What chemical test can differentiate secondary and tertiary alcohols?

Or you can try the very simple Lucas' test using zinc chloride and concentrated hydrochloric acid:
$$\ce{ROH + HCl -> RCl + H2O}$$

It works best with tertiary alkohols, slow with secondary alcohols and not at all with primary alcohols.


What is the reason behind the 2% increase in silver metal composition just after melting the raw silver alloy (55% silver)?

Zinc is used in some silver alloys; Zn would be easy to lose into dross or vaporization.


Raman scattering and "electron and proton center" of molecule

I guess for homogeneous read homo-nuclear. He is trying to explain how the electric field of the radiation induces a dipole by shoving the electrons around, hence reference to polarisability: this relates to the standard description. This induced oscillating field then radiates, hence scattering. The electron and proton centre is when the field is zero, i.e. whatever that distribution is in the molecule in no external field


Free radicals vs Hund's rule contradiction?

The first statement in the question is not correct. Formation of electron pairs is not favorable. That would lead to putting two electrons in the same orbital and there is some effect from charge-charge repulsion.

Electrons end up paired in bonding or stabilized orbitals because there is space for a second electron, and the overall stabilization that would result from putting a second electron in is greater than the destabilization of getting that electron from somewhere else.


Predicting molecular geometry from a formula

It will depend on the number of lone electron pairs as well.
For example:
$BCl_3$ will be triangular planar as it has no lone electron pairs.
$NH_3$ however will be triangular based-pyramidal because there are 3 bonded pairs of electrons and one lone electron pair.
Basically, it comes down to the number of electron pairs (both bonded and unbonded) around the central atom.

Two electron pairs - Linear

Three electron pairs - Triangular planar

Four electron pairs - Tetrahedral

Five electron pairs - Triangular based bi-pyramidal (it's like triangular planar, but with one on top and one below, or two triangular pyramids with their base stuck together)

Six electron pairs - Octahedral (Square based bi-pyramidal)

Those geometries include the lone electron pairs. For the molecular geometry, you will then need to exclude the electron pair.
E.g. $H_2O$ : central atom O. O has 4 electron pairs: 2 bonded, 2 lone. The shape, including lone electron pairs is Tetrahedral. Ignore the lone pairs now, you will be left with a bent molecular geometry.

Blue is oxygen;
Yellow is hydrogen;
White/Grey are lone/unbonded electron pairs.
I hope this makes sense. :)


Photodissociated iodine laser and population inversion

You are making a chemical laser. A laser depend on keeping a positive difference in population between the two levels involved (upper >lower), and of course feedback for a laser oscillator. It does not matter how many I* you make as long as the lasing transition does not go the the lowest ground state level. You must make the transition end at a higher level in the ground state I atoms that has no (i.e. insignificant) thermal population, thus a make a 3 level laser meaning that the laser radiation cannot pump  this lower level to the higher level from which lasing occurs.  You must of course keep producing I* atoms.


Coffee and tap water reaction - results in gray tint

The tap water might be slightly hard.  Hard water has a high content of minerals such as calcium or magnesium which makes the water more alkaline. Alkaline water will draw tannin out of the coffee much quicker than pure water.

Filtering the water with something like activated carbon filter then an ion exchange filter will lower the mineral content, thus lowering the pH. With a lower pH the dissolution of tannin will be slower.

Slowing the dissolution of tannin relative to more soluble portions like caffeine is part of the reason for steeping tea and coffee for only a limited amount of time.  More steep time, or more alkalinity in the steeping solution,  releases more tannin.  More tannin makes the coffee more anti-nutritional, more bitter, and it darkens the color.


How can I interpret the colour of flames in terms of spectral series?

The Balmer series relates to the electronic transitions in the hydrogen spectrum. In other elements it only requires that the energy difference between different excited states corresponds to a wavelength in the visible spectrum. These will clearly differ depending on the effective nuclear charge of the elemnt concerned.


Can I make a container of (kind of) perpetually boiling water?

You can buy a hand-heat powered boiler. But something must provide energy to keep the bubbles forming, and remove it to condense the gas back to liquid, or it would all be in the vapor state. In this toy, the heat of your palm provides the input energy, and ambient air cools the vapor in a condenser.

BTW, this demo won't work if the room temperature is exceedingly high, because the liquid (probably an ethanol-water mix) won't condense and the vapor pressure becomes too high for bubbles to form in the remaining liquid.


The answer to this question hinges on two key questions:

What kind of dust are we dealing with?
How is the dust arranged (i.e. what is your experimental setup)?

Which Dust?

Firstly, I'm going to assume you meant domestic dust (since there are more kinds of dust than you'd imagine).

I quote Wikipedia:

  Dust in homes, offices, and other human environments contains small amounts of plant pollen, human and animal hairs, textile fibers, paper fibers, minerals from outdoor soil, human skin cells, burnt meteorite particles, and many other materials which may be found in the local environment.

Arrangement

The arrangement of the dust particles is key. Are we considering an aerosol, or are we carefully laying a line of dust on the table and connecting it with two electrodes?

Aerosol - We can safely assume that the conductance of air has a far higher impact on the measurement than the conductance of the particles in the aerosol (the dust). So here, dust does not conduct electricity. Note: I ignore the "conducting" via static electricity processes here, but they may play a role. I do not consider that "classical" conductance.

Line of Dust - Organic material is usually insulating, as is our skin (if it wasn't for the moisture that is needed for, you know, living), paper, textiles, hair and so on. So the main constituents of domestic dust does not conduct electricity. So also here, we will not observe conductance.

There is one exception to the above, where conductance is a possibility: Metal dust. This is kind of cheating, but consider the waste of some metalworking process, in which metal dust (or powder) is generated. This dust, being composed of metals and metal oxides, would likely conduct electricity (when laying it out in a line, still not as an aerosol).


Does this set of quantum numbers have a unique solution?

You haven't gotten anything wrong, you are entirely correct!

A set of quantum numbers can be used to identify the specific electron in the sub-atomic world of an atom, and what path it occupies in three dimension.

$\ell = 2$ refers to the $d$ orbital. Now how many electrons can a d orbital contain? It has 5 "slots" because we can go from -2, -1, 0, 1, 2. These represent the orientations, each one holding up to two electrons at different spins (up or down, -1/2 or +1/2). We therefore know that $\ell = 2$ refers to a possible 10 different unique sets of quantum numbers, because each set could represent a specific electron, of which there could be up to 10 in the d orbitals, only 6 for p orbitals, etc.

The principal quantum number ($n$): Has only integral values (1, 2, 3, etc) starting at 1. Think of this as the power level. An electron in a 3s, 3p, 3d, etc orbital is in the third power level. It represents the numbers before the letter in the electron configuration such as 1s$^2$2s$^2$2p$^6$ so in other words, it represents what level the electron is on.

The angular momentum quantum number ($\ell$): goes only from 0 to $n-1$, also uses integral values like $n$. Think of this as the letter after the number in the electron configuration such has 1s$^2$2s$^2$2p$^6$ so in other words it represents which orbital we are talking about on that power level. 0 represents $s$, 1 for $p$, 2 for $d$, 3 for $f$ etc.

The magnetic quantum number ($m_\ell$): Goes from $-\ell$ to $\ell$ including 0. Think of this value as representing a particular orientation of an orbital. The number of orientations will vary depending on the orbital. s orbitals are spherical and can only have one orientation, s orbitals are referred to by saying $\ell = 0$. If we said $\ell = 1$ we would be talking about the p orbital, which has -1, 0, and 1 orientations (3 total). This can be thought of as the x, y, and z axis. $d$ orbitals have 5 possibilities, $f$ has 7, etc. I like to refer to each of these orientations as a "slot" which can hold two electrons.

The value for $m_s$ is used to identify the spin either up or down, but not both.


Determining order of reaction when concentration of reactant is not given

If the stoichiochemistry is as you present it:

$$\ce{A -> P + Q}$$

Then if at $t=0$, $[\ce{A}]=[\ce{A}]_0$ and $[\ce{Q}]=0$, then because one equivalent of $\ce{Q}$ is formed for each equivalent of $\ce{A}$ consumed, your method of determining $[\ce{A}]_t$ must be correct.

$$[\ce{A}]_t = [\ce{A}]_0 - [\ce{Q}]_t$$

Perhaps the error your instructor noted came in applying your relationship or interpreting your graph.


Number of electrons transferred & Gibbs free energy

You should use $12$ electrons. Another way to think of it is when you calculate the $\Delta G$ of the first reaction, that value will be units of energy per mole of reaction. So, since you doubled the moles of reaction, the $\Delta G$ would double.


Gaussian and Surface Reactions - Doable?

You did go to Gaussian's Documentation on the subject, right?

In principle, it should work.  In practice, you should consider using massive firepower, and the plane-wave codes (NWChem, abinit, VASP, etc. etc.)

Also, I question whether any of these codes properly model bond breaking on a surface, i.e.  metal-catalyzed reactions.  Big simulations don't guarantee correct wave functions.

I should point out that surface reactions with plane waves are also difficult. At least they used to be.  There are/were issues with Ewald summation having to be done in 3D, and so the surface normal dimension used to have to be HUGE to avoid image effects.

This is a tough business.


Where does this equation for the activity in the gas phase equilibrium between N₂, H₂ and NH₃ come from?

The activity of a molecule $\ce{M}$ as a function of its concentration in the solution can be written as

\begin{equation}
  a_{\ce{M}} = \frac{\gamma_{\ce{M}} c_{\ce{M}}}{c^0} \ ,
\end{equation}

where $c^0$ is the standard concentration and $\gamma_{\ce{M}}$ is the (dimensionless) activity coefficient (see also: here and Wikipedia).
If you have a sufficiently dilute solution so that non-ideal contributions become very small, the activity coefficient can be approximated by 1, i.e. $\gamma_{\ce{M}} \approx 1$.
So, what you're left with for the activity under these conditions is

\begin{equation}
  a_{\ce{M}} \approx \frac{c_{\ce{M}}}{c^0} \ ,
\end{equation}

Now, the concentration is defined by $c = \frac{n}{V}$ and with the ideal gas law $\frac{n}{V} = \frac{p}{RT}$ you get:

\begin{align}
  a_{\ce{M}} &\approx \frac{c_{\ce{M}}}{c^0} = \frac{\frac{n_{\ce{M}}}{V}}{\frac{n^0}{V}} \\
  &= \frac{\frac{n_{\ce{M}}}{V}}{\frac{p^{0}}{RT}} = \frac{n_{\ce{M}} R T}{p^{0} V}
\end{align}

where $p^{0}$ is the standard pressure.


How do repulsive solute interactions affect the van't Hoff factor?

I think that your first probabalistic argument is correct, i.e that as the solute interaction increases the effective number of them is reduced. In the second case, as the interaction become repulsive, the effective number of them cannot increase over the maximum number there. They are restricted in their configurational entropy because the repulsion  means that the chance of finding one next to another is greatly reduced. Thus the probabilistic aligns with your stat-mech one.

You can calculate the entropy of mixing in an ideal solution then in a 'regular' one in which the components interact with one another.

In the ideal solution a statistical argument based on number of ways of placing particles on a grid,  leads to the entropy being given by

$$\Delta S_{mix}= -R[n_1\ln(x_1)+n_2\ln(x_2)]$$

for $n_1,\,n_2$ moles and $x_1,\,x_2$ mole fractions of species 1 and 2.

In a solution in which there is interaction between (equal sized) molecules the free energy is

$$\Delta G_{mix} = R[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2E$$

where E is is an energy of mixing. This can be calculated by considering that one solute molecule interacts with energy $\epsilon _{aa}$, the solvent with energy $\epsilon _{bs}$ and between solute and solvent with $\epsilon _{ab}$. Again a statistical argument based on molecules being on lattice sites means that nearest neighbour interactions of the type a-a  and b-b occur with probability $x_a^2$, and $x_b^2$ respectively and between a and b as $2x_ax_b$ and the 2 arise because a_b and b-a interactions are counted.

The total mixed energy is then $E_{tot} \approx (\epsilon_{aa}N_a^2+ 2\epsilon_{aa}N_aN_b+\epsilon_{bb}N_b^2 )$$

but the change on mixing is this energy less that due to the two components unmixed which is $\approx (N_a\epsilon_{aa}+N_b\epsilon_{bb})$

which means that the extra mixing energy is $E = zN_0(2\epsilon_{ab}-\epsilon_{aa}-\epsilon_{bb})$

where $z$ is related to the coordination number on the lattice.

The entropy on mixing is then

$$\Delta S_{mix}= -R[n_1\ln(x_1)+n_2\ln(x_2)] -(n_1+n_2)x_1x_2\frac{\partial w}{\partial T}$$

where the second term $\displaystyle \Delta S^E =-(n_1+n_2)x_1x_2\frac{\partial w}{\partial T}$ is a positive number about $0.2R$ J /mol/K for benzene/cyclohexane and $0.01R$ for benzene /$\ce{CCl4}$

Changing $\epsilon_{bb}$ as you suggest will still increase or decrease $\Delta S^E$ acting through the derivative.


What is ideal water hardness for brewing coffee?

According to a discussion I had on this matter with a friend in the coffee processing industry, you should definitely consider yourself lucky for your naturally hard water.  We live in a region with very soft water and he pre-treats all of his coffee-making water with a bed of ground granite (as a physical substrate) and powdered gypsum $\ce{(CaSO4 *2H2O)}$.  For individual use he suggests mineral water, defined in Wikipedia as:

  "The U.S. Food and Drug Administration classifies mineral water as
  water containing at least 250 parts per million total dissolved solids
  (TDS), originating from a geologically and physically protected
  underground water source."

The "taste testing" conferences attended by my friend have rigorously determined that DI water gives the "blandest" taste and water high in calcium and sulfur containing compounds to be the best.

Unfortunately, the chemical basis for this is poorly understood.  It could certainly be pH related, and that my friend's treatment bed just raises the pH which in turn better extracts the desirable components of coffee relative to the undesirable.  The same could be true of the mineral water and hard water in general should have a higher pH than soft water.


Your intuition that the enzyme binding more strongly to substrate will be slower is correct, this is called ground-state stabilization.  I think where you might be making an error is how you handle the factor of 25.  The factor of 25 probably refers to the ratio of equilibrium constants between E and ES for the two enzymes, not the ratio of binding energies.  So you need to start with delta G = - RT ln Keq, and compare for 2 different Keq's.  Then use your equation to estimate the difference in rate constants.


How to know which kind of distortion takes place due to Jahn-Teller effect

If you read and watched videos, you probably already know that the Jahn-Teller effect causes stabilisation in the case of unevenly populated orbitals. Reducing the symmetry of the system renders the orbitals no longer symmetrically equivalent thereby creating an energy difference used to the compound’s advantage.

Let’s separate two cases: 1) octahedral complexes with $\mathrm{d^4}$ and $\mathrm{d^9}$ configuration, in which the $\mathrm{e_g}$ orbitals are unevenly populated; and 2) all the other cases.

Uneven $\mathbf{e_g}$ population

In an octahedral complex with four or nine d-electrons, there will be either one or three within the $\mathrm{e_g}$ orbitals ($\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$). It is unfavourable for orbitals of the same energy (as symmetrically equivalent orbitals have) to have different populations so the system distorts itself. The distortion will always be along a single axis and per defintion this axis is always the $z$ axis. Thereby, ligands on the $z$-axis (negative charges in the most simple crystal field model) move away from the metal centre stabilising any orbital with a contribution in $z$ direction and most significantly the $\mathrm{d}_{z^2}$ one.

The effect is greatest in this case because the $\mathrm{e_g}$ orbitals actually point towards the ligands thus rendering them most susceptible to distance effects.

All other cases

In all other cases, the effect is much weaker. This is mainly because no other orbitals are known to point directly at the ligands; even for tetrahedral complexes the $\mathrm{t_2}$ orbitals point in-between them. Therefore, moving a ligand away slightly has much less effect on the electronics of the system.

What exactly happens really depends on the system itself and how it can be stabilised. If you are speaking of an octahedral complex, it will usually distort itself to $D_{\mathrm{4h}}$, i.e. equivalently to the former case because that is the distortion that keeps the highest possible symmetry. In that case again, one would say that the ligands in $z$ direction are moved away. Otherwise you need to figure it out by drawing, thinking and guessing or by studying crystal structures.

Tl;dr

Instead of memorising a list remember the first case and $z$-distortion. Ignore all the other cases for being weak.


Uncertainty Principle in the Ground State of a Rigid Rotor

As far as the uncertainty relation between $L_{z} = -i\hbar \partial/\partial\phi$ and $\phi$ is concerned, there is a simple way to understand why we cannot use the commutation relation $[\partial/\partial\phi,\, \phi] = 1$, which leads to the problematic inequality $\Delta L_z \Delta\phi \ge \hbar/2$.

Remember that the variable $\phi$ is defined on a circle, i.e., $\phi = 0$ and $\phi=2\pi$ represent the same point. A function of the form $\phi\psi(\phi)$, where $\psi(\phi)$ is $2\pi$-periodic, has a discontinuous jump by the amount $-2\pi \psi(0)$ at $\phi = 0$ (or $2\pi$). Hence, when applying $\partial/\partial\phi$ to $\phi\psi(\phi)$, one should make sure to differentiate the discontinuity as well, which leads to an additional term: $-2\pi \psi(0)$ times a Dirac delta function. Then, the correct commutation relation should be $[\partial/\partial\phi,\, \phi]\,\psi(\phi) = \psi(\phi) - 2\pi\psi(0)\delta(\phi)$, and we have the modified uncertainty principle
$$
\Delta L_z \Delta\phi \ge \frac{1}{2}\big|\langle[L_{z}, \phi]\rangle\big| = \frac{\hbar}{2}\Big|\langle\psi|[\partial/\partial\phi,\, \phi]|\psi\rangle\Big| = \frac{\hbar}{2}[ 1- 2\pi|\psi(0)|^2].
$$
For a normalized angular momentum eigenfunction $\psi_{m}(\phi) = e^{im\phi}/\sqrt{2\pi}$, the right-hand side of the above yieids zero. This is exactly what should be the case when the uncertainties in the angular momentum and the angle are respectively zero and $2\pi$.


How to calculate partition coefficients?

Equilibrium are really dependent on the chemical activities of the species not their concentrations. So an overriding assumption for both parts is that the activities of the chemical species is the same as the concentration. That is a reasonable for these solutions. As a rule of thumb, for solutions with concentrations more than 0.1 molar the assumption is dicey.

I think you're very close to the right solution for part 1, but I'd state it a bit differently.

Since pH is given, $\ce{[H^+]}$ in the aqueous phase is determined by

    $[\ce{H^+}] = 10^{-\mathrm{pH}} = 10^{-3.21} = 6.2 \cdot 10^{-4}~\mathrm{mol\,dm^{-3}}$

      The mantissa of the pH only has two significant figures, so should the $\ce{[H^+]}$

We know that in the aqueous layer the charges have to balance so $\ce{[H^+] = [A^-] + [OH^-]}$, but $\ce{[A^-] >> [OH^-]}$ so we can assume that $\ce{[H^+] = [A^-]}$
Using $\ce{[H^+] = [A^-]}$ the $K_\mathrm{a}$ equation can be solved for the $\ce{[HA]}$ present in aqueous layer which gives $0.038~\mathrm{mol\,  dm^{-3}}$.

$$\ce{[HA]} = \dfrac{\ce{[H^+][A^-]}}{1\times10^{-5}} = \dfrac{(6.2\times10^{-4})^2}{1\times10^{-5}} = 0.038$$

Next, the $\ce{[A-]}$ in the aqueous phase is insignificant compared to the $\ce{[HA]}$ in the aqueous phase, so the moles of $\ce{HA}$ present in the organic layer can be determined by subtracting the moles of $\ce{HA}$ present in aqueous layer (0.038*0.5=0.019)from the initial moles of $\ce{HA}$ (0.057*0.50 = 0.029), which results in a final concentration of $0.020~\mathrm{mol\,  dm^{-3}}$ of $\ce{HA}$ in the organic layer.
Finally, $K_\mathrm{D} =
   \dfrac{[\ce{HA}_\mathrm{aqueous~layer}]}{[\ce{HA}_\mathrm{organic~layer}]}
   = \dfrac{0.038~\mathrm{mol\,  dm^{-3}}}{0.020~\mathrm{mol\,  dm^{-3}}} = 1.9$

    The $\ce{[HA]}$ in both layers is known to two significant figures so the $K_D$ value should have two significant figures too.

For Question 2.

The reaction between $\ce{HA}$ and $\ce{NaOH}$ in aqueous solution is, as you noted,

$$\ce{HA + NaOH -> Na^+ + A^- + H2O}$$

Assume none of $\ce{NaA}$ migrates into the organic phase.
Assume that no $\ce{NaA}$ forms in the aquaeous phase either, which is to say that both the $\ce{Na^+}$ and $\ce{A^-}$ ions are completely solvated.
Since $\ce{[Na^+]}= 0.027$ and the volume is 0.5000 $\mathrm{dm}^{-3}$, the number of moles of $\ce{NaOH}$ present is $0.0135$.
Started with 0.0295 moles of HA total, so the solution will be acidic.
Since the charges must balance $\ce{[Na^+] + [H^+] = [A^-] + [OH^-] }$, but $\ce{[A^-] >> [OH^-] }$ and $\ce{[Na^+] >> [H^+]}$ so $\ce{[Na^+] \approx [A^-] }$ which means that there are $0.0135$ moles of $\ce{[A^-]}$ in the aqueous solution, and 0.016 moles of $\ce{HA}$ split between the aqueous phase and the organic phase.
From above $0.016 = \ce{0.5[HA]_{aq} + 0.5[HA]_{org}}$ and using $K_D$ we can determine $\ce{[HA]_{aq} = 0.021}$ and $\ce{[HA]_{org} = 0.011}$
We can use the $K_a$ to solve for $\ce{[H^+]}$

\begin{align}
1\times10^{-5} &= \dfrac{\ce{[H^+][A-]}}{\ce{[HA]}}\\
\ce{[H^+]} &= \dfrac{1\times 10^{-5}\ce{[HA]}}{\ce{[A^-]}} = \dfrac{(1\times 10^{-5})(0.021)}{0.027} = 7.7\times10^{-6}\\
\mathrm{pH} &= 5.11
\end{align}


Ideal gas constant applied to water, lead, and other non-gases

It is a very good question: Why does R appear everywhere in equations even though it is apparently related to or derived from the ideal gas law? Yes historically R was derived from the limiting gas law in the 1830s.

The beauty of R is that it is related to two physical constants called the Boltzmann constant and Avogadro's constant. Read here

https://en.wikipedia.org/wiki/Gas_constant#Relationship_with_the_Boltzmann_constant

The Boltzmann constant appears everywhere in thermodynamics, semiconductors and statistical mechanics: https://en.wikipedia.org/wiki/Boltzmann_constant

  "The name is a relic of the argon gas chamber used to determine energy
  and temperature at different pressures and temperatures, but it
  applies to every thermodynamic system."

I feel this is nothing but fiction. Not sure why people make up stories. William B Jensen traces the history of R in the Journal of Chemical Education.


Energy balance in battery thermodynamics

The answer to this problem explains how to arrive at

$$ dQ_p = - dw_{ele} +  dG + TdS $$

This expression is generally true for electrical work at constant T and p. From it the differential equation in the OP can be derived by taking the time derivative on both sides and making appropriate substitutions.

  $\frac{\mathrm dG}{\mathrm dt}$ can be expanded with dT:  $\frac{\mathrm dT}{\mathrm dt}S = -\frac{\mathrm dT}{\mathrm dt}\frac{\mathrm dG}{\mathrm dT}$

But remember that you defined

$G=H-TS$

and you should use that when computing $\frac{\mathrm dG}{\mathrm dt}$. But introducing that just results in a circular argument (a lot of terms cancelling to no use).

  to get to the equation at the very beginning of this post, it is necessary to set $E_0$ and $dE_0/dT$ constant

That's ok. $E_0$ is the voltage when no current is running (that is, battery is not being depleted, as there is no reaction, and therefore composition is constant over time), and is thus constant with respect to time.

  But this makes no sense to me, since the whole purpose of this calculation is the temperature increase with time during cycling a cell.

That's tricky and I do not have a complete answer. The equation derived is for a specific T. To solve for the time-dependence of the temperature evidently requires additional steps, presumably incorporating expressions for the heat capacity.


Titration is in the question because ascorbic acid is likely the principal, if not only, reducing agent.

  Using iodine

For example, [this site] describes a titration using the iodine-starch redox indicator (the presence of iodine and starch together creates a blue/black color).

The active oxidant is $\ce{I3-}$, formed from $\ce{I2}$ and $\ce{I-}$:

$$\ce{I2 + I- <=> I3-}$$

The redox reaction then is:

$$\ce{C6H8O6 + I3- ->C6H6O6 + 3I- + 2H+}$$

Add a small amount of a saturated solution of soluble starch to your soft drink sample and then titrate against a known solution that is equimolar in $\ce{NaI}$ or \$\ce{KI}$ and $\ce{I2}$. When the dark color persists in your soft drink sample, you have your end point.

This site provides a full procedure (PDF).

  Using $\ce{KMnO4}$

Potassium permanganate is its own indicator, since $\ce{KMnO4}$ solutions are purple and the reduced version is colorless (technically an insoluble brown solid).

If your soft drink is dark-colored, you may want to dilute it so that you can see the color change.

The redox equation is:

$$\ce{2MnO4- +6H+ 5C6H8O6-> 2Mn^{2+} + 8H2O}$$

  Spectrophotometrically

Ascorbic acid absorbs UV light (but then so does other things in your soft drink, like the preservative sodium benzoate). However, you could use ascorbic acid's reducing power to turn on an absorbance in the visible. Here is one example. This paper may be behind a paywall, but the abstract contains the essential idea. In this paper, the authors describe a system using copper(II) and phenanthroline(phen). Ascorbic acid reduces copper(II) to (copper(I), which creates a coordination complex with phenanthroline that has $\lambda_{\text{max}}=460 \ \text{nm}$. The authors claim a detection limit of 40 $\mu\text{g/L}$, which is pretty good (40 parts per million).

$$\ce{2Cu^{2+} + C6H8O6 -> 2Cu+ + C6H6O6 +2H+}$$
$$\ce{Cu+ + 3phen ->Cu(phen)3+}$$

  Another colorimetric assay

The Folin-Ciocalteu assay is another nice colorimetric assay. Reducing agents react with the phosphomolybdate and phosphotungstate compounds in the reagent mixture and produce a blue color. This method needs to be calibrated against some ascorbic acid standards because the redox chemistry is complex and not fully understood. The absorbance max of the reduced FC reagent is in the 670ish range, if I remember correctly.


If dark surfaces absorb/store more light/energy than light surfaces, then does dark skin, on average, store energy from light most efficiently?

Melanin is a chemical compound(s) found in the skin and is responsible for skin pigmentation.  The darker the skin tone, the more melanin is present.  In light-toned-skin, melanin is produced as part of the tanning process when the skin is exposed to UV radiation.  Melanin is a very effective filter for UV radiation.  The more melanin present in the skin, the less light is available for light-catalyzed chemical processes that occur in the skin.

As Dissenter noted in his comment, light energy is not stored per se in the human body.  Rather light catalyzes certain reactions in the skin and the products of these reactions can be stored (at least for a while) for use in the body.

Vitamin D production is an example of an extremely important light-mediated process that begins in the skin.  Vitamin D is thought to play a role in a large number of regulatory processes throughout the body ranging from bone health to cancer.  Because melanin is such an effective filter of sunlight, non-white populations are at an increased risk of vitamin D deficiency.


Can Le Chatelier's principle be derived?

Yes. To prove Le Chatelier's principle doesn't require a general theory of chemical kinetics, just an understanding of (thermodynamic!) fluctuations and response. As usual, Callen's Thermodynamics and an Introduction to Thermostatistics is a good reference. Ch. 8.5 is relevant, and I shall simply recapitulate Callen's development below.

(Incidentally, I think this principle is closer to Newton's third law of action and reaction, but in general thermodynamic laws have a very different flavor from mechanical laws because of their underlying structure. Thermodynamics deals with nebulous relationships; mechanics with definite equations of motion.)

In fact, we will prove an extra statement, the Le Chatelier-Braun principle. This principle states that the secondary effects induced by a perturbation also serve to curtail it. This augments the Le Chatelier principle, that the primary effect induced by a perturbation serves to curtail it.

I will steal Callen's example of a system immersed within a pressure and temperature reservoir, with diathermal walls and a movable domain wall with which to control its volume. The wall is moved slightly outward, causing a positive volume change $\text{d}V$. The primary effect is the decrease in pressure of the system, which leads to a driving force to decrease the system's volume. A secondary effect is the change in temperature $\text{d}T$ resulting from this change in volume, $$\text{d}T = \left(\frac{\partial T}{\partial V}\right)_S\text{d}V = -\frac{T\alpha}{Nc_v\kappa_T}\text{d}V.$$ The prefactor is unimportant; we care only about the sign of the result. All variables are positive except for $\alpha$, which is of variable sign. For now, we can assume it positive. The reduction of temperature in the system then drives heat flow into it, which will itself affect the pressure of the system: $$\text{d}P = \left(\frac{\partial P}{\partial S}\right)_V\frac{\text{d}Q}{T} = \frac{\alpha}{NT^2c_v\kappa_T}\text{d}Q.$$ This change in pressure is positive and diminishes the effect of the original perturbation. The same result is obtained if one takes $\alpha$ negative.

Let us consider a thermodynamic system with first law $$\text{d}U = f_1\,\text{d}X_1 + f_2\,\text{d}X_2.$$ It is coupled to a reservoir, itself with first law $$\text{d}U' = f_1'\,\text{d}X_1' + f_2'\,\text{d}X_2' = -f_1'\,\text{d}X_1 - f_2'\,\text{d}X_2,$$ noting that $\text{d}X_i' = -\text{d}X_i$, because $X_i+X_i'$ is assumed fixed. A perturbation $\text{d}X_1^\text{pert}$ drives fluctuations $$\text{d}f_1^\text{fluc} = \frac{\partial f_1}{\partial X_1}\text{d}X_1^\text{pert} \quad \text{and} \quad \text{d}f_2^\text{fluc} = \frac{\partial f_2}{\partial X_1}\text{d}X_1^\text{pert}.$$ These fluctuations in intensive quantities themselves lead to responses $\text{d}X_1^\text{resp}$ and $\text{d}X_2^\text{resp}$. The signs of these responses can be determined by minimizing the total energy of the system and reservoir given the initial perturbation, $$\text{d}(U+U') = (f_1-f_1')\,\text{d}X_1^\text{resp} + (f_2-f_2')\,\text{d}X_2^\text{resp} = \text{d}f_1^\text{fluc}\text{d}X_1^\text{resp} + \text{d}f_2^\text{fluc}\text{d}X_2^\text{resp} \leq 0.$$ Because $X_1$ and $X_2$ are independent variables, each term in the sum must be negative, and we have $$\text{d}f_1^\text{fluc}\text{d}X_1^\text{resp} \leq 0 \quad \text{and} \quad \text{d}f_2^\text{fluc}\text{d}X_2^\text{resp} \leq 0.$$ The first equation yields $$0 \geq \text{d}f_1^\text{fluc}\text{d}X_1^\text{resp} \quad \Longleftrightarrow \quad 0 \geq \text{d}f_1^\text{fluc}\frac{\partial f_1}{\partial X_1}\text{d}X_1^\text{resp} = \text{d}f_1^\text{fluc}\text{d}f_1^{\text{resp}(1)},$$ where we have multiplied both sides of the inequality by $\partial f_1/\partial X_1$, which must be positive by stability. $\text{d}f_1^{\text{resp}(1)}$ is the response of $f_1$ to the fluctuation due to $X_1$ only, and so we have the Le Chatelier principle. The second equation yields $$0 \geq \text{d}f_2^\text{fluc}\text{d}X_2^\text{resp} = \frac{\partial f_2}{\partial X_1}\text{d}X_1^\text{pert}\text{d}X_2^\text{resp},$$ and hence that $$0 \geq \frac{\partial f_1}{\partial X_1}\text{d}X_1^\text{pert}\frac{\partial f_1}{\partial X_2}\text{d}X_2^\text{resp} = \text{d}f_1^\text{fluc}\text{d}f_1^{\text{resp}(2)},$$ where we have both used a Maxwell relation and multiplied both sides of the inequality by $\partial f_1/\partial X_1$. This is the Le Chatelier-Braun principle.


Would the mixture of vinegar and baking soda make a cleaning agent?

Well, cleaning is a very broad subject. This is a question which could make a great Science Fair Project. Because vinegar is just a dilute solution of acetic acid (ignoring trace compounds present), and because both baking soda and sodium acetate are pretty soluble, it wouldn't probably be practical to try to make a saturated acidic solution of BS. That is, any precipitates will be made from alkaline solution, under normal circumstances.

There's two questions which would need to be answered:

Does fresh V+BS slurry act the same way as old V+BS slurry?
Are either better than water + BS?

Well, hmm. To be honest, IDK, but I'd guess "maybe" for both (which is why it would be interesting to see what happens). Let me explain:

Acetic acid dissolves soap scum (mostly K, & Ca salts of fatty acids) and dissolves water hardness (mostly Ca & Mg carbonates) So starting off getting rid of some or all of that junk could definitely be a good thing (and acetic acid is only going to be around for a very short time if BS is present in excess). But we know that acids attack metals, so leaving vinegar on metal isn't a good idea...and guess what? Once the BS fully reacts, the acid will be gone! I can't think of any other plausible reason that V+BS would be better than W+BS.

One other potential reason would be if the vinegar changed the abrasiveness of the BS, making it more effective (which is another thing you could test for), but I doubt it would. (possible but not quite plausible, if you know what I mean). Anyway, what you've got left is W+BS after the reaction, so what's the worst it can do?

I do know for sure that most cleaners (with the exception of many metal cleaners) are alkaline because higher pH tends to dissolve (and/or lift-off (which means form a colloid of or weaken the adhesion to the surface-to-be-cleaned of)) organics better than acidic solutions. Cleanser, which is usually just mostly silica, certainly is more abrasive than BS, but that also means it will tend to scratch (or fog) more surfaces. I don't know, but I'll bet most cleansers are alkaline for the reason mentioned above. (alkaline solutions are NOT more slippery, but they form soaps out of your skin, and those soaps are more slippery. So, the slipperiness you feel when you touch an alkaline (basic) solution is just your skin dissolving :)


Both statements on the behaviour of $\ce{Cr2O7^{2-}}$ are true, but you are confusing two different reactions.

The formation of chromium trioxide from (sodium) dichromate in sulfuric acd is not a redox reaction. There is no partner that could be oxidized by $\ce{Cr2O7^{2-}}$. Concentrated sulfuric acid is an oxidant itself and metals like copper are dissolved (oxidized to $\ce{Cu^{2+}}$) while not hydrogen is generated.

Remember also that sulfuric acid is a strong, oxidizing and dehydrating acid. This effect is used in the production of $\ce{CrO3}$, which can be considered the anhydride of chromic acid ($\ce{H2CrO4}$). Note that chromium in $\ce{Cr2O7^{2-}}$, $\ce{H2CrO4}$, and $\ce{CrO3}$ always has the same oxidation state.

The second reaction that you mentioned does happen too, and this is indeed part of a redox reaction in which $\ce{Cr(VI)}$ is reduced to $\ce{Cr(III)}$. However, this can only happen when a suitable reaction partner is present, which in turn is oxidized by $\ce{Cr2O7^{2-}}$.


Electrical and Thermal conductivity of electrum alloys

This reference may be a good place to get started:

Thermoelectric power and thermal conductivity in the silver-gold alloy system from 3-300°K

https://doi.org/10.1080/14786437008228219


What is a simple formula to approximate the change in relative humidity as temperature rises/falls?

If you want to convert the %RH from one temperature to another you can use one of various parameterizations, for instance the Antoine equation:

  range: 273. - 303 K,
  Bridgeman and Aldrich, 1964,
  https://webbook.nist.gov/cgi/inchi?ID=C7732185&Mask=4#Thermo-Phase

  A = 5.40221;
  B = -1838.675;
  C =   -31.737;

  log(P) = A + B/(C+T);

Then obtaining the $RH_2$ at $T_2$ from that at $T_1$ is as simple as
$RH_2 = RH_1 P_1/P_2$, where $P_1$ and $P_2$ are the saturation vapor pressures of water computed with the equation above, assuming the ambient pressure (and therefore the partial pressure of water) is constant.


Why is there no such substance as p-phthalic anhydride

First lets take a look at o-pthalic acid and it's anhydride.

Now as you can see the ring formed in case of o-pthalic anhydride is sterically feasible. There is not very high angle strain and the aromaticity is still intact. Now, take a look at p-pthalic acid.

Now, if you try to dehydrate this molecule as you see there is a very big steric problem. To accomplish this dehydration, the molecule must take a Boat form, which looks something like this, (I am using image of cyclo-hexane just to show you the structure, benzene ring will never do this)

But to accomplish this conformation the benzene ring has to loose it's aromaticity and also create a very unstable system with the pi bond densities repelling each other. So, as this conformation is not Thermodynamically feasible, anhydride formation doesn't take place.


Does resorcinol give positive iodoform test?

There is an equilibrium between the keto enol forms, however small the keto form may be as a contributor. The rest is as you say. The precipitation of iodoform drives the equilibrium further to the products, as per Le Chatelier's principle.Of course, it is not in all cases where the equilibrium is controlled by such factors. The net gibbs energy change for the reaction must be negative.


When will organometallics act as base or nucleophile?

Grignard is a base as well as an nucleophile. When there's a strong electron deficient group it'll attack there, when there' a very acidic group it'll attack there. For other bases it depends upon their relative basicity and nucleophilicity. BAsicity is thermodynamically related while nucleophiicity is kinetically decide. They're silent about each other. You need to practice and gather examples and an experience too :D


Is exposure to amyloid beta proteins dangerous?

Amyloid proteins form insoluble aggregates and they cannot pass through the dermis. The aggregates can be (and frequently are, in the intracellular context) resistant to digestion but they would not be absorbed from the gut. Moreover, these peptides are not "toxic" per se. It is their property of forming aggregates and plaques that gives rise to the pathology observed in the associated neurodegenerative diseases. It is important to note that neural tissue does not regenerate like many other tissues in the body and that is one reason why neural tissue is vulnerable.

So unless you inject some of these proteins inside your cranium, it is unlikely that you will suffer from any problems beacause of Aβ exposure.


How do we know a reaction is in equilibrium?

By definition, the macroscopic observables do not change once equilibrium is reached, and experiments that probe the microscopic domain are usually expensive. I use an analogy to show the microscopic changes:

The model

Take two beakers A and B, and fill one or both with some water (and add some food color for better visibility), and use two straws F and R to continuously transfer from beaker A to beaker B (with straw F) and from beaker B to beaker A (with straw R). The transfer is by having the straw touch the bottom, and then covering the top with your index finger or thumb.

At that point I ask students what they expect will happen if I continue for the rest of the class period. At one point, the level of liquid does not change anymore even though I am still transferring liquid. This is the state of dynamic equilibrium.

To make it more interesting, take straws of unequal diameter (or two vs one straw, like in the looped video).

The analogy explained

The liquid level in the two beakers represent the concentration of reactant A and product B. The diameter of the straws represent the rate constant of the forward reaction (straw F) and the reverse reaction (straw R).

$$\ce{A <=>[k_f][k_r] B}$$

Limitations

There is only one reactant and one product. The reaction will always be first order.

Avoiding misconceptions

If you take a single straw to transfer back and forth, students might think that equilibrium is like a pendulum. If you take straws of equal diameter, they might think equilibrium means that concentrations are equal (rather than correctly, rates are equal).

  Every video just claims that there is a forward AND backwards reaction but doesn't really explain why and is also something I've taken for granted

Starting with just A, you learn that the A molecules have the capability of turning into B. Starting with just B, you learn that the B molecules have the capability of turning into A. If equilibrium were static, someone would "have to tell" the molecules to stop reacting. With the dynamic equilibrium model (which has support from expensive molecular-level experiments), molecules keep on doing what they always did - react every now and then. There is no requirement for some kind of switch in behavior, it just is an emergent property of having both forward and reverse reactions going on and influencing concentrations of species.


Inductive vs conjugation effects in ester hydrolysis

Clearly, you have not read benzene halogenation before. In the electrophilic substitution of halogens in a benzene ring, the halogen, taking Cl for example, deactivates the ring precisely because its inductive effect of withdrawing electron density from the benzene ring outweighs the resonance effect of Cl donating its lone pair into the ring. It is experimentally proven that chlorobenzenes are less reactive, and you can understand the reason why inductive effect outweighs the resonance effect as the Cl atom, being highly electronegative, is "not willing" to readily donate its lone pair of electrons into the ring; High nuclear charge causes Cl atom to hold tightly to its valence electrons, thus they are not available for donation. Also explains why Cl- ion is a weak conjugate base and weak nucleophile.


Absinthe is a strongly alcoholic beverage containing anethole, which is insoluble in water and very soluble in ethyl alcohol.  When water is added to the absinthe, the alcohol becomes too dilute to keep the anethole dissolved, and it appears milky. When more ethyl alcohol is added in the form of cognac, the solution again becomes strong enough in alcohol to dissolve the anethole and the solution becomes clear again.

The phrase "stab the fat" is a bit odd, but I'll do my best to explain. Anethole is a fat-like or fat-loving oil, meaning it dissolves well in other fatty sorts of compounds, but not in water (which is lipohpobic, or fat-avoiding/fat-fearing). Ethyl alcohol has two parts to it. One end of it (labeled (fat) in the chemical formula in the question) is lipophilic (attracted to fat) and the other end is hydrophilic (attracted to water). If there is enough alcohol in the mixture, the lipophilic end of the alcohol molecules can surround the anethole ("stabbing the fat") which leaves the other end of the alcohol free to be dissolved in the water (and also any excess hydrophilic portions of the alcohol).

I hope the "stab the fat" explanation was clear. Don't hesitate to ask for specific clarifications in the comments below.


There is no simple and affordable way to obtain fine aluminium powder chemically. As permeakra said, in the absence of a ball mill, it's just easier to buy some.

Note that a finely dispersed aluminum powder obtained commercially is always going to have low level of impurities such as $\ce{Fe, Mn, Si, Mn}$ (from the grinder's balls) and traces of wax/fats on the surface (added by manufacturer on purpose to prevent adhesion and separate particles from each other).

As for the chemical/physicochemical methods, there is a great overview included in the US5885321A patent, among which there are the following:

exploding of aluminum wire in a vacuum by a high electric current (particles' size varies a lot);
condensation of vaporized aluminum in a current of inert gas (very energy-inefficient);
decomposition of aluminium carbonyls (there is no experimental evidence for any of these to exist);
reduction of halides in solution (issue with separation of the particles from the salts);
CVD of aluminium thin films by decomposition of tertiary amine complex of aluminum hydride (complex multistep process with low yield and expensive and volatile precursors);
decomposition of the dialkylaluminum hydride in $\ce{H2}$ atmosphere at elevated temperatures (up to $\pu{260 °C}$) with a catalyst ($\ce{Ti, Zr, Hf, Ln}$) and a tertiary amine (no particle size control, questionable efficiency);
decomposition of diethylhydridoaluminum or diisobutylhydridoaluminum in diisopropyl ether or triethylamine at $\pu{90 °C}$ to $\pu{185 °C}$ with titanium isopropoxide catalyst.

The patent itself proposes

  decomposing alane-adducts in organic solvents under an inert atmosphere to provide highly uniform particles selectably sized from about $\pu{65 nm}$ to about $\pu{500 nm}$

where adducts are

  trialkyl amines and tetramethylethylenediamine, ethers and other aromatic amines;

and conditions such as

  atmospheric pressure and at temperatures as low as $\pu{50 °C}$ with xylene solvent, ... , toluene, dioxane, and tetramethylethylenediamine.

As a catalyst, titanium halide, amide and alkoxide are used. Authors also provide a complete table of the systems tested:

References

Higa, K. T.; Johnson, C. E.; Hollins, R. A. Preparation of Fine Aluminum Powders by Solution Methods. US5885321A, March 23, 1999. (PDF)


There reaction environment would have to be acidic to protonate the amide, the amide is not natively electrophilic enough to be attacked by the amine.

A possible point of concern is 'won't the nuclophilic amine be protonated?'  And the answer is absolutely.  But the protonation is a reversible reaction, and at equilbrium with a catalytic amount of acid (almost always TSA is used in catalytic quantities), the proton will be exchanged with various, some of which will lead to the condensation resulting in product.


What are appropriate techniques to ensure homogeneous solidification of an agar-based aqueous solution?

Let me suggest two avenues :

Vertical set up à la potter's wheel: At school we once did something related to your aim: using an old $\pu{75 rpm}$ record player, obviously no longer needed, with a beaker containing previously molten bees wax in the centre.  Under constant rotation (rather than the revolution by a centrifuge), the surface of the wax formed a nice meniscus which frooze by the slow cooling. While I am not aware if the alread-gel particles would keep nicely floating in your suspension, until all is solidified, it may be worth to giv it a try.  Centering of the whole ensemble is important (unbalance and spill), especially if the sample to be generated should really become as large as a human head.  Scaling up appears possible (larger beaker / bucket, and potter's wheel).

Horizontal set up: If you were interested in a smalller sample, a roller mixer would be an alternative:

(screen photo)

A wide mouth bottle (like the still cylindrical $\pu{1 L}$ polymer centrifuge bottles used in biochem) would be filled by immersion with liquid and -- still immersed in solution -- closed with the lid to ensure absence of air bubbles in the bottle.  After external cleaning of the bottle, they would be layed along the rolls.  Subsequently, you would turn on and crank up the rolling and hopefully the gel would form sooner than the liquid dropping out from the bottles.  (One may orient the bottles with the lids facing you, and place a bucket underneath them to collect potential spill.)  You may find such a roller mixer in biochem groups cultivating their own cell cultures.


Reactions which are impossible theoretically

If the $20$ J you mention is the average energy this will correspond to a certain temperature. However, molecules have a distribution of energies, caused by their random motion in solution or in the gas phase, so that some have far more energy than the average and some far less. The relative populations of vibrational and rotational energy levels a molecule has are determined by the Boltzmann (exponential) distribution. This means that at a given temperature some molecules are more likely to react than others. If the $100$ J you mention corresponds to a temperature at which the reaction proceeds with a reasonable rate constant then if you only supply $20$ J the rate constant will be $e^{-5} \approx 7.10^{-3}$ times smaller. This may be a small enough rate constant to suspect that the reaction is not proceeding, but its just a matter of waiting !

(If you have a temperature gradient across your sample then of course the reaction rate constant will also vary across the sample being greatest where the temperature is greatest.)


How does dilution affect result of titration of jam?

You can do this at any dilution of both titrate and titrant, as long as you "keep the books" up to date when calculating the results you should be fine. I would probably only dilute the titrate though, unless you have good reasons. (If the dilution of jam is to make it less viscous then that is a good reason for doing it)

Whenever I needed to perform titration something of unknown pH and contents (and I had a large supply of the solutions) I would do a low resolution "sweep" with high concentrates and low tech (old school burettes are usually the fastest, valve fully open and stirring at... vigorous). I'd do two or three of those, just to pinpoint the area of interest. (Sometimes there are several, but then you just do the below for all of them). Assuming one specific point of interest was at 100% addition of titrate, I would add 80% of the titrant to a sample of titrate, then do another titration at 10x dilution of titrant. Three fast samples. Repeat the above, and now at 100x dilution i'd put it in the autotitrator and get some graphs.


Is there a difference between imidazolidinone and imidazolidone?

According to the NIST ChemBook, 2-imidazolidinone and 2-imidazolidone are synonymous. The official IUPAC name is imidazolidin-2-one. 1,3-ethyleneurea is also another name for this same molecule:

But, as you note, tiny changes in spelling can sometimes be much more tricky than this. To take an example from this family: imidazolidinone and imidazolinone (note that the later ends in –inone, not –idone) are ketones formed from imidazolidine and imidazoline, respectively. Imidazolidine and imidazoline are partly saturated and fully saturated imidazole derivatives:

Imidazolidine is a proper IUPAC name, but the official name of imidazoline is 4,5-dihydro-1H-imidazole.


In the mass spectrum of niacin (3-pyridine-carboxylic acid), what ion corresponds to the peak at m/z = 105?

From the general interpretation of electron ionization mass spectra, it would seem logical to report the $m/z$ 105 ion as a loss of $\ce{H_{2}O}$, thus $\ce{[C_{6}H_{3}NO]^{\bullet +}}$. As correctly suggested in the question, the peak at $m/z$ 106 is likely arising from a loss of $\ce{OH^{\bullet}}$ through an $\alpha$-cleavage of the $\ce{C-O}$ bond.

So the issue is: why would niacin (or nicotinic acid) undergo this $\ce{H_{2}O}$ loss? It is not observed for other similar molecules such as benzoic acid, which starts by a $\ce{OH^{\bullet}}$ loss. I did find two relevant publications in the literature on this point. Neeter and Nibbering[1] performed $\ce{D}$ labelling experiments, showing that the transferred hydrogen atom originates mostly from the $\alpha$ position (ortho from both the carboxylic acid and the nitrogen atom). Opitz[2] measured the appearance energy (AE) for $m/z$ 105 at 10.94 eV, more than 0.6 eV lower than the AE for $m/z$ 106 at 11.58 eV. He also suggests, based on metastable dissociation data showing loss of CO to $m/z$ 77, two possible structures for the $m/z$ 105 ion. So to answer the question: (a) formation of the $m/z$ 105 ion is thermodynamically favored compared to the formation of the $m/z$ 106 ion.

Based on this (scarce) literature evidence, I would suggest, based on labelling experiments, that some proton transfer occurs between the $\alpha$ cycle hydrogen and the hydroxyl group. This exchange can either move back and forth, or lead to a loss of water through formation of a cyclic structure.

References

Neeter, R.; Nibbering, N. M. M. Mass spectrometry of pyridine compounds. II: Hydrogen exchange in the molecular ions of isonicotinic and nicotinic acid as a function of internal energy. Org. Mass Spectrom. 1971, 5 (6), 735–742. DOI: 10.1002/oms.1210050612.
Opitz, J. Electron-impact ionization of benzoic acid, nicotinic acid and their n-butyl esters: an approach to regioselective proton affinities derived from ionization and appearance energy data. Int. J. Mass Spectrom. 2007, 265 (1), 1–14. DOI: 10.1016/j.ijms.2007.04.014.


Computing two-electron integrals with an STO-3G basis set

Actually there is a mistake in the analtical expression in Cook's Book. On his web page he has a pdf with the corrected verison

http://spider.shef.ac.uk/

Maybe this solves your problem, but I would also recommend to implement the Obara-Saika Scheme or rys-Quadrature since they are really much more efficient. If your are programming in Python, you might have a look at the PyQuante project, which implements all this stuff.
Concerning Obara-Saika you might also read about the Head-Gordon Pople Scheme. It is in principial an adapted version of Obara-Saika which reduces the FLOP count.


Why does a Diels–Alder reaction not occur here?

Firstly, Why doesn't Diels-Alder reaction happen here?

On first sight the reaction seem completely feasible as both the diene and the dienophile are good species that could undergo Diels-Alder, but no. It could be proven by just analyzing the products (the actual one and the expected one).

Resonance. The property that literally defines organic stability. The DA product has not got even a trace of the stabilization that the actual product receives. The actual product gets to experience extended resonance due to the benzene ring that is formed.
The second factor being that the actual product has got TWO (yes...two) aromatic rings fused together, each having 6 electron resonance.
Steric hindrance. The DA product experiences Steric hindrance while the reaction happens due to the bulky dithiane group. While the other path is free from this.
Strain. Consider the carbon that is holding the dithiane group in both the products. In the DA product the carbon is $sp_3$ while the other product has $sp_2$ (perfect 120 degrees required, and 120 degrees received). In fact, all carbons in the actual product are $sp_2$, other than the ones in the dithiane group obviously.

All of these factors encourage the reaction to proceed in the second path which leads to an amazing yield of 62%.

P.S. I kinda feel stupid after typing "actual product" like 10 times.

Now, the MECHANISM (ohh yeahh!)

Let's start with para-benzoquinone. It is an $\alpha-\beta ketone$. Ergo, the $\beta$ hydrogen is acidic. So, lets prepare it for the incoming nucleophile.

umm...something's missing. Yeah!

OK. Looks better.

Now that we are done with this, we will get to the other reactant. Observe that the methoxy group on the $sp_2$ makes it a good nucleophile. But there is a problem, the dithiane group is too bulky to be drawn again and again. Therefore,

Properties of 'G' :

*bulky, must be kept away

*observing the product, where it is intact, we can say that it is useless for the reaction.

Coming back to the reaction, this nucleophile attacks our first molecule.

Here the (*) marked carbon is activated due to the methoxy group, hence the attack is at that site.

Now the carbocation can stabilized using proton transfer onto the oxygen.

The encouraging factor for this is the achievement of an aromatic ring in the molecule.

Observe the (*) marked carbon in the above image. We see that it is strained due to the thiane group present on + the methoxy group. Therefore the methoxy could abstract the proton from the adjacent carbon and exit as methanol.

Yet again, another encouraging factor for this exit is the formation of another aromatic ring.

Replacing the 'G" with it's true form we have...(drum roll)


Why does chlorine have a higher boiling point than hydrogen chloride?

This is because $\ce{Cl2}$ has close to double the mass and is also a larger molecule compared to $\ce{HCl}$. You were correct in saying that London dispersion forces are weaker but because of $\ce{Cl2}$s size they overcome the dipole-dipole forces in $\ce{HCl}$.


For the E1 pathway, molecule A will react faster since the intermediate is more stable than from molecule B. According to this web site, the major product for A is 1,2-dimethylcyclohexene; for B, it is 1-methylcyclohexene.

For the E2 pathway, both reactions will have nearly the same rate. Depending on the orientation of the α-methyl group (cis or trans to Br), the major product is the same for cis, but for trans it will be 2,3-dimethylcyclohexene for A and 2-methylcyclohexene for B.


Using an organic example, as you move from ethylene to butadiene to hexatriene, you have 1, 2 and 3 bonding and antibonding pi molecular orbitals respectively.  If you continue this chain extension and wind up with a very large polyene, the energy difference between the all of the bonding (or antibonding) orbitals becomes small and instead of having discreet molecular orbitals, you can imagine them "fuzzing" out into a "band" of molecular orbitals.  All of the bonding orbitals would become what is termed the valence band, while the antibonding orbitals would comprise the conduction band.  Just as ethylene is neutral, so is the large polyene.  If you excite one of ethylene's pi electrons to the antibonding molecular orbital you have created an excited state which is still neutral.  Scientists working with the solid state would use different terms to describe this process.  They would say that the antibonding orbital contains 2 holes and the boding orbital 2 electrons. When the electron is promoted to the excited state a hole is also promoted (holes have lower energy the further they are from the nucleus) from the antibonding orbital to the boding orbital.  Just like the ground state, the excited state molecule is still neutral, the electron and hole are just in different orbitals from where they started.


Resonance energy is a fictitious value, it cannot be measured, it is always only estimated. The definition in the gold book[1] clearly states this:

The difference in potential energy between the actual molecular entity and the contributing structure of lowest potential energy. The resonance energy cannot be measured, but only estimated, since contributing structures are not observable molecular entities.

The arising question is how to obtain the energy of the hypothetical contributing structure. One way to do it is to use tabulated (averaged) values for 'normal' bonds to estimate the total energy of the hypothetical molecule.[2]
Another way to do it is to extrapolate from measured hydrogenation energies. This was famously done for benzene. For example, the hydrogenation energy of cyclohexene to cyclohexane is about $\pu{110 kJ mol-1}$. This is the energy to hydrogenate one double bond. In the hypothetical Kekulé structure of benzene you'd expect three of them, giving you a theoretical hydrogenation energy of about $\pu{330 kJ mol-1}$. The measured hydrogenation energy of benzene is about $\pu{200 kJ mol-1}$. This difference of $\pu{130 kJ mol-1}$ can now be taken as the stabilisation due to delocalisation, i.e. the resonance energy.[3]
There is no universal way of doing this, because none of the ways can be proven to be right; it's not real after all. Therefore you may find vastly different values for the resonance energies. It is better to think about resonance as a phenomenological, qualitative argument of mostly pedagogical value. If you are comparing values of hydrogenation (something that can be measured) against extrapolated values (or expectation values) then it is unnecessary to obscure this comparison by giving it another label.

Resonance Energy in: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. DOI 10.1351/goldbook.R05333

You can find such values for example in this Chemistry LibreTexts article on Bond Energies.

Values are loosely converted to SI units from the article on Chemistry LibreTexts Structure and Resonance Energy of Benzene: A First Look at Aromaticity.


What component(s) of instant coffee make it deliquescent?

The behavior strongly depends on the temperature and humidity in which the coffee was kept and on the specific type of coffee.

From Tea and coffee powders, M. Huang, M. Zhang, in Handbook of Food Powders, 2013:

  Being hygroscopic, instant coffee particles are susceptible to the action of moisture – that is, they absorb moisture from the air. If the moisture content increases to 7–8%, the powder or granules can become a paste or solid mass (Clarke, 1987).

Clarke (R.J. Clarke, in Encyclopedia of Food Sciences and Nutrition (Second Edition), 2003) writes:

  Instant coffee is relatively hygroscopic, easily picking up moisture from the atmosphere and caking at about 7–8% w/w moisture content. For example, to keep instant coffee below 5% (dry basis) moisture content, the relative humidity of the air with which it is in contact must be below 35–40%, though the precise value depends upon the nature of the instant coffee in question, primarily due to differences in porosity of the particular particles made. It is necessary, therefore, that jars of instant coffee be well sealed prior to sale.

My answer is admittedly incomplete. As pointed out in comments to the post and in the above quotes, porosity (characterized by e.g. the mean particle surface-to-volume ratio) influences condensation, but so do the intrinsic chemical properties of the particle surface.

The chapter by Huang and Zhang referenced above provides the following table (Table 20.1) detailing the composition of instant coffee powders:

  Constituent   Content(% dry weight basis)

  Minerals                  7.6–14.6

  Reducing sugars (glucose) 3.2–13.1

  Galactomannan             2.4–10.5

  Low molecular organic acids   12

  Brown pigments             15–28

  Caffeine                   2.5–5.4

  Trigonelline               1.56–2.65

  Source: Belitz et al. (2009).

Simple sugars are often hygroscopic as may be organic acid compounds, and caffeine. In particular the presence of glucose (part of "invert sugar") could account for much of the absorption of water. In addition, minerals such as various potassium salts (which make up much of the mineral content of processed coffee beans) are known to be deliquescent.

The Belitz reference appears to be: Belitz et al. (2009). Coffee, tea, cocoa. In Food Chemistry (4th Edition), ed. H.-D. Belitz, W. Grosch, and P. Schieberle, 938-970. Germany: Springer. However haven't accessed the original pubs due to paywalls.


The equilibrium constant of a reaction only involving pure solid or liquid

Concentration of  pure solids and liquids is conventionally taken to be 1. Since reaction involves only solids and liquids the equilibrium constant will be 1


Why is the alkene a more stable species in this case?

I don’t think that your interpretation is what the author meant. I think that they are comparing the reactivity (or inverse stability) of the alkene to that of the carbocation. I think we can both agree that a carbocation is a much more reactive species than an alkene.

When it comes to comparing the products of a nucleophilic substitution to those of an elimination, I don’t think there’s an easy call that can be made. It depends far too much on variable factors such as the steric congestion around the reactivity site, the substitution pattern of the double bond, the size and shape of the nucleophile etc.

In the case of this specific reaction, however, we can make that call. The only mildly nucleophilic species is the hydrogen sulphate anion $\ce{HSO4-}$ deriving from deprotonated sulphuric acid. Hydrogen sulphate has three equivalent oxygens that each carry one third of the negative charge, and they are also practically symmetry-equivalent (not ideally, though, since the proton will angle off). The high oxidation state of the central sulphur atom reduces the electron density on the oxygens. Each of these oxygens is therefore a very bad nucleophile — compare to typical sulphonic acid residues such as tosylate or mesylate which are equally terrible nucleophiles. This should make a nucleophilic attack highly unlikely.

A second thing to remember is that sigma bond does not equal sigma bond. Usually when discussing sigma bonds, $\ce{C-C}$ or $\ce{C-H}$ bonds are meant which are very unpolar, have a bond dissociation energy of something around $\pu{400 kJ/mol}$ and are thus very hard to dissociate. Here, we would be making a $\ce{C-O}$ bond. But not any kind of $\ce{C-O}$ bond: the oxygen belongs to a sulphate residue. This makes the $\ce{C-O}$ bond very weak and very easily cleaved, especially since sulphate is a very good leaving group.

Therefore, in this specific case we can actually make the call and say that the elimination product is more stable than a product of a nucleophilic substitution.

One more thing: you do write cleavage of a sigma bond. But what is actually happening is akin to a deprotonation: a proton leaves leaving its electrons behind in a type of acid-base reaction. As you may know, these are typically very rapid as protons have an easy time ‘jumping’ from one molecule to another if the overall conditions for acid-base reactions are favourable.


The negative absorbance values in the baseline measurement

Negative absorbance in a double beam instrument has no physical meaning because absorbance, by definition, cannot be negative. It ranges from $0$ to $\infty$.
Instrumentally, all that means is that somehow the sample cuvet is transmitting more light than the reference cuvet. In other words, the measured absorbance of the sample is less than than measured absorbance of the reference cell.  Usually this can happen when the cuvets are not optically matched (i.e., they may look identical physically but their optical transmission characteristics are very slightly different).
Try swapping the reference cuvet with the sample cuvet and see if the readings remain negative again. If you are getting large negative absorbance values with the same solvent, try another cuvet set and if the problem persists, the spectrophotometer must be checked. However, if the negative values are quite small such -0.001 or -0.0002, this is tolerable and you can subtract the baseline from the measured signal in order to correct this systematic error.


What exactly is a 'chiral centre' in a carbon compound?

Skipping axial and helical chirality altogether, no to point #2: there are examples of atom centred chirality without a carbon atom in the centre.  The careful oxidation of thioethers may yield chiral sulfoxides:

(Han et al., Chem. Soc. Rev. 2018, 47, 1307-1350, doi 10.1039/C6CS00703A).
Here, chiral requires just the two carbon substitutents to differ, because the third is oxygen, and the fourth the lone electron pair.  Thus, applying CIP rules as usual where the lone electron pair has the lowest priority, DMSO is not chiral, contrary to methyl phenylsulfoxide which is.
Just as with stereogenic centres around carbon, deprotonations on the adjacent prochiral carbon atom may proceed with preference for one hydrogen over the other:

(Solladié, Synthesis 1981, 185-196, doi 10.1055/s-1981-29378).


Can alkyl fluorides be formed by the Hunsdiecker reaction?

The classic reaction of silver salt of fatty acid and halogen to produce alkyl halide was first studied by Alexander Boronin in 1861. But the reaction was generalized by Heinz Hunsdiecker due to which the reaction is named Hunsdiecker reaction. Hunsdiecker used bromine to make methyl bromide. Chlorine was then used to make alkyl chloride. When iodine was used, then ester was formed. This reaction was called Simmonini reaction. Fluorine was not yet discovered then and hence alkyl fluoride was not made through this reaction (fluorine was discovered in 1886 by Henry Moissan).

But there are some speculation discussed among the chemist community whether Boronin actually used some fluorine containing reagent to make alkyl fluoride. It is discussed in this forum that Boronin might have used "acidic potassium fluoride" or $\ce{KHF2}$/potassium fluoride($\ce{KF}$) to make benzoyl fluoride in 1862 but there is no report or scientific paper on this reaction making it a speculated information. (Note: fluorine was discovered by electrolyzing $\ce{KHF2}$).

Fluorine containing compounds were made through Hunsdiecker reaction though proper alkyl fluoride synthesis through this reaction is yet to be reported:

Perfluoroalkyl iodides were made through this reaction:
$$\ce{R_fCO2Ag + I2 ->[\Delta] R_fI + CO2 + AgI}$$
This compound is used to make numerous fuctionalized fluorocarbons and
first Grignard reaction of this class.
Perfluoroalkyl diiodides has also been prepared by heating perfluorocarboxylic acid in presence of iodine. It has been used to make diacids having ether linkage.
Fluorinated ester of type $\ce{RFCO2CF2RF′}$ has been was obtained from reaction of silver 3,6,9-trioxa-F-undecanoate and iodine with a diluent at 130° C. Substitution products, $\ce{RFCOY}$ (Y = nucleophile) and pyrolysis products, $\ce{RFCOF}$ and $\ce{RF′COF}$ were isolated and characterized. It appears probable that the acyl hypoiodite and the iodo-F-alkane reacted by an $\ce{S_Ei -}$ type process to give the completely fluorinated ester.
Fluoro-functionalized graphene oxide  is made via the Hunsdiecker reaction. It is made by treating of carboxylated graphene oxide with selectfluor at 90 °C for 10 h under an atmosphere of nitrogen, using silver nitrate as a catalyst.

References:

Organofluorine Chemistry: Principles and Commercial Applications
edited by R.E. Banks, B.E. Smart, J.C. Tatlow - page 18
Well-Architectured Fluoropolymers: Synthesis, Properties and Applications
By Bruno Ameduri, Bernard Boutevin - page 108
Brace, N. O. Some approaches to the synthesis of fluorinated alcohols and esters. I. Completely fluorinated esters from the hunsdiecker reaction of silver F-alkanoates with iodine. J. Fluorine Chem. 1981, 18 (4), 515–524. DOI: 10.1016/S0022-1139(00)82668-8.
Xing, R.; Li, Y.; Yu, H. Preparation of fluoro-functionalized graphene oxide via the Hunsdiecker reaction. Chem. Commun. 2016, 52 (2), 390–393. DOI: 10.1039/c5cc08252h.


Derivation of relation between acid and base equilibrium constants

$K_\mathrm{a}$ and $K_\mathrm{b}$ should not have a term for the concentration of water inside.
When you dissolve $x$ moles of $\ce{NH3}$ in water (doesn't matter how much $x$ is), both relations

$$K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]} \tag{1}$$

as well as

$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} \tag{2}$$

must simultaneously hold true. You agree that both of these relations must independently hold no matter what concentration of $\ce{NH3}$ (or $\ce{NH4+}$, or $\ce{H+}$, or $\ce{OH-}$) there is, right? That is why they are called equilibrium constants, after all.

So, in any given single solution with one well-defined value of $[\ce{NH3}]$, one well-defined value of $[\ce{NH4+}]$, ... both equations must simultaneously hold. You can then cancel out the terms as necessary to obtain

$$K_\mathrm{a}K_\mathrm{b} = K_\mathrm{w} \tag{3}$$

If you use two different solutions, with different concentrations of all the relevant species, the relation still holds true. However, the maths will be much harder because for example if you have solution 1 with pH = 1, and solution 2 with pH = 10, then $[\ce{H+}]_\mathrm{sol,1}[\ce{OH-}]_\mathrm{sol,2}$ is no longer equal to $K_\mathrm{w}$. The relation $[\ce{H+}][\ce{OH-}] = K_\mathrm{w}$ is only necessarily true if we are talking about one single solution.


There is a shell structure within the nucleus also. So the neutron/proton ratio varies from 1:1 for the light elements to 1.5 for the heavy elements.

  Note: I ripped off the image from another website. Not sure where they got it from...


H is enthalpy.

By definition H = U + PV (where U is internal energy, P is pressure and V is volume).

For question 1, your work says:

  (250) x (4.184) x (26.26-24.85)

  = 1474.86 kJ

Because you failed to show units in your work, you are off by a factor of 1000.

But the main reason your answer is different from the desired answer is you have calculated the amount of heat released by 0.152 g reacting, whereas the desired answer comes from calculating the heat released by 3 moles of A reacting, considering change in enthalpy of the system to be negative when heat is released to the surroundings.

For question 2:

Answer ii is wrong because the monoatomic forms of hydrogen and bromine are not the standard states of the elements.  Answer iii is wrong because the equation is written for 2 moles being formed rather than 1 mole.


Percent difference: What to place on the denominator?

The experimental value is the one you measured/calculated yourself, from your own experiments. The theoretical value is the one that you obtained from some third party, to which you are comparing your experimental value.

So, the number on the bottle is the theoretical, and the number from your data and calculations is the experimental.


Oscillator Strength and Absorption Intensity in TDDFT

I'll address the last part of your question. I think the first part

What formulae relates the transition dipole moments, the oscillator
strength and the observed absorption relative intensity?

of your question could stand on its own.
The oscillator strength is connected to the transition dipole moment vector via
$$
f_{if} = \frac{2}{3}\omega_{if}|\vec \mu_{if}|^2
$$
in atomic units,i.e. $m_e=\hbar =1$, $\hbar \omega_{if}=(E_f-E_i)$. The equation in SI units can be found on Wikipedia. $i,f$ stand for the initial and final electronic state involved in the transition.
For spontaneous emission, you have to multiply $f_{if}$ by an additional factor of $\omega_{if}^2$ since spontaneous emission is proportional to $\omega_{if}^3|\mu_{if}|^2$.
The oscillator strength is already a dimensionless relative quantity and the ratio of different oscillator strengths should correspond to the ratio's in an absorption spectrum measurement. Just normalize everything by the highest value if you are going for a  plot that is comparable with a relative intensities plot where the peaks are also  normalized .
It is important to note that the calculation only describes a purely electronic transition and yields only a single line at the energy difference of the involved adiabatic electronic states. The result of a real measurement is not a single line and includes many effects that are not described by a simple excited electronic states calculation.
But the calculation gives you an estimate where the center of the peaks should be and how intense they should be. A simple TD-DFT calculation does not give you information about the detailed lineshape. The lineshape is typically assumed to have some form and then added by broadening a stick spectrum based on the oscillator strenghts with a spectral lineshape function. The width of the lineshape function is then adjusted to match the experimental measurement manually.  Common lineshape functions are here.


Assigning labels and answering questions using given DSC graph

As example, Skoog / Leary's principles in instrumental analysis contained a section about DSC in the chapter of thermoanalysis.  Your university library likely has some of them or other books about the topic, too.

Then, beside monitoring glas transitions, melting, decomposition (while heating) or solidification / crystallisation (while cooling) I suggest to attribute the other phase transitions with (in the best case: simultaneous) visual observation of the sample (heat stage microscopy, like here) at these critical temperatures.  As these transitions take time, from a practical perspective, a heating rate of $\pu{10 K/min}$ is maybe rather quick.

After reading "liquid crystals" in the title, I thought a cholestic phase would be mentioned, too, but if your data support its absence, ok.  On the other hand, one of PerkinElmer's brochures includes this representation

(source)

which might be useful to complement the structural description of your liquid crystals.


How are alkanes, alkenes, and alkynes homologous series since they have no functional group attached?

For the linear alkanes, the internal methylene groups $\left(-\ce{CH2}-\right)$ are considered to be the functional group defining the homologous series:

$$
\ce{H3C-\underline{(\ce{CH2})_n}-CH3}~,~~n\geq 0 \tag{n-alkanes}
$$

Alternatively, this series is defined by the absence of any particular functional group other than methyl and methylene.

Ethane, $\ce{C2H6}$ is the member of the series for $n=0$. Technically, one could exclude methane, $\ce{CH4}$, from the series, since it doesn't have two capping methyl groups. However, to my knowledge it's usually considered part of the series as a special case.

For the 1-alkenes and 1-alkynes, the double and triple bonded carbons are the functional groups defining the series:

$$
\ce{H3C-(CH2)_n-\underline{\ce{CH=CH2}}} ~,~~n\geq 0 \tag{1-alkenes}
$$

$$
\ce{H3C-(CH2)_n-\underline{\ce{C#CH}}} ~,~~n\geq 0 \tag{1-alkynes}
$$

Parallel to the alkanes, the shortest "official" members of the two series are technically propene $\left(\ce{H3C-CH=CH2}\right)$ and propyne $\left(\ce{H3C-C#CH}\right)$, respectively. As with methane above, though, ethylene/ethene $\left(\ce{H2C=CH2}\right)$ and acetylene/ethyne $\left(\ce{HC#CH}\right)$ are typically grandfathered in.

The case of non-terminal linear alkenes/alkynes is less straightforward. Depending on one's philosophy, they can either be lumped in with the terminal alkene/alkyne series, or they can be considered as distinct homologous series:

$$
\ce{H3C-(CH2)_n-\underline{\ce{CH=CH-CH3}}}~,~~n\geq 0 \tag{2-alkenes}
$$

$$
\ce{H3C-(CH2)_n-\underline{\ce{C#C-CH2-CH3}}}~,~~n\geq 1 \tag{3-alkynes}
$$

$$
etc.
$$

Note that $n\geq 1$ is strictly required for the 3-alkynes, since $n=0$ would give 2-pentyne $\left(\ce{H3C-C#C-CH2CH3}\right)$, a 2-alkyne. This requirement is general: e.g., for the x-alkenes, $n \geq x-2$ must hold for $x\geq 3$.

Branched alkanes/alkenes/alkynes complicate things even further. There are at least two common alkane series that I can think of off the top of my head:

$$
\ce{H3C-(CH2)_n-\underline{\ce{CH(CH3)2}}} ~,~~n\geq 0 \tag{iso-alkanes}
$$

$$
\ce{H3C-(CH2)_n-\underline{\ce{C(CH3)3}}} ~,~~n\geq 0 \tag{neo-alkanes}
$$

You can mix and match branching, double bonds and triple bonds in countless ways to make new homologous series. I would personally argue you could define a homologous series to contain multiple variable-length segments:

$$
\ce{\underline{\ce{(H3C)2C=C}}-(CH2)_n-\underline{\ce{C#C}}-(CH2)_m-\underline{\ce{CH(CH3)}}-(CH2)_r-\underline{\ce{CH=CH2}}} \tag{?}
$$

Of course, the practical utility of any correspondence of properties among the various species in such a convoluted homologous series may be ... limited.


You seem to be confusing the elements with the pure simple substances of the elements. In atomic terms, both the oxygen atom and the sulfur atom are paramagnetic, since they contain two unpaired electrons in the ground state. However, the substances oxygen ($\ce{O2}$) and sulfur (generally intended as $\ce{S8}$) are not the same as the atoms.

In the simple picture of Lewis structures, both substances have all atoms forming two bonds, completing the octet rule and therefore without unpaired electrons. This would imply that both are diamagnetic. However, simple valence bond theory is inadequate to explain the bonding in oxygen. Simple molecular orbital theory suggests that there are in fact two unpaired electrons in the $\ce{O2}$ molecule. This gives rise to paramagnetism in the oxygen molecule, also.


Change in volume of two beakers on reaching equilibrium

Okay, so you know that the vapor pressure of the pure water is some value, $p_w$. And you know from Raoult's Law that a solution with a nonvolatile solute of mole fraction $\chi_s$ will have a vapor pressure $p=p_w \chi_w$, where $\chi_w = 1- \chi_s$, so $p=p_w(1-\chi_s)$.

Since $\chi_s <1$, then $p < p_w$. How do you use this information?

Well, as the system approaches equilibrium, the two beakers will want to exchange water through the vapor phase to equilibrate the vapor pressures. What will that look like when it happens?  Can it happen?

By this logic, you can determine where the water is in your system. However, you cannot determine the volume(s) in the beakers without the following additional information:

Volume of the enclosing vessel - helps determine what mass of water remains in the gas phase
Temperature of the system - determines initial vapor pressure of pure water
Concentration of the sugar solution.
The non-ideal behavior of the sugar-water solution's volume.


How would you know if the light emitted is of one wavelength or if it is a mixture of two or more wavelengths?

The green light from copper flame is sadly not a single wavelength. It is very good question though because not many students even think about it. In order to "see" what wavelengths are present in a light source, you either need a spectroscope (old school way) or a monochromator with a detector (modern way). Of course you can search about Bunsen-Kirchoff spectroscope. It uses a prism to disperse the light falling on the prism.
Depending on how much interest and money you have you can easily construct a spectroscope in a shoebox (several published articles out there). Ten years ago or so I used to make CD diffraction spectroscopes in shoeboxes or cardboard boxes for seeking answers to queries like yours.
If you want spent some money ($400-500) you can buy a high quality pocket spectroscope with a wavelength scale . Then you can enjoy all light sources (except the Sun). It is fascinating to see and explore so many phenomena. You will find out by looking at the moon that it has some dark lines (never look at the Sun from a spectroscope). I am not suggesting to buy third rate products from Amazon because twice I purchased from them and the wavelengths were all wrong!
The monochromator is a fancier product. I never had the courage to buy one. They are very expensive. The most common design is a Czerny-Turner design. I am sure Prof. EdV had even fancier products which we have never fathomed :-)
This picture shows white light entering a slit, it falls on a mirror which makes all the rays parallel. The mirror sends the light to a high quality diffraction grating, the reflection diffraction grating separates wavelengths in space. The light from reflection grating falls on a focusing mirror, and the mirror focuses lies on the exit slit. The exit slit has a detector. You would slowly rotate the grating. If there are multiple wavelengths in the light source (say a flame) the detector will record a spike every time a different wavelength reaches the detector.


Boron: borates, borides.  $\ce{MgB2}$ is the highest temperature BCS superconductor by a large margin.

Carbon: carbonates, carbides, hexacoordinate carbon in complexes, organolithiums, Grignard reagents, carbenium ions ($\ce{Ph3C+}$).  $\ce{Al4C3}$ hydrolyzes to methane.

Silicon: silicates, aluminosilicates, silicides.  Cook up silica gel in ethylene glycol plus potassium hydroxide (NOT in glass) to get pentacoordinate siliconate,

http://patentimages.storage.googleapis.com/WO1994020507A1/imgf000014_0001.png

You asked the second half in the Physics Stack.


In an isothermal reversible process the total entropy change of piston and energy reservoirs (surroundings) must be zero, $\Delta S_T = 0$. Thus the entropy of the expanding gas may increase but only if it is compensated for by a decrease in entropy of the reservoirs. So no contradiction of second law. (The heat absorbed from the surroundings is $T\Delta S $ and if this is positive more work can be done by a chemical reaction than the heat of reaction.)

In the second question you will realise that you can turn all the work into heat but not all the heat into work. Its entropy again.


According to Markovnikov rule and stability of carbocation which is most likely to form if HI "excess" is not given?

There are three reasons as far as I can see as to why the 1,2 product is favored. There may be more.

Steric hindrance
The iodine atom is large in size compared to the carbon atom. Having two iodine atoms on successive carbon atoms in a hydrocarbon is quite unstable. In fact, should two successive carbon atoms possess iodine substituents, they usually undergo elimination reactions. And here the 1,1 product would have two iodine groups on the same carbon atom!

Inductive effect of halogens
While it is true that halogen atoms have a positive mesomeric effect, this effect is almost always dominated by their negative inductive effect. If you see a halogen atom, you can usually neglect it's +M effects in favor of it's −I effects.

Carbocation stability
The carbocation formed in the process of formation of the 1,2 product is more stable (due to hyperconjugation) than the one formed in the process of formation of the 1,1 product (again, due to the −I effect of iodine).

Either way,  1,2-diiodopropane is more stable than  1,1-diiodopropane.


Can density functional approximations including dispersion correction perform worse than without it?

tldr; I'll check the detailed results, but the overall answer is that dispersion correction is absolutely worth it

We just finished a benchmark over ~700 molecules with up to 10 conformers, so ~6700 single points at the DLPNO-CCSD(T) / def2-TZVP level. I will post a link to the preprint when it's available soon.

In the meantime, here's a quick summary for the effects of dispersion corrections:

  Comparing un-corrected PBE, B3LYP, and ωB97X single-point energies to DLPNO-CCSD(T) illustrate a significant effect. The uncorrected median R2 values drop by ~0.12, and the median Spearman correlations drop by ~0.08. For example, the median R2 of B3LYP / TZ drops from 0.920 to 0.706 without the D3BJ dispersion correction.

The metric here is the correlation between a method (i.e. B3LYP) with the DLPNO-CCSD(T) energies.

I'll look to see if there's ever a case where the dispersion correction does worse. But there are many, many papers illustrating that conformer energetics are dominated by non-bonded interactions .. which are improved with dispersion-corrected density functional methods.

Your methodology seems solid, although if you want good energy evaluations of different conformers, I'd consider either RI-MP2 or DLPNO-CCSD(T) single-points. Our timings in the paper show that RI-MP2 can be about as fast as hybrid DFT methods, but more accurate at ranking conformers.


A couple of thoughts:

There are mainly 3 different definitions of acids and bases:

The Arrhenius formulation: Acid = proton donor in water,  Based = hydroxide ion donor in water
The Brønsted-Lowry formulation: Acid = potential proton donor, Base = potential proton acceptor
The Lewis formulation: Acid = Electron pair acceptor, Base = Electron pair donor

The Lewis definition is the most general, followed by the BL and finally by the Arrhenius definition.

You are right, for example HCl is considered a strong acid in water, but it is very weakly acidic in glacial acetic acid. If you follow the Lewis definition a large number of reactions will be considered acid base reactions. An exception are free-radical based reactions

There are many compounds that are amphoteric, but it really also depends on the definition you use.


How large an energy difference in Hartree can be used to distinguish two different structures?

I wouldn't use energy as a general criteria to discriminate conformations; a root mean squared deviation (RMSD) of superimposed structures would be a better metric. This is because two very different conformations may have very similar energy. For example, if you perform a relaxed energy scan of one of the dihedrals of cyclopentane, you get nearly the same energy for most of the dihedral angles, bu very different ring conformations.

If you found an energy difference of 0.0000001 a.u. and the superimposition of the conformations is nearly perfect, I'd say they are in the same minimum of the energy surface.


Write the combustion equation of an organic compound

combustion is a reaction using O2 to burn the other reactant and will result in CO2 and H2O

$\ce {\_\_\ C5H10O + \_\_\ O2 -> \_\_\ CO2 + \_\_\ H2O}$

then just balance
so..

$\ce {C5H10O + 7O2 -> 5CO2 + 5H2O}$


Are molecules with bond orders less than one stable?

Molecules possessing a bond of order below 1 can be perfectly stable, in the sense that their resulting molecular structure lies in an energetic potential well. Strictly speaking it is enough that at $T=\pu{0 K}$ and in the absence of any interactions with matter or fields, the molecule will not spontaneously disassemble. However, there is no need to go quite so far to protect such a molecule from decomposing; there are examples of species which are chemically important in regular laboratory conditions.

All other things being equal, it is true that species with bond orders below 1 are relatively unstable. This is mainly because the fractional bond is comparatively weak (requiring comparatively little activation energy to break, i.e. a smaller $E_\mathrm{a}$), and because in most cases the molecule can react with other substances in such a way as to form products with all covalent bonds of bond order 1 or higher (increasing the exergonicity of most reactions, i.e. a more negative $\Delta_\mathrm{r}G$).

Since the kinetic barrier for reaction is comparatively low, and the thermodynamic drive for reaction is comparatively high, species with bond order below 1 tend to need some extra protection against the world to stay in one piece. For example, diborane and trimethylaluminium are compounds possessing bonds of order 0.5, and while being indefinitely stable when pure, they spontaneously ignite in air from exposure to oxygen and moisture. As rightly pointed out in the comments, noble gas compounds require fractional bond orders, yet several compounds can be obtained and stored (especially xenon compounds), though they tend to be sensitive to moisture and heating. Boron is also responsible for a very interesting class of compounds in which many boron atoms attach to  each other in cage-like structures with very complicated bonding, in which fractional order bonds are involved. Some of the larger and more symmetric structures can be relatively stable, especially with adequate substituents.

In space, there's really not much of anything around, so you might expect to find some molecules with fractional bond orders floating about. Indeed, one can find the trihydrogen cation, which is actually one of the most common ions in the Universe!


Powdered sugar is often used in chemistry demonstrations, in the presence of oxygen it is highly combustable (there have been significant explosions and fires at sugar processing plants in the past due to this).

The sugar is high in energy, and the fine powdered variety has a greater surface area with which to interact with oxygen, producing the burst of energy needed. The sugar by itself however, is probably not enough to get any propellant effect, you'd need to add some kind of oxygen source. A quick google suggests mixtures of powdered sugar and potassium nitrate (link).

Dry gelatine (the powdered version) is essentially just sugar plus, as you noted, animal extracts or vegetarian friendly alternatives, so the same should apply, though using actual sugar is probably a lot cheaper.

Edit:  As pointed out by @MaxW in the comments below, and as eluded to in my opening comment, igniting a fuel in the presence of an oxygen source can be highly explosive. Be sure to take necessary safety precautions (work on a small scale, away from buildings. Ignite from a distance, and be sure any spectators are also out of harms way etc.) and also ensure that you're not breaking any laws in your area.


Can ethyl alcohol fumes be suppressed by a household air purifier/filter?

No home air purifier removes alcohol vapor from the air. There are a number of ways it can be eliminated, though, depending on quantity, purity and other factors.

Bakeries have to comply with clean air standards for VOC (volatile organic compounds), primarily ethanol, which can cause smog. One way the alcohol is removed is by catalytic oxidation, similar to the way an automobile catalytic converter works. Even hot copper is sufficient to oxidize alcohol.

Another way to remove ethanol is by condensation, which allows the alcohol to be saved for resale, or by washing (scrubbing) the oven gases with water.

Sadly, though this regulation of ethanol emission is necessary, it also removes the sublime aroma of baking bread from the vicinity. sigh...


Can aniline be made with the method described in Dr Stone?

Believe it or not, aniline was actually isolated from coal tar way back in 1834. A chemist called Friedlieb Runge isolated aniline along with phenol from coal tar. He discovered that aniline produced a beautiful blue color when treated with chloride of lime (calcium hypochlorite). He named it cyanol/kyanol and that was the birth of the first aniline dye but it was made a dye by the exhaustive work of another chemist, William Perkin.
Perkin distilled the components of coal tar like benzene, aniline, toluene, anthracene, phenol, naphthalene and other aromatic hydrocarbons. He first worked around toluene to make allyl toluidine which he then transform to quinine by oxidizing it with a mixture potassium dichromate and sulfuric acid. He was left with a useless reddish-black powder. He then took aniline and repeated the experiment. He was left with a black gooey substance. Disappointed by the results, he took the test tube and washed it off but then came out a beautiful looking purple residue which hung off at the wiping cloth. He named the dye "mauveine". Perkin was so amazed by the result that he started working with other distilled products hoping to make other dyes$\ce{^{[4]}}$ . His discovery revolutionized the dye industry. Perkin proposed the following statement:

Benzene, toluene, and other components of coal tar were colorless
because they absorbed ultraviolet light undetectable by the human eye.
But if those aromatic hydrocarbons were treated with an acid to create aniline or another amine, after some additional steps the
newly synthesized molecules very efficiently absorbed light particles
from specific wavelengths in the visible spectrum.

So, Senku (the main character of Dr. Stone) being well-versed in chemistry knows what he's talking about.
Notes and references

https://trc-leiden.nl/trc-needles/materials/dyes/aniline
https://www.scientificamerican.com/article/toms-river-excerpt-on-aniline-dye/
https://www.cavemanchemistry.com/cavebook/chtar.html
Magentas from toluene, reds from anthracene, pinks from phenol, and indigos from naphthalene.
Dr. Stone is an amazing anime. I recommend fellow readers to check it out.


How do we predict the order of melting points of group 1 and group 2 halides?

For group 1 halides polarisation is not a dominating factor(except for lithium) since the size of cations is relatively larger than any other cation therefore we must consider lattice energy for determining melting point. As interionic distance increases lattice energy decreases hence melting point decreases down the group.
Lithium is an exception because of its very small size that brings high degree of polarisation. you may want to refer to lother meyer's curve on atomic volumes of elements to get relative idea about the size of the elements.


How come one of the hydroxides in the phosphate group are more acidic than the other?

Because the rendering is not really accurate.  It would have been better to render the two dissociation constants without trying to associate them with specific hydrogen atoms.

Given the neutral phosphate group with both hydroxyl hydrogens attached, either one has an equal probability of being removed as a proton, corresponding to the lower of the two $\mathrm{p}K_\mathrm{a}$ values.  The drawing simply picked one of the protons at random.  Then with that proton removed and the functional group now negatively charged, the second hydroxyl proton has become harder to remove and that is what the higher $\mathrm{p}K_\mathrm{a}$ represents.