London Forces and Liquifying Noble Gases

I'd say that because noble gases are non-polar atoms, the only force that applies to them is the London force.

If you want a liquid phase, you need to have forces between the atoms / molecules that constitute this liquid (forces too strong would lead to solid state).

So because London force is the only one that applies on noble gases molecules, they are necessary to liquify to gases.


Final pressure of an ideal gas mixture, when two separate gases are put into one vessel

Note that since $1\ \mathrm{cm^3} = 1\ \mathrm{mL}$, I will talk in mL.

I doubt that you are supposed to use the ideal gas law for this problem since no temperature is specified, and moles are not talked about. Instead, let's look at a few instances of Boyle's Law ($P_1V_1=P_2V_2$ -- for comparing the pressures and volumes of gasses).

The gasses are, indeed, unreactive, and, according to Kinetic-Molecular Theory, have nearly identical properties where gas laws are concerned. Let's find the partial pressure of each gas with Boyle's Law and sum them with Dalton's Law of Partial Pressures ($P_\mathrm{tot} = P_1 + P_2 +\dots$).

Note that to make the units cancel, I will use $1\ \mathrm L = 1000\ \mathrm{mL}$.

$$720\ \mathrm{mmHg} \times 200\ \mathrm{mL} = x\ \mathrm{mmHg} \times 1000\ \mathrm{mL}$$

$$\frac{720\ \mathrm{mmHg} \times 200\ \mathrm{mL}}{1000\ \mathrm{mL}} = x\ \mathrm{mmHg} = 144\ \mathrm{mmHg}$$

$$750\ \mathrm{mmHg} \times 400\ \mathrm{mL} = x\ \mathrm{mmHg} \times 1000\ \mathrm{mL}$$

$$\frac{750\ \mathrm{mmHg} \times 400\ \mathrm{mL}}{1000\ \mathrm{mL}} = x\ \mathrm{mmHg} = 300\ \mathrm{mmHg}$$

$$144\ \mathrm{mmHg} + 300\ \mathrm{mmHg} = P_\mathrm{tot} = 444\ \mathrm{mmHg}$$

444 mmHg is the final pressure in the container.


An equilibrium reaction in the presence of other fast reactions -- is it necessary also to consider the individual rate constants?

You're correct. If other reactions are occurring that are fast on the time scale of the equilibrium reactions, then you must treat the forward and reverse reactions separately, including by using their separate rate constants.

"Equilibrium" is a mathematical model that never actually perfectly matches reality.  In order for most systems to actually be at "perfect equilibrium" we would have to let them sit and react for an infinite amount of time. Of course, practically speaking, that's not necessary: equilibrium is reached to within measurement precision in quite reasonable amounts of time for many reactions. What's necessary in order for the concept of 'equilibrium' to be useful in analyzing a reaction system is for the time scale of those reactions to be "very short" (or, equivalently, for the rate of those reactions to be "very fast") when compared to all other reactions of interest in the system.

In other words, when we write the following system of reactions:

$$
\ce{A + B <=>[K_1] C + D} \\ ~ \\
\ce{C + E ->[k_2] P}
$$

what we're really writing is:

$$
\ce{A + B \underset{k_{-1}}{\overset{k_1}{\rightleftarrows}} C + D} \\ ~ \\
\ce{C + E ->[k_2] P} \\ ~ \\
\begin{align}
k_1\ce{[A][B]} &\gg k_2\ce{[C][E]} \\
k_{-1}\ce{[C][D]} &\gg k_2\ce{[C][E]} \\
K_1 &= {k_1 \over k_{-1}}
\end{align}
$$

The relative values of the forward and reverse rate constants can vary, with their ratio described by $K_1$, but at all points through the reacting process of interest, both the forward and reverse equilibrium reactions are assumed to be much faster than the final, product-forming reaction. If this assumption doesn't hold, then we can't treat the first reaction as "just" an equilibrium reaction.


Alkali metals in general are extremely reactive metals. They are highly reducing in nature. Hence, they react with water forming their corresponding hydroxides evolving dihydrogen gas.

In case of sodium - as cited in your question - its reduction potential is $\pu{-2.7109 V}$ hence it easily reduces hydrogen.

Also, as the density of sodium is quite low it floats on the water. It immediately reacts with water to form a white trail of $\ce{NaOH}$ which further gets dissolved to give a colourless solution.

This reaction is highly exothermic evolving a lot of heat which is sufficient enough to melt sodium due to its low melting point.

Sodium moves in water as the dihydrogen gas formed below the waterline pushes sodium. If sodium is trapped in the container the increasing temperature could induce dihydrogen to catch fire.


Are there two compound resins that tolerate over 1000°C for glass/carbon fiber?

In an oxygen-containing environment, no carbon-based polymer (which includes all epoxies) that I know of can tolerate 1000 C, at least not without significant degradation in mechanical strength.

There are two reasons for this:

One is that the carbon-carbon bonds will start to break down in large numbers at higher temperatures, and since the carbon-carbon to carbon-oxygen reaction is exothermic, if oxygen is around there will be no going back upon cooling.

The second is that even if the polymer doesn't chemically degrade, it will undergo a physical phase transition - it will melt, or at least soften. This can be compensated for by reinforcing with fiber or particles to make a composite. However, if it softens enough, a fiber composite will "delaminate" - the physical bonds holding the resin onto the reinforcment will come apart, and the composite will no longer be a composite.

Here is an interesting study that talks about both mechanisms of degradation, and lists several systems with their correpsonding melt and thermal decomposition temperatures. Polytetrafluoroethylene has the highest thermal decomposition temperature in this study, at 775 K (502 C).

Non-carbon polymers might be an option, although the tradeoff is that they are weaker and composite options are a lot more limited. Silicon-based polymers generally withstand higher temperatures than carbon analogues, but I wasn't able to find any that can approach 1000 C. Fluoro-polymers are another possibility - as you can see from the reference, they tend to have much higher stability in terms of thermal decomposition. There might be a fluoro-epoxy resin out there. One thing to keep in mind, though, is that the products of fluoro-polymer decomposition are particularly toxic and reactive, and so the applications where they could safely be used at high temperatures are probably very, very limited.

Edit in response to questions

  One question regarding the "oxygen around", would this also include if layers of epoxy were applied as a extra layer ontop of the carbon or would it need to be a complete vacuum in order to prohibit oxidization?

If you had a layered carbon/polymer composite, the chemical reaction with the oxygen wouldn't start at the polymer-carbon interface, it would start at the surface exposed to oxygen. However, since the reaction is exothermic, once it starts burning, it would go quickly. The carbon would burn too at these temperatures. An inert atmosphere would prevent that reaction, but it would still thermally decompose - the difference would be that hopefully the decomposition wouldn't be as exothermic, and therefore would go more slowly. Now for epoxies, they have oxygen in the polymer structure. At high temperatures, that oxygen will react and the polymer will break apart whether there is an inert atmosphere or not.


Assuming the $a$ is large enough to accommodate a BCC lattice with spheres of radius $r$, we may define

$V_s = \frac{4}{3}\pi r^3$ is the volume of the sphere

$V_c = a^3$ is the volume of the cube.

in a BCC lattice, there are 8 corner points, and one central. Because the lattice points on the corners are defined with respect to the center of each of the spheres, we know that each of the corner spheres will contribute $\frac{1}{8}$ of their total volume to the occupied volume of the unit cell. In addition, the sphere at the center will contribute its entire volume to the occupied volume. We may then write the total volume occupied in the unit cell as

$V_{occ} = 8\left(\frac{1}{8}\right)V_s + V_s =2V_s$

We know that the total volume of the unit cell is that of the cube, meaning that the unoccupied (empty) volume of the unit cell may be written

$V_{unocc} = V_c - V_{occ} = a^3 - \frac{8}{3}\pi r^3$

Wherefore we may write the empty fraction as

$\frac{V_{unocc}}{V_c} = \frac{a^3 - \frac{8}{3}\pi r^3}{a^3} = 1 - \frac{8\pi r^3}{3a^3}$


Why are the ozonides of heavier elements more stable?

Potassium and heavier alkali metals ozonides are formed by treating with ozone, or by treating the alkali metal hydroxide with ozone. They are very sensitive explosives that have to be handled at low temperatures in an atmosphere consisting of an inert gas.

Lithium and sodium ozonide are extremely unstable and must be prepared by low-temperature ion exchange starting from $\ce{CsO3}$. Sodium ozonide, $\ce{NaO3}$, which is prone to decomposition into $\ce{NaOH}$ and $\ce{NaO2}$, was previously thought to be impossible to obtain in pure form. However, with the help of cryptands and methylamine, pure $\ce{NaO3}$ may be obtained as red crystals isostructural to $\ce{NaNO2}$.

As pointed out by user2617804, lithium and sodium ozonide are extremely unstable because the ozonide ion is big for the tiny sodium and lithium ions so it is stretched around it due the distributed charge on the ozonide ion- puts it under large internal stresses.

See wikipedia for more details.


Taking volume contraction into account when mixing water with ethanol

Measuring out volumes can sometimes be easier than dealing with mass, but in this case, converting the calculations to mass-centric can make the process easier. (I'll use your symbols, but they will now be in mass units (kg). If you desire percentages of ethanol by volume, you still need to convert to mass units because of the volume contraction!))

First, calculate the total mass  in kg of the volume of solution needed; divide the volume by the density of the specific ethanol-water solution. You can look up this density in a handbook or Google it. One simple table is at https://www.engineeringtoolbox.com/ethanol-water-mixture-density-d_2162.html, although many tables are available, even to tenths of a percent.

Second, calculate the mass of 96% ethanol needed:  = (∗/100)×1/0.96. The ratio / will be higher than , the % desired, because the ethanol is only 96%. Convert this to volume in liters, if you desire, by dividing by 0.80138, the density of 96% ethanol.

Third, calculate the amount of pure water needed:  =  − .   / will be less than 100 -   because of the 4% water already in the ethanol. But the total mass will give you the volume required, and the % ethanol will be correct - BY MASS.

Commercial isopropanol solutions are frequently sold as 70% by volume; this calculates out to 64.7% by weight. Your procedure, if you desire to make variable % solutions BY VOLUME, will need to do a calculation before you even do the first step: convert the % ethanol by volume to % ethanol by mass.


For all radioactive decay (or other nuclear reaction) of a nuclide into other nuclides, the atomic number $Z$ and mass number $A$ need to be conserved.

$$\ce{_{Z_1}^{A_1}X -> _{Z_2}^{A_2}Q + _{Z_3}^{A_3}R }$$
$$Z_1 = Z_2 + Z_3$$
$$A_1 = A_2 + A_3$$

Additionally, the charges must be conserved. If $\ce{_{Z_3}^{A_3}R }$ is an alpha particle $\ce{_2^2\alpha^{2+}}$ or $\ce{_2^2He^{2+}}$, then:

$$\ce{_{Z}^{A}X -> _{Z-2}^{A-2}Q^{2-} +_2^2He^{2+}}$$

Neutrons decay into protons and electrons (beta particles). Both particles are needed for the math to work. And in order for the math to work, you need to know that the atomic number $Z$ is $0$ for a neutron (no protons), and $1$ for a proton (1 proton, or the $\ce{H+}$ ion). The electron does not have an atomic number. In fact, it is not that $Z$ and charges must be conserved separately. The sum of atomic number and charge is conserved.

$$\ce{_0^1n -> _1^1p+ + _{0}^0e-}$$

For the decay of $\ce{^{14}_6C}$ into $\ce{_7^{14}N}$, we first have a neutron decay into a proton and an electron, which is captured by the newly formed $\ce{_7^{14}N^+}$. The nitrogen nuclide is initially a cation because carbon only had six electrons, and nitrogen needs seven.

$$\ce{_6^{14}C -> _7^{14}N+ +e- -> _7^{14}N}$$


Is alkene reaction with hydrogen to form an alkane an addition reaction?

Yes, the hydrogenation reaction of alkenes is typically classified as an addition reaction. See here and there, for example.


Why is the calomel electrode widely preferred despite its toxicity?

[…] Why and how chemist are still able to perform experiments using Calomel electrodes despite its toxicity? […]

I mostly used non-aqueous silver electrodes as a reference for my measurements in acetonitrile.

But I wouldn't be afraid to use a saturated calomel electrode (SCE) when necessary. We have recently discussed the toxicity of mercury metal and I think it is largely overrated.

As far as $\ce{Hg2Cl2}$ (calomel), the other mercury component in the SCE is concerned, have a look as the wikipedia article that you have cited yourself. The solubility is extremely low ($\mathrm{2.3\ mg\cdot l^{-1}}$) and the ${\mathrm{LD_{50}}}$ is $\mathrm{210\ mg\cdot kg^{-1}}$ (rat, oral uptake). This isn't overly dramatic either, when compared to the infamous potassium cyanide (${\mathrm{LD_{50}}}$ $\mathrm{5\ mg\cdot kg^{-1}}$, rat, oral uptake).

Actually, calomel had medical uses :)

In an article on Venereal disease prevention, the author writes:

  […] the absence of therapeutic modalities spurred the United States Armed Services to utilize Credé's solution for urethral irrigation and locally applied calomel ointment for prophylactic purposes in World War I […]

As far as the toxicity is concerned, both mercury metal and mercury(I)chloride are by no means comparable with dimethylmercury.


How to find probability in different basis representations?

In quantum mechanics observables are represented by operators and the outcome of a given measurement can only be an eigenvalue of this operator. After the measurement, the state of the system will be projected to the eigenstate of the operator corresponding to the measured eigenvalue (consider for simplicity the case where you don't have more than one eigenstate corresponding to one eigenvalue).

Now, in you case, the probability $|4/5|^2$ corresponds to the following:
$$
P(\text{-}a) = |\langle \text{-}a|\psi\rangle|^2.
$$
This is the probability of projecting (by measurement) the state of the system into $|-a\rangle$. This corresponds to computing the probability of obtaining the eigenvalue $\text{-}a$ as a result of your measurement. If you want to know the probability of being in a give energy state, you have to compute the following probability
$$
P(E) = |\langle E|\psi\rangle|^2.
$$
where $|E\rangle$ is the state with energy $E$ (eigenstate of the Hamiltonian). Therefore, in order to compute the braket $\langle E|\psi\rangle$ you need to write $|\psi\rangle $ in the basis of Hamiltonian's eigenstates: in this basis, the solution is trivial.


Why do the left- and right-handed enantiomers of a chiral molecule react differently?

With other chiral molecules, the transition states are necessarily diastereomeric. Those transition states thus have different energies, which implies that rates will be different. Also, the products with other chiral reactants are diastereomeric, so equilibrium mixtures will be different.


Page 11699 of this paper has one approach to what you want: basically, you compute the relative Gibbs free energies of the possible stable conformers, assume they can interconvert, and use a Boltzmann distribution to work out what the percentages are.

  First, we selected a reference structure $R$ for a given species (neutral or protonated).

  Next, for every structure $M$ other than the reference structure $R$ we determined the
  equilibrium constant $K_M$ $$K_M = \frac{[M]}{[R]}$$ from the difference in Gibbs free energies for $M$ and $R$. The fraction of $M$ in the equilibrated sample is given by:

  $$x_M = K_M/\left(1 + K_1 + K_2 + ...\right)$$

  where the sum in the denominator goes through all structures for a
  given species. The fraction of $R$ in the sample is:

  $$x_R = 1/\left(1 + K_1 + K_2 + ...\right)$$

They don't include it in the paper, but the way to calculate an equilibrium constant $K$ from a difference in Gibbs free energy $\Delta G$ is:

$$ \Delta G = RT \ln K $$

where $R$ is the molar gas constant, and $T$ is the temperature (in Kelvin).

The tricky part is then working out what your stable conformers are, generally using some form of either guesswork or conformation sampling algorithm. They use a genetic algorithm of some sort in the paper I linked to, but given that methylcyclohexane isn't that complex a molecule, you might want to just take a look at an energy diagram and discussion of conformers of cyclohexane for some hints.

If you're getting deep into molecular dynamics, you could also use some sort of algorithm designed to use dynamic simulation to sample a wide area on a potential surface, like metadynamics or umbrella sampling, to find your conformers, and probably calculate your free energies at the same time.

You can use a static method to calculate the free energies instead, though -- the method you use to do this will depend on what sort of accuracy you're after, and may vary from a classical forcefield technique through Density Functional Theory to coupled-cluster theory (in ascending order of computational expense).


Is it correct to say that active carbon is an enhancer of either acid or base?

Active carbon is neither an acid, nor a base. It cannot. It does not react directly in water and in acidic or basic solutions.  But it may have a catalytic action on the solutes. It may accelerate the rate of chemical reactions. That is probably what you have observed when mixing sodium bicarbonate in water. Without knowing it, you must have increased the rate of the forward reaction : $$\ce{HCO_3^- <=> CO_2 + OH^{-}}$$ and this reaction is favored if the air in the room contains a low concentration of $\ce{CO_2}$, according to Le Chatelier's principle. As a consequence, the pH of the solution has slightly increased, due to this supplement of $\ce{OH^-}$ ions.


Have I identified the IUPAC name of this ketone correctly?

Your assessment is correct. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

  (…)
  16. Ketones
  17. Hydroxy compounds (includes alcohols)
  (…)

However, locants are placed immediately in front of the part of the name to which they refer. Therefore, the correct name for the compound that is given in the question is 1-hydroxy-4-methylpentan-3-one.


why "amu" is used as a unit in relative atomic mass?

$$
\pu{1 u} = \pu{1.660539040(20) \cdot 10^{-27} kg}
$$

The u (amu is the old unit name) is $1/12$ of the weight of an $^{12}\ce C$ atom.

The way the u is chosen ensures that all core and atom masses are multiples of 1($\pm $0.1) u. The u is a count of nucleons in an atom.

Until the 60s, chemists used $1/16$ of the mass of a average atom in a oxygen-isotope-mixture and physicist used $1/16$ of the mass of a $^{16}\ce O$ atom.

It's really just about choosing what is the best to measure and have a consistent result with. So you would chose an element with a very consistent isotope ratio (isotope-pure $^{12}\ce C$ e.g.) and one where you can measure weights easily.


Can plastic leak into water from a HDPE "2" plastic container at room temperature?

HDPE
has a molecular weight more than several hundreds of thousands,
"leaking" is plainly impossible.
This taste is the tase of stale water, nobody knows
what that taste is actually.
Maybe its just the taste of not cold enough water,
lacking some dissolved air, or some bacteriae
develop after contact with air and light.
(google for pseudomonas)
The only substance which ever got into the jug prior
to the water is the air (or some other gas?) used to
inflate the jug in production.
But that cause would impair the taste of the water in
the fridge as well.

BTW why do You drink such botteled water?
What about the tap water in your town/city/village?


What does (sol) in a chemical equation mean?

The state (sol) or (solv.) stands for "solvated". Since you are dissolving $\ce{HF}$ in liquid $\ce{SbF5}$, there is no water in that system (which would cause hydrolysis). So it is not correct to use (aq), since it is a nonaqueous system.

Here are some other examples:

Autoprotolysis of liquid ammonia:
$$\ce{2NH3 (l) <=> NH4+ (sol) + NH2- (sol)}$$
Autoprotolysis of nitrosyl chloride:
$$\ce{NOCl (l) <=>  NO+ (sol) + Cl- (sol)}$$
Dissolving sodium metal in liquid ammonia creates solvated electrons:
$$\ce{Na (s) <=>  Na+ (sol) + e- (sol)}$$


Finding stationary solutions and verifying differential equations

Your differential equations are correct.

From $\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_1[\ce{A}][\ce{B}]+k_2[\ce{A}] = 0$ follow the stationary solutions $[\ce{A}] = 0$ or $[\ce{B}] = \frac{k_2}{k_1}$.

Reaction (2) obeys a first order decay law. Its half-life is $t_{\frac{1}{2}} = \mathrm{ln}(2) / k_2$.

The image below shows how $[\ce{B}]$ reaches the stationary state soon after the reaction has started ($k_1 = 0.3, k_2 = 0.2$).


Amine salts are:

charged
polar
hydrogen-bonding (unless they're quaternary)

There some cross-over between those categories, but all will increase solubility in water, a polar solvent, and decrease solubility in a nonpolar solvent, like ether.

That said, if you put very long oily carbon chains on an amine salt, you will decrease the solubility in water, and increase the solubility in nonpolar organic solvents.


How can I determine if there will be a reorganization in organic reaction?

Predicting? Not really. You can use sophisticated quantum mechanical calculations to explore the energetics of the reaction channels, and apply let's say trainsition state theory to get dynamical informations. This method is cumbersome, time-consuming and still far from being exact - but this is as close to predicting reaction pathways as it gets.

Now, in organic chemistry, you usually go with your "gut". I will try to present a way which organic chemist can think. When you have seen a lot of mechanisms, then you can guess whether reorganization happens or not. It also helps if you enumerate the circumstances that can affect the reaction. Let's see some of these:

Steric factors

In order to have a rearrangement, you obviously need one part of the molecule (the "jumping" part) to have access in space to the receiving part of the molecule. This requirement  can be satisfied if we let only a hydrogen atom jump, or we let the groups jump only a single bond. Now there are a lot of exceptions of this rule: for example in Benzilic acid rearrangement you can have phenyl groups mitigating. The idea that the groups have to be close to each other is also not always true, because you can have "jumps" between different molecules too. However, if the target part of the molecule is extremly hindered, then no rearrangement can be expected.

Electronic effects

Now, if we look at the electronic requirements of a rearrangement, then we have two different things to look for. One type of reactions can be called "orbital-controlled", because the mechanism depends mostly on the frontier orbital shapes and overlap of the reactants - in this case the rearrangements. A good example of these types of reactions are the sigmatropic processes, like Cope rearrangement.

If we assume that the orbitals are alright - which is not a huge assumption if we consider hydrogen transfer - then we have to look at the free energy profile of the reaction. This means we have to check how much energy we can "win" by rearranging the reactants. In the case you mentioned with the carbocations, this is equal to checking how stable is the carbocation. If it is instable (high energy), then we can "win" a lot by rearranging the molecule to the lower energy - by mitigating a hydrogen atom let's say. The stability of the carbocation then depends on the order of the cation and the electronic effects of different substituents. The general idea is that if you do not have your charge well distributed over many atoms, then you have a high energy - for example, CH3+ cation is extremly instable. If you have a lot of groups to "help" carrying the charge, either by inductive effects, or by resonance, then the carbocation stabilizes, making the rearrangement less favoured.

It is important to note that the things I have mentioned are mostly about the energetics of the reaction. Organic chemists seldom have a way to say anything about the kinetics of a particular reaciton. There are tricks like Hammett's relations and tranisiton state theory, but most of the time, they simply are not applicable. In our case with the hydrogen transfer, we could also look for quantum effects, which is completely out of organic chemistry's predictive reach.


Resonance is a concept that originates from the incapability of the original Lewis structure to describe delocalised systems. The formulation of these resonance structures are a representation of extreme states that are possible. The true bonding pattern lies between all possible resonance structures.

Consider benzene, where we, among others, consider two major resonance structures. In fact the electron structure is well known to be a delocalised $\pi$ system, therefore better described by the right figure.

In this particular case, electrons appear to be "pushed" from one bond to another. This is not the case, as electrons are occupying one (or more degenerate) orbitals the whole time.

So when it comes to resonance structures, there are no clear rules, what is allowed and what is not allowed, however it has to be a possible representation of the actual bonding pattern. Resonance is a mere concept of describing a very complicated bonding pattern in simpler terms.


This is a heat transfer related problem. You can certainly do that. For solid medium heat transfer you need to use Fourier heat transfer equation
\[q=-k\cdot\nabla T\]
Here, k= material's conductivity and $\nabla T$ = 1st derivative of the Temp. along the driving force or heat
For Liquid and gas phase you should use law for thermal convection. And for vacuum medium you should use radiation law. There are lots of ways to do this but from your problem statement i think it's a numerical grid type problem. For your better understanding, i would suggest 'Engineering heat transfer' by william s janna (I may be biased here because i like this book) but you can certainly use any other standard heat transfer book or internet.


Can PBE (and LDA) actually be a better choice sometimes?

Of course, they can. This is in fact one of the major problems with DFT: there is no systematic way of improving a functional, so we never know a priori which will work better for some particular problem. The only thing we know is that some functionals perform systematically better that the others for some cases (see a bit more on that at the very end).

The choice of a functional (and even more generally the level of theory) is dictated by which property has to be calculated with which accuracy and is usually based on extensive benchmark studies. A rather quick search by "magnetic coupling dft benchmark" terms found few papers claiming that hybrids indeed work worse than pure functionals for that particular property. For instance, here, it is explicitly claimed in the abstract that

  The inclusion of exact exchange does not lead to improvements but
  worsens results drastically.

Exactly the same thing happens in your case, and this is not a big surprise for a number of reasons.

First, hybrid functionals are primarily better for energetics, and not just for any property. Many of hybrid functionals were even crafted with that particular aim in mind, i.e. to improve the description of energetics. But for other properties hybrid functionals might provide almost no improvement comparing to pure functionals and the results can indeed be even worse.
Secondly, the might be some cancelation of errors out there. In some cases you can get lucky and have a better results using a lower level of theory with a smaller basis set due to cancelation of errors.
Finally, benchmark has to be extensive. When it is said that some functionals perform better, the meaning is that they perform better systematically, i.e. for a wide variety of cases at average they spit out "better" numbers. There is absolutely no guarantee that these functionals will perform better for each and every case.

So, I recommend to:

Always start from the literature search for benchmark studies on the property of interest which were already done and reported. They very likely exist and can be of a great help.
On the basis of extensive benchmarks choose just a few combinations of methods and basis sets which are known to perform systematically well for the property of interest and benchmark them on your system, or, even better, on a number of similar systems which you might want (or even planned already) to study in the future.
Pick one or two methods and stick to them.


Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral?

The geometry of the complex changes going from $\ce{[NiCl4]^2-}$ to $\ce{[PdCl4]^2-}$. Clearly this cannot be due to any change in the ligand since it is the same in both cases. It is the other factor, the metal, that leads to the difference.

Consider the splitting of the $\mathrm{d}$ orbitals in a generic $\mathrm{d^8}$ complex. If it were to adopt a square planar geometry, the electrons will be stabilised (with respect to a tetrahedral complex) as they are placed in orbitals of lower energy. However, this comes at a cost: two of the electrons, which were originally unpaired, are now paired:

$\hspace{30 mm}$

(source: Wikipedia)

and on top of that, there are additional steric repulsions introduced by moving the ligands closer together. (This is, to some extent, a throwback to VSEPR theory, which simply says that the ligands would rather adopt a tetrahedral arrangement because of their mutual repulsion.)

We can label these two factors as $\Delta E$ (stabilisation derived from occupation of lower-energy orbitals) and $P + S$ (pairing energy + steric repulsions) respectively. One can see that:

If $\Delta E > P + S$, then the complex will be square planar;
If $\Delta E < P + S$, then the complex will be tetrahedral.

This is analogous to deciding whether an octahedral complex adopts a high- or low-spin configuration; where the crystal field splitting parameter $\Delta_\mathrm{O}$, also called $10~\mathrm{Dq}$ in older literature, plays the same role as $\Delta E$ does above. (However, unlike $\Delta_\mathrm{O}$, there is no simple graphical way to represent $\Delta E$ on a diagram since multiple orbitals are changed in energy.)

The two main factors that differentiate the Ni(II) complex from the Pd(II) complex are therefore:

The $\mathrm{4d}$ orbitals of $\ce{Pd^2+}$ are more radially diffuse (i.e. bigger) and therefore form stronger overlap with the orbitals on $\ce{Cl-}$, as compared to the $\mathrm{3d}$ orbitals of $\ce{Ni^2+}$. This leads to larger splitting of the $\mathrm{d}$ orbitals - or the relevant MOs (I'm too lazy to check their symmetry labels) - and hence a larger $\Delta E$.
Again because the $\mathrm{d}$ orbitals are more diffuse, the pairing energy $P$ is smaller in the palladium complex (it basically costs less energy to stuff them into the same orbital).

Together, these two factors ensure that practically all $\mathrm{4d}$ and $\mathrm{5d}$ $\mathrm{d^8}$ $\ce{ML4}$ complexes adopt a square planar geometry, even if the ligand is not a strong-field ligand. Other examples of such square planar complexes are $\ce{[PtCl4]^2-}$ and $\ce{[AuCl4]^-}$.

In fact, I am digressing here, but the same factors also cause the octahedral complexes to be almost invariably low-spin. For example, the $\mathrm{d^5}$ complex $\ce{[Fe(ox)3]^3-}$ has five unpaired electrons, indicating a high-spin configuration, $\mathrm{(t_{2g})^3(e_g)^2}$. However, $\ce{[Ru(ox)3]^3-}$ has one unpaired electron, which implies a low-spin configuration $\mathrm{(t_{2g})^5}$.


Why is a ketone not capable of hydrogen bonding among themselves?

Short answer: the ketone oxygen can participate in hydrogen bonding

The compound you mention, methoxymethane, is an ether not a ketone.

Ethers don't hydrogen bond very well because the oxygen isn't very polarized.

On the other hand, the carbonyl in a ketone is polarized.  We can draw resonance structures that show this polarization.

(image source)

The negative polarization on the carbonyl oxygen allows it to play the role of hydrogen bond acceptor as illustrated in the following drawing.

(image source)

Edit: If you are asking if the hydrogens on the carbon adjacent to the carbonyl carbon (or any other carbon in the backbone of a simple ketone) can participate as the donor in a hydrogen bond, then the answer to this question is "no". These $\ce{C-H}$ bonds are typically less  acidic and less polar than the $\ce{O-H}$ bond in water.

Therefore, while the oxygen in pentanone can participate in hydrogen bonding with donor hydrogens in other molecules, the hydrogens in pentanone cannot participate in hydrogen bonding.  Pentanone cannot hydrogen bond with itself.


Is there a way to faciliate remembering which are alpha/beta hemiacetals of sugars?

When you draw the Haworth projection, the alpha anomer is written with the -OH group pointing downwards (see the image), when you write the symbol alpha, you start by writing it downwards. For the beta anomer the -OH group is written upwards and when you write the symbol beta, you start writing it upwards. This mnemonic applies only to D-monosaccharides. For L-monosaccharides the opposite applies.

Actually, a better definition is that the alpha isomer has the -OH group pointing in the direction opposite to the -CH2OH group (on carbon 5), the opposite applies for the beta isomer, therefore for L-monosaccharides the -CH2OH group should point downwards and the alpha isomer will have the -OH group pointing upwards.


Can atomic partial charges be measured in molecules experimentally?

Let me put it this way. There are no atoms in molecules. There is just a continuous cloud of electron density, and as you correctly pointed out, we have no way to pinpoint where one atom ends and another starts. Or rather, we have many ways to do that, all slightly different and all inherently arbitrary. (My personal favorite is the concept of Bader charges, but that's a matter of taste.)

So yes, the very idea of partial atomic charges is unavoidably vague.


Ullmann's Reaction (similar to Wurtz) - aryl chloride with deactivating groups - stability of free radical

Your assumption that electron-withdrawing groups destabilize radicals is wrong. Radicals are in fact stabilized by both electron-withdrawing and electron-donating groups and even more so if both types are attached to the same carbon atom (captodative effect).

I modified an image from Ian Fleming's Molecular Orbitals where you have the proof. In the image you can see that the combination of the carbonyl group's (which is electron-withdrawing) pi bonding/antibonding pair of orbitals with the p orbital of the radical results in an overall decrease in energy:


What is the enol form of 4-methylcyclohex-2-en-1-one?

There are two possible enol forms of this ketone. The first one is:

This is the one you drew, which shows the α hydrogen being abstracted. It is perfectly valid.

The other one is:

Note how the alkene next to the carbonyl group "extends" the range of the carbonyl group. This allows the γ hydrogen (i.e. the one three carbons away) to be removed, and then you can "push" electrons into the carbonyl group through the alkene group. In general, this is a property of α,β-unsaturated ketones (i.e. ketones which have a double bond between the α and β carbons).

Note that in this enol form, the methyl substituent is now in the same plane as the C=C next to it. This means that, if you were to reverse the tautomerisation to get back the keto form, there's an equal chance of the hydrogen coming in from above the plane and from below the plane, and racemisation would occur. In practice, there is a rapid equilibrium between keto and enol forms, and this means that if you were to dissolve this compound in any kind of acid (or base), the racemisation would occur pretty quickly.

Here, I have drawn the mechanisms for acid-catalysed tautomerisation. I assume you are now able to draw the mechanisms for the reaction in base (which in general forms an enolate instead of an enol).

At this stage, you might wonder why we cannot abstract the other α hydrogen. There's a couple of reasons. Firstly, it's attached to a $\mathrm{sp^2}$ carbon, and the C-H bond is therefore somewhat stronger. Secondly, even if you could take off that proton and form the "enol", you would have two double bonds next to each other. That's a huge problem because those double bonds want to be linear, but don't quite have the space to do so since there's a ring:

Regarding the stability of the two legitimate enol forms, I would not say that they are any more stable than the keto form. In general, the keto form is more stable. There are cases in which the enol form is relatively stabilised, most often when there is aromaticity or extra conjugation in the enol form (other factors such as solvent will also influence the position of equilibrium):

However, that is not the case here. The starting keto form is as conjugated as any enol you can form.


White snow-like powder below sodium hydroxide solution

Sodium hydroxide is easily contaminated with sodium carbonate. The free base continuously reacts with atmospheric carbon dioxide to form the salt.When a relatively concentrated solution of NaOH is prepared, sodium carbonate slowly precipitates out as a white gelatinous mass.
a) How long did you expose it to air?
b) What is the expected concentration of the base?
c) What are your electrodes made of?


Assigning 1H NMR spectrum and predicting peak coupling patterns for 2H-1-benzopyran-2-one

Don't you have an integral for this spectrum? This would show you clearly that you have two sets of overlapping signals, just as you have surmised.

Looking at the numbering for your molecule described:

You are correct with the assignment for the vinyl protons. Proton E will be strongly downfield shifted, and F only moderately shifted from the base double bond chemical shift of 5.25. These values could be determined using additive tables found in lots of NMR textbooks.

The proton shifts for the aromatic ring can also be roughly calculated using additive tables, but what you have are indeed overlapping signals; Protons A and C are at ~7.3 and protons B and D are at ~7.55 ppm.

To predict the splitting patterns you anticipate for the aromatic system, consider first the general couplings you'd expect:

ortho coupling ~7-8 Hz
meta coupling ~2-3 Hz
para coupling ~1Hz or less

Let's ignore the para couplings for simplicity - they are not often seen.

So,

Proton A will have one ortho coupling and one meta coupling, so a will be a dd
Proton B will have two ortho couplings and one meta coupling, so will be a ddd
Proton C will have two ortho couplings and one meta coupling, so will be a ddd
Proton D will have one ortho couplings and one meta coupling, so will be a dd

Let's now build in double bond :

Protons E and F have a cis coupling (~10 Hz).
Proton F will appear as a doublet, but proton E is likely to have a homoallylic type coupling to proton D (4J), splitting both of these signals to an additional doublet with a coupling of about 1-2 Hz. This means that proton D may be a ddd, and proton E a dd if this coupling is observable.

So that should be the expected splitting pattern from this molecule. Now to justify the appearance of the peaks. Overlapping signals can often make things difficult to interpret. Without knowing what frequency this spectrum was recorded at, you also need to consider whether there are second order effects starting to manifest, certainly with the meta couplings. There will definitely be tenting of the peaks giving rise to distorted peak intensities from what would normally be expected.

So, I wouldn't be devoting too much time to the interpretation of overlapping signals like these. One thing I will say is that if you label the peaks as (say) a ddd, then you should be able to report all couplings for each d. Otherwise, if you cannot report couplings (and this means couplings, not splittings) report them as a multiplet (or in this case overlapping multiplets).

Complex splitting patterns of second order and overlapping signals can be analysed and simulated, but it takes a lot of work.You can try drawing out a splitting diagram for each signal using the approximate couplings for ortho and meta couplings, and then start overlapping through constructive addition to see what you get. It gets very complicated very quickly.


Why do silver nitrate and sodium hydroxide react to produce silver(I) oxide?

Silver in silver oxide is no more oxidized than in $\ce{AgNO3}$. So you should ask yourself the same question earlier, even before the reaction. Yes, noble metals are somewhat resistant to oxidation. But still they can be oxidized, and thus can form compounds, of which $\ce{AgNO3}$ and $\ce{Ag2O}$ are the examples.

Even gold can be oxidized, though that requires some extreme measures.

As to the formation of oxide instead of hydroxide, there is no clear-cut distinction between Ag and the rest of metals. Many insoluble hydroxides are not quite stable. $\ce{Cu(OH)2}$, for example, would slowly lose water and turn into oxide; likewise, $\ce{Fe(OH)3}$ turns into $\ce{FeO(OH)}$. Silver does the same, just a little faster.


Calculating molarity of a gas using ideal gas equation

Hydrogen at one atmosphere pressure and 298 K is certainly not a one molar solution. However, you can still calculate the concentration with the ideal gas. Note that this is a gaseous solution and since it is a pure gas, the solute is also the solvent.
$$
c(\ce{H2}) = \frac{n(\ce{H2})}{V(\ce{H2})}
           = \frac{p(\ce{H2})}{\mathcal{R}T}
           \approx 40.9~\mathrm{mol\,m^{-3}}
           \approx 0.04~\mathrm{mol\,L^{-1}}
$$


How many molecules does it take to have a phase?

This is actually an active area of research for water clusters. In principle, for $\ce{(H2O)_{n}}$ there should be a "melting" phase transition, much like for ice to liquid water.
So, in principle, if you had an accurate enough theoretical method, you could do molecular dynamics and see when the melting point of the cluster matches bulk water.
Defining a line between liquid and gas seems tricky to me, but I don't do statistical mechanics, and perhaps there are molecular dynamics techniques for defining a boiling point.
Defining the solid-liquid phase transition is much easier, because the radial distribution function of the system can be monitored to see the transition (i.e., the density changes).
At the moment, I don't think the best dynamics methods get the melting transitions accurately yet, but I think n~20 is about right. I'll revise this when I get a better answer.
EDIT: It turns out that 21 seems to be the number for water drops.
"How many water molecules are needed to solvate one?"
Chem. Sci. 2021

The minimal structure eventually responsible for proper solvation is made of a total of 21 water molecules and includes two complete solvation shells, of which the whole first one is tetrahedrally coordinated to the central molecule.


Why does nose grease knock down the head on beer and soda?

Any oil should work.

Naturally occurring surfactants, such as lecithin, make bubbles or foam because one end of the detergent molecule is hydrophilic while the opposing end is hydrophobic (not rabietic, though). Without oil, the hydrophobic end prefers air to water, encapsulating the air. With oil added, though, the water-repellent end is attracted to the oil, forming micelles, instead.

You could put vegetable oil or mineral oil on your finger, but then the server might wonder why you stick your finger in the beer.

You might also try making a beer-powered boat! Tell your server, and ask for a free beer to prove that this works.


Between 3-chlorophenol and 3-fluorophenol, which one is more acidic?

Experimentally, the $\mathrm pK_\mathrm a$ values for 3-chlorophenol and 3-fluorophenol appear to be (perhaps unexpectedly) equivalent.

The table below1 shows the measured $\mathrm p K_\mathrm a$ values of substituted phenols in dimethyl sulfoxide (DMSO):

$$\begin{array}{ccc} \hline
\text{substituent} & \sigma & \mathrm p K_\mathrm a \\ \hline
\ce{H} & 0 & 18.03 \pm 0.04, \color{silver}{16.4} \\
\ce{p-CH3} & -0.170 & 18.86 \pm 0.05 \\
\ce{m-CH3} & -0.069 & 18.23 \pm 0.01 \\
\ce{p-Cl} & 0.227 & 16.74 \pm 0.03 \\
\color{red}{\ce{m-F}} & \color{red}{0.337} & \color{red}{15.88 \pm 0.07} \\
\color{red}{\ce{m-Cl}} & \color{red}{0.373} & \color{red}{15.83 \pm 0.05} \\
\ce{m-CN} & 0.56 & 14.76 \pm 0.05 \\
\ce{m-NO2} & 0.710 & 14.39 \pm 0.12, \color{silver}{13.75} \\
\hline
\end{array} \\
\small\text{(Greyed out values are taken from ref 2.)}$$

In general, halides have slightly unusual effects on aromatic rings. Despite being highly electronegative (and deactivating), they actually donate into the pi system via their lone pair.

This donation means that the phenoxide generated upon deprotonation (i.e. the dissociation being measured by the $\mathrm pK_\mathrm a$ value) isn't as stabilised as it would be with a purely withdrawing group (such as the nitrile or the nitro groups).

References

Bordwell, F. G.; McCallum, R. J.; Olmstead, W. N. Acidities and hydrogen bonding of phenols in dimethyl sulfoxide. J. Org. Chem. 1984, 49 (8), 1424–1427. DOI: 10.1021/jo00182a020.
Chantooni, M. K., Jr.; Kolthoff, I. M. Comparison of substituent effects on dissociation and conjugation of phenols with those of carboxylic acids in acetonitrile, N,N-dimethylformamide, and dimethyl sulfoxide. J. Phys. Chem. 1976, 80 (12), 1306–1310. DOI: 10.1021/j100553a009.


Order of −I effect among ethenyl and 1-propenyl groups

I would say yes, the -CH3 group adds a little +I effect, making the -CH=CH-CH3 a little less electron withdrawing compared to the vinyl.


When it comes to acidity, stability of the counter-ion is one of the main considerations. The nitro-group acts to stabilize the negative charge of the resulting nitrobenzoate by pulling the negative charge onto itself.

In the case of 2-nitrobenzoic acid, it is much nearer and therefore able to more effectively 'remove' the negative charge from the carboxylate group. In the case of 4-nitrobenzoic acid, it's still able to do so but less effectively as it is further away.

The hydrogen bonding effect that you describe does exist in 2-nitrobenzoic acid but this does not act to make it less acidic. While one might reason that the hydrogen bonding "holds the proton in place" and therefore makes it harder to leave, the presence of the hydrogen also means that both the nitro and the carboxylate group are highly unlikely to be simultaneously coplanar with the benzene ring (it's not obvious from the image because there's no bond between O and H). The energetics in this case are likely to favor the release of H+ in exchange for a coplanar configuration.


Methods for dissolving Scotch/Kaptop/etc. Tape for Electron Microscopy?

Generally you use Acetone. The following is taken from
Enoki, Toshiaki, and Tsuneya Andō. 2018. Physics and Chemistry of Graphene : Graphene to Nanographene p. 94

The Scotch tape method is very common in the fabrication of
graphene devices, which are mainly used in research into the
fundamental properties of graphene. It can produce high-quality
graphene with lateral sizes ≤ 0.1 mm; however, controlling the size
and position of the graphene is almost impossible. The following is
the typical procedure of the Scotch tape method:

(i) Place a few flakes of graphite (≈ 1 mm) on the adhesive side
of a plastic sticky tape with tweezers (Fig. 3.1(a)). Scotch tape
(3M) and Nitto tape are commonly used. Natural graphite,
Kish graphite, or HOPG (highly oriented pyrolytic graphite) is
usually used for the starting material.

(ii) Fold the adhesive side to sandwich the graphite flakes and
press the tape firmly (Fig. 3.1(b)).

(iii) Peel the tape apart slowly, so that the graphite flakes are
cleaved and attached on both sides of the tape (Fig. 3.1(c)).

(iv) Repeat the second and the third steps with slightly shifting
the fold line, so that the graphite flakes do not overlap (Fig.
3.1(d)).

(v) Stop repeating the cleavage when graphite flakes spread over
the sticky tape (Fig. 3.1(e)).

(vi) Prepare a silicon substrate with a silicon dioxide layer on
the top surface (Fig. 3.1(f)). Place address markers on the
surface in advance using photolithography so that one can
easily locate the position of a graphene flake in an image.
These markers are indispensable for further microfabrication
on graphene. An example of the address markers is shown in
Fig. 3.2

(vii) Attach the adhesive side of the tape with the graphite flakes to
a silicon substrate, and gently rub the surface to remove any
air between the substrate and the tape (Fig. 3.1(g)).

(viii) Slowly peel the tape off the substrate (for example, more than
2 min for a 1 cm substrate) (Fig. 3.1(h)). Not only graphite
flakes but also some adhesive remains on the substrate. The
latter can be removed by submerging the substrate in acetone
for a few seconds.

It is also reported that in the scotch tape method, not only graphene flakes but also a large amount of adhesive is attached to the substrate. You get rid of it immersing the substrate in Acetone. Other methods are reported in the reference too.

About adhesion
There are lots of aspect about adhesion. About scotch tape the most interesting answer you can find here "https://www.scientificamerican.com/article/what-exactly-is-the-physi/" is the following

"The simplest answer that I can give to the question is that pressure-sensitive adhesives (which are polymers) are 'tacky' or 'sticky' because they are essentially very high viscosity liquids that also have some elastic characteristics--in technical terms, they are 'viscoelastic.' This property means that they exhibit some of the characteristics of liquids, and so they will 'wet' a surface to which they are pressed. But then, because of their elasticity, they will resist separation when stressed. Thus, 'stickiness' is strictly a physical (viscoelastic) phenomenon, not a chemical one."


What is the heat energy required to perform electrolysis equivalent?

I won't do any rib poking, but there is some flawed thinking here.  Certainly not off topic though.

Nomenclature: First, when you pass a current through water, you are not breaking the hydrogen bonds, which are intermolecular forces, but (a) breaking the bond between hydrogen and oxygen and (b) forming molecular oxygen and molecular hydrogen, $\ce{2H2O -> 2H2 + O2}$, and this reaction can be broken down into two half reactions: $$\ce{4H2O + 4 e- -> 2H2 + 4OH-}$$ $$\ce{2H2O -> O2 + 4H+ + 4e-}$$

Potentials: The 1.2 V you refer to likely arises from the standard reduction potential of oxygen, which is the reverse of the second reaction above.  I won't go in to the thermodynamics here, but suffice it to say that a brief look at a standard reduction potential table indicates that the reduction of water (first equation) has a potential of -0.83 V and the oxidation of water has a potential of -1.23 V for a grand total of ~ -2 V. (Note, these values are pH dependent and standard reduction potential tables report values at a pH of 0.)

Electrochemistry When you apply an overpotential to a system (the potential beyond what is needed to get the oxidation/reduction reactions to go forward) then two things will happen (a) the reaction will go faster and (b) heat will be generated.  This heat would be analogous to heat due to friction in a physical system and is considered wasted energy.

Thermochemistry Finally, we get to the answer to your question, which is "probably not, but the way you asked it, yes"  (hunh?)  Recall the nomenclature issue above.  Adding heat to water will break the hydrogen bonds in water eventually leading to the water changing from a liquid to a gaseous state.  However, adding heat to water would not result in the oxidation/reduction reactions outlined above because an important ingredient is missing: electrons.


Value of mixing ratio and rate constant in Chapman mechanisn

The entire stratosphere lies within the "homosphere", where the composition of gases is dominated by mixing processes and the atmosphere is thus homogenous.  This reaches up to about 100 km, whereas the stratosphere doesn't go any higher than about 50 km.


If it was about the number of protons in the nucleus, then Mercury would hardly be liquid at room temperature, would it?

Actually, it is about the density of free electrons, i.e. the density of the "electron gas" that makes up the metallic bonding. Obviously, this density has to be the negative of the ions' charge, so you might equivalently say it's about magnesium's (2+) charge, though that doesn't make much sense physically unless you also consider the conduction electrons, as the ions themselves of course repulse each other.


Gas chromatography flame ionization detector (FID) - why hydrogen gas?

The key to the answer is understanding how FID works.

The hydrogen flame has a minimal flame ionisation, what is needed for the low signal baseline.

Incoming organic molecules from the HPGC column create in the flame a lot of ions and increase the flame electric conductivity.

Using alternatives causing higher ionisation would decrease FID sensitivity that is not great even at ideal conditions, compared to e.g. ECD.


The major isotopes of chlorine are 35 and 37.  If 4 chlorines are present in a molecule, then the following isotopic combinations of chlorine are possible

35, 35, 35, 35
35, 35, 35, 37
35, 35, 37, 37
35, 37, 37, 37
37, 37, 37, 37

So there will be 5 molecular ion peaks.  You could calculate their relative heights by factoring in the relative isotopic proportions, along with the number of permutations of each of the above combinations.


The strength or weakness of an acid is independent of its concentration. A strong acid dissociates completely in solution, so $\ce{HCl}$ at any molarity would be considered strong, and $\ce{HCOOH}$ (for example) at any molarity would be weak. It is more enlightening to consider $K_a$ values, which are essentially the equilibrium constants for the ionization of an acid $\ce{HA}$.
$$K_a=\frac{[\ce{H}^+][\ce{A}^-]}{[\ce{HA}]}$$
This value will either be very small (<<1) or very large (>>1000). Acids with very small $K_a$ are considered weak, and acids with very large $K_a$ are considered strong (the $K_a$ of strong acids is usually not stated and assumed to be infinite).

Addendum: The $K_a$ value is concentration-independent because you are only interested in how well the acid dissociates from its molecular form. If you had $0.0000001\ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCl}$, practically every molecule of $\ce{H-Cl}$ would dissociate into $\ce{H^+}$ and $\ce{Cl^-}$ no matter how many there are. But if you had $12 \ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCOOH}$ (methanoic acid), only a small proportion of the $\ce{HCOOH}$ would dissociate into  $\ce{H^+}$ and $\ce{HCOO}^-$, irrespective of how many you had to start.


Rebar for concrete is frequently stored in heaps, and rusts. It actually develops a stronger bond if it is not too rusty, because the rebar surface is rougher and better able to adhere to the hardened cement. It's interesting to see a pile of old rebar with a fresh delivery on top - silvery, not a trace of rust - on top of a pile of rebar that looks almost disgustingly rusted. And then you think, yeah, but the stuff on the bottom will have a better bond.
The key word in the observation about the fences is "sometimes".
I'm going to guess that the maker of the fence in the top picture also had a pile (or two) of iron bars, and some were rusty from being stored long or outdoors, or were a teensy bit different from some other pile of iron bars. This fence maker, knowing how his fence weathers, has to make it as uniform as possible, so he takes one bar from one pile, then a bar from the other pile... He prepares them identically, primes them identically, paints them identically... But then nature takes over and the most corrodable bar develops a spot, which spreads until it is all corroded-looking, before the more corrosion-resistant bar has even gotten started!
And then you come along and take a picture just before the other bars begin to corrode! This experiment is in the initial stages, not yet at equilibrium. Check back on the fence every year or two - I believe things will even out in time.


Your description is almost correct - note that when sodium metal burns in air, it forms primarily sodium peroxide in addition to sodium oxide. Sodium peroxide will react with hydrochloric acid to give sodium chloride and hydrogen peroxide. Heating will decompose that to water plus oxygen. You should still be able to get the sodium chloride product that you expected.


Part of the explanation is that orthophosphate forms complexes -or even precipitate, traces of heavy metal ions present in the solutions. Many of these ions, like manganese and platin, will act as catalysts for the degradation of hydrogenperoxide. The phosphonic acid HEDP (Etidronic acid) is another example of a stabilizer.


The Nernst equation for the anode is :$$E_a = E°_a + \frac{RT}{F}ln{\frac{[Li^+]_e [C_6]_s}{[LiC_6]_s}}$$ where the concentrations $[C_6]_s$ and $[LiC_6]_s$ are  defined in the solid phase of the anode, and not in the electrolyte. The concentration $[Li^+]_e$ is defined in the electrolyte. When the anode is working, the concentration $[LiC_6]_s$ decreases in the solid phase and the concentration $[C_6]_s$ increases.

The same reasoning may be done on the Nernst equation at the cathode :  $$E_c = E°_c + \frac{RT}{F}· ln{\frac{[Li^+]_e[CoO_2]_{s'}}{[LiCoO_2]_{s'}}}$$ where $[CoO_2]_{s'}$ is the concentration of $CoO_2$ in the solid phase of the cathode, and $[LiCoO_2]_{s'}$ is the concentration of $LiCoO_2$ in the solid phase of the cathode.

The overall voltage of the battery $\Delta E$ can be written :  $$\Delta E = E°_c - E°_a + \frac{RT}{F} · ln\frac{[LiC_6]_s [CoO_2]_{s'}}{[C_6]_s [LiCoO_2]_{s'}}$$ As a consequence, both $[LiC_6]_s$ and $ [CoO_2]_{s'}$ decrease when the battery is working. So the overall voltage decreases.


Why is the water reduction of oxygen favored in acidic conditions?

If you look at the half equations, you will see when you add them together, you get 1.23 V, 1.22 V, and 1.23 V.

As the mixture becomes more basic, it is more difficult for the oxygen to grab $H^+$ ions and easier for $H_2$ to lose electrons to generate $H^+$ ions.

The mechanism probably involves these combinations: $O_2, H_2O_2, H^+, HO_2^-, H_3O_2^-, H_2O, and\ HO^-$.


Entropy change of system in adiabatic irreversible process

The change in entropy is the reversible heat flow divided by $T$.  So if the process is irreversible, you can't use the irreversible process's heat flow to find the entropy change.


Separating sodium sulfate from chromium trioxide solution

The overall reaction is:
$$\ce{H2SO4 + Na2Cr2O7 → 2 CrO3 + Na2SO4 + H2O}$$

How to separate chromium trioxide and sodium sulfate from solution?

Since chromium trioxide is soluble, it exist as chromic acid in the solution. So, in the reaction pot, there are compounds like  $\ce{H2CrO4, H2Cr2O7}$ along with our target compound $\ce{CrO3}$, all in equilibrium. But with the addition of a large excess of concentrated sulfuric acid, water is withdrawn from the hydrated forms and the chromium trioxide separates from the solution in the shape of red needles. Successively, the reaction pot is made to stand overnight for the chromium trioxide to precipitate slowly in form of needles and then filtered off from the solution using Buchner filter. The synthesis procedure of chromium trioxide can be found in prepchem.com:

100 grams of sodium dichromate are dissolved in 250 ml of water and
the obtained solution is filtered. 200 ml of concentrated sulfuric
acid are dropwise added with constant stirring until a slight
permanent precipitate of chromium trioxide is formed. The mixture is
cooled for half an hour or longer, then slowly, while stirring, 200 ml
of concentrated sulfuric acid are added dropwise. The reaction mixture
is left to stand over night in order that the crystal meal may become
somewhat coarser. The larger crystals (and purer) of chromium trioxide
could be obtained by heating the solution with crystal meal with
stirring to $\pu{100 ^{\circ}C}$ and allowing to cool slowly. If this process is
repeated once or twice, a more satisfactory product will be obtained.
The obtained crystals of chromium trioxide are filtered with a Buchner
funnel with a sintered glass disc instead of the usual paper filter
(chromium trioxide is a strong oxidizing agent and will oxidize
organic material).  After draining the crystals completely and
pressing the surface and treating with 15 ml of concentrated (65-70%)
nitric acid. The nitric acid is removed by suction and the operation
is repeated twice with 10 ml of nitric acid each time. Finally, the
red crystals of chromium trioxide are dried by suction as much as
possible, transferred in a dry evaporating dish and by gentle heating
the remaining nitric acid is removed. Dried product is stored in a
stoppered bottle. Yield is 60%.

Note: chromium trioxide is toxic, corrosive, a powerful oxidizer and a potent carcinogenic. Sulfuric acid is corrosive and can burn your skin if you are careless. Handle those chemicals with care. Perform this synthesis only after you become a qualified chemist. This answer is only for educational purpose only.


Very interesting question! The key word you are looking for is planar chirality. In trans-cyclooctene, the polymethylene bridge can either go "in front of" or go "behind" the plane of the double bond, assuming you fix the double bond and the two hydrogens in place.

As pointed out by @jerepierre, they are considered different molecules due to a high-energy barrier which prevents the interconversion. Cyclooctene is the first cycloalkene to have both stable cis- and trans- isomers. The chain in trans-cyclooctene is not long enough to swing over the double bond. As the chain gets longer, the energy barrier to rotation decreases.

Here are the two mirror images of trans-cyclooctene (image source: own work).

These two molecules are mirror images of each other but are not superimposable. Therefore trans-cyclooctene is chiral despite not having a chiral center.

Source

Edit: this is the first time I learned that a chiral molecule does not have to have a chiral center. After making some searches online, I feel a need to expand the answer to clarify the concept of chirality.

A chiral molecule is one that has a non-superimposable mirror image. Mathematically a molecule is chiral if it is not symmetric under an improper rotation. Chirality arises due to:

point chirality: typically a carbon center with four different substituents;
axial chirality: such as allenes with different substituents on each carbon (see this question);
planar chirality: such as the case of trans-cyclooctene;
inherent chirality: due to the presence of a curvature in a structure that would be devoid of symmetry axes in any bidimensional representation, such as fullerenes.


State solid synthesis of LiMn2O4 mechanism

$\ce{LiMn2O4}$ has been made by rotary kilning of $\ce{Li2CO3}$ (or $\ce{LiOH}$ or $\ce{LiOH.H2O}$) + $\ce{MnO2}$. I made hundreds of grams on a small scale and 1000 pounds on a larger scale about 25 years ago.

The fine powders are added to a rotary kiln (similar to making Portland cement) and heated to about 800 degrees C. The kiln rotates, stirring the mix; the lithium compound melts and reacts with the $\ce{MnO2}$ as the powdery mass gets mushed around and around. The product was used as the cathode material in a lithium battery.

The oxidation state of the manganese drops from +4 to +3.5. The significant process is that oxygen gets lost, so to balance the equation, take one lithium, two manganese and four oxygens on the product side to make $\ce{LiMn2O4}$; everything else from the raw material side goes away.

Perhaps $\ce{LiMn2O4}$ can be made in water from dissolved salts, but gaseous $\ce{O2}$ would not be a convenient reagent.


As Wikpedia Article says ,the criteria for a molecule to be aromatic are --

the molecule must follow Hückel's rule, having 4n+2 electrons in the delocalized, conjugated p-orbital cloud;
the molecule must be able to be planar;
the molecule must be cyclic; and,
every atom in the ring must be able to participate in delocalizing the electrons by having a p-orbital or an unshared pair of electrons.

When we are talking about aromaticity , we consider the electrons inside the ring. You are right that the styrene has 8 pi electrons but when checking for Huckel's rule, we only consider the electrons inside the ring. The alkenyl group is just an attachment to the stable aromatic compound which might take part in resonance but wouldn't affect the compound's aromaticity.


Applying the Solubility Guidelines is AgClO4 soluble or insoluble?

Different guidelines will say slightly different things to convey the same idea. Many I've seen don't explicitly mention silver, but still produce the same result for most compounds. I don't have that book but if it's anything like these, there's probably another line that says salts of anions like nitrate and chlorate are soluble.

This table is worded a bit better to make the possibility of exceptions more explicit:

  Salts containing Group I elements are soluble ($\ce{Li+, Na+, K+, Cs+, Rb+}$). Exceptions to this rule are rare. Salts containing the ammonium
  ion ($\ce{NH4+}$) are also soluble.
  Salts containing nitrate ion ($\ce{NO3-}$) are generally soluble.
  Salts containing $\ce{Cl^-, Br^-, I^-}$ are generally soluble. Important exceptions to this rule are halide salts of $\ce{Ag^+, Pb^{2+}}$, and $\ce{(Hg2)^{2+}}$.
  Thus, $\ce{AgCl, PbBr2}$, and $\ce{Hg2Cl2}$ are all insoluble.
  Most silver salts are insoluble. $\ce{AgNO3}$ and $\ce{Ag(C2H3O2)}$ are common soluble salts of silver; virtually anything else is insoluble.
  Most sulfate salts are soluble. Important exceptions to this rule include $\ce{BaSO4, PbSO4, Ag2SO4}$ and $\ce{SrSO4}$ .
  Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements
  ($\ce{Ca, Sr,Ba}$) are slightly soluble. Hydroxide salts of transition
  metals and $\ce{Al^{3+}}$ are insoluble. Thus, $\ce{Fe(OH)3, Al(OH)3, Co(OH)2}$ are not
  soluble.
  Most sulfides of transition metals are highly insoluble. Thus, $\ce{CdS, FeS, ZnS, Ag2S}$ are all insoluble. Arsenic, antimony, bismuth, and lead
  sulfides are also insoluble.
  Carbonates are frequently insoluble. Group II carbonates ($\ce{Ca, Sr, Ba}$) are insoluble. Some other insoluble carbonates include $\ce{FeCO3}$
  and $\ce{PbCO3}$.
  Chromates are frequently insoluble. Examples: $\ce{PbCrO4, BaCrO4}$
  Phosphates are frequently insoluble. Examples: $\ce{Ca3(PO4)2, Ag3PO4}$
  Fluorides are frequently insoluble. Examples: $\ce{BaF2, MgF2 PbF2}$.


$\ce{HCl{(l)}}$ is not the correct notation in this case. Arrhenius acids and bases (chemicals that interchange $\ce{H^+}$ with water to form $\ce{H_3O^+}$ or $\ce{OH^-}$) break apart in water, forming ions. In your case, there is an equilibrium $\ce{HCl{(aq)} + H_2O{(l)}<=> H_3O^+{(aq)} + Cl^-{(aq)}}$ which very strongly lies to the right; virtually all of the $\ce{HCl{(aq)}}$ is broken apart in this way (which is why $\ce{HCl}$ is known as a strong acid).

Now, not everything dissolved in water has to dissociate into ions. For example, glucose $\ce{(C_6H_12O_6{(s)})}$ dissolves in water to form $\ce{C_6H_12O_6{(aq)}}$. Weak acids are acids that do not entirely dissociate into ions (for example, $\ce{CH_3COOH{(aq)} + H_2O{(l)} <=> CH_3COO^-{(aq)} + H_3O^+{(aq)}}$ still has some undissociated $\ce{CH_3COOH{(aq)}}$ in solution.)

A final point is that some dissolved species can, in fact, be solvents on their own. In this case, the definition of whether the species is aqueous or liquid is not well-defined and usually depends on the context. For example, a 1:4 mixture of $\ce{MeOH}$ and $\ce{H_2O}$ is typically written as $\ce{MeOH{(l)} + {H_2O{(l)}}}$, but a solution where a small amount of $\ce{MeOH}$ (maybe used as a reactant) would be denoted $\ce{MeOH{(aq)}}$. This has to do with the definition of the thermodynamic activities of the species and if you have questions about this, there's probably someone else on here who is a better expert than I to explain.


How to find the largest cyclic substructure with RDKit?

In the RDKit Cookbook is an example how to find the indices of ring systems.
RDKit Cookbook Example
Having the indices, it is simple to find the largest ring.
from rdkit import Chem
from rdkit.Chem import Draw
from rdkit.Chem.Draw import IPythonConsole
IPythonConsole.molSize = (400, 300)
IPythonConsole.drawOptions.addAtomIndices = True

m = Chem.MolFromSmiles('C1=CC2C(C=C1)=c1ccccc1=C2C1C=C(c2ccccc2)c2cccc3c4ccccc4nc1c23')
m

# the cook book example
def GetRingSystems(mol, includeSpiro=False):
    ri = mol.GetRingInfo()
    systems = []
    for ring in ri.AtomRings():
        ringAts = set(ring)
        nSystems = []
        for system in systems:
            nInCommon = len(ringAts.intersection(system))
            if nInCommon and (includeSpiro or nInCommon>1):
                ringAts = ringAts.union(system)
            else:
                nSystems.append(system)
        nSystems.append(ringAts)
        systems = nSystems
    return systems

ringSys = GetRingSystems(m)
ringSys

[{16, 17, 18, 19, 20, 21},
 {13, 14, 15, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35},
 {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}]

Highlighting the largest ring for proof.
highlightAtoms = max(ringSys, key=len)
Draw.MolToImage(m, size=(400,300), highlightAtoms=highlightAtoms)


How can we confirm the number of protons in an atom?

1) While not easy, it is possible to obtain mass-spectrum of full range of ions, including fully ionized. By number of peaks with Z/E corresponding as X, X/2, X/3 ... X/n it is possible to ensure that an element has exactly n electrons and protons. It is still difficult to ensure full ionization of heavier atoms, so the method is not applicable for heavier elements. This, however, is the only direct method I can think of. Should work for carbon, though.

2) Various X-ray-derived spectral methods. While not providing direct evidence, they do provide information on electronic shell structure, that can be compared with theoretical calculations.


pH value of HF, hydrogen fluoride or hydrofluoric acid?

I was able to get some rough pH indicators that are allowed in our lab (cleanroom environment). No clue why a supplier like Honeywell has completely wrong data.

Apparently some sources are completely off, doing the calculation based -log(1/H+) gives the correct values in the end, tested for 50, 10, 5, 1 and 0.5%.
for a quick calculation see:
http://www.endmemo.com/chem/phcal.php


Forming ester from amide using protonated alcohol as catalyst

There are several points that I'd throw out:

As a technicality, you probably can't buy protonated alcohols; you'd need to use the alcohol plus an acid like HCl in a non-aqueous solvent.

Now, once the amine drops off, it will mop up your catalytic acid, so you had probably better use at least stoichiometric acid, and possibly quite a bit more.

Amides are more stable than esters; you know that already. So, the reaction thermodynamics are not in your favour to begin with. Thus, in order to get a decent yield, you probably also need to use a big excess of alcohol (Le Chatelier's principle: more alcohol → more ester).

You'd probably also need to heat it quite a bit to get the reaction going. (The same is true for hydrolysis with aqueous acid.)

All in all, probably you could get it to work, but it might not necessarily be fun, and if your reactants had any other sensitive functional groups on it they would pretty much be destroyed by heating in strong acid.
In general, this is a tough reaction, and I suspect that if you could find a general way to do this cleanly then it would be quite a substantial achievement. Recently, Hie et al. have developed a Ni-catalysed procedure (Nature 2015, 524 (7563), 79–83), but it only works on aromatic  amides, so isn't applicable to your example substrate. (Essentially, both the group attached to the carbonyl carbon, as well as one of the groups on nitrogen, have to be aromatic. Despite that obvious limitation in scope, it's still in Nature, which maybe suggests something about the difficulty of the transformation.)


The two protons Hb and Hc are also chemically inequivalent, because they have a different chemical environment with the methyl group, i.e. they don´t have the same chemical shift.

In a chiral molecule, the protons in a methylene group are rarely really chemically equivalent. There is usually a preferred orientation.


What is the simplest formula for the mineral beryl which contains 5.03% Be, 10.4% Al, 35.5% Si and 53.58% O?

Firstly the "correct" answer seems to be wrong, which is not an uncommon occurrence, so please do not always take answers at the back of books to be correct. Wikipedia lists the formula of beryl as $\ce{Be3Al2(SiO3)6}$ which is much more in accordance with what you have done.

That aside, there are two things you need to correct.

The percentages that you have in your first step are different from those quoted in the title. For example, for silicon, you divided 31.35 by 32, but the title says that there is 35.5% silicon. Please, be more careful when you are pressing numbers into a calculator or the computer. The same issue exists for aluminium and oxygen.
In your second step, oxygen, you seem to have done a calculation pertaining to the first step? You meant to do 3.3/0.558, correct?

Fix those and you should obtain the correct answer - and not the one in the back of whatever text you are using.


Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?

The answer is, as @orthocresol mentioned, due to two parts.
Initially, if we are assuming the ligands are point charges (crystal field theory) then in octahedral complexes, the ligands are more closely associated with the d orbitals, as shown in this diagram:
But, in tetrahedral complexes, the ligands are not so closely associated with the d orbitals, as shown in this diagram:
By associating the d orbitals seen in the first diagram, with the tetrahedral point charges in the second diagram, you can see how close the point charges are to the d orbitals in the octahedral case compared to tetrahedral.
Also as otho mentioned, octahedral has 6 ligands interacting with 5 d orbitals, whereas tetrahedral has only 4 ligands interacting with 5 d orbitals. Octahedral has the more negative charge due to more electrons, therefore more repulsion.


The borosilicate tube gives a very broad signal, the signals from the boron compounds you're measuring should be much sharper. If your samples are concentrated enough and have reasonably sharp lines, that should be sufficient to distinguish them from the broad background signal from the NMR tube.

You could of course also measure an identical NMR tube with the same amount of solvent, but without your boron sample. Comparing this with your real sample should make it easy to identify which signals are background from the tube.

And of course you could use an NMR tube made from a different material.


If you are given ΔH for this reaction which occurs at constant pressure, what additional information you need to determine ΔE?

Because:

1) your question assumes that the volumes of the solids ($\ce{Al}$ and $\ce{AlCl3}$) are negligible compared to the volume of the $\ce{Cl2}$ gas. This means that $\Delta V \approx V(\ce{Cl2})$.

2) your question assumes that chlorine gas behaves ideally. This means that $V(\ce{Cl2}) = \frac{nRT}{p}$.

In the end, it means you only need two of the three variables ($p$, $T$, $V$) in order to obtain all the information you want.

I would however be very careful about using molar quantities, in particular $V_m$ (since $p$ and $T$ are intensive properties), if the question is asking you to find $\Delta H$ and $\Delta E$ in terms of $\text{kJ mol}^{-1}$.


How come we can't use the equivalence point equation of NV=NV in this problem?

My question is, since are looking at the equivalence point in a titration, why can't we use $N_\text{acid} \times \text{Volume}_\text{acid} = N_\text{base} \times \text{Volume}_\text{base}$, since equivalents of acids and bases are the same at the equivalence point? If we can, how would we do it?

First of all, I would suggest that you stick to the original method using moles and proper mole ratios.

Please forget about this normality equation because it is only historically important and rarely anyone uses this concept in 2020 for research or academic work. Note, industrial analytical calculations still use this equation for very routine work. Still if you want to know why this normality equation fails in your case is as follows.

0) The $N_aV_a = N_bV_b$ relation holds if and only if all the concentration units are in normality, where $a$ and $b$ can be acid-base or oxidizing-reducing agent pair. It will also work for precipitation titrations etc. Corollary: This equation fails in titration calculations when the units are in molarity.

(i) This is a redox titration not an acid-base titration.

(ii) Let us change the subscripts for the sake for indexing, $$N_\text{ox}V_\text{ox}=N_\text{red}V_\text{red}$$

You have $\pu{0.01 M}$ $\ce{KIO3}$, what will be its concentration in normality units?

For that you need to see how many electrons are gained in the half-reaction...

$\ce{KIO3}\longrightarrow\ce{I2}$; Balance the equation and see how many electrons are involved in the redox process. See the example here
Balancing this reaction

Multiply the number of electrons involved in the half reaction by the molar concentration to get normal concentration of potassium iodate.

Apply the above equation and you will get the correct result but in normality units i.e. thiosulfate concentration in normality units.


Example of chemical reaction with a given form for the kinetic equations

A chemist friend found an answer for the simplest case:
$$
\begin{cases}
\frac{\partial u}{\partial t} - d_u \Delta u = \tilde{k}_1 v - \tilde{k}_{-1} u - \tilde{k}_2 uv \\
\frac{\partial v}{\partial t} - d_v \Delta v = \tilde{k}_{-1} u - \tilde{k}_1 v - \tilde{k}_2 uv
\end{cases}
$$

$u$ is the concentration of ethenol;
$v$ is the concentration of ethanal;
the tautomerization of ethenol and ethanal gives the first-order part of the reaction;
the aldol reaction of ethanal with ethenol, giving the 3-hydroxybutanal, gives the second order part.

This is definitely a satisfying answer.

Nevertheless I’m still interested in other answers, especially with autocatalysis of either $u$ or $v$ (that would decrease the first-order consumption of it) or with autoreactions (that would give the term $-k_i u^2$ or $-k_i v^2$). As a matter of fact, the French Wikipedia claims that an aldol reaction of ethanal with itself in presence of a base can also produce 3-hydroxybutanal, so that the same example would also give a term $-k v^2$? But maybe these two aldol reactions correspond in fact to the same elementary mechanism?


According to Electronic Structure of NO2 Studied by Photoelectron and Vacuum-uv Spectroscopy and
Gaussian Orbital Calculations J. Chem. Phys. 53, 705 (1970) :

The highest molecular orbital $4a_1$ is occupied by the 1 unpaired electron.

The electron population of this orbital is (see table VI):

0.53 on the N atom (0.16 2s, 0.37 2pz)

0.24 on each O atom  (0.24 pz)

Experimentally, Oxides and Oxyions of the Non-metals. Part II.
C02- and NO2 of the Chemical Society (1962): 2873-2880 collects values for the partition of unpaired electron density among N and O atomic orbitals.

The most purely experimental values are:

N 2s  0.094
N 2pz 0.364
N 2px 0.054
O 2pz 0.33

and

N 2s 0.103
N 2pz 0.471
O 2pz 0.33

(where the values for O are both O atoms added together)

There is extensive discussion of the various Lewis structures of NO2 from a dimerization point of view (that the opposite spin electrons should be on the atoms that form the dimer bond) starting on page 90 if the 2015 book Bonding in Electron-Rich Molecules: Qualitative Valence-Bond Approach via Increase-valence Structures , and also elsewhere in the book from a monomer point of view.

Overall, it is clear theoretically and experimentally that the unpaired electron is delocalized, having significant density on both N and the two O atoms.  It is wrong to say that a particular one of the Lewis structures is "correct".


Differences in using molten NaCl compared to aq. NaCl in a galvanic cell having copper anode and cathode

A galvanic cell with the same electrodes ( copper ones in this case ) is usable as a galvanic cell only if cathodic and anodic electrolyte differs in the concentration of electroactive components.
E.g. if we can use 2 different concentrations of $\ce{CuSO4}$ solution, separating them by a diaphragm or a salt bridge. The electrode with more concentrated solution around would become cathode with the more positive potential and vice versa.
Putting 2 copper electrodes into homogeneous solution of $\ce{NaCl}$ ( or molten one ) does not form a galvanic cell, unless some minor undefined differences create some residual voltage.
In case you did mean an electrolytic cell with moltan salt instead, anodic copper would be dissolving.
At the cathode, only metallic sodium would depose initially. Since some copper ions reach the cathode, copper would start priority deposition. Sodium deposition would depend on the applied electrode potential.


What current and voltage are best for electrolysis rust removal?

In order to answer the question we also need how large a part you are electrolyzing.

The current density in $mA/cm^2$  is the key number. For electrolysis on stainless steel, I would say you need to stay less than about $ 1~mA/cm^2$ to stay safe.

The voltage is hard to gauge. In any two electrode system like yours, the electrochemical potential is only one part of the total potential. The total potential includes contributions from the resistance of the solution, resistance of the contacts, overpotential of the counter reaction, etc... The current should be a decent gauge of what you are seeing.


How is "the surrounding" and its temperature defined to calculate entropy change during a reaction?

Thermodynamics deals with changes between systems at equilibrium, which is to say that the initial and final states of a process are regarded as equilibrium states. As explained in the comments, the surroundings are usually defined as an ideal reservoir of infinite size and thus infinite heat capacity and constant temperature, clearly an approximation, but the key point remains that stated in the previous sentence.

It does not make sense to define an equilibrium diathermal state in which the surroundings and the system have a different temperature. Since  in the stated problem the system undergoes exchange of heat with its surroundings, and the system is at an initial and final T of 263 K after the transition, then the surroundings have to be at 263 K throughout.

One potential point of confusion regarding this type of problem concerns the metastable nature of the supercooled state. This is not a true thermodynamic equilibrium but is regarded as sufficiently stable to be regarded as such. An additional point of confusion may be that heat is transferred to the surroundings even though system and surroundings are at the same T. This is possible because the surroundings have infinite size and heat capacity. One way to envision such surroundings is as a reservoir of a substance with a melting point of 263 K.


Interpretation of 1H-NMR spectra for Z-ochtodenol

Let’s get the obvious out of the way first: Yes, the light blue protons should probably be the allylic ones since their shift is so similar to the other allylic protons’ shifts. (The green ones, that is; the red ones also experience deshielding from their neighbour oxygen.)

Then also, the report of the spectrum has been done badly. It should have included the field used. It makes a big difference if a spectrum is measured at $90~\mathrm{MHz}$ or at $700~\mathrm{MHz}$ — it may well be that weaker couplings are unobservable in lower field spectra but clear in higher ones. If I had to guess, I would assume a rather weak magnet since the smallest coupling reported is $6~\mathrm{Hz}$ — and they don’t even include a further significant figure.

Therefore, concerning your first bullet point: It is entirely possible that those protons do couple only the spectrum had too low a resolution to show it. However, remember that coupling should always be considered a possibility. If you check out the Karplus curve, you will see that for an angle of $\approx 90^\circ$ vicinial coupling effectively becomes zero. Same things may happen with long-range couplings: They may be absent with no apparant reason to the organic chemist since we cannot see the molecule’s exact orientation.

A ‘broad multiplet’ makes no sense to me whatsoever. In my opinion, multiplet alone should be enough. And I also strongly suspect an error in the book here: That is the shift where the four missing question-marked hydrogens should appear at. They should have badly overlapping, possibly even complex spins since each one is magnetically unequivalent and we are dealing with a cyclohexane system. This line should be, in my opinion:

  $1.35-1.70$ (4H, m)

And that also explains why three protons are missing. The pink ones are the only ones that would allow more protons in the same spot.

All in all that last bit is an error that really shouldn’t have happened in a textbook exercise. I do hope there is an erratum somewhere.


Approximate time for pure water pH to equilibrate in open air?

Assume you have poured your distilled water in a 1 liter container.

We write the flux of $n_{\ce{CO2}}$ between air towards your container, as
$$
J_{\ce{CO2}}=\frac{dn_{\ce{CO2}}}{dt}\times\frac{1}{A}=\frac{d[\ce{H2CO3}]^{*}}{dt}\times\frac{V}{A}
$$
where $n_{\ce{CO2}}$ is the number of ${\ce{CO2}}$ moles in water, $A$ the water surface and $V$ the water volume.
Assuming $\ce{CO2}$ atmospheric pressure equal to $10^{-3.5}\,atm$, we obtain the concentration $[\ce{CO2}]_{air}$ approximately equal to $10^{-5}\,M$ (the latter value obtained via $PV=nRT$).

We call $[\ce{H2CO3}]^{*}$ the sum of $[{\ce{CO2}}]_{water}$ and $[\ce{H2CO3}]_{water}$ concentrations in water: at the beginning, we set it equal to $0$. Note that $[\ce{H2CO3}]^{*}$ could be reasonably written as:

$$
[\ce{H2CO3}]^{*}\approx[{\ce{CO2}}]_{water}
$$

indicating that the majority of carbon dioxide in water is not converted into carbonic acid, being the kinetics for such a conversion very slow, as pointed out by @Nicolau Saker Neto, .

By expressing $J_{\ce{CO2}}$ as:

$$
J_{\ce{CO2}}=\frac{D_{{\ce{CO2}}}}{Z}\times ([\ce{CO2}]_{air}-[\ce{H2CO3}]^{*})
$$

with $D_{\ce{CO2}}$ the diffusion coefficient of ${\ce{CO2}}$ equal to $7.2\times10^{-4}\,dm^{2}h^{-1}$and $Z$ the thickness of superficial layer through which the exchange occurs. For the latter, we take $40\times10^{-5}\,dm$.

Putting previous equations together (with $A/V=1\,dm^{-1}$), the following is obtained:

$$
\frac{d[\ce{H2CO3}]^{*}}{dt}=1.8\times10^{-5}-1.8\times[\ce{H2CO3}]^{*}
$$
which yields:

$$
[\ce{H2CO3}]^{*}=1\times10^{-5}(1-e^{-1.8t})
$$

The graph below shows that about $2$ hours later, a plateau is established, for $[\ce{H2CO3}]^{*}= 1\times10^{-5} M$.

By considering the acid-base equilibrium ($pK_{a}=6.3$):
$$
\text{H}_{2}\text{C}\text{O}_{3}^{*} \ce{<=> HCO3- + H+}
$$
we obtain a $\text{pH}=5.7$, at the plateau.

UPDATE 1: the value for $Z$ has of course a big impact on time. A larger value will increase the time needed to reach the plateau. I have used $40\times10^{-5}\,dm$, considered a "typical value" by the authors of this book (sorry in French, page 181, 3rd edition). Any more accurately derived value for $Z$ is welcome !

UPDATE 2: I used the notation $\text{H}_{2}\text{C}\text{O}_{3}^{*}$ to represent the two species $\ce{CO2}$ and $\ce{H2CO3}$, in water.


Derivation of the relation between temperature and pressure for an irreversible adiabatic expansion

When you first posted this question I was very busy, but now I've found the time to look at it again and can give you an answer:

As I already mentioned in one of my comments your equation is slightly incorrect. It should actually be

\begin{align}
  T_2=\left[\frac{c_{V} + R\frac{p_2}{p_1}}{c_{p}}\right]T_1
\end{align}

Now, I will come to the derivation:
You should always start out by identifying what you already know about the system. You know the expansion will be performed adiabatically, so there can be not heat exchange with the environment, i.e.

\begin{align}
  \delta Q = 0 \ .
\end{align}

This means that from the first law of thermodynamics you get:

\begin{align}
  \mathrm{d} U = \delta Q + \delta W = \delta W
\end{align}

where $U$ is the internal energy and $W$ is the work.
Furthermore, you know that the expansion process is irreversible. If the only work performed is the pressure-volume work, where the system has to expand against a constant external pressure, which is $p_2$ in your equation, then this work is given by:

\begin{align}
  \delta W = - p_{2} \mathrm{d} V \qquad \text{where:} \qquad p_{2} = \text{const.} \neq f(V)
\end{align}

So, together with the first law of thermodynamics this yields:

\begin{align}
  \mathrm{d} U = - p_{2} \mathrm{d} V \ .  \tag{1}
\end{align}

Now, you need to assume that the expanding gas is an ideal gas. For ideal gases you know that

\begin{align}
  \mathrm{d} U &= \underbrace{\frac{\partial U}{\partial T}}_{= \, C_{V}} \mathrm{d} T + \underbrace{\frac{\partial U}{\partial V}}_{= \, \Pi \, = \, 0} \mathrm{d} V \\
  \mathrm{d} U &= C_{V} \mathrm{d} T \tag{2}
\end{align}

where $T$ is the temperature, $C_{V}$ is the heat capacity at constant volume, and $\Pi$ is the internal pressure, which is equal to zero for an ideal gas.
Other relations that hold for ideal gases are the ideal gas law

\begin{align}
  p V = n R T
\end{align}

where $n$ is the amount of the gas and $R$ is the universal gas constant,
and

\begin{align}
  c_{p} - c_{V} = R  \tag{3}
\end{align}

where $c_{p} = \frac{C_{p}}{n}$ and $c_{V} = \frac{C_{V}}{n}$ are the molar heat capacities at constant pressure and constant volume, respectively.

Now, you just have to combine equations $(1)$ and $(2)$:

\begin{align}
  - p_{2} \mathrm{d} V = C_{V} \mathrm{d} T \ ,
\end{align}

integrate the result (under the assumption that $C_{V}$ is temperature independent)

\begin{align}
  - p_{2} \int^{V_2}_{V_1} \mathrm{d} V &= C_{V} \int^{T_2}_{T_1} \mathrm{d} T \\
  - p_{2} \left( V_2 - V_1 \right) &= C_{V} \left( T_2 - T_1 \right) \ ,
\end{align}

use the ideal gas law $V = \frac{n R T}{p}$ to substitute $p$ for $V$

\begin{align}
  - n R p_{2} \left( \frac{T_2}{p_2} - \frac{T_1}{p_1} \right) &= C_{V} \left( T_2 - T_1 \right) \\
  - n R \left( T_2 - \frac{p_2}{p_1} T_1 \right) &= C_{V} \left( T_2 - T_1 \right) \ ,
\end{align}

introduce the molar heat capacity via $C_{V} = n c_{V}$

\begin{align}
  - R \left( T_2 - \frac{p_2}{p_1} T_1 \right) &= c_{V} \left( T_2 - T_1 \right) \ ,
\end{align}

and separate $T_1$ and $T_2$

\begin{align}
  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 \left( c_{V} + R \right) \ .
\end{align}

Finally, you can use equation $(3)$ to get the desired result:

\begin{align}
  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 \bigl(\underbrace{c_{V} + R}_{= \, c_{p}} \bigr) \\
  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 c_{p} \\
  \Rightarrow \ T_2 &=\left[\frac{c_{V} + R\frac{p_2}{p_1}}{c_p}\right]T_1 \ .
\end{align}


Indeed, it should be inversely proportional. In case you divide the copper concentration by the nickel concentration, the sign of the RT/nF part should be plus and not minus. If you divide the product of the reduction by the precursor concentration, you have to use minus.

The temperature has a great effect on the current, especially in the "Cottrell-determined" area. The Nernst equation, which you use in this case, is true for a near-equilibrium environment with fast electrode kinetics. If you drive the potential too high, mass transport can become so slow (in comparison to electron transfer) that an increase in potential will not yield a higher current flow as predicted by Butler-Volmer kinetics.


Confusion about the charge of electrons while orbiting through bohr's energy shells

The main confusion I can see in your question is charge vs. energy. Energy is the capacity to do work. Energy comes in many forms. Kinetic energy is the energy of motion. Potential energy is the energy of the system due to its arrangement. Electrons in Bohr shells have energies dependent on the mass of the electron, which is constant, the charge of the electron (constant), the distance from the nucleus (fixed at certain allowable values), and some fundamental constants. Thus, an electron in a shell has unchanging energy. The only way for the electron to change its energy is to move to a different shell. $\Delta E=0$ for an electron in a Bohr shell; $E\neq0$.

An electron in a Bohr shell certainly has kinetic energy because it has orbital angular motion. An electron in a Bohr shell certainly has potential energy because it is being held apart from the oppositely charged nucleus. The total energy of the electron in a Bohr shell ($E_n$) is a function of the principal quantum number, $n$. $Z$ is the number of protons (1 for hydrogen, the only atom for which the Bohr model works) and $R_E$ is the Rydberg energy (related to the Rydberg constant by $R_H=R_E/hc$).

$$E_n=-\frac{Z^2 R_E}{n^2}$$

The Rydberg energy is the smash up of some other more fundamental constants, like the mass of an electron ($m_e = 9.109\times10^{-34} \text{ kg}$), the fundamental unit of charge ($e=1.602\times 10^{-19} \ \text{C}$), Coulomb's constant ($k_e = 8.987 \times 10^9 \ \ce{N m^2 C^{-2}}$ also $k_e=(4\pi\varepsilon_0)^{-1}$, where $\varepsilon_0=8.854\times 10^{-12} \ \text{F/m}$ is the permittivity of a vacuum), and the Bohr radius ($a_0 = 5.29\times 10^{-11} \ \text{m}$). The Bohr radius is itself equal to a mashup of other constants  $a_0=\hbar/(m_e c \alpha)$, where $c=2.998\times10^8 \ \text{m/s}$ is the speed of light in a vacuum, $\hbar=1.054\times10^{-34} \ \text{J s}$ is the reduced Planck constant, and $\alpha=7.297\times^{-3}$ is the fine-structure constant. Eventually, $R_E$ simplifies down to a relativistic energy:

$$R_E = \frac{k_e e^2}{2a_o}= \frac{1}{2}m_e c^2 \alpha$$

Electric charge is something else. It is a unit of electrostatic interaction with other charged particles. The electric charge of an object influences the energy of that object in the presence of an electric field, see Coulomb's Law. The (magnitude of) charge on an electron ($e$) is part of the Rydberg energy, so it influences the total energy of the electron.

However, charge$\ \neq\ $energy.


Dielectric constant and dissolution of ionic salts

The dissolving of a substance into a solvent and the avoidance of contact ion pairing depends on several things. The dissolving relates to the change in Gibbs free energy between the crystaline solid and the solution of the cations and the anions from a salt. A Born-Haber cycle can be used to explain this.

You should be aware of the fact that the Debye-Huckel and extended Debye-Huckel equations for the activity coefficent for ions in solutions have terms in them for the relative dielectric constant of the liquid.

The higher the activity coefficent then the more the ions are able to move freely of each other. The extended Debye-Huckel equation can be modified into the Bromley equation by adding another term for ion-ion interactions.

When considering the effect of changing the relative dielectric constant.

I think that one of the best examples is sodium bromide, in water sodium bromide is a strong electroyte where the cations and anions are well separated. But in acetone where the relative dielectric constant is lower the same salt is a "weak electroyte" where the cations and anions tend to associate with each other.

I think that the formation of contact ion pairs is the first step when going from a perfect solution of ions to a crystaline ionic solid. But the water does more than act as a simple dielectric. What the water molecules will do is to bind to the ions. In particular the water molecules will coordinate as Lewis bases to the cations and then enlarge the size of the cations. This binding of waters can act as a bridge between the cation and the water.


Difference between polarity and polarizability

Polarity and polarisability are almost entirely unrelated.

Polarity refers to the distribution of partial electric charge in molecules (or even materials). Or, more technically, the degree to which a molecule has a dipole moment (which relates to its inherent electric field). Some molecules, like benzene, are non-polar as they have no net dipole moment; others, like nitromethane, have a strong dipole moment. Polarity is a fixed property of the molecule that doesn't depend on the external field.

Polarisability refers to the degree to which the electron clouds in a molecule or atom can be influenced by an external electric field. Everything, polar or not, has a polarisability. Crudely (i.e. simplifying a lot) you can think of it as the degree to which electrons are held tightly in the atom or molecule. So xenon atoms are fairly polarisable compared to helium atoms as their electron cloud is more spread out and less tightly bound. Water is very polar but a lot less polarisable than hexane which is non-polar. Bulk substances have complicated polarisability that depends on the crystal structure and may be inhomogeneous.

Polarity is about the ground state electron distribution; polarisability is about how that distribution changes when subject to an external electric field.


It has nothing to do with aromaticity. It is related to this answer.

Why does benzaldehyde not respond to Fehling's test?

Any aldehyde without an alpha hydrogen will not give Fehling's test.
Acrolein also doesn't have an alpha hydrogen, so it will not give Fehling's test.