"Strength". Hmm, what firstly comes to mind is you want a comparison of lattice energies (or more precisely, lattice enthalpies)
Lattice energy is a [crude] way of estimating how strong the electrostatic attractions are between ions in an ionic compound. Let's take the formation of $\ce{NaCl}$ into account, as I see it's the most common "ionic" compound that teachers start teaching the lesson with.
According to Born-Haber cycle this is the reaction that results in the formation of pure table salt:
$$\ce{Na+(g) + Cl- (g) \rightarrow NaCl(s)}$$
Forming the salt crystals is exothermic and this is called the lattice enthalpy of formation.

Note that some textbooks mean to define lattice energy as the energy that is obtained by the thermodynamic system in the process of turning the solid compound to its gaseous ions.

To make a long story short, the more the lattice enthalpy is, the stronger "bonds" you get.
There are two main factors that affect the lattice formation enthalpy:

Charge of the ions
Radius of the ions

Charge of the ions:

Source
Look at the diagram above. $\ce{NaCl}$ is known to release approximately 822 kilojoules upon formation, while $\ce{MgO}$ releases 3890 KJ. And what does that mean? It means that Magnesium Oxide is way more "stable" than Sodium Chloride. So, complete the rest:

The more the charge of the ions is in an ionic compound, the _____ is its lattice enthalpy and the ______ its stability.

Radius of the ions:

Source
As you see, and could've imagined if you were on the right track, bigger ions (literally) mean weaker attraction. I wanted the answer to be fairly complete, so I did an addition to the main point that answers your question. Thus, there's no need of expanding this.
In your case:
You might as well have reached a conclusion. Mg ion has a 2+ charge, while Na has only a single electron missing. The anion is pretty much the same, $\ce{HCO3-}$ (hydrogen carbonate), and its priorly covalent nature shouldn't mislead you in doubting your answer. There is a case of comparison of the charge of the ions, and I would say Magnesium ion wins.
In case of additional study, you can visit:
http://www.chemguide.co.uk/physical/energetics/lattice.html
http://www.knockhardy.org.uk/sci_htm_files/08le.pdf
http://www.rsc.org/learn-chemistry/resources/gridlocks/downloads/answers/Born%20Haber%20NaCl%20ans.pdf

Addendum:
The charge and the radius of ions are by no means the only factors that affect the lattice enthalpy. It's just that the other factors are considered "trivial" by current textbooks.
The lattice enthalpy and lattice energy aren't the same. They do have close meanings, though.


What type of compound/element might be in these glowing tiles

Glow in the dark materials are materials that exhibit phosphorescence. The most common materials used for this effect are zinc sulfide and strontium aluminate, doped with various elements that affect the colour of luminescence.

Phosphorescence is similar to fluorescence, where an electron is excited to a higher energy state, but instead of dropping back immediately by emitting a photon, an intersystem crossing event occurs where the electron's spin flips resulting in a triplet state (S=0 to S=1 in the figure). Because the electron has the same spin as the other electron it left behind in the ground state, the transition to the ground state is forbidden and thus occurs much more slowly. In practice, a given material usually exhibits both fluorescence and phosphorescence. Materials that are very phosphorescent have a high probability of intersystem crossing so that when light is shined on them, a large population builds up in triplet states that will release light slowly. What makes a given material have a high intersystem crossing probability is pretty complicated, but it tends to be higher in molecules with large atoms—even things like fused silica glasses are phosphorescent, but very weakly.

Source


Separating plant extract from ethanol solution

Free evaporation or using a water bath ?
Both ethanol and water evaporate at any temperature, just with the exponential-like rate. ( Washed clothes will dry up even at freezing temperatures ).
If living in area with currently cold weather, consider using evaporation on the central heating radiator, parhaps partially isolated.
Free evaporation could be a safe bet, but slow evaporation can support oxidation of dissolved substances, so I would try both free and heated ways.
I remember that in my young-chemist days, I had bought a water stream vacuum pump, based on the Bernoulli equation.


How long does it take for nitric oxide to turn into dinitrogen and dioxygen?

Theoretically, it will decompose, but the reaction is rather slow, on a human timescale. One experiment ran for almost forty years (J. Phys. Chem. 1958, 62 (3), 360–361), from 1918 to 1957, without noticeable decomposition.

That said, in aqueous solutions it breaks down more rapidly. In vivo, the half-life is a few seconds.

It is amazing to me that a free radical is so stable!


Diels Alder reaction with maleic anhydride. Reactivity of different dienes

a) The compound can rotate into a trans configuration that is not suitable for Diels Alder

b) The compound is locked into a cis configuration that is favorable. Furthermore, a carbon group is slightly electron donating

c) The compound has an electron withdrawing group.

d) What compound is that? It's compounds labelled b) and c)

Comment: Normal demand Diels Alder is enhanced by an electron withdrawing group in the dienophile and an electron donating group in the diene. Look at reaction mechanism tab (http://en.wikipedia.org/wiki/Diels%E2%80%93Alder_reaction)

I am not sure as to why that is. I have forgotten.


How to find partial pressures of gases only given their percentages?

Pressures of gasses are additive. For example, 1 atm X plus 1 atm Y makes 2 atm total pressure. Because all the components of a mixed gas are at the same temperature and pressure, the fraction of each component by volume is also the same fraction of the total pressure.


What happens if you put a piece of copper into a solution containing Al3+ ions?

You are right with copper does not react with aluminium ions.

But as a side reaction, copper may get slowly oxidized by oxygen and dissolve in mildly acidic solution of aluminium salt.

Such a thing may happen, if you decalcify copper heating spiral by vinegar and let it stay overnight.

$$\ce{ 2 Cu + 4 CH3COOH + O2 -> 2 (CH3COO)2Cu  + 2H2O  }$$

Something similar can happen in aluminium salt solutions as well, because of hydrolysis.

$$\begin{align}
\ce{ [Al(H2O)6]^3+ + H2O &<=> [Al(H2O)5(OH)]^2+ + H+}  \\
\ce{ 2 Cu  + O2 + 4 H+ &->  2 Cu^2+ + 2 H2O} \\
\ce{ 2 Cu  + O2 + 2 H2O &->  2 Cu(OH)2 v} \\
\ce{ 2 Cu  + O2 +  H2O + CO2 &->   Cu(OH)2\cdot CuCO3 v} \\
\end{align}$$


No serious mining operation would just dump a third of their desired product.
 The question does say that the reactions are "part of the manufacture of copper." This does not preclude other reactions. With the given reactions alone we are not able to handle the surplus of $\ce{Cu2S}$.
$$\ce{\color{\red}{4CuFeS2} + 9 1/2 O2 + 4 SiO2 \rightarrow \color{\red}{Cu2S + Cu2O} + 2SO2 + 4FeSiO3}  \tag{1}$$
$$\ce{\color{\red}{2} Cu2S + 2 Cu2O \rightarrow 6Cu + SO2 +\color{\red}{Cu2S}}\tag{2}$$
However we can infer from the first reaction that the sulfide can react with oxygen to form the needed oxide to balance reduction and oxidation.
$$\ce{ Cu2S + 1 1/2O2 ->  Cu2O + SO2}\tag{3}$$
The second equation can now become:
$$\ce{\color{\red}{2} Cu2S + 2 Cu2O +\color{\red}{1 1/2 O2} \rightarrow 8 Cu + 2 SO2 \tag{4}}$$
Thus in the first step $\ce{Cu2O \text{and} Cu2S}$ are produced according to quation 1.  In the second step they are reduced with aditional oxygen according to equation 4. To pove they balance first double equation 1.
$$\ce{8CuFeS2 + 19 O2 + 8 SiO2 \rightarrow \color{\red}{2Cu2S + 2Cu2O} + 4 SO2 + 8 FeSiO3}  \tag{5.1}$$
Add $\ce{1 1/2 O2}$ to each side
$$\ce{8CuFeS2 + \color{\red}{20 1/2 O2} + 8 SiO2 \rightarrow \color{\red}{2Cu2S + 2Cu2O + 1 1/2 O2 }+ 4 SO2  + 8 FeSiO3}  \tag{5.2}$$
Substitute the red products with equation 4 reactants.
$$\ce{8CuFeS2 + \color{\red}{20 1/2 O2} + 8 SiO2 \rightarrow \color{\red}{8 Cu + 2 SO2}+ 4 SO2  + 8 FeSiO3}  \tag{5.3}$$
This Gives a final balanced equation of
$$\ce{8CuFeS2 + 20 1/2 O2 + 8 SiO2 \rightarrow 8 Cu + 6 SO2  + 8 FeSiO3}  \tag{5.4}$$
As you can see on a molar eatio $\ce{CuFeS2\text{:}Cu}$ is 1:1 after accounting for the additional oxygen needed.


Most effective means of removing coffee stains from carpet

Oxidise it using a mild oxidizing agent like $\ce{H2O2}$ (hydrogen peroxide). When it undergoes oxidation coloured compounds get decolorized. Good luck on cleaning it.


Inflection point, slope, and equivalence point in a titration

Your intuition is not sufficient in this matter. The equivalence point is the steepest point of any titration curve, however weak or strong is the acid and the base.


What is the exact reason that compounds such as carbon dioxide cannot interact with other compounds through dipole-dipole interactions?

As the other answers have indicated, CO2 has no NET dipole moment. However it does have two dipoles pointed in opposite directions (as OP keeps mentioning). This means that CO2 can possibly interact through higher moments, such as the quadrupole moment. The mathematical procedure behind this is known as the multipole expansion. It is important to note however, the higher order multipoles get weaker and weaker by a factor of 1/R approximately, so quadrupole interactions are usually ignored for example.


As long as there is less than 100% relative humidity, the ice will sublime to some extent.

Just in the past week it has been in the -12 °C to -5 °C range in my area, and the snow and ice on the ground did sublime, on the order of 1 cm per day, more in the sun less in the shade.

For quantitative information see "Sublimation from a seasonal snowpack at a continental,
mid-latitude alpine site" Hydrol. Process. 13, 1781-1797 (1999).


A liquid that forms a surface membrane that prevents evaporation but allows oxygen to pass through

I don't think you or your cells want any strange low molecular weight substance in your experiment.

You can buy polymethylpentene (PMP) foil, this stuff has a rather low density compared to PE and other polyolefines, and therefore a high permeability for gases (10x higher than PE for oxygen, and PE is already not exactly tight against gases, x20 higher permeability than PVC for example). 50µm films are readily available, you find them online in a minute.


The composition of Silver tarnish is Silver sulfide, to quote from Wikipedia:

Silver sulfide is an inorganic compound with the formula $\ce{Ag2S}$. A dense black solid, it is the only sulfide of silver. It is useful as a photosensitizer in photography. It constitutes the tarnish that forms over time on silverware and other silver objects. Silver sulfide is insoluble in most solvents, but is degraded by strong acids.

The last sentence is important as Silver cleaning is not accomplished with harsh chemicals, but via electrochemistry which leads to the hydrogen atom radical formation, which then reduces the $\ce{Ag2S}$ liberating $\ce{Ag}$ and $\ce{H2S}$.  Here is a short summary of likely mechanics:
$\ce{Al -> Al3+ + 3 e-}$
$\ce{H2O ⇌ H+ + OH-}$
$\ce{HCO3- ⇌ H+ + CO3(2-)}$
$\ce{H+ + e- ⇌  •H }$
Effectively, $\ce{•H}$ can operate as a $\ce{(e-, H+)}$ pair:
$\ce{Ag2S + 2 •H -> 2 Ag + H2S (g)}$
Here is a confirming source, which states to quote:

The nascent hydrogen may also act as a reducing agent as it evolves from the metal surface.

And also, per the same source, the formation reaction of Ag2S can be reversed:

In that case silver suffers attack by dissolved oxygen and present sulphide species according to Reaction (5):
$\ce{4Ag (s) + 2HS- (aq) + O2(g) ⇌ 2Ag2S (s) + 2 OH- }$ (5)

So working in air, it is advised to employ a base to drive the above equilibrium to the left. Also, in alkaline conditions expect the reaction:
$\ce{H2S + 2 OH- -> 2 H2O + S(s)}$
Which is also noted in the referenced source, to quote:

However, after the cleaning of heavily tarnished samples (immersed for 30 and 60 min) some small quantities of sulfur (~0.3wt.%) are detected on the surface by EDS analysis, and the colour is changed to light yellowish.

[EDIT] With particular reference to the thiourea system, it is described in this hydrometallurgy source.  A cited reaction equation is:
$\ce{ Ag_2S + (SC(NH)(NH)2)2 + 2 H^+ + 4 SC(NH2)_2 -> 2 Ag(SC(NH2){_2}){_3}^{+} + S_o }$
Silver metal can then be recovered with reductive electrolysis.


How does strength of boric acid solution increase in presence of salicylic acid?

Equilibrium will be far to the right, as a stable six membered ring is formed. The proton speaks for itself.

References: Queen, A. The kinetics of the reaction of boric acid with salicylic acid. Can. J. Chem. 1977, 55 (16), 3035–3039 DOI: 10.1139/v77-421.


Comparison of carbocation stability in bicyclo compounds

I think the third option as greatest number of resonance structures can be alluded to it.
The following may suffice:

In the first one, the carbocation is isolated except for the presence of a single double bond in conjugation beside it. The oxygen and the double bond beside it play no role in stablizing it.
In the second one the lone pair on oxygen, the double bond and the carbocation are in conjugation. Hence, it is somewhat stable.
In three, similar conjugation like two is present, but is it extended over another double bond i.e. the lone pair of oxygen, two double bonds and the carbocation are all in conjugation and an extended resonance comes into play. Thus, it is more stable than two, since more number of resonance structures may be drawn for it than 2.
The fourth one is the least stable among them, as no conjugation or resonace structures may be constructed. Moreover the I- effect of the neighbouring oxygen tends to destablise the carbocation.

Hence, the order of stablity:
3>2>1>4
An arrow-pushing representation of resonance in the third option can be made as follows:

Hope it helps.


Dehydration of alcohol with Lewis acid aluminum trichloride

In the reaction $\ce{{}^{$i$}Pr-OH + H2O ->[\ce{AlCl3}] C3H6 + 2H2O}$ the $\ce{H2O}$ on the reactant side and one of the $\ce{H2O}$s on the product side cancel, so you get $\ce{{}^{$i$}Pr-OH ->[\ce{AlCl3}] C3H6 + H2O}$.

I think the reaction would also work with other Lewis acids. But it will rather proceed through the $\mathrm{E}1$ mechanism in any case. And it is likely that you'll get $\mathrm{S}_\mathrm{N}1$ side reactions.


What are the primary, secondary and tertiary physicochemical properties of ionic liquids?

The definition of generations of physicochemical properties can be attributed to Hugh et al. [1] as recapitulated by e.g. Ferraz et al. [2]:

  The search for new and different ILs has led to the progressive development and application of three generations of ILs: 1) The focus of the first generation was mainly on their unique intrinsic physical and chemical properties, such as density, viscosity, conductivity, solubility,and high thermal and chemical stability. 2) The second generation of ILs offered the potential to tune some of these physical and chemical properties, allowing the formation of “task-specific ionic liquids” which can have application as lubricants, energetic materials (in the case of selective separation and extraction processes), and as more environmentally friendly(greener) reaction solvents, among others. 3) The third andmost recent generation of ILs involve active pharmaceutical ingredients (API), which are being used to produce ILs with biological activity.

The passage quoted in the OP stems from an article by Zakrewsky et al. [3] and refers to a study on a third generation IL. Third generation characteristics would in this case include those granting the IL antibacterial, nontoxic and transdermal delivery properties, that is allowing solubilization of drug compounds, minimizing damage to human cells, and disupting bacterial biofilms. In this case, "secondary" properties were not targeted, but these might include (as per the above quote) reduced environmental toxicity.

References

Hough, W. L.; Smiglak, M.; Rodríguez, H.; Swatloski, R. P.; Spear, S. K.; Daly, D. T.; Pernak, J.; Grisel, J. E.; Carliss, R. D.; Soutullo, M. D.; et al. The Third Evolution of Ionic Liquids: Active Pharmaceutical Ingredients. New J. Chem. 2007, 31 (8), 1429–1436. https://doi.org/10.1039/B706677P.
Ferraz, R.; Branco, L. C.; Prudêncio, C.; Noronha, J. P.; Petrovski, Ž. Ionic Liquids as Active Pharmaceutical Ingredients. Chem. Med. Chem. 2011, 6 (6), 975–985. https://doi.org/10.1002/cmdc.201100082.
Zakrewsky, M.; Lovejoy, K. S.; Kern, T. L.; Miller, T. E.; Le, V.; Nagy, A.; Goumas, A. M.; Iyer, R. S.; Sesto, R. E. D.; Koppisch, A. T.; et al. Ionic Liquids as a Class of Materials for Transdermal Delivery and Pathogen Neutralization. PNAS 2014, 111 (37), 13313–13318. https://doi.org/10.1073/pnas.1403995111.


The $K_p$'s here are those for the standard formation reactions.
If you go to the following link, you will see $\Delta G^o_f$ values for $\ce{CO2(g)}$  and $\ce{CO(g)}$, in $100 \,\mathrm{K}$ increments, up to $1500 \,\mathrm{K}$:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwjbt56Z9IbvAhXVjp4KHbD-BS0QFjAJegQIAhAD&url=https%3A%2F%2Flink.springer.com%2Fcontent%2Fpdf%2Fbbm%253A978-3-319-41616-8%252F1.pdf&usg=AOvVaw3rJgF0nzpqXtq9ML9qvGaV
Thus, doing a simple linear extrapolation to $1600 \,\mathrm{K}$ gives us the following:
$$\Delta G^o_f,\ce{CO_2(g)} = -396.309 \mathrm{\frac{kJ}{mol}}$$
$$\Delta G^o_f, \ce{CO(g)} =  -252.3298 \mathrm{\frac{kJ}{mol}}$$
Solving for $\ln K_p$ (using $\Delta G^o = -RT \ln K_p$) gives:
$$\ln K_p,\ce{CO_2(g)} = 29.7906$$
$$\ln K_p, \ce{CO(g)} = 18.9677$$
Dividing by $\ln(10)$ to convert from the natural logarithm to the base-10 logarithm gives:
$$\log K_p,\ce{CO_2(g)} = 12.940$$
$$\log K_p, \ce{CO(g)} = 8.238$$
[$\Delta G^o_f$ for $\ce{CO(g)}$ changes more rapidly with temperature than that for $\ce{CO2(g)}$, so my extrapolation for the former may have a larger error, which could explain the small discrepancy vs. the value you were given.]
In addition, if you appropriately combine the formation reactions to give the net reaction (as you would for a Hess's Law problem), you will see that you get the supplied equation for $\log K_p$ for the net reaction (I'll leave this as an exercise for you).


How to find how much there is of a compound given the following information?

Maybe I'm just too tired right now, but didn't you mention that the absolute amount of magnesium oxide doesn't change during aqueous workup?

Isn't thus the amount of 3.22 g $\ce{MgO}$ exactly what was formed initially?

Wouldn't it be more reasonable to let the students calculate how much magnesium metal was burned initially, taking the $\ce{N2}$/$\ce{O2}$ ratio in air and the stoichiometry of the magnesium nitride into account? ;)


Equation for reaction of tin and iodine to produce tin(IV) iodide

The balanced chemical equation is independent of the amounts of each reactant added. In this case we can easily deduce the equation to be: $$\ce{Sn + 2I2 -> SnI4}$$

What the given masses tell us is that there is excess tin being added. We can see this because the $\ce{Sn:I2}$ ratio added is $1:1.5$ whereas the stoichiometric ratio is $1:2$ and so there is less iodine than is required to completely react with the added tin - or in other words there is an excess of tin.


Is 1,2-dihydronaphthalene "partially aromatic"?

Aromaticity is a quite difficult concept but in this case we can answer the question by simply looking at similar compounds and experimental data.

What you got here is a styrol derivative, the pi-system (which we need to look at if we are talking about aromaticity) is the same. So if this compound is not aromatic then styrol shouldn't be either.

But styrol is in fact aromatic (or better: the benzene ring in styrol is). It reacts like an aromatic compound should, it got the right shift in NMR, the bond lengths are in accordance with aromaticity and so is the UV/VIS spectrum.

So a conjugated double bond clearly doesn't make a benzene (significantly) less aromatic.


Reverse Cannizzaro reaction to make aldehydes?

No, this is not practicable because aldehydes are higher in free energy than either alcohols or acids.  Thus, the equilibrium constant of the Cannizaro reaction lies strongly against the aldehydes.

The case of formaldehyde may be even more extreme than the example reaction you gave, but it is what I could find data for.

The eQuilibrator database lists the equilibrium constant for the reaction of

$$\ce{2 HCHO + H2O <=> CH3OH + HCOOH}$$

as $$K'_{eq} = 3.4 × 10^{13}$$

This is for each reactant at 1 mM in concentration, at pH 7 and an ionic strength of 0.1 mol / L.  Thus, you'd need a teramole or two of formaldehyde to get one mole of acid/alcohol.  The thermodynamics for other aldehydes may be slightly more "favorable" but you aren't going to get 13 orders of magnitude of favorability for any "regular" pair of aldehyde and acid/alcohol.


Search for the latest IChO tests and preparatory problems. You can find quite a few interesting problems.

Here are some tests:
http://www.iuventa.sk/files/documents/5_icho/past%20competition%20tasks/volume3-icho41-45.pdf

http://icho2013.chem.msu.ru/materials/Preparatory_problems_IChO_2013.pdf (problems 18-21)

http://icho2014.hus.edu.vn/Preparatory-Problems/0001/01/Preparatory-Problems-224.html (problems 18-26)


Approximating the concentration of ions when a salt dissociates

Really, your question seems to be a math question.

You need to solve the cubic equation:

$$\mathrm{9.0\cdot10^{-12}=[0.100+}2x]^{2}[x]$$

Cubic equations are exactly solvable; however, it is common to use an approximate method.

Whether or not it is appropriate to approximate 0.100 +2x as 0.100 depends upon how small x is and how precise an answer you need.

Once you get an approximate solution for x, you could recalculate 0.100 +2x and use that value to re-solve the problem. By repeating this process you can get arbitrary close to the correct value by iteration.


In the phase diagram of a pure substance there are regions of a single phase e.g. solid, liquid or vapour. The boundary lines you refer to are where the substance exits as $2$ two phases, say solid and vapour in equilibrium with one another.

The Gibbs phase rule states that the number of degrees of freedom, F,  which is the number of (intensive) variables that can be varied is given by
$$F = C-P+2$$
where C is the number of components, $1$ for a pure substance, and P the number of phases, which is $1$ anywhere in the diagram except on a boundary line. Therefore the number of degrees of freedom is $2$, (except on a boundary line) which means that pressure and temperature can both be varied independently.

The boundary lines indicates the condition of P and T where there are two phases in equilibrium, say solid and vapour. As there are $2$ phases there is now only $1$ degree of freedom so that temperature and pressure are no longer able to be varied independently. Thus on the liquid /vapour curve, if the pressure is increased some vapour is condensed (heat is given out) and the temperature increases to a new value but one that that is still on the boundary line, i.e. vapour is still in equilibrium with the liquid. Along any boundary curve the two phases are in equilibrium and so, e.g., the rate of melting is equal to rate of freezing. (Any point along the liquid/vapour boundary line can be considered to be boiling, but we define the normal boiling point to be the boiling temperature at $1$ atm pressure.)

At the triple point, where the three boundary curves meet, there are no degrees of freedom and this point is fixed. At and above the critical point the density of the liquid and vapour are the same and the liquid's meniscus  disappears. The super-critical fluid looks very much like the swirling foam that one sees for a short while on pouring a bottle of Guinness into a glass.

The boundary lines can be calculated for a substance using the Clausius and Clausius-Clapeyron equations.


It is risky to predict just by looking at the structure, of course there is the ring size effect and one could easily conclude that $\ce{[Rh(Ph2PCH2CH2PPh2)3]3+}$ is the most thermodynamically stable compound since you would have a 5-membered ring instead of a 4-membered as in $\ce{Rh(Ph2PCH2PPh2)3]3+}$. However, there is also the contribution of the electronic effect, which is the increase or decrease in electron density of the metal ion caused by the distribution of ligands around the metal center. The combination of electronic and steric effects was reviewed by C.A Tolman in 1977 for monodentate ligands and he proposed a way of measuring these effects experimentally and quantitatively, they are the so-called Tolman's parameters (Tolman, C. A. (1977). "Steric effects of phosphorus ligands in organometallic chemistry and homogeneous catalysis". Chem. Rev. 77 (3): 313–348.)

For bidentate ligands such as DPPE (2 $\ce{-CH2 -}$ groups between phosphorus atoms) or DPPM (1 $\ce{-CH2 -}$  group between phosphorus atoms) another important parameter is the bite angle "β" (the circular sector angle between the metal and the two phosphorus atoms). For example, in square-planar and octahedral complexes the ideal bite angle is 90º , deviations from this value will electronically destabilize the complex (KUMAR, M.  et al. Importance of Long-Range Noncovalent Interactions in the Regioselectivity of Rhodium-Xantphos-Catalyzed Hydroformylation. Organometallics, v. 34, n. 6, p. 1062-1073, Mar 2015.).

Bidentate phosphines with a natural bite angle of 120° preferentially occupy diequatorial sites in a trigonal bipyramidal complex whereas a bidentate phosphine with a natural bite angle of 90° preferentially occupy apical-equatorial positions (Casey, C. P.; Whiteker, G. T.; Melville, M. G.; Petrovich, L. M.; Gavney, J. A.; Powell, D. R. (1992). "Diphosphines with natural bite angles near 120° increase selectivity for n-aldehyde formation in rhodium-catalyzed hydroformylation". J. Am. Chem. Soc. 114 (2): 5535–5543.).

DPPM has a bite angle of 78º while DPPE has a bite angle of 85º (closer to 90º), so considering only the angular aspect, the $\ce{Rh-DPPE}$ complex would be more stable (Ligand Bite Angle Effects in Metal-catalyzed $\ce{C−C}$ Bond Formation
Piet W. N. M. van Leeuwen,, Paul C. J. Kamer,Joost N. H. Reek, and, and Peter Dierkes, Chemical Reviews 2000 100 (8), 2741-2770)*. Electronically, I assume that the 2  $\ce{-CH2 -}$ groups of DPPE could donate more electronic density to the P atoms than DPPM, which would somehow strengthen the $\ce{Rh-P}$ bonds.

However, it is the combination of sterical + electronic effects that can unequivocally decide which of the two complexes are more stable, and this would only be possible using computational chemistry or experimentally, though the above can be used when discussing about it on a statement or paper.


Why does the titanium go to the tungsten filament in the crystal bar process?

By "gaseous titanium", do you mean "titanium tetraiodide ($\ce{TiI4}$) vapor"?
Van-Arkel De-Boer process is a purification process of titanium and zirconium. Basically, what happens is that the impure metal, let's say titanium is heated in iodine environment at a temperature of $\ce{250 ^\circ C}$ to form volatile titanium tetraiodide ($\ce{TiI4}$) vapor. The impurities are left behind, as they do not react with iodine. This vapor is then  passed over a hot tungsten filament at $\ce{1400 ^\circ C}$ for which the vapor gets decomposed and pure titanium is deposited on the filament and is removed. The iodine is reused. The overall reaction is:
$$\ce{Ti _{(impure)} + 2I2 ->[250 ^\circ C] TiI4 ->[1400 ^\circ C][W filament] Ti_{(pure)} + 2I2}$$
The thermodynamics of this reaction is discussed in Buttonwood's comment.


TL;DR: Naphthalene anion radical won't tolerate a presence of unshielded alkali metal cation nor in solution, neither in crystal. Reduction of naphthalene is possible strictly under air- and moisture free conditions by alkali-metals in the presence of coordinating solvents (cryptands). In this case solvent-separated cations don't form coordination bonds with naphthalenide radical.
On the other hand, further reduced (dianion) naphthalene readily forms various compounds not only with alkali metals, but also with lanthanides, typically $\eta^4$-complexes.

Various studies on stabilization of radical anions, especially in solid state, show that complexation (preferably with polar aprotic solvent or polydentate ligands) of counter-cations is essential:
$$\ce{C_nH_m + [M^0]_{\infty} + $x$ Solv + $n$ L <=> [M+(Solv_x|L_n)](C_nH_m^{.-})},$$
where $\ce{[M^0]_{\infty}}$ - any metal in oxidation state $0$ as a bulk reactant, $\ce{Solv}$ and $\ce{L}$ - solvent and ligand, correspondingly.
If cation is optimally solvated (complexated/sterically protected), radical counter anions $\ce{C_nH_m^{.-}}$ with extended $\pi$ systems are preferred because they lack extensively charged centers which favor contact ion formation (1). For example, crystals of bis(diglyme-O,O',O'')-sodium naphthalenide radical
$\ce{[Na+(diglume)2](C10H8^{.-})}$ ([Na]) (1) has been isolated based on a principle of increasing enthalpy (calcd. based on crystal structures):
$$\Delta H (\ce{[Na+(THF)6]}) = \pu{-587 kJ mol^{-1}}$$
$$\Delta H (\ce{[Na+(DME)3]}) = \pu{-671 kJ mol^{-1}}$$
$$\Delta H (\ce{[Na+(diglume)2]}) = \pu{-677.3 kJ mol^{-1}}$$
Under certain conditions, the reduction of the naphthalene can proceed further to the naphthalene dianion, $\ce{C10H8^{2-}}$. For example, a complex with radical ion $\ce{[Li+(TMEDA)2](C10H8^{.-})}$ ([Li]) can reversibly be transformed into naphthalene dianion-complex $\ce{[Li+(TMEDA)]2(C10H8^{2-})}$ (2):

Of course, bigger cations require better shielding, commonly resulting in a usage of bulkier solvents or supramolecular assemblies, such as ([2,2,2]-cryptand)-potassium naphthalenide radical, $\ce{[K+(crypt-222)](C10H8^{.-})}$ ([K]) (3), in comparison to previously mentioned [Li] and [Na] naphthalenides (in the following crystal structures sphere-packing model is applied to metal atoms, and the rest is represented with sticks for simplicity):

It is also worth mentioning that naphthalene dianion forms numerous complexes with rare earth elements (4), e.g. with [Y] (see picture): $\ce{CpYC10H8(DME)}$, ($\eta^5$-Cyclopentadienyl)-(1,2-dimethoxyethane)-($2\eta^1:\eta^2(2\sigma,\pi)$-naphthalenide)-yttrium(III) with a $\ce{C6}$ ring being bent by $26.1^\circ$:

$\ce{La}$ and $\ce{Eu}$ complexes with bridged $\mu-\eta^4:\eta^4$-naphthalene are also obtained ($\ce{[LaI2(THF)3]2(C10H8)}$, $\ce{[EuI(DME)]2(C10H8)}$).
(1) Bock, H.; Arad, C.; Näther, C.; Havlas, Z. J. Chem. Soc., Chem. Commun. 1995, 23, 2393–2394. DOI: 10.1039/C39950002393
(2) Melero, C.; Guijarro, A.; Yus, M. Dalton Trans. 2009, 0 (8), 1286–1289. DOI: 10.1039/B821119C
(3) Rosokha, S. V.; Kochi, J. K. J. Org. Chem. 2006, 71 (25), 9357–9365. DOI: 10.1021/jo061695a
(4) Protchenko, A. V.; Zakharov, L. N.; Fukin, G. K.; Struchkov, V. T.; Bochkareva, M. N. Russ Chem Bull 1996, 45 (4), 950–953. DOI: 10.1007/BF01431330


Ring contraction when cyclohexene oxide is treated with methyl Grignard

Sara Jane: Lest you miss the point of the stimulating discussion between Waylander and Zhe, I would like to sum up their thoughts with a diagram. Your first clue should have been the opening line: "Recall that epoxides rearrange with Lewis acid...". It is common that catalytic Lewis acids rearrange 2,3-dialkyl oxiranes [epoxides] to ketones. In the case of cyclohexane epoxide (1), rearrangement to ring-contracted cyclopentane carboxaldehyde (4) occurs. In the illustration I have used CH3MgBr as the Lewis acid. Alternatively, one might use the other components of the Schlenk equilibrium, i.e., (CH3)2Mg or MgBr2. The rearrangement is shown as two separate steps (2 and 3) although it is likely to be a concerted process.

The aldehyde 4 in the presence of the Grignard reagent will undergo addition. Given sufficient Grignard reagent, addition to the carbonyl group will occur to afford the secondary alcohol 6. The only ways that the aldehyde 4 can be isolated is if there is insufficient methyl magnesium bromide or that the Grignard reagent acts as a base and forms the unreactive enolate of the aldehyde. The aldehyde 4 would be produced upon aqueous workup. While Grignard reagents with β-hydrogens, e.g., isopropyl and ethyl Grignard reagents, are capable of reduction of aldehydes and ketones, methyl Grignard reagents are incapable of this reaction. Therefore, the primary alcohol 9 may be assumed to be a misprint in place of secondary alcohol 6.
The oxidation of alcohol 6 to ketone 7 is straightforward. The final step is the formation of the alkene 8 (most likely of the (E)-configuration) via the McMurry coupling.


How do derive the equation for relative velocity of gas molecules

Your equation $v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$ is incorrect.

$v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ (your last equation) is also incorrect.

Instead $v_{rel}=(\frac{16KT}{\pi m})^{\frac12}$ and $v_{rel}=(\frac{8KT}{\pi \frac m2})^{\frac12}$ would both be correct.

The need for reduced mass is for when the particles are not the same mass, usually in the context of considering collisions between two different molecules or atoms.

$\mu = m_1m_2/(m_1 + m_2)$ by definition.  If $m_1 = m_2$ then $\mu = m/2$

When there are two different masses, there are two different mean velocities.

To see where the equation with the reduced mass comes from, look at appendix 1.4 of Kinetic Theory of Gases, along with pages 1-30 to 1-34.


Can Silicon be used as anode for Li-Ion batteries

Generally, heavier atoms means lower specific stored charge per mass.

Specific energy per mass depends of the cell redox systems, but also on the above.

Silicium, in contrary to carbon, does not form hexagonal planar structure with delocalized electrons like graphite, but prefers the diamond-like structure.

The Li-ion cell anode is formed by lithium-graphite intercalate, for which does not exist Silicium analogue. If it had existed and the cell had been possible to create, the major difference would have been being heavier for the same mAh capacity.

Silicon anodes (thanks @andselisk  for the link) are possible, but not as analogue of the graphene structure  of graphite. The huge problem with them is material swelling in the charge state, limiting the anode life to few cycles. Few percents of silicium together with carbon can be used to increase somewhat the graphite anode capacity.


I think that the main effect of salt in cooking vegetables is to act osmotically to draw water out of the plant cells, reducing the volume of the veggies and hence the time required to cook the vegetables. Perhaps this also changes the flavor and texture in beneficial ways as well. Example: If you soak/cook dry beans in fresh water, they tend to burst open, but remain whole in salted water.


What is the evolutionary reason for the existence of rings in molecules?

I noticed that most molecules (or at least drug-like molecules) have rings, but I was wondering: from an evolutionary perspective, what does it mean?

First note: Delete the ‘most molecules’ bit. The numbers are insane and nigh uncountable, but non-cyclic molecules are similarly common as cyclic ones are.

From an evolutionary persective this means: nothing.

Drugs are designed to inhibit certain enzymes as specifically as possible. The enzyme has a shape that happens to best fulfil it’s purpose according to evolutionary principles. (my draft had ‘design’ here for a second, but it would be the wrong word choice. Nature and evolution do not design.) This will render an active pocket having a certain orientation, size and hydrophobic pattern that drug designers attempt to mimick. If a (poly-) cyclic molecule fits well, it fits well. If a linear molecule fits better it fits better. Think of it as trial and error.


What are the parameters of protonation of a molecule?

We are taking about electrophilicity of ions.
It basically depends upon:
1. charge on the ion(+)
2. density of plus charge on it

So comparing $\ce{H+}$ and $\ce{CH3+}$, the former has a higher plus charge density so that's why it will be more attracted towards 'O'.


How do noble gases bond with themselves?

Noble gases usually do not form strong bonds between their atoms - it takes fair amount of energy to dimerise them into excimers, but those are short-lived excited molecules. Thanks to excitation, shells of the atoms aren't closed and they react, but very quickly they lose energy and become separate atoms. On the other hand there are many stable molecules created by heavier noble gases (mainly xenon) with other elements.

As mentioned in comment, you may heard about detection of so called van der Waals molecules of helium, which aren't "true" molecules, but very weakly bound pairs of atoms. In fact helium vdW dimer may have weakest bond even among them, and it was an achievement to observe it.


IUPAC nomenclature of 1,2,3,4-tetramethylcyclobutane

The Cahn-Ingold-Prelog (CIP) assignments of the configurations of the two tetramethylcyclobutanes above are accomplished as follows:
1)  Tetramethylcyclobutane 1: This isomer has all of the methyl groups on the same face of the ring. There are two mirror planes passing through C1-C3 and C2-C4. Numbering the cis-isomer is arbitrary in that there is a 4-fold axis of symmetry present. The common methods for R/S-configurations do not apply. Because the ring carbons lie in symmetry planes, they are stereogenic and achirotopic requiring r/s-assignments. When an assignment is made to one carbon, the same assignment may be applied to the remaining three.
  Temporary assignments for 1a are made to C2, C3 and C4 (r0/s0). Priorities are assigned at each of these carbons such that C1 is the highest priority group.  Returning to C1 the priorities are r0>s0>CH3>H. C1 is of the s-configuration. Digraph 1b illustrates the assignment to C1 in a different way. The horizontal line represents the plane of the ring with C2, C3 and C4 above the plane of the ring and the dotted line (C1) signifies the carbon under assignment. Since r0 precedes s0, C4 precedes C2. C1 is of the s-configuration.

2) Tetramethylcyclobutane 2: This stereoisomer has ring carbons that are both stereogenic and chirotopic. Here R/S assignments apply. There is a plane of symmetry passing through the C1-4 and C2-3 bonds while a two-fold axis of rotation exists between the C1-2 and C3-4 bonds. Applying the methodology of 1a to 2a, C1 is assigned the R-configuration. Referring to digraph 2b, R0R0 precedes R0S0 [like(lk)>unlike(ul)]. Thus, C2 precedes C4. C1 is of the R-configuration.
One may apply this method to the remaining three ring carbons but there is a shortcut. Because C1 mirrors C4, C4 is of the S-configuration. The rotational axis shows that both C1 and C2 are of the R-configuration while C3 and C4 are both of the S-configuration. The numbering system requires that the lowest numbered atoms, if possible, be labeled R.


Finding the sensitivity of zero-order ultra sensitivity and its parameter dependence

I found the answer.
First, we need to find balanced equation for reaction. Then express the concentration of $xˆst$ in the steady state through E.
Then we can arrive to this kind of solution for the value of E.
$\displaystyle{E\frac{x_0 - \frac{x_o}{2}}{K + x_0 - \frac{x_0}{2}}} - {\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}} = 0.$
$\displaystyle{E\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}} - {\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}}= 0. $
E = 1


Electrochemical series - when not at standard conditions

While the electrochemical series does assume standard-state conditions, we will show that in certain cases it will also work for non-standard-state conditions. One recalls that the Nernst equation is given by $$E = E^\circ - \frac{RT}{nF}\,\ln Q,$$ and can be used to determine electrochemical cell potentials when the cells are not at standard state. This works directly for the second reaction you list, as $Q = \ce{[Sn^2+]}/\ce{[Cu^2+]} = 1$ and $\ln Q = 0$, so the second term in the Nernst equation vanishes, but will not work for the first reaction, for which $Q = \ce{[Sn^2+]}/\ce{[Ag+]^2} \neq 1$.

In this case, it is reasonable to assume that one knows both (i) that the correction due to the second term is relatively small, and (ii) that $\ce{Ag}$ and $\ce{Cu}$ are relatively inert metals, and strongly favor being in their elemental form relative to other metals. The quoted result then follows.


When we heat a borax crystal, why does it first expand and then contract?

I've never seen this reaction, but it sounds quite entertaining. And I learned a couple of new words in looking at the literature to explain this phenomenon.

Intumescence: The property of a material to swell when heated. Borax is intumescent.
"Calcination is a thermal treatment process in presence of air applied to ores and other solid materials to bring about a thermal decomposition, phase transition, or removal of a volatile fraction. The calcination process normally takes place at temperatures below the melting point of the product materials."

The mechanism for the calcination of borax quite complex and is covered in some detail in this article. Borax decahydrate (BDH), $\ce{Na2B4O7.10H2O}$ is easily dehydrated to Borax pentahydrate (BPH), $\ce{Na2B4O7.5H2O}$ without much visible change. The actual form of the borate ion in both BDH and BPH is $\ce{B4O5(OH)4^2-}$ with 2 of the water molecules actually incorporated as hydroxyl groups in the ion. (The other 2 hydroxyl groups come for the water hydrogens and 2 oxygens from the $\ce{B4O7^2-}$.) That means that the last two water molecules can only be removed by molecular decomposition. So in BPH, you actually have only 3 "free" (i.e., easy to remove) water molecules.

So here's what happens. The borax decahydrate dehydrates smoothly with minimal density changes (no swelling) until you get to borax pentahydrate. At this point, the BPH dissolves  in its own water of dehydration and the "blossoming" you see is from the solution erupting through the partially dehydrated crystal surface as it's heated. The density of the borax changes dramatically as it goes from the pentahydrate (about $0.85\ \mathrm{g/cm^3}$) to the dihydrate  (about $0.14\ \mathrm{g/cm^3}$) which is why it ends up looking "whipped".

To do the final transformation to a tiny bead, the remaining 2 waters of hydration (those tied up as hydroxide ions) are driven off as the $\ce{B4O5(OH)4^2-}$ decomposes to give metaborate, $\ce{BO2^2-}$ and boric anhydride, $\ce{B2O3}$. $$\ce{B4O5(OH)4^2- -> 2 BO2 + B2O3 + 2H2O}$$

Fascinating.


Fluorination of Carbonyl Compounds using Molybdenum Hexafluoride

Molybdenum(VI) fluoride can be used as a fluorinating agent though not as popular as the selective DAST and Deoxyfluor as @Waylander noted. Molybdenum(VI) fluoride fall under the category of electrophilic fluorinating agent where it acts as a Lewis acid. The reaction is run in an inert solvent like $\ce{CH2Cl2}$ and a catalyst like boron trifluoride is used. The following is the reaction procedure taken from an e-book$\ce{^{[1]}}$. Also yield of various gem-difluorocarbonyl compounds from corresponding carbonyl compounds are listed:

  A solution of $\ce{MoF6}$ (102g, 0.49g) in anhydrous $\ce{CH2Cl2}$(500mL) is cooled at 20-30 C. While passing a gentle flow of $\ce{BF3}$ through the solution, the carbonyl compound(0.5g) is added in small portion. The mixture is stirred while warming upto rt. Anhyd. $\ce{NaF}$ and silica gel were added. After filtration, the solvent is distilled in vacuo. The residue is distilled/purified by chromatography.

Further, it is written that:

  In general, the yields are not that high due to the sensitivity of the
  reaction towards electronic and steric factors. Yield increase with
  the elctronic withdrawing group of the substituent of the substrate
  and decrease with the steric hindrance of the carbonyl group.
  $\ce{MoF6}$ attacks ether, amino, hydroxy, carboxy as well as
  $\ce{C=C}$ bonds. Ester, amide, cyano, nitro are inert towards the
  reagent. Halo substituent are not exchanged for fluorine.

The general reaction is given by$\ce{^{[2]}}$:

If a cyclic compound is used then R1 and R2 will be fused together.

If fluorination is proceeded at high temperature then catalyst is not needed. Aromatic carboxylic acids are converted into into gem-trifluoro derivatives without any catalyst.
Also, Aryl trifluoromethyl derivatives can also be obtained under less drastic conditions by fluorination of aromatic acid chlorides.

Now, the same reaction can be achieved by using DAST or Deoxylfluor and they are the more popular fluorinating reagent than $\ce{MoF6}$. Their reaction mechanism has been studied extensively. The mechanism is more or less similar than the mechanism achieved by $\ce{MoF6}$. The following is an explanation of the mechanism$\ce{^{[3]}}$(emphasis mine):

  The electronic effect of the substituent exerted a significant
  influence on the reactivities of benzophenones. The presence of an
  electron-withdrawing group is believed to facilitate the nucleophilic
  attack of the fluoride anion to the carbonyl group, causing more
  facile cleavage of the carbon–oxygen double bond. Among the
  benzophenone substrates examined, 4-nitrobenzophenone exhibited the
  highest reactivity and the corresponding gem-difluorinated product was
  obtained in 91% yield[...]

Notes and References

Houben-Weyl Methods of Organic Chemistry Vol. E 10b/1, 4th Edition Supplement: "Organo-Fluorine Compounds - Synthesis of Fluorinated Compounds I, Transformations of Fluorinated Compounds" by Bernd Baasner and Georg Thieme Verlag, 14-May-2014 (link)
http://reag.paperplane.io/00002051.htm#2
Ying Chang, Amit Tewari, Avi-Izak Adi, Chul sung Bae "Direct nucleophilic fluorination of carbonyl groups of benzophenones and benzils with Deoxofluor" Volume 64, Issue 42, 13 October 2008, Pages 9837-9842
DOI: https://doi.org/10.1016/j.tet.2008.08.009
https://www.alfa.com/media/docs/FluorinatingAgents.pdf
Direct carbonyl group attack can be avoided and α-fluorination of ketones can be achieved by using highly selective fluorinating agent, NFSI. α-fluorination of cyclohexanone has been observed.

Enantioselective Organocatalytic α-Fluorination of Cyclic Ketones
Piotr Kwiatkowski, Teresa D. Beeson, Jay C. Conrad, and David W. C. MacMillan
Journal of the American Chemical Society 2011 133 (6), 1738-1741
DOI: 10.1021/ja111163u


What is the physical quantity/magnitude of pH?

I disagree with the answer that says "pH measures acidity or basicity" (no matter what is in Wikipedia).

pH is the negative of the log of the hydrogen ion (H+) activity. Activity is a thermodynamic concept that is related to but not the same as concentration. Activity is similar to chemical potential. The same concentration of hydrogen ions will have different activities, depending on what else is in solution. In weak solutions, the activity coefficient that relates the concentration to the activity depends mainly on the charges of the other ions in solution and not on the particular ion species. But at higher amounts of other species, the particular chemicals and chemical interactions become increasingly important.

On the other hand, acidity is the ability to provide hydrogen ions to a reaction and considers not only the hydrogen activity but also the amount of acid-base reaction that can occur. You may have heard of pH 'buffers'. These are solutions that are designed to have a balance of chemical species so that the pH is very stable, even if small amounts of acid or base are added or if the ionic strength of the solution changes. If you add a bit of base to the buffer solution, the solution reequilibrates to provide acid to react with the base. You have to add a fair bit of base before the buffer capacity is used up and the pH changes very much. Then, with further addition, the pH will change a very quickly until it approaches the pH of the base that is being added. So a pH buffer has high acidity and alkalinity, keeping the pH stable.

In groundwater geochemistry, acidity is measured by reacting the solution with strong base until the acid is essentially used up. This is done by titration either to a fixed endpoint pH or to the inflection point in the curve of base added vs pH curve.

Bottom line, pH doesn't measure acidity. In terms of the original question, I would say:

Variable: pH, Quantity: negative log of the hydrogen ion activity, Units: dimensionless pH units.


When is the activity coefficient greater than 1?

The classic case of having activity coefficient greater than 1 is the solution of very hydrophobic organic solvents in water. For example, if you have toluene or benzene in water, these molecules do not prefer to stay in water. This type of work was a popular research topic in the 50-70s.
Note that here we are talking about neutral solution so the concept of ordered lining up does not arise.
There are plenty of papers on this topic. Just search activity coefficient of benzene in aqueous solutions.


Why a TS must have only one negative frequency

Well, let's just look at the simple case you mention. If we are at a transition state, then we are at a saddle point on the potential energy surface (PES). As you say, there are two directions where we go down in energy and two where we go up in energy which are along or perpendicular to the gradient vector.

Note that to confirm we are at a saddle point, in mathematics one could do the second-derivative test. Or, in chemistry (and in math) in order to do this, we construct the Hessian matrix and find its determinant. So, let our PES be given by a function $f(x,y)$. Then, the Hessian takes the form,
$$
\begin{bmatrix}
f_{xx} & f_{xy} \\
f_{xy} & f_{yy} \\
\end{bmatrix}
$$
where $f_{xx}$ denotes the second partial derivative of $f$, both with respect to $x$ and likewise for the rest of the terms. Notice that $f_{xy}=f_{yx}$, so this matrix is symmetric and hence all its eigenvalues are real. Also note that sometimes people will write this in terms of the force constants instead of the second-derivatives because a harmonic force-constant is a second derivative.

So, if we take the determinant of this matrix, its sign tells us about whether the function is concave up or concave down at the point we are interested in. This determinant is given by,
$$
f_{xx}f_{yy}-(f_{xy})^2.
$$
The sign of this actually tells us information about the eigenvalues of the matrix because the determinant of a matrix is equal to the product of its eigenvalues.
Notice that in order to find the eigenvalues of this matrix we would solve the following equation:
$$
(f_{xx}-\lambda)(f_{yy}-\lambda)-(f_{xy})^2=0
$$
Or, after rearranging,
$$
\lambda^2-\lambda(f_{xx}+f_{yy})+f_{xx}f_{yy}-f_{xy}^2=0
$$

These eigenvalues can be related to the frequencies of the vibrations if we first mass-weighted each second-derivative. All I want to show here is that for this case there are only two eigenvalues, and these eigenvalues do not have to do with the number of directions to move towards a minimum. One of the vibrations you can imagine as being the mode which in two directions have an increase in energy. So, along this mode, the molecule stays at the transition state and hence there is a physical frequency associated with this motion. If the second derivative test is worth anything, then the overall sign of this determinant will be negative at a transition state and hence there will be one and only one negative eigenvalue. Note that this only works at critical points, so that's why at the transition state there must be only one negative imaginary frequency (or negative eigenvalue).

Because for most every molecule we have more than two modes to worry about, you could imagine splitting a much larger matrix into many 2-d matrices and looking at two modes at a time. From this idea it is clear that a transition state need not be limited to having only one negative frequency. Rather, at second-order saddle points, (i.e. a saddle point where following the eigenvector corresponding to a negative eigenvalue leads you to another saddle point) there will be two negative imaginary eigenvalues.

To summarize:

The number of imaginary frequencies is the number of negative eigenvalues of the Hessian matrix (the matrix of all possible mixed-second-partial derivatives)
There can be more than one negative frequency at a higher-order saddle point, but physically speaking these are much less common than transition states with only one imaginary frequency
One cannot determine the number of imaginary frequencies by thinking about the number of paths to a minimum. You could think about the number of modes which do not kick you off of a saddle point if you prefer this.

Edit:

I mistakenly referred to negative frequencies which is not correct (as alluded to by a commenter). The eigenvalues of this matrix correspond to the force constants (second derivatives with respect to displacement coordinates), so one gets negative force constants, not negative frequencies at a saddle point. Because the force constant for one mode is negative at a transition state, the "frequency" of this mode is evaluated as $\nu=\sqrt{\frac{k}{\mu}}$ as usual, and the frequency is imaginary. So, one is looking for either one negative force constant or one imaginary frequency, not a negative frequency. Sorry I missed that.


I'm sorry for the wrong answer initially given; you might consider to unaccept this one.

I'm not sure whether this has ever been examined in detail. It is likely that nobody bothered to collect the (volatile) reaction products of hydrazine.

According to a procedure provided by the group of Rodney Ruoff (DOI), the reduction was performed in an open flask equipped with a reflux condenser. An aqueous suspension GO was first sonicated (to support exfoliation) until it became clear, then hydrazine hydrate was added and the mixture refluxed at 100 °C for 24 hours. During this time, the reduction product precipitated as a black solid.

No remarks on what actually happened to the hydrazine were given in this article.


How was water formed if oxygen and hydrogen exist as molecules?

Water does not have two oxygen atoms (has one) and one hydrogen (has two) hence the molecular formula of water is H2O. Water forms when there is enough energy to break the molecular bonds within the Hydrogen and Oxygen diatoms, they then combine and form water. As Mathew said, look at the Standard Gibbs free energy of formation. It will explain that H2O is much more favourable to be formed than H2O2.


Simplest way to convert glucose into water and carbon dioxide other than burning?

I'd suggest you read into the Fenton process. Using an iron catalyst with hydrogen peroxide will break down (potentially) any organic matter into $\ce{CO2}$ and $\ce{H2O}$ within minutes or hours. The only drawback from your requirements is the pH, since I think this is only down in acidic solution, but you could do a little research to find out more.


Can the alkyl-oxygen bond in lactones be cleaved in an SN2 reaction?

Yes, it is possible. If you consider the chemistry of a four-membered lactone (a β-lactone), it is able to act as an alkylating agent, as nucleophiles can attack the C–O bond in an SN2 fashion.

One example is in the synthesis of a series of β-substituted amino acids (J. Am. Chem. Soc. 1985, 107 (24), 7105–7109). An intramolecular Mitsunobu reaction of Cbz-protected serine 1 forms the β-lactone 2 which is then opened by a variety of nucleophiles. Here, benzylthiolate is used as an example, giving N-Cbz-S-benzylcysteine 3 as the product:

The authors note that:

  Only relatively “hard” nucleophiles like ammonia and methoxide attack the carbonyl to give acyl-oxygen cleavage. In earlier reports serine β-lactones were used mainly for synthesis of seryl peptides by attack of amines at the carbonyl. However, with ammonia as the nucleophile, simply altering the solvent [from MeCN to THF] inverts the 3:1 amide/amine ratio.

The following article has a wide selection of more examples (see references 6–8): J. Org. Chem. 2002, 67 (14), 4680–4683.


Why are DOM lithiations sometimes conducted at −78 °C and othertimes conducted at elevated temperatures?

The reason a given set of conditions used in any given reaction is likely fairly arbitrary unless you're following a paper where the conditions have been extensively optimised (for instance in an OPR&D paper where a process group has rigorously screened) - as has been discussed here before, −78 °C  is a convenient temperature to run reactions, so if it works at −78 °C, its unlikely anyone will ever bother to warm it up.

That said, in Clayden's Organolithiums: Selectivity for Synthesis (which is probably the most comprehensive reference of R–Li chemistry I can think of), a table of directing groups and representative conditions is presented. I've included the table below, but the actual text has a discussion of the various groups and a discussion of the representative conditions used. As a general guide, it seems that temperature is largely dictated by:

the activating/deactivating nature of the directing group
the sterics of the directing group
the stability of the directing group, with low temperatures required in the case of electrophilic groups like the carbamate to prevent nucleophilic addition that would occur at higher temperature

As a general point, as temperatures get higher, the stabilisation of the anion by the directing group becomes less important to the overall energy and as such at higher temperatures, one would (generally) expect selectivity to drop.

Summary of directing groups used in DoM reactions with typical reaction temperatures. Taken from Clayden, J. Organolithiums: Selectivity for Synthesis; Pergamon: Oxford, U.K., 2002; p 34.


Cis/trans Proline vs. Double bond character

The change from cis to trans (called isomerization) is not spontaneous. There are proteins called prolyl isomerases (see Wikipedia Prolyl Isomerase) that assist in the isomerization process. This happens when the protein is being folded into its final secondary and tertiary structures.


Why is osmium the densest known element?

The density of an element is related to how many atoms can be placed in a given volume and the weight of the nuclei.  Therefore, the smaller the atomic radius of an atom and the higher the atomic number of the nucleus, the greater the density of the element. The very small atomic radius of osmium results in a small metal-metal separation.  This small atomic separation along with osmium's relatively high atomic number gives rise to osmium's high density.

The small atomic radius can be attributed to the following factors:

f orbitals are very diffuse and therefore result in poor screening of the electrons further out.  In the case of osmium ($\ce{[Xe] 4f^{14} 5d^6 6s^2}$) the poor screening by its 4f orbitals leads to a contraction in the n=5 and n=6 orbitals.
Due to osmium's high atomic number, relativistic effects come into play.  These relativistic effects have been discussed previously on SE Chem for gold and mercury.  A general discussion as well as the mathematics describing the relativistic effect can be found in these earlier posts.  Basically, for heavy nuclei the s electrons must circulate at relativistic speeds in order to maintain a stable orbit (not fall into the nucleus).  Therefore the electron mass increases and the s orbital radius decreases (the p orbital radius also decreases, but to a lesser degree).

The orbital contraction caused by these two effects leads to a small atomic radius for osmium.  Therefore, short metal-metal bonds result.  This is reflected in the small unit cell volume for osmium (27.96 cubic angstroms).  For comparison the unit cell volume for lead is 121.3 cubic angstroms! As a result, more osmium atoms can pack into a given volume compared to other elements

Osmium's relatively high atomic number along with its small atomic radius, as explained above, gives rise to osmium's high density.


Bent's rule for Sulfur trioxide and Sulfur dioxide

Here are structures of dominant resonance structure (only one of 3 and 2 differencing only by rotation) for sulfur tri- and dioxide. Despite common misconception, d-orbitals apaprantely play very insignificant role in chemistry of p-block elements, so the structures include formal charges.

As can be seen, the average order of sulfur-oxigen bond lowers with move from di- to trioxide from $1.5$ to $1.(3)$. However, average formal charges grows from $+1$ to $+2$ on sulfur and $-0.5$ to $-0.(6)$,  on oxygen, thus greatly increasing ionic bonding (roughly from $1*0.5 = 0.5$ to $2*0.(6)=1.(3)$) Generally, ionic bonding greatly outperforms covalent bonding, and here charges grown considerably so bond shortening may be attributed to that. The point, however, is, that in this particular structure more than one effect is in play, and their effects outperform whatever small effects are considered when Bent's rule is applied.


Product formed under ethoxide/ethanol and heating?

You are correct in your prediction for this first reaction:

The second is simply an $\mathrm E2$ β-elimination reaction, which forms product I:


Is this a spontaneous or a non-spontaneous process?

Think like this, if the process occurs it is spontaneous.

Without the heating the process is not spontaneous because it does not occur.

With the warm up, you've modified the system. It is this system that is spontaneous (with heating).

Your last statement is right for the unheated system.

Notice, there are two distinct situations.


The half-reaction stops primarily because it's still producing/consuming electrons, and there's only so many free electrons an electrode (the local environment) can accept/supply.

Just like chemical species, electrons have an equivalent to concentration. (Whereas the generalization of the concentration is the "chemical potential" of the species, the "concentration" of electrons is related to the electrical potential or voltage). If the "concentration" of electrons gets too high or drops too low, the reaction rate of the half reaction will drop off - just like it does if the concentration of a chemical product/reactant gets too high/drops too low. In normal operation, the external conductor allows electrons to move, renormalizing the "concentration" of electrons and allowing the reaction to proceed further.

Note that if you have a poorly made half-cell, you may indeed have a pathway for the electrons to travel to the other half cell, or to react in some other (half-)reaction. If this occurs, then your battery will indeed continue to run even when disconnected, draining its stored chemical energy. But decently made batteries control for this, and the only way for the half cells to receive/dispose of electrons is through the external circuit.


Devoting sufficient time on Google didn't help at all.

Hmmm... I found this PDF in just a few seconds, where we read the following.

The origin of the name microcosmic salt.
Beads of sodium metaphosphate
saturated with some oxides cool to a clear glass. Small crystals grow
in the glass on repeated heating and cooling. A microscope with a 80
to 100 x magnification is needed to inspect these crystals with
various morphologies. Such beads resemble a microcosmos. Some authors
tried to use the effect for analytical purposes in the 19th century,
but these efforts failed. Borax beads sometimes show the same effect.

Microcosmos here is understood (as usual) as a "small world", a "world in miniature".


Decarboxylation of cyclohex-1-ene-1,2-dicarboxylic acid by electrolysis method

Turns out a similar reaction has been performed,   the reaction proceeds via the following mechanism

  SOURCE

The cyclohexyne dimerizes to 1,2,3,4,5,6,7,8-octahydrobiphenylene. Then further rearranges. So I would say that electrolysis is feasible with rearrangement as depicted in the picture .
 The second compound(1,2,3,4,5,6,7,8-octahydrobiphenylene) which you've drawn is more stable since very unstable compounds (cyclohexyne) do not form even more unstable compounds. The second compound will not be the final product/major product owing to its high instability due to anti aromaticity.

PS
 I haven't delved into the details regarding the instability of cyclohexyne and 1,2,3,4,5,6,7,8-octahydrobiphenylene since they have been well covered in Gaurang's answer


How to optimize the molecular geometry under PyMol?

Disclaimer: I don't typically use PyMol, so I only made a quick look at the Optimize.py script.

You're right that the problem is with the bond orders. The Optimize.py script uses a PDB file as an intermediate representation that it sends to Open Babel for force field optimization and other properties.

The problem is that PDB does not support bond orders. So yes, you don't get double-bond information back, and I can't verify if Open Babel is getting bond orders from PyMol and Optimize.py.

I looked through the PyMol commands and it seems as if there's a way to write an MDL Molfile (which would have bond orders) but not read one - PDB is the only option for reading and writing in a PyMol script.

My suggestions:

Either push the PyMol developers to support other formats (e.g., Sybyl .mol2 or MDL Mol) in the scripting commands.
Use another program like Avogadro to draw and optimize the geometry and export to read into PyMol later.

(Disclaimer: I'm an Avogadro and Open Babel developer, but there are many programs that could be used.)


Determination of solubility equilibrium using galvanic cell reactions

The leftmost and rightmost parts of the galvanic cell notation are supposed to be electrodes, not electrolytes (no KBr (aq) and the likes of it there). What is more, the double vertical line represents the interface where the junction potential has been eliminated ((c) Atkins), i.e. salt bridge, which in this case is the interface between electrolytes. The left hand side of the cell is anode, where the oxidation occurs, and the silver ions are reduced at the cathode, which is written on the right hand side. It seems to me that the correct notation is:

Ag (s) | AgBr (s) | KBr (aq) || AgNO3 (aq) | Ag (s)

In turn, the cell reaction is $\ce{Ag+ + Br- -> AgBr}$, and the EMF of this reaction is +0.704 V. It is not surprising that the equilibrium constant of this reaction is much larger than unity, otherwise silver bromide wouldn't be so stable and so insoluble. What you're missing is that the solubility product, which you want to calculate, is an equilibrium constant of the inverse process - not the formation, but the dissolution of $\ce{AgBr}$. So take the inverse of the equilibrium constant which you obtained from the galvanic cell EMF, and you'll get $7.9\cdot10^{-13}$, which is pretty close to the reference values.


Is there a metal that melts and solidifies quickly?

The melting points of some eutectic mixtures are tabulated in an article by French.1 Some options for low-melting point metals/mixtures are:

Gallium-indium eutectic, mp 15.7 °C (Sigma-Aldrich catalogue)
Galinstan, mp –19 °C (Wikipedia article)
Various gallium-based Indalloy® alloys, mp between 7.6 and 29.78 °C (corporate page)
A general search for "fusible alloys" brings up several sources with many choices:

Wikipedia page
Rotometals sales page
Belmont sales page
Bolton Metal sales page

Reference

French, S. J. Melting Points of Eutectics. Ind. Eng. Chem. 1936, 28 (1), 111–113. DOI: 10.1021/ie50313a026.


Calculate the concentration of the deprotonated adipic acid anion

I'll follow your lead and represent adipic acid as $\ce{H2A}$ for the sake of simplicity.

Equations we know:

(I) o.100 = $\ce{[H2A] + [HA^{-}] + [A^{2-}]}$

(II) $\dfrac{\ce{[H^+] [HA^{-}]}}{\ce{[H2A]}} = 3.9 * 10^{−5}$

(III) $\dfrac{\ce{[H^+] [A^{-2}]}}{\ce{[HA^{-}]}} = 3.9 * 10^{−6}$

(IV) $\ce{[H^+]} = \ce{[HA^{-}] + 2[A^{2-}]}$

There are four equations in four unknowns so the system does have a solution. I could find any way to simply all the equations into anything which seems reasonable for an exact solution.

The way to solve this problem is to solve for the first ionization and assume that the second ionization has an insignificant contribution. A check can be done with equation (IV) to verify that  $\ce{[HA^{-}] \gg 2[A^{2-}]}$.

I am going to show an excessive number of significant figures in the answers so that various methods of calculation can be compared. The acid constants are only given to 2 significant figures so that is the limiting number of significant figures that can justified.

Quick and dirty

Let $\ce{[H^+] = [HA^{-}]}$ and let $\ce{[HA]} = 0.100$

then from (II)

$\ce{[H^{+}]^2 = [HA^{-}]^2 =} \sqrt{(0.100)(3.9E-5)}$

$\ce{[H^{+}] = [HA^{-}] =} 1.97E-3$

thus $[\ce{HA}] \approx 0.1 - 0.00197 = 0.09803$

from equation (III)

(V) $\ce{[A^{-2}]} = \dfrac{\ce{[HA^{-}]}}{\ce{[H^{+}]}}3.9E-6 = 3.9E-6$

So indeed $\ce{[HA^{-}] \gg [A^{-2}]}$

This method can be made somewhat better by noting

                  x*(0.1/0.10197)
   HA    0.10000  0.09807
   HA-   0.00197  0.00193
         -------  -------
         0.10197  0.10000

Cleaner solution

The Quick and dirty is a bit dodgy since $\ce{[H2A]}$ was assigned a value of 0.100 to calculate, then reassigned a value of 0.09803. Since the pK constant is only given to two significant figures this isn't really wrong, but a more consistent approach can be easily calculated.

Let $\ce{[H^+] = [HA^{-}]} = x$ and let $\ce{[HA]} = 0.100 - x$, and also let's use $k_2$. Substituting into equation (III):

$x^2 = (0.100 -x)k_2$

$x^2 + k_2x -0.1k_2$

This is a quadratic in x so roots are:

$x = \dfrac{-k_2 \pm \sqrt{k_2^2 - 4(1)(-0.1k_2)}}{2(1)}$

The negative root doesn't work so the positive one is the only one. Also $k_2^2$ is so small that it can be neglected so;

$x = \dfrac{-k_2 + \sqrt{0.4k_2}}{2} =0.00186$

$\ce{[H^+] = [HA^{-}]} = 0.00186$

$\ce{[HA]} = 0.100 - 0.00186 = 0.09814 $

Again from equation (V)

$\ce{[A^{-2}]} = \dfrac{\ce{[HA^{-}]}}{\ce{[H^{+}]}}3.9E-6 = 3.9E-6$

Exact Solution

Solve equation (II) for $\ce{[HA^{-}]}$ in terms of $\ce{[H2A]}$ and $\ce{[H^+]}$ to get:

$\ce{[HA^{-}]} = \dfrac{k_1\ce{[H2A]}}{\ce{[H^+]}}$

and multiply (II) and (III) to solve for $\ce{[HA^{2-}]}$ in terms of $\ce{[H2A]}$ and $\ce{[H^+]}$ to get:

$\ce{[A^{2-}]} = \dfrac{k_1k_2\ce{[H2A]}}{\ce{[H^+]^2}}$

Both of the above can be substituted into the left hand side of the following three equations to get the values shown on the right.

$\dfrac{\ce{[H2A]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{[H^+]^2}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}}$

$\dfrac{\ce{[HA^{-}]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{k1[H^+]}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}} $

$\dfrac{\ce{[A^{2-}]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{k_1k_2[H^+]}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}} $

Being a glutton for punishment I created a spreadsheet with the calculations. After some manual manipulation I was able to determine the "exact solution" to a totally ridiculous number of decimal places.

$\ce{[\ce{H+}]} \text{ }= 0.00195932639309$

$\ce{[\ce{H2A}]}~~= 0.09804455814272$
$\ce{[\ce{HA-}]} ~= 0.00195155732146$
$\ce{[\ce{A^{2-}}]}~~= 0.00000388453582$
$\text{sum} \quad = 0.10000000000000$

$\ce{2[\ce{A^{2-}}] = [\ce{H+}] - [\ce{HA-}]} = 0.00000776907164$

$\ce{K_{a1}}$ from data = $3.9E-5$
$\ce{K_{a2}}$ from data = $3.9E-5$

$\ce{[H^+]} = \ce{[HA^{-}] + 2[A^{2-}]} = 0.00195932639309$


MO Diagram "greatest stabilisation when orbitals are close in energy"

Fundamentally it's mathematics. The energies of the MOs are the eigenvalues of the matrix $$\begin{pmatrix}\alpha_1 & \beta \\ \beta & \alpha_2 \end{pmatrix},$$ where $(\alpha_1,\alpha_2)$ are the energies of the two constituent orbitals and $\beta$ is (loosely speaking) the overlap between them.

For the same size of $\beta$, if $\alpha_1 \approx \alpha_2$, then the eigenvalues can differ quite a lot from $(\alpha_1,\alpha_2)$. On the other hand, if $\alpha_1$ is very different from $\alpha_2$, then the eigenvalues will be closer to $(\alpha_1,\alpha_2)$.

As an illustration consider the matrix

$$\begin{pmatrix}1.1 & 0.5 \\ 0.5 & 1 \end{pmatrix}$$

This has eigenvalues of $1.55$ and $0.55$, which are relatively distant from the "original energies" of $1.1$ and $1$. On the other hand, the matrix

$$\begin{pmatrix}3 & 0.5 \\ 0.5 & 1 \end{pmatrix}$$

has eigenvalues $3.12$ and $0.88$, which are closer to $3$ and $1$.

Taking a more abstract perspective, the eigenvalues of $$\begin{pmatrix}\alpha_1 & \beta \\ \beta & \alpha_2 \end{pmatrix}$$ are

$$\lambda_\pm = \frac{\alpha_1 + \alpha_2}{2} \pm \frac{\sqrt{(\alpha_1 - \alpha_2)^2 + 4\beta^2}}{2}$$

and you can play around with this expression to gain some insight. For example, in the limit where  $(\alpha_1 - \alpha_2)^2 \gg 4\beta^2$ (corresponding to a large energy difference between the original interacting orbitals), the eigenvalues reduce to

$$\lambda_\pm \to \frac{\alpha_1 + \alpha_2}{2} \pm \frac{\alpha_1 - \alpha_2}{2}$$

which are simply $\alpha_1$ and $\alpha_2$.


Work done by system of ideal gas in isobaric expansion

At the interface between the gas and its surroundings, by Newton's 3rd law, the pressure exerted by the gas on its surroundings is equal to the pressure exerted by the surroundings on the gas.  So we can use either.  For a reversible expansion, the gas pressure is given by the ideal gas law.  But the ideal gas law only applies at thermodynamic equilibrium and during a reversible process (which consists of a continuous sequence of thermodynamics equilibrium states).  For an irreversible expansion, the ideal gas law doesn't work (i.e., gives the wrong answer), and we can't use it to get the gas pressure at its interface with the surroundings (where the work is being done).  But, if we know the surroundings pressure during the irreversible expansion, we can still calculate the work (because, at the interface between the gas and surroundings), the two will match.  So using the external pressure is always a valid approach to getting the work.


What would be the product of following elimination reaction?

The first thing you notice is an alcoholic base reacting with a molecule. The base looks for the most acidic hydrogen in the molecule. Since the no.of moles of $alc.KOH$ is not mentioned, we go for both the acidic hydrogens at the adjacent positions to the carbonyl functional group. Once the acidic hydrogens are abstracted, it's a simple elimination of $Cl$ that will happen.

I agree with the comment made by @Waylander as there are plenty of examples with three-membered rings. Well, (c) is not possible but you can say (a) is a possibility. Of course (a) forms too by a simple elimination reaction but the most probable one is the abstraction of acidic hydrogen in the molecule. The stability of [3]rings can be brought about by banana bonds.


How to calculate entropy change for a gas reaction using $S^\circ$?

Just to make sure that I get your question correctly; the process in question is a chemical reaction that takes place at a temperature and pressure close enough to the reference temperature and pressure (those used for the tabulated values) that any changes to the tabulated values are neglible? (I assume this, as this is what you have done in the enthalpy calculation)

In that case, it is probably easier that you would expect - you can use (almost) the same approach as you used to calculate $\Delta H$ as both entropy and enthalpy are state functions. That is:

$\Delta_{rx} S = \sum_i \nu_i S^\circ_i$

where $S^\circ_i$ is the tabulated value for each of the components. This is described on Wikipedia.

If the temperature deviates sufficiently from the reference temperature used then you should adjust the entropy change accordingly

$\Delta S_{T_1 \rightarrow T_2} = \int_{T_1}^{T_2}\frac{C_V}{T}dT=C_V\ln(T_2/T_1)$

where the last equality holds if $C_V$ can be assumed to stay constant in the interval $T_1 \rightarrow T_2$. A similar adjustment can be made if the pressure, volume etc. changes during the course of the reaction. Again, as the entropy is a state function, it is enough to adjust the total change in entropy for the reaction.

Hope this helps! Please let me know if I am not being sufficiently clear, and I'll try to elaborate things a bit.


Database with organic reaction mechanisms (curly arrows)

As far as free online resources go, I've found the following useful:

Organic Reactions (organic-chemistry.org)
Name-reaction.com
List of organic reactions (en.wikipedia.org)

I would also strongly recommend acquiring a copy of March's Advanced Organic Chemistry at some point. I think it's fair to say that it's widely regarded as one of the most essential reference works in the field of organic chemistry. It's concise, more comprehensive than one would've thought possible, beautifully organized, and contains thousands of primary literature references. While not every reaction is presented with an accompanying mechanism, many are, and useful exposition is typically provided.

Another book you might find useful is Strategic Applications of Named Reactions in Organic Synthesis. More than a mere reference text, it includes an overview describing important applications for the reactions it covers, examples of uses in practical syntheses, as well as comprehensive stepwise mechanisms.


Does pyrite retain the magnetic properties of its element iron?

Pyrite is not magnetic. Some related minerals are, but only weakly, not as strong as iron, so the test will likely fail anyway. Magnetism is not an inherent property of an element, but that of a compound. Most compounds of iron are not magnetic (though some are); on the other hand, there are organic magnets made out of totally non-magnetic elements.

Think of something based on chemical properties instead. Pyrite is a salt of a metal and a weak acid. Gold is a metal, of the kind that would not react with acids. See where this gets you.

As for the other properties of the elements, some would be inherited by their compounds and some (arguably, most of them) would not.


Under $E_2$ reaction conditions, the MeO$^-$ product will be

while the Me$_3$CO$^-$ product will be

.

This is due to the bulkiness of Me$_3$CO$^-$ being unable to reach the hydrogen at the substituted carbon so instead it goes to the primary position instead.


Well, you are right, the synthesis of your second compound is described in DOI:10.1021/ja038384r:


Software for technical calculation of chemical reactors

Years ago a colleague of mine mentioned ASPEN as a typical software (bundle) used to model chemical plants.  Despite this and this lack of examples how the print-out will look like -- and I did not work with it, either.


Indirectly Measuring Diffusion of NO2 into Water

NO2 is a free stable radical, better express as $\ce{.NO2}$, and being a radical, does not directly react with H2O, or H+ or OH-.

The reaction mechanism likely proceed via a self-reaction creating nitrogen tetroxide:

$\ce{.NO2 + .NO2 <=> N2O4 }$

In a diluting medium of air, depending on the concentration of nitrogen dioxide, this reaction is kinetically expected to be slow.

When it does occur, it can be followed by:

$\ce{N2O4 +  H2O -> HNO2 + HNO3 }$

Source: See Reactions (9) and (10) on Page 28 of this paper.

Now, one can ignore the intermediate reaction mechanics, for simplicity, however, this does present a problem for some, as yourselves, as clearly the rate determining step is not even cited.

Also, if you review the paper, there are other reaction equilibriums that you may have to take into consideration. As a sidebar, what the paper does not mention is that aqueous nitrite is a photo-sensitive producing products, so you may wish to control the amount of lab light exposure as well.


The general approach to solve acido-basic equilibrium problems is:

respecting charge balance and charge conservation
respecting mass conservation
respecting dissociation equilibrium constants for acids, bases and water itself.

Generally, it leads to a set of nonlinear equations for a set of independent variables. The application of a substitution method usually results to a polynomial equation for $\ce{[H+]}$ of the second up to the forth order. The latter is hard to solve analytically.
Fortunately, in many scenarios, there can be made justified simplifications, like

neglecting water autodissociation
neglecting concentration of one substance form wrt the other one or their sum.

That largely simplifies an equation to be solved.
E.g. to calculate $\mathrm{pH}$ of solution of acetic acid, ammonia, or sodium carbonate,
we can consider none of $\ce{H+}$, resp. $\ce{OH-}$ are created by the water autoionization and just negligible part of the compound underwent the acido-basic reaction.
E.g. for sodium acetate hydrolysis
$$K_\mathrm{b}=\frac{\ce{[HA][OH-]}}{\ce{[A-]}}\approx
\frac{{\ce{[OH-]}}^2}{\ce{[A-]}}\approx \frac{{\ce{[OH-]}}^2}{c_0}
 $$


What is the total number of isomers, considering both structural and stereoisomers of cyclic ethers with the molecular formula C4H8O?

I would not include the last structure, even if it was clever of you to think about it.

I think that the term "cyclic ether" restricts the scope to "cyclic compounds which have an ether moiety inside the ring". Your last compound would be considered as a linear ether with an alkyl and a cycloalkyl substituents.

Oh and your 6th structure does not have 3 isomers but just 2 because the (R,R) an the (S,S) isomers are actually equivalent by rotation of the molecule. This compound is said to be "meso".


Can I deduct heteronuclear van der waals parameters from homonuclear parameters in a forcefield?

Found the answer myself. In case someone else stumbles over this, the combining rules for Lennard-Jones parameters of van der Waals interactions are:

Distance parameter:
$r_{XY}=\frac{1}{2}(r_{XX}+r_{YY})$ or $\sigma_{XY}=\frac{1}{2}(\sigma_{XX}+\sigma_{YY})$

Depth parameter:
$\varepsilon_{XY}=\sqrt{\varepsilon_{XX}\varepsilon_{YY}}$


Which of these reduction reagents would me most appropriate to yield this product?

1) As you say*, the ketone must be protected.  Protect it. Any propagating naughtiness is irrelevant to the next step.
2) Reduce the ester. Beat on it.
3) Deprotect the ketone.

*Note that you begin with an alpha-ketoester.  In the old Morrison and Boyd that was a French joke.  LAH would have the same problem.  How the mighty have fallen.


Why are acids and bases are defined as both donors and acceptors of species in the acid-base theories?

These are primarily old theories that were kind of evidence or observation based. These are not completely correct theories as there are may loop holes in them. For eg. many types of compounds with significant acidic and basic characters cannot be predicted by these theories. Also, to a certain extent the Bronsted and Lewis concepts are complementary.Wikipedia says - ""

  A Lewis base is often a Brønsted–Lowry base as it can donate a pair of electrons to H+; the proton is a Lewis acid as it can accept a pair of electrons. The conjugate base of a Brønsted–Lowry acid is also a Lewis base as loss of H+ from the acid leaves those electrons which were used for the A—H bond as a lone pair on the conjugate base.

and

  In 1923, Lewis wrote An acid substance is one which can employ an electron lone pair from another molecule in completing the stable group of one of its own atoms. The Brønsted–Lowry acid–base theory was published in the same year. The two theories are distinct but complementary.

Basically, proton is a $H^+$ ion and we all know from our simple understanding that compounds that give $H^+$ or proton are acids and those that accept $H^+$ or proton are bases.


Solution based measurement of Solvent-Accessible Surface Area of macromolecules

As you noted, the solvent-accessible surface area is a geometric concept, and as such is most easily calculated using computational geometry tools from a given molecular (or crystalline) structure. However you obtain this atomistic structure (X-ray, neutrons, molecular simulation, NMR, …) is of course of significant importance to the resulting area, but the calculation method will be the same.

There are, however, a few studies which have attempted to link the solvent-accessible surface area to directly measurable properties, and in particular to spectroscopic properties. A quick search of the literature reveals three examples:

Amide H/2H Exchange MALDI-TOF Mass Spectrometry
UV Raman directly on the exposed residues
Using diazirine as a probe


Can plastic be used as fuel for vehicles?

There are many ways to recycle different types of plastic, and conversion to liquid fuels suitable for internal combustion engines is currently done on an industrial scale. The key to converting the carbon contained in long-chain polymers that make up the plastics is called pyrolysis, which is essentially a way to thermally decompose these products into shorter chain hydrocarbons suitable for use as liquid (or even gaseous or solid) fuel.  According to this Wikipedia article:

Anhydrous pyrolysis can also be used to produce liquid fuel similar to diesel from plastic waste, with a higher cetane value and lower sulfur content than traditional diesel.

An example of a current commercial waste-plastic-to-fuel operation is the Canadian corportion Plastic2Oil, Inc., who claims to "convert waste plastic to ultra clean oil" in their 250,000 gallon fuel production and blending facility.
The following is an excerpt from the description of the patent-pending process used by Plastic2Oil:

The processor accepts unwashed, unsorted waste plastics, composites and commingled materials. Although many sources of feedstock are available, we are focusing initially on post-commercial and industrial sources, since these are readily available in large supply, and present a cost-effective solution for companies. We currently have several companies providing plastic feedstock.
The hopper is loaded with up to 4,000 lbs. of plastic feedstock per machine per hour using a forklift. The plastic is loaded into the pre-melt reactor by a continuous conveyor between the hopper and the reactor. The plastic is then heated using the off-gases produced by the process. After the plastic has been liquified in the pre-melt reactor it passes through a solids-liquids separator before going into the main reactor.
In the main reactor, the liquified plastic hydrocarbons are cracked into various shorter hydrocarbon chains and exit in a gaseous state. Plastic2Oil's proprietary catalyst and unique process engineering enables us to capture an average of 86% of the hydrocarbon content of plastic. The Petcoke residue produced at this stage (approx. 2-4%) remains in the processor chamber and is automatically removed while the processor is in operation.
From the main reactor, the gases that are fuel oil and diesel are condensed and separated, then proceed into temporary fuel tanks. All of the gaseous "light fractions" or off-gases (approx. 10-12% of process output), such as methane, ethane, butane and propane, exit the system and are used to fuel the furnaces.

Note: I have no affiliation whatsoever with Plastic2Oil or their industry. I chose their company as an example largely at random, or more accurately, due to their very usable and easily found website.


The good people at the American Academy of Pediatric Dentistry have a number of useful resources available to shed light on questions such as yours.

I direct you to a manuscript entitled Microbial contamination of toothbrushes and their decontamination in which it is written:

  Storage conditions of toothbrushes are an important factor
  for bacterial survival. Dayoub et al. and Meier et al. reported
  that the number of microorganisms in the toothbrushes kept
  in aerated conditions was lower than in toothbrushes stored in plastic bags. Several authors have reported that bacterial contamination can be reduced by washing toothbrushes after use, and drying in aerated conditions. Caudry et al. reported that a wet environment increases bacterial growth and cross contamination. Therefore, as time increases between one toothbrushing and another, more microorganism development can occur in the toothbrushes a wet/moisture environment.

The references cited in the above passage are detailed in the linked PDF. Happy brushing!


What are the molecular requirements for condensation polymerization?

Monomers must contain a functional group at each side (at least bifunctional) and these functions  have to react with each other (like amine and carboxylic acid to yield amide group or alcohol and carboxylic acid to yield ester group,...) .

  We can start with two types of monomers that hold identical functional groups at each side.

One example of that is the reaction of 1,6-diaminohexane with hexandioic acid to yield Nylon-6,6.

  We can also start from one type of monomers with two different functional groups at each side.

One example of that is the reaction of 11-aminoundecanoic acid to yield Nylon-11.

The condition that the reaction should give off a small molecule is no longer valid. The counter example is the formation of polyurethane by reaction of diisocyanates and diols without giving off  small molecules. We prefer name this type of polymerization "a step growth polymerization" in opposite to "chain growth polymerization" (known before as polyaddition).


What is the difference between an industrial and a laboratory production method?

This answer will be separate in few parts. In the first one I will just make a review of the scope between lab an industrial scale. In the other parts, you’ll find some things more deeply explained and exemplified. In any case, if something isn’t clear or the English grammar is wrong, please ask me a question and point it (the English mistake) out.

Part I - Short Overview

A laboratory method can perfectly be adapted at industrial scale however there are many things to take in consideration.

Let say you are working in a lab to perform the synthesis of a molecule. In the basics we will assume you will need a reactor (round-bottom flask), and a distillation column to separate your compound from the others (solvent, remaining reactants, possible side products).

If you do all that stuff at a laboratory scale, even if your reaction is super exothermic even involving very hazardous chemicals, you'll always have the possibility to do it safely or at least having the possibility to stop the reaction by cooling it down rapidly.

I would mention at this point that the ratio between the volume and the surface of an object isn't constant. For example, if you multiply by 10 the length of the sides of a cube, its volume rises by a factor 1000 whereas its surface by a factor 100. You can find more interesting information here.

So, if you make your reaction at laboratory scale, let say under few litre, even if it might be dangerous, you’ll control easily and readily the temperature inside your flask or small reactor.  So, at industrial scale, let say over 1 cube meter (in general I would say 5 to 50 cube meters is the average, but some reactors are bigger than 1000 cube meter), you must take in consideration safety and security details as well as the sizing of your operation. I recommend you to read here a really nice overview about most of the kinds of the reactors used in chemical plants with many details related to them.

Now I bet you start to understand the big difference. But let me go a bit deeper on that part. If you want to size a chemical operation, you must first take in consideration the hazards of all of your chemicals as well as their fundamental properties (ebullition temperature, state at a given temperature (gas, liquid, solid, supercritical fluid ,  …), the possible interactions between to compounds or more.

After that you’ll need to know about the kinetics of your reaction, is it fast or slow?, you’ll also need to know about the thermodynamics of all the chemicals to make a prediction for safety (see the next part) and to complete you’ll have to take in consideration if you have a system with only liquids, with gas and liquids or even harder with gas, liquid and solid (hydrogenation (using hydrogen gas) of a liquid organic molecule (let say an oil) catalysed by a given catalysor (a solid) (see the third part).

For years now engineers and researchers, developed methods to compute thermodynamic and kinetic properties as well as many different methods for sizing many different basic or very complex operation (distillation, crystallization, multiphase reactors, drying, precipitation, …) (and I’ll give when I’ll have time as many details as I can in a fourth part about that point). So now it exits a lot of extrapolations and models made from the experiments because of the too complex existing situations to make nice calculus involving mathematics and physics learned at school. This article gives you an idea of all the possibilities of having nightmare while sizing an operation at industrial scale.

Part II - Predicting the safety of a reaction

The Semenov diagrams

So, to give some concrete applications of what I said before, I will show you a very basic thing that safety engineers will do to know more about the danger of the reaction. They will use a Semenov-diagram (or plot).
The first idea is to consider your reaction having a zero-order kinetics, i.e. the kinetics doesn’t involve the concentration of the reactants. It’s a big assumption but it gives you really a good idea of how dangerous it can be. The second idea is, as your reactor will be quite big, to consider it as adiabatic. It’s not true at all of course because you wouldn’t be able to heat or cool it, but it allows you to know what could happen in the worst case.
Doing an energy balance on your system you’ll get:
$$Q_{acc} = k_0 \cdot C_0 \cdot \exp\left(\frac{-E_a}{RT}\right)\cdot \left(-\Delta_r H\right)-UA\cdot (T_r-T_c)$$
Where $Q$ represent the amount of heat generated by the reaction, $k_0$ is the pre-exponential factor of the Arrhenius’s law, $C$ is the concentration, $E_a$ is the activation energy, $R$ is the constant of ideal gases, $U$ is the global exchange coefficient of your reactor and $A$ the area of heat exchange.
Now if you plot this equation which tells you the amount of energy liberate by the whole process at a given temperature with the plot only $UA\cdot (T_r-T_c)$, you’ll get something like that,

reference

If you consider the blue dotted line crossing the red plot, the blue curve tells you about how much heat you can evacuate and the red one tells you about what the whole process liberates. The point A is the stable point as even if you go under in temperature your reaction will still stay in control, however the B point is the point from which you have instability. It’s clear to understand than between these two points you can control the temperature of your reactor.

Keep in mind the heat you can evacuate is a linear function of the temperature $UA\cdot (T_r-T_c)$ and the heat produced by the whole operation is exponential with the temperature as shown by Arrhenius.

Predict the dangerous nature of a reaction

Another thing you can do easily is to predict the kind of danger of your reaction (not dangerous, a little dangerous, dangerous, extremely dangerous) using the worst case (adiabatic reactor).

Then you have rises of the temperature equal to the amount of heat produced by the reaction. Then you have,
$$C_p \cdot \Delta T_{adiab} = [reactants]_0 \cdot V \cdot (- \Delta_r H)$$
Where $C_p$ is the heat capacity of your mixture, $[A]_0$ means the concentration of A at the beginning (you want to know what happens if all of it reacts), and V is the volume of the mixture.

You can then predict the variation of temperature knowing all the needed parameters. I would say if the rising of temperature is below 50 °C your reaction is quite safe; if you find something more than 100 °C you’ll have to find either another solution (a different reactor, …) or add a lot of security systems (emergency open valves, a possibility to almost instantly flush your reactor content in a inhibitor mixture under the reactor, …).

Part III - What if I have a multiphase system?

Well this is the case most of the time. It’s not common to have a system in which you only have soluble fluids (except for gases). If you are performing a treatment of your affluents or your effluents (because you can’t use row compounds and you can’t throw away your potential toxic, corrosive, … mixture in the nature) you will often have an exchange between a gas and a liquid. If you are doing a reaction involving a catalysor then you will probably have three phases if it’s a hydrogenation or a fluidized-bed reactor for a cracking operation or something else. You’ll have only one phase for vapocracking, polymerization (not in emulsion and relatives) and urea production for example.

Now in these cases you have to take in consideration of possibility of having an exchange of chemicals (the ones you want at least) between the two phases or three phases. If you have a liquid/gas system, the situation is common and even if a bit hard sometime you can get good approximation and correlation using the two films theory see that I wasn’t really confident with it some time ago.

If you use a catalysor, you have to predict the possibility of the species to go inside it so that they can react otherwise nothing will happen. Then you have to understand how your catalyst works and how to avoid the possibility of blocking it or poisoning it [ref] as in most of the cases catalysors are like sponge solids. I found two articles pdf1 and pdf2 if you want to got deeper on how complex it is.

So I hope this very long answer help you more than you need. I will also use it as a reference for some more questions in the future. You may say that all these parameters can also be taken in consideration even at a lab scale. I will anwser yes in the sense that you will never have all the data you want and you'll have to do experiment to extrapolate back to the industrial scale. But I will moderate that if you are only doing stuff in a lab scale it doesn't cost the same as when you are involving tons of chemicals and it's way more secure so even if you are wrong in something you can afford it much more easily.

  I will point out that in most of the cases the chemical plants are
  sized as the lab plants a company may have meaning that the lab plants
  are made by extrapolating down the data (size of the reactors, ratios
  of length especially, ...) so that when you do your experiment in the
  lab you can then extrapolate back using your own models and then be
  more close to the reality with really tiny variations.