Can oxidizing agents be used to increase the combustion efficiency of waste oil?

It is safer to keep the process as simple as possible without the addition of perchlorates or peroxides to the oil.  Perchlorates are salts with low solubility in oil; hydrogen peroxide equally is a polar compound with little affinity to form a solution with oil.  Thus, the salts could clog the fuel delivery, or in case of $\ce{H2O2}$ simply yield a biphasic liquid mixture if stored in the same container as the fuel.  It equally would increase the potential hazard to run such a combustion engine; perchlorates are heat sensitive and could explode when you don't want them to do so. In addition, over time residues of salts and salt residues facilitate corrosion (already a problem for the transportation of crude oil); here, the elevated temperatures could amplify the problem e.g., at the burner's nozzle, or as deposits in the fire box.
Thus, stay with the delivery of liquid oil and air.  If your are lucky in terms of viscosity of the oil, you still may obtain an aerosol but not the fine mist of air and fuel as characteristic for Otto engines, slow motion at 2:56 min onward).  More often, however, similar to oil driven heaters, you will rely on special burners running which bring pressurized air via fans into contact with a fine film of oil:

(credit)


So, I ask, what are the trends leading to the brittleness of a metal

Brittle materials absorb very little energy before fracturing. The only trend that can be linked to this is the strength of bonding in the metal (Other factors like crystal structure and purity do not have any trends and so there are a lot of exceptions). This also means that the trend in brittleness is not very regular since it is complex and depends on too many other factors.

  why are these trends there? (usually trends can be explained in terms of simpler things like atomic size/sheilding/etc)

Since brittleness cannot be explained by a very simple trend, I will explain what reasons I have found.

The stronger the bonding in the metal, the more difficult it is to change its structure and thus it tends to break instead of deforming. Here's a graph (melting point vs. Group number) that gives a general measure of strength of metallic bonding in all the transition elements of the first three series.

According to the graph, we predict that Tungsten (Period 6 Group VIB) should the most brittle and this is true for impure tungsten. Again, this successfully predicts that Silver, Gold and Copper (Group IB) are least brittle (and so most malleable and ductile). Also Lanthanum (Period 6 Group IIIB) is predicted to be quite malleable and this is also true.

Note that this also explains the high malleability of alkali metals (the graph is not very clear at that point).

There may be much more exceptions to this trend when examined closely so take this only as an approximation.

Crystal structures also play an important role in brittleness but they cannot be explained by a trend.

Purity of a metal tends to increase malleability and decrease brittleness because there are less points at which a fracture can take place.


When converting between a hydrated electrolyte amount in mass and in milliequivalents (meq), why are the water molecules taken into account?

When you want to obtain a solution of $\ce{CaCl2}$ in water, you can do it by dissolving anhydrous compound $\ce{CaCl2}$, or  the hydrated compound $\ce{CaCl2·2H2O}$. It does not matter which one to choose, provided you add enough water to compensate the difference and obtain at the end the wanted total volume.
If you want to obtain $1$ mL of a final solution that has to be $4$ meQ/ml, you must dissolve $4$ mEq of $\ce{CaCl2}$ or $4$ mEq of $\ce{CaCl2·2H2O}$ in $1$ mL . It does not matter which one you use. The only requirement is that the number of Equivalent has to be $4$ mEq. But the Equivalent weight  $\ce{CaCl2}$  is $111/2 = 55.5 $ g, and the equivalent weight of $\ce{CaCl2·6H2O}$ is $73.5$ g
To get the final solution with $\ce{CaCl2}$ you must choose to dissolve $4$ mEq $\ce{CaCl2}$, which weighs $\ce{(4·55.5}$ mg) = $222$ mg $\ce{CaCl2}$.
To get the same final solution with $\ce{CaCl2·6H2O}$ you must choose to dissolve $4$ mEq $\ce{CaCl2·6H2O}$, which weighs $\ce{(4·73.5}$ mg) = $294$ mg $\ce{CaCl2}$,
Do you understand ?


Energy required for obtaining required mass of aluminium from alumina via electrolysis

Required energy $E$ can be found as a product between voltage $U,$ current $I$ and time $t$:

$$E = UIt\label{eqn:1}\tag{1}$$

Unknown multiplier $It$ can be found using united Faraday law:

$$m = \frac{MIt}{zF}\quad\implies\quad It = \frac{mzF}{M}\label{eqn:2}\tag{2}$$

where $m$ is the mass of the substance liberated at the electrode; $M$ is the molar mass; $z$ is the number of electrons transferred per ion; $F$ is Faraday constant.
Considering aluminium is produced via Hall–Héroult process or similar, at the cathode the following reaction takes place:

$$\ce{Al^3+ + 3 e− → Al}$$

and $z = 3.$
Plugging \eqref{eqn:2} into \eqref{eqn:1} allows to finally find the energy:

$$
\begin{align}
E &= \frac{mzUF}{M}\\
  &= \frac{\pu{20 g} × 3 × \pu{5 V} × \pu{96485.33 C mol-1}}{\pu{26.98 g mol-1}}\\
  &= \pu{1.07 MJ}\tag{3}
\end{align}
$$


Reaction between free radicals and water

Bromine radical reacts with water

$$\ce{Br* + H2O -> HBr + *OH}$$

OH radical reacts with bromine

$$\ce{*OH + Br2 -> BrOH + Br*}$$

to give a new bromine radical, and so on, until the bromine is totally consumed.


Why are strong acids and bases not suitable as primary standards?

strong Bases in general are not suitable as they will react with $\ce{CO2}$ from the air. This is actually more problematic than hygroscopicity, as it means that you also cannot store diluted $\ce{NaOH}$ solution (openly) as standard.
Instead, $\ce{Na2CO3}$ or $\ce{KHCO3}$ are much better alternatives (for strong acids).
The acids are either a gas ($\ce{HCl}$) or are produced from gas. I don't see how you can produce a conentration reliably enough for use as primary standard.
even if you would manage to produce saturated $\ce{HCl}$, it's much more difficult and less reliable than known alternatives (see below).
but that wouldn't work e.g. for $\ce{H2SO4}$ as $\ce{SO3}$ is soluble in $\ce{H2SO4}$ (Oleum).
$\ce{HNO3}$: redox reactions are not only not particularly useful, but actually harmful as they use up acid equivalents:
$\ce{NO3- + 2e- + 2 H+ -> NO2- + H2O}$
also here, a solid alternative (KH-Phthalate) is available. If you don't have that, you may get away with oxalic acid which is still better than the acids suggested here.


How does salt concentration affect corrosion rate of iron?

The involvement of water accounts for the fact that rusting occurs much more rapidly in moist conditions as compared to a dry environment such as a desert. Many other factors affect the rate of corrosion. For example the presence of salt greatly enhances the rusting of metals. This is due to the fact that the dissolved salt increases the conductivity of the aqueous solution formed at the surface of the metal and enhances the rate of electrochemical corrosion. This is one reason why iron or steel tend to corrode much more quickly when exposed to salt (such as that used to melt snow or ice on roads) or moist salty air near the ocean.

Water is the enabler of fast oxidation of iron so freshwater will also cause rust. However, salt water is a very good conductor (lots of dissociated ions) and so there are a number of electrolysis reactions that tremendously accelerate corrosion in salt water.
For example if you have iron in contact with salt water and also in contact with another metal such as aluminum (also in contact with the water) you effectively get a battery which drives very fast corrosion processes.
This effect can be reversed by using a metal (like zinc) which causes the current to be reversed and in effect the zinc corrodes rapidly, protecting the iron. This is the principle of a 'Cal-rod' which is used to slow the rusting of hot water tanks


I can't post a comment, so this is not a true answer but I think can help. If I've understand what you mean you want to perform a Temperature-programmed reduction (TPR) followed by a temperature-programmed oxidation (TPO). In this case I think this old book can help: Temperature-programmed Reduction for Solid Materials Characterization. I don't know much about single crystal but I think that many things from TiO2 studies of nanomaterials can be consider valid for your case, this online book can be helpful to understand all the variables.


Is Beer-Lambert law also valid for fluorescence and luminescence?

No the Beer Lambert law  ($I_{trans}=I_0\exp(-\epsilon[C]L)$) does not apply as this shows that the transmitted light intensity decreases exponentially with concentration as it passes through a solution.

Fluorescence/luminescence is directly proportional to the concentration of molecules excited provided that the solution  is dilute so that dimers or excimers are not formed and that self-absorption of the fluorescence light by another part of the same solution does not occur.


organic chemistry of alkyl halides and alcohols

You are correct that copper metal is a catalyst for alcohol dissociation to aldehyde + $\ce{H2}$.  The reaction of ethanol, for example, can be written as:

$$ \ce{CH3CH2OH  ->[\ce{Cu}~\textrm{metal}][300~^{\circ} \textrm{C}] CH3CHO + H2} $$

Your question about halo-substituted alcohols is very interesting, because those species are very unstable.  If I understand you correctly, you are not referring to halohydrins such as 2-chloroethanol, which are species that have a halo substituent beta to the alcoholic carbon, but rather to species such as 1-chloroethanol where the halo and hydroxyl substituents are on the same carbon atom.
alpha halo alcohols are very unstable, and tend to eliminate halo-acid and form the corresponding carbonyl compound.  For example, gaseous chloromethanol spontaneously eliminates hydrogen chloride to form formaldehyde at 296 K (~23 °C) with a half-life of about 300 seconds.
Thus I think the best answer to your question is that:

Halo alcohols like 1-chloroethanol or chloromethanol are already oxidized to the same degree as carbonyl compounds.
Even before reacting them with copper metal at 300 °C, they are likely to decompose to the carbonyl compound via elimination of hydrogen halide.
Unlike the reaction of simple alcohols on hot copper metal, this reaction is not a redox dissociation, but is a simple elimination reaction.$%edit$


How to avoid the appearance of extra hydrogens from lattice packing of PDB file?

This is a bug, which is clear.

My guess is that you don't get $\ce{CH6}$ but that there's some disorder, so the transformations put two near-identical $\ce{CH3}$ groups on top of each other. Thus, one C becomes two, and each gains hydrogens.

There's usually a "slop" factor to detect near-equivalent atoms, but it's clearly failing.


Why doesn't pure water conduct when it dissociates?

Pure water does conduct as pointed out by the above comments. But it has very less dissociation hence there are very few ions to conduct electricity.
you may have heard that pH of pure water is 7. This statement implies that the concentration of H+ ion is 10^-7. Since it is a neutral solution concentration of OH- ion is also 10^-7. As such the conducting property of water is very less.


One elder reference is Brauer's Handbook of Preparative Inorganic Chemistry, which states for this compound

"In the method of Wöhler and Streicher, $\ce{IrO2}$ is prepared from green $\ce{IrCl3}$ which can readily oxidized in a stream of $\ce{O2}$ at 600$^\circ$C giving blue-black  $\ce{IrO2}$.
The oxidation of fine iridium powder in a stream of air or oxygen does not give $\ce{IrO2}$ quantitatively. [...]"

which is accessible in archive.org's public scanned version here, p. 1590 (book's counting) / p. 1636 (the website's counting) of the 1963-edition by Academic Press.  The age of the edition should neither defer you from using it as a reference, nor inhibit you to look up additional references.


Do polyfluorosulfuric acids like H2SO2F4 exist?

As the comments imply, no dice.  You can, of course, make a six-coordinate sulfur compound with the formula $\ce{SF6}$.  But trying it with a mixture of oxide, hydroxide and fluoride ligands instead of just fluoride presents the opportunity to evolve $\ce{HF}$ leaving the sulfur with a lower coordination number.  There is a five-coordinate compound $\ce{SOF4}$, but otherwise you should expect the sulfur to get down to four-coordination.  So the only stable protic acids you can get with one sulfur atom and oxide, hydroxide and fluoride ligands are plain old $\ce{H2SO4}$ and the more strongly acidic $\ce{HSO3F}$.


Delocalization of π-electrons in nitrate ion

Unfortunately the key to understanding the delocalisation of electrons lies in understanding rudimentary molecular orbital theory. It is also necessary to understand that hybridisation is a mathematical concept which can be used to describe bonding. It is no necessity to have bonding.

Your explanation of benzene is the most common description, but it is somewhat incomplete. First, you are right, one can consider the carbon atoms having three (roughly) sp2 hybrid orbitals, giving rise to two σ bonds to the neighbouring carbons and one σ bond to hydrogen. The remaining 6 perpendicular p orbitals form the delocalised π system. These orbitals form 6 new orbitals and due to symmetry constraints, i.e. D6h, three of them are bonding and three of them are antibonding. In the π system, there are six electrons, so only the bonding orbitals will be occupied. This makes the extraordinary stability of these compounds. The following picture schematises this on the left side, while on the right side these are the orbitals obtained by a quantum chemical calculation. (At the DF-BP86/def2-SVP level of theory.)

The nitrate ion $\ce{NO3-}$ is what I like to call a Y-aromat since it has some similarity in the electronic structure. The bonding situation is very comparable to benzene.
The resonance structures, however, look much more complicated.

In quite good approximation, you can regard the nitrogen as having three sp2 hybrid orbitals (and one p orbital) and the oxygen as (roughly) having two sp (and 2 p) orbitals. You follow the same scheme, therefore you obtain four equivalent nitrogen oxygen σ bonds. Each oxygen has one lone pair in the in-plane p orbital and one lone pair in the sp hybrid. The remaining out-of-plane p orbitals of nitrogen and oxygen build up the delocalised π system. This time there are only four orbitals. Because of symmetry constraints, i.e. D3h, one is bonding, two are (close enough to) non-bonding, and one is antibonding.* With the remaining six electrons you only fill the bonding and non-bonding orbitals. The concept is similar. Below you find the schemes like the ones above for benzene.

Since the carbonate anion $\ce{CO3^{2-}}$ is isoelectronic to nitrate, approximately the same orbital picture will result.

* Strictly speaking, there are no non-bonding orbitals. The classification can only be with respect to a bonding axis, and there are only two options: 1. there is no nodal plane (or surface) perpendicular to that bonding axis; it is then called bonding. 2. There is a nodal plane (or surface) perpendicular to that bonding axis; it is the called antibonding. A molecular orbital can be bonding respectively to one bonding axis, and antibonding with respect to another.
For example, the E" orbitals of nitrate have this feature, since the nodal surface curves through the nitrogen atom. The above depicted orbital on the left is bonding with respect to one of the $\ce{N-O}$ bonds and antibonding towards the other. (It is also bonding with respect to a $\ce{O\bond{~}O}$ interaction.) The orbital on the right is exactly opposite (orthogonal) to that.


Question is, what is "better"? Faster, or higher melting enthalpy, or degree of crystallinity? The methyl group is bulkier and thus hard to get into the right backbone conformation, and the OH group is polar, as you've noted, leading to a melting point of PVA of 230°C (iPP ~175°C).

But then, it all depends on the actual tacticity, which is not specified. The isotactic species have a totally different crystal structure than the syndiotactic, and sample with low tacticity (PVA!) might still have another one. That leads to totally different crystallisation kinetics.

So while a lot speaks for "A", I say there can be no definite answer.


What is the proper description/name of this amide/ester structure?

In IUPAC speak it is a carbamic acid methyl ester.

The IUPAC name for the compound drawn according to Chemspider structure search is Carbamic acid, N-[5-(propylthio)-1H-benzimidazol-2-yl]-, methyl ester  1

Most working chemists would call it a methyl carbamate


Dissociation equilibria of ammonium bifluoride in water

You need to write out all the possible equilibria just like you do in the case of a multiprotic acid like Phosphoric acid. You will also need to have a pKa for every equation. The bifluoride ion will obviously dissociate in water


What's the relationship between Van der Waals forces and hydrophobic interactions?

Van der Waals forces are the attractions/repulsions - the forces - between molecules or atoms, other than attractions like ionic attractions, and covalent attractions. These forces are:

Keesom Effect - This is an effect caused by two polar atoms interacting with each other. Two permanent dipoles are involved, meaning the molecules/atoms involved are polar. This may be attractive or repulsive, depending on the dipoles involved.
Deybe force - This is an effect caused between a molecule with is polar, and one that is not. Because the one that is polar affects the electrons on the second (non-polar) molecule, it creates an attractive force between the two molecules.
London dispersal effect - This is a force that acts between two non-polar molecules/atoms. Because the electrons around each molecule/atom repel each other, it creates a redistribution of charge, inducing an instantaneous dipole moment. This is the force that acts in liquid noble gases, to make them liquid, or between methane molecules in liquid form.

Hydrophobic forces are caused because molecules like ethane and other hydrocarbons - petrol etc. and oils like sunflower oil etc. are non-polar, and therefore don't "like" being dissolved in a polar solvent like water. In water, there are hydrogen bonds between the individual molecules, which is why water is a liquid - the van der Waals forces are far too small in water to bind the water as a liquid at room temperature. If we introduce non-polar substances into the water - polar solvent - it upsets these hydrogen bonds, and creates an increase in enthalpy because of this. Therefore the lowest energy state is for the hydrophobic hydrocarbons to separate themselves from the water.

The relationship between van der waals forces and hydrophobic interactions is that the van der waals act to bind the hydrophobe - non-polar substance - together, to separate from the polar solvent/water, and these contribute to the energy needed to separate the two substances. The seperation causes a decrease in the entropy of the system. To counter this decrease in entropy, there must be some decrease in enthalpy. Because the hydrophobe disrupts the hydrogen bonding in water, when they separate, the hydrogen bonding then causes an decrease in enthalpy, because of the favorable interactions. This is also where the van der Waals forces come in. They are favourable interactions, so cause a decrease in enthalpy, and this helps the separation of the hydrophobe and the water into two separate phases, because it makes the separation more energetically favourable. They are not necessary, but make separation more favourable, because the enthalpy change is greater for the separation process. This makes it more favourable.


Is the deprotonation of phenol exothermic?

Any bond dissociation reaction is endothermic, hence the termin 'bond'. Heterolytic dissociation (into ions) is more endothermic than homolytic. Strong acids are not exception.

However, the words above are about molecules in vacuum. When the molecules are in a media, interaction with media may stabilize ions strongly. This is why salts dissolve and acids/bases dissociate.

In case of phenol, dissociation in vacuum is definitely endothermic. Dissociation in water is also endothermic, since the dissociation constant is small. (1) However, deprotonation implies that protons moves to another acceptor, and in this case it depends entirely on the acceptor. Similarly, in nonwater media dissociation of phenol also may be exothermic.

(1) Dissociation constant is tied to dissociation free energy with law
$\ln K_e = - \Delta G / RT$ . Since entropy in reaction dissociation reactions is definitely increasing, the only reason for $\Delta G$ to be unfavorable is enthalpy.


This is what Encyclopaedia Britannica definition of a solution in chemistry:

  Solution, in chemistry, a homogeneous mixture of two or more substances in relative amounts that can be varied continuously up to what is called the limit of solubility. The term solution is commonly applied to the liquid state of matter, but solutions of gases and solids are possible. Air, for example, is a solution consisting chiefly of oxygen and nitrogen with trace amounts of several other gases, and brass is a solution composed of copper and zinc.

A solution consists of solutes (at least one) and a solvent. The solute is define as the substance that is dissolved in the solvent. In other words, for solutions with components in the same phase, the substance(s) present in lower concentration are solutes while the major substance present in highest abundance is the solvent. For example, 190-proof alcohol is a liquid-liquid solution, which is the mixture of 95% ethanol and 5% water by volume. The major substance in this mixture is ethanol. Therefore, it is the solvent. The 5% water is the solute. Brass is a solid-solid solution of copper and zinc metals. General composition of brass is 65% copper and 35% zinc by weight. Hence, in brass, copper is the solvent and zinc is the solute. Air is a solution of multi-gases. In air, 72% nitrogen and 20% oxygen, and rest is other gases including $\ce{CO2}$ and $\ce{H2}$. Thus, the solvent of air is $\ce{N2}$ while $\ce{O2}$ is one of its solutes. Saline solution, on the other hand, is a solid-liquid solution. In a saline solution, solid salt ($\ce{NaCl}$) is the solute dissolved in water, which is the solvent (liquid).


GROMACS won't be able to calculate the enthalpy of mixing directly but using the energy function in GROMACS you will be able to get energy, pressure, and volume of the simulations.

Going through the paper that you posted it looks like the ran multiple simulation boxes for the different components and mixtures. They did this because you will not be able to get U,P, and V for individual components in GROMACS. So in order to calculate the enthalpy of mixing for a 2 component mixture they ran 3 simulations: one of the mixture, and one of each component by themselves.

Once you have the simulation boxes for the different components and mixtures it should be pretty straight forward to calculate the enthalpy of mixing.

Just a side note: I am always a little nervous calculating thermodynamic expressions using pressure from GROMACS because the pressure will fluctuate so wildly. So if you're going to make sure you're careful about your equilibration run and your NPT equilibration run as well.

Below are the GROMACS manual for energy which you would use to get U,P, and V from your simulation. The next is a tutorial for an NPT run from Virginia Tech and their tutorials are really well written and should help as a starting point and finally the last is a chat between a user and GROMACS developer discussing pressure fluctuations in NPT.

http://manual.gromacs.org/documentation/current/onlinehelp/gmx-energy.html

http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin/gmx-tutorials/lysozyme/07_equil2.html

https://mailman-1.sys.kth.se/pipermail/gromacs.org_gmx-users/2011-May/061370.html


Why sugar solution/sugarcane juice becomes sour when kept for certain time?

This is not a complete answer but it should give you a head start.

I suspect two things: Fermentation and Caramelization.
In any fermentation process, sugar will turn into acid. It is observed from this answer from cooking.SE that at a temperature of 35-45 °C, fermentation occurs. If any sugar laden food is kept at this temperature, bacterial or yeast colony would flourish giving you the fermentation effects and making your food acrid. Fermentation process is a two-step process. First, glucose is turned to ethanol by Saccharomyces cerevisae, also known as Brewer's yeast. This ethanol is then converted to acetic acid by Acetobacter aceti, giving you the sour effect. This two-step process is called aerobic fermentation.

Caramelization generally occurs at higher temperatures, at 150-170 °C where sugars are broken down to simpler products and turning sugar into a brown-black mixture of products. This reaction is also called Maillard reaction. At this temperature, the reaction happens almost instantly but at lower temperature (35-45 °C), this reaction will takes hours, days or even months to finish it but it will speed up if there is moisture where it will take a few hours. Note that caramelization process is very complex as there are various steps. The finished product is a brown-black goo which is actually a mixture of several products. But if put it in a simplified way, sucrose breaks into glucose and fructose which further breaks into acetic acid, maltol, furan and many other products. This acetic acid contributes to the sour taste. So, in a hot-humid day, sugar can turn sour.


What causes the colour when conducting flame tests on solid salts?

There is a very detailed discussion of what cause the emission lines in flames with $\ce{NaCl}$ in The Origin of the Spectra (ironically by P. Darwin Foote et al.) 1922, starting around page 165.
They are saying that for $\ce{NaCl}$ most of the transitions are from atomic $\ce{Na}$, the $\ce{NaCl}$ having vaporized, and Na+ then captured an electron, and a small fraction of the transitions are from the act itself of Na+ capturing the electron.
They calculate that a Na atom emits 2000 D-line quanta per second.
They also calculate the fraction of Na atoms with electrons in the $3p$ state is $\ce{10^{-11}}$ at 1000K, rising to $\ce{10^{-7}}$ at 1500K. (Note: the book says "$2p$", but $2p$ in Fowler notation is $3p$ in modern notation)
Also, they say the vapor pressure of $\ce{NaCl}$ in a Bunsen flame is on the order of $\ce{10^{-6} \ mmHg}$.
So the color is from vaporized $\ce{NaCl}$ (vapor phase $\ce{NaCl}$ being molecular), that has dissociated and formed atomic $\ce{Na}$.
Also in The Absorption of Light by Flames Containing Sodium Phys. Rev. 38, 699 (emphasis added):

The dissociation probably occurs in two or more steps. Possibly the
molecules break up into sodium and chlorine ions which then lose their
charges becoming uncharged atoms. There may also be one or more
intermediate steps in the recombination.  But certainly the sodium
giving the light which we are studying consists of uncharged atoms, as
is evident from its spectrum. Evidence that there are comparatively
few charged sodium atoms in the flame is given by a study of the
conductivity of flames. Wilson, (p. 85) for example, estimates that
only 1.6 percent of the sodium in a fame is ionized even when the
concentration is small.
It also seems reasonable to assume that at the
edge of the flame there is more complete dissociation of the salt than
in the center, and that this is the cause of the difference between
the concentration ratios obtained with light from the center and from
the edges. A fairly definite proof that there is more dissociation at
the edges is given by the appearances of a fame into which copper
chloride is being sprayed. Such a fame gives almost no green color
except at the edges, where it is a bright green. This is the more
noticeable if there is a small amount of sodium impurity in the copper
chloride. The center of the flame is then yellow with a distinct border
of green.
The greater dissociation at the surface is no doubt in some way due to the greater amount of air there, but whether it is caused by a higher temperature
or by a difference in chemical action can not at present be stated.


Dissolving sand (or something similar) in a bottle of water?

To clarify: it sounds like you want a layer of insoluble sediment, so that if you shake the bottle, the water looks cloudy. Maybe it settles back out quickly, or maybe the water stays cloudy.

Then, you add something and shake the bottle again. This time, the sediment disappears.

There are lots of possibilities here. One that comes immediately to mind is lime (calcium hydroxide), which is only slightly soluble in water. Add acid (vinegar if you're timid and don't mind adding a large quantity, hydrochloric acid if you want fast results), and it will be converted to a highly soluble salt.

Chalk (calcium carbonate) would work, too, but it would fizz (produce bubbles of gaseous carbon dioxide) when you added the acid.

If you can say a little more about the context of your demonstration, we might be able to provide better ideas. Is this actually about solubility, or is it a metaphor for something else?


Why doesn't the Brady's test (2,4-DNPH) work on carboxylic acids?

2,4-Dinitrophenylhydrazine (2,4-DNP or 2,4-DNPH) reacts readily with aldehydes and ketones via a condensation reaction (the lone pair of electrons on the terminal amino group in 2,4-DNPH makes it a strong nucleophile and the condensation starts by the nucleophilic 2,4-DNPH attacking the electrophilic carbonyl carbon) to produce the corresponding hydrazone.  The hydrazine is usually a brightly colored yellow, orange or red compound, so the reaction is often used to test for the presences of an aldehyde or ketone

2,4-DNPH does not react with amides, esters or carboxylic acids.  As shown below for the case of an ester, an extra resonance structure can be drawn for these 3 types of compounds as compared to a ketone.  This extra resonance structure delocalizes some of the positive charge away from the carbonyl carbon onto the adjacent hetero-atom (oxygen in the case of the ester or carboxylic acid, nitrogen in the case of an amide).  This makes the carbonyl carbon less electrophilic in these compounds, and consequently attack by the nucleophilic 2,4-DNPH is less favored.


If you count electrons and determine the formal charge on each atom, you find that in structure #1, the negative charge is on the oxygen.  Do the same exercise for structure #2 and you find that the negative charge is on nitrogen.  Since oxygen is more electronegative then nitrogen, the negative charge is more stable when its on the oxygen atom.  Therefore, structure #1 is more stable (preferred) than structure #2.

Structures #3-#5 are even less stable.  They either violate the octet rule (#4, #5) and / or have charges (sometimes multiple) on more than one atom (#3-#5).


Is there a chemical reaction which could give off enough heat underwater to cook something?

Interesting prompt!
There are plenty of reactions which could be performed underwater to release a large amount of energy. In fact, many substances react vigorously with water itself. You may have heard about how alkali metals such as sodium are not compatible with water, and there's a good reason for it!
Explosive reactions aren't particularly suited to cooking food, though, and a sustained flame would be much more useful. Interestingly enough, there was legitimate research performed on making underwater flames, for a very specific purpose: the Olympic torch!
For the 2000 Summer Olympics in Australia, during the classic Olympic torch relay before the start of the games, the torch was actually carried ablaze underwater for three minutes at the Great Barrier Reef. According to Wikipedia, the usual fuel for the burning torch is a mixture of liquefied propane/butane, which burns using oxygen gas in the air as an oxidizer (more details of a torch's construction can be found here). There is very little dissolved oxygen in ocean water, however, so the fuel/oxidizer mixture had to be changed. From the BBC article:

The special flare was developed after nine months by a team of chemists and engineers at Melbourne pyrotechnics company Pains Wessex.
The company's managing director Charles Tegner said designing the torch was a challenge - not only to produce a flare to burn underwater at such a depth, but to burn like the Olympic torch flame as well.
"It had to be clearly visible," he said. "Such flares don't normally exist."
The flare's chemical composition, pressed into a steel tube, produces sufficient oxygen and nitrogen to maintain a very hot flame.
The flare burns so fiercely at more than 2,000 degrees Celsius that this creates enough pressure to keep the water from entering the tube.
Its intensity is produced by a mix of oxygen-generating chemicals as well as the combustible element magnesium in a finely powdered form

I haven't found the detailed composition of the fuel/oxidizer mixture, but this is already a very good hint. Magnesium is an alkali earth metal which burns vigorously with a bright white flame in many conditions, be it in pure oxygen, air, pure nitrogen, or even pure carbon dioxide, and it will also react with water when hot producing hydrogen gas, much like sodium (this is why reactive metal fires are so hazardous; don't even think about using a $\ce{CO2}$ or water extinguisher on them, that just feeds the fire!). In this case, the pyrotechnicians decided to mix finely pulverized magnesium metal with a solid oxidizer, very likely something such as potassium chlorate, $\ce{KClO3}$, which decomposes under heat releasing oxygen gas. The flame was lit in air until the reaction between the magnesium and solid oxidizer started going, and then it was safe to submerge the torch while the reactants lasted. The stunt has in fact been repeated since.
To add a few more options to this answer, there is a general class of reactions with water called hydrolysis reactions, some of which are quite energetic. Here, the starting material tears apart water molecules directly and uses the fragments to modify its structure to a more stable form. For example, this is what happens when phosphorous pentoxide is allowed to come into contact with water. This avidity for water makes phosphorous pentoxide a useful material in a laboratory, as it is capable of scavenging trace amounts of water from a liquid or gas. However, these types of reactions already tend to release much less energy per gram of reactant, compared to combustion reactions.
Lastly, I'll cite one more thing. Even without any chemical reaction, the dissolution of some solids in water is capable of releasing some heat. This happens when starting material has a weaker attraction to itself than to water molecules. This property can be exploited in disposable heat packs. These types of processes usually release even less heat per gram of starting material than combustion or hydrolysis, but at least the process is conceivably easily reversed, provided energy is given to dry the material again.
Therefore, it is possible to get a lot of heat from reactions underwater, though it is quite a more elaborate process than starting a fire in air.
Edit: Now that I saw the original question at Worldbuilding.SE, I realize part of it wondered about how cooking would be done in earlier stages of technological development. In this case, I'm hard-pressed to think of a chemical reaction which happens using materials found easily underwater and releases a lot of energy. The difficulty is great enough that it has actually been suggested previously as a general argument for why technologically advanced or even sentient life would be unlikely to develop underwater, as any hot thermal processing of food or materials would be severely hindered. Except for incredible strokes of luck, hydrothermal vents on the sea floor sound like the most reliable heat source which could be accessed by a primitive civilization.


Help with molecular dynamics and computational modelling to check validity of hypothesis regarding SN2 reaction mechanisms

Your hypothesis concerning a "bouncing" contribution to the reaction kinetics would be incredibly difficult to observe or quantify even with the best supercomputers.

The likelihood of a molecule to react in a manner like the SN2 reaction is actually an incredibly rare event at the molecular time scale (femtoseconds 10^-15 s). This may sound contradictory since these reactions are commonplace and occur readily under appropriate conditions (e.g. a few hours heating). However, the two timescales are separated by over 15 orders of magnitude. Many reactions which do approach the transition state can also revert back to the reactant, or in other words abort. So you'd have to run the incredibly long simulations thousands of times in order to see only a small number of reactions. This is the meaning of the pre-factor in the Arrhenius rate  equation

$k = Aexp(\frac{-\Delta G^\ddagger}{k_B T})$

Nevertheless, some of your intuition is correct. A molecule would indeed require momentum along the transition vector in order to overcome the highest energy point (the transition state) and pass over to the product. Otherwise it would return to reactant. Some of this velocity may naturally come from bouncing off a nearby solvent molecule.

Extra Information:

The topic concerning the relationship of transition states and molecular dynamics is actually very complex and a major challenge in the field of chemistry.  The challenge essentially boils down 1) the size of the system and 2) the time length of the simulation.

Fortunately for most practical cases we can simply ignore dynamical effects and study isolated because it is known that solvent interactions are generally weak and reactions closely follow the minimum energy paths.Therefore, the free energy at the transition state ($G^{\ddagger}$) can be approximated in solution using the Harmonic approximation and implicit solvent models like PCM. If you'd like to see how some of these calculations are done, please check out my GitHub on the growing string method (ref 1). These are much simpler to model and actually give a lot of information. Usually the transmission coefficient is assumed to be 1. or 0.5.

If you're interested in some more advanced considerations of the relationship between dynamics and reaction barriers I'd recommend the following references: ref 2,3

References

"Single-ended transition state finding with the growing string method" Journal of Computational Chemistry 2015, 36, 601–611
"QM/MM Protocol for Direct Molecular Dynamics of Chemical Reactions in Solution: The Water-Accelerated Diels–Alder Reaction" J. Chem. Theory Comput. 2015, 11, 12, 5606-5612
https://cims.nyu.edu/~eve2/string.htm


Does nitrogen tetrafluoride act as an acid or a base in its reaction with water?

Neither. The reaction is NOT one of ion exchange; it is a redox reaction. Thus, $\ce{NF4+}$ works as electron acceptor here, like this:

$$\ce{NF4+ + 2e- -> NF3 + F-}$$

Since the reaction mechanism is unclear, it is possible that some of its stages may have acid-base character (for example, $\ce{F-}$ sometimes acts as a base), but $\ce{NF4+}$ certainly isn't acting primarily as base or acid here.


N.B.: the advice below mostly applies to organic compounds. If you're dealing with inorganic salts, or solutions thereof, the approach will be very different.

For a solid mixture, I would take an IR first. That should allow you to rule out potentially useless chemical tests by identifying the functional groups present in your mixture without wasting any of the sample. This would also give you some indication whether acid/base extraction is likely to be of any use. After that, it seems logical to test solubility in a wide range of solvents and at different temperatures. If you can find one that dissolves your mixture only partially at low temperature, but completely at high temperature, then you can likely separate the components immediately by recrystallization. Judicious use of a centrifuge can help if the crystals form a fine suspension.

In the case of a liquid mixture, I would again take an IR initially. In some cases, one component of the mixture can be chemically separated, and IR results would again be useful in determining the viability of that approach. Ketones and aldehydes, for example, will react to form semicarbazones that can be recrystallized and then identified by melting point. As far as physical methods of separation, standard are distillation and fractional freezing.


Why are intensive properties not additive?

Its because intensive properties like temperature, refractive index, density are same for all the portions of the container of gas. You can't say that if the temperature of one portion is x and and other portion is also x, then the total temperature is $2x$.

But extensive properties, they are different for different portions of the container like mass, volume, length. So they can be added.

It doesn't matter if the intensive property is a ratio of two extensive properties. If two extensive properties are in ratio then if first extensive property changes then the second one should change.


pH of lemon juice and cola is around 2-3. As long as you don't go below this point, drinking should be fine.

It is different if you consider fish living in such water. pH from several ppm would not drop below 5, which is close to rainwater, but significantly lower than normal river or sea water (6-7 and 7-8 respectively). You should consult specialized literature in the case.


How could mass increase when sulfuric acid is added to calcium carbonate?

There are several contributing factors to this outlier observation. First of all, when scientists did mass based experiments, even hundreds of years ago, they recorded mass up to four decimal places. The mass of CaCO3 should have been recorded to 0.1 mg for example 5.0612 g. When you say that mass increases, we do not know the details of mass change. In short, never record mass with one significant figure.

Secondly, how did you measure volumes of 20 mL acid? If you were using cylinders, they are never accurate. There is significant error in volume transfer. With a relatively viscous solution such as 2M H2SO4, if you transfer 20 mL, the transferred volume via cylinder is usually less than 20 mL.

This all about good laboratory practice. Coming to chemistry: when CaCO3 reacts with H2SO4, if forms an insoluble salt called calcium sulfate (CaSO4). With concentrated 2 M sulfuric acid, once it comes in contact with CaCO3, it may form a coating of CaSO4, which would slow down the rate of reaction. Did you see some unreacted solid mass?

Also conc. sulfuric acid is hygroscopic.


Recrystallization and purification techniques for getting a pure sample

Recrystallization is a purification method because it is a slow selective generation of a crystal framework, (mostly) free of trapped impurities. It is only effective as a purification method if done properly. Poor recrystallization techniques; rapid cooling, dramatic changes in solubility with temperature, bad choice of solvent, over evaporation of solvent, can all produce crystals of low purity which are not much better than the starting crop.

Impurities in crystals can occur via three mechanisms;

inclusion: impurities occupy a lattice site within the crystal framework; that is, they become part of a defective crystal unit themselves
occlusion: impurities become physically trapped within the crystal framework. This can be a common occurence for some solvent molecules
adsorption: impurities form a weakly bound interaction to the crystal surface.

The faster the crystallization takes place, the greater the chances these mechanisms have of occurring.  Crystals are specific odd-shaped units that need to pack in nicely to their neighbours to form a good crystal. This takes time. A new molecule may attach to the growing lattice, but if it is not the right fit (such as because it is an impurity), it will leave in favour of a better suited fit attaching to the lattice. I may be showing my age here, but we used to liken crystal growth to a game of Tetris. Heck you can still play it! Done slowly, there is time to rotate all the pieces and get the perfect fit. As you speed things up, it becomes harder to get the right fit, holes are left, and before long the whole crystal goes to the dogs. That is essentially what is happening with rapid crystal growth. It's why you get smaller crystals and greater levels of impurities.

Of course, there other methods for purifying solids which are effective also, if used correctly. Precipitation, sublimation and fractional crystallization are all methods for purifying solids, which can give high levels of purity, matching purities obtained be recrystallization.


If the membrane is permeable to both gases, then both gases will become uniformly distributed between the two containers, and the final pressure will be uniform.  So, if the initial pressures in the two containers are $p_1$ and $p_2$, the initial number of moles in the two containers are $$n_1=\frac{p_1V}{RT}$$ and $$n_2=\frac{p_2V}{RT}$$  The total number of moles does not change, so the final pressure is $$p=\frac{(n_1+n_2)RT}{2V}=\frac{p_1+p_2}{2}$$The final mole fractions in the combined container are $p_1/(p_1+p_2)$, and $p_2/(p_1+p_2)$.  The final partial pressures are $p_1/2$ and $p_2/2$.


For the first subquestion:

Using unit analysis, the answer is very quickly found: The unit of the slope $m$, assuming dimensionless values for the absorbance, is
$$[m] = \frac{[\Delta y]}{[\Delta x]} = \frac{[\Delta A]}{[\Delta t]} = \frac{1}{\text{min}}$$

Now you need a volume in the denominator as well, so dividing by a volume seems like a good idea. Since you need the change in absorption per $\pu{mL}$ of $\ce{1X}$ solution, you divide the slope of the $\ce{1X}$ reaction by $\pu{0.24 mL}$ for the correct value.

For the subquestion with which you seem to have the most trouble: simply apply $c=n/V$.

You will find the amount of substance changed via
$$ \Delta n = \Delta c\times V = \Delta c \times \pu{1.2 mL}$$


Why is 1,2,3,4,5,8-hexahydronaphthalene more stable than 1,4,4a,5,8,8a-hexahydronaphthalene?

Disclaimer:
This post might be hard to read for some people. Some may even consider this a rant. You have been warned.
This answer will question the validity of the used models, as well as the basis for such exercises. (It will also point to a fundamental flaw of this specific exercise.) If you are a student, you might learn concepts that will later turn out to be inaccurate or outdated. This answer should give you some general context, but it will probably not be able to help you in an exam.

Introduction
I have previously stated in the comments how much I hate these exercises, I even called them really, really big garbage. There are several reasons for that. I am going to try to convince you step by step why I dislike these exercises, this one in particular.
Stability is not a very meaningful concept
There is a reason why there is no easy definition for stability. It is too fuzzy a concept with too many different connotations that basically everybody could understand something different. This is an ultimate flaw, and for these exercises, some kind of understanding of 'stability' is implied, but probably never properly defined.
Let's start with something tangible, one possible (and accepted) definition of stable; one that is so trivial you might have not even thought about it. Stable means something exists. In more official terms, i.e. in the gold book, it is called structural stability (10.1351/goldbook.ST07109):

Within the Born–Oppenheimer (adiabatic) approximation, this is associated with an energy minimum on a potential energy surface. Structural stability implies that any change in the coordinates of the nuclei can only increase the total energy.

I'm not absolutely happy with the above definition, as it can be read that only global minima are stable. If you start from a local minima, some changes in the coordinates will lead to a decrease in total energy; that is how different conformations work and eventually also reactions. So the changes implied by this definition should be reasonably small.
Obviously, this definition does not at all help us with the exercise. Both compounds are stable according to this definition, but we cannot say which one is 'more stable'.
From this we could go ahead and rank molecules according to their total energy. As long as we would be looking at isomers, this could actually work, but another question is how meaningful this approach is. I won't and cannot go into any detail or discussion how to obtain the total energy; for the purpose of this post just assume every molecule does have an exact total energy.
In the end the total energy is a number associated with a molecule. It will not tell you anything about the chemistry that molecule does. Almost all tangible and practical information is not contained in this number. Sure it offers an exact and consistent way of sorting things, but –in my opinion– an utterly useless one.
It is also problematic in a semantic way: you are using one term to describe a phenomenon which already has a name.
This definition also will not help in answering the question, because estimating this value is hard. Quantum Chemistry has plenty of options to estimate or calculate the expectation value of these compounds. Some are simple, based on empirical data, e.g. semi-empirical methods, some are a bit more complex, e.g. Hartree-Fock, Density Functional Approximations, some are very complex, e.g. Configuration Interaction, Coupled Cluster. Almost all of them you cannot do on a piece of paper in a reasonable timeframe.
That leads us to the point where we need another definition. The gold book actually has another definition for us (10.1351/goldbook.S05900):

As applied to chemical species, the term expresses a thermodynamic property, which is quantitatively measured by relative molar standard Gibbs energies. A chemical species $\ce{A}$ is more stable than its isomer $\ce{B}$ if $\Delta_\mathrm{r}G^\circ > 0$ for the (real or hypothetical) reaction $$\ce{A -> B}$$ under standard conditions. If for the two reactions:
\begin{align}
\ce{P &-> X + Y} & (\Delta_\mathrm{r}G^\circ_1)\\
\ce{Q &-> X + Z} & (\Delta_\mathrm{r}G^\circ_2)
\end{align}
$\Delta_\mathrm{r}G^\circ_1 > \Delta_\mathrm{r}G^\circ_2$, $\ce{P}$ is more stable relative to the product $\ce{Y}$ than is $\ce{Q}$ relative to $\ce{Z}$. Both in qualitative and quantitative usage the term stable is therefore always used in reference to some explicitly stated or implicitly assumed standard. The term should not be used as a synonym for unreactive or 'less reactive' since this confuses thermodynamics and kinetics. A relatively more stable chemical species may be more reactive than some reference species towards a given reaction partner.

This brings us back to the original dilemma, the implied understanding of stability. While we certainly can estimate the hypothetical relative molar standard Gibbs energy of the reaction, it actually won't tell us much about the chemistry. At least this definition will give us some handling on the exercise, i.e. the problem becomes finding 'numbers'.
Sometimes, but this is very wrong, stability is used as masquerading a different concept altogether, a kinetic concept. Stability may then wrongly be defined as the lower likelihood to react, in other words a lower reactivity means more stability. Now this brings back the already discussed semantic problem, but this time it is actually worse. There is also no reason to not use the proper terminology. The gold book defines reactive (reactivity) as follows (10.1351/goldbook.R05180):

As applied to a chemical species, the term expresses a kinetic property. A species is said to be more reactive or to have a higher reactivity in some given context than some other (reference) species if it has a larger rate constant for a specified elementary reaction. The term has meaning only by reference to some explicitly stated or implicitly assumed set of conditions. It is not to be used for reactions or reaction patterns of compounds in general. The term is also more loosely used as a phenomenological description not restricted to elementary reactions. When applied in this sense the property under consideration may reflect not only rate, but also equilibrium, constants.

So this is pretty much linked to the activation barrier. This is surely nothing that is asked about by the exercise, especially since no reaction conditions are mentioned. However, it felt necessary to be included here, since quite a few people will actually mix up the terms or even use them synonymously.
You might come across terms like air-stable which means you can handle a substance in a normal atmosphere without having to worry that it would decompose or react. This is exactly the problem that I mentioned before.
Another point worth mentioning is that some people use 'stable' to indicate whether a species is long-lived; this is also a wrong usage according to the above definitions. It becomes especially problematic when it is used synonymously with unreactive or even inert; but these are discussions for a later time.
In conclusion: There are two correct ways to define stable/stability. (1) Stable either means something exists as a local minimum on the potential energy surface. (2) As a relative concept, A is more stable than B if it costs energy to transform A into B.
Hyperconjugation and the problem of counting interactions
The gold book again gives us a definition for hyperconjugation (10.1351/goldbook.H02924), but I will only quote a portion:

In the formalism that separates bonds into σ and π types, hyperconjugation is the interaction of σ-bonds (e.g. C–H, C–C, etc.) with a π network. This interaction is customarily illustrated by contributing structures, [...] sometimes said to be an example of 'heterovalent' or 'sacrificial hyperconjugation', so named because the contributing structures contains one two-electron bond less than the normal Lewis formula [...].

The important takeaway from this is, that this is only valid within a formalism that separates into σ and π bonds. This concept does not (in this form) exist in Molecular Orbital Theory.
As previously mentioned, there are various ways to describe electronic structure. Many of them require more than the back of an envelope. Those descriptions are basically modeling the electron density at given nuclear positions; these are just mathematical tools to reproduce observables.
As such, hyperconjugation interactions are also just mathematical models and you cannot measure their stabilising factor. All numbers you might find online or in textbooks, are pretty much partitioning schemes of experimental data.
As a very simple example: You can measure the hydrogenation enthalpy of ethene and propene. You can then assume that the π-bond will (in first approximation) behave equally. The resulting energy difference can then be interpreted as the stabilising effect of hyperconjugation. You can do that for thousands of molecules and find some average energy value for these interactions.
That certainly is a valid scientific approach; the problem comes then if you neglect all the conditions that lead to this value and take it as a given constant.
It should be painfully obvious, that different σ-bonds will interact differently with other bonds. There is no reasonable justification to assume that a hyperconjugation interaction will always have the same stabilising effect. It should therefore come as no surprise that a carbon-hydrogen σ-bond located between two carbon-carbon π-bonds will have a totally different stabilising effect than a terminal carbon-hydrogen σ-bond.
Unfortunately, this is taught as some kind of holy grail of organic chemistry. Counting hyperconjugation contributors is at best a zeroth order approximation, an educated guess, a ballpark figure. Surely, you can use it to do a rough estimation, but you should always be aware about how imprecise this is.
It is also important that for a complete description of the electronic structure, you always need all resonance contributors, n.b. hyperconjugation is a more specialised form of resonance. Since these are hypothetical structures, they cannot be assigned an energy and they cannot be weighted that way. It really does need more involved methods to accurately describe this.
Focusing on this criterion alone also only ensures that you are neglecting any and all other possible influences. Dispersion will certainly have a different effect in one molecule than it will have in the other. There will be other mesomeric, or inductive effects; the conformational space will also be of concern.
So while $\pi \approx 3$ is a good guess to roughly estimate the size of a round window, you probably shouldn't use it to built your house.
The problem is the question
Now that we have gotten all of the general problems out of the way, there was one problem pointed out previously in the comments by user55119:

There is a cis/trans issue with the second structure. You might explore the heats of hydrogenation/combustion of hexenes that bear the substitution of the double bond. Welcome to ChemSE!

As if we hadn't already enough problems, the question is already ambiguous by specifying the molecules. In my personal opinion, this shows that the instructor did not spend enough time actually posing the exercise, really thinking about it. It shows that this exercise is targeted exactly at this specific counting exercise. To be honest, I would not have been surprised if the approximation had yielded a completely different result.
To be on the safe side, I performed a couple of calculations. First I ran a conformational analysis with CREST on the xtb level of theory, then I optimised the lowest structures with RI-BP86/def2-SVP in Orca. These are the relative electronic energies. (I did not bother with thermal corrections, since they are not accurate enough for this level, and qualitatively will probably not change the general order.
(1)          1-2-3-4-5-8-hexahydronaphtalene      0.0 kJ/mol
(2)    cis-1-4-4a-5-8-8a-hexahydronaphtalene     48.8 kJ/mol
(3)  trans-1-4-4a-5-8-8a-hexahydronaphtalene     26.8 kJ/mol

You can see from these numbers, especially between 2 and 3 where the number of hyperconjugation interactions is the same, that there are quite large differences. So using hydrogenation or combustion values is probably a better approach to this kind of question.
Additionally, those numbers don't tell you much about the molecules themselves or the chemistry they most likely will perform. In general I'd expect 1,4-cyclohexadiene derivatives to readily undergo oxidation to an aromatic species, but that is besides the point.
My personal opinion is that these counting models oversimplify chemistry beyond the point of usefulness. They do not further any understanding of molecular and electronic structure. It turns useful and educational models into formulaic use cases, which are applied without further understanding.


How can it be certain that some bonds are 180 degrees?

If we look at the molecule we see it is symmetrical. The forces acting upon the 2 hydrogen atoms are identical. As you know, when hydrogen forms a covalent bond with carbon, it's electron is drawn more strongly by the carbon than by the hydrogen, which leaves the hydrogen with a net positive charge (carbon is more electronegative than hydrogen).

Also, the 3 covalent bonds between the carbons are kind of bowed outward in the middle. The 3 bonds between the carbons constitute the most negative part of the molecule.

Naturally, the positive and negative charged moieties are drawn toward each other as much as possible. But, given that the force of attraction is equal on all sides of the hydrogen, it will be aligned equidistant on all sides to the negative triple bond.

I don't know if I am painting a good word picture here, but the forces exerted on the hydrogens place them exactly on a straight line, i.e. their bonds are 180 degrees.

Of course there are other forces in play, namely the forces of neighboring molecules. However, they will also align themselves as much as possible so as to minimize the repulsion of having hydrogen atoms close to one another and to maximize proximity of hydrogen atoms to the triple bonds of neighboring molecules. This arrangement will be preserved as much as possible as the molecules move about. So, the net forces upon the hydrogen-carbon bonds will always force those bonds to remain 180 degrees.


Algebra of conserved quantities for chemical reactions

What I think you are asking about is the number of thermodynamic parameters which characterize a chemical reaction. Those parameters are used to make quantitative predictions of the final state of the reactive system (at least in the context I'm familiar with, i.e. chemical engineering); for example, the amount of heat exchanged with the reactor's environment, the composition of the mixture at equilibrium, it's material phase, overall density, etc.

In classical thermodynamics, the way to address this problem is through the potentials defined on the system of interest, and studying equilibrium as the extremum of one these. The chosen potential depends on the particular configuration of the system under study; and in all cases their relation with equilibrium is a consequence of the Second Law. For example, if the system is restricted to evolve at constant volume and entropy, the minimized potential is the internal energy; if it evolves at constant energy and volume (an isolated simple compressible system) it maximizes entropy; etc.

From a practical point of view of natural systems and industrial processes -which evolve at nearly constant temperature and pressure fixed by the environment- the involved minimized potential is the Gibbs free energy. That the minimization of $G$ yields an equilibrium state for a closed system at constant pressure and temperature is straightforward (it can be found in any physical chemistry textbook). Moreover, when we say here that the system at it's minum $G$ is at equilibrium, we are saying it's at overall equilibrium: it's simultaneously at mechanical equilibrium (because of null pressure gradients), thermal equilibrium (because of null temperature gradients) and in chemical equilibrium (because of null chemical potential gradients); so there's no "force" (in a generalized sense) to drive any further process.

So, going back to your question, which are the independent variables of a reactive system as such? Well, suppose that in your system you have $P$ chemical species which can be both reactives or products of $Q$ reactions taking place in a monophase. For every species, the balance on the number of moles of each one yields:

$n_{i}=n_{0i}+\sum_{j=1}^Q{\nu_{i}^j\cdot \xi_{j}}$

Where $\xi_{j}$ is the extent of the j-th reaction.

Then, we can define the chemical potential as the partial molar Gibbs free energy per species:

$\mu_{i}=\displaystyle\frac{\partial G}{\partial n_{i}}$

And also, given that $G$ is an extensive property (or better phrased, a homogeneous function of first degree in mass) by Euler's Theorem it follows that:

$G=\sum_{i=1}^P\mu_{i}n_{i}=\sum_{i=1}^P\mu_{i}\cdot (n_{0i}+\sum_{j=1}^Q{\nu_{i}^j\cdot \xi_{j}})$

Every chemical potential can be expressed as the sum of it's specific Gibbs free energy of it's pure form, at the temperature and pressure of the mixture, plus a term which ponderates the influence of concentration and chemical affinity of each particular species towards the rest of the present compounds which constitute the mixture (for example, the solvent matrix, if one species is so concentrated that it can be considered so). The latter term contains an adimensional parameter known as the activity of the species:

$\mu_{i}=\mu_{0i}(P,T)+RTln(a_{i})$

The zero index in the first term just means that it's referred to a pure system. This specific free energies can be related to the formation enthalpies and entropies which are tabulated through:

$\mu_{0i}=h_{0i}(P,T)-T\cdot s_{0i}(P,T)\\\mu_{0i}=\left\{{h_{fi}^0(P_{0},T_{0})+\displaystyle\int_{(P_{0},T_{0})}^{(P,T)}dh_{0i}}\right\}-T\cdot \left\{{s_{fi}^0(P_{0},T_{0})+\displaystyle\int_{(P_{0},T_{0})}^{(P,T)}ds_{0i}}\right\}$

Where $h_{f}^0(P_{0},T_{0})$ and $s_{f}^0(P_{0},T_{0})$ are the standard formation enthalpy and entropy, respectively. The integrals evaluate the variation in enthalpy and entropy of the pure compounds due to a difference in temperature and pressure in reference to standard values (25 ºC and 1 bar). The evaluation of the integrals in terms of measurable properties must be done using Maxwell relations and definitions of potentials:

$dh_{0i}=\displaystyle\frac{\partial h_{0i}}{\partial {T}}dT+\displaystyle\frac{\partial h_{0i}}{\partial {P}}dP=C_{p}dT+\frac{dP}{\rho}\\ds_{0i}=\displaystyle\frac{\partial s_{0i}}{\partial {T}}dT+\displaystyle\frac{\partial s_{0i}}{\partial {P}}dP=Cp\frac{dT}{T}-\frac{\alpha}{\rho}dP$

What about the activities? These are non-linear functions of the composition, temperature and pressure of the mixture, and usually that non-linearity is expressed through an activity coefficient (it may be the case that further coefficients are introduced theoretically, but all they do is to "partition" the contributions to non-ideality into a product of these, and so here I talk about a unique coefficient without loss of generality). So:

$a_{i}=x_i\cdot\gamma_{i}(P,T,x_1,x_2,...,x_P)$

And:

$x_i=\displaystyle\frac{n_i}{\sum_{k=1}^P n_k}=\displaystyle\frac{n_{0i}+\sum_{j=1}^Q{\nu_{i}^j\cdot \xi_{j}}}{\sum_{k=1}^P\left\{{n_{0k}+\sum_{j=1}^Q{\nu_{k}^j\cdot \xi_{j}}}\right\} }$

Finally, the (enormously complicated) expression for G turns out to be dependent of the variables:

$G=G(P,T,h_{fi}^0,s_{fi}^0,Cp_{i},\rho_{i},\alpha_{i},n_{0i},\xi_j) \ \ \ \ \ \ i: 1 < i < P \ \ \textrm{and}\ \ \ j: 1 < j < Q$

Those are, a priori, the "independent quantities" involved in the overall reaction's equilibrium. Once you've fixed the temperature, pressure, the original amount of substances and get the intensive properties stated for all the chemical species present in the system along it's evolution, G becomes only a function of the extents of the reactions:

$G=G(\xi_j) \ \ \ \ \ \  j: 1 < j < Q$

And the equilibrium is reached when $G$ is at minimum; that's a problem solvable through optimization. The values of the extent of every reaction at equilibrium is enough to calculate every other useful parameter of the system: it's composition, the heat involved, the change in overall volume, etc.

Summary: the conserved quantities in a chemical reaction are physical constraints (conservation of total mass, balance of moles according to stecheometry, conservation of energy exchanged with the environment, strict increase in entropy, etc) but that's not enough information to uniquely evaluate the equilibrium parameters of a general reaction (or reactions). Most of the expressions I left in an undefined closed form in this answer (such as the activity coefficients, all of them functions of compositions) must be experimentally correlated, or modelled with some theoretical formalisms, generally highly complex. The standard thermodynamic potentials might be tabulated though, as it's the case with the intensive properties of pure substances (specific heats, thermal expansion coefficients, etc).


What Would Hypochlorite Ions Prefer to React with, Hydrogen Ions or Ozone?

First of all, hydronium ion, $\ce{H3O+}$ is basically a proton in aqueous solution. A proton undergoes hydrogen bond in aqueous solution to forms various forms of hydrated proton. But for simplicity, we denote $\ce{H+}$  which is considered an acid. Now, UV light dissociates hydronium ion in ozone atmosphere into oxygen, singlet oxygen, proton and water. So, if hypochlorite  salt or its corresponding acid is introduced, it might react with these. Chloride ion in form of salts is very much stable and is unlikely to react with any of the dissociated products.

So, first of all, any hypochlorite salt or hypochloric acid added will dissociate into component ion in presence of UV light. Then will react with the following dissociated products:

Singlet oxygen

$$\ce{ClO- + 2O ⇄ ClO3-}$$

(Source, considered hypochloric acid)

Proton

$$\ce{2HClO + 2 H+ + 2  e{−} ⇌ Cl2(g) + 2H2O}$$

Water is present as moisture so it is unable to attack hypochlorite ion readily. Also oxygen is present in minute amount(mainly singlet oxygen is present), so it cannot oxidize hypochlorite ion.


A couple of things, first, inden-1-ol tends to tautomerize into the corresponding 1,3-dicarbonyl.  Second, when adding TsCl to an alcohol, the group formed is -OTs not -OTMS, the latter is a silyl ether.  The reaction with addition of TsCl requires a base such as pyridine, and because of the presence of a 1,3-dicarbonyl, rather than being added to the alcohol, it will be added to the carbon alpha to both the carbonyls.  Third, the reaction of Grignard reagent should be avoided due to loss of yield because of interaction with the carbonyl.  Fourth, the inden-1-ol formation is done in the presence of a base (to form enolates) which is then followed by an acid wash(to form enols which tautomerize to ketones), your mechanism seems okay if enolization occurs instead of tautomerization.  Finally, the correct structure of Dimetindene is below:

https://en.wikipedia.org/wiki/File:Dimetindene.svg

I will briefly explain how I would do it from the indenol you described.  First, would be a Wittig reaction using 1-(pyridin-2-yl)ethylenetriphenylphosphorane, (which can be prepared from 2-acetylpyridine) and the indenol tautomer.  This will yield a beta-gamma unsaturated ketone.  This product can undergo a photochemically induced [1,3] hydride shift to form an alpha-beta unsaturated ketone.  This can be subject to Wolff-Kishner conditions to form Dimetindene.


Why is p-toluidine more basic than m-toluidine?

In p-toluidine you are not actually considering the hyper conjugating structures due to which the compound becomes more basic than m-toluidine which only has it's +I effect and I would like to suggest you that as far as I know resonance effect "usually" overpowers the other effect!


how to calculate the initial concentration of NaOH in a mixture?

As @MaxW pointed out, if you assume that the volumes are additive implies that your total volume is 40 mL. Now, remember the definition of molar concentration:
$$C=\frac{\text{moles solute}}{\text{total volume}}=\frac{n}{V}$$

That means that the concentration and the volume are inversely proportional, thus if initially, you have a $\text{0.0400 } \frac{\text{mol}}{\text{L}}$ in 20 mL and then you increase the volume twice, then your concentration decrease to the half of initial value. In numbers:
$$\frac{\overset{\text{this are your moles of solute}}{\overbrace{\frac{\text{0.0400 mol}}{\text{1 L}}\times{\text{20 mL}}}}}{\underset{\text{this is your final volume}}{\underbrace{\text{40 mL}}}}=\text{0.0400 }\frac{\text{mol}}{\text{L}}\times\frac{1}{2}=\text{0.0200 }\frac{\text{mol}}{\text{L}} $$


Same set of carboxylate reagents yielding two different products with an alkylide

Reaction 2 is wrong. Reaction 1 is correct.

Later in the book (q87), another question is asked with the same set of reagents (carboxylic acid and methyllithium). Reaction 1 has been used in that as well.

Reaction 2 is wrong because, as ralk912 noted above, charge conservation has to be followed. If methyl lithium reacts with a negatively charged species, the product has to be negatively charged as well. If the ester is neutral, then lithium must be negatively charged, and would associate with the $\ce{Li+}$ formed in step E to form $\ce{Li2}$, which is only observed in gas phase.

Not to forget that there simply isn't any mechanism wherein a carboxylate would react with an alkylide to form an ester. It does not simply occur. Reaction 2 is wrong.

Waylander has suggested a good reason for a misprint i.e. the reagent $\ce{CH3Li}$ might have been intended to be $\ce{CH3I}$ instead. Then, it would make sense for the carboxylate anion to attack an alkyl halide, and displace an iodide ion (which could easily attach with the $\ce{Li+}$ ion to form lithium iodide) forming the methyl ester.

user55119 has confirmed that the product of the step F is cyclohexylmethyl ketone, and that it is obtained by reaction 1 only. The book containing this is "Margret Jorgensen in Organic Reactions."


What is the difference between the roles of a surfactant and a stabilizer in the process of nanocrystal growth?

CTAB is a cationic surfactant, SDS is an anion surfactant.

These surfactants act to selectively stabilize ionic faces on the crystal as it grows, causing those faces to grow more rapidly. This leads to a change in geometry or crystal shape. CTAB causes one morphology being cationic, while SDS will cause another, its head ion having the opposite charge. Further, a mixture of the two will cause an intermediary morphology.

For instance, CTAB will cause an aqueous solution of thioacetamide and Lead(II) acetate to form PbS nanocrystals in a octahedron shape, while SDS will cause them to form a dendritic fern like shape, while a CTAB/SDS mixture (5:1) will cause a stellated octahedron or star shape to form.


You are correct about the absolute numbering of $\ce{CH2=CH-C#CH}$. By the IUPAC priority rules, the alkene has the lower absolute numbering.
However, numbering can also be relative. In your case, the relative numbering 1,2 refer to either of the following:

relative position from the end of the chain, where the chain end is #1
relative position from the functional group you care about (where that group is #1)
suggesting that a reaction occurs on neighboring positions in a molecule.

That being said, the definition given by Brown et al. is imprecise. A better definition is one of the following:

A terminal alkyne is an alkyne in which one (or both) of the two alkyne carbons is at the end of a chain.
A terminal alkyne is an alkyne $\ce{RC#CR}$ in which one or both $\ce{R}=\ce{H}$


What is the behavior of a Zwitterion during ion exchange chromatography?

I would expect it to depend on the pH of the mobile phase. The charge of your stationary phase and whether you want the azide to elute will determine what charge you want on the azide. Given that, you should set the pH of the mobile phase either acidic or basic enough to ensure that the azide is fully charged in the direction that you want.

From Wikipedia:

  Proteins have numerous functional groups that can have both positive and negative charges. Ion exchange chromatography separates proteins according to their net charge, which is dependent on the composition of the mobile phase. By adjusting the pH or the ionic concentration of the mobile phase, various protein molecules can be separated. For example, if a protein has a net positive charge at pH 7, then it will bind to a column of  negatively-charged beads, whereas a negatively charged protein would not. By changing the pH so that the net charge on the protein is negative, it too will be eluted.

If changing the pH is problematic, apparently increasing the ionic strength can help shield the azide's charge from the column's charge:

  Elution by changing the ionic strength of the mobile phase is a more subtle effect - it works as ions from the mobile phase will interact with the immobilized ions in preference over those on the stationary phase. This "shields" the stationary phase from the protein, (and vice versa) and allows the protein to elute.


Distillation: temperature not reaching azeotrope boiling point

I think your major mistake is using a simple distillation method, which may not give accurate reading for azeotropic distillation. The early distillate may not have enough cooling to get to expected azeotrope temperature of aqueous $\ce{HBr}$. If you have followed the work of Carriére et Cerveau (Ref.1), you may able to assume approximate composition of your fractions. Although their work cannot be regarded as of high precision as the composition was determined by titration (Ref.2), they have given an insight to $\ce{HBr}$ distillation. Their work has clearly showed aqueous $\ce{HBr}$ has been yielding a "high-boiling" azeotrope (boiling temperature of azeotrope is higher than boiling points of either of pure components). For example, the $\ce{HBr}$ composition of the early distillate collected at $\pu{110 ^\circ C}$ was reported as 50.8%. The temperature had increased to $\pu{126 ^\circ C}$ maximum (47.5% $\ce{HBr}$), and then declined passed  $\pu{111.5 ^\circ C}$, at which, the $\ce{HBr}$ composition was found to be 30.8%. Since your initial $\ce{HBr}$ concentration was around 30%, I’d say your distillate is already at downward slope and never have able to reach constant boiling azeotrope temperature. If you had used fractional distillation setup with long enough column (and measured atmospheric pressure accurately), you may have able to collect some azeotrope. For your convenience, I listed published data of azeotrope boiling point, density, and composition at different pressure (measured by pressure control apparatus):

$$
\begin{array}{ccc}
\text{Reference} &\text{Pressure} & \text{Boiling Point} & \text{Density at}\: \pu{25 ^\circ C} & \text{Composition, % HBr}\\\hline
\text{Carriére et Cerveau}^1 & \pu{760 mmHg} & \pu{126 ^\circ C} & - & 47.5\\
\text{Ewing and Shadduck}^3 & \pu{760 mmHg} & \pu{125 ^\circ C} & - & 47.795 \pm0.033\\
\text{Bonner et al.}^2 & \pu{760\pm0.5 mmHg} & \pu{124.3\pm0.02 ^\circ C} & - & 47.63\pm0.01\\
\text{Bonner et al.}^2 & \pu{100 mmHg} & \pu{74.12 ^\circ C} & 1.5116 & 49.80\\
\text{Bonner et al.}^2 & \pu{200 mmHg} & \pu{90.35 ^\circ C} & 1.5030 & 49.28\\
\text{Bonner et al.}^2 & \pu{300 mmHg} & \pu{99.91 ^\circ C} & 1.4961 & 48.83\\
\text{Bonner et al.}^2 & \pu{400 mmHg} & \pu{107.00 ^\circ C} & 1.4908 & 48.47\\
\text{Bonner et al.}^2 & \pu{500 mmHg} & \pu{112.94 ^\circ C} & 1.4866 & 48.19\\
\text{Bonner et al.}^2 & \pu{600 mmHg} & \pu{117.82 ^\circ C} & 1.4832 & 47.95\\
\text{Bonner et al.}^2 & \pu{700 mmHg} & \pu{122.00 ^\circ C} & 1.4802 & 47.74\\
\text{Bonner et al.}^2 & \pu{800 mmHg} & \pu{125.79 ^\circ C} & 1.4775 & 47.56\\
\text{Bonner et al.}^2 & \pu{900 mmHg} & \pu{129.13 ^\circ C} & 1.4752 & 47.40\\
\text{Bonner et al.}^2 & \pu{1000 mmHg} & \pu{132.12 ^\circ C} & 1.4733 & 47.27\\
\text{Bonner et al.}^2 & \pu{1100 mmHg} & \pu{134.80 ^\circ C} & 1.4716 & 47.14\\
\text{Bonner et al.}^2 & \pu{1200 mmHg} & \pu{137.34 ^\circ C} & 1.4700 & 47.03\\
\end{array}
$$

References:

MM. E. Carriére and Cerveau, Compt. rend. Acad. Sci. 1923, 177, 46-48 (https://gallica.bnf.fr/ark:/12148/bpt6k3130n/f50.image).
W. D. Bonner, L. G. Bonner, F. J. Gurney, “Azeotropic Hydrobromic Acid Solutions at Pressures of 100 mm to 1200 mm,” J. Am. Chem. Soc. 1933, 55(4), 1406-1409 (DOI: 10.1021/ja01331a012).
D. T. Ewing, H. A. Shadduck, “The Composition of a Constant-Boiling Solution of Hydrogen Bromide in Water,” J. Am. Chem. Soc. 1925, 47(7), 1901–1904 (DOI: 10.1021/ja01684a017).


Are all homogeneous mixtures just solutions?

Homogeneous and heterogeneous classifications are scale-dependent. What we mean when we say a mixture is homogeneous is that there are no visible phase differences on the scale of interest.

For example, milk (when homogenized) and lotions are both examples of colloidal suspensions. There are no phase differences at the human scale, but neither one is a solution. On a more microscopic scale, you could call the suspended micelles phase boundaries, and in that case might not call the mixture homogeneous.

"Solution" always implies mixing on the molecular level (single-phase).

To summarize:

  Solutions are always homogeneous mixtures, but homogeneous mixtures are not always solutions.


Why can't small trans-cycloalkenes exist?

Because you're stuck in the ring structure.

The trans isomer has branches that, in their natural state, would want to rest antiparallel to one another. But in double bond 6 membered ring structure (like in cholesterol), these bonds would need to be twisted a lot so that they could sit in the octagonal ring structure. Given how a double bond is rigid and will already resist rotation, this theoretical trans isomer will have such a high energy requirement for it to exist that it will simply refuse to do so.


Which metals with high density do not react with mercury?

Here is a link to a chart of all the elements in the Periodic table and their densities.  Iridium at $22.65~\mathrm{g/cm^3}$ is the densest of all the elements with Osmium a close second at $22.61~\mathrm{g/cm^3}$.  Sigma-Aldrich sells both of these metals (Iridium \$126.50 for 0.5 gram; Osmium \$195 for 1 gram).

  the highest density material between 10 17 range that is less than
  mercury and does not react with mercury

Mercury has a density of $13.534~\mathrm{g/cm^3}$.  It sounds like you're asking for something with a density right below mercury that doesn't react with mercury. Rhodium at $12.45~\mathrm{g/cm^3}$ is probably the most readily available element with a density less than mercury.  As to rhodium's reaction with mercury, an alloy is formed at 175 C.  If you're working at room temperature, especially for shorter periods of time (like hours) you'd probably be OK, but it really depends on your specific application.


Solubility curve for sodium tetraborate in water

The solubility curve of borax is taken from here.

We can infer that solubility of borax at 100°C is around 520g/L. It is observed that borax in water exhibits supersaturation . When the solution is at 100°C and then you drop at 40°C, then a clear solution is obtained when amount of borax is 60 grams per litre. More than that, there is a tendency of the solution to overflow. See ref.1 for the more detailed explanation of supersaturation and ref.2 for the dissociation reactions happening in water. First the tetraborate ion is formed and it is a weak base in solution and thus reacts with additional water to form boric acid, a weak acid and hydroxide ion.

$$\ce{[Na2B4O5(OH)4].8H2O <=> 2Na+ + B4O5(OH)4^{2-} + 8H2O}$$

$$\ce{B4O5(OH)4^{2-} + 5H2O <=> 4H3BO3 + 2H2O}$$

Reference

https://www.enotes.com/homework-help/how-much-borax-should-dissolve-hot-water-order-721345
https://www.google.co.in/url?sa=t&source=web&rct=j&url=https://www.smc.edu/AcademicPrograms/PhysicalSciences/Documents/Chemistry_12_Experiments/Solubility%2520and%2520Borax.pdf&ved=2ahUKEwjr5b6W1tXdAhWRbisKHb1rBUAQFjAXegQIBBAB&usg=AOvVaw3c9IA0jqEROPS534lSedA8


Why does water spills on our shirt change its colour?

This is not so much a change (physical and especially not chemical), it is more about perception of the diffuse reflection and refraction of light. According to Australia's ABC Science page Darker when wet:

  When your clothes are wet, all those air gaps (between the fibres) get filled with water. So when light hits a wet patch it's moving in and out of water and fibres, not air and fibres. The refractive index of fabric is a lot closer to that of water than it is to air, so the light doesn't change speed quite as dramatically going between water and fibres. And that means it bends at a smaller angle when it goes from one to the other.

This is summarised in the diagram below:

The information on the linked page is from Professor David Jamieson from the School of Physics at The University of Melbourne, Australia.

Another explanation according to Dr. Christopher Baird of the website Science Questions with Surprising Answers is that we perceive a darkening due to the fabric becoming more transparent - an explanation from the website:

  Spill some water on your white shirt and the situation changes. In the
  spot that is wet, the water fills all the little pockets in the fabric
  that used to be filled with air. In terms of its interaction with
  light, water behaves very similar to fiber material. Adding water to
  fabric effectively removes the multiple reflecting surfaces (at least
  as far as the light is concerned) and turns it back optically to a
  solid chunk of material, which is transparent.

Essentially, what we are perceiving when the fabric appears to darken, is due to the colour of what is underneath - view the cloth with a light source behind it, it appears 'lighter', due to being more transparent.


If A is a stronger acid than B then is B necessarily a stronger base than A?

In general this is not true.

For instance, water is a stronger acid than ethane and is also a stronger base than ethane.


I will ignore the IR as I always do and work only with the NMR.

The ten aryl protons are probably two phenyl groups ($\ce{C6H5-{}}$). That leaves $\ce{C4H6O2}$ and takes care of eight of the nine double-bond equivalents.
The two triplets are very likely (examining the coupling constant would confirm that) part of a $\ce{-CH2-CH2-{}}$ fragment. On either side of these is either an oxygen or a quarternary carbon as there are no additional couplings.
The final singlet points to a $\ce{-CH2-{}}$ group with the same constraints as above.
We still have $\ce{CO2}$ and one double-bond equivalent to take care of. One functional group particularly suggests itself. Then, we have to play connect the fragments.

   $\ce{CO2}$ is likely an ester functionality since we do not have a carboxylic hydrogen to spare.

   One of the two triplet $\ce{CH2}$ is shifted downfield rather strongly. This suggests that it is directly bound to the ester group oxygen, giving $\ce{-CH2-CH2-OC(=O)-{}}$.

   Both of the remaining $\ce{CH2}$ groups will be benzylic but the singlet is more deshielded than the triplet. That suggests that the singlet $\ce{CH2}$ is not only benzylic but also α to the ester.

   Therefore, the final result is $\ce{C6H5-CH2-CH2-OC(=O)-CH2-C6H5}$.


Is there a general way to predict products of a chemical reaction

Often times it is possible due to previous data. Trends appear and you can make deductions from such trends. Reactions may not happen at all. There may be deviations from the trends as well. The deviations may be for unknown reasons or there may be unproven theories that are included in textbooks. Inorganic chemistry reactions are not as clear cut as organic chemistry reactions. You don't have to memorize all the reactions. Know the trends, conduct the experiment, and if your prediction fails, try to think hard about the conditions in which the experiment took place. Sometimes the explanation may lay in the container that you used for the reaction, or it may be stoichiometry.

As in all sciences, you need experiments.


sp2 hybridized carbon versus sp3 hybridized silicon

Carbon — and all the other elements of the second period, for that matter — is exceptionally good at forming double bonds. A double bond is formed if there is a σ bond and a π bond. Forming a σ bond is not hard: You have two orbitals coming from two different atoms pointing at each other. Forming a π bond is much harder, since the orbitals are pointing orthogonally away from the bond-axis. Thus, the overlap of π bonds is often a lot weaker than for σ bonds.

This gets more important if you consider third-period elements rather than second-period ones. They have a total of eight more core electrons making them much larger in size and the orbitals more diffuse. A σ bond is still formed easily, there will be the point when two diffuse orbitals are close enough to interact in a σ fashion. To arrive at favourable interactions for a π bond would require the atoms to come much closer than what would be comfortable; unfavourable interactions between the core electrons would already be a thing.

In fact, I was taught in my Concepts of Inorganic Chemistry class, that disilene $\ce{Si2H4}$, the heavier homologue of ethene, does not form a classical π bond. Instead, it can be understood as a par of two singlett silene fragments $\ce{SiH2}$ which approach each other so that the empty p-orbital of one silene can interact with the lone pair in an $\mathrm{sp^2}$ orbital of the other silene fragment. Obviously, the symmetry is lower and the bond is much weaker.

So the reason behind it is that $\ce{Si=Si}$ double bonds are hard to create and this gets even harder for heavier homologues while $\ce{C=C}$ double bonds are reasonably stable and easy to generate.


Is m-cresol or p-cresol more reactive towards electrophilic substitution?

To start, it is important that you understand that the methyl and hydroxy groups are activating only in the ortho and para positions. For example, if I were to substitute plain phenol, which contains an activating hydroxy group, which position would my electrophile (in this a bromine electrophile) add to? You can see that the intermediary sigma complex involves a positive charge. When the bromine attacks at the ortho or para (I did not include a diagram for the para attack, but it is similar to the ortho attack) positions, it is stabilized by an additional resonance structure provided by the oxygen lone pair. In the case of the meta attack, as you can see in the diagram below, this does not happen, and the intermediate is left with one less resonance structure.

Most activating groups will have the same directing effects as shown with the activating hydroxyl group above (para-ortho directors), so a methyl group, which contains no lone pair (and thus no additional resonance structures) will still stabilize the intermediate by donating electron density in much the same way as if an additional resonance structure was present.

Now getting back the para and meta cresol, the meta cresol contains two activating groups

Both of these activating groups activate the same positions, so this combination of activating effect will make the meta cresol more reactive than para-cresol. In para-cresol, the two groups do not work to activate the same positions, so, no matter which carbon is substituted, it will not be activated by both groups simultaneously like in meta-cresol.


Urea ($\ce{H2N-CO-NH2}$) is the quintessential amide. It is extremely soluble in water, and you could say it is a hydrogen donor and acceptor in water.
However, if you replace one of the $\ce{-NH2}$ groups with an alkane or other organic group, solubility is reduced, depending on the size of the group.
Amides generally have higher boiling points and melting points than related alcohols and acids - because of their unique ability to be internal hydrogen donors and acceptors. This donor-acceptor capability makes their solid state more stable and less able to be solvated to a greater degree in water. You could think of the internal bonds being like a dimerization, where the organic chain effect is doubled, while the double $\ce{-CO-NH2}$ interaction gains little benefit from solution in water.


Is there a soluble salt, or compound that when dissolved in water results in a solution with density less than pure water?

No, you can't dissolve a salt to make a lower density solution.

Instead of thinking about salt ions just fitting into water molecules, think about it as if the salt ions are letting everything fit together better. The water molecules are achieving their current density because of the presence of hydrogen bonding. When you add the ions you introduce better bonds between the charged ions and the polar water molecules. These bonds allow everything to fit together tighter leading to a rise in density.

This decrease in overall bond length is a key to the spontaneity of the solvation reaction. If you introduce an ion that increases the bond length you would see no solubility because they system would be more energetically stable as a bulk solid and water with hydrogen bonding.

The polyelectrolyte you bring up raises a good point, but it is not an exception to this rule. Instead, the polyelectrolyte is so large that it "solvates" into water to form a gel. This new material is not really a solution so much as just a new material with water in it. Hydrated salts would be another example.

This should be the case for all solute-solvent systems. However, this is not to say what you want can't be done. You may have to looks at some reversible reaction or cis/trans type switching material. A simple solute-solvent system won't get it done.


Actually, the more the polarization of a bond, the stronger it is, since both the atoms are drawn close together due to electrostatic attraction, decreasing the bond length and increasing the bond stability (think about ionic bonds. Your reasoning would dictate them to be very weak, while we know how strong they are!). Actually, I would predict $\ce{CH3-CF3}$ to have a really small bond length (the bond is probably the strongest, would someone confirm this please?), since the fluorinated carbon has quite a positive charge, while the other one is relatively negatively charged (not negative in the absolute, just relative to the other carbon).

All the remaining options have equal number of fluorines on either carbon, differing only in their number. These three are tricky. Consider $\ce{CF3-CF3}$. Both of its carbons are severely deficient in electrons. Each badly needs electrons, and since it can't argue with fluorine (fluorine is an unreasonable bully), so both carbons try to drag the shared pair of electrons towards itself. This ultimately results in them both coming closer to each other. As we reduce the number of fluorines on each carbon, their desparation is reduced, thus increasing the bond length. Thus the answer is (B).

Edit: If @pentavalentcarbon's calculations of bond lengths are correct, then my reasoning was too simplistic, and indeed it seems so now. So if we dig deeper, $\ce{CH2F-CH2F}$ could have the second smallest bond length and $\ce{CF3-CF3}$. Why? Well, it is because of hyperconjugation, similar to the gauche effect (read it up to get better understanding of the gauche effect, although I explain it in short here also).

So, what happens is that in the case of $\ce{CH2F-CH2F}$, the bonding C-H molecular orbital (MO) overlaps with the antibonding C-F MO, transferring some of its electron density to the C-F MO (see diagram). This interaction is stabilizing (since delocalization of electrons stabilizes the molecule), and also introduces slight double bond character in the C-C sigma bond, thus serving to shorten it.

Now in $\ce{CH2F-CH2F}$, there will be two such interactions, but then there are also two such interactions in $\ce{CHF2-CHF2}$! There is a difference though. In the latter case, there are two fluorines on each carbon, which heavily draw electron density away from the carbon through the sigma bonds. This decreases the sigma bond density in the C-H bond, which being poor in electrons, gives relatively low stabilization when overlapping with the antibonding C-F MO. So $\ce{CHF2-CHF2}$ has a slightly higher bond length than $\ce{CH2F-CH2F}$.

Using this same reasoning, we can see why $\ce{CH3-CF3}$ has the smallest bond length (three favourable hyperconjugational interactions!) and $\ce{CF3-CF3}$ has the longest bond length (no hyperconjugation).


(Note that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’.)
Simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl, methoxy, and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.
For example, ‘1,2-dibromo-’ is considered to begin with ‘b’; ‘isopropyl’ is considered to begin with ‘i’.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order in your first example corresponds to the name x-isopropyl-y-methoxy-z-methylbenzene.
In your second example, the correct alphanumerical order corresponds to the name x-isopropyl-y-methylcyclohexane.
Furthermore,

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g.
Therefore, your first example is named as 1-isopropyl-2-methoxy-4-methylbenzene rather than 4-isopropyl-3-methoxy-1-methylbenzene since the locant set ‘1,2,4’ is lower than ‘1,3,4’.

Your second example is named as 1-isopropyl-4-methylcyclohexane rather than 4-isopropyl-1-methylcyclohexane since isopropyl is cited first as a prefix in the name.