What type of Organic Substance is used in LCD

Liquid crystals are usually rigid rod-like molecules that tend to show some degree of intermolecular order even when notionally liquids (there is a good summary on Wikipedia). This happens because the rigid polarisable parts of the molecules can have relatively strong directional interactions even in the liquid phase.

Some of the structures they form can rotate the plane of polarised light You can get some intuition about why knowing that polarising lenses consist of highly ordered molecules whose interaction with light is highly direction dependent. In normal liquids the molecular orientations are effectively random so, even in each molecule interacts with light depending on its orientation, this effect washes out due to the randomness of the overall structure. Liquid crystals form structures where there is the possibility or large-scale order allowing macro-effects on polarisation (just like the rigid structure in a polaroid film).

The other thing that makes liquid crystals useful is that the details of the ordering in the liquid can be affected by external electric fields, especially in thin films of the compounds. In principle this is possible when any molecule is polar (that is it has a net electrical dipole). Of course, making an effective display requires a lot of fine tuning of the electrical, mechanical and chemical structure, but the basic details are simply that the molecules are affected by electrical fields and this allows some control of the macro-structure which, in turn, alters the way the liquid crystal interacts with light. the effect is not "turned on" by the field, just altered and that is enough to make a viable display technology.

Liquid crystals have a long history. The first compounds known to show liquid crystal behaviour is cholesterol benzoate, the first molecule shown below, which was dicoveredn in 1888. The other structures show some actual molecules from the early days of LCD displays to modern developments. For more examples of the history there is a great summary from Merck (one of the major manufacturers) in this slideshow.


Only for solids that have regular and repeating pattern of constituent particles we define shape and geometry.

So, on the basis of order of arrangement of constituent particles solids are classified into crystalline and amorphous.The former has regular pattern of arrangement of particles which repeat itself periodically over the entire crystal (as shown below is the two dimensional structure of quartz)

Amorphous solids consist of particles of irregular shape(as shown below is 2d structure of quarts glass.

If the three dimensional arrangement of constituent particles of crystalline solid is represented diagramatically in which each particle is depicted as a point, the arrangement is called crystal lattice

Unit cell is the smallest portion of crystal lattice which when repeated in all directions generate the entire crystal (see below). There are only 14 possible crystal lattices called Bravais Lattices( after a French mathematician who first described them.

On the basis of unit cell parameters (axial distances and axial angles)
we get the following seven primitive unit cell which defines the shape of solid.

COORDINATION NUMBER
this term is used in crystallography in a somewhat different sense than it is in coordination compounds.

The number of nearest neighbors (or touching particles) that a particle has in a crystal is called its coordination number. All atoms in bravais lattice have the same coordination number.

In a simple or primitive unit cell, around any given atom there would be six equally spaced nearest neighbours.Thus the coordination no=6.

Any other particle in crystal lattice will be at a greater distance from the particle under consideration. This also apply to your NaCl crystal.


Is the reaction of silver nitrate and copper endothermic or exothermic?

You may already know that the world is not about enthalpy but about Gibbs free energy. A reaction is sponanteous if the change Gibbs free energy is negative.

$$\Delta G = \Delta H - T \cdot \Delta S$$

As you see, enthalpy and entropy play into this. Looking at your reaction, we note that we are taking two ions out of a solution to release one into solution. Compare the following, slightly different description:

$$\ce{2 Ag+ (aq) + Cu (s) <=>> 2 Ag (s) + Cu^2+ (aq)}$$

From an entropic point of view, this is bad. Rather than having two dissolved ions, we have one on the products’ side. Therefore, we can assume $\Delta S < 0$, therefore $T \cdot \Delta S < 0$. Negating it gives a positive value.

But the reaction is spontaneous, so $\Delta G < 0$. This can only be true if $\Delta H$ is sufficiently negative to counteract the loss of entropy. Thus we conclude that $\Delta H < 0$ and the reaction is exothermic.


Comparing SN2 reaction rate for phenacyl chloride and methyl chloride

First, let me point out that a rate difference of 500 is really not that large.  There are solvolysis reactions with relative rate differences on the order of $\mathrm{10^{10}}$ or greater (1).

The transition state for an $\mathrm{S_{N}2}$ reaction involves hypercoordinate (or hypervalent) bonding.  The transition state is, more or less, a trigonal bipyramid structure and the bonding at the central carbon atom ($\ce{Nu-C-X}$) involves a 3 center-4 electron bond (see the above link and links therein for more detail on this concept). With 5 electron pairs being shared by this central carbon, anything that removes electron density from the central carbon atom will prove beneficial and accelerate the reaction.

$\mathrm{S_{N}2}$ Reaction Pathway

In your molecule #2, a benzoyl group is attached to the central carbon involved in the $\mathrm{S_{N}2}$ process.  You are correct that such a group is larger than a hydrogen and should decrease the reaction rate due to steric effects. However the size of the benzoyl group is not exceptionally large and the steric effect will likely be small. More importantly, the benzoyl group will also exert an electronic effect. The central carbon in an $\mathrm{S_{N}2}$ reaction is $\mathrm{sp^2}$ hybridized [e.g. leading to a trigonal bipyramid structure as mentioned above] and has a $\mathrm{p}$ orbital.  This $\mathrm{p}$ orbital can interact with the adjacent carbonyl group through resonance.  This resonance effect will tend to remove (delocalize) electron density from the central carbon atom.  Also, the carbonyl group will inductively remove electron density from the central carbon atom.  These electronic effects will remove electron density from the central carbon atom thereby accelerating the reaction as observed.

(1) For example, the relative rates of solvolysis of 7-tosyloxynorbornane and anti-7-tosyloxynorbornene in acetic acid are $\mathrm{1:10^{11}}$


Are there ion combinations which can't be separated using the ion exchange resin method?

Ion Exchange resins are not selective.  You are essentially just statistically overpowering the situation by providing many more times the concentration of the ion you want compared to the ion you are 'extracting'.

There are some situations where chelation is so strong you can't exchange because the other ion is less stable in said position, say for example in crown ethers.

So the answer to your question, is no, there are not specific ions that can't be exchanged as a general rule, but there are ion pair combinations which won't exchange.


My take on the possible chemistry starting with the electrolysis of salt water, which can produce some chlorine. The latter can further react with water as follows:
$\ce{Cl2 + H2O <=> H+ + Cl- + HOCl}$
The created Hypochlorous acid may react with cuprous (here present as a $\ce{Cu2O}$ on the copper metal electrode). A fenton-type reaction based on $\ce{HOCl}$ could then proceed as follows for $\mathrm{pH > 5}$:

$\ce{Cu(I) (s) + HOCl (aq) -> Cu(II) (aq) + ^.OH (aq) + Cl- (aq) }$

The dissolving of the $\ce{Cu2O}$ results in a de facto cleaning of the copper metal surface, which per a reference to quote:

Copper is a beautiful burnished-gold color when it is clean and well maintained, but like all metals, copper can become discolored when exposed to air and water.

For background on the fenton-type reaction, see "Fenton chemistry in biology and medicine" by Josef Prousek, to quote reaction (15) on page 2330, to quote:

For $\ce{Fe(II)}$ and $\ce{Cu(I)}$, this situation can be generally depicted as follows [20,39]
$\ce{ Fe(II)/Cu(I) + HOX -> Fe(III)/Cu(II) + HO^. + X- }$ (15)
where $\ce{X = Cl, ONO}$, and $\ce{SCN}$.


What happens if you mix two insoluble liquids but with the same density?

There is no "or" in the first place. An emulsion is a two-phased system, it is just that one phase comes in a form of small droplets. Small as they are, however, they are pretty big from molecular point of view.

Now to the point. If you stir the liquids vigorously, you'll have an emulsion. Short of that, they will attempt to minimize the surface, and the shape with minimal surface is a sphere. Indeed, that's what you will end up with. It's been done before. Look up the Plateau's oil drop experiment: they put oil in a water/ethanol mixture with ratio tuned so as to match the density of oil, and behold, it just floats there in a huge spherical drop.

If both liquids get in contact with the wall, the geometry might become more complicated. The phase separation surface will develop certain contact angle where it touches the wall, and the rest of it will form some kind of minimal surface, which can be surprisingly diverse.

So it goes.


Crystal field splitting of other orbitals

According to Crystal Field Theory, d-orbitals of central metal atom splits due to non uniform repulsion of ligands around the metal atom. Why isn't p-orbital splitting discussed?

To be blunt, the reason nobody mentions p-orbital splitting or non-splitting is that neither the teacher nor the learner has ever understood the crystal field theory properly in its true depth (including myself).  They just propagate what they saw in standard textbooks, repeat the same in class, the students pass never to see CFT again. The story ends. If that student ever becomes a teacher, the story continues.

You are mixing apples and oranges. Crystal field theory has nothing to do with hybridization. The original work of Hans Bethe, the man behind crystal field theory, wrote a 72-paged highly theoretical paper in Annalen der Physik in German. The translations are available and I quote the abstract where he does clarify what happens to p-orbitals. How many can claim that that they read this paper completely before teaching CFT. I cannot. The abstract clarifies your misconception.

  The influence of an electric field of prescribed symmetry (crystalline field) on an atom is treated wave - mechanically . The terms of the atom split up in a way that depends on the symmetry of the field and on the angular momentum I (or J) of the atom . No splitting of s terms occurs , and p terms are not split up in fields of cubic symmetry . For the case in which the individual electrons of the atom can he treated separately ( interaction inside the atom turned off) the elgenfunctions of zeroth approximation are stated for every term in the crystal; from these there follows a concentration of the electron density along the symmetry axes of the crystal which is characteristic of the term. - The magnitude of the term splitting is of the order of some hundreds of cm-1. - For tetragonal symmetry, a quantitative measure of the departure from cubic symmetry can be defined, which determines uniquely the most stable arrangement of electrons In the crystal.


At what temperature does carbon dioxide decompose into carbon and oxygen?

Disclaimer: I am too lazy to check whether the reaction $\ce{CO2(g) <=> CO(g) + 1/2 O2(g)}$ is more favourable.

There is no single tipping point where all carbon dioxide will suddenly be converted to carbon and oxygen. As you heat carbon dioxide, the percentage of oxygen present would gradually increase.

You're interested in the reaction

$$\ce{CO2(g) <=> C(graphite) + O2(g)}$$

for which, at $298~\mathrm{K}$, $\Delta_\mathrm{r} H^\circ = +393.51~\mathrm{kJ~mol^{-1}}$ and $\Delta_\mathrm{r} S^\circ = -2.862~\mathrm{J~K^{-1}~mol^{-1}}$ (data from Atkins, Physical Chemistry 10th ed., pp 974-7). The equilibrium constant for this reaction is remarkably simple:

$$K = \exp{\left(\frac{-\Delta_\mathrm{r}G^\circ}{RT}\right)} = \frac{a_{\ce{O2}}}{a_{\ce{CO2}}}$$

and if we assume ideality, then both activities are proportional to the partial pressures, which are in turn proportional to the number of moles of both substances:

$$K \approx \frac{n_{\ce{O2}}}{n_{\ce{CO2}}}$$

Therefore if $K = 100$, then you have a hundred times more oxygen in your system than carbon dioxide. If $K = 1$, then you have equal amounts. The pressure and volume of the system do not matter in this interpretation of the value of the equilibrium constant. Our goal is now to find out the variation of $K$ with temperature.

The variation of $\Delta_\mathrm{r} H^\circ$ with temperature is given by Kirchhoff's law, and to simplify things, let's assume that the heat capacities at constant pressure are independent of temperature over the temperature range we are interested in:

$$\begin{array}{cc}
\hline
\text{Compound} & C_{p,\mathrm{m}}^\circ\text{ / }\mathrm{J~K^{-1}~mol^{-1}} \\
\hline
\ce{C(graphite)} & 8.527 \\
\ce{O2(g)} & 29.355 \\
\ce{CO2(g)} & 37.11 \\ \hline
\end{array}$$

$$\Delta_\mathrm{r}H^\circ(T) = +393.51~\mathrm{kJ~mol^{-1}} + (0.77~\mathrm{J~K^{-1}~mol^{-1}})(T - 298~\mathrm{K})$$

One can derive a similar expression for the entropy change, which is "left as an exercise for the reader":

$$\Delta_\mathrm{r}S^\circ(T) = -2.862~\mathrm{J~K^{-1}~mol^{-1}} + (0.77~\mathrm{J~K^{-1}~mol^{-1}})\left[\ln{\left(\frac{T}{298~\mathrm{K}}\right)}\right]$$

All that remains is to calculate the Gibbs free energy change

$$\Delta_\mathrm{r}G^\circ = \Delta_\mathrm{r}H^\circ - T\Delta_\mathrm{r}S^\circ$$

and the equilibrium constant

$$K = \exp{\left(\frac{-\Delta_\mathrm{r}G^\circ}{RT}\right)}$$

You can do this easily with a spreadsheet. I've thrown together some very messy Mathematica code and I'm sure someone will come along and tell me that there's a much more concise way of doing it, but:

Manipulate[{Power[10,logT],Exp[((393510+(0.77*(Power[10,logT]-298)))-(Power[10,logT]*(-2.862+(0.77*(Log[Power[10,logT]/298])))))/(-8.31446*Power[10,logT])]},{logT,2.47422,12}]

The first number in the curly brackets gives you the value of $T$ in Kelvins, and the second gives you the value of $K$ at that temperature $T$. Here are some of the numbers:

$$\begin{array}{cc}
\hline
T\text{ / K} & K \\
\hline
298 & 7.513 \times 10^{-70} \\
10^3 & 2.073 \times 10^{-21} \\
10^4 & 0.007895 \\
10^5 & 0.6899 \\
10^6 & 1.307 \\
10^9 & 2.598 \\
10^{12} & 4.9257 \\
\hline
\end{array}$$

$K = 1$ when $T = 2.52 \times 10^5~\mathrm{K}$.

This is well above the temperature at which carbon vaporises ($4827~\mathrm{^\circ C}$ according to Wikipedia). If you redo the calculations using instead the equation

$$\ce{CO2(g) <=> C(g) + O2(g)}$$

then you get these new expressions (thermodynamic data again from Atkins):

$$\Delta_\mathrm{r}H^\circ(T) = +1110.19~\mathrm{kJ~mol^{-1}} + (13.083~\mathrm{J~K^{-1}~mol^{-1}})(T - 298~\mathrm{K})$$

$$\Delta_\mathrm{r}S^\circ(T) = +149.498~\mathrm{J~K^{-1}~mol^{-1}} + (13.083~\mathrm{J~K^{-1}~mol^{-1}})\left[\ln{\left(\frac{T}{298~\mathrm{K}}\right)}\right]$$

and much more favourable numbers:

$$\begin{array}{cc}
\hline
T\text{ / K} & K \\
\hline
298 & 1.637 \times 10^{-187} \\
1000 & 1.470 \times 10^{-50} \\
2000 & 3.417 \times 10^{-21} \\
5000 & 0.003130 \\
6000 & 0.351794 \\
7000 & 10.65 \\
10000 & 5594.31 \\
\hline
\end{array}$$

This time, $K = 1$ at $T = 6276~\mathrm{K}$. You can play around with the code here:

Manipulate[{T,Exp[((1110190+(13.083*(T-298)))-(T*(149.498+(13.083*(Log[T/298])))))/(-8.31446*T)]},{T,298,10000}]

Why does $K$ for this reaction increase so drastically with temperature? It is likely to be because of the entropy contribution: because the entropy change is much more positive, increasing the temperature makes the term $-T\Delta_\mathrm{r}S^\circ$ much more negative, lowering $\Delta_\mathrm{r}G^\circ$ and increasing $K$.


What exactly is a reducing equivalent and what does it do?

The number of "equivalents" of a reagent refers to how much of the reagent there is relative to another reactant of interest. To use a concrete example, if we are interested in the $\mathrm{S_N2}$ transformation $$\ce{CH3Cl -> CH3Br},$$ a possible way of bringing this about would be to treat the starting material (methyl chloride) with a certain amount of bromide ion. If we carry out the reaction on 0.5 mol of methyl chloride, in theory we would only need 0.5 mol of bromide ion, because of the stoichiometry of the reaction:

$$\ce{CH3Cl + Br- -> CH3Br + Cl-}$$

In this case, 0.5 mol of bromide ion would be referred to as being "1 equivalent" of bromide (relative to methyl chloride, which is the reactant of interest).

We could also say that for the complete combustion of hexane, we would (theoretically) need "9.5 equivalents" of oxygen gas:

$$\ce{2C6H14 + 19O2 ->12CO2 + 14H2O}$$

In the case of a redox half-reaction, you usually have electrons in either the reactants or the products:

$$\ce{Fe^3+ + e- -> Fe^2+}$$

(this highly simplified reaction is relevant in, e.g., cytochrome c's role as an electron carrier in oxidative phosphorylation). Under biological conditions, this is rather unrealistic, as electrons don't actually exist in aqueous solution, nor is there any electrolysis going on. So, you need something else to act as a supplier of electrons, i.e. a reducing agent, such as NADH.

$$\ce{NADH + H+ -> NAD+ + 2H+ + 2e-}$$

In this case, because 1 mol of NADH provides 2 mol of "electrons", you could say that NADH acts as two reducing equivalents. Here, NADH is acting as a reducing agent, and is itself oxidised to NAD. [I don't agree with the Wikipedia page about hydride ion being one reducing equivalent; it should be two.]


Is the boundary of an atom well defined?

No, the boundary of an atom is not well defined. According Quantum mechanics the electron density around an atoms decreases exponentially, thus getting asymptotically close to zero.
Considering atoms as rigid spheres is only an approximation. The criteria used to decide on the actual value of the atomic radius is arbitrary. You can for example choose the expectation value $\langle\hat r\rangle$ of the electronic wavefunction. For the Hydrogren atom, this results in the bohr radius $a_0$. But you might as well choose the maximum of the electron density ($\frac{3}{2}a_0$ for $\ce{H}$), or something based on experiments.
Now the aggregate state is something completely different and not directly  related to atomic radius and whether they are rigid sphere or not. Solid means the molecules/atoms of a substance are fixed in their positions. They can vibrate, but do not change their average position with respect to each other. In gases molecules may move freely. And a liquid is something in between, you have some order if you look at a small group of molecules, but not on a larger scale as for solids.


What makes the reaction a zero-order reaction, is it being dependent only on the concentration of one of the elements.

In a system made by a reagent and a catalyst, we expect the reaction to behave in a zero-order fashion when the concentration of the reactant varies in a linear fashion during the course of the reaction.
You could write an equation in the form: $rate = k$ and, implying that only the reactant A varies in concentration, $[A]=[A_0]-kt$.

Now, during a reaction you know that the concentration of the catalyst remains constant.

What you are asking is: what happens if I vary the concentration of the catalyst? Assuming that you are still in zero-order conditions and you run another reaction with a larger amount of catalyst, then you will obtain another equation rate (another value for k).

But that will not make the curve [A] vs t non linear (that is, non zero-order).
More catalyst means a larger rate constant, but doesn't imply a variation in kinetic behaviour of the reaction.

Still, if you were to evaluate the variation of rate in respect to the catalyst's concentration, then you would obtain a reaction rate that is still of order 0 in respect to A, and of order non-zero in respect to the catalyst.

Maybe your book was just pointing out that a set of reactions in which the same amount of catalyst is used, could be studied with a zero-order approach.


Why does the emission wavelength of sodium appear to be an exception despite the trend for ionisation enthalpies among alkali metals?

On the NIST website you can look up the transitions corresponding to the wavelengths you list:

$$
\begin{array}{l c c c c}
\hline
\text{Element} & λ/\pu{nm}& |i\rangle & |f\rangle & \text{I.E./cm$^{-1}$}\\
\hline
\ce{Li} & 670.8 & 1s^22s & 1s^22p & 43\,487\\
\ce{Na} & 589.2 & 2p^63s & 2p^63p & 41\,449\\
\ce{K} & 766.5 & 3p^64s & 3p^64p & 35\,010\\
\ce{Rb} & 780.0 & 4p^65s & 4p^65p & 33\,691 \\
\ce{Cs} & 455.5 & 5p^66s & 5p^67p & 31\,406\\
\hline
\end{array}
$$

You see that the transition in Cs that you give in the table corresponds to a $7p\leftarrow 6s$ transition and that for consistency you should compare it with a $6p\leftarrow 6s$ transition which has a wavelength around 860 nm (no $d$ electrons involved).

The term values of the electronic states of multi-electron atoms are relatively well described by Rydberg's formula

$$
\tilde{\nu}_{n,\ell}=\text{I.E.} - \frac{\mathcal{R}_M}{(n-\delta_\ell)^2},
$$
with $\mathcal{R}_M$ the mass-corrected Rydberg constant and where the quantum defect $\delta_\ell$ depends on the orbital angular momentum of the electron $\ell$ and on the properties of the ion core. The transition $n'p\leftarrow n''s$ can therefore be approximated by
$$
\tilde{\nu}_{n',\ell=1}-\tilde{\nu}_{n'',\ell=0}=\frac{\mathcal{R}_M}{(n''-\delta_{\ell=0})}-\frac{\mathcal{R}_M}{(n'-\delta_{\ell=1})},
$$
and thus depend on the quantum defects for the $s$ and $p$ states. The quantum defects depend on the scattering properties of the Rydberg electron with the ion core and it is difficult to predict their values for different elements. It is therefore difficult to make quantitative predictions. Also for the heavier elements (higher nuclear charge $Z$), relativistic effects become more important, which influences the electronic structure.

Note that these lowest states are not really Rydberg states and the Rydberg formula is a bad approximation. For high values of $n$ the Rydberg formula gets much more accurate. In fact, it allows for a very accurate determination of the ionization energies of these elements.


How to find trace methanol content in fermented fruit/distillate?

Methanol, formed by degradation of pectines from the skin of fruits, as well as higher alcohols is volatile and does not show significant absorption in the UV.

Consequently, gas chromatography is the method of choice to determine these components in the fermentation mix.


Where does the energy of solar radiation hitting the atmosphere go?

It is broadly true that some solar radiation ends up heating the atmosphere. But it isn't true that the atmosphere absorbs most of the heat emitted by the sun.

Broadly there are several ways that radiation is absorbed:

Some of it (but not much of it) is absorbed directly by certain molecules in the atmosphere (carbon dioxide and water vapour are biggies), directly heating them.
Some of it is reflected by clouds and ice and other parts of the earth and reemitted to space (quite  a lot of it)
Some radiation is absorbed by plants and used to drive photosynthesis (important but not a lot of absorbed radiation)
A lot of it is absorbed by the surface of the earth (land and sea) which is the part that contributes most of the warming effect of sunlight (of course this warmer sea and earth then warms the atmosphere above it either directly be contact or by being absorbed before it is transmitted back into space, though a lot of it is emitted to space).

So, in the end, a lot of the radiation ends up as heat, but mostly via heating of earth and sea not by directly heating the atmosphere. On the other hand, the atmosphere does end up much warmer as a result. And the absorption by the atmosphere does make the difference between the earth being very cold or very habitable as a planet. The details of this are very complicated which is why people build very big computer models when they want to understand global warming which depends of the complicated details of how the atmosphere absorbs fractions of the heat.


A theoretical chemistry question about elements and their presence over time

Is it possible that in these times, the amount of energy would permit atoms larger than we have observed in nature or been able to create in the lab?

Quite possibly, but we do not need to go back in time. In fact, the early moments after the Big Bang are the least likely time to find very heavy elements, as the primary elements produced initially are believed to have been almost exclusively hydrogen and helium.
Since those early moments, the universe has certainly expanded, but there are still places where energy and mass are highly concentrated, such as in supernovas and black holes. Whether or not very heavy elements are transiently created in these places is an open question.
The problem is that elements heavier than uranium tend to be unstable, so they break apart (nuclear fission) very quickly into lighter elements. In order to make very heavy elements, you need to combine two less heavy elements (nuclear fusion), but it's hard to get two heavy atoms together to fuse if they keep falling apart themselves. A good discussion of the possibility of very heavy element creation in stars is
Goriely, S. and Martinez Pinedo, G. (2015) The Production of transuranium elements by the r-process nucleosynthesis. Nuclear Physics A 944:158. DOI:10.1016/j.nuclphysa.2015.07.020

If we then travelled to a far future, hundreds of millions of years from now, when the energy of the universe is much more spread out, and atoms have had more time to decay and lose electrons, wouldn't there be, at some point, a time where we no longer observe certain heavier elements in nature?

The answer to this question relates to the first answer. Since heavy atoms are constantly being synthesized within stars, the elements that are decaying get replaced. But as the number of stars decreases over time, the rate of synthesis will necessarily decrease as well, and elements that are susceptible to fission will slowly decrease in abundance. The time scale for this, however, is much longer than even hundreds of millions of years. Our Sun, for example, is expected to last about another 4.5 billion years, and the final dying out of all of the stars in the universe is expected to take somewhere between 1 and 100 trilliion years. I encourage you to read up on ideas about the "heat death" of the universe if this is a topic you are interested in.

At either of these points in time wouldn't you expect to find more ionic elements in nature?

Since it takes energy to separate opposite charges from one another, ionization happens with high energy, so places like stars where there is a lot of energy will have a lot of ionic matter. Likewise, when high energy emissions from those places interact with other matter (like cosmic rays hitting the Earth's atmosphere), ionization will take place. So you're right that when the universe's energy is more concentrated, there were probably more ions.
In the distant future, the opposite is true. Since the universe decays to a low energy state, and ionization takes energy, we expect that oppositely charged matter will combine to make neutral atoms and molecules as the universe gets older. Some have theorized, however, that the expansion of the universe will mean that matter is so spread out that oppositely charged ions get too far away from each other to interact, so it is conceivable that the final state of the universe is separated ions, but that is just speculation.
And as noted in the comments, you may get a much better answer over at physics.SE.


Does a unit cell have to contain a whole number of atoms?

Crystallographic background

A unit cell cannot contain parts of atoms from a formula unit, otherwise it would not be a geometric unit of repeatability.
The simplest condition arising from translational symmetry is that the total number of atoms present in the unit cell must be either equal to or be a multiple of the number of atoms in the chemical formula, which is an equivalent to the integer number of formula units per unit cell: $Z \in \mathbb{N}.$

Note, however, that the asymmetric unit (crystallographically independent region) contains the $Z/n$th part of the atoms of the formula unit, where $n$ is the order of the crystallographic point group.
Thus, the asymmetric region may contain an integer number of formula units, one formula unit or its fractional part, meaning that some atoms located on symmetry elements, are "divided" across several asymmetric regions.

Data precision

It appears you are using the correct formula for the crystallographic density $ρ$ (I strongly encourage not to use $d$ as a symbol for density as in crystallography $d$ is reserved for spacing between planes):

$$ρ = \frac{MZ}{N_\mathrm{A}V}$$

where $M$ is molar mass, $N_\mathrm{A}$ is Avogadro's number and $V$ is the unit cell volume.
Although you can use the exact value for $N_\mathrm{A} = \pu{6.02214076E23 mol-1}$ and a precise tabulated value $M(\ce{FeO}) = \pu{71.844 g mol-1},$ note that you are given both $a$ and $d$ (experimental data) to the single significant figure, which prohibits your answer from being more precise than these parameters, and you must present a rounded number $Z = 4$ and not $Z = 4.18,$ which doesn't make either crystallographic or mathematical sense.

In real life situation there always will be a systematic difference between the measured and crystallographic densities due to the presence of defects, and the deviation of $Z$ from the integer value is often due to statistically filling of the proper point group by the atoms of a different kind.


What are the standard math prerequisites for physical chemistry?

In my experience, the most helpful advanced math courses you can take for Physical Chemistry are Differential Equations and Linear Algebra. Multivariate Calculus is helpful, but not as helpful as Diff. Eq. would be, especially in the beginning. Linear Algebra is very helpful for quantum chemistry, Diff. Eq. is also helpful for quantum, but more so for statistical mechanics. In the comments you mentioned several different Diff. Eq. options - as @permeakra said, taking all of them is a good idea, but ODEs and PDEs will probably be the most useful.


Is it possible to have Volume(solution) < Volume(solvent before adding solute)?

Gosh, sure. According to various on-line info (of unknown quality) a 2% by wt. solution of NaOH in water has a density of 1.0207 at 20°C. This means 100 g of it has a volume of 97.97 ml. At 20°C water has a density of 0.9982 meaning that 98 grams of water has a volume of 98.18 ml.


Influence of sodium cyanoborohydride on C=N bonds in DNA bases

Sodium cyanoborohydride is a very selective reducing agent. Another (safer) alternative would be sodium triacetoxyborohydride. But this doesn't really matter as far as $\ce{C=N}$ bonds of DNA bases are concerned since they are very stable.


What is the difference between radioactive equilibrium and transient equilibrium?

I guess you have been caught in a semantic trap. Adjectives radioactive and transient describe  two different aspects of the same thing.
It could be as well written as "radioactive transient equilibrium". Or, even better as "radioactive steady state", as it is not true equilibrium.
Radioactive describes its nature . It may not be radioactive,by can be thermal, chemical, radiative, mechanical etc .
Transient describes its dynamics. It need not to be transient ( temporary, also called  a steady state ), but it can be a permanent, timeless, true  equilibrium.


Why doesn't fluoride ion have highest nucleophilic power (nucleophilicity)?

Nucleophilicity of an ion does not solely depend on the number of lone pairs it has. The solvent in which the ions are dissolved matters, too.

If you are using a polar protic solvent, then flouride ion would behave as a very bad nucleophile (it would be solvated in the case of water; water will form hydrogen bonds with it).

Among the nucleophiles you have stated, $\ce{R3C}$ anion would be the best nucleophile in a polar protic solvent. However, if you use a polar aprotic solvent, flouride ion would behave as a very good nucleophile (because aprotic solvents do not solvate anions).

Remember, nucleophilicity is a kinetic concept and basicity is a thermodynamic one. Your reasoning for comparing the nucleophilicities was based somewhat on basicity of the ions. (This would work well in an aprotic solvent.)

In very layman terms:

Nucleophilicity: How good is something at reacting (doesn't concern stability of the ion)
Basicity: How badly something would want to react (concerns stabilities)


The color change results from the reversible oxidation-reduction reaction of the methylene blue indicator.  In alkaline solutions, glucose is oxidized to D-gluconic acid or alpha-D-gluconolactone.

$\ce{HOCH2(CHOH)4CHO + 3OH- -> HOCH2(CHOH)4CO2 + 2H2O + 2e-}$

In the course of this reaction, methylene blue is reduced from the blue (oxidized) form to the colorless (reduced) form.

Shaking the flask dissolves $\ce{O2}$ in the solution, which oxidizes the indicator back to the blue (oxidized) form.


I'm looking for a substance that can be used as an adhesive, but that will burn without producing any ash

Nitrocellulose(flash paper). Using acetone or something you get a gel. Caution recommended.


For the "shotgun" approach, saliva samples ($\pu{1 mL}$) were pre-fractionated with the use of Millipore ultracentrifuge filters (Millipore Corp., Billerica, MA, USA). Individual fractions were treated with $\pu{10 mM}$ dithreitol for $\pu{30 min}$, and then $\pu{50 mM}$ iodoacetamide for $\pu{30 min}$.

This reaction sequence in Ref.1 has a clear intention: First, break disulfide bonds in saliva proteins using a thiol exchanging reagent and then trap the recently formed thiol groups by iodoacetamide (Ref.2). According to Ref.2:

  Alkylation of cysteine thiols with iodo, bromo or chloro substituted acetic acid or acetamide is a classic approach that has been exploited since the 1930′s.

Thus, it is safe to assume that this second reaction is intended to trap thiol group as quickly as possible (The relative reaction rates between glutathione and these halogenated acetates are $100:60:1$ for iodo-, bromo- and chloro-acetates respectively; Ref.3).

Also keep in mind that disulfide reduction to corresponding thiol is accomplished primarily by thiol exchange type reagents such as dithiothreitol (DTT) or 2-mercaptoethanol (ME). Thus, it is also safe to assume that dithreitol mentioned in Ref.1 is definitely a thiol exchanging reagent, possibly DTT. Therefore, I think it is a chain mistake as M. Farooq suggested in his comment.

References:

S. Hu, Y. Li, J. Wang, Y. Xie, K. Tjon, L. Wolinsky, R. R. O. Loo, J. A. Loo, D. T. Wong, "Human Saliva Proteome and Transcriptome," J. Dent. Res. 2006, 85(12), 1129–1133 (DOI: 10.1177/154405910608501212).
Jakob R. Winther, Colin Thorpe, "Quantification of Thiols and Disulfides," Biochimica et Biophysica Acta 2014, 1840(2), 838-846 (https://doi.org/10.1016/j.bbagen.2013.03.031).
Frank Dickens, "CLII. Interaction of halogenacetates and SH compounds. The reaction of halogenacetic acids with glutathione and cysteine. The mechanism of iodoacetate poisoning of glyoxalase," Biochem J. 1933, 27(4), 1141–1151 (doi: 10.1042/bj0271141).


Heat given off from an electrochemical cell compared to mixing reactants

I think the problem here is related to what each of these terms means, in a fundamental sense.

$\mathrm{d}S$ is the entropy of the system

$\frac{-\mathrm{d}q}{T}$ is the entropy of the surroundings - the heat change of the surroundings divided by the temperature.

$\mathrm{d}H$ is the enthalpy of the system - it is equal to $\mathrm{d}q$ only when the internal energy change occurs at constant pressure, and when there is no form of work being done besides PV work. In an electrochemical cell, moving the electrons is work, and so we can't leave it out of the total energy balance.

In other words, you can't just use $\mathrm{d}q = \Delta H$ for the 1st question because $\mathrm{d}q_p \neq\Delta H$ when you are doing any work besides PV work.

Instead you are using the Clausius inequality, which is just another way of stating the second law: the entropy of the universe must increase for any process. In this case, the universe is the system plus the surroundings, and so:

$$\Delta S - \frac{\mathrm{d}q}{T} \geq 0$$

Solving that for dq gives you the heat absorbed by the surroundings, which is equal to temperature times the entropy change of the system.

For a process that is only (possibly) doing PV work, at constant pressure, $\Delta H = \mathrm{d}q_p$, and so we can just use that. This would be an example of using the 1st law at constant pressure, when there is no other type of work besides PV.

In your case, you used:
\begin{align}
\Delta G &= \Delta H - T\,\mathrm{d}S\\
\Delta H &= \Delta G + T\,\mathrm{d}S\\
\end{align}

This works because the equation for $\Delta G$ is derived directly from the Clausius inequality at constant pressure. In other words, you are using the 2nd law to derive the same result.

Edit

To answer your last questions, and questions from the comments:

  why is $T\,\mathrm{d}S$ less than $\mathrm{d}q$ in my example? I don't really get this bit

It's not - but you are dropping the signs and that makes it hard to see. If you leave the signs in:

\begin{align}
T\,\mathrm{d}S &\geq \mathrm{d}q\\
(298~\mathrm{K})\left( -61.6~\mathrm{\frac{J}{K\cdot mol}}\right) &\geq -70~\mathrm{\frac{kJ}{mol}}\\
-18~\mathrm{kJ} &\geq -70~\mathrm{kJ}\\
\end{align}

Which is true, although it is easier to see if you divide by -1:

$$18~\mathrm{kJ} \leq 70~\mathrm{kJ}$$

  The entropy divided by the temperature gives $234.9~\mathrm{JK^{−1}mol^{−1}}$ which is not the value for entropy as it should be according to the Clausius inequality given that the change in entropy is equal to the heat given off in part (b). Why is this so?

Here I assume you meant "enthalpy divided by temperature." The second reaction is not carried out in a reversible manner, so we have no guarantee that $\mathrm{d}S = \mathrm{d}q/T$. All we know is that it will be greater than or equal to $\mathrm{d}q/T$, and if you plug in the numbers, it is.

Remember that all of this is built into the $\Delta G$ equation as well, which is why we use it when doing "regular" chemical reactions at constant T & P.


Dirac notation when bra and ket aren't vectors

Wave function for a quantum system is a vector in a Hilbert space, which is a special kind of an inner product space. Specifically, for a spin-0 particle constrained to move in 1 spatial dimension the wave function $\psi(x)$ is an element of $L^2(\mathbf{R})$ space, i.e. the space of functions that are square-integrable on the whole real line $\mathbf{R} = (-\infty;+\infty)$. And the inner product between any two elements $\vec{\psi}_{i} = \psi_i(x)$ and $\vec{\psi}_{j} = \psi_j(x)$ of $L^2(\mathbf{R})$ space is indeed defined as follows,
$$
\newcommand{\inner}[2]{\langle{#1},{#2}\rangle}
\newcommand{\ket}[1]{\,\lvert{#1}\rangle}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
\inner{\vec{\psi}_{i}}{\vec{\psi}_{j}} := \int\limits_{-\infty}^{+\infty} \psi_i^*(x) \psi_j(x)\, \mathrm{d} x \, .
$$

Just note here that, first, it is the matter of convention in which argument the inner product is linear: in physical literature it is often defined to be linear in the second argument, while in mathematically oriented literature it is usually defined to be linear in the first argument. Consequently, mathematicians usually define the inner product between any two elements $\psi_i$ and $\psi_j$ of $L^2(\mathbf{R})$ space in the following way,
$$
\inner{\vec{\psi}_{i}}{\vec{\psi}_{j}} := \int\limits_{-\infty}^{+\infty} \psi_i(x) \psi_j^*(x)\, \mathrm{d} x \, ,
$$
as it is in fact done in some Wikipedia articles referenced above.

And secondly, physicists usually prefer to use even a different notation, the so-called Dirac bra-ket notation, in which elements of a Hilbert space are denoted as $\ket{\psi_i}$ and $\ket{\psi_j}$ (rather than $\vec{\psi}_{i}$ and $\vec{\psi}_{j}$ as it is usually done in linear algebra), while their inner product is designated as $\braket{\psi_i}{\psi_j}$ (rather than $\inner{\vec{\psi}_{i}}{\vec{\psi}_{j}}$).


The use of the word "manifold" in most of the quantum chemistry literature is purely as a synonym for multiple or different. Usually it appears when the number may be very large. Nothing more complicated than that.

For a simpler example than coupled cluster, consider configuration interaction with only single excitations (CIS). One of the key quantities is the set of single excitation amplitudes, $t_{ia}$ or $t_{i}^{a}$ or $k_{ai}\hat{a}^{\dagger}\hat{i}$ depending on your preference for notation. $i$ is an occupied MO index, and $a$ is a virtual MO index. We might say that the set $\{t\}$ forms a manifold, and the number quickly becomes large as you increase the basis set size: there will be $N_{ov} = N_{\text{occ}} \times N_{\text{virt}}$ single excitation amplitudes.


Double titration also often refers to Gilson's titration method (also known as "Gilman color tests" or "Gilman double titration") for determining the concentration of organometallic compounds in hydrocarbon solvents.
At the first stage the total concentration of the base in the solution is determined.
At the second stage the residual amount of alkali after the organometallic compound reacts with the halide is established, allowing to calculate the concentration of metalloorganic compound. [1]

For the double titrations, a 3 ml aliquot of the solution to be analyzed was
withdrawn by pipette using a suction bulb and added to 10 ml of ether.
The solutionwas hydrolyzed with 10 ml of distilled water and titrated with standard acid, using phenolphthalein as indicator, to give the total present.
A second 3 ml aliquot was withdrawn and added 10 of ether, containing 1 ml of the halide, under an atmosphere of nitrogen.
The solution was swirled gently and allowed to stand for 2 min, after which it was hydrolyzed with distilled water and titrated immediately with standard acid, again using phenolphthalein as indicator.
In titrating the latter solution the end point is easily overstepped, since the
aqueous layer decolorizes before the ether layer. Vigorous shaking near the end point
is recommended.
There is the possibility of interference from hydrolysis of the halide, so that the titration should be carried out within a reasonable length of time.

An updated cumulative summary for the method can be found in Wakefield's Organolithium Methods [2, pp. 17–18]:

For detecting the presence of an organolithium compound (for example for checking whether any remains at the finish of a reaction) Gilman's Colour Test 1 is convenient:
$$
\begin{align}
\ce{(4-Me2NC6H4)2C=O + RLi &-> (4-Me2NC6H4)2C(R)OLi}\\
\ce{ &->[(i) H2O][(ii) I2, AcOH] [(4-Me2NC6H4)2CR]+ I-}
\end{align}
$$
A sample (0.5—1 ml) of the test solution is added to an equal volume of a I solution of Michler's ketone, 4.4'-bis(dimethylamino)benzophenone, in dry benzene.
A few drops of water are added, and the mixture is shaken.*
A few drops of an 0.2% solution of iodine in glacial acetic acid are added,
resulting in the formation of a blue or green colour.
Gaidis proposed two modifications, to distinguish between alkyl- and
aryllithium compounds (15).
(a) Following hydrolysis, the solution is buffered to pH 9 by the addition of 20% aqueous catechol (1—2 ml).
A few drops Of iodine in benzene are added and the solution is shaken.
A green to blue colour in the organic layer indicates the presence of an aryllithium compound.
The mixture is then acidified with acetic acid to reveal the presence Of an alkyllithium compound.
(b) Following hydrolysis the solution is acidified with acetic acid (5—15
drops).
20% aqueous sodium bisulphite is added; aryl- but not alkyllithium
compounds give a green colour.
Gilman's Colour Test 2 (16) which also distinguishes alkyl- from aryllithium compounds, depends on the rapid metal—halogen exchange of the former with 4-bromo-N,N-dimethylaniline:
$$
\begin{align}
\ce{4-BrC6H4NMe2 + RLi &<=>> 4-Li2C6H4NMe2 + RBr}\\
\ce{4-Li2C6H4NMe2 &->[(i) Ph2CO][(ii) H3O+] [Ph2CC6H4NMe2]+}
\end{align}
$$
The solution to be tested (0.5-1 ml) is added to an equal volume of a solution of 4-bromo-N,N-dimethylaniline in dry benzene.
A 15% solution of benzophenone in dry benzene (1 ml) is added.
After a few seconds, water is added and then the mixture is acidified with concentrated hydrochloric acid.
A positive result is indicated by a red colour in the aqueous layer.

* Aerial oxidation sometimes leads to formation of the blue-green at this stage.

References

Gilman, H.; Cartledge, F. K.; Sim, S.-Y. The Analysis of Organic-Substituted Group IVB Lithium Compounds. Journal of Organometallic Chemistry 1963, 1 (1), 8–14. DOI: 10.1016/S0022-328X(00)80039-X.
Wakefield, B. J. Organolithium Methods; Best synthetic methods; Academic Press: London; San Diego, 1988. ISBN 978-0-12-730940-8.


When Z>1 are repulsive forces dominant or are attractive forces dominant?

Your arguments for attractive forces are not quite correct, particularly the first one, the other two are partly correct. The points for repulsive forces are more accurate but not completely correct.

The intermolecular potential energy is fixed by the type of molecule and varies with separation between molecules (e.g. a Lennard-Jones potential). The potential well formed by the interaction is short-range. It is effectively zero at about four to six molecular diameters decreases to negative values as the molecules approach  one another (negative energy is attractive) but rapidly becomes positive ( repulsive) when the molecules are effectively in contact.The potential is unaffected by temperature and pressure.

The ratio $p\overline{V}/RT$ is called the compressibility ratio Z and is unity for an ideal gas. ($\overline V$ is the molar volume $V/n$). Ideal gases have no intermolecular potential, or size. Real gases differ primarily in their size and magnitude of their intermolecular potential energy and so deviations from $Z=1$ are to be expected. At all temperatures and pressures the molecules have an attractive potential that tries to pair up molecules, and repulsion on collision when the charge on the electrons in each molecule force them to repel one another.

It is the competition between attraction, repulsion and kinetic energy at a give temperature that determines the shape of the $Z ~ vs. p$ plot.

For molecules at low temperature the line of the graph $Z= p\overline{V}/RT$ vs. p dips down further below $Z=1$ than it does for the same molecules at higher temperatures. This is because at low temperatures the  molecules have less kinetic energy ( so move more slowly) and are more influenced by attractive forces than at higher temperatures. This makes the actual volume of gas smaller than the ideal (as a larger fraction of molecules are paired up), which in turn makes Z smaller.

Similarly, at a given temperature, a larger negative Z is seen in gases with a larger attractive interaction, say $\ce{CO2,  CH4 or ethene}$ compared to those with smaller attraction, e.g. $\ce{He or H2}$ .

At high temperature the molecules have more kinetic energy and so attractive interactions are relatively less important, and repulsion more so as the kinetic energy is larger. This makes the actual volume greater than the ideal and $Z>1$.


How does oil on the surface of water prevent rust?

Well, I went and searched for the solubility of oxygen in water and oil, and found this summary paper on the NIST web site: "The Solubility of Oxygen and Ozone in Liquids" by Battino, Rettich and Tominaga, J. Phys. Chem.Ref. Data., vol 12, no. 2, 1983.

Conveniently, the paper gives solubility data for oxygen in both water and olive oil.  The solubility is given both as molar fractions, and also as the Ostwald coefficient $L$, defined as the volume of gas absorbed per volume of absorbing liquid.

Given the very different molar volumes of water and olive oil, I believe the latter provides a much more useful comparison here.  Comparing tables 1 (water; note the $10^2$ factor in the header) and 25 (olive oil) in the paper, we see that the Ostwald solubility coefficient for oxygen in water is about 2,000 times higher than in olive oil:

 Temp. (K) | Temp. (°C) | L (water) | L (olive oil)
-----------+------------+-----------+---------------
    298.15 |         25 |     310.4 |        0.1269
    308.15 |         35 |     276.6 |        0.1326
    318.15 |         45 |     254.5 |        0.1383
    328.15 |         55 |     240.9 |        0.1441

So, yes, oxygen is a lot more soluble in water than in oil.  But why?

Well, if this web site is to be believed, it's because of the strong hydrogen bonding between water molecules, which causes even liquid water to maintain some degree of an ice-like lattice structure, with gaps that small gas molecules like $\ce{O2}$ can easily slip into.  Once there, the gas becomes solvated, with the strongly polar water molecules inducing a temporary dipole in the $\ce{O2}$ molecule and thus creating an attractive Debye interaction.  The overall solvation process is exothermic (i.e. releases heat), explaining why the solubility of oxygen in water decreases with temperature.


Arrow pushing: why does the O become positive when its lone pair becomes a double bond?

Ah, the confusion that layman’s terms can cause. You are assuming sharing to mean ‘both have the same electrons as before, they just share a little bit of them’ — but that’s not really the nature of a chemical bond let alone the idea behind formal charges. Rather, once a molecule forms the electrons are delocalised across the entire molecule and the Lewis structures we draw are a mere approximation that makes things easier for our brains.

When determining things such as formal charges, you should stick exactly to the book and perform it strictly as directed. To determine a formal charge, all bonds must be formally cleaved homolytically, i.e. oxygen only gets two electrons back from the double bond (and another from the other single bond). Then count the resulting electrons and compare to the initial state. Since you will arrive at five electrons for oxygen (which is one less than six), it gets a single, formal positive charge. And indeed physically, there is less negative charge density on oxygen than there was before. (But note that not all formal charges correspond well with actual charge density.)

This is a distinct process from determining oxidation states, whereby you must cleave the bonds heterolytically with the electronegative atom taking it all.

In none of these formal processes you can say ‘but it was originally oxygen’s electrons …’ You cannot make electrons wear pink and yellow hats (not even during Winterbash) and thus you cannot distinguish them.


Dependence of EMF of a cell on stoichiometry

The electromotive force of a redox reaction is not dependent on the coefficients in the balanced equation as it is an intensive property (i.e. a property that does not depend on the size of the system). Thus, your reasoning is incorrect.

You can read more about electrochemistry here: https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Basics_of_Electrochemistry

Also, I am wondering if your textbook is actually correct because $\ce{Ag2+ + e- -> 2Ag}$ and $\ce{Ag+ + e- -> Ag}$ are definitely different. Thus, the standard reduction potential of silver (I) should not be used. Instead, the standard reduction potential of $\ce{Ag2+}$ should be used. Someone who is more knowledgeable in this area please correct me if I am wrong.


Many factors contribute to the solubility of a compound in water. In this case, the main difference between the compounds is the bonds’ strength.

Fluoride is very electronegative, and thus the bonding forces between $\ce{Ca^2+}$ and each fluoride anions are very strong, they are tightly held together in the crystal form. A few water molecules do not have enough strength to get in between and dissolve it.

Chloride is less electronegative, thus $\ce{CaCl2}$ has weaker bonds, then comes $\ce{CaBr2}$ and at last $\ce{CaI2}$, the weakest of them all. You may also expect increase in bond length depending on the crystal structure, which may also contribute to the solubility properties of the given compound. So weaker the bond strength, greater the solubility – at least that’s the expected.

In fact, $\ce{CaCl2}$ is more soluble than $\ce{CaF2}$, and $\ce{CaBr2}$ is even more – as expected. However, I did some research and the Wikipedia article on $\ce{CaI2}$ claims it’s less soluble than $\ce{CaBr2}$, which is not the expected, but it does not provide a source for that information. It might be true, though, since there are a variety of factors that may or may not contribute to it’s solubility in water, such as it’s crystal structure and the particular characteristics of iodide.

You can do the experiment yourself to find out which one is more soluble, since all solubility facts and numbers we know come from experiments. I couldn’t find anything on the solubility of $\ce{CaI2}$ in my CRC Handbook of Chemistry and Physics either, so I can’t provide you with a precise answer on why it’s less soluble than $\ce{CaBr2}$ (if in fact it is), but I hope you can now understand some of the factors that make $\ce{CaF2}$ the least soluble compound on that list.


(1) determine at which side substance A will diffuse out of the membrane

    should I answer at the right side or that the substance stays within the membrane?

Neither, because the concentration is higher in the membrane than either side, the substance will flow out of both sides initially at least. After a while the material will flow from left to right (high concentration to low concentration) and converge towards steady state.

  (2) chose one of the following profiles below which represents the concentration of a substance A at steady-state.

First thing - When doing diffusion problems immediately refer to Fick's laws of diffusion.

Fick's first law assumes a steady state system and is given as:

$$J = -c D\frac{dx}{dz}\quad \text{or more simply} \quad J = -D\frac{\Delta C(x)}{\Delta z} \quad \text{or more generally}\quad J = -cD\ \nabla x$$

Where $J$ is flux (mols through an cross-sectional area per time), $c$ is total concentration, $C(x)$ is the concentration at point $x$, $D$ is the diffusion constant, $x$ is the mole fraction and $z$ is position.

For steady state to occur, $J$ must by definition be constant through the entire membrane and thus $-D\frac{\Delta C(x)}{\Delta z}$ must be constant through the entire interface. Since $D$ is already a fixed constant, we can assume $\frac{\Delta C(x)}{\Delta z}$ must also be constant. Thus the plot with a constant slope is the steady state profile which is B.

Q.E.D.


why are Ca(OH)₂ and Al(OH)₃ less soluble than Na(OH) in water

$\ce{OH}$ is small anion and bonds effectively with small cations. Thus, in general, the bigger (and less specific charge) of the cation, the higher is solubility of the corresponding base in polar solvents, meaning that $\ce{Ca(OH)_2}$ and $\ce{Al(OH)_3}$ having much smaller and more charged cations are almost covalent in nature and thus, do not dissociate.

In general, solubility of salt in water is determined by energy balance between breaking of the ionic bonds and hydratation of anions formed. Thus, the most soluble salts are usually formed by small and big ions, that cannot bond effectively, but at least one of them can be hydrated effectively.

Please, note: solubility for molecular compounds, like sugars and some non-dissociating compounds that looks like salts (like $\ce{Cu(CH_3COO)_2*H_2O}$) is an entirely different matter and determined by relative strength of intermolecular interactions (read: the molecule have to be able to form hydrogen bonds)


What could these letters "S" in red circles mean in a biochemical diagram?

It almost certainly refers to the amino acid (residue) serine. The numbers (300, 333, 351) refer to the residue numbers. The COOH at the end signifies the C-terminus of the protein (subunit).
The black P's next to it are phosphate groups; both protein kinase A and C ("PKA" and "PKC") are involved in control of protein activity precisely via phosphorylation of serine residues in proteins (amongst others).

https://en.wikipedia.org/wiki/Protein_kinase_A
https://en.wikipedia.org/wiki/Protein_kinase_C


What is a suitable acid to remove limewash from masonry brick

Glycolic acid (hydroxyacetic, HOCH2CO2H) is a mild organic acid, not very expensive, and the calcium salt is quite soluble. It is essentially a non-smelly vinegar! It has been used for brick washing.

One issue is that whereas HCl will dissolve CaCO3 until the HCl is all gone, the pH of glycolic acid solution increases as it dissolves CaCO3, so the dissolution reaction slows down and finally stops at maybe 50% completion.

Citric acid and tartaric acid are very soluble, but their calcium salts are not. Calcium citrate and tartrate may come down as fine crystals in the pores of the brick and be difficult to remove. Calcium formate is moderately soluble and so would be useful, but formic acid is a vesicant, so it must be used with as much or more care than HCl.

Muriatic acid is readily available at 32% and 20% concentrations. Diluting by a factor of 50 would give a pH about 2, which is safer for people and brick, but that's a lot of water to clean a brick wall.


Does the Battery Bounce Test actually work?

Yes.

The battery bounce test kind of works. You can tell whether the battery still has a charge greater than 90%, but bounce hight levels off at about 50% and doesn't go higher. This means that it is not clear whether the battery is completely discharged when you observe a bounce.

Why?

Steingart et al. just recently published a paper on the subject, and through scientific rigor could confirm this claim. The reason they give has  to do with how an alkaline battery is built:

(Source)

As you can see, a large part of the inside is devoted to the zinc anode. During the discharge, a redox reaction takes place to produce the current:
$$ \ce{Zn + 2MnO2 -> ZnO + Mn2O3} $$
The zinc (usually in some gel form at first) is slowly oxidized. As a result of this oxidization, crosslinks between the small zinc granules are formed, which make the whole center more rigid. As such, its "bounce" (or coefficient of restitution) becomes greater, because the downward kinetic energy cannot be fully absorbed by the gel anymore.

If you are interested in more details, check out the article they published. If you don't have access you can read a summary given on the Princeton University News web page, where you also can see an animation of a series of discharged batteries bouncing to different heights.


Which has greater dipole moment: cis-1,2-dichloroethylene or 1,1-dichloroethylene?

Dipole moment is a vector entity. Any applying dipole moment can be replaced by two perpendicular components:

For convenience to identify, I put $a$, $b$, $c$, and $d$ as the magnitudes of applying dipole moments of relevant $\ce{C-Cl}$ bonds. Yet they are theoretically identical: $a = b = c = d = \mu$.

Let's consider cis-1,2-dichloroethene first. The horizontal components of vectors $a$ and $b$, $a \sin 30^\circ$ and $-b \sin 30^\circ$, will cancel each other (negative sing is because of opposite direction of the vector). However, vertical components are in the same direction, thus they add together: $a \cos 30^\circ + b \cos 30^\circ = 2\mu \cos 30^\circ = 2\mu \times \frac{\sqrt{3}}{2} = \sqrt{3}\mu$. This is the net dipole moment of cis-1,2-dichloroethene.

Now, consider 1,1-dichloroethene. The vertical components of vectors $c$ and $d$, $-c \cos 30^\circ$ and $d \cos 30^\circ$, will cancel each other (again, negative sing is because of opposite direction of the vector). Here, horizontal components are in the same direction, thus they add together: $c \sin 30^\circ + d \sin 30^\circ = 2\mu \sin 30^\circ = 2\mu \times \frac{1}{2} = \mu$. This is the net dipole moment of 1,1-dichloroethene.

Therefore it is clearly, $\sqrt{3}\mu \gt \mu$. Realistically, $\mu_{1,1}$ is smaller than $\mu_{1,2}$ because of increasing electron density of two chlorine atom attached carbon in 1,1-dichloroethene compared to that of one chlorine atom attached carbon in 1,2-dichloroethene. This can be easily show using MaxW's values: $\mu_{1,1} = \mu = \pu{1.3 D}$ and hence, $\mu_{1,2} = \sqrt{3}\mu = \pu{1.3 D} \times \sqrt{3} = \pu{2.25 D}$, which is actually $\pu{1.9 D}$.


Pauli-forbidden term symbols for atomic carbon

Okay, actually I managed to find somewhat of an explanation myself.

We could construct a state with $M_L = 1$ and $M_S = 0$, for example:
$$\begin{align}
\psi = \frac{1}{2}\left[\mathrm{2p_1}(1)\mathrm{2p_0}(2) + \mathrm{2p_0}(1)\mathrm{2p_1}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right]
\end{align}$$

where $\mathrm{2p}_m$ represents the 2p orbital with $m_l = m$.

However, such a combination of $(M_L,M_S)$ can only occur once, since the two electrons are indistinguishable. In other words, the following two other wavefunctions are not permissible because permuting the two electrons changes the wavefunction:
$$\begin{align}
\psi &= \frac{1}{\sqrt{2}}\left[\mathrm{2p_1}(1)\mathrm{2p_0}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right] \\
\psi &= \frac{1}{\sqrt{2}}\left[\mathrm{2p_0}(1)\mathrm{2p_1}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right]
\end{align}$$

and the $^1D$ state, which has $S = 0 \Rightarrow M_S = 0$ and $L = 2 \Rightarrow M_L = 2,1,0,-1,-2$, already has to encompass one state with such a combination of $(M_L,M_S)$. Therefore, a state with $(M_L,M_S) = (1,0)$ cannot contribute to a $^1P$ term.

It is logically not the strongest argument (sufficiency vs necessity), but it is kind of what I was looking for. If anybody can provide another viewpoint though I will be glad to accept the answer.


Calculate concentration gradients in FPLC

Percentage of both A and B would change 70% from 24 to 144mL, with A from 75% to 5% and B from 25% up to 95%. The change rate would be 70% during the 120mL period or $0.70/120$ per mL. At any given point $x$ mL, between 24 and 144mL, concentration of $\ce{NaCl}$ in $\ce{mol/L}$ is:  $$\ce{[NaCl]} = (0.75 - 0.7\times (x-24)/120)\times0.2 + (0.25 + 0.7\times (x-24)/120)\times1.6$$

For a 15mL solution between 102 and 117mL, the concentration should be the average of the two points when x = 102, and 117mL.


Mixtures are never linear. Mixing $50$ mL water + $50$ mL ethanol gives a $96$ mL mixture. The masses are additive, but not the volumes. I agree the discrepancy is not huge. If you can tolerate a $10$% error, you may consider that the volumes are additive. Anyway take care of the fact that, usually percentage are indicated in mass, not in volume.


Is constant pressure heat capacity = entropy during adiabatic reversible expansion of perfect gas?

Inspired by Chet Miller.
The fallacy in the argument is that, ΔG = 0  because the process is reversible. This is not true and so leading to wrong final result.
ΔG is a signpost for spontaneity of a process only if the process occurs through constant temperature and pressure. Likewise, that ΔG = 0 means reversibility, is true only if the process happens through constant temperature and constant pressure.
Adiabatic expansion clearly involves change in temperature, so ΔG is not a signpost for spontaneity of adiabatic change, and hence ΔG ≠ 0 for reversible adiabatic expansion.


Is there a liquid that looks like water but boils at a low temperature?

The following list gives some boiling points

I've added the suggestions of Ben Norris and Todd Minehardt.

Common organic solvents

Pentane: 36 °C.
Diethylether: 35 °C.
MTBE: 55 °C.
Acetone: 56 °C.

Please read the Material Safety Data Sheets (MSDS), pay attention to the hazard statements and decide whether you (still) want to put your bare hands into these solvents!

Refrigerants

Dibromodifluoromethane: 23 °C
Trichlorofluoromethane: 24 °C
2,2-Dichloro-1,1,1-trifluoroethane: 28 °C
1,1-Dichloro-1-fluoroethane: 32 °C


When you heat the copper wire in a flame, it is oxidized on the surface to copper (II) oxide:

$$\ce{2Cu(s) + O2(g) -> 2CuO(s)}$$

Then when you mix it with the halide and heat it, the higher reduction potential makes it displace the oxygen.

Here's an example with sodium chloride:

$$\ce{CuO(s) + 2NaCl(s) -> CuCl2(g) + Na2O}$$

Copper halides are volatile (except for fluorides); the gas is hot enough to push some electrons into an excited state, and as they cool a tiny bit, the electrons drop back into their rest state and emit the excess energy in the form of a photon with a wavelength corresponding to the color green.


Solubility of Potassium Dichromate in Water

What happened was very simple.

I've been duped.

The seller sold me (intentionally or not) sodium dichromate and not potassium dichromate, or at least very contaminated with sodium dichromate.

According to online source (https://en.wikipedia.org/wiki/Solubility_table#S), the solubility of sodium dichromate in water is about 198g/100ml of water at 30°C, and a very high 415g/100ml at 100°C.

That explains clearly why everything dissolved promptly while the water was only slightly warm (40°C? Didn't come to mind put a thermometer in since I was expecting it to dissolve only near 100°C), since 500ml of solution would be able to dissolve 970g (my raw material) at only 30°C.

Also, when I crystallized some of it overnight on a cold (~15°C?) water bath, the solubility would drop to about 175g/100ml, causing the precipitation of about 100g of salt.

Finally, to settle matters about the presence of sodium dichromate, I dropped some saturated potassium chloride on a small sample of the solution in a test tube, and observed the instant formation of an orange precipitate.

The same occurred when I picked some of the crystals I separated before and made a small amount of saturated solution - therefore the crystals were made of sodium chromate and not dichromate - since they have much higher solubility they shouldn't've precipitated at all before potassium dichromate.

Also, I picked some potassium dichromate of known procedence and made some saturated solution and added a few drops of a saturated solution of potassium chloride, and nothing happened.

In the end, everything said, for what I intend to make, sodium dichromate or potassium dichromate are interchangeable after adjusting quantities, but I'll have to execute extra steps to purify it from this solution.

As an piece of advice (for me and everyone else), I'd test any new potassium/sodium dichromate from unknown sources before doing anything with it, simply by making some saturated solution and adding saturated potassium chloride.

I'll have to devise a procedure for separating one from another given unknown proportions of either.

Thanks for everyone time.


How does Walter White make pure crystal meth using a non-stereospecific reaction?

Intriguing question.

First, the best yield would be achieved by selectively producing one enantiomer instead of the other. In this case, White wants D-methamphetamine (powerful psychoactive drug), not L-methamphetamine (Vicks Vapor Inhaler). Reaction processes designed to do this are known as "asymmetric synthesis" reactions, because they favor production of one enantiomer over the other.

The pseudoephedrine method for methamphetamine employs one of the more common methods of asymmetric synthesis, called "chiral pool resolution". As you state, starting with an enantiomerically-pure sample of a chiral reagent (pseudoephedrine) as the starting point allows you to preserve the chirality of the finished product, provided the chiral point is not part of any "leaving group" during the reaction. However, again as you show, phenylacetone is achiral, and so the P2P process cannot take advantage of this method.

There are other methods of asymmetric synthesis, however none of them seem applicable to the chemistry shown or described on TV either; none of the reagents or catalysts mentioned would work as chiral catalysts, nor are they bio- or organocatalysts. Metal complexes with chiral ligands can be used to selectively catalyze production of one enantiomer, however the aluminum-mercury amalgam is again achiral. I don't remember any mention of using organocatalysis or biocatalysis, but these are possible.

The remaining route, then, is chiral resolution; let the reaction produce the 50-50 split, then separate the two enantiomers by some means of reactionary and/or physical chemistry. This seems to be the way it works in the real world. The advantage is that most of the methods are pretty cheap and easy; the disadvantage is that your maximum possible yield is 50% (unless you can then run a racemization reaction on the undesireable half to "reshuffle" the chirality of that half; then your yield increases by 50% of the last increase each time you run this step on the undesirable product).

In the case of methamphetamine, this resolution is among the easiest, because methamphetamine forms a "racemic conglomerate" when crystallized. This means, for the non-chemists, that each enantiomer molecule prefers to crystallize with others of the same chiral species, so as the solution cools and the solvent is evaporated off, the D-methamphetamine will form one set of homogeneous crystals and the L-methamphetamine will form another set. This means that all White has to do is slow the evaporation of solvent and subsequent cooling of the pan, letting the largest possible crystals form. Then, the only remaining trick is identifying which crystals have which enantiomer (and as these crystals are translucent and "optically active", observing the polarization pattern of light shone through the crystals will identify which are which).


I've collected rocks and tumble polished rocks. It is not possible that just sitting in bleach water tumble polished the stones. You can leave the stones in bleach water till you die and they still won't get polished. The only explanation is that you bought polished stones and put them into the bleach water. Over a period of time you just forgot what type of stones that you started with.

The other point is that "river rocks" (rocks smoothed in a river...) will look much smoother when wet because the water wets the surface and "hides" some of the surface roughness. So dry Petoskey Stones just look sort of mottled gray stones, but when wet the coral pattern shows nicely.


What does "reaction coordinate" mean? What does it mean when a reaction has TWO reaction coordinates?

The reaction coordinate is the progress of a reaction from reactants to products with various intermediates and transition states in between. It is an abstraction. It has no relation to time. Rather it is the progress of bond-forming and bond-breaking reaction steps. The free energy change of partially formed and partially broken bonds cannot be measured. The free energies of transition states (local maxima) cannot be directly measured. These states are transient and cannot be directly observed. For a simple reaction with few mechanistic steps, the activation energy (transition state energy) can be calculated from kinetics measurements. The free energy of intermediates can often be determined directly or indirectly if they are long-lived. Most of these diagrams are assembled from a few measured data points and either extrapolation or theoretical computation.

If two reaction coordinate axes are presented, this often means that there are two separate pathways the reaction can take, and each pathway is presented on one of the two reaction coordinate axes. Perhaps the reaction involves two different steps, but those steps need not happen in the same order every time. As a generic example, consider nucleophilic substitution reactions in organic chemistry: A generic nucleophile ($\ce{Nu-}$) reacts with an organic molecule R-X, where X is a leaving group, and R is the rest of the molecule (which is unchanged or nearly unchanged). The overall reaction would be

$$
\ce{Nu- + R-X -> Nu-R + X-}
$$

This process involves two steps: forming the bond between Nu and R, and breaking the bond between R and X. These two steps can happen in three different orders: 1) the R-X bond breaks first and the R-Nu bond forms second (this is the dissociative or $\mathrm{S_{N}1}$ mechanism); 2) The R-Nu bond forms first and the R-X bond breaks second (this is the associative or $\mathrm{S_{N}Ac}$ mechanism); and 3) The R-X bond breaks and the R-Nu bond forms at the same time (this is the intermediate or $\mathrm{S_{N}2}$ mechanism). All three of these cases occur in reactions that are covered in any undergraduate organic chemistry text. These three processes would look like the following. The items in parentheses are transition states with partially formed/broken bonds ($\ce{\bond{....}}$).

Dissociative:

Step 1: $\ce{Nu- + R-X -> (Nu- + R\bond{....}X) -> Nu- + R+ + X-}$
Step 2: $\ce{Nu- + R+ + X- -> (Nu\bond{....}R + X-) -> Nu-R + X-}$

Associative:

Step 1: $\ce{Nu- + R-X -> (Nu\bond{....}R-X) -> [Nu-R-X]-}$
Step 2: $\ce{[Nu-R-X]- -> (Nu-R\bond{....}X) -> Nu-R + X-}$

Intermediate:

Only step: $\ce{Nu- + R-X -> (Nu\bond{...}R\bond{....}X) -> Nu-R + X-}$

A reaction coordinate energy diagram for a substitution process could then be graphed over two different reaction coordinates: breaking the R-X bond and forming the R-Nu bond.  For ease of two dimensional representation, these are often presented as contour diagrams. Various trajectories could be taken through this space based on what real species R, Nu, and X represent. For example, if R+ is a stabilized cation, then the dissociative pathway becomes more likely. These plots are called More O'Ferrall-Jencks plots. This link will take you to the wikipedia article, which is not a very good article. The references in that article to the papers by R. A. More O'Ferrall and W. P. Jencks are probably better places to go to find out more. Most advanced undergraduate or graduate level textbooks of physical organic chemistry, particularly Advanced Organic Chemistry, Part A: Structure and Mechanisms by Carey and Sundberg, will cover this topic, at least as it relates to organic chemistry.


Sodium cyanide is slowly decomposed by water, but the implication is that it can still be used over short time periods provided that acids are kept away and proper safety practices are followed.  See here.


Why is this molecule considered a cis isomer of 2-bromobut-2-ene?

With the same reasoning as provided here, I recommend that you name this compound (2 E)-2-bromobut-2-ene.

Using (cis/trans) may be useful around an isolated double bond with only two substitutents. While in the example presented by you the two methyl groups are on the same side of the double bond, they are not the only ones to be considered here. Triple and quadruple substituted dienes should be named in accordance to the CIP rules only.

Edit by 02-Jan-2019:
Agreeing with a comment by @Loong, the previously suggested name, "(E)-bromo-2-butene", was replaced by "(2 E)-2-bromobut-2-ene".  This edit removes simultaneously the ambiguity about the position of the bromine, as well as the double bond.  The interested is referred to the illustrated examples of rule R-7.1.2 of the corresponding Blue Book here.


Is CH₃CH(CH₂CH₃)₂ the same as CH₃CH(CH₃CH₂)₂

In chemistry, as in any "language", there is what must be said to be correct: it is the 1st way your molecule is written.
Your second way of writing it is not wrong but it does not correspond to the globally institutionalized standard. The goal for any language is to be understood no matter where you are.
The conventions indicate that it should be written $\ce{CH3-CH-(CH2-CH3)2}$ which does not induce any ambiguity


Why is the the rate of a reaction measured in concentration?

Simply, because if you used an absolute amount (moles, for instance), by changing the quantity of reagents used, you couldn't use the existing data (or you would still need the volumes, to calculate the concentrations, in order to scale up/down the existing data, and fit them with your reaction).

Since the kinetics doesn't change with volumes, but with concentrations (for instance, if you use the same concentrations of reagents in 100 ml, or in 1 liter, the reaction, at least in theory, doesn't change) it's much easier and useful to express it in terms of concentrations.


Why is the Lucas test not recommended to differentiate higher alcohols?

The problem is that the alcohols themselves might be insoluble. Ethanol is soluble, yes, but as the alkyl chain length increases, the hydrophobic nature due to the chain starts dominating over whatever hydrophilicity the $\ce{-OH}$ group had been providing for lower homologues.

From source [1]:

  .. any alcohol that is insoluble in the reagent might appear to be giving a positive 3° test.

The paper goes on to properly quantify several things:

  Based on the results of this study, using 1 drop of alcohol added to 10 drops of reagent, it can be said that, at the very least, all saturated acyclic monofunctional alcohols having six or fewer carbons are soluble in the Lucas reagent.

It is clear that the relative amounts of the sample and the reagent become important in such borderline cases (~6 carbon long chains). If too much alcohol is added, it may cause an illusion of positive lucas test.

From source [1]: (emphasis mine)

  That statement may not be true when 3 or 4 drops are used
  per 10 drops of reagent. In fact, sometimes when 10 drops of
  reagent are added to 3 or 4 drops of some of the alcohols, an
  interesting event occurs after a delay of a couple of minutes
  or so. Apparently, some undissolved alcohol that has been
  clinging to the bottom of the test tube suddenly breaks loose
  and, by streaming to the surface of the very dense Lucas
  reagent, gives a momentary illusion of a positive test. The
  real cloudiness, alkyl chloride formation, appears later—all
  the more reason to use a high reagent/alcohol ratio.

References:

[1] "A study of the Lucas test"
R. A. Kjonaas and B. A. Riedford
J. Chem. Ed. 1991 68 (8), 704
DOI: 10.1021/ed068p704


Search for the last step of an altered Raschig process reaction mechanism

It can't be

$$\ce{R-NH-NH-R + 2R-NHCl -> R-N=N-R + 2R-NH2Cl}$$

The reason is that while the atoms are balanced, the electrons aren't. Going from $\ce{R-NH-NH-R}$ to $\ce{R-N=N-R}$ involves the loss of two protons and two electrons, but the $\ce{R-NHCl}$ are really only picking up two protons. (Nitrogens with four substituents are positively charged.)

When thinking of mechanisms, it may help to explicitly split the final listed reaction into two steps.

5a. $\ce{H-NH-NH-H + NH2Cl -> H-N=N-H + NH4Cl}$

5b. $\ce{H-N=N-H + NH2Cl -> N2 + NH4Cl}$

That is, each chloramine is involved in pulling one dihydrogen (two protons and two electrons) off of the diazo. - The addition of a "hydride" (a proton and two electrons) to the nitrogen kicks the chlorine off as a negatively charged ion, and the remaining proton protonates the lone pair on the nitrogen.

The analogous last step for the $\ce{R-NH2}$ case might be:

$$\ce{R-NH-NH-R + R-NHCl -> R-N=N-R + R-NH3Cl}$$

This is equivalent to reducing the $\ce{N-Cl}$ covalent bond to $\ce{R-NH2 + HCl}$ followed by an acid/base reaction.


Unknown white precipitate in electrolysis of water

The slight bluish tinge to that white precipitate makes it look to me more like copper(II) hydroxide than anything to do with the hardness minerals, especially since you haven't observed it with other anode metals thus far.

One way to test this would be to make a whole bunch of the precipitate, by running the current for a long time, and then taking a small sample of the liquid with precipitate in it and adding an equal volume of distilled vinegar (5% acetic acid).  If it is indeed $\ce{Cu(OH)2}$, I think it should dissolve to form a clear blue solution of copper(II) acetate.

The only problem might be if 5% acetic acid isn't able drop the pH far enough, in which case you'd have to track down a more concentrated acid. One choice might be $\ce{HCl}$, which is available at, e.g., aquarium supply stores under its historical name of muriatic acid. You'd then get a solution of copper(II) chloride, the color of which will vary depending on the $\ce{Cl-}$ concentration. Another option, per MaxW's comment, would be pickling vinegar (10% acetic acid) -- in fact, a quick search just revealed that Amazon has several vinegar options available in higher acidities.


Silver identification of the group I of cations

Both formulas for diamminesilver(I) chloride essentially mean the same. Writing it as

$$\ce{Ag(NH_3)_{2(aq)}^+ + Cl^{-}_{(aq)}}$$ emphasizes that the compound is dissolved in water as discrete ions (the cationic silver complex and chloride anions). One can also write $\ce{Ag(NH_3)_2Cl}$ instead because it is known that the compound exists in the aforementioned way in aqueous solution. In the latter case, it would still be more precise to use brackets to indicate the complex, i.e. $[\ce{Ag(NH3)2}]\ce{Cl}$, so that it becomes clear that the two $\ce{NH3}$ are the ligands of silver.


What is the difference between quantities reported as optical rotation and circular birefringence?

It gets very complicated, there are many papers and books on the topic and this little table, from Jensen et al. 1, gives a glimpse of the anisotropic effects:

An extremely short answer is this: when linearly polarized light is incident on an optically active medium, it can be thought of as consisting of equal components of right circularly polarized (RCP) and left circularly polarized (LCP) light. The indices of refraction for these are unequal, so RCP and LCP propagate at different speeds through the medium. This is circular birefringence. Upon exiting, it is linearly polarized light that exits, not the RCP and LCP fictive components. But the differential transit speed results in a rotation of the plane of polarization of the exiting linearly polarized light. This is optical activity, so optical activity is circular birefringence. The longer the optical path, the more the rotation and so on.
This is a truly vast topic, so I recommend not jumping directly into the deep end. Shurcliff's wonderful old book 2, maybe still findable, is extremely readable and a delight: the Poincare sphere is genius from a true genius! The book by Kliger et al. 3 is a good starting point and not insanely hard to understand. The paper by Jensen et al. 1 is rewarding, but definitely not easy reading.
There are many more papers, books, animations, and so on from which to learn. But be aware that notations are all over the place and reconciling equations and terminology can be very frustrating.
References:

H.P. Jensen, J.A. Schellman, T. Troxell, “Modulation Techniques in Polarization Spectroscopy”, Applied Spectroscopy 32 (1978) 192-200 (doi.org/10.1366/000370278774331567).

W.A. Shurcliff, Polarized Light, Harvard University Press,
Cambridge, MA, 1962.  For example on archive.org.

Kliger, D. S.; Lewis, J. W.; Randall, C. E. Polarized Light in Optics
and Spectroscopy, 1st ed.; Academic Press: Boston, 1990.  To borrow for  example from archive.org.

EDIT: In response to the OP's comment, I am adding two of my tables.

Circular birefringence (optical activity)
$$
\begin{align}
[φ] &≡ \text{molar optical activity}~(\pu{deg/cm M}) \\
\mathrm{CB} &≡ 2π(n_- - n_+)l/λ ≡ 2[φ]cl × π/180
\end{align}
$$

Circular dichroism
$$
\begin{align}
[θ] &≡ \text{molar ellipticity}~(\pu{deg/cm M}) \\
\mathrm{CD} &≡ \ln 10 (ε_- - ε_+)cl/2 ≡ 2[θ]cl × π/180 \\
g &≡ \text{dissymetry ratio} ≡ Δε/ε ≡ ΔA/A \\
ε &≡ \text{mean molar absorptivity} = (ε_- + ε_+)/2
\end{align}
$$

Complex angle of rotation, $χ$
$$
\begin{align}
χ &≡ χ' + iχ'' ≡ \mathrm{CB}/2 + i\mathrm{CD}/2 = φ + iθ \\
Δε &= 4π[θ]/180\ln 10 ≅ 0.03032[θ]
\end{align}
$$

Definitions of symbols and expressions
$$
\begin{align}
\mathrm{RCP} &≡ \text{right circularly polarized light (‘+’ subscript)} \\
\mathrm{LCP} &≡ \text{left circularly polarized light (‘−’ subscript)} \\
ε_± &≡ \text{molar absorptivities for RCP and LCP}~(\pu{L/mol cm}) \\
ε &≡ (ε_+ + ε_-)/2 ≡ \text{mean molar absorptivity}~(\pu{L/mol cm}) \\
A_± &≡ \text{absorbances for RCP and LCP} \\
A &≡ (A_+ + A_-)/2 ≡ \text{mean absorbance (true)} \\
n_± &≡ \text{(real) indices of refraction for RCP and LCP} \\
c &≡ \text{concentration}~(\pu{M}) \\
l &≡ \text{path length}~(\pu{cm}) \\
λ &≡ \text{wavelength}~(\pu{cm}) = 10^{-7} × \text{wavelength}~(\pu{nm}) \\
n &≡ (n_+ + n_-)/2 ≡ \text{mean index of refraction} \\
A_l &≡ \ln 10 (A_+ + A_-)/2 ≡ \text{mean absorbance} \\
η &≡ 2πnl/λ ≡ \text{phase (radians)} \\
[θ] &≡ \text{molar ellipticity}~(\pu{degree/cm M}) \\
[φ] &≡ \text{molar optical activity}~(\pu{degree/cm M}) \\
\mathrm{CD} &≡ \ln 10 (ε_- - ε_+)cl/2 ≡ 2[θ]cl(π/180) = 2θ \\
\mathrm{CB} &≡ 2π(n_- - n_+)l/λ ≡ 2[φ]cl(π/180) = 2φ \\
a &≡ \mathrm{CD}^2 - \mathrm{CB}^2 + \mathrm{LD}^2 - \mathrm{LB}^2 + \mathrm{LD'}^2 - \mathrm{LB'}^2  \\
b &≡ 2\mathrm{CD} × \mathrm{CB} + 2\mathrm{LD} × \mathrm{LB} + 2\mathrm{LD'} × \mathrm{LB'} \\B &≡ \left[(a^2 + b^2)^{0.5} + a\right]^{0.5}/2\sqrt{2} \\
C &≡ \left[(a^2 + b^2)^{0.5} - a\right]^{0.5}/2\sqrt{2} \\
Q &≡ B + iC
\end{align}
$$

The expressions for a, b, B, C, and Q are used in the general anisotropic medium Jones optical calculus and Mueller optical calculus matrices. These two matrices are denoted by J(GAM) and M(GAM), respectively, and are given in the paper by Jensen et al.


Since Al2O3 is very stable, powerful acids such as HF and chromic acid are used to remove the oxide surface. Since you're working with fine particles and wish to use a gas, rather than liquid, you'll need to consider what will become of the removed oxides. Gaseous HF would leave fluorides behind. Perhaps atomic hydrogen might work, but you'd have to remove the resultant water before the Al reoxidizes.

It is easier to prepare powdered aluminum in an inert atmosphere from bulk aluminum. There are a number of sites showing how to grind aluminum using anything from a file to a coffee grinder; you'd need to adapt this to run in argon, perhaps with a glove-box.

Sintering might be difficult, since Al is so soft that pore spaces would close.


How does the acidity of soft drinks not affect certain drugs, such as opiates? What about drugs and stomach acid?

Many drugs are not very soluble in water what is a common property for a lot of organic substances, including Codeine. Therefor it is very common to turn them into their hydrochlorid which is simply $\text{Drug}\ce{.HCl}$. [4] E.g., Crystal Meth got it's name from the crystals that the hydrochloride forms.

$\hskip1.7in$
Source: Wikimedia Commons

Other examples are, e.g., Cocaine, Cetirizine, hydroxyzine, and many more.

As your substance is either the hydrochloride or would turn into the hydrochloride in stomach there should not be any destructive influence on the structure of the drug. In fact it even promotes the absorption.

On the other hand, the better solubility of the corresponding hydrochloride is a relatively big problem for those drugs that should not get absorbed in the stomach. They need to be coated to "survive" the stomach acidity to reach the intestine where they should get absorbed.


NIST Webbook Chemistry is a reference which I would recommend as a first (as in initial) resource to consult for enthalpies of small organic molecular and inorganic materials.  The entry about methanol lists for one temperature the value in question, the experimental method of acquisition, and the primary literature reference.  The later may contain more data than the one tabulated here, too.

The references might be old, but in a scientific library you will find assistance to figure out who cited these publications which in turn may lead you to more recent determinations.


Photocatalysis - reactions caused by light. The term seems to be predominately related to free radicals.

This is obviously a misnomer. The photon is consumed which isn't true for a "real catalyst."

I know that lasers have been used to create excited molecular states which then react. I investigated this for reactions in gas phase of different atomic isotopes of a chemical.

I think if you get any more specific than that then it degenerates into "commonly accepted names" for certain reactions. Sort of like you just have to know what acetic acid is. There is no way to "figure out" what the name means if you don't just know.


What is the reason that tertiary alcohol pKa is higher than primary alcohol pKa?

The short answer is that the steric hindrance of the tert butyl alcohol makes it hard to add an protonate the hydroxy group.

Based on this lecture from the vanderbilt university. The answer is due to steric hindrance:

The hydroxy group of the tert butyl alcohol is intramolecular stabilized by hydrogen bonds (2nd image), between the methyl hydrogens and the oxygen. In the other hand, i.e. the hydroxy group of methanol is intermolecular stabilized with another methanol molecules ("Behave more like a water") (3th image). Therefore, the equilibrium $\ce{ROH}\rightleftharpoons\ce{RO^-}+\ce{H^+}$ is more displaced to the right with the methanol than with the tert-butil alcohol.

The protonation of ter butyl alcohol is slower than methanol (Figure 4th and 5th), for the same reason: steric hindrance.


What tools are used to predict the products of a chemical reaction?

I don't know for sure, but I suspect these two reagents would lead to the same product.  They are both superacids, and extremely harsh.

Whichever is a better Bronsted acid (proton donor) in benzene probably leads to a faster reaction, or better conversion.  But who knows?  Maybe the reactions are so acidic (two plus charges in a small organic molecule!) that they are just about the same.

There may be some superacid expert lurking on chemistry.se who knows this answer, but I think the only way to know is to do the reaction.

You may also want to write the corresponding author of the paper. They may already know the answer. Or, they might be flattered and may find the question interesting enough to go to the lab and do the experiment.

(I don't think computational chemistry will help here.)


What is the difference between a positive and negative electron orbital?

There are no such things as positive or negative electron orbitals. The number in the subscript is $m$, known as magnetic quantum number and running from $−\ell$ to $\ell$. It doesn't tell you much about the atom. It is a property of the orbital, not of the atom. Every atom has all these orbitals at once (some of them occupied and some empty); also, there are versions of each for different $n$.

In a free atom, all orbitals with the same $n,\;\ell$ and different $m$ have the same energy, so this number does not mean much for an orbital either (at least unless you put the atom in a magnetic field).

Moreover, the pictures are fake anyway. Orbitals which are the true eigenfunctions of $m$ (that is, correspond to specific $m$) are complex-valued and thus can't be properly represented in a plot. As for the orbitals featured on this diagram (and all over the textbooks), these are the linear combinations of complex orbitals for different $m$, hence each of them does not correspond to any particular $m$. It's not that they are wrong per se (absolutely not!), it's just that you can't point at them and tell which is $p_1$ and which is $p_{-1}$.


Lennard-Jones diagram - what is on the x-axis?

Ahh, silly me. The x-coordinate is of course just the distance between the $\mathrm A$ atoms / $\mathrm A_2$ molecule and the surface, to which they bind. This diagram doesn't show the interaction between $\mathrm A$ atoms but between them and the surface. I will close this question as solved.


How much heat is required to dehydrate a hydrate?

According to Wikipedia, copper(II) sulfate will dehydrate from penta- to trihydrate at 63 °C, then to monohydrate at 109 °C and then finally to the anhydrate at 200 °C. It melts at 150 °C, but won't decompose into ions until 650 °C which is beyond most benchtop rigs (your Pyrex flask would give way at about 500 °C).
The required heat of decomposition, as you can see, does vary for each hydrate (especially for the same salt), but doesn't always have to be greater than the boiling point; water has a significant vapor pressure at nearly all ambient Earth temperatures (even at freezing), and so in a dry environment, water will evaporate rapidly at temperatures well below boiling.


Why does substance that sublime not come to liquid state?

From a thermodynamic point of view there are two competing possibilities for condensation of a vapor:

(1) vapor $\ce{->}$ liquid, with some negative $\Delta H$

(2) vapor $\ce{->}$ solid, with an absolutely larger negative $\Delta H$

In the absence of large quantum mechanical effects (see here), the second process with a more negative enthalpy change must become favorable at a low enough temperature above absolute zero; the actual temperature where this happens  depends on the material.  So when the condensation is done at a low enough temperature, it goes directly from vapor to solid, and the reverse process of evaporation then also misses the liquid.


Baking soda is essentially sodium hydrogen carbonate, which should not be expected to react with oil because of its poor solubility. $\ce{NaHCO3}$ melts at 50 °C and starts to decompose slowly to sodium carbonate, carbon dioxide and water.

$$\ce{2NaHCO3 ->~ Na2CO3 + H2O + CO2\uparrow}$$

By the release of carbon dioxide, $\ce{NaHCO3}$ acts as a raising agent when used in baking. Under the conditions of deep frying (150-200 °C), the decomposition is fast as the reaction is entropically favored (generation of gases). $\ce{CO2}$ and $\ce{H2O}$ escape from the baking mixture, leaving sodium carbonate behind.

The ingestion of small amounts of $\ce{Na2CO3}$, which are generated from typically used amounts of baking soda, should not be a health problem. Acidic ingredients in the baking mixture will convert sodium carbonate into the sodium salts of the respective acids, carbon dioxide and water. Higher amounts of sodium carbonate, stemming from the use of too much baking soda, would cause an unpleasant metallic taste of the baked food. Sodium carbonate is actually used as a food additive (E 500) and in toothpastes as a foaming agent and abrasive.


The alcohol displaces water gradually through the process of osmosis. To prevent rupture of cells from the osmotic pressure, samples are immersed in a series of baths of increasing alcohol concentration.

After immersion in 95% or stronger alcohol, it is sometimes the practice to displace the alcohol with toluene or other non-polar solvent so that the specimen can be permanently mounted with paraffin or balsam.


What's the reddish powder that oozes out of a black rock?

Is your rock magnetic?  It might not be, but what you are reporting sounds like iron oxide (i.e. rust), and so your rock could be an iron ore like taconite, or even magnetite.  Such an ore could look like carbon.  Even if not magnetic, it could still be an iron ore.


Why is CuI2 not stable in solution while other copper(II) halide salts are?

It is common knowledge that copper (II) iodide ($\ce{CuI2}$) does not exist. Thus, your question "why is $\ce{CuI2}$ not stable in solution?" is clear on that, but unlike this previous question, you wasn't clear about what happens (or what you have observed). I have no knowledge of what would happen when some one add $\ce{KI}$ solution to $\ce{CuSO4}$ solution. Yet, I can speculate what would happen by considering few redox half reactions:

$$\ce{I2 + 2e- <=> 2I-} \qquad E^\circ = \pu{0.536 V} \tag{1}$$
$$\ce{Cu^2+ + 2e- <=> Cu} \qquad E^\circ = \pu{0.342 V} \tag{2}$$
$$\ce{Cu^2+ + e- <=> Cu+} \qquad E^\circ = \pu{0.153 V} \tag{3}$$

For convenience, let;s rewrite equation $(1)$ in oxidation format:

$$\ce{2I- <=> I2 + 2e-} \qquad E^\circ = \pu{-0.536 V} \tag{4}$$

Sum of $(2)$ and $(4)$ would gives the appropriate redox equation:

$$\ce{Cu^2+ + 2I- <=> I2 + Cu_{(s)}} \qquad E_\mathrm{EMF}^\circ = \pu{-0.194 V} \tag{5}$$

Similarly, sum of $(3)\times 2$ and $(4)$ would gives the appropriate redox equation:

$$\ce{2Cu^2+ + 2I- <=> I2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-0.383 V} \tag{6}$$

The excess $\ce{I-}$ would react with $\ce{Cu+}$ and $\ce{CuI}$ would precipitate as a result:

$$\ce{Cu+ + I- -> CuI_{(s)}} \qquad K_\mathrm{sp} = 1 \times 10^{-12} \tag{7}$$

Therefore, albeit forward reactions of equations $(5)$ and $(6)$ are not thermodynamically favorable (e.g., for equation $(5)$, $K_\mathrm{eq} \approx 2.7 \times 10^{-7}$), both forward reactions persist and they would be driven to completion because of one product separate as a solid (Le chatelier principle).

Let's now consider oxidation half reactions of other three halides:

$$\ce{2F- <=> F2 + 2e-} \qquad E^\circ = \pu{-2.866 V} \tag{8}$$
$$\ce{2Cl- <=> Cl2 + 2e-} \qquad E^\circ = \pu{-1.358 V} \tag{9}$$
$$\ce{2Br- <=> Br2 + 2e-} \qquad E^\circ = \pu{-1.087 V} \tag{10}$$

If $\ce{Cu^2+ -> Cu+}$ reduction happens, the respective three redox reactions are:

$$\ce{2Cu^2+ + 2F- <=> F2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-2.713 V} \tag{11}$$
$$\ce{2Cu^2+ + 2Cl- <=> Cl2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-1.205 V} \tag{12}$$
$$\ce{2Cu^2+ + 2Br- <=> Br2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-0.934 V} \tag{13}$$

And if $\ce{Cu^2+ -> Cu}$ reduction happens, the respective three redox reactions are:

$$\ce{Cu^2+ + 2F- <=> F2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-2.524 V} \tag{14}$$
$$\ce{Cu^2+ + 2Cl- <=> Cl2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-1.016 V} \tag{15}$$
$$\ce{Cu^2+ + 2Br- <=> Br2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-0.745 V} \tag{16}$$

Since values of $E_\mathrm{EMF}^\circ$s are so high negatives, these reactions are practically not happen. For example, $K_\mathrm{eq}$s are in the range of $6.14 \times 10^{-26}$ (for eq. $(16)$) and $1.55 \times 10^{-92}$ (for eq. $(11)$). Also keep in mind that, contrary to $\ce{CuI}$, $\ce{CuBr}$ and $\ce{CuCl}$ are also water soluble and $\ce{CuCl}$ is water soluble as well as hygroscopic.