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On the directing nature of these groups in electrophilic aromatic substitution
These three groups are listed in our workbook. I don't know the answer and our teacher isn't sure either:

(each has a free valency through which it is attached to a benzene ring):

Diphenylhalomethyl
Difluoromethyl
Dichloromethyl

Now, I've read and understood this question - How does hyperconjugation lead to the directing properties of alkyl group? Therefore, I think that the groups 2,3 are deactivating (due to -I effect of fluorines and chlorines) yet o/p directing due to hyperconjugation with the H atom. On the other hand, the first group should be meta directing since it has only -I effect and no +/-M or hyperconjugation.
I googled for the complete list of directing groups but I could not find these types of groups.

Is my reasoning - and hence answer - correct? Or is there another answer, and why?

Is 1 a deactivator meta director? 2 deactivator , o/p? 3 deactivator meta? If any of those is correct let me know. I might be thinking on the rightish track.

@AvnishKabaj I don't know the answer :(

I agree with you, on the first group being a meta directing group. But I believe that the degree of hyperconjugation in the second and the third group is very low (due to a single hydrogen atom). Carbon is quite electrophilic in both the cases and hence they must behave like meta directing groups just like CF3.

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TINKER molecular modeling extra forces - energy unit
I use the TINKER molecular modeling software package (version 7.1.2) for computational physical chemistry. There is an extra file (extra1.f) included for the user to edit and incorporate their own unique forces/energies to the dynamics of the molecules. The energy that I want to add contains the Boltzmann constant, but I am having trouble determining if it is correct to use the units of calorie (kcal/mole/K) or joule (kJ/mole/K). The source code seems to use both units in different ways. The default Boltzmann parameter in the user guide and source code is 0.831446215 (kB) in g*Ang2/ps2/mole/K, but the energy totals are all listed in kcal/mol K. Which unit is correct to use in the extra1 file?

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Phospors and Phosphorus
I'm not a chemist.
I understand that phosphorus comes in multiple forms, one of which is white, volatile, and highly toxic.  I know that phosphors come in multiple chemical forms, two of which are zinc sulfide and strontium aluminate for glow-in-the-dark products like light sticks, etc, many of which kids use.
In doing some research for a magazine article I'm writing, I read a sentence that said "Chemists have been able to mimic phosphors …" and the word "mimic" leaves me wondering if this was done because original phosphors were (and still can be) from phosphorus?

Phosphors made out of phosphorus are about as impossible these days as heroin cough syrup or radium beauty powder.

For those voting to close, I think you are making a harsh and unfair decision. This is a perfectly good question from a non-chemist and deserves a good answer not closure.

Phosphers have never been made with phosphorous just like lead pencils have never been made with lead. The words have a common origin but there is very little chemical relationship.
Phosphorus is a Latin name for the planet Venus and, figuratively, for any glowing object. Early chemists in the 17th century used the word to refer to any substance which glowed spontaneously or when heated. They characterised and named several different types of phosphoruses. When the element phosphorus was isolated from urine in the late 17th century it was called “English Phosphorus” or “Phosphorus of Urine” and later simply “Phosphorus”.
The older meaning of “phosphorus” and “phosphor” continued into the 19th century and, in the early 20th century, “phosphor” began to be used with more scientific rigor to refer to a material exhibiting phosphorescence.
Phosphorus (the element) is not phosphorescent; it emits light only insomuch as it spontaneously ignites in air.
Graphite (a.k.a. plumbago) was used as a substitute for black lead hence it was the "lead" in the pencil.

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Mechanism for conversion of gem-dihalide to carbonyl compound
How does nucleophilic substitution work in a compound consisting of two leaving groups?

I understand that hydroxide ions will attack the carbon centre (which bears a partial positive charge) and cause a nucleophilic substitution reaction.
What I'm unsure of is whether the reaction proceeds via

The hydroxyl groups replacing the leaving groups one at a time (double $\mathrm{S_N2}$).
After the replacement of the first hydroxyl group, the repulsions between the electron pairs of oxygen and of chlorine cause the chlorine, a better leaving group, to leave, causing formation of a cation. ($\mathrm{S_N2}$ followed by $\mathrm{S_N1}$).
Both bonds forming at once (a transition state with six bonds to the center), although this seems very unlikely.

I don't think it needs to go through the geminal diol - simply one SN2 followed by oxygen lone pair kicking chloride out is sufficient.

But if it does, it will go one by one.


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What is the grey film/black sludge from mercury? How to remove it?
How to remove it? Any other simple method apart from distillation.
I have a small container of mercury that I’ve collected over the past decade. I opened the container after a few years now and I noticed that there is a grey coating on the surface of mercury. Also there is a black powdery deposit at the bottom. It sticks to the bottom and sides of the plastic container.

There is also chance that the mercury you stored was contaminated with other metals. In this case they might react with air, forming oxide powder. If you are lucky, they might dissolve in acid. However, the resulting solution must be assumed to contain mercury, i.e. considered a toxic waste.

Out of idle curiosity, what did you decide to do with the mercury? No need to answer if you prefer not!

@EdV No problem sure I can tell you. I eventually figured out what was the issue. So, about 2 months after I had posted this, I got another small batch of mercury and in the same type of glass bottle container. The bottle used to store both the contaminated one and the new one was the same and its laboratory type. The new one is exactly how I had stored it. Well maybe a little less shiny due to mild oxidation. But the other one still sticky and It coats on the glass walls and the base of the container...contd

The reason was my friend who was playing around with it while I was showing him, had transferred it to a watch glass to take a better look. But unknown to me he had used some aluminium rod to touch its surface and then later we had transferred it back to the bottle. He didn't know about the amalgamation process. I guess I might distil it off someday to get pure Hg.

Ah, aluminum reacts vigorously with mercury, once the oxide coating on the aluminum is breached. In WW II, the allies actually considered doing a covert operation to sabotage German aircraft by putting mercury amalgam on the fuselages. So that explains the  stuff that that you are seeing. Thanks for the information!

Yeah, no wonder mercury is a big no-no in airplanes! One person, I know had asked me last year if I could give him a few drops of the mercury to kill a tree in his backyard since he said it was legal only to cut if the tree was dead. I'm aware that if you drill a foot deep hole into the tree or near it's root and inject some mercury or sulfuric acid. The tree would be dead pretty soon. Mercury kills it a lot faster. Anyway I wasn't very comfortable with the idea and told him to get some KNO3 from a store & inject and get it done. I didn't want to get into trouble if he did something stupid lol

First, I would advise against using mercury in a home lab due to the ease which scattered droplets fall into cracks and evaporate.
That said, if distillation is out, then try bulb-pipetting from the center of the mass of mercury to avoid some of the surface contamination. Avoid dropping the mercury when lifting the pipette, since $\ce{Hg}$ is so dense it tends to run out as soon as lifted. Ideally, work in a hood with a large tray with raised sides underneath the apparatus.
If purity is important, electrolysis will yield a higher-grade product... at the risk of working with even more toxic compounds!
Better, get some gallium if you want to play with a liquid metal, and don't need $\ce{Hg}$ to make Grignard reagent or other esoteric substance.


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Determining pH of immunoprecipitation buffer recipe
I'm following a protocol for extraction of proteins from yeast, and I came across this recipe for immunoprecipitation buffer:

$\pu{50 mM}$ HEPES-$\ce{KOH}$, pH 6.8, $\pu{150 mM}$ $\ce{KOAc}$, $\pu{2 mM}$ $\ce{MgOAc}$, $\pu{1 mM}$ $\ce{CaCl2}$, $15\%$ glycerol.

Does this mean I bring the whole solution to pH 6.8, or do I make the HEPES at pH 6.8, and then add the remaining reagents?


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What is the difference between BET and Langmuir model for calculating the specific surface area?
Brunauer–Emmett–Teller (BET) surface area is measured by specific surface area of the adsorbent. BET surface area is measured by using non-corrosive gases like $\ce{N2}$, $\ce{CO2}$, $\ce{Ar}$ etc. BET surface area depends on size and number of gas molecules adsorbed. Langmuir surface area depends on adsorption capacity of the adsorbent.
Why is Langmuir surface area always higher than BET surface area?

Langmuir adsorption model considers that only a monolayer of adsorbate can be formed on top the surface.
In contrast, BET model considers multilayer formation.
Then, if (as generally occurs) multilayers are formed, BET is able to consider that they are not being formed by directly covering the surface, but by covering extra layers. Langmuir will consider that multilayers can only form as the first adsorbed layer, thus resulting in an exaggerate (unrealistic) surface area.
You can clearly see the differences in the Wikipedia page for Langmuir adsorption.

As correctly identified, the Langmuir model assumes a monolayer of adsorbed molecules. The adsorption cross section of an adsorbate depends on numerous factors, not just the molecular size. Furthermore, the observed specific surface area is scale variant, similar to the length of a coastline. A phenomenon known as fractality. That mean that the measured surface area depends on the size of the adsorbate molecule.

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Why do many organic halogen compounds smell or taste sweet?
Is it just coincidence that many organic halogen compounds (especially chlorinated ones) tend to either taste or smell sweet? Examples include dichloromethane, vinyl chloride, chloroform, and sucralose. This chart also shows that many low-carbon haloalkanes have a sweet smell. If this isn't coincidence, what is the biochemical basis for this phenomenon?

Come to think of it, "sweet smell" is a misnomer in itself. Pure sugar doesn't smell much of anything, and sweet-smelling chemicals typically aren't sweet to the taste.

The taste of halogenated organic compounds mainly depends on the halogen atoms involved. Study showed that halogenation of aliphatic series imparted sweetness and halogenation of aromatic compounds increased/decreased the intensity of sweetness/bitteness. This increase/decrease of intensity is due to the efficacy of halogen atoms which is directly related to electron withdrawing power (Hammett $\pu{\sigma}$ value). This is called AH-B theory of sweetness$\ce{^{[3]}}$. Effect of halogen substituent on aromatic compounds to impart bitterness was in opposite trend when compared to effect of halogen substituent on aliphatic compounds to impart sweetness. For example, halogenation of  aminonitrobenzene compounds impart sweetness$\ce{^{[1]}}$. Nitroaniline has a relative sweetness of 40 and substituting halogen at para position respect to nitro group increases sweetness (chlorine: $\pu{x 400}$, bromine:  $\pu{x 800}$, iodine:  $\pu{x 1250}$).

Commercially, chlorine played an important role in imparting sweetness in organic compounds. One instance is halogenation of saccharin. Substituting fluorine at para position imparts sweetness but a bitter aftertaste. Substituting chlorine imparts sweet but with accentuated bitterness. Substituting bromine enhances sweetness but it is extremely bitter. Substituting iodine loses is sweetness leaving a slight bitter aftertaste. Chlorination of other halogen compounds like sucrose and its derivatives, tryptophan, saccharides, lactose, maltose has also been studied where sweetness/bitterness increases by many folds$\ce{^{[2]}}$.
Notes and references

Taste by Lloyd M. Beidler
Taste Chemistry by R.S. Shallenberger
Robert Shallenberger and Terry Acree proposed the AH-B theory of
sweetness. Simply put, they proposed that to be sweet, a compound
must contain a hydrogen bond donor (AH) and a Lewis base (B)
separated by about 0.3 nanometres. According to this theory, the
AH-B unit of a sweetener binds with a corresponding AH-B unit on the
biological sweetness receptor to produce the sensation of sweetness. (Wikipedia)


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How does the smell of a compound come about, and is it possible to define a smell?
Colour - and eyesight in general - arises because objects reflect/transmit certain wavelengths of colour, which is detected by our eyes.
On the other hand, what gives rise to smell? Is there a branch of chemistry associated with this? Is it possible to define the smell of any substance, using some parameters analogous to wavelength in the case of sight?
We know that $\ce{H2S}$ has a foul odor, but distilled $\ce{H2O}$ does not have any strong smell. Does $\ce{H2Se}$ have a similar foul smell?

Related: https://chemistry.stackexchange.com/questions/80977/what-is-smellhow-come-2-things-have-the-same-smell

I have rewritten your question based on a few of your comments, etc. Hopefully this will stop the unclear votes, but if you think I have changed your meaning, just rollback. I personally think this could be somewhat on the broad side (you're asking multiple questions at once, albeit related), but it seems that it can be somewhat answered, so I'll leave it to the community to decide.

This might be a good place to start: https://en.wikipedia.org/wiki/Vibration_theory_of_olfaction

Just a (slightly pedantic) note about colour/vision: as I alluded to in my comment to Ivan's answer it's nowhere near as simple as just wavelengths and intensities. See the Checker shadow illusion for example (there's something similar directly involving colours, but I can't locate an example at the moment). Wavelengths are a property of light; colour is a property of the brain.

As a sensation, olfaction does not seem to possess the same status as, say, vision. Most biologists, indeed most people not directly involved with fragrances or flavours seem to think that odour sensation is “subjective” and not necessarily shared by others.
What makes an odourant?
The general requirements for an odourant are that it should be volatile,
hydrophobic and have a molecular weight less than approximately 300 daltons.

The first two requirements make physical sense, for the molecule has
  to reach the nose and may need to cross membranes. The size
  requirement appears to be a biological constraint. A further
  indication that the size limit has something to do with the
  chemoreception mechanism comes from the fact that specific anosmias
  become more frequent as molecular size increases.

To be sure, vapor pressure (volatility) falls rapidly with molecular size, but
that cannot be the reason why larger molecules have no smell, since some of the
strongest odourants (e.g. some steroids) are large molecules. Additionally, the cut-off is
very sharp indeed e.g substitution of the slightly larger silicon atom for a
carbon in a benzenoid musk causes it to become odourless.

Comparison of molecular size between a benzenoid musk (left) derived from
acetophenone and its sila counterpart ( right) in which the central carbon atom in the t-butyl
groups has been replaced with Si. The carbon musk is a strong odourant, the sila musk odourless.
Attempts have been made to accommodate discrepant structure-odour relations by a process known as conformational analysis. This involves exploring the space of conformations adopted by the odourant molecule when deformed away from its energy minimum.
Odour descriptors and odour profiles
Odour descriptors are the words that come to mind when smelling a substance.
The more generally understood the words are, the more useful they are as descriptors. In practice, it is easy for any observer, after a little training, to use the standard descriptors of fragrance chemistry. Example of descriptors include musky, camphoraceous  etc
Smelling chemical groups
A fact that has, in our opinion, received too little attention from olfaction researchers is the ability of humans to detect the presence of functional groups with great reliability.
The case of thiols ($\ce{-SH}$) is familiar, but groups ($\ce{NO2}$), aldehydes ($\ce{C=O(H)}$), can be reliably identified once the odour character the functional group character confers is known. When nitriles are used as chemically stable replacement for aldehydes, they impart a metallic character to any smell: cumin nitrile smells like metallic cumin (cuminaldehyde), citronellyl nitrile smells like metallic lemongrass (citronellal), and nonadienylnitrile smells like metallic cucumber (nonadienal). Oximes give a green-camphoraceous character, isonitriles a flat metallic character of great power and unpleasantness, nitro groups a sweet-ethereal character,etc.
Here are some odour categories and their representative molecules, chosen to illustrate
structural diversity:
Musk

Musk odour descriptors might be “smooth clean, sweet and powdery”. The
  molecules that possess this odour character are exceptionally diverse
  in structure. Macrocyclic musks contain a 15-18 carbon cycle closed
  either by a carbonyl or by a lactone and smell similar but fresher and
  more natural, often with fruity overtones (cyclopentadecanolide,
  ambrettolide). Nitro musks, discovered originally as a byproduct of
  explosives chemistry, smell sweeter and are reminiscent of
  old-fashioned barbershop smells.

Representatives from five chemical classes which yield musk odors. 1 androst-16-en-3a-
ol, a steroid musk. 2: ambrettolide, a macrocyclic musk. 3: Musk Bauer, a nitro musk. 4: Tonalid, a
tetralin musk. 4: Traseolide, a indane musk.
Ambergris

Originally derived from concretions spat out by whales and aged in the
  sun, ambergris odorants smell nothing like natural ambergris tincture,
  which has a weak animalic marine smell. The smell of ambergris
  odorants was once aptly described to us by a chemist-perfumer as
  “glorified isopropanol”. Ambergris odourants provide an interesting
  combination of very closely related smells with widely different
  structures: amberketal, timberol, karanal and cedramber

Two ambergris odorants, timberol (left) and cedramber (right)
Bitter almonds
This easily-recognized category is interesting  because it includes a small molecule (HCN) which, however, is perceived by a large fraction of observers  to smell as metallic not almond-like to. Benzaldehyde, nitrobenzene,trans-2-hexenal (but see above) are good examples.

The complexity of structure-odour relations, and the fact that the three dimensional structure of the receptor site is unknown, make it very difficult to apply conventional quantitative structure activity relationships.
Plausible theories of odour

Many theories of Structure-odour relations (SORs) have been proposed
  in the past (reviewed in Moncrieff, 1951) but advances in biological
  understanding, not least the discovery of odourant receptors, have
  gradually ruled them out. There appears to be two possible types of
  SOR theory left standing:

Shape-based theories: Odotopes

Most enzyme-substrate and receptor-ligand binding relies on molecular
  recognition between protein and ligand. Recognition depends on
  interactions that can be either attractive or repulsive (Davies and
  Timms 1998). All attractive chemical interactions are ultimately
  electrostatic in nature whether they occur between fixed charges,
  dipoles, induced dipoles or atoms able to form weak electron bonds
  (e.g. hydrogen bonds).
Repulsive interactions can be electrostatic or
  quantum-mechanical (electron shell exchange repulsion). Almost every
  change in molecular structure (with some exceptions which will
  described below) alters the set of surface features capable of forming
  such attractive or repulsive interactions, and thus affects what we
  loosely call molecular shape.

Recently, both  in vivo and  vitro studies have shown that, generally receptors respond to more than one odourant, suggesting that they detect the presence not of the whole molecule but of a partial structural feature thereof, hence odotopes.

According to odotope theory the smell of a molecule is then due to the pattern,
  i.e. the relative excitation of a number N of receptors to which it binds.

Ethyl citronellyl oxalate, a molecule possessing a macrocyclic musk odour but linearin shape. Right: a macrocyclic musk, cyclopentadecanolide. Shape-based theories assume that the linear musk assumes a conformation close to that of the macrocyclic when binding to the receptor, hence the similarity in odour.
Vibration theories

The idea that the nose operates as a vibrational spectroscope was
  first proposed by Dyson (1938) and later taken up and refined by
  Wright (1982). What makes it attractive in principle is that
  vibrational spectra share three properties with human olfaction.

No two molecular spectra are exactly alike, particularly in the aptly named
“fingerprint region”.
Many functional groups are easily identified by their specific
vibrational frequencies.
System utilizing a physical property as basic as vibration will be ready for never-before-smelt molecules, i.e. does not depend on a repertory of existing or expected structures. In that sense, it does not rely on molecular recognition.

Remarkably, even bonds between atoms can be detected: the acetylenic C-C triple bond of –ynes imparts a isothiocyanate-like mustard-like smell to molecules which is clearly recognizable, for example in acetylene and in methyloctynoate.
Functional groups as odotopes
An odotope theory can explain these regularities only by assuming that the
functional group is an odotope. In the older structure-odour literature, this used to be
described as electronic factors (as opposed to steric). The idea was that, given that many
functional groups were similar in size, the recognition mechanism must somehow be
sensitive to the fine structure of the electron distribution (orbital energies, charge
density, etc) of the functional group.
However this proposition has some shortfalls;
Consider for instance the SH group in, say methanethiol. Alcohols never smell of
sulfur, whereas thiols always do. What could make the SH infallibly distinctive as an
odotope, as compared to the OH group? Partial charge, bond length, bond angle and
atom size are somewhat different between $\ce{R–SH}$ and $\ce{R–OH}$, but it is hard to see how
these can be detected with absolute reliability by, say, an aminoacid side chain in the
presence of thermal motion.

Replacing a C=C bond with a sulfur atom does not change odour character, suggesting that “electronic” properties of sulfur are not sufficient for molecular recognition.
Functional groups and vibrational theory
By contrast, the distinctive smell of functional groups is a natural feature of a
vibrational theory. Above 1800 wavenumbers, IR absorption lines are diagnostic of the
stretch frequencies of diatomic functional groups.

The clearest example so far is that of boranes. The terminal B-H bond
  in boranes has a stretch frequency whose range overlaps with that of
  thiols. Turin (1996) therefore predicted that boranes should smell
  sulfuraceous, despite the complete absence of similarity, both
  structurally and chemically, between boron and sulfur.

A comparison between borane and thiol smells is best made using decaborane. Decaborane smells strongly of boiled onion, a typical SH smell. Other, less stable boranes share this sulfuraceous smell character;

The dependence of the sulfuraceous character on molecular vibrations
  and atomic partial charges, as predicted by a vibrational theory.
  Decaborane (left) smells sulfuraceous, and its terminal B-H bonds have
  a stretch frequency ˜ 2500 wavenumbers. In triethylamine-borane
  (middle), the B-H stretch is shifed to 2300 wavenumbers and the
  sulfuraceous smell is no longer present. In p-carborane (right) the
  near-neutral partial charges make the SH bond odourless.

In summary it it could said there is still more work needed on study of structure-odour chemistry to have conclusive evidence on the best theory, currently vibrational theory is evidently successful at explaining the fact that we smell functional groups even when sterically hindered, and in accounting for differences in smell between isotopes, while the odotope theory explains little.
References
Structure-odour relations: a modern perspective: Luca Turin et al. [Available online: https://pubs.acs.org/doi/abs/10.1021/cr950068a]

Smell is what you feel when molecules of some compound (and not any kind of radiation, mind you) touch the olfactory nerves deep inside your nose. It is just another property of a compound, like molecular weight, or melting point, or color, only different.

Unlike molecular weight, it can't be calculated from molecular formula by simple rules.
Unlike molecular weight or melting point, it can't even be adequately described with a single number.
Unlike all of the above, it can't be reliably predicted for a new compound, though much can be done by the way of analogy.

$\ce{H2Se}$ smells worse than $\ce{H2S}$, and $\ce{H2Te}$ is worse still. A disturbing characteristic of the latter, given by an anonymous expert in the field, I'd better keep hidden so as not to offend the public taste:

 Smell it, and you'll vomit farther than you can see.

Is there anyone who is researching on this subject?

There are guys who make artificial food smell [almost] like real, but I don't know if they have some sort of deep theory behind it.

What is the secret behind it?

I just told you I don't know what's behind it. Maybe there is no theory at all. Maybe they just decompose the natural smell into components, identify them one by one, produce the pure compounds, mix them and see how it feels.

As I (believe I) understand it (neither a food scientist nor chemist), perhaps even more so than for colour, smell is "defined" more by the physiology of noses than any inherent property of a chemical. Just as the wavelength of light plays a part in our sensation of colour (but does not totally define it), there will be physical properties of certain chemicals that might affect how they smell (IIRC things like angles of bonds; how well a molecule "fits" receptors), but these properties are not the complete picture.

Aside:  There are reports of "smell" at the initiation of a round of radiation therapy.  Example.  I doubt this is olfaction (which you call "smell" in your first sentence), but do not find that the mechanism has been studied.  This phantosmia experience has been reported.

The molecular mechanism of smell is still unknown. What can be said is that molecules of all sorts are usually identified in proteins (probably of many different types) and that the effect this detection has is amplified, possibly by passage of ions through a membrane, and then interpreted in the brain which we experience as a sensation. Why we experience different smells from different molecules or non at all for some, is more likely to be answered by reference to Darwin's theory of evolution.


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What actually is electrode potential?
I was thinking about how Daniell cell works and got into a serious doubt.
Let's consider a half cell of Zn. In standard conditions, it'll have an electrode potential of -0.76V with respect to SHE. If I now increase the zinc ion concentration, I know from Nernst equation that it's electrode potential (still with respect to SHE) should become less negative. Intuitively also, it's tendency to get reduced will be more due to Le Chatelier's principle, so more reduction potential as predicted by Nernst equation.
So far so good.
But, if I see it like this, then the results seem to get flawed:
When I increase zinc ion concentration, in the new equilibrium state that will be attained, the absolute potential of the electrolyte will be greater (due to more positive (zinc) ions), and a little bit (safe to neglect it's consequences) of negative charge on Zn electrode will be decreased, with a little deposition of Zn.
Zn2+ + 2e-  <--->  Zn
So the net effect would be of electrolyte solution becoming more positive. Now, the definition of  "reduction" potential is the (absolute) potential difference between the reduced and the oxidised species (along with some additive constant that makes SHE's electrode potential 0V, but it's of no concern here as we're seeing where the electrode potential (of Zn) increases or decreases, NOT its absolute value); hence, E = V(Zn electrode) - V(Zn2+ in electrolyte). So this should get more negative as the potential of electrolyte increases. (Potential of the electrode also becomes more negative, but its effect is marginal.)
So what is the flaw in this perspective?


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Do chemical garden coral formations leach toxins?
Short version, I want to make a coral garden with sodium silicate and salts, and I would like to put that in my fish tank, but I'm concerned about the fish getting poisoned.
https://en.wikipedia.org/wiki/Chemical_garden
https://www.thoughtco.com/make-your-own-magic-rocks-607653
The above pages describe the chemicals needed. The articles say that once the corals have formed, they can be rinsed to remove the sodium silicate, and kept in fresh water. Most of the articles I've read say the same thing, in different ways. But none of them say whether the coral degrades, or if it is stable.
Are there experiments that I can do to test this? How can I be sure?

I'd rather not do this, even after rinsing.

"Chemical garden" is experiment for chemistry shows not for decoration of fish tanks.

@paracetamol Thanks for the spelling fix! I've learned something :)

Experiments to test if toxins leach from silicates do exist; however, these experiments involve dangerous chemicals.  For that reason these tests should be left to qualified lab personnel and not described here.  Performing these experiments are expensive and would need to be performed on every combination of methodologies to make a "chemical garden".  Because so many combinations exists, quality control on such an experiment would be difficult.
Experiment aside, a "chemical garden" in a fish tank is a bad idea.  While silica is chemically insoluble,  it can be converted biologically into orthosilicic acid.  This form of silica is soluble. This process in nature is called biological weathering.
Chromium, lead, and perhaps some of the other compounds are carcinogenic to humans.  These compounds, even in an immobilized form, are not worth the risk and should be nowhere near a fish tank.


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How can I calculate the concentration of ethanol vapor in an airstream?
How can I calculate the concentration of ethanol vapor in an airstream?
The steady state generation flow rate (of the vapor) is 625 ft^3/min.  The ambient temp is 50-60 F (we can use avg temp 55 F).
These vapors would come from liquid ethanol.
What more information do I need to calculate this concentration in kg/m^3 or ppm?
I apologize as I do not have a thorough chemistry background.

Would you not need to know the rate of evaporation from the liquid ethanol? If the evaporation  rate is very slow  this will limit the amount of ethanol in the airstream, if high it may be the equilibrium vapour pressure.

Are you saying that ethanol vapor is being generated at a rate of 625 $\ce{ft^3}$/min or that liquid ethanol is being generated at a rate of 625 $\ce{ft^3}$/min.

@J.Ari Ethanol vapor.

@porphyrin  ok, how shall I find the evaporation rate of the ethanol at this temp?

So if you know how much ethanol vapor you are generating you need to know what else is in the stream and measure the amount of the stream. Calculating a concentration at that point becomes trivial.

You can't. Evaporation rate is not a thermodynamic property. It depends on the specifics of the process it is occurring in. The vapor pressure at a given temperature is determined AT EQUILIBRIUM - meaning the process/history is unimportant. The rate that a specific process occurs is obviously process dependent. It is true that various organics have had cited evaporation rates. These rates are generally a weight ratio relative to some standard liquid (ethyl ether, acetone, benzene, etc.). The reason they are relative is because no two investigators are likely to have sufficiently similar experimental processes to state weight loss on an absolute basis and get the same results. I'm just saying the same thing in another way; that your absolute weight loss per unit time is dependent on the specifics of your process. You've two realistic options: 1. find some Chemical Engineering (etc.) literature which has measured that for a "sufficiently similar" process and interpolate. 2. Measure it yourself. There are organic vapor monitoring devices, weight loss or tracer methods available. (There are other ways as well, depending on the accuracy required and the processes involved.) Note: a simple example of the difficulty in determination of evaporation rate involves a wet surface. The evaporation rate from that surface will clearly depend on the air flow across the surface, even at a constant temperature. It will also depend on the humidity of the air. Demonstrating that the rate depends on air flow proves that the relationship is going to be complex since only the air in contact with the surface will influence it, while air flow rates are for the bulk flow through a volume (not across a surface).


text:
Why does adding a charged group to an organic molecule decrease its stability?
As I understand it, during glycolysis, glucose is converted into fructose 1,6-bisphosphate which greatly destabilizes the molecule. This is so it can be divided into pyruvates more easily.
Now I don't understand why adding charged groups such as phosphates destabilize molecules, what actually happens to the molecule when a charged group associates with it? Why does it become destabilized?

The quick and general answer is electrostatic repulsion. Fructose-6-phosphate is already a dianion. Phosphorylating it a second time will mean adding another two negative charges to the dianion. Like charges repel each other, so the two dianionic phosphate groups are pushed apart electrostatically.
However, you cannot generalise this. Sometimes, the additional negative charge may serve to better balance charges overall. It could also be so far away from an existing negative charge that electrostatic interactions are minimal. Finally, there are many ions and positively charged organic molecules dissolved in the cell which altogether stabilise negative charges relatively with respect to a simple solution of fructose-1,6-bisphosphate. (The same can be said for the stabilisation of positive charges.)
Finally, please note that the additional negative charge does not really assist the fragmentation of fructose-1,6-bisphosphate. The reaction — a retro-aldol — is enzymatic and catalysed principally by a lysine and aspartate residue. The former is used to convert the keto functionality into an imine which has a greater tendency to form an enamine while the latter serves as a proton shuttle. The phosphate groups are merely there so the enzyme recognises the correct molecule; modifying the enzyme’s structure to allow for unphosphorylated fructose to be consumed would not alter the reaction rate significantly.

What actually is destabilization ? What happens to the molecule after destabilization?

@HariharanV I didn’t use the term once, so I don’t entirely understand why you’re asking me, but: essentially, it means a molecule being somehow given more internal energy. What happens depends on the molecule and is way too broad to answer here.


text:
Which term for heat energy in calorimetry is correct in scientific usage?
In calorimetry several terms are interchangeable:

Lower Calorific Value
Lower Heating Value
Net Calorific Value
Net Heating Value

The following terms also appear to have identical meanings:

Higher Calorific Value
Higher Heating Value
Gross Calorific Value
Gross Heating Value
Gross Energy

So far I can think of several factors affecting which term should be used.

In the days of the early discoveries of calorimetry, Latin was just
as much of a research language as English, so interchangeability of
the terms calorimetric (which derives from the Latin word calor
meaning heat) and heating would be understood.
The word value sounds like a non-specific reference to a number.
The unit Calorie is not part of the SI standard though it is derived
from SI units.  In the context of describing the energy required to
raise the temperature of water it may be appropriate.

It seems like gross or net heat energy would be the best term for scientific usage.  Does any standard or consensus exists which defines which of these terms are best to use?

I haven't ever seen calorific used. Are you sure you don't mean caloric? Caloric value is pretty common though mainly used for food calories.

It might also help if you give an example sentence where you want to include one of these times. "We find that the ___ of cubane is ..."

Pulling random sentences from an internet search. ...  The higher heating value of anthracite coal was tested and found to be 32.50 MJ/kg.  ...  The biogas from the digester has a calorific value of 4,500 to 5,000 K. cal/m³.  ...  LOL .. add caloric to the pile of words ...  "The current system 'provides only an estimate of the energy content of foods,' but 'determining the real caloric value of a food item requires actual feeding experiments,' one speaker said.

An explanation of differences connotation may work for an answer. Higher heating value is associated with fuel energy content.


text:
Solubitity of FeS
I have some questions related to a problem of multiple equilibria and solubility. I have to calculate the solubility of FeS in pure water considering that the second stage of hydrolisis of $\ce{S^2-}$, producing $\ce{H2S}$, can not be neglected.
For this problem the book that I was consulting gives me the values of the dissociation constants for $\ce{H2S}$ ($K_\mathrm{a1} = \pu{1.0e-7}$ and $K_\mathrm{a2} = \pu{1.2e-13}$) and the solubility product of $\ce{FeS}$ ($K_\mathrm{sp} = \pu{8e-9}$). I proposed the following reactions:
\begin{align}
\ce{FeS &<=> Fe^2+ + S^2-}\tag{R1}\\
\ce{H2S + H2O &<=> HS- + H3O+}\tag{R2}\\
\ce{HS- + H2O &<=> S^2- + H3O+}\tag{R3}\\
\ce{2H2O &<=> OH- + H3O+}\tag{R4}
\end{align}
From these ones I got the mass and charge balance equations:
\begin{align}
[\ce{Fe^2+}] &= [\ce{HS-}] + [\ce{S^2-}] + [\ce{H2S}]\tag{1}\\
2[\ce{Fe^2+}] + [\ce{H3O+}] &= 2[\ce{S^2-}] + [\ce{HS-}] + [\ce{OH-}]\tag{2}
\end{align}
I also get other equations from the equilibrium of the acid in the two stages of its dissociation:
\begin{align}
K_\mathrm{a1} &= \frac{[\ce{HS-}][\ce{H3O+}]}{[\ce{H2S}]}\tag{3}\\
K_\mathrm{a1} &= \frac{[\ce{S^2-}][\ce{H3O+}]}{[\ce{HS-}]}\tag{4}
\end{align}
The solubility product of salt and the ionic product of water:
\begin{align}
K_\mathrm{sp} &= [\ce{Fe^2+}][\ce{S^2-}]\tag{5}\\
K_\mathrm{w} &= [\ce{H3O+}][\ce{OH-}]\tag{6}
\end{align}
I have six equation in total. So my doubt is: how to simplify these equations? Well, what relation could I get from the mass or charge balance equation in order to simplify the process of resolution?
In my book the answer is $\pu{4e-7}$, but I don't get the necessary relation to resolve all of this problem.
I would really appreciate some advice about it.

This exercise isn't particularly good. Ka2 is way off, and isn't appropriate to describe protonation of sulfur bound to iron. Also check out http://www.sciencedirect.com/science/article/pii/S0016703706001992


text:
Safest Methylating Agent
I am planning on doing some synthesis in the near future and due to safety purposes I am looking for a methylating agent that is both effective and safe. The substrate is a deprotonated amine and the solvent is water. I understand that DNA methylation is involved in the toxicity of the compounds, but there should still be ways to minimize risk. Ideally, the compound should be a powder that is soluble in water. Can any ideas that work for this substrate and solvent be thrown on the table?
I was planning on using methyl iodide solely as a last resort.

Would this work?

AirHuff- I'm not sure I'd go with formaldehyde as being non-toxic. To the OP, most methylation agents aren't compatible with water as a solvent, they're too reactive

Also, a deprotonated amine in water isn't really a deprotonated amine.

It is an aromatic amine and the solution will be basicified by k2co3, also means that the reaction proposed doesn't work (no imine formation), sorry should have specified more.

Is there any reason for the water.

Cheap solvent, safe compared to other solvents, good for separating unwanted compounds.

Would a solvent such as 2-propanol work in that case for methylation?

What you are looking for does not — to the best of my knowledge — exist and likely won’t ever exist.
Methylating agents are usually classified as toxic as they are usually able to penetrate cells and nuclei and methylate DNA bases — although they will also methylate most nucleophiles on the way there. This can lead to all sorts of things will the shadow of cancer overarching it all. Off the top of my head there are a few methylating agents that might not penetrate the skin such as methyl Meerwein salt. These, however, are so reactive that they just react before they reach the skin.
Secondly, methylating agents will likely never work in water. Water itself is a nucleophile and a (Brønsted) acid. If you use an electrophilic methylating agent such as methyl iodide or methyl Meerwein salt, water is able to nucleophilically attack these to generate methanol and then dimethyl ether, thereby deactivating your methylating agent. If you choose to use a nucleophilic methylating agent such as methyllithium instead, water will supply a proton to liberate methane (and lithium hydroxide in the case of methyllithium) instead. So practically no methylating agent will survive in aquaeous solution.
Finally, your proposed reaction pathway is skewed from the beginning: you suggest using a deprotonated amine in water as a reactant. However, the $\mathrm{p}K_\mathrm{a}$ of amines is typically around $30$, depending on the substituents. ($\mathrm{p}K_\mathrm{a}(\ce{LDA}) \approx 35; \mathrm{p}K_\mathrm{a}(\ce{LiHMDS}) \approx 25$; source: Evans $\mathrm{p}K_\mathrm{a}$ table.) On the other hand, the $\mathrm{p}K_\mathrm{a}$ of water is $14$ or $15.7$, depending on the source you consult. This is a difference of $10$ to $20$ logarithmic units depending on the type of amine — meaning that amines are practically never deprotonated in aquaeous solution.[1]
If you mistyped and are using a deprotonated amide, things won’t get that much better ($\mathrm{p}K_\mathrm{a}(\ce{RCONH2}) \approx 25$).
Therefore, you will likely have to give up your plan and go for a more traditional methylation method.

[1]: Note you might argue that the free base, $\ce{R3N}$ is enough, so all you want to do is lower the $\mathrm{pH}$ of water to below $\approx 10$ so that the protonated form $\ce{R3NH+}$ is neglegible. However, at these $\mathrm{pH}$ values, water is also deprotonated to a non-neglegible amount and a hydroxide ion is a better nucleophile than a free amine base.


text:
Ozone-Air Paradox
One day, I decided to calculate the density of air, so I searched for the individual densities of the 3 main elements that forms breathable air. I found the following data, at $\pu{0^\circ C}$ and $\pu{100 kPa}$:
\begin{array}{lrS}
\text{Component} &\% &\rho,\,\pu{g cm-3}\\ \hline
\text{Nitrogen}  &78 &0.001251          \\
\text{Oxygen}    &21 &0.00142897        \\
\text{Argon}     &1  &0.001784
\end{array}
I'm calculating the density of $\pu{1 cm3}$ of air. Individual masses in $\pu{1 cm3}$ of air:
\begin{align}
&\text{Oxygen}   &\pu{0.0003000837 g}\\
&\text{Nitrogen} &\pu{0.00097578 g}  \\
&\text{Argon}    &\pu{0.000001784 g} \\
\end{align}
The final density is equal to $\pu{0.0012937037 g cm-3}$
Then I realized that the density of ozone should be lighter due to the ozone layer to be able to exist, otherwise, the ozone gas, as it is fairly more dense than air, would pass directly through the atmosphere. The problem is that the density of ozone gas at the same conditions (STP), have a density of $\pu{0.00214 g cm-3}$, which is $65.41654785\%$ heavier than air.
With the context explained, my question is: if is really a ozone layer up there in the atmosphere, how it is there, being heavier than the base gas mix that is below it?

The best way of saying thanks is to upvote the answer (if you think it is a good answer) and accept it (if it helped you!). Relevant help centre section

The atmosphere is not well-separated into layers of density; the most prominent reason for that is wind that mixes the different layers.
However, even aside from that there is a reason for ozone to be found in the ozone layer primarily and not much elsewhere: simply because it is produced there by UV radiation and because it breaks down fast enough to not come all the way down.


text:
Precision of non-adiabatic couplings
Let me ask a question about the differentiation of Non-Adiabatic Couplings (NACs), specifically the terms $D_{IJ}$ and $d_{ij}$.
According to the Pyxaid paper, the set of $D_{IJ}$ computed via Slater Determinant (SD) basis are equal to the set of $d_{ij}$ from a one-electron orbital basis; so $D_{IJ}$ is computed numerically by differentiating $d_{ij}$. According to Libra-X, however, $D_{IJ}$ is directly differentiated. $D_{IJ}$ in the latter case seems less precise than that in the former because $D_{IJ}$ = $d_{ij}$ * (non-unity couplings of one electron orbital basis).
How important is the difference?

Can you add links or references to the Pyxaid and Libra-X papers? It's difficult to answer the question without these.

Coincidentally I just came across this link, it's a starting point: https://groups.google.com/forum/#!forum/quantum-dynamics-hub

I'll ask this question in the google forum you showed me. Thank you!


text:
What is the Si–N–C bond angle in silyl isothiocyanate?
In silyl isothiocyanate, supposedly, the value of the bond angle $x$ is $180^\circ$.

I don't know how to figure this out. There is a lone pair on nitrogen and two sigma bonds. Shouldn't it have sp2- or sp3-hybrid orbitals and thus $x$ be smaller than $180^\circ$?

The answer that you are supposed to give is presumably something along the lines of "dπ–pπ" bonding between silicon and nitrogen. This essentially means that the lone pair of nitrogen is involved with backbonding into a silicon 3d orbital, leading to a linear geometry at nitrogen (i.e. the bond angle is 180°). This can be represented with the following resonance structures:

To be very clear, this is very likely incorrect. The extent of d-orbital involvement in main group compounds is usually tiny. If there is any backbonding, the acceptor orbital is probably a Si–H σ* orbital rather than a d-orbital (cf. backbonding from transition metal to phosphine).
Electron diffraction studies of H3SiNCS show that the Si–N–C bond angle is actually 163.8°.1 A computational study (TZVP + MP2 level of theory) found 165.91°.2. In fact, the linear structure above is not an energy minimum, but rather a transition state for the interconversion of the two energy minima:3

I don't actually know how one can predict this a priori without access to a computer, but clearly this system is not amenable to bond angle prediction with VSEPR theory, and invoking the so-called "Si–N dπ–pπ bonding" doesn't give the correct answer either.

References

Glidewell, C.; Robiette, A. G.; Sheldrick, G. M. Electron diffraction study of large amplitude vibrations in H3SiNCO and H3SiNCS. Chem. Phys. Lett. 1972, 16 (3), 526–529. DOI: 10.1016/0009-2614(72)80416-0.
Palmer, M. H.; Nelson, A. D. The structures of the azido-, isocyanato- and isothiocyanato-derivatives of methane and silane and their derivatives. A comparison of ab initio with experimental results. J. Mol. Struct. 2004, 689 (1–2), 161–173. DOI: 10.1016/j.molstruc.2003.11.008.
Palmer, M. H.; Guest, M. F. The methyl, silyl and germyl esters of the pseudo-halogen acids. A comparison of structural data from experimental and ab initio studies. Chem. Phys. Lett. 1992, 196 (1–2), 183–190. DOI: 10.1016/0009-2614(92)85951-6.

Despite the bond angles being measurably less than $180°$, the assumption of $sp$ hybridization remains quite accurate. See this answer with respect to the central atom in carbon suboxide.


text:
Ammonia for cleaning - what is the interaction between ammonia and oils?
Household ammonia is recognized as being an excellent cleaner for grease and oil. My question is, how does this work? Does ammonia merely put the oils into a suspension, like a detergent does? Or is there a more significant chemical reaction between the two - along the lines of how enzymes affect oils?

There is no reaction, just the detergent action.

I can't verify your answer, but if you can offer backup, you should make this an answer. Thank you for replying!

I found this, a bit of a long winded video but has some relevant  information.  -  https://www.nsf.gov/news/special_reports/chemistrynow/chem_cleaners.jsp

The comment that there is no reaction, other than a detergent action, is, I am pretty confident, incorrect. There is apparently a chemical reaction between the ammonia and oils, with at least some of the oil being converted to amides. Unfortunately, I don't have enough chemistry knowledge to know what this means in plain English. Starting a bounty. Relevant sources: https://naldc.nal.usda.gov/download/31213/PDF, https://aocs.onlinelibrary.wiley.com/doi/10.1007/BF02640174, and a commercial application https://www.ossmaine.com/wp-content/uploads/2017/08/Oil-in-Ammonia-Introduction.pdf

@CorvusB, Are you specifically talking about Windex? The so called ammonia solutions are not pure ammonia solutions. They have alcohols and surfactants.

https://chemistry.stackexchange.com/questions/81977/ammonia-for-cleaning-what-is-the-interaction-between-ammonia-and-oils

@M.Farooq Nope, I'm not referring to commercial cleaning mixtures, such as Windex. I am referring to household ammonia, aka ammonium hydroxide, and/or aqeous ammonia.

I agree with Ivan Neretin. But, Why are bases so good dissolving oil? Even much better than a apolar solvent.

@JoaquinTorrens, Ivan Neretin's comment is, I am quite certain, wrong. A detergent puts oils in emulsion. Eventually the oils will return to an unemulsified state. This is why there are "fatbergs" in sewer and septic systems. Based on the USDA document, the reaction with ammonia converts some of the oils into something else. James Gaidis' answer to the question may have gotten one part right: there might be some saponification. I'm lousy at chemistry, so I don't know. But I'm good at practical applications, and I know making soap alters the input ingredients.

https://chemistry.stackexchange.com/questions/81977/ammonia-for-cleaning-what-is-the-interaction-between-ammonia-and-oils

LOL, I find it curious that readers are linking directly back to my question in the comment section! Not sure what they are trying to say. Circular non-logic!

@KalleMP While the link provided is dead, a simple search turned up the document. And, yes, it does provide pertinent info. The doc's reference links are also bad, but also findable, and also relevant. Combined with two bits of info from my old chem teacher, and I think I have a valid answer. Thank you.

Ammonia is a surfactant, like detergents, but it also reacts chemically with some quantities of oils to convert them into other things. In addition, as a cleaning agent, unlike detergents and soap, ammonia has anti-microbial action.
Ammonia is a historically notable cleaning agent. It was readily available, prior to the advent of the industrial era, as it is a natural by-product of the breakdown of animal (and human) waste (1). Aging urine produces a liquid with a high ammonia content. Ammonia has significant antimicrobial properties (2) as well as anti-grease/oil properties. As an anti-oil agent, it has both surfactant and transmutational  properties (3)(4)(5). In plainer English, that means that ammonia emulsifies grease and oils, like soap and detergents, so that the mess can be washed away with water, and it changes the nature of some of the oil.
As a surfactant, ammonia acts like soap and detergents, meaning they pick up and disperse the oil into tiny droplets in the water (emulsification), which then can be rinsed away. Soap and detergents are a relatively unstable emulsifier. Which means that, as the original soap or detergent breaks down, the oils come out of emulsification, and regroup. And, voila, you get your fat back - just in a different location - and probably with lots of other impurities. Ammonia, as a surfactant, does the same thing: i.e. it acts to emulsify the oil. And, when it breaks down, the emulsion ends.
However, ammonia also chemically interacts with the oil, in significant amounts, in a process called ammonolysis, in which it also has catalytic effects. In this process it converts some of the grease and oils into other substances (e.g. amines), which no longer have the same properties as the original grease or oils. (4)(5) Ammonia does this latter through ammonolysis, in which it also may act as a catalyst. This is the part of ammonia's action that is similar to saponification, in which the original substances are altered, and no longer exist in their original form. How much of the oil is converted into amines depends on the temperature of the interaction, the type and characteristics of the oils, etc. However, this change is stable, and permanent, in that the end products do not revert to being oils.
(1) history:
Smithsonian Magazine: The science behind historic uses of urine
(2)Ammonia is not currently recognized as a primary antimicrobial product, as, for example, is bleach (chlorine). Chlorine, my example, is far more effective as an antimicrobial. However, in ammonia's favor, is the fact that ammonia is far more effective in this regard than soap or detergents alone. Please note that, on the linked chart, bleach (chlorine) is referred to by its chemical name: sodium hypochlorite.
 Antimicrobial Spectrum of Disinfectants
(3) Surfactants:
Dawn Chemical: Chemistry of Cleaning
Essential Chemical Industry: Surfactants
(4) Ammonolysis (subset of Solvolysis)
(5) Catalytic Effects in the Ammonolysis of Vegetable Oils

Common food fats and oils usually contain some free fatty acid1. The alkalinity of household ammonia serves to form a soap which will emulsify the main body of the fat/oil, if enough elbow grease (i.e. vigorous agitation with a scrub brush or wash rag) is applied. Other alkalis do as well - better, if you consider breathable ambient air to be desirable.
Some brands of commercial household ammonia already contain fatty acid soaps and exhibit a cloudy appearance.
It also seems to me that a dilute solution of washing soda ($\ce{Na2CO3}$) would clean quite well, and ammonia should do no better. But when the odor of $\ce{NH3}$ disappears, and the oily/fatty surface is now clean, it is as if there are now two indicators of cleanliness: no odor and no oiliness, therefore the surface must certainly be clean, because all evidence of the cleaning process has been removed. It must be a psychological thing, like foaming with detergents.
Reference:

Analytical approaches for the assessment of free fatty acids in oils and fats
S. A. Mahesar, S. T. H. Sherazi, Abdul Rauf Khaskheli, Aftab A. Kandhro, Siraj uddina, Anal. Methods, 2014 ,6, 4956-4963, DOI: 10.1039/C4AY00344F

Nice try, but I'm pretty sure this is not a good answer. There are, I believe, a number of issues in the answer, beginning with the reference to FFA. The relevance of that link is not established here. 2nd, washing sodas are also traditionally noted cleaning agents, so the alkalinity aspect may be pertinent. However, saponification (soap-making), is not, I think, an emulsification, as the initial ingredients are altered in the process. Also, it is not necessary to apply effort with ammonia to remove grease. However, emulsification IS the question here, as that is all that detergents do to oil.

Continuing the above comment. The ammonia in household ammonia is volatile - it readily evaporates - leaving the water behind. This is both the source of the odor, and the reason the odor dissipates. However, the cleaning effect is most certainly not psychological, and in most circumstances is more powerful for dissolving and softening grease and oil than a cleaning soda. Sodas do work because of alkalinity, but primarily as an adjunct to detergents or soaps. One can demonstrate that there is a difference in action by soaking a greasy pot in one or the other overnight. Cont. . .

The pot soaked in household ammonia may well be cleaner outright, and any remaining oils will be softened so that the mentioned elbow grease is far more effective than previously. Thanks for trying!

I wonder how rapidly the ammonia leaves the aqueous solution compared to how fast the water itself evaporates, since NH3 is so soluble in H2O. The odor threshold of NH3 is <50 ppm, but 4% is 40,000 ppm, so there's lots to go around!

I rather imagine there is some dependence on surface area and air motion, but an open pot containing ammonia will be only H20 in less than 24 hours or very nearly so. That's an original quantity of NH3 equal to or less than one liter. There will be little or no odor at that point, but I haven't the tools to analyze the remaining water. However, I'm pretty sure the water will still be alkaline at that point, so there must be something remaining besides H2O.

Interesting observation. Odor and pH vs time would be an interesting graph. It could be that household ammonia is made with a concentration only a little above mildly detectable, so it can lose some NH3, yet still be an effective cleaner.  The extra NH3 in there is to let the consumer know it's aggressive.     HCl at >20% smells strongly acidic; <20%, not so strong. The degree of odor is not linear wrt concentration (at least for HCl in water).

Once household ammonia has sat out in the open air and loses the odor, it is no longer effective for cleaning, although I have not measured the difference. I am reasonably sure, from practical experience, though, that this is the case.

I have determined that this answer, while it does contain some key points, is not a good answer. There is an interaction of ammonia and the oils, and it is, in this regard, similar to saponification. Detergents and soaps put oils and dirt in suspension, rather than significantly changing their chemical nature. I will attempt to answer the question myself after the bounty dies.


text:
Is this rearrangement in carbocation possible?
Generally, in the following type of reactions, positive charge on $\alpha$-Carbon causes ring expansion, but in the following example positive charge is formed on $\beta$-Carbon and the cation is cyclo-propyl methyl.

So is the above rearrangement even practically possible?

For what it's worth, there are no reports of this reaction in the scientific literature.

yes,  the rearrangement in carbocation is possible depending upon the condition of reaction and the stability of the new carbocation formed
 eg- the general order of stabilty for carbocation is 3degree >2degree > 1degree
NOTE- the stability of a carbocation depends on aromaticity resonance hyperconjucation inductive effect and sometime ring expansion also occurs


text:
Mechanism for bromoetherification of olefinic alcohol

Propose a mechanism for the following reaction showing the relevant stereochemistry of intermediates and the final product.

I am confused as to what is going on here. I guess there's some sort of neighboring group participation after the formation of the bromonium intermediate. From the molecular formula of the product, I deduce the mechanism seems to proceed via an addition and subsequent elimination. But then again I'm not very sure.

As you already know, treatment of an alkene with aqueous bromine leads to halohydrin formation. This results when the intermediate bromonium ion is trapped by the solvent, water:

Generally, water is a lousy nucleophile, but this process can happen because there is simply so much water around. In our case, the alcohol in the same molecule acts as the nucleophile, instead of water which may or may not be present. Alcohols aren't great nucleophiles, but what this has going for it is that the process (called bromoetherification, because you form a bromo+ether) is intramolecular.

As you can see in the scheme above, there are two possibilities in the formation of the bromonium ion: either on the top face of the alkene, or on the bottom face. There is probably a slight bias because of the presence of the substituent on the ring, but it turns out that this probably does not matter.
When the alcohol attacks the bromonium, it is an $\mathrm{S_N2}$ process, which requires the oxygen to attack carbon from behind the C–Br bond. Now, if the bromonium ion forms on the top face, this can't happen, because it's tough for the oxygen to loop all the way to below the plane of the alkene. Only if the bromonium ion is formed on the bottom face can the oxygen attack. Thankfully, bromonium formation is pretty much reversible, so even if the bromonium ion is initially formed on the top face, it can revert to starting material.

Extra
This bromoetherification reaction doesn't seem to have been carried out on this particular substrate - perhaps it's too simple - but it is a thing. Here's a more impressive example (Org. Lett. 2004, 6, 2961):

It also has a somewhat more popular cousin, called bromolactonisation, where the nucleophile is a carboxylate instead of an alcohol. Here's an example, taken from E. J. Corey's prostaglandin synthesis. The halogen used is different ($\ce{KI3}$ acts as a source of $\ce{I2}$, so this is an iodolactonisation) and the nucleophile is different, but the mechanism is generally the same - except that $\ce{NaHCO3}$ deprotonates the carboxylic acid before it attacks the bromonium. Can you work it out?

See also: Facial selectivity of iodonium formation in an iodolactonisation for a related question on iodolactonisation.


text:
Does Wurtz-Fittig-reaction involves Sn1 or Sn2?
Following mechanism is commonly proposed for the Wurtz-Fittig-Reaction:

Picture from Wikipedia
The red marked reaction steps is a nucleophilic substitution (SN). I suppose that this is a SN1 reaction, with an intermediate carbocation, rather than a SN2 reaction. In my opinion the resulting carbocation is good enough stabilized throughout the aromatic ring and therefore SN1 should be preferred, moreover because of the steric hindrance of the aromatic ring.
Is this assumption right? Thanks a lot for your help.

I’m kind of torn between upvoting this question because you actually thought about it and downvoting because your thoughts are entirely in the wrong direction. Notice that the phenyl cation is not stabilised by resonance and not delocalised over the ring. Quite contrary to your expectations, it is very reactive.

@Jan Ok thanks, the orientation and type of the orbitals I totaly forgot. This article explains this problem further. Therefore it cannot be a Sn1.
But a Sn2 neither (massive steric problems).
So I  assume that there is a total different mechanism. Maybe nevertheless a formation of a Phenyl cation, but through a different mechanism (abstraction the Br as a Bromid through the carban-Anion)?

My best guess is that it’ll be akin to an $\mathrm{S_NAr}$ reaction.

Note that the picture you linked to isn't present on the corresponding English Wikipedia article as of today.

Aryl carbons cannot go through SN1 nor SN2 substitution reactions. SNAr involves the creation of benzyne intermediate. The mechanism is this:

The benzyne has sp hybridized carbons, which should be linear, but forced to be bent. Therefore it can be attacked by a nucleophile, and become sp2 which isn't linear.
Notice that the carbanion can also attack the adjacent carbon. It's a special phenomena for this type of reaction. The mechanism was confirmed by carbon-14 labeling (Organic Chemistry/Paula Bruice, sixth edition, page 692).

"The mechanism was probably confirmed by carbon-13 labeling." You should have a source for this, imo - otherwise it just sounds like speculation.

I added a source.

Correct me if this is wrong, but isn't this the mechanism for elimination-addition of benzene, not $\ce{S_NAr}$?

The title of page 690 is: "Nucleophilic Aromatic Substitution: An Addition-Elimination Mechanism", and in page 691 it clearly says that this type of reaction is called SNAr ("the reaction is called an SNAr reaction" [substitution nucleophlic aromatic]"). In page 692 the title is "Nucleophlic Aromatic Subsitution: An Elimination Addition Mechanism That Forms A Benzyne Intermediate". It's simply a unique case of SNAr.


text:
Conservation rule in closed systems
In a closed system the reacting species interconvert but do not leave the system. Therefore the amounts of different species in the system would be bound by a conservation rule.

A simple example:
  $$\ce{A -> B}$$

Here the conservation rule would be: $\ce{A + B = constant}$, which says that at any point of time during the reaction the sum of the amount of both the species would be constant.

A more complex example
  $$\ce{2A -> B}$$

Here, the conservation rule would be: $\ce{A + 2B = constant}$.
Let's take a more complex system:

$$\begin{align}
\ce{A &-> B} \\
\ce{2B &-> C} \\
\ce{C &-> D + E} \\
\ce{2A + C &-> F}
\end{align}$$

From empirical analysis I can deduce that the conservation rule would be: $$\ce{A + B + 2C + D + E + 4F = constant}$$
However, I am not able to deduce a mathematical equation that gives these co-efficients. I assume that the stoichiometry matrix may give a clue but I am not sure. The stoichiometry matrix for the above system is:
$$\mathbf{S}=\begin{bmatrix}
-1 & 0 & 0 & -2 \\
1 & -2 & 0 & 0 \\
0 & 1 & -1 & -1 \\
0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}$$
This is relatively easier to figure out empirically as there is only one starting reactant (A). For reaction systems with multiple starting reactants (as shown below), the deduction of conservation rule(s) becomes even more difficult.
$$\ce{aA + bB -> cC + dD}$$
The problem seems quite elementary but I am not able to solve it. Am I missing something obvious?

Let us continue this discussion in chat.

With 6 components and 4 reactions, you must have two conservation rules (that is, if the reactions are independent, which they are).
Mathematically, we find ourselves in a 6D space (as there are 6 components), we move in a 4D subspace defined by the 4 reactions, and we need a basis for the 2D subspace normal to it, which is precisely where we can't go. The basis vectors will be our conservation laws.
You may use any CAS or your own hands to solve the problem. In Wolfram Mathematica, this is done as follows:
NullSpace[{
{1, -1, 0, 0, 0, 0},
{0, 2, -1, 0, 0, 0},
{0, 0, 1, -1, -1, 0},
{2, 0, 1, 0, 0, -1}
}]

As expected, the answer includes two vectors. One is $\{1, 1, 2, 2, 0, 4\}$. Another is $\{0, 0, 0, -1, 1, 0\}$, which translates to $\rm [D] - [E] = const$ ($0$ if there were no initial amounts), which is obviously true, since both are produced in the same reaction and never spent. That's your second conservation law.
BTW, the sum of the two vectors corresponds to the conservation law you already know. A basis can be chosen in a multitude of ways, after all.
That's what you do in any problem of this sort. That's linear algebra; it is about as simple and robust as a cleaving axe. It will never run out of gas, or freeze because its usage license has expired, or refuse to chop the incompatible firewood. It just works.
So you prepare the stoichiometry matrix (which you already did; I just transposed it), find the normal subspace, translate the basis back to your compounds. That would be it. The notion of time is irrelevant. The notion of differential equations is irrelevant. The notion of limiting reactants is irrelevant. It is just vectors and spaces. They do not even know you apply them to chemistry, nor do they need to.
So it goes.


text:
Why is edta in its calcium complex still a tetraanion at pH 10?
$\ce{edta}$ has different forms according to the pH. For example if the pH is $7{-}10$, $\ce{HY^3-}$, whereas if the pH is $11{-}14$ its form is $\ce{Y^4-}$.
But I've come across an exercise where it says that we titrate $\ce{CaCl2}$ with $\ce{edta}$ and the pH is $13$, and then we titrate it again with $\ce{edta}$ but the pH now is $10$.
The reactions that are given are the following:
$$\ce{CaCl2 + Na2H2Y <=>  Na2[CaY] + 2HCl}$$
Why is the $\ce{edta}$ given as $\ce{Na2H2Y}$ both times? Shouldn't its form be different at different pHs?

The $\mathrm{pH}$ discussions only make sense in the way you are using them for free $\ce{edta}$. Free $\ce{edta^4-}$ can indeed accept up to six protons depending on the acidity of the solvent; the most common forms are probably the neutral compound $\ce{H4edta}$ and the disodium salt $\ce{Na2H2edta}$. For these species, a $\mathrm{pH}$ equilibrium exists as you know it for polyprotic acids and only for these species is the discussion you’re beginning your post with correct.
Once $\ce{edta}$ forms a coordination compound, e.g. with calcium, things change. Six atoms of $\ce{edta}$ — four oxygens and to nitrogens — are now octahedrally surrounding the central metal ion. The complex will look something like this:

The calcium ion is of course calcium(II), the $\ce{edta}$ ligand carries four negative charges resulting in the complex anion being a dianion $\ce{[Ca(edta)]^2-}$. Technically, this species can still accept up to four protons, but only the carbonyl oxygens can now be protonated to some extent. If you do, you will encounter a very acidic oxygen-centred cation which will want to get rid of the extraneous proton again. Note that the two sites which I intuitively expect to be the most basic, the two nitrogen atoms, cannot accept any more protons since they are already tetracoordinated.
Thus, once you form a complex from free $\ce{edta}$, the equilibrium situation changes completely. We now no longer may consider $\ce{edta}$ by itself but must consider the complex as a whole; the remaining acid-base equilibrium is
$$\ce{[Ca(edta)]^2- + H+ <<=> [Ca(Hedta)]-}\tag{1}$$
As I mentioned, protonating the ligand will lead to a very acidic species (akin to protonating a protonated carboxylic acid) and therefore the equilibrium is shifted to the left.
So even though the $\ce{pH}$ values may tell you that you are dealing with a tri- or tetraanion of free $\ce{edta}$ in solution, the complex has quite different acidity constants.


text:
Water electrolysis - what is happening to an iron anode?
So I made an experiment to find a good electrolyte for a water electrolysis.
I tried citric acid, which turned out to not produce any gases at the anode, I tried sodium hydroxide, which turned out to be the very best, and I tried sodium carbonate.
The $\ce{Na2CO3}$ solution was about 2 g/l.
As my electrodes I decided to try some nails, very likely made of iron or some iron alloy.
So I hooked up my power source, just two nine-volt-batteries in series, and started the electrolysis.
The kathode (negative terminal) bubbled vigorously, indicating the production of hydrogen gas.
The anode (positive terminal) however bubbled very little and started decomposing, gray-green flakes started falling off and sinking to the bottom. This decomposition started as soon as the power was supplied.
This struck me, because I was expecting oxygen to be liberated at the anode!
When I took my anode out, it was at first a dirty green or green-gray, but it quickly turned a rusty brown.
So in conclusion, the kathode produced hydrogen with no sign of wear when on the other hand the anode made little gas and decomposed into probably some iron compound, which turned into probably iron oxide (I'm judging by the color here) on prolonged contact with air.
So, what could it be? I thought it might be iron hydroxide, but that'd be brown. This however was green.

May you please explain your downvote so I can improve in the future?

This might be iron(II) hydroxide - judged from it's green color and the capability to turn brown (Oxidation to Fe(III)) on contact with air.

On an anode made of iron, some of the removed electrons may be used oxidizing iron metal to ferrous iron(II) ions, or oxidizing iron(II) ions to iron(III), rather than bubbling off oxygen gas.  Pushing more current through the cell is likely to result in gas evolution alongside the iron oxidation, since all three of these processes will run simultaneously.
$$\ce{Fe(metal) -> Fe^{2+}(aq) + 2 e^- }$$
$$\ce{Fe^{2+}(aq) -> Fe^{3+}(aq) + e^- }$$
$$\ce{2H_2O(liquid) -> O_2 (gas) + 4H^+ (aq) + 4e^- }$$
Ferrous iron(II) ion solutions are a pale green, but rapidly oxidize in air to vivid red-brown ferric iron(III).  Iron oxides and hydroxides form precipitates and slimes in a variety of colors including black (magnetite), red (iron(III) oxide), and yellow (iron(III) hydroxide).

With Regards to Colors of Fe2+ and Fe3+, here is an interesting video. And especially interesting since Sodium Carbonate is often used as electrolyte for electrolysis.


text:
Clausius–Clapeyron relation for states of water and vapor
I've seen many questions about this topic on this site, but I still have some issues trying to understand this aspect. Here's what I believe I understood
looking at the phase diagram for a pure substance (water):

If you put water in a container with a piston pushing $\pu{1 atm}$ directly on the water with no extra space for inert gases, it will stay 100% liquid as at room temperature and $\pu{1 atm}$ liquid is the stable phase.
If the water is put at room temperature in a closed box where the volume is more than the volume of the water (creating a vacuum), the molecules will evaporate in a number so that the water vapor will be at the equilibrium pressure corresponding to room temperature, which can be calculated by the phase diagram.

I can't understand why the air (and its pressure) doesn't work like a piston, making sure no water is in gas state at $\pu{1 atm}$.
In particular, there was this exercise where a piston with $p = p_\mathrm{tot}$ was pushing down on a mixture of vapor $\ce{A}$ and inert gas above pure liquid $\ce{Ar}$, so that:
\begin{align}
p_\mathrm{tot} &= p_\mathrm{inert} + p_\mathrm{a}\\
\text{Equilibrium:}\qquad \ce{A(l, $p_\mathrm{tot}$) &= Inert(g, $p_\mathrm{inert}$) + Vapor(g, $p_\mathrm{a}$)}
\end{align}
Why doesn't the vapor $\ce{A}$ 'feel' the $p_\mathrm{tot}$ and get condensed into liquid $\ce{A}$?
Was it because $p_\mathrm{tot}$ is less than the vapour pressure of pure $\ce{A}$ at that temperature? If it was $p_\mathrm{tot} = p_\mathrm{vap}$ ($T = T_\mathrm{exercise}$), then the equilibium would be
$$\ce{A(pure, l, $p_\mathrm{tot}, T$) = Inert(g, $T, p_\mathrm{tot}$)?}$$
This doesn't look like an everyday case to me though where water is in gas form even if the pressure is $\pu{1 atm}$.

The answer may be somewhat counterintuitive. Yes, an inert gas does very much exert pressure on a solid or liquid. The problem is that the chemical potential of the substance in the gas phase is not affected by the inert gas (if the gas is assumed ideal).
Assume you have a substance (liquid or solid) in a rigid isothermal but otherwise empty box. The result is as expected evaporation or sublimation of the substance until the pressure of the vapour has increased to an equilibrium value (the saturation vapor pressure). We assume there is plenty of substance so that some remains in the condensed phase while the rest fills the box as vapour. What drives evaporation/sublimation? The fact that the chemical potential of the substance is (initially) lower in the vapour. Once equilibrium is achieved the chemical potential is the same in both phases.
Now assume an alternate scenario: in addition to the substance (solid or liquid) there is now some inert gas in the box. As suspected this gas exerts a pressure on the liquid and thereby affects its chemical potential (free energy). But the free energy of the substance in the gas phase is unaffected by the inert gas (assuming the gases behave ideally). And here the somewhat counterintuitive result: the substance again evaporates/sublimes until reaching the saturation vapour pressure, but precisely because the inert gas exerts a pressure on the condensed phase and increases its chemical potential in this phase, but does not alter the chemical potential of the vapor, it increases the saturation vapor pressure. The increased vapour pressure raises the chemical potential of the vapour until it matches the heightened potential of the condensed phase.
And the short answer to the question is that the inability of vapor A to 'feel' the total pressure including the inert gas is part of the definition of an ideal or perfect gas, which fails if the pressure is raised sufficiently.

Nice description. It should be noted that, for the inert gas to work as you describe, it must also be insoluble in the liquid. To the extent it dissolves in the liquid, it would decrease the liquid's mole fraction which, by itself, will decrease the chemical potential of the liquid, thus lowering its vapor pressure.  Hence you would have two competing effects from the inert gas, one which raises the chemical potential of the liquid, and one which lowers it.

@theorist Good point, I guess I should do a calculation to see what Henry's law coefficients and pressures would be necessary for the effect of solubility to become important.

@BuckThorn So for the gas phase the chemical potential is function of "partial" and not total pressure?

@Anton if the gas is assumed ideal, yes. Otherwise a correction factor in the form of an activity coefficient is necessary.

Consider the situation from a more microscopic/statistical mechanics perspective: In the liquid water there are quite strong intermolecular forces from all sides, at the water surface the forces are only from "one side" and in the gas the intermolecular forces are negligible.
Remember that the energy is not distributed evenly amongst the individual molecules, but instead some are faster and some are slower. (We shall ignore  vibrational modes and such for the sake of simplicity.)
When a certain molecule at the surface of the water has a kinetic energy that is higher than the potential energy from the intermolecular forces; it can overcome the latter and fly away from its neighbors, because the intermolecular forces fall off rapidly, it is essentially moving freely now.
Similarly, a currently free molecule might have a lower than average speed and become "stuck" at the surface of the water, typically transferring some momentum to the body of water.

After a while, a dynamical equilibrium depending on the velocity distribution (i.e. temperature) and the amount of water vapour per volume (partial pressure) will develop. A higher temperature increases the amount of molecules that randomly have a high enough energy  to overcome intermolecular forces; whereas the amount of water vapour increases the likelyhood of a water molecules hitting the surface.
That said likelyhood is independent¹ of the pressure of other gasses is not quite easy to see though; for example, one might argue that the mean-free-path is heavily reduced when the ambient pressure is higher. To speak of the vapour pressure however, our system (i.e. slightly moist air and slightly aerated water) must be in a thermodynamic equilibrium². Clearly in equilibrium  water molecules are spread evenly across all the space available for the gasses, well mixed with the others constituents of the air.³ (Also clearly, I hope) this is so regardless of the pressure/density that the inert gasses have.
I hope these considerations make clear why only the partial pressure is relevant for evaporation.
For very readable explanations of this and similar topics I'd recommend the textbook Thermal Physics by Stephen and Katherine Blundell.

¹ Or almost independent; at very high pressures there are intermolecular forces in the gas complicating things. The ideal gas assumption is very accurate in the troposphere though.
² If you have e.g. a test tube that is open to the atmosphere with  a bit of water at the bottom, you would have a gradient with saturated air at the bottom and drier air at the top of the tube. The steepness of that gradient does slightly depend on pressure. Personally considerations like these confused me greatly when first learning about partial pressures and such.
³ Gravity would make for slight gradients, but let's ignore that for now.

I can't understand why the air (and its pressure) doesn't work like a piston, making sure no water is in gas state at 1 atm

It is because water vapor and air mix. If you place a thin foil on the water surface (e.g. saran wrap), the air will work like a piston, pushing on the foil, which pushes on the liquid water and on any molecule that tries to escape.

If the water is put at room temperature in a closed box where the volume is more than the volume of the water (creating a vacuum), the molecules will evaporate in a number so that the water vapor will be at the equilibrium pressure corresponding to room temperature, which can be calculated by the phase diagram.

Yes, and now you could say the water vapor pushes on the liquid so that it does not further evaporate, but it is not a good model. Instead, molecules from the liquid phase are free to evaporate, but the same number of molecules from the gas phase condensate at the same time.
Pressure of the liquid
Where the pressure of the liquid is relevant (and the piston model works no matter whether the pressure is created by a piston, an inert gas or water vapor) is for the inside of the liquid. If the vapor pressure is lower than the pressure of the system, no bubbles form (no boiling occurs). If the vapor pressure is higher, bubbles will form, either displacing the liquid water ("boiling") or increasing the pressure to match the vapor pressure (still no boiling water).


text:
Discrepancies in calculating free energy values listed in Stumm and Morgan
I am hoping someone can help me with a clarifying a calculation on a fundamental thermodynamic understanding of reactions. Below is Table 2.5 from Stumm and Morgan's Aquatic Chemistry.

Table 2.5. Influence of Pressure and Temperature on the Energies of Water Ionization a
  $$
\begin{array}{rrr}
\hline
p (\pu{bars}) &\log K_\mathrm{w} &\Delta G_\mathrm{w}^\circ (\pu{kJ mol-1})\\
\hline
    &T = \pu{298.17 K} &\\
1 &-14.00 & 79.94\\
200 & -13.92 & 79.48\\
400 & -13.84 & 79.02\\
600 & -13.77 & 78.63\\
800 & -13.70 & 78.22\\
1000 & -13.63 & 77.83\\
\hline
T (\pu{^\circ C}) &\log K_\mathrm{w} &\Delta G_\mathrm{w}^\circ (\pu{kJ mol-1})\\
\hline
&p = \pu{1 bar} &\\
0 & -14.93 & 85.25\\
10 & -14.53 & 82.97\\
20 & -14.17 & 80.91\\
30 & -13.83 & 78.97\\
50 & -13.26 & 75.71\\
\hline
\end{array}
$$
  aData from Harned and Owen (1958).

I'd like to focus on the second portion of the table, where temperature is varied. $K_\mathrm{w}$ and $\Delta G_\mathrm{w}^\circ$ values are listed, but using the given $K_\mathrm{w}$ values, I calculate different $\Delta G_\mathrm{w}^\circ$ values than what the table offers. Here is a sample calculation for $\pu{50^\circ C}$:
$$K_\mathrm{w} = 10^{-13.26} = 5.49\cdot10^{-14}$$
$$\Delta G_\mathrm{w}^\circ = -RT\ln K$$
$$\Delta G_\mathrm{w}^\circ = -\pu{8.314e-3 kJ mol-1 K-1} \cdot \pu{323.15 K} \ln (\pu{5.49e-14}) = \pu{82.03 kJ mol-1}$$
The table lists $\Delta G_\mathrm{w}^\circ = \pu{75.71 kJ mol-1}$
Not only is this answer substantially different, but the value is heading in the opposite direction of where my calculations would head (more positive vs more negative). Unfortunately, I don't have access to the original source (Harned and Owen).
Can someone please help me to understand my error?

TL;DR: It appears that OP is right and has estimated standard Gibbs free energy correctly. The latter book accurately fetched data from the original source, but the calculations were carried out based on a false premise that $\Delta G^\circ$ refers to $\pu{25^\circ C}$ exclusively.

First, a general disclaimer. The uppercase circle ($^\circ$) generally denotes standard state. According to IUPAC, temperature is not declared within, but many authors attribute (negligently) $\pu{298.15 K}$ to Gibbs standard free energy $\Delta G^\circ$. Also check out What is the difference between ∆G and ∆G°?
For the given equilibrium one can use
$$\Delta G_T^\circ = -2.303RT\log K_T, \tag{1}$$
and at $\pu{298.15 K}$ this equation can be rewritten as follows, assuming $[\Delta G_{298.15}^\circ] = \pu{kJ mol-1}$:
$$\Delta G_{298.15}^\circ = -5.71\log K_{298.15}. \tag{2}$$

Now to the references. I didn't find the 3rd edition of The Physical Chemistry of Electrolyte Solutions by Harned and Owen (1958), but I do have an earlier version from 1943 [1]. In chapter 15 "The Ionization and Thermodynamic Properties of Weak Electrolytes" [1, p. 485], table 15-1-1 accumulates values for $K_\mathrm{w}$ determined electrometrically:

In subsequent investigations, the cells without liquid junction, were employed in certain combinations for the determination of the ionization constant itself, as well as $m_\ce{H}m_\ce{OH}$ and $\gamma_\ce{H}\gamma_\ce{OH}/a_\ce{H2O}$ salt solutions.
  \begin{align}
&\ce{H2 | HX($m$) | AgX-Ag} \tag{I}\\
&\ce{H2 | MOH($m_1$), MX($m_2$) | AgX-Ag} \tag{II}\\
&\ce{H2 | HX($m_1$), MX($m_2$) | AgX-Ag} \tag{III}
\end{align}
$$-\log K'_\mathrm{w} \equiv \frac{(\ce{E_{II} - \ce{E_{III}}})F}{2.303RT} -
 \log\frac{(m_\ce{H}m_\ce{X})_\mathrm{III}}{(m_\ce{X}/m_\ce{OH})_\mathrm{II}} + \\ + \frac{3.629 \times 10^6}{(DT)^{3/2}}\sqrt{\mu} = -\log K_\mathrm{w} + f(\mu) \tag{15-1-10}$$

So, Stumm and Morgan [2, p. 57] correctly adapted the data from table 15-1-1:
\begin{array}{rr|rr}
T (\pu{^\circ C}) &K_\mathrm{w} \times 10^{-14} {^{[1]}} &\log K_\mathrm{w}{^\text{calcd. [1]}} &\log K_\mathrm{w} {^{[2]}}\\
\hline
0 & 0.1139 &-14.9435 &-14.93\\
10 & 0.292 &-14.5346 &-14.53\\
20 & 0.6809 &-14.1669 &-14.17\\
30 & 1.469 &-13.8330 &-13.83\\
50 & 5.474 &-13.2617 &-13.26
\end{array}
But, what Stumm and Morgan did is they calculated the standard free energy exclusively at $\pu{25^\circ C}$ using the $K_\mathrm{w}$ values for other temperatures; disregarding the fact that equation (2) is only valid for $K_{298.15}$, but it was applied to all values as follows:
$$\Delta G_{298.15}^\circ {^{[2]}} = -5.71\log K_T, \tag{2a}$$
which is $\color{red}{\text{incorrect}}$. The latter column is populated with appropriate values calculated with (1), as it was done by OP:
\begin{array}{rr|rr}
T (\pu{^\circ C}) &\log K_\mathrm{w}{^\text{calcd. [1]}} &\color{red}{\Delta G_{298.15}^\circ {^{[2], (2a)}} (\pu{kJ mol-1})} &\Delta G_T^\circ {^{(1)}} (\pu{kJ mol-1})\\
\hline
0 & 0.1139 &\color{red}{85.29} &78.14\\
10 & 0.292 &\color{red}{82.96} &78.79\\
20 & 0.6809 &\color{red}{80.86} &79.50\\
30 & 1.469 &\color{red}{78.95} &80.28\\
50 & 5.474 &\color{red}{75.69} &82.04
\end{array}
Bibliography

Harned, H. S.; Owen, B. B. The Physical Chemistry of Electrolyte Solutions; American Chemical Society Monograph Series; Reinhold Publishing Corporation: New York, 1943; Vol. 95.
Stumm, W.; Morgan, J. J. Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters, 3rd ed.; Environmental science and technology; Wiley: New York, 1996.

Thank you for both your post edits (still learning mathjax) and your very detailed answer. I can sleep better knowing this information!

@prof.kvothe No prob at all, no one should ever stop learning:)

Indeed, when I plot the new data as a function of temperature I get a slope of 0.0808 and an intercept of 55.81, which both match nicely to the S and H values calculated from std (298 K) tables (-0.08066 and 55.81, respectively).


text:
Discrepancy observed for standard free energy change with temperature
My understanding of standard free energy is that it is only valid at the specified temperature, which is typically 25$^\circ$C. Thus, if T changes, so does $\Delta$G$^\circ$$_{rxn}$ according to this equation:
$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}$$
(assuming $\Delta$H$^\circ_{rxn}$ and $\Delta$S$^\circ_{rxn}$ are constant with temperature, which is a common assumption I'm finding in texts)
For example, $\Delta$G$^\circ_{rxn}$ at 25$^\circ$C would be different than at 250$^\circ$C. My confusion is with regards to $\Delta G^\circ_{rxn}$'s relationship to chemical equilibrium (or reactions) as:
$$\Delta G^\circ_{rxn}=-RT~ln~K$$
I've seen resources on the internet treat this oppositely, which has caused me great confusion. Let's take the Haber Bosch as an example, where we want to find K at 250$^\circ$C:
$$\ce{N2 + 3H2 <=> 2NH3}$$
From standard tables, at 25$^\circ$C (all kJ, mole, and K):
$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1306$$
$${N_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$
$${NH_3}: \Delta G^\circ=-16.48, \Delta H^\circ=-46.1, \Delta S^\circ=0.192$$
For this reaction:
$$\Delta G^\circ_{rxn}=2(-16.48)-(0+0)= -32.96~kJ/mole$$
$$\Delta H^\circ_{rxn}=2(-46.1)-(0+0)= -92.2~kJ/mole$$
$$\Delta S^\circ_{rxn}=2(0.192)-[(0.1915)+3(0.1306)= -0.1993~kJ/mole~K$$
First approach (how I think it is):
1. Calculate $\Delta$G$^\circ$ at 250$^\circ$C:
$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}=-92.2-(523.15 K)(-0.1993)=12.06~kJ/mole$$
2. Calculate K (assuming kJ, mole, and K for R and other values):
$$K=e^{-\Delta G/RT}=e^{-12.06/[(8.314E-3)(523.15 K)]}=0.062$$
Second approach (also commonly seen):
1. Calculate K using $\Delta$G$^\circ$ from tables, i.e., 25$^\circ$C, but inserting a new T:
$$K=e^{-\Delta G/RT}=e^{32.96/[(8.314E-3)(523.15)]}=1954.6$$
Edit: I want to also add that this reaction is more favorable at lower temperatures, so Approach 1 makes sense and Approach 2 does not, even though many treat problems this way. Also, I get $K=10^{5.77}$ at 25$^\circ$C, so the decreasing K matches up.
Question: Which one do you consider correct and why is there so much confusion and discrepancy in online tutorials? If this first approach is correct, is the error in the second common because of the misinterpretation of "standard"?
And I'd like to add: do the rules follow for $\Delta G_{rxn}$ and non-equilibrium as well? That is:
$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~Q$$

you're right that the error in the second approach is due to the misinterpretation of the "standard" term... Although the second method seems right at first glance, the error is that the $\Delta G^o$ is not the value to be used at this temperature

To expand on what @Abhigyan has said, the standard enthalpy and entropy of formation (and thus free energy of formation) are functions of temperature.  This certainly has to be taken into account.  One approach for including the temperature dependence uses the Van't Hopf equation, and the other approach is to determine the changes in H and S with temperature.

@Chester Miller. Thank you. So, why do some texts claim that std change in enthalpy and entropy dependence on temperature can be ignored for calculations of standard delG at different temperatures? What is their basis for this?

I guess it's OK if the temperature change is not very large.  But we do know the enthalpy and entropy are functions of temperature.

Here is an example where someone took the second approach: https://www.thoughtco.com/free-energy-and-pressure-example-problem-609491

Here you have a related questions and useful answers https://chemistry.stackexchange.com/q/37257/43021


text:
Can anyone explain this condensation pattern?
What is the hair-like polymer that condenses on the caps of the plastic cell vial? Inside vial are cells in a mixture of FBS and DMSO. The vials were frozen slowly in a special isopropanol-filled container (isopropanol does not touch the vials) to $\pu{-80^\circ C}$, then put on dry ice to move to a liquid nitrogen cryogenic freezer.
The hair-like crystals appear to be water, but bend toward my finger when nearby, similar to static attraction. I suspect that it has something to do with DMSO/isopropanol leaking out, and the very low temperatures that cause this strange condensation pattern, but I'm not sure.

Those strains might also be DMSO or DMSO-water-isopropanol mixture. If you could determine the melting point more precisely, it would be easier to answer. Also, assuming you work in a decent lab, maybe do some HPLC analysis in a spare time to be 100% sure what that is.

@andselisk Thank you for your response. I was planning on doing an LCMS/HPLC on it sometime to be sure. Just wondering if anyone knew off the top of their heads :)


text:
Mono-oxidized and dioxidized tryptophan
I'm translating a Russian text in which there is the following diagram of tryptophan degradation:

Further on in the text, hydroxytryptophan is called "the mono-oxidized form" (моноокисленная форма) and dihydroxytryptophan is called "the dioxidized form" (диокисленная форма). Is this correct and would this be understood by the readers if I translate it as-is?

First, you can use Google Ngram viewer to check out how often and in which context proposed words and their combination is used/has been used in the literature.
Second, I feel that "mono-oxidized" is overly redundant and just "oxidized" will deliver. Even in Russian you would rarely say so ("моноокисленный" or "единожды окисленный" is usually seen as an unnecessary complication in most cases). For "dioxidized" there are plenty examples when it is used in a similar context both in inorganic and bio chemistry, so it should a safe usage from what I see.
There is also an another option to say "two-stage oxidation" ("окисленный по двум ступеням") instead of "dioxidized" if you want to underline that the process has been carried out step by step.

Thank you! I googled for "monooxidized tryptophan". I'm not sure how Ngram could help me in this case. I was not sure whether the addition of the -OH group could be called "mono/di/oxidation", because I last self-studied chemistry in April.. zero free time.

I googled that too and the first link was this paper where "mono-oxidized (+16) and dioxidized (+32) form of tryptophan" expression has been used. So again, I see both mono-oxidized and dioxidized valid, though subjectively the mono prefix is not necessary.


text:
Ethanol discolored yellow when stored with a rubber stopper. Forming fine white precipitate when added to water
I stored some Klean-Strip denatured alcohol in a test tube and capped it with a rubber stopper. I did this in preparation for a chemistry demonstration. When I opened my kit a few days later at the event, I was surprised to find that the ethanol in the tube had turned a definite urine yellow. It burned just fine but let off nearly imperceptible wisps of smoke. Even more interesting was when I added water into the tube, which caused an almost instant white precipitate to form. I had to improvise the demonstration and play it off as turning flammable urine into semen, which it closely represented (because it was meant to be a comedic demonstration for a mature audience). The next few days I ran an experiment of using the same denatured ethanol in a jar over the black rubber stopper. By the end of the experiment, the solution exhibited the same properties and characteristics as the accidental batch. Now, not even to save my own skin, I cannot come up with a reasonable explanation for why the ethanol turns yellow, what causes the discoloration, and how does it turn milky when water is added.
Why does this happen. Chemically, what is happening. And would this be a good indicator for a non-anhydrous chemical? For example, I can add this to methanol or NaOH, measure the "turned whitedness" and determine whether or not that chemical is truly anhydrous?

That's kinda strange; I guess it dissolved some low molecular stuff from rubber, but still...

It’s not easy to know unless you decide to throw the full set of analytical chemistry at your yellowish ethanol. But we do have a few clues.
It is most likely that something from the rubber stopper got dissolved in the ethanol. This solution turned out to be yellowish so we can assume the compound itself to be yellow to brown-ish. Dissolution either came by the ethanol soaking the stopper while the flask was lying down or by evaporation/condensation.
We also cannot determine from the information you have given what the precipitate is that forms when you add water. But my first guess would be ions from the water being less soluble in an ethanol-water mixture and precipitating. If that is the case, it would be made up mainly of sodium chloride. It could also be that some component of the ethanol is insoluble in water and precipitates out. Considering your alcohol is denatured, it might also be the additive that makes it undrinkable. However, once again there are too many variables and you need the full set of analytical chemistry thrown at the precipitate to find out what it is.
Most certainly whatever is happening is not a good indicator of anything.


text:
Quantum Chemistry, quantum physics
In both classical and quantum mechanics knowledge of the present state of an isolated system allows it's future state to be calculated. If yes then why if not then why ?

yes and no: it is possible to find chaos from some starting conditions for a physical system and predictable behaviour from others. It is a huge subject. In a physical system  an example is double and triple pendulums and in a chemical one the Schlogl equations, $\ce{Y +2X= 3X;   Z = X}$ . See 'From Calculus to Chaos' by D. Acheson (Publ OUP) and Nonlinear Dynamics and Chaos, by S. Strogatz (Westview Press) for many examples, including the Lorenz equations, turbulence and forced oscillations. The Poincare-Bendixon theorem is used to predict when chaos will occur and when it will not.