How to refer to this method of mixing solids?

I think the closest term would be recrystalisation though this is normally used to dissolve one component (the desired one) to allow the removal of impurities, so I would be hesitant to use it but it is used here to describe the rocket candy example you used: http://www.jamesyawn.net/rcandy/index.htm

With regards to your second example, baking a cake is actually a series of interesting and complex chemical reactions and structural changes. The New Zealand institute of chemistry wrote a nice summary of the chemistry of baking that can be found here: http://nzic.org.nz/ChemProcesses/food/6D.pdf

Hope that helps


Michael reaction or nucleophilic attack at the alpha carbon of an alpha-beta-unsaturated ketone?

An $\alpha,\beta$-unsaturated ketone is electrodeficient at the $\beta$ position. This can be seen if you draw the resonance structures of such a molecule.

The $\beta$ carbon is thus a good site for nucleophilic attack. But, as you know, carbonyls are also prone to nucleophilic attack. To discriminate between the two, you need to look at how the reaction is controlled, either thermodynamically or kinetically.

In a kinetically controlled reaction, the product that is formed fastest predominates. In a thermodynamically controlled reaction, the predominant product is the energetically favored one.

A Michael addition is a 1-4 addition, where a nucleophile attacks the $\beta$ carbon, and produces the thermodynamically favored product. On the other hand, a 1-2 reaction (on the carbonyl) gives the kinetic product, and is obtained at low temperatures.

Why is the 1-4 product thermodynamically more stable? Because the resulting product benefits from keto-enol tautomerism, which results in lowering the energy of the system. Usually, the more resonance forms a compound has, the more its electrons are delocalized, the more stable it is.

Draw the resonance forms of the 1-4 and 1-2 products, and see.

You asked for specific affinities of different organometallics in 1-4/1-2 additions. My knowledge is that organocuprates ($\mathrm{R-CuLi}$) will perform Michael additions, and that organolithians seem to prefer 1-2 addition. Also, according to this source, Grignard reagents do not seem to have a preference.

My take on this is that the cuprate is less reactive, and therefore can form the thermodynamic product, whereas the lithium reagent is so destabilized that it reacts right away.


The 'long chain band' described in your textbook (Pavia) is actually a vibrational mode in which the terminal methyl groups are 'rocking'.

This rocking mode is observed in open chain alkanes,  -(CH2)n-, where the number of methylene units is greater than 4.

Polyethylene is the 'classic' examples, the IR spectra and corresponding vibrational modes are given below:

Why do we see this band?

The reason we see any given IR band is as a consequence of the symmetry of the molecule, with only certain molecular vibrations IR active (we don't, for example, observe the C-C vibrations directly, they aren't active). Your question, more broadly, then becomes 'how to predict the IR active vibrational modes for a given molecule?'.

For a full explanation of the IR spectra in this particular case, using molecular symmetry and group theory, you might try reading J. Chem. Phys 1956, 25, 549.

Why don't we see the band in cyclic alkanes?

In cyclic alkanes, the methyl groups have been constrained to form one continuous ring of methylene groups. As discussed above, the band at ~725 cm-1 is caused by the terminal methyl groups of longer chain alkanes, hence, removing these methyl groups removes their corresponding vibrational modes.


16 steps total

Longest linear sequence of 9 steps

Synthesis of Precursor 1:

The hydroformylation of allylamine, Rhodium catalysis (drew the product wrong, it has one more carbon please ignore it for now, I drew the right structure in the one-pot step)
Diazotization of product, occurs immediately before addition to other precursors in one-pot step later on (produces precursor 1)

Synthesis of Precursors 2 & 3

Hydroformylation of methylacrolein, Rhodium catalysis (produces precursor 2)
Protection of aldehyde via hemithioacetal
Protection of ketone through dioxolane
Deprotection of aldehyde by base (produces precursor 3)

Formation of grignard reagent by reaction with magensium of bromoethanal diethyl acetal

One pot reaction, diazo thermolysis resulting in carbene substitution followed by aldol
Weak acid dehydration to form alpha-beta unsaturated ketone
Wittig reaction to establish alkene
Dehydration using Rhenium Catalyst with good selectivity, click here for the paper
Reduction of aldehyde (special reducing agent used just in case free ketone exists)

Deprotection of the ketone (loss of ethane-1,2-diol)
Grignard reaction followed by dehydration to form the ketone of the product
Reduction with lithium aluminum hydride (not shown) to get product, Plaunotol


Electrolysis using electrodes of metals whose ions are in solution

The reason why silver dissolves (or water is oxidized) is well known. The reason why platinum does not dissolve is more elusive. Thanks to Maurice's comment for pointing this out.
Let's assume your solution is 1M in AgNO3 and 1M in acid ($H^+$). The potential required to oxidize silver is 0.80 V; the voltage required to split water (oxidize it) is 1.23 V, but there are complications of overvoltage, mostly with hydrogen, but also with oxygen. To sum up: it is easier to grab an electron from silver than to pump another hydrogen ion into solution. AHA! How about reducing the acid concentration, or not adding any at all? (Of course, the reason for adding the acid was to make the silver dissolve easier.) If the pH is increased so that hydrogen ion concentration is reduced to give a lower oxidation potential than silver, then water will be oxidized, hydrogen ion will be produced, and eventually, silver will become more easily oxidized. This electrolysis wil go thru a phase where hydrogen and silver are codeposited, probably making a porous, weak deposit of silver.
For platinum to yield 2 electrons requires 1.188 V. Although this is less than required for oxidizing water, overvoltage effects due to oxygen and hydrogen are apparently important. Although platinum can be electroplated with a platinum anode, apparently the process fails to work well; older methods of platinum plating were kept refreshed with additions of platinum salts rather than relying on dissolution of the anode (Ref 1). More modern methods use a pulsed electropotential, but their success seems to be from a solving difficulties at the cathode rather than getting the anode to dissolve (Ref 2). They deposit a monolayer of platinum attached to a monolayer of hydrogen, then oxidize off the hydrogen and repeat. The electrode potentials strongly suggest that increasing the cell potential, if you had a platinum anode, would result in dissolving platinum metal, but the catalytic reactivity of platinum with respect to H2O2 suggests that maybe the oxidation reaction occurs in the water layer, and the platinum acts merely as an electronic conductor.
Ref 1. https://www.technology.matthey.com/article/25/1/32-41/
Ref 2. https://www.chemistryworld.com/news/platinum-plating-at-the-flick-of-a-switch-/5720.article


I have asked experts -> I am right. The common convention is to label the physical electrodes as cathode/anode based on discharge even though that will be backwards when the battery is charging. That is a poor, confusing, and illogical naming convention, but it is what it is.

The more clear and logical naming convention would be to name the physical electrodes either positive or negative since those labels apply in both charge/discharge directions. Some people use this logical naming convention, but the illogical convention seems more prevalent.


Okay, to find the density of the gas, to need to know its mass and its volume. So to do this, lets say we have one mole of this substance. So its mass will be $16.04~\mathrm g$. Now all we need to do is find its volume.

This can be done by rearranging the ideal gas equation: $$V = \frac{nRT}{p}$$ Now all we have to do is plug in the values. However you have be careful that you use the write units. This is a common mistake, especially when using the ideal gas law however it is good to see that you have used the correct units: $$\begin{align}V &= \mathrm{\frac{1~mol\times0.08206~L~ atm~ mol^{-1}~K^{-1}\times336.25~K}{1.05~atm} }\\
&=\mathrm{26.28~L}\end{align}$$

As you can see, all the units cancel nicely so that you get the final answer in litres which is the correct unit for volume.

Now, density can be calculated: $$\rho =\mathrm{\frac{16.04~g}{26.28~L} = 0.61~g~L^{-1}}$$


Wikipedia gives the following solubilities for the noble gases in water at $25^\circ \text C$ in $\mathrm{cm^3~kg^{-1}}$:

$\mathrm{He: 8.61~~~Ne: 10.5~~~ Ar: 33.6~~~ Kr: 59.4 ~~~Xe: 108.1 ~~~Rn: 230}$

Why does solubility of noble gases increase down the group?

The intermolecular interactions of noble gases are in the form of London dispersion forces, which are strongly influenced by the polarizability of the atom which increases as the atomic radius increases. This can be thought of as resulting from the electrons being 'less tightly bound' to the nucleus and the atomic radius increases. Atomic radius increases down the group therefore solubility also increases. For more on polarizability see here.

Why is radon not the most soluble noble gas?

Radon (and other heavy radioactive elements) are often ignored when talking about trends because they are uncommon. Radon is in fact the most soluble noble gas but the half lives of its isotopes are on the order of hours or days so it is not really very useful.


What are the oxidation states of galium and arsenic in GaAs semiconductor?

In general, gallium can have +1 and +3 oxidation states, though +1 is a very rare condition. In its turn arsenic possible oxidation states are -3, +3 and +5. As it is stated in Wikipedia, gallium oxidation state in gallium arsenide is +3. Once again, arsenic has -3 oxidation state in arsenides or intermetallic compounds. Therefore, gallium arsenide oxidation states are +3 for Ga and -3 for As. This is logical because the electronegativity of As is higher then Ga. As for Simon Sze, he is an electrical engineer and physicist and not a chemist. From the mathematical point of view (-3)(+3) is the same thing as (-1)(+1).


This isn't something you could know except by experiment, but boric acid doesn't exist as solely $\ce{H3BO3}$ in solution. In addition to being an Arrhenius acid, $\ce{H3BO3}$ is also a Lewis acid. This means it is capable of accepting a lone pair, and will undergo the following reactions in aqueous solution:

$$\ce{B(OH)3 + H2O <=> B(OH)4- +  H+}\tag1\label{Eq1}\\$$

$$\ce{4B(OH)4- + 2H+ <=> B4O7^{2−} + 9H2O}\tag2\label{Eq2}\\$$

By combining \eqref{Eq1} and \eqref{Eq2}, we get the full reaction:

$$\ce{4B(OH)3 <=> B4O7^{2−} + 2H+ + 5H2O }\tag3\label{Eq3}\\$$

Boric acid does not produce tetraborate exclusively, however, and many other species species are also present in solution. The system is actually so complex that it remains difficult to characterize, one paper mentioning that there are some ten different equilibrium reactions in solution$^{[1]}$. In performing an acid-base calculation, however, it is okay to simply treat it as a solution of $\ce{H2B4O7}$, neutralizing according to the equation:

$$\ce{H2B4O7 + 2NaOH <=> Na2B4O7 + 2H2O}\tag4\label{Eq4}\\$$

Titration confirms that $\ce{H3BO3}$ reacts with $\ce{NaOH}$ in a $\text{2:1}$ ratio, and crystallization of the product confirms $\ce{Na2B4O7}$ is produced.

$[1]$ Trejo G., Frausto-Reyes C., Gama S.C., Meas Y., Orozco G. Raman Study of Benzylideneacetone on Silver. Int. J. Electrochem. Sci. 2012, 7, 8436–8443.


Plastics Design Library series include tabulated parameters for the variety of polymers.
One of such parameters is a PDL number rating from 0 to 9:

0: solvent dissolved disintegrated
1: decomposition
2: severe distortion; oxidizer and plasticizer deteriorated
…
9: highest resistance, no change

In the table below I assembled data on thermoplastics chemically unstable towards various oils with PDL numbers below 3:

$$
\begin{array}{lrcll}
\hline
\textbf{Polymer} & \boldsymbol{T/\pu{°C}} & \textbf{PDL} & \textbf{Oil} & \textbf{Resistance Note} \\
\hline
\text{CAB} & 20 & 0 & \text{spearmint oil} & \text{Limited Resistance} \\
\hline
\text{LDPE} & 20 & 1 & \text{aniseed oil} & \text{Not resistant; tensile strength at yield and elongation at break greatly reduced} \\
\hline
\text{HDPE} & 20 & 2 & \text{camphor oil} & \text{Not resistant} \\
\hline
\text{PP} & 100 & 2 & \text{two-stroke engine oils} & \text{Unsatisfactory/severe effect} \\
  & 20 & 2 & \text{turpentine oil} & \text{Not resistant} \\
  & 100 & 2 & \text{vaseline oil} & \text{Unsatisfactory/severe effect} \\
\hline
\text{LLDPE} & 20 & 1 & \text{aniseed oil} & \text{Not resistant; tensile strength at yield and elongation at break greatly reduced} \\
\hline
\text{PS} & 22 & 2 & \text{clove oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{lemon peel and oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{orange peel and oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{pine needle oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{spearmint oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
\hline
\text{SAN} & 23 & 2 & \text{citronella oil} & \text{Severe attack; softened in few hrs} \\
  & 23 & 2 & \text{clove oil} & \text{Severe attack; softened in few hrs} \\
  & 22 & 2 & \text{lemon peel and oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 52 & 2 & \text{pine needle oil} & \text{Severe attack; softened in few hrs} \\
  & 23 & 2 & \text{spearmint oil} & \text{Severe attack; softened in few hrs} \\
\hline
\text{ABS} & 52 & 2 & \text{pine needle oil} & \text{Severe attack; softened in few hrs}  \\
  & 23 & 2 & \text{spearmint oil} & \text{Severe attack; softened in few hrs} \\
\hline
\end{array}
$$

Acronyms

$$
\begin{array}{ll}
\text{CAB} & \text{Cellulose Acetate Butyrate} \\
\text{LDPE} & \text{Low Density Polyethylene} \\
\text{HDPE} & \text{Polyethylene} \\
\text{PP} & \text{Polypropylene} \\
\text{LLDPE} & \text{Linear Low Density Polyethylene} \\
\text{PS} & \text{Polystyrene} \\
\text{SAN} & \text{Styrene Acrylonitrile Copolymer} \\
\text{ABS} & \text{Acrylonitrile Butadiene Styrene}
\end{array}
$$

References

Chemical Resistance of Thermoplastics; Woishnis, W. A., Ebnesajjad, S., Eds.; Plastics Design Library; William Andrew: Norwich, N.Y, 2012.


A Lewis acid can accept a pair of electrons from a Lewis base. The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid.


I would not expect to find any relationship.  The potential you are referring to involves breaking metallic bonds and ionizing the resulting atoms while conductivity involves electrons moving through the solid metal.  Copper's is -.34 and it is between gold and silver in conductivity.  Aluminum's is 1.66 and it is #4 behind gold in conductivity.


What kind of isomerism does Ni(PPh3)2Cl2 have and what is their IUPAC name?

A metal phosphine complex, dichlorobis(triphenylphosphine)nickel(II) exists in two isomeric forms (blue and red) based on how you prepared it (Wikipedia). The blue isomer, which is paramagnetic can be prepared by treating hydrated nickel chloride ($\ce{NiCl2.6H2O}$) with triphenylphosphine in alcohols or glacial acetic acid. When allowed to crystallize from chlorinated solvents (e.g., dicloroethane (Ref.1), dicloromethane (Ref.2)), the tetrahedral isomer converts to the square planar red isomer, which is trans and diamagnetic:

References:

Benedetto Corain, Bruno Longato, Roberto Angeletti, Giovanni Valle, “trans-[Dichlorobis(triphenylphosphine)nickel(II)]•$\ce{(C2H4Cl2)2}$: a clathrate of the allogon of venanzi's tetrahedral complex,” Inorganica Chimica Acta 1985, 104(1), 15–18 (https://doi.org/10.1016/S0020-1693(00)83780-9).
A. S. Batsanov,  J. A. K. Howard, “trans-Dichlorobis(triphenylphosphine)nickel(II) bis(dichloromethane) solvate: redetermination at $\pu{120 K}$,” Acta Cryst. 2001, E57, m308–m309 (https://doi.org/10.1107/S1600536801008741).


Does rice protein have a high concentration of arsenic?

The US FDA reported arsenic testing of 12 rice protein powders as follows:

  333863609,
  Dietary Supplements,
  Rice Protein Powders,
  Plant-Based Energizer:
  0.1

  333511892,
  Dietary Supplements,
  Rice Protein Powders,
  Protein Energizer:
  0.8

  333511937,
  Dietary Supplements,
  Rice Protein Powders,
  Protein Energizer:
  0.9

  334620956,
  Dietary Supplements,
  Rice Protein Powders,
  Raw Vegan Chocolate:
  0.9

  334094485,
  Dietary Supplements,
  Rice Protein Powders
  Plant-Based Drink Mix:
  1.0

  333511880,
  Dietary Supplements,
  Rice Protein Powders,
  Raw Protein:
  1.1

  333687971,
  Dietary Supplements,
  Rice Protein Powders,
  High Protein Energy Meal:
  1.4

  333687931,
  Dietary Supplements,
  Rice Protein Powders,
  High Protein Energy Meal:
  1.6

  334093871,
  Dietary Supplements,
  Rice Protein Powders,
  Blend of protein, fruits & veggies:
  1.6

  334094067,
  Dietary Supplements,
  Rice Protein Powders,
  Chocolate Plant Protein Shake:
  3.7

  334093956,
  Dietary Supplements,
  Rice Protein Powders,
  Blend of protein, fruits & veggies:
  4.4

  333688312,
  Dietary Supplements,
  Rice Protein Powders,
  Organic Meal Replacement:
  5.7

Where the units are micrograms of inorganic arsenic per serving.


Does the slippery feeling of bases always appear at certain pH?

There probably is no one pH level where things get slippery. Capacity to react as a base has something to do with it.

You can make ammonia and sodium hydroxide solutions both with pH 11, but the sodium hydroxide solution is only 0.001 molar and thus cannot react with much oil on the skin.  Ammonia solution has a lot more base available to react with those fatty acids.  The weak base "hides" most of its reactivity in water alone but reveals when there are fatty acids to neutralize, so it's slip sliding away.

Probably anything that we consider basic and is sufficiently concentrated will slip in the fingers.


Confusion about electron arrangement in elements

Basically, it depends on subshell. If you google something like "subshells", you'll see that s-subshell can hold 2 electrons, p - 6, d - 10, f - 14, g - 18. It's very unlikely that you'll need subshells f and g, since you're school student.
Look at this picture (about Aufbau principle).
It shows you the order of orbitals. Each s orbital holds up to 2 electrons, p - 6, so on. Fill them with all your electrons in given order.
For example, you have 12 electrons. The order (using Aufbau principle) is 1s-2s-2p-3s-3p-4s-3d...
1s holds 2 electrons (12-2 = 10), 2s holds 2 (10-2=8), 2p - 6 (8-6=2), 3 s - 2 (that's all). We see that 1st shell has 2, 2nd shell has 8, 3rd shell has 2, but 3rd is not full (since you still have empty 3p and 3d subshells).
So let's check. n is for number of shell. $2*n^2$ is for maximum number of electrons that shell can hold. For 1st, $2*1^2=2$, as we've seen. For 2nd, $2*2^2=8$, as we've seen. For 10th shell, the maximum is 200 electrons.
[edited]
Use Aufbau principle as well as info about subshells. Let's look at Y(39 electrons), which is in G3 P5. So we expect 5 shell to be outermost, and 3 valence electrons. Not "electrons in outer shell", but valence. For school course of chemistry, outer shell electrons are usually referred to as "valence", but these concepts are not always interchangeable.
Back to our example, Y has 1s-2electrons, 2s-2,2p-6,3s-2,3p-6,4s-2,3d-10,4p-6,5s-2,4d-1 (sum - 39). So yes, "maximum shell" is 5, as expected. But why do we have 1 electron on the outermost subshell, when we need to have 3? It's because valence electrons are those on 5s and 4d (2+1=3). Usually d subshell electrons are also considered "valence", when you have 1-10 of them (if total number of electrons, starting with 4d electrons is 18 or less). I highly recommend you to use periodic table with 18 groups, it's much more convenient for these purposes.


Is the largest number of molecules in 36 g of water or 54 g of dinitrogen pentaoxide?

I'm stuck, because both of them are having two moles of each of the respective molecules.

Are you sure?

\begin{align}
M(\ce{H2O})  &= \pu{18 g mol^-1} &
\implies \pu{36 g} &\mathop{\hat{=}} \pu{2 mol} \\
M(\ce{N2O5}) &= \pu{108 g mol^-1} &
\implies \pu{54 g} &\mathop{\hat{=}} \pu{0.5 mol} \\
\end{align}


What causes triboluminescence and what process occurs to produce light?

I'm only a first year chemical engineering student, and we haven't really covered this topic in our course yet, but I will try and answer your question to the best of my knowledge.

(I am sure people more qualified me would provide better responses.)

When we apply certain stresses to a material such as a crystal, we separate charges (which is why it usually is asymmetric... the presence of anistropy allows charge separation.) These charges then can cause electric discharge, which ionizes the surrounding air and releases a flash of light.

There exist symmetric crystals which display this phenomenon as well... and it is suggested that (probably) impurities exist within such crystals which break the symmetry.

Anyway, this is my understanding of the phenomenon, and I don't claim is perfect in any way, but hopefully is helpful to you.


Using stoichiometry to split a complex mineral

Is it possible to separate out $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ into simply the $\ce{(Mg_{x}Cu_{1-x})(CO3)}$ part that reacts with the dilute sulfuric acid to produce the carbon dioxide, the moles of which I know?

$4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{CO2}$ implies $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$.
$4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ is 0.8167 g.
Then, considering the atomic weights of C, O and H, subtract the contribution of these elements from 0.8167 g to determine the mass of $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2}$.
Then divide by 2 to get the mass of $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{Mg_{x}Cu_{1-x}}$;
Then use the atomic weights of Mg and Cu to solve for $x$.


Is there a possible distortion of XeF6 from Oh point group to reduced symmetry?

Now, there are a few things that must be stated very clearly. Firstly, it pays to be very careful when you are talking about point groups. It's careless (and confusing to others) to write $\mathrm{T_{2u}}$ when you mean $\mathrm{T_{1u}}$, or to write $\mathrm{A_{g}}$ when you mean $\mathrm{A_{1g}}$, so please pay attention to this next time, regardless of whether you are writing online or submitting a piece of work. Good communication in science has to be precise and clear!

Secondly, let's use this reduced descent of symmetry table (I have removed some columns and some rows, which aren't relevant now)

$$\begin{array}{ccc} \hline
O_\mathrm h & T_\mathrm d & D_\mathrm{3d} \\
\hline
\mathrm{A_{1g}} & \mathrm{A_1} & \mathrm{A_{1g}} \\
\mathrm{T_{1u}} & \mathrm{T_2} & \mathrm{A_{2u} + E_u} \\ \hline
\end{array}$$

This tells you that if you decrease the symmetry of the system from $O_\mathrm h$ to $T_\mathrm d$, then anything that previously transformed as $\mathrm{A_{1g}}$ will now transform as $\mathrm{A_1}$. Likewise, anything that previously transformed as $\mathrm{T_{1u}}$ will now transform as $\mathrm{T_2}$.

If you are trying to look for potential mixing of orbitals in the lowered symmetry, then you have to look what the irreps in the lowered symmetry are. From what you have written, I sense that you're trying to "look for an $\mathrm{A_{1g}}$ component", which isn't what you should be looking for, because something that transformed as $\mathrm{A_{1g}}$ in $O_\mathrm h$ symmetry may not necessarily transform as "$\mathrm{A_{1g}}$" in the lower symmetry. Point in case: $T_\mathrm d$.

If you are considering the viability of a distortion to $T_\mathrm d$, then you should be looking for an $\mathrm{A_1}$ component in the second row, not $\mathrm{A_{1g}}$. Of course, $\mathrm{T_2}$ does not contain $\mathrm{A_1}$, so the distortion to $T_\mathrm{d}$ isn't favourable.

Now, in $C_\mathrm{3v}$ the correlations are $\mathrm{A_{1g}} \to \mathrm{A_1}$ and $\mathrm{T_{1u}} \to \mathrm{A_1} + \mathrm{E}$. If we add this column to the reduced table above

$$\begin{array}{cccc} \hline
O_\mathrm h & T_\mathrm d & D_\mathrm{3d} & C_\mathrm{3v} \\
\hline
\mathrm{A_{1g}} & \mathrm{A_1} & \mathrm{A_{1g}} & \color{red}{\mathrm{A_1}} \\
\mathrm{T_{1u}} & \mathrm{T_2} & \mathrm{A_{2u} + E_u} & \color{blue}{\mathrm{A_1 + E}} \\ \hline
\end{array}$$

then it is clear that because the $\color{blue}{\text{blue}}$ $\mathrm{A_1 + E}$ shares one irrep in common with the $\color{red}{\text{red}}$ $\mathrm{A_1}$ component, the HOMO–LUMO mixing is allowed in $C_\mathrm{3v}$ symmetry, and this provides a qualitative argument for why distortion to $C_\mathrm{3v}$ symmetry is favoured.

Again, to drive home the point, you shouldn't be looking for whether the $\color{blue}{\text{blue}}$ bit contains $\mathbf{A_{1g}}$. There is no $\mathrm{A_{1g}}$ irrep in $C_\mathrm{3v}$, so of course you won't find it! You should be asking whether the $\color{blue}{\text{blue}}$ bit contains the $\color{red}{\text{red}}$ bit.

How to get those correlations, is quite a fair question. Well, I got it from this diagram in Orbital Interactions in Chemistry, 2nd ed. by Albright et al. The correlation diagram 14.4 should be self-explanatory, hopefully (you can ignore the pictures of the MOs, 14.5 and 14.6).

But if you want to obtain it without flipping through every book in the library, this is how you need to do it. Because high-symmetry point groups like $O_\mathrm h$ contain tens and tens of subgroups, books don't usually list every single possible correlation to lower symmetry, since there are far too many possibilities. The table you were given has a bunch of omitted correlations, listed under "other subgroups" (the implication being, "other subgroups that I couldn't be bothered to put in"). You'll notice that $C_\mathrm{3v}$ is one of those "other subgroups" that weren't explicitly given.

The strategy therefore is to carry out the correlation in two steps. First you go from $O_\mathrm h$ to $D_\mathrm{3d}$, and then you go from $D_\mathrm{3d}$ to $C_\mathrm{3v}$. The tables you are using don't provide that, but a more complete list of tables is provided in the Appendixes of Molecular Vibrations by Wilson et al. The $D_\mathrm{3d}$ to $C_\mathrm{3v}$ correlation is given as (again I removed the irrelevant rows)

$$\begin{array}{cc} \hline
D_\mathrm{3d} & C_\mathrm{3v} \\ \hline
\mathrm{A_{1g}} & \mathrm{A_1} \\
\mathrm{A_{2u}} & \mathrm{A_1} \\
\mathrm{E_u} & \mathrm{E} \\ \hline
\end{array}$$

so now the correlation from $O_\mathrm h$ to $C_\mathrm{3v}$ is easy: $\mathrm{A_{1g}}$ in $O_\mathrm h$ goes to $\mathrm{A_{1g}}$ in $D_\mathrm{3d}$, which in turn goes to $\mathrm{A_1}$ in $C_\mathrm{3v}$. Likewise, $\mathrm{T_{1u}}$ in $O_\mathrm h$ goes to $\mathrm{A_{2u} + E_u}$ in $D_\mathrm{3d}$, which in turn goes to $\mathrm{A_1 + E}$ in $C_\mathrm{3v}$.


First to clear up some terms. Glass transition just means that the intermolecular bonds are weak enough for the material to continue crystallizing—it is not the melting temperature.  This can be accompanied by softening of the material but that does not mean the polymer will flow.  Thermoplastics can flow at elevated temperature but thermosets (which includes cross-linked thermoplastics) do not by definition.

Epoxy has a glass transition due to a loss of crystallinity but is not usually shapeable unless formulated to or poorly prepared.  PVA is a thermoplastic that melts around $\pu{200^\circ C}$  which is also the temperature that it starts to pyrolyze and char. With careful temperature control you can melt and form PVA and there are 3D filaments dissolvable in water based on PVA.  Without careful control, you decompose the plastic before melting it.


Limiting the minimum percentage of oxygen in atmosphere maintained by an oxygen absorber / scavenger?

Oxygen absorbers have the purpose to prevent food spoilage by oxidation (of fatty acids, flavors, etc.), so the desire is to remove oxygen as much as possible and also to bind small amounts which may diffuse into the package upon storage.

In order to achieve a certain defined oxygen containing atmosphere, consider flooding the package with an artificial gas mixture, e.g. 5% oxygen, 95% nitrogen. Should be easy to obtain from scuba dive shops who are filling bottles with homemade nitrox blends if they are willing to give you a custom mixture (which in this case is not suitable for diving).

Or just mix air and nitrogen or carbon dioxide (from a soda machine) by adjusting the flow.


Why are potentials of cells with magnesium electrode always lower than expected?

At least one significant reason is that the voltage you "expect" is probably for a standard state of one mole per liter of $\ce{H+}$, i.e. at pH 0.  Neither the magnesium acetate nor the plain (tap?) water you used for the experiments come close to this pH. Also, the standard state for $\ce{H2}$ would be one bar of atmospheric pressure.  Probably this condition was not achieved either.

You can calculate the expected potential at any pH using the Nernst Equation.

The half-cell reactions in your system are

$\ce{Mg^0 -> +2e- +Mg^{+2}}$ (this has a potential of 2.372 V not 2.7 V as a mistakenly wrote in a prev. version of this answer)
$\ce{2H^+ +2e- -> H2}$

The total reaction is thus $\ce{Mg^0 +2H+ -> Mg^{+2} + H2}$

So applying the Nernst equation to a one molar magnesium solution at pH 7 (i.e. $\ce{H+}$ concentrations of $10^{-7}$ mol/L) would give:

$E_\text{cell} = E^{\ominus}_\text{cell} - \frac{RT}{zF} \ln Q$
$E_\text{cell} = 2.4 \text{ V} - \frac{RT}{zF} \ln \frac{1}{(10^{-7})^2 }$
$E_\text{cell} = 2.4 \text{ V} - 0.41 \text{ V} = 1.9 \text{ V}$

Obviously, this correction gives you a predicted voltage that is still higher than what you measure closer what you measured.  But you reported using acid and getting a higher potential, so it's clear that the pH effect matters for your system.  But something else must be going on too.  Some other possibilities for other minor effects that tweak the voltage a bit more:

Hydrogen partial pressure is unaccounted for
overpotential
your cell is not open but is conducting current
dirty electrodes etc.
other unaccounted for electrochemical reactions (copper oxidation, etc.)


References to draw 3D molecules with directionality of non-bonding electron pairs and p-orbitals

As it is already pointed out in the comments, what you are learning for is hybridisation and some basic MO theory. Do not go too far into the MO direction, because you will get into somewhere completely different place, and you will hard time to make the connections.

So aftre you reviewed hybridization: Once you can identify in a molecule the sp1, sp2 and sp3 carbons, you should be able to orient the hybrid orbitals. For example in the case of sp2 carbon of the lower image, it is in the HHCC plane. In case of sp1 there are two, for sp2 one un-hybridized p orbitals. These unhybridized orbitals should be perpendicular to the hybridized ones: perpendicular to the axis of sp1 or the plane of sp2.


Any side effects to using calcium chloride as a desiccant?

The only issue with $\ce{CaCl2}$ I've encountered is that eventually it pulls in so much water that it dissolves.

To keep dust down and to prevent the $\ce{CaCl2}$ solution from sloshing around, place it in a plastic container with a tightly-fitting sponge atop.

$\ce{CaCl2}$ is relatively non-toxic (it's used to make pickles), is inexpensive and is readily available, in cold climates, for melting snow and ice.


In the case of weak acids dissociating like strong ones when they are highly dilute, there is a limit to how far this can go.  If you were to try to measure acetic acid dissociation at $10^{-n}$ molar for $n=4,5,6,7,8,...$ at ambient temperature you would at first rise nicely towards $100$% as you might expect, but eventually level off at only about $99.4$%.
What stops you is the autoionization of water.  As you approach infinite dilution the hydrogen ion concentration approaches not zero but the autoionized value of $10^{-7}$ molar.  So the ratio of dissociated acid to undissociated acid levels off at $K_a/10^{-7}$.  Accordingly the limiting degree of dissociation is
$1/[1+(10^{-7}/K_a)]=K_a/(K_a+10^{-7})$
For acetic acid with $K_a\approx1.8×10^{-5}$ this limit is about $99.4$%, but for a much weaker acid such as a typical phenol or hydro-cyanic acid the dissociation is much lower even at extremely low concentrations.  When $K_a<10^{-7}$ the limiting dissociation is less than $50$%; you need a stronger base than water to coax dissociation as a major reaction in aqueous solution.


Why is the pressure value altered in this gas law numerical?

Your tutor is a master of confusion :-D

Would the exercise be easier if the pressure was given in stones per square acre? ;-)

Write down the equation for the ideal gas law again. Pay attention to the units!

Don't let the mercury fool you ;-)


Total valence electrons in a rhodium complex

This is not a benzene ligand ($\ce{C6H6}$), but rather a phenyl ligand ($\ce{C6H5}$).

In the benzene complex, all the carbons have their hydrogens. The pi electrons of the ring act as donors to the metal. On the other hand, in the phenyl complex, the ring carbon bonded to the metal does not have its hydrogen. That carbon is formally negatively charged and the lone pair on carbon acts as a donor to the metal.

Aryl ligands, just like alkyl ligands, count as 1-electron donors within the "covalent" electron-counting system (which you seem to be using, since you counted H and Cl as 1e donors as well).

Therefore the total electron count is $9 + 1 + 1 + 2 +2 + 1 = 16$.


Conceptual thoughts about mixing two immiscible liquids (not on surface tension):

A reflux condenser used in the ochem lab equally has a circulation of about 2 L/min water (reference from an advertisement, which doesn't look too much off-range)

After the two liquids are mixed, what application will follow?  Do you target cleaning an engine, or deployment for food processing?  What are intended storage conditions once the two are mixed (e.g., time of storage / transport, temperature of storage / use)?  Altogether with the other liquid you want to mix with water, this may narrow your choice of the surfactant.

Mixing the two continuously may be achieved with a tube reactor with two inlets, one outlet, and a screw-shaped inner geometry

(reference, section «Liquid-Liquid & Liquid-Gas Mixers»)
Depending on parameters of your liquids (e.g., density, viscosity, flow rate, key word Reynolds number) you may enter the regime of turbulent flow until you divide the liquid into tiny droplets which increases the surface accessible for mixing.  The addition of surfactant may help to achieve the intended mixing, too.

(loc. cit.)

The formation of foam however is the result of an entrapment of a gas.  Whipping cream with a whisk for example intentionally aerates the cream.  Liquids like water may boil (i.e., generation of gas) because of sudden pressure changes while passing a propeller blade at a flow rate too high (cavitation), too.
Thus, mixing water and your other liquid in a tube reactor like the one mentioned above, in absence of air and at the right hydrodynamic state may prevent the formation of foam.  (You mau fight foam by ultrasound (example), though.)


According to IUPAC definition:

  ‘carbohydrate’ includes monosaccharides, oligosaccharides and polysaccharides as well as substances derived from monosaccharides by reduction of the carbonyl group (alditols), by oxidation of one or more terminal groups to carboxylic acids, or by replacement of one or more hydroxy group(s) by a hydrogen atom, an amino group, a thiol group or similar heteroatomic groups. It also includes derivatives of these compounds [...]

But according to Wikibooks

  the term is generally understood in the biochemistry sense, which excludes compounds with only one or two carbons. Natural saccharides are generally built of simple carbohydrates called monosaccharides with general formula $\ce{(CH2O)_n}$ where $n$ is three or more [...]

Because formaldehyde and glycolaldehyde (not a true sugar) has one and two carbon respectively in its structures we can exclude both as a carbohydrate.


Rate of dehydration for butan-2-ol and butane-2,3-diol

For this question, given the lack of initial information, I'm making the following assumptions:

Butane-2,3-diol dehydrates to give butan-2-one
2-butanol dehydrates to give but-2-ene

Butane-2,3-diol dehydration is nicely described in this paper from Emerson1:

Aqueous 2,3-butanediol was shown to react in a pseudo-first-order reaction in the presence of sulfuric acid to form methyl ethyl ketone. The reaction could be described by a pinacol rearrangement mechanism and had an activation energy of 36 kcal/mol.

Finding the data for dehydration of 2-butanol proved more elusive. One study2 conducted dehydration rate experiments for secondary alcohols at 250 °C and 40 bar pressure and obtained a rate constant for 3-hexanol to be $k = \pu{(0.067 \pm 0.021) h^{-1}}$. Assuming this rate would be similar to 2-butanol, we can find the activation energy via the arrhenius equation, which comes out to be around 48 kJ/mol, which is around 11.52 kcal/mol. Another source3 conducted the same experiment over a copper-chromite catalyst, and obtained an activation energy of 13 kcal/mol. This means that 2-butanol will dehydrate faster than butane-2,3-diol.
From a conceptual standpoint, this makes sense: The pinacol rearrangement mechanism would require a hydride shift, which would be the RDS and would take more time and energy than a simple carbocation formation. However, if you're comparing absolute reaction rates, a large number of factors come into play, and given the right conditions, butane-2,3-diol can dehydrate faster than butan-2-ol.
References:

Emerson, Richard R., et al. “Kinetics of Dehydration of Aqueous 2,3-Butanediol to Methyl Ethyl Ketone.” Industrial & Engineering Chemistry Product Research and Development, vol. 21, no. 3, 1982, pp. 473–77. doi:10.1021/i300007a025
Bockisch, Christiana, et al. “Kinetics and Mechanisms of Dehydration of Secondary Alcohols Under Hydrothermal Conditions.” ACS Earth and Space Chemistry, vol. 2, no. 8, 2018, pp. 821–32. doi:10.1021/acsearthspacechem.8b00030.
Joseph Sires Cantrell, "Rate and activation energy of the first order dehydration of 2-butanol over a copper chromite catalyst", 1954, source


How is electron-impact mass spectrometry data formally reported?

Third edition (latest offline edition) of the ACS Style Guide (currently known as The ACS Guide to Scholarly Communication) has a section devoted to reporting analytical data, including mass-spectrometry and EIMS in particular [1, p. 275]:

Mass Spectrometry
MS $m/z$ (relative intensity): $238.2058~(44.8\%),$ $195.1487~(100\%),$ $153.1034 ~(21.2\%).$
GC–MS $m/z$ (% relative intensity, ion): $202~(9,~\ce{M} + 4),$ $200~(32,~\ce{M} + 2),$ $198~(23,~\ce{M+}),$ $142~(35,~\ce{M} - 2\ce{CO}),$ $321~(95,~\ce{M} - \ce{Me}),$ $415~(\ce{M+} - \ce{Cl}).$
HRMS–FAB $(m/z):$ $[\ce{M} + \ce{H}]^+$ calcd for $\ce{C21H38N4O6S},$ $475.259;$ found, $475.256.$
EIMS $(\pu{70 eV})$ $m/z:$ $\ce{M+}~420~(15),$ $241~(15),$ $201~(59),$ $135~(14),$ $69~(23).$
Abbreviations: $m/z$ is the mass-to-charge ratio, $\ce{M}$ is the molecular weight of the molecule itself, $\ce{M+}$ is the molecular ion, HRMS is high-resolution mass spec-
trometry, FAB is fast atom bombardment, and EIMS is electron-impact mass spectrometry.

The recommendations for ACS journals also align with the general recommendations given in RSC Experimental data policy — Guide to the presentation of experimental data (accessed 2021-04-22):

Mass spectrometry data
Given in the form: $m/z~183~(\ce{M+},~41\%),$ $168~(38),$ $154~(9),$ $138~(31)$ etc. The molecular ion may be specified as shown if desired. Relative intensities in parentheses (% only included once). Other assignments may be included in the form $m/z~152~(33,~\ce{M} − \ce{CH3CONH2}).$ Metastable peaks may be listed as: $\ce{M^*}~160~(189\to 174),$ $147~(176\to 161),$ etc. The type of spectrum (field desorption, electron impact, etc) should be indicated. Exact masses quoted for identification purposes should be accurate to within $\pu{5 ppm}$ (EI and CI) or $\pu{10 ppm}$ (FAB or LSIMS).

If you are a $\mathrm\LaTeX$ user, check out chemmacros package (TeX.SE) and its spectroscopy module (PDF) which assists with the input of various experimental data.
Reference

The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. ISBN 978-0-8412-3999-9.


We usually don't use the real gas $C_V(T,p)$ directly in calculating the change in U in going from thermodynamics equilibrium state A to thermodynamic equilibrium state B, if both these states are beyond the ideal gas region.  Starting from state A, what we do first is use the real-gas p-v-T equation to determine the change in U in going from the initial pressure $p_A$ to a low pressure (or large specific volume) in the ideal gas region $p^*$ at constant temperature $T_A$.  Then, our second step is to hold the pressure constant at pressure $p^*$ and raise the temperature from $T_A$ to $T_B$, using the ideal gas heat capacity to determine the change in internal energy for this step (since we are now in the ideal gas region).  Finally, we hold the temperature constant at $T_B$ and raise the pressure from $p^*$ to $p_B$ using the real-gas p-v-T equation again to determine the change in U.  The sum of the changes in U for these three steps adds up to the change in U in going from $(T_A,p_A)$ to $(T_B,p_B)$.


How to calculate the vapor pressure of water?

It doesn't matter if there is a container.  I'm picturing bubbling the nitrogen through the water, for example the water is in a flask, and there is a hose going down into the water, maybe a frit at the end of the hose.

The important assumption is that nitrogen reaches equillibrium with the water, which is a big assumption.

But adding 1.2 grams of water would increase the total volume by about 1 L.  I would use the total volume to answer the question, not 48L, but it will only change the answer by about 2%.


Why is fluorine more reactive than chlorine?

It depends on what reaction you are talking about.

In the context of organic SN1 reactions, the reactivity is $\ce{R-F}$ (least) $\ce{< R-Cl < R-Br < R-I}$ (most). [1] This reflects the leaving group ability, with larger atoms better able to accommodate the extra electron.
In the context of radical homolysis of $\ce{X-X}$ the order of reactivity is $\ce{F2 < Cl2 < Br2 < I2}$ as the sigma bonds become weaker as you move down the period.[2]
In polar aprotic solvents (DMSO) nucleophilicity follows $\ce{F- > Cl- > Br- > I-}$. This order reverses in polar protic solvents due to solvation: J. Chem. Educ., 1970, 47 (6), p 473
Transition metal $\ce{MX_n}$ bond enthalpies have been measured empirically and calculated: J. Chem. Phys., 1957, 26 (6), 1644–1647.. It would be inappropriate to over-generalize, as any discussion here needs to also account for metal oxidation states, HSAB theory, electronic configurations, and ligand-field theory

References:

Clayden, Greeves and Warren Organic Chemistry, p. 348.
Clayden, Greeves and Warren Organic Chemistry, p. 971.


What non-poisonous chemicals can get rid of SLS effectively?

SLS is highly soluble in water (100g/L)

Therefore, in the specific case of your teeth having lots of bubbles around them after brushing, you should simply gargle your mouth with water to remove the excess SLS.


The air in the balloon will be saturated (with water vapour) and the air outside not. As a result there will be a vapour density (or partial pressure) gradient between the inside and outside of the balloon that will drive the mass flow of water vapour through the balloon and to the outside.


What is the meaning of scores when comparing reaction classes in the KEGG database

And does [the score] signify the bonding strength of how probably it is for a pair of compounds to be an R pair?

No, it has nothing to do with that. It is not about how probably a particular reaction is.

When comparing RCs (reaction classes), the scores in the table of related reaction classes are just a measure on how similar these reactions are.

RC00064 apparently is an N-acylation or the formation of a peptide bond. RC00197 with a score of 1.0 is a N-formylation and considered very similar.

Change the cutoff to a lower value, such as 0.4, and compare the results at the end of the table with RC00064. You'll find dissimilar reaction classes like RC02518, the oxidation of a selenol to a selenic acid.


Verification of variants of the Ebullioscopic formula

This answer expands upon my first comment:

  I am familiar with the version with $\Delta_{\text{vap}}H$, and in the derivation somewhere I remember that the Gibbs-Helmholtz equation is invoked and that is the origin of this term.

I have tried to perform the derivation I alluded to, and it is given:

Assuming a binary mixture, A denotes the solvent and B denotes the solute. Superscript (*) implies pure solvent without solute (e.g $T^*$ is boiling point of pure solvent). Also, I assume the enthalpy of vaporisation doesn't change over the temperature range we are interested in.At equilibrium the chemical potential obeys:

$$\mu^*_A(g) = \mu^*_A(l) + RT\ln(\chi_A)$$
$$ \ln(\chi_A) = \frac{\mu^*_A(g)-\mu^*_A(l)}{RT} = \frac{\Delta_{\text{vap}}G}{RT}$$

Differentiating with respect to $T$ and invoking the Gibbs-Helmholtz equation

$$\frac{\mathrm{d}\ln(\chi_A)}{\mathrm{d}T} = \frac{\mathrm{d}(\Delta_{\text{vap}}G/T)}{R\mathrm{d}T} = -\frac{\Delta_{\text{vap}}H}{RT^2}$$

Integrating from $\chi_A = 1 \ \text{to an arbitrary value}$

$$\int_{0}^{\ln(\chi_A} \mathrm{d}\ln(\chi_A) = \frac{-1}{R}\int_{T^*}^{T}\frac{\Delta_{\text{vap}}H}{T^2}\mathrm{d}T$$

$$\ln(\chi_A) = \ln(1-\chi_B) = \frac{\Delta_{\text{vap}}H}{R}\left( \frac{1}{T}-\frac{1}{T^*}\right)$$

For dilute solutions, $\chi_B < 1$ so the approximate expression

$$\chi_B =\frac{\Delta_{\text{vap}}H}{R}\left( \frac{1}{T^*}-\frac{1}{T}\right)$$

Since, $T \approx T^*$
$$\left( \frac{1}{T^*}-\frac{1}{T}\right) = \frac{T-T^*}{TT^*} \approx \frac{\Delta T_b}{T^{*2}}$$

Now, we can rearrange to get the desired expression,

$$ \Delta T_b = \overbrace{\left(\frac{RT^{*2}}{\Delta_{\text{vap}}H}\right)}^{K_b} \chi_B$$

  Moreover, $\Delta_{\text{vap}}H$ and $L_v$ have the same physical meaning, and only the dimensions are different ($\mathrm{kJ \ mol^{−1}}$ vs. $\mathrm{kJ\ kg^{−1}}$  The multiplicative factors that result from the conversion between the two, can just be absorbed into the constant itself,I believe. So I am guessing the tabulated values might differ based on definition. Plus, I think it is an empirical parameter anyway.

Note, that the relation given above is in terms of molar fraction however it is easier to work with molalities when you are doing experiments. The same relation can thus be reformulated

$$\Delta T_b = \tilde{K_b}b$$

and this $\tilde{K_b}$ has the definition which involves $L_v$, I believe. (just look at the dimension in both the expressions and you'll see).


Why can't helium be solidified at 'ordinary' pressures?

When the temperature of helium gas is decreased to about 5.2 K, a phase transition to ordinary liquid helium ($\ce{He}$ I) occurs. The behavior of this liquid phase is normal and identical to any other liquid phase.

As temperature is decreased more (at moderate pressures), helium does not solidify. In fact, it undergoes a phase change to a second liquid phase known as $\ce{He}$ II at a temperature of approximately 2.17 K. As shown in the phase diagram below, the $\ce{He}$ II phase persists until absolute zero. Due to quantum effects, $\ce{He}$ II has remarkably unique properties. It is considered a superfluid.

For more details, please click here. Pressure is required to produce solid helium (25 atmospheres or more), because of the small mass and extremely weak forces between the helium atoms.


As per the official IUPAC guidelines, namely Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
     (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
     (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The directly relevant part here is rule (g), which states that the substituent cited  first gets the lowest locant. Since substituents are cited in alphabetical order, this indirectly means that the lowest locant is assigned to the substituent which comes first alphabetically, i.e. 'bromo'.
Note that this is only rule (g), and that in general rules (a)–(f) should be applied first (their full text can likely be found elsewhere on Stack Exchange). In this case, rules (a)–(d) are irrelevant and rules (e)–(f) do not distinguish between the two proposed numbering systems, so (g) is used as the tiebreaker.
In any case, the preferred IUPAC name of the above molecule is (2E)-2-bromo-1,4-difluoro-3-iodobut-2-ene. [Technically, the locant of the double bond must be specified with the stereodescriptor, hence (2E)- instead of just (E)-, although in practice this is often ignored when there is only one double bond.]


A process is thermodynamically reversible if it is essentially at equilibrium. Specifically, the system and its surroundings stay infinitesimally close to equilibrium with each other throughout a reversible process. Small changes in intensive variables of the system are perfectly balanced by changes in those variables in the surroundings. For example, $T_{\text{system}} = T_{\text{surroundings}}$ if heat can be exchanged reversibly; any temperature difference will result in an irreversible exchange of heat.
What's special about reversible heat transfer? When a tiny amount of heat $dq_{\text{rev}}$ is released by the system during a reversible process, the surroundings will absorb exactly that amount of heat, no matter what path the process takes.

If you have an irreversible process, the amount of heat (and work) depends on how the system and surroundings change during the process.
What's the difference between reversible and irreversible heat? The heat transferred by an irreversible process $dq_{\text{irrev}}$ is less than $dq_{\text{rev}}$, where $dq_{\text{rev}}$ is the heat that would have been transferred if the process had been performed reversibly. The difference between the two is the amount of work you'd have to do on the system to reverse the effects of the irreversible process.

How do we know that? Energy $U$ is a state function, so it'll be the same whichever way we do the process:
$$dU = dq_{\text{irrev}} + dw_{\text{irrev}} = dq_{\text{rev}} + dw_{\text{rev}}$$
so the difference between the two heats is
$$dq_{\text{rev}} - dq_{\text{irrev}} = dw_{\text{irrev}} - dw_{\text{rev}}$$

The differences on either side are positive, so $$dq_{rev} > dq_{irrev}$$
This means that an irreversible process will always absorb less heat from the surroundings and perform less work on the surroundings than it would have if the process had been done reversibly.
Why $dS = dq_{\text{rev}}/T$? With a reversible process, $$\left(\frac{dq_{\text{rev}}}{T}\right)_{\text{system}} = -\left(\frac{dq_{\text{rev}}}{T}\right)_{\text{surroundings}}$$
so if we define $dS \equiv dq_{\text{rev}}/T$, then we have $dS_{\text{universe}} = dS_{\text{system}} + dS_{\text{surroundings}} = 0$.

Consider what would happen in an irreversible process that took the system from state A to state B. The inequality $dq_{\text{rev}} > dq_{\text{irrev}}$ together with the definition of entropy in terms of reversible heats gives
$$\Delta S = S_B - S_A = \int_A^B\frac{dq_{\text{rev}}}{T} > \int_A^B\frac{dq_{\text{irrev}}}{T}$$ for any such process. This also implies that for any cyclic process (e. g. A to B back to A), we have $$\oint\frac{dq_{\text{irrev}}}{T} \lt 0, \ \text{and}\  \Delta S = \oint\frac{dq_{\text{rev}}}{T} = 0$$
This shows that $q_{\text{rev}}/T$ is a state function (it's zero for any cyclic process) but  $q_{\text{irrev}}/T$ is not a state function. If entropy is to be a state function, then, you can't define it as simply $q/T$; you've got to use $\Delta S = q_{\text{rev}}/T$, or for the differential form $dS = dq_{\text{rev}}/T$.
How could we relate the entropy change for an irreversible process to $dq_{\text{irrev}}$? From the difference between $q_{rev}$ and $dq_{irrev}$ above and the definition of entropy, we have
$$dS = \frac{dq_{\text{rev}}}{T} = \frac{dq_{\text{irrev}}}{T} + \frac{dw_{\text{irrev}}-dw_{\text{rev}}}{T}$$

TL;DR: There IS no TL;DR; you're asking some important questions here, and the answers need to be made carefully.


What is the meaning of a decimal in a subscript in a chemical formulas?

In the case of complex glasses such as $\ce{Co_{28.6}Fe_{12.4}Ta_{4.3}B_{8.7}O46}$, Instead of normalizing the subscripts to have integer subscripts they normalized them by molar percent such that the sum of subscripts is $100\%$  ($28.6+12.4+4.3+8.7+46=100$)

This is fairly common practice in materials science. For many complicated materials such as this, it's hard to gain useful meaning from the chemical formula if the subscripts are normalized to integers so they are normalized to 100. People reading this formula now can compare it to similar materials they have seen and future materials are also more comparable (e.g. if they substituted the tantalum with niobium).


Contribution of Tetrahedral and Octahedral voids in HCP

I cannot find anywhere..

Couldn't find? You can figure it out on your own, with me, right now.

Disclaimer: By "hexagonal unit cell" I assume you mean hexagonal prism, which comprises of 3 primitive hexagonal Bravais lattice.

Octahedral Voids (OhV)

The one on the left displays octahedral voids(visibly surrounded by 6 atoms).
The OhV is at the center of the highlighted octahedron, and lies on the plane containing four of the six lattice points in the octahedron.
Note that these lattice points belong to the unit cell, and the OhV must hence lie inside the unit cell.

This is the same for the other 5, amounting to a total of 6 OhV per HCP unit cell.

Being completely inside, its contribution is taken as 1.

BONUS: If you remember, No. of atoms per unit cell is 6, which
tallies with the no. of OhV given by the formula $N_{OhV}=N$ where
$N$ represents no. of atoms per unit cell.

Tetrahedral Voids (ThV)

I'm going to be frank: it IS not as easy as locating OhV.
The figure above absolutely freaked me out when I tried to visualize ThV's. I'm not keen on recommending that.
Instead, this 3D rendition should help immensely.

Type 1 ThV

The Type 1 Thv's are the ones inside in unit cell completely(so again, contribution=1).
Notice that I have marked 2 tetrahedrons. Also note that each tetrahedron we form will tell us about the ThV its existence creates.

RED: Note that there will be another 2 like this one in the lower half. That makes a total of 6 Red-type ThV in the entire HCP unit cell.
GREEN: Note that there will be a total of 2 of the green-type ThV, the other one belonging to the inverted tetrahedron just below it.

That gives us a total of $6+2=8$ Type 1 ThV.

Type 2

Let's do a little recap to see which ThVs have we counted so far(displayed in red)

I constructed a 3D model of 3 HCPs, which is required to visualize the remaining.

Now I will mark the ThV forming lattice points in red.

As you can see, lattice points from 3 unit cells are involved.

A quick analysis tells us that there will be 2 such Tetrahedrons for every vertical edge.
A quicker analysis tells us that since each ThV so formed is present in 3 HCP unit cells, or is "shared" by them, Its contribution must be $\frac 13$.

We count the number of such tetrahedrons to be 12 per unit cell; with contributions, ThV Type 2 will be $12*\frac 13$ $=4$.

Ultimately, we get:

ThV type 1= 8

ThV type 2= 4

Total ThV= 12.

Which brilliantly tallies with the formula for number of ThV($N_{ThV}=2N$).

I hope managed to enlighten you.

TL;DR

OhV contribution is 1, no issues there.
ThV does not have a single "contribution", but rather could be 1 or
$\frac 13$ depending on which ThV you're talking about.


How can I balance the following equation atomically and electrically?

Deal with the two half equations separately and then combine them.

Starting with the oxalate equation: $$\ce{C2O4^{2-} -> CO2}$$

Balance the atoms: $$\ce{C2O4^{2-} -> 2CO2}$$

Now balance the charge by adding electrons: $$\ce{C2O4^{2-} -> CO2 + 2e-}$$

Now for the manganese dioxide reduction: $$\ce{MnO2 -> Mn^{2+}}$$

We can balance the half equations by adding water, hydrogen ions or electrons. Since water is the only one which contains oxygen we should add this first: $$\ce{MnO2 -> Mn^{2+} + 2H2O}$$

Now add hydrogen ions to balance the hydrogens: $$\ce{MnO2 +4H+ -> Mn^{2+} + 2H2O}$$

Finally add electrons the balance the charge: $$\ce{MnO2 +4H+ + 2e- -> Mn^{2+} + 2H2O}$$

Now combine the equations to cancel all the electrons. In this case there are two on both sides so we don't need to multiply any of the equations by anything.

$$\ce{MnO2 +4H+ + C2O4^{2-} -> Mn^{2+} + 2CO2 + 2H2O}$$


For starters, I think most chemists would agree that a chemical change is different from a physical change by involving the formation or breaking of chemical bonds, or changes to the charge, coordination, or bonding state of some of the atoms involved. A chemical change can be reversible, however, so reversibility is not a determining factor. Either a physical change or a chemical change could result in a change of color, or not--as the case may be.

In your examples, the absorption of water by a disposable diaper, as well as the color changing of liquid crystal-based displays (like the beer temperature indicator, thermometer, and mood ring) are clearly physical changes. The liquid crystals change their optical properties due to the physical alignment of the molecules, and not because of making or breaking of bonds. On the other hand, the pregnancy test and blood detection test probably involve chemical changes to produce a color readout (sometimes initiated by a relatively loose binding between an antigen and an antibody, or formation of a metal complex). The photochromic lenses turn dark as a result of a chemical change induced by light (according to Wikipedia, anyway--this is not my area of expertise).


Is it possible to recycle an alkaline battery?

Yes, in theory. Everyready has 4% recycled materiel in some of its batteries.

  The application I am thinking about is that in 3rd world countries often access to western markets is limited. You can't just pop into a store and buy batteries. Would it be possible for a small factory to be in these places and the locals could bring their batteries in for recycling?

No. The chemistry and processing involved would be far to complicated for some sort of rudimentary factory. There is also the economic factor. Even with a well educated work force such a factory would be very expensive to run and need a massive number of used batteries to recycle.

Part of the expense that recycling would offset is the cost of disposing of the used batteries. For example NiCd batteries in a landfill are bad news because of the toxicity of cadmium. Third world countries generally have much poorer environmental regulations than the developed countries.

To answer the additional question...

  The reactions for an alkaline battery are the conversion of zinc to zinc oxide and MnO2 to Mn2O3. So, to recycle a battery of this type you would have to reverse these reactions. Is that easy to do?

No it isn't easy. You can't just mechanically separate the zinc component of an used battery from the manganese component. You'd have do something like dissolve the mess with acid then separate and process the two elements.


Geometry of the transition metal complex in solid dipotassium bis(oxalato) copper(II) tetrahydrate

Yes, square brackets are used to denote a coordination sphere, but they may sometimes be used incorrectly. Ideally you want to check the crystal structure, luckily there is one [1]:

The $\ce{Cu(II)}$ atom has square-planar geometry with two bidentate oxalate ligands. Each $\ce{K+}$ cation is coordinated by eight oxygen atoms from one bidentate and two unidentate oxalate groups and four water molecules.

Fig. 1. (a) Crystal structures of $\ce{[K2Cu(ox)2 · 4 H2O]}$ (1) and (b) $\ce{[K2Cu(ox)2 · 2 H2O]}$ (2) with atom numbering scheme. The thermal ellipsoids are drawn at 50% probability.

Reference

Fan, J.; Sun, W.-Y.; Okamura, T.; Yu, K.-B.; Ueyama, N. The X-Ray Crystal Structural Characterization of Dipotassium Bisoxalato Copper(II) Tetrahydrate, $\ce{[K2Cu(Ox)2·4 H2O]}$ $(\ce{Ox}$ = oxalate Dianion). Inorganica Chimica Acta 2001, 319 (1), 240–246. DOI: 10.1016/S0020-1693(01)00466-2.


Car Parinello advantages and disadvantages

Car Parinello is an Molecular Dynamics (MD) model. The MD models of chemicals I've worked with use a projection of the expected wave-function using classical mechanics to approximate what the atoms and molecules in the model are doing. Essentially assigning a force constant based on distance between the atoms in the system to approximate the energies involved in the simulation. This kind of simulation can give semi-qualitative numbers to energies of association or dynamics of a reaction in some cases where the motions of all the atoms are essentially classical in nature. However, if there is proton motion involved in the reaction coordinate or if you would like chemically meaningful energies of association MD models will not be workable for your goals. MD models can, typically, only get a "feel" for whether or not an association is possible or if a reaction could take place. And if there is a quantum mechanically allowed and classically forbidden step, say proton tunneling, the model will fail.

So MD models are great for very large or distributed systems. For example modelling a solute in solvent or large solid systems, both of which are beyond most ab initio Quantum Mechanics (QM) calculations due to the number of individual calculation steps needed for convergence. There are some "high level" (nit-picky) calculations that require precision beyond what a classically built model can give which is why DFT and ab initio calculations are still run today.

Now I have not ever used Car Parinello in my work, (I'm part of one of those nit-picky groups) so I cannot give any real numbers for the errors in this method as compared to other MD simulations. I would, however, say that the writer of the Wiki page is not un-biased (neither am I, see above regarding nit-picky group) and is trying to show how great this method is compared to other MD simulations.  This particular MD simulation seems to be using some SCF calculations and long range basis set functions for outer shell electrons and so I think it is mainly for use in solid ionic systems rather than the small isolated molecules my group typically models.

For DFT (or essentially all modern QM calculations) the electronic wave function is approximated using basis functions. These are pre-programmed functions that can be mixed together with little computational effort to approximate the true wave function. The basis sets used by Car Parinello are plane wave basis function which means the wave function repeats at long range and allows (forces) assignment of electrons far removed from one another. This means that a self consistent state is quickly reached for solid crystalline systems.

My personal bias (use a grain of salt for this section) is against such basis sets due to the fact that it forces a pseudo-crystalline pattern on the system (at least at long range). For most chemically interesting systems it is the oddities on a surface or the internal structure which give rise to chemical change. And so, this model may fail to represent these systems well due to the choice of basis set. Once again, not having used this model I cannot say it would fail to represent these interesting features, I just doubt it would give reasonable energies or wave functions after completing the calculations.


How does pH change during the electrolysis of water?

The pH will decrease, because magnesium sulphate is acidic.

As the electrolysis proceeds, the magnesium sulphate becomes more concentrated.


How can a decomposition reaction be exothermic?

It just can be so.  Look for something that is favored by cooling instead of heating.
In the case of decomposition of something to pure elements that is exceedingly rare, but decomposition involving other compounds offers more possibilities.  During hot processing of steel (in the hot rolled coil, actually) we can get this reaction which is favored by cooling below 570°C:
Wustite --> Magnetite + Iron
$4 \ce{Fe_{1-x}O} \rightarrow \ce{Fe_3O_4}+(1-4x)\ce{Fe}$
$x$ typically 0.05-0.10.
See the phase diagram here noting what happens to the $\ce{Fe_{1-x}O}$ phase on cooling.


How different are molecule with same connectivity

In my experience, calculated parameters like topological indices are not viewed as significant if considered alone. It's a strategy for finding trends or correlations and, subsequently, candidates from  large volumes of chemical structures. Many studies will publish trends they find in a target chemical groups which are not meant to apply to every chemical group.

An example of how all these seemly useless data is useful - see the Merck Kaggle Challenge.
https://www.kaggle.com/c/MerckActivity

http://blog.kaggle.com/2012/11/01/deep-learning-how-i-did-it-merck-1st-place-interview/

Statistical significance is harder to comment on. I work with between 8 and 20 million data points and the idea of how to define statistical significance is evolving in cheminformatics as it evolves in big data analytics. It gets much more complicated because standard methods, such as the t-test, fail.

NOTE
I just realized there might be some fundamental confusion. Mol files include atomic information as do most graph based chemical representations. They're stored as properties of the nodes. An algorithm can choose to use or ignore this information. Not all topological parameters ignore atomic information.


Why is phosphorous acid diprotic and not triprotic?

The two isomers you're referring to both exist and are in equilibrium:

When only one or two protons are removed, resonance stabilizes the anions formed, e.g. the $\ce{HPHO3^-}$ anion:

The trianion (shown below) doesn't have the benefit of resonance stabilization. This makes removing the third proton much harder.


NaCl water electrolysis without membrane, how measure the free chlorine concentration?

The best way for determining the active chlorine amount ($\ce{ClO^-}$) in a bleach is to dilute it with water, then to add Potassium iodide $\ce{KI}$, and enough HCl solution to make it acidic. It produces $\ce{I_2}$ by the equation $$\ce{ClO^- + 2 H^+ + 2 I^- -> I_2 + H_2O}$$ Then the $\ce{I_2}$ so produced is titrated by a standard solution of sodium thiosulfate $\ce{Na_2S_2O_3}$, according to : $$\ce{I_2 + 2S_2O_3^{2-} -> 2 I^- + S_4O6^2-}$$

A quicker (but not so precise) way is to add 2 mL acetone to 50 mL bleach in a insulated flask, and to measure the change of temperature due to the reaction. The reaction produces chloroform $\ce{CHCl_3}$ and acetate. After one minute, the temperature increases to a stable final value. The change of temperature is proportional to the amount of chlorine. If $\Delta T $= 10°C, the concentration of active Cl is 5.6%. Generally speaking, the concentration of active Chlorine is : 0,56% x $\Delta T$


How much "cleaner" does summer fuel burn?

Gasoline, diesel fuel, and ethanol-gasoline blends are adjusted seasonally and geographically to ensure proper starting and performance. Winter blends contain higher levels of high vapor pressure compounds like butane to ensure the fuel is sufficiently volatile to ignite in cold temperatures. This in turn makes the gasoline cheaper to produce as it does not have to be as pure as the summer blend$^{[1]}$.

Conversely, summer blends contain lower levels of high vapor pressure liquids to ensure that the fuel doesn't evaporate away in the heat and pollute the atmosphere. This is why summer blends are marginally cleaner than winter blends. A study on winter versus summer emissions near tunnels form the University of Wisconsin collected the following data on non-methane hydrocarbon emissions$^{[2]}$:

As for fuel efficiency, winter-grade gasoline yields $1.7$% less energy on average than the summer. This may seem like a considerable amount, but energy content can vary widely from station to station in either season from $3.4$-$4.8$%$^{[3]}$.

$^{[1]}$ The Oil Drum
$^{[2]}$Univeristy of Wisconsin, Summer and winter nonmethane hydrocarbon emissions from on-road motor vehicles in the Midwestern United States
$^{[3]}$ EPA, Fuel Economy Impact Analysis of RFG


Sum of spins of electrons in a wavefunction

I think the reason you're getting confused is because when we talk about "spin", we actually are referring to two separate observable values, the z-component of the spin (properly designated $\hat{S}_z$) and the square of the magnitude of the three-component spin vector (properly designated as $\hat{S}^2$).

The commonly used values of $+\frac12$ and $-\frac12$ are shorthand for $+\frac12 \hbar$ and $-\frac12\hbar$, the only values possible to measure for $\hat{S}_z$.

Because of quantum uncertainty requirements, we can know the values of both the z-component and the square of the magnitude of spin at the same time, but we cannot know either the x- or y-component while we know the z-component. As a result, we can only say that the spin vector lies somewhere on a the surface of a cone centered on the  z-axis that expands upwards from the origin ($+\frac12$) or downwards from the origin ($-\frac12$).

For two uncorrelated spins (like in a triplet state), the z-components either add (2 positive) or (2 negatives) or cancel (1 positive and 1 negative), but that only means that the net z-component is 1, 0, or -1. Even if $\hat{S}_z=0$, there is still a net spin from the x- and y-components. The value of the magnitude (or technically its square) comes from $\hat{S}^2$, which is conserved as 1 for all three possibilities. That's the S value given in your notes. For the $\hat{S}_z=0$ case, that means the spin vector lies in the x-y plane as one of the possible radial vectors of a circle with radius 1 centered on the origin in that plane.

For the correlated spins (singlet state), the magnitude squared is 0, which occurs because the two spin vectors exactly oppose each other, and all three components cancel out, not just $\hat{S}_z$, though we have no measure of the x- and y- components that are cancelling. We only know the resulting magnitude is 0.


Rationalising the reactivity of phosphoryl chloride

AFAIK it's not a particularly common choice, but Vilsmeier reagents are sometimes used to convert alcohols to chlorides (see e.g. Tetrahedron Lett. 1992, 33 (34), 4901–4904). This explains the requirement for DMF/DMAc as solvent.

A reasonable mechanism is as follows:

(see also Org. Lett. 2002, 4 (4), 553–555 for a similar mechanism)


Thermionic Emission Conservation of Energy

I was mistaken on the principle that governs vacuum tubes. It is not that the signal is being amplified by thermionic emission; rather, the radio wave is being used as a gate to allow or block electrons from being received. Thermions are being emitted from a coil using electricity, emitting at a higher current, modulated by the radio wave as a control grid (preventing or allowing the thermions from being collected at the anode).


Aluminum chlorohydrate (ACH) is a group of specific aluminum salts having the general formula $\ce{Al_nCl_{(3n-m)}(OH)_m}$ (Wikipedia). It is used in cosmetics as an antiperspirant and as a coagulant in water purification. Specifically in wastewater treatment processes, ACH is used as a coagulant to remove dissolved organic matter and colloidal particles present in suspension. In his paper on Water Treatment Coagulants, Peter Gebbie describes what coagulants do (Ref.1):

  With aluminium-based coagulants, the metal ion is hydrolyzed to form aluminium hydroxide floc as well as hydrogen ions. The hydrogen ions will react with the alkalinity of the water and in the process, decrease the pH of the water as can be seen from Equation [...].  Similarly for ACH, which is described as being a pre-hydrolyzed coagulant, the following reaction takes place:
  $$\ce{Al2(OH)5Cl -> Al2(OH)5+ + Cl- + H2O -> 2Al(OH)3 + H+ + Cl- }$$
  Less hydrogen ions are produced with ACH, reflecting the hydroxylated nature of this compound.

There it says it all: ACH is more basic than other aluminum salts because it is pre-hydroxylated.

Aluminum chlorohydrate is one of the most common active ingredients in commercial antiperspirants. The variation most commonly used in deodorants and antiperspirants is $\ce{Al2Cl(OH)5}$, similar to the one used in water purification.

Note: Your statement of "The pH value of ACH is about 4" is evidently not correct. However, W. R. Grace & Company (producer of solution form of ACH) listed solubility of ACH as $\pu{1000 g/L}$ at $\pu{20 ^\circ C}$ and $\mathrm{pH} = 3.3$, the optimum condition.

Reference:

Peter Gebbie, “An Operator’s Guide to Water Treatment Coagulants,” 31st Annual Water Industry Workshop – Operations Skills; University Central Queensland Campus: Rockhampton, Australia, 4 to 6 July, 2006, pp. 14–20 (http://wioa.org.au/conference_papers/06_qld/documents/PeterGebbie.pdf).


What are standard controllers for automatic pH control?

Automatic pH control is not that easy as is shown by the fact that automatic titrators, the equipment that you want to build, are relatively expensive.
Besides titration, automatic titrators usually have also a pH stat function, which is for maintaining a certain fixed pH.

In first place, note that to  have a good control, typically you must know which type of system you are working with.

Normally, to maintain a constant pH only one reagent is added, either acid or base.
This is because chemical systems typically become more acidic or more alkaline when they proceed, but do not oscillate.
When you add both reagents, they usually start to compete: a bit of acid is added, it passes the set point, base is added, it passes the set point in the other direction, etc.
This happens especially when the response of the system is slow and the mixing is not good (shaker plate is not a good idea, use a stirrer instead).
The way to solve this is by adjusting very well the concentrations of the reagents, the delay times, addition rates, etc., and this is different in every system.

You also should know in which part of the pH curve you are working to set the concentration of the reagents and the rate of addition.
The pH curve has three very different zones, a vertical part and two horizontal parts.

Depending on where you are, the conditions must be different.
Note also that the inflection point becomes less obvious when weak acids or bases are involved.

Finally, you might need to integrate also a temperature compensation since, depending on the $\Delta\epsilon$ you have set, the pH change with temperature will be to high.

In short, a PID might work in some specific system, but for a general purpose system you will probably need some more sophisticated control like a PLC.


What is the meaning of the "-osic" suffix on a metal (e.g., cobaltosic oxide)?

In the course of searching for examples of usage of the "-osic" suffix, I think I found the answer to my own question.  From the last link in the question, a web store selling pigments (emphasis added):

  The source of natural black oxide usually is from an iron ore called magnetite. Magnetite, also known as lodestone, is a ferrous ferric oxide (ferrosic oxide) that is a heavy, black color and opaque.

So, it would appear that the suffix indicates a compound of mixed oxidation state: a combination of -ous and -ic forms of the metal.


If the hydroxide ion is such a poor leaving group then why does water dissociate?

Only 1 in 10,000,000 water molecules dissociate.

HI, HBr and HCl fully dissociate, and I-, Br-, and Cl- and are much better leaving groups.


Why can't two Silicon tetraoxide tetrahedrons share an edge or a face?

In solid-state chemistry, we know a certain set of rules called the Pauling rules. One of them states that sharing faces or edges is unfavorable. Imagine your silicate-tetrahedra to share faces. That would equal the closest distance between two highly charged silicon cations. They both have a charge of +4, so there is really some repulsion. Same goes for the edges which is also quite close.

If you increase the overall size of the tetrahedron however by exchanging oxide by sulfide you will end up in the silicon(IV)sulfide structure which is essentially long chains of tetrahedra that share edges. Here by introducing a much larger anion the distance between the two silicon centers is increased and edges can be shared.


Why the optimization in gaussian shows no bond with some atoms in diborane?

I wouldn't put too much faith in the representation of bonding in GaussView, it's simply determined by bond distances and often it doesn't show the bond at all for some unknown reason. If your geometry is good, then you can use your knowledge and intuition of bonding to recall the true structure.


Now, does that imply that if the value of $|\Psi|^2$ is maximum at a
  point, the electron is most likely to be found at that point?

It certainly does imply something similar to what OP is saying, but not quite. As always in quantum theory one has to be very careful with his or her wordings. Namely, we are talking about a continuous random variable here, so the only probability which makes sense is that of it falling within a particular range of values, i.e. probability of finding an electron in some infinitesimal element of volume $\mathrm{d}V$ centered at a particular location $r$ and not on this single point. Still, if you divide the whole space into volume elements of an identical (infinitesimally small) size, the bigger the value of $|\Psi|^2$ is, the bigger is the above mentioned probability.

Now to the actual findings. So, for an electron occupying the $\mathrm{3d}_{z^2}$ orbital, the probability of finding it in an infinitesimal element of volume $\mathrm{d}V$ centered at a particular location $r$ is given as $|\Psi|^2 \mathrm{d}V$, where $|\Psi|^2$ is the probability density. We analyze the probability density
$$
|\Psi|^2
=
N^2 \sigma^4 \mathrm{e}^{-2\sigma/3} (3\cos^2\theta-1)^2 \, ,
$$
or, more precisely only its angular $(3\cos^2\theta-1)^2$ part. So, when does this expression have its maximum value? The global maxima are at $\theta = 0$ and $\theta = \pi$, i.e. alongside $z$-axis, plus there are local maxima at $\theta = \pi/2$ and $\theta = 3\pi/2$, i.e. in a plane perpendicular to $z$-axis. 1

From the discussion with OP, it seems it must be noted that terms "probability density" and "probability" are defined with respect to a particular random event. Above we analyzed the probability density of finding an electron in an infinitesimal element of volume $\mathrm{d}V$ centered at a point $r$. The radial distribution function arises when considering a probability of a different event, namely, the probability of finding an electron between two concentric spheres of radii $r$ and $r + \mathrm{d}r$. We start from the probability of finding an electron in infinitesimal volume element $\mathrm{d}V$ at a point $r$
$$
  |\Psi|^2 \mathrm{d}V
  =
  \underbrace{R_{nl}^2(r) |Y_l^m(\theta, \phi)|^2}_{|\Psi|^2}
  \underbrace{r^2 \mathrm{d}r \sin\theta \mathrm{d}\theta \mathrm{d}\phi}_{\mathrm{d}V} \, ,
$$
and integrate out $\theta$ and $\phi$ dependences
$$
  \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi}
  |\Psi|^2 \mathrm{d}V
  =
  \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi}
  R_{nl}^2(r) |Y_l^m(\theta, \phi)|^2
  r^2 \mathrm{d}r \sin\theta \mathrm{d}\theta \mathrm{d}\phi \, .
$$
And since the spherical harmonics $Y_l^m$ are normalized (to unity)
$$
  \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi}
  |Y_l^m(\theta, \phi)|^2 \sin\theta \mathrm{d}\theta \mathrm{d}\phi
  =
  1 \, ,
$$
for the probability of finding an electron between two concentric spheres of radii $r$ and $r + \mathrm{d}r$ we simply get
$$
  R_{nl}^2(r) r^2 \mathrm{d}r \, .
$$
The corresponding probability density is, of course, $R_{nl}^2(r) r^2$. Such analysis is different from what we did above for the angular $\theta$ dependence: we simply ignored the radial and the other angular dependence instead of integrating them out. This ignorance makes our analysis a bit rough, but not worthless: at least we were able to identify some general trends in the angular $\theta$ dependence.

1) See Wolfram Alpha for details.


Why would a reaction favor one isotope over another?

Chemically speaking compounds containing isotopes are very similar but not exactly similar

Broadly when compounds contain different isotopes of the same element (say different isotopes of oxygen) their reactivity is the same. But this is a little over broad. There are subtle and (usually) small differences and these can show up as small differences in chemical reactivity.

To get an idea why consider the simple physical properties of water and heavy water. In heavy water the two hydrogens are replaced by deuterium (hydrogen with an extra neutron in the nucleus). A heavy water molecule is about 10% heavier than a normal water molecule (molecular mass ~20 rather than ~18). As a result the physical properties are a little different: bp is about 2°C higher, mp about 4°C higher and the liquid is about 10% denser. So careful techniques can separate the two (by repeated fractional distillation, for example).

The slight differences also appear in chemical reactions. The reasons why are more complex than the explanation of simple physical effects. But one hint that they exist is simply that the vibrational frequency of a bond depends on the mass of the atoms making up the bond. Isotopes have different masses so there will be some difference with different isotopes. And bonds vibrating are part of the mechanism of reactions involving that bond so the reactions will have different rates. The detailed theory is pretty complex (see this) and the size of the effect depends on the detailed mechanism of the reaction.

Reaction rates for some reactions involving bonds to hydrogen can differ by large factors (because deuterium is twice as heavy as hydrogen) and are sometimes 10 fold. Most other isotopes have only small relative mass differences compared to their parent element and the net rate differences are often only small single-digit percentages. This results in some reactions concentrating some isotopes over others (eg 18O vs 16O. But these can be large enough to detect by mass spectroscopy.

So, despite their chemical similarity, some reactions and some physical processes will favour one isotope over another (though often only very slightly). If you understand the key reaction mechanism, the small differences in the concentrations can tell something about the conditions of the reaction (temperature dependence, for example).


Finding the density of nitrogen at 0 degrees Celsius and 1 atmosphere

Using the ideal gas law
\begin{align}
  pV &= nRT, &
  \frac nV &= \frac{p}{RT},
\end{align}
where
\begin{align}
  T &= \pu{273 K},\\
  p &= \pu{1 atm},\\
  R &= \pu{0.0821 L atm mol-1 K-1},\\
  M(\ce{N2}) &= \pu{28.0 g mol-1},
\end{align}
we get
\begin{align}
  \frac{n}{V} &=
    \frac{\pu{1 atm}}{\pu{0.0821 L atm//mol K} \times\pu{273 K}},\\
  \frac{n}{V} &= \pu{0.0446 mol//L},
\end{align}
and finally
$$
\rho(\ce{N2})
  = \pu{0.0446  mol//L} \times \pu{28  g//mol}
  = \pu{1.2  g//L}.
$$
Therefore answer (d) is correct.


This picture is quite bad, since there are patterns inside the gray areas which shouldn't be there but the gray areas are the orbitals. You could think of it like writing down the position of an electron over and over again and every dot is a position where you found the electron. More dots means it was in that area more often (=higher probability of finding it there).

An electron, theoretically, can be everywhere except the nodes. So an electron from an atom right in front of you could be lightyears away. The chance is very, very, very low for that to happen but it is possible. So for having a 100% chance of finding the electron you have to look in the whole universe, of course it has to be somewhere. That's not very useful at all, since every electrons can be found somewhere in the universe so we don't work with 100% chance but rather with 95% ( or 99% or 99.9%,..) since that's already a probability of finding the electron very, very close to the core of the atom and it is there most of the time.


There is some minor argument as far as nomenclature goes for these lipids. Most sources I can find, including IUPAC have glycerides as only including esters of glycerol with fatty acids. Monoglycerides have a single fatty acid ester with two free OH groups, diglycerides have two fatty acid esters, and triglycerides have three. Phosphoglycerides are defined separately as esterified derivatives of phosphatidic acids, so you can think of them like glycerides where for one of the substituents, instead of a fatty acid, some derivative of phosphoric acid is present. However, the way the two terms are defined, they don't encompass any common molecules, i.e. there are no molecules that are both glycerides and phosphoglicerides and neither group is a subset of the other.

So: mono-, di-, and triglycerides are subsets of glycerides while phosphoglycerides are a different group altogether, though structurally similar.


What is the difference between an ionic compound and a salt?

The GCSE answer is.
An ionic compound is one in which the bonding between the atoms is ionic, for example sodium chloride is an ionic compound. It is formed from sodium cations and chloride anions.
There are substances which can be regarded as being ionic which are not normally viewed as salts. For example some metal oxides such as potassium oxide are probably best described as being ionic solids but in this case the compound is formed from oxygen and potassium.
A salt is the product of reacting an acid and a base together. Many salts are ionic. For example if you were to react hydrochloric acid with zinc and then to dehydrate the product (avoiding hydrolysis) you would end up with zinc chloride.
The PhD chemists view of the matter
One great problem in chemistry is that there are very few perfectly ionic compounds. I hold the view that the more chemistry you know the more one becomes aware of the fact that simple explanations do not work well.
The great problem in my mind is that a great spectrum of bonds exist from perfectly ionic to simple unpolarised covalent bonds. All cations have some ability to pull electron density towards themselves. A simple anion such as a chloride should in the idea world be a sphere of electron density.
The electrostatic attraction of the electron density by a cation on an anion pulls electron density towards the cation. This can be regarded as leading to some covalent character of the bond existing. Thus as a result I hold the view that there are very few true perfectly ionic compounds.
One example would be tetraphenylphosphonium tetraphenylborate which has two very large ions, the interaction between these ions is close to a perfect ionic attraction. Here is a diagram of the tetraphenyl phosphonium cation.

We can see here that the ion has the positive charge deeply buried inside it, here is the solvent accessible surface, I have rendered it in colours to represent the charges at points on the solvent accessible surface.

The atom locations can be seen in the following picture to help you understand what is happening.

Even the tetramethyl phosphonium cation is able to isolate the positive charge away from the solvent accessible surface, in contrast the trimethyl sulfonium cation (Me3S+) presents to the outside world (the solvent accessible surface) part of the ion which is quite positive.

We can also take a molecular orbital view of bonding, a lot of coordination chemistry of transition metals is considered with the ligand field model. The ligand field model is based on the crystal field model which treats all bonding as ionic. However the same results can be obtained for d block metals using molecular orbital theory.
It is also interesting to note for the lanthanides that they do not have strong ligand field splitting effects, this is due to the fact that their f orbitals are close to the centre of the atom and do not interact well with ligands. In many compounds often the bonding of lanthanides is viewed as being ionic.
If we consider for a moment palladium chloride then it is clear that the attraction of the chlorides to the palladium cations results in the formation of covalent bonds. If we move back from palladium to a metal which is less able to form bonds to chloride (cobalt).
Then if we have anhydrous cobalt chloride, we have in the anhydrous solid a Co-Cl distance of 2.453 Å. If we dissolve cobalt chloride in water then the water molecules will react with the cobalt(II) centres to form hexaaqua complexes of cobalt [Co(H2O)6]2+ which have a larger radius than cobalt(II) cations. These are less able to electrostatically attract chloride anions and the bonding of the chloride anions to the [Co(H2O)6]2+ cations will become closer to being a simple electrostatic (ionic) attraction / bonding than the bonding between the chlorides and the cobalt in the anhydrous cobalt chloride.
Here is a picture of the crystal structure of anhydrous cobalt chloride. (H. Grime and J.A. Santos, Zeitschrift Fur Kristallographie, 1934, 88, 136-141) You can see how the cobalt has a octahedral coordination environment in which six chloride anions form bonds to the cobalt. Keep in mind that the chloride anion is a ligand with a weaker ligand field than water. As a result the cobalt adsorbs red light in the form of anhydrous cobalt chloride.

By the way the solid is a layered solid, the solid can be thought of as being like a series of slices of bread (chlorines) with the cobalt as the jam.

By the way the distance between cobalt and chloride in the [CoCl4]2- complex is 2.276 Å, which is a bit shorter than the distance in the anhydrous solid.


Can I solubilize lemon essence in water somehow?

Some hints are provided by Wikipedia:Lemon liqueur:

To produce the Lemon liqueur requires sugar, water, lemon zest,
mixing in organic salutes  liquor, and time to mature. Lemon zest is soaked in high proof neutral spirits to extract from it the lemon oil (an essential oil). The extraction is then diluted with simple syrup.

In other words, lemon zest is drawn into water by mixing in water-soluble organic compounds, in this case sugar and (ethyl) alcohol.  If you want it to be nonalcoholic, maybe just the sugar will do.
During late autumn I make a cranberry jam by boiling the cranberries with oranges, sugar and water.  In the presence of the sugar and organic material from the oranges and cranberries and with the application of heat, orange zest is extracted into the jam.  The jam is too dark and thick to see whether the zest remains in solution when the jam is cooled, but the extraction of the zest is evident by taste.


clarification on background limits in electrochemistry

the background limits ‘are the potentials where the cathodic and anodic currents start to flow at a working electrode when it is immersed in a solution containing only an electrolyte added to decrease the solution resistance (a supporting electrolyte).

Bard's book is notoriously complex. Some bright scientists are not good at communication. Richard Feynman was an exception.

All Bard is saying that the background limits define the potential window [see flat line in your figure] where electrolysis does not occur on the working electrode. The background limits are dependent on the electrolyte used, it is also dependent on the solvent and the electrode material as well.

Now keep in mind that modern potential sweep experiments there are three electrodes in a cell. There is a reference electrode, in your case, an Ag/AgBr electrode. Then you have a working electrode, Pt/H+ and then the writer is quiet about the third one, which is called the counter electrode. Most of the time it is understood that it is their. It could be another Pt electrode but not necessarily a hydrogen electrode i.e., nobody is bubbling hydrogen gas over the counter electrode.

Now the reference electrode has no role in the cell. There is no redox reaction taking place at the reference electrode because the circuitry is such that no or very little current passes through the reference electrode. Electrochemists don't care what is happening at the counter electrode. The potential difference is measured between the reference electrode and the working electrode. This is the only purpose of the reference electrode.

When you are sweeping (=continuously changing) the potential of the working electrode with respect to the reference electrode, it can become a cathode or an anode during scanning of the potential. You can scan in any direction. You are right, isolated reduction or oxidation cannot occur. If the working electrode was a cathode at a particular moment, the counter electrode is the anode and vice versa.


If one rearranges the ideal gas law equation, you can obtain the following (assuming $n$ and $T$ are non-zero):

$$\frac{PV}{nT} = R$$

$R$ is a constant, and there are in fact infinitely many possible sets of values $(P, V, n, T)$ that satisfy the equation. Let $(P_1, V_1, n_1, T_1)$ denote one such set, and let $(P_2, V_2, n_2, T_2)$ denote a second one. Then, we have:

$$\frac{P_1V_1}{n_1T_1} = R$$
$$\frac{P_2V_2}{n_2T_2} = R$$

So, by the transitive property:

$$\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}$$

You can let $n$ and $T$ be constant as well (as in your specific example), but the principal is the same and leads to an analogous result, since any product and/or ratio of constants will itself be a constant.


Does an increase of area lead to more collisons in a confined jar of gas?

Pressure is force over an area, or as an equation you might state that as:

$$P=\frac{F}{A}$$

Which can be rearranged as

$$F=P \cdot A$$

You are correct in saying that if the number of molecules are held constant, that the force would be over a larger area and therefore decreased. As you increase the area but keep the number of molecules constant, the pressure decreases proportionally, and no change in force is seen, but what Feynman was talking about was keeping the amount of molecules per cubic centimeter constant, or in other words keeping pressure constant. This way as area is increased, you see a proportional increase in force.


qualitatively: Why are thioesters more reactive than regular esters?

http://guweb2.gonzaga.edu/faculty/cronk/CHEM440pub/thioesters.html

"thioester cannot make as much of a contribution because the required overlap between carbon's 2p orbital and the sulfur 3p orbital is ineffective. The large difference in energy between these two orbitals means they cannot "mix" as well as the 2p orbitals of oxygen and carbon."


It is because of the very small difference in electronegativity between carbon and hydrogen.This is the reason why alkanes are very less acidic,do not show electrophillic substitutions etc.


The definition of peptide, peptide bond and protein

Two or more amino acids form a peptide. The peptide has the repeated form $\ce{-CO-NH-C(\alpha)R - }$ where R is an amino acid side-chain. The peptide bond is the amide bond N-C. The $\ce{CONH}$ group is planar. A peptide chain can be artificially constructed in the lab from any of the amino acids, its length and composition is then for you to determine.

A protein is a naturally occurring chain of amino acids folded so as to perform some function. Usually a protein has many amino acids because it has to perform a function in its organism. I'm not sure what the number of amino acids in the smallest protein is, but 20 seems about right. This number is determined by the protein's function, it can be many hundreds. A part of a protein's  backbone is also called a peptide chain.


Analysis of mixture by precipitation and complexometric titrations

Let $x$ be the amount of $\ce{MgCl2}$ in the original mixture, and $y$ be the amount of $\ce{NaCl}$.

Both of these salts dissociate completely in aqueous solution to give their constituent ions:

$$\begin{align}
\ce{MgCl2 (s) -> Mg^2+ (aq) + 2 Cl- (aq)} \\
\ce{NaCl (s) -> Na+ (aq) + Cl- (aq)}
\end{align}$$

Therefore, the total number of moles of chloride ions in the mixture is:

$$\begin{align}
\eta_{\ce{Cl-},\text{total}} &= \eta_{\ce{Cl-}\text{(from }\ce{MgCl2})} + \eta_{\ce{Cl-}\text{(from }\ce{NaCl})} \\
&= 2\eta_{\ce{MgCl2}} + \eta_{\ce{NaCl}} \\
&= 2x + y
\end{align}$$

$\ce{Ag+}$ ions react with $\ce{Cl-}$ ions in a $1:1$ stoichiometric ratio:

$$\ce{Ag+ (aq) + Cl- (aq) -> AgCl (s)}$$

Therefore, we have $\eta_{\ce{Ag+}} = \eta_{\ce{Cl-},\text{total}} = 2x + y$.

$\ce{EDTA}$ only complexes magnesium, and again in a $1:1$ stoichiometric ratio:

$$\ce{Mg^2+ (aq) + edta^4- (aq) -> [Mg(edta)]^2- (aq)}$$

Therefore, $\eta_{\ce{EDTA}} = \eta_{\ce{Mg^2+}} = \eta_{\ce{MgCl2}} = x$.

From your titration, you would have determined both values $\eta_{\ce{Ag+}}$ and $\eta_{\ce{EDTA}}$. Therefore, you have a system of two simultaneous equations in two unknowns, which is extremely simple to solve:

$$\begin{align}
\eta_{\ce{Ag+}} &= 2x + y \\
\eta_{\ce{EDTA}} &= x
\end{align}$$