What is the relationship between enthalpy of atomisation and enthalpy of sublimation?

tl; dr: In the case of a metal, sublimation and atomization can be regarded as synonymous when atomization refers to the process of converting the solid into a dilute atomic gas. In particular, the end points (states) of the process should be identical.

There are usually conventions on what you call "enthalpy of X". However the official source of chemical nomenclature - IUPAC - has not bothered to post a definition for enthalpy of atomization (or atomization enthalpy) in its Gold Book. There are entries for atomize:

  To subdivide a liquid into very small particles; methods include: impact with a jet of gas, use of a spinning disk generator, vibrating orifice generator, etc.

and for atomization (in analytical flame spectroscopy):

  The conversion of volatilized analyte into free atoms

These definitions do not refer to the same process, which leads to some ambiguity. Are you speaking of a transition from a condensed into the gas phase that preserves molecular structure, or of a chemical process that converts a sample in some phase into an atomic gas?

The wikipedia meanwhile has an entry for enthalpy of atomization which provides the following definition:

  The enthalpy of atomization (also atomisation in British English) is the enthalpy change that accompanies the total separation of all atoms in a chemical substance (either a chemical element or a chemical compound)

There is an entry in the IUPAC gold book for bond energy (theoretical):

  The energy required to break a given type of bond between atoms in certain valence
  states. An averaged bond energy is commonly derived by dissecting the heat of atomization of a molecule into contributions of individual bonds. For molecules with localized bonds, the heats of atomization (formation) are usually well approximated by the sum of pertinent averaged bond energies.

This reference to "heat of atomization" (which is not otherwise defined in the Gold Book) is consistent with that in the wikipedia.

If you apply this last definition then yes, the sublimation enthalpy and the atomization enthalpy of a solid metal should be identical, since both refer to the process:

$$\ce{M(s)->M(g)}$$

(assuming identical P,T end points). Note that sublimation enthalpy means standard sublimation enthalpy when referring to the process at standard pressure (1 bar) and a fixed temperature. Similarly for atomization enthalpy.

As regards the question:

  The $E^θ(\ce{M^{2+}}/\ce{M})$ value for copper is positive ($\pu{+0.34 V}$). What is possibly the reason for this? (Hint: consider its high $Δ_aH_V$ and low $Δ_{hyd}H_V$)

Different uses of the word "atomization" render the meaning of "enthalpy of atomization" ambiguous. This is why I open my answer with an exposition on the definition of "atomization". Your problem refers to an ideal solution, but I assume here atomization refers to sublimation since reference to hydration would not seem to make sense otherwise. The question asks to break down the reaction into steps including atomization (sublimation), ionization and solvation. Ignoring the ionization step, a larger atomization enthalpy and smaller hydration enthalpy is consistent with the metal being more stable than the solubilized ion, ie the reduced form is favored (relative to standard hydrogen electrode).

A libre text and the accompanying diagram explain some of this


Get the molar mass of an unknown liquid in a dissolution

As long as you work with sufficently low concentrations, you do not need the molecular mass of your unkown.  Indeed, this was (is?) your initial question.

Freezeing point depression (salting the footwalk to melt ice in the winter), elevation of the boiling point (adding salt to cooking water that increases it's boiling temperature, too) and osmotic pressure are colligative properties.  What you need is to look up the crysoscopic constant, for water (same page) of $1.853 \mathrm{~K·kg/mol}$.

On the other side, if you already state the equation of $0.16$ moles corresponding to $10\mathrm{~g}$.

Despite advent of mass spectroscopy, such a determination may be still helpful today.


How are the toxic components in a mixture identified?

You have many questions nested together. Whenever you have a mixture + identification of components in your mind, bring chromatography to your mind. You must be familiar with paper chromatography. Same idea, but applied to more sophisticated instruments.
When you have a completely unknown mixture, you start a so-called screening procedure with a high-performance liquid chromatograph and a mass spectrometer. The HPLC separates the components, and the mass spectrometer identifies the molecules from the fragmentation pattern and molecular mass.
You start to see how many components are there in the HPLC-MS chromatogram. Let us say you had natural products and you see 100 peaks (which is not unusual at all).
Later you scale up the process, go at a preparative scale and start collecting fractions of mixtures. You perform biological tests on them, and the relevant fraction is further studied by more advanced tools such as recycling the HPLC-NMR system.
The question you ask is equivalent to a decent Ph.D. project in natural product isolation and characterization followed by investigating their biological activity. In pharmaceutical companies, HPLCs run day and night with a typical HPLC may have 10-15 columns for screening as to which column under which mobile phase will produce the best separation.
I did not check how fagopyrins were discovered, but the process is general. Wiki gave a hint on the first German reference Fagopyrin, ein photodynamischer Farbstoff aus Buchweizen (Fagopyrum esculentum) = Fagopyrin, a photodynamic dye out of buckwheat, locked behind paywall. This procedure is from 1950s, when HPLCs did not exist but the technique of chromatography was very well known.

Wir haben aus einem Methanolextrakt von getrockneten Buchweizenblüten
durch fraktionierte Verteilung und chromatographische Adsorption einen
photodynamisch wirksamen Farbstoff in Form eines dunkelroten, fast
schwarzen mikrokrystallinen Pulvers isoliert, der im folgenden als
Fagopyrin bezeichnet ist.
We have isolated from a methanol extract of dried buckwheat flowers by
fractional partitioning and chromatographic adsorption a
photodynamically active dye in the form of a dark red, almost black
microcrystalline powder, hereafter referred to as fagopyrin.


Why butane exist as a liquid in a lighter ?

Now, am I right to say that the vapour pressure is equal to the
  pressure of the butane gases in the small container, that's why they
  undergo a phase change immediately?

You seem to have answered your own question.

The pressure inside the lighter container is indeed the vapour pressure of the mixture of n-butane and iso-butane that it contains, at ambient pressure.

Here is the pressure-temperature dependence for iso-butane:

(Source)

The high vapour pressure inside means that when you open the valve, a gaseous mixture of both isomers will come streaming out and some of the liquid mixture will spontaneously evaporate to replace it.


Can 99% isopropyl alcohol be stored in a clear container?

I would not expect that isopropanol would evaporate appreciably from a properly sealed container. Rather the main purpose of storing isopropanol out of light is to prevent peroxide formation.  Isopropanol can form peroxides and when distilled they can concentrate and explode. A clear container allows more light to pass and thus more peroxides to form.


Nomenclature of an organic compound with benzene and double bonds

Since the compound that is given in the question doesn’t have a principal characteristic group (that would be expressed as a suffix and that would determine the senior parent structure), the ring has seniority over the chain when selecting the preferred IUPAC name. In general nomenclature and depending on the context, however, the chain may be favoured to recognize the unsaturated acyclic structure.

The corresponding subsection in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Method (1): (penta-3,4-dien-2-yl)benzene (preferred IUPAC name; ring preferred to chain)

Method (2): 4-phenylpenta-1,2-diene (unsaturated acyclic structure may be recognized)

Therefore, the preferred IUPAC name for the compound that is given in the question is (penta-3,4-dien-2-yl)benzene.

Note that parentheses are used around substituent prefixes (here: “penta-3,4-dien-2-yl”) to separate locants referring to different structural elements.

Also note that locants are placed immediately before that part of the name to which they relate (i.e. “penta-1,2-diene” and not “1,2-pentadiene”).


E, Z designation - what to do when substituents are identical on one side of the double bond?

When both the substituents are identical on one side of the double bond, the compound does not show geometrical isomerism.
There is no point in using E-Z nomenclature in that case.


Two monoprotic acids vs. one diprotic acid

Your intuition is right, because a potentiometric sensor (a $\mathrm{pH}$ probe in your case) does not know the source of hydrogren ions, it only senses the presence of hydrogen ions. The same is true for any other detection mode. All we care is $\ce{H+ + OH- -> H2O}$.

It can come from two monoprotic acids or it can come from a diprotic acid. In order to make your wish true, $\mathrm{p}K_\mathrm{a}$ must be less than $\mathrm{p}K_\mathrm{b}$. They cannot vary randomly and they cannot be equal.

However intuition is not enough to prove a point. You have to pick numbers and simulate. Here is useful online tool to calculate titration curve and see how well they match.

pH Titration Curve Generator

Choose to add $\ce{HA}$ or $\ce{H2A}$ and $\ce{NaOH}$ as a base.


The cleavage that you have shown is NOT favourable. The dipole of the carbon double bond O points towards the oxygen , not the carbon. The bond tends to shift onto the oxygen giving  it a negative charge , while the carbon a positive charge. This, however is anti-aromatic so the molecule tends to do a Diels-Alder reaction with itself.

The bond will not cleave the way as you have mentioned . Just for the sake of aromaticity, a carbon double bond oxygen will not cleave in such a manner where the carbon contains the negative charge and the oxygen a positive charge. An oxygen with positive charge is extremely unstable(as it has a high electronegativity) and will show a strong inductive effect which will pull the electrons from the carbon(which is way less electronegative than oxygen) thus reverting to carbon double bond oxygen structure. Aromaticity can't be obtained as oxygen would just pull those elctrons from the carbon atom.

However let us consider the case of cyclopropanone. Here, the molecule becomes aromatic due to the shifting of the double bond onto the oxygen (oxygen gains a negative charge while carbon , positive).


Calculating the potential of an electrochemical cell (Nernst equation)

If you are not sure about the use of the parameter Q, go back to Nernst's law for each electrode, namely $$\ce{E_{M^{2+}/M} = \pu{ E°_{M^{2+}/M}} + \pu{0.0296 V} \times  \log[M^{2+}]}$$ Here the redox potentials of the two half-cells are, according to Nernst's law :$$\ce{E_{Cu} = \pu{+ 0.339 V} + \pu{0.0296 V} \times  \log0.10} = \pu{0.309 V}$$  $$\ce{E_{Fe} = \pu{-0.44 V} + \pu{0.0296 V} \times  \log0.0030} = \pu{- 0.515 V}$$
So the overall performance of the cell is $$\ce{E_{cell} = E_{Cu} - E_{Fe}}=  \pu{0.309 V} - \pu{(-0.515 V)  =  0.824 V}$$


What is the benefit of using Am-241 in smoke detectors even though it is a radioactive element?

The usage in ionizing smoke detectors requires a radioactive isotope to work.  In addition to a sufficient half-life to make a smoke detector with a suitable service life, Am-241 emits mostly alpha radiation (which is used for the ionizing mechanism) with relatively little of the more hazardous and useless gamma radiation.  See the Applications section in  https://en.m.wikipedia.org/wiki/Americium.


Based on your provided context, your book's "answers" are indeed misleading and non-answers.

Clearly I would think both (c) and (d) are spontaneous processes. For (c), the iron will naturally align itself in the magnetic field. For (d), I know the reverse reaction is non-spontaneous (table salt does not normally disassociate), so the forward reaction must be spontaneous.
The entropy of the universe increases for any spontaneous process. If we assume the entropy of the system is negative (like the book does) then the entropy of the surroundings must be positive and greater.
I agree that it makes sense that the entropy of the system is negative. The iron filings are moving to a more ordered state in (c). For (d) there are more micro-states originally since there were more moles of reactants.

These statements are correct. As you've argued, the change in entropy of the surroundings must be positive for these reactions to be spontaneous. The surroundings constitute a thermal reservoir, so the change in entropy is simply proportional to the heat evolved during the process, which implies that heat must be transferred from the system to the surroundings.
For the formation of sodium chloride, this is quite intuitive: sodium chloride is very stable and has a large, negative enthalpy of formation. The case of magnetization is less obvious, but it can be shown (see Ex 17.4 in Blundell and Blundell's Concepts in Thermal Physics) that an isothermal increase in the magnetic field leads to evolution of heat into the surroundings, and the same reasoning as above applies.


Carbocation Rearrangement - Till which neighbouring carbon is it possible?

Consider that in the situation you present, it would be much easier to lose the proton to form the alkene immediately, than to migrate the positive charge until a position where ring-opening is viable.

Not to say it is impossible, but the required 'steps' to get to that point are very improbable to the point that it doesn't happen.


Electronegativity and nucleophilicity are different concepts. Electronegativity refers to how strongly an atom pulls bonding electrons towards itself. Nucleophilicity refers to ability of atom to donate a lone pair of electrons. Different factors affect nucleophilicity, such as orbital size, charge of ion, size of molecule, hybridisation etc etc.


The statements are not contradictory when read both carefully. One is about overlapping and other is about strength. Also, you might be getting confused that more overlap leads to more strong bond but the actual statement is for a particular type orbital(s, p, d, etc.) greater extent of overlapping leads to more strong bond.


Suggestion for a Synthesis of Paracetamol — plausible?

Your route sounds possible, yes. (Aniline is commercially available.) There would very likely be side reactions of the amide under Baeyer-Villiger conditions but nothing exceptionally out of control—except you lose a little bit of yield.

There seems to have been a little race concerning the shortest and most environmentally favourable route to paracetamol. To the best of my knowledge, and assuming that their NMR spectra are unaltered and directly from the reaction published, Joncour et al. have the shortest route in 1 (one) step starting from (also commercially available) hydroquinone and utilising ammonium acetate in acetic acid.[1]

However, for the industry to decide to use a non-established method, that new method must offer a significant advantage over what is presently used. ‘Significant advantage’ typically means much safer or much cheaper. The industrial paracetamol synthesis as employed by Hoechst–Celanese starts off from phenol (extremely inexpesive), does a Friedel–Crafts type acylation with acetic anhydride and $\ce{HF}$ (cheap materials), transforms the ketone into a ketoxime (cheap $\ce{NH2OH}$) and finally uses catalytic acids ($\ce{CF3COOH, SOCl2}$; the latter being rather cheap) to induce a Beckmann rearrangement.

Viewed zealously, this is technically already a more optimised synthesis than the one you propose—it is a step shorter and uses even cheaper reagents. There is probably not much room left for optimisation along these traditional lines.

Referene:

[1]: R. Joncour, N. Duguet, E. Métay, A. Ferreira, M. Lemaire, Green Chem. 2014, 16, 2997. DOI: 10.1039/C4GC00166D. The openly accessable supplementary information with experimental details and the NMR spectrum is available on the same page.


Identifying phycobiliprotein based on wavelength

Its maximum wavelength is 620.6 (nm)

I'm afraid that this is pretty much a link-only answer, but based on   Spectroscopic Properties of C-Phycocyanin and of Its $\alpha$ and $\beta$ Subunits, published in J. Biol. Chem., 1973, 248, 5679-5685 (PDF) there's a good chance that the phycobiliprotein is C-Phycocyanin.


Is paper or water the stationary phase in paper chromatography?

The stationary phase in chromatography is the one that doesn’t move according to the eyes of a macroscopic (i.e. human researcher) observer. (That complicated way to put it was to prevent anybody raising any relativism arguments.)

Obviously, the paper does not move through the water but the water does through the paper. You should discard the five-ish sources that state water be the stationary phase as they obviously have no clue about chromatography.


Concept of heat of reaction in a isolated system

When chemical bonds are broken potential energy increases.  When they are made potential energy decreases. A loss in PE raises kinetic energy and a gain in PE lowers KE.  Since all reactions involve bond breaking and bond making the reaction will involve an increase or decrease in PE and a decrease or increase in KE.  When KE of the system increases, a system in contact with the surroundings will lose energy as heat (exothermic).  When KE of the system decreases a system in contact with the surroundings will gain energy as heat (endothermic).  In an isolated system "exothermic" reactions just raise temperature and "endothermic" reactions just lower temperature.


Yes, it’s coincidental. There is no expectation that prime (or near-prime) atomic weight means anything for the element itself. In fact, your affirmation is not even true: on the list given by Wikipedia, the first two examples are:

$\ce{Au_75 Si_25}$, your assertion is true
“an alloy of 77.5% palladium, 6% copper, and 16.5% silicon”, and your assertion isn’t met for copper

Let's go further. For a given element, how likely is it that its atomic weight, rounded, is within ±1 of a prime number? Well, atomic weights are relatively low integers, and prime numbers are plenty numerous among small integers, so it is relatively frequent. The answer can be computed with Mathematica:

weights = {#, Round@ElementData[#, "AtomicWeight"]} & /@ ElementData[All];
f[i_] := PrimeQ[i] || PrimeQ[i - 1] || PrimeQ[i + 1]
res = {f[#[[2]]], #[[1]]} & /@ weights
Select[res, ! #[[1]] &][[All, 2]]

74 out of the first 118 elements turn out to satisfy your “near-prime atomic weight” condition, which is 63%. This seems high, but it's not: if you do the same check, not only for atomic weights, but for all integers below 200, 61% of them are near-prime:

Total@Boole@Select[f /@ Range[1, 200], # &]


Does the explosion of TNT include oxidation?

3 nitrogroups of TNT are rich of oxygen, but the aromatic ring has only carbon and hydrogen atoms.

Therefore oxygen of nitrogroups oxidizes carbon and hydrogen atoms.

There is no principle difference between oxidation and reduction agens being separate molecules, or being different parts of the same molecule.

Note that TNT, in contrary to e.g nitroglycerine, is oxygen deficient, so the amount of available oxygen is not enough for complete oxidation to water and carbon dioxide.

TNT:

$$\ce{2 C7H5N3O6 -> 3 N2 + 5 H2O + 7 CO + 7 C}$$
or
$$\ce{2 C7H5N3O6 -> 3 N2 + 5 H2 + 12 CO + 2 C}$$

Nitroglycerine

$$\ce{4 C3H5N3O9 -> 6 N2 + 10 H2O + 12 CO2 +  O2 }$$


What test can be used to differ amide and amine?

Easiest way is to use TLC(thin layer chromatography) plate and stain with ninhydrin. Ninhydrin colors primary and secondary amines and amides only with vigorous heating when amide bond breaks.


Would magnesium and calcium hydroxide undergo a single replacement reaction?

The key part of the question is given sufficient energy.  The difference in electronegativity between Ca and Mg is ~0.3 eV. The relationship of eV:K is about 1:12,000. Given that, you should be able to calculate a temperature at which the reaction $\ce{Ca(OH)2 + Mg -> Mg(OH)2 + Ca}$ is favored over the reverse.

However, temperature is based on the random motion of particles, so the energy distribution follows a Gaussian curve. Some Mg atoms are sure to have enough energy to knock loose some Ca atoms at room temperature.

Furthermore, you can force the suggested reaction by making use of relative solubilities. Find a solvent (not water, since magnesium and calcium are far more electronegative than hydrogen, and would decompose water to form hydroxides of both metals) in which $\ce{Ca(OH)2}$ is more soluble than $\ce{Mg(OH)2}$. Precipitation will "lock up" the $\ce{Mg(OH)2}$ as it forms, little by little, forcing the reaction to proceed even though it is endothermic.

Of course, you could also force this to proceed in molten salts by driving it with an electric current.


Does adding prefixes to SI units make them no longer SI units?

Yes, when prefixes are used with coherent SI units, the resulting units are still SI units; however, they are no longer coherent.

  The metre per second, symbol m/s, is the coherent SI unit of speed. The kilometre per second, km/s, the centimetre per second, cm/s, and the millimetre per second, mm/s, are also SI units, but they are not coherent SI units.

BIPM: The International System of Units (SI)


The ideal gas equation of state assumes all of the particles are not interacting. This is a bad assumption when particles are close together, such as in a liquid. However as particles get further apart this becomes a more realistic model since particles collide less and less often.

In general low pressure and high temperature leads to conditions where the ideal gas equation will work. We see this in combustion reactions actually. You can model a gasifier fairly well with the ideal gas equation of state because of how high the temperatures are.

At high temperatures, molecules move with higher energy, so they bounce of walls harder. Thus to attain a given pressure, you need fewer molecules if they are hitting the walls harder. Thus, at a higher temperature, there are fewer molecules in your vessel, making them further apart, which makes them more ideal.

In terms of molar volumes (I prefer pressure but won't begrudge your textbook), yes as the molar volume gets bigger that means the same number of particles take up more space, meaning they are farther apart from each other. In the extreme limit they will become infinitely far apart and be exactly like an ideal gas. In practice the molar volume does not have to be extreme for the ideal gas to start modeling the system well.


Importance of zinc blende and wurzite crystal structures for semiconductors?

As suggested I can try to summarize the basic idea in an answer although I wouldn't say it solves the bounty. I'm also unsure why you even set the bounty for it.

For this I'll quote the book that I found a few days ago. As I own the missing pages now as well the answer seems a bit more elaborate then before. As the book is in German I will translate the text.

  The formation of the band-structure of typical element semi-conductors like diamond (C), Si and Ge was already shown in chapter 7.3. By mixing s- and p-wavefunctions we'll get a tetrahedral bonding orbital ($sp^3$), that will lead to a splitting into a bonding and an anti-bonding orbital in the range of the equilibrium-bonding distance. The bonding orbitals form the valence band and the anti-bonding orbitals form the conducting band. The distribution of the four s- and p-electrons into bonding and anti-bonding orbitals leads to a completely filled valence band while the conducting band remains empty.
  The result is therefore an insulator like diamond or, if the energy gap between the valence and conducting band is smaller, to semi-conductors like Si or Ge.

I'll skip the next pages that deal with the formation of the band-structure and the symmetry aspects as I have no experience in these topics and continue with the $III-V$-semi-conductors.

  As the formation of $sp^3$ hybrid orbitals in the chemical bonding of Si and Ge seem to be the a substancial part of the semi-conducting properties it seems reasonable that other materials with a tetrahedral crystal structure, i.e. $sp^3$ hybridisation, should show semi-conducting properties as well.

At this point I will switch to basic solid state chemistry as the book doesn't elaborate this part well enough. As solid state chemist I often have examples like these with my students. A while ago we were discussing the silver-mobility in high-temperature $AgI$, which shows quite some conductitiy through defects and the low temperature modification. At this point I would typically ask for reasons why $AgI$ might be problematic. Typical answer that I expect are:

HSAB-concept  with soft $Ag^+$ and soft $I^-$
A covalent three-dimensional tetrahedral network with ZnS-structure

For the second part they either check these facts before we talk about it or they can use the things they have learned in lectures. One of them being the Grimm-Sommerfeld rule. Basically, compounds, that form between two elements to the left and to the right of another element will likely adopt the structure of said element. So for a group $IV$ element like the silicon structure, it can be $III-V$ or $II-VI$ or $I-VII$. It's one of these rules that are taught in the very first semester but people forget about it although it's super useful for compounds you don't know.

Therefore my idea would be here that if, as the book mentions, $sp^3$ hybridisations and covalent, tetrahedral structures as the main source for the semi-conducting properties it's only natural that $III-V$ compounds show a similar effect. Those bands are basically molecular orbitals. If you start from a di-atomic molecule and increase the amount of bonding partners you will get more and more orbitals that move closer and closer together until you have bands. If let's say Ge has a certain band-gap and you change to GaAs, or course the single contributions to the HOMO and LUMO and later to the bands will be different but not that far away from Ge, so we expect a comparable band-gap.

Source: H. Ibach, H. Lüth, Festkörperphysik: Einführung in die Grundlagen, 7th edition, Springer, 2008.


Comparing the Lattice energy of different compounds

You might want to look at

https://scilearn.sydney.edu.au/fychemistry/prelab/e12.shtml

and the related page

https://scilearn.sydney.edu.au/fychemistry/calculators/lattice_energy.shtml

which has a nice lattice energy calculator that allows you to play with the parameters and see how the lattice energy varies, but I'll try to summarise those the argument below.

Here we are dealing with the ionic model - everything is totally ionic, there is total charge separation, all binding is electrostatic. Thus our system consists of a set of point charges. We can write the energy of such a system as

$$
E=N_AM \frac{z_1 z_2 e^2}{4 \pi \epsilon_0 r}
$$

where $N_A$ is the Avogardro constant, $z_1$ the charge on the first ion, $z_2$ the charge on the second, r is the closest separation of the ions and M the Madelung constant. Now note the Madelung constant only depends upon the structure of the crystal - all Sodium Chloride type crystals, whether it be NaCl, KF, or FeO, all use the same value off M. Looking at the formula we can see that FOR A GIVEN STRUCTURE (i.e. with the same M)

High charges on the ions mean high lattice energy
Small separation means high lattice energy

Thus point 2 addresses point 2 in your question, K is bigger than Li, hence the separation is bigger in KF than LiF, hence KF has a lower lattice energy than LiF. The first point explains why MgO has a higher lattice energy than NaF. However it doesn't cover your first example, as $\ce{AlF3}$ and MgO have different structures (and actually I don't think it is a very good question).

Comparing different structures is very difficult. About the best we can do is the electrostatic term from the Kapustinskii equation (https://en.wikipedia.org/wiki/Kapustinskii_equation)
$$
U_{L}={K}\cdot {\frac {\nu \cdot |z^{+}|\cdot |z^{-}|}{r^{+}+r^{-}}}
$$
This is an approximation to the above. Here K is a constant which is INDEPENDENT of structure and $\nu$ is the number of ions in the empirical formula, 2 for MgO, 4 for $\ce{AlF_3}$. Thus assuming ${r^{+}+r^{-}}$ is the same for both of these $\ce{AlF3}$ has a bigger lattice energy because $\nu\cdot |z^{+}|\cdot |z^{-}|$ is bigger - For $\ce{AlF3}$ it is 4*3*1=12, while for MgO it is 2*2*2=8. What this is really saying is that the $\ce{AlF3}$ structure has a much larger Madelung constant than the NaCl one, enough to overcome the charge differences, and thus it has a higher lattice energy.


What organic solvents are suitable to use with potassium permanganate?

It sounds like you need a solvent that 1) does not react with KMnO4 and 2) is immiscible with water.

As @Janice DelMar said in her comment, some of the lower-chlorinated hydrocarbons (methylene chloride, 1,2-dichloroethane, 1,1,2-trichloroethane) should work. If you can live with a high boiling point solvent, you might consider chlorobenzene.

Other solvents that fit these criteria include most alkanes (pentane, hexane, cyclohexane, petroleum ether, etc.), benzene (but not toluene), and diethyl ether.

A quick test to determine compatibility of your solvent with KMnO4 involves a TLC stain. Make (or borrow) a batch of KMnO4 stain as described here. Dip a TLC plate into your solvent and then dip it into the stain with minimal drying between. If the plate turns brown (MnO2), then your solvent was oxidized and is no good. If the plate stays purple, then your solvent is resistant.


How to calculate vibrational spectra of organic molecules?

The usual approach is:

Find the equilibrium geometry (within the Born-Oppenheimer app.).
Expand the energy in Taylor series till second order term. The first order will be 0 because of the minimum energy condition.
Include this expression in the Hamiltonian.
Generate a coordinate transformation to get normal coordinates.
Relate second order derivative of the potential energy with the force constant of Hook law (Using second order term we can identify it with a harmonic oscillator). Then get the harmonic frequencies.
The software employed should give also the intensity for each frequency.
Plot intensity vs. frequency/energy/wave length. You'll get only "points". To transform them to a continuous function you can use a Lorentzian like function.

Details can be found in most computational quantum chemistry books.


Deuterium is already obtained by electrolysis of water. Deuterium is a little less reactive than Hydrogen, So when we repeatedly electrolyze water, we get heavy water. Then we can obtain deuterium from that.


Magnetic nature of tetraamminedichlorocobalt(III) chloride

The ammine ligand is not really a strong-field ligand. It is not a strong π acceptor which is characteristic of strong-field ligands such as the nitrosyl cation or carbonyl. Instead, it is a weak π donor. Quite unlike other strong-field ligands such as cyanide, there are a great number of high-spin complexes with ammine ligands. Ammine (and aqua) ligands are better termed intermediate field. They can stabilise low-spin complexes but they typically also support high-spin environments.

Understanding when ligands cause the electrons to pair is very non-trivial but essentially guesswork. This is exemplified by some complexes being isolated in both the high-spin and low-spin states as is the case in $\ce{[Fe(ptz)6]^2+}$ ($\ce{ptz}$ being 1-propyl-1H-tetrazol-κN4). Furthermore, the metal has even more influence on the question with higher oxidation states typically stabilising low-spin configurations while lower oxidation states often favour high-spin complexes. The only clear cases are 4d and 5d transition metals which can be safely assumed low-spin unless stated otherwise.

With that said, cobalt(III) complexes feature a combination of a high (-ish) oxidation state of the central metal combined with a $\mathrm d^6$ environment which generally favours low-spin a bit more as it allows for maximum electronic stabilisation. Therefore, cobalt(III) complexes can be guessed as low-spin if they are not entirely made up of weak-field ligands. Ultimately, however, the experiment or calculation will have to answer.


Assuming that "atom type" already accounts for the actual nuclei and only refers to the different chemical character, I can think of examples of influence from a distance.

The one with the smallest distance is the acidity of hydrogen attached to carbon. When we compare the hydrogens compared to the central carbon in Propane and Propionaldehyde, we observe a large difference.
Comparing any carbon of Benzene with one of the inner carbons of Butadiene, we observe a large difference in reactivity towards electrophiles.
The same comparison for a carbon of Benzene and the carbon in the 4 position of Pyridine yields a drop in nucleophilicity for the latter.


How do we indicate a volume fraction like 0.05 in procedure descriptions?

I think it would be wise to include units, because how would you distinguish between mass and volume fractions? Options 2 and 3 that you present are both seen in the scientific literature and text books. My opinion is that option 3 is the least ambiguous way to present both the numeric value and the context (i.e. the units).


Which one is more acidic – 3,5-dinitrophenol or 4-nitrophenol

As Zhe's comment points out, intuition does not count for much here, unless you have enough actual experience working with these systems to know what is more acidic. There are some things which cannot be predicted a priori using simplified models.

The only way to find out is to look up the data. The acidity constants of some of these phenols are tabulated in J. Org. Chem. 1998, 63 (19), 6680–6688. 4-Nitrophenol has a $\mathrm pK_\mathrm a$ of $7.14$, and 3,5-dinitrophenol a $\mathrm pK_\mathrm a$ of $6.73$, in aqueous solution.


Arrange in order increasing reactivity towards nucleophilic substitution

The reactivity of the SN2 reaction is determined by the stability of its transition state. More stable the transition state (hereby referred to as TS), lower will the activation energy be for the reaction, and therefore, the reactivity increases.
Let's first have a look at the transition state of an SN2 reaction.$^{[1]}$

What you see inside the square brackets is the transition state, Nu represents the incoming nucleophile, while X represents the outgoing leaving group (both bearing a partial negative charge).
In this transition state, the carbon atom in the middle has a p orbital that shares one pair of electrons between the old and the new bonds. This suggests that the transition state for an SN2 reaction has a more or less planar carbon atom at the centre with the nucleophile and the leaving group arranged at $180$° to each other.
The point to be noted here is that the p-type orbital of carbon in the TS stays directly perpendicular to the plane formed by its three substituents, which is the same as the plane of the ring (in this case, all the three hydrogens shown in red). This means that sterically bulky groups which are present in the plane of the ring, and directly protrude towards the site of reaction (which is still perpendicular to this plane) and therefore will have a much bigger effect in inhibiting the entry of the incoming nucleophile as compared to those outside it.
This will be a general idea for the stability of TS (and hence the reactivity towards the SN2 reaction) in other cases as well.
Now, although all the options have adjacent benzyl groups, it might be worthwhile to see how benzyl and allyl groups stabilize the TS. These compounds react rapidly by the SN2 mechanism because the π system of the adjacent double bond can stabilize the transition state by conjugation.
The p orbital at the reaction center (shown below in brown, and corresponding to the brown orbital in the previous diagram) has to make two partial bonds with only two electrons — it is electron deficient, and so any additional electron density it can gather from an adjacent π system will stabilize the transition state and increase the rate of the reaction. This is shown here for allyl bromide:$^{[2]}$

The benzyl group acts in much the same way using the π system of the benzene ring for conjugation with the p orbital in the transition state. This stabilization will be present in all the four options,and I am making no attempt to compare them here
So, the geometry of the transition state and steric consideration will put S at the end due to the highly directed bulkiness of the di-ortho substituted benzene group, while P will jump ahead due to the absence of such clashes. S will still have benzylic interactions, but the incoming nucleophile will be hindered by the methyl group which points towards the perpendicular p orbital
Out of R and Q, R might achieve favourability over Q because again, a steric clash in the plane of substituents by a benzene ring having a methyl group directed towards the site of reactivity (the p-orbital perpendicular to plane of the ring) will generate much more hindrance than a single methyl group smaller than o-methyl benzene, which is directed away from the site.
Finally, another small reason that may tip the scales in the favour of R could be an interaction between the σ $\ce{C-H}$ bonds and the p orbital of carbon, in a kind of hyper-conjugative fashion similar to what happens in a carbocation (although this interaction would most probably be weak) as follows:

This weak σ $\ce{C-H -> p+}$ might lead to a lowering of the energy of the p-orbital, and thereby slightly increase the feasibility of attack by lowering the activation energy.
The visualization for this interaction may be seen in a similar fashion as hyperconjugation for a carbocation:$^{[3]}$

Such an interaction is not quite feasible in the TS of Q, where any interaction with the ortho-substituted methyl group will require complete delocalization over the benzene ring, which is not very probable as the TS is a short-lived species, and is not a full-fledged carbocation (especially since the leaving group is chloride, which is a good leaving group and doesn't lead to excessive charge development by a very early or late departure)
Hence, you get the order P>R>Q>S.
References:
$[1]$ Organic Chemistry Second edition, Jonathan Clayden, Nick Greeves and Stuart Warren,
ISBN $9780199270293$ Nucleophilic substitution at the carbonyl group, [p.$340$]
$[2]$ Organic Chemistry Second edition, Jonathan Clayden, Nick Greeves and Stuart Warren, Nucleophilic substitution at the carbonyl group, [p.$341$]
$[3]$ Molecular orbitals and organic chemical reactions, Fleming


Electrolysis of water - why not pure water?

The terrible conductivity of water can explain this.

Pure water is a really bad electric medium, since it has a low tendency to ionize by itself (we know the low concentrations of ions involved by looking at the Kw).

This means that the voltage needed to get a current flow from anode to cathode would be really high, with consequences both in term of safety and energetic waste.

A salt dissolved provides ions, which make the diluted solution a good electric medium, and allow a current to easily flow between anode and cathode.


Soxhlet extraction condenser - addition of compound to refluxing setup

In my opinion the intention is to have a very slow addition of the reagent to the hydride. The compound is dissolved in ether, so a Soxhlet is of no use since it is for solids.

What they are possibly referring to is a liquid-liquid extraction apparatus, specifically the one for extracting with solvents lighter than water:

The saturated solution of the compound is then placed in the thimble and when reflux starts, the solvent is condensed, falls into the tube and returns to the flask carrying over the compound. Normally the tube goes to the very bottom of the thimble. In fact, usually it is a loose tube with a wide opening on the top. How to know that the addition is complete is another story. If the compound is not colored, it will not be easy without opening the system.  In any case, maintaining reflux for a long time will ensure that everything goes into the flask.

I suppose that for some reason it is difficult to perform a slow addition through a funnel in this case. It would be, however, better to use a syringe pump but not everybody has one of those.


Calculation of the cell potential of a lemon battery experiment

I am afraid that nobody can predict the true potential of Zn/ Cu lemon battery. It is not a standard system. The electrode potentials you see in the published tables are measured or calculated under unknown or controlled conditions and one of the basic requirement is that the metal must be in equilibrium with its own ions.

The lemon battery, although looks very simple, it is too complicated to deal with.

When you insert the electrodes into the lemon, there are no metallic ions to begin with, i.e., the concentration of Zn$^{2+}$ and Cu$^{2+}$ is essentially zero. Just using plug & chug approach for the Nernst equation will give you infinite potential difference. This is not what is experimentally observed.

1b. Even if we assume that some metal dissolves around in a few seconds, still the metal concentration is not high enough to allow a reliable use of Nernst equation.

Citric acid is not only an acid, but it has complexing/chelating property as well.
Any voltage which comes close the calculated potential from redox tables is a mere coincidence.

I think the proper way to determine the potentials generated by contact potentials. This is the realm of physics. Two dissimilar metals, when brought in contact, will generate a potential difference. See my previous answer Electrochemical reactions in a salt water battery


Why is heroin a more potent drug than morphine, despite having a similar structure?

The answer to this question lies in the pharmacokinetics of these two drugs: the acetyl groups cause Heroin to be 200 times more lipid soluble than Morphine1. Another example of a common drug would be Fentanyl, which is 580 times more lipid soluble than morphine, with a structure very similar to that of Morphine.
Increased lipid solubility allows these drugs to pass more rapidly across the blood-brain barrier into the Central Nervous System, giving them more analgesic potency and a more rapid onset of action than morphine (the peak blood level of heroin when used intravenously is around 5 minutes2, whereas that of fentanyl is 1.7 minutes3).
As for the sociological aspects, the high potency of these drugs acts as a positive feedback loop. The heroin/fentanyl dose is smaller than morphine, which is preferable for drug smugglers, as they have to smuggle less of the drug, reducing their risk. This explains the high potency and usage of heroin as a recreational drug as compared to morphine.
References:

Mahesh Trivedi, Shafee Shaikh, Carl Gwinnut: "Pharmacology of Opioids", Updates in Anaesthesia, 2007, pp. 118-124. Link.
Oelhaf RC, Azadfard M. Heroin Toxicity. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; Jan 2020. Available from: https://www.ncbi.nlm.nih.gov/books/NBK430736/
Fentanyl Citrate Injection, USP. FDA Label.


What is the minimum distance of Rubidium and Iod ions in a centered cube?

The distance between $\ce{Rb^{+}}$ and $\ce{I^{-}}$ in the fcc lattice would be $(d(\ce{Rb^{+}})/2) + (d(\ce{I^{-}})/2) = 3.76 \times 10^{-10} \mathrm{m}$


What is the origin of the Quadratic factor in calculation of XRD peaks?

Reciprocal space

One way to derive the formula is to use reciprocal space coordinates. The reciprocal space lattice has three unit cell vectors, a*, b* and c*. The diffraction vector d* is given by:

$$d^* = h a^* + k b^* + h c^*$$

The d-spacing is given by the reciprocal length of d*. The relationship between direct space and reciprocal space unit cell vectors is

$$a \cdot a^* = 1$$
$$b \cdot a^* = 0$$
$$c \cdot a^* = 0$$

$$a \cdot b^* = 0$$
$$b \cdot b^* = 1$$
$$c \cdot b^* = 0$$

$$a \cdot c^* = 0$$
$$b \cdot c^* = 0$$
$$c \cdot c^* = 1$$

So to get the d-spacing, we have to figure out the length of $d^*$. For a cubic system, the reciprocal space unit vectors all have the lenght of 1/a, and they are all at right angles. To get the length for the cubic system, we use Pythagoras' theorem (or the length of a vector in Cartesian space): It is the square root of the coordinates squared.

So:

$$|d^*| = |a^*| \sqrt{h^2 + k^2 + l^2}$$

To get to your formula, substitute 1/d for $d^*$ and 1/a for $a^*$. For the hexagonal system, it is a bit different because the angles are not all 90 degrees, so there will be a trigonometric factor (sin 120 or cos 120 or such) in the respective formula.

Direct space

One plane will go through the origin, the next through a/h, b/k and c/l. In the cubic case, a vector normal to the plane is (h, k, l). To get the distance of the plane from the origin, you can take any point in the plane and form the dot vector with the normalized vector (h, k, l). For example,

$$ d = \frac{(h, k, l)}{\text{length} (h, k, l)} \cdot (1/h, 0, 0)$$

For a cubic system, this will simplify to the expression for the d-spacing. The term $h^2 + k^2 + l^2$ is from the length of the (h, k, l) vector.

(h k 0), no vectors

Both derivations above use vectors. Here is a derivation for the 2-dimensional case using no vectors.

Let the origin be point A. In a cubic system, the plane hk0 intersects the x-axis at a/h (point B) and the y-axis at b/k (point C). Our goal is to calculate the altitude. We use that the area of the triangle can be calculated two ways, using the perpendicular sides or using the hypothenuse and altitude:

$$\text{Area} = \frac{1}{2} * \frac{a}{h} * \frac{b}{k} = \frac{1}{2} * \text{hypothenuse} * \text{altitude}$$

We can get the length of the hypothenuse from Pythogoras:

$$\text{hypothenuse} = \sqrt{(\frac{a}{h})^2 + (\frac{b}{k})^2}$$

Then, we can solve for the altitude:

$$\text{altitude} = \frac{\frac{a}{h} * \frac{b}{k}}{\text{hypothenuse}}$$

$$ = \frac{\frac{a}{h} * \frac{b}{k}}{\sqrt{(\frac{a}{h})^2 + (\frac{b}{k})^2}}$$

With a = b and simplifying, we get the formula the OP asked for (in two dimensions).


I don't like to reinvent wheel -- if you go to Wikipedia's  Metal carbonyl and follow just two 1st literature position, you'll find your answer concerning bridging.

For MO calculations and a general scheme applicable to systems featuring “dative” MM bonds and non-compensating, semi-bridging carbonyl ligands, see: R. D. Barr, T. B. Marder, A. G.
Orpen, I. D.Williams, J. Chem. Soc. Chem. Commun. 1984, 112–114


How to find the molar mass of a compound given mass, vol, temp, and pressure?

Sometimes (too often) answer keys are wrong. In this case we've now triple checked the calculation, and you, Jan (as implied in his comment) and myself have all come up with a molar mass of $\pu{257 g/mol}$. At this point, we have to say the answer key is wrong and you are correct.

Your answer is $\pu{1.71}$ times that given by the key, and neither that nor any other comparison appears to correlate to any obvious error on their part. It simply appears to be a misplaced or incorrectly entered answer in the key, end of story and move on. And good job getting the correct answer ;)


That's very common, actually.

https://en.wikipedia.org/wiki/Polymorphism_(materials_science)#Ostwald's_rule

A substance melts, but the recrystallisation from the iostropic melt can be kinetically inhibited. If you cool further down, a less stable form of the same substance, with lower melting point, can crystallise, and re-form over time, into the original phase.

This (according to Ostwalds rule) happens practically every time a suitable, less stable polymorph exists. Or let's rather say there is a general tendency for it to happen.

Another, very common example (to which Ostwalds rule also loosely applies) are polymers. Because of their chain structure, crystallisation (if possible at all) is much inhibited and usually never occurs until you cool a good ten K or so below the equillibrium melting point.


How much worse is methane than CO2 as greenhouse gas?

Let's take a look at this graph just so we have a reference.

This shows the IR absorption bands of various gases found in the atmosphere. For this answer, let's focus in on the absorption bands of $\ce{H2O}, {CO_2}$, and $\ce{CH_4}$.

First, the answer to why methane is considered a worse greenhouse gas than carbon dioxide is largely because methane absorbs in regions where carbon dioxide does not absorb at all. That is, methane absorbs different wavelengths of light than carbon dioxide.

Another thing to note is that methane's absorption peaks overlap with some of the absorption peaks of water vapor. This is an interesting thing to note because the strength of methane's absorptions in those overlapping regions are much stronger than the absorptions of water.

So, we have two things which contribute to why methane is sometimes called a worse greenhouse gas than carbon dioxide. First, it absorbs in areas where $\ce{CO2}$ does not absorb. Second, these absorptions are very strong.

Something you might be interested in, however, is the fact that there are some people who make the argument that methane essentially doesn't matter as a greenhouse gas. This would be due to the overlap in absorption between water vapor and methane. For instance, methane has an average concentration of 1.7 ppm (parts per million) in earth's atmosphere. Compare this with the fact that the earth's atmosphere can be composed of 1% water vapor at sea level and .4% across the whole atmosphere on average.

Thus, the amount of water vapor in the atmosphere is much more than that of methane at any given time. So, it can be argued that we do not need care much about methane as a greenhouse has because water vapor will do all the IR absorption that methane would have done.

I don't know enough about atmospheric chemistry to really say if I agree or disagree with those claims. One thing I do know, however, which is interesting is the fact that the idea of global warming would seem to present more of a problem with water vapor than methane or carbon dioxide because as temperature increases, the ability of the atmosphere to hold water vapor increases which would accelerate the warming from water vapor, which would increase the amount of water vapor the atmosphere can hold...

Also, I believe that the metric which says methane is more dangerous than carbon dioxide is a comparison which does not account for other things which might absorb that light. In other words, the figure you quote do not account for the fact that water vapor might make methane irrelevant as a greenhouse gas.

Interesting Read: http://www.nasa.gov/topics/earth/features/vapor_warming.html


What would be the mechanism for this condensation reaction?

Your 2-methyl-1,3-benzoxazole is a masked acetic acid.

Remember that one way to synthesize oxazoles actually starts from aldehydes (and uses TosMIC).

So, yes, the reaction is an aldol condensation where the $\ce{CH}$-active component is carboxylic acid in disguise.


What is role of palladium in hydrogenation reactions?

The general catalysis cycle of hydrogenation reaction is pretty simple and AFAIK well documented. It consists of for reactions, all well documented in metalloorganic chemistry.

Hydrogen activation. It comes in many flavors, but idea is the same: $\ce{H-H}$ bond is broken and instead $\ce{M-H}$ bonds are formed.
Alkene bonding with a metal center
Alkene insertion in $\ce{M-H}$ bond, forming $\ce{R-M}$ alkyl bond
Elimination of alkane from $\ce{R-M}$ and $\ce{M-H}$ fragments.

This general process can occur on mononuclear or polynuclear complexes and on metal surface. However, when selecting particular metal for the process it is important that the intermediate particles were neither too stable, or they will cover catalyst completely and remain there, nor too unstable or the process won't happen. It so happens, than Ni, Pd, Pt are in the sweet spot with Pd being one of the best. They are not, however, the only hydrogenation catalysts in existence, they are simply the ones with high activity and simple composition. For example, $\ce{Ni}$-promoted $\ce{WS2}$ catalystalso has significant hydrogenation activity.


What is the name of the crinkly plastic used in clear parts of envelopes?

It usually is BOPS: Biaxially oriented polystyrene.


Why does a six-membered ring contract in this SN1 reaction?

A cursory look through several organic chemistry texts (Clayden, Carey, March, Vollhardt) as well as SciFinder doesn't reveal any instances of the same reaction used in the Khan Academy video - this would imply that either the reaction is made up to illustrate a point (albeit not very well) or comes from somewhere more obscure (historical literature in a language other than English).

SN1 vs SN2

The first question that it would be useful to answer is whether the reaction proceeds via an SN1 or SN2 pathway, or, to phrase this a more accurate way, to what extent is the halide dissociated at the time when the rearrangement occurs.

In general, secondary halides can undergo both SN1 or SN2 reactions depending on the conditions used (substitutions at secondary centres are always hard to predict, and often go with neither complete inversion nor complete racemisation). In March's Advanced Organic Chemistry,  the chapter on cationic  rearrangements summarises:

  The SN1-type process occurs mostly when B is a tertiary atom or has one aryl group and at least one other alkyl or aryl group. In other cases, the SN2-type process is more likely.

Stereoelectronics

If we accept that the reaction goes via an SN2-like pathway, we must then acknowledge the stereoelectronic requirement that the migrating group must be able to interact with the C-Br σ* orbital.

This requirement immediately rules out the migration of the exocyclic methyl group, leaving only the internal C-C bond to migrate (assuming that bromine adopts the equatorial position and the ring sits on a chair-like conformation - both of which are reasonable assumptions).

Migrating group ability

Even ignoring the steroeelectronic issues, migratory aptitude would favour rearrangement of the internal C-C bond as it is marginally more able to stabilise the developing positive charge during the rearrangement.

tl;dr

In the absence of the original literature, it's hard to say with certainty which group migrates (though it would be an easy experiment to carry out....). That said, following the general trends for rearrangements/migrations found in undergraduate organic chemistry texts, it would seem most likely that the rearrangement takes place as shown rather than the competing 1,2-methyl shift (we're also ignoring the possibility of a proton elimination to give an alkene).


Why doesn't sulfate get electrolyzed in sulfuric acid electrolysis?

To analyse this kind of problem, the first step should be identifying all the species you have in solution that could take part in any electrolytic reaction. These are:

$$\begin{array}{}\ce{SO4^2-} & \ce{H+} & \ce{H2O} & \ce{Cu}\end{array}$$

I did not add graphite to this list since it generally never participates in electrolytic processes. The next step would be to identify what can be reduced and what can be oxidised. Thankfully, all species are either in their lowest or highest common oxidation state. See the list below with oxidation states added:

$$\begin{array}{cccc}\ce{\overset{{+VI}}{S}\overset{-II}{O4}^2-} & \overset{\mathrm{+I}}{\ce{H+}} & \ce{\overset{{+I}}{H2}\overset{-II}{O}} & \overset{\pm 0}{\ce{Cu}}\end{array}$$

Obviously, the only candidates for oxidation are oxygen or copper, and the only candidates for reduction are sulfur or hydrogen. I’ll concentrate on the oxidation first, as the equations are simpler. We need to find a redox pair for each component and then decide which is more likely. For copper, the only real possibilty is generating copper(II), while for oxygen the only real possibility is generating oxygen gas. See equations $(1)$ and $(2)$:

$$\begin{align}\ce{Cu &-> Cu^2+ + 2 e-}&&E^0 = +0.337~\mathrm{V}\tag{1}\\
\ce{2H2O &-> O2 + 4 e- + 4 H+}&&E^0 = +1.229~\mathrm{V}\tag{2}\end{align}$$

I added the standard electrode potentials $E^0$, too. We can easily compare which oxidation is more likely: the smaller potential will act first, thus, copper is oxidised. Now on to the reductions. For hydrogen, it is naturally the known equation. What about sulfur in sulfate? Well, the most obvious choice is assuming $\ce{SO3^2-}$ as a reduction target. See equations $(3)$ and $(4)$:

$$\begin{align}\ce{2H+ + 2e- &-> H2 ^}&&E^0 = \pm 0~\mathrm{V}\tag{3}\\
\ce{SO4^2- + 2 e- + 2 H+ &-> SO3^2- + H2O}&&E^0 = +0.17~\mathrm{V}\tag{4}\end{align}$$

The reduction potential is actually given for the target $\ce{SO2 (aq)}$; however, that has the same oxidation state and is in equilibrium with sulfite, so they are comparable. Here, the more positive potential should actually react first. However, using the graphite electrode typically favours hydrogen evolution via overvoltage mechanisms. This is, unfortunately, something you will have to memorise.

Summing up:

There are only four reactions that can happen
of these, two can be chosen as more favourable by their respective electrode potentials
one choice needs to be adjusted due to overvoltage.


How does lime prevent soda lime glass from dissolving in water?

The addition of lime does indeed form calcium silicate, but it doesn't serve the purpose you mentioned. Silica of itself isn't water soluble, nor soda glass.

The difference is in the crystalline structure. Silica ($\ce{SiO2}$) consists of silicon atoms bridged by oxygen atoms, forming a periodic crystal with no net charges on any atom.

However when some alkali is introduced ($\ce{Na2O}$), the additional oxygen atoms remain charged, and cannot link silicon atoms together; those form silicate anions. The silica breaks and depolimerizes, and the remaining $\ce{Na+}$ fill holes in the structure to maintain charge balance. Thus the solid melts at lower temperature with reduced viscosity, it's "more fluid".

Lime contains calcium, which bonds more strongly to silicate anions than sodium. It also decreases viscosity of molten glass because of the non bridging oxygen, but in addition increases hardness and tensile strength of the glass in solid state.


Resonance and reductions with sodium borohydride

Sodium borohydride is definitely strong enough to reduce an acetophenone derivative. It is likely that the resonance stabilization does reduce the reactivity of the ketone. It's not often that a sodium borohydride reduction requires 35 hours. In the cases that I have run, usually less than one hour is sufficient.

I did some quick searching of whether an isolated ketone could be reduced in the presence of an acetophenone, and I did not find an adequate example of a case where only the isolated ketone was reduced. That says to me that the reactivity difference is not that great. Perhaps someone else has come across this though.


How can the density of water be maximised?

Clerici solution is a solution of equal parts of thallium formate and thallium malonate in water, density is 4.25 g/ml,     Thoulets solution is solution of potassium-mercury iodide in water with density 3.2 g/ml. Don't even think of using them.

Sodium polytungstate $\ce{Na6[H2W12O40]}$ is a relatively safe compound and allows densities up to 3.1 g/ml. $\ce{ZnCl2}$ allows densities up to ~ 1.9 g/ml, but has acidic solutions.

Unfortunately, common nontoxic salts usually either are not dense enough or not soluble enough to make really heavy liquids, but densities around 1.1-1.2 g/ml are quite easy to achieve even with common salt. Note, however, that concentrated salt solutions 'siphons' water from human body through damaged skin and mucous membranes, and to some extent even through undamaged skin, which may lead to desiccation, trauma of mucous membranes and very unpleasant sensations.


Collecting my comment thoughts together, there is an early demethylation step in this synthesis. Structure 1 is the result of carboxylation while methylation provides ester N. Friedel-Crafts acylations of anisoles at the ortho-position can undergo demethylation via aluminum chelation. Magnesium chelate 2 is a candidate for demethylation by iodide ion. Phenol O undergoes thermal dehydration catalyzed by the acidic phenolic hydroxyl group. Steps P --> Q are straightforward. Step R --> S methylates the phenol hydroxyl and excess KOH insures that any ester that is formed is saponified. Clemmensen reduction of S and remethylation afford T.
Addendum: The OP's comment reveals that T is a carboxylic acid, not a methyl ester. Presumably the Clemmensen reduction effected some demethylation which is the reason for the last methylation step.


Assumption behind Gibbs energy and maximum work

The whole point of Gibbs free energy is to quantify a bound on possible non-expansion work at constant p and T. That non-expansion work is implicit as part of the term "dU" which in the reversible case can be written as

$$dU = -pdV + TdS + dw_{\text{non-exp,rev}}$$

The $-pdV$ term is the expansion work if the process is carried out reversibly. You remove that contribution by adding a  $pdV$  term:

$$dU + pdV= -pdV + TdS + dw_{\text{non-exp,rev}} + pdV = TdS + dw_{\text{non-exp,rev}}$$

Finally if you subtract the entropy term you are left with the maximum possible non-expansion work (obtained when the process is performed in a reversible fashion):

$$dU + pdV - TdS= dw_{\text{non-exp,rev}}$$


1.) The mirror plane is orthogonal to the green line drawn between the respective center of the two enantiomers.

2.) The rotation is around the red axis. The red marked carbon atom shows the rotation angle.


How to purify water with activated carbon?

The main reason activated carbon is used in water treatment is that its large surface area is useful for adsorbing volatile organic contaminants and chlorine. It is not effective at removing inorganic ions (dissolved minerals, including fluoride), but these are removed quite well by distillation. I think distillation first would make more sense—distillation will remove less volatile contaminants so that less has to be adsorbed by the carbon.

Tap water has to meet certain standards (that depend on your area) for dissolved minerals and volatile organic compounds, so depending on the end use of the water, removing the small amount of remaining contaminants may not be necessary. Formerly, double-distilled water was the standard for making highly purified water, but certain applications are more or less tolerant of certain types of contaminants, and double-distillation requires a lot of energy, so it has largely been replaced for many purposes.

Consumer water filters usually contain filters to remove particulates, ion exchange resin to remove inorganic ions, and activated carbon to improve the taste of the water, and as far as purification for laboratory use, combinations of filtration, deionization, reverse osmosis, ultrafiltration, etc. are common these days. Different processes are better at removing some types of contaminants than others, so combinations are used. Distillation of tap water followed by treatment by activated carbon will produce quite pure water by consumer standards, but might not be pure enough for say semiconductor microfabrication or biomedical applications.


What happens to Aluminium in Autoclaved Aerated Concrete

$$\ce{2Al + 7Ca(OH)2 -> Ca[Al(OH)4]2 + 3H2 ^ + 6CaO}$$
or maybe
$$\ce{2Al + Ca(OH)2 + 6H2O -> Ca[Al(OH)4]2 + 3H2 ^}.$$
I think you'd get basically (ha) aluminium hydroxide.  This is effective because bases can dissolve the protective aluminium oxide layer that usually protects aluminium metal from corrosion.  The concrete is foamed by the hydrogen.


Intermolecular interaction between aniline and dichloromethane

According to this NIST publication (caution: link leads directly to a biggish PDF file) the dipole moment in the gas phase of dichloromethane and aniline are quite similar (1.6 and 1.5 D respectively).  These are respectable dipole moments, and neither molecule is very big, so offhand I would say an argument can be made that the dipole-dipole interaction is most important, particularly because it's orienting.

Note by the way that rating dispersion forces as the weakest should be done cautiously. It works for small molecules, but since dispersion forces scale with the number of electrons in a molecule and no other interaction does, once you get above a fairly modest size of molecule dispersion forces (along with excluded volume forces) dominate.  That's why eicosane is a solid at room temperature while water is a liquid.


How to assign priorities around the asymmetric carbon with its substituents having multiple bonds?

I would like to add a diagram to supplement @Raven's response and add a caveat about flipping structures to make R,S-assignments. The C's are the phantom carbons from the double and triple bonds. Working outward from the center of chirality, the first shell (in red) does not allow a decision to be made. At the second shell (in blue) the tertiary-butyl group drops out of the race and becomes the third priority group ahead of the hydrogen (4th priority). The third shell (in green) assigns the benzene ring as the top priority because the acetylene group has "run out of room" and must settle for second priority.
Flipping structures often result in drawing the enantiomer when one tries to place the lowest priority group in the rear. Leave the structure where it is and use your hands to determine chirality. Point your thumb from the chiral center to the lowest priority group. Curve your fingers so that they point in the direction 1 --> 2 --> 3. If you come up with 3 --> 2 --> 1, you are using the wrong hand! In this case your right hand (Rectus) will do the job.


No chart I'm aware of.  However, an informal and incomplete survey indicates most chemists to this by trial and error.

For pipe-type stir bars in flat bottom flasks/beakers, and you want a 'fast' stir, generally you want one roughly two thirds the size of your magnet arm to minimum the motor moving ahead of the magnet spin and 'flinging' it around.  Obviously this is not possible if your flask is much smaller than the motor.  If your stir bar is too close to the size of your vessel, you have 'catching' issues. with the sizes.

For egg-type stir bars in round bottom flasks, the concavity of the surface will help, but not eliminate 'flinging'.If you need a fast stir, you need a bigger stir bar.  Personally, I've never needed larger than 3/4" because of my use preference, (also do not typically use roundbottoms larger than 1L).

Ultimately, however, remember stir bars are for mild to moderate stirring.  If you need vigorous, you are better served with a mechanical stirrer.  Additionally, if you are using very large (5L+) flasks, mechanical stirring is best.


How is spiro[4.4]nonatetraene antiaromatic

The orbital interaction between the 2 butadiene units in spiro[4.4]nonatetraene is referred to as spiroconjugation.  Note that the 2 rings attached through the common or spiro carbon are perpendicular to one another.  Each cyclopentadiene ring contains a 1,3-butadiene unit.  In the figure below, I've drawn in the molecular orbitals for the highest occupied molecular orbital ($\Psi_2$) of 1,3-butadiene.

It is a bit easier to analyze the situation using a Newman projection.  In the Newman projection on the right I only show the 4 carbons (and their P-orbitals) that are directly attached to the central spiro carbon, since these atoms and orbitals are the only ones close enough to interact between one ring and the other.  Notice how all of the through space interactions between nearby P-orbitals is stabilizing (indicated by arrows).  That is, all of the overlaps are construcive (bonding}, either + + or - -.  This indicates that $\Psi_2$ has the correct symmetry to interact with itself.  Look at the molecular orbitals (MO) for 1,3-butadiene and convince yourself that $\Psi_1$ and $\Psi_3$ do not have the appropriate symmetry to interact with $\Psi_2$ (e.g. there are no + + overlaps between the carbons attached to the spiro carbon, only + - overlaps that cancel out; and energetically, $\Psi_4$ is to far away to interact.  When the two $\Psi_2$ MOs interact, as usual, new MOs are formed; one is stabilized and one is destabilized and the destabilized orbital is always slightly more destabilized than the stabilized orbital is stabilized.  The net result in this case is that spiroconjugation produces a (small) net destabilization and the molecule is termed antiaromatic in a spiroconjugation sense.

On the other hand, consider spiro[2.4]heptatriene. Now the only interaction of consequence (e.g. where the interactions are of the ++ or -- type rather than +-; said differently, only $\Psi_2$ of the diene and $\ce{\pi_^{*}}$ of the ethylene have the appropriate symmetry to interact) occurs between the HOMO of the diene ($\Psi_2$) and the LUMO of the ethylene in the 3-membered ring.

(image source)

In this case, as the orbital mixing diagram indicates, there is a net stabilization of the pi electrons that results.  Spiro[2.4]heptatriene is said to be aromatic.

Spiroconjugated compounds containing 4n+2 pi electrons will turn out to be stabilized (aromatic), while those with 4n pi electrons will be destabilized (antiaromatic).  Remember, these effects are small, but they can be seen in UV spectral shifts and even in solvolysis reaction rates of related compounds.  If you google spiroconjugation or spiroaromaticity you can find links to the original articles that appeared in the late 60's, as well as some more recent information on the subject.


Does carbonated water change over the course of several days, or was my container leaky?

The difference is in what is leaking. Gas leaks out a lot easier than fluid. The container is rated not to leak liquid, not that it won't leak air. So if you turned it upside down with 35 psi inside (a carbonated drink) the fluid won't leak out.

So turn the Klean Kanteen flip-top growler upside down. ;-)


Why is silver nitrate used for cyanide titration?

Due to the hardness of copper ions relative to silver ions, copper ions will bind hydroxide more readily in the presence of cyanide than silver ions do. Once ligated by $\ce{OH^-}$ ions, $\ce{Cu^{1+}}$ will oxidize to $\ce{Cu}^{2+}.$

The softer acid, $\ce{Ag^{1+}}$, doesn't do this nearly as quickly, especially not in the presence of a softer ligand like cyanide.


Product of treating 1,3‐dibromocyclobutane with alcoholic KOH at elevated temperatures

The two equivalents $\ce{HBr}$ are eliminated from two different molecules, one from the starting material and one from an intermediate. In the first step, $\ce{HBr}$ is eliminated under cleavage of the cyclobutane ring, which is favored by the release of ring strain. Elimination of $\ce{HBr}$ from 1-bromo-1,3-butadiene finally yields but-1-en-3-yne.


In High Performance Liquid Chromatography, why are ratios of solvents used?

Not sure if I can fulfill Ed's "real answer". I like the word you used for tuning solvent polarity- a knob. Modern students may understand this better.

There is no theoretical restriction in chromatography to use multiple solvents or a single solvent. For example in gas chromatography, you always use a pure gas. The reason for using a mixture of solvents in liquid chromatography originates from the concept of "general elution problem of chromatography" which basically says that there is no single column or a single mobile phase which can separate everything. Not only separation scientists use a single mixture composition of mobile phase (isocratic), they can also continuously change the mobile phase composition as a function of time. This is called gradient elution.

When you do real separations, you have no idea beforehand about the solubility of analytes injected into the column. If these injected solutes are not soluble in a pure solvent they will permanently absorb on the column. A real sample may contain 30-40 compounds. Say a petroleum sample, literally have thousands and thousands of different molecules. Now think yourself: will they all dissolve in 100% hexane?


pH calculation - adding distilled water after titration

So I presume you can work out the initial H-H equation. I got pH 5.8? might be worth checking.
Then use the equation $$c_1 \cdot V_1 = c_2 \cdot V_2$$
Where $c_1 = [\ce{H+}]$

$[\ce{H+}] = 10^{-\mathrm{pH}}$

so I got $[\ce{H+}] =10^{-5.8} = 1.58489 \times 10^{-6}$

Now we can rearrange our original equation: $(c_1 \cdot V_1)/V_2 = c_2$

$c_1=1.585\times10^{-6}$

$V_1=120\ \mathrm{ml}=0.12\ \mathrm L$

$V_2=(120\ \mathrm{ml}+880\ \mathrm{ml})= 1\ \mathrm L$

$c_2=1.902\times10^{-7}$

$\mathrm{pH}=-\log[\ce{H+}]$

$\mathrm{pH}= -\log(1.902\times10^{-7})$

$\mathrm{pH}= 6.72$

Hope this helps, Its been a long time since I did anything like this so let me know how it goes and if this is correct. cheers.


Why is potassium cation harder than copper?

$\ce{Cu+}$ has a kernel of $\ce{K+}$, with an extra 10 protons in the nucleus and a $\mathrm{3d^{10}}$ electron shell around it. $\ce{Cu+}$ is smaller because it has more positive charge pulling those electrons in. But size isn't everything!

The first ionization potential of potassium ($\pu{4.34 V}$) is much lower than the first ionization potential of copper ($\pu{7.73 V}$). But from there on, the situation changes: Second ionization potential: $\pu{31.63 V}$ for $\ce{K}$ vs $\pu{20.29 V}$ for $\ce{Cu}$; third ionization potential: $\pu{45.72 V}$ for $\ce{K}$ vs $\pu{36.83 V}$ for $\ce{Cu}$; fourth ionization potential: $\pu{60.91 V}$ for $\ce{K}$ vs $\pu{55.2 V}$ for $\ce{Cu}$; and fifth ionization potential: $\pu{82.66 V}$ for $\ce{K}$ vs $\pu{79.9 V}$ for $\ce{Cu}$. It is not till the sixth ionization potential that potassium is again easier to ionize than copper: $\pu{100.0 V}$ for $\ce{K}$ vs $\pu{103 V}$ for $\ce{Cu}$.

So, although the smaller size of the $\ce{Cu+}$ ion suggests that it would be harder (after all, it's stuffed with more electrons!), it turns out that the $\ce{Cu+}$ ion has a softer (i.e. more easily ionized, therefore more easily polarized) outer skin.

Perhaps this is not predictable unless you take into account more than just ionic radii, but at least it is explainable.


Which salt will best be able to buffer the addition of HCl?

To pick the best salt in this case, we should pick the least acidic cation and the most basic anion.
While sodium cannot dissociate into $\ce{H^+}$ ions, ammonium can, so the ammonium salts will be more acidic than their sodium counterparts.
In addition to being the most basic ($\mathrm{p}K_\mathrm{b}(\ce{CO3^{2-}})=3.67$,  $\mathrm{p}K_\mathrm{b}(\ce{CH3COO^{-}})=9.24$, $\mathrm{p}K_\mathrm{b}(\ce{Cl-})=20.3$), carbonate is also a diacidic base whose conjugate acid is also a stronger base than the other two anions ($\mathrm{p}K_\mathrm{b}(\ce{HCO3-})=7.63$).
So, sodium carbonate is the best choice.


Balmer's formula and lagurre polynomials

When solving Schrodinger equation different potentials are included depending on the problem at hand, for example, particle in a box, harmonic oscillator, H atom etc. Then boundary conditions are applied; for a particle in a box, for example, the wavefunction has to be zero at either end of the box and this ensures that the solutions are quantised, i.e the wavefunction depends both on position x and an integer n and the energy also depends on an integer.

The different solutions to different types of differential equations were worked out many many years ago, and often by French mathematicians consequently the polynomial solutions obtained often carry their names, Laguerre, Hermite, Legendre and Chebychev (although he was Russian). These polynomials all depend on position x and one or more integers. The Hermite polynomials, for example, which form solutions to the harmonic oscillator, are often labelled as $H_n(x)$ where different polynomials are found with  $n=0,1,2 ...$, e.g. $H_0(x)=1, H_1(x)=1, H_2(x)=4x^2-2 \cdots$.

In solving the H atom the radial and angular solutions are separated and each solved independently. The associated Laguerre polynomials ($L_n(x)$) are used in finding the radial solutions, and the angular solutions  depend on the spherical harmonic polynomials which themselves depend on the associated Legendre polynomials $P_l^m(\cos(\theta)$.

(Notes. The polynomials are often described by recursion formulae meaning that once the first couple are known the rest are easily found. The associated polynomials are found by repeated differentiation of the polynomial, via Rodrigues' formulae. The polynomials are orthogonal.)


7-Hydroxymitragynine

The extraction and crystallization of the parent alkaloid, mitragynine, from dried leaves of Mitragyna speciosa (Kratom), has been described in great detail by A. H. Beckett, E. J. Shellard and A. N. Tackie in Planta Med., 1965, 13, 241-246 (DOI).

(Interestingly, the procedure also furnished speciofoline, a spiroindolo indolizidine alkaloid.)

Leonurine

A first extraction was seeming reported by S. Kubota and S. Nakajima in Nippon Yakubutsuqaku Zasshi, 1930, 153. I don't have a DOI and the source is most likely in Japanese.

I don't have personal experience with the extraction of guanidines from biological material and would probably rather try to synthesize leonurine, following the procedures described in Tetrahedron, 1969, 25, 5155-5161 (DOI).

General guidelines

I have given some hints on extraction in an answer to a question on the use of TLC in alkaloid identification.


Why do atoms stay nearer to each other even when delocalized electron moves away

There is a significant quantum probability a delocalized electron is there, what in average attracts atoms together, compared to the state there was no such an electron.

E.g. for benzene, the distance is shorter than for a single bond, but longer than for a double bond, being effectively 1.5 bond.


Which form of chromium oxide is formed naturally in the atmosphere?

An investigation of the phase diagram reveals that the only stable chrome oxide near room temperature is $\ce{Cr2O3}$.  $\ce{Cr3O4}$ appears, but only near 2000K.  There is no stable chrome analogue to $\ce{FeO}$ on the phase diagram.

Thus, the one stable oxide that will form under normal atmospheric conditions is $\ce{Cr2O3}$. However, note that it is easy to find mixed metal-oxygen compounds with high mutual solubility.

If you are interested in the thermodynamics of the $\ce{Cr-O}$ system, you could start with M. Kowalski and P.J. Spencer, Calphad 19(3) 229-243 (1995), which covers the $\ce{Cr-O, Fe-O}$ and $\ce{Ni-O}$ systems


The algebraic problem in the equilibrium of weak acids and bases

This sort of derivation is a just painful, but straightforward. As Poutnik mentioned it is just a series of equations where you eliminate variables by substitution.
So start with the basics.

$\ce{HA <=> H+ + A-}\tag{A}$
$\ce{BOH <=> B+ + OH-}\tag{B}$
$\ce{H2O <=> H+ + OH-}\tag{C}$
Given:

$\mathrm{C_A}$ is the concentration of the HA reagent
$\mathrm{V_A}$ is the volume of the HA reagent added to the solution
$\mathrm{C_B}$ is the concentration of the BOH reagent
$\mathrm{V_A}$ is the volume of the BOH reagent added to the solution

So there are six unknowns: $\ce{[HA], [A-], [BOH], [B+], [H+]}$ and $\ce{[OH-]}$
Thus we need 6 independent equations to solve the problem.
Step 1
$\mathrm{K_a} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}\tag{1-1}$
$\mathrm{K_b} = \dfrac{\ce{[B+][OH-]}}{\ce{[BOH]}}\tag{2-1}$
$\mathrm{K_w} = \ce{[H+][OH-]}\tag{3-1}$
$\ce{[H+] + [B+] = [OH-] + [A-]}\tag{4-1}$
$\ce{([HA] + [A-])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_A\cdot C_A}}\tag{5-1}$
$\ce{([BOH] + [B+])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_B\cdot C_B}}\tag{6-1}$
So now for the 6 equations just eliminate equations one at a time.
Step 2
So for example we want to know the final equation to only have the unknown $\ce{[H+]}$.  Rearrange equation (3-1) to:
$\ce{[OH-]} = \dfrac{\mathrm{K_w}}{\ce{[H+]}}\tag{3-2}$
Now in equations (1-1), (2-1), (4-1), (5-1), and (6-1) substitute for $\ce{[OH-]}$ and equation (3-1) is eliminated.  So now you're left with 5 equations and 5 unknowns.
$\mathrm{K_a} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}\tag{1-2}$
$\mathrm{K_b} = \dfrac{\ce{[B+]\cdot \mathrm{K_w}}}{\ce{[BOH][H+]}}\tag{2-2}$
$\ce{[H+] + [B+] = \dfrac{\mathrm{K_w}}{\ce{[H+]}} + [A-]}\tag{4-2}$
$\ce{([HA] + [A-])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_A\cdot C_A}}\tag{5-2}$
$\ce{([BOH] + [B+])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_B\cdot C_B}}\tag{6-2}$
Obviously as you continue it gets more and more tedious, but the process is straightforward.

The point is that you have 6 unknowns and 7 equations below. So all of the equations can't be independent. With a problem that is going to be this tedious I like to write out everything explicitly so that:
(1) I don't get confused
(2) So that I can check my work step by step.


You picked a difficult question because this question involves acid-base reactions. It is no longer a simple "double displacement" reaction where the cations and anions "swap partners".

  If I were to combine solutions of $\ce{(NH4)2CO3}$ and $\ce{HNO3}$ in water, the only ionic compounds that could form are $\ce{H2CO3}$ and $\ce{NH4NO3}$ right?

No, not really. You have to realise that the protonation of $\ce{CO3^2-}$ occurs in two stages: firstly, to $\ce{HCO3-}$, and then only to $\ce{H2CO3}$. Also, the $\ce{NH4+}$ ion is acidic, and can react with water to form $\ce{NH3}$.

On top of that, $\ce{H2CO3}$ will spontaneously decompose into $\ce{H2O}$ and $\ce{CO2}$, the latter of which is released as a gas.

Since these species are all soluble in water, and soluble ionic compounds exist as ions in water, a more appropriate way to think about this question is "what species are present", not "what ionic compounds are present". Species is just a general term that includes ions as well as molecules (since, for example, $\ce{CO2}$ and $\ce{NH3}$ are not "ions").

  But since both are soluble no precipitate would form correct?

That is correct.

  So does that mean that no reaction occurred?

No, there is a reaction happening because of the $\ce{CO2}$ gas formed. You will observe a lot of effervescence if you were to mix these two together.

However, if this question did not involve $\ce{CO2}$... Let's say you are reacting $\ce{NaCl}$ with $\ce{KBr}$. Since all the possible products ($\ce{NaBr}$ and $\ce{KCl}$) are soluble, you could say that in this case, no reaction would occur.

  Does it also mean that an ion like $\ce{NO3-}$ will never react in an aqueous solution because it is always soluble in water?

No. You could say it will never be precipitated out. However, $\ce{NO3-}$ can also react in other ways apart from precipitation. For example, it can be used as an oxidising agent. One famous example is the dissolution of a copper coin in concentrated nitric acid.


I found a very old book that used this notation to describe a plot where $X$ many cubic centimeters of $\ce{HCl}$ are added to a solution.


Why does fluorine stabilise a carbocation?

Fluorine belongs to the same period as carbon. Carbocation is electron deficient which has its two $\mathrm{2p}$ orbitals vacant. One electron of carbon in $\mathrm{2p}$ orbital goes into the formation of $\ce{C-F}$ σ bond. $\ce{F}$ has $\mathrm{2p}$ orbital containing a lone pair which can be donated to the $\mathrm{2p}$ vacant orbitals of carbocation. Same size $\mathrm{2p}$ orbitals of carbon and fluorine overlap effectively to form a π bond.


Resonance is a remnant of valence bond theory, which is necessary because it is impossible to describe delocalised bonding within a localised bonding scheme. Resonance structures are not real. One Lewis structure describes one electronic configuration with fully localised electrons. For many molecules it is completely sufficient to describe the bonding situation with one such structure in a first order approximation. For many other molecules this is completely insufficient. For $\ce{NO2-}$ this is the case, hence you need a superposition of more than one structure to get closer to a qualitative description. This caveat is not necessary in molecular orbital theory, since this is designed to treat delocalisation.

Nitrite anion

One question therefore arises from the above: Are the $\ce{N-O}$ bond lengths equal because there is resonance, or is there resonance because the $\ce{N-O}$ bond lengths are equal?

This is obviously more of a philosophical question, but it shows the dilemma of the whole concept. The bonding of any given molecule is a result of many different factors; to name but a few: repulsion of the nuclei, repulsion between the electrons, attraction between electrons and nuclei.

For $\ce{NO2-}$ there is one key question, that gives us a very important clue as to why the bond lengths are equal: Why should one oxygen atom be different from the other? In fact the molecule is symmetric (C2v), therefore the bond lengths must be equal.
If you try to write a Lewis structure that respects this symmetry, you will end up with an electron sextet on nitrogen (and a formal charge of +2), which certainly can only be an insufficient approximation. If you write a Lewis structure with fewer formal charges, keeping symmetry, you will notice that nitrogen will have 10 electrons. That would mean that it would have to use d-orbitals, which simply are not available for nitrogen (more on that another time). There is another way to write a Lewis structure, however, this one has to break the symmetry, and you'll notice that there is a second one, which is equal. In order to keep the symmetry you introduce resonance, to state, that the "true" electronic structure is a superposition of (at least) those two configurations.

TL;DR: The $\ce{N-O}$ bonds in the nitrite anion are equal, because the oxygens are equal.

Nitrous acid

It is false to say that in $\ce{HONO}$ is no resonance. And it is also wrong to say that a lack of resonance makes the bonds unequal. If the solution manual claims this, it is wrong.

  Why can $\ce{HNO2}$ not have 2 possible structures where the single and double bonds switch between the two oxygens?

These are valid resonance structures of nitrous acid. You have to keep in mind though, that resonance does not change the geometry of a molecule (or molecular structure). The contribution of the Lewis structure with fewer formal charges to the total wave function will be much higher than the contribution of the other structure in this case. This is not a general statement, there are molecules where the charge separated Lewis structure has a higher contribution to the wave function.

By analogy to the anion case, there is one key question, that gives us a clue about the molecular structure: If you bind a proton to one oxygen, are both oxygen really equal anymore? In this case the symmetry of the molecule is reduced to Cs. Therefore the bonds, cannot be equal.
For this molecule you will have no problem writing a single Lewis structure that satisfies every imposed criteria.
To answer the initial question: The bonds in $\ce{HONO}$ are unequal, because the proton is bound to one of the oxygens. It is a very polar, but covalent bond. The electron density of the $\ce{N-O}$ bond is therefore shifted towards the $\ce{H-O}$ bond, and simultaneously reduced for the $\ce{N-O}$ bond when compared to the anion. Therefore the $\ce{HO-N}$ bond is longer than the $\ce{N=O}$ bond (I implied the most contributing Lewis structure here).

TL;DR: The $\ce{N-O}$ bonds in the nitrous acid molecule are unequal, because the oxygens are unequal.

There frequently is the notion of a most stable resonance structure. While this is very popular, it is completely wrong. I again refer to the question What is resonance, and are resonance structures real? Resonance is used to describe the electronic structure of a molecule, there is no change in the molecular structure allowed. It is necessary to understand, that only the description in terms of all possible resonance structures is in itself correct, i.e. the theoretical limit of valence bond theory, everything else is an approximation. However, not all resonance structures will have the same contribution to the wave function. Whenever someone speaks of the most stable resonance structure, it is likely that they mean the resonance structure with the highest contribution to the wave function. But even if you know what is meant, it is still wrong to speak of a most stable resonance structure, it must be corrected.

(I currently don't have the time to provide a full fledged answer, so instead I was trying to give some pointers in the right direction. I might come back to amend it whenever I have more time.)


Why does mercury not form amalgam with iron?

Iron does not form an amalgam readily, hence the use of iron electrodes in mercury switches.  Apparently, iron does form alloys with mercury under specific conditions; see http://www.madsci.org/posts/archives/2011-04/1304143502.Ch.r.html.

By contrast, mercury spilled on gold (e.g. gold-alloy jewelry) soaks in immediately, discoloring the gold, and migrates further into the bulk metal in a few hours, thus restoring the color.


Reaction of aqueous sodium carbonate with aluminum foil

It is rather:

$$\ce{2 Al + 6 H2O + 2 OH- -> 2 [Al(OH)4]^- + 3 H2 ^}$$where $\ce{OH-}$ comes either from hydroxide dissociation, either from carbonate hydrolysis.

$$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$

The reaction with carbonate would gradually slow down as the carbonate/bicarbonate buffer will kick in.

Initial $\mathrm{pH}$ is ( see notes below line ):

$$\mathrm{pH_{init}}=14 - 0.5 ( \mathrm{p} K_\mathrm{b2} - \log{ [  \ce{Na2CO3} ]} )\\=  7 + 0.5 ( \mathrm{p} K_\mathrm{a2} + \log{ [  \ce{Na2CO3} ]} ) $$

because $\mathrm{p} K_\mathrm{a2} +
  \mathrm{p} K_\mathrm{b2} =   \mathrm{p} K_\mathrm{w} = 14 $ ( at $\mathrm{ 25\ ^{°}C}$ )

As $ \mathrm{p} K_\mathrm{a2} = 10.329$:

$$\mathrm{pH_{init}}=12.17 + 0.5 \cdot  \log{ [  \ce{Na2CO3} ]} $$

..and when some $\ce{OH-}$ is spent and $\ce{HCO3^-}$ is formed:

$$\mathrm{pH}=10.329 + \log \frac {[  \ce{CO3^2-} ]} {[  \ce{HCO3^-} ]}$$

From Henderson–Hasselbalch equation

$$\ce{pH} = \ce{p}K_\ce{a} + \log \left( \frac{[\ce{A^-}]}{[\ce{HA}]} \right)  $$

If there is supposed all $[\ce{H+}]$ comes from acid dissociation and majority of acid is not dissociated, then  :

$$\ce{pH}   \overset{cca}=
\ce{p}K_\ce{a} -   \ce{pH} - \log   {[\ce{HA}]_\ce{total}}
$$

$$\ce{pH} \overset{cca}=  \frac12 (
\ce{p}K_\ce{a} - \log   {[\ce{HA}]_\ce{total}} )$$

Alternatively

$$\ce{pOH} = \ce{p}K_\ce{b} + \log \left( \frac{[\ce{BH+}]}{[\ce{B}]} \right) \\  \overset{cca}=
\ce{p}K_\ce{b} -   \ce{pOH} - \log   {[\ce{B}]_\ce{total}}
$$

Then

$$\ce{pOH} \overset{cca}=  \frac12 (
\ce{p}K_\ce{b} - \log   {[\ce{B}]_\ce{total}} )$$

$$14 - \ce{pH} \overset{cca}=  \frac12 ( 14 -
\ce{p}K_\ce{a} - \log   {[\ce{B}]_\ce{total}} )$$

$$\ce{pH} \overset{cca}=  7 + \frac12 ( \ce{p}K_\ce{a} + \log   {[\ce{B}]_\ce{total}} )$$


In water both equilibrium constants will be extremely small; the concentrations of $\ce{O^2-}$ or $\ce{H4O^2+}$ ions in water are essentially zero.

According to Cotton and Wilkinson's Advanced Inorganic Chemistry, for the hydrolysis of solid oxide ions you have

$$\ce{O^2- (s) + H2O (l) -> 2 OH- (aq)} \qquad \qquad K_b > 10^{22}$$

so you can never have any significant concentration of oxide ions in aqueous solution. As DavePhD points out, $K_\mathrm{a}$ for the hydroxide ion is less than $K_\mathrm{w}/K_\mathrm{b} = 10^{-36}$.

Now, $\ce{H3O+}$ may have some capacity to act as a base in media like $\ce{HF:SbF5}$ (see this paper on the role of $\ce{H4O^2+}$ in isotopic exchange reactions between hydronium ions). $\ce{H4O^2+}$ can exist in sulfolane solution, too (see this paper).

If $\ce{H4O^2+}$ exists in water at all, it's probably going to actually be two protons bridged and hydrated by a lot of waters, rather than an actual $\ce{H4O^2+}$ ion. It will have a much shorter lifetime than even a hydronium ion. The molecule can exist in theory, though, and its electronic structure has been studied for at least half a century (see Rosenfeld's 1964 paper including SCF calculations on $\ce{H4O^2+}$).


Why is ΔS the same regardless of whether a process is irreversible or reversible?

In an irreversible process, we can't use the integral of dQ/T to calculate the change in entropy between the initial and final states of the system, because there is entropy being generated within the system (as a result of mechanical energy dissipation to internal energy) which the integral does not account for.  However, since entropy is a function of state, we can focus on the exact same initial and final states of the system, and devise a path between these two states that does not involve dissipation of mechanical energy, and calculate the integral of dQ/T for this alternate path.  Such a path is a reversible path, and the integral along this (or any other) reversible path gives us the change in entropy.


Re-read the statement carefully. It says that the separation of opposite charges ("charge separation") destabilizes the resonating structures. The second phrase ("charge delocalisation") refers to the delocalisation of the same charge around a resonating structure.

For example, here's an example of how "charge delocalisation" increases stability:

(source)
Notice the same positive charge is delocalised throughout the conjugated system. It reduces - the unit positive charge on a single carbon atom - to a partial positive charge over several carbon atoms in the resonance hybrid, thus increasing the stability of the molecule.

Here's an example of "charge separation":

which destabilises the molecule. Remember that uncharged structures are more stable than charged structures.

I hope it helps!


Not every model is perfect. Rutherford's model suffered from the problem of electromagnetic radiation but answered important questions about the structure of atoms by showing the existence of a positively charged nucleus.

Bohr improved on this by realising that if electrons could only have certain energies then he could explain other things such as the spectra of hydrogen. As far as I am aware he did not attempt to explain why electrons are fixed in these orbits - this was only explained later by the advent of quantum mechanics. However, this does not mean that Bohr's model was useless since it accurately predicts useful things.

We do not have to know everything about why a model works in order for it to be useful. Einstein did not know why the speed of light was constant but he assumed that it was (based on all the available experimental evidence) and derived special relativity, which is an extremely useful theory. Nowadays we have quantum mechanics and general relativity which we know are incompatible but they are both exceptionally useful theories so we do not ignore them.


Mechanism of the formation of peracetic acid

You are not too far off. It is somewhat of a mixture of the two mechanisms you proposed. The carbonyl oxygen is much more basic than the non-carbonyl oxygen and will be protonated preferentially. Then hydrogen peroxide can attack, and once the tetrahedral intermediate collapses, deprotonation yields peracetic acid.


Kinetic Theory of Gases and molecular velocities of gases

Inelastic collisions are such collisions, that dissipate mechanical energy to other energy forms. It may be e.g. acoustic energy, light, thermal energy ( in macro context only ) or internal energy of an object.

The change of the internal energy in molecular context can be creation/breaking of chemical bonds, or changes of electron energy states. In usual conditions of stable enough gases, these events are very negligible, as average kinetic energy is much smaller than the energy needed to such a change.

Be aware that even fully elastic collision generally leads to a change of the mechanical energy of an object, in whole, or its translational, vibrational and rotational parts.

Imagine 2 identical balls moving at the same speed on a colliding course. After the collision, they have still identical speed if and only if their collision was symmetric. Otherwise 1 ball will slow down and the other ball will speed up.

The justification of the kinetic gas theory is the best shown on very good agreement of prediction and observed behavior of permanent gasses at normal conditions.


Can superoxide ion act as a ligand in a similar fashion as peroxide?

Here the oxygen "molecule" in its complex with hemoglobin is rendered as a superoxide ligand.  But remember, electrons are shared between the metal and the ligand; so there may be some latitude in how you can apportion them to determine "oxidation states".


#4 is a collection vessel of some sort.

The ground glass fitting would have connected to some other ground glass fitting. A set of the ears would be on the connecting piece of glassware also. Then springs would hold the two pieces together.