Using quantiative analysis to find empirical and molecular formula

The problem you gave doesn't quite work out, so I searched it up; did you perhaps mean $0.115 \space \text{g}$ of the sample were used in finding the sulfur content (you said $0.15 \space \text{g}$ instead).

If you did mean $0.115 \space \text{g}$, then we can figure out that the molecular formula of the unknown compound is $\text{CaC}_2\text{N}_2\text{S}_2$. Let me explain:

First, let's try figuring out how many moles of $\text{Ca}$ there are for each mole of the unknown compound, using the information that $0.250\space \text{g}$ of the compound was used to form $0.16 \space \text{g}$ of CaCO3. Converting everything to moles, this means we used $(0.250\space \text{g})\space/\space (156 \space \text{g/mol})=0.00160 \space \text{mol}$ of the unknown compound to get $(0.16\space \text{g})\space/\space (100.0869 \space \text{g/mol})=0.00160\space \text{mol CaCO}_3$, or $0.00160\space \text{mol Ca}$. This means that one mole of the unknown compound will contain $(0.00160\space \text{mol Ca})\space/\space (0.00160 \space \text{mol Unknown Compound}) \space\times\space (1 \space\text{mol Unknown Compound})=1\space\text{mol Ca}$

We can do this for each of the other procedures to get that for each mole of the unknown substance, we have $1\space\text{mol Ca}$, $2\space\text{mol S}$, and $2\space\text{mol N}$. Now, we can just subtract the masses of $\text{Ca}$, $\text{S}$, and $\text{N}$ from the molar mass of the unknown compound to get the mass of $\text{C}$ in the compound. For each mole of the unknown compound, we get that we have $\text{156 g}-\text{40.08 g}-2(\text{14.01 g})-2(\text{32.06 g})=23.8 \space\text{g}$ of carbon. This is just $(23.8\space \text{g})\space/\space (12.01 \space \text{g/mol})=1.98\space \text{mol} \approx 2 \space \text{mol}$ carbon per mole of unknown compound.
So, since we have $1\space\text{mol Ca}$, $2\space\text{mol S}$,  $2\space\text{mol N}$, and $2\space\text{mol C}$ for each mole of the unknown compound, the unknown compound must be $\boxed{\text{CaC}_2\text{N}_2\text{S}_2}$.


Hydrogen peroxide / sodium or potassium permanganate: reaction type

Learning from the comments and from looking at this video (https://www.youtube.com/watch?v=9S6v-NMMnB4), there are two reactions:

$$\ce{3H2O2 +2KMnO4 -> 3O2 + 2MnO2 + 2K+ + 2OH- + 2H2O}\tag{1}$$

$$\ce{2H2O2 ->[MnO2] O2 + 2H2O}\tag{2}$$

The first reaction creates the catalyst $\ce{MnO2}$, the second generates oxygen gas even after the permanganate is used up. If you look closely at the video, you will see bubbles (oxygen) coming out of the solution even after the solution is colorless (i.e. permanganate has completely reacted).

For some more chemistry, I also recommend this video by Prof. Martyn: https://www.youtube.com/watch?v=OLUyeCC-2Ko

  Can someone please enlighten me as to why some people would consider this reaction catalytic with the presence of the permanganate the catalyst?

This is not accurate. What is true is that you need much less permanganate than hydrogen peroxide because only a little is required to make the catalyst, which then allows the reaction to proceed without being used up.


Reaction of dimethyl sulfate with Protocatechualdehyde

When one equivalent of dimethyl sulfate reacts with 3,4-dihydroxybenzaldehyde (protocatechualdehyde), there are two nucleophilic sites in the molecule, namely 3- and 4-hydroxy-groups. If one equivalent of strong base such as $\ce{NaOH}$ is used, it will deprotonate the most acetic group between them, in this case, 4-hydroxy-group. For example, $\mathrm{p}K_\mathrm{a}$ of 4-hydroxybenzaldehyde is $7.61$ while $\mathrm{p}K_\mathrm{a}$ of 3-hydroxybenzaldehyde is $8.98$ at $\pu{25 ^\circ C}$ (Ref.1). Thus, it is safe to assume that product K is 3-hydroxy-4-methoxybenzaldehyde (isovanillin).

However, the regioselectivity of the synthesis (when two nucleophilic sites are present) can be controlled by the number of base equivalents in this kind of alkylation steps (Ref.2). Thus, in excess $\ce{NaOH}$, both phenolic hydroxy groups of 3,4-dihydroxybenzaldehyde would be deprotonated and methylation would occur at 3-phenolate, because it is the most reactive at this point between two (basically, negative charge of 4-phenolate should be largely delocalized within aromatic nucleus with benzaldehyde group than that of 3-phenolate without direct resonance effect from benzaldehyde group).  Therefore, the major product M in this case would be 4-hydroxy-3-methoxybenzaldehyde (vanillin). You can probably compare with methylation of ascorbic acid in different conditions.

Note: Some literature mentioned that when one equivalent of $\ce{NaOH}$ is used, reaction is must faster and tends to use both methyl groups of dimethyl sulfate. Thus, I won't surprised if the given answer for K is 3,4-dimethoxybenzaldehyde.

References:

E. P. Serjeant, Boyd Dempsey, Editors, IUPAC chemical data series no. 23: Ionization Constants of Organic Acids in Aqueous Solution; Pergamon: Oxford, United Kingdom, 1979 (Values extracted from PDF here).
Tobias Schneider, Vladimir Kubyshkin, Nediljko Budisa, "Synthesis of a Photo‐Caged DOPA Derivative by Selective Alkylation of 3,4‐Dihydroxybenzaldehyde," Eur. J. Org. Chem. 2018, (18), 2053-2063 (https://doi.org/10.1002/ejoc.201701749).


Interpreting from phase diagram whether O2 will float or sink

Yes, we can do that.

In fact, it is only the sign of the slope that matters. Say, you stand right below the gas/liquid curve; you apply a little more pressure, and you step from gas to liquid. How does that correspond to Le Chatelier's principle? Well, just like that: the system responds to increased pressure by decreasing its own volume, hence the liquid must be denser that the gas (which we of course already knew from independent sources).

Guess what?

The same reasoning applies verbatim to the liquid/solid curve, since it is sloped in the same direction.

As to the intermolecular forces, I think the usual hand-waving along the lines of "less freedom $\to$ less motion $\to$ lower volume" will do, and even that is redundant. Having solid denser than liquid is normal. If anything, it is the opposite situation that would require an explanation.


Is there an explanation for the tendency of silicon compounds to react
  with water?

Reaction with water, oxygen, etc., the answer is the same and has to do with the strength of the $\ce{Si-O}$ bond. Let's compare the following two reactions:

$$\ce{CH4 + \frac{1}{2}O2 -> CH3OH}$$
$$\ce{SiH4 + \frac{1}{2}O2 -> SiH3OH}$$

In both reactions we break an $\ce{X-H}$ bond and an $\ce{O-O}$ bond, while we make an $\ce{X-O}$ bond and an $\ce{O-H}$ bond.  The $\ce{O-O}$ bond breaking and the $\ce{O-H}$ bond making are common to both reactions and therefore cancel out.  Let's compare the only differences, the $\ce{X-H}$ bond breaking and the $\ce{X-O}$ bond making.

The following Table contains the bond energies for these bonds.

\begin{array}{|c|c|c|c|} \hline
\ce{X} & \text{X-H Bond Strength} & \text{X-O Bond Strength} & \text{Overall Change}  \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline
\ce{C} & 99 & 86 & -13\\ \hline
\ce{Si} & 75 & 110 & 35\\ \hline
\end{array}

Looking at the numbers it becomes clear that the $\ce{Si-H}$ bond is a lot weaker than the $\ce{C-H}$ bond, and the $\ce{Si-O}$ bond is a lot stronger than the $\ce{C-O}$ bond.  Consequently, the reaction involving silicon is more exothermic than the analogous reaction involving carbon.  This greater stability in the $\ce{Si-O}$ product is reflected in the transition state leading to it.  It lowers the energy of the transition state and consequently the reaction involving silicon is much more facile than the same reaction involving carbon.


Ratio of conjugate bases of a tricarboxylic acid

Let's use citric acid as an example. Although the $pK_a$ values of citric acid might not necessarily be the same as those above, it will be useful for demonstration.
$$\ce{C6H8O7 + H2O <=> C6H7O7- + H3O+ <=> C6H6O7^2- + H3O+ <=> C6H5O7^3- + H3O+}$$
Throughout the polyprotic acid titration, two protons are donated. The way this problem is solved is you have to use successive $\mathrm pK_\mathrm a$ values and concentrations.
Primary proton transfer equilibrium:
$$\ce{C6H8O7 + H2O <=> C6H7O7- + H3O+} $$
\begin{array} {|c|c|c|c|c|}
\hline
C_\mathrm{initial} & 0.20 & - & 0 & 0 \\
\hline
\Delta C & -x & - & +x & +x\\
\hline
C_\mathrm{final}& 0.20 - x & - & x & x \\
\hline
\end{array}
$$K_\mathrm {a_1} = 0.01 = \frac{[\ce{C6H7O7-}][\ce{H3O+}]}{[\ce{C6H8O7}]} $$
$$ K_\mathrm{a1} = 0.01 = \frac{x^2}{0.20-x} $$
Unfortunately, we can't disregard the x in the denominator so we have to solve the quadratic:
$$ x = 0.04 \ \mathrm{or} \ x = -0.05$$
We disregard our negative root. From $[\ce{C6H7O7}]^- = [\ce{H3O+}]$:
$$[\ce{C6H7O7}-] = \pu{0.04 mol L-1}$$
Now we use our second $\mathrm pK_\mathrm a$ to find the concentration of $\ce{C6H6O7^{2-}}$. Since $K_\mathrm {a_2} << K_\mathrm {a_1}$, we can assume that the hydronium ion concentration is unchanged by the second deportonation.
$$\ce{C6H7O7- + H2O <=> C6H6O7^2- + H3O+} $$
\begin{array} {|c|c|c|c|c|}
\hline
C_\mathrm{initial} & 0.04 & - & 0 & 0.04 \\
\hline
\Delta C & -x & - & +x & +x\\
\hline
C_\mathrm{final} & 0.04 - x & - & x & 0.04 + x \\
\hline
\end{array}
$$ K_\mathrm {a_2} = .00001 = \frac{x(0.04 + x)}{0.04-x} $$
$$ x = -0.04002 \ \mathrm{or} \ x = 9.995 \times 10^{-6} $$
Ignoring the negative root:
$$ [\ce{C6H6O7^2-}] = \pu{9.995 \times 10^{-6} mol L-1}$$
Finally, assuming that the concentration of hydronium ions and the conjugate base of citric acid after the 2nd deprotonation is unaffected by additional deprotonation, we can write up our equilibrium table again:
\begin{array} {|c|c|c|c|c|}
\hline\
C_\mathrm{initial}& 9.995*10^{-6} & - & 0 & 0.04 \\
\hline
\Delta C & -x & - & +x & +x\\
\hline
C_\mathrm{final} & 9.995*10^{-6} - x & - & x & 0.04 + x \\
\hline
\end{array}
$$ K_\mathrm {a_3} = .00000001 = \frac{x(0.04 + x)}{9.995 \cdot 10^{-6} -x} $$
$$ x = -0.04002 \ \mathrm{or} \ x = 2.49875 \times 10^{-12}$$
$$ [\ce{C6H5O7^3-}] = \pu{2.49875 \times 10^{-12} mol L-1}$$


Stability of dialkyl substituted alkenes

Following is an image of hyperconjugative effect. We can see a $\ce{\sigma}$-orbital delocalising electron pair with an empty $\ce{\pi^\ast}$-orbital.

The explanation to your problem can be given by deciding whether the effect is "centralized" or "decentralized". Let me explain it by an example.

A quick scene:
Suppose, you and your friend is sitting across a boat ($\ce{C_{you}=C_{friend}}$) in the middle of a river, and both of you have to reach on either bank soon (a process similar to delocalization, probably). Assume the methyl groups to be oars you need to move across the river.

In $\ce{R2C_{you}=C_{friend}H2}$, both the oars are with you. So, only you can row the boat and hence both of you can reach to the bank soon. Technically, the $\ce{\beta-H}$'s are at one carbon, so the effect can happen only on it, hence "centralized" and super-fast.
But, in $\ce{RHC_{you}=C_{friend}HR}$, both of you have the oars. So, both of you can row the boat and therefore both of you will reach to either bank a bit later. Technically, the $\ce{\beta-H}$'s are distributed equally among the carbons, so the effect can happen on either side, hence "decentralized" and not as fast as in the previous case.
As per your concern on steric factors, you may know that "alkyl" is referred to a straight hydrocarbon chain, and it isn't as bulky as to create "steric hindrance".
I guess that makes some sense :)

EDIT: The above stuffs explains how is first isomer the most stable. I'd previously said that "steric effects" doesn't matter, but in case of cis and trans isomers, it matters and it's the only thing that differentiate them (along with polarity). Due to this, cis isomer is least stable among the three.
Reference table of $\ce{\Delta H^°_f}$: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/heats-formation.html#Alkenes (in #C = 4)


The short answer is that the nuclear spin and electronic spin are separate effects. So an electronic radical ($\ce{N2H+}$) does not affect the nuclear spin of the protons or the nitrogen ($\ce{^14N}$ has nuclear spin 1).

So on a basic level, the net charge does not affect the overall nuclear spins.

The somewhat longer answer is that the nuclear spin states depend on the local chemical environment. This means that we can "see" different proton nuclear spins based on the molecular structure.

If there's are unpaired electrons, paramagnetic NMR requires understanding the coupling of the electron spin and the nuclear spin. It's possible, but not as straight-forward as typical diamagnetic NMR (for example $\ce{^1H}$ peaks may spread over 200-250 ppm, unlike ~8-12 ppm for normal protons).

Two main effects in paramagnetic NMR are pseudo-contact and contact shifts.

  The effect of the contact term arises from transfer of spin polarization to the observed nucleus. Spin polarization is a consequence of the very strong electron-nuclear (NMR detected) interaction. This coupling, also known by EPR spectroscopists as hyperfine coupling, is on the order of MHz, vs the usual internuclear (J) coupling observed in conventional NMR spectra, which are on the order of a few Hz. This difference reflects the large magnetic moment of an electron (−1.00 μB), which is much greater than any nuclear magnetic moment (e.g. for $\ce{^1H}$: 1.52×10−3 μB).


This is a part answer - an annotated short list of references regarding burst nucleation and mechanisms of nanoparticle growth.

Model of nanocrystal formation in solution by burst nucleation and diffusional growth (Robb and Privman, 2007), in this article the authors

  present a model with the assumptions of instantaneous thermalization below the critical nucleus size and irreversible diffusive growth above the critical size

A main feature of this article is that

  for the first time formulates LaMer's explanation of burst nucleation in a manner allowing quantitative calculations.

This is followed up by one of the authors the next year with the article Diffusional Nucleation of Nanocrystals and Their Self-Assembly into Uniform Colloids (Privman, 2008), where he

  review theoretical explanation of mechanisms of control of uniformity in growth of nanosize particles and colloids.

An explanatory diagram of LaMer's explanation (referred to in the 1st article) is shown below:

The image is from the article Controlled synthesis of colloidal silver nanoparticles in organic solutions: empirical rules for nucleation engineering (Sun, 2012) - unfortunately, the article is behind a paywall.

However, a sample chapter explaining the process in full entitled Growth of Nanocrystals in Solution (Viswanatha and Sarma), provides a greater detail about the processes involved with nanoparticle growth.


Is this the true expression for isochoric enthalpy of non-ideal gases?

The derivation for the first part of your question is shown below. I'm not sure what you are trying to do in the second part.

The heat capacity is the heat required to raise the temperature of  1 mole of substance by 1 K. Suppose  for the moment that heat capacity C is constant over a temperature range $T_1 $ to $T_2$, then the heat q required to raise 1 mole of substance  over this temperature range is
$$q=C(T_2-T_1) = C\Delta T$$
If the heat capacity is not constant, which is generally the case as real molecules have vibrational and rotational energy levels then,
$$ q=\int _{T_1} ^{T_2}C(T)dT$$

where in many cases the heat capacity can be approximated as $C(T)=a+bT+cT^{-2}$. For a gas the partition function can be calculated, if spectroscopic values are known, and $C(T)$ calculated that way.

The heat absorbed depends on the path taken to go between the two end points so that C is not well defined. To make it well defined either T or P must be constant.

At constant volume, $q_V=C_V\Delta T$ and the work done as expansion against external pressure is zero, thus from the first law with internal energy as U,
$$ \Delta U = C_V\Delta T $$

and a similar integral as described above is used when the temperature range is too large for $C_V$ to be considered a constant. (This equation shows also, in the limit of small changes,  that heat capacity is the slope of internal energy vs. temperature.)

In the case of constant pressure $q_P=C_P\Delta T$. The change in enthalpy is
$$\Delta H = C_P\Delta T$$
and so in general
$$ \Delta H=\int _{T_1} ^{T_2}C_P(T)dT$$

To demonstrate that $q_P=\Delta H$ we note that in this case the volume is not constant and so the heat transferred at constant pressure $q_P$  is related to internal energy change $\Delta U$ as
$$ \Delta U = q_P-P\Delta V$$
where $P\Delta V$ is the work done. Rearranging and expressing the $\Delta$'s directly gives
$$ q_P= (U_2+PV_2) - (U_1 +PV_1)$$
from which using the definition of enthalpy $H=U+PV$ gives $q_P=H_2-H_1$ and shows that $q_P=\Delta H$.


Empirical data

If you analyze crystal structure data for short intermolecular distances between hydrogen atoms and potential acceptor atoms, you will find a distribution of these angles. The simplest explanation for this distribution is that there is a favorable interaction that is strongest for the most commonly observed angles. The fit between empirical data and computational results is fairly good, e.g. https://www.researchgate.net/publication/242862021_Distribution_of_hydrogen_bond_angles_in_molecular_crystals

Dipole-dipole interactions

If you treat hydrogen bonds as pure dipole-dipole interactions (Q-Y ... H-X), you expect the bond to be directional. After all, you maximize the interaction by close approach of the negative partial charge on the acceptor and the positive partial charge on the donor (i.e. the hydrogen atom in a polar bond). However, for this model the angular dependence should be symmetrical, i.e. there should be a similar decrease in bonds strength when you rotate the donor or the acceptor away from a colinear arrangement. The "rules" for hydrogen bonds say otherwise. It is more important that the angle at the hydrogen atom is maximal (Y ... H-X) than that the angle at the acceptor atom is maximal (Q-Y ... H).

Covalent contribution

There is a covalent contribution to the hydrogen bond that explains why a linear arrangement around the hydrogen atoms is favored. The strongest hydrogen bonds are those where hydrogen bond and covalent bond with hydrogen become interchangable ("symmetric hydrogen bonds"). In his 2002 review, Steiner wrote:

  All hydrogen bonds can be considered as incipient proton transfer reactions, and for strong hydrogen bonds, this reaction can be in a very advanced state.

There is some controversy whether the hydrogen bond has covalent character, and some question how to describe it theoretically (as a 3 center 4 electron bond?). There is, however, strong evidence from NMR that there is indeed a covalent contribution because nuclei couple in a through-bond fashion across hydrogen bonds (e.g. this review).


Charge separation and stability in resonating structures

The answer is 1) when positive and negative charges on C atoms are adjacent.This is because in the second Resonating Structure the charges have to be seprated and as both charges are opposite in nature they will attract each other and make the process of seprating them difficult making it unstabilized.


Spices are absorbed or adsorbed in food?

The answer varies with the type of food and the type of condiment. Both are very complicated mixtures, so a same combination of food and condiment can display varying levels of adsorption/absorption depending on the region/compound being analyzed, but in general, I expect that while all of chemical/physical adsorption and chemical/physical absorption happen to some extent (cooking is quite a messy collection of physical and chemical interactions!), either physical absorption is the most prevalent, or none of them happens to the largest part of the spices.

First of all, adsorption is a surface phenomenon, which fundamentally limits how much matter can be considered adsorbed onto an object. Adsorbed films are usually very thin (probably not much more than a few nanometers thick, often a single molecule thick). Regular macroscopic objects have very little surface area relative to their volume, so it takes a comparatively tiny amount of substance to completely cover the object in a thin adsorbed layer, compared to say, activated charcoal. Thus we can likely exclude adsorption as the main factor.

It's also important to consider that most food is cooked with some amount of solvent (water or oil/fat), and in fact solvents come out of the food while it's being cooked. This means that a lot of the spices can very well end up as part of a liquid (albeit often quite concentrated) solution, which remains stuck on the surface simply due to its high viscosity and significant friction with the uneven surface of the food. I bet this is what goes on in most cases.


How are reactants ad-/ -desorbed during Haber process?

All chemical catalysis can be grouped into 1 of 2 categories: homogeneous or heterogeneous catalysis. Homogeneous catalysts are ones that are in the same phase (i.e. solid, liquid, gas BUT USUALLY IN SOLUTION) as the reactants they are helping to react. Heterogeneous catalysts are in a different phase than the reactants they are working on. Industrial processes (like the Haber-Bosch process for making ammonia) frequently use heterogeneous catalysts due to their comparatively inexpensive nature, and there ease of handling on a large scale. Doing it this way can cause some added problems, namely the reactants need to bind to the catalyst's surface (adsorption) for the catalyst to work. In the H-B process this is gaseous N2 and H2 binding to a mainly iron catalyst. To help this process along the iron is made with a high surface area with holes and such (think like the texture of a sponge except on a much smaller scale). After the reaction happens the products then need to release from the catalyst (desorption) so they can be separated from the catalyst and be collected.  The active site simply refers to the place on the catalyst where the reaction happens. Adsorption is helped by the high temperature and massive pressures in the H-B process, as the gasses are flowed through a chamber at high pressure containing the solid heterogeneous catalyst. Not all of the molecules stick and not all that do stick react, so the underacted gas is recycled and passed over the catalyst again. Long story long...the very question you asked won the Gerhard Ertl the 2007 Nobel Prize in Chemistry.


The name supercritical fluid is fancy but don't be confused by thinking that it this would a very odd looking state of material. Supercritical CO2 is just a state of CO2 at a certain T and P, so it looks very much like gaseous CO2 i.e., it is a transparent clear "fluid" whose density and viscosity is higher than a gas but lesser than that of the liquid. If you are able to see an interface in those transparent cells (e.g. here https://www.youtube.com/watch?v=GEr3NxsPTOA), it means that the liquid phase and the gas phases exist in the system. We are able to see the interface (boundary) when the refractive index is different from the container walls, and the surroundings. In short do not expect a new form of matter in supercritical state. It is just like thinking what does a plasma look like? It very much appears a like an ordinary flame which is extremely hot.

Response to the comment "is supercritical fluid a special types of a liquid ? if no, why then some SCFs used as a solvent ? what if a small pot of liquid heated inside a large pressure chamber to its critical point? would it expanding fill the whole chamber like a gas or stay in its container like a liquid ?"

I am afraid you are mixing several concepts here. The key question is, how does one distinguish a liquid phase and a gas phase? Supercritical fluid is not a liquid because its density and viscosity are not close to liquids, nor it has a definite boiling or melting point (a liquid does, right?). Think of supercritical fluid as a state or condition of a gas where it is above its critical temperature and pressure. That is all. We cannot say supercritical carbon dioxide has a boiling point of x Celsius or a freezing point of y Celsius.

Coming to your second part on SFC extraction: First, let us talk about gases. Can the gases act as a solvent or carrier of molecules? The answer is yes, and this is the very basis of gas chromatography. How do you detect odors? The gas molecules are carrying the odor causing molecules from a far away place to your nose i.e., gases also have some capability of "dissolving" substances. In the same vein, if I use a temperature of Tc, and a pressure Pc for carbon dioxide (or any gas), we can use it for extraction. It just turns out that supercritical carbon dioxide has higher dissolving power than gaseous carbon dioxide and certainly less than liquid carbon dioxide.

Third point: Let us do an imaginary experiment: If you have a sealed container of fixed volume containing nothing but carbon dioxide at very low temperature. Also assume a pressure gauge is also attached. You have filled in enough carbon dioxide in the chamber that the gauge shows a value of 85 atm. Now you start heating the chamber until the temperature of the chamber reaches 35 Celsius. Imagine you can see the insides of that chamber with a camera: You will see nothing because there will be no liquid in that chamber. It will look empty to your eyes. Given the fact you know that an empty glass is not empty, it is filled with gases (nitrogen, oxygen etc). In the same way, our chamber is also filled with a carbon dioxide which our eyes cannot see. We will still call this carbon dioxide as supercritical carbon dioxide because we know from our experimental settings that we are above critical temperature and pressure of carbon dioxide. You can go higher temperature to 45 Celsius, still you are in supercritical state. Thus in a phase diagram, there is a supercritical region.

Don't let the word fluid confuse you. Since we cannot call a gas above its critical temperature and pressure a typical gas, nor we can call it a liquid, it is best to call it a fluid. Hope that clarifies your confusion.


What is the process involved in performing a mass spectrometry experiment?

This is hard to answer, because in my experience mass spectrometry isn't that time consuming at all. The experiment itself doesn't take that long. In this experiment the sample is ionized and charged fragments are accelerated in an electrical or magnetic field. Depending on their mass/charge ratio they will end up on specific locations in the detector.

The process around the experiment itself is dependent on your specific lab, but typically the mass spectrometer is operated by only a few people that do all the samples of the entire department. Because of that, there will often be a queue which could be several days. Additionally, not all mass spectra can be digitally identified as coming from a certain chemical, especially for larger molecules there are often various options that require human interference for the identification. In our lab this human interpretation is left to the person owning the sample, but if that part is also done by the mass spectrometer operators (I know in some labs this is the case) then I can imagine that this last step will be time-limiting.

To summarize: the experiment is short, the procedure around it (dependent on your specific lab) can take quite some time.


Safest way to conduct a reaction with bromine under reflux conditions

Check if you may skip using elemental bromine as starting material altogether.  An alternative approach may be running the bromination in a heterogenic, biphasic setup with the couple $\ce{NaBrO3/NaHSO3}$ in water (e.g., 1998JOrgChem6023, 2000OrgProcRes Dev.30), or $\ce{NaBr}$ / sodium perborate (e.g., 1998SynthCommun925), or pyridinium tribomide (e.g.,2005JChemEduc306), or bromodimethylsulfonium bromide prepared by mixing aqueous $\ce{HBr}$ with DMSO (1997JOC4321, often at rt).  They share a slower release of bromine, and because of this, lower amount and concentration of reactive bromine and attenuate the hassle with corrosive gases.
If staying with liquid bromine:

Trapping unconsumed bromine with thiosulfate per se is a good idea, but the reaction of bromine gas with solid $\ce{Na2S2O3}$ in the drying tube is both incomplete, bears the potential that the drying tube gets congested, and tends to be messy to clean.  You could end up with a closed vessel, which isn't safe; and even dangerous as you generate HBr in stoichiometric amounts.  Thus passing the exhaust gas across scrubbers filled with aqueous thiosulfate solution has a much higher chance of win.  Don't forget the safety flask (different reaction, but same principle).

The the normal balloons from the toy shop does break under bromine, as well as HBr gas too quickly.

Beside @Mathew Mahindaratne's comment to dissolve bromine in chloroform, using acetic acid as solvent provides you a wider margin of temperature.  (If your substrate can withstand the acid.)


Uses of Fourier / Laplace transforms in chemistry (apart from spectroscopy)

For the principle «what is small, gets large; what is large, gets small» when applying Fourier transformation and inverse Fourier transformation, it is one important tool in crystallography to relate between the regular arrangement of molecules in crystals at Angstrom scale, and the intensity recordings on film (old school) and electronic detector (very often area CCD) where signals are much wider apart from each other.  (This is often is referred to the relationship between direct and reciprocal space, too.)  Thus, scrutinizing the diffraction pattern's symmetry from special directions (figure b) below) allows the trained eye to relate to possible symmetries of the crystalline sample (a) , and the ratio of vector lengths describing the smallest unit of a crystal structure, the unit cell.  Today, assessing position and intensities of these data is typically assisted by a computer.

(credit, open access)
As shown by Tsutaoka et al. (Eur. J. Phys. 35 (2014) 055021, doi 10.1088/0143-0807/35/5/055021), it is possible to simulate such patterns with a laser pointer passing two film gratings slightly rotated to each other; the 2D crystal is then build by the points where the two gratings S1 and S2 intersect each other (note the vector ratio $\vec{a} / \vec{b}$ vs. $\vec{a^*} / \vec{b^*}$, and angles enclosed), e.g.

ImageMagick is one program allowing you to experiment with FT on 2D images but recommend to consult it altogether with the above mentioned open access publication by Aubert and Lecomte (J. Appl. Cryst. (2007) 40, 1153-1165, doi 10.1107/S0021889807043622) for additional examples and background.


Which set of reactants is better for preparation of 1-methoxy-4-nitrobenzene?

Drastic conditions for reaction 1? Not really. Reflux the reactants in $\ce{MeOH}$ and the reaction should go smoothly and in high yield.

Reaction 2 - two difficulties.
The anion of 4-nitrophenol is very stable and, because of the electron withdrawing power of the nitro group, it is not very nucleophilic. Second difficulty is that bromomethane has a boiling point of 3C so the reaction mixture has to be kept cold to keep it in the reaction mixture. This does not help you overcome the poor nucleophilicity of your phenoxide.

Without running the two reactions it is guesswork to predict which will give the better outcome, but I favour the SNAr.


Why does Bismuth only create a +3 ion instead of a +5 ion?

Isn't it just that +5 is a lot of (positive) charge for a single cation? Which other (transition) metals do form a $\ce{M^{5+}}$ cation? Just out of my mind, I can't think of any.

Typically, these high oxidation states exist in the form of oxo-anions ($\ce{CrO4-}$, $\ce{MnO4-}$, $\ce{VO4^{3-}}$, etc.) and bismuth is no exception here: $\ce{Bi(V)}$ exists as bismutate ($\ce{BiO3-})$.


How can one find products when one knows the reagents?

If I went into specifics, this answer would span a few thousand books. Instead, I'll give an overview:
There are many ways of determining the products given the reagents:
Actually reacting the reactants
One can, of course, actually make the reaction take place. Once you've done that, you need to identify the contents of the resulting "soup". A few methods for this  (You may have to separate the products first) :
NMR Spectroscopy
Here, a nifty technique is used to determine the magnetic properties of a compound. From these properties, one can make reasonable guesses at the substructures existing in a molecule, and check against a list of possible products. This is a very powerful technique, and usually gives good results.
Qualitative analysis
Here, we use various chemical reagents on the products, and analyse these products, making conclusions about the nature of the original products.
Microscopy
Here, using an STM, one figures out the structure of the molecule. This isn't used that often, but is an interesting technique.
There are many other such techniques, with varying accuracies and applicabilities.
Predicting the products in the absence of experiment
A lot of chemical research is into improving its (already vast) predictive power. The end-game of most sciences, after all, is to be able to predict stuff. So, in many cases, we can just predict how the reaction proceeds, without actually making the reaction occur in a lab.
Figuring out a reaction mechanism
This generally occurs in organic reactions. Here, you break the reaction into elementary steps. Each step can be explained by looking at electronic attraction/repulsion, as well as the increase in stability (which can generally be estimated by taking into account the inductive, mesomeric, hyperconjugation, and steric effects, though other things may effect the overall stability as well). From these steps, one gets a proposed mechanism, which may be verified.
Knowing solubilities and other numerical values
In ionic/inorganic reactions, the path a reaction takes generally hinges on the solubility of its final products. This is sometimes the case for other parameters as well. Knowing various numerical values lets one compare two possible reaction paths
Computational chemistry
Here, one uses the equations of quantum mechanics (Schrodinger's equation and a few of its cronies), and solves them for the given problem using numerical methods.

The particular reaction you have given
This one can be easily explained by resonance. In $\ce{CH3COOH}$, the $\ce{H}$ is loosely bonded, as the $\ce{C=O}$ bond resonates with the $\ce{C-O}$ bond. This makes it easy for the $\ce{H}$ to float away as $\ce{H+}$, leaving behind a pretty stable $\ce{CH3COO-}$ (made stable by resonance of the negatively charged lone pair over the oxygen atoms).
So far, we know that $$\ce{CH3COOH -> CH3COO- + H+}$$
Now we know that it is behaving like an acid. It is extremely well known that water forms a Hydronium ($\ce{H3O+}$) ion by self-dissociation, and this is accelerated by making the medium acidic (All free $\ce{H+}$s find the nearest $\ce{H2O}$ molecule and form $\ce{H3O+}$). $\ce{H3O+}$ is stable due to resonance, again.
So we now know that $$\ce{H+ + H2O -> H3O+}$$
Put the two reactions together and you get $$\ce{CH3COOH + H2O -> H3O+ + CH3COO-}$$
Though, in this case, the historical method of finding out the reaction may have been different -- acetic acid has been well known and studied for a very long time.


Is turning graphite carbon into diamond industrially doable?

It has been done since 1954 and has been a commercial success for some time. Many products now have diamond-like coatings, produced comparatively inexpensively through vapor deposition without need for high pressure.

BTW, in English "burning" means to set on fire, oxidize; I believe you mean "heating graphite", and, as your graph shows, pressure is needed as well as heat.


Abstraction of a proton in cycloheptatriene

The pKa for a vinylic proton is ~ 43, the pKa for propene is ~ 40, and the pKa for 1,4-pentadiene is ~ 35.  Consequently it's not the vinylic protons that are the most acidic, it's the allylic ones.  Now the question whether tropylium anion is aromatic/antiaromatic is complicated.


When ammonium chloride is hydrolyzing, there is established equilibrium

$$\ce{NH4+  + H2O <<=> NH3 + H2O+}$$

with $$\mathrm{pH}=7-\frac 12 ( 4.75 + \log {c_{\ce{NH4Cl}}} )$$

$$\mathrm{pOH}=14 - \mathrm{pH}=7+\frac 12 ( 4.75 + \log {c_{\ce{NH4Cl}}} )$$

Concentration of ammonia is several orders higher than concentration of hydroxide ions, with origin of water dissociation, so no ammonia hydroxide there.

When ammonia is dissolved, there is established equilibrium

$$\ce{NH3  + H2O <<=> NH4+ + OH-}$$

with $$\mathrm{pH}=14-\frac 12 ( 4.75 - \log {c_{\ce{NH3}}} )$$

Equimolar concentrations of $\ce{NH4+ + OH-}$ are sometimes formally called ammonium hydroxide, but using this term is deprecated and it cannot be isolated.


Removing prenylated phenolics from clothing, shoes, and upholstery

As a starting point, I shall assume that the irritant compounds are freely exposed to the environment (which is not necessarily the case, as I mention further on). As such, one of the conceivably simplest ways to remove them is to wash them off, i.e. disperse them in a solvent and discard the liquid mixture. For several reasons (abundance, extensive solvation ability, non-toxicity, etc.) water is a good choice.

First of all, how well would pure water work at dispersing the prenylated phenolics? Probably not too great. The representative parent compounds of the irritants, hydroquinone and p-hydroxybenzoic acid have modest solubilities in water (59 g and 5 g per 1 L, respectively), and the addition of farnesyl chains in the compounds would quite likely cause a severe reduction in water solubility, unless the farnesyl chains are highly substituted with hydrophilic groups.

Of course, there is no substance that is entirely insoluble in any solvent, though it is very common for certain solutes to require impractical amounts of solvent to disperse completely. It could well be that a washing machine's load of water (of order 100 L) is actually capable of dissolving the farnesylated toxins, especially because natural products are generally present in rather small quantities. However, there are some techniques which can possibly help the process. I list a few in the following:

Use hot water - The majority of substances become more soluble in hot solvents because dissolution very often increases the entropy of a system ($\Delta S>0$), and hence from the Gibbs' free energy relation $\Delta G=\Delta H - T \Delta S$, an increase in temperature leads to a larger decrease in free energy, making the process more favourable. Of course, this only really works well if the water is hot throughout the soaking/washing process; adding hot water and letting it cool off will just cause part of the solute to precipitate out of solution again.
Use a surfactant (detergent) - Since the offending compounds potentially have large hydrophobic chains, they can conceivably removed by the same process we use to wash away oils and fats. An amphipatic molecule contains a hydrophobic part (such as the hydrocarbon chain of sodium dodecylbenzenesulphonate) responsible for interacting with the oils and a hydrophilic part (in SDBS, the sulphonate group), which is well solvated by water. Thus, the surfactant bridges the interaction between oil and water, allowing effective dispersion.
Use a basic solution - In this specific case, the undesirable compounds have some acidic character. In particular, the hydrogen atom in the -COOH (carboxyl) group of the p-hydroxybenzoic acid derivatives are about as acidic as acetic acid. The -OH (hydroxyl) groups in both p-hydroxybenzoic acid and hydroquinone are also slightly acidic, though weaker by some 5-6 orders of magnitude. Thus, mixing in a basic solution may allow an acid-base reaction to take place, forming salts of the organic compounds with significantly increased solubility in water. Many washing powders/liquids are already slightly alkaline, but it is also easy to prepare basic solutions separately. A cheap and very safe base would be sodium carbonate, $\ce{Na2CO3}$, though a few tens of grams may be required to prepare a decent volume of basic medium for soaking. Sodium bicarbonate, $\ce{NaHCO3}$, also works though it is a weaker base and would require a higher amount to reach a given pH value. Caustic soda, $\ce{NaOH}$, is yet another cheap option, but care must be taken to not add too much, as it is a strong base (in contrast to the previous two). I don't know how resistant different fabrics are but, very approximately, any more than a couple of grams of lye per litre of water risks partially hydrolysing the fabric and weakening its structural integrity. Use at your own risk!
Use a bleaching agent (chlorinated or non-chlorinated) - Bleaches work by partially oxidizing organic matter, breaking organic molecules into smaller chunks and making each piece more hydrophilic (due to added oxygenated functionalities). Even if the compounds aren't removed entirely, the reaction products may be much less harmful.
Miscellaneous - You suggest exposure to sunlight. Light is certainly capable of decomposing organic substances, especially UV light, but there's no guarantee that the exposure required to eliminate the irritants won't also break down fabric and bleach dyes. Waiting is also probably not very efficient, as there is no guarantee the irritant compounds are unstable enough to undergo spontaneous decomposition in ambient conditions at a significant rate.

Unfortunately, all this discussion is of limited use if the initial assumption (freely accessible irritants) is incorrect. It could be that the compounds are stored in cavities in the plant hair's surface or interior, and the hairs themselves are insoluble and impermeable to water, effectively locking the toxins away. In this case, the compounds will likely be removed only slowly, as the hairs are progressively worn out by abrasion and other processes. Thus a physical removal method may be preferable. I have heard that adhesives (sticky tape, waxing strips, etc.) can be successfully used to peel out painful hairs. This is also a useful option for surfaces which cannot be washed easily (such as car upholstery).

In practice, I am not sure how this will all work out. It's been a few days since I wrote my comments, so perhaps you've had time to test something out? If so, I would be curious to hear some feedback!


Is the 'central Tin atom' in crystalline tin oxide really offset like this?

From http://www.crystallography.net/cod/1000062.html one can see that Sn sits in the middle, without any displacement, and mimics the classic Rutile-type structure:

I checked original Wikimedia image, and according to metadata

  CIF retrieved from The American Mineralogist Crystal Structure Database

is no longer available. I also failed to discover this source:

  X-ray crystallographic data from:
  R. W. G. Wyckoff (1963) Second edition. Interscience Publishers, New York, New York. Crystal Structures 1, 239-444

Ideally, if you have *.cif file, it might be worth checking occupancies for the atoms you are interested in. Probably there is a 50-50 disordered Sn atom about center of unit cell, and you only observe a single site. As long as this is not the case, I would rely on the available COD data from 1971 instead.


Why does increase in pressure favor formation of water in water-ice equilibrium (at 273K)?

To answer this question, it is best to look at a phase diagram of $\ce{H2O}$.
Note the axes are kilopascal and celsius, but we know that $0^{\circ}\text{C}=273\text{K}$ and $101\text{kPa}=1\text{atm}$. We can see that if the pressure goes up at a constant temperature, the system will move from the phase boundary into the liquid phase.

The generally cited reason for why this occurs is that hydrogen bond network of ice makes it less dense than liquid water, and since higher pressure will favor the more dense state, the ice will melt to become liquid. This leads to a lower overall system volume.


How would you assign the chemical shift for those 4 peaks?

Since there is overlap between some of the resonances I would run a COSY experiment to resolve the peaks and find correlations, and make assignments on that basis. A number of things can complicate assignment. First, the CHO group can be in axial or equatorial positions, leading to two overlapping subspectra under slow exchange. In addition you have different shifts for geminal Hs, in equatorial/axial positions, separated by ~0.5 ppm (equatorial further downfield), with resolution into two peaks depending on the exchange time (e.g. temperature). The equatorial/axial H are coupled with a relatively strong J-coupling (~12 Hz). The chemically equivalent hydrogens are not magnetically equivalent, but this is not very important since long distance couplings are small.


Is the density of solute same as the density of solution?

The density of the solution isn't needed for either wt/wt% nor vol/vol%.

wt% = (grams solute)/(grams solution)

vol% = (vol solute)/(vol solution)

For vol% it only really makes sense if mixing two liquids like ethanol and water.


Grignard reagents react with epoxides but dissolve in THF, what happens in oxetane (four membered ether ring)?

In a 1979 paper entitled Copper-catalysed reactions of Grignard reagents with epoxides and oxetane, Huynh, Derguini-Boumechal, and Linstrumelle report that the copper-catalysed ring opening of oxetane by various Grignard reagents gives the alcohols "in acceptable yields":

$$ \begin{array}{cccc}
\ce{R} & \ce{X} & \text{Solvent} & \text{Yield (%)} \\
\hline
n\text{-}\ce{C4H9} & \ce{Cl} & \text{Ether} & 75 \\
\ce{C6H5} & \ce{Br} & \text{THF} & 52 \\
\ce{CH2=CH-CH2} & \ce{Br} & \text{Ether} & 50 \\
\ce{(CH3)2C=CH-CH2} & \ce{Cl} & \text{THF} & 50
\end{array} $$


I completely agree with Gonçalo Justino's answer, but would like to add another shade to it.
In principle there are quite a few possible arrangements thinkable for this molecule - at least in the gas phase. In this thought it simply refers to the fact that they are local minima on the potential energy surface. The most stable of all is the one that has been discussed here before. Hypochlorous acid is also one of the compounds that disrupt the ozone cycle. Because concentrations in the stratosphere are low, we can find most possible arrangements of the three atoms. It should also be noted, that these reactions usually occur via a radical pathway.
Structure of $\ce{HOCl}$
The bonding situation in these "weird" compounds is not obvious at all, classifying it as a purely covalent bound is equally as wrong as calling it ionic.
Interestingly the $\ce{O-H}$ bond is much more covalent than the $\ce{O-Cl}$ bond, when we look at the most stable, i.e. bent, conformation. I am not going into detail about possible linear intermediates, as they are only very short lived species.
The analysis of a DF-BP86/def2-SVP calculation with the quantum theory of atoms in molecules (QTAIM) shows us this somewhat unexpected behaviour.

What we see here is that the bond between the hydrogen and the oxygen is strongly covalent as the electron density is $\rho_\mathrm{BCP}(\ce{O-H})=0.34~\mathrm{e\,a.\!u.^{-3}}$ and the Laplacian has a high negative value, $\nabla^2\rho_\mathrm{BCP}(\ce{O-H})=-2.02$, i.e. much charge accumulation. On the other hand the chlorine oxygen bond has quite some ionic character. The electron density is $\rho_\mathrm{BCP}(\ce{O-Cl})=0.18~\mathrm{e\,a.\!u.^{-3}}$, while the Laplacian is slightly positive, $\nabla^2\rho_\mathrm{BCP}(\ce{O-Cl})=+0.01$, i.e. low charge depletion. This makes this bond about half ionic and half covalent. (An ideal ionic bond would have an electron density close to zero and a much larger positive Laplacian value.)
From this analysis it makes sense, that in the gas phase the molecule more likely dissociates via $$\ce{HOCl ->[h\nu] HO + Cl}.$$
In aqueous solution there are quite a few more equilibria at work and the reaction and stability of this compound is highly pH dependent. In general this compound will decompose whenever it gets the chance.
Structure of $\ce{OClH}$
This structure is about $61.5~\mathrm{kcal\,mol^{-1}}$ higher in energy and therefore quite unstable. But that does not mean it does not exist.
The interesting point here is that both bonds are about the same strength and both can be considered predominantly covalent. The oxygen chlorine bond, $\rho_\mathrm{BCP}(\ce{O-Cl})=0.25~\mathrm{e\,a.\!u.^{-3}}$, has a little higher electron density than the chlorine hydrogen bond, $\rho_\mathrm{BCP}(\ce{Cl-H})=0.21~\mathrm{e\,a.\!u.^{-3}}$. Both have a negative Laplacian, $\nabla^2\rho_\mathrm{BCP}(\ce{O-Cl})=-0.19$, $\nabla^2\rho_\mathrm{BCP}(\ce{Cl-H})=-0.57$, i.e. charge accumulation.

Other possible structures of $\ce{HOCl}$
Any other structure, for example linear or bent $\ce{OHCl}$, are no minima on the potential energy surface. They might exist as excited or transition states, but they are incredibly short lived. One only has to care about these species when dealing with complex gas phase reactions. (There is now way they could be stable in solution.)
Conclusion
While it is theoretically possible to have the oxygen in a terminal position, it is certainly not the ground state. Another aspect to think about is also, how accurately can a Lewis structure represent the molecule. Especially when it comes to small molecules one always has to take it with a grain of salt as most of the theories brake down at this level.
For up to highschool level of chemistry, however, the answer to this question is the one given by so many others before.

If that did not bore you to death, might I suggest additional literature: Kirk A. Peterson, Sergei Skokov and Joel M. Bowman; J. Chem. Phys. 1999, 111, 7446. (mirror)


The order is correct but this is my procedure.
Stronger the bond smaller the length


Are triglycerides soluble in polar solvents?

Polar is not a black-and-white distinction. There are weakly polar compounds (like ether), medium polar compounds (dichloromethane) and strongly polar compounds (dimethyl sulfoxide). Fats (i.e. triglycerides) can dissolve in weakly and maybe moderately polar solvents but not in strongly polar ones.

While a dipole can be induced in triglycerides to some extend, most of the long nonpolar side chain can only give weak induced dipoles. Thus, weakly polar solvents may still dissolve these by dipole–induced dipole interactions but the triglycerides are just too nonpolar for strongly polar solvents.

Water is one of the most polar solvents and additionally able to hydrogen bond. While triglycerides can accept hydrogen bonds around the glycerin moiety, they are just not polarisable enough for water to act. Hence, water prefers to form its own phase.


Half equivalence point is exactly what it sounds like. It is the point where the volume added is half of what it will be at the equivalence point.

There are 3 cases

Strong Acid vs Strong Base: Here one can simply apply law of equivalence and find amount of $\ce{H+}$ in the solution. Similar method for Strong base vs Strong Acid.
Weak Acid vs Strong Base: Here one can apply Henderson-Hasselbalch equation as a general result one gets pH = pKa.
Weak Base vs Strong Acid: This case is similar to WA vs SB and one can easily use Henderson-Hasselbalch for bases again one gets pOH = pKb


The larger $\ce{Cs+}$ ion may interact with more water molecules at once than $\ce{Li+}$, but each interaction is weaker. The prevailing interaction between water molecules and ions are ion-permanent dipole interactions (or just ion-dipole), which are proportional in strength to the amount of charge on the ion and the partial charge on the dipole, and inversely proportional to some power of the distance between the centres of the charges (via the shell theorem). Hence, the closer the water molecules get to the ion, the stronger the interaction. Since $\ce{Cs+}$ is very large, the charge centres don't come so close, and the interaction is relatively weak. $\ce{Li+}$ is much smaller, so there is strong solvation. For sufficiently small, highly charged ions, the first solvation shell comes so close to the ion that it allows covalent bonding to take place.

You may find this succinctly described as $\ce{Cs+}$ having a lower charge density than $\ce{Li+}$, which is about the same thing.


Does freezing point depression depend on the charge of the solvated ions?

First let's gather the data from the CRC Handbook of Chemistry and Physics, 92nd edition (mol/L is moles per liter). The data below is reported as molarity rather than molality since the problem is posed in molarity. However in a solution as dilute as 0.1 M the difference shouldn't be large.

         wt%      mol/L      Delta-F
  HCl     0       0        0.00
          0.5     0.137   -0.49
          1       0.275   -0.99
          2       0.553   -2.08
          3       0.833   -3.28
          4       1.117   -4.58
          5       1.403   -5.98
  least squares fit Delta-f = = -3.4683*[HCL] - 0.5624*[HCL]^2
  [HCL] =0.1 M, Delta-f = -0.35

         wt%      mol/L      Delta-F
  NaBr    0       0.00     0.00
          0.5     0.049   -0.17
          1       0.098   -0.34
          2       0.198   -0.69
          3       0.301   -1.04
          4       0.405   -1.39
          5       0.512   -1.76
  least squares fit Delta-f = = -3.4823*[NaBr] + 0.0930*[NaBr]^2
  [NaBr] =0.1 M, Delta-f = -0.35

         wt%      mol/L      Delta-F
  KNO3   0        0        0.00
         0.5      0.05    -0.17
         1        0.099   -0.33
         2        0.2     -0.64
         3        0.302   -0.94
         4        0.405   -1.22
         5        0.509   -1.50
  least squares fit Delta-f = = -3.4264*[KNO3] + 0.9923*[KNO3]^2
  [KNO3] =0.1 M, Delta-f = -0.33

         wt%      mol/L      Delta-F
  MgSO4  0        0        0.00
         0.5      0.042    -0.10
         1        0.084    -0.19
         2        0.169    -0.36
         3        0.257    -0.52
         4        0.346    -0.69
         5        0.437    -0.87
  least squares fit Delta-f = = -2.3466*[MgSO4] + 0.9726*[MgSO4]^2
  [MgSO4] =0.1 M, Delta-f = -0.22

So there is a difference. 0.1 molar $\ce{MgSO4}$ has a freezing point depression of 0.22 degrees Celsius whereas the other 0.1 molar salts are about 0.35 degrees.

As far as the "answer" goes the problem is very poorly worded. First the problem states to "assume an ideal solution," then poses the problem in such a manor that some sort of non-ideal behavior must be assumed to obtain an answer. "Ideal behavior" for electrolytes only occurs in very dilute solutions. A 0.1 molar solution is much too concentrated for ideal electrolyte behavior.

Without any "lookup" of data the only solution seems to be some hand-waving about $\ce{MgSO4}$ being the only salt with divalent ions so the effective dissociation is smaller because of forming ion-pairs in solution.


You have now given yourself some information to go on with, and indeed have identified most of the important fragments. Now, the tricky bit is putting them all together! Let's work through what you've got.

Let's first assume that the sample is pure, and the integrals are correct to within reasonable error. This then gives you 5 obvious environments with ratio 3:3:4:1:1. Environments with 4 equivalent protons are not that common, so we might suspect there is some overlap here, and that we have more than one group. So, I count 12 protons.

The peak at 0.9 integrates to 3 - it is most likely a methyl group, and is closely resembling a triplet, so we can make the guess that it is next to a CH2. (It is in fact not a true triplet, and is immediately recognisable as a terminal methyl for a longer alkyl chain - see this link for more information). But for now let's call it a triplet.

At 1.2, we also have a -CH3 (integrates as 3), which is a doublet, so must be next to a -CH.

The multiplet mess around 1.4 must contain at least one -CH2 (to couple to the methyl at 0.9), which will leave us with another -CH2, or perhaps two -CH groups.

That broad peak at 1.72 is characteristic of an alcohol peak. It is also indicative of a sample which, if not run almost neat, is very concentrated and in a dry solvent. Its integral is not quite correct, but probably because of some excess water in the sample. But it is broad, and in the correct region for an alcohol. You could confirm this by doing a D2O shake, and seeing this peak disappear.

And, at 3.8 you have a -CH which is coupled to more than one thing. You have reasonably proposed that it is adjacent to a -CH3 and -CH2- Its chemical shift suggests that it is near some electronegative group or deshielding group. Knowing you have an alcohol might help narrow the possibilities here. The chemical shift is not indicative of a ring on its own. An alkyl ring would still be to higher field.

So, let's summarise what you know:

CH3CH2-
CH3-CH(OH)-CH2-

I think there is perhaps only one way to connect these fragments....
The key to these types of questions is that once you propose a structure, you can then go back and rationalise what you see to test the structure. Sometimes, you might have to come up with several structures before you get the correct one. But it usually best to start with the simplest, most probable and start eliminating from that.

If you still have problems with understanding any of this, ask some more, and we can go from there.


Why is it lead(II) chromate and not lead(II) chromate(VI) tetraoxide?

Chromate ($\ce{CrO4^2-}$)is a poly atomic ion which is partially defined by the oxidation state of chromium.  Its like the sulfate ($\ce{SO4^2-}$), phosphate ($\ce{PO4^3-}$), nitrate ($\ce{NO3-}$), permanganate($\ce{MnO4-}$), and carbonate ($\ce{CO3-}$) polyatomic ions. For each ion the oxidation state of the center ion is defined therefore there is no reason to state it.  You could in principle call lead (ii) chromate lead(ii) chromium(vi) oxide or tetroxide, but this is less informative of the structure of the dissolved species.


How is Gibbs free energy related to stability?

The thermodynamic data refers to formation of the given compounds from the elements in their standard states. For instance for $ \ce{ ZnO(s)}$  it refers to the reaction
$ \ce{ Zn(s) +1/2O2(g) ->ZnO(s)} \tag{1}$
with each element or compound present at a pressure of 1 bar and (unless noted otherwise) a temperature of 298.15 K.
A more negative value of $\Delta G^\circ$ means greater product stability. This follows from the relationship
$$\Delta G^\circ = -RT\log K \tag{2}$$
where K is the equilibrium constant for the reaction at that temperature. An increased value of K implies an increased amount of products relative to reagents. By the relation in equation (2) a larger value of K also means a more negative value of $\Delta G^\circ$.
By definition a more stable compound decomposes less readily. Decomposition here refers to the reverse of reaction (1).
When comparing the tabulated oxides, a more stable oxide has a more negative free energy of formation. Therefore $ \ce{ZnO(s)}$ is the most stable of the oxides relative to oxygen and the reduced metal. Of the metals the mercury(II) oxide would require the least energy to reduce.
Note that for all of the reactions here
$$K = p^\circ/p_\ce{O2}$$
where $p^\circ=\pu{1 bar}$ (assuming the pressure is low). When starting from metal oxides, allowed to decompose until at equilibrium, a less stable oxide is at equilibrium with a higher partial pressure of oxygen (smaller K) due to more of the oxide decomposing. Therefore imagine you have an evacuated box containing the oxide of one metal and an amount of another metal in pure (reduced) form. If the equilibrium partial pressure $p_\ce{O2}(\text{eq})$ for the oxide is higher than $p_\ce{O2}(\text{eq})$ for the oxide of the second metal, then (if you wait long enough) all of the oxide will be reduced and the second metal oxidized until either the original oxide or pure metal is exhausted.
The enthalpies of formation refer to how much heat evolves when  1 mole of the oxide is formed from the reagents at constant T and p, with all compounds in their standard states. The more negative the value the more heat is released (the reaction is more exothermic). The oxidation of zinc is the most exothermic reaction (per mole of product formed) of those shown in the table.


What's the difference between the two equations for work?

Aside from a sign convention artifact, they're the same thing.

Pressure is force per unit area:

$$
 P = {F\over A}
$$

If a surface with a given area is moved through a displacement $\Delta x$, then the volume of displacement is the product of $A$ and $\Delta x$:

$$
\Delta V = A \Delta x
$$

So, the $P\Delta V$ work formula can be rewritten:

$$
w = -P\Delta V = -P\left(A \Delta x\right) =
  - \left({F \over A}\right)\left(A \Delta x\right) =
  - F \Delta x
$$

There are a lot of assumptions embedded in the above, and you have to be a lot more careful than I was about defining positive and negative quantities correctly, but on a high level that's all there is to it.


Why do electron donating groups stabilize free radicals?

Throughout this answer, I'm assuming that you mean carbon free radicals, as you tagged this with organic chemistry.

A free radical is actually more than just a molecule. It is somewhere between an ion and a stable molecule, but has no formal charge. However, it is highly reactive, as you mentioned. You've also identified the right reason for the need of electrons, it is the octet rule, at least at an elementary level. They behave as electrophiles, in order to complete their octet.

A general free radical of carbon has 7 valence electrons, (like the methyl, ethyl free radical etc.) and while other free radicals of carbon do exist (like carbene, benzyne, and a few I don't know about), most of these are hypo octet, i.e. they have fewer electrons than necessary to complete their octet.

Hence, they get stabilised by electron releasing groups, which "quench" the requirement of electrons for the free radicals. This can happen via induction, or by resonance,or even by hyperconjugation (which is basically the same as resonance).

I hope this answers your question.


Are polar bond stronger than non-polar bonds?

The argument is circular: electronegativity is a measure of the tendency of atoms to attract electrons. Therefore electrons are attracted more strongly to electronegative atoms. More attractive has no more meaning than "bound more strongly" and a measure of binding strength can be regarded as a mixture of ionization energy and electron affinity, that is, an electronegativity parameter as defined by Mulliken.
Pauling's original definition of electronegativity relied on the idea that the bond dissociation energy contains contributions from a QM interaction plus an additional purely electrostatic interaction (a classical Coulombic interaction not involving a QM description of chemical bonds). In homonuclear diatomic molecules the electrostatic term is not important because the atoms "tug" at the electrons with equal strength, and so only a QM term contributes to bonding. For heteronuclear diatomics one can estimate the magnitude of the contribution of the QM term as an average of the terms in the two homonuclear molecules. Subtracting this average from the dissociation energy for the heteronuclear diatomic then results in an estimate of an electrostatic contribution to bonding. The electrostatic contribution is argued to be due to the different electron attraction in the two atoms, which is encoded in Pauling's electronegativity parameter. This is summarized by the equation for the heat of formation postulated by Pauling (in $\pu{kJ/mol}$):
$$Q_f = 100 (\chi_A-\chi_B)^2$$
The additional electrostatic term enhances the binding energy:

The greater the separation of two elements on the electronegativity scale, the greater is the strength of the bond between them. The extra stability is the resonance energy between the normal covalent structure and the ionic structure.

The excerpt comes "from the horses mouth" (Pauling's General Chemistry textbook). If the explanation is not satisfying I recommend looking into more advanced theories.


What is the electronic configuration of Fe(II) ion?

When researching the explanation for this phenomenon, I found this:

  It is natural to question why one or two electrons are usually pushed into a higher energy orbital. The answer is because $\mathrm{3d}$ orbitals are more compact than $\mathrm{4s}$, and as a result any electrons entering $\mathrm{3d}$ orbitals will experience greater mutual repulsion. The slightly unsettling feature is that although the relevant $\mathrm{s}$ orbital can relieve such additional electron-electron repulsion, different atoms do not always make full use of this form of sheltering because the situation is more complicated than just described. One thing to consider is that nuclear charge increases as we move through the atoms, and there is a complicated set of interactions between the electrons and the nucleus as well as between the electrons themselves. This is what ultimately produces an electronic configuration and, contrary to what some educators may wish for, there is no simple qualitative rule of thumb that can cope with this complicated situation.

On a semi-related note, according to one source I found (the source sites many others as well), it is in fact the $\mathrm{3d}$ electrons that fill first, followed by the $\mathrm{4s}$ electrons. This more satisfactorily explains why the $\mathrm{s}$ electrons are first lost in the ionization of the $\mathrm{d}$-block transition metals.


What are Degenerate Rotational Energy Levels Physically?

I have two questions...

  Why does a molecule "gain potential energy" when it rotates? Does it
  want to stop rotating for some reason?

  What do degenerate energy levels correspond to physically in terms of
  the molecule's rotation?

Do you remember the Newton's law of motion? A body in motion will always remain in motion until and unless there is a force acting on it. So you can imagine, that once a force caused the molecule to rotate, the molecule will continue to rotate forever. As a crude analogy can you recall the time when the Earth stopped spinning on its axis? It is spinning for billions of years. In the same way you can think that the isolated molecule, once excited, will rotate for a pretty long time (on a spectroscopic time scale).

You can think of degenerate rotation as a a way of physical rotation of a molecule in such a way that if you calculate the rotational energy in each case, it turns out to be the same number.

Third point, as pointed out by orthocresol, rotational energy is kinetic energy not potential energy. So no need to invoke potential energy.

Sometimes, we should not actually try to visualize what is happening physically at very small molecular levels. This is where mathematics enters. Many results originate out of pure mathematical calculations which can be hard to imagine mentally. For instance, Bohr's got correct H2 spectrum results by assuming the electron is rotating around the proton. The question is it actually rotating like a planet around a proton? A short story is worth sharing which will help you in future. Feynman (Physics Nobel) used to think a lot about of photons as a student and discussed them with his father. When he finished his PhD in physics, he realized, he still could not explain what a photon is. Sorry I cannot recall the reference. But mathematics has no problem. Yet on the web we can "see" photons in pictures as balls, which are obviously pure but meaningless imaginations.


How to calculate mass of nanomaterial grown on a substrate

One possible option for thin film thickness measurement is by Energy Dispersive Spectroscopy (EDS) in a scanning electron microscope (SEM).  There exist commercial packages for this such as LayerProbe by Oxford and STRATAGem (I have no financial interest in Oxford or SAMx).  If your campus/research facility has an SEM with an EDS (many do), but not a commercial package, you can still work it out yourself.  If I were tackling this in my own lab, I would probably do something along the following lines:

Option #1:

Obtain a spectrum from a known thickness of carbon on your Cu substrate.  This can be any method such as sputtering or evaporation so long as the thickness is known by other means.  Most geology labs have the ability to do this kind of deposition since they need reproducible carbon thicknesses on their samples.

For example, geo labs will typically place a polished brass object next to the sample to be coated in an evaporative coater.  The change in the color of the brass corresponds to the thickness of the carbon: e.g. blue is 250 Å but bluish green is 300 Å (Kerrick, D. M., et al., 1973).  So something like 25 % accuracy is possible this way.

Then you measure that thickness in the SEM using an exact beam current, size, distance from the objective lens.  Then you measure your graphene in the SEM using the exact same conditions.  If your sample is less than about 200 nm thick, then the number of counts/second is proportional to the number of carbon atoms present in the layer.  Using the mass for your standard layer, (assume a density of graphite), then you can compute the mass of your graphene.

If your thickness is more than about 200 nm, then you may need to do an absorption correction – i.e. the photons are being absorbed on their way out and your count rate is no longer proportional to the number of atoms.  You can compute attenuation coefficients using the CXRO calculator (and a couple other places if you google).  Then you can do a thin film correction.  At 200 nm, your error from ignoring the thin film correction is probably about 5–10 %, which is likely a bit less than the error in the determination of the standard film using the color method.  So you can think with that for your sample.

Option #2:

You can measure the mass directly using an oscillating crystal.  Commercial systems comprise a quartz crystal with electronics set to find the resonant frequency.  You place the quartz crystal in your chamber alongside your Cu substrate.  As your reaction happens, it grows both onto your Cu and onto the quartz crystal.  The resonant frequency of the crystal changes as mass is added.  These are sold commercially, though a bit pricey.  For example, at TedPella (again, no financial interest).  The resolution is typically billed as 1 Å thickness resolution (assuming you know the density of your material.)

For your application, you would have to be sure that the graphite grows on your monitor crystal the same as on your Cu substrate.  That may require checking in an SEM, and it may require prepping the crystal somehow.  Hence, I offer this as suggestion #2.

Finally, you could also figure this out using a focused ion beam and sectioning the sample from the side.  If this is too complex, there also exist labs where one can hire them to do this kind of measurement.

I hope one of these solutions works for you!

Ref:

Kerrick, D. M., Eminhizer, L. B., & Villaume, J. F. (1973). The Role of Carbon Film Thickness in Electron Microprobe Analysis. American Mineralogist, 58, 920–925.


Calculating the ionic polarizability from the Clausius-Mossoti relation

Yep, you missed quite an important part: Clausius-Mossoti equation in the form you've written is limited  and holds true for the dielectric that consists of individual particles. However, the problem says you are dealing with a crystal, which, by the way, is most likely sodium chloride $\ce{NaCl}$, judging from its dielectric constant and refractive index.

In the first Clausius-Mossoti equation you've written $\alpha$ consists of two main components:

electronic polarizability $\alpha_\mathrm{e}$ caused by the displacement of the electron shell of the atom relative to the nucleus under the action of the field;
ionic polarizability $\alpha_\mathrm{i}$ caused by the displacement of ions with respect to other ions.

For the ionic crystal Clausius-Mossoti equation can be written as

$$\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} = \frac{N}{3\varepsilon_0}(\alpha_\mathrm{e^+} + \alpha_\mathrm{e^-} + \alpha_\mathrm{i})\, ,\label{eq:1}\tag{1}$$

where $\alpha_\mathrm{e^+}$ and $\alpha_\mathrm{e^-}$ are electron polarizabilities of positive and negative ions in crystal lattice. Their sum can be determined using a Lorentz–Lorenz equation, which is derived from the fact that the effect of an external electromagnetic field with relatively high frequencies corresponding to the visible and ultraviolet range of the spectrum results in the displacement of only the electron shells relative to atomic nuclei, while the more massive particles (atoms and ions) do not have time to shift from the occupied positions. Correspondingly, only electronic polarization contributes to the polarization of the medium (e.g. $\alpha_\mathrm{i}\approx0$), and the refractive index is related to the electron polarizability of the particles by the Lorentz-Lorentz formula:

$$\frac{n^2 - 1}{n^2 + 2} = \frac{N}{3\varepsilon_0}(\alpha_\mathrm{e^+} + \alpha_\mathrm{e^-})\label{eq:2}\tag{2}$$

Subtracting \eqref{eq:2} from \eqref{eq:1}, one eliminates the electronic constituent from the equation:

$$\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} - \frac{n^2 - 1}{n^2 + 2} = \frac{N}{3\varepsilon_0}\alpha_\mathrm{i}\tag{3}$$

and ionic polarizability is actually determined as follows:

$$
\begin{align}
\alpha_\mathrm{i} &= \frac{3\varepsilon_0}{N}\left(\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} - \frac{n^2 - 1}{n^2 + 2}\right)\\
&= \frac{3\cdot\pu{8.85e-12 F m-1}}{\pu{2.24e28 m-3}}\left(\frac{5.62 - 1}{5.62 + 2} - \frac{1.5^2 - 1}{1.5^2 + 2}\right)\\
&= \pu{3.7e-40 F m2}\tag{4}
\end{align}
$$


Molecular Orbital Theory and No. of bonds

Overall bond order equals number of electrons that occupy bonding orbitals minus numer of electrons that occupy antibonding orbitals divided by two.

Now you can apply the same concept indivudually:

For s molecular orbitals, both electron pairs "cancel out".

For p molecular orbitals, you have 2 electrons in sigma bonding orbitals and none in sigma antibinding orbitals, which means (2-0)/2 = 1 sigma bond.
Also, you have 4 electrons within pi bonding orbitals and only 2 electrons within pi antibonding orbitals, which means (4-2)/2 = 1 pi bond.
This yields the result that you expected.


Nucleophillic addition reaction in the presence of an acid catalyst on a molecule with an alkene

The acid will protonate the O of the carbonyl carbon, so a positive charge is generated. The oxygen of the alcohol with the free pair of electrons will act as nucleophile. So the alcohol is added to the cycle and a pair of electrons of the double bond of the carbonyl will compensate the positive charge generated. As the medium is acid, in the second step the hydroxyl group is protonated, so it is generated a water molecule, the alcohol attacks again at the same carbon and the water is released to the medium.

This is why this process is used to protect carbonyl groups. The reaction is reversible so if you add water, the reaction direction is reversed and you get the keto form again.


As a simple answer one might say the more nodes the higher the orbitals energy, i.e. the less stable it is. But you can have orbitals with the same number of nodes and different energies.

In general the is a node rule in quantum mechanics, which states that the ground state of a system has no nodes and higher excited states have more nodes.
Two common examples are the particle in a box and the harmonic oscillator. Both links contain the equations for the eigenfunctions and their energy as well as their plots.
A simplified example dealing with orbitals would be Hückel MO Theory.

Now for atomic orbitals the story is a bit more complicated. First of all, orbitals are not the electronic states of an atom or molecule, but something you can approximately decompose the electronic state into. Furthermore we have not 1 but 3 quantum numbers (5 if you include spin) defining an orbital: $n$, $l$ and $m$. The node rule for atomic orbitals is:

$n-l-1$ radial nodes
$l$ angular nodes:

$|m|$ nodal planes shaped like a cone around $z$-axis
$l-|m|$ planar nodal planes perpendicular to $xy$-plane

So we have $n-1$ nodes in total. The energy of an atomic orbital mainly depends on the quantum number $n$:
\begin{equation}
  E_n \propto -\frac{Z}{n^2}
\end{equation}
where $Z$ is the atomic charge. So higher $n$ means more nodes and higher (less negative) energy, which is in agreement with the general node rule for $n$. This also agrees with the quantum numbers $l$ and $m$, as they neither change the orbital energy (it only depends on $n$) nor the total number of nodes. (This is considering an atom with one electron only).

However, the orbital energies furthermore depend on the coulomb interaction with other electrons. This is what leads to $2p$ orbitals usually having slightly higher energy than $2s$ orbitals, but they still have the same number of total nodes, they are just differently distributed between radial and angular part of the orbital.

Also note, that you cannot directly compare orbitals of same quantum numbers between different elements, as the nuclear charge $Z$ goes into the orbital energy as well. For example the energy of the $1s$ orbital in the Hydrogen atom is $-0.5\ E_{\rm h}$. In a Carbon atom this would be something more like $-16\ E_{\rm h}$ for $1s$, while $2s$ with around $-1.2\ E_{\rm h}$ would be energetically much closer to the $1s$ Hydrogen orbital.


Dissociation degree hell, Ostwald's law of dilution

You are on the right track according to the directions given to you ( = the problem statement), but the directions themselves are wrong.

Say you had an equilibrium with $\ce{{[H^+][A^-]\over[HA]}=K}$. You dilute it five-fold and tell your particles to wait a bit with dissociation until you are done with the math. That makes all concentrations 5 times smaller, and consequently, the LHS is now $1\over5$ of what it used to be. Now you tell them to increase their dissociation by a factor of 4. In doing so, each of $\ce{[H+]}$ and $\ce{[A-]}$ increases four-fold, thus growing the LHS to $16\over5$ of its initial value, or more than 3 times greater than $K$, and they do that at the expense of the remaining $\ce{[HA]}$, decreasing the denominator and increasing the LHS further yet. Just how is it going to be the same $\ce{K}$ again?

I suppose it is some subtlety of wording that we are misinterpreting here. It might have been lost in translation, but can't we at least make a reasonable guess? Could it be that the 4-fold increase is compared to the initial state, that is, to $\alpha_0$ and not to $\alpha_1$? No, that would not suffice. Still, there might be other possibilities...


How to convert cell diagram to a chemical equation?

Assuming you meant $\ce{MnO^-4}$ and not $\ce{MnO^{4-}}$, your reduction half reaction is incorrect. Because according to your equation $\ce{Mn^{2+}}$ is being oxidised to $\ce{MnO^-4}$ rather than being reduced.
The correct half cell reactions would be:
$$
\ce{Zn -> Zn^{2+} + 2e-}\\
\ce{MnO^-4 + 8H+ + 5e- -> Mn^{2+} + 4H2O}\\
$$
And so the overall balanced reaction would come out to be:
$$\ce{5Zn + 2MnO^-4 + 16H+ -> 5Zn^{2+} + 2Mn^{2+} + 8H2O}$$


Well, I can think of these reasons that can justify the hybridization in ammonia molecule.

Bond angle: If the molecule did not have hybrid orbits, and instead had unhybridized p-orbitals taking part in the bond formation, then the bond angle between the orbitals would be 90 degrees. And as for the real situation the bond angle is nearly 107 which makes the molecule more stable, decreasing the bond pair-bond pair and bond pair-lone pair repulsion.
Energy: Recalling the definition of hybridization, it is the mixing of the atomic orbitals having slightly different energies to form new orbitals that have equal energies. This stabilizes the molecule. As for ammonia molecule due to hybridization the energies of the lone pair of electrons and the bond pair of electrons becomes almost equal, thereby increasing the stability of the molecule.
Geometry: Thinking about the spatial arrangement of the atoms in the molecule of ammonia if there is no hybridization in the molecule then the size of the orbital containing lone pair of electrons would be different from that of the orbitals containing bond pair of electrons. Also if hybridization would not be taking place, explaining the geometry of the molecule, that is it's trigonal pyramidal shape, would not be possible.


Automated Identification of Isomeric Relationship from 3D Structure?

You might become interested in OpenBabel.  Written in C++, it offers bindings to Python (ref).  It looks like stereochemistry is covered more in detail (here, here) than the obchiral command (here).

OpenEye's Chemistry Toolkit, on the other hand, equally capable to read numerous chemical file types (ref), offers several approaches to detect stereochemistry (examples).  Written in Python, it may offer an alternative starting point.  Perhaps you qualify for one of their academic licenses (ref).


General notation for one of the d-orbitals

Generally you are free in your choice of coordinate system to describe a certain problem and calling the d-orbital with $l=2$, $m_l=0$  either $\mathrm{d}_{3z^2-r^2}$, $\mathrm{d}_{3x^2-r^2}$ or $\mathrm{d}_{3y^2-r^2}$ depends on this choice but by convention the $z$-direction is usually chosen to be the system's preferred direction, i.e. it points along the main symmetry of the system.

Let's move to $d$ orbitals in crystals: Usually, metal $d$ orbitals are not "properties" of the crystal itself but "belong" to the local metal centers. Thus, the $z$-direction along which a $\mathrm{d}_{3z^2}$ orbital points refers to the preferred direction of the metal ion in its local environment, e.g. in a Jahn-Teller distorted Perowskite $\ce{ABO3}$ the $\mathrm{d}_{3z^2}$ orbital of metal ion $\ce{B}$ always points along the direction of the elongated/compressed $\ce{B-O}$ bonds within the local $\ce{BO6}$ octahedron or quadratic bipyramid. So, even though in $\ce{LaMnO3}$ you have differently tilted $\ce{MnO3}$ octahedra that have different preferred directions and thus differently oriented $\mathrm{d}_{3z^2}$ orbitals, you still refer to all of them as "$\mathrm{d}_{3z^2}$" and everyone you talk to about this will know which orbitals you mean and where they will point in the different local octahedra in the crystal. That is pretty much the status quo.

 $\hspace{2em}$

Now, what you want to do, is talk about the $\mathrm{d}$ orbitals not in the context of their local coordinate system but in the context of a more global coordinate system related to the whole crystal, e.g. declare a certain crystal plane to be the $x$,$y$-plane and name the $\mathrm{d}$ orbitals according to this choice to let their name reflect their orientation within the plane.

Of course, you can do this and it has been done here pretty much as you suggested  where the authors chose notations like "$\mathrm{d}_{3\sigma^2 - r^{2}}$ orbital ($\sigma = x, y , z$)" or "$\mathrm{d}_{3(x, y)^2 - r^{2}}$ orbitals". But be aware that this is a kind of notation that needs to be explained beforehand when you use it because it refers to an unconventional choice of coordinate system as a frame of reference for the $\mathrm{d}$ orbital labels.

TL;DR: Yes, you can use a general notation like you suggested but make sure to explain which coordinate system you refer to because otherwise you might cause confusion.


How does one test for lead in dry paint?

This site is not specialized and reserved to professional chemists. On the contrary. A lot of teenagers are using it every day.
Now if you want to identify lead in a dry paint, you may burn it in the flame of a Bunsen burner, and save the ashes. They contain a mixture $\ce{Pb + PbO + Pb3O4}$. Then dissolve the ashes in nitric acid (concentrated $\ce{HNO3}$ diluted with its own volume of water). Filtrate the insoluble part. Dilute the filtrate still more, by adding an equal volume of water. This gives you a lead nitrate solution, with which you carry out the following tests. Add some $\pu{1 M}$ $\ce{HCl}$ to a part of the obtained solution : it should produce a white precipitate of $\ce{PbCl2}$, if the solution contains some lead ions. Or add some $\ce{KI}$ solution : You would obtain a yellow precipitate of $\ce{PbI2}$, if the solution contains lead. The reactions are :
$$\ce{Pb(NO3)2 + 2 HCl -> PbCl2(s) + 2 HNO3}$$
$$\ce{Pb(NO3)2 + 2 KI -> PbI2(s) + 2 KNO3}$$
A further proof is obtained by heating the solution containing the yellow deposit of $\ce{PbI2}$, because this substance gets redissolved at $\pu{100°C}$. And after slow cooling, $\ce{PbI2}$ will precipitate back and it looks like wonderful golden plates, floating in the liquid. It is amazing. If these precipitate do not occur, there is no lead in your paint.


First off, let's think of the light as photons, instead of waves. That makes it clear that yes, there are two photon and multi-photon processes. As you surmise, the selection rules for these are much more complicated than a single-photon process.

Your main question is about "how exact" an absorption needs to be. In reality, it depends on the system.

Atomic Absorption Spectra

For atoms, there are no vibrational or rotational transitions, only electronic (or perhaps nuclear). So atomic spectra are indeed very tight lines, e.g., the Rydberg formula.

Diatomic Molecules

Now you have one vibration and some rotational levels. The latter are important, since you can go up or down some rotational levels with a vibrational transition. (And if you have an electronic transition, you can usually go up vibrational and/or rotational levels.)

So the spectra is a bit more complex than single lines. For example you can do rotational-vibrational spectroscopy.

In this case, the individual transitions are still pretty sharp - not completely lines, but $0.3\:\mathrm{cm^{-1}}$ probably won't match any particular transition.

Larger Molecules

Now you have very small spacing in the rotational lines, so if you take a vibrational absorption spectra (e.g., IR spectra) you'll see that most molecules have some width to the absorption peaks, so $\pm 0.3\:\mathrm{cm^{-1}}$ will still fit within the peak, just not as strongly.


This answer to this question applies to your query. Many physical properties depend on the intermolecular forces experienced in the liquid or solid. These intermolecular forces are hydrogen bonding (strongest), dipole-dipole attractions (in polar molecules), and London dispersion forces (weakest of the forces when comparing molecules of the same size, but dependent on the number of electrons in the substance). Stronger intermolecular forces would make the substance less volatile.

As noted in the first response, methanol is more volatile than ethanol. Methanol and ethanol would both have both hydrogen bonding (a relatively strong type of dipole-dipole attraction) and London dispersion forces. Because ethanol has more electrons (because it is a bigger molecule, but not necessarily because it is a heavier molecule) it would have more London dispersion forces in comparison to methanol, so, with stronger intermolecular forces, it would be less volatile. So, this example follows the general guidelines--figure out the intermolecular forces present in the molecules you want to compare and the molecule with the weakest cumulative intermolecular forces will be the most volatile.

Intramolecular bonding (covalent bonds, when you're talking about molecules) does not affect physical properties.


Software for visualization of 3D models of chemical structures with atomic orbitals

As porphyrin mentions in the comments, have a look at Jmol.

It indeed does the rendering as in your original version, but with the many options (see this mo command documentation) and particularly interactivity provided by the scripting, I think you may pedagogically achieve what you are looking for. Proteopedia is a nice website where the efficiently use Jmol's scripting feature for educational purposes (try the green links).


Pure Hydrogen and Dry Hydrogen Difference

It depends what you (or your field) is considering what an impurity is for a given substance. For example, if you want really pure methane, ethane will became an impurity. However, if you want gaseous alkane(s), there is no real difference between methane and ethane, so ethane is no longer an impurity.

On the other side, moisture can be your enemy, so pure methane (without ethane) can be considered bad if it contains 3% water whereas a mixture of methane and ethane (50/50) could be considered better because it contains only 0.1% water.

In your case, "pure hydrogen" would mean that "it is composed almost entirely of hydrogen gas" (but it can contain water) whereas "dry hydrogen" would rather mean that the gas does not contain water (but it can contain other molecules as well, but not too much)


How does polarimetry account for the orientation of the molecules?

Short answer: The light beam interacts with all of the randomly aligned molecules in its path.  There are roughly $\ce{6 x 10^{23}}$ molecules in a mole of material. If we were to dissolve a micro mole of this compound in 10 ml (1 deciliter is a typical polarimetry sample size) of solvent we would have $\ce{6 x 10^{17}}$ molecules in our sample cell.  Since the molecules are randomly oriented, there are enough molecules so that the polarized light beam will sample all possible orientations of the molecule with respect to the light beam many, many times over.  This allows for measurements to be made in an accurate and statistically reproducible manner.

Detailed answer: A beam of light has associated with it an electric and magnetic field.  In simple polarimetry experiments it is the electric field (or vector) associated with the light beam that is of interest.  Let's say the beam is propagating in the Z direction, then the propagating electric vector lies in an infinite number of planes that contain the Z axis.  When we plane-polarize the electric vector associated with the light beam, we remove all of the planes that the electric vector resides in except for one.

The next important point is to realize that a plane polarized electric vector associated with a beam of light can also be represented by (is equivalent to) two circularly polarized beams (or vectors), rotating in opposite directions (clockwise and counter-clockwise), each one-half the intensity of the plane polarized electric vector.  What we wind up with is two helices, one right-handed and one left-handed, traveling through the sample in our polarimeter.

Let's say our polarimeter sample contains a significant concentration of chiral molecules in solution.  Although the chiral molecules are randomly aligned with respect to the light beam, each chiral molecule will interact with the right-handed or clockwise rotating helical electric vector differently than it will interact with the left-handed or counter-clockwise rotating helical electric vector.  Our sample must contain a large number of molecules so that, despite all of the different molecular alignments between our molecules and the beam, when averaged over the entire sample, the difference in interaction of the right- and left-handed helices with our chiral sample will be repeatable.  It is not hard to have a significant concentration of molecules because a reasonably long sample path length is used.

The unequal (dissymmetric) interaction between our chiral sample and the right- and left-handed helices results in a rotation of the plane of polarization of the emerging light beam from its original position - and this angle of rotation is what the polarimeter measures.


Will mixing methyl 2-cyanoacrylate and cellulose make the cellulose harden quickly?

The working principle of super glue, based on 2-cyano acrylate, is an anionic polymerisation reaction which in principle triggered by any nucleophile.  Water conveniently may be such a nucleophile (the basic anion of water, hydroxyde $\ce{OH^-}$ were even better) which is why the container's mouth originally is not only closed by a cap, but by a thin metal plate.

Cotton consists mainly of cellulose, and cellulose itself

(source)

posesses many hydroxyl groups which like water may trigger the polymerisation.  Since there are so many that are easily accessible, the process is very fast, and proceeds under liberation of much heat (i.e., an exothermic reaction).  Because of this and because the acrylate may generate a strong 3D polymer, one should not wear cloths made of cotton-based fabric, nor gloves of such.

While it is new for me to learn that ionic liquids are able to dissolve cellulose, it might be that such an unmoderated polymerisation "runs away"; i.e. already proceeds at a high rate and, because of the liberated heat and high local concentration of the nucleophiles and monomers, will become even faster, eventually is no longer manageable.  Beside potential splatter of liquid-as-such (thermal threat), the polymerisation reactions could proceed further even on and with the skin of an operator (chemical threat) if not protected properly.  Be warned, as each of the two are potentially dangerous, in addition to the potential hazard that working with ionic liquids, and chemicals in general, represent.

Wikipedia hints to this potentially relevant Critical Review doi 10.1039/C2CS15311D in Chem. Soc. Rev., 2012, 41, 1519-1537.


Rate of reaction of silver salts of carboxylic acids having different alkyl groups towards Hunsdiecker Reaction

If the free radical is less stable, it will react faster.
So once the system is started, the propagation (step 4) will proceed faster for primary radicals.
That means that it is the limiting step and that step 3 is faster than 4 for any type of R.


Why does the -OH group in β-naphthol direct the incoming diazonium salt towards the ortho position?

It's sort of a process of elimination.  The electrophilic ion could potentially attack anywhere, but:

The hydroxyl group is activating, so the electrophile will prefer that ring.  You are left with three open positions.
One of those three is meta to the hydroxyl group and so less favored, as activating substituents are generally ortho/para directing.  Hydroxyl follows that rule.
Finally -- electrophilic substitution in a naphthalene system is kinetically favored next to the other ring.  See here for example.

We have one position that's on the same ring as hydroxyl, ortho or para to that hydroxyl group, and next to the other ring.  That is the position shown in the book.


Are radioactive isotopes of silver dangerous?

A glance through the table of the isotopes in the venerable CRC handbook reveals that the longest lived radioactive isotope of Ag is $Ag^{108m}$, made by neutron capture by $Ag^{107}$.  this has a listed half life of "> 5y", but the capture cross section is only $35\pm5$ barns, pretty small for thermal neutrons.  So, it would be really hard to get a lot of radioactive silver.

Next, all isotopes listed decay by $\beta, \beta^{+}$, EC, or IT (electron capture of internal transitions for the metastable nuclei), with decay energies ranging from 93 keV ($Ag^{107m} \rightarrow Ag^{107}$) to 6MeV ($Ag^{116}$).  Those are ionizing radiation, which would increase your probability of getting cancer.

The 'nasty radiation poisoning' things are actually nasty primarily because of being heavy metals, the radiation comes along for the ride, so to speak.  The normal Uranium isotopes are very long lived, so emit very little radiation per unit time.  Polonium, the scary poison of choice in stories and perhaps by the KGB, in contrast, has a short lifetime (138 days for $Po^{210}$), and spews lots of ~5MeV $\alpha$ particles, heavily damaging tissue (much more so than gammas).

Given the difficulty in making the stuff, assuming you would try $Ag^{108m}$, (small cross sections, can't chemically separate from the stable $Ag^{107}$, etc., your best bet would just be acute heavy metal poisoning from Ag salts (which may or may not be highly toxic). Or, using a silver candlestick to bludgeon the poor victim to death.


I'd like to model some compounds with DFTB, but some elements don't have parameter files in the official site. Is there a workaround?

I recommend you try out xtb by the Grimme group. It is a state-of-the-art Tight Binding program parameterized for elements up to Radon (already built-in). There is a small caveat: the only precompiled binary they distribute is for Linux (but the program is open source, so you can compile it yourself if you want.)

And yes, it is really fast.


In meta hydroxy benzaldehyde which position (w.r.t hydroxy group) is more prone to electrophilic attack?

This is known as the ortho effect. According to it if an ortho/para and meta directing group are meta to each other, then the incoming electrophile prefers to come on ortho of meta directing group.

See the reference.