Photodissociation is a unimolecular reaction triggered by absorption of light. You start with one molecule, a bond (or more) breaks and you end up with two (or more) products. In water you can get $\ce{OH}$ and $\ce{H}$ (etc), usually in excited electronic states. You can not get the products of the electrolysis.

Water electrolysis is a more complex reaction. The final balance is:
$\ce{2H2O -> 2H2 + O2}$,
so you break four bonds and make three bonds.

Both reactions are not spontaneous (in the electronic ground state of water). The net balance is not energetically favorable (you start with more energy in bonds than what you end up). However, the way you add this energy is also different. In photodissociation the energy is given by light; in electrolysis the energy is given by forcing an electron current.


Why is the exo product formed in the Diels–Alder reaction between furan and maleic anhydride?

Great question!  It turns out that the rate of formation of the "expected" endo product is actually ~500 times faster than the rate of formation of the exo product. However, the Diels–Alder is a reversible reaction. In this case, the exo product is thermodynamically favored over the endo product by about $\pu{1.9 kcal/mol}$.

So, even though the predicted endo product is the kinetically favored product (and formed first), thermodynamics eventually takes over and the more stable exo adduct is what you wind up with. More details are given in J. Org. Chem. 1978, 43 (3), 518. What a nice example of competing kinetic and thermodynamic pathways.


Do perfect gases react when placed in a container whose volume can vary?

You have yourself all wrapped around the axle with odd conditions.

  Suppose that $n$ moles of four generic perfect gases...

This is a really odd way to specify the initial concentrations of four gases. Say I have 5 moles total. There are four gases, so I have four unknowns and only two equations (total moles and equilibrium equation) which is unsolvable.

This is also ambiguous. It is $n$ moles total, or $n$ moles for each gas?

  Now, if temperature and pressure are constant, then so will the density of the gases be. But if the density is constant, then the concentration of the gases will be constant as well, ...

Since you're holding temperature and pressure constant, and given the reaction:
$$\ce{A(g) + B(g)<=>C(g) + D(g)}$$
then the density will always be the same. Even if you have all A and B which are totally converted to C and D, the density of the gas mixture will be the same (assuming ideal gas behavior which you stipulated).

If at a constant volume and the reaction had been:
$$\ce{2A(g) + B(g)<=>C(g) + D(g)}$$
then the density would change as the equilibrium shifted.

But since you stipulated constant temperature and pressure then the volume would also have to change as $\ce{2A(g) + B(g)}$ was converted to $\ce{C(g) + D(g)}$ to maintain constant pressure, so the overall density of the gas remains constant.

  But if the concentrations are constant, then at the outset of the reaction, when there are n moles of each gas, the reaction is already at equilibrium, because the reaction quotient $Q=K_c$.

Let's break this down. If the concentrations are constant, then the reaction is already at equilibrium.

It does not necessarily follow that when there are $n$ moles of each gas then the reaction is at equilibrium. That would only be true if $K_c = 1$, but $K_c$ has not been specified.

  If not, how will the number of moles of each gas vary with time?

Ok, let's assume $n$ moles of each of the four gases. You'd also need to provide the rate constants for the forward and reverse reactions to calculate concentrations over time. Without the rate constants it isn't clear if the equilibrium will be established in seconds or years.


You should have read your explanation material more carefully! It clearly states that $A_{AL}2$ has not even been observed in acid hydrolysis due to water being a poor nucleophile in an $S_N2$ process. As for $A_{AL}1$, that only happens when the esterified radical can leave as a stable carbocation which, in this case methyl is a very very unstable carbocation.


molar solubility of CaF2 at pH3 given molar solubility at pH7 and pKa of HF

The $K_{sp}$ of $\ce{CaF2}$ which you found out as $3.7 \times 10^{-11}$ is indeed correct, and after that essentially you have to solve two equations.

Now, suppose at $pH =3$, (i.e. $\ce{[H+] = 10^{-3}}$),  $\ce{CaF2}$ dissociates as $\ce{Ca^2+}$ and $\ce{F-}$ and their concentration in the solution are as follows, $$\ce{CaF2 <=> Ca^2+ + 2F-}$$ $\ce{[Ca^2+ ] = S }$ mol/L. and $\ce{[F-] = 2S -}$ $x$ mol/L. ,as, this $x$ mol/L. of $\ce{F-}$ is in equilibrium with $\ce{HF}$, as, $$\ce{H+ + F- <=> HF}$$ This equilibrium satisfies the equation $$\ce{\frac{[H^+][F^-]}{[HF]}= 10^{-3.17 }   => \frac{10^{-3}(2S - x)}{x} = 10^{-3.17}  ...(i)}$$.

And another equation which you have to solve is just regarding $K_{sp}$, i.e $$S(2S-x)^2 = 3.7 \times 10^{-11} .....(ii)$$ From first equation, you will get $\frac{2S}{x} = 1.676$ ; if you substitute for $x$ in the second equation by this, you can now solve for $S$ and the answer will be $S= 3.845 \times 10^{-4}$.

Thus , the factor will be $\frac{3.845 \times 10^{-4}}{2.1 \times 10^{-4}} = 1.83$


The clotting in a yoghurt is indeed caused by the lactic bacteria producing lactic acid and therefore lowering the pH of the milk. However in this particular case (yoghurt making) this is not due to the denaturation of the milk protein, they are in fact intact.
When the pH drop in the milk, it get near the isoelectric point (IEP) of the casein (milk protein) therefore making them partialy neutral (or completly if the pH reach it).
It weaken the electrostatic repulsion between the casein.
Consequently when the pH get closer to the IEP, casein solubility decrease and often precipitate.

And this is reversible, if you add caustic soda or another base, the yoghurt or any sour milk will go back to being liquid (well not perfect in some case because the bacteria often produce exopolysaccharide which thicken the milk).


R/S configuration of bridging carbon in bicyclic system

Several days ago I offered a solution to the R/S assignment of the bridgehead carbons of (1R,2S,4R)-2-bromo-2-methylbicyclo[2.2.2]octane 1. During the middle of the night I realized that the method was faulty because the solution would not accommodate the (1S,2R,4S) enantiomer  3 but would rather predict (1R,2R,4R) which clearly is a non-existent diastereomer. The shortcoming lay in not applying, in a heirarchical sense, CIP rule 4b (RR/SS>RS/SR) over rule 5 (R>S).
Focusing on bromide 1, the configuration of C2 is "S" with Br>C1>C3>C9. Structure 2 is structure 1 from a different perspective to which we will refer during the ensuing discussion. To determine the configurations of the bridgehead carbons, C1 and C4, a digraph is the method of choice. In the "Digraph of C1" one starts at the non-duplicate atom C1 and traces the six possible paths back to duplicate atoms C(1) that are each attached to three phantom atoms of atomic number zero. The vertical chain has the top priority while the hydrogen (not shown) at C1 has the lowest priority. The two horizontal chains are mirror images and they must be assigned to the second and third priorities. In these chains C2 has been predetermined as having the S-configuration. Locate C4 in the left chain. It has the R-configuration because C3>C8>C5. Locate these atoms in structure 2 and convince yourself. [I use my hands. Point your right thumb from C4 to H and your fingers will point from C3 to C8 to C5]. In the right hand chain C4 has the S-configuration because C3>C5>C8. Check these priorities in structure 2.
Now the left hand chain is designated as RS and the right hand chain as SS. Since RR/SS>RS/SR, the order of carbon atoms attached to non-duplicate C1 is C2>C6>C7. Locate these atoms in structure 2 and convince yourself that C1 does indeed have the R-configuration!
Look at "Digraph of C4". Now that you are versed in the method, the priorities around non-duplicate C4 are C3>C8>C5 because C3-chain>SS>RS>H. C4 also is of the R-configuration! <continued>

The digraphs below confirm the assignments to the enantiomer 3 (4), (1S,2R,4S)-2-bromo-2-methylbicyclo[2.2.2]octane. I'll leave the analysis to the reader.


Acid-catalyzed hydrolysis of diphenyl malonate

If the question is "What are the products of hydrolysis of diphenyl malonate?", then my answer would be 2 equivalents of phenol and one equivalent of malonic acid.

If the question is "What are the products of diphenyl malonate when heated in an acid aqueous solution?", then my answer would be 2 equivalents of phenol, one equivalent of acetic acid, and one equivalent of carbon dioxide.

I see it sort of as a semantic question, since the hydrolysis of an ester gives an alcohol and a carboxylic acid, as you suggest. I don't consider decarboxylation to be a hydrolysis reaction, literally defined as cleavage of bonds by addition of water. Water isn't incorporated into the products of the decarboxylation step.


Evaporation is an endothermic phenomenon, i.e. it absorbs heat in order to proceed. Acetone is a volatile solvent (it evaporates easily) so it absorbs much heat when evaporating, and your skin gets colder because of that. That is what you are feeling. So acetone is not "colder", it just cools down your skin more easily when evaporating.

Ether (aka diethyl ether) is even more powerful at this.

On another note, while acetone is a pretty safe solvent, it can damage your skin because it can take away skin's lipids and you get "dry hands". In France, your employer would be forced to provide you with protecting gloves, e.g. latex gloves.


Is it wise to use ECPs on light atoms for electronic structure calculations?

I think you will find that the time savings are minimal. Modern integral screening will eliminate virtually all inter-atom core-core integrals anyway and if you only have a 1s core then the cost of evaluating intra-atom core-valence integrals will be comparable to evaluating the ECP-core interaction.


Do the products in exothermic reactions have higher kinetic energy than the products? Do they also have lower chemical energy?

In line with the comment of orthocresol, these potential energy diagrams only display the change in the chemical potential as a function of the reaction coordinate.

To clarify: Let's consider the theoretical situation of a reaction mixture in complete vacuum that undergoes a reaction towards a product state. If we assume that the reaction is exothermic and there is no loss of energy due to irradiation, then this means that the products are heated up by an amount of energy that is gained from the loss in chemical potential (i.e. the exothermic reaction). In other words, if you would plot the chemical potential + the kinetic energy (the two terms combined) as a function of the reaction coordinate, you would obtain a straight line due to the law of conservation of energy, because chemical energy is converted to kinetic energy (heat).


What's the difference between atomic orbitals and energy levels?

Neon's atomic number. But then what are the atomic orbitals, like s,
  p, d, and f. Where do they come in? Are they completely different from
  the energy level?

Atomic theory was developed in stages almost a century ago. Bohr's model is the simplest version of hydrogen atom. General chemistry textbooks write rarely mention the Sommerfeld's model from which the concept of orbital angular momentum originated on elliptical paths (recall that a circle of Bohr's model is a special ellipse). Sommerfeld model model explained more features of the hydrogen spectrum. After Sommerfeld the next person was Schrodinger who proposed the H atom in terms of probabilities of finding electrons rather than fixed orbits.

In short, s, p, d, f come into play in more advanced models of Sommerfeld. Bohr was the first one to talk about electronic configurations for multielectron atoms.

Sommerfeld' model


What is the meaning of an oxide being 'Higher' or 'Lower'?

I think it generally means that the average oxidation state of the metallic element in the oxide is higher or lower. You can also say that the mass percentage of oxygen in the oxide is higher or lower.


Show that decomposition of AsH3 follows first order kinetics

In a first order reaction $\ce{A \rightarrow B + C}$ with rate constant k the rate equation is
$$ {\frac{d\ce{[A]}}{dt} = - k\ce{[A]} } $$
which  if integrated to give the concentration vs time where the initial concentration is $\ce{[A_0]}$ produces
 $$\ln\left(\frac{\ce{[A]}}{\ce{[A_0]}}\right)=-kt$$ or $$ \ce{[A]= [A_0]}e^{-kt}$$

Thus a log of the ratio concentrations or pressures (or any measurement proportional to concentration) vs time should be a straight line if the reaction is first order.

In your case you are observing the product which appears as the $\ce{AsH3}$ disappears therefore
$$ {\frac{d\ce{[B]}}{dt} = + k\ce{[A]} } $$

You can substitute for $\ce{[A]}$ and integrate again to get $\ce{[B]}$.

$$\ce{[B]-[B_0]=[A_0]}(1-e^{-kt})$$
However the total pressure is measured which is the sum of $\ce{2[A] + 3[B]}$ and if $\ce{[B_0] =0}$ then

$$\ce{3[B] + 2[A] =3[A_0]}(1-e^{-kt}) + \ce{2[A_0]}e^{-kt}$$
if the measured pressure is proportional to sum of concentrations with $p_0$ as initial pressure then

$$p = p_0(3-e^{-kt})$$

At $t=0$ the initial pressure is $2p_0$ which is the value given in the table. At very long times the pressure becomes constant at $3p_0$ and calculating $3p_0 - p$, rearranging and taking logs produces

$$\ln\left(\frac{3p_0 - p}{p_0}\right) =  -kt $$

which when plotted is the 'normal' equation for a first order equation.


Can molten Strontium and Barium Hydroxides be electrolysed to give the metals?

Expanding on @Oscar Lanzi comment: first, all the hydrogen (and half the oxygen) would be electrolyzed before then alkaline earth metals.  At that point, if kept liquid, then the $\ce{BaO}$ and $\ce{SrO}$ remaining would be split. That would not be easy, as the melting point of $\ce{BaO}$ is 2,196 K and that of $\ce{SrO}$ is 2,804K (though a eutectic mix of the oxides would be a bit lower).

[It's more efficient to dissolve the oxides in a halide salt. The electrolysis of $\ce{CaO}$ in $\ce{CaCl2}$ has been explored here and here.


HCl rota-vibrational lab: ab initio simulation of rotation constants

Disclaimer:

I never used Gaussian, so I am just guessing about the program.

Coriolis coupling:

Coriolis coupling couples bending modes with rotation and there is no bending in diatomic molecules.

There is no Coriolis coupling in Gaussian for HCl, because there can't be one. The relevant output could also be NaN, it is simply not defined.

Alpha "matrix":

For diatomic molecules you have two degenerate rotations and one stretching mode. Hence your alpha "matrix" is actually well represented by a scalar and used as such in your equations. For this reason one would expect only one alpha value.

That there is written -0.000 and not 0.000 is an indication for a truncated negative number. In addition other values are properly set to NaN. I think that your one reported alpha value is simply close to zero. You can try a verbose output to verify this.

Solution:

If you want to get a non zero alpha
I would recommend a molecule with a shallower potential hyper surface, i.e. a lower force constant. Or you could use this example to discuss floating point arithmetics and truncation with your students.


Why do methane molecules have 15 degrees of freedom?

15 is there because you have 5 atoms which can move independently, and we live in a 3D space, and $3\ \times\ 5\ =\ 15$.

3 of these are translational degrees of freedom, which you seem to agree with, so I'll just leave it at that.

3 are for rotations, because the methane molecule is not linear. Axes of symmetry are irrelevant: indeed, you may rotate methane around the axis of symmetry as well as around any other axis. Being or not being diatomic is hardly relevant either, except for the fact that all diatomic molecules are necessarily linear; however, the converse is not true.

Subtracting 3 and 3 from 15, we get 9, and that's how many vibrational degrees of freedom we have, because they can't be anything else. To visualize them is another story, and a great deal more difficult. Counting bonds doesn't help much. The separation of degrees into kinetic and potential is not a thing at all. You don't have to do that anyway. Write $15\ -\ 3\ -\ 3\ =\ 9$ and call it a day.


Question about null space balancing method

Fortunately this issue was just caused by a small oversight. In the midst of so many coefficients, you accidentally skipped the second nitrogen atom in $\ce{NH4NO3}$, which changes the fifth element in the fourth row from 1 to 2. The correct composition matrix is then:
$$\begin{bmatrix}
3 & 8 & 1 & 12 & 4 & 2 \\
1 & 0 & 0 & 1 & 0 & 0 \\
4 & 4 & 3 & 40 & 3 & 1 \\
0 & 2 & 1 & 3 & \color{red}{\textbf{2}} & 0 \\
0 & 1 & 0 & 12 & 0 & 0
\end{bmatrix}$$
After complementing with the bottom row $\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$ and inverting the matrix, we obtain:
$$\frac{1}{96}\begin{bmatrix}
3 & 79 & 2 & -9 & -14 & -8 \\
36 & -204 & 24 & -108 & -72 & -96 \\
39 & -477 & 90 & -213 & -246 & -168 \\
-3 & 17 & -2 & 9 & 14 & 8 \\
-51 & 417 & -66 & 249 & 174 & 168 \\
0 & 0 & 0 & 0 & 0 & 96
\end{bmatrix}$$
Finally, transposing the rightmost column and dividing its coefficients by 8, the result is  $\begin{bmatrix} -1 & -12 & -21 & 1 & 21 & 12 \end{bmatrix}$, exactly as expected.
Thank you for the linked article! Hopefully others may find it useful. If you're into linear algebra applied to chemistry, I recommend you take a peek into chemical reaction network theory!


Does selenium sulfate corrode or otherwise react with metals?

Yes, selenium sulfide is well known and it is used in anti-dandruff shampoo but other sulfur-containing selenium compounds are not really well known. Selenium sulfate, properly selenium(II) sulfate, $\ce{SeSO4}$ is unknown to the best of my knowledge. The analogous tellurium compound has been synthesized recently, but still it is not a proper sulfate compound, rather it is actually an oxide-sulfate mixture type of compound.

In case you are wondering whether selenium sulfate will be obtained by reacting selenium with sulfuric acid, then you are wrong. Selenium and sulfuric acid would react to form a mixture of acids (selenous and selenic acid) but they would all disproportionate to selenium dioxide, sulfur dioxide and water.

$$\ce{Se + 2H2SO4 ->[\Delta] SeO2 + 2SO2 + 2H2O}$$

Selenium sulfite or properly selenium sulfoxide, $\ce{SeSO3}$, is however known.

So, selenium sulfate is a fictitious compound and thus used in the cartoon.


Can aldehydes be reduced to alcohols efficiently using LiAlH4 (excluding cost effectiveness)?

Yes, $\ce{LiAlH4}$ can be used for the reduction of an aldehyde to an alcohol. Compared to $\ce{NaBH4}$ it is a stronger reducing agent due to the electronegativity differences (Pauling electronegativity of $\ce{Al}$ is 1.61 & $\ce{B}$ is 2.04) making the metal-H bond more polar in case of $\ce{LiAlH4}$. Therefore, it reacts very violently with water and other acidic group containing compounds, and careful addition of $\ce{LiAlH4}$ is required.

The mechanism is rather straightforward, with alkoxide formation and protonation steps to get the final product.

In the lithium aluminum hydride reduction, water is usually added in a second step, whereas in the sodium borohydride reduction the methanol solvent system achieves this protonation immediately.


Every $\boxed{^{14}\textrm{Si}}$ atom is surrounded by 4 $\boxed{^8\textrm{O}}$ atoms.   On the other hand, every $\boxed{^8\textrm{O}}$ atom is surrounded by 2 $\boxed{^{14}\textrm{Si}}$ atoms.

Suppose that the number of $\textrm{Si-O}$ bonds is $n$, then the number of $\boxed{^{14}\textrm{Si}}$ atoms is $n/4$, while the number of $\boxed{^8\textrm{O}}$ atoms is $n/2$.  Therefore you arrive at the formula $\textrm{Si}\textrm{O}_2$.


The Nernst equation only technically applies when the system is in electrochemical equilibrium—when there's no net current flow, but there's actually an exchange current. A system in which the reaction can only happen in one direction because there are no products wouldn't apply. Source:    Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008.

What I'm not sure about is the validity of ∆G values calculated from $∆G=∆G°+RT\ln Q$ at the  limits where the system is all products or all reactants. ±∞ don't seem to be meaningful values, but the system should quickly move away from these boundary values, even if only by a small amount. These chemical thermodynamics methods are well-behaved when near equilibrium, but well away from equilibrium, they can produce invalid results.


Why does a 57.15% ABV spirit (ethanol+water) have a density of 923 kg per cubic meter?

There is nothing wrong with your calculations: but some liquids don't play nice
Your calculations are, likely, correct.
But liquids, unlike ideal gases, don't necessarily obey simple rules when mixed. In particular the two components, ethanol and water, strongly interact with each other when mixed and those interactions mean that the volumes of the pure liquids are not additive when they are mixed.
The effect is well known and a chart of it looks like this:

This is a fairly common behaviour for liquid mixtures where the components of the mixture interact. But the effect for ethanol/water is particularly strong and of practical significance both to chemists, industry and the tax authorities.


Which functional group gets reduced first?

I believe you understand $\ce{LiAlH4}$ is stronger base than $\ce{NaBH4}$. In the reduction of 2 to 1, $\ce{NaBH4}$ failed to reduce ester function at all, but reduce 3-keto function to 3-hydroxy function. That result tells us that isolated carbonyl group is easy to reduce than that in ester (not even to relevant aldehyde). Thus, you can conclude that isolated carbonyl function reduced first before hard to reduce ester function in the presence of stronger reducing agent such as $\ce{LiAlH4}$, which can able to reduce both.


I have been doing quite a lot of reading recently about the theory of atoms in molecules and think that I have semi-satisfactory answers to your questions, so I will share them here and perhaps others will add complementary answers as there actually seems to be quite an extensive literature on this subject, but there is still a lot of debate over these topics.

First, a quick note on how the charge density, $\rho$, can be used in the theory of atoms in molecules (AIM) to determine chemically intuitive properties. This is usually done by the definition of an index. For instance, the first index one might think to define is the electron number index, $N(\Omega)$
, where $\Omega$ is the space defined by the critical points surrounding a single nucleus. Essentially AIM claims that $\Omega$ and the very idea of an atom are interchangeable, so $N(\Omega)$ is conceptually identical to answering how many electrons are associated with a partiular atom. Hence, the AIM definition of the charge on an atom follows as,
$$
Q_{\Omega_i}=Z_{\Omega_i}-N(\Omega_i)
$$
where $Z_{\Omega_i}$ is the charge of the nucleus of atom $i$ in the basin $\Omega_i$.

Not surprisingly, $N(\Omega_i)$ is defined as[1],
$$
N(\Omega_i)=\int_{\Omega}\rho(\textbf{r})d\tau
$$
which is clearly just the charge density in a particular basin. This is interchangeable with a number density of electrons by division by the electric charge $e$.

These indices are always defined in terms of the charge density. The reason these are called indices, rather than something more exact, is because we are really computing a quantity which dimensionally satisfies the property we are interested in, and then interpreting it as the thing itself. The ambiguity comes from the idea of a basin, $\Omega$, which is mathematically well-defined but is not guaranteed to be physically exact for any reason.

  How dependent are computed charges using the quantum theory of atoms in molecules on the used level of theory?

Ref. [2] is a review of the AIM theory with a specific emphasis on the electron localization functions (ELFs) to describe aromaticity. More on these ELFs in a moment. They present the calculation of the index $N(\Omega)$ for several diatomic molecules using DFT, HF, and CISD calculations. The total charge in each of the basins stays more or less the same for all of these methods. The value of $N(\Omega)$ generally only changes in the first or second decimal place for the calculations they present (so tenths to hundredths of an electron). The trend when Coulomb correlation is included is that $N(\Omega)$ decreases for what would traditionally be considered the more electronegative atom. That is to say that coulomb correlation makes the bonds more covalent and less ionic. For mostly ionic systems such as $\ce{LiF}$, the addition of coulomb correlation has a small effect because the electrons are already mostly localized in the basins [2]. All of this is to say that the actual charges computed based on $N(\Omega)$ seem to be relatively insensitive to coulomb correlation. This makes sense based on the fact that the primary features of the spatial distribution of $\rho$ will be dictated by the requirement of antisymmetrization of the wavefunction which is, of course, included in a HF description.

That covers the question of charges based on $N(\Omega)$, but this does not actually give us a picture of how electrons are shared between atoms. In other words, how localized and how delocalized are the electrons? In order to answer this question, ref. [3] provides definitons of both localization and delocalization indices, which are unique because the localization index describes localization of electrons to a single basin, while the delocalization index describes the sharing of electrons with either a specific basin, or with all other basin in a molecule.

While $N(\Omega)$ is not greatly affected by coulomb correlation, the localization index, $\lambda(A)$, and the delocalization index, $\delta(A,B)$ are. For instance, $\delta(A,B)$ which is a measure of the sharing of electrons between the atoms $A$ and $B$, changes by almost $1.0$ when going from the HF description of $\ce{N2}$ to the CISD description of $\ce{N2}$. In general, for homodiatomics, coulomb correlation has the effect of increasing the density in individual basins. And, consistent with the statement above about charge, the localization index for polar diatomics tends to increase the charge on the less electronegative element when a correlated wavefunction is used.

I would highly recommend reading ref. [3] as a place to gain some intuition about the behavior of these different indices.

  Do calculated charges converge to a specific value with higher level of theory?

I have not been able to find any explicit discussion of this, but of course any method which approaches the exact wavefunction, and hence the exact density, must have the same limiting value for any index defined in terms of the density. This is really the power of the AIM model. It is completely quantum mechanical in the sense that the exact wavefunction will yield the exact basins. Whether or not these basins mean what AIM wants them to mean, however, is what people debate.

Another related point is how the various indices are affected by the particular basis set used. I have seen a couple papers which mention this, and note that the results really do not depend much on the particular basis set chosen[1]. The reason which is speculated for this behavior is that the partitioning used in AIM is in real space rather than in the Hilbert space to which the basis set belongs. Unsurprisingly, however, the indices are sensitive to the basis set when the density itself is very sensitive to the basis set. Open-shell molecules seem to be more complicated on this front [2].

  Are charges from calculated electron densities comparable to charges from measured electron densities?

The primary purpose of ref. [1] is to compare the value of properties determined from AIM and from a density determined by x-ray diffraction of p-nitroaniline. This molecule was chosen, I believe, because it has a large dipole moment.

One complication in using the experimental density is that it is not as simple as just measuring the density and then having the whole scalar field in front of you. Rather, from what I understand, you measure the diffraction data and then use certain theoretical models to translate this diffraction data into a charge density. Thus, there is ambiguity in what charge density you actually measure. Ref. [1] uses the UMM and KRMM multipole formalisms which are discussed in this wikipedia article. The point of bringing this up is that the answer for the charges actually depends on which method you use to construct the experimental density.

Ref. [1] indicates that the KRMM multipole method which uses the AIM analysis of charges agrees much more closely with an AIM analysis of the charges from a theoretical density found from periodic-HF and periodic-DFT calculations (this is all solid phase). The p-HF and moreso the p-DFT charges agree very well with experimental charges which are calculated from KRMM using the AIM method. See Table 2 of [1].

The dipole moments also agree very well, but only after various refinements are made to both the partitioning of the experimental density as well as the use of the KRMM method. See Table 4 of Ref. [1].

One point which was mentioned in this book[4] (the relevant chapter is available on Google Books, ch. 3) is that the location of bond critical points, saddle points on the density between two atoms, can be very different for theoretical and experimental densities. Thus, properties derived from the basins can, in principle, be very different for the densities. It seems, however, that despite this problem of locating the bond critical point accurately, properties which are derived from the theoretical and experimental basins are not very different. This is because in the region of the saddle point, the density is usually quite flat, thus shifting the bond critical point does not have much effect on how much density is in each basin.

  Are there more reliable methods to obtain partial charges?

Only briefly, I will point out that various population analyses are always possible to determine charges on atoms. The answer to this really depends on what you mean by reliable. In a sense, AIM is the only reliable method of determining charges on atoms in molecules because it is the only method I am aware of which bothers to define what an atom in a molecule actually is. Additionally, it is reliable in the sense that it exists in a completely self-contained mathematical framework. Population analyses, however, can be very dependent on the particular basis set used, and also are not invariant to unitary transformations of the orbitals. In this sense they are very unreliable.

Mulliken charges are a common form of finding charges from a population analysis, but these are known to very basis set dependent. A better scheme, I think, is the CHELPG scheme which fits the molecular electrostatic potential of a system and hence conserves the dipole moment when it assigns charges to atoms. This is a very nice property for a population analysis. Nonetheless, it will be sensitive to method and basis set depending on how much the electrostatic potential changes upon changing either of these.

I think that generally the AIM charges and charges from population analyses will not be too different, but one notable exception is given in ref. [4] on page 162 where it is noted that the AIM charge on boron in $\ce{(H3BNH3)2}$ is positive, $q(\Omega_{\ce{B}})=2.15$, while the Mulliken population analysis gives a negative charge of $-0.26$. So that's a very stark difference. I guess I would trust AIM, but I don't really know.

As a concluding point, everything I have discussed above calculates properties from indices based on $\rho$, but there is no reason to limit the definition of indices to the density. Rather, some authors have defined indices, for physically motivated reasons, based on the gradient of the density, $\nabla\rho$ and the Laplacian of the density, $\nabla^2\rho$.

An interesting application of one of these based on the gradient is for some alkanediols where there is no bond critical point between the two $\ce{O-H}$ groups despite the obvious presence of an internal hydrogen bond [5].

Ref. [5] is a really cool paper. Definitely give it a read.

[1] Volkov, A., Gatti, C., Abramov, Y., & Coppens, P. (2000). Evaluation of net atomic charges and atomic and molecular electrostatic moments through topological analysis of the experimental charge density. Acta Crystallographica Section A: Foundations of Crystallography, 56(3), 252-258.

[2] Poater, J., Duran, M., Sola, M., & Silvi, B. (2005). Theoretical evaluation of electron delocalization in aromatic molecules by means of atoms in molecules (AIM) and electron localization function (ELF) topological approaches. Chemical reviews, 105(10), 3911-3947.

[3] Fradera, X., Austen, M. A., & Bader, R. F. (1999). The Lewis model and beyond. The Journal of Physical Chemistry A, 103(2), 304-314.

[4] Popelier, P. L. A., Aicken, F. M., & O’Brien, S. E. (2000). Atoms in molecules. Chemical Modelling: Applications and Theory, 1, 143À198.

[5] Lane, J. R., Contreras-García, J., Piquemal, J. P., Miller, B. J., & Kjaergaard, H. G. (2013). Are bond critical points really critical for hydrogen bonding?. Journal of chemical theory and computation, 9(8), 3263-3266.


The particular reaction is of acetylene ($\ce{C2H2}$) and ethanol ($\ce{C2H5OH}$) to yield ethyl vinyl ether (IUPAC name ethoxyethene). It's one example of Reppe chemistry, a fairly diverse set of reactions, often operating on alkenes and alkynes (though certainly not exclusively). The specific reaction in the question is a vinylization reaction, though it was Reppe's carbonylation reactions which were probably of greater practical (read: economic) import historically, as they were famously used in the Monsanto process and in production of various acrylates and other acrylic acid derivatives.

The vinylization reactions could still be synthetically useful, as they provide an extremely simple alternative route to vinyl ethers without having to proceed via elimination, substitution, coupling, olefination, or substrate-specific oxidation reactions.

I'm not aware of any specific uses for ethyl vinyl ether, however it could certainly be used as a building block in any number of organic syntheses.


I suspect one might utilize the fact that all three compounds – cyclohexanol $\ce{C6H12O}$, acetone $\ce{C3H6O}$ and pentanal $\ce{C5H10O}$ – share the same empirical formula $\ce{(CH2)_xO}$, which is also applicable to the mixture of reactants, and so instead of writing a set reaction equations

\begin{align}
\ce{C6H12O + 8.5 O2 & -> 6 H2O + 6 CO2} \tag{R1}\\
\ce{C3H6O + 4 O2 & -> 3 H2O + 3 CO2} \tag{R2}\\
\ce{C5H10O + 7 O2 & -> 5 H2O + 5 CO2} \tag{R3}
\end{align}

one can avoid finding volume fraction $\phi_i$ of each individual component and use a single reaction scheme such as

$$\ce{(CH2)_xO + $\frac{3x-1}{2}$ O2 -> x H2O + x CO2} \tag{R4}$$

and find relation between average molecular weight $\bar{M}$ and $x$ solely from (R4). Lets denote the mass and the total amount of gas mixture components as $m$ and $n$, respectively. Then according to (R4) total amount of gas mixture is $n = \frac{n(\ce{H2O})}{x}$, and the average molecular weight is established as

$$\bar{M} = \frac{m}{n} = \frac{x \cdot m}{n(\ce{H2O})} = \frac{x \cdot m \cdot M(\ce{H2O})}{m(\ce{H2O})} = \frac{x \cdot \pu{2 g} \cdot \pu{18.015 g mol-1}}{\pu{1.998 g}} = (18.033 x) \,\pu{g mol-1} \tag{1}$$

At the same time using molecular weights of the elements:

$$\bar{M} = x \cdot (M(\ce{C}) + 2M(\ce{H})) + M(\ce{O}) = (14.027 x + 15.999)\,\pu{g mol-1} \tag{2}$$

Equating (1) and (2):

$$18.033 x = 14.027x + 15.999 \tag{3}$$

and solving the resulting equation (3) for $x$, one finds that $x = 3.994$. Solving either (1) or (2) with obtained $x$ yields in average molecular weight $\bar{M} = \pu{72.023 g mol-1}$.


The nitroxyl anion is isoelectronic to oxygen. It has pretty much the same orbitals as oxygen does and it also features the same number of electrons. The principal difference is the orbital shape: since oxygen and nitrogen are two different atoms with a different number of protons, there is no plane of symmetry bisecting the $\ce{N=O}$ bond axis (in group theory terms: the point group is $C_{\infty \mathrm v}$ rather than $D_{\infty \mathrm{h}}$). This difference means that bonding orbitals will be more oxygen-centred while antibonding ones will be more nitrogen-centred.

However, if you take a look at the molecular orbital scheme, the key orbitals are identical to those in oxygen. The highest occupied molecular orbitals are two symmetry-equivalent π* orbitals and the unoccupied σ* orbital is higher in energy. Since the two π* orbitals are symmetry equivalent, there is no way the molecule can pair the spins; the highest energy electrons will be parallel and occupy one of the π* orbitals each.

Considering that we have exactly the same situation as oxygen in the key orbitals, this also means that $\ce{NO-}$ features exactly the same states. Most notably, the ground state is a triplet $\ce{^3NO-}$ state, while a higher-energy singlet $\ce{^1NO-}$ state also exists — entirely analogously to the singlet $\ce{^1O2}$ and triplet $\ce{^3O2}$ states of dioxygen.

When calling the nitroxyl anion a radical, one should remember that it is, like the oxygen molecule, more accurately described as a diradical. Its properties are distinct from those of monoradicals such as $\ce{NO}$, which feature a doublet state.


Number of stereoisomers of 1-(2-methylcyclohexyl)ethan-1-ol

The comments on the question are incorrect [edit: the offending comments seem to be gone now], as is the answer given in the book. In general, answers printed at the back of a book should never be taken as the absolute truth. There are typos in all books. Even well-known and well-written textbooks have many errata (usually corrected in subsequent printings).

As you said, there are three chiral centres and therefore 8 stereoisomers.

There are no other elements of symmetry present in this molecule which render any of those 8 stereoisomers equivalent.


Is there a simple way to get the circular dichroism of a molecule from its structure?

Is there a simple way to get the circular dichroism of a molecule from
  it's structure?

Short Answer:  Yes, for many molecules if you know the molecule's structure, then you can predict the shape of the optical rotary dispersion (ORD) or circular dichroism (CD) curve.  Conversely, and perhaps more importantly, if you know the shape of the ORD or CD curve, then you can predict the molecule's structure, in particular, the absolute configuration at chiral centers in the molecule. There are many empirical techniques, such as the "Octant" rule.  These methods are based on numerous experimental observations that have been codified into sets of rules for different classes of compounds.  These empirical rules can be used to determine the absolute configuration of cyclic ketones, conjugated dienes, alpha, beta-unsaturated ketones, twisted biphenyls as well as other systems.

Long Answer: When polarized light is passed through a chiral sample and the light is absorbed, there are two physical phenomenon that can be measured.  One is called the circular birefringence and it measures the difference in the refractive indices ($\ce{n_{R}, n_{L}}$) of the sample when a beam of right and left circularly polarized light is passed through the sample.  When this difference is plotted against wavelength an ORD curve results.  The other parameter is called circular dichroism and it measures the difference in the molar extinction coefficients ($\ce{\epsilon_{R}, \epsilon_{L}}$) when right and left circularly polarized light passes through the sample.  When this difference is plotted against wavelength a CD curve results.  The dependence of these two phenomenon on wavelength is termed the Cotton effect.

The following figure shows the uv-vis absorption (blue line), ORD curve (green line) and CD curve (red line) for the two enantiomers of camphor sulfonic acid.  Note how the ORD and CD curves for one enantiomer are mirror images of the same curves for the other enantiomer.  It is the carbonyl chromophore's $\ce{n->\pi^{\ast}}$ absorption that is responsible for the curves seen in this image.  In ORD spectra there are typically 2 maxima, one positive and one negative.  The positive maxima is a "peak", while the negative maxima is termed a "trough". If, when moving from longer to shorter wavelength, a peak is encountered first, then the ORD curve is said to be positive (+).  Conversely, if a trough is encountered first, then the ORD curve is negative (-).

image source

Chromophores can be divided into two groups, symmetric or inherently dissymmetric.  A carbonyl, nitro and olefin are examples of groups that, by themselves, are symmetric.  They may be perturbed by asymmetry on other atoms around them, but alone, by themselves, they possess no inherent asymmetry.  On the other hand, groups such as helicenes, twisted biphenyls and unsaturated ketones possess a twist or handedness; such molecules are inherently dissymmetric.  The intensity of an ORD or CD curve from a molecule containing an inherently dissymmetric chromophore is usually much greater than the peak intensity from a molecule containing a symmetric chromophore.

With that information as background, we can now apply these concepts to the examination of real molecules.  One of the most studied classes of molecules where these concepts have been applied are the substituted cyclohexanones.  In the early 60's  Djerassi, Moffitt, Woodward, Moscowitz and Klyne (some of the greats!) developed an empirical set of rules to predict the absolute configuration of chiral centers in substituted cyclohexanones.  The rule became known as the "Octant" rule and came about because of the work they were doing with complex natural products that often contained a substituted cyclohexanone ring.  They would run a reaction on a pure enantiomer of a cyclohexanone containing natural product and wanted a method to quickly determine if a chiral carbon in the molecule retained its absolute configuration or if the configuration was inverted in the reaction.  After comparing ORD curves from countless cyclohexanone containing natural products, they were able to compile their observations into the Octant rule.

As its name implies, the Octant rule divides the region around a cyclohexanone carbonyl into 8 regions or octants.  As the picture below (left part of figure) illustrates, the octants are generated by bisecting the carbonyl group with 3 planes.  Two of the planes contain the carbonyl group.  One divides the carbonyl into top and bottom regions and the other divides the carbonyl into right and left regions.  The third plane actually passes through the middle of the carbonyl and is perpendicular to the 2 other planes.  This third plane divides the molecule into front and back regions.  There are a few molecules that do have substituents in the front region, but most molecules do not, so we can usually simplify our analysis by only considering the 4 rear octants.  Each octant is also assigned a plus or minus symbol telling us whether we would expect a substituent in the octant to contribute in a positive or negative manner to the ORD or CD curves.

image source

The figure on the right shows how a cyclohexanone would be placed in the octant diagram and how its projection appears as a rectangle when sighted down the carbonyl group.

Let's examine the case of 3-methyl cyclohexanone.  First let’s draw the possible conformers of both S- and R-3-methylcyclohexanone

Now let’s take the 2 possible conformations for R-3-methylcyclohexanone (axial and equatorial methyl group) and project them onto the octant diagram.  We see that the equatorial conformation has a positive contribution to the Cotton effect, while the axial conformer has a negative contribution.  At equilibrium, the molecule exists predominantly in the conformation with the methyl group equatorial, therefore we expect the positive Cotton effect from this conformer to outweigh the negative contribution to the Cotton effect from the axial conformer.  Therefore, we expect a positive ORD and CD curve for R-3-methylcyclohexanone.  Similar analysis leads to the expectation of a negative ORD and CD curve for the S enantiomer.  Indeed, experimentally it is found that the ORD curve for R-3-methylcyclohexanone is positive and negative for S-3-methylcyclohexanone.  The full name for the R enantiomer is R-(+)-3-methylcyclohexanone

image source

More detail on the 3-cyclohexanone example, as well a nice, concise review of ORD can be found here.

Books have been written on this subject so there is much that I have left out.  For those who would like to explore further here are a few phrases that, in addition to the previous link, can help you get started:  Cotton effect, ORD biological applications, and magnetic circular dichroism


Sodium peroxoborate can be prepare using two methods.

Electrolysis of sodium borate with some sodium carbonate
Oxidation of sodium meta borate using $\ce{H2O2}$

Heating borax strongly will give a mixture of sodium meta borate and boric anhydride.

If you can remove boric anhydride from the transparent glass like mixture, you can then add $\ce{H2O2}$

$$\ce{2NaBO2 + 2H2O2 + 6H2O -> Na2[(OH)2B(O - O)2B(OH)2].6H2O}$$


NMR calculations with NWChem quantum chemistry program

Those calculations are tricky because some functionals work well, others don't, some basis sets work well while even a better basis might give bad results and so on.

We did some studies in the past using the GIAO Method in G09 with B3LYP and 6-311+G(2d,p) als basis for the NMR calculations and M06-2X/6-311++G(d,p)for geometry optimization. And that worked quite well. However, we always used solvated systems (PCM) which seems to be the common way to do it. Try to reoptimize in solvent and run the NMR calculation in solvent if that's possible in NWChem.


Why are the t2 orbitals above the e orbitals in a tetrahedral complex using the crystal field theory?

Take into account that $\mathrm{e_g}$ and $\mathrm{t_{2g}}$ (or $\mathrm{e}$ and $\mathrm{t_2}$) are symmetry terms, and they make sense in the context of a certain symmetry. In other words, they do not describe something that is implicit to the orbitals, but, rather, how are orbitals affected by an environment with a certain symmetry.

In crystal field theory, we look only at the electrostatic effect of a certain distribution of charges around the nucleus on the orbitals; the discussion is primarily qualitative and geometric in nature.

In an environment with spheric symmetry (such as an isolated atom), all 5 d-orbitals are degenerate; since electrostatics are the same in any radial direction, and the set of 5 d-orbitals are only distinguished by their angular part (how are they distributed in the solid angle around the nucleus), they are equivalent. [Incidentally, this would be true not only for an isolated atom but also for an atom surrounded by a spherically symmetric charge distribution, such as a uniform charge distribution a certain distance away from the nucleus; however, I can't think of a real system that would have exact spherical symmetry.]

As soon as you introduce point charges around the nucleus, however, symmetry will break down. All 5 d-orbitals will no longer interact equally with the point charge; some orbitals will be closer to the point charge, some will be further away; and, most importantly from the point of view of symmetry, perhaps several orbitals will interact equally with the point charges.

This may be easier to visualize with the three $\mathrm p_x$, $\mathrm p_y$, $\mathrm p_z$ orbitals: if there are two point charges along the $z$ axis, it is clear that the $\mathrm p_z$ will interact more strongly with them than the $\mathrm p_x$ and $\mathrm p_y$ orbitals, but, crucially, $\mathrm p_x$ and $\mathrm p_y$ will interact exactly the same way with the charges; their interactions with the point charges are symmetry-equivalent.

In the extreme case where the distribution of point charges around the atom has no symmetry, all five d-orbitals would in general have different energies, because their interactions with the charges would be different; but if our distribution of charges is symmetric (as happens in both the octahedral and the tetrahedral case), we need to look at how the orbitals behave under the symmetry.

If you do that (and Jan's excellent answer explains what happens with the Cartesian axes under both symmetries), you will see that in the octahedral case two orbitals behave the same way (and are destabilised to the same extent), while another set of three orbitals behave equally (and are stabilised to the same extent). Looking further at their symmetry properties, these sets of two and three orbitals behave in the ways we describe with $\mathrm{e_g}$ and $\mathrm{t_{2g}}$, respectively. In the tetrahedral case, per contrast, we find that two orbitals are equally stabilised while three orbitals are equally destabilised; the symmetry label that fits their behaviour is $\mathrm e$ and $\mathrm t_2$. [In both cases, $\mathrm e$/$\mathrm{e_g}$ is a symmetry label that corresponds to two elements, and $\mathrm t_2$/$\mathrm{t_{2g}}$ corresponds to three.] These "blocs" of orbitals are degenerate (i.e. these two or three orbitals all have the same energy because they interact similarly with the point charges in the environment), while the energy of these blocs is not the same - that's what we call "splitting".

Note that the temptation to think that there are two "$\mathrm e$" d-orbitals and three "$\mathrm t_2$" d-orbitals in an atom is to be avoided: there is nothing intrinsic in the orbitals that produces this particular splitting into two blocs; rather, it's a product of the symmetry of the environment. The same 5 orbitals, if put in a square planar environment, do not form blocs of 2 and 3 orbitals, but 2, 1, 1 and 1; in a trigonal bipyramidal environment, they split into 2, 2, and 1; and so on.

So it just happens that two orbitals are stabilised in a tetrahedral environment (and thus are labelled $\mathrm e$) and three are destabilised (thus labelled $\mathrm t_2$), while it's the opposite in an octahedral environment.

[As a thought exercise: in crystal field theory, we assume that we are dealing with a central cation surrounded by anions, because that's the problem for which CFT was developed. But, what would happen if the point charges around the atoms were positive instead of negative? You'll find that you end up with stabilised $\mathrm{e_g}$ (octahedral) and $\mathrm t_2$ (tetrahedral) and destabilised $\mathrm{t_{2g}}$ (octahedral) and $\mathrm e$ (tetrahedral), exactly the opposite splitting but the same symmetry behaviour.]

There are rigorous mathematical ways to deduce the symmetry labels of d-orbitals (or any orbital, for that matter) under an environment of a certain symmetry, as provided by point group theory, which involve symmetry tables and so on. Most likely, your textbook will include a discussion of the derivation for at least the octahedral case.


A dehydrating agent acting as an hydrating agent?

Reaction with water is not always a hydration reaction.

A hydration reaction is one in which all of the atoms of the water molecule get added to the reactant to form a product:

$$\ce{MX + H2O -> MH2XO}$$

$$\ce{C2H4 + H2O}\overset{\ce{H+}}{\longrightarrow}\ce{C2H6O}$$

$$\ce{P4O10 + 6H2O -> 4H3PO4}$$

The reaction of water with phosphorous pentachloride is more properly a substitution or partial metathesis reaction. The hydrogen atoms in water are not incorporated into the product:

$$\ce{PCl5 + H2O -> POCl3 + 2HCl}$$

This reaction is difficult to control because $\ce{POCl3}$ is reactive with water also,
and under the right conditions can yield phosphoric acid or even $\ce{P4O10}$.

$$\ce{POCl3 + 3H2O -> H3PO4 + 3HCl}$$
$$\ce{4POCl3 + 6H2O -> P4O10 + 12HCl}$$

The reaction of $\ce{P4O10}$ and $\ce{PCl5}$ is thus also a partial metathesis. It is more obvious when you balance the equation. The oxygen and chlorine atoms are rearranging between the phosphorous atoms.

$$\ce{P4O10 + 6PCl5 -> 10POCl3}$$


CH4 + O2 => CO2 + 2H2, what's wrong with it?

The reaction as you state it is correct only if there will react only one molecule of oxygen. But the reaction describes burning of methane which is supposed to be in the presence of excess of oxygen. Then not only methane is burnt, but also the arised hydrogen.

So in "first" step:

$\ce{CH_4 + O_2 -> CO_2 + 2H_2}$

but then the hydrogen will be also combusted in reaction :

$\ce{2H_2 + O_2 -> 2H_2O}$

so the overall reaction will be:

$\ce{CH_4 + 2O_2 -> CO_2 + 2H_2O}$

As was pointed out in the others answers the real mechanism of a methane combustion at a low oxygen pressure is more complicated:

49th AIAA Aerospace Sciences Meeting including the New Horizons Forum and Aerospace Exposition, 4-7 January 2011, Orlando, Florida, http://enu.kz/repository/2011/AIAA-2011-94.pdf

another:

Intermediate Species Proﬁles in Low-Pressure Methane/Oxygen Flames Inhibited by 2-H Heptaﬂuoropropane: Comparison of Experimental Data with Kinetic Modeling, http://www.nist.gov/el/fire_research/upload/Williams-Intermediate-Species-profiles-in-Low-pressure-Methaneoxygen-Flames-inhibited-by-2-heptafluoropropane.pdf


Does Intercalation and Deintercalation of Lithium ions occurs in Half Cell?

When you're using a lithium metal anode there is no intercalation really, just reduction of the lithium ions to lithium metal:

$$\ce{Li^+ + e^- -> Li^0_{(s)}} $$

The cathode reaction is still the same.


The ACS Style Guide suggests using $\ce{L}$ [1, p. 186] in APPENDIX 10-2 Abbreviations, Acronyms, and Symbols:

  \begin{array}{ll}
[\ldots] & \\
\small{L} & \text{absolute conﬁguration} \\
\mathrm{L} & \text{leucine} \\
 & \text{ligand} \\
 & \text{liter} \\
\mathrm{L_I} & \text {spectral line} \\
[\ldots] &
\end{array}

IUPAC "Green Book" Quantities, units, and symbols in physical chemistry also defines $\ce{L}$ [2, p. 160] in chapter 9 ABBREVIATIONS AND ACRONYMS:

  \begin{array}{ll}
[\ldots] & \\
\mathrm{KS} & \text{Kohn-Sham} \\ \\
\mathrm{L} & \text{ligand} \\
\mathrm{L2TOFMS} & \text {laser desorption laser photoionization time-of-flight mass spectroscopy} \\
[\ldots] &
\end{array}

Glossary of Class Names of Organic Compounds and Reactivity Intermediates Based on Structure (IUPAC Recommendations 1995) uses $\ce{L}$ [3, p. 1349]. For example, see this entry:

  metal-carbyne complexes:
  Metal complexes of the type $\ce{RCML_n}$, ($\ce{M}$ = metal, $\ce{L}$ = ligand) in which formally a carbyne is coordinated to a metal. E.g.

$\mathrm{Lig}$ and other notations are not mentioned in the above sources as suitable ligand notations.

Bibliography

The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. DOI 10.1021/bk-2006-STYG.
IUPAC “Green Book” Quantities, units, and symbols in physical chemistry, 3rd ed.; Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Moss, G. P.; Smith, P. A. S.; Tavernier, D. Pure and Applied Chemistry 1995, 67 (8–9), 1307–1375 DOI: 10.1351/pac199567081307.


Why is the dissociation reaction of acetic acid in water initially endothermic and then exothermic?

I have found a paper which does this same experiment and provides a discussion of this phenomenon[1]. I will begin by just quoting from the discussion section:

  It has been stated in the introductory section that a
  maximum in the formal Cp° for a pair of thermally
  stable free ions can be ascribed to an endothermic
  change in the solvent shell around one of the ions.
  However, the physical description of that change can
  cover a wide range of phenomena. At one extreme,
  the change might be a genuine chemical reaction: for
  instance, an octahydrate becomes a hexahydrate. At
  the other extreme, it might be merely an artifact resulting from a peculiar energy level diagram for a single
  species.

Intuitively, one would expect the reaction to stay either endothermic or exothermic with only small changes due to the relative thermal populations of various vibrational modes in the reactants and products. This picture is more relevant to the gas phase. Thus, the authors are stating that this kind of large change in the behavior of a reaction must be related to the environment which each ion experiences, namely, the solvation environment.

The authors go on to give diagrams for each of the cases they describe above. They stress that in the case where a genuine chemical change occurs, you are essentially moving from populating one set of states to another set of states entirely.

On the other hand, and I suspect this is more commonly the case when water is the solvent, the density of states might increase quite markedly as energy increases. That is the solvent may begin exciting librational modes quite strongly or even move to a regime where the solvents can more or less freely rotate. This behavior of the solvent would be very sensitive to temperature. The authors are very careful to note that each of these situations can result in the same thermodynamic outcome, so one can determine which is happening on the basis of thermodynamic data alone. Nowadays, we could easily run a molecular dynamics simulation or carry out some kind of 2-dimensional IR experiment and extract meaningful kinetic information to supplement the thermodynamics.

So, to summarize, this large change in enthalpy which crosses from exothermic to endothermic is almost certainly due to large changes in the solvation environment and not large changes with respect to the reactants and products.

References:

[1]: Leung, C. S., & Grunwald, E. (1970). Temperature dependence of. DELTA.-Cp. deg. for the self-ionization of water and for the acid dissociation of acetic acid and benzoic acid in water. The Journal of Physical Chemistry, 74(4), 687-696.


Does solubility of acetic acid depend on the pH?

You are correct in both cases, though your reasoning isn't quite on target.

Regardless of the pH, retinol will not be protonated or deprotonated to any significant degree (which would greatly affect it's solubility if it were). It's just a nonpolar molecule that will stay that way regardless of pH.

Acetic acid also would not be protonated to any significant degree at low pH, but that's a mute point as it's fully miscible in water anyway.

The one situation in which you would see deprotonation of the compound is that of acetic acid at high pH. For the most part, both acetic acid and the acetate ion produced by the basic conditions should remain highly soluble. It is possible however, that if you added enough strong base, say NaOH, to a concentrated solution of acetic acid, you could eventually precipitate sodium acetate.

Summary, TL;DR:
You are correct that the answer to both questions is no. Acetic acid is miscible with water (so it's solubility definitely can't be increased) and it's solubility simply won't be affected by pH. If retinol could be protonated or deprotonated, that would increase it's solubility, but it can't be protonated or deprotonated to any significant degree so it is insoluble in water, regardless of pH.


Why isn't aluminium involved in biological processes?

One argument put forward has been that aluminum is very poorly bioavailable, moreso than many other elements.  Aluminum oxide is very insoluble in water.  In addition, any dissolved aluminum that does form in seawater is likely to be precipitated by silicic acid, forming hydroxyaluminosilicates.

From Chris Exeter's 2009 article in Trends in Biochemical Sciences:

  But how has the by far most abundant metal in the Earth's crust remained hidden from biochemical evolution? There are powerful arguments, many of which influenced Darwin's own thinking [15], which identify natural selection as acting upon geochemistry as it acts upon biochemistry. I have argued previously that the lithospheric cycling of aluminium, from the rain-fuelled dissolution of mountains through to the subduction of sedimentary aluminium and its re-emergence in mountain building, depends upon the ‘natural selection’ of increasingly insoluble mineral phases of the metal [7]. The success of this abiotic cycle is reflected in the observation that less than 0.001% of cycled aluminium enters and passes through the biotic cycle. In addition, only an insignificant fraction of the aluminium entering the biotic cycle, living things, is biologically reactive. However, my own understanding of such an explanation of how life on Earth evolved in the absence of biologically available aluminium was arrived at by a somewhat serendipitous route! In studying the acute toxicity of aluminium in Atlantic salmon I discovered that the aqueous form of silicon, silicic acid, protected against the toxicity of aluminium [16]. Subsequent work showed that protection was afforded through the formation of hydroxyaluminosilicates (HAS) [17] which, intriguingly, are one of the sparingly soluble secondary mineral phases of the abiotic cycling of aluminium! The discovery that silicic acid was a geochemical control of the biological availability of aluminium, though now seemingly obvious in hindsight, was a seminal moment in my understanding of the bioinorganic chemistry of aluminium, and although it helped me to understand the non-selection of aluminium in biochemical evolution, it also provided me with a missing link in the wider understanding of the biological essentiality of silicon.

Dr. Exeter is one of the few scholars who appears to have written in depth about this issue.  Thus, perhaps it is fair to say that (a) your question doesn't have a definitive answer, but (b) the poorly accessible nature of aluminum over geological time due to its interaction with and precipitation by silicic acid is the leading hypothesis.

It's worth noting that when aluminum is artificially introduced into metalloenzymes in place of naturally occuring metals, the resulting alumino-enzymes can retain activity, as a 1999 article in JACS by Merkx & Averill shows.


Calculation of Degree of Dissociation from Ostwald's Dilution Law

The OP asked a simple question to which I gave a complex answer. My daughter called such answers "Dad answers".

The OP essentially asked "If the dilute solution approximation for $\alpha$ is used when calculating, how do you know how good the approximation is?"

In his answer user porphyrin focused on the pertinent idea.

Dilute Solution Approximation

Now if we assume that $(1-\alpha) \approx 1$ meaning that very little of the compound dissociated, then we get the standard approximation, $\alpha'$:

$$  \alpha^{'} \approx \sqrt{\dfrac{K_d}{c_{0}}}\tag{1}$$

Exact Solution

The equation below is an exact solution even though it doesn't allow $\alpha$ to be calculated directly. You would have to guess at $\alpha$ values and iterate until you found a solution.

$$\dfrac{\alpha}{\sqrt{1 - \alpha}} = \sqrt{\dfrac{K_d}{c_{0}}}\tag{2}$$

Estimated % Error

Now let's calculate an estimated % error as:

$$\pu{Est \% Error} = 100 \times \dfrac{\dfrac{\alpha'}{\sqrt{1 - \alpha'}} - \alpha'}{\alpha'}\tag{3}$$

Quadratic Solution

The exact value of $\alpha$ can be calculated directly using a quadratic equation. The solution is given below.

$$\alpha = - \dfrac{K_d}{2\cdot c_0} + \sqrt{\left(\dfrac{K_d}{2\cdot c_0}\right)^2 + \dfrac{K_d}{c_0}}\tag{4}$$

Data Table

\begin{array}{|c|c|c|c|c|} \hline
(\alpha^{'})^2=K_d/c_0 & \alpha (quadratic) & approx & true\ \% \ error & est.\ \% \ error \\ \hline
1.00000000 & 0.61803399 & 1.00000000 & 61.80339887 &  \\ \hline
0.81000000 & 0.58192705 & 0.90000000 & 54.65856100 & 216.22776602 \\ \hline
0.64000000 & 0.54162637 & 0.80000000 & 47.70329614 & 123.60679775 \\ \hline
0.49000000 & 0.49663670 & 0.70000000 & 40.94810050 & 82.57418584 \\ \hline
0.36000000 & 0.44641839 & 0.60000000 & 34.40306509 & 58.11388301 \\ \hline
0.25000000 & 0.39038820 & 0.50000000 & 28.07764064 & 41.42135624 \\ \hline
0.16000000 & 0.32792156 & 0.40000000 & 21.98039027 & 29.09944487 \\ \hline
0.09000000 & 0.25835623 & 0.30000000 & 16.11874208 & 19.52286093 \\ \hline
0.04000000 & 0.18099751 & 0.20000000 & 10.49875621 & 11.80339887 \\ \hline
0.01000000 & 0.09512492 & 0.10000000 & 5.12492197 & 5.40925534 \\ \hline
0.00810000 & 0.08604108 & 0.09000000 & 4.60119879 & 4.82848367 \\ \hline
0.00640000 & 0.07686397 & 0.08000000 & 4.07996803 & 4.25720703 \\ \hline
0.00490000 & 0.06759286 & 0.07000000 & 3.56123125 & 3.69516947 \\ \hline
0.00360000 & 0.05822699 & 0.06000000 & 3.04498988 & 3.14212463 \\ \hline
0.00250000 & 0.04876562 & 0.05000000 & 2.53124512 & 2.59783521 \\ \hline
0.00160000 & 0.03920800 & 0.04000000 & 2.01999800 & 2.06207262 \\ \hline
0.00090000 & 0.02955337 & 0.03000000 & 1.51124937 & 1.53461651 \\ \hline
0.00040000 & 0.01980100 & 0.02000000 & 1.00499988 & 1.01525446 \\ \hline
0.00010000 & 0.00995012 & 0.01000000 & 0.50124999 & 0.50378153 \\ \hline
0.000001 & 0.0009995 & 0.00100000 & 0.05001250 & 0.05003753 \\ \hline
0.00000001 & 9.9995E-05 & 0.00010000 & 0.00500012 & 0.00500038 \\ \hline
1E-10 & 9.99995E-06 & 0.00001000 & 0.00050000 & 0.00050000 \\ \hline
\end{array}

So the Est % Error from equation (3) works quite well.

When the Est % Error is 5% or less, $\alpha' \le 0.1$, the Est % Error is quite good.
So for the approximate equation when $\alpha^{'} = 0.1$ you get 1 significant figure, 0.01 gets two significant figures and so on. The improvement doesn't continue forever of course - this is chemistry not math. To get three significant figures or more you'd have to be a fastidious experimenter.
For the approximate equation when $\alpha' = 0.7$ the estimated error is 82.6% which is much too large, but an 82.6% error still tells you that the  approximation $\alpha^{'}$ is no good.
To make user M. Farooq's point in another way, by the time you calculate the approximation from equation (1) and the Est % Error from equation (3), you've pretty much done enough work to calculate the exact answer using equation (4). So why bother with the approximation?

EDIT A comment by user Buck Thorn "I never thought about the usefulness of the estimate to bracket the true value" made me think.

It turns out that we can bracket the true value $\alpha$ using a simple calculation of the approximate value $\alpha'$:

$$\alpha'\cdot(1-\alpha') \lt \alpha < \alpha'\tag{5}$$

\begin{array}{|c|c|c|c|} \hline
(\alpha^{'})^2=K_d/c_0 & \alpha (quadratic) & \alpha'\cdot(1-\alpha') & \alpha' \\ \hline
1.0000000000 & 0.6180339887 & 0.0000000000 & 1.0000000000 \\ \hline
0.8100000000 & 0.5819270490 & 0.0900000000 & 0.9000000000 \\ \hline
0.6400000000 & 0.5416263691 & 0.1600000000 & 0.8000000000 \\ \hline
0.4900000000 & 0.4966367035 & 0.2100000000 & 0.7000000000 \\ \hline
0.3600000000 & 0.4464183905 & 0.2400000000 & 0.6000000000 \\ \hline
0.2500000000 & 0.3903882032 & 0.2500000000 & 0.5000000000 \\ \hline
0.1600000000 & 0.3279215611 & 0.2400000000 & 0.4000000000 \\ \hline
0.0900000000 & 0.2583562262 & 0.2100000000 & 0.3000000000 \\ \hline
0.0400000000 & 0.1809975124 & 0.1600000000 & 0.2000000000 \\ \hline
0.0100000000 & 0.0951249220 & 0.0900000000 & 0.1000000000 \\ \hline
0.0081000000 & 0.0860410789 & 0.0819000000 & 0.0900000000 \\ \hline
0.0064000000 & 0.0768639744 & 0.0736000000 & 0.0800000000 \\ \hline
0.0049000000 & 0.0675928619 & 0.0651000000 & 0.0700000000 \\ \hline
0.0036000000 & 0.0582269939 & 0.0564000000 & 0.0600000000 \\ \hline
0.0025000000 & 0.0487656226 & 0.0475000000 & 0.0500000000 \\ \hline
0.0016000000 & 0.0392079992 & 0.0384000000 & 0.0400000000 \\ \hline
0.0009000000 & 0.0295533748 & 0.0291000000 & 0.0300000000 \\ \hline
0.0004000000 & 0.0198010000 & 0.0196000000 & 0.0200000000 \\ \hline
1.000000E-04 & 0.0099501250 & 0.0099000000 & 0.0100000000 \\ \hline
1.000000E-06 & 0.0009995001 & 0.0009990000 & 0.0010000000 \\ \hline
1.000000E-08 & 0.0000999950 & 0.0000999900 & 0.0001000000 \\ \hline
1.000000E-10 & 9.999950E-06 & 9.999900E-06 & 0.0000100000 \\ \hline
\end{array}


Is there an easy way to detect methanol in spirits (which may have been adulterated)?

Well that depens on your possibilities.

a) Destillation / Boiling Point (easy, unsafe)

b) Refractive index (depends on impurities)

c) GC (best, very expensive)

d) 1H NMR


Complexes formed by lanthanides and actinides

After a lot of digging around, I finally found something about this in J.D. Lee's Concise Inorganic Chemistry. I thought I'd post whatever I found over here anyway since it's very interesting:

Answer to question 2

Complex formation by lanthanides is different from that of actinides. In lanthanides, the $\mathrm{4f}$ orbitals are well shielded by the larger $\mathrm{5d}$ and $\mathrm{6s}$ orbitals and are deep inside the atom. So $\mathrm{f}$ orbitals do not participate in any bonding and complex formation is similar to that of transition metals. However in the actinides, the $\mathrm{5f}$ orbitals extend outwards and participate in bonding much more easily making the interactions much more complex than in the transition metals.

Answer to question 3

Higher coordination numbers are apparently extremely common among f-block elements (strange right?).
Octahedral (6) and tetrahedral (4) structures are very rare except when bulky ligands are present. Most common coordination numbers among lanthanides are

7 (Capped trigonal prismatic) yttrium acetylacetonate hydrate (Yttrium seems to be grouped with the f-block elements due to similar properties)
8 (Square antiprismatic and Dodecahedral) cerium acetylacetonate and holmium tropolonate
9 (Tri-capped trigonal prismatic) nonaaquaneodymium(III) complex $\ce{[Nd(H2O)9]^3+}$

10 (very complex) and 12 (icosahedral) are seen only in the larger elements like cerium and thorium with small ligands like $\ce{NO3-}$ and $\ce{SO4^2-}$

The actinides also commonly form exotic structures like chained tricapped trigonal prismatic in $\ce{[ThF8]^4-}$ and $\ce{[PaF7]^3-}$ distorted cubic in $\ce{[PaF8]^3-}$

The text mentions that the nature of bonding in $\ce{[Ce(NO3)6]^2-}$ is still not understood because it would imply bond orders of less than 1 or participation of f orbitals.

Could anyone explain anything about the first question since there is nothing mentioned on the stability of these compounds.


What is a “spectator mode” in this analysis of molecular vibration modes

The paper in question is Ab Initio Trajectory Surface-Hopping Study on Ultrafast Deactivation Process of Thiophene. From a quick Google Scholar search that returns 187 hits, it is clear that the expression “spectator mode” for a harmonic vibration mode is not a standard terminology, but is nonetheless used by plenty of authors.

As to the meaning, I think you got it from the everyday meaning of “spectator”: they are vibration modes that do not participate in the phenomenon of interest, i.e. that are not coupled to the chemical or physical process studied (which can be a chemical reaction, an adsorption process or a photophysical process).


What happens in the reaction of ammonium dichromate and mercury(II) thiocyanate?

This is a combination of two reactions.  The first is the ammonium dichromate vocano and the second is called Pharaoh's snakes.

The volcano

The reaction occurring here is: $$\ce{(NH4)2Cr2O7(s) -> Cr2O3(s) + N2(g) + 4 H2O(g)}$$  The reference above provides several pictures and a description of the chemistry.

Pharaoh's snakes

Again, the reference above describes what is happening, so only a summary of the chemical reactions is given here:
$$\ce{2 Hg(SCN)2->2 HgS + CS2 + C3N4}$$ $$\ce{CS2 + 3O2 -> CO2 + 2SO2}$$ $$\ce{2C3N4->3(CN)2 + N2}$$ $$\ce{HgS + O2->Hg + SO2}$$

Pharaoh's snakes is a rather hazardous demonstration and should not be performed without the proper ventilation, protective gear and waste disposal equipment.


Will Raman photon counts/Spectral resolution increase if laser beam becomes parallel?

Spectral resolution is not dependent on the laser intensity. It is determined by the state of the sample ( gas, liquid or solid) and more importantly by the components of the instrument itself. For example, what is the slit width anf the number of lines on the grating? What type of detector is there? If it is charged coupled device (CCD) then the spectral resolution will also be dependent on it.


Why is the reaction between hydroxide and water reversible, when the reaction between oxygen ions and water is not?

Correct, all reactions are reversible, just to different extents. Think about the combustion reaction in your vehicle. Is that reaction reversible? Sure, in that nothing is 100% irreversible. The reversing of the combustion just might take a while. But it's not reversible to any significant extent in the short and medium-term; i.e. you're not going to be re-capturing your combusted gasoline and be putting that back together as usable fuel for your vehicle.

Also, about the reaction of oxide ion with water - that reaction goes practically 100% because oxide ion is a very strong base. The strongest base supported in high concentration by a medium is the conjugate base of the medium. (Leveling effect). For water, that would be hydroxide ion (remember that conjugate bases lose a proton). Oxide ion is the conjugate base of hydroxide ion. We would expect oxide ion to be an even stronger base than hydroxide ion because of oxide ion's even higher negative charge density - we're talking about a negative two charge here (versus negative one for hydroxide) and we're missing a hydrogen proton (which only further decreases the size of the oxide ion and further increases its negative charge density). If you think about this Coulombically, then of course oxide ion is going to be a stronger base than hydroxide ion - opposite charges attract (recall that water has two partial positive charges/dipoles stabilized by its hydrogen atoms). F=kq1q2/r^2 and one of the q's here is massive.

Also about how the products are written - it doesn't matter how the products are written. Either way you are creating two HO- (hydroxide ions). Write hydroxide with the negative next to the O (so either -OH or HO- are acceptable) because that's what you're doing - heterolytic clevage of the O-H bond in H-O-H, which (in this case) creates a negative formal charge that is mostly stabilized by the oxygen atom (NOT the hydrogen atom). Plus it's the hydroxide ion not the oxyhydride ion (whatever the hell that is).

On second thought, when writing -OH, make sure the "-" is clearly a superscript, lest people think you're talking about a hydroxyl group.


Is FA 18:2 undifferentiated the same as linoleic acid?

It's part of an old nomenclature used in the fatty acid industry.  FA 18:2 translates to F atty A cid with 18 carbon atoms and 2 double bonds.  You can read more about these "Lipid numbers" here.  So to answer your question, there are number of fatty acids with 18 carbons and 2 double bonds, linoleic acid is just one of them.  FA 18:2 includes linoleic acid, but is not limited to just linoleic acid.


Activated Carbon - Desorption in sunlight?

Your guess is right: the mechanism of charcoal regeneration is undoubtably temperature-induced desorption. Gas molecules are adsorbed on the large surface area of activated carbon by a mechanism of physisorption (or physical sorption) caused by weak nonspecific interactions (van der Waals forces) between gas molecules and the carbon surface. Unlike chemisorption, this process is reversible and, since it involves only weak interactions, it can actually be reversed by relatively small increases of temperature.

Wikipedia gives a specific example:

  The activated carbon can adsorb a large amount of methanol vapours in ambient temperature and desorb it at a higher temperature (around 100 degrees Celsius). During the daytime, the sunshine irradiates the collector, so the collector is heated up and the methanol is desorbed from the activated carbon. In desorption, the liquid methanol adsorbed in the charcoal heats up and vaporizes.

You will surely notice the resemblance between this text and your case.


What makes a ligand stronger than another?

One definite factor is the ability of the ligand to stabilize its lone pair. You observe that the cyanide ion is a better ligand than the nitrogen ligand. Let's think about this. The lone pair that reacts on the cyanide ion is the lone pair on the carbon atom. Not the lone pair on the nitrogen atom.

Carbon is less electronegative than nitrogen. Thus carbon is less able to stabilize its valence electron density, which in turns enhances the reactivity of its valence electron density through increased negative charge density.

Same with sulfide and oxide anions. Sulfur is less electronegative than oxygen, and thus less able to stabilize valence electron density. Hence the much greater stability of silver sulfide as opposed to silver oxide.

Other competing factors include charge density; the Lewis acid the ligand is coordinating to, medium (solvent), etc. For example the oxide anion is more basic than the sulfide anion in water solvent; i.e. the oxide anion forms a stronger bond with the hydrogen proton in water. But we just saw that silver sulfide is much more stable than silver oxide.


Why do gases need to be cooled to liquefy/solidify?

What your book states is not generally true. Two counter-examples:

Ammonia ($\ce{NH3}$) will liquefy at room temperature and a pressure of approximately $\pu{10bar}$. (CRC Handbook of Chemistry and Physics 44th ed; information as cited on Wikipedia’s data page).
Butane ($\ce{C4H10}$) will liquefy at room temperature and a pressure of just over $\pu{2bar}$. (Same source as cited on Wikipedia’s data page).

Especially the second is well-known to all those who have a butane lighter of transparent material. If it is half-full, a clear distinction between the liquid and gaseous phases can be seen inside at room temperature. Naturally, the inside of the lighter must be under positive pressure for gas to flow out if a flame is required.

On the other hand, there are a great many gases that cannot be liquefied at room temperature by applying pressure. One of these is nitrogen ($\ce{N2}$). In this case, applying enough pressure will make the sample transform directly from the gaseous to a solid phase.

The difference between nitrogen and ammonia/butane is their thermodynamic data, most importantly their critical point. The critical point defines the last position of the liquid/solid separation of a phase diagram. If the temperature is higher than the critical temperature, no liquid phase will be observed under any compression. Likewise, if the pressure is higher than the critical pressure, a substance will not liquefy even under extreme cooling (it will go straight to the solid phase).

The critical points of the compounds mentioned herein are:

$$\begin{array}{lrr}\hline
\text{Compound} & \vartheta_\text{crit} [\mathrm{^\circ C}] & p_\text{crit} [bar] \\
\hline
\ce{NH3} & 132.4 & 112.8\phantom{0} \\
\ce{C4H10} & 152\phantom{.0} & 37.96\\
\ce{N2} & -146.9 & 33.90\\ \hline \end{array}$$

As you can see, the critical pressures of all compounds are well above standard pressure (hence they can be liquefied under reduced temperature). Nitrogen’s critical temperature is below room temperature, so it cannot be liquefied by compression; the others can.


Why can Wurtz reaction be not used for preparation of alkanes with odd number of carbon atoms?

It's always good to think of what would probably go on in the reaction vessel. Here's a mechanism of the Wurtz reaction:

First, one electron from the metal is transferred to the halogen to produce a metal halide and an alkyl radical.

$\ce{R–X + M → R. + M+X−}$

The alkyl radical then accepts an electron from another metal atom to form an alkyl anion and the metal becomes cationic. This intermediate has been isolated in several cases.

$\ce{R. + M → R^-M+}$

The last step is a $\ce{S_N2}$ reaction; the nucleophilic carbon of the alkyl anion displaces the halide, forming a new carbon-carbon covalent bond

$\ce{R−M+ + R–X → R–R + M+X−}$

Simple. Now, let's take a look at an example.

If we start with a single alkyl halide, in this case a methyl halide,
$\ce{2 CH3-X + Na ->[][-2NaX] CH3-CH3 }$

Now, let's consider two dissimilar alkyl halides, ethyl and methyl halide:
3 possible reactions take place, giving a statistical distribution of products.

$\ce{2CH3-X + Na ->[][-2NaX] X-CH3}$

$\ce{2CH3-CH2-X + Na ->[][-2NaX] CH3-CH2-CH2-CH3}$

$\ce{CH3-X + X-CH2-CH3 + Na ->[][-2NaX] CH3-CH2-CH3}$

Sure, you've produced an alkane with an odd number of carbon atoms, but at the same time you have a mixture of alkanes (this is not really a selective reaction), that you will have a hard time separating.


Reducing sugar among the given compounds

The English wikipedia page on reducing sugars starts with

  A reducing sugar is any sugar that is capable of acting as a reducing agent because it has a free aldehyde group or a free ketone group. All monosaccharides are reducing sugars [...]

Unfortunately, the monosaccharides tend to hide the aldehyde and keto groups by undergoing intramolecular reactions between a carbonyl group and a hydroxy group. This reaction leads to a cyclic hemiacetal.

In order to figure out which of the structures in your question is a monosaccharide in disguise, we need to spot the hemiacetal moiety. With other words:

Which of the structures can open again to show a carbonyl group?

We can definitely rule out (C)! This is a hydroxymethyl-substituted cyclohexane with some hydroxy and a methoxy group.

What about (D)? Close, but no cigar. This is an acetal, which won't open that easy.

I'll leave the last two examples to you ;-)


What defines the specific enthalpy of atomization?

ChemGuide says that 'The standard atomisation enthalpy is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state.'

For example,
\begin{array}{rc}
\ce{1/2 Cl2 (g) -> Cl (g)} & \Delta H_{a}^{0} = +122~\mathrm{kJ/mol} \\
\ce{1/2 Br2 (l) -> Br (g)} & \Delta H_{a}^{0} = +112~\mathrm{kJ/mol} \\
\ce{Na (s) -> Na (g)} & \Delta H_{a}^{0} = +107~\mathrm{kJ/mol}
\end{array}


Catalyst and Inhibitors home experiments

You might look up the decomposition of hydrogen peroxide, $\ce{H2O2}$. The 3% solution sold as antiseptic should be safe to use at home, as would be some readily available catalysts and inhibitors.
Horseradish might be used for a catalyst... but don't eat it in the lab.
To measure speed of reaction, measure gas evolution with an inverted, liquid-filled graduated cylinder or small measuring cup over time.
You can have fun burning things (carefully) in the evolved $\ce{O2}$, too.


Software for predicting chemical reactions

For predicting chemical reaction you should use classical mechanics to take in account the simple interactions between the molecules  and quantum mechanical for all the aspects concerning the molecule itself and complex interactions.
Knowing the principal parameter such as temperature and pressure that could describe the system with the classical mechanics (kinetic energy etc. etc.) is not enough for predicting the reaction with ab initio methods.
Adding the quantum mechanical contribute is mandatory and quite hard to do. A amateur chemist do not have neither the skills nor the computation power require to make such calculations. For the moment you can't do that with any software but the last Nobel was won by three theoretical chemists who are dealing with this topic. Have a look here for more information.

In your case where the reactions are well known you can use a chemical reactions database like the NIST one.
This is a simple query:

And this is the...


How to derive the Gibbs-Helmholtz equation?

You need to know that

$$\require{begingroup} \begingroup \newcommand{\d}[0]{\mathrm{d}} \d G = V\,\d p - S\,\d T$$

from which you can determine that

$$\newcommand{\pdiff}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}} \pdiff{G}{p}{T} = V \qquad \pdiff{G}{T}{p} = -S$$

Therefore (by the quotient rule)

$$\begin{align}\pdiff{}{T}{p}\left(\frac{G}{T}\right) &= \frac{T(\partial G/\partial T)_p - G(\partial T/\partial T)_p}
{T^2} \\[8pt]
&= \frac{T(-S) - G(1)}{T^2} \\[8pt]
&= \frac{-TS-G}{T^2} \\[8pt]
&= -\frac{H}{T^2}
\end{align}$$

as desired (since $G = H - TS$).


Any classification like chemical or a physical change, homogeneous or heterogeneous substance, will eventually fail at one point. From an analytical or physical chemistry's perspective, homogeneous substance is a substance, in a single phase, whose chemical composition is uniform. A heterogeneous substance has different composition from place to place like your garden soil.

Why do we treat a water+water vapour system as heterogeneous in spite of similar molecules present? Also if there exists an equilibrium in the two phases would we still not call them homogeneous?

It is a bad example to label water+ water vapor system as heterogeneous- but with respect to what? It is better to call it a two phase system. You have to define heterogeneous with respect to some property.
For example, if we take tap water in a glass, we can say the liquid (water) is homogeneous with respect to its chemical composition. Let us put some ice cubes in water, now it is a heterogeneous phase, although chemically it is all water.
More formally, from the International Union of Pure and Applied Chemistry's Gold Book:

homogeneity (in analytical chemistry): The degree to which a property
or a constituent is uniformly distributed throughout a quantity of
material. A material may be homogeneous with respect to one analyte or
property but heterogeneous with respect to another.

You can see the subtilty above in italics.


Regarding reversible phase transition of a real gas

$$\Delta S_\mathrm{sys} =\frac{\Delta H_\mathrm{vap}}{T_b}$$
is the entropy change of the system.
$$\frac {\Delta G}{T_\mathrm{b}} = \Delta S_\mathrm{tot} = \Delta S_\mathrm{ext} + \Delta S_\mathrm{sys} = \Delta S_\mathrm{ext} + \frac{\Delta H_\mathrm{vap}}{T_b} = 0$$
is proportional to the total entropy change.


When precessing nuclei absorb electromagnetic radiations, am I supposed to consider the wave nature or particle nature of light?

The nuclear spin states in an atom are excited by the radio frequency radiation and one spin state moves to the other when a photon is absorbed.

The precession part is not so much about individual spins but the magnetisation of the sample which is the vectorial addition of all the nuclear magnetic dipoles; precession is motion about the permanent magnetic field. (Think of the magnetisation rather like a vertical rotating pendulum if you want.) As circular polarised rf is used the spins are excited coherently and all add up 'in phase', hence a large magnetisation is produced that then undergoes precession. Loosing this phase in the magnetisation corresponds the the T2 decay time and loosing population excess above equilibrium to the T1 time. It happens that the precession frequency is the same as the transition frequency.

(An individual spin angular momentum is a complicated nutation type motion as it experiences a force from the rf field which is at right-angles to he permanent magnetic field.)


How does charge density affect the rate of a decomposition?

Increase in charge density increases the polarizing capacity of the atom, decreasing the bond strength, thus facilitating easy decomposition


When the interior gunpowder charge explodes, it blows the stars in all directions and ignites them.  They then burn very rapidly (perhaps explosively) giving the star effect exactly because the stars are a mixture of fuel and oxidizer.  If the stars were all fuel (carbon), they would be essentially charcoal briquettes.  In that case the center would explode, throwing (possibly) burning charcoal chunks in all directions.  This could be a truly novel approach to barbecue...


There are all sorts of resources for picking the best gloves for handling any given type of chemical. This resource describes latex as being good for protecting against bases, alcohols, dilute water solutions, and fair against aldehydes, ketones.

Indeed if you check the Glove Type and Chemical Use table towards the bottom of the page, you can see that latex fairs pretty well against most acids. These ratings give you an idea of how long it takes to penetrate the glove when exposed to that chemical. As you can also see from this table, neoprene gloves are slightly better for handling acids, and though it's not shown on the table, PVC gloves are also good.


Are there ways to fully dissociate a complex salt?

Let's consider the formation constant of this ion complex at $25^\circ\mathrm{C}$:
$$\ce{Fe^{2+} + 6CN- <=> [Fe(CN)6]^4-}\qquad K_\mathrm{f}=10^{+24}$$
This means that, the equilibrium is highly displaced towards the formation of the  complex ion. The ion complex is very stable.

So, if we want to displace this equilibrium towards the dissociation of the complex ion, we have to add an adequate quantities of  a metallic ion or a ligand that form with $\ce{CN^{-}}$ or $\ce{Fe^{2+}}$, respectively more stable complexes.


The only way you can know for sure is to analyze the reaction using oxidation numbers, and figuring out which elements lose/gain electrons. Here is a link explaining how to assign oxidation numbers in a reaction.

As an example, consider the reaction, what is the oxidation number of sulfur in this particular molecule, $S_2O_3$. According to these rules, which are in order of priority, the oxidation number of oxygen is $-2$, unless it is bonded with hydrogen or fluorine, which in this case, it is not. Given that we have three oxygen atoms in this molecule, the total oxidation of oxygen is $-6$, which must be offset by two sulfur atoms so the total oxidation number is $0$. Obviously, the oxidation number of a single sulfur must then be $+3$, so that the molecule is neutral.

In any reaction, you would essentially have to do this for every molecule on both the products and reactants side in order to determine which atoms are losing/gaining electrons.


What happens to water under high pressures without possibility of escape?

What happens to water under high pressures without the possibility of escape?
It depends on what you mean for high pressure and on the temperature, however, the water phase diagram can help you to understand what will happen. This is from Wikipedia User Cmglee:

You can see that at high-pressure water assumes a solid form (ice): you will have ice X at 100 GPa, and what is labelled as hexagonal ice XI in the diagram but is actually superionic water or ice XVIII at 1 TPa (temperature range 0 to 650 K).  These sorts of ices have different lattice and internal energies.
What would happen to water as pressures increase towards infinity without the possibility of the water escaping confinement?
Unfortunately, chemists don't deal with infinity, basically, however, there will be a point where the water molecule won't longer exist all the bonds will break.  After that probably you will have something called electron-degenerate matter have a look here, here is all about quantum mechanics, Pauli exclusion principle and black holes so on more Physic S.E. stuff.

Could triple points in water play a role in keeping temperatures low under high pressures via fluid thermodynamics?
You can also note that there are other triple points like (100 k, 62GPa) this however doesn't directly affect the properties of water but are the properties of water that determine where are the triple points. In a close system, kinetic energy is constant so I think that fluid thermodynamics doesn't matter in this case is more related to equilibria.
More info about water phase diagrams here.


Heating of Enamel coated cooking utensils and health hazards

I unfortunately don't have time to dig through these articles in great detail, but they should help get you started. Feel free to comment with more questions.

The overall impression I got from my brief search is that there is definitely a potential health hazard depending on how the coloring was made, but in most modern countries, there are  regulations in place to prevent those kinds of problems. Enough about what I think though, let's get into the references.

According to Wikipedia, the coating on most cookware is vitreous enamel, which is made by heating a mixture of glass powder and pigment to a very high temperature and letting it melt onto a surface (much like glazing in pottery).

The glass itself is pretty much harmless, so the main danger comes from the added pigments. Here, we're stuck, because it really does depend on what was used to make the pigment. If it's a zinc pigment, you're probably fine. If it was cadmium based, you might be kinda screwed.

I've found references to several articles which describe experiments trying to find out what the leach rate of metals from the glass into a solution is (usually by boiling in vinegar), but I can't hunt down the fulltexts right now, aside from the one open-access article from the NIH. Here's what I have:

Full text:Lead, Cadmium and Cobalt (Pb, Cd, and Co) Leaching of Glass-Clay Containers by pH Effect of Food
While this study was on clay containers, it should theoretically apply to glass-on-metal as well. Keep in mind that the containers seem to have been acquired from local craftsmen in Mexico, so the results won't necessarily apply to stuff you find on the supermarket shelves.

Abstract: Hot leaching of ceramic and enameled cookware: collaborative study.
This is what you're looking for, but I can't find the full article, and the authors seem to have been more interested in establishing a method than finding the health effects. If you can find the full paper and cross-reference their results with acceptable limits of exposure (maybe from the EPA), that'd probably be your best bet.

The second link also has other articles on similar problems. If you can't find your information from these two articles, maybe you can find it by following the links on PubMed. Good luck!


You are mostly right. The horizontal axis represents the distance from the nucleus. The maximum of the electron density of 1s orbitals is in fact at the nucleus. This is not impossible, remember that electron density can be interpreted as the probability that an appropriate experiment will detect the electron there. In fact, electron density "at" the nucleus is one of the major ways NMR spectroscopy works.


Resonance condition for absorption of energy

This is indeed the basis of modern Fourier transform NMR (the technique that all modern spectrometers use).

The molecule is irradiated with a range of frequencies and the response detected. Application of some clever maths (the Fourier transform) then allows the individual signals to be observed.

You can think of this a bit like bashing every key on a piano and recording the noise. Application of Fourier transform would then allow you to get back the sound for the individual keys.


Reaction of Iron(III) with hydroxide ion

I wouldn't expect the reactions to yield different products but rather different particle size. The higher concentrated $\ce{NaOH}$ should hereby produce smaller particles†.

The "solution" in the left vial is probably rather a colloid of microscopically dispersed iron(III) hydroxide. It should show the Tyndall effect if you send a beam of light through it. However, the particles will aggregate over time to form larger particles that also precipitate, as seen in the second image.

† Higher concentration means higher supersaturation. Higher supersaturation leads to a higher nucleation rate, so more and thus smaller crystals are being formed.


Best way to remove Aluminium from "silvered" glass?

Heating the bulb in a dilute NaOH solution would probably work, yes. Something that might be more 'interesting', and potentially safer (with less hydrogen) though, is to acquire a bit of gallium.

Gallium infiltration is a really fascinating phenomenon. Simply melt a little bit of gallium and dunk your bulb in it, and watch as the aluminum becomes as soft as wet tissue paper. Gallium also melts at a temperature very slightly above that of room temperature, meaning that the the chance of the glass bulb cracking because of heating would be very minimal.


What is the difference between ignition temperature and flash point?

From Wikipedia:

Autoignition Temperature: the lowest temperature at which a substance spontaneously ignites without a source of ignition (such as a flame or spark).

Flash Point: the lowest temperature at which the vapours of a (volatile) material will ignite, with an ignition source.

Ignition temperature means the same thing as autoignition temperature according to Merriam Webster.


Highest known is about 4.25 (all
densities in g/cm${}^3$) at room temperature:
http://en.wikipedia.org/wiki/Clerici_solution

If you're willing to allow heating it comes close to 5 because the solubility of the salts increases with temperature:
http://books.google.com/books?id=kaa2qeFRXmUC&pg=PA99

My favorite runner-up is the Thoulet solution mentioned at
http://en.wikipedia.org/wiki/Potassium_tetraiodomercurate(II) but it
only reaches around 3.20.

Lowest would depend on what you mean by "water solution". If you mix a
drop of water with a gallon of a very light liquid such as diethyl
ether (which is not very miscible with water but does dissolve some
-- hence the need to dry it in the lab over sodium or molecular
sieves), then the answer is the density of the liquid (ether = 0.73).
If water has to be the majority component I'd expect the
minimum to be very close to the density of water, since mixtures tend to decrease a bit in volume. (The density of water is 1.0 at 4${}^\circ$, close to 1.0 at room temperature, 0.94 near the boiling point).