<h2>Solution 1:</h2>
I agree with @andselisk that this question is quite broad. I will focus on two specific questions asked

<blockquote>
 The only equation I know for x-ray diffraction is Bragg's Law but is this the only equation used to interpret the data? [...] How do you translate the spots on a detector to electron density plots using Braggs law? 
</blockquote>

Apart from Bragg's law (which tells you where the diffraction spots are for a known orientation of a crystal with know unit cell), it was also known that real space (electron density) and reciprocal space (diffraction pattern) are related by 3D Fourier transform. For structures containing one or two atoms, just knowing the unit cell parameters and the symmetry is enough to get the entire structure. For anything slightly more complicated, the Fourier transform of amplitudes or later, of intensities (Patterson methods, http://reference.iucr.org/dictionary/Patterson_methods), had to be used. The first crystal structure analyses were of crystals with centrosymmetry, where the phase problem is easier to solve.

<blockquote>
 My question is what is the general process for doing this without computer software?
</blockquote>

Diffraction data was measured on film, with gray-scales to assess intensity of signals. To calculate a Fourier transform, pre-computed tables were used, such as the Beevers-Lipson strips. As Andselisk commented, Fourier transform was used late in the 20s, and initially for problems that were one- or two-dimensional. There is a nice account by P.P. Ewald available here: https://www.iucr.org/publ/50yearsofxraydiffraction/full-text/structure-analysis
This was written at a time when the myoglobin structure had just been solved, after "50 years of X-ray diffraction".

<h2>Solution 2:</h2>
Not just in the '20s but up to the '90s at least, the d-spacings were estimated by hand measurements of diffractometer peaks or film lines and applying the Bragg formula. d-spacings were then ranked from the most intense down to the least intense. Starting with the 2 most intense d-spacing values - and allowing +/- 0.02 angstrom error margin to each - a process of searching the Hanawalt Search Manuals was done till a chemically plausible phase with these two values was found. If no plausible phase was found for these two most intense d-spacings, the second value was replaced with each of the third/fourth/fifth/etc most intense d-spacing value and another search made till a chemically plausible phase was found. Once located, the remaining less intense d-spacings listed for this phase were compared with the remaining calculated d-spacings and if found this was supporting evidence for this phase. Remaining d-spacings were relisted according to intensity and the process repeated. It always seemed daunting at first but, with a little initial guidance from a more experienced person (usually another postgrad student) and with a shortlist of all plausible phases plus their d-spacings beside you, it got easy enough in metallurgy and ceramics. It must have been much harder in biochemistry but I guess those working in that métier had their own shortcuts.

For more detailed work, e.g. for residual stress calcs, grain texture effects, part-crystalline/part-amorphous phases, exploration of short-range ordering in amorphous phases, etc one referred to texts by <a rel="nofollow noreferrer">Cullity and <a rel="nofollow noreferrer">Azaroff. Beyond that you had to look at papers by other researchers, talk to x-ray lab technicians and above all try to look at the problem in the way that was most natural to oneself or most amenable to the nature of the problem.

Chemistry - How were x-ray diffraction patterns deciphered before computers?

<h2>Solution 1:</h2>
The $\mathrm{pH}$ of pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{CO2}}= 10^{-3.5}\ \mathrm{atm}$) can be calculated as follows.

$$[\ce{H2CO3^*}]=K_\mathrm H\cdot p_{\ce{CO2}}$$

where $[\ce{H2CO3^*}]$ is the total analytical concentration of dissolved $\ce{CO2}$, i.e. $[\ce{H2CO3^*}]=[\ce{CO2(aq)}]+[\ce{H2CO3}]$, and 
$K_\mathrm H= 3.39\times10^{-2}\ \mathrm{mol\ l^{-1}\ atm^{-1}}$ is Henry's law constant for $\ce{CO2}$.

$$\begin{align}
\log[\ce{H2CO3^*}]&amp;=\log K_\mathrm H+\log p_{\ce{CO2}}\\
&amp;=-1.5-3.5\\
&amp;=-5.0
\end{align}$$

The commonly used first acid dissociation constant of carbonic acid $\mathrm pK_{\mathrm a1}=6.3$ (at $25\ \mathrm{^\circ C}$) actually is a composite constant that includes both the hydration reaction 
$$\ce{H2O + CO2(aq) &lt;=&gt; H2CO3}$$
and the protolysis of true $\ce{H2CO3}$
$$\ce{H2CO3 &lt;=&gt; H+ + HCO3-}$$
For a weak acid
$$\begin{align}
\log[\ce{H+}]&amp;\approx\frac12\left(\log K_\mathrm a+\log[\ce{H2CO3^*}]\right)\\
&amp;=\frac12\left(-6.3-5.0\right)\\
&amp;=-5.65\\
\mathrm{pH}&amp;=5.65
\end{align}$$

Thus, pure rain in equilibrium with the atmosphere has about $\mathrm{pH}=5.65$. Any acid rain with lower $\mathrm{pH}$ would be caused by additional acids.

<h2>Solution 2:</h2>
You are forgetting an important component of the air: carbon dioxide. When it dissolves in pure water (=rain water), it makes it acidic. It is not considered that harmful.

Acid rain has a negative connotation; it is mainly caused by anthropogenic activities. The low pH of acid rain is due to sulfur oxides and nitrogen oxides and it is indeed below 5.7.

IUPACs definition "Rain with pH values &lt; about 5; commonly results from acids formed from pollutants. 'Pure' rain water equilibrated with atmospheric CO2 and naturally occurring acids in relatively clean air usually has a pH&gt;5."

<hr>
<h2>Solution 3:</h2>
Under atmospheric pressure, dissolved carbon dioxide can reach an equilibrium state in water that yields a pH of as low as 5.7

Chemistry - Why is it considered acid rain with pH <5.6?

<h2>Solution 1:</h2>
Karsten's answer is excellent, but here is a figure that shows the mathematics involved:

<img src="https://i.stack.imgur.com/GHIY6l.jpg" alt="Tetrahedral angle">

The central atom (green) is at the center of the cube, the four other atoms (purple) are at alternating vertices and the geometry should be clear.

Alternatively, if you orient the molecule so one peripheral (purple) atom is directly "above" the central (green) atom, then each of the other three atoms is just $1 - \theta$ ($\approx 70.52877940 ^\circ$) away from being directly "under" the central atom, so each contributes $\cos(1 - \theta)$ times the bond dipole. This is the downward component of the bond dipole from one of the lower atoms.

But $\cos(1 - \theta) = 1/3$, so this is simply (bond dipole)/3, and there are three of these lower atoms, so the three downward components exactly balance the one upward component.

<h2>Solution 2:</h2>
Orientation 1

Consider the orientation of methane below. The partial positive charges are distributed equally around the central atom. Comparing the left with the right side of the molecule, the top left and top right cancel each other out, and the other two are in the middle with the carbon. Comparing the top with the bottom side of the molecule, the two top hydrogen positions cancel out with the two bottom hydrogen positions. Comparing the front with the back side of the molecule, the bottom front and the bottom back hydrogen positions cancel each other out and the other two are in the plane of the paper.

For a polar molecule, the positive partial charges have to be separated from the negative ones along a direction (along the overall dipole moment, which is a vector). Here, the positive partial charges are on the outside and the negative partial charges are on the inside.

<img src="https://i.stack.imgur.com/mgU1J.gif" alt="enter image description here">

If you like vectors, you could also say the two top bond dipole moments added up point straight upwards, while the two bottom dipole moments added up point straight downwards. If you add those two up, you get a net dipole of zero.

If you like symmetry arguments, there are six mirror planes along the H-C-H planes (the left-right and the front-back are easy to see), so there can't be a dipole. The argument is as follows: If there were a dipole moment, for example left to right, and I apply the left-right mirror plane, it would have to switch directions, but the molecule is still the same. Because of this contradiction, the dipole moment has to be zero in that direction. The same argument goes for the three-fold and two-fold axis in the molecule.

Orientation 2

<blockquote>
 Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to 0. 
</blockquote>

The upper H atom is straight up, while the downward H atoms go at an angle, with the component down being 1/3 a bond length. So in that orientation, it also cancels out, but is more difficult to believe.

<img src="https://i.stack.imgur.com/GRRG1t.png" alt="enter image description here">

Orientation 3

In this orientation (used for Fisher projections in sugar chemistry), you can see the symmetry nicely as well one up, one down; one left, one right; two front, two back.

<img src="https://i.stack.imgur.com/xFCmyt.png" alt="enter image description here">

Orientation 4

Maybe my favorite: two up, two down, two left, two right, two front, two back - that is symmetric.

<img src="https://i.stack.imgur.com/ivvHHt.png" alt="enter image description here">

<hr>
<h2>Solution 3:</h2>
<blockquote>
 "In the beginning we saw that [methane molecules] are 3D structures with one H being at the top and other 3 at the bottom, with the bond dipole moment directed from each H towards C. Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to zero."
</blockquote>

So what happens if you turn the molecule around so that one of the other H atoms is now at the top?

Remember that a tetrahedron is symmetric, so that it looks the same when viewed from any of its four corners. So if the tetrahedral methane molecule had a non-zero dipole moment pointing towards one of the hydrogens, by symmetry it would also have to have an equal dipole moment pointing towards all of them. But then it wouldn't be a dipole, but an octupole — a dipole can, by definition, only point in one direction.

Chemistry - Why does a tetrahedral molecule like methane have a dipole moment of zero?

<h2>Solution 1:</h2>
One argument put forward has been that aluminum is very poorly bioavailable, moreso than many other elements. Aluminum oxide is very insoluble in water. In addition, any dissolved aluminum that does form in seawater is likely to be precipitated by silicic acid, forming hydroxyaluminosilicates. 

From Chris Exeter's 2009 article in Trends in Biochemical Sciences:

<blockquote>
 But how has the by far most abundant metal in the Earth's crust remained hidden from biochemical evolution? There are powerful arguments, many of which influenced Darwin's own thinking [15], which identify natural selection as acting upon geochemistry as it acts upon biochemistry. I have argued previously that the lithospheric cycling of aluminium, from the rain-fuelled dissolution of mountains through to the subduction of sedimentary aluminium and its re-emergence in mountain building, depends upon the ‘natural selection’ of increasingly insoluble mineral phases of the metal [7]. The success of this abiotic cycle is reflected in the observation that less than 0.001% of cycled aluminium enters and passes through the biotic cycle. In addition, only an insignificant fraction of the aluminium entering the biotic cycle, living things, is biologically reactive. However, my own understanding of such an explanation of how life on Earth evolved in the absence of biologically available aluminium was arrived at by a somewhat serendipitous route! In studying the acute toxicity of aluminium in Atlantic salmon I discovered that the aqueous form of silicon, silicic acid, protected against the toxicity of aluminium [16]. Subsequent work showed that protection was afforded through the formation of hydroxyaluminosilicates (HAS) [17] which, intriguingly, are one of the sparingly soluble secondary mineral phases of the abiotic cycling of aluminium! The discovery that silicic acid was a geochemical control of the biological availability of aluminium, though now seemingly obvious in hindsight, was a seminal moment in my understanding of the bioinorganic chemistry of aluminium, and although it helped me to understand the non-selection of aluminium in biochemical evolution, it also provided me with a missing link in the wider understanding of the biological essentiality of silicon.
</blockquote>

Dr. Exeter is one of the few scholars who appears to have written in depth about this issue. Thus, perhaps it is fair to say that (a) your question doesn't have a definitive answer, but (b) the poorly accessible nature of aluminum over geological time due to its interaction with and precipitation by silicic acid is the leading hypothesis.

It's worth noting that when aluminum is artificially introduced into metalloenzymes in place of naturally occuring metals, the resulting alumino-enzymes can retain activity, as a 1999 article in JACS by Merkx &amp; Averill shows.

<h2>Solution 2:</h2>
The metal’s trivalency is certainly not an issue. Iron and cobalt form trivalent compounds (in the $\mathrm{+III}$ oxidation state) in many of their biologically relevant complexes. In fact, in metalloprotein surroundings the actual ‘valency’ (oxidation state) is not so much of a deciding factor; what matters is the number of ligands and the required rigidity of the resulting complex. While some metals prefer octahedral complexes and others tetrahedral ones, many can take either shape and switching between these is inherent to catalytic activity of many metalloproteins. Aluminium has an all-empty d shell and in its compounds is spherically symmetric (only core shells filled) so it should allow for a flexible coordination environment—like zinc.

One thing that aluminium certainly cannot become is a redox catalyst: in compounds it is practically always $\ce{Al^3+}$. But zinc is likewise practically always $\ce{Zn^2+}$ and yet used in a number of (non-redox) metalloproteins. So that’s not the issue. In fact, some enzymes’ catalytic activities might (as a thought experiment) be increased when zinc is replaced by aluminium because the latter has a higher charge and thus might be able to better catalyse hydroxide generation. On the other hand, aluminium typically displays slower ligand exchange rates than other comparable metal ions (exemplified by the fact that it is precipitated as a hydroxide when its solutions are alkalised with ammonia) reducing its catalytic utility.

Reading that, one might well be inspired to think of aluminium as a structural metal, assisting in the correct folding of proteins. Again, there is nothing intrinsic to aluminium stopping it from doing this and it should fare comparable to other redox-innocent metals.

Obviously, the enzymes living organisms use nowadays are optimised towards using non-aluminium metals, so replacing one with aluminium will usually result in a less active catalyst or less-stabilised structure. But if aluminium had been used, enzyme structures would have evolved accordingly to allow for the effective use of aluminium in active cores and/or as structural metals.

As already hinted at by the other answer, aluminium has a low bioavailability. In fact, at $\mathrm{pH\,7}$ aluminium is predominantly precipitated as $\ce{Al(OH)3}$ but partially redissolving as $\ce{[Al(OH)4]-}$. At slightly lower pH values—typical of the surroundings of most living beings—it is almost entirely precipitated as $\ce{Al(OH)3}$ so practically biounavailable. Of course, specialised ligands can bring it into solution but these need to be evolved first. The problem of silicates further complicates bioavailability.

In conclusiong, I will agree with the already posted answer that bioavailability, especially low solubility at typical pH values are the most likely answer; when compared to e.g. iron, the latter’s redox abilities make it much more attractive for use in enzymes.

Chemistry - Why isn't aluminium involved in biological processes?

<h2>Solution 1:</h2>
First, hypochlorite ($\ce{ClO−}$) is not a strong oxidizing agent compared to other chloride oxyanions, at least in acidic medium. Under standard conditions in acidic medium, chlorous acid ($\ce{HClO2}$) is the best oxidizing agent among them due to the largest positive number for standard electrode potential:

$$\begin{align}
\ce{HClO2 + 2H+ + 2 e- &amp;&lt;=&gt; HClO + H2O} &amp;\quad E^\circ &amp;= \pu{1.645 V} \label{rxn:1}\tag{1}\\
\ce{HClO2 + 3 H+ + 3 e- &amp;&lt;=&gt; 1/2 Cl2 + 2 H2O} &amp;\quad E^\circ &amp;= \pu{1.628 V} \label{rxn:2}\tag{2}\\
\ce{HClO2 + 3 H+ + 4 e- &amp;&lt;=&gt; Cl– + 2 H2O} &amp;\quad E^\circ &amp;= \pu{1.570 V} \label{rxn:3}\tag{3}\\
\end{align}$$

Keep in mind that in the reduction half-reaction \eqref{rxn:1}, $\ce{HClO2}$ reduced to $\ce{HClO}$ with its largest positive standard electrode potential $(E^\circ = \pu{1.645 V})$ among chlorine oxoacids. For comparison, we should look at half-reactions of which each of these two reagents reducing to the same product. Thus, look at half-reactions \eqref{rxn:2} versus \eqref{rxn:4}, and \eqref{rxn:3} versus \eqref{rxn:5} where $\ce{HClO2}$ shows its superiority over $\ce{HClO}$:

$$\begin{align}
\ce{HClO + H+ + e- &amp;&lt;=&gt; 1/2 Cl2 + H2O} &amp;\quad E^\circ &amp;= \pu{1.611 V} \label{rxn:4}\tag{4}\\
\ce{HClO + H+ + 2 e- &amp;&lt;=&gt; Cl– + H2O} &amp;\quad E^\circ &amp;= \pu{1.482 V} \label{rxn:5}\tag{5}\\
\end{align}$$

Perchlorate ($\ce{ClO4-}$) and chlorate ($\ce{ClO3-}$) ions, on the other hand, are also good oxidizing agents, but not as strong as $\ce{HClO2}$ or $\ce{HClO}$, yet closed. This fact is clear when compared to the half-reactions where they directly reduce to $\ce{Cl2}$ or $\ce{Cl-}$ similar to $\ce{HClO2}$ and $\ce{HClO}$:

$$\begin{align}
\ce{ClO4- + 8H+ + 7e- &amp;&lt;=&gt; 1/2 Cl2 + 4H2O} &amp;\quad E^\circ &amp;= \pu{1.39 V} \label{rxn:6}\tag{6}\\
\ce{ClO4- + 8H+ + 8e- &amp;&lt;=&gt; Cl– + 4H2O} &amp;\quad E^\circ &amp;= \pu{1.389 V} \label{rxn:7}\tag{7}\\
\ce{ClO3- + 6H+ + 5e- &amp;&lt;=&gt; 1/2 Cl2 + 3H2O} &amp;\quad E^\circ &amp;= \pu{1.47 V} \label{rxn:8}\tag{8}\\
\ce{ClO3- + 6H+ + 6e- &amp;&lt;=&gt; Cl– + 3H2O} &amp;\quad E^\circ &amp;= \pu{1.451 V} \label{rxn:9}\tag{9}\\
\end{align}$$

However, the standard reduction potentials OP is mentioning is for different reduction half-reactions $\ce{ClO3-}$ and $\ce{ClO4-}$ involved (\eqref{rxn:12} and \eqref{rxn:10}, respectively):

$$\begin{align}
\ce{ClO4- + 2H+ + 2e- &amp;&lt;=&gt; HClO3 + H2O} &amp;\quad E^\circ &amp;= \pu{1.189 V} \label{rxn:10}\tag{10}\\
\ce{ClO3- + 2H+ + e- &amp;&lt;=&gt; ClO2 + H2O} &amp;\quad E^\circ &amp;= \pu{1.152 V} \label{rxn:11}\tag{11}\\
\ce{ClO3- + 3H+ + 2e- &amp;&lt;=&gt; HClO2 + H2O} &amp;\quad E^\circ &amp;= \pu{1.214 V} \label{rxn:12}\tag{12}\\
\end{align}$$

Accordingly, when consider Latimer diagrams $(\ce{ClO4- -&gt; ClO3- -&gt; HClO2 -&gt; HClO -&gt; Cl2 -&gt; Cl-}),$ you'd see, $\ce{HClO2}$ cannot oxidizes $\ce{ClO3-},$ although it is the strongest oxidizing agent. However, $\ce{HClO2}$ can oxidize another species, which $\ce{ClO3-}$ cannot oxidize.

Note that all reduction potential values are from: http://depa.fquim.unam.mx/amyd/archivero/TablasdepotencialesREDOX_26700.pdf (Petr Vanẏsek), which cited the following references:

<ol>
<li>G. Milazzo, S. Caroli, V. K. Sharma, Tables of Standard Electrode Potentials; Wiley, Chichester, 1978.</li>
<li>A. J. Bard, R. Parsons, and J. Jordan, Standard Potentials in Aqueous Solutions; Marcel Dekker, New York, 1985 (Series: Monographs in electroanalytical chemistry and electrochemistry).</li>
<li>S. G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at $\pu{298.15 K}$,” J. Phys. Chem. Ref. Data 1989, 18(1), 1-21 (https://doi.org/10.1063/1.555839).</li>
</ol>

<h2>Solution 2:</h2>
Besides the the thermodynamic aspects discussed by Matthew, consider the kinetic aspects. Oxidation by a chlorine oxyanion involves displacement of oxygen from its bond with the chlorine. Such a displacement, in a protic solvent such as water, must involve protonation of the oxygen:
<ul>
<li>Oxygen without the proton would have to be displaced as oxide ion, which would be a high-energy intermediate in any protic solvent.
</li>
<li>In this article from the University of Utah, molecular orbitals for some phosphorous-oxygen species are discussed. These show that oxygen can back-donate its $p$ electrons to phosphorous, strengthening the bond. A similar back-donation to chlorine would make the oxygen harder to displace in a redox reaction, but protonation would cut down on this back-donation.
</li>
</ul>
Given these considerations, oxidation by a chlorine oxyanion is kinetically favored if the oxygen can act as a base and thus form a species that is more easily displaced than a bare oxide ion. Hypochlorite ion does that relatively well, whereas in higher oxyanions the needed basic character weakens rapidly unless we force the issue with a strongly acidic solution.

Chemistry - Why is hypochlorite a stronger oxidizing agent than other oxychlorides?

<h2>Solution 1:</h2>
Most likely, mercury was used as a colorant, specifically red. This list shows mercury providing a red color in flame tests, and the same would presumably be seen in fireworks.

Several other, less dangerous elements also give red or reddish colors. This site identifies lithium as providing the color in red fireworks, and also mentions lithium carbonate specifically.

Be safe this 4th, handle fireworks with care if you do so at all!

<h2>Solution 2:</h2>
In "Fireworks, the Art, Science, and Technique", Takeo Shimizu gives a very good explanation of color production. For blue, the color-producing firework chemical is thought to be copper monochloride. If there is no free chlorine in the flame, there can be no blue color.The discovery of colors produced by chlorine along with other elements belongs to Chertier, who described the phenomena in 1836.In years past, mercurous chloride, also known as calomel, was used. Calomel decomposes at ordinary flame temperatures into chlorine and mercury metal vapor. Along with potassium chlorate ,mercurous chloride(calomel) was used as the oxidizer in older blue firework star compositions, calomel was often included to insure that plenty of chlorine was available.
Today we know that mercury is a very bad actor in the environment, witnessed by the birth defects in Japan resulting from mercury poisoning, not to mention the deaths!
<h2>Making Pharaoh's Snakes</h2>
<img src="https://i.stack.imgur.com/yP2dL.png" alt="enter image description here">
Tomasz Szymborski, Creative Commons License
Ignite a small pile of mercury(II) thiocyanate, $\ce{Hg(SCN)2}$.Mercury thiocyanate is an insoluble white solid which can be purchased as a reagent or can be obtained as a precipitate by reacting mercury(II) chloride or mercury(II) nitrate with potassium thiocyanate (All mercury compounds are toxic and probably banned for this reason).
<h2>Pharaoh's Snakes Chemical Reaction</h2>
<ul>
<li>Igniting mercury(II) thiocyanate causes it to decompose into an insoluble brown mass that is primarily carbon nitride, $\ce{C3N4}$. Mercury(II) sulfide and carbon disulfide are also produced.</li>
</ul>
$$\ce{2Hg(SCN)2 → 2HgS + CS2 + C3N4}$$
<ul>
<li>Flammable carbon disulfide combusts to carbon(IV) oxide and sulfur(IV) oxide:</li>
</ul>
$$\ce{CS2 + 3O2 → CO2 + 2SO2}$$
<ul>
<li>The heated $\ce{C3N4}$ partially breaks down to form nitrogen gas and cyanogen:
$$\ce{C3N4 → 3(CN)2 + N2}$$</li>
<li>Mercury(II) sulfide reacts with oxygen to form mercury vapor and sulfur dioxide. If the reaction is performed inside a container, you will be able to observe a gray mercury film coating its interior surface.</li>
</ul>
$$\ce{HgS + O2 → Hg + SO2}$$

Chemistry - What purpose does mercury dichloride have in fireworks?

<h2>Solution 1:</h2>
How do I get into method development?

As was said in the comments, one would join a group that works in this field and learn from them. From up close, there is some derivation from previous results and some trial and error involved. And yes, a PhD-sized investment is typically required.

How do I get into method implementation?

It depends on how deep you wish to go. There are open-source projects to look at and contribute to. There are publicly available exercises on implementing HF and beyond. I would not advise to implement DFT from scratch because the numerical integration grid is both critical and difficult. Same is true for the integrals, use a library for them.

How do I learn to use Program X?

It is helpful to learn one program suite well and learn others later. The reason is that many concepts are similar between the programs. Learning one program comes down to locating good worked examples, reading the manual, attending a workshop if possible. One may for instance attempt to reproduce data from a DFT benchmark, since the supporting material usually has a detailed listing, possibly down to absolute energies.

<h2>Solution 2:</h2>
The Psi4 project recently published a set of Python notebooks and tutorials on a wide variety of methods, called Psi4NumPy- J. Chem. Theory Comput. 2018 14(7) 3504-3511 - 

All the code is available as open source on Psi4NumPy GitHub:

<ul>
<li>Hartree Fock including RHF vs. UHF, DIIS, etc.</li>
<li>Density Functional - including grids</li>
<li>Properties</li>
<li>Geometry Optimization</li>
</ul>

Generally these are well-written and give significant extra details. Moreover, they're live code with Psi4, so you can test out the actual implementation.

Beyond the tutorials, there are scripts indicating implementations of coupled cluster, MP2, response theory, etc.

I think anyone starting out or who wants more details with a readable introduction should take a look.

Chemistry - Developing Computational Chemistry Software

Every adsorbent has a saturation capacity depending on its active surface area. The idea to replace carbon filter is not only due to the saturation capacity but imagine fungi, bacteria may be adsorbed as well. It is worth replacing.

Chemistry - Dangers of using activated carbon past its lifespan

<h2>Solution 1:</h2>
Your intuition is on point. Pretty much all the "lifehacks" praising cleaning abilities of vinegar and sodium bicarbonate are ineffective. The reaction appears vigorous due to the formation of carbon dioxide gas, but what is left is a diluted solution of sodium acetate:
$$\ce{CH3COOH(aq) + NaHCO3(aq) -&gt; NaCH3COO(aq) + H2O(l) + CO2(g)}$$
Sodium acetate doesn't have any pronounced cleaning abilities. It doesn't act like a surfactant, doesn't drastically increase solubility of the impurities, or participate in complexation helping to dissolve metal-containing compounds.
As for unclogging abilities—yes, this is something that my grandmother used to use, but these days there are much better alternatives.
The idea was to pour some hot water into the system and then put a couple of spoons of baking soda into the drain creating warm concentrated solution in the siphon.
This solution would also be mildly basic due to hydrolysis and help to dissolve some fat and thin hair:
$$\ce{HCO3^-(aq) + H2O(l) &lt;=&gt; H2CO3(aq) + OH-(aq)}$$
After some time (approx. 30 min) it would make sense to add some vinegar hoping the reaction would cause mixing up the constituents of the clog and excessive pressure would help to push the waste further down the drain.

<h2>Solution 2:</h2>
Better is using washing soda alone first ( instead of baking soda ), pouredby boiling water and kept there overnight. Next day vinegar may be used to make some bubbles desintegrating the clogging.
Washing soda has cleaning ability by itself and has better solubility than baking soda. Adding vinegar later is not to produce some extra powerful chemical cleaner, but it's effect is mechanical. It is producing CO2 bubbles and pressure to tear and push clogging, hopefully weakened by prior soda action.
Detergents then make the clog desintegration easier.
An even stronger cleaner is sodium hydroxide, that is often used as well. But is cannot be used alone, without soda, to produce CO2 with vinegar.
Note that majority of clogging is caused by protein a/o fat rich materials, like hairs, that are chemically dissolved in alkaline solutions.

Chemistry - Cleaning with vinegar and sodium bicarbonate

<blockquote>
First of all, how can a "half" sigma bond exist?
</blockquote>
Usually, you expect double bonds to be shorter and stronger than the corresponding single bonds, and triple bonds even shorter and stronger. The OP already mentioned bond-orders of 1.5 that occur for conjugate double bond systems, and those have properties in between single and double bonds.
Perhaps the most famous case for a "half" sigma bond is the hydrogen molecule ion, $\ce{H2+}$. It has a bond distance of 100 pm (compare 74 pm for dihydrogen) and a bond energy of 2.77 eV (compare 4.52 eV for dihydrogen). Source: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm
So in terms of comparing the bonding in the hydrogen molecule ion with the hydrogen molecule, it makes sense to talk of a partial bond in terms of bond length and strength. It might make sense to talk of "half" a sigma bonds because the single bond has 2 electrons, but in this case there is only one.
The reason this case is famous, at least in courses introducing quantum chemistry, is that it is a single-electron system, so easier to work with theoretically and computationally than the dinitrogen cation.
<blockquote>
Then I wrote down the molecular configuration using MOT. I noticed that the last electron in N2+ goes into σ2p orbital.
</blockquote>
Here is the molecular orbital diagram, and a depiction of the highest occupied molecular orbital.
<img src="https://i.stack.imgur.com/3FEQO.jpg" alt="enter image description here">
Source: https://www.chemtube3d.com/orbitalsnitrogen/
As you can see, this orbital has two nodes. In comparison, the orbitals corresponding to the pi bonds have only one node. Using arguments similar to the ordering of orbitals in Hueckel theory, you would expect the sigma orbital to be higher in energy.
However, comparing the energies of states in all diatomics, you see that it is more complicated (the difference between the energy of 2s and 2p changes across the 2nd period elements, and with it the amount of mixing between 2s and 2p of same symmetry):
<img src="https://i.stack.imgur.com/f1UGI.jpg" alt="enter image description here">
Source: https://chem.libretexts.org/Courses/Heartland_Community_College/HCC%3A_Chem_161/9%3A_Molecular_Geometry_and_Bond_Theory/9.8%3A_M.O._Theory_and_the_Period_2_Diatomic_Molecules
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[OP in comments] However why would the loss of electron be from a stable bond pair? Why cant it be one of the lone electrons? Also why does it have to be a sigma bond and not a pi bond, if at all the loss is not from one of the lone pairs?
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You are using the valence bond model (or the Lewis structure) to "name" electrons. There is no 1:1 mapping of "lone pairs" to the molecular orbital diagram. It does say that the two combination of 2s orbitals "cancel each other out", i.e. are non-bonding; however, the shape of these orbitals does not match the valence bond model, where sp-orbitals pointing away from the center of the molecule are counted as lone pairs.

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