Is it true that human cells grow in levorotation?

I remember reading this in a popular science magazine, some 20 years ago: that in most of the living organisms, cells grow in levorotation and very few grow in dextrorotation.

There is a kernel of truth in what you wrote, but either the science fiction magazine or your memory has distorted things a fair bit.  Cells do not grow "in levorotation" or "in dextrorotation".  Chirality is not a property of cells, it is a property of the individual molecules that comprise the cell.  Many different types of molecules are essential for life and are found in all living cells.

Two examples of these ubiquitous, essential molecules are carbohydrates and amino acids.  In all organisms on Earth, most carbohydrates are found in what is known as the D configuration.  The "D" stands for "dextrorotatory" but the nomenclature is confusing.  For carbohydrates, the configuration is defined as relative to glyceraldehyde, a simple sugar.  Thus D-fructose or D-glucose have the same configuration as D-glyceraldehyde, but only D-glyceraldehyde is guaranteed to be dextrorotatory.

Amino acids in life on Earth are predominatly "L", meaning they share an absolute configuration with L-glyceraldehyde.

  Is that true, at least for humans?

So humans, like most life on Earth, is made of a mixture of levorotatory and dextrorotatory molecules.  Some types of the molecules in humans are predominantly dextrorotatory and some are levorotatory.


The difficulty with creating an ionic bond glass is that creating true glasses depends on a combination of directional bonding between atoms (or molecules) and a rate of cooling that doesn't give those bonds time to move into optimal positions. The result is a bit like piling together sticky spheres that become locked before they can find an optimal stacking arrangement.

Pure ionic bonds in contrast are more like slippery spheres that are attracted towards each others' centers, but don't have any particular preferences about which parts of their surfaces make contact.

Given that, the most likely result by far from very rapid cooling of a molten ionic solid will be to produce micro or nanocrystalline ionic solids that may superficially resemble glasses, but which under X-ray or neutron diffraction crystallography would prove actually to have crystal structure at a very fine scale.

With that said, you nonetheless can come up with a recipe for trying to make a truly glassy ionic solid, especially since there's no such thing as a pure ionic bond. You would want to cool a very thin layer of molten salt at at an incredibly high rate, at least millions of degrees per second if I recall glassy metal cooling rates rightly. For ionic, you would also want the result to be pretty close to absolute zero, since whatever degree of "directional stickiness" you can get out of ionic solids is likely to be less even than metallic bonds.

Finally, if you really want to try to make such a thing as a serious project, you of course must check the literature in detail. My rule of thumb is that if you dig deep enough, there are almost no decently plausible ideas that you can think of in the physical sciences that someone has not at least written a paper on in some journal, and maybe even done some experiments.

Still, what you are asking is an intriguing idea for a real experiment, and it wasn't that many years ago that both metallic glasses and Penrose-tiled quasicrystals were new ideas.


How to quantify the adsorption affinity of gases?

The term you are looking for is "adsorption enthalpy". However calculating that value will depend on a variety of factors mainly the absorbate and substrate, but it is not that simple. The surface structure of substrates matters and a $[111]$ oriented crystallographic face will have a different adsorption enthalpy with a given absorbate that say a $[001]$ crystallographic face.

I suggest you study this more, namely the Langmuir adsorption model and Brunauer–Emmett–Teller Theory (BET). Especially since BET is a popular technique to measure adsorption.


The answer hings on the colligative properties of solutes. In essence solutes lower the vapor pressure of the solvent. NaCl splits into two ions (i.e. two solute "particles") and lowers the vapor pressure of water more than urea for which the whole molecule dissolves as one solute "particle."

If there was no liquid the Hg level would be higher than any of the three tubes shown since there would be a pure vacuum above the Hg.

The scale of the relative differences in drawing is obviously greatly exaggerated.


For Acetone or IPA we are more than happy to keep it in normal squeeze bottles which are not air tight so the bottle you've got should be fine. I believe you're more likely to see it evaporate first than to see it get 'noticeably weaker' (I assume you're using technical grade acetone since it's just for cleaning). There should be no need to dump the acetone/IPA in the bottles on a weekly basis - although acetone is cheap its unnecessary to waste money like that!

Along the lines of trying to figure out what could be the cause of the experience of your staff, I doubt it has anything to do with the acetone itself because you say that they can use the same chemical on two different days and have two different experiences. So it's either they're imagining it (let's assume they are not), or it is the dirt that has to be cleaned that is the issue.

I've experienced it in the lab before that sometimes it does seem more difficult to wash out and clean glassware despite using the same chemicals - and this usually has something to do with how long the 'dirt' has been sitting in my flask or how it got there in the first place (slow precipitation/crystallization, crashed out, solvent dried out and left overnight, etc.). Without knowing what your product is and what it is you're trying to clean out, I can't offer much else other than checking out the consistency of the incoming product that needs to be cleaned. It could be a sign that the manufacturer's process is contaminated on some days.


Energetic stability in relation to α-hydrogens

Apparently the intent of the question is to assess the stability of the three alkenes. Evaluation of the heats of formations provides useful information. The NIST site provides heats of formation and hydrogenation all of which are shown in black in the diagram. Computed values appear in red. Because hydrogenation of A, B and C all provide ethylcyclopentane, the greater the heat of hydrogenation the less stable the alkene. Alternatively, the less negative the heat of formation, the less stable. Thus, the order of stability is A < C < B.


How to calculate the pressure of an ideal gas that undergoes isothermal expansion?

I am going to use the physics conventions of work according to which, work done by the gas is positive. So if you want the answers as per chemistry conventions than just flip the sign of the work done, keeping the magnitude same.
Part (a): Reversible Isothermal Expansion
\begin{align}
\Delta U_a &=nC_v\Delta T\\
\Delta U_a &=\pu{0 J}\\
\end{align}
\begin{align}
\Delta H_a &=nC_p\Delta T\\
\Delta H_a &=\pu{0 J}\\
\end{align}
\begin{align}
W_a&= \int_{V_i}^{V_f} P.dV\\
W_a&= \int_{V_i}^{V_f}  \frac{nRT}{V}.dV\\
W_a&= nRT\ln\frac{V_f}{V_i}\\
W_a&= \pu{2.35 \times 8.314 \times 300 \times \ln3 J}\\
W_a&= \pu{6439.37 J}\\
\end{align}
\begin{align}
q_a&= W_a+ \Delta U_a\\
q_a&= \pu{(6439.37 + 0)J}\\
q_a&= \pu{6439.37 J}\\
\end{align}
By Boyle's Law,
\begin{align}
P_iV_i&=P_fV_f\\
P_f&=P_i\times \frac{V_i}{V_f}\\
P_f&= \pu{\frac{1750}{3} kPa}\\
P_f&= \pu{583.33 kPa}\\
\end{align}
Part (b): Isochoric Heating**
\begin{align}
\Delta U_b &=nC_v\Delta T\\
\Delta U_b &=\pu{2.35 \times 1.5 \times 8.314 \times 78 J}\\
\Delta U_b &=\pu{2285.93 J}\\
\end{align}
\begin{align}
\Delta H_b &=nC_p\Delta T\\
\Delta H_b &=\pu{2.35 \times 2.5 \times 8.314 \times 78 J}\\
\Delta H_b &=\pu{3809.89 J}\\
\end{align}
\begin{align}
W_b&= \int_{V_i}^{V_f} P.dV\\
W_b&= \pu{0 J}\\
\end{align}
\begin{align}
q_b&= W_b+ \Delta U_b\\
q_b &= \pu{(0 + 2285.93)J}\\
q_b&= \pu{2285.93 J}\\
\end{align}
By Gay-Lussac's Law,
\begin{align}
\frac{P_i}{T_i}&=\frac{P_f}{T_f}\\
P_f&=P_i\times \frac{T_f}{T_i}\\
P_f&= \pu{\frac{1750}{3} \times \frac{378}{300} kPa}\\
P_f&= \pu{735 kPa}\\
\end{align}
Overall Process:
\begin{align}
\Delta U&= \Delta U_a+ \Delta U_b\\
\Delta U &= \pu{(0 + 2285.93)J}\\
\Delta U&= \pu{2285.93 J}\\
\end{align}
\begin{align}
\Delta H&= \Delta H_a+ \Delta H_b\\
\Delta H&= \pu{(0 + 3809.89)J}\\
\Delta H&= \pu{3809.89 J}\\
\end{align}
\begin{align}
W&= W_a+ W_b\\
W&= \pu{(6439.37 + 0)J}\\
W&= \pu{6439.37 J}\\
\end{align}
\begin{align}
q&= q_a+ q_b\\
q&= \pu{(6439.37 + 2285.93)J}\\
q&= \pu{8725.30 J}\\
\end{align}


How does a lone pair of a central atom affect the dipole moment?

How does lone pair of a central atom affect the dipole moment?

There is no single answer to your question, let me explain.  Unlike a typical covalent bond where the electrons are shared between two nuclei and the electron density is spread out over the entire bond, in a lone pair the electrons are not shared and the electron density is more localized around the atom that has the lone pair of electrons.  This increased electron density could lead to a more significant contribution from the lone pair electrons to the molecular dipole moment than from electrons spread out more diffusely in a covalent bond.  Next we must understand the directionality of the lone pair of electrons.  Consider the two molecules pictured below, ammonia and phosphine.  The molecules appear to be very similar, they are in the same column in the Periodic Table.

However in ammonia the $\ce{H-N-H}$ angle is around 107 degrees and the molecule is roughly $\ce{sp^3}$ hybridized, the lone pair and the 3 $\ce{N-H}$ bonds roughly pointing towards the corners of a tetrahedron.  You can see that in this case (as shown by the arrows, the "arrowhead" end representing the negative end of a dipole), the lone pair on nitrogen will make a contribution to the molecular dipole moment.  Next, let's examine phosphine.  The $\ce{H-P-H}$ angle is around 90 degrees and the molecule can be viewed as being unhybridized, the lone pair is an $\ce{s}$ orbital and the 3 $\ce{P-H}$ bonds are constructed from phosphorous $\ce{p}$ orbitals.  You can see that in this case, the lone pair on phosphorous, due to its spherical symmetry will not make a contribution to the overall molecular dipole moment.

So in summary, a lone pair of electrons can make a significant contribution to the magnitude of a molecular dipole moment due to the fact that they are more localized than bonding electrons and consequently there is a high electron density.  But, the directionality (or lack thereof) of the lone pair must also be assessed, since a lack of directionality may preclude it from making a significant contribution to the overall molecular dipole moment.


How can I determine the boiling point of a compound using simple distillation?

Do you mean that the substance could be pure methanol, ethanol or 1-propanol?

In that case there is a simple way to distinguish between the four options. A pure substance will have a T-V curve that consists of 2 straight lines, 1 running from $T=T_{start}$ to $T=T_{b}$ at $V=0$ and one running from $V=0$ to $V=V_{total}$ at $T=T_{b}$. Probably you will have some impurities which cause some small deviations from this perfect hook, but that will be the general shape. Determining the boiling temperature is just a matter of checking where the curve goes straight up. Depending on the temperature at which this happens you can identify methanol ($T_{b}\approx65$), ethanol ($T_{b}\approx78$) or 1-propanol ($T_{b}\approx97$).

The azeotropic mixture makes it a bit tricky, because the boiling points of 1-propanol and water are so close together, but the curve should show a distinct difference not going vertically up at a certain temperature.


Yes, you are right. One electron from 3p orbital is unpaired and excited to a 3d orbital. The hybridization of sulfur atom is sp2 hence a lone pair and two bond pairs(due to sigma bonding) reside in these hybrid orbitals.

The unpaired electrons are 3p and 3d hybridized orbitals are used in pi bonding with oxygen's unhybridized 2p orbitals. Hence two pi bonds are formed which are p-p pi and d-p pi bonds. Because of delocalisation these bonds are identical and have properties intermediate to the two different pi bonds.

To understand how the electron is able to make a jump with high energy difference, you will have to understand that d orbitals usually participate in bonding only when central atom is joined to many highly electronegative atoms. Because of polarization, there is a slight positive charge on the sulfur atom due to which orbital contract. This contraction is very very significant in d orbitals. Because of this electrons can be excited to the less energetic and tightly held d orbital. Also, the bonds formed will have a greater extent of overlap because d orbitals are not that diffused anymore. This is the reason that PF5 exists but PH5 does not.

Also, note that we cannot use dz^2 orbital for pi bonds as its lobes have the same phase and hence net overlap during pi bond will become zero.


Cleaning silver - which method is better?

Abrading the silver sulfide removes it (and some silver underneath the tarnish). Electrolytic cleaning actually restores the silver previously part of the tarnish. In addition, it cleans crevices inaccessible to manual polishing.

There are disadvantages, though:

If the article has niello inlay, it could be damaged.
If other materials, such as parchment or glue, are part of the article, they would be damaged.
If the article is a valuable antique, cleaning may reduce its value.


I think the most intuitive way to think about this is rates of reaction. A reaction involving 2 molecules in the gas phase only happens when they collide. At higher pressures there will be more collisions between molecules.

This means that the reaction with the most molecules will be sped up more, so if a reaction has 2 molecules reacting to give 1 molecule of product, the forward reaction will be sped up more than the reverse reaction by increasing the pressure.

To do the kinetic calculations properly will require all the mechanistic steps involved. The general principle is that that at equilibrium, both reactions are happening at the same rate, but when you increase the pressure you will speed up the reaction which has more moles of gas, resulting in that reaction happening faster than the other reaction, shifting the equilibrium until the rates of the two reactions are balanced again.


How to know when electrons will be added onto Lewis Structure from surrounding atoms to become ion?

I think I know where your confusion comes from. Firstly, I would want to note an important point:

Covalent bonds consist of pairs of electrons shared by two atoms, and bind the atoms in a fixed orientation. Chemwiki.UCDavis

This is the characteristic that makes them different from ionic bonds. The surrounding atom will not become an ion in its full sense.
I would want to describe a way of drawing Lewis electron dot structures that will avoid any confusions.
$\ce{H2O}$:
We have two Hydrogen atoms and an Oxygen atom to begin with. They have this lewis structure:

Source 1
Follow these instructions for "normal" molecules (the ones that abide octet in a perfect sense):

Compute all of the electrons that are in the valence shells of separate atoms. For instance, an O atom has 6 es in the valence shell and each of H atoms have one. So you get: $$6+(2*1)=8$$
I prefer to call this $\text{n}_v$. V represents valence shell.
Compute all of the electrons the atoms will have after the compound has been formed. That is, when all of them abide octet. In the instance of water, O will have eight (just like other octet abiding atoms) and H atoms, as they are exceptional, will have two. So you get:
$$8+2*2=12$$
I prefer to call this $\text{n}_f$. F represents "final". You can name it whatever you want.
If you want to know how many electrons are bonding and are shared between atoms, subtract $\text{n}_v$ from $\text{n}_f$:
$$\text{n}_f - \text{n}_v \rightarrow 12-8=4$$
For an ease in calculation, I call this $\text{n}_b$. As you might have just figured out,
If you want to know how many es are nonbonding and "belong" to a single atom, subtract $\text{n}_b$ from $\text{n}_v$ to get:
$$\text{n}_v - \text{n}_b \rightarrow 8-4=4$$
Call this whatever you want. I rather calling it $\text{n}_{nb}$.

With the information you just gathered you can draw the lewis structure for most of the covalent compounds. In the case of water, 4 bonding es mean two covalent bonds, right? Plus, H has no other electrons than it's only one. So O has 4 nonbonding electrons. Thus, you may come to something like this:

Source 2
$\ce{NH4+}$
In ions, we have to take the charge of the ion into an account. Positive charge means lesser electrons than a neutral species, and negative charge means more es in a comparison with a neutral covalent compound. Thus, we should add to or subtract from $\text{n}_v$.
So let's get on with this thing's lewis structure:

$\text{n}_v = 5+(4*1) -1=8$ (Notice (-1) that was because of a 1+ ion charge.)
$\text{n}_f = 8 + (4*2)=16$
$\text{n}_b = 16-8=8$
$\text{n}_{nb} = 8-8=0$

There is an important thing to note here. Nitrogen has 5 es in valence shell, but ends up having four bonds. This is because an H atom had lost it's electron to become n ion AND, with an empty orbital. So, N shares two electrons instead of one which is described as dative covalent bonding. We get this structure in accordance to the calculations and thoughts:

Source 4
Finally, $\ce{NH2-}$
Now this case is easily solvable through the explanations for ammonium.

$\text{n}_v = 5 + 2 +1=8$ The +1 is because of the charge of the ion.
$\text{n}_f = 8 +2*2=12$
$\text{n}_b = 12-8=4$
$\text{n}_{nb} = 8-4=4$

Put some thinking, and you should get this (ignore the other ones if you like):

Source 5


What is chemical decomposition in the context of a crystal or amorphous solid?

When it comes to non-molecular solids such as glass, metal or diamond, "a single entity" is in fact the formula unit, the concept identical to a "molecule" for, well, molecular compounds. Both formula unit and molecule are the smallest portion of a substance that preserves its chemical properties. As long as the formula unit remains the same, there is no chemical decomposition to speak of.

However, this doesn't mean that there is no chemical processes that can be initiated mechanically. An entire branch, mechanochemistry deals with them. As for laser ablation, I think you are basically introducing a secondary phase, a thin film of another substance (for example, metal oxide) on the surface.

One more practical example I can think of: for a single-crystal diffraction experiment one has to use a 3D crystal (in a sense that it shouldn't be a 2D plate for the proper absorpton correction) with linear dimensions not exceeding fractions of a millimeter. For the probe preparation a crystallographer often has to literally cut the crystals with a special scalpel under the microscope. However, the resulting crystal structure doesn't depend on how many cuts and in what direction the crystallographer did. The x-rays though, can alter the structure of some sensitive molecular crystals (often biological samples, proteins), but this is another story.


What are the possible magnetic quantum numbers of silver?

Silver has electronic configuration as [Kr] 4d10 5s1 .
The s orbital ( 5s ) has only one electron. And the s orbital here has n = 5 (as it is in the fifth period) and l = 0 .
The s orbital is symmetric in space (sphere) which has only one orientation and hence has no "different" orientations  each of which is represented by a particular value of ml. So ml can take only one value and that is 0.( If this was a p orbital which could take 3 different orientations, ml would take 3 values -> -1,0,1. In general an orbital with azimuthal quantum number l, ml would take 2l+1 values of ml  ranging from -l to +l )
The quantum number ms represents the spin. It can either be +1/2 or -1/2.
Hence ,  ml =  0 amd ms = +1/2 or -1/2.
Hope this helps!


The absolute values of enthalpy H at a single state point are meaningless. It is only the difference between two different state points that matter. Thus, the value for a single state point can be set to any arbitrary value. Many handbooks set the arbitrary state point so that the values of these properties are positive for most liquid or gas states.

Apparently, WolframAlpha is setting enthalpy H to zero for the saturated liquid at the normal boiling point (NBP) for acetone but not for water and ethanol.

For ethanol, WolframAlpha is setting specific enthalpy to h = 200 kJ/kg for the saturated liquid at 0 °C (IIR reference state).

For water, the triple point is selected as the reference state, where the internal energy U of saturated liquid is assigned a zero value. Accordingly, the enthalpy H = U + pV of liquid water at the triple point is slightly larger than zero, at a specific enthalpy of about h = 0.6115 J/kg.


How are clathrate hydrates formed in the lab?

I don't know about some of the clathrate hydrates you mentioned, but I've recently been reading some papers about both bromine and chlorine clathrate hydrates, and I imagine the procedure would be very similar for the systems you've mentioned.

I'll summarize from Spectroscopic Signatures of Halogens in Clathrate Hydrate Cages. 1. Bromine.

Basically, they're procedure is as follows. Get some really pure water and really pure bromine solution. Make a solution of bromine and water and place it in a quartz cell (cuvette). Place dry ice above the solution and the solution will begin to crystallize on the quartz cell wall. After it is clear that the solution has condensed/frozen onto the the side of the wall, place the cell into around $3^\circ C$ water. This is because at this temperature, both water and bromine will be liquid, but a bromine clathrate hydrate will be solid.

Thus, one can isolate the crystal by simply melting off the water and bromine which are not in a clathrate hydrate cage-type structure.

I believe this same sort of procedure could easily be applied to many of the other guest molecules you mention, however, because in general clathrate hydrates are going to have a higher melting point than ice. For instance, carbon dioxide clathrate melts at about $10^\circ C$ according to wikipedia.

I would suspect that methane clathrates and other clathrates with more weakly interacting guests would melt at a temperature very close to ice, so the procedure might be more complicated or sensitive. Indeed, that paper I cited above make some bromine-doped clathrates of dichloromethane, which is fairly insoluble in water, so a slightly different procedure is chosen.

One thing I find is that you can make these clathrate hydrates at around the melting point of ice, but then people frequently cool the clathrates down to very low temperatures to conserve the structures and keep the gases trapped in the cage, not necessarily because the this is needed in the formation of the clathrates.

After reading a bit more, I believe that a nearly identical procedure can be used for weakly interacting gases like methane, but the process is done under high pressure of the gas desired inside the cage which obviously shifts around the temperatures but doesn't change the idea presented above.


On Electrochemical Cells and Salt Bridges

You have it all backwards. Liquid-liquid junction does not exist at all in your supposed meaning. Solid-liquid junction does, but it is very much not what you think it is, to the point that you might just as well assume it does not exist either.

The complete story would hardly fit in an answer (and, moreover, it is explained in any decent textbook on the subject), so I'll focus on the details that are especially wrong. First of all, the double electric layer, being electrically charged, creates an electric field of its own, and thus shields the rest of the solution from the electrodes and their field. Second, the electric forces are really, really strong (like a thousand times stronger than you seem to think, and even stronger than that), and therefore the said layer at any point contains but a very tiny portion of metal ions. That portion suffices for creating an electric field which would completely negate the field of electrodes. The vast majority of metal ions float freely inside the solution, having no idea at all that they are being electrolyzed.

As the ions build up, the shielding becomes ever more perfect. Dissolve some $\ce{NaCl}$ in a glass of water without any electrodes or anything; the ions would float freely in the solution. Do they feel mutual attraction? Well, in a way, they do. Do they assemble into strange walls across the glass? I've never seen anything like that. Now, the ions in you cell (the majority that does not participate in the double layer on either side) feel exactly the same way as ions in that glass.

Electrochemical cell, if set up properly, will give current until one of the electrodes is completely eaten up. The need for a salt bridge has absolutely nothing to do with electric effects. Rather, it serves to prevent the mixing of solutions, which are presumably different and better left that way, for fear that otherwise they might react with the electrodes directly.

As for the cell with separate spaces and no bridge, it will stop working right away, much like the light bulb connected with a single wire to one pole of a battery.


Copper substituting silver in Tollen's reagent

For the particular reaction where you get a silver mirror with an aldehyde, I doubt it'll work for copper, due to the reduction potentials. I'll take acetaldehyde $\ce{<->}$ acetate as an example here, but I believe that the reduction potentials of most aldehydes will be similar. The nitrate is a spectator ion here, so we need not consider its presence.

$$E^o_{\ce{Ag^+,Ag}}=+0.80$$
$$E^o_{\ce{Cu^{2+},Cu}}=+0.34$$

$$E^o_{\ce{CH3COO^{-},CH3CHO}}=+0.580$$

Our reaction involves the metal ($\ce{Ag}$ or $\ce{Cu}$ being reduced, so it must have a higher reduction potential than the acetate/acetaldehyde pair. Of the two metals, only silver has a higher SRP, thus the reaction involving silver nitrate is spontaneous--whereas the copper nitrate reaction is  infeasible.

Of course, you could find some substrate other than an aldehyde which reacts with copper nitrate to achieve the effect you want--though I'll have to look up some stuff to figure this out. I think the Tollen's reaction is a single step everyday redox reaction, which means that we just need to look for a stronger reducing agent which doesn't harm the glass.

Actually, it looks like the standard process for plating nonconductors with metals is to give them a conducting layer by dipping them in something, and then using electroplating. Not sure about this, though.


Why doesn't water wash off the coronavirus alone?

Because your hands are covered in a thick crust of dirt, dust and fats, in which the virus can hide.

But anyway still nobody has been able to show that this virus is regularily (or even rarely) transmitted via smear infection. (If someone knows about any evidence to the contrary, please come forward. And yes, some other diseases, especially bacterial infections, do even mainly take this route. That´s how you usually get the "shits".)

It can be detected on surfaces after hours via PCR, however PCR is as sensitive to broken down pieces of a virus particle as it is to the fresh, contagious specimen. Obviously the RNA pieces also of a destroyed virus would not evaporate, so the result of these investigations proves nothing but the extremely obvious. Newspapers are spreading it as a piece of clickbait and tinder, causing people to buy lots of "sanitizers".

Anyway, use soap, leave the disinfectants to the health care people who absolutely, direly need them, and be done.


Precipitation of AgCl from the tap water solution of the group 2 chloride

Yes the amount of $\ce{AgCl}$ formed will be more due to more $\ce{Cl-}$ ions present in the tap water. They are mainly because of some salts already present in water, and due to chlorination.


Why does naphthalene have a higher melting point than both water and diphenylamine?

In comparisons such as this water needs to be excluded as it is exceptional in almost every respect due to its highly directional and strong hydrogen bonds.

As a broad generalisation the melting of solids is more dependant on the nature of the repulsive forces between molecules than on their attractive ones. Experimentally the enthalpy of melting ( Latent Heat or just Heat of Fusion) is approximately ten times less than the corresponding heat of vaporisation which indicates that repulsive forces play a greater role in melting than in vaporisation: in a solid molecules tend to be slightly closer together than in a liquid, thus may experience larger repulsive forces. Van-der-Waals forces tend to have weak orientational dependence, but repulsive forces do not as they are short range and  depend on the asymmetric shape of the molecules and can have a large effect.

Thus melting points are rather dependent on the geometry of the molecules, if they can pack well into a lattice then a high melting point is expected. If the molecule has a dipole this can cause both attractive and repulsive interactions depending on the relative positioning of molecules; head to tail can be attractive if molecules are face to face, but repulsive if in line one behind the other and vice versa if their relative orientation is changed.

In the case of naphthalene and the amine, clearly the naphthalene can stack nicely and so increase its attractive interaction relative to repulsive ones, whereas the amine is twisted and cannot so easily pack meaning that repulsion can be stronger and attractive interactions weaker (even with dipole) and melting points lower. As an example consider the  $\ce{C6H14}$ isomers, n-hexane mp -95 C, 2-methyl-pentane mp -154 C but boiling points are 69 & 60 C respectively and $\ce{C8H18}$ isomers, n-octane mp -57, 2-me-heptane mp -120 and bp 126 & 119 C respectively.

( Attractive forces largely determine the enthalpy of vaporisation, and by Trouton's rule their boiling points. The enthalpy of vaporisation is closely related to the (attractive) cohesive energy in a liquid.)


Gibbs free energy can be expressed as a function of P,T and n but are enthalpy and internal energy also (partially) functions of n?

Yes, you can define chemical potentials as partial molar enthalpies, or partial molar internal energies, or partial molar Helmholz free energies.  (Sorry for my initial confusion on this point ... see also my answer on your other question on this topic, here).

The definitions as partial molar enthalpies and partial molar internal energies are not particularly useful, since they involve working at constant entropy, which is practically impossible, but they are still mathematically valid.

The definition of chemical potential as partial molar Helmholz free energy at constant temperature and volume may actually have some practical applications, (e,g. perhaps for reactions that evolve gases in closed containers? although I have never come across any.


Where does the extra electron come from in the Lewis Dot Electron structure of say for example nitrate ion?

Unfortunately your Lewis structure for the nitrate ion, $\ce{NO3-}$, is wrong. You did not include any charges and as it is an ion there is at least one. Let's first have a look at how an ion may be formed and then hopefully you see, that there is no mystery at all about it.

Take for example ordinary table salt, which has the chemical formula $\ce{NaCl}$. If you dissolve it in water the lattice will be broken and the ions will by hydrated and are now able to move more freely.
$$\ce{NaCl_{(s)} -> Na+{}_{(aq)} + Cl- {}_{(aq)}}$$

The same happens when you dissolve potassium nitrate, which releases the nitrate ion:
$$\ce{KNO3_{(s)} -> K+{}_{(aq)} + NO3- {}_{(aq)}}$$
Here you can see, that the nitrate ion has an extra electron which it has taken from the potassium. See ionic bonding for more information.

For the Lewis structure of the nitrate ion you first have to count all the valence electrons of the involved atoms:
\begin{array}{lclr}
1 &\times &\ce{N} & 5\\
3 &\times &\ce{O} & 6\\
1 &\times &\text{charge} & 1\\\hline
  &       &              & 24\\
\end{array}

In the Lewis structure you have to distribute 24 electrons in a fashion, that you do not violate the rule of two, which is better known as the octet rule. Therefore your structure may include a nitrogen oxygen double bond, a positive formal charge on nitrogen and a negative formal charge at each of the singly bound oxygens.
Another representation is that all oxygens are singly bound and therefore the nitrogen has a $+2$ formal charge and each of the oxygens has a negative formal charge. The problem with this structure is that nitrogen does not have a full octet. Nevertheless, it is a strong contributor to the actual electronic structure. The bonding situation is much more complicated than can be expressed through a Lewis structure. The concept of resonance structures has to at least be included. Better in this case would be molecular orbital theory and or valence bond theory. The molecule itself is a y-aromatic system, the charge is delocalised through the extensive $\pi$ system. The blue structure is an attempt of showing this feature.


Can dust or fingerprints interfere in ultra-violet spectroscopy

You can deal with this through background correction of your spectra, which is a very common practice in spectrometry and implemented in any good spectrometry software.

In the most general sense, background correction consists of recording the spectrum of the spectrometer path without the analyte, and then subtracting that background spectrum from the spectrum of the analyte. In that sense, background correction is a form of difference spectroscopy. The spectrum you obtain from this ideally contains only the peaks attributable to the analyte.

A concrete example is subtracting the absorption spectrum of a solvent in a cuvette from an analyte. You would run a cuvette with solvent but without the analyte and use that as your background, followed by a cuvette with the analyte. Subtracting the first spectrum from the second would yield the signal attributable only to the analyte, assuming that the cuvettes, solvents and atmosphere between the two runs are negligibly different.
Obviously, there are a lot of potential confounds here that need to be accounted for:

Cuvettes have to be made of the same material, have the same dimensions (path length is important!) Cuvettes and such are often designed to be maximally transparent to radiation in the region of the spectrum of interest to maximise sensitivity. This is why they sell uber-expensive quartz cuvettes for UV/vis.
Solvents/substrates have to be of identical composition between runs. You also don't want a solvent or substrate that absorbs strongly in the spectral region of interest for similar reasons to the above.
Cuvettes have to be homogeneously clean - if one is dirty, greasy or scuffed and scatters light more than the other, it may ruin the results.
The atmosphere has to be the same between runs - I've heard stories about quantitative IR spectroscopy of $\ce{C=O}$ stretches being thrown out by an increase in $\ce{CO2}$ due to people breathing (how rude!). You can understand that researchers who look at atmospheric $\ce{CO2}$ concentration need very careful sampling and analysis protocols for reasons similar to this.


Your professor's premise of dissolution is only true if the process is incongruent melting which by the way is still melting. Whereby the process proceeds as:

$$\ce{KAl(SO4)2.12H2O_{(s)} ->[\Delta][T1] KAl(SO4)2~_{(s)} + 12H2O_{(l)}\tag{$\beta$ to $\alpha$ + liquid}}$$
Then
$$\ce{KAl(SO4)2~_{(s)} ->[\Delta][T_2 > T_1] KAl(SO4)2~_{(aq.)}\tag{$\alpha$ to liquid}}$$

This concept is outlined in the generic binary phase diagram below where $\alpha$ is anhydrous potash alum, $\beta$ is the hydrated form, and liquid is potash alum solution. Above a certain temperature $\beta$ will decompose into $\alpha$ and liquid. As the temperature rises more potash alum will dissolve until eventually it will all be liquid again.

Figure 1. Binary Phase Diagram with Peritectic melting

However if it is a direct conversion from liquid to solid then it is congruent melting and your professor is wrong. This can occur either as melting an eutectic (shown in Figure 2) or congruent melting of a compound (shown in Figure 2) and proceeds directly as follows.

$$\ce{KAl(SO4)2.12H2O_{(s)} ->[\Delta][92.5^\circ C] KAl(SO4)2~_{(aq.)} + 12H2O_{(l)}\tag{Interionic Melting}}$$
$$\ce{Sn_{(s)} + Pb_{(s)} ->[\Delta][183^\circ C] Sn_{(l)} + Pb_{(l)}\tag{Eutectic}}$$
$$\ce{Ti5Si3_{(s)} ->[\Delta][2130^\circ C] 5Ti_{(l)} + 3Si_{(l)}\tag{Intermetallic Melting}}$$

Figure 2. Binary Pahse Diagram with Eutectic.

Figure 3. $\ce{Si\! -Ti}$ Binary Bhase Diagram with Compound at $\ce{Ti5Si3}$ That Undergoes Congruent Melting.

Usually differential scanning calorimetry is used to determine which melting regime you are in indicated by a range of temperature causing a spike in heating rate corresponding to incongruent melting or a single temperature where heating spikes which would indicate congruent melting.

In this case, however, we can confidently assume that $\ce{KAl(SO4)2.12H2O}$ is under going incongruent melting since it melts above the temperature of water and below the temperature of the anhydrate and there are no other hydrates that it forms, the process is the peritectic melting thus the process your professor describes is correct, but it is still considered a melting process.


Why some compounds form different crystal structure at different temperature?

You are just entering into the field of Polymorphism. I couldnt find any source for Why polymorphism occurs but you can think about it as Energetically favorable structure at a particular temperature.

Related - Why polymorphs form?


Why six C atoms are usually seen in cyclic compounds?

It is all about minimizing the energy of a molecule.
In the case of carbon, the only molecule that adopts a perfect hexagonal geometry in its ground state is benzene (and its derivatives that possess a 6-fold rotational axis).  In this case the hexagonal geometry is adopted because all of the carbons are $\ce{sp^2}$ hybridized.  The ideal geometry (lowest energy) for an $\ce{sp^2}$ hybridized carbon involves three 120 degree bond angles around the carbon atom.  This can be accomplished in the case of benzene by placing 6 carbon atoms in a plane and connecting them as shown to produce the benzene structure.

Other carbon-based molecules may project as hexagons, but they are not hexagons when viewed in 3 dimensions.  For example, cyclohexane is not flat like benzene, but rather typically exists in what is called a "chair conformation" in its ground state where the C-C-C angle is around 109.5 degrees.

why not octagon?

Some other regular polygon shapes can be found in molecules.  For example, the cyclopentadiene anion exists as a regular pentagon, and the cyclooctatetraene dianion exists as a regular octagon.  Both of these molecules contain only $\ce{sp^2}$ hybridized carbons.

As mentioned above, these atoms prefer to exist in a geometry with 3 equivalent 120 degree bond angles.  However, the internal angle in a regular pentagon and a regular octagon are 108 and 135 degrees respectively.  Since these angles deviate from the 120 degree ideal, these molecules will be strained, unlike benzene.

What other ring shapes are preferred, low-energy minima?

We've already discussed the $\ce{sp^2}$ case where the preferred 120 degree angle is found in benzene which exists as a hexagonal structure.  $\ce{sp}$ hybridized carbon is most stable when the X-C($\ce{sp}$)-X bond angle is 180 degrees.  Clearly small or medium size rings comprised solely of $\ce{sp}$ hybridized carbon cannot exist as stable entities.  In the case of $\ce{sp^3}$ hybridized carbon, the closer we get to the ideal tetrahedral angle (an angle whose cosine = -1/3, ca. 109.47 degrees) the lower the energy of the molecule.  We've already mentioned cyclohexane and while it's C-C-C angle is close to the ideal tetrahedral angle, it is not exactly the tetrahedral angle (because each of its carbon atoms does not exist in a perfect tetrahedral environment).  If we fuse a few more cyclohexane rings onto cyclohexane, we can produce the molecule adamantane with 4 cyclohexane faces all in the chair geometry.

There is less strain energy in adamantane (per carbon) than in cyclohexane.  The geometry at each carbon atom is closer to the ideal tetrahedral geometry, but still not quite there.  How can we further elaborate adamantane into a ring system where all carbons have the ideal tetrahedral angle? Does adamantane look vaguely familiar?  By continuing to add more and more cyclohexane rings we build the following ring structure...

... Diamond, where all of the $\ce{sp^3}$ carbon atoms are in the ideal tetrahedral environment, an extremely stable arrangement.
EDIT: response to OPs comment

So, the lesser energy state must be the reason that after six number
of atoms, five is the common one, right?

Kinda.  Here's a table showing estimated ring strains for saturated hydrocarbon rings of various sizes.

[Image source  Also, see this link for a discussion about why the ring strain for cycloheptadecane (-3.4 Kcal/mol) is suspect and probably should be higher.]
5- and 7-membered saturated hydrocarbon rings have similar strain energies, yet 5-membered rings are more common.  We must consider the free energy of the process, enthalpy and entropy changes. In addition to strain (an enthalpic factor) we must also consider entropy.  In ring forming reactions it is entropically more difficult for a 7-membered chain to bring both ends together for closure to a ring, then it would be for a smaller 5-membered chain.


How to calculate pH of a solution with two dominant equilibria?

The first step is to write down the variables of interest, and the equations that relate them.  If there exactly as many (non-redundant) equations as variables, then the system might be solvable.

In this case, the variables are:

$\ce{[H+]}$
$\ce{[X^{-}]}$
$\ce{[Y^{-}]}$
$\ce{[HX]}$
$\ce{[HY]}$
$\ce{[OH^{-}]}$

The equations are:

water dissociation equilibrium: $K_w=\ce{[H+][OH^{-}]}$
HX dissociation equilibrium $K_{a,x}=\frac{\ce{[X^{-}][H+]}}{\ce{[HX]}}$
HY dissociation equilbrium $K_{a,y}=\frac{\ce{[Y^{-}][H+]}}{\ce{[HY]}}$
charge balance: $\ce{[H+]}=\ce{[OH^{-}] + [X^{-}] + [Y^{-}]}$
X mass balance: $\ce{[X^{-}] + [HX]}=\text{constant}=0.1\text{ molar}$
Y mass balance: $\ce{[Y^{-}] + [HY]}=\text{constant}=0.1\text{ molar}$

So, "all" you have to do is solve those equations for the six unknown variables.  That isn't easy to do with pen and paper, but it can be done.  Or you could use a computer to numerically solve the equations since you know all the numbers $K_w$, $K_{a,x}$, $K_{a,y}$, and the two different 0.1 molar total concentrations.

Once you solve the equations for $\ce{[H+]}$, the pH is just calculated from $pH = -\log_{10}\ce{[H+]}$.


How can we find the eV value of molecular orbital diagrams for a given atom?

Exact values of MO energy are rarely given as they are difficult to obtain in anything other than relatively trivial systems. When they are given, it is often not even clear whether they are experimentally derived (Photoelectron spectroscopy, as mentioned by orthocresol), or computationally derived (by calculations to approximate the wave functions and their properties). For this reason, most MO diagrams presented are qualitative, derived from our knowledge of symmetry, orbital energy etc.

There is a book, The Organic Chemist's Book of Orbitals (Salem and Jorgensen) which gives the MO's of many simple, and slightly more complex  molecules along with computationally derived energies (in eV).

For anything more complicated, you could also try searching the literature, however this would probably just turn up very specific examples that people have calculated to support their research.

If you need specific energies for molecules, you might look into computing them yourself. Simple software such as Spartan does an acceptable job at computing MO energy, with a variety of computational methods available depending on how much computing power you have.


1. Origins of the Houk model
The initial idea behind the Houk model was infact provided by Kishi, who proposed that alkenes with a chiral centre on the adjacent carbon adopted a 'reactive conformation' in which the small group eclipsed the alkene in order to minimise unfavourable steric interactions.[1]

Fig1: Kishi's "reactant conformer" model, taken from ref [1]
Houk took the 'reactive conformation' from the Kishi studies and modelled it computationally, showing that the lowest energy ground state conformation for alkenes with substituents at the allylic position was indeed the one in which the small group eclipsed the alkene, as proposed.[*]
2. Allylic strain
The Houk model works largely on steric grounds, balancing A1,2  strain and A1,3 strain, as can be seen from the figure below:

Fig2: The reactive conformation, taken from Organic Synthesis- Strategy & Control, Wyatt and Warren (Wiley)
The 'reactive conformation' is the middle structure (now days often known as the Houk conformation), in which A1,2  strain and A1,3 strain is minimal. In the left hand structure A1,2  strain is present, in the form of a H-H clash (doesn't look too significant, but raises the energy of the conformation significantly enough, similar to in the conformations of ethane). In the right hand structure A1,3  strain is present, this A1,3 strain is far more significant than A1,2, making this conformation sufficiently high in energy as to be barely populated at room temperature.
Based on the arguments above, it's worth pointing out that the Houk model works best for cis alkenes, with trans alkenes often giving poorer selectivity:[2]

Fig3: cis and trans alkenes reacting via the Houk model, taken from Organic Synthesis- Strategy & Control, Wyatt and Warren (Wiley)
This result is consistent with the fact that in a trans alkene, A1,3 strain is already minimised by virtue of the olefin geometry, meaning that two possible 'low energy' conformers exist which have neither A1,3 strain nor significant A1,2 strain.
3. Reactions not following the Houk model
Not all reactions for which the Houk model could be used give the expected 'Houk product'. The most common of these (and indeed the one in Clayden that you mention in the question as not following the Houk model) is when the starting material is an allylic alcohol.

Fig4: Reactions not following the Houk model due to directing effects, taken from Organic Chemistry, Clayden and Warren (Oxford University Press)
In the example above, the hydroxyl group is acting as a directing group, guiding epoxidation with the mCPBA to the same face (the steric argument, which would always predict epoxidation away from a bulky group, does not apply here).

Notes and references
[1]: Houk, K. N.; Rondan, N. G.; Wu, Y.-D.; Metz, J. T.; Paddon-Row, M. N. Theoretical studies of diasteroselective hydroborations. Tetrahedron 1984, 40 (12), 2257–2274. DOI: 10.1016/0040-4020(84)80009-5.
[2]: Vedejs, E.; McClure, C. K. Hyperconjugative effects of allylic substituents are not important in osmylations. J. Am. Chem. Soc. 1986, 108 (5), 1094–1096. DOI: 10.1021/ja00265a048.
[*]: The Houk group has been instrumental in providing a bridge between theoretical predictions and experimental results, providing weight to proposed models, and giving confidence in being able to look at a reaction and predict what product it may be.


Why does elemental Iodine form a brown solution with Ethanol but not with Isopropyl alcohol?

While iodine is readily soluble in ethanol and isopropanol, it is only soluble to about 300 ppm in water. I think it's pretty clear that the 30% water in your isopropanol solution was plenty to thoroughly inhibit the iodine's solubility.


Finding the equivalent weights of the oxide and the metal when reduced by hydrogen

Equivalents are part and parcel of Indian chemical education. I don't think college chemistry is going to be updated soon anytime there. I recently heard a complaint by a company to the chemistry chair in an American University. They said that your students do not understand normality! Please teach these concepts to your undergraduate students. Of course, normality concept sounds Greek to modern students.
Anyway let us assume we are in early 19th century and we are doing a classical experiment of chemical analysis. It is not a bad exercise, then. Balanced equations existed then, so we can write:
$$\ce{M_xO_y + yH2 -> M + yH2O}$$
833 mL of hydrogen gas corresponds to 0.833/22.4 = 0.0372 moles of hydrogen. This concept existed in early 19th century.
So what is the definition of an equivalent weight of anything? It is the weight that reacts with 1 equivalent weight of hydrogen (which is half its molecular weight).
Your first step is wrong. Realize that
No. of Equivalents of $\ce{H2}$ in 833 mL are = 2 x 0.0372.
This implies that equivalents of metal oxide which are present in the reaction are also 2 x0.0372.
Equivalent weight of the metal oxide is then = 1.80 g/ 0.0744 = 24.19 or as per your textbook 24.2
Since the equivalent weight of oxygen is 8, the equivalent of metal is 24.19-8= 16.19.
Now, there is no metal whose atomic weight matches 16.19. So what if the metal is divalent = 32 (this is sulfur, not a metal). Is it possible that this is trivalent metal like Ti so  the atomic weight of that M is 16.19 x3= 48.57 which matches with titanium.
One can do back calculation to double check this using moles. I leave that for an interested reader.


Which one of the following is NOT a haloalkane?

I have went through each option:
A) Freon is $\ce{CCl2F2},$ which is derived from methane
B) Teflon is a polymer of $\ce{-F2C-CF2 -}$ (or) $\ce{(C2F4)_n},$ derivative of 1,2-difloroethene. Here, carbon is saturated, which means no double or triple bond is present.
C) Iodoform is $\ce{CHI3},$ which again is a derivative of methane.
D) Vinyl chloride is unsaturated thus it does not classify as an haloalkane.


Understanding effect of ph on solubility of salt

Let assume we have the following equilibria:
$$\ce{MX \rightleftharpoons M+ + X-}$$
$$\ce{HX \rightleftharpoons H+ + X-}$$
The expression for their respective equilibrium constants are the following:
$$\mathrm{K_{sp} = [M^+][X-]}$$
$$\mathrm{K_a = \frac{[H^+][X^-]}{[HX]}}$$
We also know the following (let S be solubility):
$$\mathrm{S = [M^+] = [X^-]_{tot}}$$
Now we just need one more equation before we can start solving for S which is a mass balance. Since all the $\mathrm{X^-}$ that is produced either reacts with water to form $\mathrm{HX}$ or remains as $\mathrm{X^-}$, we get the following equation :
$$\mathrm{[X^-]_{tot} = [X^-] + [HX]}$$
Now lets put $\mathrm{[X^-]}$ and $\mathrm{[HX]}$ in terms of $\mathrm{[M^+]}$:
$$\mathrm{[X^-] = \frac{K_{sp}}{[M^+]}}$$
$$\mathrm{[HX] = \frac{[H^+][X^-]}{K_a} = \frac{K_{sp}[H^+]}{K_a[M+]}}$$
Now let plugs those values into our mass balance and replace $\mathrm{[X^-]_{tot}}$ and $\mathrm{[M^+]}$ with S:
$$\mathrm{[X^-]_{tot} = \frac{K_{sp}}{[M^+]} + \frac{K_{sp}[H^+]}{K_a[M+]}}$$
$$\mathrm{S = \frac{K_{sp}}{S} + \frac{K_{sp}[H^+]}{K_a\times S}}$$
$$\mathrm{S^2 = K_{sp} + \frac{K_{sp}[H^+]}{K_a}}$$
$$\mathrm{S = \sqrt{K_{sp} + \frac{K_{sp}[H^+]}{K_a}}}$$
This gives you the exact same expression as the one on the sheet but hopefully this is much clearer for you.


Acid catalysed mechanism of the addition of alcohols to alkenes

The short version: You are correct, the schematic drawing is not the correct mechanism and it does also not provide the most likely path. For a training exercise it is close enough though. It shows the most important steps and it conserves stoichiometry.

The depicted scheme does not show a reaction mechanism in the sense of the definition given in the goldbook. A reaction mechanism always also has to include transition states, that lead to the respective products. It is rather a reaction sequence, that shows the most important steps. In this sense, the stoichiometry has to be conserved and in this sense it is as correct as it could be.

Noteworthy is, that the catalytically active species is not sulfuric acid, but instead a mixture of hydronium and protonated methanol, $\ce{H3+O/H2+OCH3}$. Sulfuric acid is such a strong acid, that it is capable of protonating any solvent that has a lone pair, even itself via autoprotolysis (in concentrated $98\%$ solution). $$\ce{2H2SO4 <=> [H3SO4]+ + [HSO4]- <=> H3+O + [HS2O7]- <=> H2O + H2S2O7}$$
The use of concentrated sulfuric in organic synthesis is usually not advisable, since its dehydration property can quickly decompose many organic compounds that contain hydroxyl groups. The use of diluted sulfuric acid (either in water or alcohols) is therefore much more common.

It is only logical, that sulfuric acid will protonate methanol, which can be considered a preceding equilibrium. $$\ce{H2SO4 + HOCH3 <=> H2+OCH3 + [HSO4]- }$$
Since methanol is also the solvent and the acid is only available in catalytic amounts, it is feasible to assume, that the equilibrium lies entirely on the right.

It is also important, that the last step must be also in equilibrium, since it is only the protonation/deprotonation of an ether. Here again we have much more methanol than product, so that the equilibrium will most likely be entirely on the product side.

A more descriptive mechanism would be the following:

For simplicity and sufficiently accurate bond formation and breaking is indicated by arrows, omitting the transition states. Resulting a slightly simplified picture. However, the preceding equilibrium - as you noted - should not be omitted. This formulation of the mechanism is particularly important for many simple organic reactions, as it focusses entirely on the reactants and products (and to a minor extent on the reactive intermediates).

When it comes to more complex reactions, e.g. with asymmetric catalysts it is often much better to include the transition states.

From this it can also be seen, that the choice of sulfuric acid is somewhat arbitrary, as most acids would do the same. However, sulfuric acid is often used, since bisulfate and sulfate are very weak nucleophiles.

Additional note
In all schemes only the most likely path is shown. Any possible side reactions, like concerted [2+2] cycloaddition, have been omitted.


Benzofuran is named by the position of the oxygen, hence 2-Benzofuran.  (below is 1-benzofuran)

The =O s are then added on the 1 and 3 carbons, hence, 1-3-dione

Compare, cyclopentan-1,3-dione

The end result is

2-benzofuran-1,3-dione


Regarding why your ceramic material did not get coated nicely on the fiberglass, I think there's a problem of adherence between the fibers and the the ceramic. You need to prepare the surface of the fibers by etching for example before coating. They are often shielded.
Regarding why your  sol-gel can't be calcinated at $500°\mathrm{C}$. You need to figure out the proper temperature of calcination by monitoring the phase appearance of the ceramic materials by XRD diffractometry. You start with the XRD diffractogram of your as-prepared sample(normally, it should be amorphous). Then, you treat your sample at different  temperatures within different periods of times (you have two parameters to find: Temperature and Time). Unless you are working on a published article that has addressed all these issues.


You are neglecting the hydronium ions produced during dissociation of acetic acid.

We can consider the following sequence of events when we add more acetic acid to solution. In reality they occur simultaneously, but it's easier to think about it sequentially.

We have more than the equilibrium concentration of acetic acid, so some of it dissociates to produce $\ce{H+}$ and acetate ion.
Now we have more than the equilibrium concentration of acetate ion, so some of it reacts with $\ce{H+}$ to form acetic acid again.

Your argument is that step 2 makes the solution more basic, but you are neglecting the effect of step 1. Every extra acetate ion produced in step 1 also involves the production of a $\ce{H+}$. Furthermore, step 2 has to occur to a lesser degree than step 1, because step 2 consumes the acteate ions produced in step 1. The acidity of the solution must therefore go up as a whole.*

*Of course it's also possible that steps 1 and 2 both occur at nearly the same rates, and then the pH of the solution doesn't really change much as a result. This is what happens in a buffer.


Here's a breakdown: First, you identify that ethyne is the limiting reagent and would be consumed completely. Second, you correctly calculated the volume of oxygen left that is $450\ \mathrm{cm^3}$. $20\ \mathrm{cm^3}$ of ethyne will completely react to produce $40\ \mathrm{cm^3}$ of carbon dioxide.

Thus, the total gas volume when the reaction reaches completion is $490\ \mathrm{cm^3}$.


Formation of tetrahydrofuran in the reaction of n-butanol with lead tetraacetate

The reaction of lead tetraacetate, $\ce{Pb(OAc)4}$ with alkanols, $\ce{R-OH}$, is believed to proceed via an alkoxy-lead acetate, $\ce{R-O-Pb(OAc)3}$.

In this intermediate, the bond between the $\ce{Pb}$ atom and the oxygen of the alkanol cleaves homolytically, this yields an alkoxy radical, $\ce{R-O^.}$ and $\ce{Pb^.(OAc)3}$.

If the alkoxy radical has a hydrogen atom in the $\delta$ position, intramolecular hydrogen atom abstraction may occur.

By this $\ce{CH3CH2CH2CH2O^.}$ is converted to a carbon-centered radical, $\ce{^.CH2CH2CH2CH2OH}$.

The later is oxidized by $\ce{Pb^.(OAc)3}$, which leads to $\ce{Pb(OAc)2}$
and the cation $\ce{^+CH2CH2CH2CH2OH}$.

This cation undergoes cyclization to furnish tetrahydrofuran.


Why do liquids and solids have constant concentrations?

If you think about what concentration, it usually refers to the molar concentration in $\mathrm{mol/l}$ (at least in the field of chemistry). To calculate this concentration we need to find how many moles of a certain chemical are in 1 liter of the liquid or solid.

I will start with an example and describe the general principle afterwards. Let's say we have pure water at room temperature and atmospheric pressure. This means that the density is $998\,\mathrm{g/l}$. We also know that the molar mass of water, consisting of 2 hydrogen and 1 oxygen, is $18.016\,\mathrm{g/mol}$. If you look at the units of the molar mass and the density you can already see that we can combine these into something that has units $\ce{mol/l}$. If we do that we get that the concentration of water is $\frac{998\, \mathrm{g/l}}{18.016\,\mathrm{g/mol}}=55.4\,\mathrm{mol/l}$. So the concentration of water is $55.4\,\mathrm{mol/l}$. Notice that I have only used the density and the molar mass of water, which are (at least for equal temperature and pressure) constant.

In general you can do this for any solid or liquid (or gas) and find the concentration as $c=\frac{\rho}{M}$, where $\rho$ is the density, $M$ the molar mass and $c$ the molar concentration. For a given chemical, the molar mass is constant.

Now about the ‘fact’ that the molar concentration is constant for liquids and solids: strictly speaking it is not. That is because the density of liquids and solids is temperature and pressure dependent. The reason that the molar concentration is often called constant is twofold:

the density of liquids and solids has a much weaker dependence on the temperature than gases do therefore it can be regarded as approximately constant
if you look at solutions, the variation of concentration can be
orders of magnitude, whereas for e.g. water the concentration
variation when going from $0$ to $100\,\mathrm{^\circ C}$ is only a few percent.
Therefore, again, the concentration of the liquid is approximately
constant.


Does combustion occur more easily at higher oxygen concentrations or just higher oxygen partial pressure?

When you're dealing with questions like "how well does this burn", you are intrinsically dealing with the reaction rate of the combustion reaction.

When looking at chemical reaction rates, they are typically proportional to the "chemical activity" of the species reacting. (There's complications due to the stoichiometry of the reaction, but those corrections are confined solely due to the nature (formula) of the reaction, rather than any particular concentration/pressure conditions.) The definition of chemical activity can get complex in certain cases, but for most normal reaction conditions, the "chemical activity" of a reactant is proportional to its molar concentration - or in other words, how many molecules of the reactant are in a given volume.

For solution reactions, we typically use the concentration directly, but for gaseous reactants, we can take advantage of the ideal gas law and realize that at a fixed temperature, the concentration of a substance is proportional to it's partial pressure. Which means the chemical activity of a particular gaseous reactant will be proportional to its partial pressure.

So in determining the reaction rates of such things as combustion, it's the partial pressures of oxygen and other reacting gases that matter, rather than the total pressure. Absent any side reactions, or considerations such as thermal transfer, adding or removing an inert gas should not affect the rate of reactions.

(But note for combustion, which is highly exothermic, thermal transfer effects can't necessarily be neglected. The ability of the inert gas to either carry away excess heat - or act as a blanket to keep heat near the reaction point - could substantially change how quickly a fire burns in practice. But I'd expect it to be a complex relationship based on air circulation, convection considerations, etc.)


Can I write $\ce{X + 2 e- -> X^2-}?$ Why can't I start with $\ce{X}$ instead of $\ce{X^2+}?$

The question says that the metals are $\ce{A(s), B(s), C(s) and D(s)}$. That means the letters stand for the uncharged element, not for something more arbitrary. In the table, the letters are used again, in the following way (shown for A):

$$\ce{A(NO3)2(aq)}$$

Because nitrate has a negative charge, the metal A forms an ion in water with a positive 2 charge ($\ce{A^2+}$).

$$\ce{A(NO3)2(s) -> A^2+(aq) + 2NO3-(aq)}$$

With the same argument, you can figure out that B also forms a divalent cation in solution, while C and D are monovalent. There is no flexibility of having $\ce{A}$ stand for an ion, or for the formation of a metal anion; you should stick to the species and symbols in the text and table of the question.

Examples where letters stand for something else

In acid/base chemistry, A and B are used in the following way:

$$\ce{AH <=> A- + H+}$$

$$\ce{B + H+ <=> BH+}$$

If you look closely, $\ce{B}$ stands for a base (with conjugate acid $\ce{BH+}$) and - which might cause confusion - $\ce{AH}$ stands for an acid (with conjugate base $\ce{A-}$). Neither A nor B have to be a single atom, and their charge could be anything. However, $\ce{B}$ and $\ce{A-}$ have one charge less than $\ce{BH+}$ and $\ce{AH}$, respectively.

In redox (bio)chemistry, $\ce{NADH and NAD+}$ are used. Again, the charges are to be interpreted as relative, and the relationship between the two is given by:

$$\ce{NAD+ + H+ + 2e- <=> NADH}$$


Why are there drastic changes sometimes in radius of isoelectronic ions?

The whole concept of ionic radius is not sharply defined one. Note, that the same is true for the notion of atomic radius, as well as for any other kind of radius or any other notion of size. In general, the notion of size of an object looses its usual (i.e. classical) meaning in the microscopic realm since objects there do not have well-defined physical boundaries. So from the very beginning one should not expect a perfect trend in a not so well defined property.

Anyway, values of the ionic radii are based on crystallographic data, but what is actually determined in X-ray crystallography is the distance between two ions, not their radii. Then it is basically up to you how to divide this distance into the radii of ions in question. But whatever way the division is done, the ionic radius of an ions is not a fixed, perfectly defined property, since for the same ion, the radius can differ in different crystal lattices due certain variables such as coordination number and electron spin.

For instance, you mentioned the ionic radius of 138 pm for $\ce{K+}$. But this value is for 6-coordinate compounds only while for 8- and 12-coordinate compounds of $\ce{K+}$ the ionic radius is 151 pm and 160 pm respectively. Same story for $\ce{Ca^2+}$: 100 pm, 112 pm, and 135 pm for 6-, 8-, and 12-coordinate compounds respectively.1

Interestingly, for anions there is a little variation in the ionic radius depending on coordination number, so that for many of them only one value independent of coordination number is specified. To be honest I could not come up with a quick explanation why, so it feels like I have to dig into the matter deeper. But I'm quite reluctant to do so, since, as I said in the beginning, the whole concept of ionic radius looks suspicious, so I expect to find just some obscure rule arriving out of the blue which will explain this little variation in the anionic radius relatively to cationic ones.

Slater's rules to the rescue!

Now, I guess, I know which rule can be used to explain this drastic change in ionic radii. Let us examine the application of the good-ol' Slater's rules to the matter. These rules allows us to calculate the effective nuclear charge  $Z_{\mathrm{eff}}$, which is the net positive charge experienced by an electron in a multi-electron atom.
$$
    Z_{\mathrm{eff}} = Z - s \, ,
$$
where $s$ is the screening constant. Slater's rules provide a way to calculate the values for the screening constants, and consequently, to obtain the values of effective nuclear charges.

All the ions $\ce{S^2-}$, $\ce{Cl-}$, $\ce{K+}$, $\ce{Ca^2+}$ are isoelectronic: the electron configuration of them is that of $\ce{Ar}$, i.e. $\mathrm{[Ne] 3s^2 3p^6}$. Thus, we have to compare effective nuclear charges for $\mathrm{3p}$ electrons in these ions. And as a first approximation we could compare effective nuclear charges for $\mathrm{3p}$ electrons in the corresponding atoms. Now without going into the details (listing the actual Slater's rules & applying them) I will just leave here the final numbers, effective nuclear charges for $\mathrm{3p}$ electrons for the corresponding atoms:

$$
\begin{array}{c|cccc}
     & \ce{S} & \ce{Cl} & \ce{K} & \ce{Ca} \\
\hline
Z & 16 & 17 & 19 & 20 \\
Z_{\mathrm{eff}}(\mathrm{3p}) & 5.482 & 6.116 & 7.726 & 8.658 \\
r_{\mathrm{ion}} & 182 & 181 & 138 & 100 \\
\end{array}
$$

Now you can see that despite a relatively small change in the nuclear charge going from $\ce{Cl}$ to $\ce{K}$
$$
   (19-17)/17 * 100 \% \approx 11.8 \% \, ,
$$
there is a rather drastic change in effective nuclear charge
$$
   (7.726-6.116)/6.116 * 100 \% \approx 26.3 \% \, .
$$
And the drastic change in ionic radius is exactly in line with this drastic change in effective nuclear charge
$$
   (138-181)/181 * 100 \% \approx -23.8 \% \, .
$$
To summarize: the effective nuclear charge for $\mathrm{3p}$ electrons increases by about 26.3% when going from $\ce{Cl}$ to $\ce{K}$ and the ionic radius accordingly decreases by 23.8%.

Note that we used effective nuclear charges of the atoms just as the first approximation for that of the ions. For anions an additional electron will occupy an outer orbital and there will be an increased electron-electron repulsion. This will push the electrons further apart and increase the shielding, so that $Z_{\mathrm{eff}}$ for anions will be smaller than for the corresponding atoms. For cations the effect of the removal of an electron is the opposite: when an electron is lost, electron-electron repulsion and therefore, shielding decreases, so that $Z_{\mathrm{eff}}$ for cations will be bigger than for the corresponding atoms. It would not change the above mentioned trend though; in fact, it will make it even more prominent.

1 Ionic radii values are taken from here.


I don't think it is safe or realistic to do that without extensive training and equipment, but yes, historically sodium metal has been made by chemical reduction of sodium salts at high temperature and condensing sodium vapor produced.  For example, sodium carbonate was heated with charcoal.  Even though equillibrium does not favor sodium metal, the fact that the small fraction of sodium produced is removed takes advantage of Le Chatelier's principle to keep making more sodium.


Why does equilibrium favour weak acid or weak base?

Here is a simple intuitive explanation. In your above example, there is actually 2 reactions occurring, the forward and the reverse reaction. Let us consider the forward reaction. Here, both $\ce{NaOH}$ and $\ce{HCl}$ are strong bases and acids. What this means is that HCl has a strong ability to donate a proton and NaOH has a strong ability to accept protons. This means that when NaOH and HCl react together, the reaction will basically go to completion. So you will end up with up with very little NaOH  and HCl and the equilibrium will favour the products.

Now lets look at the reverse reaction which is: $$\ce{Na+ + Cl- + H2O -> NaOH + HCl}$$ Here $\ce{Cl-}$ is the conjugate base of HCl which is  a strong acid. This means that $\ce{Cl-}$ is a really weak base, meaning that it is pretty bad at pulling protons from molecules. Also water is the conjugate acid of NaOH, which is a strong base. This means that water is a really weak acid, meaning that it is also pretty bad at kicking out one of its protons. This means that in this reaction, barely any of the products will react and you will mainly end up with NaCl and water.

As you can see in both reaction the equilibrium will favour the weak acid and base as they aren't able to react that well together. This effect can also be explained by using equilibrium constants but I won't go into that.


Order of reaction / Rate-determining step in enzyme catalysed reactions

The rate determining step doesn't have anything to do with any of the factors you mentioned. It's not even true that the rate determining step can be determined directly from the rate of $$\ce{\mathrm{ES}->\mathrm{EP}}$$
The correct definition is that it is the biggest barrier from a stable species (reactant or intermediate) to any subsequent transition state along the reaction coordinate.

Your comment about the apparent rate of the reaction is unrelated. The rate-determining step is an intrinsic property of the reaction and its mechanism. It is unaffected by concentration, though concentration effects may change which mechanism out of several is predominant.


Why are transition metal hydroxides insoluble?

The rule of thumb for hydroxide salts is actually that all hydroxides are insoluble save those of the alkali metals and the heavier alkaline earth metals (namely $\ce{Ca^{2+}}$, $\ce{Sr}^{2+}$, and $\ce{Ba}^{2+}$). There are many resources you can look up to check the solubility rules of certain ions, this one, for example.

Before researching this, I suspected that the solubility of the salts with a certain anion might be determined by the cation's control over its valence electrons, which is measurable through ionization energy. Sure enough, when you compare the first ionization energies of various metals, you find that those with the lowest first ionization energies are those that are soluble hydroxides.

Alkali Metals

The salts of all alkali metals are completely soluble because they have relatively little control over the outermost electron orbiting them, in addition to this being the only electron they must lose in order to attain an octet structure.

Alkaline Earth Metals

For the alkaline earth metals, the complete outermost $\mathrm{s}$ orbital must be lost in order to attain an octet structure. Because the orbital is complete, and among other reasons, the first ionization will require more energy than the alkali metal in the same period. In addition to this, the second electron must then also be lost. This means that overall, it requires much more energy to ionize this group than the alkali metals. This explains the insolubility of $\ce{Be(OH)2}$ and $\ce{Mg(OH)2}$.

The solubility of the hydroxides of the metals lower in the alkaline earth metal group can be explained by the shielding effect, which effectively says that the ability of a positive nucleus to attract electrons decreases as the amount of orbitals surrounding the nucleus decrease.

Transition Metals

The insolubility of the hydroxides of the transition metals can be explained both by the increasingly large ionization energies that result from successively removing electrons (or negatively charged hydroxide molecules in this case) from the metal cation, and the greater effective nuclear charge on the electrons across a period.


Do SALC-AOs really belong to their symmetry species?

To show that your two functions do represent the $\mathrm{E}$ irreducible representation, we can approach the problem algebraically.

We start by representing the operators of the $C_{3v}$ point group in the basis of the AOs. The identity element is easy in any basis:
$$E=\pmatrix{1 &0 &0\\0 &1 &0\\0 &0 &1\\}$$

We can see that applying this operation to $e_1$ and $e_2$ will just return these functions, so $\chi(E)=2$. We can also represent this operation in the basis of $a_1$, $e_1$, and $e_2$, which will be important for explaining the remaining operations.

For the $C_3$ rotation, we can notice that this just has the effect of moving each AO to the next site (i.e. $\phi_1\to\phi_2$, $\phi_2\to\phi_3$, and $\phi_3\to\phi_1$), so we can write it in matrix form as
$$C_3=\pmatrix{0 &1 &0\\0 &0 &1\\1 &0 &0\\}$$

Lets see what happens when we apply this to $e_1$:
$$C_3e_1=\frac{1}{\sqrt{6}}\pmatrix{0 &1 &0\\0 &0 &1\\1 &0 &0\\}\pmatrix{2\\-1\\-1}=\frac{1}{\sqrt{6}}\pmatrix{-1\\-1\\2}=e_1'$$

Writing $e_1'$ in the basis of the original functions, we obtain:
$$e_1'=0a_1+\frac{-1}{2}e_1+\frac{-\sqrt{3}}{4}e_2$$

Doing the same with $e_2$:
$$e_2'=0a_1+\frac{-\sqrt{3}}{4}e_1+\frac{-1}{2}e_2$$

We can then take the trace of this matrix (or, to simplify, the 2x2 submatrix spanned by $e_1$ and $e_2$) to get the character of $C_3$:
$$\chi(C_3)=\mathrm{Tr}\pmatrix{\frac{-1}{2} & \frac{-\sqrt{3}}{4} \\ \frac{-\sqrt{3}}{4} & \frac{-1}{2}}=-1$$

You can go through the same sort of exercise with $\sigma_v$ (determine form in AO basis, apply to each function, write resulting functions in original function basis, take the trace to get the character).


Why doesn't fast-scan cyclic voltammetry work with metal electrodes?

It actually does work. Carbon fiber electrodes are just more useful for the neuro work that fast scans are popular for these days.

Here is an example from the literature of a metal electrode used up to $3*10^6$ V/s , Amatore et. al. JEAC


Start with a reading about the types of errors that arise in spectroscopy. Clearly photometric error $\delta I_{pe}$ is a device uncertainty. The suggestion is, its magnitude depends on the number of "beams". What changes as the number of beams increases? Path length? Number of gratings or optical components? What else? How do each of those factors contribute to an analytical expression for photometric uncertainty?

At the end, a reasonable result might suggest that, as you double the number of beams, you double the relative uncertainty in signal $\delta I_{pe} / I$, where $I$ is the signal intensity. To obtain the same absolute confidence, you therefore have to double the measurement intensity $I$.


How to create the equivalent of luminol for a compound?

Probably not easily, especially if the concentrations are really low. Luminol + $\ce{H2O2}$ only works with blood because it's a catalytic reaction, i.e. it doesn't take much iron to create a lot of chemiluminescence. I'm not an organic chemist, but I'm not aware of any sort of similar reaction specific for this class of compounds. I did see this reaction: http://onlinelibrary.wiley.com/doi/10.1002/(SICI)1521-4168(19980501)21:5%3C315::AID-JHRC315%3E3.0.CO;2-V/abstract Unfortunately, it's not open access and I can't share my institutional access, but it's basically a photo-generated ruthenium complex that reacts with tertiary amines and indoles which might be applicable here, though I'm guessing is not sufficiently sensitive. It's not terribly selective so you might run into interference problems and in that paper they use it for HPLC  detection so it may not be bright enough to see with the naked eye. (I have seen similar chemiluminescent compounds before and they're pretty dim, even with the lights out)

There are ways to produce visible changes in response to small concentrations of an analyte: LSPR, ELISA, etc. but they're a lot of work to implement for new analytes and probably too expensive for looking at a large area. Chemical imaging based on mid-IR or Raman spectroscopy may be applicable, though as far as I know, they're not geared for measuring large areas.

It seems that directly visualizing pheromone traces is a non-trivial task. Groups seem to do things like use image analysis (http://www.ploscompbiol.org/article/info:doi/10.1371/journal.pcbi.1002592) to estimate the levels of pheromones. I would like to have seen a comparison of their model with actual pheromone measurements and the way I would propose to do it is to take surface samples after the experiment. If a big sheet of paper was used as the surface, you could cut it up in a grid and extract the pheromone for analysis by LC-MS or whatever (or maybe even used directly as a paperspray emitter for MS). An ambient-ionization mass spec technique like LTP or DART would also probably be a good way to construct a surface map of pheromone concentrations.

Briefly: There are certainly ways of measuring trace amounts of things like pheromones, but a simple CSI-esque naked eye visualization probably isn't easily achievable.


Why can UV light initiate a reaction between hydrogen and chlorine gas?

Before going into the mechanism of this reaction, I suggest you look up free radical mechanism, as this reaction takes place through that.

$\ce{Cl-Cl}$ bond in $\ce{Cl2}$ is weak enough to be broken by mere UV rays (present in sunlight), and hence they undergo homolytic cleavage (the resultant products are $\ce{Cl}$ atoms, not ions) to form two $\ce{Cl}$ free radicals. Now these $\ce{Cl}$ free radicals are extremely reactive (due to one free electron that it can share in a covalent bond) and hence attack hydrogen gas.

Now is the time you consider energy lowering. $\ce{H-Cl}$ bond is lot more stronger than $\ce{H2}$, and hence the formation of $\ce{H-Cl}$ bond formation will be favored instead of $\ce{H-H}$. Therefore, $\ce{Cl}$ free radical attacks $\ce{H2}$ gas to break the bond and form $\ce{HCl}$. The other $\ce{H}$ atom hence liberated forms a bond with the other $\ce{Cl}$ free radical.


DCC coupling of N-hydroxysuccinimide and 2,2-dimethoxyacetic acid

If peaks for "some impurities" are appear around $\delta \: \pu{1.8 ppm}$, I assume you have residual THF in your crude product. The other multiplet of THF is probably burried under $\delta \: \pu{3.43 ppm}$ resonance. That would confirmed by your impurity integration, which should be equal to 1H unit equivalence. If I'm right, your 6.36H unit equivalence is actually overlap one of $\ce{CH3}$-resonance and one of $\ce{CH2}$-resonance.

On the other hand, the two resonances at $\delta \: \pu{3.43 ppm}$ and $\delta \: \pu{2.73 ppm}$ would very well be representing two $\ce{CH3}$-groups and (two $\ce{CH2}$-groups + part of THF impurity), respectively, as well.


Regioselective enolate formation, choosing major and minor products

A major factor contributing to the stability of the enolate ion formed as the major product (F) is the stability associated with extended conjugation or extended delocalisation via resonance. Due to this, terming (F) as the major product seems justified.
If the product were to be a kinetically-controlled one (KCP), then the presence of three acidic alpha hydrogen atoms could've caused faster formation of product (E) ; it still would not have been the more stable product, but since the major product in KCP-type reactions is controlled more by the kinetics , rather than the thermodynamics, we would possibly expect to see (E) in a larger yield in that case.
Note: Strictly speaking, the problem should mention conditions that can help us to unambiguously determine which pathway (KCP vs. TCP) should be chosen. But a practical advice, garnered from the experience of working problems which seldom perfectly mention the reaction conditions, is that unless mentioned otherwise, it is always wise to first consider the thermodynamically stable product.


Is a compound composed out of gold, manganese, xenon, krypton, and oxalate possible?

No, this is not possible.

Actually, if I would have to think of the most unlikely chemical conceivable, that would be it. Let's see why:

Krypton is a noble gas that doesn't bond to anything. All of the known krypton compounds can be counted on one hand, and most of them contain fluorine. Putting krypton in a large molecule like this just can't be.
This is the same for xenon. Even though it's slightly more reactive than krypton, it still has problems forming bonds with other stuff and even when it does it's extremely unstable.
Gold is another noble element that doesn't like bonding to stuff.
Permanaganate is a very strong oxidiser. This is not something you want to have around organic stuff (such as your oxalate). They are simply not stable together.

I don't know how you derived the data for it, but it seems completely unreasonable. A boiling point of 7538384392 K? How about no way? Even if you somehow could create this compound (god knows how), I would expect it to be so unstable it would decompose completely at ambient temperature. And by "ambient" I mean ambient on a Kuiper belt object.


Molecular kinetic interpretation of pressure

The derivation can be obtained without having to explicitly consider the time of impact. Suppose the molecules have kinetic energy $E=mc^2/2$ where velocity is $c$ and mass $m$. The molecules are in a fixed box of volume $v=xyz$. The molecule can move in any direction and so has components $u, \;v,\; w$ along each axis meaning that $c^2=u^2+v^2+w^2$ but as no direction is special (the gas fills the volume uniformly) then we can let $c^2=3u^2$.

The pressure on the vessel's walls is due to the change in momentum and for a wall collision after moving along the $\pm x$ direction is $2mu$. A molecule will on average hit opposite walls $u/2x$ times per second when moving in the $\pm x$ direction. By Newtons second law the rate of change in momentum equals force, thus a molecule exerts a force $2mu \cdot u/2x=mu^2/x$ on the face of the container perpendicular to $x$ which has area $yz$. As pressure is force /area and as the collision area is $yz$ then $p=mu^2/(xyz)\equiv mu^2/V$. Finally, substituting for $u$ and allowing for $N$ molecules the pressure is $p=Nmc^2/3V$ i.e pressure is twice the kinetic energy divided by three times the volume, $\displaystyle p=\frac{2E}{3V}$


Relationship between vapour pressure and saturated vapour pressure

If you're at equilibrium, you're saturated.   If you're saturated, you're at equilibrium.  The end.

If things aren't at equilibrium, the situation is different.  Consider a dish of water in the middle of the Sahara desert.  The amount of water in the atomsphere in the Sahara desert is generally nonzero, say in the range of 1 $\mathrm g~{kg}^{-1}$.  Even though the water vapor pressure (even in the Sahara!) is nonzero, it is way less than "saturated".  Therefore it isn't at equilibrium, and the pan of water quickly evaporates.


In a copper wire, how many electrons does each atom give away?

This is a number which can be measured via the Hall effect. This reference give the Hall coefficient as $-5.4\times10^{-11}\,\mathrm{m^3/C}$ for a number density of charge carriers as
$$n_\mathrm e=\frac1{\left(-5.4\times10^{-11}\,\mathrm{m^3/C}\right)\left(-1.602\times10^{-19}\,\mathrm C\right)}=1.16\times10^{29}/\mathrm m^3$$
The number density of copper ions is
$$n_\ce{Cu}=8920\,\frac{\mathrm{kg}}{\mathrm{m^3}}\times\frac{1000\,\mathrm g}{\mathrm{kg}}\times\frac{1\,\mathrm{mol}}{63.546\,\mathrm g}\times\frac{6.022\times10^{23}}{\mathrm{mol}}=8.45\times10^{28}/\mathrm m^3$$
So that works out to about $1.37$ charge carriers per ion.


How to chemically postpone a chemical reaction

In general the answer to your question is no, you cannot postpone the reaction once the reactants are actually in contact.

Reaction rate usually decreases with temperature.  Thus it is possible, but not certain, that if you freeze a reaction mixture down to liquid helium temperatures the reaction would stop.  This isn't really practical in general.


This is something of a standard reaction stochiometry question, although the situation is a bit more involved than the ones you normally see in textbooks.

The first step is to work out what the chemical reaction we're talking about actually is. One complication is that "kerosene" doesn't name a single molecular species, but is instead a mix of hydrocarbons. But the hydrocarbons are similar enough that we can probably approximate the mixture as a "typical" hydrocarbon of roughly the right type and mass. Wikipedia approximates kerosene as dodecane, and even gives a combustion reaction (saving us the effort of balancing one):

$$\ce{2 C12H26(l) + 37 O2(g) → 24 CO2(g) + 26 H2O(g)}$$

Now that we have stoichiometry, we just need to convert the quantities of reactants involved to numbers of molecules (moles). We're dealing with rates of consumption for both reactant, so let's just run the numbers for a fixed period of time, say 1 minute.

We have an amount of kerosene we want to burn ($\ce{X\; mL*s^{-1} * 60s}$) and a conversion from volume to mass (the density), and all we need to get the "moles of kerosene" would be the molecular weight of kerosene. Since we're approximating kerosene as dodecane, use the number for dodecane ($170.34\;  \ce{g*mol^{−1}}$).

Using the knowledge from the balanced chemical reaction equation that 2 moles of dodecane require 37 moles of oxygen to combust completely, we can determine how many moles of oxygen we need to burn the given quantity of kerosene.

Now we need to convert moles of oxygen to a volume of air. If we assume oxygen and air is an ideal gas (a safe assumption under normal conditions), we can use the ideal gas law to interconvert moles ($n$) into volume ($V$): $PV = nRT$, where $P$ is the pressure, $T$ is the temperature, and $R$ is the ideal gas constant (about $8.314\;\ce{L*kPa*K^{−1}*mol^{−1}}$). Or, if we're dealing with standard temperature and pressure (about 1 atmosphere at sea level and about room temperature), then 1 mol of ideal gas takes up about $22.4\;\ce{L}$. So we can find out what volume of oxygen we need.

But we're not dealing with pure oxygen, so we have to divide by 0.21 to get the volume of air which would contain that volume of oxygen. (This works because the 21% is a volumetric percentage.) The final little bit is to convert the number of liters of air used per minute into cubic feet used per minute.

So there's really no calculus involved - it's just lining up the conversion factors and multiplying through. The tricky bit is probably figuring out the chemical reaction formula you want to use (e.g. the assumption of dodecane). It also helps to know that things are easiest if you do most of the calculations in SI (metric) units, and only convert to non-SI at the end.

One final caveat is to keep in mind that we assumed that kerosene was well represented as dodecane. If your kerosene is substantially different from dodecane, the reaction will have to be adjusted. We also likely assumed that the input air was at standard temperature and pressure, and was dry (the 21% value is for dry air). Warmer/cooler air, air at altitude, and humid air will all change the calculation. Also keep in mind that the calculation is for combustion under ideal conditions - incomplete mixing and non-ideal combustion issues might mean you need more air in order to ensure complete combustion.


Why is lead sulfide found in nature, whereas lead oxide is less common?

Lead(IV) and sulfide ions are large and polarizable, making them soft in the context of hard/soft acid/base theory (HSAB), which states that ions with similar hardness will form stronger bonds.  The oxide ion is hard and therefore has a tendency to form weaker interactions with Pb(IV).

HSAB theory has been discussed on Chem.SE in this question and this answer.


Which conformer of beta D-glucose is more stable?

I also know that the anomeric effect makes the axial one favored as well.

The anomeric effect does stabilize an axial group, but only at the anomeric position. Other axial groups are not similarly stabilized. Thus $\alpha$-D-glucose is more stable than you think it should be based on steric considerations alone.

Knowing this, which is the more stable?

PS - very pretty chairs.


In an isothermal process, how can the change in internal energy be 0?

The internal energy of an ideal gas is simply $$ U = \alpha nRT,$$
where $$\alpha = \frac{\text{degrees of freedom}}{2}$$
So, in an isothermal process,
$$
\Delta T = 0 \Longrightarrow \Delta U = \alpha nR\Delta T = 0,$$
and likewise any
$$\left(\frac{\partial U}{\partial P}\right)_T = \left(\frac{\partial U}{\partial V}\right)_T = 0.$$


What is dielectric constant and how does it explain the reason why water is one of the best solvents

The dielectric constant $\epsilon$ (known also as permittivity) is a measure of the extent to which a substance is polarized under an applied (external) electric field. Polarization amounts to net separation of charge across the substance. A "dielectric" is the common name used to refer to a substance placed between the plates of a capacitor and used to store electrical energy in the form of polarization, that is, as charges separated across the material. A capacitor is the non-chemical analogue of a voltaic cell, that is, a form of storing electrical energy.

A substance with a high dielectric constant is easily polarized. Polarization allows countercharges to be placed around an ion, resulting in Coulombic interactions between solvent and ion, promoting solubilization of the ion by competing with interionic interactions. Similarly, a polar solvent - one with a high dielectric constant (relative permittivity $\epsilon_r>\approx15$) - will solubilize polar molecules by forming stabilizing interactions with the solute that compete with solute-solute interactions.

Water is a good solvent for polar molecules for the above-mentioned reasons. Water is sometimes referred to as the "universal" solvent, but it is generally a good solvent for polar solutes, not apolar ones.


Is phosphoric acid to be considered a hazardous waste?

Phosphoric acid is a "food" additive because its acidity prevents the growth of microorganisms.  By itself it can cause burns, and high levels of phosphate in the diet have been linked to weaker bones (greater risk of fractures).

If you pour it down the drain, its acidity should not be the issue.  You could neutralize it with baking soda, and it will tend to become neutralized and buffered in nature on its own, but I think the primary issue is that it will yield phosphate ions which act as a fertilizer/nutrient source for algae growth and can cause algal blooms in ponds and streams that can dominate the ecosystem and choke out plant and animal diversity.

For that reason, don't pour it down the drain even though it will likely become neutralized as it acts on the clog.

If you just need to manage it, neutralize it with baking soda, and boil off the water, or let it evaporate to yield sodium phospate salt.  We don't want the sodium phosphate to enter directly into the local watershed, but you could put it on houseplants.  I don't know if phosphate fertilizer is allowed where you live, but if it is, you could put it on your lawn.  It PROBABLY would also be OK to dispose of the solid Sodium Phosphate in the trash.


Reason 1:

while $T\,\mathrm dS=\mathrm dQ$ for a reversible process, this does not mean $T\Delta S = Q$

Reason 2:

The term "reversible reaction" is unrelated to "reversible process".

Lame Reason 3:

Because your "always" statement does not repeat the "constant pressure" requirement assumed for $Q=\Delta H$


What does the Sm5Ge4 orthorhombic structure look like?

You can look at the $\text{Sm}_5 \text{Ge}_4$ crystal structure at Materials Project here.


Why do we observe signal coalescence during dynamic NMR?

The basic explanation is quite simple: a spectrum reports the rate of precession of a collection (ensemble) of nuclear spins. Precession can be pictured as orbital motion about a common axis, like earth rotating about the sun (merely for purposes of illustration). Now assume you have two groups of spins (equal number in each), spins in each group precessing at a fixed rate, that is, spins in each group precess coherently. Since the NMR signal is an average over all of the spins, and you have two populations with distinct precession rates, you see two independent peaks in the spectrum.

If on the other hand there is fast exchange (fast to be defined later) between the spins in the two populations, each spin ends up with an average precession rate, the mean of the starting values of the populations. In that case, the NMR spectrum contains only one peak. The spins have been mixed and each population is no longer pure, the original populations have lost coherence.

In the intermediate regime, when exchange rate is not so fast that the populations are thoroughly averaged, the spins will inhabit mostly one population but their rates will deviate from the pure initial rate due to sampling of the rate of the other population. This jumping between populations leads to a loss of coherence, and the NMR spectrum will show broadened peaks. These eventually coalesce into a broad central peak with a precession rate equal to the average over the two starting populations. The condition for coalescence (fast exchange) is that the exchange between populations happen rapidly relative to the precession rate about the axis, thereby blurring the precession rate of individual spins.


Is photosynthesis endothermic and exothermic at the same time? ... somehow?

Consider that when $1$ molecule of glucose forms, this happens at the cost of $6$ molecules of  $\ce{CO2}$ and $6$ molecules of $\ce{H2O}$ :

$
\ce{6CO2(g) + 6H2O(l) -> C6H12O6(s) + 6O2(g)}
$

Considering that at $25\ \mathrm{^\circ C}$ and  $1\ \text{atm}$

$
\Delta H_{\text{f},\ce{CO2}}^\circ=-393.5\ \text{kJ/mol}
$

$
\Delta H_{\text{f},\ce{H2O}}^\circ=-285.8\ \text{kJ/mol}
$

$
\Delta H_{\text{f},\ce{C6H12O6}}^\circ=-1273.3\ \text{kJ/mol}
$

You get, by summing the enthalpies of formation for each of the involved species and considering that for $\ce{O2}$ that enthalpy is zero:

$
\Delta H_\text{r}^\circ=6\times393.5\ \text{kJ/mol}+6\times285.8\ \text{kJ/mol}-1273.3\ \text{kJ/mol}=2802.5\ \text{kJ/mol}
$

Note that, for $\ce{CO2}$ and $\ce{H2O}$, the change of sign is required, since you "lose" them, while you form glucose.

So, finally $\Delta H_\text{r}^\circ$ (heat to be absorbed to form one mole of glucose) is positive, indicating that photosynthesis is indeed endothermic.


What is the mechanism for reduction of alkylmercuric compounds by hydrides?

The reducing agent NaBH4 is used for this reaction because it is highly selective. If H2SO4 were used in place of NaBH4 in this reaction the molecule would be dehydrated and we would probably see some HgSO4 form as well. The main thing to remember here is that a reducing agent has much more selectivity than an acid solution.


Could thin layer chromatography separate enantiomers?

The reason is simple: the physical properties of chiral pairs are exactly the same.

So, for any given compound that is chiral, the melting point, the solubility, the partitioning between liquid and solid phase in a chromatography column etc. are all exactly the same for the pairs of chiral molecules. Simple physical processes can't distinguish the two molecules. The solvents and solid components in chromatography are rarely chiral, so they can't tell the enantiomers apart.

The only things than can are other chiral things. If the solid phase of a chromatography column were chiral (or the solvent, though that is a lot less common) then the column could separate the enantiomers.

Think of shaking hands. It is natural that a right handed person shakes hand with the right hand of another person but the interaction is radically different if they to shake hands with the other persons left hand. The interaction of chiral things (like hands) depends on chirality. But if you bump fists instead (fists being closer to a sphere which has no handedness) then it matters a lot less which hand your friend uses.