Is there a way to tell which one is the racemic or meso compound with 1H-NMR if all you had was the structure and spectra?

Hydrobenzoin has 3 stereoisomers: meso, R,R and S,S. You have not included the S,S. Also, your Newman projections might make sense based on steric grounds, but largely ignore electronic factors such as pi-stacking and hydrogen bonding. A very quick run through the Avogadro Bable geometry optimisation has these molecules optimising with the phenyl rings somewhat gauche.

While there are many methods in NMR that can assist in determining relative and absolute stereochemistry, there is no method that offers universal application. Generally it is not possible to identify stereoisomers from first principles in non-rigid skeletons. This is because conformers are rarely locked into a single position. Properties such as spin coupling, and nOe are useful, and the method by Murata is considered groundbreaking in this area. It relies on measurable  3J(HH) and 2,3J(CH).
Derivatisation, database comparison, measurement of residual dipolar couplings in aligned media (liquid crystals), and computational methods. Determination of absolute configuration usually involves complexation with a enantiomerically pure reagent.

And I'm not confident that your method for assigning cis/trans geometry based on chemical shifts is reliable.


I could not find a definitive mechanism for 1,4-dioxane, but this is the mechanism for the formation of THF peroxide is likely very similar

diagram from here


$\ce{F2}$ will produce  $2\space \ce{F-}$ ions.

You can write the balanced half cell reactions as follows:

$\ce{F2 + 2e- ->2F-}$

$\ce{Li -> Li+ + e-}$

Multiply the second equation by 2 to get $\ce{2Li -> 2Li+ + 2e-}$. This is to ensure that the final equation does not have any electrons on either side.

Now add this equation with the fluorine one.

You'll get $\ce{2Li + F2 -> 2Li+ + 2F-}$ which you can write as $\ce{2Li + F2 -> 2LiF}$


The issue here is that Schrödinger's equation doesn't yield us a "solution of 90% probability". It doesn't yield a probability at all. It yields a wave function, $\psi(x,y,z,t)$, which has units $\frac{1}{\sqrt{m^3s}}$. To get a probability density, we square its norm: $|\psi|^2=\psi^*\psi$. Note: even this is not a probability. A probability is dimensionless, whereas this has units $\frac{1}{{m^3s}}$. We can represent the probability of the particle described by $\psi$ being in the range (x, y, z, t) and (x+dx, y+dy, z+dz, t+dt) by multiplying the norm squared by dxdydz*dt: $P(x,y,z,t) = |\psi(x,y,z,t)|^2dxdydzdt$. Of course, these increments are infinitesimal, so this alone would evaluate to zero. However, this approach allows you to integrate over a range of x,y,z,and t:
$P(x+\Delta{x},y+\Delta{y},z+\Delta{z},t+\Delta{t}) = \int{dx\int{dy\int{dz\int{dt|\psi(x,y,z,t)|^2}}}}$.
A better understanding of this will help you understand the issues with your second question. However, we can also rely on the Pauli exclusion principle to assert that there cannot be two Fermions that share the same set of quantum numbers (note that some of the quantum numbers may be shared, but not all simultaneously.


If we start heating a substance, does it melt when we break its intermolecular bonds?

I'm somewhat befuddled as to the confusion. I think it is perhaps the problem is in the phrase "intermolecular weak bonds."

If you look at a diamond crystal there aren't any diamond "molecules." Rather the whole crystal is one big "molecule."

Likewise we write the chemical formula for sodium chloride as $\ce{NaCl}$, but in solid and liquid sodium chloride there are no sodium chloride molecules. True sodium chloride molecules can only exist in the gaseous state.

If you look at $\ce{H2O}$ the hydrogen atoms are extremely labile. So the two hydrogen atoms and the one oxygen atom that were a molecule when the ice froze might not be part of the same molecule when the ice melts.

So to fit the overall criteria of "intermolecular weak bonds," we need a molecule which has stronger intramolecular bonds than intermolecular bonds. Say methane for example. So when methane freezes the same molecule that freezes in the solid will be the same molecule that is freed when the solid methane melts.

So in the case of methane adding energy in the form of of heat to solid methane will eventually reach the melting point of methane. At that point the weaker intermolecular bonds of the crystal break and the bulk solid turns into liquid methane.

So the overall notion is that the solid form exists as colder  temperatures than the liquid form and that the liquid form exists at cold temperatures that the gaseous form. So generally we take solid and heat it. When (IF!) it turns to a liquid then the temperature of the transition is the melting point. Likewise if the liquid can be turned into a vapor then the temperature of the transition is the boiling point.  Thus solid methane and sodium chloride will both "melt," but for different reasons.


How is ammonia used in production or processing of steel?

Likely, for nitriding http://en.wikipedia.org/wiki/Nitriding . Essentially, ammonia is used as nitrogen donor. Nitrogen atoms diffuse into steel, producing hardened, corrosion-resistant surface layer.


Determining the Energy Density of a Battery given the anode and cathode

Cathode and anode will have different specific charge capacities. These capacities can be made equal by tailoring the thickness of the electrodes. The question is asking you to make sure that the charge capacity ratio is equal to one, which means you will need to find what anode thickness you must have for the capacities match (they give you cathode thickness of 100 microns)
This is the equation to find the energy density of a battery: $$E=VC/m$$
$E$ is energy density in Wh/kg, $V$ is nominal voltage in volts, $C$ is capacity in Ah and $m$ is mass in kg. Why are you multiplying this equation with thickness and porosity?
You were given some electrode geometric parameters and you will have to find the others. Remember, you are looking for the electrodes masses so that you can compute energy density (hint: start by looking into densities of your materials).
Keep in mind that not all electrodes will be porous (for example for Li metal you can assume porosity is 0%). Also, you will have to report results in $J/kg$, so make sure you keep track of your units and convert if necessary.


Completely chemically inert metal alloy?

Due to electronegativities no alloy would be thermodynamically stable in the presence of fluorine (F$_2$). Even in an alloy the constituent metal atoms are either "sharing" electrons or receiving/donating electrons. Fluorine is the element with the highest known electronegativity and would "pry" the electrons from the electron-accepting metal atoms.


Let's have a look at the definitions of dew:

  dew (the free dictionary)

  Water droplets condensed from the air, usually at night, onto cool surfaces.
  Something moist, fresh, pure, or renewing: "The timely dew of sleep / ... inclines / Our eye-lids" (John Milton).
  Moisture, as in the form of tears or perspiration, that appears in small drops.

Therefore dew is water in the condensed phase, i.e. liquid. Water vapour on the other hand is the gaseous phase. They are definitely not the same.


What causes polarization in noble gases?

Although this applies to all noble gases, consider helium. There are two electrons moving around the nucleus. The movement of those electrons is symmetrical and during a defined amount of time, the atom is not polarised. However, the momentary positions of those electrons can polarise the atom. This can happen when both electrons are relatively close, on the same one side of the atom. In this case a momentary dipole is formed, since there is separation of two charges, nucleus being positive and the pair of electrons being the negative pole.


One way you can do this is by looking at the packing structure of the metal.

As an example, if you look at Wikipedia, you see that Tungsten has a body-centered cubic crystal structure. This means that in each unit cell, there are going to be two Tungsten atoms. We can then predict the density of a perfect Tungsten crystal lattice using some geometry and unit conversion.

First off, I'll give you an equation which you can prove to yourself quite easily so I won't go into that. Density of a crystal is: $$\rho=\frac{n*M}{N_A*V}$$

Where, $n$ is the number of atoms in the unit cell, $M$ is the molar mass of the atom, $N_A$ is Avogadro's number, $V$ is the volume of the unit cell.

So, for Tungsten this comes out to be $$\rho=\frac{2*183.83 g*mol^{-1}}{6.022*10^{23}*(\frac{4*139*10^{-10}cm}{\sqrt{3}})^3}=18.45\frac{g}{cm^3}$$

The experimental density of Tungsten is $19.33 \frac{g}{cm^3}$.

The answer is usually a little better than that, but still quite close.

The only information you need to do this calculation which isn't on a periodic table is the packing structure and the atomic radius.

Something which is noteworthy is the atomic-packing-factor, $APF$, which comes from finding the ratio of volume of the atoms to volume of the unit cell and represents how much space the atoms fill in the cube, or how efficient the structure is at packing.

For the body-centered cubic (BCC), $$APF=\frac{2*\frac{4}{3}\pi r^3}{(\frac{4r}{\sqrt{3}})^3}=0.68$$

That means BCC, takes op 68% of the total available space per unit cell for equal sized spheres.

Check out this link if you want more info on that.

So, to answer the actual question which how do we find a trend with all this, we now know that density depends on radius, which we already have a trend for, molar mass, which also has a very simple trend, and packing structure, which is the real unknown.

There is this from this page,

  In the resonating valence bond theory, the factors that determine the choice of one from among alternative crystal structures of a metal or intermetallic compound revolve around the energy of resonance of bonds among interatomic positions. It is clear that some modes of resonance would make larger contributions (be more mechanically stable than others), and that in particular a simple ratio of number of bonds to number of positions would be exceptional. The resulting principle is that a special stability is associated with the simplest ratios or "bond numbers": 1/2, 1/3, 2/3, 1/4, 3/4, etc. The choice of structure and the value of the axial ratio (which determines the relative bond lengths) are thus a result of the effort of an atom to use its valency in the formation of stable bonds with simple fractional bond numbers.
  which I don't actually understand but seems to explain why certain lattices are chosen.

Basically, using the fact that radius decreases going right and molecular weight increases going right, we would predict that density would increase uniformly across the periodic table for elemental metals, except that various metals pack in different ways. Hexagonal Close Packed is the most efficient packing system, so I wouldn't be surprised to find that being associated with many high density metals.

I hope that gives a good idea of how there is sort of a trend, but also why no trend is truly there.

EDIT:

To figure out which has the highest density, I would start by figuring out which pack in a Hexagonal Close-Packed structure as that is the most efficient packing structure with an $APF$=.74


Iron on reaction with HCl forms Fe(II) salts and not Fe(III) salts because even if Fe(+3) is formed, as we can see in electrochemical series Fe(+2) will be favoured ie. H2 which gets can reduce Fe(+3) to Fe(+2). So thermodynamically Fe(+2) salt is favoured. If you want to make Fe(+3) salt then you can react Fe and Cl2 directly.

  How H2 gets its reducing property?

H2 gas will react with FeCl3 to form FeCl2 and HCl. I don't know the exact mechanism but it can be explained well on the basis thermodynamic data available.


Why is there a difference in the colours of K₂Cr₂O₇ and K₂CrO₄?

I'm really excited for this because I get to reference the almighty color wheel!! Fair warning, this answer is much more qualitative than quantitative, but that's more interesting sometimes anyways.

If you look at the structure of the chromate and dichromate ions next to each other (see here for structures: https://en.wikipedia.org/wiki/Chromate_and_dichromate), the only major difference between the two is that the Cr-O bond joining the two chromate ions (missing an oxygen) is now a single bond. This means that bond will vibrate at a lower frequency, and because frequency and wavelength are inversely related, that bond will absorb a longer wavelength of light.

Now to the color wheel! It is a general chemistry (often unexplained) fact that the color we see is the complementary color of the wavelength of a bond's vibration. Thus, in the case of the chromate ion, we see yellow, and across from yellow is the purple-ish region. That means, if one of the bonds in the chromate ion, and thus two of the bonds in the dichromate ion, were absorbing a longer wavelength like we said earlier, on average we would expect something just longer than purple-ish, like blue, to be absorbed.

The complementary color of blue is red slash orange, and that is in fact the color we see in the dichromate ion!

At the heart of all this is the principle that the colors we see are those wavelengths of light which on average are not absorbed by a large number (on the order of Avogadro's number) of molecules.

An approach like this will only be reliable for very similar molecules like the two we have here.


4-Bromophenol vs 4-aminophenol pKa in water comparison confusion

Answer to this question is, as orthocresol pointed out, simpler than what you think. As you know, $\ce{-Br}$ is deactivate the ring (even though it is o,p-directing) while $\ce{-NH2}$ is great ring activator. What that means is regardless of the mesomeric effect, compared to $\ce{-H}$, $\ce{-Br}$ is overall electron withdrawing and $\ce{-NH2}$ is electron donating (remember, $\ce{-NH2}$ is a Lewis base).

Said that, I'd say basically, $\mathrm{pH}$ of a phenol ($\mathrm{p}K_\mathrm{a} = 9.95$) is depended on how stable phenolate ion is after it donated $\ce{H+}$. Electron withdrawing groups stabilize phenolate ion (e.g., $\ce{-NO2}$ group, p-nitrophenol: $\mathrm{p}K_\mathrm{a} = 7.1$) while electron donating groups destabilize it (e.g., $\ce{-OCH3}$ group, p-methoxyphenol: $\mathrm{p}K_\mathrm{a} = 10.2$). Thus, one can expect that p-bromophenol should have a lower $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more acidic than phenol) while  p-aminophenol should have a higher $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more basic than phenol). This is true in reality. Actual $\mathrm{p}K_\mathrm{a}$ values of p-bromophenol  and  p-aminophenol are $9.34$ and $10.3$, respectively ($9.34 \lt 9.95 \lt 10.3$).

Sources for $\mathrm{p}K_\mathrm{a}$ values: Evans pKa Table and Wikipedia.


Why does opened wine develop the sour taste and smell of vinegar?

The ethanol (alcohol content) of the wine, is oxidised, by the oxygen in the air into ethanoic acid. Vinegar is dilute acetic acid which is where it gets its smell.

The reaction happens in to stages,

The ethanol is partially oxidised to ethanal
The ethanal is further oxidised to ethanoic acid


Difference between electropositivity and electronegativity

Background:
Electropositive  implies having tendency to lose electrons to form positive ions. Electronegative implies having tendency to gain electrons to form negative ions. Electronegativity is a concept used to describe the tendency of an atom to attract electrons in a covalent bond.

  Why is the electronegative value of hydrogen greater than that of sodium?

This is because sodium after losing an electron gets its octet configuration, but a hydrogen atom after losing an electron will not get the configuration of helium but it can get helium configuration if it accepts an extra electron or forms a covalent bond.

So this implies that sodium is more electropositive that hydrogen, even though they belong to same group.

Sometimes "more electronegative" is used to describe higher electronegativity and "more electropositive" is used to describe lower electronegativity. In this sense, elements are less electronegative (or more electropositive) as you go down any group in the periodic table. However, the difference in electronegativity between hydrogen and sodium is exceptionally large, and correlates with the very different chemistry these two elements show.


What is the role of acid in keto enol tautomerism

In acid-catalysed keto-enol tautomerism, a lone pair of electrons is removed from the oxygen of carbonyl group by the acid.

If enol is the starting compound, markovnikov's addition of $\ce{H+}$ occurs.


Ionization energy of neon vs its cationic counterpart

Does this anything to do with size of the atom shrinking?

Yes, precisely; this is one way of putting it. Another way is to say that because one electron has been removed, the remaining electrons are less shielded from the nuclear charge: therefore, the effective nuclear charge increases, and the remaining electrons are harder to remove.

  Is this because cationic counterparts always have more ionization energy than their neutral counterparts because it becomes successively more difficult to remove electrons from atoms on successive removals?

Yes. This is true for every element in the Periodic Table, with no exceptions, and it doesn't matter what the electronic configuration is.

If all else were equal, then it would indeed be harder to remove an electron from a noble gas configuration than a non-noble gas configuration. However, that is a very big if! Between $\ce{Ne}$ and $\ce{Ne+}$, the electron configurations are indeed different, but all else is not equal: as discussed previously, the effective nuclear charge in $\ce{Ne+}$ is greater than that in $\ce{Ne}$.


How can I calculate the boiling temperature of a pure substance as a function of pressure?

If you know the latent heat of vaporization at one temperature, you should be able to determine the latent heat of vaporization at any other temperature from Hess' law and knowledge of the heat capacities of the liquid and vapor as a function of temperature.  You may also have to include a pressure contribution as the critical point is approached.


Cs Point Group's "Horizontal" Reflection Plane

There is always a principal rotation axis, even if it's only a $C_1$ axis. As there are an infinite number of these one can always be found perpendicular to the reflection plane, and as horizontal planes take precedence over veritcal ones, the operation is listed as $\sigma_h$.


Short Cut Method to Calculating pH of Polyprotic Acid?

The assumption that the second (and higher) deprotonation steps are negligible for the calculation of the pH of a polyprotic acid is valid for citric acid.

You can compare this by calculating the exact solution and then stepwise neglect deprotonation steps.

(For the sake of simplicity I write $c(\ce{H+})=x$ and $k_{a,n}=k_n$.)

Exact: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}$$
Neglecting $k_3$: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x k_1+2 k_1 k_2}{x^2+x k_1+k_1 k_2}$$
Neglecting $k_2$ and $k_3$: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{k_1}{x+k_1}$$

Solving those equations give the following pH values (with super unrealistic high precision):

Exact:               1.2469291050971560
w/o $k_3$:           1.2469291065371300
w/o $k_2$ and $k_3$: 1.2470654318802320

As a pH can only be measured precisely within something like $\pm \left(10^{-2}\ldots10^{-3}\right)~\mathrm{pH}$, the calculated pH can be rounded to be 1.247 in every singly case. No matter which deprotonation step was neglected.

So for your case it is negligible. For more general cases I have to look it up and will add it to my answer within the next days.


Why are 1,3-diols protected with aldehydes but 1,2-diols are protected with ketones?

Let's look at the protection of a typical 1,3-diol with benzaldehyde and acetone:

The theory is that the formation of the six-membered cyclic acetal is less favourable with acetone than it is for benzaldehyde, because with acetone you cannot avoid having one methyl group being axial.

The next-most stable thing after a six-membered ring is a five-membered ring, which can be formed by the protection of a 1,2-diol instead of a 1,3-diol. So, acetone happily reacts with those 1,2-diols to form five-membered cyclic acetals, whereas benzaldehyde is more selective for 1,3-diols since it can enjoy the stability of a six-membered ring.

When there is a choice, like in glucose, you get rather different results:

However, I am not sure how much this principle extends to protection of compounds with only two hydroxyl groups. I am guessing it should be possible to protect 1,3-diols with acetone too - after all, a methyl group is hardly the worst axial substituent to have.


What effect does Baking Soda and Citric Acid have on a soft drink?

Actually if we care just about the reaction between Baking Soda ($\ce{NaHCO3}$) and Citric Acid ($\ce{C6H8O7}$) the following reaction should ensure :

$$\ce{3NaHCO3(aq) + C6H8O7(aq) -> 3 H2O (l) + 3 CO2(g) + Na3C6H5O7(aq)}$$

So basically the reaction should produce water, carbon dioxide and trisodium citrate (as known as E331 : a food additive)


Why does the binding of DMSO to metal via the S atom (k-S) strengthen the S=O bond?

The confusion seems to arise from the incorrect statement that the S-O bond has pi character and that there exist $\pi$ and $\pi^*$ molecular orbitals in DMSO. Instead, DMSO (despite the way it is often drawn) is most accurately depicted as having three single bonds and one lone pair on the the S atom (clearly visible as a dominant element in the HOMO shown in the question). The trigonal pyramidal geometry (also shown in the question) is consistent with this description.
The S-O bond is a dative bond, meaning that the S atom provides both electrons rather than having one electron come from each atom. That results in a formal positive charge on S and negative charge on the O. The O therefore is best represented as having one bond (to S) and three lone pairs, similar to the oxygen atom in an alkoxide ion. Because of the charge separation, the S-O bond also has a strong ionic component.
[Aside: so-called hypervalent sulfur compounds are typically drawn with double bonds that are actually combined sigma + ionic bonds rather than sigma + pi bonds. DMSO is not the only one. Keep this in mind in the future when trying to understand properties of sulfur-containing molecules.]
We now consider what happens when DMSO interacts with a transition metal. Similar to the O of an alkoxide ion, the O atom in DMSO can interact with metals as both a $\sigma$ and $\pi$ donor because of the multiple lone pairs. Both the $\sigma$ and $\pi$ donations result in decreased electron density on O, reducing its charge and weakening the S+/O- ionic interaction. The weaker ionic interaction means that the bond overall weakens, consistent with the decrease in the observed stretching frequency. There is no need for an S-O $\pi$ interaction to exist in order for this to happen.
If instead the S atom interacts with the metal, it does so primarily as a $\sigma$ donor via its one lone pair. There may be capacity for a $\pi$ acceptor interaction with $\sigma^*$ orbitals, as has been proposed for ligands such as triphenylphosphine, but if this occurs, it is with the $\sigma^*$ orbital associated with the S-C bonds, not the S-O bond, consistent with the LUMO shown in the question, which has a large contribution from the py orbital on S (ie the p orbital in the plane of the C atoms). Thus there is no weakening of the S-O $\sigma$ interaction due to $\pi$ acceptance.
As long as the $\sigma$ donation is stronger than the $\pi$ acceptance (which is what we expect), there is a net loss of electron density from S, which further increases the positive charge on the already electron deficient S atom. Increased charge means an increased strength in the ionic S+/O- interaction, so the S-O bond overall is stronger, consistent with the observed increase in the S-O stretching frequency.


How to determine electrolyte concentration in solutions?

Electrolytes turn water into a conductive solution.  Conductivity is affected by how free the electrons are to move.  Conductivity (which is the reciprocal of resistivity) is affected by by concentration, temperature, and how much charge the ions carry.  Conductivity alone will give an equation with both ions and concentrations; however, this equation will have multiple solutions. For some fun understanding this phenomena look at this this link.

Conductivity is different than voltage.  Voltage in this situation is likely referring to a half-cell reaction with a standard hydrogen electrode in a galvanic cell.  Using this approach has the same issue with a multiple solution equation.

Electrolytes are typically ions.  Ion concentration is measured with ion chromatography.  Ion chromatography involves separating the ions from the solution based on various properties to perform a measurement. It is explained here, here, and here.

When electrolytes of drinks are measured it is important to include if the electrolytes are bio-available.  Bio-available electrolytes depend on the activity performed.  An activity such as sports uses a different set of electrolytes than sitting.

Testing bio-available electrolytes requires more work than analytic chemistry. Analytic results both the drink and bodily fluids of a population doing a certain activity are need to draw conclusions about the electrolyte composition of a drink.

Reading a patent for a sports drink ions include $\ce{Na+}$, $\ce{K+}$, $\ce{Mg^2+}$, $\ce{Cl-}$, $\ce{SO4^2-}$, $\ce{PO4^2-}$.  The patent also includes citrate, sucrose, dextrose, ascorbic acid and pyridoxine as these are known to affect the uptake of electrolytes.

If you have access to a university library the values of these compounds for orange juice are available.


Are pi bond like like interactions possible when substituents in a compound are not directly connected?

First, let's get to the root of this problem, then I'll double back to your question. We are asked which of the two compounds is most acidic. As a side note there are various definitions of acidity, but note that broadly these theories describe acidity as linked to propensity to accept electron pairs in some form rather than general instability.
Considering neither II nor III have any immediately acidic functionalities (alcohol, carboxylic acid, sulfonic acid), we identify the alpha carbons of the carbonyl functionalities in each molecules as the most acidic hydrogens with pka ~20-25 (since all other hydrogens are alkyl (pKa ~50) and cannot form a stable conjugate base upon deprotonation).
With that in mind, the real difference between II and III is the identity of the carbonyl group. Comparing II to III we see II is an ester, with the carbonyl center bonded to an alkoxyl substituent and an alkyl substituent. Comparatively, III is a ketone, with its carbonyl center bound to to alkyl substituents. Recognizing that O lone pairs can hybridize to conjugate with the carbonyl center were C atoms cannot, we predict III to be more acidic.
The reasoning for this can be approached from a couple of angles, so here's two:

O lone pairs present in II destabilize the enol formed by deprotonation, making it more likely to donate electrons (I.E has more basic character than III), thus making its conjugate acid form less acidic (since weaker acids have conjugate bases that are stronger bases and vice versa)

O lone pairs can form an additional bond to the carbonyl carbon in a octet-satisfied resonance contributor, which greatly reduces the contribution of the analogous hyperconjugative resonance contributor with dissociation of H+. Note that this hyperconjugative contributor is a rather poor description in a thermodynamic sense, but it does play an effective stand in for a mechanism with a basic species.

Now to your actual question: You are correct with your analysis of the pi system of II, but I would note that your description is using atomic orbitals which can become very confusing in the world of ochem. In the hand-wavy hybridization terms, each oxygen would adopt an sp2 bonding orbital hybridization with unhybridized p orbitals participating in the pi-system. However, these pi-interactions can only occur if sequential atoms have unhybridized p orbitals.
This is due to the fact that the stability of pi systems arises from overlap of p orbitals. When two p orbitals are close enough to overlap, their wavefunctions can recombine into  relatively lower energy (bonding) and relatively higher energy (antibonding) wavefunctions which become orbitals. This same recombination can occur for any number of p-orbitals, with the sole condition being that overlap actually occurs.
This can only have a fully stabilizing effect when atoms are ~1.20-1.5 angstroms apart as indicated by spectroscopic bond lengths. Overlap is feasible between single-bonded substituents (Like C-O in II), but impossible once the continuous system of unhybridized p orbitals is broken.
Thus while the O in II can directly conjugate its electrons with the carbonyl pi system by "rehybridizing" from the general expected sp3 hybridization to stabilize the system, in III there are no adjacent unhybridized p orbitals to interact with the carbonyl pi system. The only electronic interactions present between the substituent (cyclic ether) and the carbonyl system would be inductive/field effects.
While one could argue that a conformer of III could possibly involve delocalization of O p electrons into the carbonyl system, I would respond by noting that such a conformer

Would have to have an unhybridized O p-orbital in order to be involved with pi interactions
Would have to have a torsional angle Θ between the oxygen and the carbonyl group of Θ=/=0 in order for overlap to actually occur (Distance between orbitals r<~1.5).

However, this would mean that the oxygen's unhybridized p orbital and the carbonyl pi system would have different symmetry elements, meaning such a pi-interaction would be symmetry-forbidden
That last bit might be jumping the shark a little, but I hope this helps!


You'd think that reaction of an alkyl grignard with a (primary) alkyl bromide/iodide gave all sorts of grignards via metal/halogen exchange.  I guess that's what your professor means by transmetallation.  You'd also get some elimination products.  Certainly that's what would happen with alkyllithiums.

That's why for carbon-carbon bond formation reactions you'd use cuprates.  Reduced basicity, increased nucleophilicity, all that.


What determines "polarizability" of an element?

Fluorine is not very polarizable because it is small. Its electrons, therefore, are all close together. A polarized atom has shoved all its electrons to one side. Since they are close together in Fluorine, the negative-negative repulsion is too large for that. In another halide such as Iodine, whose valence electrons are much farther from each other, much better polarization can be achieved.

More polarizability does NOT always mean better or weak bonding. Generally, polarizable atoms like to bond to each other, and non-polarizable atoms like to bond to other non-polarizable atoms. For more info, see:

http://www.adichemistry.com/inorganic/cochem/hsab/hard-soft-acid-base-theory.html


The water referred to in the article was being stored in a reservoir after being treated with chlorine.  It is the combination of chlorine and sunlight that facilitates oxidation from bromide to bromate.

So, no, this is not applicable to natural bodies of water, only bromide containing water which has been artificially treated with chlorine.

See Bromate ion formation in dark chlorination and ultraviolet/chlorination processes for bromide-containing water for more information.


There is a old paper which discuss the mechanism of decomposition of bleach. It says the major product is chlorate($\ce{ClO3-}$)

  The decomposition of sodiun hypochlorite has been re-examined. The
  results show that Foerster and Dolch's mechanism of the decomposition
  to chlorate and chloride is correct; they postulated a slow
  bimolecular reaction to chlorite and chloride, followed by a faster
  reaction of chlorite with more hypochlorite. [...] The rates are such
  that at 40 C. A solution of sodium hypochlorite will contain about lyo
  as much chlorite as hypochlorite.[...]. No signs of specific catalytic
  effects of sodium chloride, hydroxide, or carbonate could be observed,
  and it seems probable that earlier reports of this were due to
  variations in ionic strength. The decomposition to chloride and oxygen
  has been measured and is a unimolecular reaction, which is possibly,
  but not certainly, uncatalyzed.[...]

  $$\ce{2NaOCl -> NaClO2 + NaCl}$$
  $$\ce{NaOCl + NaClO2 -> + NaClO3 + NaCl}$$
  $$\ce{2OCl- -> 2Cl- + O2}$$

But since this paper is in the 50's, most of the information would  be outdated or misleading. So, here is the result.

Wikipedia says:

  It decomposes (autoxidizes) when heated, forming sodium chlorate    and sodium chloride:
     $$\ce{3 NaOCl(aq) → NaClO3(aq) + 2 NaCl(aq)}$$
  Sodium hypochlorite decomposes with increasing temperature and under    the influence of light and such metals as copper, nickel, or
  cobalt:
     $$\ce{2 OCl⁻(aq) → 2 Cl⁻(aq) + O2(g)}$$

There are two pathways of decomposition of bleach. One is forming chlorate and other one is forming chloride. But the 1st reaction is dominant one i.e chlorate is the major product. But the formation of chlorate actually requires heat(to be precise 303-323 K).

The 2nd reaction also requires heat(343 K) and also some catalysts:

  Typical metals for nickel and copper are in the range of 50 ppb or
  less when sub-micron filtration is provided but many times, levels in
  the range of 300-400 ppb for nickel and 50-60 ppm for copper can be in
  the product which will create $\ce{O2}$ and increased decomposition.(source)

So, the conclustion is that the formation of chlorate or chloride do require heat. But there are 2 reaction that indeed happens when bleach is remained untouched for a long time i.e reaction with carbon dioxide and moisture.

  If not properly stored in airtight containers, sodium hypochlorite
  reacts with carbon dioxide to form sodium carbonate:

  $$\ce{2 NaOCl(aq) + CO2(g) → Na2CO3(aq) + Cl2(g)}$$

  $$\ce{NaClO + CO2 + H2O → NaHCO3 + HClO}$$

  Sodium hypochlorite react with carbon dioxide and water (moisture) to
  produce sodium bicarbonate and hypochlorous acid. Sodium hypochlorite -
  saturated solution. The reaction proceeds at room temperature.(chemiday)

Hypochlorous acid is somewhat unstable so it might decompose to give chlorine fumes or further react with hypochlorite to form chlorate.

$$\ce{OCl- + 2HOCl -> ClO3- + 2HCl}$$


You need two half reactions for the total redox reaction: one reduction and one oxidation. The convention is that we provide all potentials as standard reduction potentials. If you need an oxidation potential, you're going to need to reverse the reduction half reaction and negate the sign standard potential.

This means that the total potential of the cell is the raw sum of the oxidation and reduction potentials. In your specific example, you want:
$$\ce{Ga^{3+} + 3e- -> Ga}, E_{red}^{\varnothing} = -0.560\,\mathrm{V}$$
$$\ce{Fe -> Fe^{2+} + 2e-}, E_{ox}^{\varnothing} = +0.409\,\mathrm{V}$$

The EMF of the cell is the sum:

$$E_{cell}^{\varnothing} = E_{red}^{\varnothing} + E_{ox}^{\varnothing}$$

And this value is negative.

Now, as a shortcut, you can identify which of the two reduction half reactions needs to be reversed to make an oxidation, and you can add that value--negated--to the potential for the reduction half reaction, as above.

What your teachers or book meant was the following. I've changed the names of the variables so that they don't conflict with the standard notation (which is why we're talking past each other).

$$E_{cell}^{\varnothing} = E_{red,1}^{\varnothing} - E_{red,2}^{\varnothing}$$

where $E_{red,1}^{\varnothing}$ is the standard reduction potential for the reduction half reaction and $E_{red,2}^{\varnothing}$ is the standard reduction potential for the reduction that corresponds to the oxidation half-reaction. This is hellishly confusing, which why I think you are confused. The key here is that

$$E_{\mathrm{cathode}} = E_{red,1}^{\varnothing} = E_{red}^{\varnothing}$$

but

$$E_{\mathrm{anode}} = E_{red,2}^{\varnothing} = -E_{ox}^{\varnothing}$$

Note also that ordinarily, for a spontaneous reaction, iron would constitute the cathode of the cell because it would be the reduction. However, in this case, because the reaction is non-spontaneous, the iron half reaction is actually going to be an oxidation, so that means it's actually going to be the anode. This is the other factor you're confused about. You're trying to apply shortcuts for a spontaneous cell to a non-spontaneous one. Try to understand what the underlying principle is.


G3 is generally a term used to describe a particular computational model used in calculating molecular bonding in quantum chemistry.  The G refers to Gaussian and 3 is just a third iteration of including additionally complex terms to the molecular orbital calculations being used.  There's a brief discussion on wikipedia, but you'll find more extensive discussion of these types of calculations in Quantum Chemistry text books or directly from the researcher's works just an example, a G3 Theory site.
Aromaticity can actually be measured quantitatively via Aromatic ring current and calculated with [NICS] nucleus-independent chemical shift.  The wikipedia article included a link to a paper describing the quantitative scale.  It's just a way of saying how aromatic is the molecule compared to another (call it a standard / perfect example molecule.)  The problem with such relative standards is there may always be one that exceeds the standard, so you could potentially end up with results larger than 100%.


Which d-orbitals split according to CFT?

The $d$ orbitals that split are the $(n-1)$ $d$-orbitals because they are the relevant ones that do chemistry (because they have the electrons).

If the orbitals are degenerate, it doesn't matter whether they're filled or not, the energy level is the same for all the degenerate states. So it also doesn't matter if they're filled or not, the energy states will split up according to CFT.

The compound only "feels" this raising or lowering of energy depending on how many electrons are filled in into the different orbitals after splitting.

To maybe clarify some confusion: The existence of an orbital doesn't mean that the electron has to be in there, it is only a mathematical construct.


The reason why carbon dioxide is a gas and silicon dioxide is a solid is because their chemical structures are different.

Carbon dioxide is a linear structure with two double bonds between carbon and oxygen. It is a small molecule and non-polar with only weak bonds between the molecules. Hence it is a gas.

Silicon dioxide is not formed of small molecules. It consists of an infinite array of silicons where each silicon is bonded to four separate oxygens (and each oxygen is shared between two silicons). This creates a strong refractory solid (glass and sand are mostly silicon dioxide aka silica). So the same apparent overall formula doesn't describe the actual structure of the compounds at all. But the structures explain the difference in behaviour.

Of course this doesn't explain why silicon prefers to bond with four oxygens when carbon prefers just two. This is not completely simple and results from the relative bond strengths of carbon-oxygen bonds, carbon-oxygen double bonds and the equivalent bonds for silicon and oxygen. The simple version is that silicon oxygen bonds are strong relative to their double-bond equivalents whereas carbon-oxygen double bonds are strong relative to their single bond equivalents. Or, more precisely, if we could make a carbon-oxygen network solid with the equivalent structure to silica, it would tend to fall apart into carbon dioxide. If we could make silicon dioxide molecules, they would react with the release of energy to give silica.

Deeper explanations would need to look at why the relative strengths of double and single bonds turn out that way, but that would get into molecular quantum mechanics and would not be much more useful as an explanation.

The simplest explanation is the fact that the structures are different.


How do lone electrons impact the charge of an ion?

The two red dots indicate the presence of two electrons, not the absence, if there were no electrons there it would be balanced. The colours of the electrons in the image indicate their origin: blue electrons originate with oxygen, black from sulfur and red from an outside source (eg. Potassium or H2).
Ions have a charge because they have more or fewer electrons than the number of available protons. Sulfate has gained two extra electrons.
The arrows in the second image indicate a 'dative covalent bond' where the electrons in the covalent bond all come from one atom rather than being shared, while the lines indicate a normal covalent bond (ie. an electron from each atom. The extra electrons added to the oxygen atoms to form a full orbital level in the normal covalent bond are responsible for the extra charge.

There's a nice visual depiction here: http://chem-net.blogspot.co.uk/2013/02/electron-dot-structure-of-sulfate-ion.html


Electrostatic potential map and local ionization map

The excellent MultiWFN is able to produce both functions and luckily, a reference for it has explanations for both as well:

Electrostatic potential

According to the reference above,

  $$V_{tot}(r)
    = V_{nu}(r) + V_{ele}(r)
      = \sum_A \frac{Z_A}{|r - R_A|} - \int \frac{\rho(r')}{|r - r'|} dr'$$
  This function measures the electrostatic interaction between a unit point charge placed at $r$ and the system of interest.
  A positive (negative) value implies that current position is dominated
  by nuclear (electronic) charges.
  Molecular ESP has been widely used for prediction of nucleophilic and electrophilic sites for a long time.
  It is also valuable in studying hydrogen bonds, halogen bonds, molecular recognitions, and the inter-molecular interaction of aromatics.

Average local ionisation energy

Again, according to the reference above,

  $$\bar{I}(r)
    = \frac{\sum_i \rho_i(r) |\epsilon_i|}{\rho(r)}$$
  where $\rho_i(r)$ and $\epsilon_i$ are the electron density function and orbital energy of the $i$th molecular orbital, respectively.
  This function has many uses, for example, reproducing atomic shell structure, measuring electronegativity, quantifying local polarizability and hardness, and predicting sites for electrophilic or radical attack.

Thus, while the electrostatic potential represents the hypothetical electrostatic interaction energy between the molecule and a positive unit point charge at $r$, the average local ionisation energy is the orbital density-weighted absolute orbital energy average.

Finally, resonance/delocalization/electronegativity are built in the method you use to produce the electron density $\rho(r)$ (edit: a nice discussion about it can be found here).


Nomenclature - Common vs IUPAC Names and Alphabetical Priorities

No, 2,2-dimethyl-7-(1-methylethyl)bicyclo[3.2.0]heptane is not an acceptable name.  Like the rule you quoted says, start numbering at a bridgehead carbon and first pass through the largest ring (in this case the 5-membered ring) and number that ring so that a substituent will have the lowest number.  So this molecule would be named 2-isopropyl-7,7-dimethyl-bicyclo[3.2.0]heptane.  Usage of "1-methylethyl" is not acceptable.  If you weren't going to use the term "isopropyl", you could call the fragment "2-propyl". So you could write 2-(2-propyl)-7,7-dimethyl-bicyclo[3.2.0]heptane, but the term "isopropyl" is much more common usage.

Edit: After reading Greg E's comment, I looked around the internet and found that "1-methylethyl" is both correct and the official IUPAC name.  Here is a paraphrasing of the relevant IUPAC rule...

These groups (those attached off the main chain) are always numbered so that the point of attachment to the rest of the molecule is C1, with the longest continuous chain beginning from the point of attachment taken as the parent.

Then all sites I visited would add something like this...

"For historical reasons, some of the simpler branched-chain alkyl groups also have nonsystematic (or common, or trivial) names."

Most of these sites seemed to (strongly) favor the trivial name.  As one site put it, "Common names are used because they are easier to say, shorter, and save paper
and ink in scientific publications when they have to be used repeatedly in manuscripts."


Use of combinations to determine number of isomers

For the following I've numbered the positions on ethylene where the four halogen atoms could go. (The Carbon-Carbon double bond doesn't "spin.") It doesn't matter which halogen goes to 1, so let's make it Fluorine. If Fluorine were in any other position then you could flip or rotate the molecule to get Fluorine to position 1. Now there are three choices left for position 2, which leaves 2 choices for position 3. So the number of possible isomers is:

$3 \times 2 = 6$

Note that if you just willy nilly drew molecules then you could draw 4! or 24 permutations. However when you rotate and flip the molecules, then only 6 unique molecules exist.


Yes, it is detailed here and in numerous other places. It refers to a class of reactions, and does not mean "join."

From the above-cited page on named reactions:

  "Click Chemistry" is a term that was introduced by K. B. Sharpless in 2001 to describe reactions that are high yielding, wide in scope, create only byproducts that can be removed without chromatography, are stereospecific, simple to perform, and can be conducted in easily removable or benign solvents.

Sigma-Aldrich has a primer where they state:

  The traditional process of drug discovery based on natural secondary metabolites has often been slow, costly, and labor-intensive. Even with the advent of combinatorial chemistry and high-throughput screening in the past two decades, the generation of leads is dependent on the reliability of the individual reactions to construct the new molecular framework.
  Click chemistry is a newer approach to the synthesis of drug-like molecules that can accelerate the drug discovery process by utilizing a few practical and reliable reactions. Sharpless and coworkers defined what makes a click reaction as one that is wide in scope and easy to perform, uses only readily available reagents, and is insensitive to oxygen and water. In fact, in several instances water is the ideal reaction solvent, providing the best yields and highest rates. Reaction work-up and purification uses benign solvents and avoids chromatography.

A pretty comprehensive overview is given here, in a manuscript entitled Click Chemistry, a Powerful Tool for Pharmaceutical Sciences. Hein, Liu, and Wang. Pharm Res. 2008 Oct; 25(10): 2216–2230.


What is the mechanism of endothermic dissolution?

Since the temperature of the solid is now slightly lower than the temperature of the surrounding water, heat will be transferred from the water to the solid until the temperatures are equal

It is fine to think that way as a rough draft in rationalizing this process. The only issue is that temperature is a macroscopic quantity, and mechanism tends to be a particular (molecular) concept. So you could say that the individual solid particles (e.g. ions) will have slightly lower kinetic energy than average, and collisions with individual solvent molecules will balance this out.
Or you could start with solvent molecules near the solid-liquid interface that already have a higher kinetic energy than the average become part of the solid, thus lowering the average kinetic energy of the remaining solute molecules.
Or you could say that the water molecules solvating the ions use parts of their kinetic energy in the desolvation process, directly lowering the temperature of the solvent.
Because all the species have a distribution of kinetic energies around the average, all these scenarios are correct.

As a result, the temperature of the remaining solid will drop a little.

I don't think the bulk of the solid will have a lower average kinetic energy, so I don't think the temperature of the solid will drop faster than that of the solution. I would think that there is a temperature gradient between the solid-liquid interface on the one hand, and the bulk solid and liquid on the other hand. This is maintained as long as there is a net transfer from solution to solid. Once equilibrium is reached, of course, this will include thermal equilibrium.

this energy will be sufficient to overcome the 4 kJ/mol barrier

"Barrier" sounds like an activation energy, something you can step over. Here, I would rather talk of "energetic cost" or "step up", something that you don't get back. The solvation energy is higher than the lattice energy, so there is a cost for taking the ions out of solution and adding them to the ionic solid.


Radicals are so reactive because they require so much energy to form. When we speak about radical reactivity, "more reactive" generally means a more exothermic hydrogen atom abstraction step. The more exothermic this step, the more the transition state resembles the reactants (Hammond's postulate), with a correspondingly lower character of the carbon-centered radical product. This makes the reaction less sensitive to the stability of the carbon-centered radical.

This being said, some radicals do have an activation energy, and some don't. The $E_a$ is essentially zero for all hydrogen atom abstractions by fluorine atom, and the free energy barrier arises solely from the pre-exponential factor of the Arrhenius equation ($k=Ae^{-E_{a}/(RT)}$), which is near 13 for the halogenations listed below.

The activation energies for abstraction by chlorine atoms are also exceedingly small, but there is a slight preference for more highly substituted $\ce{C-H}$ bonds. Lastly, the activation energies for abstraction by bromine atoms are substantial, clearly showing a preference for abstraction of 3°, 2°, and 1° $\ce{C-H}$ bonds in that order. It is because of these differences in activation energies that we observe the varying regioselectivities of the different halogenation reactions$^{[1]}$.

$[1]$ Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science: Sausalito, CA, 2006.


Determine concentration Using Beer's law

I don't know anything about Red 40, but you will need the extinction coefficient.  Perhaps you determined this from a plot of concentration vs absorbance in the lab?

I am going to attempt to answer you question with reference to NADH, which has an accepted extinction coefficient of 6220 M-1.cm-1 at 340 nm.

Say I have a solution of NADH in a test-tube, I take 4 ml of this and dilute with water to 10 ml.  I then take this diluted solution and measure its absorbance at 340 nm in a spectrophotometer using a cuvette with a path-length of 1 cm.  The value I obtain is 0.4.
(Edit.  In the above measurement, I take care to 'blank' the spectrophotometer against water).

What is the concentration of NADH in the original solution?

From the Beer-Lambert law (where c is concentration)

0.4 = 6220 x c x 1

Therefore

c = 0.64 x 10-6 M  = 0.64 µ M

But the solution on which the measurement was made was
diluted 4 in 10 or 1 in 2.5

Therefore the concentration of NADH in the original solution is

c = 2.5 x 0.64 x 10-6 M = 160.77 x 10-6 M = 160.77 µ M

So, the concentration of NADH in the original solution is about 161 µ M (micro-Molar).

I am making a number of assumptions here, the main one being that the spectrophotometer obeys the Beer-Lambert law in the absorbance region of interest.  You should determine that this is so by measuring  the absorbance of series of NADH solutions and make sure that the resulting graph of absorbance vs concentration is linear and through the origin.

At the very least, doubling the concentration of NADH should exactly double the absorbance (0.8) and halving it should give exactly half the absorbance (0.2).  In addition, measurement of water (or whatever you are 'blanking against) should give zero - that is the spectrophotometer should be properly 'blanked'.  If not you will need to subtract this value.

So what happens if we do not dilute our original sample, but just measure the absorbance at 340nm directly?

From Beer-Lambert

A = 6220 x 160.77 x 10-6 = 1

The original solution is predicted to have an absorbance of 1, but IMO even with modern spectrophotometers you are 'pushing your luck' to assume that Beer-Lambert is obeyed at such a high absorbance.  A 'rule-of-thumb' is to keep the absorbance below 0.8.

One final comment.  It is very difficult to accurately determine an extinction coefficient from absorbance versus concentration plots, unless you accurately know the concentration of the original material.  If this figure relies on a 'dry-weight' measurement, you will need to be sure that the compound is pure, for a start.

Reading your question again, I think we need a bit more experimental detail, and if you could supply it, I'll have another go, maybe.

NADH is widely used in the biochemical sciences where (among many other things) the relatively high extinction coefficient of 6220 M-1.cm-1 is exploited in continuous enzyme assays.  If an enzyme produces or consumes NADH (or NADPH) then it is (usually) very easy to design an assay.

EDIT

It would appear that the accepted extintion coefficient (Molar Absorption Coefficient) for Allura Red AC (Red 40) is about 25 000 M-1.cm-1 (or 25 000 mol-1 L cm-1) at about 505 nm. The best reference I can get it here. Azo dyes tend to have very high extinction coeffficients, so this 'looks about right'

The diluted solution has an absorbance of 0.283

Taking the above as the extinction coefficient, and in addition assuming the measurement was made in a cuvette with a path length of 1 cm (l =1), one can calculate from Beer-Lambert that the concentration of Allura Red in the diluted solution is about 11.32 x 10-6 M.

The dilution factor is 4 in 10, or 1 in 2.5.  Therefore, the concentration of the original solution is about 28.3 x 10-6 M.


Electrochromic Materials, electrode setup

I will answer your last question first, this is very material dependent. Not all materials will be electrochromic. It is a very specific effect where the material has different oxidation states that have different absorbance profiles(colors if you will). The most common ones are conducting polymers (https://en.wikipedia.org/wiki/Conductive_polymer) since these materials have different oxidation states that have vastly different electronic structure.

Two electrode electrochemistry can get pretty hard to understand. Two compare your setups 1 and 2, the main differences are the electrode area and the electrode distance. Other than that, the two are the same. So, yes, electrochromism will happen in setup2 as well. However, it is hard to predict exactly where it will happen and to what extent. It all depends on the material you select, it's resistance and doping characteristics. In either setup, it will start at the electrode and work it's way inward, obviously, setup 1 is more advantageous with the larger electrode area and shorter inter-electrode distance.


Enthalpy and Internal Energy for Isothermal Expansion

You are right that an in in ideal gas, internal energy is a function of temperature only, and that in this problem, temperature is not changing.  However, I think you are confused about how broadly the ideal gas law applies to this problem.

The question states that the ideal gas law applies to the water vapor.  But the question is about a phase change of water.  Let's break down some of the components of the problem.  In the question we have:

Liquid water.  The ideal gas law does not apply to liquid water.
Water vapor.  The ideal gas law does apply.
A phase change of liquid water to water vapor.  $\ce{H2O(l) <=> H2O(g)}$  The ideal gas law does not apply to the process of the phase change, simply because processes are not gases and cannot be modeled by the ideal gas law.

Thus only one of three "components" of the problem is an ideal gas.

As a look at any reasonable steam table will tell you, the internal energy of water vapor is higher than the internal energy of liquid water.  This difference is the heat of vaporization (at constant volume).


Periodic Trends of Zeff and Electronegativity

The Slater shielding rules give more quantitative information about the amount of shielding.

Put simply, valence electrons (as you go across a row) do not shield very well, roughly 35% of a full electron. So when you go from B to C to N, you keep increasing the nuclear charge by one proton, but the electrons don't fully shield the nucleus.

Core electrons, as suggested in comments and other answers do shield fully, since they're closer to the nucleus.

So yes, the shielding for Li and Na are similar. As you mention, the valence electron in Na is further from the nucleus, and that does effect the ionization energy and electronegativity as you suggest.


Farnesyl Diphosphate: Chemical commercial synthesis

Farnesol is commercially available as a distillate of essential oil or synthetically derived from Nerolidol. These chemicals are commonly used in the fragrance industry, as such efficient synthetic processes should exist.

Here's a potential synthesis route for Farnesol (1) from the starting material Nerolidol (2). Farnesol is then phosphorylated into Farnesyl diphosphate.

Nerolidol (2) is treated with anhydrous acetic acid under basic conditions to produce Farnesol (1). From The Total Synthesis of Natural Products
edited by John ApSimon, page 201. Also referenced in US1663817A.

If we assume that isopentenyl pyrophosphate is similar to farnesyl diphosphate (1 isoprene unit vs. 3 isoprene units), then this technique "An Improved Synthesis Of Isopentenyl Pyrophosphate" can theoretically be used.

The double phosphorylation has stated yields from 32%-51%. Let's assume the yield of Nerolidol to Farnesol is 70%. Given the price of Neroidiol is 39 USD/kg†, the least efficient price would be about 174 USD per kilogram. The price of Farnesol is 129 USD/kg†, and Geraniol is 19 USD/kg†. (If one would like to synthesize Geraniol into Nerolidiol, steps are given on page 201 above.)

This is pure conjecture, but should most likely work given that Farnesol is similar to both Geraniol and Isopentenol. Also the reason I included the step from Nerolidol to Farnesol is because Farnesol is too expensive compared to the price of Nerolidol despite being one synthetic step away.

†Prices are quoted from Vigon International, a North American reseller.


Why is magnesium hydroxide not hygroscopic?

Try looking at Inorganic Chemistry in Aqueous Solution by Jack Barrett around page 41 on google books or elsewhere.

Entropy of hydration of Mg2+ is very negative compare to Na+.

According to Barrett, more highly charged ions restrict the movement of water molecules of hydration more.  Also, the smaller the radius of the ion, the more restricted the movement of the hydrating waters is.  Both the number of waters whose movement is restricted, and the degree to which they are restricted, is affected by charge and radius.

A complete analysis would consider both the entropy and enthalpy contributions.


Any way to see "cubic-active" vibrational modes in molecules?

The product $xyz$ is an odd function and as such should produce the same selection rules as 1 photon (electric dipole) transitions. Thus CO has a 1 photon transition at 146.7 nm in the vacuum uv that can be more easily reached with a 3-photon transition with a pulsed laser at 440 nm. ( Three photon transitions are also used in bio-imaging where a ir photon is used so as not the excite the sample. Also, as the transition is multiphoton the focus area is smaller than a 1 photon transition and this improves spatial resolution).

As the product $xyz$ can be thought of as $xy$ multiplied by $yz$ then this is like a double Raman transition. Raman involves a change in polarisability. Polarisability is proportional to a volume but the projection for the radiation is 2D hence $xy$ type products in the point group tables. A  Raman type transition may occur between 4 levels. Levels 1 and 4 are  'real' and 2 and 3 'virtual'. The steps are 1 up to 3 , 3 down to 2, and 2 up to 4. This has been demonstrated in doppler free geometry between Na $3^2S_{1/2}\to 3^2P_{1/2}$.


Should binding constants be unitless when deriving fractional occupancy equations from reactions?

The equilibrium constant does not need to be unitless, because it is depending on its definition. See goldbook:

  Equilibrium Constant
  Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always unitless, as it is defined differently (goldbook)

  Standard Equilibrium Constant $K$, $K^\circ$ (Synonym: thermodynamic equilibrium constant)
  Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{\mathcal{R}T}\right\}$$  where $\Delta_rG^\circ$  is the standard reaction Gibbs energy, $\mathcal{R}$  the gas constant and $T$  the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.

In your example both would work if you straighten your definitions, involving concentrations
\begin{aligned}
K_A \cdot c(\ce{X}) \cdot c^n(\ce{A}) &= c(\ce{X_{A}}),\\
y=\frac{c(\ce{X_A})}{c(\ce{X})+c(\ce{X_{A}})} &= \frac{K_A\cdot c^n(\ce{A})}{1 + K_A\cdot c^n(\ce{A})},
\end{aligned}

or activities
\begin{aligned}
K^\circ_A \cdot a(\ce{X}) \cdot a^n(\ce{A}) &= a(\ce{X_{A}}),\\
y=\frac{a(\ce{X_A})}{a(\ce{X})+a(\ce{X_{A}})} &= \frac{K^\circ_A\cdot a^n(\ce{A})}{1 + K^\circ_A\cdot a^n(\ce{A})}.
\end{aligned}

In both cases $y$ remains unitless.

In experimental chemistry it is much easier to observe concentrations instead of activities. Therefore to fist named constant is more often in use. A straightforward use of the thermodynamical equilibrium constant might sometimes prove a little tricky.

For reasonable dilutions one can simply substitute activity with a unitless concentration as given through
$$a(\ce{Y})=\gamma\frac{c(\ce{Y})}{c^\circ},$$
with $\gamma\approx1$ for $c(\ce{Y})\to0$ and $c^\circ=1\:\mathrm{mol/L}$.


Negative temperature as limiting case for two-level system

I think you would need to describe the derivative $\frac{\partial U}{\partial S}$ at that point, which should be infinity by your logic since the entropy is at a maximum with respect to internal energy ($\frac{\partial S}{\partial U}=0$). This is obtained using $$\frac{\partial U}{\partial S}=\frac{1}{\frac{\partial S}{\partial U}}$$


Do molecules with a hexagonal planar geometry exist?

I think it's nearly impossible to find or synthesize a "canonical" complex with hexagonal molecular geometry, but in the field of host-guest supramolecular chemistry there are numerous examples of "unusual" geometry. Probably the most well-established class of such compounds are torands ("hosts") incorporating alkali metal cations ("guests").

Check out, for example: (1) Bell, T. W.; Cragg, P. J.; Drew, M. G. B.; Firestone, A.; Kwok, D.-I. A. Angew. Chem. Int. Ed. Engl. 1992, 31 (3), 345–347, DOI 10.1002/anie.199203451.

Here is an example of the structure with potassium ion from (1), Tri-n-butyltorand-potassium picrate clathrate, which I quickly sketched in Olex2:

Top view:

Side view:

Unit cell and packing:


How to identify the repeating unit in a polymer?

It's not that the "double bond breaks," it's that the vinyl C=C bond reacts during the radical addition polymerization to create polyvinyl acetate.

I'd disagree that the repeat unit is $\ce{CH3OCO}$ since that ignores the backbone entirely. Instead, if you look at the Wikipedia article (or any other picture of polyvinyl acetate), I'd call the repeat unit something like $\ce{-(CH2CH(OC=OCH3)-)}$ or something like that.


Correct.

For the non-liquifying gases (room temperature is far above their critical temperature) knowing the pressure is enough, if you know the volume of your cylinders. (You need a better formula than $pV=nRT$ to calculate the amount of gas at those high pressures. https://en.wikipedia.org/wiki/Van_der_Waals_equation, virial expansion, ...)

Actually the van der Waals equation is not useful, because, although it gives a good  idea why the ideal gas law breaks, it gives very wrong numbers esp. for high pressures. BWR or BWRS are supposed to work well, at the cost of using eight or eleven constants.

For those gases which are partially liquified ($\ce{CO2}$ at $57\ \mathrm{bar}$, $\ce{H2S}$ at $12\ \mathrm{bar}$, $\ce{SO2}$ at $3.5\ \mathrm{bar}$), you need to know the weight (and tare weight) of the cylinders, of course.


Why is pyridine a monodentate ligand, but pyrrole a tridentate ligand?

From what I gathered in the comments you confused a few terms. Both compounds are monodentate, i.e. they can only coordinate one metal at a time. However, pyridine usually donates one electron pair, while pyrrole can donate three (to one metal) — I haven’t seen a structure where pyrrole does that, though.

The reason is pretty simple: Pyrrole is much like the cyclopentadienyl anion except that one carbon atom is replaced by nitrogen. If it wants to coordinate in any way, it will have to do that like a slice of bread from its side, forming a $\eta ^5$ complex. Much like ferrocene in the picture below.

Pyridine, on the other hand, has no hydrogen on the nitrogen atom; instead it has a lone pair pointing outwards, but still in the plane of the ring. This allows pyridine to donate two electrons in a $\sigma$-like fashion to a metal centre, $\eta ^1$-style. Of course, pyridine could also perform a sandwich-like coordination much like benzene sometimes does. However, the direct two-electron $\sigma$ donation will be much stronger and therefore largely outweigh a sandwich-type donation in most compounds.


Stability of complexes based on number of rings

When polydentate ligands binds to a metal in a ring form, it is called chelate ring and the corresponding complex is called chelate complex.
Experimentally, it is observed that metal complexes of polydentate ligands with more rings are significantly more stable than the corresponding complexes of similar ligands but with less rings. Because, the larger the number of ring closures to a metal atom, the more stable the compound. This phenomenon is called the chelate effect.
But, do note that ring number is not the only factor to determine the stability of a complex. There are several other factors like nature of central metal ion or the ligands, ring size, ionic size. For this question, let's consider ring size.
The more the ring size, the less stable is the complex. The entropy of complex is changed if the size of chelating ring is increased or decreased. Four membered ring compounds are unstable, whereas five-membered are more stable because of negligible ring strain. Likewise, ligands with a flexible organic backbone like ethylenediamine are significantly more stable than complexes with six-membered chelate rings, which are in turn much more stable than complexes with four- or seven-membered rings.
Refer to the following articles for more details:

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Stability_of_Metal_Complexes_and_Chelation
https://old.amu.ac.in/emp/studym/99996027.pdf


Difference between thermodynamic and kinetic stability

To understand the difference between kinetic and thermodynamic stability, you first have to understand potential energy surfaces, and how they are related to the state of a system.

A potential energy surface is a representation of the potential energy of a system as a function of one or more of the other dimensions of a system. Most commonly, the other dimensions are spatial. Potential energy surfaces for chemical systems are usually very complex and hard to draw and visualize. Fortunately, we can make life easier by starting with simple 2-d models, and then extend that understanding to the generalized N-d case.

So, we will start with the easiest type of potential energy to understand: gravitational potential energy. This is easy for us because we live on Earth and are affected by it every day. We have developed an intuitive sense that things tend to move from higher places to lower places, if given the opportunity. For example, if I show you this picture:

You can guess that the rock is eventually going to roll downhill, and eventually come to rest at the bottom of the valley.

However, you also intuitively know that it is not going to move unless something moves it. In other words, it needs some kinetic energy to get going.

I could make it even harder for the rock to get moving by changing the surface a little bit:

Now it is really obvious that the rock isn't going anywhere until it gains enough kinetic energy to overcome the little hill between the valley it is in, and the deeper valley to the right.

We call the first valley a local minimum in the potential energy surface. In mathematical terms, this means that the first derivative of potential energy with respect to position is zero:

$$\frac{\mathrm dE}{\mathrm dx} = 0$$

and the second derivative is positive:

$$\frac{\mathrm d^2E}{\mathrm dx^2} \gt 0$$

In other words, the slope is zero and the shape is concave up (or convex).

The deeper valley to the right is the global minimum (at least as far as we can tell). It has the same mathematical properties, but the magnitude of the energy is lower – the valley is deeper.

If you put all of this together, (and can tolerate a little anthropomorphization) you could say that the rock wants to get to the global minimum, but whether or not it can get there is determined by the amount of kinetic energy it has.

It needs at least enough kinetic energy to overcome all of the local maxima along the path between its current local minimum and the global minimum.

If it doesn't have enough kinetic energy to move out of its current position, we say that it is kinetically stable or kinetically trapped. If it has reached the global minimum, we say it is thermodynamically stable.

To apply this concept to chemical systems, we have to change the potential energy that we use to describe the system. Gravitational potential energy is too weak to play much of a role at the molecular level. For large systems of reacting molecules, we instead look at one of several thermodynamic potential energies. The one we choose depends on which state variables are constant. For macroscopic chemical reactions, there is usually a constant number of particles, constant temperature, and either constant pressure or volume (NPT or NVT), and so we use the Gibbs Free Energy ($G$ for NPT systems) or the Helmholtz Free Energy ($A$ for NVT systems).

Each of these is a thermodynamic potential under the appropriate conditions, which means that it does the same thing that gravitational potential energy does: it allows us to predict where the system will go, if it gets the opportunity to do so.

For kinetic energy, we don't have to change much - the main difference between the kinetic energy of a rock on a hill and the kinetic energy of a large collection of molecules is how we measure it. For single particles, we can measure it using the velocity, but for large groups of molecules, we have to measure it using temperature. In other words, increasing the temperature increases the kinetic energy of all molecules in a system.

If we can describe the thermodynamic potential energy of a system in different states, we can figure out whether a transition between two states is thermodynamically favorable – we can calculate whether the potential energy would increase, decrease, or stay the same.

If we look at all accessible states and decide that the one we are in has the lowest thermodynamic potential energy, then we are in a thermodynamically stable state.

In your example using methane gas, we can look at Gibbs free energy for the reactants and products and decide that the products are more thermodynamically stable than the reactants, and therefore methane gas in the presence of oxygen at 1 atm and 298 K is thermodynamically unstable.

However, you would have to wait a very long time for methane to react without some outside help. The reason is that the transition states along the lowest-energy reaction path have a much higher thermodynamic potential energy than the average kinetic energy of the reactants. The reactants are kinetically trapped - or stable just because they are stuck in a local minimum. The minimum amount of energy that you would need to provide in the form of heat (a lit match) to overcome that barrier is called the activation energy.

We can apply this to lots of other systems as well. One of the most famous and still extensively researched examples is glasses.

Glasses are interesting because they are examples of kinetic stability in physical phases. Usually, phase changes are governed by thermodynamic stability. In glassy solids, the molecules would have a lower potential energy if they were arranged in a crystalline structure, but because they don't have the energy needed to get out of the local minimum, they are "stuck" with a liquid-like disordered structure, even though the phase is a solid.


Why do PBr3 and PCl3 have larger bond angles than PH3?

The larger angles can simply be explained as a result of repulsion between the larger atoms of $\ce{Br}$ and $\ce{Cl}$. Hydrogen atoms in $\ce{PH3},$ as the are so small, experience less repulsion as compared to $\ce{Br}$ atoms in $\ce{PBr3}$ or $\ce{Cl}$ atoms in $\ce{PCl3},$ therefore the larger bond angles in $\ce{PBr3}$ and $\ce{PCl3}.$


by definition a nucleophile, is a lone pair donor. phile meaning 'loving'. i.e its attracted to species which are electron deficient. the term is used to describe a species in an organic mechanism.a  bronsted base, is a proton accepter. Mostly in organic chemistry, you draw curly arrows, going from the nucleophile to electrophiles,most usually a electron deficient carbon. so it acts more as a Lewis base. which is a lone pair donor. however there are protons floating around in most organic reactions, also very electronegative atoms such the halogens(leaving groups). which pick up protons so i guess by defintion you could say they act as bonsted bases as well as nucleophiles. However the term nucleophile is mostly reserved for the species acting on the electrophile in an organic mechanism.


Prefixes -  I think the prefixes neo, iso etc. are used only in common names. There's a different system in IUPAC for naming things having a branched chain.
Aromatic Compounds - The only aromatic compound that I know of having same common and IUPAC name is benzene.


Find the maximum temperature attained by the gas during the change

The ideal gas law $pV = nRT$ allows us to relate one of $p$, $V$, and $T$ to the two other variables, and is the main equation to work with here.

Rephrasing the original question to make it more suggestive, we are interested in maximizing $T(p,V)$ subject to the constraint $p(V)$, a linear function in $V$ readily found from the conditions specified in the problem. Substituting $p(V)$ for $p$ to get $T = T(p(V),V) = T(V)$, we have reduced the problem to one amenable to simple calculus techniques, and the answer follows from taking a derivative.

Remark. In general, extremization problems are not so easy, and we will need to use the more powerful tool of Lagrange multipliers.


The kinetic energy $E_\text{k}$ equals the work $W$ done on an object:
$$E_\text{k}=W$$
The work $W$ is the result of a force $F$ on a distance $s$. If $F$ is constant, the work $W$ is equal to the force $F$ multiplied by the distance $s$:
$$W=Fs$$
The force $F$ on an object is equal to the mass $m$ of that object multiplied by the acceleration $a$:
$$F=ma$$
For uniform acceleration $a$, the final velocity $v$ is
$$v=at$$
and the distance $s$ is
$$s=\frac{1}{2}vt=\frac{1}{2}at^2$$

Therefore,
$$\begin{align}
E_\text{k}&=W\\
&=Fs\\
&=mas\\
&=ma\cdot\frac{1}{2}vt=ma\cdot\frac{1}{2}at^2=\frac{1}{2}m(at)^2\\
&=\frac{1}{2}mv^2
\end{align}$$


Predicting relative solubilites of salts without Ksp values

To a certain extent, you can predict the relative solubilities of two salts, especially when there is a large difference in solubility. The existing answer mentions Fajans' rules and obviously it is quite easy to predict which of $\ce{NaCl}$ and $\ce{AgI}$ is more soluble.

But I'm inclined to be pessimistic and say no, this is not generally applicable.

You're perhaps on the right track. The dissolution of the salt arises from a couple of measurable quantities: the lattice enthalpy of the salt, and hydration enthalpies of the ions. You could measure both, and try to come up with a trend.

However that's often far easier said than done. The trends are sometimes quite subtle, and even if you have all the relevant data and limit yourself to a "simple" system like the alkali metal halides, you'll find that the interplay between two factors leads to a very delicate balance, and the resulting trend is quite complex. I don't think it's something that we can predict solely in our heads without doing any measurements.

For example, if you simply compare $\ce{NaF}$ and $\ce{NaI}$, $\ce{NaI}$ has a smaller lattice enthalpy (favours dissolution) but the $\ce{I-}$ ion also has a smaller hydration enthalpy (disfavours dissolution). So which factor wins out? There's no simple heuristic or rule of thumb that applies here, not any that I know of, at least.

How about two obviously more different salts? Let's try $\ce{NaF}$ and $\ce{MgF2}$. $\ce{NaF}$ has a smaller lattice enthalpy (favours dissolution) but also the sum of the hydration enthalpies is smaller than $\ce{MgF2}$ (more ions and $\ce{Mg^2+}$ is more charged than $\ce{Na+}$ - which disfavours dissolution of $\ce{NaF}$). Again we have the same dilemma.

In fact, that's just the enthalpic changes, so only half the story. There's also entropy to care about, which can either be positive or negative (and it's extremely difficult to predict this!).


Working with wave number equations, use trial-and-error?

[…] my teacher told me to continue using a trial and error method […]

But nobody told you to do it manually once you have understood how to adjust the Rydberg constant for $\ce{He+}$ and use the formula to get the energies for the given  transistions, in order to compare them with the $\ce{H_{\alpha}}$ line (656.3 nm):

import scipy.constants

transitions = ((7,5), (5,3), (6,4), (4,2))

def get_he_line_nm(n_upper, n_lower):
    '''calculate the energy for a line in a He+ spectrum and return
       it in nanometer'''

    energy = 4 * scipy.constants.Rydberg * (pow(n_lower, -2) - pow(n_upper, -2))

    return 1.0/(scipy.constants.nano * energy)
    # energy in nm

for upper, lower  in transitions:
    wavelength = get_he_line_nm(upper, lower)
    print('{0} -> {1} : {2:7.2f} nm'.format(upper, lower, wavelength))

will give the output

7 -> 5 : 1162.81 nm
5 -> 3 :  320.37 nm
6 -> 4 :  656.11 nm
4 -> 2 :  121.50 nm


Arrange in the order of their reactivity of SN2 reaction

The order you've given is absolutely correct with respect to SN2, I believe, the question needs to be authenticated once again, as the order for SN1 would be the exact reverse of this, which is as given in the answer to your question.


According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the ‘N’ is a locant just as the ‘2’:

  P-14.3 LOCANTS

  P-14.3.1 Types of locants

  Traditional types of locants are arabic numbers, for example, 1, 2, 3; primed locants, for example, 1′, 1‴, 2″; locants including a lower case Roman letter, for example, 3a, 3b; italicized Roman letters, for example, O, N, P; Greek letters, for example, ‘α-, β-, γ-’ and compound locants, for example, ‘1(10)’ and ‘5(17)’.

Note that ‘N’ is lower than ‘2’:

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
  (…)
  Italic capital and lower-case letter locants are lower than Greek letter locants, which, in turn, are lower than numerals.

Identical simple substituents are indicated by multiplicative prefixes, such as ‘di’, ‘tri’, etc. The current IUPAC recommendations do not contain any rule that would stipulate that such substitutents should be grouped. By way of comparison, similar examples given in the Blue Book include names such as N,N,2-trimethyl-3-{4-methyl-3-[2-methyl-3-(methylamino)-3-oxopropyl]phenyl}propanamide.

Therefore, the preferred IUPAC name (PIN) for the compound given in the question is N,N,2-trimethylpropan-1-amine.


Confusion between electronegativity and electron affinity

I wouldn't shoot this down so quickly - consider, e.g. Mulliken electronegativity, defined as the average of the electron affinity and ionisation energy. If you think about it, electronegativity depends not only on how strongly an atom pulls other electrons to itself, but also how willing it is to give up its own electrons. The former can be somewhat represented by the EA, the latter by the IE.

One point that hasn't been mentioned is that electronegativity is not a physical parameter which can be measured. It is a human construct, which is often helpful in explaining chemical reactivity, but at the end of the day it is not something which can be derived from physics. Therefore, there is no real reason to expect that there should be any one-to-one correlation between the EN and the EA (unless the EN scale is explicitly defined to follow the EA). For example, if you use the more common Pauling scale (defined by bond dissociation energies), there's no real reason why the trend in EN should match the trend in EA exactly.

Broadly speaking, the trends match (because of what you've said) - e.g. both EN and EA generally increase going across a period. However, the small details like chlorine vs. fluorine won't necessarily match.


You can etch ITO on PTE by mean of conc. HCl solution without problems. The treatment duration depends on the ITO thickness but is limited to minutes.

Just make few samples (residuals from cutting etc will suffice to make this test) to find out how long it takes to fully remove ITO.

No reference need it as I do etching of ITO coated PTE electrodes since years.

For sake of completeness let me add that I never investigated PTE itself prior and after the etching so I do not know if there are changes which, for instance, may interfere with long (very long) stability etc.

In my case it suffices that PTE maintains its transparency and still behave as a robust but flexible substrate. It seems it could be your case, too.

Just to be pedantic. Obviously after etching the foil must be washed thoroughly.


Shouldn't 1 mole of any ideal gas at open space occupy more than 22.4 L volume at S.T.P. due to diffusion?

Here is the calculation for the volume occupied by one mole ($n$) of an ideal gas at STP (temperature $T = 273.15\ \mathrm K$, pressure $P = 1\ \mathrm{bar}$):

gas constant: $R = 0.08314\ \mathrm{L\ bar\ mol^{-1}\ K^{-1}}$
ideal gas law: $PV = nRT$

$$\begin{align}
V &= \frac{nRT}{P}\\[6pt]
&= \mathrm{\frac{1\ mol\ \times\ 0.08314\ L\ bar\ mol^{-1}\ K^{-1} \times\ 273.15\ K}{1\ bar}}\\[6pt]
&= 22.71\ \mathrm L
\end{align}$$

Another way of looking at this of course is that any 22.71 liter chunk of particulate-free air, of arbitrary shape, will contain about 1 mole of air. Molecules of the different gases composing the air will diffuse in to and out of this region, but it will always contain about 1 mol of air at STP.


What exactly happens while washing clothes

When you wash a piece of cloth, even without soap, it's better to agitate. The vibration, friction, etc. will help speed up the process of mixing the dirt/grease with water.

Even the very definition of "agitation" uses the example of washing. From Vocabulary.com

  Agitation is the act of stirring things up, like the agitation of a washing machine that moves the water, detergent, and clothes around and gets the dirt out

And as you rightly pointed out from another question, "the froth has little or no effect on the detergent action".


What are aryl chlorides currently used for?

Synthetic handles: for organometallic couplings (Sonogashira, Suzuki...), there is a selectivity ($\ce{aryl-I} > \ce{aryl-OTf} >\ce{aryl-Br} >\ce{aryl-Cl}$) that can be crucial in the total synthesis. You can make the other position ($\ce{aryl-I}$ or whatever) react first, then use modified conditions suitable for aryl chloride couplings (plenty of examples on this website). Or make an halogen exchange before coupling your chloro- position.
Common functionalities: halogen bonds can change binding affinities. Even if they are weaker than hydrogen bonds, they are specific (see C. Bissantz et al., J. Med. Chem., 2010, 53 (14), 5061–5084.). They can make interactions with electrophiles, nucleophiles, or with themselves (ex: $\ce{C-X...O}$ $\ce{sp^2}$ or $\ce{X...X}$). One example of molecule designed with chloroaryl moieties is
represented here (see Furet et al., Bioorg. Med. Chem. Lett.
2012, 22 (10), 3498–3502, which shows the docking model). This is an inhibitor which blocks the binding pocket of the
regulator protein MDM2. The 6-chloroindolyl moiety enables to fill a
subpocket (TRP 23), while the p-chlorophenyl fills another one (Leu 26). The chlorine-chlorine interactions between the 6-chloroindolyl moiety and the triptophan residue in the protein was a key point in the design of this inhibitor, strongly enhancing the binding affinities.


What kind of compound absorption the 365nm UV light on TLC

Your question is almost impossible to answer - it simply is too broad.

Excitation spectra of fluorescent organic compounds typically aren't sharp lines but consist of one (or more) bands. Even if the maximum of the excitation spectrum isn't located at $\lambda = \textrm{365 nm}$, monochromatic irradiation at that wavelength will excite a lot of different compounds (and even compound classes) which then deactivate by fluorescence.
Your light source does not emit monochromatic radiation! Hand lamps used to inspect TLCs typically have coated low-pressure mercury lamps. While the maximum of the radiation density might be centred around 360 nm, the full with at half maximum is - roughly estimated - about 30 nm or more. This means that there's even more compounds that might get excited by the emission of the hand lamp.

In summary, there's no exhaustive answer to your question, but just to name some compounds that are potential candidates:

Diphenylacetylene (1)
1,6-Diphenylhexatriene (DPH, 2)
4',6-diamidino-2-phenylindole (DAPI, 3)
6,8-difluoro-7-hydroxy-4-methylcoumarin (DIFMU, 4)


It has to do with the fact that $\ce{Mn^{2+}}$ is a high-spin $d^5$ ion (in most cases) which is afforded special stability, as all frontier $d$ orbitals are half-filled. This reduces affinity for more covalent (soft) interactions that would involve disrupting the $d^5$ state by addition of electrons. So, the character is more ionic.

Editing after some additional searching: This also explains why ions such as $\ce{Fe^{3+}}$ are also hard. However, it seems to contradict in cases like organocuprates, where it would seem that the formation of $\ce{Cu(I)}$ would result in a similarly stabilized $d^{10}$ ion (same with silver). In these cases, is it simply because polarizability of transition metals in that region outweigh the $d^{10}$ stabilization?


Why are some weak acid equilibrium approximations invalid in very dilute solutions?

The above simplified formula considers 2 simplifying assumptions:

Concentration of hydrogen ions from water autodissociation can be neglected, so $[\ce{H+}] \approx [\ce{A-}]$ .
Concentration of an acid $c_0$ , in relation to its $\mathrm{p}K_\mathrm{a}$, is high enough for the fraction of ionized acid to be negligible, so $c_0 \approx [\ce{HA}]$ .

Especially the second assumption is broken for too dilute weak acid solutions. For more diluted solutions, higher fraction of an acid is ionized to match the equilibrium equation.
if we consider $$K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}}$$
then if we dilute the acid 4 times, concentration of the conjugate base decreases just 2 times, so the fraction of the ionized acid form increases 2 times as well.
It also means the stronger the acid is, the higher concentration must be . As for the same concentration, stronger acids are more ionized and $c_0 \approx [\ce{HA}]$ causes a bigger error.
The rule of thumb is, for the formula to be acceptable, this inequality should be true: $$ - \log {c_0} < \mathrm{p}K_\mathrm{a} - 2$$
so maximally about 1% of an acid is ionized.
The first assumption is broken mainly for too weak acids. E.g. if $\mathrm{p}K_\mathrm{a} = 10$, practically all $\ce{H+}$ comes from water autodissociation and the equation fails.


Explain on the basis of valence bond theory that tetracyanonickelate(II) ion with square planar structure is diamagnetic

@M.L has already provided the solution as per VBT, but that answer seems to have some flaws when it comes to CFT (looks like he has corrected the mistakes now). So, I would like to take some time to answer this question using CFT.

According to crystal field theory, the degeneracy of the orbitals of a same orbital is lost in presence of a ligand and the pattern in which they are distorted depends upon the nature of the ligand. The spectrochemical series helps us understand the nature of a ligand.
In general there are two types of ligands, strong field and weak field. Strong field ligands have a greater power to distort the degeneracy of the sub-orbitals whereas weak field have a weaker power of distortion.
As per the spectrochemical series, $\ce{CN-}$ is a strong field ligand and thus creates a greater distortion in the degeneracy of $\mathrm d$ orbitals.
The central metal ion here is the $\ce{Ni^2+}$ with its $\mathrm d^8$ electronic configuration.
A $\mathrm d^8$ system prefers a square planar approach of ligands if the ligands are strong field. In such a condition the the degeneracy is disrupted and the energy diagram of the $\mathrm d$ orbitals is as below:

Now when trying to fill this diagram with electrons we have to take into consideration that an electron in the $\mathrm d_{x^2-y^2}$ orbital can pair up with the electron in $\mathrm d_{xy}$ in case of a strong field ligand. This is because the extent of splitting done by a strong field ligand is so high that it can overcome the energy gained due to pairing of electrons.
Thus the energy diagram filled with electrons looks as following:

Now the ligands donate their lone pairs in the $3\mathrm d_{x^2-y^2}$, $4\mathrm s$, $4\mathrm p_x$ and $4\mathrm p_y$ sub-orbitals.
In reality, no hybridization takes place but if forced to say the hybridization of this complex, we can say $\mathrm{dsp}^2$ because of the orbitals involved in the $\sigma$ bonds between the metal and ligands.
As you can see the energy diagram does not contain any unpaired electron and thus the complex is diamagnetic.


Naturally occurring molybdenum disulfide can occur in either of two crystal systems, hexagonal or "trigonal"; the latter is also known as rhombohedral (see here).

In both cases the coordination geometry around molybdenum has only threefold rotational symmetry, but in the hexagonal form the crystal will add a center of symmetry to become hexagonal.  Such a conversion is seen in other cases such as hexagonal close-packed metals like magnesium and zinc, which actually have only threefold symmetry about any axis perpendicular to the basal plane and passing through an atom, yet the Crystal's are hexagonal.  In general, when you have a unit cell with $D_{3h}$ symmetry -- a mirror plane perpendicular to the threefold exis -- you will get hexagonal crystals in both molybdenum disulfide and hexagonal close-packed metals.

The rhombohedral crystal, with a different arrangement of mirror planes (the point group symmetry is $D_{3d}$ instead of $D_{3h}$, and $D_{3d}$ has no mirror plane perpendicular to the threefold axis), has the required center of symmetry without going from threefold to sixfold rotational symmetry.


Equilibrium Constants For Phenol Tautomers

I believe that's because in the first step, phenol regains its aromaticity which makes the $K_{eq}$ substantially ($1000$ times) larger than that in the second step where it loses its aromaticity.


Splitting of $d$ orbitals when ligands approach central metal ion

The picture you pose is only a part. The full picture is posted below.

This solves your doubt. The splitting does not occur about the original level, i.e. not about the energy level the d-orbital has in complete absence of the ligand's electrostatic field. But, the splitting occurs about a hypothetical Barycentre. This is the energy level of the d-orbitals (still degenerate) in a hypothetical spherical electric field (with spherical symmetry). Due to the symmetry of the hypothetical barycentre field, the d-orbitals remain degenerate but at an energy higher than in the absence of the ligands. Now, the $t_{2g}$ group which is lowered in case of an octahedral field, is still above the original free ion energy level in the complete absence of the ligands.

EDIT
By conservation of energy, we mean that the energy is conserved during splitting. The very process of splitting occurs symmetrically about the barycentre, so that the net amount by which $t_{2g}$ is lowered ($3\times 0.4\Delta_o$) is equal to the amount by which $e_g$ is raised ($2\times 0.6\Delta_o$). But the net energy in absence of field and in presence of field need not be same for energy conservation to hold. Whatever the rise in the net energy (from free ionic to barycentre) will be taken from the kinetic energy (or potential) of the approaching ligands.


Factors affecting mass of a gas-evolving reaction mixture over time

I. False – because the magnesium has a higher surface area in this reaction, the rate will proceed faster, producing a steeper slope in the graph; however, the final mass of the solution will be the same in both experiments. We would note that the lines converge when each system reaches equilibrium.

II. True – the higher concentration of $\ce{HCl}$ means more $\ce{H2}$ gas will be lost from the system at equilibrium. Thus, the line would flatten beneath line 1.

III. False – a lower temperature would mean the average kinetic energy of molecules in the system would be lower; therefore, a smaller fraction of the molecules will have sufficient energy to react with others. In this case, we would note an elongated curve that converges with the first.


Is there any kind of reaction with two types of reactants (gases) and one product (also gas) such that the total volume after the reaction increases?

Decomposition of ozone:

$\ce{O3 + O3 -> 3 O2}$

Technically, not 3 different gases, but the introduction of isotopes may be judged as in compliant (and yes, contrived).


Is peroxynitrous acid an isomer of nitric acid?

Keeping aside its stability and considering only their structure and structural formula, is peroxynitrous acid an isomer of nitric acid?

Yes.

It seems that peroxynitrous acid is a very, very unstable compound and I hardly doubt that it has been synthesised in pure, stable form. This video shows the preparation of peroxynitrous acid but it readily converts into nitric acid and so the peroxynitrous acid yield is very poor.

Indeed, it is impossible to isolate peroxynitrous acid $\left(\ce{ONOOH}\right)$ in aqueous solution, due to the rapid decomposition to nitric acid. The yield is not just poor, it is zero.
However, the conjugate base of peroxynitrous acid, peroxynitrite $\left(\ce{ONOO^-}\right)$, happens to be extremely stable. (Apropos the question, peroxynitrite is an isomer of nitrate.) So, as long as a synthetic approach is devised that keeps the product strictly in its anionic/alkaline form, one can actually produce peroxynitrite solutions and salts and work with them in experiments.  In fact, I synthesized sodium peroxynitrite solutions myself as part of my Ph.D. research.  You can read about the procedure in Section 4.2.4 of my thesis (free for download).  A short literature review of various $\ce{ONOO-}$ synthesis methods is also included there.
Briefly, I loaded aqueous solutions of (1) $\ce{NaNO2}$, (2) $\ce{H2O2}$ plus $\ce{HCl}$, and (3) $\ce{NaOH}$ into syringes fitted onto a syringe pump.  I then connected the syringes to a tubing network that first mixed the acidified $\ce{H2O2}$ with the $\ce{NaNO2}$, forming $\ce{ONOOH}$, and then quickly mixed the product solution with the $\ce{NaOH}$.  The alkali "caught" the $\ce{ONOOH}$ while a significant concentration still remained, providing a bright yellow solution of alkaline peroxynitrite.  After decomposing residual $\ce{H2O2}$ by treatment with $\ce{MnO2}$, these solutions held their concentrations for a couple of years when stored at $-80\,^\circ\mathrm C$.


What is the precise definition of maximum work in the context of electrochemistry and thermodynamics?

"Maximum Additional Work that can be extracted at constant pressure and temperature" to be precise, which may include electrical work, etc.  It means the work besides the expansion work.

The negative sign signifies the sign of the charge of the electron, since electrical work is proportional to the charge.