Do metals form covalent bonds besides ionic and cordinate bond?

The first thing that should be said is that there's no difference between a coordinate bond (dative bond) and an ordinary covalent bond. Yes, the electrons "come from different places"; but the molecule doesn't actually know this, nor does it care. Once a covalent bond is formed, it is a covalent bond, regardless of where the electrons "come from".

Some of the examples you have given can equally be described as coordinate bonds or covalent bonds. For example, $\ce{MnO4-}$ could be viewed as four $\ce{O^2-}$ ligands forming coordinate bonds to a central $\ce{Mn^7+}$. Or, you could also just view them as plain old Mn=O covalent bonds. It doesn't actually matter.

Having established that there is no real difference between coordinate bonds and covalent bonds, the only real question is how can you tell a covalent bond apart from an ionic bond. The answer is mostly to do with polarisability. If you imagine an ionic bond between a strongly polarising cation ($\ce{Al^3+}$) and a polarisable anion ($\ce{Cl-}$), the cation will pull over some of the anion's electron density: this is exactly what is meant by "polarising". This changes the nature of bonding from an ionic model (where the bonding electrons reside entirely on one atom) to a covalent model (where the bonding electrons are somewhere in between the two atoms). You'll notice that all the cations you've listed: $\ce{Li+}$, $\ce{Be^2+}$, $\ce{Al^3+}$, (formally) $\ce{Cr^6+}$, and $\ce{Mn^7+}$, are very highly polarising. The first three have a very small size, and the last two have exceptionally large positive charges.

That said, covalent bonds between two transition metals are reasonably common, too: a simple example is $\ce{Mn2(CO)10}$ (Wikipedia), which is essentially $\ce{(OC)5Mn-Mn(CO)5}$ with a single bond between the two.

The bond order between two metals can go up to slightly ridiculous numbers. Cr–Cr and Re–Re quadruple bonds are well-known in $\ce{Cr2(OAc)4(H2O)2}$ and $\ce{[Re2Cl8]2-}$ respectively. Even quintuple and sextuple bonds between metals have been described (although some of these accounts are not without controversy).

How can you predict this, though? It's not trivial at all, and in truth you probably don't need to care about these exotic examples until you have enough experience to understand the bonding.


reaction with Barium Chromate, Potassium nitrate

If you used distilled water to prepare the solutions of potassium chromate and barium nitrate, it is more likely that you got a yellow precipitate of $\ce{BaCrO4}$ in a yellow solution of excess $\ce{K2CrO4}$. The yellow color originates from $\ce{CrO4^{2-}}$, and a small amount of finely divided precipitate may look white at first sight (namely when the yellow color of the solution is more intense), when it is actually pale-yellow (it can look like this).


Why …carbaldehyde?
It's similar to $\ce{C6H11-COOH}$ cyclohexanecarboxylic acid (not cyclohexanoic acid). It's the case where the functional group's carbon is not included in the parent structure. (Such case are also nitriles/carbonitriles, $\ce{R-CN}$)
Why not …hexyl… or rather …cyclohexyl… ?
This (wrong name) would be constructed similarly to a different, non-preferred type of nomenclature, a radical name, which names e.g. ethanol as ethyl alcohol.
There are more than two aldehydic group in the compound.
They are not. Alhedyde group is ${\ce{-CHO}}$. The other one, $\ce{>C=O}$ is an oxo group (ketone). The group for suffix is chosen by the pre-defined priority, ${\ce{-CHO}}$ > $\ce{>C=O}$.
Why 2-bromo-6-oxo-… and not 6-bromo-2-oxo-… ?
The groups named by prefixes (after highest priority - suffix group is assigned lowest locant possible) are not numbered by their priority, but such that the assigned locant set is lowest possible. If there are more equivalent possibilities, then they are numbered in the alphabetically sorted order (which is used in the name).


Chemoselectivity in reduction of 3′-nitroacetophenone

The key question is how the reduction by the two reagents proceeds.

For tetraborohydride the reduction proceeds as a standard nucleophilic addition of hydride ion to carbonyl group.

For tin/HCl the reduction proceeds as a series of one-electron transfers from metal surface mixed with protonation. It seems that the lowest free orbital in the compound belongs mostly to nitro group (which makes sense as it has a strong π-withdrawing effect, stronger than carbonyl group), so it is reduced to a nitroso group and then further via a hydroxylamine to an amine. It is possible to stop the reduction at intermediate stages under some conditions.

So the question is, why borohydride does not reduce the nitro group. It should be able to do so, as a strong reducing agent which is able to reduce iron from solutions. A possible answer is that extremely small size of nitro group with a net negative charge on the nitro group prevents formation of a stable adduct with hydride ion.


You can tell by the dark gray appearance that it is not vanadium chrome, but rather an alloy of hardened steel.  If it were made of vanadium chrome, it would not have broken, and it would also have a silver appearance to it.  Density may help tell if you are close, but with alloys the density does not necessarily conform to the rule of mixtures.


How does salt push out the water from vegetables, meat, etc.?

It is called "Osmosis" and the applications of this are many. Since you are a self-identified "noob of chemistry" I'll not go too far into the specifics of why this happens but try to explain on a laymans basis.

As you probably know, water is a solvent. It can dissolve (some) solids so that no solid remains, only liquid. This usually has some effect on the solute (water in this case). Lets say you have two glasses of water - one of which has some salt dissolved in it. Nature would very much prefer if they were at the same concentration of salt. Nature wishes for 2 glasses of water with $\frac{some}{2}$ salt in each.

So, when there is an apple, with water inside and some salt outside, nature would like there to be about the same concentration of salt inside and outside. So she (nature) will provide a driving force for taking water out of the apple and into the salt. The skin of the apple is a semi-permeable membrane: only water can pass. So nature will drive the water through the skin to the salt. And she will keep at it until the apple is so empty of water that it is so hard to find water in the apple that the forces are in balance.

The same thing applies to sugar. Sugar is also one of these compounds that can dissolve in water and will affect the water in the same way.

For a more detailed explanation I'd google "Osmosis". This interesting phenomenon is responsible for many things, wonderful and useful. You can actually pump water with it. Nature will "pull" water against the force of gravity in order to balance out the concentrations. (This is basically how a salt power plant works)


Does sulfur have various allotropes at room temperature?

"Exists" is an awfully tricky word: it can be understood in a thermodynamic sense or in a kinetic (=everyday) sense, and the meanings are sometimes quite different. Ditto for "stable".

Thermodynamically, there exists only one stable allotrope of any element at any given conditions. All other allotropes are not allowed to exist and must eventually transform to that one. This is true, but the problem is that the transformation may take forever.

Now, your book on sulfur was using this meaning, but your test was not. It was meant to be understood in a kinetic sense, where multiple allotropes may coexist all right. In particular, the high-temperature monoclinic sulfur can be cooled down to room temperature and still remain monoclinic. I can attest to that; I used to hold it with my own hands, as did numerous other chemists. That's what people call metastability.

There is also plastic sulfur and other forms, but that's another story. And as you already know, there is oxygen vs ozone, and white phosphorus vs red, so hydrogen is really the only element here that does not have multiple allotropes.

So it goes.


Why is the amu of chlorine-35 less than 35?

You need to account for the energy released when nucleons and electrons come together and form a Cl-35 atom. Its called the Binding Energy.

This sort of equation can help to explain :

(Rest Mass Energy of Individual nucleons,electrons*) - (Various Binding Energies) = (Rest Mass Energy of Natural)

*edit


Adding cyanide vs nitrite for nucleophilic substitution and why does it change when we use silver salts for it

Potassium cyanide ($\ce{KCN}$) is a predominantly ionic compound ($\ce{K+}$ is Hard Acid while $\ce{CN-}$ is Soft base) and would dissociate completely to give cyanide ions in solution. Cyanide ion ($\ce{CN-}$) is considered an ambient nucleophile, which means that the negative charge can associate with either carbon or nitrogen. However, the negative charge of cyanide ion concentrated on carbon than nitrogen. Thus, reaction with haloalkanes yields cyanides (nitriles) since $\ce{C-C}$ bond is thermodynamically more stable than $\ce{C-N}$ bond (simplest explanation).

Silver cyanide ($\ce{AgCN}$) is largely covalent and it does not dissociate easily to give $\ce{CN-}$ due to soft-soft interaction ($\ce{Ag+}$ is Soft Acid and $\ce{CN-}$ is Soft base). Thus, negative charge of $\ce{CN-}$ ion is mostly available at nitrogen, and hence the attacking centre of the nucleophile will be $\ce{N}$-atom. As a result, you get nucleophiluc substitution of haloalkanes to give isocyanides (isonitrile).

Same argument can be given for nitrite formation with $\ce{KNO2}$ and nitro formation with $\ce{AgNO2}$. In this case, $\ce{KNO2}$ is a predominantly ionic compound ($\ce{K+^-O-N=O}$) where one of oxygen atoms of nitrite ion has negative charge. Therefore, the nucleophilic attack of $\ce{NO2-}$ on haloalkane would result alkyl nitrite ($\ce{R-O-N=O}$). Silver nitrite ($\ce{AgNO2}$) is, on the other hand, largely covalent and would not dissociate easily to give free $\ce{NO2-}$ ions. As a results, the lone pair on nitrogen acts as an attacking site, forming nitroalkanes (high electronegativity of oxygen compared to nitrogen prevented it from donating its lone pair, as explainnation given by Clayden).


Best reagents state for air free chemistry. Choosing halide for grignard synthesis

The chloride can be a bit slow to react, particularly if the aryl halide is hindered. You might need a small amount of iodine to get the reaction started. (Maybe you ran into this already?)

Once you form the Grignard, they should be identical for most applications.

Personally, I would just go with iodide because the chloride is probably a bit less reactive, and frankly, weighing out substances that are almost solid is extremely annoying. Strongly prefer the actual solid.

The main reason I would elect the chloride over the iodide in this case is if the cost of the chloride were much lower, and even then only if the reactivity weren't seriously impeded by using the chloride.


As Tan Yong Boon stated, it is Bent’s Rule with which we can explain the lower bond angle of $\ce{NF3}$ when compared to $\ce{NH3}$. The rule as stated by Henry Bent:

  Atomic s character concentrates in orbitals directed towards electropositive substituents.

Fluorine is more electronegative that hydrogen, and the $\ce{N−F}$ bond would have greater p character than the $\ce{N−H}$ bond. And more s character leads to large bond angles. Thus, the bond angle is greater in $\ce{NH3}$ than in $\ce{NF3}$.
Now, consider $\ce{NCl3}$. Clearly, $\ce{Cl}$ atom islarger in size than the central atom, nitrogen. Hence the higher bond angle here is due to the steric crowding caused by $\ce{Cl}$ atoms.(More pronounced than the electrongativity of $\ce{Cl}$ atom). They repel eachother and hence bond angle increases. This is not the case with small $\ce{H}$ atoms around the nitrogen in $\ce{NH3}$. Therefore, the $\ce{Cl-N-Cl}$ bond angle of $\ce{NCl3}$ is greater than $\ce{H-N-H}$ bond angle of $\ce{NH3}$.


Do all the precipitates undergo an equilibrium with its respective ions, why can't they dissociate fully?

In the two examples you mentioned, we have an equilibrium between the solid and the small quantity of the solid that is soluble in water and completely dissociated:
$$\ce{Ca3(PO4)2_{(s)}<=> 3Ca^2+_{(aq)}  + 2 PO4^{3-}_{(aq)} }$$
$$\ce{AgCl_{(s)}<=> Ag^+(aq)_{(aq)}  + Cl^{-}_{(aq)} }$$

I think it's wrong to write a solid forward reaction.
It's an heterogeneous equilibrium between the solid state (the precipitate) and the aqueous state (ions constituents of the precipitate). The constant of this equilibrium is the solubility product $K_{\mathrm{sp}}$ of the precipitate.


Does the relationship equation between standard cell potential and equilibrium constant violate potential's intensive properties?

No, there is no violation of "intensivity".  The reason is that $K_{eq}$ depends on $n$, and changes in one cancel the other out.

For example, consider the electrolysis of water:

$$\ce{2H2O_{(l)} -> 2H2_{(g)} + O2_{(g)}}$$

The equilibrium constant for this reaction is $K_{1}=\frac{[\ce{H2}]^2 [\ce{O2}]}{1}$ and if you wrote out each electrochemical half reaction separately, $n$ would be 4.

Now consider this reaction:

$$\ce{4H2O_{(l)} -> 4H2_{(g)} + 2O2_{(g)}}$$

The equilibrium constant is now $K_{2}=\frac{[\ce{H2}]^4 [\ce{O2}]^2}{1}=(K_1)^2$.  If you wrote out the half-reactions for this reaction, $n$ would be 8, twice as big.  But the $\ln K_{eq}$ term would also be twice as big, since $\ln K_2 = \ln{(K_1)^2} = 2 \ln K_1$.


Does electrolysis of aqueous tin(II) chloride produce Cl₂, SnCl₄ or both?

Let's try to look at those electrons. C - cathode, A - anode.

$$
\begin{align}
&\mathrm{C(-):} &\ce{Sn^2+ + 2e- &→ Sn^0} \\
&\mathrm{A(+):} &\ce{2Cl^- &→ Cl2^0 + 2e-}
\end{align}
$$

So, your second reaction is right. But the first is wrong: $\ce{Sn}$ is reduced, but there are no oxidation. You should use

$$\ce{Sn + 2 Cl2 → SnCl4}$$

But I should mention that this reaction is for $\pu{115 °C}$, so don't worry, it won't go for your conditions.

Another way is

$$\ce{SnCl2 + Cl2 → SnCl4}$$

It requires a lot of work, bubbling $\ce{Cl2}$ through aqueous solution and distillation in the end, so you also shouldn't worry.

You can basically think that all you $\ce{Sn}$ that becomes a part of new structure goes into solid $\ce{Sn}$. It's acceptable since synthesis of $\ce{SnCl4}$ is not an easy thing.


Determining reaction order in equilibrium reactions

The difficulty with problems in kinetics is that you can envision a pool of intermediates such that, while B is consumed, you don't see significant formation of C, so that C is a poor reporter of the consumption rate of B. Concepts such as intermediates, autocatalysis, feedback, steady-states, lag times can be important here. Things can get complicated pretty quickly.

The initial forward rate, in the absence of a significant accumulation of reaction intermediates, can for simplicity be written as

$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_\mathrm{fwd}[\ce{A}]^a[\ce{B}]^b$$

Again, this assumes a negligible accumulation of intermediates whose reverse reaction may lead back to formation of B, as in a process under steady state.

For this to be useful in the determination of $k_\mathrm{f}$ it helps to satisfy two conditions:

$k_\mathrm{fwd}[\ce{A}]^a[\ce{B}]^b \gg k_\mathrm{rev}[intermediates]^m$
$[\ce{A}]^a\approx \text{const}$

These conditions are essential to guarantee a decay of the concentration of $\ce{B}$ with a rate constant $k_\mathrm{fwd}[\ce{A}]^a$. The answer to your second question is no unless the mechanism is simple and conditions 1 and 2 hold (small amount of product), that is, if the reverse reactions from intermediates or products are not important.

So, it is not essential that the measurements come from the beginning as long as the mechanism is elementary (no rate limiting steps leading to accumulation of intermediates), the reverse reaction is insignificant in rate and the concentration of A greatly exceeds that of B throughout your measurement time.


What you have to understand about resonance structures is that they do not actually exist in the molecule. The separate resonance structures are never formed on their own but they can be used to represent a delocalised bonding situation, more accurately described using molecular orbitals.

In the case of propenal the $\pi$ MOs result from the overlap of four $p$ orbitals on the four atoms in the chain. The two occupied $\pi$ MOs show how most of the electron density is between C1 and C2, and C3 and the oxygen but there is also significant electron density between C2 and C3. This reflects the different positions of the double bonds shown in the resonance contributors of propenal.

http://www.chemtube3d.com/orbitalsacrolein.htm


Determining equilibrium concentrations from initial conditions and equilibrium constant

The text

  4.0 = [x]2 so x = 0.85 [1.0 – x][2.0 – x]

appears to be an error. The proper equation is:

$$4  = \frac{x^2}{(2-x)(1-x)}$$

The 0.85 is one of two roots of the quadratic equation you'll solve, using

$$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$

after you simplify the first expression, above.

As for the rest of your question, please expand: What does

  the 3.00

mean?


I agree with M. Farooq, a test that might work is cutting small pieces of thermal paper and placing them in a tube with grain alcohol or a facsimile, and allowing them to soak for a few hours, then send the sample to a lab and ask them to test for bisphenol A/S. If you can, it's just better to ask the company.
Edit: you can just soak them in grain alcohol. It's hard to do a specific test for bisphenols, but I found one for phenols in general.
Procedure:

Prepare a dilute solution of ferric chloride in grain alcohol.
Using an eyedropper, add a few drops of the solution to a teaspoon of your solution to test.
If a color change is observed, phenols are present.

This is not ideal, but it's the best you can do.


How does light emission depend on temperature by fluorescence?

This phenomenon is thermoluminescence. The incoming photons excite electrons of the material to a higher level, a metastable state, which fall back to a lower state and emit photons. The metastable state may last for only nano- or picoseconds, in which case it is termed fluorescence. If the electrons need a bit of a "nudge" and hang around for a few seconds or hours, it is termed phosphorescence.

In the extreme case of some materials, the metastable state can last for millenia, which can be useful for dating the age of the item. For example, when pottery is fired, atoms in a metastable state relax. From then on, energy accumulates from cosmic rays and radioactive decay of nearby minerals such as thorium or potassium. By heating the ceramic and measuring the amount of light emitted, one can approximate the age of the pottery.

Ceramics need to be heated to a few hundred Celsius, but your glass works in just warm water. There are similar infrared detection cards with phosphors that are charged using visible or UV light, and that emit visible light when triggered by low-energy IR.


Are biological macromolecules organic or inorganic compounds?

The definition of "organic" is somewhat vague: it can mean containing carbon, or it can mean coming from living things, particularly as defined by the United States Department of Agriculture.

To further confuse the issue, some carbon compounds, such as sodium carbonate or tungsten carbide, are sometimes defined as inorganic.

So state a particular definition, and examine those compounds according to that definition.


XAS spectroscopy - Differences in magnitude of 2p SOC and 3d SOC?

The energy of a spin orbit level for a '1 electron' or 'hydrogenic' atom such as sodium is given by

$$E_{so}=\frac{AZ^4}{n^3(l(l+1)(l+1/2))}\frac{j(j+1-l(l+1)-s(s+1)}{2}$$

where $A$ is a collection of constants and $j,\,l,\,s$ quantum numbers derived from the term symbols for the levels involved.

The second fraction in the equation is the angular momentum term, the first due to $\langle1/r^3 \rangle$ where $r$ is the distance from the nucleus.

For a D electron you might have $j$ quantum numbers $j$=5/2 and 3/2 with $l$=2 and $s$=1/2 and for a P,  $j$=3/2 and 1/2 with $l$=1 and $s$=1/2 thus the separation of the D level pair is less than that of the P.

Most good textbooks will have a derivation of this equation and other corrections such as due to relativity etc.


Dissociation behaviour of polyelectrolytes

Do a thought experiment: Into your polycationic molecule, insert very large spacers in between the amino groups so that they are far away from each other in terms of covalent connection and in terms of spatial distance. Now, break a bond in the middle of each spacer. Do you think this will influence the protonation state of the amino groups?

If the $\mathrm{pH} = \mathrm{p}K_\mathrm{A}$ for your $100^2$ independent amino groups, half of them will be protonated and half deprotonated. If you randomly pick 100 amino groups and turn them into a molecule by repairing the spacers in our thought experiment, you get a distribution of charges (a Gaussian bell curve centered on 50).

If you lower your pH by one unit, about 91% of amino groups will be protonated and 9% deprotonated, and again there will be a distribution of charges for the polycationic molecule/ion.

If, instead of taking a snapshot at a given time, you ask about average protonation state over a couple of milliseconds, you will have a molecular (or ionic, I guess) charge of 50+ for pH 6.5, and about 91+ for pH 5.5.

If the amino groups are not separated by spacers, they will "talk to each other", like the carboxylate groups in citric acid, for example. For citric acid, three pKa values are given, 3.13, 4.76 and 6.40, even though each group has similar chemistry and two are identical. So in that case, things are more complicated. It also gets more complicated if you bundle species with identical charge when figuring out pKa values theoretically (for citrate, there is one neutral species and one 3- species, but three 1- and three 2- species).


Nature produces cis fats exclusively (someone may prove me wrong with some weird example.) This is why unsaturated fats such as vegetable oil have lower melting point than saturated fats of the same MW (the cis double bonds lead to a more compact molecule which gets less tangled with its neighbours.)

The reason nature produces cis bonds is due to the enzymatic mechanism in which these bonds are produced. Therefore this is down to reaction kinetics, and as often happens the product which is not the most thermodynamically stable of all possible products is produced. It just has to be more stable than the reactants.

When fat is artificially hydrogenated with a metal catalyst (for example nickel) to increase its viscosity/raise its melting point (to turn vegetable oil into margarine) some of the cis double bonds that escape being hydrogenated are able to change from cis to the more thermodynamically stable trans. These trans fats are unnatural and are therefore widely claimed to be unhealthy.


Anthocyanins as pH-indicators: stop the titration at red, purple or blue?

If you titrate some hydrogen bromide solution to some sodium hydroxide [...]

It's probably a bad idea to start with the anthocyane in strongly alkaline solution. Like other pyrylium compounds, flavylium cations do not tolerate (strongly) akaline conditions very well.

The nucleophilic opening of the pyrylium ring to a chalcone is irreversible under the conditions of the titration.


Why is NaBr only a reducing agent? Can you tell if something is a reducing/oxidizing/both agent?

In the presence of something reasonably oxidizing, it could form $\ce{Br2}$ and reduce whatever oxidizes it. It could also oxidize to $\ce{NaBrO_{x}}$.

Those other things are not going to easily oxidize because that would put a formal positive charge on a highly electronegative atom. In the case of $\ce{HIO4}$, it would increase the charge on iodine from $+7$. This would be very hard.


Are all fatty acids monocarboxylic acids?

You are asking two separate questions.

Not all fatty acids are typically found in fats and oils (common name for storage lipids, which is also somewhat a common name...)  Identifying fatty acids that have an additional charged tail does not inherently mean you'll find them in your fats and oils - fatty acids do have other roles. There may be exceptions, but they're not going to be common.

From what information you have given us, the larger question your teacher is driving you towards is to think about what the physical nature of fats and oils must be given their biophysical properties and how that correlates with their biological role, and secondly what the physical nature of fatty acids would be that drives those properties.

Considered macroscopically, your storage lipids will remain neutral, and pack together. If you added charge to your storage lipids in any significant amounts, you would turn them into something that behaves like a membrane lipid.


Vacuum distillation is employed when the boiling point of the substance is too high (400 Celsius for example). Under reduced pressure, the boiling point will be reduced to the range that is easier to handle.
Steam distillation take advantage of azeotropic effect, water steam is bubbled into your liquid (aromatic compound for example) and carry out the mixture of water vapor and your liquid vapor. This technique can also avoid handling the high boiling point distillation because the azeotropic boiling point is usually lower than pure liquid.


Why is Aluminium Chloride a Lewis Acid if it's an ionic compound?

if it's an ionic compound

Actually, it is debatable. $\ce{Al^{3+}}$ is a very small cation and its formation is strongly unfavored. In fact, this cation, just like many others, may exist only in solvates, like $\ce{[Al(OH2)6]^{3+}}$, or unchemical environment.

In particular case of anhydrous $\ce{AlCl3}$ the molecule is a strong Lewis acid, strong enough to coordinate with lone pair of another similar molecule, so it forms either dimer $\ce{Al2Cl6}$ with two chlorine bridges, donating one pair to both Al ions, or polymer, with Al atoms exhibiting octahedral coordination. Unlike many ionic compounds, melted $\ce{AlCl3}$ has poor conductivity.

This, however, does no apply to $\ce{AlCl3.6H2O}$ which is, again, a ionic compound $\ce{[Al(OH2)6]^{3+} (Cl^{-})3}$ and does not exhibit any Lewis acidity (though it does exhibit strong Brønsted acidity, i.e. tendency for hydrolysis)

This said, however, I have to clarify: ionicity of compound does not prevent particular ion from being Lewis acid/base. In fact, cations often are excellent Lewis acids. The only thing needed for a particle to be a Lewis acid is presence of low  vacant orbital, and sometimes presence of such orbital is not derivable from octet rule. For example, $\ce{SnCl4}$ is Lewis acid, though in it $\ce{Sn}$ has full octet. The orbital in question is describable in terms of MO theory which I strongly encourage to study.

The most often considered Lewis acid is hydrogen cation. In fact, this Lewis acid is so strong, that it never occurs free, it is always bound to Lewis base (again, except unchemical environment). Another common Lewis acid is $\ce{Na^+}$. This cation is quite weak Lewis acid, but it is sterically accessible, and usually is paired with anion, which is usually a Lewis base. This said, despite mostly ionic nature of $\ce{NaCl}$, there is considerable covalent bonding in it. (of course, on the other hand there is such anion as $\ce{BPh4^-}$ which is not a lewis base of considerable strength, so its salts are almost purely ionic in nature)


Is there a very stinky chemical solution that isn't a health hazard?

My vote would be for butyric acid and related short-chain fatty acids with between 3 and ~8 carbons (isobutyric acid, valeric acid, isovaleric acid, caproic acid, etc.)

In pure form these chemicals would be dangerous acids, but in dilute aqueous solution buffered at pH 5 or so, they are not very corrosive or toxic but are still pretty stinky.

Skatole could be another example.  See the Wikipedia article for malodorants for some more ideas.


How are the π bonds arranged in Osmium Tetroxide?

Your conception of $\pi$-bonds is a bit too restrictive. A $\pi$-bond is a bond in which there is one node along the internuclear axis. So, yes, two side-on $p$-orbitals do form a $\pi$ bond because there is one node along the internuclear axis (the one separating the bottom lobe from the top lobe). A $\sigma$-bond is a bond with no nodes (i.e. cylindrically symmetric), a $\delta$-bond has two nodes, etc. It's important to note that these symmetry labels work best in molecules where each bond axis either exactly lines up with or is exactly orthogonal to all the others (like diatomics and octahedral metal complexes) but we can work in other molecules with a little more effort.

Now, in osmium, the valence orbitals most involved in bonding are actually the $d$-orbitals. Unfortunately, because the molecule has tetrahedral symmetry and the bond axes are not all orthogonal or parallel, it's less easy to see how the oxygen and osmium orbitals come together to form $\pi$ bonds, and we need to resort to MO theory to get a qualitatively correct answer. If you don't know MO theory, the most basic way of approaching the following is to think of all the ligands and metal orbitals combining (much like the simple $p-p$ overlap in your question) and we label each combination by a label that has to do with the symmetries (rotations and reflections) of the combination of atomic orbitals.

Let's start with a qualitative MO diagram from this answer:

Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs. Orbital energies here are qualitative (and in fact the order is wrong here for $\ce{OsO4}$ which has occupied orbitals energies $t_1>t_2>a_1>t_2>e$).

The oxygen $p$ orbitals that point toward the metal and will make up the $\sigma$ bonds form the $a+t_2$ orbitals. The oxygen $p$ orbitals that make up the $\pi$ bonds are the $e+t_1+t_2$ orbitals. The osmium $d$ orbitals are $e+t_2$. Since only orbitals of the same symmetry can mix, the $e$ orbitals are purely $\pi$-bonding while the $t_2$ orbitals are a mix of $\pi$ and $\sigma$ bonding.

So your picture is sort of correct for the oxygens but they have to be combined in specific ways in order to overlap with the osmium. Here is a picture of one of the $\pi$ bonds in this molecule:

You can see the $p$ orbitals on each oxygen, the node along the bond axis and the osmium $d$ orbital at the center taking part.

Here is a purely $\sigma$ orbital:

In this case, the Os $s$ orbital participates with the oxygen $p$ orbitals pointing towards the center.

And here are two views of one of those weird $t_2$ orbitals, which is a mix of both $\sigma$ and $\pi$ in tetrahedral symmetry:

You can see that the oxygen $p$ orbitals clearly have a node near the O, but as you get towards the metal center the node bends off-axis. Meanwhile, near the metal, the nodes are not oriented near the bond axes at all. Very confusing!

In tetrahedral symmetry, some orbitals are clearly $\pi$-type, some are clearly $\sigma$-type, and some are a mix and therefore not a well-defined $\sigma$ or $\pi$.

EDIT: The careful reader will note that I ignored the $t_1$ orbitals. These are $p$ orbitals on the the oxygens that have no osmium orbitals to bond with and so are non-bonding. Here is a picture of one of these non bonding orbitals.

You can clearly see all the oxygen $p$ orbitals, but note that the osmium center has no orbital.


Most probably some hydrocarbon like butane = big flame. A quick way to check is if the can makes a sound when you shake it. If it does then it's most likely hydrocarbon. It would have been less confusing if it said "compressed gas".


The structure of 1-(butan-2-yl)-4-tert-butylbenzene is displayed below. What is -yl on this compound and when do I use it?

The "yl" refers to removing a hydrogen atom from a molecule making a univalent group.

Butane is

$\ce{CH3-CH2-CH2-CH3}$

To make butan-2-yl take a hydrogen atom off carbon 2:

$\ce{CH3-CH-CH2-CH3}$

Then butan-2-yl X would be

$\ce{CH3-CH(X)-CH2-CH3}$.


The accepted answer has nice pictures, but has a couple of small factual inaccuracies in the last paragraph. Here's a bit more maths to illustrate what's going on.
Atomic orbitals, which are one-electron wavefunctions, are split into two components: the radial and angular wavefunctions
$$\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_{lm}(\theta,\phi)$$
so-called because they only have radial ($r$) and angular ($\theta$,$\phi$) dependencies respectively. If either of these two components are zero, the total wavefunction is zero and the probability density there (given by $\psi^*\psi$) is also zero.
A radial node occurs when the radial wavefunction is equal to zero. Since the radial wavefunction only depends on $r$, this means that each radial node corresponds to one particular value of $r$. (The radial wavefunction may be zero when $r = 0$ or $r \to \infty$, but these are not counted as radial nodes.)
Analogously, an angular node is simply a region where the angular wavefunction is zero.* In the case of the p-orbitals, this is a plane. However, angular nodes do not necessarily have to be planes, as we'll see later.
The number of radial and angular nodes is dictated by the forms of the wavefunctions, which are derived by solving the Schrodinger equation. For a given orbital with quantum numbers $(n,l)$, there are $n-l-1$ radial nodes and $l$ angular nodes, as previously described.
An example
Let's take a look at one of the hydrogen 3d wavefunctions with $(n,l) = (3,2)$. $a_0$ is the Bohr radius of the hydrogen atom. We would expect $3-2-1 = 0$ radial nodes and $2$ angular nodes.
$$\mathrm{3d}_{z^2} = \underbrace{\left[\frac{4}{81\sqrt{30}}a_0^{-3/2}\left(\frac{r}{a_0}\right)^2 \exp{\left(-\frac{r}{3a_0}\right)}\right]}_{\text{radial: }R_{32}} \underbrace{\left[\sqrt{\frac{5}{16\pi}}(3\cos^2\theta - 1)\right]}_{\text{angular: }Y_{20}}$$
All the individual terms in the radial wavefunction can never be zero (excluding the cases of $r = 0$ or $r \to \infty$ as I described earlier). Therefore, this orbital has no radial nodes. Surprise, surprise.
The angular nodes are more interesting. The angular wavefunction vanishes when $3 \cos^2\theta - 1 = 0$. Since $\theta$ takes values between $0^\circ$ and $180^\circ$, this corresponds to the two solutions $\theta = 54.7^\circ, 125.3^\circ$. Both of these solutions are angular nodes. This is how they look like:

The dotted lines are the angular nodes. They are not planes, but rather cones. They correspond to one particular value of $\theta$, which in spherical coordinates is the angle made with the positive $z$-axis; I have marked these angles on the diagram.
If you can obtain the forms of the wavefunctions, then it is easy to find the radial nodes. Mark Winter at Sheffield has a great website for this; just click on the orbital you want on the left, then "Equations" near the top-right corner.

If we stick to complex atomic orbitals, which are simultaneous eigenfunctions of $H$, $L^2$ and $L_z$, then the dependency on $\phi$ is always of the form $e^{\pm im\phi}$, which can never be zero, so angular nodes never arise due to $\phi$-dependency. However, radial and angular nodes are most commonly discussed in the context of real atomic orbitals, obtained by linear combination of the spherical harmonics. These do have angular nodes that depend on both $\theta$ and $\phi$, but they are often much simpler to express in terms of Cartesian coordinates $(x,y,z)$ (obvious examples being the $\mathrm{2p}_x$ and $\mathrm{2p}_y$ orbitals). The $\mathrm{3d}_{z^2}$ orbital is an exception that was deliberately chose as an example because its angular nodes are not planes.


Why would the loss of H₂ to space mean that the Earth's geochemistry would be based on acid/base reactions?

The statement you cite contains a reference to “How Earth’s atmosphere evolved to an oxic state: A status report”, D. C. Catling and M. W. Claire, Earth and Planetary Science Letters 237 (2005) 1–20. This paper in turns give some more details on this assertion, which I'll summarize below.

It all turns around the question of how reducing the prebiotic atmosphere was. The paper I cited above contains this quote:

  before life existed the atmosphere contained mainly $\ce{N2}$, with negligible $\ce{O2}$, probably more $\ce{CO2}$ than today, and only minor levels of reducing gases such as $\ce{H2}$.

So, in the absence of widely available gases such as oxidative $\ce{O2}$ or reducing $\ce{H2}$, oxygen was mainly present as $\ce{H2O}$ and $\ce{CO2}$. This means redox reactions are not big players in the overall chemical scheme of the periodic atmosphere, and acide-base reactions dominate (by default).


Which one is more basic: acetamide or aniline?

You can consider this in the following way.

If amide undergoes protonation, it forms a positively charged nitrogen atom next to the strong electron-withdrawing group. This destabilises the molecule, so most probably it will not act as a base. Instead, it will act as a moderate acid with a strong base, since conjugate anion will be stabilized by the adjacent carbonyl group.

In the case of aniline, the phenyl group is not a strong withdrawing group. Still, the phenyl group withdraws electron density from the nitrogen atom through resonance. However, anilines can be protonated by strong acids. You can observe that during the formation of a diazonium salt, where aniline is protonated by strong acids like $\ce{HCl}$.


Wohler synthesis: magnesium cyanate + ammonium chloride?

The reason Wohler used silver cyanate is because silver chloride precipitates out, leaving ammonium cyanate, which (in a separate step) was thermally converted to urea. If you used magnesium cyanate, there would be no precipitate since magnesium chloride is very soluble in water. However, if you evaporated the water and heated the mixture, you probably would end up making some urea. But it would not be pure.


I see no special meaning there, aside of the common English. It means eventual metals present in the solution as side components or impurities, aside of calcium. Like alkali metals, magnesium, heavy metals as iron, manganese, copper atc.


Stability of diboron(B2) and dilithium(Li2)

Firstly, the use of molecular orbital diagrams allows you to predict the electronic configuration, if a molecule would exist or not (as you say, evaluating the bond order), and some magnetic properties. Moreover, if you have "the same molecule" as O$_2$ and its ion  O$_2^-$, etc, you can say the higher bond order, the bigger stability.

But if the bond order is the same (your case), you can not talk about stability using those diagrams.

Coming back to the question, looking the dissociation energy (the energy relative to separated atoms) for B$_2$, you will see it is bigger in module.
So it will be more stable.
I do not have a qualitative way to estimate which of those molecules is more stable.

The way to know it theoretically is using electronic structure programmes.


Why interconversion in tautomers is so rapid?

If you are using a protic solvent (the most common situation when you're dealing with polar functional groups), then you are "cheating" because the solvent can generate its own proton transfers, making the reaction look rapid.


Entropy change during irreversible process

The answers to your first and third questions are "yes."  Here is a tutorial on how to calculate the change in entropy of the system or the surroundings for an irreversible process:  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/. The reference also provides worked examples.  Note that, in considering entropy changes for both the system and the surroundings from an irreversible process, one needs to first separate the system from the surroundings and then subject each of them separately to a reversible path.  These separate reversible paths do not have to match.
It simplifies things a little to think about how entropy of a system can change.  There are only two ways:  entropy transfer across the boundary of the system by heat flow in and out at the boundary temperature, and entropy generation (due to irreversibility) within the system.  In a reversible process, only the first mechanism is present.  In both cases, the transfer across the boundary is $\int{dq/T_B}$ where $T_B$ is the boundary temperature.
Regarding questions 2 and 4 (involving ideal constant temperature reservoirs), these are somewhat limiting cases.  Here, if you imagine a very thin conductive slab between the two reservoirs (with high temperature at one face of the slab and low temperature at the other face), then for the irreversible process involving the transfer of heat between the reservoirs, all the entropy generation would take place within the intervening slab, while the reservoirs would each separately experience only entropy transfer with the slab.  In the limit where the slab becomes infinitesimally thin, we would obtain the system you have envisioned, but all the entropy would still have been generated at the interface, and each reservoir would only experience entropy exchange with the interface.


House-hold chemistry experiments for a high-school student

My favorite "textbook" on home-chemistry is the Golden Book of Chemistry Experiments (Robert Brent).

It's a neat little book, and since it was published so long ago (1960s) it's not really available as a (literal) book these days. But fear not, because the PDF version can be easily found online!

A reliable place to download it from is archives.org.

Small warning:

I wouldn't recommend this to young children because quite a few experiments/projects mentioned there make use of an open flame (that's pretty much the only reason...). So be sure to conduct them in an open space far from anything inflammable.

Maybe this is the book/resource you're looking for? ;)

(A spot of light reading)

However, as the gents in the comments section have pointed out, conducting home-experiments won't (necessarily) make you a "Chemist". But I highly recommend you experiment (safety is important too) as frequent as you can, because over time it can firmly root a scientific bent of mind in you.

Intelligence, inquisitiveness and experience (and empathy) are highly prized values in the scientific world.

Besides, as you experiment at home (while you are still a high-school student) you'll soon stumble upon an extremely important idea:

No real experiment can be perfectly replicated

And you can only come to truly appreciate this fact when you experiment yourself...a lot.

No matter what experiment you do, no matter how perfect your set-up or procedure is, every time you conduct the same experiment you will never get the exact same result. Right now this "idea" may not seem like much (as a high-schooler, you might initially find this unbelievable), but its implications are profound and far-reaching.

That the fact that replicating experiments perfectly is impossible will soon lead you to another piece of insight: Probablity & Statistics rule science.

(End of light reading session)

Stemming from @andselisk's idea (in the comments under your post), you might find the following Youtube channels interesting (I'll try to add more when I get the time):

Nurdrage   <--- Highly recommended
NileRed
Cody'sLab
Doug's lab

The next two are largely DIY channels, but some of their videos include projects that count as "Chemistry":

NightHawkinLight
The King of Random

 Thus armed with this knowledge, you shall sally forth and become a chemist (of sorts...)!


Is there a thermodynamic driving force for racemisation?

Thermodynamics is not usually helpful in understanding racemisation: think mechanisms and kinetics instead

The thing about enantiomers is that, from a thermodynamic point of view, they are the same so any process is not being driven by differences in the energy between the molecules.

What matters, if racemisation is to occur, is that there is some accessible pathway that allows interconversion. For example, in acidic or basic ethanol solution, (3R)-3-phenyl-2-butanone will racemise via the (achiral) enol form of the molecule. There is a small amount of the enol present which interconverts to the (chiral) molecule but without remembering which chiral molecule the enol was formed from. This will, ultimately, give a racemic mixture. If no such mechanism existed the molecule would not interconvert. What matters is that some such pathway exists.

The notorious medicine thalidomide is an interesting example. One enantiomer is a useful therapeutic, the other a dangerous teratogen. But making a pure enantiomer doesn't help as the two forms interconvert in the body (by a complicated mechanism).

These conversions are not driven by the different stabilities of the end products but by the inability of non-chiral intermediate molecules in an equilibrium with the chiral starting materials to remember the chirality of their progenitor. This, in effect, increases entropy. But that entropic force is only possible if there is an available reaction mechanism that passes through a non-chiral form allowing the increased disorder that results from a chiral starting material turning into a racemic mixture.


What are the reactions occurring during the titration of a carbon dioxide contaminated water sample with diluted sodium hydroxide solution?

Your thinking is in principle correct, let me just add a few more things, that may help you further.

What you are effectively titrating is carbonic acid, or more precise the hydrogen carbonate ion. Carbonic acid is not very stable in water, an most of the carbon dioxide stays the way it is, i.e. dissolved and interaction via van der Waals forces. The main equilibria are as follows (and derivations thereof):
$$\begin{align}
\ce{CO2 + 2H2O &<=> HCO3- + H3O+}\tag1\label{major}\\
\ce{HCO3- + H2O &<=> H2CO3 + OH-}.\tag2\label{minor}\\
\end{align}$$
In a contaminated water sample $\eqref{major}$ is by far the more predominant reaction.

When you now add sodium hydroxide solution you shift $\eqref{major}$ to the right side, as you are removing hydrogen carbonate ions via
$$\ce{HCO3- + OH- <=> CO3^2- + H2O}\tag3\label{conversion}$$

As you correctly identified, phenolphtalein is a weak acid by itself and the colour change happens via
$$\ce{H2In + 2OH- <=> In^2- + 2H2O}.\tag4\label{indicator}$$

Once you have added enough sodium hydroxide solution to completely shift the equilibrium $\eqref{conversion}$ to the right, you will start deprotonating the indicator molecule via $\eqref{indicator}$.

It is important to understand, that all these reactions happen at the same time and they keep on happening. This is the basic characteristic of a dynamic equilibrium. Since you are changing the composition of these equilibria the overall composition changes and the system adjusts, see Le Chatelier's principle.


Sublimation of Iodine from Iodine Povidone

A procedure is available on Erowid for this process.

  Make or buy a coldfinger (see below), boil your fluid to absolute dryness, without burning the dry restant, then hang your coldfinger in an appropriate Erlenmeyer, and heat gently the now dry restant. Make your coldfinger fat-free with acetone first (on the outside where the I2 will condense). The I2 will sublimate from the solid form into the gas form, without getting a fluid first (that's sublimation), and the gaseous I2 will resublime from the gaseous form into the solid form against the cold wall from the coldfinger. A little bit will of course also resublime against the upper part of the Erlenmeyer, but that you can also scrape off later, when all your 10% I2 hangs on the cold glass walls.


It seems that the method to circumvent this problem is indeed to use a protecting group to protect one end of the alkyl chain, which is the end to be converted to a bromide after the Wittig reaction.

As Waylander aptly mentioned, another problem which comes with using an alkyl dibromide is that there may be an intramolecular substitution reaction, in which the anionic carbon in the phosphonium ylide reacts with the other end of the alkyl dibromide.

To circumvent this problem, we could use an alkyl bromide with an alcohol functional group at the other end of the chain. This alcohol group can then be protected using various protecting groups, such as a tetrahydropyran ether or trimethylsilyl ether. After the Wittig reaction, the compound can then be deprotected using acidic hydrolysis. After which, the alcohol functional group can then be converted to a bromide using $\ce {PBr3}$. The desired compound is thus obtained.


The iodine atom in $\ce{ICl2}$ is what is termed hypercoordinated (or hypervalent). Basically, this means that the 3 atoms combine to produce 3 molecular orbitals (the 3-center 4-electron model). There is no need to invoke d-orbitals.

In neutral $\ce{ICl2}$ there are three electrons spread across the two $\ce{I-Cl}$ bondss. Placing those electrons in our molecular orbitals, 2 go into the bonding orbital and one into the nonbonding orbital.

In $\ce{ICl2^{+}}$ we have removed the one nonbonding electron.  As a consequence there will be less electron-electron repulsion and only the bonding molecular orbital is occupied, because of this we would expect shorter bond lengths in $\ce{ICl2^{+}}$ compared to $\ce{ICl2}$.

In $\ce{ICl2^{-}}$ we have added another electron and it will go into the nonbonding orbital.  This will increase the electron-electron repulsion in the molecule without increasing bonding, because of this we would expect longer bond lengths in $\ce{ICl2^{-}}$ compared to $\ce{ICl2}$.

These arguments tell us that we would have shorter bond lengths in $\ce{ICl2^{+}}$ compared to $\ce{ICl2^{-}}$.


Is the pressure in an HPLC system the same throughout?

I figured this out. The pressure displayed is between pump and column inlet. There's a pressure gradient from the column inlet to the outlet, so the pressure reaching the detector is significantly less than the readout.


Is there a compound that contains water as a subcomponent?

Off the top of my head: Coordination complexes that have water ligated to a (transition) metal atom.

See corresponding Wikipedia article
A familiar example would be (hydrated) copper sulfate $\ce{CuSO4.5H2O}$, which has 4 water molecules ligated to the copper atom. (The fifth water molecule mentioned in the chemical formula is associated copper sulfate via hydrogen bonds)

 The picture is sources from the Wikipedia article for copper sulfate, and the diagram is sourced from this Quora post

If you're new to the idea of complex ions, this chemguide article offers a good intro to the topic.


Mixing naphthalene, formaldehyde and sulfuric acid (Marquis reagent + naphthalene)

Marquis Reagent (1) is a spot test for alkaloids, first reported in 1896. The original agent was a mixture of 2 drops of 40% formaldehyde and 3 milliliters of concentrated sulfuric acid and was used to detect certain (e.g., opium) alkaloids and distinguish between them. The signature of the alkaloid is both the initial color produced and the sequence of color changes with time. Morphine reacts with Marquis' Reagent to give a purple to violet color. Two molecules of morphine and two of formaldehyde are proposed to condense in the presence of sulfuric acid to the dimeric product which is protonated to the oxoniumcarbenium salt.
The key reaction is protonation of formaldehyde, which attacks an aromatic ring, ortho to the activating phenol group, producing a hydroxymethyl derivative which is immediately protonated and dehydroxylated to give a benzylic cation which attacks another aromatic ring. And it happens again at ortho positions to close a new ring.

Now the ingredients in your old school days reaction (naphthalene, sulfuric acid and formaldehyde) are used to make what was a common concrete admixture. Kao Soap (2), who made Mighty 150 (“Naphthalenesulonic acid-formaldehyde condensate sodium salt”) offered it to my company. We were making a similar material for paint dispersion and promoted our material for concrete strengthening by water reduction.
The condensate is a polymer and is made by first sulfonating napthalene in the beta position with concentrated sulfuric acid. (The alpha sulfonate is the first product, but quickly rearranges to the more stable beta isomer.) Then formalin, 40%, is added, slowly. The formaldehyde protonates, attacks the alpha position of the (still activated) unsulfonated ring (the sulfonated ring is deactivated), forms a hydroxy methyl aryl group, is immediately protonated, dehydrated and attacks another naphthalene sulfonate in the alpha position of the unsulfonated ring.
$H_2SO_4$ + $H_2CO$ --> $H_2COH^+$ +$HSO_4^-$
$H_2COH^+$ + $HO_3SNap$ --> $HO_3SNapCH_2OH$ + $H^+$ -->
$HO_3SNapCH_2^+$
$HO_3SNapCH_2^+$ + $HO_3SNap$ --> $HO_3SNapCH_2NapSO_3H$
This methylene group is the first link of a polymer chain. The other alpha positions (one on each unsulfonated ring) are still open and will be attacked if there is enough formaldehyde. Controlling the amount of formaldehyde added is a method of controlling ultimate molecular weight of the polymer, which affects its dispersing power because of the ability of longer polymer chains to adsorb more strongly and repel other particles without deabsorption of the polymer.
Controlling the rate of formaldehyde addition is key to preventing runaway polymerization. The addition of formaldehyde at near room temperature lowers the temperature of the 500-1000 gallon batch; however, the batch temperature must be raised to cause polymerization, which is exothermic - sulfuric acid generates heat as it hydrates - and the batch viscosity is already very high, so the heat cannot be quickly removed.
There is a possibility of a runaway reaction in your mixture: the naphthalene is not even sulfonated, so both rings are activated (relative to benzene). If you did get 2-methoxynaphthalene as an intermediate, the sulfuric acid would protonate the methoxy oxygen, generating a benzylic cation, and water and heat of hydration, and a polymer with chains coming out of up to 4 alpha positions per naphthalene. Now the typical naphthalene sulfonate condensates (the acids and the sodium salts) are brown (light to dark), but I could imagine a polymer coming out with a black color.
This approach to 2-methoxynaphthalene seems destined to fail, because the protonation of 2-methoxynaphthalene is easier than  protonation of formaldehyde. But organic chemists frequently generate tars.
Ref.1 https://en.wikipedia.org/wiki/Marquis_reagent
Ref.2 https://chemical.kao.com/en/products/B0012344_en/?region=gl


I think you mean "wash resin 3 times with 10 vols of the extract buffer".

Vol means volume strength. It tells you how much of a certain element in milliliters will be RELEASED from a certain substance. In this case, your buffer is supposed to release 10 vol of something. That's all it means.

However, I am surprised about the usage of vol in this context. Volume strength is usually a term directly associated with hydrogen peroxide, and dealing with the release of a certain volume of oxygen.


I think steric effects may play a role in that chlorine atoms are smaller than bromine atoms. The bromine atomic radius is about 0.15 Angstroms larger than the chlorine atomic radius. The larger bond angle can be in part explained by the need to accommodate the larger atoms.


Do hydrogen peroxide stabilizers affect oxidizing ability

Shouldn't hinder your oxidation much, just keep in mind they will still likely be around when your reaction is finished.


What is the first ionization energy of the hydride ion?

You probably don't look at the "ionization energy" for hydride. Instead, you have the reverse reaction:

$\ce{H + e- -> H-}$

You'd measure the enthalpy change here as the electron affinity (EA) of the hydrogen atom. The Wikipedia data page says $72.8\ \mathrm{kJ}/\mathrm{mol}$.

You're basically asking for the opposite of that, so the enthalpy change has the same magnitude but opposite sign. Note that this is not an ionization energy because you're not creating an ion.


Spontaneous/Non-Spontaneous Reactions and Reversible Reactions

"Reversible reactions constitute a limiting case between spontaneous and non-spontaneous processes." So what does this mean?

To me this sentence is meaningless gobbledygook and should be ignored.  That is, this sentence is unhelpful and trying to understand it will just make you more confused and less able to master chemistry.  Here is a short summary version of the points I make in the "Advanced discussion" below:

All reactions can be viewed as reversible from a mathematical standpoint, if the reacting system is "big enough".  Reactions that are highly spontaneous in the forward direction mean that at equilibrium, the number of "reactant" molecules will be very very small (but not zero!).
Many times, "big enough" would mean astronomically big, such as bigger than the Earth.  In these cases, reactions can be regarded as "irreversible", although there is no strict, universally agreed upon boundary between "reversible" and "irreversible".
Whether a reaction is spontaneous isn't that much related to whether it is "reversible".  Don't worry about understanding the connection between these concepts until you have understood each concept independently.

Advanced discussion.

The meaning of spontaneous

Spontaneous processes have negative Gibbs free energy changes ($dG < 0$) and also result in positive entropy changes in the universe ($dS_{univ} > 0$).  However, $dG$ is a function of state.
For chemical reactions, this means that $dG$ depends on the concentration of the reactants and products, i.e.

$$dG = dG^\circ + RT \ln Q$$

where $Q$ is the reaction quotient.

Thus, if we start off with all "product" and no reactant, then the "reverse" direction of the reaction will be spontaneous.  But if we start off with all reactants and no product, then the forward direction is spontaneous.

The meaning of reversible

The short answer is that for chemical reactions, "reversible" does not have a precise thermodynamic definition.

$$dG = dG^\circ + RT \ln Q = -RT \ln K + RT \ln Q = RT \ln \frac{Q}{K}$$

where $K$ is the equilibrium constant.  People will usually call reactions with a "large" equilibrium constant irreversible and those with a "small" equilibrium constant reversible.  But the exact definition of "large" and "small" isn't well-defined.  Many people might say $\frac{Q}{K} \leq .01$ represents an "irreversible" reaction.  For chemical reactions with very large equilibrium constants, the size of the system required to ensure that an equilibrium can be reached can get very large, as explained here.


What is diamagnetic in diamagnetic dilution?

The term diamagnetic dilution (in general) implies that the material used for dilution has all its electrons paired, so we will not see its response in an electron spin resonance experiment or any other experiment which can sense unpaired spins.


Why is a C–D bond stronger than a C–H bond?

Physics is better able to answer "how" questions than "why" questions, but here goes. The quantum mechanical description of the C–D system versus the C–H system gives the former a lower zero-point energy, which is the minimum energy the quantum system can attain. A good conceptual model of this is to consider the C–X system as two masses connected by a spring. When X = D, the system will vibrate more slowly than when X = H simply due to mass.

Since energy is proportional to the frequency of vibration, the energy of the C–D system is less (lower). This table lists the C–D bond dissociation energy as $\pu{341.4 kJ/mol}$ and C–H as $\pu{338 kJ/mol}$. Since this is the energy to break the bond, the C–D bond is stronger.

As far as your question about hyperconjugation; I'm surprised that C–H would have greater hyperconjugation than C–D, but it is not something that I know anything about. I guess I could speculate and wave my hands around and say that since the C–D bond is stronger, it keeps the electrons "closer" and so more confined to the sigma orbital, but that is just hot air.

Here's something which claims that the reason D does less hyperconjugation is because it vibrates less (see slide 15). So, it seems (if this is correct) that it's the smaller bond deformations that reduce the hyperconjugation for deuterated systems. Here is a pretty clear explanation of the kinetic isotope effect.


Half-equations for the combustion of sulfur dioxide

Orthocresol already sufficiently pointed out, that half-reactions are a book-keeping tool for electrons. When talking about electrolysis there might be coincidental overlap between half-reactions and the overall mechanism, but as the wording is supposed to indicate, this is not necessarily true.
Many reactions look simple at first glance, but on the second look they appear as highly complex systems of different equilibria. Actually the most simple looking reactions are often the ones that are painstakingly complex.

  I actually mean the real half equations that express how combustion is made in reality. I thought that we don't need water and acid to perform it. Am I mistaking?

Since you have been asking about "the real half-reactions" I thought I give a little more insight into the mechanism.

I found a fairly old publication on the mechanisms of the formation of sulfur oxides during combustion.[1]
For the purpose of illustrating the complexity this shall be enough, but I am fairly certain there are more modern approaches and kinetic and mechanistic studies available today.

The reaction itself proceeds only at moderate temperatures (400-600 °C) and is slightly exothermic.
$$\ce{SO2 + 1/2O2 <=> SO3 (g) + $99.0$~kJ}$$
Because the reaction proceeds quite slowly, catalysts are necessary. Here comes into play what you find in Orthocresol's answer: you need a species that provides you with oxygen(-2) ions. In praxis you use $\ce{V2O5}$ catalysts, e.g. in the synthesis of sulfuric acid.[2]
Sulfurtrioxide is very hydroscopic, which indicates, that the reactions that you proposed cannot happen, since it would hydrolyse immediately in the presence of water.
$$\ce{SO3 + H2O -> H2SO4(l)}$$

Sulfurtrioxide has many modifications, most commonly it exists as $\ce{(SO3)3}$ units in solid phase. Condensation is also one of the driving forces of this reaction, in the gas phase the equilibrium between $\ce{SO2}$, $\ce{O2}$, and $\ce{SO3}$ would not be entirely on the product side.
There is also a variety of different other sulfuroxides that are possibly involved in this complex equilibrium. This is due to the fact, that sulfurtrioxide is a powerful oxidiser.

Let's have a glance at a few reactions that are involved in the formation. The reaction itself most likely proceeds via a radical pathway in the gas phase, with $M$ being a catalyst of some sort (it can be the wall or some inert molecule):
$$\begin{align}
\ce{SO2 + O2 &<=> SO3 + O}\\
\ce{SO2 + O + $M$ &<=> SO3 + $M$}
\end{align}$$

One important reaction that disturbs the equilibrium is
$$\ce{SO_{$n$} + O <=> SO_{$n-1$} + O2}.$$

I also found a publication about the mechanism in the troposphere and things get even more complicated there.[3] The reason for that is dilution, a reasonable amount of molecules in excited states, and various other species involved. This topic became interesting because of air pollution and acid rain.
I am only giving a few examples and leaving out a lot of details, as stated previously, this is for illustrative purposes.
$$\begin{align}
\ce{SO2 + O2 &<=> SO4}\\
\ce{SO4 + O2 &<=> SO3 + O3}\\
\ce{SO2 + SO2 &<=> SO3 + SO}\\
\ce{SO2 + O3 &<=> SO3 + O2}\\
&\text{etc.}
\end{align}$$

As soon as you consider species that are usually available in the troposphere, e.g. $\ce{N2}$, $\ce{NO}$, $\ce{H2}$, $\ce{CO}$, $\ce{CO2}$, etc, things get really complicated.

Conclusion

You cannot write half reaction equations for this equilibrium. Half reactions in general do not represent the reaction mechanism; they are a just a tool to generate overall reaction equations.
The mechanism of the reaction is significantly more complex than only two half-reactions.
You are not mistaken, you don't need water to perform the combustion, however, you need some sort of catalyst, that is able to absorb the excess bonding energy.

References

Arthur Levy, Earl L. Merryman, William Thomas Reid, Environ. Sci. Technol., 1970, 4 (8), 653–662.
Arnold F. Holleman, Nils Wiberg, Egon Wiberg: Lehrbuch der Anorganischen Chemie. 102. Auflage. De Gruyter: 2008. (in German)
English version: Nils Wiberg, A. F. Holleman, Egon Wiberg  (Eds.): Holleman-Wiberg's Inorganic Chemistry. 1st Edition. Academic press: 2001. (Amazon, Google books)
Jack G. Calvert, Fu Su, Jan W. Bottenheim, Otto P. Strausz, Atmospheric Environ., 1978, 12 (1-3), 197-226.


Assigning absolute configuration when determining crystal structure isn't possible

This question is possibly a little broad- there are entire books dedicated to methods of structure determination and assignment of stereochemistry.

The tl;dr answer is that there is no single method by which absolute stereochemisty can be assigned unambiguously, it depends on the nature of the natural product (terpene, polyketide, polypeptide, etc) and the exact structural features.

Some common methods which are often found in isolation papers:

1. Degradation

Historically, one of the most common ways of identifying new natural products was to attempt to degrade them to known compounds. If you imagine taking a new peptide natural product (this is actually a bad example as most peptides are made up of natural amino acids), you could easily hydrolyse it back to the constituent amino-acids (or smaller peptide chains) for which rotational data is available.

2. Derivatisation to make crystals

If your material doesn't crystallise, make small structural modifications to make it crystalline.

Being able to do this somewhat depends on knowing the 2D structure (by NMR). Once you identify the functional groups present, you can begin to modify the structure in order to obtain something crystalline (to do X-ray analysis on).

There are too many possibilities to go through in detail, but you could, for instance, take a carboxylic acid and form an ester, or some kind of salt.

This method is often carried out in the pharmaceutical industry, where they want their active pharmaceuticals to be in the solid state in a known form (different forms have different properties, such as solubility, so great lengths are taken to obtain uniform material).

3. Derivatisation for NMR studies

If the molecule of interest has an alcohol present, its possible to form esters with a special kind of carboxylic acid, this is the Mosher ester assignment.

The alcohol of interest is reaction with two enantiomeric acids (the Mosher acids), to form a pair of diasteromers. The Mosher esters shield the molecule with the phenyl group, causing shifts to change on the nearby protons around the chiral alcohol. By considering the relative changes, the assignment may be assigned.  This method is often found in structure elucidation papers, however there are some instances where it can give misleading results, so shouldn't really ever be used on its own without any other evidence.

It's worth noting that the Mosher esters themselves are often crystalline, allowing a crystal structure to be obtained in addition to the assignment by the NMR analysis discussed above.

4. Computational work

Computation is one of the newer methods of assigning absolute stereochemistry, so new in-fact that the FDA (the people who approve drugs) don't accept computational assignment on its own, unless accompanied by some other, more traditional method, of assigning configuration.

In theory, there are methods to predict optical rotation, or other properties such as circular dichroism, which can then be compared to the real experimental values. In practice, this rarely works well for complex molecules as there are too many variables involved which effect the computational result.


Calculate molar mass of a compound from a given molecular mass

To get the molar mass of your mystery compound, make use of the fact that there are $N_{\rm A}$ (Avogadro's number) of molecules of it in one mole of it.

To solve the problem, we multiply the molecular mass (given) by Avogadro's number (known) to get the mass of one mole of the mystery compound:

$${\rm molar \;mass} = N_{\rm A}\times 2.19\times 10^{-22}\,{\rm g} = (6.022\times 10^{23}\,{\rm mol}^{-1}) \times (2.19\times 10^{-22}\,{\rm g}) = 131.88\;{{\rm g}\over{\rm mol}}$$


How does one determine whether a chemical requires hazardous waste disposal?

You may find information from the US Federal government relating to this questions at the following sites:
https://www.epa.gov/hwgenerators/steps-complying-regulations-hazardous-waste
https://www.epa.gov/rcra/resource-conservation-and-recovery-act-rcra-regulations

Furthermore, your organization (theoretically) has a local waste discharge permit that it must adhere to as well. If chemical waste is not common where you work, you or the responsible party should review the permit requirements and the operations taking place.


How many electrons can each shell in an atom hold?

$$\text{Electron Capacity for a shell} = 2n^2$$

where, n represents the Principal Quantum Number, a.k.a. energy level.

Keep in mind that an energy level need not be completely filled before electrons begin to fill the next level because beyond the fourth energy level, you would get into theoretical subshells.

See this.


Why are excess nuclei required to produce an NMR signal?

"A signal is produced when a radio frequency causes one state (could be either α or β) to invert"

This isn't correct. It really isn't, even though it's often incorrectly described that way. The reality is more complicated, and it involves understanding QM as well as the fact that single spins do not necessarily exist in $\alpha$ or $\beta$ states (any superposition of them is allowed). See my answer here for more details.

Without going into too much detail, population differences* can be used to generate coherences, which are responsible for generating the detected signal. If we construct a generalised wavefunction, with two different coefficients of the two states, viz.

$$|\psi\rangle = c_\alpha|\alpha\rangle + c_\beta |\beta\rangle$$

then coherences can be (loosely) understood as an oscillation of the coefficients $c_\alpha$ and $c_\beta$. For example we could have

$$\begin{align}
c_\alpha &= \cos(\omega t + \phi) \\
c_\beta &= \sin(\omega t + \phi)
\end{align}$$

which mean that the coefficients, and hence the wavefunction $|\psi\rangle$, is oscillating with an associated frequency $\omega$ and a phase $\phi$. This generates a signal in the final spectrum at the frequency $\omega$. This sort of oscillation has nothing to do with individual spins flipping from $|\alpha\rangle$ to $|\beta\rangle$. Indeed, in NMR, individual spins have no meaning at all. Only the bulk behaviour of the $\sim10^{20}$ spins in your sample matters.

How does this relate to population differences? For a coherence to be generated, there is an additional requirement that across the entire ensemble of spins, there is a correlation between spins in the phase $\phi$.

A decent analogy would be having a collection of ($\sim10^{20}$) analog clocks on a wall. Each of these is oscillating at a rate of one revolution per 12 hours, since each clock returns to the same time every 12 hours. If you started all the clocks at the same time, then they would (largely) remain in sync with each other, and you could have a pretty good idea of what time it is right now. On the other hand, if they all started out of sync with each other, then you would have zero hope of telling what time it was.

This is the real key to understanding why there is a need for population differences. Without a population difference, you can generate oscillating spins, but each of them has a completely random phase and so there is no correlation between them. Consequently, the net signal sums to zero.

I don't want to go deeper because it will get very, very long, and also because it has been covered in a way that is better than I can ever do. See: James Keeler's book at Resources for learning Chemistry.

* As I've said before, populations should not be interpreted as the actual number of particles in a specific eigenstate, but rather the ensemble average of the coefficient corresponding to the specific eigenstate in the wavefunction. In other words, if each spin has a state $|\psi\rangle = c_\alpha|\alpha\rangle + c_\beta|\beta\rangle$, then we have $n_\alpha = \overline{c_\alpha^* c_\alpha}$ where the bar indicates an average over all spins in the ensemble.


Difference in calculated pH and the real pH of a phosphate buffer?

The flaw in your calculations is the assumption that there exists a pK that is a constant independent of concentration.

In reality, pK is a function of ionic strength.

At infinite dilution (zero ionic strength) the pK is 7.199 as you assumed.

At 0.1M phosphate concentration, the pK is 6.81.

See Phosphate Buffers which cites to A. A. Green, "The Preparation of Acetate and Phosphate Buffer Solutions of Known PH and Ionic Strength", J. Am. Chem. Soc. 55, 2331, (1933).


It is not only the diene itself which is stabilised by resonance, but on electrophilic addition the carbocation formed is also stabilised by resonance if the diene is conjugated. Thus, forming the intermediate for a conjugated diene is easier than forming the intermediate for the alkene, even though the starting compound itself is more stable in case of conjugated dienes and therefore conjugated dienes react faster.


For complex or non stoichiometric ( many metal oxides ) compounds, it makes sense to calculate the average oxidation number that can be a rational number.
This can be useful if we are to enumerate chemical reactions with non-stoichiometric metal oxides, or if just a summary elemental composition is known, like hydrocarbon mixtures.
Oxidation numbers of particular atoms are integer numbers, with exceptions when   extra electrons ( or theirs deficit ) are shared among different number of atoms. The particular example of this are superoxides $\ce{M^{I}O2}$.


How one atom interact with another atom is an difficult question and requires quite some calculations, even for a system with only two hydrogen atoms.  Basically, this is done by solving the Schrödinger equation for the system.  The answer will also depend on the electronics state of the system, especially the spin state.

If the system is in a triplet electronic state, the two atom will experience a repulsive force that approach infinity as you pull the two atoms closer, and that force approaches zero as the distance approaches infinity.

If the system is in a singlet state, the force is zero at the equilibrium bond length of $H_2$ atom(~0.74Å).  As the distance gets smaller the atoms experience an increasing repulsive force which approaches infinity at 0.  As the two atoms are pulled apart from the equilibrium there is a decreasing attractive force that approaches 0 at infinity.


Limitations of extremes for the exchange-correlation functional - DFT

One way to begin to gain some insight in this is to consider the work done on benchmarking density functionals. I shamelessly plug the work done by my old group, which can be found here. The GMTKN30 data include results for Hartree-Fock and LDA functionals with large basis sets. The description of the several benchmark sets involved and the analysis in the actual paper will illustrate many of the shortcomings and strengths of these approaches (and why better, more pragmatical DFT approaches are more widely used). While I have no access to the GMTKN55 paper, I trust that it contains references to other benchmark papers that may actually deal with crystals.


Can the pyrophoric tendencies of a pyrophoric substance be suppressed in a pressurized environment?

Depending on the pyrophoric substance, there can be differeing effects from changing air pressure. Strange as it may seem, sometimes combustion is slowed at very high pressure. For example, ethane, which reacts explosively with oxygen at STP, oxidizes slowly in a series of steps at ~100 atmospheres (~107 Pascals).

That said, it depends on the substances involved, pressure and presence of catalysts or other substances. Aluminum in a ball mill produces pyrophoric dust, but adding graphite during milling suppresses ignition.


Reaction between a carbene and n-pentane

n-Hexane is one of the 3 products formed. Carbenes are high-energy, reactive intermediates.  Due to their extreme reactivity they usually react indiscriminately. In the case of methylene (the carbene in your diagram), it will insert into each of the different $\ce{C-H}$ bonds that are present in n-pentane.  Because of its reactivity, the 3 products will be formed in a statistical ratio that is roughly equivalent to the number of different hydrogens present in the starting n-pentane, 6:4:2.

If you'd like to learn more about carbenes, for example

singlet and triplet carbenes
their addition to carbon-carbon double bonds

then take a look at these earlier answers - ref_1, ref_2


What is the canonical reference for studying stereochemistry?

The classical book on stereochemistry is Eliel: Stereochemistry of Organic Compounds, first edition appeared 1962.

You will find overwhelming amount of information there. I would recommend it, regardless your age and level. It is so well written, that it is worth at least of fast reading over most important parts.
You should find it in any reasonable library, no need to get the latest edition, even the older ones should cover what you need.


What are the differences between diesel and unleaded gasoline?

Diesel engines vs. gasoline engines

Diesel engines don't rely on spark plugs, but they still work by igniting the fuel to generate a force that moves a cylinder in the engine.  The air/fuel mixture in a diesel engine is ignited by the compression of the same cylinder during the "compression" stroke.  That is, like most gasoline engines in cars, diesel engines are four-stroke engines.

The compression stroke, in a gasoline engine or in a diesel engine, causes heating of the cylinder contents as the compression happens.  Diesel engines are designed to have a more severe compression of the cylinder contents during the compression stroke.  This results in much higher heating of the contents; in fact the temperature becomes so high that the ignition of the fuel and air becomes spontaneous.

You may ask, where does the energy to power the compression stroke come from?  It comes from inertia in the engine as it is running.  Another way to say it is that the previous power stroke provides the energy for the compression stroke.  (The power stroke is when the exploding air fuel mixture pushes the cylinder out, doing useful work.)  There is so much work released in the power stroke that there is enough to move the car forward as well as provide inertia for the next compression stroke.

Here are some more web links with helpful explanations:

Wikipedia:

  The diesel engine (also known as a compression-ignition or 'CI' engine) is an internal combustion engine in which ignition of the fuel that has been injected into the combustion chamber is initiated by the high temperature which a gas achieves when greatly compressed (adiabatic compression). This contrasts with spark-ignition engines such as a petrol engine (gasoline engine) or gas engine (using a gaseous fuel as opposed to gasoline), which use a spark plug to ignite an air-fuel mixture.
  The diesel engine has the highest thermal efficiency of any standard internal or external combustion engine due to its very high compression ratio and inherent lean burn which enables heat dissipation by the excess air.

How stuff works:

  In theory, diesel engines and gasoline engines are quite similar. They are both internal combustion engines designed to convert the chemical energy available in fuel into mechanical energy. This mechanical energy moves pistons up and down inside cylinders. The pistons are connected to a crankshaft, and the up-and-down motion of the pistons, known as linear motion, creates the rotary motion needed to turn the wheels of a car forward.
  Both diesel engines and gasoline engines covert fuel into energy through a series of small explosions or combustions. The major difference between diesel and gasoline is the way these explosions happen. In a gasoline engine, fuel is mixed with air, compressed by pistons and ignited by sparks from spark plugs. In a diesel engine, however, the air is compressed first, and then the fuel is injected. Because air heats up when it's compressed, the fuel ignites.

Be sure to watch the animation at How Stuff Works.  It's quite helpful!

Diesel fuel vs. gasoline

In diesel engines, premature combustion can be a problem.  This happens when the fuel ignites "too early" in the compression phase, before maximum temperature and thus maximum efficiency can be obtained.  This means that by design, diesel fuel is a bit harder to ignite than gasoline.  It's made of slightly less volatile hydrocarbons.

As noted by @steveverrill in the comments, the above paragraph was incorrect.  A better explanation (straight from his comment) is

  Diesel is harder to ignite at ambient temp due to its lower volatility / higher flash point (approx 52 °C vs -43 °C). This means it's difficult to form an explosive mixture which could be ignited by a spark (which is a local area at much higher than ambient temp.) But inside an engine both diesel & gasoline are fully evaporated. Under these conditions diesel is easier to ignite due to its lower autoignition temperature (approx 265 °C vs 280 °C). Autoignition temperature is tuned by additives to help ensure diesel can be ignited just by compression, whereas gasoline cannot ignite without a spark.

Wikipedia again

  Petroleum-derived diesel is composed of about 75% saturated hydrocarbons (primarily paraffins including $n$, $iso$, and cycloparaffins), and 25% aromatic hydrocarbons (including naphthalenes and alkylbenzenes). The average chemical formula for common diesel fuel is $\ce{C12H23}$, ranging approximately from $\ce{C10H20}$ to $\ce{C15H28}$.

Wikipedia on gasoline

  The bulk of a typical gasoline consists of hydrocarbons with between 4 and 12 carbon atoms per molecule (commonly referred to as C4-C12). It is a mixture of paraffins (alkanes), cycloalkanes (naphthenes), and olefins (alkenes).

Thus, diesel fuel contains mostly hydrocarbon molecules that have on average 12 or so carbon atoms, while gasoline contains hydrocarbons that contain on average 8 or so carbon atoms.  The heavier, bigger molecules in diesel fuel means it evaporates more slowly at ambient conditions, requires more heating (i.e. more compression) before it spontaneously ignites.


Oxidation state of metal ion in unit cell

Edge positions are those at the centre of an edge of the cube; i.e. in the middle between two closest corner positions. As such, there are 12 edge positions and each belongs to 4 unit cells giving us 3 metal ions.

The way the nitride positions are described is basically a verbose way of saying cubic close-packed or face-centred cubic. Therefore, we can redescribe the cation positions as ‘all octahedral voids except for the one in the centre of the unit cell’. Since there are 4 atoms per unit cell in ccp and 4 octahedral voids, that again gives us a $3:4$ ratio and thus $\ce{M3^{IV}N4}$.


If chemical bonding is quantum mechanic, is saying they bend and vibrate like springs wrong?

That's a good question, but remember, an analogy is just that: comparing dissimilar things on the basis of structure or function. Though there may not be a physical similarity, the analogy is effective in describing what happens, and can even be used to calculate outcomes, within limits.

In a similar way, the concept of a mass and a spring are used to describe an electrical R-L-C tuned circuit. Both have resonant frequencies and other properties that can be calculated from the physical properties of the components.

One of the difficulties in describing quantum phenomena is that until the physical properties are measured, they may be indeterminate. An electron is not in a specific place, but has a probability of being detected in a specific place. So though the analogy is not physically comparable, it helps understanding of the chemical bond, and can even produce useful results, such as IR spectra, and its use is valid.

This begs the question, though; perhaps there is another analogy that you might find more useful... Any suggestions from others are welcome!


As Ivan explained, a glass is by definition amorphous. However, you can take a quartz crystal (silicon dioxide, which is the main ingredient of glass) if it is of high purity and optical quality, and machine it into a transparent object such as a window pane.


In general, malleability is determined by how easily metal crystals can be moved around. Anything that gets in the way, such as another substance blocking crystal movement, makes the metal less malleable.

For example, pure iron is malleable, but add some carbon to molten iron, and the carbon precipitates out on cooling, blocking crystal movement and creating much harder and more brittle carbon steel. Think of it as a two-phase material.


Does 3,6-di-X-cyclohexa-1,4-diene have stereo centers?

Good point. The (c) compound has two tetrahedral stereogenic carbon atoms, with four “substituents”

$\ce{-H}$,
$\ce{-X}$,
$\ce{-CH=CH-\overset{\text{*}}{C}HX-C=C-\bond{...}}$,
and the same one, $\ce{-CH=CH-\overset{\text{*}}{C}HX-C=C-\bond{...}}$ but from the other side, thus with the opposite configuration, or its mirror image.

(‘*’ denotes (the other) stereogenic carbon atom)

They are called pseudo-asymmetric carbon atoms.

A different relative configuration at these two pseudochiral centers leads to these two ‘cis’ and ‘trans’ isomers.

The definition of stereogenic center (for more complete one, see e.g. this answer) is fulfilled.

(Maybe this topic is too early, for this textbook stage, and the author did not notice this complication.)


What is the purpose of the Carothers equation?

It is an equation useful for describing the behaviour of a large system, not predicting the fate of a few monomers.
Polymerization is a stochastic process—if we only have 80 monomers, the resulting polymer molecules will vary wildly in length if we perform the experiment several times (and logically, could not ever have an average degree of polymerization of 100), and the equation could not provide much insight into the results.
However, 80 monomers is a very tiny quantity.
If we take one of the monomers used to make PET, terephthalic acid, at 166.13 g/mol works out to around $2\times 10^{-23}\ \textrm{kg}$ for 80 molecules, a quantity that is almost impossible to actually measure out so it doesn't really matter that the equation doesn't work.
At a more practical scale, with even 1 g of the monomer, we have ~$10^{21}$ monomers and we will see the average behaviour of the system that the equation describes.


Enzymes always catalyze reactions in both directions (all catalysts do). When an enzyme is first discovered and named, it is not always clear which direction is physiologically relevant (and sometimes both are).

There is a nice answer on StackExchange Biology talking about the enzyme pyruvate kinase, which acts as a phosphatase (i.e. in the opposite direction than the name suggests) in glycolysis. In the lab, you can study the reaction in both directions, depending on the concentrations of species present.

For the class of enzymes you are talking about, the same idea applies. A lyase can work in the other direction, as a synthase, and vice versa. If you want to classify enzymes in terms of mechanism and structure, there is no reason to separate the lyases from the synthases. They are expected to be highly similar or identical (if we are just using two names for the same enzyme).


How does mixing lemon juice in hooka water affect the composition of the mixture?

The acidic water is probably protonating nicotine, causing it to carry a positive charge. Once charged, the resulting nicotinium cation will not be volatile, but rather, highly soluble. Based on this, I would expect that acidifying the water causes a much lower amount of nicotine  to be pulled through the exit hose (because it is trapped in the acidic water).


Why Calcium Oxalate is insoluble in acetic acid?

Calcium oxide, hydroxide and carbonate react with acetic acid to form calcium acetate. Calcium oxalate does not react.

Edit: In real life, you'll of course will end up with a little oxalic acid and calcium acetate, as calcium oxalate is very slightly soluble and then you'll mix all available ions.


What is the most convenient way to prepare ferrous oxide (FeO) in the laboratory?

In appears that the synthetic procedure for preparation of iron(II) oxide from iron(II) oxalate has been first described by Günther et al. [1] and subsequently summarized in Brauer's Handbook of Preparative Inorganic Chemistry [2, p. 1497]:

  Iron (II) Oxide

  I.

  $$\ce{\underset{143.8}{FeC2O4} = \underset{71.8}{FeO} + \underset{28.0}{CO} + \underset{44.0}{CO2}}$$

  Thermal decomposition of $\ce{FeC2O4}$ yields pure $\ce{FeO}$ only under specific conditions.
  The decomposition is carried out in a quartz vessel (Fig. 332) whose lower section is kept at $\pu{850 °C}$ by means of an electric furnace.
  The joint is surrounded by a water-cooled lead coil or a rubber hose.
  The nascent gases should be removed as quickly as possible; for this reason, the reactor is connected to two parallel mercury pumps and a good forepump; the gas is carried into two liquid-nitrogen-cooled traps containing activated charcoal.

  Fig. 332. Preparation of iron (II) oxide.

  The starting $\ce{FeC2O4}$ $(\pu{0.5 - 0.8 g.})$ is placed in the small bulb above the quartz vessel, and the water of crystallization is completely vaporized by heating in vacuum for 12 hours at $\pu{200 °C}.$
  The bulb is turned in the joint, and the $\ce{FeC2O4}$ drops into the heated lower section of the reactor where it is rapidly decomposed to $\ce{FeO},$ $\ce{CO}$ and $\ce{CO2}$ (the decomposition is complete in about 20 seconds). The product $\ce{FeO}$ is retained by a quartz wool plug, which must be loose enough to prevent a buildup of pressure during the decomposition.

  The furnace is now removed and the hot quartz tube is chilled as rapidly as possible in cold water, since $\ce{FeO}$ is unstable in the range of $\pu{300-560 °C}$ and decomposes according to:

  $$\ce{4 FeO = Fe3O4 + Fe}$$

  (this decomposition proceeds most rapidly at about $\pu{480 °C},$ but
  ceases below $\pu{300 °C}).$
  The above procedure yields a jet-black product, readily soluble in dilute acids; it is rapidly oxidized in air, but does not ignite.

Alternatively, iron(III) oxide can be reduced by iron to iron(II) oxide [2, p. 1498]:

  II. The preparation from stoichiometric quantities of commercial $\ce{Fe2O3}$ and reduced iron can also be recommended.
  The mixture and a few drops of water are sealed into a preevacuated quartz tube, heated for about three days at $\pu{900 °C},$ and quenched in cold water.

$$\ce{Fe + Fe2O3 ->[\pu{900 °C}] 3 FeO}$$

Iron(II) oxide can also be produced from iron(III) oxide reduction by carbon monoxide at the temperature range between $\pu{500 °C}$ and $\pu{600 °C}$ (adapted from [3, p. 415]):

$$
\begin{align}
\ce{3 Fe2O3 + CO &->[\pu{400 °C}] 2 Fe3O4 + CO2} \\
\ce{Fe2O3 + CO &->[\pu{500-600 °C}] 2 FeO + CO2} \\
\ce{Fe2O3 + 3 CO &->[\pu{700 °C}] 2 Fe + 3 CO2}
\end{align}
$$

as well as by thermal decomposition of magnetite [3, p. 415]:

$$\ce{2 Fe3O4 ->[>\pu{1538 °C}] 6 FeO + O2}$$

References

Günther, P. L.; Rehaag, H. Über Die Thermische Zersetzung von Oxalaten II. Mitteilung. Darstellung von Reinem Ferrooxyd. Z. Anorg. Allg. Chem. 1939, 243 (1), 60–68. https://doi.org/10/cqzgk8. (in German)
Handbook of Preparative Inorganic Chemistry, 2nd ed.; Brauer, G., Ed.; Academic Press: New York; London, 1965; Vol. 2. (Archive.org)
R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


Why does denaturation of albumen protein cause pH of solution to increase?

The main component of egg white, ovalbumin, contains 93 negatively charged residues (Asp + Glu) and 68 positively charged residues (Arg + Lys), as well as 14 His. If you make the solvent less polar by adding ethanol to water, they will tend to form the neutral species (Arg+ and Lys+ will lose a proton, and Glu- and Asp- will pick up a proton).
If you had the protein unfolded already, you would expect that the solution gets more basic if the effect on all charged amino acids is the same because there are more negatively charged ones.
To test whether the observed effect has to do with denaturation of the proteins or with changes in protonation state of side chains already exposed to water, you could measure the pH of amino acids (preferably with a modification of the main chain carboxylic acid and amino functions so only the side chains show acid/base reactions) at different ethanol concentrations.


SI Units for Isothermal Incompressibility and Specific heat capacity at constant volume

The units of a partial derivative $\displaystyle\left(\frac{\partial Y}{\partial X}\right)_Z$ are the units of $Y$ divided by the units of $X$.  The units of $Z$ don't matter.

Isothermal compressibility is $-\displaystyle\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$, so the SI units are reciprocal pressure, $\rm Pa^{-1}$ or $\rm m^2\ N^{-1}$.
Heat capacity at constant volume is $ \displaystyle\left(\frac{\partial E}{\partial T}\right)_V$, so its SI units are energy/temperature or $\rm J\ K^{-1}$. When you say specific heat capacity, that means per unit mass, so the specific heat capacity at constant volume is $\rm J\ kg^{-1} K^{-1}$.


What does the ZZE tag in an Chemical Table file refer to?

The current version of the CTFile spec can be downloaded for free from Accelrys. Based on the files you have, you'll want to look at the "V2000" section.

Short answer: I believe [M  ZZE] is a vendor-specific property key, possibly introduced by ACD Labs, given it's the package cited in the header on line 2. The CTFile format spec for V2000 files says nothing about an "M  ZZE" property key.

The CTFile spec says that property keys that are not recognized by parsing software are to be ignored. This makes vendor-specific extensions feasible.

You seem to be on the right track in your interpretation. The Galactobuxin entry has this line in the properties block:

  M  ZZE  1  31  23

Breaking this down:

{M  ZZE} is the mystery properties key.

{1} is the number of data entries on the line.

{31} is the 1-based index of either an atom or bond.

Atom 31 is one of the rather boring methoxy groups.

On the other hand, bond 31 is a real problem child:

This file is abusing the meaning of 'wedge bonds' by trying to use them to indicate chair conformation perspective, rather that what they're intended for - flat stereoprojection. My guess is that the file is trying to annotate this bond.

The only question left is - what does {23} refer to?

Bond 23 is unremarkable. Atom 23 is one of the origins of a wonky wedge bond. But it's not clear why it should be associated with bond 31.

My best guess is that the software that created this file knew it was abusing wedge bonds and wanted to include a flag on bond 31 so that software reading the file wouldn't misinterpret.

Given that you're seeing the same {23} in every file, this could be a code to specially process bond 31's stereo flag.


Is (high) pressure required to form (quartz) crystals?

Pressure does not affect the crystal structure (unless it gets insanely great, so that stishovite formation kicks in; that's not going to happen in an autoclave, though). Pressure, however, does affect the temperature (indirectly), and hence solubility, and hence the rate of crystal formation. There is no lower threshold, that's why you keep getting wildly different estimates for the "required" pressure. It is just that the higher you get, the faster your crystals grow. They would grow at normal atmospheric pressure as well, just very slow (so that only your great-grandchildren may possibly be able to see them).

I suggest that you aim at something less extreme. Calcite, maybe?


Mixing 1 cubic meter of HCl and NaOH: Temperature rise of a metal container?

You have 1 mole/liter of each reactant.  How many liters of each solution do you have?  That gives you the number of moles of reactants. They will create way less than $2\cdot 10^6$ moles of water-you were counting the water that was already there.  Once you have the right number of moles of reactant, that is the number of moles of water produced.  Multiplying by your heat per mole gives $Q$.  Then divide by $mc_p$ to get $\Delta T$


How to calculate the pH of a buffer after HCl was added?

For (a), the Henderson-Hasselbalch equation, $$\ce{pH} = \mathrm{p}K_a + \log([\ce{A^{-}}]/[\ce{HA}]),$$ comes in handy. Because your molarities and volumes of the acid and its conjugate base are equal, this indeed reduces to simply $\ce{pH} = -\log(6.3 \cdot 10^{-5})$.

For (b), the volume of $\ce{HCl}$ added is required, as the concentration of the solution alone is not sufficient information. The standard practice is to assume that $\ce{HCl}$ (being a strong acid) reacts fully with the conjugate base in your buffer solution to produce an equal amount of the conjugate acid (i.e., if $x$ moles of $\ce{A^{-}}$ are consumed by $\ce{HCl}$, $x$ moles of conjugate acid $\ce{HA}$ are produced). Therefore, you can use the Henderson-Hasselbalch equation to recalculate the $\ce{pH}$, subtracting the moles of $\ce{HCl}$ added from your conjugate base, and adding that some number of moles to your conjugate acid.


All metals form basic oxides.

That's true. I cannot remember metal, that does not form basic oxide

  Metallic oxides are basic in character.

Not necessary. Polonium is not the only one metal, forming acidic oxides. Mn, Cr and some others do it as well.

Oxide is considered acidic, when it forms hydroxide, that is acidic (well, it is simplification, but let's stick with it for a while). Hydroxide is acidic, when it can lose proton, giving stable anion. Acidity of hydroxide and metallicity of element formed it does not connect any way. Acidity depends of ability of remaining part of molecule to stabilize negative charge, which usually correlate with $\ce{O/OH}$ ratio and electronegativity of central atom. So, $\ce{H6TeO6}$ is quite weak acid and $\ce{H2CrO4}$ or $\ce{HMnO4}$ are quite quite strong acids.