How to make pyramid-shaped NaCl salt crystals

TL;DR: Use 10-20% urea solution for octahedral crystals and a water-ethanol mix for pyramidal crystals.

There is no universal method to predict and to control crystal habits. Very often the experimental results are explained with molecular dynamics; there are models (Wulff condition, Donnay–Harker correlation, Hartman–Perdok theory, Hartman–Bennema attachment energy) that can only be used to represent vapor-grown crystals. Crystal morphology is vastly dependent on many factors:

phase it's growing from (gas, liquid, melt);
growing rate (temperature);
concentration of the mother solution;
ionic strength of the solution;
presence of impurities (habit modifiers);
solvent etc.

Rock salt can take one of the three morphologies dictated by its FCC lattice with a strong tendency to prefer cubic crystals:

Figure 1. Morphologies of $\ce{NaCl}$ crystal denoted with Miller planes: cube $\{100\}$; octahedron $\{111\}$; rhomb-dodecahedron $\{110\}$.

There are numerous literature sources providing experimental information on crystal growth. For sodium chloride there is a recent review by Aquilano et al. [1], not only providing a brief description of common habit modifiers, but also giving an insight on to why one crystal direction is preferred:

  The morphology of natural halite is largely dominated by the cube $\{100\}$; the $\{110\}$ and $\{111\}$ forms are seldom present. When crystals grow by evaporation from
  aqueous solution the perfection of the cube faces is usually lost; owing to the high values of the supersaturation, hopper shaped $\{100\}$ faces appear along with dendritic
  branches developing in the $❬111❭$ directions. This is the main reason of the halite tendency to cake. Hence, the best way to face this problem is to search for additives
  able to favor the appearance of the $\{111\}$ form, thus avoiding the morphological instability such as the “hazardous” dendritic growth.

Here is a brief summary for the non-standard crystal shapes of rock salt and corresponding modifiers:

Octahedra. Formamide; urea; sodium tetraphosphate $\ce{Na6P4O13}$; cadmium chloride $\ce{CdCl2}$; polymer additive containing an amide functional group.

Dendrites and needles. Hexacyanoferrate(II) $\ce{Fe(CN)6^4-}$; polyvinyl alcohol.

Pyramides. Water-ethanol solution; growth from a supersaturated solution [2].

Presence of heavy metals ($\ce{Pb^2+}$, $\ce{Cd^2+}$) generally promotes formation of large single crystals instead of many small crystals. For more practical details, photos of crystals and manuals you can refer to Crystal Growing Wiki (originally in Russian, but now most articles are translated to English).

References

Aquilano, D.; Otálora, F.; Pastero, L.; García-Ruiz, J. M. Three Study Cases of Growth Morphology in Minerals: Halite, Calcite and Gypsum. Progress in Crystal Growth and Characterization of Materials 2016, 62 (2), 227–251. https://doi.org/10.1016/j.pcrysgrow.2016.04.012.
Desarnaud, J.; Derluyn, H.; Carmeliet, J.; Bonn, D.; Shahidzadeh, N. Hopper Growth of Salt Crystals. J Phys Chem Lett 2018, 9 (11), 2961–2966. https://doi.org/10.1021/acs.jpclett.8b01082.


Why constant volume combustion gives higher energy than constant pressure?

Combustions of organic stuffs produce a greater number of gaseous moles than in the reactants. When done at constant pressures, a part of the energy produced by the combustion must be used for repelling the outer atmosphere. This is a work and it costs some enegy. This is why a combustion at constant pressure produces less energy than at constant volume.


What are inner core electrons and how do they influence chemical and physical properties?

Inner core electrons do have a very strong affect on the chemistry of elements,for example inner d orbitals have a very poor shielding effect thus the outer electrons are bounded more strongly to the nucleus thus decreasing their reactivity and oxidation states.You will come across inert pair effect,stability of lower oxidation states down the group etc because of effect of inner core electrons.


Molecular geometry of BaH2 according to VSEPR theory

There is a d-Orbital-Contributions to HOMO in d0 MX2-Complexes.
All higher‐level calculations show SrH2 and BaH2 to be bent with angles of ∼140° and 120°, respectively, while CaH2 is linear with a flat potential‐energy surface for the bending motion.
The use of a core‐polarization potential together with the 2‐valence‐electron pseudopotential approach allows an investigation of the relative importance of core‐polarization vs direct d‐orbital bonding participation as reasons for the bent structures. The calculations strongly suggest that both contribute to the bending in SrH2 and BaH2. Thus, the bent structure of [BaH2] may be considered to arise from a stabilization of the b2 HOMO upon bending by mixing in metal dyz character in second-order perturbation theory.
Reference:
https://core.ac.uk/download/pdf/232777714.pdf
http://www.chimdocet-inorganica.it/SITO_ESERCIZI/Complementi/COMP1/VSEPREccezioni.pdf
This observation has been rationalized successfully by an increased involvement of (n - 1)d-orbitals in the bonding as M gets larger,19-21
and by the influence of core-polarization interactions between the M and X sites.2 Elements in the d-block have relatively high atomic masses and they tend to have stereochemically inactive electron pairs.


Not necessarily. The vapor pressure of a liquid (or solid) is the pressure of the vapor in equilibrium with its condensed phase in a closed system. The boiling point of a liquid at a given pressure is the temperature at which the vapor pressure is equal to the given pressure. This means the vapor pressure is a function of the temperature,
$ P^* = f(T) $, and the boiling point is a function of the pressure, $ T_{bp} = f(P) $.

So your question is a bit vague. In general, yes, liquids with higher vapor pressures will have lower boiling points. In any first year chemical engineering course, students are introduced to the Antoine Equation, an emperical relation between vapor pressure and temperature. Plotting the Antoine equation for a set of liquids lets one visualize the relationship between vapor pressure and temperature very quickly; these plots are called Cox charts. For example, examine the Cox chart below:

Now look at the relationship between ethanol and benzene. At 50 F, benzene has a higher vapor pressure than ethanol. At pressures below ~10 psi, the general trend holds and benzene has a lower boiling point. However, the lines for benzene and ethanol cross at ~10 psi, so above ~10 psi benzene actually has a higher boiling point than ethanol.

So in general, yes, liquids with higher vapor pressures will have lower boiling points. However, the question is vague and exceptions do exist.


Copper(III) nitrate cannot be obtained from aqueous nitric acid, and likely doesn't exist.
Reaction carried under strongly oxidative conditions between $\ce{Cu(NO3)2}$ and fuming $\ce{HNO3}$ yields in nitrosyl copper(II) trinitrate $\ce{[NO+][Cu(NO3)3−]},$ [1, 2] sometimes written as adduct $\ce{Cu(NO3)2 · N2O4},$ which is contradictory to the crystal structure consisting of copper(II) centers octahedrally coordinated by nitrate groups and the voids in cross-linking mesh of octahedra are occupied by nitrosyl ions [3]:

Figure 1. Crystal structure of nitrosyl copper(II) trinitrate $\ce{[NO+][Cu(NO3)3−]}$ (ICSD-422033). Color code: $\color{#3050F8}{\Large\bullet}~\ce{N}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$.
According to Wiberg's Anorganische Chemie [4, p. 1707], oxidation of $\ce{Cu^2+}$ to $\ce{Cu^3+}$ is possible only with strong oxidation agents $(E^\circ(\ce{Cu^2+/CuO+}) = \pu{+1.8 V}),$ and copper(III) is highly unstable in water and has a tendency to oxidize the medium $(E^\circ(\ce{H2O/O2}) = \pu{+1.23 V}).$
The reduction potential or $\ce{Cu^3+}$ can, however, be reduced down to $\pu{+0.45 V}$ by coordinating it with appropriate strongly donating ligands (which nitrate really isn't).
Examples of such complexes are:

$\ce{K3CuF6}$ (oxidant: fluorine $\ce{F2}$; $\ce{CuF3}$ is unknown);
$\ce{K5[Cu(IO6H)2]},$ $\ce{K5[Cu(TeO6H2)2]},$ $\ce{Na5[Cu(TeO6H2)2]}$ (oxidant: peroxodisulfate $\ce{S2O8^2-}$ in strongly alkaline medium);
$\ce{Na[Cu(OH)4]},$ $\ce{KCuO2},$ $\ce{Ba(CuO2)2}$ (oxidant: hypobromite $\ce{BrO-}$/oxygen $\ce{O2}$; $\ce{Cu2O3}$ is unknown).
Cu(III) complex of 2-ethoxy-5,10,15,20-tetrapentafluorophenyl-3,7-diaza-21,22-dicarbaporphyrin (Cu(III)-$\ce{N2CP}$, a doubly N-confused porphyrin) [5]
A variety of Cu(III) peptide complexes [6, pp. 213–223]

References

Znamenkov, K. O.; Morozov, I. V.; Troyanov, S. I. Synthesis and Crystal Structure of Copper(II) Nitrato Complexes $\ce{NO[Cu(NO3)3]},$ $\ce{Na2[Cu(NO3)4]},$ and $\ce{Ag2[Cu(NO3)4]}.$ Russian Journal of Inorganic Chemistry 2004, 49 (2), 172–179. (in Russian)
Gagelmann, S.; Rieß, K.; Wickleder, M. S. Metal Oxidation with $\ce{N2O5}$: The Nitrosylium Nitrates $\ce{(NO)Cu(NO3)3},$ $\ce{(NO)2[Zn(NO3)4]}$ and $\ce{(NO)6[Ni4(NO3)12](NO3)2·(HNO3)}.$ Eur. J. Inorg. Chem. 2011, 2011 (33), 5160–5166. https://doi.org/10.1002/ejic.201100565.
Blackwell, L. J.; King, T. J.; Morris, A. Crystal Structure of the Copper Nitrate–Dinitrogen Tetroxide Adduct $\ce{Cu(NO3)2 · N2O4}.$ J. Chem. Soc., Chem. Commun. 1973, No. 17, 644–644. https://doi.org/10.1039/C39730000644.
Wiberg, E.; Wiberg, N.; Holleman, A. F. Anorganische Chemie. Nebengruppenelemente, Lanthanoide, Actinoide, Transactinoide, 103. Auflage.; De Gruyter: Berlin; Boston, 2017; Vol. 2. ISBN 978-3-11-049573-7. (in German)
Furuta, H.; Maeda, H.; Osuka, A. Doubly N-Confused Porphyrin:  A New Complexing Agent Capable of Stabilizing Higher Oxidation States. J. Am. Chem. Soc. 2000, 122 (5), 803–807. https://doi.org/10.1021/ja992679g.
Karlin, K. D.; Tyeklár, Z. Bioinorganic Chemistry of Copper; Springer Netherlands: Dordrecht, 1993.


Why does the energy required to break a bond decrease as the number of bonds increase?

You are considering all the bonds equivalent and that's where you are making mistake.The energy required to break second bond will depend upon several factors such as repulsion, configuration, also note that there are different types of bond formed (sigma and pi bonds) and we know that energy required  to break a sigma bond is greater than that of pi bond.


$\ce{Mg^2+}$ forms complexes with 4-(2-Pyridylazo)resorcinol that can be detected spectrophotometrically: Simultaneous spectrophotometric determination of calcium and magnesium in water, described by E. Gómez, J.M. Estela, V. Cerdà in Analytica Chim. Acta 1991, 249, 513-518.

(DOI: 10.1016/S0003-2670(00)83027-5)


Replicating an illustration for the crystal structure of ferrihydrite

I think the software used could be Diamond (http://www.crystalimpact.com/diamond/)


In 1928, Malaprade demonstrated that periodic acid reacted with ethylene glycol to produce iodic acid and formaldehyde (Ref.1). Hence, the oxidation of adjacent diols with periodic acid or its salt in aqueous solution is now generally known as the Malaprade reaction, the mechanism of which is depicted below:

The reaction proceeds faster under acidic conditions and can be applied to higher polyhydric alcohols, which behave similar to ethylene glycol. The necessity for the reaction is to have 1,2-dihydroxy functionality $\left(\ce{>C^1(OH)-C^2(OH)<}\right)$, and upon exposer to periodic acid, the middle $\ce{C^1\!-C^2}$ bond would oxidatively cleave to give two compounds with gem-diol functionality $\left(\ce{>C^1(OH)2 \ and \ (OH)2C^2<}\right)$, which are essentially carbonyl groups in aqueous solutions. The reaction has also been further extended to the cleavage of α‐hydroxy carbonyl compounds, 1,2‐dicarbonyl compounds, α‐amino alcohols, α‐amino acids, and polyhydroxy alcohols since then, and successfully applied for structural analysis specifically of sugars:

(source, Ref.2)
As shown in the figure, each $\ce{C-C}$ oxidative cleavage of $\ce{C-C}$ bond gives $\ce{OH}$ group to each carbon. For example, 1,2-bond chevage between $\ce{-CH(OH)-CHO}$ gives aldehyde carbon another $\ce{OH}$ group, making it formic acid as shown in the diagram. Meanwhile, 1,2- and 2,3-bond cleavages gives $\ce{C}$2 carbon two extra $\ce{OH}$ groups so that resultant molecule is formic acid again. Thus, complete oxidation of $\pu{1 mol}$ of D-glucose gives $\pu{5 mol}$ of formic acid and $\pu{1 mol}$ of formaldehyde.
The requirement is to have 1,2-diol functionality. So, how ketone and aldehyde with $\alpha$-hydroxy groups get oxidized? In aqueous solutions, ketones and aldehydes are in equilibrium with their corresponding gem-diols, thus providing 1,2-diol feature.
The 1,2-diol does not have to be cis-orientation. The trans-diol will also get oxidized but with slower rate (see here and Ref.3).
References:

Zerong Wang, "Malaprade Reaction (Malaprade Oxidation)," In Comprehensive Organic Name Reactions and Reagents; John Wiley & Sons, Inc.: New York, NY, 2010 (https://doi.org/10.1002/9780470638859.conrr406). ISBN: 9780471704508.
Fathia Mohammed Ibrahim,  Mubark Elsayed Osman, "Elucidation of Sugars Structure through Periodic Acid Oxidation Cleavage," International Journal of Science and Research (IJSR) 2018, 7(1), 1152-1155 (https://www.ijsr.net/search_index_results_paperid.php?id=6121702)(PDF)
G. J. Buist, C. A. Bunton, J. H. Miles, “149. The mechanism of oxidation of α-glycols by periodic acid. Part V. Cyclohexane-1 : 2-diols,” J. Chem. Soc. 1959, 743-748 (https://doi.org/10.1039/JR9590000743).


You are correct, but to make it a little more clear you can include the assumed "atom" in the denominator of amu:

$$
\begin{align}
m_{\ce{C}^{12}} &= \pu{12amu atom^-1} \\ \\
m_{\ce{C}^{12}} &= \pu{12g mol^-1} \\ \\
\pu{12amu atom^-1} &= \pu{12g mol^-1} \\ \\
\pu{1amu atom^-1} &= \pu{1g mol^-1}
\end{align}
$$

In other words, the ratio of amu/atom is the same as the ratio of g/mol. The definitions of amu and moles were intentionally chosen to make that happen (I'm surprised your teacher didn't explain this, actually). This allows us to easily relate masses at the atomic scale to masses at the macroscopic scale.

To check this, look at the mass of an amu when converted to grams:

$\pu{1amu}= \pu{1.6605E-24 g}$

Now divide one gram by one mole:

$\pu{1g mol^-1}= \frac{\pu{1 g}}{\pu{6.022E23 atom}} = \pu{1.6605E-24 g atom^-1}$

It's the same number! Therefore:

$\pu{1g mol^-1}= \pu{ 1 amu atom^-1}$


How can I reach the two minima connected by a first-order saddle point (transition state) using the Gaussian Software

how can I use the EF option of Gaussian to perform an optimization along an imaginary mode starting from a transition structure?

I have never used the EF mode of Gaussian 09, but...

  Is there a more elegant way, circumventing manual structural modifications, to yield the two minima connected by the TS?

I believe the standard way of proving that your TS is the TS you want is by an intrinsic reaction coordinate calculation:

http://gaussian.com/irc/

Make sure to check out the "options" tab. A decent write-up on a small system is here:

http://www.cup.uni-muenchen.de/ch/compchem/geom/irc5.html

Basically, you need to restart the calculation using the checkpoint (chk) file of your TS calculation, and as you calculated the vibrational spectrum, you can use "irc=rcfc" to read the force constants from the restart file. By default the IRC calculation does 10 points in both directions of the "reaction coordinate" starting from the TS, you can change this using MaxPoints=n, and you can also specify which direction you want to study using other keywords listed in the "options" tab on the Gaussian website.

Opening the logfile in Gaussview, you can see the IRC profile - the same way you can look at the optimization profile of an "opt" calculation. I'd follow up on the IRC by optimizing each of the end-points of the profile using a standard "opt" calculation, just to make sure.

Hope that helps.


In English, French and German, the names are less systematic. Here are some examples from English:

\begin{array}{|c|c|c|}
\hline
\text{Formula} &  \text{Name} & \text{Oxidation #}  \\
\hline
\ce{FeCl2 }& \text{ferrous chloride, iron(II) chloride} &  +2\\
\hline
\ce{FeCl3 }& \text{ferric chloride, iron (III) chloride} &  +3\\
\hline
\ce{N2O}& \text{nitrous oxide} &  +1 \\
\hline
\ce{NO}& \text{nitric oxide} &  +2\\
\hline
\ce{HNO2}& \text{nitrous acid} &  +3\\
\hline
\ce{HNO3}& \text{nitric acid} &  +5\\
\hline
\ce{NO2-}& \text{nitrite} &  +3\\
\hline
\ce{NO3-}& \text{nitrate} &  +5\\
\hline
\ce{N^3-}& \text{nitride} &  -3\\
\hline
\ce{CrO}& \text{chromous oxide} &  +2\\
\hline
\ce{CrO3}& \text{chromic oxide, chromium trioxide} &  +6\\
\hline
\ce{Cr2O3}& \text{chromium (III) oxide} &  +3\\
\hline
\ce{H2CrO4}& \text{chromic acid} &  +6\\
\hline
\ce{HCl}& \text{hydrochloric acid} &  -1\\
\hline
\ce{HClO}& \text{hypochlorous acid} &  +1\\
\hline
\ce{HClO2}& \text{chlorous acid} &  +3\\
\hline
\ce{HClO3}& \text{chloric acid} &  +5\\
\hline
\ce{HClO4}& \text{perchloric acid} &  +7\\
\hline
\ce{Al2O3}& \text{aluminum oxide} &  +3\\
\hline
\ce{CaO}& \text{calcium oxide} &  +2\\
\hline
\ce{Li2O}& \text{lithium oxide} &  +1\\
\hline
\end{array}

I think this is what happens if names of substances were chosen before the concept of oxidation state (and electrons etc) existed. It would be nice to have something more systematic, but that would mess up all the older literature.

It would be fascinating (to me, at least) to see the Slovak names in another column of this table. Also, I am glad that chemical formulae exist that are universal in any language.


Why does the outside of a plastic cooking oil bottle get sticky over time?

Some drying oils, such as linseed oil, react with oxygen in air to polymerize, eventually forming a hard film. When the oil is in the midst of change, i.e. short-chain polymers, it's a sticky mess (don't touch wet paint).

Most cooking oils are unsaturated. Having double bonds makes the oil easier to digest and less likely to deposit in the circulatory plumbing. It also makes the oil more chemically reactive. When in a closed bottle, it reacts fairly slowly with residual air, eventually becoming rancid. Spread over the surface of a bottle, the oil is exposed to more air and oxidizes and polymerizes more rapidly.

An experiment to test this would be to coat two pieces of the plastic used for bottle, usually PETE, one with cooking oil (unsaturated) and one with mineral oil (mostly saturated) and check if the cooking oil becomes sticky, while the mineral oil should not.

One other factor, though, is that oils will cause dust in the air to adhere, so that might also cause some stickiness. Place the test swatches in a container covered with paper towels to keep off dust.


It is both experimental and theoretical. Electrons have angular momentum, and sometimes we want to discuss whether it lines up with another direction (ie typically the direction an electromagnetic field, since electrons are charged their angular momentum makes them magnetism and that interacts with electromagnetic field).

How should we talk about the direction of the angular momentum? We can say "up" and "down" or "forwards" and "backwards", and sometimes we do that. But if we want to use it in calculations, it is easier to give them numbers. So what numbers should we give? They are opposite directions so really it makes sense to use opposite numbers. So what opposite numbers should we use? We can use any, so why not use a gap of "1" between them? So what opposite numbers have a gap of 1 between them? 1/2 and -1/2.

There are some other detailed reasons why 1/2 is a convenient number, but it is in my view fair to say we just want to represent two opposite directions in calculations, and the numbers 1/2 and -1/2 suit this well.


Difference between polypeptides and polyamides

Your guess is correct. Polyamides are polymers where the repeating units are linked by amide bonds. Polypeptides are a specific type of polyamide where the repeating units are amino acids.


Question about a line from Rosenfeld's paper on optical rotation

This answer was suggested by Ed V. It resolves the apparent problem nicely and readers with some fluency in the subject would probably not find the transition from $\lambda$ (medium) to $\lambda$ (vacuum) too troublesome.

We begin with $c = \nu \lambda_o$ in a vacuum and $\frac{c}{n} =\nu\lambda $ in a medium, in which $n$ is th refractive index.

In equation (42) of Rosenfeld's paper we have (using $\lambda n =\lambda_o$):

$$\Delta n =  \frac{2\pi\rho}{\lambda n}= \frac{2\pi\rho}{\lambda_o}.\hspace 74 mm
 ~~~(42)$$

Then in equation (43) above we get $\lambda_o^2$ in the denominator:

$$\theta = \frac{2\pi\nu}{c}\cdot \frac{\Delta n}{2}=\frac{2\pi\nu}{c}\cdot\frac{\pi\rho}{\lambda_o}=\frac{2\pi^2\nu\rho}{c\lambda_o}=\frac{2\pi^2\rho}{\lambda_o^2}\hspace 30 mm (43) $$

and finally substitution of (42) into (43) gives

$$\theta = \frac{2\pi\nu}{c}\cdot \frac{2\pi\rho}{2\lambda n}= \frac{2\pi^2\rho}{\lambda_o\lambda n}=\frac{2\pi^2\rho}{\lambda_o^2}. $$


How do I arrive at the mass of solvent from a molar fraction concentration?

Both your teacher and your answer are wrong!

There are two steps to this problem.  First, we need to convert between mass fractions and mole fractions.  Second, we need to convert from a per-amount basis to a per-volume basis.  "Per-amount" is my made up word for all units such as mass fraction, mole fraction, molality, etc that are expressed per amount (whether moles or mass) of substance (whether total or solvent).  "Per-volume" units include things like molarity, grams per liter, etc., where the basis is the volume of solution.  Converting from an amount basis to a volume basis requires knowing the density of the solution!

Step by step:

Converting to mass fraction

$$ 0.325 \frac{\mathrm{mol\;\ce{H2SO4}}}{\mathrm{mol\;total}} \Rightarrow  \frac{0.325\;\mathrm{mol\;\ce{H2SO4}}}{(1-0.325)\;\mathrm{mol\;\ce{H2O}}}$$

$$\frac{0.325\;\mathrm{mol\;\ce{H2SO4}}}{(1-0.325)\;\mathrm{mol\;\ce{H2O}}} \times \frac{98.1\mathrm{\frac{g\;\ce{H2SO4}}{mol\;\ce{H2SO4}}}}{18\mathrm{\frac{g\;\ce{H2O}}{mol\;\ce{H2O}}}}=2.62\mathrm{\frac{g\;\ce{H2SO4}}{g\;\ce{H2O}}}\Rightarrow \frac{2.62}{2.62+1}\mathrm{\frac{g\;\ce{H2SO4}}{g\;total}}=0.724\mathrm{\frac{g\;\ce{H2SO4}}{g\;total}}\Rightarrow (1-0.724)\mathrm{\frac{g\;\ce{H2O}}{g\;total}}=0.276\mathrm{\frac{g\;\ce{H2O}}{g\;total}}$$

This step is pretty easy to do using the information in the problem.  There are 72.4 grams of sulfuric acid present in 100 g of the solution.  So far, so good.

Converting to a per volume basis

But at this point the problem gets very tricky:

  ...what is the mass of water (in grams) in 100 mL of solution?

It says 100 mL, not 100 g.  This makes the problem much harder.  This also makes your teacher's answer wrong:

  Teacher is not convinced with this answer and says its wrong because the total must be 100g

The total doesn't need to be 100 g because we are apparently dealing with 100 mL of solution.  Depending on the density of the solution, the total mass will be more or less than 100 g.

According to Wikipedia, sulfuric acid at a mass fraction of 0.7 has a density of 1.60 kg/L and at a mass fraction of 0.78 has a density of 1.70 kg/L.  Let's suppose a mass fraction of 0.724 has a density of ~1.65 kg/L.

$$0.276\mathrm{\frac{g\;\ce{H2O}}{g\;total}}\times\frac{1650\mathrm{\;g\;total}}{\mathrm{L}}= 46\mathrm{\frac{g\;\ce{H2O}}{L}}$$

With the final number, now it is easy to see that if there are 430 grams of water per liter, then in 100 mL there are 46 grams of water.


Typical amines ($\ce{NR3}$) are roughly $\ce{sp^3}$ hybridized.  A chiral amine must undergo nitrogen inversion in order to racemize. The transition state for nitrogen inversion is a planar $\ce{sp^2}$ structure.  So in order to invert the ligands attached to nitrogen we must move from an $\ce{sp^3}$ ground state with ca. 109 degree bond angles to an $\ce{sp^2}$ transition state with 120 degree bond angles.

On the other hand typical $\ce{PR3}$ compounds are roughly unhybridized with 90 degree bond angles in the ground state.  In order to racemize they must also invert at phosphorous which again requires an $\ce{sp^2}$ transition state with 120 degree bond angles.  Changing the geometry from 90 to 120 degrees in the case of phosphorous requires more energy that a distortion from 109 to 120 degrees in the case of nitrogen.

P.S. Put a chiral nitrogen in a 3-membered ring and chirality can be preserved because you've now made it so much more difficult to expand the 60 degree bond angle in the ring to 120 degrees.


Surprisingly enough, the answer is yes, though it would be extremely difficult in practice to reduce the size of a molecule as large as $\ce{O_2}$ by the method I will describe here.

First, I must be clear that "shrinking" as seen in many science-fiction stories is completely impossible. Size is not some arbitrarily determined parameter which you can change with a ray gun!

Though the concept of size is a little fuzzy at atomic/subatomic scales, we know it is the valence electrons who are responsible for determining the distance scales in atoms and molecules (since they're the farthest from the nuclei). After solving the Schrödinger equation for a system, one finds that the electrons are concentrated in regions around the nuclei called orbitals. A meaningful definition of size for these orbitals implies that they are around 100 picometres or more across, conveniently described as twice the Bohr radius.

Usually, this would be the limit for any atom or molecule. However, nature has a rather curious quirk. We have discovered that all atoms in the Universe involved in chemistry are made of electrons, protons and neutrons (the latter two being themselves made of different combinations of up quarks and down quarks). However, there are more subatomic particles out there, and there is nothing to stop (exotic) atoms being made out of these other subatomic particles, as long as they are combined in the right way.

Physics doesn't yet understand why, but the fundamental subatomic particles in the Universe come in three generations. That is, there are groups of particles which behave exactly the same (have the same spin, electric charge, among other parameters), but differ only in their mass. In particular, electrons are just one type in a group of three types of particles called leptons. The other types are called muons and taus (or tauons). Again, all leptons share exactly the same properties, except for mass; the muon is approximately 200 times heavier than the electron, while the tau is around 3500 times heavier. We don't naturally see atoms containing these heavier cousins of the electron because they decay very quickly and are hence much rarer. Even though the muon is one of the most stable subatomic particles, its mean lifetime is a mere $2.2\ \mu s$, and the tau is much more unstable still. In the case of the muon, however, it is long-lived enough to produce it artificially and get some chemistry done.

If one were to mix muons with normal atomic nuclei in an appropriate chamber, then for a fleeting moment it would be possible to make atoms and molecules where all the electrons are replaced with muons. The resulting atoms/molecules would have muon orbitals instead of electron orbitals. By solving the Schrödinger equation for muons instead of electrons (changing $m_e$ for $m_{\mu}\approx200\ m_e$), the Bohr radius becomes around 200 times smaller and as a direct consequence the muon orbitals are smaller by approximately the same factor. Thus, an oxygen atom where all electrons have been replaced by muons will be about 1/200th the size, but will display almost exactly the same chemistry; two such exotic atoms would be expected to combine exactly as normal oxygen atoms do, creating an $\ce{O_2}$ molecule with sixteen muons instead of electrons, with an expected bond length of around $0.6\ pm$ instead of the usual $121\ pm$.

This may all seem theoretical, but muonic atoms have already been made in the laboratory before. Given the difficulties in producing and manipulating muons in their short lifespan, the only atoms produced as of yet seem to be muonic hydrogen (where the single electron is replaced by a muon) and quite recently muonic helium where only one of two electrons was replaced with a muon. In the latter case, the mix of electrons and muons makes the atom behave chemically quite differently from regular helium or doubly-muonic helium.


Styrene and HClO are going to make a racemic mixture, in which case the epoxide will also be racemic, i.e. there should be no optical rotation.

An alternative would be to use the Jacobsen epoxidation, the Shi epoxidation or some other enantioselective epoxidation.


Are these two space groups in a group–subgroup relationship?

The Pmma is a minimal non-isomorphic supergroup of Pmm2; or you can say that Pmm2 is one of maximal non-isomorphic sub-groups of Pmma.

How to solve the question:

(1) fast and dirty: take International Tables for Crystallography vol. A go to your group (in mine edition Pmma starts on p.274). On second page describing space group you can find Maximal non-isomorphic subgroups. The third listed is [2] Pmm2 (25) 1; 2; 7; 8.

You can see that the operation in  #50 and #25 have a lot of similarities. No.1 you can ignore as it's [x,y,z] but 2, 7 and 8 for #50 are [-x+.5,-y,z], [x,-y,z] and [-x+.5,y,z]. As one can see the #25 has following general operators: (2) [-x,-y,z] (3) [x,-y,z] and (4) [-x,y,z].

(2) The hard way: Mathematics is explain in the same volume in chapter 13.1. Isomorphic subgroups. Of course the knowledge of group theory and matrix calculus is to your advantage ;-)


Why is the electronegativity of hydrogen 2.20?

Interestingly, the first paper from Pauling on electronegativity "The Nature of the Chemical Bond. IV. The Energy of Single Bonds and The Relative Electronegativity Of Atoms" published in 1932 gives a different scale with H = 0.0 and F = 2.0!

Later, he changed to 2.1 for H and 4.0 for F, e.g. http://scarc.library.oregonstate.edu/coll/pauling/bond/notes/sci5.001.14-notes-01.html

If we look through his book, on p. 90 we find the following text:

  Following these there are given the values obtained by adding 2.05 to them...

The resulting numbers are then rounded to one decimal, from H = 2.05 to H = 2.1 and F = 4.0.

His notes are online (e.g., here) but there's no obvious explanation I can find yet.

My hunch is that he started with H = 0.0, and realized that the alkali metals would be negative numbers. The shifted scale gave Li ~ 1.0 and F ~ 4.0, which seem like "nice" numbers.

I tried to hunt down the reason for the move to 2.2 as the standard for hydrogen. The first reference seems to be by Huggins (J. Am. Chem. Soc., 1953, 75 (17), pp 4123–4126) but the paper itself doesn't really explain the reasoning.

Wikipedia suggests a paper by Allred, but that cites the Huggins paper for the 2.2 number.

Presumably more could be found with a very careful search through the papers of Pauling and Huggins, but I've never heard any clear explanations and the documents online don't give them either.


Regioselectivity of Birch reduction (electron donating groups)

The explanation lies in the distribution of electron density in the intermediate radical anions that appear during the Birch reduction.
I use the following reactions as guiding examples:

Electron-withdrawing groups stabilize electron density at the ipso and para positions through conjugation and so the negative charge will mainly be found in these positions; subsequent protonation occurs para.

Electron-donating groups destabilize a negative charge (when compared to simple benzene) through conjugation and this effect would be strongest at the ipso and para positions.

Thus the negative charge will mainly be found at the ortho and meta position.


Why does Cl- corrode aluminum containers?

In general, it should not. Under the right conditions, i.e. acidic conditions at high temperature on a freshly exposed (e.g. scratched) $\ce{Al}$ surface, you could get some oxidation to $\ce{AlCl3}$.  One industrial process for producing $\ce{AlCl3}$ is via the reaction:

$$\ce{2 Al + 6 HCl → 2 AlCl3 + 3 H2}$$

at $\mathrm{600-800^oC}$. It would of course happen much more slowly, if at all, at lower temperatures and lower acidity.

The necessity of a freshly scratched surface is to expose the aluminum metal itself, as aluminum pretty rapidly forms a thin, inert oxide layer upon exposure to air or water.

In other words, in the absence of this perfect storm of conditions, aluminum cans should be very resistant to corrosion by $\ce{NaCl}$.


How do the def2-SV, def2-SV(P), and def2-SVP basis sets differ?

Def2-SVP has the polarisation functions on all atoms, -SV(P) does not have these functions on the hydrogen atoms and -SV does without them.
In future, you can check on the EMSL Basis Set Exchange It wouldn't give you an answer explicitly, but you can view the basis set for different atoms and see how the different sets vary.
A side note - If your 'complex molecules' aren't too large, consider the larger Def2-TZVP basis set. Errors at the Def2-SVP level can be several kcal which may significantly impact your study. As an alternative, consider the PC-n family of basis sets from Frank Jensen. If the Def2-TZVP set is too large, the double-zeta PC-1 might work well.


What is the composition of the oil-like fluid that seeps out of dry-cells over time?

There are various kinds of dry cells.I will give some examples as to what is that oil-like fluid:

In zinc-carbon cell:

  Zinc carbon cells have a short shelf life as the zinc is attacked by
  ammonium chloride. The zinc container becomes thinner as the cell is
  used, because zinc metal is oxidized to zinc ions. When the zinc case
  thins enough, zinc chloride begins to leak out of the battery. The old
  dry cell is not leak proof and becomes very sticky as the paste leaks
  through the holes in the zinc case. The zinc casing in the dry cell
  gets thinner even when the cell is not being used, because the
  ammonium chloride inside the battery reacts with the zinc. An
  "inside-out" form with a carbon cup and zinc vanes on the interior,
  while more leak resistant, has not been made since the 1960s.

In alkaline cells:

  Technically, that white fluffy corrosion that develops at the ends of
  the battery (most often the negative end) is called potassium
  carbonate.

  All batteries will slowly gradually self-discharge over time. This
  will occur whether they are setting on the shelf (a much slower
  process) or installed in a device (which often occurs much quicker) –
  and dead batteries will eventually leak.

  Very high temperatures can also cause batteries to rupture and leak
  (e.g. in hot car during the summer).

  The “alkaline” of the battery is potassium hydroxide (it’s the alkali
  equivalent of acid’s hydrochloric acid), and this will leak out,
  forming a white “fluff” of potassium carbonate, typically on the
  negative end of the battery cell – because apparently the positive end
  is vented better.

So the reason for the cell-corrosion varies with the type of the cell.

Read the Wikipedia page to know about the composition of electrolytes of different dry cells.https://en.wikipedia.org/wiki/Dry_cell


Why total heat of hydrogenation of 1,3-cyclohexadiene is more than that of benzene?

Yes. The relatively smaller Heat of hydrogenation (HOH) for benzene as compared to that for 1,3-cyclohexadiene is due to the aromaticity of the first. Analyzing the thermochemistry is indeed among the first and perhaps more intuitive ways to present and quantifies aromaticity itself.

A) Cycloexene HOH = -120 kJ/mol

B) 1,4-Cycloexadiene HOH = -240 kJ/mol

C) 1,3-Cyclohexadiene HOH = -232 kJ/mol

D) Benzene HOH = -208 kJ/mol

While for B the HOH is about double as compared to that of cyclohexane and thus is according to the assertion in the question, the conjugation in C does results in a "lower than expected" value.

The 8 kJ/mol difference is the energy of resonance of the two conjugated double bonds.

The situation is even more striking for D as this difference from "an expected value" amounts now to about 152 kJ/mol. This is again a measure of stability and its distinctively high value is a manifestation of  aromaticity.

The HOH of benzene is even less than that of cyclohexadienes. This shouldn't come as surprise though, as for there are not really double bonds.

In other words, while hydrogenation of individual double bond is an exothermic reaction, hydrogenation of benzene to cycloexadienes is an endothermic process.

Once you disrupt aromaticity, then you can think again in terms of individual double bonds (although delocalisation may occur to little extent, as in C). And this fact should answer the question "why HOH of benzene is higher than that of cyclopentene?", too.


Spin of electrons about it's own axis, release of energy & COLLAPSE of atom?

You are confusing macroscopic spin with the spin angular momentum of electron here. Electron is a point entity and does not posses a macroscopic spin (i.e. it does not spin on its axis).
 It does have angular momentum which is expressed as 'spin angular momentum' which is of-course quantized.

This link   by  John Rennie will be useful for understanding : The electron isn't spinning on it's own axis


Thermal properties of 301 stainless steel

301 is the lowest alloy content of the "18-8" family. The slightly lower nickel ( compared to 304) makes the austenite less stable so when cold rolled it work hardens more than 304. It is still basically austenitic so does not become brittle at low temperatures. So it was chosen for the higher strength ( when cold worked)  at all temperatures , relative to 304. Austenite is non-magnetic but as 301 is cold rolled it becomes somewhat magnetic depending on the amount of cold work. Presumable caused by formation of tiny grains of martensite ( the magnetic , strong, crystal of the iron family). The small martensite grains block the slip planes in the austenite matric making it stronger.


How appropriate/adaptable is HPLC for in situ sample analysis on Titan?

It is a very good question because this is an active area of research. Having heard talks from the actual people who are making analytical instruments for space analysis, I can mention the actual issues before coming to your actual question so that you are aware of the latest problems. As far as I know, HPLCs have not been sent, but GC systems have been successfully used.

Key problems

(i) Weight. It costs an enormous amount (in millions) of money per kg to send anything in space. Routine HPLCs are very heavy. We need two heavy duty persons to lift the stack of standard equipment.

(ii) Latest HPLCs can now be reduced to the size of a small brief case. Check capillary chromatography. Also check HPLC on a chip. It is the size of a hand palm.

(iii) Elecric power consumption: Think how much power is consumed by the motors in the pump (huge) and other analytical equipment including the mass spec. You don't have lot electrical power in space. This the biggest problem besides weight issues.

(iv) Think of excessive radiation. Can the plastics and equipment tolerate huge amount of radiation from the Sun and elsewhere.

(v) Temperatures: Think of temperature extremes with liquid as solvents.

(vi) There is no HPLC column (or GC column) in the world which can separate everything. May be you would attach at least 7-8 columns at once and see which one separates the sample (having a system with multiple columns at once is commercially available. This is done by pharma people)

  Our motivation for adopting HPLC is that we can keep all substances in
  their original phase for analysis.

Yes, HPLC is a non-destructive technique. Mass spectrometer is not. However, UV-detection is useless for unknowns because you have no clue about their absorption spectra. Even if you have an absorption spectrum, it provides very little information about the full structure. Only mass spectrometer can be used in case of complete unknowns. Keep in mind that a typical HPLC C-18 column can only separate hydrophobic compounds under ideal conditions. The sole job HPLC or GC is to simplify the mass spectrum of a sample. Imagine you inject a sample containing 50 analytes into a MS. Imagine the complicated mess. With an HPLC, only compound will reach the detector one at a time. Mass spectrum is very simple then.

  My question then is, is there a way to use HPLC to identify unknown compounds suspended in the lake without disturbing the state of the sample? Any advice would be appreciated.

Short answer no. HPLC cannot detect or separate suspended matter. If you meant substances dissolved in the lake, yes, then it is possible. Every sample has to be carefully filtered to remove any particles.

Since you have two chemists, why don't you just do a real experiment on lake water.


Why BH3 its classified as a "Molecular Hydride" and AlH3 its a "Intermediary Hydride"

Alane is not a molecule: https://en.wikipedia.org/wiki/Aluminium_hydride You can't isolate molecules easily because they polymerize.

Borane on the other hand is known to exist as a dimer: https://en.wikipedia.org/wiki/Diborane

The difference in bonding is likely due to "hypervalency" (or similar concept with different terminology) in aluminum.


Often in chemistry "facts" depend on perspective. You're right that essentially H goes from -1 to 0 and As from 3+ to 0. But in general redox reaction involve two different chemical species. In this case it is better to think of the reaction as a decomposition reaction rather than a redox reaction.


How to determine interaction of photon with a hydrocarbon Bixin

In a first step I would widen the available search criteria.  While accessing the English entry of wikipedia, collection of the chemical name, CAS number, smiles code (copy/paste), pubchem and chemspider ID are presented.  These identifiers may become helpful in the subsequent searches in other databases.  Nor would I hesitate (as long as my web browser allows proper display .and. there are at least some words barely understood in reading) to access the other versions (Spanish, Portuguese, ...), too.

In the public libraries of US' NIH, just the keyword "bixin" yields 150 (pmc) or 116 (pubmed) hits, that after some narrowing yield entries like this abstract in Spectrochimica Acta or this about triplet states of bixin, published in J. Agric. Food Chem.

While the British chemspider does not yield spectra directly, the literature references listed by this database include even surprising finds like doi 10.1039/C4CS00309H (2015ChemSocRev3244) about Vegetable-based dye-sensitized solar cells.

Do not omit collections like researchgate, as here, or figshare. If your institute has subscriber access, chemical databases like Reaxys (Elsevier) or Scifinder Scholar (ACS) may indicate primary literature including spectroscopic data, too.  No longer chemistry/physics centred, a search in Thomson Reuters' Web of Science, and more engineering-oriented Scopus (by Elsevier) may round the picture.


How does adding HCl convert AgCl into a sol?

What is the mechanism behind this?

During peptization, the precipitate adsorbs one of the ions of the electrolyte on its surface. This causes the development of positive or negative charge on precipitates, which ultimately break up into smaller particles of the size of a colloid. Note that the presence of equal and similar charges on colloidal particles is largely responsible in providing stability to the colloidal solution, because the repulsive forces between similarly charged particles prevent them from coalescing or aggregating when they come closer to one another. Thus, they remain in dispersion medium as stable colloidal sols.

  And why does having a common ion matter?

The sol particles acquire positive or negative charge by preferential adsorption of positive or negative ions. When two or more ions are present in the dispersion medium, preferential adsorption of the ion common to the colloidal particle usually takes place.


Antiaromaticity is a concept describing an explicitly fictional situation, no molecule is antiaromatic, because all molecules will react to avoid the situation.

So the answer to the implied question is that cyclobutadiene has very little resonance because the molecule is prevented from being antiaromatic (by the nature of the universe and its physics).

Elaboration:
To avoid this situation, butadiene is very reactive; it will undergo an intermolecular Diels-Alder reaction avoid this antiaromatic state.  When reaction is prevented something much more interesting happens.  When frozen in an argon matrix, x-ray crystallography of cyclobutadiene indicates the molecule is highly rectangular, not square.  For dienyl resonance, there is some changing of bond lengths.  In butadiene, the 2,3 bond shortens.  If this were to occur in cyclobutadiene, then there would be degeneracy in the orbitals leading to a Frost diagram that would describe an antiaromatic state.  So if reaction is prevented, the molecule distorts its own geometry.

The fictional resonance structures you would draw are double bonds on the top and bottom, and double bonds on the sides.  However, generally, one might assume there is no resonance for this molecule.


In an Electrotlytic cell using MgSO4 electrolyte, Cu(OH)2 and Mg(OH)2 are visible products, Where Does SO4 go?

It forms $CuSO_4$, which is soluble in water. $Cu(OH)_2, CuOH$ and $Mg(OH)_2$ prefer to precipitate, while $CuSO_4$ just floating there, completely invisible. Actually, even when ppl try to obtain $CuSO_4$ in such electrolytic cells, they simply collect $Cu(OH)_2$, add $H_2SO_4$ and crystallize sulfate. So it is there, but there is no way you can obtain it directly.

[remark] You can also try to crystallize it from solution, since it's there. But maybe you won't succeed.


Color of Anhydrous Transition Metal Salt

A ‘mere ionic salt’, when inspected properly, turns out to be nothing different from a coordination compound. One side (the cation) is electron deficient while the other side (the anion) has spare electrons, so there will be some kind of coordination happening from one to the other. Indeed, all salts adopt crystal structures so that the favourable coordination interactions are maximised.

For anhydrous cobalt(II) chloride, I would immediately expect a closest packing with cobalt ions sitting in tetrahedral voids, occuping ¼ of them to give $\ce{[CoCl4]}$ tetrahedrons where each chloride coordinates two neighbouring cobalt ions. Analogously for chromium(III) chloride, I would expect chromium to occupy one-third of the anion structure’s octahedral voids, giving rise to $\ce{[CrCl6]}$ octahedrons where again each chloride coordinates two chromiums. Both are thus clearly coordination compounds. And going by what I know about $\ce{Cu^{II}}$, that salt will probably be a set of square-planar $\ce{[CuCl4]}$ fragments, again where each chloride coordinates two coppers.

Note that all of these are rough guesses (I wouldn’t even call them educated) since you can only predict solid-state (ionic) structures so far a priori. Oftentimes you can arrive at multiple possibilities and even more times you are surprised at what is actually adopted but then retrospectively you can usually explain the structure somehow with coordination.


Reaction of 2,4,6-triaminobenzoic acid in a protic medium and heat

My initial guess for the product was Phloroglucinol, or 1,3,5-trihydroxybenzene1,2

Belskaya et al.3 says that 1,3,5-triaminobenzene can be formed, but it can eventually be hydrolyzed to Phloroglucinol

The sources mention decarboxylation as the mechanism. Note that it is the trihydroxylamine 10 that is getting decarboxylated here. The mechanism becomes clearer if you consider the imino tautomers 6 or 7 with the acid group attached and then dehydrate.
In short, Phloroglucinol should be the final product. If that is not one of the options, then 1,3,5-triaminobenzene is the answer you should select.
References:

“PHLOROGLUCINOL.” Organic Syntheses, vol. 9, 1929, p. 74. doi:10.15227/orgsyn.009.0074
Tewari, K.S and Vishnoi, N.K : A textbook of organic chemistry, 4th edition.
Olga B.Belskaya, Valentin P.Talsi, Roman M.Mironenko, Vladimir A.Rodionov, Sergey V.Sysolyatin, Vladimir A.Likholobov, "Transformation pathways of 2,4,6-trinitrobenzoic acid in the aqueous-phase hydrogenation over Pd/C catalyst", Journal of Molecular Catalysis. doi: 10.1016/j.molcata.2016.04.014


Are the coefficients and exponents of contracted gaussian same for cartesian and pure form?

They are the same.

Your equation can be rewritten as

$$ \phi_\mathbf l^\text{CGF}(r,\Omega)=\frac{x^{l_x}y^{l_y}z^{l_z}}{r^l}\sum_ns_nr^l\exp(-\alpha_nr^2)\equiv\Phi_\mathbf l(\Omega)R_l(r) $$

where $l=l_x+l_y+l_z$ is the total angular momentum, $\Phi_\mathbf l(\Omega)$ is the angular part, and $R_l(r)$ is the radial part, and they are separable not only in mathematical form, but also for practical purposes. (This discussion omits the norm, which is not separable to radial and angular contributions.) When specifying a Gaussian basis set in quantum chemistry, one talks only about the radial part, independent of the angular part. The angular part is really considered an implementation detail, because its form usually doesn't have any significant effect on the results.

For completeness, two most common angular forms are Cartesian:

$$ \Phi_\mathbf l(\Omega)=\frac{x^{l_x}y^{l_y}z^{l_z}}{r^l} $$

and spherical harmonic:

$$ \Phi_\mathbf l(\Omega)=Y_l^m(\theta,\varphi) $$


How to calculate the solubility of lead (II) carbonate in water, given that bicarbonate will also form?

List known reactions and equilibrium constants

\begin{align}
\ce{PbCO3 &<--> Pb^2+ + CO3^2-} \quad \quad \quad \quad \quad K_\mathrm{sp} = 1.5\times10^{-15}\\
\ce{HCO3- &<--> H+ + CO3^2-} \quad \quad \quad \quad \quad \quad K_\mathrm{a2} = 4.7\times10^{-11}\\
\ce{CO3^2- + H2O &<--> HCO3- + OH-} \quad \quad \quad \quad \quad K_\mathrm{b2} = \dfrac{K_\mathrm{w}}{K_\mathrm{a2}} = 2.13 \times 10^{-4}\\
\end{align}

Start off by just evaluating quick and dirty

Assume $\ce{[Pb^{2+}] = [CO3^{2-}]}$, $\ce{[Pb^{2+}] = \sqrt{1.5\times10^{-15}} = 0.387 \times10^{-7}}$

OK, the assumption is no good and this is going to be fairly nasty...

The pH is close to the autoionization of water
$\ce{CO3^{2-}}$ is a strong base so $\ce{HCO3^{-}}$ can't be ignored.

GIVEN:

Starting off with pure water so $\ce{[H+] = [OH-] } = 1\times 10^{-7}$ before $\ce{PbCO3}$ is dissolved.
$\ce{[CO3^{2-}] = 0}$ and $\ce{[HCO3^-] = 0}$ before $\ce{PbCO3}$ is dissolved.

WORK

Know that

The final pH > 7 due to the formation of $\ce{HCO3^{-}}$
$\ce{[Pb^{2+}] = [CO3^{2-}] +[HCO3^-]}$
$\ce{[OH^-]_{final} - [OH^-]_{water} = [HCO3^-] }$ or $\ce{[OH^-]_{final} = 1.00\times 10^{-7} + [HCO3^-] }$

For the ratio $ \dfrac{\ce{[CO3^{2-}]}}{\ce{[HCO3^-]}}$ we can use the expression of the $K_\mathrm{a2}$ equilibrium to find that:

At pH 7 ratio = 0.00047
At pH 8 ratio = 0.0047
At pH 9 ratio = 0.047

So it would seem that $\ce{[HCO3^{2-}] >> [CO3^-]}$ is a reasonable assumption, which also means that $\ce{[Pb^{2+}] \approx [HCO3^-]}$.

Now we have the two equilibrium equations

$K_\mathrm{sp} = 1.5\times10^{-15} = \ce{[Pb^{2+}][CO3^{2-}]}$

$K_\mathrm{b2} = \dfrac{K_\mathrm{w}}{K_\mathrm{a2}} = \dfrac{\ce{[HCO3^-][OH-]}}{\ce{[CO3^{2-}]}}$

which we can multiply together to get

$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} = \ce{[Pb^{2+}][HCO3^{-}][OH^-]}$

APPROXIMATE CUBIC SOLUTION

if we let $x = \ce{[Pb^{2+}]}$ then since $x = \ce{[Pb^{2+}] \approx [HCO3^-]}$ and $\ce{[OH-] = [HCO3^-] + 1.00\times10^{-7}}$ we have

$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} =(x)(x)(x + 1\times10^{-7}) =x^3 + (1\times10^{-7})x^2$

Now if we multiply both sides by $10^{21}$ we are trying to solve:

$319.149 = x^3 + x^2$

neglecting the square term we get $x=\sqrt[3]{319.149} = 6.834$, or $\ce{[Pb^{2+}] = 6.834\times10^{-7}}$.

But looking at our cubic equation again

$319.149 = x^3 + x^2 \text{ =?= } 6.834^3 + 6.834^2 = 365.821 $

which is too far off since we need values to about 1%. Solving the cubic by iteration we get $x = 6.516$

    Using 2 guesses then interpolation
    6.834           365.876
    6.000           252.000
    6.492           315.758
    6.517           319.257
    6.516           319.116
    Check
    $319.149 = x^3 + x^2 \text{ =?= } 6.516^3 + 6.516^2 = 319.116 $

which gives:

$\begin{align}
\ce{[Pb^{2+}]} &= 6.516 \times10^{-7}\\
\ce{[HCO3^{-}]} &= 6.516 \times10^{-7}\\
\ce{[OH^{-}]} &= 7.516 \times10^{-7}\\
\end{align}$

We can calculate

$\ce{[CO3^{2-}]} = \dfrac{K_\mathrm{sp}}{\ce{[Pb^{2+}]}} = 0.023\times10^{-7} $

$\ce{[H+]} = \dfrac{K_\mathrm{w}}{\ce{[OH-]}} = 0.133\times10^{-7}$

CHECK:

$\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^{-}]}} = 4.695\times10^{-11}$

So to the correct number of significant figures we have

$\begin{align}
\ce{[Pb^{2+}]} &= 6.5 \times10^{-7}\\
\ce{[CO3^{2-}]} &= 0.023 \times10^{-7}\\
\ce{[HCO3^{-}]} &= 6.5 \times10^{-7}\\
\ce{[H^{+}]} &= 0.13 \times10^{-7}\text{,}\quad pH = 7.88\\
\ce{[OH^{-}]} &= 7.5 \times10^{-7}
\end{align}$

EXACT FORMULIC SOLUTION

Starting with the formula

$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} = \ce{[Pb^{2+}][HCO3^{-}][OH^-]}$

We let $x = \ce{[Pb^{2+}]}$ as before.

$\ce{[HCO3^-] = [Pb^{2+}] - [CO3^{2-}] =} x - \dfrac{K_\mathrm{sp}}{x} = \dfrac{x^2 - K_\mathrm{sp}}{x}$

$\ce{[OH-] = [HCO3^-]} + \sqrt{K_\mathrm{w}} = \dfrac{x^2 - K_\mathrm{sp}}{x} + \sqrt{K_\mathrm{w}} = \dfrac{x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}}{x}$

$\begin{align}
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} &= (x)(\dfrac{x^2 - K_\mathrm{sp}}{x})(\dfrac{x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}}{x})\\
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}}x &=(x^2 - K_\mathrm{sp})(x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}) \\
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}}x &= x^4 + (\sqrt{K_\mathrm{w}})x^3 - 2K_\mathrm{sp}x^2 - K_\mathrm{sp}\sqrt{K_\mathrm{w}}x + K^2_\mathrm{sp}\\
0 &= x^4 + (\sqrt{K_\mathrm{w}})x^3 - 2K_\mathrm{sp}x^2 - \dfrac{K_\mathrm{w}K_\mathrm{sp}+ K_\mathrm{sp}K_\mathrm{a2}\sqrt{K_\mathrm{w}}}{K_\mathrm{a2}}x + K^2_\mathrm{sp}\\
\end{align}$

$x = 6.531198\times10^{-07}$, and other values same as Excel also...

EXACT SOLUTION WITH EXCEL

Solved first by using an excel spreadsheet (with too many significant figures, but extra digits help show that the calculations are consistent)...*

$\begin{align}
\ce{[Pb^{2+}]} &= 6.531199 \times10^{-7}\\
\ce{[CO3^{2-}]} &= 0.022967 \times10^{-7}\\
\ce{[HCO3^{-}]} &= 6.508231 \times10^{-7}\\
\ce{[H^{+}]} &= 0.133187 \times10^{-7}\text{,}\quad pH = 7.8755376\\
\ce{[OH^{-}]} &= 7.508230 \times10^{-7}
\end{align}$

CHECKS:

$\ce{[H^{+}][OH-] = 0.9999986\times 10^{-14}}$

$\ce{[Pb^{2+}][CO3^{2-}]} = 1.500020\times10^{-15} $

$\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^{-}]}} = 4.700057\times10^{-11}$


The term "Essigäther" is actually the German name for ethyl acetate i.e Essig = "vinegar" + Äther = "ether". "vinegar" becomes acetic, hence ethyl acetate becomes "acetic ether". (etymoline.com)

But, ethyl acetate is an ester. So, why is it named "acetic ether"?

According to Leopold Gmelin, ester was called "oxy-acids ether" or "ether of third type". The original German text is:

b. Ester oder sauerstoffsäure Aetherarten. Ethers du troisième genre.
Viele mineralische und organische Sauerstoffsäuren treten mit einer
Alkohol-Art unter Ausscheidung von Wasser zu neutralen flüchtigen
ätherischen Verbindungen zusammen, welche man als gepaarte
Verbindungen von Alkohol und Säuren-Wasser oder, nach der
Radicaltheorie, als Salze betrachten kann, in welchen eine Säure mit
einem Aether verbunden ist.

which translates to:

b. Ester or oxy-acid ethers. Ethers of the third type.
Many mineral and organic acids containing oxygen combine with an
alcohol upon elimination of water to [form] neutral, volatile ether
compounds, which one can view as coupled compounds of alcohol and
acid-water, or, according to the theory of radicals, as salts in which
an acid is bonded with an ether. (italics mine)

You see the name "ether" was used to define an ester but this usage has become obsolete to avoid confusion.
Other compound classes originally named ethers
Gmelin reports in his 1848 Handbook of Chemistry that French chemists use the terms ether of the first, second and third type to refer to ethers, thio- and halo-alkanes, and esters, respectively. While the modern term ether refers to compounds with a specific functional group, the term ether was historically used to refer to a variety of volatile substances.

The names ether and ester are currently used, while the name "afer" for the thio- and halo-alkanes has not caught on.
Source: https://www.google.com/books/edition/Hand_book_of_Chemistry/U4sMAAAAYAAJ?hl=en&gbpv=1&bsq=Ethers%20du%20troisi%C3%A8me%20genre
Reference

Leopold Gmelin, Handbuch der Chemie, vol. 4: Handbuch der organischen Chemie (vol. 1) (Heidelberg, Baden (Germany): Karl Winter, 1848)


What is the molecular structure of the coating on cast iron cookware known as seasoning?

Some hints:

if you mean that "what exactly" in the original sense of the words,
then the answer is: nobody knows and will ever know.
If you are content with some more vague answer, then
look at the pyrolysis chemistry occuring from 150 °C
up with fats, proteins and starch/sugars, and all that
at once. This is a vast field of chemistry, and all
is gives to a trained chemist is a idea of the structures
of that coating.
Bonds to iron will not occur, first because on cast iron and
on steel pans there is always a layer of iron oxide,
on cast iron  you find a glaze formed by fusion of the
molds finish layer (often containing graphite and clay)


Ethanamide has a higher boiling point than ethanoic acid since amide has two hydogen atoms it can form more Hydrogen bonds than ethanoic acid and their molar masses are 45gmol-1 and 60gmol-1 which makes the difference brought by the London forces somewhat negligible .
you may suppose Ethanoic anhydride should be having a lower boiling point than ethanoic acid and ethanamide since it only has London forces and dipole dipole attractions.But ehanoic anhydride(molar mass-106gmol-1) has a higher molar mass than Ethanoic acid(molar mass-60gmol-1) making the London forces in ehatnoic anhydride more significant.
Hence we have
ethanamide(222 °C)>Ethanoic anhydride(140°C)>Ethanoic acid(118°C).


heat capacity at constant volume and internal energy

Yes, you can apply the formula $\Delta U=nC_V\Delta T$ for an ideal gas in a container whether or not the process is isochoric.

This is because the change in internal energy for an ideal gas undergoing some mechanical process is mainly affected by the change in kinetic energy of the gas. It does not depend on the change of  volume of the gas. So as long as the temperature difference is same the change in internal energy will be same. Heat capacity is the amount of heat required to raise the temperature of a substance by 1 unit so $\Delta U$ is directly proportional to heat capacity and $\Delta T$ along with the number of moles of gas.

When the process is carried with constant volume only then $\Delta W$ is $0$ so in that case $\Delta Q = \Delta U$ whereas $\Delta U =nC_V\Delta T$ holds good always for an ideal gas under some mechanical process.


Why is methane's molecular formula conventionally "CH₄", while water is "H₂O" (among others)?

The choice of whether to write binary hydrides with the hydrogen first or second depends on whether that hydrogen is considered acidic, with water marking the delineation point.

By convention, if a binary hydride is more acidic than water, then they are written in the form $\ce{H}_{\text{n}}\ce{X}$. If a binary hydride is less acidic than water, they are written as $\ce{YH}_{\text{n}}$. It so happens that this changeover neatly divides the periodic table. Binary hydrides from groups 16 and 17 (VIA and VIIA) are written with the hydrogen atoms first: $\ce{H2O,~H2S,~H2Se,~H2Te,~HF,~HCl,~HBr,~HI}$.

Binary hydrides from groups 1 (IA) through 15 (VIA) are all written with the hydrogen atoms last, regardless of whether the hydride is ionic $(\ce{LiH,~NaH,~KH,~CaH2})$ or molecular $(\ce{BH3,~CH4,~SiH4,~NH3,~PH3})$.

This notation is in keeping with that for more complex compounds. If the compound is an acid, the acidic hydrogen atoms are put at the from of the formula, while nonacidic hydrogen atoms are placed in the main part of the formula. For example, the hydrogen atoms in phosphoric acid $(\ce{H3PO4})$ are considered acidic and the hydrogen atoms in octane $(\ce{C8H18})$ are not. Acetic acid $(\ce{HC2H3O2})$ has one hydrogen atom that is considered acidic and three that are not, this the formula $\ce{HC2H3O2}$ is considered more helpful than $\ce{C2H4O2}$, although both are "correct" by different sets of rules.


Domestic cleaning products and types of solvents

There are two ways of cleaners. Ones that will "grab on to" the dirt and bacteria to wash it away e.g. soap and shampoo. Then there are ones that will dissolve the dirt and kill the bacteria such as bleaches. They won't do anything dangerous together, they'll just not work.


Chemical Equilibrium - Le Chatelier's principle

For the case you're describing, having left-over product species will likely not cause the reverse reaction to increase.

When you define a reaction equation like: $\ce{aA + bB <=> cC + dD}$, you are explicitly defining A and B as 'reactants' and C and D as 'products'. In the scenario you've described, just because you have excess reactants because you have a limiting concentration of either A or B, doesn't mean the concentration of products will increase. This is because we can think of concentration of products as $[Products](t) = [C](t) + [D](t)$ and not $[Products](t) = [C](t) + [D](t) + [A](t) + [B](t)$.

--

A nice definition of Le Chatlier's principle is the following: "A change in one of the variables that describe a system at equilibrium produces a shift in the position of the equilibrium that counteracts the effect of this change."

If one is to get a little more technical with this, equilibrium is a function of many factors, primarily Temperature (T), Pressure (P), and mixture species (n). This is typically described by thermodynamics by the state property, Gibbs Free Energy, or $\Delta G$.

For the purposes of an example to quantify Le Chatlier's principle, let's look at the differential form of Gibbs Free Energy: $dG = -SdT +VdP + \sum^{n}_{i=1}\mu_idn_i$. If we consider a case where temperature isn't changing (isothermal), and where pressure isn't changing (isobaric), then the Gibbs Free Energy equation is affected by changes in system composition: $dG = \sum^{n}_{i=1}\mu_idn_i$. For an equilibrium case, which means that $\Delta G = 0$, it can be shown that if you expand this expression out for a reaction mechanism (like the one mentioned above), you end up with an expression like: $\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$. This particular equation is defined for standard conditions (0 degC, 1 atm).

--

We're almost to an answer, let's look a little closer at $K_{eq}$. Let's use a gas phase system $\ce{aA + bB <=> cC + dD}$ as an example. If you look up the derivation of the $\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$ in any standard thermodynamics textbook, you will see that $K_{eq}$ is actually equal to the ratio of the mol fractions of each species multiplied by the system pressure: $K_{eq} = \frac{y_C^c y_D^d}{y_A^a y_B^b} P^{c + d - a - b}$. If one assumes that the pressure is fixed and that $a = b = c = d = 1$, then this expression would reduce down to: $K_{eq} = \frac{y_C y_D}{y_A y_B}$.

You can further add detail to this expression by defining the mol fraction in terms of the extent of reaction $\xi$ for a reaction system.
Where, $n_i = n_{io} + v_in_{io}\xi$--which essentially says that the mols of a given species is equal to inital mols +/- the extent to which that chemical is generated or reacted away (based on the specified chemical of reference $n_{io}$). If you plug this into the expression for $K_{eq}$ you can see that now the equilibrium conversion of a reaction system is dependent on the amount of mols of each species. For the system described above, the total mols $N_{total}$ will cancel out of the expression leaving:

$K_{eq} = \frac{(n_{Ao} + v_An_{io})(n_{Bo} + v_Bn_{io})}{(n_{Co} + v_CN_{io})(n_{Do} + v_Dn_{io})} = exp(-\frac{\Delta G^o_{T_o}}{RT_o})$

For a given temperature and pressure, the Gibbs Free Energy ($\Delta G$) will be a fixed value. Therefore the conditions in the equation (composition of the mixture of reaction system) must balance in such a way that the equilibrium conversion $\xi$ will satisfy the condition. You can see here that as one either varies products (C,D) or reactants (A,B) that the solution for $\xi$ will change. You can generally show that an excess in either products or reactants will shift the system to maximum equilibrium conversion $\xi$. And as such, if one suddenly added or removed products or reactants you would cause a shift in the equilibrium conversion $\xi$.


How does change in entropy influence the tendency of a reaction to occur?

Entropy ($S$) is a measure of the disorder in a system.

In a closed system, entropy always increases over time. In an open system, energy can be added to a system to cause a decrease in entropy, but this is not necessarily a spontaneous reaction.

A reaction will be spontaneous if the change in free energy, $\Delta G$, is negative.

Quantitatively,
if entropy increases, and the reaction enthalpy is exothermic $(\Delta H<0)$ or weakly endothermic ($\Delta H>0$ and $\Delta H < T\Delta S$), the reaction becomes spontaneous according to the free energy equation.

Since the observable universe is essentially a closed system, entropy will continue to increase until the universe essentially ceases to exist.

This way an increase or decrease(sometimes) in the randomness can cause a reaction to occur.

Here is a table relating to spontaneity and signs of $\Delta G$:

Image Source


"Thermodynamics an engineering approach, Cengel and Boles" defines a pure substance as one that has the same chemical composition throughout.

This is the correct definition of a pure substance. However, air, regardless of phase, is not a pure substance.

All matter is categorized as either a "pure" substance or a mixture. The word "pure" in front of "substance" is unnecessary, since the definition of substance implies purity. A substance is a sample that is chemically uniform in composition. In other words, all particles of that sample are the same chemically (whether they are atoms or molecules or ionic lattices or what have you). Substances cannot be separated into simpler components by any physical process (i.e. without breaking chemical bonds). Mixtures are samples that contain two or more substances. All mixtures can be physically separated, although the means to do so may be neither easy nor practical.

Air is a mixture of nitrogen ($\ce{N2}$), oxygen ($\ce{O2}$), carbon dioxide ($\ce{CO2}$), argon ($\ce{Ar}$), water vapor, and traces of other gases. Each of these gases separately is a substance. Air can be easily (but not in the comfort of your own home) separated into its constituent parts. Cooling air will cause the constituent components to condense out of it at progressively lower temperature: first water vapor, and then $\ce{CO2}$ (195 K), followed by $\ce{Kr}$ (120 K), $\ce{O2}$ and $\ce{CH4}$ (90 K), $\ce{Ar}$ (87 K), $\ce{N2}$, $\ce{Ne}$ (27 K), $\ce{H2}$ (20 K), and $\ce{He}$ (4 K).

As each substance condenses out of air it can be physically separated from the remaining gas, reducing the complexity of the mixture. Each substance, if returned to the gaseous state separate from the original mixture is physically and chemically distinct from the original mixture (or whatever is left of it). Such physical separations do not need to be easy to make air a mixture, they need to be possible.

A pure substance does not exhibit this behavior. When pure water is cooled, it undergoes a phase change to ice, which is physically separable from liquid water. However, the solid and liquid samples are still water. You have not reduced the complexity of the sample, just changed its form. If you allow the separated solid water to melt, it will be physically and chemically indistinguishable from the original water sample. Pure water cannot be separated into other chemically distinct substances without performing a chemical reaction, which means breaking and forming bonds between atoms.

The majority of matter that we interact with in our everyday lives is comprised of mixtures.

We encounter a few pure substances in the kitchen. Table sugar is a single compound (sucrose). So is table salt (sodium chloride), baking soda (sodium bicarbonate), and cream of tartar (potassium bitartrate). Water is a pure substance, so long as it has been purified. Most water that we encounter has other substances dissolved in it.

Gold (as long as it is 24K) is a pure elemental substance. So is diamond (carbon), graphite (also carbon), and the neon in neon lights. All minerals are pure substances, including asbestos. Glass is a pure substance. Most precious stones are pure substances.


Why can we not use water instead of blank control solution in enzyme catalysis with Peroxidase experiment?

It is because the water blank absorbs so much light that the usable scale is compressed, thus, causing a poor accuracy.

For water, the absorption coefficients $\alpha\ [\mathrm{m^{-1}}]$ at $\lambda$ = 680, 580, 535, 480 and 380 nm are 0.47, 0.09, 0.045, 0.013 and 0.011.

(Source: R. M. Pope, E. S. Fry, Applied Optics, 1997, 36, 8710-8723)

That is not a lot.

On the other hand, you never know about impurities in your buffers or other components (apart from your probe).

Therefore, you usually do not measure against "just the solvent", but against the solution of "everything except the probe" to record the complete background.


Some data on size of hydrocarbons and boiling point

I found complete data on the butane/octane pair but not the propane/hexane pair, so I am using those as illustration (I think they are taken at a temperature near the normal boiling point, respectively).

$$
\begin{array}{|c|c|c|c|c|}
\hline
\text{substance} &  \Delta H_{vap} & \Delta S_{vap}  & T_b & \text{ref}\\
\hline
\text{butane} & \pu{22.4 kJ/mol}  &  \pu{82.30 J/mol K}& 273 K & [1]\\
\hline
\text{octane} & \pu{39.1  kJ/mol}  &  \pu{98.3  J/mol K}& 398.7 K & [2]\\
\hline
\end{array}
$$

Ref. 1: https://webbook.nist.gov/cgi/cbook.cgi?ID=C106978&Mask=4

Ref. 2: https://file.scirp.org/pdf/OALibJ_2016091314302030.pdf

The enthalpy of vaporization is determined by measuring equilibrium vapor pressure over a range of temperatures (Clausius Clapeyron equation). Then, the entropy of vaporization is calculated from the boiling point and the enthalpy of vaporization ($\Delta G_{vap} = 0$ at the boiling point):

$$ \Delta S_{vap} = \frac{\Delta H_{vap}}{T_b}$$

  What is the difference between pulling apart a pair of propane twice, versus one pair of hexane?

As far as enthalpy is concerned, there is not much difference. You have to disrupt about the same dispersion forces removing one molecule of hexane from liquid hexane or removing two molecules of propane.

However, if enthalpy were the only aspect you had to consider, liquids would never turn into gases because you lose all interactions in the gas phase. Experience shows that if you increase the temperature sufficiently, at one point you disrupt the interactions between molecules in a liquid and get a gas.

  Why would the total amount of energy needed differ?

What determines whether a process happens or not is not the change in enthalpy alone, but the change in Gibbs energy. The Gibbs energy contains an enthalpy component and a highly temperature-dependent entropy component. The entropy change of a molecule from the liquid to the gas state is about the same for butane and octane (also see Trouton's rule), but the enthalpy change is different by a factor of two. That's why the two substances have different boiling points.

$$ \Delta G_{vap} = \Delta H_{vap} - T_b \Delta S_{vap} = 0$$

$$ T_b = \frac{\Delta H_{vap}}{\Delta S_{vap}}$$

If you ask for the analysis for evaporating a kilogram each of butane and octane, the enthalpy change for the entire sample would be similar, but the entropy change for the entire sample would be off by a factor of about 2. This is because the molar entropy of vaporization is similar for butane and octane, but a kilogram of butane contains about twice the number of molecules compared to a kilogram of octane.

  It's seems to go against the concept of pressure.

This is a bit of a puzzling statement. There are two things that require energy when a liquid turns into a gas at constant pressure. The first is disrupting the intermolecular forces in the liquid. The second is pushing against the pressure to increase the volume of the sample as liquid turns into gas. The latter is abbreviated as P-V-work. If you look at internal energy changes, you have to consider it. If you look at enthalpy changes, however, it is taken out of the equation because enthalpy is defined as
$$ H = U + P V$$
As we are discussing Gibbs energy and enthalpy, we don't have to worry about P-V-work. ("It is not part of the bill").


How to determine which of two metals will have the higher temperature after applying the same amount of thermal energy?

By definition, the heat capacity is the energy required for a body to be warmed by $\pu{1 K}$. For the specific heat capacity it's the same but normalized by the mass. The higher the heat capacity is, the more energy will be required to warm it.

The metal $\mathrm{X}$ having the lowest heat capacity will have the higher temperature if it receives the same amount of energy than the metal $\mathrm{Z}$.


Hydrogen molecule potential energy graph

Normally, when the atoms are separated by a huge amount (infinity) we take this energy to be zero. Bringing the atoms closer to one another has an attractive potential and the bond has formed on reaching the lowest point on your curve.  Moving the atoms closer together increases the potential energy because here we push electrons closer to one another and this energy increases rapidly  with each small decrease in separation due to repulsion of similar charges. The slope of the curve gives the force on the atoms, so at small separation they experience a large force and repel one another. At large separation, where the slope is positive, there is a smaller attractive force.


Why do *tert*-butyl ethers in diluted sulfuric acid undergo elimation instead of solvolysis?

It isn't necessarily the same situation as the haloalkane in water. In the sulfuric acid case, there is a large amount of acid present, and as you rightly say the alcohol and water are similar leaving groups, so the SN1 product is easily decomposed back to the tertiary carbocation. As a result, the mechanism tends towards the thermodynamic product, which apparently is the alkene (probably for entropic reasons.

Compare this with the alkyl halide in water. Here, the halide is a much better leaving group than water. Additionally, there is a low concentration of acid in solution (at most 1 equivalent, from the generation of HX). This means that loss of water from the SN1 product is much less favourable.

That said, I'm sure you'd get a mixture of E1 and SN1 products from the haloalkane reaction you describe, I merely wanted to point out that the ether and haloalkane cases aren't as similar as you may have presumed.


The reaction of iron with sulfur to produce Iron(II) sulfide is inconvenient (for me)

Not an answer per se (I lack sufficient reputation to comment): are these reactions in aqueous solution, or dry gases? Just a suggestion, but by omitting the charges on these species and not writing them as half-reactions you are obscuring the electron exchange in each. Thus in your first example (iron reacting with chlorine gas), you take metallic Fe, whose oxidation half-reaction can be written as
$$\ce{Fe^0 -> Fe^3+ + 3e-}$$
The equivalent for reduction of chlorine would be
$$\ce{2e- + Cl2_{(g)} -> 2Cl-}$$
Normalizing half reactions to electrons transferred ($6$), adding and simplifying stoichiometry gives

$$\ce{2Fe^0 + 3Cl2 -> 2FeCl3}$$

Thus chlorine is the electron acceptor (strong oxidizer) and iron the donor. In your reaction (2), iron is again a donor (this time of only $2$ electrons), hydrogen is the acceptor, with chloride ion just along for the ride:

$$\ce{Fe^0 -> Fe^2+ + 2e-}$$
$$\ce{e- + H+ -> \tfrac{1}{2}H2} $$

In contrast to these simple reactions, those involving sulfur are exceeding complex because of the diversity of species that can form. In your reaction to form iron monosulfide, one possible way for Fe to be oxidized but native sulfur (zero-valent S actually present as stable ring, $\mathrm{S}_8$) to be both reduced and oxidized is via a disproportionation reaction, with water reduction to hydrogen consuming additional electrons in an anoxic environment, thus yielding both FeS and iron sulfate:
$$\ce{Fe^0 + \tfrac{1}{8} S8 + 2H2O -> \tfrac{1}{2}FeSO4 + \tfrac{1}{2}FeS + 2H2}$$

Amended answer:
If you're heating metallic iron and S to significant temperatures, the elemental sulfur will form a gas, $\mathrm{S}_{n}(g)$ ($n$ indicates which T-dependent allotrope is present). To synthesize FeS however, I think you would need an oxygen-free system. If $\mathrm{O_2}$ is initially present, consumption could be accomplished by sacrificing some of the reactants via in situ oxidation of the metal, or production of $\mathrm{SO_2}$. Then this can proceed directly, using the remaining $\mathrm{S}_{n}(g)$, e.g.:
$$ \tfrac{1}{2}\mathrm{S_2}(g) + \mathrm{Fe}(s) = \mathrm{FeS}(s) $$
I believe the ratio of S/Fe of the product will increase with temperature, and at sufficiently high temperature reach 2 ($\mathrm{FeS_2}$, pyrite).


Is there a way to contain fluorine gas for long term so that it can be visually observed?

As @JonCuster mentions, some materials are pretty resistant to fluorine gas at room temperature.

But out of curiosity, I checked Theodore Gray's website. He's made an effort to have a "periodic table" table with actual elements. He also has a beautiful book, called (naturally) The Elements.

  Almost anything placed in the path of a stream of fluorine gas will spontaneously burst into flame. This includes things like, oh, say glass, steel, and other things not normally thought of as flammable. It is, therefore, fairly difficult to have a sample of it in an element collection. I used to have a statement here that there was no transparent container that could hold fluorine for any length of time, but I was corrected by a man who has figured out how to do it.

There's more here but I'll summarize. Basically, the problem with glass is the $\ce{Si-O-H}$ bonds, which will auto-catalytically react with $\ce{F2}$ to give $\ce{HF}$, and that etches the glass. So with a high quality quartz tube, a lot of annealing, and a fluorocarbon grease to coat the glass and minimize reactivity, it works.

Interestingly, I always had the idea that fluorine gas was a light blue-green. Instead, it's brownish.


What is the influence of glass type on silver mirror making?

It is not so much the glass, it is the contact between the substrate and the reflecting layer.
Some of the problems with the silver/glucose approach, like their optical imperfections is related to the bad adhesion of elemental silver to glass.  The «normal glass mirror» in the bathroom thus uses a tin layer between the cleansed glass, and the subsequently added silver (mostly) responsible for the reflection.  Discovery channel once  aired brief documentary about this; the more interesting part may be still seen on youtube here.
Modern mirrors for telescopes and optical instruments in general do not use silver, but aluminum; the reflective surface is now not behind the glass, but on top (first surface mirrors).  One of the reasons to use aluminum instead of silver is that silver darkens over time (think about the formation silver sulfide, already on cutlery), while aluminum oxide forming on top of aluminum still is transparent enough (for visible light) that these mirrors are aluminized about once every two years only.  Willi Koorts shows the procedure in a about 20 min video here on small scale: washing the mirror, remove of Al by KOH, washing, deposit of Al in high (real high) vacuum.  (This shows the operations at a larger, $\pu{8 m}$ scale.)


The Mechanism of the reaction
The reaction of an alcohol with thionyl chloride under base-free conditions is one of the most significant examples of the SNi mechanism (DN+ANDe to give it it’s correct IUPAC designation, which is perhaps more instructive in this case).
There are two steps to the SNi mechanism, as shown below (figure taken from Bruckner, Organic Mechanisms [1]):

Step 1: The alcohol reacts with thionyl chloride to afford an alkyl chlorosulfite, which can often be isolated.[2] At this stage, the oxymethine stereocentre hasn’t been touched– it’s the second step that will determine the stereochemistry.
Step 2: The chlorine is delivered intra-molecularly, with extrusion of sulfur dioxide gas. Since the sulfur is being delivered from the same face that the hydroxyl was on, the reaction occurs with retention overall.  This can be seen in the dotted lines in the figure above.
Two things are worth noting at this stage:

If pyridine is added to the reaction mixture, we see overall inversion via an SN2 mechanism.[2] This occurs because the pyridine interacts with the intermediate alkyl chlorosulfite formed, liberating a free chloride (nucleophilic) and hence allowing the substitution to take place  inter-molecularly. This can be seen in the solid lines in the figure above.

The second step in which I described as happening intra-molecularly doesn’t quite happen intra-molecularly. In reality, the C-O bond begins to fragment in an SN1 fashion, before the chloride attacks. The reason why we still see retention (rather than racemisation as is common with  SN1) is due to a phenomenon called contact ion pairs.[#]

Carbocation rearrangement
The simple answer to your question is that at no stage of the reaction is it thought that there is any free carbocation concentration, making rearrangement less likely on kinetic grounds (along with elimination, which could equally be argued by your logic).
The ion pair theory is such that the carbocation remains closely associated with the leaving group, allowing the nucleophilic attack to occur rapidly without the need for an additional collision to take place.
[1]: Bruckner, R. Organic Mechanisms- Reactions, Stereochemistry and Synthesis; Springer: Berlin, 2007
[2]: J. Am. Chem. Soc. 1952, 74, 308
[#]: You can read about this in any advanced organic chemistry text such as Carey or March, and is a bit too long of an explanation for me to give satisfactorily here


How to calculate mass of calcium fluoride that will dissolve in sodium fluoride solution?

In the $\pu{500 cm^3}$ solution there are is already $\ce{F^-}$ dissolved coming from the $\ce{NaF}$ $\pu{0.10 mol dm^{-3}}$ that was already there. This means that the solubility of $\ce{CaF2}$ is reduced due to the common ion effect.

NB $\ce{NaF}$ can be considered completely dissociated because it has a very high $K_\mathrm{sp};$ therefore the initial concentration of $\ce{F^-}$ is $\pu{0.10 mol dm^{-3}}.$

$$
\begin{align}
[\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\
x(2x+\color{red}{\pu{0.10 mol dm^{-3}}})^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\
\end{align}
$$

The issue  here is that to find $x$ you need to solve a cubic equation…

However, $K_\mathrm{sp}$ of $\ce{CaF2}$ is very small and $x$, that is its solubility, must be much less than $\pu{0.10 mol dm^{-3}}!$ This allows you to make the approximation that $2x + \pu{0.10 mol dm^{-3}} ≈ \pu{0.10 mol dm^{-3}}.$

$$
\begin{align}
[\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\
x(\color{red}{\pu{0.10 mol dm^{-3}}})^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\
x &= \pu{1.46E-8 mol dm^{-3}}
\end{align}
$$

Solubility $s$ of $\ce{CaF2}$:

$$s(\ce{CaF2}) = xM(\ce{CaF2}) = \pu{1.46E-8 mol dm^{-3}}\times\pu{78.1 g mol-1} = \pu{1.14E-6 g dm^{-3}}.$$

Finally, the mass of $\ce{CaF2}$ dissolved in $\pu{500 cm^3}$ would be

$$m (\ce{CaF2}) = s(\ce{CaF2})\times V = \pu{1.14E-6 g dm^{-3}}\times\pu{500 cm^3} =\pu{5.7E-7 g}.$$


Why does the chlorination of indene occur with syn selectivity in heptane?

Key

The reaction is run in heptane, a non-polar solvent.  A non-polar solvent will not stabilize or support ions as well as a polar solvent.

Background

When the $\ce{Cl2}$ begins to interact with the indene an initial ion pair is formed.  In the case of chlorine, it is not always clear whether

a cyclic halonium ion (the 3-membered structure on the bottom left)
an open ion (the structure on the bottom right)
or an equilibrium between the two ions occurs

[Side Note: In electrophilic bromination reactions involving alkenes, bromonium ions are very common intermediates usually leading to trans addition.  However, chlorine isn't as polarizable as bromine and in this particular case with indene, the open ion is extremely stable due to its benzylic nature. Hence the involvement of only a chloronium ion is questionable in this case.]

Analysis

As mentioned above, ions are not very well stabilized in the non-polar solvent. As a result

the two ions (one positive, one negative) will tend to stay close together  (tight ion pair) and stabilize each other
and further due to this lack of stabilization they will have a short lifetime leading to quick collapse to products.

Both of these factors (chlorines located close to one another, short lifetime doesn't allow much movement of the second chlorine) favor syn collapse - addition of the second chlorine from the same side that the first chlorine is on.

If we were to repeat the reaction in a polar solvent that would allow the ions to separate and have a longer lifetime, more trans product from anti addition would be expected.


Is liquid hydrogen rocket propeller pollution free?

I'll address you questions in order:

1) Yes, that is basically the correct reaction, although not quite balanced;  here is how I would write it:

$$\ce{2H2(l) + O2(l) -> 2H_2O(g)}$$

2) Ideally, if there were no side reactions, then yes, this particular reaction is "polution free", and the only product is water.

3) You are correct in your thinking that in the end this does not represent a net pollution-free energy source.

The reason for this doesn't lie in the reaction taking place at the rocket, but in how they produced the hydrogen in the first place. Unfortunately, there are no significant deposits of $\ce{H2}$ on our planet waiting to be pumped out and used as fuel. So we  have to make it, which requires engergy.

This means that the $\ce{H2}$ is not a net energy source, rather it is essentially acting as a battery, storing the energy from some other source. So, you can think of the actual pollutants of the $\ce{H2}$ rocket fuel as being the products from whatever process went into generating it in the first place.  This largely depends on what part of the world the hydrogen generation took place. They could have burned coal or other fossil fuel to produce the energy, or they could have built a dam to generate hydroelectric power, etc.  Any of the negative aspects of producing a particular type of energy would be considered a negative product of the $\ce{H2}$ production.


Single step transformation of a carboxylic acid into an aldehyde

$\ce{Li, RNH2 (R=Me, Et)/NH4Cl/H3O+}$

Burgstahler, A. W.; Worden, L. R.; Lewis, T. B. Direct Reduction of Carboxylic Acids to Aldehydes by Lithium in Ethylamine. J. Org. Chem. 1963, 28 (10), 2918–2919.
Bedenbaugh, A. O.; Bedenbaugh, J. H.; Bergin, W. A.; Adkins, J. D. Synthesis of aldehydes and secondary amines from carboxylic acids via imines. J. Am. Chem. Soc. 1970, 92 (19), 5774–5775.

$\ce{BH3.SMe2/PCC}$

Brown, H. C.; Rao, C. G.; Kulkarni, S. U. A Convenient Conversion of Carboxylic Acids into Aldehydes. Synthesis 1979, 1979 (9), 704–705.

$\ce{(CH3)2CHC(CH3)2BH2}$

Brown, H. C.; Heim, P.; Yoon, N. M. Selective reductions. XVII. Reaction of thexylborane in tetrahydrofuran with selected organic compounds containing representative functional groups. Comparison of the reducing characteristics of diborane and its alkyl derivatives. J. Org. Chem. 1972, 37 (19), 2942–2950.

$\ce{(CH3)2CHC(CH3)2BHCl.SMe2}$

Brown, H. C.; Cha, J. S.; Nazer, B.; Yoon, N. M. Exceptionally facile reduction of acyclic and alicyclic carboxylic acids to aldehydes by thexylchloroborane-dimethyl sulfide. J. Am. Chem. Soc. 1984, 106 (25), 8001–8002.
Brown, H. C.; Nazer, B.; Cha, J. S.; Sikorski, J. A. Selective reductions. 38. Reaction of thexylchloroborane-methyl sulfide complex in methylene chloride with selected organic compounds containing representative functional groups. Comparison of the reducing characteristics of thexylchloroborane, thexylborane, and diborane. J. Org. Chem. 1986, 51 (26), 5264–5270.
Brown, H. C.; Cha, J. S.; Yoon, N. M.; Nazer, B. Selective reductions. 39. Partial reduction of carboxylic acids with thexylchloroborane-methyl sulfide. A direct and simple aldehyde synthesis. J. Org. Chem. 1987, 52 (24), 5400–5406.

$\ce{(CH3)2CHC(CH3)2BHBr.SMe2}$

Cha, J. S.; Kim, J. E.; Lee, K. W. Facile reduction of saturated and unsaturated carboxylic acids and their salts to aldehydes by thexylbromoborane-dimethyl sulfide. J. Org. Chem. 1987, 52 (22), 5030–5032.
Cha, J. S.; Kim, J. E.; Oh, S. Y.; Lee, J. C.; Lee, K. W. Direct conversion of saturated and unsaturated carboxylic acids into aldehydes by thexylbromoborane-dimethyl sulfide. Tetrahedron Letters 1987, 28 (21), 2389–2392.

9-BBN/Li 9-BBN

Cha, J. S.; Kim, J. E.; Oh, S. Y.; Kim, J. D. One-pot conversion of carboxylic acids to aldehydes through treatment of acyloxy-9-borabicyclo[3.3.1]nonanes with lithium 9-boratabicyclo[3.3.1]nonane. Tetrahedron Letters 1987, 28 (39), 4575–4578.

9-BBN/t-BuLi/9-BBN

Cha, J. S.; Kim, J. E.; Kim, Y. S. One-pot transformation of carboxylic acids into aldehydes through stepwise treatment of acycloxy-9-borabicyclo[3.3.1]nonanes with Full-size image (<1 K)-butyllithium and 9-borabicyclo [3.3.1]nonane. Tetrahedron Letters 1987, 28 (49), 6231–6234.

DIBAL-H

Zakharkin, L. I.; Khorlina, I. M. Zh. Obshch. Khim. 1964 34, 1029; J. Gen. Chem. USSR 1964 34, 1021.

bis(N-methylpiperazinyl)aluminum hydride

Muraki, M.; Mukaiyama, T. Aminoaluminum hydride as new reducing agents. I. Selective reduction of carboxylic acids to aldehydes. Chemistry Letters 1974, 3 (12), 1447–1450.
Hubert, T. D.; Eyman, D. P.; Wiemer, D. F. A convenient synthesis of bis(N-methylpiperazinyl)aluminum hydride: a reagent for the reduction of carboxylic acids to aldehydes. J. Org. Chem. 1984, 49 (12), 2279–2281.

activated silyl carboxylates

Corriu, R. J. P.; Lanneau G. F.; Perrot M. The one-pot conversion of carboxylic acids to aldehydes via activated silyl carboxylates. Tetrahedron Letters 1987, 28 (34), 3941–3944.

titanium-catalyzed Grignard

Sato, F.; Jinbo, T.; Sato, M. The Reduction of Carboxylic Acids to Aldehydes by Dichlorobis[π-cyclopentadienyl]titanium-Catalyzed Grignard Reactions. Synthesis 1981, 1981 (11), 871

N,N-dimethylchloromethyleniminium chloride and lithium tri-t-butoxyaluminum hydride

Fujisawa, T.; Mori, T.; Tsuge, S.; Sato, T. Direct and chemoselective conversion of carboxylic acids into aldehydes. Tetrahedron Letters 1983, 24 (14), 1543–1546.

$o\text{-}\ce{HSC6H4OH, POCl3, HClO4/LiAlH4/H2O, HgCl2}$

Costa, L.; Degani, I.; Fochi, R.; Tundo, P. Pentaatomie heteroaromatic cations. Note III. A Convenient Synthesis of Aldehydes from Carboxylic Acids via 2-Substituted 1,3-Benzoxathiolium Perchlorates. J. Heterocyclic Chem. 1974, 11, 943–948.

$o\text{-}\ce{(NH2)C6H4, PPA/NaOet/MeI/NaBH4}$ or $\ce{LiAlH4/H3O+}$

Craig, J. C.; Ekwurire, N. N.; Fu, C. C.; Walker, K. A. M. Conversion of Carboxylic Acids into Aldehydes and their C-1 or C-2 Deuteriated Derivatives. Synthesis 1981, 1981 (4), 303–305.


Why carbon nanotubes can act as a good electrical conductor?

I am sorry to say, but your question contains several false assumptions, most importantly:

Nanotubes are good conductors:

No, not all of them. Certain types have metallic/semi-metallic conductance, others are semiconductors. Single wall carbon nanotubes are generally indexed with a so-called chirality index (m,n) that tells you how the carbon sheet is wrapped up into a tube. See wikipage.
Nanotubes behave exactly like 3D nanoparticles in terms of band structure:

No, nanosheets (graphene) and nanotubes behave in different ways and your description of quantum confinements don't work in that way in 1D infinite (nanotubes) and 2D infinite (graphene) cases.

You can derive an approximate band structure of graphene/nanotubes using tight-binding models, which in most cases predicts correctly, whether the structure is metallic or semiconducting. See simulation of nanotube band structure.


Calculating the zero-point energy for the C2N2 molecule

The adiabatic electronic energy only contains the energy of the electrons. You are missing the energy that the molecule has in form of the vibrational groundstate of the nuclei. This energy can be calculated trivially in the harmonic approximation. Within the harmonic approximation we can define normal modes and each mode corresponds to a 1D harmonic oscillator with its respective normal mode frequency. The eigenenergy of a harmonic oscillator is given by
$$
E_{n} = \hbar \omega(n+1/2), \ n\geq0
$$
Each normal mode of your molecule contributes thus such an energy when you include vibrational energy. The zero point energy refers to the system where each mode is in the groundstate i.e. $n_i=0 $, where $i$ is an index for the normalmode. The zero-point energy  correction is then simply
$$
E_{ZP} = \sum_i^{3N-5} \hbar\omega_i 1/2
$$
where $N$ is the number of atoms and $3N-5$ is the number of normal modes in a linear molecule. In a nonlinear molecule you would have $3N-6$ normal modes. $n$ is the quantum number of the oscillator state.
In your case, you have some degenerate eigenvalues due to symmetry. The $\pi_u$ and $\pi_g$ modes are actually sets of two degenerate modes, which means that their frequency would appear twice in the sum that i denoted. This also works out when we consider your example, we have the following number of modes
$$3\cdot 4 -5=7$$. Since we have to sets of doubly degenerate modes we see only 5 distinct frequencies although there are 7 normal modes.
Plugin all thatt into the formula yields,
$$
E_{ZP} = 1/2(\hbar \omega_1+\hbar \omega_2+\hbar\omega_3 + 2\hbar \omega_4 + 2\hbar \omega_5)
$$
with $\omega_i = 2\pi \nu_i.
$
It's sometimes a bit vague what people mean with zero point energy. Some people use the term for the correction that i defined above, but technically it should include also the electronic adiabatic energy. Which means that you have to add the electronic state energy to the $E_{ZP}$ value.
One might also want to include the contribution due to the rotational states of the molecule, but we can't do that without more information.


Calculating bond dissociation enthalpy of F-F bond

If you can't wrap your head around getting the answer directly, just calculate the enthalpy of reaction for the two bond dissociation energies you are considering.

Assume bond dissociation energy is 141 kJ/mol

$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot \pu{141 kJ/mol} - 6 \cdot \pu{272 kJ/mol} =  \text{negative value}$$

Assume bond dissociation energy is 423 kJ/mol

$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot \pu{423 kJ/mol} - 6 \cdot \pu{272 kJ/mol} = \text{positive value}$$

So it can't be 423 kJ/mol because the reaction is exothermic. If you complete the math for a bond energy of 141 kJ/mol, the numerical answer is -263 kJ/mol for the reaction enthalpy.

What if you actually wanted to estimate a bond energy?

In this case, you could call the bond energy $x$, and solve for it:

$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot x - 6 \cdot \pu{272 kJ/mol} = \pu{-263 kJ/mol}$$

Solving for $x$ would give you the bond energy.