Why do the alkali metals form different products upon combustion in air?

Since I will deal with all of the alkali metals in this answer, I think the question should also be broadened. There is no point in covering one single metal (sodium) without touching the others since it is the trend going down the group that we are interested in.

All thermodynamic data is taken from Prof. M. Hayward's lecture notes at Oxford.

So, firstly, some data. There is an increase in ionic radius going down the group, which should not be surprising:

$$\begin{array}{cc}
\hline
\ce{M} & \text{Ionic radius of }\ce{M+}\text{ / pm} \\
\hline
\ce{Li} & 76 \\
\ce{Na} & 102 \\
\ce{K} & 138 \\
\ce{Rb} & 152 \\
\ce{Cs} & 167 \\
\hline
\end{array}$$

This leads to an decrease in the magnitude of the lattice enthalpies of the Group I superoxides, peroxides, and oxides, going down the group:

$$\begin{array}{cccc}
\hline
\ce{M} & \Delta H_\mathrm{L}(\ce{MO2})\mathrm{~/~kJ~mol^{-1}} & \Delta H_\mathrm{L}(\ce{M2O2})\mathrm{~/~kJ~mol^{-1}} & \Delta H_\mathrm{L}(\ce{M2O})\mathrm{~/~kJ~mol^{-1}} \\
\hline
\ce{Li} & -960 & -2748 & -3295 \\
\ce{Na} & -860 & -2475 & -2909 \\
\ce{K} & -752 & -2175 & -2503 \\
\ce{Rb} & -717 & -2077 & -2375 \\
\ce{Cs} & -683 & -1981 & -2250 \\
\hline
\end{array}$$

As for why there is an increase in magnitude going across the table, we have to look at the factors controlling the magnitude of the lattice enthalpy:

$$\Delta H_\mathrm{L} \propto \frac{\nu z_+ z_-}{r_+ + r_-} \tag{1}$$

where $\nu$ is the number of ions in one formula unit, $z_+$ and $z_-$ are the charge numbers on the cation and anion, and $r_+$ and $r_-$ are the ionic radii:

$$\begin{array}{ccccc}
\hline
\text{Formula unit} & \nu & z_+ & z_- & \text{Numerator in eq. (1)} & r_-\text{ / pm} \\
\hline
\ce{MO2} & 2 & 1 & 1 & 2 & 149 \\
\ce{M2O2} & 3 & 1 & 2 & 6 & 159 \\
\ce{M2O} & 3 & 1 & 2 & 6 & 120 \\
\hline
\end{array}$$

The lattice energies of the peroxides and oxides are roughly 3 times those of the corresponding superoxides, because of the larger numerator. The lattice energies of the oxides have a  slightly larger magnitude than those of the corresponding peroxides, because of the smaller anionic radius.

Just looking at the lattice enthalpies, we might think that all metal cations would form the oxides. However, this approach is flawed because it does not take into consideration the energy cost of forming the three different anions from molecular dioxygen. Recall that the lattice enthalpy is defined as $\Delta H$ for the reaction

$$\ce{M+ (g) + X- (g) -> MX (s)}$$

However when we burn a metal in oxygen we are starting from $\ce{M}$ and $\ce{O2}$. So, we have to figure out the energy needed to get from $\ce{M}$ to $\ce{M+}$, and from $\ce{O2}$ to the relevant anion ($\ce{O2-}$, $\ce{O2^2-}$, or $\ce{O^2-}$).

The analysis here is complicated by the fact that the charges on the anions are not the same. Therefore, the reaction

$$\ce{M(s) + O2 (g) -> MO2(s)}$$

cannot be directly compared with

$$\ce{2M(s) + O2(g) -> M2O2(s)}$$

In order to "standardise" the equations, we will consider the reactions per mole of metal. One can loosely think of the combustion reaction as "releasing" some kind of energy within the metal; the most favourable reaction will be that which "releases" the most energy per mole of metal. Therefore, the three reactions we are considering are:

$$\begin{align}
\ce{M (s) + O2 (g) &-> MO2 (s)} & \Delta H_1 \\
\ce{M (s) + 1/2 O2 (g) &-> 1/2 M2O2 (s)} & \Delta H_2  \\
\ce{M (s) + 1/4 O2 (g) &-> 1/2 M2O (s)} & \Delta H_3  \\
\end{align}$$

Now, we construct Hess cycles for all three reactions.

Superoxides
$$\require{AMScd}\begin{CD}
\color{blue}{\ce{M(s) + O2(g)}} @>{\Large\color{blue}{\Delta H_1}}>> \color{blue}{\ce{MO2(s)}} \\
@V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\Delta H_\mathrm{L}(\ce{MO2})}A \\
\ce{M+(g) + e- + O2(g)} @>>{\Large\Delta H_\mathrm{f}(\ce{O2^-})}> \ce{M+(g) + O2^-(g)}
\end{CD}$$

Peroxides
$$\require{AMScd}\begin{CD}
\color{blue}{\ce{M(s) + 1/2O2(g)}} @>{\Large\color{blue}{\Delta H_2}}>> \color{blue}{\ce{1/2M2O2(s)}} \\
@V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\ce{1/2}\Delta H_\mathrm{L}(\ce{M2O2})}A \\
\ce{M+(g) + e- + 1/2O2(g)} @>>{\Large\ce{1/2}\Delta H_\mathrm{f}(\ce{O2^2-})}> \ce{M+(g) + 1/2O2^2-(g)}
\end{CD}$$

Oxides
$$\require{AMScd}\begin{CD}
\color{blue}{\ce{M(s) + 1/4O2(g)}} @>{\Large\color{blue}{\Delta H_3}}>> \color{blue}{\ce{1/2M2O(s)}} \\
@V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\ce{1/2}\Delta H_\mathrm{L}(\ce{M2O})}A \\
\ce{M+(g) + e- + 1/4O2(g)} @>>{\Large\ce{1/2}\Delta H_\mathrm{f}(\ce{O^2-})}> \ce{M+(g) + 1/2O^2-(g)}
\end{CD}$$

Now we need more data. $\Delta H_\mathrm{f}(\ce{M+})$ is simply the sum of the atomisation energy and first ionisation energy:

$$\begin{array}{cc}
\hline
\ce{M} & \Delta H_\mathrm{f}(\ce{M+})\text{ / }\mathrm{kJ~mol^{-1}} \\
\hline
\ce{Li} & 679 \\
\ce{Na} & 603 \\
\ce{K} & 508 \\
\ce{Rb} & 484 \\
\ce{Cs} & 472 \\
\hline
\end{array}$$

and the enthalpies of formation of the anions are

$$\begin{array}{cc}
\hline
\ce{X} & \Delta H_\mathrm{f}(\ce{X})\text{ / }\mathrm{kJ~mol^{-1}} \\
\hline
\ce{O2-} & -105 \\
\ce{O2^2-} & +520 \\
\ce{O^2-} & +1020 \\
\hline
\end{array}$$

Why is the enthalpy of formation of $\ce{O^2-}$ so large? The answer is that, to get from $\ce{1/2 O2}$ to $\ce{O^2-}$, you need to first break the $\ce{O=O}$ bond, then add two electrons to the oxygen. Furthermore, the second electron affinity is often an unfavourable process. For the other two anions, you don't need to break the $\ce{O=O}$ bond.

So now, we can see a trend at work already. Going from the superoxides to peroxides to oxides, the more negative lattice enthalpy favours the formation of the oxide. However, the increasing heat of formation of the anion favours the formation of the superoxide.

When do each of these two factors win out? Well, when lattice enthalpies are comparatively large, we would expect the lattice enthalpy factor to outweigh the heat of formation of the anion. Lattice enthalpies are large precisely when the cation is small, and therefore lithium forms the oxide when heated in oxygen. However, with caesium, lattice enthalpies are smaller, less significant, and the heat of formation of the anion wins out; caesium therefore forms the superoxide.

The trend is of course not black and white. Going down the group from lithium to caesium, we might guess that perhaps there are one or two elements that form the intermediate peroxide. That element is sodium. You could say that the larger lattice energies of sodium salts sufficiently compensate for the formation of the peroxide ion, but aren't enough to compensate for the formation of the oxide ion.

I leave you with the last bunch of numbers, which tabulate the values of $\Delta H_1$ through $\Delta H_3$ for all the elements (all values in $\mathrm{kJ~mol^{-1}}$). You can actually calculate these yourself by plugging the data above into the Hess cycles. It seems that the data is a little different from that given in the Chemguide screenshot you have, but the conclusion is the same, so I'll ignore that:

$$\begin{array}{cccc}
\hline
\ce{M} & \Delta H_1\text{ (superoxide)} & \Delta H_2\text{ (peroxide)} & \Delta H_3\text{ (oxide)} \\
\hline
\ce{Li} & -386 & -435 & \mathbf{-459} \\
\ce{Na} & -362 & \mathbf{-375} & -342 \\
\ce{K}  & \mathbf{-349} & -320 & -234 \\
\ce{Rb} & \mathbf{-338} & -295 & -194 \\
\ce{Cs} & \mathbf{-316} & -259 & -143 \\
\hline
\end{array}$$

As described earlier, the salt with the most negative enthalpy of formation will be preferentially formed. These are bolded in the table.


Relating change in internal energy and to that in enthalpy

You are right. B is the correct choice.

$$\Delta H = \Delta E + \Delta PV$$

Since $\Delta E $ is more negative than $\Delta H$, $$\Delta H > \Delta E$$.
Hence $$\Delta PV>0$$

$\Delta E \space $has to be negative. So option A is ruled out.

Among the other choices, B is the only one where number of gaseous moles are increasing from left to right, leading to a positive value of $\Delta V$ and hence $\Delta PV$.


I can see at Gompertz function they use the formula below. See the link for Graphs of Gompertz curves, showing the effect of varying one of a,b,c while keeping the others constant

$$f(t)=a\mathrm{e}^{-b\mathrm{e}^{-ct}}$$
where

''a'' is an asymptote, since $ \lim_{t \to \infty} a\mathrm{e}^{-b\mathrm{e}^{-ct }}=a\mathrm{e}^0=a$
''b'' sets the displacement along the ''x''-axis (translates the graph to the left or right). When ''b'' = log(2), f(0) = a/2, also called the halfway point.
''c'' sets the growth rate (''y'' scaling)
''e'' is Euler's Number (''e'' = 2.71828...)

The halfway point is found by solving $f(t) = a/2$ for $t$.
$$t_{hwp} = -\frac{\ln(\ln(2)/b)}{c}$$

Regardless of if the equation $$x(t) = x_\mathrm{max}\exp(-a\exp(b\log t)).$$ in the question is the true one, analyzing the equation as is, being aware ift has different shape than the above.:
It can be rewritten as $$x(t)=x_\mathrm{max} \cdot  \exp{\left(-a \cdot t^b\right)}$$
If $\ce{b=1}$, the equation reduces to the classical equation of the kinetic of the first order and the half-time $t_{1/2}=\frac{\ln 2}{a}$, where an asymptote at any point in time crosses x axis after $t_{1/2}$.
$$x(t)=x_\mathrm{max} \cdot  \exp{\left(-a \cdot t \right)}$$
The parameter $\mathrm{b}$ affects the time evolution of the diffusion kinetic model..
$b = 1$ keeps the relative decrease with time constant.
$b \gt 1$ accelerates the relative decrease with time.
$b \lt 1$ decelerates the relative decrease with time.
Illustrative table about the $b$ value influence.
$\begin{array}{|c|c|c|c|}  \hline
 & b & b & b \\  \hline \hline
t & 0.8 & 1 &1.2 \\
0 & 100 & 100 & 100 \\
1 & 50.0 & 50.0 & 50.0 \\
2 & 29.9 & 25.0 & 20.3 \\
3 & 18.8 & 12.5 & 7.5 \\  \hline
\end{array}$
$t$ was padded by small value to avoid log 0. Results were rounded.
$a$ was set to $\ln{2}$ to have $t_{1/2}=1$ for $b=1$.


Has a non-carbon-based form of life been discovered since 2010?

Just because an organism might need a nutrient, whether arsenic or cobalt (which homo sapiens needs for making RBC's) does not mean that the organism is not carbon-based. What had been announced was that an organism used arsenic to replace phosphorus, not carbon, which has since been amended.

There is also the question whether the GFAJ-1 microbe is a facultative or obligate user of arsenic, or simply excludes arsenic. Though the organism can survive high concentrations of arsenic, it appears to do so by discriminating between As and P and using the phosphorus in its metabolism.

There are organisms such as Sulfurospirillum arsenophilum and S. barnesii that are at least facultative As users, deriving energy from oxidation or reduction of arsenic compounds, but the arsenic is not replacing carbon, much less phosphorus, in their biochemistry.


why don't carbons use sp3 instead of sp2 orbitals to make double bonds?

Only one electron can be in an orbital at a time. There are three p orbitals, one in each of the x, y, and z axes. A pi bond is the overlap of one p oribital from each atom. Like the p orbital, a pi bond also can be rotated from the x axis to the y or z to produce a different unique pi bond. Because a sigma bond has rotational symmetry, rotating it does not give a unique bond. Therefore, you cannot have two of them between the same pair of atoms.

Edit: It appears this question has been asked before and there are much better answers than mine (Why can there not be more than one sigma bond in a set of bonds?)


Rather

$$\ce{2 NaOH(aq) + CO2(g) -> Na2CO3 (aq) + H2O}$$

and analogically on the solid $\ce{NaOH}$.

Bicarbonate in aqueous solutions cannot survive the excess of hydroxide, forming carbonate.

$$\ce{HCO3^{-}\ (aq) + OH- (aq)<=>> CO3^{2-}(aq) + H2O}$$

The same is further enforced by dehydration effect of solid hydroxide.

$$\ce{NaHCO3(s) + NaOH(s) -> Na2CO3(s) + H2O}$$

Only in excess of $\ce{CO2}$ Is formed bicarbonate:

$$\ce{CO3^{2-}(aq) + CO2(aq) + H2O <=>> 2 HCO3^- (aq) }$$


First off, may I say that I applaud your decision to test this through an experiment. It's rare to see that than I would like.

Now, on to the matter at hand. It's fairly well known from industrial chemistry that non-polar solvents degrade latex quite heavily.

I work with latex seals a lot, and the hexanes we use routinely break the seals down in under a day. Of course, if you're lubricating your condoms with hexanes, you're a) an idiot or b) absolutely insane.

A paper I managed to find suggests that there really isn't too much direct data on condoms, and it muses that the warnings might have arisen from industry, where nonpolar solvents decidedly do degrade latex.

To find out, they did a burst experiment with condoms that had been treated with various oils. Glycerol and Vaseline-treated condoms showed a very, very minor decrease in strength, while mineral oil/baby oil-treated ones burst at less than 10% of the volume of an untreated condom.

They also found that 10 month-old condoms have half the burst volume of 1-month old ones, so you could argue that using 1-month-old condoms that have been slathered in Vaseline is still much safer than using older ones.

As for the actual chemistry of the weakening, I honestly don't know. If I were to hazard a guess, I would note that the latex looks like a bunch of ethylenes glued together,

so my guess would be that the solvents get between the chains and force them apart, weakening them. For this to happen, the solvent must be nonpolar, but still small enough to slip between the chains of the polymer.

That's probably why vaseline and canola oil don't have much of an effect---they're just too big to fit between the chains. Again though, I don't know for sure, so don't quote me on this last paragraph.


Finding required drug concentration given Kdiss

Chemical reaction is

$$\ce{D1 + E <=> ED1}$$

Initial concentrations: $[\ce{D1}]_\textrm{tot}; \quad [\ce{E}]_\textrm{tot}; \quad 0$.

If $X$ percent of $E$ "reacts", then stoichiometry tells us that:

Concentrations at equilibrium: $[\ce{D1}]_\textrm{tot} (1- X) ;\quad [\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X;  \quad [\ce{D1}]_\textrm{tot} X$.

The equilibrium expression is

$$K_\textrm{diss} = \frac{ [E][D_1]}{[ED_1]}$$

Therefore

$$K_\textrm{diss} = \frac{ [\ce{D1}]_\textrm{tot} (1- X) ([\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X)}{[\ce{D1}]_\textrm{tot} X}$$

Which we can simplify to

$$K_\textrm{diss} = \frac{(1- X)}{X} ([\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X)$$

So you can solve this for $X = 0.90$, $[\ce{E}]_\textrm{tot} = 2.00 \ \mu M$ and $K = 1.00 \ nM$.


I have never worked with potassium personally, but have experience with other alkali metals.   You are spot on about potassium superoxide.  I have direct knowledge of an explosion that occurred when potassium metal was stored for an extended period of time in mineral oil.  I would therefore be extremely cautious if you intend to store it in this manner.  Factors such as length of storage time, permeability of your container seals and storage temperature can all affect the amount of superoxide that forms.  Based on the evidence I have seen, it is not safe to assume that there will be no transport of oxygen through a mineral oil barrier.  The only way I would work with this metal is in an inert glovebox, period.  That is my own preference, of course, and I always tend to err on the side of caution.


How to determine the most reactive hydroxyl group of α-ᴅ-glucopyranose?

$\ce{EtOH}$ is a rather bad nucleophile and we have the assistance of a lone pair of the neighboring oxygen atom (the positive charge can hence be delocalized over the carbon and oxygen) which greatly speeds up $\ce{S_{N}1}$ in comparison to $\ce{S_{N}2}$. So what will probably happen is protonation by $\ce{HCl}$ of the alcohol next to the ether oxygen followed by nucleophilic ($\ce{S_{N}1}$) attack of $\ce{EtOH}$ on the formed (delocalized) secondary carbocation. In comparison to the possible reactions at the other secondary centers (having the formulation of the question in mind) this primary reaction will be fast enough to justify the ignoring of them.


Why there will be only limited shells in a particular atom?

Imagine a universe that is not expanding, and contains only one large star and one small planet. It is technically possible to put the two objects into orbit no matter how far apart they are. At some very large distance, the gravitational attraction is so small that it is nearly negligible, but if given infinite time you will be able to trace the orbits for the planet-star system. If, however, you introduce more planets and stars close to the planet, then the influence of the star that is very far away will become irrelevant to the motion of the planet.

We can see a similar case when we look at an electron around an atom. Each time you excite it, you move the electron into a higher-level orbital that is more weakly bound. Eventually you get the electron into orbitals that are nearly degenerate and asymptotically approach zero binding energy. If you have a universe that contains only the electron and the atom, and get rid of the vacuum fluctuations, then yes, it is theoretically possible to put an electron into an orbital with arbitrarily high level of n.

In practice, at some point the electron becomes so weakly bound that electromagnetic forces in the environment will dominate and take the electron away from the atom, ionizing it.

It is experimentally possible to isolate an atom from the environment quite well and then excite the electron to a pretty high n without ionizing it. An atom that is excited in this way is called a Rydberg atom.


Why is there such a violent reaction when NaCl is formed?

Sodium's single unpaired electron is partially occupying a much higher energy orbital than the unfilled orbital on the chlorine which will be occupied when the two chemicals are mixed. The excess electronic energy generated will rapidly become thermal energy giving a "violent" but rarely explosive reaction (as there is no gas generation only thermal expansion).


What's the relation between molarity and molality?

$W$ and $V_{\text{sol}}$ can be related via the density of solution. We have, mass of solution = mass of solute + mass of solvent.
Hence,
$$ M_{\text{sol}} = W + M’ $$

Let us assume the density of the solution to be $\rho$. Since, density is equal to mass over volume, we have,
 $$ \rho=  \frac{M_{\text{sol}}}{V_{\text{sol}}} = \frac{W + M’}{V_{\text{sol}}} $$
Thus, you have relation between $W$ and $V_{\text{sol}}$.


Removing aluminum oxide before galvanic current and potential measurements

sure it's going to affect the test so you have to remove the oxide layer
Generally, we adopt the method of alkali corrosion to remove the oxide layer on the surface of the aluminum.The purpose of alkali corrosion is removing the oxide layer on the surface of the aluminum. (Note: Technological process of the method is as follows: under normal temperature, 20% nitric acid; the process parameters is about 0.5 minutes. Pre-degrease--degrease--rinsing--rinsing--alkali corrosion--rinsing--rinsing--bright dipping--rinsing--rinsing--pure rinsing. The purpose of bright dipping is to remove surface residual black ash after alkali corrosion.Temperature:50-60 degrees; time:0.5-1 minutes.The purpose of degreasing is to remove grease on the surface of the aluminum.The process parameters is 0.5-1 minutes. You can use acid to wash it


Extraction of hydrogen and oxygen from water

No, it won’t burn immediately. As your electrodes are generally separated by some distance, you should be able to isolate the produced oxygen and hydrogen gas by catching the respective bubble streams under a submerged test tube or the like. Alone, neither oxygen nor hydrogen will burn, however combining the oxygen and hydrogen gas will produce an explosive mixture. Mixing the hydrogen with air will also produce an explosive mixture as the atmosphere is about 22 % oxygen.

To get the hydrogen to burn, you need to supply the necessary activation energy, which can be supplied by a spark or flame. After this, the reaction will be self-sustaining due to its exothermicity. There are in fact entirely self-contained apparatus which can be purchased from scientific teaching shops that can electrolytically split water and then combust the hydrogen and oxygen using a piezoelectric ignition system, the ensuing tiny explosion causing water to spurt out of a small hole. Wikipedia mentions that the autoignition temperature of hydrogen in air is 500 degrees Celsius.

On a small scale, this reaction is quite safe, however you should always take safety precautions such as wearing protective eyewear and having fire-suppression equipment at hand.

Here’s an illustration of a reasonable water-splitting setup. Remember that your water should contain an electrolyte such as dissolved salt. Keep in mind that if your potential difference is high enough you will start to produce small quantities of chlorine gas and aqueous sodium hydroxide, both of which are hazardous in high concentrations.

I should draw attention to the fact that the $\ce{H2}$ headspace is about twice the volume of the $\ce{O2}$ headspace. This comes down to the stoichiometry of the reaction, and also means that the $\ce{H2}$ volume will increase at twice the rate of $\ce{O2}$. The wires I’ve illustrated connect to the terminals of a battery or lab PSU.

Addendum: Here’s a nice guide on how to construct and safely operate a water-splitting/combusting apparatus using household equipment, aimed at secondary school students.


Assuming 'change' means modification of composition, the answer is Digital Thermometer because even if temperature is linked to atmosphere composition it does not provide you any information about it.

Satelite can embed LIDAR system which will provide you crude qualitative (spectrum) and quantitative (intensities) informations on atmosphere composition, but it requires complex mathematical processing and relies on weak hypothesis;
Mass spectrography is probably the most accurate way because it is highly qualitative (retention time) and quantitative (ionization current);
Distillation will allow you to qualitatively 'separate' (boiling point) and 'quantify' (mass flow rate) distinct components.

Obviously temperature measurement is not able to provide any informations of this kind. Further more, temperature gradient in atmosphere is not only dominated by its composition, there is a lot of other phenomena involved in it.


The differences between integrals evaluations of contracted an uncontracted basis functions

A lot of stuff is going on in this question. So for clarity let's make this answer very explicit. We can try to consider the two cases of using an uncontracted basis and a contracted basis to clarify the differences. As a case let's look at STO-$2$G for $\mathrm{H_2}$. In the STO-$2$G basis set $\mathrm{H_2}$ would normally be described by two basis functions each consisting of two primitives.

Uncontracted case

If we use the STO-$2$G basis set in the case of using uncontracted Gaussian basis functions, each hydrogen will be described by two basis functions. Our overlap matrix will therefore be of the form:

$$ S=
\begin{bmatrix}
\left<\phi_1^\mathrm{H_1}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_4^\mathrm{H_2}\right.\right> \\
\left<\phi_2^\mathrm{H_1}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_2^\mathrm{H_1}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_2^\mathrm{H_1}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_2^\mathrm{H_1}\left|\phi_4^\mathrm{H_2}\right.\right>  \\
\left<\phi_3^\mathrm{H_2}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_3^\mathrm{H_2}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_3^\mathrm{H_2}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_3^\mathrm{H_2}\left|\phi_4^\mathrm{H_2}\right.\right>  \\
\left<\phi_4^\mathrm{H_2}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_4^\mathrm{H_2}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_4^\mathrm{H_2}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_4^\mathrm{H_2}\left|\phi_4^\mathrm{H_2}\right.\right>
\end{bmatrix}
$$

In this notation we have that:

$$\left<\phi_i^\mathrm{H_k}\left|\phi_j^\mathrm{H_l}\right.\right> =
N_iN_j\int_{-\infty}^\infty P_i\exp\left(-\alpha_ir_i^2\right)P_j\exp\left(-\alpha_jr_j^2\right) d\tau$$

With $P$ being a function of the angular momentum, in cartesian coordinates:

$$P=x^ly^mz^n$$

It can be seen that $P=1$ for s-functions. As we can see we have 4 basis functions in total (two on each hydrogen). You wrote something about, you would expect that in the case of uncontracted that we would have ""[number of electrons]x[number of basis functions] matrix"". As we can see this is not the case. Right now our matrix is in the atomic basis. When we solve the Hartree-Fock equations and find the coefficients that will take us to a molecular basis, we will see that the orbitals that are occupied in the atomic orbital basis, will have a contribution to occupied orbitals in the molecular orbital basis, and are therefore still important.

Contracted case

Lets now take a look at how the STO-$2$G basis set will be used in the case of contracted basis functions. Now our overlap matrix will take the following form:

$$ S=
\begin{bmatrix}
\left<\phi_1^\mathrm{H_1}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_2^\mathrm{H_2}\right.\right> \\
\left<\phi_2^\mathrm{H_2}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_2^\mathrm{H_2}\left|\phi_2^\mathrm{H_2}\right.\right>
\end{bmatrix}
$$

At first glance the overlap matrix look just like the one in the uncontracted case, but smaller. What has changed is how the elements are defined:

$$\left<\phi_i^\mathrm{H_k}\left|\phi_j^\mathrm{H_l}\right.\right> =
\sum_I\sum_J c_Ic_JN_IN_J\int_{-\infty}^\infty P_I\exp\left(-\alpha_Ir_I^2\right)P_J\exp\left(-\alpha_Jr_J^2\right) d\tau$$

We can see that we now sum over all the primitives included in the contraction. To make it more clear, let's write out how the second element in the overlap matrix looks like. Now remember that each of the basis functions consists of two primitives:

$$
\begin{eqnarray}
\left<\phi_1^\mathrm{H_1}\left|\phi_2^\mathrm{H_2}\right.\right>
&=& c_{11}c_{21}N_{11}N_{21}\int_{-\infty}^\infty \exp\left(-\alpha_{11}r_{11}^2\right)\exp\left(-\alpha_{21}r_{21}^2\right) d\tau \\
&+&c_{11}c_{22}N_{11}N_{22}\int_{-\infty}^\infty \exp\left(-\alpha_{11}r_{11}^2\right)\exp\left(-\alpha_{22}r_{22}^2\right) d\tau \\
&+&c_{12}c_{21}N_{12}N_{21}\int_{-\infty}^\infty \exp\left(-\alpha_{12}r_{12}^2\right)\exp\left(-\alpha_{21}r_{21}^2\right) d\tau \\
&+&c_{12}c_{22}N_{12}N_{22}\int_{-\infty}^\infty\exp\left(-\alpha_{12}r_{12}^2\right)\exp\left(-\alpha_{22}r_{22}^2\right) d\tau
\end{eqnarray}
$$

In this notation the first number refers to the basis function and the second refers to the primitive. As we can see each element of the overlap matrix in the contracted basis is made from four integrals. Even though our overlap matrix is four times as small compared to that of the uncontracted basis, we still have to evaluate the same number of primitive integrals. In this case we have investigated here, where we used STO-$2$G, we have no unoccupied orbitals. In general, this will not be the case, most basis sets (also when used in the contracted basis) will have unoccupied orbitals.

Other comments

In your question I can see that you mention molecular orbitals. Just to be clear. When we do the integrals we are in the atomic orbital basis. The molecular orbital basis is the one we find when we solve the Hartree-Fock equations. I.e. the molecular orbitals have nothing to do with uncontracted vs contrancted basis functions, though we will get more molecular orbitals if we use uncontracted basis functions, because we will get the same number of molecular orbitals as we had atomic orbitals.

Regarding how to implement this in Mathematica I have no idea. Though as my understanding of how Mathematica is ought to be used, I would recommend that you would implement your integral code in a language like Python. For inspiration you can take a look at this guide, which implements all the integrals with contracted basis functions.


What is the oxidation number of the oxygen in BaO2

"oxygen always -2 except in peroxides"

This exception is in effect here, because $\ce{BaO2}$ is a peroxide consisting of $\ce{Ba^{2+}}$ and $\ce{O2^{2-}}$ ions. The oxidation number of Ba is +II, and the oxidation number of each of the oxygens in the peroxide anion is -I. This fits with the charge of the peroxide anion ($2 \times -1 = -2$), and as $\ce{BaO2}$ is a neutral compound, the sum of all oxidation numbers is 0. Moreover, there are even more exceptions to the rule of thumb cited above, for example, the superoxide radical anion $\ce{O2-}$ with a fractional oxidation numer of $-\frac{1}{2}$, or the dioxygenyl cation $\ce{O2+}$ with a formal oxidation number of $+\frac{1}{2}$ for each oxygen.

In contrast to peroxides, "simple ionic compounds" of oxygen are oxides, which contain the anion $\ce{O^{2-}}$. The charge of the oxide anion is $2-$, so the oxidation number of oxygen in those compounds is -II.


How to figure out which phase on a phase diagram will be the most dense?

The solid phase is more dense than the liquid phase. The line that separates solid and liquids bends right.

If the solid phase is less dense than the liquid phase, the line that separates solid and liquids bends left.


Do atoms react to fill outer shell or 8 valence electrons?

Excellent question I must say.
To middleschoolers they really don't teach advanced concepts regarding orbitals. But since this question came to your mind, I guess you're capable of handling what I have to say next.
Unlike what's taught to you, shells don't exist just like that, they exist as a collection of subshells. Subshells are given names such as $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, and $\mathrm{f}$. The number of subshells a shell has is given by it's principal quantum number, or simply the shell number. In this way, the 1st shell has only one subshell, which is $\mathrm{1s}$. The second shell has two subshells; $\mathrm{2s}$ and $\mathrm{2p}$. The third has three subshells; $\mathrm{3s}$, $\mathrm{3p}$, and $\mathrm{3d}$.
Every subshell is further composed of orbitals. Orbitals are different from orbits, orbits have electrons in a precisely defined path, while orbitals show a cloud of probability of where the electrons could exist. Each orbital holds two electrons.The orbitals also have exciting shapes:
The $\mathrm{s}$ subshell consists of just one orbital, the $\mathrm{s}$ orbital. It's a spherical cloud of electron probability:

The $\mathrm{p}$ subshell consists of three orbitals, $\mathrm{p}_x$, $\mathrm{p}_y$, and $\mathrm{p}_z$. They are dumb-bell shaped, oriented along the $x$, $y$, and $z$ axes respectively:

The $\mathrm{d}$ subshell has five orbitals, $\mathrm{d}_{xy}$, $\mathrm{d}_{yz}$, $\mathrm{d}_{zx}$, $\mathrm{d}_{x^2+y^2}$, and $\mathrm{d}_{z^2}$. They have interesting shapes.

The $\mathrm{s}$ subshell can hold two electrons, the $\mathrm{p}$ subshell can hold six electrons,the $\mathrm{d}$ subshell can hold ten electrons, and the $\mathrm{f}$ subshell can hold fourteen electrons.
In middleschool, they teach you to fill the lower electrons first, and move to the top, if you want to find the electronic configuration of a given element. Unfortunately, that's not entirely correct. Infact, considering the subshell concepts, you must use the Aufbau Principle. The rule tells us to fill the electrons in the order given in the picture below:

(source: boundless-cdn.com)
If the picture wasn't clear, the order is $\mathrm{1s}$, $\mathrm{2s}$, $\mathrm{2p}$, $\mathrm{3s}$, $\mathrm{3p}$, $\mathrm{4s}$, $\mathrm{3d}$, $\mathrm{4p}$, $\mathrm{5s}$, $\mathrm{4d}$, $\mathrm{5p}$, $\mathrm{6s}$,...
Do you notice that $\mathrm{4s}$ comes before $\mathrm{3d}$? Thats why you don't get those ten extra electrons of $\mathrm{3d}$, since the $\mathrm{4s}$ subshell is being filled for the elements potassium and calcium.


Most contributing resonance structure of phosphoric acid

Both structures are not the main contributing structures. In fact, the first structure is outright wrong and should be marked as an error as the electrons of phosphorus do not add up and the non-protonated oxygen is implied to be a radical.
However, the correct most contributing structure can be derived from the first structure simply by adding formal charges: a formal positive charge on phosphorus and a formal negative charge on the non-protonated oxygen. This structure satisfies the stability conditions well:

All main group elements have octets (dublets for hydrogen)
The second proposal violates this most important principle

The number of formal charges is minimalised
We cannot go any lower without violating the more important first point

The type of formal charges matches the electronegativity difference
And also we cannot distribute them differently

The distance between formal charges is minimal
But again we have no choice.


Do most elements generally exist in a planet during early formation, or is it a small subset?

Elements are not formed through chemical reactions or in planets (except through nuclear decay) . They are formed in stars through nuclear reactions.
Your heavier elements can't even be formed in your typical star. It takes a supernova.
So pretty much everything the planet is made of is there from the start.


To understand the pH rate profile of the Aspirin hydrolysis, we have to look at all the possible mechanisms through which the hydrolysis may take place.
A good rule of thumb for interpreting a pH rate profile is: going from left to right, when the line curves upward (i.e. the rate increases suddenly), a new mechanism is dominating the reaction. When the line curves downward (i.e. the rate drops suddenly), that generally indicates that there is a change in ionisation/dissociation or that the rate-limiting step has changed for the same mechanism.
If you look at the graph for Aspirin, you will see that the graph bends up two times, around pH 2.5 and pH 8.5. So, there must be 3 different mechanisms throughout the whole pH range. (There is also a downward bend at pH ~5, that's not important at the moment).
So you can divide the graph into 3 segments (separated by blue line):

We need 3 mechanisms—(A), (B) and (C). Let's deal with them one at a time.

(A)

At pH < ~2.5 i.e. strongly acidic enviornment, the main components in the solution are the protonated form of aspirin (pKa of aspirin ~ 3.5), and $\ce{H3O+}$. A mechanism can be drawn:

This is the usual acidic ester hydrolysis mechanism. I haven't drawn the last few steps for brevity.
The rate of the reaction is: $$\ce{r=k_1[aspirin][H3O+]}$$ $$\mathrm{usually\,written\, as,}\;\; \ce{r=k_{obs}[aspirin]}$$
Which means, $\ce{k_{obs}=k_1[H3O+]} \Rightarrow \ce{log(k_{obs})=log(k_1) +log([H3O+])} \Rightarrow \ce{log(k_{obs})=log(k_1) - pH}$. You can immediately see why the graph decreases linearly as pH increases.

(B)

When pH > ~2.5, the solution contains both the protonated and the deprotonated form of aspirin, along with a lower concentration of $\ce{H3O+}$ as the solution is not strongly acidic anymore. As the rate of the reaction has increases suddenly, that must mean there is another mechanism in effect now, which overshadows the previous acid catalysed mechanism.
This part is difficult, because multiple mechanisms can be drawn, and it is difficult to distinguish which one is actually happening. Here the graph bends down once, and that happens around pH 4, close to the pKa of aspirin(3.5) which should clue you in to the fact that the mechanism here is somehow affected by the deprotonation of aspirin.
The rate rises at first, until pH 4 and then it stays almost constant until pH 8.5 (where (C) starts). This indicates that the mechanism is not catalysed by acid or base (because then the graph can't stay flat against changing pH), and that the mechanism involves the deprotonated form of aspirin (because once all the aspirin is deprotonated, the graph becomes flat).
One possible mechanism:

Note that we can't draw a mechanism where water attacks the ester directly, because water is not a good enough nucleophile, and also because the anion has to participate in the reaction somehow, as I said earlier.
Another possible(*) mechanism would be direct attack of the carboxyl anion on the ester:

Here an anhydride intermediate is formed, which is then hydrolysed as usual, and I have not drawn that part of the mechanism.
(*) I remember hearing in a lecture, that this mechanism via anhydride has been proven to be wrong, based on isotope substitution studies. I can't find a reference right now, so I will leave the mechanism in the answer, but I am pretty sure the first mechanism is the correct one.
Here, $\ce{[aspirin-]}$ and $\ce{[aspirin-H]}$ represents the deprotonated and protonated form of aspirin, respectively. The rate of the reaction is: $$\ce{r=k2[aspirin-]}$$ But the way the experimental measurements are taken means that the rate is expressed in terms of the total aspirin concentration:
$$\ce{r=k_{obs}[aspirin]}$$
From the acid dissociation equilibrium, we can write:$$\ce{[aspirin]=[aspirin-H] +[aspirin-]=\frac{[aspirin-][H3O+]}{K_a} +[aspirin-]=[aspirin-]\left( 1 +\frac{[H3O+]}{K_a}\right)}$$
We can substitute into the first equation: $$\ce{r=k2[aspirin-]= \frac{k2}{\left( 1 +\frac{[H3O+]}{K_a}\right)}[aspirin]}$$
Which gives:
$$\ce{k_{obs}=\frac{k2}{\left( 1 +\frac{[H3O+]}{K_a}\right)}}$$
In the region pH ~2.5 to 4, $\mathrm{k_{obs}}$ increases as $\ce{[H3O+]}$ decreases. After pH ~4, $\ce{[H3O+]<<K_a}$ so, $\mathrm{k_{obs}\approx k_2}$ and the graph becomes almost flat.

(C)

After pH 8.5, the main components of the solution are the deprotonated form of aspirin and $\ce{OH-}$. So, the base-catalysed mechanism becomes dominant.

As all the aspirin is in deprotonated form, we can write
$$\ce{r=k_3[aspirin][OH-]}$$
Which means, $\ce{k_{obs}=k_3[OH-]} \Rightarrow \ce{log(k_{obs})=log(k_3) - 14 + pH}$
So the graph rises linearly with increasing pH.
[P.S. The solvent does not appear in the rate equations.]


How can carbon dioxide be converted into carbon and oxygen?

In my opinion, the catalytic, solar-driven conversion of carbon dioxide to methanol, formic acid, etc. is much more interesting and promising, but since Enrico asked for the conversion of carbon dioxide to carbon itself:

The group around Yutaka Tamaura was/is active in this field. In one of their earlier publications,[1] they heated magnetite ($\ce{Fe3O4}$) at 290 °C for 4 hours in a stream of hydrogen to yield a material which turned out to be stable at room temperature under nitrogen. This material, $\ce{Fe_{3+\delta}O4}$ $(\delta=0.127)$, i.e. the metastable cation-excess magnetite is able to incorporate oxygen in the form of $\ce{O^2-}$.

Under a $\ce{CO2}$ atmosphere, the oxygen-deficient material converted to "ordinary" $\ce{Fe3O4}$ with carbon deposited on the surface.

This remarkable reaction however is not catalytic, but a short recherche showed that the authors have published a tad more in this field. Maybe somebody else finds a a report on a catalytic conversion among their publications.

Tamaura, Y.; Tahata, M. Complete reduction of carbon dioxide to carbon using cation-excess magnetite. Nature 1990, 346 (6281), 255–256. DOI: 10.1038/346255a0.


Use of significant figures in addition and subtraction

No. The answer isn't accurate enough. Say, for example I wish to add two numbers: $1.23$ and $2.367$.

I proceed like this, but I've no idea what the third digit after the decimal in $1.23$  is and so I represent it as $?$.

$$\begin{align*} 1 &.23? \\ + 2 &.367 \\ \hline 3 &.59? \\ \hline  \end{align*}$$

This is why the end result is rounded off with the number of digits equal to the number having the fewest number of digits you initially set off with.


Correct term/word for bias in calculations?

I believe the fundamental notion that you're looking for is a "systematic error in the data analysis".

A "statistical bias" would imply some sort of sampling problem which doesn't seem to be the case from your description.

An "analytical bias" would imply a different/better laboratory technique which again doesn't seem to be the case. So although the overall results are biased, you don't need to redo the experiment itself but rather reanalyze the data using the proper calculations.


What is the reason behind neglecting & putting the overlap integrals $0$ in Simple Hückel Theory?

What is(are) the reason(s) behind making the approximation of putting
  all the overlap integrals 0?

I am not 100% sure here, but using such a simplification, the mathematics becomes significantly easier. In principle, Hückel theory would still work if there would be overlap. In fact, it is possible to create a completely new basis by applying a transformation to the basis set by which a new set of orthogonal basis functions is created. The advantage of setting $\hat{S}$ (the overlap matrix) to $\hat{I}$ (the identity matrix) is that the eigenvectors of the matrix equation can be directly interpreted in terms of the $\pi$ orbitals in the system. You do not have to back-transform your solution as is the case in the Hatree-Fock method.

  Why did Hückel work only on π orbital? Doesn't his calculation work
  for σ-bonding?

Only $\pi$ electron molecular orbitals were included because these determine the properties of the molecules that Hückel was studying (conjugated molecules). The $\sigma$ electrons were ignored. This is known as $\sigma$-$\pi$ separability and is justified by the orthogonality of $\sigma$ and $\pi$ orbitals in planar molecules. As a consequence though, the Hückel method is in principle limited to planar systems.


Material that absorbs only in NIR and emits in visible

Such materials not only exist but are common

Most people will have seen an actual demonstration of materials that exploit the phenomenon of a substance that absorbs NIR light and emits visible light. Most Green laser-pointers are based on their use.

These usually rely on a semiconductor laser outputting near-infra-red (NIR) light at 1063nm wavelength which then passes through a crystal of KDP (mono potassium phosphate) or KTP (potassium titanyl phosphate) which emits green light at 532nm (which is the second harmonic of the original laser frequency). The mechanism involves non-linear optical processes (which usually require a fairly high intensity of light so won't happen as easily as phosphorescence, for example).


Actually an organometallic scandium(I) compound is known, from P.L. Arnold at al. (1996).  As with organometallic compounds generally, the metal is covalently bonded, and the use of $3d$ orbitals as valence orbitals (there are 14 shared valence electrons associated with each scandium atom) in this and similar compounds favors including scandium as a transition metal.

In predominantly ionic settings, however, it is true that scandium(I) would be unlikely.  Because scandium is only at the beginning of the transition metals it has relatively low ionization energies for its valence electrons, and this low energy cost tends to be overcome by increasing lattice or solvation energy associated with higher ion charges.  With ionically bonded scandium the balance is all in favor of more ionization and more electrostatic attraction, whereas with titanium the slightly higher ionization energies and the fact that a noble gas configuration requires four rather than three positive charges makes lower charge ions more accessible.


What is the probability of 3 possible products, between two chemical species?

Well, if you assume the rates are known and the reactions' order follows from stoechiometry (e.g. if they are elementary reactions), you can put the chemical kinetics into simple equations:

$$\frac{\mathrm da}{\mathrm dt} = -k_1 a(t)^2 - k_3 a(t)b(t) $$
$$\frac{\mathrm db}{\mathrm dt} = -k_2 b(t)^2 - k_3 a(t)b(t) $$

($t$ here being time, not temperature).

Knowing initial amounts or concentrations $a_0=a(t=0)$ and $b_0=b(t=0)$, you can pretty much integrate  the system to find out what happens.

Edit — solving this system for $k_1=k_2=k_3$ yields the quantities of AA, AB and BB at infinite time to be as follows:

$$aa = \frac{a_0^2}{a_0+b_0}$$
$$ab = \frac{a_0 b_0}{a_0+b_0}$$
$$bb = \frac{b_0^2}{a_0+b_0}$$


Does hydrogen bonding only occur between water molecules?

NO, Water is not only the example for hydrogen bonding. It's most common example for basics.

Hydrogen bonds occur in DNA, proteins, polymers, etc. But for introductory level it is easy to explain with water.

DNA: Hydrogen bonding between guanine and cytosine, one of two types of base pairs in DNA.

Polymer: Para-aramid structure

Images: Wikipedia


Let's explain this using an example. Consider a cell in which an aqueous solution of sodium chloride is electrolyzed. The $\ce{Na^+}$ ions migrate toward the negative electrode and the $\ce{Cl^-}$ ions migrate toward the positive electrode. But indeed at the anode, there are  two substances that can be oxidized at the anode: $\ce{Cl^-}$ ions and water molecules.
$$\ce{2Cl^-   \rightarrow Cl_2 + 2e^-}$$ $$\ce{2H_2O   \rightarrow O_2 + 4H^+ + 4e^-}$$
The standard-state potentials for these half-reactions are $E^0_{ox}= -1.36 V$ for the first couple and  $E^0_{ox}= -1.23 V$ for the second couple.

At first glance, it would seem easier to oxidize water ($E^0_{ox}= -1.23 V$) than $\ce{Cl^-}$ ions ($E^0_{ox}= -1.36 V$).

It is worth noting, however, that the cell is never allowed to reach standard-state conditions. The solution is typically 25% $\ce{NaCl}$ by mass, which significantly decreases the potential required to oxidize the $\ce{Cl^-}$ ion. The pH of the cell is also kept very high, which decreases the oxidation potential for water. The deciding factor is a phenomenon known as overvoltage, which is the extra voltage that must be applied to a reaction to get it to occur at the rate at which it would occur in an ideal system.

Under ideal conditions, a potential of 1.23 volts is large enough to oxidize water to $\ce{O_2}$ gas. Under real conditions, however, it can take a much larger voltage to initiate this reaction. (The overvoltage for the oxidation of water can be as large as 1 volt.) By carefully choosing the electrode to maximize the overvoltage for the oxidation of water and then carefully controlling the potential at which the cell operates, we can ensure that only chlorine is produced in this reaction.


The analogy with a proton is actually a good one if you are careful to remember that an electron is nearly 2000 times lighter than a proton. What does that mean? It means that despite the fact that an electron is very "small", the electron is actually going to be very large because lighter particles will tend to spread out and have a much more diffuse wavefunction. For instance, if we just take a case where the environment sets up a more or less constant potential for the electron to sit in, then it becomes very clear that a proton will spend much more time towards the bottom of the well than the electron will. Of course the environment wouldn't set up the same potential for both of them because this potential is determined by interactions with the electron or proton itself.

Ok so that's a basic picture. Essentially, you have to think of the solvated electron as being a very quantum mechanical ion if you wish to think of it as an ion.

Before we can get to full solvation, however, it is useful to understand the behavior of an excess electron in simpler systems. One good example, because they're studied all the time and are usually fairly representative of what might be going in the liquid phase, is a water cluster. In Sommerfeld and Jordan's paper[1] studying the behavior of excess electrons on water clusters, they provide the following wonderful image:

This gives a great picture of what a solvated electron could look like. The point is that it can look like a lot of things. If you wish, you can think of these isosurfaces as being an "orbital" that the excess electron is free to move around in. The difference is that we draw orbitals as 90% probability densities, only these are drawn at a constant volume, and hence don't all contain the same "percentage of the electron". In fact many of these surfaces contain less than 50% of the charge density, which demonstrates these electrons are very diffuse!

Now, the arrangements where the electron is primarily inside the cluster is going to resemble a solvated electron the most. It is worth noting structures such as 20c and 20d. In these structures, the electron interacts electrostatically with the free $\ce{O-H}$ bonds. In structures such as 13a and 24b, the electron is actually bound by the dipole field of the cluster (these clusters all have large dipole moments). Because these electrons are bound by dipole interactions, they will be quite diffuse.

One way in which electrons like these are definitely not like a proton in solution is that correlation effects become very important for the behavior of these electrons. Indeed, some of the structures shown above only give a negative binding energy when correlation is included at the CI level. (See 1 for details on all this.)

Some more interesting papers on electrons bound to water clusters are 2 and [3].

So, this sets us up for thinking about fully solvated electrons (sticking with water because it's where the most work is done since water is very important). As it happens, our orital analogy applies quite nicely in liquid water. Below is the little cover art from a paper hot off the presses (2017) by Ambrosio et al.. [4]

They actually show that the solvated electron in water is bound strongly enough to have multiple electronic states (they match with experiment quite well), and they can be seen to be s-like and p-like. That is, one of them is roughly spherical while the other has a node with a smaller lobe. Again, keeping our analogy with a proton in mind, these electrons are much larger than how we think of a proton in water. Both, however, induce a local structure in the water molecules to stabilize the excess charge.

The fact this aqueous electron has at least two electronic states tells us already that it is probably bound more strongly than the electrons in the sodium/liquid ammonia solution (at least for the concentrated solution). The difference being that excess electrons in water are going to be very dilute (there's not many of them), while the sodium/ammonia ones are quite concentrated when we see them.

These solvated electrons in water are actually quite mobile, as we might expect. See for instance[5], in which it is shown that a surface-bound electron actually has almost the same energy as a fully solvated electron. That doesn't really make sense to me besides that the stuff from the water clusters might be relevant. That is, for this to be true, you would expect that the dominant interactions at the surface would be different than in the bulk. I'd guess dipole-bound at the surface and primarily dispersion and other correlation effects in the bulk.

As one last way to beat the proton analogy to death, we can maybe understand the mobility of the solvated electron by noting that protons are also very mobile. That is, if I drop $\ce{DCl}$ into a solution of $\ce{H2O}$, I will eventually end up with a bunch of $\ce{HDO}$ molecules and some $\ce{D+}$ and mostly $\ce{H+}$. This is to say that the proton which I drop in and say "that's my proton" is going to get attached to a water molecule and some other hydrogen is going to become the proton, and so on.

The big difference for the solvated electron is that electrons are indistinguishable from each other, so I can't say that this is the solvated electron and those are the valence electrons of the surrounding solvent. Rather, we have to picture the solvated electron as just being part of the wavefunction describing the electrons of all nearby waters (well really the whole solution but let's not get carried away). Maybe this is where we start talking about solvated muons...

In any case, if you use google scholar there is a ton of reading on solvated electrons, so feel free to check it out, and hopefully this helps you have a picture for whatever people might be working on.

Sorry I could only speak about water. In general an aqueous electron might behave quite differently in other solvents, but it's definitely always going to be quite quantum mechanical. The fact it can be bound by purely correlation effects tells us what kind of beast we're dealing with.

[1]: Sommerfeld, T., & Jordan, K. D. (2006). Electron binding motifs of (H2O) n-clusters. Journal of the American Chemical Society, 128(17), 5828-5833.

2: Sommerfeld, T., DeFusco, A., & Jordan, K. D. (2008). Model potential approaches for describing the interaction of excess electrons with water clusters: Incorporation of long-range correlation effects. The Journal of Physical Chemistry A, 112(44), 11021-11035.

[3]: Jordan, K. D., & Wang, F. (2003). Theory of dipole-bound anions. Annual review of physical chemistry, 54(1), 367-396.

[4]: Ambrosio, F., Miceli, G., & Pasquarello, A. (2017). Electronic Levels of Excess Electrons in Liquid Water. The Journal of Physical Chemistry Letters, 8(9), 2055-2059.

[5]: Coons, M. P., You, Z. Q., & Herbert, J. M. (2016). The hydrated electron at the surface of neat liquid water appears to be indistinguishable from the bulk species. Journal of the American Chemical Society, 138(34), 10879-10886.


How does one determine the probability of H⁺ to H₂O in pure water?

The answer to this question was surprisingly difficult to find because all modern ways to measure pH use reference solutions of known pH and thus rely on prior knowledge of the $H^+$ activity of said reference solutions.

Finally I found this on wikipedia:

  it is possible to measure the concentration of hydrogen ions directly, if the electrode is calibrated in terms of hydrogen ion concentrations. One way to do this, which has been used extensively, is to titrate a solution of known concentration of a strong acid with a solution of known concentration of strong alkaline in the presence of a relatively high concentration of background electrolyte. Since the concentrations of acid and alkaline are known, it is easy to calculate the concentration of hydrogen ions so that the measured potential can be correlated with concentrations. The calibration is usually carried out using a Gran plot.[9] The calibration yields a value for the standard electrode potential, $E^0$, and a slope factor, f, so that the Nernst equation in the form
  $$E = E^0 + f\frac{2.303RT}{F} \log[\mbox{H}^+]$$
  can be used to derive hydrogen ion concentrations from experimental measurements of E. The slope factor, f, is usually slightly less than one. A slope factor of less than 0.95 indicates that the electrode is not functioning correctly. The presence of background electrolyte ensures that the hydrogen ion activity coefficient is effectively constant during the titration. As it is constant, its value can be set to one by defining the standard state as being the solution containing the background electrolyte. Thus, the effect of using this procedure is to make activity equal to the numerical value of concentration.

This means that the precise cocentration of $H^+$ is actually unknown what is known is it's chemical activity which has been defined as equal to the concentration for a certain standard solution.


For H$_2$ there are only two electrons and thus you can only have singlet (electrons with opposite spin projection) and triplet states (electrons with equal spin projection).

The spin wave function of the singlet state is a linear combination (a superposition state) of the first electron with spin projection $\alpha$ and the second electron with spin projection $\beta$ and the opposite situation. Both spin states in the quantum superposition have opposite phase:
$$|\uparrow\downarrow\rangle -|\downarrow\uparrow \rangle$$
If the electrons are in the same molecular orbital then (by the Pauli principle) only the singlet state exists. If they are filling $\pi$ molecular orbitals then they could be in different degenerate orbitals (with angular momentum projection on the molecular axis $M_z = \pm \hbar$) and thus both singlet and triplet terms would exist.

There are three degenerate states belonging to the triplet term, that is, there are three wave functions that have equal energy. There spin components are:
$$|\uparrow \uparrow \rangle$$
$$|\downarrow \downarrow \rangle$$
$$|\uparrow \downarrow\rangle + | \downarrow\uparrow \rangle$$

So you can see that it doesn't matter if you fill one molecular orbital with one electron in spin $\alpha$ and a different molecular orbital with the electron with $\alpha$ spin or you do so with both $\beta$ and $\beta$ spins. However you can not put both electrons with the same spin projection in the same molecular orbital. That is against the Pauli principle, that is, against the anti-symmetry of the electronic wave function.


Using real wave functions for chemical bond explanation

Chemists use the linear combinations of hydrogen atom wavefunctions which are real because this is simply much more convenient to think about. Notice that for hydrogen, all the orbitals for a given $n$ are degenerate, so taking these linear combinations does not affect the energy, and thus thinking about the shapes of these orbitals is potentially more useful than thinking about the shape of something which is complex.

You are correct that taking the real-valued wavefunctions causes them to no longer be eigenfunctions of $L_z$. This is a bit frustrating, but in practice we never use these orbitals for anything, so it does not really matter. I used to think organic chemists took the shape of orbitals a bit too seriously (and they maybe do), but the notion of the shape of an orbital is actually recovered in a more rigorous manner.

That is, molecular orbital theory is just a qualitative form of Hartree-Fock (HF) theory. In HF, the Fock matrix is unitarily invariant, so we are free to transform our orbitals by any unitary rotation. These orbitals can be written as functions in $\mathbb{R}^3$ as desired. Thus, there exists schemes by which molecules orbitals can be "localized" and they often take the shape of orbitals which look a whole lot like the real-valued hydrogen atom orbitals you mention.

Thus, the use of these real-valued orbitals as a notion of shape is justified after the fact from HF theory and beyond.

In practice, this never really matters if one is being rigorous because any attempt at saying exactly which orbitals are involved in a bond is thwarted by unitarily transforming those orbitals. Thus, many chemists use the shape of an orbital as a heuristic because thus far, it has proven to be a very useful conceptual tool. I think pretty much all chemists know this is not completely correct, but neither is anything in physics.

Basically, the shape of orbitals provide a useful model.


Here is another equivalent way of thinking about the problem. The mass of oxygen on the old scale is 16 amu. The mass of oxygen on the new scale is 15.994 amu. When you divide them:

$$\frac{16\ \textrm{old}}{15.9994\ \textrm{new}} = 1.0000375 \frac{\textrm{old}}{\textrm{new}}$$

Basically, taking a ratio of any value in the old scale to the equivalent value in the new scale (or vice versa) gives you a conversion factor. So for every new amu, there is 1.0000375 old amus. The question is asking you to convert 12 new amus to old amus. Just by looking at the units, it should be obvious what to do:

$$\frac{16\ \textrm{old}}{15.9994\ \textrm{new}} * 12\ \textrm{new}=12.00045\ \textrm{old}$$

If you explicitly label old and new, it's easy to see that the new units cancel out.

Incidentally, if you keep track of your significant figures, you can see why this redefinition of the amu didn't cause too many problems for chemists—even to five sig figs, the change is negligible.


Friedel-Crafts alkylation of five-membered heterocycles

Alkylation of the five-membered rings is perfectly possible - nobody says it cannot happen. But there are two problems:

Alkylation further activates the ring, making it more reactive - this is exactly the same problem that makes alkylation an unattractive reaction in benzene rings
The Lewis acid can, and will, lead to polymerisation of electron-rich rings such as the five-membered heterocycles.

With acylation, the first point is moot, and if you choose an appropriate catalyst you can avoid polymerisation. So acylation can be used in certain cases (I'm fairly sure that simply chucking in an acyl chloride with $\ce{AlCl3}$ will not be a clean reaction at all though).

From Joule & Mills, Heterocyclic Chemistry 5th ed., p 300 (chapter on pyrrole):

  Mono-C-alkylation of pyrroles cannot be achieved by direct reaction with simple alkyl halides, either alone or with a Lewis-acid catalyst, for example pyrrole does not react with methyl iodide below 100 °C; above about 150 °C, a series of reactions occurs leading to a complex mixture made up mostly of polymeric material together with some poly-methylated pyrroles. The more reactive allyl bromide reacts with pyrrole at room temperature, but mixtures of mono- to tetra-allyl-pyrroles together with oligomers and polymers are obtained.

p 327 (chapter on thiophene):

  Alkylation occurs readily, but is rarely of preparative use, an exception being the efficient 2,5-bis-t-butylation of thiophene. (J. Org. Chem., 1984, 49, 4161. DOI:10.1021/jo00196a012)

Presumably the steric bulk of the t-butyl group prevents further reactions.

p 350 (chapter on furan):

  Traditional Friedel–Crafts alkylation is not generally practicable in the furan series, partly because of catalyst-induced polymerisation and partly because of polyalkylation.


How do I make a solution of spinach to titrate with pottasium permanganate?

The best approach is to ask the teacher again after doing some "homework" (i.e. reading and searching). Extracting and then titrating oxalic acid/oxalate is not that easy from spinach. You will need a lot of spinach to extract a titratable amount. Oxalic acid can be extracted by boiling spinach in 0.1 M HCl. However, tons of other redox active substances will also appear in the extract. Direct titration with KMnO4 will give an extra-ordinary large percentage, which is of course erroneous. After extraction, you need to precipitate oxalic acid via some cation. After isolation of the precipitate you would need to bring back oxalic acid into solution, one can do a redox titration. In real-world nobody would use this approach but as an academic project, it could be a good exercise. Again ask your teacher if she/he is willing to go to this extent.


In free radical substitution of propane with halogens, how do you know which carbon ends up with an unpaired electron after fission of C-H bond?

This is to do with the relative stabilities of the free radicals produced. Here is a schematic diagram showing the two possibilities for propene (similar mechanism, also with two free radical possibilities like propane):

As you can see, a more stable and less stable free radical is produced. The stability depends on how many carbons are attached to the carbon with the unpaired electron. In the diagram, the free radical where two carbons are attached is more stable than the free radical with one carbon attached.

As a general rule, the stability will be greatest in tertiary free radicals, then in secondary, and least stability in primary.

This is due to the inductive effect of carbons bonded to the carbon with the unpaired electron:

Carbons bonded to the free radical carbon are able to 'move' some of their electron density over to the free radical carbon, stabilising it. This is due to something called sigma conjugation, which we needn't go into.

But in summary, both free radicals are formed, as seen in the first diagram. Both free radicals go onto generate products, but much more come from the more stable species. Therefore, in fact both carbons possible in propane end up with the unpaired electron, but the route involving the secondary free radical species is more successful and gives the major product. The route involving the primary intermediate gives a minor product.


Why is Kw the same in acid, water, and alkali?

You are correct that the concentration of $\ce{[H+]}$ and $\ce{[OH-]}$ both vary based on the the composition (acid/alkaline) of the solution, but the remarkable thing is that their product does not. When $\ce{[H+]}$ goes up, $\ce{[OH-]}$ goes down, in the same proportion.

The reason for this is that the Kw represents the equilibrium constant of the following reaction, and any excess (lack) of $\ce{[H+]}$ or $\ce{[OH-]}$ will react by LeChatelier's principle to reestablish equilibrium:

$$\ce{ H2O <=> H+ + OH-}$$

We can write the standard equilibrium constant equation as follows

$$K_{eq} = \frac{\ce{[H+][OH-]}}{\ce{[H2O]}}$$

For aqueous solutions, the concentration of water is a fixed constant1, so we can multiply both sides by it, and fold it into the equilibrium constant2. Therefore we can write:

$$K_{w} = \ce{[H+][OH-]}$$

So just as standard equilibrium constants do not differ with the concentrations of chemical species in the solution, but can vary with temperature, so too does $K_{w}$ not vary based on the concentration of chemical species, but does depend on the temperature of the solution.

1 Technically, the values in the equilibrium constant equation are activities rather than concentrations (see here and ref 1 therein for a rather technical discussion), a distinction we normally ignore, as activities are proportional to concentration in dilute solutions. Regardless, the activity of water in most dilute aqueous solutions can be considered a constant, so it doesn't change the argument.

2 Note this is only true in a solution which is almost entirely water, with just a little bit of solute dissolved. If you had a water/ethanol mixture, or a very concentrated solution of a highly soluble salt, then the water activity can't be considered a fixed constant, and you'd have to start making some adjustments based on it. (This is where the distinction between "concentration" and "activity" come into play - these adjustments are more complicated than just dividing by the nominal concentration of water.)


How to read the used volume from a burette?

It depends ;-)

The scale on a burette typically is ascending from top to bottom. With other words: zero is close to where you fill your $\ce{NaOH}$ solution in, the largest value is close to the valve. The total amount added (to your vinegar) is the difference between the final and the initial reading.

But if your burette was initially filled up to the zero mark, this would just mean to subtract zero from the final reading (which is just the final reading).


In preparing sulphonilamide, why is aniline acylated if the acyl group has to be removed later?

The main problem encountered during electrophilic substitution
reactions of aromatic amines is that of their very high reactivity.
Substitution tends to occur at ortho- and para-positions. If we
have to prepare monosubstituted aniline derivative, how can
the activating effect of –NH2
 group be controlled ?

This can be done by protecting the -NH2 group by acetylation with acetic
anhydride, then carrying out the desired substitution followed
by hydrolysis of the substituted amide to the substituted amine

The lone pair of electrons on nitrogen of acetanilide interacts
with oxygen atom due to resonance as shown below:

Hence, the lone pair of electrons on nitrogen is less available for
donation to benzene ring by resonance. Therefore, activating
effect of –NHCOCH3
group is less than that of amino group.

Moreover, in the
strongly acidic medium, aniline is protonated to form the
anilinium ion which is meta directing.

However, by protecting the –NH2
 group by acetylation reaction
with acetic anhydride, the reaction can be controlled
and the p derivative can be obtained as the major product.

source: word to word from
NCERT(INDIA) CLASS 12 chemistry ;)
http://ncert.nic.in/NCERTS/l/lech204.pdf


What should be the oxidation state of oxygen in HOF (hypofluorous acid)?

Until the (recent) redefinition of the IUPAC, the concept of oxidation states was not as well defined as one would expect. I have discussed the issues of the old version and outlined the new definition in more detail in an answer to Electronegativity Considerations in Assigning Oxidation States.

When you apply the official pre-2016 definition (via the Internet Archive) from the IUPAC gold book, then you have to assign $\text{OS}(\ce{H}) = +1$ as it is not a compound with a metal, $\text{OS}(\ce{O}) = -2$ as it is no peroxide compound, and leaving the disturbing $\text{OS}(\ce{F})= \color\red{+1}$.

Going with the on electronegativity based alternative description, you will still assign $\text{OS}(\ce{H}) = +1$ as it has the lowest electronegativity. Then you assign $\text{OS}(\ce{F})= -1$, because of the highest electronegativity. This leaves a oxidation state of $\text{OS}(\ce{O}) = 0$.

This assignment is matched by the 2016 definition, the summary version of which is: The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds. The definition refers to the use of Allen electronegativities (see Pure Appl. Chem. 2016, 88 (8), 831–839 for more detail). Consequently the heteronuclear approximation yields:

The true charges are of course something completely different. Based on a DF-BP86/def2-SVP calculation I ran some population analyses with Multiwfn 3.4.1 (a newer version is available). The NBO charges are taken from a prior version of this answer, but I since have lost access to the program and the files used.

\begin{array}{lrrr}\hline
\text{Method}    & \ce{H} & \ce{O} & \ce{F} \\\hline
\text{Hirshfeld} & +0.18  & -0.10  & -0.09  \\
\text{ADCH}      & +0.38  & -0.29  & -0.09  \\
\text{VDD}       & +0.18  & -0.10  & -0.09  \\
\text{Mulliken}  & +0.20  & -0.06  & -0.15  \\
\text{Löwdin}    & +0.09  & +0.01  & -0.10  \\
\text{Becke}     & +0.38  & -0.21  & -0.17  \\
\text{ADC Becke} & +0.38  & -0.29  & -0.09  \\
\text{CM5}       & +0.35  & -0.26  & -0.08  \\\hline
\text{QTAIM}     & +0.61  & -0.44  & -0.17  \\\hline
\text{NBO}       & +0.45  & -0.32  & -0.12  \\\hline
\end{array}


The Lewis structure most closely resembling reality consists of two resonance structures: the first one posted in the question and its mirror image. The reason is that the octet rule is observed this way (no hybridization of d-orbitals for main group chemistry necessary or possible) as well as the symmetry of the molecule: the two bonds are identical.


Improving copper contact and preventing oxidation

Though I have not tried them DeoxIT products claim to do what you want, available as liquid and spray.

If anyone has used these, please add a comment as to effectiveness.


Confusing definition of mol when considering second order rate equations

I think the sentence cited above in the wikipedia article is very confusing.
But the wiki article about the Avogadro constant helped me resolve my confusion, which defines $mol$ via the Avogadro constant:
$$N_A = \left\{N_A\right\}[N_A] = 6.022\times10^{23} \text{mol}^{-1}.$$
Hence
$$k_{on} = k^*_{on}N_A = \left\{k^*_{on}\right\}\left\{N_A\right\}\frac{l}{\frac{1}{[N_A]}s}=\left\{k^*_{on}\right\}\left\{N_A\right\}\frac{l}{mol~s}$$
This definition is also given on the hompege of the National Institute of Standards and Technology (NIST), see here.


Can ingredients in Tiger Balm cause any harmful effects on PVC figures?

According to this Wikipedia entry on Tiger Balm, there is camphor (among other chemicals) in the formulation of this product. This chart of chemical resistance recommendations for PVC indicates that it is not recommended that PVC should be exposed to camphor. Therefore, it is probably best to reduce or eliminate contact of Tiger Balm with the figures.

As for the paint cracking or chipping as a result of exposure to Tiger Balm, that might depend in part on the composition of the paint, and that cannot be addressed unless you can provide that information.


is water a non-metallic oxide? and hence acidic in nature?

When textbooks say that water is "neutral," textbooks mean that the water meets this condition:

$\ce{[H_3O^+]} = [\ce{HO^-}]$.

That's all.

Water can be neutral, in that its hydronium ion concentration equals its hydroxide ion concentration. Water can also be acidic if there is carbon dioxide dissolved in the water. Carbon dioxide can dissolve in water simply by letting the water come into contact with the atmosphere. Water can be basic too if you put a basic solute in water.

Water itself can also act as a weak Brønsted acid by donating a proton. This however is not a favorable process.

Water can also act as a Lewis acid because the oxygen in water withdraws electron density from the hydrogens, giving each hydrogen a partial positive charge. This in turns makes the hydrogens electrophilic - fitting the definition of Lewis acidity.

I think what your textbook is getting at is Lewis acidity. Consider carbon dioxide. This doesn't fit the Brønsted definition of acidity because it has no hydrogen protons to donate. So consider the partial charges on the carbon dioxide molecule. There is a positive partial charge on the carbon because the oxygens are more electronegative than carbon and have withdrawn electron density from the carbon. This makes the carbon electrophilic. And hence carbon dioxide as a whole can be considered a Lewis acid (but do understand that it's the carbon specifically in carbon dioxide acting as a Lewis acid).

So in general, your textbook says that non-metallic oxides tend to be acidic (in the Lewis sense) because the oxygens withdraw electron density from the non-metal (remember that oxygen is one of the most electronegative elements). This in turns creates partial positive charge(s) and makes the centers of these partial positive charges electrophilic.

Also, you could say that non-metallic oxides are precursors to Brønsted acids; consider this Lewis acid/base reaction. Remember that the central conceit of Lewis acid/base chemistry (coordination chemistry) is electron flow; nucleophiles attack electrophiles (not the other way around; that would be the fatal conceit).

So consider this reaction:

$\ce{SO_3 + H_2O -> H_2SO_4}$

We have sulfur with an electronegativity (EN) of 2.6. Oxygen has an electronegativity of 3.5. The difference is 0.9. The bond is clearly polar covalent. In addition, we have not one but three oxygens isolating electron density from the sulfur. Therefore, the sulfur most definitely has a partial positive charge and the oxygens partial negative charges. We say the oxygens are nucleophilic. And back to water, we have already concluded the hydrogen atoms in water are electrophilic due to their positive partial chage. The sulfur trioxide will attack the hydrogens and form the sulfuric acid molecule - a strong Brønsted acid (its first ionization is complete in water solution).


The lithium-ion batteries do not contain pure metallic lithium, nor do they contain any metallic alloy of lithium. They contain various intercalation compounds which are relatively harmless by themselves.

When a charged battery is short-circuited (by throwing it into the water or otherwise), it may react quite violently, but that's another story. Any battery would do the same, more or less. It's about physics, not chemistry (I mean, chemical reactions would occur as well, but they don't explain the explosive part; they would occur during the normal operation all the same).


How or Why does diesel/kerosene have a much higher 'flash point' but lower auto/self-ignition point than gasoline/petrol?

Flash point and autoignition temperature measure different things: gasoline and diesel are optimised for different ignition conditions
First, the standard definitions.
Flash point
Flash point is the lowest temperature at which a liquid can gives off vapor to form an ignitable mixture in air near the surface of the liquid.
Autoignition point
The autoignition temperature or kindling point of a substance is the lowest temperature in which it spontaneously ignites in a normal atmosphere without an external source of ignition...
The first point to make on the differences between diesel and gasoline is that the liquids are designed for different properties in engines.
Petrol/gasoline engines need fuels that vaporise easily to give a mixture that is ignited by a spark. So they want a mixture that is easy to ignite but that does not explode spontaneously in the engine (which is bad for the engine and upsets the control of when the exploding fuel happens). So a low flash point is good but a low autoignition point is bad.
Diesel engines work differently. The explosion is not triggered by a spark but by compression of the fuel mixture which heats it and leads to a spontaneous explosion in the mixture driven by the heat of compression. So a diesel engine wants a specific autoignition temperature and one that is low enough to create reliable fuel ignition.
So the properties of gasoline and diesel are there by design as that is how the engines they feed are meant to work.
The chemical properties that create these properties are, broadly, different mixtures of hydrocarbons. Gasoline tends to be more volatile and contains far more unsaturation (aromatics and branched hydrocarbons). Volatility is necessary so the mixture can be vaporized easily to make a flammable mixture. Diesel tends to have many more straight chain hydrocarbons which are more susceptible to attack by oxygens in the air. Diesel also tends to be much less volatile than gasoline as volatility is much less important to creating an effective mixture in the engine. This has the side effect of increasing the flash point for diesel.
So design intent and chemical properties make the metrics different.


How to find out the state symbol of a compound by looking at its chemical formula?

You can't do this by:

  just by looking at the chemical formula

But instead look up the properties of the compounds in question (Wikipedia is a good source of data).

$\text{RbOH}$ is highly soluble in water, so assuming there's an excess of water present, it will end up being dissolved in the water. So its state symbol will be $\text{(aq)}$.

Hydrogen on the other hand is a gas (at room temperature and above) and highly insoluble in water: it will leave the reaction as gas and its state symbol will be $\text{(g)}$.

So without knowing the physical/chemical properties of the reagents/reaction products, the state symbols cannot be determined.

Note that the state symbols are often idealisations: solid or liquid compounds are rarely completely soluble or completely insoluble. The same is true for gases. Even water is a case in point: liquid, $\text{(l)}$, at room temperature but with significant vapour pressure.


Specific kind of plastic that won't make fabrics smell?

It is likely a plasticizer in polyethylene bins, rather than polyethylene itself, that has an odor. Storing the garments in clear polyethylene plastic bags (perhaps inside the bins), which have different plasticizers, should reduce the odor -- and some bags are designed to compress clothing with use of a vacuum cleaner, saving space.

Another option is to get an inexpensive wooden (cedar lined?) chest, or a cloth storage bag. Advantage: breathable. Disadvantage: does not protect from moisture.

BTW, some of the plasticizers include polybutylene, dioctyl  phthalate, glyceryl tribenzoate, polyethylene glycol, sunflower oil etc. Handbook of Plasticizers Ch. 11, G. Wypych. Perhaps rancid sunflower oil is the odor?


Le Chatelier's Principle: Temperature dependence of endothermic and exothermic reactions (chemical equilbrium)

In a reaction, there is an exothermic and a endothermic side. Increasing the temperature will favour the endothermic side.

So if your reaction is exothermic, that means it must release heat if it shifts to the left. Then, since the left is exothermic, it shifts to the right (endothermic) to minimise this disturbance.

If your reaction is endothermic, that means it must absorb heat if it shits to the left. Then, since the left is endothermic, if you increase the temperature it shifts to the left (endothermic) to minimise this disturbance.

The above is much easier to understand if you think of 'heat' as a product or reactant. In an exothermic reaction, heat is on the product side of the equation. So if you increase this component, to minimise this disturbance you must shift the equilibrium to the LEFT to decrease the 'heat' amount - and vice versa.


What you call a molecular structure is in reality an asymmetric unit. These two terms are not related, and in general are not interchangeable. The same way $\ce{Pb3O4}$ is not a molecular formula, but a formula unit.

All pictures you presented are correct in terms that they do represent an asymmetric unit, but actually an IUCr-recommended asymmetric unit (can be generated with Olex2's command standardise) looks like this:

$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#575961}{\Large\bullet}~\ce{Pb}$.

... and with $Z = 4$ these are packed in a unit cell with $Pbam$ space group accordingly:

The crystal structure itself can be represented as infinite layers of cross-linked chains made of $\ce{[PbO6]}$ octahedra, which are stacked via $\ce{- O-Pb-O -}$ bridges:

Crystallographic data is taken from ICSD #9753 [1].

Gavarri, J. R.; Weigel, D.; Hewat, A. W. Journal of Solid State Chemistry 1978, 23 (3), 327–339. DOI 10.1016/0022-4596(78)90081-6.


Periodic table groups - which grouping is "right"?

There is no "most right classification scheme" for the elements of the periodic table, worth memorizing above all others.  This is because there is no single, 'universal' set of criteria that effectively, unarguably, and neatly divides the entire periodic table into groups. Depending on the properties of interest, different classifications/groupings are more or less appropriate.


Firstly I recommend you to understand the concept of "Magnetic Susceptibility". My source for this is Miessler,Tarr Inorganic Chemistry Book 2nd edition.

Consider Ni2+, its electronic structure is 4s03d8. This is a d8 ion. Then it has 8 electrons in d orbitals. d orbitals have 5 degenerate level of orbitals.

Also you should know gyromagnetic ratio which is the ratio of magnetic dipole moment to angular momentum of the system. (We got a system where quantum mechanics may be obeyed).

So, considering orbitals and electrons that fill these orbitals we got two magnetic moment here. Spin magnetic moment(sum of it is S) and angular magnetic moment(sum of it is L).
I don't know properly write all calculations with codes. So I attached an image.


What is the structure of the hypothetical xenon octafluoride?

There does seem to be a good body of evidence, including your reference, for the existence of xenon octafluoride.

Regarding it's structure, according to VSEPR theory as tabulated here, the structure is square anti prism:


Methods to prepare lactic acid from acetaldehyde

A fairly simple two step synthesis involves nucleophilic addition of $\ce{HCN}$ to the carbonyl group and then aqueous hydrolysis of the nitrile to yield the carboxylic acid product.


Why does Nitrogen gas appear in the equation for making producer gas?

As the comments allude, the coefficient is 4 to represent the approximate 4:1 ratio of oxygen to nitrogen in air (more precisely 21:78 or 7:26).  This added nitrogen must be heated along with the oxygen and carbon.  From a stoichiometric perspective, nitrogen is just a spectator and could be excluded, but from a thermal/total system perspective as nitrogen increases the specific heat of the system requiring more heat to be inputted for heating the mixture which subsequently requires more heat dumping to cool the system, reducing the efficiency of the system compared to a system of pure reactants.


Smallest nearest neighbor distance among elements

Both FCC and HCP packings have nearest neighbours as close as possible (in nature at least, the simple cube is even closer but doesn't exist). Their nearest neighbour distances in terms of $a/R$ are $2\sqrt{2}$. Here $a$ is the length of a side of the unit cell and $R$ is the radius of the atom the cell consists of.

When you are looking for the smallest nearest neighbour distance this means that you are looking for the smallest $a$ in an FCC or HPC packing. You can translate this into: you are looking for the atom which forms an FCC or HPC packing and has the smallest value for $2\sqrt{2}R$ so for $R$.

Now, I don't know all the options but for example $\ce{Cu, Au, Ag}$ form FCC structures and $\ce{Zn}$ forms an HCP structure (see e.g. here). You can look up the atomic radii and find that Zinc has the smallest: $122$ pm and will thus have a nearest neighbour distance in the lattice of $345$ pm.


At what position in the distillation apparatus does the temperature match the boiling point of the distilled substance?

The temperature in area A can well exceed B all right, especially if the distillation adapter is long or if there is a fractioning column or another condenser before the three-way adapter.
However, no target component is going to make it through the condenser unless its vapors reach the side arm (the one with the joint to condenser).

Apparently, it's only possible if the temperature of that area coincides with the boiling point of the target, that's why the thermocouple or thermometer probe is installed exactly at that area (B).

As you rightfully suggested the temperatures in areas C and D are arbitrary: it only should be above the m.p. to prevent clogging and less than b.p. so that the target is not evaporating away.

P. S. there is a vacuum adapter shown on the picture (area C), so there also might be distillation at lower pressure, which would also lower the b.p. of the mix constituents.
Since the question doesn't mention its usage, atmospheric pressure is assumed.


What are the chemical reactions behind fire?

Even though fire is one of the Greek classical elements, it is the only one that is not matter in our current understanding. What we experience as fire is the energy (in the form of light and heat) given off by the exothermic combustion of (usually) organic materials, like wood. The large amounts of thermal energy released by combustion often cause the gases in the fire to incandesce, that is, some of the kinetic energy produced by the fire is converted into electromagnetic radiation that we see (visible) and feel (infrared).

The chemical reaction for combustion is pretty simple. Take, for example, the combustion of n-octane $\ce{C8H18}$, which produces more than 5 MJ of energy per mole during combustion (from webbook.nist.gov).

$$\ce{2C8H18 +25O2->16CO2 +18H2O} \space\space \Delta_cH^o=-5430 \text{ kJ/mol} $$

The combustion of wood is more complex. The majority of the organic mass of dry dead wood is lignin and cellulose. Lignin is a highly-cross-linked copolymer of p-coumaryl alcohol, $\ce{C9H10O2}$, coniferyl alcohol, $\ce{C10H12O3}$, and sinapyl alcohol, $\ce{C11H14O4}$. Cellulose is a linear polymer of glucose, with formula $\ce{(C6H10O5)}_n$.

Different species will have different ratios of lignin to cellulose, different cross-link densities in the lignin, and different ratios of coumaryl to coniferyl to sinapyl alcohols. The formula for lignin could be expressed then as $\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z$.

The equation for the combustion reaction for cellulose is:
$$\ce{(C6H10O5)}_n +6n\ce{O2->}+6n\ce{CO2}+5n\ce{H2O} $$

Since lignin is more complex, its combustion equation is more complex:
$$2[\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z]+(21x+23y+25z)\ce{O2}\\ \ce{->}(18x+20y+22z)\ce{CO2}+(10x+12y+14z)\ce{H2O}$$

And assuming the variable composition of wood as a ratio (A:B) of cellulose to lignin, the overall combustion reaction becomes the following monstrosity:

$$A\ce{(C6H10O5)}_n +2B[\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z]+[6An+B(21x+23y+25z)]\ce{O2}\\ \ce{->}[6An+B (18x+20y+22z)]\ce{CO2}+[5An+B(10x+12y+14z)]\ce{H2O}$$

Given the number of variables ($A, B, n, x, y, z$), the $\Delta_cH^o$ for this reaction is difficult to determine (but not impossible, assuming we know some thermochemical reference data). Whatever its value, it is exothermic enough to induce incandescence.


How can I obtain crystal-clear glycerin?

You can buy high-purity glycerin from numerous vendors online. High-purity grades will be as crystal-clear as you can possibly get.

For instance, check out the premium grade glycerin product from Sigma-Aldrich which should be suitable for your needs but is quite expensive.

There are less expensive (but pure enough for your needs) options, so be sure to explore the other grades available from other vendors. For example, you can secure a gallon of Kosher USP food grade vegetable-derived glycerin for $14.99 from Amazon. This product (or similar) is adequate for your needs at a better price point that lab-grade reagents.


How was the dissociation constant of water determined?

I'd dare to say that it was first determined in experiments on electrical conductivity around 1890 by the physical chemist Friedrich Kohlrausch.

In Die Wasserstoffionenkonzentration. Ihre Bedeutung für die Biologie und die Methoden ihrer Messung (Leonor Michaelis, 1922, ISBN: 978-3-642-88800-7) the author writes

  F. Kohlrausch und A. Heydweiller kamen durch wiederholte Destillation des Wassers unter ganz besonderen Kautelen zu einem Grenzwert der Leitfähigkeit des Wassers, der sich durch weitere Reinigungsversuche nicht weiter herabdrücken ließ und daher als die eigene wahre Leitfähigkeit des Wassers gedeutet werden mußte. Sie stellten auf diese Weise die Dissoziation des Wassers zum erstenmal zahlenmaßig fest. Später wurden dann mehrere andere Methoden zur Dissoziation des Wassers gefunden, die wir bald kennenlernen werden, und diese bestätigten die Resultate von Kohlrausch.

This roughly translates to:
Upon repeated destillation of water under extreme precautions, F. Kohlrausch and A. Heydweiller eventually reached a limit for the conductivity […]. This way, they were the first to quantitatively determine the dissociation of water[…]

In a footnote to this paragraph a reference is given:

F. Kohlrausch und A. Heydweiller, Zeitschr. f. physikal. Chemie, 14, 317 (1894).


In HF there is a hydrogen bond which always outweighs all other factors when deciding properties related to stability. H-Bonded molecules are very stable


Is it possible that diamond can be grown from aqueous solutions?

The reaction you propose would be formally equivalent to the protection of aldehydes with 1,3-propanedithiol to form the correspondent dithiane. There are many organic reactions in which aldehydes participate this way, but all of them have something in common: the other species(lets call it donor) must have a free pair of electrons to attack the carbon atom in the aldehyde. In case this pair is not present, but the donor has an acidic hydrogen, we could use a strong base to remove it and form in situ the pair of electrons. (See aldol reactions)

Here's the problem: diamonds have no free pairs of electrons. And terminal hydrogens in a diamond lattice are by no means acidic, hence can't be removed by any base to the best of my knowledge. So there's no way a diamond will react with aldehydes.


How does one properly do this alkali melt? (Pfleger's indigo synthesis)

On academia, I found a reading of Industrial Dyes: Chemistry, Properties, Applications by Wiley (2003) that on first view sheds some light on the processes and chemistry.  Looks like an interesting secondary reference, rather than a Organic Synthesis.  I took some screen photos from there

and

and

however, the concluding bibliography may turn out sour

as many of these references are likely only in German.

Perhaps one may trace either in these, or more recent work (than 1890) citing them provide a more detailed procedure.

Addition: On archive, I found a "medium old" book about dye chemistry by Fierz, initially published in 1949, that comes pretty close to your intended alkaline melt; except omitting sodium amide:

Perhaps other old books dedicated to this topic equally tell more about these secrets, than modern textbooks?


Why does 1,3-dihydroxybenzene have a higher boiling point than 1,2-dihydroxybenzene?

Being more volatile implies that the molecules should be set free from mutual interactions. As 1,2-dihydroxybenzene has intramolecular H bonding, it is easy for it to exist independently, but for 1,3-dihydroxybenzene there might be intermolecular H bonding (interaction between two molecules); hence, it is a bit more difficult to break it apart, thus having a higher boiling point than its counterpart.


Proving a thermodynamic relation between $(\partial H/\partial T)_p$ and $(\partial U/\partial T)_V$

$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$

I would start from

$$ dH = \pd HTp dT + \pd HpT dp$$

This gives rise to

$$ \pd HTV = \pd HTp  + \pd HpT \pd pTV$$

in which you can recognize some components of the solution. Since

$$dH = dU +PdV + VdP $$

so that

$$ \pd HTV = \pd UTV  + V \pd pTV$$

it follows that

$$ \pd UTV  + V \pd pTV = \pd HTp  + \pd HpT \pd pTV$$

which is readily rearranged to obtain the desired equation.
$\endgroup$


Are there any major exceptions when comparing electron affinity?

This exception rule is actually orbital filling rule. For two electrons to be in same orbital they need to have different spins (Pauli exclusion principal). This electron pairing requires additional energy and thus it is easier to add electrons if there are free orbitals. When element has a half-filled p sublevel all 3 orbitals have one electron and pairing takes place (difference between energy levels of 2p and 3s is greater than electron pairing energy).

Electron pairing effects have significant impact to physical properties of coordination complexes (like color and magnetic properties).


What makes Gorilla Glass more durable with each generation?

Are there other ways of "chemical strengthening" besides cationic exchange that can be utilized to strengthen the glass, keeping it flexible and less fragile at the same time?

In a word: No.

To understand why ion exchange strengthens glass, you have to understand why the ion exchange makes the glass harder.  The process ion exchange hardening, its done at a temperature that allows ion diffusion, but disallows a reconfiguration of the glass structure or relaxation of bonds (below Tg). When sodium atoms are replaced by potassium in glass, the potassium ions occupy a site that is sized for sodium; this mismatch creates the compression that in turn gives the glass its scratch resistance. Thus the only way to chemically harden the glass is to substitute larger ions into the glass.

Now an astute person might be thinking:

  Are there other ways of "chemical strengthening" besides cationic exchange... What about anion exchange?

To this yes in theory if you could substitute the anions without rearranging the structure then yes anion exchange would be another way. But as Yogi Berra once said "in theory there's no difference between theory and practice, but in practice there is." The problem with anion exchange is you would be breaking and thus relaxing the bonds of atoms that you need to remain in place, thus no strengthening would occur.

I suppose you could also modify the surface to make a layer of silicon nitride or carbide and call that anion exchange, but would it this strengthening glass or changing the material altogether?

One way in which you could increase glass strength though still predominately a chemical process is to substitute sodium for lithium (or hydrogen), anneal the glass to allow the bonds to relax to form sites sized for lithium then substitute potassium back in where the lithium is to make it even stronger.

As far as what make each generation of GG stronger, I think it is base glass improvement such as modifying the aluminum content or making the glass thinner or making the ion diffusion more local to the surface and preventing the potassium from penetration too far.


Option C is correct.

You could start by looking up the definition of a coordinate bond.

Here's a tiny excerpt from the above link:

  A coordinate bond is a kind of 2-center, 2-electron covalent bond in which the two electrons derive from the same atom.

In your case, as you have correctly noted yourself, the $\ce{NH3}$ molecule will have an "lone pair" of electrons. However, your judgement that $\ce{HF}$ has none is quite incorrect– the fluorine atom has $3$ lone pairs. (not that it's very relevant to the question, but you can look up the structure of $\ce{HF}$ here)

Since the $\ce{NH3}$ is a lewis base, its lone pair of electrons would abstract the $\ce{H}$ from $\ce{HF}$.

$\ce{NH4+}$ has four equivalent bonds. You could call one of them the "dative" bond, but there's really no way to distinguish them from each others in general cases.

The concept of dative bond comes here just because the bond between nitrogen and the hydrogen from $\ce{HF}$ involves 2 electrons (like all single bonds), both of which are contributed by the nitrogen atom (unlike most single bonds).

(Image source)


How to properly model a mechanochemical reaction of barium titanate and copper(II) complexes?

Reaction rates can be calculated from the Arrhenius equation $k=Ae^{E_a/RT}$ where $E_a$ is the activation energy found at the transition state.

Please check out the Reference 1 for calculating transition states under force. Following is an excerpt from the paper (Ref.1):

  As a force is delivered to the mechanophore, the energy provided by the work can be quantified by $\Delta E =−F\Delta x$ where $\Delta E$ is the change in energy, $F$ is the amount of force applied, and $\Delta x$ is the change in distance over which the force is applied ($\Delta x$  in the examples of this article will correspond to the distance between the two atoms where the tensile force is applied…

We have recently developed a python version of the code.

Force can also be added along bonds to simulate ultrasound and we have some work in preparation showing this for polymers. Essentially ultrasound creates bubbles which pop and drag the polymer in opposite directions.

References:

Allison G. Roessler, Paul M. Zimmerman, “Examining the Ways to Bend and Break Reaction Pathways Using Mechanochemistry,” J. Phys. Chem. C 2018, 122(12), 6996-7004 (https://doi.org/10.1021/acs.jpcc.8b00467).