The spontaneous emission of light by a substance not resulting from heat is luminescence (a type of cold-body radiation).  $\ce{^1}$

A few types of luminescence are:

Bioluminescence: Made by living creatures such as fireflies, glow-worms, and many marine creatures.
Chemoluminescence: made by a chemical reaction. Glow sticks work this way.
Electroluminescence: made by passing electricity through something like in a light emitting diode (LED).
Photoluminescence: made by shining light at "luminous" (phosphorescent) paints.
Röntgenoluminescence: made by shining X-rays at things. (The curious name comes from Wilhelm Röntgen (1845–1923), the discoverer of X-rays.)
Sonoluminescence: made by passing energetic sound waves through liquids.
Thermoluminescence: made when photons are emitted from hot materials.
Riboluminescence: made by rubbing, scratching, or physically deforming crystals.
Phosphorescence: photoluminescence as a result of triplet-singlet electronic relaxation (typical lifetime: milliseconds to hours)

A chemical absorbs and then re-emits a photon of light, where the energy of the photons matches the available energy states and allowed transitions of the chemical takes place. In phosphorescence, the electron that absorbs the photon undergoes intersystem crossing $\ce{^2}$  into a triplet state.

The excited in triplet state returns to the lower energy singlet state. Such transitions, are kinetically unfavored and progress at a slow rate. Most phosphorescent compounds are fast emitters with triplet lifetimes in milliseconds. If triplet lifetimes are minutes or even hours, very slow degradation occurs.$\ce{^3}$ $\ce{^4}$ In phosphorescent materials, there's a delay between them absorbing energy and giving out light.These substances are so-called "glow-in-the-dark" materials.

To make glow-in-the-dark toys manufacturers mix zinc sulfide or strontium aluminate with plastic giving in glow-in-the-dark toys.$\ce{^5}$ $\ce{^6}$

Fluorescence

Fluorescent materials produce light instantly, when the excited atoms return to normal, in as little as a hundred thousandth of a second.$\ce{^4}$ $\ce{^7}$

Chemoluminescence

In the case of chemiluminescence, two chemicals interact which giving energy as a byproduct in the form of light( heatless glow).

When glow sticks are bent, an inner glass tube is broken releasing hydrogen peroxide. Hydrogen peroxide reacts with diphenyl oxalate giving 1,2dioxetanedione, which is unstable. It decomposes to carbon dioxide releasing energy. The energy is absorbed by electrons in dye molecule, that falls down to ground state releasing energy in the form of light.$\ce{^8}$ $\ce{^9}$

Chemiluminescence can happen naturally as well. In such instances — like fireflies — it is bioluminescence.

Some creatures, such as fireflies and jellyfish, contain chemicals within them that cause them to glow. Some of these creatures glow for protection, camouflage or to attract mates.

Ex 1.Bobtail squid $\ce{^12}$ $\ce{^13}$

Ex.2 Mycena lampadis ( mushrooms )$\ce{^14}$

The principal chemical reaction in bioluminescence involves some light-emitting molecule and an enzyme, generally called the luciferin and the luciferase.$\ce{^10}$ $\ce{^11}$ $\ce{^15}$

Electrochemiluminescence $\ce{^16}$is a special form of chemiluminescence in which the light-emitting chemiluminescent reaction is preceded by an electrochemical reaction.$\ce{^17}$

References

1 https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Map%3A_Principles_of_Instrumental_Analysis_(Skoog_et_al.)/15%3A_Molecular_Luminescence_Spectrometry

2 https://www.shsu.edu/chm_tgc/chemilumdir/JABLONSKI.html

3 https://en.wikipedia.org/wiki/Intersystem_crossing

4 https://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/lumin1.htm

5 http://www.phys.ubbcluj.ro/~dana.maniu/OS/BS_7.pdf

6 https://science.howstuffworks.com/innovation/science-questions/question388.htm

6 https://www.wonderopolis.org/wonder/how-do-things-glow-in-the-dark

[7] https://en.wikipedia.org/wiki/Fluorescence#Lifetime

[8] https://www.compoundchem.com/2014/10/14/glowsticks/

[9] https://en.wikipedia.org/wiki/Glow_stick#Fluorophores_used

[10] https://en.wikipedia.org/wiki/Bioluminescence#Chemical_mechanism

[11] https://www.nationalgeographic.org/encyclopedia/bioluminescence/

[12] https://www.nationalgeographic.org/encyclopedia/bioluminescence/

[13] https://en.wikipedia.org/wiki/File:Sepiola_atlantica.jpg

[14]https://www.timeforkids.com/g34/bioluminescence-2/

[15] http://photobiology.info/Branchini2.html

[16] https://www.sciencedirect.com/topics/chemistry/electrochemiluminescence

[17] https://www.sciencedirect.com/science/article/pii/S0008622317304943


What chemical will dissolve the plastic coating on circuit boards

This seems to be a popular question judging from some Google queries.

Printed circuit boards are made of a variety of materials, and from what I understand, the specifics may vary somewhat based on the specific needs of the product and the manufacturer. For example, in some cases, I know the PCBs are coated with special anti-static coatings to prevent delicate RAM and CPU chips from being damaged (e.g., this article found via Google among others).

So I doubt there's a definite solution - more likely you'll need to try several different things.

That said, there's a recent article that looks very interesting about separating parts by dissolving and swelling the plastic resins with DMSO: "Dissolution of Brominated Epoxy Resins by Dimethyl Sulfoxide To Separate Waste Printed Circuit Boards" Environ. Sci. Technol., 2013, 47 (6), pp 2654–2660.

Certainly DMSO is considered a near-universal solvent and will swell many polymers. So I'd probably try that first.


You're replicating the natural weathering process of granite.
Granite is a rock comprising the minerals feldspar, quartz, and mica.
There is a continuous solid-solution ternary diagram here that illustrates the various feldspars as a function of calcium, potassium, and sodium content: the end members are K-spar (potassium variety), albite (sodium variety), and anorthite (calcium variety).
The weathering process is complex and involves so-called incongruent dissolution, where one component in the solid state reacts to form one or more nother solid state components. This is why you see granitic feldspar weathering faster than the quartz or mica components out in nature.
You are reacting various minerals (the exact composition of your feldspars and micas would be determined by XRD or a similar method) with oxygen, carbon dioxide, and water from the atmosphere as well as any waters of hydration that may be in your micas, which are sheet silicates from a general class of silicates known as phyllosilicates.
Back to your query about crumbling: you're dehydrating the micas (muscovite on the light-colored end, biotite on the dark-colored end) if they are hydrated at all, but mostly you're making various clays from the feldspars via the incongruent dissolution pathway. Those new minerals you're making are the crumbly ones, mostly clays and maybe some brittle mica scraps. But it's mostly clay.
Generally speaking, carbon dioxide and water form an aqueous solution of carbonic acid species which reacts in the presence of oxygen and a feldspar to form a clay and aqueous calcium and carbonate ions. There are many, many complex reactions.
Here is an excellent write-up about anorthite hydrolysis, specifically regarding the region around Egersund, Norway. In it, the author does some modeling using PHREEQC and gives specific examples of some of the weathering reactions. Here is one of the reactions that takes a mixed albanitic-anorthitic feldspar to kaolinite:
$$\ce{2NaCaAl3Si5O16 +
      8CO2 +
      9H2O =
      2Ca^2+ +
      2Na^+ +
      6HCO3- +
      4SiO2 +
      3~Al2Si2O5(OH)4 +
      2CO2}$$
High temperature speeds things up, and also will dictate what products you get. Same for water and carbon dioxide, of course.
In summary: you are making mostly crumbly clay from granite by speeding-up the chemical weathering process of (mostly) the feldspar component(s). The particular composition of the clay varies with the entire system composition.
This is a very complex sub-field of study in aqueous geochemistry. An excellent reference (which I used in my grad school days and beyond) is Stumm and Morgan.


Why is kerosene preferred over other non-polar solvents to store sodium in?

Kerosene is a fraction from the distillation of petroleum, the one with boiling point 180° to 230°, containing hydrocarbons from $\ce{C_11}$ to $\ce{C_12}$. In reaction to what has been said in the commentaries (below the question), I note that mineral oil is a cloudy expression whereas kerosene seems to be the precise name of the more or less broader base material of this cloud.

The best means to store alkali metals, which are all strong reductants, is an alkane because they are inert (as indicated by the name “paraffin”) and cannot be reduced.$^1$ The reason why one does not take the lower boiling fraction (gasoline 40° to 180°, $\ce{C_6}$ to $\ce{C_10}$), cyclohexane (b.p. 81°) or hexane (b.p. 69°) is obvious: it would be too evaporable, which is dangerous since its evaporation when someone forgets to close the bottle sufficiently, would lead to spontaneous ignition of the alkali metal (especially in the case of potassium) by contact with moist air. (Additionally, cyclohexane would anyway be unnecessary expensive.) The reason why the higher boiling fraction (light gas oil, 230 to 305°, $\ce{C_13}$ to $\ce{C_17}$) is not used is that this is much more greasy than kerosene.

It is strange that in the question benzene and carbon disulfide are proposed since they are both more expensive, very toxic and their b.p. is very low. — In the case of carbon disulfide this would be a paradigm of a very dangerous choice for three reasons: 1. concerning the boiling point (46°, extremely fiery), 2. concerning its extreme toxicity and 3. because it is not inert against reductants.

Footnote:

$^1$ In the case of lithium, which is much less reactive than the other alkali metals, the container should ideally be filled level with the top since it is swimming on the kerosene. But anyway the kerosene stays over as a protecting film on the lithium, which obstructs its reaction with moist air, even if it is not submerged.


Changing iron(III) acetate to iron(II) acetate

You need a reducing agent that will reduce iron(III) back to iron(II). Actually finely distrubuted iron could be the best reagent to do this, as it will not introduce any contaminants. The reaction taking place is 2 Fe(III) + Fe = 3 Fe(II).

Of course the iron used should not be rusty. I am not sure that you can easily get hold of fine iron powder, but I would give a try boiling the vinegary solution with fresh iron shavings or steel wool and as little contact with air as possible. The color change should be visible to greenish from the red-brown of the iron(III) acetate.

Actually, during dissolution of Fe in acetic acid you should get iron(II) acetate predominantly until the Fe runs out, but usually scrap iron is covered by iron(III) oxides/hydroxides, and these also dissolve in the vinegar. When the iron(II) solution is in contact with oxygen in air, it oxidises rather quickly. It is a good idea to start with a batch that was not left standing for weeks.

Other reducing agents work quicker than iron (e.g. tin(II) salts or zinc powder), but these introduce other products that would interfere with your dyeing.


In the solid state of ammonium chloride, the nitrogen and hydrogen atoms are connected by covalent bonds, forming the ammonium ion. The bond length is about 98 pm. Each ammonium ion is surrounded by eight chloride ions (and in turn, each chloride ion is surrounded by eight ammonium ions). The distance between a chloride and the closest hydrogen atoms is about 237 pm. In contrast, the bond length in HCl is 127 pm.

In the reaction $$\ce{NH4Cl(s) -> NH3(g) + HCl(g)},$$

you lose the ionic interactions, and ammonium gets deprotonated while chloride gets protonated (lose an N-H bond and gain a H-Cl bond). The change in entropy for the reaction is positive, so it is favored at high temperature.

We might think that high temperature makes the H-Cl bond stronger and the N-H bond weaker. The bond dissociation energy for N-H is 390 kJ/mol while that of H-Cl is 432 kJ/mol. However, bond dissociation energies are not directly relevant because we are not transferring a hydrogen atom (making $\ce{NH3+ + HCl-}$). Instead, we are transferring a proton, so the pKa of ammonium and hydrogen chloride are more relevant but not directly either because this does not happen in aqueous solution.

The key is that the products are in the gas phase. If the products of the decomposition were ions and had opposite charges, they would not remain in the gas phase. In aqueous solution, you get $$\ce{NH4Cl(s) -> NH4+(aq) + Cl-(aq)}$$ because water is good at solvating ions. In the gas phase, the most stable combination of the atoms at hand is $\ce{HCl + NH3}$.


Why did my lemon and penny battery fail?

By not removing every last trace of copper from the sanded-side of the pennies, I created a kind of short-circuit that discouraged the charges from moving through the stack of pennies.

This is the main problem. Until all of the copper is gone from the zinc side of each penny, there is little reason for the two halves of the electrochemical reaction between zinc and copper not to just take place locally.  Your battery likely had one or more ionic short-circuits, just as you say, on these incompletely sanded surfaces.

  Since the cardboard was not very absorbent, it might have been particularly porous either and this might have prevented sufficiently free flow of particles between each penny in the stack.

This is unlikely.  As long as the cardboard was soaked through with the lemon juice, it should function fine as an ion-transport medium.

  By using a relatively low number of pennies, I guess I set it up to have a relatively low voltage. Maybe I could have overcome problems 1 or 2 if I used more pennies.

This is also unlikely. With five junctions (one between each neighboring pair of pennies), at a theoretical $1.1~\mathrm V$ per junction your stack voltage should have been at least about $5~\mathrm V$. This should be plenty to light a simple LED.


Rocket propellants: accuracy of vapor pressure estimation techniques

I would like to determine the vapor pressures of a couple of substances (rocket propellants) over a significant temperature range within a ~10% error margin.

The best way to determine vapor pressures is to measure them.  Is that not an option for you?

  It is important that the approximation results in a closed representation of the vapor pressure curve , therefore I would like to avoid methods that have points of discontinuity (e.g. Antoine equation?).

The Antoine equation isn't discontinuous by nature, it's only discontinuous if you use two different parameter sets for a single substance in order to widen the temperature range of applicability of the equation.

  What are your experiences with vapor pressure approximation? Could you provide some tips on how to improve my results?

I would recommend the Antoine equation as a first step beyond the Clausius Clapeyron equation.  Unlike the CC equation, the Antoine equation does not assume that the enthalpy of vaporization is independent of temperature.  It also provides a single, closed-form representation of the vapor pressure dependence on temperature.

Unfortunately, the Antoine equation requires compound-specific parameters.

However, many of the other methods you report using, such as the Lee-Kesler and Ambrose-Walton methods, also require compound specific parameters: specifically, the critical temperature and pressure of the compound of interest.  Even the Clausius-Clapeyron equation requires an (assumed constant) enthalpy of vaporization, which is also a compound-specific parameter.

Thus, the Antoine equation is probably the best place to start.  As Wikipedia shows, it can have accuracy better than 10% deviation for some compounds.  Ideally you would measure vapor pressures for your two compounds and then fit the data to the Antoine equation.  If measuring the vapor pressure of your compound is not possible, you could start by examining the Antoine equation predictions for a series of compounds that approximate your unknowns in chemical structure and molecular weight.  However, with this latter approach, I doubt error under 10% will be possible.

Antoine equation parameters for your compounds of interest

UDMH has Antoine parameters available from NIST.
The case of dinitrogen tetroxide is complicated by the equilibrium $\ce{N2O4 <=> 2 NO2}$.  It is unusually difficult to fit vapor pressure data to.
Hydrazine has Antoine parameters available from NIST.
In general NIST has Antoine parameters for a variety of substances.


Finding molecular formula with limited information given

Ionic solids like NaCl, KOH, KBr, PbCl2 and KBrO3 have ratios of individual elements, but the "molecule", if you could use that term to describe it, would be the whole crystal, because all the ions are attracted to nearby ions, ad infinitum.

The white solid could be written as "K23.4% Br47.8% O28.8%", and every time this experiment is run, the same product with the same composition will be obtained. And it will behave exactly the same as the commercial product called potassium bromate with a label saying KBrO3. We prefer the simpler formula.

The numbers given for the weight of PbCl2 and the product KBrO3 do not define a specific reaction, because other reaction products and excess reactants are not specified. And to be realistic, it seems that the filtrate was not evaporated to dryness, because all these other components are not obtained. It seems that this question is posed not just to get you to divide 3 percentages by 3 atomic weights, but to decide what numbers are important to answering the question. The mole ratio between PbCl2 and KBrO3 is unimportant (it's 3) - you got the answer at KBrO3 and should stop there.

Remember this riddle: What has 4 legs and a tail, barks like a dog and concrete?
Answer: A dog. (I just threw in the concrete to make it hard.)

A common test technique of teachers is to throw in extra information to see if you know how to select the correct information to use.


Charge, multiplicity, level of theory and coordination coordinates of gold compleaxes

There are various software to draw cartesian coordinates. Gaussian has its own GaussView, for example. I believe MolView or Avogadro are also popular. You can draw the molecule in any of these, then use the export option to create your coordinate file (export to Gaussian input file or xyz file). Take advantage of the fragment options for things like phenyl ligands. Also, depending on your goals, it's sometimes helpful to align your molecular axis with a principle axis (e.g. z-axis) for easier reading later, if you're using the calculation for more than just basic output parameters.

Furthermore, you can try to find the molecule in an x-ray database and adjust the geometry yourself to approximately match, which can pay dividends later by drastically simplifying the Gaussian calculation, particularly if you are using a larger basis set. You at least want to make sure you're drawing the correct stereoisomer, because the calculation may not correct this on its own.

The charge and multiplicity are manually defined in the input file for Gaussian. I believe you type in the values in the line just above the input coordinates if I recall correctly. You should check the official manual for sample inputs. There's much more syntax to be observed for an input file.


Why does Tyndall scaterring produce conical shape?

It is simple.
Light always travels in a straight line and if a lens is present it would diverge or converge ( in a straight line ) accordingly.

The question is why no other shape like maybe a hemisphere or a paraboloid ?

Now let me ask the question, have you seen such thing?
Check out this diagram to make everything clear.

If the beam of light increases in width ( diverging beam) linearly with distance. It is because all the particles scatter equally in all directions and some of them reach your eyes and this reveals the shape of the beam. The adjacent particles does the same and so on.
Therefore you can also see the light getting more and more diffused linearly as more and more light is scattered.

It also depends on the refractive index of the medium where the incident light ray will either move closer to the normal or move further. Again this would lead to a conically diverging beam.

EDIT

  This is to be noted that the colloidal solution does not affect the shape of the beam of light ( converging or diverging ).

This is also evident from your diagram.

From your question I feel that you think that the colloidal solution is responsible for the cone shape, it is not so. It is simply the geometry of the light rays

What the solution does is scatter the light in all directions and makes the "beam" visible.  A straight beam of light in perfect vacuum would not be visible as a stalk of light ( that we see when light enters a dark room thought a window )


What is the smell of a glass of water in contact with fresh air?

I do have a very similar experience! What the OP calls the mollusk smell, I call it slightly rotting "seaweedy" smell. It has been a very unpleasant experience for us in our new home.

Similar to OP's experience, this seems to be more intense with glasses and ceramics right after the newly washed dishes get exposed to air when the windows are open. But I also detect it with plastic cups as noted below.

In order to eliminate other variables such as dishwasher problems and tap water quality, I bought bottled water and disposable plastic cups. I did the following experiment.

WITH CLOSED WINDOWS

Water from bottle poured in a previously unused plastic cup, sitting for 15 minutes : No smell

Water from bottle poured in a cup and couple of sips taken, sitting for for 15 minutes : No smell

WITH OPEN WINDOWS

Water from bottle in a previously unused plastic cup, sitting for 15 minutes : No smell

Water from bottle in a cup and couple of sips taken, sitting for 5 minutes : Seaweedy smell

So I'm about to conclude that something in the outside air combines with the saliva (and/or chemicals/enyzmes from the detergents) to produce the seeweedy smell! The same thing happens with the newly washed dishes out of the dishwasher, and hand washed dishes after they get exposed to air with the windows open.

Yes, I know this doesn't make any sense to people that haven't experienced it before! It sounds like nonsense, but it's quite annoying every time I take sips from water cup.

I'm almost on the verge of calling up the EPA about it. There must be something in the air, but what...


Why did I get these inconsistent experimental results in silanization?

Rubbing is a very inconsistent process, and will introduce a lot of variation. I would suggest using a volatile non-polar solvent such as acetone, hexanes, or ethyl acetate to remove residual kerosene and allowing to air dry.


Why is the solubility of calcium sulfate so low even though it is a salt?

Solubility of calcium sulphate is quite high, compared to many much less soluble salts, like barium sulphate, calcium phosphate or calcium fluoride.

Most of insoluble minerals are salts.

Insoluble/limited solubility salts have as crystals lower Gibbs energy than dissolved, leading to solution being thermodynamically  unfavourable.

That is related to mutual relation of energy of crystal lattice and of hydration of ions, and also to dissolution entropy change.


Order of reaction for hydrogen ions being ignored

Imagine you are doing a process which takes 'n' steps. Label the steps : 1,2,3..n, now consider the time it takes to do step 1 , step -2 and so on. Now, wouldn't you agree that the time taken to complete process would be entirely dependent on the step which takes the most time(slowest step) to complete?
So, in your example the hydrogen ions are in one of the faster steps. So, if the concentration of hydrogen is 'enough' such that once you come to the step that their ample hydrogens to fuel the reaction, then it would seem that rate is independent of hydrogen concentration ( I say seem because there may be minute effects of it on the rate)


Explain the enolisation process of 5,8-dihydroxynaphthalene-1,4-dione

There is no need for any catalyst apart from acid or base. Also, the $\ce{^{13}C}$ is just there to tell you that it is not the same molecule. Also, this is not a zwitterion, it is tautomerism since there is pi bond movement involved.
Don't deprotonate any carbons. They're all $\ce{sp^2}$ and won't take kindly to deprotonation. Just push electrons from one oxygen to the other through the conjugated system.
Try doing this one first. In acid, your general tautomerism mechanism would look like this:

Protonate the carbonyl group.
Push electrons from phenol oxygen.
Deprotonate the carbonyl formed on the other side.

In base, it would be:

Deprotonate the phenol.
Push electrons from phenoxide oxygen.
Reprotonate the phenoxide formed on the other side (strictly this shouldn't happen, but whatever).

If you can do this one, the mechanism for your original question just consists of the same thing repeated. You can "fill in" the parts of the molecule that I omitted to see what I mean.


This chemistry was first described in this 1960 JACS paper here.

The first step is cleavage of the $\ce{Br-CCl3}$ bond to give $\ce{Br^.}$ and $\ce{^.CCl3}$.
The second step is $\ce{Br^.}$ abstracting a proton from toluene to give $\ce{HBr}$ and a benzyl radical.
The third step is the $\ce{^.CCl3}$ radical abstracting a proton from toluene to give chloroform and a benzyl radical.
The fourth step is a benzyl radical abstracting $\ce{Br}$ from $\ce{Br-CCl3}$ to give benzyl bromide and $\ce{^.CCl3}$.
The fifth step is two $\ce{^.CCl3}$ radicals quenching to give hexachloroethane $\ce{Cl3C-CCl3}$.

So overall the products are $\ce{HBr}$, Benzyl bromide, Chloroform, and Hexachloroethane.
Reactions 3 and 4 are the major portion of the reaction so equimolar amounts of benzyl bromide and chloroform are formed along with small amounts of $\ce{HBr}$ and hexachloroethane (approx 20:1).


Phosphate $\ce{PO4^3-}$ is an anion of the phosphoric acid $\ce{H3PO4}$. Term phosphate comes from a salt, for instance a sodium phosphate $\ce{Na3PO4}$. $\ce{Na+}$ are bound with a $\ce{PO4^3-}$ through ionic bonds. No electron sharing.
To annotate specific moieties of a molecule skeleton, a chemical group term is used. The phosphate has donated one pair of electrons on the oxygen to have a covalent bond with other molecule, for instance sugar in the image, and becomes a phosphate group in the molecule. Since the electron pair does not sit on the oxygen any more, but is shared with carbon, the phosphate charge is lowered to −2.


meth vs eth

[OP] Why is '-ethane' in 'methane'?

This is a coincidence.
Methyl is

ultimately from Greek methy "wine" + hylē "wood.

Source: https://www.etymonline.com.
The terminology was created by Dumas and Péligot in 1834 to distinguish wood alcohol from ethanol, and was published first in the French language:

Source: https://gallica.bnf.fr/ark:/12148/bpt6k6569005x/f15.item
Ethyl is from

Greek aithēr "upper air; bright, purer air; the sky" (opposed to aēr "the lower air"), from aithein "to burn, shine," from PIE *aidh- "to burn"

Source: https://www.etymonline.com.
Liebig had argued in 1834 that what now is known to be $\ce{-C2H5}$ is a "radical", i.e. a reoccuring group in molecules such as diethyl ether, ethanol and ethyl ethanoate, and gave it the name ethyl.

Source: https://babel.hathitrust.org/cgi/pt?id=uva.x002457887&view=1up&seq=30
Using the ending -yl for both the methyl and the ethyl group is a suggestion by Berzelius from 1835 (the correct atomic weight of carbon was not know yet, so the chemical formulas are off by a factor of two):

Source: https://books.google.com/books?id=1DM1AAAAcAAJ&pg=PA376#v=onepage&q&f=false, see Wikipedia entry on Ethyl group.
methane vs ethane
That these both end in -ane is not a coincidence - all alkanes end in -ane. As stated in M. Farooq's answer, this nomenclature was introduced by A. W. Hofmann in 1865.
Acknowledgements
Maurice's answer inspired me to find the first work by Dumas that led to the methyl terminology. I have edited my answer to incorporate information from the excellent answers by M. Farooq and Jan and the insightful comments. In trying to figure out the sequence of events in the 1830s, I made use of this account by Frederick E. Ziegler.
References
Dumas, J.-B, Mémoire sur l'esprit de bois et sur les divers composés éthérés qui en proviennent. lu à l'Académie des Sciences les 27 Octobre et 3 novembre 1834, Paris, 1834
Jean-Baptiste Dumas et Eugène Péligot, Mémoire sur l’Esprit de Bois et sur les divers Composés Ethérés qui en proviennent, Annales de chimie et de physique, 58 (1835) p. 5-74
Justus Liebig (1834) "Ueber die Constitution des Aethers und seiner Verbindungen" (On the composition of ethers and their compounds), Annalen der Pharmacie, 9 : 1–39
Jacob Berzelius, Årsberättelsen om framsteg i fysik och kemi [Annual report on progress in physics and chemistry] (Stockholm, Sweden: P.A. Norstedt & Söner, 1835), p. 376
A. W. Hofmann 1866, in Monatsbericht der Königl. Preuss. Akad. der Wissenschaften zu Berlin, 1865 653


Zero filling leading to increases in resolution (NMR)

The key idea behind zero padding or zero filling is that the "resolution" or the step size of the frequency axis in the Discrete Fourier Transform is dependent on the number of points you have in the time domain. Zero filling is done before doing the Fourier transform. Thus after the FFT the resolution appears to be improved because you have more points in the spectrum and the spacing between each point appears to be decreased.
This is a result of a beautiful theorem (although this is casually mentioned in most textbooks), but never shown as a theorem. I found in Cooley's original paper from 1967 [1, p. 80].
Cooley & Tukey are the key persons who made Discrete Fourier Transform (DFT) possible by computers for lowly mortals like us, otherwise it was an elitist subject among the mathematicians. Here is a snippet of it. It took me several years to find an original paper which showed this theorem. However, math researchers told me this was known in 1754. Full discussion here: History of Integral Transform Theorem
References

Cooley, J.; Lewis, P.; Welch, P. Application of the Fast Fourier Transform to Computation of Fourier Integrals, Fourier Series, and Convolution Integrals. IEEE Transactions on Audio and Electroacoustics 1967, 15 (2), 79–84. https://doi.org/10/bmw8z8.


Positive or Negative Anode/Cathode in Electrolytic/Galvanic Cell

The anode is the electrode where the oxidation reaction

\begin{align}
 \ce{Red -> Ox + e-}
\end{align}

takes place while the cathode is the electrode where the reduction reaction

\begin{align}
 \ce{Ox + e- -> Red}
\end{align}

takes place. That's how cathode and anode are defined.

Galvanic cell

Now, in a galvanic cell the reaction proceeds without an external potential helping it along. Since at the anode you have the oxidation reaction which produces electrons you get a build-up of negative charge in the course of the reaction until electrochemical equilibrium is reached. Thus the anode is negative.

At the cathode, on the other hand, you have the reduction reaction which consumes electrons (leaving behind positive (metal) ions at the electrode) and thus leads to a build-up of positive charge in the course of the reaction until electrochemical equilibrium is reached. Thus the cathode is positive.

Electrolytic cell

In an electrolytic cell, you apply an external potential to enforce the reaction to go in the opposite direction. Now the reasoning is reversed. At the negative electrode where you have produced a high electron potential via an external voltage source electrons are "pushed out" of the electrode, thereby reducing the oxidized species $\ce{Ox}$, because the electron energy level inside the electrode (Fermi Level) is higher than the energy level of the LUMO of $\ce{Ox}$ and the electrons can lower their energy by occupying this orbital - you have very reactive electrons so to speak. So the negative electrode will be the one where the reduction reaction will take place and thus it's the cathode.

At the positive electrode where you have produced a low electron potential via an external voltage source electrons are "sucked into" the electrode leaving behind the the reduced species $\ce{Red}$ because the electron energy level inside the electrode (Fermi Level) is lower than the energy level of the HOMO of $\ce{Red}$. So the positive electrode will be the one where the oxidation reaction will take place and thus it's the anode.

A tale of electrons and waterfalls

Since there is some confusion concerning the principles on which an electrolysis works, I'll try a metaphor to explain it. Electrons flow from a region of high potential to a region of low potential much like water falls down a waterfall or flows down an inclined plane. The reason is the same: water and electrons can lower their energy this way. Now the external voltage source acts like two big rivers connected to waterfalls: one at a high altitude that leads towards a waterfall - that would be the minus pole - and one at a low altitude that leads away from a waterfall - that would be the plus pole. The electrodes would be like the points of the river shortly before or after the waterfalls in this picture: the cathode is like the edge of a waterfall where the water drops down and the anode is like the point where the water drops into.

Ok, what happens at the electrolysis reaction? At the cathode, you have the high altitude situation. So the electrons flow to the "edge of their waterfall". They want to "fall down" because behind them the river is pushing towards the edge exerting some kind of "pressure". But where can they fall down to? The other electrode is separated from them by the solution and usually a diaphragm. But there are $\ce{Ox}$ molecules that have empty states that lie energetically below that of the electrode. Those empty states are like small ponds lying at a lower altitude where a little bit of the water from the river can fall into. So every time such an $\ce{Ox}$ molecule comes near the electrode an electron takes the opportunity to jump to it and reduce it to $\ce{Red}$. But that does not mean that the electrode is suddenly missing an electron because the river is replacing the "pushed out" electron immediately. And the voltage source (the source of the river) can't run dry of electrons because it gets its electrons from the power socket.

Now the anode: At the anode, you have the low altitude situation. So here the river lies lower than everything else. Now you can imagine the HOMO-states of the $\ce{Red}$ molecules as small barrier lakes lying at a higher altitude than our river. When a $\ce{Red}$ molecule comes close to the electrode it is like someone opening the floodgates of the barrier lake's dam. The electrons flow from the HOMO into the electrode thus creating an $\ce{Ox}$ molecule. But the electrons don't stay in the electrode, so to speak, they are carried away by the river. And since the river is such a vast entity (lots of water) and usually flows into an ocean, the little "water" that is added to it doesn't change the river much. It stays the same, unaltered so that everytime a floodgate gets opened the water from the barrier lake will drop the same distance.


How does the work that won the 2012 Sustainable Chemistry Award contribute to sustainable chemistry?

His work is about developping new catalysts based on copper and iron, to replace to traditional catalysts based on palladium. Copper and iron are both very common elements in nature, while palladium is considered a high supply risk (see the 2012 British Geological Survey risk list for details).

The new RSC Visual Elements Periodic Table can be used to check this kind of information.


Why does a silane group react with hydroxyl groups?

It is well known that silicon forms very strong bonds with oxygen (think sand, $\ce{SiO2}$). Hence, any time one can convert a $\ce{Si-H}$ bond into $\ce{Si-OH}$, it will happen.

I believe that silanes are extremely air sensitive due to the moisture content.


How are reaction enthalpy and entropy affected by temperature

For each species you could use Kirchoff's equation to calculate the change in enthalpy with temperature, in its integrated form (at constant pressure) this is $\displaystyle \Delta H_2-\Delta H_1=\int_{T_1}^{T_2}\Delta C_p dT$. The heat capacity $C_p$ changes with temperature also, but for small temperature changes it may be considered to be effectively constant, then $\displaystyle \Delta H_2-\Delta H_1=\Delta C_p(T_2-T_1)$, if not, the heat capacity can be approximated by $C_p=a+bT+cT^2 \cdots$ and integrated but the constants should be determined experimentally.

The entropy is zero at zero Kelvin for a pure substance so must increase with temperature and as $\displaystyle S_T=\int_0^TC_p/T dT$ you can calculate the change in entropy between two temperature.

However, there are phase changes as the temperature increases (solid, liquid, gas) then there are discontinuities in the heat capacity and the entropy becomes $\displaystyle S_T=\int_0^TC_p/T dT+\Delta H_{fus}/T_{fus} + \Delta H_{vap}/T_{vap}$. The terms with phase changes cancel out in $\Delta S$ if your species are in one phase, i.e. all reactants are in the gas phase.

(If you are familiar with partition functions in Statistical Mechanics then thermodynamic properties of molecules can, in principle, be calculated directly.)


How many stereoisomers are possible for the compound in picture?

Follow these steps:

Identify all possible sources of stereochemistry. You got pretty far.
Determine how many isomers can derive from each. I think you went wrong here.
Do you have symmetry? Reduce accordingly.
Multiply all the numbers together.

In your molecule:

Double bonds:

the internal double bond in the residue on the top right
the cumulene, which also shows isomerism (E/Z for uneven double bond counts, aR/aS for even ones).

Asymmetric carbons:

the top methyl group
the bottom methyl group
the cumulene-containing side-chain
the other unsaturated side-chain.

Every one of these can be either R/S or E/Z.
The molecule does not have any elements of symmetry, so we cannot reduce.
All we have are factors of 2; two from the double bonds, another four from the asymmetric carbons. This leads us to:
$$n(\text{isomers}) = 2^{(2 + 4)} = 2^6 = 64$$


Is 1-methoxy-2-methylpropane a positional isomer of 2-methoxy-2-methylpropane?

2-methoxy-2-methylpropane is a synonym for MTBE (Methyl tertiary-butyl ether). The name "propane" is a little bit missleading, because it comes from the standardized IUPAC nomenclature system, which names the  ether-oxygen together with the smaller alkyl group an "alkoxy" group with the suffix "-oxy" (in our case: methoxy) and takes the larger alkyl part as its base identifier (here propane) when describing ethers. So MTBE is referred as an ether with all its chemical properties.

Now to the isomerism: Structural isomers have the same chemical formula but different constitutions (arrangement of atoms, bonds, groups). So they definitely must be constitutional isomers.

Constitutional isomerism is divided into 5 subcategories:

Chain isomerism
Positional isomerism
Functional isomerism
metamerism
tautomerism

As you already mentioned: they are metamers. Thats because different alkyl rests are attached to the same functional group (ether). This is the first reason why they are not likely to be positional isomers. And the other reasons might come from the "moving substituent" (alkoxy group), which has a carbon atom (the carbon structure has to be constant within positional isomers, therefore moving the -OCH3 could mean a difference) and is no functional group.


Reduction of copper oxide by hydrogen in the lab to copper

As the system cools, the gases will contract. Unless a positive flow of hydrogen (or some other inert gas) is maintained during the cooling process, this will result in room air being pulled into your tube. The oxygen present in air would then oxidize some of $\ce{Cu}$ back to $\ce{CuO}$, so long as some regions within the tube remained hot enough for this reaction to take place (at least $\pu{300^oC}$ or so).


Why is freezing point depression greater than boiling point elevation for equimolar substances?

To find an answer, let us take a look at how the cryoscopic constant (for decrease in melting point) and the ebullioscopic constant (for increase of boiling point) are defined:

Please note that the Kelvin-scale is used for temperature here. The boiling point is always above the melting point (we keep pressure constant to keep the example simple).

For most substances, the enthalpy of vaporization is greater than the enthalpy of fusion. It follows that the denominator for K_eb is greater, resulting in a smaller K_eb.

This means that the effect of greater enthalpy of vaporization compared with enthalpy of fusion outweighs the effect of boiling temperature being greater than the melting temperature (even though it's the square of absolute temperature).


Reason behind more Reactivity of Unsaturated Hydrocarbons

The unsaturated regions are electron-rich and typically allow for electrophilic attack, such as by halogens or protons. Saturated hydrocarbons usually react only with radicals (radical halogenation). C-H activation is often hard in saturated hydrocarbons.

While double and triple bonds are stronger than a single bond when looking at complete cleavage of a bond, one notices that the bond strength of a double bond is less than twice the bond strength of a single bond (with other factors being comparable, there are surely some exceptions). When a double bond reacts, it typically is changed into a single bond, not fully cleaved. Similar arguments hold for triple bonds.


How to calculate the ratio of rate constants of competing reactions?


How can an α,β-unsaturated ketone tautomerise?

If you have an α,β-double bond next to your carbonyl, a γ hydrogen can also take part in tautomerism. Take crotonaldehyde $\ce{H3C-CH=CH-CHO}$:

$$\ce{O=CH-CH=CH-CH3 <<=> HO-CH=CH-CH=CH2}$$

In general, a double bond is able to ‘extend’ the mesomeric properties of a carbonyl compound across two additional atoms. This is known as vinylogous; crotonaldehyde can form a vinylogous enol.

In the specific case of 6,6-dimethylcyclohex-2-en-1-one, these are the two tautomers:


How to determine the calorimeter constant from its heat capacity

To give a precise answer, the following assumptions are necessary and must be clear:

bomb calorimeter works at constant volume ($V=const$);
both water and calorimeter itself are at thermodynamic equilibrium before the experiment and during the measurement, in particular their temperatures $T_w$ and $T_c$ are equal before the experiment and during the measurement;
the system is compound by calorimeter itself plus water;
the system is an isolated one;
pressure is 1 bar.

Initially the system is at temperature $T_1$. Let's imagine that an object at $T_o>T_1$ is put inside the chamber of the calorimeter. Temperature of the system increases and, once reached thermodynamic equilibrium, it stops at a precise value $T_2$.

Since $V=const$, heat transferred from object to system is:
\begin{equation}
Q_V=\Delta U=\Delta U_{calorimeter}+\Delta U_{water}=(mc_V\Delta T)_c+(mc_V\Delta T)_w
\end{equation}
where $\Delta T_c=\Delta T_w=T_2-T_1$.

We know that heat capacity at constant volume is defined as:
\begin{equation}
C_V=\left(\frac{\partial U}{\partial T}\right)_V\approx \left(\frac{\Delta U}{\Delta T}\right)_V
\end{equation}
So, reshaping the first equation, we obtain:
\begin{equation}
C_V=\frac{\Delta U}{\Delta T}=(mc_V)_c+(mc_V)_w=(C_V)_c+(\rho Vc_V)_w
\end{equation}
Adding the following data:

$\rho_w=1000\;kg/m^3$;
$(c_V(300\;K,1\;bar))_w\approx 4.134\;J/(kg\;K)$ (source: Perry's Chemical Engineers' Handbook)

and carrying out the conversion: $V=600\;mL=6\times10^{-4}\;m^3$, we obtain finally:
\begin{equation}
C_V=787\;J/K=0.787\;kJ/K
\end{equation}
So right answer is A.


Why is the ortho isomer a major product in the nitration of toluene?

A methyl-group is not t-butyl, so it isn't all that crowded.
There are two ortho-positions.
There is also the possibility of an ipso-attack with the nitronium donor attacking the position of the methyl group, then shifting to the ortho-position.

Given all that without any orientation effect, the ratio between ortho and para should be around 3/1 - 2/1. It is commonly observed around 3/2, so the crowding does have an effect.


Why did my beer slowly froth up after removing from freezer?

Melting takes time.
As your beer is thawing, temperature remains roughly constant. Then the beverage will slowly assume ambient temperature.
Solubility of gases (e.g., $\ce{CO2}$) in common solvents usually decreases with warming. This is because the breaking of solvent$-$solvent interactions are not compensated by new interactions between solute and solvent. In other words, solubility of gases is often exothermic.

The bottom of the beer might have warmed more quickly, causing a buildup of carbon dioxide under a layer of still-frozen beer. Reaching some critical value for pressure, the top popped. According to Henry's law, immediately after even more gas bubbles are formed.

Effects due to positive thermal expansion of beer are probably rather small but these would only aggravate any overflow.

For further reading on similar subjects, see

Does shaking increase the pressure inside a bottle of champagne?,
Plateau–Rayleigh instability.


Oxidation of Carbon atom in Alkene, Alkyne and Alkanes of functional groups?

Nomenclature priorities are in order of the degree of oxidation of the carbon atom.

This may be true for the heteroatom functional groups, but not for alkenes, alkynes, and alkanes.

Consider that each $\ce{C-H}$ bond contributes $-1$ to the oxidation state and each $\ce{C-C}$ bon regardless of order contributes $0$ to the oxidation state of carbon.

Thus:

hydrocarbon   formula   functional group   oxidation number of C
ethane        CH3CH3    alkane             -3
ethene        CH2CH2    alkene             -2
ethyne        CHCH      alkyne             -1

  Alkynes have a higher oxidation number than alkenes.

Alkenes react better in oxidation reactions than alkynes because alkenes are less oxidized than alkynes.

The organic chemistry textbook that I teach from explains that alkenes have higher nomenclature priority than alkynes because alkenes are more common. I cannot find a more authoritative reason/source.


Which one of displayed formula below does not represent an isomer of heptane?

The structure in the question and option (D) are the same thing. Note that a branching as apparently seems is not a real branch on the principle chain that would give an isomer. Basically (D) is same as that given in the question.


General justification for steady state approximation

No very satisfying justifications are available unless you do a case-by-case analysis. But PSSA does not come from common sense. In general, pseudostationary state approximation (PSSA) is justified by the theory of differential equations.

In fact, the pseudostationary state concentrations represent the zeroth approximation
to a set of particular solutions of the correct chemical kinetic equations, which are called principal solutions. After the initial period, the concentrations of the intermediates approach their principal solution rather than their pseudostationary state values.

Therefore, the accuracy or usefulness of the pseudostationary state assumption depends upon the magnitude of the differences between the principal solution and the pseudostationary state concentrations.

OK, that sounds useless. But after doing some maths (like finding the principle solutions), some (obviously very ambiguous) rules are indeed justified mathematically.

  Induction period should be short. (As you can see PSSA solutions cannot satisfy the initial conditions)
  Destruction of the intermediates should be fast.

For example, in $\ce{A ->[$k_1$] B ->[$k_2$] C}$ if $k_2\gg k_1$ then the conditions above are met while if $k_1\gg k_2$, they aren't.

By the way, PSSA originates from Michaelis-Menton theory for enzyme kinetics. PSSA in it is well justified while the conditions above are not necessarily met. Consider

$$
\ce{E + S <=>[$k_1,k_{-1}$] ES ->[$k_2$] P}.
$$

Let $[\ce{S}]=s,[\ce{E}]=e,[\ce{ES}]=c,[\ce{P}]=p,e+c:=e_0,s+p+c:=s_0$. Furthermore, let $\tau=k_1e_0t,u(\tau)=s/s_0,v(\tau)=c/e_0$. The kinetic equations can be written as

$$
\begin{cases}
\frac{\mathrm du}{\mathrm d\tau}=-u+(u+K-\lambda)v\\
\color{red}{\varepsilon}\frac{\mathrm dv}{\mathrm d\tau}=u-(u+K)v
\end{cases}
$$

where  $\lambda=k_2/\left(k_1s_0\right),K=\left(k_{-1}+k_2\right)/\left(k_1s_0\right),\color{red}{\varepsilon=e_0/s_0\ll 1}$. Therefore, PSSA can be applied (i.e. setting $u-(u+K)v=0$, justified by perturbation methods) without loss of much accuracy. This example indicates that more often than not, a case-by-case analysis is required.

Finally, I think to use PSSA safely, one should do some experiments to check (numerical simulations, real experiments, etc.).


Fischer to Haworth projection: which carbon connects to the oxygen?

It just is. There are pyranose and furanose forms for all aldoses and hexoses (given they are long enough), including glucose:

$\hspace{3cm}$

You will just have to remember which connections correspond to which sugars. You can read more about anomers ($\alpha$ vs. $\beta$) here, and more about carbohydrate chirality (D vs. L) here.


Stereochemistry of product in Diels-Alder reaction

1) The regiochemistry of the product can be predicted by drawing resonance structures of the diene and dienophile, allowing us to assign partial charges to the reactive centers. Doing that shows us that 2 and a are partially positive while 5 and b are partially negative. We'd then predict that 5 attacks a and b attacks 2. This answer has a figure showing prediction of regiochemistry this way.

2) We would predict that the product is the endo- product, which would have the bridging ester on the opposite side of the cyclohexene relative to the methyl ketone. This has to do with the endo-rule, which is classically explained by secondary orbital overlap between the the lobe of the HOMO on carbon 4 interacting with the lobe of the LUMO on carbon c. Keep in mind that this trend does exist, but mixtures of endo and exo are common.


Is there a hypervalent molecule which has no contribution of d-orbitals at all?


I don't think that question is trivial and except for what you have already deduced, I would not have an idea how to solve that without employing computational chemistry.

The problems already start that I suspect that the proton from position 9 readily exchanges to position 7 in any kind of polar solvent (and in absence why not with itself).

Well from that point on forward, I'd say you need at least a lot of acid to protonate 9, as the π system is aromatic and will most likely not react first.

After that you could probably deduce from the approximate bond angles which of the nitrogen would have the largest s- and p- character, then rank them from highest to lowest p contribution. The one with the most will probably be the HOMO and therefore react first. To be honest though, this is just guessing.
I have absolutely no clue what the lesson with this exercise it, except for: "They're pretty much the same."

Let's have a closer look anyway. I have computed the neutral molecule at the DF-B97D3/def2-TZVPP level of theory, and used NBO6.0 for the partial charges:

As expected, there is not much of a difference; I'd even go as far and say they are the same. Position 9 is a little bit more negative, because it has a hydrogen to draw from.

Next up, let's have a look at the highest molecular orbital, as we would assume to protonate there:

Again, there is only little to no difference. Or in numbers (contributions > 3%):

Alpha occ 31 OE=-0.217 is
N1-p=0.3635 N1-s=0.0382
N3-p=0.2551
N7-p=0.0667
C5-p=0.0563
C4-p=0.0473
C2-p=0.0452
C6-p=0.0382

From this we would conclude that since N1 has the largest contribution, the proton is most likely to go there.

Therefore I also calculated all the protonated species, and their relative energies are:
\begin{array}{lr}\hline
\text{Position} & \Delta G(\pu{298.15 K}, \pu{1 atm})/(\pu{kJ/mol})\\\hline
1  &     0.0  \\
3  &    41.2  \\
7  &    26.1  \\
9  &   196.8  \\\hline
\end{array}

From this we conclude that the most likely position to be protonated is N1, followed closely by N7, and also N3. Off the charts is, as expected, N9 as it breaks the aromaticity.

TL;DR: Order of protonation $1 > 7 > 3 > 9$, calculated at DF-B97D3/def2-TZVPP.


Why can some ions have different valence?

An empty, half-filled or fully-filled electron shell is especially stable.  $\ce{Fe}$ metal is [Ar] 3d6 4s2.  $\ce{Fe^{2+}}$ emptiess the 4s-shell.  $\ce{Fe^{3+}}$ give that plus a half-filled 3d-shell.  Vanadium II, III, IV, V.  Now try chloride, hypochlorite, chlorite, chlorate, and perchlorate.

The most stable oxidation state depends on the medium and added ligands, if any.  Cu(I) is a Lewis soft acid. It is stabilized by soft bases like pyridine, $\ce{MeCN}$, olefins, $\ce{I^{-}}$.  $\ce{Cu^{2+}}$ is a hard Lewis acid.  It is stabilized by hard bases like ammonia, water, $\ce{Cl^{-}}$.  Dissolve $\ce{CuCl2}$ in pyridine or $\ce{MeCN}$ and you have a powerful oxidizing agent. Dissolve a $\ce{Cu^{1+}}$ salt in water and it disproportionates into $\ce{Cu}$ metal and $\ce{Cu^{2+}}$.  Put $\ce{Cu}$ metal and $\ce{Cu^{2+}}$ into $\ce{MeCN}$ and you get $\ce{Cu^{1+}}$.


Distribution coeffecient without concentrations

Molar concentration can be expressed via mass $m$ and volume $V$ all right:

$$C_i = \frac{n_i}{V_i} = \frac{m_i}{M_iV_i}$$

Analyte doesn't change during the extraction and its molar mass $M$ remains the same $(M_i = \text{const})$ so the distribution coefficient $D$ can be rewritten as such:

$$D = \frac{C_\mathrm{s}}{C_\mathrm{w}} = \frac{m_\mathrm{s}V_\mathrm{w}}{m_\mathrm{w}V_\mathrm{s}}$$

where "s" refers to the solvent phase and "w" to aqueous phase.
At equilibrium

$$m_\mathrm{w} = m_0 - m_\mathrm{s}$$

where $m_0$ is the initial mass of the analyte.
Finally, the distribution coefficient is

$$D = \frac{m_\mathrm{s}V_\mathrm{w}}{(m_0 - m_\mathrm{s})V_\mathrm{s}} = \frac{\pu{0.0376 g}\cdot\pu{150 mL}}{(\pu{0.192 g} - \pu{0.0376 g})\cdot\pu{50 mL}} = 0.731$$


Why is sodium chloride edible whereas both sodium and chlorine are highly reactive?

Sodium chloride is edible. But elemental sodium and chlorine are highly reactive and poisonous. What is the difference?

Solid sodium chloride is an ionic compound. This means it consists of positively charged sodium ions ($\ce{Na+}$) and negatively charged chloride ions ($\ce{Cl-}$) arranged in a cubic lattice.

On the other hand, sodium metal and chlorine gas are elements. $\ce{Na}$ has an electron configuration of $\mathrm{[Ne]~3s^1}$ and so it really wants to lose that single electron to form a full outer shell as a $\ce{Na+}$ ion. This makes it highly reactive towards species which can easily accept an electron such as $\ce{Cl2}$, $\ce{H2O}$ etc. $\ce{Na+}$ has a full outer shell, being isoelectronic with neon, and so it is quite happy to stay as an ion and not react. Chlorine, with an electron configuration of $\mathrm{[Ne]~3s^2~3p^5}$, really wants to gain a single electron to get a full outer shell, isoelectronic with argon, as the $\ce{Cl-}$ ion. This makes it highly reactive towards species which can easily donate an electron, such as $\ce{Na}$.

  is there any difference in eating sodium chloride in powder form and after dissolving in water?

Not really. Sodium chloride dissolved in water consists of hydrated $\ce{Na+}$ and $\ce{Cl-}$ ions. If you eat solid sodium chloride then it rapidly dissolves in the water in your body, so the effect is the same. Interestingly though, solid salt tastes quite nice, but salty water tends to taste horrible.


Take this generic reaction representing the Michaelis-Menten enzyme kinetic:

You can derive the Michaelis-Manten constant $K_m$:
 (for derivation see https://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics)

The quasi-steady state hypothesis was used, therefore the rate of formation and breakdown of $ES$ are the same; this is a good approximation if the enzyme concentration is much less than the substrate concentration $[S]$ or $K_m$ or both. That means that $K_m$ can be extremely small (almost zero) and the Michaelis-Menten equation still be valid proven that the enzyme concentration is much less than  $[S]$.
$K_m$ is very small if:
$k_r$ and $k_{cat}$ are both very small compared to $k_f$, thus if the complex $ES$ is formed very quickly compared to its breakdown.
In this case the reaction proceeds at its maximum velocity $V_{max}=k_{cat}*E_0$, with $E_0=[ES]+[E]$:

About the theoretical limit on how small $K_m$ can be there are two considerations:
 First, $k_f$ cannot be cannot be faster than the diffusion-controlled encounter of an enzyme and its substrate. This means that $k_f$ cannot be higher than $10^9 s^{-1} M^{-1}$  (ref.https://www.ncbi.nlm.nih.gov/books/NBK22430/#:~:text=This%20rate%20cannot%20be%20faster,s%2D1%20M%2D1.)
Second, $k_{cat}$ cannot be too small or the product $P$ formed is produced at a rate not suitable for the cell survival.
Generally, I haven't encountered $K_m$ smaller than $10^{-6}M$ in biological enzymes (e.g Triosephosphate isomerase, TPI) but possibly there are. In TPI, $ k_f$ ∼ $10^9 s^{-1} M^{-1}$, therefore  $k_r$  ∼ $k_{cat}$  ∼ $10^{3}s^{-1}$ that is still quite high. It would be very interesting to know if there are enzymes with an even lower $K_m$. Very intresting question.


Does the absolute value of enthalpy have any physical significance?

The absolute value of enthalpy has direct relation to the mass of the system:

$$H = U + pV = mc^2 + pV$$

with the second term mostly negligible. If the system accepts thermal energy Q, its mass change is $\Delta m = \frac {Q}{c^2}$

For most of enthalpy applications, the absolute $H$ value is irrelevant.

The absolute value interpretation is the total energy stored in the system at the constant pressure, consisting of

the equivalent energy to the mass of contained matter
work done by reaching the system volume against the external pressure.

This energy is formally extractable from the system by annihilation and shrinking to zero volume.


The term "thermally" is used here to contrast with a photochemical isomerization. In the chemistry of vision, 11-cis retinal absorbs light and is isomerized to the all-trans isomer. This is a photochemical process because it involves the absorption of a photon to promote one electron into an antibonding orbital. In this excited state, the geometry can rearrange to the more stable all-trans form before the electron relaxes back to the ground state.

In a "thermal" process, no photons are absorbed, no electrons are promoted, and no excited states are needed. The molecule takes advantage of ambient or supplied kinetic energy to reorganize.


Does Chemical Formula of Matter Influence on how would it be Charged?

Firstly: Suppose we rub glass ($\ce{SiO2}$) against paper (which is made from cellulose $\ce{(C6H10O5)_{$n$}}$). We receive positively charged glass and negatively charged paper. Is that because of total ionization energy of molecule stronger than glass or crystal system too?

Although atoms and molecules like to be uncharged there is a difference between the electron affinity and the ionization energy. Electron affinity is a measure of willing a substance is to accept an extra electron. Ionization energy is a measure of how willing a substance is to donate electrons.

When the paper is rubbed against the glass the surfaces of the two substances come into physical contact. So the electron affinity of the paper is greater than the ionization energy of the glass rod, so electrons are stripped from the glass and are transferred to the paper. Enough electrons are transferred to balance the affinity/ionization relationship. This is known as the Triboelectric effect.

To be totally honest the Triboelectric effect is not well understood. There are a lot of enigmatic interactions in how the surfaces contact, how many times the surfaces contact when rubbing, how specific contact points my heat up, or how some physical transfer may take place. Read the Wikipedia article for more details on these points.

    Secondly: When we rub atoms against other atoms, do electrons accelerate each other by the same electrostatic charge and change their orbitals to higher orbits due to synchrotron radiation they emit. Or is it because of the fact that protons are more close are ionizing electrons by electrostatic attraction? Am I correct?

There are a number of erroneous concepts in the setup/question. The first explains the phenomena.

Synchrotron radiation is electromagnetic radiation and has absolutely nothing to do with the electrostatic experiment that you are discussing.

The electrostatic experiment that you are discussing does not "accelerate electrons." So rubbing a glass rod with paper doesn't make a particle accelerator.

Neither the paper nor the glass has bare protons floating around.


How would I measure the Carbon Dioxide content in Coca Cola over a period of time?

Measure the change in mass over time of the remaining liquid. Though some water will also evaporate, you can control for that by keeping the humidity near 100%.

If you have to be precise, collect the outgassed $\ce{CO2}$ in a liquid-nitrogen cold trap. Check the mass of the condensate, which should equal that lost from the soda. Check purity, to be really finicky.


Find pH and Kw for water with given values of conductivity

We know,

$$\lambda_c = \frac{(\text{Specific Conductivity})}{1000 \times c}$$

$$\alpha\lambda_\infty = \lambda_c$$

(where $c$ is concentration in moles per litres of the electrolyte and $\alpha$ is degree of dissociation of the electrolyte)

$\lambda_{\infty}$ of water (the electrolyte) is the sum of $\lambda_{\infty}$ of $\ce{H+}$ and $\ce{OH-}$.

From equations above, one can easily find  the the value of $c\alpha$ is $10^{-7}$, which is the concentration of $\ce{H+}$.


1) Cr3+ has 3 electrons in 3d orbital. Obviously, it is a tetrahedral splitting, so configuration will be eg2 t2g1 which shows it is paramagnetic.

2) Here you need to know that for Co3+ all complexes are low spin except [CoF6]3- and [Co(H2O)3F3]. So, the complex given will be low spin with all 6 electrons paired & thus it will be diamagnetic.

3) Cu2+ has 3d9 configuration. So, it can never be diamagnetic as one electron will always be unpaired.

4) What you thought is right. But I would like to add that NH3 doesn't always behave like a weak field or strong field ligand. It depends on many factors which can't be explained using CFT.

Hope it helps!


The reason stated in the literature is fairly straightforward. However, actually tracking down the original source is harder than it looks.

Going back to Phys. Rev. B22, 115 (1980) one finds in the opening paragraph the statement

  Mixed Zn ferrites of the type Zn$_{x}$M$_{3-x}$O$_{4}$ where M
  is a magnetic ion have been the subject of numerous
  investigations in the past, especially with regard to
  their magnetic properties.$^{1}$ The metallic ions occupy
  the tetrahedral A or octahedral B sites in the spinel
  crystal structure. The Zn ions preferentially occupy
  the A sites because of their tendency to form covalent
  bonds involving sp' orbitals.

Reference number 1, on the first sentence, is a book [J. Smit and H.P.J. Wihn, Ferrites (Wiley, New York, 1959)]. The statement on $sp3$ bonding is not backed up by a reference. I can only assume that it was considered common knowledge in the spinel community in 1980.


Naproxene syntheses: electrophilic aromatic substitution on activated naphthalene

No other byproducts are shown

I haven't checked the original Syntex patent (if you've ever written a patent you know that sometimes important details are omitted), but it seems odd to me that those two isomers are claimed to be the most significant isomers formed. As you point out, the 2-methoxy-5-acetoxy isomer is not expected to be formed in significant amounts based on electrophilic substitution patterns in substituted naphthalenes (Note: here is a nice concise reference that shows how to apply the same concepts used to predict which isomers will  predominate in electrophilic substitution with benzene to substituted naphthalenes).

In this abstract the Friedel-Crafts acetylation of 2-methoxynaphthalene is studied.  By changing solvents they can shift the major product from 2-methoxy-3-acetoxy to 2-methoxy-6-acetoxynaphthalene.  These two isomers are among those expected and the experiment shows that these two isomers are close enough in energy that small factors can significantly impact the product distribution.  Two other factors to keep in mind

the 6-acetoxy isomer corresponds to "para" substitution
When running a Friedel-Crafts with oxygenated (hydroxyl, methoxy) aromatics, the aluminum chloride will (depending to some degree upon solvent) complex with the oxygen.  In the case at hand this could make the methoxy a bulky substituent disfavoring reaction at the ortho positions.

  But again, why in pos. 6? Why not in pos. 3, in ortho to the
  activating substituent?

Careful low temperature mono-bromination of phenol shows that para-substitution is strongly favored (>95%, reference).  If additional bromine is available a second bromine will be substituted at the ortho position. It seems that the same thing is happening here in your 2-naphthol example.

Like you, I might not have predicted these outcomes, but that's why we run experiments.

  are these reactions done under kinetic or thermodynamic control?

Sorry, without more information (like product composition as a function of time, temperature, etc) I can't answer that.


Actually solubility $x$ is given as a function of pressure $p$ and henry's constant $K_H$:
$$p=K_Hx$$

  My book says that higher the value of K at a given pressure lower is the solubility of the gas in the liquid. I don't understand how.

Since at a given pressure, $p$ is constant, if you increase $K_H$, logically solubility $x$ would decrease.


As I remember I have read somewhere that $\ce{Al3BC}$ is konwn to have Boron in -5 oxidation state.

you can also reffer to this PDF page no. 139
PDF IN GERMAN


How does the polarizability of an apolar compound influence properties such as its boiling temperature?

In non-polar liquids, the intermolecular forces are known as London dispersion forces (or dipole induced dipole interaction). This is the weakest interaction that exists between molecules.
The strength of this interaction depends on polarisability of non-polar molecule. As an example, let's take the following examples:
$$
\begin{array}{lr|lr|lr|lr}
\hline
\text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} \\
\hline
\ce{CO2}  & -78.5 & \ce{CCl4} & 76.7  & \ce{CBr4}  & 189.5 & \ce{CI4}   & 329.2 \\
\ce{CS2}  & 46.3  & \ce{CBr4} & 189.5 & \ce{SiCl4} & 57.7  & \ce{SiBr4} & 153.0\\
\ce{CSe2} & 125.5 & \ce{CI4}  & 329.2 &            &       &                 \\
\hline
\end{array}
$$
In these molecules, polarisability depends on polarisability of anions $(\alpha),$ which depends as
$$\alpha \propto q \times r,$$
where $q$ is the charge and $r$ is the ionic radius.
As charge on each anion is same, hence we can conclude that, greater the radius, greater is the polarisability, and thus greater is London dispersion forces.
Therefore, greater boiling point.


The shorthand way that we draw structures in organic chemistry, with implicit hydrogens, leads us to often forget that the hydrogens are there, or to neglect considering them in our analyses. In electrophilic aromatic substitution (EAS) reactions, the generally accepted mechanism involves attack of the aromatic ring on the electrophile to give a Wheland intermediate, which then loses a proton from the $\mathrm{sp^3}$-hybridized ring carbon to restore aromaticity. Ipso substitution, then, is really nothing special - we simply substitute a position on the aromatic ring that bears a non-hydrogen (e.g., $\ce{X}$) group, presumably via an analogous mechanism. For this to be viable, both steps of the mechanism need to be favorable:

$\ce{X}$ needs to enhance, or at least not greatly reduce, the nucleophilicity of the carbon it is attached to. For example, inductive effects would great disfavor ipso attack at groups like $\ce{F}$ or $\ce{CF3}$.
$\ce{X+}$ should be a reasonable species to lose in the second step, to regain aromaticity, or there should be some mechanism by which $\ce{X}$ can be lost, with electrons flowing back towards the aromatic ring.

Sulfonic acids and carboxylic acids are rather electron-withdrawing, but if you drill down into the details of these ipso-substitution reactions, many are actually conducted on the conjugate base (i.e. sulfonate or carboxylate), which also provides a pathway for loss of a neutral group in the second step, i.e. $\ce{SO3}$ or $\ce{CO2}$. Alternatively, harsh conditions are used, e.g., sulfonation/desulfonation.

It turns out that ipso attack is more common than it first appears, because the product of the initial ipso attack can re-arrange. So, for example, nitration of p-cresol involves 40% attack at the ipso carbon bearing the methyl group, followed by acid-catalyzed rearrangement to the expected product.[1] There is more discussion here.

Closer to the original thrust of your question - there are some groups that are even better at being substituted in an EAS reaction than a proton. Trialkylsilyl and trialkylstannyl groups are often used. Both $\ce{Si}$ and $\ce{Sn}$ are less electronegative than carbon and highly polarizable, making the ipso carbon a good nucleophile. These groups also help to stabilize the Wheland intermediate through hyperconjugation. Finally, they both have energetically-favorable pathways to leave with electron density flowing back towards the aromatic ring. For these reasons, ipso substitution at $\ce{SiR3}$ or $\ce{SnR3}$ substituted carbons can often be achieved rapidly, under mild conditions, with high regioselectivity. This is often helpful in specialist applications, like radiolabelling.

I haven't specifically addressed each of the original three numbered questions, but they can be tackled using the principles above. Ipso substitution is possible for any group - it's a competition, $\ce{H}$ vs $\ce{X}$. Is ipso substitution affected by reaction conditions? I'm sure it could be - factors like protonation states, hard vs soft electrophiles, reversibility, etc. would have an influence.


Did you not carry out some research?

Well, you can find information about it in Wikipedia, or even dictionaries like Merriam-Webster or Dictionary Online.

You will see that pH is calculated through:

$$pH = -\log(a_{\ce{H+}})$$

Or which can most of the time be stated as:

$$pH = -\log(\ce{[H+]})$$

$a_{\ce{H+}}$ is the hydrogen ion activity, the 'effective concentration' of hydrogen ions. Without active hydrogen ions, you don't have any pH.

You might find that the pH scale is from 0 to 14, but you can also have circumstances where you have negative pH and pH above 14. It depends on the extent of dissociation of an acid or base to produce active hydrogen ions, or hydroxide ions in the case of bases.

Since I mentioned hydroxide ions, it might also be worth nothing that:

$$14 = pH + pOH$$

And that basically, a pH of 1 is equivalent to a pOH of 13.


Actually, not all atoms are stable, at least not in the gas phase, which is where we measure the electron affinity. You should first read Why do atoms "want" to have a full outer shell? for detailed elaboration.

Now, electron affinity involves the gain of an electron to an isolated gaseous atom, but not all atoms actually want to gain an electron, i.e., they are already pretty stable. This is precisely the reason why elements like the noble gases or nitrogen have negative electron affinity i.e. you need to supply energy for them to accept an electron.

Also read: Are there any major exceptions when comparing electron affinity?

So, in short: "Atoms are stable so..." not always, in fact, such atoms are in the minority. "...how can they pull electrons..." read the first link. "how do they have any electron affinity at all?" we can always supply excess energy to make an atom accept an electron, forcibly. So, the only difference for stable vs non-stable atoms would be the sign of electron affinity.


Relating a compound's radii and unit cell dimensions

The ratio of the radius of the cation to the radius of the anion is called radius ratio. The ratio of the radius of cation and the radius of anion i.e, radius ratio ($r_+/r_-$) plays an important role in determining the structures of ionic solids and coordination number of ions. The possible coordination numbers and structural arrangements of anions around  cations for different $r_+/r_-$ values are given below:


I'm not sure about why it's not matching but when I carry out substructure matches in rdkit I use SMARTS instead https://www.daylight.com/dayhtml/doc/theory/theory.smarts.html

Here is what I would have done.

from rdkit import Chem

smiles_list = ['C12=CC=CC=C1C3=CC=CC4=C3C2=CC=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=CC=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C=C4', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CCCCC7=C6', 'C12=CC=CC=C1C3=CC=C4C5=C(C=CC2=C35)C6=C4C=C7CC8=CC=CC=C8CC7=C6', 'c1ccccc1']

pattern = Chem.MolFromSmarts('[#6]12~[#6][#6]~[#6][#6]~[#6]1[#6]3~[#6][#6]~[#6][#6]4~[#6]3[#6]2~[#6][#6]~[#6]4')
for idx,smiles in enumerate(smiles_list):
    m = Chem.MolFromSmiles(smiles)
    print("Structure {}: pattern found {}".format(idx+1,m.HasSubstructMatch(pattern)))

and the output

Structure 1: pattern found True
Structure 2: pattern found True
Structure 3: pattern found True
Structure 4: pattern found True
Structure 5: pattern found True
Structure 6: pattern found False

I've added a Benzene smiles to show that it's not just matching everything. The SMARTS pattern I made probably could be better I think. I hope this helps anyway.


Are electron orbital orientations filled in any particular order?

No, you cannot say that there is a particular order in which the shells are filled.
Relation between $\mathrm{p_x,p_y,p_z}$ with $\mathbb{m_l}$ values
The magentic quantum number ${m_l}$ represents the projection of the angular momentum $\vec{l}$ of an electron in an orbital, on an axis, usually taken to be the z-axis. The orbitals, and their $\mathrm{m_l}$ values, come from the solution of the Schroedinger's equation for a Hydrogen atom.
Now, the problem is, the solution for the orbitals with ${l=1}$ (i.e. p orbitals) produces one real wavefunction (corresponding to $m_l=0$), and two other complex wavefunctions (corresponding to $m_l=1$ and $m_l=-1$). If the main axis is taken to be z, then the $m_l=0$ orbital has the usual dumb-bell type shape and points in the z-direction, so it is the $\mathrm{p_z}$ orbital. However, the complex orbitals are conjugates of each other, and if you plot the real part, they both look like donuts. (You can find more on this here and here). So the other two orbitals, let's call them $\mathrm{p_1}$ and $\mathrm{p_{-1}}$, do not really correspond to anything we usually see in the chemistry textbooks.
However, wavefunctions themselves don't hold any physical meaning, because we cannot observe them. We can only observe things like electron energy, electron density etc. So, we can take linear combinations of the two complex orbitals to get two new real p orbitals, as long as it conserves the total energy. However, doing this means that the new $\mathrm{p_x}$ and $\mathrm{p_y}$ orbitals are no longer eigenfunctions of the $\hat{L_z}$ operator (i.e. they have no well defined angular momentum along the z-axis, so no well defined $m_l$ value). In most cases angular momentum is not really important, so we can safely use $\mathrm{p_x}$ and $\mathrm{p_y}$.
I suspect that there is a similar case with the d orbitals as well.
Which orbital is filled in first?
Inside one shell (e.g. 2p) the electrons are indistinguishable, so it is not possible to say which orbital is occupied by which electron. However, if a shell is not completely filled, i.e. there are unpaired electrons, then there would be an total orbital angular momentum and total spin angular momentum coming from all of those electrons combined. It is possible to identify states which have different angular momenta, as they have different energies. This forms the basis of the coupling schemes (LS coupling or jj coupling). However, you still cannot say exactly which electrons are residing in orbitals with which $m_l$ values, because multiple arrangements (microstates) can contribute to the same energy level (term).


I can't calculate the $\mathrm{pH}$ using Henderson-Hasselbalch approximation, because the concentration of acid is 0.

Yes, as commented above, sodium acetate solution cannot be considered as buffer;the result at the equilibrium , will be the same as dissolving  $\pu{0.025 mol }$ of $\ce{CH3COONa}$ in $\pu{250 ml}$ water,so you have the salt solution of $\pu{0.1M}\ce{[CH3COONa]}(c_\mathrm{\text{salt}}=\ce{[CH3COONa]_\mathrm{I}}=\pu{0.1M})$

  I don't understand the last part, why is this so? what is...

The salt ($\ce{ CH3COONa}$) is  electrically neutral substance formed by cation $\ce{Na+}$ (an acid) and  anion $\ce{ CH3COO−}$ (a base), completely dissociated or ionized in an aqueous solution as :
  $$\ce{CH3COONa   -> Na^+_\mathrm{(aq)}   + CH3COO^−_\mathrm{(aq)}}$$
a)  Since $\ce{Na+}$  is the conjugate acid of a strong base, it won't be strong enough to react with water; $\ce{Na+}$  actually spectator ion.

b)  Meanwhile, since $\ce{ CH3COO−}$ is the conjugate base of a weak acid, and  therefore strong enough to be able to hydrolyze and accept ions $\ce{ H+}$ from water, so water act as an acid leaving a hydroxide ion $\ce{OH−}$ as :
$$\ce{CH3COO^− +H2O <=>CH3COOH + OH^−}$$ $$(K_\mathrm{b(\ce{CH3COO-})}=\frac{K_\mathrm{w}}{K_\mathrm{a(\ce{CH3COOH})}}=\frac{10^{-14}}{1.76\times 10^{-5} }=5.68\times{10^{-10}})$$

c)The equilibrium equation of the hydrolysis of the conjugate base $\ce{CH3COO-}$:
$$ K_\mathrm{b} =\frac{[\ce{OH-}]_\text{equilibrium}[\ce{CH3COOH}]_\text{equilibrium}}{[\ce{CH3COO-}]_\text{equilibrium}}$$

Neglecting water autoionization and assuming :${\ce{[OH-]_\text{equilibrium}}=\ce{[CH3COOH]_\text{equilibrium}}}$, ${[\ce{CH3COO-}]_\text{equilibrium}=c_\mathrm{\text{salt}}=[\ce{CH3COONa}]_\mathrm{I}}=\pu{0.1 M}$,so:
$$ K_\mathrm{b} =\frac{[\ce{OH-}]_\text{equilibrium}^2}{c_\mathrm{\text{salt}}}$$

$$[\ce{OH-}]_\text{equilibrium}=\sqrt{  K_\mathrm{b}\cdot c_\mathrm{salt}}$$
You  can calculate $\ce{[H+]}$ -without neglecting the water autoionization  and without approximation- by substituting in the following cubic equation and solving bywolframalpha

$$\
 [\ce{H+}]^3
+( K_\mathrm{a}+[\ce{CH3COONa}]_0) [\ce{H+}]^2
  -K_\mathrm{w}[\ce{H+}] -    K_\mathrm{w} K_\mathrm{a}
  $$
$\ce{[H+]} =1.32658 ,\pu{pH}=8.877$


Di-alpha-halogenation of ketones in acidic medium

Acid Catalysed Halogenation of ketone

In the halogenation of ketones, both in acidic and basic medium, enols are reactive intermediates. After addition of first halogen to double bond, deprotanation occurs immediately (Scheme 1). The second halogen does not add$\ce{^1}$.

Scheme 1

The introduction of second Bromine is slower then the first ,since cabocation is destabilized by the electron-attracting polar effect of two Bromines (**Scheme 2).

Scheme 2

Consequently, reprotonation of the monobrominated ketone does not occur.

Base Catalysed Halogenation of ketone

The mechanism of this reaction involves the formation of an enolate ion as a reactive intermediate. After mono bromination, the enolate ion of the alpha-bromo ketone is even more stable the enolate ion of the starting ketone.(Scheme 3). Consequently, a second bromination occurs$\ce{^2}$.

Scheme 3

As discussed above, in acid catalysed halogenation  the intermediate is unstable ,hence reprotonation to give dihalo products is less (If one equivalent of Bromine is taken).

Note$\ce{^3}$ :

Comparatively, in base calalysis the intermediate (Bromo enolate anion) is quite stable leading to di and tri bromination (provided acidic H is present).

Reference

Organic Chemistry, 5th Edition  by Prof. Marc Loudon
Organic Chemistry, 5th Edition by Prof. Marc Loudon
Organic Chemistry, 2nd Edition ,Jonathan Clayden , Nick Greeves ,Stuart Warren.


Is barium sulfate soluble in dilute mineral acid?

The key point is to have an acid with such a high $\ce{H+}$ activity that in large extent shifts the below reaction to the left :
$$\ce{  HSO4-  + H2O <=> SO4^2- + H3O+ } \\ K_\mathrm{a2}=1.0 \cdot  10^{-2}$$

Then activity of sulphate ions from dissolution of $\ce{BaSO4}$ is decreased by the equilibrium reaction and the solubility increases.

$$\begin{align}
K_\mathrm{s}&=a_{\ce{Ba^2+}}\cdot a_{\ce{SO4^2-}}=1.1\times 10^{-10}\\
\frac{K_\mathrm{a2}}{a_{\ce{H+}}}&=\frac{a_{\ce{SO4^2-}}}{a_{\ce{HSO4^-}}}\\
\end{align}$$

The  effective solubility product
$$K_\mathrm{s,ef}=K_\mathrm{s} \cdot    \frac{a_{\ce{H+}}}{K_\mathrm{a2}}$$

Note that despite activity coefficients {gamma} decrease with increasing ionic strength, for concentrated acids this tendency reverses and raises very rapidly

[HCl]  Gamma
0.0005 0.975
0.01 0.904
0.1 0.796
1 0.809
2 1.01
5 2.38
10 10.44
12 17.25

Therefore, $a_{\ce{H+}}$ of 10 M HCl = 104.4.

$$K_\mathrm{s,ef}=K_\mathrm{s} \cdot    \frac{104.4}{K_\mathrm{a2}}\\=1.1\cdot 10^{-10}\cdot 1.044\cdot 10^4=1.148\cdot 10^{-6} \mathrm{~ mol^2 L^{-2}}$$

Solubility would be then $ M \cdot \sqrt{K_\mathrm{s,ef}}=233.4 \cdot 0.001072=0.25\mathrm{~g/L}$

Conclusion: The $\ce{BaSO4}$ solubility in concentrated $\ce{HCl}$ is limited, in diluted one is very low. The solubility in concentrated sulphuric acid would be much higher, as $a_{\ce{H+}}$ would be very high.


How to determine the structure of organic molecules without spectroscopy

One thing that was well known, even in the early days, was elemental analysis. It is Justus von Liebig’s contribution to have worked out the sum formulae of hundreds of natural compounds. This generally worked by combusting a given sample of a compound, capturing the vapours, absorbing the water with desiccants such as calcium chloride, absorbing carbon dioxide by potassium hydroxide (forming a carbonate), weighing the different parts and thereby calculating the mass of carbon and hydrogen in a given sample. Tests for nitrogen and sulfur also existed, but I can’t remember them off the top of my head. Oxygen content was indirectly calculated as being the remainder.

Chemists since the advent of their science had done everything to generate pure compounds. It was widely accepted even then that a pure compound was one with a defined melting or boiling point, and liquids could also be analysed by their refractive index and density. This provided a rather good library of methods for realising ‘is my compound pure?’, and ‘is it the same as $\ce{X}$?’

Another very old concept was that of valencies. By analysing samples of known and simple elemental composition, it was realised that certain elements bonded to a certain number of other elements, e.g. that a hydrogen would always require a bonding partner of value 1. Somebody (I can’t remember who but I could look it up in my lecture notes) proposed a valency of 4 for carbon. Together with a valency of 2 for oxygen, you had most pieces of the puzzle in hand.

It was recognised by Liebig and Wöhler, that isomers existed, i.e. compounds of the same elemental composition but distinct. (This is also a consequence of the ammonium cyanate experiment Wöhler performed to create urea, breaking the wall between organic and inorganic chemistry and proving vis vitalis to be inexistent — both are isomers of each other.) So there must have been something important in the structure of molecules that needed to be discovered.

A lot of further clues then came from reactions that transformed known compounds into other known compounds. For example, oxidation would transform some known compound $\ce{C3H8O}$ into acetone $\ce{C3H6O}$. This also lead to Liebig’s theory of radicals — where radicals is something a modern chemist would refer to as functional group. With these clues and radicals, the task of putting the puzzle together to pinpoint certain structures became solveable.

From then on, it all depended on whether the initial assignments were correct (they mostly were). You would go ahead either breaking down unknown compounds by known reactions after having established their functional groups or resynthesising them from known fragments in other known reactions to arrive at a final product.

Working out the difference between hexan-2-ol and hexan-3-ol would have been a none-trivial task; that is true. One variant seems to be starting from hex-2-ene and adding water to the double bond. You would get two different products and with the knowledge of structure in hand, you would realise that they had to be isomers of each other and had to be the 2- and 3-isomer respectively. How did you get hex-2-ene? Oh, maybe you managed to dehydrate hexan-2-ol into two different products, hex-1-ene and hex-2-ene. How did you get hexan-2-ol in the first place, then? Addition to hex-1-ene sounds reasonable. How to get that? Dehydration of hexan-1-ol. That final one was easily established as a special hexanol, since it could be oxidised to hexanoic acid, not only to a ketone. It is all possible but requires a lot of work. Think how lucky we are that we can record an NMR spectra within minutes (or a complete set over night) and solve the structure the next day!

It is entirely possible that for the cholesterol family, the exact ring connections were initially not known, only that there is some type of a fused ring system that can be modified in certain positions (not all!). It would have taken extra time to break down the ring system accordingly. And Emil Fischer’s work on carbohydrates probably warrants one or two questions of its own.

And then, crystallography happened, and all of a sudden was it possible to see structures. It must have seemed like the dawn of a new era.


Comparison of stability of cyclic imines

The cyclohexane-1,3,5-triimine tautomer is benzene-1,3,5-triamine, which is aromatic. This makes this trienamine form much more stable than any of the other possible tautomeric forms.


Why is the reduction by sugars more efficient in basic solutions than in acidic ones?

This is based in the underlying redox rection. If we take e.g. mannose and attempt to oxidise that, the (unbalanced) half-reaction we need is the following:

$$\ce{C6H12O6 -> C6H10O6 + 2e-}\tag{Ox1}$$

To balance it, we either need to add protons to the right-hand side (acidic medium) or negative charges to the left-hand side. Since we are asked about alkaline solutions, let’s do the latter:

$$\ce{C6H12O6 + 2 OH- -> C6H10O6 + 2 e- + 2 H2O}\tag{Ox2}$$

As you can see, the oxidation of sugars consumes hydroxide ions. If you increase the concentration of hydroxide, that will typically increase the reaction rate.

Of course, one should always check what the reduced compound is. In your case, it is always a complexated copper(II) ion and is reduced by the following equation:

$$\ce{2Cu^2+ + 2 e- + 2 OH- -> Cu2O v + H2O}\tag{Red}$$

Even here, if we set up the half-reaction, we notice that we are consuming hydroxide. Thus, the overall redox reaction is:

$$\ce{C6H12O6 + 2 Cu^2+ + 4 OH- -> C6H10O6 + Cu2O v + 3 H2O}\tag{Redox}$$

If a reaction consumes hydroxide ions, it is enhanced by basic conditions. Likewise, if a redox reaction consumes protons, it is enhanced by acidic conditions.


Why is the equilibrium constant unitless?

There are a few separate issues here to keep in mind:

$K_c$ (the equilibrium constant in terms of concentrations) is defined as
$$K_c = \prod {c_i}^{\nu_i} \tag{1}$$
Agreement between $K_c$ and the product of forward/back rate constants ($k_{\pu{fwd}}/k_{\pu{rev}}$ in the OP example) is expected only if the mechanism is correct, assuming some proportional rate law, and at equilibrium.
When dealing with gases, $K_c$ cannot generally be equated with $K_p$ (the equilibrium constant in terms of partial pressures). They may be numerically equal under some circumstances. For instance, they are numerically equal in reactions involving only ideal gases, provided the number of gaseous molecular entities remains constant during the reaction.
The (standard) equilibrium constant $K$, which simplifies to $K_a$ (under standard fugacity and some other conditions, agreements) which is defined as $$K_a = \prod a_i^{\nu_i}, \tag{2a}$$ cannot generally be equated with $K_c$. They may be numerically equal under some circumstances. $K$ itself is actually defined as follows:  $$\Delta_\pu{r} G^\circ=-RT\log(K) \tag{3}$$ whereas, strictly speaking, $K_c$ can't be used in this relation (but see below). The equilibrium constant $K$ cannot have units or this equation will not make sense.  In terms of concentrations,$$K_a = \prod \left(\frac{\gamma_ic_i}{c_i^\circ}\right)^{\nu_i} \tag{2b}$$ (To be nitpicky, one should use molality-based activities instead of amount concentration-based $c$ when saying that $K$ reduces to $K_a$).  When activity coefficients are all unity (ideal conditions), it might be that $$K_a \approx \frac{K_c}{K_c^\circ}= \prod \left(\frac{c_i}{c_i^\circ}\right)^{\nu_i} \tag{2c}$$ Strictly speaking, the equilibrium constant $K$ should remain unchanged with a change in the units used to describe concentrations. Units have no place in the constant because they cancel when ratios to reference standard states are computed (see Eq. (2)). $K_c$ and $K$ differ by the product  $$K_c^\circ = \prod ({c_i^\circ})^{\nu_i}\tag{2d}$$ (ignoring, once again, activity coefficients; standard fugacity). This difference is the source of the dangling units leading to the question in the OP.

You should therefore be careful equating $K_c$ with the ratio $k_{\pu{fwd}}/k_{\pu{rev}}$. And, assuming they are equal, the concentration equilibrium constant cannot generally be equated with $K$. However you can divide $K_c$ by $K_c^\circ$, and, assuming activity coefficients are unity, the result might be the thermodynamically correct unitless equilibrium constant (if standard fugacity), equivalent to that which appears in usual thermodynamic derivations.

Sloppy or not, associating units with an equilibrium constant is common practice. Attaching units to the equilibrium constant during reporting is a useful mnemonic when performing some computations, but it should be kept in mind that such a constant may not fulfill the requirements for computation of an associated standard free energy change, unless first modified by a ratio product of the standard concentrations.

Note that some of this is also explained in an answer to this post. IUPAC likes to use the convention standard equilibrium constant when referring to the thermodynamic equilibrium constant.


Pair of analogous periods on the periodic table

In a bit of a web search, there does not seem to be any particular names for families of periods in the Periodic Table.  However, according to WebElements,

  The f-block elements are assigned to Periods 6 (lanthanoids) and 7 (actinoids) since that is where they are located in the full, or extended, version of the periodic table.

However, these, according to the image below, only apply to the Lanthanides and Actinides, not all of periods 6 and 7.

Image source


It certainly could undergo the acid-base equilibrium you described, if there's enough energy in the system. Your problem is omitting an important piece of information, which is the temperature at which the reaction is taking place. Lithium enolates (the right compound) are usually formed at low temperatures, down to -78 C. The hope/assumption is that the aldol reaction happens faster than the acid-base reaction between the two reactants. This is certainly hard to control in this particular example, and thus the yield of the reaction is not great, according to this example mentioned in a review article on aldols:

Furthermore, it's likely that in doing this reaction, the experimentalist would form the lithium enolate separately before mixing it with the cyclohexanone. You can read about the importance of order of addition here: https://en.wikipedia.org/wiki/Aldol_reaction#Order_of_addition

Here's the mechanism, note that the problem didn't provide you with a viable source of H+.


Which C-C bond would break first, the one in ethane, or 2,2-dimethylpropane?

It does not have to do with steric effects as you suggested. Rather, the reason behind this trend is the inductive effect. Figuratively speaking, carbon side-chains attached to an central atom "push" electron density towards this atom, while hydrogen atoms don't. If you break a C-C bond and produce a carbon radical, it will be stabilized if electron density is donated towards the site of the unpaired electron, which is the case in the tert-butyl radical formed from dimethylpropane, but not in the methyl radical from ethane.


Interpretation of 4-benzoyl butyric acid H-NMR Spectrum

You should not assume that just because the coupling constants are the same that this means all the nuclei are coupled. It simply means that two sets of coupled nuclei have the same coupling constants. This is not unexpected when you consider that the nuclei are very similar - they are freely rotating alkyl chain methylene groups. And you will find that freely rotating alkyl chain methylene groups have coupling constants of around 7 Hz. This is because the 3J-coupled nuclei (called vicinal coupling) rotate through a range of conformers that have a range of coupling constant values (described by the Karplus equation) to give an average value. Different types of functional group give characteristic coupling constants. Trans-vinylic coupled protons have a coupling of around 15Hz, for instance. So, given the functionality of these alkyl methylenes, you would expect similar coupling constants.

To work out which nuclei are in fact coupled, you would use other information. A set of homonuclear decoupling experiments, or a COSY experiment would give this information clearly, although there is some very useful information in this spectrum. Your signals at 2ppm, 2.5ppm and 3ppm all show significant 'tenting'; they lean towards their coupled partner. That is, you do not get a true Pascal's triangle intensity, with the peaks closer to their coupled partner having greater intensity, as shown below:

The signals at 2.5 and 3 both lean towards the right, indicating they are coupled to something upfield (the signal at 2ppm). Similarly, the peak at 2ppm is leaning to the left, indicating it is coupled to the peaks at 2.5 and 3ppm. You could use your integrals of 2:2:2 to rationalise a -CH2-CH2-CH2- chain. And this would explain the expected splittings you see of a triplet (3ppm), triplet (2.5ppm) and triplet of triplets (2ppm). Of course, if you draw out your splitting diagram for a triplet of triplets with identical coupling constants, it will give you a pentet with intensity 1:4:6:4:1.

To answer your second question, you should try to predict what nuclei these signals represent. I have drawn out some useful props below to assist in the interpretation.

The signals between 7 and 8ppm come from your mono-substituted benzene ring, so you have three distinct environments: ortho, meta and para to the substituted position. They should integrate 2:2:1. The coupling constants for protons around an aromatic ring aren't important for this exercise, other than to say you will only see splitting due to a proton on an adjacent carbon; that is, 3J coupling. So, you have 5 aromatic nuclei in 3 environments, that integrate 2:2:1, and I can see 2 sets of signals that integrate 2:3. Clearly you have two signals overlapping at 7.5ppm. Which signals?

Let's have a think about the splitting we would expect to see from this ring. The ortho protons would be a doublet, as they are coupled to one other 3J proton only.The meta protons would be a doublet of doublets, as they are coupled to the ortho (doublet) and para (doublet) protons. And the para proton would appear as a triplet (coupled to the two meta protons).

What else can we see? The substituent on the benzene ring is a -(C=O)-R. This is a strongly electron withdrawing group, and so if you draw out your resonance structures for this, you will find that the ortho protons are deshielded compared to the meta protons, and so will be downfield of them. The para protonalso  experiences a bit of shielding, and so will also be downfield of the meta protons. So, you can assign the peak at 8ppm as your ortho proton; furthest downfield fits with your electronics adn doublet fits with your expected splitting. The overlapping peaks at 7.5 are the para proton (the left hand side of the forest); this fits with your electronics and (if you were really keen) you could assign some of these peaks to a triplet. The remaining right hand side of the peaks are your meta signals, and again you could assign this asa doublet of doublets. Of course some of the lines for the para and meta signals are overlapping (actually, possibly just one). You could resolve these signals on a higher field instrument quite easily.