How can we recreate old pharaos' perfumes?

The analysis won't be done before lunch, but neither will it be an extremely difficult process.  Certainly some of the researchers will have a fragrance industry background  and access to necessary equipment like  a tandem gas chromatograph - mass spectrometer.  Also, Pharoah's perfumes were undoubtedly less complex than todays, and all oils used in perfumes are available as "known" compounds for comparison.  Additionally, all fragrances have been "fingerprinted" making identification of the compound in a sample fairly routine and straightforward.  By "fingerprinted" I mean that if a certain pattern of compounds can be detected than they will suggest the presence of a certain oil.  For example, look at lavender oil, there are about 30 chemical components in it - just like a fingerprint.

Perhaps the base solvent has evaporated through the wall of the container and oxygen has flowed into the container.  Certainly some of the ingredients will have degraded, but the container won't be full of acetic acid.  Some amount of an alcohol may have oxidized to the corresponding ketone, aldehyde or carboxylic acid, perhaps some amount of a conjugated diene may have oxidized to the corresponding dienol; but that shouldn't make the process that much more difficult.


Where to find common inorganic/organometallic molecules?

I appreciate the suggestions but some digging found a few possibilities:

As I mentioned in the question, there's the Cambridge Structural Database, which includes over 700,000 compounds (both organic and inorganic/organometallic). It's decidedly not free, but available at many universities.
There is the "Teaching Subset" of 733 compounds from CCDC which offers a Java web view of interesting compounds, including plenty of VSEPR examples and inorganic and organometallic species. This is free to access, but as far as I can tell, not free to redistribute.
Cool Molecules from St. Olaf college, including 900+ structures from experimental data, and lots of inorganic and organometallic species (most elements are reflected), including lots of "cool" or unusual shapes (e.g., hexagonal bipyramidal).
Crystallography Open Database as mentioned by @permeakra. I was aware of this for solid-state materials, but it seems as if they've merged in data from CrystalEye, a resource which grabbed crystallography data from journal articles, including many molecular species now, and can be browsed by journal (e.g., Organometallics).

The last two resources seem the best, since the data can be freely distributed. They also don't require using Java to view the structures.


Will piperine endure heating for several days of soxhlet extractions?

So, I promised to answer this question based on my experiences if I'd remember. And I did. Here we go:

I did three soxhlet extractions, each with 175 grams of black peppercorns.

The first extraction took 6 hours and the cycles were quite slow due to the vapor pathways of the apparatus not being insulated.
The second extraction took 2 hours, again with an uninsulated apparatus.
The third extraction also took 2 hours, but the cycles were much faster because I used aluminum foil for insulating the apparatus.

Each extraction used 400 ml of 99% isopropyl alcohol of which 200 ml was distilled off after extraction. Something to note is that this reclamation for the first run took at least two times as long as it did for the two later ones.

The yields were as follows:
- Run 1: 8,6 grams of piperine
- Run 2: 6,8 grams of piperine
- Run 3: 7,5 grams of piperine

From this I concluded that with sufficient insulation, something like 3-4 hours could potentially provide the optimal efficiency and that the yields of piperine aren't significantly affected by prolonged periods of heating in an alcoholic solution.

I hope this write-up will help or be of interest to someone!


You can get an estimate of the bond strength by using Coulomb's law and periodic trends for atomic radius. $$F = k{ {q_1q_2} \over r^2}$$The charge of all these alkali metals is the same in these compounds, so the only difference in bond strength comes from the the distance between the charges. The distance between the charges is related to each atoms' atomic radius and that is an increasing trend as you move down the alkali metals.

So, if distance is increasing, the force holding the solid together is decreasing, which would justify the trend $\ce{LiH}>\ce{NaH} >\ce{KH}> \ce{CsH}$ for melting point.


Is there any difference between de-Broglie wavelength and the wavelength of normal waves?

Is de-Broglie wavelength different from wavelength of waves such as
  electromagnetic waves?

I would atually say no, there is no difference, since the wavelength is purely mathematical characteristic of a wave, the distance over which the wave's shape repeats. There is, however, an important difference between de-Broglie waves on the one side and usual physical waves (mechanical & electromagnetic) on the other:

The later consist of periodic oscillations of a physical quantity, such as, for instance, the pressure for sound waves or electrical and magnetic fields for electromagnetic waves.
The former consist of periodic oscillations of a mathematical quantity, the probability of finding a particle at some location.


Determine heat flux from temperature profile

Your first line of thinking to part a) is correct. For conductive heat transfer through rectangular layers, the heat flux needs to be constant and equivalent through all layers for steady state, or you would have a 'build up' of heat at between one layer. This would locally increase the temperature, and we wouldn't be at steady state.

For b) your second approach to a) is useful. You know the relative sizes of the temperature gradients from the sketch, and know from a) that the heat flux must be the same for each slab. Combine these two and you will find a requirement on the relative magnitudes of the conductivities.


What factors promote honey's crystallisation?

The decisive factor is the plant origin of honey. The major components ( reportedly by some beekeeper books of my passed away father ) have ratio glucose : fructose about 1:1 to 1:1.7. Glucose reportedly crystallizes more readily, triggering fructose crystallization as well. So honey closer to 1:1 ratio has stronger tendency to crystalize.(**)
Aside of already mentioned factor of temperature, another factor is obviously the ripeness of honey. Lower water content in more matured honey means higher crystallization probability.
Seeding by already crystalline honey helps to trigger or speed up crystalization.
If crystallization is already happening, or if it is intended and desired, then stirring speeds up crystallization.
Seeding and stirring leads to many small crystals, what is applied during production of the pasted honey. This form is favourite one among people. It often dissolves faster then liquid honey during drink sweatening. It is also very convenient for bread spreading, as its flow is almost frozen.
Slow spontaneous static crystallization leads to big crystals, forming mechanically tough product, requiring big force to separate a chunk of honey(*). The crystallization rate, the size of crystalsfraction of residual liquid phase and physical properties of the product seem to be composition dependent as well.
If such honey is wanted to be liquid again, it is recommended to carefully heat it up in $\pu{40 ^{\circ}C}$ warm water bath. High temperatures affect the present probiotics and enzymes. Note that honey is along fermented milk the major source of natural probiotics.

(*) My father stored honey in 4 L flasks for home-made pickles. When honey was crystallized, we often digged it up by table spoons or forks. But unless they were very strong, like heavy stainless steel ones from Soviet Union, they had tendency to bend.
(**) Infamous one is larch tree honey. It has so strong crystallization tendency it often crystalizes in the honeycomb. It is not being able to be directly centrifuged out without honeycomb destruction.


How do equilibrium shifts affect solids?

When a particular chemical process is at equilibrium, the respective rates of the forward and backward reactions are equal, meaning that no net change in the concentrations of products and reactants occurs and the composition of the reaction mixture remains stable. If $\ce{OH-}$ is added to a solution already at equilibrium, then there will be an excess of product relative to reactants and the rate of the reverse reaction will increase relative to the forward reaction until equilibrium is reestablished. This means that the ions will recombine into a crystal lattice and form a precipitate. So, to answer your first question, no, the amount of $\ce{NaOH_{(s)}}$ does not remain constant; more of it will be formed if additional ions are added to a solution already at equilibrium.

The reason why pure solids are not factored into equilibrium expressions is that they are not in fact part of the solution. Any excess precipitate, irrespective of the exact quantity, has no impact on the composition of the solution at equilibrium because it exists outside of the solution by definition. If equilibrium is disturbed due to changing conditions, then more precipitate can form, or some of the existing precipitate may dissolve back into solution. Under any given set of controlled environmental conditions, however, the product of the concentrations of the dissolved species remains constant at equilibrium, which is precisely why the equilibrium will shift if conditions or concentrations change.

For a more rigorous and mathematical explanation, you could read up on dynamic equilibria, equilibrium constants, and thermodynamic activity.


what is the difference between excitation and emission spectroscopy?

Absorption:

Light with a specific frequency is selected
This frequency is shined (radiated) onto a molecule (sample)
The amount of light that shines through the molecule is measured, ie. the detector is behind the sample (180 degrees).
Step 1 is repeated with a different frequency.

When the sample absorbs a specific frequency, less light of that frequency will shine through to the detector. The machine records this and a peak appears in the spectrum for that frequency.

Frequency is usually measured in wavelength, and a wide range of wavelengths are applicable. In UV-VIS, this can be eg. 300-800 nm, so 800 nm is measured, then 799 nm, then 798 nm...down to 300 nm. In IR, you will go into micrometers. In X-ray, you will be around 10 nm I think.

Emission:

A light with a specific frequency is selected
The sample is irradiated with this frequency.
The amount of light that shines from the molecule is measured, ie. the detector is to the side of the sample (90 degrees).
The emitted light is measured for a range of frequencies
Step 1 is repeated with a different frequency.

In both processes, the sample absorbs light. The major difference here is the detector being located to the side of the sample. This allows it to ignore the light being shined onto the molecule, and instead measure light that is emitted from the sample. Think of the incident light as a column. In absorption spectroscopy, that column is absorbed by the sample and continues to the other side, where the detector can measure it. In emission spectroscopy, you want to avoid the column of light. Think of the sample like a light bulb that shines light in all directions. The detector in fluorescence is placed to the side, avoiding the column of light and picking up the light from the sample instead. The advantage to fluorescence is you can often selectively measure a moiety (part of the molecule) with superior sensitivity, whereas in absorption impurities or stray light are more prone to interfere with the signal and also leads to higher concentrations usually being necessary.

The reason the emission wavelength detected is different than the frequency used to irradiate it is due to Stoke's Shifts.

Stoke's Shift

Stoke's Shift requires basic knowledge of the difference between electronic and vibrational states. A molecule has electronic states, S=0, 1, 2, 3...etc. Within each electronic state is a set of vibrational states, v=1,2,3.... For example, two electrons can be in the 0 electronic state (termed ground state), but one electron can be in the v=2 vibrational state, while the other electron in v=4.

This is easier to picture if you look up Stoke's Shift, or specifically the diagram on Wikipedia's page for Fluorescence.

For example, in emission spectroscopy, an electron absorbs light, being excited electronically from S=0, v=1, to S=1, v=2. Vibrational relaxations will occur before electronic relaxations. That means the electron will first relax from S=1, v=2 to S=1, v=0. After this, it relaxes to S=0, emitting a photon in the process for detection. Because it relaxed its vibrational state first, energy was lost not in the form of light (thermodynamically). As a result, the photon finally emitted is at a lower energy than the photon that had been absorbed. That's the rudimentary explanation.


The compounds given in the question contain the characteristic group $(\ce{-OH})$. Since there is only one characteristic group, the seniority order of classes is not relevant in this case; thus, the $\ce{-OH}$ substituent corresponds to the principal characteristic group that is expressed as a suffix (‘ol’).

Therefore, the preferred IUPAC name (PIN) for compound (e) is benzene-1,2,4-triol.

Various traditional names exist for hydroxy compounds. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), only the name ‘phenol’ is retained as a PIN.

  P-63.1.1.1 Only one name is retained, phenol, for $\ce{C6H5-OH}$, both as a preferred name and for general nomenclature. The structure is substitutable at any position. Locants 2, 3, and 4 are recommended, not o, m, and p.

Therefore, the PIN for compound (c) is indeed phenol.

Furthermore, according to Subsection P-63.1.1.2, the names

pyrocatechol (benzene-1,2-diol),
resorcinol (benzene-1,3-diol), and
hydroquinone (benzene-1,4-diol)

are retained but only for general nomenclature and only when unsubstituted.

Therefore, the PIN for compound (d) is the systematic name benzene-1,2-diol (the traditional name ‘pyrocatechol’ may be used in general nomenclature).


Difference between state of matter and phase

A phase is a region of space where all physical and chemical properties are uniform. This means that for example the density, the chemical composition and the temperature is equal everywhere in the region. This will also mean that the state of matter (solid, liquid, gas, etc.) is equal everywhere.

A state of matter is a form that matter can take. Examples are solid, liquid, gas, plasma. There are some others that appear under extreme conditions.

For example, nitrogen and water at 50K are both solid, so they have the same state of matter. However the chemical composition is not the same and hence they don't have the same phase.


Cycloenneatetraenyl cation, using the Greek root for nine, is certainly not antiaromatic.  A ring that large will in general be nonplanar unless strong aromatic coupling is there to drive a planar shape, and if the planar form is antiaromatic so much the better for nonplanarity.  By classical rules such a nonplanar form would be nonaromatic.

Whether this nonplanar form is Mobius aromatic is open to interpretation, or at least it seems that way given the twisted nature of the ring.  In any event Mobius aromaticity is unlikely to give the strong stabilization we see in the Huckel version because the p orbitals do not overlap perfectly.  Most likely the structure and stability of the ion  are more determined by sigma bond interactions (minimizing steric strain) than pi bond interactions.


If you read the Wikipedia page for Nernst's Heat Theorem carefully, it explains that the novelty lies not in showing (as you point out) that as $T \rightarrow 0$, $\Delta G \rightarrow \Delta H$, but rather that as   $T \rightarrow 0$, the slope of $\Delta G$ as a function of T goes to 0, that is

$$\lim_{T\rightarrow0} \left( \frac{\partial \Delta G}{\partial T} \right)_p =0$$

Then, since that slope is given by $-\Delta S$, it predicts the experimentally confirmed conclusion that as $T \rightarrow 0$, $\Delta S \rightarrow 0$,


So how exactly are these flavonoids shielding UV radiation?

Due to the structure of the flavonoids, there are many transitions to absorb the UV radiation. Also, there are many transitions to release the energy. Most of the time, it is released in 2 steps (2 photons) which equal the energy of the photon that exited the molecule.

Fluorescence emits a photon that is close to the same energy as the first photon. For quercetin, the excitation is at 440 nm and the emission is at 540 nm.


Elimination reaction of 1-bromo-2-phenylcyclopentane

There are three common elimination mechanisms: $\mathrm{E1, E2}$ and $\mathrm{E1_{cb}}$. $\mathrm{E1}$ and $\mathrm{E1_{cb}}$ both imply that an ionic intermediate is formed first; $\mathrm{E1}$ implies the same carbocationic intermediate as in an $\mathrm{S_N1}$ reaction. The $\mathrm{cb}$ in $\mathrm{E1_{cb}}$ means conjugate base, i.e. the first step is deprotonation of the compound to give an carbanionic intermediate.

$\mathrm{E1}$ reactions require good leaving groups and a moderately stabilised carbocation; this is not the case in your reactant.

$\mathrm{E1_{cb}}$ requires a very strong base and an at least mildly stabilised anion; this is not the case with your reagent and reactant.

Thus, the mechanism must be according to $\mathrm{E2}$, where the abstraction of a proton and the displacement of the leaving group happen at the same time. $\mathrm{E2}$ mechanisms require a $180^\circ$ anti-periplanar configuration of the $\ce{Br-C-C-H}$ bonds for easier attack and favour those double bonds where the hydrogen is less stericly hindered. Additionally, your reactant is trans-configured, thus meaning that there is no available proton on the higher-substituted carbon that can participate in elimination. Altogether, this leads to $\mathrm{E2}$ eliminations generally following Hofmann’s rule rather than Zaitsev’s rule. Hofmann initially observed what later became the Hofmann rule in the Hofmann elimination, which follows an $\mathrm{E2}$ mechanism.

As the ethanolic $\ce{KOH}$ is not strong enough to isomerise the double bond, the Hofmann product will be observed.


Does changing EDTA concentrations affect the amount of calcium ions in milk?

Since this post was a while ago, I'm assuming an answer has already been decided on, but I'll take a shot at answering anyway.

If I were you I would rely on concentration changes of EDTA as the independent variable since it allows for more precision in your titration. If you were to vary temperature, it would have to be the temperature of the titrant such that volume changes could be accounted for (since analyte is usually in an approximate volume of water). While volume fluctuations of titrant are theoretically possible, they are rather difficult in a laboratory setting and not exactly worth the effort unless you have lots of time and expensive equipment to control the temperatures exactly. Not only that, but volume changes of water are only approximately 5% from 0 Celcius to 100 Celcius.1 Thus, as I said, changes in [EDTA] would be more efficient and easier to manage.

Thermal Expansion of Water


Raman Spectroscopy: How to calculate their constants?

In Raman spectroscopy the difference in frequency between the exciting line and the measured signal frequencies contains the information you want. These values, made positive if necessary, give the ro-vibrational transition frequencies.

You are assuming that the molecule is vibrating with an anharmonic potential, such as a Morse potential. The energy levels are given by $E_n= \omega_e(n+1/2) - x_e\omega_e(n+1/2)^2$ (in wavenumber units) with quantum number $ n = 0, 1, ...~$, and $\omega_e$ the frequency and $x_e$ is the dimensionless anharmonicity.

In addition to the vibrational energy there is rotational energy whose levels are $E_J =B_eJ(J+1)$ for rotational constant $B_e$ (in wavenumbers) and quantum number $J=0, 1, 2,...$ and ignoring centrifugal distortion. You will also have to use different $B_e$ values for the $n=0$ and $n=1$ vibrational levels since the bond length is slightly different between (anharmonic) vibrational levels.

You know that the rotational-vibrational levels appear in groups  O, P, Q, R, S in increasing frequency and that Raman has O, Q & S bands only. In the Q band $\Delta J = 0$ between vibrational levels n to $n+1$ and is $-2$ and $+2$ for O  & S branches respectively. The Raman selection rules are $\Delta n = \pm 1$ and $\Delta J = 0, \pm 2 $.

If you look at a rotation-vibration energy level diagram for IR transitions you should by adapting this be able to work out your values for your Raman transitions.

There is a figure showing energy levels at
 https://en.wikipedia.org/wiki/Rotational%E2%80%93vibrational_spectroscopy

The Q band transition ($\Delta n =0, \Delta J =0$) in your notation is $\omega_0 = \omega_e - 2x_e \omega _e $ and the other branches calculated similarly with $\Delta J = \pm 2$ as appropriate.


Is there a household method to decolorize potentially contaminated water?

A simple test you can do is to boil a large amount of the water and see if a sediment is left in the vessel. This will tell you certain things about you contaminant.

Is it particulate? We expect so because a filter is removing the contaminant.
Is it's boiling point below that of water?

If you get a residue what is it's state? I'm expecting a solid, I dont anticipate a coloured liquid that is miscible in water with a boiling point higher than water.

Does it burn? How does it burn? Colour of flame? Further residue?

I'd also try adding a colourless surfactant to the water, does it remain coloured?


IChO problem involving the condensation point of NF3, NHF2, NH2F

The interactions do not limit to only dipole-dipole, which means positive charge is not everything.

What's special about $\ce{NF3}$? It has no hydrogens. As a result, it cannot form hydrogen bond, therefore its boiling point (or condense point) is far lower than the others'.


The ink you describe is an example of a leuco dye, which are molecules that exist in colored or non-colored states depending on what they're interacting with and under what conditions (temperature, pH, electronic environment, etc.).

For the ink system, you need 3 components: the electron-donating leuco dye molecules; the electron-accepting molecules (the developer); and a matrix of compounds that stabilize the reaction of the first 2 components for a range of temperatures (meaning the reaction is favored one way above a certain temperature, and another way below a certain temperature).

There is a simple graphical explanation from the notable pen company Pilot, part of which is below. Note that their system uses microcapsules that interact with other molecules in the ink in certain temperature ranges that result in either colored or transparent ink. In the diagram, A is the leuco dye, B is the developer, and C is the stuff that stabilizes the reaction of A and B over a temperature range.

This detailed patent application for "Thermosensitive decolorable ink composition" details the physical and chemical processes involving formulations of leuco dyes. When you add heat to the ink by erasing, the non-colored form is present. When you cool it down, you are letting the system revert to the colored form and the ink reappears.


How do I find molar mass using gas laws?

A gaseous sample of a compound has a density of $\pu{0.977 g/L}$ at $\pu{710.0 torr}$ and $\pu{100.00 ^\circ C}$. What is the molar mass of this compound?

The ideal gas law states $$pV = nRT\tag{1}\label{id-gas}$$ where $p$ is defined as the pressure, any units, $V$ is the volume in liters, $n$ is the amount of substance of gas, $R$ is the universal gas constant, and $T$ is temperature in kelvin.

The information given to us is the density of $\pu{0.977 g/L}$, a pressure of $\pu{710.0 torr}$, and temperature of $\pu{100 ^\circ C}$.

Knowing that amount of substance, $$n = \frac{M}{m},$$ where $M$ is molecular mass in $\pu{g/mol}$, and $m$ is the mass of gas in gram, we can see the unit gram cancels out and we are left with the unit mole.

However, we are given a density, $$d = \frac{m}{V} = \pu{0.977 g/L}.$$

Plugging into $\eqref{id-gas}$ the molecular mass and mass for $n$ we get:

$$ pV = \frac{M}{m}RT .$$

Isolating $m$ in density, $m = d\cdot V$, and isolating $m$ in $n = m/M\Leftrightarrow m = n\cdot M$, we can combine those two equations for $m$:

$$n\cdot M = d\cdot V$$

Isolate $n$, and plug back into our ideal gas equation $\eqref{id-gas}$:
\begin{align}
 n &= \frac{dV}{M}&
 \implies&&
 pV &= \frac{dV}{M}RT
\end{align}

Volume cancels out, and we isolate for $M$:
$$ M = d \cdot \frac{RT}{p}. $$

Plug in our known values, knowing that $\pu{1 torr} = \pu{1 mm Hg}$ and $\pu{760 mm Hg} = \pu{1 atm}$:

$$ M = \pu{0.977 g//L} \cdot \frac{
  \pu{0.082057 L * atm // mol * K}\cdot\pu{373.15K}}{
  \pu{710 torr}\cdot \pu{mmHg//torr}\cdot\pu{{1atm}//760mmHg}}
     = \pu{32.0 g//mol}$$

And our answer has three significant figures.


The molecule contains conjugated system of double bonds, being essentially a substituted styrene. The phenyl cannot therefore freely rotate and its steric requirements are difficult to estimate just by hand-waving arguments.

Correct answer needs some quantitative argument, either e.g. heats of formation or calculation.

From the second, the energies come out using MMFF94 (using Avogadro):

Z: 177.6 kJ/mol
E: 175.6 kJ/mol

Also, both the isomers rotated the styrene substantially from planarity. The difference in results is quite small and can easily change when using other method. According to the numbers I'd agree with your answer, Z is less stable.

I would not consider this a valid question in test, the answer is not straightforward.


Base catalyzed rearrangement of an alkyl substituted cyclopent-2-enone

I believe that the mechanism for rearrangement lies in a retro-aldol condensation (technically a Michael attack as a first step) followed by a new aldol condensation; both catalysed by the excess base which can both catalyse the mechanism and be the Michael attacker in step 1. My version of the mechanism is shown here:

One should be able to support the mechanism by labelling either of the ketone carbons with $\ce{^{13}C}$. (Oxygen labelling would not work since in the hydroxide conditions the oxygen will exchange rapidly via a hydrate intermediate.)

As I forgot to say and as Orthocresol correctly noted: Every step is to be seen as being in equilibrium and the overall equilibrium is shifted towards the thermodynamically most stable product: the tetrasubstituted double bond.


How can HCl be liquid if it has a boiling point at -85°C?

All halogen compounds with hydrogen ($\ce{HX}$) are gases at STP except for $\ce{HF}$, which is a liquid (b.p.: $\mathrm{19~^\circ C}$). When dissolved in water, $\ce{HX}$ dissociates to ions ($\ce{H+}$ & $\ce{X-}$), which are readily hydrated by water molecules as $\ce{H+ (aq)}$ and $\ce{X- (aq)}$, hence not exists as a gas any more.

To answer your side question, a lot of gases have this feature. For instance, sulfur trioxide ($\ce{SO3}$) is a gas but gives sulfuric acid ($\ce{H2SO4}$) when it come to contact with water. Similarly, nitric acid ($\ce{HNO3}$) can be considered as an aqueous solution of nitrogen dioxide ($\ce{NO2}$). In reality, during an industrial process, $\ce{NO2}$ react with $\ce{H2O}$ in air to give $\ce{HNO3}$.
$$\ce{4NO2(g) + O2(g) + 2H2O(l) -> 4HNO3 (aq)}$$
Exception to this trend is carbonic acid ($\ce{H2CO3}$), which decomposes under ambient conditions:
$$\ce{H2CO3(aq) -> CO2(g) + H2O(l)}$$


but also some of the chemical constituents of custard powder

Don't blame it on the custard. As long as there's

dust of a combustible material, preferably with a small particle size
sufficient oxygen in the athmosphere
an igniter, such as a spark or an open flame

there's a good chance for a devastating dust explosion.

Your combustible material can be something seemingly unspectecular, coal dust, saw dust, flour, or custard powder.

UPDATE

The incident in the BBC game most likely took place at a factory in Banbury, UK in 1981. Their traditional custard powder, known as Bird's Custard, is egg-free and mostly consists of cornflour.

  how you might use it to tell that the reaction was an explosion in a custard factory

In risk assessment for dust explosions of organic materials, product analysis rarely plays a role; the "equation" isn't interesting and it is usually assumed that all carbon ends up as $\ce{CO2}$.

In order to establish workplace safety, it is more important to determine, how "strong" a possible dust explosion might get.

A measure for the "strongness" of an explosion is the rate by which the pressure (rapidly) increases. This is expressed in the $\mathrm{K_{St}}$ value (in $\mathrm{bar\cdot m \cdot s^{-1}}$), and experimentally determined in standardized tests.

Based on the $\mathrm{K_{St}}$ value, materials of certain properties (particle size, rest humidity) are assigned to the explosion classes (ST 0 to ST 3), and suitable workplace safety measures are taken.


How does equalizing voltage for lead acid battery depend on the temperature?

Generally, the electrode reactions are both based on lead in different oxidation states. At the negative electrode, $\ce{Pb}$ is oxidized to $\ce{Pb^2+}$ during discharge.
$$\ce{Pb <=> Pb^2+ + 2 e-}$$
At the positive electrode, $\ce{Pb^4+}$ is reduced to $\ce{Pb^2+}$.
$$\ce{Pb^4+ + 2 e+ <=> Pb^2+}$$
For a classical lead–acid battery, the overall cell reaction is approximately
$$\ce{Pb + PbO2 + 2H+ + 2 HSO4- <=> 2 PbSO4 + 2 H2O}$$
As long as $\ce{Pb}$, $\ce{PbO2}$, and $\ce{PbSO4}$ are available at the electrodes, the equilibrium cell voltage depends only on the acid concentration (that's why equilibtrium cell voltage can be estimated based on measured acid density).
The dependence of the equilibrium voltage on concentration is given by the Nernst equation:
$$U_\text{cell}=\left(1.931+0.0592\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$
This equation applies to one cell; a battery, however, has six cells in a row. Thus the total voltage is
$$U_\text{battery}=6\left(1.931+\frac{RT}{ZF}\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$
where
$R$ is the gas constant,
$T$ is temperature,
$z$ is the number of electrons transferred in the cell reaction, and
$F$ is the Faraday constant.

We are only interested in the dependence of voltage $U$ on temperature $T$. All the other parameters may be taken as constants. Thus, our equation can be simplified to
$$U=\left(11.586+T\cdot k\right)\ \mathrm V$$
where $k$ is a constant.

The exact parameter values for a classical lead–acid battery might differ a bit from a modern battery with lead–carbon electrodes; however, we can calibrate our model using the data given by the manufacturer, i.e. $U=14.1\ \mathrm V$ at $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$
$$14.1\ \mathrm V=\left(11.586+298.15\ \mathrm K\cdot k\right)\ \mathrm V$$
and we find
$$k=0.008432\ \mathrm{K^{-1}}$$
Therefore, our complete equation is
$$U=\left(11.586+T\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V$$
For a new temperature of $T=0\ \mathrm{^\circ C}=273.15\ \mathrm K$, we get an estimate of
$$\begin{align}U&=\left(11.586+273.15\ \mathrm K\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V\\
&\approx13.9\ \mathrm V\end{align}$$


How to systematically find the reducible representation of a molecule? (eg: XeF4)

A way to do it is to put 3 axes ($x$, $y$ and $z$) on each atom (to represent the 3N degrees of freedom) and determine the characters for each axis.

After that, if you need the representations for the 3N-6 (3N-5 for a linear molecule) vibrational modes, you need to remove the representations for the 3 translational modes (what they call $\Gamma_{xyz}$), as well as the representations for the 3 (2 for a linear molecule) rotational modes. Those are usually stated in the character table for the space group, but you can also determine them by checking the characters for a vector.

Example with $\ce{H2O}$

Character table for point group $C_2v$

C2v | E  | C2 (z) | v(xz) | v(yz) |  lin  |    quad
-------------------------------------------------------
A1  | +1 |     +1 |    +1 |    +1 |   z   | x2, y2, z2
A2  | +1 |     +1 |    -1 |    -1 |   Rz  |     xy
B1  | +1 |     -1 |    +1 |    -1 | x, Ry |     xz
B2  | +1 |     -1 |    -1 |    +1 | y, Rx |     yz

Character table for all 3N = 9 axes

C2v | E  | C2 (z) | v(xz) | v(yz)
----------------------------------
Ox  | +1 |     -1 |    +1 |    -1
Oy  | +1 |     -1 |    -1 |    +1
Oz  | +1 |     +1 |    +1 |    +1
H1x | +1 |      0 |     0 |    -1
H1y | +1 |      0 |     0 |    +1
H1z | +1 |      0 |     0 |    +1
H2x | +1 |      0 |     0 |    -1
H2y | +1 |      0 |     0 |    +1
H2z | +1 |      0 |     0 |    +1
----------------------------------
Γtot| +9 |     -1 |    +1 |    +3

Reduction of $\Gamma_{tot}$ to irreducible representations

Using the reduction formula $n_i = \frac{1}{h}\sum_R \chi_{\Gamma_{tot}}(R) \chi_i(R)$ for each irreducible representation $i$ of the point group, we get :

A1 : $\frac{1}{4} \times (9 + (-1) + 1 + 3) = 3$
A2 : $\frac{1}{4} \times (9 + (-1) + (-1) + (-3)) = 1$
B1 : $\frac{1}{4} \times (9 + 1 + 1 + (-3)) = 2$
B2 : $\frac{1}{4} \times (9 + 1 + (-1) + 3) = 3$

Therefore, $\Gamma_{tot}$ = 3 A1 + A2 + 2 B1 + 3 B2.

We can check that there are the 3N = 9 representations that are associated with the 3N degrees of freedom of the molecule.

For degenerate representations ($\chi(E) > 1$), the representations correspond to $\chi(E)$ modes.

Isolation of the vibrational representation

Determine $\Gamma_{trans}$ and $\Gamma_{rot}$ (from the character table or by hand).

$\Gamma_{trans}$ corresponds to the linear terms in the table, $\Gamma_{rot}$ corresponds to the rotational linear terms in the table.

For $C_2v$ :

$\Gamma_{trans}$ = A1 + B1 + B2
$\Gamma_{rot}$ = A2 + B1 + B2

You can check that there are 3 modes for $\Gamma_{trans}$ and 3 (2 for a linear molecule) modes for $\Gamma_{rot}$

For degenerate representations, you will see things like $x + y$ in the character table. This means that this representation corresponds to both x and y and should be included only once

Finally, $\Gamma_{vib}$ = $\Gamma_{tot} - \Gamma_{trans} - \Gamma_{rot}$ = 2 A1 + B2.

Activity of the vibrational modes in IR and Raman

IR activity is determined by the dipole moment and a mode will be active if the irreducible representation contains a linear part (x, y or z).

Raman activity is determined by the polarisability and a mode will be active if the irreducible representation contains a quadratic part ($x^2$, $xy$, ...).

For $C_2v$, all the representations correspond to active modes in Raman and all except A2 correspond to active modes in IR.

Therefore, the 3 vibrational modes of $\ce{H2O}$ (2 A1 and B2) are active in both Raman and IR.

Two modes have the same irreducible representation: A1. That does not mean that they have the same wavenumber. For $\ce{H2O}$, there are 3
vibrational modes with distinct wavenumbers.

For degenerate representations, there will be $\chi(E)$ modes with the same wavenumber.

In groups that have an inversion center (like $D_4h$), linear and quadratic terms are mutually exclusive. Therefore, an irreducible representation will always correspond to a mode active only in IR or Raman, not both.


What is the name of this molecule? (C₈O₃H₁₃N)

If I managed to see the structure correctly it should be Methyl 1-acetylprolinate


Deducing orbital degeneracy in geometries apart from octahedral or tetrahedral

The easiest way to do it is to look at what other people before you have done.

If you look at the $D_\mathrm{3h}$ character table, you see that $(x^2 - y^2, xy)$ transform together as $E'$.

I don't think there is a way of looking at diagrams of the orbitals and concluding that they are necessarily degenerate. As you said, there isn't "enough information to assume that the two parts add up to one".

It isn't really a limitation of crystal field theory, either. Crystal field theory isn't about looking at how much orbitals overlap with each other and saying "oh, I think those look about the same, they're probably degenerate". If that was the case then in an octahedral complex you would not be able to tell why $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$ are degenerate, since the former only really overlaps with two ligands and the latter overlaps with four.

The degeneracy is predicted by symmetry, the application of which is integral to, and implicit in, crystal field theory. There is no CFT without symmetry considerations.


Transferring suspended substance from isopropyl alcohol to water?

Slowly drive off the alcohol in a vacuum or by gentle heating. Replace the lost alcohol with distilled water. Experiment with small volumes first to ensure that heat does not degrade the ink.


Elemental Potassium as both an oxidizer and a reducer

Compounds having potassium anions are referred to as potassides and are rare, usually involving K- within a cyrpt such as:

Crystalline Salts of Na- and K- (Alkalides) that Are Stable at Room Temperature J. Am. Chem. Soc. vol. 121, pages 10666-10667 describes preparation of K+(aza222)K- , where "aza222" is the compound in the above figure, but with the 6 H's replaced with methyl groups.  The compound is formed by dissolving potassium metal in methyl amine solution of the methylated aza222.


Why is propenoic acid more acidic than p-methoxybenzoic acid?

This is propenoic acid, $\mathrm{pK_a}=$4.25

This is p-methoxybenzoic acid, $\mathrm{pK_a}=$4.47

An electron withdrawing group increases the acidity of a carboxylic acid. It disperses negative charge by inductive/resonance effect and stabilizes the carboxylate ion. Thus, propenoic acid (pKa 4.25) is more acidic than p-methoxybenzoic acid (pKa 4.47).

However the molecular structures are totally different. It seems to me to be comparing apples and oranges! To me such hand waving comparisons between molecules are only meaningful if the two molecules are similar.

  The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.20.$ Why does the $\ce{-OMe}$ group make such a big effect on $\mathrm{p}K_\mathrm{a}?$

This is a more interesting question.

\begin{array}{|c|c|} \hline \hline
\text{acid} & \text{pka}  \\ \hline
\text{2-Methoxybenzoic acid} & 4.09   \\ \hline
\text{3-Methoxybenzoic acid} & 4.10  \\ \hline
\text{Benzoic acid} & 4.19  \\ \hline
\text{4-Methoxybenzoic acid} & 4.50  \\ \hline
\end{array}

So the phenyl group has about the same effect on the acidity of a carboxylic acid as observed with an ethylene group. However the substitution of other function groups, and their position on the ring can increase or decrease the pKa of the acid.

Data sources:
2-Methoxybenzoic acid

3-Methoxybenzoic acid

Benzoic acid

4-Methoxybenzoic acid


Mechanism of reaction between triphenylmethanol and acetyl chloride

I find it unlikely that the substitution proceeds via the hydroxyl group directly leaving. If there is any evidence that this reaction proceeds via direct $\mathrm{S_N1}$ of $\ce{OH-}$ by $\ce{Cl-}$, I would be very interested to see it. When an alcohol and an acyl chloride are mixed, I think it is much more probable that they will initially form the ester via nucleophilic acyl substitution:

$\hspace{27 mm}$

Presumably, this ester is rather unstable because of steric repulsions between the three phenyl groups and the ester group. So now, the $\ce{-OAc}$ group ($\ce{Ac} = \ce{COCH3}$) can leave as a relatively stable carboxylate ion to give the resonance-stabilised carbocation:

$\hspace{35 mm}$

The carbocation is then intercepted by chloride ions that were formed earlier to give the product you want. So, the second part of the reaction is a $\mathrm{S_N1}$ substitution of $\ce{OAc-}$ by $\ce{Cl-}$.

$\ce{OAc-}$ ($\mathrm{p}K_\mathrm{aH} = 4.76$) is a much better leaving group than $\ce{OH-}$ ($\mathrm{p}K_\mathrm{aH} = 15.7$).

There is another pathway possible, since $\ce{HCl}$ is formed from the initial reaction of the alcohol with the acyl chloride. If $\ce{HCl}$ is present in appreciable quantities, it can protonate the original alcohol and you could perhaps have a direct $\mathrm{S_N1}$ of $\ce{H2O}$ by $\ce{Cl-}$.

$\hspace{10 mm}$


How can I make an acidic pen to burn paper on writing on it?

You can pour anything into your pen. Is it a good idea? Most likely not. Is it dangerous? You bet! Is it a good idea to play with? NO! Whether it is legal or not to create such an object, you'll have to check with legislation in your country.

If you want paper to ignite on touch with this pen, you would need some truly strong acid (and probably undiluted). I'm certain that if this substance has the capability to burn paper, it would also corrode away the feed and nib of the pen. In the worst case, it will even eat through the ink container and leaking everywhere. In any way, this is no viable option to cut paper in a controlled fashion.

Is there a safer method? Well, there would be laser-assisted cutting of paper, but the good ol' knife or saw would still be cheaper.


What is Meant by "Eluting with 1-2% EtOAc/pentane" in Terms of Flash Chromatography?

Based on OP's question, I tentatively assumed OP is a novice to the isolation of organic compounds using column chromatography. Thus I'd explain this question with as basis as possible.

This paper OP sighted is for the synthesis of 2$H$-chromenes (2$H$-benzopyrans). They can be non-polar to polar compounds based on the substitution on benzene nucleus. The paper describes a series of relatively non-polar chromenes. When you do organic reaction, you monitor the progress of reaction with TLC analysis. During that analysis, you can determine that what solvent should be the most suitable for the particular compound elution. That's how the authours have determined several solvents such as 1% ethyl ether/hexanes and 2% ethyl acetate/pentanes, etc. for different different chromenes.

Some of these chromenes have been isolated using just one solvent. For example, evidently, they have mentioned isolation of 7-Methoxy-4-methyl-2-propyl-2$H$-chromene, using flash chromatography on silica gel, eluting with 2% diethyl ether/hexanes solvent system (this means solvent made by 2 parts of diethyl ether and 98 parts of hexanes, by volume). Note that it would also mean that they might have used certain volumes of following systems: (1) pure hexanes; (2) 1% diethyl ether/hexanes; (3) 2% diethyl ether/hexanes; etc. However, expected compound has come out with only fractions eluted with 2% diethyl ether/hexanes (remember, you supposed to analyze all the fraction by TLC), which may have combined and further analyzed by spectroscopy to confirm the structure.

As in another data set, they have mentioned isolation of 5,7-Dimethoxy-2,2-dimethyl-2$H$-chromene, using flash silica gel column, eluting with 1-2% EtOAc/pentane solvent system.It would definitely mean that they should have used certain volumes of few solvent systems: (1) pure pentane; (2) 1% EtOAc/pentane; (3) 2% EtOAc/pentane; (4) 3% EtOAc/pentane; etc. However, their expected compound has come out with fractions eluted with 1% and 2% EtOAc/pentane (concluded by the fractions analysis by TLC).

Later Note: As commented by Waylander, these type of separation is called gradient elution. By the scale of reaction, total elution must not go over $\pu{100 mL}$ of solvent for each compound.


Free database of polymers with Hansen parameters

I'm afraid that due to copyright claims the only database available is HSPiP, plus there is the most recent handbook Hansen, C. M. Hansen solubility parameters: a user’s handbook, 2nd ed.; CRC Press: Boca Raton, 2007. The book costs ten times less than a database and contains tables for solvents on pp. 345 -- 485, and for polymers on pp. 487 -- 500.


Why use glacial acetic acid in bromination of anisole?

[...] bromine is not reactive enough to brominate an aromatic without FeBr3

That's true for benzene, but the methoxy-group in anisole isn't just o/p-directing.
It is activating as well. Actually, it's that much activating that anisole will react with bromine even without the presence of a Lewis acid.

So, why glacial acetic acid? It's a polar solvent and thus nice to stabilize the Wheland intermediates. Moreover, it's easily removed in aqueous workup.

Edit 1
Does the acetic acid react with the (mostly undissociated) acetic acid? At least a polarization of bromine ($\ce{Br^{\delta+}-Br^{\delta-}...HOAc}$) is conceivable and I remember that $\ce{BrOAc}$ has been postulated as the electrophilic species.

Edit 2
I remembered some brominations on dimethoxy-substituted hydroxymethylnaphthalines I did back in the days. The reaction was performed at 60 °C in $\ce{CHCl3}$ in the presence of about 3 equivalents of anhydrous sodium acetate. Excess of bromine was reduced by addition of aqueous sodium sulfite. The yield was in the 85% range and I don't find any notes on dibrominated products.

Edit3

  And with excess Bromine you would (theoretically) react all of the anisole?

Pretty much, with 4-bromoanisole as the main product.


Simple experiment to do at home that involves acids and bases?

Measure the pH+ of various foods, and make a spreadsheet listing them.

You might also measure pH+ change with time in a food, e.g. apple juice or bread dough, with a few mg of yeast sprinkled in.


Can an organic compound have an ionic bond?

There's a large number of salts with organic anions (e.g. potassium hydrogen tartrate) or organic cations (e.g. tetrabutylammonium hexafluorophosphate)

There are even salts, in which both the anion and the cation are organic compounds with covalent bonds. This is often used in pharmaceuticals. Here the active component appears in protonated form and is served as a salt with the conjugated bases of carboxyclic acids (e.g. citrates, succinates, etc.)


How to identify the concentration of nitric acid?

It looks like you have about 40 mL of the solution. You could measure the density by using a small volumetric flask (10 or 25 mL), or even a graduated cylinder (which you could calibrate with water). Add your solution to the mark, weigh reasonably accurately, and you will have your answer (which I predict will lie between 70 and 90%).
At 20C, the density of 70% acid is 1.41; the density of 90% acid is 1.48. The difference is large enough to determine easily, and if there is any deterioration, you should be able to determine the exact concentration fairly easily from density tables, e.g.: https://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry


Is the speed of sound the fastest rate that a chemical reaction can progress through a substance?

Certainly, chemical reactions can propagate far faster than the speed of sound. If the reaction emits electromagnetic radiation, e.g. light, and is itself initiated by light, then the speed of propagation is the speed of light in that medium. See this article from Sandia National Labs, or from the Royal Society.


Physical Explanation for variation of Gibbs Free Energy with Temperature and Pressure

Probably the easiest way to picture it is that, since the Gibbs function is indeed a measure of the potential of the system to transfer energy in certain ways, these transformations affect the energy that can be transferred.

For a certain system at constant pressure and entropy (which therefore cannot absorb thermal energy from its surroundings), an increase in temperature is necessarily linked to an internal transformation - an internal work - that reduces the potential energy available to exert work on its surroundings; that increase in temperature is "using up" part of the potential reversible work as measured by the Gibbs function.

On the other hand, a system at constant volume and temperature which increases its pressure (and which, given these constrains, is therefore receiving some form of energy input from the outside; note that there isn't a pressure-volume tradeoff) will increase the potential reversible work it can exert on its surroundings, as measured by the Gibbs function.

It is important to note the implications of the constant conditions that each of those two cases set, because these constant conditions are telling you which forms of energy transfer are permitted or prohibited in each case.


Why are copper salts less soluble than zinc salts?

It is because the difference in electronegativity.
The higher the difference in electronegativity between the cation and the anion, the stronger the ionic bond.
Since we are comparing species with the same anions and different cations, we only have to look at the values of the latter ones.

cation    electronegativity   configuration
  Cu             1.9         [Ar] 3d^10 4s^1
  Zn             1.65        [Ar] 3d^10 4s^2

$\ce{Cu}$ is more electronegative so the bonds have less ionic character which makes for stronger bonds.

The same happens in the same period between $\ce{K}$ and $\ce{Ca}$, but here the situation with regard to the electronic configuration is reverse.
Potassium salts are much more soluble than calcium salts, and again it is explained by the electronegativity:

cation    electronegativity   configuration
  K              0.82           [Ar] 4s^1
  Ca             1.0            [Ar] 4s^2


After quick digging I found, that glass cleaner solutions usually contain alcohols and surfactants. This additions reduce surface tension and allows the solution to interact with glass more easily, distributing the solution over the entire surface.

The white stripes visible after water dried are usually solid salts. If the water does not contain any salts, but only compounds, that can quickly evaporate, it will not leave stripes. But it is not addition, but absence of additions that do the trick. For example, most bleacher/toilet cleaners contain hypoclorites, salts of unstalble $\ce{HOCl}$ acid and leave $\ce{NaCl}$ after evaporation

When tap water is allowed to dry, it usually leaves white stripes, but they are not visible on most surfaces

The only ideas I can get why the salts don't deposit if using a glass cleaner is that salts are removed (and if you have a cleaner not meant to be diluted, the water in it may be deionized) or to trap salt ions into organic ligands so resulted salts were not crystalline (possible, but I do not like the idea).


How to find the potential of a mixed Iron/ Cer solution?

First it helps to have an idea of what will happen. Since 1.72 V is a
lot more than 0.771 V (a difference of 1V in chemistry goes a
long way), the numbers are telling us that $\ce{Ce^{4+}}$ is a much
better oxidizer than $\ce{Fe^{3+}}$; in other words, in the presence of
enough $\ce{Ce^{4+}}$, almost all $\ce{Fe^{2+}}$ will be oxidized to $\ce{Fe^{3+}}$.
Conversely, in the presence of enough $\ce{Fe^{2+}}$, almost all $\ce{Ce^{4+}}$
will be reduced to $\ce{Ce^{3+}}$.

In equilibrium, therefore, either you'll have almost no $\ce{Fe^{2+}}$
left, or almost no $\ce{Ce^{4+}}$ left: one of the reagents will have been almost entirely consumed. To tell which one, look
at the number of moles of each ion that went into the reaction at the
moment of mixing. Keep in mind that each unit of $\ce{Fe^{2+}}$ reacts
with one unit of $\ce{Ce^{4+}}$.

So now you've figured out which is the limiting reagent (the one that gets
consumed almost entirely). Let's suppose for the sake of argument it's
the $\ce{Fe^{2+}}$. Because you know how many
moles of $\ce{Fe^{2+}}$ went in, and because all of it reacted, you know
how many moles of $\ce{Ce^{4+}}$ got turned into $\ce{Ce^{3+}}$, and you can
find the concentration of $\ce{Ce^{4+}}$ and $\ce{Ce^{3+}}$ in the
equilibrium mix. And you also know the concentration of $\ce{Fe^{3+}}$,
because essentially all the $\ce{Fe^{2+}}$ became $\ce{Fe^{3+}}$. That's
all under the assumption that $\ce{Fe^{2+}}$ is the limiting reagent.

If the limiting reagent is $\ce{Ce^{4+}}$ instead, you can use the same
logic to find the concentration of $\ce{Fe^{2+}}$, $\ce{Fe^{3+}}$ and
$\ce{Ce^{3+}}$ in the equilibrium mix.

So now you have 3 concentrations out of 4.  Finding the 4th requires
knowing the equilibrium constant, which you can calculate from the
difference in potentials. (Write the combined full reaction based on the half-reactions.)

You'll find that the 4th concentration is
very small, justifying the claim that one of the reagents
was almost entirely consumed.

If you're with me so far, please update the question with these
results, and if you still need help with finding the potential at that
point, we can take it from there...


The connection of the "R" in the ideal gas law and osmotic pressure

It's not a coincidence at all!  If you do an online search for "derivation of osmotic pressure", you'll see how $R$ enters into the derivation.

Indeed, that's one of the beauties of the van 't Hoff equation for osmotic pressure – it reveals that (under the modest simplifying assumptions of the van 't Hoff equation derivation) the osmotic pressure created by a dilute solute is equal to the pressure that solute would exert if it instead acted as an ideal gas!:

$$\text{van 't Hoff equation:  } \Pi_i = \frac{n_i}{V} R T = C_i R T$$

$$\text{ideal gas equation:   } p_i = \frac{n_i}{V} R T = C_i R T$$

Here the "$i$" refers to species $i$.


Understanding strong acid titrated with weak base graph

You are missing an important point. Conductometric titration is not just plotting the conductance after adding the titrant. One cannot plot conductance as it is, because you are significantly diluting the solution.  You have to apply a correction to take dilution into account. Once you apply a correction, the curve will remain flat because further addition of ammonia will not enhance conductivity.
See for example,

Conductometric titrations
Throughout a titration the volume of the solution is always increasing, unless the
conductance is corrected for this effect, non linear titration curves result. The
correction can be accomplished by multiplying the observed conductance either
by total volume (V+V') or by the factor (V+ V')/V, where V is the initial volume
of solution and V' is the total volume of the reagent added. The correction
presupposes that the conductivity is a linear function of dilution, this is true only
to a first approximation.


Why does cyclopropane react with bromine?

The following ring opening reaction will occour:

You are quite right about the angle strain. Because orbital interactions are not optimal in this geometry. Consider p-orbitals, then a natural bond angle would be $\theta\in [90^\circ; 180^\circ]$. A mixing of s- and p-type orbitals allows a wide range of angles $\theta\in (90^\circ,\dots, 180^\circ)$.

In cyclopropane $\ce{C3H6}$ - which you can also describe as trimethylene $\ce{(CH2)3}$ -  bonds have to be bent to overlap at all. A possible way of describing the bonding situation is regarding each $\ce{CH2}$ entity as $\mathrm{sp^2}$ hybridised. Two of these orbitals are used for $\ce{C-H}$ bonds (not shown) and one forms an inner two-electron-three-centre σ bond (left). This leaves p-orbitals to form some kind of degenerate π-like orbitals (middle, right).

This very general approach can be derived from a Walsh diagram. Schwarz et.al. {@academia.edu} and Hoffmann {@roaldhoffmann.com} described bonding quite similar and it is in quite good agreement with a calculation (BP86/cc-PVTZ, $D_\mathrm{3h}$) I have done. From this I have prepared a chart of all occupied molecular orbitals formed from valence orbitals and the LUMO. Here is a preview. Each orbital is viewed from three different angles:

Especially the symmetrical orbital 8 resembles very well the schematics. A quite rigorous approach for this theory can also be found here.

It is noteworthy - as mentioned by ron - that there is no notable increase in electron density in the centre of the ring. This may be due to the fact that there are much more orbitals having nodes in the centre than there are without.

Now bromine is known to be easily polarised $\ce{{}^{\delta+}Br-Br^{\delta-}}$ and may intercept at any point of the ring causing a bond break and relaxation to a less strained structure. It will most likely attack at the the $\pi$ type orbitals since bromine is an electrophile. The mechanism is analogous to the addition of bromine to ethene, which is nicely described at chemguide.co.uk. The essential part is the attack of the bromine at the HOMO(s).

The ring opening reaction can be reversed by adding sodium.

However, when there are bromine radicals present (UV light) then substitution will occur:
\begin{aligned}\ce{
Br2 &->[\ce{h\nu}] 2Br.\\
&+(CH2)3 -> (CH2)2(CHBr) + HBr
}\end{aligned}


in acidic media, the nucleophile tends to attack the more substituted side of the epoxide due to electronic reasons

This is an $\ce{S_{N}1}$ like reaction. First we would protonate the epoxide oxygen.  If we examine the case of an unsymmetrical epoxide and draw the resonance structures for this situation (see below) we would expect the secondary carbocation structure to count more towards the actual structure than the primary carbocation structure.  Therefore, we would expect a nucleophile to react preferentially at the tertiary epoxide carbon, quite selectively.

  in basic media, the nucleophile attacks the less substituted side of
  the epoxide due to steric reasons.

This would be a traditional $\ce{S_{N}2}$ reaction.  Attack at the least substituted (sterically most accessible) carbon would be expected.

  What if the hydride anion attacks an epoxide? I'd imagine the hydride
  anion to be quite small, and able to attack both the more substituted
  and less substituted sides of an asymmetric epoxide. What are the
  experimental results?

The hydride anion is not free, but complexed, so it has reasonable steric bulk - enough for stereoselective reactions to occur.   Further, the reaction between LAH and epoxides is quite facile, no doubt accelerated by the release of steric strain in the 3-membered epoxide ring system. A traditional $\ce{S_{N}2}$ reaction takes place.  In the figure below, the alcohol shown is formed in ca. 90% yield.  Because of the "backside" attack nature of the $\ce{S_{N}2}$ reaction axial attack is strongly preferred in cyclic systems.


How does complex ion differ from ligand complex?

To quickly pre-answer the question at the end of the body: Yes, the oxygen atoms around the chromium(VI) cause d-orbital splitting. However, chromium(VI) is a $\mathrm{d^0}$ system and thus the colour is not caused by d-electron transitions. Rather, as I highlighted in my permanganate answer, the cause of the colour is a ligand to metal charge transfer transition. For details, see there.

The question in the title basically refers to the difference between a complex such as $\ce{[CoCl4]-}$ and an ion such as $\ce{SO4^2-}$ or $\ce{CrO4^2-}$. These can be considered two extremes of a transition much like the extremes unpolar covalent bond and ionic bond when used to describe bonding.

All coordination complexes are nothing more than dative bonds, which in turn are nothing more than a special subset of weak covalent bonds. From a qualitative bonding point of view, there is really no difference between either compounds: For both you get orbital mixing, a bonding and an antibonding set of orbitals and an overall stabilisation of the system. And if you totally ignore energy differences and the like, you will find that each of the three compounds written up there have more or less the same MO scheme.

The difference from a theoretic point of view is the bond strength and the energy difference between the central atom’s and the ligands’ orbitals. For a complex ion, the orbital energies of the unbonded atoms are very similar to each other resulting in a large energy difference upon forming a bond and therefore a large difference between bonding and antibonding orbitals. For typical ligand (or coordination) complexes, the ligands have much lower orbital energies than the central metal and thus the energy gained in the bonding process is comparatively small. On this scale between $\ce{[CoCl4]-}$ and $\ce{SO4^2-}$, chromate(VI) would be somewhere just to the sulphate side of the axis’ centre.

What does this difference mean from a practical point of view? Well, one typical reaction of coordination complexes is the ligand exchange reaction: Add a better ligand into the solution and more often than not it will exchange with a worse ligand. Sometimes this results in a different geometry. $\ce{[CoCl4]-}$ can be synthesised by preparing an aquaeous cobalt(III) solution which contains $\ce{[Co(H2O)6]^3+}$ ions and adding excess chloride ions — the complex will form spontaneously. The reverse reaction is possible by adding silver nitrate which will precipitate the chloride ions, reforming the hexaaqua complex. Ligand exchange is simple.

The sulphate ion, as the extreme opposite example, has such strong $\ce{S-O}$ bonds that they will not ‘simply dissociate by offering a better ligand.’ (In fact, I don’t know if a better ligand even exists.) It is impossible to displace the oxygens to attach other ligands — and most notably, there is no good oxophilic compound known that could simply remove (i.e. precipitate) those oxygens. Note, however, that sulphuryl chloride $\ce{SO2Cl2}$ will readily react with water to form sulphuric acid (diprotonated sulphate) and hydrogen chloride, so one type of ‘ligand exchange’ is still possible.

Chromate(VI), as written earlier, is closer to sulphate than to the cobaltate complex. It will — much like permanganate — not take part in displacement reactions, the metal–oxygen bonds are pretty stable. Again note that an analogous chloride complex exists with chlorochromate $\ce{CrClO3-}$.

Note that not always will the ligand exchange I highlighted above occur easily. Ammonia should form a more stable complex with iron(III) than water. However, when adding ammonia to an iron(III) solution, iron hydroxides will precipitate rather than an ammin complex being formed. So the practical point of view is a really rough approximation that only works sometimes.


Here is the way how I got the product in my question:


Why are not all metals included in the reactivity series

It's a good question: "reactivity" may vary depending on concentration, temperature etc. but should generally closely approximate the electromotive (galvanic) series.

That said, as @Karl mentions, reactivity series I've seen usually are incomplete, often leaving out transition metals with multiple valences such as Mo and W. To fill in the missing species, look at a good table of the electromotive series.

The omission of so many metals is downright shocking, revolting. One would think a current table would have included more metals; these reactivity series seem like works in transition.


What compounds are created by mixing Styrofoam and paint thinner?

Polystyrene will be dissolved in acetone. The long polymer molecules will dissociate from each other and float around in the acetone solvent; but the polymer chains will likely not be broken down into their small styrene monomer units. If you mix acetone and polystyrene ("styrofoam"), you will simply end up with polysterene in acetone. Not significant chemical reaction is expected to take place, only the physical process of dissolution/solvation/dissolving.

I would assume such a solution is toxic and keep it far away from animals, eyes and mouths.


Does pure HCl (without water) is an acid?

If we look at the classic case of ammonia and hydrogen chloride in the gas phase, the answer is yes -- but it's more complicated than one might think.  The salt does form, but it requires either a proton-transfer medium such as water vapor or the formation of particles out of the gas phase to actually get the ions.
The acid-base reaction is formally,
$$\ce{HCl + NH3->(NH4)^+Cl^-}$$
but according to an an ab initio calculation by Cazar et al. [1] a single pair of molecules do not actually form the indicated ions.  They form instead a hydrogen-bonded species in which the would-be transferred proton remains primarily bonded to the chlorine atom.  Only with the introduction of water molecules do we find the formation of ammonium and chloride ions in the gas phase.
An empirical study by Countess et al. [2] does show the formation of ammonium chloride salt as condensed particles in anhydrous nitrogen.  They give a rate constant in terms of molecules per unit volume for formation of the monomeric gas-phase species, which is polar and attracts other monomers or reactant molecules to nucleate the particles.  It is during this polymerization phase where the ions are definitively formed; Chaban et al. [3] compute that even two ammonium chloride entities are sufficient to form an ionic structure.
References

Robert A. Cazar, Alan J. Jamka, and Fu-Ming Tao,
"Ab Initio Investigation of Proton Transfer in Ammonia−Hydrogen Chloride and the Effect of Water Molecules in the Gas Phase", J. Phys. Chem. A., 1998, 102, 26, 5117–5123.

Richard J. Countess and Julian Heicklen, "Kinetics of particle growth. II. Kinetics of the reaction of ammonia with hydrogen chloride and the growth of pariculate ammonium chloride", J. Phys. Chem. ,1973, 77, 4, 444–447.

Galina M. Chaban, R. Benny Gerber, and Kenneth C. Janda, "Transition from Hydrogen Bonding to Ionization in (HCl)n(NH3)n and (HCl)n(H2O)n Clusters:  Consequences for Anharmonic Vibrational Spectroscopy", J. Phys. Chem. A 2001, 105, 36, 8323–8332.


Oxidation of aluminum and manufacturing of electrolytic capacitors

You are performing an electrolysis.
This means that on the (+) electrode de reaction is Al $\rightarrow$ Al$^{3+}$ and on the (-) electrode the reverse occurs.

Distilled water is required to avoid the presence of chloride ions which will form chlorine at the (+) electrode and attack your aluminum foil.
Sodium bicarbonate is used to raise the pH so that the formed aluminum ions precipitate as the oxide instead of dissolving as Al$^{3+}$ ions.
The oxidation is Al $\rightarrow$ Al$^{3+}$. Then, the basic pH due to the dissolution of bicarbonate:

NaHCO$_3$ $\rightarrow$ Na$^+$ + HCO$_3^-$

HCO$_3^-$ + H$_2$O $\rightarrow$ OH$^-$ + H2CO$_3$

the aluminium oxide precipitates:

2 Al$^{3+}$ + 6 OH$^-$ $\rightarrow$ Al$_2$O$_3$ + 3 H$_2$O

The electric current forces the oxidation reaction to continue.
In air a thin layer of oxide is formed and then the rest of the metal is not in contact with the air anymore.
This is called passivation.

Since the aluminium ions are precipitating as oxide, in the (-) electrode the reduction is of sodium ions:

$\hspace{1cm}$Na$^{+}$ $\rightarrow$ Na

which reacts with water to form hydrogen and OH$^-$ so that the cycle is complete:

$\hspace{1cm}$2 Na + 2 H$_2$O $\rightarrow$ 2 Na$^{+}$ + 2 OH$^-$ + H$_2$


According to this document (https://arxiv.org/pdf/physics/0305011v1.pdf)

  "Water molecules can pass through the membranes in either direction, and they do. But because the concentration of water molecules is greater in the pure water than in the solution, there is a net flow from the pure water into the solution."
  Hydraulic equilibrium is achieved when the lower concentration of water in the high pressure side is balanced by the higher pressure and therefore higher energy pV on that side

This is the diffusion model of osmosis which is easy to understand at your level.


Miscibility of pairwise miscible liquids

So you want to know whether there is a ternary system with an "island of immiscibility". Well, it turns out that people have thought of this before; such systems have a name ("type 0 systems"), and this paper mentions water/DMSO/THF as an example.


Why do polarizable nucleophiles favor SN2 over E2?

Polarisable Nucleophiles increase the rate of an Sn2-type reaction but not an E2 reaction because the more polarisable a base is, lower the electronegativity and charge density (conversely, higher ionic radius) it has.

Now, we know that the basic strength depends on these factors( in a period, we check the electronegativity and in a group, the ionic radius).

As, the Rate Determining Step of an E2 reaction involves the attack of a base, lower basic strength would mean that the reaction would be slower. Hence, more polatisable bases means that E2 would be slower.


Are there more endothermic or exothermic reactions?

Answering "does every reaction have a reverse reaction?" (are there more endothermic or exothermic reactions):

You've actually hit on a topic people have been exploring for nearly a century:

  Lewis. A New Principle of Equilibrium. PNAS, 1925, 11(3),179-183

    The Law of Entire Equilibrium — Thus I am led to propose a law which in its general form is not deducible from thermodynamics, but proves to be compatible with the laws of thermodynamics in all cases where a comparison is possible. It may be called the law of entire equilibrium, and may be stated as follows. Corresponding to every individual process there is a reverse process, and in a state of equilibrium the average rate of every process is equal to the average rate of its reverse process. The rate at which one group or set of groups L goes over into another group or set of groups L' is the same as the rate at which the groups L' go over into the groups L. Moreover if there are various paths by which the first process occurs, there is an equal number of paths by which the second process occurs, and the rate is the same in both directions along every path. This will be true no matter how detailed are the specifications which define the several groups and the several paths.

The implications of this concept (and your question) have been well-elucidated by SE user Curt F. in two posts:

  Is every chemical reaction in equilibrium?

    Commenters suggest that "irreversible" reactions do not have an equilibrium. This is true, but tautological. In the real world, all reactions are reversible, at least to a (perhaps vanishingly small) degree. To say otherwise would violate microscopic reversibility. A reaction that was 100% irreverible would have an equilibrium constant of infinity. But if =∞, then Δ∘=−ln would turn into Δ∘=−∞. So to get infinite energy we would just have to use 100% irreversible reactions! Hopefully the problems with the idea of "irreversible" reactions are becoming apparent.

  Spontaneous/Non-Spontaneous Reactions and Reversible Reactions

    All reactions can be viewed as reversible from a mathematical standpoint, if the reacting system is "big enough". Reactions that are highly spontaneous in the forward direction mean that at equilibrium, the number of "reactant" molecules will be very very small (but not zero!).
    Many times, "big enough" would mean astronomically big, such as bigger than the Earth. In these cases, reactions can be regarded as "irreversible", although there is no strict, universally agreed upon boundary between "reversible" and "irreversible".

This should implicitly answer your question, as a 'reversed' exothermic reaction is an endothermic reaction. If every reaction is fundamentally reversible (on some scale, whether or not it's reasonable, as Curt F. describes), then there are an equal number of exothermic and endothermic reactions.

Answering "which occurs more frequently: exothermic or endothermic reactions?" (are there more endothermic or exothermic reactions?):

This is an interesting way to look at things. Considering the ever-increasing entropy of the universe and the favorable relationship between entropy and heat, I would guess that there are more exothermic reactions. Exothermic reactions represent energetically favorable transformations and are more spontaneous than endothermic reactions, so it stands to reason that there are "more exothermic reactions."

  Roduner &  Radhakrishnan. In command of non-equilibrium. Chem. Soc. Rev., 2016, 45,2768-2784

    2.7 Heat death: the ultimate end of an evolving universe

      From the second law of thermodynamics we know that the entropy of the universe will increase for all spontaneous processes. Since the total possible reversible heat flow is limited by everything having the same temperature, and when no chemical or nuclear reactions are available anymore which can produce further heat, entropy will reach a maximum but will not go to infinity. This extrapolated end of a universe that evolves further in time has been dubbed heat death. But how will this final state look like? This is a question that is complex to answer. Considering an extremely simplified model universe helps to get a rough idea.

      We first consider a closed system consisting of 1 mol of glucose and 6 mol of molecular oxygen at 25°C and 1 bar pressure (see Fig. 18a). The equilibrium state at the same temperature and pressure consists of the system after complete combustion to 6 mol of CO2 and 6 mol of liquid H2O. The heat of reaction of 2805 kJ will be dissipated from the system...

    5 Concluding remarks

      Heat engines produce work that is typically used to drive processes which do not occur spontaneously. This is achieved by letting the machine operate between two temperatures, where energy is provided to the system at the high temperature and partly converted to work. This can be done only when an unavoidable remaining fraction of the energy input is degraded to lower temperature and dissipated as waste heat. It is shown here that this principle is not limited to heat engines but holds more generally for many other processes in which a subsystem of the universe is to be transformed to higher order, as in the separation by distillation or crystallisation, photosynthesis, solar cells, or self-assembly...

      The simple model universe of Fig. 18, the combustion of glucose, predicted that the final state of the universe will be much warmer compared with today and consists of a higher fraction of atmospheric CO2 and H2O at the cost of O2. One might therefore ask whether the global warming that we see on our planet actually reflects that we are noticeably approaching the heat death of the universe as a consequence of combustion of fossil fuels with corresponding heat and entropy production. The answer is straightforward: the direction of the effect is correct, but it does not quantitatively reflect the approach of our planet to heat death. The solar heat incident on our atmosphere is about 5 orders of magnitude more than the total energy consumption by the human population. CO2 and water produced by combustion just increase the mismatch between incident and reflected solar energy by increasing the greenhouse effect of the atmosphere, thereby amplifying the direct effect by a considerable factor.

If we're to believe that heat death is a logical conclusion of the second law of thermodynamics, then there must be more exothermic reactions to result in an ultimately stagnant system where all available energy has been dispersed as heat and atomic components are at their lowest possible energy states — this outcome, by definition, represents an excess of exothermic reactions.


it smells bad, and the beer bottle's metallic label is slowly corroding.

That's, unfortunately, very common. At one of my former labs 'acid' cabinet  was corroded from within. For this reason it is recommended to keep such bottles under fume hood.

AFAIK, common stoppers are made from polyethylene or polypropylene. At one of my former labs concentrated hydrochloric acid was stored in stock plastic canister, similar to that used to store petrol (on the other hand, both nitric and sulfuric acids were stored in 20l glass bottles, even though the stoppers were made from same polymer).  I HIGHLY recommend to by (or order by mail) dedicated chemical bottles with caps, though. They are usually fairly cheap and pretty universal.


Is the electronegativity value ever used in calculations, or is it just a number assigned to each element?

True, with values of electronegativity in hand (and beside the by Pauling, there are others by Rochow, Mulliken, etc. to name some scales) you describe bond polarity. Beyond this number, infrared vibrational frequencies, chemical shifts in NMR spectroscopy and isomer shifts in Mössbauer spectroscopy are mentioned by the corresponding wikipedia article.


I will assume that the y-axis in your graph is temperature, and that the pressure of the system is fixed.

Consider the physical system: a mixture of two species in two phases is at equilibrium. Gibbs' phase rule tells us that we have two degrees of freedom, which can be chosen from the set $\{P, T, X_B^{liq}, X_B^\text{solid}\}$. By fixing $P$ (first degree of freedom) and letting $T$ (second degree of freedom) vary over a prescribed range, we produce the functions $X_B^{liq} = X_B^{liq}(T)$ and $X_B^\text{solid} = X_B^\text{solid}(T)$, and this is exactly what is represented by a phase diagram.


Chloride ion does not react with the water to any significant extent. To understand this, we must first accept that all chemical reactions are equilibriums. This means that all chemical reactions are reversible to some extent. Some reactions aren't very reversible at all (at least not in the short or medium term) - such as the combustion reaction in your car's gasoline engine. But if you give it a few million years, the carbon dioxide released by your car's combustion process will be fixed (biological term) by plants and these plants might eventually turn into fossil fuels again.

So we've established that all chemical reactions are equilibriums. Now we can consider the case of the chloride ion and why this reaction does not happen, as you say, or does not happen to a significant extent, as I say:

$\ce{Cl^- + H_3^+O -> HCl + HO^-}$

The reason is that if the above reaction happened to any significant extent, then this would make HCl a weak acid. Strong acids are defined to dissociate nearly completely in water. HCl is a strong acid by any account. Think of it as a waterfall. This reaction represents going down a waterfall (this reaction is large extent; heck, it's practically 100%):

$\ce{HCl + H_2O -> Cl^- + H_3^+O}$

So if the above reaction is the going down part of the water fall, its reverse reaction must represent the going up part of the waterfall. Going up the waterfall is hard no matter how you slice it; the only difference is whether you're trying to go up a small waterfall or Niagara falls. The reaction of hydrochloric acid with water is akin to going down Niagara falls - this is a favorable reaction. This is basically a one-way street.

Therefore, chloride ion can't affect solution pH; if chloride ion could somehow react again to form more hydrochloric acid then we'd violate the definition of a strong acid - nearly complete dissociation. This is what makes hydrochloric acid a strong acid!

And this is empirically true. Chloride ion can't even be protonated effectively by the strongest acid in a solution of hydrochloric acid, which is hydronium ion, so there is no way that you can expect chloride ion to be protonated by water (which is only about a trillion times weaker than hydronium ion as an acid). Thus, your solution of ammonium chloride has a pH controlled not by the chloride ion but by the ammonium ion.


How does liquid mercury conduct electricity?

Keep in mind that "stabilizing" orbitals does not "prevent" electrons from being shared, only that they would be shared to a smaller extent.  The very fact that mercury is a liquid and not a gas shows that there is some bonding in mercury, even if it is weaker than the bonding in most metals.  So, it is conductive the same way other metals are, just less conductive.  You can look up the conductivity of mercury, and you will find it to be lower than most metals.


...equilibria symbol which is one line rather than two with 'half arrows' at either end

I'm pretty confident you mean "⥋" arrow within the reaction scheme for enzyme $\ce{E}$ and inhibitor $\ce{I}$ which I tried to recreate from your description:

$$\ce{E + I ⥋ \underset{\rm loose\\bond}{EI} <=> \underset{\rm strong\\bond}{EI'}}$$

I've seen this arrow in literature before, and in general it's been also utilized to denote equilibrium. This notation is not officially supported by IUPAC standards. However, there is a couple of references tangentially mentioning it.

From the article "The Chemical Equation Part I: Simple Reactions" [1, p. 189] (emphasis mine):

  Sometimes reactants interact quickly, but the reaction never goes to completion because the reverse reaction is competing with the forward reaction, and the system ends up in a state of chemical equilibrium. A pair of arrows pointing in both directions indicates an equilibrium system, but the second arrow need not be there. Even when two arrows are used, the equation still cannot identify the point of equilibrium.

There is also a recent outstandingly well-written review with the title that speaks for itself: "Chemistry: A Panoply of Arrows" [2, p. 592], which pretty much says the same (emphasis mine):

  Van't Hoff's double arrow was modified in 1902 by Hugh Marshall,... who removed the inner barb of each arrow. This created the half-headed double arrows, ⇌, which have since been further simplified to ⥋. In justifying his proposal, he noted that single arrows were being used in organic chemistry
  “to indicate merely the stages and methods by which a substance can be produced from some other substance as a starting point. In such cases, as no attention is paid by the bye products (sic), there is nothing of the nature of an equation
  involved.”

My humble opinion is that authors' intent was to visually separate those two processes, e.g. due to exceptional difference in rates, or maybe they wanted to denote conformational change aside from the chemical equilibrium.

Notes

"⥋" arrow belongs to the Supplemental Arrows-B Unicode block under the name "Left barb down right barb up harpoon" and is invoked via U+294B.
This and many other symbols you are having troubles identifying can be recognized by services like Shapecatcher or Detexify.
I also asked a question on Mathematics.SX hoping that someone could provide more definitive answer regarding the proper usage of this arrow (no answer yet).

References

Kolb, D. Journal of Chemical Education 1978, 55 (3), 184. DOI: 10.1021/ed055p184.
Alvarez, S. Angewandte Chemie International Edition 2012, 51 (3), 590–600. DOI: 10.1002/anie.201101767.


Major form of methyl 3-hydroxy pent-2-enoate?

quick googling shows that for a very similar compound ethyl acetoacetate enol and ketoforms are in equilibrium with 10% of enol form. However, this equilibrium is influenced by solvent significantly. Actually, the percentage shows that the difference in stability between these two forms is veeery little.

The difference may be attributed to the difference in stability of the bonds, i.e.   C=O + C-H is better then C=C + C-O + O-H. However, internal hydrogen bonding DOES exist and participates in stability of enol form. For example, for hexan-1,3-one, the percentage of enol form is much lower.


How do we recognize the internal plane of symmetry in meso compounds?

I see two issues here:

Rotational conformations do not change whether or not the molecule is meso. For meso compounds, the presence of the symmetry element means that for each conformation that is chiral, we are guaranteed to have the enatiomeric conformation with equal probability. Therefore, the net optical rotation is zero.
Fischer projections are no doubt a usual way to represent chiral molecules, but in general, we use Fischer projections most heavily in the case of carbohydrates. The reason is because carbohydrates have shared homology where we care about the specific placement of hydroxyl groups. In general, they are not a great visualization tool. Wedge and dashes are much better. In this case, for a simple two carbon system, where the two carbons have identical connectivity, you could just as well have assigned Cahn-Ingold-Prelog absolute stereochemistry to each carbon to see that the molecule is meso.


You are correct that there is no difference between the last fourth and last molecules shown.

However, the thermodynamic driving force behind the Claisen condensation is the formation of the highly stable enolate of the $\beta$ ketoester, or in your case $\beta$ diketone.

In order to form this stable enolate the original ester must have at least two $\alpha$ hydrogens so that there is at least one left at the end to deprotonate. Therefore the Claisen condensation will not work with esters having only one $\alpha$ hydrogen (and obviously not with esters with no $\alpha$ hydrogens).

Once you have completed the reaction to yield the enolate product, it can be acidified as shown in your reaction scheme to yield the neutral molecule.


13C-NMR of aniline: Why are carbons that are further from the amine group more downfield than carbons that are closer?

Although the nitrogen atom is more electronegative than carbon, nitrogen will readily share its electron pair, allowing one to draw a resonance structure where the nitrogen is donating electron density into the ring. Therefore, the donating of the lone pair of electrons overcomes the inductive effects arising from the difference in electronegativity.


What are the oxidation states of sulfur in the tetrathionate ion?

Consider the structure of tetrathionate. The two central sulfurs each have two lone pairs and are assigned half of the electrons from the two bonds they make, since the electrons of bonds between atoms of the same element must be distributed evenly (due to there being, by definition, an electronegativity difference of zero between two atoms of the same element). A neutral sulfur atom has six valence electrons, so the oxidation state of the central sulfurs can be calculated as follows:

$$6 - 4 - \frac{1}{2}(4) = 0$$

That is, six electrons in neutral sulfur, minus four from the lone pairs, minus half of the four sulfur-sulfur bonding electrons, gives zero.

The terminal sulfurs, on the other hand, have no lone pairs, and all the electrons from the sulfur-oxygen bonds are assigned to oxygen, since it's the more electronegative element. The only electrons assigned to those sulfurs are half of those from the sulfur-sulfur single bonds. Hence, their oxidation state is:

$$6 - \frac{1}{2}(2) = +5$$

That is, six electrons in neutral sulfur, minus half of the two sulfur-sulfur bonding electrons, giving an oxidation state of +5.


Reactions of alkynes with sodium/ammonia vs sodium amide

Sodium in ammonia ($\ce{Na/NH3}$) and sodium amide ($\ce{NaNH2}$) are very different sets of conditions, despite superficially looking similar!

Sodium metal itself, $\ce{Na}$, is a one-electron reducing agent – it is oxidised to $\ce{Na+}$ in the process. When it reduces neutral hydrocarbons, negatively charged species are formed, which pick up a proton from the ammonia solvent. Overall, $\mathbf{Na/NH_3}$ is a way of reducing some organic compounds (including alkynes).

Internal alkynes $\ce{R-C#C-R}$ can be reduced to trans-alkenes using $\ce{Na/NH3}$. Specifically, solvated electrons are created which add to the triple bond as shown. Ammonia acts as a proton source to protonate the carbanions formed. The vinyl anion intermediate 4 can interconvert between a cis and trans geometry, and the trans geometry is preferred as this minimises steric repulsion between $\ce{R^1}$ and $\ce{R^2}$. When this anion is protonated, the trans-alkene is formed. This is a nice synthetic procedure for the preparation of trans-alkenes, complementary to Lindlar hydrogenation which produces cis-alkenes from alkynes.

On the other hand, in sodium amide $\ce{NaNH2}$, sodium is already in the +1 oxidation state and is no longer a reducing agent. However, the amide ion $\ce{NH2-}$ is a very strong base (the $\mathrm pK_\mathrm a$ of ammonia, $\ce{NH3}$, is 38). So $\mathbf{NaNH_2}$ is a way of deprotonating some organic compounds (including terminal alkynes).

The proton attached to the terminal carbon in a terminal alkyne is acidic enough ($\mathrm pK_\mathrm a \sim 25$) to react with sodium amide, forming the corresponding carbanion 7. Such carbanions are good nucleophiles and can be used in many synthetic procedures involving nucleophiles, such as addition to a carbonyl compound, as illustrated here.


The $\Delta V$ is supposed to be the volume of the vapor minus the volume of the liquid.  The volume of the liquid is only 0.025 liters, which they neglect.

I don't know where they get the 12470.6 from.  I'm guessing this is an error.

(0.082)(373)(25)/18=42.48. You solve it basically the way you said.

Δ(PV)=42.48 liter−atm =1029 cal=42.48 liter-at = 1029 cal

So, ΔU=13500−1029=12471 cal


When two materials precipitate at the same time, you'll often get an extremely intimate mix of the two. I'm assuming that the notation you've got here means a precipitate consisting of a mixture of both nickel (II) ions, iron (II) ions, and sulfide ions, or a mix of nickel and iron sulfides (NiS & FeS).

Iron and nickel (II) have similar ionic radii and charge densities, so it should be possible to create crystal domains containing both metal ions interchangeably, in any ratio.


Acid catalyst decarboxylation of 1-carboxyl-tryptoline

Your suggested mechanism should be acceptable (need two more steps, protonation and deprotonation, see last two steps of my scheme below), if you have isolated the carboxylic acid product (compound $\bf{1}$ in my scheme). If compound $\bf{1}$ is not isolable, I suggest following path would be taken to decarboxylation:

Your reaction is an analog to Pictet Spengler reaction (Wikipedia). When the intermediate $\bf{I}$ is formed during the reaction as depicted in the mechanism, 1,2-hydride transfer is possible to give more stable carbocation $\bf{II}$, which is resonance stabilized $(\bf{\ce{II <-> III}})$. The intermediate $\bf{III}$ can easily undergo decarboxylation to give some what stable enamine product, $\bf{2}$. In acidic medium, this enamine $\bf{2}$ would undergo protonation to give the intermediate $\bf{IV}$, which would then deprotonated to gain aromaticity and give thermodynamically stable indole product, $\bf{3}$ (the intended product).


Theoretical comparison of critical constants of real gases

In essence regarding this problem
$$T_\mathrm c \propto \frac{a}{b}$$
$$P_\mathrm c \propto \frac{a}{b^2}$$
$$V_\mathrm c \propto {b}$$

Values of $a$:

$\ce{NH3} \to 4.225\ \mathrm{l^2\ bar/mol^2}$
$\ce{CO2} \to 3.640\ \mathrm{l^2\ bar/mol^2}$
$\ce{H2O} \to 5.536\ \mathrm{l^2\ bar/mol^2}$.

This trend as justified by the OP through hydrogen bonding is absolutely correct.

Values of $b$:

$\ce{NH3} \to 0.0371\ \mathrm{l/mol}$
$\ce{CO2} \to 0.04267\ \mathrm{l/mol}$
$\ce{H2O} \to 0.03049\ \mathrm{l/mol}$.

$b$ or effective molecular size is essentially calculated by multiplying the size of a molecule by four. The trend over here is obviously determined by molecular size only.

Now:

The critical temperature and critical pressure are in direct /inverse proportion with $a$ as well as $b$ .An interplay of two factors. Predicting the orders of both of these cases become extremely difficult  without data. The opposing orders of $P_\mathrm c$ and $T_\mathrm c$ are seen as the order of proportion changes i.e $P_\mathrm c$ is inversely proportional to the square of $b$ and $T_\mathrm c$ is inversely proportional to $b$.
Trends in $V_\mathrm c$ on the other hand can be safely predicted by estimating the size of molecules as it only dependent upon one variable that is $b$.

All data is from this Wikipedia page


Raman Spectrum of carbon chains: How to identify the vibrational modes

When you perform a IR or Raman calculation, the primary step is the diagonalization of the mass-weighted Hessian:

$$
\begin{align}
H_{ij,ab}^{\mathrm{mw}} &= \frac{1}{\sqrt{m_{i}m_{j}}} \frac{\partial^{2} E}{\partial a_{i} \partial b_{j}} \\
\mathbf{Hq} &= \lambda \mathbf{q}
\end{align}
$$

where $H_{ij,ab}^{\mathrm{mw}}$ is a $3N\times3N$ Hermitian matrix, $i,j$ are indices running over the atoms, $m_{i}$ is the mass of atom $i$, and $a,b \in x,y,z$ (the $3N$ Cartesian molecular coordinates). The eigenvalue $\lambda_{i}$, multiplied by some conversion factors, is a frequency in the harmonic oscillator approximation, and the corresponding eigenvector $\vec{q}_{i}$ is the normal mode, the collective atomic displacements, for that frequency.

(A calculation of Raman intensities involves extra work, namely the calculation of $\frac{\partial^{3} E}{\partial \varepsilon_{\alpha} \partial \varepsilon_{\beta} \partial a_{i}}$, but the normal modes are identical.)

Aside from performing literature searches for normal mode frequency assignments on related systems, a good alternative is the visualization of these normal modes. Many outputs from frequency calculations can be opened in Avogadro:

Clicking on a frequency in the right-hand side pane (only IR intensities are displayed) can display both the displacement vectors and an animation of the displacement (see the lower right corner box). Avogadro is able to read this portion of the NWChem output file:

          -------------------------------------------------
          NORMAL MODE EIGENVECTORS IN CARTESIAN COORDINATES
          -------------------------------------------------
             (Projected Frequencies expressed in cm-1)
...
                    7           8           9          10          11          12

 P.Frequency      252.66      366.45      606.76      773.89      790.45      808.06

           1     0.06344     0.00000    -0.02127     0.00000     0.00000     0.00096
           2    -0.10753     0.00000    -0.09540     0.00000     0.00000    -0.00247
           3     0.00000    -0.03639     0.00000    -0.01614     0.01014     0.00000
           4    -0.08820     0.00000    -0.09326     0.00000     0.00000     0.00127
           5     0.02291     0.00000    -0.01064     0.00000     0.00000    -0.00238
           6     0.00000     0.00615     0.00000     0.14073    -0.07465     0.00000
           7    -0.13091     0.00000     0.17533     0.00000     0.00000     0.05512
           8     0.01443     0.00000     0.09744     0.00000     0.00000     0.01777
           9     0.00000     0.19419     0.00000    -0.08894     0.11256     0.00000
          10     0.11294     0.00000    -0.04324     0.00000     0.00000    -0.12834
          11     0.07728     0.00000     0.04956     0.00000     0.00000    -0.02839
          12     0.00000    -0.10132     0.00000    -0.04711    -0.12489     0.00000
          13     0.29675     0.00000    -0.23722     0.00000     0.00000     0.83429
          14     0.12532     0.00000     0.00136     0.00000     0.00000     0.22209
          15     0.00000    -0.33078     0.00000     0.53191     0.69392     0.00000
          16     0.24449     0.00000     0.20104     0.00000     0.00000     0.01418
          17    -0.04794     0.00000    -0.02443     0.00000     0.00000     0.00175
          18     0.00000    -0.37449     0.00000    -0.49495     0.26769     0.00000
          19     0.03319     0.00000    -0.05962     0.00000     0.00000    -0.00066
          20    -0.29762     0.00000    -0.33571     0.00000     0.00000    -0.01410
          21     0.00000     0.25757     0.00000     0.27323    -0.15332     0.00000
          22    -0.06568     0.00000    -0.11340     0.00000     0.00000    -0.00250
          23     0.13575     0.00000    -0.12891     0.00000     0.00000    -0.02550
          24     0.00000    -0.29795     0.00000    -0.17367     0.10673     0.00000
...

So, to find the C-C stretching mode (or any vibration of interest), clicking through the list in Avogadro is more convenient than eyeballing the displacements in the output file. With the advent of accurate computational methods, the assignment of peaks in experimental spectra is usually aided by calculations identical to these. It's even possible to plot the computed spectra with some artificial broadening:


They are not coupling with the aldehyde proton. Because the three protons around the ring are locked in such close proximity to one another, they will all couple each other. Let us consider these protons:

$\hspace{6cm}$

As by the $(n+1)$ rule of multiplicity, a proton will couple with each nearby, chemically unique proton and be split into $(n+1)$ peaks. These split signals can then also be split into $(n+1)$ more peaks by other nearby, chemically unique proton. Since all three protons are in the same region of space around the ring, they all couple each other, splitting each proton's signal into $(1+1)(1+1)=4$ peaks, making each signal a doublet of doublets (a quartet).

If you'd like a detailed analysis of the NMR of furfanal, you can read about it here.


Does sheer amount affect reaction rates?

If you define "completion" by a certain percentage of conversion then no, they will behave exactly the same, provided that things like cooling and stirring are the same in both systems.