Is there a danger in electrolyzing NaCl?

If the cell is mixed, you would make bleach, sodium hypochlorite and some free chlorine (and probably chew up your anode). Since the solution is not very basic, hypochlorite would probably disproportionate to chlorite, and that possibly to chlorate.

One mole of electrons is one faraday, about 96 500 coulombs. One ampere is one coulomb per second. Five minutes passes $(5\ \mathrm{min})(60\ \mathrm{s/min})(10\ \mathrm{C/s}) = 3000\ \mathrm{C}$ or $0.031\ \mathrm{F}$ in a reaction that requires two moles input for each mole of chlorine. $0.01554\ \mathrm{mol}$ chlorine or $350\ \mathrm{ml}$ at STP, tops. It wouldn’t be good for you.


Is oxygen above the critical point always supercritical fluid? Would it still appear to roughly follow the ideal gas law?

Yes.  Any fluid with a temperature is above critical temperature and the pressure above the critical pressure is by defintion a supercritical fluid.  Don't be mislead by all the claims that supercritical fluids are special and wonky with all sorts of amazing, bizarre properties.  This is true of some supercritical fluids near the critical point, but the farther you go away from the critical point, the more like a regular gas supercritical fluids become.  (The only exception here is that at gargantuan extremes of P and/or T, eventually fluids will become plasmas, metallic solids, black holes, etc.  But as long as it is a fluid of the same substance [no chemical reactions], then it will be supercritical.)
Yes, supercritical fluids can often be modeled by the ideal gas law.  A good look into why would take into the theory of corresponding states.  This theory says that most gases will most often have similar compressibilities $Z = \frac{PV}{nRT}$ when they are in "corresponding" states, which means that at similar temperature and pressure relative to each gas's critical point.  There are charts that you can use (approximately) for many, many real gases.  Things to realize:

The compressibility factor $Z$ is a measure of non-ideality.  Ideal gases always have a compressibility factor of 1.0.
You can use the chart to find approximate compressibility factors for many, many real gases if you know that gas's critical temperature $T_c$ and critical pressure $P_c$.  The "reduced temperature" is just $\frac{T}{T_c}$ and likewise for pressure.
If you check out the generalized compressibility factor chart, look at the dependence on reduced temperature!  At a reduced temperature of 2.0, the line is nearly flat at $Z = 1.0$.  This is an indication that the gas behaves nearly ideally.  It takes pretty close approaches of $\frac{T}{T_c}$ to 1.0 to get strong non-ideality.
Oxygen's critical temperature of 154 K means that a reduced temperature of 2.0 just requires 309 K, about ambient temperature.  At these temperatures, oxygen will be fairly ideal, regardless of temperature, up to reduced pressures of 5 or 8, which correspond to 250 to 400 bar or so.


Explaining the Relative Energies of Various Vibrational Modes

I don't know how to explain the bending mode, but there is a nice mathematical explanation for the difference in stretching modes.

Take $\ce{CO2}$, carbon dioxide, as an example. I'll use B3LYP/6-31G(d,p) values. $\nu_{1}$, the symmetric stretch, is 1372 cm$^{-1}$, and $\nu_{3}$, the asymmetric stretch, is at 2436 cm$^{-1}$. Now consider $\ce{CO}$, a single carbonyl group (the molecule won't work for this example). Looking at an IR table, the average stretch will be about 1750 cm$^{-1}$. $\ce{CO2}$ looks a lot like two $\ce{CO}$ bonds that are linear and share a carbon. But, clearly $\ce{CO2}$ doesn't have a doubly-degenerate stretch mode of any kind at that frequency, so something is causing the local modes to not be identical to the normal modes. Because both local modes have the same symmetry, they are allowed to mix with some coupling between them, just like molecular orbitals. A local mode is a vibration that is completely uncoupled from all other modes; its motion does not influence the motion of other atoms and vice versa.

Set up a Hamiltonian for a degenerate two-level system with some coupling between them. Keeping with a Huckel-like picture, I'll use $\alpha$ for the energy level of each degenerate local mode and $\beta$ for the coupling between them.

$$
\hat{H} = \begin{pmatrix}
\alpha & \beta \\
\beta & \alpha
\end{pmatrix}
$$

Obtaining the normal modes and their energies is equivalent to diagonalizing this Hamiltonian, where the eigenvalues are the normal mode energies and the eigenvectors are the normal modes themselves as linear combinations of the local modes. Doing so gives the normal mode energies:

$$
\alpha - \beta, \, \alpha + \beta
$$

and the normal modes themselves:

$$
\begin{pmatrix}
-1 \\
1
\end{pmatrix},\,
\begin{pmatrix}
1 \\
1
\end{pmatrix}
$$

By convention in Huckel theory, $\beta$ is usually negative, so the 2nd eigenvalue is the one lower in energy.

If you take the dot product between each normal mode eigenvector and the vector of local modes:

$$
\begin{pmatrix}
\phi_{1} \\
\phi_{2}
\end{pmatrix}
$$

out falls the symmetric and antisymmetric combinations leading to the symmetric and asymmetric stretches. Leaving out the normalization factor of $1/\sqrt{2}$:

$$
(-\phi_{1} + \phi_{2}), \, (\phi_{1} + \phi_{2})
$$

Setting $\alpha = 1750$ cm$^{-1}$ and $\beta = -500$ cm$^{-1}$ gives a symmetric stretching frequency of 1250 cm$^{-1}$ and an asymmetric stretching frequency of 2250 cm$^{-1}$. Not bad for a model with a single parameter.

This is how the energy levels (eigenvalues) split with increasing the coupling between the initially degenerate local modes into distinct normal modes:

Here is a link to the Jupyter Notebook containing the math I did. This is a simplified (math-wise) version of what we did in a paper, since the conceptual point is identical.

This sort of description is how any two-level time-independent system works. See how the two spin energy levels of a free electron change in energy when an external magnetic field is applied (Zeeman effect):


The original solution contains the square planar $\ce{Pt(II)}$ complex ion $\ce{[Pt(NH3)4]^2+}$. When sodium iodide is added to the solution, one or more $\ce{NH3}$ molecules are replaced by $\ce{I-}$ ions in a ligand exchange reaction. In particular, replacing two $\ce{NH3}$ molecules with $\ce{I-}$ ions yields the neutral complex $\ce{[PtI2(NH3)2]}$, which is poorly soluble in water, and therefore a precipitate is formed.

Remarkably, two isomers of square planar $\ce{[PtI2(NH3)2]}$ exist: $\ce{$cis${-}[PtI2(NH3)2]}$ and $\ce{$trans${-}[PtI2(NH3)2]}$.
When starting from $\ce{[Pt(NH3)4]^2+}$, the first $\ce{I-}$ can replace any of the four equivalent $\ce{NH3}$ ligands. Because of the larger trans effect of $\ce{I-}$ compared to $\ce{NH3}$, the second $\ce{NH3}$ is preferably replaced trans to the first $\ce{I-}$ ligand and therefore the trans product $\ce{$trans${-}[PtI2(NH3)2]}$ is obtained.
(Note that, if one starts from $\ce{[PtI4]^2- }$ and replaces two $\ce{I-}$ ligands with $\ce{NH3}$, the cis product $\ce{$cis${-}[PtI2(NH3)2]}$ is obtained instead.)


What's the solid product of the reaction of liquid ammonia with excess sulfur?

I don't think this question has a simple answer.

This paper's first page (all I can see without a subscription) indicates that sulfur can dissolve in liquid ammonia, producing hydrogen sulfide (as the question correctly states) and... complicated stuff: S7N-, which disproportionates into S4N- and "an" ammonium polysulfide (maybe (NH4)2S6, maybe another composition).

The paper's first page also refers to the system's light sensitivity, so if you don't keep it in the dark, you may get different products.


Which proton is more acidic in this compound?

You could start by drawing the resonance structure of the naphthalene ring, this gives a compound in which the conditions on the two nitrogen atoms are interchanged, thus both nitrogen atoms are equivalent. As $\ce{N-H}$ bonded protons are more acidic than $\ce{C=C-H}$ protons and $\ce{N-C-H}$ protons, the answer must be the hydrogen atom bonded to the nitrogen atom, as both nitrogen atoms are equivalent, either of the hydrogen atoms is a valid answer (as they are equivalent).


Do individual crystals resulting from evaporating a mixture of two ionic compounds contain both cations/anions within the crystal lattice?

Crystals with randomly intermixed ions only occur when the ions have very similar diameter (really the same spatial charge distribution in its exterior, but alkali ions are in very good approximation spherical) and same charge. That would be called a single phase solid solution.

However, $\ce{Na+}$ and $\ce{K+}$ have quite different diameters ($r=102$ resp. $\pu{138 pm}$). Your mixed solution will crystallise as NaCl with a low amount of potassium ions and KCl with some potassium replaced by sodium ions. In other words, your isotropic solution decomposes into a solid two-phase mixture.

The two substances will surely have a different tendency (also temperature dependent) to include such defect ions, and the two compounds have different solubility, so one will precipitate first, so the concentrations in the residual solution changes greatly during the process, and consequently also the defect concentration in the single crystals.


Is it wrong to write a coefficient of 1 in a chemical equation?

It is not wrong, per se, but it goes against conventions.
These rules are in place to simplify how we write formulas and equations, etc.
If we did write the $1$ all the time, then perhaps we would write it for a chemical formula, and water would be $\ce{H2O1}$, for instance.
If you want to write it as a reminder and placeholder that is totally fine.
A teacher might say that they have a rule, or that they want you to follow standard convention. You should do this if requested. When I taught, I would offer to students that it is OK if they want to do it to get in good habit of always writing coefficient.
All that being said, IUPAC guidelines say to write the species with no coefficient if the coefficient is one. So, technically it is not to the standard, but does not make it less clear, so use it in your work, but if you are submitting as classwork follow the protocol of the class.
Below is cited a brief note to this from the The ACS Style Guide[1].

➤ Specify the number of each species (molecules, atoms, ions, etc.) of reactants and products by a numeral written on the line and closed up to the symbol.
$$\ce{2Al + 6NaOH -> 2Na3AlO3 + 3H2 }$$
➤ To indicate the aqueous, solid, liquid, or gas state, use the appropriate abbreviations on the line, in parentheses, and with no space preceding them.
$$\ce{Ag(s) + H+(aq) + Cl–(aq) -> AgCl(s) + ½H2(g)}\\
  \ce{4FeS(s) + 7O2(g) -> 2Fe2O3(s) + 4SO2(g)}$$

Within these few lines, it is stated to write a numeral, but then it is omitted when it is $1$.
House rules prevail, as this is commonly accepted practice.
Reference:
[1]: The ACS Style Guide: Effective Communication of Scientific Information; Coghill, A. M., Garson, L. R., Eds.; Oxford University Press: New York, NY, 2006(pdf).


Electrophilic Aromatic Substitution: Are protonated OH-groups ortho-/para- or meta-directing?

O-protonated phenols direct in meta, according to an article on that topic. As it occurs, phenols can be, depending on acid, protonated on oxygen, but also on carbon, and even diprotonated (but not twice on the same atom), when most powerful mixtures of HF and SbF5 are used. O-protonated forms can be brominated, hydroxylated and alkilated. C-protonated forms don't posses such reactivity and are essentially Meisenheimer's complexes.


Significant figures when standard deviation is 0

Various sources of error are inherent in measurements. The so-called Type A evaluation, i.e. the statistical analysis of series of observations (e.g. calculation of the standard deviation and the standard error of the mean), can be used to evaluate the uncertainty arising from random effects.

However, this evaluation does not include the uncertainty arising from systematic effects. For example, if the concentration of the titrant (the solution of a known concentration that is added during the titration) has a small unknown error and the same titrant is used for all titrations, each of the individual results includes the same systematic error. Or, if the burette has a small unknown error and the same burette is used for all titrations, again each of the individual results includes the same systematic error. You can neither eliminate such errors nor reduce the caused uncertainty by making additional trials, and you cannot evaluate the caused uncertainty by using statistical analysis (i.e. by calculating the standard deviation and the standard error of the mean) of the results. The uncertainty arising from such systematic effects has to be evaluated by other means. The so-called Type B evaluation is usually based on scientific judgment using all of the relevant information available on the possible variability of the results. Finally, you can combine all calculated or assumed uncertainties by using the rules for error propagation.

For example, the uncertainty arising from the unknown error of the burette may be estimated using the manufacturer’s specifications as follows:
Usually, you estimate (interpolate) the measurement between the graduation divisions of the burette. You might be able to estimate the measurement to one tenth of the smallest division. However, volumetric instruments, such as graduated cylinders, volumetric flasks, bulb pipettes, graduated pipettes, and burettes, – like any other measuring instrument – have a limited accuracy. The dimensions, graduating divisions, and accuracy limits of volumetric instruments are standardized. For example, the requirements of burettes are specified in the international standard ISO 385. Two classes of accuracy are specified: Class A or AS for the higher grade and Class B for the lower grade. The nominal capacity, graduation divisions, and accuracy limits are shown in the following tables.

$$
\textbf{Burettes, Classes A and AS} \\
\begin{array}{lll}
\hline
\text{Nominal capacity} & \text{Graduation divisions} & \text{Accuracy limits} \\
\mathrm{ml} & \mathrm{ml} & \pm\ \mathrm{ml} \\
\hline
\hphantom{00}1 & 0.01 & 0.006 \\
\hphantom{00}2 & 0.01 & 0.01 \\
\hphantom{00}5 & 0.01 & 0.01 \\
\hphantom{00}5 & 0.02 & 0.01 \\
\hphantom{0}10 & 0.02 & 0.02 \\
\hphantom{0}10 & 0.05 & 0.03 \\
\hphantom{0}25 & 0.05 & 0.03 \\
\hphantom{0}25 & 0.10 & 0.05 \\
\hphantom{0}50 & 0.10 & 0.05 \\
100 & 0.20 & 0.10 \\
\hline
\end{array}$$

$$
\textbf{Burettes, Class B}\\
\begin{array}{lll}
\hline
\text{Nominal capacity} & \text{Graduation divisions} & \text{Accuracy limits} \\
\mathrm{ml} & \mathrm{ml} & \pm\ \mathrm{ml} \\
\hline
\hphantom{00}1 & 0.01 & 0.01 \\
\hphantom{00}2 & 0.01 & 0.02 \\
\hphantom{00}5 & 0.01 & 0.02 \\
\hphantom{00}5 & 0.02 & 0.02 \\
\hphantom{0}10 & 0.02 & 0.05 \\
\hphantom{0}10 & 0.05 & 0.05 \\
\hphantom{0}25 & 0.05 & 0.05 \\
\hphantom{0}25 & 0.10 & 0.10 \\
\hphantom{0}50 & 0.10 & 0.10 \\
100 & 0.20 & 0.20 \\
\hline
\end{array}$$

For example, a burette of Class AS with a nominal capacity of 50 ml has graduation divisions of 0.10 ml. By interpolation, you might be able to estimate the measurement to one tenth of the division, i.e. to 0.01 ml. However, the accuracy limits of the burette are ±0.05 ml (at any point on the scale).


My memory tells me that KOH is more soluble in ethanol than NaOH. The chemistry is the same, just more efficient at the higher concentration.


Major product of radical fluorination vs bromination

There is nearly no thermodynamic aspect to this reaction. The activation energy for any alkane hydrogen abstraction by a fluorine radical is essentially 0$^{[1]}$. This means that there is an equally likely chance of fluorinating at any given position. In this case, we need only know the proportions of homotopic hydrogen. Homotopic hydrogen are labelled with the same colors in the molecule:

The proportion in decreasing order is 9:6:1:1:1. That means that we should expect about the following yield of monofluorinated products:

As an aside, you would never realistically expect to obtain a monofluorinated product. The fluorination of alkanes is violent, extremely fast, and only gets faster upon fluorination.

$^{[1]}$ Chemistry StackExchange, Why are free radicals so reactive?


Chemical kinetics of a reaction, rate constant, and activation energy

The rate constant and the activation energy tell us how fast the reaction proceeds (moles/sec), not whether it is spontaneous. The higher the temperature, the higher the rate constant and, for specified concentrations of reactants, the faster the reaction. The rate of a reaction always increases with increasing temperature (higher rate of collisions), so the activation energy is always positive.  As the temperature increases, the rate constant approaches the pre-exponential factor (E/T approaches zero). For some reactions, the activation energy is very low, so even at lower temperatures, the rate constant approaches the pre-exponential factor. The activation energy has a weak dependence on temperature, and, in practice, this temperature dependence is usually  neglected.


Why opposite of electrophilic is not Protophilic

Because not all "nucleophilic" agents react equally well with all types of relatively exposed nuclei.  Phosphines are very unwilling to extract a proton but quite nucleophilic if the available electronic vacancy is in carbon or a transition metal.


CIP Rules - How to deal with a double bonded nitrogen?

I have taken your functionality and I have incorporated it into structure 1 that has stereochemistry. The two double bonds in 1 are shown in the CIP shorthand of structure 2. C3 and C4 are non-duplicate atoms and they are attached to duplicate carbon atoms (4) and (3), respectively. Similarly, in the C=N double bond C1 is attached to the duplicate red nitrogen (N) while the non-duplicate red nitrogen is attached to duplicate carbon (1). Both non-duplicate and duplicate atoms of the same element have the same atomic number, however, duplicate atoms are attached to three phantom atoms of atomic number zero. In addition, all atoms (C,N,O,Cl,etc.) are treated as though they are quadrivalent.
In the digraph the blue numbers are the successive spheres emanating from the stereogenic center C2. In sphere 1 hydrogen has the lowest priority while C5 has the next lowest priority owing to the 3 H's in sphere 2. C1 is attached to N,(N),N in sphere 2, having the highest priority, while C3 is connected to 4,(4),H in sphere 2. The overall priorities are C1>C3>C5>H.
Now to the gist of your question. While N=N=(N), in sphere 3, (N) is attached to 0,0,0. (N) has the lowest priority of the three nitrogens. Of the remaining two nitrogens, C,C,0 = C,(1),0 in sphere 3 but this tie is broken in sphere 4 where H,H,H/H,H,H has priority over H,H,H/0,0,0. The nitrogen priorities are black N>red N>(N).


ClO2 chlorine dioxide how to know if it is still active or not ? by color ? by smell ? by reaction?

If you start with $\pu{100 mL}$ stock solution at $\pu{100 ppm}$ concentration, and dilute this by a factor of 10 to yield $\pu{1 L}$, then you are correct, the later one has a concentration of $\pu{10 ppm}$.  Keep both in a cool place.
As to check if the solution still is active, or not, and both stock and diluted solution at the expected concentration (which will decrease already if the solution is exposed to hot summer temperatures), you may use test strips you immerse dip into the solution, dry at room temperature, and compare against a scale.  An example were the following

(source)
Depending on the manufacturer, the scale of concentration to test may vary.


Why are only sigma bonds used in determining the geometry and bond angle in VSEPR theory? Why aren't pi bonds used?

VSEPR is used to predict highly idealized geometries corresponding to Platonic solids or combinations thereof (e.g., the trigonal bipyramidal geometry being essentially the result of fusing two faces of a tetrahedron). The basic rationale behind its use, and the reason for its success in many situations, is the recognition that electrons, being like-charged, repel one another electrostatically. Hence, as a first approximation, it's reasonable to assume that the most stable spatial configuration of any molecule would have the atoms spaced as uniformly and as distant from one another as possible. VSEPR achieves this by placing the central atom of a simple molecule (or some molecular sub-structure) at the center of a geometrically perfect polyhedron having as many vertices as non-central atoms and lone pairs, and then placing the non-central atoms and lone pairs at those vertices.

The reason that $\sigma$ bonds (and lone pairs) determine the geometry is that they form the basic skeleton of the molecule. Recall that $\sigma$ bonds are formed by head-on overlap of atomic orbitals, meaning that they are oriented along the imaginary axis connecting two atomic nuclei, and hence concentrate electron density in the region directly between the two nuclei. $\pi$ bonds, on the other hand, are essentially orthogonal to the $\sigma$ bond skeleton, and are substantially weaker. Moreover, $\pi$ bonds do not exist in isolation, meaning any $\pi$ bond between two given atoms is always formed secondarily to the $\sigma$ bond between said atoms. As such, $\pi$ bonds do not alter the basic idealized geometry of a molecule as dictated by $\sigma$ bonding, although in practice, because they do actually introduce additional electron density and require closer orbital overlap, bond multiplicity affects bond length and bond angles (as does, for that matter, the size of the atoms involved). The magnitude of those deviations from the idealized VSEPR geometry could, in principle, be quite large, although most often those deviations are slight enough that VSEPR remains useful as a first approximation.


Why is ionization potential considered a periodic property?

The ionization energy is described as the energy required to remove an electron from an atom in the gaseous phase. Apart from some exceptions, it is possible to identify a general trend in ionization energies within the periodic table: moving from left to right in a period the ionization energy increases, whereas it decreases from top to bottom in a group.


Why do we need to add Al in excess for the reaction between a nitrate and a base to give ammonia?

Hydroxide and nitrate alone do not react  with each other.

Their reaction with aluminium is the part of the classical quantitative determination of nitrate content by distillation of produced ammonia and by the back acid titration.

Nitrite + ammonium ions interfere, but the former can be eliminated due their high reactivity, the latter can be determined and/or eliminated by prior distillation without aluminium.

Aluminium acts as a reduction agents, reducing nitrates to ammonia, forming alkali aluminates.

$$\ce{3 NO3-(aq) + 8 Al(s) + 5 OH-(aq) + 18 H2O(l) -> 3 NH3(g) + 8 [Al(OH)4]-(aq)}$$

Generally, aluminium need not to be in an excess. But if there is the desire the all nitrate ions react quantitatively, aluminium and hydroxide must be in excess.

Usually, the Devarda's alloy is used, instead of pure aluminium, containing also copper and zinc.


Does clove oil/eugenol interact with nylon type plastics?

Nylon 612, primarily used for filaments and toothbrushes, is resistant to clove oil at 20 °C [1, p. 2034]:

Nylon 12, used for coatings, is also resistant to clove oil at 20 °C [1, p. 1829]:

The tech data on nylon at newmantools.com also suggests there is no effect of clove oil on nylon in general.

References

Chemical Resistance of Thermoplastics; Woishnis, W. A., Ebnesajjad, S., Eds.; PDL handbook series; William Andrew: Norwich, N.Y, 2012. ISBN 978-1-4557-7896-6.


Should I be worried about the waste from "sugar rockets" and other potassium nitrate fireworks?

Potassium nitrate is transformed into potassium nitrite only when heated alone. And it is not really toxic, as it is added in ham to prevent its too early oxidation in air.  When heated with charcoal, sulfur or any other oxydable substance, potassium nitrate is transformed into $\ce{K2CO3}$ or $\ce{K2S}$. For example, black powder is a mixture $\ce{KNO3 + C + S}$ that reacts violently according to ; $$\ce{2KNO3 + 3 C + S -> K2S + 3 CO2 + N2}$$ And if the explosion is made in contact with air, the hot $\ce{K2S}$ is quickly oxidized into sulfate according to $\ce{K2S + 2 O2 -> K2SO4}$
Sugar rockets are made of ordinary sugar $\ce{C_{12}H_{22}O_{11}}$ and $\ce{KNO3}$. This mixture reacts according to the two following equations : $$\ce{C_{12}H_{22}O_{11} -> 12 C + 11 H2O}$$  $$\ce{4 KNO3 + 5 C -> 2 K2CO3 + 2 N2 + 3 CO2}$$
So the ashes of any combustion reaction are mostly made of $\ce{K2CO3}$ which was the main constituant of the wood ashes. These ashes was used in the Middle Age for washing clothes by our grand-grand-mothers. They are not considered as toxic


Why does an increase in Kinetic Energy increase the likelihood of successful collisions?

The kinetic energy of a particle is an alternative way of measuring how fast it moves (the mass of the particle also matters but for a given type of particle of a given mass the only thing that matters is how fast it is moving).

Potential energy has nothing to do with how far particles are apart (unless they are very close) and is pretty irrelevant in this situation.

In a gas or liquid, molecules are moving around a lot and banging into each other a lot. The speeds of the molecules are distributed statistically but their average speed is related to temperature. Whether a reaction happens when they collide depends on whether the molecules hit each other with enough energy to cause a reaction to happen (the orientation of the molecules might also matter but there are usually enough collisions in many different orientations that we don't think about this explicitly other than as part of the rate of the reaction).

In general, more energy in the collisions will lead to more reactions as more molecules will have enough energy to make the reaction happen (rather than just bounce off each other). More energy means more molecules have faster speeds. So, in general, more kinetic energy leads to a higher probability that any given collision will lead to a reaction.


Azeotropes and separation by distillation?

You seem to have the right idea.

To elaborate, take the specific example of the positive azeotrope of (roughly) 96% ethanol and 4% water. If you begin with a mixture containing less than 96% ethanol, distillation will result in a distillate more abundant in ethanol and nearer to the azeotropic composition. That implies the residue will necessarily be proportionally lower in ethanol. If you continuously repeat the distillation procedure using the remaining pot residue, it will approach 100% water. On the other hand, if you begin with a concentration above 96% ethanol, the resulting distillate will actually be lower in ethanol (and therefore nearer to the azeotropic point), while the residue will actually become more concentrated in ethanol. Hence, in theory, with many repeated distillations you can approach purity in the residue when dealing with positive azeotropes, and the ultimate composition of the residue will depend on which side of the curve you're on to start with.

When dealing with negative azeotropes, the situation is analogous but the particulars are basically reversed. Take the example of hydrofluoric acid and water, which form an azeotrope at approximately 37% HF. If you initially begin distilling a mixture comprising less than 37% HF, you'll obtain a distillate more dilute in HF, while the residue will contain more HF and be nearer to the azeotrope. If you were to collect the distillate and repeat the distillation, you would again end up with an even more dilute distillate, with the concentration of the distillate approaching 100% pure water. If you begin distilling a mixture containing more than 37% HF, then the residue will end up being more dilute as it approaches the azeotrope, while the distillate will actually be richer in HF. If you repeat this process many times over, you can eventually obtain a distillate approaching 100% pure HF (in theory).

So, in summary, you're correct to suggest that one of the components can be isolated with something approaching complete purity from an azeotropic mixture, assuming a sufficient volume to enable an adequate number of simple distillation cycles or theoretical plates in fractional distillation. Of course, the problem is that you really have no choice in determining which of the two components you can isolate, as that depends entirely on the starting composition of the mixture. In practice, you can often separate azeotropeic mixtures by modifying the distillation process in various ways.


Which of the following para-substituted benzyl alcohols reacts fastest with HBr?

Inductive effects will play much more of a role in electrophilic aromatic substitution. Here, positive charge on a benzylic position is being delocalized to the aromatic ring, and the question is asking which group will best stabilize this positive charge (via resonance):

Between $\ce{-H}$ (neutral), $\ce{-NO2}$ (which would destabilize the positive charge), $\ce{-Cl}$ (poor at stabilizing positive charge), and $\ce{-OCH3}$ (good at stabilizing positive charge), $\ce{-OCH3}$ is the clear winner.


Okay, I think I found an answer. It's a set of four papers which were published consecutively in Physical Review 36 (1930). Interestingly, each was submitted on June 23rd, 1930, which makes me wonder about the back story.

The first comes from General Motors Corporation Research Laboratories in Detroit (Kettering, Shutts, and Andrews). The other three papers are out of Johns Hopkins (Donald Andrews, from Chemistry, and Yates, from Mathematics). I suppose Andrews could have graduated from Hopkins and found that GMCRL was working on the same thing; or perhaps he had a dual appointment.

As a side note, there appears to be a patent suggesting that L. W. Shutts invented shock absorbers, so it makes immediate sense that he would propose the spring law as an explanation.

So Donald Andrews (or at least Johns Hopkins) is the obvious link between GMCRL and Robert C. Yates. Did Andrews know Yates? Was someone about to be scooped? The answer to that is probably beyond the scope of this question.


Permanent suspension of cocoa powder into water

Sounds like you need a food-grade surfactant. Maybe you could try lecithin?

The problem is that cocoa powder is mostly made of stuff that is not soluble in water (hydrophobic). So, to create the emulsion, you will need to "package" the powder molecules in something they are soluble in. You could either use a surfactant directly, or build up in layers using fat molecules and proteins, with a surfactant around the whole package.

Lecithin would be a good starting choice since it is FDA approved and used in lots of products already.

I would try this:

Add lecithin in gradually larger amounts until the suspension worked (you will have to try different heat/pressure and emulsification treatments - consider high-pressure homogenization as well.)
Add some whole milk (we are looking for the fat molecules and proteins) and see if you can find an acceptable milk/lecithin ratio that tasted good and that was still stable.

If that doesn't work, your other option is to modify the cocoa powder granule surface so that it is hydrophilic. That means you will need to add polar functional groups. One way to do that is with an acid (citric, for example), but you will add a sour taste which might be hard to get rid of. There is a patent for an approach using enzymes, but I don't know which one they used (it just refers to "starch degrading," "protein degrading," and "cell-wall degrading" enzymes), or how well it works.


We describe an object, which does not exist, but is very real, called and 'orbital'.  It is a region of space where an electron is 'likely' to be.  It is the result of charge field interactions between positively charged nuclei.  Normally we only discuss orbitals in the context of  1-2 Angstroms distance between atomic nuclei.  Some orbitals have a character which we call 'bonding'.  A bonding orbital, occupied by one or two electrons might be thought of as a 'bond'.  However, bonds also do not exist, even if they are very real.

It is somewhat nonsensical to describe where in an orbital the electron is.  If you must, the electron may be described as having a location of the entire orbital simultaneously, with vary weights to all of those locations, which are averaged.  This needs to be thought of in the concept of 'wavefunction', which is a mathematical function describing the distributions of these weights.  In the case of a polar covalent bond, the wavefunction more heavily favors locations near the more electron negative nucleus, and so the orbital 'concentrates' more of its charge near one end of the 'bond' and results in what we call a partial charge.

The reason we get squeamish about describing 'when' and 'where' is that the Heisenberg uncertainty principle renders the uncertainties energy and momentum of the system to be absurdly large when dealing with systems that are roughly the size of a chemical 'bond'.

At this point, nothing should be cleared up with regard to the original question.

The reality of the situation is that using when and where in this context is very likely to lead to predictions that will lead one astray when predicting physical phenomena.  As bizarre as quantum mechanics is, it has been shown to be highly predictive of the bizarre world of atoms and molecules, and so you have to try to figure out how to not ask questions from a classical  physics framework.


Relationship between Van der Waals forces and the boiling point

We talk about the forces between molecules.  The Van der Waals force represents the summation of all attractive and repulsive forces between molecules, other than those due to electrostatic charges or actual bonds.  This summation most often turns out to be a weak attractive force.  Typically the magnitude of the Van der Waals force is small compared to electrostatic or bonding forces and (again typically) only apply at relatively short range.  So they usually serve to weakly hold molecules together.  The stronger the Van der Waals force, the more strongly molecules will be held together and the more energy will be needed to be put into the system to break them apart from each other.  In the case of boiling a liquid, you are really talking about breaking molecules apart from one another.  So in the case of boiling, the weaker the Van der Waals attraction between molecules, less energy will needed to break them apart from each other and allow them to boil.
Here is a link to a Wikipedia article on Van der Waals forces if you'd like to read more about them.


How to convert 1,4 pentadiene to 1,3 pentadiene?

My first instinctive answer would be to use a precious metal catalyst. Palladium usually displays this sort of activity. However, this article should cover most methods from basic and metallic to photochemical.
https://www.thieme-connect.de/media/10.1055-b-003-125756/lookinside/10-1055-b-003-125756_chapter012-1.jpg


Yes, when bonds are formed, the hybridization can change in a carbon atom.
I think it’s important to remember something. Hybridization is not a real phenomena that occurs in reality. It is a model we use to simplify the complex bonding that occurs between atoms.
So, yes hybridization of the carbon will change, as it is not concrete and can change as bonds are made.


Why is it that neither silver nor copper react with a strong acid?

This will involve some degree of hand-waving. Let me first limit the analysis to the transition metals so that more effective comparisons can be made.

The reaction between a metal and an acid requires the metal to be oxidized, i.e. lose electrons to the acid, creating positively-charged metal ions in solution. The general idea is that as you go along the transition metals in a period from left to right, there is an increase in effective nuclear charge felt by the valence $ns$ electrons (due to the additional $(n-1)d$ subshell electrons poorly shielding the $ns$ electrons from the nucleus) making the valence electrons harder to remove. This trend can be seen to some extent in the increase in ionization energies of the elements, for example, and also in the standard reduction potentials for metal cations of same charge (they become more positive, in other words making oxidation and hence attack by acid more difficult). Thus, metals at the left of the transition metals tend to be more easily attacked by acids than the metals at the right.

There is a rather interesting exception though: zinc strongly breaks the trend, being far easier to oxidize than copper, the element before it, even though zinc has a much higher ionization energy, clearly indicating its electrons are held tighter (cadmium also exhibits this anomaly, to a milder extent). What gives? The thing is that the tendency for a metal to oxidize is the result of several combined factors, one of which is also how strongly bound the solid metal is in the first place. If a metal contains atoms which are strongly bonded to each other, then oxidation tends to be more unfavourable as it would require these bonds to break. When comparing copper and zinc, it is clear that the latter has far weaker metallic bonding in the solid (their melting/boiling points are $1360\ \rm{K}$ / $2835\ \rm{K}$ and $695\ \rm{K}$ / $1180\ \rm{K}$, respectively). Thus, if you include the energy necessary to separate the atoms from the solid before ionizing them, it turns out that the process is easier for zinc than copper.

Comparing different rows of the transition metals is a bit less clear, in part because not all elements have ions of the same charge which can be directly compared. The general tendency is that the metals become tougher to oxidize as you go down the rows. For the transition metals in the sixth and seventh period this is probably a consequence of lanthanide/actinide contraction and relativistic effects, which decrease the energy of the valence $ns$ orbitals.

Now for some other elements. Magnesium is very reactive towards acids because it is both a metal containing relatively weak bonds and because its ionization energy is comparatively low, being an alkaline earth element. For tin, the metallic bonding is not weak, but its ionization energy is not too high, so it will oxidize in acids, though not as strongly. Lead is somewhat less reactive to acids, possibly because its oxidation tends to stop at +2 instead of tin's +4, creating a significantly softer cation which isn't as well solvated by water.


Can we prepare alkyl chloride or alkyl bromide by Finkelstein reaction?

The thing is that you have to start with some compound $\ce{R-X}$ where $\ce{X}$ is a good leaving group. The quality of the leaving group is often thermodynamically defined, and in the Finkelstein reaction that you described, the energy motor for the whole thing is the insolubility of the salt that is produced.

I don't see what favourable starting material you could use for this in order to prepare other haloalkanes, especially since there are other ways for doing that.

That being said, it would in principle be possible to do
$$\ce{R-X + NaY -> R-Y + NaX v}$$
as long as the solubility of $\ce{NaY}$ is larger than the solubility of $\ce{NaX}$.


Is N-methylimidazole an unusual solvent, or are there more common substitutes for it?

http://www.chemicalbook.com/ChemicalProductProperty_EN_CB1316726.htm
http://www.lookchem.com/cas-616/616-47-7.html
http://www.guidechem.com/cas-616/616-47-7.html  2000 RMB = $(USD)320
http://ahjinao.en.alibaba.com/product/595605676-209523668/Time_honored_Supplier_of_N_Methylimidazole_CAS_No_616_47_7_.html
http://www.alibaba.com/countrysearch/CN/n--methylimidazole-supplier.html
http://www.buyersguidechem.com/chemical_supplier/N-Methylimidazole.php

Info and suppliers,  Note that a Chinese contract is often not the legal or philosophical equivalent of a Western contract.


How to deposit a durable layer of sensing molecules in a PVC matrix on the tip of an optical fiber?

We have no idea how to do this, since we dont have any step/by/step instruction, reaction time, environment we need to maintain for this.

Let's redefine the general task:

How to deposit a layer of 2,4,5-triphenylimidazole (lophine) on the tip of an optical fibre?

Requirements:

The layer must be permanent/durable and does not wear or wash off during measurements of the nitrate concentration in drinking water.

The workflow therefore includes:

How to prepare the fibre?
How to deposit one (or several) layers of lophine?
How to cure the layers to make sure that they do not wear/wash off?

As far as the preparation of the fibre is concerned, your primary paper (Optics and Lasers in Engineering, 2014, 60, 38-43) gives some hints on page 39:

  To deposit the sensitive layer on
  the optical fiber, the coating was removed in a 1 cm section, and
  the exposed cladding was sterilized. Isopropyl Alcohol Level was
  used for cleaning the optical fiber in special cleaning cloths and
  immediately the LI–PVC was introduced [...].

Apparently, lophine is deposited on the fibre in a PVC matrix and you have cited the composition of the mixture used.

How is this actually done?

An earlier publication by the same authors gives some directions! It's reference 19 in the paper that you cited!

In Sensors, 2008, 8, 847-859, the describe authors describe the dipping method in chapter 2.2. and provide further references.

In summary, the prepared fibre is

dipped into the solution (lophine, PVC, TBP, THF) and then
cured by placing the fibre in an oven for 15 minutes at 80 °C.

These two steps are repeated as often as necessary to obtain the required total layer thickness.


Reacting sodium metal with aqueous sodium hydroxide, what would happen?

Consider first the solvation of sodium metal in water to give a sodium ion and a hydrated electron.
$$\ce{Na(s) \xrightarrow{\ce{H_2O}} Na^+(aq) + e^-(aq) }$$
The hydrated electron will then act as the reactive species to reduce water to hydrogen atoms and hydroxide anions.
$$\ce{e^-(aq) + H_2O(l) \rightarrow H(aq) + OH^-(aq)}$$
Note that two of these hydrogens come together to form the diatomic hydrogen gas.
$$\ce{2H(aq) \rightarrow H_2(g)}$$
When we multiply the coefficients in the first two equations by two, we are able to consider the net ionic equation through the elimination of the reactive intermediates.
$$\ce{2Na(s) + 2H_2O(l) \rightarrow 2Na^+(aq) + 2OH^-(aq) + H_2(g)}$$
We see that the dissolution of sodium metal in aqueous sodium hydroxide would be affected by Le Chatelier's principle due to the presence of the sodium hydroxide. We would expect there to be a reduction of solubility and rate of dissolution. The greater the initial concentration of sodium hydroxide, the more appreciable the effect.
Given that a solution of sodium hydroxide at $\ce{25 ^{\circ}C}$ is saturated at $\ce{25 M}$, a $\ce{1 M}$ solution would not be expected to have an appreciable effect. Hydrogen gas would remain a product, evidenced by the net equation.
As far as the reaction of molten sodium and molten sodium hydroxide, I expect that is what they were referring to on the Wiki page for sodium oxide when they gave this equation.
$$\ce{2NaOH + 2Na \rightarrow 2Na_2O + H_2}$$
The source given is Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. ISBN 978-0-08-037941-8.
The solubility of sodium hydroxide was resourced from the Wiki page for sodium hydroxide. The source given appears to be Haynes, CRC Handbook of Chemistry and Physics.
The first set of equations were sourced from Walker, David C. Production of Hydrated Electrons. Canadian Journal of Chemistry. Volume 44. 1966.


How to calculate the rate of reaction of Sodium and Chlorine

Rate laws are determined via experimental data. Reactant concentrations are held constant while others are modified in order to determine the effect of each reactant's concentration on a reaction. For this reason, one cannot determine the rate law from just a stoichiometric equation.
$$\ce{2Na + Cl2 -> 2NaCl}$$

We can propose reaction mechanisms to make an educated guess of the rate law:
\begin{align}
\ce{Na + Cl2 &-> NaCl + Cl}&&\text{(slow step)}\\
\ce{Na + Cl &-> NaCl}&&\text{(fast step)}\\
\end{align}

Using Hess' law, we can combine these to make sure it equals the original equation.
$$\require{cancel}
\begin{align}
\ce{  && Na + Cl2 &-> NaCl + \cancel{Cl}}\\
\ce{+ && Na + \cancel{Cl} &-> NaCl} \\\hline
\ce{  && 2Na + Cl2 &-> 2NaCl} \\
\end{align}$$

The coefficients of the slowest elementary step in reaction mechanism determine the exponents of the rate law. Hence:
$$\text{Rate}=k[\ce{Na}]^{1}[\ce{Cl2}]^{1}$$

Or just
$$\text{Rate}=k[\ce{Na}][\ce{Cl2}].$$


What would be the name of this compound?

I think it is Cubane ($\ce{C8H8}$), a synthetic hydrocarbon molecule (Wikipedia), which has synthesized using following scheme:

I am confident about that because your teacher had said that there wasn't a central carbon atom (assumingly). See following illstration:


What does β stand for in β-potassium zirconate?

They signifies one polymorph (the beta polymorph) of barium zirconate.

Allotropy is the property of some chemical elements or compounds to exist in two or more different forms, in the same physical state at different temperature. Now, technically they are the same element or compounds but have different crystal structure at changing temperature. So, in order to name them, alpha, beta, gamma etc. are used in the prefixes to depict them as they are the allotropes formed at different temperature phases.. That helps in their differentiation.

For e.g. iron has 5 allotrope namely:

ferrite (α-iron): the ferromagnetic form at T < 770 °C (the Curie point, TC); BCC crystal structure
β-iron: T = 770 – 912 °C; the paramagnetic form of α-iron; same structure as α
austenite (γ-iron): 912 – 1,394 °C; FCC crystal structure
δ-iron: 1,394 – 1,538 °C; forms from cooling down molten iron; same structure as α
ε-iron (also called hexaferrum): forms at high pressures; HCP crystal structure

There are several more examples of such type of nomenclature of elements. You can find them in the wikipedia article of allotrope(see table).

Also, not only elements form allotropes. Compounds also form allotropes but have a different name i.e polymorphs. Generally, binary metal oxides are polymorphs. Polymorphs have different temperature phases in which one form of crystal structure exist. These phases are depicted by giving greek alphabet at the prefixes like α-phase, β-phase etc. Polymorphs or allotropes of compounds are given that greek alphabet prefixes according to that phase in which one form of that compound exist.

For e.g. iron oxide $\ce{Fe2O3}$ has 4 phases.

α-phase: at ambient conditions: has Corundum-type Rhombohedral (R3c) like structure
β-phase: below 773 K: has Body centered cubic (Ia3) like structure
γ-phase: up to 933 K: has Cubic, spinel structure (Fd3m) like structure
ε-phase: has Rhombic (Pna21) like structure

So, iron oxide $\ce{Fe2O3}$ has 4 polymorphs : $\ce{\alpha-Fe2O3, \beta-Fe2O3, \gamma-Fe2O3, \epsilon-Fe2O3}$.

For more information you can check this question(asked by myself) or this question that describes how polymorphs is formed.


Heating organic compound in a stream of nitrogen gas

This is a strangely written text. In a practical setting, the way to get oxygen concentration is by sending an organic sample to a lab for an ultimate analysis.  The ultimate analysis chemistry is explained elsewhere.

The highlighted text seems to describe anoxic pyrolysis of organic matter well above 450°C (for coals this temperature may be higher).  Pyrolysis proceeds as follows.

$$\ce{Organics ->[heat]  syngas + tar + char  }$$

If any molecular oxygen were present in the system these two reactions would very quickly happen.
$$\ce{Organics + O_2 ->[heat]  2CO }$$
$$\ce{2H^+ + O_2 ->[heat]  H_2O }$$

The syngas contains elemental oxygen mostly in the form of $CO$ and $H_2O$, however, other gasses like $CO_2$, $H_2S$, and perhaps even other hydrocarbons may be present.  For the sygas to have most of the elemental oxygen as opposed to the char or tar phase high temperatures exceeding 450°C is important.

If the syngas containing the volatilized matter is passed over a bed of hot coal around 900°C then several things happen.

Complex particles break down (hence the lysis in pyrolysis.)
The gas stream picks up elemental carbon from the bed of hot coal.
$$\ce{CO_2 + C ->[heat]  2CO}$$
$$\ce{H_2O + C ->[heat]  H_2 + CO}$$

By this point, if the temperature is hot enough, all but perhaps an insignificant fraction of the elemental oxygen will convert into $CO$.

Testing for $CO$ is done with the iodine pentoxide test.


I wondered about that for a long time, but Ben Norris and BigGenius have the truth of it: the question is about the bond dipole moment, otherwise the answer would make sense. Here's why.
Molecular dipoles

Sulfur trioxide has planar trigonal geometry, and its molecular dipole is zero. Similarly, molecular $\ce{SiO2}$ has the same geometry as $\ce{CO2}$, i.e. linear, and thus its molecular dipole is zero.

$\ce{Cl2O7}$ is bent at its central oxygen atom:

and thus has a non-zero dipole moment. While $\ce{P2O5}$ generally occurs as its “dimer” $\ce{P4O10}$, it is possible to write a molecular structure for $\ce{P2O5}$ which then strongly resembles that of $\ce{Cl2O7}$, being bent at the central O atom. It thus also has non-zero molecular dipole.

Finally, ordering between $\ce{Cl2O7}$ and $\ce{P2O5}$: because the P is less electronegative than Cl, the partial charges borne by the P atoms will be larger than those of the Cl atoms, and thus I expect the molecular dipole of $\ce{P2O5}$ to be greater than that of $\ce{Cl2O7}$.

Bond dipoles
If the question is talking about the $\ce{X–O}$ (or $\ce{X=O}$) bond dipole in the series where X = S, Si, P or Cl, it becomes easier to understand. For one thing, because you're asked about $\ce{X–O}$ bonds where X spans a row of the periodic table, which is typical of this sort of problems. The answer would then be that bond dipole moment get bigger when the electronegativity difference of X and O becomes larger (the sharing of electrons gets more asymmetric). This leads to the answer you gave.


The classification we use to distinguish bonds in other chemical compounds will obviously be valid for the classification of oxides.

They can be classified as covalent or ionic.


...According to my opinion it should be Oxygen as it is more electronegative than Nitrogen...

It's correct that oxygen is more electronegative than nitrogen. However, this makes an oxygen group a poorer electron donating group than the corresponding nitrogen group. Electronegativity means that the atom is less willing to give up the electrons. Nitrogen is more willing to give up (donate) its electrons because it is less electronegative.


From Wikipedia:

An alum is a type of chemical compound, usually a hydrated double sulfate salt of aluminum with the general formula $\ce{XAl(SO4)2 · 12 H2O},$ where $\ce{X}$ is a monovalent cation such as potassium or ammonium.

There are also chromium alums and ferric alums, but the color in your sample does not seem to suggest either of them.
The alums melt at relatively low temperatures, because they already have a lot of water in the crystal structure, so ammonium alum and potassium alum can not be differentiated on the basis of melting points. Solubilities are fairly close, too:
$$
\begin{array}{llc}
\hline
\text{Compound} & \text{mp}/\pu{°C} & \text{Solubility}/\% \\
\hline
\ce{(NH4)Al(SO4)2 · 12 H2O} & 93 & 15 \\
\ce{KAl(SO4)2 · 12 H2O} & 92.5 & 11 \\
\hline
\end{array}
$$
If there is no violet color to a flame, there may be no potassium, but the test can be sensitive and may be indeterminate. Eliminate the possibility that the alum is an ammonium salt by dissolving some in water and adding a strong alkali, then sniffing it gently. If you smell ammonia, you have an ammonia alum; it will still work as a flocculant for muddy water, but is different chemically. A strong alkali readily available at home, well, at the grocery store anyway, is sodium carbonate (washing soda).


Why 1 mole of H2 occupied the same volume occupied by 1 mole of O2?

The volume occupied by a gas has no relationship to the size of the molecules of the gas
This is one of the principles of the gas laws in chemistry. It is based on observation: in the early days of chemistry it is what chemists observed to be true. And it is a very important chemical law.  But, of course, that doesn't explain why it is true.
There are several ways to gain some appreciation of this without too much theory. consider water as a liquid compared to water as a gas (I'm simplifying a bit but not by much). One mole of liquid water occupies about 18mL of space and this space is largely dominated by the size of the water molecule (again reality is a little more complex as the specific way the molecules interact makes a small difference to the volume occupied). As water vapour (the same substance but as a gas) 1 mole of water occupies over 22L of space (ignoring the fact that water is not a gas at STP, but bear with me as this is the space it would occupy if it were). That is well about 1200 times more space. This implies that very little of the space occupied has anything to do with how big the water molecule is.
The same broad idea applies to all gases.
So why does the molecular size not matter? A detailed understanding requires a fair few calculations about how gas molecules behave. But, the bottom line involves thinking about the molecular cause of pressure and the meaning of temperature. Pressure is caused by gas molecules bouncing off the walls of a vessel. Temperature is the average energy of those molecules which is a function of their speed and their mass. Doing the calculations shows the relationship between volume, pressure and temperature. This calculation explains the properties of gases but the volume of the molecules has no part.
Strictly speaking the ideal gas laws assume that the size of molecules don't matter. But this is a good approximation when the gases are far away from the point where they become liquids. The simplified theory matches actual observations well. And the fact that, in typical gases, the volume occupied by the molecules is less than 0.1% of the volume occupied by the gas shows why the molecular volume is usually irrelevant.


Is it incorrect to use Fisher Projections for carbons that don't have stereocenters?

According to IUPAC it is not incorrect for using it to depict carbohydrates being stereogenic or not. However here is an excerpt from the link listed below:

  The use of
  Fischer projections in non-carbohydrates is problematic. Any drawings
  of this type are necessarily ambiguous since it is unclear whether the
  author of the drawing intended to make a Fischer projection or simply
  happened to draw a diagram in that orientation. Molecules depicted in
  the style of Fischer projections are likely to be recognized as such
  when they contain at least two acyclic stereo- genic centers oriented
  vertically, although such interpretation is not guaranteed. Molecules
  containing a single stereogenic center pose even greater problems
  since it is often convenient to separate the four ligands by equal
  angles whether or not a single stereoisomer is intended. Accordingly,
  although Fischer projections of carbohydrates are acceptable, similar
  depictions of non-carbohydrates should be used only when the viewer of
  the diagram is certain to understand that the diagram was intended to
  represent a Fischer projection. Fischer projections should be avoided
  where ambiguity is possible, including when used in computer
  applications.

See more here (page 37-38):

http://www.iupac.org/publications/pac/2006/pdf/7810x1897.pdf


When does a piston stop in an irreversible process?

You're right.  The piston can gain kinetic energy and overshoot its equilibrium position, but then, the force of the gas on the piston will drop until the piston comes back the other way.  The piston will undergo an oscillation back and forth, but this oscillation will be damped as a result of viscous dissipation stresses in the gas.  Eventually, the oscillation will decay until the piston stops.  At this point, the gas pressure will be in equilibrium with the lower pressure outside.
If I do a force balance (Newton's 2nd law) on the piston, I get
$$F_\mathrm{g} - p_0 A = m \frac{\mathrm{d}v}{\mathrm{d}t},$$
where $F_\mathrm{g}$ is the force the gas exerts on the piston, $p_0$ is the outside pressure, $A$ is the piston cross sectional area, $m$ is the mass of the piston, and $v$ is the piston velocity.  If I multiply this equation by the velocity of the piston $v = \mathrm{d}x/\mathrm{d}t$, and integrate with respect to time, I obtain:
$$
W_\mathrm{g}(t)
  = \int\limits_0^{x(t)} F_\mathrm{g}\,\mathrm{d}x
  = p_0(V(t) - V(0)) + m\frac{v(t)^2}{2},
$$
where $W_\mathrm{g}(t)$ is the work done by the gas on the piston (its surroundings) up to time $t$, $V(t)$ is the gas volume at time $t$, and $m\frac{v(t)^2}{2}$ is the kinetic energy of the piston at time $t$. At infinite time, the piston is no longer moving, so its kinetic energy is zero at final equilibrium. Therefore, at final equilibrium, the work done by the gas is just
$$W_\mathrm{g}(\infty) = p_0\Delta V,
\text{ where } \Delta V = V(\infty)-V_0.$$


Given the whitening observed at the cathode, one potential explanation is that calcium salts were precipitating there, forming a layer blocking passage of part of the current.
I usually write the cathodic hydrogen-evolution reaction in neutral solution as:
$$
\ce{2 H2O + 2 e- -> H2 + 2OH-}
$$
Thus, the pH in the vicinity of the cathode will be considerably greater than 7 during the electrolysis process, due to the $\ce{OH-}$ formed. Since you're using a calcium-based electrolyte, it seems likely that some combination of $\ce{CaO}$/$\ce{Ca(OH)2}$ is precipitating in that alkaline surface layer. Once formed, in my experience such solids will dissolve only slowly in neutral aqueous solution.
If possible, I'd recommend trying the experiment again but in a different electrolyte, such as $\ce{KNO3}$ or $\ce{NaNO3}$; those alkali cations shouldn't exhibit any precipitation.


Active mass of solids shouldn't necessarily be unity

Rather, active mass of solids need not necessarily be unity, but arbitrarily is unity with advantage. You do not have to know the molar mass nor density.
The common thermodynamic expressions like $K = \frac{[C][D]}{[A][B]}$
or kinetic ones like $\frac{\mathrm{d}[C]}{\mathrm{d}t} = k \cdot [A] \cdot [B]$
should be more exactly written as $K = \frac{a_\mathrm{C}] \cdot a_\mathrm{D}}{a_\mathrm{A} \cdot a_\mathrm{B}}$ or $\frac{\mathrm{d}[C]}{\mathrm{d}t} = k \cdot a_\mathrm{A} \cdot \cdot a_\mathrm{B}$
where $a_\mathrm{X}$ is unitless quantity called activity, defined by the chemical potential of the substance $ \mu_\mathrm{A} = \left( \frac {\partial G}{\partial n_\mathrm{A}} \right)_{T,p}$ by formula:
$$\mu_\mathrm{A} = RT \ln {a_\mathrm{A, unitless}}$$
If we implement the standard chemical potential, we can define unitless activity as ratio of the activity with unit and the standard activity with the unit value.
$$\mu_\mathrm{A} = \mu_\mathrm{A}^{\circ} + RT \ln \left( \frac {a_\mathrm{A}}{a_\mathrm{A}^{\circ}} \right)$$
where $a_\mathrm{A}^{\circ}$ is the activity at the standard chemical potential $\mu_\mathrm{A}^{\circ}$. We can consider the activity $a_\mathrm{A} = x\ \text{"whatever unit"}$ and the standard activity $a_\mathrm{A}^{\circ} = 1 \ \text{"whatever unit"}$.
If we arbitrarily choose $a^{\circ} = \pu{1 mol/L}$, then $a$ then more or less closely follow the respective concentration [A].
Chemical potential of the pure solid substance $\mu_\mathrm{A}^{\circ}$ does not depend on the amount of substance, so we arbitrarily choose $a^{\circ} = x_\mathrm{A}^{\circ}=1.0$ and $a$ then more or less follows the  molar fraction of substance $x_\mathrm{A}$.


Why do we use hydrochloric in the test for sulfate ions but nitric acid for halide ions?

The link above which I used for reference mentions that for the test
  for sulfate anions, hydrochloric acid (HCl) is used to ensure that
  there is no presence of carbonate ions. For the test for halide ions
  (F−, Cl−, Br−, I−, etc), nitric acid (HNO3) is used instead, for the
  same purpose.

Good question, but this is not a universal practice. In reality, the choice of acid depends on the actual sample. Qualitative tests, like these, have become merely academic exercises due the to advances in instrumental analysis these days. As you said, one cannot add HCl to destroy carbonates in a sample which is to be tested for halide ions. Dilute nitric is the only choice, because you would not use sulfuric acid in the case of halides because (i) your test sample may contain ions which will precipitate with sulfate (Ba, Ag, Pb, Ca etc.). Concentrated nitric acid would be a poor choice as well because it will oxidize iodide to free iodine.

An excellent book, which is far better than any online resource, is Vogel's Qualitative Inorganic Analysis. It should be available in any good library and it will give an in depth view of qualitative analysis.


Is the breaking of bonds by using energy a prerequisite for new bonds to be made with the involved atoms?

A: No, it is possible for a bond to form between two species without first breaking a bond. But it is more common for reactions to occur with a tradeoff between breaking old bonds and forming new ones.

If so, is this because the more energy there is, the more the atoms will move/vibrate, eventually pulling the bond apart?

A: Adding energy in the form of heat can lead to breaking of bonds, which in turn can lead to formation of new bonds.

Otherwise, can bonds be broken by a new bond sort of pulling it apart in some way?

A: You can have a "displacement" reaction where one atom replaces another in a structure: the bond to the new atom forms at the same time as the other bond is broken. That process has a lower energy barrier between reactants and products, relative to breaking an existing bond first and then forming a new bond later.

Is the strength/length of the bond what determines the amount of energy needed to break it? I'm also unsure of what the strength or length of a bond is based on.

A: The strength of a bond is the same as the amount of energy needed to break it (in the absence of other reactants). Bond length tends to correlate inversely with bond strength, as for example in going from a single to a double to a triple bond between carbon atoms. Bond strength depends on a number of factors--bond strength tends to be greater between two atoms of about the same size. Bonds between two electronegative atoms like oxygen or chlorine are weaker than a bond between two carbon atoms.

Am I correct in thinking that bond order is about the number of shared electrons in a covalent bond? But what about ionic bonds? Is it the amount of electrons "taken" by the more electronegative atom?

A: You are correct here in your understanding of bond order. But the term "bond order" does not apply to so-called ionic bonds -- for example in sodium chloride -- where an electron is completely donated from sodium to chlorine and does not participate in keeping the atoms together. Rather, it is the electrostatic attraction between the sodium and chloride ions that keep them together.


Sodium carbonate titration in presence of phenolphthalein

Suppose you want to find out the amount of sodium carbonate  in a solution of sodium hydroxide. Let's use phenolphthalein as the indicator and HCl for titrating the mixture.

The first reaction that takes place is $$\ce{NaOH + HCl -> NaCl + H2O}$$
The pH at this point isn't alkaline enough for phenolphthalein to become colourless, since there's still base ($\ce{Na2CO3}$) present in the solution.

Then $\ce{Na2CO3}$ reacts. When all the $\ce{CO3^2-}$ becomes $\ce{HCO3^-}$, the pH is given by:

$$\mathrm{pH}=\frac{\mathrm{p}K_{\mathrm{a}1} + \mathrm{p}K_{\mathrm{a}2}}{2}$$

which is approximately $8.3$. (The above expression is for amphiprotic salts.) Phenolphthalein changes colour at approximately pH $8.2$.

So when the colour of phenolphthalein disappears, it means that all the carbonate is converted to bicarbonate. So if $x$ is the volume of HCl required to neutralize NaOH and $y$ is the volume of HCl required for complete neutralization, of carbonate, then phenolphthalein changes colour when volume of HCl used is $\displaystyle x+\frac{y}{2}$.

Remember that for complete neutralization, two moles of HCl is required for every mole of carbonate. For half neutralization only one mole of HCl is required. (Which is half the amount required for complete reaction).


Do maxima in Gibbs energy also correspond to equilibrium positions?

G vs. $\xi$ does not have a maximum

If all reactants and products are pure liquids and solids, G vs. $\xi$ is linear. If some of the species are in mixtures, the entropy of mixing is responsible for the "sagging" shape of the curve. If the curve has an extreme value, it will be a minimum.

Analogy to mechanics

In mechanics, this situation would be called metastability. Giving a nudge in either direction will allow the system to reach lower potential energy.


This question requires some additional specification before it can be answered.

Propanediol is ambiguous as a chemical name. There is 1,2-propanediol and 1,3-propanediol, and these are different compounds (namely, structural isomers).

Though either way, if you check references, you will see that both compounds are miscible with water, so the solubility is infinite.

See: 1,2-propanediol and 1,3-propanediol.

The term "oil" simply refers to the fact that it is a viscous liquid at room temperature.

Unfortunately, generalized solubility is hard to determine. 1,2-propanediol appears to be miscible in several different polar solvents, so you might be able to dissolve in something anhydrous.


You need to define what you mean by "stronger".  Are you talking about resisting deflection (and thus rolling losses)?  Yield?  Break?  Impact?  Abrasion?  Lifespan?  Ability to withstand heating from its deflection as the vehicle rolls?  Also, any material can "withstand the weight of a car" if you use enough of it, do you know what sort of geometry you're talking about?

Lastly, the material properties of polymers depend a lot on the manufacturing processes and the details of the raw materials (chain lengths, branching, monomer ratios, etc) and additives.  A good example is PET.  Check out the difference in tensile strength between simple thermoformed PET and BoPET (Mylar), for example; they're vastly higher for BoPET.  And the reason is because it's elongated to near breaking at lower temperatures - thus orienting the polymer - and then given a heat treatment to relax it into its new shape.

I think you need to define your problem statement better.  Think about what properties you want, your geometry, and how you plan to manufacture it.  Also, what are the parameters you're trying to optimize?


Does SiCl2 exist and its possible structure

Via Wikipedia, $\ce{CCl2}$, where carbon is in the +2 oxidation state, does exist but it is fleeting due to its high reactivity.

I was able to find a reference on the analogous dichlorosilylene, and I also found some information on difluorosilylene, also known as silicon difluoride, both of which have silicon in the +2 state and react similarly to a carbene.

So yes, both $\ce{CCl2}$ and $\ce{SiCl2}$ exist where the central atom in the +2 oxidation state.

Tin(II) chloride fill its octet by way of either  bridging chloride bonds in the crystal structure or hydration bonds in water (see Wikipedia images). The structure of solid $\ce{GeCl2}$ isn't given, but I would expect something similar. I'll update if I find anything.


Why aren't lower aromatic organic compounds taught?

Hueckel's rule (i.e. that aromatic compounds have $4n+2$ $\pi$ electrons) places limits on aromaticity in cycles with less than 6 carbons.

As there are only 3 cases to look at, let's enumerate them:

5 carbons: cyclopentadiene ($\ce{C5H6}$) is not aromatic, but the cyclopentadienide anion ($\ce{[C5H5]^{-}}$) is. This species is well known in organometallic chemistry, due to its propensity to act as an $\eta^5$ ligand and form sandwich complexes with transition metals. The most famous of these is no doubt ferrocene, where two cyclopentadienide ligands sandwich an $\ce{Fe^{2+}}$ ion. This ion is didactically problematic and in my experience not covered in high school or early undergraduate curricula because most of its interest lies in its sandwich bonding mode with metal ions, which cannot be easily reconciled with the simplified VB/VSEPR/hybridisation theory used to teach introductory organic chemistry. The same goes for the under-loved tropylium cation ($\ce{[C7H7]^+}$), which has 7 carbons.

4 carbons: cyclobutadiene is antiaromatic, and as such only exists at very low temperatures. If I recall correctly, it possesses two unusually long bonds and two unusually short bonds, quite at odds with the bond equivalency found in benzene. This molecule would have to lose 2 e- to become aromatic. It does so happen that the cyclobutadienyl ligand is known in $\eta^{4}$ metal complexes such as $\ce{(C4H4)Fe(CO)3}$ (1) - I was not aware of these complexes until an expert in the field pointed them out to me.

3 carbons: I was not sure about this, but apparently the cyclopropenyl cation is a thing and is aromatic. Cotton (2) also mentions this being known as a $\eta^3$ ligand. I suspect that it is not taught because it's too weird.

In conclusion:

The smaller aromatic rings are principally known as polyhaptic ligands to metals due to their non-neutrality. This places them outside of the comfort zone of the theories of molecular structure and bonding used in introductory organic chemistry, and as such they are usually left until later.

(1) Cotton, F.A., Wilkinson, G.; Advanced Inorganic Chemistry; Wiley Interscience, 5e, (1988), p. 1170

(2) Ibid, p. 1168


Would Al ions still be reducd to Al metal if it was aqueous soltion?

The Bayer process does not produce aluminium. It produces sodium aluminate, and finally aluminum oxide, which is later used in the Hall–Héroult process (without water) to produce aluminium metal.

As @Mithoron implied, aluminium will displace the hydrogen from water, so cannot be made in an aqueous environment.


Amides will react with three equivalents n-butyllithium to give nitriles via geminal lithium oxyimide intermediates (J. Org. Chem., 1967, 32 (11)). n-butyllithium is a strong enough base to deprotonate the molecule twice at nitrogen and once alpha to the carbonyl group. Grignard reagents may well behave similarly.

I do not think this is a particularly useful transformation though because n-butyllithium is a rather indiscriminate reagent. There are a wide range of other methods available - see this paper if you're interested.

In the special case of a N,O-dimethylhydroxyamide, you can make ketones using either.


Consider that there may be other reactions also going on, not just acid-base chemistry. So, for instance in the Claisen reaction of ethyl acetate, if methoxide was used instead of ethoxide then you may get methyl acetate formed as well.


Will ground glass joints relieve pressure non-catastrophically?

Ground-glass joints are notorious for seizing at the wrong time, particularly if reactants are alkaline.

If pressure might change, I'd add some relief mechanism, such as a water-trap or rupture disk. Without knowing your setup, something as simple as a balloon might work, if not attacked by reactants.


What's the meaning of the $-\equiv-$-like bond symbol?

This is a triple bond between two $\ce{C}$ atoms, i.e. $\ce{-C#C -}$. It is drawn that way because the carbon atoms in a triple bond are $\mathrm{sp}$-hybridised and thus exhibit a linear arrangement.


Partial pressure can be imagined to be the pressure of a gas in the absence of other gases.

Le Chatelier's principle states:

  When a system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

Hence when the pressure of $\ce{Cl2}$ is reduced by 75% in the $\ce{POCl3_{(g)} <=> POCl_{(g)} + Cl2_{(g)}}$ system, the system naturally acts to counteract the change: by increasing the pressure of $\ce{Cl2}$.

Using the ideal gas law, $pv = nRT$ so pressure is directly proportionate to the number of moles of $\ce{Cl2}$ present. The position of equilibrium therefore shifts to the right, favouring the forward reaction that produces more $\ce{Cl2}$ (though not as much as to increase partial pressure of $\ce{Cl2}$ to its former value). This would mean it also consumes more $\ce{POCl3}$.

As for more $\ce{POCl}$ being consumed, if all other factors remain constant I'm not sure why this would happen.


Galvanic cells/redox - confused regarding diagram

I think the arrow for $\ce{O2}$ just means there's oxygen gas in that part of the apparatus.

The reason B is wrong is that conversion of hydrogen peroxide to oxygen gas and water is the spontaneous reaction at these conditions. Assuming SLC implies 1 atm of pure oxygen, then everything is under standard conditions so we don't need to bring Nernst into it. The half reactions are as follows with the corresponding potentials.

$$ \ce{H2O2 + 2H+ + 2e- -> 2H2O} (\Delta E^{\circ} = +1.78) $$
$$ \ce{H2O2 -> O2 + 2H+ + 2e-} (\Delta E^{\circ} = -0.7) $$

so the net reaction is
$$ \ce{2H2O2 -> O2 + 2H2O} (\Delta E^{\circ} = +1.08) $$

Since $\Delta E^{\circ} $ is positive, this reaction is spontaneous.

The fact that this reaction doesn't happen means that there's a high kinetic barrier. I think that's what A is getting at, though I have a little bit of an issue with how it's worded. Regardless it's still better than B.


Can organomercury compounds synthesize accidentally from household cleaning products?

Unfortunately, you seem to have already broken some of the protocols of cleaning up a broken CFL according to the EPA's website.

According to this site:

  Vacuuming is not recommended unless broken glass remains after all other cleanup steps have been taken.  Vacuuming could spread mercury-containing powder or mercury vapor.

Also according to data from a study of the effect of the breakage of CFLs performed by the Maine Department of Environmental Protection:

  Mercury concentration in the study room air often exceeds the Maine Ambient Air Guideline (MAAG) of 300 nanograms per cubic meter (ng/m3) for some period of time, with short excursions over 25,000 ng/m3, sometimes over 50,000 ng/m3, and possibly over 100,000 ng/m3 from the breakage of a single compact fluorescent lamp.

and also:

  Vacuuming a carpet where a lamp has broken and been visibly cleaned up, even weeks after the cleanup, can elevate the mercury readings over the MAAG in an un-vented room.

At the end of the report, it states that although mercury vapor is extremely toxic:

  ...it is unclear what the exact health risks are from exposure to low levels of elemental mercury...

It seems as though the best course of action you could take now would be to ventilate the room and if possible your house extremely well for several hours, and to avoid vacuuming any more.


Is there an iron–iron bond in diiron nonacarbonyl?

tl;dr: There probably isn’t an Fe–Fe bond. We should probably stop drawing an Fe-Fe bond. Textbooks and lecturers should really stop teaching students that there is an Fe-Fe bond. But all that said, there isn’t one single satisfactory explanation for why there isn’t one…

Despite a lot of studies (and despite $\ce{[Fe2(CO)9]}$ being one of the earliest examples of a molecule with symmetrically bridging carbonyls) , the nature of bonding in $\ce{[Fe2(CO)9]}$ is still a little hard to get a clear picture of.
Whilst at a research level, most academics now accept that there is no significant bonding interaction between the two metal centres, this is not echoed at the undergraduate level where the Fe–Fe bonding model is still often taught.
Most textbooks either avoid the issue entirely (never providing a structure), or show the Fe–Fe bond (with some footnote that it might not be correct), and even those that do acknowledge the lack of the metal-metal bond (Elschenbroich being a good example) don’t really discuss how this works in practice, or how one would go about electron counting (a source of many a nasty tutorial problem).
Historical context
The simple answer is that there was never really any strong evidence for the presence of the Fe–Fe bond. As is often the case, it was adopted by several (well known) chemists at the time, and subsequent publications (and especially textbooks) propagated the idea, largely due to its convenience with respect to electron counting.
The earliest data that really looked at the nature of the Fe–Fe bond in $\ce{[Fe2(CO)9]}$  can be attributed to a crystal structure measured by Powell (2.46 Å)[1](later corrected by Cotton (2.53 Å)).[2]
Although these values didn’t necessarily prove the existence of an Fe–Fe bond (you can’t see a bond with X-ray, only the atoms, from which you can infer bonding), its presence was necessary to fulfil the 18-electron rule, and the values obtained were well in-keeping with other M–M bond lengths known.
The issue with the above argument (as convenient as it is) is that it’s impossible to tell whether the 2.53 Å represents a true bonding interaction or whether it’s simply a consequence of the bridging Fe–C(O)–Fe bonds pulling the metals close together. The iron itself is relatively contracted, so unlike in some other bridged systems, the steric problem of bringing the two metals together is relatively minimal.
The Beginning of Change
Although the Fe–Fe bond was fairly widely adopted, not all were in agreement.
Roald Hoffmann (of Woodward–Hoffmann fame) published several papers which appeared to contradict the presence of an Fe–Fe bond, showing that the interaction was in-fact “antibonding and repulsive”.[3]
Though not fully quantitative (simple Hückel calculations), Hoffmann was able to construct qualitative molecular orbital diagrams using molecular symmetry and isolobal/electronic fragments (from well behaved molecules) to piece together the likely bonding in $\ce{[Fe2(CO)9]}$. To date, his findings are largely still valid (though exact energies have now been calculated with far more accuracy).
Quantitative Computational Studies[#]
Modern computational studies agree that overall, the Fe–Fe interaction is net anti-bonding (though some bonding interactions have been found) with no genuine covalent Fe–Fe bond, as was initially proposed.
In Liddle’s Molecular Metal-Metal Bonds, McGrady provides a qualitative  molecular orbital diagram for the complex. The σ and σ* orbitals, as well as the δπ and δπ* orbitals (sandwiched between the former pair), are fully occupied and provide no net bonding character. However, the population of the Fe–Fe antibonding πδ* orbitals leads to a formal bond order of –2.[4]

This diagram correlates well to the results in one of the earliest computational studies on this system comparing various methods (DFT, MP2, SCF) by Reinhold.[5]
In the study, various properties of the molecules are calculated (M–M bond length, M–C bond lengths, bond angles etc.) and this data is compared to known data from crystal structures/other experimental work. They determine that only DFT gives accurate results (casting some doubt on earlier computational work using less rigorous methods).[##]
The molecular orbital diagram constructed (in agreement with the one from Ref [4] above) crucially shows a negative overlap of the Fe–Fe orbitals is observed, suggesting a direct repulsion. To quote directly from the paper:

in the iron systems, with more contracted metal orbitals, the direct metal–metal repulsion is relatively weak while the metal–bridge bonds are sufficiently strong

Other computational studies have looked to calculate properties of the complex, such as vibrational frequencies of any proposed Fe–Fe bond, or the NMR shieldings that would be present on the ligands with and without the presence of the Fe–Fe bond. This work is referenced in Comprehensive Organometallic Synthesis II (Vol. 6), but doesn’t seem to have lead to very useful results.
Conclusion
The take home message from this is that there really is unlikely to be an Fe–Fe bond in the way it is commonly taught to undergraduates, and certainly not for the purposes of electron counting.
A nice possible explanation is given by McGrady,[4] which provides some explanation for why the 18-electron rule isn’t important in this case (and at a level that an undergraduate could understand):

There are, in fact, many parallels between 34-electron $\ce{[Fe2(CO)9]}$ and square-planar 16-electron species, not least the tendency of both to react with nucleophiles. The breakdown of the 18-electron rule in square-planar complexes occurs because the vacant pz orbital fails to find an appropriate symmetry match among the linear combinations of ligand-based orbitals, and so is not used, either to form a bond or to accommodate metal-based lone pairs of electrons. [...]
In $\ce{[Fe2(CO)9]}$, there are two vacant $\mathrm{a_2''}$-symmetric metal-based combinations (primarily linear combinations of 4s and 4pz) but only one occupied ligand-based combination of the same symmetry. Thus, one of the two metal-based orbitals remains “unused” in the same sense that the pz orbital remains unused in square-planar species. This simple symmetry-based analysis suggests that there is no need to invoke an Fe–Fe bond in $\ce{[Fe2(CO)9]}$ simply because the metal centers do not conform to the 18-electron rule.

References & Notes
[1]: Powell, H.M.; Ewens, R.V.G J. Chem. Soc 1939, 286
[2]: Cotton, F.A.; Troup, J.M. J. Chem. Soc. Dalton Trans. 1974, 800
[3]: Lauher, J.W.; Elian, M.; Summerville, R.H.; Hoffman, R. J. Am. Chem. Soc., 1976, 98, 3219
[#]: Full disclosure, I’ve read most of the papers that will follow at some point whilst trying to answer this in the past, but (big but) I’m not a computational chemist. I’m hoping someone else may be able to shed light on how valid these computational results actually are.
[4]: McGrady, J.E. in Molecular Metal–Metal Bonds; Liddle, S., Eds.; Wiley-VCH: Weinheim, 2015
[5]: Hunstock, E.; Mealli, C.; Jose-Calhorda, M.; Reinhold, J. Inorg. Chem. 1999, 38, 5053. Note that this isn’t a complete geometry optimisation, they start with some known parameters. New J. Chem. 1994, 18, 465 is a paper supposedly attempting the complete optimisation from scratch, but I can’t find it anywhere right now to access it
[##]: Even now there are a lot of issues with accurately calculating properties of organometallic complexes… it takes a vast vast amount of computational time due to the need for basis sets far expanded compared to those developed for simple organic systems


Shakespeare apparently said "What’s in a name? That which we call a rose, by any other word would smell as sweet." Rule no. 1, don't trust the internet until and unless you have some idea about the credentials of the website (and this comes with experience).

The usage of the term transition metal is nothing but historical. You can stick to IUPACs definition (https://goldbook.iupac.org/terms/view/T06456)

  An element whose atom has an incomplete d sub-shell, or which can give rise to cations with an incomplete d sub-shell.

By that zinc should not be a transition element. The definition has nothing to do with paired electrons as you wrote in the comments.

Look at the history of word, this is from 18th century from the unabridged OED.

  transition element  n. Chemistry  [after Russian perexodnyj èlement (see transitional element n. at transitional adj. and n. Compounds)] (originally, now historical) any of the nine metallic elements forming group VIII of Mendeleev's periodic table; (now) any of a large class of metallic elements occupying the central block of the periodic table (groups IV b–VIII, I b, and II b, or 4–12, the d-block), which are characterized by partly filled d orbitals, commonly show variable valency and an ability to form coordination compounds, and form many coloured compounds; often extended to include also those elements having partly filled f orbitals (the lanthanides and actinides).


Oxygen can exist in two states, the singlet and triplet state. The triplet state is the ground state whereas the singlet state is the higher energy state. In the triplet state the highest energy electrons are unpaired whereas in the singlet state the highest energy electrons are paired.
Now, take a look at the MO diagrams of both states in Wikipedia. You'll be able to tell that the singlet state does not obey Hund's first rule.
Now lets look at the Lewis diagram for a oxygen molecule:

As you can see, all of the electrons in this diagram are paired. The oxygen represented in the diagram above is an oxygen that has two bonds and two lone pairs. The only MO state that has lone pairs is the singlet state.
Now on to "why does being paired in the Lewis structure equate to being paired in MO?" The reason is that paired electrons in Lewis diagrams represent paired electrons. The concept of paired electrons does not depend whether you are using MO theory or Lewis diagrams. Paired electrons are electrons that are in the same orbital.
Now, the next question. "Why are the name of MOs like σ and π? Doesn't σ and π get used when talking about σ and π bonds? Are σ orbitals and σ bonds correlated in anyway?" Sigma bonds and sigma orbitals refer to the same thing. Molecular orbitals are a way to represent the bonds formed by two atoms.
Sidenote: Mulliken first proposed the sigma, pi, delta notation according to this journal article..


Is (S)-3-hydroxypalmitoyl-CoA a hydroxy fatty acyl-CoA?

This just looks like an arbitrary distinction they made in the ontology between 'short chain' acyl-CoAs and 'long chain'. As you can see, the 3-hydroxypalmitoyl chain is somewhat longer (16C) than the others. It stands to reason that the creators of the ontology have decided to subdivide hydroxy fatty acyl-CoAs into those with shorter hydroxyl fatty chains:

hydroxyoctanoyl:   8C
hydroxydecanoyl:  10C
hydroxylauroyl:   12C

and long:

hydroxypalmitoyl: 16C

As to where the dividing line is, the hydroxytetradecanoyl-CoA (14C) homologue is not 'long chain', whilst there is no entry for hydroxypentadecanoyl-CoA. The octadecanoyl homologue belongs to its own class (of 3-hydroxyoctadecanoyl-CoAs, of which there are only two conceivable enantiomeric members), and there does not appear to be icosanoyl homologue in the database. There may be other members of the 'long chain' family if you search around, who knows?

In short, it appears to be arbitrary.


Well, tetrahydrocannabinol, you said it correct only the correct reason/explanation is hyperconjugation only.
More the number of (alpha) hydrogen atoms on a C=C [isolated as in alkenes or in resonance as in benzene], more the possibility of hyperconjugation.

In benzene, this will shift the electron density from C-H bond of CH3 group towards the ring, making it electron rich.

Contrary to what could have been thought from +I effect of corresponding alkyl groups on the ring, hyperconjugation is more in CH3.

Since hyperconjugation involves complete shifting of electron from valence shell of atoms to the other, it is more "effective" that INDUCTIVE which just pushes electron little more towards more EN (more correctly, towards group with more -I) atom.


Which diagram shows the effect of catalysis on chemical equilibrium?

If (as suggested in the comments) the y-axis is supposed to represent concentration, such that the upper curve is "concentration of $\mathrm{A+B}$" and the lower curve is "concentration of $\mathrm{C+D}$", then Figure 1 is correct, albeit horribly mislabeled.

Your Figure 2 is correct if we want the y-axis to represent a rate-of-conversion instead, i.e., if the upper curve is "number of parcels of $\mathrm{A+B}$ being converted to $\mathrm{C+D}$ per second", and the lower curve is vice-versa. As you point out, a catalyst increases the conversion rate, and does so in both directions. So, exactly as you have drawn in Figure 2, those catalyzed rate curves should both be higher than their uncatalyzed siblings.

("Rate of reaction" is an awfully ambiguous term here – since at equilibrium, the rate of reaction must be zero!)


Why do atoms generally become smaller as one moves left to right across a period?

As you move from left to right across a period, the number of protons in the nucleus increases.  The electrons are thus attracted to the nucleus more strongly, and the atomic radius is smaller (this attraction is much stronger than the relatively weak repulsion between electrons).

As you move down a column, there are more protons, but there are also more complete energy levels below the valence electrons.  These lower energy levels shield the valence electrons from the attractive effects of the atom's nucleus, so the atomic radius gets larger.


Does conductivity change in a solution due to rust?

This is an interesting question. As iron rusts, you will have a couple of reactions. First, as you expect is the oxidation of iron:

$\ce{Fe ->Fe^2+ + 2e^-}$ or $\ce{Fe ->Fe^3+ + 3e^-}$

Then, either one of these may precipitate as their oxide, or hydroxide or their chloride.

On the flip side, somewhere else on the same piece of nail, there has to be a reduction. This may be one of:

$\ce {2H+ +2e- -> H2} $ or $\ce{2O2 + 4e- +4H+ ->2H2O}$

The overall reaction will probably be a combination of the above depending on the various rates of reaction. A priori, I would say it is very hard to predict which mechanism would be operative and therefore it is hard to predict which way conductivity would go.

Another point is even if the overall mechanism was such that there was some imbalance that would affect the conductivity, that would be very hard to measure. The changes would be a very small fraction of the total conductivity which would make it a hard measurement.

The typical way this is measured is "weight loss", which involves cleaning the corrosion products off the metal specimen and measuring the weight after given time periods. You can find lots of cleaning procedures and details of weight loss measurements over the net.


Why only the unstable resonating structure proceeds in the reaction?

Imagine a case where the resonance form with the negative charge localized on the oxygen went forward and attacked the second carbonyl:  That charge needs to go somewhere so the double bond is broken and the electrons go up to the second oxygen (the one that started as the second carbonyl). Those electrons would look to come back and reform the double bond, and in doing so would force one of the groups to leave. And the bond that leaves is the one you just made. This reaction happens with both atoms acting as a nucleophile except only one (the carbanion) goes forward and doesn't immediately leave again.


Correlation between 1H and 13C shifts - coincidence or not?

Maybe it would help to comprehend the coincidental changes in 1H and 13C chemical shift by understanding the origin of chemical shift. In both cases, very crudely speaking, the influence of ring currents and electronegative elements/groups cause a downfield (away from zero) shift. The factor of 20 quoted is a function of the number of electrons (usually outer ones) involved in forming the bond. So the trend will also be true for other elements e.g. 15N where aromatic N compounds i.e. pyridine have a larger downfield shift.


Acetic(vinegar) and citric acid as the rust remover

To answer your first question, no, you would mainly just get dilution of the stronger acid (citric) with the weaker (acetic). If you used a citrus fruit as the source of citric acid, it's important to note that there is a fair bit of variation in acidity from one citrus juice to another. The best bet would be lemon or lime juice, which have among the lowest pKa values (measurement of acidity, lower is more acidic) of the common citrus juices.  If you just used something like orange juice, you might notice significant improvement by switching to lemon or lime juice.

Your second question probably is the right way to go. Rust is fairly soluble in dilute solutions of strong acids like hydrochloric acid (the acids you have used so far are weak acids).  You can get consumer-grade hydrochloric acid from a number of sources, the best probably being an aquarium or pond shop. These products usually have names like "pH down" or something indicating that they are acidic. What you want to do is look at the ingredients for a product that is just dilute hydrochloric acid, without a bunch of other stuff. Be sure to read and follow all of the precautions on the container, particularly with respect to wearing eye protection.

Another alternative source for dilute hydrochloric acid would be rust-cleaning products from a hardware store. These typically have a detergent of some sort mixed in with the acid, which is OK, and this may actually be your best solution depending on availability where you live.


Why are Friedel Crafts reaction not possible for molecules less activated than benzene?

Technically, the teacher's claim might be slightly modified.  Deactivated benzenoid aromatics do fail to undergo Friedel-Crafts reactions, but according to a summary description from the University of Calgary:

Deactivated benzenes are not reactive to Friedel-Crafts conditions, the benzene needs to be as or more reactive than a mono-halobenzene.

So, a monohalogenated ring which is only weakly deactivated, can still react.  However, stronger deactivation, for example with an acyl group, would kill the F-C reaction.
This property can actually be quite useful when we need to do the F-C reaction selectively.   Using acyl as the deactivating group, we can acylate an aromatic ring and be sure the reaction occurs only once.  The ring becomes too deactivated to react further.  In principle this can then lead to a monoalkylated product with good yield by subsequently reducing the acyl group, whereas direct alkykation which activates the ring could give polyalkylated products.


Conductance of the mixture of two solutions with given resistance values

There is an excellent treatment of electrolytic conductivity in Wikipedia. However, the OP’s main question involved a specific equation, pertaining to the conductance of a mixture of equal volumes of two electrolyte solutions, and this was not specifically discussed in the Wikipedia article:

$$\text{conductance} = \frac{1}{2}\left(\frac{1}{R_{1}} + \frac{1}{R_{2}}\right)$$

For simplicity, assume the electrolyte concentrations in the two solutions are in the “Goldilocks” zone: not so high that activities and fancy theory are needed, but not so low that the conductivity contribution due to water’s autoionization has to be taken into account. Neither solution is concentrated, so the ions just bounce around like they would in the unmixed solutions. This would certainly be incorrect in relatively concentrated solutions, but that is not the case at hand.

In this idealized scenario, a solution’s conductivity, i.e., specific conductance, is directly proportional to the mobile charge carrier (i.e., ion) concentrations in the solution. The conductance of a solution, denoted by $G,$ is directly proportional to the solution’s conductivity. The resistance of the solution, denoted by $R,$ is simply the reciprocal of the conductance and, likewise, the resistivity, i.e., specific resistance, is the reciprocal of the conductivity.

Now consider two electrolyte solutions that can be mixed without causing any chemical reaction. In particular, no precipitate forms, no gas is evolved, no complexes form, and so on. Measurements on the first solution provide values of $G_1$ and $R_1$, while measurements on the second solution provide values of $G_2$ and $R_2$. If equal volumes of the two solutions were well mixed, and a conductivity measurement was performed on the mixed solution, the conductivity of the mixed solution, $G_\mathrm{mix}$, would be

$$G_\mathrm{mix} = \frac{G_1}{2} + \frac{G_2}{2} = \frac{1}{2}(G_1 + G_2)$$

since each solution, when mixed with an equal volume of the other solution, had its constituent ion initial concentrations halved. But the total conductance is just the sum of the two conductances, $G_1 = 1/R_1$, and $G_2 = 1/R_2$, so

$$G_\mathrm{mix} = G_\mathrm{tot} = \frac{G_1}{2} + \frac{G_2}{2} = \frac{1}{2}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$

Like all approximations, this one is only as good as the foundational assumptions upon which it is predicated.


Can there be two IUPAC names of a compound?

Generally, since many compounds can have two or more names in accordance with several methods recommended by IUPAC, a compound may be named correctly in more ways than one. Therefore, current IUPAC recommendations include the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including the many retained names.

In particular for primary amines, the rules given in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) are as follows.

  P-62.2.1.2 Primary amines, $\ce{R-NH2}$, are systematically named in the following ways:

  (1) by adding the suffix ‘amine’ to the name of the parent hydride;

  (2) by adding the name of the substituent group $\ce{R-{}}$ to the parent hydride ‘azane’;

  (3) by adding the name of the substituent group $\ce{R-{}}$ to the term ‘amine’ used as a preselected parent hydride name for NH3; this method is used only with monoamines.

  (…)

  Method (1) leads to preferred IUPAC names.

Since, according to the current recommendations, the substituent group $\ce{(CH3)2CH-{}}$ can be named ‘propan-2-yl’, ‘isopropyl’, or ‘1-methylethyl’ (where ‘propan-2-yl’ is the preferred prefix), the compound given in the question can have the following seven names in accordance with methods recommended by IUPAC. The PIN is ‘propan-2-amine’.

Method (1):
propan-2-amine (PIN)

Method (2):
propan-2-ylazane (propan-2-yl is the preferred prefix)
(1-methylethyl)azane
isopropylazane

Method (3):
propan-2-ylamine (propan-2-yl is the preferred prefix)
1-methylethylamine
isopropylamine