What is the oxidation state of nitrogen in N2O5?

It is +5 .Even by the structure.

Notice that it is a resonance structure of:

Now one of the bonds are coordinate bonds where nitrogen donates BOTH its electrons to oxygen (otherwise, oxygen cannot have a single bond without a negative charge and no peroxy type linkage).

So, while determining oxidation state from the structure :

O.N=+2+2+1=+5

References:
https://en.wikipedia.org/wiki/Dinitrogen_pentoxide


Heat lost in evaporation at constant temperature

Not a complete answer, but this should help clarify things for you:

You wrote "since the mass is decreasing during evaporation, both mass and heat would change".  But note that the change in mass per se does not affect the needed heat flow.  For instance, consider pure water.  Here, the change in mass would not affect the rate at which heat is lost due to evaporation (assuming other conditions, like surface area, temperature, the partial pressure of water vapor in the air above the water, wind speed, etc., don't change), and thus would not affect the heat flow needed to maintain the temperature.  This is because you're not trying to change the temperature of the water (in which case its mass would be relevant), you're just trying to maintain an energy balance of heat flow in and heat flow out.

Hence, regardless of the volume of water, for a specific evaporation rate, the heat flow needed to maintain constant temperature would be the same, since the latter is determined only by the rate at which heat is lost due to evaporation.

However, since you are working with salt water solutions, their concentration would increase with time, which would decrease the evaporation rate and thus decrease the heat flow needed to maintain the temperature.

So you have two choices:

(1) Just calculate the heat flow as a function of concentration (i.e., for each solution, determine the initial heat flow needed).  If the volume of water is large relative to the surface area, treating the concentration as a constant could be a good approximation for short periods of time.

(2) Use the evaporation rate as a function of concentration to determine the concentration as a function of time. I.e., derive a time-dependent expression. This is trickier.

I'm not sure from your question which you want to do, since your concern seemed to be the change in the total mass of the water, not the concentration. Note also that the equation you found in the linked article doesn't include a variable for concentration.


I have to write a Python program relating with molecular weight. Any insight?

It's pretty simple actually. The molecular weight of a compound is the sum of the atomic weights of all its atoms. The same numbers apply to a single atom as for a mole of that atom, but at the single-molecule level the unit of measure is the atomic mass unit or amu (aka the dalton or Da). The number of grams per mole of an element and the number of daltons that a single atom of the element weighs are intrinsically linked due to the definition of what a mole of something is (the amount of a pure chemical substance that weighs the same amount in grams as the mass of one atom or molecule of the substance in daltons).

So, methane (CH4), for instance, has a molecular weight of 1.0079 x 4 + 12.011 = 16.0426 Da. A mole of this gas would weigh 16.0426 grams (which is - interesting science fact - lighter by almost half than that of N2 or O2 gases making up most of the atmosphere, so a balloon of it will float). So, your Python script would simply prompt for the number of hydrogens (accepting integer values only), carbons (ditto) and oxygens (ditto), and plug them into the expression:

total_mass = num_hyd * 1.0079 + num_carb * 12.011 + num_oxy * 15.9994

... before returning total_mass as the answer.

Theoretically, the energy contained as enthalpy in the bonds between atoms contributes to a molecule's mass by general relativity and the first law of thermodynamics. In practice, the amount of energy inherent in even the least stable bonds (which produce the most net energy when they decompose to form more stable ones) is a rounding error when you run it through E=mc2, and so chemists don't bother. Nuclear physicists, who actually change the atoms themselves, do care, because the equation works in reverse for them; they get lots of energy from very small changes in the atomic masses of the reactants vs the products.


How is a phase equilibrium defined for a one-component system?

Melting is a process. A process is not a state. Since only states can be "at equilibrium", it doesn't make sense to speak of a real melting substance as being in an equilibrium state. However, melting is deemed reversible if performed infinitely slowly, and during such a transition the system passes through a series of equilibrium (or "quasi-equilibrium") states. But such a process is an idealization.  In reality, you can't "melt ice isothermally". Real transitions are therefore irreversible because system and surroundings are not actually in thermal equilibrium during heat transfer$^\dagger$.
Note that if you place in contact two of the intermediate states sampled during the transition neither will change, that is, each remains in its original equilibrium state.
The second definition encompasses the first one (equality of chemical potential across phases). The second definition therefore better captures what is meant by phase equilibrium. The conditions for phase equilibrium are those used to derive the phase rule, namely mechanical, chemical and thermal balance between phases. A phase is a region of uniform composition, temperature and pressure.

$^\dagger$ We say that it is possible to melt a pure substance using an ideal reversible process that leads the system infinitely slowly through a series of equilibrium states (each intermediate stage in that reversible process is an equilibrium state).  In order to move between intermediate states, an infinitesimally small amount of heat has to be transferred from the infinitesimally warmer surroundings (exploiting, say, something like a Maxwell demon). In what is a logical sleight of hand, we say that the temperature difference is so small however, that the system and surroundings are effectively isothermal. This does not reflect what happens in a real process.


Poly(methyl methacrylate), PMMA withstands acids and bases of low concentration. So it should be safe for electrolysis of water.


They differ by definition, which states that work function is difference in energy between Fermi level (work required to add one electron to the body) and vacuum level (energy of an electron that drifts in vacuum, free from any material), while ionization energy is difference in energy between HOMO (highest occupied molecular orbital) and our old friend, vacuum level.

It's clear that electrons that occupy orbitals (HOMO energy level) are lower in energy than Fermi level (the boundary between occupied and unoccupied states in a continuum of states). Below is the picture which you might find useful. Also check for the source of picture: it has good and detailed explanation about your next possible questions on topic.

[source]: Antoine Kahn. Fermi level, work function and vacuum level. DOI: 10.1039/C5MH00160A (Focus) Mater. Horiz., 2016, 3, 7-10.


How do you figure out fuel specific impulse non-experimentally?

The relevant property is the temperature of the burnt fuel when it leaves the rocket engine nozzle, i.e. when it stops exchanging impulse with the engine.

That temperature you can convert into a velocity distribution of the gas particles, multiply with their molar mass, and that's it. (somewhat less than that of course, because no all gas particles are flying in the exact same direction. depends on the nozzle type, among other influences)

The temperature you can estimate to some extent from the reaction enthalpy, but of course you have radiative and conductive losses. Which is where this becomes an engineering problem.


Identifying Nucleophiles and Electrophiles

Nucleophilicity and Electrophilicity are no absolute measurable quantities - they are based on an per entry basis and usually in reference to some kind of reaction. So basically you are right in assuming that polar molecules have both properties. In the special case of water these properties also stem from $$\ce{H2O <=> {}^{-}OH + H3+O}$$

But let us have a look at some molecules:

$\ce{CO}$ pretty good nucleophile especially towards transition metals. It has quite a small dipole towards the carbon, but a massive overload of electron density due to the $\sigma$-type HOMO
$\ce{CO2}$ also quite polar bonds, but no (permanent) dipole - still quite a potent electrophile given the right nucleophile
$\ce{SO3}$ has no permanent dipole (but a transitional one) although quite polar bonds. However, it is only a strong electrophile.
$\ce{NH3}$ quite polar, hence a reasonably good nucleophile. But it is also almost always way too strong a base to act as an electrophile
$\ce{NH(Et)2}$ not so different from ammonia, quite polar, but yet a terrible nucleophile, because it is too big
$\ce{BH3}$ not very polar (no permanent dipole) but quite a good electrophile
$\ce{I-}$ a monopol, very polarisable, good nucleophile, good leaving group.
...

The consideration of the strength of nucleophiles and electrophiles lies in many variables. It might be polarity, local electron density, molecular orbitals, polarisabilities. It is often highly dependent on the solvent and the other reactants. If you add $\ce{I-}$ to a acyl chloride you will certainly have a substitution to some extend:
$$\ce{R-COCl + I- <=> R-COI + Cl-}$$
If you add hydroxide ions, then you will certainly end up with the acid:
$$\ce{R-COCl + I- + {}^{-}OH <=> R-COOH + I- + Cl-}$$
Look at it nucleophilicity or electrophilicity from a case to case basis.


What is the correct structure for trans-3,4-dichlorocyclopentene?

The issue here really is that the name 'trans-3,4-dichlorocyclopentene' is not really correct, as it is unspecific. It lacks stereo-descriptors, and the locant for the double bond.

Assuming that the double bond starts at the first carbon, which is usually the case if it is omitted, there are two enatiomers still possible:

(3R,4R)-3,4-Dichlorocyclopent-1-ene

See for details: National Center for Biotechnology Information. PubChem Compound Database; CID=12430197.
(3S,4S)-3,4-Dichlorocyclopent-1-ene

See for details: National Center for Biotechnology Information. PubChem Compound Database; CID=131015828.

So the structure you have drawn as 1 is (3R,4R)-3,4-dichlorocyclopent-1-ene.

The structure you have drawn as 2 doesn't really exist, as the double bond is planar, but ignoring that it would be 1,2-dichlorocyclopent-1-ene.

See for details: National Center for Biotechnology Information. PubChem Compound Database; CID=10606793.


Calculation of the lowest energy conformation with a webbased solution or freeware

Avogadro can do a lot of this for you. It might not be precisely what you want but the molecular mechanics from Open Babel integrated into it can do at least some of the tasks you mentioned. It's free and has a very intuitive and smooth user interface. A real pleasure. The downside is that it's poorly documented.

To play around with conformations, go to Extensions -> Molecular Mechanics. To view the generated conformers (from the Conformer Search option) go to View -> Properties -> Conformer Properties...


Is it possible for the forward and reverse reactions to have different reaction orders?

Why is this strange?

Consider:

$$\ce{2A <=>[$k_1$][$k_{-1}$] A_2}$$

Forward rate is second order: $k_1[\ce{A}]^{2}$.

Reverse rate is first order $k_{-1}[\ce{A2}]$.


Can I melt Barium Titanate powder into a cylinder and preserve the high dielectric?

As you know, barium titanate is a piezoelectric ceramic materials. It has its property as a piezoelectric material due to the specific spatial arrangements of its constituting atoms. Melting this material, means that it will lose its crystalline  structure upon heating (knowing also that its Curie Temperature is $T_C=120^o\mathrm{C}$, i.e. above this temperature it loses its ferroelectric property).
On the other hand, processing and shaping ceramic materials are often performed during the preparation of these materials, not after. Heating at temperatures around $T_C=1600^o\mathrm{C}$ for processing ceramics is not a common procedure and it is expensive.
If you want to prepare cylinders and discs from a powder of barium titanate, you can add it to a polymer matrix (wax, epoxy resin,..) in an appropriate mold using adequate processing techniques like injection molding (heating below Curie Temperature). Of course, the resulting dielectric constant of the composite material is less than the one of pure  barium titanate.


Experimental measurment of thiosulphuric acid decomposition kinetics

Measuring conductivity over time might do the trick (7 ions (8 charge equivalents) in, 4 ions/charge equivalents out).

Regarding spectrometry: you could deliberately make use of the light scattering caused by the precipitating sulphur. Should be somewhere in the 600+ nm range. It's a standard method in biological sciences to follow the growth of bacteria.


Calculating Vapor Pressure from Delta G/ Kp

This topic came up the other day but in regards to the evaporation of water.  You might find my answer to that question helpful.

Here's how I would solve the problem.

Explicitly note standard state used and assumptions.  I will assume:

Methanol vapor is an ideal gas
Standard states for all species are 1 bar of partial pressure 25 °C.  (This assumption matters a lot.  The meaning of the $^\circ$ symbol in $\Delta G^\circ$ can't be understood without it.
The free energy of methanol liquid does not vary with pressure, i.e. $G_{liq} = G_{liq}^\circ$ at all pressures.  I'm fairly confident that this assumption is equivalent to assuming methanol liquid is an incompressible fluid.

Note the equilibrium condition.  At equilibrium,

$$\Delta G = 0 = G_{vap} - G_{liq}$$

Substitute the pressure-dependent free energy of an ideal gas

$$ 0 = G_{vap} - G_{liq} = \left [G_{vap}^\circ + RT \ln{\frac{p}{p^\circ}}\right ] - G_{liq}$$

Use the assumption of incompressibility of liquid methanol that we made above

$$0 = \left [G_{vap}^\circ + RT \ln{\frac{p}{p^\circ}}\right ] - G_{liq}^\circ$$

Rearrange and substitute

$$0 = \left [G_{vap}^\circ - G_{liq}^\circ \right ] + RT \ln{\frac{p}{p^\circ}}$$

$$0 = \Delta G^\circ  + RT \ln{\frac{p}{p^\circ}}$$

$$0 = 4.3 \frac{\textrm{kJ}}{\textrm{mol}} + RT \ln \frac{p}{p^\circ}$$

Solve for $p$

$$\ln{\frac{p}{p^\circ}} = \frac{-4.3 \frac{\textrm{kJ}}{\textrm{mol}}}{RT} $$

$$p = p^\circ e^{\frac{-4300 \textrm{ J/mol}}{RT}}$$

$$p = p^\circ e^{\frac{-4300}{8.314\cdot 298}}=  p^\circ e^{-1.73} = 0.176~p^\circ  $$

Figure out what $p^\circ$ is.  Just like it does for $\Delta G$, adding a $^\circ$ indicates that we are referring to a standard state.  Thus $p^\circ$ refers to the partial pressure of methanol vapor at its standard state.  This is almost always defined as one atmosphere (or in more modern times, one bar, which is nearly but not exactly the same thing) of pressure. Assuming that the standard state is the usual one, then $p^\circ$ is one bar, which is equal to 750 mmHg.  This is what we have assumed in step 1.  If that is the case, then $p = 0.176~ p^\circ=0.176\cdot750~\textrm{mmHg}=134~\textrm{mmHg}$.

  How would it know that this unitless quantity is atm and not pascal or mmHg?

Hopefully my answer above helps you answer your question.  It has to do with the common use of $\circ$ to indicate a standard state of one bar of partial pressure for all gases.


Yes free $\ce{H+}$ ions, protons, really exist.

Protons are constantly emanating from the sun and reaching Earth.

The proton flux is continuously monitored by satellite.

However, in a solution such as water, instead of bare $\ce{H+}$ ions, they are $\ce{H3O+}$ or larger ions such as $\ce{H5O2+}$ or $\ce{H9O4+}$.

When $\ce{HCl}$ dissolves, the proton is transferred to a water molecule to form $\ce{H3O+}$.

$\ce{HCl}$ only has 17% ionic character according to Pauling's The Nature of the Chemical Bond, so it is covalent, but reacts with water to form separate ions in solution.

For more information see Eigen or Zundel Ion: News from Calculated and
Experimental Photoelectron Spectroscopy:

  The appearance
  of the hydronium ion was and is
  currently debated as several possible
  structures, three of them being $\ce{H3O+}$
  (the Eigen cation), $\ce{H5O2+}$ (the Zundel
  cation) and $\ce{H9O4+}$ (the strongly solvated
  Eigen cation)

and

Both Zundel and Eigen Isomers Contribute to the IR Spectrum of the Gas-Phase $\ce{H9O4+}$ Cluster:

  The “Eigen cation”, $\ce{H3O+(H2O)3}$, is the most prevalent protonated water structure in the liquid phase


Friedel-Crafts Alkylation and hybridization

If anything this seems to be more favorable as an sp carbocation is likely less stable than an sp3 carbocation.

That's the problem, sp2 and sp hybridized carbocations are less stable than sp3. Those carbocations are difficult to form and will likely undergo rearrangement faster than attack of the aromatic ring.

Modern organic synthesis has much better ways of making sp-sp2 and sp2-sp2 carbon-carbon bonds by palladium catalyzed cross-coupling.


Spectroscopic Determination of an Equilibrium Constant

You must first calculate dilution factors. Assuming the volumes are additive, we can calculate dilution using $M_1V_1=M_2V_2$ equation. Also, assuming $\ce{Fe(NO3)3}$ and $\ce{KSCN}$ are each completely dissociated, we can say:

$$\ce{[Fe(NO3)3] = [Fe^3+]} \text{ and } \ce{[KSCN] = [SCN-]}$$

Thus, initial $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ are $0.20$ and $\pu{2.0 \times 10^{-4} M}$, respectively.

In test tube one: $\pu{5.0 mL}$ of $\pu{0.20 M} \; \ce{Fe(NO3)3}$ and $\pu{5.0 mL}$ of $\pu{2.0 × 10^{−4} M} \; \ce{KSCN}$ are added together, thus dilution factor is 1:2. Therefore, using $M_1V_1=M_2V_2$ equation you can calculate initial concentrations of $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ after the addition (but before they react with each other) as follows:

$$\ce{[Fe^3+]} = \pu{\frac{0.2 \times 5.0}{10.0} M} = \pu{0.10 M} \text{ and } \ce{[SCN-]} = \pu{\frac{2.0 × 10^{−4}  \times 5.0}{10.0} M} = \pu{1.0 × 10^{−4} M}$$

As a matter of fact, since the original solution of $\ce{KSCN}$ did not undergo any dilution maneuver before final 1:1 dilution, $\ce{[SCN-]}$ of all 6 test tubes are same, and its value is $\pu{1.0 × 10^{−4} M}$.

On the other hand, the original solution of $\ce{Fe(NO3)3}$ has undergone serial dilution of $\pu{10 mL}$ to $\pu{25 mL}$ (hence dilution factor is $10/25$ or $2/5$). Therefore, if original concentration of each solution is $x$ before add together in test tubes, the final $\ce{[Fe^3+]}_a$ before add with $\ce{[SCN-]}$ is:

$$\ce{[Fe^3+]}_a = \frac{x \times 2.0}{5.0} \: \pu{M}$$

Thus, the final $\ce{[Fe^3+]}_a$ before add with $\ce{[SCN-]}$:

$$\ce{[Fe^3+]}_2 = \pu{\frac{0.20 \times 2.0}{5.0} M}= \pu{0.08 M}, \, \ce{[Fe^3+]}_3 = \pu{\frac{0.08 \times 2.0}{5.0} M}= \pu{0.032 M}, \,  \ce{[Fe^3+]}_4 = \pu{\frac{0.032 \times 2.0}{5.0} M}= \pu{0.0128 M}, \,  \ce{[Fe^3+]}_5 = \pu{\frac{0.0128 \times 2.0}{5.0} M}= \pu{0.00512 M}, \,  \ce{[Fe^3+]}_6 = \pu{\frac{0.00512 \times 2.0}{5.0} M}= \pu{0.002048 M}$$

Finally, those concentrations were further diluted in 1:1 ratio in test tubes, and therefore, initial $\ce{[Fe^3+]}$ are half of those $\ce{[Fe^3+]}_a$ values. I tabulated all of them accordingly in following table, as answer to your question 1:

$$
\begin{array}{c|cc}
\text{Test Tube #} &\text{Initial $\ce{[Fe^3+]}$} &\text{Initial $\ce{[SCN-]}$} \\\hline
1 & \pu{0.10 M} & \pu{1.0 × 10^{−4} M}\\
2 & \pu{0.04 M} & \pu{1.0 × 10^{−4} M}\\
3 & \pu{0.016 M} & \pu{1.0 × 10^{−4} M}\\
4 & \pu{0.0064 M} & \pu{1.0 × 10^{−4} M}\\
5 & \pu{0.00256 M} & \pu{1.0 × 10^{−4} M}\\
6 & \pu{0.001024 M} & \pu{1.0 × 10^{−4} M}\\\hline
\end{array}
$$

There are 7 different values in the table. But you have given one
relevant answer, and it is also incorrect.
Find the answer for $\ce{[SCN-]}$ in test tube 4 from the table, which is
identical in all 6 test tubes.
The concern equilibrium is: $$\ce{Fe^3+ + SCN- <=> FeSCN^2+}$$ Reaction mole ratio of $\ce{[Fe^3+]}$ : $\ce{[SCN-]}$ : $\ce{[FeSCN^2+]}$ is 1 : 1 : 1. If equilibrium concentration of $\ce{[FeSCN^2+]}$ in test tube 2 is $\ce{ 8.3×10^{−5} M}$, the equilibrium concentration of $\ce{[Fe^3+]}$ in test tube 2 is $\ce{ (0.04 - 8.3×10^{−5}) M}$ and that of $\ce{[SCN-]}$ in test tube 2 is $\ce{ (1.0×10^{−4} - 8.3×10^{−5}) M}$. Now, you may able to calculate equilibrium constent for the reaction using following equation:

$$K_\mathrm{eq} =  \frac{\ce{[FeSCN^2+]}}{\ce{[Fe^3+]}\ce{[SCN-]}}$$


A solution is cloudy at lower concentration and becomes clear at higher concentration. Why?

At very dilute concentrations, the solvating power of water overcomes the natural tendency of surfactant molecules to agglomerate into a separate phase. The shear numerical excess of water molecules just about totally dissociates the bulk surfactant and dissolves it as separate molecules. So a mixture of surfactant and water can be a true solution - clear and homogeneous.
However, as the concentration of surfactant increases, aggregates form (micelles) which can absorb oleophilic molecules (e.g., the polymers in the formula). With sufficient surfactant, the polymer is dispersed enough and into small enough micelles to appear clear. In the second graph, the knee of the curve is at the CMC; at this point, the aqueous phase is saturated with single molecules and some are beginning to stick together, starting to be able to contain polymer.

To get to the point: Your floor cleaner needs to be concentrated enough to work well. This suggests that there is enough surfactant to exert a solvent effect on the polymer, dispersing the chains, separating the molecules, making the smallest dimension significantly less than a wavelength of light. The original floor cleaner product could appear clear to the eye. Addition of a little of the product to a lot of water reduces the micelle formation of the surfactants by favoring dissolution of single surfactant molecules: the polymer chains then have fewer surfactant molecules to surround them. Polymer molecules escape, agglomerate and become visible - but only a little, since you can reverse the process.
A more delicate detection method is to use the Tyndall effect: in a darkened room, shine a narrow beam of light through a liquid (say from a single LED flashlight). The beam may be visible, much like a flashlight beam in a smoky room, and can detect some materials at levels in the ppm range (e.g., milk in water).

I suspect your floor cleaner would be represented by the middle test tube in the series of five. Small micelles are present but the liquid seems clear, and will seem to have a small or negligible Tyndall effect. The polymers will be dissolved in the many micelles, which are too small yet to be very visible. When you dilute, you go to the left one step, the surfactant becomes more dispersed into single molecules, exerting less solvent power on the polymer. Either the micelles  grow so large as to become visible, or actually collapse, allowing the polymer molecules to agglomerate.
Adding formula back to the mixture increases the surfactant (the solvent power), redisperses the polymer molecules into tiny micelles, and the liquid becomes clear again. The key (hypothesis) is that the pure polymer dispersion would be cloudy by virtue of its large particle size (~1 micron?), but a high enough surfactant concentration makes the formula essentially clear by generating micelles much smaller than the wavelength of light.


What is it that is "symmetry-adapted" about symmetry-adapted perturbation theory?

The "symmetry" in symmetry-adapted perturbation theory refers to the anti-symmetry of the wave function with respect to electron exchange:

$$\Psi(\mathbf{r}_1,\mathbf{r}_2) = -\Psi(\mathbf{r}_2,\mathbf{r}_1) $$

The most usual way to enforce antisymmetry in electronic structure theory is to enforce orthogonality between spin-orbitals.  So SAPT is intermolecular perturbation theory that includes terms that depend on the overlap between orbitals in different molecules.

Intermolecular perturbation theory gives rise to concepts like eletrostatic interactions, polarization, and dispersion.  With SAPT you then get additional concepts like exchange repulsion (aka steric or Pauli repulsion).


Is oxidation always thermodynamically favorable?

As the term redox implies you have an oxidation and a reduction in every redox reaction.
It makes no sense to just talk about a half reaction, but you have to look at the cumulated energy difference of educts and products.
Now you have to remember two things:

You can easily convert between the energy and potential difference of a redox reaction by multiplying with the exchanged charges:
$$\Delta G= Uzq$$
$U$ is the potential (usually measured in Volt), $z$ the number of exchanged elementary charges and $q$ the elementary charge (usually measured in Coulomb).
Energy differences are transitive. That means:
$$\Delta G (A\rightarrow C) = \Delta G (A \rightarrow B) + \Delta G (B \rightarrow C)$$
There is no absolute zero point for the energy in chemical reactions. You can only talk about differences.

What chemists did, was to define a Standard electrode potential.
With this it is possible to measure half reactions compared to this standard. Because of sloppy naming, people call this the potential of the half reaction. But strictly speaking it is just the difference to the standard electrode potential. Since differences are transitive that is not a problem when working with potentials.

This means: The oxidation can be unfavoured compared to the standard electrode potential if the reduction is even more favoured.


Why can't you achieve the equilibrium exactly in the middle (K=1) when you add enough heat/energy?

There are indeed reactions where the equilibrium constant passes through unity as a function of temperature.  The water-gas shift reaction

$\ce{CO + H2O <=> CO2 + H2}$

has an equilibrium constant that gradually decreases with increasing temperature (it's slightly exothermic as written), and the constant is close to $1$ at $\pu{800°C}$.


Determining whether a compound is ionic or molecular in nature

In principle, when studying a structure like HCl, CH3COOH, or any other compound, the first thing to do is drawing the symbol of the atoms in a reasonable geometry on a sheet of paper. Then consider the outer electron layers : put enough points around each symbol to get a reasonable Lewis structure. For example, one point near $\ce{H}$ and six points around $\ce{O}$. Then try to join atoms with their neighbors to make covalencies (or doublets). When all atoms are joined to neighbors, count the number of electrons around each atom, taking into account that the two electrons included in a covalence between A and B  are at the periphery of both atoms A and B. Check all atoms. If an atom respects the octet law, don't modify it : its structure is covalent. If an atom A does not respect octet law, try to remove one electron from its outer layer, and send this electron to the B atom, which becomes negatively charged. The covalency disappears from the region between A and B, and the bond AB becomes ionic with a positive charge on the A atom, which becomes $\ce{A+}$, and a negative charge on $\ce{B}$ which becomes $\ce{B-}$.
Using this method, you will see that $\ce{HCl}$ and $\ce{CH3COOH}$ are both made of covalencies, because all their atoms are respecting the octet rule. You will also see that $\ce{NaCl}$ may first be drawn in a covalent structure. But then the sodium atom does not fit the octet rule. It will if an electron is removed from the sodium atom, to be given to the chlorine atom, producing then $\ce{Na+}$ and $\ce{Cl-}$ ions.
Hydroxides contains two sorts of bonds. Hydroxides like $\ce{NaOH}$ are made of covalent bonds between $\ce{O}$ and $\ce{H}$ atoms, and of ionic bonds between $\ce{Na^+}$ and $\ce{O-}$.
Oxygenated acids are a special case. They may get two sorts of bonds between $\ce{A}$ and $\ce{B}$. For example sulfuric acid $\ce{H2SO4}$ can be drawn with this method to give first a covalent structure with two bonds $\ce{H-O}$, two bonds $\ce{O-S}$, and two double bonds $\ce{S=O}$. This is OK for all atoms except the central Sulfur atom, which is surrounded by six doublets or six covalences. At this point, some chemists are thinking that it is a case of hyper valence. Some think that the octet rule may be obtained by removing the second bond from each double bond from the sulfur atom, and sending it to the outer Oxygen atom. This makes a central sulfur atom charged $2+$ (respecting the octet rule) and two oxygen atoms charged $1-$. Both theories can be discussed.
The only problem with this reasoning is the fact that it does not work properly  for atoms of the third column. For example, $\ce{BF3}$ should be ionic making $\ce{B^{3+}}$ and $\ce{F-}$ ions. Indeed $\ce{BF3}$ remains covalent, and it is hard to justify. Maybe, on may admit that Boron atom prefers to stay covalent rather than to loose three electrons. Three is probably too much.


If PQ, T2U2, RS2 and VW4 have the same Ksp, which has the lowest solubility in moles of solute per litre?

Easier just to migrate my comment into an answer. Your assumption
$$\ce{[PQ]} > \ce{[VW4]}$$ is not valid based on
$$\ce{[PQ]^{2}} = 256\ce{[VW4]}^{5}$$

For $y = a^{x}$, if $a>1$, larger $x$ leads to larger $y$. But for $a<1$, larger $x$ leads to smaller $y$, and the question indicates that solubilities are between zero and one, i.e., small.

If you plug in $K_{\mathrm{sp}}= 10^{-10}$, you get $\ce{[PQ]}=10^{-5}$ and $\ce{[VW4]} = 3.3\cdot 10^{-3}$.

As I said in my comment, I believe that this is a very poorly designed question: the constraints are too broad and there is a typo in the key. Someone was very sloppy...


Trends in solubility of group 2 nitrates

Solubility is one of the present scientific problems that nobody is able to explain  thoroughly. As you are speaking of Calcium compounds, look at the series of Calcium compounds made with halogens. $\ce{CaCl2, CaBr2, CaI2}$ are so soluble in water that they can be dissolved in less than their weight of water. $\ce{CaF2}$ should be similar. It is not. On the contrary, $\ce{CaF2}$ is one of the most insoluble compounds, because it forms the mineral called fluorite, which is one of the main sources of Fluorine atoms on Earth. If it would have been at least partly soluble, it would have been washed away by the rains a long time ago. Nobody is able to explain such a difference of solubility.

There are plenty of theories explaining the solubilities of some groups of substances using electronegativities, ionic or covalent radius, and other parameters, which work pretty well with a lot of substances. But there are always exceptions. The final explanation remains to be discovered.


Mechanisms involved in iodination of salicylamide protocol

The order of events is probably:

  Protonation of the hypochlorite ion

$$\ce{ClO- + H3O+ <=> ClOH + H2O}$$

  Nucleophilic displacement of chloride by iodide

$$\ce{I- + ClOH \phantom{0}<=> Cl- + IOH}$$

  Protonation of hypoiodous acid

$$\phantom{000}\ce{IOH + H3O+ <=> IOH2+ + H2O}$$

  Fragmentation of the $\ce{IOH2+}$ cation

$$\phantom{00000}\ce{IOH2+ <=> I+ + H2O}$$

Alternatively, the second step could have been protonation of hyochlorous acid, followed by displacement of water by iodide, and then fragmentation of iodine monochloride:

$$
\begin{align}
\ce{ClOH + H3O+ &<=> ClOH2+ + H2O}\\
\ce{I- + ClOH2+ &<=> ICl + H2O}\\
\ce{ICl &<=> I+ + Cl-}
\end{align}$$


How to decide anode and cathode form electrochemical series?

If one element A is above B in the standard electrode potential series, then the one with greater oxidising power or reducing agent(ie,B) is anode. While the one with greater reducing power or oxidising agent (ie, A) is cathode.

Standard Reduction Potential Table


Given your hypothesis that the three $\mathrm{p}K_\mathrm{a}$'s are dependent on one another and that experimentally you do not seem to be able to identify which histidines are protonated and which are not at any given time, it would seem that the simplest model to use is that of a triprotic acid such as citric acid. The triple HH equation model you used is only appropriate for three completely independent $\mathrm{p}K_\mathrm{a}$'s (such that the $\mathrm{p}K_\mathrm{a}$ at site 1 is unchanged by the protonation state of site 2).

In the dependent $\mathrm{p}K_\mathrm{a}$ model, we have four species (vs the eight in your model): $\ce{H3P},$ $\ce{H2P},$ $\ce{HP}$ and $\ce{P},$ where "$\ce{P}$" is the protein. There are three "apparent $\mathrm{p}K_\mathrm{a}$'s" none of which represents a true $\mathrm{p}K_\mathrm{a}$ of a specific site. They instead correlate to the combination of possible reactions that give a change in the overall protonation state. For example, the first $\mathrm{p}K_\mathrm{a}$ represents the change in state from $\ce{H3P}$ to $\ce{H2P},$ regardless of which of the three sites loses a proton.

We can then write four equations starting with mass balance:

$$
\begin{align}
[\text{total protein}] &= [\ce{H3P}] + [\ce{H2P}] + [\ce{HP}] + [\ce{P}]\tag{1}\\
K_\mathrm{a1} &= \frac{[\ce{H}][\ce{H2P}]}{[\ce{H3P}]}\tag{2}\\
K_\mathrm{a2} &= \frac{[\ce{H}][\ce{HP}]}{[\ce{H2P}]}\tag{3}\\
K_\mathrm{a3} &= \frac{[\ce{H}][\ce{P}]}{[\ce{HP}]}\tag{4}
\end{align}
$$

We can also define the fraction of sites unprotonated as

$$\phi = \frac{3[\ce{P}]+2[\ce{HP}]+[\ce{H2P}]}{3[\text{total protein}]}\tag{5}$$

It's relatively straightforward to use the four above equations to get expressions for each of the species concentrations in terms of $[\ce{H}]$ and $[\text{total protein}]$ only. Substituting those expressions into the bottom equation will give you an equation to which to fit your data.

If you post your data in tabular form, others can play around with different fitting approaches and give feedback.


Chemical elements or compounds with yellowish white color

Aluminium(III) chloride - often described as white but samples is sometimes contaminated with iron(III) chloride giving it a light yellow color
Antimony(V) oxide - pale yellow solid
Bismuth(III) oxide - light yellow solid
Rhodium(III) oxide - lemon yellow
Cerium(IV) oxide - pale yellow-white compound
Dysprosium(III) nitrate - white to pale yellow
Holmium(III) oxide- Color depends on lighting condition. In daylight, it is tannish light yellow; Under trichromatic light, it is fiery orange red (Thanks @EdV)
Samarium(III) oxide - light yellow
Silver(I) iodide - samples can have a range of color(white to pale yellow to yellow) due to impurity and photosensitization
Selenium(IV) chloride - white to faint yellow volatile solid
Gold(III) chloride - pale yellow to yellow
Potassium Telluride - pale yellow
Thallium(III) bromide - pale yellow
Some vanadate salts are light yellow

Bromium nitrate is pale yellow but not confirmed. Silver(I) bromide and lead(II) bromide are light yellow and white cream colored respectively.
Above data source: http://lanthanumkchemistry.over-blog.com/article-the-colors-of-chemicals-complexes-104821449.html


Specific reduction of certain functional groups

1.The selective reduction of the carboxylic group is explained in this article. Link:https://www.ch.imperial.ac.uk/rzepa/blog/?p=5114. It could reduce esters but is not favourable at all and difficult.So if u only have an ester grp and its treated with bh3, then you can reduce it. However if both carboxylic and ester grps are present, carboxylic acid gets reduced .

Regarding reduction with h2/pt and h2/pd-c not much can be said. The thing is at higher temperatures and pressures(extreme conditions)both could reduce esters. Generally esters are NOT reduced via cataytic hydrogenation (it is just not favourable).However reduction has been performed in mild conditions via other complexes like ruthenium.(https://pubs.acs.org/doi/pdfplus/10.1021/op4003278


Why doesn't the bidentate ligand form dative bonding three times?

Orthocresol already pointed you towards the Jahn-Teller effect. The effect is there in $\ce{[Cu(H2O)6]^2+}$, the hexaaquacopper(II) cation, but more pronounced in the tetraammindiaquacopper(II) cation $\ce{[Cu(NH3)4(H2O)2]^2+}$. Compare an image of the complex as shown below:

Even the naked eye can see that the copper-oxygen bonds are much longer than the copper-nitrogen ones. Measuring the distances in crystal structures containing that cation gives $203~\mathrm{pm}$ for the $\ce{Cu-N}$ bonds and $251~\mathrm{pm}$ for $\ce{Cu-O}$.[1] Why is this the case? To answer that we need to refer to the molecular orbital scheme of a typical octahedral complex:[2]

The three lowest groups of orbitals containing six MOs in total are typically populated by the ligand lone pairs which form the coordinate bond. The next set of two groups — five orbitals in total — are the metal d-orbitals that can (or cannot, in case of $\mathrm{t_{2g}}$) interact with the ligands. For copper(II), we need to fill in nine metal d-electrons meaning that the $\mathrm{t_{2g}}$ level is fully populated while $\mathrm{e_g}$ has three electrons in two orbitals — a denatured state.

This denatured state is not energy-optimal. Energy could be released if we somehow manage to unequalise the $\mathrm{e_g}$ orbitals, give them different energies and have the lower-energetic one be fully populated. Remember that the $\mathrm{e_g}$ orbitals are $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$. In a gedankenexperiment, remove the two ligands in $z$-direction symmetrically away from the metal centre. Any orbital that has a $z$ contribution should now be stabilised — most notably, that includes $\mathrm{d}_{z^2}$. The orbital scheme looks slightly different, the complex’ symmetry is no longer $O_\mathrm{h}$ but $D_\mathrm{4h}$ but we have gained energy by being able to differentiate between the $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$ orbitals. This is the Jahn-Teller effect.

Consequences of the effect include that ammin ligands preferentially coordinate copper(II) equatorially, and that the tetraammindiaqua complex is the preferred one in high aquaeous ammonia concentrations. Hexaammincopper(II) can be prepared only in liquid ammonia to the best of my knowledge.

Now let’s consider using $\ce{en}$ (ethylenediamine) rather than $\ce{NH3}$. The bridge between the two amino groups is long enough to bridge across one edge of the octahedron but not any further. Replacing all the equatorial ligands with $\ce{en}$ makes sense because we will get another very symmetric complex $\ce{[Cu(en)2(H2O)2]^2+}$, and because these are the ones that are replaced first. If we now wanted to replace the two other water molecules, we would create a rather distorted $\ce{[Cu(en)3]^2+}$ complex, where two $\ce{en}$ molecules have one short, strong and one long, weak bond. This is an unfavourable condition, since the former shorter, stronger bonds have to be weakened to get there. In short, $\ce{en}$ behaves like ammonia and only occupies the equatorial positions, not the axial ones. A consequence again of Jahn-Teller.

[1]: Data taken from Professor Klüfers’ internet scriptum for his general and inorganic chemistry course at the LMU Munich (section 23.4).
[2]: First presented in this answer and originally taken from Professor Klüfers’ coordination chemistry course.


Is there a way to convert aniline to phenol without diazotizing it?

Benzenediazonium salt is so versatile that it can be used to make many compounds. That's why, diazotization reaction is so popular and useful. The reaction conditions is not that harsh and is easily achievable. Coming to the reaction, diazotization is a straight one-way procedure to achieve aniline to phenol conversion. If you talk about other ways of converting aniline to phenol, the reaction conditions for such ways are generally harsh and hence not done practically to obtain phenol. In addition, there are so many steps, that the conversion is practically not feasible.
One of the ways of converting aniline to phenol without diazotizing is to convert aniline to acetanilide and then hydrogenolyze it to make benzene which can be then used to make phenol.

Step 1

Step 2

The reaction is taken from a paper1. The paper discussed the use of $\ce{NiMo}$ based catalyst along with novel neo-microporous composite material HSM/ZSM-5 as support. The reaction result is compared with other supports like $\ce{\gamma-Al2O3}$ and lone ZSM-5 and HSM system.

Step 3

Reference

Sajedeh Jafarian, Ahmad Tavasoli, Hasan Nikkhah, "Catalytic hydrotreating of pyro-oil derived from green microalgae spirulina the (Arthrospira) plantensis over $\ce{NiMo}$ catalysts impregnated over a novel hybrid support," International Journal of Hydrogen Energy 2019, 44(36), 19855-19867 (https://doi.org/10.1016/j.ijhydene.2019.05.182) 44. 10.1016/j.ijhydene.2019.05.182(PDF).


Delta G for Standard Hydrogen Electrode reaction

You cannot apply the $\Delta{G}$ equation to a single electrode potential. It can be applied to a cell though so if the hydrogen electrode is connected to another electrode (say copper dipped in copper sulfate solution) then you can find the free energy.

It's better to remember the formula as:

$\Delta{G} = -nFE_{cell}^o$

Where

$E_{cell}^o = E_{cathode}^o - E_{anode}^o$ (using all standard reduction potentials)

That way you won't accidentally apply it anywhere else.

In this case, for $\ce{\Delta G}$ to be zero you would have to take Hydrogen electrodes for cathode as well as anode.

A More Complicated Explanation

Just in case you're curious about the reasons behind this (Since I was also very confused about this when I first learned Electrochemistry. It's ok if you don't understand this),

It is very difficult to measure the absolute potential of a single electrode and so the standard reduction potentials you are using are actually based on a reference system.

This is actually very similar to the system of atomic masses. The atomic masses you use in calculations are not the actual masses of the elements in grams but they are measured in unified atomic mass units. One unified atomic mass unit is defined as 1/12th the mass of a carbon-12 atom.

[Note: Now of course, after the mole was made a fundamental SI unit, the unified atomic mass unit has become equivalent to $1 g/mol$. This is just for the sake of explanation.]

So if you come across a situation where you have to use the mass of the atom (maybe some nuclear chemistry calculations) you can't use those normal atomic masses but you would have to convert it into kg (One unified atomic mass unit is about $1.6\times10^{-27}$ kg).

Similarly, the standard reduction potentials are actually measured with reference to a standard electrode (Whose reduction potential is arbitrarily taken to be zero) which is the standard hydrogen electrode.

This standard reduction potential of an electrode can be measured easily by connecting the electrode to a standard hydrogen electrode (SHE) and then measuring the potential difference using a voltmeter and applying the $E_{cell}^o$ equation. (If $H^+$ is being reduced to $H_2$, then $E_{cathode}^o$ is zero otherwise $E_{anode}^o$ is zero).

So this just means that when you say $E^o$ of the SHE is zero, you are actually saying that "the potential of the SHE with respect to itself is zero" which is quite obvious. That doesn't mean that the potential of the electrode itself is zero.

That is why you cannot apply the first equation in this case. But this does not apply to cells. If we assume the absolute potential of the SHE to be some $x$, then,

$E_{abs}^o = E^o + x$

Then,
$E^o_{cell} = E_{abs}(cathode) - E_{abs}(anode)$

$= E^o_{cathode} + x - [E^o_{anode} + x]$

$E^o_{cell} = E_{cathode}^o - E_{anode}^o$

The absolute potential of the SHE cancels out in the difference.


Firstly, some reactions have a very low probability, secondly, the probability depends on temperature (activation energy!), and thirdly, diffusion is not exacly a fast process, and takes time even in a gas phase. Once the reaction progresses, the remaining educt particles have an increasingly hard time finding each other even if you thoroughly mixed everything at the beginning.


Can every conductor be electroplated using copper and sulphuric acid?

While you can use basically any conductive material as an electrode to reduce the metal ions in solution, to actually plate the electrode with the metal, the metal has to be able to attach to the surface. If you have contaminants like surfactants, it can prevent deposition. The trouble with using aluminum as a cathode is that it readily forms an oxide layer that can be difficult to stick things to. I understand that it's possible to electroplate copper onto aluminum, but it's usually done with copper cyanide baths, so you may find it easier to plate copper onto steel or nickel. The other issue is that reduction of the copper ion is not the only possible reaction at the cathode. If your voltage is too negative, you may get significant hydrogen evolution and the bubbles will prevent a solid film from adhering properly. (which looks like it may be the problem in your photo) What you want is to deposit very slowly by keeping your voltage down so that you minimize gas evolution, though you may still have trouble with an aluminum cathode.

As to the sorts of things that can be electroplated, many metals have similar behaviour where it's possible to produce quite high quality films of a second metal over top. Other metals like mercury and bismuth tend to form alloys or intermetallic compounds. It is certainly possible to deposit thin metal films on materials like indium tin oxide, graphite or glassy carbon, thought adherence can be a problem.


I hope this is not a homework question. Suppose your polymer has $n$ reapeating units and capped with methyl groups at the end as you described. Thus, it should looks like following figure:

Thus it has $n$ $\ce{(-CH2CH2-)}$ units and $2$ $\ce{(-CH3)}$ units. Thus your $\ce{(-O-CH2-)}$ signal is accounted for $4n$ protons while $\ce{(-O-CH3)}$ signal accounted for only $6$ protons. Thus,
$$\frac{4n}{6} = \frac{20}{1}$$
You may find the value of $n$ from this equation and you can calculate the molecular weight of the polymer in hand.


Why is hydrochloric acid often sold as 33% concentration?

The user ssavec prompted me to research further on my own and post an answer.

It seems the limits are practical ones.

For hydrochloric acid — $\ce{HCl}$ a practical limit is 38% with absolute
limit around $40~\%$ and commercial concentrations ranging from 30 to 35%
for convenience in transportation infrastructure.
For nitric acid — $\ce{HNO3}$ the $68~\%$ limit is the maximum that can be
reached by direct distillation, higher concentration up to $98~\%$
(fuming nitric acid) needs alternative dehydration or synthesis techniques.
For sulfuric acid — $\ce{H2SO4}$, which can be distilled to $98.3~\%$ (and can be used
to concentrate nitric acid) it is sold at concentrations that are
directly suited for the intended use ranging from $20~\%$ to $80~\%$. With concentrations around $30~\%$ common for lead-acid battery electrolyte preparation.

So it seems the limits are all practical but for different reasons, one was close to the absolute limit for efficient transport.  The second was at the direct distillation limits and the last one was to suit the most popular consumer applications.

The spread in the numbers had caused me to wonder on their relation but it seems there was no direct relation to the concentrations we see for consumer sale.


Radicals are electron-deficient compounds. Therefore, according to Bent’s rule a radical should always populate an atomic orbital with a maximised $\mathrm p$ contribution. Ignoring 2 and 4 for a second, the radical occupies an $\mathrm{sp^2}$ orbital in 3 and 5 while it occupies an $\mathrm{sp^3}$ orbital in 1.[1] This already gives us $3\approx5<1$.

In a next step, let’s consider 2 and 4. These, as you noticed, are adjacent to a π bond, so the radical can occupy a $\mathrm p$ type orbital and participate in π resonance. This is the main way to differentiate 1, 2 and 4, since all three could essentially adopt a planar configuration with the radical confined to a $\mathrm p$ orbital — yet in 2 it can also resonance with the allyl system and in 4 it can resonate across two double bonds. Thus, we now intermediately have $3\approx5<1<2<4$.

All that remains is to determine the difference between 3 and 5, which is not trivial. In 3, you have a single double bond. This structure can distort a little bit, increasing the bond angle on the radical-bearing carbon towards something closer to $\mathrm{sp}$. The freely rotatable single bonds making up most of the ring allow this distortion. The phenyl radical can distort much less, since its geometry is basically already defined from the six-membered aromatic ring. Thus, we may assume 3 to be more stable than 5. Therefore: $5<3<1<2<4$.

Note:

[1]: In fact, rehybridisation is also possible in 1, giving the radical a pure $\mathrm p$ orbital and making the arrangement of substituents planar around the radical centre.


If double and triple bonds are stronger than single bonds then why are unsaturated hydrocarbons less stable than saturated hydrocarbons?

When I was a first year, this concept was hard for me to grasp as well. The idea is that, even though the bond itself (double or triple) is stronger than a single bond as a whole, we must consider the components of a double or triple bond. A double bond consists of one sigma and one pi bond.

Let's take ethene for example. The $\ce{C=C}$ double bond has one sigma and one pi bond. The sigma exists because of two, sp2 hybridized orbitals overlapping head-to-head. This type of covalent bonding is very strong. But since it is only sp2, there is one p orbital leftover that was not hybridized. So, each carbon takes their left over p orbital, and forms a pi bond from the p orbitals overlapping side-to-side (the p orbitals are parallel). The unhybridized p orbitals are perpendicular to the sp2 orbitals.

Pi bonds, because of the way they are constructed, are much weaker than a sigma bond due to side-to-side overlap, versus head-to-head. Having said, that pi bond can be easily cleaved by an incoming group. This properties of the pi bond is what makes an unsaturated hydrocarbon so much more reactive and unstable compared to a saturated hydrocarbon.


Can someone intuitively explain the reason for the units of entropy (J/K )?

The units of of energy over temperature (e.g. J/K) used for entropy in the thermodynamic definition follow from a historical association with heat transfer under temperature gradients, in other words, the definitions of temperature and entropy are intertwined, with entropy being the more fundamental property. Entropy can be thought of as a potential and temperature (or its inverse, rather) as a generalized force associated with displacements along energy dimensions in the entropy potential. The (inverse) temperature is a measure of the effect of changes in the amount of energy on the entropy of the system, as the following definition suggests:

$$\frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)_V$$

However if you start from a statistical mechanical definition of the entropy,

$$S=k_\mathrm{B}\log\Omega$$

then it is easy to see that a unitless definition might be just as well suited:

$$S_{\mathrm{unitless}}=\log\Omega$$

However, in the absence of the Boltzmann constant you need some other way to relate entropy and energy (e.g. heat transfer) in our conventional system of units. You can of course subsume $k_\mathrm{B}$ into the definition of a new temperature scale:

$$\frac{1}{T_\mathrm{new}}=\left(\frac{\partial S_\mathrm{unitless}}{\partial U}\right)_V$$

where the old and new temperature scales are related as $T_\mathrm{new}=k_\mathrm{B}T$. The new temperature scale then has units of energy.


Lost track of the hydrogen in step 3 of the Krebs cycle

Your problem stems from the statement
"I would say this process entails the loss of two hydrogens from Isocitrate, and the gain of those hydrogens in NADH/H+."
This statement is incorrect. In isocitrate, the carboxylic acid functional groups are deprotonated, as would be the case at neutral pH. The oxidation reaction thus involves loss of only $\ce{H-}$ to NAD. There is no loss of $\ce{H+}$, because that has already been removed.
If you start with isocitric acid, which is isocitrate in its neutral form with $\ce{-CO2H}$ instead of $\ce{-CO2-}$, then you will need to remove $\ce{H-}$ and $\ce{H+}$ to accomplish the reaction and your above statement will be correct.
The textbook image you showed takes a confusing approach of implicitly associating an $\ce{H+}$ with isocitrate even though it is not actually on the molecule, so that "NADH/H+" is indicated as a product even though only NADH has actually been produced in the reaction shown.


How's the IUPAC name of basketane determined?

Validation
Let us build this alien step by step:

We will build bicyclo[$4$.$4$.$0$]decane.
We will build tricyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$]decane.
We will build tetracyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$]decane.
We will build pentacyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$.$\mathrm{0^{4,7}}$]decane.

Bicyclo[$4$.$4$.$0$]decane looks like this:

The part "decane" means that there are $10$ carbons and the molecule is saturated, just as indicated.
The part "bicyclo" means that there are two rings. One ring is $1$-$2$-$3$-$4$-$5$-$6$ and the other is $1$-$6$-$7$-$8$-$9$-$10$.
The part [$4$.$4$.$0$] is based on $\ce{C}1$ and $\ce{C}6$:

The first $4$ refers to the first link between $\ce{C}1$ and $\ce{C}6$, namely $2$-$3$-$4$-$5$ which contains $4$ carbons.
The second $4$ refers to the second link between $\ce{C}1$ and $\ce{C}6$, namely $7$-$8$-$9$-$10$ which contains $4$ carbons
The $0$ refers to the third link between $\ce{C}1$ and $\ce{C}6$ which is just a line with $0$ carbons between.

Tricyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$]decane looks like this:

It is now "tricyclo" because there are three rings: $1$-$2$-$5$-$6$, $2$-$3$-$4$-$5$, $1$-$6$-$7$-$8$-$9$-$10$.
The extra part $0^{2,5}$ means that $\ce{C}2$ and $\ce{C}5$ are linked by $0$ extra carbons.

Tetracyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$]decane looks like this:

(from now on, a strong intuition on topology is required)
The new part $\mathrm{0^{3,8}}$ refers to the new linkage between $\ce{C}3$ and $\ce{C}8$ with $0$ carbons between.

Finally, pentacyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$.$\mathrm{0^{4,7}}$]decane:

Now you might ask: how on earth is that basketane? Let me show you how:

Not convinced? Here is the proof:

$1$ is connected to $2$, $6$, $10$.
$2$ is connected to $1$, $3$, $5$.
$3$ is connected to $2$, $4$, $8$.
$4$ is connected to $3$, $5$, $7$.
$5$ is connected to $2$, $4$, $6$.
$6$ is connected to $1$, $5$, $7$.
$7$ is connected to $4$, $6$, $8$.
$8$ is connected to $3$, $7$, $9$.
$9$ is connected to $8$, $10$.
$10$ is connected to $1$, $9$.

In the above two pictures, you can verify the connectivities, and they match exactly.

Derivation
Firstly, we try to find the largest ring that can contain the most carbon atoms. In this case, it is already indicated above (trace the ring $1$-$2$-$3$-$4$-$5$-$6$-$7$-$8$-$9$-$10$).
Then, you might ask, "How do we judge which numbering is correct?"
Curiously, the 1993 IUPAC Recommendations do not contain the criteria, but the 1979 IUPAC Recommendations do. Rule A-$32.31$, Nomenclature of Organic Chemistry, Sections A, B, C, D, E, F, and H. Pergamon Press, Oxford, 1979. Copyright 1979 IUPAC.:

$32.31$ - When there is a choice, the following criteria are considered in turn until a decision is made:
(a) The main ring shall contain as many carbon atoms as possible, two of which must serve as bridgeheads for the main bridge.
(b) The main bridge shall be as large as possible.
(c) The main ring shall be divided as symmetrically as possible by the main bridge.
(d) The superscripts locating the other bridges shall be as small as possible (in the sense indicated in Rule A-$2.2$).

Let us look at the first rule:

(a) The main ring shall contain as many carbon atoms as possible, two of which must serve as bridgeheads for the main bridge.

This means that we should maximize the sum of the first two numbers and minimize the sum of the other numbers.
Also, this means that we need to choose two carbons which are each connected to three carbon atoms.
For example, this would be wrong:

Incorrect name: pentacyclo[$\color{red}{4}$.$\color{orange}{2}$.$\color{green}{2}$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$.$\mathrm{0^{4,7}}$]
This name is incorrect because the first two numbers, $\color{red}{4}$ and $\color{orange}{2}$, add up to $6$, which is smaller than $8$ as derived from the correct name.

Let us look at the second rule:

(b) The main bridge shall be as large as possible.

The "main bridge" refers to the third number, which is to be maximized.
This is not applicable here, as every number beyond the third is zero.

Let us look at the third rule:

(c) The main ring shall be divided as symmetrically as possible by the main bridge.

This means, that although the first number is to be larger than the second, the difference between the first two numbers is to be minimized.
For example, this would be wrong:

Incorrect name: pentacyclo[$\color{red}{6}$.$\color{orange}{2}$.$\color{green}{0}$.$\mathrm{0^{2,7}}$.$\mathrm{0^{3,10}}$.$\mathrm{0^{6,9}}$]
This name is incorrect because the first two numbers, $\color{red}{6}$ and $\color{orange}{2}$, have a difference of $4$, while the difference between the first two numbers in the correct name is $0$.

Let us look at the fourth rule:

(d) The superscripts locating the other bridges shall be as small as possible (in the sense indicated in Rule A-$2.2$).

This is what A-$2.2$ says:

[...] When series of locants containing the same number of terms are compared term by term, that series is "lowest" which contains the lowest number on the occasion of the first difference. [...]

For example, this would be wrong:

Incorrect name: pentacyclo[$\color{red}{4}$.$\color{orange}{4}$.$\color{green}{0}$.$\mathrm{0^{2,9}}$.$\mathrm{0^{5,8}}$.$\mathrm{0^{7,10}}$]
Let us compare the locants in the incorrect version and the locants in the correct version:

Incorrect: $2,9,5,8,7,10$
Correct: $2,5,3,8,4,7$

The first difference is $9$ vs $5$, in which the correct version is smaller.

Conclusion
Therefore, we have arrived at the correct systemic IUPAC name for basketane:

Pentacyclo[$\color{red}{4}$.$\color{orange}{4}$.$\color{green}{0}$.$\color{purple}{\mathrm{0^{2,5}}}$.$\color{blue}{\mathrm{0^{3,8}}}$.$\color{pink}{\mathrm{0^{4,7}}}$]decane.


Reactivity order of electrophilic addition

For the electrophilic addition to carbondouble bonds, the $\pi$- electron cloud double bonds first attack the electrophile and then a carbocation is created adjacent to the carbon where the electrophile is added. Then that carbocation is attacked by the nucleophile to form the addition product. $$\ce{R-CH=CH2 + E -Nu -> R-CH^+-CH2-E -> R-CH(Nu)-CH2E}$$The above path is the general electrophillic addition reaction in the case of alkenes. The carbocation position may vary depending upon the $\ce{R}$ group.

When the $\ce{R}$ is $\ce{CH3-O}$, there is a positive mesomeric effect (+M or +R effect) of the group due to which  when the carbocation is formed in the adjacent carbon, it gets  high stabilisation due to the +R effect, which in turn decreases the energy of the transition state of carbocation heavily, and thus the reaction rate becomes very high. That's why, the rate of electrophilic addition is very high in this case is very high, i.e it is the most easy.
When the $\ce{R}$ is $\ce{F}$, there is a significantly stronger +R effect due to $\ce{2p\pi -2p\pi}$ overlap. So, the carbocation gets a slight stabilisation due to this +R effect but there is also a -I effect in case of $\ce{F}$ which is the factor of destabilisation of carbocation. That's why, ease is relatively lower due to significant +R but stronger -I effect.
When the $\ce{R}$ is $\ce{Cl}$, there is a very weak +R effect (vinyl chloride has a weak  effect of +R, which is observed from the lesser dipole moment due to lesser charge separation  than ethyl chloride, where there is no +R), but dominant -I effect. So, in this case carbocation is more destabilised. So, the ease of addition is decreased even further.
When the $\ce{R}$ is $\ce{NO2}$, there is -I effect and also a dominant -R effect or -M effect. So the $\pi$ -electron cloud gets delocalised with the nitro group and thus due to decreased $\pi$-electron cloud density, it becomes difficult for it to even attack the electrophile. So, fractions of molecules forming the carbocation is extremely less, which actually increases the activation energy of the reaction. Thus, the reaction becomes very difficult with nitroethylene.

Consideering the overall effects described above, the ease of electrophilic addition should be,$$\ce{CH3 -O - CH=CH2 > F - CH=CH2 > Cl - CH=CH2 > NO2 - CH=CH2 }$$


In ESI-MS, why do we assume that the positive charge on the analyte must come from protonation from the solvent?

You might be asking two questions here.

(1) Can the high electric field strength induce electron detachment in the vapor? No. Normal electrospray voltages are not high enough to do so. Air, for instance, has a breakdown voltage of around 30 kV/cm, and liquids and aerosol particles are ionized only at even higher electric field strengths. Electrospray voltages are around one-tenth of that.

(2) Can the positive potential at the capillary walls induce oxidation of $M$ to $M^+$? In principle, yes, but only if there are no other species present with higher oxidation potentials than $M$ -- such as the metal walls of the capillary.


You could remove the bulk of the sodium chloride (and most other salts) by evaporation and crystallization. Most of the impurities you are looking for are going to stay in solution or suspension when you filter off the salts.

You could also use dialysis to remove the remaining salts: this requires a bag or length of thin-walled tubing made of a common polymer such as cellulose acetate. Tie off one end (the bottom), put the concentrate from the above step into the tubing, and tie off the top end. Then place it in a bucket of distilled water. The salts and other small molecules will diffuse out into the water, whereas the higher molecular weight materials like toxins will stay in the bag. You may need to change the water in the bucket a few times to get rid of all the salts.

You will still need some chemistry to figure out what substances remain in the bag. With seawater, it could be a very complex mix. However, looking at it with a fluorescent light might show something interesting. Good luck, and hope you get inspired to study chemistry!


$1$)
Let's admit first that the reaction rate is proportional to the amount of reactants, in the usual way. Whatever the order $1$ or $2$ of the reaction, the rate constant ${k_{1} }$ or ${k_2}$ will not change if the amount of reactant is increased. But the rate of reaction ${r = - dn/dt}$ does increase when the amount of reactant ${n}$ increases, whatever the order.
For $1$st order, the rate is : ${r = - dn/dt = k_1 n}$. So the rate ${r}$ increases by increasing the amount of reactant ${n}$.
For $2$nd order, the rate of rection is : ${r = - dn/dt = k_2 n^2}$. Here too, the rate of reaction increases when the amount of reactant $n$ increases.
$2)$ As the reactant is a solid, the reaction rate may not be proportional to the amount of reactant (in mole). Let's suppose that it is proportional to the surface of the grains. Even so, the surface increases by increasing the amount of powder. As a consequence, the reaction rate does also increase.


What kind of 'product' is the 'product' in 'ionic product of water'?

Per Zhe's comment, in the specific context of the term ionic product of water, the word product is used in the sense of the result of the mathematical operation of multiplication:

  The ionic product of water equals the concentration of $\ce{H+}$ times the concentration of $\ce{OH-}$:
  $$K_\mathrm w = \ce{[H+][OH-]}$$

(Strictly, per MaxReid's comment, the ionic product water is of course the product of the activities of $\ce{H+}$ and $\ce{OH-}$, but concentrations are commonly used in routine/non-rigorous calculations as they provide good accuracy for dilute solutions.)

This sense of the word product applies any time one is referencing the mathematical description of the equilibrium among the ionic forms of a given species. This usage also includes the solubility product (again with the above caveat about activities):

  The solubility product of silver chloride equals the concentration of $\ce{Ag+}$ times the concentration of $\ce{Cl-}$:
  $$K_\mathrm{sp} = \ce{[Ag+][Cl-]}$$

As an aside, note that the main exception to this terminology is when the expression describing the distribution of species requires division by one or more quantities. In this case, the term to use is quotient, for which I expect little ambiguity exists:

  The reaction quotient of the reaction
  $$\ce{A + B->C + D}$$
  is
  $$ Q = {\ce{[C][D]}\over \ce{[A][B]}}$$

However, to be complete: chemistry does also use the word product in the other sense you describe, as in something being produced by a reaction.  Due to this, the word product in the phrase the ionic product(s) of X could also bear this meaning.  For example:

  The ionic products of the aqueous decomposition of dinitrogen tetroxide at neutral $\mathrm{pH}$ are nitrite and nitrate ion:
  $$\ce{N2O4(aq) + 2H2O -> NO2-(aq) + NO3-(aq) + 2H3O+}$$

In general, I think the difference boils down to whether one is talking about the ionic distribution of a specific species of interest, as opposed to the transformation that occurs in a particular reaction of interest.

As I noted above, though: In the specific case you are asking about, product is used in the sense of the result obtained from multiplication.


How to measure chlorine content using ORP sensor?

For a school project like yours, you can use iodometric titration. Iodometric titration is one of standard methods of water examination. The amount of free chlorine in tap water is really small on the order of < 1 parts per million therefore you need a lot of sample. One would typically take 500 mL of tap water (or more), add potassium iodide, and titrate with 0.01 M sodium thiosulfate. You can use an ORP and plot a potentiometric titration curve.
Consult your teacher for this option. He/she should explain to you the details.
There are spectrophotometric methods but the reagents may not be available in your school.


Why HF plus H2O is not a buffer solution?

What are you talking about?
$\ce{HF}$ in water solution is acting like moderately strong acid. It's because:

$$\ce{HF + H2O \rightleftharpoons H3O+ + F-}$$
$$\ce{H3O+ + F- \rightleftharpoons [H3O]F}$$
which is undissociated.Then:
$$\ce{F- + HF \rightleftharpoons HF2-}$$

As you can see the large affinity of $\ce{F^-}$ to $\ce{H^+}$ won't allow it to dissociate.
The $K_a(HF)(6.6x10^{-4})$ is strongly favoring reactants.
So the $\ce{HF}$ in $\ce{H2O}$ isn't buffer because there isn't almost any $\ce{F-}$ to be considered as conjugated acid. So if you add some acid there will not be enough $\ce{F-}$ for it's neutralisation.
So if you add some acid or base it will react with water and equilibrium constant will be changed and that will lead to drastic change of pH.

So the buffer is solution which consist of:

salt of strong acid and weak base and the weak base (for example $\ce{NH_{4}Cl}$ and $\ce{HN3}$ ).
salt of weak acid and strong base and the weak acit (for example  $\ce{CH3COONa}$ and $\ce{CH3COOH}$).

Hope i helped you. If you have any other questions please ask.


Converting benzoic acid into meta-fluorobenzoic acid

Two steps should be enough ;-)

Nitrate benzoic acid. In J. Chem. Educ., 1967, 44, 464
(DOI), the authors added benzoic acid to fuming nitric acid at 0 °C. 3-nitrobenzoic acid is the major isomer (~ 75%). Recrystallize from ethanol.

Your product is an electron-deficient arene and $\ce{-NO2}$ is a decent nucleofuge. Try to replace it by fluorine in a nucleophilic aromatic substitution!

In the case of 4-nitrobenzoic acid, the transformation to 4-fluorobenzoic acid has been described to work with a yield of >95% in a room temperature reaction using DMSO as a solvent and anhydrous tretrabutylammonium fluoride (TBAF) as the reagent. (DOI)


Why did the red colour in a flame test for strontium disappear and appear again on longer heating?

Good, careful observation! Here's my assumption of what happened:

The $\ce{SrCl2}$ solution on the wire boiled violently, releasing a large amount of the salt into the flame.
Some remaining anhydrous $\ce{SrCl2}$ stuck to the nichrome after all water boiled off, and melted when the wire reached ~$\pu{874^\circ C}$, coating the loop with a thin layer of the salt.
Additional heating, after melting the salt, went into raising its temperature until it approached the boiling point, ~$\pu{1250 ^\circ C}$. Nichrome can withstand ~$\pu{1400 ^\circ C}$ before melting, but you don't even need to reach the b.p of $\ce{SrCl2}$ to have a bit evaporate and color the flame. Ergo, the flame turned red again.

By the way, this implies you could melt $\ce{SrCl2}$ in a ceramic crucible, insert two nichrome wires connected to a battery, and electrolyze the molten salt to see a bit of strontium (which will catch fire as soon as it rises to the surface, with its characteristic flame color). Do this only under a teacher's supervision.


What is the difference between 'particles', 'molecules' and 'chemical species'?

A chemical species consists of chemically identical atoms, molecules or ions (more generally molecular entities) and is identified as a single species in a spectroscopic experiment. The latter part of the definition is important, because, for example, a compound which is identified as a single chemical species by NMR spectroscopy can still contain different enantiomers (that are separate molecular entities) which cannot be distinguished by an achiral spectroscopic method.

A substance is matter which has a specific and distinguishable chemical composition and specific properties. From this follows that every pure element and chemical compound is a substance. For example, pure $\ce{Fe}$ and $\ce{CH4}$ are substances, while an aqueous solution of $\ce{NaCl}$ and gasoline are not. The latter two are a mixture of two ($\ce{NaCl}$ and $\ce{H2O}$) or more different substances and have therefore properties which depend on the composition.

Particles can be defined as microscopic constituents of matter, like atoms or nuclei. The term can also refer to even smaller constituents like electrons, protons, or neutrons, which are called elementary particles. It is also used in the definition of an atom, the smallest particle which is characteristic for a chemical element. Atoms with the same number of protons $Z$ (and equal number of electrons) represent the same chemical element. Isotopes are atoms of the same chemical element which have different mass numbers, i.e. the same number of protons $Z$, but different numbers of neutrons in the nucleus. In the three isotopes of hydrogen, for example, the nuclei contain different number of neutrons (0, 1 and 2).

A salt is a chemical compound which consists of cations and anions. It should be distinguished from a zwitterionic compound, which contains charged but overall neutral molecules as building blocks. An example for the latter is trimethylamine N-oxide.


Heat up the soft tubing to make it easier to fit over the connection. Most labs have a heat gun that is used for this (amongst other uses). A hairdryer will serve, or put the end of the tubing in hot water. Many glass connection have what is called an olive which you have to push the tubing over while hot, when it cools it contracts giving a good seal (particularly important for the water lines of condensers which may be left overnight)


The "rules" for LCAOs in Molecular Orbital Theory

You are right to be questioning the validity of this method, and I congratulate you for doing so. This is actually an extremely important skill, that differentiates the best students from the rest.

The variational method, like many methods from physical chemistry, is a method of approximation (a model) to what really happens. There exist a whole hierarchy of methods for computing the molecular orbitals of molecules (which are themselves models, being the stationary states of the Schrödinger equation), including at the top end post-Hartree-Fock theory and density functional theory. These methods provide quantitative information about molecular orbitals, but also require serious computer power (which is generally unavailable to undergraduate students). However, all of the basic physics can be explored and understood with simpler models that can be solved on a few sheets of paper, such as LCAO theory, hence why we teach them to undergraduates.

As regards to your question, we do indeed optimise a linear combination. We start off by assuming that the total molecular orbital wavefunction can be approximated using a linear set of atomic orbitals:

$$|\Psi\rangle = c_1|\phi_1\rangle + c_2|\phi_2\rangle + \cdots + c_n|\phi_n\rangle$$

We then need to find the coefficients. The variational principle (also known as the Rayleigh-Ritz method) states that the coefficients that give the best approximation to the wavefunction will minimize the energy, given by

$$\mathcal{E} = \frac{\langle\Psi|\hat{H}|\Psi\rangle}{\langle\Psi|\Psi\rangle}$$

Now, the second part of your question involves how to compute the two terms in this fraction. Without going into massive mathematical detail, the LCAO method can be recast into a matrix problem rather than an integral problem using the atomic orbitals as a form of basis vector. In the simplest case (Hückel theory) we assume that they are normalized, such that the denominator is always 1 or 0 (see later). The problem now is mostly how to determine the numerator.

In brief, each element of the Hamiltonian matrix is given by

$$H_{ij} = \langle\phi_i|\hat{H}|\phi_j\rangle$$

The variational principle applied to this matrix implies that the optimized energies are the eigenvalues of the Hamiltonian matrix, and the coefficients are given by the eigenvectors. Since an $n$ x $n$ matrix has precisely $n$ eigenvalues, this implies the "number of molecular orbitals" rule. Notice that the reason for the +/- combinations is by our own design (LCAO). We could have picked more complex trial functions by assumption, but this would make computation far more difficult.

It most certainly is true that in order for there to be a significant interaction, two orbitals must be close in energy. The detailed reasons are complex, but essentially it comes down to the size of the $\langle\phi_i|\hat{H}|\phi_j\rangle$. Orbitals that are far apart in energy have small values of this term hence interact weakly. This doesn't necessarily mean they can't form molecular orbitals, however, but these effects are negligible and are not important in understanding the chemistry, so are neglected in simple models (although are often included in some of the most complex modern methods).

The reason that only atomic orbitals of the same symmetry give molecular orbitals is the overlap integral $\langle\phi|\phi\rangle$. This is the total sum (integral) of the product of the AOs. This is zero for different symmetries. This must be non-zero to give an MO. Consider a $\mathrm{2p_z}$-orbital and a $\mathrm{1s}$-orbital for instance (for illustration purposes only!). The s-orbital is spherically symmetric, with all points the same sign. The $\mathrm{p_z}$-orbital has a "dumbbell" shape, with equal areas of different signs above and below the $xy$-plane. Therefore, the total sum of the product is zero. No overlap = no interaction.


Why is propanoic acid weaker than butanoic acid?

As a trend is usually a linear fit through experimental values it does neither mean that there are no minor variations between the values nor that there is a constant rise between values but only that one can see a difference between earlier and later values of that measured property.

And besides all the answers on the actual influence of the longer alkyl chain on the acidity of the carboxylic group there are no big differences between the acidities for those small alkanoic acids between chain lengths of two to ten carbons.

However, there is a small overall trend $\left(+0.02~\mathrm{pK_a}\right)$ which can be seen in the following graph.

So one is a little bit stronger than the other but it is not very much.

One could start to argue why exactly those two differ "so much" in strength but looking at those nine values above and the "noise" around the trend I guess that makes no sense.


Melting points of benzoic and stearic acids

Puzzles are more puzzling when the pieces almost fit together, but for some unexplained reason, just don't fit. In other words, a comparison should be made between two almost  (but not quite) similar things.
In this case, stearic and benzoic acids are quite different. They have some carbons (quite different, and not even the same number!), and one similar group of four atoms ($-COOH$) - not very much similarity.
How about we compare the melting points of octadecane and stearic acid - and their structures? There's a lot of similarity there, and a little dissimilarity. Octadecane melts at 28-30 C (Ref. 1) vs. stearic acid at 69.3 C (Ref. 2). Obviously the difference is caused by the added stickiness of the carboxyl groups for one another, because they form dimers in the crystal, a structural feature (Ref. 3) lacking in the paraffin.

That establishes that a carboxyl group raises the melting point of a paraffin chain. An x-ray structure for $C_{24}H_{50}$ (octadecane not available) looks just like the one for stearic acid, but without the carboxyls (Ref. 4).

Now let's compare benzoic acid with toluene. The numbers of carbons are the same, and we just convert one $CH_3$ group to a $COOH$, as in the previous comparison. Toluene stacks in the solid (Ref. 5) with energies about 3-4 kcal/mol, which isn't much, considering that toluene melts at -95 C. But benzoic acid stacks nicely (Ref. 6) and melts at 122 C.

Now there's a puzzle: why does a carboxyl group raise the mp of a paraffin by 40 degrees, but raises the mp of an aromatic ring by 217 degrees? Looking at the formulas, looking at the drawings of a molecule (or two) doesn't help. Looking at actual data (because those little molecules have a mind of their own, and we have a hard time thinking like they do) like x-ray crystal data, suggests that putting a little planar group ($COOH)$ onto a long chain might strengthen intermolecular bonding a little, but extending a ring's planarity and adding in some hydrogen bonding, which amplifies the extent of planarity could do some powerful intermolecular stickiness.
I guess the wiggly straight chains don't have such good contact with their neighbors as the benzoic acid dimers do. Who would have guessed, without the crystal structure data?
Will puzzles never cease? If octadecane melts at 28-30 C, and its mp is raised by turning a methyl into a carboxyl, you might wonder what a phenyl group would do. Apparently, not much. The mp of 1-phenyloctadecane is 29 C (Ref. 7).
Ref.1 https://en.wikipedia.org/wiki/Octadecane
Ref.2 https://en.wikipedia.org/wiki/Stearic_acid
Ref. 3 https://www.researchgate.net/figure/Crystal-packing-of-left-the-A-super-14-and-right-the-A-1-13-two_fig1_233992144
Ref. 4 https://www.researchgate.net/figure/Crystal-structure-of-C-24-H-50_fig1_270344720
Ref. 5 https://www.x-mol.com/paper/6024417
Ref. 6 https://www.researchgate.net/figure/Selected-intermolecular-contacts-in-the-benzoic-acid-crystal-viewing-the-dimers-from-the_fig2_51049984
Ref. 7 https://pubchem.ncbi.nlm.nih.gov/compound/1-Phenyloctadecane#section=Experimental-Properties


Reasonably complete infrared spectral database with data in CSV form?

First off, I would think you could fairly easily design some data extraction tools for the AIST database. After all, the X and Y axes are standardized (i.e., $4000-400 \mathrm{cm}^{-1}$).

Beyond AIST, you can find machine-readable JCAMP files from the NIST WebBook. There are supposedly ~16,000 spectra, although you'd have to script some way to download them one-by-one.

I think a better bet would be to purchase a database:

Bio-Rad sells an IR database. A quick look now indicates that it has ~235,000 spectra and I'm sure it's in an accessible format.
NIST sells a smaller database ~5000 compounds in JCAMP format.

A quick search also turned up:

IR Spectra.com - claiming ~140,000 spectra in various formats.


Yes, but not practically.

The reduction of silica and carbon dioxide to silicon carbide is a conducted at high temperatures (commonly higher than 1500 °C, reaching 2500 °C). The energy required to reach these temperatures is immense. Furthermore, this is done in a nearly pure carbon dioxide atmosphere. I remind you that our atmosphere has about 400 ppm of carbon dioxide.

Silica is not abundant as you think. The main source for silica, sand, is actually quite rare and many countries are experiencing environmental damage due to illegal or legal mining of sand. Sand is commonly host to fragile and unique ecosystems. You can mine sandstone, but you will need to crush it to sand. Pulverising is the most energy intensive phase of any mining operation. For silicon carbide to form efficiently, it has to be crushed to a fine powder. Sand size is not enough. This is extremely energy intensive.

This is going to be an extremely expensive undertaking, and since the source of most of our energy is still fossil fuels, then this would be counter productive.


Correlating the decrease in orbital radius and penetrating power with the increase in l value

tl;dr

The statement that "the orbital radius slightly decreases with the increase of $l$ value" is true only for orbitals with the same value of $n$ (shell). The penetration of the nucleus by an electron is measured by the relative electron density, which depends on both shell ($n$) and subshell ($l$) of an electron in an atom.

What is Orbital Penetration:

  Penetration describes the proximity of electrons in an orbital
  to the nucleus. Electrons which experience greater penetration
  experience less shielding and therefore experience a larger Effective
  Nuclear Charge ($Z_\text{eff}$), but shield other electrons more effectively. Electrons in different orbitals have different
  wavefunctions and therefore different distributions around the
  nucleus. However, contrary to what many think, penetration is not the
  outer electrons penetrating through the shield of the core electrons.
  It is actually just how well the electrons feel the nucleus. This is
  similar to the idea of outer electrons penetrating, but not the same.
  They are not the same because the core electrons have more penetration
  than the outer electrons since they (the core electrons) feel the
  strongest pull.

Clarification

As you can see from the definition above, the penetration of a nucleus by an electron is measured by the relative electron density near the nucleus of an atom. It is essentially, how effectively electrons can get close to the nucleus.  For instance, below are the electron probability densities for s orbitals and a p orbital.

Nodes are regions of zero electron probability (white areas),
The orange color corresponds to regions of space where the phase of the wave function is positive, and the blue color corresponds to regions of space where the phase of the wave function is negative.

In a multi-electron system, the penetration of the nucleus by an electron is measured by the relative electron density near the nucleus of an atom for each shell and subshell of an electron. For example, the 2s electron is penetrating the nucleus of an atom more than the 2p electron, because the 2s has more electron density near the nucleus than the 2p electron. The penetration power of an electron, in a multi-electron atom, is dependent on the values of both the shell (n) and subshell (l) of an electron in an atom.

For the same shell value (n) the penetrating power of an electron follows this trend in subshells:

$$\ce{s > p > d > f}$$

And for different values of shell (n) and subshell (l), penetrating power of an electron follows this trend:

$$\ce{1s > 2s > 2p > 3s > 3p > 4s > 3d > 4p > 5s > 4d > 5p > 6s > 4f ...}$$


Why is vulcanization called "vulcanization"?

According to Thomas Hancock's 1857 Personal narrative of the origin and progress of the caoutchouc or india-rubber manufacture in England

  In preparing to carry out the new process on a
  large scale, it appeared desirable to give the mate-
  rial a more definite name than " the change,"
  which I had adopted for the purposes of my speci-
  fication; and, whilst discussing the subject amongst
  my friends, Mr. Brockedon proposed the term
  " Vulcanization," and, as no better suggested
  itself, I determined to adopt it, and it is now
  known by that name both in Europe and America.
  It owes its derivation to the Vulcan of mythology,
  as in some degree representing the employment of
  sulphur and heat, with which that mythological personage was supposed to be familiar. I shall,
  therefore, of course use the term in the remainder
  of my narrative.


How can a bond be more stable and more polar at the same time?

The important thing here to keep in mind is that the bond strength is not a spherical chicken in vacuum, meaning that we need to keep the entire process in mind when analyzing all the factors that can influence it. Although the fluorine has a higher electronegativity, it is significantly smaller than Cl, Br, or I (same goes to their corresponding ions). And without going too deep into intricacies, $\ce{F-}$ is $way$ less stable than $\ce{I-}$ and thus, using the OChem language, a much better nucleophile. Due to it's inability to stabilize the charge, it holds onto the hydrogen with all it's might (pardon me for anthropomorphizing the atoms).

SO, when you're thinking about breaking $\ce{H-F}$ bond, you gotta take into consideration what happens with the products. In the example with organic chemistry and leaving groups that you have pointed out, fluorine is the worst leaving group for the precisely the same reason--it cannot stabilize the negative charge well, so it doesn't "want" to leave. However, if you try to do a, say, $S_N2$ reaction in a protic solution (not the best way to go around, but doable), then $\ce{F-}$ will be a much better leaving group than $\ce{I-}$, b/c $\ce{F-}$ will instantly "catch" a hydrogen from the solution and stop being nucleophilic, which according to Le'Chatelie'r principle shifts the equilibrium to where $\ce{F-}$ is a leaving group.


What is the quantum mechanical explanation of X-ray diffraction?

The description given in your question pretty much explains things. I would put it in slightly different words which may or may not help, which is that the x-ray photon induces a dipole moment in the atom (a field induced dipole) which then radiates and so the x-ray is scattered. This is pretty much what happens in scattering  by visible photons and we must assume that the photon is distant in energy from any transitions. (In Compton scattering part of the x-ray energy is imparted to the electron, so the frequency of the scattered x-ray is changed).

The field induced dipole moment is proportional to the polarisability which for a single bound electron is
$$ \alpha = \frac{e^2}{m}\left ( \frac{1}{\omega_0^2-\omega^2} \right )$$

where $\omega_0$ is the resonance frequency of the bound electron and $\omega$ the x-ray frequency and so $\omega >> \omega_0$ which in the limit gives
$$\alpha \rightarrow -\frac{e^2}{m\omega^2} $$
Thus the calculation reduces to understanding the QM calculation for the polarisability. See for example Atkins & Friedmann 'Molecular Quantum Mechanics' Ch $12$.

(  By considering that the x-ray is a plane wave the intensity of scattered radiation relative to that incident is given by
$$ \frac{I}{I_0}= \frac{e^4}{m^2c^4R^2}P(\theta)$$

where R is the distance from the scattering centre to the detector, m the electron mass and $\theta$ the scattering angle and $P(\theta)=(1+\cos^2(\theta))/2$ is called the polarisation factor.  This formula is sometimes named after Thomson. The $m^{-2}$ shows why significant scattering occurs only from electrons and not from protons. The book by Flygare 'Molecular Structure and Dynamics' section 8.5 gives a detailed description of x-ray scattering in general and details such as calculating atomic scattering factors. )


Why are condensation reactions endergonic?

One of the most simple biosynthetic reactions that generates ‘oligomers’ from ‘monomers’ (between inverted commas because that description is not fully correct here) is the ester formation used to synthesise lipids. It can be broken down to:

$$\ce{R-O-H + H'-O'-C(=O)-R' -> R-O-C(=O)-R' + H'-O'-H}$$

With each important bond highlighted and the oxygen and hydrogen atoms made distinguishable by primes. A simplistic way of deriving standard reaction enthalpies is to use Hess’ law and add up the bond dissociation enthalpies of bonds that are broken and subtract that from the bond dissociation enthalpies of bonds that are formed. Doing that in this case gives:

Broken:

$\ce{C-O'}$
$\ce{O-H}$

Formed:

$\ce{C-O}$
$\ce{O'-H}$

Since the bonds being formed and those being broken are identical, at first approximation $\Delta_\mathrm{r} H^0 \approx 0$, meaning the reaction depends entirely on the entropic term.

On the entropic side of things, you are creating one huge, dissolved entity out of two not-as-huge dissolved entities. Making an exact call here is difficult, as it depends on a number of things, most notably how large the hydration sphere is for all three of these compounds. If each of them has a hydration sphere of approximately the same diameter, the product side will have more entropy since less water molecules are needed for an identical hydration sphere (akin to the hydrophobic effect). The opposite effect is also possible, though.

I would like to challenge the ‘all condensations are …’ It seems to simplistic to me. Many condensation reactions are probably endergonic but likewise I expect a few to be overall exergonic. The cell just takes no chances and uses an $\ce{ATP}$-dependent pathway in all cases to ensure the outcome it desires.


It looks to me like tubes used to distribute gas flow.

If you work with substances that would react with the oxygen in air, you prevent that by working under an inert gas such as nitrogen, i.e. you fill all the flasks containing your substance with nitrogen gas, keeping up a steady flow of the nitrogen in order to prevent any from air leaking in.

The nitrogen is usually provided through a valve in the fume hood, and then distributed through the tubes you have there. The left end in your picture is connected to the valve, and on the ends on the right side you can attach tubes leading to your reaction flasks. The fibreglass is there to prevent air from getting into the nitrogen distributing system while there's nothing attached to the tube.

The ends of the glass tubes look weird to me though, so I might be entirely wrong.


Crystalline structures of inorganic materials aren't usually represented by diagrams, but rather by 3D renderings. First, you need to find a crystallographic structure file, which are usually in CIF format. A Google search leads to this one for cummingtonite.

Then, you need some molecular visualization software. Here, I use CrystalMaker, which can create nice photorealistic 3D images, and has a simple tool to record animations. The result is here:

($\ce{Fe}$ is pink, $\ce{Mg}$ is yellow, $\ce{Si}$ is blue; both $\ce{Mg}$ and $\ce{Si}$ are at the center of the tetrahedra, and oxygen atoms correspond to all the tetrahedra edges)


Activation Energy in Acid/Base reactions

Proton transfer, especially in aqueous media, has an activation barrier that is easily surmounted at room temperature or higher.  Such reactions are said to be diffusion controlled (the reaction happens at the rate of proton diffusion through the media) or spontaneous.  In such cases thermodynamics will control the process.   That said, I'm sure there are cases where the base the proton is transferring to, is, for example, surrounded by bulky ligands substantially increasing the activation energy for the process.


Why are spherical compounds with a half filled HOMO concidered aromatic?

OK, let's get this straight. There is no such thing as spherical aromaticity. The so-called Hirsch's rule is sheer nonsense and never works.

Now don't get me wrong, the premises were sound. When we have dozens of atoms arranged kind of uniformly on the surface of a sphere, the whole thing starts to look mighty like continuous - reminiscent of a huge "atom" consisting of surface only, without any depth. Naturally, the eigenfunctions (AKA molecular orbitals) of such a system would be the pure angular functions without the radial part, and their energy will be determined by the angular quantum number $\ell$. The bottommost orbital would be like $s$ (a sphere of one sign), then would go three $p$-like orbitals (each with one change of sign), then five $d$-like orbitals, and so on. Look at the lowest energy levels of $C_{60}$ (pictured), or any other fullerene, for that matter; they would go just like that - one, three, five... (With any other fullerene, the correspondence would be approximate; with $C_{60}$, because of its high symmetry, it is exact). And of course, when we count the electrons up to the $n$'th level, we get $2n^2$.

Sounds good, does it?

Well, it just does not work. There are too many crude approximations on top of each other. Most importantly, when we get all the way up to the HOMO, our eigenfunctions start changing sign too often, so we can't any longer pretend that the surface of a fullerene is continuous. Consequently, the pattern breaks and all conclusions fall apart.

One might point out that pretty much any rule in chemistry is like that, including the Hückel's rule itself. It is also built of crude approximations, isn't it?

That's right, it is, but there is a huge difference, too. Hückel's rule works. Hirsch's rule doesn't. The former predicts the outstanding stability of benzene, and not-so-great stability of cyclooctatetraene, and that cyclopentadienyl would prefer to be an anion, and so on - and they all obey. The latter predicts... well, see for yourself.

What are the magic numbers of electrons for the presumed spherical aromaticity? $2,\,8,\,18,\,32,\,50,\,72,\,98$ and so on. The first five are too small for a fullerene, so ignore them. 60 is not magic, so one would expect it to be unstable. WRONG. It is pretty stable. The paper you mentioned says it is strained. Now what was that about? Of course it is strained, as are all fullerenes and nanotubes; if you don't strain that $sp^2$ carbon, it would just form a flat sheet known as graphene. But what in the world does this have to do with [anti]aromaticity? Think again of the old good Hückel's aromaticity. Think of the cyclopropenyl cation; is it aromatic? Positively yes (pun intended). And yet it is like ten times more strained than fullerene. Aromaticity and strain are totally unrelated things; if people bring up one of them to judge the other, they are either clueless or deliberately misleading, even if they managed to get a paper published in a peer-reviewed journal.

Antiaromaticity, if anything, is about having a half-filled HOMO and thus being prone to Jahn-Teller distortion, radical reactions, and other unpleasant things. Is this the case with $C_{60}$? Look again at those energy levels. Look at the completely filled HOMO, and the completely empty LUMO, and that huge gap between them. Look and think twice before calling this guy antiaromatic.

Now comes the "magic" number 72. That would be $C_{72}$. Does it exist? Well, yeah, it kinda does, like any $C_n$ where $n$ is even and $\ge70$. Is it particularly stable? No, it does not seem so; it is way less abundant than the smaller $C_{70}$ and the bigger $C_{76}$, which themselves are way less abundant than $C_{60}$. Also, starting at $C_{76}$, the fullerenes would have multiple isomers, and as soon as $C_{80}$, we find an example where one isomer is unstable for electroniс reasons (antiaromatic indeed, if you'd like) and the other is stable. So the stability is not determined by the number of electrons alone, after all.

Now what would you call a seemingly reasonable rule which works $\bf{0\%}$ of the time?


Quantifying reactivity in substrates and electrophiles

One large body of work aiming to quantify electrophilicity was accumulated by the Mayr group. Numerous reagents were tested against typical substrates, recording the kinetic rate constants allowed to attribute electrophilicity / nucleophilicity along the equation of

$$ \log k (\pu{20 ^\circ{}C}) = s_N (N + E) $$

with
$E$ = electrophilicity parameter
$N$ = nucleophilicity parameter
$s_N$ = nucleophile-specific sensitivity parameter

Beside the selection of publication mentioned on the group page, work was invested to create a map, and a public database with currently "1075 nucleophiles and 276 electrophiles".


Why do only unsaturated hydrocarbons undergo addition reactions?

To what would you add given a saturated hydrocarbon? You must substitute, abstract, or displace.

Benzene is aromatic — different pattern of reactivity.  It photochlorinates.  It UV reacts with maleic anhydride, then adds another mole thermally.  Palladium catalysis will swap appropriate substituents.


Yes, water is ionized to some extent in all phases. One study even extends to 1,000 K!

A concise discussion of the change in ionization with temperature at Chemguide also explains the variation of pH of neutral water with temperature. [The concept of pH 7 being neutral is based on Kw (in mol2 dm-6) of ~10-14 at ~300 K.]

In general, the higher the temperature, the more a compound is ionized.