The title is actually a different question than the post, which contains several questions, but I'll answer all of them since they are closely related.

Why are optically active compounds abundant in nature?

Nature needs some kind of complexity to work. To build a living system you need hundreds, thousands or even millions of different molecules, just a few won't do it. If you just draw random molecules you'll end up with a lot of chiral ones. Unless you are sticking to linear chains without functional groups or aromatic (flat) systems you will almost always end up with chiral molecules. It actually would be surprising if there would be no chirality.

Given that many optically active compounds are extracted from nature, why do organisms go through pain of seperating enantiomers for their consumption ?

As Zhe already said in the comment: they don't need to separate those. All the animals and plants basically use only one enantiomer.

Why do optical isomers are found in nature that are hard to seperate in a lab?

Well, you are right. Enantiomers have the same physical properties unless they are in a chiral environment. Organisms on earth are able to produce only one enantiomer because the whole environment is chiral which means both enantiomers have different physical properties.

If we want to separate enantiomers in the lab we are using the same system. We take a chiral phase for chromatography (those are quite expensive) which means both enantiomers behave differently.

We are also able to synthesize enantiomerically pure compounds if we use a chiral environment during synthesis. But this is often expensive and we aren't nearly as good as biological systems doing this.


How can I properly calculate the isoelectric point (pI) of amino acids?

Isoelectric point of an amino acid is the $\mathrm{pH}$ at which the molecule carries no net charge[1]. It can be calculated by the average of the relevant $\mathrm pK_\mathrm a$ values as you have mentioned.
Your confusion seems to stem from choosing the relevant $\mathrm pK_\mathrm a$ values. For this we should refer to the titration curve of the amino acid.
For a neutral amino acid[2]:

From the curve we can infer that the $\mathrm{pI}$ is simply the average of the two $\mathrm pK_\mathrm a$ values of the carboxylic acid and the amino group.
For a basic amino acid[2]:

From the curve we can infer that the $\mathrm{pI}$ is simply the average of the two $\mathrm pK_\mathrm a$ values of the two amino groups. The $\mathrm pK_\mathrm a$ of the carboxylic acid group is not relevant.
For an acidic amino acid[3]:

From the curve we can infer that the $\mathrm{pI}$ is simply the average of the two $\mathrm pK_\mathrm a$ values of the two carboxylic acid groups. The $\mathrm pK_\mathrm a$ of the amino group is not relevant.
Here are examples for all three cases:

References:

Wikipedia

Titration curves for neutral and basic amino acids

Titration curve for acidic amino acids


What is the purpose of atomisation in atomic absorption spectroscopy?

Atomic absorption spectroscopy is a quantitative measure of the elements in the analyte.
We atomise the analyte so that the absorption of light corresponds to the characteristics of the elements.
Otherwise how we can predict the fundamental elements when these are still part of some molecules?
We measure the absorbance and refer to the Beer-Lambert law.


You have already ruled out the equations b, c, and e.

  But I'm not sure how to compare the remaining two reactions.

Have a look at equation a again.

The reaction of $\ce{P4O10}$ with water generates phosphoric acid ($\ce{H3PO4}$), which is a fairly strong three-protonic acid ($\mathrm{pK_{a1} = 2.16; pK_{a2} = 7.21}$).

As a result, subsequent deprotonation of phosphoric acid
$$\ce{H3PO4 + H2O <=> H3O+ + H2PO4-}$$

will pull the equilibrium in equation a to the right.

Hydrogen sulfide ($\ce{H2S}$), on the other hand, is a rather weak acid with $\mathrm{pK_{a} = 7.0}$. Consequently, you will find a significant amount of undissociated hydrogen sulfide in aqueous solution. Therefore, the correct answer in the comparison is indeed d.


Hofmann rule and deviation from E2 mechanism

The difference between the different eliminations like E1, E2 or Hoffmann elimination is the proper choice of the base and proper solvent medium.

Let's take your example i.e. 2-bromo-2-methylbutane, and choose different bases and reaction media for observing differences between E1, E2 and Hoffmann elimination.

Why not E1?

First let's consider the reaction as:

$$\ce{2-bromo-2-methylbutane + EtO^- (in EtOH) -> }$$

Clearly, this reaction is conducted in a highly polar solvent where the carbocation - formed when $\ce{Br^-}$ leaves - gets stabilised. The base then abstracts the hydrogen of $\ce{C3}$ to form 2-methylbut-2-ene by an E1 mechanism.

If you use the same substrate but this time in a highly non-polar solvent, the possibility of E1 decreases as carbocation formation won't get extra stability. But as the base is not sterically hindered, it can still abstract hydrogen to form the Saytzeff product, but there is a case where it can also form the Hoffmann product.

Why not E2?

The E2 mechanism follows an "anti-periplanar orientation of hydrogen and leaving group". Suppose the reactant's stereochemistry is such that there is no anti-$\ce{H}$ on the more substituted adjacent carbon for forming the Saytzeff product, but there is an anti-$\ce{H}$ on a less substituted adjacent carbon, which will lead to the Hoffmann product. In that case, the major product can be the less substituted alkene due to stereospecificity of E2 reaction.

Why E1cb?

Now consider your above given case. Here, the base $\ce{t-BuO^-K+}$ is sterically hindered and is a very strong base. So, prior to any carbocation formation, this base will abstract hydrogen from the less hindered position, and form a carbanion. Also, a less substituted carbanion is more stable than a more substituted one (due to a smaller +I effect). Both these effects lead to formation of a less substituted alkene via an E1cb mechanism.

So, depending upon which product is needed as a major one and the proper stereochemistry of reactant, you should choose whether to go with an E1, E2, or E1cb mechanism by choosing the proper base (bulky or less hindered) and proper solvent (polar or non-polar).


Does the addition of water to N2H4 produce N2H4

Yes, equations must be balanced, thus the only correct way to write this is:

$$\ce{N2H4 + H2O <=> N2H5+ + OH-}$$

Easy to remember: Nitrogen typically has a free electron pair which can act as a Lewis base for a proton.

If your textbook says $4$, it is an error in your book. Good job on spotting it and realising it.


Can a reaction with positive Gibbs free energy change ever occur?

Yes, $\Delta G = -RT\ln K$.

If a reaction is endothermic with a negative entropy change, $\Delta G$ is positive.

That $\Delta G$ is positive only means that $K<1$.

$\Delta G$ would need to approach infinity for $K$ to approach zero.

Therefore, at equilibrium, there will always be some products in principle no matter how unfavorable the reaction.

As a practical matter, the reaction would need to be only slightly endothermic and having only a slight negative entropy change to get a significant amount of product.

However, if the small amount of product is continuously removed from the system, as for example by precipitation from a liquid phase, or gas bubbling out of a liquid phase, or by liquid-liquid extraction, one can keep the reaction going forward.


In aminobenzoic acid, do we have three different pKa's?

The two acidity constants in 4-aminobenzoic acid are due to the loss of $\ce{H^+}$ from the protonated form of 4-aminobenzoic acid (the 4-carboxyphenylammonium cation), transforming it into 4-aminobenzoic acid, and then a further loss of $\ce{H^+}$ turning it into the 4-aminobenzoate anion.

The first two ionizations have a $\mathrm{p}K_\mathrm{a}$ of around 5, while the next ionization quite likely has a $\mathrm{p}K_\mathrm{a}$ well below ~25 (compare with other substituted anilines), so you won't find any appreciable amount of deprotonated nitrogen atoms in an aqueous solution. Groups with N−H bonds are far less acidic than their O−H counterparts due to the lower electronegativity of the nitrogen atom, which is less capable of stabilizing the extra negative charge.


What are the units for m in the average kinetic energy formula?

$$\ce{Kinetic ~Energy ~= ~1/2 (mass~of ~the~object~ [kg]) * (velocity~of~the~object~[m/sec])^2}$$


Stability order of octahydrobiphenylenes

While solving other related questions, I came up with a idea that

  More is the number of $\alpha$ hydrogen present(hydrogen attached to carbon which is adjacent to a carbon containing double bond ) more stable is the compound.

In the question $Z$ has maximum 10 $\alpha$ hydrogen so it is the most stable.$X$ and $Y$ both have 8 $\alpha$ hydrogen.But $Y$ is more stable since conjugated double bond in short non aromatic cyclic system is highly disfavoured.

Hence the correct stability order is $Z>Y>X$

Side note- I am still not satisfied with my second argument.Any other explanations or thought will be highly appreciated.


Reaction of Silicon monoxide with alcohols

I doubt very much its likely to occur.  Reducing an alcohol to a alkane is not trivial. strong reducing agents such as $\ce{LiAH4}$ cannot reduce alcohols, and to my knowledge $\ce{TiO}$ cannot reduce alcohols which would be an even better canidate that $\ce{SiO}$.


How does chromium help stainless steel prevent rusting?

TL;DR
    Note that the passive layer forms on the surface, there needn't be any change to lattice constant. Chromium needn't migrate , the Cr present on the surface will form the layer to protect it. The key point is how the layer develops from say a single-atom layer of oxide to the usual/maximum width by migration of electron and oxygen in the oxide layer. The process at molecular and atomic level is more complex than you think.

From wikipeddia article on Passivation

  There has been much interest in determining the mechanisms which describe how the thickness of the oxide layer on a material increases with time. Some of the important issues include: the relative volume of the oxide compared to the parent metal, the mechanism by which oxygen diffuses through the metal oxide to the metal - oxide interface and the relative chemical potential for the oxide to form. Boundaries between micro grains, if the oxide layer is crystalline, form an important pathway for oxygen to reach the unoxidized metal below. For this reason, vitreous oxide coatings – which lack grain boundaries – can retard oxidation. The conditions necessary (but not sufficient) for passivation are recorded in Pourbaix diagrams.

  Some corrosion inhibitors help the formation of a passivation layer on the surface of the metals to which they are applied. Some compounds, dissolving in solutions (chromates , molybdates) form non-reactive and low solubility films on metal surfaces.

The mechanism

  All passivating‐type inhibitors are oxidizing agents. However, not all oxidizing agents are inhibitors. A passivating ‐type inhibitor functions by producing local‐action current which anodically polarizes a metal into the passive potential region and thereby provides the means for obtaining a noble mixed potential. This mechanism is independent of whether passivity is caused primarily by oxide or adsorbed films.

  The major factors which determine whether a particular system will exhibit passivity are:
  (a) reversible potential of the redox system created by the oxidizing inhibitor;

  (b) electrochemical parameters for reduction of the inhibitor on the metal surface (exchange current, Tafel slope, and limiting diffusion current), and

  (c) critical anodic current and Flade potential(The potential of a passive metal immediately preceding a final step fall from the passive to the active region.) of the metal. If oxygen is present in the system, some chemicals may function by changing its reduction kinetics.

  It is further shown that the amount of inhibitor found associated with the surface is not necessarily related to adsorption. Data are presented for passivation of stainless steel and titanium. The mechanism is discussed in terms of various oxidizing agents including oxygen, chromate, molybdate, and pertechnetate.

Some theories I could find for passivation are Kinetic Theory of Inhibition and Passivation in Electrochemical Reactions and Scaling theory in a model of corrosion and passivation


All the special rules you need for naming of such dispiro cycloalkane systems can be found in one chapter of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-24.2.2 Linear polyspiro alicyclic ring systems
Polyspiro parent hydrides consisting of unbranched assemblies of three or more saturated cycloalkane rings are named using the nondetachable prefixes ‘dispiro’, ‘trispiro’, etc., according to the number of spiro atoms present, cited in front of the name of the acyclic hydrocarbon that has the same number of skeletal atoms. The von Baeyer spiro descriptor indicates the number of carbon atoms linking the spiro atoms by arabic numbers that are cited in order, starting at the smaller terminal ring if one is smaller, and proceeding consecutively, always by the shorter path, to the other terminal ring through each spiro atom and then back to the first spiro atom; the numbers are separated by full stops and enclosed in square brackets. The compound is numbered in the order in which the numbers of the von Baeyer spiro descriptor are cited, including spiro atoms when encountered for the first time. Each time a spiro atom is reached for the second time its locant, which has already been assigned, is cited as a superscript number to the number of the preceding linking atoms.
(…)

Following these rules step by step, you will find that the name for the compound that is given in the question is dispiro[3.0.45.14]decane.


Why do we need to incubate this system for 18 hours at 60 degrees Celsius?

Until you post the original source procedure, my answer is going to be generic.

  When I am repeating an experiment I found in the literature, why do I need to do [insert step that seems unnecessary, difficult, or just plain long]?

The short answer is, because it works.

You need to incubate the slides at 60 oC for 18 hours because that's what the author of the paper found to be necessary to guarantee that the slides were properly PEGylated. It's possible that the author of your procedure is also using a procedure that he or she found in the literature, and the the rationale for the specific incubation procedure will require some digging.

The longer answer is, the reaction probably needs some incubation to run to completion. The reaction between the PEG and the ITO surface is not instantaneous. If you want the coating to be good, you need to give it time to cure, and mild heating can accelerate the process. While 18 hours seems like a long time, it would likely take interminably longer at room temperature. Could you heat at a higher temperature for less time? Probably, but I would wait until you are certain you can make the existing procedure work before messing with it. That way you have some successfully coated slides to compare to.

The 18 hour period has one of two origins. The second is more likely.

The developer of the procedure maximized the efficiency of the process: He/she carefully experimented with the conditions until he/she discovered the exact time/temperature combination that maximized the efficiency of the reaction - getting the best coating in the shortest time.
The developer of the procedure maximized the convenience of the process: He/she determined 1) that the coated slide needed to cure for several hours (3 or 4) at 60 oC,  2) that higher temperatures were detrimental to the quality and integrity of the coating, and 3) prolonged heating at 60 oC is not detrimental. In this case, the researcher puts the slides into the oven in the afternoon to cure overnight and picks up the cured slides the next morning. 4 PM on Monday afternoon to 10 AM on Tuesday morning is 18 hours. As a synthetic chemist, I often wrote down '18 hours', when I really meant 'overnight'. The reaction may have been done at 6 hours or 8 hours, but 18 hours is more convenient. I set up the procedure one afternoon a few hours before I left for the day so that I could make sure it was stable. I worked it up the next morning.


Originally, both of us, andselisk and I, thought you are taking about a plain which goes through all four groups ($\ce{H-H-Cl-OH}$ squire). It is a forbidden plane when it come to symmetry of a compound, if it had been the case. Fortunately it is not the case. We later realized after Mithoron's comment that the plane you are talking about actually goes through $\ce{C^*-C^*-Cl-OH}$ squire instead (asterisks indicate chiral $\ce{C}$s). That is an allowed plane for symmetry concern. Yet again, you are wrong about it's been a plane of symmetry. For your convenience, I draw all four possible isomers of 3‐chlorobutan‐2‐ol and relevant sawhorse structures showing the plane of your concern:

I tried to put the plane as it is a mirror on a plane where $\ce{C^*-C^*-Cl-OH}$ squire overlap. The particular stereoisomer representing both Fischer projection and sawhorse structure (with your 'plane of symmetry') is $(2R,3S)$-3‐chlorobutan‐2‐ol, so I drew that particular isomer first. Now look at my sawhorse structure below that particular structure. The $\ce{C^*-C^*-Cl-OH}$ squire is on the plane so that those groups and atoms do not count on symmetry. The only countable groups are two methyl groups and two hydrogen residues. If you consider the plane as a mirror (which dissects each $\ce{H-C^*-CH3}$ angle), you would see both $\ce{CH3}$-groups are in the front of the mirror and hydrogen residues are inside of the mirror (resembling mirror images). As a result, we can conclude that the mirror is not a plane of symmetry because each $\ce{H}$ atom can't be the mirror image of adjacent $\ce{CH3}$ group. Therefore, both carbon atoms marked with $\color{red}{red}$ $\color{green}{green}$ circles are chiral centers. As a result, there are 4 stereoisomers for 3‐chlorobutan‐2‐ol: $(2R,3S)$-, $(2S,3R)$-, $(2R,3R)$-, and $(2S,3S)$- where first two and last two isomers are enanthiomers of each other (as indicated in the diagram).


EDTA titrations of singly charged cations

Titrations involving chelating of metal ions with ethylenediaminetetraacetic acid (EDTA) is an extremely useful analytical tool and a very well studied system in analytical chemistry. It can be bind to virtually any metal ion, evidently even to alkali metals. For instance, sodium ion complexes with EDTA in certain situations (Ref.1). The abstract of the reference states that:

The crystal structures of tetrasodium (1,2-ethanediyl-dinitrilo)tetraacetate pentahydrate and disodium dihydrogen (1,2-ethanediyldinitrilo)tetraacetate dihydrate have been determined. $\ce{Na}$ ions prefer to act as bridges between different $\mathrm{edta}$ ligands and only in the tetrasodium salt is the $\mathrm{edta}$ ligand hexadentate to one $\ce{Na}$ ion. $\ce{Na}$ ions display seven-coordination, when coordinated to one $\mathrm{edta}$ ligand (distorted pentagonal bipyramid), and six-coordination, when linked to several $\mathrm{edta}$ ligands (from distorted octahedral to trapezoidal bipyramidal geometry). An $\ce{Na}$ ion in the tetrasodium salt shows five-coordination, because the sixth site is blocked by the $\mathrm{edta}$ ligand. $\ce{H}$ atoms in the disodium salt are linked to amine groups. Only these two salts can be obtained on varying the $\mathrm{pH}$ of an aqueous solution containing $\ce{Na+}$ and ethylenediaminetetraacetic acid. The first is obtained around $\mathrm{pH}$ 2-5 and the second around $\mathrm{pH}$ 12.

Nonetheless, there are no evident in the literature that EDTA titrations can be done to determine alkali metal concentration (agreeing with M. Farooq in his comment). As a matter of fact, disodium ethylenediaminetetraacetate (dihydrate) salt is the EDTA source it has been almost always used in most of EDTA titrations to determine  a large number of metal cations including alkaline earth and transition metals. Keep in mind that $K_f$ values of alkali metal complexation with EDTA are very law compared to other metals (Ref.2):
$$
\begin{array}{|c|c|c|c|}
\hline
\text{Cation} & K_f & \text{Cation} & K_f & \text{Cation} & K_f & \text{Cation} & K_f\\
\hline
\ce{Li+} & 6.17 \times 10^2 & \ce{Be^2+} & 1.58 \times 10^9 & \ce{Cr^2+} & 3.98 \times 10^{13} & \ce{Sc^3+} & 1.26 \times 10^{23}\\
\ce{Na+} & 4.57 \times 10^1 & \ce{Mg^2+} & 6.17 \times 10^8 & \ce{Mn^2+} & 7.41 \times 10^{13} & \ce{Y^3+} & 1.23 \times 10^{18}\\
\ce{K+} & 6.31 \times 10^0 & \ce{Ca^2+} & 4.90 \times 10^{10} &\ce{Fe^2+} & 2.09 \times 10^{14} & \ce{La^3+} & 3.16 \times 10^{15}\\
\ce{Ag+} & 2.09 \times 10^7 & \ce{Sr^2+} & 5.37 \times 10^8 & \ce{Co^2+} & 2.04 \times 10^{16} & \ce{In^3+} & 1.00 \times 10^{25} \\
\ce{Tl+} & 3.47 \times 10^6 & \ce{Ba^2+} &  7.24 \times 10^7 & \ce{Ni^2+} & 4.17 \times 10^{18} & \ce{Bi^3+} & 6.31 \times 10^{27} \\
\hline
\end{array}
$$
This lack of complexation of alkali metals with even strong chelator such as EDTA can be attributed to their lower Lewis acidity (EDTA is a strong Lewis base because it contains both amino and carboxylate groups).
Apart from alkali metal cations, singly charged transition metal cations form  complexes with chelates and can be determine by relevant titrations. For example, concentration of silver(I) ions ($\ce{Ag+}$) can be determined by cyanide titration (a monodentate):
$$\ce{Ag+ + 2CN- -> [:\!N#C-Ag-C#N\!:]-}$$
The fairly high $K_f$ values for those singly charge cations (e.g., see the values for $\ce{Ag+}$ and $\ce{Tl+}$ in the table above) other than alkali metals suggest that they were able to involve in complexation titrations (They are compatible with alkaline earth metals).
References:

M. Font-Bardia, X. Solans, M. Font-Altaba, “Sodium ion complexes with ethylenediaminetetraacetic acid,” Acta Crystallographica Section C 1993, C49, 1452-1456 (https://doi.org/10.1107/S0108270193001350).
Arthur Earl Martell, Robert M Smith, Editors, In Critical Stability Constants, Volume 1; Plenum Press: New York, NY, 1974, pp. 204-211 (ISBN-13: 978-0306352119).


Have the STRUCTURES of synthesized anthropomorphic molecules (nanoputians) been verified experimentally?

The first article wikipedia's entry refers to (here) is not only a publication behind a paywall, but with freely available supporting information, too.  This in principle allows to replicate the work.

Browsing through the latter provides all / most / many details of how building blocks were prepared, how they were joined; including both numerical and graphical display of 1H and (broadband decoupled) 13C NMR spectral data.  Like the following, found on page 22 of the SI:

and 13C NMR (same page, below)

Perhaps (i.e. the following is a speculation) the "main" article in the journal includes additional spectroscopic data (UV, IR; MS) or characterisation (melting point, combustion analysis, etc.) as the article stretches from page 8750 to 8766.

To have a more rounded perception of this work done, it would be interesting to look at the echo by others in the community.  Which was the context this publication was cited?  Did the original researchers continue to work on these compounds, and which context?  In which ways other researchers than the authors echoed this work?  Because ACS's page indicates the publication was noticed in freely accessible media, it were worth to look up about this article in SciFinder, too.


Do you get a new chemical compound if you mix sodium hydroxide and water?

According to Oxford's dictionary of chemistry "A compound is a mixture of two or more elements in a fixed ratio".
So if $\ce{NaOH}$ (in solid state) is added to water, it just dissociates into its constituent ions that are $\ce{Na+}$ and $\ce{OH-}$. You can add any ratios of two until you cannot add any more.


Counting number of acidic-H in a molecule

An aldehyde or a ketone with an alpha hydrogen forms a carbanion that resonates to enolate form. This leads to two canonical structures that are in resonance. The example below is a polyphenol compound extracted from turmeric, called curcumin.

source: Biomaterials. 2010 May;31(14):4179-85. doi: 10.1016/j.biomaterials.2010.01.142. Epub 2010 Feb 23. (https://www.ncbi.nlm.nih.gov/pubmed/20181392)

The hydrogens between two keto groups are most acidic. This is due to formation of two equivalent enol structures stabilized by hydrogen bonding. The structural requirement would be planarity of atoms involved in hydrogen bonding.

Below are 3-D structures of the given compound in the question above.

Structure without hydrogens

Bridgehead carbon is 4. Keto groups are at 2 , 5 and 8 (shaded in grey). Keto groups on 5 and 8 are not in the same plane . Hence Hydrogen on 4 , 22 (see figure with hydrogens) does not form stable hydrogen bonded enol structures as discussed in the above example.

Therefore, hydrogen at 8th one (pointing with red arrow) may not be acidic.

Note: 3-D structures have been drawn on Chem3Dpro

Structure with hydrogens

Structure with hydrogens (2nd view)


Why does soda explode if artificial sweetener or other powders are added?

Artificial sweetener powder causes nucleation of the carbonation in the soda/pop/soda-pop. The carbon dioxide concentration in the soda is higher than the equilibrium concentration for the partial pressure of $\ce{CO2}$ in the atmosphere thus over time the soda will off-gas $\ce{CO2}$ and if allowed to sit long enough will become "flat".

The off-gassing is spontaneous but not instantaneous, thus it is subject to kinetic rates. To form a bubble, enough $\ce{CO2}$ molecules need to coalesce at one time to form a large enough bubble to be stable and not redissolved in the soda (nucleating).

When a rough surface is introduced, the number of $\ce{CO2}$ molecules that must come together to form a stable bubble is far less that the pure drink, so the off-gassing occurs much faster causing the observed "explosion".

You can observe this when certain glasses (ussually beer glasses) etch a small ring in the glass, creating a rough surface and the formation of bubbles at the ring.  you can also observe this if you put a larger rough object in the soda. This could be a rock, a piece of wood (like a tooth pick), or even a Mentos.

To solve this you might be able to dissolve it in a small amount of water first. This may not completely solve issue though, as most sweeteners use silicon oxide or titanium oxide powder to prevent clumping, neither of which are soluble and will still nucleate bubbles.


What exactly is Indium's ability to stabilize non-ferrous metals?

All I did was click on the reference from the History wikipedia link above for the sentence you were talking about. The sentence comes from a paragraph of a manuscript which is found here: http://pubs.acs.org/doi/abs/10.1021/ed011p270

Note the bolded part, that was what is meant.

" Its[Indium's] identification as well as its name came from the beautiful indigo-blue line which it so prominently displays in its spectrum. Its commercial development came as the result of the knowledge that it should be a valuable stabilizer of non-ferrous metals much as chromium is a stabilizer of the ferrous metals. "

Above is quote from:

French, Sidney J. (1934). "A story of indium". Journal of Chemical Education 11 (5): 270. Bibcode:1934JChEd..11..270F. doi:10.1021/ed011p270.


Why should I acidify twice in the procedure for qualitative analysis of chloride anions?

Playing around with pH values, going from acidic to basic and back to acidic is a major element of the separation methods used in traditional analytic inorganic chemistry. It takes advantage of the facts that both acids and bases were easy to acquire even in the early days of chemistry and that the pH can be monitered well by indicators which had also been known for a long time (even though in the past they might have used different indicators than nowadays). Making sure the pH is at a correct value at a certain time is also important in this cycle.
(Because the Markdown syntax does not allow calling enumerated lists with lowercase letters, I’ll use numbers instead to refer to the different steps.)

You have a solution of potentially many different ions. For the sake of this answer, let’s assume chloride, bromide, nitrate and carbonate. If you measure the pH of this solution, you will notice it is slightly basic due to the carbonate. If you were to add silver nitrate immediately, silver carbonate $\ce{Ag2CO3}$ would also precipitate, complicating later analyses.
Therefore, you add an acid (whether it is acetic or nitric acid does not matter. It should not be hydrochloric acid for obvious reasons and it should not be too strong.) until the pH of the solution is acidic (i.e. at least 5).

You can now add your silver nitrate. Only chloride and bromide will precipitate as $\ce{AgCl}$ and $\ce{AgBr}$, respectively.

By centrifuging, you now have only the silver chloride and bromide precipitates remaining since you are discarding the supernatant. Due to residual amounts of whichever acid you used (acetic acid in the protocol), the damp precipitate is still acidic. However, most of the original acid will have remained in the supernatant and has been poured away.

Acidic residues remain even after washing albeit to a lesser extent. We can basically consider the precipitate neutral now. This is important for step 5.

Adding diluted ammonia (not $25~\%$ concentrated) will dissolve $\ce{AgCl}$ as $\ce{[Ag(NH3)2]Cl}$ while leaving the $\ce{AgBr}$ untouched. Due to the ammonia you are adding, the pH of the resulting solution is basic. Any remaining residues of acid from step 1 have been neutralised now at the latest.

Centrifuging makes sure to remove any remaining dispersed silver bromide. The supernatant should contain $\ce{[Ag(NH3)2]Cl}$ in solution.

Now we want to reprecipitate the silver chloride to confirm we actually had it in the precipitate. (Note we cannot be certain that silver chloride was present if silver bromide has precipitated out.) We do this by reacidifying the solution until all ammonia is protonated to give ammonium and the ammine ligands therefore dissociate from the $\ce{[Ag(NH3)2]+}$ complex. This causes silver chloride to reprecipitate.

In case you did not add enough acid you are here reminded to do so. Failing to add sufficient amount of acid would lead to the silver ions remaning complexed in solution and you incorrectly assuming there be no chloride in your unknown sample.


The simplest kinds of soap are just a fatty acid chain with something more ionic on the end - something like the ubiquitous sodium laureth sulphate:

When it's in bar form, it behaves a bit like a solid block of plastic, because chemically speaking, those long chains of carbon are quite a lot like a common plastic such as polythene. These chains attract each other and repel and exclude water, keeping the block together. The sulphate group on the end is hydrophilic, though - it's chemically stable surrounded by water. This means that when you add water, it infiltrates and teases apart the structure of strands, getting to the ionic ends on each chain and forming a "solvation shell" of waters around each one. There's no real chemical reaction here - as you say, it's a physical process, which is why rubbing the soap helps - you're applying shear to the structure and helping to separate the strands.

Once they're separated from the bulk, the soap forms tiny bubbles called "micelles", which have the chains pointed inwards towards each other and the ionic sulphate groups meeting the water at the outside.


Plot of Arrhenius Equation with temperature

I believe that this should have been the graph...(please omit the
  negative side of x axis), so concave downwards. Why is there a
  discrepancy?

Your analysis and your graph are correct.  Let's examine why the discrepancy arises.  I've drawn two graphs below.  The top one is similar to your graph, only I'm examining a smaller range on the x-axis.  My lower graph examines an even smaller range and it is a range close to x=0.  You can see that this lower graph appears to be "concave-up" just like the Wikipedia graph you referenced.

Here's a third graph where I changed the numerator from -1 to -100, notice how it moves the curve further to the right along the x-axis.

Depending upon 1) what numbers we put in the numerator and denominator of our exponential equation and 2) depending upon how we scale our axes, we can dramatically change how our graph appears.  There will always be a relatively flat part to this curve located close to x=0, then there will be a concave-up section, followed by a concave-down section, and finally another relatively flat section further out on the x-axis.  You happened to choose a large range (up to x=60-70) for the x-axis of your graph.  This compressed your graph, so it did not show the first two parts of the curve (the relatively flat part located close to x=0 and the concave-up section) clearly.


The association of entropy with disorder is anthropocentric. "Disorder" is a concept derived from our experience of the world. The association with "disorder" is clearer once we explain what we mean by "order". The following definition among those provided by Merriam-Webster most closely fits the intended meaning:

a regular or harmonious arrangement

Since regularity always implies lower entropy, all else being equal, it is therefore fair to associate "order" with lower entropy.
The association with "order" or regularity also syncs with the concept of entropy according to Boltzmann's statistical mechanical definition ($S= k_\mathrm B \log \Omega$). $\Omega$, the number of microstates available to the system, can be quantified using the entropy. More possible unique microstates implies a higher entropy. Greater regularity implies more constraints regarding arrangement of the system and therefore less possible microstates. Solids are usually more regular and therefore have lower entropy than fluid states at the same T. Same when comparing gases and liquids. The possible microstates increases from solid to liquid to gas.
This also jives with the informational-content definition. You can use less information (use a more compact description) to describe an orderly (regular) system. You need more information to describe all possible arrangements of molecules in a gas or liquid than in a solid. Think of entropy as measuring the length of the recipe required to build all possible arrangements of the system.


Relating collision theory to rate of reaction

It is interesting to see you creating your own approach. The comments above provide good responses. I would recommend looking at Buttonwoods suggestion regarding collision theory and cross-sections. However, it appears none have answered your question directly, so I will provide my response to it here.

Question: "Now the molecules of the first substance must bump to the molecules of the second substance with the appropriate orientation. How can I add this to my type?"

My Response: You recognize that orientation is important when it comes to chemical reactions. There are many ways to handle this, I will provide three short suggestions and you can consider them or use them to come up with something else.

Arrhenius (constant) approach: The pre-exponential factor, $\alpha$, in the following Arrhenius equation absorbs two important concepts of chemical reactivity - 1) the frequency of collisions, 2) the orientation of the molecules.

$$u = \alpha e^{-E_a/RT}$$

Note, I changed from the conventional notation in order to match and not confuse with your own. In fact, $\alpha$ is defined as the frequency of properly oriented collisions. So a simple constant multiplier that accounts for orientation can help modify your equation.

Boolean logic: You can possibly simplify things further by using a simple selection function that tells you whether the molecules are in proper orientation. Here is the simplest you could use:

$$f(x) =
\begin{cases}
1 \text{ if proper orientation}\\
0 \text{ if not proper orientation}
\end{cases}$$

Functional dependence: Last mentioned here would be the most creative approach to come up with a function that describes the orientation of your molecules. You will have to come up with the functional dependencies you need (i.e. a reaction coordinate) to help solve your problem. Here is an example of a reaction between atomic hydrogen with molecular hydrogen to make the same species using sinusoidal function where the independent variable (reaction coordinate) is the angle of the H-H bond relative to the hydrogen atom.

$$\text{Reaction:    } H-H + H^{\boldsymbol{\cdot}} \to H^{\boldsymbol{\cdot}} + H-H$$
$$\text{Function:    } f(\theta) = \frac{1}{2}\left[\cos\left(\frac{\theta}{2}\right)+1\right]$$

The interpretation of such a function as this would be 100% likely to react if the $H$ atom is in the plane of the bond with $H_2$ while 0% likely to react if $H_2$ is vertical. This isn't an exact reality, but just an example.


Nernst equation is the only way of solving this problem. Let's consider the example of the reduction of permanganate. The half-reaction is:
$$\ce{MnO4-  +  8 H+  +  5 e-  ->  Mn^2+  +  4 H2O}$$

The potential can be written as:

$$E  =  E^\circ  +  \frac{0.059}{5}\log \left(\frac{\ce{[MnO4-][H+]^8}}{\ce{[Mn^2+]}} \right)$$

If you want to see explicitly the $\mathrm{pH}$, you can rewrite the preceding expression, which gives:

$$E  =  E^\circ  +  0.012 \log \left(\frac{\ce{[MnO4^-]}}{\ce{[Mn^2+]}}\right) - \left(\frac{8\times 0.059}{5} \right) \mathrm{pH}$$

or :
$$E  =  E^\circ  +  0.012 \log \left(\frac{\ce{[MnO4^-]}}{\ce{[Mn^2+]}}\right) - 0.094 \: \mathrm{pH}$$


The easiest way to grow diamond through chemical vapor deposition (CVD) is on a heated filament, which just takes a few watts.

You'd need a low-pressure chamber (~30 torr, or ~0.04 atm) with gas-tight fittings for electricity, feed-gas of ~1% $\ce{CH4}$ and ~99% $\ce{H2}$ and $\ce{Cu}$, $\ce{Ti}$, $\ce{W}$ or $\ce{Ta}$ filaments. Since this is a flow-though reactor, you'll also need to dispose of the flammable feed-gas without letting it accumulate near the vacuum pump.

N.B. This is not a simple project one could do at home, since you'd need vacuum equipment, pressure regulators, high-pressure bottles* of these highly flammable gases, and a vacuum-vessel which could implode. If you can get a knowledgeable instructor involved who has access to laboratory facilities and a thorough understanding of safety procedures, though, it's not impossible to accomplish.

* One of the greatest dangers would be a gas bottle falling, taking off like a rocket and leaving explosive gas in its wake!


Oh my God! I hope I help you in time. I just spent hours dealing with the same issue, and I finally figured it out. Basically, OpenChrom will only read CSVs formatted in a very specific way, so you will have to change your .csv files to meet these requirements:

The first three columns MUST be have the headers 'RT(milliseconds)', 'RT(minutes) - NOT USED BY IMPORT', and 'RI', exactly as I have written them.
The second column ('RT(minutes) - NOT USED BY IMPORT') does not need any values; as its name suggests, it is not used by the import, but it still needs to be there! You will have to convert your times to milliseconds to populate the first column. The third column ('RI') is for retention indices, if you do not have them you must populate the entire column with zeros (0.0).
Every subsequent column must have an integer header representing the mass of the signal (e.g. '176' for m/z 176).
No trailing comma at the end of the line! E.g. a line of values should look like

  1,2,3,4

  5,6,7,8

and NOT

  1,2,3,4,

  5,6,7,8,

In Excel, the only way for my files to save like this is if I save them as CSV (MS-DOS).

I hope this helps!


Why is there an exponent 4 after the brackets in sp3?

The electron configuration of carbon is canonically written as 1s22s22p2. What you now can do is invoke orbital hybridization and combine the electrons in the 2s and 2p orbitals. Then you get sp3 hybrid orbitals with a total of 4 electrons in them. The 1s electrons stay where they are. To mark the new hybrid orbitals as separate, you put brackets around them.

You then end up with an electronic configuration of 1s2(2sp3)4.


It would have been helpful if you have said what kind of polymer-based nanocomposites you would like to create. Yet, I found a good article for the fabrication of poly dimethylsiloxane/carbon nanofiber (CNF)-based nanocomposites. The schematic diagram of experimental procedure is illustrated in following figure (from the Ref. 1).

The experimental section in Ref.1 describes the step-wise procedure for the fabrication. It also provides a useful few more references for polymer-based nanocomposite formulations. I also included another reference (Ref.2), which may be helpful to you.

References:

S. Chowdhury, M. Olima, Y. Liu, M. Saha, J. Bergman, T. Robison, Poly dimethylsiloxane/carbon nanofiber nanocomposites: fabrication and characterization of electrical and thermal properties, International Journal of Smart and Nano Materials, 2016, 7(4), 236-247 (https://doi.org/10.1080/19475411.2016.1269027).
H. Azimi, F H. Tezel, J. Thibault, Effect of embedded activated carbon nanoparticles on the performance of polydimethylsiloxane (PDMS) membrane for pervaporation separation of butanol, J. Chem. Technol. Biotechnol., 2017, 92(12), 2901–2911 (https://onlinelibrary.wiley.com/doi/pdf/10.1002/jctb.5306).


How to compare viscosity of alcohol to water?

A common experiment would be to have two concentric cylinders. The outer one can rotate almost freely (with some slight and constant retardation/breaking applied), while the inner one is driven by a motor, like so. Between them is a layer of the liquid that you wish to determine the viscosity of. The speed of the outer cylinder is a measure for the viscosity.

However, since you want to explain this to first-graders, you might wanna start with something more obvious than the difference in viscosity between ethanol and water. How about the difference in viscosity between water and honey? As a demonstration experiment, you could use a funnel in which you pour an equal amount of either liquid, and then measure the time it takes until it has passed through the funnel. This should be more tangible and easier to understand. You could also prepare beakers with water, honey and maybe peanut butter, and have your students run a spatula through all of them. That way they can experience the difference in force that is necessary to move the spatula first hand.


There are two ways of rationalising this. One way is to look up or qualitatively determine the frontier orbitals (HOMO and LUMO) of anthracene, the other is to think about the stability of various radical anions.

But first, a quick note on the Birch reduction and its mechanism. As you may know, the $\ce{Na/NH3}$ system effectively works as a single electron transfer reagent. The electron is added to the π system’s LUMO, and the resulting radical anion is trapped by a proton (donated from a proton source; e.g. added ethanol) to give a radical. This radical is still electron-deficient and thus susceptible to another single electron transfer reduction resulting in a second anion which again is captured by a proton resulting in the reaction product.

Frontier orbital analysis

A great resource for frontier orbitals is the WOrbit page of Hans-Ulrich Wagner from the LMU Munich. Among the systems whose orbitals you can look at is, luckily, anthracene. For posteriority, I have included the HOMO and LUMO pictures in the image below.

Figure 1: LUMO and HOMO of anthracene as calculated on the WOrbit page linked above.

The LUMO is of importance; after all, we are adding an electron. We see that the MO is most localised on carbons 9 and 10 of the central ring — thus, this is where the first electron will be added in a localised picture, and this is where the first proton will capture.

Scheme 1: First reduction sequence.

As I drew in scheme 1, this gives us the radical 3, of which I have only shown one resonance structure. Applying the same analysis for the second step will tell us that the second electron will be added exactly at that position where I have drawn the radical (data not shown). Thus, we find that the central ring is reduced as depicted in scheme 2.

Scheme 2: Second single electron transfer and subsequent reduction of radical 3 to 9,10-dihydroanthracene 5.

Stability of radical anions

If we consider the $+I$ effect on substituted carbons, we see that we only need to consider the three structures shown in scheme 3 below; reversing charge and radical in structure 2 would result in the negative charge being on the tertiary carbon which is much less favourable when compared to the secondary one.

Scheme 3: Possible radical anions of anthracene.

At first, it looks like 6 and 7 might be better stabilised structures: 6 effectively represents a methylnaphthalene radical and 7 a naphthylallyl radical. However, if we take a closer look at 2 we see that it can be rearranged in a manner shown in scheme 1 above. If you consider the above drawn structure 3 as a radical anion (i.e. before protonation), notice that the radical is now a diphenylmethyl radical. This diphenylmethyl radical wins in favourability against both naphthyl derivatives.

As a general rule, if you can preserve the aromaticity of a benzene ring the structure is ‘better’ than one where you can’t. And two benzene rings as are present in structures 3 and 4 (and 5) are better than the naphthyl rings of structures 6 and 7. Thus, it is most favourable to reduce the central ring.

Note, by the way, that depending on the conditions employed, 5 is not inert towards further Birch reduction. Mehda, Sen and Ramesh report a triple Birch reduction of anthracene leading to the very interesting tricyclic triene 8 in $80~\%$ yield (scheme 4).[1]

Scheme 4: Synthesis of 1,4,5,8,9,10-hexahydroanthracene.[1]

Reference:

[1]: G. Mehta, S. Sen, S. S. Ramesh, Eur. J. Org. Chem. 2007, 423. DOI: 10.1002/ejoc.200600610.


Scandium actually does have oxidation states lower than +3.  Wikipedia reports a double chloride $\ce{CsScCl3}$ where scandium is clearly in the oxidation state +2.

The stability of oxidation states in transition metals depends on the balance between ionization energy on the one hand, and binding energy due to either ionic or covalent bonds on the other.   In the case of scandium the third ionization energy is low because all three valence electrons are held rather loosely, being in diffuse orbitals that are shielded from most of the nuclear charge by the argon core.  Thus it's easy to get from the +2 to the +3 oxidation state because of the greater binding energy enabled by the latter.  Later first-row transition elements like iron have more tightly bound $3d$ electrons and thus higher third ionization energies, making the +2 oxidation state more favorable.


What makes pentane, hexane, heptane, and octane dangerous, while butane and nonane aren't?

Pentane, hexane, heptane, and octane are labelled with the ‘environment’ hazard pictogram (GHS09) since they are considered toxic or very toxic to aquatic life with long-lasting effects (H411 or H410) because of their toxicity to the water flea Daphnia magna in standardized tests.

Nevertheless, nonane is not completely harmless; it still may cause long-lasting harmful effects to aquatic life (H413). However, a pictogram is not required for this hazard category.

According to a comparative study, the toxicity of these substances is mainly determined by their solubility in water and their organism–water partition coefficient.


Although the direct complete oxidation of methanol (or formaldehyde or formic acid, on that matter) to carbon dioxide ($\ce{CO2}$) using  oxidizing reagents such as $\ce{KMnO4}$  or $\ce{K2Cr2O7}$  is hard to find in literature, electrochemical oxidation of methanol to $\ce{CO2}$ through formaldehyde and formic acid is one of the concepts used in fuel cell research (Ref.1). The extensive use of simple organic compounds such as methanol, formaldehyde, and formic acid as fuels is because they have several advantages, likes of being easy to store and handle, and having possess a high energy density of the order of $\pu{kWh/kg}$. Aside from that, one would also expect they should have the simplest and most straightforward reaction mechanisms of all the possible organic fuels because of their simple structure. Yet, in their review article (Ref.1), Parsons and VanderNoot revealed that the oxidation of methanol to $\ce{CO2}$, involves $\ce{6 e-}$ so the oxidation must occur in several steps with several products or intermediates. The product resulting from complete oxidation of methanol (and other individual organic fuels such as formaldehyde and formic acid) is of course $\ce{CO2}$. The reactions of all these three fuels have been studied on a variety of electrodes, and with a wide spectrum of experimental conditions to optimize the outcome (most of those references are sited in Ref.1). Keep in mind that the studies involving methanol are outweighed those of formaldehyde and formic acid, simply due to the fact that its complete oxidation to $\ce{CO2}$, generates 6 electrons compared to 4 and 2 for formaldehyde and formic acid, respectively. Electrode potentials of these reactions are still debatable, and Koppenol and Rush have collected and calculated some reduction potentials for $\ce{CO2/CO2^{.-}}$ couple and plotted them in a Frost diagram (Ref.2).
We noted above that the electrochemical oxidation of methanol to $\ce{CO2}$ involves $\ce{6 e-}$.  As expected, the oxidation occurs in several steps with several products or intermediates, and, for example, mass spectral measurements has showed that both $\ce{H2CO}$ and $\ce{HCO2H}$ were produced (Ref.3).  Their conclution was listed as:

(a) At room temperature, a stationary decomposition of methanol takes place on a platinum catalyst, with formation of formaldehyde and formic acid;
(b) $\ce{HCO2H}$ reacts with $\ce{CH3OH}$ to form $\ce{HCOOCH3}$ (methyl formate);
(c) $\ce{HCO2H}$ is also produced during electrolysis, and is an intermediate in the oxidation to $\ce{CO2}$;
(d) $\ce{H2CO}$ reacts with $\ce{CH3OH}$ to form $\ce{CH2(OCH3)2}$ (methylal), but, contrary to the results of other authors, $\ce{H2CO}$ is not an intermediate of electrolysis.

Accordingly, you’d see even under electrochemical conditions, this conversion is not straightforward. Ref.1 also proposed a step-wise mechanism to explain all those intermediate and product, involving main byproduct, carbon monoxide ($\ce{CO}$), which is believed to be poisoning the $\ce{Pt}$ electrode in most cases.
References:

R. Parsons, T. VanderNoot, “The oxidation of small organic molecules: A survey of recent fuel cell related research,” Journal of Electroanalytical Chemistry and Interfacial Electrochemistry 1988, 257(1–2), 9-45 (https://doi.org/10.1016/0022-0728(88)87028-1).
W. H. Koppenol, J. D. Rush, “Reduction Potential of the $\ce{CO2/CO2^{.-}}$ Couple. A Comparison with Other $\ce{C1}$ Radicals,” J. Phys. Chem. 1987, 91(16), 4429-4430 (DOI: 10.1021/j100300a045).
T. Iwasita, W. Vielstich, “On-line mass spectroscopy of volatile products during methanol oxidation at platinum in acid solutions,” Journal of Electroanalytical Chemistry and Interfacial Electrochemistry 1986, 201(2), 403-408 (https://doi.org/10.1016/0022-0728(86)80064-X).


Energy level of molecular orbitals in heteronuclear diatomic compounds

This question requires idea of the Linear Combination of Atomic Orbitals (LCAO). It involves the following equation:
$\psi_n = \sum\limits_{i} c_{ni}\phi_{i} = c_{n1}\phi_{1} + c_{n2}\phi_{2} + ... c_{ni}\phi_{i}$
Here $\psi_n$ represents the wavefunction for the resulting molecular orbital, $\phi_{i}$ represents the atomic orbitals that contribute to the bond, and $c_{ni}$ represents the weight coefficient that can basically tell us how much of each atomic orbital is involved in the molecular orbital. If we look at the molecular orbital diagram for carbon monoxide, you are right, the bonding pi orbitals are lower in energy than the bonding sigma orbital:

The reason for this is due to the contributions from carbon. This comes from the relative energies of the atomic orbitals. From photoelectron spectroscopy, the energies of the carbon 2p orbitals and 2s can interact with the oxygen 2p orbitals because they are close enough in energy while the 2s oxygen orbital is a little too far to interact with the 2p orbitals of carbon. See image below

The important thing to note here is that the 2s orbital can interact as well. This means when we solve for the wave function using LCAO and the equation above, carbon can give a large contribution because two of its atomic orbitals interact, and it turned out it did. The same thing can be done for NO (which in fact is also similar to CO where the pi bonding orders are lower energy than the sigma bonding ones) and other molecules. However, it gets extremely complicated with larger and larger molecules.
I got this information from this public PowerPoint  https://www.chem.uci.edu/~lawm/10-9.pdf which come from lectures from UCIrvine which were released: https://www.youtube.com/watch?v=1CH6CeORL_g&list=PLaLOVNqqD-2GDVq2xFUuAltjIXPgQNNTL&index=7&ab_channel=UCIOpen


It means "the liquid that swims above", from the Latin terms super = above and natare = to swim.
So it is the liquid that remains on top.

The subnatant would be the liquid at the bottom, in case you should centrifuge two liquid phases.

The term you name, nato, means son in Latin.


What would an ammonium sulfide fire look like?

If your container is supposed to contain solid ammonium sulfide $\ce{(NH4)2S}$, it better be cooled to -20 °C or below.

(Typically, ammonium sulfide is only used in aquous solution to precipitate metal cations of cobalt, nickel and others as their sulfides in the Kationentrennungsgang).

Long before the solid ammonium sulfide reaches room temperature, it will start to decompose and ammonia and hydrogen sulfide are liberated.


What does bond polarity have to do with acid strength?

This can be extrapolated to why polar substances ionize more often in water than non-polar substances. This is because water too is polar. Always remember "like dissolves like." Polar dissolves polar. This is because the positive end of the water molecule pulls at the negative end of the solute while the negative end of the water molecules pull at the positive end of the solute. In a non-polar molecule this would obviously not happen. Hence, in the case of acids, polar acids will dissociate more and hence more hydrogen atoms will be released into the solution.


Unit derivation concerning Nernst equation

M. Farooq have given good description of how to analyse units in a equation. Therefore, your assumption of the unit of $\dfrac{RT}{nF}$ should be Volts is correct. However, your question is not to assume but to derive. What's that means is unit conversion. Let's see the units of all four variables:

$R$ is a constant: $\pu{8.314 J mol-1 K-1}$
$F$ is also a constant: $\approx \pu{96485  C mol-1}$
$T$ is a variable: $\pu{K}$
$n$ is number of total electrons in the redox reaction involved, and hence a variable but unit less.

Now lets see the unit conversion:
$$\frac{RT}{nF} = \frac{\pu{J mol-1 K-1}\times \pu{K}}{\pu{C mol-1}} = \frac{\pu{J}}{\pu{C}} $$
By definition (Ohm's law):
$$ V = \frac{\text{potential energy}}{\text{Charge}} = \frac{\pu{J}}{\pu{C}} $$
$$\therefore \:\; \frac{RT}{nF} = \frac{\pu{J}}{\pu{C}}= \pu{V} $$


The first step toward answering this question is understanding the extent to which the statement "Group 11 metals are unreactive" is or isn't true.

As stated in How do plants respond to copper deficiency? Plant Signaling and Behavior vol. 3, pages 231-232:

  The transition metal copper is essential for all organisms.

It is not just that copper is present in every living thing.  Copper is reacting in every living thing. Copper ions cycle between the +1 and +2 oxidation states in Plastocyanin,  Copper/zinc superoxide dismutase and cytochrome oxidase. So as you are saying copper is unreactive, it as reacting in cytochrome oxidase in every mitochondrion of your body, and in Cu/Zn superoxide dismutase in the cytosol of cells through your body.

One way to quantify the reactivity of metal ions in through reduction potential, here listed for reduction of the ion to the zero oxidation state metal.  (limiting to d-block metals at least as unreactive as Cu)

\begin{array}{c @{:} c} \\
\color{blue}{\text{Ion}} & \color{blue}{\text{Reduction potential}} \\\hline
\mathbf{\color{red}{Cu^{2+}}} & 0.3419  \\\hline
\ce{Tc^2+} & 0.400   \\\hline
\ce{Ru^2+} & 0.455  \\\hline
\ce{Cu+} & 0.521 \\ \hline
\ce{Rh+} & 0.600  \\
\ce{Rh^2+} & 0.600  \\
\ce{Rh^3+} & 0.758  \\ \hline
\mathbf{\color{red}{Ag^+}} & 0.7996   \\\hline
\ce{Os^2+} & 0.85  \\\hline
\ce{Hg^2+} & 0.851  \\\hline
\ce{Pd^2+} & 0.951   \\\hline
\ce{Ir^3+} & 1.156  \\\hline
\ce{Pt^2+} & 1.18   \\\hline
\mathbf{\color{red}{Au^{3+}}} & 1.498  \\
\ce{Au+} & 1.692  \\
\end{array}

Looking at this list, we see that there are a substantial number of d-block metals that are less reactive than copper.  A few are less reactive than silver, and gold is the least reactive on the basis standard reduction potential.

The overall trend in d-block metals, is down and to the right in the block is less reactive. The exception is that, for a given row, the group 12 metal is more reactive than the group 11 metal, but due to the formation of a 2+ ion, not a 1+ ion.

Next we should look at ionization energies of d-block metals to see how much of the reduction potential corresponds to properties of an individual atom, versus bulk or solvent effects.

Looking at first-ionization energies, for each row of the d-block, the highest ionization energy is always the group 12 element (Zn, Cd, Hg). This corresponds well with the fact the Zn, Cd and Hg do not form +1 ions.  The group 11 elements have significantly lower first-ionization energies, and do form +1 ions.

Then looking at second-ionization energies, group 11 has the highest second-ionization energies for every row of the d-block!

Much of the reactive/unreactivity of d-block is explained by the increase of nuclear charge to the right and down the d-block.  The complete d-shell of neutral group 11 atoms shields the valence s electron well enough that it may be removed to react.  However, to lose 2 electrons, the group 11 elements would need to lose a d-electron from a completed d-shell, which is difficult.  For group 12, two s valance electrons can be lost, and the complete d-shell maintained.


What is the difference between the Bohr model and the quantum model?

The Bohr model is a comparatively simple model, whereas the quantum description is part of the huge theory of quantum mechanics, so to list all the differences including implications would need to explain quantum mechanics itself.

But I think quite a good starting point would be to say, the Bohr model has electrons travel on certain, specific, classical trajectories, picked out by certain rules. The quantum model does not have electron trajectories. Instead electrons are quite different, they are spread out a bit like clouds of gas, in areas called atomic orbitals. I would recommend reading about atomic orbitals, the Wikipedia page is quite interesting. You will see very quickly they are quite different from the Bohr model.


Can you model intermolecular interactions by using the Born-Oppenheimer approximation to get a potential function for nuclei?

The Born-Oppenheimer approximation is very used in ab initio molecular dynamics (AIMD). When the Born-Oppenheimer approximation is used, it is more specifically called Born-Oppenheimer molecular dynamics (BOMD). Various software packages have implemented this method, here is a list of a few of them:

CP2K
CPMD
Gaussian
Orca

Other types of AIMD is also used, including Car-Parrinello molecular dymanics (CPMD) and Ehrenfest molecular dynamics.

As always it is important to remember that the Born-Oppenheimer approximation is only a good approximation if the relaxation time of the electronic structure is much faster than that of the movement of the nucleus.


Home science optical absorption test for approximate caffeine quantity in coffee?

Yes..., except that for "...in coffee?" it's No.
AAI Solutions' Measuring Caffeine Concentration; Applied Analytics Application Note No. AN-019 front artwork certainly looks hopeful...

(click for larger)
After all an Optical Multichannel Analyzer (OMA) in this case, to a home scientist, looks like some light source, a grating, three photodiodes and some tedious calibration, this is really for extracted caffeine in supercritical CO2.
However, from Background in

As different literatures indicated spectrophotometric determination of caffeine is also reported as preferred method of determination such as UV–Vis spectrophotometry because of its relatively low cost, rapidity, high accuracy and reproducibility. But UV–Vis spectrophotometric method cannot be used directly for determination of caffeine in coffee beans extracted with water owing to the matrix effect of UV–Vis absorbing substances in the sample matrix [9]. In aqueous solution of coffee beans it was observed that there is spectral interference from caffeine and chlorogenic acid in the wavelength regions of 200–500 nm. Yet this method requires the extraction of caffeine from the aqueous solution of coffee beans using dichloromethane for the spectroscopic determination. This is necessary since the caffeine spectrum is overlapped with other compounds found in coffee. Hence, the use of dichloromethane limits the wider application of UV–Vis method.
[9] Genetic diversity and correlation of bean caffeine content with cup quality and green bean physical characteristics in coffee (Coffea arabica L.)

In simple language: Coffee is full of all kinds of molecules besides caffein, so there will likely be many broad spectral features in the UV interfering with any simple way to measure one particular molecule's concentration.
You're going to have to resort to some kind of specific extraction. If not supercritical CO2 mentioned above, then something like this:
SPECTROPHOTOMETERS: How can I measure the caffeine content in drinks? with 16 items of apparatus, two reagents including somewhat dangerous to "home scientists" Dichloromethane
or like this: Jenway's 73 SERIES SPECTROPHOTOMETER Application note: A09-010A; The quantitative determination of caffeine in beverages and soft drinks using UV wavelength spectroscopy again with the dichloromethane!
Bottom line
Try not to drink so much coffee that you need to worry about all of this. Listen to "Sir Professor" Martyn Poliakoff's and switch to tea. (note, I'm also the OP)


Euclidean distance between atoms using RDKit

The issue was that I had to embed the molecule to have a 3D conformation. So I added following line:

AllChem.EmbedMolecule(mol)

Feel free to answer with better suggestions and alternatives.


One possibility is that chocolate has a lower thermal conductivity than water-based foods, which make up much of the food in our freezers.

If you open your freezer and touch a cardboard food container, then touch a metal ice tray, which one feels colder? They are both the same temperature, but the metal tray feels colder because it conducts heat better than the cardboard box, so the heat gets transferred out of your hand quicker and it feels colder.

Water is a relatively good conductor of heat and is present in large quantities in many foods. Chocolate is largely made up of fatty organics, which do not conduct heat as well. It is thus very possible that chocolate has a lower thermal conductivity than many frozen foods and therefore doesn't feel as cold to the touch in comparison.


What happens when an alkyne reacts with ammoniacal Cu2Cl2?

Terminal alkynes (including acetylene) form a $\color{red}{\text{red precipitate}}$ of copper acetylides when it is passed through ammoniacal cuprous chloride solution:
$$\ce{R-C#C-H + CuCl + NH3 -> R-C#C-Cu + NH4Cl}$$
However, in the presence of oxygen, this reaction proceeds further to dimerize the corresponding alkyne. This reaction, the oldest homo-coupling of terminal alkyne reaction,  was first discovered by Carl Andreas Glaser in 1869 (Ref.1, 2), hence the reaction is named Glaser-coupling after him:
$$\ce{2R-C#C-Cu ->[O2][NH3/EtOH] R-C#C-C#C-R }$$
The actual Glaser's reaction was (ref.2):
$$\ce{2Ph-C#C-H ->[CuCl/O2][NH3/EtOH] Ph-C#C-C#C-Ph}$$
References:

Carl Glaser, "Beiträge zur Kenntniss des Acetenylbenzols," Berichte der deutschen chemischen Gesellschaft 1869, 2(1), 422–424 (https://doi.org/10.1002/cber.186900201183).
Chao Liu, "Chapter 1: Introduction," In Lecture Notes in Chemistry 102: Transition Metal Catalyzed Oxidative Cross-Coupling Reactions;  Aiwen Lei, Editor; Springer-Verlag GmbH DE (Part of Springer Nature): Berlin, Germany, 2019, pp. 1-10 (ISBN: 978-3-662-58102-5).


Unknown Mass in Dalton's Law of Partial Pressures

Assuming that the total pressure $p_\text{total}=2068.44\ \mathrm{kPa}$ is equal to the sum of the partial pressures of the individual gases in accordance with Dalton’s law, the partial pressure of air $p_\text{air}$ can be calculated as follows:

$$\begin{align}
p_\text{total}&=p_\text{air}+p_{\ce{CO2}}\\[6pt]
p_\text{air}&=p_\text{total}-p_{\ce{CO2}}\\[6pt]
&=2068.44\ \mathrm{kPa}-344.74\ \mathrm{kPa}\\[6pt]
&=1723.70\ \mathrm{kPa}
\end{align}$$

Since molar mass $M$ is defined as

$$M=\frac mn$$

where $m$ is mass and $n$ is amount of substance, the ideal gas law

$$p\cdot V = n\cdot R\cdot T$$

where $V$ is volume, $R$ is the molar gas constant, and $T$ is temperature, can be rewritten as

$$\begin{align}
p\cdot V&=\frac{m\cdot R\cdot T}M\\[6pt]
\frac V{R\cdot T}&=\frac m{M\cdot p}
\end{align}$$

This equation applies to air
$$\frac V{R\cdot T}=\frac {m_\text{air}}{M_\text{air}\cdot p_\text{air}}$$
where $m_\text{air}=0.454\ \mathrm{kg}$ is the given mass of air and $M_\text{air}=28.97\ \mathrm{g\ mol^{-1}}$ is the average molar mass of air, as well as to $\ce{CO2}$
$$\frac V{R\cdot T}=\frac {m_{\ce{CO2}}}{M_{\ce{CO2}}\cdot p_{\ce{CO2}}}$$
where $m_{\ce{CO2}}$ is the unknown mass of $\ce{CO2}$ and $M_{\ce{CO2}}=44.01\ \mathrm{g\ mol^{-1}}$ is the molar mass of $\ce{CO2}$.

Therefore,

$$\begin{align}
\frac {m_{\ce{CO2}}}{M_{\ce{CO2}}\cdot p_{\ce{CO2}}}&=\frac {m_\text{air}}{M_\text{air}\cdot p_\text{air}}\\[6pt]
m_{\ce{CO2}}&=\frac {M_{\ce{CO2}}\cdot p_{\ce{CO2}}\cdot m_\text{air}}{M_\text{air}\cdot p_\text{air}}\\[6pt]
&=\frac {44.01\ \mathrm{g\ mol^{-1}}\times344.74\ \mathrm{kPa}\times0.454\ \mathrm{kg}}{28.97\ \mathrm{g\ mol^{-1}}\times1723.70\ \mathrm{kPa}}\\[6pt]
&=0.138\ \mathrm{kg}
\end{align}$$

Note that the given value for the volume $V=85\ \mathrm l$ is not required for the calculation of $m_\text{air}$. Thus, the uncertainty of the volume $V=85\ \mathrm l$ (with only two significant digits) does not affect the uncertainty of the result.


How does the Wave-Mechanical model reconcile with the particle definition of an electron?

Wave particle duality is a fundamental idea in the quantum mechanics required to understand the behaviour of fundamental particles

One of the key ideas in quantum mechanics is that particles fundamental behaviour cannot be explained if we assume their "real" behaviour is Newtonian (or billiard-ball like). The behaviour of particles like electrons depends on how we measure them and sometimes they appear wave-like sometimes particle like.

This is very important in chemistry as it leads to good explanations of chemical bonding and the shapes of molecules. Electrons in atoms and molecules, for example, were originally thought to behave like a mini solar system where they orbited the nucleus. But this explanation doesn't work well to explain much in chemical bonding or atomic behaviour.

Schrödinger developed a better way to think about their behaviour with his famous wave equation. This describes the shapes of electron clouds around atoms. One way to think of the output of this equation is that it describes clouds of probability describing the possible locations of an electron in an atomic orbital. Those orbitals are the "standing waves" of electrons bound to atoms and they have specific definite shapes (related to spherical harmonics which are the 3D equivalent of the standing waves in a vibrating violin string).

This description of the properties of electrons is the core of our understanding of chemical bonding and the shapes of molecules.

What are electrons? Quantum mechanics doesn't tell us what they really are: it just tells us how they behave when we make certain measurements. And, in chemistry, the measurements that matter mostly give results where the wave-like behaviour is what matters.


A ball mill can get you to sub-micron particle size with a minimal capital investment - less than 500 USD on eBay.


Conservation of angular momentum in rotational spectroscopy

Sorry for the late answer, but I could not find the time to write something earlier. A concept that you encounter often in quantum mechanics is the so-called coupled and uncoupled basis. Your question is a variation on that theme: before the absorption the uncoupled representation is the more natural where the angular momentum of the photon and the angular momentum of the molecule (and their projections) are good quantum numbers, whereas after the absorption the coupled representation is more suitable and the total angular momentum of the system (and it's projection) is a good quantum number. In Dirac's braket notation we have

$$
|jm_j\rangle |j_\text{ph}m_{j_{\text{ph}}}\rangle
$$
for the uncoupled representation and
$$
|JM_J\rangle
$$
for the coupled representation.

Your question concerns the conservation of the total angular momentum of the system $\hat{J}^2$ and in the coupled representation the expectation values are easily obtained because $J$ is a good quantum number
$$
\hat{J}^2|JM_J\rangle=\hbar^2 J(J+1)|JM_J\rangle
$$

In the uncoupled representation the total angular momentum is given by $\hat{J}^2=\left (  \hat{j}+\hat{j}_\text{ph} \right)^2=\hat{j}^2+\hat{j}_\text{ph}^2+2\hat{j}\cdot \hat{j}_\text{ph}$

The first two terms of this expression are again easy, but the thrid term is a bit tricky. Using the ladder operators you can write it as
$$
2\hat{j}\cdot \hat{j}_\text{ph}=2(\hat{j}_x\hat{j}_{\text{ph},x}+\hat{j}_y\hat{j}_{\text{ph},y}+\hat{j}_z\hat{j}_{\text{ph},z})=\hat{j}_+\hat{j}_{\text{ph},-}+\hat{j}_-\hat{j}_{\text{ph},+}+2\hat{j}_z\hat{j}_{\text{ph},z}
$$
where
$$
\hat{j}_\pm|jm_j\rangle=\hbar\sqrt{j(j+1)-m(m\pm1)}|jm_j\pm1\rangle
$$
and it thus couples states of different $m_j$. Let us, for example, couple $|jm_j\rangle=|1m_j\rangle$ and $|j_\text{ph}m_{j_\text{ph}}\rangle=|1m_{j_\text{ph}}\rangle$ so that we have $(2j+1)(2j_\text{ph}+1)=9$ basis states:
$$
|1,-1\rangle|1,-1\rangle,\\|1,-1\rangle|1,0\rangle,\\|1,-1\rangle|1,1\rangle,\\
|1,0\rangle|1,-1\rangle,\\|1,0\rangle|1,0\rangle,\\|1,0\rangle|1,1\rangle,\\
|1,1\rangle|1,-1\rangle,\\|1,1\rangle|1,0\rangle,\\|1,1\rangle|1,1\rangle
$$

We determine all expectation values of the total angular momentum operator using the 9 basis states which gives us a total of 81 values that we can order in a 9x9 matrix (in units of $\hbar^2$)

6   0   0   0   0   0   0   0   0
0   4   0   2   0   0   0   0   0
0   0   2   0   2   0   0   0   0
0   2   0   4   0   0   0   0   0
0   0   2   0   4   0   2   0   0
0   0   0   0   0   4   0   2   0
0   0   0   0   2   0   2   0   0
0   0   0   0   0   2   0   4   0
0   0   0   0   0   0   0   0   6

We can diagonalize this matrix and find the eigenvalues 0, 2, 2, 2, 6, 6, 6, 6, 6. These are the same eigenvalues (including the $2J+1$ degeneracy) of $\hat{J}^2$ in the coupled representation!

So the expectation value of the total angular momentum is the same in both cases, the difference is that in the uncoupled case the matrix has off-diagonal elements and it is a bit more work to find its eigenvalues. In the coupled case, the matrix for $\hat{J}^2$ is diagonal and you get the expectation values directly. You might ask if there is an easier way to convert from the coupled to the uncoupled case and back. As you might have guessed there is such a way and it involves the so-called Clebsch-Gordan coefficients.


Why isn't the ionic radius of Mg2+ bigger than F-?

These two species are isoelectronic (have the same number of electrons). In fact, they're electronically identical with the except that the magnesium ion has 3 more protons. Since all other electronic factors are equal, the net effect is that the three extra protons pull all of the electrons in closer to the nucleus. I think you understand this part.

Electronegativity is not correctly used in this context. One way to think about it is that you should only be using that concept on a neutral atom. The fact the fluoride anion is negative and magnesium cation is positive means you've already applied electronegativity to the system in order to ionize them.


Ortho-para directing and meta directing groups: Activating or deactivating the aromatic ring?

I'm not sure if G is an activating or deactivating compound.

You may want to look at the relative order of activation/deactivation of some substituents,
(source: www.chem.ucalgary.ca)

As per the image, $\ce{-NR2}$ is a strongly activating group, while $\ce{-CONH2}$ is a moderately deactivating group (bcoz, it lies somewhere in between $\ce{-COOH}$ and $\ce{-COCl}$). Hence, the net effect would be activating.

How can resonance structures help to explain the reactivity of the compound?

Resonance structures can tell whether the substituent is o,p -directing or m -directing. It can't say anything about rate of reaction.

How do we draw resonance structures that has two substituents such as G?

Just like you've drawn for a single substituent. The only difference is that the other substituent has replaced one of the $\ce{H}$ from the resonance structure drawn with single one. So, the other may participate in delocalization.

For F, does the lone pair on N affect the resonance?

Obviously, but not significantly, as it will participate in resonance too.

I think Br will attach para to N because N is a strong activating group and it will not go to between the two substituents because it is crowded, is this true?

Yes, but it is partially true.
para product (wrt. $\ce{-NR2}$) will be major, however, ortho product (wrt. $\ce{-NR2}$) will be also present as the minor one.


How difficult is it to produce diatomic hydrogen?

Diatomic hydrogen is hydrogen’s stable elemental modification under standard temperature and pressure — so if you generate hydrogen you will be generating diatomic hydrogen.

The simplest way to do this with household means is probably to use acetic acid concentrate (the stronger the better) and zinc powder. (Solid zinc works but much slower.) The reaction may be sluggish; if that is the case, very carefully and slowly heat the acetic acid. Capture the hydrogen in some glass, cup or test tube held upside down (hydrogen is the least dense gas with each molecule only weighing $2~\mathrm{u}$ (atomic mass units) so it will always rise).

If you have collected some amounts of hydrogen gas you can proove its presence. This was a common experiment in my high school chemistry courses so I’m going to tell you how to do it. However, exercise caution! If you didn’t do this at school yourself or didn’t watch someone do it before, I would advise you to not attempt it. Only do this with very small amounts of hydrogen, no more than a test tube of $2~\mathrm{cm}$ diameter. Make sure the test tube is full (by collecting enough hydrogen). Carefully bring the opening towards a candle flame. Do not use a pocket lighter where your thumb is right next to the flame for safety reasons. The hydrogen should burn with a faint ‘pop’ creating water. Considering how slurrish the reaction of zinc and acetic acid probably is, it may be too slow to actually collect enough hydrogen. Don’t attempt the burning if that is the case.

The reason for being cautious is the following: $2\,:\,1$ mixtures of hydrogen and oxygen are explosive. And with explosive I mean that they give a very loud bang. Luckily, air is mainly nitrogen so the probability of accidentally creating an explosive mixture is rather low. However, if you did, you will regret it.

Finally, on the topic of ‘hydrogen-enriched water’: Hydrogen, like all other gases, is not well soluble in water. However, it is hardly present in standard water since it is almost non-abundant in the atmosphere so it cannot dissolve. (Carbon dioxide, nitrogen and oxygen are far more abundant and therefore spontaneously dissolve in water to a certain extent — especially carbon dioxide which, in fact, reacts.) It is possible to create a hydrogen-saturated aquaeous solution or something less hydrogen-containing and sell it as ‘hydrogen-enriched water’. It would have no positive or negative effects on you or your surrounding. (Unless maybe your voice got slightly lighter in the same way as when you breathe helium — however, the amount of hydrogen is very likely too low for this to happen.) So it sounds like my dream of life: Making money with other people’s stupidity. It’s a shame that I’m not ruthless enough to do such a thing …


How can ionic bonding in lanthanides occur without valence orbitals available for overlap?

Lanthanide atoms also have $5d$ and $6s$ valence orbitals available for bonding, and these can overlap effectively with ligands.  Whether this overlap leads to a bonding orbital encompassing both metal and ligand (covalent bond) or transfer if the electrons to a much low-energy ligand orbital (ionic bond), a bond is formed either way.  The bond may not be entirely ionic but is entirely (or nearly so) derived from the $5d$ and $6s$ orbitals of the lanthanide atom instead of the $4f$ orbitals.
Lanthanides do tend to form complexes with high coordination numbers, exceeding the coordination number of six commonly seen with $d$-block transition metal complexes.  From here:

Lanthanides share many similar characteristics. These characteristics include the following:
...
Adoption of coordination numbers greater than 6 (usually 8-9) in compounds
...
A preference for more electronegative elements (such as O or F) binding
...

The coordination number thereby exceeds the number of separate covalent bonds that can be formed from $5d$ and $6s$ valence orbitals alone, so a limited number of covalent bonds must be de-localized over the many linkages.  This would lead to the bonds defaulting towards a more predominantly ionic mode, requiring the ligand orbitals to be low-energy and associated with highly electronegative ligand atoms (also noted in the above excerpt).


I'm wondering why an E1cb reaction needs a poor leaving group like -OH attached to one of the carbons.

In an E1cb reaction, the rate-determining step (RDS) is the $\ce {C-H}$ bond cleavage. According to Carey & Sundberg (2007), p. 552, for the irreversible E1cb reaction (when the hydrogen abstracted is very acidic), the rate of the reaction may increase in the following order of leaving group: $\ce {I^- < Br^- < Cl^- < F^-}$. Why does the reaction rate increase as the leaving group becomes "lousier"?

From kinetics, we known that the rate of a reaction is determined by the slowest step, the RDS. In the E1cb, the RDS involves removal of a proton to form a carbanion. This step would be accelerated if the $\ce {C-H}$ bond is weakened and when the carbanion experiences greater stabilisation of the negative charge. Poor leaving groups, such as $\ce {OH^-}$ and $\ce {F^-}$, are often able to provide inductive stabilisation of the negative charge, e.g. $\ce {-OH}$ and $\ce {-F}$ substituents are electron-withdrawing in nature. This would thus help to speed up the RDS and thus, the overall reaction rate would be faster with these poor leaving groups.

Note: The explanation for this is similar to that for this question: Why are fluorides more reactive in nucleophilic aromatic substitutions than bromides?

  Paraphrased: If this is the case, can E1cb reactions still involve good leaving groups, such as $\ce {Br^-}$?

On the same page, the authors also conclude that:

  E1cb reactions generally require both carbanion stabilization and a relatively poor leaving group

If the leaving group becomes a better one, e.g. if the $\ce {F^-}$ becomes $\ce {I^-}$, then the $\ce {C-X}$ cleavage may be more advanced than the $\ce {C-H}$ bond cleavage, resulting in the reaction becoming E2, E1-like or even E1. Thus, if good leaving groups are present, likely the elimination will not proceed by E1cb but by the other pathways.

Update

I just realised someone had a similar question answered already: Why must the leaving group in E1cb be poor? You can refer to that answer for a different perspective. But ultimately, the essence of the correct explanation is well-contained in my answer above.

Reference

Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.


How does metallic bonding in alkaline earth metals work?

Metallic bonding is essentially delocalisation of covalent bonds to the logical extreme and cannot be considered as "two-centre, two-electron". The metallic lattice is a network of cations immersed in a "sea" of electrons. As the highest energy orbital is the spherical s, there is no directional influence and hence why beryllium has the efficient hexagonal close packed structure.


First I have to ask about the premise of your question- why are you teaching yourself chemistry from a book on heterocyclic compounds?  The rest of your question suggests you've skipped the basics and jumped to fairly advanced stuff.  There are many good introductory chemistry that will walk you through problems similar to this one that will be a far easier starting point than heterocyclic compounds.

Assuming your general question is "how do I do calculations on oxidation reactions?", you fill find whole chapters and even books on the topic.  I would highly recommend starting with something like "What is the balanced reaction of combusting propane (C3H8) in pure oxygen?"  You will need this kind of basis to be able to work up problems like the one you ask about with many possible products and spectator species.

This will get you started: http://en.wikipedia.org/wiki/Combustion#Chemical_equations with answer to the first question.  In general though, pick up a used introduction to chemistry book up off Amazon for $10 and start with those walk throughs.

Hope this helps answer your real question and please let me know if I misunderstood and you had a more specific question about this particular problem.

EDIT: For the yield questions, you can use the the balanced equation and molecular weights to come up with a yield.

You'll have something like:
$$yield=stoichiometricRatio*\frac{MW_{product}}{MW_{thiazole}}$$

$$\frac{g_{product}}{g_{thiozole}}=\frac{mol_{product}}{mol_{thiozole}}*\frac{g_{product}}{mol_{product}}*\frac{mol_{thiazole}}{g_{thiazole}}$$

If they give you a mass (or mass flow rate) of thiazole, you could further convert this to a mass (or mass flow rate) of products.


How can the concentration of a solid be constant during a reaction?

In physical chemistry one does not use the concentration of a chemical but its activity (see https://en.m.wikipedia.org/wiki/Thermodynamic_activity)

In a solid the standard condition is the pure chemical. In case of the solid block of aluminum it is therefore in its standard condition which leads to the fact that the activity is 1 (the chemical potential is equal to the chemical potential in the standard condition as the block of aluminium is in its standard condition - see the link I provided for the definition of the activity).

Therfore it doesn't need to be considered in calculations like the law of mass action as multiplying or diving by 1 does not affect the result.

However in "real chemistry" one does not calculate with activities too often and uses concentrations instead and in order to apply the same trick one simply says that the concentration is constant.

So in the end it's just a result of sloppiness because one doesn't want to work with activities.


Why laboratory glassware shatter in big pieces?

Kitchen glassware is made to contain large internal stresses, so it shatters into small, non-dangerous pieces. That is hard to achieve with Duran etc., because that stuff has a small thermal expansion coefficient, just so that it does not break under large temperature changes.

I expect that the kind of kitchen glassware that is supposed to withstand large temperature changes also breaks into rather large pieces.

And yes, glass is highly elastic. Elasticity is effectively the only mechanical property it does possess, before breaking. ;-)


Preference for tin or iron in the reduction of nitrobenzene

The book's reasoning is correct. Fe/HCl gives $\ce{FeCl2}$ on reaction and it gets hydrolyzed by steam vapors to produces more hydrochloric acid and hydrogen to push the reaction forward thus making the reaction self-sustaining to produce more aniline from nitrobenzene.

$$\ce{Fe + 2HCl ->[160°C,CH3OH] FeCl2 + H2}$$

$$\ce{3FeCl2 + 4H2O(steam) ->[Hydrolysis] Fe3O4 + 6HCl + H2 }$$

whereas $\ce{SnCl2}$ gives $\ce{Sn(OH)Cl}$ and 1 mole of hydrogen chloride which is insufficient and it gets exhausted after sometime.

The other thing is cost factor as @Waylander said. Iron scrap is way cheaper than tin.

Both Sn/HCl and Fe/HCl reduction(Bechamp reduction) is replaced by catalytic hydrogenation due to the latter being more efficient and more productive in industial production. Fe/HCl reduction is now an academic interest and is studied as a route to produce iron oxide based pigments.


Ethanol solution, soaked over night of APTEOS will work.  You won't get great coverage, but it will be likely good enough for what you want to do.

The subsequent chemistry is a bit more touchy.  The problem is acid byproduct which forms ammonium salt with the amine functions that is difficult to remove.  This leaves latent acid catalyst at your surface.

You are better served forming the amide from your APTEOS and coating with that product.


The effect of air pressure on the melting point

For most substances, higher pressure (or air pressure, in your case) will cause the melting temperature to go up. To think about it intuitively, imagine that you have a certain solid. Melting it would increase the volume of that substance because liquids take more space than solids. If you increase the pressure, it becomes harder for that transformation to occur. It's harder to change into a state that needs more volume if you are under pressure! So, you need more energy, hence temperature, to melt.

Notice that this relation is actually reverse in water. Because ice takes more volume than liquid water, it's actually easier to melt it at high pressure and it requires less heat. Try taking an ice cube and apply pressure on it with a knife, see how it liquifies at the contact point.


Estimating standard Gibbs free energy of a molecule

You will need to calculate the partition functions for each type of motion and then calculate the thermodynamic quantities from these.

As to internal rotation, if the energy for this is far greater than $k_BT$, such as with a C=C bond, then treat it as a torsion and use this frequency. If the barrier to rotation is small compared to $k_BT$ then treat it as a free rotor but reduce the partition function by n if there are n barriers. In the intermediate case you will have to find the barrier (say by spectroscopy) and fit the data to a model of the potential (usually of a cosine form)and then numerically solving the Schrodinger eqn. to find energy levels.

If your temperature is so high that cis-trans reaction occurs then you should probably have to treat this as a mixture.


I have read that $\ce{NH3}$ acts as an SFL with a $\ce{M^3+}$ metal ion and acts as a WFL with a $\ce{M^2+}$ metal ion.

(I assume that SFL = strong-field ligand and WFL = weak field ligand.)
No, this is not correct. Ammonia is neither a true strong-field ligand nor a true weak-field ligand. Instead, it can form high-spin and low-spin complexes depending on the properties of the metal. Thus, it is best to term it a medium-field ligand.
It is also not true that all metal(III) ammine complexes be low-spin while all metal(II) ammine complexes be high-spin. However, among the many different factors that influence the spin state of a complex, it is true that a higher oxidation state tends to favour (note the wording) a low-spin configuration.

It also states that when $\ce{NH3}$ forms a complex with the $\ce{M^2+}$ ion, its stability constant is $10^{11}$.

It would be very weird if the stability constants of all $\ce{[M(NH3)6]^2+}$ complexes have the same value. Think about it: the metals are different while everything else is the same. There has to be variation. The same is true for the aquacomplexes.

So, if that is true then $\ce{[Ni(H2O)6]^2+}$ should be more stable than $\ce{[Ni(NH3)6]^2+}$ as Ni is in the $\mathrm{+II}$ oxidation state.

You can’t say much about the stability of the two a priori. What you can say is that both of these complexes should be relatively inert (although not as inert as, say, $\ce{[Co(NH3)6]^3+}$) from a simple CFSE estimation.

The crux of the question is that we must arrange the complexes in correct order for the absorption wavelength in the visible region

This has absolutely nothing to do with the stability of the complexes or their stability constant; only with their electronic structure and the energy difference between the orbital an electron is excited from and the orbital an electron is excited into. Furthermore, there has to be an error in $\ce{[Ni(NO2)6]^2+}$ – I find it hard to see a stable complex involving six stable radicals around it.
As for the order the question asks for, this is where the spectrochemical series actually comes in and in handy. Determine which orbitals are involved, what effect a greater field split has on these, find the ligands on the spectrochemical series, determine their relative field splits and translate that all into an photon wavelength. You will soon notice that the wavelength of the hexaaquanickel(II) complex must be longer than the wavelength of hexaamminenickel(II) which gives you your answer by inferring.


How can I recognize aromatic rings in a connection table?

I will reiterate what I have written in a comment earlier, as of why I find this a bit problematic (especially for a cheminformatics approach).

There still is no clear cut definition of aromaticity, there even is an (at the time of writing) upcoming conference Aromaticity 2018 to get more clarity on the matter.

There are a couple of questions on the site, where I have written an answer for which demonstrate the issue. I don't want to be shamelessly promoting my own posts, but I also don't want to repeat myself. I think one of the crucial examples is How to explain (non-/anti-) aromaticity in fulvene with the help of resonance structures?

From examples like this, it seems clear that from structure and Kekule form alone it appears impossible to determine whether a ring has aromatic character or not. Mithoron stated in the comments something quite categorical, which I find in its extend debatable; but I wouldn't want to dismiss it either:

  I wouldn't say 2H-indene is aromatic, much like 1,2-benzoquinone also isn't.

Take the first as an example, it has a high C-H acidity (higher than indene), because the resulting structure would have a fully conjugated system very closely related to naphthalene. This structure is undoubtedly more aromatic (handwavy, but if you could quantify it, it probably would), so any kind of structural diagram, yes/no-parameter would misrepresent that.

  Given a connection table for a molecule that contains any number of six-membered rings (possibly fused, spiro, etc.) and is represented in Kekule form, can one write an algorithm to determine which rings are aromatic and which are not? Atomic coordinates are available if necessary.

I'm afraid the answer to this question is no.

What you could do, and where you'll probably be right many times is using the bond lengths as an indicator. The bonds in such rings should all be very, very similar (maybe ±10%), an given Pekka Pyykkö's data (pdf from his homepage) fall somewhere around 140 pm. Couple that with the ring-finder and I would assume it's going to be close many times.


I poked around on the internet and found a lot of sites use the Pauling electronegativity scale (link 1, link 2).  For carbon the value is 2.55 and for bromine 2.96.  Not quite a 0.5 difference, but closer - and this "rule" is somewhat arbitrary to begin with.  Also the dipole moment of bromobenzene is around 1.7 D (link 3), certainly not insignificant.  I too would have said bromobenzene is polar.  It is not strongly polar, whether it is "a little" polar, "mildly" polar, "somewhat" polar, etc. seems to be the question - but that's just semantics.  Maybe the reason many people consider the $\ce{C-Br}$ bond to be polar is because it has a very observable effect on electrophilic aromatic substitution reaction rates.  Specifically, why should the relative rate for electrophilic aromatic substitution at the meta position in bromobenzene be slower that the rate in benzene itself unless you invoke the inductive effects caused by a polar $\ce{C-Br}$ bond?


Your concentration is 0.615 M. The M stands for molar concentration, which is the number of moles per liter of solution. Thus, your concentration of 0.615 M (mol/L) means that in one liter of solution there will be 0.615 moles of solute (KCl). Similarly, a solution with concentration of 0.615 g/L would have 0.615 grams of solute per 1 liter of solution.
A mole is just like any other unit for amount, except that it is built on a very large number. Since, for example, there at $6.02\times 10^{23}$ atoms of chlorine in 35.45 grams of chlorine, we generally do not discuss numbers of atoms (or molecules) as a useful measurable amount. How much would a dozen atoms of chlorine weigh? Much less than any of our balances or scales could measure!
One mole of stuff is an amount of stuff equal to the number of carbon atoms in 12 grams of $\ce{^{12}C}$, most common isotope of carbon. This number happens to be Avogadro's Number - $6.02\times 10^{23}$, which is really large! This is also the number of atoms or molecules present in a number of grams equal to one formula weight of that substance, so it is a convenient number. We know that when we weigh out 12.01 grams of carbon, 1.008 grams of hydrogen, 15.99 grams of oxygen, etc., we know that we always have the same number of atoms in each sample, because each is one mole. We tend not to worry too much about the large number hiding in the definition of one mole.
Now, if you have one mole of a compound, like $\ce{KCl}$, then you know that there is one mole of the $\ce{KCl}$ formula unit in that sample. Since each $\ce{KCl}$ formula unit has one $\ce{K+}$ and one $\ce{Cl-}$, then each mole of $\ce{KCl}$ has one mole of $\ce{K+}$ ions and one mole of $\ce{Cl-}$ ions.
Now, you have your 0.615 M solution, and you only have 230 mL of it. How much $\ce{KCl}$, $\ce{K+}$, and $\ce{Cl-}$ is in that sample? Well, you would have 0.615 moles of each in 1 liter, but you do not have one liter. You have 230 mL. Since there are 1000 mL in a liter, we can do a little maths:
$$230\text{ mL}\times \dfrac{1 \text{ L}}{1000\text{ mL}}\times\dfrac{0.615\text{ mol KCl}}{1\text{ L}}=0.141 \text{ mol KCl}\equiv 0.141 \text{ mol}\,\ce{K+}\equiv 0.141 \text{ mol}\,\ce{Cl-}$$


Pseudoelementary reactions in the Briggs-Rauscher oscillatory reaction mechanism

What do they mean when they say the reactions are pseudoelementary?

Here is a portion of the first page:

So for the reaction (I added an H+ to the product to balance it)
$$\ce{IO3- + HIO2 + 2H+ <=> 2IO2 + H2O + H+},$$

they assume that the two protons attach in a fast reversible step (not sure to which of the reactants), and then there is a bimolecular reaction between the two iodine-containing intermediates X and Y:

$$\ce{IO3- + HIO2 + 2H+ <=>[fast] X + Y -> 2IO2 + H2O + H+},$$

The rate law for this set of reactions taken together is the one given in the question.

  Are reactions involving the collision of four or three molecules more common than I thought?

No, they are rare (and slow). They could have written out the separate elementary reactions, but they probably also did not know where the protons attach and in what order, and they are not claiming to get the mechanism right. They just want to show a plausible set of reactions and rate constants that leads to the observed oscillations.

  When they write down the equations for these reactions they write them as if the stoichiometric coefficients were the order with respect to each reagent, which seems to imply that the reactions are somehow elementary, which further confuses me.

It does not imply that. From a rate law alone you can't distinguish between mechanisms. Take these two mechanisms for example:

$$\ce{ A + H+ <=>[fast] AH+}$$
$$\ce{ AH+ ->[slow] B + H+}$$

on the one hand and
$$\ce{ A + H+ -> B}$$

on the other hand. Both will be first order in $A$ and in $\ce{H+}$.


What are the bond angles in a normal trigonal planar arrangement? Treat the center atom as in H C H as a trigonal planar arrangement with the middle (carbon) atom in the center with a third atom attached on the left over position (another carbon). What angle maximizes their separation?

Your other question about the plane of the molecule is a bit more interesting. The pi bonds between the carbons ensure that the molecule is not rotating in space. To maximize the distance between the terminal hydrogen atoms, one set should be aligned up and down, while the other two on the opposite side should be rotated 90° in space, otherwise they would ellipse one another. In other words, one pair is horizontal while the opposite is vertical for example, depending how you want to imagine it.