text:
Do conjugated polymers absorb very low energy photons?
I was reading that in conjugated compounds, the extent of conjugation decides what wavelength photon it can absorb. Longer the conjugation "chain", lesser the photon energy required. So, molecules like lycopene
with 11 conjugated bonds is able to absorb in the blue range, appearing red. If this is the case, consider the polymer poly-acetylene

which has a conjugated "chain" spanning "thousands". Does this mean, this molecule would absorb infra red and lower energy light?

In principle yes, longer conjugated regions absorb at longer wavelengths. The simplest model when thinking about this is to consider a particle in a box, or on a ring if considering cyclic conjugation as in the chlorophyll's. However, in very long polymers, such as polyacetylene, the conjugation length is in fact rather small, approx 10 to 20 double bonds. Nanometer thick  films appear brown and absorb below 800 nm ish whereas if the conjugation length were very large, as per the chemical structure you draw, the molecules would absorb to longer wavelengths and appear black.
This limited conjugation is due to two things, first the molecules are  twisted and bent in forming the polymer and these distortions alter the electronic properties just enough to shorten the conjugation length. Second even when this twisting etc. were removed there are two ways of arranging the double bonds depending how you start. At their junction these two forms form what is called a 'solition' or 'topological kink'  and this is  a non-bonding electron between two regions of conjugation. This also limits conjugation.


text:
How do I calculate the amount of polyprotic acid and base needed to achieve a certain buffer pH?
e.g. Citric Acid and Trisodium Citrate
$\ce{H3C6H5O7 <=>[ka_1] H2C6H5O7- + H+ <=>[ka_2] HC6H5O7^2- + H+ <=>[ka_3] C6H5O7^3- + H+}$
Using the definition of Ka and substituting for intermediates, I can get the relation:
$$\ce{[C6H5O7^3-] = (ka1\times ka2\times ka3\times [H3C6H5O7])/[H+]^3}$$
Taking the negative log of both sides gives:
$$\mathrm{-\log[}\ce{C6H5O7^3-}\mathrm{] = pka1+pka2+pka3-\log[\ce{H3C6H5O7}]-3\times pH}$$
$$\mathrm{\log(\frac{[\ce{H3C6H5O7}]}{[\ce{C6H5O7^3-}]}) = pka1+pka2+pka3-3\times pH}$$
$$\mathrm{\frac{[\ce{H3C6H5O7}]}{[\ce{C6H5O7^3-}]} = 10^{(pka1 + pka2 + pka3 - 3\times pH)}}$$
$$\mathrm{[\ce{H3C6H5O7}] = 10^{(pka1+pka2+pka3-3\times pH)} \times [\ce{C6H5O7^3-}]}$$
For Citric Acid's pkas of $3.13,~4.76,~6.4$, this equation implies that you need Trisodium Citrate at 5000x the concentration of Citric Acid to get a pH of 6, which makes no sense as that's essentially pure Sodium Citrate, which has a pH in the $8-9$ range. What am I doing wrong here?

Welcome to Chemistry.SE!  Please note that formulas can be better expressed with $\ce{}$ for chemical formulas/equations, $\mathrm{}$ for math term/equations, and $\pu$ for units.  Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies.

HINT - Look at the stoichiometry. At pH 3.13 citric acid equals sodium citrate. At pH 4.76 sodium citrate equals disodium citrate. At pH 6.4 disodium citrate equals trisodium citrate.

Nothing is wrong with your equations; you are simply misinterpreting your calculated result. At a pH of 6, sodium citrate is present at 5000-fold the concentration of citric acid, but these are not the only species in solution, and sodium citrate is not the dominant species in solution.
The hydrogen citrate dianion is, and is thus the main contributor to the pH of the solution.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Would standing next to a pool of mercury do you any harm?
I was watching this clip from the 2017 film The Mummy in which the characters stand next to a pool of mercury which was being used as part of an elaborate prison.
Several comments under this video imagine that standing next to a pool of mercury would be very dangerous, but I can't see any reason to think that a significant amount of vapour would be coming from the pool so it seems like the mercury just being there in liquid form would be quite harmless. But maybe I shouldn't go robbing any graves any time soon.
The automatic warning dialogue told me the question appears quite subjective. So, let's say you're standing next to the pool for no more than ten minutes and we'll define harm as requiring medical intervention or (if chronic) as having a demonstrable, long-term health effect, such as erethism. I have no idea how large the pool is in the video, but for the purposes of the question, we'll say it's a circular pool of 10 metres in diameter.

The expression "mad as a hatter" came from real exposure problems. But would you die instantly, no. Would you necessarily die from the mercury eventually, no. However the salient fact is that mercury certainly won't do you any good, so it is prudent to limit your exposure.

The data on this is...frustrating to find. There's wildly different toxicity thresholds for organic and inorganic mercury, and it seems very few actual toxicity tests. I found a CDC reference that points out that over 80% of mercury in vapor can be absorbed by the lungs, and that in a room at 25C, the saturation for mercury vapor is about 18mg/m^3. Next step would be to try to find a source on toxicity (this book seems like a candidate, but paywalled) to see if this could seriously injure you in 10 minutes.

I won't comment on health but that sarcophagus should have amalgamated.

It depends. But there is little short term risk.
Mercury metal isn't very toxic. You can even ingest it without significant short or long term effects. But mercury vapour is nasty and you don't want to be continuously exposed to any significant amount of that (see this answer for more detail). And some mercury compounds are pretty nasty too.
Of course mercury metal emits mercury vapour so there will always be some in any room with exposed mercury. But the reason why this isn't an immediate worry is that it takes time to absorb enough vapour for it to become notably toxic. Chemists are used to handling liquid mercury in the open lab but are a lot less comfortable if it has been spilled and not cleaned up. This didn't used to be so which is one of the reasons we know the dangers: chemists who were exposed to vapour over years did suffer toxic effects. Mercury miners suffered similar long term toxic effects from the vapour. In modern labs we make sure the vapour doesn't build up by good ventilation and making sure any spills are cleaned up quickly so exposed mercury doesn't hang around emitting vapour.
So would you be in danger standing next to a pool of mercury? No, if you didn't do it for long. Or, if the ventilation in the room were good and the vapour wasn't allowed to build up. Even then, though, modern safety standards wouldn't be comfortable having pools of mercury exposed to the atmosphere: you wouldn't want to regularly work in that room with the pool.
In short, standing next to a pool of mercury even in a badly ventilated room wouldn't cause immediate harm; working next to that pool for months or years would certainly cause you harm.

Just an addendum to Matt`s answer.
The main problem in the research lab was not the short term exposure, but it accidental spillage. Even a small spill is quite time consuming to clean and it is easy that small amounts of mercury acumulates leading to unsuspected long term exposure.
In Physical Chemistry labs. mercury was commonly used to seal vacuum systems, measure pressure or in mercury diffusion pumps. Fortunately, the technology has render these techniques obsolete.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Reaction of chloroform, resorcinol and strong base
A common reaction used to identify resorcinol (benzene-1,3-diol) in the laboratory is to react it with chloroform in presence of 30% sodium hydroxide (NaOH). A red coloured solution (or precipitate) is produced. I have searched many books and websites, but none of them give any clue about the actual reaction, its products, far less the reaction mechanism.
The reaction is as follows:
$$\ce{Resorcinol + Chloroform + 30\% NaOH -> Red Colour}$$
So any ideas about what reaction is taking place?

https://www.quora.com/Can-resorcinol-give-Iodoform-test#

@SoumikDas no, no, in this reaction, iodine or any other halogen is not added.

Well, turns out to be, the reaction proceeds to product a red colored species.

Reaction with chloroform in the alkaline medium
Dissolve 0.1 g in 1 ml of water R, add 1 ml of strong sodium hydroxide
  solution R and 0.1 ml of chloroform R, heat and allow to cool. An
  intense, deep-red colour develops which becomes pale yellow on the addition of a slight excess of hydrochloric acid R.

@OscarLanzi probably. I could not find enough evidence to prove that.

Do you know the mechanism of this reaction? I cannot figure it out.

You're probably looking at a two-part reaction.  First chloroform+sodium hydroxide imparts a carbonyl group ortho to a phenolic hydroxyl group (see here).  Then this carbonyl group electrophilically attacks a second resorcinol molecule to give the dimeric structure.


text:
Why does formic acid give positive Fehling's test?
$$\ce{HCOOH ->[\text{Fehling's solution}] CO2 + H2O}$$
The points discussed in a previous question do not account for the fact that formic acid also gives positive Fehling's test since it is not enolisable. Then what is the reason/mechanism for the reaction?

I thought that the exact mechanism is not known in case of both fehlings and tollens.

Formic acid is something like an aldehyde, it has a hydrogen atom attached to a carbonyl group.

Because Fehling's solution oxidizes enolizable aldehydes to carboxylic acids, it does not preclude all other oxidations. Draw cupric formate and move formate electrons toward copper.

Here is a short simple available answer provided online by Prachi Sawant in 2016 to quote:

Formic acid has both aldehydic (-CHO)  and carboxylic (-COOH) functional groups.
  Hence, it gives both Tollen's and Fehling's test positive.
Reaction of formic acid with Tollen's reagent:
$\ce{HCOOH + 2[Ag(NH3)2]+ +2OH- → 2Ag + CO2  + 2H2O  + 4NH3 }$
Reaction of formic acid with Fehling's reagent:
$\ce{HCOOH + 4OH- + 2Cu^{2+}(in complex)  → CO2 + 3H2O + Cu2O }$

Now, here is my attempt on the possible underlying mechanic, which is likely complex. I start with the above net reaction, which in reality, is probably not completely accurate, but nevertheless, may provide some insight into actual half-cell reactions, beginning with the reduction of the organic cupric complex:
$\ce{Cu^{2+} +  e^-  ->  Cu^+ }$
where the insoluble Cu2O (the test indicator) is formed. The corresponding other half-cell is less clear accept for the liberation of an electron:
$\ce{COOH^- -> .HOCO + e^- }$
which may speculatively may be followed by:
$\ce{.HOCO + .HOCO -> H2O + CO + CO2 }$ (speculation)
And/or, this paper favoring non-conventional ionized species in the context of formic acid as possessing more stability as demonstrated, for example, by:
$\ce{HCOOH -> [.HCOOH]^+ + e^- }$
With its source cited below, followed by:
$\ce{[.HCOOH]^+ -> H2O .CO^+ }$  (a distonic ion, per 2nd cited source)
$\ce{OH^- + H2O .CO^+ -> H2O + COOH^- }$ (speculation)
To quote a cited source above :

Rate constants for the radical-induced hydrogen abstraction from formic acid, HCOOH, are presented here. Only those reactions leading to the formation of $\ce{.HOCO}$ were investigated.

Per a source: 'The gas phase chemistry of the formic acid radical cation $\ce{.[HCOOH]^+}$ Mechanism for exchange of the hydrogen atoms: a quantum chemical investigation', to quote:

Earlier labelling experiments had indicated that prior to the loss of H the hydrogen atoms in ionized formic acid, $\ce{.HCOOH^+}$, can become positionally equivalent via an unknown pathway. Using ab initio molecular orbital calculations we have located a minimum energy reaction pathway for hydrogen exchange reactions which takes place close to the dissociation level and proceeds via flexible ion/dipole complexes. Along this path three equilibrium structures of different atom connectivity are found: $\ce{.HCOOH^+}$, 1, ionized formic acid; two forms of the surprisingly stable ion/dipole complex $\ce{.HOHCO^+}$, 2, 3; and the distonic ion $\ce{H2O CO.^+}$, 4. Ion 1 is separated from ion 2 by a barrier of 142 kJ mol−1 and this barrier corresponds to a 2A′ → 2A″ surface crossing. The barrier for the proton transfer 3 → 4 is only 25 kJ mol−1 relative to 3; the transformation $\ce{.HCOOH^+ -> H2O .CO^+}$, formally a 1,2-hydrogen shift, takes place via the ion/dipole complex $\ce{HO· HCO^+}$, $\ce{H2O .C^+}$ is the reacting configuration for decarbonylation and this reaction is calculated to take place at the dissociation limit in full agreement with experiment.

Where ‘decarbonylation’ can also be effected by several transition metal complexes, (source), as in the current context with an organic copper complex.

According to some people:
If you look carefully, you will find both the aldehyde radical and the carboxyl radical in formic acid:

The red line shows the aldehyde functional group character and the blue line indicates the carboxyl functional group character.
As a result, it has mainly characteristics of carboxyl compound, but sometimes shows carbonyl character as well which happens in case of tollen's reagent and fehling's solution reacting with it.
You can identify it by doing the 2:4-DNPH test to which the formic acid doesn't respond.
But to me and most other people:
This reaction certainly is given by aldehydes, it is not specific for this class of compound since many other acids, such as tartaric or citric, similarly reduce silver nitrate. The reduction of mercuric chloride to mercurous chloride by formates likewise offers no support for the aldehyde theory, since aldehydes in general, including formaldehyde, cannot effect this change. Actually, formates give none of the reactions which characterize an aldehyde; thus they fail to restore the colour to Schiff's reagent, produce no reaction with 2.4 dinitrophenylhydrazine hydrochloride, and above all do not reduce Fehling's solution, which is probably the most specific of all aldehyde reagents.

They’re not radicals; at best, they are functional groups. But formic acid really doesn’t undergo any typical aldehyde reactions, so calling the carbonyl group an aldehyde is missing the point …


text:
Excluded Volume in van Der Waals equation
I was just reading over the EOS topic when where I stumbled upon excluded volume.
Although the pictorial representation of the hard sphere model clearly shows that the excluded volume is 4 times the volume occupied by molecules, but what i am not unable to understand is that why do we take the sphere with diameter equal to one of the molecules in the first place.

So, my question is that why do we have to take the whole spehere..

I guess this has something to do with viewing the molecules in 3-dimensions, but i am unable to visualize.

Draw the exclusion volume centered on one of the particles, rather than encompassing the two. In your picture, take the lower right particle as A and the upper left particle as B. Draw your circle concentrically around A.  Then it becomes clearer that the center of B can not penetrate into that circle. The center of particle B can never enter that volume, one considers this as particle B can never occupy that volume.

We assume that the molecules are identical hard spheres with diameter $d$. The excluded volume is emphatically not the volume of the sphere, but rather the volume excluded to the other particles by the presence of that sphere.
When two particles collide, they touch each other at some point on their surface, and the distance between the center of one molecule to that of the other is $d$. Indeed, two particles cannot get closer than a distance $d$ to each other, and hence the excluded volume of each particle is the volume of a sphere of diameter $d$—-we must take into account the radii of both colliding spheres.
Edit: This suggests that the excluded volume should be eight times as large as the original volume of a sphere, not four times. However, a collision event involves two particles, and this excluded volume is shared, in a sense, by both particles. We must therefore divide by a factor of two, and that gives the resulting multiplier of four.

I knew that part, whats confusing me is that why do we take a whole sphere? In the second picture, cant a third molecule (shown in red) occupy that space? shouldn't the area enclosed with solid black line equal excluded volume? Thanks

@ShahMHasan, your picture is misleading, and is not the picture that my post describes. You should think of the excluded volume as a sphere of radius $d$ centered about one of the particles.

i know my picture isn't correct, the question is: why is it misleading?

@ShahMHasan, from my earlier comment, because you should think of the excluded volume as a sphere of radius d centered about one of the particles.

A sphere with twice the diameter of a molecule should have 8 times the volume, shouldn't it?

@IvanNeretin, yes, sorry, I've neglected to address that point. A collision event involves two particles, and this excluded volume is shared, in a sense, by both particles. We must therefore divide by a factor of two, and that gives the resulting multiplier of four.

One more thing please, why do we not consider collisions greater than two? Or does that give the same result?

@ShahMHasan - Just two particles colliding is considered because it is much more probable than three particles colliding. However three particles can collide, so the simplification isn't entirely accurate.

If the probability that a pair of balls overlaps is $8v/V$ where $v$ is the ball volume, and $V$ is the container volume, the probability that they don't overlap is
$$1 - 8v/V$$
Given $N$ balls, there are $N(N-1)/2$ combinations of pairs. The probability that none of them overlap is approximately
$$(1 - 8v/V)^{N(N - 1)/2} \sim (1 - 4vN/V)^N$$
Multiplying by $V^N$ as in calculation of partition functions, one gets
$$(V-4vN)^N,$$
which looks like an excluded volume of $4v,$ not $8v.$


text:
Why can't NH5 form?
In ammonia a nitrogen atom forms 3 covalent bonds with hydrogen atoms and has one lone pair which can be used in a dative covalent bond with another hydrogen atom to form an ammonium ion. Instead of using both electrons for one dative covalent bond why can't they both individually be used for 2 separate covalent bonds forming NH5?

This violates the octet rule.

As the person whose answer supposedly answers this question, I really don't see how this is a duplicate, the obvious difference being that nitrogen is a Period 2 element whereas sulfur is Period 3. Furthermore, $\ce{NF5}$ is just as hypothetical as $\ce{NH5}$, so the argument I used doesn't apply here. So I have reopened this. Please think and look at the questions carefully before closing.

@orthocresol I would say that, besides the comments about size effects for period 2 elements, the currently accepted answer here doesn't differ much from the reasoning given in your prior answer. I definitely see now that it is not a duplicate, but I think at the very least it covers some of the groundwork for this question.

I have pursued such questions regarding the hypervalency of particular molecules before and thus, I would like to provide some insights to the matter.
A common superficial answer that one can provide to answer the question is that there are ten electrons around the central nitrogen atom in $\ce {NH5}$ and thus, it cannot exists based on the octet rule. And we do know that the octet rule must be strictly abided by for the period 2 elements.
However, let us look at the deeper reasons why period 2 elements behave in this way by looking at the characteristics of hypervalent molecules.
One very crucial reason why particular hypervalent molecules and molecular ions, such as $\ce {PCl5}$ and $\ce {SO4^{2-}}$, can exist is due to the presence of electronegative ligands around the central atom which can draw electron density away, reducing the inter-electronic repulsion in the molecule and hence, increasing the stability of the molecule significantly. Due to the significant ionic character resulting from forming multiple polar covalent bonds, there is actually significant ionic character within the molecule. Computational studies have shown that some of these central atoms are, in fact, bearing significant charge, with some even having a greater charge than +1. Although you may expect that central atoms in hypervalent molecules have more than 8 electrons, the reality is that they likely have less! This was shown in a study conducted by Noury, Silvi and Gillespie.[1] This goes to show how electron-withdrawing are the electronegative ligands.
Another important factor to consider is sterics. Simply put, we cannot put multiple large atoms around a small atom. Atoms of the elements of period 2 may not have reached the "critical size" for it to accommodate more atoms around it while period 3 elements seem to be able to accommodate more atoms much more easily. This is the likely reason why the molecule $\ce{NF5}$ does not exist. While the electronegative fluorine ligands may help to reduce electron density accumulating in nitrogen, the space constraints simply are too limiting. You can read more about the molecule and its associated instability at its Wikipedia page. The references provided on the page are also very insightful.
I hope it is now clear why $\ce {NH5}$ cannot exists based on considerations of the two factors above.
References

Noury, S.; Silvi, B.; Gillespie, R. J. Chemical Bonding in Hypervalent Molecules:  Is the Octet Rule Relevant?. Inorg. Chem. 2002, 41 (8), 2164–2172. DOI: 10.1021/ic011003v.

Nice. Regarding the positive charge on the central atom, it makes perfect sense too that the Period 2 elements - which are more electronegative - don't quite tolerate it as well.

Just a hint: the term hypervalent is misleading (one could say deprecated) and such molecules should better be referred to as hypercoordinated instead.


text:
Mechanism for conversion of imide to ester with NaH

What is the mechanism of this reaction? My proposed mechanism is shown below, would it work?

Sodium hydride is a non-nucleophilic base.

Joe kent: Try forming the alkoxide with NaH first and then transfer the CO2Et group either intra- or intermolecularly. The amide ion that is left is ready for alkylation. Try it and see what you get.

Not sure the sodium hydride is intended as,a nucleophile.  It could react with the benzoyl bromide.

NaH is a strong base but a bad nucleophile. This is one possible mechanism.

Correct about NaH but I can’t agree with that SN2...


text:
Separating 2 chemicals
Is it possible to  separate two chemicals with slightly different melting points? Can this be done with the use of a lab freezer? As MP is -44 and The other is -61 - -59 °C.
Will it solidify these two substances separately? Or each becoming a separate layer?

Depends, see: Eutectic

The process of separating chemicals by their melting points is called fractional freezing. In zone refining, it is used to produce some of the most pure chemicals, for example Si 99.999999999% (11N, meaning 11 9's) pure.
However, your question per se is far too general to be answered. Some substances can be more easily separated than others. For example, an attempt to separate $\ce{DHO}$ and $\ce{D2O}$ from $\ce{H2O}$ did not succeed under the test conditions.

Physicochemical Information of unwanted chemical is
Boiling point 119 - 121 °C (1013 hPa)
Density 0.913 g/cm3 (20 °C)
Explosion limit 1.8 %(V)
Flash point 18 °C
Ignition temperature 329 °C
Melting Point -61 - -59 °C
Vapor pressure 21.3 hPa (20 °C)
Solubility 5.1 g/l.

And the desired in is.
Molar mass
86.09 g·mol−1
Appearance
Colorless liquid
Density
1.1286 g/mL (15 °C), 1.1296 g/mL (20 °C)
Melting point
−43.53 °C (−46.35 °F; 229.62 K)
Boiling point
204 °C (399 °F; 477 K)
log P
-0.76[3]
Acidity (pKa)
4.5
Refractive index (nD)
1.435, 1.4341 (20 °C)
Viscosity
1.7 cp (25 °C

What if I buy a HPLC Chromatography machine. Will this work?

With the difference in boiling points you quote (120C  v 204C) distillation loooks a good bet.


text:
neutralization of a strong acid with a strong base
When a 1M H2SO4 is mixed with an equal volume of 1M NaOH, is the mixture acidic, basic, or neutral?

Welcome to Chemistry.SE!  Please note that formulas can be better expressed with $\ce{}$ for chemical formulas/equations, $\mathrm{}$ for math term/equations, and $\pu$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies.

Since the sulfuric acid is diprotic, the reaction of sulfuric acid with sodium hydroxide has a 1:2 ratio, with twice as much sodium hydroxide being used up in the reaction. The neutralization reaction would be:
$\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}$
After that reaction is complete, approximately half of the sulfuric acid remains, and you would be left with an acidic solution.
Edit: As was pointed out below, I failed to balance my sodium hydroxide.

Balance sodium hydroxide :$\ce{2NaOH}$

Since hydroxide ion is a monoprotic base and sulphuric acid is a diprotic acid
, there are actually two equations involved.
Because we have an equimolar amount of acid and base, the products of the first reaction are sodium bisulfate and water;-
$$\ce{H2SO4 + NaOH -> NaHSO4 + H2O}$$


text:
Evaporation of isopropyl alcohol below its boiling point

Would total evaporation of isopropanol (isopropyl alcohol) at room temperature leave residue? And, why it is able to evaporate although it does not reach its boiling point?

Why does water evaporate below the boiling point?

Residue left behind is not the alcohol but impurities dissolved in or mixed with the alcohol so pure isopropyl will not leave residue.
The reason it is able to evaporate below its boiling point is the same reason that all other liquids can evaporate below their boiling points. A liquid such as isopropyl alcohol will have a certain property called vapor pressure. Since the molecules are mostly free to move around, some move faster than others and are able to escape the fluid's surface. This will continue to happen until either the partial pressure of isopropyl alcohol in the surrounding area is equal to the alcohol's vapor pressure or there is no longer any more alcohol to evaporate. This pressure reaches this equilibrium because at the vapor pressure, there are just as many molecules of alcohol colliding with and joining the liquid alcohol as there are leaving it.
The boiling point need not necessarily be reached for all of a liquid to evaporate. Instead what the boiling point means is that once at that temperature, any more energy will go to evaporating more molecules faster rather than increasing the temperature.
So to summarize, liquids will evaporate until there is either no more or until the vapor pressure is reached.

Pure isopropyl alcohol would normally evaporate completely at room temperature in our standard atmosphere. If there is a residue, it is due to dissolved or suspended impurities, including anything it may have dissolved from the surface that it is evaporating on. Like other volatile liquids, it evaporates because the molecules have enough kinetic energy at room temperature to overcome the forces (attractions) that otherwise hold the molecules in liquid form. In addition to their own kinetic energy, additional energy will come from air molecules impacting the molecules of alcohol on the surface. Room temperature or any temperature measures the average kinetic energy of the molecules. Some of the molecules have a high enough energy to create a vapor pressure at room temperature.

One additional source of temporary "residue" on a surface after the evaporation of a volatile liquid can be condensed water droplets. Water will condense out of humid air onto some surfaces which lose heat to the volatile liquid as it evaporates. In time this too will evaporate as heat enters the surface from the surroundings. I believe the vapour pressure of the liquid must be greater than that of water (under the conditions of the experiment) for this to happen.


text:
Predicting products of a reaction and balancing
2BiOCl + NaOH = BiOH + NaCl.
This is the reaction where I am trying to get Bismuth Hydroxide as one product and salt. My problem is what is the state of Bismuth Hydroxide? Whether it is BiOH or Bi (OH)2 or Bi(OH)3. How do I know which one I will get?

You know that from the valence of Bi, and that you deduce by looking at BiOCl.

Well, the reaction actually produces $\ce{BiO.OH}$. According to this site:

It reacts with alkalis, especially when the solutions are
  concentrated. It is probable that with a dilute solution of potassium
  hydroxide a reversible reaction takes place according to the equation
$$\ce{BiOCl + KOH <=> BiO.OH + KCl}$$
With concentrated solutions the action is much more complex.

Some sources states that the reaction produces Bismuth(III) oxide. In conclusion, the reaction may produces different species depending on concentration of alkali, reaction conditions, amount of bismuth oxychloride etc.


text:
Can sucralose form a syrup?
(Almost) Everyone is familiar with making simple syrup from table sugar and water.
I was wondering if pure sucralose is also capable of forming a syrup.
I wanted to do an experiment myself, but to my consternation,

Splenda is not pure sucralose.
pure sucralose is not readily available to me

So, I would like to hear if anyone has tried this experiment, or reasons why sucralose would not form a syrup if such is the case. Pointers to references would be nice.
P.S. I do not have a deep chemistry background, so simple explanations (or even pictures) will be very much appreciated.

Syrup is just a very concentrated solution of sugar.
Because of the high concentration (for instance 65% w/w for USP Syrup, the concentration that avoids bacterial growth) the viscosity increases, and the solution gets syrupy.
In the case of sucralose, the solubility is much lower than that of sugar.
That is normally not an issue, since the sweetening potency is about 600 times higher, at moderate concentrations (80 ppm sucralose have the same potency as 5% sugar).
But to make a syrup, you need a very high concentration.
With a solubility of 28.3 g/100 mL, the viscosity will, probably, be much lower than that of sucrose syrup.
To compare, note that the viscosity of a 30% sugar solution is 3.21 mPa•s, while that is a 60% sugar solution is 56.76 mPa•s.
Take into account that sucralose differs from sucrose only in the exchange of three $\ce{-OH}$ for $\ce{-Cl}$.


text:
How to predict ease of oxidation or reduction for a given oxidising or reducing agent?
I have never understood which agent will be able to oxidize or reduce which type of reactants and yield which type of products will be formed. For example I want to know about DIBAL-H and how is it able to reduce some groups but not carboxylic acid. Here is what all I want to know:

What are the criteria for judging which agent will oxidize or reduce?
What will be the condition in which it will be able to oxidize or reduce a given molecule.
What type of products will be formed?

Note: I am in high school and thus need an explanation which I will be able to understand.

Are you looking for a qualitative or quantitative criteria? Reduction potentials would give an easy quantitative measure, but qualitative measures are much more specific to the systems you are studying. @Pan

I am more interested in the qualitative measure @Tyberius

This goes to the heart of organic chemistry and most of it is just about learning the properties of the most common reagents, then understanding where to find the detailed information. A good Organic chemistry text book is the starting point, then there are lists to help, for examples list of oxidising agents, list of reducing agents or entire series of books on Organic chemistry reagents e.g. Fieser & Fieser which a good library should have. For individual transformations databases such as Reaxys or Scifinder are searchable by structure and can find the conditions under which a transformation has previously been done. If the transformation has not previously been reported then you have to use your judgement as to which conditions to try, that is where the real art of synthesis comes in. It is important to note that solvent and temperature play a huge role in such transformations. Reactions that proceed in one solvent may give a different outcome in another.
Dibal-H does reduce carboxylic acids, but require multiple equivalents and elevated temperature or additional reagents such as TMSiCl paper here

Thank you for answer. But i think there is no intuitive for me to say if or not oxidation will happen at my high school level.

Probably not, just learn the reactions of the most common reagents and those will provide the answers at the high school level.


text:
Why does carbon not form 3 pi bonds?
Why doesn't carbon form  3 $\pi$ bonds? If s-s overlap is possible between 2 carbon atoms,then it will be possible for 3 p orbitals to overlap laterally .

Carbon in graphite does participate in three $\pi$ bonds all right.

@IvanNeretin But not using all 3 p-orbitals, which seems to be what the OP wants.

https://chemistry.stackexchange.com/questions/35002/can-carbon-have-quadruple-bonds

Possible duplicate of Bonding in diatomic C2, a carbon-carbon quadruple bond?

No, this is not a duplicate. Related, maybe, but definitely not the same questions. If I have to, I will reopen this.

Okay so say you have two carbon atoms and you want to make a sigma bond with their s orbitals. They won't want to since their orbitals are filled but VSEPR theory tells us that carbon can rearrange and hybridize its orbitals to allow such bonds to happen. Normally you'd have your 3 pi orbitals along the 3 spatial axis, but the s orbital, holding 2 electrons, can hybridize with the p orbital that has no electrons to get two sp orbitals that have one electron each. This sp orbital will be along an axis we'll call z. This leaves us with two p orbitals orthogonal to our sp orbitals which may overlap laterally to give two pi bonds.
If we have two p orbitals overlapping to make two pi bonds, the third p orbital leaves us a problem. If we don't hybridize the orbitals, the third p orbital is empty and cannot make a bond. Even if you had a suitable lewis acid to try to give it some electrons, the bond would't be a pi bond since the orbital overlap would be along the inter nuclear axis. If we do hybridized then we end up with the situation described earlier.
Perhaps there could be a strange configuration where the inter nuclear axis is equiangular from the three base axis and the empty p orbital is filled by some electron donor and the bonding p orbitals overlap at right angles. I wouldn't really call it a pi bond at that point and anything in that position would likely collapse into a more stable state.
Basically the geometry of the p orbitals means that we can really only have two of them overlapping at any one time in the basic sense of a pi bond. This is assuming we are not counting the cases where a single p orbital participates in bonding with more than one atom such as in graphite, though I wouldn't really call that three distinct bonds.

VSEPR theory does not tell us that carbon hybridises its orbitals. VSEPR only lets us guess the molecular structure, then we would find the best fitting molecular orbitals.


text:
How can you tell when a molecule forms a lattice
I came across the molecule SiO2 and didn't realize that in fact, it forms a lattice. How could I have found out that, in fact, it forms a lattice?

By the fact that it is a solid.

@A.K. SiO2 is easily supercooled into a glassy state. Not the point.

There is no easy way to guess or predict the answer in general, but experience will give you some clues
The way to tell whether something is a molecule or a covalent solid (Made from a lattice) is to look up the structure. When you first learn chemistry, some things have just have to be learned. But when you learn a few things, some patterns will emerge and you will get better at predicting the properties of other compounds.
The structure will usually tell you the answer. Other physical properties of the compound can hint at the structure. Carbon dioxide is a gas so you should be able to predict it is a small molecule; silicon dioxide is a refractory solid: this is a big clue that it is not a small molecule but is, in fact, a network solid mad from a lattice of Si-O bonds. Nitrogen and Oxygen are gases, so most likely small molecules; graphite and diamond (both forms of pure carbon) are solids so  out likely networks made from large lattices.
There isn't really a simple way to predict this, but once you have learned a few examples, patterns will emerge.

SiO2 is usually not a molecule, but just the chemical formula indicating the relative composition of quartz: Two oxygen atoms for every silicon atom.
In the solid state, it forms huge (macroscopic) network "molecules", which can form a crystalline lattice (e.g. quartz), or an amorphous glass. They slowly become smaller upon heating after crossing the melting point (1713°C), until you reach the boiling point (>2200°C), where you probably get a monomolecular gas of $\ce{SiO_2}$ molecules.


text:
Proof that BF3 exists?
What I know is, elements with octet are stable, But In BF3, Boran doesn't get octet, So, How come BF3 molecule exist? If it exists, How do we know whether it is BF3 or not? Is there any proof that shows BF3 exists?

https://www.mathesongas.com/pdfs/products/Boron-Trifluoride-Pure-Gas.pdf

Do you mean, for example, how do we know that BF3 exists in a monomeric form and not as a hypothetical B2F6 dimer which would be octet-complete (analogous to Al2Cl6)?

If the dimer were a thing, at low temperatures where there's no scrambling, you should see 1 peak by fluorine NMR instead of 2.

Possible duplicate of https://chemistry.stackexchange.com/questions/57784/has-anyone-even-taken-a-picture-of-a-molecule-to-confirm-the-geometry-predicted

https://chemistry.stackexchange.com/questions/32240/how-can-an-inexperienced-chemist-determine-the-chemical-structure-of-a-molecule https://chemistry.stackexchange.com/questions/51626/what-is-the-correct-way-to-verify-a-structures-geometry-for-example-for-benzen

Boron trifluoride, as described in Wikipedia is proved to exist the way everything in chemistry is ultimately proved, by experiment.  The octet rule is really only a model that describes the electronic structure of some but not all molecules.  Boron trifluoride is one of those (many, actually) molecules for which the octet rule is not really correct.  The actual electronic structure is not just a single Lewis structure but a combination of multiple contributing structures, some of which do satisfy the octet rule and some of which don't.

The octet rule is not really a model either; it's a rationalisation of an observation. It's been shown for many compounds to work, and on that basis it was suggested it works for many more. Also the octet rule is an upper limit.

Yes $\ce{BF_3}$ does exist. In fact its a Lewis acid . Since it doesn't have its octet complete it is called as hypovalent. Also there is Back bonding happening in $\ce{BF_3}$ as Fluorine has electrons to donate and Boron has empty orbital available and the back bonding is of the type $\ce{2Pπ-2Pπ}$ due to which there is partial double bond character and hence it helps in making  boron less electrophilic thereby increasing its stability. Here are some properties of boron trifluoride -https://en.m.wikipedia.org/wiki/Boron_trifluoride

What you said is theroetically possible, But where can I get real existence proof?, I mean, Can you show me the existence of BF3 not in theoretical way.??(like with photos, Composition of a Bf3 compound or anything but theory?)

Please see edited answer

To make your answer better I'd add some of the known physical properties and synthetic routes. Maybe even references to commercial sources. Heck, you can buy it from chemical suppliers.

@Matt_black the OP exactly wants some proof of existance of Borontrifluoride so if you can then please add some proif in the answer


text:
Do carboxylic acids form hydrates like aldehydes and ketones?
Having learned that aldehydes exist in equilibrium with their hydrate form -
Do carboxylic acids do the same in a water solvent and form a hydrate with 3 $\ce{OH}$ groups and one alkyl group?

No, they don't.

@Mithoron Any reason for it?

It breaks conjugation.

Such hydrates are called orthocarboxylic acids https://chemistry.stackexchange.com/questions/75588/why-are-organic-ortho-acids-unstable https://en.wikipedia.org/wiki/Ortho_acid

Let me go against the previous answers, and say that they "do" form hydrates.
What you should consider is that the equilibrium towards such hydrates is extremely unfavorable, making them virtually non-existent.
There have been, anyway, clever ways of "blocking" those forms as orthoesters. If I recall correctly, Corey used a protecting group of the OBO family in the synthesis of a natural product.
A more recent paper, which describes a protecting group of that nature, is this:
https://pubs.acs.org/doi/abs/10.1021/ol047493n?journalCode=orlef7
Anyway, in normal mechanism-writing, that hydrate is not considered, being highly unfavored and unlikely to form.

Ortho esters are well known (i.e. trimethylorthoformate). Ortho acids are not known


text:
Are crown ethers "recycled", typically?
I know that crown ethers have the interesting property of strongly complexing alkali metals, which allows them to be used in organic solvents (eg. as "dissolving-activity enhancing agents" for potassium permanganate or potassium tert-butylate).
Having never used such molecules myself, I was wondering: is there a procedure employed in order to remove the complexed metal, hence allowing a recycling of that ether?
I am sorry if the question might be trivial, but I couldn't find an aswer online.

In my experience it is rare to try to recover chemicals used in the lab. I've only known it to be done with precious metals, silver gold and platinum.

You can always recover the lighter ones by (vacuum) destillation. Most chemistry labs would just scale down the reaction until recycling in no longer worthwhile. But, if you need liters of the stuff, one would sure think about recycling. Mind, worthwile would mean that over a year, it would save thousands of Euros or dollars. Because you already cost ten times as much. ;-)

Given that crown ethers are very expensive (edit: apparently not quite so), I imagine if they were economically recyclable the procedure would be widely divulged, but it doesn't seem to be the case. That said, the Wikipedia article for 18-crown-6 says that it can be dried using NaK, which destroys any water and forms the electride [K(18-crown-6)]Na. This effectively "spends" the crown ether, so it makes no sense to use this process if there weren't a way to recover the metal-free crown...

Crown ethers are typically not recycled because they remain in the organic phase where normally also all the byproducts are (so the purification is very complicated) and they are still complexing the alkali metal after the reaction.
You want to recycle them to again complex some alkali metal, but then you have first to break down the previous complex, which is not straightforward.
To recycle crown ethers you must use a different approach.
The trick is to immobilize them on some solid support.
Typically, you have some crown ether functionalized monomer and polymerize it, or you have some reactive polymer and introduce the crown ether at the end.
A review can be found here.
These solid phase crown ethers are used in the same way as the better known ion exchange resins.
The problem is that you will have to synthesize them, because they are not readily available for purchase.


text:
What could possibly cause a "melting" reaction between vinyl and plastic?
This question was posted in Electronics. Something is really odd about it! What could possibly be causing this "melting" reaction between a common flexible covering (vinyl, apparently) on test lead clips and the plastic there?

a couple of breadboards had been lying in a box with a rubbered alligator wire sandwiched between. They've been there for maybe a month untouched. No other chemical or electricity has been present and only ambient heat (hot summer but 30°C tops here in Sweden). I'm pretty shocked that the plastic of these two different styles of breadboards both got so messed up simply by touching this rubber:

This is a very common alligator clip and very common electronics breadboard, often used together. I'm wondering whether the asker is telling us everything.

Although I understand that you are likely not to have that information, it would be interesting to know which polymers are involved here :)

@The_Vinz Indeed! I was hoping that a chemist here would ask that question there.

Check on the breadboard, there should be a recycling symbol somewhere on its back. Google for the number in the recycling symbol, it tells you the polymer class it's made of, and report here.

It is probably a consequence of the plasticiser not the polymer
It is hard to be sure what has happened here without checking the details of which polymers are involved which the picture can't make clear. But we can make some intelligent guesses based on commonly used polymers.
First: the cable isn't made from natural rubber. Most likely it is made from plasticised PVC. This is a good and cheap cable material with good insulating properties and doesn't decay like rubber from interacting with atmospheric oxygen. This involves a lot of plasticiser, usually something like a phthalate. The cable might contain 10-40% of plasticiser by weight.
Depending on the exact polymer formulation and the specific plasticiser, the plasticiser can migrate into other materials if they are in contact for long periods of time. When I used to work in a lab, for example, the cheap PVC tuning we used in low pressure gas lines would sometimes accumulate a thick liquid in the lower parts of the tubing loops if used for years: this liquid was plasticiser that had migrated out of the PVC over years of use. This migration is worse when other volatiles are in contact with the polymer.
This may be what is happening in this case. In warm conditions the plasticiser in the PVC might migrate and damage other plastics. It isn't possible to tell from the picture what the breadboards are made from but ABS is most likely in a good quality breadboard, though acrylic, polystyrene or polycarbonate are all possible. All of these polymers can be affected by the migration of plasticiser from PVC as this study shows (though it depends on the specific plasticiser). Good choice of materials for the board and plasticiser can minimise this problem but cheap breadboards and cables are likely to have worse problems.
In short, this is the likely problem here.


text:
Is chitin actually protein?
Recently insects are featured as a protein rich source for human nutrition. That humans can really digest chitin through chitinase enzym has been only recently confirmed.
But, does the chitin shell of their body actually count to the protein family?
Wikipedia does not state a clear yes or no. It says it is derivative of glucose, and may be compared by function to keratin protein. I'd really like to exclude any ambuguity and undersrand why.

It is a carbohydrate. See: https://answers.yahoo.com/question/index?qid=20130512114358AArmD5V&guccounter=1

But how does not qualify to be such?

Wikipedia can be a pretty bad place to look for credible information, perhaps the reason college lecturers discourage students to use it. Anyway, strictly speaking chitin is a polysacharride which contains amino sugars (which doesn't necessarily make it qualify to be classified as protein)

To say, "[i]n terms of function, [chitin] may be compared to the protein keratin" does not remotely imply that chitin is a protein. The implication is that despite functional similarities chitin is not a protein. There is no ambiguity.

Chitin and protein are completely unrelated.
The only common thing is that they are polymers.
Chitin is a polymer of amino sugars while protein is a polymer of amino acids.
Both monomers are very different and are not converted one to the other.
In fact, chitin, like cellulose, is not fragmented by animals, so it is not absorbed.
Moreover, chitin is a homopolymer, so even if it where absorbed and transformed to amino acids, it would yield only one (or maybe a few ones) but all the others would have to be obtained through a different source.
So chitin is not a source of protein or amino acids.


text:
Shouldn't all isothermal processes be adiabatic?
We know, $\mathrm{d}H = nC_p\mathrm{d}T$ which implies $\mathrm{d}H = f(\mathrm{d}T)$ using  $\mathrm{d}H = \mathrm{d}U + \mathrm{d}PV$ and $\mathrm{d}q = \mathrm{d}U + \mathrm{d}w$ we get $\mathrm{d}H=\mathrm{d}q$
In isothermal process, $\mathrm{d}T = 0$ implies $\mathrm{d}H=0$  since $\mathrm{d}H=\mathrm{d}q$, $\mathrm{d}q = 0$
for a reaction to be at constant temperature (unless it is a case of free expansion), there must be an exchange of heat, but $q = 0$ here.
Same result can be obtained using $\mathrm{q} = ms(\mathrm{d}T)$
Where is the flaw in my reasoning?

adiabatic means no exchange of heat. isothermal means no temperature change over time for the system. For a reaction involving a release of enthalpy, for isothermicity to be true, heat must be drained. Hence, the reaction cannot be adiabatic. Only isenthalpic reactions can concurrently be adiabatic and isothermal. That makes for a tiny selection of reactions.

The enthalpy change for reacting species is a function not just of temperature. In chemical reactions, bonds are broken and new bonds form.  There is an enthalpy change associated with this as well.  So the enthalpy of the reacting mixture depends both on the temperature and on the amounts of the various chemical species present. For a reaction occurring at constant pressure, the heat Q is still equal to the change in enthalpy of the reacting mixture, even if the temperature is also held constant.

There is a very serious flaw in your reasoning.
You have written that $\mathrm{dH = dU + d(PV)}$, which is absolutely correct but, you should remember that $\mathrm{d(PV) = (PdV + VdP) \neq dW}$, because $\mathrm{dW}$ is only equal to $\mathrm{PdV}$. So, your  logic that $\mathrm{dH = dQ }$ is only true at constant pressure where $\mathrm{dP = 0}$. In general case, it is not at all true. In general, $\mathrm{dH = dQ + VdP}$. So, in Isothermal process, surely $\mathrm{dH = 0}$, which means $\mathrm{dQ = -VdP}$, but in Isothermal processes, $\mathrm{dT=0 = d(PV)}$, which implies $\mathrm{PdV = -VdP}$. So, we have, $\mathrm{dQ = PdV}$ which doesn't say that it should be adiabetic always.

Reactions on earth occur at const. pressure so all the reactions if isothermal are also adiabatic? But isn't this equation only valid for processes Not reactions?

The enthalpy change for reacting species is a function not just of temperature.  It changes with the amounts of the various chemical species.  For a reaction occurring isothermally, the heat Q is still equal to the change in enthalpy of the reacting mixture.

@ChesterMiller I get it but what's wrong in Soumik Das's answer?

@quantised. It doesn't take into account chemical reactions, which was part of the original gist of your question.  Even for an isothermal process, $\Delta H$ is not zero if chemical reaction occurs.


text:
Inorganic reactions related to nascent hydrogen

Nascent hydrogen is prepared by
(A) $\ce{Na}$ and $\ce{C2H5OH}$
(B) $\ce{Al}$ and $\ce{NaOH}$
(C) $\ce{Zn}$ and dil. $\ce{H2SO4}$
(D) All of these

My attempt
I know that option (A) & option (C) are correct option. But don't know about option (C) and in the answer all of them are correct.
Can someone provide the complete reaction about option (B) and its important information such as similar reactions or mechanism?

Why downvote? I don't understand?!

Welcome to Chemistry.SE! Questions are the central focus of this site, but we ask that questions demonstrate an attempt to solve/understand the question to maintain quality.  Please take a minute to look over the help center and tour page to better understand our guidelines and question policies.

Zinc and sulfuric acid react to form nascent hydrogen. From chem-guide.blogspot.com (emphasis mine):

[…] if ordinary hydrogen is passed through acidified $\ce{KMnO4}$ it does not get decolourised. However, if zinc pieces are added to the same solution, bubbles of hydrogen rise up through the solution and the colour is discharged due to the reduction of $\ce{KMnO4}$ by nascent hydrogen.
$$
\begin{align}
\ce{KMnO4 + \underset{molecular}{H2} + H2SO4 &-> \text{No reaction}} \\
\ce{Zn + H2SO4 &-> ZnSO4 + \underset{nascent hydrogen}{2H}} |\times 5 \\
\ce{2 kMnO4 + H2SO4 + 10 [H] &-> K2SO4 + 2 MnSO4 + 8 H2O} \\
\hline
\ce{5 Zn + 8 H2SO4 + 2 KMnO4 &-> 5 ZnSO4(aq) + K2SO4 + 2 MnSO4 + 8 H2O}
\end{align}
$$

See this youtube video for more information.
As for the reaction given in option (B), see previous SE questions: this, this and this.


text:
redox signaling molecules and electrolysis
There is a product whose presentation screams of snake oil, allegedly consisting of nothing but electrically treated salt water. The product is claimed to be chock full of "redox signaling molecules" that have lots of health benefits.
I am not a chemist, and I certainly don't want to promote this (un-named) product; I just want a different perspective on it.  My understanding is that electrolysis of salt water produces some mix of $\ce{H_2, O_2, Na, Cl, NaOH}$, and $\ce{NaClO, NaClO_3}$ depending on the configuration of the reactor, with possible inclusion of other substances due to decomposition of the electrodes.
A search on PubMed turned up several papers describing positive effects (in mice) of consuming water containing dissolved $\ce{H_2}$. I haven't found anything indicating that there are health benefits from consuming the other electrolysis products of salt water.
My question is this:  Could any of the other substances produced by electrolysis of $\ce{NaCl}$ in water ($\ce{O2, Na,Cl, NaOH,}$ and $\ce{NaClO, NaClO_3}$) act as "redox signaling molecules" as claimed by the manufacturer?

Welcome to Chemistry.SE!  Please note that formulas can be better expressed with $\ce{}$ for chemical formulas/equations, $\mathrm{}$ for math term/equations, and $\pu$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies.

@orthocresol, the term "redox signaling molecules" is used in cell biology, e.g.,[https://pdfs.semanticscholar.org/c997/ef50e63b833d60410b6ca0b4e0c838aa930a.pdf], and seems to refer primarily to small molecules containing oxygen, produced by mitochondria and diffusing to neighboring cells.  Haven't been able to find an explicit definition yet.

Without seeing the ad, or knowing the product, I'd say that it is snake oil. The electrolysis of NaCl in water will yield bleach -- hardly a healthful substance to be drinking, unless you're using a dilute solution to disinfect the water.

@S.McGrew you are right and I had deleted my hasty comment - although not before you saw it! Sadly, a google search for redox signalling molecules seems to give more pseudoscience than science...


text:
Understanding energy level splitting of an octahedral complex from spin-orbit coupling, 1st and 2nd order zeeman effects
While studying notes from a couple years ago, I encountered a diagram that I don't have further information on but would like to understand.
While I believe I managed to find sources to guide me through the details of the calculation, there are a few conceptual questions I had, which are summarized at the bottom.
The diagram is for a $d^1$ octahedral complex under a magnetic field:

a) ligand field splitting
b) Spin-Orbit coupling
c) First-order Zeeman splitting
d) second-order Zeeman splitting
a)
The ligand-field splitting is straightforward to me.
b)
The spin-orbit coupling I believe I understand.
The $E_g$ level does not have an electron, and therefore does not undergo any spin-orbit coupling.
For the $t_{2g}$ energy level splitting, I followed the procedure covered on pages 9 and 10 here (the formula is also given on the wiki). The formula for spin-orbit coupling is:
$$\Delta E=\frac{B}{2}[j(j+1)-l(l+1)-s(s+1)]$$
For a $^2D$ term, $l=2$ and $s=\frac{1}{2}$. Then $\Delta E$ would be:
$$j=\frac{5}{2}\therefore \Delta E=\frac{3}{2}B$$.
$$j=\frac{3}{2}\therefore \Delta E=-B$$
I assume in the diagram, $\zeta=B$.

Why does the diagram obtain different coefficients ($-\frac{1}{2}\zeta$ and $\zeta$)? Did I do this wrong?

c)
I think I might have understood this one, using the following link. I'll be referencing equations from this link for these two sections. I'm substituting the Bohr Magneton, $\mu _B$ with simply $\mu$, as it renders better this way, and replacing $B_{ext}$ with $B$ for the same reason.
In deriving equation 8.15 from the link, I am confused what is meant by the line

let the hermitian, unperturbed hamiltonian $H_0^\dagger =H_0$ "act to the left" on $\langle n^0|$

I have never used an operator to "act to the left." I've always applied it toward the right.
Assuming it somehow works out to $E_n^0$, I understand how theorem 8.1 is obtained:
$$E_n^1=\langle n^0|H_I|n^0\rangle$$
They apply the above theorem in equation 8.25 with $H_I=H_z$, where
$$H_z=\frac{\mu}{\hbar}(L_z+2S_z)B$$
to obtain:
$$\Delta E =\mu (m_l+2m_s)B$$
In my diagram above, $m_l=2$ and $m_s=\frac{1}{2}$, so
$$\Delta E = \mu (2\pm 1)B=3\mu  B-\mu B$$
$$= 2\mu B$$
That seems to match the diagram (where $B=H$).

Why does the $E_g$ level split into two double degenerate levels, and the ground state $\Gamma _8$ (4) not split at all?

d)
For second-order splitting, the linked PDF provided this formula (theorem 8.3):
$$E_n^{(2)}=\sum_{m\ne n}\frac{\left| \langle m^0|H_I|n^0\rangle \right| ^2}{E_m^0-E_n^0}$$
I was unsure how this relation was true:
$$(\langle n^0|H_I|m^0\rangle )(\langle m^0|H_I|n^0\rangle )=\left| \langle m^0|H_I|n^0\rangle \right| ^2 $$
but otherwise followed the derivation. Accordingly, the numerator indicates the 2nd order zeeman correction is simply the square of the first order correction; however, the denominator I did not follow, namely:
$$E_m^0-E_n^0=3\zeta$$

Where does this $3\zeta$ come from?
Similar to the first-order splitting, why do the energy levels experience different types of splitting?

summary of questions

In section b), why are my splitting coefficients different from the diagram?

In section c) of the diagram, why is the highest-lying energy level in the diagram split into two double degenerate levels, the middle-lying energy level split into two singly degenerate levels, and the lowest-lying energy level not split at all? I assume the answer to this question will also explain why the second-order splittings in section d) are also inconsistent with each other.

In section d) of the diagram, how does the $3\zeta$ term in the denominator for the energy splitting come about?

What is responsible for certain energy levels being doubly degenerate after following splitting from the Zeeman Effect?

Where does the term symbol splitting according to the projection of total angular momentum along the axis of the magnetic field, ie. $j_z=j,j-1,...-j$ fit in here? Aren't these also separated in a magnetic field?

this might be a bit too broad due to the number of questions. I think they would be better split up and linked together.

To answer part b it is necessary to consider the symmetry properties in an octahedral field. The spin orbit Hamiltonian is $H_i=\zeta\sum_i\bar l_i\cdot \bar s_i$ where $\zeta$ is a constant characteristic of the radial part of the d orbitals. The orbital angular momentum, which may be written in the form (for the $x$ component) as $\displaystyle l_x =-i\hbar \left(y\frac{\partial }{\partial z}-z\frac{\partial}{\partial y}\right)$ (and similarly for $l_y$ and $l_x$ by rotating subscripts) transforms as does the $T_{1g}$ irreducible representation in an $O_h$ field.
If a term has an orbital symmetry $X$ it will only be susceptible to spin orbit coupling if the symmetry product $X\cdot T_{1g}\cdot X$ contains the totally symmetric representation $A_{1g}$. In other notation $\langle i|H |j\rangle \ne 0$ or $\int \psi_i H \psi_f d\tau \ne 0$). Thus the $E_g$ levels are unaffected in point group $O_h$ because $E_g \cdot T_{1g}\cdot E_g$ does not contain $A_{1g}$ but the $T_{2g}$ interact ($T_{2g} \cdot T_{1g}\cdot T_{2g}$ does  contain $A_{1g}$).
Notes: It is assumed that the potential energy effect of the ligands is only small compared to that of the central field of the ion, thus the spin-orbit coupling is a small perturbation on the energies.
The full derivation is complex and involves using double symmetry groups.
The free ions have SO couplings $^2\mathrm D_{3/2} =-3\zeta_d/2$ and $^2\mathrm D_{5/2} =\zeta_d$. (The formula you use if for the free ion).  In the octahedral field the calculation is rather complicated but the result is that $T_{2g}$ levels are split by $3\zeta/2$ in the ratio you show.
In the remaining parts of your question similar symmetry arguments should indicate what levels are split, but only a detailed calculation will produce the energies that they are split by.


text:
Why do electrons in sigma bonds have zero orbital angular momentum about the internuclear axis?
As per the title, why do electrons in σ-bonds have zero orbital angular momentum about the internuclear axis?
As far as I know, the outermost electrons participate in bonding. Suppose that the 3p electrons are the outermost and form a σ-bond. Initially, their individual orbital angular momenta are
$$|\vec{l}| = \sqrt{l(l+1)}\hbar = \sqrt{3}\hbar$$
since they have the orbital angular momentum quantum number $l = 1$. How can this become zero after forming a σ-bond?

The rules to get quantum numbers in diatomics are called "Wigner-Witmer-Rules". The original paper (https://doi.org/10.1007/BF01400247 in german) should have an explanation, I just don't have time right now to read it.

My quantum is slightly rusty, but I think this is how it goes (I would appreciate any corrections or clarifications). We're talking about diatomic molecules here. As you no doubt know, σ-orbitals are invariant about rotation around the internuclear axis: that is to say, you can rotate the molecule as much as you like, but the sigma orbital will still look the same. This is true whether the sigma orbital is made from s, p, or d orbitals.
If we define the $z$-axis to be the internuclear axis and $\phi$ to be the angle around this axis (the azimuthal angle), then we can also say that σ-orbitals have no dependence on the angle $\phi$.

In a diatomic molecule, because the orbital angular momentum (OAM) operator $\hat{L}$ does not commute with the Hamiltonian, the quantum number $l$ is no longer a 'good quantum number': that is to say, the eigenstates of the Hamiltonian (which are the MOs) are no longer simultaneously eigenstates of $\hat{L}$, so the MOs do not have a specific value of $l$ that can be assigned to them. However, for a one-electron molecule (e.g. $\ce{H2+}$) it turns out that $\hat{L}_z$ still commutes with the Hamiltonian, where the $z$-axis is the internuclear axis (or molecular axis). This operator represents the projection of the OAM onto the internuclear axis, and the corresponding quantum number is denoted $m_L$ or $\lambda$ (depending on where you read).
Now, let's denote the wavefunction of the σ-orbital by $\psi$. We know that the eigenvalues of $\hat{L}_z$ are unit multiples of $\hbar$, or in other words, $\hat{L}_z\psi = m_L\hbar\psi$. [Note that this is no different to $\hat{L_z}\psi = m_l\hbar\psi$ in an atom, except that we use a capital L because the context is different, i.e. we are talking about a diatomic molecule.]
It turns out that in spherical coordinates, $\hat{L}_z$ is equal to $-\mathrm i \hbar (\partial/\partial \phi)$. Since we've established that $\psi$ has no dependence on $\phi$, the partial derivative $(\partial \psi/\partial \phi)$ is zero and hence $\hat{L}_z\psi = 0$, meaning that the quantum number $m_L = 0$, as desired.
It doesn't matter what the constituent atomic orbitals are. However, I thought a bit of clarification would be useful. For the atomic p-orbitals, you are correct that these possess $l = 1$. However, note that $l$ is a quantum number which characterises the magnitude of the angular momentum, whereas $\lambda$, or $m_L$, is a projection quantum number. That is to say, it is possible to "add" two atomic orbitals with $l = 1$ and have the projection cancel out.
If you think about the projection of the orbital angular momenta in the original p-orbitals, they are of course $m_l = (+1,0,-1)$. However, these are the complex p-orbitals which are obtained as eigenstates of $\hat{L}_z$. The real p-orbitals, i.e. the dumbbell-shaped p-orbitals that you and I think about when constructing sigma bonds, are either $m_l = 0$ (i.e. the $\mathrm p_z$ orbital) or are linear combinations of $m_l = \pm 1$ (i.e. the $\mathrm p_x$ and $\mathrm p_y$ orbitals). So, it's not at all counterintuitive that when you add these together to form sigma bonds, the projection $m_L$ is simply zero.


text:
How to quickly predict the group of elements with atomic numbers 117 and 120?

The elements Z (atomic number) = 117 and 120 have not yet been discovered. In which group would you place these elements?

I tried to write their electronic configurations but it was too tedious and I do not remember the Aufbau principle to that extent ($\ce{5g, 6f}$...).
Is there any quicker way to solve this question?

Till Uuo (atomic number=118 ) the places in periodic table ate known.So you start placing the 119th one in 1st group and so on so that you the shape of periodic table is not disturbed. But i am not sure

One way you can go about doing this is learning the number of elements present in each period. The first period has 2, the second and the third have 8 each, the fourth and the fifth have 18 each, the sixth and the seventh have 32 elements each. Since the noble gases are present at the end of the period, you can now calculate their atomic numbers which are - 2, 10, 18, 36, 54, 86 and 118 (2+8=10, 10+8=18, 18+18=36, 36+18=54, 54+32=86, 86+32=118). Now you can place Z-117 in the Halogen group(17) just before the noble gas, and 120 in the Alkaline earth metals group(2).

117 has been made years ago, and it even has a name since 2015.

Ditto element 118.

Elements 119 and 120 should be more or less typical alkali and alkaline earth metals. But the increasingly important relativistic effects make things more complicated. There is actually a wikipedia article about elements 119 and beyond: https://en.wikipedia.org/wiki/Extended_periodic_table

Given :
Atomic number z=117
Number of protons=117
electronic configuration is [Rn]5f¹⁴6d¹⁰7s²7p⁵
Belongs to group 17-- Halogen Family
11) atomic number =z=120
number of protons= 120
so, electronic configuration is [Uuo]8s²
element belongs to group 2 ---Alkali earth family

You would place Z = 120 in the second group, Alkali Earth Metals, as that element completely fills the 8s orbital. Z = 117 is Tennessine, which is a group 17 Halogen and has already been discovered.


text:
How does the equilibrium shift when concentration of reactant and product are increased simulatenously
Before answering my question, take a look at this example
$$\ce{2NOCl <=> 2NO + Cl2}$$
Let's suppose I have $\pu{2 mol/L}~\ce{NOCl},~\pu{2 mol/L}~\ce{NO}$, and $\pu{1 mol/L}~\ce{Cl2}$ in a closed system. If we were to calculate the equilibrium constant, it would be
$$K_c = \frac{(2)^2(1)}{(2)^2}$$
Now suppose I increase the concentration of both $\ce{NOCl}$ and $\ce{NO}$ by $\pu{1 mol/L}$. The equilibrium will stay the same,
$$K_c = \frac{(2+1)^2(1)}{(2+1)^2};$$
however, if I were to instead increase the concentration of both $\ce{NOCl}$ and $\ce{Cl2}$ by $\pu{1 mol/L}$, the equilibrium will shift to the product side
$$K_c = \frac{(2)^2(1+1)}{(2+1)^2}$$
Can someone explain this phenomenon in terms of rate of reaction or in terms of molecules?

Can someone explain this phenomenon in terms of rate of reaction or in terms of molecules?

In terms of rate of reaction, at equilibrium the forward rate is proportional to the concentrations of the reactant (as if this were an elementary reaction). Likewise, at equilibrium the reverse rate is proportional to the concentrations of the products (as this were an elementary reaction). If you change the concentration of a single reactant and a single product by the same factor (i.e. multiply by the same number, not add the same number) and if the stoichiometric factors are the same (i.e. the order of reaction in this special case), and the reactants are sufficiently dilute, the system will still be at equilibrium. In the second change described by the OP, neither is the case (one concentration increases by 50%, the other doubles, and the exponents are different).
In terms of thermodynamics, if the Gibbs free energy of the reactants is changed by the same amount (i.e. add or subtract a number) the system will remain at equilibrium. This will happen if one species on either side is changed by the same factor because the Gibbs free energy changes with the logarithm of the activity of a species. Again, this only works if the stoichiometric coefficient of the two species are the same.
Nomenclature K vs Q
In the equations shown by the OP, different concentrations are substituted into the equilibrium constant expression even when the reaction is not at equilibrium. The correct name for these expresssions is reaction quotient Q. If Q matches K, the system is at equilibrium. In the examples provided, Q stays the same in one case (still in equilibrium) and changes in the other case (no longer in equilibrium). What the equilibrium would be needs not to be calculated. Knowing that Q is unequal to K is sufficient for showing that the system is in fact not in equilibrium.

In an equilibrium, the equilibrium constant remains the same.
When you change the concentration of some component, the whole system shifts to new concentrations to maintain this constant.
When one mole each of $\ce{NOCl}$ and $\ce{NO}$ are added,  let us suppose that the equilibrium is shifted to the right and some $\ce{Cl2}$ is formed.
The new concentration of $\ce{NOCl}$ will now be $3-x$, the concentration of $\ce{NO}$ will become $3+x$ and the one of $\ce{Cl2}$, $1+\frac{x}{2}$.
The final concentrations can be calculated by solving the cubic equation:
$$K_c=\frac{(3+x)^2 (1+\frac{x}{2})}{(3-x)^2}$$
Note that if in reality the equilibrium does shift to the left instead of the right, the value of $x$ will be negative.
In the other case, the equilibrium will shift to the left, and the equation to be solved is:
$$K_c=\frac{(2-x)^2 (1-\frac{x}{2})}{(3+x)^2}$$
Again, a negative value of $x$ will mean that the equilibrium shifts to the right instead of going to the left as has been supposed.

When a reaction shifts to the right or left direction, it means that the reaction is trying to remove the equilibrium stress. So removing or adding reagent or product in a reaction at equilibrium will not change the value of the constant.
You said you would increase the amount of substance of $\ce{NOCl}$ and $\ce{NO}$, but you did not add it into $\ce{NO}$ concentration.
Adding products and reagents concentration would change the constant's value, but not shift the equilibrium towards any side.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". Unfortunately your answer is a bit ambiguous, if not wrong. If the equilibrium doesn't shift, the constant's value stays the same, too.

(-1) The equilibrium constant $K_c$ does not change... it is the reaction quotient $Q_c$ that changes.


text:
Are p‐xylylene and fulvene aromatic?

Are the following structures aromatic?

I have not seen this type of question anywhere so I do not know the answer.
This is just a problem that has been roaming around in my head that I may have seen somewhere.
My approach: since aromaticity leads to stability the carbon-carbon bond in both structure should be polar leading to aromaticity like this:

But this is just a thought and I do not know if it is right.
Is any of the two structures aromatic?

Related: How to explain (non-/anti-) aromaticity in fulvene with the help of resonance structures?

First off, the polarized form drawn for fulvene is wrong.  To make this contribution aromatic, draw a positive charge on the pendant carbon and a negative charge on the ring.

To decide if a molecule is aromatic or not you must know some rules-

The Molecule Must Be Cyclic and the system should be conjugated.
The molecule must follow Huckle's rule i.e. it should have (4n+2) Pi Electrons participating in resonance.
3.The molecule must be planar.

For details see-https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/Chapter_15%3A_Benzene_and_Aromaticity/15.03_Aromaticity_and_the_Huckel_4n___2_Rule

Now if you try to draw the resonance structure for your 1st molecule then you will see that 4 electrons are participating in resonance at a time. So the system is not fully conjugated. It will be a cross conjugated system.So both the structures you have mentioned are non aromatic.

Stability is a result of aromaticity. You can not comment that if a compound is stable then it is due to aromaticity.

Here is an image showing resonance in para xylylene.

It can be seen from the resonating structures that some of the resonance structures are not aromatic making the molecule non-aromatic in total.
Any recommendations on making the image better are welcomed.

I still do not understand why the overall molecule in 1st problem is not conjugated. Can you please elaborate on that.

I have edited the answer if its still unclear do let me know.

It is clear now but i think the 3rd structure should be the most contributing structure. Since that structure is highly contributing, shouldn't the molecule exhibit its properties?

Molecule will exhibit the properties of the resonance hybrid and not of any one of its contributing structures.

@Groverkss This will give you the most appropriate idea on it https://chemistry.stackexchange.com/questions/22100/is-the-aromaticity-broken-in-some-resonance-structures-of-para-nitro-aminobenzen


text:
Equilibrium for very small amounts of reactants
Is there any difference in the concept of equilibrium when it comes down to dealing with extremely small amounts of reactants?
Say we have $$\ce{A + B <=> C + D}$$
 and $K_c$ is $100,000$. If we only have $10$ atoms of each $\ce{A}$ and $\ce{B}$ in a beaker, we would expect the container to be totally full of $\ce{C}$ and $\ce{D}$. But $K_c$ is not infinite, so there must be some unreacted $\ce{A}$ and $\ce{B}$. However, if there was even one atom of reactants, the reaction would quickly shift back to satisfy the equilibrium.
Does the concept of equilibrium only apply to macro-sized reactions? Is there an extendable concept for the micro scale?

Equilibrium constants are more related to thermodynamic functions then numbers of molecules.

https://chemistry.stackexchange.com/a/50496/9961

Can you estimate the minimum number of molecules of A and B you should have initially so that, with an equilibrium constant of 100,000, the statistics at equilibrium would be valid?

Thermodynamics describe an average state for a large number of items. However, for a time, that state will vary from that average. So if you take a snapshot of the system at an instant, it would likely not be exactly as equilibrium would predict, yet over time that would be the average state.
"The larger your sample size, the smaller the standard deviation" from the expected value. However, by increasing the number of samples over time, even a small number of molecules would behave as expected.
However, the question is not unreasonable, in that it describes the problem of reconciling macroscopic behavior with quantum effects. Boltzmann explored this issue, and Planck introduced quanta to harmonize statistical thermodynamics with observed black-body radiation.

If you observe a small number of reactants and for a short time then their behaviour will not conform to that of many millions as there will be some variation simply due to the small number of reactants and the order in which they react with one another etc. From statistics we know that  the fewer the number of measurements the larger is the standard deviation of that measurement.
However, if you repeat the experiment on your small number of molecules many thousands of times then the average behaviour is recovered. This means that the time average is the same as the ensemble average and is expressed in the Ergodic Hypothesis. This is sometimes expressed as 'an ensemble of particles (atoms ,molecules) will pass through all the states available to it given sufficient time'.

Thermodynamics is only applicable to the systems of size in between astronomic and microscopic scales. Systems with a small number of particles can spontaneously abandon
equilibrium state due to fluctuations. Statistical physics tells us [1] that the relative magnitude of the fluctuations of thermodynamic quantities $x$ (e.g. pressure, temperature, volume, internal energy, entropy etc.) is inversely proportional to the root of the number of particles in the system:
$$
\frac{
\sqrt{\left\langle\left(\Delta x \right)^2\right\rangle}
}{
\left\langle x \right\rangle
}
\sim
\frac{1}{\sqrt{N}}
$$
Lets arbitrarily set the detection limit for experimental determination of the relative change of $x$ to optimistic value of $10^{-8}$. In this case we would need at least $N = (10^{-8})^{-2} = 10^{16}$ particles belonging to the thermodynamic system before we can speak of equilibrium at all.
References

Landau, L. D.; Lifshitz, E. M. Statistical Physics, 3rd Edition, Part 1: Vol. 5; Butterworth-Heinemann: Amsterdam u.a, 1980.


text:
double replacement
$$\ce{KNO3(aq) + BaCl2(aq) -> KCl(aq) + Ba(NO3)2(aq)}$$
I found products but I noticed in product that both of them are in the aqueous phase. So actually no reaction occurs?
Does a reaction need at least a precipitate to occur to be double replacement?

Yes, precipitates are necessary. Otherwise all ions remain ions, and there is no reaction.

Or some other way that two of the ions are removed together. E.G. 2 ZnS (s) + HCl (aq) -> ZnCl2 (aq) + H2S (g)

Welcome to Chemistry.SE!  Please note that formulas can be better expressed with $\ce{}$ for chemical formulas/equations, $\mathrm{}$ for math term/equations, and $\pu$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies.

@A.K., I like welcome comments, but this meta post is probably more helpful than the hidden points of editing one. Not all maths should be in \mathrm{} too, although there can be subtleties there...

The reason the reaction you drew isn't a double replacement reaction is that the aqueous compounds listed don't have their ions actually bound together in solution, but instead the individual ions are solvated by shells of water molecules. If you write the net ionic equation for your reaction, you get $$\ce{K+(aq) +NO3-(aq) + Ba^{2+}(aq) +2Cl-(aq) -> K+(aq) +NO3-(aq) + Ba^{2+}(aq) +2Cl-(aq)}$$
where you can see that we haven't actually created any new compounds in going from reactants to products.
If one of the products were a precipitate, it then becomes a true reaction, as you start with four kinds of ions in solution, and end with a solid formed from two of the ions and a solution containing the other two.


text:
Gibbs Free Energy: Exothermic, low temperature & reduced system entropy - such reaction is spontaneous or not?
According to $\ce{\Delta G^\mathrm{o}= \Delta H^\mathrm{o} -T\Delta S^\mathrm{o}}$, if a reaction
a) is exothermic,
b) reduces system entropy, and
c) the temperature is low,
then it is spontaneous.
Consider the combustion of hydrogen gas, with $$2\ce{H2}+\ce{O2}\to 2\ce{H2O}.$$ This reaction is exothermic, and the system's entropy decreases. So does it mean that:
1) this reaction is spontaneous at low temperatures?
2) this reaction is not spontaneous at high temperatures?
If so, it seems counter-intuitive that placing the reagents in a high-temperature environment will not cause spontaneous combustion. Can you explain why this should be the case?
I really appreciate your help!

I no expert in thermochemistry, but from the equation, low temperature means a neglictible impact of  −TΔS on the free energy. So if the reaction is exothermmic, the reaction would be spontaneous.
At high temperature, the kinetics are higher, but the entropy loss of the conbustion would be a lot more important considering the second part of the equation.

Yes, this leads to my 3rd question: seems unintuitive that under High Temperature, the loss of entropy is amplified by High Temperature so much that the reaction is not spontaneous (because it seems that in High Temperature, this combustion should occur with even more ease). Any idea?

Why don't you just use the van't Hopf equation to calculate the equilibrium constant for the reaction of hydrogen and oxygen to produce water as a function of temperature to see quantitatively how this plays out?

Even at 1000 K, the equilibrium constant is still huge.


text:
How should you name the solid-solution compound ScY(SeO4)3?
How should you name the compound $\ce{ScY(SeO4)3}$? There are 2 cations, and I know that $\ce{SeO4}$ is a polyatomic ion with $-2$ charge. However, I am unsure how to proceed.

This is similar to a double salt, but because of the chemical similarities between scandium and yttrium, $\ce{ScY(SeO4)3}$ is a solid solution in this case. And the answers is you just name the cations in alphabetical order, then the anions like any normal salt. Thus the name is scandium yttrium selenate.
If the ratio between cations had a large difference such as $\ce{Sc_{1.9}Y_{0.1}(SeO4)3}$ It might be called yttrium-doped scandium selenate, but could still correctly be called just scandium yttrium selenate too.


text:
What's the difference between FFKM and PTFE?
These two fluorocarbon polymers seem to me very similar, the only difference I see is that FFKM is elastomer while PTFE aka Teflon is more of hard type of plastic.
Isn't FFKM just PTFE with lower average molecular weight, that is the average molecule is shorter chain so its softer? Is the production of these two polymers vastly different or is the process mostly same?

Teflon (PTFE) and Kalrez (FFKM) are different polymers.
PTFE is made from tetrafluoroethylene, and contains only carbon and fluor.
The material is stiff.
FFKM is made from tetrafluoroethylene, vinylidene fluoride and perfluoromethylvinylether, so besides carbon and fluor, it also contains hydrogen and oxygen.
These additional monomers make the polymer an elastomer.

Is it possible to make elastomer from just carbon and fluorine atoms?

I am not sure. Parofluor, another trademark FFKM polymer, does not use vinylidene fluoride, so it does not contain hydrogen. But oxygen is still there. There are elastomers made with only carbon and hydrogen containing monomers so, theoretically, it should be possible by using the equivalent perfuorinated compounds.

Kalrez is much more common in industrial applications.

FFKMs don't contain any hydrogen. FKMs does. TFE/HFP (FEP) copolymers aren't elastomeric due to the high Tg (by far above room temperature)

Would you mind spelling out the acronyms?

PTFE stands for PolyTetraFluoroEthylene. FFKM isn't an acronym. Its definition is given by ASTM D1418: "Perfluorinated rubbers of the polymethylene type
having all fluoro, perfluoroalkyl, or perfluoroalkoxy substituent
groups on the polymer chain"


text:
How does protium fusion produce helium?
In stellar nucleosynthesis, how is helium produced from a protium atom that has no neutrons?

It is a good question, but I think it might be better off being regarded as nuclear physics. Nuclear chemistry is chemistry of radioactive and nuclear materials and some other things. It has been argued that it is "the study of radioactivity using chemistry and the study of chemistry using radioactivity".

In the process of stellar nucleosynthesis, the helium comes from deuterium ($\ce{^2H}$) produced through the Proton–proton chain reaction (P-P 1).
In P-P 1, two proton ($\ce{^1H^+}$) atoms fuse to form one deuterium atom, as well as a positron ($\ce{e^+}$) and a electron neutrino ($\ce{v_e}$).

One deuterium atom and one protium atom fuse, forming one helion ($\ce{^3He}$) atom and releasing a photon as a gamma ray ($\ce{Y}$).

This is followed by the fusion of two helion atoms, creating an alpha particle ($\ce{a^2+}$ / $\ce{He^2+}$).

Here's a diagram from Wikimedia Commons depicting the process:


text:
Why is FEP transparent and PTFE not?
fluorinated ethylene propylene (FEP) is close relative of poly(tetra-fluoroethylene) (PTFE), both only contain fluorine and carbon atoms,both have very similiar properties yet FEP is relatively transparent while PTFE is opaque, why?

Otherwise clear plastics lose their transparency from crystallization of the polymer.  The crystalline phases have different refractive indices and when a sufficient number form the light becomes so scattered that the material is no longer transparent. Note this is an extensive property and if a material is thick enough it will become opaque, while a thin film of the same material could be translucent.
FEP lacks the regularity in polymer structure that PTFE has, and thus does not crystallize as readily. Since the crystallite density is lower, the polymer is more amorphous and the medium does not scatter as much and thus appears more transparent.

The transparency of a plastic does not depend on its nature but on how it is made.
To get a transparent plastic it must be cooled very fast when it is in molten state.
If the cooling is too slow, the polymer begins to order it self in more or less crystalline domains.
When it is not isotropic anymore, it is also not transparent.

I cant imagine what non-isotropic plastic is.What is the difference between normal isotropic and non-isotropic plastic?


text:
How do we know that we have isolated a substance?
In The Historical Background of Chemistry by Henry M. Leicester page 131, the author says:

Carl Wilhelm Scheele (1742-1786) [...] isolated many organic compounds, including hydrogen cyanide, lactic, citric, and malic acids, and glycerol.

But how did Scheele know that those substances were isolated?
According to me, we can’t be sure of that without doing some very high-tech analyses.

Who says he was sure?

There is another consideration. What "pure" meant in 1780 versus what "pure" means today. Even today we can't isolate absolutely pure compounds.

@Mithoron leaving the philosophical debate on certainty aside, one has to be sure enough to claim to have a substance isolated. the answer is in @MaxW‘s comment and the answer: the chemical rigor was much less and there was still sufficient evidence.

While modern tools make this easier, there are plenty of physical characteristics that help identify that a substance is pure
Modern tools like NMR, X-ray structure determination and others make identifying pure substances and their structures relatively easy, there are plenty of other characteristics than can support the identification of a pure substance.
Pure compounds tend to have sharply defined boiling points or melting points, for example. So if you can generate crystals, that is a good start. Crystallisation from solution, for example, tends to yield relatively pure compounds and this can be checked by measuring the melting point. Repeated fractional crystallisation can confirm this if the melting point doesn't change. For liquids you can distill them and observe the boiling point. Fractional distillation can separate mixtures and you can check that the results boil at a single temperature. Gases also have physical properties that can be checked after fractional distillation or other separation techniques.
Here is an example where attention to detail yielded the discovery of a new element. Rayleigh and Ramsay were experimenting with the separation of gases and noticed that nitrogen isolated from air (by removing all the other known gases) seemed to have different properties to nitrogen formed from chemical reactions (it was about 0.5% denser, for example). By careful fractional distillation, Ramsay was able to isolate a much denser gas from the nitrogen which we now call Argon. This was verified to be an element by observing its discharge spectrum which had new lines in it not present for other elements. So careful separation techniques and carful observation of physical properties could yield a new substance with different properties to nitrogen.
Before we had modern instrumentation this careful application of known chemical and physical techniques was the basis for the development of the laws of chemistry and the identification of new pure substances. It takes more time, but it works.

Another example for a discovery by attention to detail: Rainer Ludwig Claisen discovered his well known rearrangement when he tried to find the melting point of his newly synthesized crystals of allyl 1-naphthyl ether (1-allyloxynaphthalene).


text:
Why is cooking not potentially lethal?
Even the simplest ingredients used in cooking, such as a banana, contain hundreds of chemical compounds. When we cook we mix several of these ingredients and apply heat, which makes a perfect environment for chemical reactions between the various compounds. Actually cooking is the art of initiating the right chemical reactions.
For a typical recipe there are billions of possible reactions between the initial compounds and there are many possible recipes with different initial compounds.
Why are none of the chemical products of this vast space lethal enough to harm a human? In other words why aren't there ingredients that must not be mixed?

Our bodies are also made of thousands of chemical compounds. How is that not lethal?

@IvanNeretin Our body evolutionary evolved to work. Read my question carfully, It is not about the amount of ingredients. It is about the vast space of possible chemical products in cooking due to the number of compounds involved, i.e. that a single toxic product could be created.

Well, then consider this: we also evolved to consume cooked meat. Then again, why would a random product be acutely toxic? Most chemicals aren't.

Life as we know it on Earth has a fairly limited palate of chemistries. And, those chemistries are shared broadly across all of life. While some organisms have evolved to create hazardous materials (cyanide, snake venoms, ...), cooking tends to break up complex molecules into simpler ones. And, often these simpler ones are the building blocks used by all life.

Those molecules may not be that reactive as you think, also, the main reaction pathways are the degradation of macromolecules, falling apart to their units

Doing cooking wrongly is most certainly lethal. Cooking often kills pathogens in food (many people are poisoned by undercooked chicken). It is also essential for some foods because it breaks down poisonous ingredients in the raw material (cassava has to be thoroughly treated and cooked or you will be poisoned by its cyanide content; red kidney beans that haven't been boiled will also kill you).

Cooking is potentially lethal.
Just like how going out for a walk is potentially lethal (You could get run over by a car, maybe develop a stroke, or get hit by a meteorite).

Cooking or cookery is the art, technology, science and craft of preparing food for consumption.
 From this Wikipedia article on Cooking

The basic idea behind cooking, as a whole, (regardless of the actual physical process) is to either make certain inedible (or indigestible) items  edible (digestible), or otherwise unsavory items more palatable.
Sometimes we have toxic substances that are rendered "safe", through the reactions that the cooking process affords. Other times, the opposite happens.

Now, does cooking (boiling, frying, baking) something generate toxicants?
Almost certainly yes. But the amount and type of toxicant(s) produced is what determines the overall lethality of the item you've cooked.
The potato chips/fries we all hold so dear almost invariably contain acrolein and acrylamides, however in trace amounts. If, you chose to let the potatoes simmer for a little too long, then you'll better be able to taste the wrath of those substances.
It is the dose, that maketh the poison (as I've discussed in a previous answer).

The cooking process does, as you point out, initiate a gamut of reactions: Some simply break down the food into products you can better digest, absorb and assimilate... while others lead to products that aren't as useful.

During cooking many harmful compounds are formed, but they are not immediately lethal.
However, long exposure to the same harmful compounds can be lethal in the end.
Therefore it is important to have a broad food basis, with many different raw materials and different cooking procedures.
In that way, you are less exposed to the same harmful compounds all the time.
The alternative is to not eat anything, but that is more lethal in the short term.


text:
Are polyhydroquinones considered to be polyphenols?
Looking at the Wikipedia article for polyphenol, I was struck by how the monomer units shown are all substituted meta or ortho for the hydroxyl groups.
https://en.wikipedia.org/wiki/Polyphenol#Chemical_structure_and_synthesis
I assumed, naively, that polyhydroquinones would be considered polyphenols, but I am uncertain now. From what I can see, there is no IUPAC definition of polyphenols, so I am interested to know what commonly accepted criteria (and what contentious aspects, if any) are there for polyphenols.

The WBBSH definition states that polyphenols are defined as:

generally moderately water-soluble compounds
with molecular weight of 500–4000 Da
with >12 phenolic hydroxyl groups
with 5–7 aromatic rings per 1000 Da

This is the most widely accepted definition. However, due to the increase in research, there is a proposed Quideau definition stating
“The term "polyphenol" should be used to define compounds exclusively derived from the shikimate/phenylpropanoid and/or the polyketide pathway, featuring more than one phenolic unit and deprived of nitrogen-based functions.”
However, this is not an official definition (yet)

Sorry for the Wikipedia link, I have yet to follow it to its source

I read all these criteria on the Wikipedia page as well, but yeah, I don't know why they are picked as the criteria for polyphenols when it seems like a larger class of chemicals would fit...

@selkie222 apparently it links to a book by Haslam himself that I do not have time to check out from my university library


text:
Why does water diffuse from a lower solute concentration to a higher one?
I'm studying biology and this came up in the context of passive transport. I understand why, in general a gas moves down a concentration gradient (the random movement of gas particles and probability). But I don't understand why the concentration of some solute (ions, salt, etc) would affect the movement of water. Why does water move from an area with a lower concentration of solute to an area with a higher concentration of solute?

In general chemical reactions have a forward reaction and a backwards reaction. The equilibrium is established where the two rates of reaction are equal.

The driving force is the same. Usually, in the cases you're describing, the solute is prevented from moving, so the water moves instead. Whereas in the general case, both the solvent and the solute freely disperse to provide a relatively uniform concentration.

It might help to think of a lower concentration of water in the area of a higher concentration of solute

Think of it this way: water moves both ways, but there is more of it on one side, hence the flow from that side to another will be greater than the reverse flow.

There are two possibilities. The first is that it is simple dilution, but for more likely is that osmosis is occurring. In this case the concentrated solution and the water, or a  dilute solution, are separated by a semi-permeable membrane, i.e. one that allows passage of small water molecule but not large solute ones.  To lower the overall energy (as represented as the chemical potential) water moves through the membrane from the dilute solution to the more concentrated one. In a cell this can lead to the cell swelling and possibly bursting if left uncontrolled. There is some more explanation in the answer to this question Entropy as the driving force for osmosis


text:
Pigmentation of aromates
So, they try to color naftalene, nitrobenzeen and benzene. Only one of them colors. They can say with certainty that this is nitrobenzeen. Right/wrong and why?
Sorry if my English isn't that great, not my native language!!
Thanks a lot x

It is indeed nitrobenzene.
You really should look at the compounds, literally, because chemistry is ultimately an experimental science.
That being so, you can still note that the conjugated nitro group with its electronegative atoms gives relatively low-lying vacant orbitals into which electrons can be excited.  These orbitals can be accessed by absorbing short wavelength (violet/blue) visible light leading to a yellow color.