Deciding the order of heat of combustion of isomeric alkanes

Branched-chain alkanes have lower values of ΔcH⊖ than straight-chain alkanes of the same number of carbon atoms, and so can be seen to be somewhat more stable. https://en.wikipedia.org/wiki/Alkane#Branched_alkanes

The heats of formation are given in https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation  and go from -40.0 kcal/mol for a to -41.8 for b to -42.5 for c to -44.5 for d. Compound d has the largest heat of formation, therefore will have the smallest heat of combustion.

The explanation is complicated. Sometimes you have to wonder why teachers spend so much time on effects that are so minor. The data are interesting, in a trivial sort of way. Show the data; it's interesting; move on.

Highly branched alkanes may stretch the bond angle. This can cause steric hindrance, and can substantially increase the reactivity. "Substantially". That's interesting. In addition, the reactivity of highly branched alkanes can be used to increase the octane rating of gasoline, because the stability of highly branched free radicals slows down the combustion and avoids explosions in the cylinder (pinging). Substantial is the key word, whether it it is reactivity or stability. The difference between 40 and 44.5 is hardly substantial. I would call that difference about half of substantial.

Wikipedia goes on: "However, in general and perhaps surprisingly, when branching is not extensive, branched alkanes are actually more thermodynamically stable than their linear (or less branched) isomers. For example, the highly branched 2,2,3,3-tetramethylbutane is about 1.9 kcal/mol more stable than its linear isomer, n-octane. Due to the subtlety of this effect, the exact reasons for this rule have been vigorously debated in the chemical literature and is yet unsettled. Several explanations, including stabilization of branched alkanes by electron correlation, destabilization of linear alkanes by steric repulsion, stabilization by neutral hyperconjugation, and/or electrostatic effects have been advanced as possibilities. The controversy is related to the question of whether the traditional explanation of hyperconjugation is the primary factor governing the stability of alkyl radicals." Surprisingly subtle and unsettled. Not really a great teaching topic.

The question regularly pops up in secondary school exams. Why? The reasoning seems unsettled. How to answer the question? The teacher must have given the answer if the teacher is at all competent and meant to provide the reasoning; so regurgitate the correct answer without reserving a special place in your memory for it. If the teacher did not provide the answer, then you must do your own research, in advance of the test - first, to discover what kind of questions your teacher is likely to ask (can you search out previous year's tests?) and second, to research the possible answers.

If you do well at this, you may be asked to teach the next class.


How is the ln k term in the linearised Arrhenius equation acceptable?

The reason is because of sloppy notation. This sort of logarithmic plot is widespread in chemistry, though, and isn't limited to the Arrhenius equation, so do not expect it to change very much in the near future. You will have to get used to it.

With the Arrhenius equation

$$k = A \exp\left(-\frac{E_\mathrm a}{RT}\right)$$

you cannot directly take logarithms on both sides, because of the inherent units in $k$ and $A$. First you would have to divide through by a constant which we could call $k^\circ$:

$$\frac{k}{k^\circ} = \frac{A}{k^\circ} \, \exp\left(-\frac{E_\mathrm a}{RT}\right)$$

Now we can write

$$\ln \left( \frac{k}{k^\circ} \right) = \ln \left( \frac{A}{k^\circ} \right) - \frac{E_\mathrm a}{RT}$$

and as long as $k$, $k^\circ$, and $A$ are expressed in the same units, the resulting arguments of the logarithms will not have units. Note that the actual value of $k^\circ$ does not matter: you will of course get a different y-intercept, but the final values of $A$ and $E_\mathrm a$ that you get out of the Arrhenius plot will not be affected.

The division by $k^\circ$ is often implicit. When people write $\ln k$, what they usually mean is $\ln(k/k^\circ)$ where $k^\circ$ has the numerical value of $1$ and the units of whatever $k$ is in. Once you have the y-intercept, you take the exponential of that and tack the units back on to get $A$.

Although it is not rigorous, from a practical standpoint, this generally does not lead to much confusion as long as the author is consistent with their usage of units. Another reason for omission is that the actual value of $k^\circ$ has no physical effect, as previously discussed, and an overly pedantic focus on writing $k/k^\circ$ and $A/k^\circ$ can distract one from the chemical principles behind these equations, especially for more complicated equations which have many more symbols.


Is a discharged battery in a state of chemical equilibrium?

Post-script: After recently encountering a related post I re-read my answer here and realized I could have provided a much clearer and more useful definition of equilibrium (rather than my somewhat philosophical earlier answer) that encompasses the different cases you present: in thermodynamics, a system is in an equilibrium state when no lower free energy states are accessible.

Therefore, the two beakers, just like the exhausted battery, are at equilibrium, because there are no accessible lower energy states under the defined conditions.

You can ask yourself: if the universe is expanding (that is, if it is in a dynamic state), is thermodynamics truly ever applicable? The answer is that in science we need to take approximations, during measurement or when extracting information from data. So it is in thermodynamics, which works well enough even though on long enough scales the approximations fail (on timescales of the "in the long run we are all dead" sort).

This is of course a tricky point. We often need to distinguish between "real equilibrium states" and "kinetically trapped or metastable states". However it is useful to invoke a "duck-typing" rule applicable to thermodynamics problems: if the measured properties of the system remain constant and homogeneous during the interval of the experiment, then the system can be considered to be at equilibrium during that interval. This definition is particularly valuable since thermodynamics is an empirical science. Caveat: that doesn't mean you can be sure that the system is in the state you meant it to be in (!), since if you wait long enough you might end up in another state (which might be the one you actually wanted the system to be in!). Perhaps more important is reproducibility: can you put the system into the desired state, reproducibly?

To summarize: the concept of equilibrium, like that of reversibility, is subject to the practical condition that no variations be observed across the system or over time during the course of the experiment, when the system is in the proposed state.

As regards a battery, if you don't see a change in the potential, you can regard the system as being "at equilibrium", at least temporarily! Again, the definition of equilibrium is not as strict as we'd like to think. The materials in the battery, left exposed to the environment long enough, will rust and fall apart, leading eventually to  other "equilibrium" states.


Can separated and purified substances be obtained from a chromatogram?

Generally, all chromatography is done on silica gel. It's the exact same procedure for TLC (thin layer chromatography), except that instead of paper you have a thin layer of glass covered with silica. What can be done in that case is scraping the spots, then extracting the product with organic solvents and filtration. There exists plates made for this, which are thicker than just for analysis.

Of course this is not viable for large amounts of products. In that case, we use a column of silica (see the image below, taken from https://bitesizebio.com/29947/basics-chromatography-column/).

This time, gravity makes the solvent go through the column and come out at the other end. Compressed air is often used to make this process go much faster.

You will however think "how do you reveal the products IN the column?" As the products aren't always colored, you can't really tell when what you want is coming out. Therefore, everything is collected in tubes of the appropriate size and revealed afterwards. If everything goes right, there is no overlap between your products and you get everything out of the column. If there is an overlap, you have to either do the column again for the mixed fractions or just discard those and keep the tubes containing only the pure product you want.

This technique is used pretty much all the time, as it can separate anything (as long as it doesn't degrade on silica) and can be adjusted by changing which solvent or mix of solvents you use.


Why does chlorine oxidise iron to iron(III), not iron(II)?

The enthalpy of formation ($\Delta H_\mathrm{f}^\circ$) of $\ce{FeCl3}$ is $\pu{-399.49 kJ mol-1}$, while the $\Delta H_\mathrm{f}^\circ$ of $\ce{FeCl2}$ is $\pu{-341.79 kJ mol-1}$. This means that $\ce{FeCl3}$ is $\pu{57.7 kJ mol-1}$ more stable than $\ce{FeCl2}$, a considerable amount. This means that it is more thermodynamically favorable for $\ce{FeCl3}$ to form than $\ce{FeCl2}$, likely due to the larger lattice energy.

Furthermore, in the +2 oxidation state, one electron remains paired in the $\mathrm{3d}$ orbital. When $\ce{Fe}$ is in the +3 oxidation state, however, it has a half filled $\mathrm{3d}$ orbital, a state which is known to be particularly stable, which you can read about further here.


What do '(6S)' and '5' mean in (6S)-5-formyltetrahydrofolate?

I've redrawn a truncated version of the title molecule below with the "5-formyl" group in place.  I've also inserted small numbers so you can see how the base system is numbered.  So "5-formyl" means that a formyl group ($\ce{-CHO}$) is attached to the "5" position.  The 6-position has 4 different substituents attached to it so it is a chiral carbon atom.  The R-S nomenclature system is used to describe the 2 mirror image possibilities at chiral centers.  The enantiomer you've drawn is the "6S" stereoisomer.


Why does the energy level of a particle in a box decrease with length?

In quantum mechanics, particles are waves, and as with any wave the longer the wavelength the less energy is in the wave.  You don't need the box to see this; infrared light has less energy in the photon than ultraviolet (the former makes you warm but the latter can damage your skin) and a slow moving free electron (lower energy) has a shorter associated wavelength than a fast-moving one (higher energy).  In the particle/box problem, the possible wavelengths are set by the dimensions of the box; a bigger box allows longer wavelengths -> lower energies.


Which it is the oxidation state of indium and phosphorus?

I've estimated net atomic charges in rocksalt type InP by simple HF calculation in CRYSTAL06 in random basis.
Lattice parameter of 5.54 A is taken from:

K Kabita et al. First principle study on pressure-induced electronic structure and elastic properties of indium phosphide (InP). Indian J. Phys. (2015), DOI 10.1007/s12648-015-0701-0.

Results: http://pastebin.com/zdb8js8Z

Total atomic charges are 47.2861139, 16.7138861, so partial charges are:

In: +1.71
 P: -1.71

Given that oxidation state is an hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, the answer is probably better estimated as +2/-2, not 1.


Reaction of hydrazine and hydroxylamines with carboxylic acids

For simplicity, I will consider the reaction of acetic acid with ammonia, because the same principles apply. The two alternatives are:
$$
\underbrace{\ce{CH3C(O)NH2}}_{\mathrm{amide}} + \ce{H2O} \leftarrow \ce{CH3C(O)OH} + \ce{NH3}\rightarrow \underbrace{\ce{CH3C(NH)OH}}_{\mathrm{imidic \ acid}} + \ce{H2O}
$$
where the group connected via a double bond has been set in parentheses. The two products are connected by protonation/deprotonation in a tautomeric equilibrium, so the question boils down to thermodynamic stability. For most cases, we expect the C-O double bond/C-N single bond combination to be more stable than the C-N double bond/C-O single bond combination.


How to write the last cell reaction for the following cell?

The species do cancel out. However, the concentrations of the species do not. It is a bit of a matter of preference as to how you express this. The way I would write it is:

$$\ce{Zn^{2+}_{(aq)} ($c = 0.01\text{ M}$) + Zn_{(s)} <=> Zn^{2+}_{(aq)} ($c = 0.1\text{ M}$) + Zn_{(s)}}$$

for which $E = E^0 - \frac{RT}{2F}\,\ln\!\left(\frac{0.1}{0.01}\right) = -0.030 \text{ V}$ at 298 K, since $E^0 = 0 \text{ V}$. A negative value of $E$ agrees with the reverse reaction above being spontaneous, which is of course what you would expect since $0.1 > 0.01$.


What is distorted square pyramidal structure of a compound?

Going by a standard argumentation, the $\ce{[XeF5]+}$ cation would be expected to have two different bond types: a classical 2-electron-2-centre bond to the axial fluorine and a pair of two 4-electron-3-centre bonds making up the base of the pyramid with the equatorial fluorines. This structure is shown in figure 1.

Figure 1: Hypothetical perfect structure of $\ce{[XeF5]+}$. Image taken from Wikipedia, where a full list of authors is available.

The actual structure of $\ce{[XeF5]+}$, reported by Leary et al.,[1] is not perfect, though. The axial fluorine atom does not sit at the centre of the pyramid, destroying the perfect $C_\mathrm{4v}$ point group. This is probably an effect of core orbitals on the central xenon atom (I cannot access the paper where I am right now). Because the pyramid no longer has four equal faces, it is distorted.

Reference:
K. Leary, D. H. Templeton, A. Zalkin, N. Bartlett, Inorg. Chem. 1973, 12, 1726. DOI: 10.1021/ic50126a004.


Calculating ΔG at the extremes of reaction extent

After a lot of help, I have the following to suggest as an answer:

Imagine a reaction with $\Delta_\mathrm{r} G^{\circ} = -1000Jmol^{-1}$ at 298K. Using the following equations: $$\Delta_\mathrm{r} G^{\circ} = - RT\ln K$$ $$e^{-\frac{\Delta_\mathrm{r} G^{\circ}}{RT}} = K$$
This would give a K value of 1.50, which indicates the Q value on the curve where G is at a minimum (favoring products slightly in the equilibrium mixture). Using another equation (valid e.g. for a reaction $\ce{A(aq) <=> B(aq)}$): $$ξ = \frac{Q}{1+Q}$$

Gives the ξ value (1.50/2.50 = 0.600) which is the x coordinate of the minimum on the graph, where the x axis runs from ξ = 0 (pure reactants) to ξ = 1 (pure products). This graph is pretty similar to the one posted in the question:

Using the equation:$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^{\circ} + RT\ln Q$$

And substituting in:$$Q = \frac{ξ}{1-ξ}$$

Gives us:$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^{\circ} + RT\ln \frac{ξ}{1-ξ} $$

or:

$$\frac{\delta G}{\delta \xi} = \Delta_\mathrm{r} G^{\circ} + RT\ln \frac{ξ}{1-ξ} $$

$\Delta_\mathrm{r} G^{\circ}$ is a constant even as G changes as the value of ξ changes, so are R and T. This differential equation can be separated and integrated to gives us the original function of how G depends on ξ:

$$\begin{align} \int \delta G = \Delta_\mathrm{r} G^{\circ}ξ + RT \int \ln \frac{ξ}{1-ξ}\delta \xi \end{align}$$

$$\begin{align} G = \Delta_\mathrm{r} G^{\circ}ξ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1 + C)\end{align}$$

Graphing this on desmos gives us the correct graph for how G varies with ξ.

The graph shows the correct ξ value for the equilibrium point and also the difference in G between the "pure" reactants and products. All of this is as predicted by the theory.

If you trace the line along the domain of x towards zero (or 1) the y values increase until the value becomes undefined, but I have zoomed in until my brain hurt and the value never seems to shoot upwards (as I thought the maths predicted). L'Hopital's rule seems to prevent this from happening.

I can't pretend that I fully understand the maths (I don't), maybe others' comments will help here. If anyone wants to use the desmos calculator to produce their own plots (you can change T or $\Delta_\mathrm{r} G^{\circ}$ to other values to see the effect) please do so, I may use this in teaching the material in the future.

I credit @orthocresol @Night Writer and @Karsten Theis with helping me. Also Darrell Bach and my students Nick Kleiderer and Sylver Riddell.

****** Addition to deal with limits ******

The limits of G at each end of the domain of ξ are below.
(For ease of reading I will use x = ξ)

$$\begin{align} G = \Delta_\mathrm{r} G^{\circ}ξ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1 + C)\end{align}$$
$$ G(x) = \Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)$$

Limit of G when $x \to 0 ^+$ :

$$ \lim_{x \to 0 ^+} G(x) = \lim_{x \to 0 ^+} [\Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)] $$

$$\lim_{x \to 0 ^+} G(x) = \lim_{x \to 0 ^+} [\Delta_\mathrm{r} G^{\circ}x - RT + RTC] + \lim_{x \to 0 ^+} [RT(x \ln (x) + (1-x) \ln (1-x))] $$

Note (1):   $ \lim_{x \to 0 ^+}[(1-x) \ln (1-x)] = 0 $

Note (2):   $\lim_{x \to 0 ^+} [x \ln (x)] = 0$   (by l'Hopital's rule)

$$\lim_{x \to 0 ^+} G(x) = - RT + RTC + RT(0+0) $$

$$\lim_{x \to 0 ^+} G(x) = - RT + RTC $$

$$\lim_{x \to 0 ^+} G(x) = RT(C-1) $$

Limit of G when $x \to 1 ^-$ :

$$ \lim_{x \to 1 ^-} G(x) = \lim_{x \to 1 ^-} [\Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)] $$

$$\lim_{x \to 1 ^-} G(x) = \lim_{x \to 1 ^-} [\Delta_\mathrm{r} G^{\circ}x - RT + RTC] + \lim_{x \to 1 ^-} [RT(x \ln (x) + (1-x) \ln (1-x))] $$

Note (1):   $ \lim_{x \to 1 ^-}[(1-x) \ln (1-x)] = 0 $ (by l'Hopital's rule)

Note (2):   $\lim_{x \to 1 ^-} [x \ln (x)] = 0$

$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} - RT + RTC + RT(0+0) $$

$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} - RT + RTC $$

$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} + RT(C-1) $$

These limits to G show that there is a value to G as you approach pure reactants and products along the curve of G vs ξ. The algebra shows some interesting points, which agree with other parts of the theory of equilibria and state functions:

The curve of G vs ξ can be thought of as the sum of a simple line between the G of reactants and the G of products $$G = \Delta_\mathrm{r} G^{\circ}ξ + RT(C-1)$$

for $C = 0$

$$G = \Delta_\mathrm{r} G^{\circ}ξ - RT$$

and a second function (which is always negative) that deals with the Gibbs free energy of mixing $$ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1)$$

This is seen as a dip from the line, reducing the Gibbs free energy in the mixed system. This disappears as ξ approaches 0 or 1, as the system is becoming pure reactants or products:

For a reaction where K > 1 and $\Delta_\mathrm{r} G^{\circ} <0$ (such as the graphs in this example) the pure products are lower in G than the reactants, but the equilibrium mixture is lower still
$\Delta_\mathrm{r} G^{\circ}$ refers to the tangent of the curve at standard conditions ($Q=1$ and $ξ = 0.5$) and also to the slope of the straight line that connects $G_{reactants}$ and $G_{products}$. As ξ changes by 1 from reactants to products, $\Delta_\mathrm{r} G^{\circ}$ is also the absolute $\Delta G$ between pure reactants and products


Reverse Image Searching a ChemDraw Figure

The best tool I have come across so far to do this is this one: https://cactus.nci.nih.gov/osra/
I was also reading about this other one, but can't find the link to the tool itself.


How can I calculate the charge distribution of a water molecule?

Two suggestions: MolCalc: Calculate Properties > Polarity and Solvation gives you a 3D surface plot with the electrostatic potential superimposed.

Another suggestion is Avogadro/GAMESS/MacMolPlt. See for example here


Are all 'dextrorotatory' sugars in biology actually 'd' or '+' in chirality?

Dextrorotatory = d = (+). Levorotatory = l = (-). These terms describe the absolute sense of rotation in an optically active substance.
D and L are older terms for the sense of chirality that can be traced back to dextro- or levorotation of glyeraldehyde. The actual compound may have the same or opposite sense of rotation, but the assignment is determined by analogy. (Cf. Fisher/Rosanoff).
R and S describe absolute stereochemistry of a center, so a specific molecule with multiple chiral centers cannot be sufficiently described with a single R or S, but require one for each stereocenter. This designation is part of the more modern (and widely accepted) Cahn-Ingold-Prelog system, which has largely been adopted by IUPAC.
Note that while D and L might be sufficient to describe a sugar, anything more complicated than a 3-carbon aldose or 4-carbon ketose is not likely to be properly described with a single R or S because there are multiple stereocenters.


Which is more soluble in water and why: K₂HPO₄ or CaHPO₄?

The charge and size of an ion influences both lattice stability and stability of the hydrated ion.

The less stable the lattice the more soluble the ion (less energy required to disrupt the lattice).  The more highly charged and smaller the ion is, the larger are the electrostatic repulsions within the lattice which will act to destabilize the lattice.
The more stable the solvated ion, the more soluble the ion.  The higher the charge on the ion and the smaller its radius, the more stable the solvated ion.

The size of the potassium ion is around 152 pm, the size of the calcium ion is around 114 pm.  So the calcium ion is both the smaller and the more highly charged of the two.  Therefor we would expect the calcium ion to have the less stable lattice and be more highly solvated (stabilized) when in solution.  Both of these factors would lead us to expect $\ce{CaHPO4}$ to be more soluble than $\ce{K2HPO4}$


Molecular differences in the HB graphite hardness scale

Pencil leads are not comprised of pure graphite: they're made from graphite and a varying amount of clay, which in turn is what determines the hardness of the lead. Thus, you're not actually comparing graphite to graphite but rather different mixtures of graphite and clay.

According to one article published by Jetpens.com (a vendor of writing implements) entitled Picking the Perfect Pencil Hardness Grade:

  Graphite is the key component of modern day pencil lead—in fact, lead is a misnomer as it's actually made up of a mixture of graphite and clay. The formulation of this mixture determines its lead grade. The higher the proportion of graphite content relative to clay the lead has, the softer and darker the lead will be, and vice versa.

And Wikipedia tells us:

  The various graphite pencil grades are achieved by altering the proportion of graphite to clay: the more clay the harder the pencil.

The difference in hardness is attributable to the amount of clay mixed in with graphite in your different pencil leads. There is no difference in the graphite at the molecular level.


How does higher carbonyl compounds disolve in water?

Sugar contains a lot of hydroxyl groups. So due to hydrogen bonds, sugar can readily dissolves in water.


How to change CO₂ to a less polluting gas?

It makes no difference where you grab the $\ce{CO2}$.  Don't burden the car.

Plant mammoth sterile forests with gene-gineered high lignin wood for building.  Or, deep bury the trunks in the Sahara Desert of the Empty Quarter.  RuBisCO is the slowest, least efficient enzyme on Earth, the first carbon fixation step in photosynthesis.  Gene-gineer a faster, more efficient (e.g, not poisoned by oxygen) RuBisCO for the forests.

http://www.cea-jaea-collaboration.net/literature/cea/ICCAP%20Nice%202007%20NaCO2finalpaper.pdf

A reactor for your tailpipe.


how is there no centre of symmetry in this case?

One way to quickly get a visualization without building a 3D model (having a model is more instructive, though):

Go to https://chemapps.stolaf.edu/jmol/jsmol/simple.htm
Click on Load by name and type "(2S,3S)-Butane-2,3-diol"
Click on console and type "rotate branch @2 @4 180" to get a conformation that is easy to compare with the Fisher projection.


The procedure to convert methane to longer hydrocarbons at low temperature is one of the so-called Holy Grails in catalysis (http://pubs.acs.org/doi/abs/10.1021/ar00051a001). The person who is able to invent a process wherein this is possible (at an efficient rate) would be a very good candidate for a Nobel Prize.

Why is it then so difficult to convert methane to longer hydrocarbons? Methane is a highly stable molecule. In order to form a longer hydrocarbon, methane first has to be broken down to smaller CH$_{x}$ fragments. The first dehydrogenation step of methane (to CH$_{3}$) has a very high activation energy and even using a good catalyst, this is not feasible at an appreciable rate under 100 degrees celcius.

Industrially, this process is done by first creating synthesis gas (CO + H$_{2}$) from methane by partial oxidation using pure oxygen (which is cryogenically separated from air). Thereafter, the synthesis gas is compressed to roughly 20 atm of pressure and converted to very long hydrocarbons (waxes) in the Fischer-Tropsch process. In turn, these waxes are cracked to form diesel fuels and lubricants.

I am afraid that there is no easy way of converting methane to longer hydrocarbons. Though please keep trying because basically the whole world would benefit if you succeed! :-)


The difference is quite huge. Firstly, the deposition will occur on both electrodes (if possible). Secondly, if the frequency is high enough, there will be no deposition at all as only polarization and depolarization effect take place. The frequency effects mass, transport and electrical double layer. Depending on the electrolyte composition, the dissolution (if possible) will reduce the deposition. Usually pulse electroplating is used.  The problematic is huge, please see some articles for more info:

An Overview of Pulse Plating - Norman M. Osero

and

Pulse and pulse reverse plating—Conceptual, advantages and applications, M.S. Chandrasekar, Malathy Pushpavanam


Is specific rotation defined for mixtures of enantiomers or meso compounds?

Nothing stops you from taking the equation and calculating a specific rotation of a mixture. Indeed, as far as I am aware when establishing a new asymmetric reaction, the e.e. is measured by something like chiral HPLC (which will unambiguously give two peaks that can be integrated and compared) while the specific rotation is measured for the resulting mixture of enantiomers as is. So the paper will say something like $99~\%$ yield ($98~\%\ \text{e.e.}$) but below there will be a standard calculation of $[\alpha]_\mathrm D^\pu{20^\circ C}$.

So obviously you can measure a specific rotation of the racemate or meso compound but whatever value you measure should fall into the error of the polarimeter so that the final value of $[\alpha]_\mathrm D^\pu{20^\circ C}=0$.


You have already mostly answered the question yourself ;)

A heating block, made from a material with a high thermal conductivity, helps to keep (a lot of) samples distributed over a larger area (several vials, Eppendorf tubes, etc.) at exactly the same temperature even if the primary heat source is punctual or just a layer of heating wires at the bottom of the instrument.


When you are considering redox reactions, you should get familiar with electrochemical series (Ref.1). Oxidizing and reducing characters of certain reagent also depends on the conditions  used as well. For example, according to Ref.1, $\ce{HClO}$ in acidic medium:
$$\ce{HClO + H+ + e- <=>  1/2Cl2 + H2O} ~~~~~~~~ \mathrm{E}^\circ = \pu{+1.594 V} $$
On the other hand, $\ce{ClO-}$ in basic or neutral medium:
$$\ce{ClO- + H2O + 2e- <=>  Cl- + 2OH-} ~~~~~~~~ \mathrm{E}^\circ = \pu{+0.890 V} $$

Consider the following half reaction:
$$\ce{Br2 + 2e- <=>  Br- } ~~~~~~~~ \mathrm{E}^\circ = \pu{+1.098 V} $$

Now, you can say that $\ce{HClO}$ in acidic medium spontaneously oxidizes $\ce{Br-}$ to $\ce{Br2}$, but $\ce{ClO-}$ in basic or neutral medium cannot spontaneously oxidize $\ce{Br-}$ to $\ce{Br2}$.

In similar way, you may also say that $\ce{HClO}$ in acidic medium is a better oxidizing agent than that of $\ce{ClO4-}$ in acidic medium, because:
$$\ce{ClO4- + 8H+ + 7e- <=>  1/2Cl2 + 4H2O} ~~~~~~~~ \mathrm{E}^\circ = \pu{+1.392 V} $$

Reference:

Steven G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at 298.15 K,” Journal of Physical and Chemical Reference Data 1989, 18(1), 1–21 (https://doi.org/10.1063/1.555839).


Measuring pH of pasty solutions/mixtures

Diluting the sample does alter the pH, but it does so in a predictable way.

When we are measuring pH, we are measuring the concentration of hydrogen ions (or hydornium ions, or whatever ) via the negative logarithm:

$$\mathrm{pH} = -\log_{10}\left([\ce{H+}]\right)$$

Diluting a solution has a predictable effect on concentration, since the total amount (in moles) of $\ce{H+}$ will not change by the addition of more solvent. This relationship is summarized by the formula (where $c$ represents concentration, $v$ represents volume, and $1$ and $2$ represent the solution before and after dilution, respectively:

$$c_1v_1 =c_2v_2$$

Let's take an example. Suppose you have 1.000 mL of a viscous acidic aqueous solution. The viscosity of the solution will impede easy measurement of pH because of the kinetics of mass transport. If you dilute this solution with water to a total volume of 10.00 mL, you now have a less viscous solution with the same number of moles of $\ce{H+}$, because sources of $\ce{H+}$ were neither added nor removed.

If this new diluted solution has $\mathrm{pH = 3.03}$, then $$[\ce{H+}] = 10^{\mathrm{-pH}}=10^{-3.03} = 9.33\times 10^{-4}\ \mathrm{M}$$.

The concentration of the original solution can now be determined:

$$c_1 = \dfrac{c_2 v_2}{v_1} = \dfrac{\mathrm{9.33\times 19^{-4}\ M)(10.00\ mL)}}{\mathrm{1.000\ mL}}=9.33\times 10^{-3}\ \mathrm{M}$$

The pH of the original solution is:

$$\mathrm{pH = -\log_{10}(9.33\times 10^{-3}) = 2.03}$$

Note, that since I chose to dilute by a factor of 10, the pH of the dilute solution is 1 pH unit higher than the original solution.

This approach works for solutions of strong acids and strong bases and will work over much of the pH range, but may get a little awkward around neutral. The approach breaks down for weak acids and bases since changing the concentration changes the position of equilibrium. For weak acids, diluting likely results in additional ionization events. However, even without knowing the acid/base and its equilibrium constant, you could perform multiple dilutions and do some curve-fitting and get back to the original pH.


The difference between CASSCF and MCSCF methods

Both methods optimize orbitals and CI coefficients. MCSCF is the more general term, while CASSCF is a special type of MCSCF.

As the name Multi-Configurational Self Consistent Field already implies, it combines the Configuration Interaction method with the SCF method (Hartree-Fock). Since this combination increases complexity of the wave function, one usually applies restrictions to both sets of parameters (CI and orbital coefficients). For example core orbitals may be kept fixed (frozen) or only a small subset of configurations is considered.

Regarding the selection of configurations, there are various possible approaches. For example:

hand selecting configurations using chemical intuition or some other a priori knowledge
selecting a subset of orbitals, which should at least cover the HOMO and LUMO. In this active space one considers all possible configurations, hence Complete Active Space Self Consistent Field (CASSCF)
restricting configurations only to certain excitation levels (e.g. Singles and Doubles) within the active space: Restricted Avtive Space Self Consistent Field (RASSCF)

So CASSCF is a special variant of MCSCF, therefore there is not really a difference between the two. Although not every MCSCF is necessarily  a CASSCF, the later is the most common type.


Nitrogen is more electronegative than hydrogen hence N will pull electrons towards itself. Hence not a good candidate to donate electrons!

Yet it makes bonds with hydrogen, as do other electronegative elements (think of hydrogen chloride and water). So polar bonds form in all kind of molecules.

  lets take the case of Oxygen O=O, here each Oxygen atom has 2 lone pairs, and going by the above logic these lone pairs should also participate in bond formation and create an Oxygen polymer(!!)

The concept of stability makes sense when comparing two states with the same number of atoms (e.g. ammonia and hydrogen ion vs. ammonium ion). For elemental oxygen to polymerize, each $\ce{O2}$ molecule would bring in electrons. For a molecule to act as base, a lone pair turns into a bond with hydrogen (where hydrogen provides zero electrons and the base provides two). This does not "break the octet" of the base, and give the hydrogen access to two electrons (completing its duet).

  Hence with the above points I still could not see why Nitrogen has to accept proton

It does not have to. It is a weak base. If you add ammonia to neutral water, most of it remains undissociated (the pH goes up, though, because some of it does dissociate).

  [devils advocate question:] How can I use electronegativity to predict something about acid/base chemistry?

The trick is to compare green apples with red apples, not oranges with bananas. The easiest comparison is isoelectronic species, such as:

$$\ce{HF vs OH-}$$
Hydroxide is a better base than hydrogen fluoride.

$$\ce{H2O vs NH2-}$$
Azanide (had to look up the name) is a better base than water.

$$\ce{H3O+ vs NH3 vs CH3-}$$
The methyl anion is a better base than ammmonia, which is a better base than hydronium.

Or you can compare compounds by exchanging O for NH or CH2.

$$\ce{H2O vs NH3 vs CH4}$$

$$\ce{CH3-OH vs CH3-NH2 vs CH3-CH3}$$

$$\ce{CH2=O vs CH2=NH vs CH2=CH2}$$

This is more complicated because the carbon atoms lack a lone pair while nitrogen and oxygen have one. So you can compare nitrogen and oxygen using an electronegativity argument, but have to exclude carbon because it can't act as a base without the lone pair (well, $\ce{CH5+}$ does sort of exist but is called a super-acid, so $\ce{CH4}$ is hardly a base).

Hydrides

Here is a set of examples from OpenStax Chemistry 2e showing some of the trends across the periodic table. Again, $\ce{CH4 and SiH4}$ don't have lone pairs, so they can't be bases, even though the trend suggests they might be the strongest base in the series. The reason that water is described as neutral is not some special property of water but because the pKa is typically defined in aqueous solution, and a neutral pH references the hydroxide and hydronium concentration in pure water. If liquid ammonia were our solvent of choice, that would be the "neutral" case.


What should be the molarity of HCl and H2SO4 acid so that they can be considered safe to touch?

According to the Globally Harmonised System (GHS), hydrochloric acid with a concentration of c ≥ 25 % and sulfuric acid with a concentration of c ≥ 15 % are labelled "causes severe skin burns and eye damage" (H314) and are classified based on standard animal test data in skin corrosion Sub-category 1B ("Corrosive responses in at least one animal following exposure > 3 min and ≤ 1 h and observations ≤ 14 days"). "Skin corrosion refers to the production of irreversible damage to skin; namely, visible necrosis through the epidermis and into the dermis occuring after exposure to a substance or mixture."

Hydrochloric acid with a concentration of 10 % ≤ c < 25 % and sulfuric acid with a concentration of 5 % ≤ c < 15 % are labelled "causes skin irritation" (H315) and are classified based on standard animal test data in skin irritation Category 2. "A substance is irritant to skin when it produces rversible damage to the skin following its application for up to 4 hours."

Concentrations well below these values may be considered safe for the exposure of skin to the substance.

Note that the natural skin surface pH is below 5.


Comparison of alkali and alkaline earth melting points - MO theory

From "The Nature of the Interatomic Forces in Metals" by Linus Pauling

  This qualitative description of the interactions
  in the metal is compatible with quantum mechanical treatments which have been given the
  of
  problem, and it leads to an understanding
  such properties as the ratio of about 1. 5 of
  crystal energy of alkali metals to bond energy of
  their diatomic molecules (the increase being the
  contribution of the resonance energy), and the
  increase in interatomic distance by about 15
  percent from the diatomic molecule to the
  crystal.
  The alkaline earth metals, by assuming the
  configuration nsnp, are able to form twice as
  many bonds as the alkalis. Similarly the suc-
  ceeding elements in the periodic table can form
  bonds in increasing number.

You are right on the fact that alkaline earth metals should not bond if we only considered their highest atomic energy levels. The dissociation energy for diatomic clusters of alkaline earth metals is low.

However, the disassociation energy gets higher for tetra-atomic as shown in this computational study: http://www.sciencedirect.com/science/article/pii/0039602885902328

Now, the reason why alkaline earth metals have higher melting points to be due to the p orbitals. The previous study relates p hybridization with bond strength. And so does this study: http://scitation.aip.org/content/aip/journal/jcp/77/8/10.1063/1.444313

Both journals state that p orbitals are involved in bonding. And the more involved the p orbitals in bonding, the stronger the bond. This trend is shown for alkaline earth metals. Beryllium, which has the highest melting point, also has more p orbital character. Also, take a look at the density of states of alkali metals and alkaline earth metals:
http://www.colorado.edu/engineering/MCEN/MCEN5024/

You can see that not all p orbitals are entirely above the s orbitals.

So, to answer your question. The reason why alkaline earth metals have higher melting points is due to p orbitals. As Pauli states in his review, the np orbitals "hybridize" with the s orbitals. This allows for more bonds to form.

I think that p orbitals participate in bonding in alkali metals. But I couldn't find information that compares p orbital "hybridization" of alkali and alkaline earth metals directly.


This seems to be a very specific case. I would say that you should start with a standard Ni electroplating solution and see the result. You could then adjust the parameters of your plating operation and the concentrations of the additives in the solution.

However, I have a bigger concern. How stable is your liquid crystal template under applied potential in an electrolyte? Before trying to deposit Ni, you should try to do simple electrochemistry find the potential window where your template stays intact.


What molecule is mostly responsible for the smell of petrol/gasoline when I refill my car?

As you state, aromatic compounds such as benzene, naphthalene, xylene and toluene have noticeable odor (ergo "aromatic"). In addition, odorous oganosulfur compounds may remain, though in Europe and North America effort is made to remove them (kerosene and heating oils may have higher amounts of oganosulfur contaminants).

In the US, Methyl tert-butyl ether (MTBE) was a common strong-smelling additive that raised octane number and reduced CO emission, but it has been largely phased-out because of it propensity to leak into and damage aquifers.

A Safety Data Sheet list some other chemicals that may have a noticable odor, such as ethanol (added for the same purpose as MBTE).

A trained nose can detect and identify any or all of these compounds, so it really depends on the particular petrol you buy. As a rough guide, the aromatics may have a "sweet' or fruity smell to some people, though naphthalene has the smell of moth-flakes, MBTE a "burning" odor, and sulfur compounds may smell like rotten garlic.


How would I know that a compound and its isomers are enantiomers?

I am assuming that your question is this:
"How to check whether two given compounds are enantiomers of each other?"

First, you check for structural or geometrical isomerism.
If there isn't any, you check whether the compounds are diastereomers. (Check whether they are not the mirror images of each other.)
If they are not diastereomers, you have two possibilities:

Either the compounds are enantiomers of each other.
Or they represent the same compound.

You could try this approach:
First note the groups attached to the chirality centre. The Carbon is obviously an sp3 hybridized atom with four different groups attached to it.
Interchanging any two groups attached to the chirality centre produces the chiral compound's enantiomer.
So you could try interchanging the positions of the groups attached to the chirality centre until you get the other compound. If the number of swaps is even, the two compounds are identical. If odd, they are enantiomers of each other.


Reactivate dried up ink of ballpoint pens

As ink typically consists of non-polar dyes dissolved in a non-polar liquid, I would recommend using a straightened paper clip and a drop of acetone to clear out the dried ink clogging up the ink cartridge. This works by agitating the dried ink, as well as by replenishing the ink's solvent. Just be careful to not make too much of a mess!

Another method I have seen work before is heating the end of the pen under a flame, which melts the ink clogging the tip. For this method, my advice would be to not hold the pen in one place for too long, lest you risk melting the plastic.


Ascorbic acid is considered an antioxidant because it can supply an electron to a species with an unpaired electron (a "radical", labeled "X" in the below diagram) to stabilize the radical. Of course, since ascorbic acid has now lost an electron, it is itself a radical - but it is a very stable radical due to resonance. Below I have included a diagram of this process below, from UC Davis.

Note how the we can "electron-push" to get the unpaired electron in the middle structure to move to the oxygen on the other side of the ring. This resonance delocalization of the unpaired electron in the ascorbyl radical makes it much more favorable to have the unpaired electron in the ascorbyl radical rather than the X.

Because radicals normally poach electrons from surrounding molecules to pair with their lone electron, they normally cause a chain reaction where their "victim" goes on to pull electrons from another neighboring molecule (oxidation), and so on and so forth. However, because the ascorbyl radical is stable, it is less likely to poach an electron from another molecule in its environment, preventing the chain reaction from occurring. Thus, it is an antioxidant.


How does dimethyl sulfate react with water to produce methanol?

The mechanism of methyl transfer of $\ce{Me2SO4}$ is generally thought to be an $\mathrm{S_N2}$ reaction. Thus, water would attack the methyl group, releasing $\ce{MeO-SO3-}$, a good leaving group generating methanolium $\ce{H3C-OH2+}$. The proton is later transferred back to a methyl sulphate to give methyl hydrogen sulphate $\ce{HSO4Me}$ and methanol.

One could imagine the proton from methanolium to initially be transferred to water to generate hydronium. That’s actually pretty probable and it will be hard to tell where it will end up. Water, methanol and hydrogen sulphate (and therefore likely methyl sulphate, too) all have similar $\mathrm{p}K_\mathrm{b}$ values and are equally eager to displace it/non-eager to accept it. adding some base, e.g. sodium carbonate should provide a good ‘resting point’ for such an unhappy proton.

The question is why water can do this even though it is a lousy nucleophile. Well, it turns out that for one, $\ce{MeSO4-}$ is a great leaving group enhancing the reaction rate and for two, there is nothing else around if you mix just $\ce{Me2SO4}$ and water that would be a better nucleophile. ($\ce{OH-}$ is present in too low concentrations to actually count.) Thus, the two reagents have the choice of doing nothing or slowly doing something thermodynamically favourable (= reacting). Kinetics dictate that it must happen slowly, but it can happen.


Thermochemistry: forms of energy & pressure

1.Does pressure decrease with time?

The ideal gas law tells us that
$$PV=nRT$$
or
$$\mathrm{P=\frac{[nRT]}{[V]}}$$

If the temperature, number of moles or volume does not change, then the pressure will not change.

  If pressure is caused by the force of gas molecules hitting the inside
  surface of a container, don't they transfer energy and eventually slow
  down enough to totally reduce pressure?

Yes, that's correct.  If energy is transferred out of the system, then the temperature will fall and the pressure will decrease.  But if everything is already at the same temperature, there will be no energy flow in or out and the pressure will remain constant.

  after the pressure is created, doesn't it require continuous energy to
  maintain pressure against a constant force

No, there is no requirement that energy must be supplied to maintain the pressure.  Again, if the gas, the vessel and the surroundings are all at the same temperature the pressure will not change.  However, if the entire system is not at thermal equilibrium, then heat will flow so as to put things at equilibrium and the temperature will change.


Why is the aqueous solution of KMnO4 slightly basic?

The strength of permanganic acid that you quote, combined with that of potassium hydroxide as a base, would guarantee that pure potassium permanganate is neutral in aqueous solution. But commercially prepared potassium permanganate is made in the presence of alkali, the use of potassium instead of sodium arising from the fact that the reaction scheme does not work with the latter.
$\ce{2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O}$ (in molten KOH solvent; reactants are fused together)
$\ce{2 K2MnO4 + 2 H2O → 2 KMnO4 + 2 KOH + H2}$ (electrolysis in alkaline media; a direct chemical reaction would convert some material back to $\ce{MnO2}$)
Samples are of course purified but must be assumed to contain some leftover alkali from this preparation.


How can a Zn-Cu galvanic cell have a voltage of 1v in stead of 1.1v

You can get a smaller potential if you alter the concentration of one or both ion solutions away from ideal 1M concentration. This behavior is described by the Nernst equation:

$$E_{cell}=E_{cell}^{o} - \frac{RT}{zF}lnQ$$

where $E_{cell}$ is the actual potential in the cell, $E_{cell}^o$ is the potential of the cell under ideal conditions, R is the ideal gas constant, T is the temperature, F is Faraday's constant, z is the number of electrons in the reduction oxidation reaction and Q is the reaction quotient. By altering Q to be greater than 1, the actual voltage of the cell dips below the ideal value.


Without knowing what dissolved species are causing the hardness, it's a little difficult to know what to advise. Performing qualitative analysis would likely help you a lot.

You describe using lime softening (i.e., using hydroxides) which can help in some cases. Usually the resulting hydroxide precipitates are pretty hard to filter, requiring a membrane or a very good filtration system.

As Wikipedia notes, there are a few other reliable techniques:

reverse osmosis
distillation
ion exchange resins
chelating ligands

While the first three are definitely the best in terms of ultimate water quality, an easy thing to supplement lime softening would be to add various chelating ligands. For example, after you carefully filter the water from lime softening, maybe add citric acid?

In any case, I'd suggest you do some sort of qualitative analysis (and pH) testing as you go. This is the best way to know what steps you'll still need.


Are entries ever removed from the Merck Index?

Without several print editions side-by-side, I restrain my answer to the electronic version. This public link or here regarding the online version based on the 14th edition may be read like it may be the case.  To quote:

 "The 14th Edition also contains "retired" records from earlier editions of The Merck Index which still contain valuable information but are no longer being updated with new information."

Of course, beside RSC you may contact them (Merck in Rahway, NJ), too.


There are a few ways to do nanoparticle tracking, although I'm not sure they will exactly apply to your system.

Dynamic light scattering or DLS has been used to get the distribution of nanoparticle sizes in solution/suspension over time. Generally, the data gives information on the ensemble, not single-particle tracking.

Nanoparticle tracking analysis or NTA also relies on scattered light, but gives frame-to-frame analysis of each particle's trajectories.

There are other related techniques... Another approach is to create an Anti-Brownian ELectrokinetic (ABEL) trap - to monitor the particle or protein dynamics and apply forces to push it back. This technique was developed by the Moerner lab at Stanford, for example Acc. Chem. Res., 2012, 45 (11), pp 1955–1964


is there any mnemonic for the AXE method?

I think you shouldn't abuse with mnemonic. In this case you should start with this rule of VSPR:

each atom in a molecule will be positioned so that there is minimal
repulsion between the valence electrons of that atom.

Now you should begin to train your three dimensional thinking!
This is the method I've adopted to learn AXE, of course everyone has his own but maybe it could help to you to find yours:
Learn to draw the $AX_nE_0$ series:
This for me was the first step: learn to draw the basic geometry 0 lone pair. This is quite intuitive knowing the previous law. Begin with the $A$ then add the $X$s. Note that the linear module could help you to remember it.

Add 1,2,3 lone pairs series
This is the next step. I've brutally modified the wikipedia table to show how you can easily do this step from the Basic geometry 0 lone pair. Simply begin to substitute $X$ with $E$: one $E$ for 1 lone pair, two for 2 lone pair...
Nomenclature
The last step is to add the nomenclature. First of all you should take in account that the lone pair ($E$) don't appear as a part of the final shape from which the nomenclature is derived. If not considering the $E$ group you obtain a planar molecule you have to add the word planar after the name, if you obtain a pyramidal shape you should use pyramidal after the name of the planar pyramid's base.
So imagine that $E$ groups are invisible and try use this method if you  can't find it only with pyramids try to figure out which of the following figures is more close to the actual shape of the molecule.

The first two are platonic solids  (Octhaedron, and Tetrahedron) then there is a T for a T-shaped, and a bent figure. In the original wikipedia table respective molecule configuration is already rotated to the right position, I think you can do it on your own, with your 3D thinking.
For the last, seesaw shaped, there is a powerful "mem":

Here is an old photo of two boys in a $SF_4$ molecule.Good ol' times!


Because the enol and ketone aren't stable, equilibrium happens; therefore the enol has a chance of losing a hydrogen ion and because of that, the enol becomes an enolate.

Also keep in mind that an acid is a molecule that can donate a proton.
The enolate is a base and it's gathering/accepting a proton from the hydrogen.

I assume this is what you meant when you asked the question.


How does a limiting reactant relate to the concentration of ions in solution?

Of course, the concentration of $\ce{Na^+}$ is $0.2\ce{M}$, because it is a spectator ion. It doesn't participate in the  main reaction between  $\ce{Pb^2+}$ and $\ce{SO4^2-}$.
The concentration of $\ce{Na^+}$ is $0.1 \times 2=0.2 \ce{M}$ as there are two ions of sodium in sodium sulfate.

The number of mols  of ion nitrate, as a spectator ion, is $2\times1.5/331.2= 0.00906 \ce{mol}$.
The concentration of ion nitrate is  $ 0.00906/0.125=0.0725 \ce{M}$.

Let's go back to the main reaction between  $\ce{Pb^2+}$ and $\ce{SO4^2-}$: The limiting ion is $\ce{Pb^2+}$ as the number of moles of lead nitrate  $1.5/331.2=0.00453 \ce{mol}$ is smaller than the number of moles of sodium sulfate  $0.0125 \ce{mol}$.
So, after the reaction, there  are no  $\ce{Pb^2+}$ ions left in solution.

As for the remaining $\ce{SO4^2-}$; its concentration is:  $(0.0125-0.00453)/0.125=0.06376 \ce{M}$


Where can I find (citable) FTIR spectra of CuCl and CuCl2?

If your institution holds a subscription to Elsevier's database Reaxys, than you may find, based on chemical name, molecular formula, or CAS number (the later for example from a catalogue of a supplier of chemicals) links to primary literature in the section of spectroscopic properties. Reaxys tries to cover both organic materials (formerly deposited in the Beilstein database) and inorganic / metalorganic materials (formerly deposited in the Gmelin database).  Otherwise, Scifinder from the ACS equally points towards primary literatures, too.

Beside the databases behind a paywall, freely accessible databases like the NIST webbook may be worth to consult; the entry about $\ce{CuCl2}$ here
mentions in the references already one with IR data of gaseous $\ce{CuCl2}$ (here).


Can carbon and titanium form an ionic bond?

Yes, carbon can form ionic bonds with Titanium. $\ce{TiC}$ is titanium carbide, an ionic compound with $\ce{C^{4-}}$ (the carbide anion) and $\ce{Ti^4+}$ (titanium cation). It has the same crystal structure as $\ce{NaCl}$ (which probably has to do with the $\ce{1\!\!:\!1}$ ratio of anion to cation).

This is a space filling model of titanium carbide:

Here is sodium chloride for a crystal comparison (click the image for a larger version):

The second model (the $\ce{NaCl}$) shows the ionic bonds as covalent bonds but still, both titanium carbide and sodium chloride have a cubic lattice structure.

However, carbon does not often form ionic bonds with metals since it prefers to be covalent (much like oxygen). In fact carbon has much more covalent character than oxygen does which is why oxides form easily (just $\ce{O2 + element}$ and usually, oxides will form) but carbides don't (a catalyst is needed to form carbides and some oxides like gold oxide).


You can look up the molecule on chemspider, where you have a little applet for the 3D structure. Or you can download a coordinate file from NIST and view it in a molecular viewer, like Avogadro. Or keep on reading for some deeper insight.

As Klaus already pointed out, if VSEPR is a valid concept, one would arrive at the conclusion, that the molecule is trigonal pyramidal, like ammonia. I performed a DF-BP86/def2-SVP calculation on this to analyse the electronic structure a bit more.

One reason why VSEPR does fail (a lot) is that lone pairs are more stable in orbitals with high s character. Another point that is often observed is that orbitals with higher p character are directed to more electronegative elements (Bent's Rule).[1,2] As a result of that, the bond angle between electronegative elements gets smaller. This can be observed here, too. From the ideal sp³ arrangement we would expect an 109.5° angle, but here it is about 5° smaller. As a consequence also the hydrogen-nitrogen-fluorine angle has to decrease. In this case the bond angle becomes much smaller because of an intramolecular hydrogen bond.[3]
The NBO analysis consequently builds the same picture:
$$\begin{array}{llll}\hline
\text{type}&\text{main contribution}&\text{other contributions}\\\hline
\sigma\text{-Bd}~\ce{N-F}& 64\%~\ce{F}~(14\% s + 86\% p) & 36\%~\ce{N}~(15\% s + 85\% p) \\
\sigma\text{-Bd}~\ce{N-H}& 68\%~\ce{N}~(25\% s + 75\% p) & 32\%~\ce{H}~(100\% s) \\
\sigma\text{-LP}~\ce{N} & 100\%~\ce{N}~(52\% s + 48\% p) &  \\
\sigma\text{-LP}~\ce{F} & 100\%~\ce{F}~(75\% s + 25\% p) &  \\
\sigma\text{-LP}~\ce{F} & 100\%~\ce{F}~(10\% s + 90\% p) &  \\
\sigma\text{-LP}~\ce{F} & 97\%~\ce{F}~(100\% p) &  2\%~\ce{N}~(7\% s + 93\% p) & 1\%~\ce{F2}~(15\% s + 85\% p)\\\hline
\end{array}$$

The planar (trigonal) structure on this level of theory is about 33 kcal/mol higher in energy. It is the transition state for the nitrogen inversion, which is therefore very slow.

Notes:

Bent's rule on wikipedia
Bent's rule on our network: Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?
The bond length is much smaller than the sum of the corresponding van der Waals radii, suggesting a hydrogen bond. Compare A. Bondi J. Phys. Chem., 1964, 68 (3), pp 441–451.
$r_{\mathrm{vdW}}(\ce{HF})=r_{\mathrm{vdW}}(\ce{H})+r_{\mathrm{vdW}}(\ce{F})= (106 + 140)~\mathrm{pm}= 246~\mathrm{pm}$


Which are the radicals containing phosphorus in ATP?

You have almost answered yourself completely.

'Phosphoryl’ in current IUPAC Nomenclature of organic chemistry, preferred names (2013):

  P-67.1.4.1.1.2 Fundamental acyl groups for substituent derived from mononuclear noncarbon oxoacids

  (…)

  $\ce{-P(O)<}$
  phosphoryl

Example compound with “hydroxyphosphoryl” group:

$\ce{C2H5-P(O)(OH)-CH2COOH}$
[ethyl(hydroxy)phosphoryl]acetic acid

(software generated, in the book there is an example without hydroxy parens: O-({[(2R)-2,3-bis(octadecanoyloxy)propoxy]hydroxyphosphoryl})-L-serine)

‘Phosphono´:

  P-67.1.4.1.1.1 Retained names for substituent groups derived from mononuclear noncarbon oxoacids

  (…)

  $\ce{-P(O)(OH)2}$
  phosphono (preselected prefix)

(There's no “phosphonyl” if someone wonders, there's however different group with similar name, phosphonoyl $\ce{-PH(O)\bond{-}}$.)

Example:

$\ce{(HO)2P(O)-CH2COOH}$
phosphonoacetic acid

And, I think that the correct layout for the name provided should be rather:
{[(2R,3S,4R,5R)-5-(6-aminopurin-9-yl)-3,4-dihydroxyoxolan-2-yl]methoxy}(hydroxy)phosphoryl phosphono hydrogen phosphate

This name is somewhat overcomplicated. ‘Triphosphate’-based name could be used, see

  P-67 Mononuclear and polynuclear noncarbon acids and their functional replacement analogues as functional parents for naming organic compounds

  (…)

  P-67.2 Di- and polynuclear noncarbon oxoacids

  (…)

  P-67.2.1 Preselected names

  (…)

  $\ce{(HO)2P(O)-O-P(O)(OH)-O-P(O)(OH)2}$
  triphosporic acid (preselected name)

and

  P-106 Nucleotides

  (…)

  P-106.2 Nucleotide diphosphates and triphosphates

  Diphosphate, triphosphate, etc. esters of nucleosides are named by citing a phrase such as diphosphate, after the name of the nucleoside. The presence of hydrogen atoms on the diphosphate, triphosphate, etc. component of the molecule is indicated by the words ‘hydrogen’, ‘dihydrogen’, etc. Parentheses are used to avoid ambiguity.

  Examples:

  uridine 5’-(tetrahydrogen triphosphate)

(but this is natural compounds section, which is more permissive)

Seemingly apparent reason why ‘triphosphate’ is currently not used in software generated systematic names is that there's an ambiguity in binding the residue to the first (last), or the middle phosphorus' oxygen (positional isomers), and no numbering system for non-carbonic acids is defined.

However it is defined (P-67.2.2.1), with esters examples (P-67.2.5.2). The numbering is from-to central atoms including the oxygen connectors, so the terminal phosphorus atoms get numbers 1, 3, 5; so (generalized) names for the two isomeric alkyl triphosphates (alkyl on terminal or middle phosphate unit, respectively), are

O1-alkyl tetrahydrogen triphosphate
O3-alkyl tetrahydrogen triphosphate

which might be confusing, especially in the latter case. So this can be the main reason, why these are not used.

Let's use it anyway (with a minor improvement – indicated hydrogen):
O1-[(2R,3S,4R,5R)-5-(6-amino-9H-purin-9-yl)-3,4-dihydroxyoxolan-2-yl] tetrahydrogen triphosphate.

It should be OK to omit ‘tetrahydrogen’, as phosphates are ionized in physiological conditions.

(Not speaking of yet better names like adenosine 5’-triphosphate or ATP.)


How can the hybridisation schemes of transition metal complexes be determined?

Firstly, note that hybridisation theory as applied to transition metal complexes is an incorrect, flawed theory. It is an attempt to rationalise experimental observation (e.g. geometries), but in the process it invokes models of bonding which are highly unrealistic and demonstrably untrue. See: Why is it wrong to use the concept of hybridization for transition metal complexes?

Now, with that disclaimer out of the way, let's examine how that flawed theory works. In traditional hybridisation theory, tetrahedral and square planar 4-coordinate complexes are considered to be $\mathrm{sp^3}$ and $\mathrm{dsp^2}$ respectively. Since $\ce{[Ni(CO)4]}$ is tetrahedral and $\ce{[Ni(CN)4]^2-}$ is square planar, one can assign the hybridisations accordingly. If you have a textbook that still mentions this outdated theory, it should have a table telling you what hybridisation each geometry corresponds to. Here is an example from Housecroft's Inorganic Chemistry 4th ed. (p 667). To their credit, there is a subsequent section debunking this theory.

These are generally based on post hoc justifications. Essentially, somebody looked at the geometry and then tried to piece together which orbitals could be combined to form such a geometry. So, I do not think there is a good way to predict them a priori. This formula you have dug up does not work very well, even for main group elements. For transition metals the factors deciding geometries are much more complex, and simply cannot be captured in one line of mathematics.

The real problem with the approach above is that you need to know the geometry before you can assign a hybridisation. (Don't even try to do it the other way round - it simply does not work because hybridisation is an incomplete theory. It is like trying to predict wave-particle duality using classical mechanics.)

If you cannot predict the geometry of the complex, then you cannot predict the hybridisation. You could argue that cyanide is a strong field ligand and in conjunction with a $\mathrm{d^8}$ metal, a square planar geometry is preferred. On the other hand, in $\ce{[Ni(CO)4]}$, the central atom is $\mathrm{d^{10}}$ and hence even with a strong-field ligand like carbon monoxide, a tetrahedral geometry is favoured, simply based on steric grounds.

However, this does nothing but introduce more questions. The term "strong-field" has no meaning outside of molecular orbital theory. Hybridisation theory has no inkling of what strong- or weak-field ligands are, and consequently cannot explain why $\mathrm{d^8}$ Ni(II) only prefers to go square planar with strong-field ligands. On top of that, it is not at all obvious from hybridisation theory how the electronic configuration of the central metal should favour a particular geometry. It's simply not possible for your formula to explain any of this.

So, to answer your questions directly:

  Is there a systematic and logical way of knowing hybridisation of coordination compounds?

Yes. You use a more advanced theory (MOT) to predict the geometry, and then you use the table above to find the hybridisation. That's a step backwards though, you already used MOT and there's no reason why you should go back to hybridisation.

  Why this formula does not work for coordination compounds?

Because hybridisation theory is an incomplete theory and cannot capture the subtle factors influencing geometry of coordination compounds.

Main group compounds

I guess I will elaborate a bit more on main group compounds. The formula you quoted seems to work for main group compounds, but let's get two things out of the way.

It gives the correct hybridisations as espoused by old theory. $\ce{SF6}$ is not $\mathrm{sp^3d^2}$ hybridised because d-orbital involvement in bonding is negligible. So, this formula does not actually work.
It does not predict the hybridisation without predicting the geometry.

Point 1 is important, but let's not go into that now. Point 2 may seem confusing to you. Clearly it is predicting the hybridisation, simply from the number of valence electrons, etc. etc.?

The truth is that this formula only "works" because these quantities (number of valence electrons, number of bond pairs, ...) allow one to predict the geometry, using VSEPR theory. For example, the structure of $\ce{CH4}$ is tetrahedral, and VSEPR theory is sufficient to predict that. Once the geometry has been established as tetrahedral, the hybridisation can then be correctly assigned as $\mathrm{sp^3}$.

So, this formula is skipping a step in between. In fact, the correct sequence is to predict the geometry using VSEPR theory, and then obtain the hybridisation. Now, with transition metal complexes, VSEPR does not work. Thus, it is no surprise that that formula also does not work.


Good collection of images showing periodicity of the Table?

I have an entire chapter regarding periodicity in my textbook. Have a look at the images in it:

Atomic Radius:

Ionization Enthalpies:

A summary of periodic trends:

These are the pictures in my textbook. As a bonus, I'll even add a few examples from our day-to-day life:

Alkali Metals:
Lithium:

Sodium:

Potassium:

Rubidium: (its enclosed in glass because it spontaneously combusts in air)

Cesium:

You can notice that properties of elements periodically repeat. All the alkali metals have silvery appearance, with cesium having a little bit of yellow-golden in it. They are soft, and have exceptionally high reactivity.

Halogens:

Fluorine:

Chlorine:

Bromine:

Iodine:

You can see that all the halogens are quite colored, and most have low melting and boiling points. Thats why fluorine and chlorine are gases, while bromine is a liquid with thick vapors. Iodine, however is a solid due to higher attractions due to increased molecular mass. All the halogens are quite reactive, with fluorine and chlorine violently so.

Transition Metals (a few examples):

Iron:

Cobalt:

Nickel:

Chromium:

You can see how the transition metals are quite similar to each other. They are lustrous in appearance, are quite hard, and also display quite a few catalytic properties. For example, iron is used in the manufacture of ammonia, nickel is used to hydrogenate vegetables oils, palladium is used in catalytic converters, et cetera.


If you took the 100 mL sample of air, and took out everything but the oxygen, then the amount of oxygen trapped would fill 21 mL at STP.

One way to think of it is, even though the oxygen would expand to fill the container, its pressure would need to reduce according to Boyle's law: P1V1 = P2V2. This is where the concept of Partial Pressure comes from. But if you then compressed that oxygen so that P = 1 atm (since you specified initial pressure to be 1 atm), the oxygen would take up 21 mL of volume.


Do eta-4 benzene ligands participate in Diels Alder or other cycloadditions?

Coordinated benzene as a dienophile

It would be logical to suggest that an η4-benzene ligand could behave as a dienophile. The reason why benzene is characteristically unreactive, both as a dienophile and as a diene, is because of its aromaticity; if aromaticity is lost, one would expect it to become more reactive.

However, to the best of my knowledge (and SciFinder-fu), such reactivity does not seem to have been achieved. :(

In the rest of this answer I'll try to summarise some of the research that is (somewhat) related to what you were asking about.

Benzene as a dienophile - in general

Benzene itself (and some substituted benzenes) can act as a dienophile under some conditions:

with 1,2,4,5-tetrazines which act as electron-poor dienes (i.e. inverse demand Diels–Alder).1 Presumably, the reaction is favoured due to the loss of $\ce{N2}$ gas.

Under high pressure.2

Intramolecularly, when constrained to be in close proximity with a reactive diene.3

With incredibly reactive dienophiles (cyclobutadienes).4

(This list is probably not exhaustive.)

Coordinated benzene, but as a diene

In ref 4, Inagaki et al. also summarised the ways in which benzene has been activated towards Diels–Alder reactions:

  Currently, there are only a limited number of options to activate benzene sufficiently, including coordination of a transition metal complex to benzene, naphthalene with small resonance energy and the use of benzene derivatives with a sterically induced strained geometry [...] (emphasis mine)

The only reference given for the bolded sentence is the same article that Klaus has linked in his answer.5 It describes the reaction of η2-benzene as a diene, with a hypothesis very similar to what you have written in the question:

  In the proper coordination environment, both osmium and rhenium can form stable complexes with arenes in which only two of the six carbons are coordinated. Once complexed in a dihapto fashion, the uncoordinated portion of the arene more closely resembles a conjugated diene. We therefore hypothesized that the complexation should increase the propensity of arenes to undergo cycloaddition reactions.

In this reaction scheme taken from the paper, the carbons numbered 1 through 4 are acting as the diene component in a Diels–Alder reaction with N-methylmaleimide:

References

Seitz, G.; Hoferichter, R.; Mohr, R. Benzol und linear anellierte Arene als Dienophile in der Diels-Alder-Reaktion mit inversem Elektronenbedarf. Angew. Chem. 1987, 99 (4), 345–346. DOI: 10.1002/ange.19870990413.
Jarre, W.; Bieniek, D.; Korte, F. Benzene as Dienophile in the Diels-Alder Reaction. Angew. Chem. Int. Ed. 1975, 14 (3), 181–182. DOI: 10.1002/anie.197501812.
Saito, K.; Omura, Y.; Maekawa, E.; Gassman, P. G. The Addition of 1,3-Diphenylisobenzofuran to Cyclooctatetraene and to Dimethyl Cyclooctatetraene-1,2-dicarboxylate. The Benzene Ring as a Dienophile in an Intramolecular Diels–Alder Reaction. Bull. Chem. Soc. Jpn. 1990, 63 (2), 395–400. DOI: 10.1246/bcsj.63.395.
Inagaki, Y.; Nakamoto, M.; Sekiguchi, A. A Diels–Alder super diene breaking benzene into C2H2 and C4H4 units. Nat. Commun. 2014, 5. DOI: 10.1038/ncomms4018.
Chordia, M. D.; Smith, P. L.; Meiere, S. H.; Sabat, M.; Harman, W. D. A Facile Diels−Alder Reaction with Benzene: Synthesis of the Bicyclo[2.2.2]octene Skeleton Promoted by Rhenium. J. Am. Chem. Soc. 2001, 123 (43), 10756–10757. DOI: 10.1021/ja011689q.


Should data tables be made in pen or pencil?

In the real world, chemists have a lab journal to document what they did; Good Scientific Practice guidelines state that results should be written there immediately, readably, non-removably, directly and a few other adverbs. ‘Great’ scientific practice or patent-safe documentation would require a witness.

In universities, I have heard of a trend of more and more lab journals being replaced by computer files. In the labs I have worked in, this was not the case and the paper lab journal remained the way to go. Apparantly, if you want to document something in a patent-safe manner, a paper book is mandatory.

Since the entries into a lab journal must not be removable, the only option you have is ink; usually, ball-point pens (biros) are used.

However, the wording of your question (‘chemistry lab’ and ‘data table’) makes me think you could still be a student. In that case, the only advice you should take is that of your teacher/advisor/…


Is iron in the brown ring compound in a +1 oxidation state?

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

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

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


I agree with Jan. The "coffee molecules", probably mostly alkaloids like caffeine, have larger and heavier structures than water molecules. I would expect most have dipole-dipole intermolecular forces, and (minor) hydrogen bonding may be present for some.

When you heat the solution with thermal energy, when we zoom in to the molecular scale we see this energy us actually kinetic. The molecules move faster, and some at the surface break free of their attractions to their neighbors. Despite water being held together by the strong hydrogen bonding intermolecular force, water molecules will be first and most to gain enough energy to pop out of the liquid solution. As Jan said, there are many more water molecules that coffee solute molecules. Also, the coffee solute's greater weights will make them harder to bring up to speed (though as a general rule of thumb IMFs like H-bonds actually play a more significant role than mass).

So those water molecules that pop out are far from their neighbors, so they constitute what we call a gas or vapor. There is much more water-gas above the solution than coffee-molecule-gas. This water vapor exerts a pressure based on how many molecules there are (and depending on kinetic energy / temperature), and we call this pressure from the gas above the coffee the "vapor pressure of the solution". The vapor pressure of water, despite being low relative to many organic liquids, is far higher than the vapor pressure of the molecules from coffee.

Think about the coffee residue in your pot, or the grounds - do they evaporate? Essentially no. There is essentially no vapor pressure, they are too bound by their IMFs to pop out and become gas (at temperatures a human usually encounters anyway). But water has a high vapor pressure, it is easily evaporated.

So when the gaseous water molecules encounter cooler surfaces, they will transfer some kinetic energy to the surface and transition from energetic gas state to lower-energy liquid state. This is the meaning of "condense". No coffee molecules condense, because none were in the gas phase to begin with (essentially).

Jan had a great point that these concepts are the foundation of the incredibly important process of distillation.


This phase diagram has one critical point - it is where the line separating the liquid and gas comes to an end - it is now one fluid. So, you can go from liquid to gas (or vice versa) by swinging T,P around the critical point without undergoing a first order phase transition.

The points with arrows pointing to them are triple points, which are also special points, but different from the critical point. Movement away from a triple point means that one (or more) of the phases have to undergo a first order phase transition to the stable phase(s).

Thermodynamically, these are very different things.


Physical properties of geometric isomers

In the case of fumaric acid, intermolecular hydrogen bonds are the reason beyond its higher melting point compared with the melting point of maleic acid ( where intramolecular hydrogen bonds dominate).


You left out the adjective “reversible.”  The entropy change for an adiabatic reversible process is zero.  In an adiabatic reversible process, the temperature also changes.


Pipetting strong acid with plastic material

This depends on type of plastic. You definitely can if plastic is perfluoropolyetylene (Teflon). You probably can if plastic is polyethylene or polypropylene, but both will degrade in long term (polypropylene will do it faster). You should not, if plastic is PET or other polyester, or polystyrene.

It is probably most reliable to find which plastic you dealing with and search for 'chemical resistance chart' for this plastic.


2 problems with elimination-addition of halobenzene using NH2-

1. Electrophilicity of the triple bond
Hard-soft acid-base theory ties in with this question. The general trends in nucleophile-electrophile reactivity show that hard (highly charge dense) nucleophiles have a tendency to react well with hard electrophiles, and soft (low charge density) nucleophiles with soft electrophiles. This relationship is due to two competing factors; electrostatics and overlap of frontier orbitals. Hard-hard interactions lend their stability to electrostatic attraction, while soft-soft interactions rely on a good HOMO-LUMO overlap.
A π-bond is generally considered a soft species that doesn't reply on electrostatics very strongly. Good soft nucleophiles have relatively high HOMO's (highest occupied molecular orbitals) and good soft electrophiles have relatively low LUMO's (lowest unoccupied molecular orbitals). The third bond in benzyne is abnormal in that it doesn't involve overlap of 2 unhybridised p-orbitals, but 2 hybridised $sp^2$ orbitals that aren't completely parallel. From Clayden Organic Chemistry:

This will, as expected, result in a very weak orbital overlap. As a result, the split in energy between the resulting bonding and antibonding orbital is very small, giving a very low LUMO which can easily accept a nucleophile's electrons. From here:

This is the source of the triple bond's electrophilicity.
2. The final protonation step
(Image from Clayden Organic Chemistry)

Your assumption is correct that the anion is very unstable. According to this source, , benzene has a $\mathrm{p}K_\mathrm{a}$ of 43. This is due to the negative charge being situated in an $\pu{sp^2}$ orbital, which is highly charge dense, with no delocalisation into the ring (since it is not a part of the π-system). Ammonia has a $\mathrm{p}K_\mathrm{a}$ of 38, so its conjugate base, $\ce{NH2^-}$, is an extremely good base. This value is still lower than benzene's, so the benzene anion is still less stable. By a quick calculation, this gives the reaction above an equilibrium constant of roughly $\pu{10^5}$, heavily in favour of ammonia deprotonation.


When using Gaussian to calculate NMR, what's the default solvent and frequency?

While Greg's answer describes sufficiently what the program does, I would like to extend a little on his remark. (This should have been a comment, but it was too long.)
The HF/STO-3G level of theory is what you call the minimised ab initio approach. It is unreliable in many ways, but it is fast. However, even geometries of well known and/or calculated molecules may be wrong. In short, it fails more often than it is correct. The problems start with HF, since it does not describe correlation sufficiently. (It only describes Pauli correlation, but this is exact.) The next problem is the basis set: It substantially lacks basis functions. It is called the minimal basis for a reason. Using this setup for NMR calculations is worse than guessing.
The first thing to do would be to increase the basis set. Maybe move away from Pople basis all together. A few basis set families come to mind: Dunning's augmented family, aug-cc-pVDZ and larger, Ahlrichs and Weigend's def2 family, def2-SVP and larger, and many others, see EMSL.
Calculating NMR tensors is always an accuracy versus time tradeoff. And it might be complicated to find the best (in the sense of most efficient) setup. The best way is to start a geometry optimisation on quite a low level. For reasonably large molecules a semi-empirical model like pm6 can be used to scan for configurational space.
Increase the level of theory to pure DFT. You can use density fitting here to speed up your calculation: e.g. #P BP86/Def2SVP/W06 DenFit.
Validate your geometries with a higher basis set, e.g. #P BP86/Def2TZVPP/W06 DenFit and different functionals, e.g. TPSS, PBE1PBE, M06, etc.
Validate your geometries with the dispersion correction of Grimme, e.g. BP86 EmpiricalDispersion=GD3BJ, PBE1PBE EmpiricalDispersion=GD3BJ, B2PLYPD3, etc.
Apply solvent corrections for your system, e.g. SCRF=(Solvent=Chloroform). Make sure to run single point calculations first and try optimisations later.
If you have fully converged geometries at DFT level, you might want to check basic properties with ab initio techniques. Most appropriate would be perturbation theory, e.g. MP2.
After you have done all the above you can go ahead and calculate NMR tensors. It may be combined with solvent approximations. It requires a very well converged structure, a local minimum on the potential energy hyper surface. (Check that with a frequency calculation, e.g. FREQ.) You can set it up like this sample input:
%oldchk=filenameOfPreviousOptimization.chk
%chk=newCheckpoitfile.chk
#P MP2/def2TZVPP
SCRF(solvent=chloroform)
NMR(GIAO)
geom=AllCheck guess=read

Be advised that that computing NMR data with only one method can lead to many misinterpretations. It is very important to verify your structures and tensor on a variety of methods. If you cannot afford MP2 with a large basis, then you should at least check 4 or 5 DFT methods. It is also important to know, that including solvent effects is still something where the theory is not exact. If you have a system that changes a lot with the solvent, then you should consider explicit treatment. Then you might want to dive into molecular modelling, molecular dynamics, MM/QM, ONIOM, etc. pp.


Both can be formed. It depends on the amount of water added:
\begin{align}
&&\ce{XeF6 + H2O  &->  XeOF4 + 2HF}\\
&\text{or}&
\ce{XeF6 + 2H2O &-> XeO2F2 + 4HF}\\
&\text{or even}&
\ce{XeF6 + 3H2O &-> XeO3 + 6HF}
\end{align}


What is the difference in the reducing power of various ionic hydrides?

This is a very generalized answer but I'll go for it anyhow, who knows how it'll turn out to help (if it does add to what you already know, I suspect you may already know about what I'm going to type....)

The metrics you're considering are 'Reactivity' and (since you brought up LiAlH) 'Reducing character'. Clump these together, and it turns out what you're looking for is the ability for these hydrides to donate/release their hydrogen.

Now I've reasons to believe you already know the 'reactivity' trend for p-block hydrides (correct me if I'm wrong). Figuring out the reducing character of the s-block ionic hydrides involves similar logic.

In the p-block hydrides, you've probably learn't that as you move down the group, the E-H bond length increases (I use 'E' to denote p-block elements). If the bond length increases, the easier it is to cleave, therefore the hydrogen is released more readily. Here's an analogy if that bit seems a bit hard to grasp. Think of the E-H bond as a pencil. The longer it is, the easier it is to break; the shorter it gets the more difficult it is to break (try it......P.S-this credits for this analogy goes to my Chem. Teacher).

In a nutshell, for the p-block hydrides, as you move down the group the tendency to donate hydrogen increases.

Now back to the s-block. You can't apply the logic you used for p-block hydrides here (at least not directly ;) ).

At first glance, the main impediment you'd see is "s-block hydrides are ionic but p-block hydrides are covalent, so I can't figure out the trend the same way...". You'd be correct, but if you trying extrapolating (in reverse) the logic you use for p-block hydrides, see what happens.....

Since they're all ionic hydrides, and since you want the reactivity trend, what you'd want to do is compare the covalent nature of each s-block hydride.

As you'd already know, an ionic bond isn't 100% ionic, and the same can be said for covalent bonds. You call them ionic or covalent, if the ionic or covalent nature of a bond is relatively significant.

So by using Fajan's rules (assuming you know those as well) you can figure out that LiH is the most 'covalent' of the ionic s-block hydrides, and therefore more prone to donating hydrogen thereby acting as a reducing agent. This Wikipedia page just confirms that...

As you move down Groups 1 and 2, the reducing character of the s-block hydrides decreases.

Also LiAlH and NaBH are much, much stronger reducing agents than plain ole' s-block hydrides.


Analytical solution for kinetics of bimolecular reaction

Assuming that the bimolecular chemical reaction $\ce{A + B ->[\kappa] C}$ has mass action kinetics, we have the following pair of coupled ODEs

$$\begin{array}{rl} \dot a &= - \kappa \, a \, b\\ \dot b &= - \kappa \, a \, b\end{array}$$

where $\kappa > 0$ is the rate constant, $a := [\ce{A}]$ and $b := [\ce{B}]$. Since $\dot a = \dot b$, we have $\frac{\mathrm d}{\mathrm d t} \left( a - b \right) = 0$ and, thus, integrating, we obtain

$$a (t) - b (t) = a_0 - b_0$$

where $a_0 > 0$ and $b_0 > 0$ are the initial concentrations. Since $b (t) = a (t) - (a_0  - b_0)$, the 1st ODE can be decoupled from the 2nd, as follows

$$\dot a = - \kappa \, a \, \left( a - (a_0  - b_0) \right)$$

which can be rewritten in the form

$$\frac{\mathrm d a}{a \, \left( a - (a_0  - b_0) \right)} = - \kappa \, \mathrm d t$$

Assuming that $a_0 \neq b_0$, we have the following partial fraction expansion

$$\left( \frac{1}{a - (a_0  - b_0)} - \frac{1}{a} \right) \mathrm d a = - \kappa \, (a_0  - b_0) \, \mathrm d t$$

Integrating, we obtain

$$\ln \left( \frac{a (t) - (a_0  - b_0)}{a_0 - (a_0  - b_0)} \right) - \ln \left( \frac{a (t)}{a_0} \right) = - \kappa \, (a_0  - b_0) \, t$$

which can be rewritten as follows

$$\ln \left( \frac{a (t) - (a_0  - b_0)}{a (t)} \right) = \ln \left( \frac{b_0}{a_0} \right) - \kappa \, (a_0  - b_0) \, t$$

Exponentiating both sides, we obtain

$$\frac{a (t) - (a_0  - b_0)}{a (t)} = \frac{b (t)}{a (t)} = \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)$$

and, eventually, we obtain

$$\boxed{\begin{array}{rl} &\\ a (t) &= \dfrac{a_0  - b_0}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}\\\\ b (t) &= \dfrac{(a_0  - b_0) \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}\\ & \end{array}}$$

Taking the limit,

$$\lim_{t \to \infty} a (t) = \begin{cases} a_0 - b_0 & \text{if } a_0 > b_0\\\\ 0 & \text{if } a_0 < b_0\end{cases}$$

$$\\$$

$$\lim_{t \to \infty} b (t) = \begin{cases} 0 & \text{if } a_0 > b_0\\\\  b_0 - a_0 & \text{if } a_0 < b_0\end{cases}$$

What if $a_0 = b_0$?

Previously, we assumed that $a_0 \neq b_0$. If $a_0 = b_0$, then

$$\frac{\mathrm d a}{a \, \left( a - (a_0  - b_0) \right)} = - \kappa \, \mathrm d t$$

becomes

$$-\frac{\mathrm d a}{a^2} = \kappa \, \mathrm d t$$

Integrating, we obtain

$$\frac{1}{a (t)} - \frac{1}{a_0} = \kappa \, t$$

and, eventually, we obtain

$$\boxed{ a (t) = \frac{a_0}{1 + a_0 \, \kappa \, t} = b (t)} $$

In this case, both reactants are eventually exhausted

$$\lim_{t \to \infty} a (t) = \lim_{t \to \infty} b (t) = 0$$


Why is the formation of double helix between non complementary sequences disfavoured?

If you take a look at the base pairings, as shown below, you can see that adenine and thymine form two hydrogen bonds when they are matched, while guanine and cytosine form three hydrogen bonds when they are matched. If you try to match an adenine with cytosine or guanine with thymine, the orientation of hydrogens and lone pairs will not line up.

For example, guanine has two hydrogens and one lone pair available for hydrogen bonding, while cytosine has one hydrogen and two lone pairs available for hydrogen bonding. They fit! However, if you try to match adenine (with one hydrogen and one lone pair available for hydrogen bonding) with cytosine, the elements will not match or line up, resulting in an energetically less-favorable interaction.

Summary: A base on one strand of DNA requires another, specific base to complement the exposed hydrogens and lone pairs to form the most most energetically favorable interactions.


What are the oxidation states in Borohydride (BH4-)

I would have expected the oxidation states to be +III for B and -I for
  H.

You're right, boron is in the (+3) oxidation state in these equations.

In $\ce{B(OH)_3}$ each $\ce{OH}$ is (-1) and the boron is (+3), overall the molecule is neutral

In $\ce{BH_{4}^{-}}$ each $\ce{H}$ is (-1) and the boron is (+3), overall the ion is (-1)


White creamy substance (not calcium) observed when boiling water is filtered through coffee paper filter

I assure you it is calcium carbonate.

The particles are so small they do not just sink to the bottom like a stone. Instead, they aggregate at the nearest phase boundary. All small impurities do that, unless they have a very low interfacial tension with water. It's energetically much more favourable than sinking.

Over time, this creamy substance aggregates further, up to a point where the aggregates are large and dense enough to dissociate from the surface, and sink to the bottom.


On heating in the Earth's atmosphere, can magnesium react with nitrogen to form magnesium nitride?

A large pile of grey magnesium powder, when lit in air, produces a smouldering pile which cools down to reveal a crusty white solid of magnesium oxide. However, if you break apart the mound, you can find something quite strange in the middle - a clearly brownish powder that wasn't there before.
Seeing is believing! The author of the video also has a clever idea to identify the brown solid. By adding water and placing some moist pH paper above the puddle, it clearly shows the transfer of some alkaline substance across the gap. This is ammonia gas, $\ce{NH3}$, whose presence is explained by the hydrolysis of magnesium nitride:
$$\ce{Mg3N2(s) + 6H2O(l) -> 3 Mg(OH)2(aq) + 2 NH3(g)}$$
It is important that the pH paper not come in direct contact with the water used to hydrolyze the magnesium oxide, as $\ce{Mg(OH)2}$ is itself also basic, and could also be formed by reaction with either $\ce{MgO}$ or $\ce{Mg}$ directly. Only $\ce{Mg3N2}$ produces a basic gas which forms an alkaline solution in water.
As you can see, magnesium metal does react directly with molecular nitrogen ($\ce{N2}$) when burned in air. However, the reaction is thermodynamically and kinetically less favourable than the reaction with molecular oxygen ($\ce{O2}$). This is almost certainly due to the extreme strength of the bond between nitrogen atoms in molecular $\ce{N2}$, whose bond dissociation energy of $\mathrm{945\ kJ\ mol^{-1}}$ is one of the strongest in all of chemistry, second only to the bond in carbon monoxide. For comparison, the bond dissociation energy of molecular $\ce{O2}$ is drastically lower, at $\mathrm{498\ kJ\ mol^{-1}}$.
So why did the Chem13 magazine article referenced in Aniruddha Deb's answer not find any magnesium nitride? It is likely that 1 g of magnesium metal is far too little for the experiment run under their conditions. It takes a significant amount of "sacrificial" magnesium to completely consume the oxygen in its surroundings. Only once practically all the oxygen is consumed (and while the pile of magnesium is still hot enough from the reaction between magnesium and oxygen) will the remaining magnesium metal react with the nitrogen in air. Alternatively, the reaction would have to be performed in an oxygen-free environment. Magnesium metal is such a strong reductant that many substances can act as an oxidant for it, including pure $\ce{CO2}$ (also shown in the video above) and water (never put out a magnesium fire with water!).


Can metal oxides react with water to form a hydroxide, then that with HCl to make a salt?

tl;dr: Yes, depending on extent of reaction

The group 1 and (group 2) hydroxides for example: $\ce{NaOH,  KOH, RbOH, CsOH, Ba(OH)2, Ca(OH)2}$ are strong bases, being essentially fully ionized in aqueous solution; (although some are weaker $\ce{LiOH}$, $\ce{Mg(OH)2}$) . These should react with $\ce{HCl}$ - a strong acid forming water and a neutralisation product a typical corresponding salt $\ce{MCl}$ or $\ce{MCl2}$ .

Being soluble in aqueous envronment implies the salt is also obtained in this phase, so the question will be on extration method to use, in order to get this salt product in solid form.

Why I said depending on extent of reaction, is beacuse some metal oxides e.g $\ce{MgO}$ hardly form the hydroxide in the first place, and also insolubility with render the second reaction effectively out of question.


Configuration of endocyclic double bond in Vitamin A

The endocyclic double bond of vitamin A has a (Z) configuration. Here’s why:

First, let’s determine priorities of the atoms connected on the top side of it:

C

C (ring)
C (methyl 1)
C (methyl 2)

C

C
(C) (ghost atom; due to the double bond)
H

So the ‘ring side’ — the atom on the left in the way it is drawn at the moment — has a higher priority than the side chain (C at (1.3) beats H at (2.3)).

What about the bottom side?

C (ring)

C
H
H

C (methyl group)

H
H
H

Here again the ‘ring side’ has a higher priority than the methyl group: (1.1) carbon beats (2.1) hydrogen.

Both left-hand sides have a higher priority than their corresponding right-hand side. Thus, the double bond is (Z) configured.


Which chemical should I use to wash soot precipitate from cast iron?

Depends on your resources and how delicate the object is (is it an antique?).

You can put it in a furnace or self-cleaning oven to burn-off the soot. Determining the correct temperature for this will be an art.
You could put it in a furnace under a hydrogen atmosphere.  the hydrogen will react with the carbon to form hydrocarbon gases and it definitely won't oxidize the iron.  You can also use carbon dioxide which will for carbon monoxide with the soot.  You may not have this sort of equipment though.
You can clean the object in an ultrasonicating bath. These are widely available and the best part is that you will not require chemicals to remove the soot.  This method may take a while but it is the most gentle.


At what acid concentration does potassium metabisulfite react?

Potassium metabisulfite ($\ce{K2S2O5}$) is a white crystalline powder, which also known as potassium pyrosulfite. It is mainly used as an antiseptic for winemaking, process engraving, and as a source for sulfurous acid. It also acts as a potent antioxidant, and hence use as a food preservative (known as E224). For example, potassium metabisulfite is sometimes added to lemon juice as a preservative (Wikipedia). Lemon juice contains citric acid ($\mathrm{p}K_{\mathrm{a1}}= 3.13$) and ascorbic acid (Vitamin C; $\mathrm{p}K_{\mathrm{a1}}= 4.17$) as ingredients, and hence, has $\mathrm{pH}$ ranging from 2 to 3. For that reason alone, your concern about reaction of $\ce{K2S2O5}$ with citric acid or a mixture of acids within $\mathrm{pH}$ 3-4 has no base (a good news).

However, a major reason for addition of potassium metabisulfite ($\ce{K2S2O5}$) as a food preservative is to have sulfur dioxide ($\ce{SO2}$) as an active ingredient. Its melting point is listed as $\pu{190 ^{\circ}C}$, but it decomposes to two compounds:
$$\ce{K2S2O5(s)  -> K2SO3(s) + SO2(g)}$$
It is also in equilibrium with $\ce{K2O}$ and $\ce{SO2}$ when contacted with water (Ref.1):
$$\ce{K2S2O5(aq)  -> K2O(aq) + 2SO2(aq)}$$
In the aqueous media, there is an equilibrium between molecular sulfur dioxide ($\ce{SO2}$; E220) and the anions, bisulfite ($\ce{HSO3-}$) and sulfite ($\ce{SO3^2-}$), which are collectively and conventionally defined as ‘free $\ce{SO2}$’ (Ref.1). The reference also states that:

  The active substance is sulfur dioxide (($\ce{SO2}$, E 220) which in the present application is generated in situ from potassium metabisulfite ($\ce{K2S2O5}$, E 224).

During their investigation:

  The ANS Panel noted that sulfur dioxide, bisulfite and sulfite ions coexist in solution in a series of equilibria and that a read across between the different sulfites used as food additives is possible.

In general, following equilibria exist in aqueous solution of $\ce{S2O5^2-}$ (potassium or sodium):
$$\ce{S2O5^2- + H2O  <=> 2 HSO3-}$$
$$\ce{HSO3- + H2O  <=> 2 SO3^2- + H3O+}$$
$$\ce{HSO3- + H3O+  <=> SO2(H2O) + H2O}$$
Where $\ce{SO2(H2O)}$ is called hydrated $\ce{SO2}$.

A final note: Iowa State University has listed following values as desired 'free $\ce{SO2}$’ by $\mathrm{pH}$ of wine:

$$
\begin{array}{ccc}
\hline\
\text{pH of wine} & '\text{free }\ce{SO2}', \pu{ppm} & & \text{pH of wine} & '\text{free }\ce{SO2}', \pu{ppm} \\\hline
2.9 & 11  &   & 3.5 & 40  \\
3.0 & 13  &   & 3.6 & 50  \\
3.1 & 16  &   & 3.7 & 63  \\
3.2 & 21 &  & 3.8 & 79  \\
3.3 & 26 & & 3.9 & 99 \\
3.4 & 32 & & 4.0 & 125  \\\hline
\end{array}
$$

References:

Claudia Bolognesi, Laurence Castle, et al. (Panel members), EFSA CEF Panel (EFSA Panel on Food Contact Materials, Enzymes, Flavourings and Processing Aids), “Safety assessment of the active substance potassium metabisulfite, for use in active food contact materials,” EFSA Journal 2016, 14(5), 4465, 8 pp. ( doi: 10.2903/j.efsa.2016.4465).


Why is it important to minimize impurities when doping a semiconductor?

Semiconductors are based on impurities.  They just call it doping.  The addition of group 3 (Boron group) or group 5 (Nitrogen group) impurities create a hole or an extra electrno, respectively.  These form charge carriers which are necessary for a semiconductor to be used in a circuit.  However, the dopants need to be added in highly controlled and precise amounts to get the necessary mobility.  The silicon also needs to be highly pure for regular electronic response, for regular electronic properties over arbitrary distances.  Another important consideration is that the dopant allows some degree of tuning of the electrical conductivity of the semiconductor.

Also, highly pure crystalline silicon  fractures nicely along the crystal planes, meaning it can be cut with high precision.  Since most mass-produced electronic processors are made on 12- or 16-inch silicon wafers with dozens of devices manufactured at once, and initially all physically connected, the ability to precisely excise each tiny circuit is critical for high through put and yield in manufacturing.


Number of stereoisomers of polysubstituted cubane

I hope I am not too late.

All carbons are $\mathrm{sp^3}$-hybridised and three bonds of every carbon are fixed (one corner shares three bonds), so fourth one has only one place in space to maintain an $\mathrm{sp^3}$ tetrahedron. This one is optically active, so this and its enantiomer are the only two possible isomers.

Now you get why we don't use the $2^n$ formula. And similar thing happens in a bridgehead position.


How is it that earth's inner core is still solid?

The melting point of the vast majority of materials increases as more pressure is applied. At https://www.researchgate.net/figure/312054885_fig1_Figure-231-Experimental-pressure-temperature-phase-diagram-of-pure-iron-a is a phase diagram showing how this variation occurs for pure iron:

From https://physics.stackexchange.com/questions/184032/what-is-the-pressure-at-the-center-of-the-earth-or-a-neutron-star the pressure at the centre of the Earth is 365GPa. At this pressure we can see from the graph that the melting point of pure iron is well over 5000k. Thus, given the experimental uncertainties and the that the core is not pure iron, it is easily conceivable that the core of the Earth is solid, or at least a large proportion of it is.