Do companies recover carbon dioxide and water from the purification step of cryogenic air separation, and how?

It is unlikely that this is an economically feasible $\ce{CO2}$ recovery method, which means commercial gas producers probably don't bother to isolate $\ce{CO2}$ this way (surely not water). The removal of $\ce{CO2}$ and water is performed with a pre-purification unit (PPU) that contains an adsorbing material such as alumina or molecular sieve. When the material nears saturation with impurities, the PPU needs to be taken off-line for regeneration. This is done by flowing warm $\ce{N2}$ over the adsorbent to heat it. It certainly seems possible to attempt capture of the $\ce{CO2}$ at this stage. Water is likely to be released from the adsorbent after the $\ce{CO2}$, allowing in principle a stream containing CO2 and N2 to be isolated. However isolation of $\ce{CO2}$ from this stream is expensive. As explained in a brochure from the CO2 Capture Project:

$\ce{CO2}$ can be separated from other gases by cooling and condensation. Cryogenic separation is widely used commercially for streams that already have high  $\ce{CO2}$ concentrations (typically >90%) but it is not used for more dilute  $\ce{CO2}$ streams.
A major disadvantage of cryogenic separation of  $\ce{CO2}$ is the amount of energy required to provide the refrigeration  necessary  for  the  process,  particularly  for  dilute  gas  streams. Another  disadvantage  is that some components, such as water, have to be removed before the gas stream is cooled, to avoid blockages.
Cryogenic separation has the advantage that it enables direct production of liquid  $\ce{CO2}$, which is needed for certain transport options, such as transport by ship. Cryogenics would normally only be applied to high concentration, high pressure gases, such as in pre-combustion capture processes or oxygen fired combustion.

This is complemented by information provided by Universal Industrial Gases Inc.:

Producing carbon dioxide as a commercial product requires that it be recovered and purified from a relatively high-volume, CO2-rich gas stream, generally a stream which is created as an unavoidable byproduct of a large-scale chemical production process or some form of biological process.
In almost all cases, carbon dioxide which is captured and purified for commercial applications would be vented to the atmosphere at the production point if it was not recovered for transport and beneficial use at other locations.
The most common operations from which commercially-produced carbon dioxide is recovered are industrial plants which produce hydrogen or ammonia from natural gas, coal, or other hydrocarbon feedstock, and large-volume fermentation operations in which plant products are made into ethanol for human consumption, automotive fuel, or industrial use.  Breweries producing beer from various grain products are a traditional source.  Corn-to-ethanol plants have been the most rapidly growing source of feed gas for CO2 recovery. CO2-rich natural gas reservoirs found in underground formations found primarily in the western United States and in Canada are another source of recoverable carbon dioxide.


Peroxide in acidic conditions being reacted with iron sulphate mechanism

At least for the second part of the product formation, you have pretty much given the answer yourself.

The methylcyclopentane moiety of the products results from two different 5-exo-trig radical ring closures, that is from the intramolecular addition of two different radicals to $\ce{R-\color{\red}{C}H=CH2}$.

How are these two different radicals generated?

Have a look at the starting materials! Isn't the combination of the hydroperoxide $\ce{R-OOH}$ and $\ce{FeSO4}$ very similar to the conditions of a Fenton reaction?

Let's therefore postulate the following: $\ce{R-OOH + Fe^2+ -> R-O* +\quad{}OH-  + Fe^3+}$

The resulting 1-methylcylohexan-1-ol derived radical can form methyl ketones by cleaving two different $\ce{C-C}$ bonds; the resulting carbon-centered radicals then give rise to the different cyclization products.


Alright, so I've worked out my answer (thanks to @user55119). The Reduction step was:

  $$\ce{ 2H2O_{(l)} + 2e^- -> 2OH^-_{(aq)} + H_2_{(g)}}\quad\text{(Reduction)}$$


Relationship between Energy Level and electron position

The term "position in their orbital" is incorrect. The position of the electron is smeared out over the whole orbital.

I guess, you are still thinking about the electron as a particle, which moves around in the space defined by the orbital. That's wrong!

The orbital IS the electron (at least for one-electron systems; for many-electron systems this is just approximately true). The "probability density" is just about the likelihood of a measurement, if you would do one. Then the electron suddenly behaves like a particle, and needs to decide on a certain position.

One might try to make many of such measurements with very small time steps in between them. Any attempt to connect them to some continuous path or trajectory would fail. The electron would just appear to be randomly jumping around, popping up at different locations, no matter how short the time steps between the measurements.

During a transition to a higher state, the electron does not move. It just changes its wavefunction, instantaneously when absorbing the photon. And as the probability density depends on the orbital (aka wavefunction), the probability density changes as well. There is no need to consider the exact position, as in quantum mechanics we cannot grab it. It is smeared out. It is at every position at the same time (just with varying probabilities).


What substance is used for staining brass?

This site regarding patina forming solutions might help:  http://www.sciencecompany.com/-W160.aspx#1


Molecular orbital diagram for nitrogen monoxide, the nitrosyl cation and the nitrosyl anion

As usual with these simple molecules: It's not as simple as it seems. Here it actually doubles down quite a bit more.
The short and very dis-satisfactory answer is: It is very, very complicated. For this reason, I am reluctant to present a molecular orbital diagram. However, I will provide calculated MOs. While it may be tempting to just look at the pictures, the rationalisation behind it is more important. What will follow now are a few attempts at that rationalisation.

In the nitric oxide series, $\ce{NO+}$ is surprisingly the simplest. Why is that surprising? Because it is isoelectronic with $\ce{CO}$; and this molecule is one of the most complicated ones there are. You can read a bit more about it here: How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?
I do compare $\ce{NO+}$ more with $\ce{CO}$ rather than $\ce{N2}$ to which it is also isoelectronic, because of the asymmetry introduced by the different nuclei. The molecule is the simplest to explain, because it is a singlet ground state. All electrons are neatly paired, and spacial symmetry can easily be matched with the degeneracy of the orbitals.
The π-orbitals are not the HOMO, instead it is the σ lone pair at nitrogen.

The next more complicated is $\ce{NO-}$, which is isoelectronic to $\ce{O2}$. In this case it is hard to find a less symmetric isoelectronic version. It is well known, that the ground state is a triplet because of the degeneracy of the π-orbitals. While we would expect the orbital ordering to be the same as in $\ce{NO+}$, it is not. This is mainly due to the fact, that occupied orbitals behave differently than unoccupied orbitals.
In some way, because of the paramagnetic character, we have a situation wich cannot be described by a single molecular orbital description. You need at least two: one for α-electrons (or spin up) and one for β-electrons (or spin down). The good thing however is, that both of these descriptions may still be matched with spatial symmetry.
The bonding π orbitals become higher in energy than the σ lone pair because the anti-bonding counterparts are partially filled.

The most complicated molecule in the series is the radical $\ce{NO}$. There is absolutely no simple description. Because of the unpaired electron, you need at least two descriptions again. However, this is not enough because now one of the spin-state description cannot match the spatial symmetry. This is because one electron cannot occupy two orbitals at the same time in an MO description. Because of that (from a technical point of view) the orbital degeneracy has to be broken. Physically this is not possible.
The molecule has to be treated with a multi-reference approach, i.e. one MO description for every possible configuration. It is probably safe to assume that the orbitals of π symmetry don't split that much and are higher in energy than the σ lone pair.

Below you can find the calculated MO. The level of theory used for $\ce{NO+}$ was RMP2/def2-QZVPP, because this is equivalent to the level the other molecules were treated, which is ROMP2//UMP2/def2-QZVPP. One can clearly see how the orbital symmetry for $\ce{NO}$ breaks the degeneracy of the pi orbitals.
For $\ce{NO+}$ MOs 5,6 and 8,9 are degenerated. For $\ce{NO-}$ MOs 6,7 and 8,9 are degenerated.
Colour code: blue/orange, α, β-occupied; purple/yellow-orange, α/β-occupied; red/yellow, unoccupied; contour value, 0.1.

With a little more literature research (prior to answering the question) one would have found the following publication:
Renato P. Orenha and Sérgio E. Galembeck. Molecular Orbitals of $\ce{NO}$, $\ce{NO+}$, and $\ce{NO–}$: A Computational Quantum Chemistry Experiment. J. Chem. Educ. 2014, 91 (7), 1064–1069.


Does an airborne chemical stimulant exist?

Regardless of whether the stimulant is actually gaseous at standard temperature and pressure, it can be made airborne as an aerosol simply by spraying. If the room to be filled with vapour is sufficiently small then it is no problem to just spray a solution of your stimulant and pump it inside. But even gym hall sizes should be fine with adequate technique applied.


I remember my days when I started confusing myself in the exact same way. The issue I believe is that you haven't been explained the details of how all this happens.
Evaporation:
All molecules in a liquid have various amount of energy. But the ones at the top, the ones that form the surface - they have the maximum energy in them because they are present there by overcoming any forces of attraction. (The other way to imagine this would be, the layer of molecules at the surface will have mostly downward forces of attraction on them compared to the distributed force of attraction from all directions on a bulk molecule.) Now these surface molecules have the tendency to escape. These molecules will individually leave the liquid by getting rid of the forces of attraction because it is easier for them compared to bulk molecules. And when individual molecules leave the liquid, it is called evaporation. Although this is a form of vapourisation, it doesn't need a constant temperature where the change of state occurs. They are just individual molecules that leave the liquid and escape and cause a vapour over the liquid.
Boiling:
But in boiling, a lot of constant heat supply, constant temperature, constant pressure. All this is required. And when this is obtained, the liquid changes its state and becomes a gas. This gas forms bubbles and escapes from the bulk of the liquid as well and not only from the surface as in case of evaporation.


Alkane, alkene, alkyne boiling point comparison

Disclaimer: All of this "jazz" will be about reaching a mere rule-of-thumb. You can't just compare whole families of organic compounds with each other. There are more factors to consider than below, mostly based on isomerism notions. However, as most of the A grade exams emphasize on the lighter aliphatic compounds, we can understand each other here. :)

It's all about polarizability.

  Polarizability is the ability for a molecule to be polarized.

When determining (aka comparing) the boiling points of different molecular substances, intermolecular forces known as London Dispersion Forces are at work here. Which means, these are the forces that are overcome when the boiling occurs. (See here for example)

Source

London forces get stronger with an increase in volume, and that's because the polarizability of the molecule increases. (See the answer to this recent question)

Alkanes vs. Alkenes

In their simplest form (where no substitution etc. has occurred) alkanes tend to have very close boiling points to alkenes.

  The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids.

  Boiling points of alkenes depends on  more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules.

  \begin{array}{|c|c|}\hline
\text{Compound} & \text{Boiling point / }^\circ\mathrm{C} \\ \hline
\text{Ethene} & -104 \\ \hline
\text{Propene} & -47 \\ \hline
\textit{trans}\text{-2-Butene} & 0.9 \\ \hline
\textit{cis}\text{-2-Butene} & 3.7 \\ \hline
\textit{trans}\text{-1,2-dichlorobutene} & 155 \\ \hline
\textit{cis}\text{-1,2-dichlorobutene} & 152 \\ \hline
\text{1-Pentene} & 30 \\ \hline
\textit{trans}\text{-2-Pentene} & 36 \\ \hline
\textit{cis}\text{-2-Pentene} & 37 \\ \hline
\text{1-Heptene} & 115 \\ \hline
\text{3-Octene} & 122 \\ \hline
\text{3-Nonene} & 147 \\ \hline
\text{5-Decene} & 170 \\ \hline
\end{array}
  In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons.

Alkanes vs. Alkynes

As explained, since there is a bigger volume to an alkane than its corresponding alkyne (i.e. with the same number of carbons) the alkane should have a higher boiling point. However, there's something else in play here:

Alkynes, have a TRIPLE bond!

I currently can think of two things that happen as a result of this:

London Dispersion Forces are in relation with distance. Usually, this relation is $r^{-6}$. (See here) The triple bond allows two alkynes to get closer. The closer they are, the more the electron densities are polarised, and thus the stronger the forces are.
Electrons in $\pi$ bonds are more polarizable$^{10}$.

These two factors overcome the slight difference of volume here. As a result, you have higher boiling points for alkynes than alkanes, generally.

\begin{array}{|c|c|}\hline
\text{Compound} & \text{Boiling point / }^\circ\mathrm{C} \\ \hline
\text{Ethyne} & -84^{[1]} \\ \hline
\text{Propyne} & -23.2^{[2]} \\ \hline
\text{2-Butyne} & 27^{[3]} \\ \hline
\text{1,4-Dichloro-2-butyne} & 165.5^{[4]} \\ \hline
\text{1-Pentyne} & 40.2^{[5]} \\ \hline
\text{2-Heptyne} & 112\text{–}113^{[6]} \\ \hline
\text{3-Octyne} & 133^{[7]} \\ \hline
\text{3-Nonyne} & 157.1^{[8]} \\ \hline
\text{5-Decyne} & 177\text{–}178^{[9]} \\ \hline
\end{array}
1: http://en.wikipedia.org/wiki/Acetylene
2: http://en.wikipedia.org/wiki/Propyne
3: http://en.wikipedia.org/wiki/2-Butyne
4: http://www.lookchem.com/1-4-Dichloro-2-butyne/
5: http://en.wikipedia.org/wiki/1-Pentyne
6: http://www.chemsynthesis.com/base/chemical-structure-17405.html
7: http://www.chemspider.com/Chemical-Structure.76541.html
8: http://www.thegoodscentscompany.com/data/rw1120961.html
9: http://www.chemsynthesis.com/base/chemical-structure-3310.html
10: https://chemistry.stackexchange.com/a/27531/5026

Conclusion: We can't fully determine the boiling points of the whole class of alkanes, alkenes and alkynes. However, for the lighter hydrocarbons, comparing the boiling points, you get:
$$\text{Alkynes > Alkanes > Alkenes}$$


By definition, for a cell to be galvanic, you would need the difference $\ce{E^0_(cathode) - E^0_(anode)}$ to have a positive value. So yes, the standard potential for the anode to have a more negative potential than the cathode.

Lower and higher are terms that are best avoided when talking about electrochemistry. More positive and more negative are unambigious.


Why is gold so popular in nanotechnology?

First, let's look at gold's properties.

Gold is the most malleable of all metals. This malleability alone is very useful in aiding scientists to create such small gold nanoparticles.
Gold also reflects infrared radiation very well and is an excellent conductor of heat and electricity.
Generally, gold is not easily affected by oxygen and is generally nonreactive with most acids and bases. This property serves well especially in the human body where there are an abundant of acidic and basic fluids and the circulatory system is enveloped with oxygen. Because gold is generally unaffected by most acids, bases, and oxygen, there is more assurance for scientists that their golden nanoparticles for medicinal uses would not react with other things before reaching their specific target.
Going back to gold's ability to reflect infrared radiation, an essential reason why physicians, scientists, and researchers utilize nanotechnology is to track down certain cells (cancer, etc), molecules, or processes. Thus, it is important to use such a metal that can reflect radiation to tell them where their targeted molecules are or where their targets are going. In essence, gold nanoparticles are suitable for "staining," or showing contrasts in cells more visible, such as in tissue samples. This can provide higher spatial resolutions for labeling applications.
Gold, as a nice heat conductor, can also absorb heat very well, as I mentioned earlier, and in these biotechnologies, strong absorption of light energy is also another crucial aspect for allowing gold to then dissipate this absorbed light energy in its surroundings so that these heated areas can also be tracked and help with drug delivery purposes.

Here are some references:

Instant insight: A golden future
Nanotechnology Research into the Uses of Gold Nanoparticles


References for Transition State Theory and Quantum Transtion State Theory

There are many sources that you can check out, in particular the reviews of Truhlar:

Current Status of Transition-State Theory by Truhlar et al.

This article was used in my undergraduate course of chemical kinetics, so I think that it could be a good source. Also we used:

Variational transition-state theory by Truhlar and Garrett

Maybe they are a bit outdated, but they are good as start points. The abstract of this newer article make it seems to be useful for you

Transition state theory: Variational formulation, dynamical corrections, and error estimates by Vanden-Eijnden and Tal

To deeply understand the theory statistical physics is a must. A good source (that also deals with TST) is the Ralph H. Fowler's book.


How can a waterproofing material be "water-based"?

One might need to have more information about the product you've posted to say what occurs in the specific example. But, the general concept could be similar to Latex (water-based) paint. Where the water proofing chemical is dispersed by an emulsion, and once the water evaporates, the proofing material is restructured by its, now unhindered, intermolecular forces. These forces might be strong enough, once the proper structure exists, to resist water coming back and trying to interrupt the interactions.

Below is the Wikipedia explanation for my example of latex paint.

  The term "latex" in the context of paint in the USA simply means an aqueous dispersion; latex rubber from the rubber tree is not an ingredient. These dispersions are prepared by emulsion polymerization. Such paints cure by a process called coalescence where first the water, and then the trace, or coalescing, solvent, evaporate and draw together and soften the binder particles and fuse them together into irreversibly bound networked structures, so that the paint cannot redissolve in the solvent/water that originally carried it.


Theoretically you can simulate any reaction an a computer, provided it is big enough of course.

Practically however, this should be close to impossible. Here are some thoughts on how to approach that and their related problems.

First, one would have to model a complex reaction mixture including the educts $\ce{H2O}$, $\ce{CH4}$, $\ce{NH3}$, and $\ce{H2}$. Simulating those alone, without considering reactions, should be possible using molecular mechanics. This should scale up to a couple of thousands of these simple molecules.

Modelling all the possible reactions requires quantum mechanical approaches. But this does not scale anywhere close to macroscopic quantities.

One could model all the individual possible reactions one at a time. Then it is possible to compare how fast they are and how much energy is required. From that one could deduce how likely a certain reaction is, and feed this into the molecular mechanics calculation mentioned above.

First one would have to figure out all possible product of the reaction mixture, including all intermediates. This alone should yield a huge number of candidate compounds. But maybe one could try to focus just on a single product and one path there. Based on experimental results, pick a target compound of interest and a possible reaction path/mechanism. Experiments should give information on both, products and intermediates.

But even simulating a predefined reaction path (i.e. the educts, intermediates and final product) is a challenge. Different relative positions of the atoms in a reaction need to be considered (the energy landscape along the reaction coordinate). Ab initio molecular dynamics (AIMD) is an approach to do that. But there are still a number of approximations involved. This is costly and of limited accuracy.

We also need to consider the spark from the electrodes. This will lead to electronically excited state and ionic intermediates. This adds quite a lot of complexity to the already difficult task.

In summary, tackling the whole thing at once is impossible. One would have to break it down into many small steps, maybe then it becomes feasible to focus only on very specific aspects.


Reason for inconsistency in specific gravity of calcium choride added to water

When you say, "of products", do you really mean "of the solution"? If so:

The solvent mass and the solute mass add up to the mass of the solution.

But solvent volume and solute volume often don't add up to the volume of the solution. This is because the forces between molecules in the solution are different from the forces between molecules in the pure solute and the pure solvent.

In your solution, the calcium ions and chloride ions attract water molecules around them. The water molecules are packed more tightly than they would be if the ions weren't there. That can increase the density of the solution relative to the total volume of the unmixed solvent and solute. Another effect is that the ions in solution are packed less tightly than they are in solid $\rm CaCl_2$.

You can see which effect wins by dumping some calcium chloride into a volumetric flask. Fill the flask up to the mark with water, and then mix until all the calcium chloride dissolves. The solution probably doesn't quite come up to the mark anymore, because the calcium and chloride ions have drawn the water molecules in around themselves, making the solution denser than it would be otherwise.

More detail than you may want below:

The volume of the solution $V$ is given by the following equation:

$$V = n_1 \overline{V}_1 + n_1 \overline{V}_2$$

where $n_1$ and $n_2$ are moles of calcium chloride and moles of water in the solution, and $\overline{V}_1$ and $\overline{V}_2$ are the partial molar volumes of the calcium chloride and the water.

These partial molar volumes are actually derivatives of the volume of the solution with respect to moles of calcium chloride ($\overline{V}_1 = \partial V/\partial n_1$) and with respect to water ($\overline{V}_2 = \partial V/\partial n_2$). They are related to each other (when one rises, the other falls) and they vary with the composition and temperature of the solution in a rather complicated way.


How much does a light heat the air around it?

During a period of $\Delta t = 1\ \mathrm h =3600\ \mathrm s$, the light bulb converts electric power of $P = 60\ \mathrm W$ to light and heat. Assuming that none of this energy escapes from the room, the entire power is eventually converted to heat:

$$\begin{align}
Q &= P \cdot \Delta t \\
&= 60\ \mathrm W \times 3600\ \mathrm s \\
&= 216\ \mathrm{kJ}
\end{align}$$

According to REFPROP (NIST Standard Reference Database 23, Version 9.0), the relevant parameter values of air at the assumed initial temperature of $T_0 = 20\ \mathrm{^\circ C}$ and pressure $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$ are a density of $\rho_0 = 1.2046\ \mathrm{kg/m^3}$ and a specific heat capacity at constant pressure of $c_{p,0} = 1.0061\ \mathrm{kJ/(kg\ K)}$.

The initial volume is given as $V_0 = 85\ \mathrm{m^3}$. The corresponding mass is

$$\begin{align}
m &= V_0 \cdot \rho_0 \\
&= 85\ \mathrm{m^3} \times 1.2046\ \mathrm{kg/m^3} \\
&= 102.391\ \mathrm{kg}
\end{align}$$

Ignoring the heat capacity of the walls and inventory, the temperature of the air is approximately increased by

$$\begin{align}
\Delta T &= \frac QC \\
&= \frac Q{m \cdot c_{p,0}} \\
&= \frac{216\ \mathrm{kJ}}{102.391\ \mathrm{kg} \times 1.0061\ \mathrm{kJ/(kg\ K)}} \\
&= 2.1\ \mathrm{K}
\end{align}$$

By way of comparison, a better result may be obtained using the enthalpy of air, since the heat capacity of air is not constant. In this case, however, the difference is insignificant because the temperature change is small. The initial specific enthalpy of air is $h_0 = 419.40\ \mathrm{kJ/kg}$. Thus, the enthalpy is

$$\begin{align}
H_0 &= m \cdot h_0 \\
&= 102.391\ \mathrm{kg} \times 419.40\ \mathrm{kJ/kg} \\
&= 42\,943\ \mathrm{kJ}
\end{align}$$

At constant pressure, the change in enthalpy corresponds to the heat transferred to the air:

$$\begin{align}
H_1 &= H_0 + \Delta H \\
&= H_0 + Q \\
&= 42\,943\ \mathrm{kJ} + 216\ \mathrm{kJ} \\
&= 43\,159\ \mathrm{kJ}
\end{align}$$

Thus, the final specific enthalpy is

$$\begin{align}
h_1 &= \frac{H_1}{m} \\
&= \frac{43\,159\ \mathrm{kJ}}{102.391\ \mathrm{kg}} \\
&= 421.51\ \mathrm{kJ/kg}
\end{align}$$

At a pressure of $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$, this specific enthalpy corresponds to a temperature of $T = 22.1\ \mathrm{^\circ C}$.

Note that, at the final temperature of $T = 22.1\ \mathrm{^\circ C}$ and the pressure of $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$, the new density of air amounts to $\rho_1 = 1.1960\ \mathrm{kg/m^3}$. Thus, the new volume is

$$\begin{align}
V_1 &= \frac{m}{\rho_1} \\
&= \frac{102.391\ \mathrm{kg}}{1.1960\ \mathrm{kg/m^3}} \\
&= 85.6\ \mathrm{m^3}
\end{align}$$

Therefore, a small amount of air must escape from the room.

Finally, if the heat from a $60\ \mathrm W$ light bulb really is a problem, you might want to leave the room since the human body generates about $100\ \mathrm W$.


The key to understanding this type of question is that the equilibrium (saturation) vapor pressure of a pure liquid is to a first approximation (that is, assuming ideal behavior) independent of the presence of other gases, such as (in this case) oxygen. Therefore, you can think of $\pu{355 Torr}$ as being always the pressure of the vapor when in equilibrium with the pure liquid at $T=\pu{80 ^\circ C}$. For equilibrium to be established requires that there is (more than) enough substance to fill the volume with vapor at that pressure. The minimum amount of substance which will fill the volume with vapor at that pressure is, under the assumed ideal conditions, given by the ideal gas law: $$n=\frac{pV}{RT_b}$$where in the present problem $T_b=\pu{80 ^\circ C}$, $p=\pu{355 Torr}$, and V is the volume occupied by the vapor, which changes from $V<\pu{100 mL}$ to $V<\pu{50 mL}$. In the present problem you don't know how much of the water is initially a liquid, only that the inequality $V<\pu{100 mL}$ is satisfied by the vapor. During the volume reduction some of the vapor condenses into additional liquid in order to retain the equilibrium pressure. The oxygen is treated here as a spectator, although it contributes to the total pressure in the vessel. In practice it also alters the vapor pressure of the liquid, although here by a negligible amount.


Why do self-replicating molecules dissociate?

Well, allow me to answer since I worked on self-replication during my master’s thesis.

The thing about all these associationen is that there is always an equilibrium to be considered. That is also true for chemical reactions: Technically, every reaction is at an equilibrium of some kind.

$$\ce{A + B <=> C + D}$$

or in the special case of associations:

$$\ce{A + B + AB <=> A-B-AB}$$

For nomal bond-creating and bond-breaking reactions, the energy involved in the whole process is relatively large — breaking a carbon-carbon bond requires approximately $400\,\mathrm{\frac{kJ}{mol}}$! However, if we are merely talking about association as in complexation, the energies involved (i.e. gained or lost) are rather small. This es especially also true for the activation energy, meaning that forming and dissociation of such complexes are rather fast and at equilibrium the whole time. It is important to realise that the bonds formed for the self-replicating complex are all intermolecular and only the bond between the fragments is a covalent, intramolecular bond.

Let’s take a look at an exemplary self-replicating complex:[1]

On the right half of this picture we have the two fragments, on the left we see the template. All solid lines are actual (intramolecular) bonds while all dashed lines are intermolecular associations — in this case hydrogen bonds. The intermolecular dashed interactions build up and break down rapidly with low energy barriers. The new intramolecular bond is formed between carbon and nitrogen on the aromatic rings (not formed on the right, formed on the left) and represents the energy liberated in the reaction. This one is only formed, not broken.

To form the bond, the two fragments need to linger on the template long enough. That is generally no problem, despite rapid association and dissociation.

In fact, form too stable of a complex (i.e. one that won’t dissociate quickly) and you have effectively eliminated the complex’ ability to self-replicate.

[1] A. Terford, G. von Kiedrowski, Angew. Chem. 1992, 104, 626. DOI: 10.1002/ange.19921040523
In English: Angew. Chem. Int. Ed. 1992, 31, 654. DOI: 10.1002/anie.199206541


Salts that are more hydrophilic than NaCl

As per DavePhD's answer, the contact angle of NaCl with water is zero, so we need another measure.

I think the word you are looking for is hygroscopic. http://en.wikipedia.org/wiki/Hygroscopy Substances which are so hygroscopic that they dissolve in the water they absorb from the atmosphere are said to be deliquescent, although the property of deliquesence in itself does not directly indicate the strength of a substance as a drying agent.

The strength would best be measured as the equilibrium vapour pressure at a certain temperature, or alternatively, the temperature required to dehydrate: i.e. the temperature at which the vapour pressure of water vapour above the hydrated form of the substance becomes equal to atmospheric.

Cheap substances which are hygroscopic find applications as desiccants

Here is a list of some common ones http://en.wikipedia.org/wiki/List_of_desiccants. It includes the workhorse laboratory desiccants: silica gel for use in desiccators where products are dried under vacuum; molecular sieves, for drying solvents; CaCl2 for drying gases; and MgSO4 and CaSO4 in their anhydrous form for a quick drying of organic solutions. I would assume (though I'm not sure) that all of these would be stronger than NaCl.

Unfortunately the really strong ones are missing, so here's a list of mine. The weaker ones above tend to act by physical absorption, or by changing from anhydrous crystals to hydrated crystals. Stronger ones tend to react chemically.

Concentrated sulfuric acid Strong enough to dehydrate sugar to "charcoal." One of the reasons for the affinity is the energy released when H2SO4 proponates H2O, forming HSO4- + H3O+. More of a curiousity than a desiccant. Not widely used as a desiccant because of its acidity.

Phosphorous pentoxide Reacts with water to form various different polymeric phosphoric acids, ultimately ending up with 2P2O5 + 6H2O= 4H3PO4 though strength is gradually reduced as water is absorbed. Commonly used in the lab when a strong desiccant is required.

Calcium Oxide The reaction CaO + H2O = Ca(OH)2 has been proposed for large scale energy storage (the storage occurs in the reverse direction), but so far has been used only for smaller applications. As the calcium species are basic, exposure to air causes absorption of CO2, which then irreversibly degrades the material to bicarbonate and carbonate (unless heated to a much higher temperature.)

Sodium Metallic sodium, extruded into wire using a press to increase its surface area, is the best way to ensure absolutely dry solvents.


Does rate of auto-ionization of water change after adding a strong acid?

Ideal solutions
If we pretend for a moment that all the solutions under consideration are ideal, and that we have already reach equilibrium, we can consider the following reaction:
$$\ce{H2O(l) + H2O(l) <=> HO-(aq) + H3O+(aq)}$$
As Zhe said in the comments, the concentration of the reactants don't change much (even if you have $\pu{1 M}$ sodium hydroxide or $\pu{1 M}$ hydrochloric acid), so the forward rate at equilibrium is the same at any pH (again, assuming an ideal solution). That means the reverse rate is also the same at any pH. Because the product of the concentrations of hydroxide and hydronium ions is constant, the reverse rate constant does not depend on the pH.
Considering ionic strength and activity coefficients
The above argument does not work in a real solution with high concentrations of dissolved ions. If the pH is far from neutral, the concentration of ions (and counter-ions is high), and equilibrium concentrations (and with that the rates) will depend on the solutes present in solution.
Molecular level
At the molecular level, the auto-ionization of water is a special case because hydronium and hydroxide ions seem to diffuse faster than other ions (Grotthuss mechanism). At neutral pH, both species are rare. At acidic pH, there are more hydronium ions, but fewer hydroxide ions, so collisions are also rare (and reverse for basic). A high ionic strength would make it easier for ion pairs to form and separate, while a low ionic strength would be advantageous for the combining of hydronium and hydroxide.


Two may be as many protons that a Bronsted-Lowry base as strong as hydroxide ion can remove.  Building up more than two negative charges than that on the ion is a tough job.  Possibly a stronger base, such as amide ion in ammonia, could do that.

That the basicity of the medium could impact how many protons can dissociate is nicely demonstrated by citric acid.  Ordinarily it is triprotic in aqueous solution, but spectroscopic evidence indicates that at high alkali concentration (pH>14.4) a fourth, alcoholic proton can be removed.  There would be no doubt with a base such as amide in ammonia.


Under physiological conditions (aqueous solution, pH near neutral), amino acids don't react with each other directly, and the equilibrium lies on the side of peptide bond hydrolysis, not formation.

  When amino acids react with each other, how do you know which one reacts using it's carboxylic group and which one with it's amine group?

In the cell, amino acids first react with ATP, then form an ester with RNA, and then their amino group forms a peptide bond with the growing amino acid chain. The last step is catalyzed by ribosomes that select the next amino acid to be added using messenger RNA as a template.

  For example the dipeptide val-ser, between the amino acids valine and serine, does valine's amino group react or does it's carboxylic group with serine's respective group?

In your example, valine and serine could form either valyl-serine or seryl-valine. As Tyberius writes in the comments, peptides are written so that in each peptide link, the amino acid on the left of the link provides the carboxylic acid group and the one of the right of the link provides the amino group. The very first amino acid (the one at the amino terminus) carries an amino groups that has not undergone peptide bond formation, and the last amino acid (the one at the carboxy terminus) carries a carboxylate group that has not (yet) undergone peptide bond formation.


How can antibonding orbitals be more antibonding than bonding orbitals are bonding?

Mathematical Explanation

When examining the linear combination of atomic orbitals (LCAO) for the $\ce{H2+}$ molecular ion, we get two different energy levels, $E_+$ and $E_-$ depending on the coefficients of the atomic orbitals. The energies of the two different MO's are:
$$\begin{align}
E_+ &= E_\text{1s} + \frac{j_0}{R} - \frac{j' + k'}{1+S} \\
E_- &= E_\text{1s} + \frac{j_0}{R} - \frac{j' - k'}{1-S}
\end{align} $$

Note that $j_0 = \frac{e^2}{4\pi\varepsilon_0}$, $R$ is the internuclear distance, $S=\int \chi_\text{A}^* \chi_\text{B}\,\text{d}V$ the overlap integral, $j'$ is a coulombic contribution to the energy and $k'$ is a contribution to the resonance integral, and it does not have a classical analogue. $j'$ and $k'$ are both positive and $j' > k'$. You'll note that $j'-k' > 0$.

This is why the energy levels of $E_+$ and $E_-$ are not symmetrical with respect to the energy level of $E_\text{1s}$.

Intuitive Explanation

The intuitive explanation goes along the following line: Imagine two hydrogen nuclei that slowly get closer to each other, and at some point start mixing their orbitals. Now, one very important interaction is the coulomb force between those two nuclei, which gets larger the closer the nuclei come together. As a consequence of this, the energies of the molecular orbitals get shifted upwards, which is what creates the asymmetric image that we have for these energy levels.

Basically, you have two positively charged nuclei getting closer to each other. Now you have two options:

Stick some electrons between them.
Don't stick some electrons between them.

If you follow through with option 1, you'll diminish the coulomb forces between the two nuclei somewhat in favor of electron-nucleus attraction. If you go with method 2 (remember that the $\sigma^*_\text{1s}$ MO has a node between the two nuclei), the nuclei feel each other's repulsive forces more strongly.

Further Information

I highly recommend the following book, from which most of the information above stems:

P. Atkins and R. Friedman: Molecular Quantum Mechanics, $5^\text{th}$ ed. Oxford University Press, 2011.


Is home synthesis of cellulose acetate possible, and why?

The bottom line is that this would be a very bad idea to try due to safety reasons alone, and is not realistically feasible using only household materials regardless.

Glacial (100%) acetic acid is dangerous if not handled properly and requires an appropriate fume hood at a minimum. You also are not going to get to glacial acetic acid from vinegar, and water will kill this reaction.

Acetic anhydride is a reagent that you just won't be able to obtain or synthesize yourself. Even if you did, it would be hydrolyzed by the water in your impure acetic acid.

Sulfuric acid is commonly used as a catalyst, and while you could obtain this, it should absolutely not be handled by someone without appropriate equipment and training in it's safe usage.

So again, this is just not a feasible project and there would be serious dangers along the way if you tried. I like your creative thinking process and I hope you will pursue some other, safer, home science project.


Alkene production by dehydrohalogenation

The fundamental reason lies in the difference between the action of a base and a nucleophile.

There are two types of mechanism through which substitution proceeds:-

SN1 Reaction - Here, the substrate undergoes hetrolytic cleavage to produce a carbocation and an anion. The thing which solvents do is to stabilise the newly formed ions. A Polar solvent(One case-Water) stabilises the intermediate stage more than the reactant stage because the reactant state forms weak dipole-dipole bonds with the solvent where as the intermediate stage has full fledged proper charges on then to interact with the solvent and forms strong ionic bonds.
SN2 Reaction - Same thing but because of the changes in mechanism type of solvent changes but the essence is same.

This tells us that the solvent is crucial to generate a strong electric field which can help the reaction proceed further by ionising the substrate whereas a base is not so capable of handling this type of work

In elimination reaction, the solvent generates a strong electric field around the substrate which in turn helps in the cleavage of C-H bond so that it can stabilise the carbocation(in E1 mechanism) or help the transition state to get rid of the leaving group(In E2 mechanism).

Basic action is a matter of equilibrium which shows how readily it can accept protons and go forward whereas nucleophilicity is all about the rate how fast it can go which in turn shoes us how eager it is to form bond or get rid of its negative charge.


How to avoid explosion when there is a leakage of LPG gas and coal gas/town gas?

Aside of open flame and operating electronic devices, another danger hard to eliminate is static electricity, mostly due synthetic material of clothes, or rubber soles of shoes.
The safest and the most available prevention choice at emergency scenario is intense ventilation. All others are rather hope it would be enough, as it is hard to eliminate or not to forget all possible ignition sources.
Be aware that switching devices off may be more dangerous than keeping them on, as interrupting high voltage  AC circuit usually creates sparks, unless this is addressed by parallel capacitors. This is typical for devices transforming AC to  low voltage, like most low power electronics.

Coal gas is of little use these days, with major fuel components  carbon monoxide and hydrogen.
Much more common is natural gas, based on methane, or LPG, based on butane (typically 80%) and propane.
The oxidant is in all cases aerial oxygen.


It is caused by physical and chemical reactions in the cake batter of two types:

1) When heated and mixed with water, starches present in the flour will undergo a process called gelatinization in which water is absorbed in the structure. This will cause the batter to 'set' from a liquid into a solid.

2) Additionally, reactions of the proteins from the egg that you added can also work to 'set' the batter. At first the proteins will unfold (denature) after which they start to form bigger proteins through cross-linking reactions that turn the liquid into a solid.

The baking soda in the batter turns into sodium carbonate, water and carbon dioxide. The latter will make your cake nice and fluffy.


How can I remove a PEG shampoo from a natural horn-comb without damaging it?

Soap, water and a dash of elbow grease. PEGs are all surfactants which will wash away with water and if they don't then an alcohol-water mix (vodka) will certainly remove any PEG residue, probably along with some of the natural oils in the horn comb that has built up over time and usage.

This may leave the comb with whitened patches or streaks on it.  These are not a PEG residue but fine cracks in the horn, interstices in the layers of keratin that the horn is made of, that used to be filled with oil which the PEG shampoo has removed.  Any number of natural oils will improve the appearance of the horn, including, but not limited to, natural scalp oils (i.e. use the comb for its intended purpose), linseed oil (a favourite of cricketers for their bats), castor oil, beeswax, olive oil (a bit sticky) or a light application of any proprietary furniture polish.  Choose something unscented if you don't want Mom to find out. And start with the least viscous options you have, these will achieve greater penetration of the fine cracks in the horn and "seal" them with a wax if possible.

I've seen antique tortoiseshell respond very badly to cleaning with detergents, but the damage is really illusory and can be restored with the gentle application of oils and waxes.


What workflow can I use on to extract cartesian coordinates of many molecules from both .spartan and .out (Gaussian) files?

The only tool that I know for reading Spartan files (besides Spartan) is Jmol. With some programming or scripting, you will probably be able to save conformers to other files in a workflow - I don't think there is a pre-existing script.

There are multiple open source tools that read Gaussian output files. I would suggest either:

Open Babel - GUI, command-line and Python
cclib - using Python

For example to extract a bunch of Gaussian files to XYZ Cartesian:

obabel *.log -oxyz -m

The -m flag indicates "multiple" and the -oxyz indicates that you want XYZ-format files. More is in the manual.


Sodium chloride is colorless because the electrons are tightly bond tho the sodium cation and the chloride anion. Visible light has not enough energy to excite the electrons to higher energy levels. So there's no absorption or emission of visible light of a certain wavelength and therefore no color.
I wouldn't say that the majority of salts is colorless, since the more complex the composition of a salt is the more likely it is not colorless.


Theoretical yields and determining the number of moles in a reaction

You are correct in your thinking.

First, knowing the mass of the compounds, use the molar mass relationship for each (g/mol) to convert them all to moles. This is because when it comes to reactions, compounds react on a mole or number basis.
This lets you know the "starting amounts" of $\ce{H_2O}$ and $\ce{CO_2}$. Now, our forum has a policy against providing direct answers to homework, so I'll give examples. If you hypothetically had, say, 12 moles of both water and carbon dioxide, then using the numbers in the balanced chemical equation, they can fully react to make 2 moles of glucose. Thus, 2 moles is your theoretical yield. However, in another case, if you had 9 moles of water and 14 moles of carbon dioxide, the $\ce{H_2O}$ limits the amount of glucose you can produce, and it is the limiting reactant. (14 moles of $\ce{CO_2}$ would require 14 moles of $\ce{H_2O}$. There will be some $\ce{CO_2}$ left over because you are limited by only 9 moles of $\ce{H_2O}$). In this hypothetical case, the theoretical yield would be 1.5 moles of glucose. (Your problem may or may not involve limiting/excess reactants).

And if you're wondering, it's called "theoretical" yield because if you were carrying out this reaction in real life, some inefficiencies in combustion would yield less than the calculated amount. Hope this helps.


Why does the anode solution contain Sn2+ in a Sn-Cu voltaic cell?

A voltaic circuit needs to be a complete circuit first an foremost. So, to answer your 2 main concerns, think of a voltaic cell as a electric circuit:

  In one half cell, we have a $\ce{Sn}$ anode in water connected to a wire.

Water is not a good conductor of electricity, dissolved ions are needed, such as the $\ce{Sn^{2+}}$ ions are needed to facilitate the movement of the electric current in the first place. Also, the presence of the ions allows the overall redox reaction be spontaneous to provide the electric current (ChemWiki).

  The wire is submersed in the other half cell which is simply $\ce{Cu^{2+}}$ solution

According to the ChemWiki Voltaic Cells, the metals plates provide a surface for the reactions to occur on, a wire just submersed into the solution provides a very small surface area for the reaction to occur.

A diagram from the UC Davis ChemWiki page Electrochemical Cells helps explain this (the example uses zinc instead of tin, but the principle is the same):


Where are they stored? (How does tea plant make them? I'm looking for the reactions that occur in the leaf)

Both citrate and oxalate are associated with ATP production via the citric acid cycle. As is obvious from the name, citrate is a critical component of the process. Oxalate is a hydrolysis product of oxaloacetic acid, which is also part of the cycle. Both of these ions are present within the plant cells.

  How and why do they get released from tea leaf? Since you don't usually see any "big" differences between tea leaf and a soaked tea leaf? (I'm not looking for the how of solution)

If your tea leaves are crushed or broken in any way, this really is just simple dissolution. If you have whole tea leaves, the oxalate and citrate are migrating into and out of the plant cells. The cell wall is permeable, so the ions (or corresponding acids) can simply flow through it. The cell membrane is semipermeable, and the diffusion of the ions across it is a biologically controlled process that I honestly don't have the background to explain.

  What makes the color of water change? Why does the color vary with pH alteration?

The pH and color are not related properties. The pH is affected by the citric acid and oxalic acid associated with the anions previously discussed, as well as other minor components that are likely present. The color comes from some of the many other compounds that are extracted into the water, which number in the hundreds, if not thousands. Some of these compounds are inherently colored - no reaction is required to produce color in the solution.


Determination of pKa by absorbance and pH of buffer solutions

Make the plot $10^{\mathrm{pH}}(A_{\ce{L-}} - A)$ versus $A$,
the adjusted equation will be of the form
$$
10^{\mathrm{pH}}(A_{\ce{L-}} - A)
  = 10^{\mathrm{p}K_\mathrm{a}}A
  - 10^{\mathrm{p}K_\mathrm{a}}A_{\ce{HL}},
$$
the slope is $10^{\mathrm{p}K_\mathrm{a}}$, that is, $\mathrm{p}K_\mathrm{a} = \log_{10}(\text{slope})$.


Finding the concentration from a dilution

Solving one out of the four would be good enough to understand the concept behind this. Therefore, I shall explain the first experiment and the concentrations of the components in the solution produced.
In experiment 1 all four are added in equal quantities. Finding the total volume, we get the total volume
$$V_\text{tot} = \pu{4 ml}.$$
Assuming the amount of $i$th solute to be denoted by $n_i,$ its molarity $c_i$ can be found as
$$c_i = \frac{n_i}{V_\mathrm{tot}}.$$
Now, we add $\pu{1 ml}$ of $\pu{2.0 M}$ acetone. This is equivalent to saying we add $1/1000$ of $\pu{2 mol}$ of acetone. Therefore we add $\pu{2 mmol}$ (millimoles) of acetone.
This means that we have $\pu{2 mmol}$ of acetone in $\pu{4 mL}$ of water. Using the formula for molarity we get that acetone is in $\pu{0.5 M}$ in solution.
Hope you can follow on for the other parts as well, the concept remains the same.


What is the significance of Sphingolipids in human?

There are multiple known functions of sphingolipids and their metabolites in humans, although the roll played in the structure of myelin might be the most important and well understood.

However, according to this Wikipedia article, sphingolipids also play a roll in cell recognition and signalling, among other specialized functions. Additionally, some of it's metabolites have been shown to perform numerous functions. From the above article:

  Recently, simple sphingolipid metabolites, such as ceramide and sphingosine-1-phosphate, have been shown to be important mediators in the signaling cascades involved in apoptosis, proliferation, stress responses, necrosis, inflammation, autophagy, senescence, and differentiation.

The bottom line answer to your question is that there are many other functions of sphingolipids in humans, particularly if you include it's metabolites.  The article cited above is well referenced and should serve as a good starting point to investigate (or just enumerate) the other functions of sphingolipids in humans.


Finding molarity of hyposulfite solution used in titration

Your method in using the equivalent concept is correct, but the miscorrect in your solution is in determining the n-factor.

Refer to the above equations and the calculation N-factorPoutnik
N-factor Is the change of atom oxidation state multiplied by the number of atoms changing this state.
Because the iodine is just an intermediate product to be titrated by $\ce{Na2S2O3}$, the oxidation number change is by  $6$.Therefore, n-factor of  $\ce{KIO3}$ is $6$.So, the equivalent mass is M/6.And , n-factor of  $\ce{Na2S2O3}$ is $1$.So, the equivalent mass of $\ce{Na2S2O3}$ is equal to its molar mass.
$$
\begin{array}{cc}
\ce{&KIO3 &+ 5KI + 3H2SO4 -> &3I2 +&3K2SO4 + 3H2O}\\
&\frac{0.6}{214}~\mathrm{meq} & &\frac{0.6}{214}~\mathrm{meq}
\end{array}
$$
$$
\begin{array}{cc}
\ce{&I2 &+ &2 Na2S2O3 &-> 2 NaI + Na2S4O6}\\
&\frac{0.6}{214}~\mathrm{meq} & &\frac{0.6}{214}~\mathrm{meq}
\end{array}
$$
$$∴\frac{0.6}{214} = 44\times10^{-3}\times x\times 1 \to n\text{-factor of hypo}$$
$$∴x \approx{\pu{0.064 M}}$$


In what way could benzoin give Tollen's test?

Of your three "thoughts", you are correct that an aldol reaction is not an option. Not only is the product "crowded", but the reaction is reversible. [BTW, an aldol condensation occurs when water is eliminated from the initial aldol product. In the case of a benzoin aldol product, elimination of water is impossible.]

Your second premise is a good one. Benzoin (1) in the presence of alkaline Tollens' reagent, can be enolized (blue arrows) and the hydroxyl group deprotonated (red arrows). The enediolate 2 is a prime candidate for oxidation by Ag+. This species is the dianion of the enediol 6 tautomer of benzoin. The enediolate 2 is formed in the acyloin condensation of ethyl benzoate (7).
A one-electron oxidation of the enediolate 2 produces resonance stabilized radical anion 3. Protonation of 3 may occur at this point but a second one-electron oxidation produces the diradical 4, which is benzil (5). Under more vigorous alkaline conditions, Liebig's benzylic acid rearrangement of benzil occurs to afford benzylic acid (8), which itself is known to oxidize to benzophenone (9) under a variety of oxidative decarboxylation conditions. Neither of these reactions, i.e., the formation of 8 and 9, is expected to occur under the mild conditions of the Tollens' oxidation. Accordingly, your third idea is unlikely under the reaction conditions.


Which is more amphoteric: zinc or copper hydroxide?

Zinc hydroxide is more amphoteric as quantitatively investigated in Effect of pH, Concentration and Temperature on Copper and
Zinc Hydroxide Formation/Precipitation in Solution

In summary, in the case of Cu(OH)2, only at extremely low concentrations and very high pH can a third OH- be coordinated.

  At higher pH values, copper
  hydroxide, Cu(OH)2, is the dominant species up to pH 12.3 where the copper ion
  Cu(OH)3
  - forms according to Equation 3. At higher copper concentrations, solid
  Cu(OH)2 is formed and precipitates out of solution at copper concentrations above the
  solubility product of copper hydroxide at 1×10-8 M. It is important to note that the
  domain of stability of solid Cu(OH)2 is expanding to lower and higher pH values with
  increasing copper concentration.

On the other hand:

  Above pH 11.4, Zn(OH)3
  - forms according to Equation 7. At
  higher zinc concentrations, solid Zn(OH)2 forms and precipitates out of solution at zinc
  concentrations above the solubility product of zinc hydroxide of 1×10-5 M.

So qualitatively both display similar amphoteric behavior in the limit of infinite dilution, but at a appreciably non-zero concentration zinc is more amphoteric.

The article says it is getting the equilibrium constant values from Mineral Equilibria, Low Temperature and Pressure.

An alternative source of this information is Hydrolysis of cations. Formation constants and standard free energies of formation of hydroxy complexes Inorg. Chem., 1983, 22 (16), pp 2297–2305, which gives quantitative values for OP equations 1, 2 and 3 as well as the corresponding equations for zinc, in terms of Gibbs energy of formation and in terms of cumulative equilibrium constants.

Simplifying the general equation in the paper to only mononuclear (one metal atom) complexes:

$$K_y$$
$$\ce{M^{n+} + yH2O <=> M(OH)_y^{n-y} + yH+}$$

$\ce{Cu^{2+}}$ :
$pK_1 = 7.96$
$pK_2 = 16.26$
$pK_3 = 26.7$
$pK_4 = 39.6$

$\ce{Zn^{2+}}$ :
$pK_1 = 8.96$
$pK_2 = 16.9$
$pK_3 = 28.4$
$pK_4 = 41.2$

(These are all experimental values, calculated values are also given)

So really, without considering the solid phase (solubility), Cu2+ and Zn2+ seem very similar, with Cu2+ showing slightly more acidity.

Another study, which does consider the solid phases, is Zinc Hydroxide: Solubility Product and Hydroxy-complex Stability
Constants from 12.5-75 [degrees] C Can. J. Chem. 53, 3841.

This study approaches the problem with 5 equilibria involving the solid phase. (These authors use the symbol "c" for the solid phase.)

$\ce{Zn(OH)2(c) <=> Zn(OH)+ + OH-}$
$K_1 = [\ce{Zn(OH)+}][\ce{OH-}] = 2.54 \times 10^{-11}$

$\ce{Zn(OH)2(c) <=> Zn(OH)2(aq)}$
$K_2 = [\ce{Zn(OH)2}] = 2.62 \times 10^{-6}$

$\ce{Zn(OH)2(c) +OH- <=> Zn(OH)3-}$
$K_3 = [\ce{Zn(OH)3-}]/[\ce{OH-}] = 1.32 \times 10^{-3}$

$\ce{Zn(OH)2(c) +2OH- <=> Zn(OH)4^{2-}}$
$K_4 = [\ce{Zn(OH)4^{2-}}]/[\ce{OH-}]^2 = 6.47 \times 10^{-2}$

$\ce{Zn(OH)2(c) <=> Zn^{2+}} + \ce{2OH-}$
$K_{sp} = [\ce{Zn^{2+}}][\ce{OH-}]^2 = 1.74 \times 10^{-17}$

This study (see Fig. 1) shows that aqueous $\ce{Zn(OH)2}$ is only the major aqueous species in the pH range 9-11, having acted as an acid or base outside this range.

Instead, if I were just given the information in the OP, meaning the two supposed Ka values, as some kind of test or homework question,  I would say copper is more amphoteric.  My reasoning would be, for zinc, just to remove one proton, you need to get up to almost pH 10.  For two more protons to be removed, as required to get to Zn(OH)2 and lose yet another proton to act as an acid, seems impossible because I would expect pKas to be spaced apart and if you would need to approach or go beyond pH 14 to observe the acidic behavior, that isn't reasonably considered amphoteric.

Note that EXAFS Investigations of Zn(II) in Concentrated Aqueous Hydroxide Solutions J. Phys. Chem. 1995, 99, 11967-11973 finds that rather than being 6-coordinate as indicated in the OP, 4 hydroxides coordinate Zn2+ tetrahedrally with no water ligands.


Ghiorso, Sikkeland, Larsh, and Latimer of the Lawrence Berkeley Laboratory reported the discovery of element 103 in 1961. It is correct that the discoverers suggested the name lawrencium, symbol Lw, for the new element in honor of E. O. Lawrence, inventor of the cyclotron and founder of the Lawrence Berkeley Laboratory. The original proposal can be found in Ghiorso, A.; Sikkeland, T., Larsh, A. E.; Latimer, R. M. New Element, Lawrencium, Atomic Number 103. Phys. Rev. Lett. 1961, 6 (9), 473–475:

  In honor of the late Ernest O. Lawrence, we respectfully suggest that the new element be named lawrencium with the symbol Lw.

At the XXII Conference of the IUPAC in London 1963, the name was accepted but the symbol was changed to Lr. According to Comptes rendus / XXII Conference, London 5th to 9th July, 1963, the reason was simply the difficulty with the letter “w”:

  Following the observation by H. Remy that “w” is an uncommon letter in many languages and difficult to pronounce it was proposed to alter the spelling to Laurentium. The Chairman emphasized that the Commission had no right to modify the spelling of the name, but that it could be pronounced according to the custom of the language. To make the name more acceptable it was decided to recommend the chemical symbol Lr and not Lw for element 103.


What noble metal is most oxidation resistant?

Ruthenium, Rhodium, Palladium, Osmium, Iridium, Platinum, Gold

  ... I will narrow this question down to oxidation by diatomic oxygen in air at room and elevated temperatures.

According to this criterion, gold is the most oxidation resistant of the above metals.

First of all, gold oxide cannot be obtained by reaction of gold and oxygen ($\ce{O2}$) at standard pressure. Gold oxide ($\ce{Au2O3}$) decomposes to gold and oxygen above $\pu{160°C}$.

Ruthenium reacts with oxygen (air) above $\pu{700°C}$ to give $\ce{RuO4}$. Ruthenium powder burns in air.
Rhodium reacts with oxygen (air) above $\pu{700°C}$ to give $\ce{Rh2O3}$.
Palladium powder burns in air.
Osmium powder reacts with oxygen (air) at room temperature to give $\ce{OsO4}$.
Iridium reacts with oxygen at high temperatures to give $\ce{IrO2}$ or $\ce{IrO3}$. Iridium powder burns in air.
Platinum powder burns in air.


Accepted book enthalpies are in the units $\mathrm{kJ/mol}$.

Simply convert $\mathrm J$ to $\mathrm{kJ}$ by dividing by $1000$. Since $1\ \mathrm{kJ} = 1000~\mathrm{J}$.

The moles of dilute aquaous solutions is best determined by volume (L) times concentration (mol/L).$$n = V \times c$$ You would use the moles of the product. If there is, in fact, "$0.2\ \mathrm{mol}$ of the reactants present" then there would be $0.2\ \mathrm{mol}$ of the products present. since the chemical reactions are $\ce{1 mol + 1 mol -> 1 mol + 1 mol}$.  (You are not interested in the number of atoms per mole, so you would not use Avogadro's number.)

$$q= C_\mathrm{sol} m\Delta T$$ where $C_\mathrm{sol}= 4.06~\mathrm{J/(g \ ^\circ C)}$ (Specific heat capacity for dilute aqueous solutions)

$$q= C_\mathrm{sol} m\Delta T / n_\mathrm{product}$$  ($n = \text{moles product}$)

Remember that Hess's Law states that

  if a reaction can be expressed as the sum of two or more reactions, then the heat flow for the overall reaction is equal to the sum of the heat flows.

Hence, the sum of the first 2 reactions should be equal to the 3rd reaction.

$\displaystyle \ce{NaOH + HCl -> NaCl + H2O}$
$\displaystyle \ce{NH4OH + NaCl -> NaOH + NH4Cl}$  (Reversed and Enthalpy sign changed)
$\displaystyle\ce{HCl + NH4OH -> NH4Cl + H2O}$

Note how $\ce{NaOH}$ and $\ce{NaCl}$ cross out leaving the third reaction.


How is the position of a peak on an IR spectrum determined?

Generally, the width of IR absorption peaks depends on the environment of the target molecule. Intermolecular forces alter vibrational modes.

If intermolecular attractions are fairly strong (like the hydrogen bonding present in liquid ethanol), the number and strength of interactions can vary greatly from molecule to molecule, causing the vibrational modes observed in an IR absorption spectrum to widen.

As for the reported width of the peak, the convention is to report the width at 1/2 the height of the peak.


The ratio of Cu to EDTA is the number of copper ions per EDTA ion. In this case there is one EDTA ion for every copper ion so the ratio is 1:1.

Copper bonds with six electron pairs on EDTA to form the complex shown below where the central metal species is copper(II):


Is there a theory behind selecting elements that may be successful in potential superconductors?

Suffice to say, nobody really knows, and a Nobel prize surely awaits the first people who can explain the relationship between chemical composition and high-temperature superconductivity.


I do not feel that the word catalyst applies to photochemistry.

Photochemistry allows to access much higher energy states and thus reach reaction pathways that are otherwise not possible. Photons do not stabilize transition states (how would you rationalize the interaction?).

Catalysts on the other hand are focused on lowering barriers so that reactions proceed via lower energy transition states.

Because catalysts only affect the transition states, catalysts cannot affect the equilibrium position of a reaction. For example, if you had a catalyst that allowed easier conversion between (E)- and (Z)- double bonds, you could not use the catalyst to selectively provide the one that is less favorable.

On the other hand, with photochemistry, we do not establish an equilibrium because the higher energy state is not necessarily accessible via microscopic reversibility. And indeed, we can selectively use photochemistry to provide the less favorable species as is well know in the case of stilbenes.

EDIT:

Thanks to airhuff for pointing out that the requirement that a catalyst not be consumed also implies that we should not think about photons as catalysts. Hard to imagine a situation where you would absorb a photon and release one that is the same energy...


How to redefine internal pressure in terms of κ and α?

$$dV=\left(\frac{\partial V}{\partial T}\right)_pdT+\left(\frac{\partial V}{\partial p}\right)_Tdp$$So at constant volume dV = 0 and $$\left(\frac{\partial p}{\partial T}\right)_V=-\frac{\left(\frac{\partial V}{\partial T}\right)_p}{\left(\frac{\partial V}{\partial p}\right)_T}$$

The rest is straightforward.


What will be the product of this reaction?

This reaction is one of a special kind. The one and onlyend product is 3-hydroxy,3- methylcyclohexan-1-one. As the Grignard reagent is in excess, and it is an $\alpha$- $\beta$ unsaturated ketone, there is a possibility of undergoing two different type of reaction mechanisms. But Surprisingly, both will give the same ultimate end product after two steps.   The major mechanism through which the reaction should undergo, is the direct nucleophilic attack of the grignard reagent at the electrophilic carbonyl carbon. As, Grignard reagent is a hard base , it's attack to hard acid centre of carbonyl carbon is more preffered. So, first it will undergo this addition after which it acidic hydrolysis is done, which cleaves the ether and forms an enol which tautomerises to give the final end product ketone. The reaction scheme is shown below,

Now, the minor mechanism is the process of conjugated addition of the methyl group to that $\alpha$-$\beta$ unsaturated ketone. But again after tautomerisation and  hydrolysis it actually forms the same compound, which is 3-hydroxy,3-methyl cyclohexan-1-one. The reaction scheme is shown below,

So, it is seen that however the presence of excess reagent may interfere with the reaction mechanism but the end product is $100 $$\%$ $3$-hydroxy,$3$-methylcyclohexan-$1$-one.


Volume required to dilute solution for a pH change

Since the amount of substance stays equal, you can calculate the volume of your final solution $V_\mathrm{end}$, form your start concentration $c_\mathrm{start}(\ce{OH-})$ and your initial volume $V_\mathrm{start}$ and your target concentration $c_\mathrm{end}(\ce{OH-})$. Your working equation should be:

\begin{align}
c_\mathrm{start}(\ce{OH-})\cdot V_\mathrm{start}
&= c_\mathrm{end}(\ce{OH-})\cdot V_\mathrm{end} \\
V_\mathrm{end} &= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start}
\end{align}

You can work out the start concentration for the initial pH and the target concentration from your final pH. Given (at $T=298.15~\mathrm{K}$, $p=1~\mathrm{atm}$):

\begin{align}
\ce{pH}_\mathrm{start} &= 13 \\
c_\mathrm{start}(\ce{OH-}) &= 10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.1~\mathrm{mol/L} \\
V_\mathrm{start} &= 0.1~\mathrm{L} \\
\ce{pH}_\mathrm{end} &=11 \\
c_\mathrm{end}(\ce{OH-}) &=10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.001~\mathrm{mol/L}
\end{align}

Hence

$$
V_\mathrm{end}= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start} = 10~\mathrm{L}
$$

and therefore

$$
\Delta V = V_\mathrm{end}-V_\mathrm{start} = 9.9~\mathrm{L}= 9900~\mathrm{mL}.
$$


Comparison of acidic strength between nitric acid and nitrous acid

The correct terminology is "cross conjugation" and "extended conjugation" (but the latter makes more sense in the context of more than two pi bonds, as in penta-1,3,5-triene).

Now, you have correctly identified that the nitrate anion has cross conjugation while the other has simple conjugation. However, you've incorrectly concluded one of them to be more stable. The conclusion is incorrect because there is a much dominant factor at play instead.

The three canonical resonating structures of the nitrate anion are "equivalent" i.e. wherever the negative charge may delocalise, the new resonance structure is exactly the same as the original one. Therefore, all the resonance structures are "equivalent" in their potential energy. This leads to a much rapid delocalisation of the lone pair, between the three oxygen atoms, leading to a rapid charge dispersal.

Both nitrate and nitrite anions have equivalent resonating structures. But, in the former, the negative charge is dispersed over three oxygen atoms (instead of only two in nitrite). From this, we conclude that the nitrate anion is much more stable than the nitrite anion.

Therefore, the next time you apply the fact that cross conjugation imparts lesser stability than extended conjugation, do look out for this additional dominant factor of "equivalent resonance"!


Why doesn't copper react with hydrochloric acid while the other metals do?

You might want to look up some terms, such as

noble and less noble metals
reduction potential
galvanic series

Here, reaction means that

hydrogen gas is formed
the metal is dissolved

In order to form hydrogen, protons need to be reduced to hydrogen atoms which then combine to $\ce{H2}$.

$$\ce{2 H+ + 2 e- -> H2}$$

The metal serves as an electron donor and is oxidized, e.g.

$$\ce{Zn -> Zn^{2+} + 2 e- }$$

The more noble a metal is, the more reluctant it is to lose electrons. This is the case for copper, which is therefore not oxidized under these conditions.


To choose a more basic example, let us consider the particle in a box with

\begin{equation}
V(x)={\begin{cases}0 & 0\lt x \lt L\\\infty & \text{otherwise}\end{cases}}
\end{equation}

To solve this we use the Schrödinger Equation

\begin{equation}
-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)=E\psi(x)
\end{equation}

subject to the boundary conditions
\begin{equation}
\psi(0)=\psi(L)=0.
\end{equation}

The origin of these boundary conditions is, that the wave function needs to be a) continuous and and b) zero outside of the potential well, since probability goes to zero for an infinite potential.

As an ansatz for the wave function we take the solution of the free particle, i.e. planes waves:
\begin{equation}
\psi(x) = \exp\left(-ikx\right)
\end{equation}
with
\begin{equation}
k^2=\frac{2mE}{\hbar^2}.
\end{equation}
What we now have is a periodic wave function, with period $L$ (or an integer multiple of that) due to the boundary condition, hence
\begin{equation}
kL = 2\pi n.
\end{equation}
At this point we have introduce the integer quantum number $n>0$. With this we can now write down wave function and the corresponding energy. Both depend on $n$ and are therefore quantized.

So the origin of the quantized nature here are the boundary conditions. The quantum number naturally show up in the solution of the Schrödinger Equation. Obviously this can happen for other system, such as the Hydrogen atom as well.

As a final side note: The Schrödinger equation does not always lead to quantized solutions, it depends on the system. There are for example continuous states when considering the dissociation of molecules.


Question about change in pressure and change in temperature for system at equilibrium

For a general idea, see Le Chatelier's principle

  When any system at equilibrium for a long period of time is subjected to change in concentration, temperature, volume, or pressure,
  (1) the system changes to a new equilibrium and
  (2) this change partly counteracts the applied change.

For the changes of $p, V$, read about equilibrium constant $K$ and Reaction quotient $Q$ and their relation.

Notice, how the Reaction quotient changes with increased pressure/decreased volume, if the reaction increases the molar amount of gases.
And how it changes if it decreases them.

For the former case, increasing pressure increases Q and the equilibrium shifts backwards to make Q  equal to K again.

For the latter case, increasing pressure decreases Q and the equilibrium shifts forwards to make Q equal to K again.

For the temperature effect, see van't Hoff equation:

  $$\frac{d}{dT} \ln K_\mathrm{eq} = \frac{\Delta H^\ominus}{RT^2}$$

Endothermic reaction direction has higher activation energy than the exothermic direction.

As the consequence,

the reaction rate of the endothermic reaction direction

raises with temperature faster

than the reaction rate of the exothermic reaction direction.

Faster endothermic then exothermic reaction means the equilibrium constant shifts at higher temperature in favour of the endothermic reaction.


Synthesizing 2-bromoaniline from bromobenzene

It will be much quicker to run the direct nitration of bromobenzene and isolate the ortho nitro product by chromatography. It is the minor nitration product, but your overall yield will be better by cutting out two steps which will give only moderate yields.
Nitration and chromatography procedure here 1


Can we prepare aryl halides by heating with halogen acids?

The partial double bond character of the C-O bond in phenol becomes obvious from the resonance structures. Contributions from the structures with $\ce{C=O}$ double bond explain why the character of the C-O bond is in between single and double bond, and also show that oxygen lone pairs are less available for protonation to form $\ce{Ph-OH2+}$.

Protonated phenols are therefore fairly acidic, with a $pK_a$ around -6. $\ce{HBr}$ and $\ce{HI}$ have significantly lower $pK_a$ values (-9 and -10, source), so they would indeed be able to protonate phenol. However, $\ce{H2O}$ is unlikely to be eliminated from this compound due to the instability of the resulting phenyl cation. Its positive charge cannot be delocalized into the aromatic system as it resides in an empty $sp^2$ hybrid orbital which is perpendicular to the delocalized $\pi$ bonds (source). Formation of the phenyl cation by elimination therefore requires much better leaving groups like $\ce{N2}$, which can be eliminated from aryl diazonium cations.

Substitution of the protonated OH group by halides with an $S_N2$-like mechanism is hindered by Coulomb repulsion between the $\pi$ electron cloud of the benzene ring and the lone pairs of the nucleophile. This reaction only works with aromatic compounds in which the electron density in the $\pi$ system is decreased by multiple electron-withdrawing substituents, like nitro groups.


Relativistic effects in element 137 (Feinmanium) and above

Yes, the value 137 occurs even considering Einstein's theory of special relativity.

137 comes from considering Sommerfeld or Dirac relativistic theories, when the nucleus is modeled as a point.

See equation 1 of A new method for solving the Z > 137 problem and for determination of energy levels of hydrogen-like atoms


Loss of aromaticity vs low entropy of activation

There are several factors we can point to before resorting to entropy of activation.

1)  At the end of the "chain" is displacement of an iodide ion.  Iodine, with its diffuse valence orbitals, forms rather weak bonds with carbon making this displacement thermodynamically and kinetically favorable.  Presumably the nucleophile that re-opens the cyclopropane ring forms a stronger bond with the carbon.

2)  We may no longer have an aromatic ring, but a carbonyl group flanked by two conjugating carbon-carbon bonds still has plenty of resonance stabilization.  We certainly do not lose all that much stabilization.  Moreover...

3)  Upon further review, there is still an aromatic contribution after all.  The cyclopropane ring is well known to stabilize carbocations by hyperconjugation, and with the carbonyl oxygen tending to withdraw electrons we may argue that the quinone-like ring has some carbocation character.  Adding a pair of pi electrons to this ring through hyperconjugation will then give aromatic contributing structures in which the cyclopropane ring is positively charged and the oxygen regains its negative charge.

All told, we don't have to reach for "entropy of activation".  The reaction is better off than it looks just in terms of energy.


Does battery self discharge when connected?

Great question! I studied battery chemistry as part of my graduate work, and I think the answer depends on the mechanism of self-discharge.

I think that by discharging the battery through an external circuit, you will generally cut off the self-discharge mechanisms (depending on the discharge rate), so if you add the discharge rate and the self-discharge rate, you will probably underestimate the battery life. You should also keep in mind that the rate of self-discharge depends on the state-of-charge for the battery, and is generally worse at high state of charge. Thus, even the manufacturer specification for self-discharge rate isn't a constant.


Grignard reagent will act as a base or a nucleophile

As most of us chemists have understood and experienced, the Grignard reagent have undergone the many reactions with suitable substrates. Yet, the most familiar and probably the most used reaction is the addition to a carbonyl group to give $2^\circ$- or $3^\circ$-alcohol. The Grignard reagents' affinity for a corbonyl group in the presence of an acidic $\alpha$-$\ce{H}$ has been documented here. However, if at least $\ce{OH}$ group is present in the molecule, acid base reaction would be take over. For example, the usual Grignard reaction would not happen if your glassware is wet or not properly dried.

Albeit they were rarely published, it is well known fact that some side reactions with available acidic $\alpha$-$\ce{H}$ and some reduction reactions have been taken place during original Grignard reactions. These side reactions are frequently complicated the intended reactions, which may replace some or all of the intended addition reaction, reducing expected percent yield (Ref.1). Most importantly, the conversion of an alkyl halide to a deuterium labelled hydrocarbon has been achieved via Grignard reagent preparation (Figure A):

Keep in mind that these labelling are selective based on the alkyl halide. Recently, there was a method developed to selective labeling by using organozinc reagents (Ref.2). Although these are not conventional Grignard reagents, organozincs are close relatives (Figure B). Another example for Grignard reagents acting as a base is a method to prepare terminal alkynyl-Grignard reagents (Figure C).

Some times during Grignard preparation, you might experience no Grignard was evidently formed because you have loosen your alkyl halide reagent while substrate ketone was recovered unchanged. That is because, if your ketone is bulky like diisopropyl ketone (2,4-dimethylpent-3-one), following reaction may happen (look under an example for Grignard acting as a base):

If your Grignard reagent have an active $\alpha$-$\ce{H}$ to give, it can also acts as an reducing agent as in above case of Grignard acting as a reducing agent. This is the case of reaction between benzophenone and specific Grignard reagent (isobutyl magnesium bromde or 3-methylbutylmagnesium bromide) illustrated in Ref.1. The main products from that reaction were dibenzylcarbinol and isobutene.

References:

G. E. Dunn, John Warkentin, "An Isotopic Study of the Reducing Action of the Grignard Reagent," Canadian Journal of Chemistry 1956, 34(1), 13841–13857 (https://doi.org/10.1139/v56-007).
Aiyou Xia, Xin Xie, Xiaoping Hu, Wei Xu, Yuanhong Liu, "Dehalogenative Deuteration of Unactivated Alkyl Halides Using $\ce{D2O}$ as the Deuterium Source," J. Org. Chem. 2019, 84(21), 13841–13857 (https://doi.org/10.1021/acs.joc.9b02026).


Explain the selectivity in bromination of this compound with two substituted benzene rings

Because that is not the most electron rich aromatic. The ring at other end of the molecule is, because it is an aniline. The first, and probably the second, bromination will go there. If you keep throwing in Bromine you will eventually get bromination of the Acetophenone ring.


The free energies of formation of the elements in their standard states is zero, by definition.  You can look up the (variously reported) standard enthalpy of formation of the product, −204.6 kJ/mol (exothermic!).  It is roughly the binding energy of the crystal lattice less the ionization energies of the inputs.  So, by the numbers, do you obtain spalerite or wurtzite?  ZnS has crystal polymorphs and polytopes with differently populated unit cells

$\ce{Zn_2S_2}$ Wurtzite-2H
$\ce{Zn_4S_4}$ Wurtzite-4H and Sphalerite
$\ce{Zn_6S_6}$ Wurtzite-6H
$\ce{Zn_15S_15}$ Wurtzite-15R


How to determine number of structural isomers?

It isn't easy but it is an interesting research topic

Determining the number of possible structures for a given range of chemical formulae isn't simple even for saturated hydrocarbons. The number of possible structural isomers rises rapidly with the number of carbons and soon exceeds your ability to enumerate or identify the options by hand. Wikipedia, for example, lists the numbers of isomers and stereoisomers for molecules with up to 120 carbons. But the counts are getting silly even at 10 carbons where there are 75 isomers and 136 stereoisomers.

It has been an interesting research topic in computational chemistry and mathematics. This old paper (pdf), for example, list some formulae for simple hydrocarbons among other simple series. Part of the interest arises because of the relationship to the mathematics of graph theory (it seems that chemistry has inspired some new ideas in this field of mathematics partially because enumerating possible isomers of hydrocarbons is strongly related to drawing certain simple trees which is intuitively obvious if you use the standard chemical convention of drawing just the carbon backbone and ignoring hydrogens).

You can look up the answers on the fascinating mathematics site OEIS (the online encyclopaedia of integer sequences). The sequence for simple hydrocarbons is here.

But the mathematical approach oversimplifies things from the point of view of real-world chemistry. Mathematical trees are idealised abstract objects that ignore real-world chemical constraints like the fact that atoms take up space in three dimensions. This means that some structures that can be drawn cannot exist in the real world because the atoms are too crowded and cannot physically exist without enough strain to cause them to fall apart.

Luckily, computation chemists have also studied this. There is, unfortunately, no obvious shortcut other than trying to create models of the possible structures and testing them to see if they are too strained to exist. The first two isomers that are too crowded are for 16 and 17 carbons and have these structures:

If you have any intuition of the space filling view of these, you should be able to see why they are problematic. A research group at Cambridge University has produced an applet to enumerate the physically possible isomers for a given number of carbons which is available here if your Java settings allow it. The results are discussed in a paper available in the Journal of Chemical Information and Modelling.


According to this abstract$\mathrm{^{[1]}}$, both aerobic and anaerobic processes play a role in the natural degradation of PCB's.

  Studies have identified two distinct biological processes capable of
  biotransforming polychlorinated biphenyls (PCBs): aerobic oxidative
  processes and anaerobic reductive processes. It is now known that
  these two complementary activities are occurring naturally in the
  environment. Anaerobic PCB dechlorination, responsible for the
  conversion of highly chlorinated PCBs to lightly chlorinated
  ortho-enriched congeners, has been documented extensively in the
  Hudson River and has been observed at many other sites throughout the
  world. The products from this anaerobic process are readily degradable
  by a wide range of aerobic bacteria, and it has now been shown that
  this process is occurring in surficial sediments in the Hudson River.
  The widespread anaerobic dechlorination of PCBs that has been observed
  in many river and marine sediments results in reduction of both the
  potential risk from and potential exposure to PCBs. The reductions in
  potential risk include reduced dioxinlike toxicity and reduced
  carcinogenicity. The reduced PCB exposure realized upon dechlorination
  is manifested by reduced bioaccumulation in the food chain and by the
  increased anaerobic degradability of these products.

1) Aerobic and anaerobic PCB biodegradation in the environment, Environ Health Perspect. 1995 Jun; 103(Suppl 5): 97–99


Why does the corrosion of a steel nail develop iron ions in the ends of the nail and hydroxide in the middle of the nail?

Very good question and thought provoking too. It appears that the nail somehow knows or is it is able to decide which part will be a cathode and which part will be an anode! Why is that so? Why one end was not a cathode and the other end an anode, just the one shown in the cartoon taken from here Imaging metal corrosion

So the fact you should keep in mind is that corrosion is superficially a random process just like cancer. Corrosion comes from corrodere "to gnaw to bits, wear away," it can start from any defect on the metal (scratch, impurity, etc.). However once it starts, just like cancer, it decides its propagation (i.e., the flow of electrons), hence the direction of cathode or an anode. It is likely that in your case the nail tip and the head had more defects than the center, so corrosion started at the tips. The pattern you are seeing is not universal or a fundamentally symmetric phenomenon.

Let us now bend a nail, create stress or defect in the center. Now the corrosion also starts in the center. Have a look at this gallery. Corrosion gallery


On the addition of HCl to a conjugated diene

Benzylic carbocation is formed as it is stable due to resonance.
So Cl attaches to benzylic position.
More substituted alkenes are more stable so the terminal double bond changes its position.
(Also this isn’t free radical reaction)


Determining rate constants for parallel reactions from experimental data

One way you could get $k_1$ and $k_2$ is to first find their sum using your equation for $[A](t)$ at a known $t$.

Once you have done that, you can solve for $k_1$ using your equation for $[B](t)$ at a known $t$, since you now have the denominator of your first factor. You could alternatively find $k_2$ using $[C](t)$.

Once you have $k_1$, then $k_2$ is easily found from the sum.


Why can toluene be chlorinated at either the benzylic position or the ring depending on the conditions?

The halogenation of toluene is possible by two mechanisms. In German, they are referred to by the shorthands KKK and SSS, where KKK stands for Katalysator, Kälte, Kern (catalyst, cold and core) while SSS is Sonne, Siedehitze, Seitenkette (sun, sizzling heat and side chain).

At low temperatures and in the presence of an activating catalyst such as $\ce{FeCl3}$, halogenation of toluene will typically occur as phenyl ring halogenation following the electrophilic aromatic substitution mechanism.

Scheme 1: Catalytic chlorination of toluene with $\ce{Fe^3+}$ as a catalyst. ortho- and para-products expected; (para not shown) mesomeric structures of the intermediate not shown.

Und higher temperatures and irradiation, halogenation takes place according to a radical chain mechanism.

Scheme 2: Radical chlorination of toluene under irradiation. Multiple chlorination possible. Mesomeric structures of the benzyl radical not shown.

The electrophilic aromatic substitution is, of course, not as easily possible on the benzylic carbon; we would somehow need a free electron pair to appear there — not likely.

On the other hand, if a chlorine radical attempted to remove a hydrogen of the phenyl ring, that would result in a radical that cannot be delocalised (the $\ce{C-H}$ bond is orthogonal to the π-system) which is therefore very unstable. Another way of looking at it is to consider the $\ce{C–H}$ bond strengths: weaker bonds are more easily cleaved homolytically, as is the case in free-radical halogenation. Aryl $\ce{C–H}$ bonds are roughly $\pu{100 kJ/mol}$ stronger than benzylic $\ce{C–H}$ bonds (Wikipedia), so radical abstraction preferentially occurs at the benzylic position.


Are compounds which are used to seed clouds like silver(I) iodide hygroscopic?

Silver iodide is not hygroscopic. It is used to seed clouds, because when the atmosphere is saturated with water vapor, liquid water should start condensing. But to do this, it needs a seed, a place when the first molecules will start to condense. No water molecule feels that it should be the first one to show other molecules where to be condensed. But, as silver iodide has the same hexagonal shape and same dimensions as snow, it may make believe it is an ice crystal. And the neighboring water molecules will be mistaken, follow the example and deposit on it. The common salt cannot do this. It does not look like ice.


How does polydimethylsiloxane stick to glass?

If you ask a chemist about PDMS and its adhesion to glass, she will immediately think of the work done in constructing micro-reactors.  This is because one method of manufacturing micro-reactors is to stick a patterned piece of PDMS to a piece of glass; the PDMS is patterned such that it has reactor channels that are sealed when it is stuck to the glass.  The glass is usually a micro-scope slide or something similar.

With the above in mind, in order to answer your question I began to search through the literature which describes the preparation of micro-reactors from glass and PDMS.  What they were saying was that in order to form a bond between the molecules within the glass and the molecules within the polymer (PDMS) the glass had to be pre-treated,

  "most of the research papers mention the
  use of oxygen plasma for developing chemical (siloxane) bonds be-
  tween the participating surfaces"

from "Studies on Surface Wettability of Poly(Dimethyl)Siloxane (PDMS) and Glass Under Oxygen-Plasma Treatment and Correlation With Bond Strength", DOI: 10.1109/JMEMS.2005.844746.

Since the manner in which your material is applied does not have any sophisticated pre-treatment then the only alternative method of adhesion (as pointed out earlier in this thread) is Van der Waals forces.

If this isn't quite the answer you expected, at least it is a good start.


How does % purity on household products work?

For example, toilet bowl cleaner is 9.5% HCl.

First off, do you see "w/v" mentioned after those figures, i.e- $\ce{HCl}$ 9.5% (w/v)?

  Does this mean that for every 90.5 moles of H2O (or other stuff) there is 9.5 moles of HCl? Does it mean that for every 90.5 g of H2O there are 9.5 g of HCl?

Even if you don't, I'd still expect the concentration to be expressed in the form: $x$ grams of $\ce{HCl}$ for every $y$ units of water. As for whether water's mass or volume's considered, it would depend on the regulations for such products (which I'm unfamiliar with).

However, I wouldn't be surprised if it did turn out to be a w/v quantity (which is quite common in pharmaceutics).

  Kinda weird that we don't know if this is weight or molarity.

Like I said, I'm going to throw my weight behind "weight", since it's far more common than molarity outside of a lab (come to think of it, I've never seen a household product's ingredient concentration expressed in molars).

Not exactly important, but since you mentioned "molarity", here's a handy formula to convert something in % w/v to molars:

$$\text{Molarity} = \frac{{10}{(w/v)}{(d)}}{M}$$

Where $\ce{d}$ is the density of the solution, $\ce{M}$ is the molar mass of the solute and $\ce{w/v}$ is the mass concentration of the solute.


How to convert benzamide to benzylamine?

The more reactive $\ce{LiAlH4}$ is required for the reduction of the amide to the corresponding amine (source). The carbonyl carbon of an amide is less electrophilic compared to a ketone, because electron density is donated by the amide nitrogen via resonance.

Bromine and hydroxide react with an carboxylic acid amide to yield an amine with one fewer carbon atom (Hofmann degradation). $\ce{Br2}$ and $\ce{KOH}$ react to form $\ce{KOBr}$ in situ, which transforms the amide into an intermediate isocyanate ($\ce{R-N=C=O}$). Hydrolysis of the isocyanate yields a carbamic acid $\ce{R-NH-COOH}$, which decomposes to the amine $\ce{R-NH2}$ and $\ce{CO2}$. As described in this paper, the reaction of $\ce{PCl5}$ with carboxylic acid amides produces amidochlorides $\ce{RCCl2NH2}$ as intermediates, which break down to the corresponding nitrile $\ce{RC#N}$ and $\ce{HCl}$.


How is bonding in the trihydrogen cation and triatomic hydrogen possible?

Here are the vital parts from the links about $\ce{H3}$ in the question:

  The molecule can only exist in an excited state [...] the electronic state for a trihydrogen cation with an electron delocalized around it is a Rydberg state.

and from second-level link

  Rydberg states of the neutral triatomic hydrogen molecule can be viewed as
  an electron attached to a tightly bound $\ce{H3+}$ core

Basically, at least some states can be viewed as an $\ce{H3+}$ ion working as a nucleus in a bigger hydrogen-like atom. If said 'nucleus' catches the electron, the molecule immediately dissociates. So, let's focus for a bit on $\ce{H3+}$. In this ion a two-electron three-center bond is formed. Such bonding is actually pretty common, even if (almost never) mentioned in school. As long as orbitals of several atoms can overlap effectively and have fitting symmetry, they can form a single highly symmetrical bonding orbital, that can accept two electrons. In this case the bonding order between atoms is fractional, so in $D_\mathrm{3h}$ (triangular) $\ce{H3+}$, each of the three $\ce{H-H}$ bonds has bond order $1/3$. This kind of bonding plays a critical role in the chemistry of boranes (and Group 3 elements in general, maybe except thallium) and some families of compounds of d-block elements.

To obtain more intimate and correct understanding you should invest in molecular orbital theory. Your link for $\ce{H3+}$ contains a comprehensive description of bonding in terms of said theory.


Copper chloride color in electrolysis of salty water

The color of the solution is probably due to the presence of blue copper(II) ions ($\ce{Cu^2+}$) and greenish tetrachlorocuprate(II) ions ($\ce{[CuCl4]^2-})$.

The change in color comes from the equilibrium

$$\ce{Cu^2+ + 4 Cl- <=> [CuCl4]^2-}$$

Low concentrations of chloride favor the formation of the blue copper ions. High concentrations of chloride favor the formation of the greenish tetrachlorocuprate ions.

If a significant amount of chlorine had been created you should have observed bubbles at the anode and the typical chlorine smell.


Why can primary cells not be recharged effectively?

Galvanic/electrolyic refers to how the cell is being run. If you have an electrochemical cell, and you connect a load to it, and it produces a spontaneous current, it's a galvanic cell. If instead you connect a voltage source, and the current from the voltage source induces a reaction, it's an electrolytic cell. There's no requirement that a galvanic cell be made with two beakers and a salt bridge, that's just a very common setup for introductory chemistry experiments using the Daniell cell.

The defining characteristic of a secondary battery is that the net "electrolytic" reaction is the reverse of the net "galvanic" reaction. So a secondary battery is a galvanic cell when it's discharging and an electrolytic cell when it's charging. In a primary battery, whatever reaction happens during charge doesn't return the battery to its initial state. A common reason for this is that the galvanic reaction produces insoluble products, so the concentration is really low by the electrode. Then when you reverse the current and run it as an electrolytic cell, something else reacts instead, like water electrolyzing.


Naming the spin system present in ethyl butanoate

By Pople Nomenclature convention, coupled nuclei for which the chemical shift difference is much greater than the coupling between them (say, greater than 3-5 times) are labelled with letters alphabetically well separated (AX, AM etc). Labelling nuclei with adjacent letters of the alphabet (AB, MN, XY etc) indicates that the chemical shifts are close compared to their coupling, and this means they are typically strongly coupled ie cannot easily be analysed using first order techniques.
As for so many NMR related questions on this site, I refer you to Hans Reich's valuable online resource for further reading.

So, what to call your system. Clearly, this is an example for teaching purposes only. You have correctly identified that you have 5 non-equivalent sites with relative populations 3:2:2:2:3. If you cannot work out, or do not know what the expected approximate chemical shifts would be for each of these guys I can tell you they are well enough separated not to worry about second order coupling.

So now you need to do is determine whether you have any magnetic inequivalence in your system. Again, have a read of this useful page to understand how to test for this, but I think you are fairly safe with this case that there are no nasty surprises.

All that is left is to assign your spin labels to your system. So for the propyl group, none of the chemical shifts are close (compared to coupling) so your labels should be well separated; so not ABC, but something like AJM. Likewise for the ethyl chain; something like RX. I'll leave you to work out how to put the two together.


In Graham's law, is the rate of effusion only dependent on molar mass?

Molar mass is just one of the factors affecting effusion. Graham's law states that everything else remaining same, the rate of effusion of a gas is inversely proportional to $\sqrt M$. In other words, Graham's law describes the relation of the rate of effusion with only one of the factors involved in determining the overall rate.

For the general rate of effusion of a gas at pressure $P$, and the hole area $A$ at a temperature $T$, using kinetic molecular theory, it turns out to be: (derivation on Physics.SE)

$$r=\frac{PA}{\sqrt{2\pi MRT}}$$

Now, in your question, the balloon with two moles will have a greater pressure (more moles in the same volume) and therefore it will have a greater rate of effusion for the same hole area and temperature and molecular mass.