Significance of Temperature in Gibbs Free Energy

Change in the Gibbs free energy ($\Delta$$\ce{G}$) of a reaction at any temperature can be determined by the equation <$$\ce{$\Delta$G = $\Delta$G^0 + RTlnQ_c (Q_p if the reactants and products are gaseous)}$$ Here, $\Delta$$\ce{G^0}$ is the standard Gibbs-free energy change of the reaction which is determined by,$$\ce{$\Delta$G^0 = $\Sigma$a_iG^0_i_{products} - $\Sigma$b_iG^0_i_{reactants}}$$ where $\ce{a_i}$s and $\ce{b_i}$s are proper stoichiometric coefficients. Now, $\ce{G^0}$ of products or reactants are determined by producing 1 mole of ech of them from their corresponding constituting elements in their standard states and at a FIXED TEMPERATURE(generally at $\ce{298.15K}$). So,  $\ce{$\Delta$G^0}$ is determined also at that particular temperature,So, it is independent of Temperature.  But,when we have to calculate $\ce{$\Delta$G}$ for any reaction at any temperature($T$) by the above given relationship , we have to know the temperature of the reaction condition as you can see clearly, $\ce{$\Delta$G}$ is directly  dependent on $\ce T$.That's why, it's important to know the temperature.


In the photoelectron spectroscopy (PES) experiment a sample is irradiated with photons of differing energy and the kinetic energy of any ejected electrons is measured.  By knowing the energy of the incoming photon and the kinetic energy of any ejected electrons, we can calculate the energy required to ionize the electron as the difference between these two recorded values. An electron will only be ejected if the photon energy is equal to or greater than the energy required to ionize an electron in the molecule.  See the section entitled "Photoelectron Spectroscopy" in this link if you would like to read more.

In the case of methane, all of the bonds are equivalent as required by symmetry.  There is only one $\ce{C-H}$ bond strength for methane.

When we run the PES experiment with methane we see 3 signals (as pictured in your link; two of the signals are due to photoionization of electrons from the $\ce{C-H}$ bond, and the third signal, at much higher photon energy, is due to photoionization of the carbon 1s electrons).  We know that the 4 bonds in methane can be described by 8 orbitals, 4 bonding, 4 antibonding.  We can transform these 8 localized orbitals into 8 molecular orbitals; the 4 bonding MO's are pictured in the link you provided.  Notice from the picture in your link that 3 of the 4 MO's are degenerate, but there is also a different 4th MO.  All of these bonding MO's are filled and ionization can occur from either one of the 2 types - hence 2 signals from the occupied $\ce{C-H}$ MO's (in the PES experiment, these are not sequential ionizations, e.g. these are not first and second ionizations; rather they are independent ionizations that will occur only when our incoming photon energy matches the energy of the orbital containing electrons).  During the PES experiment with methane, no bonds are broken, rather an electron from an orbital is ejected and the methyl radical cation is formed.


Two electron integral evaluation over primitive Gaussians without recurrence relations

As you mentioned, most implementations such as Obara–Saika (OS), McMurchie–Davidson (MD), Head-Gordon–Pople (HGP), and many others use a series of horizontal and vertical recurrence relations to build up angular momentum and transfer it among each of the four centers. These are analytic solutions, up until the need to solve the Boys function, however we could debate as to whether or not they are closed-form expressions by terminology. They are closed-form in the sense that each part of the recursive expansion does terminate properly with $(ss|ss)$, given in OS equation 44.

In general, any recursive function can be rewritten to be non-recursive using loops of sums and or products, though it may involve considerable effort. In that sense, the recurrence relationships are quite powerful and avoid lots of "boilerplate". If you are interested in a non-recursive expression, there is the earlier Taketa–Huzinaga–O-ohata (THO) paper that, while not very descriptive, includes explicit working expressions for 2-electron and 1-electron (overlap, kinetic energy, and nuclear-electron attraction) integrals. The most relevant expression is

\begin{align*}
\text{ERI} &= \frac{2\pi^{2}}{\gamma_{1}\gamma_{2}} \left( \frac{\pi}{\gamma_{1} + \gamma_{2}} \right)^{1/2} \exp \left\{ - \frac{\alpha_{1}\alpha_{2}\overline{AB}^{2}}{\gamma_{1}} - \frac{\alpha_{3}\alpha_{4}\overline{CD}^{2}}{\gamma_{2}} \right\} \\
&\times \sum_{i_{1}i_{2}r_{1}r_{2}u} B_{i_{1}i_{2}r_{1}r_{2}u}(l_{1}, l_{2}, A_{x}, B_{x}, P_{x}, \gamma_{1} | l_{3}, l_{4}, C_{x}, D_{x}, Q_{x}, \gamma_{2}) \\
&\times \sum_{j_{1}j_{2}s_{1}s_{2}v} B_{j_{1}j_{2}s_{1}s_{2}v}(m_{1}, m_{2}, A_{y}, B_{y}, P_{y}, \gamma_{1} | m_{3}, m_{4}, C_{y}, D_{y}, Q_{y}, \gamma_{2}) \\
&\times \sum_{k_{1}k_{2}t_{1}t_{2}w} B_{k_{1}k_{2}t_{1}t_{2}w}(n_{1}, n_{2}, A_{z}, B_{z}, P_{z}, \gamma_{1} | n_{3}, n_{4}, C_{z}, D_{z}, Q_{z}, \gamma_{2}) F_{\nu}(\overline{PQ}^{2}/4\delta),
\label{tho-2.22-1}\tag{THO eq. 2.22 part 1}
\end{align*}

with

\begin{align*}
&B_{i_{1}i_{2}r_{1}r_{2}u}(l_{1}, l_{2}, A_{x}, B_{x}, P_{x}, \gamma_{1} | l_{3}, l_{4}, C_{x}, D_{x}, Q_{x}, \gamma_{2}) \\
&\quad = (-)^{i_{2}} f_{i_{1}}(l_{1}, l_{2}, \overline{PA_{x}}, \overline{PB_{x}}) f_{i_{2}}(l_{3}, l_{4}, \overline{QC_{x}}, \overline{QD_{x}}) \\
&\qquad \times \frac{i_{1}!i_{2}!}{(4\gamma_{1})^{i_{1}}(4\gamma_{2})^{i_{2}}\delta^{i_{1} + i_{2}}} \cdot \frac{4\gamma_{1})^{r_{1}}(4\gamma_{2})^{r_{2}}\delta^{r_{1} + r_{2}}}{r_{1}!r_{2}!(i_{1}-2r_{1})!(i_{2}-2r_{2})!} \\
&\qquad \times [i_{1} + i_{2} - 2(r_{1} + r_{2})]! \frac{(-)^{u} p_{x}^{i_{1} + i_{2} - 2(r_{1} + r_{2}) - 2u} \delta^{u}}{u! \{ i_{1} + i_{2} - 2(r_{1} + r_{2} - 2u)\}!},
\label{tho-2.22-2}\tag{THO eq. 2.22 part 2}
\end{align*}

and intermediate quantities are

\begin{equation}
\nu = i_{1} + i_{2} + j_{1} + j_{2} + k_{1} + k_{2} - 2 (r_{1} + r_{2} + s_{1} + s_{2} + t_{1} + t_{2}) - u - v - w,
\end{equation}

\begin{align*}
\gamma_{1} &= \alpha_{1} + \alpha_{2},~\gamma_{2}\text{ is similar}\\
\delta &= \frac{1}{4\gamma_{1}} + \frac{1}{4\gamma_{2}} \\
\mathbf{P} &= \frac{\alpha_{1} \mathbf{A} + \alpha_{2} \mathbf{B}}{\alpha_{1} + \alpha_{2}},~\mathbf{Q}\text{ is similar} \\
\mathbf{p} &= \mathbf{Q} - \mathbf{P} \\
\overline{AB} &= A - B
\end{align*}

\begin{equation}
F_{\nu}(t) = \int_{0}^{1} u^{2\nu} \exp(-tu^{2}) \mathop{du},~\text{the Boys function.} \tag{THO eq. 2.11}
\end{equation}

$(l, m, n)$ are the usual polynomial exponents leading to s-, p-, and d-type functions, $(\mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D})$ are the usual basis function positions, the $\alpha$ are the basis function exponents, and $f_{j}(l, m, a, b)$ is the coefficient of $x^{j}$ in the expansion of $(x + a)^{l} (x + b)^{m}$.

  In \eqref{tho-2.22-1} the summations with respect to the indices $i_{1}, i_{2}, r_{1}, r_{2}, u$ extend from 0 to $l_{1} + l_{2}$, $l_{3} + l_{4}$, $[i_{1}/2]$, $[i_{2}/2]$ and $[(i_{1} + i_{2})/2 - r_{1} - r_{2}]$ respectively. The ranges of $(j_{1}, j_{2}, s_{1}, s_{2}, v)$ or $(k_{1}, k_{2}, t_{1}, t_{2}, w)$ may be found in the same way.

Regarding the indices used in the summations:

If the index $i_{1}$ runs from 0 to $l_{1} + l_{2}$, and $i_{2}$ runs from 0 to $l_{3} + l_{4}$ we can deduce that $i_{1}$ must cover all possible values of the total angular momentum along the x-component for the bra, and $i_{2}$ goes up to the total angular momentum along the x-component for the ket. The indices $j,k$ match with $m,n$, meaning the y- and z-component.
$r,s,t$ are not to be confused with distances themselves, like how we usually define $r_{A} = |\mathbf{r}_{A}| = |\mathbf{r} - \mathbf{A}|$. Both $r,s,t$ and $u,v,w$ are also related to the three Cartesian coordinates, but in a more complicated manner than $i,j,k$; they appear to originate from the closed-form expressions for the Hermite polynomials.

Their implementation is slightly modified, with the necessary expressions in equations 3.3, 3.4, and 3.5, though the above can be implemented directly as well. A quick search revealed this code that implements all the THO integrals in C++. It is worth commenting that in real packages, the Boys function is precomputed from interpolation tables, rather than from the gamma and incomplete gamma functions.


Diagonal line going through groups IIIA-VIA on the periodic table

The red line is the dividing line between metals and non-metals. It is rather arbitrary; as Mendeleev himself wrote: “It is...impossible to draw a strict line of demarcation between metals and nonmetals, there being many intermediate substances.” Most modern periodic tables do not feature it.

The elements with a blue background are the metalloids. Against, it's a rather arbitrary choice, but it corresponds to elements that have some properties of metals. Wikipedia describes them in the following way:

  Physically, metalloids usually have a metallic appearance but they are brittle and only fair conductors of electricity; chemically, they mostly behave as (weak) nonmetals. They can, however, form alloys with metals. Ordinarily, most of the other physical and chemical properties of metalloids are intermediate in nature


When dispensing liquid nitrogen, why does the surface get wet rather than frozen?

Your "first thought" is basically right, with one clarification. The addition of "so much new water" to the existing ice results in the formation of liquid water due to the latent heat released upon condensation. This results in melting of the existing frost and thus the transitory existence of liquid water.

Also you can rule out oxygen condensation.  Firstly there is some insulation between the 77 K liquid nitrogen and the atmosphere, so the surface is not likely to cool all the way to the 90 K boiling point of oxygen.  More importantly, oxygen would not condense from the air at 77 K anyway.  You have to think in terms of saturation vapor pressures, and the vapor pressure of oxygen in our air is only about 0.21 atmospheres.

By analogy, albeit an extreme one, the temperature in the coldest regions of our atmosphere drops below the sublimation point of carbon dioxide.  Does that cause the formation of dry-ice clouds?  Not even close. The sublimation point of carbon dioxide in this example, like the boiling point of oxygen in your example, is only relevant at 1 atmosphere of pressure of that compound, and in neither case does our atmosphere contain 1 atmosphere of either of those compounds.


First of all, How can you assume that, $\Delta S$ is positive in any exothermic reaction? Secondly, the relation you have written $\Delta G = -RT\ln K_\text{eq}$ is not at all correct. The correct relation is $\Delta G = \Delta G^\circ + RT\ln Q_c$ (or $Q_p$ if the reactants are gaseous). Now at equilibrium $\Delta G= 0$ and $Q = K$. So, the correct relation is

$$\Delta G^\circ = -RT\ln K_\text{eq}$$

And, there are obviously differences between $\Delta G$ and $\Delta G^\circ$. So, the reasoning you have given is not totally correct.

The correct reasoning comes from Van't Hoff Equation, which states $$\frac{\mathrm d}{\mathrm dT}(\ln K_{\text{eq}}) = \frac{\Delta H}{RT^2}$$

Integrating both sides, we have

$$\int_{K_1}^{K_2}\mathrm d(\ln K_{\text{eq}} ) = \int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\,\mathrm dT$$

So, we have $$\ln\frac{K_2}{K_1}= \frac{\Delta H}{R}\left(\frac{1}{T_1} -\frac{1}{T_2}\right)$$ Thus we see that increasing the temperature from $T_1$ to $T_2$, the equilibrium constant changes. Here $\Delta H < 0$ (exothermic reaction) and $\displaystyle\frac{1}{T_1}- \frac{1}{T_2} > 0$ so, $\displaystyle \ln\left(\frac{K_2}{K_1}\right) < 0$, which means $\displaystyle {\frac{K_2}{K_1}< 1}$, which implies $K_2 < K_1$. So, the equilibrium constant decreases.

Note: Without any mathematical equation also, the same result can be deduced. In an exothermic reaction, the products are formed along with heat production, which is the definition of exothermic reaction. That means, temperature of the system will increase due to this heat production. Now if you increase the temperature, the system will obviously want to go in the direction where this temperature minimisation is possible, which is actually the backward reaction. So, the reaction shifts backwards, and thus the equilibrium constant decreases as now the product concentration has become lesser and reactant concentration has become larger.


Is this approach to finding density in crystal lattice acceptable?

Yes of course, if you do the math right, that limes gives the exact same number.

Don't forget to look out for atoms that are completely within the unit cell.


Oxidizer/reducer in decomposition of potassium chlorate

If I remember my introductory lab course correctly, there are two reactions going on, one after the other. First, $\ce{ClO3-}$ disproportions according to the following equation:

$$\ce{4 ClO3- -> Cl- + 3 ClO4-}$$

Which has these two half-equations:

$$\ce{ClO3- + 6 e- + 6 H+ -> Cl- + 3 H2O}\\
3(\ce{ClO3- + H2O -> ClO4- + 2 e- + 2 H+})$$

Where the formally required water is merely spectator.

Only upon further heating does the perchlorate anion decompose according to:

$$\ce{ClO4- -> Cl- + 2 O2}$$

Which again can be broken down to:

$$\ce{ClO4- + 8 e- + 8 H+ -> Cl- + 4 H2O}\\
2(\ce{2H2O -> O2 + 4 e- + 4 H+})$$

And again formally requires water as a spectator.

Note that it is not at all unusual for one entity to be both; the simplest example is probably the disproportion of hydrogen peroxide according to:

$$\ce{2 H2O2 -> O2 + 2 H2O}$$

Where one molecule is reduced by another of the same type.


Why does fluorine act as a leaving group in organophosphates?

I am reluctant to post this as an answer because it is not really backed up by much research, but since nobody has said anything yet, here's my theory:

  It is probably to do with the fact that the covalently bound sarin (which necessarily involves kicking out either $\ce{RO-}$ or $\ce{F-}$) mimics the transition state for acetylcholine hydrolysis too well, such that the expulsion of one of those two leaving groups is much more favoured than in aqueous solution.

Acetylcholinesterase (AChE) is very similar to a typical serine protease, and serine proteases are very well-studied. I am going to postulate that the mechanism of acetylcholine (ACh) hydrolysis in AChE is exactly the same as it is in a typical serine protease. So first, let's look at what the enzyme does normally. It's nothing special, just a nucleophilic acyl substitution:

(There's an additional glutamate residue, analogous to the aspartate in a serine protease, that fixes the conformation of His440, but it's not directly relevant to the mechanism. Also, there's an anionic site which binds to the positively charged quaternary nitrogen, but that's also irrelevant.)

The acyl-enzyme intermediate then goes on to be hydrolysed by water (not relevant here).

The reason why the enzyme can catalyse the reaction is because it stabilises the transition state for the ester hydrolysis, which means that the activation energy for hydrolysis is decreased and the rate is increased. (Strictly, it provides an alternative hydrolysis pathway in which the activation energy is lower - but the point is that since the catalysed hydrolysis is mechanistically similar to the uncatalysed hydrolysis, the enzyme must play a role in stabilising a tetrahedral intermediate or transition state.)

The transition state looks something like this (partial charges omitted):

Sarin is a potent AChE inhibitor because it covalently binds to the active serine residue as in the left-hand side of this diagram:

You can see that structurally, this is very similar to the transition state shown above. If the enzyme stabilises the TS above, this covalent binding of sarin to the enzyme must also be extremely stable. I do not pretend to know exactly how it is stabilised, but it must be related to the specific 3-D conformation of the enzyme. If you do not kick out the fluoride ion (right-hand side), then it does not quite resemble the transition state so well.

What this means is that there is a strong driving force for the fluoride ion to be expelled, in order for it to reach that stable, 4-coordinate phosphorus form.

That leaves one question that I feel is worth addressing, which is why fluoride is expelled instead of isopropoxide. Honestly, I do not know. As pointed out by jerepierre, fluoride is certainly a better leaving group than isopropoxide ($\mathrm{p}K_\mathrm{a}(\ce{HF}) = 3.17$, $\mathrm{p}K_\mathrm{a}(\ce{^{i}PrOH}) = 16.5$ - quite a large difference), and it might very well just be that. On the other hand, as with most biological systems, it is difficult to give a full explanation of why things happen the way they do, and applying (relatively) simple chemistry heuristics might not be the full picture. As to what other factors might influence this, I can only guess that is that there might be some hydrogen bonding arrangement that requires the specific size of the oxygen in the $\ce{-OR}$ group over the size of the fluorine (in the catalysis TS, there is an oxygen there as well). Essentially, I am saying that the LH diagram resembles the TS more than the RH diagram (don't quote me on it though...):


How is the the law of chemical equivalence defined and what are its limitations?

Basically the law of equivalence wants you to balance the equivalents that are involved in your reaction. I'd like you to view you reactions in two parts, the oxidation and reduction halves. I see:

$$\begin{align} \ce{I- -> I2} \tag{oxidation} \\
\ce{IO3- -> I2}\tag{reduction} \end{align}$$

In other words, we've just discovered a (reverse disproportionation?) comproportionation reaction. The change in the oxidation state of iodine in the oxidation part is $-1 \to 0 $ ($\Delta_{o.s.} = 1$) and for the reduction half it's $+5 \to 0$ ($\Delta_{o.s.} = 5$). So far so good.

Usually, the questioner would provide you with the number of moles of each species and you're done with number of moles times the n-factor. Nevertheless, let's try to give it a shot. I'm balancing this equation:

$$\ce{1KIO3 + 5KI + 3H2SO4 \to 3I2 + 3K2SO4 + 3H2O}$$

This means for $1$ mole of $\ce{KIO3}$ to react, you'll need $5$ moles of $\ce{KI}$. As this is a proportion, converting it to an absolute value, I'd say you've $1n$ moles of $\ce{KIO3}$ and $5n$ moles of $\ce{KI}$.

So if I try to equate the gram-equivalents, I'll have $1n \times 5 = 5n \times 1$. In other words, $\text{Eq.}_{\ce{KIO3}} = \text{Eq.}_{\ce{KI}}$

  I've heard that the law of chemical equivalence states that the gram equivalence of each of the reactants equals the gram equivalence of each of the products.

Law of equivalence is even more fun, for a balanced chemical reaction:

$$ \ce{aA + bB -> cC + dD}$$

Equivalents of every species will be equal. Mathematically,

Equivalents of A = Equivalents of B = Equivalents of C = Equivalents of D

will hold.

  Why aren't the gram equivalents of $\ce{H2SO4}$ equal to that of water?

You need to strip your equation of the spectating species. For instance $\ce{H2SO4}$ is primarily here because it acts as a medium-provider, it makes the solution acidic. Neither $\ce{SO4^2-}$ nor $\ce{H+}$ underwent any oxidation/reduction so applying the equivalent concept makes no sense in this case.


Molar concentration of water and pKa value

For most practical purposes, we assume that the aqueous solution is dilute and that the concentration of water is constant, that is, $[\ce{H2O}] \approx 55~\mathrm{M}$. This is the standard that is used to define the ionization constant of water, $K_\mathrm{w}$.

If the equation $1\times 10^{-14}~\mathrm{M^2} = [\ce{H3O+}][\ce{OH-}]$ didn't hold (at $25~\mathrm{^\circ C}$) then we would have to come up with other equations for acid-base equilibrium. Even for strong acid/base dissociations we approximate the equilibrium equation with full dissociation (i.e. one forward arrow).

I recommend reading pp 12-15 of the following if you want a more theoretical treatment of strong acid/base dissociation where activities are used: http://www.chem1.com/acad/pdf/c1xacid2.pdf


Denaturing of Proteins and Nucleic Acids - Effect of Temperature (Heating vs Cooling)

Complimentary nucleic acids are held together mainly by hydrogen bonding, so I'm not sure if they are able to be cold denatured, but cold denaturation is a well known phenomenon for proteins. Here is a decent review of the processes involved. In short, most proteins' tertiary structure is dominated by hydrophobic interactions and the energy barrier to unfolding a protein is generally quite small. The stabilizing effect of the hydrophobic interactions decreases with temperature and at a certain point, is insufficient to keep the protein folded. For many proteins, the cold denaturation temperature is below 0 °C, so it's difficult to study. Generally, studies on these proteins are done by adding chemical denaturants or changing the pressure of the system to shift this temperature higher, having the risk of conflating this deliberate denaturation with the temperature effect. Of course, not all proteins are stabilized this way and things like metallothioneins probably don't denature in this way as their tertiary structure is maintained by coordinating metal ions.


Finding is the concentration of H⁺ in a benzoate buffer solution after adding HCl

Initially, it's true that you have the following equilibrium between sodium benzoate and benzoic acid:
$$\ce{C6H5COOH + H2O <=> C6H5COO- + H3O+_\mathrm{(aq)}}$$
When you add  $\ce{HCl}$ to your buffer solution, sodium benzoate reacts with the acid, and the aforementioned equilibrium is displaced to the left:
$$\ce{C6H5COO- + H3O+_\mathrm{(aq)}  <=>  C6H5COOH + H2O }$$
Now, the equilibrium constant of this reaction is:
$$K=\frac{1}{K_ a}=0.16 \times 10^5>10^3$$
The reaction is quantitative (total). So, we can rewrite it:
$$\ce{C6H5COO- + H3O+_\mathrm{(aq)} ->  C6H5COOH + H2O }$$
Now, we have to construct an ICE table:
Assuming that we didn't dilute the solution when we added $\ce{HCl}$ (this point is not clear in your question): $$\ce{[C6H5COOH]_0} = \pu{0.12 M}$$
$$\ce{[C6H5COO^-]_0} =  \pu{0.21 M}$$
$$\ce{[H^+]_0} = (\frac{0.050}{1.3} + 10^{-4.4}) \mathrm{M}\approx 0.0385 \mathrm{M}$$
As you can clearly seen $\ce{[H^+]_0}$ is the lowest, and so ion hydronium is the limiting reactant.
At the end of the reaction, we have:
$$\ce{[C6H5COO^-]_\mathrm{(eq)}} = (0.21-0.0385) = \pu{0.1715 M}$$
$$\ce{[C6H5COOH]_\mathrm{(eq)}} = (0.12+0.0385) = \pu{0.1585 M}$$
$$\ce{[H^+]_\mathrm{(eq)}} = \epsilon$$
Substituting all these numbers in the expression of $K$
$$K = \frac{ 0.1585}{0.1715\epsilon}=0.16 \times 10^5$$
By solving the above equation, we find:
$$\ce{[H^+]} = \epsilon = \pu{5.82 \times 10^{-5} M}$$
$$\mathrm{pH} = 4.23$$


Can precipitation reactions attain equilibrium?

The statement is inaccurate. A better statement would be something like:

A chemical equilibrium can not be established if one of the products is continuously removed (i.e. its concentration always decreases).

If one of the products is a gas and can mix with earth's entire atmosphere, that would make its concentration (or partial pressure) drop to almost zero. In order to establish an equilibrium, however, all concentrations have to be non-zero. Otherwise, the equilibrium constant will not be equal to the reaction quotient, and there will continue to be a net reaction.
In case of the precipitation, the solute is not continuously removed. At some point, the solution will no longer be supersaturated, so then the concentration of solute will be constant.

My teacher told me about about a reaction
$\ce{Ca(HCO3)2(aq) + heat -> CaCO3 +CO2 +H2O}$
and on cooling the solution and bubbling CO2(g)
through it the CaCO3(ppt.) reacts back to form calcium bi-carbonate.

Once you change the temperature, you don't expect the reaction to stay at equilibrium. So this does not support either the original statement or the opposite statement.

But I want to know if precipitation reactions can attain equilibrium. Why does the book state otherwise?

Yes, it can. Here is an example:
$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}$$
The equilibrium constant expression for this reaction is:
$$K = \frac{1}{[\ce{Ag+(aq)}][\ce{Cl-(aq)}]}$$
(I could have written the chemical equation in the other direction, and then the equilibrium constant would have been the solubility product.) So the product is a solid, and its concentration does not change. The silver and chloride ion concentrations will drop until the equilibrium has been reached.
Here is another set of reactions:
$$\ce{CoCl4^2-(aq) + 6 H2O(l) <=> Co(H2O)6^2+(aq) + 4 Cl-(aq)}\tag{1}$$
$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}\tag{2}$$
Here, the equilibrium of the first reaction will be far on the right side because the second reaction removes a lot (but not all) chloride ions from the solution if the silver ions are in excess of the chloride ions. Still, the system will reach equilibrium.

Why does the book state otherwise?

I don't know. Maybe they consider an equilibrium where one of the species is at extremely low concentration not an equilibrium. Or maybe it is a typo.
It is great to have a critical mindset when reading books and hearing people talk about chemistry. Even if all the statements are true, it helps you learn.
edit

I want to know if in a reaction like
$$\ce{AgNO3(aq) + NaCl(aq)\to AgCl(s,ppt) + NaNO3(aq)}$$
the precipitate $\ce{AgCl(s,ppt)}$ can react with $\ce{NaNO3(aq)}$ in the solution to give the reactants?

Yes. I would rather write it like this, though:
$$\ce{Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) <=> AgCl(s) + Na+(aq) + NO3-(aq)}$$
and then leave out the spectator ions:
$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}$$
The following "reaction" might serve as illustration why leaving the spectator ions in the equation is not a good idea in preparation for writing the equilibrium constant expression:
$$\ce{KCl(aq) + NaNO3(aq) <=> KNO3(aq) + NaCl(aq)}$$
For this "reaction", the concentration of species is undefined. Writing the equilibrium constant would be strange because how can I tell apart $\ce{KCl(aq) + NaNO3(aq)}$ from $\ce{KNO3(aq) + NaCl(aq)}$ if they fully ionize in solution? If I show the ions separately, it becomes clear that the "reaction" is not a reaction:
$$\ce{K+(aq) + Na+(aq) + NO3-(aq) + Cl-(aq) <=> \mathrm{the\ same???}}$$

I've seen this and this but couldn't understand.

Those two Q&A's are quite confusing, I agree.


Atoms at the edge of a crystal that have an unsatisfied valence are said to have "dangling bonds."  Many elements, in addition to carbon, can have dangling bonds.  Dangling bonds is a subject of current interest because of the impact these structures can have on semiconductor properties.

These dangling bonds are very similar to free radicals, except since they are immobilized in a solid, they are somewhat less reactive than free radicals in solution. Nonetheless, they can react with whatever materials they are exposed to, such as hydrogen, water vapor, oxygen, etc. In addition, if there is a neighboring dangling bond then they can both react with one another to form a bond and satisfy there valence.

When carbon or silicon surfaces are prepared under clean room conditions, the dangling bonds can persist.  In the semiconductor industry this clean room preparation technique is followed by bringing in a doping gas in order to purposefully alter the electronic band structure of the substrate material.

Since unpaired electrons have magnetic properties, in carbon (or any other element) nanostructures where there is a lot more surface area to volume, the concentration of dangling bonds is much higher.  Consequently, the unpaired electrons in the dangling bonds confer magnetic properties on these materials that are large enough to be easily  detectable and to manipulate.

Finally, since dangling bonds represent a non-equilibrium situation, surfaces containing dangling bonds undergo a relaxation or reshaping that is referred to as "surface reconstruction."

Here is a full-text article that should help you get started:
1) Structure of the diamond 111 surface: Single-dangling-bond versus triple-dangling-bond face


What is the chemistry behind this "petrol from air" technology?

Step one: trap $\ce{CO2}$. This can be done using specially selected amines, like triethanolamine. Aminosilicones (a common chemical used in hair conditioner), can also do this.

Step two: release trapped $\ce{CO2}$. In case of triethanolamine you'll have to boil mixture.

Step three: mix $\ce{CO2}$ with hydrogen (hydrogen is produced by electrolysis of water). Volume proportion should be somewhere 1:3.

Step four: threat the mixture with proper catalyst. This step may be divided into two stages: reverse water-shift reaction  and some variant of Fischer–Tropsch process with some type of alcohol synthesis as very interesting candidate. There are some multi-functional catalysts that can perform both stages at once.

Another variation on step 4 is the Sabatier process, which involves generating methane from $\ce{CO2}$ and hydrogen.  Typically the catalyst is Rainy nickel (powder).

The required energy in form of electrical current is probably from 1.5 to 3 times of heat of combustion of resulting fuel, which is in turn at least twice more that can be converted into actual work again.


Unique calling code for a chemical reaction

I don't know about Godel numbering specifically (though this is interesting, my being a math guy at heart), but there are existing textual systems for encoding molecules (not reactions that I know of).  From these, you may be able to encode reactions.

Two examples are InChI and SMILES.  These are Wikipedia links.


The karat is a unit used specifically for describing the purity of gold alloys:

Various ways of expressing fineness have been used and two remain in common use: millesimal fineness expressed in units of parts per 1,000 and karats used only for gold. Karats measure the parts per 24, so that 18 karat = 18/24 = 75% and 24 karat gold is considered 100% gold.
The karat (not carat as a unit of mass) (symbol: K or kt) (US) or carat (symbol: C) is a unit of purity for gold alloys.

Note that in the U.S., the symbol K or kt is used; in your locale, the symbol C and the word carat might be used, from which confusion might arise when talking about the unit of mass carat.
The carat is a unit of mass used for, well, telling us what the mass of gemstones are:

The carat (ct) is a unit of mass equal to 200 mg (0.2 g; 0.007055 oz) and is used for measuring gemstones and pearls. The current definition, sometimes known as the metric carat, was adopted in 1907 at the Fourth General Conference on Weights and Measures, and soon afterwards in many countries around the world. The carat is divisible into one hundred points of two milligrams each. Other subdivisions, and slightly different mass values, have been used in the past in different locations.


Are there any tricks I can use to pull SO2F2 out of the air?

Sulphuryl Fluoride is chemically fairly stable, considering it's a pesticide - toxic effects are caused by chemical reactions which means they can't be inert.

However, according to the Vikane documentation, it's not UV stable. UV lighting is fairly widely available, but TBH a sunny day will do far more.


Final update, all earlier edits incorporated.

Groundrules: Considering compounds with:

only carbon and hydrogen
only 4 bonds to carbon

There are 37 isomers without considering trans isomers; 49 when trans isomers are included.  Also, many of these compounds seem extremely unstable and therefore unlikely to exist.

Note to self: check back in 20 years and see how many of the unlikely ones have been detected.

Names by row:

butane, isobutane
but-1-ene, but-2-ene, 2-methylpropene
buta-1,3-diene, buta-1,2-diene, buta-1,2,3-triene
but-1-yne, but-2-yne, but-1-ene-3-yne, buta-1,3-diyne
methylcyclopropane, 2-methylcyclopropene, 1-methylcyclopropene, methylenecyclopropane, methylenecyclopropene, methlycyclopropadiene
methylcyclopropyne, methlenecyclopropyne
cyclobutane, cyclobutene, cyclobuta-1,2-diene, cyclobuta-1,3-diene, cyclobutatriene, cyclobutatetraene, cyclobutyne, cyclobutenyne, cyclobutadiyne
bicyclobutane, bicyclobut-1(3)-ene, bicyclobut-1(2)-ene, bicyclobuta-1,3-diene, bicyclobuta-1,2-diene
tetrahedrane, tetrahedrene, tetrahedradiene


Would the change in enthalpy change of chemical reaction in space be considered as change in internel enrgy?

Well, in principle yes, for example if you had a blob of liquid floating in empty space and wanted  to consider the whole blob as a thermodynamic system.

But it may be worth mentioning that wouldn't be a typical use of thermodynamics in astrophysics.  Usually you'd consider your thermodynamic system to be some small part of a larger system and the larger system would be a thermal and volume reservoir, so T and P would both be nonzero.

For example, it's common to consider the thermodynamics of stars, but what you do is consider a small packet of gas to be your system, and the rest of the star to be surroundings that establish a constant T and P for your packet.


In the nonideal solution, why can't one of the solute be positive deviation and the other be negative deviation?

For multiple component systems we use activity coefficient $\gamma$ in modified Roult's law as:
$$K_i=\gamma \frac{P_i^\star}P $$

and $\gamma$ is defined in the terms of $f$ : fugacity or fictitious pressure for a real gas.

You should read further about "Vapour Liquid Equilibrium" for multiple components.

Maybe:  http://www.journalrepository.org/media/journals/BJAST_5/2012/Dec/1355741742-Olaleke312012BJAST1789.pdf. You can see there is no reason why it can't have deviation in multiple directions.


Does bleach react with disinfectants containing quaternary ammonium salts?

Calling poison control first was the best thing you could do! Thumbs up!

Please note that we're usually reluctant to give medical advice - we don't want to get you or anyone in your house get killed or harmed either.

You might want to call poison control again and/or send them the safety data sheet of the product.

The following is just additional information to give you some direction on why it is probably safe to use the mold remover on the floor previously treated with bleach.

The College of Veterinary Medicine at Cornell University has published Guidelines for the Use of Bleach. According to this document,

  Bleach can safely be mixed with quaternary ammonium compounds that contain detergents, such as A-33 and Roccal to create a single solution with both cleaning and disinfecting properties.

So, I'd dare to say that the combined use, as well as subsequent treatment of the floor is safe, but if I were you, I would search again confirmation with official poison control rather than trusting some guy from the internet ;-)


Do ions still "conduct electricity" if a physical connection made by aqueous electrolytes between the anode and the cathode is nonexistent?

If a cathode (negative pole) is placed alone in an ionic solution (without any anode in the solution), it will attract cations from the solution, and discharge maybe one or two or one thousand of these cations. This will last maybe 1 nanosecond, and then it will stop. The reason is that this number of atoms is extremely small, so small that the neutral atoms which are deposited on the cathode cannot be seen or weighed. And then the cathodic discharge will stop, because the whole solution has lost positive charges (cations), but has not lost any negative charges (anions). As a consequence, the whole solution becomes negatively charged. This negative charge increases with the time and prevents the cathode from bringing new electrons necessary for discharging further cations. The electrolysis will stop. The electrode is said to be "polarized".


Hybrid orbitals forming molecular orbitals

It is still LCAO-MO theory, but just dumbed down a lot. The difference is that, instead of feeding the "pure" atomic orbitals into the LCAO mechanism, you carry out an additional mathematical step in order to get orbitals that have nice directional properties, and you feed those into the LCAO mechanism instead.

Let's talk about a simple example, methane. You could hybridise the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals on carbon to obtain a set of four $\mathrm{sp^3}$ hybrids. If you now try to derive the symmetry of these $\mathrm{sp^3}$ hybrids, you will realise they transform as $A_1 + T_2$. And if you didn't bother doing the hybridisation step, guess what are the symmetries of the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals? The $\mathrm{2s}$ orbital transforms as $A_1$ and the $\mathrm{2p}$ orbitals transform as $T_2$. In the end, you get roughly the same picture through MOT regardless of whether you use hybridised orbitals or not. (Note that the $4\mathrm{sp^3}$ in the second diagram does not indicate a principal quantum number of $4$, but rather that there are four such orbitals. I should have changed it, but I was lazy.)

There is a difference between this and conventional VB theory. Conventional VB theory tells us that each of the $\mathrm{sp^3}$ orbitals overlaps with one of the hydrogen $\mathrm{1s}$ orbitals to form four identical bonds. Once you draw a MO diagram, complete with symmetries, you are entering the realm of MO theory.

However, part of the reason why that example worked so well is that methane has a central atom. When you come to larger molecules, and have to piece together different fragments of the molecule, there is a difference between LCAO using the hybridised and unhybridised orbitals. The best example is the very first MO diagram that you drew - I cannot tell what it is, but I think it is that of an alkene. So let's consider ethene, $\ce{C2H4}$.

In fact, the diagram is "wrong". There are 8 electrons in total in the constituent AOs, but the MOs only have 4 electrons. This is not because of any elementary mistake by your teacher. It is because two of the $\mathrm{sp^2}$ orbitals on each carbon do not point towards each other; they point towards the hydrogens instead. Therefore, they are assumed to be involved only in C-H bonding and not in C-C bonding. Strictly speaking, they should therefore not be shown on the MO diagram. Nevertheless, it is this assumption that gives us some insight into the difference between this "hybridisation-LCAO" treatment and normal LCAO treatment using the pure AOs.

We'll first look at the "real" MOs of ethene (image taken from Fleming, Molecular Orbitals and Organic Chemical Reactions):

$\hspace{15 mm}$

We can try and draw some parallels between your "simplistic" MO diagram and this "proper" MO diagram here. In the simplistic diagram, the C-C $\sigma$ bonding arises from overlap of the two $\mathrm{sp^2}$ hybrid orbitals. So, this is likely to somewhat correspond to the $\pi_x$ MO above, which kind of looks like what you would expect from constructive overlap of $\mathrm{p}$ orbitals on carbon pointing towards each other. However, there's likely to be a bit of the $\sigma_s$ MO mixed into it, because you are using $\mathrm{sp^2}$ hybrid orbitals which also have the $\mathrm{s}$ orbital mixed in. The other filled MOs are not likely to contribute since they are antibonding with respect to the two carbons. In fact, they are mostly there to provide the additional C-H bonding required to make the molecule stable, which is exactly what is neglected in the simplistic diagram.

However, the $\pi$ bonding orbital in your simplistic diagram is a resounding success. It's exactly the same as the $\pi_z$ MO in the real diagram! And the $\pi^*$ antibonding orbital in the simplistic diagram is also exactly the same as the $\pi_z^*$ orbital in the real diagram! This leads us to the main point: we can do this much simpler MO treatment using hybrid orbitals and get a sufficiently accurate MO picture, without having to go through a bunch of symmetry arguments to determine which AOs overlap and which AOs don't. Furthermore, guess which are the most important MOs that control the reactivity of ethene? Exactly; it's the two $\pi$ orbitals.

Now that I've gotten that out of the way, I can proceed to answer your specific questions:

  I was confused by this because I thought that VB and MO theories were two separate theories.

What I wrote above mostly applies to this confusion. The crux of the matter is, hybridisation can be applied to VB theory and it is actually in this context that you first meet hybridisation, but it is not exclusive to VB theory. You can do a little mix-and-match in order to simplify your life, without sacrificing too much accuracy.

  Are molecular orbitals formed from regular (s,p,d, etc.) atomic orbitals, or are they also formed from the hybrid orbitals?

It is your choice. One leads to a more accurate picture, but it becomes extremely tedious as the molecule becomes more complex.

  And if hybrid orbitals do combine to form molecular orbitals, are the electrons still in between the two bonded atoms or are they delocalized like in the MO theory?

The electrons are localised. This is most obvious from the difference in the $\sigma$ orbitals shown above. Full delocalisation a la "proper" MO theory would imply that there are MOs which lead to bonding over the whole molecule, i.e. the $\sigma_s$ orbital.

  Then why do you still call it MO theory if the electrons are localised?!??!!??!

You didn't ask this question, but I anticipated you might wonder about it, since I've been going on and on about MOT only to end up saying that the electrons are localised. It's still LCAO-MO theory, just that the basis set of orbitals chosen is different, which leads to a different form of the wavefunction which happens to be localised between two nuclei. In terms of the electron density distribution, it is the same qualitative picture that is painted by VB theory. But, remember that VB theory has no idea what an antibonding orbital is, or that it even exists. So, what you have done is: you have actually arrived at the same result (localised electron density) by another means (MOT).

Disclaimer: There are people who will debate exactly what "VB" and "MO" theories are. Wavefunctions exist in VB theory as well, except that they are constructed differently, and one might very well say that the simplistic MO theory above is, in fact, VB theory. Nevertheless, I am sticking to the introductory, qualitative notions of VB and MO theory here.


I don't know the answer for sure, but will take a guess based on what I read at the link provided by Mithoron.

Non-chemists outside the laboratory, and persons who have to handle and work with those chemicals would certainly benefit from prilling to reduce dusting which could be detrimental to their health in breathing and on the mucous membranes.

Gardeners use Ammonium, potassium and calcium nitrates in fertilizers and stump removers, and do not always wear eye protection or dust masks. Thus prilling helps to mitigate exposure to these chemicals.


The hydrocarbon tail is considered nonpolar mainly due to

The Pauling electronegativity of Carbon is 2.55 and Hydrogen is 2.2. Due to the small difference, .35, and the definition of a polar bond, C-H bonds are considered nonpolar.

Indeed, the molecule methane, $\ce{CH4}$ has a dipole moment of 0 D (debyes). I think part of your confusion arises from the picture you provided. Skeletal drawings are 2-d in nature and do not convey information about 3-d. For instance, each carbon has it's hydrogens arranged in a tetrahedral structure, with bond angles 109.5 degrees. The symmetry of this cancels all the vectors contributing to polarity.

If you look at the picture of methane above, and consider the top most hydrogen as the next carbon in the hydrocarbon tail, you can see that in reality, the bottom 3 hydrogens do experience a net dipole moment towards the carbon, and this increased by having another carbon further up in the tail. Relatively speaking, this distance is rather large from carbon 1 to carbon 2. We can say that the vector that arises from this is negligible, and indeed, there is a net dipole moment anyways towards the oxygen at the polar head.


You essentially make cottage cheese in the process of making cheddar cheese. But cottage cheese is about 80% water. The "cheese" solids are soluble in oil, but not in water. So cheddar cheese will "melt" but cottage cheese won't.


Why is this organic compound trans, and not cis?

You are right, there is no difference between the two molecules that result if you rotate 180 degrees about the $\ce{C=C}$ double bond - they are the same exact molecule.  So for this molecule, the terms "cis" and "trans" would not be used, they have no meaning here.  The name of the molecule is simply 3-ethyl-3-hexene.  If the on-line course says otherwise, it is wrong.


Why is the recipe of Coca Cola still a secret?

There are different angles this question can be answered:
Chemical point of view:
A full analysis of a totally unknown mixture is painful and extremely costly. It is always helpful to know how many components you are looking for; what types etc. In this case, it is not enough to analyse the elemental composition or some pure elements, but Coke contains a lot of natural products and mixtures like caramel. Imagine how you identify (and correctly describe the production parameters) caramel in a dilute solution...  Also, liquids with extremely high concentrations of sugar and different acids are unpleasant for analysis, as you often have to separate these main component, so you can identify only small components.
That being said, it is not an impossible task; there are many methods to identify e.g. natural products based on DNA traces. But consider another factor:
Business point of view:
The recipe of Coca Cola itself worth nothing.
Most probably any competent soda maker can make a drink that 99% of the consumers cannot distinguish from Coke by the taste (and could do that decades before). However they don't do because no one would buy a drink that tastes like Coca Cola, but made by others.
The value is in the brand. People buy a fake Rolex if they cannot afford a real. But anyone can afford a Coke- why would buy a fake? If you are in beverage business, it is an imperative to make a drink that is somehow DIFFERENT than the other ones!


Your predictions are right; the molecule has more basic character.

The drug itself is practically insoluble in water, because the big organic molecule is near it's isoelectric point at the pure water's pH. However by protonating or deprotonating it and thus making it charged it's solubility is greatly enchanced.

This is also the case with those medical ready-for-use infusions:

  ...CIPRO IV solutions are available as sterile 0.2% ready-for-
  use infusion solutions in 5% Dextrose Injection. CIPRO IV contains lactic acid as a solubilizing agent and hydrochloric acid for pH adjustment. The pH range for the 0.2% ready-for-use infusion solutions is 3.5 to 4.6.

Reference


Enthalpy is slightly affected by isotope composition, but mostly below enthalpy measurement errors.

Generally, the standard enthalpy is considered for natural mixture of stable isotopes, unless considered for particular isotope explicitly.


Is the reaction of calcium carbide with water a Brønsted-Lowry acid-base reaction or Lewis acid-base reaction?

We may regard the calcium hydroxide and calcium carbide  as essentially ionic except for the covalent bonding within the anions.  So there is no Lewis reaction involving the calcium.

That leaves the proton transfer, which can be described fully by the B-L theory.  You actually have two reactions, successively forming $\ce{C2H-}$ and then $\ce{C2H2}$.


Mechanism of the reaction of an alkene with excess HI in presence of CCl4

According to [1], 1,2-diiodopropane reacts with hydrogen iodide to give 2-iodopropane.

So the reaction path might be

$$\ce{CH2=CH\bond{-}CH2\bond{-}I + HI -> CH3\bond{-}CHI\bond{-}CH2\bond{-}I}$$

$$\ce{CH3\bond{-}CHI\bond{-}CH2\bond{-}I + HI -> CH3\bond{-}CHI\bond{-}CH3 + I2}$$

The mechanism of the second part is unclear to me. It may be elimination of $\ce{I2}$ followed by addition of $\ce{HI}$.

1 Th. M. Schmitt, Analysis of Surfactants, Second Edition, p. 67


Reaction mechanism of glycerol with hydrogen iodide acid

The iodine isn't changing positions, it's donating a pair of electrons to stabilize the charge by forming an iodonium ion similar to the bromonium ion formed during the halogenation of an alkene shown below. You can see that the bonded pair of electrons was not affected by the step. It looks like they confused the reaction by short-cutting the reaction step.


Why are alpha-carbonyl relatively less stable?

The carbon atom of the carbonyl group has a partial positive charge due to electron withdrawal by the electronegative oxygen atom. An $S_{N}1$ mechanism would result in an intermediate with a positive charge next to a partial positive charge, which would be high in energy.


How do the pKa and pH indicate to the proper environment of protonation?

The sentence immediately before the bolded one indicates that the $\mathrm pK_\mathrm a$ value of the ligand bound to the enzyme was determined to be $9.1$. It also indicates that the $\mathrm pK_\mathrm a$ value o the free ligand is 2.5 logarithmic units less, i.e. $\mathrm pK_\mathrm a(\ce{Lig}) \approx 6.6$.

The human cell is well-buffered to ensure that the pH stays within a very narrow range. There are enough buffering compounds around, that we never need to use any Henderson-Hasselbalch equation or other difficult calculation methods to determine the protonation state of a given compound. We can simply examine its $\mathrm pK_\mathrm a$ value; if that is lower than the cell’s pH value, the compound is deprotonated. In all other cases, it is not.

The $\mathrm pK_\mathrm a$ value of the free ligand indicates it should be deprotonated in the cell. However, upon binding to the protein with the $\mathrm pK_\mathrm a$ value rising by $2.5$  logarithmic units, the ligand suddenly becomes protonated. This is what the bolded sentence implies.


Why do we measure the initial rate of a reaction?

First, note that one does not always measure initial rates. It is fairly common to fit an entire reaction curve (or nearly entire), but it requires a little bit more math. It is also common to measure an "initial" rate that is not quite initial. That is, it is the rate after a burst phase has occurred, and the system has settled into something close to a steady state.

That said, measuring initial rates is done for two practical reasons.

1) we can approximate the product concentration as 0. For a reaction that is at all reversible or product-inhibited, the math is much easier if [P]=0.

2) we can often approximate the concentration of one (or more) of the reactants as fixed. For example, if you measure the rate only for an amount of time required for a few percent of the reactant to be consumed, you can get a pretty good fit to the data by assuming that reactant is fixed at its initial concentration. That makes it easy to produce plots of the initial rate vs the concentration of that reactant.

Both of these factors make it much easier to determine basic rate laws and the approximate value of the associated rate constants.


How can we say whether or not an organic compound is soluble?

The traditional rule of thumb has always been that 1 oxygen can dissolve 4 carbons.
In this case, the compound is $\ce{C8H8O2}$ which makes an oxygen:carbon ratio of 1:4, so it should be soluble in water.

The source of this rule of thumb is Dr. Pedro J. Victory, SJ, who taught us Organic Chemistry. That was a long time back but I recall this rule of him.
The context was liquid-liquid extraction.

I do not remember if he added a corollary for the case of esters, where you should count only one of the oxygens of every ester group.
We perform extractions with ethyl acetate all the time and there the ratio is 1:2.  Very probably, he did add that.

To support this rule I have searched for some solubilities at PubChem. On basis of this data, the threshold for soluble (as asked in a comment by @Zhe) would be around 10% (100 mg/mL).

                    mg/mL T (C)
acetone             completely soluble
methyl ethyl ketone 223    25
3-pentanone          43    25
2-pentanone          46    25
isopropanol         completely soluble
1-butanol            63    25
2-butanol           181    25
1-pentanol           22    25
2-pentanol           45    25
3-pentanol           55    25
methyl ethyl ether  230    25
diethyl ether        60    25
methylacetate       243    25
ethylacetate         80    25
n-propylacetate      19    20


How Can I Test Something for the Presence of Iodine?

Just keep in mind that only elemental iodine will cause a blue-black coloration. Nobody add elemental iodine to a supplement just like nobody can eat chlorine out of table salt (NaCl). The typical compound they add is KIO3, which will not color starch right away, until and unless you reduce it to KI and then oxidize KI to free I2. In any case, search natural foods which contain iodine.


These two isomers are enantiomers, i.e. mirror images, of each other:

$\hspace{50 mm}$

This one has a plane of symmetry through the molecule and is therefore meso:

$\hspace{60 mm}$


Do undissolved salts in saturated solvents enter equilibrium with the dissolved salts?

Yes, they enter into a constant equilibrium. Just as we write equilibrium expressions for insoluble salts as equilibrium expressions, like $\ce{CaF2} \leftrightharpoons \ce{Ca}^{2+} + \ce{2F-}$, when you reach the saturation point, solute particles are in dynamic equilibrium between aqueous and solid phase.


Migration of ethyl group in pinacol rearrangement

The two possible compounds resulting from the pinacol rearrangement of your substrate, 3,4-dimethylhexane-3,4-diol, are as follows.

A quick Reaxys search throws up a paper in which this pinacol rearrangement was investigated.1 I quote directly:

  Four different investigators have studied the pinacol-pinacolone rearrangement of 3,4-dimethyl-3,4-hexanediol. It was finally demonstrated by Nybergh that, contrary to the conclusions reached by the three previous investigators, the rearrangement involved the preferential migration of an ethyl group rather than a methyl group.

The paper describes how the authors carried out the reaction and analysed the mixture to determine the ratio of the two products shown above. In 1956, there was much less technology than we enjoy now, so no NMR and no HPLC. It turns out that they had to hydrogenate the products to make a mixture of the corresponding alcohols and figure out the ratio by IR spectra. The result: $70\mathrm{-}73\%$ of 3-ethyl-3-methylpentan-2-ol (corresponding to the ethyl shift), depending on the form of the starting pinacol used.

Unfortunately, I can't access the (very old) journals that claim that the Me shift happens preferentially.

References

(1) Reeve, W.; Karickhoff, M. The Pinacol Rearrangement of meso- and dl-3,4-Dimethyl-3,4-hexanediol. J. Am. Chem. Soc. 1956, 78 (23), 6053–6054. DOI: 10.1021/ja01604a028.


Is activated carbon classified as organic or inorganic?

While nomenclature is of particular interest to organic chemists to specify an exact compound, the classification of X into broad category Y or Z isn't a precise science, and not really of practical use.  The article cites a textbook by Seager to this effect, stating

  The distinction between "organic" and "inorganic" carbon compounds, while "useful in organizing the vast subject of chemistry... is somewhat arbitrary"

Even if you find a source that says "charcoal is (in)organic", you may just as well find one stating the opposite.  Just like the coal from which it may have been produced, it was once biomass and decidedly organic, but so was graphite and diamond, or CO2 and CO32−. I think it's overly pedantic and unproductive to try to come up with definitive judgements for these decidedly edge case scenarios.

After all, it's just a chemical on the shelf, what one does with it is far more relevant.  I don't use it on a daily basis, but it seems more like a tool than a reagent.  The fact it contains carbon seems beside the point; it's value isn't in the chemical composition but rather its extraordinary adsorptive properties.


Stoichiometry of alkane-perchlorate reaction

I think that there is some $\ce{H2O}$ as product:
$$\ce{2Mg(ClO4)2 + 4CH4 -> 2MgCl2 + 4CO2 + 8H2O}%edit$$


Why are metal oxides basic and non-metals' acidic or neutral

We have learnt that metallic oxides are basic nature. This is because the oxides of metals like $\ce{MgO}$ or $\ce{CaO}$ or $\ce{Na2O}$ form hydroxides in their aqueous solutions.
For example:
$$\ce{K2O + H2O -> 2KOH}$$
$$\ce{MgO + H2O -> Mg(OH)2}$$

But when non-metallic oxides dissociate into their constituent ions in water, they give $\ce{H+}$ ions in their aqueous solutions forming acids
For example:
$$\ce{CO2 + H2O -> H2CO3}$$
$$\ce{SO3 + H2O -> H2SO4}$$


In fact, you can see $C_2$ and other symmetry elements for yourself in 3D by opening the CIF [1, COD-2005680] with Mercury (free, available for Windows, Linux, MacOS):

Figure 1.
Unit cell of crystalline sulfuric acid $\ce{H2SO4}$.
Green lines parallel to crystallographic axis $b$ depict $C_2$ rotation axes.

This can be achieved by drawing the unit cell content with Calculate → Packing/Slicing…  and adding symmetry elements of interest via Display → Symmetry Elements….

Figure 2.
Single molecular unit $\ce{H2SO4}$ in two projections with the $C_2$ placed in drawing plane (left) and perpendicular to drawing plane (right).

References

Kemnitz, E.; Werner, C.; Trojanov, S. Reinvestigation of Crystalline Sulfuric Acid and Oxonium Hydrogensulfate. Acta Crystallographica Section C Crystal Structure Communications 1996, 52 (11), 2665–2668. https://doi.org/10.1107/S0108270196006749.


Designing a mathematics course for 1st year chemistry students

At least in the US, most undergraduate chemistry curricula do not go deep enough into the math of QM that you need to spend a great deal of time on it. Students that do take more advanced QM classes can learn the math at that time. I also am not sure it's safe to assume that all incoming students at that level have a good understanding of 1D calculus, though most will have had some exposure.
With that in mind, here are topics that I'd include (not in any specific order), structured around the relevant applications:

basic calculus review of integration and differentiation in 1D - main intro level application is chemical kinetics. Also applies to intro quantum.

setting up and solving systems of equations, especially including iterative solving methods - intro level application is solving systems of multiple equilibria by mass and charge balance

a module on statistics with a focus on dealing with experimental error and error propagation. Include methods of curve fitting and associated pitfalls. - applies to lab component of curriculum. Probably one of the more commonly overlooked aspects of math in chemistry

Fourier transform and other techniques involved in spectroscopy

Basics of symmetry operations and group theory


What exactly is the 'shuttle effect' in sulfur batteries?

OP's title indicated OP is looking for the real mechanism of  the Shuttle Effect in $\ce{Li-S}$ batteries. Yet, OP is asking different questions in the text body. I'd try to address as much as possible in this answer.

A most recent review article (Ref.1) states that:

  Lithium-sulfur ($\ce{Li-S}$) batteries are one of the most promising batteries in the future due to its high theoretical specific capacity ($\pu{1675 mAh g−1}$) and energy density ($\pu{2600 Wh kg−1}$). However, the severe capacity fading caused by shuttle effect of polysulfide needs to be addressed before the practical application of $\ce{Li-S}$ batteries.

To understand what is the Shuttle Effect, OP need to first understand the working chemical mechanism of $\ce{Li-S}$ batteries, and can take a head stat in that direction with this article about Lithium–Sulfur Battery, which gives a fair description of the chemistry involve in the battery. The same Wikipedia article simply described the Shuttle Effect as follows:

  The lithium polysulfide $\ce{Li2S}_x$ ($6 < x < 8$) is highly soluble in the common electrolytes used for $\ce{Li-S}$  batteries. They are formed and leaked from the cathode and they diffuse to the anode, where they are reduced to short-chain polysulfide and diffuse back to the cathode where long-chain polysulfide are formed again. This process results in the continuous leakage of active material from the cathode, lithium corrosion, low coulombic efficiency, and low battery life.

The formation of polysulfide shuttle effect is classified into five steps in Ref.1:

  Formation of long-chain polysulfide,
  Detaching of polysulfide from sulfur host,
  Dissolution of polysulfide into electrolyte,
  Migration of polysulfide toward lithium anode side, and
  Reaction between polysulfide and lithium anode.

If interested, OP may read Ref.1 and Ref.2, which thoroughly discussed the mechanism and possibilities of how to prevent the Shuttle Effect in depth. For example, one such techniques is modifying the separator, which has been done successfully recently (Ref.3):

References:

Wenchen Ren, Wei Ma, Shufen Zhang, Bingtao Tang, "Recent advances in shuttle effect inhibition for lithium sulfur batteries," Energy Storage Materials 2019, 23, 707-732 (https://doi.org/10.1016/j.ensm.2019.02.022).
Chao Deng, Zhuowen Wang, Shengping Wang, Jingxian Yu, "Inhibition of polysulfide diffusion in lithium–sulfur batteries: mechanism and improvement strategies," J. Mater. Chem. A 2019, 7(20), 12381-12413 (https://doi.org/10.1039/C9TA00535H).
Yanpeng Fan, Zhihui Niu, Fei Zhang, Rui Zhang, Yu Zhao, Guang Lu, "Suppressing the Shuttle Effect in Lithium–Sulfur Batteries by a UiO-66-Modified Polypropylene Separator," ACS Omega 2019, 4(6), 10328-10335 (https://doi.org/10.1021/acsomega.9b00884).


Bromination on the aromatic ring vs aliphatic chain

The key in this reaction is ketone in propiophenone is acting as Lewis base to $\ce{AlCl3}$ Lewis acid. If $\ce{AlCl3}$ is in catalytic amount, it activate $\alpha$-$\ce{H}$ in side chain sufficiently enough (similar to acid would do to a carbonyl compound) and progress to produce $\alpha$-substituted product. If $\ce{AlCl3}$ is in excess, it seemingly deactivates aliphatic side chain completely (Ref.1), and ring substitution predominates to give $meta$-substituted product exclusively as a result (Ref.2 and 3):

The reaction with excess $\ce{AlCl3}$ should necessarily be done in solvent free conditions, and excess $\ce{AlCl3}$ is needed such that one equivalent is used to complex with carbonyl oxygen and another equivalent is used to activate $\ce{Br2}$ molecule. As a result, benzene nucleus would be substituted and 3-bromoacetophenone would be obtained exclusively (Ref.1). The crude liquid product is purified as an oil after fractional distillation in 70-75% yield (Ref.2). The reaction with trace amount of $\ce{AlCl3}$ has done in dry ether and solid crude yield has been 88-96%. When recrystallized by methanol, the yield has been reduced to 64-66% (Ref.3). Another stark different between these two reactions are addition order. In the excess reaction, acetophenone has been added to $\ce{AlCl3}$ while in the catalytic reaction, $\ce{AlCl3}$ has been added to etherial acetophenone solution.

References:

D. E. Pearson, H. W. Pope, W. W. Hargrove, W. E. Stamper, "The Swamping Catalyst Effect. II. Nuclear Halogenation of Aromatic Aldehydes and Ketones," J. Org. Chem. 1958, 23(10), 1412–1419 (https://doi.org/10.1021/jo01104a003).
D. E. Pearson, H. W. Pope, W. W. Hargrove (Checked by B. C. McKusick, D. W. Wiley), "3-Bromoacetophenone," Org. Synth. 1960, 40, 7 (DOI: 10.15227/orgsyn.040.0007) (Organic Syntheses, Coll. Vol. 5, p.117 (1973)).
R. M. Cowper, L. H. Davidson (Checked by Lee Irvin Smith, E. W. Kaiser), "Phenacyl Bromide ($\alpha$-Bromoacetophenone)," Org. Synth. 1939, 19, 24 (DOI: 10.15227/orgsyn.019.0024)(Organic Syntheses, Coll. Vol. 2, p.480 (1944)).


Why does ethene react explosively with fluorine?

The equation gives only the reactants and final products and tells us nothing about the mechanism; stoichiometric equations rarely do.This is an explosive reaction so it will proceed by many steps usually involving radicals. To get feedback, and so explosion, one step must be of the branched chain form $$\ce{R_1^. +  M -> R_2^. + R_3^.}$$ where $\ce{R^.}$ is a radical and $\ce{M}$ a reactant species (e.g. $\ce{CH2=CH2}$). As two radicals are produced for one reactant radical a chain reaction ensues, the reaction rate accelerates and so bang!

Any explosive reaction has initiation, propagation, branching and termination steps, some also have an inhibition step. The initiation step in your reaction probably forms the radical $\ce{FCH2-CH2^.}$ although I don't know the mechanism.


Deviation of real gases from ideal behaviour

The ideal gas law can be derived using kinetic molecular theory by making two very important assumptions that are not true for real gases:

There is no attractive force between gas molecules
The gas molecules have no volume

Think about how the PV curve would change when you remove one or both of these conditions. Then think about which types of molecules the assumption of "no attractive force" would be more important for.


It is almost always possible to draw resonance structures. In case of $\ce{CO2}$ you could imagine a resonance structure in which the carbon doesn't have an electron octet and a positive charge and one of the oxygen atoms carrying a negative charge.
You could also draw one in which the carbon is only connected via a single bound to each oxygen and carrying two positive charges while the two oxygens carry a negative charge each.

You might come up with more of those structures.

So the question you really wanna ask is whether these resonance structures have any real contribution to the actual bonding state of $\ce{CO2}$ and for that you will see that all of the described strucutures are highly unfavourable due to their energetic state. Therefore the normal Lewis-notation of $\ce{CO2}$ decribes the real bonding state pretty well as all other resonance structures don't contribute much.


Nitrogen and an unknown gas is leaked into a common vessel... The gaseous mixture contains 0.4 moles of nitrogen. Molecular mass of the other gas is?

Calculate total moles using the ideal gas law and solving for $n$:

$$n = {4.15\;{\rm bar}\cdot 3\;{\rm L}\over {0.0831\;{\rm L}\,{\rm bar}\,{\rm K}^{-1}\,{\rm mol}^{-1}}\cdot 300\;{\rm K}} = 0.5\;{\rm mol}$$

Given 0.4 mol $\ce{N2}$, moles of unknown gas = 0.5 mol - 0.4 mol = 0.1 mol.

Rate of effusion of $\ce{N2} = r_{\ce{N2}} = 0.4\;{\rm mol}/10\;{\rm min} = 0.04\;{\rm mol}\cdot{\rm min}^{-1}$.

Rate of effusion of unknown ${\rm X} = r_{\rm X} = 0.1\;{\rm mol}/10\;{\rm min} = 0.01\;{\rm mol}\cdot{\rm min}^{-1}$.

Using Graham's law of diffusion:

$${r_{\ce{N2}}\over r_{\rm X}} = \sqrt{M_{\rm X}\over M_{\ce{N2}}}$$

we solve for $M_{\rm X}$:

$$M_{\rm X} = {M_{\ce{N2}}\times (r_{\ce{N2}})^{2}\over (r_{\rm X})^{2}}$$

and plugging in numbers we get the molecular mass of the unknown gas:

$$M_{\rm X} = {(28\,{\rm g}\cdot{\rm mol}^{-1})\times (0.04)^{2}\over (0.01)^{2}} = 448\,{{\rm g}\over{\rm mol}}$$

where the units on the effusion rates have been omitted for clarity and because they cancel.


This is one of the many reasons why hybridisation including d-orbitals fails for main-group elements.

Xenon in $\ce{XeF6}$ is not hybridised at all. Instead of invoking populated core d-orbitals or energetically removed d-orbitals (remember the aufbau principle: the next shell’s s-orbital has a lower energy than the d-orbitals you are proposing to include in hybridisation!) xenon just offers its three p-orbitals $\mathrm{p}_x, \mathrm{p}_y$ and $\mathrm{p}_z$ for four-electron-three-centre bonds. These 4e3c bonds can be understood using the following two mesomeric structures:

$$\ce{F^-\bond{...}Xe^+-F <-> F-Xe^+\bond{...}F-}$$

Each $\ce{Xe-F}$ bond has a bond order of ½, and for each fluorine there is another with a bond angle $\angle(\ce{F-Xe-F}) \approx 180^\circ$ as part of the same 4e3c bond.

Also note that this means that xenon’s lone pair is comfortably located in the $\mathrm{5s}$ orbital.


Why does electrolysis take so much time?

It depends: The rate of electrolysis is limited either by the electric source or by the ion concentration. I am inexperienced with electrolysis but I imagine that quality may increase as the the process slows, much like crystals grow with more imperfections when the rate of crystallization is fast.

There is no such thing as an electrolyte solution composed of positive ions and electrons.

  "Activity coefficients of single ions cannot be measured
  experimentally because an electrolyte solution must contain both
  positively charged ions and negatively charged ions." -Debye–Hückel
  theory

Electroplating systems tend to use an electrode source that replenishes the solvent ions plated onto the target electrode.


Can Molisch test be used to separate monosaccharides from polysaccharides?

No. Mono and poly saccharides both give a positive test.


You're right and wrong.  Your right that pyruvate anion has a higher enol content than lactate anion, at least that would be my guess (pyruvate anion has 3 enolizable hydrogens, while lactate anion has one and needs to destroy the carboxylate resonance in order to create the carbonyl required for enolization...read on).

But I think that lactic acid could enolize, after all esters can enolize.  I suspect that the enol content in the lactate carboxylate anion would be non-measurable - you would need to disrupt the carboxylate anion resonance in order to generate the carbonyl required for enolization.  If we use 36 kcal/mol as an estimate of the carboxylate resonance energy and say that for every 1.4 kcal/mol we can shift an equilibrium by a power of 10, then we might expect 1 enol molecule in every $10^{26}$ lactate carboxylate molecules.  In the lactic acid itself, we have no carboxylate resonance to disrupt, the enol content would be higher.


Does benzene's resonance structure allow it to enter DNA?

I will start my answer with a preface that the website linked to in the question is a pseudoscience website (and I am glad that it has vanished from the face of this earth, only accessible via the Wayback Machine). These people use scientific-sounding jargon that sounds impressive to the lay reader, but any actual scientist will know that it is simply rubbish. The objective is usually to fear monger, or to promote some product of theirs that has no special properties. Unfortunately, these seem to be becoming more and more prevalent recently.

However, the premise of the question is sound: benzene itself is known to cause cancer in humans. It is listed as a Class 1 carcinogen, meaning that there is evidence for carcinogenicity in humans.

As with many things in molecular biology, we do not actually know the full mechanism by which benzene causes cancer. However, there have been a number of papers written on the topic: some of them are linked at the bottom of this post.

Exactly which metabolite of benzene causes cancer is not known. It is likely to be a combination of them. The first step is oxidation of benzene by the enzyme cytochrome P450 2E1 (CYP2E1) to benzene oxide, which exists in equilibrium with oxepin (via a 6π electrocyclic reaction). The cytochrome P450 enzymes are found in the liver, and their role is generally to insert oxygen atoms into molecules, making them more polar and water-soluble, so that they can be excreted. For more information about this process, see Drug metabolism on Wikipedia.

These undergo further reactions, catalysed by various other enzymes, to give a large range of metabolites:

In particular,

The benzoquinones have been shown to inhibit DNA topoisomerase II, an enzyme that makes temporary cuts in double-stranded DNA in order to "unwind" DNA that has been entangled. It plays a key role in many cellular processes such as DNA replication, DNA repair, and chromosome segregation (during cell division); therefore, inhibition may lead to chromosome breakage or failure to segregate.
The same quinones can undergo a process called redox cycling, where they undergo an enzymatic reaction in which a single electron is added to them to form a radical anion. These species are then released, and react with molecular oxygen $\ce{O2}$ to give the superoxide anion, $\ce{O2^-}$... and then the process repeats itself. The buildup of $\ce{O2^-}$ (and other reactive oxygen species) leads to oxidative stress and DNA damage.
(E,E)-muconaldehyde has been recently shown to form an adduct with two molecules of deoxyguanosine, i.e. the guanine bases in DNA. Here, R represents the rest of the deoxyribose sugar. Intra- or inter-chain links can be formed via this method, which then lead to inaccurate replication or chromosomal aberrations.
A mechanism was proposed in reference 3. It is not reproduced here but it is not difficult to imagine how such a reaction might happen: nitrogen atoms in guanine are nucleophilic, and literally every carbon in muconaldehyde is electrophilic.

The literature on the topic contains much more information than I can write in here. Reference 4 is a relatively recent review on the topic, which would be a decent starting point to find further information.

Regardless of what mechanism it is, one thing is certain: benzene itself and its molecular properties are not likely to be the cause of its carcinogenicity. It is almost certain that multiple metabolites of benzene are the culprits.

References

Chen, H.; Eastmond, D. A. Topoisomerase inhibition by phenolic metabolites: a potential mechanism for benzene's clastogenic effects. Carcinogenesis (Oxford) 1995, 16 (10), 2301–2307. DOI:10.1093/carcin/16.10.2301
Rappaport, S. M.; Kim, S.; Lan, Q.; Li, G.; Vermeulen, R.; Waidyanatha, S.; Zhang, L.; Yin, S.; Smith, M. T.; Rothman, N. Human benzene metabolism following occupational and environmental exposures. Chem.-Biol. Interact. 2010, 184 (1–2), 189–195. DOI:10.1016/j.cbi.2009.12.017.
Harris, C. M.; Stec, D. F.; Christov, P. P.; Kozekov, I. D.; Rizzo, C. J.; Harris, T. M. Deoxyguanosine Forms a Bis-Adduct with E,E-Muconaldehyde, an Oxidative Metabolite of Benzene: Implications for the Carcinogenicity of Benzene. Chem. Res. Toxicol. 2011, 24 (11), 1944–1956. DOI:10.1021/tx2002838.
Hartwig, A. The role of DNA repair in benzene-induced carcinogenesis. Chem.-Biol. Interact. 2010, 184 (1–2), 269–272. DOI:10.1016/j.cbi.2009.12.029.


Where does the rate equation come from? (r = k [A]^x [B]^y)

A to the power of x or B to the power of y are telling you how the concentration of that reactant will affect the rate of reaction. For example, if you find that doubling the concentration of reactant A quadrupled the reaction rate than A could be raised to the power of 2. If doubling the concentration of B doubled the rate of reaction than you could raise B to the power of 1. K is the Arrhenius equation and its value is taking into account the frequency that the particles are colliding, the amount that are colliding with the right orientation and the energy required for the reaction to take place. K = Ae^(-Ea/RT) where A is the product of the frequency of the colliding particles and the amount that have the correct orientation. e is 2.718281828 and is raised to the power of negative the activation energy divided by RT where R is the ideal gas constant with a value of 8.314 and T is the temperature. You could use algebra to solve for A if you have the other values.


How this tricky Wagner-Meerwein rearrangement occurs?

My literal back of the envelope solution. Notable points:

I invoked a nonclassical carbocation in the first step of the rearrangement. This was the easiest way to explain the stereochemistry of the new acetate group.
This is a reaction from 1959-1960. It's almost certainly racemic. I think the product as drawn is actually the enantiomer of the one that you would expect from the reactant.


The IUPAC periodic table (2016-pdf) is good because it quantifies the accuracy of the natural abundance atomic weights, but it doesn't give electron configuration.

The NIST periodic table has electron configuration, and probably is closest to what you want.

The NIST table is also good from a copyright point of view, because you can reproduce it without worrying about copyright infrigment, as US government works are generally not subject to copyright protection.


Does a bulky nuclephile react faster by Michael or direct addition?

First, you got wrong thinking Michael addition is faster than the direct addition. Actually, Michael addition (thermodinamic control) is comparatively slower than the direct addition (kinetic control) here.

Second, to my understanding, Bürgi-Dunitz angle (Ref.1) has nothing to do here (may be it affect both additions, you never knows). However, there are several factors control the outcome.  The way that nucleophiles react depends on three factors:

The conditions of the reaction;
The nature of the $\alpha,\beta$-unsaturated carbonyl compound; and
The type of the nucleophile.

An example for the conditions of the reaction: The reaction of but-3-en-2-one and $\ce{KCN}$  in the presence of an acid catalyst gives 1,1-cyanohydrin (2-cyanobut-3-en-2-ol) at $\pu{10-20 ^\circ C}$ by 1,2-addition (kinetic product), but would give 4-cyanobuta-2-one at $\pu{80 ^\circ C}$ by 1,4-addition (thermodynamic product). Thus, it is clear that the Direct addition to the carbonyl group (which forms at low temperature) is faster than conjugate addition (forms at higher temperature), giving cyanohydrin. However, that reaction is reversible and, as a result, converted to more stable cyanobutanone at high temperatute, which is essentially an irreversible reaction.

Why is direct addition (1,2-addition) faster than conjugate addition (1,4-addition)?: Simple molecular orbital calculation shows both, the carbon atom of the $\ce{C=O}$ group and the carbon atom $\beta$ to it, carry some positive charges. However, the carbon atom of the $\ce{C=O}$ group carries comparatively more charge, so electrostatic attraction for the charged nucleophiles will encourage it to attack the carbonyl group directly rather than undergo conjugate addition.

On the other hand, with direct 1,2-addition, we gain a new $\ce{C–C}$ $\sigma$-bond, but lose the $\ce{C=O}$ $\pi$-bond and keep the $\ce{C=C}$ $\pi$-bond. Meantime, with conjugate 1,4-addition, we still gain a new $\ce{C–C}$ $\sigma$-bond, but lose the $\ce{C=C}$ $\pi$-bond instead of $\ce{C=O}$ $\pi$-bond, and keep the $\ce{C=O}$ $\pi$-bond instead of $\ce{C=C}$ $\pi$-bond. Overall, the $\ce{C=O}$ $\pi$-bonds ($\pu{369 kJmol-1}$) are stronger than $\ce{C=C}$ $\pi$-bonds ($\pu{280 kJmol-1}$), hence the conjugate addition gives comparatively more stable product.

An example for the nature of the $\alpha,\beta$-unsaturated carbonyl compound: The reaction of 2-butenal (an $\alpha,\beta$-unsaturated aldehyde) and butyllithium in THF at $\pu{-70 ^\circ C}$ gives oct-2-en-4-ol by 1,2-addition exclusively. Yet, under the identical conditions, N,N-dimethyl-2-butenamide (an $\alpha,\beta$-unsaturated amide) would give N,N-dimethyl-2-butylbutanamide by 1,4-addition exclusively.

An example for the type of the nucleophile (Hard or Soft): The reaction of 4-methylpent-3-en-2-one (an $\alpha,\beta$-unsaturated ketone) and thiophenol at $\pu{25 ^\circ C}$ gives 4-methyl-4-thiophenoxypent-2-one exclusively by 1,4-addition (Remarkably, no acid or base catalyst is needed for this reaction,which would be needed for the alcohol additions, instead).

  Here, it has been explained that the attraction between nucleophiles and electrophiles is governed by two related interactions: (1) Electrostatic attraction between positive and negative charges, and (2) Orbital overlap between the HOMO of the nucleophile and the LUMO of the electrophile. Successful reactions usually result from a combination of both, but sometimes reactivity can be dominated by one or the other. The dominant factor, either it be electrostatic or orbital control, depends on the nucleophile and electrophile involved. Nucleophiles containing small, electonegative atoms (such as $\ce{O}$ or $\ce{Cl}$) tend to react under predominantly electrostatic control, while nuclophiles containing larger atoms (including the sulfur of thiols) are predominantly subject to control by orbital overlap. The terms hard and soft have been coined to describe these two types of reagents. Hard nucleophiles are typically from the early rows of the periodic table and have higher charge density, while soft nucleophiles are from the later rows of the periodic table—they are either uncharged or have larger atoms with higher-energy, more diffuse orbitals (Ref.2).

Few examples:

$$
\begin{array}{c|c}
\hline
\text{Hard nucleophiles} & \text{Borderline nucleophiles} & \text{Soft nucleophiles} \\
\hline
 \ce{F-, \mathbf{HO^-, RO^-}, SO4^2-, Cl-} & \ce{N3-, \mathbf{CN^-}} & \ce{\mathbf{I^-, RS^-}, RSe-, S^2-}\\
\ce{\mathbf{H_2O, ROH}, ROR', RCOR'} & \ce{\mathbf{RNH_2, RR'NH}} & \ce{\mathbf{RSH, RSR'}, R3P}\\
\ce{\mathbf{NH_3, RMgBr, RLi}} & \ce{\mathbf{Br^-}} & \text{Alkenes, aromatic rings}\\
\hline
\end{array}
$$

Electrophiles can also classified in this manner. For example, $\ce{H+}$ is a very hard electrophile because its small size and charged. Bromine $\ce{Br2}$ is, on the other hand, a soft electrophile due to its large size and having no charge.

In general, hard nucleophiles prefer to react with hard electrophiles, and soft nucleophiles with soft electrophiles.

Overall, Hard/Soft reactivity:

Reactions of hard species are dominated by charges and electrostatic effects
Reactions of soft species are dominated by orbital effects
Hard nucleophiles tend to react well with hard electrophiles
Soft nucleophiles tend to react well with soft electrophiles

References:

H. Bürgi, J. Dunitz, J. Lehn, G. Wipff, "Stereochemistry of reaction paths at carbonyl centres", Tetrahedron 1974, 30(12), 1563-1572 (http://dx.doi.org/10.1016/S0040-4020(01)90678-7).
Jonathan Clayden, Nick Greeves, Stuart Warren, Organic Chemistry; 2nd Edn; Oxford University Press: Oxford, United Kingdom, 2012.


Derivation of van 't Hoff equation for temperature dependence of equilibrium constant

The equation linking $\Delta H^\circ$ and $K$ is called the van 't Hoff equation. Since Philipp's comment on your question already links to a very thorough discussion of where the equation $\Delta G^\circ = -RT\ln{K}$ comes from, I won't repeat it.

The definition of the Gibbs free energy, $G$, is $G = H - TS$. Using $\mathrm dG = V\,\mathrm dp - S\,\mathrm dT$ we obtain the Maxwell relation

$$\left(\frac{\partial G}{\partial T}\right) = -S$$

and hence the Gibbs–Helmholtz equation (derivation here)

$$\left(\frac{\partial (G/T)}{\partial T}\right) = -\frac{H}{T^2} \quad \Leftrightarrow \quad \left(\frac{\partial (\Delta G^\circ/T)}{\partial T}\right) = -\frac{\Delta H^\circ}{T^2}$$

Since $\ln K = -\Delta G^\circ/RT$, we have

$$\frac{\mathrm d(\ln{K})}{\mathrm dT} = -\frac{1}{R}\frac{\mathrm d}{\mathrm dT}\left(\frac{\Delta G^\circ}{T}\right) = \frac{\Delta H^\circ}{RT^2}$$

This is the differential form of the van 't Hoff equation; it's not the most useful thing to us though because it only tells you the slope of a plot of $\ln{K}$ against $T$ at one given point. We usually separate the variables and integrate with respect to both sides:

$$\int_{\ln{K_1}}^{\ln{K_2}}\!\mathrm d(\ln{K}) = \int_{T_1}^{T_2}\!\frac{\Delta H^\circ}{RT^2}\,\mathrm dT$$

$$\ln{K_2} - \ln{K_1} = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2} \right) $$

So, if you know the equilibrium constant $K_1$ at a certain temperature $T_1$ and you want to find the equilibrium constant $K_2$ at a different temperature $T_2$, you can just plug in your values into the equation and solve for $K_2$.

Note that this equation supports what you know of Le Chatelier's principle; if the reaction is exothermic, $\Delta H^\circ < 0$, and if you increase the temperature from $T_1$ to $T_2 > T_1$ then $(1/T_1 - 1/T_2) > 0$. The RHS of the equation is therefore negative, and that means that $\ln{K_2} < \ln{K_1} \Rightarrow K_2 < K_1$ which implies that the equilibrium position has shifted to the left.

Note that the last step (the integration) makes the assumption that $\Delta H^\circ$ is a constant over the temperature range $T_1$ to $T_2$. Note that this is, in general, not true but if the temperature range isn't too huge you will get pretty accurate results from the use of this equation.


AFAIK, production scale processes are highly optimized, producing just traces of $\ce{CO}$. There traces are either not managed, or can by passed through catalyzators.
On the other hand, burning coke with limited air or oxygen $\ce{2 C(s) + O2(g) -> 2 CO(s)}$ or passing water vapour through burning coke $\ce{C(s) + H2O(g) -> CO(g) + H2(g)}$, are used intentionally, to produce $\ce{CO}$, eventually with hydrogen, for further synthesis.


Is there a way to tell what the coordination of a complex ion will be without knowing the ligand?

There is no way you can predict this a priori. For example, cobalt(III) forms an octahedral hexaaquacomplex $\ce{[Co(H2O)6]^3+}$. If you add chloride to the solution, you will form a tetrahedral tetrachloridocomplex $\ce{[CoCl4]-}$. This was a pure ligand exchange reaction as shown in the equation below.

$$\ce{[Co(H2O)6]^3+ + 4 Cl- <=>> [CoCl4]- + 6 H2O}$$


It will never happen. You can not hydrolyze a salt to any strong acid (nitric acid in your case) or base because the reverse reaction is predominant.


How can the thermal conductivity of a binary gas mixture be estimated?

I'm thinking you could use the Fourier equation for heat flux

$$
\phi =K \frac{S\theta}{e} \,\, ,
$$
where

$\phi$ = Heat flux
$K$ = Termal conductivity coeficient in units of  $kcal  s^{-1}\ m^{-2}\ K^{-1} /m^{-1}$
$S$ = Area of the section you're dealing $m^{2}$
$θ$ = Difference of temperature $K$
$e$ = Thickness in $m$.

And use for the resulting thermal conductivity the fraction

$$
K_{r}=\frac{\sum K_{i}J_{i}(M_{i})^{1/3}}{\sum J_{i}(M_{i})^{1/3}} \,\, ,
$$
where:

$K_{r}$ = Resulting termal conductivity of the moisture;
$K_{i}$ = Termal conductivity of each gas;
$j_{i}$ = Molar fraction of each gas;
$M_{i}$ = molar mass of each gas.

I think it'll work with some 2% of standard deviation. If you're able to read in portuguese, here is the link of my reference:

Reference

And here is the link of the Book cited by @Chester Miller.

I hope I hope I was Helpfull!


A chemist would say that the reaction proceeds through an "aromatic transition state". The bridged, chloronium ion transition state drawn in your question involves 2 electrons.  No electrons from the electrophile were used, just 2 electrons from the pi-bond that it added to (draw the resonance structures for the bridged intermediate, each has 2 electrons in the $\ce{C-Cl}$ bond and no electrons in the carbocation p-orbital).  If a transition state contains 4n+2 electrons, then it is said to be aromatic. This case has n=0.

Here is another example. The top line in the figure below shows a suprafacial 1,5 hydrogen shift.  6 electrons are involved here, so the transition state is also referred to as aromatic (4n+2, n=1).

image source (look far down on the page)

Like the bridged halonium ion, this reaction is also very common because it too has an energetically favored (aromatic) transition state.

The second line in the figure depicts a suprafacial 1,3 hydrogen shift.  4 electrons are involved here, so the transition state does not fit the 4n+2 rule.  The transition state is said to be anti-aromatic.  Because it is disfavored energetically (anti-aromatic), the suprafacial 1,3 hydrogen shift does not occur (at least not via a low-energy concerted pathway).

Extra Credit: I don't want to jump to far ahead, but let me briefly say, that what I've written above applies to Hückel transition states, opposite rules apply to Möbius transition states.  See this earlier answer, and references therein, for an explanation of the Hückel and Möbius concepts.


What is the principle of diffuse reflectance spectroscopy behind band gap measurement?

Reflectance spectroscopy is very closely related to UV/Vis spectroscopy, in that both of these techniques use visible light to excite valence electrons to empty orbitals. The difference in these techniques is that in UV/Vis spectroscopy one measures the relative change of transmittance of light as it passes through a solution, whereas in diffuse reflectance, one measures the relative change in the amount of reflected light off of a surface.

A solution that is completely clear and colorless has essentially 100% transmission of all visible wavelengths of light, which means that it does not contain any dissolved components that have (allowed) electronic transitions over that energy range. By the same line of reasoning, a white powder effectively reflects 100% of all visible wavelengths of light that interacts with it.

However, if the material has electronic energy levels that are separated by an energy in the visible region, then it may absorb some of light energy to move electrons from the filled energy level (valence band) into this empty level (conduction band). This causes an relative decrease in the amount of light at that particular energy, relative to a reference source. In other words, the % transmission/reflectance will decrease.

Your data suggests that the separation between the filled valence band and the empty conduction band is around 2.76 eV. Because when light of that energy or higher interacts with the material, it is absorbed resulting in an excitation of the electrons from the valence bands into the conduction band. This lowers the relative reflectance of the material (i.e. lower % transmittance).

I don't this that entirely answered all your questions, but I hope it helps a little.


How can I get xyz coordinates of atoms of a unit cell structure from CIF format files?

CIF contains coordinates for the atoms in asymmetric unit only (what Ivan called "the independent part" in the comments), plus the set of symmetry operation to fill the remaining space with those atoms.
What you want to do is to open CIF in any viewer supporting that format, make sure the unit cell is packed, e.g. the symmetry operations are applied along all three crystallographic axes $a, b, c$ from $0$ to $1$ and export the data to *.xyz format.
There are several programs that are capable of doing this.
For consistency I suggest using crystallographic data by Meagher [1] for both minerals at 25 °C from AMCSD:

Pyrope CIF
Grossular CIF

CSD Mercury

As Martin suggested, Mercury (Free, available on Windows, Linux, MacOS) is one of the most user-friendly and reliable crystallographic viewers.

Open the program and load CIF (I loaded pyrope.cif).
An asymmetric unit pops up:

Go to Calculate → Packing/Slicing… and under Packing area tick Pack option, and under Include Atoms area select … that Fit.

Now you have the content of the entire unit cell:

Save the data as XYZ by clicking File → Save As… → Save as type: XMol files (*.xyz).
This produces a file with the coordinates of all atoms of the unit cell.

Unfortunately, I'm not aware of a good method to eliminate atoms located on the periodic boundary with Mercury as packing coordinates can only be changed by a 0.1 Å step, which is rather crude.
However, there is a workaround I'm aware of with another program, VESTA.

VESTA

VESTA (free, available on Windows, Linux, MacOS) is yet another functional crystallographic viewer.

Open the program and load CIF (again, I loaded pyrope.cif).
By default VESTA not only recognizes atomic connectivity, but also populates the unit cell and the first coordination spheres of all atoms on top of that.

Since the goal is to cut off the content of neighboring unit cells and the boundary atoms, the interatomic bonds should be removed.
Go to Edit → Bonds… and delete all [three] bonds from the list.

Now we have the same atomic set as in Mercury after packing with all the boundary atoms present.

To eliminate the boundary atoms, you can decrease the boundary box by a tiny fraction.
On Style tab click Boundary… button and adjust upper and lower limits for  all three axes ($x_\mathrm{min}$, $x_\mathrm{max}$, $y_\mathrm{min}$, $y_\mathrm{max}$, $z_\mathrm{min}$, $z_\mathrm{max}$) making the box smaller by 0.001 Å on each margin.

After applying new boundaries you get the unit cell content without boundary atoms.

Save the data as XYZ by clicking File → Save As… → Save as type: XYZ file (*.xyz).

References

Meagher, E. P. The Crystal Structures of Pyrope and Grossularite at Elevated Temperatures. American Mineralogist 1975, 60 (3–4), 218–228.


High or low spin complex of Zn2+ and Ti3+

Are you familiar with ligand field stabilisation energy? This approach answers your question. Interaction of the central metal ion with the surrounding ligands causes splitting of the $d$ orbitals into an essentially upper and lower energy level, the form of the splitting is rather complex and depends directly upon the symmetry point group that the ligand complex belongs to.

Either way, high spin complexes only form if the reduction in inter electron repulsion energy is greater than the energy splitting of the orbitals, with the definition of high spin being a none aufbau filling of the split orbitals. It does not make sense to describe the electronic structure of a $\ce{Zn^2+}$ ion in a complex as high or low spin, there is only one configuration available as there are 10 $d$ electrons present, and the 5 $d$ orbitals split to form 5 new orbitals of varying degeneracies (dependent upon the point group of the complex and the symmetry of the coordination environment).

$\ce{Ti^2+}$ has 2 valence d electrons available to be involved in bonding with the ligand environment, however there is no coordination environment that I am aware of (The splitting would have to be extremely small and other thermodynamic factors such as the ionic contribution to the Gibbs energy of complex formation would mean that the complex would never be observed anyway - such a low splitting would only be the result of very poor overlap which would be caused by large separation between the central ion and the ligands, resulting in a lower ionic term) that causes splitting where the magnitude of the splitting is less than the repulsion term invoked when both electrons occupy the same molecular orbital, be it degenerate or not.

The $\ce{Ti^3+}$ ion has only 1 $d$ electron, it will definitely reside in the lowest available molecular orbital. However, the first excited states, ignoring hyperfine splitting, of both titanium ions will be high spin - the magnitude of the splitting usually corresponds to wavelengths of light in the visible region of the spectrum, hence many transition metal complexes are coloured.


Chemistry is determined entirely by electron interactions. Since the number of electrons in an atom is determined by the number of protons in the nucleus, the atomic number (number of protons) is what defines the chemical behavior of a given element. This means that in terms of chemical properties, isotones would have no similarities unless they happened to be in the same group (column) of the periodic table, but that would be a coincidence and would have nothing to do with the number of neutrons.

As DavePhD and ron said, the nuclear stability could be similar, since nuclear stability has a lot to do with the total number of nucleons (neutrons + protons) as well as the relative proportion of protons to neutrons. The wikipedia article on isotones discusses stability in a little more detail.


There are a few reasons. Some of the most important are:

-Oxalic acid dihydrate isn't hygroscopic (it doesn't absorb water from the atmosphere, whereas sulfuric acid does). If a substance can absorb water from the atmosphere, then when you weigh it out, you won't be able to accurately calculate the number of moles used because you don't know how much is water and how much is your substance. This won't be the case with oxalic acid dihydrate.

It's chemically stable when stored.
As Waylander said (in his comment elsewhere), it's a solid so is easily weighed out.
It has a moderately high molecular mass of $\pu{126 g/mol}$ (as the dihydrate) so any error in the mass weighed corresponds to a relatively small error in the calculated amount in $\pu{moles}$.

As a result, oxalic acid has become a primary standard: So you can use it to make a solution of accurately known concentration to determine the concentration of a solution of sodium hydroxide (by titration).


How do I obtain Chrysoidine from Benzene?

I think aniline yellow is likely to nitrate ortho to the aniline giving you the wrong structure. Also the Sn/HCl reduction may well reduce the azo double bond.

Here is a route from benzene.
From benzene a simple nitration gives nitrobenzene. Divide your nitrobenzene, reduce one portion to aniline by hydrogenation over Pd or Pt catalyst. Nitrate the second portion to 1,3-dinitrobenzene procedure here. Reduce the 1,3-dinitrobenzene by hydrogenation over Pd or Pt catalyst to give m-phenylenediamine. Diazotise your aniline to the diazonium and react this with the phenylenediamine. This route was used industrially to prepare Chrysoidine in the 19th century reference here.


I've done some research about it, and have found that the salts of some metals can be used for the intended purpose (for instance, copper sulphate burns green).

You're right! Various melts salts are known to colourize a flame. In the lab and/or for demonstration purposes, this is typically achieved by holding a magnesium oxide stick with a bit of the salt attached into the flame of a Bunsen Burner. There's a wikipedia page on Flame tests that shows some possible effects.

    I use kerosene as my fuel[…]

Here's a problem. The salts you could use soluble in water, but insoluble in a solvent as unpolar as kerosene! Consequently, you might want to give a flammable, but more polar solvent a try.

Is it conceivable to change the flammable solvent used in your performances? I reckon that you might achieve the desired effect using ethanol or methanol. With respect to the toxicity of both solvents, I'd definitely try ethanol first!