The recipe linked to the question is not very detailed. It does not even say to beat the egg whites (which you did and which needs to be done). Here is a snippet of another recipe:

  You just need a sugar thermometer so you can cook the sugar until it reaches a temperature of 113°C/235°F. Once the sugar syrup reaches that temperature, immediately take it off the heat – or you’ll risk bringing the sugar up to ‘hard-ball’ stage – and carefully drizzle it into the foamy egg whites, mixing continuously. You do remember the foamy egg whites, right? I bet you do. Drizzle the hot (!) syrup slowly into the egg whites, making sure not to pour it directly onto the whisk attachment of your mixer, or the syrup could splatter against the sides of the bowl or into your face.

Source: https://thetoughcookie.com/2014/09/11/different-kinds-meringue-make-italian-meringue/

I think the key is "drizzle into, mixing continuously". And here is a picture of what it should look like at that stage (from the same source):

  [OP} But since I don't understand the chemistry of egg whites I am not sure how to figure it out.

There is some chemistry and some physics here. You want to add a lot of sugar and little water. You need the sugar in solution so you can add it quickly, and if the syrup cools down, it is no longer a liquid. On the other hand, you don't want the egg white to heat up too much, so that's why you drizzle in the syrup while mixing vigorously. That way, the syrup cools down while being diluted with the water in the beaten egg whites, effectively removing water from the bubbles surrounded by the protein from the egg white, making the foam more stable.

If everything else was correct (water content of syrup, using bowls that are not greasy, correct ratio of egg whites to syrup), I think the key is to drizzle the syrup into the egg white where it gets mixed right away. You describe adding it to the side of the bowl, but the recipe does not say that. If I were to experiment with the recipe, this step would be my focus.


I had the same problem as well, until I sketched out the actual molecules.

When you have one mole of $\ce{KOH}$, you have one mole of the collective "molecule".

If you were to break it down into $\ce{K}$ and $\ce{OH}$, you would get 1 mole of $\ce{K}$, and 1 mole of $\ce{OH}$. This is because you need 1 mole of each to make 1 mole of $\ce{KOH}$.

It's kind of like if you have one hundred keyboards, you would get one hundred 'Enter' keys, one hundred 'Up arrow key' and so on...


What does it mean to 'fuse' ores and why do we use alkali metals in the process?

"To fuse" is another word for "to melt" (e.g. "heat of fusion"). Specifically, if you say you want to fuse two materials, you melt them in the hope that they will mix.

In this case, you melt the carbonate, and hope that the chromite will dissolve in it. Because e.g. $\ce{Cr2O3}$ has a melting point of $\pu{2435 °C}$, chromite ($\ce{Fe(II)Cr2O4}$) typically around $\pu{2140 °C}$. That is out of reach with a common Bunsen burner, and even if you did reach it, it would still be the same insoluble mess after cooling down.

When your chromite is dissolved in alkali carbonate, the mixture easily dissolves in $\ce{HCl}$, and you can properly run your wet analytics.

Soldering for example is a very similar process: Your soldering iron cannot nearly melt copper, but the molten tin/lead/etc. fuses (partly dissolves/melts) with the copper surface, and you get a continous metallic connection.


Deciphering stereochemical notation in electrocylization reaction

You are both correct and incorrect. (And this entirely without superimposition of quantum states!)

You are taking an achiral molecule and performing a reaction in achiral media to form a chiral product. Symmetry dictates that the reaction must be racemic, so there must be equal amounts of upward-pointing and downward-pointing carboxylates.

However, the drawing is also incomplete. It does not make sense to discuss the reaction without labelling the entire stereochemistry at the newly formed bond. It is important to realise that it generates only two of the four possible stereoisomers: the two syn ones ((S,R) and (R,S)). You should add an upwards-pointing hydrogen to the right-hand carbon of the new bond.

Judging by your comments, you are only interested in the arrow direction and resulting movement. Well, the arrows are not clear. They are 2D on a 2D plane and trying to signify a 3D movement. You could grab the curved bit of the arrow like a handle and:

either pull it upwards — then you have a rotation which would move the carboxylate below your screen’s plane and create the downward-pointing product ((R) on the α-carboxylate carbon),
or push it downwards — then you have the rotation which would move the carboxylate above the screen’s plane and create the upward-pointing product ((S) on the α-carboxylate carbon).

So the arrows’ directions are arbitrary, we do not know if the are shaped below or above the plane.


Questions about intramolecular interactions in AMBER forcefield

As for your first question, this site states that 1-4-interactions are scaled by a factor of 0.5, which I suppose implies 1-2 and 1-3 interactions being excluded.

The answer to your second question is yes, as can be seen here (section TORSIONS). The potential is divided (and thus scaled) by Ntors.


Can I perform a stable isotope analysis at home?

I am assuming that you wish to measure isotopes of carbon.  It is unknown to me how you will extract sample from a bird feather that is suitable for measurement in a spectrometer, and does not taint the isotope ratios. The spectrometers that can distinguish between isotopes are generally quite sensitive and complicated. So the literal answer to your question is "NO, you can not do it at home with a DIY spectrometer."

But, you might be able to purchase a spectrometer. You might be able to differentiate carbon isotopes in carbonyl compounds with infrared spectroscopy. (But don't count on it, unless you are somehow making a gaseous product such as $\ce{CO}$ or $\ce{CO2}$.)

The usual way of measuring carbon isotopes is via mass spectrometry. You can probably get a crude measurement with a machine that costs $20,000 (on eBay) but then you must get it calibrated and keep it supplied with carrier gas, etc.  In short, it will bankrupt you.

User permeakra has the best idea: send a sample to a lab.  But, I am not sure of any lab will allow you to send samples through their spectrometer, not knowing anything about your reputation or the volume of business that they might expect to get from you.


What does this example mean? Slow cooling of 40% Sn alloy from 800°C to 600°C: L → L and γ → L, γ, and ε → L and ε

Consider the 40 at% (Sn) composition at 800 °C: it's a liquid.
When you cool it down at around 650 °C you enter a two-phase field: the solid gamma forms (Cu-enriched) and the liquid enriches in Sn. As you further cool down the gamma phase will continue to form, but with progressively higher content of Sn
You cool down at 640 °C and you enter a new different two-phase field: the liquid partially reacts with the gamma phase forming the solid epsilon phase. If you further cool down, the ε phase will continue to form from the liquid, but its composition will progressively enrich in Cu.
Just below 415 °C you enter into another two-phase field: the liquid will solidify into ε and η phase.
That's what occurs by cooling down to 300 °C, but if you further would cool down below 189 °C you will enter another two-phase field; in this case the ε and η' phases will form.
The amounts of the different phases at the different $T$ are obtained by applying the lever rule.
Take care, I described what happened from the thermodynamic point of view (that is what is represented in a phase diagram); the kinetics can remarkably change the things.


Orbital hybridisation is, like most other topics bordering quantum mechanics, just a model of reality (in fact, I would go as far as to say like all topics ever thought of, but lets not get too philosophical). As such, there are limits to its usefulness, so models should never be thought of as a "true picture" of anything.

Models are always simplified and in the end "wrong", but sometimes they are useful. Through models we can gain insight and make new predictions that can be falsified or not and as such we advance our understanding (and the most learned even advance science as a whole). Without models I would not understand anything, but I considert it counter productive for myself to think of them as representing some "truth".

$$\text{Observation} \rightarrow \text{interpolation} / \text{modelling} \rightarrow \text{prediction} \rightarrow \text{experiment} \rightarrow \text{observation}$$


Cis/trans interconversion of crotyl-boron compounds

The lack of an octet and the empty p-orbital on boron make trialkyl boron compounds quite reactive.  If we replace one of the alkyl ligands with an alkoxy group (borinate), we can now draw a resonance structure where oxygen shares one of its lone pairs with boron.  In such a resonance structure the boron atom has achieved an octet and the empty p-orbital on boron has been filled.  Consequently borinates are less reactive than the corresponding boranes.  Boronates, with two alkoxy substituents, have an even higher contribution from the resonance structure with the $\ce{B^{-}=O^+}$ double bond and are even less reactive.


Steric factors are more important here. The positive charge resides on the N and the nucleophile attacks the least hindered site hence the product is as drawn.


Instead of doing the whole derivation, I will provide the following two hints usefull in the derivation:

Use the quotient rule (as is was already done by OP).
$$
  \left({\frac {f}{g}}\right)'={\frac {f'g-fg'}{g^{2}}} \, ,
$$
where in this particular case $f = \sum_{i,j} c_i c_j H_{ij}$ and $g = \sum_{i,j} c_i c_j S_{ij}$.
Then, when taking the derivatives of the above mentioned $f$ and $g$, take into account that for each and every $c_i c_j X_{ij}$ term (where $X$ is either $H$ or $S$) there are only three possibilities:

$i = k$ in which case $\partial (c_i c_j X_{ij})/\partial c_k = c_j X_{kj}$;
$j = k$ in which case $\partial (c_i c_j X_{ij})/\partial c_k = c_i X_{ik}$;
$i \neq k$ and $j \neq k$ in which case $\partial (c_i c_j X_{ij})/\partial c_k = 0$.


Thermodynamics of DNA double helix formation?

The entropy of a system decreases when 2 single stranded DNA molecules come together and form a double stranded molecule.

Are you sure?

I would argue that association of two single-stranded DNA strands to form a double-strand is energetically favourable — for entropic reasons. This should be a combination of

the hydrophobic effect
the liberation of a few water molecules per base that are no longer needed in hydrogen bonding.

Point 2 is explained quickly. Upon association of two single strands, the characteristic Watson-Crick base pairs form and hydrogen bonds are established between the DNA bases. These hydrogen-bond donors and acceptors were previously associated with water molecules keeping them close to the DNA single strands and thus rather ordered. These water molecules are now liberated; and since there are at least two per base, this partial process is entropically favourable.

Point 1 may not be as clear as point 2, but it again boils down to the fact that water molecules are liberated from a rather well-defined structure. Large parts of DNA — the area around the bases — are not as hydrophilic as other parts of the molecule or water itself. While they still are somewhat polar, their interaction with water is not ideal. Thus, the ‘surface’ water molecules will form a kind of defined boundary around the molecule. This boundary becomes lesser overall if the surface to volume ratio decreases. And upon association, you have almost double the volume in practically the same surface (i.e. you have halfed your effective surface). Therefore again, a large number of water molecules are liberated.

If DNA association is entropically favoured, there is no need to invoke any heat.


You made $K_b$ = 0. That is where you were wrong. Taking approximation of Neglecting $\alpha$ with respect to 1 only gives us advantage of not solving the quadratic equation.

From there you can write $\alpha = √(K_b/C)$ here you can see $\alpha$ comes inversely proportional to $√C_o$ or we can say directly proportional to $√V$. From here you can deduce if $V$ approaches infinity $\alpha$ approaches 1. And that's the law.


I'm pretty sure this doesn't follow Le Chatelier's Principle

Le Chatelier's Principle does not directly say what happens to concentration ratios.  Nor does it directly compare conditions after a disturbance with those before.  It says what happens only after a disturbance has been applied.

Here, the disturbance is adding one mole of A per liter.  When equilibrium is re-established you have only an additional one-half mole of A per liter, the rest having reacted to form B.  The initial addition of A has been partially offset, which is exactly what Le Chatelier's Principle says.


Are such high oxidation states as reported by Vojovodic et al. possible?

Oxidation states are just numbers, a bookkeeping tool for chemistry. These hardly ever correspond to anything observable. Point in case: hypofluorous acid $\ce{HOF}$, see for example the oxidation state of oxygen and of fluorine in the confines of the definition. Maybe also of interst in that regard is the introduction to "oxidation state"/"oxidation number" in general.

Therefore instead of asking whether the oxidation states are stable, one should ask if the bulk/ molecular structure is stable.

When it comes to computationally aided catalyst design, you can pretty much use the entire periodic table to play with. It is a matter of interpreting the results that count, and obviously if experiment can reproduce such predictions. Where you start becomes a matter of taste and starting with perovskites is as good a guess as any.

The authors of the publication make quite an effort, calculating hundreds of potential catalysts. However, they also indicate problems within their methodology. Quoting from the supporting information:

  3 Oxygen evolution data
  [...]
  \begin{array}{lcl}
\text{formula} & \text{values of }\Delta G, \eta, \dots & \text{warnings}\\
\hline
\ce{MgBaO3} & [\cdots] & \text{a,b}\\
\ce{NaLaO3} & [\cdots] & \text{a,b}\\
\ce{CaBO3}  & [\cdots] & \text{b}\\
&\vdots&\\
\hline
\end{array}
  a: One or more runs did not converge or failed in Quantum Espresso, missing adsorbates are reconstructed from successful runs using scaling
  b: Changes in atomic positions of greater than 5.5 angstrom found in the slab or adsorbate during optimization, typically indicates structural instability
  [...]

And there are a lot more unstable species.

TL;DR: Basically, in these extreme cases, where you would observe abnormal oxidation states, the calculation also predicted that the bulk/ molecular structure is not stable.


Why Fe(CO)5 exists, but Fe(NH3)5 doesn't?

While we cannot be sure that $\ce{Fe(NH3)5}$ does not exist, the bonding between ligand and metal offers a good explanation for why it is expected to be less stable than $\ce{Fe(CO)5}$.

In metal carbonyls like $\ce{Fe(CO)5}$, the bonding between CO and the metal consists of three single bonds, 1 $\sigma$ and 2 $\pi$ bonds. The $\sigma$ bond is formed by overlap of the carbon $sp$ hybrid orbital (which contains the lone pair) with an empty $d$ orbital of the metal. Electron density is donated from the orbital of CO into the metal orbital, so CO acts as a $\sigma$ donor. The two other $\pi$ bonds are formed by overlap of a filled metal $d$ orbital with a pair of $\pi^*$ molecular orbitals of CO. Electron density is donated from the metal orbital into the molecular orbitals of CO, so CO acts as a $\pi$ acceptor.

For the $\pi$ bonding to occur, the metal is required to have sufficient $d$ electrons, and thus preferably a lower oxidation state. This explains the relatively low oxidation state of iron (0) in $\ce{Fe(CO)5}$. The combination of one $\sigma$ and 2 $\pi$ bonds gives the M-CO bond a partial triple-bond character, while the C-O triple bond is weakened by the transfer of electron density into the $\pi^*$ orbitals (reference).

Because of its ability to serve as a $\sigma$ donor and $\pi$ acceptor simultaneously, CO is a stronger ligand than $\ce{NH3}$ which can only act as a $\sigma$ donor with its lone pair in the nitrogen $sp^3$ hybrid orbital. $\ce{NH3}$ lacks a $\pi$ system and therefore cannot act as a $\pi$ acceptor. Because $\pi$ backbonding is crucial for stabilization of a metal center in low oxidation state, ammonia is unlikely to form stable ammin complexes which are analogous to the corresponding metal carbonyls.


Is it possible to make a disaccharide from two fructose monomers? If so, what is its name?

in agave plants the typical sugars are fructans (fructose-polimers) like inulin, levan, agavin - with sometimes really complex structures comparable to starch. As starch breaks down into glucose, maltose, etc. you have also a need for break-down process e.g. in the tequila/mezcal industry which is usually a thermic hidrolisis (no enzymes like in beer). This hidrolisis is never complete so you have dimers and trimers (or higher) fructose-sugars, which are even metabolizable by yeast-strains. They are referred to as nistose or kestose (https://pubchem.ncbi.nlm.nih.gov/compound/1-Kestose#section=Top).
I hope this answer helps you to finally resolve your question - check out also user55119 coment.


Whose concentration is more 1 molar or 1 molal?

To convert from molal to molar, one generally needs to know the density of solution as well as the formula weight of solute.  Suppose there is a single solute $x$ in water, and that $x$ has a formula weight of $f_x$ g/mol.  Suppose also that the solution of $x$ has a density of of $\rho$.
Let the molarity of $x$ in the solution be $C_x$ mol/L.  Then:

the mass concentration of $x$ in g/L is $C_x f_x$.
the weight fraction of $x$ in grams of $x$ per total gram of solution is thus $w_x = \frac{C_x f_x}{\rho}$.  The weight fraction $w_x$ must be a number between 0 and 1.
The weight ratio of $x$ in grams of $x$ per gram of water is $\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}} = \frac{w_x}{1-w_x}$.
The molality is simply the weight ratio divided by the formula weight, $\frac{1000}{f_x}\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}}$.  The factor of 1000 is because molality is moles per kg of solvent, not moles per g.
That last expression can be written as  $\frac{1000}{f_x}\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}} = \frac{1000}{f_x} \frac{w_x}{1-w_x}$

This last formula shows that molality values will always be higher than molarity values.  Because $w_x$ has to be between zero and 1, then the molality formula will divide by a smaller number than then molarity formula.  One molar water "dissolved" in water (say we are dissolving distilled water in tap water, or dissolving isotopically labeled water into regular water) has a molality of around 1.02.

My approach is:

Convert 1M into molality and then compare if it is less or more than 1m. (?)

Good approach.

The density of water is 1kg per L.

True, but as noted in the comments, the density of solutions in water can be either higher or lower than the density of pure water.

When solute is added to it, density would definitely increase(?)

Not necessarily true.

And therefore 1L of that solution will weigh more than 1 kg.

Not true, and this is the point where I think your logic mislead you.  To make one liter of a 1 molar solution, less than one liter of pure water is usually required.  This is because part of the liter of solution is taken up by solute.  So not all of the liter needs to come from pure water.  And since water has a density of 1 kg/L, if we need less than one liter of pure water, we need less than 1 kg of pure water.  This then means that our one mole of solute is being dissolved by less than kg of solvent: thus the molality of a one-molar solution is greater than one.
The neat thing is that this is true regardless of the density or formula weight of the species being considered.


Why is the base-centered orthorhombic crystal lattice a unique crystal system?

All the centered Bravais lattices are redundant in the sense that it is possible to use a primitive cell of smaller volume instead. However, this often means that crystal symmetries are no longer aligned with the cell axes. For the orthorhombic system, you either have 2-fold rotation or screw axis parallel to the cell edges, or mirror planes or glide planes perpendicular to the cell edges. If you switch to a primitive lattice for the centered cases, these symmetry elements will no longer be aligned in this simple way (they might be on diagonals, making the symmetry operations harder to figure out).
So it is a convention, and one that is useful.


Why does magnesium bromide transfers from a carbon atom to nitrogen in piperidine?

There is no step 1 and step 2 in this reaction. Both are consecutive and second one is fast because it is an acid-base reaction:

Note that approximate $\mathrm{p}K_\mathrm{a}$ values of ring $\ce{C-H}$ and $\ce{N-H}$ are 50 and 40, respectively. Thus, as soon as Grignard reagent is formed, it exchanges relatively acidic protons of $\ce{N-H}$ fast (Remember, Grignard is also a strong base). To visualize the exchange process, I included schematic representation at the bottom of the image.


What is the difference between wet paint, dry paint, and in between those stages?

Ullmann’s Encyclopedia of Industrial Chemistry lists numerous components of commercial paints:

dyes, pigments, extenders
organic solvents, water, coalescing agents
binders
plasticizers
paint additives (levelling agents, film-formation promoters, wetting agents, dispersants, antisettling agents, antifoaming agents, catalysts, antifloating and antiflooding agents, antiskinning agents, matting agents, neutralizing agents, thickening agents, preservatives, corrosion inhibitors)

The application method, drying and hardening behaviour of a paint is essentially determined by the binder. Binders are macromolecular products. The low molecular mass binders include alkyd resins, phenolic resins, polyisocyanates, and epoxy resins. These binders must be chemically hardened after application to produce high molecular mass cross-linked macromolecules in order to form acceptable films. The higher molcular mass binders include cellulose nitrate, polyacrylate, and vinyl chloride copolymers, which are readily suitable for film formation.

As the paint dries, a firmly bonded film is formed. Drying and film formation take place physically and chemically. Physical drying involves evaporation of the solvents (organic solvents and/or water). This is the main process for paints with high molecular mass polymer binders. Chemically drying paints contain binder components with a relatively low molecular mass that react together on drying to form cross-linked macromolecules. This process can occur by polymerization, polyaddition, or polycondensation.

In practice, physical drying and the various mechanisms of chemical drying can proceed concurrently or consecutively, depending on the composition of the binder system.


Vapour pressure of a liquid subjected to pressure

Here's how it works.  The fugacity of the liquid at temperature T and total pressure P is equal to the equilibrium vapor pressure at T times the Poynting correction (assuming that the vapor phase behaves as an ideal gas mixture).  The fugacity of the same species in the vapor is equal to the partial pressure of the substance in the gas phase.  The vapor and liquid are at equilibrium at temperature T and total pressure P if the fugacity of the substance in the liquid is equal to the fugacity of the substance in the vapor phase.


Why do DMSO and acetone have such radically different melting temperatures?

The first thing to remember is that boiling and melting points, while seemingly simple, are incredibly hard to predict. Yes, many teachers and professors give their pupils and students a number of simple prediction rules, but there is a lot of intrinsic balance and experience associated with applying them that they may often seem unintuitive.

Comparing DMSO and acetone, the first striking difference is the replacement of carbon with sulfur. This may seem like a small change but results in radical differences. For example, compare propane with dimethyl sulfide; or the latter with dimethyl ether. $\ce{Me2S}$ is a liquid with a boiling point of around $40~\mathrm{^\circ C}$, the others are gases at room temperature and standard pressure.

There are two factors that explain the difference between $\ce{Me2S}$ and propane/methyl ether — one does better in explaining the difference to propane the other does much better in explaining the difference to dimethyl ether.

Polarity. Sulfur is more electronegative than carbon, thus the $\ce{Me2S}$ molecule is slightly polar while propane is essentially nonpolar. Since all of these molecules exhibit (approximate) $C_\mathrm{2v}$ geometry, the polarity of individual bonds does not cancel out. However, one would expect $\ce{Me2O}$ to be more polar and thus have an even higher boiling/melting point.
London interactions. These are often simplified as ‘rising with molecular weight’, but in fact, the key factor is not molecular weight (which is a nuclear property) but polarisablility of electrons. Sulfur, being larger, having an additional shell and thus being much more diffuse than both carbon and oxygen is much more polarisable. Thus, the significant difference in the boiling and melting points of $\ce{Me2S}$ and $\ce{Me2O}$ can be attributed mainly to the stronger London interactions, made possible by adding eight (core) electrons and thereby moving the valence electrons further away from the core.

Both factors play a role in the difference of melting points of DMSO and acetone, although there is even more. While acetone has a trigonal planar structure featuring a $\ce{C=O}$ double bond, DMSO is trigonal pyramidal with an additional lone pair and formal positive charge on sulfur, to give a $\ce{S^+-O^-}$ single bond. While this feature reduces the London interactions somewhat — sulfur’s higher oxidation state means that the electrons are more stable and the orbitals appear smaller at the same cutoff value — it increases polarity significantly.

Indeed, it is better to compare similar molecules; isopropanol’s structure is much closer to DMSO’s than acetone’s is.


Where does the energy required for excitation of electrons in hybridization come from?

Ab initio hybridisation is a concept taught in schools to help understand how carbon with its one s- and three p-orbitals can achieve a tetrahedral environment. As part of this process, it is required to formally excite an electron.

Formally is the key word here. The entire process is a formal process that has practically nothing to do with real-life atoms (especially because these real-life atoms rather rarely end up in a vacuum without anything around them as the theory assumes). Thus, in real life this formal energy is not needed.


Reaction between phenol and nitrous acid

What the reference says

The reference gives only following figure to support their theory. A weakness of this mechanism is requiring a high energy intermediate like sp2-carbocation.

An alternative proposal

This alternative avoids high-energy intermediates and thus appears more likely:


Pulsed NMR change in magnetisation vector

In the nmr experiment the sample is placed into an intense but static magnetic field. By the (nuclear) Zeeman effect the spin energy levels that are degenerate in the absence of a magnetic field become split in energy and thus more spins reside in the lower energy levels that in higher one. The number of any level is given by the Boltzmann distribution at the prevailing temperature. The total spin angular momentum of the sample is called the magnetization. To make a measurement a magnetic field from a radio frequency source is applied at right angles to the large, static field. This tips the magnetization into the x-y plane making measurements possible. I try to explain how and what happens below.

The spins have a magnetic moment $\bar m$, which is a vector quantity meaning it has direction and magnitude, and as a result the static field causes magnetic induction $\bar B$ in the sample. This is also a vector. The magnetic induction (confusingly but commonly also called the magnetic field) is induced in the sample by the the large static magnetic field often labelled $H$.

If all the individual spin moments are added up these form the magnetization $ \bar M$. This is a magnetic field and is also a vector and so has both magnitude and direction.    All nuclear spin magnetic moments experiences a torque ( a rotational motion generating angular momentum ) in a magnetic field and is given by classical electromagnetism as $\bar m \times \bar B$ where $\times $ means this is a vector cross product; $\bar M \times \bar B=|M\|B|\sin(\theta)$ where $\theta$ is the angle between the vectors.  The change in angular momentum produced is given by $\displaystyle \frac{d\bar L}{dt}=\bar m\times B$. As the angular momentum is given by $\bar m=\gamma \bar L $,   where $\gamma$ is the magnetogyric ratio (these are known and for H is $27.75\cdot 10^{7}$ rad/T/s), then

$$ \frac{d\bar M}{dt}=\gamma \bar M\times B \tag{1}$$

This equation can be broken down into components in x, y and z directions. The z direction is conventionally taken as the direction along which $\bar B$ lies, and this has no x or y components. The three equations produced are

$$\frac{d M_x}{dt}=\gamma m_xB;\quad \frac{d M_y}{dt}=-\gamma m_yB; \quad \frac{d M_z}{dt}=0  \tag{2}$$

If there is no external perturbation then these equations show that the magnetization remains along the z axis and as we place the detector along the x or y axis then no signal can be observed. This is  because the magnetic field of the magnetization has to 'cut the coils' of the detector as so induce a signal that can be measured. It clearly cannot do this if it has magnitude only along the z axis.

The equations (2) can be solved if some initial conditions at time zero are chosen and physically important solutions are found only when the x and y components are not initially zero. In that case the solutions show that then magnetization in x and and y are proportional to $\sin(\omega_0t)$ and $\cos(\omega_0t)$ respectively where $\omega_0= \gamma B_0$ where $B_0$ is the z component of $\bar B$. The Larmor frequency is $\nu_0=\omega_0/(2\pi)$ and is the frequency measured in an nmr experiment.

As these solutions require that the magnetization has to be in the x-y plane this  condition has to be satisfied and is done so by exciting the sample with rf radiation directed along the x (or y) axis. This radiation has a magnetic  field that is weak compared to $\bar B$ and of magnitude $B_1$; its frequency we will call $\omega_r$. Classically the magnetic field applies a torque to the magnetization and rotates it away from the z axis.  In a molecular (quantum) explanation, if the rf energy is resonant with the energy gap, formed by the large static field $B$, then spins are excited from the lower to upper level and this changes both the size and direction of the magnetization $\bar M$.

In a classical  explanation the  polarised rf field $B_1$ is considered to be composed to two oppositely rotating fields. One moves in the same sense as the magnetization precesses, and so interacts with it, the other moves in the opposite sense and has little interaction and can be ignored. The experiment is now such that the spins experience both a static magnetic field $\bar B \equiv B_0$ in the z direction and a rotating field $B_1$ in the x-y plane. Clearly this is now a very complex problem but can be simplified if one assumes that the axes x and y rotate about z at the frequency of the $B_1$ magnetic field. This is called the 'rotating frame'. The result is that the equations similar to (1) are produced but with $B_{x,y,z} $ replaced by $B_{eff(x,y,z)}$ where $\displaystyle B_{eff}^2=(B_0-\frac{\omega_r}{\gamma})^2+B_1^2$.


What are the metamers of penta-1,2-diene?

Yes. Consider metamer to be a different word for isomer, as noted by Lighthart:

  The term metamer is equivalent to isomer...As an aside, I have never heard the term metamer before, I expect it is archaic.

And, according to Dictionary.com, consider a diene to be:

  a hydrocarbon that contains two carbon-to-carbon double bonds in its molecules.

There are too many diene isomers to be listed here ($\ce{C5H8}$ has many atoms), but you can begin by changing the numbers in the name to generate different Lewis structures.

Penta-1,2-diene:

Penta-1,4-diene:

This is just one structure. Try cis vs. trans, position isomers, etc. You may think of many more.


How to solve this dimensional analysis/stoichometry problem

Figured it out, thanks to @M. Farooq's help.
I first did some cross multiplication of 250/24 times X/15 to find out that in 15 hours, there would be 156.25 L of methane. This is equal to 86.56 grams of methane. Using that, I found out there were 5.393 moles of methane, and multiplying that by 2 (because for every one methane it makes 2 waters) I found that there would be 10.79 mol of water when the reaction was done. This, times the molar mass of water, gives me 194.4 grams of water after the explosion.


Is HMgX also a Grignard Reagent like RMgX?

According to this reference $\ce{HMgX}$ compounds are a thing.  You can't make them with magnesium +$\ce{HX}$, of course (what actually happens?), but you can get them from magnesium halide salts plus an "active" form of magnesium hydride in THF.  They act as hydride ion sources for reaction with $\ce{AlH3}$ and $\ce{BH3}$ (forming mixed tetrahydridoaluminate and tetrahydridoborate salts, respectively), so we may reasonably expect the same sort of hydride transfer with organic substrates.  Such a hydride transfer would be analogous to conventional Grignard reagents transferring their alkyl or aryl anion moieties.


Are electron shells probability distributions?

Short Answer:

No.

Long Answer:

First, strictly speaking, the orbitals themselves in the quantum mechanical sense are not probability distributions.  They are eigenfunctions $\Psi_i$ of the Hamiltonian as defined by the time-independent Schroedinger equation $H\Psi_i=E_i\Psi_i$.

The probability distribution function $p(\vec r)$ for electrons is generated by summing the products of the eigenfunctions and their complex conjugates for all occupied eigenfunctions at a given 3-D point $\vec r$ and dividing by the total number $N$ of electrons in the system:

$$
p(\vec r) = \frac{1}{N}\sum_{occ}{\Psi_i^*(\vec r)\Psi_i(\vec r)}
$$

Note that $p(\vec r)$ is simply $\frac{1}{N}$ times the electron density function $\rho(\vec r) = \sum_{occ}{\Psi_i^*(\vec r)\Psi_i(\vec r)}$.

To be fair, prior to a detailed introduction to the quantum mechanical description of electronic structure, the distinction between the electronic wavefunction $\Psi(\vec r)$ and the electron density function $\rho(\vec r)$ is typically not clearly established in chemistry instruction, so your having conflated them is quite understandable.

Second, again strictly speaking, electron shells are not probability functions.  While Pushkar Soni is correct that the principal quantum number for a given electron approximately indicates the 'electron shell' to which it is assigned, the indistinguishability of electrons (among other factors) in general prohibits unambiguous translation of quantum number information into the real-space function $\rho(\vec r)$.  The best we can do is to interpret electron shells as corresponding to certain features of the structure of $\rho(\vec r)$. Offhand, I know of at least two ways of qualitatively identifying 'electron shells' from $\rho(\vec r)$.

First, as Geoff Hutchinson notes, the radial distribution function $r^2\rho(r)$ shows local maxima that can be interpreted as electron shells.  Note that  $\rho(r)$ in the r.d.f here is a function only of $r$, the distance from an isolated atom/nucleus; the angular dependencies (on $\theta$ and $\varphi$) have been integrated out.

Second, Bader's QTAIM describes how the Laplacian of the electron density, $\nabla^2\rho(\vec r)$, exhibits minima in 3-D space at positions corresponding to the notion of electron shells.


Does the acid strength of hydrochloric and hydrofluoric acid correlate with the electronegativity of the halides?

You are correct, HCl is a stronger acid than HF. Fluorine is both more electronegative and smaller than chlorine. Because fluorine is more electronegative, the bond between it and the hydrogen is more polar, meaning that the proton would need to overcome a larger coulomb force to separate from the fluorine. In addition, fluorine is a smaller atom, meaning that the addition of an electron has a larger effect on the net charge of the atom. This both makes the fluoride anion (the conjugate base) less stable and it also - once again - increases the coulomb force between the proton and the fluorine. In addition, the coulomb force is even stronger because there is a smaller distance between the two atoms. As a general rule of thumb, as you move down the halogens, the acids become stronger and they form more stable conjugate bases.


Can we predict with any reliability whether a chemical is carcinogenic in people?

One of the great problems is that there is no single mechanism by which a chemical carcinogen causes the damage which leads to cancer. There are a range of different mechanisms, I will mention some of them.

One mechanism is alkylation of DNA, the attachment of alkyl groups to DNA will prevent it being read correctly. In some cases the damage can be repared while in other cased the damage will cause the body to misread one base pair as another. There is an interesting problem with electrophilic alkylation agents, if they are too reactive towards water then they will not survive to reach the DNA.

There is an interesting paper in which this idea is discussed E.W. Vogel and M.J.M. Nivard, Mutation Research, 1994, volume 305, pages 13-32 considers it at length. If we consider the ratio of the rate constant for the reaction of an alkylation agent with thiosulfate and acetate, then as the ratio of these reaction rates changes in favour of the thiosulfate reaction something known as the Swain Scott constant (s) also increases. The Swain Scott constant is the ratio between reaction at two different sites (7-alkylguanine/O6-alkylguanine) in DNA. In general the higher the value of the s constant the stronger a carcinogen is. In the paper by A. Barbin and H. Bartsch in Mutation Research, 1989, volume 215, page 95 to 106 a series of directly alkylating carcinogens a good but not perfect relatioship is seen between the s constant and how carconogenic they are in rodents.

So it appears for one class of carcinogen that it is possible by doing some simple chemical kinetics experiments that it is possible to make an assessment of how carcinogenic a substance.

One of the great problems is that many "carcinogens" are in fact "precarcinogens" which require metabolic activation to make them carcinogenic. For example hydrazine reacts with formaldehyde in the liver before being oxidized to form diazomethane which then methylates the liver. This is why acute large hydrazine exposures in humans can cause liver failure.

This brings us to the Ames test, this is a test using bacteria which measures how able a substance is to alter DNA. The idea is that cancer is caused by altering the DNA of normal cells. So something which is able to randomly alter the DNA of cells is able to induce cancer. The great problem with the Ames test is that while it is a good test. There are both false positives and false negatives if it is used as a test for carcinogenic potential.

The normal Ames test protocol includes some sterile extract of rat liver to provide some metabolic activation. This will increase the ability of things like polycyclic aromatic hydrocarbons to alter the DNA of the bacteria.

If we consider an inorganic example, the chromate anion is known to be carcinogenic while chromium(III) is not carcinogenic. To understand this we need to know something about chromium biochemistry.

It is true that chromium(III) can form some very stable complexes which are very slow to react with DNA. But the chromium(III) is unable to reach the DNA in the cell. However chromate is able to pass through the sulfate transport channels in both the cell and nuclear membranes to allow it to reach the nucloplasm. There it is reduced to chromium(III) which then damages the DNA. As a result we can make a judegment about chromate and sulfate.

Sulfate is unable to react in the nucloplasm to form something which will form long lasting complexes with DNA, so even while sulfate is able to reach the DNA by the same route as chromate it is clear that sulfate is unable to act as a carcinogen by the same mechanism as chromate.

Also permanganate or magnanate is a stronger oxidant than chromate, I suspect that as it would be less able to survive the journey through the body to the DNA it is less able to incude cancer. This is likely to be one of the the reasons why permanganate is far less carcinogenic than chromate. Plenty of evidence exists for chromate being carcinogenic but little if any evidence exists for permanaganate.

Also the complexes of manganese with DNA are likely to be shorter lived than those of chromium. This is becuase chromium(III) complexes are slow to exchange ligands while manganese(II) are rather quick to exchange ligands. This is due to the fact that chromium(III) complexes are d3 while manganese(II) are d4.


what the solution of this problem, negative absorbance values in UV-vis spectrum

The negative values can arise due to two major reasons: You did not mention that whether you are using a single beam instrument or a double beam instrument.
In a double beam instrument, absorbance of the sample and the blank is continuously measured. In a single beam spectrophotometer, A(blank) is recorded first and later subtracted. In each case, the output value at each wavelength is:
Abs (measured) = A(sample)- A (blank)

If at any region of the spectrum A(blank) is larger you get slightly negative absorbance values. This could be due to several reasons including refractive index differences. Under ideal cases, the cuvet which contains the blank is same as cuvet containing the sample or both of them should be optically matched i.e. they should show identical absorbance readings at given wavelengths.

Another reason to see random negative values is that noise is a part of any instrumental analysis. Eliminating all sources of noises, you still get stuck with shot noise, which originates due to statistical counting of photons. That is the very nature of detection especially in spectroscopic measurements- perhaps the ultimate limit in analytical sciences.


Why is Ma*Va=Mb*Vb still valid for weak acids/bases?

We normally titrate a weak acid by a strong base, or  weak base by a strong acid, to ensure that the reaction of titration is total (quantitative). We don't titrate weak acid by a weak base (and vice-versa).
I will calculate the equilibrium constant of the titration of  weak acid $\ce{HA}$ by a strong base $\ce{OH-}$:
 $$\ce{HA + OH- <=>A- + H2O}$$ The equilibrium constant is $$K=10^{(\mathrm{p}K_w -\mathrm{p}K_a)}$$ For common weak acids (acetic acid $\mathrm{p}K_a=4.7$ ): $$K>> 10^3$$
So, the equation of titration (neutralization) must be written:$$\ce{HA + OH- ->A- + H2O}$$
At the equivalence point: $$n(\ce{HA})=n(\ce{OH-})$$


StrainViz is the program that might be useful [1].

Reference

Colwell, C. E.; Price, T. W.; Stauch, T.; Jasti, R. Strain Visualization for Strained Macrocycles. Chem. Sci. 2020, 11 (15), 3923–3930. DOI: 10.1039/D0SC00629G.


Magnesium Bicarbonate Water Reaction With MSM - Methylsulfonylmethane

On the reactions happening

It helps to have clear view of what actually happens at what point of the experiment. Before you add anything to your mineral water, $\ce{CO2}$ and water are in the following equilibrium (I am explicitly ignoring all other ions present, even if they are not spectator ions because they too can take part in acid-base equilibria):

$$\ce{CO2 (g) + H2O <=> CO2 (aq) + H2O <=> H2CO3 <=>\\ <=> HCO3- + H+ <=> CO3^2- + 2 H+}\tag{1}$$
$$\ce{CO2 (g) + 3 H2O <=> CO2 (aq) + H2O <=> H2CO3 + 2 H2O <=> \\ <=> HCO3- + H3O+ <=> CO3^2- + 2 H3O+}\tag{1'}$$

(Equation (1') is merely an equvalen version of equation (1) with extra water molecules to show that naked $\ce{H+}$ does not exist in solution.)

Under equilibrium conditions, most of the $\ce{CO2}$ is either dissolved ($\ce{(aq)}$) or present as minute gas bubbles ($\ce{(g)}$) and the amount of carbonic acid is rather low. Adding $\ce{Mg(OH)2}$, an Arrhenius and Brønsted base, shifts these equilibria rightwards by consuming protons:

$$\ce{Mg(OH)2 (s) + 2 H+ <=> Mg^2+ (aq) + 2 OH- (aq) + 2 H+ <=>> Mg^2+ (aq) + 2 H2O}\tag{2}$$

Since reaction (2) will actively push equilibrium (1) rightwards, gaseous $\ce{CO2}$ is dissolved and then reacted, lowering the bottle’s overall pressure. Now you are adding methylsulphonylmethane. This compound can act as an Arrhenius and Brønsted acid by the following equation:

$$\ce{H3C-SO2-CH3 <=> H3C-SO2-CH2- + H+}\tag{3}$$

Hence, it is able to shift equilibrium (1) back leftwards, counteracting the force of reaction (2). The acidity of dimethyl sulphone (as it is also called) is due to the strong $-I$ effect of the sulphone group, which has a formal $\ce{S^2+}$ atom in its centre. The resulting anion can be understood in a similar way as phosphorus ylides can.

Since equation (3) pushes the equilibrium (1) leftwards, it again becomes more favourable to release $\ce{CO2}$ into the surroundings, i.e. form gaseous $\ce{CO2}$, i.e. increase the bottle’s pressure.

On the safety aspects

You should never drink anything that has come into contact with laboratory grade chemicals. These are not tested to be safe for consumption, they are tested and controlled only to be safe for use in a lab (and hence in chemical reactions). There is at least one urban legend of a fatal accident happening due to mixed-up chemical bottles.

That taken care of, it seems that dimethyl sulphone be a relatively safe additive. According to Wikipedia, its $\mathrm{LD_{50}}$ value is larger than $17.5~\mathrm{g / kg}\text{ body mass}$, making it a safe chemical. It is sold as a dietary supplement, showing that the risk seems to be very low. As such, it will probably be safe to drink the soda after addition of both compounds, provided the pH value is where it should be (i.e. dimethyl sulphone has been added sufficiently to counteract the pH increase induced by $\ce{Mg(OH)2}$.)


Persian blue salt is formed by evaporating sea water, as are most if not all salt deposits. In the geosciences, we call salt deposits like this evaporites. Persian blue salt is sodium chloride (the mineral name is halite).
According to this description:

Blue salt is one of the rarest salts known to mankind, and is extracted from a salt mine in the northern province of Semnan in Persia.
The intriguing blue color occurs during the forming of the salt's crystalline structure, as intense pressure is exerted on the salt deposits. The individual crystals fracture the light in an unusual way and the resulting blue (which is caused by an optical illusion), becomes visible.


working with ratios and RMS to find variables

As I said solve using variables first and simplify as much as you can, then plug in the numbers. From the Maxwell-Boltzmann Distribution:

$v_p = \sqrt{\frac{2RT}{M}}$

$v_{rms} = \sqrt{\frac{3}{2}}v_p$

where:

$v_p$ most probably velocity
$v_{rms}$ rms velocity
R universal gas constant
T temperature
M molecular mass

so let:

$v_{p,He}$ - most probably velocity for He
$v_{p,U}$ - most probably velocity for $\ce{UF6}$
$v_{rms,He}$ - rms velocity for He
$v_{rms,U}$ - rms probably velocity for $\ce{UF6}$

we're given:
 $v_{rms,He} = v_{rms,U}$

but $v_{rms,He} = \sqrt{\frac{3}{2}}v_{p,He}$
and $v_{rms,U} = \sqrt{\frac{3}{2}}v_{p,U}$
so:
$\sqrt{\frac{3}{2}}v_{p,He} = \sqrt{\frac{3}{2}}v_{p,U}$
which can be simplified to:
$v_{p,He} = v_{p,U}$

Now substituting from the Maxwell-Boltzmann equation for $v_p$ for both He and $\ce{UF6}$ we have:

$\sqrt{\frac{2RT_{He}}{M_{He}} } =  \sqrt{\frac{2RT_U}{M_U}}$

simplying
$\frac{T_{He}}{M_{He}} = \frac{T_{U}}{M_{U}}$

rearranging:
$\frac{T_{U}}{T_{He}} = \frac{M_{U}}{M_{He}} = \frac{352.02}{4.003} = 87.94$

Superfluous information is that $\ce{UF6}$ decomposes about $1800^o \text{ K}$ and is a solid below about $330^o \text{ K}$. Helium liquefies about $4.2^o \text{ K}$. So this only works if Helium is in range $4.2^o \text{ K}$ to $18.4^o \text{ K}$ which would mean that the uranium hexfluoride is between $369^o \text{ K}$ to $1800^o \text{ K}$.


Why is the 2s orbital lower in energy than the 2p orbital when the electrons in 2s are usually farther from the nucleus?

I think your question implicates another question (which is also mentioned in some comments here), namely: Why are all energy eigenvalues of states with a different angular momentum quantum number $\ell$ but with the same principal quantum number $n$ (e.g., $\mathrm{3s}$, $\mathrm{3p}$, $\mathrm{3d}$) degenerate in the hydrogen atom but non-degenerate in multi-electron atoms?
Although AcidFlask already gave a good answer (mostly on the non-degeneracy part) I will try to eleborate on it from my point of view and give some additional information.
I will split my answer in three parts: The first will address the $\ell$-degeneracy in the hydrogen atom, in the second I will try to explain why this degeneracy is lifted, and in the third I will try to reason why $\mathrm{3s}$ states are lower in energy than $\mathrm{3p}$ states (which are in turn lower in energy than $\mathrm{3d}$ states).
$\ell$-degeneracy of the hydrogen atoms energy eigenvalues
The non-relativistic electron in a hydrogen atom experiences a potential that is analogous to the Kepler problem known from classical mechanics.
This potential (aka Kepler potential) has the form $\frac{\kappa}{r}$, where $r$ is the distance between the nucleus and the electron, and $\kappa$ is a proportionality constant.
Now, it is known from physics that symmetries of a system lead to conserved quantities (Noether Theorem).
For example from the rotational symmetry of the Kepler potential follows the conservation of the angular momentum, which is characterized by $\ell$. But while the length of the angular momentum vector is fixed by $\ell$ there are still different possibilities for the orientation of its $z$-component, characterized by the magnetic quantum number $m$, which are all energetically equivalent as long as the system maintains its rotational symmetry.
So, the rotational symmetry leads to the $m$-degeneracy of the energy eigenvalues for the hydrogen atom.
Analogously, the $\ell$-degeneracy of the hydrogen atoms energy eigenvalues can also be traced back to a symmetry, the $SO(4)$ symmetry.
The system's $SO(4)$ symmetry is not a geometric symmetry like the one explored before but a so called dynamical symmetry which follows from the form of the Schroedinger equation for the Kepler potential.
(It corresponds to rotations in a four-dimensional cartesian space. Note that these rotations do not operate in some physical space.)
This dynamical symmetry conserves the Laplace-Runge-Lenz vector $\hat{\vec{M}}$ and it can be shown that this conserved quantity leads to the $\ell$-independent energy spectrum with $E \propto \frac{1}{n^2}$. (A detailed derivation, though in German, can be found here.)
Why is the $\ell$-degeneracy of the energy eigenvalues lifted in multi-electron atoms?
As the $m$-degeneracy of the hydrogen atom's energy eigenvalues can be broken by destroying the system's spherical symmetry, e.g., by applying a magnetic field, the $\ell$ degeneracy is lifted as soon as the potential appearing in the Hamilton operator deviates from the pure $\frac{\kappa}{r}$ form.
This is certainly the case for multielectron atoms since the outer electrons are screened from the nuclear Coulomb attraction by the inner electrons and the strength of the screening depends on their distance from the nucleus.
(Other factors, like spin and relativistic effects, also lead to a lifting of the $\ell$-degeneracy even in the hydrogen atom.)
Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?
Two effects are important here:

The centrifugal force puts an "energy penalty" onto states with higher angular momentum.${}^{1}$ So, a higher $\ell$ value implies a stronger centrifugal force, that pushes electrons away from the nucleus.

The concept of centrifugal force can be seen in the radial Schroedinger equation for the radial part $R(r)$ of the wave function $\Psi(r, \theta, \varphi) = R(r) Y_{\ell,m} (\theta, \varphi )$
\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \underbrace{ \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} } } - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0
\end{equation}
\begin{equation}
{}^{= ~ V^{\ell}_{\mathrm{cf}} (r)} \qquad \qquad
\end{equation}
The radial part experiences an additional $\ell$-dependent potential $V^{\ell}_{\mathrm{cf}} (r)$ that pushes the electrons away from the nucleus.

Core repulsion (Pauli repulsion), on the other hand, puts an "energy penalty" on states with a lower angular momentum. That is because the core repulsion acts only between electrons with the same angular momentum${}^{1}$. So it acts stronger on the low-angular momentum states since there are more core shells with lower angular momentum.

Core repulsion is due to the condition that the wave functions must be orthogonal which in turn is a consequence of the Pauli principle. Because states with different $\ell$ values are already orthogonal by their angular motion, there is no Pauli repulsion between those states. However, states with the same $\ell$ value feel an additional effect from core orthogonalization.

The "accidental" $\ell$-degeneracy of the hydrogen atom can be described as a balance
between centrifugal force and core repulsion, that both act against the nuclear Coulomb
attraction.
In the real atom the balance between centrifugal force and core repulsion is broken,
The core electrons are contracted compared to the outer electrons because there are less inner electron-shells screening the nuclear attraction from the core shells than from the valence electrons.
Since the inner electron shells are more contracted than the outer ones, the core repulsion is weakened whereas the effects due to the centrifugal force remain unchanged. The reduced core repulsion in turn stabilizes the states with lower angular momenta, i.e. lower $\ell$ values. So, $\mathrm{3s}$ states are lower in energy than $\mathrm{3p}$ states which are in turn lower in energy than $\mathrm{3d}$ states.
Of course, one has to be careful when using results of the hydrogen atom to describe effects in multielectron atoms as AcidFlask mentioned. But since only a qualitative description is needed this might be justifiable.
I hope this somewhat lengthy answer is helpful. If something is wrong with my arguments I'm happy to discuss those points.


Thermodynamic and kinetic products - carbocation stability

Background

First off, let's discuss what we mean by "kinetic" and "thermodynamic" products.  The kinetic product is the first formed product, the one that forms the fastest, the one with the largest rate constant. The rate constant, k, can be described using the Arrhenius equation
$$\mathrm{k=Ae^{\frac{-Ea}{RT}}}$$
where A is the pre-exponential factor, a measure of the probability of a successful collision. Ea is the activation energy, the energy required to surmount the barrier to reaction, R is the gas constant and T is the temperature in Kelvin.

Usually, in a series of related compounds like we have here, the A factor (which is a very large number to begin with, so would need to change by a large amount to alter the course of a reaction) is assumed to remain relatively unchanged.  Consequently the activation energy is viewed as the controlling factor in terms of which reaction pathway is favored.

The thermodynamic product is the product with the lowest energy of formation, the most stable product.

In a reaction where several products can be formed the thermodynamic product could be the same as the kinetic product, or it could be different.

Two examples to further clarify kinetic and thermodynamic products can be found in this earlier answer describing the Diels-Alder reaction between cyclopentadiene and maleic anhydride. The endo isomer is formed first - it is the kinetic product, but if the reaction is run under conditions permitting equilibration, then ultimately the more stable exo product will be predominate. Another example can be found in the sulfonation of naphthalene where naphthalene-1-sulfonic acid is formed initially (the kinetic product), but upon heating will equilibrate and rearrange to the more stable 2-isomer.

A key point here is that in order for the kinetic product to rearrange to the thermodynamic product, the reaction must be reversible under the reaction conditions.

Back to your question

  could one also argue that the far left product is the kinetic product?

Technically, according to your diagram, the answer is "no". There are no equilibrium arrows in your reaction diagram, therefore if the far left product were the kinetic product it would be observed.

But maybe you inadvertently left out the equilibrium arrows, what would the answer be if you included them? All 3 possible products are rather analogous, so for starters we would expect similar A factors in the 3 pathways and look to differences in the activation energies to decide which is the preferred pathway (kinetic product). Your product on the left requires a tertiary carbocation as its intermediate, the other two products require a resonance-stabilized allylic carbocation which is further stabilized by a methyl substituent.  The latter carbocation is lower in energy (resonance stabilization plus inductive and hyperconjugative stabilization by the methyl group) and so the kinetic product will derive from it.

Of the two allyl resonance structures you've drawn, the one on the right will carry a bit more importance in terms of describing the actual allyl carbocation due to the methyl stabilization described above. Hence I would expect the kinetic product to involve attack at the methylated allyl carbon leading to your far right product being the kinetic product.

Attack at the other allyl terminus produces a product with a methyl group on the double bond. We know from heats of formation that alkylation of a double bond stabilizes the system (because electrons are moving from higher [$\ce{sp^3}$] to lower [$\ce{sp^2}$] energy). Hence, your middle product is a possible  thermodynamic product.  Your molecule on the far-left also contains a tri-substituted double bond, it may also appear as a thermodynamic product.

Whether or not equilibration between the kinetic and thermodynamic products occurs depends on the reaction conditions.  Assuming equilibration is possible, then short reaction times at low temperatures will favor formation of the kinetic product, while higher temperatures and longer reaction times will facilitate transformation of the kinetic product into the thermodynamic product.


Can thermite be lit while mixed into butane? If not is there a flammable liquid that would work?

Thermite is a solid-solid reaction that I think would be greatly inhibited by any sort of intermediate. In either case, the smoke, the dust and the copious amounts of nitrous oxides produced will not make your asthma any better off than if you were to pick up a shovel.

And then there will likely be other safety-, legal-, environment- and relationship-with-neighbours -related issues that alone would make this a very bad idea.


Question 1:

You are mixing the sign conventions for work from physics and chemistry. Sadly, they are not the same. In chemistry, the work done by the system on the surroundings is negative to signify energy leaving the system. Any transfer of energy (heat, work, etc.) from the system to the surroundings is denoted with a negative sign to ensure the energy change of the system is negative (energy is lost). Chemistry is concerned ultimately only with the system (usually the chemical reaction under investigation), and not the surroundings (apparatus, equipment, experimenter, laboratory, and the rest of the universe).

Thus, for pressure-volume work:

$$W=-p\Delta V$$
$$\delta W=-pdV$$

We need this negative sign. The system does pressure-volume work on the surroundings when the system expands (think a piston in an automotive engine) so $dV>0$. Without the negative sign, we would assume that the system gains energy and the surroundings lose energy when the system expands.

Thus, in the chemistry formulation of the first law (assuming only pressure-volume work) is:

$$dU=\delta Q + \delta W =TdS - pdV$$

In physics, some consideration is given to the surroundings (and in some cases the surroundings may be more interesting than the system, such as in the proper definition of entropy). Especially in the early days of physics, work leaving the system was useful in that it could be utilized. Thus, in physics $W=p\Delta V$ means work is being done by the system on the surroundings. Thus, the first law as formulated in physics is:

$$dU=\delta Q - \delta W = TdS -pdV$$

Note however, regardless of the sign convention in the definition of $W$ as a function of $p$ and $\Delta V$, we still end up with $$dU=TdS-pdV$$

The sign of $W_{\text{max}}$ in chemistry needs to be negative to follow the same conditions of chemistry. Thus, the equation in your chemistry textbook is correct. An equivalent definition in physics would be:

$$W_{\text{max}}=nFE_{\text{cell}}$$
$$\Delta G_\text{cell}=-W_\text{max}=-nFE_\text{cell}$$

In chemistry the negative sign goes in the definition of work. In physics the negative sign goes in front of the work term. Don't mix and match conventions, and you will get consistent answers.

  Question 2:

$\Delta G = 0$ is the definition of equilibrium not the conditions for reversibility. A reaction can be reversible (and most electrochemical reactions are) and still be far from equilibrium at the initial conditions.

  Question 3:

Temperature is not required to be constant here. Also, note that the term $W_\text{max}$ refers to the maximum possible electrochemical work. Usually less electrochemical work is done because heat is also generated (which changes the temperature). Also, most electrochemical systems are not closed. Even a battery is not a closed system (by the strict physics definition). It's hard to maintain constant pressure in a closed system.

  Question 4:

Technically the formulation of the first law that you used is explicitly for pressure-volume work done by closed gaseous systems. As most redox reactions occur in solution in open systems, that formulation is invalid for a different reason.

  Question 5:

Given that your initial equation is correct, then for a spontaneous reaction:

$$\Delta G < 0$$
$$\therefore -nFE_\text{cell} < 0$$
$$nFE_\text{cell} > 0$$
$$E_\text{cell} > 0$$
$$E_\text{red} - E_\text{ox} > 0$$


What are some good references on conductometry as an analytical technique?

I am certainly no expert in the field (in fact, I know nothing about it) but I imagine this 225+ page text will prove to be useful for you:
Oscillometry and Conductometry: International Series of Monographs in Analytical Chemistry, Volume 21 by E. Pungor is available via Google Books. You might be able to find a used copy of this book somewhere (it was published in 1965).


Water has hydrogen bonding in it. Hydrogen bonding is some kind of intermolecular force (a tutorial and the wikipedia page) that is usually seen in molecules that have $\ce{OH}$, $\ce{NH}$ or $\ce{FH}$ somewhere in their structure.
How does it happen?
Hydrogen atom is really small (atomic radius: About 37 pm) When it bonds with some very electronegative species, like O atom, (in the case of water), the electron is highly absorbed by the more electronegative. That results in a relatively high concentration of positive charge. The other O atom in water is attracted to H and this is the basis of hydrogen bonding.
So...?
Hydrogen bonding in water is far too weak to be compared with intramolecular bonds like covalent or ionic bonds, but it's strong enough that to be broken, it will require lots of energy. That's why water boils at $100 ° C$.
Many know the reason of water's high specific heat to be hydrogen bonding. Partially, it's related to that: In giving heat to water, some heat is "spent" on loosening the hydrogen bonds rather than increasing water's kinetic energy. The argument is valid, and I haven't seen anyone bringing evidence as to its being wrong, but another reason, which was the one hyperphysics gave, relies on the motion of water.
Expanding the topic isn't necessary, and it will add up to confusion. In short, the water molecule is nonlinear. That will result in three rotational degrees of freedom that allows it to have an awful lots of rotational motions! (More types of motions are a cause of an additional bonus to the specific heat)
Liquid ammonia has a higher specific heat, just for the same reason.
Even a very easier reason to understand is the water's relatively low molar mass: $\frac{18 g}{mol}$. What should that mean? More moles of water per one kilogram.
As a summary, the reasons are:

The heated water will contribute much of the heat to loosening, bending or breaking the hydrogen bonds.
Water had three rotational degrees of freedom. In addition to vibration, rotations happen a lot to water molecules. This will result in a higher heat capacity.
Specific heat capacity is defined as the amount of heat required per unit mass to increase the temperature by a degrees celsius. The relatively low molar mass of water allows more moles of it to be there in a mass unit (either kg or g)

As a side note, the exceptional high specific heat is not the only weird property of water. See here for example.


Iron(II) reduces nitrite to nitrogen monoxide which forms a brown pentaaquanitrosyliron(II) complex with the iron(II) ions.

$$\ce{NO2^{-} + Fe^{2+} + 2H+ -> NO + Fe^{3+} + H2O}$$

$$\ce{[Fe(H2O)6]^{2+} + NO -> [Fe(H2O)5NO]^{2+} + H2O}$$

Iron(II) can also reduce nitrate to nitrogen monoxide. This is used in the brown ring test to determine the presence of nitrate in aqueous solutions.


Why is in the reaction of cyclic unsaturated aldehydes with hydrazine and hydrogen peroxide the aldehyde preserved?

The reaction of $\ce{H2O2}$ with hydrazine forms diazene, $\ce{HN=NH}$.  This reacts in a syn fashion to deliver both hydrogens to the same side of the alkene.


Latimer diagram for a metal in basic solution

Let's look at how you would add in the water and hydroxide ions for the $\ce{MO4^{3-}->MO(OH)}$ reaction.

Step 1:  Start with the given metal species, with the metal oxidation states included:

$\ce{M^VO4^{3-}->M^{III}O(OH)}$

Step 2:  Add the electrons according to the oxidation states.  Here the oxidation state on the metal drops by two, so there must be two electrons on the left.  Ignore the fact that the overall charges may be unbalanced, that comes next.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH)}$

Step 3:  Now we are ready to attack the charge balance.  Add the hydroxide ions, which do not couple with the electron exchange, to balance the charges.  Counting out the charges we have going into this step we need five hydroxide ions on the right.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH) + 5OH^-}$

Step 4:  Finally add water to balance hydrogen; if you rendered the oxidation states and did the math correctly, the oxygen balance falls into place simultaneously.  With six hydrogen atoms to balance we use three molecules of water to get the final answer for this reduction:

$\ce{M^VO4^{3-} + 3H2O + 2e^- ->M^{III}O(OH) + 5OH^-}$

Can you now do the remaining reductions?


The fact that the book is using $\Delta G$ as the relevant state variable for free energy implies this is being carried out at constant T and p.  So, proceeding:

  "However, this implies that we are getting usable work from increases in entropy...."

Yes, that's correct.  The positive $\Delta S$ lowers $\Delta G$

  "However, this implies that we are getting usable work from increases in entropy – meaning that the muscles are somehow absorbing heat energy from the environment."

No, that's incorrect.  The heat flow here is purely due to $\Delta H$, and it is flowing out of the system.

What you're probably thinking is that, at constant T (which we have here), $$\Delta S = \frac{q}{T},$$
and thus, since $\Delta S >0 $,  q must be $> 0$ as well, i.e., that the  positive $\Delta S$ means heat is flowing into the system.

This is wrong.  It is not the case that $\Delta S = \frac{q}{T}$.  Rather:

$$\Delta S = \frac{q_{rev}}{T},$$

and this reaction is not happening reversibly.  Instead,  $\Delta S$  is determined from the difference between $S_{products}$ and  $S_{reactants}$, and each of these individual values are calculated from:

$$S(T') =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$$

[For more on this, see my answer here:  Infinite Increase in Entropy when Energy added to Absolute Zero ]

In summary, in a chemical reaction at constant pressure (and with no non-pV work),

$$\text{đ}q_{sys}=dH_{sys}$$

Thus it is the sign of $\Delta H$, not $\Delta S$, that determines the direction of heat flow.

Finally, the answer to (d) is that, as thermal energy is generated by the reaction, it heats up the system, and the 2nd law of thermodynamics says that thermal energy naturally flows from hotter to cooler regions.  So there is heat flow into the surroundings until the reaction mixture returns to the temperature of the surroundings.


Why do we use external pressure when calculating work done by gas?

The ideal gas law (or other equation of state) is valid only when a gas is in thermodynamic equilibrium.  In a rapid irreversible expansion or compression, the pressure depends not only on the volume (and temperature) but also on the rate of change of volume.  This is because a gas has a viscous property which acts to dissipate mechanical energy and convert it to internal energy. In a reversible process, where the rate of change of volume is negligible, the ideal gas law can be used.  But the ideal gas law (or any other equation of state for that matter), cannot be used to determine the pressure (or more precisely, the compressive stress) in an irreversible process.

At the interface between the gas and its surroundings (say the inside face of a piston) where the gas is doing work, by Newton's 3rd law, the pressure of the gas matches the "external pressure," and, if this external pressure is known, the amount of work that the gas does on its surroundings can still be determined. In the "vertical piston" example, if one does a force balance on the piston during the process, one obtains:
$$P_{gas,I}=\frac{Mg}{A}+P_{atm}+\frac{M}{A}\frac{d^2V}{dt^2}=P_{ext}$$ where $P_{gas,I}$ is that gas pressure at the interface with the piston (the subscript I is present because, in an irreversible process, the pressure of the gas within the cylinder may not even be uniform), and the mass times acceleration of the piston is included. The work done by the gas on its surroundings during the process is obtained by integrating this equation between the initial and final equilibrium states to yield$$W=\int{P_{gas,I}dV}=Mg\Delta h+P_{atm}\Delta V=\int{P_{ext}dV}$$This equation acknowledges that, as a result of the viscous property of the gas, the kinetic energy of the piston has been damped out by the time the system reaches its final state.


Why is the effect of beta-branching neglected in this comparison of SN2 reaction rates?

Yes, the beta branching will hinder the nucleophile a bit, but the substrate is still a primary halide with 2 hydrogens on the carbon that is attacked. The steric effect of having a methyl group and a hydrogen on that carbon vs two hydrogens is very much larger.


Does lead carry the negative charge in plumbane?

$\ce{Pb^{4+}}$ is very unstable due to the inert pair effect, wherein for p-block elements lower oxidation state becomes stabler as we move down the group. Also, the main factor is the electronegativity of lead being more than hydrogen, which you stated in your question (although they are very close). Hence the oxidation states would not be flipped.


How do I have to interpret the input to the OCC flag in Molpro?

You are correct in your interpretation of the OCC flag (that command corresponds to $3$ electrons in $\mathrm{a_g}$ orbitals, $1$ in $\mathrm{b_{3u}}$, $1$ in $\mathrm{b_{2u}}$, and $2$ in $\mathrm{b_{1u}}$). (Because these are orbitals, not overall wavefuction symmetries we use lowercase letters)

Your confusion seems to be arising from visualizing the orbitals which you are including. Remember that these representations refer to the symmetry of the molecular orbital, not atomic orbitals. $\mathrm{a_g}$ refers to the set of orbitals which have no phase change under all of the symmetry operations, so that corresponds to the $\sigma(\ce{1s})$, $\sigma(\ce{2s})$, and $\sigma(\ce{p_z})$, orbitals. Similarly, $\mathrm{b_{1u}}$ refers to the filled $\sigma^*(\ce{1s})$, $\sigma^*(\ce{2s})$. Finally the $\mathrm{b_{3u}}$, and $\mathrm{b_{2u}}$ orbitals refer to the filled $\pi(\ce{p_x})$ and $\pi(\ce{p_y})$ orbitals.

As we know, the occupation of $\ce{N2}$ is:
$$
  \ce{\sigma(1s)^2 \sigma(2s)^2 \sigma^*(1s)^2 \sigma^*(2s)^2
  \pi(p_x)^2 \pi(p_y)^2 \sigma(p_z)^2},
$$
and using the $D_\mathrm{2h}$ point group we can rewrite this as:
$$\mathrm{1a_g^2\ 2a_g^2\ 1b_{1u}^2 \ 2b_{1u}^2 \ 1b_{3u}^2\ 2b_{3u}^2\ 3a_g^2}.$$
Hopefully, this rewriting of the occupation pattern in the chosen symmetry makes the choice of orbitals clear.

To really understand which orbitals fall under which irreducible representation you need to visualize them (or draw them in the case of a dimer) and then pull out the character table for the point group.

You can find a $D_\mathrm{2h}$ character table on the homepage of Dr. Achim Gelessus (Jacobs University Bremen).


What's the smallest Hoberman sphere possible?

Hoberman sphere shares the same topological features being an icosidodecahedron with a family of polyoxometalates called Keplerates $\ce{\{(\color{teal}{Mo|W})_72\color{orange}{M}_30\}}$ ($\ce{\color{orange}{M} = Fe,Cr,V}$, anionic residues and alkali metal cations are removed for clarity) ([1, 2]):

Also, there is a cluster $\ce{[\color{blue}{La}_20\color{green}{Ni}_30(IDA)30(CO3)6(NO3)6(OH)30(H2O)12](CO3)6 * 72H2O}$ (IDA - iminodiacetate $\ce{C4H5NO4}$) consisting of an icosidodecahedron of Ni(II) Ions
encapsulating an inner La(III)-dodecahedron [3]:

The problem is, these structures only resemble the symmetry, but not the expandability of Hoberman sphere. Currently there are many attempts of creating nanomachines with mechanically active molecular assemblies, but I didn't find any working models close to Hoberman structures (yet).

Also, some "spherical" viruses such as Cowpea chlorotic mottle virus (CCMV) with a structure of truncated icosahedron, which is significantly larger than Keplerates ($\pu{28 nm}$ (CCMV) vs. $\pu{3.2 nm}$ (average Keplerate)), can undergo swelling (pH < 6.5) and expansion (pH > 6.5), and show pH and metal ion dependent polymorphism [4, 5]

  as a result of deprotonation of carboxyl moieties at the inter-capsomere contacts, leading to electrostatic repulsion that opens out the structure but falls short of complete disassembly through preservation of interwoven
  carboxyl/protein links between capsomeres.

Don't forget that "there's plenty of room at the bottom", so Hoberman sphere can possibly be a real working entity at molecular scale.

Bibliography

Botar, B.; Ellern, A.; Hermann, R.; Kögerler, P. Angewandte Chemie International Edition 2009, 48 (48), 9080–9083. DOI 10.1002/anie.200903541.
Schäffer, C.; Merca, A.; Bögge, H.; Todea, A. M.; Kistler, M. L.; Liu, T.; Thouvenot, R.; Gouzerh, P.; Müller, A. Angewandte Chemie 2009, 121 (1), 155–159. DOI 10.1002/ange.200804496.
Kong, X.-J.; Ren, Y.-P.; Long, L.-S.; Zheng, Z.; Huang, R.-B.; Zheng, L.-S. J. Am. Chem. Soc. 2007, 129 (22), 7016–7017. DOI 10.1021/ja0726198.
Speir, J. A.; Munshi, S.; Wang, G.; Baker, T. S.; Johnson, J. E. Structure 1995, 3 (1), 63–78. DOI 10.1016/S0969-2126(01)00135-6.
Kovacs, F.; Tarnai, T.; Guest, S. D.; Fowler, P. W. Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences 2004, 460 (2051), 3191–3202. DOI 10.1098/rspa.2004.1344.


What are the subfields of chemistry that work very closely with biology?

In addition to jonsca's link of chemistry subfields, Wikipedia also has a list of biochem subfields. Find what research interests you, perhaps not based on a course you've taken.

You can look up what professors are doing in their respective fields of research by visiting faculty research group sites online. Check out their publications, current projects, probably would be good to join a lab too. (I'm blindly assuming you're interested in research and/or graduate studies)

The list on Wikipedia first lists (for whatever reason these are "main branches" but I'm surprised biotech is categorized under "other branches"...):

Animal Biochemistry
Plant Biochemistry
Molecular Biology
Cell Biology
Metabolism
Immunology
Genetics
Enzymology

Getting into more or less random scopes of research topics which may or may not be within the above fields:

Biotechnology,
Bioluminescence,
Molecular chemistry,
Enzymatic Chemistry,
Genetic engineering,
Pharmaceuticals,
Endocrinology,
Hematology,
Nutrition,
Photosynthesis,
Environmental Chem,
Toxicology
(Probably many, many others...)


Which calcium salt is good enough to make calcium benzoate?

Thanks to @Shoubhik who pointed out that calcium hydroxide is good enough to make calcium benzoate. To support this claim, I found a patent which indeed describes that calcium hydroxide was used to make calcium benzoate (See comments). Here is the summary of the patent:

  The invention relates to a processing method for synthesizing calcium
  benzoate at low temperature. The method comprises the following steps:
  heating and stirring 65-85kg of industrial-grade benzoic acid, 28-32kg
  of calcium hydroxide, 1-3kg of catalysts comprising triethanol
  distearate, ethylene distearamide and ethylene bislauramide and
  300-500kg of clear water in a reaction kettle at 333-363 K to react
  for 2-4 hours; and then adding 5-15kg of separant, stirring for half
  an hour, filtering and drying. The method provided by the invention
  can be used for directly synthesizing calcium benzoate, thereby
  lowering the equipment investment, reducing the working procedures,
  lowering the labor intensity of workers, increasing the output,
  enhancing the quality, lowering the cost and preventing the
  environmental pollution.


Why is the probability of a didentate ligand to form a chelate with a central atom, high?

Speaking in general, when a molecule is in a "energetic" environment (in this case, for instance, the energy comes from the temperature of the room, about 298 kelvins away from the "zero energy" point), the molecule moves A LOT. It bends, rotates, and changes its conformation.
When EDTA has no complexed metal, its "two arms" just keep moving randomly, until they catch a cation. When one of the arms "catches" a cation, the arms keep moving randomly. But this time, it's much more likely that the other arm will find the cation, being it so close (the cation won't move too far: the other "arm" keeps it close!).
When both arms find the cation, the rigid complex is formed.

To make a stupid example, think about a person with two magnets on his hands, moving randomly in a room in the hope of capturing with both his hands a metal ball: the "first arm" has to search around all the room, but the second one only has to find the first one


Why can't this mixture be used as a buffer solution?

You have an example of an alkaline buffer solutions here. They are commonly made from a weak base ($\ce{NH3}$ aqueous solution) and one of its salts ($\ce{NH4Cl}$) and work in a little simplified picture as follows.

Ammonia is a weak base, so the position of the equilibrium for the following reaction is well to the left
$$
    \ce{NH3_{(aq)} + H2O_{(aq)} <<=> NH4+_{(aq)} + OH^{-}_{(aq)}} \, ,
$$
and once you add some $\ce{NH4Cl}$ it pushes the equilibrium of the reaction above even more to the left (see Le Chatelier's Principle). As the result you have both lots of ammonia (due to the equilibrium of the above reaction being well to the left) and of ammonium ions (coming from $\ce{NH4Cl}$). And you will still have some hydroxide ions due to the reaction above, so that solution will be alkaline. The actual $pH$ depends on in which molar proportions the ingredients were mixed. Usually, $\ce{NH3_{(aq)}}$ and $\ce{NH4Cl}$ mixed in equal molar proportions, the solution would have a pH of 9.25.

When some acid is added to such buffer it will mostly be removed by the following process
$$
    \ce{NH3_{(aq)} + H+_{(aq)} <=>> NH4+_{(aq)}} \, ,
$$
and when some base is added it will mostly be consumed by the following reaction with ammonium ions
$$
    \ce{NH4+_{(aq)} + OH^{-}_{(aq)} <=>> NH3_{(aq)} + H2O_{(aq)}} \, ,
$$
where in both cases the equilibrium is well to the right.

So, as you see, both the weak base and its salt should present in the solution. Thus, there should be some excess of weak base when you mix it with a strong acid, otherwise there will be no weak base at the end of neutralization reaction. The idea that one should have double amount of a weak base comes from the fact that, as I mentioned, usually a weak base and its salt are mixed in equal molar proportions.

$%edit$


Why does the sulfone ring in tazobactam open when the lactam is hydrolyzed?

Here's a draft:
First, read this paper. Considering we're under alkaline conditions (blood pH between 7.35-7.45) we'll take their alkaline mechanism proposal.
Mechanism

This is obviously incomplete, with good reason. Without knowing what the active site of the enzyme looks like (except for Ser70), it's difficult to devise of a mechanism. I could imagine another alkoxy attacking the sp3 carbon between S and N, and shifting two electrons to the sulfur thus generating your SO2- in one step, but that would require a proper active site.
The geometry of the molecule inside the active site of the enzyme is obviously optimal for this reaction to proceed. I'd like to know more. For some reason my school doesn't give me access to this paper.


Markovnikov's addition to halogenated alkene?

Markovnikov's rule is usually stated in terms of subsitution on the alkene, for example:

  In the addition of HX to an alkene, the hydrogen atom always attaches at the less substituted carbon atom (the carbon atom which already has more hydrogen atoms).

This version of the rule works well most of the time. It works in your case. The hydrogen atom of $\ce{HCl}$ end up on the carbon atom that already has two hydrogen atoms. The chlorine atom ends up on the other carbon atom:

$$\ce{CH2=CHCl +} {\bf \ce{H-Cl}} \ce{-> {\bf H}-CH2-CHCl-{\bf Cl} }$$

A better version of the rule has to do with the stability of the carbocation intermediate (and can be generalized to electrophiles other than $\ce{H+}$.

  An alkene reacts with electrophiles in such a way as the electrophile attaches to the carbon atom which will result in the more stable carbocation.

The order of stability for simple carbocations is $3^\circ >2^\circ >1^\circ >\text{methyl}$, however there can always be more complex carbocations (de)stabilized by heteroatoms and/or resonance. The original wording of Markovnikov's rule may fail in these more complex cases.

For example, the addition of $\ce{HX}$ to $\ce{CH3O-CH=C(CH3)2}$. In this case the major product (before it decomposes into the aldehyde) is $\ce{CH3O-CHCl-CH(CH3)2}$. Two possible carbocations form, and the one where the proton attaches at the more substituted position is the one that is more stable.

First carbocation, $2^\circ$, but stabilized by resonance (more stable):

$$\ce{CH3-O-CH+ -CH(CH3)2 <-> CH3-O+ =CH-CH(CH3)2}$$

Second carbocation, $3^\circ$, but not stabilized by resonance (less stable):

$$\ce{CH3-O-CH2-C+(CH3)2}$$

The first carbocation goes on to form the product. Can you extend this to what happens to your molecule?


A true annihilation peak at $E = 511\ \mathrm{keV}$ is caused by the detection of individual annihilation photons. The source of such annihilation photons can be:

1) Pair production in the surrounding material caused by high-energy gamma radiation
If the gamma energy is large enough to make pair production in the surrounding material (typically in the high-$Z$ shielding material surrounding the detector) relevant, the gamma quantum (a high-energy photon) may disappear and be replaced by an electron and a positron. The electron and positron may travel a few millimetres before losing their kinetic energy to the absorbing medium. When its kinetic energy becomes low, the positron may combine with an electron in the absorbing medium. Then both disappear and are replaced by two annihilation photons of $E = 511\ \mathrm{keV}$ each.

2) Positron emission of the source
If the source consists of a radionuclide that decays by positron emission (β+ decay), and the positron is stopped and undergoes annihilation before it reaches the detector (i.e. the positron is stopped within the source volume or in the material that is encapsulating the source), this region acts as a source of annihilation photons of $E = 511\ \mathrm{keV}$ each.

If one annihilation photon reaches the detector and is absorbed within the detector, an annihilation peak appears in the spectrum at an energy of $E = 511\ \mathrm{keV}$.

Furthermore, a peak at $E = 511\ \mathrm{keV}$ could be caused by mere coincidence due to

1) Coincidence summing
If two gamma rays are detected approximately at the same time, the spectrum will show only one count at the sum of their gamma energies. If the sum amounts to approximately $E = 511\ \mathrm{keV}$, the resulting peak would look like an annihilation peak.

2) Double escape
If the gamma energy is large enough to make pair production in the detector material relevant, the photon may disappear and be replaced by an electron and a positron. The electron and positron may travel a few millimetres within the detector before losing their kinetic energy and finally combining with an electron. Then both disappear and are replaced by two annihilation photons. If one annihilation photon escapes from the detector without interaction within the detector, a single escape peak appears in the spectrum at an energy of $511\ \mathrm{keV}$ below the full-energy peak. If both annihilation photons escape, a double escape peak appears at an energy of $1.02\ \mathrm{MeV}$ below the full-energy peak.
If the remaining energy is approximately $E = 511\ \mathrm{keV}$, the resulting escape peak would look like an annihilation peak. This effect is not possible for single escape peaks since their energies are always well above $E = 511\ \mathrm{keV}$, but it is theoretically possible for double escape peaks.


What is the cause for thermodynamic reversible and irreversible process?

Let me start by reiterating what @Phillip said in his comment: this is not a good definition for an irreversible processes - a better definition is:

  An irreversible process is one in which the net entropy generation term is greater than zero.

All real processes are irreversible. The "process speed" comes into play when we want to make approximations for idealized systems. It turns out that one way to convert a process into a theoretically reversible one is to set the condition that it moves very slowly between thermodynamic states, so that at any given step, it is in equilibrium. The reason that this isn't a good definition for a reversible process is that there are cases where a rapid change is reversible, and cases where a very slow change is irreversible. It's just that in general use, the "very slow process" approximation is commonly used to obtain a reversible process.

  Which equilibrium are they speaking of?

Typically they mean equilibrium between the system and surroundings, but it can also mean within the system itself. Equilibrium is another way of saying "maximum entropy" - the matter and energy are as "spread out" as they are going to get. If the entropy is maximized, no more can be generated. In "real life" this means the system is well-mixed and that matter and thermal energy has dissipated throughout the system - there are no temperature or concentration/density gradients. This is easier to visualize using particles instead of energy - imagine an ideal gas in one side of a partitioned chamber, where the partition has a hole in it to allow the molecules to cross into the other section:

This is an example of an internal equilibrium - the particles cannot cross the system boundary, but they can reach an equilibrium within the system. An example of a system in equilibrium with its surroundings would be a container filled with water at room temperature - there are no thermal gradients across the container boundary.

  Why does rapidity of the change prevent it from attaining equilibrium (how does an infinitesimal change in a reversible process help it to reach equilibrium)?

Whenever a system is perturbed, meaning moved away from an equilibrium state, it takes some amount of time to reach a new equilibrium. This amount of time is called the relaxation time. Different systems have different relaxation times, and if you make changes quickly enough, the system won't have time to relax, and therefore won't be in equilibrium. Using a piston in a cylinder as an example, if you compress the piston by a very small amount, then the pressure inside will be in equilibrium with the pressure outside almost immediately.

  What do they mean by driving and opposing force?

Using the piston as an example again, in this case the driving force would be the external pressure on the piston. The opposing force would be the pressure of the gas inside the cylinder. For a very slow process, at any given time the outside pressure and inside pressure would be the same - the volume would change to make sure this is true. For a very rapid change, the gas inside the cylinder would take longer to reach equilibrium, and so the pressure outside would not equal the pressure inside - the driving and opposing forces would not be equal.

It is important to remember that these are approximations that are intended to allow us to convert really difficult (usually unsolvable) differential equations that describe non-equilibrium or dynamic systems into something that is solvable. The trade-off is that we go from a realistic, fully dynamic system to one with constraints - we treat it like a series of tiny jumps between systems that are in equilibrium. In many cases, these approximations don't affect the accuracy of the solution. However, whether or not it will is determined by the time scale of the change relative to the relaxation time of the system.


Why do spin isomers of hydrogen (ortho and para hydrogen) change their nuclear spin with temperature variance?

The effect is due to the symmetry properties of the rotational energy levels and those of the nuclear spin. The change with temperature is, as usual, traced back to the Boltzmann distribution.

In homonuclear diatomic molecules such as $\ce{H2}$ it is necessary to consider the spin of the overall wavefunction to exchange of nuclei. If this is $\Psi = \psi_t\psi_v\psi_R\psi_n$. The translational wavefunction $\psi_t$ depends on the coordinates of centre of mass, the vibration only on separation of nuclei. Thus we need only consider the product of rotational and nuclear wavefunctions, $\Psi = \psi_R\psi_n$. As there are identical nuclei the degeneracy is $g=2s_n+1$ with nuclear spin $s_n$ ($=1/2$) for hydrogen atoms.

As the product wavefunction $\Psi$ is neither odd nor even, it is necessary to ‘symmetrise’ by making the symmetric form in terms of states $\psi_1$ and $\psi_2$ for particle $(1)$ and $(2)$, as these also satisfy Schrodinger’s equation.

$$\psi_S=\psi_{1}(1)\psi_{2}(2)+\psi_{1}(2)\psi_{2}(1)$$

and asymmetric form

$$\psi_A=\psi_{1}(1)\psi_{2}(2)-\psi_{1}(2)\psi_{n}(1)$$

The number of symmetric functions is $g_n(g_n+1)/2$ and asymmetric $g_n(g_n-1)/2$ making a total of $g_n^2$.

The rotational wavefunctions are symmetric for even J rotational states and asymmteric for odd rotational quantum numbers J. As the product $\psi_R\psi_n$ must be asymmetrical for a molecule with an odd mass number for even J only asymmetrical nuclear states are accessible and  and vice versa.

The partition function for odd mass number is
$$Z= \frac{g_n(g_n-1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n+1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) $$

and for $\ce{H2}$ the partition function is
$$Z= 1\sum_{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) + 3\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) $$

From the equilibrium distribution equation the ratio of number of ortho- to para- molecules is therefore

$$\frac{N_{ortho}}{N_{para}} = \frac{ 3\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT))  } {   \sum_{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT))} $$

As the temperature changes so does this ratio; as the temperature tends to infinity the ratio tends to a limit of 3 and to zero as the temperature tends to zero.

For any spin degeneracy the ratio in the limit of very high temperature is
$$\frac{N_{ortho}}{N_{para}}=\frac{g_n+1}{g_n-1}$$
and for $\ce{D2}$ with spin of $1$ and $g_n=3 $ the ratio is $2$.

The figure shows the ratio for hydrogen.


Why is the Vitamin B complex, a "complex"?

It is essentially exactly what your teacher says: the term vitamin B complex is used because the structures assigned a name ‘vitamin Bx’ don’t really have anything in common although they are all somewhat important.

As to why nobody ever uses vitamin K complex or vitamin D complex: well, at least in my country those vitamins are very under-represented when compared to ABCE. ABCE vitamins are those often seen as the most important, the most valuable or similar.[citation needed] They are often advertised as being included in supermarket products. The advertising industry probably doesn’t want to ‘establish’ yet another thing they could add but would be rather cost-intensive.


The issue at hand is competition between the intermolecular forces that pull the solvent together and the forces that cause the solvent to adhere to the surface. In this case, the solvent's interactions with itself are greater, causing the ink to bead. Since the amount of ink that is actually drawn out when writing is very small, the beads of ink are tiny and very hard to see.


Why Frenkel defect does not impact the density of the solid?

The sources may not explicitly say this, but I suspect that they consider mass density to be a macroscopic property, to be measured from at least milligrams and cubic millimetres.

If the volume considered is small enough, then yes, one would see local variations of the density. However, if you make the volume even smaller, you would see huge variations in the density as additional nuclei are included. (Similar statements will be true for Frenkel defect concentrations.) On this level, the definition of mass density simply breaks down and is no longer very useful.


Formaldehyde, $\ce{H_2CO}$, has the following structure

There is no $\ce{OH}$ group, just a central carbon atom with a single bond to each of two hydrogens and a double bond to oxygen.


Are the stereoisomers of nucleosides actually anomers?

The C1 position of carbohydrates and carbohydrate derivatives is generally referred to as the anomeric center and compounds differing only in their configuration at this position are called anomers. In nucleosides, the 1'-position may not be part of a hemiacetal or acetal functional group, but it is at the same oxidation state. The IUPAC Gold Book defines anomers as:

  Diastereoisomers of glycosides, hemiacetals or related cyclic forms of sugars, or related molecules differing in configuration only at C-1 of an aldose, C-2 of a 2-ketose, etc.

The fact that the definition includes "glycosides" and "related cyclic forms of sugars, or related molecules" points towards why it is acceptable to describe the 1'-position as the anomeric center. Furthermore, this position in nucleosides is at the same oxidation state as a hemiacetal or acetal. Therefore, there are structural and chemical reactivity similarities between these functional groups and the 1'-position of nucleosides.

The correct name for this R1C(OR2)(NR3R4)R5 functional group is a hemiaminal ether. Hemiaminals have a free hydroxyl group: R1C(OH)(NR2R3)R4.

Also of note is that the anomeric effect refers to the preference for polar groups to be axial on pyranose derivatives, where the polar group may be not only oxygen substituents (eg. hemiacetals, acetals), but also other derivatives such as halides (eg. glycosyl halides). In all these cases, C1 would be referred to as the anomeric center.


Reconciling multiple inorganic properties for the purposes of learning these elements

first of all take one group a day.
right whatever you have to learn and then try to identify some relationship. Then reread the material and write down whatever you remember . Now look what you miss.
everyday take a new group and do the same.
Also write whatever you remember of the previous groups you have learned.


What is the official file extension for SMILES and SMARTS files

I don't think I've ever seen anything proposed as "official," though I admit I don't use Daylight's official toolkit.

Open Babel accepts:

.smi
.smiles
.can (ideally canonical SMILES, but that matters more for writing, not reading)

I haven't seen many SMARTS files, but I would suspect both .sma and .smarts are out there "in the wild."

Incidentally, this is probably something to bring up with the OpenSMILES effort. I know there's an OpenSMARTS specification started as well.