Kopecký et.al. indeed utter the word unstable, but not in the sense of its recommended definitions, which were established later.

  Although the parent compound, isoindole (I), has resisted isolation, the existence of this unstable species has been shown by trapping with dienophiles (i).

The problem here is that stability always requires a point of reference, which is not given in the quote above.
I would assume that the authors use unstable (or instability) as a way to say the compound cannot be isolated. They go on and justify why this compound behaves in this way, describing what I would say is its reactivity. While their argument is sound and I believe correct, the choice of their wording is a little bit ambiguous (in today's terms).
We clearly need to distinguish a compounds reactivity (towards a certain reaction) from its stability within a common set of parameters. As such the following statements are correct, but neither of them is a justification for the other.

2H-Isoindole is a less stable $\ce{C8NH7}$ isomer than 1H-indole.
2H-Isoindole is more reactive towards dienophiles than 1H-indole.

While statement one is easy enough to prove computationally, it is a lot harder to explain or analyse. Quick calculations on the DF-M06L/def2-TZVPP level of theory gives the following hypothetical reaction energy:
$$\begin{align}
\ce{1$H$-indole &-> 2$H$-isoindole} & \Delta_\mathrm{r}G^\circ &= +38.6~\pu{kJ mol^-1}
\end{align}$$

Unfortunately Kopecký et.al. offer no explanation for this, because they are more or less focused on explaining why it is "unstable", by which they most likely mean highly reactive or not isolatable. They are concerned with the reaction paths that lead away from this species and find that 1H-isoindole plays a big (probably most important) role in this.

  The [...] electron deficiency for position 1 of the isoindolenine [...] leads us to conclude leads us to conclude that the reactive species responsible for the instability of this series of compounds probably is the isoindolenine [...].
  (Note that the numbering differs from the IUPAC name and the position referred to as 1 is actually 3.)

I think this analysis is spot on a holds up to modern standards of computational methodology. The 1H-tautomer indeed opens up a very big variety of decomposition and reaction pathways, and it is readily accessible at room temperature (DF-M06L/def2-TZVPP):
$$\begin{align}
\ce{2$H$-isoindole &-> 1$H$-isoindole} & \Delta_\mathrm{r}G^\circ &= +3.4~\pu{kJ mol^-1}
\end{align}$$

The reason for this is - and this is the point where Joule and Mills get it completely wrong - the retention (not the creation) of the aromatic system in the benzene moiety. The chemical behaviour (reactivity towards cycloadditions) is not determined by the lack of a 'complete' benzene ring. But that is us slipping into statement 2, which we are not quite ready yet.
Let us first dissect the latter statement in the same book (p.447), which is equally as wrong:

  Isoindole, benzo[c]thiophene and isobenzofuran are much less stable than their isomers, indole, and benzo[b]thiophene and benzo[b]furan.

Although they offer no appreciable effort to prove this statement they are probably correct. At least the calculations for the $\ce{C8NH7}$ system confirm that. They even use the correct (or recommended) definition of stable in this case.

  This is undoubtedly associated with their lower aromaticity, which can be appreciated qualitatively by noting that in these [c]-systems, the six-membered ring is not a complete benzenoid unit.

Again, they offer no proof for that other than a hand-wavy argument based on a single Lewis-structure. If we include resonance into the mix then we have an at least as reasonable argument as this to disprove it.

If we have a look at the molecular orbitals of the compounds, we will see that there is more to this than just one resonance description. I have pulled out two characteristic molecular orbitals for example, but the complete set can be found here. On the left side (top) we have 1H-indole and on the right side (bottom) we have 2H-isoindole. Apart from a light shift of the electron density due to the loosened symmetry in indole, the orbitals are pretty much the same.

When we take the 1H-isoinole isomer into the mix, we see that the aromatic system of the benzene ring is retained completely.

Aromaticity is still not a fully understood concept and as a result there is not really a rigorous concept that can be applied (see the definitions below). There will probably be debate about it for the next decades. (Hopefully my bounty will pay off and we can find out more in this question.)
As a result we don't really have a handle on judging whether or not 2H-isoindole has a lower aromaticity than 1H-indole. And according to Kopecký's paper, with available methods at this time, this cannot be backed up with theoretical results.

While I hoped to find some experimental evidence for the existence of isoindole due to the general progress within the last fifty years, I was still unable to find a single source.

I'll skip over the rest of the quote because the stated facts are certainly accurate, just the reasoning with the 'incomplete' benzene ring is false.

Let's look at statement two for a while.
The reactivity of 2H-isoindole towards dienophiles is very easily explained looking at the HOMO of the compound; for comparison the HOMO of butadiene on the right (bottom).

The very easy rationale is that the HOMO is of the right symmetry for a Diels-Alder reaction, which additionally retains the aromatic character of the benzene moiety.
Now a look at the HOMO (left/ top) of 1H-indole will tell us that the only viable attack can be carried out on the 4,7-positions, leaving a much smaller aromatic moiety, i.e. pyrrole, than isoindole. The only way for an inverse electron demand  Diels-Alder reaction would also be in the benzene moiety as the LUMO (right/ bottom) shows.

This is why one molecule reacts readily, while another has to be forced (really, really hard) or reacts as the dienophile via 'the double bond in the pyrrole system'.

After all that information the question still remains, or rather the question has become

"Why is isoindole less stable than indole?"

As I said before, this is really not easy to interpret. In general, lifting symmetry restrictions results in a more stable configuration, as fewer boundary conditions equal more flexibility and that usually gives a lower energy.
One aspect is certainly also that the lone pair of the nitrogen will be better delocalised in indole as there are more carbon atoms in proximity. This is actually something visible in the molecular orbitals.
Apart from that, neither geometries differ much from idealised benzene ($\mathbf{d}(\ce{C-C})=138.7~\pu{pm}$) or pyrrole($\mathbf{d}(\ce{C-N})=136.7~\pu{pm}$, $\mathbf{d}(\ce{C-C'})=137.1~\pu{pm}$, $\mathbf{d}(\ce{C'-C'})=141.4~\pu{pm}$, picture), so that a significant distortion energy is probably not a large contributing factor.

Other than that, I don't know. I also cannot explain, why the other aromatic $\ce{C8NH7}$ isomers are even less stable than isoindole. That would probably require more extensive calculations and analyses.
\begin{array}{lr}\hline
\text{Isomer}& \Delta E_\mathrm{el}/\pu{kJ mol^-1} & \Delta G/\pu{kJ mol^-1}\\\hline
\text{1$H$-indole} & 0.0 & 0.0\\
\text{2$H$-isoindole} & 37.3 & 38.6\\
\text{indolizine} & 52.0 & 50.6\\
\text{1$H$-cyclopenta[$b$]pyridine} & 83.4 & 82.4\\
\text{2$H$-cyclopenta[$c$]pyridine} & 79.7 & 79.1\\\hline
\end{array}

  By this logic, I would expect other carbons in this situation, $\ce{C_{ar}-\color{\red}{C}=N}$, to also be very electron-deficient. However, aromatic Schiff bases seem pretty common, for example salen. What am I missing?

It is true that the carbon atoms in question will be very electron deficient, and they will probably be a good attach point for nucleophiles. In the salen ligand the nitrogens are probably stabilised by intramolecular hydrogen bonds, and while in complexes the entropy factor should make is very stable, too.

On another note, the isoindole moiety is also  very, very common. Just think heme ligands and complexes.

One final thought: Carbonic acid decomposes rapidly in water, however, it is seldom questioned, that the molecule itself is stable under certain conditions.

My personal preferred definition for stable:  An certain arrangement of atoms that forms a local minimum on a potential energy hypersurface and therefore could theoretically probed is stable.

Helpful definitions from the IUPAC gold book

  unstable
  The opposite of stable, i.e. the chemical species concerned has a higher molar Gibbs energy than some assumed standard. The term should not be used in place of reactive or transient, although more reactive or transient species are frequently also more unstable. (Very unstable chemical species tend to undergo exothermic unimolecular decompositions. Variations in the structure of the related chemical species of this kind generally affect the energy of the transition states for these decompositions less than they affect the stability of the decomposing chemical species. Low stability may therefore parallel a relatively high rate of unimolecular decomposition.)

  stable
  As applied to chemical species, the term expresses a thermodynamic property, which is quantitatively measured by relative molar standard Gibbs energies. A chemical species A is more stable than its isomer B if $\Delta_\mathrm{r}G^\circ > 0$ for the (real or hypothetical) reaction $$\ce{A -> B},$$
  under standard conditions. If for the two reactions:
  \begin{align}
\ce{P &-> X + Y} & (\Delta_\mathrm{r}G_1^\circ)\\
\ce{Q &-> X + Z} & (\Delta_\mathrm{r}G_1^\circ)\\
\end{align}
  $\Delta_\mathrm{r}G_1^\circ > \Delta_\mathrm{r}G_2^\circ$, P is more stable relative to the product Y than is Q relative to Z. Both in qualitative and quantitative usage the term stable is therefore always used in reference to some explicitly stated or implicitly assumed standard. The term should not be used as a synonym for unreactive or 'less reactive' since this confuses thermodynamics and kinetics. A relatively more stable chemical species may be more reactive than some reference species towards a given reaction partner.

  aromatic

  In the traditional sense, 'having a chemistry typified by benzene'.
  A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.
  See also: Hückel (4n + 2) rule, Möbius aromaticity
  The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

  aromaticity
  The concept of spatial and electronic structure of cyclic molecular systems displaying the effects of cyclic electron delocalization which provide for their enhanced thermodynamic stability (relative to acyclic structural analogues) and tendency to retain the structural type in the course of chemical transformations. A quantitative assessment of the degree of aromaticity is given by the value of the resonance energy. It may also be evaluated by the energies of relevant isodesmic and homodesmotic reactions. Along with energetic criteria of aromaticity, important and complementary are also a structural criterion (the lesser the alternation of bond lengths in the rings, the greater is the aromaticity of the molecule) and a magnetic criterion (existence of the diamagnetic ring current induced in a conjugated cyclic molecule by an external magnetic field and manifested by an exaltation and anisotropy of magnetic susceptibility). Although originally introduced for characterization of peculiar properties of cyclic conjugated hydrocarbons and their ions, the concept of aromaticity has been extended to their homoderivatives (see homoaromaticity), conjugated heterocyclic compounds (heteroaromaticity), saturated cyclic compounds (σ-aromaticity) as well as to three-dimensional organic and organometallic compounds (three-dimensional aromaticity). A common feature of the electronic structure inherent in all aromatic molecules is the close nature of their valence electron shells, i.e., double electron occupation of all bonding MOs with all antibonding and delocalized nonbonding MOs unfilled. The notion of aromaticity is applied also to transition states.


Metal reagent in dehalogenation of vicinal dihalides

You mean something like the following right
$$
\mathrm{R-X ~+~ Na \longrightarrow R\cdot ~+ ~NaX}
$$
and we add a proton source like an alcohole to transfer an H atom to $\mathrm{R\cdot}$:
$$
\mathrm{R\cdot~ + ~R'-H} \longrightarrow \mathrm{R-H}
$$
Well, its in principle possible. But a) you don't want an explosion right?, so be careful as this will probably be very exothermic b) it will probably be really messy as side products like
$$
2~\mathrm{R\cdot} \longrightarrow \mathrm{R-R}
$$
can be formed. Also the whole approach is not really tolerant to pretty much any functional group and will just tear the substrate apart in many cases. The other alternatives (e.g. Zinc, $\mathrm{SmI_2}$) are just much milder, easier to use and much more versatile. But sure, try it out have fun (less than 0.5 g sodium should still be relatively harmless) ; )


Influence of the neighbouring molecules in a crystal at the example of XeF4

There are two single crystal-structure investigations done at room temperature, both published in 1963. Both suggest that $\ce{XeF4}$ crystallizes in monoclinic system ($P 1 2_1/n 1$) and preserves square-planar geometry in solid phase; however, the angle $\ce{F-Xe-F}$ slightly deviates from $90°$.

Figure 1. Packing of $\ce{XeF4}$ molecules. Color code: $\color{#90E050}{\Large\bullet}~\ce{F}$;$\color{#429EB0}{\Large\bullet}~\ce{Xe}$.
Ibers and Hamilton [1] published a structure with an $R$-factor of $11\%$ (ICSD #27467)

The molecule has symmetry $D_\mathrm{4h}$ within the limits of error of this study, although such symmetry is not required by the space group. The angle $\ce{F1-Xe-F2}$ is $\pu{86 ± 30°}$ and the two independent distances $\ce{Xe}$ to $\ce{F}$ do not differ significantly and have an average value of $\pu{1.92 ± 0.03 Å}$ (Fig. 1). The shortest intermolecular $\ce{F...F}$ contact is $\pu{2.95 Å}$. This and all other intermolecular distances are reasonable; the molecule fills space well, providing further confirmation of the formula $\ce{XeF4}$. Since all of the intermolecular $\ce{F...F}$ contacts in this structure exceed slightly the expected van der Waals contact of $\pu{2.7 Å}$ it is not surprising that a second phase of $\ce{XeF4}$, with a density some $10$ percent higher than that of the present structure, exists (2). The structure of this denser phase is at present unknown.

Fig. 1. The structure of the $\ce{XeF4}$ molecule in the solid.

Templeton et al. [2] did a subsequent XRD experiment and presented improved crystallographic data (ICSD #26626) with $R$-factor of $8.6\%$ and stated that

The structure consists of a molecular packing of square planar molecules of $\ce{XeF4}$.

With a comment on [1]:

Ibers and Hamilton have deduced two structures by refinement of data with $h + k + l$ even. These data do not permit determination of the relative signs of the two $y$ coordinates. One of these two structures is in approximate agreement with our result.

References

Ibers, J. A.; Hamilton, W. C. Xenon Tetrafluoride: Crystal Structure. Science 1963, 139 (3550), 106–107. https://doi.org/10.1126/science.139.3550.106.
Templeton, D. H.; Zalkin, A.; Forrester, J. D.; Williamson, S. M. Crystal and Molecular Structure of Xenon Tetrafluoride. Journal of the American Chemical Society 1963, 85 (2), 242–242. https://doi.org/10.1021/ja00885a038.


Now a days, sulfonephthaleins can be prepared by reaction of readily available saccharin and the desired plenol compound (Ref.1 & Ref.2). In this method, active reagent, sulfobenzoic anhydride will be prepared in situ as depicted in following diagram:

In this method, it's described the preparation of phenolsulfonephthalein (Phenol Red). If you use thymol instead of phenol, you will get thymolsulfonephthalein, which can be subsequently brominated by bromine/acetic acid to obtain dibromothymolsulfonephthalein (Bromothymol Blue). This procedure is described in Ref.2.

References:

V. H. Tillu, D. K. Dumbre, H. B. Borate, R. D. Wakharkar, V. R. Choudhary, "Solvent-Free One-Pot Synthesis of Sulfonephthaleins from Saccharin and Phenols," Synthetic Communications 2012, 42(8), 1101-1107 (https://doi.org/10.1080/00397911.2010.535946).
B. S. Rao, J. B. Puschett, B. M. Karandikar, K. Matyjaszewski, "Synthesis of Functional Bromothymol Blue Dyes for Surface Attachment to Optical Fibers," Dyes and Pigments 1991, 16(1), 27-34 (https://doi.org/10.1016/0143-7208(91)87018-I).


That diagram is misleading because it’s technically in reference to bis(hexamethylbenzene)ruthenium(0). This is a famous example by the late and great E.O. Fischer. This Ru(0) complex is 18e, three methyl resonances were observed for the bent ring at −10 °C. Crystallography later confirmed this. This structure is surprising to the new student, perhaps, because the classic 18e bis(benzene)chromium is sandwiched between two planar arenes.

This is not an organic chemistry question. It is deep in the realm of organometallic chemistry, with a proper explanation requiring a molecular orbital energy diagram and then one has to invoke a Jahn Teller instability argument that requires lowering symmetry from $D_{6d}$ or $D_{6h}$. Suffice it to say, stability is increased by changing the hapticity.


Response factor calculation for gas chromatography

This is a bit convoluted.

A simply dimensional analysis shows that your approach doesn't work. The units of $O$ are unit area. The conversion factor $f$ has units moles per unit area. The variable $k$ is in moles. If you divide the two equations you get a relative sensitivity of $f_2$ as compared to $f_1$, but the value is dimensionless. So each compound has its own response factor. So for two unknowns you have:

$f_1 = \dfrac{O_1}{k_1}$
$f_2 = \dfrac{O_2}{k_2}$

The twist here is the GC typically uses a standard either on a run before the unknown or spiking the unknown sample. So the standard is known to be $x$ ppm and you'll get some area from the run. The hope is that if the gain of the detector drifts them the area of the standard will change too by the same amount. So a 10% gain in area means that the detector gain has drifted up by 10%. Thus measuring area relative to the standard compensates for the drift.

The sensitivity changes a bit from day to day because the ionization in the detector changes due to aging, the gas flow rate isn't 100.0000% the same each time, the column ages and so on.

So the ratio of two unknowns is meaningless. You ratio each of the unknowns to a standard.


Why does ester hydrolysis occur in basic medium

In basic medium, there is a huge amount of $\ce{OH-}$ and they will deprotonate the carboxylic acid as soon as it forms.
$$\ce{RCOOR' + OH- <=> RCOOH + R'O-}$$
$$\ce{R'O- + H2O <=> R'OH + OH-}$$
$$\ce{RCOOH + OH- -> RCOO- + H2O}$$
The last deprotonation step is almost irreversible and consumes the base. This is why, an excess of base is usually added for ester hydrolysis.
The $\ce{RCOO-}$ is a negative ion, so it repels any nucleophile that comes close to it, so nucleophilic attack cannot happen, and the reverse reaction is prevented (as Maurice has mentioned in the comments). In essence it is this last step that drives the reaction to the product side.


What is the reason why protons and electrons do not collide?

They do collide.  But it's not like cars crashing and smashing each other up on the highway.  Their quantum mechanical wavefunctions overlap and the electrons do in face settle in with wavefunctions that are centered on the nucleus.  But the low mass of the electrons, the Heisenberg uncertainty principle (which has its greatest effect on low-mass particles), and the Pauli exclusion principle (electrons can't all pile up in the most stable quantum state)  cause the electron wavefunctions to get diffused out around the nucleus.  Instead of getting smashed up, the electrons basically surround snd embed the atomic nucleus.


Why is finding [H+] when given [OH-]= 3.57e-5 M wrong when I use -log[OH-]?

Well You have to note that you cannot simply "e" everything because the base of log in chemistry is base 10 while you assumed that it was all base e. Ie ln(x). in order to get x, we use e, while if we say log(x) we have to assume that it is base 10 which is the general norm


Which out of the two has greater bond angle? Water or diethylether.Why?

You are mostly correct with your reasoning , and the $\ce{C-O-C}$ bond angle was indeed expected to be lesser than that of $\ce{H-O-H}$ bond angle of water . But it's all theoretical. Experimentally the opposite was observed and was explained on the basis of steric hindrance or steric repulsion of the two bulky $\ce{C2H5}$ groups.

In case of $\ce{(C2H5)2O}$, the positive deviation of the bond angle can be explained on the basis of steric crowding of two ethyl groups. They repel each other and as a result , they do not let the lone pairs of Oxygen compress the bond angle more .

$\ce{H-O-H}$ bond angle is found to be $104.5^\circ$ whereas the $\ce{C-O-C}$ bond angle in diethyl ether is found to be roughly close to $110.5^\circ$. It can be attributed to the steric repulsion that trumps the bond pair-bond pair and lone pair-lone pair repulsion in this case.


Temperature Dependent Specific Heat of Nitrogen

The function looks like its easy to differentiate or integrate. Should I be doing something involving calculus?

Yes, exactly!  The equation you should be using is the differential form of the enthalpy equation you are already using.

$$dH = m~C_p(T)~dT$$

$$\int_{H_1}^{H_2} dH = m \int_{T_1}^{T_2} C_p(T)~dT  $$

As you can verify, if heat capacity is independent of temperature, i.e. a constant, the equation reduces to

$$H_2 - H_1 = m~C_p~\left (T_2 - T_1 \right ) \implies \Delta H = m~C_p\Delta T$$

This is the equation you tried to use originally.  However, since we derived this equation assuming a heat capacity that is independent of temperature, and your question explicitly involves dependence on the temperature, we should not use the already integrated equation.  Instead you'll have to do the integration yourself by substitution of temperature-dependent equation.


Can the direction of electron flow in a voltaic cell be reversed?

Can going to non-standard conditions reverse the cell potential of a voltaic cell?

I have already answered that question, in the affirmative, for a simple tin and lead galvanic cell, here: https://chemistry.stackexchange.com/a/116734/79678.

What follows is an elaboration with some specific values and illustrative figures. In fig. 1, the tin and lead voltaic cell is shown:

In fig. 1, the tin and lead ion concentrations are each exactly $\pu{1 M}$ and the cell potential is $\pu{+0.014 V}$, with tin as the anode. All necessary equations are shown in the figure and the usual simplifying assumptions are made: unity activity coefficients, zero cell resistance, etc. The electron flow is from $\ce{Sn}$ to $\ce{Pb}$.

In the next figure, the lead ion concentration is $\pu{0.3363 M}$ while the tin ion concentration remains at exactly $\pu{1 M}$. This results in zero cell potential, as shown in fig. 2:

Since the cell potential is zero, it is a moot point to call tin the anode: the cell is not doing (net) electrochemistry. There is no net electron flow.

In the final figure, the lead ion concentration is exactly $\pu{0.1 M}$ while the tin ion concentration remains at exactly $\pu{1 M}$. This results in a cell potential of $\pu{+0.016 V}$, as shown in the figure:

But now lead is the anode and tin is the cathode. The electron flow is from $\ce{Pb}$ to $\ce{Sn}$.

So changing the ion concentration to non-standard values can reverse the electron flow direction. But this was an especially easy case because the two metal electrodes had standard potentials that were relatively close. For the Daniell cell, with $\ce{Zn}$ and $\ce{Cu}$, it would not be feasible.


All nucleophiles are Lewis bases. They are just electron pair donor.

A molecule or ion is called a "base" (or, “Brønsted base”) when it interact with proton, it is called a "nucleophile" when it interact with other atom centers, which can be carbon, nitrogen, sulfur and so on. So the row "base" and "nucleophile" may overlap. For example, $\ce{OH-}$ is a base when it take proton off an acid, and it can also be a nucleophile when it hydrolyzes ester (nucleophilic attack to carbonyl group).

The broad nucleophilicity definition is how good the nucleophile interact the positive or partial positive atom center. When the atom center is proton, nucleophilicity is then basicity. Because proton center is hard and usually less steric hindred, a bulky and hard (for hardness $O$ > $N$ > $S$ center) electron donor will be likely to prefer proton over the softer and more hindred non-proton center. These electron donors are called Non-nucleophilic bases. Potassium tert-Butoxide is a good example.


Why do atoms bond? Why are unfilled valence shells energetically unstable?

The idea of a "full" outer shell is a bit of a misnomer, for instance 3rd row has 3d orbitals but doesn't bother to occupy them until 4th row. Does Argon have a full outer shell, even though it has a $3d^0$ configuration?

Generally there's a large energy gap between the p-orbitals in one row and the s-orbitals in the next row, because of the increase in shielding and such factors, which acts as a kind of natural barrier, or a step, to adding any more electrons than what appears to be a 'full' valance shell.

But within any row, the amount of shielding and associated factors (penetration, nuclear charge, etc) remain largely similar, so if the valence shell isn't 'full' there is still sufficient attraction between the electron and a neutral atom that a negative ion may be stable.  Adding more than this, the energy gap is too large to overcome under usual circumstances so it stops at a 'full' shell.

There are always exceptions.  Noble gases lower down can form compounds even though they are "stable" "full" outer shells, and quirks like some transition metals having $s^1d^{10}$ configuration go to show that some of the ideas about electron filling and stable shells are rules of thumb, each individual case really needs to have its energy levels calculated to explain them.


Is it possible to synthesis adenine, thymine, cytosine, or guanine at home?

Based on jonsca's comment, extract DNA from something. The protocol will involve detergent and isopropol alcohol. The details are available in several places and is fairly simple, but you won't get much DNA per onion, so be ready to prepare a lot of them. To do the degradation, there are several things you could try. If your detergent contains nucleases ( many contain enzymes but do a poor job of labeling which enzymes ) you could mix that in and react it at 37C for a while. Can also try high pH solutions, if you happen to have NaOH around. If you do the DNA extraction differently, you can get the RNA instead of DNA. RNA is very unstable at high pH, and RNAse enzymes are EVERYWHERE. So the RNA should degrade into nucleotides pretty readily if you just use sloppy lab technique. When I extract RNA in the lab, I use phenol/chloroform to pull the proteins and genomic DNA away from the RNA, which stays in the aqueous layer. You probably don't have access to those at home, but cooking oils may be hydrophobic enough to pull it off. Also, it helps to have a centrifuge.

Don't ask me how to purify the nucleotides from your solution and ask me even less about how to verify what you have at home. Verification would require something like a UV-Vis or HPLC or NMR. Good luck getting those at home.


You have to study existed architectures. First, it is necessary to distinguish between photo-conductive and photovoltaic elements. In photovoltaic elements one needs a mechanism of separation of positive and negative charges formed under an irradiation of the semiconductor. Even a simple p-n junction can do this. But, this is all about efficiency. The solar cell has to absorb as much light as possible without reflections, and, at the same time, the charge transport inside should be without losses that is impossible in thick structures. Also we have to care about the absorption spectrum. To absorb in all spectrum of the solar radiation, people make cascade structures with different materials absorbing in different ranges of spectrum.


How to calculate the concentrations of the species in the carbonate equilibrium from a titration with hydrochloric acid?

You have correctly identified the amount of substance $\ce{HCl}$ which were necessary to neutralize all $\ce{HCO3-}$ and $\ce{CO3^2-}$ in the sample. With the given sample volume of $\pu{10 cm^3}$, we can calculate the concentration of both ions in the sample:
$$
\frac{\pu{4.4E-3 mol}}{\pu{10 cm^3}}
= \pu{4.4E-4 mol//cm^3}
= [\ce{HCO3-}]+[\ce{CO3^{2-}}]
$$

Using the given ratio of $0.958$, we can express $[\ce{CO3^{2-}}]$ in terms of $[\ce{HCO3^{-}}]$:
\begin{gather}
[\ce{CO3^2-}]
  = 0.958\times[\ce{HCO3-}]\\
\pu{4.4E-4 mol//cm^3}
  = [\ce{HCO3-}] + [\ce{CO3^2-}]
  = [\ce{HCO3-}] + 0.958\times[\ce{HCO3-}]
  = 1.958[\ce{HCO3-}]\\
[\ce{HCO3-}]
  = \pu{2.25E-4 mol//cm3}
\end{gather}

We then use the calculated hydrogen carbonate concentration and the ratio to find $[\ce{CO3^2-}]$:
$$[\ce{CO3^2-}] = 0.958\times[\ce{HCO3-}] = \pu{2.15E-4 mol//cm3}$$


Effective nuclear charge and electron repulsion

Is the effective nuclear charge just a semiempirical value fitted to approximate the total potential energy of an electron in an orbital, or is there another meaning behind the number?

No, that's pretty much it. The Wikipedia entry has a partial periodic table with some $Z_\mathrm{eff}$ values. If you go further down to larger atoms, it becomes obvious that $Z_\mathrm{eff}$ decreases quite a lot towards the outer shells, so you could argue that the concept has its merits in capturing a quite significant effect.


Fourier transform for spectroscopy spectra?

As Mad Scientist said, many spectroscopy methods are already based on Fourier transform of a time-dependent observable. (This is not true of the basic spectrophotometry measurements, where you directly measure the transmittance as a function of frequency.) NMR is a method based on Fourier transform, as is FTIR.

Regarding NMR signal processing, it's a very wide field, which includes interesting research ranging from academic interest to software development. So, it's hard to answer your question other than saying “yes, of course, further processing of NMR signals is sometime possible”. One thing you have to realize is: while you have probably learnt proton NMR during your studies, this is only one NMR technique, and the simplest of them all. There are plenty of different NMR techniques, depending on the type of excitation and sequence of RF pulses, many of them being multidimensional. Each method can use a variety of signal processing techniques.


Change in internal energy for isothermal process

It must not be an ideal gas.  This is irrespective or whether the process is reversible or irreversible.   That is the only possible conclusion.  Otherwise the amount of heat received would adjust  until it was equal to the work done.


Why are smoke odors persistent in fabrics?

Why is the smell of smoke so persistent, while many other strong smells are not?
In fact this is a tricky question, because you can't make such a comparison. According to IUPAC Gold book Smoke is:

An aerosol originating from combustion, thermal decomposition or
thermal evaporation. Its particles may be solid (magnesium oxide
smoke) or liquid (tobacco smoke).

So it is a fluid colloidal system in which the "the dispersed phase is a solid, a liquid or a mixture of both and the continuous phase is a gas" others strong smells that you smell during your every day life, are mainly already gasses. The interaction areosol-material an a gas-material is completely different.

Interaction between fabric and gasses
We can decompose the interaction in two different phenomena:

gas trapped inside the little voids inside your fabric
gas adsorbed in the fibers surface

The first is governed by Fick law of diffusion this process reach the equilibrium very quickly e.g. if you enter in a room with a gas after a couple of minutes the concentration of gas inside the voids of your fabric will be the same as the outside concentration.
The second is more hard to determine, however is more slower because the adsoprtion occur if there are free sites where the molecule can be placed, and this is a competitive process between all the others gasses present in larger quantities in the atmosphere, that already occupy these sites.
However if you stay a day in a room with a gas is possible the some gas molecule will be ad-sorbed in your fabric. This interaction is more persistent, however I think that the amount of substance is quite negligible.
So interaction between gasses and fabric is not very persistent.

Interaction between aerosol and fabric
This is completely different because colloidal suspensions are unstable. Aerosol are high energy system that tend to lower their energy forming bigger particles or bonds with other materials. So the deposition of these particles over the surface of others materials will be most of the times a spontaneous process. The high surface of the fabrics indeed make it easier.
So what happen when a particulates/aerosols settles on the fabric? This is not  gas trapped inside the little voids of your fabric or gas adsorbed in the fibers surface is a mixture of liquids and solids: many big (relatively!) "clusters" of molecules spread all over the fabric, from the tiny pores to the fiber's ravines. What matter here is the vapor pressure of the substance  you got this equilibrium (for a liquid):
$$\ce{Molecule_{(l)} <=> Molecule_{(g)} }$$ and the analogous with $_{s}$ instead of "l" for the sublimation. For both of these equilibria only the molecules in the liquid-gas or solid-gas interface can evaporate\sublimate, so the cluster surface is very important!

The case of Syringol

In the case of Syringol you have a solid that melt at 50–57 °C. So the sublimation is the only path that allows you to smell it. For most solids the vapor pressure is very low so the equilibrium will keep going to the right for a long long time, releasing syringol, until the source end.


How can sodium chloride melt ice or keep it frozen?

The apparent contradiction is resolved as follows: in both cases, addition of salt makes ice melt faster, which is important in the street and negligible in a beer box (or an old style cold bath used in the lab). But in doing so, it also causes the temperature of the ice and its surroundings to go well below $0^\circ\rm C$, which is important in the lab and negligible in the street.


It's a little hard to tell from this angle, but that looks like a pretty sensible representation of the axial methyl geometry given on the website you linked to.

I'm not entirely sure what you mean by "push to the center of the molecule", though.

ETA: These might help a little more: this isn't exactly the most accurate geometry, but it's a pretty good representation.


Why is the minus sign (-) not allowed in reaction equations?

Don't get too caught up in notational rigidity. You're "allowed" to use any notation you want, if it gets you to the correct answer. If you really want, go ahead and use $\ce{Na\ - \  e^{-} -> Na+}$, or go all out and use $\ce{-Na^{+}\ - \  e^{-} -> - Na}$. However, as Richard Feynman discovered while learning trigonometry, if you start using too much obscure/uncommon/personal notation, you get into trouble showing others your work, which makes doing science hard. Standardizing notation is just a way to save time and headaches later on.

The most basic chemical equations actually carry very little chemical information, to the point that it's easy to write reactions which are chemically outrageous. One of their most fundamental purposes is simply to establish atom conservation (which is why so much time is spent on teaching how to "balance" reactions). Thus, chemical equations are always equivalent to a system of linear equations, which is why you can treat them like such and add/subtract molecules to both sides, multiply by a number, add two different reactions together, etc. Using negative stoichiometric coefficients would thus be a very trivial extension of this behaviour, and the equation would remain completely valid.

Let's look at an example. Take the intermolecular dehydration of ethanol to produce diethyl ether:

$$\ce{2 C2H5OH -> C2H5OC2H5 + H2O}$$

The equation is just as valid if you subtract a water molecule from each side:

$$\ce{2 C2H5OH\ -\ H2O -> C2H5OC2H5}$$

Not many people would write the above equation. However, we do have another type of standardized notation which is often used, and which involves a negative stoichiometric coefficient. Using this notation, the equation is written as:

$$\ce{2 C2H5OH ->[\ce{- H2O}] C2H5OC2H5}$$

The water molecule has a legitimate negative stoichiometric coefficient of -1, and again the equation is exactly as valid as all of the ones above. Why is it that negative stoichiometric coefficients are rarely written in the reactants or products, but common when written atop the reaction arrow? I have no idea, but ultimately it makes no difference. That's just how notation happened to become standardized, and if everyone understands it, that's fine.


Making metal powders using electrolysis - chosing the right electrolyte

Fine copper powder can be prepared using thermal decomposition of copper citrate (can be prepared by mixing dirt cheap citric acid, three equivalents of a base like potassium hydroxide and copper sulphate). I once made pyrophoric (!) copper powder using this technique. There are mentions of preparation of iron powder same way. Anyway, it is much more practical than electrolysis for big batches.

Frankly speaking, fine aluminium flakes are widely employed in various paints and are dirt cheap. They should be available in shops for artists. There is no reason to make them at home (and aluminium reacts with water once cleaned of oxide film)

Magnesium is probably the worst. It is quite costly and active and making fine powder is dangerous. It also reacts with water and alcohols, so making it employing electrolysis is out of question. There are mentions of home smelting of aluminium and magnesium, producing magnalium. Magnalium is brittle and can be easily milled, and it is active enough so fine milling is usually not needed. While fancy, proper pyrotechnics can be made without magnesium with easy, aluminium is usually much more preferable. Magnesium often produce huge volumes of white smoke, a habit aluminium doesn't have.

Regardless of above, I strongly encourage to be cautions when working with metal powders, especially magnesium.


Why is sulfur dioxide more stable to UV radiation than ozone?

I'm going to attempt to answer this from three points of view.
Qualitative considerations and MO theory:
First, why isn't ozone linear? You can imagine $\ce{O3}$ like a $\ce{CO2}$ molecule that lacks two electrons. This lack of electrons will result in less $s$-character, and less bonding "attitude". That's why ozone is bent.
Still, the repulsion in its electrons ($\ce{O}$ is a small central atom, very small) will want it to be free. Oxygen's van der Waals radius is 1.5 $\mathrm{\overset{\circ}A}$. Thus, $\ce{O3}$ has a bond length of $\approx 1.2\ \mathrm{\overset{\circ}A}$. That's really cramped. Electrons aren't happy with this and will try to get free as soon as $E_a$ and other conditions are provided.
Now compare that with sulfur's 1.8 $\mathrm{\overset{\circ}A}$ radius; which, results in a $\approx 1.43\ \mathrm{\overset{\circ}A}$ bond length. Less cramped, the electrons aren't as much complaining as they do in $\ce{O3}$'s case. We can't get really further as these two molecules aren't isoelectronic.
_____________________________________________
Approach using classic high school knowledge:
It's because oxygen has an arbitrarily positive formal charge. Take a look at $\ce{O3}$'s electron-dot diagram:
$\hspace{18ex}$
$\hspace{45ex}$ $\small{Source}$
The oxygen atom, which has an electronegativity of $3.6$ in Allen scale, has formed $sp^2$ orbitals. This, a very electronegative atom with a positive formal charge, is highly unfavorable for the molecule, and acts very "non-bonding".
The sulfur, on the other hand, has an electronegativity of $\approx 2.5$. It will also get a positive formal charge. But since it's less electronegative, it bears it.
_____________________________________________
Thermodynamics and kinetics:
Ozone decays into diatomic oxygen under certain conditions:
$$\ce{2O3 -> 3O2}$$
Three moles of oxygen will be produced per two moles of ozone. The measured enthalpy of formation for ozone is $142~ \mathrm{kJ/mol}$, while it's 0 for diatomic oxygen. So
$$\mathsf{\Delta H_f~of ~the~ products - \Delta H_f ~of ~the ~reactants} = \mathsf{reaction~ enthalpy}$$
$$ 0 - 142 = \fbox{-142} ~\mathrm{kJ/mol}$$
Furthermore, this reaction will increase entropy, as two moles of a gas produce another three moles. Thus, it's thermodynamically highly favorable.
Above that, its energy of activation is a measly $\approx 32 \mathrm{~kJ/mol}$, which is a very small kinetic barrier.
Thus, you can also conclude that $\ce{O3}$ is exceptionally "inclined to decompose". Also, note that ozone is a very strong oxidant; while $\ce{SO2}$ is a weaker one.


Lowering the melting point of water with salt

You should write independent equilibrium expressions for the chemical potential of each species.

In a freezing point depression problem the solvent in frozen form is at equilibrium with the solvent in liquid form, so you can write that
$$\mu(\ce{H2O(s))}=\mu(\ce{H2O(sol))}$$
Next, by Raoult's law (assuming it holds), the chemical potential of solvent in the solution (the right-hand term) differs from that of the pure liquid by the term $RT\log(\chi_1)$ where $\chi_1$ is the mole fraction of solvent, so that
$$\mu(\ce{H2O(s))}=\mu(\ce{H2O(l))} + RT\log(\chi_1)$$
This is just the beginning. Next you need to recognize that the standard Gibbs free energy change for melting is just $\Delta_m G^\circ=\mu(\ce{H2O(l))}-\mu(\ce{H2O(s))}$ (assuming the solution is at $\pu{1 bar}$ or approx normal pressure) and invoke the Gibbs-Helmholtz equation in differential form:

$$\left(\frac{\partial(\Delta G /T)}{\partial (1/T)}\right)_p=\Delta H$$

so that

$$\left(\frac{\partial\log(\chi_1)}{\partial (1/T)}\right)_p=-\frac{\Delta_m H^\circ}{R}$$

Now you integrate the van't Hoff equation from $\chi_1 = 1$ (pure solvent) to $\chi_1 = 1-\chi_2<1$, while assuming $\Delta_m H^\circ$ to be constant over this range of $T$. Finally, assuming $T\approx T_m$, expressing $\chi_1$ in terms of solute molality, and assuming the solute concentration is low, results in the freezing point equation
$$\Delta T = K_m m_2 $$
where $K_m$ is the cryoscopic constant
$$K_m=\frac{RT_m^2M_2}{1000 \Delta_m H^\circ}$$
Here $M_2$ is the molar mass of the solute.
You can use this last expression to compute the molality of solute $m_2$ required to depress the freezing temperature by $\Delta T$.

The process of dissolving the solute is not important (whatever happens before it is fully dissolved doesn't matter as long as it is soluble). Also, it is assumed an ideal solution forms, which means interactions of the solvent with solute, and between solvent molecules are equally strong. And the salt remains in solution (none is incorporated into the solid phase). Note also that liquid solvent remains in equilibrium with solid solvent at the melting point $T_m$, and that we assume that the solute concentration is dilute (this is important to derive the last expression, but Raoult's law may hold even for substantial solute concentrations).


Gelatination of solution in saponification

Sodium stearate is not very soluble in water (above about 2%, it becomes viscous but not a firm gel). Ethanol can reduce the viscosity at first, but sodium stearate is not soluble in ethanol either, so an excess of ethanol causes a precipitation of solid that firms up the aqueous gel. More water  dissolves that precipitate and restores the fluidity.
The solubility and rheology of fatty acid soap solutions depends on temperature, concentration, solvents and the structure and purity of the fatty acid group. Stearate is very different from oleate, and both are different from laurate.


Spontaneous explosion of a bottle containing alcohol and milk

The bottle can't have exploded for "no actual reason"; there may have been no apparent reason, but everything does indeed happen for a reason, at least at this level. But this isn't English.SE...

My bet is on a combination of things. As with many things in everyday chemistry, the substances involved are never pure enough to pinpoint one exact cause. Case in point, milk is an emulsion containing a significant percentage of fats and proteins. So, possible causes include:

Hygroscopy and acidity of the alcohol causing the milk to curdle, by taking the water from the milk to form an alcohol solution that the complex milk proteins and fats precipitate out of due to lower solubility. This may produce a "greater than the sum" volume of product by leaving these proteins and fats in a porous gel (basically cottage cheese, though usually a stronger acid than ethanol is used in cheesemaking).
Esterification. Milk, as stated, contains fats including fatty acids. Add alcohol, and the fatty acids react with the alcohol to produce esters (neutral fats). The byproduct is water, formed from the acid stripping the alcohol's hydroxyl group. Again, the products of this reaction can take up more volume than the reactants.
Biological reduction. I actually do not think the concentration of alcohol in this solution would be antiseptic; if the alcohol is pure then we're only talking about a 20% solution by volume, and that would have to be a lab-distilled product like Everclear; "natural" distillation in open air only gets you to about 150 proof, and using even 100-proof vodka gives this mixture only a 10% concentration by volume of ethanol. According to at least one source, ethanol needs to be concentrated to at least a 65% solution by volume to be effective as an antiseptic. So, to a bacteria's point of view, the alcohol is an overworked police officer; it'll slow down bacterial growth, but there's way too much fat and sugar here for the alcohol to be much of a deterrent. Consumption of organic fuel by cellular metabolism, as a rule, produces CO2 gas, and in a closed container at room temperature, well, you get the picture.


Comparing radii in lithium, beryllium, magnesium, aluminium and sodium ions

Many people may think that orbitals, or ‘shells’ essentially work like planet orbits around the sun; i.e. no matter how many electrons populate ‘Mercury’, ‘Venus’ and ‘Earth’ or how many protons are in the ‘Sun’ Venus’ orbit will always have the same dimension.

This is very much not so when studing atomic orbitals. The principle of hydrogen-like orbitals (which are the only orbitals that can be exactly calculated with the Born-Oppenheimer–approximation) is that we are dealing with an electron free to rotate and translate on a potential energy surface which is formed by the electrostatic potential of a proton or a nucleus with a charge of $z+$. Commonly, taking the 1s orbital as an example, there is a distance from said nucleus where the probability of locating the electron in the sphere defined by that radius is larger than a certian cutoff value, e.g. $90~\%$. If we go from a hydrogen-like system to a $\ce{He+}$ cation, we are doubling the charge thus strongly increasing the attractive force between nucleus and electron. As such, the $90~\%$-radius is much closer to the nucleus than before.

This holds true with limitations when no longer observing hydrogen-like atoms but actual multi-electron atoms: the more protons in the nucleus the more contracted the orbitals become. While that effect in itself is rather small, the second one is substantial: the higher an ion’s charge is, the more contracted its orbitals become. Therefore, $\ce{Be^2+}$, which is even considered to be the smallest atom or atomic ion known (yes, smaller than hydrogen) is substantially smaller than lithium.

When directly comparing two atom sizes of elements that are above and below each other in the periodic table, naturally under otherwise identical conditions the lower one will be larger; even though it has at least 2 (lithium) more protons it has also started populating an additional shell which is further removed from the nucleus. Similarly, if you compare $\ce{Li+}$ and $\ce{Mg+}$, magnesium will still be substantially larger. (In this case, the additional remaining electron in magnesium’s 3s orbital is the deal breaker but note that the same argument applies to $\ce{C+}$ and $\ce{P+}$ where no shell is completely empty.

As noted before, however, if we substract another electron, i.e. if we compare the isolobal $\ce{Li+}$ and $\ce{Mg^2+}$, the latter’s orbitals shrink substantially. By empirical observations scientists have discovered what is known as the diagonal relationship: two ions that are top-left bottom-right diagonally placed in the periodic table, whereby the bottom-right one’s charge is greater by one, have approximately the same radius.


Isolating calcium acetate from water with electrolysis

How to get rid of water to collect calcium acetate?

Evaporate it.

How to do it in an environmentally responsible way?

Electricity from the outlet cost you money and probably used some
combination of coal, natural, gas, and or nuclear to produce it.
Bunsen burner, but it likely uses a fuel that was taken from the ground, made from a crop that that competes against food production for land.
Solar cells to produce electricity. It uses renewable energy but
is relatively expensive and electricity is a high value energy
compared to solar thermal. Using commercial grade solar cells only
converts about 17% of the energy in sunlight into electricity that is then converted into heat to evaporate the water.
Solar thermal is cheap and easy to make with household items like an aluminum or soup can and black spray paint. Further more, it has the potential to be more than twice as efficient as solar cells at providing energy you need to evaporate water.

Electrolysis:

If you insist on electrolysis for experimental reasons, calcium acetate is a salt of acetic acid and calcium from the calcium carbonate in the egg shells. As a salt, like all salts, calcium acetate will dissolve in water forming ions that will conduct electricity. So, there is no need for any electrolyte other than water and calcium acetate.

Furthermore, your end products will be something other than calcium acetate and would depend in part on your anode and cathode, temperature, and applied voltage.


Why No Change in Polarizability in Asymmetric Stretch?

Please explain using plain english (without any rules or tables) why is that in asymmetric stretching like the following, there is no change in the size, shape or orientation of the electron cloud that surrounds the molecule?

There is a change. The size and shape of the electron cloud does change for the individual bonds affected by the asymmetric stretch. Polarizability relates to the ease with which electrons can be moved from their equilibrium position.  Electrons in weaker (longer) bonds are more easily displaced than electrons in stronger (shorter) bonds.
From your diagram, I believe that you are discussing the case of a linear $\ce{AB2}$ molecule. Picture the asymmetric stretch in our $\ce{AB2}$ molecule with the two $\ce{B}$ atoms remaining stationary and the $\ce{A}$ atom oscillating back and forth. Simultaneously one $\ce{A-B}$ bond becomes longer and the other shorter.  One becomes more polarizable, the other less polarizable.  In the linear $\ce{AB2}$case these changes in polarizability are equal and opposite, hence for the molecule overall they cancel out and there is no change in overall polarizability (for example, see here).

In order for a molecule to be Raman active, there must be a change in the polarizability

Yes, but it must be a change in polarizability for the entire molecule, not just an individual bond. Further, keep in mind the  Principle of Mutual Exclusion. It states that, for centrosymmetric molecules (molecules with a center of symmetry, your linear $\ce{AB2}$ is an example), vibrations that are IR active are Raman inactive, and vice versa.
The asymmetric stretch in such a molecule changes the dipole moment in the molecule from zero to non-zero; hence the asymmetric stretch is IR active. Therefor by the Principle of Mutual Exclusion, the asymmetric stretch must be Raman inactive.  Even though the asymmetric stretch changes the polarizabilty of some bonds in the molecule, the polarizability of the whole molecule does not change (as explained above) and the asymmetric stretch is Raman inactive as predicted.
The links I posted in the comments show how these concepts can be applied in the linear $\ce{AB2}$ examples of $\ce{CO2}$ and $\ce{XeF2}$.


Why carbon monoxide reacts with hydrogen to give different products in the presence of different catalysts

The thermodynamics are always the same. If you go from carbon monoxide to methane, there is only one final enthalpy, and if you have the choice of going to methane or methanol the thermodynamically preferred product is always the same.

The difference indeed less within the different pathways and the activation barriers needed to get from one intermediate to the next. This isn’t really thermodynamics, it is mainly kinetics (although you can argue about thermodynamically preferred intermediates). The pathways are often very complex and entire papers revolve around how one reaction is catalysed on such a solid state catalyst, so please forgive me for not looking them up; but the general point still stands: one catalyst will lead to intermediates where the $\ce{C-O}$ bond is easily broken (and thus give methane), the other will not (and thus give methanol).


A very handy diagram and explanation of photoconductivity is shown below, from the American Physical Society page Viewpoint: Holey Intrinsic Photoconductivity:

The caption explains:

  Excitation process leading to photoconductivity in a condensed matter system (upper panel) and its cold-atom counterpart (lower panel). In the condensed matter system, incoming laser light ($hv$) excites an electron to the valence band, leaving a positively charged hole behind. In the cold-atom analogy, the optical lattice is tuned to allow an atom to move into the excited band, also leaving a hole behind.

From the Wikipedia page Photoconductivity, states that some 'classic' examples of photoconductive materials (asides from selenium) include:

The polymer polyvinylcarbazole
Lead sulfide

Some more examples exist in the paper Organic photoconductive materials: recent trends and developments include the chemical classes:

Phthalocyanines
Squaraines
Azo pigments
Perylene pigments


Why a catalyst is not consumed in a reaction when all steps are reversible?

By saying Catalysts are not consumed by reactions. is meant there is no stoichimetric ratio to reactants, consuming catalysts and forming from them catalytically inactive compound.
By other words, some of reaction steps regenerates the original form of a catalyst, consumed by a prior step, so the net consumption is negligible.
The transition between active states of a catalyst is not considered a catalyst consumption, as it does not affect its catalytic activity.
Catalysis means the catalyst is reversibly switching between its 2 – chemically different – active states, producing an alternative and faster reaction path. Each form catalyzes one of the mutually opposite direactions of the net reaction. It is like a kind of a ferry to move compounds between reactant and product shores, bypassing a faraway bridge. The ferry is not consumed if ends at the opposite shore.

Note that things get little complicated, if some of products or intermediates catalyse a part or whole net reaction. Not being particularly aware of the intermediate case, but for the product, the famous case is oxidation of oxalate by permanganate in acidic environment:
$$\ce{5(COO)2^2-(aq) + 2 MnO4-(aq) + 16 H3O+(aq) -> \\ ->10 CO2(g) + 2 Mn^2+(aq) + 24 H2O}$$
$\ce{Mn^2+(aq)}$ as the product acts as a catalyst, forming catalytic system $\ce{Mn^3+(aq)/Mn^2+(aq)}$ with a similar redox potential as $\ce{MnO4-(aq)/Mn^2+(aq)}$.
$$\ce{4 Mn^2+(aq) + MnO4-(aq) + 8 H3O+(aq) <=> 5 Mn^3+(aq) + 12 H2O}$$
$$\ce{(COO)2^2-(aq) + 2 Mn^3+(aq)  -> 2 CO2(g) + 2 Mn^2+(aq) }$$
The effective result is an autocatalytic reaction. There is sometimes applied a small amount of Mn^{+II} salt to speed up the reaction, or it is waited until initial addition of $\ce{KMnO4}$ creates enough $\ce{Mn^2+}$.


Most efficient way to purify household vinegar: distillation or freezing?

There is no acetic acid/water azeotrope, so distillation can produce glacial (pure) acetic acid.


Why are hydrogen ions always associated with another molecule?

The way I understand it is (and my understanding is by no means perfect, or complete), as you pointed out correctly: a hydrogen ion is in fact a proton. The proton is a "bare charge" and as you rightly said, "tiny". a
This makes it extremely reactive (in a sense), and thus in a chemical system of any sort would immediately seek out and associate with the electron clouds of a surrounding molecule.

For instance, in an aqueous solution it forms $\ce{H_3O+}$ (hydronium ion, which are further solvated by water molecules and form a manner of aggregates).

An extremely interesting way of looking at it was pointed out in the comments (just putting it here, so no one misses it)

  An interesting way to thinking about the reactivity of a bare proton is to calculate its charge density and compare with other monocations, such as Li+. This will make it clear why free protons will abstract electrons from just about anything. –  Nicolau Saker Neto

Do note (as far as I know, someone might correct me if I am wrong) free protons are fairly stable (by free I mean not bound to electrons or other nucleons) and usually don't decay into other particles. They exist in several naturally occurring situations, high energy plasmas come to mind.


Synthesis of phosphonate salt not working

Looking at the procedure (and considering the structure of the desired product) it appears that its boiling point is 142C at 27 mm (not at atmospheric pressure). So you should try a vacuum distillation to purify your product. Heating it to try to distil it at 1 atm is probably causing decomposition. Also, be sure to protect the reaction mixture and the final product from atmospheric moisture.


Yes! There is a simple possibility. Take one racemate and do what Pasteur did:

be lucky, in that both enantiomers crystallise separately;
be lucky again in that they form macroscopically distinguishable crystals (in this case probably: mirror images);
collect the different enantiomers. Voilà, you have enantiopure substances and can use them in synthesis.

(Thankfully, using tartaric acid immediately gives access to the Sharpless asymmetric epoxidation.)

Once you have any single enantiopure compound, you are able to generate further ones. For example, reacting a racemate with an enantiopure compound in a way that preserves the respective stereocentres will result in diastereomeric products. Diastereomers can theoretically always be separated (it may be harder for some and some may interconvert under ambient conditions but the theory stands). You could also generate diastereomeric salts, by taking an enantiopure anion (say (R)), and a racemic cation. You will receive the diastereomers (R,S) and (R,R) which can then be separated and so on and so forth.


What are the different types of catalysts used in catalytic hydrogenation of alkynes?

The "Lindlar catalyst" (Pd/CaCO3) is probably the most common way of effecting stereoselective reduction of alkynes to (Z)-alkenes. It is fairly accessible and the reaction is easy to run. But it is not the only way: there are other catalyst systems. One of the more notable ones is probably the Brown catalyst "P2-Ni" (Wikipedia).
There is a nice review on alkyne semihydrogenation here: Chem. Rev. 2013, 113 (3), 1313–1350, which covers both heterogeneous and homogeneous methods, as well as methods involving other hydrogen sources apart from H2 gas itself. I recommend it to interested readers and especially anybody considering carrying out such a reduction.

The Lindlar hydrogenation of an alkyne forms (Z)-alkenes stereoselectively. Syn hydrogenation of the (Z)-alkene gives product (c), not product (a).

Note that in the fully reduced alkane, there is free rotation about the central C–C bond. So, having both D atoms pointing "out of the plane" is irrelevant, as you can easily rotate to a conformation in which one D atom is pointing in the plane. However, the configuration of the product cannot be changed, and that is the difference between options (a) and (c).


If Alkynes can be reduced to Alkenes using Sodium in liquid ammonia, why can't Alkenes be reduced to Alkanes?

As shown in your diagram, when the initial electron is transferred to the alkyne a radical anion is formed; the transferred electron pairs with an electron from the alkyne to form an anion at one end of the molecule. These two electrons are in an $\ce{sp^2}$ orbital.  If you try to repeat this process with an alkene, you wind up with the paired electrons in an $\ce{sp^3}$ orbital.  An $\ce{sp^2}$ is lower in energy than an $\ce{sp^3}$ orbital since it contains more s character and an s orbital is lower in energy than a p orbital.  The electron solvated in ammonia has enough energy to transfer into a lower energy $\ce{sp^2}$ orbital, but not enough energy to transfer into a higher energy $\ce{sp^3}$ orbital.


The positive terminal of a battery is where 'conventional current' comes out of the battery and flows through an external circuit, returning to the battery via the negative terminal. Since electrons are (by convention) negatively charged, the negative terminal of a battery is where electrons are 'released' by the battery.

NOTE: Of course, if it is a secondary battery which is 'recharging', electrons will 'enter' the battery at the terminal marked negative.


Calculating Compressibility factor from the Van der Waals' Gas equation

$$Z = \frac{V^{\mathrm{r}}_{\mathrm{m}}}{V^{\circ}_{\mathrm{m}}}\tag{1}\label{1}$$

$V^{\mathrm{r}}_{\mathrm{m}} = \text{Volume of 1 mol real gas.}$

$V^{\circ}_{\mathrm{m}} = \text{Volume of 1 mol perfect gas.}$

The van der Waals equation is $$ \left( P^{\text{r}} + \frac{an^2}{(V^{\text{r}})^2}\right) (V^{\text{r}} - nb ) = nRT \tag2\label{2}$$

Now, consider you have a container containing a $1~\mathrm{mol}$ real gas. You know its pressure $P^{\text{r}}$, volume $V^{\text{r}}$ and temperature $T$ and you wish to find the compressibility factor of this gas.

So, for calculating $Z$, we know the real volume of the gas, i.e $V^{\mathrm{r}}$ and now we need to calculate  $V^{\circ}$ which is the volume it should have occupied if it behaved like a perfect gas, i.e. it obeyed the perfect gas law.

Thus, $$V^{\circ} = \frac{RT}{P^{\text{r}}}$$

Therefore, substituting this value of $V^{\circ}$ in $\eqref{1}$ $$Z = \frac{V^{\mathrm{r}} P^{\text{r}}}{RT}\tag3\label{3}$$

From $\eqref{2}$, by rearranging the terms, we get $$P^{\text{r}} = \frac{RT}{(V^{\text{r}} - b)} - \frac{a}{(V^{\text{r}})^2} \tag4\label{4}$$

Substituting this value of $P^{\text{r}}$ in $\eqref{3}$, we get $$Z = \frac{V^{\text{r}}}{(V^{\text{r}} - b)} - \frac{a}{V^{\text{r}}(RT)} \tag5\label{5}$$


Enthalpy of formation and bond energies of CH2(OH)2?

Firstly, regarding the bond energy, you will have to specify exactly what bond you are talking about, you can't ask for the bond energy of a compound because compounds have many bonds. This doesn't apply for diatomics (for example, you can ask what the bond energy of $\ce{N2}$ is; although that is loose usage of terminology, the meaning is clear). But methanediol is obviously not a diatomic.

Secondly, from what I understand, you need to find the enthalpy change for the hydration of formaldehyde: $\ce{CH2O_{\;(aq)} + H2O_{\;(l)} -> CH2(OH)2_{\;(aq)}}$ and your approach was to find the enthalpies of formation of the individual compounds. That's a good idea and more often than not, it works because  enthalpies of formation are widely tabulated and easy to find. But it's often worthwhile just searching for the enthalpy change of the reaction you are interested in, especially if it is a relatively well-known or simple reaction (like this one). Anyway, this recent paper gives the value of $\Delta _{\text{r}}H$ for the reaction above as $-39.0 \text{ kJ mol}^{-1}$.


Determine reaction concentration at chemical equilibrium with multiple reactions

Answer already found by myself. Short conclusion:

To calculate variations of the input concentrations at chemical equilibrium 2 major steps are needed.

1) Calculate the equilibrium constants $K_x$
for every reaction $x$ at given concentrations. Solve that system of x equations (details in 2nd edit) with a numerical solver of your choice.

EDIT: No numerical solver is needed, since the equations simplify to:

$\ce{K_1 - \frac{CN}{C*N} = 0}$

$\ce{K_2 - \frac{CO}{C*O} = 0}$

which is very interesting, since this simplifies the whole problem a lot. It also means that at chemical equilibrium a reaction does not care if a reactant is depleted via another reaction e.g., is bound within another product, or if the (initial) concentration of that reactant is just lowered to achieve the same amount in equilibrium. No reaction describes the connection between $\ce{CO}$ and $\ce{CN}$ e.g., $\ce{CO + N <=> CN + O}$, so basically the concentration of $\ce{CO}$ and $\ce{CN}$ ARE DECOUPLED (looking at the decoupled equilibrium constants), although they seem to be coupled via their common reactant ($\ce{C}$) e.g., changing $\ce{N}$ (or $\ce{O}$), changes the amount of produced $\ce{CN}$ (or $\ce{CO}$), changes the amount of $\ce{C}$ and further the amount of the other product $\ce{CO}$ (or $\ce{CN}$) .

This gives rise to a new question: How would a 'forced coupling' between $\ce{CO}$ and $\ce{CN}$ via $\ce{CO + N <=> CN + O}$ influence the concentration of both products?

Since we are at chemical equilibrium and have already taken 'indirect coupling via $\ce{C}$' into account, I guess that introducing a triangular-like reaction scheme would only introduce a redundant equation and won't give new information. Maybe I will test for that and share my results.

EDIT: Testing showed no difference in using a triangular-like reaction scheme (including the reaction above). As stated above, the 'new' reaction/equation does not contain new information. The 'new' equation can be written as a linear combination of the other 2 reaction equations. Therefore, one of these 3 reaction equations can be canceled/eliminated since 1) we need $n$ equations for $n$ variables and 2) an equation consisting of a linear combination of other equations cancels out (leads to a zero result) when solving the system of equations.

This means we only need to consider (in this case) the 3 equations for mass/element conservation and 2 reaction equations for the 5 variables (concentration of 3 reactants and 2 products).

2) Adjust/variate the input abundance and use another system of $z$
equations (conversation of mass/element and reaction equations - details in the edit of the question) to calculate the new concentrations at equilibrium.

Note: Make sure to always have $n$ equations for $n$ variables and don't mess them up.


You CAN use R=8.31 but it has the (standard) unit of (m3 Pa K−1 mol−1), simply convert your volume from liter (aka dm3) to meter cube (m3) and the pressure will be in Pascal (Pa)

If you use R=62.36, which has a unit of (L mmHg K−1   mol−1), as long as your   volume is in liter, the pressure will be in mmHg
But the question ask for mmH2O.

Pressure is simply force on an area (F/A), weight on an area (mg/A), density * volume on a area (rho*V /A) = density * height (rho*h)

Thus P = density (Hg) * h (Hg) = density (water) * h (water)

h (water) =  (13.6 / 1.00) * h(Hg)


Why is only one lone pair in imidazole delocalised?

The correct answer to this question is more direct, and not listed in your items. The non-bonding electron pair in nitrogen 3 is in an orbital perpendicular to the π-bonding p orbitals of all other atoms in the imidazole ring. Thus, it does not have the appropriate geometry to overlap with other orbitals forming π-bonds, and does not participate in any resonance effects.

This effect of orbital geometry is often invoked to explain the difference in basicity between pyridine and pyrrole. In the former, the nitrogen lone pair is of type 3 and does not participate in resonance, and so it has a basicity comparable to regular amines and imines. However, in the latter, the lone pair is of type 1 and does participate in the aromaticity of the ring. Protonation of the nitrogen atom would require localization of the electron pair on the nitrogen, breaking most of the resonance effect. This is energetically unfavourable, so pyrrole displays a much lower basicity than expected for amines or imines.

This section on pyridine has a little more information on why the type A lone pair does not delocalize, including a helpful picture. Notice that the lone pair is in an $\mathrm{sp^2}$ orbital parallel to the plane of the ring, while the π-bonded system is perpendicular.

I can make the following observations on your conjectures:

1) I don't think this applies, actually. The fact that orbital overlap is disallowed in the first place precludes the possibility of even thinking what would happen to the geometry of the molecule if the electron pair in nitrogen 3 formed a π-bond; it just doesn't.

However, if the molecule had a boron atom instead of carbon in position 2 (with an empty $\mathrm{sp^2}$ orbital pointing in the same direction as the full $\mathrm{sp^2}$ orbital in nitrogen 3), then I'm not sure what would happen. A more simple example of this type of structure would be 1,2-azaborine (of which only the 1,2-dihydro derivative seems to have been isolated as of yet).

I could imagine canonical structures in which the electron pair on nitrogen links with the boron atom, forming an additional bond (analogous to benzyne, which is isoelectronic with 1,2-azaborine). However, to form an extra N–B π-bond, then either the bond will be a $\pi(\mathrm{sp^2}\text{–}\mathrm{sp^2})$ bond (which is unusual as far as I know), or both the nitrogen and boron atoms would have to partially rehybridize into a mix of $\mathrm{sp^2}$ and $\mathrm{sp}$ configurations to allow a $\pi(\text{p–p})$ bond, which would likely imply an increase in bond angle and a huge increase in ring strain. Neither option seems particularly good, so I suspect such canonical structures would contribute little to the actual bonding picture in the boron-substituted ring.

2) The nature of the nitrogen atom which participates in resonance is actually something between $\mathrm{sp^2}$ and $\mathrm{sp^3}$. The concept of hybridization isn't very adept at explaining resonance naturally.

I'm also not happy with how much my arguments rely on hybridization, but I don't know other effective means to explain what's going on.


How to calculate the enthalpy of reaction using Gaussian 09?

In principle your approach is correct. There are a few pitfalls though.

Quantum chemistry is usually not well equipped handling ions, especially anions, within reactions, hence the evaluation of these reaction energies will probably contain some errors. This is mostly due to the fact that you are technically calculating everything in the gas phase, and apply thermal corrections later. In well behaved systems, e.g. organic compounds without charges, that does not matter much. But even then we try to exploit error compensation with (for example) isodesmic reactions.

If you are treating the ions separately, you are neglecting all possible interactions between them, too. That might lead to additional errors. One way to approximate this is to calculate the ion pair together. That might not always lead to the correct result since solvents will also play a role here. Obviously you could also include these.

Another problem for your system is probably conformational space. Especially the phosphine is somewhat flexible and will adopt a few conformations (isolated as well as complexed). Here it is also important to find the global minimum, to evaluate the energies correctly.

It is advisable to look for similar systems in the literature and see what they have done. You may also find benchmark studies that will give you more information on how well your methodology performs for such systems. There is no general recipe for success; it always involves a lot of hands-on work. Eventually you will want to compare your calculated results to experimental ones to verify your findings.


Is there any diprotoic acid where both pka's are less than 0?

Ethanedisulfonic acid does indeed have $pK_{a1}=-2.06, pK_{a2}=-1.46$.
Using polycyano substutution can also work.  This paper reports the table below, note that the tetracyano hydroquinone "makes it" (in the excited state produced by irradiation).


Is there a quantum chemistry package which calculates a second-order wavefunction?

As @Wildcat already mentioned, here is a Gaussian 09 input that will generate the MP3 density, which should correspond to the full second-order wavefunction, thanks to Wigner's $2n+1$ rule:

%chk=g09_mp3.chk
%mem=2gb
#p mp3(full)/sto-3g density=current pop=full output=wfn

wavefunction

-1 1
O     -2.0481500778    0.0447984581   -0.0601937089
H     -1.1514551262    0.4243541138   -0.0806303110
H     -1.8059176361   -0.8898508193    0.0217239645
Cl     1.2314805228   -0.2273431057   -0.0015462666

g09_mp3.wfn

The MO coefficients corresponding to the fully relaxed wavefunction $\Psi$ are also dumped to the *.wfn file in addition to the normal output, which I think can be parsed with cclib.


This has everything to do with heat conduction, Setting aside freezing or melting for the time being, the two things that are happening in combination are (1) heat conduction and  (2) the capacity of a parcel of solid to store thermal energy by its temperature changing.  Heat conduction is driven by a spatial gradient in temperature, with heat flowing from a hotter region to a colder region.
So if the outside surface temperature of an object is suddenly increased, the interior temperatures will not change at all initially, and the temperature gradient will be zero everywhere, except at the surface, at which it will be very high. This will cause heat to begin flowing into the very outer layer of the object.  The heat that flows into this outer layer will cause its temperature to begin rising (to store at least part of the heat).  This temperature rise will then create a temperature gradient for heat to begin flowing into the next layer inward, where its temperature will begin rising.  This kind of sequence continues until, later, the temperature at the very center of the object begins to change.  So, as a result of conductive heat transfer, the temperature rise propagates like a wave traveling inward into the object.
This can all be described mathematically and modeled by the so-called transient heat conduction equation.


Comparison of evaporation rate of liquids

Evaporation rates are used to compare solvents for different reasons and under various conditions. One condition would be under vacuum in a rotary evaporator, to remove a low-boiling solvent to assist crystallization of a solute. Another condition would be in a paint, where you desire a higher boiler to allow enough time to spread the paint without brush strokes.
A practical way to compare evaporation rates of solvents is to put a small amount into a dish and weigh it as it evaporates.  Pure compounds will have a fairly sharp disappearance and therefore a sharp time to evaporate; mixtures like kerosene or gasoline will evaporate fast at first, then slower and slower. This time, or the rate for 90% evaporation, can be compared with another solvent under the same conditions.
Evaporation rate is a commonly listed on Safety Data Sheets, and in other tables, generally compared to butyl acetate or ether. One table is at https://www.dispersetech.com/content/11-properties-of-common-solvents.
Many tables of solvent characteristics do not include evaporation rate, but do show vapor pressure at room temperature, e.g., https://sites.google.com/site/miller00828/in/solvent-polarity-table. Vapor pressure at some temperature can be a good indicator of evaporation rate, but a very strong factor in evaporation is wind, or blowing across the solvent surface. Water is particularly complicated because of relative humidity, but it and wind and temperature have been investigated for reservoirs: https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/97WR02405.
Chemical distributors often list all the properties of all the solvents they deal with - and sometimes of materials they don't deal with.


Can Michaelis-Menten constant be zero in the Michaelis-Menten Equation, or is this a non-zero constant?

It does not make sense for the constant to be zero. The constant is usually expressed as follows:

$$ K_M = \frac{k_{\mathrm{reverse}} + k_{\mathrm{catalytic}}}{k_{\mathrm{forward}}} $$

Since the $k$ values are strictly non-negative, the constant would only be zero if $k_{\mathrm{reverse}} = k_{\mathrm{catalytic}} = 0$. In theory, $k_{\mathrm{reverse}}$ could just be $0$ (but that would not be called a Michaelis-Menten model if I want to nitpick), but if $k_{\mathrm{catalytic}}=0$ then you have no overall reaction at all.

Another way to see the same thing is to recall that the Michaelis-Menten constant has the same numeric value as the substrate concentration at the half of the maximum reaction rate. Can this be zero? It would not make any sense to me.

In any case, if you would extend your question to contain information about the epidemic model you would like to create, I am sure a lot of people (including me) would love to help finding a good reaction kinetics analogue for it.


$\ce{Ag+}$ reacts with Sodium Rhodizonate to form a Black colored compound - but only in a neutral or weak-acidic environment.

$\ce{Pb^{2+}}$ should give you a Blue-Violet color (or Scarlet if we're dealing with a weak-acidic environment); therefore, I don't think you're dealing with a lead contamination.

source: http://pubs.acs.org/doi/abs/10.1021/i560110a034?journalCode=iecac0


Bond angle in water vs in oxygen difluoride

Water has a bond angle of 105º and oxygen diflouride has a bond angle of 103.8º, which cannot be explained by repulsion since fluorine is larger than hydrogen$[1]$. This is phenomenon is, however, explained by Bent's rule:

  Bent's rule is that in a molecule, a central atom bonded to multiple groups will hybridise so that orbitals with more s character are directed towards electropositive groups, while orbitals with more p character will be directed towards groups that are more electronegative.

p orbitals are oriented 90º from each other, so orbitals with more p character have smaller angles between neighboring atoms. The effects of this phenomenon can also demonstrated by comparing bond lengths.

$[1]$ Bent, H. A. Chemical Reviews 1961, 61 (3), 288.


Regioselectivity in the formation of an imidazo[1,2-a]pyrimidine

I propose the route in the following scheme. I don’t see any unfavourable intermediates. In short, the first step would be formation of a Schiff base with the extracyclic amino group and the aldehyde. This — in one of many orientations as shown below — places the bromide nicely so that one of the pyrimidine nitrogens can attack it nucleophilicly. Upon displacement, an intermediate nitrogen-centred cation is generated; loss of a proton and switching of electrons regenerates aromaticity and gives the desired product.

Scheme 1: Proposed reaction scheme. The attacking nitrogen $\mathrm{sp^2}$ orbital in step 2 is drawn explicitly.

We can draw structures of the Schiff base intermediate that would allow either of the pyrimidine nitrogens to attack. However, the top one experiences a stronger $-I$ effect by the neighbouring $\ce{CF3}$ group hence its electron density and nucleophilicity should be reduced. The para-nitrogen does not experience this inductive effect since it greatly weakens with greater distance.

The ortho-nitrogen’s lone pair is indeed affected since inductive effects affect both σ and π bonds while mesomeric effects mainly affect π electrons only.


Question about calculating enthalpy change

The approach to estimating the enthalpy change in a reaction by using bond dissociation energies assumes homolytic cleavage of the bonds: each atom gets one electron from the covalent bond.

The bond dissociation energy of the $\ce{H-H}$ bond is reported as 436 kJ/mol.  The relevant reaction is:

$$\ce{H2(g) -> 2H.(g)}$$

Note that this reaction is not the same as $\ce{H2 -> 2H+ + 2e-}$. The homolytic cleavage   reaction produces hydrogen atoms, not hydrogen ions. If we look up the standard enthalpy of formation for hydrogen atoms, it is +218 kJ/mol. Since the standard enthalpy of formation for $\ce{H2}$ is defined as zero, using the enthalpies of formation we also get $\Delta H = +436$ kJ/mol for this reaction.

It is generally hard to use bond dissociation energies for heterolytic reactions (reactions that produce ions from molecules).


How does an ice bath change the rate of an reaction?

Placing the reaction vessel in ice will lower the temperature of the reaction and thus slow down the reaction. Temperature is a measure of the average kinetic energy of the molecules, so lowering the temperature lowers their kinetic energy. In order for reactants to form products, they must collide with a certain amount of energy. Thus, at a lower temperature it will take longer for the same amount of product to be formed.

You also need to keep in mind that at lower temperatures the equilibrium constant will change. Since this is an exothermic reaction, at lower temperatures there will be more product formed (Le Chatelier's principle). There will be more moles of oxygen present in the iced reaction. However, you're measuring the volume of oxygen, and a colder gas will have a lower volume (for the same number of moles of gas). So you've got two opposing trends: the ice will cause more oxygen to be produced (which would normally correlate with a larger volume) but the colder temperature will also cause the gas to shrink. Without more information, such as the temperature of the two reactions, I don't think you can determine whether you will wind up with a larger or smaller volume, so it would be hard to determine the shape of the graph for the colder reaction.


How does a combustion process produce carbon monoxide?

If combustion occurs in oxygen, (at least) the following reactions are relevant:
$$\ce{C + O2 -> CO2} $$
$$\ce{C + 1/2 O2 -> CO} $$

You can favor the first reaction by having more oxygen in the gas phase, since it required twice $\ce{O2}$ as the first.

However, one of the most important factors is the temperature: high temperature always favors the $\ce{CO}$ product, since reaction wants to proceed fast and react with any available oxygen.
You can clearly see these dependences in the Figure 6 of this article, at low temperatures the CO/CO2 ratio is low, but quickly increases by heating.

More detailed information is easily found in some classical papers, a full PhD thesis on it, or even calculations with modern quantum chemistry methods.


Your questions involve a lot of chemical concepts.

An oxydo-reduction is reaction where electrons are exchanged between species including or not molecular rearrangement.

You totally can write down any salt in the following way, emphasizing ion charges:

$\mathrm{NaCl}_{(s)} \equiv \mathrm{Na^+Cl^-}_{(s)}$

As long as you specify the phase, because when writing down an ionic form the chemist might assume that it is solvated into water. To underline it you may write the following:

$\mathrm{Na^+Cl^-}_{(aq)}$

Further more solid generally have numerous possible distinct phases, therefore you may need to specify which crystalline structure the salt have.

The following equation is explicit:

$2\mathrm{Na}^{(0)}_{(s)} + \mathrm{Cl_{2(g)}} = 2\mathrm{Na^+Cl^-}_{(s)}$

And underlines that an electron is withdrawn from metallic sodium in order to produce chlorine. Anyway, this equation is global balance this it hides many details of the complete and complex chemical process, especially the formation of the crystalline structure. And it does not provide any information on conditions of reaction.


Is there an inexpensive solvent I can use to dissolve thermal paste without scrubbing?

First the sad part - there is no magic compound that will dissolve thermal paste like a champ and leave your electronics alone. If that was no "sine non qua" as they say, we could have suggested a varied assortement of highly aggressive halogenated solvents which we would mix with something else for the extra bang, some hydrogen peroxide perhaps, or the good old fuming nitric acid. Things would happen, rapidly.

As for all-round no-nonsense solvents, isopropyl alcohol is a big favourite, it is fairly environmentally friendly especially if you consider your hands and your tabletop as part of the environment. And of course it solvates a wide range of icky substances that needs removal. As a small step up the agressive solvent ladder would be to consider acetone, "citrus solvent" (basically oils and acid of lemons, it kicks ass) or acetic acid based solvents. But none of them will magically clean the surface, you'll have to use some elbow grease. Or thumb grease, I guess.


Why is Electrochemistry often the method of choice for determining Ksp values?

Perhaps a simple example will illustrate how a solubility product constant, $\ce{K_{sp}}$, may be determined via measurement of an appropriate voltaic cell potential. Consider the estimation of the $\ce{K_{sp}}$ value for the following equilibrium:

$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq) \quad K = K_{sp}} \tag{1}$$

This equilibrium lies far to the left because AgCl has very poor solubility in water and, as a direct consequence, the concentrations of silver ions and chloride ions are very low. Therefore, the $\ce{K_{sp}}$ value is small, i.e., much less than unity, and a direct measurement of the $\ce{K_{sp}}$ value is difficult due to the low solubility of AgCl.

However, we have the following two half-reactions, with their standard reduction potentials (from standard tables):

$$\ce{Ag+ +  e−  <=> Ag(s)} \quad E^\circ_\mathrm{red} = \pu{+0.7996 V} \tag{2}$$

$$\ce{AgCl(s) + e− <=> Ag(s) + Cl−} \quad  E^\circ_\mathrm{red} = \pu{+0.22233 V} \tag{3}$$

Half-reaction (2) has the more positive standard reduction potential, hence more negative $\ce{\Delta G}$, so it will be the spontaneous reduction half-reaction at the cathode in a voltaic cell that can be constructed. Reversing half-reaction (3) yields the oxidation half-reaction at the anode in the constructed voltaic cell:

$$\ce{Ag(s) + Cl− <=> AgCl(s) + e−}  \tag{4}$$

So the voltaic cell notation is

$$\ce{Ag(s)|Cl^-(aq;\pu{1 M})||Ag^+(aq;\pu{1 M})|Ag(s)}$$

and the net cell reaction is obtained by adding half-reactions (2) and (4), resulting in

$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s) \quad K = 1/K_{sp}} \tag{5}$$

This equilibrium lies far to the right because it is simply the reverse of half-reaction (1). Therefore, the $\ce{K}$ value is large. Indeed, it is simply the reciprocal of half-reaction (1)'s $\ce{K_{sp}}$ value. This is consistent with the fact that mixing aqueous solutions containing silver ions and chloride ions immediately results in spontaneous formation of AgCl solid that precipitates out of solution.

The Nernst equation, where $\ce{Q}$ is the reaction quotient, is

$$ {E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{R T}{n F} \ln(Q) } \tag{6}$$

The natural logarithm term is basically a correction term that takes into account the fact that voltaic cells are usually not at standard state, i.e., they are not dead yet. This is accomplished using $\ce{Q}$. At standard state, $\ce{Q} = 1$, and at equilibrium, $\ce{Q} = K$, where $\ce{K}$ is the equilibrium constant. In the present case, $\ce{K}$ pertains to equation (5).

Now, at equilibrium (and ignoring units), $\ce{\Delta G_{cell} = 0, E_{cell} = 0 and Q = K}$, so

$$E^\circ_\mathrm{cell} = \frac{R T}{n F} \ln(K) = {E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = \pu{+0.7996 V} -(\pu{+0.22233 V}) = \pu{+0.5773 V}} \tag{7}$$

Note that the standard cell potential is defined as

$$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag{8}$$

With $\pu{T = 298.15 K}$ and $\pu{n = 1}$, $\pu{RT/nF \approx 0.02569 V}$, so equation (7) can be solved for $\ce{K}$:

$$\mathrm{K = e^{nfE^\circ _{cell} /RT} \approx e^{+0.5773V/+0.02569V} = 5.74 \times 10^{9} \tag{9}}$$

Hence $$\pu{K_{sp} = 1/K = 1.74 \times 10^{-10}} \tag{10}$$
which is close to typically quoted values in the literature.


What Are Examples of Reducing Sugar in Mailliard Reaction?

Quoting from medium.com

  If it wasn’t for Maillard, we would not have juicy seared steak, deep fried chicken wings, nor grilled sausages. All these browning, aromas and flavor bursts are a result of simple sugars reacting together with proteins. However, Maillard is not the only reaction occurring between these different molecules in food.

This is one of the chemical reaction that whether you have an interest in chemistry or not, we all carry out on a regular, maybe even in daily basis.The Maillard Reaction${^1}$ causes two types of browning of foods: enzymatic & non-enzymatic browning. Non-enzymatic browning can again be split in two types: the Maillard reaction and caramelization${^2}$.

The Proteins contained in meat and other comestible goods are usually rich in the amino acids arginine and lysine. The side chain func- tional groups of Arg and Lys react with reducing carbohydrates such as glucose or lactose to form Amadori reaction products${^3}$.

Maillard reaction chemistry

The Maillard reaction is a chemical reaction between an amino acid and a reducing sugar, usually requiring the addition of heat.The Maillard reaction is a ‘non-enzymatic browning' reaction involving reduced sugars with compounds possessing free amino groups. A reactive sugar, such as glucose(a reducing sugar) can react with amino groups in amino acids, peptides, and proteins as well as with other molecules that contain free amino groups${^4}$.

Why a Reducing sugar?

Freshly prepared solutions of glucose in water gradually change by the opening of the pyranose ring to the free aldehyde form. This reaction is a reversal of a hemiacetal formation reaction. Reducing sugars behave like glucose${^5}$.

Following are the steps involved.

Initial reaction between a reducing sugar and an amino group forms an unstable Schiff base${^6}$.

The Schiff base slowly rearranges to form Amadori products${^7}$.

Next, oxidative decomposition of Amadori products can lead to the formation of a wide range of reactive carbonyl and dicarbonyl compounds.The complex reactions are summarised in figure below${^7}$.

Different amino acids produce different amounts of browning.

The Maillard reaction is responsible for many colors and flavors in foodstuffs${^8}$.

  The Maillard reaction does not just make food taste better. It has a science involved in it. Understanding the key facts allows you to manipulate the process of cooking according to what your taste buds crave for. It is the difference between blindly following a recipe and being able to make a recipe work for your taste${^9}$.

UPDATE

  The Maillard reaction occurs most readily between temperatures 300°F-500°F. When the meat is cooked, the outside temperature becomes hotter than the inside, which triggers the reaction. Protein molecules in the meat are bonded in coils and when heat is applied, those bonds begin to break and the coils begin to unwind. Most of the water content in the muscle fibers then leak out. Reactive carbonyl groups of the sugars react with the nucleophilic amino acids, which produces glycosylamine and water. Pentose sugars react more than hexoses

The pentose sugars might be part of Nucleotide of DNA and RNA(In case no external reducing source is added).

That said, Reducing sugar free xylose, has been reported in onion, strawberries, prunes, apples, pears, grapes, juniper berries, barley malts, brewhouse worts, maple syrup, asparagus, the white and the yolk of eggs, corn, tomatoes, apricots, bamboo shoots, potatoes, beans, alfalfa, beer, and mangoes${ ^{10}}$.

Most of the carbohydrate of the carrot was found to be in the form of soluble reducing substance(You could try as additives for steak prepration)${^{11}}$.

References

${^1}$:https://www.compoundchem.com/2015/01/27/maillardreaction/

${^2}$:https://foodcrumbles.com/maillard-reaction-mechanism-hard-core-chemistry/

${^3}$:https://www.iris-biotech.de/global/knowledge-base/custom-synthesis/maillard-reaction-products

${^4}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx

${^5}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx

${^6}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx

${^7}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx

${^8}$:https://www.scienceofcooking.com/maillard_reaction.htm

${^9}$:https://medium.com/@vijinimallawaarachchi/maillard-reaction-the-science-of-browning-flavoring-and-aroma-1605133960ae

${^{10}}$:https://www.researchgate.net/publication/268009365_Free_Sugars_in_Fruits_and_Vegetables

${^{11}}:$http://www.jbc.org/content/46/3/537.full.pdf


Why is Enthalpy of Compound at Some Temperature Equal to its Enthalpy of Formation at the Same Temperature?

They are not equal.  The enthalpy of a compound at temperature T is relative to the pure elements at the reference temperature $T^0$.  So it is equal to the standard heat of formation at the reference temperature plus $C_p(T-T^0)$.


Is there a better way balancing reaction equations than systematic trial and error?

When using matrices, you are really just solving a system of equations. You can use any technique you know to solve the system of equations that a chemical equation represents. Using matrices is only one technique that may be useful for complicated equations or if you have a calculator/MATLAB on hand. In the given example, it would be easier to solve it directly with guess and check.
Writing the system of equations
In your example
$$\ce{C3H8 + O2 -> CO2 + H2O},$$
write in some variables as coefficients:
$$\ce{a C3H8 + b O2 -> c CO2 + d H2O}.$$
Now, we can write out a system of equations based on the stoichiometry of each element on either side of the equation. In other words, write equations that express the number of each element on the left hand side to the number of each element on the right hand side:
\begin{align}
3a &= c\\
8a &= 2d &\Longleftrightarrow&& 4a &= d\\
2b &= 2c + d
\end{align}
Finally, we have to choose a value for one variable to substitute in. This will be the fourth and final equation. We want integer coefficients here, so we will choose a substitution that gives us all integer coefficients. Otherwise, just multiply your coefficients at the end so that they are all integers. Here, I'll let $a = 1$.
Solving the system
Substitution
Substituting the first two equations into the third gives:
\begin{align}
2b &= 6a + 4a = 10a &\Longleftrightarrow&& b &= 5a
\end{align}
\begin{align}
a &= 1\\
b &= 5\\
c &= 3\\
d &= 4
\end{align}
And the balanced equation is
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}.$$
Using matrices
To answer the question though, let's do it with matrices. I won't go through the whole explanation of solving the system but I'll show you how to build the matrices and get an answer. Remember, we are just representing the same system of equations from before, but with matrices now!
We need to build two matrices, $A$ and $b$. Reorganize the equations we wrote previously so that all of the variables are on the left hand side and a constant is on the right hand side. Also, notice I added in our equation let $a = 1$:
\begin{align}
1a + 0b + 0c + 0d &= 1\\
3a + 0b -1c + 0d &= 0\\
0a + 2b -2c - 1d &= 0\\
4a + 0b + 0c - 1d &= 0
\end{align}
These equations are written like this:
\begin{align}
A &=
\begin{bmatrix}
1 & 0 & 0 & 0\\
3 & 0 & -1 & 0\\
0 & 2 & -2 & -1\\
4 & 0 & 0 & -1\\
\end{bmatrix}\\
b &=
\begin{bmatrix}
1\\
0\\
0\\
0\\
\end{bmatrix}
\end{align}
Notice that $b$ is just a column of the constants on the right hand sides of the equations, in the same order as they were given. The order is important here!
There are a few ways to solve this manually or in a calculator/MATLAB. If you have a TI-8x calculator or similar, you could use its rref function (this transforms an augmented matrix into reduced row echelon form, I'm sure google could help you out) or you could multiply $A^{-1}b = x$. In MATLAB, just do: A\b. This gave me:
$$
x =
\begin{bmatrix}
1 \\ 5 \\ 3 \\ 4
\end{bmatrix}
$$
Each row in this final matrix $x$ is the value for $a$, $b$, $c$, $d$ in that order, giving the balanced chemical equation
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}.$$


More the number of resonance structures (alias mesomers) more is the stability of the carbocation so benzylic carbocation with 4 mesomers is more stable than 2 structures of the allylic carbocation.


Reason for usage of Metal chlorides and Platinum or Nichrome wires in flame test

Metal chlorides are usually much easier volatilized than oxides or sulfates. They melt or sublime at temperatures much lower than 800°C. So they are volatilized in any flame like Bunsen burners.

Platinum is not attacked by hot chlorides or acids, and is not oxidized in hot air. So it can be used for a great number of consecutive tests in the flame. It will never wear out in the flame.