Why do Organolithium or Grignard reagents act as nucleophiles and not as bases with aldehydes and ketones

In fact, both reagents you noted here have quite complex structure and are not nucleophiles at all: both are electrophiles, because metallic atom here has too little neighbors to draw electrons from.

Let me explain, using Grignard reagent $\ce{EtMgCl}$ . In reality it has complex structure with $\ce{Mg}$ atom coordinated to alkyl fragment and two diethyl ether molecules. When reacts, it coordinates first with most nucleophilic atom. In case of alcohol it is oxygen of $\ce{OH}$ fragment, and of course it  reacts with closest easily migrating atom: hydrogen, producing alkane. In case of ketone it coordinates with $\ce{CO}$ group, and alkyl group attacks closest electrophile : carbon from $\ce{CO}$ group, whose electrophility was just boosted.

If right compound, effectively bounding to metal atom, is added and no other way of reactions is readily available, both reagents can work as bases. For example, organolithium compounds can replace hydrogen atoms in alkynes, alkenes and arenes as you noted earlier.


Food doesn't produce urea, your body produces urea from the nitrogen content of the food you eat (mostly comes from proteins). So you can eat e.g honey, which contains minimal, if any, amount of nitrogen.


Are certain logP values eluted when 100% methanol is used as the solvent in a C18 column?

HPLC is often used to determine lipophilicity in industry, as the original methods to determine logP by partitioning between octanol and water was not particularly amenable to high-throughput assay.

Lipophilicity values measured by HPLC are often reported as cLogP / cLogD (where c is relating to the fact that the value has been determined by some form of chromatography).

Experimentally, a C18 column is used with an acetonitrile/water solvent system, buffered appropriately. Unlike with the octanol/water procedure, different companies run these HPLC assays differently, and as such they aren't as comparable (i.e. if you ran 20 compounds through a GSK assay, and the same 20 compounds through a Pfizer assay, the precise values for cLogD obtained in each case would vary, even if the general trend remained the same).

In terms of you doing this yourself, you'd first need to find a training set of compounds with known and reliable values. If you know paracetamol comes of at x mins and has a  literature logP of y, and ibuprofen comes off at z mins and has a literature logP of w, you can interpolate. When you then run an unknown sample, its retention time using your method can be used to predict the cLogP/D (keep in mind that you'd need a far greater training set than 2 compounds).


Is there a heatformed vinyl-like plastic that is hard as shellac when cold?

I found a web article, "The Origin & Many Uses of Shellac" by R.J. Wakeman, to be particularly illuminating and well-written source of information on the composition and manufacture of shellac records, for instance it explains:

  Emile Berliner’s first disc records were pressed in celluloid (1894) followed by the use of vulcanized rubber (1895).(36) In 1896 Duranoid shellac records were developed and became the standard medium.(11) The formula used for making shellac records varied from firm to firm and from decade to decade; it was usually a company secret. The shellac component represented approximately five-eighths of the material used for the better quality records.(33) Essential components would include shellac, ground rock, carbon black, and cotton floc.

The article lists a few patents among the references that you could quickly search for additional preparation methods, although I found them to be somewhat evasive with regard to details, preferring to amplify the scope of the application by making general claims. The number of patents describing the manufacture of records with shellac and competing resins with a variety of additives is considerable. There is therefore no universal recipe for which to search. What common properties do the formulations share? The key is that the shellac serves as a binder, and that strength and hardness, to protect against wear of the record by the steel needle, to actively wear the needle into shape, or to prevent shattering, is largely conferred by the ground mineral filler and other additives such as abrasives and plasticizers. Apparently the fineness of the ground rock and the quality of the shellac are important.

  Abrasives were frequently added to the record matrix to help wear the point of the needle to conform to the shape of the record grooves

and

  When a needle enters the groove of a 78 rpm shellac record the needle point should quickly abrade to fit the record groove.

Regarding the preparation of shellac records, Wakeman writes:

  To mold shellac records, powdered shellac, ground stone, carbon black, and cotton floc were measured by a formula and mixed in a revolving drum or Banbury mixer.(16) The shellac needed to be finely ground for thorough mixing with and coverage of the other components. High quality orange shellac with low wax content and few impurities was the standard.(16) One report states the shellac was ground to 80 mesh, the fillers to 200 mesh, and the coloring matter to less than 0.4 micron in size. The mixture was steam heated, causing the shellac to melt and form a thick dough which was then passed onto heated rollers and formed into a long roll. The roll was cut into sections (called “biscuits” in some factories) with adequate material to form a 10 or 12-inch record.(8,31)

So in answer to the question, "what alternatives exist to shellac with similar properties?" you would like to replace shellac with other resins, while retaining the other essential additives, such as abrasives. You would not be the first to attempt to do so. As Wakeman explains:

  After the war shellac was combined with other resins to form softer and less brittle 78 rpm records.(16) Shellac was reduced in the matrix to form 15 to 30% of the mix.(16) Vulcanite (also called ebonite), a hard rubber, was sometimes included in the mix, as were Vinsol (a pine tree resin) and Manila gum (16).

Another important property is time to harden. I looked up Durium, a relatively simple shellac competitor. Durium was patented as a fast-hardening resorcinol-formaldehyde resin. Production of Durium records was faster since, unlike shellac, Durium hardened quickly and records were fabricated by pressing rather than moulding. While it has properties that represent a significant advantage over shellac (price, hardening speed, method of impression), unfortunately it also requires working with a variety of unpleasant chemicals.

For an early vinyl patent, developed during shellac shortages in WWII, see US patent 2335986. For a more recent patent providing a number of formulations alternative to shellac, see US patent 2787602.

Regarding characterization of shellac composites, this is a difficult topic to address, since the properties can be expected to depend strongly on the specific additives. I did find an old reference quoting a value for the modulus of elasticity of processed shellac as on the order of $\pu{10 \times 10^3 kg/cm2}$. You have many choices regarding a resin to match the required strength and softening temperature.  You may want to contact a company such as Ensinger plastics for guidance on selection of a thermoplastic composite.


When looking at the corresponding Wikipedia article for prochirality you will find out, that your reasoning is correct for the hydrogenation. Of course if you have a different nucleophile, then the situation would change, as the Cahn-Ingold-Prelog identifiers would change.

See also prochirality in the IUPAC goldbook.


How is the bond length calculated from the total electronic energy?

The geometry of a system is what is known (or given) and a corresponding energy is computed.  If you have an energy then you should have the geometry (or in your case, bond length).

The wave function, $\Psi$, in the Schrodiner equation is simply a function of all the positions of the particles within the system (nuclei and electrons).  Typically chemists invoke the Born-Oppenheimer approximation which assumes that the nuclei are stationary (frozen in space) with respect to the very 'fast-moving' electrons.  Therefore, we freeze the nuclei in space and compute the energy of this stationary state.  Move the nuclei in your molecule a little bit and you will get a new energy.

As Wildcat was explaining, optimization procedures essentially moves the nuclei in your molecule around in space, computing an energy for each configuration. The goal of this type of routine is to find an energy that is lower than all the rest.  Take, for example, the $\ce{H2}$ molecule.  It contains one bond length and therefore the only coordinate we need to consider in an 'optimization procedure' is the internuclear separation of the two $\ce{H}$ atoms.  It is the chemist which will choose the initial positions of these atoms.  Say we initially set these atoms to be 200 pm apart.  We compute an energy.  Now, can we do better?  (i.e. can we perturb this geometry in such a way that gives rise to a lower energy?)  Well we can certainly try other bond lengths so we now set our bond length to 150 pm and compute a new energy.  Lo and behold we see that this new energy is lower than the first!  We keep trying different bond lengths until we reach a point where the energy increases no matter what we do with the separation (whether we increase it or decrease it).  Once we are at this point, we have found the lowest energy configuration of $\ce{H2}$.  Of course optimization routines are smarter than doing a brute-force approach to sampling all the possible quantities each geometrical parameter can take on.  Furthermore, second-derivatives are essential for characterizing the nature of the stationary state that was determined (i.e. is the stationary state a true minimum, a transition state or some higher order saddle point?).


What is the difference between resonance and π conjugation

You might be confused but not missing anything big or fundamental.
A pi-conjugated system is one in which delocalisation of pi-electrons can take place. It is all.
When effective, pi-conjugation is a subset, or a type, of delocalisation than can be described by resonance.
However:
to be strict, conjugation refers to bonds, and specifically to an alternating single and double bonds structure. It is just a positional nomenclature.
It turns out that in cases and in different extent the structure with such conjugated alternating single and double bonds does not represent the real molecule right because delocalisation takes place.
While the terms are linked, a not very accurate use of them leads to their treatment as if they are synonyms. That is what I have more or less done above, but just using a more nuanced term as pi-conjugation, thereby referring to electrons rather than bonds.
This is as to say that a sequence of alternating single and double bonds (as in at least one limiting form) is conjugated by definition. The conjugation (would be better saying delocalisation, here) can eventually/effectively be so true to the point of erasing the original meaning, i.e. in fact there are even not localized conjugated bonds.
Whit all the above in mind, a perfect organic metal with equalised bonds would be seen as "the king" of conjugated polymers.
Finally, within the context and dictionary of  VB the above delocalisation is described by resonance, This again lead to the use of delocalisation and resonance as interchangeable terms.


Direction of arrows in electrocyclic reactions

The answer is rather bland and probably not what you want to hear. Electrocyclic reactions are called electrocyclic because they are cyclic. (Please don’t see cyclic circular reasoning in there.) They proceed in a cyclic, i.e. fully symmetric fashion. You simply cannot draw one arrow direction and claim it is the most logical or the only one possible. You could draw these arrows as you drew them, but you could also formulate a radical-based mechanism (each double bond splits into two radicals which combine with the two of the other double bond) at no loss.

Thus, there is no way to say whether left or right. The reaction happens; in a cyclic fashion.


How can we get the toxicity level of a plant without using animals?

Short answer
You can't!
Longer answer

Toxicity is not defined for a plant (or animal), but for specific compounds which are either produced or accumulated in these "toxic organisms". The concentration of toxic compounds may vary during the vegetation cycle and can depend, particularly in the case of accumulation, on the growth conditions.

Toxicity of particular compounds is typically given against specific endpoints, such as daphnids, rats, etc.

You asked for "the toxicity level". Which one do you mean?
$\mathrm{LD_{50}}$, $\mathrm{LC_{50}}$, NOAEL/NOAEC, NOEL/NOEC, DNEL,...

If you cannot measure in vivo for whatever reasons, use established QSAR models for specific endpoints.


Are there any substances that react very slowly with water to create heat?

Hand- and foot-warmers are made containing iron filings, with a bit of paper pulp and salt, and sometimes a sealed water packet. These rely on the (surprisingly rapid) rusting of iron in contact with air, catalyzed by the water and salt (the pulp is just to hold the water and keep it in contact with iron). This reaction is a fairly safe way to melt a bit of snow, with some caveats:

It's more expensive than using freezing point depression of salts such as $\ce{CaCl2}$, $\ce{Mg(C2H3O2)2}$, or $\ce{NaCl}$ (table salt). Even adulterated wine containing ethylene glycol has been used for runway deicing.
The rust will eventually leave brown streaks on any surface below the deicer, such as the walls of a building if used to melt snow on the roof.


Do alkyl halides with an alpha nitrogen atom undergo SN1 reaction?

The reactivity of haloalkyl ethers are well documented (e.g., they were originally used to protect $\ce{-OH}$ groups during total syntheses). For instance, chloromethyl methyl ether had been used to protect alcohol and phenol $\ce{-OH}$s as $\ce{-OCH2OCH3}$ or commonly known as $\ce{-OMOM}$. They should be sealed and stored in freezer to protect from moisture because they are susceptible to decompose by reacting with water, showing their reactivity. They are also well known carcinogens. An early review on $\alpha$-haloalkyl ethers can be found in Chemical Reviews (1955)([Ref. 1]).

A recent example in Ref. 2 shows in situ preparation of $\alpha$-halo ethers to protect $\ce{R-OH}$ groups, in order to prevent researchers from exposure to carcinogenic $\alpha$-halo ethers (see Figure below):

For example for carcinogenic activity of a $\alpha$-halo ether: Human health assessment information on bis(chloromethyl)ether can be found on the IRIS website (https://www.epa.gov/iris). IRIS stands for Integrated Risk Information System, a part of EPA’s mission to protect human health and the environment.

Based on the high reactivity of  $\alpha$-halo ethers, one can expect $\alpha$-halo amine would have been even higher reactivity than them. To my knowledge, they do not exist, mainly because, they'd undergo elimination quickly to give relevant  Schiff bases (see below):
$$\ce{-> [R-CH(Cl)-NHR1] -> R-CH=NR1 + HCl} \ \mathrm {or}$$
 $$\ce{-> [R-CH(Cl)-NR1R2] -> R-CH=N+R1R2Cl-}$$

[Ref. 1]: The alpha-Haloalkyl Ethers: Lawrence Summers, Chem. Rev., 1955, 55(2), 301-353 (DOI: 10.1021/cr50002a003).

[Ref. 2]: Simple, Rapid Procedure for the Synthesis of Chloromethyl Methyl Ether and Other Chloro Alkyl Ethers: M. A. Berliner, and K. Belecki, J. Org. Chem., 2005, 70(23), 9618-9621 (DOI: 10.1021/jo051344g).


Predicting bonds from properties of compounds

A is an ionic salt like NaCl.  Z is covalent and does not solvate to ions when dissolved in water, e.g., sugar, poly(ethylene glycol).  Y is polar covalent and requires solvation to ionize, e.g., HCl.


What is a molecule’s smallest set of smallest rings?

This is a tool to classify molecules with multiple rings. Here is an example from a PhD thesis:

So even though you can find six-membered and eight-membered rings in this structure, these are made up of multiple four-membered rings, and are thus not parts of the set of rings. As the picture illustrates, it is sufficient to choose five rings to show the connectivity of the atoms, so even though there are six four-membered rings, the SSSR includes only five of them. In this case, which five is ambiguous.

how do you determine what the smallest sets and rings are?

It turns out that this is ambiguous, so different software might get different results. In contrast, giving the size of all rings is unambiguous, but - according to the thesis I quoted - computationally more expensive and chemically less intuitive.

What is a SSSRs used for?

To classify cyclic molecules in cheminformatics. For a given software, two molecules are different if they have different SSSRs. Conversely, two identical molecules will have the same SSSR. This makes it easier to classify, sort, and search this class of molecules.


Is it good practise to mix double and triple zeta basis sets?

It is perfectly fine, and actually quite common, to use big basis sets for the most important atoms (perhaps at the active site of the chemical reaction) and a smaller basis set for the surrounding hydrogen atoms (for example).

The only problem you might face would be, that this complicates the basis set extrapolation to get closer to the complete basis set limit. Such an extrapolation works best if you are, for example, extrapolating from 2Z & 3Z to infinity (not a mixed extrapolation such as 2Z/3Z & 3Z/4Z to infinity). In fact you won't be able to extrapolate because you likely won't be doing calculations with a 1Z basis set, and you won't be able to do a 3Z/4Z mix (because if you could do that, then you wouldn't need to mix 2Z into your 3Z calculation!).

I do want to point out that if you need to compute "extremely accurate single point energies" then even if you could use 3Z on the whole molecule, it would not likely even get you to 1 kcal/mol precision for energy differences. So as long as "extremely accurate" means in your context "+/- 10 kcal/mol for energy differences", you are doing everything just fine (and don't need a basis set extrapolation anyway). If you are aiming for better accuracy than that, you'll probably need a basis set extrapolation (meaning you want all atoms represented at 3Z level).

In the following reference a "mixed" basis set was used;
in this paper TZP (3Z) is used for the atoms Fe, S, and Mo while SVP (2Z) is used for C, H, O, and N:

Li, Z.; Li, J.; Dattani, N. S.; Umrigar, C. J.; Chan, G. K. The electronic complexity of the ground-state of the FeMo cofactor of nitrogenase as relevant to quantum simulations. J. Chem. Phys. 2019, 150 (2), 024302. DOI: 10.1063/1.5063376.

Since the molecule is FeMoco (an iron-sulfur complex) it is likely that the Fe and S are very important (so they have a big basis set). It is also believed that the molybdenum (Mo) is important, because why else would a biological molecule have Mo in it, so it has a big basis set too. The rest have a small basis set.


Are these reaction equations for the formation of the brown ring complex correct?

According Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of
Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction
Revisited

The correct structure is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$

For many years it was thought that iron was reduced to Fe(I) and NO oxidized to NO+, based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin Fe(III) (S=5/2) antiferromagnetically couples with NO- (S=1) for an observed spin of S=3/2.

Other than the above, I agree with Dhanajay Gupta that the equations in the book are correct, that some of the Fe2+ is oxidized to Fe3+ in the first equation forming NO, and NO coordinates to remaining Fe2+ according to the second equation, but the second equation doesn't specify the charge on Fe or NO, just the overall charge of the complex.

In the first equation in the book, it should be understood that the ions are aqueous.  For example H+ doesn't mean there are bare protons, it is understood to mean $\ce{H_3O+}$ and Fe2+ is $\ce{[Fe(H2O)6]^{2+}}$ and Fe3+ is $\ce{[Fe(H2O)6]^{3+}}$.

In the second equation in the book, the inner coordination sphere is explicitly shown.

In the article cited above the mechanism of the second equation is explained. Neutral NO attacks hexaaquairon(II), forming a 7-coordinate transition state, a water leaves, and "This rate-determining displacement of coordinated water is followed by a rapid intramolecular charge-redistribution process to lead to the final $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+}}$ product".

  but why won't NO take electrons from the oxidation of Iron ($Fe^{2+}
 \rightarrow Fe^{3+} + e^-$), make $NO^-$ and attack the above and
  substitute one water molecule and give more of the brown ring complex?

To get the electron from Fe2+, NO would enter the inner coordination sphere, and when the electron transfer occurs, the brown ring complex is formed at that Fe.  If the NO- then left that Fe, it would destroy one instance of the brown ring complex.  Hypothetically, the Fe2+ could reduce NO via an outer sphere electron transfer, and then NO- coordinate Fe3+, but this is not what the article reports occurring.


The important thing is the solvent does not dissolve the substance. Then their vapour pressures sum up and they boil together.

This technique is massively used for obtaining aromatic oily essences from plants, that are mostly insoluble in water.

The destilate usually forms a stable emulsion, that gets extracted by lipophilic volatile solvent, that is later evaporated to got the substance, or a mixture of them.


Colour change of fluorescein in acidic and basic medium

The lactone form is not stable in basic medium, so the cycle opens and gives conjugation to flow and unite all four rings. The problem is you really swithched the acidic and basic forms of fluorescein, so the left one exists at basic pHs (except for the H atom at carboxylate - it goes away), and acid medium favors the existence of the right form.
You may check the forms of phenolphthaleine - they are pretty much the same.


Can I expect 18O-hydroxide when adding slight amount of NaOH-16O to 18O-water?

Yes, it will but it will be a hydrogen exchange, not an oxygen exchange.
$$\ce{2 H2^{18}O <=> \color{blue}{H3^{18}O+} + ^{18}OH-}$$
$$\ce{\color{blue}{H3^{18}O+} + ^{16}OH- <=> H2^{18}O + H2^{16}O }$$
The oxygen will mix with water and hydroxide to a near equivalent ratio of $\ce{^{18}O:^{16}O}$ in both $\ce{H2O}$ and $\ce{OH-}$ but not perfectly equivalent as isotopes do not have equivalent chemical potentials.


Can solvents like 99.9% isopropyl alcohol and/or n-hexane be "carbonated" with CO2?

The following are published[1] solubilities of $\ce{CO2}$ in several organic solvents at 1 atm and $\mathrm{20^oC}$. Solubilities are given in mL/g:

  acetone: 8.2
  ethanol: 3.6
  benzene: 2.7
  methanol: 4.1
  toluene: 3.0
  xylene: 2.3
  heptane: 2.8
  methyl acetate: 7.4
  diethyl ether: 6.3

For reference, the solubility of $\ce{CO2}$ in water is about 0.9 mL/g under the same conditions.

1) Topham S et al; Carbon Dioxide. Ullmann's Encyclopedia of Industrial Chemistry. 7th ed. (1999-2015). New York, NY: John Wiley & Sons


Why can't one make C60 with a combination of hexagons and heptagons?

Well, let's see. This is not quite a chemical question, though I wouldn't recommend moving it to Math.SE either, because they might have hard time recognizing what's $\ce{C60}$.

Now, let's imagine we're considering locally planar conjugated carbon
structures. Each carbon has to reserve one $p$ orbital for $\pi$-bonding,
which leaves it with 3 bonds. Now, the bonds tend to go in cycles; let's
confine ourselves to 5-, 6-, and 7-membered cycles, because otherwise we'll
have too much of angular strain (not that other cycles are totally impossible, though). So in effect, we are considering graphs
consisting of pentagons, hexagons, and heptagons, with 3 edges meeting at
each vertex. Let's say the numbers of these n-gons are $n_5, n_6$, and $n_7$,
correspondingly.

That's when old Euler kicks in. Polyhedra must meet a certain relation
between the number of faces (F), edges (E), and vertices (V):
$$V - E + F = 2$$
OK, what is the number of faces? Why, that's quite simple: $F=n_5+n_6+n_7$.
Then what is the number of vertices? Apparently, each pentagonal face has
five of them, and so on, so it would be $5n_5+6n_6+7n_7$. But wait, we've
counted each atom three times (for it belongs to three polygons), hence we must divide by 3: $V=(5n_5+6n_6+7n_7)/3$. By similar reasoning, the number of edges would be $5n_5+6n_6+7n_7$, if not for the fact that each edge belongs to two faces and thus is counted twice, hence $E=(5n_5+6n_6+7n_7)/2$.

All in all,
$$
V - E + F = 2\\
{5n_5+6n_6+7n_7\over3} - {5n_5+6n_6+7n_7\over2} + n_5+n_6+n_7 = 2\\
-{5n_5+6n_6+7n_7\over6} + n_5+n_6+n_7 = 2\\
{n_5-n_7\over6} = 2\\
n_5-n_7 = 12
$$

That's pretty much the size of it. You can do more or less whatever you want, as long as you have twelve more pentagons than heptagons. There is no going around it.

Now chemistry kicks in, too, and brings its own limitations. An $sp^2$ carbon wants to have its bonds in one plane, at $120^\circ$ angles. You can convince it to budge, but only so much. If you strain it more, the system becomes unstable. It turns out that two pentagons meeting at one carbon is already too much (not for an $sp^3$ carbon, though; see dodecahedrane). That's why you can't have constructions made of pentagons alone (or pentagons and heptagons, for that matter). Geometry would allow that, but chemistry wouldn't.

Let's see how that translates to actual shapes. Several fused hexagons give us coronene; it's planar.

(It is tilted sideways to give it a 3D feel. There are hydrogens attached on the outer edge; I omitted them for clarity.)

Extend it further, and you'll have an infinite flat sheet; that would be graphene.

Throw in a pentagon, and here's corannulene; it's bowl-shaped.

Extend that pattern, and you'll have fullerene.

Use a heptagon, and get circulene:

See that? It's bent indeed, but it's bent the wrong way. It's a
saddle-shaped bend. An example of negative Gaussian curvature, so to say. You can't wrap it to a sphere, no matter how hard you try.

Can we extend this further? Well, we kinda can (on paper, that is).
Moreover, it can be done in a number of ways, some leading to magnificent structures (purely hypothetical, of course).

Upd. It was brought to my attention that according to some studies, so-called carbon nanofoam might be a real example of structures of the latter kind. I'm not quite convinced (to me, until you have an X-ray diffraction study, everything is just hand-waving), but let's mention that anyway.


Colliding particles and activation energy

This is the nature of the quantum world. An action, at some level, either takes place or does not... there is no half-way state. Another example is the photoelectric effect. Planck and Einstein explained the requirement for at least a minimum energy of a photon before it can raise an electron to a higher energy level. A million photons just under that energy will not* raise an electron to a higher level, but just a single photon of the required energy (or higher) can bump up the electron.

Caveat: There is also the uncertainty principle, allowing electrons to tunnel to a higher level, or for multi-photon absorption to cause ionization, but they're not as  common as single-photon absorption and ionization.

If you're bothered by this non-intuitive behavior, you're not alone. Bohr, Bose, Boltzmann, Einstein, and others were forced to this theory through experimental results and the horrors of the ultraviolet catastrophe. Those who fail to appreciate the ultraviolet catastrophe shall be made, energetically, to walk the Planck.


What happens when you mix alum with soap?

I think you’ve made a so called “aluminium soap”. It is a compound of the form $\ce{(RCOO)3Al}$ where $\ce{RCOO-}$ is a fatty acid chain. The $\ce{Al^3+}$ ions from the alum you handled would have displaced the $\ce{Na+}$ ions in soap to form the thick gel like aluminium version of soap.

This is similar to how $\ce{Mg^2+}$ or $\ce{Ca^2+}$ ions form an insoluble scum (which consists of $\ce{(RCOO)2Mg}$ or $\ce{(RCOO)2Ca}$) when soap is used in hard water.

I assume here that you’re talking about potash alum $\ce{KAl(SO4)2.12H2O}$ as @BenNorris has mentioned in his comment above.

Any other aluminium containing alum will also have the same effect. But chrome alums (containing chromium instead of aluminium) might have a different effect on soap.

These “aluminium soaps” are used as a thickening agent in oils and greases and also in napalm (as Georg has mentioned in his comment above).


What trends affect hydrophobicity in amino acids?

Your assumption that hydrophobicity of amino acids increases with length of the carbon chain is correct. Also, a branched chain can have a lower hydrophobicity than a linear chain with the same number of carbons, because it has a decreased exposed surface area.

When polar acidic/basic functional groups are introduced, the side chain becomes more hydrophilic. For example, phenylalanine is classified as non-polar because of the hydrophobic benzyl side chain. Attaching a hydroxy group in para-position in the phenyl ring leads to tyrosine, which is more polar. This is due to its acidic phenol group which is deprotonated at physiological pH. Note that in both cases, the number and arrangement of carbon atoms in the side chain is the same.


If a temperature of a reaction is raised from 300 to 320K, what factor will the reaction rate increase by

The amount of increase is dependent on the activation energy. For a very low activation energy, there will be almost no change (this is what you obtained by removing $E_a$ and $T$). For a high activation energy, the change will be quite large. For example, with an activation energy of $110\mathrm{\frac{kJ}{mol}}$, you the rate will increase by about $16$ times. $$\frac{\exp({\frac{-E_a}{R*320}})}{\exp({\frac{-E_a}{R*300}})}=\frac{\exp({\frac{-111000\text{J}}{8.314*320}})}{\exp({\frac{-111000\text{J}}{8.314*300}})}=\frac{7.6\times10^{-19}}{4.7\times10^{-20}}=16.14$$

So there really isn't an "in general" answer. However, many text books give a crude estimate that the rate will double when the temperature is increased by $10$. This isn't in general the case, as it depends on the temperature range (a change from $300$ to $320$ is more significant that a change from $600$ to $620$) and the the value of the activation energy.


What ratio of bicarb to vinegar do I need in order for the result to be pH neutral?

Your calculation is correct (14.3 grams of 5% acetic acid completely reacts with 1 g sodium bicarbonate). But you're not interested really in "completely reacts"; you want something like pH 7.

You've got it just about right with your mix; you're at the endpoint, and this is a point where the pH will drop very rapidly with even a small addition of vinegar. So it would only take a few additional drops to make the pH go down to 7... but that may not be what you really want.

Sodium acetate has a mild salty taste by itself. So I'm not sure adding extra vinegar will be worthwhile, it'll make it a bit more vinegary.

Here's what I'd do: Mix vinegar and baking soda in the ratio you've figured out, and then see how that tastes. Make adjustments at that point according to taste, rather than by "titrating" to an arbitrary pH.


The stability constant of $\ce{Pb−EDTA}$ complex is $\ce{10^{18}}$, which is huge. The solubility product of $\ce{PbSO_4}$ is $10^{−8}$
, which cannot be compared to the effect of $\ce{EDTA}$. So the precipitate $\ce{PbSO4}$ must be soluble in $\ce{EDTA}$ solutions. These numerical data have been taken from J. G. Stark, Chemistry Data Book, John Murray, London, $1994$


How does electrolytic conductance depend on solvation?

Suppose you are in a field filled with a crowd of children and some of them are holding your hand, some are holding your shirt but you have an urge to reach from point A ( one electrode) to point B (another electrode)? In that case you effective size has increased. How easy would it be to travel from A to B, if you were all alone?

Change in conductance due to solvation is somewhat similar. When an ion wishes to travel under the influence of an electric field between two electrodes, solvent molecules like to travel with the ion, making it "bulkier", hence the ion moves with a retarded velocity.

P.S. Solvation comes from the word solvate which means "A more or less loosely bonded complex formed between a dissolved species and the solvent." It is a general term used in physical chemistry.

Even more formally, http://goldbook.iupac.org/terms/view/S05747
"Any stabilizing interaction of a solute (or solute moiety) and the solvent or a similar interaction of solvent with groups of an insoluble material (i.e. the ionic groups of an ion-exchange resin). Such interactions generally involve electrostatic forces and van der Waals forces, as well as chemically more specific effects such as hydrogen bond formation."


How can chemical energy be stored in chemical things like candle wax, gasoline,...?

Yes, energy isn't a "substance" it is the ability to do work. But chemicals can still store it.

The key to understanding how is the relative amount of energy in different chemicals. Some chemical reactions release heat because the products of the reaction are lower energy than the chemicals that react (reactions are a bit more complicated than this in reality and some are driven by changes in entropy but that isn't important for understanding simple reactions). A candle, for example, consists of a mixture of long chain hydrocarbons with a formula approximating CH2. A candle will burn in air (reacting with molecules of oxygen) to give a mixture of CO2 and H2O. This reactions releases heat (simplifying slightly) because the bonds making up the products have lower energies than the bonds making up the candle wax.

This is why we can say that chemicals "store" energy. But we should always be careful to specify the reaction involved as chemical energy is stored relative to the possible products from a reaction not in some absolute sense.


How does the proticity of an acid affect the enthalpy change of its reaction with a base?

You have to remember that enthalpy change is defined at kilojoules PER MOLE. The concentration won't affect enthalpy change because the same amount of heat is given off per mole of zinc , however the reaction will go faster with a higher concentration.


1) Oxygen CAN form 3 or 4 bonds. The most common example is $\ce{H3O+}$ ion This form of bonding is critical to understand basic nature of ammonia solutions, so for exact description of it look that part of the book.

2)Boron also can form more than 3 bonds if at least one its partner provides an electron pair for bonding, not one electron.

As a general rule I prefer to stay away from the chemistry of borates and alumina-silicates as it is ruled largely by ionic interactions and non-intuitive hydration/dehydration.

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In structural diagrams of polymeric salts 'unit' often refers to some recurring fragment... however, it is not nessesarily has same stochiometry than formal anion. In alumosilicates the anionic frame consists of $\ce{AlO4-}$ and $\ce{SiO4}$ tetrahedral units. However, they are usually connected by shared vertexes, so the exact formula of formal anion is different. Borates are a bit more complex system, since they may include both $\ce{BO3}$ and $\ce{BO4-}$ units, and oxygen is often connected to hydrogen.

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I looked into open cristallographic database. Unsurprisingly, it does NOT feature a compound $\ce{Ca2B6O11}$. However, salts are often written in 'dehydrated' form, so I looked for any borate with $\ce{Ca:B} = 1:3$ stochiometry. Several with rather complex crystal structure were found, so I left only two: one with rather simple structure of typical 'hidrated' borate anion and one with rather complex continous 'dehydrated' boron-oxygen layers. Note: the structures are refined from X-Ray , so they does not reveal hydrogen positions. Isolated oxygens of water molecules and cations are omitted.

$\ce{B3O8}$ unit from mineral Inyoite, $\ce{CaB3O3(OH)5·4(H2O)}$

$\ce{[B3O6]_{\infty}}$ layer from mineral Fabianite $\ce{CaB3O5(OH)}$

and its main building block


Why is law of mass action only applicable for elementary reaction?

The Law of Mass action states:

The rate of a reaction is directly proportional to the concentration of reactants raised to powers equal to their stoichometric coefficients

This is only valid for elementary reactions since they happen in a single step. As a result, such simple relations hold good.

But what about the more complicated ones. Certain reactions, like the oxidation of ethylbenzene to benzoic acid using hot $\ce{KMnO4}$ don't happen in a single magical step. They involve lots of reactive intermediates and transition states, so it's basically a bunch of several elementary reactions. You can't consider the whole reaction as an elementary reaction.

But you can formulate a rate equation for non elementary reactions by considering them as a bunch of elementary ones. However such formulations involve knowing the mechanism of the reaction in question, along with other data.


Does the Lewis acid have to match the halogen in electrophilic halogenation of benzene

The reason you use the respective halides for halogenation is that the halogen and Lewis acid can inter-react to form a chemical equilibrium if you don't.

$$\ce{FeBr3 + Cl2 <=> Cl+ + [FeBr3Cl]- <=> Br+ + [FeBr2Cl2]- }\\
\ce{AlCl3 + Br2 <=> Br+ + [AlBrCl3]- <=> Cl+ + [AlBr2Cl2]- }$$

Given these equilibrium reactions you can see that now you have two electrophile species which means your halogenation is not as controlled. It will work and halogenation will still occur, but now and you will have bromination and chlorination occurring in competition with each other when you only wanted one or the other.


How to calculate bicarbonate and carbonate from total alkalinity

If I understood your question correctly, you have solutions where you know there is a given amount of calcium carbonate dissolved, and would like to know the distribution of this carbonate between all the species present. This proportion is commonly refered as the alpha($\alpha$) for a given species, that varies from 0 to 1(0% - 100%). From your question, I can make some assumptions:

As a groundwater sample, any solids dissolved are very diluted, so we don't need to worry about ionic strength;
pH is not fixed;
Temperature is not fixed, but I will assume its close to room temperature;
As other components are not mentioned, I will assume all carbonate comes from calcium carbonate.

Carbonic acid, $\ce{H2CO3}$, has two ionizable hydrogens, so it may assume three forms: The free acid itself, bicarbonate ion, $\ce{HCO3-}$(first-stage ionized form) and carbonate ion $\ce{CO3^2+}$(second-stage ionized form). The respective proportions in comparison with the total concentration of calcium carbonate dissolved are $\alpha0$, $\alpha1$ and $\alpha2$. They must sum to 1(100%), as in chemical reactions matter is neither created or destroyed, only changing between forms. When the calcium carbonate dissolves, a equilibrium is established between its three forms, expressed by the respective equilibrium equations:

First stage:
$$\ce{H2O + H2CO3 <=> H3O+ + HCO3-}$$
$$K1 = \frac{\ce{[H3O+][HCO3-]}}{\ce{[H2CO3]}} \approx 4.47*10^-7 $$

Second stage:
$$\ce{H2O + HCO3- <=> H3O+ + CO3^2-}$$
$$K2 = \frac{\ce{[H3O+][CO3^2-]}}{\ce{[HCO3-]}} \approx 4.69*10^-11 $$

You can also write a equation for the overrall reaction, by sum of each stage (and multiplication of the respective equilibrium constants):
$$\ce{2H2O + H2CO3 <=> 2H3O+ + CO3^2-}$$
$$K1K2 = \frac{\ce{[H3O+]^2[CO3^2-]}}{\ce{[H2CO3]}}$$

Analysing our system, to give a full treatment, if we know the solution pH, we can calculate $\ce{[H3O+]}$. So we are left with three unknown variables, $\ce{[H2CO3]}$, $\ce{[HCO3-]}$ and $\ce{[CO3^2+]}$. But so far we have only two independent mathematical equations, for K1 and K2 (the overrall equation does't count as independent, as it's only the merging together of the other two). To solve it, we need at least one more independent equation, to match the number of unknows. What we need is the equation for the material balance of the system. As we assumed all carbonate came from calcium carbonate, we can write:
$$Cs = \ce{[CaCO3]} = \ce{[H2CO3] + [HCO3-] + [CO3^2-]}$$

Where Cs here stands for the known concentration of the salt, calcium carbonate. Now we can start replacing values taken from the equilibrium expressions into the material balance, isolating each unknow. For the bicarbonate, for example:
$$Cs = \ce{[H2CO3] + [HCO3-] + [CO3^2-]}$$
$$Cs = \ce{\frac{[HCO3-][H3O+]}{K1} + [HCO3-] + \frac{K2[HCO3-]}{[H3O+]}}$$
$$Cs = \ce{\frac{[HCO3-][H3O+]^2 + K1[HCO3-][H3O+] + K1K2[HCO3-]}{K1[H3O+]}}$$
$$\frac{\ce{[HCO3-]}}{Cs} = \ce{\frac{K1[H3O+]}{[H3O+]^2 + K1[H3O+] + K1K2}} = \alpha1$$

So we got the expression for $\alpha1$, that has a curious structure: a fraction, where the denominator is a polynomial of degree 2, and the numerator its middle term. The same procedure can be repeated to find the expressions for the alphas  of the other dissolved species. For sake of brevity, I won't do it, but the final result will be:
$$\alpha0 = \frac{\ce{[H2CO3]}}{Cs} = \ce{\frac{[H3O+]^2}{[H3O+]^2 + K1[H3O+] + K1K2}}$$
$$\alpha2 = \frac{\ce{[CO3^2-]}}{Cs} = \ce{\frac{K1K2}{[H3O+]^2 + K1[H3O+] + K1K2}}$$

Note that a interesting pattern emerges. The expressions for the remaining two species have the same structure, just changing the term that goes in the numerator.

With the expressions for all species, it's helpful to use a spreadsheet to automate the calculations for a entire range of pH values, to grasp in a visual way what happens with carbonates as pH changes. I did just that, look at the results (here the spreadsheet, to whomever wants to download and play with it):

We see that in lower pH the predominant form for carbonate is the free carbonic acid. A bit over 6 bicarbonate ion takes over, and reigns up to pH a bit over 10, from where fully ionized carbonate ion takes over. For a given pH, the concentration of each species can be computed multiplying the respective $\alpha$ by the concentration of total calcium carbonate originally present.

The plot that looks like a "XX" also allows us to see a interesting property of carbonates. Nowhere in the plot you will find a pH value where we have the three species all in significant amounts. In the lower pH region you can find both bicarbonate and carbonic acid. But carbonate only shows up when carbonic acid goes away. It's like the unconfortable situation where you have two close friends who both hate each other. In a given moment I can see you in a room talking with either friend, but I will never see you three in the same room, or both friends of yours. The dividing line is close to the pH 8.6 you mentioned in your question. However, that sad situation has a upside. It makes the problem easier to calculate.

The full treatment I gave to this problem was indeed overkill. If I have three species, but only two show up together at any given time, I can "forget" I'm dealing with a diprotic acid. I need only to see the dividing line I've found, around pH 8.6. If I'm above it, free carbonic acid concentration is zero, and I have to deal only with the pair bicarbonate/carbonate, pretending the bicarbonate anion is just a monoprotic acid. From the equilibrium, we have:
$$\ce{[H3O+]} = \frac{\ce{K2[HCO3-]}}{\ce{[CO3^2-]}}$$

Or in logarithimic form:
$$pH = pK2 + log(\frac{\ce{[HCO3-]}}{[CO3^2-]})$$

This is the old Henderson–Hasselbalch equation you surely heard about before. As we know the pH and K2, we can calculate the ratio between carbonate and bicarbonate.

In the other side, if I'm below my dividing line near 8.6, carbonate ion concentration is zero, now I have to deal only with the pair carbonic acid/bicarbonate, pretending carbonic acid is just other monoprotic acid. From the equilibrium, we have:
$$\ce{[H3O+]} = \frac{\ce{K1[H2CO3]}}{\ce{[HCO3-]}}$$

Or in logarithimic form:
$$pH = pK1 + log(\frac{\ce{[H2CO3]}}{[HCO3-]})$$

As we know the pH and K1, we can calculate the ratio between carbonic acid and bicarbonate.

If you want to study in depth such calculations, I recommend this book: Butler, James N. Ionic Equilibrium: Solubility and PH Calculations. John Wiley & Sons, 1998.


Chain transfer agents and rate of free radical polymerization

Not really.

As long as you are in solution (=unreacted monomer or solvent), it makes no difference if the reactive chain end is diffusing slow or fast. The monomer molecules come by at the same rate.

The mobile chain end would help against a local depletion of monomer, but once your mixture gets so viscous that this could happen, neither a long or a short active chain can still diffuse well (and neither can the transfer agent).

The only general difference you will see with transfer agent is that the Trommsdorff-Norish or gel effect sets in a bit later, because the overall viscosity of short chains is lower. In that sense, the transfer agent even slows down the reaction a bit. Which in that case is very welcome.

Of course the transfer agent radical can be either more reactive with the given monomer or less. (See "copolymerisation parameter".) So the chain restart can either be fast (in which case it makes practically no difference), or it can eventually be much slower. That depends on the specific transfer agent / monomer pair.


Will monosodium dihydrogen phosphate react as base or acid in water?

Monosodium dihydrogen phosphate, $\ce{H2PO4^-}$, is an amphoteric species and it will act as both an acid and a base.
These are the related chemical equations:

\begin{align}
  \ce{H2PO4- + H2O &<=> HPO4^2- + H3O+}  & \mathrm{p}K_\mathrm{a} &= 7.21\\
  \ce{H2PO4- + H2O &<=> H3PO4 + OH-}     & \mathrm{p}K_\mathrm{b} &= 11.88
\end{align}

As an example, $\pu{0.1 M}$ solution of such a salt will have a pH of about $4.7$.
In other words, the acidic behaviour dominates over the alkaline one. This is due to the value of $\mathrm{p}K_\mathrm{a}=7.21$ (related to the acidic behaviour) compared to the $\mathrm{p}K_\mathrm{b}=11.88$ (related to the alkaline behaviour).


Upon adding $\ce{SO2}$ gas to water, it shows its acidic oxide character by hydrolyzing to sulfurous acid:

$$\ce{SO2(g) + H2O(l) <=> H2SO3(aq)}$$

$\ce{H2SO3}$ is a weak diprotic acid in water, with the following acid dissociation equilibria:

$$\ce{H2SO3(aq) <=> H+(aq) + HSO3^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1}=1.4\times 10^{-2}}$$
$$\ce{HSO3{}^{-}(aq) <=> H+(aq) + SO3{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=6.3\times 10^{-8}}$$

Thus, addition of $\ce{SO2}$ gas to water will produce an acidic solution.

What would happen if air were bubbled into the solution? Air is a mixture of mostly nitrogen and oxygen. It so happens that $\ce{SO2}$ reacts with oxygen to form $\ce{SO3}$:

$$\ce{2SO2(g) + O2(g) -> 2SO3(g)}$$

Here, the sulfur atom in $\ce{SO2}$ is oxidized (oxidation number goes from +4 to +6) while the oxygen atoms in $\ce{O2}$ are reduced (oxidation number goes from 0 to -2). Now, $\ce{SO3}$ also happens to be an acidic oxide, and it hydrolyzes in water to form sulphuric acid:

$$\ce{SO3(g) + H2O(l) <=> H2SO4(aq)}$$

This time, $\ce{H2SO4}$ is a strong diprotic acid:

$$\ce{H2SO4(aq) <=>> H+(aq) + HSO4^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1} \approx 10^6}$$
$$\ce{HSO4{}^{-}(aq) <=> H+(aq) + SO4{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=1\times 10^{-2}}$$

Assuming that not much $\ce{SO2}$ is lost while bubbling air through the sulphurous acid solution, then you're effectively replacing a weak acid with a strong acid, which would cause the pH to decrease.

As a final remark, if the sulphurous acid solution were bubbled with pure nitrogen, then there would be no oxidation of $\ce{SO2}$ into $\ce{SO3}$, and after a while enough $\ce{SO2}$ will have been dragged away for the solution to become a less concentrated sulphurous acid, which would therefore have a higher pH.


How would I measure the dissolved carbon dioxide content of a liquid (water to start with)

This paper looks helpful: http://www.sciencedirect.com/science/article/pii/S0144860910000683

Quoting from the abstract:

  Many methods are available for the measurement of dissolved carbon dioxide in an aqueous environment. Standard titration is the typical field method for measuring dissolved CO2 in aquaculture systems. However, titrimetric determination of dissolved CO2 in marine water aquaculture systems is unsuitable because of the high dissolved solids, silicates, and other dissolved minerals that interfere with the determination. Other methods used to measure dissolved carbon dioxide in an aquaculture water included use of a wetted CO2 probe analyzer, standard nomographic methods, and calculation by direct measurements of the water's pH, temperature, and alkalinity. The determination of dissolved CO2 in saltwater based on partial pressure measurements and non-dispersive infra-red (NDIR) techniques with a CO2 gas analyzer are widely employed for oceanic surveys of surface ocean CO2 flux and are similar to the techniques employed with the head space unit (HSU) in this study. (...)


Constant Pressure and Temperature during Phase Change

For a pure substance, the phase rule implies that at coexistence of two phases only one intensive variable (pressure or temperature) can be altered independently, the other being dependent on the choice of the independent variable. So the answer is yes and no, you can choose to alter one of the two intensive variables and still retain two-phase equilibrium, but the choice of one intensive variable determines the value of the other. This fundamental principle derives from the conditions of thermodynamic equilibrium necessary for coexistence of the two phases. The set of $(p,T)$ points on the coexistence line can be described in terms of the Clapeyron equation $dp/dT = \Delta S/\Delta V$ and a single reference point $(p_{ref},T_{ref})$ on the line.

Consider a pure substance in a diathermal container whose volume we can alter at will, and kept at a fixed temperature. Assume we begin with the container completely filled with a condensed phase (say liquid) of the substance, that is, the volume is too small (or alternately, the pressure too high) to allow vapor to form. We have however chosen a temperature that corresponds to a point on the coexistence line at a lower pressure (or greater volume) than that currently in the vessel. You can bring the substance to an equilibrium between two phases by changing the volume until the pressure in the container equals the equilibrium vapour pressure - a fixed pressure at that temperature. You can then continue to change the volume (all the while keeping the temperature constant). The proportion of the two phases (their volumes) will readjust to accommodate the new total volume, with heat exchanging with the surroundings to keep the temperature constant, and the pressure returning to the value at equilibrium - the same vapour pressure - provided the volume change is not too big. Naturally if you increase or decrease the volume too much you will end up with a single phase and a different pressure below or above the vapour pressure.

Remember: you don't have to have two phases. But if you want to observe coexistence at a chosen temperature, you have no choice wrt the pressure that will ensure coexistence. What's cool is that the proportions of the two phases can be altered within a certain limited range of total volume so that p and T remain constant. I recommend consulting a p-V-T phase diagram and inspecting it carefully to see this graphically. The following is a slice in the p-V dimensions which I borrowed from this site:


Does isotretinoin affects contraception hormonal pills - Ethinylestradiolum, Drospirenonum

My professional knowledge of vitamin A derivatives is limited to preparative aspects and some photochemical properties.

But it seems that the German Bundesinstitut für Arzneimittel und Medizinprodukte (BfArM) provides some information in german and english.

I didn't know whether a link to the search would survive the session, so the link above is just to their website and you would need to search yourself in case you're interested.

Note: I had a look at the Fachinformationen (RTF in German) among the documents "linked".

In this publication, it is very explicitly pointed out that Isotretinoin is a TERATOGEN and efficient contraception therefore is absolutely vital (in the truest sense of the word).

It is suggested that patients use two independent methods of contraception, of which one should be a "barrier method", e.g. the use of condoms.


First order differential equation for transmembrane diffusion

The solution to your differential equation, subject to the prescribed initial conditions is $$c_{intra}=c_{extra0}(1-e^{-k_{diff}t})\tag{1}$$ under the constraint that $c_{extra}$ does not change from its initial value of $c_{extra0}$.  The true material balance constraint is $$V_{intra}c_{intra}+V_{extra}c_{extra}=V_{extra}c_{extra0}$$
which would yield a result virtually indistinguishable from Eqn. 1.

CASE WHERE $c_{extra}$ IS NOT CONSTANT

OK.  Here is the solution for the case where $c_{extra}$ is a function of time $c_{extra}(t)$, and $c_{intra}=c_{intra}(0)$ at time t = 0:

$$c_{intra}(t)=c_{intra}(0)e^{-k_{diff}t}+k_{diff}\int_0^t{e^{-k_{diff}\tau}c_{extra}(t-\tau)d\tau}\tag{2}$$
where $\tau$ is a dummy variable of integration and where $c_{extra}(t-\tau)$ represents the value of the concentration $c_{extra}$ at time $t-\tau$.  This solution reduces to the previous result if $c_{intra}(0)=0$ and $c_{extra}$ is constant.

So, to find the value of $c_{intra}$ at time t, you need to use Eqn. 2 and perform the indicated integration (usually numerically).


In an x-y diagram, the "bulging out" is demonstrating the increase in relative volatility of the components; the number of stages is a strong function of relative volatility, so an increase in relative volatility leads to less stages.


Comparing basic strength of nitrogen heterocycles

It is likely that the electron pair (from the HOMO) being donated to the acid is from a π molecular orbital. Since the π bond is rather weak, the π electrons should be rather high in energy and is thus able to interact well with the LUMO of the acid. When nitrogen atoms are incorporated into the π system, the energy of the HOMO is lowered due to the high electronegativity of N. Thus, incorporation of N atoms into the system would decrease the basicity of the molecular ion. As such, A and D would be inferior bases, compared to B and C.

Choosing between A and D, and between B and C, we would have to base our criteria for the basicity on the extent of delocalisation. In D, the π system is larger, thus greater delocalisation was achieved, with a lowering of the energy of the HOMO. Thus, D is likely to be less basic than A. Similarly, the extent of delocalisation is also larger in C, thus it would have a lower-energy HOMO, compared to B. Thus, B is likely more basic than C.

Therefore, option (2) is likely the correct order of basicity.


Have a look at this: http://www.chem.utoronto.ca/coursenotes/analsci/stats/LimDetect.html

It is possible to calculate the limit of detection from the standard error of the regression:

$$C_{\mathrm{LOD}} = \frac{3s_{y/x}}{b}$$ These values are calculated from the regression of all the data points, not the standard deviation of any subset of data.

However, in this case we have replicate blank measurements (four measurements at C=0), so we can also calculate the limit of detection as simply 3 times the standard deviation of the blank measurements:

$$y_{\mathrm{LOD}} = y_{\mathrm{blank}}+3s_{\mathrm{blank}}$$

(in instrument response units—convert to concentration from a known standard or from the regression equation)


How does increasing the ionic strength of an aqueous solution decrease the activity coefficient?

Thermodynamic activity is essentially the same the chemical potential.  At the solubility limit of a compound, the dissolved species is in equilibrium with the solid species.  For example, a saturated solution of sodium chloride is in equilibrium with solid sodium chloride crystals.  In such equilibria, the chemical potential of the dissolved sodium chloride is equal to the chemical potential of the solid sodium chloride crystals.

  If the ionic strength increases the solubility of an ionic compound, then wouldn't that mean that the concentrations of the ions that make up that ionic compound would increase, and thus the activity coefficient, therefore, must increase to show an increase in activity, otherwise known as the "effective concentration?"

Let's assume that increasing the ionic strength doesn't alter the crystals of solid sodium chloride that dissolved sodium chloride is in equilibrium with.  That means the chemical potential of the crystal is unchanged.  At the solubility limit, the dissolved sodium chloride is still in equilibrium with the crystals, and as before, this equilibrium requires that the crystals have the same chemical potential (aka thermodynamic activity) of the dissolved species.

If there is more of the dissolved species (a higher concentation), the only way the activity can stay the same is if the activity coefficient goes down.

So you were right when you wrote:

  ...the activity [is] independent of the ionic strength such that in order to keep the activity the same under higher concentrations, the activity coefficient must be smaller...

  This would then bring up the question: is the equilibrium constant, $K$, (determined by the activities of the components in the equilibrium expression) constant even under different ionic strengths, while the "concentration coefficient" (determined merely by the concentrations of the components in equilibrium with one another) does vary under different ionic strengths?

The equilibrium constant is a ratio of thermodynamic activities.  The "concentration coefficient" can vary and will not be constant any time that activity coefficients are not 1.0.  Ionic strength is one of many variables that can affect activity coefficients.  So to conclude:

Yes, the equilibrium constant $K$ is determined by the activities of the components in the equilibrium expression.
The "concentration coefficient" $K_c$ can vary as a function of ionic strength (or other variables).


Why does superglue lose strength at high temperatures?

It is the temperature because as you probably know all kinds of substances get denatured at high temperatures. Another reason could the that the rate of reaction increases after running hot water, therefore taking no effort to take it off. Hope the answer helped!


Why does electronegativity increase across a period?

The electronegativity is the tendency of an atom or a functional group to attract electrons (or electron density) towards itself.

This is because the attraction of bonding electrons by an atom increases with nuclear charge (Atomic Number) and decrease of atomic radius. Both these factors operate as we move to the right in period.

  Does it have something to do with the shielding effect of added electrons?

The electron cloud in the inner orbits act as a shield and reduces the nuclear attraction to the outer orbits. Because of the shielding effect, the tendency of nuclear attraction reduces and thus electronegativity reduces.


Do we use activities or concentrations in the rate expression?

The rate expression is always described in terms of molar concentrations. Units of the rate constant will change based upon the overall order of the reaction. This happens such that the rate is always given in M/s.


Determining the rate of stratospheric ozone destruction

Dioxygen is split in the stratosphere by UV light into oxygen atoms. This is the main source of free oxygen atoms. Their production rate depends on the intensity of the UV light.

In steady-state we can assume that the concentration of oxygen atoms is constant.

$$\frac{d[\ce{O}]}{dt} = 0  \tag{1}$$

For the above destruction mechanism we can formulate the rate laws

$$\begin{align}
\frac{d[\ce{O3}]}{dt} &= -k_1 [\ce{O3}] [\ce{M}] + k_{-1} [\ce{O}] [\ce{O2}] [\ce{M}] - k_2 [\ce{O}] [\ce{O3}] \tag{2}\\
\\
\frac{d[\ce{O}]}{dt} &= k_1 [\ce{O3}] [\ce{M}] - k_{-1} [\ce{O}] [\ce{O2}] [\ce{M}] - k_2 [\ce{O}] [\ce{O3}]  \tag{3}
\end{align}$$

Adding equation $(2)$ and $(3)$ which by $(1)$ is $0$ we get

$$\frac{d[\ce{O3}]}{dt} = -2~k_2 [\ce{O}] [\ce{O3}] \tag{4}$$

With equations $(1)$ and $(3)$ we can express the concentration of oxygen atoms as

$$[\ce{O}] = \frac{k_1 [\ce{O3}] [\ce{M}]}{k_{-1} [\ce{O2}] [\ce{M}] + k_2 [\ce{O3}]} \tag{5}$$

Substituting $(5)$ into equation $(4)$ we get

$$\frac{d[\ce{O3}]}{dt} = \frac{-2~k_1~k_2 [\ce{O3}]^2 [\ce{M}]}{k_{-1} [\ce{O2}] [\ce{M}] + k_2 [\ce{O3}]}  \tag{6}$$

So your friend seems to be right.


If you don't mind me asking, why would you expect the $\ce{Na}$ to have a higher entropy than the $\ce{NaCl}$? The entropy of a state is defined as,$$S=k_b\ln W$$where $W$ is the number of possible microstates which can make up that macrostate.

In this case, we are interested in the number of possible ways that we can arrange $\ce{Na}$ into a lattice or the number of ways we can arrange $\ce{Na}$ and $\ce{Cl}$ ions into a lattice. The latter has many, many more microstates for a given macrostate, and hence a larger entropy.

The standard entropies can be found in this table.

It is worth noting that what I have written above is not exactly the same as the standard entropy $S^\circ$ which is defined as,
$$S^\circ=\sum_{i=1}^N\int\frac{dq_i}{T}dT$$which just says that the standard entropy going from zero kelvin to 298 kelvin depends on the heat put into all $N$ atoms of a system. Thus, there is a dependence on the energy of any state the system passes through.

For the solids you're talking about, if the energetics of the systems were exactly identical, then the ratio of standard molar entropies would be the same as the ratio $\frac{\ln W_1}{\ln W_2}$. But, because sodium has much weaker bonds than sodium chloride, it has a higher entropy change per unit change in temperature than sodium chloride. This is because for weaker bonds, the atoms have greater mobility in space which increases entropy, and it is more likely for there to be lattice defects when the bonds are weak which also increases entropy.


Are there any crystals that are theoretically possible, but haven't yet been made?

When we are talking about the prediction of materials, it is generally about the structure and corresponding free energy, not the crystallization process. That being said there are a huge number of inorganic materials that predicted to be stable yet no one has synthesized them.

There are computational databases like Materials Project that contain automatically generated (theoretical) structures and their estimated free energy, solubility, magnetic properties, bandgap, stability compared to competing phases, etc.


Preventing build-up of Diethyl Ether Peroxides

There's no information regarding the stopping of formation [of ethylidene peroxide] on the page...

The required information is given implicitly. The wikipedia page states

  Diethyl ether peroxide, also known as ethylidene peroxide, [...] is a polymerization product of diethyl ether hydroperoxide

With other words:

If you prevent the formation of diethylether hydroperoxides by

storing diethylether in a brown bottle over sodium hydroxide

you will automatically prevent the formation of subsequent products, such as the ethylidene peroxide.

Note that there is a different approach, which is recommended in terms of workplace safety:

Substitute diethylether with methyl tert-butyl ether (MTBE), which does not form peroxides.


What is the difference between a chemical state and just a combination?

You seem to be confused with "state of matter"/"phase" and "chemical state".

The chemical state encompasses all the macroscopic parameters necessary to fully define a given system. For example, "pure $\ce{O2}$ at $1\:\mathrm{atm}$ and $100^{\mathrm o}\:\mathrm{C}$" is an example of chemical state. A more complicated example would be something like: "10 moles of liquid $\ce{O2}$ at $-100^{\mathrm o}\:\mathrm{C}$, mixed with 1 mole of gaseous $\ce{O2}$ at $1\:\mathrm{atm}$ and $100^{\mathrm o}\:\mathrm{C}$ and 2 moles of gaseous $\ce{N2}$ at $1\:\mathrm{atm}$ and $200^{\mathrm o}\:\mathrm{C}$" (Yes, this isn't at equilibrium)

On the other hand, the phase or sate of matter only applies to a homogeneous system. A gas is in gas phase, but a system consisting of a gas and a solid doesn't have a phase/state of matter. Note that if a gas/solid is dissolved in liquid, then the phase is liquid since solutions are homogeneous.


There are 4 structural isomeric alkenes ($\ce{C_6H_{10}}$) that can be hydrogenated to give methylcyclopentane.

1-Methycyclopentene

3-Methylcyclopentene

4-Methylcyclopentene

Methylenecyclopentane


In a battery, what happens to LiCoO2 when the lithium leaves? It becomes CoO2?

The general textbook formula for the Li ion battery is
$\ce{Li_{1-a}Ni_{1-x-y}Mn_{x}Co_{y}O2}$
, and the simplest/earliest type has $y=1$, i.e. cobalt-only. These "mixed (lithium transition metal) oxides" have a layered structure (see e.g. wikipedia), where you can relatively easily electrochemically remove (and later reinsert) a part $a$ of the Li atoms, without destroying the whole structure. You cannot increase $a$ to much more than 0.1 - 0.3 or so, depending on the exact formulation/grain size/quality/temperature/etc., before the structure does break down.
Even the "uncharged" $a=0$ mixed Li-cobalt(III) oxide gives off oxygen (!) if heated above 180°C. (It is produced by tempering at ~800°C in a pure oxygen atmosphere.) With $a>0$, this happens earlier already, and basically always leads to a runaway destruction of your whole battery setup. With higher contents of Ni and Mn, this danger becomes less pronounced. I´m not sure the Co-only variant was ever commercialised. (?)
Simply speaking, a cobalt-rich battery blows up when overcharged, and in ones with more nickel and manganese the layered structure still breaks down, and you get somewhat stable Ni/Mn(III/IV) species (and still quite a lot of overheating/general destruction). In any case this happens long, long before you reach the ideal $\ce{CoO2}$ (or generally $\ce{MO2}$) stochiometry.
If you look closely (e.g. via XRD) at the crystal structure during charging, you seemingly find that there are a number of intermediate deformed structures, and even more mixed/disordered states in between. And that of course becomes even more complicated (or washed out) with a (more or less) random arrangement of different transition metals in the structure, which will depend also on the specific synthesis route.


Proper use of vacuum pump during vacuum distillation

This pump is way overkill for a vacuum distillation. The problem is not glass implosion necessarily (round bottom jointed glassware is designed to be put under vacuum, less clear on the condenser), but rather that you may not be able to condense your product given the low vacuum. Vacuum distillation is typically done with an aspirator that gets down to 25 Torr (25,000 microns) or so. At too low a pressure, you may need a very cold receiving vessel and condenser, and depending upon the solvent dry ice slush may not even be enough. You need to determine the boiling point of your solvent at low pressures to figure this out. Your pump is for Schlenk lines and similar applications.

I don't see a reason not to use a vacuum gauge (and you are not supposed to use mercury anymore in any application), though again, depending upon the setup, it is possible to harm the gauge. You should try to monitor pressure as part of the experimental process.

Not using the capillary bubbler seems foolish, as bumping is a major problem with vacuum distillations. A capillary bubbler is a very thin tube (like when you draw glass tubing after melting it) that goes into one joint of the stillpot (drawn end in the reaction mixture). When vacuum is applied, air is forced through the tube, but because the drawn end is so narrow it only allows small bubbles into the mixture. These take the place of boiling chips which will not form bubbles as all gases are extracted from the chips via the vacuum.


Will these K-alum crystals combine into a "cohesive/well-glued" mass?

According to the Wikipedia article Potassium alum, the dodecahydrate melts at 92 to 95 °C. So if I wanted a large lump, I'd put the salt in a test tube (or some other container), dunk the container in boiling water to melt the salt, then pour the melted salt into some type of a mold to cool and solidify.


Is there an electrolyte that will halt the production of Hydrogen in water electrolysis?

The reduction of hydrogen is what the reduction potential of every other reduction potential is based off of, so:

$$\ce{2 H+_{(aq)} + 2e^{–} ->H2_{(g)}}~~~~~\Delta E^{\mathrm{o}}=0\,\mathrm{V}$$

So if a species has a has a reduction potential $>0\,\mathrm{V}$, then it will be be reduced before the naturally occurring $\ce{H+}$ ions are (these exist in solution due to the auto-protolysis of water, $\ce{H2O -> H+ + OH-}$, thereby not producing any $\ce{H2_{(g)}}$.

If your goal is to produce $\ce{O2_{(g)}}$ from the oxidation of $\ce{OH^{-}_{(aq)}}$, the reaction for this is:

$$\ce{4OH^{−}_{(aq)} -> O2_{(g)} + 2 H2O_{(l)} + 4 e^{−}}~~~~~\Delta E^{\mathrm{o}}=-0.401\,\mathrm{V}$$

Similarly, if a species has a has a oxidation potential $>-.401\,\mathrm{V}$, then it will be be oxidized before the  $\ce{OH-}$ ions, thereby not producing any $\ce{O2_{(g)}}$.

Wikipedia has a rather comprehensive standard reduction potential table, so use that to compare the reduction potential of ions. Choose your salts wisely!

To answer your other question, if this reaction continued indefinitely, a very acidic solution would result, as for every $\ce{OH-}$ that is oxidized, a $\ce{H+}$ is left in solution. Once all of the other cation from the salt is reduced, however, the $\ce{H+}$ would then start being reduced, producing $\ce{H2_{(g)}}$.


See, in an equilibrium reaction in gas phase at constant temperature, noble gases can be added in two ways―

At constant volume in the system
At constant pressure in the system.

Effect of adding noble gases at constant volume

At constant temperature and volume, if noble gas is added, total number of molecules,i.e., amount of the element increases (or you can say, 'number of moles' increases, though it's not correct, have you ever heard of number of kilograms?) in the system. As a result, also total pressure of the system increases, but there is no change in partial pressure of the reactants and products (because total pressure increases and mole fraction decreases). So, there is no effect on equilibrium by adding noble gases in a system at constant temperature and volume.

Effect of adding noble gases at constant pressure

At constant temperature and pressure, addition of noble gas to the system leads to increase in volume (cause, amount of the element increases). As a result, partial pressure of the reactants and products decreases (because total pressure constant but mole fraction decreases) as well as the sum of partial pressure of the reactants and products also decreases. Here, according to Le Chatelier's Principle, the equilibrium would be shifted in that direction where number of molecules or volume increases.


Hydrogenic Wavefunctions:
A hydrogenic wavefunction, $\psi_{nlm}(\boldsymbol{r})$, can be written as the product
$$
\psi_{nlm}(r,\theta,\phi) = R_{nl}(r) Y_{lm}(\theta,\phi)
$$
where $R_{nl}$ is a radial wavefunction and $ Y_{lm}$ is a spherical harmonic.

$R_{nl}$ is a radial wavefunction
$ Y_{lm}$ is a spherical harmonic
$n$ is the principal quantum number, $n = 1,2,3\dots$
$l$ is the angular quantum number, $l = 0=s, 1=p, 2=d,\dots,n-1$
$m$ is the magnetic quantum number $m = -l, -l+1,\dots,l-1,l$

Probability/Electron Density
The probability density, $\rho$, is equal to the square modulus of the wavefunction:
$$
\begin{align}
\rho_{nlm}(\boldsymbol{r}) &= |\psi_{nlm}(\boldsymbol{r})|^2 \\
\rho_{nlm}(r,\theta,\phi)&= |R_{nl}(r)|^2 \cdot |Y_{lm}(\theta,\phi)|^2
\end{align}
$$
The spherical harmonics $Y_{lm}$ have angular nodes through the nuclei for all values of $l$ and $m$ except $l =m =0$, so we can say that for all $l>0$ orbitals, i.e. all orbitals except s-orbitals, the nuclei will have $0$ probability density.
S-orbitals
Consider s-orbitals, in atomic units and substituting in the solution of the radial wavefunctions in generalized Laguerre polynomials $L^\alpha_k$:
$$
\begin{align}
\rho_{n00}(\boldsymbol{r})&= |R_{n0}(r)|^2 \cdot |Y_{00}(\theta,\phi)|^2\\
&= |R_{n0}(r)|^2 \frac{1}{4\pi} \\
\rho_{n00}(r)&= \frac{Z^3}{\pi n^5} \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right) \\
&= A(n) \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right)
\end{align}
$$
For all values of $n$, $L^1_{n-1}\left(0\right) = n$, so we know that the nucleus never has zero probability density for s-orbitals, and has probability density
$$
\rho_{n00}(r=0) =  \frac{Z^3}{\pi n^4}
$$
Stationary Points
But we can differentiate with respect to $r$ to find stationary points in the radial probability density.
$$
\begin{align}
\frac{d}{dr}\rho_{n00}(r)&= A(n) \frac{d}{dr} \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right) \\
&= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) +  \frac{d}{dr}L^1_{n-1}\left(2Zr/n\right) \right] \\
\text{if}\;n=1: \frac{d}{dr}\rho_{000}(r)&= -\frac{Z^4}{\pi n^6} \exp\left(-Zr/n\right) \\
&> 0 \;\forall\;r
\end{align}
$$
Hence the probability density is constantly decreasing away from the nucleus.  So as you previously assumed the maximum for probability density for the 1s orbital is at the nucleus.
$n>1$:
$$
\begin{align}
\text{if}\;n>1 : \frac{d}{dr}\rho_{n00}(r)&= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) +  \frac{d}{dr}L^1_{n-1}\left(2Zr/n\right) \right] \\
&= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) - L^2_{n-2}\left(2Zr/n\right) \right]
\end{align}
$$
Setting the expression above equal to zero and solving for $r$ will give the positions for the radial nodes and maxima inbetween for a given value of $n$.  I am not aware of a general method/expression in n.
As you asked about nucleus, we can also consider the point $r=0$:
$$
\begin{align}
\text{if}\;n>1 : \frac{d}{dr}\rho_{n00}(r)&= -  \frac{Z^3}{\pi n^5} \left[\frac{Z}{n} L^1_{n-1}\left(0\right) + L^2_{n-2}\left(0\right) \right] \\
&= -  \frac{Z^3}{\pi n^5} \left[\frac{Z}{n} n +\frac{n^2-n}{2}\right] \\
&= - \frac{Z^4}{\pi n^5} - \frac{Z^3 \left(n-1\right)}{2\pi n^4} < 0
\end{align}
$$
Conclusion/TL;DR
Hence the nuclei is always a local maximum in the probability density for an s-orbital.  You would expect the exponential decay in the radial wavefunction to dominate over the Laguerre polynomial term, so would expect local maxima further out to be smaller that the maxima at the nuclei and hence that the point of highest probability density is at the nucleus.
This makes physical sense, as classically the negatively charged electron wants to be as close as possible to the positively charged nucleus to minimise its potential energy, but the quantum mechanical nature of the electron prevents it from tumbling directly into the nucleus.


Number of electrons in pseudo wavefunction and all electron wavefunction in pseudopotentials

In pseudopotential approach the core electrons are replaced by the pseudopotential. In case of all electron calculation we obtain wavefunction for both core and valence electrons. In case of pseudopotential only the valence electron remains. So when we compare the all- electron and pseudo wavefunction we compare only the valence ones. All the wavefunctions are one electron wavefunction. The charge is basically the norm of the square of the
wavefunction. The condition of charge equivalence between the pseudo and all electron wavefunction is imposed to get correct scattering properties. This condition is known as norm conserving pseudopotential.


I have established the stereochemistry via Evans aldol addition:

Then, after protection of the hydroxyl group, the chiral auxiliary can be transformed to a Weinreb amide which will be attacked by Grignard reagent:

The methyl ketone is then transformed to an alkyne which is methylated:
Finally, hydroboration (should be very regioselective) of the alkyne will provide the trans alkenylborane which on treatment with NaOH and iodine will provide target 12:

Length (total steps and longest linear sequence): 7 steps


Do "weak-field" and "high spin" complexes refer to the same thing?

The spectrochemical series only gives info on the relative strength of the ligand field. Whether or not a metal complex is high or low spin (assuming there is an appropriate number of d electrons that both are possible) is a function of both the ligand and the metal center.
Thus, it is possible for "weak field" ligand to be in a low spin complex if the metal center favors low spin. For example, almost all complexes of second and third series transition metals are low spin, independent of ligand identities.


You can deduce the end product from considering the change in the molecular formula from start to finish. Aromatic bromination results in the loss of one hydrogen and gain of one bromine, bromination of the double bond results in no loss of hydrogen but gain of two bromines and the Sandmeyer reaction results in loss of two hydrogens. Therefore the intermediate must have fourteen hydrogens which means only one aromatic bromination has taken place.

The aromatic bromination likely produces a major product with the bromine at the ortho position shown. The amine group strongly activates the positions ortho and para to it so substitution will occur at one of these three positions. I suspect the one shown is most likely due to the bulky nature of the other attatched group, which stericically hinders attack at the other two positions.


Masamune's rule of diastereoselectivity multiplicity explained

First, we need to take a look at the possible transition states of the first reaction in your table using $\ce{9{-}BBN}$ as an enolising agent. There are two transition states shown in scheme 1: The upper, favoured and the lower, disfavoured one.

Scheme 1: Aldol reaction with $\ce{9{-}BBN}$ as an enolising agent. Both the favoured (above) and the disfavoured (below) Zimmermann-Traxler transition states are shown along with their respective products.

The difference in both transition states is in the steric clash of the α-carbon and the boron residue: If the aldehyde attacks from the si-side, the stereocentres on said α-carbon can be oriented favourably, namely with hydrogen pointing toward the most crowded area and the largest residue pointing toward the least crowded area. If the aldehyde attacks the enolate’s re-side, methyl must point towards the most crowded area to allow the largest residue to point away from crowded areas. A similar discussion can be made for the orientation of the aldehyde which could attack the enolate’s si-side with its si-side, resulting in an anti-aldol product.

It is important to note here that all four attacks are possible. However, they do not have the same probability of reacting. Based on the observed diastereoselectivity, one can say that $\mathrm{P}_R(F) \approx 92~\%$, while $\mathrm{P}_R(\bar F) \approx 8~\%$ ($R$ is the set reaction occurs and $F$ is the set favoured attack. I have subsumed the two anti-product giving attacks under $\bar F$, since they are nigh irrelevant).

Scheme 2: Aldol reaction with $\ce{(+){-}(Ipc)2BOTf}$ as the enolising agent. $\ce{Ipc}$’s methyl groups disfavour an aldehyde attack on the enolate’s re-side. This is the same as the substrate’s inherent diastereoselectivity; a matched case.

A similar case can be made for $\ce{DIP}$-mediated aldol additions in the absence of substrate diastereospecificity ($\ce{Me} = \ce{H}$). Here, we need to construct a new set $A$ meaning auxiliary-favoured attack. This auxiliary-favoured attack is shown in scheme 2, although I used the matched case from your examples, i.e. substrate diastereoselectivity is present. The auxiliary-unfavoured attack (same caveat) is shown in scheme 3.

Scheme 3: Aldol reaction with $\ce{(-){-}(Ipc)2BOTf}$ as the enolising agent. $\ce{Ipc}$’s methyl groups disfavour an aldehyde attack on the enolate’s si-side — but substrate diastereoselectivity prefers the si-side attack on the enolate (mismatched case).

Stochastically, we can consider both selectivities separately. Remember that all attack combinations are technically possible but some are just more likely to be productive. It is a good approximation to assume that $\mathrm{P}_R(A)$ and $\mathrm{P}_R(F)$ are independent of one another, which means that $\mathrm{P}_{R,A}(F) = \mathrm{P}_R(A) \times \mathrm{P}_R (F)$. Why? Well, the substrate’s diastereoselectivity is mainly based on the orientation of the residues on the α-carbon. This is pretty much independent of the orientation of $\ce{Ipc}$’s methyl groups, which cause the auxiliary’s enantioselectivity: in both methyl group orientations the same rotamer of the α-carbon has the same probability.

Having taken into account that both sets are independent, we see that we can multiply the probabilities of the different attacks to get the resulting attack’s probability. The substrate will favour (R,S) (counting from ketone to hydroxy group) in each case. To get the probability of (R,S), we multiply a large percentage (substrate selectivity) with a large percentage getting another large percentage (auxiliary selectivity). The probability for (S,R) will be a small percentage (substrate) multiplied by a small percentage (auxiliary) so even smaller. So favouring of the desired (matched) diastereomer is enhanced.

On the other hand, if the auxiliary favours (S,R), to get the probability of (R,S) we must multiply a large percantage with a small one giving a medium result. The same thing for the probability of (R,S). This means that both diastereomers have more equal chances of forming.

(A better discussion of this problem would involve relative reaction rates, as the article NotWoodward mentions does. I feel that the probability argument gets us the same conclusion without all the big kinetic calculations which I am not good at.)


Can we make glass by heating Martian Soil

The problem is sorting. Ergo, unless you bring your separator with you, the best areas for finding naturally sorted materials are in outflows such as found in Candor Chasma. Semi-translucent larger grains of quartz-rich phenocrysts might be found in useful quantities. Fusing these with a solar kiln would produce a brick of about 0.125 meters thick that would both shield from the rain of radiation and pass some of the weak Martian sunlight, easing the circadian rhythm of colonists.
Time to temperature (1700 degrees C) still makes production slow. And the loaders and other equipment need to be transported and successfully landed.

See
A Fused Regolith Structure
Ethan W. Cliffton
Proceedings of Space 90
Engineering, Construction, and Operations in Space
ASCE 1990
Albuquerque
and
Candor Chasma Camp Ethan W. Cliffton
Proceedings of Space 88
Engineering, Construction, and Operations in Space
ASCE 1988
Albuquerque


In Energy-Dispersive X-ray Spectroscopy, do higher atomic elements ever generate just K lines without L or M lines?

With a tuned X-ray source, e.g. from a synchrotron, it should, at least in theory, be possible to excite any specific line, or cause absorption at a specific wavelength, as in X-ray absorption spectroscopy.


The reactions of nitric acid could be due to 1) its acid nature, 2) its oxidizing properties, or both.

Orthocresol was correct that the substrate could predetermine which course was more likely: acid-susceptible substances (like alkalies, or alkaline materials) would respond to the acid nature; oxidizable substances would respond to the oxidizing power (like organic alcohols and sugars).

An interesting case is pure iron: in dilute nitric acid (less than 10%), it dissolves. In concentrated nitric acid, it is oxidized (passivated) and does not dissolve.

Increased temperature would also increase the likelihood of an oxidation reaction compared to an acidic reaction, because the activation energies for oxidations are usually higher.


How can oxygen atoms suddenly appear on the product side of an equation?

They probably want you to calculate the enthalpy change required to convert the reactants into the products. Another way of looking at it is the enthaply difference between the products and reactants.

Remember, $\ce{O2(g)}$ has 0 enthalpy of formation in standard state. (so do $\ce{C(s)}$ and $\ce{H2(g)}$, but we don't need that)

So, adding $\ce{2O2}$ to the reactants side won't change $\Delta H^o=H^o_{products}-H^o_{reactants}$. So calculating $\Delta H^o$ for the given "reaction" is the same as calculating it for $$\ce{C(s) + 2H2(g) +2O2 -> CO2(g) + 2H2O}$$. And $\Delta S^o$ for this reaction can be calculated very easily using enthalpies of combustion. You don't even need to use the given value for methane.

So, while the question is wrong in calling it a reaction, it is still asking us to do something reasonable -- that is, calculate $$\Delta H^o=H^o_{\ce{CO2(g) + 2H2O}} - H^o_{\ce{C(s) + 2H2(g)}}$$. Since we can subtract $H^o_{\ce{2O2(g)}}=0$ easily, the problem can be solved by taking a combustion path.


Reaction between sulfur and sodium hydroxide

The reaction products depends on the conditions(temperature, pressure etc.) and stoichiometric ratio of reactants taken. Depending on that, you will get various products in each case if you slightly change the conditions or the ratio of sodium hydroxide to sulfur taken.

$$\ce{4S + 8NaOH ->[600 C] Na2SO4 + 3Na2S + 4H2O}$$

  Sulfur react with sodium hydroxide to produce sodium sulfate, sodium
  sulfide and water. This reaction takes place at a temperature of over
  600°C.(source)

$$\ce{4S + 6NaOH → Na2S2O3 + 2Na2S + 3H2O}$$

  Sulfur react with sodium hydroxide to produce sodium thiosulfate,
  sodium sulfide and water. Sodium hydroxide - concentrated solution.
  The reaction takes place in a boiling solution.(source)

Your reaction

When you compare the three equation, you can  see that though stoichiometric ratio of the reactants in the 1st and 3rd equation is same, the moles of sulfur and sodium hydroxide  is different and thus gives different products in each case.

Also, from reaction 2 we can say that sodium thiosulfate may have disproportionate into sodium sulfite and sulfur.

$$\ce{Na2S2O3 ->[220-300 C] Na2SO3 + S}$$

and thus one mole sulfur cancel from each side and gave reaction 3 (your reaction). This site  explains the disproportionation reaction.

  Chemical reactions of sulfur compounds have been studied in equimolar $\ce{NaOH-H2O}$ melt at 100°C by voltammetry and UV spectrophotometry. Disproportionation of sulfur is fast and quantitative according to:

  $$\ce{S8 + 6OH- -> 2S3^2- + S2O3^2- + 3H2O}$$

  On the contrary, addition of sulfur to sulfide is not quantitative and
  gives $\ce{S2^2-}$ besides prevailing sulfur
  disproportionation.$\ce{SO3^2-}$ reacts with $\ce{S2^2-}$ and
  $\ce{S3^2-}$ , while $\ce{S2O3^2-}$ decomposes $\ce{S^2−}$. A
  pseudoequilibrium is observed:

  $$\ce{S^2- + S2O3^2- <=> S2^2- + SO3^2-}$$

  with concentration quotient Formula Q = $\ce{10^{-0.8}}$.


Equivalence point of titration of Sodium carbonate

When the $\ce{Na2CO3}$ is completely neutralized by the $\ce{HCl}$, the solution will be saturated with $\ce{CO2}$, so the pH will be lower than 7 - think of carbonated beverages. Although the $\ce{CO2}$ bubbles out, not all of it bubbles out - not even if stirred. If you had stopped the titration at pH = 7, the pH reduction from 11 would have been partly from $\ce{HCl}$ and partly from $\ce{H2CO3}$.


Terminology of redox reactions: which species are the oxidizing and reducing agents?

Ignoring spectator ions,

$\ce{Cl2 +2 I- ->2Cl- +I2}$

So, $\ce{Cl2}$ is the oxidant and $\ce{I-}$ is the reductant.


Is it possible to create an acid anhydride that's stable in water and if so what conditions must its structure satisfy?

This answer is years later than the original question, but I add it for anyone who come across it now or in the future...
All acid anhydride and acid chlorides are thermodynamically unstable in water.  They will eventually hydrolyze, but that time could be seconds, minutes, hours, days, or years. Moreover, in the absence of a catalyst, the rate of hydrolysis (kinetics) of any acid anhydride or any acid chloride (carboxylic, phosphoric, sulfonic, etc.) is strongly temperature dependent (hotter = faster, colder = slower, as you would expect).
For example, the classic hydrolysis of acetic anhydride is quite slow until temperatures exceed 55—60 °C.  Similarly, sulfonyl chloride hydrolysis is very slow until temperatures exceed about 70 °C.  This is in part driven by solubility - the substrate needs to be in the same phase as the water for fast rates.  Phosphoric anhydrides such as ATP or ADP can be extremely stable in water at physiological temperatures (37 °C).  The presence of an protic acid, or particularly  a nucleophilic catalyst (hydroxide, amines, thiols), can change the kinetics dramatically.  Thus, you have to pay attention to the temperature, the phasing/solubility, and the identities/properties of the other components in your reaction mixture.


How can you infer that density decreases when ice melts from the water phase diagram?

Draw an isotherm at a low enough temperature (< the "triple point temperature of 0.01°C), so that it passes through "all three" phases (ignoring the high-pressure ices that are not relevant to the question).

Note that as pressure goes up you go from solid to liquid.  Increasing pressure always favors a denser phase, so liquid is denser than solid for water.

Water is not quite unique in this respect. Silicon and germanium are two elements that do the same thing.


Limiting molar conductivity of multiple solutes

I've found a answer for my question from this question.

The way to compute the conductivity of electrolyte with multiple ion types is given from the work Pawlowicz, Rich, ( 2008), Calculating the conductivity of natural waters, Limnol. Oceanogr. Methods, 6, doi:10.4319/lom.2008.6.489.

For general case consider the system, which consist of $N_+$ number of cation types, $N_-$ number of anion types and $N_{types} = N_+ + N_-$ total number of types, fully dissolved in solvent. This system can be viewed as a weighted sum of all possible pairwise combinations between cation and anion types. Then the conductivity $\sigma,\pu{[S m-1]}$ of the solution will be determined by:
$$\sigma = \sum_{i=1}^{N_+}\sum_{j=1}^{N_-}{\frac{c_i^+ z_i c_j^- z_j}{C_{eq}} \Lambda_{eqm, \, ij}}$$
where $c_i^{\pm}$ - molar concentration of corresponding ion type, $\pu{[mol m-3]}=\pu{[mM]}$;

$z_i$ - valency of corresponding atom;

$c_i^{\pm} \cdot z_i$ - equivalent ionic concentration of corresponding ion type $\pu{[mol m-3]}=\pu{[mM]}$;

$\Lambda_{eqm, \, ij}$ - equivalent molar conductivity of binary subsystem of cation type $i$ and anion type $j$, $\pu{[S m^2 mol^{-1}]}$.

Equivalent ionic concentration $C_{eq}$ is defined as:
$$C_{eq} = \sum_i^{N_+}c_i^{+} \cdot z_i = \sum_j^{N_-}c_j^{-} \cdot z_j = \frac{1}{2}\sum_k^{N_{types}}c_k^{\pm} \cdot z_k$$

In general a binary solute dissociates according to formula:
$$\ce{A_{\nu^+}B_{\nu^-} -> {\nu^+}A^{z^+ +} + {\nu^-}B^{{z^-}-}}$$
where $\nu^+$ and $\nu^-$ are coprime numbers and represent the moles of ions  for 1 mole of solute. It follows:
$$\nu^+ = z^-$$
$$\nu^- = z^+$$

Equivalent molar conductivity $\Lambda_{eqm}$ of single binary electrolyte is defined as:
$$\Lambda_{eqm} = \frac{\sigma}{c\nu_+z^+} = \frac{\sigma}{c\nu_-z^-}$$

For a binary solute, dissolved in water at normal pressure, $\pu{25 °C}$ and infinitely diluted, the law of the independent migration of ions can be written to find a limiting equivalent molar conductivity $\Lambda_{eqm}^{0}$:
$$\Lambda_{eqm}^{0} = (\frac{\nu_+}{z^+ } \lambda_{+}^{0} + \frac{\nu_-}{z^- }\lambda_{-}^{0}) = (\frac{z^-}{z^+ } \lambda_{+}^{0} + \frac{z^+}{z^- }\lambda_{-}^{0})$$
where $\lambda^{0}$ - limiting ionic conductivity of the ion, $\pu{[S m^2 mol^{-1}]}$.

Here interaction between ions is ignored.

Change of equivalent molar conductivity from changing of the solute equivalent concentration is determined by Debye-Hückel-Onsager equation of the form:
$$\Lambda_{eqm} = \Lambda_{eqm}^0 (1-A \sqrt{I}) - B \sqrt{I} =  \Lambda_{eqm}^0 K_A - K_B$$
where $A, \, B$ - Debye–Hückel–Onsager coefficients;

$I$ - stoichiometric ionic strength;
$$I = \frac{1}{2} \sum_{k=1}^{N_{types}}c_i z_i^2$$
$$ A = \frac{z^2eF^2}{3 \pi \eta}\left(\frac{2}{\varepsilon RT}\right)^{1/2}$$
$$ B = \frac{qz^3eF}{24 \pi \varepsilon RT}\left(\frac{2}{\varepsilon RT}\right)^{1/2}$$
where $\eta$ - viscosity of solvent, $\pu{[Pa s]}$;

$\varepsilon$ - dielectric permittivity of solvent;

$q$ - coefficient depends on $z^+/z^-$ of binary solute.

The expression for the conductivity of electrolyte (solvent - water at $\pu{25 °C}$ and normal pressure, ion pairng reduction factor is ignored ($\alpha_{ij}=1$ for all $i,j$)) will be:
$$\sigma = \sum_{i=1}^{N_+}\sum_{j=1}^{N_-}{\frac{c_i^+ z_i c_j^- z_j}{C_{eq}} (\frac{z_j^-}{z_i^+} \lambda_i^0 \, K_{A, \, ij} + \frac{z_i^+}{z_j^- } \lambda_j^0 \, K_{A, \, ij} - K_{B, \,ij})}$$

In my case I have 2 cations and 2 anions, their pairwise combination give 4 different solutes, which are assumed to be completely dissociate in water. Relation between solute molar concentration and its ions molar concentrations can be established.

$\ce{NaCl -> Na+ + Cl-}$
$\ce{NaHCO3 -> Na+ + HCO3-}$
$\ce{MgCl2 -> Mg^2+ + 2Cl-}$
$\ce{Mg(HCO3)2 -> Mg^2+ + 2HCO3-}$

Conductivity of my electrolyte will be:
\begin{align*}
\sigma &= \frac{1}{[\ce{Na+}]+2 \cdot [\ce{Mg^2+}]} \cdot \\
           &\cdot \Bigl(\ce{[Na+]}  \ce{[Cl-]}       K_{A(\ce{NaCl})}        ( \lambda_{\ce{Na+}}^0 + \lambda_{\ce{Cl-}}^0      - \frac{K_{B(\ce{NaCl})}}{K_{A(\ce{NaCl})}}      )+\\
           &+             \ce{[Na+]}  \ce{[HCO3-]}  K_{A(\ce{NaHCO3})}  ( \lambda_{\ce{Na+}}^0 + \lambda_{\ce{HCO3-}}^0 - \frac{K_{B(\ce{NaHCO3})}}{K_{A(\ce{NaHCO3})}})+\\
           &+             2\ce{[Mg^2+]}  \ce{[Cl-]}       K_{A(\ce{MgCl2})}        ( \frac{1}{2}\lambda_{\ce{Mg^2+}}^0 + 2\lambda_{\ce{Cl-}}^0      - \frac{K_{B(\ce{MgCl2})}}{K_{A(\ce{MgCl2})} }      )+\\
           &+             2\ce{[Mg^2+]}  \ce{[HCO3-]}  K_{A(\ce{Mg(HCO3)2})}  ( \frac{1}{2}\lambda_{\ce{Mg^2+}}^0 + 2\lambda_{\ce{HCO3-}}^0 - \frac{ K_{B(\ce{Mg(HCO3)2})}} {K_{A(\ce{Mg(HCO3)2})}})
           \Bigr)
\end{align*}

All the values here depend from given concentration of ions or from values found in tables. Hope it is correct, please, correct me if I'm wrong.