Is there an 'easy' way to determine if a ligand is a pi acceptor, donor or neutral?

If you ask for an "easy" way to determine if a ligand is a pi acceptor/donor or none, look at the spectrochemical series. A quick-and-dirty rule (which means that it is correct most of the time but not always) is that weak ligands (iodide, bromide, hydroxide etc) are pi-donor ligands. The medium ligands (water, ammonia etc) are pi-neutral, and the strong ligands (cyanide, carbonyl, bipyridine, etc) are pi-acceptor.
But this method is some kind of reversing here. Because we normally use a ligand's pi character to guess if it is a strong or weak ligand.
If you can, ligand field theory is your best friend in this problem.
Generally, a pi-donor ligand has an extra electron pair that lies in an orbital that has approximately the same symmetry as the metal center's orbital (usually t2g). This allows the two orbitals to interact with each other and creating two new sets of orbitals: t2g and t2g* (as in the molecular orbital theory). The ligand's pi electron will reside in the  lower t2g orbital, whereas the metal's original d-electrons will reside in the t2g* one. That explains your question of the copper complexes.
In the other hand, a pi-acceptor ligand has an empty molecular orbital of low energy and approximately the same symmetry as the orbital of the metal center. In this case, it will interact with the metal's orbitals and the d-electrons of the metal will be delocalized to the ligand. These ligands usually have a highly conjugated system that greatly decreases the energy of the LUMO (bipyridine, triphenylphosphine etc). In the cases of carbon monoxide (called carbonyl as a ligand) and cyanide, they have low-energy LUMO for other reasons.
A pi-neutral ligand (also called sigma-donor ligand) does not process any of the above characteristics. For example, ammonia has only one electron pair on the nitrogen, and that pair is already used in the formation of the sigma bond, so no more pi electrons to donate. The LUMO of the ammonia is also high in energy, so no pi accepting here.
I think trimethyl phosphate is mostly a pi-neutral with the donor atom is the oxygen double-bonded to phosphorus. The molecule is not conjugated at all, and the oxygen can only donate one pair of electron since the other's orbital does not have the same symmetry.


What is the appropriate procedure for determining the quantum numbers of an electron?

1s2 2s2 2p6 3s2 3p6 4s1 is the electronic configuration of the 19th electron of chromium.

since the last shell is 4, therefore n=4,

l=0(as there is only one orbital in the s sub shell) ,

m=0 (since we're still in the s sub shell only)

and s=+1/2 or s=-1/2(arbitrarily assigned values)

note:+1/2 or -1/2 are used to distinguish between electrons as no two electrons can have the same electronic configuration, but here the question isn't that specific, so we need not bother.


Why hypochlorous acid is formed in second reaction why doesn't it proceed like the first one?
  Has it got to anything with reduction potentials? Cl and O are very close and is that reason?

Your intuition in 2. is a good guess and could answer the question mathematically but would be an indirect explanation. If you look closer at reaction 1 you will see that oxygen is oxidized by fluoride where as in reaction 2 oxygen is not oxidized. From a re-dox perspective in the first reaction your equation 1 looks something like:

$$\ce{H2O -> 2H+ + 1/2O2 + 2e-}\tag A\\
\text{and}$$
$$\ce{F2 + 2e- -> 2F-}\tag B$$

Here the fluorine molecule takes 2 electrons from the oxygen atom in an irreversible process. But for equation 2 the redox reaction is more like:
$$\ce{Cl2 <=> Cl- + Cl+}\tag C$$
Here the chlorine is oxidize by the other chloride molecule and reacts with the water to form $\ce{HCl}$ and $\ce{HOCl}$

$$\ce{Cl+ + OH- <=> HClO} \qquad\text{and}\qquad \ce{Cl- + H+ <=> HCl}\tag D$$

  side note: $\ce{HOCl}$ would more properly be presented as $\ce{HClO}$ since chlorine has a $+1$ formal charge but not $\ce{ClHO}$ since $\ce{ClO-}$ is the conjugate base.

Note that the reactions for clorine are revresable now whereas fluorine was not. Combining equations C and D give your original equation 2

$$\ce{Cl+ + Cl- + OH- +H+ \implies Cl2 + H2O <=> HCl + HOCl}\tag2$$

Now for the why. The answer comes from the electronegativities of the elements. $(X_\ce{F} = 4.0 > X_\ce{O} = 3.5 > X_\ce{Cl} = 3.2)$ as you can see fluorine's electronegativity is greater than oxygen's which allows flourine to oxidize oxygen and oxygen's electronegativity is greater than chlorine's which mean chlorine cannot oxidize it. Further Hydrogen is already oxidized and so the only thin for chlorine to oxidize is itself.

Since the other halogens also cannot oxidize oxygen, this equation can be generalized for all of the halogens that are not fluorine as well:

$$\ce{X2 + H2O -> HX^{(I-)} + HX^{(I+)}O}\tag{X = Cl, Br, I}$$


All in all this isn't a well stated problem. The molarity can be solved, but the molality can't really be solved.
The wording "with a 96% of mass" is odd. I'm not sure if this is your interpretation of the problem or if that is what is literately written in the book. I'll interpret the phrase to mean that the sulfuric acid was 96% sulfuric acid by weight.
For both problems the grams of sulfuric acid are:
$$\mathrm{grams\ }\ce{H2SO4} = 10 \times 1.84 \times 0.96 = 17.66$$
$$\mathrm{moles\ }\ce{H2SO4} = \dfrac{10 \times 1.84 \times 0.96}{98.079} = 0.1801 $$
Molarity
So the molarity, $M$, of the solution will be:
$$\mathrm{Molarity} = \dfrac{10 \times 1.84 \times 0.96}{98.079 \times 0.50} = 0.360 $$
Molality(1)
The problem with the molality calculation is that volumes are not additive. I'll admit that for two or three significant figures the calculation sort of works.
So for the 500 ml of solution, you must assume that 10 ml is sulfuric acid, the other 490 ml of volume is water. Water is assumed to be 1.00 kg/liter, so to figure out how many moles of sulfuric acid per kilogram of solvent we'll need to multiply the 0.360 moles per liter of solution by a correction factor of 1/0.980 which gives a value of 0.368 for the molality.
Molality(2)
As a second way to calculate the molality let's divide moles of $\ce{H2SO4}$ by the kilograms of water.
From above we know that the 500 ml of solution contained 0.1801 moles of $\ce{H2SO4}$.
Again we have to assume that the volumes are additive so that 10 ml of concentrated sulfuric acid and 490 ml of water yield 500 ml of solution. Given no temperature we can just assume that water is 1.00 g/ml, so 490 ml of water has a mass of 490 grams or 0.490 kilograms.
$$\mathrm{Molality} = \dfrac{0.1801}{0.490} = 0.368 $$

To perhaps beat a dead horse here let's back calculate the volumes for one liter of a similar solution. Steffen's Chemistry Pages give that a 4% solution of sulfuric acid by weight has a density of 1.0250 g/ml, 41.00 g sulfuric acid per liter and a molarity of 0.418 at 20°C.
To get 41.00 grams of sulfuric acid we'd need
$$\mathrm{ml\ conc\ sulfuric\ acid} = \dfrac{41.00}{1.0250\times 0.96} = \pu{41.7 ml}$$
But since water is 1 g/ml and the solution is 96% water
$$\mathrm{ml water} = 1000 \times 0.96 \times 1.0250 = \pu{984 ml}$$
$$41.7 + 984 = 1025.7$$


How to explain this acid-catalysed acetal rearrangement mechanistically?

You have managed to draw the correct acyclic hydrolized compound with a decent mechanism.

This is a problem that can be solved by numbering the carbon atoms. Number the carbon atoms sequentially along the chain, starting with the acetal carbon (which has to be the carbonyl group). Likewise number the unpacked acyclic molecule. Use what you get to figure out which of the three hydroxy groups are used to make the second acetal.

In general, numbering your atoms using common structural features as anchors can help unravel complicated mechanism problems.


O-H infrared band dependence of alcohol concentration in water

The easiest thing to do is look at some example spectra. All of them are taken from the Spectral Database for Organic Compounds.

This is 1-propanol, first as a liquid film and then in $\ce{CCl4}$.

In the liquid, the broad peak at $\pu{3333 cm^{-1}}$ is due to intermolecular hydrogen bonding, just as you would see in an IR spectrum of bulk pure water. The $\ce{CCl4}$ spectrum shows that as you reduce the probability of 1-propanol (or any alcohol) molecules contacting each other, the free stretch appears at $\pu{3637 cm^{-1}}$. This is similar to water in a gas-phase spectrum.

  water is explicitly made from $\ce{O-H}$ bonds whereas the $\ce{O-H}$ bond in ethanol only represents a fraction of the bonds in the molecule

Close. It isn't about the fraction of total bonds, in this case it's simply that water has two hydroxyl stretches and ethanol only has one. As water is added, the intensity of the hydrogen bonding $\ce{O-H}$ band should increase over the pure alcohol hydrogen bonding band, but we cannot know quantitatively what the increase will be without performing an experiment.

  Is there also a wave number shift that would occur too that I am missing?

Yes. In the gas phase, the water symmetric and asymmetric $\ce{O-H}$ stretches are at $\pu{3657 cm^{-1}}$ and $\pu{3756 cm^{-1}}$, which then red-shifts to $\pu{3404 cm^{-1}}$ in the bulk due to hydrogen bonding. Compare this to 1-propanol, where the "free" and hydrogen bonding values are $\pu{3637 cm^{-1}}$ and $\pu{3340 cm^{-1}}$, respectively. The water stretches are blue-shifted (higher in energy) compared to the alcohol.

Water vibrational frequency values taken from http://www1.lsbu.ac.uk/water/water_vibrational_spectrum.html.


What's the difference between demixing temperature and cloud point?

Cloud point typically refers to the LCST (or UCST) in solution. Demixing temperature usually refers to the LCST (or UCST) of a polymer mixture or copolymer in bulk.


Does corrosion follow the law of conservation of mass?

The answer to the question in the title is yes.

Iron oxide flakes away from the iron so just looking at the iron would make you think that the mass has disappeared; it's actually on the surface below, unless someone has moved it.


Why does thallium form TlF and TlF3 but aluminium only forms AlF3?

Ahh, Thallium is a very nice element. Because it's weird.

The +3 oxidation states are favorable except for the heavier elements, such as Tl, which prefer the +1 oxidation state due to its stability; this is known as the inert pair effect. Chemwiki.UCDavis
As an example in group 13 the +1 oxidation state of Tl is the most stable and Tl(III) compounds are comparatively rare. Inert pair effect, Wikipedia

So, why?
Inert pair effect was introduced to describe why the facts above are correct. However, the original explanation, provided by Sidgwick, relied on the fact that the $s$ subshell has an "inert" pair of electrons. This can't be correct, and Wikipedia uses the trend in ionization potentials of "$s$" electrons to prove it wrong:
$$\ce{In < Al < Tl < Ga}$$
A better explanation lies within the lower bond enthalpies of $\ce{MX}$ where M is one of the p-block heavy elements.
To reach a high oxidation number, elements will need to have a strong bond. They wanna bond in order to lose energy (i.e. become more stable) after all! If the bond is weak, this energy isn't provided. Thus, a heavy element is better off with being less oxidised.
See this and this. (Sorry, they might be behind a paywall)


Why is there a sudden drop in the concentration of ammonia in this graph?

In the graph the following reaction will occur: $$\ce{N2 +3H2 <=> 2NH3}$$
This reaction is an equilibrium reaction and you can see that there is exponential decay for the reactants while the product is formed. After a while the concentrations do not change any more, so the system is in equilibrium. This is at the end of the time span "a".

At the beginning of time span "b" the system is disturbed, as the ammonia is completely removed. The system is changed and the equilibrium has to form again. Hence the concentration of the reactants is further reduced while new product is formed. Like santiago already pointed out in the comments, this is known as Le Châtelier's principle. At the end of time span "b", the system is in equilibrium again.


Which of the following compounds will form significant amounts of meta product during mono-nitration reaction?

The answer to the question is (1) Aniline.

This occurs because under standard strongly acidic nitrating conditions (typically cH2SO4/HNO3) the aniline NH2 is protonated removing the ability of nitrogen lone pair to donate any electrons. Consequently the electron-withdrawing polar effect makes it m-directing. Some p-nitroaniline is formed because there is a small amount of the aniline freebase in equilibrium with the anilinium, but that is much more reactive than the anilinium.

Further reading here

Also discussed in this question  Why does nitration of aromatic amines give meta directed nitro compounds?


Good question. Enthalpy of reaction dependence on temperature is given by Kirchoff's equation (you can check it) and it is true that enthalpy of reaction changes with temperature. For most reactions if temperature change isn't too big, enthalpy of reaction won't change much so we can regard it as constant in that temperature span on which we are plotting lnK vs 1/T for sake of simplicity.


IUPAC project 2014-003-2-800 addresses this issue. However there seems to be no officially released recommendation to date.


Forms of Henry's law versus ideal gas law

I wrote out an answer to this but realized I was wrong. But I've noticed a number of things:

First off, your form of the ideal gas law has the Molar Mass inverted. If you make the substitution that $n=m/M$, then the rearrangement reads $m = MVp/RT$, you have the $M$ on the bottom.

Secondly, You are selecting a Volume of 1 Liter, so your calculation will solve for a relation of mass and pressure @ 1L, which I don't think you want.

Thirdly (?), I'm guessing that the 760 in the denominator is there so you can use a gas constant in terms of atm? Or did I miss something?

So those are my observations about your attempt with the ideal gas law. I also am concerned over the bit with the mass fraction, as I think you converted a tissue mass into an $\ce {O2}$ volume using $\ce {O2}$'s density, which would be wrong if that was the case.

Looking again at Henry's law in the $p=kC$, you can see that in order to use the standard k units available, you would need to convert the $C$ you have from $L/kg$ to $mol/L$. You can convert $L$ to $mol$ using the ratio you used earlier, and you end up with $C$ in terms of $mol/kg_{tissue}$. The trick here is that you need the density of the tissue (not of $\ce {O2}$) to convert this into the volume you're looking for in order to use the equation. I think haha.


Regioselectivity of alpha halogenation of ketones

The acidic and basic catalyzed alpha-halogenation of a carbonyl proceed through an enol and an enolate intermediate respectively.  The difference in regioselectivity arises because of the different process used to arrive at the enol/enolate in the acidic and basic cases.

In the acidic case, 1) the carbonyl oxygen is protonated and then 2) a proton is removed from the alpha carbon to form the enol.  See here for a good picture of this mechanism.  In this process, the reaction is an equilibrium so thermodynamic control will lead to formation of the most stable enol.  Therefore, a proton will be removed from the more highly substituted alpha-carbon to produce the more stable double bond (the one that is more highly substituted).

When the reaction is run under basic conditions, the process is no longer a thermodynamically controlled equilibrium, but rather proceeds under kinetic control.  The same link used above also shows the mechanism of the base-catalyzed alpha-halogenation.  The rate determining step is removal of an alpha-proton by the base.  The lowest energy proton to remove is the one on the least substituted carbon. Remember that a primary carbanion is more stable than a secondary carbanion which is more stable than a tertiary carbanion, so removing a primary proton will produce the most stable carbanion-enolate.  Hence, in the base catalyzed case the proton will be removed from the least susbstituted alpha carbon so to produce the most stable carbanion-enolate intermediate.

To summarize the acid-catalyzed halogenation proceeds under thermodynamic control to produce the most highly substituted (most stable double bond) enol.  Whereas the base-catalyzed halogenation is kinetically controlled and produces the least substituted (most stable carbanion) enolate anion.


Interestingly, IUPAC states that a "salt" is "a chemical compound consisting of an assembly of cations and anions". Under this definition, all ionic compounds are salts, and all salts are ionic compounds.

Therefore, something like sodium hydroxide ($\ce{Na+OH-}$, definitely an ionic compound) could actually be correctly called a salt. This clashes with the commonly taught high-school level definition of a salt ("the product of an acid-base reaction"), unless you consider very general definitions of acids and bases such as the Usanovich definition, whereby sodium metal $\ce{Na^0}$ is an electron donor (and therefore a base) and water is an electron acceptor (and therefore an acid).

That said, the high-school definition is too simplistic. It is common for compounds to be an acid, a base and a salt all at the same time; consider for example sodium bicarbonate ($\ce{Na+HCO3-}$). It is made of cations and anions, and therefore is definitely a salt. Furthermore, it can act as both a Brønsted–Lowry acid ($\ce{NaHCO3 + OH- -> H2O + Na+ + CO3^2-}$) and as a Brønsted–Lowry base ($\ce{NaHCO3 + H+ -> Na+ + H2CO3}$). Another amusing example is hydrazinium sulfate, a salt, acid and base, where both the cation and anion are also both acids and bases!


How would you confirm that a colorless liquid given to you is pure water?

Measurement of properties like freezing point and boiling point would give you a quick, qualitative though likely correct, answer.  The simplest to do is the measurement of the liquid's freezing point.

If you can, partially freeze your solution in a kitchen freezer, stir well, then measure the temperature while stirring.  Since the freezing point of water is relatively insensitive to atmospheric pressure, you don't have to worry about correcting for the elevation where you live.  This method is commonly used as a quick way to check the calibration of a thermometer.  Of course a measurement of ${0 ^oC}$ would mean that you likely have water.  If it doesn't freeze, assuming your freezer is cold enough to freeze water, then it's not water.  If this is what you observe, drop a small ice crystal or a grain of salt into the solution to initiate ice formation just to be sure it is not supercooled.  Also, if it doesn't freeze, measure the temperature of the solution to be sure it is at least several degrees below ${0 ^oC}$.

A shortcut, or as a further check, if you have plenty solution to spare you could take a portion and add enough crushed ice that it doesn't all melt upon stirring.  If you get the same value (${0 ^oC}$) as in the first experiment, you can be pretty confident that your solution is water.  If it were some mixture of solutions with a freezing point of ${0 ^oC}$ (unlikely), this test would give you a different measurement than in the first test due to either colligative properties if the amount of ice is small relative to the whole solution or by creating a new, bulk solution with a different melting point than the original solution.

If both of these experiments give you the same result of ${0 ^oC}$, I'll bet my house that it's water.


None of the ingredients listed would be classified as a psychotropic substance.  However, many of the compounds, when studied in mice, do have effects on brain activity and behavior characteristics.  For example,

rapeseed oil (in a mixture with other vegetable oils): "the rapeseed oil group exhibited much higher locomotor activity than that expected from the alpha-linolenate/linoleate ratio. Additionally, the rapeseed oil group exhibited unusual behavior patterns, including higher ambulation and rearing activities, faster acquisition of the water maze task and slower habituation behavior as compared with the control group. Susceptibility to pentobarbital anesthesia tended to be higher in the rapeseed oil group." abstract link
bergamot: "produces dose-dependent increases in locomotor and exploratory activity that correlate with a predominant increase in the energy in the faster frequency bands of the EEG spectrum" abstract link
linalool: "Inhaled linalool showed anxiolytic properties in the light/dark test, increased social interaction and decreased aggressive behavior" abstract link

So I am not surprised to read that when used in combination they have a noticeable affect on you.


What is the quickest way to dissolve steel?

The fastest way is dissolving the stainless steel electrochemically using $\ce{H2SO4}$ at elevated temperature where the steel block is attached as an anode. Higher current densities lead to higher removal rate.

Alternatively a $\ce{FeCl3}$ with minimal content of $\ce{HCl}$ (up to 3%) heated at 45°C or higher.


This doesn't appear to be a rare case and there are plenty $\ce{Cr^2+}$ square planar complexes.
I found 15 square planar complexes with well-determined crystal structures $(R_\mathrm{int} < 10\%)$ by applying the following geometrical restrains:

Chromium must have C.N. exactly 4 (T4).
Chromium is allowed to be coordinated by virtually any element of periodic table $\ce{El}$.
Dashed lines signify any bond type.
All four angles between neighbors should lead to nearly ideal square planar geometry: $∠(\ce{El-Cr-El}) = \pu{(90.0 ± 0.5)^\circ}$ (by allowing higher tolerance the number of structures grows in geometrical progression: for $∠(\ce{El-Cr-El}) = \pu{(90.0 ± 5.0)^\circ}$ there were nearly 100 structures).

List of compounds

CCDC ACACCS
[1]
bis(2,4-Pentanedionato)-chromium(II)
CCDC BHPZCR
[2]
bis(Dihydro-bis(1-pyrazolyl)borato) chromium(II)
CCDC CRPORT
[3]
α,β,γ,δ-Tetraphenylporphinato-chromium(II) toluene solvate
CCDC DAQDUV
[4]
hexakis(Tetrahydrofuran)-magnesium tetrakis(perfluorophenyl)-chromium(II) tetrahydrofuran solvate
CCDC DAYNEW
[5]
bis(Pyridinium) diaqua-dibromo-chromium(II) dibromide
CCDC EGINEN
[6]
bis((μ2-Chloro)-hexakis(tetrahydrofuran)-di-lithium) tetrakis(pentachlorophenyl)-chromium(II)
CCDC EQIVOP
[7]
trans-Dichloro-bis(2,6-dimethylpyridine)-chromium(II)
CCDC HARKIU
[8]
trans-bis(tris(t-Butoxy)siloxy)-bis(diethylammonio)-chromium(II)
CCDC KOMRAF
[9]
Tetrapyridine-chromium(II) bis(hexafluorophosphate) dipyridine clathrate
CCDC PITVAP
[10]
bis(Tetra-n-butylammonium) tetrakis(isothiocyanato)-chromium(II)
CCDC PUQPEY
[11]
bis(1-t-butyl-3-(di-t-butylphosphino)imidazol-2-ylidene)-dichloro-chromium
CCDC PUQPIC
[11]
dichloro-bis(1-(di-t-butylphosphino)-3-mesitylimidazol-2-ylidene)-chromium tetrahydrofuran solvate
CCDC PUQPIC
[12]
bis((Phenothiazinyl-N)-tetrahydrofuran)-chromium(II)
CCDC XEBZAF
[13]
trans-bis(4-t-Butylpyridine)-dichloro-chromium(II)
CCDC ZOYCUL
[14]
Dichloro-bis((trimethylphosphine)trimethylsilylamido)-chromium(II)

References

Cotton, F. A.; Rice, C. E.; Rice, G. W. The Crystal and Molecular Structures of Bis(2,4-Pentanedionato)Chromium. Inorganica Chimica Acta 1977, 24, 231–234. https://doi.org/10/bhjhh5.
Dapporto, P.; Mani, F.; Mealli, C. Vanadium(II) and chromium(II) Complexes with Poly(1-Pyrazolyl)Borate Ligands. Crystal and Molecular Structures of Potassium Tris[Dihydrobis(1-Pyrazolyl)Borato]Vanadium(II) Monoalcoholate and Bis[Dihydrobis(1-Pyrazolyl)Borato]chromium(II). Inorg. Chem. 1978, 17 (5), 1323–1329. https://doi.org/10/dzfdh8.
Scheidt, W. R.; Reed, C. A. Stereochemistry of the Toluene Solvate of α,β,γ,δ-Tetraphenylporphinatochromium(II). Inorg. Chem. 1978, 17 (3), 710–714. https://doi.org/10/cvqg73.
Forniés, J.; Martín, A.; Martín, L. F.; Menjón, B.; Zhen, H.; Bell, A.; Rhodes, L. F. The First Structurally Characterized Homoleptic Aryl-Manganese(III) Compound and the Corresponding Isoleptic and Isoelectronic chromium(II) Derivative. Organometallics 2005, 24 (13), 3266–3271. https://doi.org/10/bz3r8w.
Ladd, M. F. C.; Larkworthy, L. F.; Leonard, G. A.; Povey, D. C.; Tandon, S. S. Crystal and Molecular Structure of a Complex Bromide Containing Planar Diaquadibromochromium(II) Units. J. Chem. Soc., Dalton Trans. 1984, No. 10, 2351. https://doi.org/10/cj5z95.
Alonso, P. J.; Forniés, J.; García‐Monforte, M. A.; Martín, A.; Menjón, B.; Rillo, C. A New Series of Homoleptic, Paramagnetic Organochromium Derivatives: Synthesis, Characterization, and Study of Their Magnetic Properties. Chemistry – A European Journal 2002, 8 (17), 4056–4065. https://doi.org/10/cx9r27.
Gibson, V. C.; Newton, C.; Redshaw, C.; Solan, G. A.; White, A. J. P.; Williams, D. J. Low valent chromium complexes bearing N,O-chelating pyridyl-enolate ligands $\ce{[OC(Bu^t)(=2-CHN5H3Me-x)]^-}$ $(x = 3–6)$. Dalton Trans. 2003, No. 24, 4612–4617. https://doi.org/10/ffdkvr.
Terry, K. W.; Gantzel, P. K.; Tilley, T. D. chromium(II) and Chromium(III) Tri-Tert-Butoxysiloxy Complexes. Inorg. Chem. 1993, 32 (23), 5402–5404. https://doi.org/10/dfw68w.
Fochi, Giovanni.; Straehle, Joachim.; Gingl, Franz. Disproportionation of the Bis(Benzene)Chromium Cation Induced by Pyridine. Crystal and Molecular Structure of Two Tetrapyridine Complexes Containing chromium(II) and Chromium(III). Inorg. Chem. 1991, 30 (24), 4669–4671. https://doi.org/10/c3ffnj.
Larkworthy, L. F.; Leonard, G. A.; Povey, D. C.; Tandon, S. S.; Tucker, B. J.; Smith, G. W. Crystal Structures and Magnetic Behaviour of the Two Forms of Tetra-n-Butylammonium Tetraisothiocyanatochromate(II). J. Chem. Soc., Dalton Trans. 1994, No. 9, 1425. https://doi.org/10/cgtgbm.
Ai, P.; Danopoulos, A. A.; Braunstein, P. N-Phosphanyl- and N,N′-Diphosphanyl-Substituted N-Heterocyclic Carbene Chromium Complexes: Synthesis, Structures, and Catalytic Ethylene Oligomerization. Organometallics 2015, 34 (16), 4109–4116. https://doi.org/10/ggc5mx.
Edema, J. J. H.; Gambarotta, S.; Meetsma, A.; Spek, A. L.; Smeets, W. J. J.; Chiang, M. Y. chromium(II) Amides: Synthesis and Structures. J. Chem. Soc., Dalton Trans. 1993, No. 5, 789. https://doi.org/10/fcfcdg.
Cotton, F. A.; Dikarev, E. V.; Gu, J.; Herrero, S.; Modec, B. Alkylpyridine Complexes of Tungsten(II) and chromium(II). First Rotational Isomers of $\ce{W2X4L4}$ Molecules with $D_\mathrm{2h}$ and $D_2$ Symmetries. Inorg. Chem. 2000, 39 (23), 5407–5411. https://doi.org/10/c5cfr4.
Miekisch, T.; Mai, H. J.; zu Köcker, R. M.; Dehnicke, K.; Magull, J.; Goesmann, H. Silylierte Phosphanimin‐Komplexe von Chrom(II), Palladium(II) und Kupfer(II). Die Kristallstrukturen von $\ce{[CrCl2(Me3SiNPMe3)2]},$ $\ce{[PdCl2(Me3SiNPEt3)2]}$ und $\ce{[CuCl2(Me3SiNPMe3)]2}.$ Z. Anorg. Allg. Chem. 1996, 622 (3), 583–588. https://doi.org/10/bm3fq8.


Does the rate of movement of ions in a salt bridge affect the total voltage produced by a Galvanic cell?

Voltage measurement is ideally a zero current process. Therefore the details of the ions in the salt bridge and their movements will not be an issue.

In reality, voltmeters do flow some current to measure the voltages. Yet, these currents are so small that in any self-respecting salt bridge, there will not be an issue. If your salt bridge was broken for whatever reason though, then your measurement will be completely incorrect. Most voltmeters will give you an overload in that case.


What is the mechanism for the reaction of alkenes with diborane and chloramine?

Organoboranes react with chloramine to give the corresponding amine as reported by Brown et al. [1], so P is n-propylamine (propan‐1‐amine).

The mechanism is similar to the oxidation of boranes using $\ce{H2O2}/\ce{NaOH}$, but instead of migration to an oxygen atom, the alkyl chain migrates to a nitrogen atom. The $\ce{N-B}$ bond is cleaved in the same way.

Reference

Brown, H. C.; Heydkamp, W. R.; Breuer, Eli.; Murphy, W. S. The Reaction of Organoboranes with Chloramine and with Hydroxylamine-O-Sulfonic Acid. A Convenient Synthesis of Amines from Olefins via Hydroboration. J. Am. Chem. Soc. 1964, 86 (17), 3565–3566. DOI: 10.1021/ja01071a036.


Does acetoacetic ester give haloform test?

As @user55119 says it is unlikely the di-iodide is formed under iodoform conditions.
According to this paper1, acetoacetate gives only mono-iodination at the methylene centre under more powerful iodinating conditions than the standard iodoform $\ce{I2/OH-}$. The mono-iodo product in the presence of excess base will exist as the iodo-enolate effectively stopping iodination at the methyl ketone centre. Thus no iodoform product will be formed.
References

Goswami, Papori & Ali, Shahzad & Khan, Dr. M. Musawwer & Khan, Abu. (2007). Selective and effective oxone-catalysed α-iodination of ketones and 1, 3-dicarbonyl compounds in the solid state. General Papers ARKIVOC. 2007. 82-89. (Link)


How would a pH curve look like for titration of diluted weak base compared to concentrated one?

I am first going to do the ‘flip side’ of the OP’s scenario and then basically flip it back. So we start with a $\pu{0.1 M}$ aqueous solution of benzoic acid and the initial solution volume is $\pu{25 mL}$. Benzoic acid is a monoprotic weak acid with $\mathrm{p}K_\mathrm{a} = 4.20$. The initial $\mathrm{pH}$ is $2.606$. The strong base used for the titration is $\pu{0.1 M} \ \ce{NaOH}$. At the equivalence point, the $\mathrm{pH = 8.454}$ and the solution volume is $\pu{50 mL}$ ($=\pu{25 mL}$ benzoic acid solution $+ \pu{25 mL} \ \ce{NaOH}$ solution). Thus, the $\mathrm{pH}$ is simply the $\mathrm{pH}$ of $\pu{0.05 M}$ sodium benzoate solution. After all $\pu{50 mL}$ of $\ce{NaOH}$ has been added, the $\mathrm{pH} = 12.523$ and the final solution volume is $\pu{75 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{NaOH}$ has been added and $\mathrm{pH}= 4.185$.

Now assume the initial benzoic acid solution is only $\pu{0.025 M}$, but the initial volume is $\pu{100 mL}$. The total number of moles of benzoic acid is the same as before. The initial $\mathrm{pH}$ is $2.912$: the benzoic acid solution is only 25% as concentrated. At the equivalence point, the $\mathrm{pH} = 8.254$ and the solution volume is $\pu{125 mL}$ ($=\pu{100 mL}$ benzoic acid solution $+ \pu{25 mL} \ \ce{NaOH}$ solution). The equivalence point $\mathrm{pH}$ is closer to $\mathrm{pH}=7$ because the solution is simply a $\pu{0.02 M}$ sodium benzoate solution, i.e., the number of moles of sodium benzoate is now in a volume that is 2.5 times larger, so the concentration is 40% of $\pu{0.05 M}$. After all $\pu{50 mL}$ of $\ce{NaOH}$ has been added, the $\mathrm{pH}= 12.222$ and the final solution volume is $\pu{150 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{NaOH}$ has been added and $\mathrm{pH} = 4.188$, moving slightly up toward $\mathrm{pH}=7$. See the titration curves below:

Now flipping back: the OP’s scenario involving a weak base titrated with strong acid. For reasons that will become clear, assume the hypothetical weak base has $\mathrm{p}K_\mathrm{b} = 9.80$, and that the strong acid is $\pu{0.1 M} \ \ce{HCl}$ solution. Note that the $\mathrm{p}K_\mathrm{b}$ is simply $14$ (treated as exact throughout) $- 4.20$.

As before, start by assuming $\pu{25 mL}$ of $\pu{0.1 M}$ weak base. The initial $\mathrm{pH} = 14 – 2.606 = 11.394$. At the equivalence point, the $\mathrm{pH} = 5.546$ and the solution volume is $\pu{50 mL}$ ($=\pu{25 mL}$ weak base solution $+ \pu{25 mL} \ \ce{HCl}$ solution). Thus, the $\mathrm{pH}$ is simply the $\mathrm{pH}$ of a $\pu{0.05 M}$ weak base salt solution. After all $\pu{50 mL}$ of $\ce{HCl}$ has been added, the $\mathrm{pH} = 1.477$ and the final solution volume is $\pu{75 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{HCl}$ has been added and $\mathrm{pH} = 9.815$.

Now assume the initial weak base solution is only $\pu{0.025 M}$, but the initial volume is $\pu{100 mL}$. The total number of moles of weak base is the same as before. The initial $\mathrm{pH}$ is $11.088$: the weak base solution is only 25% as concentrated. At the equivalence point, the $\mathrm{pH} = 5.746$ and the solution volume is $\pu{125 mL}$ ($=\pu{100 mL}$ weak base solution $+ \pu{25 mL} \ \ce{HCl}$ solution). The equivalence point $\mathrm{pH}$ is closer to $\mathrm{pH}= 7$ because the solution is simply $\pu{0.02 M}$ weak base salt solution, i.e., the number of moles of weak base salt is now in a volume that is 2.5 times larger, so the concentration is 40% of $\pu{0.05 M}$. After all $\pu{50 mL}$ of HCl has been added, the $\mathrm{pH} = 1.778$ and the final solution volume is $\pu{150 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{HCl}$ has been added and $\mathrm{pH} = 9.812$, moving slightly down toward $\mathrm{pH}= 7$. For the titration curves, simply imagine vertically flipping the titration curves, in the figure above, about a horizontal line at $\mathrm{pH} = 7$. Or just look at the plot:

One last thing: The $\mathrm{pH}$ values at half-equivalences were close to the respective $\mathrm{p}K$ values, as expected, but any buffer can be broken by dumping it into swimming pool full of water. So dilution may matter a great deal, or relatively little, depending on the specifics.


Which alkene on heating with alkaline KMnO4 solution gives acetone and CO2?

The best approach, and good practice, is to draw out the reaction products for each case.  You should know that alkaline potassium permanganate replaces a carbon-carbon double bond with two carbon-oxygen bonds, one to each of the original double-bonded carbons.  For instance, with 2-butene we have:

$\ce{CH3-CH=CH-CH3 -> CH3-CH=O + CH3-CH=O}$

Here the product molecules are the same because the alkene is, of course, symmetric.

But wait, there's more.  The carbonyl compounds pictured in the above example are stable in the alkaline permanganate medium if they are ketones, with only carbon atoms attached to the carbonyl function.  If there are aldehydes, the carbon-hydrogen bond(s) to the carbonyl group will be oxidized further.  Thus the "acetaldehyde" identified above for 2-butene is further oxidized to break the carbon-hydrigen bond at the carbonyl group:

$\ce{CH3-CH=CH-CH3 -> 2 CH3-CH=O -> 2 CH3-C(O)-OH}$

where in the alkaline medium the acetic acid will appear as acetate ion.

As you work through your choices you find that there is one aldehyde where the carbonyl carbon is bonded only to hydrogen atoms allowing that function to be oxidized all the way to carbon dioxide (or, again accounting for the alkali, carbonate or bicarbonate ion).  Find it, then see which choice gives that along with the other given product acetone.  Bonus question:  Would the acetone hold up against further oxidation?  How would you tell?


How can a catalyst not be included in a rate equation if, by definition, it speeds up a chemical reaction?

Enzyme kinetics is one area of catalysis that has been studied extensively. The Michaelis-Menton equation being one of the central pillars and explicitly states the roles of concentration for both the substrate and enzymes.

Also, to answer your question: the rate equation is at times more interested in establishing equilibrium conditions than obtaining the time it takes to get there. The time to reach equilibrium is slightly neglected as the question "are we there yet" seems trivial when distance = rate times time.


Would atomic oxygen and atomic hydrogen result in O2 and H2 only, or some H2O too?

It would ignite and form water vapor. It's not complicated to figure out. The conditions you have described with a mixture of monoatomic hydrogen and oxygen is what is formed when a spark passes through a mixture of hydrogen and oxygen gases.


How can I permanently write on HDPE plastic?

From experience, one good way to permanently mark on HDPE is a paint marker. The following is just the first one I found online; no endorsement intended:

   Promotional image from uline.com

They tend to be stinky and a tad messy, but they usually leave clean, visible lines and are basically indelible, requiring a lot of scrubbing with alcohol, acetone, etc. to remove. No idea how well it'll survive power washing, though.


You're looking for a mordant or fixative, such as potassium alum, $\ce{KAl(SO4)2·12H2O}$.

Though I do not know how well this will work on wood, mordants are often used on textiles. Check for cotton and linen mordants in particular, since they are cellulosic material, like wood (though wood also has lignin, which may inhibit uptake of the stain).

You might ask your teacher if she thought you are too young to dye.


This is actually just one part of a much more general phenomena, which is the release of large, polycyclic aromatic compounds due to pyrolysis or combustion of organic matter. It is not specific to wood. So, to understand why these compounds form in a wood fire, or any other kind of fire, we must first understand what wood is, and then we must understand how wood burns.

Wood is biomass - that is, matter composed primarily of carbons, hydrogens, and oxygens. We can generalize here, and talk about any biomass - wood, dead or dry grass, leaves, animal carcasses, fibers, sugar, fat, and even things like coal (which is fossilized biomass with special properties). All of these things contain cellulose or similar molecules, which consists of long chains of aromatic rings (the wikipedia article on cellulose is a good place to start, and any textbook on biochemistry will give much more detail). In the context of combustion, this is called "volatile organic matter." Volatile just means that if you heat it up, it burns, gives off heat, and turns into gas.

Living things also have non-volatile and non-organic matter. As an example, people require some trace amounts of metals for biochemical functions, so humans contain some percentage of mineral matter. In the context of combustion, this material is often called "ash." If you have a fuel like wood, this ash is what's left behind after the wood burns, and composes much of what's released into the air in the form of smoke. Some fuels can have large percentages of ash (as an example, when organic matter fossilizes and forms coal, a large amount of rock or dirt interspersed with the organic material can lead to coals with very high ash content).

When you heat up organic matter, what you're doing is dumping a large amount of thermal energy into the molecules composing that matter, and after a certain point (depends on the fuel, but we're talking about 400 or 500 C) the large chains that compose organic matter begin to break apart. If this happens in the presence of oxygen, the material will also begin to oxidize, and the process is called combustion. If it happens in the absence of oxygen, the material will gasify. As the temperature gets hotter and hotter, a soup of big organic macromolecules is formed. (When I say big, I mean, BIG - hundreds of carbon, hydrogen, and oxygen atoms, and thousands of side chains and functional groups). It is in this macromolecular soup that you can find benzene and its many, many, many cousins. (These molecules are too complex to have names, and are so large that they can't be characterized exactly - instead, chemists can only statistically describe the molecules, using techniques like nuclear magnetic resonance (NMR), mass spectrometers (MBMS), or infrared spectrometers (FTIR).) This process of solid matter turning into volatile gases is called pyrolysis.

Big polycyclic molecules form when there is not enough oxygen (if there were enough oxygen, all the carbon and hydrogen would form carbon dioxide and water). They are huge and hydrophobic, and agglomerate to form soot and tar. When you see sticky brown films on cigarettes or smokers' hands or in pipes, this is caused by these macromolecules.

If you looked very closely at a wood fire, and were able to see the actual flame-wood interface, what you would see is jets of hot, flammable gases shooting out of the pores of the wood. It is these gases that ignite - not the wood itself. (Of course, as the wood heats up, and the organic matter is burned, the mineral matter is left behind, and that becomes extremely hot but does not vaporize, so you still have radiant, red-hot material.)

You can still find benzene, even in a hot fire, because flames are not perfectly mixed, and you can get pockets of cold gas interspersed in a cloud of hot gas. If it is cold enough, benzene (which can be very stable) will not be decomposed.

So the next time you watch a campfire, or smoke a cigarette, or set your dinner on fire in the oven, you should be able to better understand what you're seeing - or know where to look.


Water and ethanol are highly "miscible"; a science term that means virtually any amount of water is fully soluble in ethanol, and vice-versa.  The next thing to consider is that ethanol and gasoline are quite miscible. Think of E-10 (10% ethanol) all the way to the E-85 some petrol engines are designed to function well on (with  some loss in power output).
However, water is very poorly soluble in pure gasoline.
So adding when water to E-10 the water mixes well with the ethanol, but the H2O/ethanol mix looses solubility rapidly as more water is added.
In theory, if you set up a separation flask, basically a clear glass vessel with a petcock at the very bottom, filled it at room temp with a gallon of E-10 and added ~3 tablespoons of water, shook it all up then let it settle to separate, you could drain away the bottom (denser) layer of liquid, which would be the water/ethanol mix, an leave you with ~0.9 gallons of relatively alcohol-free petrol.
My final remark would be to ask why anyone would wish to do this as I can think of no reason in the world why someone would bother.


What's considered a "higher" level of theory in quantum mechanics calculations?

For single reference wavefunction based methods, the comparison is relatively straightforward within method "families". This is due to the nature of how these methods are formulated, usually involving a very long series that has been terminated at some point to make the calculation feasible.

You start with Hartree-Fock, which ignores electron correlation.
After that you can add perturbative corrections of increasing order (this will be the MP2, MP3, MP4, MPn family), do something called configuration interaction, truncated at increasing levels of excitation (CIS, CISD, CISDT,...,FCI), or something called the coupled cluster methods, again truncated at some level of excitation (CCSD, CCSDT, CCSDTQ,...,FCI).

In these cases the most straightforward interpretation of "higher level" would mean a less aggressive truncation of the series that includes more terms. In case of CI and CC, this almost always results in more accurate results, and both converge to the same limit, where all possible terms are included and correlation effects are exactly calculated. (of course there are errors from other approximations) The MPn perturbation series has no such strong guarantees, and this is one of the reasons why pretty much noone uses anything beyond MP4 these days. (even MP4 is seldom used)

For DFT you have something called Jacob's ladder, but things are generally more murky, as most DFT functionals have many empirical (fitted) parameters, and there is no straightforward way of comparing them.

In general, the only way to compare them, is to take a big database of either experimental or calculated (via CCSD(T) or similar high level method) data, and compare the average/maximum errors.


What is wrong in this argument that dG must always be zero?

Your textbook's derivation is done under the assumption of constant $T$, which means $T_{sys} = T_{surr} =T$.  However, this does not mean $dG_{sys}$ is always zero.  Let's start with the following:

$$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T_{sys}}+\frac{\text{đ}q_{rev, surr}}{T_{surr}}$$

Since heat flow always affects the surroundings reversibly (the surroundings are in the limit of being infinitely large, and thus heat flow affects them only infinitesimally), the reversible heat flow into the surrondings equals the actual heat flow, which is the negative of the actual heat flow into the system.  And since $T_{sys} = T_{surr} =T$, we have:

$$\frac{\text{đ}q_{rev, surr}}{T}=-\frac{\text{đ}q_{sys}}{T}$$

Which gives:

$$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T}+\frac{\text{đ}q_{rev, surr}}{T}=\frac{\text{đ}q_{rev, sys}}{T} -\frac{\text{đ}q_{sys}}{T}$$

Under the restrictions of constant $p$, and no non-$pV$ work,

$$\text{đ}q_{sys}=dH_{sys}$$

Hence:

$$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T}+\frac{\text{đ}q_{rev, surr}}{T}=\frac{\text{đ}q_{rev, sys}}{T} -\frac{\text{đ}q_{sys}}{T}=\frac{\text{đ}q_{rev, sys}}{T} -\frac{dH_{sys}}{T} = dS_{sys} - \frac{dH_{sys}}{T},$$

which is what your textbook provides.

Note, however, that

$$\frac{dH_{sys}}{T} = \frac{\text{đ}q_{sys}}{T} \ne \frac{\text{đ}q_{rev,sys}}{T}=dS_{sys},$$
unless the process is reversible, i.e., non-spontaneous, in which case we could replace the "$\ne$" with "=", and $dG_{sys}$ would equal $0$.

This connects directly to the great beauty of using $\Delta G_{sys}$ at constant $T \text{ and } p$, when there is no non-$pV$ work.  Because, if the system is at constant $T$*,
$$\Delta G_{sys} = \Delta H_{sys} - T \Delta S_{sys}$$

[*$G=H-TS => dG = dH - d(TS) = dH -TdS -SdT$; at const $T, dG= dH -TdS $; note that a constant-pressure restriction is not required to write this.]

Where we need the additional restrictions of constant $p$, and no non-$pV$ work, is for the following:

$$\Delta H_{sys} = -T \Delta S_{surr}$$

This in turn gives:

$$\Delta G_{sys} = -T \Delta S_{surr} - T \Delta S_{sys} = -T \Delta S_{univ}$$


Relative reactivity of alkyl and aryl halides in nucleophilic-substitution reactions

There are three mechanisms to discuss here: $S_N2$, $S_N1$ for $RX$ and nucleophilic aromatic substitution by addition-elimination for $ArX$:
$S_N2$ and $S_N1$:
(Image from https://www.quora.com/Why-steric-hindrance-doesnt-affect-Sn1-reaction)

$S_N2$ is concerted (one step), so this one step is rate determining, or the slowest of the mechanism. In $S_N1$, formation of the carbocation is rate determining. In both rate determining steps, the leaving group is being kicked off. This means the easier it is for the $C-X$ bond to break, the faster the reaction goes. The bond enthalpy decreases as size of halogen increases, giving the order you stated above.
Aromatic addition-elimination:
(Image from https://www.masterorganicchemistry.com/2018/08/20/nucleophilic-aromatic-substitution-nas/#seven)

In this case, step 1 is rate determining, which does not involve the leaving group departing. The halogen, however, can speed up the reaction by stabilising the negative charge formed during this step. The stronger the inductive effect of the halogen - which outweighs the +M mesomeric effect - the more it can draw the negative charge from the ring, stabilising it. The bond strength is irrelevant here, as the step during which it is cleaved is faster than the previous step regardless, and is always 'waiting around' for the first step. This gives the reverse order to the previous one.


Radical Bromination of a 1-isopropyl-4-methylcyclohexane

They are not chiral because $\ce{C}-4$ is only considered to be bonded to 3 unique groups because the ring is completely symmetrical through $\ce{C}-1$ and $\ce{C}-4$.

As for naming these compounds, the reagent is simple: cis-1-isopropyl-4-methylcyclohexane (or cis-1-(1-dimethylethyl)-4-methylcyclohexane if you're a stickler for the rules).

The products are a little tricker. Let's start with the basics: it is 1-bromo-4-isopropyl-1-methylcylohexane. Now to account for the steric relations of these substituents, label the reference substituent (labeled r) followed by a hyphen and the substituent locator number and name. The relative configuration of other substituents are then reported as cis (c) or trans (t) to the reference substituent.

This means that in the order that you present the products, their names are r-1-bromo-t-4-isopropyl-1-methylcylohexane and r-1-bromo-c-4-isopropyl-1-methylcylohexane.

[Source - (page 1060)]


Why is cyclooctateraene shaped like a "tub" rather than a "chair"?

It's all about the 3D structure of double bonds.

If we look at the tub form we see that all dihedral C-C=C-C angles are 0°. the C=C-C angles are 125°, also pretty close to the optimal value. Everything works out fine, there's basically no strain on the whole molecule.

I tried to build the chair form of all cis, but every optimization ended up in either the all cis tube form or the cis-trans-cis-trans chair form. The best structure that looks like the all cis chair form I ended up with is this:

First of all, it's pretty flat, so antiaromaticity would again play a role and the dihedral angles are also ~33° for the double bonds in front and in the back, which is energetically quite bad and unstable.

The cis-trans-cis-trans compound is definitely no monstrosity. I'm working a lot with trans-cyclooctenes and I don't see why this shouldn't be possible if a trans,trans-cycloocta-1,5-diene is possible.

Here we've got 0° for the dihedral angles for the cis bonds and 125° for the trans. Yes, this is far away from the optimal value of 180° but quite similar to what you expect from trans-cyclooctenes (around 135°). I feel like it would be quite reactive and unstable but it could definitely exist. There's just no good way to produce it I guess.

So overall: all angles in the tub form are much closer to the optimal values than in either chair form and the all cis form is quite planar.


Separate less volatile component from distilling a mixture

The more volatile compound forms greater vapour portion, compared to its portion in the distilled liquid.

It forms e.g. 15% molar fraction of vapour, if there is 10% molar fraction in liquid.

In case of deviation from the Raoult law, when there is interaction between compounds, an azeotrope with minimal  ( ethanol-water ) or maximal ( nitric acid water ) boiling point may be formed. This azeotrope acts like if it were the 3rd compound.

Within rectification columns , for each distillation stage upwards, the composition is slightly enriched with more volatile fraction.

And vice versa,  for each distillation stage downwards, the composition is slightly enriched with less volatile fraction.

If we are interested just in more volatile fractions, the column bottom is attached to the vapour source input, what is typical for labs.

If we are interested in all fractions, what is typical in oil industry, the input is injected somewhere in the middle of column, where the equilibrium composition is the most similar to the input.

The more or less volatile fraction may be an azeotrop. Components in ratio forming the azeotropic ratio cannot be separated by destination, unless other components are added.

The typical example is the azeotrope ethanol water, denying to get more than about 95% ethanol by distillation, as this azeotrope has lower boiling point than pure ethanol.

Pure ethanol can be gained by adding benzene,  forming the ternary azeotrope  benzene - ethanol - water, with even lower boiling point.


Understanding "Gay-Lussac's law of Gaseous volumes/combining volumes"

The ratio between the volumes of the reactant gases and the gaseous products can be expressed in simple whole numbers

Wikipedia is correct.

Expressed in, not expressed as.

$3/2$ can be expressed in simple whole numbers $3$ and $2$.


Freezing point depression to find molecular weight

Mass of solute $W_b=0.845\times10^{-3}\;kgm$
  Mass of solvent $W_a=4.250\times10^{-3}\;kgm$
  Depression in freezing point $\Delta T_f=1.36\;K$
  Depression constant for solvent $K_f=40\;K.kg.mol^{-1}$
  Freezing point of Camphor $T_f=129\;^\circ C$

Part I: $\Delta T_f$

Part II: Molality($m$): $$\Delta T_f=K_fm\implies m=\frac{\Delta T_f}{K_f}$$

Part III: Moles($n_b$): $$m=\frac{n_b}{W_a}\implies n_b=W_am=W_a\frac{\Delta T_f}{K_f}$$

Part IV: Molecular Mass of solvent($M_b$): $$n_b=\frac{W_b}{M_b}\implies M_b=\frac{W_b}{n_b}=\frac{W_bK_f}{W_a\Delta T_f}$$


Can you produce activated carbon from toast?

Is toast a good source of carbon is quite a broad question without considering the type of toast we're referring to. Even if do stick to a particular toast, what makes a good carbon source sounds a little too vague.

As a side note, recent researches have led to the production of a low cost multi-functional carbon foam from bread. [Source]

As a rough estimate, let's consider we're talking about white bread. 100 grams of white bread consists of 49 grams (i.e. approximately 16%) of carbohydrates. [Source] For the sake of simplicity, let's assume all of the carbohydates to be monosaccharides, specifically glucose. (This definitely isn't the case). The molecular formula of glucose is $\ce{C6H12O6}$. If we calculate the percentage-by-mass composition of carbon in glucose, we get the value as 40%. 40% of 49 grams is 19.6 gram, or in other words, bread consists of (very roughly) 19.6% carbon.

I'm unsure of whether or not that makes it a good source. Reading this article tells me that a rough estimate for the carbon content in wood (it is about some conifers but nevertheless good for rough estimates) is about 45–50%.

The production of activated carbon is undertaken generally in two ways: [Source]

Physical activation: It requires an inert atmosphere of nitrogen or argon along with very high temperatures (600–1200 °C).
Chemical activation: Strong acids, bases or salts and a lower temperature is required (450–900 °C).

Theoretically the chemical method seems appealing but in a practical sense it is not advisable to carbonize materials in a home-kitchen as the fumes evolved contain carbon-monoxide which is extremely toxic.


You can think of water as the ash from burning hydrogen: it's already given off as much energy as possible from reacting hydrogen with oxygen.

You can, however, still burn it.  You just need an even stronger oxidizer than oxygen.  There aren't many of them, but fluorine will work,

$$
\ce{2F2 + 2H2O -> 4HF + O2}
$$

as will chlorine trifluoride:

$$
\ce{ClF3 + 2H2O -> 3HF + HCl + O2}
$$


How many samples would be worth when following a desorption? How to relate with limit of detection?

I think you are confusing two concepts: the limit of quantification (LOQ) and the sensitivity of the technique.

LOQ is the lowest concentration of the analyte that can be quantified at an acceptable level of uncertainty and accuracy, and is determined (by most methods) as the calculated concentration of the blank + 10*sd of the blank measurements.

Sensitivity, on the other hand, is the method's ability to distinguish between different amounts of analyte.

There are some guidelines that do not make a differentiation between the two (such as in the FDA's Bioanalytical Method Validation guidance document), but this is generally not the case in other situations.

In your case, you will need to find an analytical method that is sensitive enough to differentiate between 0.96 (a.u.) and 0.98(a.u), that is, 0.02 units.

As for the limit of quantification, you need to ask yourself what is the lowest concentration of analyte that you want to quantify. Based on the decay plot you have shown, it seems like the lowest concentration you would encounter is above 0.8 a.u. In that case, a method with an LOQ of 0.8 a.u should be sufficient. If you want to be even more rigorous, your LOQ should be at a level that is low enough that even if it is extended by the expanded standard uncertainty of your method, it should still be low enough to quantify the lowest level of analyte you expect to encounter.


Is it possible to create a ketone using ozonolysis, when you have an alkene with no alkyl groups?

I don't think that's possible.

The general reactions scheme for ozonolysis is as follows:

Consider the case of butene:

Herein $R_1$ and  $R_3$ can be identified as methyl groups, and similarly $R_2$ and $R_4$ are plain old hydrogens. Following the reaction scheme, one would naturally expect Ethanal as a product (if you use $\ce{Zn}$, ethanoic acid may be obtained if you use something else).

I'll drop by later and write out a mechanism perhaps.


Epoxide gets opened at the least hindered side by the ethoxide (acting as a nucleophile rather than as a base in this case).
The resulting alkoxide intermediate can itself act as a nucleophile. In the middle intermediate could either kick out ethoxide (which would give back the starting material), or kick out the chloride (to give a different epoxide). Since the chloride is a better leaving group, it is this one that probably leaves (theres an element to which the whole thing is reversible away, and if it does kick out ethoxide, the ethoxide can just attack again to re-form the middle alkoxide intermediate)

The product on the right hand side is probably what the mixture would converge upon, given that chloride is  a better leaving group than ethoxide.


What is the justification for Hückel's rule?

The justification comes from molecular orbital theory. I'm going to analyze the MO diagrams for benzene (a molecule with $4n+2$ π electrons) and an MO diagram for cyclobutadiene (a molecule with $4n$ π electrons). These results generalize for all planar cyclic molecules with π orbitals on every atom.

The above π MO diagram of benzene may be obtained through Hückel theory (see also: Pi molecular orbitals of polyenes). Here, the "zero" of energy (referred to as $\alpha$ in Hückel theory) refers to the energy of an isolated carbon 2p orbital that does not participate in bonding. Hence, any orbitals below this energy are bonding in nature, and orbitals above it are antibonding.

In the benzene MO diagram above, you'll notice that all the molecular orbitals that are being filled by the π electrons are more stable than leaving the carbons unbonded. On top of that, it can be shown that the stabilisation derived from delocalised π bonding in benzene is greater than the stabilisation derived from simply forming three localised C–C π bonds, in conjugation with one another, but not delocalised over the entire ring (e.g. in hexa-1,3,5-triene). This is the condition known as aromaticity.

Let's look at the MO diagram for a hypothetical perfectly square cyclobutadiene, for contrast.

As with all molecular orbitals, the most stable orbital is filled first. However, problems start to emerge when going up to next two orbitals. First, unlike the next orbitals in sequence for benzene, these are formally nonbonding within simple Hückel theory.

It can be shown that this situation is less stable than the case for an alternative version of cyclobutadiene, where the double bonds are localised between two adjacent carbons. In practice, cyclobutadiene therefore undergoes a distortion away from a square geometry to a rectangular geometry, which has longer C–C bond lengths for the single bonds and shorter bond lengths for the now localised double bonds (see also: Cyclobutadiene - Jahn–Teller effect or not?). This is the condition known as antiaromaticity.


The short answer is because $\ce{HF}$ doesn't dissociate as much as $\ce{HCl}$.  Let's examine why this is by introducing some terms like $\ce{pH}$ and $\ce{pK_a}$.

$\ce{pH}$ does not measure the strength of a given acid, but rather the acidity of a given solution.
$\ce{pK_a}$ does measure the relative strength of an acid, look at the following equilibrium

$$\ce{H2O + HX<=>H3O^+ + X^-}$$
$$\ce{so~~~K_{eq}=\frac{[H3O^+][X^{-}]}{[H2O][HX]}}$$ but in dilute solutions the concentration of water stays constant so we can rewrite this equation as
$$\ce{K_{a}=\frac{[H3O^+][X^{-}]}{[HX]}}$$ and finally, $$\ce{pK_{a}~=~-Log_{10}K_{a}}$$ Now we can see that the more highly dissociated our acid $\ce{HX}$, the higher the concentrations of $\ce{H3O^+}$ and $\ce{X^{-}}$, and in turn the larger the value for $\ce{K_{a}}$, and the more negative the value of $\ce{pK_{a}}$.

Bottom line: highly dissociated acids have large negative $\ce{pK_{a}}$ values

Now, here is a Table from SparkNotes.  It compares the $\ce{pK_{a}s}$ (degree of ionization or strength) of the various haloacids.  You can see that as we move down the Table from $\ce{HF}$ to $\ce{HI}$ the $\ce{pK_{a}}$ becomes much more negative; our acids become much more dissociated in water.  So we see that $\ce{HF}$ is not very dissociated ($\ce{K_{a}=10^{-3.1}}$, equilibrium lies to the left). On the other hand, 1 mole of $\ce{HCl}$ will produce about 1 mole of $\ce{H^+}$ ($\ce{K_{a}=10^{+7}}$, equilibrium lies far to the right).  Note too that the $\ce{pK_{a}s}$ of these acids correlate with their bond strength which makes sense.  As the $\ce{H-X}$ bond gets weaker, it becomes easier for the acid to ionize (dissociate).


Can ideal gas law be applied to water vapour?

Yes, for many practical purposes, you may use the ideal gas law also as an approximation for water vapour.

However, you need two independent quantities to describe the state of the gas, for example temperature and pressure. The given temperature of $T=298\ \mathrm K$ alone is not enough. The ideal gas law does not give you a value for the second quantity.

For example, assuming equilibrium conditions for saturated steam at a temperature of $T=298.00\ \mathrm K$, the corresponding pressure is $p=3141.7\ \mathrm{Pa}$.

Using the ideal gas law yields a concentration of

$$\frac nV=\frac p{RT}=\frac{3141.7\ \mathrm{Pa}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times298.00\ \mathrm K}=1.2680\ \mathrm{mol\ m^{-3}}$$

The actual value for water vapour at this temperature and pressure is

$$\frac nV=1.2701\ \mathrm{mol\ m^{-3}}$$

Thus, using the ideal gas law introduces an error of about $0.2\ \%$ in this case, which could be acceptable for many practical purposes.


If you had a sealed container of aluminum powder that became exposed to air, your prospects for turning your current material back into aluminum powder are grim.

Your block probably appears like something that became clumpy when exposed to humidity (table salt, sugar, etc.), and that you aught to be able to just break it apart with a little force to get the powdery stuff again.

Unfortunately, aluminum reacts with the oxygen in air to form a very hard oxide layer, which is likely what is now binding the formerly pure aluminum powder together. In other words, you don't just have aluminum "sticking" to aluminum, but you've formed strong chemical bonds, actually creating a new compound altogether (aluminum oxide) There is likely no simple solution for returning your 150 pound blocks of mixed aluminum / aluminum oxide back into pure aluminum powder.


How are the Cahn-Ingold-Prelog rules applied to chiral molecules without a stereogenic carbon center? i.e biphenyls and allenes

Structures with axial chirality (e.g. an allene or a 2,2′,6,6′-tetrasubstituted biphenyl) with two pairs of substituents (ab and cd) are regarded as elongated tetrahedra and viewed along the chiral axis. The substituents (a, b, c, and d) are arranged in the elongated tetrahedral system.

In each pair (ab as well as cd), one higher-ranking substituent (a > b and c > d) is chosen using the sequence rule.

If the path going from a to b to c (while looking toward d) is clockwise, the chirality is described by the symbol Ra. The symbol is Sa if this path is anticlockwise.


Aluminium sulphate solubility as function of the pH

The lack of continuity for the curve indicates that the concentration of $\ce{Al2(SO4)3}$ was below the detection limit of the analytical technique being used to measure the concentration of $\ce{Al2(SO4)3}$ in the solution.

Theoretically there should be a continuous curve.

+1 for asking this as a new question.


Does the addition or multiplication rule have precedence deciding the significant figures in 5×5.364?

If this is the question in its entirety, then you should not try to 'guess' what the origin, and therefore the accuracy, of the measurement 5 is. You should take it on face value, and complete the calculation using the multiplication rule, as the equation is requiring a multiplication operation. In this case, you should have one significant figure only.

The only ambiguity you need to consider in calculating your significant figures is whether the '5' is an exact number or not. Exact numbers come about either by definition (1 hour = 3600 seconds) or by counting (5 measurements were made) and have an infinite number of significant figures. Hence, they do not affect the accuracy of your measurements. Exact numbers are usually integers (speed of light and number of centimetres in an inch are a couple of exceptions). My guess is that your question is simply a case of going through significant figures calculations, using the information provided. To avoid ambiguity in the case of a single significant figure measurement, one could consider writing 5 as 5E0, or 5x100.

If you know 5 to be an exact number (ie the question gives that information), then you would still complete the calculation using the multiplication rule, but assume infinite number of significant figures for 5. In this case you would have 4 significant figures.


What causes this magnetic moment?
  So does the dipole cause the proton to spin which generates the magnetic moment and if so what causes the tiny dipoles along axis? It seem to me there a rotation being caused by a rotation.

Spin is a property predicted when relativity theory is applied to electromagnetic theory, leading to the Dirac equation. It is an intrinsic property of subatomic particles, like charge. It explains magnetic phenomena such as NMR. It is a quantum property of NMR active nuclei (when measured it takes on discrete values). For spin-${\frac{1}{2}}$  nuclei, it is related by a multiplicative constant to the magnetic dipole moment of the nucleus. Thanks to spin, nuclei have an intrinsic and fixed magnetic moment in an external magnetic field. One component of this moment adopts a discrete orientation with respect to the applied field. Depending on the orientation, two states of different energy, usually called $\alpha$ and $\beta$, are possible for a spin-${\frac{1}{2}}$ nucleus.
We can define the resonance frequency in the presence of the external field using the equation you provide. For a spin-${\frac{1}{2}}$ nucleus it is proportional to the difference in energy between the $\alpha$ and $\beta$ states. It is the frequency at which the nuclear moment nutates about the applied field.

  Does the varying magnetic field vary the energy levels until resonance occurs?

Yes. In the CW experiment, you measure the electrical signal induced by the sample in a pick-up coil while simultaneously irradiating the sample with a second orthogonal RF coil, and sweeping an applied magnetic field. The sweep alters the resonance frequency of the spins. When this frequency equals the irradiation frequency spins are excited and transition between the two states - this is resonance, the points at which the oscillating red curve (the field sweep) crosses the origin. The spins now oscillate together coherently back and forth between $\alpha$ and $\beta$ states, yielding a fluctuating magnetic field which induces an oscillating current in the pickup coil. As the field sweeps past the resonance point, the spins begin to lose energy - to relax. Relaxation causes the signal to eventually decay to the baseline.

  Are the individual proton within the sample at different depths become out of phase with each other and the varying magnetic field is decreasing over time?

The NMR signal is an average, reflecting the behavior of a large ensemble of spins, acting coherently (nearly identically) during an experiment. The total signal is a sum over all excited spins in the sample. The applied field is as uniform across the sample as possible, so that all spins behave similarly. But in a way the spins act independently. Some oscillate faster or slower than the mean. The differences add up over time, leading to dephasing of the signal.

There is usually no spatial information in such an experiment unless special tricks are used.

  and why dose each resonance peak have a negative voltage associated with it? How dose this negative voltage tell me about what is occurring with in the sample?

This question is not clear. The spins exist in states whose coherent oscillations induce a sinusoidal magnetic field in the RF pickup coil. The electric signal induced in the pickup coil is an AC voltage, alternating between negative and positive values until it decays completely.


What is the rate-determining step in free radical halogenation?

It does seem odd as a blanket statement. However, the bond dissociation energies (BDEs) for Cl2 and Br2 are quite low, 243 and 193 kJ/mol, respectively, while BDEs for the C-H bond in alkanes are much higher (eg. 435 kJ/mol for methane). So, it's not necessarily the case that events involving only bond breaking will be the rate-determining step, particularly since, as Zhe points out, the propagation steps are bimolecular.

Bromine radicals are significantly less reactive than chlorine radicals, leading to interesting selectivity differences between chlorination and bromination reactions - https://youtu.be/c5enwHzAwmA

Edit: BDEs are taken from Klein, "Organic Chemistry", 3rd ed., Chapter 10.


This follows from the first law of thermodynamics; i.e., $$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V,$$ which identifies $U$ as a function of $S$ and $V$. If in addition we have other forms of work, such as chemical work, then we add a chemical work term $\mu\mathrm{d}N$ to the first law, and introduce an additional dependence of $U$ on $N$.

In addition, we can also replace a given independent variable by its conjugate variable, since, by thermodynamic stability, the independent variable is a monotonic function of its conjugate and vice versa. That is, the following sets of independent variables can all be used to characterize $U$: $$(S, V, N), (T, V, N), (S, P, N), (S, V, \mu), (T, P, N), (T, V, \mu), (S, P, N).$$ Note in particular that $(T, P, \mu)$ cannot be used, because these three variables, all being intensive, are not independent and are related by the Gibbs-Duhem equation.


Difference between torsion, out of plane, coplanar and perpendicular bends?

A molecule with N atoms has 3N degrees of freedom

3 translational modes

if it is linear, it will have 2 rotational modes; if non-linear 3

These leaves 3N-6 (3N-5 if linear) vibrational modes.  Vibrational modes are further subdivided into stretching, bending or torsional motions.

Stretching motions can be symmetric or asymmetric
Bending motions are often referred to as scissoring, twisting, wagging, and rocking
Torsional motions refer to rotations about bonds

This diagram illustrates these various stretching and bending vibrations

Depending on the geometric relationship between the transition moment for a specific vibrational mode and the symmetry axis for the mode, vibrational modes may be further classified as parallel or perpendicular.

Here is a link to a 31 page article (slide show) that has a lot of background and pictures on this subject. Section 4.11 (page 19) might be particularly interesting to you.

One final example that you might be interested in, formaldehyde has 4 atoms and is non-linear so it will have 6 vibrational modes.  These are all pictured on page 2 in this link.


If nothing sticks to Teflon, how does Teflon stick to frying pans?

It has to be so common a question that the answer is actually given in various places on Dupont's own website (Dupont are the makers of Teflon):

  “If nothing sticks to Teflon®, then how does Teflon® stick to a pan?"
  Nonstick coatings are applied in layers, just like paint. The first layer is the primer—and it's the special chemistry in the primer that makes it adhere to the metal surface of a pan.

And from this other webpage of theirs:

The primer (or primers, if you include the “mid coat” in the picture above) adheres to the roughened surface, often obtained by sandblasting, very strongly: it's chemisorption, and the primer chemical nature is chosen as to obtain strong bonding to both the metal surface. Then, the PTFE chain extremities create bonds with the primer. And thus, it stays put.


There are two factors at work here. One is that going down within a group of the periodic table, say from C to Si to Ge, gives you more electropositive (metallic) elements, so they behave more like metal hydrides and are easier to hydrolyze. A second factor is that with increasing atomic radius as you go down to heavier elements, the overlap between hydrogen's valence orbital(s) and those of the element will be poorer, and the resulting bonds weaker.


Difference between basicity and reducing character

Basicity is how easily can the compound give up hydrogen atoms.

Isn't this the definition for acidity instead? (assuming your "hydrogen atoms" = proton) Basicity instead refers to how easily a compound can accept protons on its lone pair, not give them up.

$\ce{BiH3}$ is a weak base, because the lone pair on bismuth is dispersed over a large surface area, as compared to that on the nitrogen in ammonia. (details here) Being diffused, the lone pair is unable to attack a proton, and hence, is a weak base.

$\ce{BiH3}$ is a better reducing agent, because on getting oxidised (losing a hydrogen), the negative charge formed on the bismuth atom is dispersed over a much larger surface area. Hence, the negative charge is stabilized much more on the bismuth atom than it would be stabilized on the nitrogen atom in ammonia.

Notice that the dispersion of the lone pair makes it more diffused, hence, less basic. While dispersion of the excess negative charge stabilizes the specifies formed after loss of a hydrogen, hence, a better reducing agent.


The difference is that phosphonic acid ($\ce{HPO(OH)2}$) is an existing, shelf-stable compound while phosphorous acid ($\ce{P(OH)3}$) is extremely short-lived if at all and immediately tautomerises to phosphonic acid.

All the -ous acids are theoretically produced in the same manner: take the -ic acid and remove one unprotonated oxygen, leaving an additional electron pair with the central atom which reduces its oxidation state by two. So while sulphuric acid is $\ce{SO2(OH)2}$, sulphurous acid is $\ce{SO(OH)2}$.

Likewise, you can always theoretically generate the -onic acids by tautomerisation from the ‘parent’ -ous acid; one proton that previously was attached to an oxygen bonds to the central atom‘s newly-formed (when coming from the -ic acid) lone pair. Thus, sulphonic acid is $\ce{HSO2(OH)}$.

In case of sulphur, while both acids have been observed and crystal structures of both have been determined to the best of my knowledge,[citation needed] the two can still tautomerise. In the case of phosphorus, only the phosphonic acid is shelf-stable and all isolated salts have been determined to be phosphonate salts.[citation needed]

In both cases, however, organic esters have been generated in both ways. Thus, you can have e.g. triethyl phosphite $\ce{P(OEt)3}$ or diethyl ethylphosphonate $\ce{EtPO(OEt)2}$.


Li2O + 2KCl =2 LiCl +K2O: Is it a feasible Reaction?

Although every reaction is reversible in nature, let's try to understand where the equilibrium might be biased, i.e. whether the reactant side favoured or the product side.

A rule of thumb while predicting feasibility is to remember, a reaction where you can get a weak acid from a strong acid is feasible.

$$\ce{Strong Acid -> Weak acid}$$

In your case we're dealing with $\ce{KCl}$ and $\ce{LiCl}$. The acidic or basic nature or salts is determined by the extent of hydrolysis of its ions. Ions from a weaker species undergo hydrolysis while ions from a stronger species don't. Another way to look at this is to compare the acidic and basic strengths of the acids and bases which formed them in the first place.
$$\ce{LiOH + HCl -> LiCl + H2O}$$
$$\ce{KOH + HCl -> KCl + H2O}$$
Notice that $\ce{HCl}$ is common to both of these species, so we have to just compare the basic strengths of $\ce{LiOH}$ and $\ce{KOH}$. Down the group (in alkali metal hydroxides) basic strength increases. So $\ce{KOH}$ is a better base. And thus $\ce{KCl}$ is a weaker acid. Since we're making a stronger acid from a weaker acid, the reaction can be termed as unfeasible.


Calculation of rupture force in a covalent bond

A force is the derivative of a potential energy:
$$\vec{F} = -\vec{\nabla} U$$
If you choose to apply a continuous force to stretch a bond until rupture, the force will need to exceed the restorative force (as described say by Hooke's law for small displacements from equilibrium) due to mutual attraction of the atoms. The maximum value of that restorative force is the maximum value of the gradient of the potential, therefore:
$$\vec{F}_{\text{rupture, continuous}} = (\vec{\nabla} U )_{\text{max}}$$
Note a quadratic potential (such as leading to Hooke's law) does not predict rupture. You would require a more accurate description of the internuclear potential to compute the rupture force.


Rodrigo de Azevedo's answer can be stated with words by saying that the rate of the reaction is the rate of disappearance of $1$ mole $\ce{O2},$ and also the rate of disappearance of $2$ moles $\ce{NO}$. That is why the coefficient $2$ appears in the law written for $\ce{NO}:$
$$\frac{d}{dt}[\ce{NO}] = - 2k[\ce{NO}]^2[\ce{O2}]$$


Boiling point can be decrased by lowering pressure over liquide. If you insist on the azeotrop preparation, find azeotrop diagram such as:

or data table such as: wiki and determine the composition that must be achieved to obtain the azeotropic mixture. Then mix it accordingly.


Working out pH of solution when excess NaOH is added

The excess NaOH is, indeed 0.005 moles. The total final volume is 30 ml + 20 ml = 50 ml = 0.05 L. Thus, the final concentration is 0.005 moles/0.05 L = 0.1 M


Can liquids be made a lot purer than by fractional distilation?

Since ethanol and the water-ethanol azetrope boil at a lower temperature than water, if you start with a lot of water with a smaller amount of ethanol  then you can "distill" (however you do it...) all the ethanol out and be left with "pure" water.

But water and ethanol form an azeotrope which is 95.5% ethanol by weight and boils at 78.1 °C. For the water-ethanol azeotrope the wt. % ethanol in the liquid is the same as the wt % of ethanol in the vapor. Thus there is no way to exploit liquid to gas phase transition to enrich the ethanol more than that.


Comparing bond angles in carbonyl dichloride and carbonyl dibromide

The reasoning you utilized for comparing the bond angles of $\ce{NH3}$, $\ce{NF3}$, and $\ce{NCl3}$, while sensible, is simplistic and works best only for a few cases where weighing the factors of steric repulsion against bond pair-bond pair repulsions is feasible. It seems that you have encountered this very problem in the case of comparing the bond angles of $\ce{COCl2}$ and $\ce{COBr2}$.
To circumvent these issues, a better method(less prone to failures) would be using Bent's rule. Now, there has been a lot of discussion on this website over what this rule actually is (see here) and why it is physically useful (see this). The Wikipedia article on the same has a wonderful intuitive discussion under the "Justification" tab for why this rule exists. However, I will not be delving into the origins of this rule, and just state outright that the basic premise of this rule is that not all hybrid orbitals are equivalent under a given hybridization scheme. The percentage s-character and p-character (that is, the amount of s-orbital or p-orbital present in a given hybrid orbital as a percentage of the whole) varies as per the electronegativities of the substituents.
A statement of this rule was provided by Henry A. Bent as follows:

Atomic s character concentrates in orbitals directed toward electropositive substituents.

The idea is that the amount of p character will increase (alternatively, the amount of s character will decrease) in those bonds which have more electronegative substituents (that is, the atom other than carbon in the bond), while simultaneously the amount of p character will decrease(alternatively, the amount of s character will increase) in bonds towards less electronegative substituents in such a manner that the net sum of percentage s and p characters over all the bonds in the molecule comes out to be $100$%. This is essentially a redistribution of the orbital energies across bonds to minimize the energy of the system. Further details on this rule can be found in the links mentioned above.
Once you are aware of this rule, applying it to your case becomes a fairly simple task. I will try to compare the $\ce{Cl-C-Cl}$ and $\ce{Br-C-Br}$ angles here, the orders for the $\ce{O-C-Cl}$ and $\ce{O-C-Br}$ angles can be easily inferred from there. From a classical hybridization perspective, the carbon in both these molecules uses $\ce{sp^2}$ hybrid orbitals for bonding, and all the three hybrid orbitals thus generated should possess equivalent atomic orbital's character ($33.33$% s character and $66.66$% p character approximately). But as per Bent's rule, since $\ce{Cl}$ and $\ce{Br}$ are both more electronegative than carbon, they will increase p character towards their direction (that is, in the $\ce{C-Cl}$ and $\ce{C-Br}$ bonds respectively). Since $\ce{Cl}$ is even more electronegative than $\ce{Br}$, the p character increase in the $\ce{C-Cl}$ bonds will be more than that what will happen in the $\ce{C-Br}$ bonds.
Hence, we can say that more p character will be directed in the two $\ce{C-Cl}$ bonds as compared to the two $\ce{C-Br}$ bonds. This directly bears a relation to the bond angle: more is the p character in the constituent bonds, lesser will be the bond angle(you can see this from a fairly easy observation: $\mathrm{sp}$ hybrid orbitals are generally at an angle of $180°$ with $50$% p character, the $\mathrm{sp^2}$ hybrid orbitals are generally at an angle of $120°$ with $66.66$% p character and the $\mathrm{sp^3}$ hybrid orbitals are generally at an angle of $109.5°$ with $75$% p character).
Hence, Bent's rule predicts that the $\ce{Cl-C-Cl}$ bond angle of $\ce{COCl2}$ should be lesser than the $\ce{Br-C-Br}$ of $\ce{COBr2}$. Experimental data says that the $\ce{Cl-C-Cl}$ bond angle of $\ce{COCl2}$ is $111.83 ± 0.11°$, while the $\ce{Br-C-Br}$ of $\ce{COBr2}$ is $112.3 ± 0.4°$. Hence, the prediction made by Bent's rule is confirmed.
Notice that the difference between the given bond angles is quite small (around $1°$). This is also somewhat expected, as the electronegativity difference between chlorine and bromine is also quite small (approximately $0.2$).
The advantage of employing this method is that you don't have to consider conflicting factors like those you stated in your approach. The rule explains these angle variations based on Coulson's theorem, from which the following relation pops out:

$$\cos \theta = \frac{S}{S-1}=\frac{P-1}{P}$$
for $\theta \in (90^\circ,180^\circ)$

($\theta$ stands for the bond angle while ${P}$ and ${S}$ stand for p character and s character respectively)
As you can clearly see, when the p character goes up, the value of $\cos \theta$ also goes up if you closely observe the simplified form $\cos \theta = 1 - \frac{1}{P}$. This means that the value of $\theta$ will decrease for the given range of $\theta$ (as $\cos \theta$ is a decreasing function for $\theta \in (90^\circ,180^\circ)$)
Hence, you can generally predict things quite accurately with Bent's rule without getting confused with contrasting factors, as they are usually taken care of while determining the atomic orbital characters.


Proof for second-order perturbation term of energy

As the unperturbed Hamiltonian is Hermitian it follows that
$$\int{ψ^{(0)*}_0}\hat {H}^{(0)}{ψ^{(2)}_0} dτ  = \left[\int{ψ^{(2)*}_0}\hat {H}^{(0)}{ψ^{(0)}_0} dτ\right]^*=\left[\int{ψ^{(2)*}_0}E^{(0)}_0{ψ^{(0)}_0} dτ\right]^*=E^{(0)}_0\int{ψ^{(0)*}_0}{ψ^{(2)}_0} dτ$$
by the definition of Hermiticity and the fact that all eigenvalues of a Hermitian operator are real.


What kind of chemical used in a chromatography run could be denoted with the letters D.H.?

Recently I had to read up on HPLC-MS technique and I encountered D.H./DH several times as an acronym for dehydrogenase, also, like in the quoted section, in the context of parabens analysis with HPLC.

For example, DH is explicitly defined as an acronym for dehydrogenase in Kastner's Protein liquid chromatography [1, p. 5] or in the paper by Zimmerling et al. [2].

Other examples for the exact dehydrogenase composition include acronyms such as ADH (alcohol dehydrogenase), FDH (formate dehydrogenase), $\small\text{D}$-LDH ($\small\text{D}$-lactate dehydrogenase), 11βDH (11β-hydroxysteroid dehydrogenase) and many more.

References

Protein Liquid Chromatography; Kastner, M., Ed.; Journal of chromatography library; Elsevier: Amsterdam; New York, 2000; Vol. 61. ISBN 978-0-444-50210-0.
Zimmerling, J.; Tischler, D.; Großmann, C.; Schlömann, M.; Oelschlägel, M. Characterization of Aldehyde Dehydrogenases Applying an Enzyme Assay with In Situ Formation of Phenylacetaldehydes. Appl Biochem Biotechnol 2017, 182 (3), 1095–1107. DOI: 10.1007/s12010-016-2384-1.


How to derive Pauli Exclusion Principle without assuming anti-symmetry?

The Pauli Exclusion principle is determined from experimental facts alone.
First, its consequences were observed and then it was incorporated in the theory. See Kaplan´s article for a more erudite answer. Also, see Towler´s exposition. Although, he dwells in foundational issues.

Important: they are not equivalent unless you consider only totally symmtric/antisymmetric functions! The antisymmetry is assumed, it cannot be derived rigorously from particle independence nor tested with experiments.

As for getting to the conclusion that the wavefunction of fermions has to be antisymmetric here is a weak argument without requiring the Pauli Exclusion principle.
You could also say that since you cannot follow the trajectory of particles therefore they are indistinguishable -- at least, if they are of the same kind. That is, the permutation of any two of them
$$ \Psi(\pmb x_1, \pmb x_2, ...) \to \Psi(\pmb x_2, \pmb x_1, ...) $$
maintains the same physical description
$$ |\Psi(\pmb x_1, \pmb x_2, ...)|^2 = |\Psi(\pmb x_2, \pmb x_1, ...)|^2.$$
The permutation can be described as the action of an operator that satisfies $P^2 = 1$.
There are only two ways to acomplish this. Either the wavefunction is symmetric (the eigenvalue of $P$ is $-1$) under the exchange or antisymmetric (eigenvalue $-1$).
$$ \Psi(\pmb x_1, \pmb x_2, ...) =  \Psi(\pmb x_2, \pmb x_1, ...) \quad \text{  Symmetric: bosons}$$
$$  \Psi(\pmb x_1, \pmb x_2, ...) = -\Psi(\pmb x_2, \pmb x_1, ...)\quad \text{Antisymmetric: fermions}$$

Therefore, if $\pmb x_1 = \pmb x_2$ then $\Psi(\pmb x_1, \pmb x_1, ...) = -\Psi(\pmb x_1, \pmb x_1, ...) = 0$. That is, Pauli Exclusion principle.

The reason as to why this argument, that is the one that was elaborated to justify Pauli Exclusion principle and is seen nowadays in most textbooks, is weak is neatly explained in the reference mentioned above.


Why aren't the R-side-chains of acidic and basic amino acids the site of polypeptide synthesis?

Chemically

If you were making a small polypeptide you'd have a protecting group on the side-chain of glutamic acid, and preferably also an activating group on the carboxyl where you want to attach it. If you're just mixing glutamate with another amino acid you'll get a mixture as you expect (including all kinds of possible oligomers).

Biologically

Protein synthesis in the ribosome actually works the other way around (the amine-end is in the ribosome, tRNAs are attached to residues at the carboxyl side). So your expected error would be the continuation of protein synthesis at another alpha amine: the lysine sidechain. Apparently the ribosome is quite good at distinguishing this 4-carbon difference.

My 2 minute hypothesis is that this is one of the reasons why there are residues with shorter acid sidechains (glutamate and aspartate) while a shorter version of lysine is not used in biological protein synthesis.

How the tRNA synthases are so specific that they don't load the amino acid with the wrong carboxyl group is a different question, but tRNAs are extremely specific enzymes in many ways (they have no problem recognizing the difference between a methyl and a hydroxyl group).


Why when I add jam to my tea it stops producing thin "membrane" on top?

The parallel process to deposition of limescale ( that is inhibited by acidic jam) is deposition of products of oxidation of epigallocatechine gallate(EGCG) and similar compounds, that are getting adsorbed on limescale particles and cup walls.

Epigallocatechin gallate (EGCG), also known as epigallocatechin-3-gallate, is the ester of epigallocatechin and gallic acid, and is a type of catechin.
It is found in high content in the dried leaves of green tea (7380 mg per 100 g), white tea(4245 mg per 100 g), and in smaller quantities, black tea (936 mg per 100 g).During black tea production, the catechins are mostly converted to theaflavins and thearubigins via polyphenol oxidases.

EGCG is a polyphenol, vulnerable to oxidation on neutral/alkalic environment, forming insoluble product of intense brown color.
Acidic environment, caused by addition of jam or lemon juice, stabilizes the compound.
Minor natural content of ascorbic acid in fruit juice is protected by abundance of organic acids by similar way.


No. ppmw means parts per million by weight (mass). This is not the same as mg/m3, which is milligrams (mass) per cubic meter (volume).

For example, in a gas of very low density (e.g. $\ce{CO2}$ at 10-6 Pa), 1 mg/m3 could exceed the weight of $\ce{CO2}$, i.e. would be greater than 900,000 ppmw.

One measure has to do with mass in a given volume, the other measure is mass in a given mass of another substance.


Strictly speaking, the answer depends on conditions and, as a consequence, states of aggregation.

Reaction (1) can be reversible if iron(II) sulfide is subsequently heated above $\pu{700 °C}$ in vacuum to prevent oxidation of the elements [1, p. 422]:

$$\ce{FeS(s) ->[\pu{700 °C}][vac] Fe(s) + S(g)}\tag{1a}$$

Reactions \eqref{rxn:2a} and \eqref{rxn:3a} are indeed irreversible if carried in aqueous solution — as you already mentioned, a poorly soluble salt is formed in both cases:

$$
\begin{align}
\ce{NaCl(aq) + AgNO3(aq) &-> AgCl(s) + NaNO3(aq)}\label{rxn:2a}\tag{2a}\\
\ce{H2SO4(aq) + Ba(OH)2(aq) &-> BaSO4(s) + 2H2O(l)}\label{rxn:3a}\tag{3a}
\end{align}
$$

Reaction \eqref{rxn:4a} is a canonical example of a reversible reaction used to study equilibrium in a gas phase:

$$\ce{H2(g) + I2(g) <=> 2 HI(g)}\label{rxn:4a}\tag{4a}$$

I'd say it's impossible to tell in advance what reaction is reversible or not without conducting an actual experiment or at least estimating free Gibbs energy.

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


Why is the following reaction not considered a redox reaction?

You are right. The question is definitely misleading. If it would ask for the mechanism, (radical) substitution would be the correct term, as a reduction does not say anything about the mechanism. But "type of reaction" can basically mean anything, and in this case both answers are correct.


Does acetone eat through Thermoplastic polyurethane and Polycarbonate

If there is no visual damage, e.g. cloudiness or crazing, the case should be OK. If there's some surface cloudiness, it might be possible to buff it out with mild abrasive, such as toothpaste or milk of magnesia.

In the future, try oil (olive, safflower, mineral etc., but not motor oil) to remove stickers. A bit slower, requiring subsequent cleanup, but safer.


Do MO's span a bond or the whole molecule?

The unfortunate answer is: both, and it is rather complicated. It can be noted at this point, however, that between two atoms connected by covalent (or even ionic bond) there are MOs.

Molecular Orbitals, the Wave Function, and the (Electron) Density

In quantum chemistry, we can define:

(spin) MO: a one-electron wave-function. Contains one electron, can be combined to with another spin MO to form what we usually think of as an MO, that is, doubly occupied.
Wave Function (WF): Typically a product of MOs, at least in the simpler approaches (Hartree-Fock [HF], common Density Functional Theory). The product is anti-symmetrized such that it obeys the Pauli principle. These products of MOs can be linearly combined to yield more accurate, complicated and expensive approaches.
Density: The expectation value of a WF when applying the density operator. Out of the three, this is the only true observable, for instance by X-ray diffraction. If you have a open shell molecule, you may be able to observe a spin-density, for instance by Electron Spin Resonance spectroscopy.

What's the point of these definitions?

They illustrate that a MO and the observable density have a few steps in between. Let's assume that someone gives us a set of MOs which yield the correct density of a molecule (as well as other correct observables). This is already a tall order, by the way. It is a mathematical fact that one can "rotate" the MOs, that is, create new MOs from linear combinations of old ones and still yield the correct density. So the set of "correct" MOs is not unique by the reasonable measure that it should give us the correct density.

OK, so there is more than one set of "correct" MOs. So what?

For one molecule/method/basis set combination, many ways of "choosing" your MOs are possible and lead to very different pictures, even when choosing reasonable contour values (say, 85%) for the visualization. An incomplete list is:

canonical MOs: these are eigenfunctions of the Fock operator, which is central to HF, but also of relevance for post-HF methods. They also preserve meaningful orbital energies (which relate to ionization potentials), but they are usually completely delocalized over the whole molecule.
localized MOs: "rotations" of the MOs have been performed to yield orbitals of (hopefully) minimal spatial extent. They are more easily interpreted, but the mathematics involved is iterative, meaning that minor changes in molecular geometry could lead to rather different local MOs. There is also more than one scheme to do this, and the results can be fairly different.
natural MOs: suppose you have performed the very accurate/very expensive calculation of full Configuration Interaction (full CI) for your molecule. The resulting orbitals are then eigenfunctions of the true Hamiltonian (in the given basis set). You will probably never encounter these due to the sheer cost of the calculation - and they may not look all that different from canonical orbitals.

And then there is the small issue of virtual/unoccupied orbitals

They do not enter the density (duh), but are often relevant to discussions of mechanisms, such as initial point of attack or stabilization. In my opinion, there is basically no way to pick a good scheme for them - canonical orbitals sort-of yield electron affinities.

Then all hope is lost. There's no way to choose.

Well, for occupied orbitals, there kinda is. One should compare orbitals (of one computational level) with each other across molecules to get to useful results. The textbooks usually do not give you this opportunity because you have no idea where those orbitals come from.

Back to OP's question: yes, canonical orbitals would extend over basically the whole molecule. You are being shown, in all likelyhood, orbitals that have been localized somehow. They are used as illustrations by the authors and that's usually not a bad idea. Take 'em with a grain of salt.


Why doesn't anyone manufacture 4.5 V cells?

Perhaps you consider the volume allotted to the batteries as equivalent to capacity. A better measure would be ampere-hours at 4.5v, or watt-hours. And weight might be a factor to consider.

An electrochemical cell with a high voltage (4.5v) that requires complicated or dilute ingredients could better fill the volume allotted, but if it runs out of power in a shorter time, it has less capacity. You could consider 6 cells in a 9v battery to be a complicated device, but it seems so simple. Three 1.5v cells in a package would give you a simple overall structure, and could give you high capacity due to the simple internal composition of each cell.

One reason that lithium ion batteries are so highly regarded is their high electrochemical capacity.
"The more energy a [rechargeable] battery can hold when charged, the less often you need to charge it. No one wants to be replacing the batteries in their wireless keyboard every week or to have their kid’s toy conk out after just 15 minutes. Whereas a common disposable AA battery’s capacity is anywhere from 800 mAh for budget brands up to 2,000 mAh for premium versions, most rechargeable batteries start at around 1,800 mAh and go up to nearly 3,000 mAh. Most trustworthy companies right now make AA batteries around 2,000 mAh. We don’t consider batteries with stated capacities lower than 1,700 mAh, which is too underpowered, or higher than 2,500 mAh, which makes us suspicious that a company is overpromising." https://thewirecutter.com/reviews/best-rechargeable-batteries/


Why doesn't separating the layers change the equilibrium point?

Draining the aqueous phase cannot change anything. If the phases are in equillibrium before (both internally and with each other), they stay that way if they are separated.
Now if you add water to the aqueous phase, the reaction in the unpolar phase (which is saturated with water, as the phases are in equillibrium) is in the first moment not affected.
However the acid, alcohol and ester have different solubilities in the aqueous phase. That means the concentrations in the unpolar phase will change a bit, and so does the reaction equillibrium.
But: This is all irrelevant, because the reaction happens under reflux. While you are standing there with the cooled mixture in the separation funnel, the reaction rate is much lower, you will never notice a change.


Examples of molecules with D4 and C9 point groups

For nearly any point group which is actually feasible (including $D_4$), look up the Symmetry gallery. Your all-cis tetra-substituted cyclobutane would be $C_{4v}$, BTW.

As for $C_{9h}$... well, I don't quite believe this is possible, but probably we may come up with something that would look almost realistic.

A substituted nanotube fragment will do, I think.

Surely it is hypothetical, so what? Cyclo-$\ce{C18}$ is hypothetical as well, and in my opinion, has lower chances to be stable. Also, its symmetry is kinda dubious.