You'll probably find a partial answer in this post. The reversible/irreversible tag refers to what is happening to the system, not the surroundings. Events in the surroundings are as a rule treated as reversible.

You should also keep in mind that heat is a signed quantity, and that there can be different conventions with regard to what is a positive/negative heat transfer. Usually chemists say q>0 when the system receives energy as heat, and when the surroundings loses energy as heat. Since heat is regarded as the sole source of change in the entropy of the surroundings (assumed to be an inifinitely large reservoir at constant T$^\ast$), we write $$dS_{surroundings} = -\frac{dq}{T}$$ for an infinitesimal heat transfer and $$\Delta S_{surroundings} = -\frac{q}{T}$$ for a finite heat transfer.

$^\ast$ An infinite reservoir can provide or accept energy in the form of heat without temperature change (or any other change in its state).


Amino acids: why is charge found by comparing pH and pKa and not pH to pKa + log [A-]/log[H+]?

The properties you listed are correct, and all of them follow from the relationship described by the Henderson-Hasselbalch equation.
The equation shows a close relationship between $\mathrm{p}K_a$  and $\mathrm{p}H$. Note that $\ce{log ([A^{-}]/[HA])}$ equals zero when $\ce{[A^{-}]/[HA]}$ equals 1, and when $\ce{log ([A^{-}]/[HA])=0}$, $\mathrm{p}K_a=\mathrm{p}H$. This means that at the equivalence point, $\mathrm{p}H$ equals $\mathrm{p}K_a$.


According to this paper$^{[1]}$, the kvas in their study started out at a pH of 6.1 prior to fermentation, then became stable at about a pH of 3.8 after 12 hours.  This is of course significantly less acidic than a cola with a pH of 2.5 (and some colas can have a pH closer to 2).

1) QUALITY CHANGES OF NATURALLY FERMENTED KVASS DURING PRODUCTION STAGES, Department of Food Technology, Faculty of Food Technology, Latvia University of Agriculture, Liela Street2, Jelgava, Latvia


Energy required to break the "delocalised electron cloud" of benzene

Let's sum it up :)

When cyclohexene is hydrogenated, $\mathrm{-120\, kJ\cdot mol^{-1}}$ is released.
When 1,3-cyclohexadiene is hydrogenated, $\mathrm{-232\, kJ\cdot mol^{-1}}$ is released. That's pretty close to 240, so it looks that the energies for the hydrogenation of the double bonds sum up. So far, we're cool.

Based on Kekule's model of alternating single and double bonds in benzene, we would now assume that the energy for hydrogenation is $\mathrm{ 3 \times -120 = -360 \, kJ\cdot mol^{-1}}$.

But as your textbook says it's only $\mathrm{-208\, kJ\cdot mol^{-1}}$.

We must conclude that Kekule was almost right, but the bonding situation in bezene is special. The energy of the molecule is much lower than that of a hypothetical cyclohexatriene. And when it's lower in energy, less is released upon hydrogenation. The difference of $\mathrm{152\, kJ\cdot mol^{-1}}$ is attributed to resonance stabilization.

Edit 1 If that's still a bit unclear with the signs of the energies, we can do some reverse experiments in mind.

Let's take cyclohexane and rip adjacent hydrogen atoms off to generate unsaturated molecules.

We need $\mathrm{120\, kJ\cdot mol^{-1}}$ to make cyclohexane.
We need $\mathrm{232\, kJ\cdot mol^{-1}}$ to make 1,3-cyclohexadiene.
If benzene would be cyclohexatriene, we would need $\mathrm{360 \, kJ\cdot mol^{-1}}$.

But we need less energy to make benzene, because it is more stable than a hypothetical cyclohexatriene. Actually, it is even more stable than 1,3-cyclohexadiene!

Edit 2

  Benzene's actual structure is stable by 152 KJ w.r.t. Kekule's structure. That 152KJ is used for breaking down the delocalised electron cloud of benzene.

Guess what? The teacher is just repeating with other words what we have already worked on our own.

Imagine that you want to convert benzene to the hypothetical Kekule structure.

It's a process that would "consume" energy, $\mathrm{152\, kJ\cdot mol^{-1}}$ to be precise.

Why is that? It's because benzene has the delocalized $\pi$-system that aromatic molecules use to have. If you want to overcome that and push the stable molecule uphill and turn in into the less stable (= not having the delocalisation) cyclohexatriene, you have to do some work.


Breaking/reducing water adhesion with an electric field

On the question relating to breaking/reducing water adhesion with an electric field, I suspect, the answer is that magnetic fields can have an impact.
Support, here is a 2008 paper published in Journal of Applied Physics, The surface tension of water under high magnetic fields, to quote in part:

The surface tension of water-air interface was measured by means of a surface-wave resonance method under magnetic field up to 10 T. It was found that the surface tension increased linearly with an increase in the square of the magnetic field.

Of note, background on the concept of surface tension, per Wikipedia, to quote:

At liquid–air interfaces, surface tension results from the greater attraction of liquid molecules to each other (due to cohesion) than to the molecules in the air (due to adhesion).
There are two primary mechanisms in play. One is an inward force on the surface molecules causing the liquid to contract.[1][2] Second is a tangential force parallel to the surface of the liquid.[2] The net effect is the liquid behaves as if its surface were covered with a stretched elastic membrane.

So, per the cited paper, surface tension increased linearly with the square of the magnetic field. Per Wikipedia, surface tension relates directly to cohesion, as opposed to adhesion.
As such, the question likely correct cites increased magnetic fields as reducing water's adhesion (as it apparently relates to promoting surface tension/cohesion).


Is Bromide anion acting as a base in reaction with ethylamine and methylbromide?

The bromide anion is acting as a counter anion to balance out the ammonium charge in step 1.

Given sufficient heat, step 2 can occur, with bromide acting as a weak base.  This is likely to produce gaseous produces which would drive the equilibrium as the gasses escape.

Mechanistically, this is possible, but synthetically, it is more likely another base would be added in a typical synthesis.


Zinc(II) hydride is not at all an ionic compound rather the elements are arranged in a one-dimensional network bonded covalently.  From wikipedia article of zinc(II) hydride:

  New evidence suggests that in zinc(II) hydride, elements form a
  one-dimensional network (polymer), being connected by covalent bonds.(...)Solid zinc(II) hydride is the irreversible
  autopolymerisation product of the molecular form, and the molecular
  form cannot be isolated in concentration.

  Molecular zinc hydride in the gas phase was found to be linear with a
  Zn-H bond length of 153.5 pm. Quantum chemical calculations predict
  the molecular form to exist in a doubly hydrogen-bridged, dimeric
  groundstate, with little or no formational energy barrier.The dimer
  can be described as di-μ-hydrido-bis(hydridozinc), according to IUPAC
  additive nomenclature

As it is not an ionic compound, it does not dissociate in aqueous solution, rather it gets hydrolyzed slowly. If you want to decompose into component element, just heat it at 90℃.

  Zinc(II) hydride slowly decomposes to metallic zinc and hydrogen at
  room temperature, with decomposition becoming rapid if it is heated
  above 90°C.

  $$\ce{ZnH2 → H2 + Zn(0)}$$

  It is readily oxidised and is sensitive to both air and moisture;
  being hydrolysed slowly by water .


A mole of atoms is the number of atoms whose total weight equals the atomic weight?

Your textbook is inaccurate.

First of all, it should be total mass, not total weight.

The following statement is correct: A mole of atoms with an atomic weight of 12 has a total mass of 12 unified atomic mass units.

But you cannot say: A mole of atoms is the number of atoms whose total mass (not weight!) is the atomic weight. (wrong!) although it might look very similar.

The atomic weight is an alternate name for relative atomic mass. It is "The ratio of the average mass of the atom to the unified atomic mass unit." (source) It has no unit.

A mole is a unit of amount. "The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles." (source)

So your textbook sentence should end: A mole of atoms is the number of atoms whose total mass is 0.012 kg if they were all carbon-12 atoms.

The unified atomic mass unit (see definition of atomic weight) is "one twelfth of the mass of a carbon-12 atom in its ground state" (source).

So, a mole of atoms is the number of atoms whose total mass is the atomic weight (relative atomic mass) multiplied by one twelfth of the mass of an carbon-12 atom.

Or, a mole of atoms is the number of atoms whose total mass is the atomic weight (relative atomic mass) multiplied by the unified atomic mass unit.

I find these statements much harder to understand than the original definition ("a mole is as many elementary entities as there are atoms in 0.012 kilogram of carbon-12").

What might have some application: One mole of atoms with the atomic weight $x$ have a total mass of $x$ amu.


What is the nature of the oxide of boron?

The most common oxide of boron is boron trioxide, $\ce{B2O3}$, which is the anhydride of boric acid:

$$\ce{H3BO3 -> HBO2 + H2O}$$
$$\ce{2HBO2 -> B2O3 + H2O}$$
$$\text{net} \ \ce{2H3BO3 -> B2O3 + 3H2O}$$

As the anhydride of an acid, boron trioxide is acidic. It reacts with water to regenerate the acid:

$$\ce{B2O3 + 3H2O -> 2H3BO3}$$

Boric acid's acidity does not derive from the dissociation of a proton (or abstraction of a proton by water), as does many acids, for example sulfuric acid:

$$\ce{H2SO4 <=> H+(aq) +HSO4- (aq)}$$
$$\ce{H2SO4 + H2P <=> H+(aq) + HSO4- (aq)}$$

Boric acid (written now as $\ce{B(OH)3}$, which is suggestive of its structure) instead reacts with water by forming a dative complex which is the active acid. A water molecule associates with the empty $p$ orbital on boron. This difference in behavior is enabled also by the electropositive character of boron. The acidic oxides of some metals behave similarly.

$$\ce{H2O + B(OH)3 <=>H2OB(OH)3 <=>H+(aq) + B(OH)4^- (aq)}$$


Esterification of a carboxylic acid and the production of water

The mechanism of the reaction doesn't involve the direct exchange of ions in the way that, e.g., a metathesis/double replacement reaction might. Take a careful look at the reaction mechanism from the Wikipedia page. Notice that one of the intermediates (the product after step three) is a tetrahedral complex with two hydroxyl ($\ce{-OH}$) groups. One of those hydroxyls is indeed the original hydroxyl of the carboxylic acid moiety, but the other one is actually formed from the protonated oxygen of the carbonyl. Either one of them could, with equal likelihood, be protonated in a subsequent step to provide the water molecule that serves as a leaving group. In the image that you linked, it's convenient to circle those specific atoms, but that shouldn't lead you to believe that the mechanism of the reaction is in fact a simple exchange of ions.

The question that you didn't ask directly, but perhaps implied, is: why is protonation of the alcohol by the carboxylic acid, followed by nucleophilic attack by the carboxylate on the carbon of the protonated alcohol, not a viable mechanism? First, a protonated alcohol is more acidic than a carboxylic acid by six or seven orders of magnitude, which means that the initial proton transfer would be slow and unfavorable. Second, the resulting ester is apt to be hydrolyzed eventually (albeit slowly without catalysis, but then the whole reaction proceeds very slowly), yielding the original carboxylic acid and alcohol.


How much oxygen is formed in this first order reaction?

If we assume that the reaction takes place in a large amount of inert gas (like nitrogen) and that the reaction is first order in $\ce{N2O5}$, we can reach the numerical answer given by neglecting the volume change (the reaction increases the number of particles, but if there is a large amount of inert gas, the volume change will have little effect on partial pressures). The path given by Maurice is fine (no reason to figure out the partial pressure first). If you convert the first-order rate constant into a half-life (18.6 h), you can quickly estimate that a bit more than half of the reactant will be gone after 20 h.

If we assume, like the OP, that you start with pure $\ce{N2O5}$, the rate law still applies, but the integrated rate law does not (because while the reaction goes on, dilution also happens through the volume increase). You could numerically integrate, or try to find the integrated rate law that includes the dilution effect. Overall, you would expect the reaction to go slower, i.e. produce a smaller amount of oxygen. Because the final volume is larger, the partial pressure will be lower (compared to the inert gas case) by even more.

The value calculated by the OP is incorrect for either case because it assumes an integrated rate law that does not apply for the assumed conditions.


It would be a total waste of energy.

Remember that the Sun is sending orders of magnitude more energy than humanity can produce. A significant fraction of that is UV, so the Sun is already breaking up more Methane than humanity could do by adding laser diodes or whatnot.

We might be able to do more at the ground, where the Sun's UV is mostly filtered.
It would still be an enormous feat to merely reach the Sun's UV level.

However, there is a big reason NOT to do it at the ground: UV does not only break up methane, but also oxygen. Part of that will recombine to ozone, which is an irritant (in the medical sense, i.e. extended-time exposure will cause damage no matter what).
(I do not know whether the oxygen bonds are stronger or weaker than the methane bonds. If methane has the weaker bonds, one could use an UV wavelength below the oxygen energy. You'd still break up bonds of any atmospheric molecule that has weaker bonds than methane, such as pheromones and whatnot; it is entirely unclear what consequences that would have.)


So I will answer my question. In Organic Chemistry Portal it is written that Hiyama coulping requires an activating agent such as fluoride ion or a base because crucial for the success of the Hiyama Coupling is the polarization of the Si-C bond. Activation of the silane with base or fluoride ions (TASF, TBAF) leading to a pentavalent silicon compound is a first necessary step. The whole mechanism is given here:

On the other hand, Sonogashira coupling doesn't require any activation and it happens preferentially in the above mentioned synthesis.


Frequency Factor of the Arrhenius Equation

The short answer is yes, we can imagine effects on the frequency factor. Without going into too much detail, the frequency factor combines two conceptual elements - the frequency of molecular collisions if reactants are present at 1 M or 1 bar (called z in collision theory) and the probability that such a collision occurs with the correct spatial orientation that a reaction is possible (called $\rho$ in collision theory).
Let's start with $\rho$. A catalyst that binds multiple reactants can do so in such a way that they are optimally spatially oriented to react. Examples can be found in enzymes which are often large enough to bind the reactants in very specific conformations. So in addition to stabilizing the transition state of the reaction (thus lowering the activation energy), they can also increase $\rho$.
Now let's consider z. This one is a bit more complicated. In homogeneous catalysis, we could imagine z actually decreasing, since we now need the reactants and the catalyst all coming together via sequential collisions. In heterogeneous catalysis, the situation is even more complicated, since the mathematics of colliding are more complicated when the participants are in different states.


Tabulated absorption spectra of greenhouse gases?

Contributing to the first half of the answer: The MPI-Mainz UV/VIS Spectral Atlas of Gaseous Molecules of Atmospheric Interest (public access here) allows you to search by chemical name, sum formula, CAS registry number, InChi key or author.

The output is a plot, annotated with the literature reference, e.g.

(reference)

It does not cover IR data to cover the second half of your question.  However, why not asking them if they in turn would know about a complementary database and letting the users of ChemSE know about their answer?

(Otherwise, if affilated with a research institution with a department of chemistry, knowing a chemical identifier like the CAS RN could help you to retrieve spectral data in Elsevier's Reaxys [rooted in Beilstein and Gmelin database], too.)


Here is an example of an entire online GoogleBook dedicated to the topic Microwave Assisted Organic Synthesis, apparently based on some 2,000 papers.
On page 3 of the cited reference, notes to quote:

Interaction between microwave radiation and solutions of polar molecule polar molecules may be adequately described using classical models, which may be derived from Maxwell's equations.

And further:

Although this analysis has concentrated on the oscillating components of the electromagnetic field of the microwave radiation,  it is also possible to get interaction between the oscillating magnetic field and the magnetic dipoles in the sample. The high-heating rates for Fe304 may be attributed to such interactions.

With respect to benefits, per page 23 to quote:

Reaction rates are generally very high and yields in many cases can be greatly improved, as competing side reactions can be minimized...

I would also add that the use of microwave-based dielectric heating is, at times, more than just fast and efficient and heating, as it can, in the presence of activated carbon, for example, introduce surface-based radicals into the reaction mix. See, for example, Generation of hydroxyl radical in aqueous solution by microwave energy using activated carbon as catalyst and its potential in removal of persistent organic substances.
As such, I described the general use of this new heating technology as microwave-assisted synthesis, and not per se, as catalytic, albeit depending on the reaction mix, it can be (per last cited reference).


As TAR86 said pure oxygen is problematic.

A little something from the Wikipedia page on oxygen toxicity

  Severe cases can result in cell damage and death, with effects most often seen in the central nervous system, lungs and eyes. Oxygen toxicity is a concern for underwater divers, those on high concentrations of supplemental oxygen (particularly premature babies), and those undergoing hyperbaric oxygen therapy.

Turns out nitrogen is also toxic at high pressure see nitrogen narcosis.

Helium is used primarily because it's cheaper as compared to other alternatives and is not narcotic.

Other gases are too heavy and they diffuse into the bloodstream.

For more information see this PDF on exotic diving gases.

Fun Fact
Argon is used but can you guess why?

   Used to inflate scuba suits, helps in insulating them.


Python command line tool to get chemical structure

There are multiple approaches in Python.

My suggestion would be to use a cheminformatics library like Open Babel or RDKit to convert from SMILES (for example) to 3D coordinates.

If you want to grab from chemical databases, I can suggest two approaches:

CIRPy - Uses the NIH chemical resolver to convert from names, SMILES, etc. into 3D structures.
Webel - This is a web-based cheminformatics tool. I've used it in the past, but I'm not sure if it's still maintained.

There are other databases, including my PQR and PubChemQC that offer QM-optimized geometries.

The catch with databases is that you might want the geometry of a molecule that's not in the database. In that case, Open Babel or RDKit is a better solution.

One other caveat. Nothing I've indicated above does a very good job with metal-containing species. If you want ferrocene, that's a trickier problem with current solutions.


What are the products formed when nitrogen dioxide reacts with water?

Both of the reactions are correct. However, the second reaction is written by 'adding' two reaction equations.

When nitrogen dioxide($\ce{NO2}$) is dissolved in water, it produces a 1:1 mixture of nitric acid($\ce{HNO3)}$ and nitrous acid(HNO2). $$\ce{2NO2(g) +H2O(l)->HNO3(aq) +HNO2(aq)}$$
i.e.$$\ce{2NO2(g) +2H2O(l)->H3O+(aq) +NO3-(aq) + HNO2(aq)}$$

However, since nitrous acid is unstable in any environment except very cold solution, it decomposes slowly into $\ce{NO}$ and $\ce{HNO3}$:
$$\ce{3HNO2(aq)->2NO(g) +H3O+(aq) + NO3-(aq)}$$

Your second reaction equation is obtained by adding these two reaction together.

Reference:

Housecroft, C. E.; Constable, E. C. Chemistry, 4th ed.; Pearson, 2010, p 777.

Note:

I wrote the $\ce{H+}$ ions as $\ce{H3O+}$, so there  is an extra water molecule on the left side of the equation, in each case.


Can lattice enthalpy only be determined by the Born-Haber cycle?

Because you can't take an ionic solid and convert it into a gas of ions. That's not a physically realistic process. You have to infer it using a Born-Haber cycle and other quantities that can be determined experimentally.


Formal charge of oxygen in protonated alcohol

I guess you need the answer using Lewis dot structure, so I used the same.


How can we accurately carbon-date samples from La Brea?

According to Radiocarbon dating of petroleum-impregnated bone from tar pits at Rancho La Brea, California. Science  30 May 1969:
Vol. 164, Issue 3883, pp. 1051-1052 :

  Extensive bone dating was not possible before a method was devised for the
  removal of the impregnating petroleum...Bone is now routinely dated from its collagen


What is the major side product of the reaction between 2-bromo-2-methylpropane and hydroxide ion?

On addition of hydroxide to t-butyl bromide there are two competing reactions, substitution and elimination. Depending on the conditions one of these reactions will predominate but both will always be present. If you are taking the $S_{N}1$ as the main reaction then the elimination ($E1$) side product will be 2-methylpropene, which is a gas at room temperatures and pressures.


Use the given information to write the formula of this compound

For a molecule AB2, 3.5g of A represents one molar fraction, and 8.00g B represents 2 molar fractions (or 4.0+4.0). Therefore, a direct ratio can be given as 3.5:4.0, or 1:1.14.  This means a molecule AnBm will give a mass ratio for A:B of n:1.14xm

For a molecule AB, for every 1g of A, you will have 1.14g of B.

For a molecule AB2, for every 1g of A, you will have 2.28g of B.

For a molecule A2B3, for every 1g of A, you will have (1.14x3/2) 1.71g of B.

etc etc

So for 5.0g of A, in order to have 17.1g of B, you will have 1.14 x (m/n) = 17.1/5

=> m/n = 17.1 / 5 / 1.14 = 3

This means m/n=3, or AB3 (alternatively it could also be A2B6)


Is sulfuric acid still hazardous after it dries?

Sulphuric acid won’t dry — meaning evaporate; what water does when one typically thinks of drying — at any noticeable rate. It will, however, given time penetrate deeper into the material if the material is porous or susceptible to degradation.

Depending on how much sulphuric acid was actually spilt, I suggest either mopping up the area with lots of water (if not much was spilt) or first semi-generously applying dry baking soda to the area. Once the baking soda has stopped generating bubbles immediately on application, you can use normal amounts of water for wiping up.

While I would love to suggest you get the sloppy housemate to do it, I think you might be disinclined because they again will do it sloppily.


Unknown salt formed from electrolysis of urine

The main components of urine are water and then urea.  Water makes up 90% to 95% of urine with urea making up about 50% of the remainder.  It's likely that your reaction is mainly the hydrolysis of water.  In that case, you are forming hydrogen gas at cathode and oxygen gas at the anode.  Also, some $\ce{NaCl}$ will be electrolyzed and form $\ce{Cl2}$.

As you stated, you are likely oxidizing copper, probably making $\ce{CuCl2}$.  Most zinc salts that you could be forming in your reaction would be white.  If urea is being electrolyzed, then you would also be forming nitrogen and carbon dioxide at the anode.  It's possible that at least some of the migrating bubbles contain $\ce{Cl2}$ gas which then could form solid $\ce{CuCl2}$.

You could do a quick test of some aspects of my answer by observing the hydrolysis of water using your system.  Just conduct your procedure exactly as you do for urine, but instead of urine use water.  The main salt present in urine is $\ce{NaCl}$, whose concentration varies quite a bit but is on the order of 3 g/L.  So just conduct your experiment on a 3 g/L $\ce{NaCl}$ in water solution and compare your results to the urine experiment.


Are these chemical reactions with amino acids possible, and if so, what are they called?

Reaction 1

The product is called carboxylic acid anhydride. If you make sure the amino groups don't react (protecting and deprotecting) and get rid of water to prevent back-reactions, this reaction is possible.

Reaction 2

The product could be called a keto-enol tautomer (here, the enol form). The keto form is much more stable, so this product - even if made as an intermediate - would isomerize to the standard peptide bond.


What does "percentage by mass" mean in the context of a compound dissolved in a solution?

is the definition of percentage by mass different than what I'm using?

Basically, the questioner wants you to calculate mass by volume percentage contrary to your assumption that the question asks for a mass by mass percentage. The existence of a lot of terms referring to the same mass percentage can be a common confusion.

Nevertheless, as a rule of thumb, you'd say, I advise you to go for the calculation of mass by volume percentage unless the questioner explicitly mentions or you've been given enough data.

  Find the percentage by mass of copper in the leach solution.

Now that you already have the concentration of $\ce{Cu^2+}$ ions with you, you just need to calculate the number of moles, then the mass and finally divide it with the volume of the mixture.

  Also, how does $\ce{H2SO4}$ react with copper compounds to form $\ce{CuSO4}$? I thought copper couldn't reduce hydrogen ions.

Oxidic sources of copper are well known for their amenability to sulfuric acid leaching, for eg. azurite, malachite, tenorite, cuprite, chryosocolla, and brochantite dissolve easily in dilute sulfuric acid. Note that too dilute concentration is refrained from since we don't want hydrous ferric oxides to precipitate (they can adversely effect copper extraction).

And as a side-note, if you remove copper sulfides and iron sulfides, the $\ce{H2SO4}$ requirement would be about five times the weight of the dissolved copper (concentrated $\ce{H2SO4}$ is a good oxidizer).

Interestingly, this question was asked in University of Cambridge international examinations October/November 2010, your question is perhaps a simplification of this:


How can the effect of a pulse sequence on a strongly coupled spin system be analysed?

The product operator formalism still applies to strongly coupled systems.  The simple "magic formula" for calculating a closed-form solution to the step-wise time evolution only applies when the terms of the Hamiltonian commute with each other, which is not the case in a strongly-coupled system.  Maybe more accurately stated: in a weakly-coupled system, the non-commuting terms can be neglected.  The other assumption that allows the removal of non-commuting terms is that the pulses are very short compared to other evolution times, so during pulse periods the free-evolution terms are omitted from the Hamiltonian.

The effects of a pulse sequence without these on such a system can still be found by computing the exponential of the matrix rather than using the shortcut.

Now a shameless plug: the SpinDrops tool (http://spindrops.org) can be used to simulate and visualize the effects of pulse sequences with arbitrary Hamiltonians. It shows an exact 1-1 mapping of coupled-spin operators to the visualization which can not be done with the vector model.  The setting 'Pure Pulses' can be disabled to include the free-evolution Hamiltonian during pulses, but incorporating strong coupling would currently need to be done by defining the Hamiltonian explicitly.


The crux of this (Russian) article is that $\ce{Cl2O5}$ is not stable enough to be isolated in a pure form, and if it were, it would be strongly acidic oxidizer.

This web site is representative of several others that simply give example reactions of the compound (the second reaction being the most common exampe):

$$\ce{NiCl2 + 3O2 -> NiO + Cl2O5}$$
$$\ce{H2O + Cl2O5 -> 2HClO3}$$

The statement in the first article and the simple fact that there is such a dearth of information about this compound suggest that it really only exists as a transient, unstable, oxidizing acid that apparently has not been isolated as a stable compound.  Furthermore, information can be found in Wikipedia and/or PubChem for $\ce{Cl2O_x}$, where $x \in 1, 2, 3, 4, 6, 7$.


Radon gas in earthquake prediction - Why Rn?

Uranium is found in low-levels in all rocks and soil.  Radon is a gaseous radioactive decay product of uranium.  As the uranium undergoes radioactive decay, radioactive radon is generated and trapped in the rocks that contain the uranium.

The earthquake theory involving radon suggests that prior to the actual quake, there is some subterranean movement where rocks are crushed, soil is uncompacted and the trapped radon is released producing a pre-quake spike in radon concentration.

So radon presents the following attributes:

it is present in all rocks and soil
if a rock is broken or if soil is disturbed, radon will be released
it is gaseous and air currents / thermal gradients will carry it up to the earth's surface producing a detectable plume
it is radioactive, this makes detecting small amounts or small changes in radon concentration relatively routine due to well-developed and very sensitive methods for detecting and accurately measuring radioactivity.
radon has a very short radioactive half-life, a bit under 4 days.  Being a gas and having a short half-life is very useful in terms of measuring radon emissions.  If a radon emission spike occurs, the gas will dissipate quickly and after about 10 half-lives (40 days) normal background levels of radioactivity will return.


Why might PM2.5 concentration increase or stay the same when SO2 is reduced?

Your research is correct: increasing $\ce{SO2}$ concentration increases $\mathrm{PM_{2.5}}$ concentration.  Gaseous $\ce{SO2}$ is photochemically oxidized to $\ce{SO3}$ which becomes sulfuric acid on contact with water vapor. This is the process for producing the atmosphere's finest aerosols, as they grow from aggregates of a few molecules.

Your final sentence is also correct: decreasing $\ce{SO2}$ concentration decreases $\mathrm{PM_{2.5}}$ concentration. The logic here is simply the reverse of that for the first statement (that increasing $\ce{SO2}$ concentration increases $\mathrm{PM_{2.5}}$).

Although the overwhelming source of atmospheric aerosols is salt particles from dried ocean spray, and there are other sources like dust storms, these produce much larger particles than those produced by the oxidation of $\ce{SO2}$. So, in a pristine ocean environment far from natural or anthropogenic $\ce{SO2}$ sources for example, you may have more larger aerosols than would be found nearer to $\ce{SO2}$ sources, so that the total particulate matter would be greater in such a region of low $\ce{SO2}$. But, the $\mathrm{PM_{2.5}}$ portion would be very low such that $\mathrm{PM_{2.5}}$ would still be decreased in this region of low $\ce{SO2}$ emission.

For the reasons stated above, the answer to your title question is that there is no significant process by which $\mathrm{PM_{2.5}}$ would increase or stay the same when $\ce{SO2}$ concentrations are reduced.


How the crystal structure varying for the same compound?

a) Your structure analysis is wrong

or

b) the publication is wrong (your looked up the wrong substance, typo in the original work, ...)

or

c) You crystallised from another solvent, at different temperature, (pressure?) and got another modification.


Why is enthalpy change = heat change only for constant pressure?

Firstly, $\int_{V_i}^{V_f} p\, dV\neq p \Delta V$ unless $p$ is constant. Hence the formula holds in that case only. Also, the definition of $H=U+pV$, hence if we wanted $\Delta H$, we should take:$$\Delta H=\Delta U +\int_{V_i}^{V_f}\,d(pV)$$
Here, of course, $d(pV)=pdV+Vdp$ using the product rule.


What causes the really bad smell of rotten potatoes?

My first guess was that it was putrescine and other polyamines---but the smell of putrefying potatoes actually comes from methyl mercaptan,  dimethyl sulfide, and dimethyl trisulfide. Check it:

Source: A. Kamiya, Y. Ose, Study on offensive odor (Report IV): A Consideration of Putrefaction and Offensive Odor of Solid Waste, Journal of Japan Society of Air Pollution, 18(5), 1983, pp  453-463.

Notice that the headspace composition changes quite a bit with time. Most of the headspace chromatography studies I found dealt with early detection of disease organisms in infected potatoes, rather than potatoes in full putrefaction mode. You'll also find many references to solanine poisoning from potatoes; solanine is a toxic glycoalkaloid, is nonvolatile, and has nothing at all to do with the foul smell and toxic gas produced by putrid potatoes.

Methyl mercaptan ($\rm CH_3SH$) has an odor described as "rotting cabbage" by ATDSR; it's one of the major contributors to the smell of farts (oh, sorry, "flatus" if we're being polite). It has an odor detection threshold as low as 1 ppb.

Dimethyl sulfide ($\rm (CH_3)_2S$) is responsible for the smell of the sea (in low concentrations); it too has a cabbagy smell.

Dimethyl trisulfide ($\rm CH_3SSSCH_3$) is present in relatively smaller amounts but it has an even stronger odor. The detection threshold is around 1 part per trillion, and it is apparently a strong insect attractant.

The gases produced by rotting potatoes are quite toxic (see Rotting potatoes in basement kill four members of Russian family).

(To keep the notes below in context, in my original answer I said that a plot point in "The Walking Dead" was that zombies couldn't smell delicious humans if they were wearing coats smeared with rotting flesh. I suggested that when the zombie apocalypse arrives, packing your pockets with putrid potatoes might work, too. Now I'm not so sure. Can zombies distinguish between polyamines and sulfur compounds? Perhaps they're stench connoisseurs.)


Why are ethers treated as substituents and not as functional groups in IUPAC nomenclature?

The priority order for functional groups is as follows:

Radicals > Anions > Cations > Zwitterionic compounds > Acids (In the order -COOH, -COO(O)H, S-derivatives, Se-derivatives, Sulphonic acids, Phosphonic acids, etc)> Anhydrides > Esters > Acid halides > Amides > Hydrazides > Imides > Nitriles > Aldehydes > Ketones > Alcohols and Phenols > Amines > Imines > hydrazine > Phosphanes > Alkene > Alkynes > Alkanes> Ethers > peroxides >  Halo, Nitro, Nitroso, Alkoxy > Oxiranes

Functional groups such as Ethers and Nitro-groups are way down the order and will not be the primary functional group more often than not, the reason behind them be named as substituents.

In your question, the alkane group is given priority over the ether as per the above-mentioned priority order for functional groups. Thus the name given is 2-Ethoxy-1,1-dimethylcyclohexane.

Let's say instead of the 2 methyl groups over there, you had a nitro group, then the ether would have become the principal functional group and the nitro group being the substituent. There's no rule saying that ethers are always to be named as substituents as per my knowledge.


Recall that SN1 heavily favors protic solvents (polar/H+). In such cases, the nucleophile to-be-leaving-group (to be Nu-) is going to be attracted to the positively charged molecules of the protic solvent (...H+) and away from the molecule it is attached to. The power of the attraction comes from the dipole between C-Nu, where Nu has the negative bit of the dipole.

At the end of this step, when the nucleophile attached group leaves, you often see "Nu-" hanging around, but really it's with the solvent, e.g. H3O+Br- , and it's fairly common to omit the solvent here and in other similar ionic situations, if I'm not mistaken.

Examples that invoke hydrogen bonding make this more apparent.

F-C...(tertiary) + H2O

The dipole here looks like (δ-)F-C...(δ+) because F is more electronegative

The F is strongly attracted to the closer H: (δ+)H-(δ-)O-H(δ+)(δ-)F-C...(δ+) (pretend the HOH is bent :)


What happens when salt water is electrolyzed with a copper anode and a copper cathode?

The chemistry here can be explained from the associated products of the electrolysis of brine. To quote an education source:

Three important chemicals, NaOH, Cl2, H2, can be obtained by electrolyzing an aqueous NaCl solution (brine). This forms the basis of the chlor-alkali industry. The diaphragm cell (also called a Hooker cell)...

As such, depending on the cell design, expect also hypochlorous acid:
$\ce{Cl2 (g) + H2O (l) <-> HCl (aq) + HOCl(aq) }$
and hypochlorite with OH- as a side product of the electrolysis of the brine:
$\ce{ OH- + HOCl -> H2O + ClO-}$
However, some of the $\ce{HOCl}$ can also engage in a Fenton-type reaction (example here) with $\ce{Cu}$ (and, to a lesser extent, $\ce{HOCl}$ acting on $\ce{Cu2O}$ resulting in some possible $\ce{CuO}$ or even a basic chloride formation) generally depicted as follows:
$\ce{Cu/Cu+ + HOCl -> Cu+/Cu++ + .OHCl-}$
where the latter radical intermediate can further break down with products dependent on the pH level (here, likely over pH 7 from $\ce{OH-}$ created in the parallel electrolysis reaction) and the chloride concentration. Source: See reactions listed in Supplement Table S1 available here by clicking on Supplement File to obtain a download.
Support: With metal spoons serving as electrodes for the electrolysis of a saltwater solution, there is clear discoloration of one of the spoon electrodes (visible at 5 minutes and 20 seconds into this science experiment video).
Bottomline in time, expect similar discoloration of a copper electrode as well from the associated electrolysis products.


Finding equilibrium constant from solubility product constant

Is it a rule that the equilibrium constant of any backwards reaction is equal to the inverse of the equilibrium constant of the forwards reaction?

Yes. The general rule actually is: For a given chemical reaction with equilibrium constant $K$:

$$\ce{aA\+bB->cC\+dD}$$

If we multiply the chemical reaction by $n$, we get $K_1$ for the following equilibrium:

$$\ce{\mathit{na} A\+\mathit{nb} B->\mathit{nc} C\+\mathit{nd} D}$$

as $K_1=K^n$.

Reversing the reaction is equivalent to taking $n=-1$, in which case we get: $K_2=\frac{1}{K}$

You must be wondering that:

  Why does the new equilibrium constant $K'=K^n$?

The answer lies in a simple observation on the Law of Mass Action. Recall that for a reaction:  $\ce{A +B —> C + D}$, $K_\mathrm c = \frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}$

Now, notice that $n\ce{A}$ can be written as $\ce{A}+\ce{A}+...+\ce{A}$ ($n$ times) Same applies on $n\ce{B}, n\ce{C}$ and $n\ce{D}$.

So, if we multiply our reaction with $n$, it actually can be rewritten as:

$$(\ce{A}+\cdots+\ce{A})+(\ce{B}+\cdots+\ce{B}) \ce{—>}(\ce{C}+\cdots+\ce{C})+(\ce{D}+\cdots+\ce{D})$$
and, thus, its new $K_\mathrm{c}'$ will be
$$K_\mathrm c' = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^n.[\ce{D}]^n}{[\ce{A}]^n.[\ce{B}]^n}=\left(\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}\right)^n=K_\mathrm{c}^n$$

Hence, proved.

PS: The previous answer proved the same result, but incorrectly. That proof has now been corrected thanks to the comment by @DavePhD below.


Enantiomers are mirror reflections, they have same physical and chemical properties. This like left and right hand.

Diastereomers have different physical and chemical properties. You can separate them on chromatography column or by crystallization.

In your example upper and bottom products are not mirror reflections.
Mirror relection of upper product looks like:

Just play with the 3D models in any drawing software to get the point.


How to light a candle with small amounts of safe to use chemicals?

Safety and ability to light a fire, as far as I understand, directly contradict each other. What starts a fire on your wish, would occasionally start a fire against your wish. The glycerol/sulfuric acid/permanganate recipe, like it or not, is probably the safest of all.

Others include:

White phosphorus, not only flammable but also severely toxic by itself.
Butyllithium and many other organometallic compounds that would ignite instantly on air; that's not quite what you want, anyway.
Hydrogen against platinum black (danger of explosion).

I definitely would not try any of these at home.


Is it possible to derive integrated rate law algebraically?

Calculus is pretty much just a shorthand for algebra (plus the whole limits concept), so the answer is technically yes.

Your teacher might be intending for you to use the pattern @Klaus Warzecha mentioned:

  For a function $f(x) = x^n$, the primitive integral is $F(x) = \vert\ln(x)\vert$ if $n=−1$, otherwise $F(x) = \frac{1}{n+1}x^{n+1}$

So in that sense, you can use algebra if you let

$$\int_{A_0}^{A_\tau} \! \frac{1}{[A]^n} \, \mathrm{d}[A] = \frac{1}{n - 1} \left[ \frac{1}{[A]^{n-1}_{\tau}} - \frac{1}{[A]^{n-1}_{0}} \right]$$

for $n \neq 1$

and

$$\int_{A_0}^{A_\tau} \! \frac{1}{[A]} \, \mathrm{d}[A] = \ln \frac{[A]_\tau}{[A]_0}$$

for $n = 1$

Note that this only works for single-species rate equations - if you have multiple species the integration result changes. Note that this also isn't really using algebra - it's memorizing and applying one of many general integration results. Which is what you do in integral calculus.

In any case, it doesn't seem fair to me to expect you to know how to derive integrated rate laws if integral calculus isn't a prerequisite for your course. If it is a prereq, then now you are seeing why. If it isn't, then what your teacher is asking you to do is, in my opinion, unrealistic.


Let's think in terms of units. Density is mass per unit volume. Not only will different solutes have different masses dissolved into a unit volume, but the volume of the resulting solution will be different for different solutes.

When people say they can dissolve a unit of mass into a unit of volume (a common measure of solubility) they are not keeping track of the final volume of the resulting solution.

Here are the expressions for the density of the two saturated solutions (we should assume that the temperature and pressure are constant so that the density of the solvent itself is constant).

$\frac{m_{A,sat}}{V_{A,sat}} = \rho_{A,sat}$

$\frac{m_{B,sat}}{V_{B,sat}} = \rho_{B,sat}$

where $m$ and $V$ are masses and volumes, and $\rho$ is the density. $A$ and $B$ are the two solutes.

So if:

$\rho_{A,sat}{}\gt{} \rho_{B,sat}$

that could mean, in one limit where

$V_{A,sat} = V_{B,sat}$,

that

$m_{A,sat}{} \gt {}m_{B,sat}$.

Which is how I'm guessing you're thinking. If the volume change of solvation was similar for two solutes, one could trust that the density of the saturated solution with the (mass) solubility.

But there's another limit where there is a large difference in the volume change of solvation, between the two solutes. Then if:

$\rho_{A,sat}{}\gt{} \rho_{B,sat}$,

as before, and

$V_{A,sat}{} \lt{} V_{B,sat}$,

then it's possible that the unit mass dissolved in the two cases was equal,

$m_{A,sat} = m_{B,sat}$,

so that the two solubilities would be equal.

So no, with the above definitions of solubility and density, one cannot assume that the density of a saturated solution corresponds to the solubility.


Why is the bond length of CO+ less than that of CO?

Unfortunately, the arguments presented by buckminst and Uncle Al aren't completely right. The MO schemes are correct but the HOMO-$\sigma$ orbital ($s_{\sigma}^{*}(5\sigma)$ in buckminst's diagramm, $\sigma_{3}$ in Uncle Al's diagram) is not antibonding but slightly bonding in character because there is some mixing with the $\ce{p}$ atomic orbitals of the right symmetry (in this respect the MO scheme of Uncle Al is better than the one of buckminst because it indicates this mixing).

So, the original problem still persists.
One possible explanation for the bond shortening after ionization is that the ionization leads to a shift of the $\ce{CO}$-electron-polarization (on ionization an electron is lost from the mostly $\ce{C}$-centered HOMO-$\sigma$ orbital and this leads to a formation of a positive partial charge on the $\ce{C}$ atom).
This strengthens the covalence of the $\ce{CO}$-bond and thus reduces the bond length (since the HOMO-$\sigma$ orbital is only slightly bonding there is not much bonding lost by taking away one of its electrons - the lost bonding is outweighed by the gain in covalency).
You can think of this strenghtening of the bond covalence in the following way: Two atomic orbitals can interact better (form stronger bonds) if their energies are close. Without the positive partial charge on $\ce{C}$ the AOs of $\ce{O}$ lie energetically a lot below the AOs of $\ce{C}$ (this is most pronounced for the $\ce{s}$ AOs which form the  HOMO-$\sigma*$ orbital). But with the positive partial charge on $\ce{C}$ the AOs of $\ce{C}$ are shifted down in energy and thus their energies are closer to the related AOs of $\ce{O}$ which leads to a stronger interaction when bonds are formed.


Why does tetrachloromethane have a higher boiling point than trichloromethane?

You also need to account for the difference in dispersion forces between the two molecules. Chlorine is much larger than hydrogen. Therefore tetrachloromethane has a larger molecular surface area which increases the intermolecular interaction strength. In this particular case, it outweighs the weak dipole interactions present in trichloromethane.


Why do we subtract the mass of one proton in mass spectrometry (negative mode)?

Interesting question, which originates from a major misconception related to pH in LCMS. Nowadays, there are dozens of ionization techniques so in the question always mention the ionization method.

Anyway, LCMS is usually done with electrospray ionization (ESI) technique and so are amino acids. First of all nobody completely knows the exact process in  electrospray ionization as yet. This is not unusual in mass spectrometry where even a common ionization technique like ESI and electron impact, the picture is not 100% clear.

Second important point is the pH in ESI does not determine whether a negative ion or a positive analyte ion will be formed. I also used to have this misconception a decade ago about pH. The final ionization process in ESI takes place in a gas phase. However it is well known that certain ion can suppress ionization very badly. Don't try to connect mobile phase pH with the state of ionization of the molecule in the gas phase.

In negative ion, you are simply observing a net negatively charged species. Nothing

For details, I'd encourage you to read "pH Effects on Electrospray Ionization Efficiency" in the Journal of the American Society for Mass Spectrometry· Go to the journal website and locate the paper with this title. I am not providing a link so that students develop the habit of searching themselves.


What is generally meant by distinct line spectrum?

Your intuition is right: all the transitions are possible.

But the language is a little vague and might just be a simplification.

The emission spectrum of hydrogen atoms contains a lot of lines which are often grouped into series (which are named Lyman, Ballmer and Paschen, for example). What groups the series is the lower energy level in the transition responsible for the line: the Lyman series are transitions to the lowest energy level; Ballmer is transitions to the second lowest level...). A good summary of all this is given here.

Your book may have been assuming you only counted the lines corresponding to the final drop to the ground state (the Lyman series) ignoring the intermediate drops to other energy levels. If not, it is clearly wrong as all the transitions are possible.


Is hybridization of the cyclopropyl anion sp5?

Background
Charles Coulson was the originator of Coulson's Theorem, a useful tool for the chemist.  It allows you to make "back of the envelope" estimations of hybridization if bond angles are known. Conversely, if one knows the hybridization from, say, $\mathrm{p}K_\text{a}$ or $J_{C^{13}-H}$ data, then the bond angle can be estimated.
The key equation is $$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

See these earlier answers for interesting examples where the theorem is applied:

estimating the hybidization in cyclopropane
angle of attack of a nucleophile at a carbonyl
the lone pairs in water are non-equivalent - the resultant hybridization of water

The Question
The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114°.  From this, and using Coulson's theorem
$$1 + \lambda^2 \cos(114^\circ) = 0$$
where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Using the equation
$$\frac{2}{1 + \lambda_{\ce{C-H}}^2} + \frac{2}{1 + \lambda_{\ce{C-C}}^2} = 1$$
(which says that summing the "s" character in all bonds at a given carbon must total to 1), we find that $\lambda_{\ce{C-C}}^2 = 3.74$, or the C–C bond is $\mathrm{sp^{3.74}}$ hybridized.
Now, if we remove a proton from cyclopropane and generate the cyclopropyl anion, we move from a situation where we had a pair of electrons shared between carbon and hydrogen in a $\ce{C-H}$ bond to a situation (the anion) where we have a pair of electrons residing entirely in an orbital on the carbon atom.  In other words, we have increased the electron density in this carbon orbital.  Bent's Rule tells us that the molecular geometry will change so as to lower the energy of this pair of electrons and that it will lower the energy of these electrons by increasing the s-character of the orbital they are in.  In order to increase the s-character in this orbital, we will take some s-character away from the 2 $\ce{C-C}$ bonds and 1 remaining $\ce{C-H}$ bond.  Hence, $$\lambda^2_{\ce{C-C}}>3.74$$ $$\lambda^2_{\ce{C-H}}>2.46$$ and $$\lambda^2_{\ce{C-electron pair}}<2.46$$
Said differently, the interorbital (not internuclear) $\ce{C-C-C}$ angle at the anionic carbon will decrease and the $\ce{C-C-H}$ angle will decrease making the substituents about the anionic carbon appear more puckered.


Benzyne formation with Grignard reagents

NAS reactions normally require strong electron withdrawing groups to facilitate the addition.  Normally these groups, nitro, carbonyl, cyanide, are preferentially good electrophiles for Grignard reagents (nucleophiles).

So no, organomagnesium reagents will not react via NAS.

You also mention the Benzyne mechanism (which is somewhat different).  In this case, where nucleophilic attack on other groups is not possible, I expect metal-halogen exchange due to aryl anions being slightly weaker bases than alkyl anions.  After said exchange benzyne formation is possible.


It does exist. The formula $\ce{C_{n}H_{2n+2}}$ is only applicable to straight chain alkanes. They can't have any degrees of unsaturation, which is defined as a multiple bond or a ring.


Are all naturally occuring reactions exothermic?

It's not necessary that all naturally occurring reactions are exothermic. The definition of Gibbs free energy is:

$$\Delta G = \Delta H - T\Delta S$$

A spontaneous reaction needs to have $\Delta G<0$, and therefore for spontaneous endothermic reactions to occur we only need that $\Delta H < T\Delta S$.


Why does HCl both precipitate and dissolve Silver?

The precipitation reaction is:
$$\ce {Ag+ + Cl- -> AgCl_{(s)}}$$
The next reaction is:
$$\ce {AgCl_{(s)} +Cl- -> AgCl2- }$$


Why does rock salt change colour when it's heated

"Rock salt" contains many impurities, in addition to its main ingredient, sodium chloride. Some trace chemicals in rock salt, such as lithium chloride or sodium carbonate, hold onto water, which might boil out, sizzling and popping, and if your specimen got hot enough, carbonates could decompose.

Even if you had used "table salt", rather than rock salt, chemicals are intentionally added, such as sodium ferrocyanide (to prevent caking), which has a yellowish color and potassium iodide (as a food supplement).

Had you heated pure sodium chloride, there should not have been a color change.


Look at the Wikipedia article on sourness for an explanation of that. But you also really need to really read a basic introduction to acid/base chemistry, because the question does not make much sense… Acids are, by definition, substances that increase the concentration of H+ ions in an aqueous solution.


Is the specific heat capacity of liquid iron constant?

Okay, let's answer bit by bit:

Measurement of heat capacity of liquids: this isn't particularly hard to do, and in fact it's much easier to obtain $C_p$ experimentally than $C_V$… however, once you know one, the other can be determined from it and some other thermodynamic properties. Different types of calorimeters are used for liquids, but the best-known is probably the Callender and Barnes apparatus.
“Does an accepted value exist?” and “Is the specific heat of liquid iron temperature-dependent?” — Well, I found a good paper which includes measurements and comparison with previously published values: Thermophysical Properties of Liquid Iron, M. Beutl, G. Pottlacher and H. Jäger, Intl. J. Thermophys., 1994, 15, 1323.

which confirms that considering that heat capacity is roughly constant in the 1800–4000 K range. The authors thus extract the constant pressure specific heat capacity as 825 J/kg/K, apparently in reasonable agreement with previous work that they cite:

“Or am I a poor researcher?” — Well, the above paper is the first result return by Google for a search for "specific heat" and "liquid iron", so I'll let you be the judge of that ;-)


galvanic corrosion in the wild: where is the reduction?

Comparing galvanic corrosion to the zinc/copper battery is a bit unfair. A fast corrosion rate is still much much slower kinetics compared to the expected rates of the electrochemical reaction in the battery.

The counter reaction in the galvanic corrosion case is almost always reduction of water. This is why it is much worse in conditions near the sea.

Galvanic corrosion rate can be measured and there is an ASTM standard(G71) to make the measurement.

To answer your comment below, I am trying to say that it is impossible to say if it will double or triple or just increase by 10%. The factors are complicated and hard to predict a priori. That is why it is a good idea to measure using the above mentioned standard.


Michaelis Menten derivation for 2 enzyme substrates

I will propose an outline for determining an expression for the rate of substrate conversion in the given case:

Set up the reaction with rate constants, assuming $k_{-2}\approx k_{-3}\approx0$: $$\ce{E + S <=>[k_1][k_{-1}] ES_1->[k_2] ES_2 ->[k_3] E + P}$$
Set up the differential equations describing the reaction, i.e. the rate of change for each component with time. The rate of substrate change, for example, will be $\frac{d[\ce{S}]}{dt}=-k_{1}[\ce{E}][\ce{S}] +k_{-1}[\ce{ES_1}]$.
Choose initial conditions and set up two equations for conservation of mass. For example, the initial concentration of enzyme must equal the sum of the concentrations of E, ES1 and ES2.
Make the pseudo-steady-state assumption (PSSA): assume that the concentrations of the intermediate complexes do not change on the time-scale of product formation, i.e. $\frac{d[\ce{ES_1}]}{dt}\approx \frac{d[\ce{ES_2}]}{dt}\approx0$.
Solve for $-r_S$, the negative rate of substrate conversion, obtaining the Michaelis-Menten expression describing the kinetics of the given situtation.


Can there be two precipitates in a double displacement reaction?

Yes, you were right in concluding that there are no spectator ions. How is that a problem? There is no commandment that says "Thou shalt not have two precipitates in one reaction". Indeed, such double precipitations are very much possible, though rare. The most famous example is probably the formation of lithopone.


How to calculate the equilibrium constant of acetic acid-hydroxide ion equilibrium?

Firstly, to derive this equilibrium constant ($K_\text{eq}$), you don't need to bother about $K_\mathrm{b}$ of $\ce{OH-}$. There is a a far simpler way of doing it.

So from the equation, $\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)}$, you can easily write, $$K_{\text{eq}} = \frac{\ce{[CH_3COO^-]}}{\ce{[CH_3COOH][OH-]}} = \frac{\ce{[CH_3COO^-]}[\ce{H^+}]}{\ce{[CH_3COOH]}} \times \frac{1}{\ce{[H^+][OH^-]}}= \frac{K_{\mathrm{a}}(\mathrm{CH_3COOH})}{K_\mathrm{w}}$$ So, here no $K_\mathrm{b}$ of $\ce{OH^-}$ arises and therefore we don't have to bother about it.

Secondly, if we want to find out $K_\mathrm{b}$ of $\ce{OH^-}$, At $\pu{25^o C}$, $\mathrm{p}K_\mathrm{w} = 14$, and from the equation of dissociation of water, $\ce{H2O(l) <=> H+(aq) + OH-(aq)}$, we can consider water as a weak acid, and $K_\mathrm{a} = \ce{[H^+][OH^-]} = K_\mathrm{w}$, so, $\mathrm{p}K_\mathrm{a}= \mathrm{p}K_\mathrm{w}$, and from the relation $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w}$ (note that in the relation the definitions are that $\mathrm{p}K_\mathrm{a}$ is the acid dissociation constant of a weak acid and $\mathrm{p}K_\mathrm{b}$ is the base dissociation constant of its conjugate base). The conjugate base of water is $\ce{OH-}$, so $\mathrm{p}K_\mathrm{b}=\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{a} = 0$ as, both are same for water. So, $K_\mathrm{w}= 10^{-\mathrm{p}K_\mathrm{a}} = 10^0 =1.$  Now, you can see why $K_\mathrm{b}$ of $\ce{OH-}$ is 1 at $\pu{25^o C}$

Thirdly, yes, $\ce{OH^-}$ can't be said as a strong base. It's a moderately strong base. And, the concept that $\ce{NaOH}$ or $\ce{KOH}$ being a strong base should imply $\ce{OH^-}$ should be strong is a totally wrong concept. There is no logic behind it. Those are ionic compounds and completely dissociates and, those are Arrhenius bases. But to become a base, $\ce{OH^-}$ should accept a proton, and hence its Brønsted base, and there can't be a comparison as such. If you have a weak Brønsted acid, its conjugate base must be strong and vice versa. But here the cases are completely different, and there can't be any direct implication.


Why are Tertiary Radicals more stable than Tertiary Carbocations?

The magnitude of the effects needs to be considered. To form a radical from isobutane, you break a C-H bond. To form t-butyl radical requires about 93 kcal/mol (390 kJ/mol) https://en.wikipedia.org/wiki/Carbon–hydrogen_bond. To form a cation, you break the C-H bond AND ionize the radical, which costs you another 6.7 +/- eV (154 kcal/mol or 646 kJ/mol http://www.colby.edu/chemistry/PChem/Hartree.html) https://webbook.nist.gov/cgi/cbook.cgi?ID=C1605738&Mask=20

So, while the t-butyl carbocation is indeed stabilized by the indictive effect, and hyperconjugation, the energy level at which this is occurring is much higher than the energy level of the radical.

Hyperconjugation is a qualitative concept, with a possible energy of about 5 kcal/mol.  Like a fudge factor, which explains all possible effects if you need it and apply it correctly, but in this comparison, its effect is negligible compared to bond-breaking and ionization.  https://en.wikipedia.org/wiki/Hyperconjugation


What is the hybridization of chlorine in vinyl chloride?

Disclaimer: An important point to remember is, that hybridisation always follows the molecular structure, it is never the cause for a certain structure. As such, hybridisation is an interpretation tool, nothing more.

As a rule of thumb, terminal atoms (excluding hydrogen) are almost always (maximally) approximately sp hybridised. While other hybridisation schemes can be applied, they are usually not a good representation.*

One popular example is water. In many texts the central oxygen is described as having (approximately) four sp3 orbitals, which essentially makes the lone pairs equivalent. That view, however, is not in agreement with the photoelectron spectrum, which clearly shows that the lone pairs are not equivalent. At this point I'd like to refer you to Michael Laing's article: "No rabbit ears on water. The structure of the water molecule: What should we tell the students?" (J. Chem. Educ. 1987, 64 (2), 124.).

The same principle described there applies to terminal atoms. Because of the local $C_\mathrm{\infty}$ symmetry the notion of three equivalent lone pairs is likely to be false. The most likely (main resonance contributor) structure for these atoms are (if hybridisation is at all feasible, i.e. mostly in the second period) one sp orbital forming the bond, one sp lone pair, two (perpendicular) p lone pairs.

In this particular example, I have calculated the molecule on the DF-BP86/def2-SVP level of theory and analysed it with the natural bond orbital theory (NBO6). Below you find the localised orbitals which best fit with the common Lewis structure of the molecule.
I have ordered the orbitals starting with the carbon-carbon σ- and carbon-chlorine σ-bonds on the bottom. Following that is the carbon-carbon π-bond. Continuing with the three carbon-hydrogen σ-bonds, and finally ending with the three chlorine lone pairs. The occupied orbitals are in blue and orange, while the corresponding virtual orbitals are next to it in red and yellow. The detailed analysis in numbers is at the end of the post.

In ast's answer there is the claim that chlorine must be sp2 hybridised because of a resonance contributor. While this explanation is extremely tempting because it easy to understand, you cannot judge from the possibility of resonance to the electronic structure. As another general rule of thumb you can remember: the lesser the orbitals are hybridised, the more likely the structure is.
The calculation also provides an analysis in terms of natural resonance theory. On the DF-BP86/def2-SVP level the contribution of the second configuration to the total electronic structure is only 6%. The main configuration contributes with about 90%, other configurations (ionic) make up the remaining 4%.

In summary, I'm afraid, but your book is incorrect, or at least incomplete. The description with two equivalent sp2 lone pairs is unnecessary complicated and does not accurately reproduce the electronic structure.

Truncated results of the NBO analysis (skipping core orbitals)

     (Occupancy)   Bond orbital / Hybrids
 ------------------ Lewis ------------------------------------------------------
   8. (1.99378) LP ( 1)Cl  6            s( 81.55%)p 0.23( 18.44%)d 0.00(  0.01%)
   9. (1.97199) LP ( 2)Cl  6            s(  0.18%)p99.99( 99.79%)d 0.18(  0.03%)
  10. (1.90237) LP ( 3)Cl  6            s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  11. (1.99642) BD ( 1) C  1- C  2
               ( 48.61%)   0.6972* C  1 s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
               ( 51.39%)   0.7169* C  2 s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  12. (1.99535) BD ( 2) C  1- C  2
               ( 49.24%)   0.7017* C  1 s( 40.16%)p 1.49( 59.76%)d 0.00(  0.08%)
               ( 50.76%)   0.7124* C  2 s( 44.47%)p 1.25( 55.47%)d 0.00(  0.06%)
  13. (1.96941) BD ( 1) C  1- H  3
               ( 61.15%)   0.7820* C  1 s( 29.81%)p 2.35( 70.15%)d 0.00(  0.04%)
               ( 38.85%)   0.6233* H  3 s( 99.91%)p 0.00(  0.09%)
  14. (1.98105) BD ( 1) C  1- H  4
               ( 61.33%)   0.7832* C  1 s( 30.09%)p 2.32( 69.87%)d 0.00(  0.04%)
               ( 38.67%)   0.6218* H  4 s( 99.91%)p 0.00(  0.09%)
  15. (1.98492) BD ( 1) C  2- H  5
               ( 62.36%)   0.7897* C  2 s( 32.79%)p 2.05( 67.18%)d 0.00(  0.03%)
               ( 37.64%)   0.6135* H  5 s( 99.89%)p 0.00(  0.11%)
  16. (1.99056) BD ( 1) C  2-Cl  6
               ( 43.14%)   0.6568* C  2 s( 22.75%)p 3.39( 77.00%)d 0.01(  0.25%)
               ( 56.86%)   0.7541*Cl  6 s( 18.31%)p 4.43( 81.12%)d 0.03(  0.57%)
 ---------------- non-Lewis ----------------------------------------------------
  17. (0.09516) BD*( 1) C  1- C  2
               ( 51.39%)   0.7169* C  1 s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
               ( 48.61%)  -0.6972* C  2 s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  18. (0.01232) BD*( 2) C  1- C  2
               ( 50.76%)   0.7124* C  1 s( 40.16%)p 1.49( 59.76%)d 0.00(  0.08%)
               ( 49.24%)  -0.7017* C  2 s( 44.47%)p 1.25( 55.47%)d 0.00(  0.06%)
  19. (0.00909) BD*( 1) C  1- H  3
               ( 38.85%)   0.6233* C  1 s( 29.81%)p 2.35( 70.15%)d 0.00(  0.04%)
               ( 61.15%)  -0.7820* H  3 s( 99.91%)p 0.00(  0.09%)
  20. (0.01032) BD*( 1) C  1- H  4
               ( 38.67%)   0.6218* C  1 s( 30.09%)p 2.32( 69.87%)d 0.00(  0.04%)
               ( 61.33%)  -0.7832* H  4 s( 99.91%)p 0.00(  0.09%)
  21. (0.02582) BD*( 1) C  2- H  5
               ( 37.64%)   0.6135* C  2 s( 32.79%)p 2.05( 67.18%)d 0.00(  0.03%)
               ( 62.36%)  -0.7897* H  5 s( 99.89%)p 0.00(  0.11%)
  22. (0.03245) BD*( 1) C  2-Cl  6
               ( 56.86%)   0.7541* C  2 s( 22.75%)p 3.39( 77.00%)d 0.01(  0.25%)
               ( 43.14%)  -0.6568*Cl  6 s( 18.31%)p 4.43( 81.12%)d 0.03(  0.57%)

Summary of Natural Population Analysis

                                     Natural Population
             Natural    ---------------------------------------------
  Atom No    Charge        Core      Valence    Rydberg      Total
 --------------------------------------------------------------------
    C  1   -0.43992      1.99995     4.42868    0.01129     6.43992
    C  2   -0.22801      1.99996     4.20978    0.01827     6.22801
    H  3    0.22758      0.00000     0.76996    0.00246     0.77242
    H  4    0.22498      0.00000     0.77165    0.00337     0.77502
    H  5    0.23463      0.00000     0.76237    0.00300     0.76537
   Cl  6   -0.01926      9.99998     7.00353    0.01575    17.01926
 ====================================================================
 * Total *  0.00000     13.99989    17.94598    0.05414    32.00000

Optimised Geometry DF-BP86/def2-SVP

6
symmetry cs ( E = -538.027155015 A.U. )
C        1.309538928      1.048660302      0.000000000
C        0.000000000      0.759346920      0.000000000
H        1.630332304      2.102388803      0.000000000
H        2.083175437      0.265507033      0.000000000
H       -0.793639761      1.523771533      0.000000000
Cl      -0.633947150     -0.867041806      0.000000000

* It is also important to understand, that atoms are never hybridised, only orbitals are. The presence of an sp hybrid orbital does not exclude the presence of a sp3 hybrid orbital.


Will the motion of particles always increase with the rise of temperature?

Indeed they are similar, but there is a difference. For solids, particles can only vibrate about their own position. Heating results in a stronger vibration. For liquids, particles can wander around the whole volume of the liquid. Heating increases the speed of the particles so that they can move more quickly inside the liquid, hence a higher diffusion constant.


Understanding the hybridization of cyclohexane?

I'm having quite a fair bit of trouble comprehending the hybridization
  of cyclohexane.

Atoms can be hybridized. Not entire molecules. Just atoms.

  I understand that this is a ringed structure with single C-C bonds,
  therefore all the carbons should have sp3 hybridization right?

Yes, all the carbons in cyclohexane can be said to be sp3 hybridized.

  I'm reading about how it not planar like benzene is, and adopts a chair/boat conformation which I am having trouble visualizing as well.

In benzene all the carbons are said to be sp2 hybridized because benzene is flat. Carbons which are sp2 are in the same plane with their ligands. Sp3 carbons are not flat with respect to all their ligands.

  why are the sp3 carbons not planar (not flat)?

You seem to be adept at identifying sp3 carbons. So I won't go over that.

Within the framework of VSEPR theory -

1) Atoms are made of charged particles, namely, protons and electrons.

2) Protons are at the core (nucleus); electrons are distributed around the protons.

3) Electrons repel each other (like charges repel).

So if we have 4 atoms arranged around a central carbon atom the lowest energy conformation or the configuration which will keep each of the 4 atoms the furthest apart is tetrahedral. The angle between the four ligands in a perfect tetrahedron is about 109.5 degrees. If we constrain ourselves to 2 dimensions and force all the atoms to exist in the same plane then the best angle of separation we can achieve is 90 degrees (imagine carbon to be at the center of a square with each corner of the square representing an atom). Clearly, 109 degrees > 90 degrees so the tetrahedral configuration is favored.


Polarity is an umbrella term which is popular among organic chemists. It is as vague as saying the weather is good today. "Good weather" may include, bright light, sunny, temperature, right humidity and so on.  Now read your own statement

Rosmarinic Acid -> TMF -> Sinensetin and I guess comparing compounds
of similar structure it manageable to compare bond types, groups etc
etc

Since you correctly use etc. etc. ...yes polarity includes et cetra et cetra. Its usage is discouraged by IUPAC.
The correct term is dipolarity which is a measure of the dipole moment (you can look them up in tables). Even that dipole moment is of no good because you have plenty of conformers and each conformer has a different dipole moment (at least in the gas phase).


Isn't 11-trans-retinal more stable than 11-cis-retinal?

Cis-retinal is less stable than the trans isomer.  The carotenoids ingested by the body are all trans.  Enzymes in the body convert them to trans-retinoic acid, then turn them into a fatty ester which is then converted by the enzyme retinoid isomerohydrolase to cis-retinal.  The higher energy state of the cis isomer makes it prone to photoisomerization back to the lower energy trans isomer.


Why is the boiling point of ethyl fluoride lower than that of hydrogen fluoride?

The three reasons that you present are not sufficient to explain the high boiling point of pure $\ce{HF}$. There is a fourth reason, that you have not mentioned : the Hydrogen bond. Molecules $\ce{HF}$ are interacting with one another through $\ce{H}$-bonds, exactly like $\ce{H_2O}$ molecules attract one another because of $\ce{H}$-bonds. In the liquid state, $\ce{HF}$ behaves as if it is made of heavier molecules, something like $\ce{(HF)_n}$, with $\ce{n}$ being between about $4$ and $7$. The boiling point of $\ce{HF}$ and of $\ce{H_2O}$ is extremely high when comparing to substances of nearly the same molar weight, like $\ce{CH_4}$ or $\ce{Ne}$. The boiling point of $\ce{H_2O}$ and $\ce{HF}$ does not depend significantly on the van der Wals forces. $\ce{H}$-bonds are much more important than van der Wals forces.

And there are practically no $\ce{H}$-bonds in ethyl fluoride, because $\ce{H}$-bonds happens only  when an $\ce{H}$ atom is attached to a strongly electronegative atom.


Combustion of saturated hydrocarbons vs combustion of unsaturated hydrocarbons

You are viewing it from the wrong perspective.
In the case of normal burning, Generally, the oxygen is the limiting reagent (therefore some of the hydrocarbons left behind ).
Let's take 1 mole of Oxygen.
So you have to take  0.285 mol of ethane (0.57 mol carbon )  and 0.4 mol of ethyne (0.8 mol of carbon).
It's quite clear ethyne has more carbon content for a given amount of oxygen. As if some hydrocarbon get left, Ethyne will give more carbon content.

https://www.youtube.com/watch?v=FJi4ODPQSE0

(Note: If concentration of oxygen is very high and enough temperature is given then both will burn with Blue Flame, as it is done in case of welding
https://www.youtube.com/watch?v=hKMDxag8yAo )

Now In the case of welding, welding is done in presence of excess of Oxygen, so there is no question of limiting reagent of oxygen.
What matters is Auto Ignition point, Ethyne has much lower auto ignition temperature that is 305 °C as compared to ethane which has 515 °C.

Of course, you pointed out above that for, given 1 mol of ethyne and ethane, ethyne need less oxygen to be burned, so it can be taken in quite concentrated form.

You can take maximum 80% Ethyne (concentration in air), it will burn.
Whereas ethane can't be burn as its concentration exceeds 12.5% (in air).

More concentrated heat implies Higher can be temperature achieved.


Mechanism of homologation of aldehyde to alkyne: Ohira–Bestmann reaction

This reaction is the Ohira–Bestmann modification1,2 of the Seyferth–Gilbert homologation.3,4
In the original Seyferth–Gilbert homologation, a diazomethane phosphonate is used to convert carbonyl compounds to alkynes with one extra carbon in the presence of potassium t-butoxide. I used acetophenone as the reactant here just to simplify things, but the same mechanism applies to your substrate.

In the last step, the diazoalkene may be equivalently thought of as a carbene. You could lose nitrogen to form a carbene, which undergoes a 1,2-alkyl shift to generate the alkyne. I have drawn it as being concerted but I am not entirely sure whether it is really concerted (as far as I know, there is still some degree of contention over the Wolff rearrangement — but feel free to correct me if I am wrong).
This procedure has some issues due to the strongly basic conditions. For example, in your substrate, the presence of t-butoxide can easily lead to racemisation α to the carbonyl group. The Ohira–Bestmann conditions are much milder and generally do not lead to racemisation.
As Waylander mentioned in a comment, you need to specify the solvent, but generally the Ohira–Bestmann version is carried out with potassium carbonate in methanol. The combination of carbonate and alcohol generates small amounts of alkoxide, which then attacks the carbonyl group in the diazophosphonate to produce the same reactive species:

However, I looked up your compound on Reaxys and it seems that the authors used caesium carbonate in t-butanol.5 I am very hesitant to suggest t-butoxide attacking a carbonyl group as a nucleophile, so I would go out on a limb and propose an alternative mechanism. The ketene generated can probably be intercepted by the t-butanol to generate t-butyl acetate. I don't have any evidence for this, however.

In general you had a lot of good ideas (Wittig-type chemistry and a Wolff rearrangement) and were mostly on the right track. The missing ingredient was probably the solvent, which is crucial for the generation of the reactive carbanion.
References

Ohira, S. Methanolysis of Dimethyl (1-Diazo-2-oxopropyl) Phosphonate: Generation of Dimethyl (Diazomethyl) Phosphonate and Reaction with Carbonyl Compounds. Synth. Commun. 1989, 19 (3–4), 561–564. DOI: 10.1080/00397918908050700.

Müller, S.; Liepold, B.; Roth, G. J.; Bestmann, H. J. An Improved One-pot Procedure for the Synthesis of Alkynes from Aldehydes. Synlett 1996, No. 6, 521–522. DOI: 10.1055/s-1996-5474.

Gilbert, J. C.; Weerasooriya, U. Elaboration of Aldehydes and Ketones to Alkynes: Improved Methodology. J. Org. Chem. 1979, 44 (26), 4997–4998. DOI: 10.1021/jo00394a063.

Kürti, L.; Czakó, B. Strategic Applications of Named Reactions in Organic Synthesis; Elsevier: Amsterdam, 2005; pp 402–403.

Brenneman, J. B.; Martin, S. F. Application of Intramolecular Enyne Metathesis to the Synthesis of Aza[4.2.1]bicyclics:  Enantiospecific Total Synthesis of (+)-Anatoxin-a. Org. Lett. 2004, 6 (8), 1329–1331. DOI: 10.1021/ol049631e.


Does NaOH react with K to form NaOK or KOH?

The question: Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$?

This is a redox reaction regardless of what species $\ce{K}$ would react with. The answer is two-fold: (a) Having potassium added to an aqueous solution of sodium hydroxide; and (b) Having potassium added to molten sodium hydroxide, therefore no water present (or trace amount).

Answer to situation (a) is potassium would essentially react with water (other ions in the solution are $\ce{Na+}$ and $\ce{OH-}$) based on standard potentials of following possible half-reactions.

$$
\begin{align}
\ce{K+ + e- &<=> K} &E^\circ &=\pu{-2.931V}&\\
\ce{Na+ + e- &<=> Na} &E^\circ &= \pu{-2.71V}&\\
\ce{2H2O + 2e- &<=> H2 + 2OH-} &E^\circ &=\pu{-0.8277V}&\\
\end{align}$$

Therefore, the only ionic species in the solution would be $(\ce{Na+ + K+ + OH-})$ in $\ce{H2O}$.

Answer to situation (b) is a little complicated than this. Nowadays, in electrochemical research, use of molten $\ce{NaOH}$ is the choice of the electrolyte. The choice of molten $\ce{NaOH}$ over molten $\ce{Na2CO3}$ is based on a number of advantages of the molten hydroxide electrolyte over others. The first advantage is its higher electrical (ionic) conductivity, which is important in the question at hand. The ionic conductivity of molten $\ce{NaOH}$ exceeds that of molten sodium carbonate by a factor of 1.5 [1]. Thus, it is safe to assume that molten $\ce{NaOH}$ functions as $\ce{Na+}$ and $\ce{OH-}$ ions in molten state. Additionally, $\ce{Ni}$ has already been shown to be a chemically stable material in a molten $\ce{NaOH}$ environment, and hence, nickel containers are used to produce anhydrous $\ce{NaOH}$. Concentrating an aqueous $\ce{NaOH}$ solution in $\ce{Ni}$ containers by heating up to $\pu{500 ^\circ C}$ will produce the desired $\ce{NaOH}$ melt.

According to [1], molecular oxygen is extremely oxidizing in fused hydroxides and reacts with hydroxide ions to produce superoxide and peroxide ions. The concentration of these ions is determined by the following equilibria (Note also that increasing the water vapor pressure would reduce the peroxide and superoxide ion content by favoring the equilibria to LHS, but it's not our concern here):

$$
\begin{align}
\ce{O2 + 4OH- &<=> 2O_2^{2-} &+& 2H2O }\\
\ce{3O2 + 4OH- &<=> 4O_2^{-} &+& 2H2O }\\
\ce{2OH- &<=> 4O^{2-} &+& H2O}
\end{align}$$

For example, see following Abstract:

  A thermodynamical discussion of the $\ce{Pt/O2}$ electrode in molten $\ce{NaOH}$ shows that the redox potential depends on $p_\mathrm{O_2}$, $p_\mathrm{H_2O}$, $a_\mathrm{O_2^{2-}}$ and $a_\mathrm{O^{2-}}$. Potential measurements have shown that the reaction $\ce{O2 + 2e− = O2^{2−}}$ is potential-controlling. From the electrode materials investigated—$\ce{Pt}$, $\ce{Au}$, $\ce{Ag}$, $\ce{Ni}$ and $\ce{Fe}$—the last one only exhibits a positive potential shift by comparison to the other metals; this shift has been interpreted in terms of an $\ce{O2^{2-}}$ impoverishment following corrosion.

  The peroxide which is required for the corrosion reactions can be formed only by chemical reaction of the melt with $\ce{O2}$ , not, however, by anodic oxidation, because peroxide oxidation essentially precedes the oxidation of hydroxyl ions yielding oxygen and water vapour. With anodic and cathodic polarization, peroxide is reacted with limiting current flow, the anodic reaction being less inhibited. Water vapour is cathodically reduced to $\ce{H2}$ and $\ce{OH-}$ —with limiting current flow—before $\ce{NaOH}$ is reduced. $\ce{Pt}$  corrosion is anodically enhanced by water vapour; in this context, current fluctuations are observed. In the pure $\ce{NaOH}$  melt the cathidic reduction may be assumed to involve formation of $\ce{H2}$ and $\ce{Na2O}$ ; not, however, $\ce{Na}$  deposition. [2]

Thus, in a molten hydroxide, one can expect ionic species such as metal cation ($\ce{Na+}$ in this case), $\ce{OH-}$, $\ce{O^{2-}}$, $\ce{O_2^{2-}}$, and $\ce{O_2^{-}}$ anions, and molecular $\ce{O2}$ present in this so-called basic melt in equilibria. Thus, when $\ce{K}$  reacts with any of these (most probably with $\ce{H2O}$ and $\ce{OH-}$ based on their concentrations at equilibrium) to oxidize to $\ce{K+}$, it will stay as mixture of $\ce{KOH}$, $\ce{K2O}$, $\ce{K2O2}$, and $\ce{KO2}$.

Example for ionic nature in molten salt solution molten $\ce{NaCl}$:

References:

Direct electrochemical power generation from carbon in fuel cells with molten sodium hydroxide: Chemical Engineering Communications, 2005, 192 (12), 1655-1670. DOI: 10.1080/009864490896241.
Elektrochemische messungen in NaOH-schmelzen: Electrochimica Acta, 1968, 13 (3), 625-643. DOI: 10.1016/0013-4686(68)87031-8.


Why is crystalline aspirin washed with chilled water in the purification process?

Because aspirin is less soluble in cold water. In fact, aspirin is not very soluble in water at all, which is why you are supposed to take it with lots of water. You are not "chilling the aspirin", you are preventing it from dissolving too much so that you have a better yield of your product. You wash it to remove impurities that do dissolve in water.


Ionic compounds dissolve in nonpolar compounds?

First, I think you have some misconceptions about electrical charge. There isn't neutral charge, it's either positive or negative, if it doesn't have either, it has no charge. Why are you seeing static charge interaction with neutral surfaces? It might be polarization, where a charged particle warps the electron cloud around a neutral atom generating partial charges around the neutral atom. Without knowing more about your experiment I can't say more.

Meanwhile I did an experiment of my own. I took 5 20mL scintillation vials, and added about 0.5g of NaCl to each. I then added 5mL of deionized water, methanol, 200 proof denatured ethanol, isopropanol, or acetone to each vial and tried to dissolve the salt. Only water dissolved all of the salt. I can't say there isn't ANY salt dissolved in the other solvents, but clearly the solubility is solvent dependent, and only water is polar enough to dissolve all the salt.


Calorimetry: Temperature Increase due to stirring of reactants

This is addressed experimentally, not through theory or computation. You run one experiment starting with the reactants, measuring the change in temperature over time. Then, you run a control with a similar solution that does not react (e.g. the product mixture), setting the mixer to the same speed. This will allow you to correct for any effect not due to the reaction. The stirring will be a minor contribution (depending on the tools used, not measurable at all). Other effects are large and always have to be corrected for (for example, if the reaction needs to be started with a spark and an ignition charge).

  What is the standard method for estimating/calculating this increase in temperature, given the rate at which the stirring occurs and all the other necessary information?

The work done to stir is based on the viscosity of the fluid. The faster you spin, the more work is necessary. In the classic experiment by Joule mentioned in the comments, a special stirring mechanism was used to have a large effect on the temperature. The experiment served to figure out the specific heat capacity of water (what energy is required to raise the temperature of a certain mass of water by a certain amount). You can read more about the experiment here.


Magnetic behaviour of complex compounds are determined on the basis of magnetic moment, rather the extent of paramagnetism is measured in terms of the magnetic moment μ.

μ = √n(n+2) BM , where n notates the number of unpaired electrons.

First lets give a look towards the number of unpaired electron each of the central metal atom has:

Here the question is asking to determine the complex compound which shows the lowest paramagnetic behaviour. As Co^3+ has no unpaired electron, its μ = 0, hence it is diamagnetic, and hence shows shows lowest paramagnetic behaviour among the above complexes.
And best of luck for jee'17.
Me too appearing the exam.

Hope it helps


Why is there sulfur in black powder / gun powder?

Black powder is a mixture of solids. As solids, they are not particularly inclined toward fast reactions between each other. Elevated temperature generally liven things up, because it allows diffusion of solids into each other. However, both carbon (charcoal) and potassium nitrate are relatively stable compounds, so even when potassium nitrate is melted, they still do not react immediately,  further heating is required.

Sulfur, on the other side, is quite easy to melt. Nitrates are not very powerful oxidizers, but potassium nitrate in particular is a low-melting compound. And at the moment both potassium nitrate and sulfur are molten, their interdiffusion becomes quite easy, so they ignite, and charcoal follows.

Summing it up: charcoal is almost impossible to melt and potassium nitrate is a relatively weak oxidizer and also hard to melt, so the mixture is hard to ignite. Sulfur is easily flammable, easy to melt, so it is much easier to react with potassium nitrate solid, and its presence makes mixture much more ignitable.


Why do batteries drain so quickly if they are forced to draw more current than they normally do?

The capacity of a cell is measured toward the terminal cut off voltage. If e.g.a double current is drained from a cell, the external cell voltage is decrease by the doubled voltage drop on the cell internal resistance: $$U_\mathrm{e}=U_0 - R \cdot I$$
That caused reaching the terminal voltage sooner then in the half time.
It is additionally leading to the ohmic internal energy wasting $$P=R \cdot I^2$$
But this energy loss affects the useful energy output, not the capacity ( directly).
Additionally, higher than optimal current causes lowering of the immediate cell open voltage. Limited rate of diffusion of electroactive components and/or of electrode reactions are not able to maintain electrode equilibrium potentials.
In result, even the $U_0$ (non linearly) decreases with the higher current and the terminal $U_e$ is reached after providing smaller total charge.


Strength of weak acids and bases: Is methane better base than ammonia?

You got it wrong. Acidic and basic strength of a compound are unrelated, about as much so as the man's name and his weight. Within the definition you and I are currently using (Brønsted–Lowry), acidity is about having a proton on a polar bond ready to ionize, while basicity is about having electron pairs ready for accepting a proton. A compound can have one and not the other, or vice versa, or both, or neither. Why should the two be related? They shouldn't and aren't. Well, there is one obvious exception: a strong acid can't be a strong base at the same time, otherwise it would turn upon itself. Except that, everything is possible.

Indeed, methane is a very weak acid, about 13 orders of magnitude weaker than ammonia (as attested by the difference in their pKa). In fact, it is one of the weakest acids among all compounds. This is not unexpected, given its almost non-polar bonds and the lack of stabilization in the anion. Also, methane is a very weak base. Forcing it to accept a proton was no small achievement in itself (see Methanium). This is not unexpected either, given its lack of lone pairs or negative charge.

Now you must be thinking that I am pulling your leg, because every kid knows that strong acid means weak base and vice versa. It is written in every textbook, said by every teacher, and that with great confidence, too. Can they all be wrong? No, they are right. How so?

Here's the trick: strong acid does indeed mean weak base, but not in the same compound. Look again at that equilibrium:
$$\ce{HA <=> H+ + A-}$$

See that? It is $\ce{HA}$ that acts as an acid here. And now look at the reverse reaction: it is $\ce{A-}$ acting as a base. Now if the equilibrium is shifted to the right, then $\ce{HA}$ is a strong acid, which inevitably means that its conjugate base $\ce{A-}$ is a weak base.

So it goes.


How to determine the product from my attempt of HOCl synthesis

Ryan, sorry your approach to generate the desired compound is not the best one. This post also shows the problems with the concept of an "accepted" answer on StackExchange. The answer is also wrong and misleading. Please consult your academic supervisor to help you do a proper literature search. I am not sure why are you doing all this, you did not explain in your two posts. First of all, there is nothing such as pure 8% NaOCl soln. This bleach is a mixture sodium hydroxide and NaOCl. What you are apparently titrating is the free alkali with acetic acid (which you should not be doing). Simultaneously, the hypochlorite ion is decomposing into other products.
The proper place to do inorganic synthesis research is the library, not Q&A site. A good reference is Kirk Othmer Encyclopedia of Chemical Technology to look up bleach and its properties. Any good library must have it.
In your first question you linked a paper "Hypochlorous Acid as a Potential Wound Care Agent" in the Journal of Burns and Wounds. This journal has a zero impact factor which means this is of dubious quality. There was a reason to avoid organic acid such as acetic acid in that paper. Have you read about disinfection by-products in water? Search this topic this too, because disinfection byproducts can be cancer-causing.


Why does this elimination reaction occur?

Conjugated dienes are significantly favorable to non-conjugated dienes (because of delocalization and hybridization, etc.), so this will be the major product. It is not to say that the other reaction cannot happen, but that a combination of factors leads to that product being major. All of chemistry occurs in an equilibrium of some degree, and the reaction above will become most stable in conjugated diene form.


Yes, according to the Arrhenius theory, acids dissociate in aqueous solution and release a proton ($\ce{H+}$). The Brønsted–Lowry defines acids  ($\ce{HA}$) and bases ($\ce{B}$) in such a way that their interaction is characterized by the exchange of a proton according to $\ce{HA + B <=> A- + HB}$.

However, this is about the reaction of molecules in which real hydrogen atoms are bound to other atoms.

To illustrate the idea, let's think about the reaction between acetic acid, $\ce{CH3COOH}$, and benzylamine, $\ce{C6H5CH2NH2}$. If we combine both liquids in our imagination, the following will happen:

$\ce{CH3COOH + C6H5CH2NH2 -> CH3COO- + C6H5CH2NH3+}$

The acid donates a proton to the amine, which serves as the base. In the course of this reaction, the acid is converted to its corresponding base, while the base is converted to its corrsponding acid.

You are confusing dissociation and proton exchange with a type of radiocative decay known as proton emission.


What is the temperature of heat generated from LPG gas?

LPG is mostly propane and butane both of which have adiabatic flame temperature near 1970 degrees C when burned in air.  See Flame Temperatures of some Common Gases.

So 300-400 degrees C is certainly reachable.


Electronic configuration of the actinides

It is difficult to calculate the energy levels of the orbitals beyond hydrogen. "The first experimental evidence for the filling of the 5f shell in actinides was obtained by McMillan and Abelson in 1940," from Wikipedia

So the answer, as you suspected, is that experimental evidence from spectroscopy helped determine the electron configuration of the lanthanides and actinides, in particular.


Effect of solvent in the reaction of alkyl halide with nitrite ion

Let me first clear out the difference between reaction of alkyl halide with $\ce{AgNO2}$ and $\ce{NaNO2/KNO2}$

The synthesis of nitro compounds from reaction of alkyl halides with alkali metal nitrites using DMSO/DMF as medium is known as the Victor meyer reation($\ce{AgNO2}$) or Kornblum reaction($\ce{NaNO2/KNO2}$). The products of these reaction are generally mixture of nitroalkanes and alkyl nitrites. But the main difference between them is that the % yield of nitroalkane is higher on reaction with $\ce{AgNO2}$ than that of $\ce{NaNO2/KNO2}$.Whereas % yield of alkyl nitrite is higher on reaction with $\ce{NaNO2/KNO2}$ than that of $\ce{AgNO2}$. The required mechanisms for the reactions are..

Now as for the reaction of $\ce{NaNO2/KNO2}$ with alcohol, there is no involvement of alkyl halide. Alkyl nitrite can also be prepared by adding concentrated $\ce{H2SO4}$ to the aqueous solution of $\ce{NaNO2/KNO2}$ and alcohol. The mechanism of this reaction is

Source books:

The Nitro Group in Organic Synthesis By Noboru Ono
Organic Chemistry of Explosives By Jai Prakash Agrawal, Robert Hodgson
Sourcebook of Advanced Organic Laboratory Preparations By Stanley R. Sandler, Wolf Karo


Confusion with usage of heat capacity in Adiabatic expansion

For an ideal gas, U is a function only of T (and not V).  So the derivative of U with respect to T is an ordinary derivative, and is independent of V.  So why do we call $C_v$ the heat capacity at constant volume, if, for an ideal gas, it doesn't even depend on volume?  It has to do with how we measure $C_v$ experimentally. If the volume is held constant, then the amount of work is zero and $$Q=\Delta U=C_v\Delta T$$  So, in the case of constant volume, we can measure $C_v$ by measuring the amount of heat added Q.  If the volume is not held constant, so that some work is done, we can't determine $C_v$ from the amount of heat added.


Determining the redox half-reactions of single displacement reactions

Let's take the first equation in the question by example:
$$\ce{Mn(OH)2 (s) + H2O2 (aq) -> Mn(OH)3- (s)}$$
In this equation ion $\ce{Mn(II)}$ is oxidised into $\ce{Mn(III)}$, and oxygen in $\ce{H2O2}$ (oxidation number -1) is reduced into ion hydroxide (oxidation number -2):
$$
\begin{align}
\ce{Mn^{2+} &-> Mn^{3+} +e^- \tag 1}\\
\ce{H_2O2 (aq) + 2e^- &-> 2OH^- \tag 2}
\end{align}
$$
If we multiply the first equation by 2 and add the equations $(1)$ and $(2)$:
$$\phantom{iiii}\ce{2Mn^{2+} + H_2O2 (aq) -> 2 Mn^{3+} + 2OH^- \tag 3}$$
By adding four ions hydroxide to the two sides of equation $(3)$:
$$\ce{2Mn^{2+} + H_2O2 (aq) +4OH^- -> Mn^{3+} + 6OH^- }\phantom{iiiiiiiiii}$$
By rearranging the equation $(3)$, we find:
$$\ce{2Mn(OH)2 (s) + H2O2 (aq) -> 2 Mn(OH)3 (s) \tag 4}\phantom{iiiiiii}$$
We do the same for the other equations.


Cram bonds are synonymous to the "wedged" bonds or, in this case, to "hollow wedged" bonds, which IUPAC [1, p. 1910] describes as follows (emphasis mine):

  There are a variety of obscure conventions for depicting arbitrary mixtures of diastereoisomers. The most common (but still obscure) of these conventions says that bold and hashed wedges should be used for absolute configuration, unwedged bold and unwedged hashed lines represent relative configuration and racemic character, while hollow wedges and dashed lines represent relative configuration only.

A bit more context is given in [2] (bond type 4):

  Central to the correct perception of a stereounit’s topography is the stylized generation of a vanishing point, realized by the graphic stereodescriptor, or
  stereobond, commonly called a “wedge” as illustrated in Figure 1 as 1a and 1b. Three such wedge symbols as in 3, 4, and 6 are in use, each of them capable of portraying the ligand Z in 1 and 2 as being located in the front (3 and 4) or behind (6) the picture plane. In an apparent search for further simplification, line symbols 5, 7, and 8 were invented as wedge substitutes.
  Lacking differentiation of their termini, line-type symbols appear, a priori, to be inherently incapable of serving such a purpose; nevertheless, they provide a useful function in the context of a convention. Accordingly, the stereogenic unit to which the line-type bond symbol is attached is considered to lie in the picture plane so that the symbol printed in bold (5) reveals its ligand Z in front of that plane, whereas the hashed symbols (7 and 8) are intended to show that the ligand is behind it. In other words, symbol 5 portrays the stereocenter at the vanishing point and hence functions such as 3 and 4.

  Figure 1. Perspective drawings of stereocenters with commonly used stereobonds.

References

Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. https://doi.org/10.1351/pac200678101897.
Maehr, H. Graphic Representation of Configuration in Two-Dimensional Space. Current Conventions, Clarifications, and Proposed Extensions. J. Chem. Inf. Comput. Sci. 2002, 42 (4), 894–902. https://doi.org/10.1021/ci025518w.


Why does humidifier put measurable PM2.5 in the air?

All tap water contains some dissolved solids, and "hard water" contains more, particularly calcium salts. There are also microorganisms found even in treated tap water, and some organisms can build up a biofilm inside the humidifier. Ultrasonic humidifiers are effective at putting these particulates into the air. However, as others stated, you may just be detecting fine water droplets from the humidifier before they evaporate.

To find out if it's water droplets or particulates from dissolved material, do a comparison test, using distilled, filtered, low-particulate water such as High Performance Liquid Chromatography (HPLC) grade water.

Clean out the humidifier thoroughly, rinsing repeatedly with a dilute acid safe for that machine (perhaps vinegar, but check the humidifier manual).
Rinse repeatedly with distilled water and finally with HPLC water.
Refill with pure HPLC water.

If this does not greatly increase the particulates above background, but tap water does, then you've found the source of the issue. If not, then you're probably detecting mostly water mist.

Caveat: Buying HPLC water is not inexpensive. If you can't easily make your own on-site, perhaps just leave the dehumidifier off.