So acids stay acids forever?

You know a company markets that diamonds are forever. Well may be not, a diamond is not thermodynamically stable. It may wish to turn back into graphite if you somehow provide the right activation energy.

So there is nothing special about acids. Acids are like other molecules some may remain stable for a long time and some may not.

An example is that of sulfuric acid. If someone leaves concentrated acid in an open beaker, in period of few months its volume will increase. How? It absorbs moisture from air.

If you leave conc. HCl bottle open, HCl gas will escape into air and the acid concentration will decrease.

Moral of the story: Nothing is permanent but change.


What is the reason behind the phenomenon of Joule-Thomson effect?

In a reversible adiabatic expansion or compression, the temperature of an ideal gas does change.

In a Joule-Thompson type of irreversible adiabatic expansion (e.g., in a closed container), the internal energy of the gas does not change.  For an ideal gas, its internal energy depends only on its temperature.  So, for an irreversible adiabatic expansion of an ideal gas in a closed container, its temperature does not change.  But, the internal energy of a real gas depends not only on its temperature but also on its specific volume (which increases in an expansion).  So, for a real gas, its temperature changes.  The Joule-Thompson effect is one measure of the deviation of a gas from ideal gas behavior.

ADDENDUM

This addresses a comment from the OP regarding the effect of specific volume on the internal energy of a real gas.

Irrespective of the Joule-Thompson effect, one can show (using a combination of the first and second laws of thermodynamics) that, for a pure real gas, liquid, or solid (or one of constant chemical composition), the variation of specific internal energy with respect to temperature and specific volume is given by:
$$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$The first term describes the variation with respect to temperature and the second term describes the variation with respect to specific volume.  For an ideal gas, the second term is equal to zero.  However, for a real gas, the second term is not equal to zero, and that means that, at constant internal energy (as in the Joule-Thompson effect), the temperature will change when the specific volume changes.  This is a direct result of the deviation from ideal gas behavior.


Metals with a $d^8$ oxidation state are stabilized by a square-planar molecular orbital configuration. In this configuration, the $\ce{d_{yz},d_{xz},d_{z^2}}$ orbitals are stabilized, while the $\ce{d_{xy}}$ orbital is slightly destabilized and the $\ce{d_{x^2-y^2}}$ is severely destabilized (however, in $d^8$ species this orbital is unoccupied). The amount of stability provided by the square planar configuration is far greater than the amount of stability provided by the low-spin $d^6$ configuration.

This is a good online powerpoint that shows the stabilization.


How much CO2 could you remove from the atmosphere with 2 trillion USD?

This is very hard to answer precisely, as there are many different carbon capture strategies, and economics at the scale required is quite different from our normal understanding. However, I'd love to see some attempts to at least get order of magnitude estimates, or sources with more in-depth analyses.

Here is an implementation of carbon capture and storage which is useful to consider for its simplicity rather than its real-world applicability, to give a sense of scale. One way to remove anthropogenic $\ce{CO2}$ is to do "inverse combustion", more specifically:

$$\ce{CO2(g) -> C(s) + O2(g)}\ \ \ \ \ \ \ \mathrm{\Delta H=+390\ kJ/mol}$$

Assume this process can be done with perfect efficiency, and that the only energy expense in the process is to drive the reaction forwards (that is, zero energy consumed in transportation, collection, construction, etc). According to this source, the amount of anthropogenic $\ce{CO2}$ emissions from 1750 to 2008 has totalled about $1250\times 10^9\ \mathrm{t_{CO_2}}$. Suppose you wish to remove all this carbon dioxide (only about half is in the atmosphere, the rest is trapped in the ocean or land) using the above process. This would require about $\mathrm{10^{22}\ J}$ of energy, which Wolfram Alpha suggests is about 50% more energy than can be retrieved from combustion of all global proven oil reserves in 2003. Put another way, this is about 20 times the world energy consumption in 2012, or 150 times world electrical energy production. This would put the cost of this process in the range of hundreds of trillions of US dollars, meaning two trillion USD barely makes a dent.

Is there some other process where two trillion USD is close to enough? I seriously doubt it.

Edit: Several comments and answers mention biological sequestration, which is a legitimate carbon capture strategy. I did not consider it, however, because its costs are far more complicated to calculate. My intention with this answer was to find a quick and comparatively simple way to attach an energy cost to carbon sequestration. Whether the monetary cost even makes sense at this scale (how do you define "monetary cost" when it's larger than world GDP?), I don't know.

But here's another amusing comparison, which should somewhat temper hopes that biological sequestration will be a magic bullet. The amount of anthropogenic carbon released between 1750 and 2008, ~350 billion metric tons of carbon, is comparable to a significant amount of the biomass on Earth (all eukaryotic life contains approximately 560 billion metric tons of carbon, multicellular life is a fraction of this). Thus, biological sequestration of the majority of anthropogenic carbon would be broadly equivalent to sacrificing all eukaryotic life on Earth (or ~10-30% of all living organisms by mass) in order to collect and bury carbon, then seeding the Earth back to its current biological state.

Over 250 years of constant stimulation to produce as much energy/goods from fossil fuels, we have released a lot of carbon.


Identifying organic compounds in Infrared spectrum

If you're analyzing an IR spectrum to find out what compound you have, you'll first want to look past the 1500/cm region to see what functional groups you have. For example, a 1700/cm peak will tell you you have a carbonyl, alcohols will have a peak around 3000/cm, etc. From there, pull up spectra of well known compounds with those functional groups, and compare your compound's 1500/cm to 500/cm region to them. This region is called the "fingerprint region" because it's practically unique for all molecules, so if the spectra match there, they're likely the same compound.


How does this ring expansion on a steroid system occur?

It is clear that the first step is ozonolysis with a slightly reductive workup, yielding two ketones. You seem to have already established that. The second step features a base and an alcohol as a solvent. The hydroxide can react with the ketones in two ways: 1) nucleophilic attack at the electrophilic carbonyl carbon 2) abstraction of the mildly acidic $\alpha$ hydrogen. With the methanol in large excess as the solvent, it could also potentially act as a nucleophile to attack the carbonyl group to eventually form hemiacetals. However, what seems to be the most preferred reaction pathway, based on thermodynamics, here is the aldol condensation to form the $\alpha$,$\beta$-unsaturated ketone. I will now go through the considerations that a problem solver could have made.

Why is the aldol condensation thermodynamically favoured over the other alternative reation pathways?

The key reason is that the formation of the conjugated enone $\pi$ system is quite energetically favoured. On the mechanism of the aldol condensation, Carey & Sundberg (2007) wrote the following, on  p. 684:

  In general, the reactions in the addition phase of both the base- and acid-catalyzed
  mechanisms are reversible. The equilibrium constant for addition is usually unfavorable
  for ketones. The equilibrium constant for the dehydration phase is usually favorable
  because of the conjugated $\alpha$,$\beta$-unsaturated carbonyl system that is formed. When the reaction conditions are sufficiently vigorous to cause dehydration, the overall
  reaction can go to completion, even if the equilibrium constant for the addition step is
  unfavorable.

What about the ketone-hemiacetal equilibrium?

The addition products (i.e. hemiacetals) formed between methanol and the ketones in the basic medium can easily be converted back to the reactants (i.e. reaction is highly reversible). The depletion of the ketone by it reacting in the aldol condensation pathway thus pushes the equilibrium towards the ketone form. However, this consideration is not even important in this case since the equilibrium constant for the ketone-hemiacetal equilibrium is much less than one.

There are numerous $\alpha$ hydrogens but which one would be removed?

To answer this question, usually we would have to know if the conditions provide thermodynamic or kinetic control. Not much information has been provided in this case so I believe it is reasonable to assume thermodynamic control since only one product would clearly dominate while there are various possible kinetic products.

In this case, the hydrogen is first abstracted from the terminal methyl group of the newly-formed ketone, forming the enolate ion. This then acts as a nucleophile attacking the carbonyl carbon of the other newly-formed ketone. Upon protonation, the $\beta$-hydroxyketone is formed. Finally, there is E1cB elimination to form the product, as shown in B.

Other addition products may result either from 1) formation of other enolates from abstracting other acidic $\alpha$ hydrogens or 2) formation of other possible addition products from the enolate attacking other carbonyls (possibly even those on other molecules). Firstly, many of the alternatives would result in ring sizes that would possess too much strain. In this case, a 4-membered ring could form if deprotonation took place at the other $\alpha$ carbon of the same ketone. Intermolecular reaction would not be as favoured here since the proximity of the groups simply make the intramolecular reaction "hard to resist".

Conclusion

I have just gone through some important considerations behind arriving at the structure B, as well as the mechanism of the aldol reaction. Many details provided may seem extraneous though... Hopefully, you are able to understand how the "ring expansion" took place now.

References

Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.


Is a non-compound molecule a chemical substance?

Found that the answer is 'Yes' in the IUPAC Gold Book definition of chemical substance:

  chemical substance

  Matter of constant composition best characterized by the entities (molecules, formula units, atoms) it is composed of. Physical properties such as density, refractive index, electric conductivity, melting point etc. characterize the chemical substance.


Is styrofoam soluble in gasoline or paint thinner?

Unless you're trying to make napalm, don't use gasoline.

Use acetone. It dissolves egg cartons within just a few minutes, and you can then take it out and reshape it (use gloves!). Set it out to dry and once the acetone has evaporated, it should be a (crude) solid.

Just ensure whatever container you're using to hold the acetone isn't polystyrene. Check to make sure recycling code 6 isn't on the container. I had a pretty good laugh at my teacher one day who poured acetone into a red solo cup and it went running all over the table.


Is there a readily available liquid that's more dense than water and insoluble?

Trichloroethylene is a common chemical solvent that has reasonably low toxicity, it has density ~1.5 g/ml. However, it slowly becomes restricted. Tetrachloroethylene also may be considered. Dichlormethan has reasonably low toxicity, but its BP is too low. I would not use the liquids mentioned without fume hood on regular basis. One-time exposure is probably OK, but repeating exposure may lead to cancer.  Chloroform may be considered, but I recommend to avoid it, it has accumulating liver toxicity. Tetrachloromethane should be avoided, it also has undesired toxicity.

Mercury MAY be considered, it is toxic if one is repeatedly exposed to vapours, but has very low vapour pressure and is reasonably cheap. But it has annoying tendency to produce small drops that are very hard to collect and move into smallest holes and accumulates in the system, so it clearly requires dedicated  workplace.

Liquids mentioned above are reasonably cheap, so if they are not restricted in your home country, you should be able to allow yourself one. Unfortunately, that's all ideas I got. Well, there are some more, but they either has unreasonable cost, or are too toxic. Bromine, for example, is toxic and, though not mixing with water, partially dissolves in it.


What happens to the methanol formed by the hydrolysis of methyl 2-hydroxybenzoate (oil of wintergreen) when synthesising salicylic acid?

The methanol plays no underlying role in the reaction

Without more detail of the reaction procedure I cannot give a complete answer, but my expectation is that after the HCl addition the reaction will be quenched by addition of water to either precipitate the salicylic acid or then be extracted into an immiscible organic solvent. The methanol will stay in the aqueous phase in either case.


Can we explain an electron's energy of a specific orbital using the
  Penetration and Shielding effect?

Yes, to some extent it is possible. Look no further than Slater's rules which specifies the values of the effective nuclear charge $Z^*$, and the effective principal quantum number $n^*$ that can be used to replace the nuclear charge $Z$ and the principal quantum number $n$ in the hydrogen-like atom orbital energy expression. The energy of each and every orbital can be approximated then by
$$
    E = - \left( \frac{Z^*}{n^{*}} \right)^{2}.
$$
But I would be careful calling this an explanation, since it is rather a rationalization. I wrote about the difference between those two here.


The explanation is that sufficiently strong Lewis acids like $\ce{SbF5}$ will form adducts even with very weak, traditionally non-coordinating Lewis bases like $\ce{SO3F-}$. The resulting anionic complexes are extremely stable, with the negative charge distributed over a large, highly polarizable molecule and inductively stabilized by the fluorine atoms. The formation of this complex, in turn, promotes the autoprotolysis equilibrium:

$$\ce{HSO3F + HSO3F <=> SO3F- + H2SO3F+}$$

The protonated species above is the most active molecule in sufficiently concentrated solutions. (In reality, the composition of these mixtures may be much more complex, with dimers of the form $\ce{Sb2F10(SO3F)-}$ and possibly more complex oligomeric structures. See, for example, this book.)

While there are some more recent and exotic examples of such Brønsted/Lewis acid mixtures, the idea is actually a fairly old one. Perhaps the most common example: sulfur trioxide in sulfuric acid gives a solution known as oleum, which contains a proportion of disulfuric acid molecules. The resulting mixture is considerably more acidic than pure sulfuric acid alone and has various uses in industrial applications as well as laboratory-scale synthetic organic chemistry.

As for why the oxygens cannot be subsituted with fluorines to yield an even stronger acid system, I can only give a speculative answer. It seems to me that a hypothetical molecule such as, e.g., $\ce{SF3OH}$ (if this is what you're imagining) would probably be very liable to decompose:

$$\ce{SF3OH -> SOF2 + HF}$$

In any case, I can find absolutely no relevant literature, which in and of itself says something about the stability of such a molecule (or perhaps my ineptitude, as the case may be).


How do I figure out the relative polarity of organic compounds?

General steps to figure out the polarity (or at least relative polarity) and subsequently the relative boiling point of compounds:

Acquire the correct geometry
Estimate dipole moments
Sum up estimated dipole moments
Check for other intermolecular interactions

Acquire Correct Geometry

This step requires some knowledge about the geometry of hydrocarbon compounds. But it seems that you already know about the tetrahedral bonding centers of saturated hydrocarbons. This is then important later on for the dipole forces. Never try to linearize the carbon backbone just to try to make sense of this, it will do more damage than good.

Estimate Dipole Moments

The dipole moments that you have to estimate will be those along bonds. There is a larger dipole moment in the ether than the hydrocarbon without heteroatoms, because the oxygen atom is strongly electronegative. As such, there will be a dipole along the $\ce{C-O}$ bond.

Sum Up Estimated Dipole Moments

These dipoles can be understood as vectors along bonds, pointing from positive to negative partial charges, and can be summed up for the total dipole moment of a molecule. The larger the absolute value of this dipole vector is, the larger the dipole-dipole interactions between molecules: Subsequently, the higher the forces between the molecules of this substance.

Check for Other Intermolecular Interactions

One other main intermolecular action is, as you already pointed out in your question, hydrogen bonding. This one is quite strong, but isn't found for every compound out there. Basically, it is when a highly polarized $\ce{H-X}$ bond can be formed, where $\ce{X}$ can be oxygen, nitrogen, chlorine or some other electronegative atom (with respect to carbon). The hydrogen that is sandwiched in the hydrogen bond has to be (formally) bonded to such an electronegative atom as well, otherwise it wouldn't be as free to create those bonds (as you also have realized, because the hydrocarbon hydrogens are really quite boring from an intermolecular interaction perspective).

Result

Based on the steps above, we can generate a table:

The more "Yes" a substance has, the higher the intermolecular forces. Based on these data, assumed by us, we can generate a list (from highest boiling point to lowest):

Ethanol
Dimethyl ether
Propane


How to show/represent an ionic bond visually?

Some examples are as follows,
$$
\ce{[Na]+ [Cl]-} \, , \\
\ce{[Mg]^2+ 2[Cl]-} \, , \\
\ce{3[Mg]^2+ 2[N]^3-} \, .
$$
Hopefully you get the idea, so a general rule is:
$$
\text{#ions}[\text{element symbol}]^{\text{magnitude of cation charge}} \text{#ions}[\text{element symbol}]^{\text{magnitude of anion charge}} \, .
$$


Strictly speaking, drawing a MO diagram is only possible for two entities.
That being said, I will attempt to show you the MOs and order them according to their energy.

At PBE0-D3/def2-TZVPD, using NWChem 6.6, I found the following minimum geometry (vacuum):

Figure 1: Minimum geometry found for Zundel cation ($\ce{[H2O...H...OH2]^{+1}}$) with natural atomic charges, angles (in degrees) and distances (in angstrom)

MOs 1 and 2 correspond to $\ce{1s^2}$ core electrons on oxygen and orbitals 3-10 give us the states for the 8 remaining valence electron pairs.
MO 11 is the lowest unoccupied molecular orbital.

Figure 2: Molecular orbital diagram for structure from Figure 1 at PBE0-D3/def2-TZVPD

Since the structure has no symmetry, no symmetry was assigned to any MO and the diagram above has simple labels only.
Energies (in hartree) can be found further below.

Below are images of each MO:

Figure 3: MO #3 (cutoff 0.04)

Figure 4: MO #4 (cutoff 0.04)

Figure 5: MO #5 (cutoff 0.04)

Figure 6: MO #6 (cutoff 0.04)

Figure 7: MO #7 (cutoff 0.04)

Figure 8: MO #8 (cutoff 0.04)

Figure 9: MO #9 (cutoff 0.04)

Figure 10: MO #10 (cutoff 0.04)

Figure 11: MO #11 (cutoff 0.4)

As a side note, we may obtain a finer view of the MOs by representing them as linear combinations of natural bonding orbitals (NBOs):

CMO: NBO Analysis of Canonical Molecular Orbitals

Leading (> 5%) NBO Contributions to Molecular Orbitals
======================================================
 MO   3 (occ): orbital energy =  -1.356442 a.u.
               0.426*[  5]: BD ( 1) O 4- H 7
               0.400*[  1]: BD ( 1) O 1- H 2
              -0.399*[  2]: BD ( 1) O 1- H 3
               0.396*[  4]: BD ( 1) O 4- H 6
               0.396*[  3]: BD ( 1) O 4- H 5
               0.336*[  9]: LP ( 2) O 1(lp)
 MO   4 (occ): orbital energy =  -1.327547 a.u.
               0.444*[  1]: BD ( 1) O 1- H 2
              -0.440*[  2]: BD ( 1) O 1- H 3
              -0.437*[  3]: BD ( 1) O 4- H 5
              -0.434*[  4]: BD ( 1) O 4- H 6
               0.280*[  9]: LP ( 2) O 1(lp)
              -0.272*[  5]: BD ( 1) O 4- H 7
 MO   5 (occ): orbital energy =  -0.859519 a.u.
              -0.606*[  5]: BD ( 1) O 4- H 7
              -0.502*[  9]: LP ( 2) O 1(lp)
               0.369*[  3]: BD ( 1) O 4- H 5
               0.359*[  1]: BD ( 1) O 1- H 2
 MO   6 (occ): orbital energy =  -0.839700 a.u.
              -0.562*[  4]: BD ( 1) O 4- H 6
               0.511*[  2]: BD ( 1) O 1- H 3
               0.471*[  3]: BD ( 1) O 4- H 5
               0.425*[  1]: BD ( 1) O 1- H 2
 MO   7 (occ): orbital energy =  -0.838909 a.u.
               0.533*[  2]: BD ( 1) O 1- H 3
               0.514*[  1]: BD ( 1) O 1- H 2
               0.481*[  4]: BD ( 1) O 4- H 6
              -0.469*[  3]: BD ( 1) O 4- H 5
 MO   8 (occ): orbital energy =  -0.696009 a.u.
              -0.549*[  9]: LP ( 2) O 1(lp)
               0.502*[  5]: BD ( 1) O 4- H 7
               0.302*[ 10]: LP ( 1) O 4(lp)
              -0.301*[  8]: LP ( 1) O 1(lp)
              -0.253*[  3]: BD ( 1) O 4- H 5
               0.244*[  1]: BD ( 1) O 1- H 2
              -0.227*[  4]: BD ( 1) O 4- H 6
 MO   9 (occ): orbital energy =  -0.636884 a.u.
               0.692*[ 10]: LP ( 1) O 4(lp)
               0.687*[  8]: LP ( 1) O 1(lp)
 MO  10 (occ): orbital energy =  -0.623778 a.u.
              -0.620*[  8]: LP ( 1) O 1(lp)
               0.614*[ 10]: LP ( 1) O 4(lp)
               0.344*[  9]: LP ( 2) O 1(lp)
              -0.319*[  5]: BD ( 1) O 4- H 7
 MO  11 (vir): orbital energy =  -0.201773 a.u.
               0.263*[121]: BD*( 1) O 1- H 2*
               0.262*[123]: BD*( 1) O 4- H 5*
              -0.257*[122]: BD*( 1) O 1- H 3*
               0.257*[124]: BD*( 1) O 4- H 6*
              -0.232*[113]: RY*( 1) H 7(ry*)

Figure 12: Atomic numbering for the NBO Analysis of Canonical Molecular Orbitals above


Lithium pyroborate ($\ce{Li2B4O7}$) is the salt of lithium ($\ce{Li}$) and pyroboric acid ($\ce{H2B4O7}$).

Nomenclature

The prefix ortho- designates an acid with the maximum number of hydroxyl ($\ce{OH}$) groups [reference]. For boron ($\ce{B}$), orthoboric acid is $\ce{H3BO3}$ (also written as $\ce{B(OH)3}$ to show that there are $3$ hydroxyl groups).

The prefix pyro- here seems to not match with the usual usage, so I will not be explaining the usual usage here. One common feature between this and the usual usage is that their molecular formulas are multiples of the ortho acid's molecular formulas, minus some water molecules:

$$\ce{4H3BO3 -> H2B4O7 + 5H2O}$$

Gallery

Some proposed structures of lithium pyroborate:

Some proposed structures of pyroboric acid:

The above are modified from two proposed structures of sodium pyroborate, with the first one referenced here and the second one referenced here.


Excited-states properties with traditional density functional theory

Conventional density functional theory (DFT) is strictly limited to describing the electronic density of ground electronic states. This is because the Hohenberg–Kohn theorems, on which it is based, are restricted to non-degenerate ground states (with no magnetic field). Also, because DFT aims at solving the time-independent Schrödinger equation (i.e. it explicitly finds stationary solutions), time-dependent phenomena are not accessible.

From the ground state density, any property of the system including excited state properties could in theory be derived given an appropriate functional[1] $X[\rho]$ that maps the ground state density $\rho$ onto the property $X$. However, in practice, no known functional exists that maps the ground state density to excitation energies or other excited-state properties. Thus, excited states and their properties are in practice unreachable by DFT: I suppose this is what is meant by Gross when stating them as “notoriously difficult to calculate”. I myself would have said impossible.

Finally, to give a short list of properties that can be described with TDDFT but not with DFT — are thus all excited state properties. In particular: bandplots/bandgaps, excitation energies, transition moments, photo absorption spectra, frequency-dependent response properties (optical & dielectric properties), etc.

Specifically, one has to be careful about band plots and band gaps, which are routinely reported from time-independent DFT calculations: they are a representation of the Kohn–Sham energies, i.e., the energies of the fictive non-interacting Kohn–Sham system, which has no physical interpretation at all.

[1] Note here that we are talking about a functional in the generic, mathematical meaning of the word. It is not the exchange–correlation functional.


What are the effects of adding water to a system at equilibrium?

The reaction $$\ce{NaCl~(s) <=>[\ce{H2O}] Na+~(aq) + Cl^{-}~(aq)}$$ is at equilibrium when approximately $359~\mathrm{g/L}$ sodium chloride are dissolved in water, i.e. the solution is saturated, or $c(\ce{Na+}) = c(\ce{Cl-}) = c_\mathrm{max}$.

If less $\ce{NaCl}$ is dissolved, it is only one phase, which is in equilibrium with itself and all solid salt is dissolved, therefore the equilibrium would be entirely on the right side. At that point adding water would not change the equilibrium at all, it would only dilute the whole system.

When the solution is saturated, there is still solid salt present, the equilibrium is not entirely on the right side. Adding water will disturb the equilibrium, i.e. $c_\mathrm{present}<c_\mathrm{max}$. Therefore more of the excess $\ce{NaCl}$ will dissolve, until the maximum concentration is reached again. If the solution is still saturated after that, nothing will have changed in your solution except for the volume.
If your solution is not saturated anymore, the equilibrium is entirely on the right side and the concentrations will decrease.


The cell potential is related to the change in Gibbs free energy by this equation:

$$
\Delta G^\circ= -nFE^{\circ}_\text{cell}
$$

where $n$ is the number of moles of electrons transferred in the balanced reaction equation and $F$ is Faraday's constant.

The UC Davis chem wiki has a brief explanation of the connection between cell potential, the reaction quotient (via the Nernst Equation), and the equilibrium constant. You should be able to find a more thorough explanation in a general chemistry textbook. My copy of Tro's Chemistry has a derivation of all of these relations, but to summarize the first relation:

Work against any potential is given by force times distance, and in the case of moving a charge, that works out to:

$$
w = -qE^{\circ}_\text{cell}
$$

Faraday's constant $F$ gives the charge in coulombs on one mole of electrons, which means the total charge is given by:

$$
q = nF
$$

Where $n$ is the number of moles of electrons.

We can substitute this back into the work equation to get:

$$
w = -nFE^{\circ}_\text{cell}
$$

Since $\Delta G^{\circ}$ is equal to the maximum amount of work that can be done by a chemical reaction,

$$
\Delta G^\circ = w = -nFE^{\circ}_\text{cell}
$$

In other words, saying that change in Gibbs free energy is negative is the same as saying that the cell potential is positive – in both cases we are simply looking at the total usable energy change and saying that "things move downhill" – the reaction proceeds in the direction that minimizes the thermodynamic potential energy of the system.


If a reactant appears on both sides of an elementary equation, can it be cancelled?

If the elementary step is indeed written

$$\ce{A + B -> A + C}$$

and assuming that you haven't just written in $\ce{A}$ for fun, i.e. $\ce{A}$ is actually a participant in the step, then yes, this would be considered bimolecular. This sort of step, featuring the same chemical species on both sides of the reaction, does pop up in (for example) the Lindemann mechanism.

With regard to kinetics, however, it is worth noting that steps of this kind can sometimes indicate a reaction that is catalytic with respect to $\ce{A}$. If that is the case, then the concentration of $\ce{A}$ is essentially constant and you would obtain pseudo-first-order kinetics.


Difference between reversible and irreversible thermodynamic process

The simplest answer in thermodynamic terms is "a reversible process is one in which the entropy generation term is zero."

  Yes, I understood that but can you explain what difference is there in the change in enthalpy, internal energy, heat, work, etc. if the process is reversible and irreversible.

For any process, all of these would be the same whether it was reversible or not. The only one that would differ would be change in entropy.


Of course they are useful. Perhaps you have not come across them yet, but being able to determine the oxidation state of an atom allows us to understand the properties of chemicals and how redox reactions work.

Let's just give two very simple examples. There are tons more. (I don't even want to go into organometallic chemistry, where being able to determine the oxidation state is incredibly important.)

Potassium permanganate, $\ce{KMnO4}$, has manganese in a +7 oxidation state. This means that it would have an electronic configuration identical to the noble gas argon: $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6}$, and therefore its $\mathrm{3d}$ orbitals are empty.

Now, most transition metal compounds are said to be coloured because of so-called "d-d* transitions". However, this can only occur if the d orbitals are partially filled - if there are no electrons in the d orbitals, like in $\ce{KMnO4}$, then there are no d-d* transitions available.

As such, the intense purple colour of $\ce{KMnO4}$ has to be explained via a different mechanism. In this case, it is explained by ligand-metal charge transfer.

Moving to organic chemistry. Let's say you have an acyl chloride, $\ce{RCOCl}$, and you want to convert it to an aldehyde, $\ce{RCHO}$. If you calculate the respective oxidation states of the carbonyl carbons, you get +3 and +1 respectively.

So, that tells you what kind of reagent you need to effect this transformation: you need a reducing agent. Hydrogen gas, $\ce{H2}$, is one such example of a reducing agent. Why? Well, that's because of oxidation states again. $\ce{H2}$ has hydrogen in an oxidation state of 0, and when hydrogen forms a bond to carbon, it has an oxidation state of +1.

Oxidation states are key to understanding why redox reactions work and why certain species are reducing agents (e.g. $\ce{LiAlH4}$: H(-1)) and why others are oxidising agents (e.g. DMSO, $\ce{(CH3)2SO}$, in the Swern oxidation).


What accounts for the high spin state of the complex Tris(acetylacetonato)iron(III)?

I will go into the theoretical details first and come to the practical answer for your exact case later; separated from the theory by a horizontal rule.

Approaching the complex theoretically, our first observation is that we are going to have six atoms coordinating our $\ce{Fe^3+}$ centre. Six ligands means that we should consider an octahedric environment as initial approximation. For the next bit of the discussion, I am going to ignore the fact that two ligands each belong to the same molecule and treat them all as $\ce{O^{\frac{1}{2}-}}$ or maybe as $\ce{OH-}$ since the latter is close enough.

The crystal field theory, our first approximation, considers these ligands to be point-shaped negative charges that approach an $\ce{Fe^3+}$ ion. The lonely iron(III) in vacuum will have five d-orbitals of degenerate energy; the approaching negative charges will destabilise all those orbitals that point towards them — for simplicity, let us assume that the corners of the octahedron are the axes of a coordinate system, and thus the destabilised orbitals are $\ce{d}_{x^2 - y^2}$ and $\ce{d}_{z^2}$. (The other orbitals point between the ligands and are therefore unaffected.) This will give rise to the most basic of all transition metal complex energy diagrams as shown in figure 1.

Figure 1: splitting of d-orbitals in an octahedral complex according to crystal field theory. The energy difference always corresponds to $10~\mathrm{Dq}$

Crystal field theory, unfortunately, gives us no information about the energy difference between the two sets of d-orbitals — this is important, since we need a significant energy difference if we want to keep our electrons paired at a lower level. Figure 1 chose to keep a high-spin complex simply for spin retaining. Therefore, the again most simplistic approach to analyse different octahedral complexes by their actual energy differences (remember that $10~\mathrm{Dq}$ or $\Delta$ is only a variable) is to assume a certain contribution both from the metal’s side and from the ligands’ side:

$$10~\mathrm{Dq} = g_\mathrm{M} \times f_\mathrm{L}$$

The $g_\mathrm{M}$ contribution depends on the metal and needs to be experimentally measured, similarly for $f_\mathrm{L}$ on the ligands’ side. $f_\mathrm{L}$ is normalised so that $f_\mathrm{L}(\ce{H2O}) = 1$. $g_\mathrm{M}(\ce{Fe^{III}}) \approx 1.40 \times 10^4 ~\mathrm{cm^{-1}}$. By knowing that $f_\mathrm{L}(\ce{OH-}) \approx 0.9$ we could calculate the energy difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ to be $1.26 \times 10^4~\mathrm{cm^{-1}}$ and compare that to other iron complexes to find out that this would mean high-spin. (Using a value for $\ce{acac}$ should change the result slightly but not much.) Note that a typical low-spin iron(III) complex such as $\ce{[Fe(CN)6]^3-}$ has the following value: $10~\mathrm{Dq} = g_\mathrm{M}(\ce{Fe^{III}}) f_\mathrm{L}(\ce{CN-}) = 1.40 \times 10^4~\mathrm{cm^{-1}} \times 1.7 = 2.38 \times 10^4~\mathrm{cm^{-1}}$ — almost double the difference.

However, if you put more research into this phenomenon, it quickly becomes highly dissatisfying. The approximation for $10~\mathrm{Dq}$ up there is no real good, because the same ligand has different $f_\mathrm{L}$ values for different complexes — and even worse, the same metal-ligand combination can sometimes be found in both high- and low-spin states resulting in wildly different values for $10~\mathrm{Dq}$: For high-spin $\ce{[Fe(ptz)6]^2+}$, $10~\mathrm{Dq} = 1.23 \times 10^4~\mathrm{cm^{-1}}$ while for low-spin $\ce{[Fe(ptz)6]^2+}$, $10~\mathrm{Dq} = 2.00 \times 10^4~\mathrm{cm^{-1}}$. And whenever simple approximations don’t adequately explain the true picture, we need to resort to molecular orbital theory.

The simplest approximation in molecular orbital theory gives a similar picture as crystal field theory does: We have a set of six s-type orbitals approaching our metal ion, and these six orbitals will form σ-bonds with the metal. The main difference is that we can better explain why the metal orbitals will split: The ligand orbitals transform as $\mathrm{a_{1g} + t_{1u} + e_g}$ which means that they will interact with the metal $\mathrm{e_g}$ orbitals. Since the ligands are typically lower in energy, we realise that what we are in fact observing are antibonding $\mathrm{e_g^*}$ orbitals on the metal. Thus, electrons in these orbitals should weaken the metal-ligand bond (order); or complexes lacking electrons in $\mathrm{e_g^*}$ should have a shorter metal-ligand distance. However, this approach is also unsatisfactory: negatively charged ligands such as $\ce{OH-}$ or $\ce{acac-}$ should have donor orbitals of higher energy, meaning more similar to those of the metal, meaning a greater splitting – which we do not observe.

Therefore, in a second approximation we need to add the ligands’ p-orbitals to the picture. These transform as $\mathrm{t_{1u} + t_{2u} + t_{1g} + t_{2g}}$ — we have an additional set of orbitals that we can linear combine with the metal’s $\mathrm{t_{2g}}$ orbitals. We now have additional ligand to metal π-bonding and can assume the metal-centred $\mathrm{t_{2g}}$ orbitals again to be antibonding ($\mathrm{t_{2g}^*}$). This again raises their energy levels and thus lowers the difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g}$, therefore lowering $10~\mathrm{Dq}$. We finally have an adequate approximation of $10~\mathrm{Dq}$.

Now to get back to high-spin versus low-spin. Remember that it is not favourable per se to pair electrons — when applying the aufbau principle, you must never forget Hund’s rule which tells you to fill orbitals of the same energy with parallel spins first before adding the first antiparallel spin/pairing the first electron. In certain cases such as chromium, it is even favourable to have $\mathrm{3d^5 4s^1}$ — all valence spins unpaired. This is no different when going from elements to compounds: A certain energy gain is always required to make electron pairing favourable. Oftentimes, this energy gain is simply not supplied as is the case for most $\mathrm{3d}$ transition metal complexes that are not $\ce{d^8}$. The high-spin iron(III) complexes are the rule, not the exception, for the $\mathrm{3d}$ metals.

To raise the energy difference, we need strong-field ligands. The above paragraphs tell us what they need to: σ-bonding must be strong to raise the energy of $\mathrm{e_g^*}$ while π-bonding (if present at all) must be weak to not raise the $\mathrm{t_{2g}}$ level. Cyanide is good at doing this because its σ-donor orbital has roughly the same energy as the metal, so the bonding will be largely covalent with a large difference.

If, by the way, we accidentally manage to turn a high-spin into a low-spin centre, as was highlighted above with the ptz-ligand, a number of follow-up changes will also occur: Since the $\mathrm{e_g^*}$ orbitals are no longer occupied, the ligand to metal bond order increases and the ligands come closer to the metal. Therefore again the energy difference will rise because the covalent bond will be strengthend. This is the reason why the energy difference suddenly jumps from $\approx 1.3 \times 10^4~\mathrm{cm^{-1}}$ to $\approx 2.0 \times 10^4~\mathrm{cm^{-1}}$ when transitioning between high- and low-spin states.

In the case of the $\ce{acac-}$ ligand, for starters we need to realise that oxygen (the donating atom) is rather electronegative and thus we can assume its orbitals to be rather low-lying energetically. Low σ-donor orbitals means less destabilising of $\mathrm{e_g^*}$ and therefore a lower field split.

Secondly, we always have p-type orbitals on oxygen that do not participate in the π-system so that they can π-bond to the metal, too — or it could even be the π-system itself bonding to the metal. In any case, $\mathrm{t_{2g}}$ will be split and the metal component, $\mathrm{t_{2g}^*}$ will be raised in energy. This further reduces the field splitting, altogether putting $\ce{acac-}$ into the weak field department.

Finally, even though we have iron(III) and even though higher oxidation states are generally more prone to give low-spin complexes, iron(III) is not one of them. It takes significantly stronger ligands than water or even ammonia to generate low-spin iron(III) complexes, so generally any anionic oxygen-donor will not be enough to overcome the pairing energy.

Abbreviations:
acac: acetoacetone (or -ate if charged)
L: ligand
M: metal
ptz: 1-propyl-1H-tetrazol-κN4.

Acknowledgement:
Most of the knowledge is taken from Prof. Klüfers’ lectures on coordination chemistry and refreshed by checking his course’s internet scriptum.


Finding compounds with B, Si, Ca, and Mg in differing amounts

This would be an exceedingly difficult problem.

My suggestion might be to use a database like the Crystallography Open Database and perform searches with the elements you indicate.

For example, the search interface ensures B, Ca, Mg, and Si are all in the compound. I get 141 matches.

It would be incredibly hard to predict compounds with arbitrary stoichiometry. There are several genetic algorithm search packages that let one attempt to find low-energy crystal structures given a formula, including:

XtalOpt
GASP
USPEX

This is, however, an incredibly active area of research. I'm not sure if any of these packages can handle arbitrary amounts of different elements, or how they would handle missing elements. (For example, the elements you describe could be found in various borosilicate minerals given that oxygen is also allowed.)


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There are many ways to synthesize copper selenide. It generally has two forms $\ce{Cu2Se}$ and $\ce{CuSe}$ but actually it is a non-stoichiometric compound. They are grown as nanoparticles for various applications like making semiconductor, optoelectronic components, sensors etc.

By direct combination of copper and selenium vapour / direct displacement of cupric salts with hydrogen selenide.(1)(2)
Using copper(II) chloride dihydrate ($\ce{CuCl2.2H2O}$) and sodium selenite ($\ce{Na2SeO3}$) in ternary oil-in-water (o/w) emulsion system. The reducing agent, sodium borohydride ($\ce{NaBH4}$) used to reduce sodium selenite source
at room temperature(3)
Using thiol ligands(4)
Using copper acetate dihydrate and hydrazine hydrate(5)

As for your reaction, chlorine will not get displaced that easily. It needs a counter ion for chloride to make bond. That's why instead of just selenium, hydrogen selenide is used. It can be better if any displacement is avoided and just direct combination of elemenys is done. See method 1.

References

http://copper.atomistry.com/cuprous_selenide.html
http://copper.atomistry.com/cupric_selenide.html
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.452.6282&rep=rep1&type=pdf
https://pubs.acs.org/doi/abs/10.1021/acsami.5b01085
http://www.asianjournalofchemistry.co.in/User/ViewFreeArticle.aspx?ArticleID=25_10_57


The Newman projections of 3-methylpentane are:

This is a staggered conformation.
Another one would be

This is also a stggered conformation and is the most stable of conformers.

You can generate other conformers by rotating the C-3 or C-4 atom along the C-3 C-4 bond axis.


Why do bubbles form and stay when I shake a bottle of tea?

Tea is may be considered as aqueous solution of many different substances extracted from solid tea leaves.  (It is not so much about «tea» as in tea, or a tisane.)
In this solution may be ingredients altering the surface tension of water sufficiently enough that air bubbles are trapped for a prolonged time.  It could be a component added to act as a preservative, which equally happens to act as a surfactant, or an ingredient intentionally added to ease formation and preserve these bubbles; like the tapioca pearls in  bubble tea.
Compounds acting as surfactants may be artificial (like CTAB) made for cleaning, or processes from natural sources, e.g., polysaccharides (like starch), or plain natural.  Washing clothes with leaves of climbing ivy, for example  works because they contain a small amount of saponins.  That a compound acts as a surfactant does not say if it is safe in food, or not.


The ability of liquid mixtures to separate is not based on density, but on polarity of each liquid. $\ce{CCl_4}$ and $\ce{C_6H_{14}}$ are both non-polar; they will mix well in solution and not separate easily. $\ce{CH_3OH}$ is polar due to presence of a polar alcohol group. Mixtures of non-polar and polar compounds will separate easily.

Density comes into play when you need to determine which of the compounds settle on top of the other. So, for (B), you would use the fact that the density of $\ce{CH_3OH}$ (0.79 g/mL) is less than $\ce{CCl_4}$ (1.59 g/mL) to say that the methanol layer would float on top of the $\ce{CCl_4}$ layer.


First, a question: what kinds of reactions are we talking about? Is it a more specialized (in this context) field like biochemistry, or more general chemical reactions? In practice, finding data for the latter is probably simpler than finding data for the former without doing any experimental work.

In general, I would use the information about Gibbs free energy of formation that can be found here (among many other places) for the reactants and products, and then calculate the equilibrium constant from

$$ K = \exp \left(\frac{-\Delta G^\circ_{\mathrm{rx}}}{RT}\right) $$

where $\Delta G_{\mathrm{rx}}$ is the change in Gibbs free energy for the reaction. Beware that if your reaction conditions - e.g. temperature or pressure - is different from standard conditions, this should be taken into account.

Hope this helps a bit!


The correct question (taken from the 2016 JEE Paper is):

Lithium Aluminium Hydride reduces epoxides1, hence (A) would not give the required product. LAH also reduces ethers to primary alcohols, which is an easier way of eliminating the option (A). $\ce{BH3}$ in THF can reduce the acid, as explained in this question, hence the requisite product would not be obtained.
$\ce{NaBH4}$ cannot reduce esters, acids or epoxides in the absence of catalysts such as $\ce{PdCl2}$ or $\ce{AlCl3}$ and diglyme, hence (C) would be a correct option here.
Can you tell what Raney Ni would do here?
Raney Ni Reduces carboxylic acid and esters at elevated temperature (eg: $\pu{350 K}$); and no reaction with epoxide. So, it can be used for selective reduction of aldehyde at room temperature ($\pu{298 K}$).
Use this resource to help yourself out and as a reference for such reactions in the future.
References:

Murphy, David K., et al. “Reduction of Epoxides. III. Lithium Aluminum Hydride and Mixed Hydride Reduction of Some Secondary-Tertiary Epoxides.” Journal of the American Chemical Society, vol. 91, no. 10, May 1969, pp. 2649–53. doi:10.1021/ja01038a042.


Is the Kekule structure of benzene completely wrong?

In benzene, all of the carbon-carbon bond lengths are equal.  Therefore, the Kekule structure shown below is an incorrect representation of benzene.

It is incorrect because it suggests that there are two different types of carbon-carbon bonds in benzene, a carbon-carbon double bond and a carbon-carbon single bond. Typically, the bond length in a carbon-carbon double bond is around 135 pm, a carbon-carbon single bond is around 147 pm long, while the carbon-carbon bond length in benzene is roughly in between at 140 pm.

Benzene is often drawn as follows to indicate that all of the carbon-carbon bonds have the same length and the pi electron system is continuous and uniformly distributed.

Sometimes, when one is discussing the resonance properties of benzene, two Kekule structures are drawn with the double headed resonance arrow between them as shown below.  This is meant to indicate that these two Kekule structures are resonance structures of benzene, and that taken together (e.g. in one Kekule structure a carbon-carbon bond is a single bond, in the other structure the same bond is a double bond; so "real" benzene is a simultaneous, superposition of these two structures where each carbon-carbon bond is in between a single and double bond), they represent the true electronic and physical structure of benzene.


Circulated Coin

I'll assume that the coin is circulated so a very gentle cleaning won't be a problem. You want to use some sort of organic solvent to loosen the glue, then very gently rub the residue off. You don't want to use any sort of silverware polish or anything abrasive when rubbing the coin.

Good solvents might be olive oil or nail polish remover. After the organic solvent wash the coin with dish washing liquid, rinse with plain water,  then dry by flipping coin over on a dry cloth or paper towel.

Uncirculated Coin

If the coin is uncirculated then you would want to use an ultrasonic bath with an organic solvent to prevent scratches on the coin.

Two tone coin after cleaning

Because the surface of the coin would have been sealed  where the tape was, it is possible that the coin will have a different color where the tape was. If that is the case then you might need to use a coin dip to get the surface looking even again. Unfortunately this will leave the surface shiny which screams that the coin has been cleaned. The coin will dull over time but never really develop a natural patina again since a truly natural patina also depends on wear.


What would be the correct mechanism for hydration of ethene?

At first I thought your mechanism was very comprehensive but upon closer inspection I think it has some flaws because in this reaction conc. sulphuric acid is used as a catalyst and it must be regenerated at the end and in the above mechanism only ethanol is generated.

This must be the mechanism shown in your book.


The description is simplified, but actually fairly accurate. You can find all the details in a specialised NMR book (other general texts usually do not go into much detail).

Quantum mechanically, you can show that the precession frequency (Larmor frequency), denoted $\omega_0$, is related to the energy difference between the up and down spin states of a spin-1/2 nucleus. The energy levels are

$$\begin{align}
E_\alpha &= +\frac{1}{2}\hbar\omega_0 \\
E_\beta &= -\frac{1}{2}\hbar\omega_0 \\
\end{align}$$

and so the transition occurs at the frequency $|E_\beta - E_\alpha|/\hbar = \omega_0$, so if we can identify how much precession is happening at the frequency $\omega_0$, that tells us how strong the corresponding signal should be in the spectrum.

The initial data that you get is in the form of a free induction decay (FID), which plots the amount of detected magnetisation against time. The process of extracting the frequencies from the raw time-domain data is accomplished by a Fourier transform.

Feel free to ask if you would like any part of this to be elaborated. I intentionally did not go into great detail, partly because that is the role of a textbook, but also partly because it can get pretty complicated very quickly.

(There is also a complicated sign convention to do with the definition of $\omega_0$, and different books define it differently. Occasionally the definition even depends on the sign of the gyromagnetic ratio $\gamma$. I would advise to not worry too much about plus/minus signs.)


Volume of NaOH needed to precipitate copper(II) but not calcium

You need to precipitate $\pu{9.9999\times 10^-4 mol}$ of $\ce{Cu^{2+}}$, hence you need to add $2(\pu{9.9999\times 10^-4 mol}) = \pu{1.99998\times 10^-3 mol}$ of $\ce{NaOH}$, this is:

$\displaystyle{V_\ce{NaOH} = \frac{n_\ce{NaOH}}{c_\ce{NaOH}} = \frac{\pu{1.99998\times 10^-3 mol}}{\pu{0.1 mol L-1}} = \pu{19.9998 mL} \approx \pu{20.0 mL}}$

There are $\pu{1\times 10^-8 mol}$ of $\ce{Cu^{2+}}$ left in solution, i.e. $\ce{[Cu^{2+}]} = \pu{8.33 \times 10^{-8} mol L^-1}$. Then, $\displaystyle{\ce{[OH-]} = \sqrt{\frac
{K_\text{sp}}{\ce{[Cu^{2+}]}}} =  \pu{1.74 \times 10^{-6} mol L-1}}$.

The final concentration of $\ce{Ca^{2+}}$ is $\pu{0.8333 mol L-1}$, so $Q_\text{ps} = \ce{[Ca^{2+}][OH-]^2} = \pu{2.52 \times 10^{-12}} < K_\text{sp}$, i.e. $\ce{Ca(OH)2}$ won't precipitate at this point.


Say your acid is monobasic: suppose its formula is $\ce{HA}$.
So, it dissociates according to following equation:
$$\ce{HA + H2O <=> H3O+ + A-}$$
Suppose its molar concentration (given) is $c$, and it dissociate $\alpha$ amount in water (assume total volume is $\pu{1.0 L}$) at given temperature at equilibrium. Thus, equilibrium concentrations of $\ce{HA, H3O+}$, and $\ce{A-}$ are $(c-\alpha)$, $\alpha$, and $\alpha$, respectively.

$$\therefore \; K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}=\frac{\alpha \cdot \alpha}{c-\alpha}$$

Since $\alpha = [\ce{H3O+}]$, $-\log \alpha = \mathrm{pH}$ (which is given) and you know $c$, which is also given, you can estimate the $\mathrm{p}K_\mathrm{a}$ (and hence $K_\mathrm{a}$ if you needed).

Late edition:

Take (-$\log$) on both side of above equation. Thus:

$$\mathrm{p}K_\mathrm{a} = -2\log \alpha + \log{(c-\alpha)}=2(\mathrm{pH})+ \log{(c-\alpha)}$$
Most of the time (in cases of $c\gt\gt\alpha$),  $\mathrm{p}K_\mathrm{a} \approx 2(\mathrm{pH})$


Can molten sugar also rotate polarized light as sugar water does?

Molecules in every state - liquid, solid and gaseous - rotate the plane of linearly polarized light - it doesn't matter, because this rotation arises from inherent asymmetry of molecules, which is causing their optical activity

More to the point, not only water solutions of sugars, but also their crystals, molten sugars and even vapor, rotate the plane of linearly polarized light. If you're asking specifically about sucrose checking this may be problematic as it's decomposing during melting, as you can read in this question and here, but I think it still can be done, under vacuum.

You can also check these related questions about chiral gases and rotation of the plane of polarization by molecules


hypo residual chlorine testing using a 10ml reagent pack

This is sort of an extended comment, not necessarily an answer.

Overall, yes I'd expect different numbers at least. How they would vary is a matter of some conjecture.

(1) The Hach DR 900 is a colorimeter. I'm not sure how large of a sample you need to fill active volume of the colorimeter cell.

(2) The ratio of water volume relative to the amount of solid reagent is changing. So the chemical equilibrium on which the calibration curve depends might shift. If you dilute 5 and 10 ml samples to 15 ml with "pure" water and take 10 ml of each of the three samples (2 dilutions plus "full strength"), then you be comparing better.

(3) Even given (2), the results seems odd. Normal drinking water should have between 0.2 and 2.0 mg/L.

Is pH of the water ok?
Does water have some unusual buffer capacity?
Reagent pack old?
Do you have some standards which you can test?


Why does the Second Law of Thermodynamics hold true?

According to second law of Thermodynamics,

During spontaneous process:

$$ ∆S_{total} > 0 \ or ∆S_{total} = 0 $$

So total entropy is positive.

But $ ∆S_{system} \ or \ ∆S_{surroundings} $ may be less than 0.

Which $ \implies $ any one of $ ∆S_{system} \ *OR* \ ∆S_{surr} $ may be less than 0.

Now, according to Big Bang theory, the Universe, soon after existence , is expanding more and more and so - its entropy is increasing.

 Why entropy should be positive to occur a spontaneous reaction?

An example first:
• Heat flow is unidirectional. That is, from higher to lower temperature.
Similar to it, all Physical or Chemical Process will tend to proceed spontaneously in one direction only(unless altered by some external agency).
Hence,
Spontaneity is the potential to proceed without assistance of external agency.

Now, Entropy means randomness or disordered behaviour.
And there is always a tendency for the system's energy to become more disordered. An example would be diffusion.
Where gases diffuse and become more chaotic. And during the expansion of our universe, its entropy increases.

So, we interpret that entropy for a spontaneous process increases, till it reaches maximum. And our universe having a positive $∆U$ and gaining entropy in the direction of spontaneity.

$∆S=0$ at equilibrium.

If $∆S$ is negative that would mean that reaction or process is going opposite to the spontaneous direction, and which is not possible in case of our universe.


Why doesn't my stainless steel Thermos get really really hot?

I have a solid steel pint glass and I can tell you that it pretty much instantly heats up to the temperature of the liquid you put inside. You probably have some sort of double-wall vacuum flask, which insulates by using a vacuum between the steel.

As you can see, there are two walls to the flask: an inner wall and an outer wall. The inner wall is probably very close to the temperature of whatever is inside the flask. However, since the two walls are separated by a vacuum (which is a fairly good insulator), the outer wall, i.e. the one that you touch, remains pleasantly near ambient temperature*.
*Unpleasantly near ambient temperature, if you've been outside in a hot summer for a while.


Think about the reaction:
$$\ce{NaCHO3(s) + CH3COOH(l) \to CO2(g) + H2O(g) + Na+(aq) + CH3COO- (aq)}$$
The resulting $\ce{CO2}$ will make the bag expand, therefore increasing the volume and the subsequent buoyant force. This force opposes the force of gravity, making it seem like the mass of the matter inside the bag has decreased.


How do I synthesise this tertiary amide from these 5 compounds?

You seem to have got the uses of the compounds right, but could not bring that knowledge together to apply it here. As you said, you first use the Grignard to add a methyl group to D

Let's call this compound K. Then, we use $\ce{NaCN}$ to convert the alcohol into a nitrile, then to a carboxylic acid rather than an amide, as this would be easier to work with for us.

Let's call this compound L. Now we are ready to start creating the actual compound. Nitrate C and then reduce it to create the amine and reduce the ketone.

Then add L to the product, followed by K. This order is followed as adding the alcohol before the acid may create a tertiary amine rather than the amide we need.


Why does automotive glass shatter into tiny pieces but glass used in our houses into shards?

Let's start out by covering the
Why?
Glasses used in automobiles are designed to break in tiny pieces.
Imagine you're driving down a dark highway hitting dangerously high speeds and then suddenly you lose control and crash.
If the glass used was not to shatter in tiny pieces rest assured you would not remain in one piece. Essentially it shatters for safety reasons.
Household glass on the is generally not involved in such drastic accidents so there is no need for it to shatter into tiny pieces .
How is it done?
From How Stuff Works

Tempered glass is created by heating and then rapidly cooling the glass to room temperature by ushering it through a system of blowers.
The surface of the glass cools much faster than the center of the glass and contracts, causing compressive stresses, while the center of the glass expands because of its temperature, producing tensile stresses. What does that mean? Imagine a piece of glass that could be pulled or stretched to a certain length (tensile stress), while being pushed down and compressed (compressive stress) simultaneously. Both the pulling and pushing stresses achieved through the heating and cooling process give tempered glass its tensile and compressive strength. The differences between these two give the glass 5 to 10 times the amount of strength it originally had.

This system holds a lot of energy due to the stress so when it breaks all of it is released causing the glass to shatter into tiny pieces
Tl;Dr
Automotive glass contains more energy than normal household glass. When it breaks all that energy is released which causes the glass to break into tiny pieces.


Why is sodium acetate a good nucleophile?

The acetate anion is actually not particularly bulky. The negative charge is located on the two oxygens which are approximately at a $120^\circ$ angle away from the only residue that has the potential to be bulky. Whatever that residues does will be ‘behind’ the carbon and hardly noticeable. However, they are still rather bad nucleophiles since the negative charge is delocalised over two oxygen atoms in the π system.

I’m actually not sure which orbital attacks nucleophilically but it could equally be an oxygen lone pair rather than the π system. If that is the case, there is really very little steric hindrance that could inhibit the reaction from proceeding. Thus, acetate is able to attack nucleophilically, as it does for example in a Mitsunobu reaction.


EM grade is a label for chemicals used in electron microscopy.


What is the point of introducing virtual orbitals in Hartree-Fock calculations?

Virtual orbitals are automatically 'given' by the solved eigenvalue equations and are necessary for solving said equations. These orbitals are important for some post-HF methods such as CI and perturbation theory and coupled-cluster theory.


Coconut oil vs Olive oil: Spray dynamics

The fluid dynamics of spray calculations depend on many conditions including the nozzle used to create the spray.  The wikipedia article on spray nozzles has a good discussion on how different liquid properties tend to affect spray. Spray information tends to be empirical, so it is not really science at play, it is engineering.

The Institute of Liquid Atomization and Spray Systems is a professional society set up around the science and engineering of spraying.  Atomization and Sprays is a periodical journal set up around this entire topic.  A internet search of 'textbook on spray and atomization' turns up many textbooks on the topic.

Plenty of manuscripts written by scientists exist providing detailed results of experiments comparing sprays.  An example would be an experiment done comparing the spray of waste cooking oils in an engine when it is used as fuel.

Summarizing a discussion of all of these experiments is too broad and out of the scope of SE.  If you are interesting in learning about sprays make sure you have some understanding of fluid dynamics.  Conduct a literature review relevant to the spray experiment you have in mind.


Can there exist any element which has an 8th principle shell?

That is certainly possible, and in fact very likely. Nuclear physics predicts an island of stable elements around an atomic number of 120 $^{[1]}$, with the two heavier of the three elements predicted to be most stable having an $\ce{8s^2}$$^{[2]}$ and an $\ce{8s^2 8p^1}$$^{[3]}$ configuration, respectively.


Isn't the dual behaviour of subatomic particles contradictory?

Take a nice simple system like a hydrogen molecule. It's true that the two electrons repel each other, but at the same time they are attracted by the positive charge of the two protons.

If we start with an $\ce{H2+}$ ion, with just a single electron, it should be obvious that the electron is tightly bound because it feels an attractive force from both the protons.

Now imagine adding the second electron to form a neutral hydrogen molecule. The second electron will be repelled by the first electron that is already present but it will be attracted by both the protons, and the attraction from the two protons wins. The binding energy of the second electron in a hydrogen molecule is about 15.4 eV, so it is even more strongly bound than the electron in a hydrogen atom.

Note that this is not related to the wave/particle duality. Wave particle duality is in any case a misguided concept since particles are quantum fields not waves or particles – they simply behave approximately like waves in some circumstances and approximately like particles in others. See this post on the Physics Stack Exchange for more on this.


It should be better to write equations as

$$\ce{RCl +H2O <=> ROH +HCl}$$

$$\ce{Fe^2+ +Zn <=>> Zn^2+ + Fe }$$

with the latter having the equilibrium shifted to far right.

But is case of high concentration of $\ce{Fe^2+}$ and very low concentration of $\ce{Zn^2+}$, the net reaction would be ongoing leftwards.

If the left side of equation below is smaller than right one, the net reaction in inclined toward iron being replaced by zinc. And vice versa.

$$E^{\circ}_{\ce{Zn/Zn^2+}}+0.0295\log {a_{\ce{Zn^2+}}}=E^{\circ}_{\ce{Fe/Fe^2+}}+0.0295\log {a_{\ce{Fe^2+}}}$$

Note that forward and backward reaction are ongoing simultaneously, having the same rate at the equilibrium.


This question unfortunately illustrates the significant shortcomings of the VSEPR model. This model is based upon classical electrostatic arguments. The molecular geometry will seek to maximize the distances between the repulsive electrons in the valence (outer) shell, that are completely localized in their chemical bonds and their lone pairs/singly occupied molecular orbitals.
This model works well in some cases. However, when quantum mechanics comes into play, it falls pretty flat, as it totally neglects (the really important) quantum effects. A better way of finding the molecular geometry is to find the geometry that minimizes the total energy of the molecular orbitals. In this case, quantum mechanical calculations find that a set of three $\mathrm{sp^2}$-hybridized molecular orbitals arising from a $D_\mathrm{3h}$ symmetry has a lower total energy than breaking the symmetry to form the pyramidal $C_\mathrm{3v}$ geometry.
However, this does not mean that all $\ce{\overset{.}{C}X3}$ radicals follow this pattern. Put an extremely electronegative substituent on in place of $\ce{H},$ e.g. in $\ce{CF3},$ and the situation changes. The lowest energy geometry is now the pyramidal one.
Sadly, quantum chemistry doesn't always care to follow general rules.


What's the bonding in the diborane(6)- hydrogen chloride dimer?

I find this question to be quite interesting as it clearly pertains to non-covalent interactions.

I have analyzed two conformations of the diborane-HCl dimer system, one with the chlorine on HCl orientated toward diborane (HCl-Dib) and one with the chlorine oriented away from the diborane (ClH-Dib).  For comparison, I have similarly analyzed the Water-HCl dimer with the chlorine oriented toward the two hydrogens on water.

Methodology

Full geometry optimizations and corresponding harmonic vibrational frequency computations were performed with Gaussian 09 using Grimme's B97D density functional along with the 6-31+G(d',p') basis set.  All geometries were characterized to C$_{2V}$ symmetry.  Symmetry adapted perturbation theory (SAPT) was implemented at the SAPT2+3(CCD) level using the Psi4 software package.  These computations implemented Dunning's heavy-aug-cc-pVTZ (haTZ) basis set (i.e. aug-cc-pVTZ for non-hydrogen atoms and cc-pVTZ for all hydrogen atoms).  Finally, natural charges were quantified using the Natural Bond Orbital (NBO) analysis routine available in Gaussian 09 (see http://nbo6.chem.wisc.edu/ for more details).

Results

Both diborane-containing complexes were minima on the B97D potential energy surface.  The H$_2$O-HCl dimer is a higher order saddle point exhibiting two imaginary modes of vibration.  The interaction energy ($E_{\mathrm{int}}$) can be quantified by attractive and repulsive contributions, the former consisting of electrostatics ($E_{\mathrm{elst}}$), induction ($E_{\mathrm{ind}}$), and dispersion ($E_{\mathrm{disp}}$) and the latter consisting of exchange ($E_{\mathrm{exch}}$). The table below summarizes the SAPT results.

Clearly, the dominant attractive force in each case is dispersion.  The ClH-Dib complex exhibits the strongest interaction energy and contains a dipole that points from the diborane monomer to the chlorine.  The HCl-Dib complex has an interaction energy that is about 1 kcal mol$^{-1}$ smaller in magnitude and exhibits a dipole pointing toward the diborane monomer.  The analogous H$_2$O-HCl system is qualitatively similar to the HCl-Dib complex in terms of the magnitudes of the components to the interaction energy.  Surprisingly, this dipole-dipole interaction is slightly weaker (by about 0.1 kcal mol$^{-1}$) than that seen in the dipole/induced-dipole system of HCl-Dib.

I have included below figures for the HCl-Dib, ClH-Dib, and H$_2$O-HCl complexes, respectively, with corresponding natural atomic charges shown on each atom.

Figure 1 - HCl-Dib

Figure 2 - ClH-Dib

Figure 3 - H$_2$O-HCl

Interestingly, when chlorine is pointed toward diborane (Figure 1), the natural atomic charges for diborane remain rather symmetrical and diborane barely exhibits an individual dipole moment.  The H$\cdots$Cl separation between the two monomers is about 3.4 Angstroms, a long-range interaction.  However, when chlorine is pointed away from diborane (Figure 2), the H$\cdots$H interaction is much smaller, exhibiting a separation of about 2.2 Angstroms.  When considering water, the separation increases once again (Figure 3) to about 3.2 Angstroms.

Conclusions

The dominate attractive force to the interaction energy for these dimer systems is dispersion, though we see a competing contribution from electrostatics in ClH-Dib (Figure 2).  Surprisingly, the orientation of the halogen significantly affects the geometry and energetics of these systems.  When chlorine is oriented toward the two hydrogens, a weaker interaction and a larger inter-monomer separation is seen.  However, when hydrogen is oriented toward the two hydrogens, the interaction energy increases in magnitude and the inter-monomer separation decreases.

Q/A

I find it more convenient to address some of Martin's comments within the answer itself (sometimes responses can become quite long).

Which compound is lower in energy?

Based on absolute energies, H$_2$O-Cl is the lowest in energy of the three dimers (by about 23 Hartrees).

How stable are they wrt decomposition?

The interaction energy will give you this information (defined as the energy difference between the interacting monomers in an optimized dimer geometry and infinitely separated monomers adopting the geometry they have in the dimer complex).  Given that the magnitude of $E_{\mathrm{int}}$ is less than 2 kcal mol$^{-1}$, I would say that these are not very 'stable'.

Have you tried asymmetric states?

No.  I do not believe breaking symmetry in the diborane-containing complexes will change the energy/geometry appreciably.

I find it quite surprising, that the BH--ClH distance is much longer (0.6A) than the sum of the respective vdW radii, while the BH--HCl distance is about the sum of the vdW radii. Still, in both cases there should be (about) no contribution from the orbitals. How do the bond lengths change in the two conformations?

In the diborane-containing complexes, the bond lengths deviate by about 0.01 Angstroms (so... no change).

Can you make a rough comparison between the B--Cl distance in the two cases?
They appear to be quite similar.

In HCl-Dib, the B$\cdots$Cl separation is about 3.79 Angstroms.  In ClH$\cdots$Dib, the B$\cdots$Cl separation is about 3.86 Angstroms.  Similar they are indeed.


Why is Ka constant when volume is increased?

$$K_\mathrm{a}=\ce{\frac{[A-][H3O+]}{[HA]}}$$

The above equation defines the $K_\mathrm{a}$, or acid dissociation constant, of an acid. The reason it is even a value worth measuring is because it is an intrinsic property of the acid, i.e. it's value does not change from sample to sample (temperature aside).

While it is true that as a solution of acid becomes more dilute, $\ce{[A-]}$, $\ce{[HA]}$ and $\ce{[H3O+]}$ all decrease, they do not all decrease by the same amount. Consider the effect of adding water to the following equilibrium:

$$\ce{HA + H2O <=>A- +H3O+}$$

The number of $\ce{HA}$ molecules will decrease, and the number of $\ce{A-}$ and $\ce{H3O+}$ molecules will increase—a conclusion we arrive at via Le Châtelier's principle. Separating the volume of water from the $K_\mathrm{a}$ equation, we get:

$$K_\mathrm{a}=\ce{\frac{mol(A-)\cdot mol(H3O+)}{mol(HA)}}\cdot\frac{1}{V}$$

What we are left with is two larger numbers over a smaller and a larger number. While it's not obvious that these numbers should change to give precisely the same quotient, I can assure you they do.


Does chemical compound with formula N₄O₁₀ exist?

PubChem database N4O10 (molecular formula) search shows that there is no known compound with this formula.

While $\ce{NO}$ (“mononitrogen monoxide”) and/or $\ce{NO2}$ (“mononitrogen dioxide”), having odd number of electrons, will dimerize or combine to form

$\ce{N2O2{: } 2NO -> ON-NO}$
$\ce{N2O4{: } 2NO2 -> O2N-NO2}$
$\ce{N2O3{: } NO + NO2 -> ON-NO2}$

(note the $\ce{N-N}$ bond),

$\ce{N2O5}$ is formed by the condensation of nitric acid $\ce{HNO3}$:

$\ce{ 2 { }O2N-OH ->[-H2O] O2N-O-NO2 }$

(note the $\ce{N-O-N}$ bond), thus can be considered nitric acid anhydride. And unlike polyphosphoric acid formation from $\ce{H3PO4}$, there are no remaining hydroxy groups for further condensation.

With non-standard formal approach, $\ce{N2O5}$ already is a dimer of hypothetical franctional-oxygen-atom-count “mononitrogen(V) oxide” $\ce{NO_{2.5}}$.


Why does raising the surrounding pressure raise the vapor pressure of a liquid?

The answer lies in that the chemical potential of the liquid increases with pressure.

Quantitatively, the chemical potential changes with a change in pressure from $p_1$ to $p_2$ as follows:
$$\Delta\mu(\textrm{l})=\int_{p_1}^{p_2}V_mdp$$

A qualitative explanation is possible by linking the change in pressure to the concept of chemical potential. The chemical potential difference is the amount of non-pV work (the "chemical" work) which the system can perform. In essence, increasing the pressure means more of the liquid has the potential to transfer into the gas state (the chemical work in question).   You might visualize the transferred molecules as being "squeezed out" of the liquid by the increasing pressure applied by their neighbours.


What is a simple ratio in which atoms combine?

The ratio of C, H and O atoms of 12:22:11 is likely to be sucrose. Glucose has a simpler ratio of 1:2:1 (even though it has 6 carbons). The usual context for presenting the theory is with inorganic compounds with no more than a 2:3 or 3:5 ratio. With this limitation, they use the words simple ratio since the numbers are small and can easily be found. The ratio of 12:22:11 could be confused with another ratio if your calculations are off by 20%.


How to find concentration of three species from given molar absorptivities

The equation you need would seem to be $\ce{A=\epsilon [C] L}$ where A is the amount of absorption (optical density), L the cell path length, your 'distance light travels' and [C] concentration. As you have 3 species absorbing at the same wavelength you have $\ce{A=(\epsilon_1[C_1] + \epsilon_2[C_2] + \epsilon_3[C_3])L}$. You know L and the $\epsilon$'s but only one of the three concentrations and therefore you do not have enough information to determine the two remaining ones.


How does current start to flow when galvanic cell is first connected?

BTW, electrons don't actually actually "flow" very fast. The drift velocity of electrons is incredibly slow, on the order of a millimeter every 5 or ten seconds in a room-temperature conductor. The propagation of current is like the particles in Newton's Cradle, where the electric field bumps each electron down the line, a much faster process. Furthermore, electrons are swept out of the conductor by the electromagnetic force, which propagates on the surface of the conductor at a large fraction of the speed of light in vacuuo, C. The fraction is the velocity factor, perhaps 0.5 to 0.75 C.
So the answer to your question is not as clear-cut as it might seem. You could say the reaction propagates at the speed of the EMF, C times the velocity factor in the connecting wire. However, one then must take into account the buildup of slow-moving ions, creating a space charge slowing things down.


Boron react with $\ce{NaOH}$ but under certain conditions. Boron exist in two form (1) crystalline form (2) amorphous form. Both forms are generally inert to chemical reactions and is unaffected by acids and alkali at normal conditions but at elevated temperature, concentrated oxidizing acids like nitric or sulfuric acids react with boron to yield boric acid. On the other hand, amorphous boron does react with fused alkali at high temperature to obtain trisodium borate:
$$\ce{2B(s) + 6NaOH(s) ->[t > 773K] 2Na3BO3(s) + 3H2(g) ^}$$
High purity, hydrated sodium metaborate(sodium tetrahydroxoborate, $\ce{Na[B(OH)4]}$) can also be produced from boron in a $\ce{NaOH-H2O}$ system. The process applied was called pressure alkaline leaching.
When fused in the presence of air at $\pu{350-400 ^\circ C}$, sodium metaborate is formed. You can find the reaction in this answer by @andselisk.
References

https://onlinelibrary.wiley.com/doi/abs/10.1002/0471743984.vse1217
Yin, B.W & Qin, S.Y & Zhang, Y.F & Zhang, Y (2014). Preparation of hydrated sodium metaborate from boron concentrate via pressure alkaline leaching. Guocheng Gongcheng Xuebao/The Chinese Journal of Process Engineering. 14. 258-265, (link)


How to determine the strongest oxyacid of chlorine?

In such questions you're supposed to know certain rules of acidity. You can find some of them on this page. In your case fourth rule, concerning oxoacids, is relevant and as you can see there, it's described on this very case. More precise formulation is so called Pauling's Rule described here. In general the more oxo groups connected to central atom, the more acidic compound.

Related: How are Pauling's rules applied to Nitric acid?


On stretching a rubberband does the entropy increase or decrease?

Think of it this way, when an elastic polymer is in its relaxed state, these molecules are all tangled up with each other and have no particular direction to them, but when you apply a force, i.e stretch the rubber band, you end up disentangling some of the polymer molecules.

We are not stretching the actual polymer molecules, just sort of changing the way they are organised. This disentanglement corresponds to a reduction in micro-states that the system occupies and thus, a decrease in entropy.

In simpler words, there's more ways for a polymer chains to jumbled up with each other (more microstates) than for them to be aligned.

More simply-- If the chain were totally stretched out, it would only have one possible conformation. By coiling itself, it increases the number of
conformations it can have, and therefore, its entropy.

A more rigorous treatment can be found if you read up about the ideal chain, which serves as a model for such systems.

Thus, assuming that no other energy exchanges takes place, the tendency of the elastic band to shrink back to its original conformation is due to something called an entropic force, i.e a tendency of the system to move to a configuration that maximises entropy.

Some Details

Note: I am not quite confident with this material and might've made some mistakes--if that be the case, I am sure someone here will correct me

Consider the following model system: The chain is made of $n$ rigid segments of size $b$; Thus, the length of the chain is $L = nb$. Also, assume that the chain has one fixed end, and one free end and the distance between the two is denoted as $r$.

Then, one can derive the probability of finding the chain ends a distance $r$  apart is given by the following Gaussian distribution:

$$ P(r,n)\mathrm dr = 4\pi r^2 \left(\frac{2nb^2\pi}{3}\right)^{\frac{-3}{2}}\exp\left(\frac{-3r^2}{2nb^2}\right)\mathrm dr$$

The classical entropy is approximately $ S = k_\mathrm b\ln(P(r,n))$

From hereon, one can go on to show that for an ideal chain, maximising its entropy means reducing the distance between its two free ends (i.e reducing $r$)


Why does bromide not react with water when methyl ammonium bromide is dissolved in water?

The key is comparing $\mathrm{p}K_\mathrm{a}$ values. A good resource for this is the Evans $\mathrm{p}K_\mathrm{a}$ table. Unfortunately, the table does not contain $\ce{CH3NH3+}$, but in the section protonated nitrogen there is a value quoted for $\ce{C2H5NH3+}$ — which is chemically similar enough to methyl ammonium to assume the $\mathrm{p}K_\mathrm{a}$ values to be essentially identical.

\begin{array}{lr}\hline
\text{compound} & \mathrm{p}K_\mathrm{a}\\ \hline
\ce{EtNH3+} & 10.6\phantom{0}\\
\ce{HBr} & -9.00 \\
\ce{H3O+} & -1.7\phantom{0} \\ \hline\end{array}

Notice how the $\mathrm{p}K_\mathrm{a}$ values differ by almost $20$ logarithmic units. $\ce{HBr}$’s $\mathrm{p}K_\mathrm{a}$ value is even lower than water’s by $7$ to $9$ logarithmic units depending on the source used. Thus, while the amine exhibits extensive acid/base chemistry in water, bromide will not be protonated to any noticeable extent. Or you could say that your final equation will be strongly favouring the reactant side.

I’ll also point you to a recent answer of mine wherein I calculated the concentration of $\ce{HF}\ (\mathrm{p}K_\mathrm{a} \approx 3.17$, so notably weaker than $\ce{HBr}$) in a 1 molar solution of $\ce{NaF}$. The case is very comparable to your’s.


The atmospheric $\ce{NO_x}$ (usually pronounced as "nox" or "knocks") concentration is the sum of the $\ce{NO}$ and $\ce{NO2}$ species.  These get combined into a single term as they each play a roll in photochemical ozone formation. $\ce{NO_x}$ can form as a byproduct of the combustion of fuels containing nitrogen, like coal, or from atmospheric nitrogen in the high-temperature environment of an internal combustion engine.

Because photochemical ozone production requires $\ce{NO_x}$, volatile organic carbon (VOC) and sunlight, the $\ce{NO_x}$ concentration is really most relevant for environments already rich in VOC that get plenty sunlight.  Modern automobiles have emission systems that have significantly reduced VOC emissions.

Depending on just where you live in Canada, $\ce{NO_x}$ levels are likely very important for photochemical ozone production during the summer, when you have plenty of natural VOC's available from local vegetation (be it forest, tundra, or any other of your rich and diverse ecosystems).

During the wintertime in Canada, or summer in places like the desert Southwest US, the atmospheric VOC concentrations are small.  In these cases it is the VOC concentration rather than that of $\ce{NO_x}$ that limits ozone production.  Of course the deserts also get plenty of the other ingredient, sunlight.

Depending on your latitude, there may be enough sunlight to produce some ozone, and due to temperature inversions the lower rate of ozone production can still lead to the buildup of ozone; so long as you have some $\ce{NO_x}$ and VOC's present.


Why does Carbon-13 NMR not have splitting patterns?

In 1H NMR, spin-spin coupling leading to multiplet structure most commonly arises due to coupling to other 1H nuclei. This is most often taught in terms of neighbouring 1H nuclei having "up" and "down" spins, both of which lead to the nucleus of interest experiencing a slightly different magnetic field, and hence having a slightly different chemical shift. While this is only a simplification, it is probably sufficient for the purposes of this question.

In 13C NMR, though, proton decoupling is often employed. As the name suggests, this removes all effects of coupling to protons (i.e. 1H). How is this achieved? Essentially, while the signal from the 13C nuclei is being recorded, the NMR spectrometer simultaneously and continually irradiates the 1H nuclei with energy. This means that each 1H nucleus is rapidly undergoing "spin-flips" between up and down spin, and therefore, what the 13C nucleus sees is actually a magnetic field that is very rapidly oscillating between the two possible values. What is actually observed in the NMR spectrum is an average of the two possible values, which is equivalent to what would be seen if there was no coupling to begin with.

Why do we do this, though?

One concern is sensitivity: when peaks are split due to spin-spin coupling, the intensity of the peak is decreased. Since 13C has a naturally very low abundance, it is of interest to prevent the peak intensity from being spread out due to coupling.
Another concern is simplicity: carbon spectra which have C–H coupling can often be extremely difficult to interpret. For an example I refer you to Glenn Facey's blogpost where an inset of a proton-coupled spectrum of toluene is shown.
A subtle point that is often neglected is the fact that the decoupling process also enhances the intensity of 13C peaks via the nuclear Overhauser effect.

All of these factors make proton-decoupled carbon spectra – or in NMR lingo, 13C{1H} spectra – the preferred experiment.

If you're interested in coupling to nuclei other than 1H:

coupling to 13C is all but invisible, considering the low natural abundance of 13C. Recall that in a 13C spectrum you are only observing 1% of all the carbons present. How many of those carbons are coupled to another 13C nucleus? Or in other words, what are the chances of finding two 13C nuclei next to each other? Pretty low is the answer – so much so that an NMR experiment designed to measure C–C couplings was given the acronym INADEQUATE.
coupling to other abundant spin-1/2 nuclei is in fact very often seen. Two simple examples are 19F and 31P, both of which are close to 100% natural abundance – in compounds containing F or P, multiplet structure will be observed in the carbon spectrum. Here's the spectrum of 1-chloro-2-fluoroethane, $\ce{FCH2CH2Cl}$, courtesy of SDBS. The two peaks on the left are from the carbon next to fluorine; coupling to fluorine leads to a 1:1 doublet being observed. Likewise, the two peaks on the right are from the carbon next to chlorine. It is still coupled to fluorine, but because of the greater distance, the coupling constant is smaller.

There is also probably a wealth of examples in organometallic chemistry (for example 103Rh is also spin-1/2), but I don't currently have the energy to find a good example.
coupling to quadrupolar nuclei (nuclei with spin > 1/2) can be seen, depending on the symmetry of the molecule and on some other physical parameters. For example, since deuterium has spin 1, the carbon peak in $\ce{CDCl3}$ is observed as a 1:1:1 triplet (a familiar sight to organic chemists). But this is not always the case: for a more in-depth explanation see, e.g. Symmetry and NMR linewidths of quadrupolar nuclei


What is the difference between enthalpy of formation and Gibbs free energy of formation?

Gibbs free energy is not more "precise." The two are different but complementary. Enthalpy and Gibbs Free Energy indicate different things. Enthalpy can tell you about the relative stabilities of the products and reactants. Gibbs free energy however can tell you about whether a reaction is spontaneous (whether a reaction will occur) under a set of specified conditions.

A lot of people will confuse exothermic - bond improving - reactions with spontaneous reactions. While it is true that a bond improving reaction is in itself quite favorable - after all - all chemical processes tend toward greater stability and lower potential energy - we must remember that there is another dimension to the problem of reaching maximum stability - namely, entropy (a measure of disorder).

Recall that all physical processes tend toward greater entropy (greater disorder) as well. Just as one's bedroom becomes more and more chaotic despite the pleas from Mom to clean one's room, physical processes tend toward chaos (disorder) as well. Gibbs free energy takes entropy into account with enthalpy and can tell us whether a process will actually go.

Consider this reaction:

$\ce{2NO_2(g) -> N_2O_4(g)}\ $

Consider only the forward reaction. The forward reaction is spontaneous under standard state conditions and 298 K - i.e. we have a negative Gibbs free energy value. Don't believe me? Nitrogen dioxide is brown. Nitrogen tetraoxide however is colorless. Put some brown nitrogen dioxide into a flask and if you observe it, you'll notice that the brown color fades somewhat as the colorless nitrogen tetraoxide is made.

The enthalpy of reaction is definitely negative - the reaction is exothermic because the reaction is a dimerization reaction - we are forming a bond here (specifically, a N-N bond) by using two nitrogen dioxide molecules (which are actually radicals, and remember that radicals are highly unstable). Nitrogen tetraoxide however is not a radical.

The entropy of reaction however is negative as well - i.e. we have an unfavorable change in entropy. Remember that $ΔS^o = S_{final} - S_{initial}$. In this case, it is clear that $S_{final} < S_{initial}$ because on the right hand side, we have one mole of gas. On the left hand (reactant) side, we have two moles of gas. There is definitely a lot more chaos (disorder) with 2 moles of gas molecules bumping around and whizzing around every single way imaginable. So we've gone from a disordered system on the left to a somewhat more ordered system on the right.

We would describe this reaction as (at 298 K and std. state conditions):

1) Spontaneous
2) Exothermic
3) Entropy lowering
4) Exergonic


What is the right way to express pH and concentrations?

This is more a question of English usage than Chemistry, but the English Language and Usage people regarded it as “proof reading” and migrated it here. So I will move my original comment here to make it an answer (until the Chemists throw the question out).

First, you really should try to answer this yourself by looking at the style used in scientific journals relevant to your work in papers from English-speaking countries with authors whose surnames suggest they may be native speakers. Several of your alternatives are grammatically correct but unneccessarily long-winded and never used.

Having said that, my opinion (as a native English-speaking biochemist with many publications) is:

A: A-1 (“a solution of pH 3”) should be used.

The use of ‘with’ alone in A-2 is unusual, A-3 is ok in speech but too verbose for a paper, A-4 with two ‘of’s is clumsy, ugly and unnecesary.

B: B-1 (“a 3 nM buffer solution”) seems best to me — brief and correct.

There is no need to expand it as in B-2, B-3 (both of which sound clumsy and I advise you not to use), or B-4. B-5 has the ugly double ‘of’ again.

If for some reason you don’t like B-1 you can name the buffer in a phrase such as “3 nM Tris-HCl Buffer”.


A good book for learning paper and thin-layer chromatography

I think an entire book on TLC/paper chromatography would be quite overkill and your best bet is on a more general practical organic chemistry book. I can recommend The Organic Chem Lab Survival Manual: A Student's Guide to Techniques by Zubrick if you are only interested in technique or need a introductory text. Else Vogel is of course the book for practical organic chemistry. TLC is covered in detail.

I'm not so sure about paper chromatography, I don't believe it is often used in the lab (at least I have never worked with it). TLC is much more common.


Total entropy change for a spontaneous reaction

The initial state of your system consists of 4 moles of $\ce{Fe(s)}$ and 3 moles of water vapor at $\pu{298 K}$ and $\ce{1 bar}$ in separate containers. The final state of your system consists of 2 moles of $\ce{Fe2O3(s)}$ and 3 moles of $\ce{H2(g)}$ at $\pu{298 K}$ and $\pu{1 bar}$ in separate containers.
The $\Delta H^\circ$ and $\Delta S^\circ$ correspond to a process for taking this system reversibly from the initial state to the final state. They don't include the changes in enthalpy and entropy of the surroundings in executing this process.
Do you have an idea of what such a process would be?


Explaining why s and p orbitals do not interact

According to my chem textbook:
"Molecular orbitals obtained from 2px and 2py orbitals are not symmetrical around the bond axis because of the positive lobes above and below the molecular plane."
Your instructor might be pointing out to the bond-symmetry thing in MO formation between pz and s orbitals extrapolated in most chemistry textbooks.
But orbital overlap is a result of constructive and destructive interference. As s orbitals are spherical, they have infinite possibilities of orbital overlap with other orbital shapes and thus even overlap with d orbitals (as explained by VSEPR: Valence Shell Electron Pair Repulsion Theory in a similar manner).
Thus s and px and s and py orbitals do overlap; if you take px to be of lower energy than py and pz, s-px overlapping occurs in boranes (formed by boron, which has one electron in 2px orbital, and hydrogen, which has one electron in 1s orbital), that often dimerize. This is also the case with methane, where px, py-s overlapping occurs, assuming that in sp3-hybridized state px, py orbitals in C bond with H 1s orbitals.


The most stable allotropic form of sulfur exists in the formula of S8. In other words, sulfur occurs naturally in S8 formula. Multiply all the stoicheomtric coefficient by 8 (other than S) to obtain the balanced chemical equation


Charge of polyatomic ions doesn't equal the charge of ions it consists of?

H-O-O-H.  The bond in the middle is covalent.  Deprotonate to $\ce{{HOO}^{-}}$ (hydroperoxide) then $\ce^{^{-}{OO}^{-}}$, peroxide dianion, $\ce{O2^{2-}}$.  You learn it by doing it, having examples, learning what to look for, and gaining understanding.  Both oxygens are in the -1 oxidation state.

Even good answers here have a legitimate diversity of explanation.  The world is complicated and dirty.  Unlike physicists (quantum gravitation, SUSY, dark matter), we can test our predictions with experiments.  Chemists are not too stinking proud to admit we were wrong, e.g., benzene displacements then palladium catalysis.  That is how we make real world progress, by discovering footnotes.

Introductory inorganic does not have wiggle room.


What would the phosphorus-NMR of trifluorodiphenylphosphorane look like?

Well, these are the fundamentals of NMR that are important:

Signals will be split if they couple to neighbouring atoms with a spin $\ne 0$ (though coupling to spin ½ nuclei is most common especially in organic NMR).
Every different environment of a coupling nucleus will produce a distinct coupling.
Depending on the number of nuclei in a given environment, the number of signals (coupling pattern) is given by the equation $2nI+ 1$, where $n$ is the number of nuclei and $I$ is that nucleus’ spin.

In your example you have correctly identified two different fluorine environments with one and two fluorines in them respectively.

The single equatorial fluorine will give a doublet because $2nI +1 = 2\ (n = 1, I = \frac{1}{2})$.

For the same reason, the axial fluorines will give a triplet: $2nI + 1 = 3\ (n =2, I = \frac{1}{2})$.

If one coupling gives a triplet and the other a doublet, the final result is either a doublet of triplets or a triplet of doublets. (The larger coupling constant i.e. the greater splitting is given first.)

The fluorine NMR would be more complex: They can couple to both phosphorus and the other fluorine. It is important to know that the axial fluorines are homotopic, i.e. the molecule can be rotated exactly exchanging their places — this means that the two give an identical signal. The equatorial one cannot be brought to overlap with the axial fluorines by any mean of symmetry; the two environments are heterotopic, i.e. they will produce different signals (and typically couple).

The axial fluorines will couple to the equatorial fluorine and phosphorus.
The equatorial fluorine will couple to the axial ones and phosphorus.

I’ll leave you to work out the coupling patterns. (Hint: The less bonds between two coupling nuclei, the larger the coupling constant usually is.)

A final note: It is due to electronic reasons that the phenyl groups occupy the equatorial positions rather than going axial. A four-electron three-centre bond is more stable for electronegative elements than electropositive ones.


This is not a simple question at all, in fact, it is an interesting insight in chemical bonding.  You make a valid assertion with your question:

  are ionic bonds just extremely polar covalent bonds?

This answer is a simplistic way of looking at it.

According to ChemGuide page Electronegativity, there is no real clear-cut 'boundary' between covalent and ionic bonding. The website describes that there is a 'spectrum of bonding' based on the electronegativity (chart shown below) of the constituent elements.

Image source: UC Davis ChemWiki

So, the greater the electronegativity, the greater the polarity or "pulling power" on the bonding electrons towards elements with a higher electrongativity.

In the following diagrams (from the ChemGuide site linked before), A and B represent arbitrary elements, and the green and red dots represent bonding electrons. Starting from a situation where the polarity is zero as the electronegativity of the constituent elements are the same (e.g. diatomic molecules $\ce{O2}$, $\ce{Cl2}$ etc), as in diagram I.

As the polarity increases, the bonding electrons move towards the more electronegative member, causing the bond to become increasingly polar (II), then eventually to the point where the difference is enough (III) for:

  To all intents and purposes, A has lost control of its electron, and B has complete control over both electrons. Ions have been formed.

I bolded the 'to all intents and purposes' to emphasise that this is a simplistic generalisation

(I)

(II)

(III)

As the lowest electronegativity is not zero (it is Francium at 0.7), there will always be an influence on the bonding electrons. So, the reality is that A does not completely lose its influence on the bonding electrons.

For example, if we were to look at a compound combining the most and least* electrongative elements (as appear in the diagram above): francium fluoride $\ce{FrF}$, even though fluorine has a far stronger electronegativity than francium, the electronegativity of the latter is not zero, so, the influence of the francium is very small, but not zero - indicating that purely ionic bonds don't exist.

As noted in the comments, cesium is likely to be less electronegative than francium, however the point made above still stands.

This is nicely summarised by an article from the University of Malta:

  Pure ionic bonding is not known to exist. All ionic compounds have a degree of
  covalent bonding. The larger the difference in electronegativity between two
  atoms, the more ionic the bond.