Szabo and Ostlund's calculation of repulsion integral on page 172 via A.41

Restricting attention to the integrals representing "self-interaction " of helium in HeH+ we have to find coefficients of the integral,
$$\int A\cdot A\frac{1}{r_{12}}\cdot A  \cdot A$$
in which for an STO-3G calculation $A=a\cdot g_1+b\cdot g_2+c\cdot g_3$ with $g_1$ a gaussian and $a$ a coefficient formed as on page 153 sec. 3.203. The coefficients will fall out as a sum of 81 terms, some of which are identical:
$$(a+b+c)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4+4a^3c+12a^2bc+12ab^2c+4b^3c+6a^2c^2+12abc^2+6b^2c^2+4ac^3+4bc^3+c^4.$$
That there are 81 terms may be checked by viewing the fortran at page 420 and 421. The expression for V1111 is formed by pre-multiplying by D(i)D(j)D(k)D(l) in which i,j,k,l range from 1 to 3. The coefficients DDDD do not appear explicitly in A.41.
The 12 terms in $ab^2c$ and so on may represent different quantities, and care must be taken to keep the accounting correct because (using notation defined below)
$$12ab^2c\cdot 2\pi^{5/2}/((a_1+a_2)(a_2+a_3)\sqrt{a_1+a_2+a_2+a_3})$$
$$\neq 12ab^2c\cdot 2\pi^{5/2}/((a_2+a_2)(a_1+a_3)\sqrt{a_1+a_2+a_2+a_3}) $$
Letting $\zeta_1 = 2.0925$ (page 170) we have:
$a_1=0.109818\cdot\zeta_1^2,~~ a_2=0.405771\cdot\zeta_1^2,~~ a_3=2.22766\cdot\zeta_1^2; $
$a =0.444635\cdot (2a_1/\pi)^{3/4},~b=0.535328\cdot(2a_2/\pi)^{3/4},~ c= 0.154329\cdot(2a_3/\pi)^{3/4}$
The sum represented in the appendix A.41 for the 2-center 2-electron integral
$$ \int He \cdot He \frac{1}{r_{12}}\cdot He\cdot He$$ is given below, and in Mathematica I get 1.307238. Substituting in the constants for hydrogen I get about 0.77466. These are Szabo and Ostlund's values on page 172. They are a good test because they avoid calculation of the error function used in cross terms and they are close to results for the integration of Slater orbitals using spherical coordinates, which are $(5/8)\cdot \zeta. $
The calculation exhibits the coefficients in A.41 explicitly and is in essence the fortran code for V1111.
$ a^4\cdot 2 \pi^{5/2}/((a_1 + a_1)\cdot(a_1 + a_1)\cdot\sqrt{4\cdot a_1})+$
$4 a^3b\cdot 2 \pi^{5/2}/((a_1 + a_1) \cdot(a_1 + a_2)\cdot\sqrt{3 a_1 + a_2})+ $
$ 2 a^2 b^2\cdot 2 \pi^{5/2}/((a_1 + a_1) (a_2 + a_2)\cdot\sqrt{2 a_1 + 2 a_2}) +$
$ 4 a^2 b^2\cdot2 \pi^{5/2}/((a_1 + a_2)\cdot(a_1 + a_2)\cdot \sqrt{2 a_1 + 2 a_2}) +$
$4 a b^3\cdot 2 \pi^{5/2}/((a_1 + a_2) (a_2 + a_2)\cdot \sqrt{a_1 + 3 a_2}) +$
$ b^4\cdot 2 \pi^{5/2}/((a_2 + a_2) (a_2 + a_2)\cdot \sqrt{2 a_2 + 2 a_2}) +$
$ 4 a^3 c\cdot 2 \pi^{5/2}/((a_1 + a_1)*(a_1 + a_3)\cdot \sqrt{3 a_1 + a_3}) +$
$ 4 a^2 b c\cdot 2 \pi^{5/2}/((a_1 + a_1)\cdot(a_2 + a_3)\cdot \sqrt{2 a_1 + a_2 + a_3}) +$
$ 8 a^2 b c\cdot 2 \pi^{5/2}/((a_1 + a_2)*(a_1 + a_3)\cdot \sqrt{2 a_1 + a_2 + a_3}) +$
$  4 a b^2 c\cdot 2 \pi^{5/2}/((a_1 + a_2) (a_2 + a_3)\cdot \sqrt{a_1 + 2 a_2 + a_3}) +$
$  8 a b^2 c\cdot 2 \pi^{5/2}/((a_2 + a_2) (a_1 + a_3)\cdot \sqrt{a_1 + 2 a_2 + a_3}) +$
$ 4 b^3 c\cdot 2 \pi^{5/2}/((a_2 + a_2)\cdot(a_2 + a_3)\cdot\sqrt{3 a_2 + a_3}) +$
$ 2 a^2 c^2\cdot 2 \pi^{5/2}/((a_1 + a_1)\cdot(a_3 + a_3)\cdot \sqrt{2 a_1 + 2 a_3}) +$
$ 4 a^2 c^2\cdot 2 \pi^{5/2}/((a_1 + a_3)\cdot(a_1 + a_3)\cdot\sqrt{2 a_1 + 2 a_3}) +$
$ 8 a b c^2\cdot 2 \pi^{5/2}/((a_1 + a_2)\cdot(a_3 + a_3)\cdot \sqrt{a_1 + a2 + 2 a3}) +$
$ 4 a b c^2\cdot 2 \pi^{5/2}/((a_1 + a_3)\cdot(a_2 + a_3)\cdot \sqrt{a_1 + a_2 + 2 a_3}) +$
$ 2 b^2 c^2\cdot 2 \pi^{5/2}/((a_2 + a_2)\cdot(a_3 + a_3)\cdot \sqrt{2 a_2 + 2 a_3}) +$
$ 4 b^2 c^2\cdot 2 \pi^{5/2}/((a_2 + a_3)\cdot(a_2 + a_3)\cdot \sqrt{2 a_2 + 2 a_3}) +$
$  4 a c^3\cdot 2 \pi^{5/2}/((a_1 + a_3)\cdot(a_3 + a_3)\cdot \sqrt{a_1 + 3 a_3}) +$
$ 4 b c^3\cdot 2 \pi^{5/2}/((a_2 + a_3)\cdot(a_3 + a_3)\cdot \sqrt{a_2 + 3 a_3}) +$
$ c^4\cdot2 \pi^{5/2}/((a_3 + a_3)\cdot(a_3 + a_3)\cdot \sqrt{4 a_3})$
Some copy-and-paste Mathematica code --don't know if it works in Wolfram Alpha.
z11 = 2.0925; z21 = 1.24;

a1 = .109818*z11^2; a2 = .405771*z11^2; a3 = 2.22766*z11^2;

a = .444365 (2 a1/Pi)^(3/4); b = .535328 (2 a2/Pi)^(3/
     4); c = .154329 (2 a3/Pi)^(3/4);

a^4*2 Pi^(5/2)/((a1 + a1)*(a1 + a1)*Sqrt[4 a1]) +
 4 a^3 b*2 Pi^(5/2)/((a1 + a1) (a1 + a2)*Sqrt[3 a1 + a2]) +
 2 a^2 b^2*2 Pi^(5/2)/((a1 + a1) (a2 + a2)*Sqrt[2 a1 + 2 a2]) +
 4 a^2 b^2*2 Pi^(5/2)/((a1 + a2)*(a1 + a2)*Sqrt[2 a1 + 2 a2]) +
 4 a b^3*2 Pi^(5/2)/((a1 + a2) (a2 + a2)*Sqrt[a1 + 3 a2]) +
 b^4*2 Pi^(5/2)/((a2 + a2) (a2 + a2)*Sqrt[2 a2 + 2 a2]) +
 4 a^3 c*2 Pi^(5/2)/((a1 + a1)*(a1 + a3)*Sqrt[3 a1 + a3]) +
 4 a^2 b c*2 Pi^(5/2)/((a1 + a1)*(a2 + a3)*Sqrt[2 a1 + a2 + a3]) +
 8 a^2 b c*2 Pi^(5/2)/((a1 + a2)*(a1 + a3)*Sqrt[2 a1 + a2 + a3]) +
  4 a b^2 c*2 Pi^(5/2)/((a1 + a2) (a2 + a3)*Sqrt[a1 + 2 a2 + a3]) +
  8 a b^2 c*2 Pi^(5/2)/((a2 + a2) (a1 + a3)*Sqrt[a1 + 2 a2 + a3]) +
 4 b^3 c*2 Pi^(5/2)/((a2 + a2)*(a2 + a3)*Sqrt[3 a2 + a3]) +
 2 a^2 c^2*2 Pi^(5/2)/((a1 + a1)*(a3 + a3)*Sqrt[2 a1 + 2 a3]) +
 4 a^2 c^2*2 Pi^(5/2)/((a1 + a3)*(a1 + a3)*Sqrt[2 a1 + 2 a3]) +
 8 a b c^2*2 Pi^(5/2)/((a1 + a2)*(a3 + a3)*Sqrt[a1 + a2 + 2 a3]) +
 4 a b c^2*2 Pi^(5/2)/((a1 + a3)*(a2 + a3)*Sqrt[a1 + a2 + 2 a3]) +
 2 b^2 c^2*2 Pi^(5/2)/((a2 + a2)*(a3 + a3)*Sqrt[2 a2 + 2 a3]) +
 4 b^2 c^2*2 Pi^(5/2)/((a2 + a3)*(a2 + a3)*Sqrt[2 a2 + 2 a3]) +
  4 a c^3*2 Pi^(5/2)/((a1 + a3)*(a3 + a3)*Sqrt[a1 + 3 a3]) +
 4 b c^3*2 Pi^(5/2)/((a2 + a3)*(a3 + a3)*Sqrt[a2 + 3 a3]) +
 c^4*2 Pi^(5/2)/((a3 + a3)*(a3 + a3)*Sqrt[4 a3])


There are multiple issues that need to be addressed before one can give a definite answer to this question. Let's start by what we are certain about:

1- The soap is not the issue (unless you haven't rinsed the barrel thoroughly with water). The remnant soap would be so dilute that should not cause any problem.

2- Since the film keeps forming there seems to be an equilibrium issue for your oil to separate. If you look into partition coefficients of the oils you should be able to estimate how much will remain in the water. Anyway, the less water you have, the more oil you can collect. (concentration factor pushes the oil out to keep the equilibrium) One idea is to add less water to your feed.

3- THEORETICALLY one way to enhance oil extraction is to use a nonpolar solvent. HOWEVER, I STRONGLY ADVISE AGAINST ACTUALLY TRYING IT AT HOME due to the flammable nature of most nonpolar solvents and health issues related to it. If you can find a nonpolar benign solvent (that is a big IF), mix it with your final water oil mixture, and then decant the water from the bottom and then distill the solvent out to get the oil as a residue.

4- Depending on the chemical structure of the oil, some of it will always remain in water. If they are not highly hydrophobic, it might be difficult to separate them from water anyway without using an assistant solvent.


Caclulating the Solubility of Fluorite in River Water

the ppm of fluorite in pure water (15.6ppm)

the ppm of fluorite in pure water is zero!

  $K_\mathrm{sp}=[\ce{Ca_{original}} + \ce{Ca_{added}}][2\ce{Ca_{added}}]^2 = 4[\ce{Ca_{original}} + \ce{Ca_{added}}][\ce{Ca_{added}}]^2$

Yes, this is correct and can be solved, you say you know $[\ce{Ca_{original}}]$ is $\pu{0.375E-3M}$ and  $K_\mathrm{sp}$ is $10^{-10.5}$ so there is one equation and one unknown, just a matter of solving a cubic equation, which can be solved exactly or by approximate methods.


If there is a solution of mix of a weak acid/base and it's alkaline/acidic salt in molar ratio 1:1, then

The solution is acidic, if $K_\mathrm{a} \gt K_\mathrm{b}$, like acetic acid + sodium acetate(alkaline base)
The solution is alkalic(ine)/basic, if $K_\mathrm{a} \lt K_\mathrm{b}$, like ammonium chloride ( acidic salt) + ammonia(base)

Note that $$\begin{align}
K_\mathrm{w}&=\ce{[H+][OH-]} \\
K_\mathrm{a}&=\frac{\ce{[H+][A-]}}
{\ce{[HA]}}  \\
K_\mathrm{b}&=\frac{\ce{[HA][OH-]}}
{\ce{[A-]}}  \\
&=\frac{\ce{[HA]  K_\mathrm{w}}}{\ce{[A-][H+]}}  \\
&=\frac{\ce{  K_\mathrm{w}}}{K_\mathrm{a}}  \\
K_\mathrm{w}&=K_\mathrm{a} \cdot K_\mathrm{b} \\
\end{align}$$

Similarly, if there is such a 1:1 (molar amount) solution of an acid and a base, that are not in the mutual relation acid/salt or base/salt, the above is valid as well.


How can I neutralise H2 gas without burning it?

Many nuclear power plants use so-called "passive autocatalytic recombiners" for the catalytic oxidization of hydrogen that could be released into the containment in the event of severe accidents. (Note that the affected units of the Fukushima-Daiichi plant did not have such equipment.) The catalytically active materials, typically platinum and/or palladium, are doted on carriers, preferably designed as catalyst plates orientated vertically to support the flow.


Reaction of Iron with Concentrated and dilute hydrochloric acid

No, $\ce{FeCl3}$ is not formed upon reaction with Conc. $\ce{HCl}$. Both dilute and concentrated $\ce{HCl}$ give $\ce{FeCl2}$ as product.

$$\ce{Fe + 2HCl_{dil} -> Fe^2+ +  2Cl^- + H2 ^}$$
$$\ce{Fe + 2HCl_{conc} -> Fe^2+ +  2Cl^- + H2 ^}$$

Source: Vogel's Qualitative Inorganic Analysis


Is there a name for the pair (c, R), where c is the stoichiometric coefficient of a reactant R in some chemical reaction?

In chemistry, often quantities get denoted by an index $i$, which then stands for some compound ($i=\text{A, B, C, ...}$).

Examples:

The stoichiometric coefficient $\nu_i$
The concentration $c_i$
The amount of substance $n_i$

The notion of grouping together the stoichiometric coefficient with the substance it uses is a novel one for me, and will most likely not have a name.

I suggest you use a name such as "coefpair" or the like and document your code accordingly.


Can I damage a biotinylated antibody in solution by vortexing or mixing too vigorously?

It is good practice to avoid vortex mixing any protein solutions, especially vigorously.  Vigorous vortex mixing can produce foaming and proteins can denature (unfold and lose activity) when they are present at an air/liquid interface (as in a foam).  This is because the hydrophobic regions of the protein will want to be in the air region and the hydrophilic regions will want to be in solution, and so unfolding may occur.
In practice, antibodies are usually quite stable as proteins go and you can often vortex them without much negative effect to your experiment. That said, if you routinely vortex proteins in the lab and come across a less stable one you will suffer degradation.  Since you will not always know in advance which proteins will be OK to vortex, it is best practice to mix gently using a pipetman (again avoid introducing air that creates a foam) or inversion, or other method. If you must vortex, then be gentle with short pulses, avoiding foaming.
I have purchased some antibodies where the datasheet specifically says not to vortex mix.


Short story

Maybe it will help to think first about a ridiculous case. Say you have a large hydrophilic solid, and a small drop of water is added on top. We would  not expect the entire solid to disappear into the liquid. Yes, there will be some adsorption and solubility here, yet most of the solid will remain undisturbed. In some sense, the solid 'does not fit' into the liquid droplet over a certain limit.

Similar things are true,  again in a certain interpretation, for more realistic situations. What does this 'fitting' mean exactly? It definitely does not mean there is not enough physical space per se. Instead as you increase the concentration of your solute, the distances between solute entities starts to decrease on average. Hence, at a certain radii the intermolecular forces become strong enough to pull the solutes together. There is often some energy barrier to this approaching, as colloid chemists will tell you, but if the end result is some energy minimum, the process will occur at some rate.

If the ratio of the amount of solvent to the amount of solute increases, that is we add more solvent, we should expect the solubility to increase. This is indeed what we observe. In summary, the amount of 'space' with respect to solubility is a function of the strength of intermolecular forces and the amount of solvent, among other things. Here you might also discover yourself that smaller particles dissolve more (not only faster!) than bigger particles.

You may compare precipitation to condensation of water in air. There is plenty of actual physical space to go around but once a certain limit is crossed (the dew point) the water will become liquid. Nitrogen and oxygen will not condense at these temperatures since their intermolecular forces are much weaker.[a]

An equally short story

Note first that for $\Delta_\pu{r} G \leq 0$ ($\Delta G$ was poor notation on my part) to be the criterion (necessary and sufficient) of spontaneity, some five conditions must be met: constant temperature, no pressure change, only expansion work is done, and the application of the thermodynamic limit. Fifthly, only one chemical reaction is assumed to take place.[b] Provided these prerequisites stand, we may look at the terms in

$$\Delta_\pu{r} H-T\Delta_\pu{r} S\le0.$$

(We should not really compare entropy directly to enthalpy because they have different dimensions.) Jumping in quickly, we might say immediately that if the solute can wriggle around more in a solution than in a solid, then certainly $\Delta_\pu{r} S>0$ for the transition $\text{solid}\to\text{solute}$. The only thing standing in our way is that damned enthalpy! Sure enough, that may be the case for some solutions but it is not the whole picture.

Since the solute entities have a solvent layer in solution, there is an entropy effect for the solvent as well. If the solvent is tightly bound to the solute, entropy decreases for this solvent entity. There is also an enthalpy contribution from solvation, not only from the act of breaking or forming bonds in a solid. I would not really dare to generalise which is the bigger contributor 'for most cases'. In general,

\begin{align}
\Delta_\pu{r} H_{\pu{solid to ideal gas}} & + \Delta_\pu{r} H_{\pu{breaking bonds in solvent}} + \Delta_\pu{r} H_{\pu{solute forming bonds with solvent}}\ + \\ &+ \Delta_\pu{r} H_{\pu{residual}} - T\cdot(\text{corresponding ‘entropy’ terms}) \le 0
\end{align}

where 'residual' signals the  solute-solute interactions in the solution.

A sentence or two on $K_{\pu{sp}}$ (with concentrations)

From the IUPAC Gold Book,

  solubility

  The analytical composition of a saturated solution, expressed in terms of the proportion of a designated solute in a designated solvent, is the solubility of that solute. The solubility may be expressed as a concentration, molality, mole fraction, mole ratio, etc.

As pointed out by Nicolau Saker Neto in a comment, we should not overemphasise the role of $K_{\pu{sp}}$ in determining solubility. Again there are cases where it is a good indicator, and situations where it fails poorly. There are assumptions which must be satisfied:

there actully exists an equilibrium $$\ce{AB(sln) <=> A+(sln) + B+(sln),}$$ that is not all of the solute "molecules" are dissociated, not really the case for strong electrolytes in a dilute solution,
that the simpification of $a \approx c$ is good enough ($a$ is activity, $c$ is some type of concentration),
provided that the above holds, the concentration of $\ce{AB(aq)}$ should be constant wrt additon of another electrolyte (in some calculations).

If these do not hold, $K_{\pu{sp}}$ (with concentrations alone) is a poor approximation for true solubility. See Denbigh, K. G. 'The Theory of Solubility Product'. J. Chem. Educ., 1941, 18 (3), 126. DOI: 10.1021/ed018p126.

[a] I have largely ignored the discussion of surface energies and surface tension, and how the interaction is often more favourable in the bulk than on the surface, and the fact that $r^3$ grows faster than $r^2$, except for small $r$ when it does not, and hence we need some seeds or nuclei to start the process, and supersaturation, and how particle sizes change etc. The example of smaller particles dissolving more is, of course, a surface tension effect.

[b] In the thermodynamic limit, the general condition of spontaneity comes from Clausius inequality. This is written as

$$\small{T\pu{d}S \ge \delta q.}$$

With the first law of thermodynamics, this is equivalent to

$$\small{T\pu{d}S\ge\pu{d}U-\delta w}$$

or

$$\small{-(\pu{d}U +p\pu{d}V - T\pu{d}S)\ge-\delta w(\text {other})}.$$

With constant temperature and pressure,

$$\small{\pu{d}(U +pV - TS)\le\delta w(\text {other})}.$$

Finally, if only expansion work is done,

$$\small{\Delta_\pu{r} G\le 0}$$

where $\small{\Delta_\pu{r} \equiv \partial / \partial \xi}$; $\small{\xi}$ is the extent of a reaction. The last result also assumes only one reaction takes place, otherwise it is no longer a necessary–sufficient condition for spontaneity.


For a 1st order reaction, how do you derive the integrated rate law for a product versus time?

If you are doing this on a molar basis, you have to be careful about assumptions like: $$[A_o] = [A]_t + [P]_t$$
In the same way that you have defined a component balance for the reactant system, you can also do so for the product system. Let's look at the concentration of B in your reaction equation $\ce{aA -> bB + cC}$.

Defining the component balance,
$$\frac{d[B]}{dt}=k[A]$$
$$[B]_o = 0, t = 0$$
This is essentially saying that the generation of $B$ is proportional to the inverse depletion of $A$ (as defined in your component balance). We know that from the solution of the A component balance that, $$[A](t) = [A]_oe^{-kt}$$ Plugging in this expression to the differential equation for $[B](t)$ yields,
$$\frac{d[B]}{dt} = k[A]_oe^{-kt}$$ Solving this differential equation will yield the following expression for $[B]$,
$$[B](t)=-[A]_oe^{-kt}+C_1$$
Using our intial condition that $[B]=0, t = 0$ we solve for our constant $C_1$ and end up with,
$$[B](t)=[A]_o(1-e^{-kt})$$
Note that this is similar to the form you had already derived, however... this is only for $B$. Since the kinetic expression will be the same assuming no initial mols of $B$ or $C$ and same stoiciometric coefficients (i.e $b = c$) then we can reasonably assume that $$[P]_t = 2[A]_o(1-e^{-kt})$$


How is the hydration state of a chemical related to its adsorption of water?

Surprisingly enough, hemihydrate reacts quickly with water to produce dehydrate (gypsum). It is due to the fact that hemihydrate is produced by heating gypsum to a high temperature. This heating produces a structure with plenty of holes or voids corresponding to the previous positions of the water molecules. Water molecules can easily fill these holes if hemihydrate is mixed with water. Anhydrous calcium sulfate is a mineral (anhydrite) that does not react easily with water. The structure of anhydrous calcium sulfate is compact and does not contain any holes between its atoms to include easily water molecules.


Extracting other mineral from salt in evaporated seawater

One material extracted from seawater is magnesium.

In the United States, magnesium is obtained principally with the Dow process, by electrolysis of fused magnesium chloride from brine and sea water [emphasis added]. A saline solution containing $\ce{Mg^{2+}}$ ions is first treated with lime (calcium oxide)[sic; calcium hydroxide would seem more likely] and the precipitated magnesium hydroxide is collected:

$\ce{Mg^{2+}(aq) + CaO(s) + H2O → Ca^{2+}(aq) + Mg(OH)2(s)}$

The hydroxide is then converted to a partial hydrate of magnesium chloride by treating the hydroxide with hydrochloric acid and heating of the product:

$\ce{Mg(OH)2(s) + 2 HCl → MgCl2(aq) + 2H2O(l)}$

The salt is then electrolyzed in the molten state. At the cathode, the Mg2+
 ion is reduced by two electrons to magnesium metal:

$\ce{Mg^{2+} + 2 e^− → Mg}$


What will be the major product of of the Wurtz reaction of 3,6-dibromocyclohexa-1,4-diene?

The answer is (B) because (B) is benzene, which is the prototypical aromatic hydrocarbon. The aromaticity of (B) increases its thermodynamic stability relative to (A), which is called Dewar benzene (mistakenly attributed to James Dewar), which is notoriously difficult to synthesis and isolate. It reverts back to benzene with a half-life of two days.

Generally, if a pathway exists for a reaction to produce an aromatic product, then the reaction likely will do so. In this particular case, the diradical intermediate that might be produced as part of the Wurtz reaction's mechanism, is the triplet excited state of benzene.


Propanoic acid reaction with potassium permanganate

Can you detail the concentrations of both the acid as well as $\ce{KMnO4}$ and also the time it took to de-colourise?

I tried an experiment however, I only have Glacial Acetic Acid, which didn't seem to react ( as the smell was still prominent).

I can suggest you a simple trick to analyse whether it is decomposing. Smell Propionic acid before adding $\ce{KMnO4}$. Then smell it after the colour of $\ce{KMnO4}$ is gone. Keep adding $\ce{KMnO4}$ till the pink colour stay(similar to titration). Then smell again. If the smell is gone then the carboxylic acid is oxidised.

There are 2 possible oxidation products of carboxylic acid:

Ketone
Carbon Dioxide

Both can be pretty obvious. Ketone will smell differently than the acid(more like nail polish remover) and Carbon Dioxide should be visible because of effervescence.


Is the concentration of negative ions always equal to that of positive ions?

The only point that $[\ce{H+}]=[\ce{A-}]$ is when the $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$ of a given acid, so indeed in most instances $[\ce{H+}] \neq [\ce{A-}]$. In addition to $\ce{A-}$, the solution also contains negatively charged $\ce{OH-}$ that balances the positive charge of the $\ce{H+}$ ions. If the acid is titrated with a strong base (e.g. $\ce{NaOH}$), the positively charged $\ce{Na+}$ ions balance the negative charge of $\ce{OH-}$ and $\ce{A-}$.

To answer the question in your title though, yes, charges are always balanced. To quote another good post on this topic:

  While it is true that solutions are not always locally electroneutral, that non-neutrality is small, and does not extend to larger scale. Thus, it does not impact chemical concentrations: you cannot separate positive and negative charges in a solution at a macroscopic level.


How to extract force when the center of mass of molecules is put in the box in GROMACS?

I solved the problem. As gmx trjconv -pbc mol option only make molecules translate, the cooridinates obtained from alad.gro can correspond to forces obtained from gmx traj -f alad.trr -s alad.tpr -n alad.ndx -of force-dump.xvg.


Are wrinkles\folds in paper reversible by heat, pressure or something else?

Once you fold a paper or crush a paper, then it was found experimentally that the internal geometry of paper developed ridges throughout inside it and these ridges inside the folded or crushed paper holds the energy you imparted by folding the paper. This energy imparted by you by folding the paper gets stored in the ridges of the paper in form of bending energy (and this bending energy is also the reason that a crumpled paper ball cannot be further compressed beyond around 90% air).

The crumpling of a piece of paper leaves localised permanent marks. This shows that, by some mechanism, the crumpling has focused the stress at some points, exceeding there the limit yield of the material and leading thus to irreversible deformations.

Indeed, a piece of crumpled paper does not usually return to its planar shape even if you iron it or apply pressure to it because along the folds, the material is often strained beyond the level of reversible elasticity.


According to the Ref.1, 5-hydroxy-1,3-dioxane (the compound of structures V and VI) both in the gaseous state and in dilute $\ce{CCl4}$ solution exists as a chair conformer, the hydroxy group in an axial position with an intramolecular hydrogen bond of the $\ce{O—H⋯O}$ type:

Accordingly, the axial conformer (V) is more stable by $\Delta G^\circ = \pu{-1.2 kcal mol−1}$). Microwave spectroscopy and the $^3J_\ce{H(5eq),OH}$ coupling constant suggest that the $\ce{OH}$ group lies in the plane of symmetry $(\ce{C_{(2)}-C_{(5)}-O}\text{-plane})$ and is a part of a bifurcated hydrogen bond to the two ring oxygen atoms (Ref.2). Therefore, it is safe to say the conformer V is more stable than conformer VI between two possible conformers of 5-hydroxy-1,3-dioxane (marked with $\color{green}{\text{green}}$ oval).
It is well known that the most stable conformer of tert-butylcyclohexane is tert-butyl in equatorial position. In a 1,4-substituted version of 1-tert-butyl-4-methylcyclohexane, since methyl group is so small compared to tert-butyl group that I is almost exclusive in equilibrium (marked with $\color{green}{\text{green}}$ oval).
However, I wasn't so sure about stability between III and IV. There is clearly no H-bonding opportunities except theoretical chemist may argue hyper conjugation with ring oxygen by MO calculations. I leave it open for any computational chemist to prove that possibility, yet my best guess is III more stable between them (marked with $\color{orange}{\text{orange}}$ oval with question mark).
References:

J. C. Jochims, Y. Kobayashi, “Bifurcated hydrogen bonds in z-2-phenyl-1,3-dioxan-5-ols,” Tetrahedron Letters 1976, 17(24), 2065-2068 (https://doi.org/10.1016/S0040-4039(00)93819-X).
Jose L. Alonso, E. Bright Wilson, “Study of an intramolecular, bifurcated hydrogen bond in 1,3-dioxan-5-ol by microwave spectroscopy,” J. Am. Chem. Soc. 1980, 102(4), 1248–1251 (https://doi.org/10.1021/ja00524a005).


How can we do amide coupling reaction in water?

A search on the phrase "Amide Formation in Water" produces a good selection of hits.
Recent papers such as this and this offer conditions.
From the abstract of 1

A general and environmentally responsible method for the formation of amide/peptide bonds in an aqueous micellar medium is described. Use of uronium salt (1-cyano-2-ethoxy-2-oxoethylidenaminooxy)dimethylaminomorpholinocarbenium hexafluorophosphate (COMU) as a coupling reagent, 2,6-lutidine, and TPGS-750-M represents mild conditions associated with these valuable types of couplings


Kinetic Energy Evaluation Integral Evaluation Program Ostlund

The off-diagonal elements $T_{\mu\nu}, (\mu \ne \nu)$ are just as necessary as the diagonal elements $T_{\mu\mu}$ to represent the kinetic energy operator in the chosen AO basis set. Substituting a few of Szabo/Ostlund's equations into each other (in my edition, 3.147-3.154), one finds
$$
F_{\mu\nu} = T_{\mu\nu} + V_{\mu\nu}^{\text{nucl}} + \sum_{\lambda\kappa} P_{\lambda\kappa}\left[ \left( \mu\nu \vert \kappa\lambda  \right) - \frac{1}{2} \left( \mu\lambda \vert \kappa \nu\right) \right]
$$
where $F$ is the Fock matrix, $V^{\text{nucl}}$ represents the nuclear attraction term, $P$ is a density matrix and $\left( \mu\nu \vert \kappa\lambda  \right)$ are two-electron repulsion integrals over the AO basis functions (denoted greek indices). Of course, your Fock matrix may be diagonal already, if you have a very special case and basis set, but in general and to perform the SCF calculation, the off-diagonal parts cannot be ignored.


Hydrogen storage using activated carbons

Short answer: yes, if you want to store the same amount of hydrogen and on condition that you do not mess with pressure.

Long answer: I'll skip some intermediate physicochemical derivations for brevity. The phenomenon we are dealing with is called adsorption. Perhaps, the simplest theoretical explanation of adsorption was given by Langmuir. Langmuir adsorption isotherm can be written as $$\theta = \frac{Kp}{1 + Kp},$$ where $\theta$ is a fraction of the surface of adsorbent (solid) covered by the adsorbate (gas) and $K$ is an equilibrium constant. It is seen that at a constant temperature the more pressure you apply to the system, the more surface is covered (i.e. the more gas is adsorbed).

But what about the temperature dependence? Since $K$ is an equilibrium constant, its temperature dependence is defined by the van't Hoff equation. Also, $K$ is directly proportional to the pressure required for a certain amount of gas to be adsorbed. In other words, when $\theta$ is constant, $\ln{K}+\ln{p}=\mathrm{const}$, transforming our van't Hoff into something like $$\left(\frac{\partial\ln{p}}{\partial(1/T)}\right)_\theta=\frac{\Delta_{ad}H^○}{R}.$$ Here, $R$ is the gas constant, and the sign of $\Delta_{ad}H^○$ - adsorption enthalpy - is negative, for adsorption is exothermic. Consequently, the slope of $\ln{p}$ vs $1/T$ line is negative. In plain English, the equation above reads like this: in order to obtain the same amount of adsorbate $\theta$ you should either decrease the temperature or increase the pressure.

If you're that interested in the topic, you may also want to check out how the sorption pumps work.


What happens during cleaning silverware?

Put them into a pot with some club soda and a piece of aluminium foil and pour over a hot water.

I am familiar with this process. I am not familiar with the other. Perhaps someone else can provide an answer for it.

  What happens chemically during these procedure - why does it work, and what are the byproducts of these reactions?

This process cleans silver by a redox process. Aluminum is a more active metal than silver. This means that aluminum metal, Al(s), will reduce silver salts (salts of Ag+). The most common type of tarnish on silver is silver oxide, Ag2O.
The reaction is:

$$\ce{3Ag2O(s) + 2Al(s) -> 3Ag(s) + Al2O3(s)}$$

The club soda - containing sodium bicarbonate - plays two roles. First it acts as a supporting electrolyte so that electrons can be conducted through the solution. Second, the carbonic acid in it dissolves the passivation layer of aluminum oxide that protects the aluminum foil. This produces aluminum carbonate, which is unstable in water and decomposes into amphoteric aluminum hydroxide, which is soluble through the formation of a number of complex ions.

$$\ce{Al2O3(s) + 3H2CO3(aq) -> Al2(CO3)2(s) + 3H2O(l)-> 2Al(OH)3 (aq) + 3CO2(g)}$$

The net reaction is then:

$$\ce{3Ag2O(s) + 2Al(s) +3H2CO3(aq) -> 3Ag(s) + 2Al(OH)3(aq) + 3CO2(g)}$$

The byproducts are carbon dioxide and aluminum hydroxide.

How toxic (if all) are the byproducts?

Not very. Aluminum salts have very low health impact on you or the environment. Additionally, you produce only very small amounts of aluminum salts this way.

How much silver is lost in such a procedure?

None (Almost). The process coverts solid silver oxide into solid silver metal, both of which are negligibly soluble in club soda (or any other aqueous solution). The $K_{sp}$ of AgOH (which is what the surface of Ag2 looks like when exposed to water) is $1.52\times10^{-8}$. All of the action happens at the surface of both metals. The only common silver salts of appreciable solubility in aqueous solutions are silver nitrate and silver sulfate.

Is this procedure "idempotent"?

Yes. This procedure converts silver oxide into metallic silver. It has no effect on metallic silver: metallic silver does not react with most acids; it cannot be reduced farther by aluminum metal; and it is not soluble in water or most aqueous solutions.


Balancing the reaction equation for the oxidation of lead(II) sulfide with ozone

This is a semi-standard example of why 'naive' balancing does not work.

Ozone is a source of atomic oxygen, producing free oxygen molecules. So, in 'mild' conditions only one oxygen per ozone molecule would react as a strong oxidizer, and the remaining molecular oxygen would require elevated temperatures to react. Thus the equation would be

$\ce{PbS + 4O3 -> PbSO4 + 4 O2}$

Similarly, in excess of reducer at elevated temperature (though why would you want it with ozone is another matter) all atoms in ozone would react. Thus, the reaction should be

$\ce{3PbS + 4O3 -> 3PbSO4 }$


How to convert -OH to terminal -SH group in nucleolipids:

An alcohol can be converted to a corresponding thiol using Lawesson's reagent (LR; 2,4-bis(4-methoxyphenyl)-1,3,2,4-dithiadiphosphetane 2,4-disulfide). However, the reaction is messy and could give you a mixture of products in addition to intended thiol, including eliminated product, alkene (Ref.1). For example, treatment of acyclic 1,4-diols with LR gives the 1,3-dienes (Ref.1).

However, the substitution versus elimination outcome with LR is depends on the structural features of the substrate (Ref.2), and hence, worth considering. The computational mechanism is depicted below:

Alternatively, you can use highly efficient Mitsunabu type reaction with diisopropyl azodicarboxylate (Ref.3). The abstract summerize the reaction as:

  Various alcohols were converted to their corresponding thiolacetates by treatment with triphenylphosphine and diisopropyl azodicarboxylate in the presence of thiolacetic acid. The overall conversion was both highly efficient (89–99% yields) and stereoselective (99.5% inversion).

References:

T. Nishio, “Direct conversion of alcohols into thiols,” J. Chem. Soc., Perkin Trans. 1 1993, (10), 1113–1117 (DOI: 10.1039/P19930001113).
M. A. Chiacchio,  L. Legnani,  P. Caramella,  T. Tejero,  P. Merino, “Pivotal Neighboring‐Group Participation in Substitution versus Elimination Reactions – Computational Evidence for Ion Pairs in the Thionation of Alcohols with Lawesson's Reagent,” Eur. J. Org. Chem. 2017, (14), 1952–1960 (https://doi.org/10.1002/ejoc.201700127).
R. P. Volante, “A new, highly efficient method for the conversion of alcohols to thiolesters and thiols,” Tetrahedron Letters 1981, 22(33), 3119–3122 (https://doi.org/10.1016/S0040-4039(01)81842-6).


Why does acidity increase going from sulfide to sulfoxide to sulfone?

The oxidation state of sulphur and its therefore increased inductive effect should be enough to explain this effect.

Besides the number of oxygens, the principle difference between the three structures is the oxidation state of sulphur that rises from $\mathrm{-II}$ to $\pm0$ to $\mathrm{+II}$. The higher an element’s oxidation state, the stronger it draws electrons towards itself since a higher oxidation state typically correlates with less electron density.

Unfortunately, we cannot draw good comparisons. Phosphorus would suggest itself, but neither the Evans table nor the Bordwell data have enough data points for scrutinisation. Furthermore, phosphorus is less electronegative than carbon, meaning the bonds would be polarised towards carbon.

Chlorine might be an option being on the other side of sulphur in the periodic table, but to the best of my knowledge no ‘chlorone’ or ‘chloroxide’ organic compounds are known let alone their acidities.

Nitrogen in its higher oxidation state (a nitro group) has a double bond which can resonate with the negative charge so it is not a good comparison. And the elements in the fourth period and below are again too electropositive in general. If anything, we may be able to draw conclusions based on the β-acidity of an ether, the α-acidity of an acetal/ketal or the α-acidity of an — if these are even stable enough to be deprotonated.


in the electrolysis of H₂SO₄ solution, when the concentration is increased why does the current increase even when OH⁻ ions reduce?

Near room temperature, considering solutions of water and sulphuric acid, the maximum conductivity is at about 30% sulfuric acid.  For pure water or pure sulfuric acid the conductivity is orders of magnitude lower.

In pure water, there are only $10^{-7}$M of $\ce{H+}$ and $\ce{OH-}$ to carry current.

When a relatively small amount of sulfuric acid  is added, it dissociates fully to $\ce{H+}$ and $\ce{HSO4-}$.  For example if you increase the concentration of sulfuric acid from 0M to 0.01 M, the number of ions in the solution increase by a factor of $10^5$.


What is the mechanism for oxidation of ketones by the Popoff's rule?

I am uncertain about the "Popoff's" rule you mention. There are two reactions that can oxidize ketones, and one seems to follow the behavior you are suggesting, but does not form a carboxylic acid. The other reaction does form carboxylic acids, but is more complex.

The Baeyer-Villager oxidation is an oxidation of ketones to esters using a peracid in the presence of a mild base:

$$\ce{CH3COCH3 + RCO3H->[\ce{Na2HPO4}] CH3CO-OCH3 + RCOOH}$$

The mechanism involves the fragmentation of one of the $\ce{C-CO}$ bonds. Unsymmetrical ketones fragment in a predictable pattern, but not always that the carbonyl remains with the smaller group. The fragment that would be a more stable carbocation (even though carbocations are not formed in this reaction) is the one to move. The migratory aptitude is tertiary alkyl > cyclohexyl > secondary alkyl, aryl > H > primary alkyl > methyl . For exampl, with 2-methyl-3-octanone, the isopropyl group moves because it is attached via a $2^\circ$ carbon:

$$\ce{(CH3)2CHCOCH2CH2CH2CH2CH3 + RCO-O-OH ->[\ce{Na2HPO4}] (CH3)2CH{\bf O}COCH2CH2CH2CH2CH3 + RCOOH}$$

Nitric acid chews ketones apart into two carboxylic acids. Concentrated $\ce{KMnO4}$ in acid also does this. These reactions are a little bit harder to find information on, since they tend to be considered uncontrollable.

The reaction goes through a series of oxidations from ketone to $\alpha$-hydroxyketone to cleaved acids through several enol intermediates.

$$\ce{RCH2COCH2R <=>[\ce{HNO3}] RCH=C(OH)CH2R ->[\ce{HNO3}] RCH(OH)COCH2R}$$
$$\ce{RCH(OH)COCH2R <=>[\ce{HNO3}] RC(OH)=C(OH)CH2R ->[\ce{HNO3}] RCO}$$

If the ketone is unsymmetrical, there is no guarantee that is will cleave predictably. For example, 2-butanone could cleave into propanoic acid and carbon dioxide or two equivalents of acetic acid.

$$\ce{CH3CH2COCH3 ->[\ce{HNO3}] n(CH3CH2COOH + CO2) + (1-n)(2CH3COOH)}$$

A balanced equation for the formation of acetic acid would look like:
$$\ce{CH3CH2COCH3 + 3NO3^- +  -> 2 CH3CO2H + 3NO2^-}$$

A balanced equation for the formation of propanoic acid and carbon dioxide would look like:
$$\ce{CH3CH2COCH3 + 4NO3^- +  -> CH3CH2COOH + CO2 + 4NO2^- + H2O}$$

The permanganate reactions are tougher to balance, since permanganate is a three electron oxidant.

Update:

When I wrote this answer, I had never heard of Popoff's rule. No textbook I own mentions this rule, and a Google search about it brings up this question as the top hit (and similar questions at other sites like ask.yahoo.com as the other hits). I now know that Aleksandr Popov published a paper in Liebigs Annalen in 1872 describing a variation of the reaction that would become the Baeyer-Village oxidation. This article is behind a paywall for me, and the first page preview confirms I would not be able to make much of it. I only know a small amount of German and the PDF sadly looks like a low quality copy of a copy of a copy.

However, from the title "Die Oxydation der Ketone als Mittel zur Bestimmung der Constitution der fetten Säuren und der Alkohole" I can parse what the paper was about. Roughly this paper was about "The oxidization of ketones - a means for determining the constitution of the fatty acids and alcohols". Thus, his method involved the oxidation to the ester and hydrolysis of said ester in one step. I don't know what reagents Popov used, but I'm pretty sure it was not a peroxyacid (since this reagent will not hydrolyze the ester). I have previously answered a question about migratory aptitude in Baeyer-Villager reactions. Since I cannot read Popov's paper, nor can I find any authoritative resource on his rule (on- or offline), I have to assume that the known migratory aptitude for the Baeyer-Village reaction is the same as Popov's rule.


Is Blue Silica Gel more harmful than the Orange ones?

I had a poke around in an effort to determine the dye used in orange silica gel and was able to find two material safety datasheets (MSDSes) for orange silica gels which both gave different empirical formulae for the dye ($\ce{C25H30ClN3}$ and $\ce{C20H14O4}$), which suggests that there are multiple formulations of orange/yellow silica gel. Both of these dyes are organic molecules (as opposed to transition metal salts), and now their possible identities are given at the bottom of this response. Suffice to say that orange silica gels were designed specifically to avoid the use of cobalt chloride, which is fairly toxic.

As a document1 from the US National Parks Service indicates, the chief danger of cobalt chloride is its environmental toxicity if it enters soil or water through improper disposal. Cobalt salts are often rather evil, and this is actually reflected in the name, which is ostensibly an alteration of 'kobold'. However as long as the beads are kept in a suitable enclosure such as a filtered dessicator plug, they should not present a hazard. Inhalation of dust from the silica gel beads, direct handling of beads or cobalt chloride-impregnated materials or complete inundation of the dessicant (which could dissolve some of the cobalt chloride) are three possible ways the cobalt chloride could present a direct hazard, but the disposal of blue silica gel-containing products in landfill and subsequent leaching of cobalt chloride is probably the chief concern.

As far as the orange beads go, as I was not able to determine the specific indicator molecule(s) I can't ascertain their toxicity, suffice to say that they are by design less harmful than cobalt chloride and appear to be generally considered safe under normal use.

P.S. (finally, after chrome crashing multiple times)

F'x points out that $\ce{[C25H30N3]^{+}Cl^{-}}$ is crystal violet, which is not super toxic but for which there is evidence of carcinogenicity2 and which is considered an environmental pollutant.

$\ce{C20H14O4}$ may very well be phenolphthalein, which is considered to be a carcinogen by some authorities, most notably appearing on the California Prop. 65 listing3.

I guess the idea is that cobalt chloride is less desirable than either of these dyes. Don't eat the beads, and wash your hands!

(1) U.S. National Park Service; Cobalt Indicating Silica Gel Health and Safety Update; 2005, pp. 1-3

(2) Littlefield, N.A.; Blackwell, B.-N.; Hewitt, C.C.; Gaylor, D.w.; Chronic toxicity and carcinogenicity studies of gentian violet in mice; Fundamental and Applied Toxicology, 2004, pp. 902-912

(3) Chemicals Known to the state to cause cancer or reproductive toxicity; State of California EPA; March 16 2012, p. 17


How does the reaction Mg + O₂ → MgO occur?

First of all, do you know that chemical equations are means to express chemical reactions?

Second, this equation represents the reaction of magnesium with oxygen. Magnesium and Oxygen are the reactants and Magnesium Oxide is the product.

The arrow indicates the direction of the reaction.

To write basic chemical equations, you only need to know your reactants, your products and how to write the symbols for them. Usually, you can 'guess' what the products can be from generic patterns or mechanism, but I don't think you will be there yet (or required to remember them) soon if you just got started on them.

To get the molecular formula or ionic formula in this case, you need to know the valencies of the elements constituting that compound. For instance here, Magnesium has valency 2, Oxygen has valency 2, so $\ce{MgO}$ is the formula for Magnesium Oxide.

For example, Sodium Oxide, Sodium has valency 1, Oxygen has valency 2, so $\ce{Na_2O}$ is the formula for Sodium Oxide.

Iron (III) Oxide becomes: $\ce{Fe_2O_3}$ (Iron valency = 3; Oxygen valency = 2)

Magnesium Chloride becomes $\ce{MgCl_2}$ (Magnesium valency = 2; Chlorine valency = 1)

Does this answer your question or were you looking for something different?


What effect does hydrogen bonding have on H NMR spectra?

Hydrogen bonding is well known to cause large changes in chemical shift values– this is evident if you look up the chemical shift of an alcohol in a data-book (for an aliphatic alcohol this is anywhere between 0.5 and ca. 6.0 ppm depending on solvent/concentration), where a large range of values are given.[1]

In general,  formation of hydrogen bonds causes shifts to move to a higher frequency (higher ppm) due to deshielding.[2] In itself this is counter-intuitive, as in a hydrogen bonding interaction the H is actually receiving electron density from the donor atom (and therefore should be shielded). In the case of NMR however, the actual electron density isn’t whats causing this, but rather the field set up from the presence of the interaction.[#]

Intramolecular vs intermolecular hydrogen bonding

Although hydrogen bonding does cause a change in chemical shift, it is often only evident (/causes an issue) with intermolecular hydrogen bonding (between two molecules). With intramolecular bonding, the system is sufficiently stable that the chemical shift of the proton is consistent (and doesn't change with concentration/solvent, much).

To give an example,[##] consider the 1H-NMR of methanol and hexafluoroacetylacetone:[3]

On the top is the spectra of the neat compounds, where hydrogen bonding is present. On the bottom, the spectra of the compounds with added tetrachloromethane (which breaks down hydrogen bonding). Whilst the intramolecular hydrogen bonding in hexafluoroacetylacetone is unchanged, we can see immediately that the chemical shift of the hydroxyl proton to move.

Applications

Hydrogen bonding in NMR may seem rather inconvenient (and indeed it can be in routine structure elucidation work where the shift keeps on moving depending on how dry the NMR solvent is on any given day), however it has played a vital role in biochemistry and supramolecular chemistry, where the change in proton shifts can be used to determine whether or not multiple components are interacting.

A classic example is the interaction between cytosine and guanidine (C pairs with G in DNA via a hydrogen bonding interaction), which can be observed by NMR when the two compounds are mixed in an equimolar ratio:

[1]: Presch, E.; Buhlmann, P.; Badertscher, M. Structure Determination of Organic Compounds- Tables of Spectral Data; Springer: Berlin, 2009

[2]: Günther, H. NMR Spectroscopy- Basic Principles, Concepts, and Applications in Chemistry; Wiley: Weinheim, 2013

[#]: This is covered in Spin Dynamics by Levitt if you have any interest in the theoretical reasoning behind this

[##]: In this other answer, i’d previously given the example of salicaldehyde, which shows a consistent intramolecular H-bond free from solvent/concentration effects

[3]: Figure reproduced from Breitmaier, E. Structure Elucidation by NMR in Organic Chemistry; Wiley: Chichester, 2002


Can an elimination reaction be visualised as the combination of the loss of a leaving group and proton transfer?

Elimination refers to the ends rather than the means. On top of that, it does not necessarily refer to the loss of a proton + a leaving group; it can in general refer to the loss of any small molecule. So, for example, a loss of CO2 is also considered an elimination, even though there is no proton there.

As far as proton + leaving group eliminations go (elimination of HX), the order in which the leaving group leaves and the proton falls off determines what type of elimination it is.

The leaving group can leave first - that is called an E1 elimination.

The proton can fall off first - that is called an E1cb elimination.

And if both of them happen at the same time, it is an E2 elimination.


Help understanding this passage on optical activity

I believe what is causing you confusion is the difference between optical activity and absolute configuration.

A molecule with a single stereocenter such as limonene and carvone have two stereoisomers, one rotates polarized light to the left (-) and the other to the right (+). Additionally, each one has its own absolute configuration in their chiral center, which can be R or S. The thing is, R and S configurations are not attached or are, anyhow, synonyms with (+) and (-) configurations. Carvone and limonene are good examples:

In carvone, the S-enantiomer rotates polarized light to the left and so it is named S-(+)-carvone. The S-enantiomer in limonene rotates polarized light to the right and so it is named S-(-)-limonene. The two nomenclature systems are different since they specify distinct properties.

I hope this answers your question. If not, I would be glad to help in anyway I can.


Using solubility product to determine the mass of the precipitate of a reaction

This is awfully closed to a homework question. Since you already have an answer (real answer is actually $\pu{4.3 g}$, see following calculations) and still insisting on how to use the solubility product information, I'd like to give some insight. Since all components are in solutions, solubilities of $\ce{AgNO3}$, $\ce{NaCl}$, and $\ce{NaNO3}$ should not be in concern. Solubility of $\ce{AgCl}$ is the only one should be concern:

$$
\begin{array}{lccc}
\ce{&2 AgNO3(aq) & + NaCl(aq) &<=> &AgCl(s) &+ & NaNO3(aq)} \\
\text{Initial}, \pu{mol} & 0.3 \times 0.10 = 0.03 & 0.5 \times 0.10 = 0.05 && 0 && 0 \\
\text{Final}, \pu{mol} & 0 & 0.05 - 0.03 = 0.02 && 0.03 && 0.03 \\
\text{Final}, \pu{mol/L}& 0 & \frac{0.02}{0.2}= 0.1 && \pu{0.03 mol} && \frac{0.03}{0.2}= 0.15 \\
\end{array}
$$

Thus, amount of $\ce{AgCl}$ precipitated $= \pu{0.03 mol} \times \pu{143.3 g/mol} = \pu{4.3 g}$

Since $K_\mathrm{sp}^\ce{AgCl} = \pu{1.77 \times 10^{-10} mol^2L^-2}$ is such a small number, it is safe to say almost all of $\ce{AgCl}$ is in solid form. However, the solubility product come to play a role if there is a question about what is the final $\ce{[Ag+]}$ in solution. Now you have to consider following equilibrium:

$$\ce{AgCl(s) + H2O <=> Ag+ + Cl-}$$

Hence, $K_\mathrm{sp}^\ce{AgCl} = [\ce{Ag+}][\ce{Cl}] = \pu{1.77 \times 10^{-10} mol^2L^-2}  \tag1$

Suppose the solubility of $[\ce{Ag+}]$ is $\alpha \ \pu{ mol\:L-1}$. Then, $[\ce{Cl-}]$ is $\alpha \ \pu{+ 0.1 mol\:L-1}$ (common ion effect). Yet, $K_\mathrm{sp}^\ce{AgCl} = \pu{1.77 \times 10^{-10} mol^2L^-2} $ is such a small number, the $\alpha$ amount from the dissolution of $\ce{AgCl}$ can be considered negligible.

Thus, from equation $(1)$:

$$K_\mathrm{sp}^\ce{AgCl} = [\ce{Ag+}][\pu{0.1 mol L-1}] = \pu{1.77 \times 10^{-10} mol^2L^-2} \\ \Rightarrow [\ce{Ag+}] = \frac{\pu{1.77 \times 10^{-10} mol^2L^-2}}{\pu{0.1 mol L-1}} = \pu{1.77 \times 10^{-9} molL-1} $$


Decomposing chemical compounds - Ease of Decomposition

It really depends on what the definition of "decompose" is in this case. If you take the definition of "decompose" to be:

  ... to break down or cause to break down into component elements or simpler constituents

then the only option would be ammonia because the others are elements whereas ammonia is a compound which can be broken down into elemental Hydrogen and Nitrogen by the Haber Process. The energetics is not of concern in this case.


Why is 2-(3-aminomethyl-2-hydroxypropyl) cyclohexan-1-one named so and not 2-(2-hydroxy-4-aminobutyl) cyclohexan-1-one?

As drawn the compound does not contain a butyl side chain with four carbon atoms; the side chain is only a propyl chain with three carbon atoms.

The propyl side chain has two substituents: a hydroxyl ($\ce{-OH}$) group and an methylamino ($\ce{-NH-CH3}$) group. Since these groups are not the principal characteristic group in this case, they have to be expressed as prefixes. Note that the correct prefix for the secondary amine is “methylamino” and not “aminomethyl” as proposed in the question.

Therefore, the correct name is 2-[2-hydroxy-3-(methylamino)propyl]cyclohexan-1-one.


Why hydrogen bonds are stronger than Van der Waals forces?

Distance
In vacuum, the calculated (classical) interaction between to aligned dipoles decreases with the square of the distance. If the dipoles can't come very close to each other, the interaction will be weak. If you treat hydrogen bonds as a special case of dipole-dipole interaction, you will find that a N-H ... O=C interaction is much stronger than a C-H ... O=C interaction. For one, the N-H group has a larger dipole moment than the C-H group. For another, the distance between the donor and acceptor are larger for the weak C-H hydrogen bond.
When the dipoles of two ethanol molecules approach, they can come fairly close because the hydrogen on the hydroxyl group lacks inner electrons. For ethanal, in contrast, the carbonyl carbon of one molecule has to come close to the oxygen atom of the other. Both atoms have filled inner shells (and more electrons in the outer shell), so that approach is limited by electronic repulsion.
Covalent character
The hydrogen bond has covalent character, explaining why it is a directional interaction. This covalent character (like all covalent bonds) can not be fully explained using classical (Coulomb interaction) arguments. So in a hydrogen bond, we have an additional attractive term that is lacking in pure dipole-dipole interactions. In extreme cases, you might think of the hydrogen bond as a 4 electron-3 center bond (the 4 electrons are the lone pair of the acceptor and the bonding electrons of the covalent bond of hydrogen with the rest of the molecule; the three atoms are the hydrogen, the electronegative atom it is covalently bound to, and the acceptor atom).
Nomenclature
Some use Van der Waals forces as an umbrella term for (permanent) dipole-dipole, dipole-induced dipole, and temporary dipole-induced dipole interactions. Others use Van der Waals forces (or dispersion forces or London interactions) as a separate and weaker kind of interaction and contrast it with the stronger dipole-dipole interaction.


Sodium and potassium produce a strange liquid, why the liquid explode when it gets contact with water?

This site provides an explanation about the mechanism of the explosion.

...causes an explosion because the metal dissolves, generating an
extreme amount of heat and transferring electrons to the water. The
dissolution step also generates steam and forms hydroxide ions and
hydrogen, which can be ignited, making the process even more
energetic.
Some researchers have puzzled over how the process can occur so
quickly, though. They recognized that the steam and hydrogen generated
early on in the reaction should form a buffer layer over the metal
surface and impede water from continuing to react.
...within a fraction of a millisecond of making contact with water,
the Na/K droplets form numerous spikes that protrude into the water.
Molecular dynamics analysis indicated that nearly instantaneous
transfer of electrons from the spikes to the water rapidly generates
positively charged alkali ions, which vigorously repel and cause a
so-called Coulomb explosion. It is the speedy manner in which that
process propagates and generates reactive metal surfaces that triggers
the overall explosion.
The researchers have figured out many of the key aspects that enable
this highly exothermic reaction to become explosive, rather than
self-quench.

It does not generally reacts with any other liquids. Hence they are stored in mineral oil.

It is highly reactive with water and is stored usually under hexane or
other hydrocarbons, or under an inert gas (usually dry nitrogen or
argon) if high purity and low levels of oxidation are
required.(Wikipedia)


Is there an electronic structure method which accounts for 3-, 4-, ..., n-electron exchange integrals?

I am posting this answer not as the definitively correct answer, but as an example of the type of thing I am talking about.

azago's answer is correct in pointing out that the Slater determinant contains within it the correct permutational symmetry of the wavefunction required by electrons. Namely, it is antisymmetric upon interchange of two orbitals. Additionally, it symmetric upon interchange of three (or any odd number of) orbitals. This is the symmetry which must be present in higher cycles of exchanges.

What I have found upon further reading is that I could have been more precise in the wording of my question. Namely, these higher-order exchange interactions are not surprising if I had made it clear that these are necessarily three-center, four-center,... integrals. Then it is more clear that HF, while having the correct symmetry to adhere to Pauli exclusion, does not include these integrals as it only deals with 2-electron integrals at the most.

So, I will discuss three contexts I have found in which the types of integrals I am thinking of show up.

Hamiltonian for a Three-Atom System:

Following along with ref. [1], the Hamiltonian for the three atom system, $\ce{H3}$ is,
$$
H=\sum_i\left(-\frac{1}{2}\nabla^2_i-\sum_{\alpha}\frac{1}{r_{\alpha i}}+\sum_{j\gt i}\frac{1}{r_{ij}} \right)+\sum_k\frac{1}{R_k}
$$
where $i$ and $j$ are labels for the three electrons, and $\alpha$ is the label for each Hydrogen atom with locations $R_1$, $R_2$, and $R_3$.

It is clear from this Hamiltonian that we have the usual coulomb terms and the usual two-electron integrals. Something weird is going to happen with the exchange terms though. There are actually three possible configurations of the atoms, but the only point I'm making here is that the matrix elements for one of the atomic configurations can be written as,
$$
\langle\psi_1|H|\psi_1\rangle=4J+4K_1-2K_2-2K_3-4K_{123}
$$
I have deviated from their notation here for consistency with current convention. These integrals are written explicitly as:
$$
\begin{align}
J&=\langle abc|H|abc\rangle \\
K_1&=\langle abc|H|bac\rangle \\
K_2&=\langle abc|H|acb\rangle \\
K_3&=\langle abc|H|cba\rangle \\
K_{123}&=\langle abc|H|cab\rangle \\
\end{align}
$$
where now it is quite clear that $K_{123}$ is a double exchange. That is, two electron labels are changed, and it is clear this is distinct from the other exchange integrals which are the usual exchanges we are used to.

I do not think this has anything to do with the fact there are three atoms present, however. Only that there are three electrons. Thus, I believe that even a complete description of Lithium would include these three-electron exchanges. Thus, the reason we do not see these integrals in HF is that they are left out of the Hamiltonian.

Explicitly-Correlated Methods:

I'll be brief here because explicitly-correlated methods are pretty complicated and I haven't spent enough time studying them to really know what's going on, but one thing which does happen is that you get matrix elements of the following form[2]:
$$
Z_{\nu w,p}^{xy,q}=\langle w_{\nu w,p}|(1-\hat{p}_{13})r_{13}^{-1}+(1-\hat{p}_{23})r_{23}^{-1}|w_{xy,q}\rangle
$$
The only important in this horrible integral is that the exchange operator $\hat{p}_{ij}$ appears with the labels of three-distinct electrons. Now, I have no idea if it's valid to interpret this as the type of integral I'm talking about, but these kinds of integrals appear all over the place in explicitly correlated methods.

Three-Body Symmetry-Adapted Perturbation Theory

Now, two-body symmetry-adapted perturbation theory (SAPT) is quite a common method, but there is a 3-body version of SAPT which can be quite a useful tool at times.

In Ref. [3], the authors present the relevant equations for 3-body SAPT, but they note that they approximate the exchange term out to the fourth power of orbital overlap. That is, for the exchange operator they do the following:
$$
P\approx 1+P_2+P_3+P_4
$$
They describe this as follows:

  In this expression, $P_2$ represents the operator which interchanges
  one electron from $X$ with one electron from $Y$ and
  therefore produces terms proportional to $S^2$. The operator
  $P_3$ includes double exchanges between three monomers involving
  three electrons. This operator, absent for a dimer,
  generates terms proportional to $S^3$. The operator $P_4$ is a
  double exchange operator which interchanges four electrons
  among the monomers and generates terms proportional to
  $S^4$. The double exchanges produced by $P_4$ are of the
  $(XY)(XY)$ or $(XY)(XZ)$ types and can be thought of as two
  exchanges of the type produced by $P_2$ occurring simultaneously.
  In contrast, double exchanges of the $(XY)(XYZ)$
  type are included in $P_5$ and double exchanges of the
  $(XYZ)(XYZ)$ type are included in $P_6$, leading to terms proportional
  to $S^5$ and $S^6$, respectively.

The whole purpose of three-body SAPT is to be able to describe the 3-body non-additivities which arise from intramolecular interactions. It has already been pointed out that the electrostatic interaction is additive up to 2-body because Coulomb's law follows the superposition principle. This then makes it clear that these higher-order exchanges involving more than two electrons generate some of the non-additive forms of coulomb correlation.

Finally, the authors of ref. [3] provide the form of these three-electron cycles I was trying to describe which can be written as:
$$
P_{111}=\sum{i\in A}\sum{i\in B}\sum{i\in C}[P_{ij}P_{jk}+P_{ik}P_{jk}]
$$
which interchanges electrons in a cycle on atoms $A$, $B$, and $C$. This operator generates all possible three-electron exchanges between three atoms. There are no intra-monomer exchanges however.

[1] Porter, R. N., & Karplus, M. (1964). Potential energy surface for H3. The Journal of Chemical Physics, 40(4), 1105-1115.

[2] Hättig, C., Klopper, W., Köhn, A., & Tew, D. P. (2011). Explicitly correlated electrons in molecules. Chemical reviews, 112(1), 4-74.

[3] Lotrich, V. F., & Szalewicz, K. (1997). Symmetry-adapted perturbation theory of three-body nonadditivity of intermolecular interaction energy. The Journal of chemical physics, 106(23), 9668-9687.


What compounds are produced when you heat CuO and sugar?

If there is enough copper oxide, all you obtain is metallic copper, $\ce{CO2}$ and water. This mixture ($\ce{CuO}$ + organic substance) was the standard way of analyzing any new organic substance in the $19$th century. The formula of the compound was derived from the weight of $\ce{CO2}$ and $\ce{H2O}$ collected. This is is still a standard question to be solved in high school chemistry courses all over the world.


Use of concentrated sulfuric acid in Fischer esterification

In fact, sulfuric acid is not a must. Many other H-acids, as well as Lewis acids and their combinations would be sufficient.

The trick here is not to get concentrated acid, but to reduce amount of water in the system. Consider the simpliest transformation, which is typically performed with a large excess of methanol and trace amounts of acid:

$$\ce{R-C(O)OH + CH3OH_{(excess)} <=>>[HCl] R-C(O)OCH3 + H2O}$$

This is an equilibrium reaction. The catalyst allows us significantly faster achieve the equilibrium state, but it does not shift it. Catalyst is equally promoting both direct and reverse reactions. A large excess of alcohol, in accordance with the law of mass action, shifts the equilibrium toward the formation of the ether and thus increases the degree of conversion of the acid into ether.

But if water is not eliminated from the system, or is initially present, hydrolysis occurs:

$$\ce{R-C(O)OCH3 + H2O_{(excess)} <=>>[HCl] R-C(O)OH + CH3OH }$$

Hydrolysis of ether will happen even faster, if a strong base is present, as in this case reaction equilibrium shifts further to the right due to the salt formation.

The bottom line is, in order for the reaction to go to the end, you either add an excess of acid (or alcohol), or remove water as it is produced. Starting with concentrated or water-free acid you are not only simplifying your life, but you are also getting better yield.


Do core electrons have molecular orbitals?

When valence atomic orbitals start to interact with orbitals on another atom or molecule (e.g. bond formation), the valence electrons respond by forming molecular orbitals with the electrons on the other atom or molecule.  The core electrons also respond to this bonding interaction.  Their energy levels and spatial distributions are perturbed.  The more deeply buried the core electrons, the smaller the perturbation - but the core electrons will also be affected by the bonding interaction.  If one were able to write out the exact wave function to describe the new molecule that was formed, the wave function would include the core electrons.  Their importance would be much less than the valence electrons, but they would be involved in describing the total bonding situation.


Surface tension, not hydrogen bonding, is the primary phenomenon which causes cereal rings such as Cheerios to adhere together while floating in milk.

The total linear perimeter of the 3-phase cereal-air-milk interface is minimized when pieces cling together.

Here are some links:

http://www.livescience.com/9350-cereal-science-floating-objects-stick.html
http://phys.org/news/2010-09-well-known-effect-breakfast-cereal-physicists.html
https://en.wikipedia.org/wiki/Cheerios_effect


Why is the attached hydrogen of 1,3-diethyl-5-(1-phenoxyethyl)benzene homotopic?

Which hydrogen is homotopic to which is rather hard to see in this compound without actually sitting down and drawing the relevant structures. Before I do this, I am going to get a few obvious points out of the way:

The compound is chiral (see the $\ce{PhO}$ group on the third side chain). This not only greatly reduces the number of enantiotopic hydrogens but also the symmetry of the molecule.
Further analysis shows that there is, in fact, no element of symmetry, i.e. the molecule’s point group is $C_1$. This means that we cannot resort to any ‘cheap hacks’ to determine homotopic hydrogens. We have to use the old-fashioned methods.
Homotopic hydrogens are those that can be transformed onto each other by rotations only (of the entire molecule or rotatable bonds) in such a way that the final depiction is identical to the initial depiction.

We have a few rotatable bonds at our disposal; I will be using the $\ce{Ph-CHXMe}$ bonds, where $\ce{X} = \ce{H, OPh}$ — i.e. I will be rotating around each carbon-carbon bond that connects a residue to the central aromatic ring. Following are the two rotation steps necessary to transform the arrow-hydrogen onto the circled hydrogen:

Scheme 1: The required rotations. The rotated bonds are shown by curly arrows; non-moving parts of the molecule are grey. The circled hydrogen (red) and the other hydrogen on the same carbon (blue) have been highlighted.

I chose to rotate both ethyl groups first and to rotate the entire benzene ring second. However, the order is not important. As we can see, the first rotation transforms the circled hydrogen behind the plane. (I need to rotate the other ethyl group, too, so that the final rotation will give me the initial product again.) The second rotation of the entire benzene ring with the two ethyl groups again puts the circled hydrogen in front of the plane.

Thus, your analysis was correct; the two hydrogens in front of the paper plane are homotopic as are the two behind it. The two hydrogens on a single carbon (red and blue in the picture) are diastereotopic as they cannot be transformed onto each other preserving the symmetry of the molecule.


Where can I find the infrared band positions for phosphorus compounds?

A good resource for IR and Raman bands is "Infrared and Raman Characteristic Group Frequencies: Tables and Charts" by George Socrates. There you can find a chart containing the positions and intensities of bands observed in the infrared spectra of phosphorus compounds (explanation: "w", "m", "s"  specify the observed intensity of the bands means whereby "w" stands for "weak", "m" for "medium", and "s" for "strong"; additionally the prefix "v" can be given meaning "very", e.g. "vs" means "very strong"):

More specifically you find the following numbers for the different bands you are looking for:

There is also another comment on $\ce{P-Cl}$ bands:

  Compounds containing the $\ce{P-Cl}$ bond have a medium-to-strong absorption
  at $605 - 435$ $\mathrm{cm}^{-1}$ due to the $\ce{P-Cl}$ stretching vibration.
  The position of the band due to the $\ce{P-X}$ ($\ce{X}$=$\ce{F}$ or $\ce{Cl}$) stretching vibration is affected by the oxidation state of the phosphorus atom. 'In the presence of
  more than one halogen atom directly attached to the phosphorus atom, two
  peaks are observed due to the asymmetric and symmetric stretching vibrations
  respectively. Difluorides of the type $\ce{-P(O)F2}$ absorb at $930-895$ $\mathrm{cm}^{-1}$ and $890-870$ $\mathrm{cm}^{-1}$


Proton affinities in machine readable format?

To my knowledge, a (Python) API for the NIST chemistry WebBook does not exist.

Rather than filling out the Search for Species Data by CAS Registry Number by hand, you might want to use mechanize.

But since you're only interested in Gas phase ion energetics data anyway, you might want to construct the relevant URLs directly.

Isoprene:

http://webbook.nist.gov/cgi/cbook.cgi?ID=C78795&Units=SI&Mask=20#Ion-Energetics

Styrene:

http://webbook.nist.gov/cgi/cbook.cgi?ID=C100425&Units=SI&Mask=20#Ion-Energetics

The parameter pattern is obvious: ID is the CAS number of the compound minus the dashes, prefixed by C.


What is the molecularity of the RDS of a zero order complex reaction?

Terminology
There are two term which you must understand and use carefully.

Molecularity - The [integer] number of reactant molecular entities that are involved in the 'microscopic chemical event' constituting an elementary reaction. (From IUPAC Gold Book)
Order of reaction - If the macroscopic (observed, empirical or phenomenological) rate of reaction (v) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $\ce{k = [A]^\alpha[B]^\beta\text{ ...}}$ (expressing in full the dependence of the rate of reaction on the concentrations A, B, ...) where $\alpha$, $\beta$ are constant exponents (independent of concentration and time) and k is independent of A and B etc. (rate constant, rate coefficient), then the reaction is said to be of order $\alpha$ with respect to A, of order $\beta$ with respect to B, ... , and of (total or overall) order $n = \alpha +\beta + ...$ The exponents $\alpha$, $\beta$, ... can be positive or negative integral or rational nonintegral numbers.

Equilibrium reaction

For a reaction like the decomposition of ammonia you could write a bimolecular reaction like:

$$\ce{2NH3 -> N2 + 3H2}\tag{1}$$

Often chemists consider such a reaction reaching equilibrium such as:

$$\ce{2NH3 <=> N2 + 3H2}\tag{2}$$

where there is a forward reaction which (naively) could be consider as the forward reaction

$$\ce{2NH3 ->[$k_\mathrm{f}$] N2 + 3H2}\tag{2.a}$$

with the forward rate given by $r_\mathrm{f} = k_\mathrm{f}[\ce{NH3}]^2$ and the reverse reaction

$$\ce{N2 + 3H2 ->[$k_\mathrm{r}$] 2NH3}\tag{2.b}$$

with the reverse rate given by $r_\mathrm{r} = k_\mathrm{r}\ce{[N2][H2]}^3$.

However, if you consider the reverse reaction, you can see that there are four molecules involved. It is highly unlikely that four molecules would ever collide in just the right way to cause a reaction. So the reverse reaction occurs in multiple steps with intermediate species involved. The exponents $x$ and $y$ would be determined experimentally and would almost certainly not be integer values. Thus more realistically:

$$r_\mathrm{r} = k_\mathrm{r}[\ce{N2}]^x[\ce{H2}]^y\tag{3}$$

Catalytic reaction

Now, back to the more salient point of the question. The decomposition of ammonia is typically not done just by heating ammonia to some high temperature to form an equilibrium, but rather it is done selectively with a catalyst and a lower temperature.

$$\ce{2NH3 ->[catalyst] N2 + 3H2}\tag{4}$$

The catalytic surface absorbs ammonia preferentially (and very quickly compared to the reaction time of the ammonia decomposition reaction) and can only absorb a small amount of ammonia. Hence the rate of decomposition doesn't depend on the concentration of ammonia so the rate of decomposition is essentially constant until almost all of the ammonia is consumed. So with a catalyst there is a zero order kinetic reaction with the equation:

$$r_\mathrm{f} = k_\mathrm{f}\tag{5}$$


To simplify slightly both Raman and IR spectroscopy show the vibrational modes of a molecule (though the techniques used to reveal these are very different).

IR spectroscopy relies on coupling between the electromagnetic field of light passing through a sample and the electric dipole of the molecule. But that absorption is only possible is the vibration in the molecule changes the molecules dipole. Consider, for example, carbon dioxide, a linear molecule with no net dipole moment. Symmetric stretches (ie both oxygens moving the same amount in opposite directions) don't change the dipole so can't be detected by IR. But asymmetric stretches or bending will change the dipole and can be detected by IR.

Raman spectroscopy (simplifying a little) relies on detecting changes to the molecules polarisability not the dipole. So the symmetric vibrations of carbon dioxide are detectable because they do change the polarisability of the molecule.

In both cases the symmetry tables tell you whether the symmetry of the vibrational mode changes the electric dipole or the polarisability of the molecule. In simple molecules physical intuition can usually tell the same thing but symmetry tables are more reliable.

A full understanding would involve a quantum mechanical description of the interaction of EM radiation with the possible vibrational modes of a molecule considering the allowed and forbidden transitions. But the physical intuition above captures the essentials without getting too mathematical.


in which way calcium promotes cell division

see its simple all you need to know is that calcium is responsible for the formation of middle lamellum which is a key structure in the cell wall formation if only karyokinesis occurs and cytokinesis fails to happen then the condition of resultant cell would become multinucleated which most plants do not have


What is the balanced equation of BH3/THF with an amide to form an amine?

The reduction of secondary amides with BH3 requires 3 equivalents, mechanism shown below image sourced from here. The borane becomes HB=O which may transform further depending on work up.


How can normalities be interchangeably used?

The fundamental idea behind gram equivalents and normality is that one gram equivalent of an acid reacts with one gram equivalent of a base. This idea eliminates the need of using molarity and thinking about mole ratios for any acid base titration.

Equal volume 1 N HCl will react with 1 N NaOH. Similarly, 1 liter of 1 N $\ce{H2SO4}$  or 1 N $\ce{H3PO4}$ will consume 1 L of 1 N NaOH.

The key calculations is that you should know how to calculate normality from a given molecular weight and a balanced equation:

For example:
$\ce{HCl}$ -> $\ce{H+}$ + $\ce{Cl-}$
gram equivalent weight= formula weight/ (No. of acidic protons) = 36/1 =36

$\ce{H2SO4}$ -> $\ce{2H+}$ + $\ce{SO4^2-}$
gram equivalent weight= formula weight/ (No. of acidic protons) = 98/2 = 49

Can you now make the connection that why "The part 1.80 N HCl can also be written as 0.9 N H2SO4" is quite wrong? because

1 N HCl = 1 N $\ce{H2SO4}$

and both solutions contain 1 gram equivalents of protons.

Yes if your teacher were comparing normality and molarity, then it is different.

Molarity calculations are different and far more easier. By definition, M of X= moles of A/ Total volume in which X is present.

A 1 M HCl just means there is one mole HCl (gas) dissolved in 1 L of solution
Similarly,
1 M $\ce{H2SO4}$ is 1 mole $\ce{H2SO4}$ dissolved in 1 L of solution.

If you were to titrate  1 L of 1 M NaOH with 1 M HCl you would require 1 L of HCl. However, since  sulfuric acid furnishes two moles of proton $\ce{H+}$ for each mole of sulfuric acid, only 500 mL would be required.

In short, 1 M HCl is equivalent to 0.5 M $\ce{H2SO4}$ in terms of titer for $\ce{OH-}$ ion.

Contrast this molarity unit with normality which is almost obsolete except in some South Asian colleges.


Why has lithium bromide a slight solubility in organic solvents?

Lithium bromide is a salt of small cation and large anion. They cannot interact effectively, so the crystal lattice is quite easy to break. On the other hand, lithium cation is small, so it can interact effectively with small negatively charged atoms, like oxygen. Indeed, lithium cation has highest hydration energy of all alkaly metal cations. Altogether, it means, that in solvents with oxygen atoms (alcohol, esters, acetone) lithium cation effectively bounds to solvent molecules, leaving crystall lattice, and bromide has to follow, resulting in some solubility of lithium bromide in solvents with negatively charged oxigen and to lesser extent, nitrogen (for example pyridine). Lithium bromide is still insoluble in organic solvents that cannot interact with lithium cation, like benzene.


Face-centered cubic (FCC, or closest-packed cubic, CCP) and hexagonal close-packed spheres have the greatest (and equal) density of packed spheres, occupying 74.048% of the actual volume. Ref 1
However, because of the different interlaying of the hexagonal, tight packing in two dimensions, the repeating unit, or unit cell, is described differently for each crystal type. While each layer in a FCC crystal is identical to every other layer, and to each layer in a HCP crystal, the different layering makes it possible to distinguish between them, and to identify repeating units. The smallest repeating unit in a FCC crystal contains 4 atoms and can be described with one size parameter. The smallest repeating unit in a HCP crystal contains 6 atoms. Although its density is the same as that of a FCC crystal, it needs 6 atoms to describe its layering. Interestingly, body-centered cubic crystals can be described with only 2 atoms per unit cell, but have a density of only 68% (they are not close-packed). Ref 2
The construction of FCC and HCP crystals is identical for two layers, but the third is differently placed. HCP has only two positions, a and b, but FCC places a third layer c so that the atoms do not lie directly over any in the first or second layer. HCP places layers as a,b,a,b,a,b...; FCC does it this way: a,b,c,a,b,c,a,b,c...

FCC and HCP unit cells are drawn to show aspects of their packing, so interatomic distances (lattice parameters) specified are between differently placed atoms. Comparing heights isn't really comparing repeating units. If you measure the distance between two adjacent atoms in FCC, you will find the same interatomic distance in HCP. However, comparing the distance between more remote atoms can give you different results. It's a bit like comparing your vertical height at age 6 months with your horizontal dimension now. They're similar, but different because you are looking with a different viewpoint.
It's fascinating to consider that crystal structures are energetically distinguishable not just by next-door atoms, but by atoms down the street by one house. Or maybe more!
Ref 1. https://en.wikipedia.org/wiki/Close-packing_of_equal_spheres
Ref 2. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Solids/Crystal_Lattice/Closest_Pack_Structures


How does alum improve the settling of particles to produce 'clean' water?

OK, so after a bit of researching this is what I've found.

First of all,why does the dust not settle in the water in the first place? Well, dust forms a kind of a suspension.

The particles are all similar and ionized, so therefore they have a similar charge (positive, I think) causing them  repel each other and stay separate.
And because they're so small, the gravitational force acting on them isn't very strong either.
Also, the dust particles, previously dispersed in the air, have a lot of kinetic energy already.

All these 3 reasons coupled cause the particles to remain suspended for so long.

So what does adding alum do? Alum has a negative charge and tends to disperse in water very fast and very well (Why so? Because it's a colloid?).

This causes it to join up with all of the offending particles and neutralize them. Now that the particles don't have any repelling charges, they tend to clump together into 'flocs'.

The increased size as well as the lack of repelling charges cause the alum particles to settle down at the bottom or rise up and float in the water.

Why do the particles clump together and settle? Apparently even this isn't as simple as it seems.

After the particles are neutralized, they clump together because of something called the London Dispersion Force which are part of the Van der Waals forces. Basically, weak intermolecular force arising from quantum-induced instantaneous polarization multipoles in molecules causes even nonpolar particles to attract each other due to the correlated movements of the electrons in interacting molecules. More info can be found inside.

As to why they then settle, I think it can be sufficiently explained by gravity.

Oh well, that's all. All for explaining something even the ancients knew.


Precautions and tips for Qualitative Inorganic Analysis of Coins

1) The coins must be dissolved (at least partially). There is a good amount of metals that won't dissolve in common  acids (hydrochloric acids), some will require nitric acid, some will require aqua regia and some will require even worse, though that's unlikely. So, most likely you will have to deal with concentrated nitric acid, that produce nitric oxide (fox's tail) when react with metals. Given that you must have fume hood. In case something undissolved remains, you can try aqua regia, and if even that fail, you have something rare, like nioubium, with no common method of wet analisys.

2) Most metals used in coins have toxic salts. So, be cautios, and avoid skin contact. Latex gloves and protective goggles are not vital, but certainly recommended unless you believe you can be accurate enough. Usual safety requirements for work with conc. acids applies as well.

3) In case of non-destructive analisys, X-ray fluorescent spectrometry may be employed. It will give accurate analysis in almost any realistic case, but requires quite costly equipment. It is not extremely costly per analysis, though.


Instead of carrying out distillation under reduced pressure at the end of the procedure to remove the ether, could I have just carried out simple distillation?

You could have performed any procedure to remove the solvent:

Simple distillation  - This would be slower than what you did and could overheat the undistilled aniline causing decomposition.
Vacuum distillation - This is what you did, perhaps even on a special apparatus called a rotary evaporator. This method is still a distillation, even if your product was not distilled. Something was distilled.
Evaporation on a hot plate (or steam bath)-

  Why is steam distillation suitable in part 3? Could I have used vacuum distillation here?

Steam distillation is suitable because the aniline is already in an aqueous suspension. If do vacuum distillation, you would need to

Separate the aniline from the aqueous mixture (extract with ether)
Remove the ether (simple distillation or reduced pressure distillation)
Vacuum distil the aniline

If you look at what you did, it is the same steps in a different order:

Distil the aniline
Separate the aniline from the aqueous mixture (extract with ether)
Remove the ether (simple distillation or reduced pressure
distillation)

Either is probably suitable. In the end, the steam distillation is easier and less hazardous than the full-blown vacuum distillation you would need.


Disparity between LeChatelier's Principle and Solutions

You misinterpret the principle. It does not say if you add more salt, more salt dissolves. More generally, the principle focuses on changes of intensive properties, that do not change with the system scaling.
The principle says if the thermodynamic activity of the solid had increased by adding more salt, more salt would have dissolved to decrease this activity. But by adding more salt, the salt activity remains constant.
You can also consider  chemical potential   $\mu = \frac{dG}{dn}$  instead of activity, as they are closely related.
By other words, thinking more salt dissolves by adding more salt to saturated solution is equally wrong as thinking more salt crystalizes by adding more of saturated solution. As the salt activity in saturated solution or in solid state does not depend on their amount, which is extensive property.


Ammonia and hydroxide ion - Lewis basicity

With respect to Brønsted basicity, definitely hydroxide is the stronger base. With respect to Lewis basicity, I think the same scenario holds. As you said, hydroxide has three lone pairs, whereas ammonia has only one lone pair. As far as the energetics of these situations are concerned, the magnitude of lone pair to lone pair repulsion is far greater than that of lone pair to bond repulsion. Thus, it would make sense for hydroxide to grab the proton instead of ammonia.


The equation is rooted in experimental observations. It is not derivable from any physical formalism, and it is not the best way to describe many reactions. Similar equations can be derived from Transition State Theory. It is important to note that the similarities are limited to the functional form.


Why can the acidic strength between H2S and HCl be explained on the basis of electronegativity but H2S and H2O can't?

You are right that electronegativity is an important consideration but another factor that is also very important is size of the atom. Although there is a decrease in electronegativity from $\ce {O}$ to $\ce {S}$, the size of the atom which the charge is mainly centred on is now larger. This allows the charge to be distributed over a larger volume of space, decreasing inter-electronic repulsions, providing greater stabilisation of the negative charge.

However, the greater size decreases charge density, decreasing the strength of ion-dipole interactions between the anion and the surrounding water molecules. Thus, the extent of solvation decreases for the larger anion.

Electronegativity is defined as the ability of the atom to attract a bonding pair of electrons. It merely serves as a proxy for the ability of the atom to stabilise negative charge. It should not be viewed as the most important factor that influences the atom's ability to stabilise negative charge.

Also, mentioned in your answer, the strength of the bond formed with the larger atom with more diffuse orbitals decreases. A bond that is more easily cleaved would facilitate bond cleavage and thus, favour the acid with the larger atom bonded to $\ce {H}$ as the stronger acid.

Consolidating the above discussion, there are really two factors on each side of the table below. Perhaps, the distribution of charge increases very significantly and the bond energy also decreases significantly while the decrease in extent of solvation and the electronegativity exhibit a relatively less significant effect. It would be great if someone is able to quantitatively show this...

\begin{array}{|c|c|c|c|} \hline
\text{Favouring acids with larger atom} & \text{Favouring acids with smaller atom} \\ \hline
\ce{Distribution of charge} & \ce{Extent of solvation}\\ \hline
\ce{Bond strength} & \ce{Electronegativity}\\ \hline
\end{array}


Electrolysis not running, maybe dependant on distance of electrodes?

Solution resistance in a two electrode cell can make a big difference as it is totally uncompensated. You may be applying 15 V between the electrodes, but the redox reaction only occurs at a very close proximity to the surface (<10 nm), so it's the interfacial potential we're concerned with, which can be significantly different than the applied voltage if the solution is highly resistive. A tight constriction in the cell can certainly increase the resistance and you should make sure you're using an electrolyte as plain water is very resistive.


problem with loops over basis sets in psi4 using python

Here is an input file that will reproduce this error:

molecule {
    b      0.0000        0.0000        0.0000
    h      0.0000        0.0000        1.0000
}

basis_sets = ["6-31G", "6-31G*", "6-31G**"]
for basis_set in basis_sets:
    set basis = $basis_set
    energy('scf')

The problem is that Psi4 sees there are files remaining in scratch that it can use, starting with the spin orbital-basis overlap integrals, but because the basis set has changed, the file is the wrong size. You'll need to clean these up after the energy call by adding clean() within the loop.


Relationship between magnitudes of forward and reverse kinetic rate constants

A chemical reaction may be a very complicated thing, involving many different paths and elementary reactions. Basically it always leads to the formation of an equilibrium, where the lowest possible energy will be the most favoured. Under certain conditions more than one state will be populated and hence there will be enough potential energy to interconvert molecules in certain states. We usually refer to that as an dynamic equilibrium.

Changes under equilibrium conditions will most likely be elementary reactions (or an infinitely complicated chain of these.)

The population of the various states seems to be static at equilibrium, since the overall change of enthalpy is zero. In a dynamic picture that only means, that forward and backward reaction cancel each other. In that sense they are the same reaction in different directions. Therefore they can be described as one trajectory on the potential energy surface. (If there is a chemical reaction involving many different elementary reactions and the equilibrium still occurs, then the mathematical description will be much more complicated as all elementary reactions will be correlated. However, the main conclusions should stay the same.)

\begin{aligned}
\Delta G_r &= \Delta G_r^\circ + \mathcal{R}T\cdot \ln K = 0\\
\Delta G_r^\circ &= − \mathcal{R}T \cdot \ln K
\end{aligned}

The equilibrium constant may be derived from the mass-action law and should be defined as a product of activities (for the standard state). For $\ce{A + B <=> C}$ this will lead to
$$K^\circ= \frac{a(\ce{C})}{a(\ce{A})\cdot a(\ce{B})}$$

The activities are proportional to concentrations (in first approximation) and are unit less ($c^\circ$ being the standard concentration $1 \:\mathrm{mol/L}$)

$$a=\gamma\frac{c}{c^{\circ}}$$

For reasonable dilutions ($c\to0\:\mathrm{mol/L}$) one can assume that the activity coefficient becomes one $\gamma\approx1$ and therefore rewrite the equilibrium constant with concentrations.
$$K^\circ= \frac{c(\ce{C})/c^\circ}{c(\ce{A})\cdot c(\ce{B})/(c^\circ)^2}$$

At equilibrium the forward and backward reaction are coupled and therefore have to reflect the equilibrium constant.
$$K^\circ= \frac{k_f}{k_b} = \frac{c_{\text{eq}}(\ce{C})/c^\circ}{c_{\text{eq}}(\ce{A})\cdot c_{\text{eq}}(\ce{B})/(c^\circ)^2}$$

As one can see, the constants also need to have the same unit. As they are at equilibrium, they are the same reaction with different populated states.

If you are moving away from the equilibrium, thing will start to get a little bit more messy. On reaction will be faster than the other and rates will not be comparable any more.


Why are alkali metal compounds like sodium hydride or sodium amide strong bases but weak nucleophiles?

You're right they're difficult to use in organic reactions because of the solubility problem you mention. However, by using crown ethers like dibenzo-18-crown-6 they can be captured and the crown ethers are subsequently dissolved so that the bases are "brought into solution" and react.


Is there a general method for working out electron configurations?

There is a semi-empirical method to fill the atomic orbitals: atomic orbitals are filled  in the order of increasing n+l. If the sum n+l is equal for two atomic orbitals, they are filled in order of increasing n.
This gives the following order for filling the orbitals:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s, 5g, 6f, 7d, 8p, and 9s


The standardized definition of the change in internal energy $U$ for a closed thermodynamic system is

$$\Delta U = Q + W$$

where $Q$ is amount of heat transferred to the system and $W$ is work done on the system (provided that no chemical reactions occur).

Therefore, heat transferred from the surroundings to the system is assigned a positive sign in the equation, whereas heat transferred from the system to the surroundings is assigned a negative sign.

Similarly, work done by the surroundings on the system is assigned a positive sign in the equation, whereas work done by the system on the surroundings is assigned a negative sign.

(Note that some books on physics or engineering still use a non-standard definition when discussing the work done by a thermodynamic heat engine.)


Relative abundance of the isotopes of iridium

An urn contains nothing but

63 white balls and
37 black balls.

You grab into the urn, pick one item, look at it and put it back.

What is the probability that you drew:

a ball
a white ball
a black ball
a red ball?


Looking for a non-toxic, environmentally safe highly colored or intense dye

Many organic dyes which are fluorescent can be carcinogenic like the famous Rhodamine dyes. You can try a couple of tricks. One of best inks, at least for demonstrations, is the yellow highlighter's ink which fluoresces bright green in water (see my photo below). If you can shine a bright white LED or a bright blue LED, it will fluoresce very strongly. You can use an optical filter in front of the camera that can block blue light and let the green and longer wavelengths in. With that you can increase the sensitivity.
Note that fluorescence intensity is directly proportional to the incident light intensity in dilute solutions.


Fractional distillation of crude oil and resulting fuel ratios

Crude oil is a complicated mixture consisting of a variety of different hydrocarbons and the particular composition is very variable depending on where it is found.

Distillation is merely a means of separating those components so it can do nothing to create more of one of them. And gasoline is, essentially, just a mixture of those components with similar properties and boiling points. There is little you can do by distillation to create more gasoline.

However refiners do have access to chemical means to alter the composition. Catalytic cracking, for example, converts constituents into different chemicals and this can allow a refinery to create more gasoline or diesel by breaking down the higher-boiling components into smaller parts. But this is much more complicated than distillation.


Am I Using the Correct Number of Significant Figures (Heat Problem)?

This is sort of tricky, but I'd say that neither 10.8 nor 11 is right!

You got
$$Q_1 = 9.7\bar22753346\ \mathrm{kcal}$$

which you rounded to $9.72\ \mathrm{ kcal}$. The bar over the first two indicates that is the significant figure and that the excess numbers can be carried in intermediate calculations but would be trimmed for the answer.

Now for the second part you got

$$Q_2 = 1.\bar134\ \mathrm{ kcal}$$

which you rounded to  $1.1\ \mathrm{ kcal}$.

Using the intermediate calculations you are calculating the sum.

$$\begin{align}
9&.7\bar22753346\\
1&.\bar134\\
Q_1+Q_2 =10&.\bar856753346 \end{align}$$

Since 9.72+ answer is good to hundredths, but 1.1134 answer is only good to the tenths, thus the answer must be rounded to the tenths. So the "best" answer using significant figures is:
$$Q_1+Q_2 = 10.9\ \mathrm{ kcal}$$

You should not round intermediate results to their "correct" number of significant figures since that leads to rounding errors in the intermediate calculations.

$$\begin{align}
9&.7\bar2\\
1&.\bar1\\
Q_1+Q_2 =10&.\bar82 \end{align}$$

Obviously carrying the ten digits in $9.7\bar22753346$ is somewhat ridiculous, so a good rule of thumb with a calculator would be to carry two extra digits in intermediate calculations. So round the intermediate number to $9.7\bar228$. The value $1.\bar134$ has two extra digits so its fine as shown.

Notice that this is exactly the same rule you used in calculating the temperature difference.

$$\begin{align}
10\bar0&.\\
-37&.\bar0\\
=6\bar3&.\end{align}$$

Had it been 37.5 degrees for body temperature, then you should have used $6\bar2.5$ in the intermediate calculations.

$$\begin{align}
10\bar0&.\\
-37&.\bar5\\
=6\bar2&.5\end{align}$$

As a matter of curiosity I used the rnd() function in Excel to simulate uniform distribution between all the values. So:

17.95 grams to 18.04999 grams of water
99.5 to 100.49999 degrees C
36.95 to 37.04999 degrees C

$\text{=(2.26)*(18+0.1*(RAND()-0.5))/4.184+(18+0.1*(RAND()-0.5))*(100+(RAND()-0.5)-(37+0.1*(RAND()-0.5)))/1000}$

The 4.184 and 2260 values were assume to be exact. Averaging over a thousand such values I got 10.85703268
with a standard deviation of 0.016370732.

So the gist is that using significant figures doesn't give as good a result as possible. The problem is in the use of 2 significant figures in the calculation of $Q_2$. For the difference between the uniform differences of 100 and 37.0 the relative standard deviation is 0.46%. However the uniform distribution of 1.05 to 1.15 has a relative standard deviation of 2.6%, much worse than what is really happening.