For a given element do heavier isotopes form stronger bonds than lighter isotopes?

Isovalent isotopes will have the same force constant. However the different masses of the isotope will affect the position of the vibrational state in its potential well. You can rationalise the difference in well depth briefly using the vibrational frequencies of a classical oscillator as the harmonic approximation to the asymmetric well for low lying states. If the stretching frequency is given by;
\begin{equation}
\nu \propto \frac{1}{\sqrt{\mu}}, \ \ \ \ \ \ \ \ where \ \ \ \ \ \ \ \mu =\frac{m_1\cdot m_2}{m_1+m_2}
\end{equation}
Then as the reduced mass increases the frequency decreases and the vibrational energy decreases $E=\hbar\omega $ with  $\omega\cdot 2\pi=\nu$. We can see that if we were to double the reduced mass then the energy would decrease by root 2.

So we need to examine the reduced masses of some simple systems, lets say the diatomics you suggest. If we let the masses of $H=1$, $D=2$, and $T=3$ then we calculate the reduced masses to be $1/2$, $2/3$ and $3/4$. The reduced mass is increasing as we go along the series and hence we expect the successive molecules to have stationary states lower in the well.

In vibrational spectroscopy this is best seen by the decrease in zero point energies for HCl and DCl. It is often easiest to take the ratios of the frequencies $\omega _{HCl}/\omega _{DCl}$, where using $m_{Cl}=35$ and H and D as above we obtain 1.395. indicating that the deuterated molecule is deeper in its well.

Edit

I have been advised to explain zero point energy. In short it is a irremovable quantity of energy from a quantum oscillator. The classical oscillator has no parallel to this. It is a manifestation of the uncertainty principle in position and momentum. If you want the truth of where it comes from you will need to consult the physics SE, but I will do my best :)

The quantum oscillator has a Hamiltonian of the classical oscillator except just replace the classical position and momentum variables with their quantum operator analogues $p\mapsto \hat p$ and $x\mapsto \hat x$.

The classical Hamiltonian (or total energy) is learnt at school to be kinetic + potential,
\begin{equation}
H=\frac{p^2}{2\mu}+\frac 12 kx^2
\end{equation}
For the reduced two body system. It is a postulate of quantum mechanics that in the position representation (just a way of formulating things) that wherever you see a $p$ and an $x$ you replace them by $\hat p$ and $\hat x$ where,
\begin{equation}
\hat p =-i\hbar \frac{\partial }{\partial x}, \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \hat x =\cdot x
\end{equation}
Therefore with these changes in the above classical expression we generate the quantum Hamiltonian of the quantum harmonic oscillator,
\begin{equation}
\hat H=\frac{\hbar^2}{2\mu}\frac{\partial ^2}{\partial x^2}+\frac 12 kx^2
\end{equation}
Where $i=\sqrt -1$ so $i^2=-1$ and $--1=1$.

It is another principle of quantum mechanics that the Schrodinger equation is the equation to solve for quantum systems, just like in classical mechanics you use Newton's second law, $F=ma$, This equation has a bit of a different shape but in essence (decoupled from time) takes the following form.
\begin{equation}
\hat H\psi =E\psi
\end{equation}
Substituting the above expression for $\hat H$ in and doing a bit of maths (wayyyy to advanced for us here) we can solve for the energy $E$ in the Schrodinger equation, when we do this we get,

\begin{equation}
E_n=(n+\frac 12)\hbar \omega
\end{equation}
What this tells us is that the vibrational levels are not continuous like the classical oscillator (above) but can only take discrete values. We see in the maths of the solution for $E$ the need for discrete integers $n$ which we call quantum numbers. These may take in this case values of $0,1,2,...$ etc etc. Each corresponding to the next adjacent energy solution.

Something interesting happens if we choose $n=0$ however. You would think that the lowest energy would be zero right? well no... quantum mechanics says we can never remove a little bit of energy from the system. (try substituting $n=0$ in the above expression and we see that the answer is non zero (1/2 $\hbar \omega$).

This is a zero point energy.

As promised here is the home-made graph of the molecular potential energy. Im sorry its not very good but honestly the best I could do with the available tools! Apologies.

The main points to follow are:

At the base of the well the graph looks like a parabola, i.e. the harmonic oscillator approximation to the molecule works best at low lying states since the anharmonicity term grows with excitation (well increasing quantum number).
Where the harmonic oscillator is valid you can see that the energy levels of are evenly spaced, each one corresponds to the quantum numbers $n$ as given above. You can calculate $\Delta E=E_{n+1}-E_{n}$ to get the difference in energy as $\frac 12\hbar \omega$. In this approximation the equilibrium bond length does not change with excitation. However we can see it does in the anharmonic case.
To answer your question, the energy levels corresponding to the heavier isotope are the blue ones since they sit deeper in the well. So you could imagine the blue ones as C-D while the red ones would be C-H. (normally you would just be looking at one species per graph, so only one set of coloured lines.


"n factor" of oxidizing or reducing agent = total number of electrons transferred per a mole of the reactant.

When more than one element in a substance is oxidized/reduced, the n-factor of the substance is taken as the sum of the n-factors of the individual elements of the substance.

This is the case here.You have  ammonium ion that undergoes oxidation reaction (n-factor= 3) and  nitrite ion that undergoes reduction reaction (n-factor= 3).
So, the n-factor of ammonium nitrite is 6.


Rate of change and kinetics order of reaction question

Indeed, the only thing you'll need is solving the first order differential equation (but before that, let me define $R_0 := \ce{[RSH]_0}$ and $C :=\ce{[Ker-S-S-Ker]}$.
Now, let's solve it assuming (that's what is given) that $R_0$ can be considered a constant.
Then, we have : $$\frac{\operatorname{d}C}{\operatorname{d}t} = -kR_0C \implies \frac{\operatorname{d}C}{\operatorname{d}t} + kR_0C = 0 \implies C(t) = \exp\left(-\frac{t}{kR_0}\right) + C_0$$ (integration constants can be determined using initial values, I won't be giving further details here on how to solve this...)
Hence, since the concentration of your molecule is an exponential function of time, it is inevitably a convex function (so only curves C and E can work).
Finally, a simple thought on the values of $k$ and $R_0$ show that curve E cannot be the answer as the decay is to "fast".
Hope this helps, if needed I can also provide a "variables separation" method of solving the differential equation.


Which element has larger atomic radius -- beryllium or xenon?

You cannot really effectively compare the atomic radii of xenon and beryllium empirically; the atoms are too far apart. Once you remove yourself from direct neighbours, it is generally not possible to estimate relative sizes of atoms.

The closest hint you can get are calculated atomic radii on Wikipedia’s data page. These show that xenon is slightly smaller ($\pu{108 pm}$ to $\pu{112pm}$) but the difference is smaller than the expected error. So, for all intents and purposes they have the same size.


Here is nice comparison of all 3 ring sizes. Octahydro-1H-inden-1-one (either cis- or trans-) carbonyl strechings fall between cyclohexanone and cyclobutanone, thus, fused ring is not a factor (it is to some extend because of sunstitution).


Is the conjugate base of every polyprotic acid amphoteric?

The conjugate base of sulfuric acid, bisulfate ion, shows little amphoteric character in water solution.  Unless you drop the pH to about 2 or below, it is only an acid, and a moderately strong one at that.


The bond lengths have little to do with the stability of the atom positioning around the central $\ce{S}$ atom. Rather, the relative sizes of the electron clouds are what should be considered when positioning these atoms. An atom of $\ce{Cl}$ has a 8 more electrons than does an atom of $\ce{F}$, so naturally its electron cloud will be larger and therefore more repulsive. The $\ce{Cl}$ should therefore be positioned in the equatorial position for the same reason you predicted that the lone pair would be present in the equatorial position--because these positions offer the least repulsion between the electron clouds of the atoms.

In the axial position, the 3 equatorial positions, present at 90 degree angles to the axial position, repel the electron cloud.

In an equatorial position, however, only the 2 axial positions are present at 90 degree angles to repel the electron cloud. The other 2 equatorial positions are positioned away from the equatorial bond in question, and repel its electron cloud only slightly. For this reason unbonded electron pairs and atoms with large electron clouds occupy these spaces first-- they offer the least repulsion and the most stable positioning.


The key here is that a system will tend towards the lowest energy state; in other words, for a process to be spontaneous, the final state must have lower energy than the initial state.

The mass of a neutron is slightly higher than the combined mass of the proton, electron and neutrino that result from beta decay, meaning that the decay will be energetically favoured. This might be easier to understand by considering the decay of a free neutron:

$$\text{n}^0 \rightarrow \text{p}^+ + \text{e}^-+\bar{\nu}_\text{e}$$

Whether this process will occur in a nuclear neutron is determined by the $N/Z$ ratio and the atomic number. In small neutron-rich nuclei, the process above will predominate, but as the nucleus gets larger, more neutrons are needed to counteract the electromagnetic repulsion between protons (which has a longer range than the strong nuclear force). Again, the reason is energy: the unbound nuclei collectively have higher energy than a nucleus, due to nuclear binding energy. Hence, for larger nuclei, neutron decay is not energetically favourable.


Reaction of acetylene with hypochlorous acid

The enol probably does tautomerize but it doesn't last. Some of the enol remains in enol form at equilibrium, and that reacts further with the hypochlorous acid.  This causes some of the keto form to revert to the enol, which then gets consumed again by the hypochlorous acid, and so on.  The selective reaction of the acid only with the enol ultimately pulls all of the keto form back to the enol and then the final, fully reacted product you see.


Why isn't hydration an oxidation reaction?

As context is always King, I will briefly mention that the original quote is about the redox chain from carboxylic acid via an aldehyde to a primary alcohol (and potentially to an alkane). In this context, the aldehyde is capable of adding water as shown in $(1)$, but this reaction does not constitute a redox process in any definition of the word (count electrons).

$$\ce{R-CH(=O) + H2O <=> R-CH(-OH)(-OH)}\tag{1}$$

In other cases, the removal or addition of water does change the oxidation states of various carbon atoms as you observed. This is not only true for the dehydration of an alcohol as in your example but e.g. also in such key reactions as the Wittig reaction or the Suzuki cross-coupling.

However, organic chemistry deals with a great number of individual synthetically useful reactions that must be classified in some way or another for humans’ benefit. In this organic chemistry classification system, only those reactions that

unilaterally oxidise a single carbon atom by means of addition of an oxygen atom or removal of hydrogen atoms; or
concurrently oxidise two carbon atoms bound to each other by the addition of oxygen or removal of hydrogen in such a manner that both oxidation states change by the same value

as oxidations. (And vice-versa for reductions. Individual exceptions most certainly will apply.) Your example is, in inorganic terms, a comproportionation which means one carbon is oxidised while the other is reduced. Not only does this not cleanly fit into ‘carbon is oxidised’ or ‘carbon is reduced’, meaning it creates a terminological mess, but it also would subsume most if not all reactions under the general redox labels which destroys the idea of classifying reactions as oxidations or reductions in organic chemistry altogether.

So: yes, in a strict (inorganic) sense it is a redox reaction, but in the interest of a more practical classification of organic reactions it is not considered one.


Rate of SN2 reaction depending on following different conditions in case of primary halides

I know this question was from awhile ago, but I'll answer anyway.

a) The key to understanding this is the concept of the transition state. Below is an image taken from Khan Academy

For an $\mathrm{S_N2}$ reaction to take place, there must be enough activation energy to reach the transition state. In this type of reaction, the transition state comprises an alpha carbon in a pseudo-$\mathrm{sp^2}$ state. The nucleophile and the leaving group are directly above and below the alpha carbon in this state, with atoms bound to the carbon in a horizontal plane.

The reason this reaction proceeds more quickly with an unsaturated beta-carbon is the pi-electrons from the unsaturated carbon stabilize the transition state. This lowers the activation energy of the reaction, making it more likely to happen with each molecular collision.

c) I'm assuming you're comparing $\ce{N, O}$ and $\ce{S to C}$. $\ce{N, O,}$ and $\ce{S}$ are more electronegative than carbon, which will impart a more positive-like characteristic on the alpha carbon, making it more susceptible to attack from the nucleophile.


how to modify a transport device with an internal combustion engine with different atmospheres

Each of the cases will strongly depend on the non-atmospheric natural
resources available on each planet.

For example in an atmosphere of $\ce{O2}$ + inert gases, it would be desirable to have deposits of hydrocarbons or $\ce{H2}$ gas to be mined and pumped into your fuel tank to burn with the atmospheric $\ce{O2}$.  In each case proposed below it is assumed that the non-atmospheric reagents would be available as deposits on the planet or that they could be synthesized in a way that they are acting as batteries, storing the energy that went into their synthesis.  That said, here are some possible simplistic solutions for the different scenarios.

Scenario 1: $\ce{CH4}$ + $\ce{N2}$ atmosphere:

Here you could simply reverse the typical internal combustion engine process used in Earth-vehicles. Rather than filling your tank with hydrocarbons to be mixed and burned with atmospheric $\ce{O2}$, you would fill your tank with (probably liquid to keep the volume small) $\ce{O2}$ to be mixed and burned with atmospheric $\ce{CH4}$.  Just as with conventional internal combustion engines on Earth, the reaction products would be $\ce{H2O}$, $\ce{CO2}$, and heat.

Scenario 2: $\ce{CO2}$ atmosphere:

As $\ce{CO2}$ is already thoroughly oxidized, you would need to fill your tank with a good reducing agent.  If there were deposits of $\ce{H2}$ gas trapped within the planet, then you would mine the $\ce{H2}$ and fill your tank with pressurized $\ce{H2}$ gas (the resources required to form and store liquid $\ce{H2}$ just wouldn't be worth the trouble, though there are other ways to concentrate $\ce{H2}$ then liberate it when needed).  The $\ce{H2}$ would then be mixed and burned with $\ce{CO2}$, ideally giving off $\ce{CH4}$, $\ce{H2O}$ and heat as products.

Scenario 3: $\ce{Ar}$ + $\ce{Cl2}$ atmosphere:

First to address is the composition of the vehicle, or at least the parts that would come into contact with the $\ce{Cl2}$ and other reactants and products, I propose a $\ce{SiO2}$ based coating for all components that would be in such contact.  For a gas-phase ignition scenario, I propose a diesel type of concept (ignition heat from compression of the mixed gases) to avoid the need for electrical ignition, metal wires, etc.

One possibility is to use the $\ce{Cl2}$ analogously to how $\ce{O2}$ is used in internal combustion engines, namely as the oxidant.  A hydrocarbon, $\ce{CH4}$ to keep it simple, would act as the reductant.  This reaction would ideally produce $\ce{CCl4}$, $\ce{HCl}$ and heat.  The atmosphere is already toxic and corrosive so the $\ce{HCl}$ exhaust is probably OK.

In the same manner as the $\ce{CO2}$ atmosphere, $\ce{H2}$ is another possibility.  In this case, the ideal reaction product is just $\ce{HCl}$.

One final possibility is worth considering here.  In all other scenarios I've only proposed fluids as fuel for the combustion.  But $\ce{Cl2}$ reacts very exothermically with $\ce{Fe}$ metal.  And the ideal product, $\ce{FeCl3}$, is also a solid.  So introducing a fine iron dust in some carrier gas could actually have the advantage of ease of collecting the exhaust do be disposed of at a dedicated treatment center.  That way we wouldn't pollute the air for the $\ce{Cl2}$-breathing natives.

Note that most of the proposed fuels here are composed of some of the most abundant elements in the universe.


Purification of Laboratory Chemicals (a generally accepted reference book for methods of purification of common chemicals) appears to agree with your procedure:

  N-Iodosuccinimide [516-12-1] M 225.0, m 200-201 °C. Crystallise it from dioxane/CCl4. It iodinates arenes in triflic acid. [Olah et. al. J. Org. Chem. 1993, 58, 3194, Beilstein 21/9 V 544.]

As you point out, carbon tetrachloride is difficult if not impossible to get hold of now, so the methodology will not work.

The most obvious thing to try would be the method for the purification of NBS, however NIS is not tolerant to recrystallisation from water and from previous attempts doesn't achieve anything.

Depending on what you're doing with the NIS, I've often had success in simply dissolving in solvent and washing with sodium thiosulfate to remove any liberated iodine. Drying and evaporation then gives cleaner, though probably not pure NIS.

Also note that NIS should be stored in the dark at reduced temperatures to prevent decomposition (which is noted by the appearance of an orange/brown decoloration).


Comparing the experimental and calculated UV/vis spectra for ethene

In short, there are two obvious problems with the setup OP uses for TD-DFT calculations:

B3LYP functional is not a good choice for TD-DFT.
6-31G(d) basis is usually too small.

At M06-2X/Def2-TZVP level I get a maximum at ~160 nm, which, taking the accuracy of the TD-DFT approximations into account, is close enough to the experimental value.


Heat energy of phosphoric acid evaporation

You need to know the latent heat of vaporization of the solution, which would be the same as that of water when phosphoric acid concentration is low.

See Fig. 10 of this bulletin for latent heat of vaporization as a function of phosphoric acid concentration, but keep in mind this is only strictly valid at the stated temperature.

If you are varying phosphoric acid concentration (concentrating phosphoric acid) you would need to integrate heat of vaporization as a function of phosphoric acid concentration over the appropriate concentration range.

Also, as phosphoric acid concentration gets very high, it is no longer valid to consider the evaporating solution to be only water, it will increasingly contain phosphoric acid as well.


How are organic compounds with radioactive atoms synthesized?

The current methods employed for synthesis fall into the following three categories1:

Chemical synthesis
Biochemical methods
Isotope exchange reactions

Chemical Synthesis:
Most chemical syntheses involving $\ce{^14 C}$ are done with Grignard reagent and $\ce{^14 CO2}$ or $\ce{^14 CN-}$. As an example, Benzene-14 is synthesized as such2:

Basic preparatory methods such as the above have been worked out long ago. An example of a more complex synthesis would be preparation of 18F-Fluorodeoxyglucose:

Biochemical methods
A few examples of these, as mentioned in (1), are:

High yields of [U-14C]sucrose, up to 70% based on $\ce{^14 CO2}$, are obtainable in detached leaves after a few hours photosynthesis using tobacco or Canna indica leaves for example.
DNA labelled either with $\ce{^14 C}$ or with $\ce{^3 H}$ is prepared by feeding E.coli with the correspondingly labelled thymine precursor. Similarly, by feeding [$\ce{^35 S}$]sulphate to E.coli or to yeast cells, isolation and hydrolysis of the proteins yields L.[$\ce{^35S}$]methionine and L-[$\ce{^35 S}$]cystine in radiochemical yields of 30% and 10% respectively.

Another example is the synthesis of C-14 labelled Phenylpyruvic acid3

Isotope exchange reactions
These are fairly common for atoms that can easily be exchanged, such as replacing $\ce{^23 Na}$ for $\ce{^24 Na}$, which was mentioned in the question. Another common mechanism for synthesis of Hydrogen isotope compounds is Hydrogen-Deuterium exchange. A similar mechanism also allows hydrogen-tritium exchange4.
Purification of radiolabelled compounds
This is mainly done through Chromatography1, as the amount of radiolabelled substrate required is generally quite small. High-performance liquid chromatography is used to separate the compounds. The chemical identity of the compounds is  confirmed by co-chromatographic behaviour with the unlabelled compound.
For further information, refer to the references (1) and (3), which go into a lot more detail with regards to the synthesis of these compounds.
References:

Anthony Evans, E. “Synthesis of Radiolabelled Compounds.” Journal of Radioanalytical Chemistry, vol. 64, no. 1–2, Mar. 1981, pp. 9–32. doi:10.1007/BF02518337.
Schmid, K., et al. “A Simple Method of Preparing Labeled Benzene.” Advances in Tracer Methodology, edited by Seymour Rothchild, Springer US, 1966, pp. 37–44. doi:10.1007/978-1-4684-8625-4_5.
Atzrodt, Jens, and John Allen. “Synthesis of Radiolabeled Compounds for Clinical Studies.” Drug Discovery and Evaluation: Methods in Clinical Pharmacology, edited by Hans Gerhard Vogel et al., Springer Berlin Heidelberg, 2011, pp. 105–18. doi:10.1007/978-3-540-89891-7_12.
Englander, S. W., and J. J. Englander. “[3] Hydrogen—Tritium Exchange.” Methods in Enzymology, vol. 49, Academic Press, 1978, pp. 24–39. doi:10.1016/S0076-6879(78)49005-6.


What is the heat needed for an isothermal expansion of gas. Why do physics and chemistry yield different answers?

The problem is that both the problem statement you presented and the chemistry solution you showed are fallacious.  The fallacy is that, for an ideal gas $V=nRT/P$, so that, if the pressure and temperature are both constant, the volume can't change.  If the volume does change, then either the pressure must change or the temperature must change.

In the "chemistry solution" you presented, the pressure is taken as constant (atmospheric).  So the temperature must change.  But if the temperature changes, then $\Delta U$ is not zero (as they assume).  So the solution is incorrect.  The correct solution, based on the 1st law, is $$\Delta U=nC_v\Delta T=Q-P_{atm}(V_2-V_1)$$or equivalently, $$\Delta H=\Delta U+P_{atm}(V_2-V_1)=nC_p(T_2-T_1)=Q$$Also, since $$n=\frac{P_{atm}V_1}{RT_1}$$we have $$\frac{T_2}{T_1}=1+\frac{QR}{C_pP_{atm}V_1}$$

In the "physics solution" you presented, they make the opposite assumption that, rather than the pressure remaining constant, the temperature remains constant.  For the temperature remaining constant, the solution you presented is correct.  But the final pressure is:  $$P_2=P_{atm}\frac{V_1}{V_2}$$


Why is a trialkylamine a better leaving group, in nucleophilic reactions, than a Cl group?

Leaving group ability refers to how well a leaving group can stay off whatever it came from.

Consider the following prototypical nucleophilic displacement reaction:

$\ce{R-LG + Nuc- <=> R-Nuc + LG-}$

There are two issues at play here:

1) Thermodynamics. How stable is the reactant with the leaving group, $\ce{R-LG}$, and the nucleophile, $\ce{Nuc-}$? How stable are the products, $\ce{R-Nuc}$ and $\ce{LG-}$? And when examining stability, we must look at not only the reactants and products but also their environments - i.e. the solvent(s).

2) Kinetics. What's the physical ability of the nucleophile to actually displace the leaving group? What's the physical ability of the leaving group to displace the nucleophile again?

Usually we're taught that leaving group ability has to do with acidity and basicity. The less basic the leaving group is, the better its ability to leave. This isn't a perfect rule of thumb, but it's acceptable.

That's the thermodynamic part of the question. If $\ce{LG-}$ weren't thermodynamically stable, then it would be inclined to attack $\ce{R-Nuc}$ and return to the status quo. (Or just grab a proton).

However, how capable is $\ce{LG-}$ of carrying out such an attack? Here we come to the kinetic part of the question. The bulkier the leaving group, the more likely it is to be sterically hindered and/or strongly solvated (trapped by solvent molecules). Trialkylamine  is a very genetic term, but when I think of amines with alkyl groups I think of quaternary ammonium compounds, which have alkyl groups that can contain upwards of 18 carbons (!). Such bulk affects the ability of the leaving group to line up properly with the substrate to perform nucleophilic attack. Nucleophilic attack often has specific steric demands; $\ce{SN2}$ requires backside attack; the approach of attacking nucleophiles with respect to flat carbonyl groups has an angle of 104.5 degrees, etc.

Here is a good summary of sterics and amine nucleophilicity.

https://books.google.com/books?id=tP4cqrYF7nwC&pg=PA231&dq=amine+nucleophilicity+steric&hl=en&sa=X&ei=nXOhVLiOD8iKgwSS5YPICg&ved=0CEUQ6AEwBA#v=onepage&q=amine%20nucleophilicity%20steric&f=false


Introducing the idea of "infinite atomicity" can be useful to highlight the importance of particle size on the properties of metallic or network covalent solids. As the surface to volume ratio of discrete particles becomes sufficiently large, properties of crystalline solids such as melting point, absorption wavelength and reactivity begin to diverge significantly from the limiting values observed for larger crystals. A pronounced change often occurs as the nanoparticle regime is entered, at ~$\pu{ 0.1 \mu m}$, which corresponds to ~$\pu{10^8 atoms/particle}$ or so, well short of infinity, but converging to that limit (presumably heading closer to infinite atomicity will bring you into the regime where gravitational forces become very important, but I am guessing your teacher did not mean this).

But to answer your question, I found the wikipedia useful. On network solids, it states that

  In a network solid there are no individual molecules, and the entire crystal or amorphous solid may be considered a macromolecule.

On atomicity it provides the following comment:

  All metals and some other elements, such as carbon, do not have a simple structure but consist of a very large and indefinite number of atoms bonded together. Their atomicity cannot be determined and is usually considered as 1.


What exactly is the parasitic white layer that forms after iron parts are treated with ammonia?

Steels are nitrided in ammonia gas at 900 to 1050 F ; It forms a very hard , very thin ( < 0.01 ") hard layer of iron nitride . An addition of aluminum alloy to the steel ( 0.5 to 1.0 % ) enhances the nitriding. Nitriding is applied typically to cutting tools like drill bits and wearing surfaces. At higher temperatures "carbo-nitriding " is done where C and N are diffused into the steel surface. The "white layer" is something seen in a metallographic sample at high magnification -100 x and higher. It is undesirable and apparently unidentified as the ASM  handbooks refer to it as only "white layer" ; they give procedures to remove it if necessary . The white layer should be thin , even compared to a 0.005 " thick layer of nitride. Plain ammonia or dissociated ammonia may be used with slightly different procedures.


Why does dimethyl dichlorosilane undergo hydrolysis to form siloxane polymers, but its carbon analog does not?

Actually a $\ce{Si-Cl}$ bond is much stronger than a $\ce{C-Cl}$ bond.  It takes about 90 kcal/mole to break a $\ce{Si-Cl}$ bond, but only around 81 kcal/mole to break a $\ce{C-Cl}$ bond.  The strengths of the corresponding $\ce{Si-O}$ (110 kcal/mole) and $\ce{C-O}$ (85 kcal/mole) bonds also need to be considered.

In the silicon reaction we need 90 kcal/mole to break the $\ce{Si-Cl}$ bond, but we get 110 kcal/mole back when we form the $\ce{Si-O}$; the reaction is exothermic by roughly 20 kcal/mole - a large amount.  In the analogous carbon case, we need 81 kcal/mole to break a $\ce{C-Cl}$ bond, and we only get 85 kcal/mole back when we form the $\ce{C-O}$ bond; the reaction is exothermic by only 4 kcal/mole - not much.  There is quite a difference in the overall driving force for the two reactions with the silicon case being energetically much more favorable than the carbon case!


Does attachment of phenyl to an alkene increases its reactivity towards electrophilic addition reaction?

Not always, and it can depend upon the groups attached to a phenyl ring, like $\ce{-OMe, -COOH, -NH2, -NO2, -X}$, etc.

The attachment of a phenyl ring to a carbon-containing double bond can affect both the stability of carbocation and the possibility of attacking with the help of the compound's $\pi$-electron cloud.

If you have only the phenyl ring with no substitution, first of all the $\pi$-electron cloud of the double bond will be in high conjugation with the phenyl ring, and thus the $\pi$-electron density will be reduced on the $\ce{C=C}$ double bond, which can reduce the possibility of attack of that double bond to a electrophile. Though the carbocation formed can be stabilised through resonance with the phenyl ring (as it is a benzyl carbocation), the step of carbocation formation can have a very high activation energy ($E_\mathrm a$) if you don't have enough stabilisation of the carbocation. But, if you have the carbocation which can stabilise through resonance with multiple benzene rings, the actual potential energy of the carbocation can be reduced to a great extent. In that case, the rate of reaction will increase.

So, actually this situation can be thought as a competition between the stability of carbocation and the decreased electron density. Generally, the stability of carbocation is the dominating effect if it is stabilised to a large extent.

The rate will be decreased even more if an electron withdrawing group is attached to the phenyl ring due to even more delocalisation of $\pi$-electron cloud, and less stabilisation of the carbocation.

If you have a strong electron donating group attached to a phenyl ring and the double bonded carbon is ortho/para w.r.t that group, that can itself exhibit the $+R$ effect, and increase electron density near the double bond through delocalisation of its own electron density through the $\pi$-bonds. In that case, the attacking of $\ce{C=C}$ double bond to a electrophile will be much easier, and the carbocation formed will be even more stable. Then, the rate of reactivity of the double bond in the electrophilic addition can increase for attachment of a substituted phenyl ring.


Reasons for low melting point of gallium

For your interest, I quote a paragraph from a book "Handbook of Nanophysics: Nanoparticles and Quantum Dots" by Klaus D. Sattler (Ed.). Section 12-8. I don't think people know the reason as to why Ga likes to exist as a dimer (as you know $\ce{S}$ is $\ce{S8}$ and so on).

  The cohesive energy of the elemental solids is defined as the
  difference between the energy per atom of the solid and the energy of
  an individual atom in isolation. However gallium is rather unique
  among metals in that in the solid state it has a very unusual
  structure—an orthorhombic 8-atom unit cell composed of dimers which
  have a bond distance 15% shorter than the next nearest neighbor
  distance. As such, it has been called a “molecular metal,” which
  really contains two distinct kinds of bonding: covalent bonds within
  the dimers which result in a pseudogap at the Fermi level, and
  metallic bonding formed by overlap of the electronic wavefunctions
  along the buckled planes perpendicular to the molecular bond of the
  dimer. As a consequence, conduction in gallium is highly anisotropic,
  with a much greater conductivity along the buckled planes than
  perpendicular to this direction (Gong et al., 1991; Heine, 1968).

  If we estimate the “metallic” cohesive energy of the solid, by
  calculating the cohesive energy relative to the dimer, we immediately
  see that gallium moves much closer to the other elements, as shown in
  Figure 12.6. In fact, in this case we are still overestimating the
  cohesive energy relative to the melting temperature, since the dimers
  in the bulk will be distorted from the free dimer geometry. This
  argument certainly explains the low melting point of gallium relative
  to other metals, and due to the anisotropy of the lattice most likely
  relates to the surface superheating of gallium that was observed very
  early on (Volmer and Schmidt, 1937).


Can a system of 2 consecutive biphenyl systems be optically active , even if any adjacent biphenyl systems aren't optically active?

The key point you have to remember is that optical activity isn't decided by a single section of the molecule but rather it is a property of the entire molecule - and while not being able to find a mirror plane is almost a guarantee, the nail in the coffin to decide that a molecule is optically active is if you cannot find an improper rotation axis (reflect and rotate).

There is no mirror plane in the tri-phenyl molecule you show (cut it into left and right the Br and I don't overlap; cut it up and down the methyl and chloride won't overlap; and so on and so forth). Hence, it is optically active.


How can I determine the highest boiling point given a list of molecules?

Q1 and Q2 have the same approach. One notices that they are homologous alkanes and picks the biggest and the smallest, having the most London dispersion/van der Waals force and least, respectively. Other forces are known not to be present between alkanes.$^1$ (There is a small difference between branched and linear alkanes, but that is negligible compared to another $\ce{CH2}$ group.)

For Q3, one sees that they are combinations of hydrogen and halogen. While van der Waals forces are also present here, they are likely to be dominated by dipole-dipole interactions and hydrogen bonding (where applicable).$^2$ Which of the two is actually determining here, does not matter, since both point to a high boiling point in $\ce{HF}$.

In Q4, one notices that the question deals in homo-atomic molecules, so there can be neither hydrogen bonding nor dipole-dipole. Thus, the more electrons, the easier it is to polarize the molecule, and the heaviest molecule wins.

To summarize: To me, the questions are more about test-taking than actual chemistry. I cannot say whether the boiling point of $\ce{C4H10}$ or of $\ce{HF}$ is higher without looking up the values or knowing them and I do not expect most chemists to. Look for which forces may apply and reason about the trends. To do so, one may have to derive the molecular structure, but I would expect that in a test, one either does not need to or has the time to do so.

$^1$ This is where practice and knowledge are indispensable, which is unfortunate for the learner.

$^2$ In this case, this is trivial, but there are cases where exact quantitative calculations would be necessary. One can construct systems in which van der Waals forces outweigh ionic repulsion.


In the nitration, why fluorobenzene has less products in the ortho position than Clorobenzene?

Halogens are deactivating groups even though they show $+M$ mesomeric effect. The inductive effect is stronger than mesomeric effect. As the halogens show $-I$ inductive effect, they are deactivating groups.

When it comes to selectivity, the mesomeric effect plays a role. As the halogens show $+M$ mesomeric effect, the resonance structure has higher electron density at ortho and para positions. Therefore, halogens are ortho-para directing.

When you take inductive effect into account for its directing properties, the para position is more favorable for electrophilic reagents because it has a higher electron density compared to the ortho position. The strength of the inductive effect decreases with distance. As the ortho position is closer to the halogen, the inductive is stronger at this position. Therefore, the major product for electrophilic substitution reactions for haloarenes is the para-substituted product.

Fluorine is more electronegative than chlorine and shows a higher tendency to withdraw electrons towards itself than chlorine. Therefore, fluorine shows a stronger inductive effect than chlorine. As fluorine reduces the electron density at the ortho position more than chlorine, you get a lesser amount of ortho substituted product with fluorine.


Forms of carbon in incomplete combustion

It heavily depends on reaction parameters (oxygen/oxidants availability, temperature, pressure, etc.)

In a free-air incomplete combustion at a (relatively) high temperature, you get mainly carbon-based molecules of various kind.

In a internal combustion engine, that usually works at high temperature and high pressure, you can also get nanoparticles of various kind, including graphite.

In a free-air, mixed, relatively cold combustion, the original small molecules, graphite and other carbon-bases compounds can further react with other molecules giving you larger molecules and/or can stick to larger particles forming nano or micro-particles of various kind. Predicting/modelling this kind of reactions can be very complicated.


Is there a difference between Ca++ and Ca(2+)?

The answer to this question is: No. There is no difference between the two, they are merely two different notations.


Why is the ionic radius of hydride so large?

It is not.
Presenting the order of ionic radii like that is useless.
Ionic radius of a given ion is not constant and mainly rely on crystal structure (coordination environment, more precisely) and used experimental techniques (crystal structure determination and theoretical calculations).
As a consequence, you can only compare homogeneous data.
Proposing the ordering based on some cherry-picked values from random sources or their averaged values is pointless at best.
This very issue is discussed by Messer [1]:

The effective ionic radius of fluoride ion is given as 1.33 Å by Pauling (8) and 1.36 Å, by Zachariasen (9), for coordination number 6. The effective ionic radius of hydride ion varies from 1.27 to 1.52 Å (10), being much more highly sensitive to the particular cation present and to the particular assumptions used in deriving the radius than the fluoride value. For the alkali metal hydrides other than lithium hydride, with the same face-centered cubic structure as the corresponding fluorides, the effective hydride radius is 1.47 Å for $\ce{Na+},$ and 1.52–1.54 Å for $\ce{K+},$ $\ce{Rb+},$ and $\ce{Cs+}.$ For $\ce{LiH},$ $\ce{CaH},$ and $\ce{BaH},$ the hydride ion radius is 1.34–1.36 Å. The closest similarities between hydrides and fluorides should thus appear here.

You can find the order of ionic radii for group 1 hydrides and halides in a more recent work by Lang and Smith [2]:

$$\ce{Li+} < \ce{H-} < \ce{Na+} < \ce{F-} < \ce{K+} < \ce{Cl-} < \ce{Rb+} < \ce{Br-} < \ce{Cs+} < \ce{I-}$$

Again, note that this doesn't only contradict your order, but it is no universal sequence and is only true for these compounds with soft-sphere model used for calculation.
References

Messer, C. E. Hydrides versus Fluorides: Structural Comparisons. Journal of Solid State Chemistry 1970, 2 (2), 144–155. DOI: 10.1016/0022-4596(70)90062-9.
Lang, P. F.; Smith, B. C. Ionic Radii for Group 1 and Group 2 Halide, Hydride, Fluoride, Oxide, Sulfide, Selenide and Telluride Crystals. Dalton Trans. 2010, 39 (33), 7786. DOI: 10.1039/c0dt00401d.


Calculating pH of weak base and strong acid: Why is K(a) not required?

Your doubt is legitimate because you are assuming that $K_{(a)}$ is a parameter that you can plug in every reaction. Actually $K_{(a)}$ is calculated using this equation in a solution of the acid with water:

$$
\ce{K_{(a)} = \frac{[H+][CN-]}{[HCN]}}
$$
That is derived from this ideal equilibrium:
$$\ce{HCN ⇌ CN- + H+}$$

Even though it reflects the strength of that specific acid in a pure water solution it's practical usage is limited when the equilibrium is not perturbated by others factors.
Anything that will interfere with this equilibrium can cause problems to your assumptions.

In your specific case, the conditions are different there is an excess of $\ce{OH-}$ in the solution due to the strong base, so every $\ce{H+}$ produced will react with the $\ce{OH-}$. Due to the Le Chatelier principle the equilibrium will respond shifting to the right: dissociating more and more $\ce{HCN}$ from producing more $\ce{H+}$.

In simple words, if one reaction is favourable the solution doesn't care about your $K_{(a)}$ it will go following the most thermodynamic favourable path to the equilibrium.

Possible practical learning aim of this question?

So if you want to have high products amount (i.e. $\ce{CN-}$) and you have cheap reactant (i.e. $\ce{KOH}$) you can use an excess of reactant to shift the equilibrium where you want.

Bonus question: what can you do if you want to have HCN?


Basic oxide are those oxides that dissolve in water to form soluble hydroxide. Basic oxides are very soluble for this reason. Acidic oxides are that oxides that dissolve in water to form strong acids. Generally, covalent oxides form acidic oxides as the element to which oxygen is bonded is electronegative. As charge increases, electronegativity increase and thus oxide become acidic. This also applies for elements having multiple oxides. As oxidation no. increases, charge increases and thus oxides become acidic. This link shares a good information on this topic:

  Since the acidity of a cation rises rapidly with charge, d-block
  elements which exhibit a wide variety of oxidation numbers may have
  one or more oxides that exhibit only basic properties and one or more
  oxides that exhibit only acidic properties. The higher the oxidation
  number the more acidic the corresponding oxide. Chromium is an example
  of such an element.

  $$
\begin{array}{|c|c|c|}\hline
\text{Oxide}&\text{Oxidation number}&\text{Category}\\\hline
\ce{CrO}&\ce{Cr^2+}&\text{basic}\\\hline
\ce{Cr2O3}&\ce{Cr^3+}&\text{amphoteric}\\\hline
\ce{CrO3}&\ce{Cr^6+}&\text{acidic}\\\hline
\end{array}
$$

The amphoteric character of chromium(III) is stated in its Wikipedia article:

  Chromium(III) oxide is amphoteric. Although insoluble in water, it
  dissolves in acid to produce hydrated chromium
  ions, $\ce{[Cr(H2O)6]^3+}$ which react with base to give salts
  of $\ce{[Cr(OH)6]^{3−}}$. It dissolves in concentrated alkali to yield
  chromite ions $(\ce{[CrO2]^{-}})$.

So, the rule of thumb is as oxidation number increases, charge on metal increases, acidic character increases. Also, basic oxide is soluble in water, acidic oxide is insoluble in water. This can be used to differentiate between acidic and basic oxide.


If all chemical processes are equilibria at least to some extent, where does the activation energy come from?

I don't think the statement that all processes are in equilibrium to some extent is true. By definition, dynamic equilibrium means that the rate of change in one direction is equal to the rate of change in the opposite direction.

This is certainly not true for all processes, otherwise there would not be a process - nothing would appear to change.

On the other hand, for any given reaction equation that is known to occur, we can say:

for a reaction at equilibrium, neither the concentration of reactants nor the concentration of products will be zero

I think that may be what you meant in your question?

If that is the case, then let's look at your example:

  Is a process with a very high activation energy (i.e. the ionization of sodium vapor into sodium ion) still considered an equilibrium process at least to some extent?

It's not an equilibrium process. However, the reaction:

$\ce{Na(g)->Na(g)+ + e-}$

could be at equilibrium. In other words, the ratio of reactants to products could be such that the Gibbs free energy is at a minimum for this temperature and pressure. A more mathematical way to say that is:

$Q = K_{eq}$

where $Q=\frac{[Na^+]}{[Na]}$ and $K_{eq}$ is the equilibrium constant under these conditions. (Brackets mean "partial pressure" - not concentration here.)

Under normal lab conditions (1 atm, 298 K), this is not thermodynamically favorable. The reaction quotient heavily favors the reactants, or another way to say it is $K_{eq}$ is really close to zero. So close you may not be able to measure it. The activation energy is extremely high (equal to the ionization energy), and so any electrons that are removed would have to get that energy from somewhere - which was your next question:

  If so, where might the energy to furnish the forward reaction come from naturally - i.e. with no human input of energy?

In an idealized, isolated system, you only have one source of energy for things like that: kinetic. Thermal motion of the atoms could conceivably put enough energy into a single atom that an electron could be "kicked off" and result in an ion. However - the odds of that happening are extremely low, given that the distribution of speeds would have a vanishingly small, although still non-zero, probability of occurring at the magnitude needed to overcome the activation energy barrier. Since the reaction is thermodynamically unfavorable, the probability that the electron would jump right back on is very high - much higher than it would be for leaving in the first place.

Now when you have $6.022 \cdot 10^{23}$ opportunities each femtosecond, very low odds can still work out. But the total amount that would be ionized at any one time is still very, very low. Possibly so low that you would have to wait for an essentially infinite amount of time to see a measurable amount of it.

In a real system, a much larger fraction would be ionized (and this may or may not be reflected in $K_{eq}$, depending on how it is measured). The reason is that there is always ionizing radiation present on earth, and there will be more or less depending on where you are located. Statistical thermodynamics doesn't take this into account, so any prediction of the relative amounts of reactants and products would not be correct. This is kind of a unique example in that respect - most equilibrium reactions, even with very slow kinetics, deal with systems where the thermal energy of the system has a much bigger effect than radiation.


Reasons for weakness of London dispersion force in graphite?

We need to look at the orbitals, where are the electrons localized?
because the hybridization of the carbons in grpahite are sp2 we have one free p orbital with one electron, which is denoted as the pz orbital.
The free pz electron then forms the resonance bond with the adjacent carbons. Therefore there are no non bonding valence electrons to contribute much for the London dispersion interactions.


How to increase the velocity of a reaction?

It should be $-E_a/RT$

Therefor, increasing the temperature, lowering the activation energy (catalysis), or increasing the concentration of reactant(s) would increase the reaction velocity.


Whatever one observes (whether it is a liquid, a gas, or a mixture thereof) is the state of maximum entropy for the particular values of the internal energy and volume the system currently has. As OP noted, this is a consequence of the second law of thermodynamics. According to this law, a system with a fixed internal energy self-adjusts its free parameters in such a way that the entropy is maximized. The free parameter in OP's example is the fraction of molecules in the liquid (or gaseous) phase.

When we say that the entropy of the gaseous phase is larger than that of the liquid phase, we are referring to the specific entropies (i.e., entropy per mole) of these phases when they coexist. When the two phases coexist, they have a common temperature $T$ and a common pressure $P$. Furthermore, their specific Gibbs free energies are the same, that is,
\begin{equation}
0 = \Delta g (T,P) = g_{\mathrm{gas}} (T,P) - g_{\mathrm{liquid}} (T,P) = \Delta u(T,P) + P \Delta v(T,P) - T \Delta s(T,P).
\end{equation}
The gaseous phase has a higher specific internal energy ($\Delta u>0$) and a larger specific volume ($\Delta v>0$) as well as a higher specific entropy ($\Delta s >0$) in such a way that the effects of the three terms ($\Delta u$, $P\Delta v$, and $T\Delta s$) cancel out.

One may still wonder why liquid-gas coexistence is ever possible. As the gaseous phase has a higher specific entropy, why doesn't the system self-adjust to have all its molecules in the gaseous phase? The Answer: What we know for sure is that if the evaporation took place under a constant temperature and pressure, the entropy would increase (because $\Delta s >0$). But this process would also increase the internal energy and the volume ($\Delta u >0$ and $\Delta v >0$). Therefore, to have more molecules in the gaseous phase but stay at the same internal energy and volume, the system has to lower its temperature and raise its pressure. This results in a decrease in the entropy,1 which counteracts its increase that would have happened if the evaporation had taken place at constant temperature and pressure. Therefore, evaporation at fixed internal energy and volume doesn't necessarily lead to the increase of the total entropy. As stated in the first paragraph, if one observes liquid-gas coexistence, this should correspond to the state of maximum entropy for the given values of internal energy and volume.

1 For any (stable) thermodynamics system, the entropy decreases as the temperature is lowered. On the other hand, no such strict rule exists for the relation between the entropy and the pressure. Still, for most systems, entropy decreases as pressure is raised. [Notice the Maxwell relation $(\partial S/\partial P)_{T} = -(\partial V/\partial T)_{P}$, meaning that the entropy decreasing with increasing pressure is equivalent to the volume increasing with increasing temperature.]


What is the difference between activity and reactivity when discussing elements?

Activity is an ordered series, showing which element will replace another element. For example, in a solution of copper sulfate in water, a piece of iron quickly becomes coated with copper as the iron displaces copper from the solution because iron is more active. No explosion took place.

$\ce{Fe(solid iron) + CuSO4 (dissolved in water) -> Cu(solid copper) + FeSO4 (dissolved)}$

Reactivity is less clearly defined. In general, it means how easily some chemical change takes place. For example, chlorine, oxygen and sodium are all very reactive, easily attaching to other elements or replacing elements in compounds. Argon and helium are not reactive, and rarely, if ever, combine with other substances. Nitrogen triiodide, $\ce{NI3}$ and acetylene, $\ce{C2H2}$, are very reactive, but in a way opposite that of sodium or oxygen: rather than attaching to other substances, forming molecules, these compounds want to break up their molecules. In fact, even touching $\ce{NI3}$ will make it decompose violently... Now there's your explosion.


I don’t know how to read or make sense of this: $\ce{CuAl6(PO4)4(OH)8.4H2O}$ the chemical formula of turquoise.

It means for every atom of copper (Cu), there are 6 atoms of aluminum, 4 phosphate groups (which each contain a phosphate atom and four oxygen atoms), 4 hydroxide groups (which each contain an oxygen and a hydrogen atom) and 4 waters (which each contain two hydrogen atoms and one oxygen atom). So the parentheses are for grouping, and the numbers are to count how many atoms or groups there are.

The structure is a crystal, so it goes on and on in all directions. That is why I was careful to say "for every atom of copper...". For other substance like carbon dioxide that are found as molecules, it is less complicated (there, the formula actually tells you the number and kind of atoms in a single molecule).

  Chemical structure of Turquoise

Some of the components in turquoise are positively charged (copper and aluminum) and some are negatively charged (hydroxide and phosphate). The building blocks are arranged that positive and negative come close together, and like charges are further apart. The entire structure is very complicated.

The picture below does not show all the atoms, but instead shows copper, aluminum and phosphate as 3D shapes (the metals and the phosphorus would be in the center of each of the shapes). The shapes have common corners occupied by oxygen atoms (in the formula, these appear as part of phosphate, hydroxide and water). The hydrogen atoms are shown at tiny spheres.

Source: http://rruff.info/uploads/MM64_905.pdf


Filtration methods at removing ions present in water

Cations and anions of water-soluble salts cannot be removed from water using only mechanical filters (which rely on particle size) such as coffee filters.

In order to remove such ions, ion exchange resins can be used. These resins consist of an insoluble matrix, e.g. polystyrene that is crosslinked with divinylbenzene to vary the porosity of the resin. To convert the resin into an ion exchanger, the polymer structure is functionalized with many functional groups. Typical functional groups are

sulfonate ($\ce{R-SO3-}$) for strong acid cation exchange resins,
quarternary ammonium (e.g. $\ce{R-N(CH3)3+}$) for strong base anion exchange resins
tertiary amine (e.g. $\ce{R-N(CH3)2}$) for weak base anion exchange resins

The counterion of new cation exchange resins used for water treatment is usually $\ce{H+}$ or $\ce{Na+}$ (other forms can be used for special applications, e.g. $\ce{Li+}$ in pressurized water reactors), i.e. $\ce{R-SO3- H+}$ or $\ce{R-SO3- Na+}$. These counterions can be replaced by another cation from the water. Thus, a cation exchange resin can remove various cations from the water and replace them with an equivalent amount of $\ce{H+}$ and $\ce{Na+}$, respectively, e.g.
$$\ce{2 (R-SO3- H+) + Ca^2+ <=> (2 R-SO3- Ca^2+) + 2 H+}$$

The counterion of new anion exchange resins is usually $\ce{OH-}$ or $\ce{Cl-}$. Thus, an anion exchange resin can remove various anions from the water and replace them with an equivalent amount of $\ce{OH-}$ and $\ce{Cl-}$, respectively.

In order to remove both kinds of ions, the water has to be passed through a cation exchanger and through an anion exchanger. Alternatively, a mixed-bed ion exchanger (which contains cation exchange resins and anion exchange resins) can be used. They are even more effective but cannot be easily regenerated.

Typical Brita water filter cartridges contain ion exchange resin beads to remove undesired ions and activated carbon to absorb other undesired compounds.


Gibbs Free energy of a reversible reaction

In general, any expression for $G$ will by necessity have a dependence on $T$, if only through the explicit incorporation of temperature in $G = H - TS$.

Looking at the Calphad (Calculation of Phase Diagram) community, they go further. Referring to A.T. Dinsdale, "SGTE Data for Pure Elements", CALPHAD 15(4) 317-425 (1991), they state, for the purposes of defining $G$ for the elements:

  The Gibbs energy is represented as a power series in terms of temperature $T$ in the form:
  $ G = a + bT + cT \ln(T) + \Sigma d_{n}T^{n}$

From the definition of $G$, $S$, and $H$ one can then get:

$S = -b-c-c\ln(T)-\Sigma nd_{n}T^{n-1}$ and

$H = a-cT-\Sigma (n-1)d_{n}T^{n}$

So, one sees, given the chosen representation for $G$, that both $H$ and $S$ should depend on $T$ unless $G$ can be represented in a very simple form (only $a$ and $b$ being non-zero). Since Dinsdale uses a more complex power series, one can rest assured that $G$, even for the elements, tends to be rich in temperature dependence.


Controversy over products of iodate and bisulfite?

Not much controversy here, IMO. There is more than one reaction involved, with slightly more complex kinetics.

As so often, Wikipedia already knows, so I will just quote the relevant part of the following article: http://en.wikipedia.org/wiki/Iodine_clock_reaction

  Iodate variation

  An alternative protocol uses a solution of iodate ion (for instance potassium iodate) to which an acidified solution (again with sulfuric acid) of sodium bisulfite is added.

  In this protocol, iodide ion is generated by the following slow reaction between the iodate and bisulfite:

  $$\ce{   IO3^{−} + 3 HSO3^{−} -> I^{−} + 3 HSO4^{−}}$$

  This is the rate determining step. The iodate in excess will oxidize the iodide generated above to form iodine:

  $$\ce{   IO3^{−} + 5 I^{−} + 6 H+ -> 3 I2 + 3 H2O}$$

  However, the iodine is reduced immediately back to iodide by the bisulfite:

  $$\ce{   I2 + HSO3^{−} + H2O → 2 I^{−} + HSO4^{−} + 2 H+}$$

  When the bisulfite is fully consumed, the iodine will survive (i.e., no reduction by the bisulfite) to form the dark blue complex with starch.


From the ideal gas equation, we get
$$PV=nRT \implies P=\frac{n}{V}RT=cRT$$
where $c$ represents the concentration.

Consider an arbitrary reaction
$$n_AA\rightarrow n_BB$$

$$K_c=\frac{c_B^{n_B}}{c_A^{n_A}}$$
$$K_p=\frac{P_B^{n_B}}{P_A^{n_A}}=\frac{(c_BRT)^{n_B}}{(c_ART)^{n_A}}=K_c \times (RT)^{\Delta n}$$

Most probably, the reaction you are talking about has $\Delta n=0$.


Equations for the oxidation reactions of tea teaflavins and EGCG with potassium permanganate

The two organic compounds are ultimately being converted to carbon dioxide, $\ce{CO2}$, and water, $\ce{H2O}$, just as they would be in complete combustion with pure oxygen gas, $\ce{O2}$, in a bomb calorimeter. Thus the two balanced combustion equations would be as follows:

$$\ce{C29H24O12 + 29 O2 -> 29 CO2 + 12 H2O} \tag{1}$$

$$\ce{C22H18O11 + 21 O2 -> 22 CO2 + 9 H2O} \tag{2}$$

So now assume that potassium permanganate, $\ce{KMnO4}$, is being used as the oxidizing agent, with manganese dioxide, $\ce{MnO2}$, as the product. Then, as per Ron's answer (https://chemistry.stackexchange.com/a/85865/79678), and assumed reasonable conditions therein, we have the following reaction:

$$\ce{4 MnO4- + 4 H+ -> 4 MnO2 + 3 O2 + 2 H2O} \tag{3}$$

Showing the spectator potassium ions, and arbitrarily assuming hydrochloric acid as the source of the hydrogen ions, yields

$$\ce{4 KMnO4 + 4 HCl -> 4 MnO2 + 3 O2 + 2 H2O + 4 KCl} \tag{4}$$

Since 29 $\ce{O2}$ and 3 $\ce{O2}$ have no common factor other than unity, multiply equation 1 by 3, equation 4 by 29, add them, cancel the equal number of $\ce{O2}$ molecules on each side and the result is

$$\ce{3 C29H24O12 + 116 KMnO4 + 116 HCl -> 87 CO2 + 116 MnO2 + 94 H2O + 116 KCl} \tag{5}$$

Since 21 $\ce{O2}$ is 7 times 3 $\ce{O2}$, use equation 2 as is, multiply equation 4 by 7, add them, cancel the equal number of $\ce{O2}$ molecules on each side and the result is

$$\ce{C22H18O11 + 28 KMnO4 + 28 HCl -> 22 CO2 + 28 MnO2 + 23 H2O + 28 KCl} \tag{6}$$


Why are tertiary carbocations so stable?

What is immediately obvious is that going from a primary to a secondary cation only produces a slight increase in rate, but the change from secondary to tertiary produces an enormous change.

Uh-Oh!

When we see a table reporting that the solvolysis rate for an ethyl and methyl compound are both equal (both 1 in your table), well that's a tip-off that something is grossly wrong.  No one would expect the methyl and ethyl carbocations to have anywhere near similar stabilities. Therefore we wouldn't expect the appropriate ethyl and methyl compounds to solvolyze (if they could) at anywhere near similar rates.

So if we accept that your table has bad data in it, then we really don't know that the change in going from primary to secondary is any different from the change on going from secondary to tertiary.

Carbocation Stability - Background

There are many factors that can affect carbocation stability (inductive effects, resonance effects, neighboring groups), but if we limit our discussion to the stability of saturated hydrocarbon carbocations then we need only consider two of these - inductive and resonance effects.

The carbocation center is roughly $sp^2$ hybridized with an empty p orbital; the alkyl substituents attached to this carbon are roughly $sp^3$ hybridized. There is a flow of electrons through the sigma bonds from the $sp^3$ carbon(s) to the more electronegative $sp^2$ carbon.  We call this an inductive effect and it will stabilize the carbocation center.

Carbocation stabilization can also result from the flow of electrons through p orbitals.  We call this a resonance effect, and in the particular case at hand it is a hyperconjugative resonance effect, as pictured in the diagram below.

(For more on hyperconjugation, particularly from an MO perspective, see the "Rotational barrier of ethane" section in this Wikipedia article.)

Prediction

The t-butyl carbocation has 3 methyl groups that can stabilize the carbocation center through inductive effects, there are also 9 beta hydrogens that can stabilize the carbocation center through hyperconjugation.

The 2-propyl carbocation has 2 methyl groups that can stabilize the carbocation center through inductive effects, there are also 6 beta hydrogens that can stabilize the carbocation center through hyperconjugation.

The ethyl carbocation has 1 methyl group that can stabilize the carbocation center through inductive effects, there are also 3 beta hydrogens that can stabilize the carbocation center through hyperconjugation.

The methyl carbocation has no inductive or hyperconjugative stabilization.

Consequently we might expect the relative carbocation stability to be

t-butyl > 2-propyl > ethyl > methyl

with the 2-propyl carbocation roughly midway between the t-butyl and ethyl carbocations.

Data

Chemistry is complex enough without adding extra unnecessary variables - like solvent. A relatively straight-forward method to compare carbocation stabilities is to measure the free energy ($\Delta G^o$) of the following equilibrium in the gas phase. The more negative the $\Delta G^o$ observed, the less stable the carbocation.

$$\ce{R^+ + H^- <=> R-H}$$

The following table shows the the results of this investigation and appears to  confirm the above predictions on carbocation stability.

\begin{array}{|c|c|} \hline
\text{carbocation} & \text{Delta} {~G^o} \\ &  \text{(kcal/m)} \\ \hline
\text{methyl} & -314 \\ \hline
\text{ethyl} & -270 \\ \hline
\text{2-propyl} & -252  \\ \hline
\text{t-butyl} & -237 \\ \hline
\end{array}

(data taken from Advanced Organic Chemistry, Part A: Structure and Mechanisms; fifth edition; Authors: Carey, Francis A., Sundberg, Richard J.; table 3.10, page 303)

The differences between the [t-butyl and 2-propyl] and the [2-propyl and ethyl] carbocations is 15 and 18 kcal/m respectively. That the 2-propyl cation is roughly in the middle of the t-butyl and ethyl carbocations is likely fortuitous since the ethyl carbocation is structurally quite different from the other two alkylated carbocations. The ethyl cation is non-classical in nature, involving a 3-center 2-electron bond with a bridging hydrogen as explained in this earlier answer.


Copper sulphate, Benedict's Reagent and albumin

Yes, it's most likely the same reaction as in the Biuret test that detects peptide bonds.


The first thing we need to do is address your chemical equations.  $\ce{MgCl2}$ dissociates in water as follows:

$$\ce{MgCl2 -> Mg^{2+} + 2Cl-}$$

Now you can see that one mole of $\ce{MgCl2}$ dissociates to give two moles of $\ce{Cl-}$. So all we have to do then is calculate the number of moles of $\ce{MgCl2}$ from the given mass and its molecular weight, then multiply that by 2 to get the number of moles of $\ce{Cl-}$:

$$\mathrm{\frac{1.9\ g\ \ce{MgCl2}}{ 95.2\ \frac{g\ \ce{MgCl2}}{mol\ \ce{MgCl2}} } = 0.0200\ mol\ \ce{MgCl2}}$$

$$\mathrm{0.0200\ mol\ \ce{MgCl2}\ *\frac{2\ mol\ \ce{Cl-}}{mol\ \ce{MgCl2}} =0.0400\ mol\ \ce{Cl-}}$$

Then since the volume is 1L, the $\ce{Cl-}$ concentration is just $\mathrm{0.040 \frac{mol}{L}}$.


How are chemistry and materials science related?

An undergraduate chemistry course here in the UK typically has a materials component to it in the form of a module or two, so you bet they're related!

It's sometimes not that fruitful to view materials science and chemistry as a different subject—much like physic vs chemistry—however we can associate certain motivations to either topic as we see fit. Please note that this is quite subjective; this view does indeed vary from person to person.

First of all, materials is more interested in the solid state than chemistry in general. We might like to look at crystal structures via x-ray crystallography (see Bravais lattices, space groups etc.) as well as the rather intriguing quasi-crystals (for which Dan Shectman won the Nobel in Chemistry in 2011). The skills gained in a materials crystallography course would no doubt help with chemistry. The converse is also true since many substances in chemistry are amenable to study via crystallography, such as proteins. You therefore may also have a crystallography module in your chemistry course—looking at molecules in the solid state not typically viewed as a 'material'. If it has appreciable electron density, you can study it by x-ray crystallography!

Another aspect of materials science looks at properties of materials, such as stress/strain curves etc.I've had such a lecture series specifically dealing with polymers in a chemistry course; one of the challenges of modern day chemistry is to invent and modify polymer architectures for specific industrial applications. It's clear then that chemistry and materials science complement each other here. Whilst materials looks at interesting properties of given materials, chemistry constructs those polymers.

Materials science can differ from chemistry in that it focuses (or at least did in my case) on the band-structure of solids more than chemistry does. So it's possible that a materials course would give you a more nuanced appreciation of semi-conductors and the like. Again, this would help overwhelmingly if you decided to go into the field of conducting polymers—a currently very active field in R&D, especially if it involves replacing rare earths.

To summarise: materials science and chemistry go hand in hand in my quick overview, however some aspects of chemistry (such as organic) view the solid-state as the 'graveyard of chemistry'—you'd be hard pressed to find reactions which occur in the solid state! If you want to find out a little more before delving into materials, then finding a quick Youtube course might help to sway you either way. In my opinion there are far more ridiculous pairings in the world than chemistry and materials!


The triphenylphosphine molecule would show resonance because there is a lone pair on phosphorus and it is in conjugation with the double bonds in benzene. But the lone pair would be in resonance only with one ring at a time because the molecule is non-planar (tetrahedral) having sp3 configuration. Also, we should consider the steric effects on the molecule. The steric hindrance produced by other two benzene rings would cause less resonance effect and the lone pair would be inhibited from going into the other two rings.
So, the resonance in this molecule would be simply like that in aniline.


What is the chemical test for identification of benzophenone?

After identification of ketone, to identify benzophenone you need to find the range of melting point using melting point apparatus which you should get in range of $48.3\ \mathrm{^\circ C}$ to $50.6\ \mathrm{^\circ C}$.

Then you can find IR of unknown(benzophenone).

Source: Prezi


The simple rules don't work with complicated examples. There are non-chiral molecules with "stereocenters", and chiral molecules lacking "stereocenters". The textbook example for your question is 1,4 dichloro cyclohexane (or 1-chloro 4-bromo cyclohexane to get even closer to your example):

William Reusch explains this and other cyclohexane cases here: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Stereoisomerism_in_Disubstituted_Cyclohexanes

  The way I was taught to find stereoisomers was to count the number of double bonds that can be E/Z and the stereocentres, and do $2^n$.

You established that this rule does not work here. If you added an oxo group to one of the ring carbons (turning it into a ketone and making the two branches of the cyclohexane different from one another), you would get the expected pattern: two stereocenters, four stereoisomers.

I found a more exact formulation of the rule on another one of William Reusch's pages:

  [William Reusch] As a general rule, a structure having $n$ chiral centers will have $2^n$ possible combinations of these centers. Depending on the overall symmetry of the molecular structure, some of these combinations may be identical, but in the absence of such identity, we would expect to find $2^n$ stereoisomers.

Source: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Compounds_with_Several_Stereogenic_Centers

So what about 1-chloro 4-bromo cyclohexane or terpin? If you use the following definition of stereocenter, both have two stereocenters:

  [Gamini Gunawardena] If the interchange of two ligands on an atom in a molecule results in a stereoisomer of the molecule, the atom is called a stereocenter or stereogenic center.

Source: https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Stereocenter

You have 4 combinations, but there are pairs of identical molecules when you consider the two chair conformations inter-converting.

Other examples where the simple $2^n$ rule does not work are the following:

Meso compounds: Stereochemistry of meso compounds
Axial chirality: M/P nomenclature for axial-chirality
Planar chirality: Why is trans-cyclooctene chiral?
Molecular knots: Molecular knots absolute configuration


Why is isobutane more thermodynamically stable than n-butane?

I would like to understand the term stability in organic chemistry.

We can speak of kinetic stability or thermodynamic stability when we discuss chemical compounds.  If a compound is kinetically stable, that indicates that it has a high barrier (large activation energy) to cross and is therefore less reactive than a compound that has a low barrier (small activation energy) to reaction.  In the following chart we can say that the reactants are kinetically less stable and the products are kinetically more stable since the barrier to the forward reaction is smaller than the barrier to the back reaction.

(image source)

If one compound has a lower heat of formation than another compound, we can say that the former compound is thermodynamically more stable than the latter compound.  Again using the above chart, we can say that the product is thermodynamically more stable than the reactant.

  We know empirically that i-butane is more stable than n-butane. But
  why is it more stable?

Steric interactions do not provide the answer (reference).

We use arguments based on carbon-hydrogen hyperconjugation to explain many things.

For example, why more highly substituted double bonds are more stable than less substituted double bonds.  Carbon-carbon hyperconjugation can also be used to explain various chemical phenomena.  In the case at hand, let's compare carbon-carbon hyperconjugated resonance structures for n-butane and 2-methylpropane.

The hyperconjugated resonance structure for butane involves a primary carbocation, while the hyperconjugated resonance structure for 2-methylpropane involves a more stable secondary carbocation.  This more stable carbon-carbon hyperconjugated resonance structure explains the increased thermodynamic stability of 2-methylpropane compared to the isomeric n-butane.


What is the number of unpaired electrons in Fe 2+?

First draw the electronic configuration for Iron.

Now since we are given with Iron(II) we take out two electrons from the outermost orbit so as to give the iron a +2 charge.

As evident from the configuration, in total there are 4 unpaired electrons.

Using the same logic you can count the number of unpaired electrons for any ion.


The second step is rate-determining. According to Wikipedia:

  Given a reaction coordinate (energy diagram), the rate determining step can be determined by taking the largest energy difference between any starting material or intermediate on the diagram and any transition state that comes after it. That transition state will then be the rate-determining step of a given reaction. The transition state with highest absolute energy may not necessarily correspond to the rate determining step.


Aluminium and zinc will produce a white precipitate due to formation of hydroxide. Both these hydroxides dissolve in excess acid or alkali due to their amphoteric character.

Manganese, nickel, cobalt hydroxides are probably soluble as they are precipitated in the 4th group as sulphides.


Let the number of atoms be n.

If the number of skeletal p electrons is 2n + 2, the structure is closo (a deltahedron with n vertices).

If the number of skeletal p electrons is 2n + 4, the structure is nido (an n+1 vertex deltahedron, with one vertex missing).

If the number of skeletal p electrons is 2n + 6, the structure is arachno (an n+2 vertex deltahedron with two vertices missing).

If the number of skeletal p electrons is 2n + 8, the structure is hypo (an n+3 vertex deltahedron with three vertices missing).

see Excursions Beyond the Zintl Border and Polyanionic Clusters and Networks of the Early p-Element Metals in the Solid State: Beyond the Zintl Boundary

For example:

Neutral Sn, Pb and Ge each contribute 2 valence p electrons, while Bi and Sb each contribute 3 valence p electrons.

$\ce{Sn_5^{2-}}$, $\ce{Pb_5^{2-}}$, $\ce{Ge_5^{2-}}$ and $\ce{Ge_9^{2-}}$ have 2n + 2 skeletal p electrons and are closo (the first three are trigonal bipyrimidal and the last is appoximately a triangular prism with square pyramid caps for the three sides).

$\ce{Sn_9^{4-}}$ and $\ce{Ge_9^{4-}}$ have 2n + 4 skeletal p electrons and are nido. (gyroelongated square bipyramid with one square pyramid apex missing)

$\ce{Bi_4^{2-}}$ and $\ce{Sb_4^{2-}}$ have 2n + 6 skeletal p electrons and are arachno (these two are square planar, an octahedron with two vertices missing).

Also, quoting Farley and Castleman's Observation of Gas-Phase Anionic Bismuth Zintl Ions J. Am. Chem. Soc. 1989, 111, 2734-2735

  Since bismuth and antimony are typically trivalent, anionic clusters
  of these elements very quickly exceed the maximum 2n + 8 electrons
  accountable under Wades rules for “hypo” compounds. In particular, the
  previously mentioned $\ce{Bi7^{3-}}$ possesses 2n + 8 valence electrons, and
  $\ce{Bi14^{6-}}$ has 2n + 20! Hence, anionic Zintl ions of these species are
  particularly intriguing due to the inability of current theories to
  account for them and lack of other corresponding metallic compounds
  with which they may be compared.

Now going back to the example $\ce{Bi_3^{3-}}$ in the orginal question, the consensus in the peer-reviewed literature is that the sturcture is bent (not a triangular ring and not linear), although the structure of the isolated ion has not been experimentally determined. The reasons for considering it to be bent are:

The structure is isoelectronic with species known to be bent in the ground state including $\ce{O3}$, $\ce{S3}$, $\ce{S2O}$, $\ce{SO2}$ and allyl anion.
Experimental determination published in Stabilization of Ozone-like [Bi3]3- and in a more recent article that it is bent in transition metal complexes. (Similar experiments for isoelectronic [P3]3- and[As3]3-also find bent structures).
Theoretical work in Zintl Anions Analogous to O3.

Though ozone and like species are found to be bent, it is also theorized that cyclic (equilateral triangular ring) ozone represents another higher local potential energy minimum, and this would be the same for $\ce{Bi_3^{3-}}$ too.

$\ce{Bi_3^{3-}}$ has 12 p electrons, which is 2n + 6, which implies arachno with respect to a trigonal pyramid.  A equilateral triangular ring structure would be consisent with removing two axial vertices or removing one axial vertex and one equatorial vertex, while the bent structure would be consistent with removing two equatorial vertices.


How to calculate the equivalent weight of KMnO4?

Refer to the definition of equivalent (IUPAC Gold Book):

equivalent entity
Entity corresponding to the transfer of a $\ce{H+}$ ion in a neutralization reaction, of an electron in a redox reaction, or to a magnitude of charge number equal to 1 in ions.

In other words, $\pu{1 equiv}$ is the amount of substance reacting with $\pu{1 mol}$ of hydrogen atom.
If hydrogen acts as a reducing or oxidizing agent, then either way $\pu{1 mol}$ of hydrogen atoms liberates or accepts $\pu{1 mol}$ of electrons:
$$
\begin{align}
\ce{0.5 H2 &→ H+ + e-}\\
\ce{0.5 H2 + e- &→ H-}
\end{align}
$$
That's why an equivalent of a redox agent is its amount which liberates or accepts $\pu{1 mol}$ of electrons upon being oxidized or reduced, respectively, assuming electrons don't exist in solution on its own for a significant period of time.
Note that $\ce{KMnO4}$ participates in redox reaction
$$\ce{2K\overset{+7}{Mn}O4 + 16 H\overset{-1}{Cl} = 2 KCl + 2 \overset{+2}{Mn}Cl2 + 8 H2O + 5 \overset{0}{Cl}_2}$$
and the half-reaction for the reduction of manganese is
$$\ce{\overset{+7}{Mn}O4- + 8 H+ + 5 e- → \overset{+2}{Mn}^2+ + 4H2O}$$
Also note that $n$-factor of permangante here is not the number of protons $\ce{H+}$, which are also used up in water formation; rather it's the number of transferred electrons, e.g. $n = 5$ and
$$M_\mathrm{equiv}(\ce{KMnO4}) = \frac{M(\ce{KMnO4})}{n} = \frac{\pu{158.03 g mol-1}}{5} = \pu{31.61 g mol-1}$$
Obviously, the equivalent mass of permanganate isn't a constant and depends on the $\mathrm{pH}$ of the reaction.
For example, in neutral medium half-reaction appears as
$$\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- → \overset{+4}{Mn}O2 + 4 OH-}$$
and there is no explicitly shown protons to count at all!
However, the $n$-factor is $3$, and the equivalent mass of permanganate would be a different value:
$$M_\mathrm{equiv}(\ce{KMnO4}) = \frac{M(\ce{KMnO4})}{n} = \frac{\pu{158.03 g mol-1}}{3} = \pu{52.68 g mol-1}$$


van der Waals equation for deviation of gases from ideal behavior

The term $-an^2/V^2$ is the reduction of pressure due to attractive or cohesive forces between molecules. This is most easily seen by writing the equation as
$\displaystyle p_\text{obs}= \frac{nRT}{V-b}-\frac{an^2}{V^2}$ .

While the constants $a$ and $b$ are considered to be empirical they do have a physical basis. As you mention $b$ accounts for the finite volume of the molecules and $a$ for the attractive force between molecules.

In the centre of the gas these attractive forces are the same in every direction and so cancel out, however, as close to the boundary as it is possible to go the forces act predominately inwards towards the bulk of the gas, rather as happens in capilliary action or surface tension. If the force between adjacent molecules is resolved into components one will lie along the boundary and the other at right angles pointing inwards.  Thus we can imagine over a long enough period of time a steady force close to the boundary and acting inwards. Thus molecules on reaching the surface are deflected both by the surface and by this inwards force.

The size of the force will be proportional to the product of the number of molecules /unit area and the size of the inward force. Both of these will be be proportional to the density ($Nm/V$) or equivalently molar concentration. The reduction in pressure can thus be written as $-a(n/V)^2$ where $a$ is a positive constant that depends on the gas and is determined only by experiment.


Vanadyl sulfate or $\ce{VOSO4}$ has $\ce{VO^2+}$ ion. It exists as a blue hydrated crystalline salt, known as acid vanadyl sulfate, $\ce{VOSO4 \cdot H2SO4 \cdot xH2O}$. If you're to heat it at $\pu{533 K}$ with concentrated sulfuric acid, you obtain anhydrous vanadyl sulfate, which is rather a greyish-green crystalline powder.¹

The ion $\ce{VO^2+}$ is more accurately written as a complex $\ce{[VO2(H2O)4]+}$.² Inductive effects of ligands and different symmetries of orbitals causes the d-orbitals to split up and thus they become non-degenerate.

The electrons possibly absorb and emit electromagnetic radiation in the visible spectrum. A rough estimate leads me to conclude that this complex would've been absorbing radiations towards the higher-wavelength (possibly red) region of the visible-spectrum thus emitting radiations corresponding to a blueish tone of light.

The colour is possibly due to d-d transition as the central metal ion ($\ce{V+4}$) has one electron in its d-orbital.

References:

B.Sc. Chemistry–II (UGC), R.L. Madan
Chemguide – Vanadium


Which is more stable — a carbanion on a vinylic position or a carbanion on a benzylic position?

One of the more accurate methods to measure carbanion stabilities is negative ion photoelectron spectroscopy.  In this process, a carbanion is placed (generated) in the gas phase (in a high vacuum).  The gas-phase carbanion is then irradiated with light.  Each frequency of light provides a specific energy ($E=h\nu$).  The frequency of the light is varied (and the energy supplied changes as a consequence of the change in frequency) until enough energy is supplied to knocked out of one of the carbanion electrons (this is called photodetachment).

$$\ce{CH3^{-} + hν -> CH3• +~~ e^{-} +~ KE_{electron}}$$

The kinetic energy of the photodetached electron is then experimentally measured and the electron affinity ($EA$) of the carbanion is given by

$$EA = hν ~~– ~~KE_{electron}$$

The $EA$ is the energy required to remove one electron from the carbanion. The larger the electron affinity, the more tightly the electron was held.  Stable carbanions hold their electron more tightly than less stable carbanions. The electron affinity of the allyl anion was measured as 0.481 eV.  For the vinylic carbanion, the EA was measured as 0.667 eV.

The fact that the vinyl carbanion is more stable than the allyl carbanion suggests that hybridization effects outweigh resonance effects in terms of stabilizing these carbanions.  That sounds reasonable and is what I would have guessed.

reference for EA measurements


Can grignard reagents react with any electrophillic sites?

To give you an idea of the nucleophilicity of the two carbons, you may have a look at $\ce{^{13}C}$ NMR data (here for example). These data tell you which carbons have less electron density (the ones which are more deshielded).

On methyl formate, you'd find $\delta_\text{COO} = 161.8$ ppm whereas $\delta_\text{OCH3} = 50.0$ ppm. This should give you a little intuition of how much the electron density is reduced on each carbon, due to the effect of the electronegative atoms. Clearly, the ester-bearing carbon is much more affected, and this accounts for the general behaviour that you encounter.

However, as Ben Norris pointed out, this is maybe a simplistic picture. In fact, it's not impossible to break preferentially the $\ce{COO - C}$ bond. This is called hydrolysis by alkyl-oxygen cleavage (see here).

To get back to the question of Grignard reagent, the problem is also explained by HSAB theory. To attack the carbon which is not bearing the ester, you would generally need a "soft" organometallic compound. I can give you an example that corresponds to what you are asking (reference here: Kawashima et al., Tetrahedron 1989, 45, 403.), but note that this is a special case. What drives the reaction is the opening of a strained ring, and the Grignard is modified by addition of copper halide:

I represented this one in particular to show the regioselectivity (breaking of bond represented in red), but the article shows plenty of examples.


Determining the mass of a substance using calibration curve produced by HPLC

Warning: This is also a homework question. However, this is more like mathematics (geometry) than chemistry, I decided to help you. Nonetheless, please keep in mind that we won't help you in all future perspectives, if you do not show your effort to solve the problem, regardless more mathematics involved.
As  M. Farooq pointed out that this problem can be solved using the answers to your previous question, specially my old-fashioned approach. Here is how you could do it. Analyze your graph thoroughly to find two different points of graph where it perfectly crossed the cross-lines:

For me, these two points are: $(35,2000)$ and $(60,3500)$. Therefore the tangent of the slope ($\alpha$) of line is:
$$\tan \alpha = \frac{3500-2000}{60-35} = \frac{1500}{25}=60$$
Now, take another arbitrary point in the graph. Say it is $(x,y)$. Then:
$$\tan \alpha = \frac{y-2000}{x-35} = 60 \ \Rightarrow \ y = 60x - 100 \tag1$$
The equation $(1)$ is the straight-line equation of the graph.
When $y = 2800$, $x= \frac{2800+100}{60} = \pu{48.33 mg/L}$. This is for the diluted sample. I think it is fair to leave you some work to do. So you can back calculate to get your answer. When I did, it came close to $\approx \pu{798 mg/can}$.
Note: The error ($3.27\%$) is due to the finding of the slope. Also there are few flaws in the question. The calibration curve is made with HPLC traces of known standard caffeine solutions (not using diluted samples of the drink). The peak area of $2800$ arbitrary units was obtained by the injection of diluted sample of the drink ($1$ to $\pu{50 mL}$ dilution).


Thought experiment for evaluating total energy

If we disintegrate a molecule one step further to isolated electrons and nuclei, will the overall energy spent be equal to negative of total energy of schrodinger equation?

By convention, yes.

Energy as a absolute quantity is not well-defined, energy must be defined with respect to an energy zero.  By convention, when we perform quantum chemical calculations our energy zero is free electrons and nuclei infinitely far apart, which we define to have zero potential energy as they cannot interact.

Also, quantum mechanically, we consider only the case at $0$ K.  We can however add on corrections to the free energy due to entropy and internal motion at finite temperatures, as well as vibration zero point energies.

How is this different from the atomisation energy?

We can see that to get from the products of atomisation to the products of "disintegration" (as you refered to it), is to remove all the electrons from the neutral atoms.  That energy is given by the sums of all the ionisation energies for each atom in the system, removing each electron one by one.  This allows us to construct a Hess cycle.

Finite temperature corrections are added to the quantum mechanical energies as the potential energy and free energy are different at finite temperature.


Yes, it would be considered homogeneous, since all of the components are in water. We don't write water as being aqueous because it would be somewhat redundant to say the water formed in water.


From this question on Physics SE, there is a linked audio track in Soundcloud here, which is what you are looking for. The OP references a jar of hair-stying gel which, upon being tapped with a mallet, produces a "reverberating" sound instead an otherwise expected "short 'tok' sound."
Note the the ringing gels are common ingredients in pomades, shampoos, and other products where rheological properties are engineered for various reasons. See this patent application for a ringing gel, wherein it is stated:

The phenomenon of ringing is a result of the manner in which the dispersed droplets are entrapped within the network structure of a gel. It has been suggested that the network within a gel consists of a rod-like or lamellar structure. Such a network could allow dispersed particles to vibrate within a gel much like a ball bounces off the walls of a box.

The practical application of the gels appears to be mainly in the cosmetics industry.


Why is the disordered CuAu structure more stable at high temperatures?

I'm not a metallurgist or even not a good physical chemist. However, I'd like to explain what's happening in $\ce{AuCu}$ alloy with temperature and any expert can intervene.
According to Ref.1, $\ce{Cu}$ and $\ce{Au}$ are both univalent group IB metals with the atomic size difference ~12%. The $\ce{AuCu}$ alloy forms the tetragonally distorted fcc lattice where alternate (00h) planes contain either $\ce{Cu}$ or $\ce{Au}$ atoms and cause a contraction in c-direction. Resulting tetragonal face-centered structure has c/a ratio of 0.92 $(\frac{c}{a} = \frac{\pu{367 pm}}{\pu{396 pm}} = 0.93)$. In the temperature range ~$\pu{380 ^\circ C}$ to $\pu{410 ^\circ C}$, the superlattice $\ce{CuAu}$-II is formed, which consist of $\ce{CuAu}$-bct unit cells with the antiphase domains along the b-direction (Ref.2). There is a lattice shift of $\frac12(a+c)$ at each five unit-cell length. The superlattice $\ce{CuAu}$-II is described as orthorhombic cell with 10 cells along one of a direction, $oI40$.

Suggested reading: Ref.3 and 4.

References:

Valentina F. Degtyareva, Nataliya S. Afonikova, "Simple Metal and Binary Alloy Phases Based on the fcc Structure: Electronic Origin of Distortions, Superlattices and Vacancies," Crystals 2017, 7(2), 34 (13 pages) (DOI: https://doi.org/10.3390/cryst7020034).
V. Ozoliņš, C. Wolverton, Alex Zunger, "Cu-Au, Ag-Au, Cu-Ag, and Ni-Au intermetallics: First-principles study of temperature-composition phase diagrams and structures," Phys. Rev. B 1998, 57(11), 6427 (DOI: https://doi.org/10.1103/PhysRevB.57.6427).
M. Sanati, L. G. Wang, Alex Zunger, "Adaptive Crystal Structures: $\ce{CuAu}$ and $\ce{NiPt}$," Phys. Rev. Lett. 2003, 90(4), 045502 (DOI: https://doi.org/10.1103/PhysRevLett.90.045502).
S. -H. Wei, A. A. Mbaye, L. G. Ferreira, Alex Zunger, "First-principles calculations of the phase diagrams of noble metals: Cu-Au, Cu-Ag, and Ag-Au," Phys. Rev. B 1987, 36(8), 4163 (DOI: https://doi.org/10.1103/PhysRevB.36.4163).


Proportionality between number of absorbed photons and optical density

Yes, your statement that "the number of absorbed photons only is proportional to the optical density at low optical densities" is correct. (I'll point out that optical density is a depreciated term for what is now referred to as absorbance.)

First let's define some terms.

  T: Transmittance, i.e. the fraction of photons not absorbed. So a
     transmittance of 0.90 would mean that 90% of the photon pass through
     the sample.

  A: Absorbance of sample which is given by the Beer–Lambert law: $\ce{A = -log_{10}(T)}$ or: $\ce{T = 10^{-A}}$

Now a bit of math swizzling...

For exponentials of $e$ there is a nice series expansion. $$ e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{3!} + ... $$ and if $x < 0 $ then: $$ e^{-x} = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \dfrac{x^4}{3!} + ... $$
now if also $|x| << 1$ then the higher order terms can be ignored and the function becomes linear where:
$$ e^{-x} \approx 1 - x $$

Now for a bit more math voodoo... Let's let $$10^x = e^y$$ so: $$y = x\mathrm{ln}(10) \approx 2.303x$$

So now substituting $\ce{e^{-2.303A}}$ for $\ce{10^{-A}}$ we get: $$\ce{T = e^{-2.303A}}$$ and thus only when $2.303A << 1$ is the Beer-Lambert law linear.

Now for the rest of the story let's define some more terms:

  $\epsilon$ is the molar attenuation coefficient of the attenuating species  in the sample;

  $c$ is the molar concentration of the attenuating species in the sample;

  $l$ is the path length of the beam of light through the material sample.

Now also accoding to the Beer-Lambert law: $\ce{A = \epsilon c l}$ so $\ce{A}$ is proportional to $\ce{c}$.


How am I to interpret the phrase "a solution of acid"?

The acidity constant is defined as the equilibrium constant for the reaction:

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

$$K_\text{a} = \dfrac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$

Note: $\ce{H3O+}$ can be considered equivalent to $\ce{H+}$

Since you have to find an equilibrium constant, (and no further info is given) you can safely assume the solution to be in equilibrium.

Plug in the various conductivities in the second equation (Kohlrausch's Law) you mentioned getting the limiting conductivity ($\lambda^o$) of the acid. This is the conductivity of the acid if it had completely dissociated into its constituent ions.

EDIT: To find the limiting conductivity of the acid using the equation, you must have the values of $\lambda^o_{\ce{H^+}}$ and $\lambda^o_{\ce{A^-}}$.

$$\lambda^o_{\ce{HA}} = \lambda^o_{\ce{H+}} + \lambda^o_{\ce{A-}}$$

The degree of dissociation of the acid $\alpha$ (The number of acid molecules which have dissociated at equilibrium per each acid molecule present originally -- it is fractional for weak acids and bases) would be the ratio of the measured conductivity of the solution ($\lambda$, which is given) to the limiting conductivity ($\lambda^o$).

$$\alpha = \dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}}$$

This alpha would be the same as the ratio of concentration of dissociated acid molecules at equilibrium (since each dissociated acid molecule gives one $\ce{H^+}$ ion, this is equal to the number of $\ce{H^+}$ ions at equilibrium) to the initial concentration of acid taken (which you have mentioned as given).

Therefore,
$$\dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}} = \dfrac{[\ce{H^+}]}{[\ce{HA}]_\text{initial}}$$

You can find $[\ce{H^+}]$ at equilibrium using this equation and then find the other concentrations using that.

Note that all this assumes that the acid is monoprotic (i.e. there is only one proton released per acid molecule). You would have to change all the equations accordingly in case more than one proton is released.


How come the same reaction can be balanced in two distinct ways?

As per Nicolau's suggestion, :

$$\{(1-2)\times\color\red{4}\}-2$$

gives

$$2\ce{XeO3}\to2\ce{Xe}+\ce{3O2}$$

Thus, I conjecture that the second and third reaction differ by the fact of decomposition of $\ce{XeO3}$

$2$ can be converted into $1$ by multiplying by $\color \red{\frac 3 2}$ and decomposing one of the $\ce{XeO3} $ molecules.

Wikipedia also says that $\ce{XeO3}$ is unstable.


Physical meaning of thermal parameters dimension (Å²)

Model of a crystal and deriving the expected diffraction image

In single-crystal X-ray crystallography, the diffraction image is due to scattering of electrons in a crystalline sample. A convenient way of describing the electron density is to first specify crystal symmetry and mean atomic positions (coordinates), and then to describe in more or less detail how atoms deviate from these mean positions (modeled by atomic displacement factors). Combining this with a spherical description of the electrons "belonging" to each atom (modeled by atomic form factors), you arrive at an electron density.

This description is useful because in the model, the electron density is the convolution of the crystal lattice points with the coordinates in the unit cell with the distribution of atoms around mean positions with the distribution of electrons around the atoms. The convolution theory states that the Fourier transform of a convolution is the product of the Fourier transforms. So in this model, you can separately Fourier-transform the different levels of the model and then multiply them to arrive at the expected diffraction image (via the structure factors).

Atomic displacement and Debye-Waller factor

The atomic displacement around a mean position may be described through a probability density function $p_k(\mathbf{u})$, where $\mathbf{u}$ is the position (3D-vector) with the mean position as origin. The Fourier transform of this function defines the Debye-Waller factor (source: equation 1.4.8):

$$T_k(\mathbf{h}) = \int p_k(\mathbf{u}) e^{2 \pi i \mathbf{h} \cdot \mathbf{u}}d^3 \mathbf{u}\tag{3}$$

Here, $\mathbf{h}$ represents coordinates in reciprocal space (just as Fourier transform of a time series gives you information in frequency space, the Fourier transform connects real space - coordinate $\mathbf{u}$ - with reciprocal space - coordinate $\mathbf{h}$).

If the displacement is modeled by a Gaussian, the integral can be determined as (source: equation 1.4.10)

$$T_k(\mathbf{h}) = e^{-2 \pi^2 \langle(\mathbf{h} \cdot \mathbf{u})^2\rangle}\tag{4}$$

This reflects that the Fourier transform of a Gaussian is a Gaussian again, and the width of one Gaussian is the inverse of the other (i.e. if the atoms are widely distributed, the signal will diminish a lot with increasing resolution). Again using the time series as an example, if a sound is played for a very short time, the pitch will be ill-defined (or if an NMR FID signal decays rapidly, the peaks will be broad).

Displacement parameters $\mathbf{U}$

The displacement parameters $\mathbf{U}$ or $\mathbf{U^{jl}}$ in the OP's equation (1) and (2) are derived from the $\langle(\mathbf{h} \cdot \mathbf{u})^2\rangle$ expression in equation (3), either direction-independent in the case of an isotropic approximation (spherical distribution around the mean) or separately along the three lattice axes (leading to six anisotropic displacement factors). I am not explicitly showing this - it gets a bit complicated because the derivation requires transformation into the sometimes non-orthogonal coordinate system based on the crystal lattice (see here). In any case, the expression is a mean square displacement (analogous to a variance of a one-dimensional function). For that reason, the dimensions are length squared and the typical units are $Å^2$. As stated in the IUCR nomenclature document:

  the elements of the tensor U have dimension (length)2 and can be directly associated with the mean-square displacements of the atom considered in the corresponding directions

and

  If the atomic pdf is assumed to be a trivariate Gaussian, the characteristic function corresponding to this pdf - by definition, its Fourier transform - can be described by the second moments of the pdf, which in the present context are called anisotropic mean-square displacements.

Second moments (variance) have dimensions that are the square of the dimension of the quantity described by the probability distribution function (pdf). In this case, we are looking at the distribution of the position (dimensions length), and the second moment will have dimensions of length squared.