<h2>Solution 1:</h2>
I think the clinical chemistry notation has not been standardized but certainly it has been inspired by chemists.
The capital M in mosM emphasizes that it is milliosMolar. Chemists use capital M to denote molarity.
On the other hand, mosm indicates milliosmolal. Chemists use small m to symbolize molality.
This notation is not universal.
From this slightly dated article on Osmolality, which is behind pay-wall
Osmolality, Harry F. Weisberg Laboratory Medicine, Volume 12, Issue 2, 1 February 1981, Pages 81–85.
<blockquote>
<h3>Terminology and Units</h3>
In 1934, Gamble, et al.,11 introduced the term ”osmillimole” which has been successively transformed into mosM (milliosmolar), mosm (milliosmolal), the noncommittal mOsm, and the present-day mosmol (milliosmole). Raoult’s law states that the lowering of the vapor pressure (one of the four ”colligative”
properties of dilute solutions) is proportional to the mole-fraction (more easily calculated from molality) of solute present in the mixture.
</blockquote>

<h2>Solution 2:</h2>
Judging from the following table consisting of compiled notations presented in offline edition of The ACS Style Guide and AMA manual of style, section 13.12 Units of&nbsp;Measure:
$$
\begin{array}{ll}
\hline
\text{Unit of measure} &amp; \text{Symbol} &amp; \text{Ref.} \\
\hline
\text{osmolar} &amp; \text{osm (also osM, Osm)} &amp; \text{[1, p. 191]} \\
\text{osmole} &amp; \text{osm} &amp; \text{[2, p. 635]} \\
\text{microosmole} &amp; \text{µOsm} &amp; \text{[2, p. 634]} \\
\text{milliosmole} &amp; \text{mOsm} &amp; \text{[2, p. 634]} \\
\hline
\end{array}
$$
symbols $\pu{Osm}$ and $\pu{osm}$ refer to the amount-of-substance of an osmotically active entity, and symbol $\pu{osM}$ refers to osmolar concentration suggesting the following relation:
$$\pu{1 osm L^-1} = \pu{1 Osm L^-1} = \pu{1 osM}$$
The letter “o” in $\pu{Osm}$ appears to be capitalized for the sake of better readability when used in conjunction with metric prefixes, and the letter “m” in in $\pu{osM}$ is likely capitalized to resemble analogy with the unit symbol for molarity (small capital letter “m”: ᴍ).
<h3>References</h3>
<ol>
<li>The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. ISBN 978-0-8412-3999-9.</li>
<li>AMA Manual of Style: A Guide for Authors and Editors, 11th ed.; American Medical Association, Ed.; Oxford University Press: New York, 2020. ISBN 978-0-19-024656-3.</li>
</ol>

Chemistry - What is the difference in mosM and mOsm units, if there is any at all?

<h2>Solution 1:</h2>
Exclude reasons one by one
Surface biological contamination: use two vessels, wipe the one of them with H2O2, hydrogen peroxide solution, leave another untreated. Check if time till smell appears is different by a few days. Bacterias can survive 20% vinegar and 100*C for a short time, some of them, if they are in a sleeping stage, it is not quite reliable method i would say.
Tap water biological contamination: use two vessels with water, add a few tablespoons H2O2 with water in one, and just water in other. Check if time till smell appears is different by a few days.
Tap water chemical contamination: buy some bottled water, use two vessels, one with tap water, another with bottled water. Check if it will be different in a few days.
Bulk material of the vessel releasing chemicals: leave water in different material vessels, one is glass, another one is your preferred one. Check if glass vessel has less smell adter a few days. Glass releases the least amount of chemicals into the water.
Dust accumulation: use two vessels, cover one and leave another one open, check if smell will appear faster in one.
Psychological factor: use two identical vessels, ask your friend to give you one or another, randomly, blindfold yourself, do a test 10 times, check if your score is above 8 correct guesses out of 10 tries.
After completing this tests you will narrow down your search significantly to give you some idea about what is going on.

<h2>Solution 2:</h2>
A few more thoughts:
<ul>
<li>fresh water taste is AFAIK associated with come $\ce{CO_2}$ being dissolved in the water. This can get lost, and make the water taste stale. 
Here's an easy experiment to check this: compare the taste of:
<ul>
<li>fresh tap water,</li>
<li>water gone stale "your way" (cooled to the temp of the fresh tap water), and</li>
<li>de-gassed fresh tap water: take water, boil it to remove dissolved $\ce{CO_2}$, cool to same temp as the other 2.</li>
</ul>
If the boiled water tastes stale like the stale water, loss of $\ce{CO_2}$ is likely the cause.
I wouldn't describe this as anything like paint-thinner or plastic or the like. But for me it is the textbook meaning of "stale" for water.
</li>
<li>Sidethought about "chlorine" smell. What we usually describe as chlorine smell in water are chloramines (mono-, di-, tri-). In first approximation, we can say that they form from e.g. from chlorine used for water treatment with amino groups from organic matter. AFAIK, our nose is more sensitive to them than to chlorine itself. I.e. the smell indicates that some organic contamination occured, but also that chlorine was there and did work - but it does not indicate whether the water is now safe from a microbiological point of view, the amount of chlorine may not have been sufficient. 
So if your tap water arrives fine and with some residual chlorine, but you then get organic matter into the water, chloramine smell may develop later on.
</li>
</ul>

Chemistry - Any idea why tap water goes stale overnight?

<h2>Solution 1:</h2>
Yes, for sufficiently strong acids, in sufficiently dilute conditions, in sufficiently basic solvents. However, things are hazier than you might expect, and depending on your definition, "clear-cut" examples of complete dissociation only become possible when you start swatting flies with nuclear bombs.
The problem is that chemistry works on equilibria, and at some point it becomes difficult to measure tiny concentrations - in a haystack of one billion straws, how can you be sure there isn't at least one needle? And in one trillion? And what about in a septillion?
<hr>
First, let us perform a qualitative investigation by considering the generic acid dissociation equilibrium in water, as follows:
$$\ce{HA(aq)-&gt;H+(aq) + A-(aq)} \quad \quad \quad \mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}}=10^{-pK_a}}$$
To simplify the following discussion, we will make the very crude approximation where we equate the thermodynamic activity of a species with its molar concentration. In other words:
$$\mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}} \approx \frac{[H^+(aq)][A^-(aq)]}{[HA(aq)]}}$$
Our objective now is to virtually eliminate the presence of undissociated $\ce{HA}$ molecules. For clarity, we rearrange the above equation:
$$\mathrm{[HA(aq)] \approx \frac{[H^+(aq)][A^-(aq)]}{K_a}}$$
Clearly, to make $\ce{[HA]}$ as small as possible, the numerator must be very small, and the denominator must be very large. Let us set a goal of making $\mathrm{[HA]&lt;10^{-27}\ mol\ L^{-1}}$, which implies less than $\mathrm{0.001}$ molecules of undissociated $\ce{HA}$ in a litre of solution.
At this point, I should mention that aqueous solutions of acids in ambient conditions must always have $\mathrm{[H^+(aq)]&gt;10^{-7}\ mol\ L^{-1}}$, thanks to the self-ionization of water. We take the best-case scenario (which physically corresponds to very dilute solutions):
$$\mathrm{10^{-27} \gtrsim \frac{10^{-7}\ [A^-(aq)]}{K_a} \quad \longrightarrow \quad 10^{-20} \gtrsim \frac{[A^-(aq)]}{K_a}}$$
We now take a typical strong acid, hydrogen chloride ($\ce{HCl}$) which has an estimated pKa of -5.9 in water. In other words:
$$\mathrm{10^{-20} \gtrsim \frac{[Cl^-(aq)]}{10^{5.9}}=\frac{[Cl^-(aq)]}{790\ 000} \quad \longrightarrow \quad [Cl^-(aq)] \lesssim 8\times 10^{-15}\ mol\ L^{-1}}$$
We see therefore that meeting the criterion of full dissociation requires preparing an aqueous $\ce{HCl}$ solution with femtomolar concentration. Essentially every solution used in a chemistry laboratory will have a higher concentration, so every $\ce{HCl}$ solution anyone has ever touched will have undissociated $\ce{HCl}$ molecules. Strictly speaking, you could say the dissociation is never complete. However, if you have a mind to make a fully dissociated solution of $\ce{HCl}$, it is physically possible.
<hr>
Just diluting acid solutions into oblivion until every molecule dissociates is kind of boring. The more exciting prospect is, of course, to increase the strength of the acid. The acidity scale goes far, far beyond aqueous solutions of $\ce{HCl}$.
Interestingly, for sufficiently strong acids (typically $\mathrm{pK_a&lt;-10}$, meaning acids at least 10 000 times stronger than aqueous $\ce{HCl}$), it is possible to isolate hydronium salts - when the strong acid is mixed with water in a 1:1 molecular ratio, a solid salt is formed, where the cation is $\ce{H3O+}$ and the anion is supplied by the conjugate base of the acid. The stoichiometry can be made exact. Surely this represents full dissociation of the acid?
Well, again, it depends on your strictness. If you actually look at the crystal structures of many of these salts, it's clear that there is a strong, close interaction between the $\ce{H3O+}$ cation and the anion, and you could argue this still does not count as complete dissociation.
That said, you can go crazy and consider some of the strongest acids we can put in a bottle, such as carborane superacids. The corresponding hydronium salt of these superacids can have extremely weak interactions between the hydronium cation and the anion. For example, when the hydronium salt $\ce{(H3O+)(CHB11Cl11^-)}$ is prepared in a benzene solution, the solid that forms actually contains three molecules of benzene surrounding the hydronium cation, which is completely detached from the anion. If this doesn't imply "complete dissociation" of the carborane superacid in water, I don't know what would.
The acidity scale goes further still than even carborane superacids. There are a few known acids which will practically always fully dissociate, no matter what. The representatives of this class are rather exotic things such as $\ce{H_3^+}$, $\ce{HN2^+}$, $\ce{HNe^+}$, $\ce{HHe^+}$, and so forth. These acids can only be detected in the gas phase, as they fully dissociate by protonating virtually anything they come into contact with (including any anion, which is why they cannot be isolated as a neutral compound).
<hr>
The last way to ensure full dissociation of an acid in a solution is to change the solvent in which it is dissolved. All things considered, water is a pretty weak base. If, for example, ammonia (am) were used as the solvent, the relevant equilibrium for $\ce{HCl}$ would be:
$$\ce{HCl(am)-&gt;H+(am) + Cl-(am)} \quad \quad \quad \mathrm{K_a(am)=\frac{a_{H^+(am)}a_{Cl^-(am)}}{a_{HCl(am)}}=10^{-pK_a(am)}}$$
If I were to give a crude guess, I'd imagine that the $\mathrm{pK_a}$ for $\ce{HCl}$ in liquid ammonia would very roughly be somewhere around -15; that is, dissociation of $\ce{HCl}$ in liquid ammonia is about a billion times easier compared to water. This is because ammonia is a stronger base than water.
You know how we were talking about hydronium salts as if they were some unusual thing? Well, the equivalent concept in ammonia would be the formation of ammonium salts. And indeed, $\ce{HCl}$ does easily form a salt with ammonia in a exact 1:1 molecular ratio, namely ammonium chloride, $\ce{NH4Cl}$. Not exactly an unusual compound.
There's nothing stopping you from using an extremely strong base as a solvent. For example, in pure liquid phosphazene superbases, even very weak acids may be fully dissociated in ambient conditions, even in the most rigorous sense.

<h2>Solution 2:</h2>
Fluoroantimonic acid is so strongly disassociating that molecules of $\ce{HSbF6}$ apparently have never been isolated. Only compounds in which a proton has been transferred to form ionic species are seen. (Replacing the fluorine with chlorine or bromine also gives an isolable compound, which may or may not be molecular depending on if water of hydration accepts a proton.) In its usual application as a solution in hydrofluoric acid solvent (the latter, by the way, is itself close to being a superacid, even without the fluoroantimonic acid solute), fluoroantimonic acid consists overwhelmingly of solvated $\ce{H_{n+1}F_n^+}$ cations and $\ce{Sb_nF_{5n+1}^-}$ anions, which equilibrate with $\ce{HF + SbF5}$ rather than with a molecular $\ce{HSbF6}$ species. When a "pure" fluoroantimonic acid compound is obtained from the solution, we still do not get association into neutral molecules:
<blockquote>
Two related products have been crystallized from HF-SbF5 mixtures, and both have been analyzed by single crystal X-ray crystallography. These salts have the formulas $\ce{[H2F+][Sb2F_{11}−]}$ and $\ce{[H3F2+][Sb2F_{11}−]}$. In both salts, the anion is $\ce{Sb2F_{11}−}$.[1] As mentioned above, $\ce{SbF_6−}$ is weakly basic; the larger anion $\ce{Sb2F_{11}−}$ is expected to be still weaker.
</blockquote>
Cited reference:
1.
Mootz, Dietrich; Bartmann, Klemens (March 1988). "The Fluoronium Ions $\ce{H2F+}$ and $\ce{H3F2+}$: Characterization by Crystal Structure Analysis". Angewandte Chemie International Edition. 27 (3): 391–392. Link

Chemistry - Do strong acids actually dissociate completely?

<h2>Solution 1:</h2>
<blockquote>
transported electrons in $\ce{NADH}$ have a higher energy than those in $\ce{FADH2}.$
</blockquote>
This is jargon describing the redox potential of the electron carrier $\ce{NADH}/\ce{NAD+}$ vs the electron carrier $\ce{FADH2}/\ce{FAD}.$ So if you compare the two reduction half reactions
$$
\begin{align}
\ce{NAD+ + H+ + 2e- &amp;-&gt; NADH} &amp;\quad E^{\circ '}_\mathrm{red} &amp;= \pu{-0.320 V}\tag{1}\\
\ce{FAD + 2H+ + 2e- &amp;-&gt; FADH2} &amp;\quad E^{\circ '}_\mathrm{red} &amp;= \pu{0.031 V}\tag{2}
\end{align}
$$
$\ce{NADH}$ is the stronger reducing agent.
<blockquote>
Could anyone come up with a definition understandable by this non-chemist (i.e. me)?
</blockquote>
Another way of saying this is that the reaction of $\ce{NADH}$ with dioxygen is more exergonic (the equilibrium lies further on the side of the products, more free energy is available from it) than the reaction of $\ce{FADH2}$ with dioxygen. For completeness, the reduction half reaction for dioxygen/water is:
$$\ce{1/2 O2 + 2 e- + 2H+ -&gt; H2O} \quad E^{\circ '}_\mathrm{red} = \pu{0.816 V}\tag{3}$$
<blockquote>
Do they mean that when these electrons are transferred to the electron carriers they are part of molecular structures in which they sit in high energy positions?
</blockquote>
It has to do with how tightly the electrons are bound in one and the other electron carrier. The OP is correct to place the term "high-energy" in quotes when referring to electrons; electrons are indistinguishable, so their energy is with respect to the bound state that they are in.

<h2>Solution 2:</h2>
@KarstenTheis made it clear by using E0 values of each half reaction. Here I describe what they actually mean by using the phrase "high energy electrons" in biochemistry texts.
Consider Ethyne and Ethane.
<img src="https://i.stack.imgur.com/aaIw2.png" alt="enter image description here">
Therefore Ethynide has a high energy electron compared to Ethanide
Another Example : H- , OH- , SH- . H- is stronger than OH- and OH- is stronger than HS- in the terms of basicity. We can say that H- has more active electrons compared to those 2. Therefore in the opinion of biology systems , its electron has more energy.(discussed more in the upcoming paragraphs)
The more thermodynamically unfavorable reaction gives out the more instable product.
In turn, the instable product participates in other reactions , in a fast rate.
High Energy Electrons do mean the unstable electrons present in the biochemicals. These biochemicals are unstable because they've missed aromaticity ,their previous octet configuration or became radicals etc. therefore they have high Gibbs energy.
Look at NAD+ , FAD and their reactions
NAD+ ---&gt; NADH
<img src="https://i.stack.imgur.com/Uba96.png" alt="NAD+">
FAD ---&gt; FADH2
<img src="https://i.stack.imgur.com/uauty.png" alt="FAD">
NAD+ loses its entire aromaticity by the reduction. This makes the reduction reaction ,to some extents, thermodynamically unfavorable for NAD+ (compared to FAD). This is why E0 of FAD is bigger than E0 of NAD+.
As I said Reduction of NAD+ is less favorable than reduction of FAD. Therefore NADH is more unstable/has high energy electrons than FADH2 and you can say that electrons in NADH are more accessible because they want to gain their aromaticity back.(not because of the radius).
So NADH is more reactant than FADH2.
EDIT: This just came to my mind. In biochemistry texts , usually they say that "covalent bond between carbon and carbon in sugars, has a lot of energy that eventually gives the energy to NADH,FADH2 and later for ATP."
This highness of energy comes from the oxidation number of carbons in sugars(e.g. glucose) . See this
<img src="https://i.stack.imgur.com/TOg5N.gif" alt="Carbon Oxidation States">
Each of them eventually participate in a CO2 by-product. The oxidation number in CO2 is +4. This results in huge movement of electrons from glucose to 6 CO2 molecules, hence the term "high energy covalent bonds in glucose". In one instance i.e. reduction of NAD+ to NADH in Step 6 of glycolysis , there is a nucleophilic substitution , HPO4(-2) acting as nucleophile and interestingly H- as leaving group.
Here, the attack of Pi and the exit of H-(carrying electrons) , oxidizes the aldehyde carbon located at the end of the molecule.
Step 6
this unstable product containing high energy electrons (H-) , reacts with NAD+ at an extremely high rate.(the more unstable the reagent , the faster the reaction). {H- = H+ + 2e-)
Hydrid acting as a leaving group , is an interesting concept. look at Cannizzaro reaction . In the Step 6 of glycolysis, Aldehyde becomes oxidized into phosphonic acid and later carboxylic acid (reducing another particle- NAD+) ,While in Cannizzaro reaction , an aldehyde becomes oxidized into a carboxylic acid ( reducing another particle- another aldehyde).

Chemistry - What are high-energy electrons?

<h2>Solution 1:</h2>
You are right heavy metals (as Pb) is a vague term. It is being discouraged and individual analysis should be reported. When someone says "I tried to find a clear standard describing this to no avail", it is a good illustration that not all information is on the webpages. It is worthwhile to search books :-)
I used to think this was an old-school analytical terminology and it turns out to be correct because it is based on classical methods. From the Food Chemicals Codex (ISBN 0-309-08866-6), which is a 1000 paged book, has the following information. Later they describe the tests as well.
Heavy Metals (as Pb)
<blockquote>
This test is designed to limit the content of common metallic
impurities colored by sulfide ion (Ag, As, Bi, Cd, Cu, Hg, Pb, Sb, Sn)
by comparing the color with a standard containing lead
(Pb) ion under the specified test conditions. It demonstrates that the
test substance is not grossly contaminated by such heavy metals, and
within the precision of the test, that it does not exceed the Heavy
Metals limit given as determined by concomitant visual comparison with
a control solution. In the specified pH range, the optimum
concentration of lead (Pb) ion for matching purposes by this method is
20 micrograms in 50 mL of solution. The most common limitation of the
Heavy Metals Test is that the color the sulfide ion produces in the
Sample Solution depends on the metals present and may not match the
color in the Lead Solution used for matching purposes. Lead sulfide is
brown, as are Ag, Bi, Cu, Hg, and Sn sulfides. While it is possible
that ions not mentioned here may also yield nonmatching colors, among
the nine common metallic impurities listed above, the sulfides with
different colors are those of As and Cd, which are yellow, and that of
Sb, which is orange.
If a yellow or orange color is observed, the following action is
indicated: Because this monograph does not include an arsenic
requirement, As should be determined. Any As found should not exceed 3
mg/kg. If these criteria are met, Cd may be a contributor to the
yellow color, so the Cd content should be determined. If an orange
color is observed, the Sb content should be determined. These
additional tests are in accord with the section on Trace Impurities in
the General Provisions of this book, as follows: ‘‘if other possible
impurities may be present, additional tests may be required, and
should be applied, as necessary, by the manufacturer, vendor, or user
to demonstrate that the substance is suitable for its intended
application.’’
</blockquote>

<h2>Solution 2:</h2>
In addition to @M.Farooq answer on the food sector, if you read heavy metals as Pb in a pharmaceutical environment, it stems from a method of the United States Pharmacopoeia, active until 2018 (same for the European, Japanese and British Pharmacopoeia as well as others).
Up until then, it was normal to analyse heavy metals as a sum parameter for many drug products. The analysis was a limit test against a lead nitrate standard, where you compare the colour.
From the USP chapter 231 of USP 39-NF 34 valid until 1. January 2018 :
<blockquote>
This test is provided to demonstrate that the content of metallic impurities that are coloured by sulphide ion, under the specified test conditions, does not exceed the Heavy metals limit specified in the individual monograph in percentage (by weight) of lead in the test substance, as determined by concomitant visual comparison with a control prepared from a Standard Lead Solution. [NOTE—Substances that typically will respond to this test are lead, mercury, bismuth, arsenic antimony, tin, cadmium, silver, copper, and molybdenum.]
</blockquote>
In addition for a lot of pharmaceutical products, special analysis for metals used in the manufacturing process were described in their monographs, such as limits of Arsenic, Iron, Lead and others.
Here is an article and a graph of the sensitivy of this method, which was quite bad.
It was a worldwide act and discussion, until finally in 2018 the International Council for Harmonisation (ICH) guidance for industry Q3D Elemental Impurities(ICH Q3D) was set in place and adopted by the regional pharmacopoeias, see FDA USP .
For the European part, here is a short "historic overview" from the European Agency.
<blockquote>
Elemental impurities in the Ph. Eur.
The Stone age : from genesis to 2008
<ul>
<li>use of non specific wet chemical test for « heavy metals »</li>
<li>limit to 10 or 20 ppm &gt; (with reference to lead)</li>
</ul>
The Bronze age : from 2008 to 2013
<ul>
<li>EMA GL on the specification limits for residues of metal catalysts or metal reagents</li>
</ul>
The Golden age : starting 2013
<ul>
<li>CHMP decision to delay the application of the EMA guideline to existing marketed products
discussions on the implementation of ICH Q3D guideline</li>
</ul>
</blockquote>
Then the chapter &lt;231&gt; was deleted form the USP and the chapters &lt;232&gt; Elemental Impurities—Limits and &lt;233&gt; Elemental Impurities—Procedures were released.
Now, elemental impurities are analysed by ICP or AAS on an risk based approach, mostly by studying, which elements are used in the manufacturing process, as part of the packaging, colouring etc and can leach into the drug, and then analysing the content.
So, in the pharmaceutical industry, heavy metals as Pb was a specific method for the analysis of heavy metals via colour comparison test. Until 2018, pharmaceutical products could be released with a Heav metal test based on a 1800 or 1900 visual colour comaprison test which excluded a lot of metals. Heavy metals "without Pb" did exist informal as a short version, for metals screenings and to talk about outdated methods.

Chemistry - Meaning of the Term "Heavy Metals" in CofA?

<h2>Solution 1:</h2>
<h2>meth vs eth</h2>
<blockquote>
[OP] Why is '-ethane' in 'methane'?
</blockquote>
This is a coincidence.
Methyl is
<blockquote>
ultimately from Greek methy "wine" + hylē "wood.
</blockquote>
Source: https://www.etymonline.com.
The terminology was created by Dumas and Péligot in 1834 to distinguish wood alcohol from ethanol, and was published first in the French language:
<img src="https://i.stack.imgur.com/sibbI.jpg" alt="https://gallica.bnf.fr/ark:/12148/bpt6k6569005x/f15.item">
Source: https://gallica.bnf.fr/ark:/12148/bpt6k6569005x/f15.item
Ethyl is from
<blockquote>
Greek aithēr "upper air; bright, purer air; the sky" (opposed to aēr "the lower air"), from aithein "to burn, shine," from PIE *aidh- "to burn"
</blockquote>
Source: https://www.etymonline.com.
Liebig had argued in 1834 that what now is known to be $\ce{-C2H5}$ is a "radical", i.e. a reoccuring group in molecules such as diethyl ether, ethanol and ethyl ethanoate, and gave it the name ethyl.
<img src="https://i.stack.imgur.com/jWuMH.jpg" alt="enter image description here">
Source: https://babel.hathitrust.org/cgi/pt?id=uva.x002457887&amp;view=1up&amp;seq=30
Using the ending -yl for both the methyl and the ethyl group is a suggestion by Berzelius from 1835 (the correct atomic weight of carbon was not know yet, so the chemical formulas are off by a factor of two):
<blockquote>
<img src="https://i.stack.imgur.com/D0Tfrm.jpg" alt="enter image description here">
</blockquote>
Source: https://books.google.com/books?id=1DM1AAAAcAAJ&amp;pg=PA376#v=onepage&amp;q&amp;f=false, see Wikipedia entry on Ethyl group.
<h2>methane vs ethane</h2>
That these both end in -ane is not a coincidence - all alkanes end in -ane. As stated in M. Farooq's answer, this nomenclature was introduced by A. W. Hofmann in 1865.
<h2>Acknowledgements</h2>
Maurice's answer inspired me to find the first work by Dumas that led to the methyl terminology. I have edited my answer to incorporate information from the excellent answers by M. Farooq and Jan and the insightful comments. In trying to figure out the sequence of events in the 1830s, I made use of this account by Frederick E. Ziegler.
<h2>References</h2>
Dumas, J.-B, Mémoire sur l'esprit de bois et sur les divers composés éthérés qui en proviennent. lu à l'Académie des Sciences les 27 Octobre et 3 novembre 1834, Paris, 1834
Jean-Baptiste Dumas et Eugène Péligot, Mémoire sur l’Esprit de Bois et sur les divers Composés Ethérés qui en proviennent, Annales de chimie et de physique, 58 (1835) p. 5-74
Justus Liebig (1834) "Ueber die Constitution des Aethers und seiner Verbindungen" (On the composition of ethers and their compounds), Annalen der Pharmacie, 9 : 1–39
Jacob Berzelius, Årsberättelsen om framsteg i fysik och kemi [Annual report on progress in physics and chemistry] (Stockholm, Sweden: P.A. Norstedt &amp; Söner, 1835), p. 376
A. W. Hofmann 1866, in Monatsbericht der Königl. Preuss. Akad. der Wissenschaften zu Berlin, 1865 653

<h2>Solution 2:</h2>
It’s worth pointing out how much of a coincidence it is that the modern ethyl and methyl are identical but for one letter. While Karsten already outlined the general bits of etymology behind these two words one can go a little bit deeper and find out how different they are.
Methane’s derivation traces itself back to Greek μέθυ (méthy) and ὕλη (hýlē with a long e) via the intermediate methyl. As the first component included Greek epsilon (ε), it was rendered e in all languages that are based on the Latin alphabet.
Ethane can be traced back directly to ethyl and was introduced to chemistry via German. However, when introduced the German spelling was actually Äthyl, sometimes (especially in the day) rendered Aethyl. The German letter ä represents a sound similar to the vowel in English bear – more open than the e spelling (which corresponds approximately to English here) would suggest. The German spelling changed sometime during the last century and nowadays the pronunciation of Ethyl and Ethan has the closed e-type vowel (while older non-chemists might still write ä and pronounce it with the more open vowel).
The Ä (or Ae) spelling in German was actually systematic, as it traces back to Latin aethēr that already featured ae; practically all Latin ae combinations are pronounced as ä in modern German. Latin borrowed this term from Greek (hence the th) where it was written αἰθήρ (aithḗr). Both the Greek and Latin terms were probably originally pronounced with an /ai/ diphthong (approximately as in high).
The term ethyl may have also entered English more directly, being derived from the already established and borrowed term ether (which came to English via French as a direct descendent of Latin aethēr; French had already dropped the leading a in spelling). The corresponding German term used to be written Äther or Aether exclusively. Nowadays, the chemical term is only Ether while the non-scientific term may be either but is more likely to remain Äther.
Thus, the two terms only began sharing most of their letters by accident and in late stages of their etymology; and had German remained a primary language of chemistry we might still be writing aethane or some derivative.

<hr>
<h2>Solution 3:</h2>
For word-history lovers...It is all gifted from the German language. Hope that satisfies the OP's curiosity.
The suffix -ane, was chosen by a chemist and it was not random or accidental. It was a well thought-out suggestion by Hofmann.
Current sense.
<blockquote>
In sense 2 after German -an (A. W. Hofmann 1866, in Monatsbericht der
Königl. Preuss. Akad. der Wissenschaften zu Berlin, 1865 653,
translated by the author in quot. 1867). In this passage, Hofmann
proposes a series of five such suffixes, one with each vowel (German
-an , -en , -in , -on , -un ), and the systematic application of these to specific types of hydrocarbons, or their analogues. Of these (and
their English equivalents), only -an -ane suffix2, -en -ene comb.
form, and -in -ine suffix5 were adopted by others. As influences on
his nomenclature, Hofmann cites (in the same paper) A. Laurent
(compare -ene comb. form) and C. Gerhard (compare discussion at -ol
suffix and -one suffix).
</blockquote>
Methane
<blockquote>
Origin: Formed within English, by derivation; modelled on a German lexical item. Etymons: meth- comb. form, -ane suffix.
Etymology: &lt; meth- comb. form + -ane suffix, after German Methan (A. W. Hofmann 1866, in Monatsbericht der Königl. Preuss. Akad. der Wissenschaften zu Berlin, 1865 653).
</blockquote>
Ethane
<blockquote>
Etymology: &lt; eth- comb. form + -ane, after German Aethan (A. W.
Hofmann 1866, in Monatsbericht der Königl. Preuss. Akad. der
Wissenschaften zu Berlin, 1865 653; also Äthan and (now usually)
Ethan).
</blockquote>
Ref: Oxford's Unabridged Dictionary (behind subscription).

Chemistry - Why is '-ethane' in 'methane'?

<h2>Solution 1:</h2>
The results of the iodolactonization example presented in this question were published in 1982 by Bartlett, et al.1 While @Aniruddha Deb has captured the essence of the reaction, there is more than one product formed in the reaction. A more detailed mechanism is offered, an issue not addressed by the original authors. 
Carbonate 1 is a racemate. No products derived from reaction at the remote 7,8-double bond were detected. Understandably, the rate of closure is faster for a 6-membered ring than a 7-membered ring. A transition state for ring closure for a reversibly formed iodonium species will be favored having the carbon chain R equatorially disposed in the transition state. Transition state 2 is favored over that of 5 because the latter encounters a gauche butane interaction. In the iodonium cation the C-C bond is shorter than the C-I bond thereby having the former of greater steric consequence than the latter. SN2 attack on the iodonium cation is favored by the more polarized carbonyl oxygen than the oxygen attached to the t-butyl group. In the latter instance, establishing a positive charge on this oxygen is counterproductive because it is adjacent to a partially positive carbonyl carbon. This objection is in keeping with why amides are preferentially protonated on oxygen and not nitrogen. The cations 3 and 6 are now poised to lose a t-butyl cation, which, given the solvent acetonitrile, may lead to Ritter intermediates besides t-butyl iodide. In the event, racemic cyclic carbonates 4 and 7 are produced in 69% yield with the cis-substituted isomer 4 predominating over the trans-isomer 7. 
<img src="https://i.stack.imgur.com/PmxPA.jpg" height="400">
<ol>
<li>P.A. Bartlett, J. D. Meadows, E. G. Brown, A. Morimoto and K. K. Jernstedt, J. Org. Chem., 1982, 47, 4013. https://doi.org/10.1021/jo00142a002</li>
</ol>

<h2>Solution 2:</h2>
Without further details, I would say that both products are possible.
The reaction that you have given here is an Iodolactonization reaction. You have correctly identified that the tert-butyl group would be removed and that the oxygen atom would prefer to form a 6-member ring. In terms of stereochemistry, however, it is tough to make a decision, as:
<ol>
<li>The stereochemistry of the ester group is not known</li>
<li>There are no more details on the reagents or reaction conditions.</li>
</ol>
Another note on the stereochemistry is that The iodine group would always be opposite to the attacking oxygen as this is an SN2-like reaction. Keeping these in mind, we can perform the reaction for the two different enantiomers of the substrate and see that both products are possible.
<img src="https://i.stack.imgur.com/vE5aP.png" alt="enter image description here">

Chemistry - Stereoselectivity of ring closure in intramolecular iodolactonisation

<h2>Solution 1:</h2>
Extensive properties are those that are directly proportional to the mass of the system, under the restriction that all intensive properties are held constant.
The restriction that "all intensive properties are held constant" means we are not changing the inherent nature of the system; rather, we are only changing its size.
For instance, if we keep all the intensive properties constant, and double the mass of the system, all the extensive properties will double.
Thus extensive properties all have the form $X=nZ$, where n is the number of particles in the system, and Z is some intensive property. [We can use n in place of mass because the restriction that all intensive properties are held constant means that the composition doesn't change.]
Consequently, if we have two extensive properties, X1 and X2, their ratio is:
$$\frac{X1}{X2}=\frac{nZ1}{nZ2}=\frac{Z1}{Z2}$$
Since the n's cancel out, the ratio of two extensive properties becomes the ratio of two intensive properties. And since intensive properties are independent of n, their ratio is also independent of n.
Based on the above definition of extensivity, volume is only extensive in 3-dimensional systems, because it's only in 3-D systems that $V\propto n$. Area is only extensive in 2-D systems, because it's only in 2-D systems that $A\propto n$.
I have never before considered resistance in the context of extensivity, but I believe the reason you are running into trouble is that resistance depends both on the size of the system and its geometry. Given that, I would conclude that resistance is neither intensive nor extensive.
You could make resistance extensive by constraining the geometry, specifically by constraining A to be constant, in which case resistance would be a 1-D extensive property that depends upon L: $R \propto L \propto n$.
However, I don't believe you could turn resistance into an extensive property by constraining L to be constant, since then you would have: $R \propto A^{-1} \propto n^{-2}$.

<h2>Solution 2:</h2>
The conductance (inverse resistance) describes a non-equilibrium property, the current transported through a medium divided by the applied voltage:
$$\frac{1}{V}\frac{dQ}{dt} = R^{-1} = \Lambda$$
where $\Lambda$ is the conductance.
You might say that an applied voltage is an intensive property independent of geometry, and could write instead
$$V = R\frac{dQ}{dt} = \frac{dQ/dt}{\Lambda}$$
Both conductance (or resistance) and current depend on the geometry (dimensions) of the sample, but clearly something cancels when dividing these extensive properties because we obtain an intensive property (voltage).
On the other hand, conductivity is often assumed to be an intrinsic property of a material independent of geometry, defined as
$$\rho = \frac{E}{J}$$
where $E$ and $J$ are the magnitude of the local electric field and the current density in the material (the vector properties $\vec E$ and $\vec J$ are here assumed collinear, more generally $\rho$ is a tensor).

Chemistry - If x and y are 2 extensive properties of a system then is x/y always intensive?

Rebar for concrete is frequently stored in heaps, and rusts. It actually develops a stronger bond if it is not too rusty, because the rebar surface is rougher and better able to adhere to the hardened cement. It's interesting to see a pile of old rebar with a fresh delivery on top - silvery, not a trace of rust - on top of a pile of rebar that looks almost disgustingly rusted. And then you think, yeah, but the stuff on the bottom will have a better bond.
The key word in the observation about the fences is "sometimes".
I'm going to guess that the maker of the fence in the top picture also had a pile (or two) of iron bars, and some were rusty from being stored long or outdoors, or were a teensy bit different from some other pile of iron bars. This fence maker, knowing how his fence weathers, has to make it as uniform as possible, so he takes one bar from one pile, then a bar from the other pile... He prepares them identically, primes them identically, paints them identically... But then nature takes over and the most corrodable bar develops a spot, which spreads until it is all corroded-looking, before the more corrosion-resistant bar has even gotten started!
And then you come along and take a picture just before the other bars begin to corrode! This experiment is in the initial stages, not yet at equilibrium. Check back on the fence every year or two - I believe things will even out in time.

Chemistry - Rusting in every other bar

<h2>Solution 1:</h2>
<blockquote>
I get that this makes sense, but I have no idea how to derive it myself. Any help would be appreciated!
</blockquote>
Good question because such questions were pursued by prominent physical chemists and Nobel laureates (Ostwald/ Nernst and Hoffmann).
The reduction of water can be tested visually and the apparatus to demonstrate it indeed named after Hoffman. Basically, it is an electrolysis of water experiment. What will see at the side connected to the negative terminal of the battery that it produces a gas, which is twice the volume as compared to the side connected to the positive terminal of the battery.
This gas which is very light, burns with a blue flame and it forms water again- so this must be hydrogen. Secondly, another interesting effect is that electrode region which is producing hydrogen becomes highly alkaline. If you put phenolphthalein indicator, the negative terminal becomes pink indicating an excess of OH(-) ions.
<blockquote>
<img src="https://i.stack.imgur.com/M844Bl.png" alt="Blockquote">
</blockquote>
So three things can convince us with the following:
$\ce{2H2O + 2e- -&gt; H2 + 2OH-}$
a) The electron rich electrode is the cathode in electrolysis cell. Water can gain electrons here.
b) The cathode becomes highly alkaline during electrolysis
c) The cathode produces hydrogen gas, which is twice in volume as compared to oxygen.
d) You can even calculate the number of electrons used to produce a given volume of hydrogen from the measured current and the time.

<h2>Solution 2:</h2>
First of all: Reduction is just a fancy way of saying "gain of electrons".
The posed question can thus be reformulated as:
What happens to water, if it gains more electrons?
In order to answer this, we should have a look at the oxidation numbers of water: Since oxygen is more electronegative, it has an oxidation number of -II, while the two hydrogen atoms have an oxidation number of +I (Roman numerals).
As we can see, oxygen is already in its most reduced state. We know this, because oxygen has 6 valence electrons, with an oxidation number of -II this makes eight thus fulfilling the octet rule.
What can be reduced (i.e. gain electrons) here, is thus the hydrogen, going from +I to 0 (electrons are negatively charged, thus reducing the oxidation number).
Hydrogen with an oxidation number of 0 is the hydrogen molecule $\ce{H2}$. In $\ce{OH-}$, both atoms still have the oxidation numbers as in water. Now we just have to balance the reaction.

<hr>
<h2>Solution 3:</h2>
Reduction processes as posed here, “rearranges” the atoms upon taking up two electrons. One way you may have been able to “derive” or take a good guess is to consider the electrolysis of water, breaking water down into $\ce{H2}$ and $\ce{O2}$. In this case there has to be both an Oxidation and a Reduction. In the hydrolysis case, the $\ce{H2}$ is the product from the reduction process. This is because the $\ce{H}$ in $\ce{H2O}$, formally has a 1+ charge. The $\ce{H}$ [both of them] in $\ce{H2}$ has a formal charge of zero. The $\ce{H+}$ of water accepted an electron and now has a charge of zero – hence a reduction. Now, knowing the reduction product, simply balance the reaction.

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