The compound in question is called D-galactose in carbohydrate chemistry. It has 4 chiral centers (C2, C3, C4, & C5): Thus, stereochemistry of them are $(2R,3S,4S,5R)$. The rule is if a compound has $n$ chiral centers, then it has maximum total of $2^n$ possible stereoisomers, if it does not have any symmetry. D-galactose does not have any symmetry within the molecule (e.g., It would have symmetry if aldehyde group, $\ce{-CHO}$ has been reduced to $\ce{-CH2OH}$). Thus, four chiral centers of D-galactose are accounted for a total of 16 stereoisomers ($2^4$). If you draw them all, you would find they are collection of 8 pairs of enantiomers as:

{$(2R,3S,4S,5R)$, $(2S,3R,4R,5S)$}
{$(2R,3R,4S,5R)$, $(2S,3S,4R,5S)$}
{$(2R,3S,4R,5R)$, $(2S,3R,4S,5S)$}
{$(2R,3S,4S,5S)$, $(2S,3R,4R,5R)$}
{$(2S,3S,4S,5R)$, $(2R,3R,4R,5S)$}
{$(2R,3R,4R,5R)$, $(2S,3S,4S,5S)$}
{$(2S,3R,4S,5R)$, $(2R,3S,4R,5S)$}
{$(2R,3R,4S,5S)$, $(2S,3S,4R,5R)$}.

That make maximum total of 8 pairs of enantiomers, meaning $2^3 = 2^{4-1} = 2^{n-1}$, where $n$ is number of chiral centers. Therefore, your formula should have been for pair of enanthiomers.


Temperature Dependence of The Conductivity of Electrolytic Solutions

One useful approximate relationship that can be used to predict the temperature dependence of the conductivity can be obtained by relating the electrical mobility to the viscosity.
For a $z_+:z_-$ electrolyte the limiting molar conductivity can be expressed as
$$\Lambda^\circ _m =  F^2\left(\frac{z_+^2\nu_+}{f_+} + \frac{z_-^2\nu_-}{f_-}  \right)$$
where the $f_i$ are frictional coefficients  for each ion, z is the ionic charge, $\nu$ is the stoichiometric coefficient, and F is the Faraday constant (for instance, for $\ce{Al2O3}$, $z_+=3$,$\nu_+=2$,$z_-=2$,and $\nu_-=3$.
The frictional coefficients can be related to the viscosity, for instance using Stokes' relation $f=6\pi \eta r$ for a sphere being driven through a uniform viscous medium with a low Reynold's number. From the temperature dependence of the viscosity, and assuming the ion radii are independent of T, you can derive a relation for the T dependence of the limiting conductivity:
$$\Lambda^\circ _m = \frac{F^2}{6\pi\eta (T)}\left(\frac{z_+^2\nu_+}{r_+} + \frac{z_-^2\nu_-}{r_-} \right)$$
The equations so far apply to dilute solutions where the viscosity is independent of the concentration of electrolyte and ion-ion interactions can be neglected.
Note that even simpler theories predict a somewhat complex dependence of the molar conductivity on temperature. One can however attempt to fit an expression inspired by those above, of the sort
$$\Lambda _m = A \times \eta (T)^{-1} $$
where A is assumed independent of T. This might work over a narrow T range and could be used to assess the importance of ion-ion interactions.


Why does hydrogen fluoride have a boiling point so much lower than that of water?

I think there are a variety of qualitative ways of looking at this:

Perhaps the most obvious is that $\ce{H2O}$ can form a greater number of hydrogen bonds due to having an equal numbers of hydrogen bond acceptors and donors. Each of the hydrogen atoms can be hydrogen bond acceptors; each of the lone pairs on the oxygen can be donors. In $\ce{HF}$ however there is only one hydrogen bond acceptor and theoretically three hydrogen bond donors. This imbalanced ratio inevitably leaves some acceptors without a donor. Imagine this as two dance parties; one party has two men and two women but the other has one man and three women. Obviously fewer bonds can be successfully formed in the latter, assuming that everything is monogamous.
As mentioned by the other poster, both O and F are very electronegative elements. F however is more EN than O, which means that F stabilizes electrons better than O. As a result, electrons localized on F are weaker hydrogen bond donors; they are already fairly well stabilized. Remember, electrons "want" to be stabilized by nuclei. If they already are fairly well stabilized then they will "feel" less of a need to be associated (and therefore stabilized) with (by) other nuclei. This suggests that the hydrogen bonding present among $\ce{HF}$ molecules might be weaker than the hydrogen bonds present among $\ce{H2O}$ molecules.

Experimental results are the gold standard, of course, as opposed to qualitative reasoning, and qualitative reasoning here can also lead us in the opposite direction; one might argue that by having three lone pairs, F has a lot on its plate, so to speak; it might be able to stabilize one lone pair of electrons very well but three lone pairs is a bigger ordeal, and perhaps big enough that the hydrogen bonds among $\ce{HF}$ molecules are stronger than those among water molecules. Again, this is all qualitative, but this is the sort of reasoning introductory chemistry teachers desire.

If we limit our thinking to just electrostatics then we might surmise that $\ce{HF}$ should have the stronger hydrogen bonds because F is more electron withdrawing and therefore hydrogen should be more positively polarized in $\ce{HF}$ as opposed to hydrogen in water. However, hydrogen bonding is more than just electrostatics. Hydrogen bonding actually has a covalent component; this, however is usually ignored by introductory treatments of chemistry. The bond angle of elements involved in a hydrogen bond is critical. The closer the elements involved in a hydrogen bond are to 180 degrees, the stronger the bond (this specific angle is the case with hydrogen bonding in water; not necessarily other molecules). If hydrogen bonding were purely electrostatic then this would not be the case; angles would not matter - only distance would.

Other issues must be explored as well, such as the number of viable hydrogen bonds and the electron-donating/releasing tendencies of the involved elements.

After scouring the web some common "explanations" that would not explain why water has a higher boiling point than $\ce{HF}$ would be:

  water can form 4 per molecule while HF can only form 2.

Incorrect because if we only look at hydrogen bonding as involving the polarity of atoms, then how can water form four hydrogen bonds per molecule? It has two positively charged hydrogens, and a negatively charged oxygen. Seems like it should only form three hydrogen bonds. One must understand that lone pairs can each be hydrogen bond donors.


Predict the product of reaction of quinone and hydrochloric acid

The product of the addition of one chloride ion (C) is easily oxidized by atmospheric oxygen (radical mechanism). The obtained 2-chloro-1,4-benzoquinone is susceptible to conjugate addition of another chloride ion, this happens chemoselectively at the more electron deficient double bond which has already a chlorine attached to one carbon atom.


Difference in mechanisms between formation of acetal and enamine

This does indeed work. One such reference includes US patent 3696080A in which propionaldehyde is reacted with dimethylamine to give the N,N-acetal as shown in the figure below.

The described reaction procedure involves cooling both reagents to $\pu{-78^\circ C}$ before addition, letting the temperature rise to around $\pu{-35^\circ C}$ at which point anhydrous potassium carbonate was added, letting the temperature rise again to about $\pu{0^\circ C}$ at which point the cooling apparatus was switched from dry ice/acetone to ice/water, and then stirring it overnight allowing the ice to melt in the process. The workup the inventors describe is again rather complex but they claim to isolate up to $\pu{73\%}$ of the bis-amino product.
SciFinder also gave a number of other references but one should serve sufficiently as a proof of concept.
It’s worth noting that $\pu{2.4 eq}$ of amine were added which probably assisted in driving the reaction to the product side.


Based on the Wikipedia article, I am inclined towards thinking that the lithium on the electrodes is neutral. The lithium ions move through the electrolyte to facilitate the two half-reactions. The electrons in the half reaction participate in closing the external circuit, so the lithium ions and electrons are moving in the same direction, except one does so inside the battery and the other outside the battery).

Onto your questions:

The lithium in the anode is neutral, not an ion.
There's no traditional bond per se, but it's stored in the energy difference between $\mathrm{LiC}_{6} + \mathrm{CoO}_{2}$ versus $\mathrm{C}_{6} + \mathrm{LiCoO}_{2}$. There is an interaction here, but I wouldn't call it a "bond."
Well, the anode or cathode isn't more energetic, that's not quite the right term. The $\mathrm{LiC}_{6}$ is more energetic than other species. It just happens that this happens to be the anode of a charged battery. Once the battery discharged, the anode is still that side, but it doesn't have more energy.
Electrical energy doesn't work that way per se. We generally say we build up a potential difference $V = \frac{\Delta E}{q}$. The extraction of energy from this difference is dependent on the circuit. If you have an incandescent light bulb, the kinetic energy of the traveling electron is converted to heat and light, for example.
When the battery is discharged, the lithium atoms have moved to the metal oxide. That's the thermodynamically more stable of the two substrates for the lithium to be adsorbed onto.

I've never looked this up before, but this is actually reall cool. We'll basically using the thermodynamics of lithium preferring to be adsorbed onto one electrode over the other. In order for the lithium atom to get there, it has to move as an ion through the electrolyte and the missing electron moves through the external circuit. I think your confusion is here: the system wants to move the lithium atom, so we'll extract something useful out of it by forcing it to move the lithium ion and the electron separately.


Problem with epoxy silane (Random results in glass silanization). What are crucial points to look at?

The triethoxysilane group has to hydrolyze first in order to bond with a silanol group of glass surface. It seems that you are cleaning the slides very well. However, I think your temperature is too low and your time of exposure is too short. The hydrolysis of ethyoxysilane group is slow.
Secondly, I guess you are posting questions on this topic for almost a year without much success. In such cases, it is always good to go back and try a control experiment and think about a new approach. The control experiment is that you try just neat glass slides and see if your antibody sticks to glass or not. Glass itself is a good adsorber. Probably you might get similar results of "fluorescence, no fluorescence and mild fluorescence". This will confirm that your silane binding procedure is doing nothing!
Second control test: Epoxy silane coated slides are commercially available. Why don't you try using them and see if your immobilization chemistry works on commercial slides or not. Schott makes commercial slides by this name: Nexterion® Slide E
Let me tell you how silanes are bonded onto silica. The reaction time is typically 10 hours under reflux conditions of toluene and other added solvents and catalysts! Sometimes a very very small amount of water is added to created polymer like bonding. There are at least several hundred papers on this topic. The surface area is silica is quite high so it is not a direct comparison with glass slides but still 3 min and room temperature is doing nothing!


Your mechanism would create a primary alkyl cation, namely the ethyl cation ($\ce{CH3CH2+}$). Primary carbocations are extremely unstable and only generated if there are stabilising effects present such as in the benzylic or allylic case — and these are then often not considered primary cations but rather allylic or benzylic ones.

Rather, the protonation of ethanol ($\ce{CH3-CH2-OH}$) creates an oxonium ion as you suggested in the first step:

$$\ce{CH3-CH2-OH + H2SO4 -> CH3-CH2-OH2+ + HSO4-}$$

The $\ce{CH2}$ carbon is then attacked from behind in a nucleophilic $\ce{S_{N}2}$ reaction liberating water in the same step. It is important to note that the attack of another ethanol molecule and the liberation of water happen synchronously; there is no first or second.

$$\ce{CH3-CH2-OH2+ + CH3-CH2-OH -> CH3-CH2-OH+-CH2-CH3 + H2O}$$

Finally, the oxygen which is now part of a protonated ether loses the proton to give the final product diethyl ether.

In the case of a nucleophilic substitution on a substrate like tert-butanol, or even better, tert-butyl bromide, one would consider the leaving group to be liberated first and a carbocation to be created; however, tertiary carbocations are orders of magnitude more stable than primary ones.


What is an appropriate way to write a recursive SMARTS that excludes trifluoromethyl groups as a potential R group?

I'm not sure if I correctly understand your intent.  My assumptions are

you have a set of molecules, e.g., as a list of SMILES strings (say FC(c1ccccc1)(F)F, CCCC(F)(F)F, CC(Cl)(Cl)Cl, and CCCO) in a file like probe.smi
you want to identify the molecules without the $\ce{-CF3}$ group

If so, I would i) search for the molecules with said group in a first step and then ii) remove this sub-set of positives from the set to know about the molecules without the $\ce{-CF3}$ group.

OpenBabel's positives: obabel -ismi probe.smi -s'C(F)(F)F' -osmi -O positives.smi.  In the present example, this yields only FC(c1ccccc1)(F)F and CCCC(F)(F)F, regardless of the different notation of the trifluormethyl group in the entries concerned.

The molecules lacking the $\ce{-CF3}$ group are just the complementary sub-set to the sub-set of the positives.  To start, one could opt for a Python script like:

#!/usr/bin/python3
"""Report molecules lacking the -CF3 group."""

def reporter():
    """Report .smi of molecules without -CF3 group."""
    # prepare lists about the molecules:
    all_mols = []  # all molecules, no discern
    positives = []  # molecules with -CF3
    negatives = []  # molecules without -CF3

    with open('probe.smi', mode='r') as source:
        for line in source:
            all_mols.append(line.strip())

    with open('positives.smi', mode='r') as source:
        for line in source:
            positives.append(line.strip())

    # truncate the set to yield the sub-set about molecules without -CF3 group:
    set_all_mols = set(all_mols)
    set_positives = set(positives)
    set_negatives = set_all_mols.difference(set_positives)

    # convert the sub-set (back) into a list of entries:
    negatives = list(set_negatives)

    with open('negatives.smi', mode='w') as newfile:
        newfile.write('\n'.join(negatives))

if __name__ == "__main__":
    reporter()

which in file negatives.smi correctly reports CC(Cl)(Cl)Cl and CCCO as molecules without the $\ce{-CF3}$ group.

Of course it would be nice to identify how to negate OpenBabel's search filter (-s'C(F)(F)F') to shorten the route traveled here.


pKa of phenols using UV-Vis spectroscopy

To quote the key points of an easy to read publication by Salgado and Vargas-Hernández (doi 10.4236/ajac.2014.517135, open access):

All starts with the dissociation equilibrium between the acid $\ce{HA}$ and its anion $\ce{A-}$ for which you write

$$\ce{HA + H2O <=> H3O+ + A-}$$

By consequence of this, the recorded total absorbance $A_\mathrm{t}$ for system, assuming only the acid and its anion absorb light in the region inspected, equates to

$$A_\mathrm{t} = d(c_{\ce{HA}} \cdot \varepsilon_\ce{HA} + c_\ce{A-} \cdot \varepsilon_\ce{A-}) = d \cdot c_\mathrm{t} \cdot \varepsilon_\mathrm{t}$$

with $c_\mathrm{t}$ as total concentration; $\varepsilon$, the absorption coefficient; and $d$, the optical path length.

For one, you record a spectrum at low pH to convert all of your sample into the acid form ($c_\mathrm{t} = c_\mathrm{HA}$, and $c_\ce{A^-} = 0$) where all absorption is due to the acid.  Equally, you record a spectrum at a pH this high that all or your sample is deprotonated (then  $c_\mathrm{t} = c_\mathrm{A^-}$).  You need then spectra between the two extrema, with some of $\ce{HA}$ in presence of some of $\ce{A-}$, because the proportion of the concentrations of acid to its anion may be expressed by

$$\frac{c_\ce{A-}}{c_\ce{HA}} = \frac{[\ce{A-}]}{[\ce{HA}]} = \frac{A_\mathrm{t} - A_\ce{HA}}{A_\ce{A-} - A_\mathrm{t}}$$

This is why you search for the isobestic point, where the absorption spectra recorded do not vary while varying the pH value of your analyte, and where $\varepsilon_\ce{HA} = \varepsilon_\ce{A^-}$.

(credit: doi 10.4236/ajac.2014.517135)

Because the equilibrium between the acid and its anion may be described by

$$\log{K_\mathrm{a}} = \log[\ce{H3O+}] + \log{\frac{[\ce{A^-}]} {[\ce{HA}]}} $$

and the definition of $pK_\mathrm{a} = - \log{K_\mathrm{a}}$ you may substitute absorptions by concentrations to yield

$$\log{\left(\frac{A_\mathrm{t} - A_\ce{HA}}  {A_\ce{A-} - A_\mathrm{t}}\right)} = \mathrm{pH} - pK_\mathrm{a} $$

Hence, plotting $\log{\left(\frac{A_\mathrm{t} - A_\ce{HA}}  {A_\ce{A-} - A_\mathrm{t}}\right)}$ in function of $\mathrm{pH}$ yields a slope with an intercept of $y = - pK_\mathrm{a}$.

(credit: loc. cit.)

Addition:
With WebPlotDigitizer it was possible to  convert above mentioned figure 8 into numbers, separate for the basic (b), the acidic (a) and the intermediate / about neuter run at pH = 7.7 (n).  (The result, a .csv file, is deposit here.)  For a wavelength of about $550\,\mathrm{nm}$ the readout of Abs equals to 1.9935 (b), 0.6502 (n) and 0.2241 (a), and the computation

$$ pK_\mathrm{a} = 7.7 - \log_{10}{\frac {0.4534 - 0.2984} {1.2121 - 0.4534}} $$
$$ pK_\mathrm{a} = 7.7 - \log_{10}{\frac {0.1550} {0.7587}} $$
$$ pK_\mathrm{a} (550\,\mathrm{nm}) = 8.39 $$

By similar token, $pK_\mathrm{a} (604\,\mathrm{nm}) = 8.20$ (with Abs of 1.9935 (b), 0.6502 (n), and 0.2241 (a)); or $pK_\mathrm{a} (432\,\mathrm{nm}) = 8.51$ (with Abs of 0.3411 (b), 0.8198 (n), and 0.8940 (a)).

To put this into perspectivive, the arithmetical mean of the four methods compared in Salgado's paper equals to $pK_\mathrm{a} = 8.277$.  Possibly, the difference between their result and the «recuperation»  here were smaller if either a) WebPlotDigitizer were used with a lesser data increment, b) using an interpolation / smoothing function among the data extracted by the program, c) figure 7 were reconstructed from their figure 4.


No, the phosphate functional group, $\ce{P(O)(OR)_3}$, itself is not acidic. Only protic phosphates are acidic. Phosphates that aren't phosphoric acids (i.e. $\ce{P(O)(OH)_{x}(OR)_{3-x}}$) are actually Lewis bases. The carbonyl oxygen is able to coordinate metals and other Lewis acids.


You have three variables, for simplicity let's call them, x, y, z each with a (standard deviation) uncertainty $\sigma_x, \sigma_y, \sigma_z$. The formula you want to find the error on is then $f=xy/z$ which will have a standard deviation $\sigma_f$ The method that always works is the  the formula that gives the square of the uncertainty
$$ \sigma_f ^2 = \left( \frac{\partial f}{\partial x}\right)^2_{y,z}.\sigma_x^2+\left( \frac{\partial f}{\partial y}\right)^2_{x,z}.\sigma_y^2+\left( \frac{\partial f}{\partial z}\right)^2_{x,y}.\sigma_z^2$$

which looks horrid and intimidating  but is very easy to use as it only involves taking the derivative of f with x, y and z in turn, while keeping the other variables constant.

In your example $\partial f/\partial x = y/z$, $\partial f/\partial y =x/z$ and $\partial f/\partial z =-xy/z^2$. Putting the parts together gives
$$\sigma_f^2= \frac{y^2}{z^2}\sigma_x^2 + \frac{x^2}{z^2}\sigma_y^2 +\frac{x^2y^2}{z^4}\sigma_z^2 $$

You can simplify this a bit but you should have all the values to plug into this. Don't forget to take the square root at the end.


Come on, you can figure out this one on your own! How do you compare acidities? Here's a helpful hint.

Armed with that information, I have to add something. I believe the mechanism as drawn above is not universal: it sure depends on the conditions and solvent used. Some other proposed mechanisms involve the enol, rather than the protonated ketone, which might be more plausible in mild conditions. In water, anyway, HBr is fully dissociated and it shouldn't be “reformed” like that. The proton will return to be solvated in the solution.


One can rarely tell a priori what the steps are, and the results obtained experimentally may seem not intuitive at all if one tries to deduce the dehydration steps solely from the crystal structure (e.g. from the arrangement of the aqua ligands and water molecules trapped within the crystal lattice).¹

TG-DTG-DTA techniques, often coupled with powder XRD, are commonly used to determine the exact process behind each step interpreting mass change over time.
Alternatively, you can deduce the possible steps from the initial structure, but this approach is less accurate.

So, for the classical textbook example of thermal decomposition of copper(II) sulfate pentahydrate the following steps of dehydration (1–2) and calcination (3–4) can be distinguished [1]:

$$
\begin{align}
\ce{CuSO4 * 5 H2O &→ CuSO4 * H2O + 4 H2O}\tag{1}\\
\ce{CuSO4 * H2O &→ CuSO4 + H2O}\tag{2}\\
\ce{2 CuSO4 &→ CuO * CuSO4 + SO3}\tag{3}\\
\ce{CuO * CuSO4 &→ 2 CuO + SO3}\tag{4}
\end{align}
$$

You could also assume the loss of four water molecules as a single step since in the crystal structure copper(II) atom has octahedral environment with four $\ce{H2O}$ and two $\ce{SO4^2-}$ ligands.
Writing chemical equation for each act of the water molecule loss is not a sin, but it makes little practical sense as the intermediate crystallohydrates are unstable and rapidly undergo further dehydration.

Notes

¹ From the crystal structure one may assume that during the first step only one water molecule is eliminated from the lattice (Figure 2a) since copper(II) is coordinated octahedrally by four water molecules in equatorial area and has two sulfate ligands in apical positions (Figure 2b).

Figure 2. Crystal structure of $\ce{CuSO4 * 5 H2O}$, data from CIF for Chalcanthite. Color scheme: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$. a: Unit cell content. b: Coordination environment and nearest neighboring water molecules of the copper(II) center.

However, this is not the case even though lattice water molecules are kept in place via H-bonds and should posses seemingly weaker interaction with the cationic part.
In reality many assumptions suitable for estimation of melting points of molecular crystals, e.g. regarding the number of bonds and their types don't work well for ionic crystals and so the dehydration patterns can only be established exactly experimentally.

References

Gadalla, A. M. Kinetics of Thermal Decomposition of $\ce{CuSO4 · 5H2O}$ to $\ce{CuO}$. International Journal of Chemical Kinetics 1984, 16 (6), 655–668. https://doi.org/10.1002/kin.550160604.


What is the correct Lewis structure of NH3O?

Technically, both the structures $\ce{H2N-OH}$ and $\ce{H3N\bond{->}O}$ may exist. However, in reality hydrogen atom is rather prone to migration and the second structure is not favorable. So, for a compound with composition $\ce{NH3O}$, the correct structure would be $\ce{H2N-OH}$. The structure of second type is stable for compound $\ce{NOF3}$ and may be observed for amine oxides like $\ce{(C2H5)3N\bond{->}O}$ or pyridine N-oxide.

Why $\ce{H2N-OH}$  and not $\ce{H3N\bond{->}O}$? One  can come with several explanations for that, but I would like to focus on one: charge distribution. In $\ce{H3N\bond{->}O}$, the nitrogen has to carry a formal positive (and actually very real) charge, while for $\ce{H2N-OH}$, the structure has no formal charges. Still, according to some sources[1] up to 20% of hydroxylamine in water exists as ammonia oxide $\ce{H3N\bond{->}O}$, probably, due to stabilization by hydrogen bonds.

Somewhat similar uncertainty may be found for phosphorus compounds (like hypophosphorous acid) and sulfur compounds (like sulfinic acids).

Kirby, A. J.; Davies, J. E.; Fox, D. J.; Hodgson, D. R. W.; Goeta, A. E.; Lima, M. F.; Priebe, J. P.; Santaballa, J. A.; Nome, F. Ammonia oxide makes up some 20% of an aqueous solution of hydroxylamine. Chem. Commun. 2010, 46 (8), 1302. DOI: 10.1039/b923742a.


What is occuring on the quantum level when a molecule rotates plane polarized light?

As mentioned in the comments, the answer to your $2^{nd}$ question regarding why optical rotation isn't simply cancelled out in solution due to random orientations is given in ManishEarth's answer to Molecular chirality and optical rotation

For your first question, I will try to explain what actually causes rotation when light impinges on an optically active molecule. For chiral molecules, it is simplifies the picture to think of each enantiomer as a screw of negative charge that is either right or left handed. Plane or linearly polarized light can always be decomposed into circularly polarized clockwise and counterclockwise components. When the light collides with the molecule, the clockwise component and counterclockwise components interact differently with the electric field of the molecule, leading one of these components to travel faster than the other, which alters the angle of the plane polarized light. A good visualization of this effect (for a whole solution rather than an individual molecule) can be found at http://cddemo.szialab.org/.

As an interesting aside, you might imagine that because we have a reasonable understanding of why optical rotation occurs, we should be able to predict, either heuristically or from a first principles calculation, the magnitude and direction of optical rotation for a molecule just based on its structure. This, thus far has not been the case. Not only are there very few ways of heuristically guessing the magnitude/direction, but quantum mechanical calculations at high levels of theory have proven unable to even consistently obtain the correct direction of the optical rotation. Given a particular chiral molecule, we still can't predict which direction, let alone magnitude, that a particular enantiomer will rotate light.

My main point in this little tangent is that even something as seemingly simple and in such widespread use as optical rotation is still not very well understood and is a difficult area of active research.


Does temperature change pressure in isochoric system?

Since it's given that ΔH > 0 (endothermic reaction) ,the reaction will proceed backward (towards the reactant). However as you know ,
Partial Pressure Of Gas= Total Pressure x Mole Fraction;
hence , even if the pressure is affected , the mole fraction increases/decreases in a way so that partial pressure remains constant.
If you calculate Kp ( Pressure Equilibrium constant) you will know that it depends on the partial pressure of the gases involved . Since partial pressure remains same no effect takes place. Hence , pressure  will have no effect on the equilibrium.

Also note that equilibrium constant is only affected by the change in temperature.
Hope this helps.

EDIT:-
In the light of the comments I would like to clarify ,

Partial pressure of a particular gas is the fraction of the gas multiplied by the total pressure which means it exerts a fraction of the total pressure. When I increase the total net pressure exerted by all gases , the fraction of moles of each gas decreases (according to the mathematical equation) thereby contributing lesser to the total pressure .

You can think of it as a ratio:-

If 1 mole X Gas exerts 10 bar pressure when total pressure is 20 bar , 0.5 moles of X gas will exert the same 10 bar pressure when total pressure is increased to 40 from 20 , thereby keeping the ratio same.


Why don’t we have a list of all compounds yet?

Because the combinatorial complexity is far to big for practical computation
One of the simplest possible cases that illustrates why this is impossible is given by restricting the choice of atoms to just carbon and hydrogen, forbidding any double bonds and disallowing any rings. This becomes equivalent to enumerating the number of possible mathematical trees of a certain structure a calculation interising in graph theory as well as chemistry.
Graphs are not exactly the same as hydrocarbons as there are spatial constraints on the molecules that can exist in 3D space and also because the molecules are in 3D, some variants with the same graph may be chemically different due to chirality (see this answer for more details). But Some calculations based on just 50 carbons put the number of possible variants at >1021 (see this wikipedia page) which is a ridiculously big number.
And this is is with just two elements from the hundred or so that could be used and it is ruling out most of the interesting structural features that could add a great deal of extra complexity. All this complexity arises from the carbon skeleton alone and it is already beyond any computational capacity we can contemplate having (just using one byte per compound would use a great deal more storage than at least one estimate of the total world capacity to store information.
So, if you still think this sounds like an easy task, you haven't done those calculations or appreciated the gargantuan scale of the resulting numbers.


I'm a little confused what OP's argument is. If you say "it should be a meta attack for methoxide group," then the question is showing an electrophilic substitution at meta-position to methoxy group. But, I understand you are wondering why substitution is at the meta-position to methoxy group rather than on ortho-position to methyl group. If that's what you meant, this is the answer to your curiosity:

The step in question is an example of Friedel-Crafts reaction. More precisely Friedel-Crafts acylation reaction. Here though, the substrate (aromatic nucleus) and the reagent (acyl chloride) are in the same molecule unlike any other typical inter-molecular reactions. In any inter-molecular Friedel-Crafts reaction involving 4-alkylanisole, substitution of acyl group mainly occurs on ortho-position to methoxy group, as you pointed out that "an oxygen atom's conjugation would be more contributing than carbon's hyperconjugation and inductive effects combined." I said here mainly because there would be some minor product from the substitution at ortho-position to alkyl group as well. For example, according to my experience, when Friedel-Crafts acylation reaction is performed in toluene and chlorobenzene, there were mixtures of products including o- and p-chloroacylbenzenes.

Now, we look at the problem in hand. This reaction come down to inter- versus intra-molecular acylation reactions. It's always intra-molecular reaction is much more faster than same type of inter-molecular reactions (here is acylation reaction of both kind) due to proximity. Thus, this phenomenon overcomes the deficiency of activation by methyl group versus methoxy group here, and gives most thermodynamically stable product, 7-methoxytetralone as the major product.

Edit: As OP's request, I include here an example for intra- vs inter-molecular acylation reaction (Ref.1):

Reference:

Yingju Xu, Mark McLaughlin, Cheng-yi Chen, Robert A. Reamer, Peter G. Dormer, Ian W. Davies, "A General Method for the Synthesis of 3,5-Diarylcyclopentenones via Friedel−Crafts Acylation of Vinyl Chlorides," J. Org. Chem. 2009, 74(14), 5100-5103 (https://doi.org/10.1021/jo900696k).


Does aluminium foil have an odour/gas off?

Here is related material I once posted:

Also, commentary in Bretherick, Volume 1, page 23-24, to quote:
A student mixed aluminium foil and drain cleaner in a soft drink bottle, which started emitting gas. Another student carried the bottle outside and was claimed to have been overcome by the toxic fumes [1]. Most drain cleaners are alkalis, so that aluminium will dissolve to produce hydrogen. The bleach that it is is suggested may have been present will produce no toxic fumes in alkali, and one would be surprised to find arsenic or antimony compounds present. If the collapse was not purely hysterical, the remaining, though remote, possibility would be phosphine. The soft drink the bottle had contained was one of the many perhaps best described as impure dilute phosphoric acid [2]. "
A related comment attributing the formation of gases like arsine and stibine from nascent hydrogen per this source.
Education on nascent hydrogen interaction is important so as to avoid toxic gases that may be inadvertently and unexpectedly created in significant amounts.

Now, in the context of extensive employment of Al foil in your home, there is a potential danger that an acid/base is attacking the foil. This could release active (nascent) hydrogen atoms.
The latter could interact, for example, with compounds in physical contact, like paint (containing perhaps zinc sulfide). This may, hypothetically, result in the release of odorous hydrogen sulfide. This is a problematic gas, but even more deadly gases could arise also (per source above).
As such, I would advise removing the Aluminum foil (or, at least areas in contact with paints,...).


How to write electronic configuration of p-orbitals?

Electrons don't pair up in orbitals before each is singly occupied by Hund's rule of maximum multiplicity.
Writing out px, py, pz fully or using p^n notation (n=no of electrons in a p orbital) is a matter of context and circumstance rather than the standard convention; you might use the 'full version' when explaining bonding through hybridization but omit it while stating the electronic configuration of an atom generally. Both notations are quite interchangeable.


What are introductory level books on chemical engineering?

I'm studying chemical engineering for more than two years and what I can tell you is that there is not really a lot of "good" books you may read as an introduction.
You have to understand that Chemical Engineering contains both "Chemical" and "Engineering". The first one refers to chemistry whereas the second refers to the way to make something. A chemical engineer who works at Metalor (which makes precious metal bars amidst other things) really has not the same problems as another one working at Air Liquide (which produce liquid gas).

Now if you are looking for what we call the basics, any chemical engineer must be good at, fluid mechanics, thermodynamics and kinetics, mathematics and informatics are a good start.

Fluid mechanics because in any plant, whatever you produce, you need to be able to size pipes, reactors, etc. It's also a must if you encounter a two-phase flow;
Thermodynamics because when you are sizing somethings you must ensure safety and security so you need to know what properties the molecules involved in your plant have. Kinetics because this is important to know the rate of a reaction and a lof other stuff related to catalysts etc.;
Mathematics because you can't do the first three without it;
Informatics because this is pretty hard to perform all the correlations and models by hand when sizing equipment and knowing how to use the common software (Aspen, Matlab) plus having some skills in coding a few lines in either Fortran/C/C++/Python is always a must. Especially if your company does not have the budget to pay the licences and Excel is limited in many ways;
Finally, mathematics and informatics are a strong requirement to design process and safety control loops efficiently and test them with data from the process before implementation.

If you speak French, the book "Le génie chimique à l'usage des chimistes" by Joseph Lieto gives a pretty good overview about Chemical Engineering. How to make a balance (of mass or energy or momentum), basic operation units (distillation and crystallisation), mass and heat transfer, thermodynamics, kinetics and process control.

The best reference I have for chemical engineering is the McGraw-Hill Chemical Engineering Series which contains more than you should know as a student. After you will need to get strong knowledge in many areas of chemistry (electrochemistry is sometimes a big plus) because as an engineer and a chemist especially if you are looking for sizing operation, you need to think about details (Production practices, HSE, Quality), which are not in the books, unfortunately.


Steady State Approximation final integral proof $\frac{d[P]}{d t}=k_{a}[A]$ yielding $[P]=\left(1-e^{-k_{a} t}\right)[A]_{0}$

There are a couple of ways to approach this problem. Bear in mind this problem requires some calculus.

$\boldsymbol{1^{st}}$ Way (repeated integration): To start, you were able to recognize the first order integrated rate law.

$$\frac{d[A]}{dt} = -k_a[A] \implies [A] = [A]_0e^{-k_at}$$

This result may be plugged into the rate equation for $P$ to obtain

$$\frac{d[P]}{dt} = k_a[A] = k_a[A]_0e^{-k_at}$$

This equation is directly integrable.

$$\int_0^t\frac{d[P]}{dt}\,dt = \int_0^tk_a[A]_0e^{-k_at}\,dt$$
$$[P] - [P]_0 = -\frac{k_a[A]_0}{k_a}\left(e^{-k_at} - 1\right)$$
$$\boxed{[P] = [A]_0\left(1-e^{-k_at}\right)}$$

The last step applies the knowledge from the problem that $[P]_0=0$.

$\boldsymbol{2^{nd}}$ Way (conservation law): orthocrestol in the comments above pointed out that the sum of the concentrations of all chemical species should equal to some constant in this problem. Call this constant $C$ and ignore $I$ due to the steady state approximation. Thus,

$$[A] + [P] = C$$

Now substitute this into your rate law for $P$.

$$\frac{d[P]}{dt} = k_a[A] = k_a(C - [P])$$

This differential equation is separable so we can use the neat tricks to integrate this equation.

$$\int_0^{[P]}\frac{d[P]}{C-[P]} = \int_0^tk_a\,dt$$
$$-\ln\left(1-\frac{[P]}{C}\right) = k_at$$
$$[P] = C\left(1-e^{-k_at}\right)$$

Almost done. Now recognize that at time zero, we can write the conservation law as

$$C = [A]_0 + [P]_0 = [A]_0 + 0 = [A]_0$$

This makes physical sense because since $C$ is a constant, this value cannot change at any time. Moreover, the total number of molecules at any time should be the same as the amount of molecules that you start with due to stoichiometry (everything is 1:1 in this particular problem). So make the substitution to get your answer.

$$\boxed{[P] = [A]_0\left(1-e^{-k_at}\right)}$$


Why is a spin change favourable in intersystem crossing?

There are two things going on here. You have conflated with "forbidden" with unfavorable. In many systems, the first excited triplet is lower in energy than the first excited singlet. Hence, the order of energies excited singlet > triplet > ground state singlet is correct and sensible. If your question is "why is this the order?" that should be a separate question.

The forbidden part of the transition has to do with the fact that, because of conservation of (electron spin) angular momentum, you can't excite directly from a singlet ground state to a triplet excited state with a photon. The forbiddenness is a statement about not being able to get there, not unfavorability. The triplet is still lower in energy and the molecule has to do some other shenanigans (often involving spin orbit coupling) to do the intersystem crossing into the lower energy triplet state. If your question is "what is the mechanism of intersystem crossing?", that should be a separate question.


Surface tension and evaporation are not as linked with each other as posed by your question. They are two distinct characteristics of liquids which generally follow direct relationship but not always. There are several other factors which you should not ignore:

Surface tension is the inherent force of liquid described by its tendency to minimize its surface area. A liquid with high molecular attraction takes account for a higher surface tension. It's temperature dependent and does not have similar slopes for all liquids. The case you provided is for standard temperature i.e. 20 C so you may get a contrary result at some other temperature. Compare: http://www.ddbst.com/en/EED/PCP/SFT_C11.php and http://www.ddbst.com/en/EED/PCP/SFT_C4.php2.
Evaporation usually deals with heat capacity, thermal energy required to overcome inter-molecular attractions and other thermal properties. Usually higher surface tension means energy required to overcome inter-molecular attractions is higher hence it may also have a higher value for the liquid having higher surface tension but it is not necessary. So liquids resembling each other in thermal properties usually form intersecting surface tension curves. i.e. one have a larger value till intersecting temperature and later on second has more. Ethanol and propanone is also such borderline case to be compared.


Software for generating a lattice from a unit cell for molecular crystals?

In addition to the presentations about CCDC Mercury and Vesta by @andselisk here and later mine about Avogadro could be extended.  As described earlier, Avogadro's capability to read .cif files about crystallographic models allows you to extend these  via the slab builder to any extent you need (GUI entry Crystallography -> Build -> Slab) and to export them either as normal .xyz file, or (via the extension tab, template assisted) in one of multiple formats recognized by quantum chemical programs (e.g., Gaussian, GAMESS-UK, MOPAC).

It equally is possible to start from scratch with a cubic unit cell of 3 Å.  Subsequently, you would alter the lattice parameters ($a, b, c; \alpha, \beta, \gamma$) and enter atoms one, after the other, either in cartesian coordinates (either Angstrom, Bohr, nm or pm), or in fractional atomic coordinates as typically in .cif files. You may wrap the atoms into the cell, and of course impose one of the space group symmetries to fill your cell, too.  You may symmetrize your constructed unit cell and reset (within tolerances you determine) into a standard setting, too. Equally than just reading a .cif,.pdb, or .xyz, these constructed unit cells may be extended with the slab builder, and equally may be exported in one of the computer chemistry formats mentioned earlier, too.


Test for Fe2+ ion in salt using potassium hexacyanidoferrate(III)

$\ce{Fe^2+}$ is oxidised to $\ce{Fe^3+}$ while $\ce{[Fe(CN)_6]^{3-}}$ is reduced to $\ce{[Fe(CN)_6]^{4-}}$. These then form the prussian blue precipitate.


Say dilute HCl is used to convert some alkaloid to salt form, then the alkaloid is dried with heat, would gaseous HCl be released in the process?

Likely yes, unless it's the limiting reagent in the reaction in question. Probably best to assume some of it will come off as gas - that's the safest approach, in my opinion.

  Is there a threshold concentration of the acid, below which, this would be acceptable from a safety point of view?

Yes, there are several metrics used by various authorities to determine what "safe" means. You'll have to decide which one (or ones) are relevant to your situation.

Here in the U.S., the Occupational Safety and Health Administration (OSHA) - which is under the Department of Labor - has published this information on hydrochloric acid exposure limits and hazards.

In that linked document, several other agencies such as the National Institute for Occupational Safety and Health (NIOSH) and the American Conference of Governmental Industrial Hygienists (ACGIH) list permissible exposure limits ranging from 3 to 7 mg/m$^{3}$; the Environmental Protection Agency (EPA) given an inhalation reference concentration (RfC) of 2 mg/m$^{3}$, detailed here as the concentration above which (I'm paraphrasing) bad things start to happen.

Good luck on your project and stay safe.


How do phase transfer catalysts bring anions to organic phase?

Actually, anions are not unstable in organic phases per se. There is no intrinsic physical property of organic phases that would make anions inherently unstable: the main problem is that most heteroatoms in organic solvents are more electronegative than carbon and therefore carry a slightly negative charge which attracts cations more easily than anions.

Most organic phases are best classified as ionophobe, however. They will not like cations or anions in their vicinity. Remember that organic phases are typically mainly hydrocarbons with the odd heteroatom, so they have a low polarisablility. Water, on the other hand, is a highly polarised dipole and able to stabilise both cations and anions easily with the appropriate side of the water molecule.

Now if we add a phase transfer catalyst to a biphasic mixture of water and an organic solvent — typical examples would be $\ce{Bu4N+ I-}$ or crown ethers — then these will readily transfer themselves into the organic phase. In the case of crown ethers, it is because they are basically already like the organic solvent: a hydrocarbon with a few heteroatoms. Crown ethers can, however, piggyback a coordinated cation at practically no extra cost: the cation is directly solvated by the crown ether’s oxygens. In the case of the tetrabutylammonium cation, the charge of the tetravalent nitrogen is somewhat shielded by the four butyl residues which interact much better with other organic residues than with water.

This means that we have a general positive charge in the organic phase due to the crown-ether solvated cations or tetrabutylammonium cations present in there. This in turn draws anions towards the cationic charge due to electrostatic interactions.

We can now also look at some basic principles explaining London interactions: generally, the more disperse the orbitals of a species are, the easier this species can be polarised and thus the better London interactions can be. Cations are species in which there is an overall deficiency of electrons so orbitals are drawn closer to the nuclei. Anions, on the other hand, have a surplus of electrons, meaning that these will repel each other more and orbitals will be more diffuse. Compare neutral atom radii to ionic radii to confirm this. The more diffuse anions will therefore have an easier time forming London interactions with the hydrocarbon solvent than the less diffuse cations.

Thus, in summary, one typically needs the ion pair to cross the phase boundary. An anion might transfer by itself but the cation needs help; thankfully, cations can be transferred easily with crown ethers or tetrabutylammonium compounds because these are more able to interact well with the solvent.


Because the electrons in the double bond C=O help to stabilize the CH3 group after it has released an H+ (it is similar in this way to what happens with carboxylic acids), the Acetyl groups are acid. Hence the pH drop.


Nitrogen trifluoride is extremely inert to hydrolysis, attack by base or acids or oxidation because of the fact that there is no d-orbitals in Nitrogen trifluoride and fluorine is far more electronegative than nitrogen. However, in the presence of base or acids, nitrogen trifluoride can however undergo hydrolysis because there is now excess of nucleophile which can attack nitrogen trifluoride.

There is a old paper which discuss the hydrolysis of nitrogen trifluoride in presence of acids and base.

  In the presence of aqueous base, however, slow hydrolysis occurs at
  100°C yielding nitrite and fluoride.

  $$\ce{NF3 + 4 OH- ->[100℃] NO2-  + 3F- + 2H2O}$$

It further says:

  This behavior differs sharply from that of nitrogen trichloride,
  which is known to give ammonia and hypochlorite under similar con-
  ditions. The latter products are readily explained in terms of
  nucleophilic attack directed at the chlorine atoms, a mechanism which
  appears reasonable in view of the fact that the electronegativities
  of $\ce{N}$ and $\ce{Cl}$ are very nearly the same and that the
  halogen may easily  expand its valence shell. Obviously, these
  considerations cannot be  applied to the nitrogen trifluoride molecule
  since fluorine is considerably more electronegative than nitrogen and
  it has no available  d orbitals. Although the nitrogen atom also has
  no free orbitals,  the relatively low electron density would at least
  offer less resistance to the approach of a nucleophile .

  The proposed intermediate $\ce{HONF2}$ would be expected to be
  unstable with  respect to the loss of $\ce{HF}$, as is apparently the
  case with the unknown  perfluoro alcohols .

  Two other reaction modes worth considering include the formation of
  an intermediate activated complex with water.


Balancing redox equation with oxygen on both sides?

It is always easy to balance redox equations by considering two half reactions. The two relevant half reactions to your redox equation are:
$$ \begin{align}
\ce{As2O3 (s) -> H3AsO4 (aq)} \\
\ce{NO3- (aq) -> N2O3 (aq)}\\
\end{align}
$$

First balance $\ce{As}$ and $\ce{N}$ in both equations. Then, balance $\ce{O}$ in both equations with water and $\ce{H}$ with $\ce{H+}$:

$$ \begin{align}
\ce{As2O3 (s) + 5 H2O (l) -> 2 H3AsO4 (aq) + 4 H+ (aq)}\\
\ce{2 NO3- (aq) + 6 H+ (aq) -> N2O3 (aq) + 3 H2O (l)}\\
\end{align}
$$

Finally, balance the charges by electrons:
$$ \begin{align}
\ce{As2O3 (s) + 5 H2O (l) -> 2 H3AsO4 (aq) + 4 H+ (aq) + 4 e-}\\
\ce{2 NO3- (aq) + 6 H+ (aq) + 4 e- -> N2O3 (aq) + 3 H2O (l)}\\
\end{align}
$$

Now add the last mass and charge balanced two equations in order to cancel electrons (this is easy, because both have 4 $\ce{e-}$s):

$$ \ce{As2O3 (s) + 2 NO3- (aq) + 2 H+ (aq) + 2 H2O (l) -> 2 H3AsO4 (aq) + N2O3 (aq)}$$
Or simply,
$$ \ce{As2O3 (s) + 2 NO3- (aq) + 2 H3O+ (aq)  -> 2 H3AsO4 (aq) + N2O3 (aq)}$$


Do molecular orbital diagrams exist for ionically bonded compounds and molecules?

Molecular Orbital theory is a simplified approach to describe chemical bonding based on symmetries and orbital overlap. In principle, it makes no assumptions on the type of bonding, i.e., covalent, ionic, Van der Waals. This distinction is more a construct to understand the strength of the bond.

In case of a covalent bond, the electrons that make up the bond are shared by both nuclei, whereas in an ionic bond the bonding originates mostly from the Coulomb interaction, that is, the electrons are mostly located on a single nucleus. In MO theory, this can be understood in terms of the interaction strength. When the atomic orbitals that make up the bond have similar energy (and same symmetry), there will be a strong interaction and the resulting MO will have character of both AOs. In other words, the electrons will be shared more or less equally by both atoms. When there is a large difference between the AOs, the resulting bonding MO will have mostly the character of the AO which is lowest in energy and as such the electrons in this MO stay mostly around this atom. This resembles an ionic bond.


Why do people consider caffeine ineffective after its half-life?

Personal experience: Some years ago, I mowed our lawn on a hot summer day. It took about 2 hours, and I sweated up a storm. To compensate, every 20 minutes or so, I came into the house and swigged down some Pepsi Cola. After I was finished, I noticed that the 2 liter bottle of Pepsi was also finished, and I was tired, so I sat down.
With nothing else going on, I noted that my heart was skipping beats. 1,2,3,skip. 1,2,3,skip. Extra systole. Nothing to get excited about. But never this skippy. So I listened, and checked, and after about 15 minutes, I got 1,2,3,4,skip. 1,2,3,4,skip. Interesting. And about every 15 minutes, my heart would go another beat before the skip. It finally got out to 14 before skipping, then went all crazy: 12, 17, 20, then finally stopped skipping.
It took 2 hours to consume the Pepsi, and about 4 hours to overcome the observable effects of the caffeine. There is certainly a non-linearity involved with the excretion and with the effect of the caffeine on my heart, but yeah, the half life says something, but there is a diminution of effect, not a go/no go.
And I never again drank that much caffeinated beverage all at once.


Question about the definition of an isotopes

Usually the term "relative atomic mass" for an element is a weighted average of the atomic masses of the various isotopes of an element, with the weighting determined by their abundance. So, since the atomic masses of various isotopes enter into this determination, your first answer is not too good.


Subsection P-44.4.1.10.1 in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads

  For the endings 'ene' and 'yne' lower locants are assigned first to the endings as a set without regard to type and then to 'ene' endings.

Since the locant set "1,3" is lower than the locant set "2,4" (see here for a discussion of what that means), the appropriate choice of locants for the above compound leads to the name pent-3-en-1-yne.

As an aside, the configuration of the double bond, if unspecified, should be labelled with the symbol Ξ (P-91.2.1.2.1).

  When the configuration is not known or must remain unspecified for lack of configurational homogeneity, the italicized symbol 'ξ' or 'Ξ' is used, preceded by the required locant. The symbol 'ξ' (small Greek letter 'xi') replaces noncapitalized CIP stereodescriptors such as 'r', 's', 'm', 'p'. The symbol 'Ξ' (capital Greek letter 'xi') replaces capitalized CIP stereodescriptors such as 'R', 'S', 'M', 'P', and nonCIP stereodescriptors 'E', 'Z', 'seqCis', and 'seqTrans'.

The preferred IUPAC name (PIN) for the compound with unspecified double bond geometry is therefore (3Ξ)-pent-3-en-1-yne.

Of course, if the configuration is specified, then the (E)/(Z) labels should be used instead.


Is it possible to make polyethylene in a high school laboratory?

Not polyethylene, but I recall doing an experiment in school where hexamethylenediamine and adipic acid were poured carefully so one was layered on top of the other.  They did not mix in bulk, but reacted at the contact surface to form a Nylon-type polymer barrier.  We poked a rod in to pull the polymer out (enabling more reaction) and wind it around the rod!


Is lithium bicarbonate an aqueous solution of lithium carbonate?

Lithium bicarbonate solution is something like a lithium carbonate solution in the sense that precipitation would produce the carbonate instead of the bicarbonate.  But what is in solution is mostly lithium and bicarbonate ions.  Precipitation is accompanied by a reaction in which the highly polarizing lithium ions react with bicarbonate:

$2\ce {Li+}+2\ce {HCO_3-}\rightarrow \ce {Li2CO_3}(s)+\ce {H2O}+\ce {CO2}$

Other alkali metals give less polarizing ions and do not react in the same way; they just precipitate the bicarbonate salt.


Exceptions to the Nitrogen Rule in mass spectrometry

The reason for the rule is that, outside of transition metal complexes, stable free radicals are relatively rare. If all electrons are paired and we are talking about molecules, not ions, there must also be an even number of protons. Thus an odd number of neutrons must be present to arrive at an odd molecular weight, and the only elements with the most abundant isotope having an odd number of neutrons are beryllium and nitrogen, although dysprosium is kinda close.

In an undergraduate lab you wouldn't be given an unknown with beryllium in it because it's so dangerously toxic, so that leaves you with nitrogen unless you have a deuterated sample.

But there are stable free radicals like nitric oxide (neurotransmitter) or chlorine dioxide (fumigant) that violate the rule. Molecular oxygen doesn't count because it's a diradical and so has an even number of electrons.

EDIT: I spoke too soon about abundances of nuclides with an odd number of neutrons. Here is a table of such nuclides with abundance $> 20\%$ taken from http://atom.kaeri.re.kr/nuchart/ :
$$\begin{array}{rl} \text{Nuclide} & \text{Abundance} \\
\hline
\sideset{^{9}}\ {Be}& 100\%\\
\sideset{^{14}}\ {N}& 99.636\%\\
\sideset{^{105}}\ {Pd}& 22.33\%\\
\sideset{^{129}}\ {Xe}& 26.4006\%\\
\sideset{^{131}}\ {Xe}& 21.232\%\\
\sideset{^{163}}\ {Dy}& 24.896\%\\
\sideset{^{167}}\ {Er}& 22.869\%\\
\sideset{^{195}}\ {Pt}& 33.78\%\\
\sideset{^{207}}\ {Pb}& 22.1\%
\end{array}
$$
Of these $\sideset{^{195}}\ {Pt}$ is the most abundant isotope of $Pt$ in addition to $\sideset{^{9}}\ {Be}$ and $\sideset{^{14}}\ N$ previously mentioned.


Why can't I reproduce this photolytic reaction of ferrocyanide?

I would put the solution under a blacklight exposure source or a equivelant such as a mercury arc lamp at close range. The referenced writing makes no specifics about the the timeframe of prolonged. This could have been 3 months of exposure to sunlight.


When is it okay to use $Q=mc\Delta T$, is this equation only for calorimetry questions?

The equation $Q = mc \Delta T$ is appropriate for calorimetry at a constant volume. $Q$ here refers to the net heat transfer of the isolated system. Change in enthalpy, $\Delta H$, on the other hand, is really the total change in energy of the system, that is:

$$\Delta H = \Delta U + P\Delta V + V\Delta P$$

Where $P$ is pressure, $V$ is volume, and $U$ is the internal energy of the system, which is the sum of the system's net heat transfer ($Q$) and the work done on the system ($W$): $\Delta U = Q + W$. If we substitute the definition of $\Delta U$ into our equation for enthalpy, we get:

$$\Delta H = Q + W + P\Delta V + V\Delta P$$

Because any mechanical pressure-volume work performed on a closed system is equal to $-P\Delta V$, the equation further reduces to:

$$\Delta H = Q + V\Delta P$$

From this it becomes apparent that $Q$ and $\Delta H$ are only equal if pressure is constant (i.e., $\Delta P = 0$). The upshot of all this is that you should not assume $\Delta H$ and $Q$ are equal unless you're explicitly instructed to assume constant volume and pressure. When a problem refers to heat, it's almost certainly referring to heat transfer, $Q$, and not enthalpy change ($\Delta H$).


There are many ways to describe chemical bonds. Copper II ion binds to some anions with ionic bonds in solids, sometimes in covalent bonds (Ref 1).
In aqueous solution, copper II ions are surrounded (hydrated) with six water molecules, with the oxygen ends pointing in, toward the positive copper ion. Water is acting as a Lewis base. Ammonia is a stronger Lewis base - after all, it is basic in water. So an ammonia molecule will replace a water molecule in a hexa-aquated copper ion. Eventually, you get $Cu(NH_3)_4)^{2+}$.$(H_2O)_2$. Oh, I forgot, sometimes we leave out the water molecules, since we just assume that they are there - everybody knows that, since the teacher has explained that (unless the teacher hasn't explained that).
In the equation of the question, only four of the actual six water molecules are shown, because they are important for describing the interaction with $NH_3$, and the other two are just a complication, because I know you would ask, why don't they get replaced too, to give $Cu(NH_3)_6^{2+}$? After all, nickel, cobalt and chromium III do! Well, zinc doesn't; it forms a tetra-ammonia complex. And iron, manganese and aluminum don't even dissolve to form soluble ammonia complexes in alkaline solution. Reference 3 suggests that it is an equilibrium situation, but there are deeper explanations which you can explore later.
Ref 1. https://aip.scitation.org/doi/10.1063/1.1677079
Ref 2. https://www.chemguide.co.uk/inorganic/transition/copper.html
Ref 3. https://www.chemguide.co.uk/inorganic/complexions/aquanh3.html#top


How slow is a ester reaction without a catalyst? (in a whiskey barrel?)

The best info I could come up with is from wikipedia (with no citations for the relevant sections)

https://en.wikipedia.org/wiki/Fischer–Speier_esterification

For reaction with an excess of one reactant and an acid catalysts, it says:

  "Typical reaction times vary from 1–10 hours at temperatures of 60-110
  °C."

Later it talks about ester reactions in wine:

  "Of course, when compared to sulfuric acid conditions, the acid
  conditions in a wine are mild, so yield is low (often in tenths or
  hundredths of a percentage point by volume) and take years for ester
  to accumulate."

Which is a bit of an Apple to Oranges comparison, and is not whiskey, but it gives an indication of 'Yes, slowwwwww'. Or something like that.


Why is are weak acid-strong base and strong acid-weak base reactions irreversible?

Yes the following reaction is reversible.

$$\ce{CH3COOH + OH-  <=>  CH3COO-  + H2O}$$

There is an equilibrium between the species $\ce{CH3COOH, OH-}$ and $\ce{CH3COO-}$ in aqueous solution.

Thus the gist is that particular $\ce{CH3COOH}$ and $\ce{CH3COO^-}$ molecules aren't stable, but rather continuously react.


Why does hydrogen burn with a pop sound?

We may agree that during the combustion, regardless if the one of wood, gasoline, or in your example, hydrogen gas, is an exothermic reaction.  In a simplified view, products of these reaction are gaseous water; in the processes burning organic matter varying relative amounts of carbon monoxide and carbon dioxide are to be considered, too.

The gaseous products formed take up a volume, much more than their starting materials in the condensed state (liquid, or solid). Because of the local heat, these gases aim to expand rapidly.  In contrast to burning wood for a bonfire, the volume offered by a test tube is confined except for the mouth of the test tube, where a decompression to reach ambient pressure may take place.  This small scale explosion initially occurs at the interface of air's oxygen at the mouth of your test tube filled with hydrogen, yet this frontier actually propagates up to the closed end of the test tube, too.  Sometimes, especially with lengthier test tubes, this may be seen from the outside, too.

The combustion of hydrogen is worth a special note, because -- as correctly spot by airhuff -- formally there is volume contraction along

$$ \ce{2H2 + O2 -> 2H2O}$$

which is more than compensated by the thermally caused expansion of the gaseous water.

Gasoline driven combustion engines work by the same principle; spark plugs locally ignite a mixture of finely dispersed fuel and air, triggering a rapid combustion (explosion) where the reaction products again aim to expand. As they operate at a larger scale as the test tube experiment, the generated sound is much more intense, and these vehicles need a silencer, too.

While your bonfire equally yields $\ce{CO2}$ and water, these products may expand easily; there is no pressure generation, hence there is no sudden expansion (at least, not at this scale), and hence we may perceive the bonfire as (in comparison to the other examples) as peacefully calm.

I suggest to take a view on videos about explosions of hydrogen gas, and an appealing demonstration how it looks in the inner of such an Otto engine here.  In the later example (the display of the combustion starts around 2:51 min and again at 6:00 min), gasoline was replaced and a high speed camera looks through a special lid, yet the example still shows nicely the propagation of this frontier of explosion.


"Isomers" are compounds that are equivalent but for a certain change in structural property. There's different types of isomers depending on which sort of structural property you need to change in order to inter-convert the two molecules.

Structural isomers are equivalent in their atomic makeup but differ based on their chemical connectivity. For example, 1-propanol and 2-propanol are structural isomers - they share the same formula ($\ce{C3H8O}$) but differ in how those atoms are connected. You can change one to the other by rearranging how the atoms are connected to each other.
Stereoisomers are identical in chemical makeup and connectivity, but differ in the spatial orientation of their atoms. There's several different subtypes, which vary based on what sort of change to the molecules can interconvert them.

E-Z isomers can be interconverted by a theoretical rotation around a double bond - something that normally doesn't happen.
Optical isomers can be interconverted by mirroring the molecule (again, something that doesn't happen)
Diastereomers can be interconverted by mirroring the arrangement around a certain subset of atoms.

Conformational isomers are like stereoisomers, but with the distinction that they differ based on changes which do happen under normal conditions. For example, by a rotation around a single bond, or by the umbrella-like inversion of a nitrogen. Different conformational isomers are normally not isolatable from each other, as they rapidly interconvert.

There's a subset of these isomers called atropisomers which would otherwise be interconvertable, but local conditions (e.g. steric hinderance) means that the interconversion is slow under standard conditions, such that the different isomeric forms can be isolated from each other.

Okay, that brings us to your question on topological isomers. These are conceptually very closely related to conformational isomers and atropisomers. Like conformational isomers, topological isomers are equivalent in chemical formula, connectivity, and even the normal stereochemical properties like E-Z and R-S designations. However, like atopisomers - and unlike most standard conformational isomers - two different topological isomers cannot interconvert on an appreciable timescale, and thus each isomer can be isolated from the other. But unlike atropisomers, the difference between two topoisomers is not due to rotation around a single bond.

Instead, the difference between two different topoisomer is due to the different topology of the molecules. Topology is a somewhat complex mathematical topic, but for our purposes here, you can think of topology as a description about the invariant way something structured on a global scale, that can't be changed by bending, twisting, folding, etc.

For example, take a rubber band. No matter how you bend, twist, fold or tie the rubber band, it's still topologically the same. (That's part of the definition of "topology".) However, if you cut the rubber band, pass it through itself and reattach it, you can get something that's called a trefoil knot. If you look locally at the rubber band, all the connectivity is the same. There isn't any twists in the trefoil that couldn't also be present in the original circular band form. But on a global scale the trefoil and the band are very different. There's no way you can bend, fold, twist, tie, etc. the two to interconvert them without cutting. Because of this, we say the trefoil and the circular band are topologically distinct, in a way the circular band and a twisted form of a circular band aren't.

This is the difference that's being expressed in the case of topological chemical isomers. All of the local structure between the two is the same, but the large-scale global structure of the molecule is such that you cannot interconvert the two isomers without breaking bonds. You can't make one into the other simply by rotating a bond -- not even by rotating a bond that is energetically disfavorable to rotate.


Predicting 260/280 of DNA or RNA from sequence?

The problem with doing this, especially with RNA, is that secondary structure can have a large impact on the absorption properties of the molecule. Interactions between bases will change the molar absorptivities in ways that may not be easily predictable. The "ideal" ratios are only guidelines and I wouldn't worry too much that a long string of a single base produces a deviation from them. If very accurate measurements are required, hydrolysis into nucleotides before measurement is usually done to avoid these interactions. For your specific case, it may be possible to construct a calibration curve of sorts to get an estimate if you can measure the tail length by some other means. So long as the tail is sort relative to the rest of the molecule, I would expect to see some predictable change in ratios.


How to distinguish first excited state and second excited state?

Unfortunately, no. If you don't want to work it out by hand, the best you can hope for is that the authors explicitly specify orbital occupations with something that looks like an electron configuration. This paper doesn't, the previous paper sort of does.

Here is a paper that doesn't have symmetry labels, but does have configurations. For example, on page 2977, the SCF reference for $\ce{CuO}$ is given as

$$
\ce{ (O 2s)^2 (O 2p\sigma)^1 (O 2p\pi)^4 (Cu 3d\sigma)^2 (Cu 3d\pi)^4 (Cu 3d\delta)^4 },
$$

where the wavefunction symmetry is determined by the singly-occupied orbital (everything else is doubly occupied). $\ce{2p\sigma}$ probably lies along the bond, which would be taken as the $z$-axis when looking at a character table. Go to the $C_{\infty v}$ table, check for $z$, and discover that it's $\Sigma^+$, which is even mentioned in the paper. So there are symmetry labels given here, but the linear groups also have Greek characters for their irrep labels...

One interesting point is that non-Abelian groups are not used in practical quantum chemistry. $C_{\infty v}$ would be reduced to $C_{2v}$, and the $^{2}\Sigma^+$ state would become $^{2}\!A_{1}$.


Hydrogen sulfide is a weak acid:
$$\begin{align}
\mathrm{p}K_{\text{a}1} &\approx 7\\
\mathrm{p}K_{\text{a}2} &\approx 13\\
\end{align}$$

Therefore, the concentration of sulfide $c_{\ce{S^2-}}$ is very low in acidic solutions:

$$c_{\ce{S^2-}} = \frac{K_{\text{a}1}\cdot K_{\text{a}2}\cdot c_{\ce{H2S}}}{c_{\ce{H+}}^2}\approx 10^{-20}\frac{c_{\ce{H2S}}}{c_{\ce{H+}}^2}$$

Besides, hydrogen sulfide is a gas. Its solubility in water is approximately $c_{\ce{H2S}}\approx 0.1\ \mathrm{mol\ l^{-1}}$.

For example, in a solution of a strong acid with $c_{\ce{H+}}=1\ \mathrm{mol\ l^{-1}}$, the concentration of sulfide may be estimated as

$$c_{\ce{S^2-}} = \frac{K_{\text{a}1}\cdot K_{\text{a}2}\cdot c_{\ce{H2S}}}{c_{\ce{H+}}^2}\approx 10^{-20}\frac{0.1\ \mathrm{mol\ l^{-1}}}{\left(1\ \mathrm{mol\ l^{-1}}\right)^2}=10^{-21}\ \mathrm{mol\ l^{-1}}$$

On the other hand, the solubility product of lead(II) sulfide is extremely low:

$$K_\text{sp} = c_{\ce{Pb^2+}} \cdot c_{\ce{S^2-}} \approx 10^{-28}\ \mathrm{mol^2\ l^{-2}}$$

Thus, the solubility of lead(II) sulfide (i.e. the maximum concentration of $\ce{Pb^2+}$) in a solution of a strong acid with $c_{\ce{H+}}=1\ \mathrm{mol\ l^{-1}}$ is

$$c_{\ce{Pb^2+}} = \frac{K_\text{sp}}{c_{\ce{S^2-}}} \approx \frac{10^{-28}\ \mathrm{mol^2\ l^{-2}}}{10^{-21}\ \mathrm{mol\ l^{-1}}} = 10^{-7}\ \mathrm{mol\ l^{-1}}$$

Therefore, lead(II) sulfide precipitates from acidic solutions and will not significantly dissolve in dilute hydrochloric acid.

However, lead(II) sulfide can be dissolved in hot hydrochloric acid with $c_{\ce{HCl}}=8\ \mathrm{mol\ l^{-1}}$ because of the formation of anionic complexes $\ce{[PbCl3]^-}$ and $\ce{[PbCl4]^2-}$.


Proper IUPAC name of a substituted cyclohexenol

I think you made a counting mistake. It is 6-methylcyclohex-2-en-1-ol


The product A, shown below, is formed by first reducing the ketone to a secondary alcohol with $\ce{LiAlH_4}$ and then by adding a methyl group to form an ether using the reagent $\ce{CH_3I}$.
$\hskip2in$
Next the grignard reaction with product A will form two products. The grignard reagent $\ce{MeMgI}$ can attack A in a $1$,$2$ or in a $1$,$4$ mode, due to the unique aromaticity of cycloheptatriene, giving a mixture of products B and C, the structures of which are shown below.
$\hskip1in$


There is no definitive equation for an incomplete combustion. Sometimes incomplete combustion of hydrocarbons may even form species such as polyaromatic compounds and other hydrocarbons. The result may differ by variation of factors such as temperature, oxygen concentration, and the hydrocarbon undergoing combustion. Basically what you're asking is like asking "is there a definitive number between 0 and 1?" You may already know that numbers between 0 and 1 is an infinite set. Incomplete combustion may not be an infinite set, because nobody proved that it is, but it's still long ways off from a definitive one equation solve it all thing.

There is only a definitive equation for complete combustion; like there is only two numbers within a set of {0, 1}.

It's easier to recognize that any combustion that is not a complete combustion as a incomplete combustion.


Do bonding and anti-bonding orbitals exist for ionic bonds?

Bonding and antibonding orbitals in a homonuclear diatomic molecule  A-B should mix due to polarization to give rise to pair of atomic orbitals with delocalized tails  for an ionic heteronuclear diatomic molecule. Now suppose A more electronegative and is the rightful anion and B the cation. The tailed atomic orbital of A is the bonding orbital(ie. A:- B°+) and the one belonging to B is the antibonding (ie. A°+ B:-). See original paper on Coulson Fischer theory.


The difference between one pot synthetic method and one step reaction

"One step" refers to a process that produces the desired product by running a single reaction--especially in a situation where there are alternative pathways to the same product involving two or more reactions in a sequence.

"One pot" refers to a process involving a series of distinct reactions carried out sequentially and in a single reaction vessel to produce the desired product without isolating and purifying the intermediates after each step.


Why is the Dulong–Petit law not applicable to the elements Be, B, C, or Si?

I refer to the same link and add a summary:

The deviation is quantum mechanical in nature. The Dulong–Petit Law is exact only if all vibrational modes are fully activated, in which case equipartition theory can be used. As every atom in a solid can be considered to be a three-dimensional harmonic oscillator, the contribution to the heat capacity is $3k_\mathrm{B}$ for one atom, or $3R$ for one mole.

However, this applies exactly only as the temperature goes to infinity. For finite temperatures we can still come close if the thermal energy is high enough versus the energy required to excite the vibrational modes:

$$k_\mathrm{B}T \gg \hbar\omega$$

For most elements – except the elements listed in the question – room temperature is high enough to come close to full excitation. For the elements in question, the vibrational frequency $\omega$ is large, particularly because of the strong bonds and relatively small masses ($\omega \propto \sqrt{k/m}$.)


Why does rdkit's `AllChem.ReactionFromSmarts.RunReactants` return a tuple of tuples?

From a tutorial I wrote on SMARTS reactions in rdkit:

  The output is a tuple of tuples. The inner tuples are there because even reactions that take only a single input molecule can result in multiple output molecules (e.g. hydrolysis). The outer tuples are there because rdkit applies your reaction incidence of the appropriate motifs in the input molecule, even if it occurs more than once.

The examples in the tutorial illustrate both cases where a single application of the reaction rule creates multiple molecules (the inner tuple(s) would have length > 1), and also cases where a reaction rule can be applied more than once to a molecule, which would lead to the outer tuples having length > 1.


According to Introduction to Atmospheric Chemistry, by Daniel J. Jacob, Princeton University Press, 1999,Chapter 10,

An important component of aircraft exhaust is nitric oxide ($\ce{NO}$) formed by oxidation of atmospheric $\ce{N2}$ at the high temperatures of the aircraft engine. In the stratosphere $\ce{NO}$ reacts rapidly with $\ce{O3}$ to produce $\ce{NO2}$, which then photolyzes:
$$\ce{NO + O3 -> NO2 + O2}$$
$$\ce{NO2 + $h\nu$ -> NO + O}$$

Note that there is no net consumption of $\ce{NO_x}$, thus it is able to act as a catalyst such that a single $\ce{NO_x}$ molecule can destroy many molecules of ozone by these reactions.
Please don't hesitate to ask for any clarifications in the comments below, particularly if I've failed to address any aspect of your question.


How to make spectra smooth by using gaussian or lorentzian function

please teach how to use gaussian or lorentzian function.

It will take several hours of discussion or 4-5 lectures. You are mixing several things.
a) Briefly, your collected spectrum is noisy. You want to remove the noise. What type of spectra is this?
b) The finite resolution of your spectrometer has nothing to do with noise. Increasing the resolution mathematically or so-called deconvolution is another story.
c) The easiest solution is to "fit" Lorentzian or Gaussians to your experimental spectrum after denoising it (step a). Experimental spectra are neither pure Gaussians nor pure Lorentzians.
Here are the steps. I am mentioning the keywords so that you can search the right terms.

Can you get hold of PeakFit software? It has a better function library than Origin.
Denoise your spectrum, using Savitsky-Golay filter.
Once excessive noise has been filtered, use a trial function such as Gaussian or Lorentzian and see which one fits better. This is checked by residuals.
In an experimental spectrum, not all peaks have the same width so you will have to use multiple Gaussians or Lorentzians to fit a segment of a spectrum.

Search "Pragmatic Introduction to Signal Processing Applications in scientific measurement 2019 edition" by Prof. Tom O'Haver. It is a free book and contains all the practical stuff which would be very useful to you.


The shape of the two bonds (see the first page of this .PDF) shows why pi bond electrons are less tightly bound. Sigma bonds are formed pretty much along the line connecting the two atomic nuclei, whereas the pi bonds are formed above and below that same line. Naturally, then, the distribution of the latter less concentrated (having more space to cover) than that of the former. Loosely stated, the shape of the pi bond distribution is more roundabout whereas the sigma bond directly touches the two nuclei which makes the electrons in it less strongly bound.
When it can, delocalization of electrons provides greater stability for the molecule because the electron density is distributed well over the delocalized regions, rather than, say, confining it in a small high-energy region surrounded by low-energy regions.

Think about resonant structures. Benzene (C6H6) is a good example. So the inverse of that is: less delocalization, less stability. Hope this helps.


Energy levels in non-hydrogen-like atoms

Consider an atom to consist of an electron orbiting a positvely charged ion core (with one or more electrons associated to it). If you excite the electron to larger and larger values of the principal quantum number $n$ the classical electron orbit becomes larger and larger as well (in fact it scales as $n^2$). At sufficiently large distance, the postively charged ion core behaves as a point charge (like a single proton) and at long range you basically have the hydrogen problem with a different mass (assuming perfect shielding of the core electrons). In order to find the overall solution to this problem, one has to connect the long-range and short range behavior of the system. One way of doing this, is is by considering the system as a collision problem at negative energy (that is, a bound system). The effect of the nonhydrogenic ion core then results in a scattering phase shift between an incommming and outgoing electron. It can be shown that the solutions to this problem have a very similar appearance as the Bohr formula, namely:

$$
E_{n,\ell}=E_\text{IE}-\frac{hcR_A}{(n-\delta_\ell)^2}
$$
where $R_A=R_\infty(1-m_e/m_A)$ is the mass-corrected Rydberg constant, $E_\text{IE}$ is the ionization energy and $\delta_\ell$ is the so-called quantum defect for electrons with orbital angular momentum $\ell$. The quantum defect is related to the phase shift induced by the core and is different for $s, p, d, f$ etc. electrons. In principle, the quantum defect is a function of the binding energy of the electron to the ionic core, but in many cases this dependence may be neglected (in particular for large $n$).

What does this mean in practice? Let us for example look at the $n$s levels of the sodium atom. We find that $R_\text{Na}=109734.697205$ cm$^{-1}$ and $E_\text{IP}(\text{Na})=41449.451$ cm$^{-1}$. The second column in the table below gives the energy (in cm$^{-1}$ from the NIST website) and the third column gives the binding energy of the electron (also in cm$^{-1}$). The last column gives the so-called effective quantum number, that is $n^*=n-\delta_\ell$, with $\ell=0$ for $s$ orbitals

 3s         0.000   -41449.451  1.63
 4s     25739.999   -15709.452  2.64
 5s     33200.673    -8248.778  3.65
 6s     36372.618    -5076.833  4.65
 7s     38012.042    -3437.409  5.65
 8s     38968.510    -2480.941  6.65
 9s     39574.850    -1874.601  7.65
 10s    39983.270    -1466.181  8.65
 11s    40271.396    -1178.055  9.65

As you can see the quantum numbers take the form of the hydrogen solution if we start counting the principal quantum number form the 3$s$ levels and take $\delta_0\approx -0.65$.

To calculate the quantum defect is difficult as it requires ab initio calculations and then converting the energy levels to effective quantum numbers from which the defects can be extracted. However, experimentaly these numbers are very convenient, and one may use this quantum-defect theory for instance to determine the ionization energy of atoms and even simple molecules very accurately by extrapolating Rydberg series.


Why does hydrogen bonding require a lone pair?

You don't need a lone pair. Water can hydrogen bond to methane, attacking the back side of one of the carbon-hydrogen bonds in the methane. In a molecular orbital the sigma bonding orbitals in the methane are sufficiently delocalized to serve in place of lone pairs, overlapping the antibonding orbitals in the water [1]:
Abstract (from [1])

Quantum chemical calculations were performed at different levels of theory (SCF, DFT, MP2, and CCSD(T)) to determine the geometry and electronic structure of the HOH···CH4 complex formed by water and methane molecules, in which water is a proton donor and methane carbon (sp 3) is an acceptor. The charge distribution on the atoms of the complex was analyzed by the CHelpG method and Hirshfeld population analysis; both methods revealed the transfer of electron charge from methane to water. According to the natural bond orbital (NBO) analysis data, the charge transfer upon complexation is caused by the interaction between the σ orbital of the axial С–H bond of methane directed along the line of the O–H···C hydrogen bridge and the antibonding σ* orbital of the О–H bond of the water molecule. Topological analysis of electron density in the HOH···CH4 complex by the AIM method showed that the parameters of the critical point of the bond between hydrogen and acceptor (carbon atom) for the O–H···C interaction are typical for Н-bonded systems (the magnitude of electron density at the critical point of the bond, the sign and value of the Laplacian). It was concluded that the intermolecular interaction in the complex can be defined as an Н bond of O–H···σ(С–H) type, whose energy was found to be 0.9 kcal/mol in MP2/aug-cc-pVQZ calculations including the basis set superposition error (BSSE).

Reference
1. A. N. Isaev,"O–H···C hydrogen bond in the methane–water complex", Russian Journal of Physical Chemistry A 90(2016), 1978–1985.


Can 1,2-dioxetanedione (active ingredient in glow sticks) undergo stimulated emission (and thus form a laser)?

The activated complex 1,2-dioxetanedione can't undergo simulated emissions as it isn't the emitting molecule even in glow-sticks: it is the source of the activation energy for other dyes.

But most emitting dyes can be used in suitably configured lasers by providing energy as photons (=light) of a higher energy and a suitable wavelength in a suitable device. Whether they would make good lasers depends on a whole bunch of other factors (like how easy it is for them to absorb light, whether the subsequent transitions are forbidden or not (can't easily get population inversion if they are not) and various quantum efficiencies of the relevant conversions in the dye molecule.)

Dye lasers have been known since 1966. They have some advantages over more conventional lasers as they are often wavelength tunable (as dyes tend to have broad emission lines) and can often achieve very short pulses. Rhodamine 6G is a common laser dye but the dyes in a glow-stick may not be as good in lasers, though, in principle, there is no reason why not.


Why does the rate-determining step determine the overall rate of reaction?

Think of a chemical reaction as a series of processes that must happen in order. If reaction $3$ is the slowest (rate determining) step of some overall reaction, then the intermediates formed in reaction $2$ just accumulate and wait for their turn to participate in the reaction. Depending on the order and kinetics of the reaction, the rate of reaction $3$ can be increased by this build-up of reactants, but there will always be an excess waiting to participate in this slow step. Once the reactants get past step $3$, they breeze on through the remaining faster steps of the reaction. Since reactions $4$ and above are fast compared to $3$, it makes little difference how fast they are.

Note that it is somewhat of an approximation to say that the overall rate is equal to that of the rate determining step. As mentioned in the example above, depending on the order and kinetics of the reaction, a buildup of intermediates from reaction $2$ could increase the rate of $3$, thus the rate of $2$ becomes somewhat relevant. Likewise, it's possible that a buildup of products from reaction $3$ could inhibit its rate depending on the rate of reaction step $4$. Wikipedia's definition of the rate-determining step of a reaction includes this mention that it is an approximation:

  In chemical kinetics, the overall rate of a reaction is often approximately determined by the slowest step, known as the rate-determining step (RDS) or rate-limiting step. For a given reaction mechanism, the prediction of the corresponding rate equation (for comparison with the experimental rate law) is often simplified by using this approximation of the rate determining step.


interconverting Van Der Waal's equation for real gases from number of molecules to density

Your analysis is correct in terms of number density.  But let's see how it plays out in terms of molar density.
Let n be the number of moles and A be Avagadro's number.  Then N=nA.  If we substitute this into your first equation, then I get $$P=\frac{n(Ak)T}{V-nAb}-\frac{aA^2n^2}{V^2}$$But, since Ak=R, we obtain:$$P=\frac{nRT}{V-nb'}-\frac{a'n^2}{V^2}$$where $a'=aA^2$ $b'=bA$.  In terms of molar density $\rho=n/A$, this becomes:$$P=\frac{\rho RT}{1-\rho b'}-a'\rho^2$$


Why does microwaving napalm make it nonflammable?

In a very similar video I found, some gentlemen microwave gasoline, the flammable component of napalm, along with aluminium foil. The aluminium sparks when exposed to the microwaves, igniting the gasoline vapors in the microwave. Once all the vapor in the microwave is combusted, the microwave must then once again fill with gasoline vapor. When it once again comes into contact with the ignition source, the vapor combusts again.

Since the gentleman in the video you used did not use an ignition source in the microwave, the gasoline in the napalm vaporized, but did not ignite. The vapor filled the microwave, and what was left was the much less flammable gelatin.


Which is more stable? A phenyl carbanion or a disubstituted vinylic carbanion?

Can the lone pair on benzene carbanion participate in resonance?

As Jori has pointed out, the carbanion lone pair is orthogonal to the pi system of the aromatic ring.  Therefore it cannot participate in resonance.  Here is a drawing that illustrates this point.

how is it more stable than vinylic carbanion?

The $\mathrm{p}K_\mathrm{a}$ approach discussed in @Jori's answer becomes problematic when we are trying to look at a small difference between two large $\mathrm{p}K_\mathrm{a}$s.  A $\mathrm{p}K_\mathrm{a}$ of $43$ means $1$ ionized molecule exists in every $10^{43}$ molecules.  Certainly difficult to measure experimentally and so a significant error is attached to these estimates. In fact, using the data from Evan's $\mathrm{p}K_\mathrm{a}$ table we find a $\mathrm{p}K_\mathrm{a}$ of $43$ for the benzene $\ce{C-H}$ proton and a $\mathrm{p}K_\mathrm{a}$ of $50$ for an ethylenic proton, suggesting that the phenyl carbanion might be more stable than the vinyl carbanion.
An alternate way to examine this question involves comparing the hybridizations of the phenyl and ethylenic $\ce{C-H}$ bonds.  Since bonds with more $\mathrm{s}$-character will stabilize electrons better than bonds with less $\mathrm{s}$-character (because the $\mathrm{s}$-orbital is lower energy than a $\mathrm{p}$-orbital), whichever bond has more $\mathrm{s}$-character should produce a more stable carbanion.
The $\ce{C-H}$ bond in benzene is exactly $\mathrm{sp^2}$ hybridized as required by the symmetry of the benzene molecule.  Experimentally, we know that the $\ce{H-C-H}$ angle in ethylene is $117^\circ$.  Using this fact along with Coulson's theorem (reference1, reference2, reference3) we can determine that the $\ce{C-H}$ bond in ethylene is $\mathrm{sp}^{2.2}$  hybridized.
The $\mathrm{sp^2}$ hybridized orbital in benzene has more s-character than the $\mathrm{sp}^{2.2}$ hybridized orbital in ethylene.  This would lead us to suspect that the phenyl carbanion is more stable than the vinyl carbanion.
Of course, once we remove these protons we expect the resulting carbanions to undergo further rehybridization in order to further stabilize their lone pair (carbanion) electrons.  It seems like a reasonable first approximation that they would both rehybridize similar amounts and the "relaxed" phenyl carbanion would remain more stable than the "relaxed" vinyl carbanion.  But in any case, the transition state leading to the phenyl carbanion will be more stable than the transition state leading to the vinyl carbanion because of the hybridization \ $\mathrm{s}$-character reasons presented above.
Edit: OP's comment on hybridization
Molecules are not just $\mathrm{sp, sp^{2}}$ or $\mathrm{sp^3}$ hybridized.  While the $\ce{C-H}$ bonds in methane are exactly $\mathrm{sp^3}$ hybridized (as required by symmetry), the $\ce{C-H}$ bonds in $\ce{CH3F}$ are not (see here for a discussion).  Bonds can have any hybridization index (the superscript to the right of "$\mathrm{p}$") between zero and infinity.  Hybridization indices are not limited to integer values, nature can mix $\mathrm{s}$ and $\mathrm{p}$ orbitals in whatever ratio is required to create the most stable bonds and molecules.


Calculate accelerating voltage required to separate two particles in a magnetic sector analyser

Baseline separating C-12 from C-13 from C-14 requires accelerator mass spectrometry,

http://en.wikipedia.org/wiki/Accelerator_mass_spectrometry

Your example requires 100 times the finesse.  I'm not optimistic for reduction to practice vs. a paper solution.  Your target could be a charge-coupled device or a channel plate amplifier.  Observe the spatial separation of signals above noise.  A Faraday cup works differently.  Time-of-flight is not optimistic.

You might diddle FT ion cyclotron resonance in a high-Q cavity.


Why does hot steam produce MgO instead of Mg(OH)2?

There are two steps in the process:

$\ce{Mg + H2O -> MgO + H2}$

and

$\ce{MgO + H2O <-> Mg(OH)2}$

The equilibrium point in the second step is dependent on the temperature. See file Equilibrium Assemblages for more information.


Minimising the formation of manganese dioxide in the titration of potassium permanganate

Adding a large excess of acid helps greatly as the reduction to $\ce{MnO2}$ only takes place under neutral conditions. $$\ce{2H2O + MnO4- + 3e- -> MnO2 + 4OH-}$$ Under acidic conditions the $\ce{MnO4-}$ is reduced all the way to $\ce{Mn^{2+}}$.$$\ce{8H+
 + MnO4- + 5e- -> Mn^{2+} + 4H2O}$$ Also swirling or stirring the solution as you are titrating makes a big difference as it stops the area where the permanganate is being added from becoming temporarily depleted of acid.Even if you do get some $\ce{MnO2}$ forming then as long as you have enough acid in the flask a few swirls should be enough to reduce the $\ce{MnO2}$ down to $\ce{Mn^{2+}}$.


Is Iodoform test restricted for only methyl ketones?

All compounds with $CH_3CO- $ or $CH_3CH(OH)-$ group answer iodoform test.


Dependence of the rate of a catalysed reaction on substrate concentration?

There is always a maximum convertion rate $V_{max}$ for every given enzyme-substrate system. In some cases the substrate itself can inhibit the enzyme's activity at high concentrations, because of the formation of some ESS-complexes, which can not dissociate into product and enzyme.

The frequent correlation is described by the Michaelis-Menten kinetics, which introduces the Michaelis constant $K_M$ as the substrate concentration at which half of all enzyme is bound to substrate $-$ and thus the reaction is running at a speed of $\frac{V_{max}}{2}$.

As you can see, the relation is described by a saturation function. But for loose approximations, where $[S] \leq K_M$ you could say its roughly proportional: $$rate = \dfrac{V_{max}}{2 \times K_M} \times [S]$$


Variation of compressiblity factor with temperature

Remember the mathematical definition of $Z$:

$$ Z = \frac{pV}{nRT}$$

Given all the $Z$ graphs I've ever seen -- the single-point interaction of all these isotherms is just coincidence (and kind of misleading). There is nothing in thermodynamics that suggests for a given gas, there exists a $p$ for which $Z$ is constant, regardless of temperature. Even for constant $V$ and $n$, you have that $Z \propto \frac{p}{T}$, so changing $T$ can in no way leave $Z$ constant.

You can find graphs that look similar even for different gases at the same $T$:

Nonetheless, there is no reason for a given material to have the same $Z$ at a given $p$ regardless of temperature:

Generally speaking, if you zoom-out far enough, and have significantly different temperatures (like 100s of Kelvin apart), some $Z$ graphs may appear to have a uniform $(p,Z)$ point, but in reality the intersections are not uniform for any physical/mathematical reason. For this reason, graphs like that can be a little misleading.


Reducing agent in preparation of CpRuCl(PPh3)2

You'd think it's the ethanol solvent that acts as the reducing agent here, and the byproduct is acetaldehyde.


Cannizzaro reaction versus Nitroaldol reaction

Is the nitroaldol reaction that much faster?

My guess would be that the nitroaldol is faster than the Cannizzaro.  The Cannizzaro reaction involves

attack by hydroxide on the carbonyl to reversibly form the tetrahedral intermediate
alignment of the tetrahedral intermediate with a second aldehyde molecule (entropically costly)
hydride transfer in a redox reaction (relatively high energy, even complexed hydride is not particularly stable) to form molecules (alcohol and carboxylic acid) that are not that different in stability than the starting material

The nitroaldol involves

generation of the nitroalkyl carbanion
attack by the carbanion on the carbonyl to form a tetrahedral intermediate
capture of a proton by the negatively charged oxygen to produce a neutral molecule
elimination of water to produce a more stable (more conjugated) molecule than we started with (e.g. energetically downhill)

The nitroaldol doesn't require any special molecular alignment after the tetrahedral intermediate is formed and is energetically downhill.

  Can the yields be so different due to the steric hindrance of the
  additional methyl group?

Further alkylation at the nitroalkyl carbanion center will not help the reaction in a steric sense just as you suggest, but also a secondary carbanion will be less stable (higher energy, more difficult to form) than a primary carbanion.

  What is the benefit of these nitrogen bases over $\ce{NaOH}$?

They are somewhat milder reagents than hydroxide and therefore less likely to produce side reactions.


No, all that does is give you back your NaOH. Its not a strong enough base to completely deprotonate methanol.
I use sodium metal in methanol, cooled in ice/water bath and stirred under nitrogen for safety.


What is the difference between the Joule and Joule–Thomson coefficients?

The Joule coefficient is $$\left(\frac{\partial T}{\partial V}\right)_U$$

Experimentally, Joule attempted to measure this value by expanding gas into an evacuated, insulated container, thus ensuring $U$ is constant.

The Joule–Thomson coefficient is $$\left(\frac{\partial T}{\partial P}\right)_H$$

Experimentally, this is realized by expanding a flow of gas in an insulted pipe from a high pressure upstream region to a lower pressure downstream region, the two regions being seperated by a porous frit.

For more information see THE JOULE AND JOULE-THOMSON EXPERIMENTS by Dr. J. B. Tatum, University of Victoria.


Is the Ocean acidification inconsistent with the Bjerrum plot?

source

I think I see the confusion. You are right that directly adding carbonate, perhaps in the form of $\ce{Na2CO3}$, would increase the pH. However, ocean acidification, as shown in your figure, is from atmospheric $\ce{CO2}$ dissolving in the ocean. The relevant equilibrium: $$\ce{CO2_{(aq)} + H2O <=> H2CO3_{(aq)} <=> HCO3^{-}_{(aq)} +H+_{(aq)}<=>CO_{3(aq)}^{2-} +H+_{(aq)}}$$

The Bjerrum plot is simply a graphical representation of this equilibrium. As you add $\ce{CO2}$, you push the equilibrium towards the right, forming $\ce{H+}$, lowering the pH. Conversely, when you add $\ce{CO3^{2-}}$, the equilibrium move towards the left, consuming some $\ce{H+}$ and raising the pH. Although adding $\ce{CO2}$ leads to an increase in the absolute concentration of carbonate, the ratio of concentrations goes the other way.

You can use this two ways:

If you have a change in pH, you can move along the x-axis to predict how the concentrations of the carbonate species change.
If you change the ratios of the species, e.g. by adding $\ce{CO2}$, you can predict how the pH will change.


Dependence of primary alcohol oxidation on presence of water and "strong" or "weak" oxidants

After looking up more information and taking the comments of user55119 into account, I've decided to post an answer myself, hoping to clearly address the main question points.

The determinant factor of the products obtained from an oxidation reaction of primary alcohols is the presence of water in the reaction medium, as it allows the formation of an aldehyde hydrate which can be further oxidized. Thus, the distinction between "weak" and "strong" oxidants is not necessarily informative (and could even be misleading) and emphasis should be placed in the reaction mechanism alongside the reaction medium.
The oxidation of primary alcohols with Cr(VI) reagents takes place through the mechanism depicted in the question or slight variations of it (depending on the reagent chosen). Carboxylic acids are obtained only through an aldehyde intermediate.

Further reading:
McMurry, John. Organic Chemistry. 8th edition.
MasterOrganicChemistry