How does the pyramidalization effect the bonding angles in a distorted octahedral environment?

So we are again talking about $\ce{BrSF5}$ but this post extends to all generally octahedral complexes/ molecules and even quadratic planar geometries. The main requirement is, that the molecule has a $C_{\mathrm{4v}}$ point group. (It may be extended to all non-linear $C_{n\mathrm{v}}\ (n>1)$ point groups.) First of all reconsider the molecular shape again.

Now admittedly the degree of pyramidalization is more than minimal, with less than $1^\circ$ deviation from the plane. However, this has a slight effect on the equatorial $\angle(\ce{F-S-F'})$ bond angle, purely to symmetric reasons.

Basic trigonometric considerations are enough to understand this phenomena. Consider the following scheme, where the peak of the pyramid is represented by $\mathbf{\ce{Y}}=\ce{S}$ and the base is span by $\mathbf{\ce{X_{eq}}}=\ce{F_{eq}}$, $\mathbf{\ce{X'_{eq}}}=\ce{F'_{eq}}$, $\mathbf{\ce{X''_{eq}}}=\ce{F''_{eq}}$ and for completeness $\mathbf{\ce{X'''_{eq}}}=\ce{F'''_{eq}}$.

So we have some basic quantities, that we can address, but in the end we can derive the connection of the two angles without caring about the other quantities, only knowing the point group.

$a$ is the $\mathbf{d}(\ce{F_{eq}\cdots{}F'_{eq}})$ distance
$d$ is the $\mathbf{d}(\ce{F_{eq}\cdots{}F''_{eq}})$ distance
$s$ is the $\mathbf{d}(\ce{S\cdots{}F_{eq}})$ distance
$h$ is the height of the pyramid and it refers to the out of plane shift of $\ce{S}$ compared to the $\ce{F_{eq}\cdots{}F'_{eq}\cdots{}F''_{eq}\cdots{}F'''_{eq}}$ plane
$h'$ is another virtual quantity, the angle bisector of $\angle(\ce{F_{eq}-S-F'_{eq}})$
$\alpha$ refers to the bonding angle $\angle(\ce{F_{eq}-S-F''_{eq}})$
$\beta$ refers to the bonding angle $\angle(\ce{F_{eq}-S-F'_{eq}})$

From Pythagoras's Theorem we can easily derive the following connection (square in the scheme):
\begin{align}
             && d^2 &= a^2 + a^2\\
  \therefore && d   &= \sqrt{2}\cdot a\tag{1}
\end{align}

So now we are going to use some trigonometric functions for the angles (look at the triangles):
\begin{align}
             && \sin\left(\frac{\alpha}{2}\right)
                    &= \frac{\frac12d}{s}\\
  \therefore && s   &= \frac{\frac12d}{\sin\left(\frac{\alpha}{2}\right)}\tag{2}
\end{align}

\begin{align}
             && \sin\left(\frac{\beta}{2}\right)
                    &= \frac{\frac12a}{s}\\
  \therefore && s   &= \frac{\frac12a}{\sin\left(\frac{\beta}{2}\right)}\tag{3}
\end{align}

Let's derive the connection between the angles by equalizing $(2)$ and $(3)$:
\begin{align}
             && \frac{\frac12d}{\sin\left(\frac{\alpha}{2}\right)}
                     &= \frac{\frac12a}{\sin\left(\frac{\beta}{2}\right)}\\
  \therefore && \sin\left(\frac{\beta}{2}\right)
                     &= \frac{a}{d}\sin\left(\frac{\alpha}{2}\right)\\
  \text{with (1)}\implies &&
                \sin\left(\frac{\beta}{2}\right)
                     &= \frac{1}{\sqrt{2}}\sin\left(\frac{\alpha}{2}\right)\\
  \equiv     && \sqrt{2} &= \frac{\sin\left(\frac{\alpha}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}\tag{4}
\end{align}

From this you can see, that the degree of pyramidalization is codependent on the equatorial bond angles. When the pyramidalization increases, $\alpha\to0$, then the equatorial bond angle has to decrease, $\beta\to0$, i.e. a constant ratio.

In the special case of $\alpha = 180^\circ$ this simplifies to $\beta=90^\circ$. In all other cases you can derive the degree of pyramidalization from the equatorial bond angles and vice versa.

Transformation of $(4)$ yields a function that displays the equatorial bond angle in dependence of the pyramidalization, with $\alpha\in~]0^\circ;180^\circ]$ or $\alpha\in~]0;\pi]$.
$$\beta = 2\cdot\arcsin\left[\frac{1}{\sqrt{2}}\cdot\sin\left(\frac{\alpha}{2}\right)\right]$$

Addendum

The out of plane height can also be derived from Pythagoras's theorem (look at right triangle):
\begin{aligned}
                 && h^2 &= s^2 + \left(\frac{d}{2}\right)^2\\
  \therefore     && h   &= \sqrt{s^2 + \left(\frac{d}{2}\right)^2}\\
  \text{with (1)}&& h &= \sqrt{s^2 + \frac{1}{2}a^2}\\
\end{aligned}

This is usually a value that is given for domed complexes, so it is nice to know how to find it.


If caffeine is an alkaloid, why is coffee acidic?

Coffee contains hundreds, if not thousands, of other compounds in addition to caffeine.  Included among these other compounds are many acids.  Many small, organic acids such as  citric, malic, lactic, pyruvic and acetic acid are present, but both quinic acid and chlorogenic acid (and their derivatives) are usually present in even higher concentration.  Phosphoric acid, an inorganic acid, is also present. The exact concentration of these various acids depends upon processing variables such as roast conditions and grind size.  Here is a link to an interesting, one-page discussion of the subject.
quinic acid

chlorogenic acid

Edit: In Season 3, episode 6 of "Breaking Bad", we meet Walt's new lab assistant Gale, who is brewing coffee using a lab glassware set-up.  Walt says that it is the best coffee he has ever tasted and Gale responds by mentioning "quinic acid" as something to be considered in brewing great coffee.


How can meta-cresol be made, given that the hydroxyl group is ortho,para-directing?

Here is a synthesis of m-cresol starting from benzene.  It achieves the desired meta orientation of the final substituents by introducing a meta-directing substituent (the aldehyde group) early-on in the synthetic scheme. In the nitration step, the meta isomer is by far the major product (o:m:p 19:72:9), but a distillation or recrystallization would be needed (or you could carry all the isomers along and separate later, but you'd probably have the widest range in boiling points at this stage with the nitro group present). The only novel step is the double Wolff-Kishner used to reduce both the aldehyde and nitro groups in a single reaction.


Why does methane have 4-different bond dissociation energies?

Bond energies are important, but you have to start somewhere. Actually, I'm surprised that the bond energies for the four dissociations are so similar. But that's a good start.
Now let's get into a bit more interesting compound: ethane. Very interesting!

The table is only part of the information on ethane given in an open access article (Ref 1). Note that the first C-H bond dissociation requires 422 kJ/mol, while the second requires 150 to 465 kJ/mol, depending on which bond you are breaking, because of what product you are making. Each line in the table is worth several paragraphs of explanation.
I admire your desire to tie up your knowledge neatly here - get a lot of string, because it's a very big package!
Ref 1. https://pubs.acs.org/doi/pdf/10.1021/acs.jpca.5b01346   Branko Ruscic, J. Phys. Chem. A 2015, 119, 28, 7810–7837


Possibility of SET mechanism of Wurtz reaction

How can the alkyl group and the halogen form free radicals when the halogen is clearly much more electronegative then the alkyl group?

To answer your first question, it is important to note that the halogen does not form free radicals in this mechanism. Instead, the driving force is the net formation of sodium halides, as illustrated in the mechanism I drew below. In the preceding sections, I will try to compare the Wurtz mechanism to a perhaps more familiar one, the Grignard mechanism. This should help elucidate some general aspects of organometallic mechanisms.

If you are familiar with the Grignard reaction, the following comparison may help clear up the involvement of radical species in the mechanism. The Wurtz reaction is in fact very similar to the Grignard reaction. Both involve the 'insertion' of a metal species in between the carbon-halogen bond. In the case of the Grignard reaction, this metal is magnesium. This allows for the formation of a simple alkyl magnesium halide; however, with the Wurtz reaction, the addition of a sodium atom in between the carbon-halogen bond results in the formation of neutral sodium halide and a carbon radical. This carbon radical will rapidly react with another equivalent of sodium metal to form a carbanionic species, which will in turn displace another halogen in an Sn2 reaction.

The Wurtz reaction is not a super useful reaction due to the formation of this reactive radical. Most chemists would opt to use the much gentler Grignard reagent. Below, I have drawn out the mechanism of the Wertz reaction of 1-bromo-3-chlorocyclobutane to form bicyclo[0.1.1]butane. In this, you can see that a carbon radical is formed. This radical must come in contact with another equivalent of sodium before it is neutralized to form a carbanionic species. For this reason, the radical will also have plenty of time to react with other parts of the molecule in unwanted ways. The Grignard reaction, on the other hand, results in the formation of a radical in the 'reduction' step; however, this is neutralized very quickly in the recombination step and will thus not react unnecessarily with other parts of the molecule. I have included a diagram of the Grignard reagent formation below.

Mechanism of the General Wertz Reaction

Mechanism for the Formation of a Grignard Reagent


Why is the principle of separation in GFC opposite to electrophoresis despite using same material for the gel?

Gel electrophoresis and gel filtration chromatography both use gel materials (though often polyacrylamide gels are used for GE and cross-linked dextran gels for GFC), but their operating principles are different.

In GE, the gel is a solid block of gel which contains only small pores for proteins to pass through.
The proteins generally have to be denatured (and given a charge) to pass through the pores as native proteins are too bulky.
The denatured proteins are like long strands of noodles being drawn through the gel and shorter proteins pass through more quickly because they don't have to contort as much to fit through all the pores—the larger the protein, the more interactions with the gel to slow it down.

However, GFC is a subset of what is known as Size Exclusion Chromatography.
SEC techniques generally also use porous gels, but instead of using a solid block of gel, the gel is made into small beads.
A similar situation to GE exists where small things can enter the pores of the gel and larger molecules suffer more resistance to pass through the gel, only now, the larger molecules have an alternative route with lower resistance—going completely around the beads.
The effect of this is that small molecules pass into the pores of the beads and slow down, while larger molecules bypass the beads entirely and travel faster (what is considered small and large depends on the gel's pores).

GE has the benefit of much better size separation, but requires denatured proteins.
SEC can handle native proteins (that are still functional on the other side of your column), but the separation is more akin to dialysis, where small and large molecules are easily separated from each other across a cutoff molecular mass, but there is otherwise poor resolution between species.


How are related the electrons magnetic dipole moments in atomic orbitals?

In my paper, mentioned above, I have had some doubts about the role of the electrons dipole moment distribution in chromium atom which is quite symmetrical if I use my octahedral model instead of s- and p-orbitals.

Now a have read on Wikipedia about the magnetic characteristics of pure chromium: "Chromium is remarkable for its magnetic properties: it is the only elemental solid which shows antiferromagnetic ordering at room temperature (and below). Above 38 °C, it transforms into a paramagnetic state." This matches perfect into my model abot octahedral structure of neon and argon shell electrons.


Why does atomic charge not balance in nuclear reactions?

There are two reasons why nuclear reactions don't consider (or often account for) charge.

The first is that charge is usually irrelevant to the reactions. If we are looking at changes in nuclei we can for all practical purposes ignore the electrons that would be associated with the nuclei under normal conditions. They are just irrelevant to the nuclear reaction taking place.

The second is that they do account for charge, but only in nuclear processes. Some radioactivity (isolated neutrons decay or any nucleus undergoing beta emission) can lead to the decay of neutrons into a proton and electron and some other minor particles. The electron clouds that would normally surround the nucleus are still irrelevant to these processes but the equation for the nuclear reaction should account for the emitted particles. When writing these equations only the nuclei (and the emitted particles) are accounted for and any net change of charge is ignored as a description of the nuclei involved implicitly describes the net nuclear charge (in the same way that many organic chemical structures don't explicitly draw every hydrogen as chemists can fill them in knowing that the total valence around carbon should add to 4).


Number of optically active compounds among the products of ozonolysis

Ozonolysis of the given compound gives $\ce{I}$ , $\ce{II}$ , $\ce{III}$ and $\ce{IV}$.

$\ce{I}$ and $\ce{II}$ are Homomers and have same configuration "S" and therefore identical (as shown in the figure).
$\ce{III}$ and $\ce{IV}$ are achiral . Therefore , the total number of optically
active compounds formed after ozonolysis is $\ce{= 1}$.


Are Tellurium Fluoride or Antimony Fluoride Ionic?

TeF4 is polymeric and low-melting.  Not ionic.  Covalent.
In general, the elements that are metalloids (on the "border" of metal and non-metal) like Te and Sb, tend to behave chemically like nonmetals (looking at their compounds).   But physically as metals (shiny, malleable, conductive).  So that's another good clue that these elements would be form covalent compounds (molecules or polymers ["network solids"]) with fluorine.
Finally even for hard core transition or p group metals, you will find that the lower oxidation count species tend to be ionic.  But as you get to tetravalent species a fair amount are molecular or polymeric (i.e. covalent) and few are ionic.  No penta or higher flourides are ionic.  And the few tetravalent ionic compounds tend to be pretty far to the left of the table (e.g. Zr)...pretty hard/electropositive atoms, not soft (i.e. polarizable) or less electropositive like those in the p group.  (To which Te and Sb are even more marginal metals.)
The Wikipedia article on Compounds of Fluorine has some good details on the concepts and good references to review articles (and Greenwood/Earnshaw Chemistry of the Elements).
https://en.wikipedia.org/wiki/Compounds_of_fluorine#Nonmetal_fluorides
P.s. Note also that it is important to specify which species of fluoride you mean (for elements below the second row).  Te forms a tetra and hexa fluoride (and some more complicated variants)--you did specific tetra, but that's not the only choice.  Also for Sb, you didn't specify which compound and both the tri and penta fluoride are well known.


What kind of optical activity requires chirality?

The meaning of optical activity in this context is the same as usual: the ability to rotate the plane of polarized light. To have it, a molecule must be chiral.

Chirality is the absence of certain type of symmetry. Since most small molecules have that symmetry, many students get the wrong perception that chirality is a property which must be obtained by having some functional group (asymmetric center, etc.), much like having a $\ce{-COOH}$ group makes a compound an acid. This is not quite true. Not only a compound may be chiral without any chiral centers, but it can also be achiral despite having some chiral centers, as are meso compounds.


Does the priority go for the longer substituent, the closer one, or one which contains a pi-bond in naming Alkenes and Alkynes?

You have already correctly identified the parent structure to be the longest chain, i.e. dodecene. As usual, the most important simplified criteria for the numbering of locants for substituent groups in such compounds are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds (first to double bonds if there is a choice)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
| (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound that is given in the question doesn’t have a principal characteristic group that is expressed as a suffix. Therefore, Rule (c) is not relevant in this case.
A low locant is assigned to the double bond according to Rule (e). In this case, this rule always results in the same locant for the double bound (the ‘ene’ ending); it doesn’t matter at which end of the chain you start counting; i.e. the name of the parent structure is dodec-6-ene.
Finally, low locants are assigned to all the substituents as a set according to Rule (f). Therefore, the systematic name is 8-ethenyl-3-ethyl-5-propyldodec-6-ene rather than ‘5-ethenyl-10-ethyl-8-propyldodec-6-ene’ since the locant set ‘3,5,8’ is lower than ‘5,8,10’.


Does the nitrogen atom move with in the molecule in an ammonia maser, or does the molecule flip?

Let me expand a little on my earlier comment. If we only consider the inversion of the ammonia molecule (we ignore rotation and vibrations other than the inversion mode), the Schrödinger equation is given by
$$
-\frac{\hbar^2}{2\mu_\text{red}}\frac{\text{d}^2\psi{z}}{\text{d}z^2}+V(z)\psi(z)=E\psi(z),
$$
with $\mu_\text{red}$ the reduced mass and $z=d_\text{NH}\sin{\alpha}$, where $d_\text{NH}$ is the NH bond distance and $\alpha$ the so-called umbrella angle. The potential V(z) is shown in the link given by porphyrin in one of his comments. Although the reduced mass does, in principle, depend on $\alpha$ (see Eq. 4 of this paper), the dependance is weak and we may consider that the hydrogen atoms move simultaneously with respect to the nitrogen atom. In this case
$$
\mu_\text{red}=\frac{3m_\text{H}m_\text{N}}{3m_\text{H}+m_\text{N}}\approx2.5\text{ u}.
$$
The tunneling does not change the center of mass (c.o.m.) of the molecule, so while the nitrogen moves up, the plane of hydrogen atoms moves down to ensure the stationary c.o.m. (and vice versa). I made the following animated gif at some point during my PhD (sorry for the watermark, but I've  had some negative experiences in the past):

The first paper - to my knowledge - that treats the tunneling of ammonia appeared already in 1932. The first application of QM tunneling was done by Gamow appeared 4 years prior to the ammonia work and solved the mystery of the huge range of observed lifetimes of alpha emitters (20 orders of magnitude).

In the 1932 paper, Dennison and Uhlenbeck also derive the magnitude of the tunneling splitting as a function of the reduced mass
$$
E_\text{inv}=\frac{h\nu_0}{\pi}\exp\left \{-\frac{2\pi}{h}\int_\limits{-z_0}^{z_0}\left [2\mu_\text{red}\left (V(z)-E \right ) \right ]^{1/2} \right \},
$$
where $\pm z_0$ are the equilibrium positions corresponding to the potential minima, $\nu_0$ is the vibration frequency in one of the two wells and $E$ is the energy of the system with respect to the potential minimum. This expression is derived from the WKB approximation and it basically is the product of the tunneling probability with the frequency with which the reduced mass 'hits' the barrier. As can be seen from the formula, the tunneling frequency scales exponentially with the square root of the reduced mass, which is also observed experimentally. The following picture is taken from my thesis and basically a copy of Fig. 2 of this paper.

If the molecule would simple rotate instead of tunnel, the change in the energy splitting would be inversely proportional to the reduced mass (via the moment of inertia) and this is clearly not observed.


How many possible nucleotide types can be created?

I don't know why RNAs with 5 nucleotides became so prominent in our biosphere but you can bet that it is probably because it was better (even slightly so) than other RNA models it may eventually have competed with. So in order to arrive at a nice hypothesis about why this thing exists, we need to understand what was the competition like and how would our 5-symbol code benefited against rivals.
Consider the setting we are discussing. We have self-replicating RNA strands competing for the building blocks in the primordial soup (and maybe in each other)  in order to use those to make more of itself. The single most important resource that these replicators will have is time and speed. RNA strands aren't terribly stable and will hydrolyze after a certain window of time, meaning that our critter have a lifespan. If an RNA strand is to work as a self-replicator it must be able to assemble copies of itself within its lifespan otherwise its code will die with it.
We know that the supply of nutrients must have been quite limited because if you put mindless replicators that can eat each other in a soup of free dinner they reproduce at an exponential rate until the supply and demands are met and the population reaches and equilibrium appropriate to the free energy available. A population in this level leaves very little free nutrients around.
So in any given pool teeming with RNA critters, life must have been hard and dinner difficult to come by, because the time of bounty, when life was rare and food was abundant was followed by "that time when everyone ate like nutjobs until all the food was in their bodies rather than out there".
We also have reason to accept that there must be an upper limit to the length of the self-replicating strands. Longer strands are more prone to hydrolysis, oxidation and radiation. Environmental hazards that may lead to mutations and the rupture of the strand, leading to the corruption of the code and potentially the loss of its self-replicating quality, effectively resulting in its death. Longer chains have a shorter lifespans and take longer to copy than longer chains.
These two things generate selective pressures that might help us explain why the amount of nucleotides isn't larger. On one side, individuals with too few symbols (shallow code) will require longer chains to hold their code. This will shorten their lifespans and make their code harder to copy, since it is longer. So going too shallow will hamper an individual's ability to reproduce by increasing the time required to make copies and shortening the lifespan. Eventually, there must be a specific chain size above which the average copy/lifespan is less than 1. With a shallow code only the simplest individuals can reproduce reliably and that leads to low genetic diversity and less adaptability.
On the other hand, individuals with too many symbols (deep code) can have very short chains which are, in theory, quicker to copy but must compete for a more varied assortment of resources and must search harder and longer to meet their nutritional needs. If the code gets too deep, the time needed to procure the neccessary building blocks will exceed the critter's lifespan, meaning that, on avarage, an individual using this code will die before it can make a single copy of itself. With deep codes we can have rich, dense complexity but its maintenance cost can be prohibitive for replicators of this type.
So here is a possible explanation. Maybe 5 nucleotides is the best bang for our buck. The longest code that would have granted our RNA ancestors enough genetic diversity to survive crises that may have wiped shallower rivals while keeping their average copies/lifespan greater than their deeper rivals.
While this might look like an appealing explanation, I urge you not to take this at face value. I have zero evidence to corroborate with any of this. Since your question was so broad, I do believe that some speculation is in order. However the example here may help you learn how to think these questions through and arrive a satisfactory hypothesis to test in a suitable laboratory. Send the link up here when you publish it, cheers.


Why can't the conjugate base of benzoic acid be stabilized by conjugation with the aromatic ring?

Here are some of the resonance structures we can draw for the carboxylate anion of benzoic acid.

You're correct that we can't delocalize the negative charge from the oxygen into the benzene ring.  Even though there is continuous p-orbital overlap between the oxygen and benzene carbons, there is just no way to push the electrons so that a negative charge winds up in the benzene ring. However, as the resonance structures show, we can move electrons the other way and place a second negative charge on the oxygens and leave a positive charge delocalized around the ortho and para positions in the benzene ring.  This is why the carboxylate anion of benzoic acid will not react with electrophiles (electrophilic aromatic substitution) at those positions in the aromatic ring.


Why does basicity of group 15 hydrides decrease down the group?

Each of these molecules has a pair of electrons in an orbital - this is termed a "lone pair" of electrons.  It is the lone pair of electrons that makes these molecules nucleophilic or basic.  As you move down the column from nitrogen to bismuth, you are placing your outermost shell of electrons, including the lone pair, in a larger and more diffuse orbital (the nitrogen lone pair is contained in the n=2 shell, while the bismuth lone pair is in the n=6 shell).  As the electron density of the lone pair is spread over a greater volume and is consequently more diffuse, the lone pair of electrons becomes less nucleophilic, less basic.

Commment: But greater the EN of the central atom , lesser would be its tendency to donate electrons right?

Response to Comment: The ENs of the central atoms are N (3.04), P (2.19), As (2.18), Sb (2.05), Bi (2.02). All of these atoms, except for nitrogen have similar ENs and I think the electron density argument is valid for them. However, in the case of nitrogen and phosphorous there is a significant EN difference that would tend to argue in the direction opposite to my answer. However there is another important difference between N and P. The lone pair in ammonia is in an sp3 orbital. In all of the other cases the central atom is essentially unhybridized (~90° H-X-H angles) and the lone pair exists in an s orbital. Therefore the lone pair electron density in ammonia (being in an sp3 orbital) is effectively increased compared to phosphine where the lone pair is in an s orbital. So even though the EN difference between N and P is significant, when hybridization differences of the central atom are taken into account the electron density argument still explains the trend in basicity.


Why is % yield in a purification table measured using activity instead of total protein?

Because total protein includes other proteins than the one desired.

"% yield" is expressing how much of the enzyme desired was actually recovered.


Finding reactants in a azo-coupling reaction whose product is given

The comments summarized the reasons well: the diazotization reaction for the initial structure would produce a mixture of $\alpha$-azo and $\beta$-azo compounds, which would give a mixture of products. Also, diazonium salts couple with activated aromatic systems. The diazonium salt shown would not couple with nitrobenzene, which is highly deactivating and does not favour electrophilic aromatic substitution.
Also, having an electron withdrawing group on the diazonium salt favours the reaction. Thus, the correct solution would have the azo group on the smaller, less activated benzene ring rather than the larger, more activated naphthalene ring.


There is not necessarily a set convention: https://en.wikipedia.org/wiki/Lattice_energy#Lattice_energy_vs_lattice_enthalpy

However, the sign must match the convention used. For example, the values in the table you provided are all positive which implies you are adding energy to break the lattice.

The definition from you prior experience would require negative values.


Identifying covalent character in ionic compounds when varying multiple ions

You will have to consider two factors here: polarising power of the cation and polarisability of the anion.
There is more than one part of Fajan’s rule(Source-J. D. Lee, Concise Inorganic Chemistry):

  A small positive ion favours covalency.
  A large negative ion favours covalency.
  Large charges on either ion, or both ions, favours covalency.
  Polarisation and hence covalency is favoured if the positive ion does not have noble has configuration.

According to these, the $\ce{Li+}$ ion is more polarising than $\ce{Na+}$ ion, and $\ce{Cl-}$ ion is more polarisable than $\ce{F-}$ ion.

Since both factors support it, $\ce{LiCl}$ is more covalent than $\ce{NaF}$.

Somtimes, like in your question, both the factors support each other. Sometimes, when they are opposite, like in the case of $\ce{LiF}$ and $\ce{NaCl}$, it is tough to determine relative covalency using Fajan’s rule.


Because it’s density is less than 1, the oak floats.

Therefore it displaced less than its full volume; some of the floating oak is above water.

Since it’s 2.68g of oak, it displaces only the volume equivalent to 2.58g of water.


While it would be possible to decarboxylate almost any carboxylic acid with appropriate conditions, simple aliphatic acids are not readily decarboxylated. Carbonyl group in beta position facilitate decarboxylation due to formation of a cyclic transition state:
Now, your example has five carboxyl group, and each has a carbonyl in beta position (carboxyl also counts as a carbonyl compound). Let's look at the picture:

First we can remove the three top carboxyls marked with arrows (A). There is a carbonyl in the ring that will aid in the decarboxylation, so we will arrive to a compound at B) (I'm not saying that the reaction will proceed that way, it is just an explanation). As I said before, one carboxyl group can facilitate the removal of the other carboxyl. But once we remove one group, we get compound C). That compound does not have any groups that would help to get rid of the last carboxyl. So there is your answer.


An example

Here is a balanced equation:

$$\ce{2 H2 + O2 -> 2 H2O}$$

Both sides of the equation show four atoms of hydrogen and two atoms of oxygen, so it is balanced. The coefficients are $\nu_{\ce{H2}} = 2$, $\nu_{\ce{O2}} = 1$ and $\nu_{\ce{H2O}} = 2$.

If 20 moles of $\ce{H2}$ and 10 moles of $\ce{O2}$ are used up in a reaction forming water (you can figure out 20 moles are formed), I can write:

$$\frac{n_{\ce{H2}}}{\nu_{\ce{H2}}} = \frac{n_{\ce{O2}}}{\nu_{\ce{O2}}},$$ where $n$ is the amount that is used up in the reaction.

I can not write:

$$n_{\ce{H2}} \cdot \nu_{\ce{H2}} = n_{\ce{H2}} \cdot \nu_{\ce{H2}},$$

as you will find when substituting the values from the example.

I should also not write:

$$2 \cdot \ce{H2} = 1 \cdot \ce{O2} $$

Hydrogen atoms are not oxygen atoms, no matter what we multiply it (nuclear reactions aside)

What about $I$ and $i$?

If you call the stoichiometric coefficients $i$ instead of $\nu$, and you want the equation below to be true:

$$i_{\ce{H2}} \cdot I_{\ce{H2}} =  i_{\ce{O2}} \cdot I_{\ce{O2}},$$

you have to make sure that $I$ is proportional to $\frac{1}{i}$, i.e. define them by

$$I_{\ce{H2}} = \frac{ \mathrm{const} }{ i_{\ce{H2}} } $$

$$I_{\ce{O2}} = \frac{ \mathrm{const} }{ i_{\ce{O2}} }$$

If you look at definitions of "number of equivalents" (a concept that is outdated), you will find that this is the case. Once that is in place, it is a matter of showing that for any non-zero $a$ or $b$,

$$\mathrm{const} \cdot \frac{a}{a} = \mathrm{const} \cdot \frac{b}{b} $$

Source: https://www.askiitians.com/iit-jee-chemistry/physical-chemistry/stoichiometry-and-redox-reactions/law-of-equivalents.aspx


Effect of pH on positive and negative charges of a zwitterion

Consider an amino acid as a zwitterion. It has both a basic amine functionality and an acidic carboxylic acid functionality. At a high pH, the acidic functionality will be largely deprotinated and the ion will have a net negative charge. Oppositely, at low pH, the amine group is largely protinated and the compound will have a net positive charge. Clearly, there is some intermediate pH at which the positive and negative charges are equal (or it's neither protinated nor deprotinated) and the net charge is zero.

To maintain charge neutrality, a charged species must have some oppositely charged ionic species associated to maintain charge neutrality. For example, under acidic conditions when the amine group is protinated, you need some anion present, like chloride for example, to be associated with the compound to maintain charge balance. This counter-ion is also known as an adherent ionic species (adherent meaning adhering to, like when the compound is an undissolved solid).


You should be adding stoichiometric amounts of diluted acid. Then distilation to remove the methanol by destilation. Alternatively can also use molecular sieves, or add sodium sulfate to dry the methanol


Tropone, or 2,4,6-cycloheptatrien-1-one, is an aromatic, non-benzenoid hydrocarbon.  If you look at the resonance structures in the drawing below you'll see that structure B

depicts a molecule with a continuous, planar loop of 7 p-orbitals that contain 6 pi-electrons.  Any planar (or near-planar), cyclic system with a continuous loop of p-orbitals is considered aromatic, or stabilized, if it contains 4n+2 pi-electrons and anti-aromatic if it contains 4n pi-electrons.  On this basis, molecules like the cyclopropenyl cation (4n+2, n=0), benzene (n=1), cycloheptatriene (tropylium) cation (n=1), cycooctatetraene dianion (n=2) would be considered aromatic systems.  Correspondingly, the cyclopropenyl anion (4n, n=1), cyclobutadiene (n=1), cyclopentadienyl cation (n=1), cyclooctatetraene (if it were a planar molecule, n=2) are considered destabilized or anti-aromatic molecules.


Why is the ROHF spin density of the propenyl radical not symmetric?

I believe your wave function is not the one you would expect in ROHF. The reason for this is guess=mix which will try to construct a broken symmetry wavefunction from the HOMO and LUMO. Why the SCF converges to that solution, and why this is stable (-116.460412224 ha), I don't know, and I have currently not the time to investigate (I will try to come back to that), but remove the keyword and you should obtain the symmetric correct solution (-116.457980797 ha). Inspect your output file, and you will notice, that the calculation is run in Cs symmetry, not C2v. (Also HF might not be the best method to show delocalisation.)


What is the pH of a solution of HCOONa and HNO3?

Yes. You could also put the molar ratio into the equation for calculating pH (then you didn't need to calculate the concentrations).

And those are not initial concentrations, but equilibrium concentrations.


Can someone explain why the electrical conductivity of magnesium chloride decreases with increasing concentration?

You simply started your data range at too high a concentration to see the initial increase.  Semantic Scholar gives data for the conductivities of several aqueous electrolyte solutions at 25°C, including magnesium chloride.  The strongly increasing portion of the magnesium chloride curve goes up to about 10%, which corresponds to about one molar (where your data happens to start).


Compound formed from concentrated sulfuric acid and sulfur

The reaction is given here.

$$\ce{S + 2H2SO4 → 3SO2 + 2H2O}$$

Sulfur react with sulfuric acid to produce sulfur dioxide and water. Sulfuric acid should be concentrated solution. The reaction takes place in a boiling solution.

$\ce{[S4]^2+}$, $\ce{[S8]^2+}$ or $\ce{[S19]^2+}$ is formed when sulfur is dissolved in oleum. The following is a paraphrase of an inorganic textbook.


How does mass spectroscopy of nicotine work?

The peak at 133 corresponds to a loss of 29 mass units (nicotine is 162).  N-CH3 has a mass of 29


How do you explain the thread sticking to an ice-cube when salt is sprinkled over it?

I conjecture that there are several things happening.

Step 3:  Addition of salt will melt the ice at its immediate point of contact. The string will naturally be surrounded by the highly concentrated brine (salt water).
Step 4:  This is enough time for the brine to get diluted, noting that there is an infinite reservoir of potential water by way of the ice cube.
Step 5:  Once the salt gets sufficiently diluted, then the comparatively large ice cube provides the necessary cold for the (now) very dilute salt water to reform as ice.

Recalling that in Step 3 the salt water surrounded the string, then ice naturally forms around the string in Step 5, hence you can pick up the ice cube with the string.


Titration of magnesium sulfate with potassium permangante

The $\ce{Fe^2+}$ concentration can be measured in this fashion since the iron(II) cation can be oxidized by a suitable oxidant, like the permanganate ion $\ce{MnO4-}$.

$$\ce{MnO4- +4H+ +3Fe^2+ -> MnO2 + 2H2O + 3Fe^3+}$$

Since permanganate is purple, and manganese dioxide is an insoluble brown solid, we have a built in indicator for redox titrations like these.

However, magnesium ion content cannot be determined this way since $\ce{Mg^2+}$ is the highest usual oxidation state for magnesium. The electron configuration for magnesium is $1s^2 2s^2 2p^6 3s^2$ or $[\ce{Ne}]3s^2$. The $\ce{Mg^2+}$ ion has the electron configuration of $\ce{[Ne]}$, which is relatively stable. Remove an additional electron from $\ce{Mg^2+}$ to get $\ce{Mg^3+}$, you get an ion with the electron configuration of $1s^2 2s^2 2p^5$, which is the same electron configuration as electron-hungry fluorine atoms, but with a nucleus containing three more protons. $\ce{Mg^2+}$ cannot be chemically oxidized further.

Now, one could use this method to determine the concentration of many other metal ions that can be oxidized, including even $\ce{Mn^2+}$.

$$\ce{3Mn^2+ + 2MnO4- + 2H2O -> 5MnO2 + 4H+}$$


How much CO2 is produced in burning of a candle?

Surely $\ce{CO2}$ will fill some of the space of used up $\ce{O2}$. Candles these days are commonly made out of paraffine, which is basically $\mathrm{C_{many}H_{many}}$. The equation may look as follows:
$$\ce{C31H64 + 47O2 -> 31CO2 + 32H2O}$$
You see that we get approximately 2 moles of $\ce{CO2}$ for 3 moles of spent $\ce{O2}$, hence the decrease in volume.

Another factor is that $\ce{CO2}$ is much better soluble in water than other components of air, so over time (though not immediately) it will probably vanish from the glass and thus free some more space.

Yet another factor is that air in the glass is quite hot while the candle is burning. As soon as it cools down, it shrinks.


Clarification on the term: coordination number

Coordination number, also called ligancy is the number of atoms that a central atom or ion holds as its nearest neighbours in a complex or coordination compound or in a crystal.

In EDTA, there are two nitrogen atoms and four oxygen atoms, which act as neighbors and bind by sharing their electrons. So, even though EDTA is itself one single molecule, there are six atoms in EDTA which are sharing their electrons. So, the iron has the coordination number of 6.

The complex looks like this:

So, basically, coordination number depends on how many atoms are directly participating in the complexing.


Order for acidic strength of hydrides across a period

The reasoning you gave for the increasing acidity of hydrides of group $15$ down the group is not the best one. You claim that less polar bonds are easy to break which is absolutely wrong. Non-polar covalent bonds are very difficult to break (think of beaking  $\ce{C-H , N-H , O-H }$ bond, which is easier to break ?)The logic can be modified as follows.

Actually the basicity of those hydrides decrease down the group, which in turn results in the increase of acidity of them. The reason for decrease in their basicity down the group is the increase in inability to donate their lone pair to accept the proton. In $\ce{NH3}$, the lone pairs recide in the $sp^3$ hybrid orbitals, whereas for other hydrides like  $\ce{PH3 , AsH3 , BiH3}$ the lone pairs recide almost in pure $s$ oritals (i.e. the hydrogen atoms are bonded to almost pure $p$ orbitals). So, as the $s$ character of the orbital where the lone pair is residing increases, it becomes closer and tightly bound to the nucleus, and thus difficult to protonate. Thus the basicity of group  $15$ hydrides decreases down the group, alternatively, the acidity increases.

Coming to the case of  $\ce{PH3 , H2S}$ and  $\ce{HCl}$, two factors govern the acidity of them. One is the  ease of bond-dissociation, and other is the solvation of the medium. When you consider the gas-phase acidity, i.e. no solvation is present, the bond-energy will only be the determining factor and bond energy is given as,  $\ce{P-H < S-H < Cl-H}$ (easily justifiable by  decreasing size of central atom), which tells us in gas phase the acidity of  $\ce{PH3}$ should be the highest. But normally when we talk about acidity, we talk about $pK_a$ in water. Whenever there is a medium, it stabilises the dissociated species via solvation (in water , this is hydration). As. the size of the conjugate base decreases across the period, the extent of hydration energy released will dominate over the bond-dissociation energy. Chloride is the smallest ion among them having higher electronic charge density, and $\Delta H_{hyd} \propto \frac{1}{r} $, It is highest for hydrolysis of  $\ce{HCl}$, making it a stronger acid among three.


Electrolysis of sodium chloride solution

The electrolysis of molten NaCl indeed gives elemental sodium and chlorine gas:

$\ce {2NaCl (l) -> 2Na (l) + Cl2 (g)}$

However, in solution (namely aqueous solution such that you are doing the electrolysis on a solution of NaCl in water), the highly reactive Na produced reacts instantly with water to give sodium hydroxide, hence why no solid Na is ever observed.


Covalent vs Ionic bond in Silicon dioxide

Because silicon and oxygen are nonmetals on the period, hence they form covalent bonds. For ionic bonding, it must be between a metal and a nonmetal.


A stable 3-phase liquid system can be created, as Spießbürger mentioned with three mutually immiscible liquids. The best one that I know of is the organic-aqueous-fluorous 3-phase system.

Fluorocarbons (hydrocarbons in which every hydrogen atom is replaced by a fluorine atom), are interestingly immiscible in both water and common organic solvents. This property has added a dimension to chemical separations using fluorous phase extractions.

If you have three paints, one oil-based, one water-based, and one fluorous-based, then you could separate them easily using a separatory funnel or similar device. You would have to make sure that the pigments in each paint was only significantly soluble in that solvent, which would take experimentation.


If hydrogen bonds are chemical bonds, and these bonds are broken when water changes phases, why doesn't this classify as a "chemical" change?

You must realize that the line between physical and chemical properties is very thin. Most bonds work via the same principle, which has mostly to do with electronegativity. When looked at in reactions it has probably also to do with the according entropy and enthalpy change. Hydrogenbonds are basically bonds based on electronegative attraction between two atoms (which is a simplified few of the reality). A phase change basically when a certain energy point has been reached: since heating op a substance is equal to giving it more kinetic energy (since heat is the same as the movement of individual atoms or molecules). When you reach a certain heat there is too much kinetic energy to form hydrogenbonds the same way it as it did before it changed phase. You probably can agree that molecules or atoms moving quickly is an obvious physical property. When you look at great detail to how hydrogenbonds work, it has something in common with many other way of two objects attracting each other (for example: the formula for the gravitional force between to large heavenly bodies, is very similar to the formula for the electrical force between an electron and a nucleus).

In simple: Many of these chemical properties seem very physical when you look at them in detail. So saying that a certain property should actually be chemical because the changing of this property seems to correspond with the changing of chemical bonds, is not a realistic claim looking at how "physical"/physic based these "chemical" bonds are.


Enthalpy (H) formula: what is the pressure volume product

Enthalpy $H$ is defined so that $\Delta H = q_P$, the heat at constant pressure. This is convenient for constant-pressure calorimetry---it makes it possible to track energy changes without considering volume changes in the system.

$$\begin{align*}\Delta U &= q + w & \text{(first law)}\\
\Delta U_P &= q_P + w_P & \text{(constant pressure)}\\
\Delta U_P &= \Delta H + w_P & \text{(definition of enthalpy)}\\
\end{align*}$$

...so the difference between $\Delta H$ and $\Delta U$ can be interpreted as minus the work for a constant pressure process.

But this doesn't answer your question for a general process. If the process is not constant pressure, we can still apply the definition of enthalpy, but the difference isn't minus the expansion work anymore. But what is it, then? How can we use something we've defined for a constant pressure process for all processes in general?

We'd like a new thermodynamic function $H$ that includes all of the information in $U$. We have $$dU = TdS - PdV$$ so the natural variables of $U$ are $S$ and $V$. We want this new function $H$ to have $P$ as a natural variable in place of $V$, so it will be convenient for constant-pressure calorimetry. Consider the following plot of $U$ vs. $V$:

The equation for the tangent line at some value of $V$ is
$$ U = \left(\frac{\partial U}{\partial V}\right)_S V + intercept$$

If we define the new function as the intercept of the tangent line shown in the graph, it will have the properties we need:

$$H \equiv U - \left(\frac{\partial U}{\partial V}\right)_S V$$

You can see that the derivative $\left(\dfrac{\partial U}{\partial V}\right)_S = -P$ from the equation for $dU$ above. Then

$$H \equiv U - (-P) V = U + PV$$

so

$$dH = dU + d(PV) = TdS - PdV + PdV + VdP = TdS + VdP$$

By adding $PV$ to $U$, we've transformed the information in $U$ (with natural variables $S$ and $V$) to a new function $H$ (with natural variables $S$ and $P$).

This trick is called a Legendre transformation, and it can be used to understand the definitions of other thermodynamic functions like $G$ and $A$ as well.


Why do Ca-Ca distances get shorter in ionic structures, as compared to the metal?

Simply put, the distance is measure from the center of the atoms. In the metal, the uncharged Ca atoms are kept at the van der waals contact distance.

In the ionic compounds the atomic radii of the Ca2+ ions have shrunk due to the loss of the outer shell electrons. Furthermore, negative ions cancel out the repelling charges and allow the Ca2+ ions to pack even closer together.


Which of the following procedures will allow the student to determine the rate constant, for the reaction?

From the graph it is clear that the reaction follows first order kinetics ($\ce{[X] = [X_0] e^{-kt}}$) where $ [X_0]$ is initial concentration. Taking $ln$ on both sides we get
$\ce{ln[X] = ln[X_0] +(-kt)}$
So if you plot a graph between $ln[X]$ and $t$ then you have the magnitude of slope as k ie the equilibrium constant.
So correct answer is A.


How accurately does (pKa1 + pKa2) / 2 estimate the pH of an amphoteric salt?

Not to make duplicate, I will refer to my answer in the mentioned Find the ph and amphiprotic salt is added to water at room temperature question.

Of the buffering

If $\mathrm{p}K_\mathrm{a1}$ is very close to $ \mathrm{p}K_\mathrm{a2}$, then the acidic salt solution is a  good double buffer $$ \ce{H2A/HA-}, \ce{HA^-/A^2-}$$

But if $\mathrm{p}K_\mathrm{a1} \lt\lt \mathrm{p}K_\mathrm{a2}$,  then both buffer systems are far from their $\mathrm{p}K_\mathrm{a}$.

As the consequence, they have  just fraction of the buffer capacity, compared to the case with close $\mathrm pK_\mathrm a$ values.

We have then a mix of 2 inferior buffers.

If e.g.the $\mathrm{p}K_\mathrm{a}$ difference = 4, the buffer capacity is roughly 25 times smaller, compared to $\mathrm pH=\mathrm pK_\mathrm {a1}$ or $\mathrm pH=\mathrm pK_\mathrm {a2}$.

( 2 orders=100, but 2 buffers ->50, and 0.5/0.5 vs 0.99:0.01 --> 25 )


Which karat gold is harder, 24 kt or 18 kt?

18 kt gold is generically harder than 24 kt gold.  Both the site you linked and another site agree on that.  Pure gold really is quite soft, and when you add things to it, it will get harder.

Now, one key to your problem with understanding the hardness of gold and its alloys is that multiple different alloys can be called "18 kt gold".  Different alloys will have different properties and hardness. Interestingly, there is a titanium/gold alloy that is ~4× harder than titanium normally is. The paper - Science Advances 2016, 2 (7), e1600319 - on it looks interesting.

Back to the topic on hand, note in the table below that there are different values for hardness for each carat of gold based on how it has been physically worked and how it has been alloyed.  Also note that all the values listed for 18 kt gold are significantly above that of 24 kt gold.

Source: Total Materia; Gold and Gold Alloys; Table 3: Mechanical Properties of Typical Gold Alloys

It might be worth a separate question of why 9 kt gold might be listed as softer than 18 kt gold, if it has more in the way of impurities.


A good description appears in the book, "Chemistry A Molecular Approach".

In the molecule of hydrogen fluoride, we know that it has a dipole moment, and fluoride is slightly negative. Ignoring this information (electronegativity difference) and we share the bonding electrons equally, the formal charge can be calculated, and both of them are determined to be 0.

In reality, we know that fluoride has the electron from the hydrogen most of the time (due to electronegativity difference). We say the hydrogen has a partial positive charge and the fluoride has a partial negative charge.

From this, it is apparent that the formal charge is different from the charge on the atoms in a molecule.


What is the geometrical structure of OF₂?

Oxygen difluoride ($\ce{OF2}$) is analogous to water ($\ce{OH2}$) both in terms of structure and hybridization.

$\ce{OF2}$ has 4 ligands around the central oxygen, 2 $\ce{O-F}$ bonds and 2 lone pairs.  Just like in water these 4 ligands assume a roughly tetrahedral geometry around the central oxygen. Again like in water, the $\ce{F-O-F}$ angle is less than the perfect tetrahedral angle (109.5°), in this case being around 103°.

Carbon disulphide ($\ce{CS2}$) is linear.  It is analogous to the 3 carbons in allene, in both compounds the central carbon is $\ce{sp}$ hybridized making the $\ce{X=C=X}$ structure linear.


Azo coupling with arenediazonium salts - why isn't nitrogen gas lost?

A simple explanation would be that the displacement of a diazonium group by oxygen to form a phenol or phenyl ether linkage typically requires heating, indicating that it is a slower process than the very fast room temperature coupling reaction at aromatic carbon to form an azo dye. Why? One explanation is that the diazonium cation and the phenol or amine being coupled are a "soft" acid and base respectively, whereas water is a "hard" base. But a substituted phenol or aniline that could not form an azo dye might well form a phenyl ether with a diazonium salt, as you propose.


Is there an algorithm that counts Joback groups in a structure?

So it looks like 'Jobeck groups' are some set of functional groups that you want to find in molecules.

Sounds a lot like substructure search, for which there are many cheminformatics tools (like SMSD, for example) however, since you mention OpenBabel:

http://open-babel.readthedocs.org/en/latest/WritePlugins/AddNewDescriptor.html

talks specifically about 'group contribution methods' and gives a simple example of how to get the counts for a set of functional groups.


Synthesis of methylenecyclohexane from cyclohexylmethanol

Whichever method of elimination you end up preferring, there is no way to directly eliminate the hydroxy group without derivatising it first in some way. The hydroxide ion is a very bad leaving group (but a rather good nucleophile) and transforming a hydroxy group into a better leaving group by protonation ($\ce{R-OH2+}$) requires acidic conditions that do not really allow the (basic) abstraction of a β hydrogen. For each of the simple options:

A hydroxide ion will never leave as part of an $\mathrm{S_N2}$ or $\mathrm{E2}$ reaction; it is too weak a leaving group.
An $\mathrm{E1}$ reaction would require strongly acidic conditions to protonate the hydroxy group; under these conditions the β hydrogen cannot be abstracted. $\mathrm{S_N1}$ would dominate.
An $\mathrm{E1cb}$ mechanism is absolutely impossible. The $\mathrm pK_\mathrm{a}$ value of the β hydrogen is around $50$ while that of the hydroxy group is around $15$. The only conjugate base being formed will be the alcoholate anion.

This means we have to resort to two-step procedures. A number of them have been published in literature in various different contexts. Some bear a name, others don’t. Some follow one of the paradigms mentioned above, others follow a completely different mechanism. Since there is only one elimination product possible, you need not worry about selectivity. I shall present a few possibilities grouped by the mechanistic type where possible but note that this list is not exhaustive and does not even begin to scratch the surface.

$\mathrm{E2}$ type mechanisms

The simplest is the one Orthocresol mentioned in the first comment: transform the alcohol into the corresponding tosylate, e.g. by adding $\ce{TsCl}$ and pyridine, then eliminate with additional potentially stronger base.

A variant of this would be to use mesyl chloride/DBU instead. Depending on how activated your proton is, a simple addition of more DBU to the reaction mixture may suffice.

Considering that your starting material is a primary alcohol, $\mathrm{E1}$-based reactions are very unlikely. Considering again the acidity of the β hydrogen, so are $\mathrm{E1cB}$; no matter which the reaction protocol.

syn-selective eliminations based on cyclic transition states

One example is the ester pyrolysis: use an acid chloride (e.g. acetyl chloride $\ce{AcCl}$) to synthesise an ester, then heat up to initiate thermal rearrangement to the corresponding acid and alkene. The transition state is cyclic, so this reaction is syn-selective.

A similar reaction is the Chugaev elimination. React the alcohol with a base and $\ce{CS2}$, then add an alkyl halide (typically iodomethane) to generate a xantogenate. This also fragmentates thermally to give an alkene and a dithiocarbonate while syn-selectively removing the β hydrogen.

And finally my favourite one: the selenoxide pyrolysis. First synthesise a selenoether with $\ce{NCSe-C6H4-NO2}$ (ortho-nitrophenyl selenocyanate) and $\ce{PMe3}$ in a Mitsunobu reaction, then oxidise the selenoether e.g. by using $\ce{H2O2}$ at room temperature to smoothly give the corresponding alkene — again via a cyclic transition state practically identical to the former two.


Osmium tetroxide reaction with cell membrane

According to the paper The chemical nature of osmium tetroxide fixation and staining of membranes by x-ray photoelectron spectroscopy[1] by White et al. (1976),

  The results support a scheme for the reaction of osmium tetroxide with tissues in which the initial reaction site is the double bonds of unsaturated lipids to form $\ce{Os(VI)}$ derivatives. Subsequent hydrolysis and further reduction yield complexes of $\ce{Os(IV)}$ and $\ce{Os(III)}$. A mixture of these three states is present in membrane specimens during microscopic observation. $\ce{Os(VI)}$ and $\ce{Os(IV)}$ could be present as osmate esters and osmium dioxide, respectively; $\ce{Os(III)}$ could be present as an oxo- or amino complex(es).

However, since $\ce{OsO4}$ reacts with biological material, the above list is probably incomplete.

Riemersma[2] suggests that negatively charged colloidally dispersed osmium dioxide hydrate is deposited at the polar ends of lipids that contain a positively charged (phosphoryl)choline moiety (lecithin, phosphatidylcholines, etc). In this case we would have a pure Coulomb interaction.

References

White DL, Andrews SB, Faller JW, Barrnett RJ, Biochim Biophys Acta, 436(3):577-92 (1976).
Jelle C. Riemersma, J. Histochem. Cytochem, 11, 436-442 (1963)


Is propanol 1-2 diol 178 the same as propylene glycol?

Yes, “propanol 1-2 diol 178” is the same compound as propylene glycol and appears to be a fabricated name used in a manual for a proprietary cleaning system by Kirmuss Audio.
According to the article from Analog Corner magazine [1, p. 22] (emphasis on the footnote):

The Kirmuss Audio KA-RC-1 comes with accessories: 60 mL of an antibacterial/antistatic/antifungal spray; a goat hair brush applicator; an optical-quality microfiber cloth; a larger microfiber cloth (decorated with rabbits) to cover the work space; a circular felt mat to support the record during hand cleaning; a combination carbon fiber/parastatic felt brush; and a bottle of stylus cleaner.
All of this costs $870—far less than at least one competing system and only
slightly more than the iSonic machine and its inferior record-spinning mechanism.
Granted, the accessories are not costly, but no matter how you figure it, Kirmuss doesn’t seem to be in this for the money: He’s not looking to lose money, but clearly he’s not trying to make a killing.
From what I’ve witnessed, he’s more concerned with waging a worldwide war against fungus.
The antibacterial/antistatic spray is identified in the manual as a “98-99% distilled water, 1-2% propanol 1-2 diol 178 mix.” That last one is propylene glycol,4 an organic solvent used in, among other things, pharmaceutical preparations and as a food additive.

4 That chemical name seems made-up, and Kirmuss at first disputed that the ingredient was nothing more than propylene glycol, but a subsequent correspondence made it clear.

Reference

Michael Fremer. The Seven “M”s of the Charles Kirmuss Vinyl Restoration System. Analog Corner, July 2019. (PDF, accessed 2020-09-03)


If you don't have any idea:

Start with the hypothesis that your starting material is the salt of some metal.
Some of the test are apparently useful to identify the cation, others tackle the anion.
Is it likely that the flame-extinguishing gas comes from the cation?
If it is not the cation, it must be the anion from which the gas is formed.
Think in the main property of diluted nitric acid. Is it an oxidant in diluted form, or just something else?

I almost completely agree with the solution given in the question, except for the last step.

The three known lead oxides are

$\ce{PbO2}$: dark brown to black, strong oxidant
$\ce{Pb3O4}$: red-orange, used as a rust-proof primer (in German: Bleimenninge)
$\ce{PbO}$: (normally) yellow

When dark $\ce{PbO2}$ is heated, it splits off oxygen to form red $\ce{Pb3O4}$, when it's heated up more, more oxygen is released and yellow $\ce{PbO}$ is formed. Properly done in a quartz tube the colours of the German flag can be produced this way.

In the puzzle, we are not starting with $\ce{PbO2}$ but with $\ce{PbCO3}$.

When $\ce{PbCO3}$(= G) is heated, $\ce{CO2}$ (= Y) is released. The questions mentions a hard glass tube, which suggests that the tube has to withstand high temperatures.

Let's assume that the reaction simply is $\ce{PbCO3 ->[\Delta] PbO + CO2 ^}$

A possible explanation for the colour change would be that at high temperatures, red $\ce{\alpha-PbO}$ is initially formed, which converts to yellow $\ce{\beta-PbO}$ upon cooling.


Hydrolysis of (chloromethyl)cyclopropane: Is but-3-en-1-ol a possible product?

I'd say, yes it is possible. The pioneering work on interconversion reactions of cyclobutyl, cyclopropylcarbinyl, and allylcarbinyl derivatives has been done by John D. Roberts and coworkers (Ref.1). In these works, they have checked reactions under both $\mathrm{S_N1}$ and $\mathrm{S_N2}$ conditions. They have observed the reactions expected to involve carbonium ion intermediates have shown interconversion, which have been correlated to factors such as carbonium ion stability, ease of interconversion, and stability of possible products. One such interconversion have been shown in the reaction of cyclopropylcarbinyl-,  cyclobutyl-, and allylcarbinylamines with nitrous acid, which gave relevant carbonium ions. Similar approach was recently taken by Arca, et al. 2014 using solid catalysts (Ref.2). They have proposed a bicyclobutonium cation transition state to explain the product formation:

The coorelation to liquid phase reaction is well explanied in their abstract:

Rearrangement of cyclopropylcarbinyl bromide over proton and ammonium-exchanged mordenite ($\ce{H}$-MOR and $\ce{NH4}$-MOR) was studied at different temperatures. The product distribution analysis revealed a similar profile to what is found in solution, with prevalence of cyclobutyl over allylcarbinyl bromide. The activation parameters showed that compared with 80% aqueous ethanol solution, the zeolites present lower enthalpy of activation and higher entropy of activation, but yielding a significantly lower $\Delta G^‡_{298}$. The results may be explained in terms of the ability of zeolites to serve as solid solvents, providing a polar nanoenvironment for ionic reactions to occur.

Therefore, we can assume the reaction in hand have also unedergone with similar mechanism, giving same three products, $\bf{1}$, $\bf{2}$, and $\bf{3}$ (probably in different ratios). The formation of  $\bf{3}$ can also explained by the possible resonance contribution as depicted in above $\color{red}{\text{Red Box}}$. Note that two open resonance structures are essentially the same, but shown because it would make a double contribution due to the symmetry.
It's noteworthy that the open-chain derivatives have been achieved from the reaction of cyclopropylcarbinyl halides with amines (Ref.3).
References:

John D. Roberts, Robert H. Mazur, “Small-Ring Compounds. IV. Interconversion Reactions of Cyclobutyl, Cyclopropylcarbinyl and Allylcarbinyl Derivatives,” J. Am. Chem. Soc. 1951, 73(6), 2509–2520 (https://doi.org/10.1021/ja01150a029).
Hugo A. Arca, Gabriel C. C. Gomesab, Claudio J. A. Mota, “Solid solvents: activation parameters for the rearrangement of cyclopropylcarbinyl bromide on mordenite zeolite,” New Journal of Chemistry 2014, 38(7), 2760-2762 (https://doi.org/10.1039/C4NJ00287C).
Michael B. Smith, Robert T. Hrubiec, Charles A. Zezza, “Reaction of amines with cyclopropylcarbinyl halides: $\mathrm{S_N2'}$ or solvolysis?,” J. Org. Chem. 1985, 50(24), 4815–4821 (https://doi.org/10.1021/jo00224a033).


Why do single, double and triple bonds repel each other equal amounts?

The way you have learnt to predict molecular geometries is called the VSEPR theory. As all theories, it makes assumptions and has a certain number of limitations. The issue you are asking about is one such limitation.
When predicting the geometry of molecules with VSEPR, one should treat multiple bonds as a single entity. However, the geometry obtained is just a first approximation, or “idealized” geometry. In fact, multiple bonds induce distortions from this ideal geometry, pretty much like lone pairs do (I assume you know that already. If not, well, check the above-linked Wikipedia page for “lone pair”.) Citing from this online course:

Multiple bonds contain a higher electronic-charge density than do single bonds, so multiple bonds also represent larger electron domains ("fatter balloons"). Consider the structure of phosgene, Cl2CO, which is shown below.

From the Lewis structure of phosgene, we might expect a trigonal planar geometry with 120°-bond angles. However, the double bond seems to act much like a nonbonding pair of electrons, reducing the ClCCl bond angle from 120° to 111°. In general, electron domains for multiple bonds exert a greater repulsive force on adjacent electron domains than do single bonds.

So: yes, lone pairs and multiple bonds give rise to deviations from the ideal VSEPR geometries. It's nice that you figured it out on your own!


Completing inorganic reactions and writing spectator ions

For B, you have correctly identified calcium phosphate as the precipitate, as it is the least soluble of all possible compounds. In the case of A, cobalt phosphate precipitates from the solution containing cobalt(II) and phosphate salts.

$$\ce{3Ca^{2+} + 2PO4^{3-} ->~ Ca3(PO4)2\downarrow}$$
$$\ce{3Co^{2+} + 2PO4^{3-} ->~ Co3(PO4)2\downarrow}$$


Why was atomic mass scale changed from Oxygen - 16 to Carbon - 12?

The mass scale has changed over time, largely due to different isotopes of the "baseline." Not surprisingly, there's a good Wikipedia article on the matter.

  In the 20th century, until the 1960s chemists and physicists used two different atomic-mass scales. The chemists used a "atomic mass unit" (amu) scale such that the natural mixture of oxygen isotopes had an atomic mass 16, while the physicists assigned the same number 16 to only the atomic mass of the most common oxygen isotope (O-16, containing eight protons and eight neutrons). However, because oxygen-17 and oxygen-18 are also present in natural oxygen this led to two different tables of atomic mass. The unified scale based on carbon-12, $\ce{^12C}$, met the physicists' need to base the scale on a pure isotope, while being numerically close to the chemists' scale.

In short, Dalton suggested $\ce{^1H}$ as the basis of the mass scale, but Ostwald pushed later for $\ce{^16O}$. Unfortunately, no one knew about isotopes yet.

The problem was that physics stuck to isotopically pure O-16 for their mass scale, and chemistry used "oxygen" (i.e., with some O-17 and O-18 present). There's 0.028% difference between those scales. That may not sound like much, but with molecules of ~200-500 amu, it starts to add up.

Using C-12, it's isotopically pure (making physics happy) very close to the previous chemistry "O-16 + O-17 + O-18" scale.

From IUPAC:

  In April 1957 at the bar in the Hotel Krasnapolski in Amsterdam, Nier suggested to Mattauch that the $\ce{^12C}$ = 12 mass scale be adopted because of carbon's use as a secondary standard in mass spectrometry. Also, $\ce{^12C}$ = 12 implied acceptable relative changes in the atomic weight scale, i.e., 42 parts-per-million (ppm) compared to 275 ppm for the $\ce{^16O}$ = 16 scale (which would not acceptable to chemists). Enthusiastically, Mattauch made a worldwide effort in the late 1950s to publicize the $\ce{^12C}$ = 12 scale and obtain the physicist's approval, while Wichers obtained the chemist's approval.


sorry that was my mistake. the correct mechanism is this


Why does hydration of alkene takes place in presence of an acid such as H3PO4 or H2SO4?

The acid is used as a catalyst. As can be seen from the mechanism below, the electron-rich double bond first attacks the positively-charged acid. The positively-charged hydronium ion is important because it is rather electrophilic and allows the nucleophilic pi bond to attack it, breaking the pi bond. This serves to initiate the reaction. Without an acid catalyst, the neutral water molecule would not be electrophilic enough for the pi bond to want to attack it, thus the reaction would not proceed.

(from http://slideplayer.com/slide/11423361/)


Predict the number and nature of bonds using Molecular Orbital Theory

I presume you can draw the homonuclear diatomic MO diagram for these two species.

As you probably already know, the s orbitals of the carbons form bonding and antibonding orbitals, both of which are filled. (The net effect of this is the “lone pairs” on each carbon.)

The p orbitals for 2 pi bonds and a sigma bond. Due to s-p mixing, the pi bonds are lower in energy and are filled first, and these 2 bonds (with their 4 electrons) are the 2 bonds we see in C2.

Add another 2 electrons to the system and you realise they fill the sigma bonding orbital, which means that the C2 2- anion has 2 pi bonds and a sigma bond (for a total bond order of 3).

Hope this helped.


Why does chloromethane have a larger dipole moment than chloroform?

Your conclusion is built on the assumption that the bond lengths will be constant. A bond’s dipole moment varies with bond length and thus for the vector addition to give the identical product, all $\ce{C-Cl}$ and all $\ce{C-H}$ bonds would have to be identical. They’re probably not.

Sutton and Brockway studied the bond lengths in chloromethane, dichloromethane and chloroform back in 1935 by electron diffraction.[1] Their results indicate that the $\ce{C-Cl}$ bond lengths don’t vary between the compounds in question as indicated in the table below. (Note that all chlorine atoms in each molecule are homotopic and thus equal bond lengths are theoretically expected.)

$$\begin{array}{lccc}\hline
\text{Compound} & \ce{CH3Cl} & \ce{CH2Cl2} & \ce{CHCl3}\\ \hline
d(\ce{C-Cl})/\mathrm{pm} & 177 & 177\phantom{^\circ} & 178\phantom{^\circ} \\
\angle(\ce{Cl-C-Cl}) & \text{n. a.} & 111^\circ & 111^\circ \\ \hline\end{array}$$

These results point towards the chlorine side of things not having an impact. But there is still the hydrogen side of things. Hydrogen atoms are generally very hard to locate in diffraction type experiments. X ray diffraction is observed if photons interact with electrons in a certain manner; hydrogen atoms do not have many electrons around them and are thus hardly seen. One needs to observe diffraction at very large angles to be able to locate hydrogens. I assume, although I don’t know, that it’s similar in electron diffraction. Indeed, the paper notes:

  […] and $l_{\ce{C-H}}$ was taken as $1.06~\mathrm{\overset{\circ}{A}}$.

The two carbons in chloromethane and chloroform are, however, very much not identical. One is surrounded by three electronegative chlorine atoms which will strongly remove electron density from it, polarising the $\ce{C-Cl}$ bonds towards each of the three chlorines. The other will have only one such withdrawing partner. The different electron densities around chlorine are well represented by the corresponding $\ce{^13C}$ NMR shifts: $\delta(\ce{CH3Cl}) \approx 22~\mathrm{ppm}$; $\delta(\ce{CHCl3}) \approx 77~\mathrm{ppm}$.[2] The hydrogen shifts are also different in similar magnitudes, showing that hydrogen in chloroform is much more polarised than in chloromethane.

These different electronics must have an influence on the $\ce{C-H}$ bond length of the two molecules. I can’t imagine two electronically such different systems to have equal bond lengths. My assumption is that $\ce{CH3Cl}$ should have the shorter $\ce{C-H}$ bond lengths, since carbon is more negatively charged while hydrogen remains positively charged; this should give a slightly greater attractive force and hydrogen, being such a small element, immediately feels these. Therefore, instead of the final dipole vector $\vec d$ being equal in both cases ($\vec d = \vec x + \vec y$), we have one lesser and one stronger vector.

$$\begin{gather}\vec d_{\ce{CH3Cl}} = \vec x + \vec y + \delta\vec y\tag{1}\\
\vec d_{\ce{CHCl3}} = \vec x + \vec y - \delta\vec y\tag{2}\end{gather}$$

References:

[1]: L. E. Sutton, L. O. Brockway, J. Am. Chem. Soc. 1935, 57, 473. DOI: 10.1021/ja01306a026.

[2]: Experimental spectra found for chloromethane/chloroform on SciFinder.


What does the first $1$ mean in $\ce{Li}_2 1 ^1\Sigma_g^+: (1s\sigma_g)^2(1s \sigma_u)^2(2s \sigma_g)^2$?

The conventional labelling of molecular states starts at X for the lowest state and A, B, C for higher ones and a, b, c, for equivalent states of the same symmetry but different multiplicity; see Herzberg spectroscopy books for many examples. These labels, however, reflect the order in which the states were discovered and so do not reflect their relative energies. It is becoming more common to precede a molecular state's symmetry label by an integer e.g. $1~^1\Sigma _g^+,~ 2~^1\Sigma _g^+,~3~^1\Sigma _g^+$ to reflect the energy ordering of states with the same symmetry. This notation reflects the way atomic levels of the same symmetry are labelled as $1$s, $2$s, $3$s, etc.

As an example, the conventional notation for the first four electronic states of  $\ce{C2}$ in order of energy is

X $^3\Pi_u$ ;  a $^1\Sigma_g^+$ ; b $^1\Pi_u$ ; A $^3\Pi_g$

which becomes

$1~ ^3\Pi_u$ ;  $2 ~^1\Sigma_g^+$ ; $1 ^1\Pi_u$ ; $3~ ^3\Pi_g$

In atoms the notation for a set of transitions starting from the 3p level in sodium is  Na $3~ ^2P_{3/2}$ .


How can electrons manage to be at the same distance from the nucleus at all positions in p, d and f orbitals?

I think you're confusing what the orbital shapes represent. The shapes are probability distributions-which means that you're most likely to find the electron somewhere within the shape.

  Electrons can only be at a fixed distance from a nucleus

This is false

  which is determined by n quantum number

The n quantum number tells you how big the orbital is-not precisely how far away the electron is.

In reality, the electron is moving around the atom, and is probably somewhere within the orbital shape.


Are the elements lanthanum and actinium considered to be in the d block or the f block of the periodic table?

Lanthanum and Actinium are generally both accepted as f-block elements. You will notice in the Royal Society's periodic table, La and Ac are both color coded to match the other f-block elements. They do this to show the beginning of both the Lanthanide and Actinide series, while maintaining a reasonable symmetry for their periodic table (that is, without "holes" in their table, similar to the first link you provided).


Second order reaction with different reactants

The general solution, if the reaction is $aA+bB \to product$, is to allow an amount $x$ to react at time $t$ then there will be concentrations $(A_0-ax)$ and $(B_0-bx)$ of species A and B respectively at time $t$ if $A_0,B_0$ are the concentrations at time zero.  The rate equation is $\displaystyle  -\frac{1}{a}\frac{d(A_0-ax)}{dt}=\frac{dx}{dt}=k(A_0-ax)(B_0-bx)$ which can be integrated ($0\to x, 0\to t)$ to give

$$ \ln\left(\frac{A_0-ax}{B_0-bx} \right)=k(A_0b-B_0a)t +\ln\left(\frac{A_0}{B_0} \right)$$

If the reaction follows this scheme plot the left hand side vs $t$ and this should produce a straight line of slope $k(A_0b-B_0a)$.

Should it happen that concentrations are initially $A_0/B_0 = a/b$ so that species are present in stoichiometric amounts then as $A_0b-B_0a=0$ the  solution above cannot be used. In this case the rate equation becomes $\displaystyle  \frac{dA}{dt}=-bkA^2$ which gives $\displaystyle \frac{1}{A}-\frac{1}{A_0}=bkt$.


What is the region of this wavelength (462 nm)?

Light with a wavelength of 462nm is in the visible light region (blue light visible by the human eye).

For certain, your lecturer was wrong that the light is in the infrared region, this occurs at much higher wavelengths closer to 700nm. More possibly he meant to say ultraviolet, which occurs somewhere below 400nm (shorter wavelengths), though even this isn't technically correct.

The Wikipedia article gives a fairly good overview of the various regions, along with a schematic describing where they come:


Usage of ammine vs amine in nomenclature

(Will do more research into ammines come February.)

Premise

Ammonia is spelled with two 'm's. The more natural derivative of the word in a linguistic sense would also have the same number 'm's. Thus whichever predates the other, ammine complexes or amines, would claim the throne.

In a structural sense, when awknowledgement of the first ammine complex came it was literally in a solution of ammonia (see more below).

Etymology of ammonia and amine

  Our travels begin in sunny Ancient Egypt. One important deity of the time was Amun, the god of wind and air along with his wife-in-mythology Amaunet. At the beginning of the Middle Kingdom of Egypt (c. $2055-$$1650$ BC), he was annointed patron saint of Thebes.  Later Amon become part of a trinity, along with gods Re and Ptah. This coincided with Thebes restored as capital city due to pharaoh Ahmose I (c. $1539-$$1514$ BC).

  Eventually Amon would become the god of gods. This is when he also goes by Amon-Re (or Ra), after uniting with sun god Re.$^{\text{[a]}}$

  Egypt's glory shone far beyond its borders. Ancient Greeks adopted their own version of Amon. There he would be called Ammon, with two 'm's.

As it so happens, near one of Amun's temples in then-Egypt is a depository of $\ce{NH4Cl}.$ From this háls ammôniakós (Greek for 'salt of Amun') we can easily extract $\ce{NH3}.$

$$\ce{NH4Cl(s) ->[t^\circ] NH3(g) +HCl(g)}.$$

It makes sense for the option with two 'm's to outdate amine. Etymonline agrees (in ~ English):

  ammonia (n.) $-$ $1799$, modern Latin, coined $1782$ by Swedish chemist Torbern Bergman ($1735$$-$$1784$) for gas obtained from sal ammoniac [----];

  amine (n.) $-$ "compound in which one of the hydrogen atoms of ammonia is replaced by a hydrocarbon radical," $1863$, from ammonia + chemical suffix -ine (2).

Note: has been expanded upon under 'Amine in chemical literature' further down.

The why

Why ammine for coordination compounds and not amine? It is probably because the study of metal ammines predates any knowledge of amines. Indeed, in $1798$ Tassaert$^{[1]}$ reported on 'ammoniacal solutions of cobalt(II) chloride [which when] allowed to stand overnight yield an orange-colored crystalline product containing six molecules of ammonia'$^{[2]}$. The compound in question was $\ce{CoCl3*6NH3}$. These and similar compounds were dubbed metal ammines$^{\text{[b]}}$.

Note that at the end of  $18^{\text{th}}$ century organic chemistry had not yet developed. The crucial and accidental evidence against vitalism$^{\text{[c]}}$ came in $1828$ when the German chemist Wöhler artificially synthesised an organic compound $-$ urea $-$ an amide with two amine qroups!

$$\ce{AgNCO + NH4Cl-> AgCl + NH4NCO \\ NH4NCO ->[t^\circ]NH_3 + HNCO <=>\underset{urea}{CO(NH2)2}}$$

First pivotal trace of amines pops up in $1842$, $44$ years after Tassaert's discovery of coordination chemistry. Russian chemist N. N. Zinin uses dihydrogen sulfide to reduce nitrobenzene$^{[3]}$$^{\text{[d]}}$:

$$\ce{C_6H5NO2 + 3H2S ->[NH3]\underset{aniline}{C6H5NH2} + 3S + 2H2O}.$$

On Zinin's research, co-founder of amine chemistry A. W. von Hofmann declares$^{[4]}$,

  If Zinin had done nothing more than to convert nitrobenzene into aniline, even then his name should be inscribed in golden letters in the history of chemistry.

Aliphatic amines were simultaneously and independently synthesised in $1849$ by chemists Wurtz of France and Hofmann of Germany. Both could be credited with the discovery of ethylamine. Hofmann's method allowed for the additional preparation of secondary and tertiary amines.$^{[3]}$

Wurtz treated ethyl isocyanate with potassium hydroxide; Hofmann applied heat to solutions of alkyl halogenides and ammonia.$^{[3]}$

$$\tag{Wurtz}\ce{C2H5N=CO + H2O->[KOH]C2H5NH2 + CO2}$$
$$\tag{Hofmann, I}\ce{R-I + NH3->[t^\circ]R-NH2*HI}$$
$$\tag{Hofmann, II}\ce{2R-I + NH3->[t^\circ]R2-NH*HI + HI}$$
$$\tag{Hofmann, III}\ce{3R-I + NH3->[t^\circ]R-N*HI + 2HI}$$

For quaternary amines, Hofmann used the following scheme:
$$\tag{Hofmann, IV}\ce{R3N + RI->R4-N*I}.^{[3]}$$

But because ammine was already reserved, they went with amine.$^{[5]}$ Furthermore, if you think about it, the naming is rather successful. Well, at least for primary amines.$^{\text{[e]}\ \text{[f]}}$

$$\text{metal ammines contain the complete}\ \ce{NH3->} \text{keep the}\ m\\ \text{pr. amines have one less hydrogen or}\ \ce{-NH2->} \text{lose the}\ m$$

Amine in chemical literature

In $1834$, German chemist Liebig optained a dirty white or buff-coloured amorphous substance by heating ammonium thiocyanate.$^{[5]\ [6]\ [7]\ [8]\ [9]\ [10]\ [11]}$

$$\tag{$\alpha$}\ce{KSCN + NH4Cl ->[t^\circ] NH4SCN + KCl}$$
$$\tag{$\beta$}\ce{NH4SCN <=> \underset{thiourea as thione}{S=C(NH2)2} <=>[t^\circ] \underset{thiourea as thiol}{HS-C(NH)NH2}}$$
$$\tag{$\gamma$}\ce{\underset{thiourea as thiol}{HS-C(NH)NH2} ->[t^\circ] NH3 + HNCS \\ \underset{thiourea as thiol}{HS-C(NH)NH2} ->[t^\circ] H2S\ \ \ + \underset{cyanamide as aminonitrile}{N#C-NH2}}$$
$$\tag{$\delta$}\ce{\underset{cyanamide as aminonitrile}{N#C-NH2} <=> \underset{cyanamide as diimine}{HN=C=NH}}$$

He named the product melam without any admitted etymology other than his own fantasy. Indeed, Liebig mentions in his original paper$^{[5]\ [6]}$,

  [These designations] are, if you like, grasped from the air (i.e. pure inventions), and serve the purpose just as well as if they were derived from the colour or one of the properties.

Liebig proceeded to boil melam with a solution of potassium hydroxide and ammonia. This yielded melamine.$^{[5]\ [6]\ [7]}$

Neither the structure of melam nor melamine were known to Liebig at the time.$^{[5]}$ Melamin(e) was most likely derived by Liebig as melam $+$ -in(e).$^{[5]}$$^\text{[g]}$

A few years before $1849$, Liebig had predicted the existance of amines.$^{[5]}$$^\text{[h]}$ Liebig called them compound-ammonias$^{[5]}$. The word amine itself was used already in $1853$$^{[12]}$ in the works of French chemist Gerhardt. Often times this was put in parentheses as an alternative name. For example$^{[12]}$,

  méthyl-ammoniaque (méthylamine),
  amyl-ammoniaque (amyl-amine).

Acceptance of the word amine was not immediate. Besides having to distinguish between amines and metal ammines, amides were too an obstacle. The restriction to retain amine for ammonia [where any] hydrogen atom [has been] successively replaced by hydrocarbon radicals was generally accepted in $1863$.$^{[5]}$

TL; DR

$$\text{Amun in Greek}+\text{nearby }\ce{NH4Cl}\ \text{depository}\ce{->[Tassaert dibs on ammine in 1798][\text{vitalism until a.l. 1828}]}\text{amine in 1863}$$

Extra information

$\text{[a]}$ According to some texts, the unification as Amon-Ra goes back to the reign of Mentuhotep II ($2055-$$2004$ BC). Amon's (or Amon-Ra's) importance declined, then increased again during Amenemhet I ($1985-$$1956$ BC), fell and rose to national prominence with Ahmose I.$^{[13]}$
$\text{[b]}$ Most of the early work was done with ammonia, and thus the resulting class of complexes were already then known as metal ammines$^{[14]}$. Developments in naming specific compounds are summarised in this diagram:

$$\text{after discoverer}\ce{->[\text{m.c.l.}]}\text{basis on colour}\ce{->[\text{m.c.l.}]}\text{count number of ammonias only}\ce{->[\text{m.c.l.}][Werner's coordination theory 1893]}\text{premodern}$$

where $\text{m.c.l.}$ is short for many compounds later. Some examples of such practice is included the table$^{[14]}$ below.

\begin{array}{|c|c|c|c|}
\hline
\mathbf{Complex} & \mathbf{Colour} & \mathbf{Name} & \mathbf{Present}\ \mathbf{formulation}\\
\hline
\ce{CoCl3*6NH3} & \text{yellow} & \color{gold}{\text{luteo}}\text{cobaltic chloride} & \ce{[Co(NH3)6]Cl3} \\
\ce{CoCl3*5NH3} & \text{purple} & \color{purple}{\text{purpure}}\text{ocobaltic chloride} & \ce{[CoCl(NH3)5]Cl2} \\
\ce{CoCl3*4NH3} & \text{green} & \color{green}{\text{praseo}}\text{cobaltic chloride} & trans\text{-}\ce{[CoCl2(NH3)4]Cl} \\
\ce{CoCl3*4NH3} & \text{violet} & \color{violet}{\text{violeo}}\text{cobaltic chloride} & cis\text{-}\ce{[CoCl2(NH3)4]Cl}  \\
\ce{CoCl3*5NH3*H2O} & \text{red} & \color{pink}{\text{roseo}}\text{cobaltic chloride} & \ce{[Co(NH3)5(OH2)]Cl3} \\
\ce{IrCl3*6NH3} & \text{white} & \color{gold}{\text{luteo}}\text{luteoiridium chloride} & \ce{[Ir(NH3)6]Cl3} \\
\hline
\end{array}

An attentful reader ought to notice that $\ce{CoCl3*6NH3}$ is yellow and $\ce{IrCl3*6NH3}$ is white. But they both have the same prefix -luteo. At first it did indeed mean yellow but later the same scheme was adopted to notate an equal number of ammonias. Crazy but it is true!$^{[14]}$ You can probably see that proper nomenclature could not get there soon enough.
$\text[c]$ In chemistry, vitalism was the idea that compounds found inside living things are fundamentally different from inorganic chemistry. Such organic molecules were believed to be impossible to synthesise outside an organism.

To captivate how little of organic chemistry was understood in the first half of the 18th century, Wöhler stated in $1835$$^{[5]}$,

  [Organic chemistry is] like a primaeval forest of the tropics, full of the most remarkable things.

Proponents of vitalism had a hard time accepting the evidence. They      argued that the process was achievable since carbamide is simply a waste product. However, luckily notable progress followed.$^{[15]}$

M. Berthelot $-$ natural fat ($1854$ doctoral thesis)
H. Kolbe, E. Frankland $-$ acedic acid ($1861$)
A. Butlerov $-$ sugars ($1861$, see Formose reaction)

${\text{[d]}}$ Aniline had been successfully separated in $1826$ by German merchant and chemist O. Underderben. Note, however, that this was not a synthesis $-$ he used dry destillation of indigo.$^{[5]\ [16]}$
${[\text{e}]}$ This was very much intentional. Wurtz recognised in $1849$ that methylamine and ethylamine could be regarded as ammonia in which one equivalent of hydrogen is replaced by methyl or ethyl.$^{[5]}$
$\text{[f]}$ Classification of primary, secondary, and tertiary amines was introduced by Gerhardt in $1856$.$^{[17]}$
$\text{[g]}$ Some authorities$^{[18]}$ claim melamine as having come from mel(am) $+$ amine. This is wrong since amines were unknown at that time.$^{[5]}$
$\text{[h]}$ From Zinin to Liebig, amines were probably grouped as alcaloïdes artificiels (French for artificial alcaloids) because the structure had not yet been proposed.$^{[19]}$

Refrences and bibliography

$[1]$ B. M. Tassaert, Ann chim. phys., 28, 92 (1798).
$[2]$ Fred Basolo, Ralph G. Pearson. Mechanisms of Inorganic Reactions. (1958). (page 2)
$[3]$ Michele Giua. Storia della Chimica. (1962) (pages 342$-$344)
$[4]$ A. W. von Hofmann. Ber. Deut. chem. Ges. (1880). 13,. 449$-$450.
$[5]$ W. E. Flood. The Origins of Chemical Names (1963) (pages xxii, 33$-$35, 37$-$38, 137$-$138)
$[6]$ J. von Liebig. 'Uber einige Stickstoff $-$ Verbindungen'. Justus Liebigs Annalen der Chemie, 10, 1, 1$-$47 (1834).
$[7]$ Bernard Bann, Samuel A. Miller. 'Melamine And Derivatives Of Melamine'. Chemical Reviews, 58 (1), 131$–$172. (1958)
$[8]$ Klaus Bretterbauer, Clemens Schwarzinger. 'Melamine Derivatives – A Review on Synthesis and Application'. Current Organic Synthesis, 9, 342$-$356 (2012).
$[9]$ Michihiro Ohta, Shinji Hirai, Hisanaga Kato, Vladimir V. Sokolov, Vladimir V. Bakovets. 'Thermal Decomposition of $\ce{NH4SCN}$ for Preparation of $\ce{Ln2S3}$ ($\ce{Ln}$ = $\ce{La}$ and $\ce{Gd}$) by Sulfurization'. Materials Transactions, vol. 50, No. 7, 1885$-$1889 (2009).
$[10]$ Zerong Daniel Wang, Motoko Yoshida, Ben George. 'Theoretical study on the thermal decomposition of thiourea'. Computational and Theoretical Chemistry, vol. 1017, 91$–$98 (2013).
$[11]$ A. Kawasaki, Y. Ogata. 'Kinetics of the formation of melamine from dicyandiamide'. Tetrahedron, vol. 22, 1267$-$1274 (1965).
$[12]$ M. Charles Gerhardt. Traité de chimie organique. First edition. (1853) (pages 8, 134, 210$-$211, 277, 396$-$397, 427, 463, 545, 551, 611, 613, 616$-$619)
$[13]$ Seppo Zetterberg. Maailma ajalugu. (2015). (pages 41$-$43)
$[14]$ Fred Basolo, Ronald Johnson. Coordination Chemistry. (1964) (pages 4$-$13)
$[15]$ H. Karik. Üldine keemia. Handbook for students. (1987) (page 223)
$[16]$ Alex Nickon, Ernest F. Silversmith. Organic Chemistry: The Name Game. (1987) (page 305)
$[17]$ M. Charles Gerhardt. Traité de chimie organique. Fourth edition. (1856) (page 592)
$[18]$ The American Heritage Dictionary of the English Language: Fourth Edition. 2000. https://web.archive.org/web/20081201105219/http://www.bartleby.com:80/61/24/M0202400.html (Jan 3, 2017)
$[19]$ Encyclopædia Universalis. 'Amines'. http://www.universalis.fr/encyclopedie/amines/ (Jan 2, 2017)


Optical isomerism regarding diastereomers

Some definitions (by IUPAC):

Diastereomer: A stereoisomer that isn't an enantiomer.
Stereoisomer: An isomer that has the same bond connectivity, but arises due to differences in the spatial arrangements of atoms.
Enantiomer: Non-superimposable mirror image.

So, diastereomers do not necessarily have to be optically active. Consider cis- and trans-1,2-difluoroethene.

They are stereoisomers (connectivity is the same, but the fluorines are on opposite sides) which are not enantiomers (they aren't mirror images of each other), which makes them diastereomers.

Neither are chiral because they are superimposable on their respective mirror images. Equivalently, the presence of the plane of symmetry (which is the plane of the molecule) makes both isomers achiral.


Does the Dieckmann Reaction proceed via the conjugated enolate of the product?

The first four steps (i.e. the main reaction) are conducted under basic conditions due to the presence of alkoxide. When that reaction is complete, the result is the deprotonated ketoester--the formation of that stable species is what drives the reaction to completion. Note that the first 3 steps are shown as equilibria, whereas the fourth step is essentially irreversible.

Later, during workup (shown above as the fifth step of the process), acid is added to neutralize that anion and form the final product.


Catalyst that combines oxygen and hydrogen

Yes

Not only is this possible it can be done while extracting electricity from the reaction. This is the key reaction in a hydrogen/oxygen fuel cell.

Fuel cells have been known for well over a century and Hydrogen oxygen fuel cells have been used commercially since the 1930s. Platinum is often used as a catalyst for the key reaction in the cells.


What an electrolyte "really" does in electrolysis?

Sometimes the electrolyte directly takes part in the chemistry, by being oxidized or reduced at an electrode. In those cases, it is easy to understand why the concentration would have an influence on the kinetics of the redox reactions (and with that, on the observed electrical current).
For water electrolysis, the "active" electrolytes are hydrogen ions and hydroxide ions. If you change the pH, you change their concentration, and would expect to affect the rate of reaction. You are asking about other electrolytes such as sodium sulfate, which don't participate in electron transfer.
To explain their role, take a look at this animated GIF. It shows electrolysis of water in the presence of universal indicator and sodium sulfate:

Source: CarolinaBiological
With the 9 volts applied to the graphite electrodes, you can see that the electrolysis causes local changes in pH (because one half reaction consumes hydrogen ions and the other hydroxide ions). It takes quite a bit of effort to mix the solution (with the plastic tube) to get the pH homogeneous again. This shows you that the diffusion of hydrogen and hydroxide ions takes a while. In the absence of other electrolytes, this would lead to local charge accumulation as well, stopping the reaction. With the electrolyte (at much higher concentration than the hydrogen and hydroxide ion), we get a pH gradient only.
This pH gradient also slows things down, and increases the voltage that needs to be applied (cell potential is 1.3 V at constant pH, but according to Nernst law, it would increase because of the pH differences). Because we are applying 9 volts, this still works.

What an electrolyte “really” does in electrolysis?

By moving in the appropriate directions, the ions of the "passive" electrolyte prevent large accumulations of charge, and allow the electrolysis to proceed at higher rates than without their presence. If you decrease the gap between electrodes down to sub-Debye length, they are no longer neccessary.


What is a typical monolayer formation time for liquid phase adsorbate on a solid surface?

Depends on concentration and other parameters. For thiols on gold, for example, a typical concentration is 1-5 mM, and formation time is 24-48 hours; "Longer assembly times tend to result in better monolayer packing." (ref)


What is the mathematical reason behind the gradual increase in osmotic pressure?

That formula for the osmotic pressure is known as the van 't Hoff law. It is described in some detail at the Wikipedia page for osmotic pressure, which also includes a brief (albeit incomplete and rather sloppy) derivation. To reach the van't Hoff law

$$\Pi = cRT$$

you have to approximate $\ln x_v$ by a Taylor series $(\ln x_v = \ln (1 - x_\mathrm{solute}) \approx -x_\mathrm{solute}$ which holds for small values of $x_\mathrm{solute})$ to get

$$\Pi = \frac{x_\mathrm{solute}RT}{V_\mathrm{m}}$$

The Wikipedia page uses $V$ to refer to the molar volume of the solvent (which should really be represented by $V_\mathrm{m}$).[1] In any case, $x_\mathrm{solute} = n_\mathrm{solute}/n_\mathrm{tot}$ and $V_\mathrm{m} \approx  V/n_\mathrm{tot}$ such that

$$\frac{x_\mathrm{solute}}{V_\mathrm{m}} \approx \frac{n_\mathrm{solute}}{V} = c_\mathrm{solute}$$

and if the solute is a 1:1 electrolyte, such as $\ce{NaCl}$, then you have to multiply by two to account for the total number of solute particles.

I am not sure about the term "effective concentration" (I personally have never seen it, and if I had to, I would probably write something like the "total concentration of solute particles") but the logic used is correct as long as you are in the regime where the van 't Hoff law is applicable i.e. very dilute solvent.

Lastly you should probably use $<$ instead of $>$ in your ordering to be clear.

[1] In the actual derivation this should strictly be the partial molar volume of the solvent

$$\overline{V}_{\!\!\mathrm{m,solvent}} = \left(\frac{\partial V}{\partial n_\mathrm{solvent}}\right)_{\!n_\mathrm{solute}}$$

but this can to a good extent be approximated by the molar volume of the pure solvent.


When you report Einstein units do you capitalize it?

The non-SI unit einstein is not accepted for use with the SI and should be avoided.

Anyway, unit names are normally printed in upright type, and they are treated like ordinary nouns. Names of units are spelled with a lower-case initial in English, except in the beginning of a sentence when a capital initial is used. This also applies to unit names that are derived from a proper name of a person such as kelvin, newton, joule, pascal, watt, hertz, coulomb, volt, farad, ohm, siemens, weber, tesla, henry, and thus also einstein.

For SI units, it is only the unit name degree Celsius that contains a capital letter. In keeping with the rule, the unit degree begins with a lower-case d and the modifier Celsius begins with an upper-case C because it is a proper name.

However, in quantity names containing a person’s name, the person’s name is spelled with a capital initial, for example, the Einstein transition probability for spontaneous emission $A_{jk}$.

These writing rules are in accordance with following references:

ISO/IEC: ISO/IEC 80000 Quantities and units (14 parts)
BIPM: The International System of Units (SI)
IUPAC: Quantities, Units, and Symbols in Physical Chemistry (Green Book)
NIST: Guide for the Use of the International System of Units (SI)
ACS: The ACS Style Guide


In an equilibrium, why are partial pressures not affected by the coefficients of the balanced equation?

One way of thinking about this is writing the original equation
$$\ce{2NOBr(g) <=> 2NO(g) + Br2(g)}$$
as
$$\ce{NOBr(g) + NOBr(g)  <=> NO(g) +NO(g) +  Br2(g)}$$
Now it should be clear you have to multiply, not add concentrations.


Although the IUPAC convention is for work done by a system to be negative for expansion, many physicists and engineers use the opposite convention. What's important is that you use the same convention throughout your calculation. You wrote that H = U + W, which is true only if work done by the system is regarded as positive. To be consistent with the IUPAC convention that you used later in the calculation, this should be instead H=U - W.

On a side note, since work represents a change in energy, not a state function, this equation should be written in terms of changes: $\Delta H = \Delta U - w$. Also note that this equation is only valid for the condition of constant pressure.


Reactivity order in hydrogenation reaction of but-2-yne and butene isomers

Monosubstituted alkene is more reactive than disubstituted because directly attached alkyl groups stabilize the system. Therefore 1 butene is more reactive.

Cis is more reactive than trans because Cis compounds are unstable. As they have similar groups on same side, it results in stearic crowding.

Thirdly, conjugated dienes are very stable beacuse they are resonance stabilized.

Thus order will be
1 butene will be most reactive, then Cis, then trans and then 1,3 butadiene..