What's the technical term for supercharging an acid with an oxidizer? And does it always work?

I don't know if it's called anything, but when you use an oxidizer in solution, you usually need a source of hydrogen ions for the oxidized oxygen to be reduced.  In the case of hydrogen peroxide, $\ce{H2O2}$, the oxygen is in the $-1$ state rather than the normal $-2$.  The usual way $\ce{H2O2}$ acts as an oxidizer is to be reduced to water in the presence of 2 hydrogen ions, usually from a strong acid:

$\ce{2H+ + 2e- + H2O2 -> 2H2O}$

Notice that the reaction consumes 2 electrons on the left, which would come from other reactants that have been oxidized (i.e. lost electrons).

The point is that a strong acid by itself has only a corrosive effect, and a highly dehydrating effect in the case of sulfuric acid.  But if you add an oxidizer, then in the presence of a fuel (like an organic compound) plus sufficient energy (say from the heat of reaction), then you are effectively getting a combustion reaction as well.  This is the sort of "combustion" that goes on inside your body to use food as fuel, for example.  I supposed this could be called "supercharged".

EDIT: So to answer your specific questions:

1) No technical term that I know of;

2) Yes, assuming the acid is strong enough to donate protons;

3) In most organic molecules, carbon is more electronegative than hydrogen, so it can be oxidized by oxygen.  And so you get a "combustion" reaction where molecular carbon is oxidized to elemental carbon (i.e. burnt).


Understanding the Jahn–Teller elongation and compression for octahedral complexes (stabilisation/destabilisation of d orbitals)

Before you consider J-T distortion, take a look at the d orbitals of an octahedral complex. Note that the orbitals that are on the axes where the ligands are are higher in energy. This is because these "d orbitals" are actually anti-bonding orbitals comprised primarily of the metal d orbitals. Because they are anti-bonding, they get higher in energy as the bonding interaction between the ligand and metal gets stronger. Thus, a longer distance between the ligand and metal (ie weaker bonding interaction) results in less destabilization.
Another way to think about it is to consider the interaction of a ligand lone pair (ie negatively charged electron density) with a filled d orbital on the metal (also negatively charged electron density). The closer they are together, the more unfavorable the interaction is.


Mixing cobalt(III) hydroxide and thallium(III) hydroxide

This should be a comment, but I decided to post it as an answer, since from my point of view, the present answers fail to shed light on the absolute simplicity of the truth.
We can state from the present answers that there will be no precipitates, but not because the concentration is smaller than an atom per liter, we don't know the volume used since it was not stated.
The simple fact is that the $[\ce{OH^-}]$ concentration stays the same, it will be $10^{-7}$ before and after the mixing of the solutions, no matter how much from each solution we will mix together, since the solubility product is so small and the dominating factor for the $[\ce{OH^-}]$ concentration comes from the autoprotolysis of water. Accepting this little knowledge, no matter what we will never reach the saturation concentration after mixing. The saturation concentration will stay the same always (in this problem), so we will just dilute the initial concentration.
Imagine we mix an equal amount of both solutions, then the before saturated concentration of the substances will be halved. So in the end in this specific case for an example the solution will be half saturated with both hydroxides. That is why there is no precipitation.
Edit: Also there is no sense talking about common ion effect, since this phenomenon arises from the charge balance equation.
$$3[\ce{Tl^{3+}}]+3[\ce{Co^{3+}}]+[\ce{H^+}]=[\ce{OH^-}]$$
As we postulated before $[\ce{H^+}]=[\ce{OH^-}]$, since the autoprotolysis is the domination factor, both of the concentration of the ions will be zero (simplified). In truth the $[\ce{OH^-}]$ concentration will be just slightly higher than the $[\ce{H^+}]$ concentration, but we can neglect this equation since we assumed the only $[\ce{OH^-}]$ concentration comes from autoprotolysis, meaning we don't have to solve a system of equations, just plugging in the value in the solubility product.


Synthesis of 2-phenoxybutanal from but-1-yne

Water, BH3, H2O2, NaOH

This reaction, an hydroboration, yelds butanal

Br2, H2SO4

This reaction, carried without light, is an alpha carbon bromination, and yelds 2-bromobutanal

Sodium phenoxide

This is a simple Williamson Ether Synthesis, which yelds the final product


When an enzyme is diluted with water, what is slowing down the rate of reaction?

[…] there's the same number of catalase molecule within an undiluted and a diluted solution. Also, seeing as water is not an enzyme inhibitor, […]

Q: Why is dating between a fixed number of guys and girls more difficult on a large square packed with nuns than in a small club (without the nuns)? ;)

A: It's more difficult to bump into each other since the girls a more far away and there are legions of nuns in between!

You are right in that the absolute amount of catalase molecules does not change upon dilution, but the concentration does!

Reactions are about collisions of the reactants and in the case of dilutions, there are more collisions with innocent bystanders than with an other reactant.


Finding/Calculating Thermal Conductivity of Magnesite

In the dissertation of K. Kratz (2006) the following values of thermal diffusivity along the crystallographic axes of magnesite are listed:

$D_{[001]} = 6.55 ~\mathrm{mm^2\,s^{-1}}$,
$D_{[100]} = 2.38 ~\mathrm{mm^2\,s^{-1}}$,
$D_{[010]} = 2.38 ~\mathrm{mm^2\,s^{-1}}$.

Ch. Clauser and E. Huenges, Thermal Conductivity of Rocks and Minerals lists the following values for the thermal conductivity of magnesite in the temperature range from 25-100°C:

$\lambda_{\perp} = 7.32 \pm 0.57 ~\mathrm{W m^{-1} K^{-1}}$
$\lambda_{\mid} = 7.86 \pm 0.17 ~\mathrm{W m^{-1} K^{-1}}$


As an example, let's say you're titrating a strong acid with a strong base using phenolphthalein as an indicator. As you initially add the base to the acid, you may not see the color change occurring because the acid so readily neutralizes the added base. This is because of the large concentration of H+. As you approach a basic pH, however, concentration of H+ decreases and the acid is not able to neutralize the base as quickly. As the base disperses in the solution more slowly, the phenolphthalein will have time to react and show a slight color change.


Chemical bonding based on hybridisation model

The usual explanation for $sp$ orbitals being more electronegative than $sp^3$ is the fact that more $s$ character means a shorter average distance from the nucleus, thus experiencing a greater effective nuclear charge. This partially explains the decreasing magnitude of the dipole moment of $\ce{CH3-CH2-Cl}$ to $\ce{CH2=CH-Cl}$ to $\ce{HC#C-Cl}$ and the increasing acidity of ethane to ethylene to acetylene.

I agree with you that the example they give doesn't really demonstrate that, since the orientation of the dipole (as you pointed out) relies on the electronegativity of the nitrogen, not the carbon. Specifically, the stability of the left resonance structure below:

$[\ce{^-N=C=Cl+  <-> N#C-Cl}]$

which clearly shows the orientation of the dipole, even if this resonance structure is only a minor contributor.

As a counterexample, chloroacetylene has a dipole with the (-) end on the chlorine, even though the carbon is still $sp$ hybridized. The corresponding resonance structure with a negative charge on C is much higher in energy than in the nitrogen case and so contributes negligibly to the overall structure, so the inductive electron-withdrawing of Cl dominates over the resonance electron-donating.


Which organic compound's NMR and IR are these? Can't find out. (Data available: Melting Point ~ 116 °C)

The integrals and shifts in the very simple NMR spectrum are a giveaway that this is a simple compound with two different types of aromatic signal, and another non-aromatic signal from a substituent, at a 2:1 peak integral ratio, hence possibly a para-disubstituted benzene ring with one proton in one substituent, none on the other.

The IR spectrum is practically identical to that in the NIST webbook for p-nitrophenol.
Some other clues:

Oop Signal at $\pu{800 cm^-1}$: para substitution
$\pu{1400-1600cm^-1}$: aromatic cc stretch, matches nicely with typical appearance for para substitution
No peak at $\pu{1700cm^-1}$: no carbonyl
Weak signals at $\pu{3000cm^-1}$: sp2 stretch, no sp3
Signals at $\pu{1350 and 1500cm^-1}$ may be nitro

Note this is a little tricky - a search would show some variation in reported spectra.

The melting point also matches reported values.

This is well in agreement with many of the comments under the original post.


Difference between quick lime, slaked lime, lime water and soda lime

Quick lime is calcium oxide, $\ce{CaO}$, in the solid state.

Slaked lime is calcium hydroxide, $\ce{Ca(OH)2}$, in the solid state.

Lime water is a dilute solution of calcium hydroxide, $\ce{Ca(OH)2}$, in water.

Soda lime is a mixture of sodium hydroxide, $\ce{NaOH}$, (also known as caustic soda) and calcium oxide, $\ce{CaO}$, (also known as quicklime).


Heat of hydrogenation of 1,3-substituted cyclohexanes

First of all, as Jan pointed out on the comment, better nomenclature for thes compounds is syn- and anti- instead of cis- and trans-, respectively. However, OP is more familiar with cis- and trans- nomenclature I'd continue it with my answer.

Each substituted cyclohexane ring (e.g., di-substituted ones here) has two chair conformers as depicted in OP's bottom diagram of methylcyclohexane (a mono-substituted one). In one conformer, the methyl group is in equatorial position (left hand structure) and the methyl group in the other one is in axial position (right hand structure). Because of two 1,3-interactions cost additional approximately $2 \times \pu{3.8 kJ/mol}$ steric strain energy, the predominant conformer is equatorial one.

Now, if you put tert-butyl group in $\ce{C3}$-equotorial position on equatorial-methylcyclohexane (left hand structure, vide supra), you get cis-1,3-substituted compound (C). On the other hand, if you put tert-butyl group in $\ce{C3}$-equatorial position of axial-methylcyclohexane (right hand structure), you get trans-1,3-substituted compound (D). Because it is too bulky, the tert-butyl group is always in equotorial position in cyclohexane ring (1,3-interaction with one hydrogen would cost additional approximately $\pu{11.4 kJ/mol}$ steric strain energy), as depicted in following diagram (The larger the substituent, the more the equatorial substituted conformer will be favored):

The most predominant cis-isomer (C) has both substituents in equatorial-positions (the left hand side structure of (C)). Other conformer (get by the ring flip; the right hand side structure of (C)) has both substituents in axial-positions. Thus, as a rule of thumb, the conformer with both substituents in equatorial positions is more stable.

On the other hand, the both trans-isomers of (D) have one substituent in an equatorial-position and the other in an axial-position. Yet, the conformer with the tert-butyl group in the equatorial-position (left hand side structure of (D)) is more stable between two, because tert-butyl group is comparatively much larger than methyl group. However, because most predominant cis-isomer (C) has both groups in the equatorial-positions, it is relatively more stable than (D), which has methyl on axial-position, making additional steric strain energy.


Are there natural conditions that could enable the formation of noble gas compounds?

As noticed in the comments, there are probably no natural conditions that would allow formation of noble gas compounds; although it should be noted that definition of 'natural' conditions should be specified. Lightning bolts surely carry enough energy to force such a reaction, but it is impossible for them to meet right mixture of gases, due to scarcity of both noble gases and super-strong oxidants and abundance of other substrates; as noted in comments (again), substrate able to create a compound with noble gas will create it with everything else first, and noble gas last.


Metals have several general properties in common (to some degree):

Luster - metals are shiny
Electrical conductivity - metals conduct electricity
Thermal conductivity - metals conduct heat
Ductility - metals can be drawn into wires
Malleability - metals can be beaten into shapes
Fusibility - metals can be melted and forged

All of these properties derive from the "metallic bond", a delocalized sharing of electrons throughout the metal lattice, which is the key necessity for displaying macroscopic metallic properties. All metals do not behave the same way for these properties. For example, gold is more malleable than iron.

Metalloids are those elements with one or more allotropes that have metallic characteristics. Few if any metalloids display all of the metallic characteristics (otherwise they would be metals!). For example, graphite is lustrous and electrically conductive, but not thermally conductive, malleable, ductile, or fusible. The metalloid with the most metallic behavior is tellurium, which is fusible and ductile.

As for compounds, many are like the metalloids - they have some properties of metals but not all. For example, silicon carbide is lustrous, but that's about it. Otherwise it is hard, brittle, and an insulator.

Polyacetylene and derivative polymers are sometimes call "synthetic metals" because of their electric conductivity and sometimes luster. However, again, they are thermally insulating. They also tend to burn instead of fuse.

Metalloid elements and compounds are not quite metallic because their electron delocalization is in only one (polyacetylene) or two (graphite) dimensions, not all three.

  Update!

Based on Snipergirl's comment, it appears that we are closer than I originally thought to preparing true synthetic metals. This highlight at phys-org suggests that supramolecular lattices of redox-coupled conjugated organic ions can exhibit metallic behavior at low temperature. The full article is at Nature Communications and has been made open access. These compounds appear to be close to achieving delocalization in all three dimensions. The behavior of this organic material is not surprising: metals and metalloids become even more metallic at cryogenic temperatures. The induced superconductivity allows for such cool things as magnetic levitation and MRI.


Where does the molar mass difference between N₂ and CO come from?

The differences in molecular mass stem from two sources:

  Nuclear binding energy

The definition of the atomic mass unit and Avogadro's number (and thus the g/mol), is the mass of one atom of the carbon-12 isotope $\ce{^{12}_6 C}$. 1 amu = $\frac{1}{12}$ the mass of one atom of $\ce{^{12}_6 C}$ and the g/mol unit is based on the definition of Avogadro's Number, which is the number of $\ce{^{12}_6 C}$ atoms in 12 grams of $\ce{^{12}_6 C}$.

Because of the differences in binding energies in the various nuclei, the monoisotopic mass of individual isotopes are not integers. For example, the mass of $\ce{^{16}_8 O}$ is 15.99491463 amu (and not 16). The monoisotopic masses of the most common isotopes of oxygen, nitrogen, and carbon is shown below:

           mass number   % absundance    isotope mass
Carbon     12            98.930          12 (defined)
Nitrogen   14            99.632          14.003074
Oxygen     16            99.757          15.99491463

  Isotopes

Generally, molecular masses (unless otherwise specified) are average molecular masses composed of average molecular masses of the elements based on their isotopes. For example, even though the most common isotope of carbon has a mass defined as 12 amu, there is another naturally occurring isotope $\ce{^{13}_6 C}$ with 1.07% natural abundance, leading to carbon having a non-integer average atomic mass of 12.01 (to 4 significant figures). There are two isotopes of nitrogen, and three isotopes of oxygen.

            Average atomic mass
Carbon      12.011
Nitrogen    14.007
Oxygen      15.999

Thus, $\ce{CO}$ has a molecular mass of $12.011 + 15.999 = 28.010$ and $\ce{N2}$ has a molecular mass of $14.007+14.007=28.014$. Even their monoisotopic masses are different:

                    average molecular mass       monoisotopic mass
dinitrogen          28.014                       28.006148
carbon monoxide     28.010                       27.99491463


In the production of white paper, the bleached pulp is often treated with optical brightening additives (OBA), that absorb in the uv and emit in the blue range. If I remember correctly, a typical example is 4,4'-diaminostilbene-2,2'-disulfonic acid.

Here's an article from The Open Textile journal (OA) in which the compound mentioned above is used as a starting material in the synthesis of another optical brightener for textiles.


Quoting from a comment of your’s:

  Hydrogen cannot make two covalent bonds since it has only one orbital in which an electron is present.

This sentence is correct for simple two-electron-two-centre bonds; i.e, if each of the two bonds going to hydrogen were to represent a 2e2c bond, hydrogen would have a double negative charge.

However, the $\ce{HF2-}$ anion does not contain two 2e2c bonds. Rather, it should be understood as a 4-electron-3-centre bond, similar to the situation in $\ce{ClF3}$. This can be understood by the following graphic:

Figure 1: Scheme of the 4-electron-3-centre bond in $\ce{HF2-}$. Image taken from Professor Klüfers’ web scriptum of his general and inorganic chemistry course.

The lowest molecular orbital, $\sigma_1$, is a fully bonding $\ce{F-H-F}$ orbital with two electrons. The second, $\sigma_2$ is nonbonding; no (accessable) hydrogen orbital exists that could complete it. The final one, $\sigma_3$, is fully antibonding. This is similar to the orbital scheme of extended linear π systems, where the number of nodal planes increases by 1 with every higher orbital. And it should be understood as such: the nodal plane of $\sigma_2$ traverses the hydrogen atom. Thus, while formally nonbonding, $\sigma_2$ is actually weakly bonding for both $\ce{H-F}$ bonds. If we calculate the overall bond order, we still count the electrons in $\sigma_2$ as nonbonding, giving a bond order of $0.5$ for each $\ce{H-F}$ bond.

The molecule can be understood in 2e2c Lewis terminology by using the following two resonance structures:

$$\ce{F-H\bond{...}F- <-> ^-F\bond{...}H-F}$$

‘Multiple bonds’ going away from a single hydrogen is not restricted to the 4-electron-3-centre bond known in $\ce{HF2-}$; they are also known and well-studied in boranes, although they are typically 2-electron-3-centre bonds in those electron-deficient compounds.


Would sodium salicylate dissolve in HCl?

HCl is in almost all cases already aqueous for common chemistry, mostly dissociated into $\ce{H^+} + \ce{Cl^-}$ because it is a strong acid, which means water is the primary solvent, not HCl. Pure HCl under ambient conditions is actually a gas (thus certainly not a pure liquid solvent).

That leaves the compound in question insoluble in the water after being protonated by $\ce{H^+}$ to form the insoluble acid molecule.

The process I mentioned in the comment is illustrated here:


How would I name O-methylations on sugars while retaining the sugar’s common name?

It is possible to name such compounds based on retained carbohydrate names in accordance with the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as follows.

P-102.5.6 Monosaccharide derivatives
P-102.5.6.1 O-Substitution
In order to maintain the integrity of structures and take advantage of retained names to imply the absolute configuration, O-substitution is allowed in carbohydrate nomenclature. Substituents replacing the hydrogen atom of an alcoholic hydroxy group of a monosaccharide or monosaccharide derivative are denoted as O-substituents. The substitution of an anomeric hydroxy group is discussed in P-102.5.6.3.2. The O-locant is not repeated for multiple substitution by the same atom or group. Number locants are used as necessary to specify the positions of substituents; they are not required for compounds fully substituted by identical atoms or groups.
P-102.5.6.1.1 O-Acetyl and O-alkyl functionalization.
For O-acyl derivatives, names with the acid component cited as a separate word ending in ‘ate’ after the monosaccharide name are preferred to names using O-acyl group prefixes. However when the ose ending is changed (e.g. to denote a glycosyl or an acid function having seniority over an ester) O-acyl prefixes are required. O-Alkyl derivatives are always expressed by prefixes.

Therefore, the name for the corrsponding D-arabinose derivative with methyl substituents replacing the hydrogen atoms of two alcoholic hydroxy group is 2,4-di-O-methyl-D-arabinose.


Why do some gases liquefy easily under pressure and normal room temperature?

The gas laws you describe, Boyle's and Charles' Laws, can be combined into one relationship known as the Ideal Gas Law. According to this law,

$$PV=nRT$$

Where

$P$ is the pressure of the gas ($\ce{Pa}$),

$V$ is the volume of the gas ($\ce{m^3}$),

$n$ is the number of moles or amount of gas ($\ce{mol}$),

$R$ is a constant ($R\sim 8.31 \ce{\frac{J}{mol.K}}$), and

$T$ is the temperature of the gas ($\ce{K}$).

To see how changing one property affects another property of the sample of gas, hold the other values constant and see what must happen.

For example, according to the law, an increase in Pressure will result in a decrease in Volume, if Temperature is held constant.
But if the Volume is instead held constant, then the Temperature will increase instead.

In effect, the ideal gas law is the combination of the other laws affecting a gas and its properties.

But this does not explain why some substances are liquids at room temperature while others are not and require cooling and/or a change of pressure.

To describe this, an understanding of what are known as Intermolecular forces must be expressed.

Between molecules of a given substance, there exist attractions. These attractions vary in strength and are what determine the pressures and temperatures needed to liquify specific chemicals.

The (usually) weakest kind of attractions are known as London Dispersion Forces. These forces are generated randomly due to the random movement of electrons about a chemical. Since the movement is random, there come times where the electron density is more negative on one side of a molecule than another. This results in a very slight charge, which can induce charges onto neighboring molecules and make them attract each other.

The next class of attractions are known as Dipole-Dipole Interactions. This is a result of a permanent charge density difference across a molecule. In these molecules, one or more atoms are much greater in their affinities for electrons and will draw the density away from the atoms with lesser affinity. This makes for a permeant slight charge on one or more sides of a molecule that are positive, while the other sides become negative. The negative ends of one molecule will attract the positive ends of neighboring molecules, resulting in a usually pretty strong attractive force.

The strongest by far class of interactions are known as Hydrogen Bonding. This is when hydrogen is bonded to en extremely electronegative atom, like oxygen or fluorine. Take water for example. The Oxygen atom greatly skews the electron density, making the oxygen end much more negative in charge than the positive hydrogen ends. This makes for very strong intermolecular attractions.

With these forces understood, you can now see that chemicals vary in the strength of their intermolecular attractions.

All that said, the weaker the intermolecular attractions, the easier it is for the molecules to break their attractions and become gaseous. Such chemicals with weak attractions require so little energy to become gaseous that at room temperature and standard pressure, the molecules act in the gas phase. To make these chemicals turn into liquids, one must either increase the pressure to make the formation of attractions more favorable or lower the temperature to lessen the energy of the molecules so that they don't break their attractive forces.

Some chemicals have extremely weak attractive forces, meaning that the temperature will have to be very low or the pressure very high to force the chemical into its liquid state.

Other chemicals, like water, exhibit such strong forces that the temperature has to be raised or the pressure lowered to vaporize them.

So yes, the gas properties of pressure, temperature, and volume do affect changes of state, but its the strength of the intermolecular attractions that are the most important and are the reasons that chemicals behave so differently under the same conditions.


How to display ligands as ball and stick and macromolecules as cartoons (if protein) / rings (if nucleic acids) in Avogadro?

This was also asked on the Avogadro discussion forum

It's possible in Avogadro 1.x and discussed in the documentation

Select part of the molecule
Click on the "options" button for a particular display type (e.g. "ball and stick")
Click on the button to only show the selected atoms


Practical differences between storing 2-electron integrals and calculating them as needed?

It depends on system size. There is a crossover point where it becomes faster to calculate the integrals as needed because as a system becomes larger and larger, the number of disk operations required to store the integrals becomes very large which greatly slows down the calculation. This is because disk operations are relatively slow compared to memory manipulation.

Thus, most software packages have an option to calculate the integrals directly and not store them. In Molpro I believe this command is gdirect.

To be more specific, there are two aspects of the storage which are slow when there are a large number of integrals. One is that disk operations themselves are slower than just doing the calculation which is only memory manipulation. The other is that the integrals are stored in some fancy way so as to minimize the amount of time it will take to find the specific integral you are looking for, but when there are a very large number of integrals, some non-negligible amount of time will go into finding the value of the integral. When these two components (plus maybe others I am missing) are larger than the time required to just do the integral calculation, then it is faster to compute the integrals directly.

I'll let someone else be more quantitative with the timing of disk operations and the lookup tables as these are more of computer science questions. But quite interesting ones.

For a useful reference on how many integrals we are talking about storing on disk, see this question.

Also as a practical guide, in my experience this crossover point was around 14 hevay atoms with a triple-zeta basis set, but this means almost nothing because this will be heavily dependent on the computing architecture you are using. I would expect the crossover point would be much higher for most supercomputing systems as I did not have much disk to work with comparatively.


How is solid titanium(II) oxide an electrical conductor?

The electron configuration of $\ce{Ti^{2+}}$ is $\mathrm{[Ar]~3d^2~4s^0}$.

$\ce{TiO}$, like e.g. $\ce{FeO}$, $\ce{CoO}$ and $\ce{NiO}$, adopts the rock salt ($\ce{NaCl}$) structure.

Due to the lower nuclear charge of $\ce{Ti}$ its 3d orbitals are less contracted than that of $\ce{Fe}$, $\ce{Co}$ and $\ce{Ni}$ which allows for overlapping of the 3d orbitals and the formation of a metal d band in $\ce{TiO}$. The d electrons of $\ce{Ti^{2+}}$ partially fill this band and lead to electrical conductivity. $\ce{FeO}$, $\ce{CoO}$ and $\ce{NiO}$ on the other hand are electrical insulators.

Note that in this case it is electrons that act as the mobile charge carriers, not the ions.


Which attack occurs first: nucleophilic acyl substitution or nucleophilic substitution?

The answer is - it depends. On the conditions, on the solvent, on the nucleophile being used, on which ester is being used. You may get selectivity, you may get attack at both sites which generates a mixture.

In the real world of chemistry you often don't know for certain until you run the experiment unless someone has published an exact procedure for the transormation which you follow.


tl;dr The main structural component of chocolate is cocoa butter, which is a blend of fatty acids (primarily oleic, palmitic, and stearic acids). Cocoa butter has multiple crystal structures, and manufacturers target a specific form which melts at around 33 °C. The fact that chocolate is about 90% sugar and cocoa butter is why chocolate is (chemically) insensitive to temperature changes.

While chocolate itself is a giant mess of hundreds of different compounds, the primary structural material of chocolate is fat. Wikipedia lists a typical composition (by mass) of about 60% sugar, 30% fats, and 10% proteins/other. So it's not terribly surprising that chocolate doesn't sublime or decompose. You probably wouldn't expect a ball of butter + sugar to suddenly decompose at 50 °C, and for the most part, chocolate is chemically pretty similar.

Now, what's interesting about cocoa butter is that it can form multiple crystal structures, and it's actually a bit tricky to get the right crystal structure out. You can read about the process in this RSC article if you'd like. The one that most chocolatiers are after is polymorph V, which is glossy and melts just below body temperature.

However, a different polymorph, polymorph, VI is the most stable form of cocoa butter, and it has a melting point that's a bit higher. It also tends to have a faint white crust on it (known as "fat bloom"). As a result, it's not as pleasant to look at, not as pleasant to eat (since it takes some chewing on to get it to melt), and generally undesirable.

Since form VI is the most stable, all chocolate eventually goes "bad" and becomes not as delicious as the maker originally intended. If you've ever bitten into an old chocolate bar and thought it was a bit waxy and unmelty, you've experienced this interconversion. I've been told by someone who did research on crystal structure kinetics that chocolate manufacturers spend lots and lots of money on research in hopes of being able to slow down or stop this process.

Now, I'd like to take stab at addressing two points in the question:

  chill the molten chocolate, and you'll still get chocolate

Yes, but is it the exact same chocolate as you got before? What if you take the molten chocolate and put it in the freezer, or, heaven forbid, drop it in liquid nitrogen? Does the resulting chocolate still melt at the same temperature? My suspicion (based on this chart) is that you'll have generated some of the lower-melting cocoa forms, and thus the chocolate will get soft and melt at a much lower temperature than previously.

  And why does milk chocolate (tend to) melt faster than dark chocolate (higher-cocoa content)?

I don't know. If I were to hazard a guess, I would say it's because milk chocolate tends to contain more sugar than dark chocolate, and so you're seeing some form of freezing-point depression. However, if it turns out that, at similar concentrations of sugar, milk chocolate still melts more quickly, I'd have no idea.


Why can't the molecularity be determined from the rate law?

The coefficients only correspond to the rate law coefficients if the reaction is elementary. Elementary reactions are ones where the reagent(s) combine through a mechanism implied by the reaction. These are simple. Many reactions aren't as simple and have multiple steps.

Check out this link, example two. The reaction is actually comprised of two reactions where the slow step determines the rate coefficients.

As you can see, you cannot just read the equation and get the coefficients. One can only determine whether a reaction is elementary (i.e. find its rate law coefficients) by experimentation.


Do I need a license or certificate to practice chemistry or what are the advantages if did have one?

Note that being a chemist doesn't require any kind of state license or the like in Germany, neither.

But buying dangerous* substances and running a chemical lab requires that you** have certain licenses and implement safety standards.
Some of these exams a chemist has to pass during studies (e.g. on toxicology and chemical substances legislation) so having a Master/Diplom in chemistry automatically means that you have those licenses.

Having this license (Chemikaliensachkunde) and being reliable (legal meaning) and of age are legal requirements you must meet in order to be allowed to buy dangerous substances. In practice however, the suppliers put another very practical additional need: they usually do mail orders only B2B and require you to have VAT/USt number.
So if I as a small-scale freelancer who fulfills all the legal requirements want to buy from say, Sigma, I don't get anything because I don't have the tax number (for my size of side-business it would be too much hassle to go through this, so I use the small business exemtion law and for the supplier it would be too much hassle to make sure the buyer fulfills the legal requirements if it isn't at least a certain business size).

As a private person, you can buy chemicals:

at the pharmacy
at specialized stores, e.g.

photography chemicals at a good photography stores
carbide at a speleology outfitter's
gases (possibly including dry ice) at a gas merchant's, etc.

you basically cannot buy dangerous substances by mail/online order (too much burocratic hassle: the seller would need to make sure of your identity and Chemikaliensachkunde etc.)
I've never tried to buy privately anything that requires the Chemikaliensachkunde e.g. at a pharmacy so I cannot tell how difficult that is in practice.

* As a rule of thumb, there are restrictions for substances (or mixtures) that are extremely flammable, oxidizing, toxic or very toxic, (suspected of being) carcinogenic, mutagenic or teratogenic (plus a bunch of listed things)

** or some employee of yours

However, you can of course work as a chemist without (much or even any) need to handle dangerous substances. For example, I'm analytical chemist / spectroscopist / chemometrician. About 90 % of my work is in front of a computer analysing measurement data, developing improved ways of doing so and writing reports/papers. At least half of the rest is in front of a computer programming instruments. I've handled exactly one dangerous substance according to chemicals legislation at two days (actually a sample of few μg I got for analysis) during the last year. The everyday stuff I handle (methanol, ethanol, isopropanol, disinfectant solution and cyclo-hexane and paracetamol=acetaminophen as calibration standards) are not legally restricted dangerous chemicals.
I do handle biohazard material far more often - but that is not what the Chemikaliensachkunde is good for. So for my work, the more relevant safety issues are biological/human material safety and laser safety... It would be quite easy not to have any chemical lab exposure (is seldom enough as it is... - but hey, what did I become chemist for?)

Also I have some colleagues doing the same type of work who are not chemists but e.g. physicists and thus do not have the Chemikaliensachkunde.

For getting a job it is obviously of advantage if you can show some kind of education/training that makes you fit for the job. For jobs in chemical industry/academia that is obviously a chemistry degree or training as chemical lab technician, but also related professions such as physics/biology/pharmacy/toxicology depending on what the job in question is.


How to compute the electron configuration of an atom?

To calculate electronic configurarion you need to know several thing first.

Electron Shell

  In chemistry and atomic physics, an electron shell, also called a principal energy level may be thought of as an orbit followed by electrons around an atom's nucleus. The closest shell to the nucleus is called the "1 shell" (also called "K shell"), followed by the "2 shell" (or "L shell"), then the "3 shell" (or "M shell"), and so on farther and farther from the nucleus. The shells correspond with the principal quantum numbers (1, 2, 3, 4 ...) or are labeled alphabetically with letters used in the X-ray notation (K, L, M, …).
  Each shell can contain only a fixed number of electrons: The 1st shell can hold up to two electrons, the 2nd shell can hold up to eight (2 + 6) electrons, the 3rd shell can hold up to 18 (2 + 6 + 10), and the 4th shell can hold up to 32 (2 + 6 + 10 + 14) and so on.

Subshells

  Each shell is composed of one or more subshells, which are themselves composed of atomic orbitals. For example, the first (K) shell has one subshell, called "1s"; the second (L) shell has two subshells, called "2s" and "2p"; the third shell has "3s", "3p", and "3d"; the fourth shell has "4s", "4p", "4d" and "4f"; the fifth shell has "5s", "5p", "5d", and "5f" and can theoretically hold more but the "5f" subshell, although occupied in actinides, is not filled in any element occurring naturally.

See 'Number of electrons in each shell' section under this article.

Atomic Orbitals

  An atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. Each orbital in an atom is characterized by a unique set of values of the three quantum numbers n, ℓ, and m, which correspond to the electron's energy, angular momentum, and an angular momentum vector component, respectively. Any orbital can be occupied by a maximum of two electrons, each with its own spin quantum number. The simple names s orbital, p orbital, d orbital and f orbital refer to orbitals with angular momentum quantum number ℓ = 0, 1, 2 and 3 respectively. These names, together with the value of n, are used to describe the electron configurations.

Understand electron configuration notation.

Electron configurations are written so as to clearly display the number of electrons in the atom as well as the number of electrons in each orbital. Each orbital is written in sequence, with the number of atoms in each orbital written in superscript to the right of the orbital name. The final electron configuration is a single string of orbital names and superscripts.
For example, here is a simple electron configuration: $1s^2 2s^2 2p^6$. This configuration shows that there are two electrons in the 1s orbital set, two electrons in the 2s orbital set, and six electrons in the 2p orbital set. 2 + 2 + 6 = 10 electrons total. This electron configuration is for an uncharged neon atom (neon's atomic number is 10.)

Memorize the order of the orbitals.

Note that orbital sets are numbered by electron shell, but ordered in terms of energy. For instance, a filled $4s^2$ is lower energy (or less potentially volatile) than a partially-filled or filled $3d^{10}$, so the 4s shell is listed first. Once you know the order of orbitals, you can simply fill them according to the number of electrons in the atom. *The order for filling orbitals is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s. *
An electron configuration for an atom with every orbital completely filled would be written: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{14} 6d^{10} 7p^6 8s^2$.

Note that the above list, if all the shells were filled, would be the electron configuration for Uuo (ununoctium), 118, the highest-numbered atom on the periodic table - so this electron configuration contains every currently known electron shell for a neutrally charged atom.

Now you can fill in the orbitals according to the number of electrons in your atom. For instance, if we want to write an electron configuration for an uncharged calcium atom, we'll begin by finding its atomic number on the periodic table. Its atomic number is 20, so we'll write a configuration for an atom with 20 electrons according to the order above.

Fill up orbitals according to the order above until you reach twenty total electrons. The 1s orbital gets two electrons, the 2s gets two, the 2p gets six, the 3s gets two, the 3p gets 6, and the 4s gets 2 (2 + 2 + 6 +2 +6 + 2 = 20.) Thus, the electron configuration for calcium is: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$.

Note: Energy level changes as you go up. For example, when you are about to go up to the 4th energy level, it becomes 4s first, then 3d. After the fourth energy level, you'll move onto the 5th where it follows the order once again. This only happens after the 3rd energy level.

For 'How to find quantumn numbers?', see this.


Does a benzene ring continue to oscillate at absolute zero?

"At absolute zero all molecular motion ceases; that is, the molecules stop bouncing into each other"

This statement is a common misconception about absolute zero and is totally false. At absolute zero, we are simply in the lowest possible energy state. That does not mean that molecular motion ceases. That would violate the uncertainty principle.


Can you calculate the properties of a substance based solely on its atomic properties?

For atomic and molecular-level properties (bond, angle, torsion energies, etc), quantum mechanics provides the equations for solving for these properties. However, there are a number of approximations that are employed, especially for larger, more complex molecules.

For bulk properties of a substance, thermodynamics and statistical mechanics provide equations for solving for a variety of material properties (critical properties, phase equilibria, various types of enthalpy, and so on). Read up on equations of state.

There is also the option of molecular simulation. In this case you are solving relatively simple, well-established equations (eg, Newton's equations of motion in molecular dynamics simulations) repeatedly in order to compute the (usually) equilibrium properties of a substance of interest.

In all cases, the accuracy varies, and validating results by comparing various methods is considered best practice. While there are "purists" out there who believe that all parameters in a model should be derived from first principles alone, in practice models are often fit to experimental data (or even data from quantum mechanical calculations) to improve accuracy.


Why doesn't LiHCO3 exist in solid state?

Claiming that a particular chemical compound "doesn't exist" is difficult to prove. The only convincing proof would be to show that neither LiHCO3 nor any of its hydrates can be isolated from (aqueous) solution. Essentially, you would have to show that using any of an almost infinite number of methods, that it couldn't be prepared at any temperature or pressure - not very likely to have ever been done, imho. So, I'll ask you for your source for such a claim, or better yet, I'll ask you to check and verify that your source - which you accept as being accurate - has made stringent efforts that demonstrate the claim as well as having published that work in peer reviewed literature. As an aside, there are plenty of compounds which require "waters of hydration" in their crystal structure. If it really doesn't exist, then the reason is simple: kinetically the reaction 2LiHCO3 → Li2CO3 + H2O + CO2 is spontaneous and energetically favored. It just seems to me that if I placed Li2CO3 in a pressure cell with water and CO2 at a billion gigapascals of pressure, I could make the bicarbonate salt, but perhaps some other structure is more likely...and anyway, generally what they mean by "doesn't exist" is that it "isn't known to exist at standard temperature and pressure".


What provides the energy allowing membrane proteins to facilitate diffusion across the membrane?

[...]how it works with out energy!

The transport proteins can change conformation in a stochastic manner. Even if the conformation opening to the extra-cellular side is a bit different in energy than the one opening to the intra-cellular side, there is no net change in energy/entropy/Gibbs energy in a complete cycle. The only thing that changes is that the cargo molecules are on the other side, a process for which the change in Gibbs energy is negative.

This argument is the same for any catalyst. If the catalyst is recovered in the same state as it was at the beginning of the reaction (and it has to be, otherwise it would not be a catalyst), the catalyst does not change the thermodynamic quantities of reactants or products. As a consequence, it speeds up the reaction, but does not influence the equilibrium constant.

If the catalyst has a high activation energy to do what it needs to do, then it is not an efficient catalyst. After all, the role of the catalyst is to lower the activation energy. So the transport proteins must be able to change from one to the other conformation both with cargo and without cargo in a facile manner. Otherwise, they could not go through the cycle of 1) picking up cargo 2) switching conformation 3) releasing cargo 4) switching conformation again.

  Which force provides with the torque to rotate [...]

Random forces associated with thermal energy are responsible for the conformational change. The binding constant for cargo has to be moderate so the the cargo dissociates easily. One the other hand, the binding constant for non-cargo has to be so low that nothing other than cargo binds.


Why do alkenes react with oxidising agents?

As per my knowledge, oxidation is loss of electron(s) by any species and an oxidising agent is an acceptor of electron(s). So, high electron density on alkenes make them likely to be attacked by oxidising agents.


Electron are added when you go to the right of a $d$ line, they have the same principal quantic number and so they are not making the atom bigger, but at the same time there is more protons and neutron resulting in more attractive force : the density increase because the volume is slightly the same but the masses of nucleus increase.

This phenomena start to be inverted at some point of the line because the "shell" (repulsive force due to other electron) of previous electron start to be important enough, the new added electron is not able to penetrate this cloud of electron and make the volume of the atom bigger, resulting in a tendency of decreasing density.

This phenomena of "shell" is explain by Slater and even have rule.

https://en.wikipedia.org/wiki/Slater%27s_rules?wprov=sfla1

EDIT : The explanation for the peak of density of the last group is explained by the lanthanide contraction. The $f$ orbitals have really low shield power and will cause a contraction of all the new added electron : it will be easier for them to penetrate the cloud of electron. This result in higher density

https://en.wikipedia.org/wiki/Lanthanide_contraction?wprov=sfla1


Virial equation of state - Taylor expansion

Here, Z is explicitly a function of temperature.

  At what expansion point do we expand?

Can you clarify what you mean? I can state that it's common to truncate this particular expansion at the second term, so we'd have:

$$\ce Z = 1 + {B_{2V}(T)\over \overline{V}}$$

The virial coefficient in this case accounts for deviation from the ideal case by accounting for attractive and repulsive forces between the particles in the system.


How does pressure influence the dehydration of salts?

Yes - the pressure will impact the dehydration temperature.
You have correctly identified that temperature and pressure are related by the Gas laws.
High pressure can also force water molecules out, as water is loosely bonded in salts.


Sulphite$(\ce{SO3^2-})$ and Barium Choride$(\ce{ BaCl2})$ or Strontium Chloride$(\ce{ SrCl2})$:
$$\ce{SO3^2- +Ba^2+ ->BaSO3 v}$$
the precipate disoolves in dilute hydrochloric acid, when sulphur dioxide evolves:
$$\ce{BaSO3 v +2H+ ->Ba^2+ +SO2 ^ +H2O}$$
On standing, the precipitate is slowly oxidises to the sulphate and is then insoluble in dilute mineral acids; this change is rapidly effected by warming with bromine water or a little concentrated nitric acid or with hydrogen peroxide:
$$\ce{2BaSO3 v +O2 ->2BaSO4 v }\\
\ce{2BaSO3 v +Br2 + H2O->BaSO4 v +2Br- +2H+}\\
\ce{2BaSO3 v +2HNO3 ->3BaSO4 v +2NO ^ +H2O}\\
\ce{2BaSO3 v +H2O2 ->BaSO4 v +H2O}$$
The solubilites at $\pu{18 ^\circ C}$ of the sulphites of $\ce{Ca,Sr,Ba}$ are respectively $1.25, 0.033, 0.022$ all in $\pu{g/l}$
Source:

Vogel’s Qualitative Inorganic Analysis for the JEE, by Svehla/Sivasankar,  7/e Paperback, 1 January 2013


What's your first miss is the definition of $50\%(w/w)$ $\ce{NaOH}$ solution (although it doesn't matter here). Actual definition is $\pu{50 g}$ of $\ce{NaOH}$ in $\pu{100 g}$ of solution. Since water is the solvent, your interpretation is correct.
Your second mistake is the lethal one. You have considered water and solid $\ce{NaOH}$ are additive. Assuming $V_\text{total} = V_\ce{NaOH} + V_\ce{H2O}$ is a mistake.
Using a given density of $\ce{NaOH}$ ($\pu{1.5298 g}$),we can calculate the volume of $\pu{1000 g}$ of solution: $\frac{\pu{1000 g}}{\pu{1.5298 g/mL}} = \pu{653.68 mL}$. Therefore, dissolving $\pu{500 g}$ of $\ce{NaOH}$ in $\pu{500 mL}$ of water (assuming temperature is $\pu{25 ^\circ C}$, hence density of water is $\pu{1.00 g/mL}$), the volume increased by only $\pu{153.68 mL}$. Do you still think volumes are additive?


Properties of starch in packaged pudding

Starch makes a glue or a paste with water. Humidity from the air may be sufficient to transform your starch into a glue. Water produces strong hydrogen bridges between starch molecules. This transforms a pile of independent starch molecules into a strong block of interconnected starch molecules.


Electronegativity Considerations in Assigning Oxidation States

In the IUPAC Recommendations 2016 the definition of oxidation state underwent a significant and comprehensive change. It does now resemble the version wich was proposed be Hans-Peter Loock and is quoted in the earlier version of this answer, parts of which are included below.

The electronegativity battle scheme is most helpful for all kinds of compounds since it is one of the most generic ways to derive oxidation states. The table represents just a cheat sheet that might be very helpful in the beginning. If you are spending most of your time with chemistry, this table will be present as some sort of muscle memory - usually referred to as chemical intuition.

The IUPAC gold book now defines oxidation state as follows:

  oxidation state
  gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation. Definition:
  OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.

  The underlying principle is that the ionic sign in an $\ce{AB}$ molecule is deduced from the electron allegiance in a LCAO-MO model: The bond’s electrons are assigned to its main atomic contributor. Homonuclear $\ce{AA}$ bonds are divided equally. In practical use, the ionic-approximation sign follows Allen electronegativities (see Source). There are two general algorithms to calculate OS:

  Algorithm of assigning bonds, which works on a Lewis formula showing all valence electrons in a molecule: OS equals the charge of an atom after its heteronuclear bonds have been assigned to the more electronegative partner (except when that partner is a reversibly bonded Lewis-acid ligand) and homonuclear bonds have been divided equally:

  Algorithm of summing bond orders: Heteronuclear-bond orders are summed at the atom as positive if that atom is the electropositive partner in a particular bond and as negative if not, and the atom’s formal charge (if any) is added to that sum, yielding the OS. This algorithm works on Lewis formulas and on bond graphs of atom connectivities for an extended solid:

  NOTE:

  Specific uses may require modified OS values: Electrochemical OS is nominally adjusted to represent a redox-active molecule or ion in Latimer or Frost diagrams. Nominal OS values may also be chosen from close alternatives for systematic-chemistry descriptions.
  Some OS may be ambiguous, typically when two or more redox-prone atoms enter bonding compromises and nearest integer values have to be chosen for the OS.
  The caveat of reversibly bonded Lewis-acid ligands originates from the simplifying use of electronegativity instead of the MO-based electron allegiance to decide the ionic sign. Typical examples are the transition-metal complexes with so called Z ligands in the CBC ligand-classification scheme (see Source).
  When used in chemical nomenclature as a symbol, the OS value is in roman numerals.
  At the introductory teaching level, prior to the bonding-based definition and algorithms: OS for an element in a chemical formula is calculated from the overall charge and postulated OS values for all the other atoms. For example, postulating $\text{OS} = +1$ for $\ce{H}$ and $−2$ for $\ce{O}$ yields correct OS in oxides, hydroxides, and acids like $\ce{H2SO4}$; with coverage extended to $\ce{H2O2}$ via decreasing priority along the sequence of the two postulates. Additional postulates may expand the range of compounds to fit a textbook’s scope.

  Source:

  Karen, P.; Mcardle, P.; Takats, J. Toward a comprehensive definition of oxidation state (IUPAC Technical Report). Pure Appl. Chem. 2014, 86 (6), 1017–1081 DOI: 10.1515/pac-2013-0505.
  Karen, P.; Mcardle, P.; Takats, J. Comprehensive definition of oxidation state (IUPAC Recommendations 2016). Pure Appl. Chem. 2016, 88 (8), 831–839 DOI: 10.1515/pac-2015-1204.
  Nomenclature of Inorganic Chemistry. IUPAC Recommendations 2005. p. 34. Available as pdf from old.iupac.org

I recommend having a look at the sources cited above, they contain a lot more information. (Note that the link from the gold book to the red book on the IUPAC website is broken.)

Prior to 2016 the IUPAC defined oxidation states in their gold book (via the Internet Archive) differently. I am including this definition, since it will be still present in prominent text books and it is what lead to the postulates given in the table of the original question.

  oxidation state (deprecated, pre-2016 definition)
  A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules:

  the oxidation state of a free element (uncombined element) is zero;
  for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;
  hydrogen has an oxidation state of $1$ and oxygen has an oxidation state of $-2$ when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of $-1$ in hydrides of active metals, e.g. $\ce{LiH}$, and oxygen has an oxidation state of $-1$ in peroxides, e.g. $\ce{H2O2}$;
  the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion. For example, the oxidation states of sulfur in $\ce{H2S}$, $\ce{S8}$ (elementary sulfur), $\ce{SO2}$, $\ce{SO3}$, and $\ce{H2SO4}$ are, respectively: $-2$, $04, $+4$, $+6$ and $+6$. The higher the oxidation state of a given atom, the greater is its degree of oxidation; the lower the oxidation state, the greater is its degree of reduction.

What we see is, that the table in the original question actually reflects some of these rules. However, this set is not generic at all and it lacks a definition for compounds that do not contain oxygen or hydrogen, are elemental or monoatomic ions. With only these rules it is impossible do determine the oxidation states for $\ce{BF3}$ and many, if not most, compounds.

The lack of the definition is very well known, but it took the IUPAC until 2016 to actually change the official set. Hans-Peter Loock proposed a much simpler concept in Expanded Definition of the Oxidation State, which resembles the currently used definition quite well.

  The oxidation state of an atom in a compound is given by the hypothetical charge of the corresponding atomic ion that is obtained by heterolytically cleaving its bonds such that the atom with the higher electronegativity in a bond is allocated all electrons in this bond. Bonds between like atoms (having the same formal charge) are cleaved homolytically.

So he basically came to the same conclusion as Dissenter in the question. Another statement from this article is

  This is not a new definition, but it predates the IUPAC rules by several decades. For example, Linus Pauling provided a similar definition of the oxidation state in his 1947 edition of General Chemistry (3).

  (3):Pauling, L. General Chemistry; Freeman: San Francisco, CA, 1947. Republished by Courier Dover Publications, 2012.


Heats of formation of neutral molecules and homolytic vs heterolytic bond dissociation in mass spectrometry

The heat of formation of a neutral molecule is important in mass spectrometry because in order to measure the characteristics of individual molecules, a mass spectrometer has to convert them to relevant ions. Once ions form, they can be moved about and manipulated by external electric and magnetic fields as depicted in following sketch of mass spectrometer (MS):

The Ion Source, the Mass Analyzer, and the Detector are the three main components of the mass spectrometer. The Ion Source, which can be considered as the heart of the spectrometer, is the one needed to ionize individual molecules. Electron Ionization (EI) is the most common ionization technique used for mass spectrometry (Note: Older literature usually refer to EI as Electron Impact, but Electron Ionization is the currently accepted term; EI works well for many gas phase molecules, but it does have some limitations as well, such as causing extensive fragmentation so that the molecular ion is not observed for many compounds). In this ionization technique, molecules of the sample are bombarded by electrons generated from a heated filament (see the insert red box in the diagram). The electrons are produced by passing a current through the filament. The amount of current controls the number of electrons emitted by the filament. An electric field accelerates these electrons across the source region to produce a beam of high energy electrons. When an analyte molecule (usually organic molecule) passes through this electron beam, a valence shell electron can be removed from the molecule to produce an ion, mostly a positive ion (Note: Ionization does not occur by electron capture, which is highly dependent upon molecular structure).

Since the ionization is produced by a single electron that is accelerated to $\pu{70 V}$, the technique is commonly referred to as $\pu{70 eV}$ EI. This high energy electron collides with a molecule (analyte), it transfer more than enough energy to over come ionization energy of the particular site (usually a electron from a lone pair of oxygen of carbonyl group or alcohol group) of the molecule, often ionizing the analyte by knocking away one of the molecular electrons (either bonding or non-bonding). The resultant ion is a radical cation. This web page describes the possibility of the fragmentation happens in mass spectrometer:

  The amount of energy transferred during this process depends upon how fast the electron (from ion source) is traveling and how close it passes to the molecule (analyte). In most $\pu{70 eV}$ EI experiments, approximately $\pu{1400 kJ/mol}$ ($\pu{15 eV}$) of energy is transferred during the ionization process. There is, however, a distribution of energy and as much as $\pu{2800 kJ/mol}$ ($\pu{30 eV}$) is transferred to some molecules. Since approximately $\pu{960 kJ/mol}$ ($\pu{10 eV}$) is required to ionize most organic compounds and a typical chemical bond energy is $\pu{290 kJ/mol}$ ($\pu{3 eV}$), extensive fragmentation is often observed in $\pu{70 eV}$ EI mass spectra. The distribution of energy transferred during ionization and the large number of fragmentation pathways results in a variety of products for a given molecule. Other electron voltages may be used to vary the amount of fragmentation produced during ionization. For most organic compounds the threshold energy for EI is about $\pu{10 eV}$. After ionization of the molecule, the residual energy from the collision may cause the molecular ion to fragment into neutral pieces and smaller fragment ions. This fragmentation can be through either homolytic cleavage or heterolytic cleavage. The molecular ion is a radical cation, but the fragment ions may either be radical cations or carbocations, depending on the nature of the neutral fragment. For example, Fragmentation of Toluene is depicted in following diagram:

Important: In Mass spectroscopy, the electron volt ($\pu{eV}$) is frequently used as an energy unit (for single molecules or atoms). Keep in mind that $\pu{1 eV} = \pu{1.6022 x 10^{-19} J}$ (per molecule or atom). Thus, $\pu{1 eV}$ (per molecule or atom) $= \pu{96.4152 kJ/mol}$.

A recent publication described why most fragmentations in mass spectroscopy preferred homolysis over hererolysis (Ref.1):

  Bond strength is an essential parameter for discussion of bonding and reactivity. While bond fission can occur by homolysis or heterolysis, for neutral compounds such as alkanes the term “bond strength” generally denotes rupture by the lowest energy homolytic pathway. For example, the $\ce{C–C}$ bond strength in ethane is listed as $\pu{359 kJ mol–1}$,$^{[Ref.2]}$ representing homolysis, which requires one-third of the energy for heterolysis ($\pu{1297 kJ mol–1}$).$^{[Ref.3]}$

In addition, Ref.4 is a great review on mass spectrometry‐based analysis of protein structure.

References:

Karlee L. Bamford, Saurabh S. Chitnis, Rhonda L. Stoddard, J. Scott McIndoe, Neil Burford, "Bond fission in monocationic frameworks: diverse fragmentation pathways for phosphinophosphonium cations," Chem Sci. 2016, 7(4), 2544–2552 (doi: 10.1039/c5sc03804a).
Donald F. McMillen, David M. Golden, "Hydrocarbon bond dissociation energies," Annual Review of Physical Chemistry 1982, 33(Oct), 493–532 (https://doi.org/10.1146/annurev.pc.33.100182.002425).
Pratim K. Chattaraj, Andres Cedillo, Robert G. Parr, Edward M. Arnett, "Appraisal of Chemical Bond Making, Bond Breaking, and Electron Transfer in Solution in the Light of the Principle of Maximum Hardness," J. Org. Chem. 1995, 60(15), 4707-4714 (https://doi.org/10.1021/jo00120a010).
Frederik Lermyte, Dirk Valkenborg, Joseph A. Loo, Frank Sobott, “Radical solutions: Principles and application of electron‐based dissociation in mass spectrometry‐based analysis of protein structure,” Mass Spectrometry Reviews 2018, 37(6), 750-771 (https://doi.org/10.1002/mas.21560).


Are there any solvents that dissolve talcum powder?

Talcum powder is made from talc, a silicate mineral with formula unit: $\ce{Mg3Si4O10(OH)2}$. Silicates tend to be pretty much insoluble in water, but it is possible to dissolve them in strong acids. I don't know about talc specifically, but many silicates can be dissolved in concentrated nitric acid. Hydrofluoric acid is also commonly used for dissolving minerals for chemical analysis. Both are very very dangerous substances that really shouldn't be used without proper training and safety equipment. I don't think there's really anything accessible for a home experimenter, unfortunately.


Understanding dissolution of CO₂ in Selexol and Rectisol

Simply, $\ce{CO_{2}}$ is dissolved in those solvents in a similar way to gases dissolved in water. The molecules do not react with the solvent but mix. The higher the pressure of the gas you pump into the liquid, the more that dissolves in it (rather than turning into bubbles and escaping). In fact, amount of gases dissolved in liquids can be measured both as a concentration (mol/L) or as a partial pressure (Pa).

Henry's law states that:

  "At a constant temperature, the amount of a given gas that dissolves
  in a given type and volume of liquid is directly proportional to the
  partial pressure of that gas in equilibrium with that liquid."

The formula is:

$p = k_{H}c$

where $p$ is partial pressure of the gas in solution, $c$ is its concentration and $k_{H}$ is a constant based on what the solute is (in this case $\ce{CO_{2}}$), what the solvent is (eg methanol) and the temperature.

The harder you push, the tighter you can pack it in.

Here is a picture:

In contrast, with the amine-based solvents, the $\ce{CO_{2}}$ actually reacts, which is why a degasification process needs to happen to release it.


Nuclear fusion reactions violate law of conservation of mass, but it has not yet been taken into account. Why?

Nuclear reactions appear to violate both the Laws of Conservation of Mass and Energy because mass is converted into energy or vice versa.

However, the concept of mass-energy equivalence that emerges as a consequence of Einstein's General Theory of Relativity makes the Laws of Conservation of Mass and Energy special limiting cases of the Law of Conservation of Mass-Energy. Since mass is now a form of energy, we can convert back and forth as long as there is no net loss between the two.

The equivalence of mass and energy comes from Einstein's famous $E=mc^2$, where $c$ is the speed of light (in a vacuum).


Reaction between hydrogen peroxide and ozone

This is basically an excerpt of the answer I have given to another quite similar question.

Ben is quite right, neither the reaction nor the mechanism is simple. It involves several different radical chains all resulting in the decomposition of ozone.[a]

It is most likely that the reaction will start with the formation of oxygen radicals in the gas phase ($\ce{O}$ atom transfer).$$\ce{O3 + O3 -> O3 + O2 + O}$$

In presence of Hydrogen donors like $\ce{H2O2}$, which "is expected to be an excellent $\ce{H}$ donor."[a] Those are forming the main chain carriers will be $\ce{HO, HO2}$.
 Finally stopping at some point by recombination of these species.
\begin{aligned}\ce{
O + H2O2 &-> OH + HO2\\
  HO + O3 &-> HO2 + O2\\
  HO2 + O3 &-> HO + 2O2\\
  2HO2 &->}\text{stable products}
\end{aligned}

In aqueous solution the decomposition is highly pH dependent, but may always work because of autoprotolysis. $$\ce{2H2O <=> H3+O + {}^{-}OH}$$

Especially important might also be protolysis of hydrogen peroxide. $$\ce{H2O2 + H2O <=> HO2- + H3+O}$$

Steahelin and Hoigné propose several hypotheses in which way ozone will be decomposed, but I will not go into absurd detail here.[b] The core assumptions involve $\ce{{}^{-}OH}$ as the initiator of a radical chain. Followed by radical (ionic) species like $\ce{.O2-, .O3-, HO2., .OH}$ resulting again in a chain reaction.
\begin{aligned}\ce{
  O3 + {}^{-}OH &-> .O2- + HO2.\\
  O3 + {}^{-}OH &-> HO2- + O2\\
  O3 + {}^{-}OH &-> .O3- + .OH\\
  HO2. &<=> H+ + .O2-\\
  O3 + HO2- &-> .OH + .O2- + O2\\
  O3 + .O2- &-> .O3- + O2\\
  .O3- + H2O &-> .OH + {}^{-}OH + O2
}\end{aligned}


Well, carbon monoxide can be created from formic acid by adding sulphuric acid which will dehydrate said formic acid:

$$\ce{HCOOH ->[\ce{H2SO4}] CO + H2O}$$

Therefore, we can imagine the reverse reaction theoretically, which would make carbon monoxide an acidic oxide. However, the forward reaction does not proceed easily and it needs both the high acidity of sulphuric acid and its strong dehydrative properties to actually work. And your question mentions using hot, concentrated sodium hydroxide to make the reverse one work.

Most oxides that are classified as acidic or basic either have a very electrophilic central atom (e.g. $\ce{CO2}$) which can be attacked by the weak nucleophile water (which in turn can then release an acidic proton), or they have a high charge density on the oxygen which allows it to abstract a proton from water directly. Carbon monoxide is neither. If you check out its molecular orbitals, you will notice that even though carbon is partially positive it has the largest HOMO contribution, meaning a proton would be more likely to attatch to the carbon side — which doesn’t want one at all. The LUMO is, luckily, also more carbon-centred, meaning nucleophilic attacks on carbon are possible. However, it is also degenerate due to the double bond so that an attack is not favoured.

Thus, the carbon monoxide molecule is one that won’t react with water at all and totally defies the concept of acidic/basic oxides.

Abbreviations:

HOMO is a widely used abbreviation for the Highest Occupied Molecular Orbital, i.e. the one with the highest energy that still contains electrons. It is usually the orbital that will attack nucleophilicly or that will be attacked electrophilicly.
LUMO is a widely used abbreviation for the Lowest Unoccupied Molecular Orbital, i.e. the virtual (unoccupied) orbital that has the lowest energy. When considering a nucleophilic attack, the attacking electrons will usually interact with the LUMO. Electrophiles attack with other molecules’ HOMO with their LUMO.


Barium hydroxide – or more exactly, the hydroxide ion – is a Brønsted-Lowry base, because in aqueous solution, it is completely dissociated into $\ce{Ba^2+}$ and $\ce{OH-}$ ions:

$$\ce{Ba(OH)2 -> Ba^2+ + 2OH-}$$

$\ce{OH-}$ is a proton acceptor, forming $\ce{H2O}$. In aqueous solution, the following reaction takes place:

$$\ce{OH- + H3O+ ->~ 2H2O}$$

Barium hydroxide thus reacts with acids in a neutralization reaction, with water and the barium salt of the respective acid as the products. For example, with hydrochloric acid:

$$\ce{Ba(OH)2 + 2HCl ->~ BaCl2 + 2H2O}$$


There is not much to say other than: your version is more or less correct. It is debateable whether the ester hydrolysis happens before or after the cleavage of azetidine (I would suspect before due to the lower overall charge). Remember that you always have to consider aquaeous work-up at the end of a reaction, which will give you the three protons you need to regenerate acrylic acid, ethanol and azetidine.

You also have another question written on your paper after the halogenation of the terminal alcohol, namely:

  What is the point of this step?

It’s not really a point. It’s just that the reaction conditions contain $\ce{HCl}$ which will protonate any strong enough base around and the secondary amine is a strong enough base. It’s more of a consequence of the reaction conditions than an actual pointful step.


What Causes there to be a Maximum Amount of Salt which can be Dissolved in Water?

When a salt dissolves in water, there is a competition between the electrostatic forces that keep the anion and cation together and the ion-dipole forces that solvate the ions.  One could write this equation as:

$$\ce{AB + q H2O \leftrightarrows A(H2O)_{m}+ + B(H2O)_{n}-}$$

where $q=m+n$. Since this is an equilibrium, we can write an equilibrium constant, which is oftentimes referred to as the solubility product: $$\ce{K_{sp}=[A(H2O)_{m}+] * [B(H2O)_{n}^{-}]}$$

Very soluble salts will have a large, but finite, $K_{sp}$.  The magnitude of the solubility product can be rationalized by comparing the energy required to break the electrostatic bonds (the lattice enthalpy) and the energy released by forming solvated ions (solvation enthalpies).  One can use a Born-Haber cycle to predict whether the dissolution of a salt will be endothermic or exothermic, and we would need a similar approach to address entropy changes before we could get a complete, quantitative understanding of the solubility.


Why does the Most Stable State of an Atom Tend to be One with Full s and p Subshells?

The major factor is electrons with the same quantum numbers $\mathrm{n}$ and $\mathrm{l}$ do not mutually shield the nucleus charge well.
Elements approaching the noble gas group in the Periodic table ( $\ce{N, O, F}$ ) get nearly  fully filled respective $\mathrm{p}$ orbitals. The effectively perceived nucleus charge grows for valence electrons. It gets progressively more difficult to ionisate these electrons, and at the same time the energy released by capturing an extra electron grows.
Elements at the opposite table side ( alkaline metals and alkaline earth metals ) have the opposite situation. They start to fill orbitals at the new, higher quantum number $\mathrm{n}$ level. The lower, now fully filled $\mathrm{p}$ orbitals shield the nucleus well. Additionally, the new $\mathrm{s}$ orbital  is farther from nucleus with lower attractive force. Both effects lead to low ionization energy of such atoms and very low affinity to extra electrons.
This leads to the octet rule, which is consequence of the fact, if chemical bonds lead to completing octets, the total electron energy is lower.
It has its limits. Ionization of electrons leads to progressively increasing ionization energy for every next electron. Similarly, accepting too many electrons leads to negative electron affinity, so the electron is released at the nearest convenience. So ions with high positive charge occur only in heavily ionisating environment, respectively the one with highly negative charge need source of electrons. Even in solid matrices there is partially covalent bond.
It may be challenging, but this explains a lot about the screening of the nucleus charge: Slater's rules


Effects of Ka on the Half-Equivalence Point

If we consider $\ce{HA}$ as a weak acid, then at the half equivalence point, $$\mathrm{p}H = \mathrm{p}K_\mathrm{a}$$ As $$\mathrm{p}H = \mathrm{p}K_\mathrm{a} + \mathrm{p[\ce{HA}]} -\mathrm{p[\ce{A-}]}$$ and for the half equivalence point,
$$\mathrm{p[\ce{HA}]} =\mathrm{p[\ce{A-}]}$$
So the higher the $\mathrm{p}K_\mathrm{a}$ is, the higher is $\mathrm{p}H$ at the half equivalence.


How to turn acetaldehyde into acetic acid using one of the five chemicals on the list?

Oxidation of aromatic aldehydes to acids by air oxidation catalysed by CuO is reported in Molecules 2008 13 948

None of the other reagents can oxidise aldehyde to acid.


Why aren't all bonds in cyclobutadiene not equal?

Strictly speaking cyclobutadiene is not antiaromatic but non aromatic, ie the same energy 4A + 4B is attained by applying the Hückel LCAO, as for two isolated double bonds, 2 times 2A + 2B.

Given that cyclobutadiene can only be isolated as non delocalized system when in matrixes and a very low T points to strain as a reason for instability of both forms, the delocalized one remaining hypothetical.

Edit for the readers not familiar with the terms: A system obeying all the requisites (cyclic, pi orbitals on each atoms) is antiaromatic when the hypothetical delocalization destabilizes it as compared to the localized system. If no energy difference arises, the system is called non aromatic.

Beware there are plenty of pages discussing non aromaticity in a trivial way, e.g. aliphatic or linear compounds. Cyclobutadiene is non aromatic within the Hückel picture. Also it seems worth to point out again that antiaromatic refers anyway to an hypothetical structure, not to an actual delocalization. A system which cannot avoid it by distorting would be unstable tout-court. Cyclobutadiene, albeit being non aromatic, is on the same line. Indeed you can find antithetical assertions about it, such as it being "the archetype of antiaromatics“ along with “it reacts as diradical“ as if the two things would be straight forward correlated. Normally a system escape antiaromaticity. Being cyclobutadiene non aromatic results in a distorted rectangular molecule, but only when it cannot escape this anyway unstable situation. See also Resonance structure of cyclobutadiene?. Although the situation of cyclobutadiene is again erroneously described as an antiaromatic case, it gives the answer. Also worth reading is the answer by Georg.


They are diastereomers.  Both of the "front" chiral carbons (the one with the bromine attached) are "S".  Rotate the one on the right 120 degrees CCW and it is identical to the one on the left.  But the "back" carbon (the one with the hydroxyl attached) is "S" on the left figure and "R" on the right figure - diastereomers.


No, nitric acid is a sufficiently hard acid to dissolve silicon metal without precipitating in low concentrations. In fact, nitric acid is used to dissolve silicon in ICP Standards. The oxide reacts slowly with water and water-based solutions - nitric acid included, but the acid does not increase the kinetics appreciably, otherwise glass would be unsuitable as a vessel for nitric acid. The element in an alloy, however, can be dissolved more easily like that in cast iron.


Minimum amount of water in pressure cooker

In a pressure vessel, the volume $V$ is approximately constant. It is given as $V = 6.2\ \mathrm l$. However, this volume includes

liquid water
steam
air
food

in unknown amounts.

The gauge pressure is given as $p_\mathrm e = 0.8\ \mathrm{bar}$; i.e. the absolute pressure is approximately $p = 1.8\ \mathrm{bar}$.

We may use so-called steam tables to look up the properties of water at the given pressure (in the following, I use parameter values taken from the REFPROP – NIST Standard Reference Database 23, Version 9.0). We may find that the saturation point (equilibrium of liquid water and steam) at a pressure of $p = 1.8\ \mathrm{bar}$ corresponds to a temperature of $T = 117\ \mathrm{^\circ C}$.
At this point, the density of liquid water is $\rho_\mathrm l = 946\ \mathrm{g/l}$ and the density of steam is $\rho_\text{steam} = 1.02\ \mathrm{g/l}$.

If we ignore the volume of the remaining liquid water, air, and food, we may consider the limiting case in which the entire volume $V$ is filled with steam. Thus, the mass $m_\text{steam}$ is given by
$$\begin{align}
m_\text{steam} &= \rho_\text{steam} \cdot V \\[3pt]
&= 1.02\ \mathrm{\frac gl} \times 6.2\ \mathrm l \\[3pt]
&= 6.3\ \mathrm g
\end{align}$$
Therefore, only $6.3\ \mathrm g$ of water (which corresponds to about $6.3\ \mathrm{ml}$ of cold water at normal pressure) are required to fill the entire volume with steam at the given pressure.

However, a significant amount of excess liquid water should remain in the cooker in order for it to function properly. The liquid water is heated in the lower part of the cooker and boils at about $117\ \mathrm{^\circ C}$. The steam is distributed in the entire volume and condenses on colder surfaces (food and in particular the walls of the cooker). The condensed water runs down and it is collected in the lower part of the cooker. This circulation is essential for the uniform distribution of heat within the cooker; hence, the cooker should not dry out. The additional amount of liquid water should be sufficient to wet all surfaces (food and cooker) thoroughly.

Up to now, we have ignored the air that remains in the cooker. At initial (cold) conditions (pressure $p_0 = 1\ \mathrm{bar}$ and temperature $T_0 =20\ \mathrm{^\circ C}$), the density of dry air is approximately $\rho_{\text{air},0} = 1.189\ \mathrm{g/l}$. If we now ignore the volume of the liquid water, steam, and food, we may consider the limiting case in which the entire volume $V$ is filled with air. The mass $m_\text{air}$ is given by
$$\begin{align}
m_\text{air} &= \rho_{\text{air},0} \cdot V \\[3pt]
&= 1.189\ \mathrm{\frac gl} \times 6.2\ \mathrm l \\[3pt]
&= 7.37\ \mathrm g
\end{align}$$
The air cannot escape. At the final pressure of $p = 1.8\ \mathrm{bar}$ and temperature of $T = 117\ \mathrm{^\circ C}$, the trapped air is compressed to a density of approximately $\rho_\text{air} = 1.607\ \mathrm{g/l}$. The corresponding new volume is
$$\begin{align}
V_\text{air} &= \frac{m_\text{air}}{\rho_\text{air}} \\[3pt]
&= \frac{7.37\ \mathrm g}{1.607\ \mathrm{g/l}} \\[3pt]
&= 4.6\ \mathrm l
\end{align}$$
We see that a large part of the volume is taken up by the trapped air. The amount of water that is actually required to create steam at the given pressure is reduced accordingly.


I think your question really arises from some confusion about what $\Delta G$ represents. In general, $\Delta X$ for a thermodynamic quantity $X$ is the change of $X$ along some process. You could make it clear by actually writing $\Delta G(\text{A}\rightarrow\text{B})$ where A and B are before and after states. (We'll note that, in the general case, $\Delta X$ depends on the path take from A to B, making this notation improper. If $X$ is a function of state, though, you're good to go.)
However, in the equation you quote:
$$\Delta G = \Delta G^0 + RT \ln Q$$
the $\Delta G$ is a free energy of reaction and should thus be denoted $\Delta_\mathrm r G$, with the correct equation being:
$$\Delta_\mathrm r G = \Delta_\mathrm r G^0 + RT \ln Q$$
The free energy of reaction is defined as $\Delta_\mathrm r G = G_{\text{products}} - G_{\text{reactants}}$.

Thus, this $\Delta_\mathrm r G$ is not the variation of $G$ over the entire reaction, which would be the $\Delta G$ of the system between the start of the reaction and the equilibrium.

PS: I think this link is the online resource I found with the clearer use and explanation of notations. Notations are important in thermodynamics!


Textbook problem with molar mass of carbon dioxide

The problem is that, when the flask is empty, it weighs $85.431\ \mathrm g$. But this weight is the weight of the empty flask plus the air included. So the first thing to do is to calculate the weight of the air, and the corresponding number of moles of air in the flask before adding $\ce{CO2}$. The volume of the air is the same as the volume of water when it is filled with water, namely $122\ \mathrm{mL}$.
The amount of air is $$n = pV/RT = \frac{122\times10^{-6}\ \mathrm{m^3}\times101325\ \mathrm{Pa}}{8.316\times300\ \mathrm{J/mol}} = 4.956\times10^{-3}\ \ \mathrm{mol}$$ The mass of this air is : $$m = 4.956\times10^{-3}\ \mathrm{mol}\times28.8\ \mathrm{g/mol} = 0.1427\ \mathrm g$$ So the mass $m_0$ of the empty flask without air is : $$m_0 = 85.431\ \mathrm g - 0.1427\ \mathrm g = 85.2883\ \mathrm g$$
A s consequence, the mass of the $\ce{CO2}$ in the flask is : $$m(\ce{CO2}) = 85.510\ \mathrm g - 85.2883\ \mathrm g = 0.2227\ \mathrm g$$
The molar mass of this $\ce{CO2}$ is: $$M = \frac{m}{n} = \frac{0.2227\ \mathrm g}{4.956\times10^{-3}\ \mathrm{mol}}= 44.8\ \mathrm{g/mol}$$
This is nearly the expected value for the molar mass of $\ce{CO2}$. If you take into account the fact that the volume $122\ \mathrm{mL}$ is known with a precision $\pm1\ \mathrm{mL}$, or $\pm1\ \%$, the final result will be defined with an uncertainty of $\pm1\ \mathrm{g/mol}$. It is: $$M(\ce{CO2}) = 44.8\ \mathrm{g/mol} \pm 1\ \mathrm{g/mol}$$


Question regarding $Z$ (Compressibility factor)

Forget about the ideal gas.  The definition of the compressibility factor is just $$Z=\frac{PV_m}{RT}$$


Modifying a TLC mobile phase in order to increase the RF values

Question: Would the hexane and ethyl acetate be considered the mobile phase/solvent?

Answer is yes: Hexane/ethyl acetate systems ($0$ to $100\%$ ethyl acetate in hexane) are frequently used mobile phases for TLC technique. However, as Waylander pointed out in his comment:

The type of mobile phase really depends on what the substrate and product are. It is worth running half a dozen trial mobile phases to see which one gives a decent separation.

Also, it is  true, as Buttonwood pointed out, a high $R_f$ value is not very useful if $\Delta R_f$ between the two compounds (or more in the mixture) of interest is small unless it increase upon change of mobile phase. For example, TLC of the mixture of benzoic acid and benzocaine (ethyl 4-aminobenzoate) shows no separation in $50\%$ ethyl acetate in hexanes. However, when mobile phase is change to $5\%$ methanol in dichloromethane, it has shown $\gt 33\%$ separation in $R_f$. So it is a trial and error method, but worth trying.


Regardless of the polarity, the electrode where oxidation takes place is the called the anode and therefore reduction must take place at the cathode. The electron flow from the anode to cathode is as shown in your top picture.

By way of an example, in an electrochemical cell, suppose that two beakers are connected by a salt bridge. In one beaker is a strip of zinc metal immersed in a $\ce{Zn(NO3)2}$ solution and in the other a strip of silver in $\ce{AgNO3}$ solution. The two metals are then connected by a wire and a current will flow. (The salt bridge supplies a return path so that the solutions remain electrically neutral). The redox of the Zn electrode is $-0.763$ V and that of the Ag $+0.799$ V. This means that electrons will flow from the zinc to the silver electrode. The zinc is oxidised to $\ce{Zn^{2+}}$ and the electrons are released into the metal and flow to the silver electrode through the wire. The zinc electrode is the anode and the silver the cathode.


Distinguishing among SN2/SN1/E1/E2 for a secondary bromide

Reaction a proceeds via an $\mathrm{E2}$ mechanism. Firstly, elimination reactions are much more favorable in this molecule because of the conjugation of the resulting structure. If the reactant had been 2-bromo-2-phenylbutane, this would have proceeded via an $\mathrm{E1}$ mechanism, as this would delocalize the positive charge around the phenyl ring. There is no such delocalization in the ejection of the $\ce{Br-}$ from 3-bromo-2-phenylbutane though—it would simply be a 2º carbocation. Instead, the delocalization of the negative charge around the phenyl ring resulting form the extraction of the $\ce{H+}$ by $\ce{OMe-}$ is more stabilizing.

$\hspace{1.5cm}$

There are no possible reactions between an alkene and $\ce{CN-}$ in DMSO. If this were a haloalkene, further elimination could occur, and if it were in $\ce{H2O}$, hydrocyanation could occur.

$\hspace{4cm}$


Rate law for breakdown of ozone to oxygen

You should write down the rate equations and set [O] as an intermediate so that $\ce{d[O]/dt} = 0$. This is the steady state approach and leads to $\displaystyle \frac{d\mathrm{[O_2]}} {dt}=2\frac{k_1k_2\mathrm{[O_3]}^2}{k_{-1}\mathrm{[O_2]}+k_2\mathrm{[O_3]}}$ which differs from your answer because when you assume a fast equilibrium you also assumed that $k_{-1} \gg k_1$. You should make this assumption after you have the full equation, if necessary, but not before you do the calculation.

To answer your last point you still have the intermediate [O] which is an unknown and so has to be eliminated as shown by using the steady state approach used above.


From the Bohr theory:
$$\large E=-\frac{Z^2e^4m}{8\epsilon_0^2n^2h^2}=-13.6\times\frac{Z^2}{n^2}\ \mathrm{eV}$$
where:

$$\begin{array}{c|c|}
& Z&\text{atomic number}&-\\
& e&\text{charge of electron}&1.60 \times 10^{-19}\ \mathrm{C}\\
& m&\text{mass of electron}&9.10 \times 10^{-31}\ \mathrm{kg}\\
& \epsilon_0&\text{electric permittivity}& 8.85 \times 10^{−12}\ \mathrm{F/m}\\
& n&\text{principal quantum number}&-\\
& h&\text{Planck's constant}&6.62 \times 10^{-34}\ \mathrm{m^2\ kg/s}\\
\end{array}$$

You can put values of $Z$ and $n$ for any shell of a hydrogenic species like $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, $\ce{B^4+}$…


How can I build the molecular orbital diagram of the complex tetraammincopper(II) (square planar structure)?

In principle, it does not matter which central metal you choose and which ligands; if the geometry is identical, the resulting MOs will be identical, too. The problem is as you state that the complex you wish to determine does not have a square-planar geometry (nor is it a tetracoordinated copper) so whatever you calculate will be off. At first, I was thinking of using a nickel(II) analogue (which would be identical save a missing electron) until I realised that $\ce{ls{-}[Ni(NH3)4]^2+}$ likely doesn’t exist and $\ce{hs{-}[Ni(NH3)6]^2+}$ is, of course, octahedral. A little bit of additional searching led me towards the square-planar complex $\ce{[Ni(\textbf{L})](ClO4)2}$, where $\mathbf{L}$ is basically a 3,10-bissubstituted 1,3,5,8,10,12-hexaazacyclotetradecane.[1] We can pretend that this perfectly fits the $D_\mathrm{4h}$ point group (it doesn’t) and that the ‘ammine ligands’ are fully identical (at least they’re symmetry-equivalent) and coordinate exactly from the coordinate axes (well, almost).

The first step is to consider the symmetry of the metal orbitals. We have five of them, and we expect them to group up into one $\mathrm{e}$-representation (two degenerate orbitals) and another three non-degenerate orbitals. A table of the irreducible represenations can be found on meta (soon™) or we can use the one from TU Braunschweig. Thankfully, it already includes the representations of the d orbitals, but for comprehensiveness you can also determine it yourself by thinking ‘does the orbital stay the same ($+1$) or invert lobes ($-1$) if I apply that symmetry operation. Here is the answer for the d orbitals:

\begin{array}{cccccccccccc}\hline
\text{Orb.} & E & 2C_4 & C_2 (z)& 2 C_2' & 2 C_2'' & i & 2 S_4 & \sigma_\mathrm{h} & 2 \sigma_\mathrm{v} & 2 \sigma_\mathrm{d} & \text{Rep.}\\ \hline
\mathrm{d}_{z^2} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \mathrm{A_{1g}}\\
\mathrm{d}_{x^2-y^2} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & \mathrm{B_{1g}} \\
\mathrm{d}_{xy} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & \mathrm{B_{2g}} \\
\mathrm{d}_{xz}/\mathrm{d}_{yz} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & \mathrm{E_g} \\
\mathrm{4s} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \mathrm{A_{1g}} \\
\mathrm{4p}_z & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & \mathrm{A_{2u}} \\
\mathrm{4p}_x/\mathrm{4p}_y & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & \mathrm{E_u} \\ \hline
\end{array}

$\mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$ must be grouped together; only as a group they will be transformed onto each other by $C_4$ or $S_4$. Always if you transform an orbital into empty space with a symmetry operation, find out which orbital was there and include that into a degenerate group.

Thus, the d orbitals transform as $\mathrm{a_{1g} + b_{1g} + b_{2g} + e_g}$. I also included the 4s ($\mathrm{a_{1g}}$) and 4p ($\mathrm{a_{2u} + e_u}$) orbitals for completeness; we’ll need them.

Next, do the same thing for the ligands. Here, we have to make use of group orbitals, i.e. group all the four ammine ligands together. Once we have the represenation of the group orbitals, we will need to reduce that to their corresponding irreducible represenations. Thankfully, we don’t have to consider whether the orbitals are pure or hybrid; we just need to consider σ-type orbitals and π-type orbitals. See the table below.

\begin{array}{cccccccccccc}\hline
\text{Orb.} & E & 2C_4 & C_2 (z)& 2 C_2' & 2 C_2'' & i & 2 S_4 & \sigma_\mathrm{h} & 2 \sigma_\mathrm{v} & 2 \sigma_\mathrm{d} & \text{Rep.}\\ \hline
4 \sigma & 4 & 0 & 0 & 2 & 0 & 0 & 0 & 4 & 2 & 0 & ? \\
8 \pi & 8 & 0 & 0 & -4 & 0 & 0 & 0 & 0 & 0 & 0 & ? \\ \hline
\end{array}

Here is where your ‘rule 5’ — or more accurately, the reduction formula — comes into play. It allows us to determine which irreducible representations a reducible one is made up of. (For smaller point groups you can often just derive it logically, but I wouldn’t do that for a group this size.) The general reduction formula is given in equation $(1)$ ($n$ is the number of times a given irreducible representation $\chi$, whose character with respect to a symmetry operation is $\chi_i$, is part of the reducible representation $\xi$. $h$ is the order of the group.) and the results for the σ and π type orbitals is given in the table below.

$$n = \frac1h \sum_i \chi_i \xi_i $$

\begin{array}{ccccccccccc}\hline
\text{Group} & \mathrm{A_{1g}} & \mathrm{A_{2g}} & \mathrm{B_{1g}} & \mathrm{B_{2g}} & \mathrm{E_g} & \mathrm{A_{1u}} & \mathrm{A_{2u}} & \mathrm{B_{1u}} & \mathrm{B_{2u}} & \mathrm{E_u} \\ \hline
4\sigma & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
8\pi & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 1 \\
\hline \end{array}

By doing this, we see that the σ-type orbitals transform as $\mathrm{a_{1g} + b_{1g} + e_u}$ while the π-type orbitals transform as $\mathrm{a_{2g} + b_{2g} + e_g + a_{2u} + b_{2u} + e_u}$. (That is all assuming I didn’t miscalculate.)

The next step is to start drawing an orbital diagram. The five d-orbitals should initially be on one level corresponding to the ionisation energy of nickel(II). The virtual 4s and 4p orbitals somewhat higher. The π-type ligand group orbital should be on a level corresponding to the ionisation energy of ammonia, the σ-type ones slightly below. I don’t have values here so I’m going to ballpark the relative energies draw a merely schematical diagram for simplicity. Any orbitals that have the same irreducible representation will mix with each other. If there are only two, these will form a bonding/antibonding pair with the antibonding orbital being relatively higher than the bonding one low. If there are more than two, they will form some kind of ladder with intermediate levels preserving the overall orbital count. Scheme 1 shows what I would end up with.

Scheme 1: Approximate orbital scheme for a fully square-planar, ideal $D_\mathrm{4h}$ $\ce{[Ni(\mathbf{L})]^2+}$ $\mathrm{d^8}$ ion.

Note that there are a number of estimates in there. For example, both $\mathrm{b_{2g}}$ and $\mathrm{e_g}$ are made up solely of d-π interactions; I can’t guess a priori which one would have a higher or lesser energy. Likewise for the relative orderings of other orbitals. The only thing clear is that any interactions stemming from the σ orbitals should be stronger than those stemming from π orbitals due to σ typically displaying better overlap.

To get back to your hypothetical $\mathrm{d^9}$ system, just add the missing electron into the $\mathrm{b_{1g}}$ orbital.

Reference:

[1]: M. P. Suh, B. Y. Shim, T.-S. Yoon, Inorg. Chem. 1994, 33, 5509. DOI: 10.1021/ic00102a026.


Mercury poisoning (also known as hydrargyria or mercurialism) is type of metal poisoning and a medical condition caused by exposure to mercury or its compounds. Mercury (chemical symbol $\ce{Hg}$) is a heavy metal occurring in several forms, all of which can produce toxic effects in high enough doses. Its zero oxidation state $\ce{Hg^0}$ exists as vapour or as liquid metal, its mercurous state $\ce{Hg2^{2+}}$ exists as inorganic salts, and its mercuric state $\ce{Hg^{2+}}$ may form either inorganic salts or organo-mercury compounds; the three groups vary in effects. Toxic effects include damage to the brain, kidneys and lungs.

Mercury poisoning can result in several diseases, including acrodynia (pink disease) and Minamata disease.

source wikipedia


Is histidine drawn wrong in these mechanisms for chymotrypsin?

Under aqueous conditions you can tautomerise the double bond, so it doesn't matter. The two nitrogens are, for all intents and purposes, equivalent.


Why do α-hydroxy ketones give Tollens' test?

The Tollens' test is done in base, which mediates tautomerization through the enolate:

The misconception that you have is that all alpha-hydroxy ketones can give a positive Tollens' test. Aldehydes give a positive test, and so only alpha-hydroxy ketones that can tautomerize to aldehydes will give a positive Tollen's test. My example above is 1-hydroxy-2-propanone, which isomerizes to 2-hydroxypropanol. Most alpha-hydroxy ketones with names that match the pattern 1-hydroxy-2-alkanone will give a positive Tollens' test.

Benzoin will not give a positive test, since it cannot isomerize to an aldehyde. What does benzoin form upon tautomerization?


I believe it is the number of carbon atoms in 12 grams.


Multiplicity of orbitals in a magnetic field

$m_l$ is the magnetic quantum number, and multiplicity refers to the number of allowed values that $m_l$ can be.


"Silicone water repellent" is barely helpful information. They are a category of materials which are used in an enormous variety of industrial, commercial and retail applications. Some products react and bind to the surface they are coating, others are simply dried onto the surface. Some cross-link, most don't. A typical consumer fabric silicone water repellent will be "penetrated" by solvents such as acetone, methy ethyl ketone (2-butanone) or toluene; mineral spirits might work. There's no way for us to determine whether these solvents can also damage the fabric or clothing item they are poured onto. Another possibility is to use a fairly harsh (alkaline) detergent, but again I can't say whether the fabric will be damaged. Many of the retail products gradually wear away, diminishing repellency slightly after each wash/dry cycle.


Dipole-quadrupole polarizability term for induced dipole

This question has been around for more than a year without an answer, so probably you have figured it out by now! But anyway, here goes. The standard literature references in this field are Theory of Molecular Fluids vol 1 by CG Gray and KE Gubbins (Clarendon Press 1984) and The Theory of Intermolecular Forces by Anthony Stone (Oxford University Press 2013) and these books explain the various terms in your equation. The dipole-quadrupole polarizability $A_{i,jk}$ may vanish by symmetry (e.g. for centrosymmetric molecules and isolated atoms) but does not vanish for a molecule like water. The corresponding term in your equation is also proportional to the local electric field gradient, so the importance of this term (relative to the other terms) depends on some of the details of the surroundings. In Appendix C of Gray and Gubbins they state

  For the cases we consider in this book the induction $A$ terms are negligible

but clearly there might be some cases in which they should not be neglected.


Silica gel is often used to absorb moisture; activated carbon is effective at odor removal; and zeolites (also called molecular sieves) come in different varieties and can be effective for both purposes.


For molecules to be enantiomers, do their bonds have to point the same way?

Place your two hands face down on a table. They are non-superimposable mirror images of one another (chirality does come from the Greek kheir, meaning "hand", after all). Now flip one of your hands over so that one palm is facing up. Are your hands any different now than before you flipped one over?

No. Your hands' orientations in space have no effect on their superimposability, as is the same for molecules.


Metals want to lose electrons and become positively charged. When the metal loses the electron, it is called oxidation. The substance that gets the electron is reduced.

Oxidation could occur at the anode if it takes electrons from a metal (i.e. $Fe^{2+}$ to $Fe^{3+}$).

Anodes are negative in voltaic cells since there are excessive electrons available to power the circuit. Cathodes give electrons away to the solution.