The turnover frequency is a rate. (It's actually the kinetic rate of the reaction in saturating substrate concentration, normalized to the amount of enzyme.)

It might be helpful to think about something like a factory that makes a certain number of widgets in a certain number of time. How would you naturally describe how fast the factory could make widgets? Most people would express how fast the factory works in terms of "widgets per hour". It's the same with enzymes - it's most natural to describe how fast the enzyme works in terms of "turnovers per second". It's just that the "turnover" is implied and is viewed as "dimensionless", so it's left off, leaving you simply with "s-1" as a unit.

(Note that even though the "turnover" is left off, it's still very important to know it's there, and to know what, exactly, one "turnover" is. In complicated reactions there is sometimes different reasonable ways to count what "one turnover" is, and different definitions could result in a 2, 3, 4, 5, etc. fold difference in value. Those vexing numeric differences become easy to explain once you realize this researcher is using the formation of one molecule of this product as "one turnover", but that researcher is using the disappearance of one molecule of that substrate as "one turnover".)

Naively, one could invert the ratio to come up with a value in time, rather than in 1/time. For example, if a factory makes 5 widgets per hour, it takes them 0.2 hours to make a widget ... sort of. One issue you run into is that that's the time on average. In a stochastic process like enzyme turnover, it's highly unlikely that any single turnover will take exactly the time specified. Some will take longer, some will be quicker. Only for a large number of turnovers in aggregate does that time value have meaning. Saying that "a turnover takes 5 s" gives a false impression about individual turnovers in a way that saying "this enzyme works at 0.2 turnovers per second" doesn't.

Additionally, in the case of multi-step reactions, saying that "a turnover takes 5 s" may be incorrect. The entire pathway may take much more that that. It's just that once the pathway has reached steady state it's only the rate limiting step which takes 5 s.  For example, take a pathway of 20 steps, 19 of which take 4 s each, but one which takes 5 s. It's going to take (19*4 + 5) = 81 seconds for any given substrate molecule to start at the beginning of the pathway and make it to the end. However, once the pathway is full and each step is handing off products to the next step, the pathway will be producing product at a rate of 1 every 5s. (This, of course, doesn't apply to enzymes where all reactions take place at the same active site, but it's a consideration in the general case, which explains why 1/s is the preferred unit for reaction rates.)


Why is the carbon dioxide we exhale transparent?

Essentially, because the carbon dioxide sublimates from solid (dry ice) to gas at a very low temperature (roughly −78 °C at 1 atm), it causes water vapour in the air to condense, causing a visible fog. Thus what you are seeing is not carbon dioxide, but rather water.

When we exhale and it is reasonably warm, the carbon dioxide expelled is roughly body temperature (36–37.5 °C) and so we do not see any fog forming.

What happens when we exhale in cold weather is a bit different. Our breath contains a high percentage of water, higher than the percentage of water that can stay in gaseous form at a low temperature. Thus the ‘extra’ water condenses into droplets.

For reference, here is the phase diagram of $\ce{CO2}$:

And here is the phase diagram of $\ce{H2O}$:

And here is the saturation fraction of water at various temperatures at 1 atm (sea level):


What's the pH of vinegar containg 5% acetic acid?

Consider the choices. Since it's acetic acid, you can rule out any pH values that indicate a basic or neutral solution. Knowing that acetic acid is a weak acid and that vinegar is a fairly dilute solution, you can also rule out any pH values that would indicate either a strong acid or a highly concentrated solution. Eliminating those impossible values leaves only one choice as the answer.

The trick with a question like this is to realize that it's multiple choice for a reason. It's intended to be a conceptual problem, not a calculation problem. In general, a good rule for multiple-choice questions like this is to first see if you can solve it using general concepts; only if you can't should you reach for you calculator.


Why does lemon juice reduce the "fish" odor of sea food — specifically fish?

There are 2 cases, both related to the acid-base reactions. Both are also partial reasons why so many fish recipes use lemon juice.
Fish, especially sea fish, naturally contain trimethylamine-N-oxide $\ce{(CH3)3N-O}$ that, after death, gets enzymatically reduced to trimethylamine $\ce{(CH3)3N}$, the source of ammonia-like fish odour.

Trimethylamine N-oxide is an osmolyte found in saltwater fishes, sharks, rays, molluscs, and crustaceans. It is considered as a protein stabilizer that may serve to counteract urea, the major osmolyte of sharks, skates and rays. It is also higher in deep-sea fishes and crustaceans, where it may counteract the protein-destabilizing effects of pressure.TMAO decomposes to trimethylamine (TMA), which is the main odorant that is characteristic of degrading seafood.
TMAO is biosynthesized from trimethylamine, which is derived from choline.

In acidic environment, TMA forms water soluble, non volatile, odourless trimethylammonium salts, what eliminates the volatile smelly trimethylamine.
$$\ce{(CH3)3N + H+ <=>> (CH3)3NH+ }$$
It is the direct analogy to the smelly volatile ammonia and odourless ammonium:
$$\ce{NH3 + H+ <=>> NH4+}$$

Another source of fishy smell/taste, especially of some sweet water species like carp or catfish, is geosmin, a polycyclic alcohol, being a derivate of decahydronaphthalene(decaline). It is released after death of some Streptomyces or cyanobacteria cells which gives a muddy or earthy taste, and the human smell sensors are extremely sensitive to it (5 ppt).
Geosmin is also responsible for smell of the soil after raining, of some beet roots and even of drinking water of surface origin. There is speculation the nose sensitivity for geosmin is a evolutionary response for the need of efficient searching for water.
Geosmin reportedly breaks down in acidic environments.


How does O-benzoyl-N-methylhydroxylamine hydrochloride oxidise this ketone?

The first step is the formation of the Me,O-Bz imine of the ketone. This tautomerises to the enamine which then does a pericyclic rearrangement with the carbonyl of the benzoyl group to install O at the alpha position. Finally  the imine hydrolyses to reveal the ketone. See Aldrich Chemfiles (2007) 7.1, page 7.


Can one figure out an optical absorption spectrum from EPR data?

No, it shouldn't be possible, based on where these spectroscopies lie in the electromagnetic spectrum:

The most common EPR experiments occur at 9.5 GHz (X band), which is in the microwave region. Even the highest-frequency experiments just barely break into the far infrared. Assuming you're doing an X band experiment, and that your system has a transition very close to the free electron $g$-values, 9.5 GHz corresponds to ~0.3 cm$^{-1}$.

For UV-vis, taking the longest wavelength of light in the visible region as 800 nm gives 12,500 cm$^{-1}$. This can be interpreted in two ways:

The EPR experiment needs ~42,000x more energy to probe even the lowest possible electronically-excited state.
Even the UV-vis transitions with the lowest possible energy will not show any fine structure, like in the sodium lines or benzene vibronic coupling. The energy scales need to be much closer. In the case of the spin-orbit-induced splitting of the sodium spectral lines, the unperturbed gap is 2.1 eV and gap between the split states is 0.0021 eV; this is 3 orders of magnitude between the energies. In the case of the benzene UV spectrum, the gap between peaks is ~10 nm, with the peak center at ~245 nm; this is 1-2 orders of magnitude:

$\hspace{20ex}$

Our EPR/UV-vis example has 5 orders of magnitude difference between the two energy scales. There is (hyper)fine structure in EPR:

which here, for DPPH in toluene is ~0.002 Tesla, or 0.35 GHz, so 1 order of magnitude.


what is the acceptable percentage for food grade ethanol?Is 95% ethanol plus 5% water consider as food grade ethanol?

The grade of ethanol does not depend on the water and ethanol percentages, but rather on what else is in there. If your mixture is truly 95-5 ethanol-water, with nothing else in it, than that would be safe to consume (accounting for the health hazards of consuming too much ethanol of course).

However, there might be traces of e.g. methanol or benzene in the ethanol, which are both highly toxic to humans and therefore cannot be present in food grade ethanol mixtures. According to this website the FCC grade and above are food grade, in particular due to some heavy metal specifications, so apparently there might also be trace heavy metals in the ethanol.


Negative potential of the oxidation of iron(II) to iron(III)

As IanB2016 mentioned in his answer, the redox potential of a half-cell reaction is always in reference to another half cell.  The usual "reference" half cell in electrochemistry and thermodynamics is the standard hydrogen electrode (SHE).  Some key features of the SHE:

the half cell reaction is $\ce{2 H+(aq) + 2e- <=> H2}$, i.e. the reductant is hydrogen gas and the oxidant is aqueous protons.
the pressure of hydrogen gas for a standard hydrogen electrode is 1 atmosphere.
the concentration of acid in the aqueous electrolyte is 1 molar, which corresponds to a pH of 0.

The oxidation potential of a species is related to its strength as a reducing agent.  Things that have a negative oxidation potential are less reducing than hydrogen gas, and things that have a positive oxidation potential are more reducing than hydrogen gas.  Lithium metal, for example, has an oxidation potential of more than three volts, i.e. $\gt+3\;\mathrm{V}$.  It is an extremely powerful reducing agent, far more powerful than hydrogen gas.

As you note, $\ce{Fe^{2+}}$ has an oxidation potential of $-0.77\;\mathrm{V}$, meaning it is a less powerful reducing agent than hydrogen gas.  This makes sense in several ways:

aqueous solutions of iron(II) salts do not spontaneously reduce aqueous protons to hydrogen gas.  If you mix iron(II) sulfate with water, for example, hydrogen bubbles do not spontaneously form.
Hydrogen gas releases more energy when burned in oxygen than iron(II) salts do when they react with oxygen.  Hydrogen reacts explosively, whereas iron(II) salts do not spontaneously explode or catch fire when they are oxidized by oxygen.

Finally, note that the oxidation potential of water (via the reaction $\ce{O2(g) + 4 H+ + 4 e− <=> 2H2O}$ at pH 0 at 1 atm of pressure) is -1.229 volts.  Water is a very weak reducing agent, which means that oxygen is a strong oxidizing agent, stronger than iron(III).  This means:

iron(III) salts do not spontaneously oxidize water to produce oxygen.
conversely, oxygen does spontaneously oxidize iron(II) to iron(III).  Solutiosn of iron(II) salts are thermodynamically unstable under aerobic (oxygen-rich) conditions.


Do alkanes evaporate faster than alcohols?

Yes, alkanes do evaporate faster than their corresponding alcohols (i.e. an alkane boils at a lower temperature than an alcohol with the same number of carbon atoms). This effect is due to hydrogen bonding between the alcohol hydroxyl groups.

Wikipedia says:

  Because of hydrogen bonding, alcohols tend to have higher boiling points than comparable hydrocarbons and ethers. The boiling point of the alcohol ethanol is 78.29 °C, compared to 69 °C for the hydrocarbon hexane (a common constituent of gasoline), and 34.6 °C for diethyl ether.


Finding rate constants for parallel reactions

By some simple mathematics, you can reach the following equations (the first of which you have reached):

$[X]=[X]_0\cdot e^{-(k_1+k_2)t}$

$[Y]=\frac{k_1}{k_1+k_2}[X]_0\cdot e^{-k_1 t}$

$[Z]=\frac{k_2}{k_1+k_2}[X]_0\cdot e^{-k_2 t}$

The $1:3$ ratio that you have arrived at is correct,  i.e. $\dfrac{k_1}{k_2}=\dfrac{1}{3}$.

Substituting this in the equation for $\ce{X}$, we get

$[X]=[X]_0\cdot e^{-4k_1 t}$

All that is left to do now is to substitute and solve, which leads to the answer:

$(k_1,k_2)\equiv(0.015\ (min)^{-1},0.045\ (min)^{-1})$

There you go!

PS: I've used $[X]_0$ for representing the initial concentration of $\ce{X}$.


Entropy of Irreversible Isothermal Compression

There is no reversible process that can take both the system and the surroundings to the same final states (simultaneously) that they reached in the irreversible process.  So, to get their changes in entropy for the irreversible process, we need to first separate the system from the surroundings and then subject each of them separately to new reversible processes (perhaps even involving a second and third "surroundings").

For the case of the original surroundings, we know that $Q_{irrev, surr}=-Q_{irrev, syst}$.  So, from the first law applied to the irreversible process $$\Delta U_{irrev, surr}=-Q_{irrev, syst}$$The alternate reversible process for the surroundings must take it between the same two states.  This means that the internal energy for the alternate reversible process must suffer the same change as for the irreversible process.  The only way this can happen is if the heat flow for the alternate reversible process is the same as for the irreversible process.  So, the entropy change for the surroundings in the irreversible process is given by $$\Delta S=-\frac{Q_{irrev, syst}}{T}$$

Here is an example of how this plays out for a specific irreversible isothermal compression or expansion.  The example works equally well for both expansion and compression.

The initial state of the system is specified by $P_1V_1=nRT$.  The irreversible process is specified by suddenly changing the external pressure on the piston from $P_1$ to $P_2$, and then allowing the gas to re-equilibrate irreversibly to its new equilibrium state with volume $V_2$ given by $V_2=P_1V_1/P_2$.  Since, during this irreversible process, the external pressure is constant, the work done by the gas on the surroundings is given by $$W=P_2(V_2-V_1)=nRT\left(1-\frac{V_1}{V_2}\right)$$This is also equal to the heat Q transferred from the surroundings to the system:  $$Q=nRT\left(1-\frac{V_1}{V_2}\right)$$So the entropy change of the system during this irreversible process is given by:  $$\Delta S_{syst}=nR\ln{(V_2/V_1)}$$And from the analysis I presented above, the change in entropy of the surroundings during this irreversible process is given by
$$\Delta S_{surr}=-nR\left(1-\frac{V_1}{V_2}\right)$$The overall change in entropy for the system plus surroundings for this irreversible process is thus given by
$$\Delta S=nR\left[\ln{(V_2/V_1)}-\left(1-\frac{V_1}{V_2}\right)\right]$$The term in brackets in this equation is positive for all values of the ratio of the volumes, whether the ratio greater than unity or less than unity.  Therefore, the Clausius inequality ($\Delta S>0$) is satisfied for both expansion and compression in this irreversible process.


Where can I find a downloadable spreadsheet of element properties?

Well, here's a CSV file that I parsed out of the JSON data provided by Paul Nathan's website, which in turn was produced from gPeriodic data in response to this question. gPeriodic is FOSS, so I can only assume that the element data contained within is good to share, though I have no idea of its provenance.

The parser uses the python json module to read the data, which I then wrote into comma separated values, which should be readable by any halfway competent spreadsheet program.

Here it is: Pastebin Link

save it as elementdata.csv and you're good to go.

Some gotchas:

This reproduces the gPeriodic data, warts and all. I make no guarantees regarding its correctness, recency, etc. I just munged the data.
Some of the data is augmented with tildes, notes about temperature/polymorph/state etc. I've left these as they are but you may need to trim them if you want to plot them as numeric values.
Argon had an atomic radius of '2-', whatever that means. I cut it.
Excel is joke software and habitually interprets numbers wrapped in parentheses as negatives, because apparently some accountants decided that surrounding numbers in parens is a more sensible option than using a minus sign. Needless to say, this is wrong. I've addressed the problem by wrapping the affected numbers in angle brackets.
There are no ionic radii in the source data. At all. Not my fault.
The columns inherit the units of the source data. I've left the units out on purpose because I wanted each column to have no spaces in the name for ease of processing in R or whatever. Furthermore, the units in the source file have some inconvenient characters from an encoding perspective.

Tried plotting Z versus covalent radius in R from this data - looks alright:

Some of the data points are missing, some are not read by R because they are wrapped in <> - pre-process to your heart's content.


What is the difference between ∆G and ∆G°?

Short answer

  Does it need to be at $25~^\circ\mathrm{C}$?

No. $\Delta_\mathrm{r} G^\circ$ can be defined at any temperature you wish to define it at, since the standard state does not prescribe a particular temperature. If you change the temperature, $\Delta_\mathrm{r} G^\circ$ will change.

  Does $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T\Delta_\mathrm{r} S^\circ$ always use $T = 298~\mathrm{K}$?

No. You use whatever temperature you are running your reaction at.

  (...maths...)

Yes, at equilibrium, $\Delta_\mathrm{r} G = 0$ and $Q = K$.

However, everything after the first bullet point is wrong. You cannot conclude that $\Delta_\mathrm{r} G^\circ = 0$, nor can you conclude that $K = 1$. The equation $\Delta_\mathrm{r} G^\circ = -RT \ln K$ does not analogously translate into $\Delta_\mathrm{r} G = -RT \ln Q$! The accurate relation is:

$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$

Setting $Q = K$ and $\Delta_\mathrm{r} G = 0$ in this equation does not tell you anything about the value of $K$. In fact, if you try doing it, all you will find out is that $\Delta_\mathrm{r} G^\circ = -RT \ln K$ - no surprises there!

Long answer

Any book that writes that $\Delta_\mathrm{r} G^\circ$ is the "special case" of $\Delta_\mathrm{r} G$ at $T = 298~\mathrm{K}$ is wrong.

The Gibbs free energy of a system is defined as follows:

$$G = H - TS$$

Under constant temperature and pressure (from now on, I will just assume constant $T$ and $p$ without stating it), all systems will seek to minimise their Gibbs free energy. Equilibrium is reached when $G$ is minimised. When $G$ is at a minimum, any infinitesimal change in $G$, i.e. $\mathrm{d}G$, will be $0$. Therefore, this is equivalent to saying that the condition for chemical equilibrium is $\mathrm{d}G = 0$.

Clearly, we need a way to relate this quantity $\mathrm{d}G$ to the actual reactants and products that are in the system. This can be done by using the Maxwell relation (see any physical chemistry text for details):

$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T + \sum_i \mu_i\,\mathrm{d}n_i$$

Under constant $T$ and $p$, $\mathrm{d}p = \mathrm{d}T = 0$ and therefore

$$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i$$

where $\mu_i$ is the chemical potential of species $i$, defined as a partial derivative:

$$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{n_{j\neq i}}$$

So, we now have a refined condition for equilibrium:

$$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i = 0$$

We can go further by noting that the values of $\mathrm{d}n_i$ for different components $i$, $j$, etc. are not unrelated. For example, if we have a reaction $i + j \longrightarrow k$, then for each mole of $i$ that is consumed, we must also use up one mole of $j$; this means that $\mathrm{d}n_i = \mathrm{d}n_j$.

This can be formalised using the idea of a stoichiometric coefficient $\nu_i$, which is defined to be positive for products and negative for reactants. For example, in the reaction

$$\ce{3H2 + N2 -> 2NH3}$$

we have $\nu_{\ce{H2}} = -3$, $\nu_{\ce{N2}} = -1$, and $\nu_{\ce{NH3}} = 2$.

By stoichiometry, if $1.5~\mathrm{mol}$ of $\ce{H2}$ is consumed, then $1~\mathrm{mol}$ of $\ce{NH3}$ has to be produced. We could write $\Delta n_{\ce{H2}} = -1.5~\mathrm{mol}$ and $\Delta n_{\ce{NH3}} = 1~\mathrm{mol}$. These quantities are proportional to their stoichiometric coefficients:

$$\frac{\Delta n_{\ce{H2}}}{\nu_{\ce{H2}}} = \frac{-1.5~\mathrm{mol}}{-3} = 0.5~\mathrm{mol} = \frac{1~\mathrm{mol}}{2} = \frac{\Delta n_{\ce{NH3}}}{\nu_{\ce{NH3}}}$$

The quantity $0.5~\mathrm{mol}$ is a constant for all chemical species $\ce{J}$ that participate in the reaction, and it is called the "extent of reaction" and denoted $\Delta \xi$ (that is the Greek letter xi). If the reaction is going forward, then $\Delta \xi$ is positive, and if the reaction is going backwards, then $\Delta \xi$ is negative. If we generalise the above result, we can write

$$\Delta \xi = \frac{\Delta n_i}{\nu_i}$$

and if we make $\Delta n_i$ smaller and smaller until it becomes an infinitesimal, then:

$$\begin{align}
\mathrm{d}\xi &= \frac{\mathrm{d}n_i}{\nu_i} \\
\mathrm{d}n_i &= \nu_i\,\mathrm{d}\xi
\end{align}$$

If we go back to our condition for equilibrium, we can substitute in the above to get:

$$\mathrm{d}G = \sum_i \mu_i\nu_i\,\mathrm{d}\xi = 0$$

Now, $\mathrm{d}\xi$ is no longer dependent on $i$, since we have established already that $\Delta \xi$ (and by extension $\mathrm{d}\xi$) is a constant for all chemical species. So, we can "divide through" by it to get:

$$\Delta_\mathrm{r} G \equiv \frac{\mathrm{d}G}{\mathrm{d}\xi} = \sum_i \mu_i\nu_i = 0$$

where $\Delta_\mathrm{r} G$ is defined to be $\mathrm{d}G/\mathrm{d}\xi$.

Note that $\Delta_\mathrm{r} G$ is an intensive property and has units of $\mathrm{kJ~mol^{-1}}$, since $\mathrm{d}\xi$ has units of $\mathrm{mol}$. This ensures that the units we use are consistent: since we know that $\Delta_\mathrm{r}G = \Delta_\mathrm{r} G^\circ + RT\ln Q$, $\Delta_\mathrm{r}G$ must have the same units as $RT$.

How do we interpret the physical significance of $\Delta_\mathrm{r} G$, or in other words, what does it even mean? There are two ways, each based on a different mathematical expression.

We have $\Delta_\mathrm{r}G = \sum \nu_i \mu_i$. This means that $\Delta_\mathrm{r}G$ is simply the difference between the chemical potentials of the products and the reactants, weighted by their stoichiometric coefficients. For the reaction $\ce{3H2 + N2 -> 2NH3}$, we have:

$$\Delta_\mathrm{r} G = \sum_i \mu_i\nu_i = 2\mu_{\ce{NH3}} - 3\mu_{\ce{H2}} -\mu_{\ce{N2}}$$

We have $\Delta_\mathrm{r}G = \mathrm{d}G/\mathrm{d}\xi$. This means that it is the slope of a curve of $G$ against $\xi$:

Note that up to this point, we have not stipulated any particular temperature, pressure, amounts of species present, or any conditions whatsoever. We have only said that the temperature and pressure must be constant.

It is important to realise that $\Delta_\mathrm{r}G$ is a well-defined quantity at all $T$, all $p$, and all possible values of $n_i, n_j, \cdots$! The shape of the curve will change when you vary the conditions. However, no matter what the curve looks like, it is always possible to find its gradient ($= \Delta_\mathrm{r}G$) at a particular point.

What exactly, then, is $\Delta G^\circ$? It is just a special case of $\Delta G$, where all the reactants and products are prepared in a standard state. According to IUPAC, the standard state is defined as:

For a gas: pure ideal gas when the pressure $p$ is equal to the standard pressure $p^\circ$.
For a liquid or solid: pure liquid or solid at $p = p^\circ$
For a solution: ideal solution when the concentration $c$ is equal to the standard concentration $c^\circ$.

$p^\circ$ is most commonly taken to be $\pu{1 bar}$, although older texts may use the value $\pu{1 atm} = \pu{1.01325 bar}$. Since 1982, IUPAC has recommended the value $\pu{1 bar}$ for the standard pressure (Pure Appl. Chem. 1982, 54 (6), 1239–1250; DOI: 10.1351/pac198254061239). However, depending on the context, a different value of $p^\circ$ may prove to be more convenient. Likewise, $c^\circ$ is most commonly – but not necessarily – taken to be $\pu{1 mol dm-3}$.

Note that in the above definitions, no temperature is specified. Therefore, by defining the standard Gibbs free energy, we are fixing a particular value of $p$, as well as particular values of $n_i, n_j, \cdots$. However, the value of $T$ is not fixed. Therefore, when stating a value of $\Delta_\mathrm rG^\circ$, it is also necessary to state the temperature which that value applies to.

When a reaction vessel is prepared with all its substances in the standard state, all the components of the system will have an activity of exactly $1$ by definition. Therefore, the reaction quotient $Q$ (which is a ratio of activities) will also be exactly equal to $1$. So, we could also say that $\Delta_\mathrm{r}G^\circ$ is the value of $\Delta_\mathrm{r}G$ when $Q = 1$.

Returning to the graph of $G$ against $\xi$ above, we note that at the left-most point, $Q = 0$ since there are only reactants; at the right-most point, $Q \to \infty$ as there are only products. As we move from left to right, $Q$ increases continuously, so there must be a point where $Q = 1$. (In general, the point where $Q = 1$ will not be the same as the equilibrium point.) Since $\Delta_\mathrm{r}G$ is the gradient of the graph, $\Delta_\mathrm{r}G^\circ$ is simply the gradient of the graph at that particular point where $Q = 1$:

The gradient of the graph, i.e. $\Delta_\mathrm{r}G$, will vary as you traverse the graph from left to right. At equilibrium, the gradient is zero, i.e. $\Delta_\mathrm{r}G = 0$. However, $\Delta_\mathrm{r}G^\circ$ refers to the gradient at that one specific point where $Q = 1$. In the example illustrated above, that specific gradient is negative, i.e. $\Delta_\mathrm{r}G^\circ < 0$.

Again, I reiterate that the temperature has nothing to do with this. If you were to change the temperature, you would get an entirely different graph of $G$ versus $\xi$. You can still find the point on that graph where $Q = 1$, and the gradient of that graph at the point where $Q = 1$ is simply $\Delta_\mathrm{r} G^\circ$ at that temperature.

We have established the qualitative relationship between $\Delta_\mathrm{r} G$ and $\Delta_\mathrm{r} G^\circ$, but it is often useful to have an exact mathematical relation.

$\Delta_\mathrm{r} G^\circ$ is exactly the same as $\Delta_\mathrm{r} G$ except for the imposition of the standard state. It follows that if we take the equation

$$\Delta_\mathrm{r} G = \sum_i \mu_i \nu_i$$

and impose the standard state, we get

$$\Delta_\mathrm{r} G^\circ = \sum_i \mu_i^\circ \nu_i$$

Thermodynamics tells us that

$$\mu_i = \mu_i^\circ + RT\ln{a_i}$$

where $a_i$ is the thermodynamic activity of species $i$. Substituting this into the expressions for $\Delta G$ and $\Delta G^\circ$ above, we obtain the result:

$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$

where the reaction quotient $Q$ is defined as

$$Q = \prod_i a_i^{\nu_i}$$

When equilibrium is reached, we necessarily have $\Delta_\mathrm{r} G = 0$ (see the discussion above). The equilibrium constant $K$ is defined to be the value of $Q$ at equilibrium. Therefore, at equilibrium, $Q = K$. Plugging this into the equation above gives us the famous equation:

$$\Delta_\mathrm{r} G^\circ = -RT\ln K$$

Again, no temperature is specified! In general, $K$ depends on the temperature as well; the relationship is given by the van 't Hoff equation.


Does dissociation of molecules into atoms change the volume of mixture?

Short answer is "not necessarily". If we isolate V in the ideal gas law, we have $$V=\frac{nRT}{P}.$$

If $n$ doubles and nothing else changes, then $V$ does also double. But the doubling of $n$ could also be offset by a doubling of $P$ or a halving of $T$, either of which would result in no change in the volume.

And yes, if the volume increases, pressure-volume work has been done, and if the volume remains constant, no pressure-volume work has been done.  That isn't the same as "no work has been done", as there are other types of work besides pressure-volume work.


Because of the absence of a lone pair, C undergoes usual $\mathrm{S_N2}$ transformation under the conditions. However, the availability of a lone pair in A undergoes the nucleophilic substitution of cyanide ion via neighboring-group participation (see above comments by Waylander and J. Deans) by the sulfur atom under the conditions (Ref.1). A very reliable mechanism is given in Ref.1 for the substitution of azide ion (work the same way for cyanide ion):

Reference:

D. D. Díaz, A. Converso, K. B. Sharpless, M. G. Finn, “2,6-Dichloro-9-thiabicyclo[3.3.1]nonane: Multigram Display of Azide and Cyanide Components on a Versatile Scaffold,” Molecules 2006, 11(4), 212–218 (doi: 10.3390/11040212).


Atomic units are the standard in almost any ab-initio (LAMMPS, NWChem, Orca, QChem) and/or classical force field (LAMMPS, MPMC, CHARMM, AMBER, GROMACS etc.) modeling code because you end up computing numbers much closer to 1.

Things can get bad very quickly for small values in python. To prevent this, I would use atomic units, and wherever arithmetic error occurs, try using the decimal type, instead of default float, via float -> string -> decimal:

from decimal import *

energy = Decimal(str(3.4280284e-19))

Here is a sample file I've used for atomic units in python:

pi = 3.14159265359
kb = 1.38064852e-23 # boltzmann / J/K
ab = 5.2917721092e-11 # bohr radius / meters
hbar = 1.054571726e-34 # planck / Js
h = hbar*2*pi
me = 9.10938291e-31 # electron rest mass / kg
e = 1.602176565e-19 # electron charge / C
ke = 8.9875517873681e9 # coulomb constant / kgm^3 s^-2 C^-2
Eh = 4.35974417e-18 # hartree energy / J
c = 299792458 # speed of light, m/s
cs = 340.29 # speed of sound, m/s
t = hbar / Eh # time / s
v = ab*Eh/hbar # velocity m/s
F = Eh/ab #force
T = Eh/kb # temp
P = Eh/(ab**3) # pressure
Ef = Eh/(e*ab) # electric field
Ep = Eh/e # electric potential
u = e*ab # electric dipole moment


Why does water diffuse from a lower solute concentration to a higher one?

There are two possibilities. The first is that it is simple dilution, but for more likely is that osmosis is occurring. In this case the concentrated solution and the water, or a  dilute solution, are separated by a semi-permeable membrane, i.e. one that allows passage of small water molecule but not large solute ones.  To lower the overall energy (as represented as the chemical potential) water moves through the membrane from the dilute solution to the more concentrated one. In a cell this can lead to the cell swelling and possibly bursting if left uncontrolled. There is some more explanation in the answer to this question Entropy as the driving force for osmosis


How does localized bonding theory and hybridisation work?

First of all, 'hybridisation' is a hypothetical concept, i.e. orbitals don't really mix to form new orbitals, but 'hybridisations' are very successful in explaining structures of molecules and their physical properties. So knowing that a compound has $\text{sp}^2$ hybridisation will just help you arrange the constituent atoms around the central atom. Physically it has no other meaning. It's just like vector cross product in Physics (for example the torque on a rotating wheel acts along the axis, i mean what does that even mean physically).

Secondly regarding how to know if an atom has a $\pi$ or a $\sigma$ bond: observing many structures(Lewis structures) and recognizing certain trends is the way out.

Some rules of the thumb that will help you draw structures are as follows:
1. Identify the central atom as the one that can make the most bonds.
2. Hydrogen makes only one bond. If Oxygen is present in the same compound, the Hydrogen is     generally attached to the Oxygen forming a $\ce{O-H}$ group.
3. Second period elements cannot expand their octets(i.e have more than 8 electrons in valence shell)
4. Mostly second period elements can make $\pi$ bonds.(That really solves part of your query). Although in some rare cases third period elements may also make $\pi$  bonds.
5. Never neglect the lone pairs(or non-bonding electrons) of the central atom.

Remember that these are only thumb rules and may be violated sometimes.


Difference between black and brown eumelanin / red and pink pheomelanin

First of all, melanines are mixtures of complex polymers, without highly ordered structures and precisely defined composition. It means that unlike e.g. polypropylene or PET, melanines are not composed of regularly repeating identical units (monomers).

In a great simplification, brown and black eumelanines are composed of units of DHI (5,6‐dihydroxy‐1H‐indole) and DHICA (5,6‐dihydroxy‐1H‐indole‐3‐carboxylic acid). The order of these units is random and the ratio of DHI to DHICA depends on the origin of the pigment. Darker eumelanines are enriched in DHICA, which form branched structures (each DHICA unit can be bonded to three other units). Light-brown eumelanines contain more DHI units. Pheomelanines are build of different units - alanylhydroxybenzothiazines.

Now to the color. The color of melanines results from the absorption of light by electrons of chromophores (i.e. fragments of a molecule that strongly absorb light at certain wavelengths). In eumelanines the chromophores are dihydroxyindole units, and in pheomelanines benzothiazine rings. The color of melanines does not result directly from how are these bonded with each other, but from how many chromophores are in a melanine molecule, how densely are they packed, and from a number of different factors (e.g. presence of metal ions, oxydation). Black eumelanines absorb more light than brown eumelanines, because they have more chromophores (they are more densly packed due to the branching of the polymer) and that is why they appear darker. And pheomelanines have different chromophores that absorb light of a bit different wavelength (blue region). Different bonding patterns affect the optical density of melanines, but not directly their color.


I'm going to side with you @TheChemist.

Your TA might be suggesting this because you need often need a negatively charged nucleophile in order to carry out nucleophilic aromatic substitutions.

Having the nitro group in the 4 position makes the your electrophile much more receptive to attack, since the negatively charged intermediate will be resonance stabilized.

Even if the one aniline does deprotonate another, the concentration of the deprotonated anilide will be so low (on the order of 5 x 10-28 mol/L in pure aniline, if the pKa is 27) that this will have no impact on the reaction.


Explaining a biochemical reaction (protein labelling, e.g. with fluorescent probes)

All reactions you gave are typically used in protein labelling, e.g. with fluorescent probes.

In order to allow further measurements on the labeled protein, the following conditions have to be met:

The bonds formed have to be stable like the peptide bonds in your protein. You don't want the probe flying around somewhere.

Proteins tend to be a bit sensitive to degradation; boiling in acetic acid is obviously not an option to run the labeling. As a consequence, the functional group to be attached needs to be activated, e.g. with good a leaving group, to allow the labelling of the protein under mild conditions. The succimidyl ester in the first reaction is a good example.


Can o-phenanthroline be used as a complexing agent for lead(II)?

Yes, there are numerous known lead(II) o-phenanthroline complexes with thiocyanate, perchlorate and bulkier organic ligands.
As for nitrate, Bytheway et al. [1] reported synthesis (from lead(II) nitrate in methanol) and crystal structure of bis((1,10-phenanthroline)-(nitrato-O,O'))-lead(II):

Figure 1. crystal structure of $\ce{[(phen)2Pb(NO3)2]}$ (CCDC: 1246661, CSD: RAQMAX).

Unfortunately, there is no UV-Vis/NIR spectra available.
Harrowfield et al. [2, p. 1030] reported that the crystals are colorless, which likely eliminates visible region of detection.

References

Bytheway, I.; Engelhardt, L.; Harrowfield, J.; Kepert, D.; Miyamae, H.; Patrick, J.; Skelton, B.; Soudi, A.; White, A. Lewis-Base Adducts of Lead(II) Compounds. XIII. Synthetic, Structural and Theoretical Studies of Some 2:1 Adducts of 1,10-Phenanthroline With Lead(II) Oxoanion Salts. Aust. J. Chem. 1996, 49 (10), 1099. DOI: 10.1071/CH9961099.
Harrowfield, J.; Miyamae, H.; Skelton, B.; Soudi, A.; White, A. Lewis-Base Adducts of Lead(II) Compounds. VII. Synthetic and Structural Studies of Some 1:1 Adducts of Linear Polyamines With Lead(II) Nitrate. Aust. J. Chem. 1996, 49 (10), 1029. DOI: 10.1071/CH9961029.


Which is more acidic acetic acid or propanoic acid?

My best guess is you are interpreting the statement "inductive effect decreases as the length of alkyl group increases" somewhat out of context.

So, let's compare formic acid to acetic acid.  Formic acid has hydrogen bonded to the carbonyl, while acetic acid has a methyl group.  The methyl group is inductively donating, and as a result acetic acid is less acidic than formic acid. $\mathrm{p}K_\mathrm{a}$ difference of $1$ full unit.

So now, let us compare methyl acetic acid (a precise, but incorrect, alternate name for propanoic acid) to acetic acid. Methyl acetic acid has an inductive donating group at the $\ce{CH2}$ carbon. As such, we expect some donation, and a lower acidity for methyl acetic acid as compared to acetic acid. True, the donating group is one group removed from the acidic site, which means the effect of the methyl group on methyl acetic acid's acidity is diminished as compared to an electron donating group directly attached to the carbonyl, so the acidity reduction will be smaller. $\mathrm{p}K_\mathrm{a}$ difference, $0.12$ units.

After you have this concept in your head, purge the abomination of nomenclature that is methyl acetic acid from your memory.

Ron has some views on hyperconjugation that may also be useful in this context, but strictly speaking, I don't find them necessary.


The relevant rule for the order of citation of detachable substituent prefixes in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-14.5 ALPHANUMERICAL ORDER
(…)
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
Alphanumerical order is applied as follows in organic nomenclature. Nonitalic Roman letters are considered first, unless used as locants or part of a compound or composite locant, for example, ‘N’ or ‘4a’ (see P-14.3), or in an isotopic descriptor. (…)

P-14.5.4 When two or more prefixes consist of identical Roman letters, priority for order of citation is given to the group that contains the lowest locant(s) at the first point of difference.

Therefore, the correct name for the compound that is shown in the question is
3-ethenyl-5-[(1E)-prop-1-en-1-yl]-4-(prop-2-en-1-yl)-2-(prop-1-yn-1-yl)cyclopentan-1-ol
since the order is ‘ethenyl’ > ‘propenyl’ > ‘propynyl’ and ‘prop-1-en-1-yl’ > ‘prop-2-en-1-yl’.
The fact that ‘4-propenyl’ is lower than ‘5-propenyl’ is irrelevant.


Resonance is not ‘dancing’. Resonance is often confused by undergraduates (and sometimes even professors) as some kind of oscillation back and forth between multiple structures. That is not the case.
When explained properly, resonance is a consequence of our unability to draw adequate depictions of what the bonding situation in the molecule truly looks like. For example, take a benzene ring. We can only draw single and double bonds. The structures should be identical whether the double bonds are in 1,3,5 or 2,4,6 positions. Our depictions predict different structures because double bonds are shorter than single ones. The actual benzene structure, however, has equal bond lengths between all carbons which could be described as 1.5-bonds. It is incorrect to think of the double bonds "resonating" between 1,3,5 and 2,4,6; rather, both depictions should be considered as two extreme pictures that by themselves cannot capture the truth.
This may be relatively easily understood in simple cases such as benzene or the cycloheptatrienyl cation, where the resonance depictions are all very similar. It is a lot harder to understand it in the case of a cation such as the tricyclopropylmethyl cation. There are a number of answers here explaining exactly which resonance forms stabilise it; it can and should also be seen as a nonclassical cation of the pentavalent carbonium type. You can, for example, check this question.
However, that does not mean that the resonance in these non-classical cations has a different nature. It is not correct to say that the resonance here is "faster" than in benzene, because resonance is simply not a oscillation or vibration between different structures: there is no "speed" or "frequency" to speak of.


Additive to an aluminum sulfate and water solution to lower pumping pressures

The needed pressure depends on liquid viscosity. The significant decrease of water solution viscosity cannot be achieved by adding of minor additives.

It could be achieved by methods that may not be applicable:

Decreasing concentration of the solute
Decreasing viscosity of solvent by different solvent composition
Increasing temperature of solution. A rule of thumb says viscosity decreases by 3% by increasing temperature by $1 ^{\circ}\mathrm{C}$.


Can rate constant depend upon the concentration of reactants (or other species involved in the reaction)?

It seems possible that you mix the terms. There is the reaction rate constant -- in your equation $k_r$, and there is the rate of reaction -- in your equation $r$ and sometimes expressed by $v$.
Depending on the order of a reaction, the overall rate of reaction may, but need not depend on the concentration of the reactants available.  There are reaction types known by experience to proceed typically one kinetic order (e.g., heterogeneous hydrogenation); however, reading the stoichiometric reaction equation often does not provide sufficient information and the rate law has to be determined experimentally.
Examples of kinetic reaction order are zero, one, two, three, but fractional orders are known, too.  It is possible to perform reactions which are of second order ($v = k_r[\ce{A}][\ce{B}]$) in a pseudo-first order regime if the concentration of reagent A is so much much higher, than the one of [B], that you detect after consumption of B still much of A.  This however assumes that the reaction still proceeds by the same pathway, thus requires experimental evidence to be supported.
This contrasts to $k_r$, which is independent from the concentration of the reagents, however depends on the temperature $T$ (keyword Boltzmann, and even more, Arrhenius equation).


How to make an aluminumoxide coating on an electrode?

Yes, it is definitely possible, but also relatively difficult as compared to electroplating zinc onto another metal.

It is a significantly important industrial process. Although aluminium itself is reactive, it develops an impervious oxide layer that quite much renders it unreactive.

  However, comparatively larger negative standard electrode potential of Al/Al(III) couple (− 1.67 V vs. SHE) precludes its deposition from the aqueous baths.

So it is usually considered impossible in aqueous (water-based) solution so is conducted in either organic liquids or fused (molten) state.

Fused: Usually a bath of an eutectic mixture of $\ce{LiBr, KBr}$ & $\ce{CsBr}$ is taken along with molten $\ce{AlBr3}$ (melts at about 94-98°C). A concentration of $80\%$ (by wt.) of $\ce{AlBr3}$ is reported to give a shiny finish.

Source $1$ reports that a satisfactory result was obtained with constant current of $1-3$ $\pu{A}$ passed for $1$ hour.

Non-aqueous solution:

From Source $2$ (emphasis mine)

  ..the commercial processes for electroplating of Al are based on organic solvents such as SIGAL (Siemens-Galvano-Aluminium) and REAL (Room-temperature Electroplated Aluminium) processes. In the SIGAL process which was developed in Siemens laboratories Al is electrodeposited at 100 °C from alkylaluminium compounds in toluene. In the REAL process which was developed in the laboratories of Philips $\ce{Al}$ is deposited at room temperature from $\ce{AlCl3}$ and $\ce{LiAlH4}$ in tetrahydrofurane.

The process must be conducted avoiding any traces of water, or the water would be electolysed first, leading to observable evolution of gas.

TL;DR: Electroplating aluminium is not something I would recommend trying at home.

References:

Aluminum electroplating on steel from a fused bromide electrolyte, P. K. Tripathy et al, Surf. Coat. Tech., Vol. 258, 15 Nov. 2014, 652-663. DOI:10.1016/j.surfcoat.2014.08.021
Electroplating of mild steel by aluminium in a first generation ionic liquid: A green alternative to commercial Al-plating in organic solvents, S. Zein El Abedin et al, Surf. Coat. Tech., Vol. 201, Issues 3–4, 5 Oct. 2006, 1352-1356. DOI:10.1016/j.surfcoat.2006.01.065


Why do we insert a numeric multiplier after group positions when naming an isomer?

take 2,3-dimethylbutane

In the case of different substituents on different or the same atom, we use a combination of the position and the name of the substituent, such as in

2-chloro-3-methoxy...
3-cyano-4-bromo..., etc.

Following this rule, 1,2-dimethoxyethane would be 1-methoxy-2-methoxyethane. Using the multiplier di, together with the positions, just shortens and simplifies the name and still is unambiguous.

The multiplier di explicitly denotes that there are two identical substituents at the given positions and not just one that bridges two centres.

An example for such a situation would be 1,4-methanonaphthalene (1). Note that even in this context, di can be used, as in 1,4:5,8-dimethanonaphthalene (2).


Protonating/basic site and Brønsted basicity orders of beta, gamma, and delta lactams

As the significance of the two contrasting factors cannot be easily evaluated, let us first look at some evidence. Wan, Modro & Yates (1980) reported the $\ce {pK_a}$ values of the conjugate acids of the various lactams, as shown in the table below:

\begin{array}{|c|c|c|c|} \hline
\text{Lactam} & pK_a\\ \hline
\ce{$\beta$, 4} &  -0.96\\ \hline
\ce{$\gamma$, 5} & -0.65 \\ \hline
\ce{$\delta$, 6} & -0.18 \\ \hline
\ce{$\epsilon$, 7} & -0.46 \\ \hline
\ce{$8$} & -0.38 \\ \hline
\ce{$9$} & -0.18 \\ \hline
\end{array}

As the ring size increases, from the 4-membered $\beta$-lactam ring to the 9-membered ring, the basicity of the lactam increases. This seems to suggest that the oxygen atom may indeed be the protonating site for these lactams, even for those with the smaller ring sizes. However, it seems that the 6-membered $\delta$-lactam ring exhibits rather anomalous basicity.

The authors explain the general trend in basicity as follows, using the two resonance structures shown:

  If the two canonical forms of the $\ce {O}$-protonated lactam are considered, the basicity will depend strongly on the ability of nitrogen to donate its pair of electrons
  to give a delocalized cation. Hence structure II will be more important in the conjugate acid than the analogous contributing structure in the neutral lactam. This will cause more serious ring strain the smaller the size of the lactam ring, and hence tend
  to reduce the basicity. As the rings become very large, such strain factors will become less important and the basicity would be expected to level off at a pK value similar to that of a typical secondary amide. Only the 6-lactam is unusual in that its base strength is much higher than expected. It is commonly observed (27) that for five-membered rings exo double bonds are preferred, while for six-membered rings endo double bonds are more stable, although these preferences are not easily rationalized. Therefore contributions from canonical form II in the protonated 6-lactam should be more important than in the $\gamma$-lactam case (and possibly the other lactams as well), leading to a higher basicity for the 6-lactam than its ring size would lead one to expect from the trends in Fig. 2.

Reference

Wan, P.; Madro, T. A. ; Yates, K. The kinetics and mechanism of acid catalysed hydrolysis of lactams. Can. J. Chem. 1980, 58, p. 2423-2432.


Are there any semiempirical methods that work with heavy alkaline earth metals?

The implementations of MNDO, AM1, and PM3 along with PM6 and PM7 in MOPAC include Sr and Ba, but not Ra. MOPAC is free for academic use, but modern versions of it are not open source. Most of these models have an open-source implementation in SPARROW (GitHub). An entirely different class of semiempirical models (GFNx) is available in the open-source XTB code (GitHub), which also includes Sr and Ba, but not Ra. The lower portions of the periodic table (lanthanides and beyond) are of less general interest to scientists and have less reference data for fitting purposes, so there is lower demand and higher difficulty in covering this portion of the periodic table. The only parameterization that I know of that covers Ra is the QUASINANO2013 parameter set (J. Chem. Theory Comput. 2013, 9 (9), 4006–4017) in DFTB, which is proprietary and only available in the commercial implementation of DFTB by SCM.

If you only need Ra to behave like a +2 cation with a slightly larger ionic radius than Ba, then you probably could modify the parameter set of one of the aforementioned models and tune the repulsion parameters of Ba to increase its ionic radius in a few reference ionic systems. That would of course take a bit of effort, in both learning what the parameters of the model mean, where they are stored in a particular simulation code, identifying some suitable reference data, and tuning one or more parameters to fit the data.

I'm sorry that there isn't a simple, affirmative answer to your question, but I hope that this helps anyways.


Does a technical umbrella term exist that covers the different kinds of biomolecules?

I don’t know what your background is. Maybe it is organic chemistry, where the term functional group is used in a very specific way and using what would seem to be a synonym instead is wrong. Maybe it is biology (taxonomy) where it makes a big difference if a ‘group’ is an order, a family, a genus or something else.

The typical primary metabolites in biochemistry — carbohydrates, peptides/proteins, nucleic acids and lipids — do not have an overarching, widely accepted technical umbrella term. You can call them groups, classes, kinds or even types and be well understood.

If I had to suggest something, I would go with classes of primary metabolites. But I wouldn’t call that a strict technical term like functional group (organic chemistry) or species (taxonomy).


Why do the two methylene protons in ethylbenzene have a chemical shift at 2-2.9?

Protons attached to a carbon that is connected to a benzene ring typically resonate in the range 2.2-2.9 ppm (see the Ar-CH range in this table).

Here is the proton nmr spectrum for ethylbenzene.  The methylene protons absorb around 2.7 ppm, right in the upper end of the expected range.

image source


I think that the question meant to test your knowledge about the polarity of simple molecules.

The fact that symmetric molecules contain identical bond polarities between the central atom and surrounding atoms, the net dipole moment usually tends to be zero, creating a nonpolar molecule.

In other words, I think symmetry in this case refers to the fact that all of the surrounding atoms are identical.


Does there exist a molecule of formule XαYβ where X and Y are elements that are also found in molecule XγYδ?

$\ce{GeF2}$ and $\ce{GeF4}$
$\ce{GeCl2}$ and $\ce{GeCl4}$
$\ce{GeBr2}$ and $\ce{GeBr4}$
$\ce{GeI2}$ and $\ce{GeI4}$
$\ce{SnF2}$ and $\ce{SnF4}$
$\ce{SnCl2}$ and $\ce{SnCl4}$
$\ce{SnBr2}$ and $\ce{SnBr4}$
$\ce{SnI2}$ and $\ce{SnI4}$
$\ce{PbF2}$ and $\ce{PbF4}$
$\ce{PbCl2}$ and $\ce{PbCl4}$
$\ce{XeF2}$ and $\ce{XeF4}$
$\ce{XeO2}$ and $\ce{XeO4}$
Silicon dihalides ($\ce{F}$, $\ce{Cl}$, $\ce{Br}$, $\ce{I}$) have also all been prepared in the gas phase, but are much less stable than the tetrahalides.


Firstly as @MaxW pointed out, using the given information, it is possible to find the shortest wavelength (or maximum frequency) X-ray only.

In an X-ray tube, electrons are accelerated in a vacuum by an electric field and shot into a piece of heavy metal (e.g., $\ce{W, Rh, Mo, Cu, Ag}$) plate. X-rays are emitted as the electrons decelerate in the metal. The output spectrum consists of a continuous spectrum of X-rays, with sharp peaks at certain energies as in the graph. The continuous spectrum is due to bremsstrahlung (German for "deceleration radiation"), while the sharp peaks are characteristic X-rays associated with the atoms in the target.
The spectrum has a sharp cutoff at low wavelength (high frequency), which is due to the limited energy of the incoming electrons (which is equal to the voltage on the tube times the electron charge). This cutoff applies to both the continuous (bremsstrahlung) spectrum and the characteristic sharp peaks, i.e. there is no X-ray of any kind beyond the cutoff. However, the cutoff is most obvious for the continuous spectrum.
Hence, the Duane-Hunt law comes handy in here, which gives the maximum frequency of X-rays that can be emitted by Bremsstrahlung in an X-ray tube by accelerating electrons through an excitation voltage $V$ into a metal target.
The maximum frequency $\nu_\text{max}$ is given by-
${\displaystyle \nu _{\text {max}}={\frac {eV}{h}},}$
which corresponds to a minimum wavelength
${\displaystyle \lambda _{\text {min}}={\frac {hc}{eV}},}$
where $h$ is Planck's constant, $e$ is the charge of the electron, and $c$ is the speed of light.
Thus putting the values given-
$$\lambda _{\text {min}}={\frac {hc}{eV}}={\frac {\pu{1240 eV nm}}{\pu{20*1000 eV}}}= \pu{0.62*10^-10 m}$$
For further references check the Wikipedia page.


Why does neighbouring group participation increase the rate of reaction?

In neighbouring group participation (NGP), an internal nucleophile first displaces a leaving group in an intramolecular sense, forming a ring (usually 3 or 5 membered) which is then opened by the incoming nucleophile intermolecularly.

  General mechanism of neighbouring group participation (taken from March, Advanced Organic Chemistry, 2007

In the scheme above, Z first displaces X to form a 3 membered ring, which is then opened by Y. Your question then becomes why is Z faster than Y at substituting X?

Collision

In order for the substitution to take place, the nucleophile must come into alignment with the leaving group. In a general sense, intramolecular reactions are usually faster than intermolecular since the two reacting groups are tethered together – there is no requirement that they collide in order to react. This is especially evident for intramolecular formation of 3 and 5 reasons, since those ring sizes are especially favourable. (Note that intramolecular reactions isn't always favourable, for example in macrocyclisations of large rings, where high dilution is needed to close the ring rather than getting intermolecular reactivity).

Entropy

The reaction between the substrate and the external nucleophile Y in an intermolecular fashion is accompanied by a decrease in entropy (two molecules are becoming one in the transition state). Whilst this also happens in the intramolecular sense (the molecule is becoming more rigid), it is significantly less than in the bimolecular case.

Its also worth pointing out that its not always 100% clear whether a molecule is reacting via an NGP type pathway, and indeed there are many cases where its been shown that both are operating at the same time, with the intra being faster than the inter, but not sufficiently rapid to prevent any direct intermolecular reaction.


How to determine the potential of a cadmium zinc cell under standard conditions?

I will like to correct you that there is one problem with your question. The correct standard reduction potential is:
\begin{align}
\ce{Cd^2+  + 2e- &<=> Cd}& E^\circ &= -0.403~\pu{V}.
\end{align}
 And another thing is, you have used symbol $E^\circ$ for oxidation potential of $\ce{Zn}$ which should not be done. You can use there just $E = 0.7628~\pu{V}$ or in a better way, you can just write it with it's reduction potential value as:
\begin{align}
\ce{Zn^2+ + 2e- &<=> Zn}& E^\circ &= -0.7628~\pu{V}.
\end{align}
Now coming to your question, for these two reactions,
\begin{align}
\ce{Zn^2+ + 2e- &<=> Zn}& E^\circ &= -0.7628~\pu{V}\tag1\label{Zn}\\
\ce{Cd^2+ + 2e- &<=> Cd}& E^\circ &= -0.403~\pu{V}\tag2\label{Cd}
\end{align}
$E_\mathrm{cell}$ of any cell reaction is given as
$$E^\circ(\text{cell}) = E^\circ(\text{reduced species}) - E^\circ(\text{oxidized species})$$
or
$$E^\circ(\text{cell}) = E^\circ(\text{cathode}) - E^\circ(\text{annode}).$$

Here as $E^\circ$ value of reaction $\eqref{Cd}$ is greater (i.e. towards more positive). Hence $\ce{Cd}$  has greater tendency to accept electrons and get reduced. So the $\ce{Zn}$ is going to give electrons and gets oxidized. Now the half cell reaction will be,
\begin{align}
\ce{Zn &<=> Zn^2+ + 2e^-}\tag{at anode}\label{anode}\\
\ce{Cd^2+ + 2e- &<=> Cd}.\tag{at cathode}\label{cathode}
\end{align}

So according to formula
\begin{align}
E^\circ(\text{cell}) &= E^\circ(\text{cathode}) - E^\circ(\text{annode})\\
E^\circ(\text{cell}) &= -0.403~\pu{V} - (-0.7628~\pu{V}) = 0.3598~\pu{V}\\
\end{align}

The correct answer is $0.3598~\pu{V}$.


What is the name of this 3 open tube and where can I purchase one?

The flask pictured in your question is simply a three necked round bottom flask. They're available commercially in a variety of sizes with different configurations of ground glass joints (personally, I've always found it easier to buy single neck round bottom flasks and then get our glass blower to append the joints where I want/need them).


Water in its pure form, i.e. $\ce{H2O}$ does not have a smell, or at least no smell that we can distinguish because the receptors in our nose (and mouth) are continuously exposed to it.

What you smell are dissolved gases and other volatile impurities. The nature of these chemicals will vary mainly by location/source of the water, and might also be a bit influenced by your specific local plumbing.

For example, in Iceland near Myvatn there is a lot of volcanic activity and the water from the tap (in particular the warm water because it can dissolve more) smells like rotten eggs due to the sulfur. Another example is that in many countries the water will smell a bit like chlorine, because chlorine is used to keep the water safe to drink (kills germs).


Electrolysis of Potassium Iodate/Hydrogen Peroxide in Briggs-Rauscher Oscillation

The experiment has been attempted. Electrolysis through the use of copper electrodes in fact "kills" the reaction. This means that the whole system is immediately turned dark-blue, and eventually made more clearer, and eventually light blue indicating the presence of copper ions.

It is suspected that not only is the sulphate ion electrolysed, but perhaps the copper iodizes the potassium and other compounds.


Start from

$$K_\mathrm{b} = \frac{\ce{[HB+]}\ce{[OH-]}}{\ce{[B]}}$$

Define

$$\alpha = \frac{\ce{[HB+]}}{\ce{[B]} + \ce{[HB+]}} $$

or

$$\frac{\ce{[HB+]}}{\ce{[B]}}=\frac{\alpha}{1-\alpha}$$

so that

$$K_\mathrm{b} = \frac{\alpha\ce{[OH-]}}{1-\alpha}$$

or

$$\alpha = \frac{K_\mathrm{b}}{K_\mathrm{b}+\ce{[OH-]}}$$

But in the limit of infinite dilution the concentration of $\ce{OH^-}$ is solely determined by the autodissociation of water (if the solution is pure unbuffered water) and

$$\ce{[OH-]} = \sqrt{K_\mathrm{w}}$$

so that

$$\alpha = \frac{K_\mathrm{b}}{K_\mathrm{b}+K_\mathrm{w}^{1/2}}$$

But at RTP $K_\mathrm{w}^{1/2}\approx 10^{-7}$ resulting in the final expression

$$\alpha = \frac{10^{7}K_\mathrm{b}}{10^{7}K_\mathrm{b}+1}$$


Boric acid is soluble in water, not oil.  Pick an emulsifier and add it to your ant poison.

Boric acid needs to remain acidic to retain its effectiveness as an ant killer.  Ants deaths due to boric acid poisoning is a topic still up for debate; however, it has been suggested boric acid is a neuro-toxin and that the acid damages one of the organs in an ants digestive system.

Peanut butter is usually slightly acidic; however, your particular mixture may be rendering the boric acid inert.  Solutions of sugar and boric acid with a high pH have been known to pass right through ants doing little harm. Perhaps using an acidic additive will increase the bait's effectiveness.

Boric acid in a 1-2% concentration may be ideal if the entire ant colony is targeted. To target the entire colony, enough poison should be used so the ant does not die before it makes it back it its colony.


What does "reaction rank" in Reaxys stand for?

For the Elsevier's Reaxys data base, there is a support page you may consult without subscription, nor log-in to the product.  Actually, this is more like a searchable directory, too, because you search for keywords, like «rank», too.

Doing so here -- time stamped 25/04/2019 -- offers insight about «How do I view my results?».  About half-way through this document, in subsection «Reactions. Order, limit or export your search results», the documentation states:

  "[Reaxys ranking] Defines the method of determining 'relevance'. To
  change the sort type, click 'Reaxys Ranking' and define another sort
  option: 'No of References', 'Reactant Availability', ‘Product
  Availability'', ‘MW of product’, ‘Yield’ or ‘Publication Year’."

And because the content of columns of numeric content (like molar mass, number of reagents, year of publication) may be sorted, this may affect the presentation of the export, too.  It equally may explain the occurrence of gaps in the values you mention.


How is iron(III) chloride changed from solid to solution?

You can prepare a solution of $\ce{FeCl3}$ by dissolving the appropriate amount in water.

Let's say you want a $0.1 \text{ M}$ solution of $\ce{FeCl3}$. You would do this by dissolving 0.1 mole of $\ce{FeCl3}$ in 1 liter of water.

$$0.1000 \text{ mol} \times \dfrac{162.2\text{ g}}{1\text{ mol}}=16.22 \text{ g}$$

You would need to dissolve 16.22 g of $\ce{FeCl3}$ in enough water to make 1 liter of solution.

This solution is not stable over time. $\ce{FeCl3}$ hydrolyzes in water. See my answer to this question. The hydrolysis can be slowed/reversed by adding $\ce{HCl}$. Since solutions of $\ce{FeCl3}$ are already acidic, there is no harm usually in adding more acid. For this 0.1 molar solution, you might choose to dissolve 16.22 g of $\ce{FeCl3}$ in enough 0.1 M HCl to make one liter of solution.


Relativistic effects and lanthanide contraction

The lanthanide contraction is caused by two effects

The 4f orbitals are very diffuse and therefore result in poor screening of the electrons further out, those in the n=5 and n=6 orbitals
relativistic effects

This Wikipedia article presents a nice discussion of the lanthanide contraction.  The article estimates that the first effect, the screening effect, is the major factor, with relativistic effects playing only a minor role (~10%) in the observed contraction.

  what I do not understand how do they match, one of them decreases the
  radius and the other one should expand it as the f orbitals expand
  right?

That's correct, the outer s- and p-orbitals contract due to the lanthanide effect, while d- and f-orbitals expand (see this SE Chem post for a more detailed description).  In the lanthanide series, the 6s electrons are further from the nucleus than the 4f electrons, therefore they (the 6s electrons) determine the radius.  A contraction of the 6s orbital radius results in a contraction of the observed lanthanide electron radius, even though the 4f electrons have moved slightly further away from the nucleus.


How is chirality center defined for an atom with 5 substituents?

Chirality depends not so much on coordination number, but on the symmetry of the substituents.  Meaning you have to pay attention to coordination geometry.  In the case of four-coordination you can have four different substituents and yet no chirality... if you have square planar instead of tetrahedron coordination.
For five-coordination you lack a mirror plane and thus have a chiral center in the following cases:

Square pyramidal geometry: if all four substituents on the base are different, or if just one pair of these basal substituents are identical and the identical pair is on adjacent vertices.  The axial substituent does not have any impact because all possible mirror planes pass through it.

Triangular bipyramid geometry: if all three equatorial substituents are different and the two axial substituents are different from each other.  You may still have chirality if one axial substituent happens to be the same as one equatorial one.


How does DMSO act as a carrier for transdermal transport of large molecules?

Quote taken from the review (1) (and which is also duplicated in Jacob's 2015 "canonical DMSO" book (2, p. 16)):

  There is some evidence to suggest that DMSO can increase diffusion
  through the
  stratum corneum by disruption of the barrier function. This probably occurs through aprotic interactions with intercellular lipids and may
  also include reversible distortion of lipid head groups that produce a
  more permeable packing arrangement.  DMSO may also play a role in
  partitioning as well by forming solvent microenvironments within the
  tissue that can effectively extract solute from vehicle. Finally, DMSO
  can have a profound solubilizing effect on less soluble agents in a
  variety of vehicles, increasing penetration simply by delivering a
  higher concentration to the membrane barrier.

alongside with extensive molecular modelling by Gurtovenko in 2007 (3) are both suggesting quick pore formation and expansion alongside with the thinning of cellular membranes:

  The reported membrane thinning and, in particular, pore formation
  provide a credible molecular-level explanation of how DMSO promotes
  permeation of molecules, particularly those of hydrophilic nature,
  through lipid membranes as observed experimentally.

From what I understand, DMSO paves the way for bigger molecules, such as steroids, making inner skin layers and underlying tissue reversibly accessible, sometimes even on nanoscale level. For the precise mechanism I think it is worth reading through the entire paper (3).

(1) Capriotti, K.; Capriotti, J. A. The Journal of clinical and aesthetic dermatology 2012, 5 (9), 24.
(2) Jacob, S. W.; Jack, C. Dimethyl sulfoxide (DMSO) in trauma and disease; CRC Press, 2015.
(3) Gurtovenko, A. A.; Anwar, J. The Journal of Physical Chemistry B 2007, 111 (35), 10453–10460. DOI 10.1021/jp073113e


Both schools of thought were prevalent, one said that the narrow end should be considered in the plane of paper, the second said that the wide end should be considered. Perhaps the first was based on a perspective flavored approach while the second mainly relied on popular drawing schemes. Excerpts from the 2006 IUPAC Recommendations1 (more information is given there):

  The least disruptive recommendation is to reject a perspective-based approach, and rather to interpret the bonds as more-abstract graphical objects. A solid wedged bond should be interpreted as projecting above the plane of the paper from the narrow end to the wide end (with the wide end being nearer to the viewer), and a hashed wedged bond should be interpreted as projecting below the plane of the paper, also from the narrow end to the wide end (with the wide end being further away from the viewer). To summarize, the narrow end of solid wedged bond or a hashed wedged bond always should point towards the stereogenic center to which it applies.

  That said, it is possible to use the other bond types in ways that are unambiguous. In particular, any type of hashed bond is unambiguous in print when connected to an atom with no other substituents (a hydrogen atom, a chlorine atom), no other non-hydrogen substituents (a hydroxy group, an amino group), or at most one other non-hydrogen substituent (an ethyl group, an ethoxy group). In each of those cases, a viewer would have no difficulty in understanding that the hashed bond projects below the plane of the paper from the other atom to that atom. Accordingly, both the unwedged hashed bond and the hashed wedged bond with reverse directionality are acceptable when used exclusively in those cases.

However, a caveat:

  However, anyone using bonds of these types should be aware that they are not the most common bond types used to represent behind-the-plane configuration. At best, an author using these types may cause a moment of confusion as a reader figures out which convention was intended. In extreme cases, a reader (or chemical software) might actually interpret the structure incorrectly—for example, by interpreting an unwedged hashed bond in the convention where it would represent relative (and not absolute) configuration, as discussed in ST 0.8 and ST-6. Fortunately, such cases are rare. The hashed edged bond (interpreted from the narrow end to the wide end) remains preferred...

Reference

Graphical representation of stereochemical configuration (IUPAC Recommendations 2006). Pure Appl. Chem. 2006, 78 (10), 1897–1970. DOI: 10.1351/pac200678101897. © IUPAC.


Actually it is saponification.  Bleach has alkali added to it, to stabilize it against decomposion to chlorine gas.

To wash your hands after contact with bleach was a wise move.


Why do different impurities create different colors in diamonds?

If you had googled a little bit you would have found this wonderful resource.

  In a yellow diamond, a few carbon atoms per million have been replaced
  by nitrogen atoms, each containing five valence electrons. The
  structure of the diamond crystal does not change significantly, but
  the extra electrons occupy a donor level.

  The nitrogen donor level energy in a diamond is large, peaking at
  about 4 eV. With a concentration of a few nitrogen atoms per million,
  instead of a clean "spike" donor level energy, the nitrogen donor
  level energy broadens into a band because of a number of complex
  factors, including thermal vibrations. This broadened donor energy
  band results in the difference between donor and conduction bands
  being as low as 2.2 eV. The most likely transition allows incident
  light with energy of 4 eV per photon to excite electrons from the
  donor level to the conduction band. However, it is possible for
  electrons to be excited to the conduction band with energy of 2.2 eV
  and upwards. This means that blue and violet light are absorbed from
  the full spectrum of light normally transmitted, and the resulting
  color is yellow.

  Unlike blue boron-doped diamonds, which conduct electricity,
  nitrogen-doped diamonds remain insulators. This is because nitrogen is
  a deep impurity: a relatively high energy (2.2 eV, as compared with
  0.4 eV for boron) is required to excite electrons from the donor energy level to the conduction band, and only a fraction of the
  available electrons are freed to carry a charge.

  An extremely rare green color can result from a higher nitrogen
  content of about 1 atom per 1000 atoms of carbon. At even higher
  nitrogen concentrations, the donor level broadens so that all visible
  light can be absorbed, resulting in a black color.

  Synthetic blue diamonds are created by adding boron as an impurity.
  Boron is trivalent, having three valence electrons; where a boron atom
  replaces a carbon atom in the diamond structure there is one fewer
  electron than usual. This missing electron, or hole, creates an
  acceptor energy level above the valence band. The boron acceptor
  energy is only 0.4 eV, so light of any energy can be absorbed during
  excitation. The boron acceptor band is broadened, and the absorption
  tapers off throughout the visible light energies, resulting in
  stronger absorption at the red end of the spectrum. At a level of one
  or a few boron atoms for every million carbon atoms, an attractive
  blue color results. Natural diamonds of this color, such as the Hope
  Diamond, are rare and highly priced.


Determining which hydrate of calcium chloride I have?

The mass of your sample decreased by 22%.

If you originally had $\ce{CaCl2.4H2O}$ and quantitatively dehydrated that to $\ce{CaCl2.H2O}$, then you should have seen a decrease in mass of 27%.

If you originally had $\ce{CaCl2.4H2O}$ and quantitatively dehydrated that to $\ce{CaCl2.2H2O}$, then you should have seen a decrease in mass of 18%.

Although it's tempting to say that you just have some mix of mono-and-dihydrates, I think that we just can't ignore the stability of your mass measurements and the temperature of your sample, which due to the radiative heating from the toaster heating coils was probably even greater than $\mathrm{350^oF}$. The most reasonable option remaining is that you originally had primarily $\ce{CaCl2.4H2O}$, but also some $\ce{CaCl2.6H2O}$ mixed in, and now you have just the monohydrate remaining.


Trying to understand the contradiction in the data of Proust/Dalton/Berzelius vs Gay-Lussac

In regards to your first question, I would think that Dalton would have a sense of the molecules' formulas based on the information at hand. Not only did Dalton have the mass ratios for these compounds, but he had also developed a measure of the relative mass of the atoms, as shown in your first document. Given this information, it would not have been difficult to compute that there were $\sim2$ atoms of nitrogen for every one of oxygen in nitrous oxide, nitric oxide was one to one, etc.

In regards to your second question, I think you are misinterpreting the contradiction. The authors proposal for the reactions was that if the compound formed was $\ce{N2O}$, it should only form one volume, since it should only form as much volume as there was volume of $\ce{O}$ originally. This is because there can only be as many molecules formed as there were atoms of the limiting reactant. The two volumes of $\ce{N2O}$ that are actually produced runs counter to this, as it would seem to require having $2$ volumes of $\ce{O}$ to start with.

For all the reactions, the products and reactants were related by whole number ratios, but not the same ratios suggested by their predicted molecular formulas.


Let $H^0$ be the heat of formation of the gas at $298~\mathrm{K}$ and $1\ \mathrm{atm}$. and let $G^0$ be the free energy of formation of the gas at these same conditions.  The enthalpy of the gas at temperature $T$ (any pressure) is then:
$$H(T)=H^0+C_p(T-298)$$  First let's get the free energy at temperature T and 1 atm.  At constant pressure, the change in free energy with respect to temperature can be calculated from:
$$\frac{d(G/T)}{dT}=-\frac{H}{T^2}$$
If we integrate this equation by parts, we obtain:

$$\frac{G(T,1~\mathrm{atm})}{T}-\frac{G^0}{298}=\frac{H(T)}{T}-\frac{H^0}{298}-\int_{298}^T{\frac{C_p}{T}dT}$$
Some algebraic manipulation of this equation yields:
$$G(T,1~\mathrm{atm})=G^0+\frac{(G^0-H^0)}{298}(T-298)+C_p(T-298)-TC_p\ln(T/298)$$
If we define $S^0=\frac{(H^0-G_0)}{298}$, this equation becomes:
$$G(T,1~\mathrm{atm})=G^0-S^0(T-298)+C_p(T-298)-TC_p\ln(T/298)$$
Adding the free energy change between $1~\mathrm{atm}$ and pressure $P$ at temperature $T$ yields:
$$G(T,P)=G^0-S^0(T-298)+C_p(T-298)-T[C_p\ln(T/298)-R\ln(P/P_0)]\tag{1}$$
where $P_0=1~\mathrm{atm}.$  The term in brackets in this equation is the same as $S(T,P)-S^0$.  If we substitute this into the previous equation, we obtain:
$$G(T,P)=G^0-S^0(T-298)+C_p(T-298)-T[S(T,P)-S^0]$$
If we combine terms in this relationship, we obtain:
$$G(T,P)-G^0=C_p(T-298)-(TS(T,P)-298S^0)$$
The term $C_p(T-298)$ is $\Delta H$, and the term $(TS(T,P)-298S^0)$ is $\Delta (TS)$.  So, as expected, the equation is the same as $$G(T,P)-G^0=\Delta H-\Delta (TS)$$
where
$$S=S^0+C_p\ln(T/298)-R\ln(P/P_0)]\tag{2}$$
Either Eqn. 1 or Eqn. 2 can be applied, depending on preference.


Reaction mechanism for the conversion of acyl hydrazide to acyl azide

You are correct acyl hydrazide can be converted into acyl azide by adding $\ce{HNO2}$ also you wrote that the $\ce{NH2}$ group of hydrazide will be diazotisated making a $\ce{N2+}$ group hence I am assuming that you know the mechanism, after diazotisation the adjacent $\ce{N}$ donate the lone pair to the $\ce{N2+}$ group making the final product as acyl azide.

The mechanism scheme is in the following picture :


Which of the following releases the smallest amount of energy upon gaining an electron?

Alright, this is similar to a question that I have answered before but I will just give you the reasoning for this phenomenon. I will not give you the answer since this appears to be a homework question.

Why an atom gains energy when you take off an electron?
Well for this, think of the attractions between the nucleus and the electrons, so when you must strip away an electron from this stability, you must add energy to do so.

Why an atom releases energy when you add an electron?
This one might be a little hard to grasp, but at the end it is not too bad. Essentially the molecule is becoming more stable. It is getting near to having a full octet. (Ignoring noble gasses). In other words the kinetic energy added by gaining the electron is released when the atom is achieving a stable state.

This is the general reasoning behind this, not too complicated, but I hope you will be able to understand it. Just, if you can't, remember the first one and think of the opposite for the second one.

Now, take a look at this question, to help you with your question even more. You will then be able to deduce the correct answer.


Stereoelectronic effects in torquoselectivity of 4π electrocyclic opening

The torquoselectivity of the above reactions is dictated by stereoelectronic effects, which (in this case) are significant enough to take precedence over steric effects. In general, electron-donating substituents (e.g. OMe, OTMS, ...) prefer to move outwards, and electron-accepting substituents (e.g. CHO) prefer to move inwards.

These can be understood by considering the frontier orbitals in the transition state of the conrotatory opening, as explained by the Houk group in earlier publications1,2 (the reactions given above were, in fact, carried out in order to provide experimental verification of their theoretical predictions). A more accurate orbital correlation diagram for the reaction is given below (modified from ref 2 to include diagrams of the relevant MOs):

A brief explanation is as follows. As the reaction proceeds, the C–C σ bond in the reactant is stretched, and consequently the σ* orbital is initially strongly destabilised (its energy increases rapidly). At a certain point near the transition state, though, it becomes sufficiently close in energy to the π* orbital such that orbital mixing can occur; this mixing leads to stabilisation and eventual "evolution" into the HOMO of butadiene.

Therefore, in the transition state, the HOMO is mostly σ-like, and the LUMO σ*-like; this was also verified computationally (details in ref 2). These frontier orbitals are of the correct symmetry to interact with pi-type orbitals of substituents. For donor substituents such as OMe, these orbitals are filled (lone pair on oxygen); for acceptor substituents such as CHO, these orbitals are empty (low-lying C=O π*).

The dashed orbitals in the following diagram represent the positions of these orbitals in the TS for both cases of outward and inward rotation. In the case of outward rotation, the substituent orbital can only interact with the lobe on C–3 which is in close spatial proximity; however, in the case of inward rotation, the substituent orbital can interact with the lobes on both C–3 and C–4.

In general, the interaction of a filled orbital and an unfilled orbital leads to stabilisation, whereas the interaction of two filled orbitals leads to destabilisation. The torquoselectivity is determined by maximisation of the former and minimisation of the latter. If the interaction with the lobe of C–4 leads to destabilisation, then inward rotation will be disfavoured, and vice versa.

For a donor substituent:

Inward rotation decreases the stabilising interaction between the donor orbital and the LUMO of the TS, as the lobe on C–4 is of the opposite phase as the lobe on the donor orbital, which leads to poorer overlap between the two orbitals.
Inward rotation increases the destabilising interaction between the donor orbital and the HOMO of the TS, as the lobe on C–4 is of the same phase as the lobe on the donor orbital, which leads to stronger overlap.

Based on these reasons, donor substituents generally prefer to rotate outwards. In the case of OH, this preference was calculated to be a whopping 16.4 kcal/mol.3

For acceptor substituents, the interaction with the LUMO is not relevant as the interaction of two unfilled orbitals does not lead to any stabilisation. However, inward rotation maximises the stabilising interaction with the HOMO, since this is enhanced by overlap with the lobe on C–4. Hence, acceptor substituents generally prefer to rotate inwards. This preference, however, is markedly smaller than in the case of donor substituents - perhaps because there is only one factor that biases the course of the reaction (as opposed to two in the case of donor substituents). Experimentally there are fewer examples of torquoselectivity with electron-accepting substituents, and for the second reaction above, the preference for the inward rotation of the formyl group was calculated to be a much more modest 4.5 kcal/mol.4

Interestingly enough, for the cyano group (-CN), a preference of 4.3 kcal/mol was calculated - for outward rotation.4 It turns out that the donor π orbitals are more important than the acceptor π* orbitals, which are relatively high in energy. The formyl group only rotates inwards because it has an exceptionally low-lying π* orbital!

References

1. Kirmse, W.; Rondan, N. G.; Houk, K. N. Stereoselective substituent effects on conrotatory electrocyclic reactions of cyclobutenes. J. Am. Chem. Soc. 1984, 106 (25), 7989–7991. DOI: 10.1021/ja00337a067.

2. Rondan, N. G.; Houk, K. N. Theory of stereoselection in conrotatory electrocyclic reactions of substituted cyclobutenes. J. Am. Chem. Soc. 1985, 107 (7), 2099–2111. DOI: 10.1021/ja00293a046.

3. Houk, K. N.; Spellmeyer, D. C.; Jefford, C. W.; Rimbault, C. G.; Wang, Y.; Miller, R. D. Electronic control of the stereoselectivities of electrocyclic reactions of cyclobutenes against incredible steric odds. J. Org. Chem. 1988, 53 (9), 2125–2127. DOI: 10.1021/jo00244a058.

4. Rudolf, K.; Spellmeyer, D. C.; Houk, K. N. Prediction and experimental verification of the stereoselective electrocyclization of 3-formylcyclobutene. J. Org. Chem. 1987, 52 (16), 3708–3710. DOI: 10.1021/jo00392a047.


Alternative mechanism for carbocation rearrangements

Scheme 1: Orbital model of a Wagner-Meerwein rearrangement.

You may think you are proposing something new, but you aren’t.

While these Wagner-Meerwein rearrangements are typically referred to as ‘hydride shifts’, that name should not be taken too literally (as in your first scheme). Instead, it is a concerted orbital interaction. There is never any hydrogen cation or hydride anion formed, hydrogen remains in the vicinity. The entire process is concerted and thus proceeds suprafacially.

Writing the central structure as a double bond with proton in the vicinity, as you did, or as ‘two cations coordinating a hydride anion’ is really only a formality. It is not an intermediate in any way, it is a transition state.

Being a transition state, by the way, means that it is a directed process. If the hydrogen went half way, it is going to proceed on (law of inertia, if you wish) and not go back. However, given a sufficiently stabilised cationic system, it may reverse at a later moment in time.


According to HSAB theory, Soft Acids have:

large atomic/ionic radius;

low or zero oxidation state bonding;

high polarizability;

low electronegativity.

The ion $\ce{Ag+}$ satisfies all the aforementioned properties.
I don't agree with you when you say it is anomalously small due to high nuclear charge and poor shielding by the d-orbitals. The effective nuclear charge has nothing to do here.

According to http://www.webelements.com/silver/atom_sizes.html: The radius of this ion is $114\ \mathrm{pm}$ when it's in 4-coordinate, tetrahedral geometry and it's $116\ \mathrm{pm}$ when  it's 4-coordinate, square-planar geometry. Please just compare these radii with the radius of  $\ce{Al^3+}$ ion : $53.5\ \mathrm{pm}$ which is considered as a hard acid.
I hope it's clear now!


It is a basic postulate of quantum mechanics. It states that it is impossible simultaneously to measure the position of a particle, say an electron, an atom or a molecule, and its momentum. It places limits on these measurements such that uncertainty in position $\Delta x$ and momentum $\Delta p$ is given by $\Delta x\Delta p \ge \hbar/2$ where (hbar) $\hbar =h/2\pi$ and $h$ is the Planck constant $6.626 \times 10^{-34}~\mathrm{J~s}$. In a classical world no such restriction occurs, but this is not what is observed experimentally.

The important part is simultaneous measurement. We can measure where an atom is to whatever precision an instrument will allow, (e.g. by xray crystallography) but we cannot in the same measurement (i.e. at same time) know what its momentum is. We can measure an atom's speed v as accurately as we like and hence momentum mv. We note its position in two places a and b, say 10 m apart, and divide distance by time to cover this distance.

The uncertainty principle places limits on the measurement of momentum mv and position that allows us to know both the position and momentum an atom (particle) has at any given instance. Ask again if this is not clear


When a liquid is evaporating or even boiling, minute drops of the liquid are formed by the mechanical disintegration of the surface of the liquid (bubble bursting, splashing or foaming). Some drops are carried away from the heated liquid by the steam (by convective and vapour flow).

The concentration of non-volatile dissolved matter in the entrained drops approximately corresponds to the concentration in the bulk liquid. Therefore, if the bulk liquid contains $\ce{NaOH}$, the entrained drops also contain $\ce{NaOH}$.

A quantitative parameter that describes this transport is the airborne release fraction (ARF). The ARF increases with temperature of the liquid and air velocity above the liquid. Measured values for liquids that are evaporating but not boiling range from $< 10^{-8}$ to $3\times10^{-5}$ (Mishima, J., L.C. Schwendiman and C.A. Radasch. November 1968. Plutonium Release Studies IV. Fractional Release From Heating Plutonium Nitrate Solutions in Flowing Air Stream, BNWL-931, Pacific Northwest Laboratory).

In principle, distillation and similar technical processes have the same problem. That is one reason why distillation shows only a limited decontamination factor (DF) even for non-volatile dissolved matter.


Contradiction in variation of conductance and conductivity on dilution

You get it wrong.

With the dilution, both conductance and conductivity ( specific conductance ) proportionally decreases, bound by the conductometer cell constant $G^* = \dfrac{\kappa }{ G}$.

What increases is molar conductivity $\Lambda_\mathrm{m}= \dfrac{\kappa }{ c}$ toward its infinite dilution value by the Kohlrausch equation $\Lambda_\text{m} =\Lambda_\text{m}^\circ - K\sqrt{c} = \alpha f_\lambda \Lambda_\text{m}^\circ$.

Note that conductance is NOT the ease with which ions move. You confuse it with electrical ion mobility, that is closely linked to the molar conductivity.

In fact conductance and conductivity have nothing to do with ions at all all.
Ionic solutions are just one special case how to manage these general current related quantities.

See conductance   and Conductivity on Wikipedia.


Bromination of 1,3,5-cyclooctatriene followed by nucleophilic substitution by excess dimethylamine

This is a quote from Solomons and Frylhe Organic Chemistry 10th Ed, Chapter 13, p.614

Reactions of this type are quite general with other conjugated dienes. Conjugated trienes often show 1,6-addition. An example is the 1,6-addition of bromine to cycloocta-1,3,5-triene.

A plausible mechanism for the bromination step would be the initial formation of a bromonium ion which is attacked distally by bromide:

Sequential nucleophilic substitution with dimethylamine then leads to the desired product.
Echoing Martin's comment on Siddharth Porwal's answer, there is a distillation step between the bromination and the substitution with dimethylamine. It is not possible for a putative bromonium ion to survive. Also, the isolation of the 1,6-dibromide also disproves the possibility of 1,2-addition mentioned in RE60K's answer.
The reaction above was originally reported by Cope et al.:[1]

Bromine (3 g) in 20 ml of chloroform was added to a stirred solution of 2 g of 1,3,5-cyclooctatriene in 20 ml of chloroform at −15 to −25° in a nitrogen atmosphere in a period of one hour. The product was distilled rapidly under reduced pressure; b.p. 95° (0.3 mm), n25ᴅ 1.5954. [...] Previous work had shown that dibromocyclooctadiene prepared by the two routes gave the same bis-dimethylaminocyclooctadiene by reaction with dimethylamine.

Reference

Cope, A. C.; Stevens, C. L.; Hochstein, F. A. Cyclic polyolefins. v. preparation of bromocyclooctadienes and 1,3,5-cyclooctatriene from 1,5-cyclooctadiene. J. Am. Chem. Soc. 1950, 72 (6), 2510–2514. DOI: 10.1021/ja01162a047.


Assigning locants when both unsaturation and substituents are present in a hydrocarbon

It is correct that the principal characteristic group (the characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name) has a higher priority for numbering than any other substituents (expressed as prefixes). It is also correct that the compound that is given in the question does not have any principal characteristic group.

However, you are missing the priority for unsaturation (i.e. multiple bonds), which is between the priority of the principal characteristic group. The most important simplified criteria for the numbering of simple acyclic compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

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

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Since the compound that is given in the question does not have any principal characteristic group, the first criterion (greater number of suffixes) given in P-14.4 (c) is not relevant in this case.

According to (e), the correct name is (2E)-4-bromo-4-methylpent-2-ene and not (3E)-2-bromo-2-methylpent-3-ene since the locant ‘2’ for the double bond is lower than ‘3’.


Electron transfer rates in Marcus Theory

There is a description of Marcus theory here How does the inverted Marcus region explain chemiluminescence? . The quantum nature can be added into the theory as necessary, it leads to a asymmetrical plot of rate constant vs free energy with the inverted region (large -$\Delta G$) decaying away more slowly than the normal region.  The equilibrium is between the solvent reorganisation with reactants and then with products. The solvent plays a controlling role in the process, the proof is that experiment demonstrates that  the rate of electron transfer can be limited by solvent motion. See Harrison et al.  Chemical Physics, Volume 116, Issue 3, 15 September 1987, Pages 429-448, and Irvine et al. ibid.  Volume 104, Issue 2, 1 May 1986, Pages 315-324. The novel aspect of Marcus theory is the role of the  solvent and this was not generally accepted until the inverted region was observed.


Figuring out the volume of air trapped inside a bottle of water

Displacement! Let the container that you have bottle in be filled to the brim inside a bigger container. Then when you breathe into the bottle, the amount of water that is displaced will be the same as the volume of the air.


Nice work! You clearly understand how factors like aromaticity/antiaromaticity and % s-character in a $\ce{C-H}$ bond can affect acidity.

You can use Coulson's Theorem to estimate the % s-character in the other molecules. In its general form Coulson's Theorem describes the relationship between the angle $\ce{X-Y-Z}$ in a molecule and the hybridization of the $\ce{Y-X}$ and $\ce{Y-Z}$ bonds according to the following equation:

$$\mathrm{1+\lambda_{XY}\lambda_{YZ}cos(\Theta_{XYZ})=0}$$

(where $\mathrm{\lambda}$ represents the square root of the hybridization index of a bond, e.g. $\mathrm{\lambda=\sqrt{3}}$ and the hybridization index $\mathrm{=3}$ for an $\mathrm{sp^3}$ hybridized bond)

In molecule III where the two $\ce{C-H}$ bonds are equivalent, Coulson's Theorem simplifies to

$$\mathrm{1+\lambda^2cos(117)=0}$$

and we find that $\mathrm{\lambda^2 = 2.2}$, or the $\ce{C-H}$ bond in III is $\mathrm{sp^{2.2}}$ hybridized.

As the hybridization index gets smaller, there is more s-character in the $\ce{C-H}$ bond which results in a more stable carbanion and consequently increased acidity. This is consistent with our thinking that, in general, acidity follows the order $\mathrm{sp > sp^2 > sp^3}$; or, the larger the $\ce{X-C-H}$ bond angle, the more acidic the $\ce{C-H}$ bond.


How to calculate the half time of a unimolecular reaction given the Arrhenius coefficient and the activation energy?

You calculated the high-pressure-limit half life, where the rate of collisions with M is so high that it has no influence on the rate constant. Now all you have to do is to consider that the pressure is 1 atm instead of at the high-pressure limit. That slows down the reaction such that the half-time is in the second instead of the micro second range.
Something similar happens with ozone. It is not very stable at sea level, but has a substantial half-time in the upper atmosphere (unless there are certain man-made substance up there).
To solve numerically, you could assume a Lindemann mechanism (see https://en.wikipedia.org/wiki/Lindemann_mechanism) and that $k_1$ is diffusion-limited (i.e. no activation energy, every collision leads to activated complex).
The textbook cited by the OP has:

Strangely, in an earlier table they show $k_0$ and $k_\infty$ as (along with pre-exponential temperature dependence $T^n$ and ?activation energy? E - units unknown):


How to determine and derive the protonation state of a functional group based on its pKa?

You're on the right track.  The Henderson-Hasselbalch equation is where this rule of thumb comes from; the key thing to remember is that, in biochemistry, your $\mathrm{pH}$ is fixed by the environment (there's always something acting as a buffer).  This is in contrast with typical H.H. problems in general chemistry where the amount of molecule and its $K_\mathrm{a}$ is changing the $\mathrm{pH}$ of the solution.

Anyhow, given that $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ are fixed values, we're free to simply compare the ratio of $[\ce{A-}]$ and $[\ce{HA}]$:

\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}\\
\log\frac{[\ce{A-}]}{[\ce{HA}]} &= \mathrm{pH} - \mathrm{p}K_\mathrm{a}\\
\frac{[\ce{A-}]}{[\ce{HA}]} &= 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}\\
\end{align}

If $\mathrm{p}K_\mathrm{a} = \mathrm{pH} + 1$ then we have $\frac{[\ce{A-}]}{[\ce{HA}]} = 10^{-1}$.  In other words, we have 10 times as much $\ce{HA}$ as we have $\ce{A-}$ (i.e., we have $\approx91\%\ \ce{HA}$) as we should expect for something that is mildly basic.

If $\mathrm{p}K_\mathrm{a} = \mathrm{pH} - 1$ then we have $\frac{[\ce{A-}]}{[\ce{HA}]} = 10^{1}$.  In other words, we have 10 times as much $\ce{A-}$ as we have $\ce{HA}$ (i.e., we have $\approx91\%\ \ce{A-}$) as we should expect for something that is mildly acidic.

Clearly, if the $\mathrm{pH}$ differs from the $\mathrm{p}K_\mathrm{a}$ by only 1 $\mathrm{pH}$ unit, we don't really have a completely protonated or deprotonated molecules, but it's certainly mostly protonated or deprotonated.  As the $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ get further apart, then we continue to change this ratio by orders of magnitude.

Let's look at a concrete example.  Physiological $\mathrm{pH}$ is 7.4 and the side chain of aspartic acid has a carboxylic acid group with a $\mathrm{p}K_\mathrm{a} = 3.9$, so we predict

$$\frac{[\ce{RCOO^-}]}{[\ce{RCOOH}]} = 10^{7.4 - 3.9} = 10^{3.5} = 3200$$

In other words, we still have some protonated sidechain, but we have >3000 times more deprotonated sidechain than protonated sidechain (did you notice that? the difference was >3 so we have >1000 difference in concentration ratio).  For all practical purposes, the aspartic acid side chain is "completely deprotonated" at physiological $\mathrm{pH}$.


Decarboxylation / Elimination type reaction

This type of reaction falls into the category of Grob fragmentation  [Wikipedia]. Conceptually, it is very similar to an E2 mechanism, where an anti-periplanar pair of electrons flows into the sigma anti-bonding orbital of a carbon-leaving group bond resulting in a new pi bond and cleavage of the carbon-leaving group bond. The primary difference between these two is that in an E2, that anti-periplanar pair of electrons comes from deprotonation, and in a Grob fragmentation, the electrons are from breaking some other sigma bond other than a C-H bond.

A syn-oriented version of this reaction can be seen in the generation of benzyne.

Also, decarboxyation is very common with a beta-carbonyl group. This doesn't exactly follow the Grob pattern, because there is no leaving group. The carbonyl plays the role of electron sink.