text:
Do alkali metals react with ammonia to form hydrogen gas?
I was recently studying about s-block alkali metals and I found the following reaction:
$$\ce{\underset{alkali\\ metal}{M\phantom{_y}} + \underset{solution}{$(x +y)$ NH3} -> \underset{ammoniated\\ metal cation}{[M(NH3)_x]+} + \underset{ammoniated\\ electron (blue)}{[e(NH3)_y]-}}.$$
It is also known that nitrogen has electronegativity of 3.0 on Pauling's scale. So, the hydrogen attached to it is an active hydrogen. Also, it shows H-bonding, so the hydrogen is an active-H.
Alkali metals react with active-H compounds to release hydrogen gas. But the reaction that takes place here is in contradiction to the aforementioned fact.
Do we have an explanation for this? Or is it a  mistake on my part?

What you have written is right, but it is just the first step.
The release and solvation of an electron is fast. Reaction of an electron with ammonia to release hydrogen $$\ce{2 e-(solv) + 2 NH3 -> H2 + 2 NH2-(solv)}$$ is slow. Deep blue (diluted) or bronze (concentrated) "sodium electride" solution in liquid ammonia slowly converts itself to the colorless ammonia solution of $\ce{NaNH2}$.

For the water case, all is extremely fast, as water is a much stronger acid than liquid ammonia. Blue-like color can be catched by 10000 fps camera, what is commented in SE  Chem Q - Reaction of potassium with water
BBC article, referring the Nature article by Pavel Jungwirth and colleagues (ref.1),

ABSTRACT: Alkali metals can react explosively with water and it is textbook knowledge that this vigorous behaviour results from heat release, steam formation and ignition of the hydrogen gas that is produced. Here we suggest that the initial process enabling the alkali metal explosion in water is, however, of a completely different nature. High-speed camera imaging of liquid drops of a sodium/potassium alloy in water reveals submillisecond formation of metal spikes that protrude from the surface of the drop. Molecular dynamics simulations demonstrate that on immersion in water there is an almost immediate release of electrons from the metal surface. The system thus quickly reaches the Rayleigh instability limit, which leads to a ‘coulomb explosion’ of the alkali metal drop. Consequently, a new metal surface in contact with water is formed, which explains why the reaction does not become self-quenched by its products, but can rather lead to explosive behaviour.

with high speed (cca 10000 fps) camera photos shows the blue colour near 0.3 ms after the metal/water contact (ref. 2). It may be rather more toward violet, it seems to me. The abstract of ref. 2 explains the blue color:

Alkali metals in water are always at the brink of explosion. Herein, we show that this vigorous reaction can be kept in a non‐exploding regime, revealing a fascinating richness of hitherto unexplored chemical processes. A combination of high‐speed camera imaging and visible/near‐infrared/infrared spectroscopy allowed us to catch and characterize the system at each stage of the reaction. After gently placing a drop of a sodium/potassium alloy on water under an inert atmosphere, the production of solvated electrons became so strong that their characteristic blue color could be observed with the naked eye. The exoergic reaction leading to the formation of hydrogen and hydroxide eventually heated the alkali metal drop such that it became glowing red, and part of the metal evaporated. As a result of the reaction, a perfectly transparent drop consisting of molten hydroxide was temporarily stabilized on water through the Leidenfrost effect, bursting spectacularly after it had cooled sufficiently.

References:

Philip E. Mason, Frank Uhlig, Václav Vaněk, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, “Coulomb explosion during the early stages of the reaction of alkali metals with water,” Nature Chemistry 2015, 7, 250–254 (https://doi.org/10.1038/nchem.2161).
Philip E. Mason, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, “A Non‐Exploding Alkali Metal Drop on Water: From Blue Solvated Electrons to Bursting Molten Hydroxide,” Angew. Chem., Internat. Edn. Engl. 2016, 55(42), 13019-13022 (https://doi.org/10.1002/anie.201605986).

So on reaction with water, aquated electron forms that reacts extremely fast with water to reduce it to Hydrogen?

Yes. Superfast cameras are even able to catch the blue.

@poutnik now I have a followup question.

@Oscar Lanzi The answer has been updated.

There's a nice picture at https://www.chemistryworld.com/podcasts/solvated-electrons/2500348.article

@Poutnik: My editing included the additional reference and its abstracts to emphasize the expected blue color. If you don't like is you may delete the addition.

Thanks Mathew, you improvement is appreciated.

What I see is the electrons showing a range if colors from bluish green to violet.  Compare with the range of colors seen here, suggesting that the alkaline-earth intercalates shown in that reference also have solvated electrons in the graphite.

Is water an acid? Isn't it by definition neutral?

It is neutral, it is an acid, releasing a proton,  and a base, capturing a proton.


text:
Chemical accuracy and room temperature
Chemical accuracy is $\pu{2E-2 eV}$ but room temperature $\pu{20 ^\circ C}$ is $\pu{2.5E-2 eV}$, so why is it not necessary to take into account thermal effects in DFT? I understood (maybe wrongly) that chemical accuracy is the precision we want on electronic energies for one atom or one molecule.

$\mathrm{k_B T}$ at room temperature is about $\mathrm{25\ meV}$.

Right, sorry ! still it's the same order of magnitude than chemical accuracy... so why it's not necessary to take into account thermal effects in DFT ?

About precission and accuracy, see this

DFT can be used to describe electronic structure (and more). It is used nowadays with different functionals, which I like to call density functional approximations (DFA). Most of them have errors beyond chemical accuracy. And there are further approximations, which could handle  thermal effects. I'm not quite sure whether I understand the question right, maybe it would be helpful if you could add some more context. In any case: Welcome to Chemistry.se!

Thanks ! It's because I read that some DFT approximations reach chemical accuracy, but I don't understand how it's possible since they don't take into account thermal effects while temperature energy is close to this precision. For instance let us think of computing the ionisation energy of an atom with between 10 and 20 electrons.

What is chemical accuracy, and why is it 20meV?

source: Stack Exchange

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

sort: votes


text:
How can I draw alpha helices in ChemDraw?
I am trying to draw a polyisocyanopeptide with different kinds of monomeric units that intramolecularly interact with one another. However, I wanted to show these connections with its carbon backbone in the helix configuration.
An example of what I mean is shown below, taken from Kouwer et al (Nature 2013, 493 (7434), 651–655). How can I draw the alpha-helix as shown in red in ChemDraw (or any other program)?

I use ChemDraw Professional 19.1.1.32. If you follow  File>Open Templates>Advanced BioDraw, you will find the black helix that I have reoriented from horizontal to vertical. The red helix was enlarged and colored red. I hope this is of help.

Alternative arrows: → (Markdown / HTML) and → / ► (Unicode)

This illustration may be a combination of two images, or simply a program I don't know about. However, there are many (3D) protein visualisation programs that can show alpha-helices such as PyMol, VMD or Yasara. Here you can ray-trace the image (i.e. transparent background) and then combine the illustration with something else, e.g. an illustration from ChemDraw to produce something like the illustration shown in the question.
If you don't have a protein to visualise in these programs, you can simply download one (with alpha helices) from the PDB database.
Example of a helix visualised in PyMol:

This answer is for ChemDoodle, but I think you can import SVG files into ChemDraw as well.
Just take an SVG image of an alpha helix and copy and paste it into ChemDoodle. Or go to File -> Insert Image and select the SVG file from your computer. It doesn't have to be SVG, but they have advantages with scaling.
Made this using this File from Wikipedia, but if you have an SVG editor, you can make your own.

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


text:
surface energy versus surface tension and contact angles
I am hoping to measure the surface tension between two fluids (salt water and isopropyl alcohol), by measuring the appropriate contact angles, and applying Young's equation: https://en.wikipedia.org/wiki/Wetting#Simplification_to_planar_geometry.2C_Young.27s_relation
As far as I can tell, the surface energy and surface tension are interchangable, however I thought that I should ask people that know better than I do, just to make sure.

Nope, both are very different things.

So, would the above measurement, together with a capillary measurement suffice to calculate the surface tension?

Surface energy has dimension J/m^2, surface tension has dimension N/m. Their relation is the famous high school formula Work = Force . distance. A liquid with surface tension e.g. $\pu{70 mN/m}$ has surface energy $\pu{70 mJ/m^2}$

Interfacial tension between two liquids implies that they are immiscible. I don't think you will get a distinct interface between isopropanol and an aqueous salt solution.

@James Gaidis I generally agree, but I had suspician isopropanol can be salted out, similarly as some other alkohols are. This amateur youtube video seems confirming the idea, resulting in 2 clearly distinguished phases..

@Poutnik Yes! That is exactly the idea! I am studying the interface waves on a two-fluid (liquid) system, and I would like to minimize the surface tension between the two fluids. Currently, I am using pure water and propylene carbonate, but the interface has a tendancy to create very tiny bubbles.
I am hoping that a dynamic equillibrium between the alcohol and the water prevents this problem. As well, propylene carbonate is a pain to work with, since most plastics seem to fall apart when exposed to it.
paper:
https://onlinelibrary.wiley.com/doi/full/10.1002/jctb.447

@Poutnik
Regarding the above example: I am tempted to say that the two quantities are the same, since the dimensions are the same, and they are related to very similar phenomena. Would you expect a capillary measurement to produce different results than the aforementioned Young equation?

Energy and torque have the same dimension as well, but they are not the same. The partial derivative of Gibb energy per surface area is surface energy, not surface tension.

@Poutnik
I see... (Good example with torque and energy by the way!)
But, what do you mean by saying that a liquid with surface tension 70 mN/m has a surface energy of 70 mN/m?

@user109527 Rather, what do you mean by that ? :-). I have written "A liquid with surface tension e.g. 70 mN/m has surface energy 70 mJ/m2". There are scenarios where one is interested in the force, and other scenarios, where one is interested in the energy. It makes more sense, if we leave simple linear and rectangular scenarios for general surfaces.

Let us continue this discussion in chat.


text:
ATP chemical structure: phosphate groups with Oxygen anions or with OH groups?
I've googled repeatedly (with multiple search queries) to try and get an answer to this question but
nothing comes up.
Embarrassingly, I seem to be the only person who is confused by:

ATP with oxygen anions.

ATP with $\ce{OH}$ groups.

I looked at the $3$D view on PubChem  - when I click 'show oxygens' they are indeed  present as per image 2.

The chemical formula for ATP is $\ce{C10\underline{\mathrm H_{16}}N5O13P3}$: $16$ hydrogens, which means image $2$ fits the bill better than image $1$.
However, image 1 is more commonly found in tutorials about ATP.
Which representation is it best to go with?

It will depend on the pH of the solution

If the pH is neutral, would image 1 (with oxygen anions) be the correct one?

well you also have a basic centre present so at neutral pH it is likely a zwitterion

I'm rather new to all this. Your terminology is way over my head. Let me rephrase my question. I'm trying to get my head round the ATP cycle in a cell. Which of the two diagrams would be the most helpful to my understanding?

All 3 of the OHs are acidic. The adenosine heterocycle is basic so it will pick up a proton from one of the OH groups at neutral pH    https://en.wikipedia.org/wiki/Zwitterion

What pH would image 1 be at?

That would be the case in basic solution beyond pH 12.

If your goal is to understand the ATP cycle in a cell, this level of detail is absolutely irrelevant and a distraction. Acidic protons can move around freely in aqueous solution, so you can include them in drawings or not. It makes no difference in the big picture of the function.

Yes, I understand that now. This question is pretty old and I've moved on since I posted it.

in vivo : ATP with oxygen anions.
in vitro : Depends on pH of solution
IN VIVO :
DNA = Deoxy Ribonucleic Acid.
DNA is acidic , because its monomers contain three phosphates bound together and this so-called triphosphoric acid is extremely acidic.
In living organisms, DNA is always accompanied by histones(positively charged proteins) or sometimes by Mg+2 complexes. Therefore the dNMPs ( deoxyribo Nucleoside Mono Phosphates) that constitute the genetic material , are negatively charged .
For rNMPs (ribo Nucleoside Mono Phosphates) the negative charged form is also more preferable.
Generally speaking in living creatures , ATP = -4 , ADP = -3 , AMP = -2 charged.
(It is worth mentioning that Pi which is a by-product in cellular energetics is HPO4(-2). )
You can see in Molecular Cell Biology Lodish et al 8th ed section 12.1 that the overall reaction of cellular respiration is:

Notice the charges.
IMPORTANT : Sometimes during the metabolism of these compounds Proton(H+) may be added to oxygens of phosphate groups only because of the mechanism of the reaction which is temporary. But usually the charges are as I mentioned before.
IN VITRO :
If you purify ATP in a solution isolated from the cell, the number of acidic hydrogens attached to the molecule is related to the pH of solution. you can predict the molar concentration of each chemical species the same way you predict the [H3PO4] , [H2PO4 -] , [HPO4 2-] , [PO4 3-] , [H+] in a solution of H3PO4 and H2O (a typical analytical chemistry question) Either by a complex utilization of Henderson–Hasselbalch equation or by Charge and Mass equations.
Also Look at this awesome picture wiki

You are mixing up ATP with monophosphoric nucleosides and thoroughly ignoring adenine's basicity which may be ignored in DNA, but should not be in ATP.

@Mithoron thank you , I'd totally forgotten about that. Still that wouldn't change anything. Pka for H3PO4 is 2.14 ,7.20 ,12.3 . It will be even more acidic in (PO4)3 form, while A/T/C/G base are aromatic bases , which as you know have weak basicity.

Weak, but many orders of magnitude stronger then water. What you picture would be in much more basic solutions then found in vivo - otherwise one, or even two nitrogens would be protonated. Also I quite doubt H-H equation would do much good in cases like this. On the bright side, as OP recently mentioned all this isn't really important in practice.


text:
How does the position of hydroxyl group in a nucleotide monomer affect the dehydration synthesis of nucleotides?
In dehydration synthesis of nucleotides, the hydrogen atom from the 3' carbon on the deoxyribose sugar of one nucleotide reacts with the hydroxyl group on the phosphate group of another nucleotide to form water. As the water molecule forms, a new covalent bond comes into existence between the two nucleotides.
Hydroxyl group can be in different positions within the phosphate group, for example:

I'm wondering if different positions of the hydroxyl group have an impact on the dehydration synthesis between nucleotide monomers. For example, when the hydroxyl group is at the bottom, will the dehydration synthesis occur at all? Also, when the hydroxyl group is on the left, does the spatial structure of the sugar-phosphate backbone have a spatial shape different from when the OH group is on the top?

No, because there are all interchangeable on a nanosecond timescale. These are just different 2D representations of an equilibrium situation.

This is not the reaction that happens in cells. You go from a triphosphate and a hydroxyl to a phophodiester and a pyrophophate. That’s not a dehydration.

The phosphate ion is of vital importance in biological systems. Its molecular geometry is tetrahedral. Thus, based on the $\mathrm{pH}$ of the system, hydroxy groups are inter-changeable with $\ce{P=O}$ form and hence all three remaining $\ce{O}$s are equivalent (see Waylander's comment). Therefore no impact on the dehydration.
However, the stereochemistry of the hydroxy group on the 3'-carbon on the deoxyribose nucleus of the nucleotide has an impact on the enzymatic dehydration process. For instance, if you change the original $(S)$-orientation of 3'-hydroxy group to $(R)$-orientation, DNA-synthase would not pick that nucleotide to synthesize intended DNA sequence.


text:
What is the influence of citrate on PEG-modified polyacrylamide hydrogels?
I've been having a puzzle explaining the experimental results with the citrate anions and PEG modified polyacrylamide:
According to the experimental data there is a hydrogel swelling whenever citrate is added to a 10mM PBS solution. However, according to the Hofmeister series, citrate is a well known salting-out agent, which would absorb water, thus make the hydrogel contract. In addition that, the swelling is diminished when the concentration of salts (ionic strength) is increased in a bulk solution. Any ideas?


text:
Why is 1,2,3,4,5,8-hexahydronaphthalene more stable than 1,4,4a,5,8,8a-hexahydronaphthalene?
Why is 1,2,3,4,5,8-hexahydronaphthalene (1) more stable than 1,4,4a,5,8,8a-hexahydronaphthalene (2)?

A follow-up question is about stability being estimated by counting the α-hydrogens.
In the first case, do we consider α-hydrogens individually for each double bond, or are common hydrogens only taken once?  If it is the former, then the answer would be 12, or else 8. Which one is correct?
If it is the latter, both have an equal number of hyperconjugating structures, i.e. H = 8. Therefore the stability of both must be equal. Is this wrong?

Symmetry? (Pretty weak argument though..)

It is not your fault, but these questions are really, really big garbage. There are zeroth order approximations, which are taught as some kind of holy grail of organic chemistry, like counting hyperconjugation contributors. This is not much more than a guess though and works accidentally some times. Also, stability is not an absolute concept; without a proper definition of the baseline, you can possibly generate any number of rankings. In this case, it is probably a good idea to ask: Which of these compounds has the higher hydration enthalpy; but you'd not use hyperconjugation for an estimate.

I think that the question should be more specified. (specific values indicating stabilities of two compounds, phase etc.)

@Patche this question came in my test.And high school organic chemistry that discuss much about mathematical values .

@Patche dear sir can you please confirm that 1st compound has $8$ hyper-conjugating H or $12$. I mean $12$ if we consider $2$ H individually wrt. to each double bond Carbon.

There is a cis/trans issue with the second structure. You might explore the heats of hydrogenation/combustion of hexenes that bear the substitution of the double bond. Welcome to ChemSE!

@Safdar I'm really sorry. but can you please tell me that  1,2,3,4,5,8 hexahydronaphthalene has 8 or 12 hyperconjugating H

@Newton Please don't use the comments section for extended discussion. I understand that you want to have an answer, but being patient is key here. This is not a discussion forum with a back-and-forth conversation style. Please take the [tour] to learn more about how this site works. As I previously said, the entire question is garbage and I am very sorry that you - a student - have to endure these terrible guessing methods. That being said, I will not endorse any count of any hyperconjugation contributors, it simply is not a reliable approximation and oversimplifies the chemistry behind it.

@user55119 Dear sir, can you give me any link or source where i can know more about.Simple Google search didn't gave me any satisfactory result.

Try this for a starter. http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/Heat.pdf. With some  effort on your part, I'd prepare an answer for you.

Number of hyperconjugating structures for a pi-bond is the same as total number of alpha-hydrogen that are associated to it.
We consider alpha-hydrogen individually for each double bond to calculate number of hyperconjugating structures.

In the first compound, there would be 12 hyperconjugating structures.
This is because alpha-hydrogen which are common to both pi-bonds hyperconjugate with both of them, contributing in two hyperconjugating structures.So total number of hyperconjugating structures will be 12.
However, if you only need to count number of alpha-H, then the answer would be given by considering the common hydrogens only once. So number of alpha-H is 8, where the 4 alpha-H at C1 and C2 contribute in 2 hyperconjugating structures each.

However, the second compound has both its pi-bonds at its ends, each having 4 hyerconjugating alpha-hydrogen interactions associated to them. So total number of hyperconjugating structures in this case are 8.
Hence, (1,2,3,4,5,8-hexahydronaphthalene) is more stable than (1,4,4a,5,8,8a-hexahydronaphthalene).

i do believe your answer but why no one pointed me this point till now, If this was so simple. Infact i was literally begging to users to know whether this structure has 8 or 12 alpha H .Are you sure? Don't get me wrong i just don't want to input any wrong info.

@Newton Read Martin's comment.. It is a gross oversimplification

@Safdar i can understand. But in exams we can't leave one of the easiest question just stating fact"gross oversimplification" .  So i just want to know total alpha H in 1st compound for exam purpose. I'm in high school btw.

This is just neglecting so much, much chemistry.

@Martin-マーチン Hyperconjugation leads to stabiity of compounds having pi-bonds. Whenever sp3 hydrogen is next to a pi-bond, the sigma orbital of CH bond overlaps with vacant pi*(antibonding)-orbital of pi-bond. This results in delocalisation of the the sigma electrons that increases the entropy, and in turn, the stability of the compound. Also on overlap with the pi* orbital, the electrons occupy a lower energy level than they were at initially. Hence, Hyperconjugation is a valid concept to determine stability of such compounds

@SoumyadwipChanda This is a handwavy explanation at best. Yes, hyperconjugation is a stabilising effect, but it has a wide range from weak to strong. Simply counting possible contributors is just a little bit better than guessing. With these particular structures there are much more factors at play. Also, stability should be seen as a relative concept and with this exercise the baseline is not defined. Your answer also is solely based in a very crude model, it does not even attempt to check whether the model produces the correct result.

@Martin-マーチン I agree to this. Of course you cannot use hyperconjugation to compare some completely different compounds. But comparing stability of compounds having similar structures, more precisely isomers, Hyperconjugation can give reliable results. However, we can never be sure of such predictions unless we have the accurate data of heat of combustion or hydrogenation or something else. The two compounds being isomers, I can surely say that hyperconjugation will have a significant effect over their stabilities.

Disclaimer:
This post might be hard to read for some people. Some may even consider this a rant. You have been warned.
This answer will question the validity of the used models, as well as the basis for such exercises. (It will also point to a fundamental flaw of this specific exercise.) If you are a student, you might learn concepts that will later turn out to be inaccurate or outdated. This answer should give you some general context, but it will probably not be able to help you in an exam.

Introduction
I have previously stated in the comments how much I hate these exercises, I even called them really, really big garbage. There are several reasons for that. I am going to try to convince you step by step why I dislike these exercises, this one in particular.
Stability is not a very meaningful concept
There is a reason why there is no easy definition for stability. It is too fuzzy a concept with too many different connotations that basically everybody could understand something different. This is an ultimate flaw, and for these exercises, some kind of understanding of 'stability' is implied, but probably never properly defined.
Let's start with something tangible, one possible (and accepted) definition of stable; one that is so trivial you might have not even thought about it. Stable means something exists. In more official terms, i.e. in the gold book, it is called structural stability (10.1351/goldbook.ST07109):

Within the Born–Oppenheimer (adiabatic) approximation, this is associated with an energy minimum on a potential energy surface. Structural stability implies that any change in the coordinates of the nuclei can only increase the total energy.

I'm not absolutely happy with the above definition, as it can be read that only global minima are stable. If you start from a local minima, some changes in the coordinates will lead to a decrease in total energy; that is how different conformations work and eventually also reactions. So the changes implied by this definition should be reasonably small.
Obviously, this definition does not at all help us with the exercise. Both compounds are stable according to this definition, but we cannot say which one is 'more stable'.
From this we could go ahead and rank molecules according to their total energy. As long as we would be looking at isomers, this could actually work, but another question is how meaningful this approach is. I won't and cannot go into any detail or discussion how to obtain the total energy; for the purpose of this post just assume every molecule does have an exact total energy.
In the end the total energy is a number associated with a molecule. It will not tell you anything about the chemistry that molecule does. Almost all tangible and practical information is not contained in this number. Sure it offers an exact and consistent way of sorting things, but –in my opinion– an utterly useless one.
It is also problematic in a semantic way: you are using one term to describe a phenomenon which already has a name.
This definition also will not help in answering the question, because estimating this value is hard. Quantum Chemistry has plenty of options to estimate or calculate the expectation value of these compounds. Some are simple, based on empirical data, e.g. semi-empirical methods, some are a bit more complex, e.g. Hartree-Fock, Density Functional Approximations, some are very complex, e.g. Configuration Interaction, Coupled Cluster. Almost all of them you cannot do on a piece of paper in a reasonable timeframe.
That leads us to the point where we need another definition. The gold book actually has another definition for us (10.1351/goldbook.S05900):

As applied to chemical species, the term expresses a thermodynamic property, which is quantitatively measured by relative molar standard Gibbs energies. A chemical species $\ce{A}$ is more stable than its isomer $\ce{B}$ if $\Delta_\mathrm{r}G^\circ > 0$ for the (real or hypothetical) reaction $$\ce{A -> B}$$ under standard conditions. If for the two reactions:
\begin{align}
\ce{P &-> X + Y} & (\Delta_\mathrm{r}G^\circ_1)\\
\ce{Q &-> X + Z} & (\Delta_\mathrm{r}G^\circ_2)
\end{align}
$\Delta_\mathrm{r}G^\circ_1 > \Delta_\mathrm{r}G^\circ_2$, $\ce{P}$ is more stable relative to the product $\ce{Y}$ than is $\ce{Q}$ relative to $\ce{Z}$. Both in qualitative and quantitative usage the term stable is therefore always used in reference to some explicitly stated or implicitly assumed standard. The term should not be used as a synonym for unreactive or 'less reactive' since this confuses thermodynamics and kinetics. A relatively more stable chemical species may be more reactive than some reference species towards a given reaction partner.

This brings us back to the original dilemma, the implied understanding of stability. While we certainly can estimate the hypothetical relative molar standard Gibbs energy of the reaction, it actually won't tell us much about the chemistry. At least this definition will give us some handling on the exercise, i.e. the problem becomes finding 'numbers'.
Sometimes, but this is very wrong, stability is used as masquerading a different concept altogether, a kinetic concept. Stability may then wrongly be defined as the lower likelihood to react, in other words a lower reactivity means more stability. Now this brings back the already discussed semantic problem, but this time it is actually worse. There is also no reason to not use the proper terminology. The gold book defines reactive (reactivity) as follows (10.1351/goldbook.R05180):

As applied to a chemical species, the term expresses a kinetic property. A species is said to be more reactive or to have a higher reactivity in some given context than some other (reference) species if it has a larger rate constant for a specified elementary reaction. The term has meaning only by reference to some explicitly stated or implicitly assumed set of conditions. It is not to be used for reactions or reaction patterns of compounds in general. The term is also more loosely used as a phenomenological description not restricted to elementary reactions. When applied in this sense the property under consideration may reflect not only rate, but also equilibrium, constants.

So this is pretty much linked to the activation barrier. This is surely nothing that is asked about by the exercise, especially since no reaction conditions are mentioned. However, it felt necessary to be included here, since quite a few people will actually mix up the terms or even use them synonymously.
You might come across terms like air-stable which means you can handle a substance in a normal atmosphere without having to worry that it would decompose or react. This is exactly the problem that I mentioned before.
Another point worth mentioning is that some people use 'stable' to indicate whether a species is long-lived; this is also a wrong usage according to the above definitions. It becomes especially problematic when it is used synonymously with unreactive or even inert; but these are discussions for a later time.
In conclusion: There are two correct ways to define stable/stability. (1) Stable either means something exists as a local minimum on the potential energy surface. (2) As a relative concept, A is more stable than B if it costs energy to transform A into B.
Hyperconjugation and the problem of counting interactions
The gold book again gives us a definition for hyperconjugation (10.1351/goldbook.H02924), but I will only quote a portion:

In the formalism that separates bonds into σ and π types, hyperconjugation is the interaction of σ-bonds (e.g. C–H, C–C, etc.) with a π network. This interaction is customarily illustrated by contributing structures, [...] sometimes said to be an example of 'heterovalent' or 'sacrificial hyperconjugation', so named because the contributing structures contains one two-electron bond less than the normal Lewis formula [...].

The important takeaway from this is, that this is only valid within a formalism that separates into σ and π bonds. This concept does not (in this form) exist in Molecular Orbital Theory.
As previously mentioned, there are various ways to describe electronic structure. Many of them require more than the back of an envelope. Those descriptions are basically modeling the electron density at given nuclear positions; these are just mathematical tools to reproduce observables.
As such, hyperconjugation interactions are also just mathematical models and you cannot measure their stabilising factor. All numbers you might find online or in textbooks, are pretty much partitioning schemes of experimental data.
As a very simple example: You can measure the hydrogenation enthalpy of ethene and propene. You can then assume that the π-bond will (in first approximation) behave equally. The resulting energy difference can then be interpreted as the stabilising effect of hyperconjugation. You can do that for thousands of molecules and find some average energy value for these interactions.
That certainly is a valid scientific approach; the problem comes then if you neglect all the conditions that lead to this value and take it as a given constant.
It should be painfully obvious, that different σ-bonds will interact differently with other bonds. There is no reasonable justification to assume that a hyperconjugation interaction will always have the same stabilising effect. It should therefore come as no surprise that a carbon-hydrogen σ-bond located between two carbon-carbon π-bonds will have a totally different stabilising effect than a terminal carbon-hydrogen σ-bond.
Unfortunately, this is taught as some kind of holy grail of organic chemistry. Counting hyperconjugation contributors is at best a zeroth order approximation, an educated guess, a ballpark figure. Surely, you can use it to do a rough estimation, but you should always be aware about how imprecise this is.
It is also important that for a complete description of the electronic structure, you always need all resonance contributors, n.b. hyperconjugation is a more specialised form of resonance. Since these are hypothetical structures, they cannot be assigned an energy and they cannot be weighted that way. It really does need more involved methods to accurately describe this.
Focusing on this criterion alone also only ensures that you are neglecting any and all other possible influences. Dispersion will certainly have a different effect in one molecule than it will have in the other. There will be other mesomeric, or inductive effects; the conformational space will also be of concern.
So while $\pi \approx 3$ is a good guess to roughly estimate the size of a round window, you probably shouldn't use it to built your house.
The problem is the question
Now that we have gotten all of the general problems out of the way, there was one problem pointed out previously in the comments by user55119:

There is a cis/trans issue with the second structure. You might explore the heats of hydrogenation/combustion of hexenes that bear the substitution of the double bond. Welcome to ChemSE!

As if we hadn't already enough problems, the question is already ambiguous by specifying the molecules. In my personal opinion, this shows that the instructor did not spend enough time actually posing the exercise, really thinking about it. It shows that this exercise is targeted exactly at this specific counting exercise. To be honest, I would not have been surprised if the approximation had yielded a completely different result.
To be on the safe side, I performed a couple of calculations. First I ran a conformational analysis with CREST on the xtb level of theory, then I optimised the lowest structures with RI-BP86/def2-SVP in Orca. These are the relative electronic energies. (I did not bother with thermal corrections, since they are not accurate enough for this level, and qualitatively will probably not change the general order.
(1)          1-2-3-4-5-8-hexahydronaphtalene      0.0 kJ/mol
(2)    cis-1-4-4a-5-8-8a-hexahydronaphtalene     48.8 kJ/mol
(3)  trans-1-4-4a-5-8-8a-hexahydronaphtalene     26.8 kJ/mol

You can see from these numbers, especially between 2 and 3 where the number of hyperconjugation interactions is the same, that there are quite large differences. So using hydrogenation or combustion values is probably a better approach to this kind of question.
Additionally, those numbers don't tell you much about the molecules themselves or the chemistry they most likely will perform. In general I'd expect 1,4-cyclohexadiene derivatives to readily undergo oxidation to an aromatic species, but that is besides the point.
My personal opinion is that these counting models oversimplify chemistry beyond the point of usefulness. They do not further any understanding of molecular and electronic structure. It turns useful and educational models into formulaic use cases, which are applied without further understanding.

@ Martin has addressed the shortcomings of assessing “stability” by counting hyperconjugating C-H bonds. If by “more stable” one means which one of the two isomers, 1,2,3,4,5,8-hexahydronaphthalene 1 or cis-1,4,4a,5,8,8a-hexahydronaphthalene 2,  has the lower heat of combustion, formation or hydrogenation, then a decision can be made. Unfortunately, there does not appear to be any thermochemical data [NIST] on these compounds to answer this question.
If one assumes, not necessarily justifiably, that each double bond in these isomers is not affected by the presence of the other double bond, then one can model a solution based on known values for the heats of hydrogenation (ΔHho) of 1,2-cis-dialkyl and tetraalkyl-substituted double bonds. The more alkyl substituted a double bond, the less heat is produced upon hydrogenation, a trend that is readily verified with data from the NIST site. The heats of formation and hydrogenation of compounds are from the same source.
The ΔHho for an unstrained 1,2-cis-double bond is -28.3 kcal/mol as is seen in the hydrogenation of cis-2-butene to n-butane and cyclohexene to cyclohexane.  The heat of hydrogenation of  2,3-dimethylbut-2-ene (tetramethylethylene) to 2,3-dimethylbutane is -25.7 kcal/mol. Using these two values, the heat of formation (ΔHfo) of 1 and 2 can be calculated given the ΔHfo of  cis-decalin 3. Accordingly, diene 1 (ΔHfo = +13.5 kcal/mol) is more “stable” than diene 2 (ΔHfo = +16.5 kcal/mol)(blue numbers) by ~3 kcal/mol.

Notice that the monohydrogenation of cyclohexa-1,4-diene liberates only -25.2 kcal/mol upon formation of cyclohexene. This implies that the two double bonds, although non-conjugated, are not isolated in their reactivity. If they were isolated and equivalent in reactivity, the ΔHfo of cyclohexa-1,4-diene would be +27.1 kcal/mol. [Parenthetically, the conjugated cyclohexa-1,3-diene has a ΔHfo 1 kcal/mol greater than cyclohexa-1,4-diene.] Diene 1 has a similar arrangement of double bonds, albeit one of them is tetrasubstituted. Assuming that the -3.1 kcal/mol correction may be applied in the case of diene 1, the ΔHho would be lowered to -50.9 kcal/mol (red numbers). Whatever the true correction may actually be, the heat of formation of diene 1 will be lower than that for diene 2.


text:
Are there any natural thermosetting polymers?
Is eggwhite a natural thermoset?
A thermoset is a material that is usually liquid, but it solidifies upon heat. Once it's set, trying to melt it by heating it will only solidify it further until it decomposes, chars, burns, and decays. Isn't that what eggwhite is? Does hardening of the egg involve cross-linking of the polymers of the eggwhite?
Cellulose is a natural polymer, but I don't think it's thermosetting. The same thing applies to polymers that make up the human body.

If you ask wether eggwhite is a thermoset, one wonders what your definition of a thermoset is.

You need to recheck your definition of thermoset Jonathan

Thermosets Directly from RRMs: Vegetable Oil Derivatives and Natural Rubber
This is a long answer... I couldnt find lesser words to express this
It may seem strange to refer to coatings based on drying oils as thermosets. Nevertheless, they fit the definition since they undergo chemical cross-linking during the curing process. Drying oils consist of triglycerides containing high levels of polyunsaturated fatty acids, such as linoleic and linolenic acid. Typical polyunsaturated fatty acids contain one (eg. linoleic acid) or two (eg. linolenic acid) “bis-allylic” methylene groups, marked clearly in the red ovals. This bis-allylic position is very susceptible to oxidation by molecular oxygen in air (a process known as “autoxidation”), which leads to formation of hydroperoxides and, ultimately, bond formation between different triglyceride molecules. A hard film is formed when sufficient cross-linking bond formation has occurred. Metal catalysts can dramatically accelerate the rate of curing. Properly formulated drying oils make excellent protective coatings (paints and stains) and durable flooring.

Alkyd resins are related technology, usually formulated as mixtures of renewable and petroleum-derived ingredients. While processes and compositions vary considerably, a typical alkyd resin is prepared by cooking at elevated temperature (>2000C) a mixture of polyunsaturated triglyceride, a polyhydric alcohol (a molecule with three or more hydroxyl groups) and a dicarboxylic acid (usually phthalic acid or anhydride) until a product of the desired viscosity is obtained. As with drying oils, alkyd resin films cure in the presence of atmospheric oxygen and can be formulated to make durable paints, varnishes, and coatings for electrical equipment. At present, phthalic acid is a petroleum-derived raw material, so about 70% of a typical alkyd resin is renewable.

The carbon-carbon double bonds in polyunsaturated natural oils can also be converted into reactive epoxy groups using epoxidizing reagents, such as peroxyacids, hydrogen peroxide or hydroperoxides with catalysts, and even certain enzymes. Academic publications and patents have described both properties and applications of epoxidized natural oils with much of the work focusing on materials from linseed and soybean oils. Epoxidized linseed and soybean oils are available commercially for applications such as coating. Acrylate-functionalized derivatives of epoxidized soybean oils are also available; they are made through addition of acrylic acid to the epoxy groups for applications such as printing inks.

Finally, natural rubber should be included in any discussion of thermosets derived directly from renewable materials. Natural rubber is polyisoprene with 100% of the carbon-carbon double bonds in the cis-configuration. The combination of C-C double bonds with allylic C-H bonds makes polyisoprene very susceptible to cross-linking by free radical reactions such as the vulcanization process, which involves a complex mixture of reactants including elemental sulfur. With proper formulation, cross-linked natural rubber can form extremely tough, resilient materials, as exemplified by tires.

It is evident that excellent protective coatings can be formulated from natural oils (eg. linseed and soybean oil) and chemically modified natural oils (alkyds and epoxidized oils). Attempts to formulate high performance thermoset materials based on natural oils as the only cross-linking agent have met with less success; although there is ample literature on the subject, the technology has provided few commercial products. This is not surprising since the cross-link density achievable with natural oils is too low to provide properties necessary for true structural applications. In the next post, we will discuss how renewable materials can be applied to high performance thermoset polymers.

Could you share the image sources? they are too small..

@Safdar I made them myself using chemspider draw. I have deleted those...

The text is unreadable

I will re-edit it later. Thanks for the suggestion

you meant 200°C not 2000°C right?


text:
What does simplest optically active alkene mean?
I am confused between whether we will call 2,3-dimethyl-cyclopropene or 3-methylpent-1-ene as simplest optically active alkene. Does “simplest” mean aliphatic, or does it mean having least number of carbon atoms?

That's kinda opinion based. Somehow, I doubt I've ever seen a definition of simplicity of compound. I could also beat your idea using deuterium and tritium instead of two methyl groups.

In chemistry, simple generally refers to compounds of low molecular masses. So simplest
optically alkene would mean the alkene with lowest molecular mass. Hence 2,3,dimethyl-cyclopropene would be your simplest optically active alkene.

Optical activity can also refer to the process. I have seen problems that refer to differing yields in terms of racemate, and from this optical activity was determined.


text:
Why is Sodium Transported Across the Cation Exchange Membrane in the Chlor Alkali Process?
In the Chlor Alkali Process, a sodium ion is transported across a cation exchange membrane to create NaOH in the Anode Chamber:

The Half reactions are:
2Cl− → Cl2 + 2e−
and
2H2O + 2e− → H2 + 2OH−
and the full reaction is:
2NaCl + 2H2O → Cl2 + H2 + 2NaOH
Why does the sodium ion get transported across the membrane? Sodium isn't involved in the anode/cathode reactions, so there's no concentration gradient driving it. I understand that the Anode reaction produces OH-, and that there is now a charge imbalance so that the sodium ion crosses in order to maintain electroneutrality. But how can this be described mathematically? I don't see how the standard mass conservation/ Nerset-Plank equations can describe this motion.

Don’t really know much about the Nerset-Planck equation, but isn’t there a term in the equation that describes flux (independent of concentration gradient) as a function of the electric field? So even if the concentration gradient is 0, the flux isn’t, since there is a charge difference between the two sides of the cell?

There is a rather interesting explanation as to why only sodium ion can pass through a cation exchange membrane and chloride ion is repelled by it.
Let us recall some properties
a) Ion exchangers are very good electrical conductors. This is well studied in the 1940s. The current is carried, not by electrons but by mobile ions in ion exchange resins. In a cation exchanger, current is carried by mobile cations.
b) Imagine in the cation exchanger, it consists of a sulfonated polymer. Let us call that R-SO3H, where H is mobile and R-SO3(-) unit is immobile.
c) Ignore the flow of the chlor-alkali cell for the time being.
d) The punch line is that the bulk solution has to be electrically neutral all the time.
Now consider an electrolytic cell which consists of NaCl solution. There is a partition of a cation exchange membrane in between. On the left there is an anode where chloride ions are being oxidized and the right, water is being reduced to hydrogen.
Right hand side:
When water is being reduced, neutral hydrogen gas is leaving the solution, leaving behind negatively charged hydroxide ions. There is a charge imbalance in the bulk which cannot exist. What has to happen? Na(+) ions must cross the cation exchanger from the left hand side to ensure there is no bulk charge imbalance.
Now chloride cannot pass through the ion exchange membrane because the R-SO3(-) has a negative charge. All it can do is to repel the negatively charged chloride ion. You can read more about Donnan effect and Donnan potential.
Left hand side:
You are losing negatively charged chloride ion as neutral chlorine gas. There is excess positive Na(+) ions there. The positive ions have to cross the membrane in order to maintain electrical neutrality in the bulk and eventually join the free hydroxide ions.
As a combined effect, the right hand side becomes a solution of NaOH (contaminated with NaCl) and the concentration of NaCl on the left hand side depletes.

Incredible answer, thank you. This is so much clearer now. Quick follow up question: Is the transport of sodium ions immediate (i.e simultaneous with the electron transfer at the anode?), or does the charge imbalance create a potential difference, and that potential difference exerts a force on the sodium, motivating it to cross the membrane? What are the rate laws governing this?

I would say it is instantaneous. Ion exchange membranes are very good electrical conductors.


text:
Étard Reaction of Ethylbenzene
What is the major product when the following compound undergoes Étard reaction?

Is it acetophenone or a terminal aldehyde.
The mechanism discussed in Mechanism for an Étard reaction (chromium complex) does not give any conclusive prediction.

Related: https://chemistry.stackexchange.com/questions/58381/mechanism-for-an-%c3%89tard-reaction-chromium-complex?noredirect=1&lq=1 (you've already mentioned) https://chemistry.stackexchange.com/questions/58371/mechanism-of-toluene-oxidation-with-cro3?noredirect=1&lq=1 https://chemistry.stackexchange.com/questions/127304/mechanism-of-etard-reaction?noredirect=1&lq=1

Apparently the Etard reaction of ethylbenzene is not as simple as with toluene. This information is from Chem. Absts. "The Etard complex of ethylbenzene (I) was prepd. in CCl4 at 0°.  Decompn. of the Etard complex of I by water yielded phenylacetaldehyde (II) and acetophenone (III) in equal proportions, along with styrene and BzH. Revue Roumaine de Chimie, 1967, 12, 1495-1501.

A mechanism has been proposed herein for the Étard oxidation of ethylbenzene that parallels the proposed oxidation of toluene with chromyl chloride (CrO2Cl2) as a [2,3]-sigmatropic rearrangement that is found on Wikipedia.[1] While such a mechanism is a convenient way to rationalize the formation of benzaldehyde, it does not consider the “Étard complex” that has a composition, in most cases, of RH/2CrO2Cl2 or greater proportion of CrO2Cl2.[2]
This mechanism is incompatible with the Étard oxidation of n-propylbenzene and ethylbenzene, both of which studies are cited on Wikipedia. The products derived from these higher homologs reveal both α- and β-
carbon oxidation.
Nenitzescu and co-workers[3] demonstrated that the Étard oxidation of ethylbenzene affords acetophenone and phenylacetaldehyde, which answers the OP’s question.  The following summary is from Chemical Abstracts:

"The Etard complex of ethylbenzene (I) was prepd. in CCl4 at 0°. Decompn. of the Etard complex of I by water yielded phenylacetaldehyde (II) and acetophenone (III) in equal proportions, along with styrene and BzH.”[3]

That n-propylbenzene 1a affords benzyl methyl ketone 2a and propiophenone 3 has been known since 1890.[4] Wiberg, et al.[5] explored the mechanism of this reaction using n-propyl-β-d2-benzene 1b. Oxidation of 1b provided benzyl methyl ketone 2b containing 60% deuterium at the α-position, which indicated a rearrangement was occurring. The presence of α-chloro-n-propylbenzene 5 was also observed. These investigators demonstrated that oxidation of propenylbenzene 4 gave rise to benzyl methyl ketone 2a.
The evidence points to either a radical process[5,6] or one that invokes hydride transfer,[2] i.e., cationic, at the benzylic carbon. There is little evidence, if any, for a [2,3]-sigmatropic rearrangement in this reaction. If there is, please apprise me.

https://en.wikipedia.org/wiki/%C3%89tard_reaction
I. Necsoiu, A. T. Balaban, I. Pascaru, E. Sliam, M. Eliam, C. D. Nenitzescu, Tetrahedron, 1963, 19, 1133.
C. N. Rentea, M. Rentea, I. Necsoiu and D. Nenitzescu,  Revue Roumaine de Chimie, 1967, 12, 1495.
W. v. Miller and G. Rhode, Chemische Berichte, 1890, 23, 1070.
K. B. Wiberg, B. Marshall and G. Foster, Tetrahedron Letters, 1962, 345.
M. Drees and T. Strassner, Inorg. Chem., 2011, 50, 5833.

Ethylbenzene yields acetophenone when it undergoes Étard reaction.
The mechanism is as follows:


text:
Nickel(II) acetylacetonate crystals coming out blue
An experiment in my class has us synthesising the acetylacetonate complex from hydrated nickel chloride being added to a solution of ammonia, acetylacetone and water. The products came out pale blue, yet according to every source I can find, the crystals should be green. Just wondering what the structure would be to accomodate the colour.

What do you mean by "the products"? Were there more than one type of crystals, or are you referring to the mother liquor and crystals? It would also be beneficial if you could add some information regarding the amounts of reactants used for the synthesis as well as upload the photo(s)—the dihydrate is a borderline case being greenish-blue.

Wikipedia states that the anhydrous Nickel(II) acetylacetonate is green. But the dehydrated compound is blue.

You obtained the dihydrate complex of AAN. If you want to obtain green crystals, you should use azeotropic evaporation to remove crystal water. Then you will have green product - anhydrous AAN.


text:
What are coexistent metals?
In ACS Sustainable Chem. Eng. 2018, 6 (10), 12542–12561, I saw the term "coexistent metals", but I could not find a definition for that on the Internet. So what does it mean?

Karapinar et al. utilized ammonium-loaded zeolite as a seeding material for precipitating calcium phosphate and suggested the possibility of using the precipitate-ammonium loaded zeolites as fertilizer. The efficiency of chemical precipitation is affected by alkalinity, organic matter content and the coexistent metals.

I see no special meaning there, aside of the common English. It means eventual metals present in the solution as side components or impurities, aside of calcium. Like alkali metals, magnesium, heavy metals as iron, manganese, copper atc.


text:
Ionic radius for tetrahedral metal–ligand complexes
The ionic radii for metal–ligand complexes that are in an octahedral coordination are easy to find. I understand that when the $\mathrm{e_g}$ orbital is filled, the ionic radius increases because the electrons repel the ligands meaning the ionic radius increases.
But I am not sure about tetrahedrally coordinated metal–ligand complexes. I think it might be the same for the first two to three electrons, but since the $\mathrm{t_{2g}}$ orbital electrons are degenerate, the $Z_\mathrm{eff}$ might be stronger “felt” by the ligands.


text:
How do we define "volume" of the gas undergoing free expansion in vacuum?
Usually volume occupied by the gas is defined as the the volume of the container, but as it expands freely (Joule's free expansion), we can't say anything about volume in a concrete sense as we don't have a rigid container. How do we think about this in a more concrete way?

We would first need the criterium what is a gas and what is vacuum.

What about $V \to \infty$?

I often see people talking about expansions in a vacuum , how do they do this without something to concretely call volume?

Particularly, it is undefined.


text:
Is elemental lead as toxic as lead compounds?
On the internet, nobody mentions about lead's oxidation state when talking about lead poisoning. I assume it's Pb(II) and Pb(IV), as chelation therapy is suggested as a treatment. Also, its accumulation in bones and teeth suggest its ionic state. But what about elemental lead? How is it transformed in human body? Is it as toxic as Pb(II) and Pb(IV) compounds?

In order for an element to be toxic, it has to be soluble in aqueous solution. If it is not soluble, it gets out of the body like sand or clay. Metallic lead is not soluble into water until it is oxidized into $\ce{Pb^{2+}}$. The main compounds of Pb(IV) are $\ce{PbO2}$ and $\ce{Pb3O4}$ which are insoluble and $\ce{PbCl4}$ which is decomposed by water. So $\ce{Pb(IV)}$ is probably not directly toxic. As a consequence, only $\ce{Pb(II)}$ is toxic

Here is a review of lead toxicity. I highly recommend to give it a thorough read.

Pb(IV), in all likelihood, will quickly get reduced to Pb(II) once it gets into a human body. So as far as poisonings go, these two can be considered equal.
Elemental lead will partially dissolve in stomach, but mostly go out undigested. So it is way less harmful than any compound of Pb(II). There were people in the old times who survived bullet wounds and used to carry bullets in their bodies for decades, with little or no apparent ill effects.
Then there is a wholly different story of the covalent compounds of Pb. They are more toxic than ionic Pb(II), and some are capable of penetrating through the skin. In particular, the notorious $\ce{Pb(C2H5)4}$ has caused our civilization a great deal of harm.
So it goes.


text:
What happens when the amount given doesn't allow for equilbrium to be achieved in a reaction?

The reaction $$\ce{A.3H2O(s) <=> A.H2O(s) + 2H2O (g)}$$ has a $K_p = \pu{400 mmHg2}$
One millimole each of $\ce{A.3H2O}$ and $\ce{A.H2O}$ are taken in a $\pu{1 l}$ container at $\pu{300 K}$
Relative humidity is $80\ \%$ and saturation water vapor pressure is $\pu{85 mmHg}$ at $\pu{300 K}$. Find the final composition of the container

My attempt
\begin{align}
K_p &=(P_\ce{H2O})^2 = 400 \\
\therefore P_\ce{H2O} &= \pu{20 mmHg}
\end{align}
Now,
\begin{array}{lc} \hline
\ce{&A.3H2O &<=> &A.H2O &+ &2H2O(g)} \\ \hline
i &1\times10^{-3} && 1 \times10^{-3} && \pu{68 mmHg} \\
c &+x && -x && -P \\
e &2\times10^{-3} && 0 && \pu{30 mmHg} \\ \hline
\end{array}
$\pu{38 mmHg}$ drops because $\pu{2 mmol}$ of water vapor was consumed
$$ P=\frac{1}{12}\frac{300\times 2\times 10^{-3}}{1} = \pu{38 mmHg}$$
The answer should have been $\pu{20 mmHg}$ pressure of water vapor but I am afraid there isn't enough reactant to make that much vapor pressure.
I asked a friend and he suggested that we can take $\pu{2 mmol}$ each of the reactants as they are solid and don't influence the $K_p$ expressions while also not influencing the ratio of the added moles.
I do not get why this can be done but this allows for the vapor pressure at equilibrium to become $\pu{20 mmHg}$ as required

I assume that the reaction would end at that point.. Don't listen to your friend.. Equilibrium matters only when you have reactants left at point of equilibrium..

But that does not satisfy the K_{p} of the reaction

This takes the assumption that the water vapour present acts as an ideal gas. Therefore $PV=nRT$ is valid.

Flaws/Misconceptions:
(1) You directly used pressure as you have in the RICE table.

Using pressure in a RICE table is not very useful since the RICE table deals with the amount of reactants and products and not pressure.

(2) The reaction doesn't reach equilibrium

The reaction reaches equilibrium as shown below. This means that the question in title doesn't need to be considered.
However, in such a case where equilibrium isn't reached, then the reaction stops due to lack of reagents to go forward. When you add more reagent it keeps moving forward until equilibrium is reached after which addition of solid reactants wouldn't cause a reaction to take place.

Now, as OP has stated, the $P_\ce{H2O}= \pu{20 mm Hg}$ at equilibrium. and it is given that relative humidity is $80\%$ for a saturation vapor pressure of $\pu{85 mm Hg}$ and so that means the intial pressure exerted by $\ce{H2O(g)}$ is:
$$P_\ce{H2O}= \frac{8}{10}\cdot 85=\pu{68 mm Hg}$$
Now using the ideal gas law, for the initial condition, we get:
\begin{align}
n_i(\ce{H2O})&= \frac{68}{76}\cdot\frac{1}{0.0821 \times 300} \\
&\approx\pu{0.00036 mol}
\end{align}
Similarly for the final case, we get:
\begin{align}
n_f(\ce{H2O})&= \frac{20}{76}\cdot\frac{1}{0.0821 \times 300} \\
&\approx\pu{0.00011 mol}
\end{align}
Therefore, $\pu{0.00025 mol}$ of $\ce{H2O}$ reacts for the condition to reach equilibrium. Now drawing the RICE table for the reaction we get:
\begin{array}{lc} \hline
\ce{&A.3H2O &<=> &A.H2O &+ &2H2O(g)} \\ \hline
i &1\times10^{-3} && 1 \times10^{-3} && 3.6 \times 10^{-4} \\
c &1.25\times10^{-4} && -1.25\times10^{-4} && -2.5\times10^{-4} \\
e &1.125\times10^{-3} && 8.75\times 10^{-4} && 1.1\times10^{-4} \\ \hline
\end{array}
So the final composition is:
\begin{array}{lc} \hline
\text{Component} &\text{Amount} \\ \hline
\ce{A.3H2O} & \pu{1.125 \times10^{-3} mol} \\
\ce{A.H2O} & \pu{0.875 \times10^{-3} mol} \\
\ce{H2O} & \pu{1.1 \times10^{-4} mol} \\ \hline
\end{array}

If the relative humidity is $80$% and the saturation water vapor pressure is $\pu{85 mmHg}$ at $\pu{300 K}$, it means that the vapor pressure in the flask is $\pu{0.8 · 85 = 68 mmHg}$. We will now show that this value is much higher than the value coming from equilibrium constant of the reaction. As this constant is : $\pu{K_p = 400 mmHg^2}$, the value of the water pressure when both hydrates are present simultaneously is, at equilibrium, $\pu{√K_p = 20 mmHg}$. So there is too much water in the gas phase, : the system is not at equilibrium. To be at equilibrium, the water pressure should be $\pu{20 mm Hg}$. It must be reduced by $\pu{\Delta p_w = 68 mm Hg - 20 mm Hg  = 48 mm Hg}$
As a consequence, the gas phase must loose a large part of its water by reacting with $\ce{A·H2O}$, so as to produce more $\ce{A·3H2O}$  and attain at the end the final vapor pressure $\pu{20 mmHg}$. The number of mole $\ce{H2O}$ to be adsorbed by $\ce{A·H2O}$ is $\pu{\Delta n_w}$ given by
$$\pu{\Delta n_w = \frac{\Delta p_w V}{RT} =  \frac{48~ mmHg}{760~ mm Hg/ atm} · 101325 Pa/atm · \frac{0.001~ m^3}{8.314~J K^{-1} mol^{-1} ~300~ K} = 2.565 10^{-3} mol}$$
The published equation states that  $1$ millimole $\ce{A·H2O}$ can adsorb $2$ millimole $\ce{H2O}$ And here the vapor contains more than $2$ millimoles water. It means that all $\ce{A·H2O}$ ($1$ millimole) will be transformed into $\ce{A·3H2O}$, which is to be added to the original amount of this trihydrate (1 millimole)
The final composition of the mixture will be :

$0$ millimole $\ce{A·H2O}$

$2$ millimole $\ce{A·3H2O}$

$2.565$ millimole - $2$ millimole  = $0.565$ millimole water in the vapor phase.


text:
Is there s-character in the HOMO of dinitrogen?

State true or false:
The highest occupied molecular orbital in nitrogen molecule has no s-character.

The answer given is that this false. Why is it so?
The highest occupied molecular orbital in $\ce{N2}$ is the sigma $2\mathrm p_z$ orbital. So it should not have any $\mathrm s$ character and hence the statement should be true.

mixing between the two $\sigma_g$ symmetry orbitals, ie "s-p mixing". Lots of info here: https://www.staff.ncl.ac.uk/bruce.tattershall/teaching/chy135/n2mo/n2.php

There is a significant portion of s-p mixing in dinitrogen.
Please see this great answer by Wildcat on Molecular orbital (MO) diagram for $\ce{N2}$ and $\ce{N2^-}$.
For the impatient, I'll quote an image from myself from How to rationalise with MO theory that CO is a two-electron donor through carbon?

I have marked these interactions with the dotted lines. To show these valence MO in all their glory, I've performed a calculation on the RI-BP86/def2-SVP level of theory with Orca.

The analysis (main atomic orbitals) of these orbitals showed, above they are shown in reverse order:
MO  3 (occ.=2.00, E=-1.022230)
  N2 (S on atom N1) [0.416], N16 (S on atom N2) [0.416]

MO  4 (occ.=2.00, E=-0.485050)
  N2 (S on atom N1) [0.435], N16 (S on atom N2) [-0.435],
  N3 (S on atom N1) [0.427], N17 (S on atom N2) [-0.427],
  N15 (S on atom N2) [-0.201], N1 (S on atom N1) [0.201],
  N5 (Px on atom N1) [0.200], N19 (Px on atom N2) [0.200],
  N8 (Px on atom N1) [-0.066], N22 (Px on atom N2) [-0.066],
  N13 (D+2 on atom N1) [-0.015], N27 (D+2 on atom N2) [0.015],
  N10 (D 0 on atom N1) [0.009], N24 (D 0 on atom N2) [-0.009]

MO  5 (occ.=2.00, E=-0.421640)
  N4 (Pz on atom N1) [-0.422], N18 (Pz on atom N2) [-0.422],
  N7 (Pz on atom N1) [0.225], N21 (Pz on atom N2) [0.225],
  N6 (Py on atom N1) [-0.164], N20 (Py on atom N2) [-0.164],
  N9 (Py on atom N1) [0.087], N23 (Py on atom N2) [0.087],
  N11 (D+1 on atom N1) [0.033], N25 (D+1 on atom N2) [-0.033],
  N14 (D-2 on atom N1) [0.013], N28 (D-2 on atom N2) [-0.013]

MO  6 (occ.=2.00, E=-0.421640)
  N6 (Py on atom N1) [-0.422], N20 (Py on atom N2) [-0.422],
  N9 (Py on atom N1) [0.225], N23 (Py on atom N2) [0.225],
  N18 (Pz on atom N2) [0.164], N4 (Pz on atom N1) [0.164],
  N7 (Pz on atom N1) [-0.087], N21 (Pz on atom N2) [-0.087],
  N14 (D-2 on atom N1) [0.033], N28 (D-2 on atom N2) [-0.033],
  N11 (D+1 on atom N1) [-0.013], N25 (D+1 on atom N2) [0.013]

MO  7 (occ.=2.00, E=-0.371870)
  N5 (Px on atom N1) [-0.452], N19 (Px on atom N2) [0.452],
  N3 (S on atom N1) [-0.299], N17 (S on atom N2) [-0.299],
  N8 (Px on atom N1) [0.194], N22 (Px on atom N2) [-0.194],
  N16 (S on atom N2) [-0.134], N2 (S on atom N1) [-0.134],
  N15 (S on atom N2) [-0.074], N1 (S on atom N1) [-0.074],
  N13 (D+2 on atom N1) [0.016], N27 (D+2 on atom N2) [0.016],
  N10 (D 0 on atom N1) [-0.009], N24 (D 0 on atom N2) [-0.009]

MO  8 (occ.=0.00, E=-0.071630)
  N7 (Pz on atom N1) [-0.528], N21 (Pz on atom N2) [0.528],
  N4 (Pz on atom N1) [0.513], N18 (Pz on atom N2) [-0.513],
  N9 (Py on atom N1) [-0.032], N23 (Py on atom N2) [0.032],
  N20 (Py on atom N2) [-0.031], N6 (Py on atom N1) [0.031],
  N11 (D+1 on atom N1) [0.013], N25 (D+1 on atom N2) [0.013],
  N14 (D-2 on atom N1) [0.001], N28 (D-2 on atom N2) [0.001]

MO  9 (occ.=0.00, E=-0.071630)
  N9 (Py on atom N1) [0.528], N23 (Py on atom N2) [-0.528],
  N6 (Py on atom N1) [-0.513], N20 (Py on atom N2) [0.513],
  N7 (Pz on atom N1) [-0.032], N21 (Pz on atom N2) [0.032],
  N18 (Pz on atom N2) [-0.031], N4 (Pz on atom N1) [0.031],
  N14 (D-2 on atom N1) [-0.013], N28 (D-2 on atom N2) [-0.013],
  N11 (D+1 on atom N1) [0.001], N25 (D+1 on atom N2) [0.001]

MO 10 (occ.=0.00, E=0.456580)
  N3 (S on atom N1) [-3.853], N17 (S on atom N2) [3.853],
  N8 (Px on atom N1) [-2.584], N22 (Px on atom N2) [-2.584],
  N16 (S on atom N2) [0.543], N2 (S on atom N1) [-0.543],
  N5 (Px on atom N1) [0.266], N19 (Px on atom N2) [0.266],
  N15 (S on atom N2) [0.162], N1 (S on atom N1) [-0.162],
  N13 (D+2 on atom N1) [0.035], N27 (D+2 on atom N2) [-0.035],
  N10 (D 0 on atom N1) [-0.020], N24 (D 0 on atom N2) [0.020]

The main concern is MO 7, which shows significant s contribution.


text:
What compounds given can oxidize Mn(2+) to MnO4(-)?

$$\ce{Mn^2+ Solution ->[reagent 'A'] MnO4^-}$$
The reagent '$\ce{A}$' can be
(A) $\ce{Pb2O3 + \text{conc.} HNO3}$
(B) $\ce{K2S2O8}$.
(C) $\ce{(NH4)2S2O8}$
(D) $\ce{K2SO4}$

All of the options can be reduced to their corresponding compounds so, Maybe all of them should be the answer. But the answer given is (A), (B), (C). Why is (D) not the answer? $\ce{SO4^2-}$ Can be reduced to $\ce{SO2}$ right?

D can be reduced, but you would need much stronger reducing agens than $\ce{Mn^2+}$ . And even if SO2 had been created somehow, it would be immmediately oxidized by $\ce{MnO4-}$.

Everything can be reduced. That's not what the question asks, though.

On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

$\ce{Pb2O3}$ does not exist. It should be replaced by $\ce{Pb3O4}$ which a useful substance for protecting iron pieces against rust.

@Maurice Hmm, without external reference, but: Less common lead oxides are:   Lead(II,IV) oxide, Pb2O3, lead sesquioxide (reddish yellow),     Pb12O19 (monoclinic, dark-brown or black crystals)

@Maurice https://pubs.rsc.org/en/content/articlelanding/1922/ct/ct9222101456#!divAbstract

@Safdar It seems to me MnO4^2- in the title may be an error, as in the body is listed MnO4-. Also, Mn^2+ needs acidic ( or at the best about neutral )pH, while MnO4^2- needs strongly alkaline pH. I am not aware of any alkaline path to produce MnO4- ( bot not saying it does not exist ).

It is not possible to create the manganate ion $\ce{MnO4^{2-}}$ from Mn(II) in basic medium with any reagent as presented in the question,

Pb2O3 is less known than some others, but is among several lead oxides listed in Wikipedia.

My bad, the question indeed asks for MnO4(-) :((

Looking at the standard electrode potentials answers your question (Note that conc. $\ce{HNO3}$ oxidizes $\ce{Pb2O3}$ to $\ce{PbO2}$) :
$$\begin{align}
\ce{PbO2 + 4H+ + 2e- &-> Pb^2+ + 2H2O } & E^0 = \pu{1.46 V} \\
\ce{S2O8^2- + 2e- &-> 2SO4- } & E^0 = \pu{2.01 V} \\
\ce{SO4^2- + 4H+ + 2e- &-> SO2 + 2H2O } & E^0 = \pu{0.17 V}
\end{align}$$
I couldn't find a reference for the couple $\ce{MnO4^2- | Mn^2+}$, but I'm guessing it would be slightly less than that of $\ce{MnO4- | Mn^2+}$, which is $\pu{1.51 V}$ (seeing that $\ce{PbO2}$ oxidizes it).
The standard electrode potential for the reduction of sulphur is less than that required to oxidize $\ce{Mn^2+}$, and hence the reaction does not proceed in the forward direction (net potential of cell is negative, which means that $\Delta G$ would be positive and cell reaction would be non-spontaneous)


text:
Are there limits to waters energy absorbtion capacity
I have a system where I heat water in a flask.
The water vapor generated is condensed and returns to the original container (not a pressurized system, but condensation system is efficient enough that there is no loss of vapor).
My goal is to have maximum amount of vapor. (The real goal is the condensate droplets that filter through plant material to extract water soluble components - Kumagawa extraction)
Question:
Is there a limit or plateau to the amount of energy the water can absorb/turn into vapor?
or
I have a 10l flask and a 1500W heating mantle. If I run the heating mantle at full power is there a possibility that the energy generated couldn't be absorbed by the water in the flask and used in turning it into vapor and thus would heat the electrical control circuits and outer casing of the heating mantle instead.

The system you describe is not absorbing energy: it is moving energy from one place to another. As long as the vapour can be condensed as fast as it is created the system will keep functioning.

Thank you for pointing out about absorption. The vapour condensating isn't the limiting factor for the moment. My concern is if there is a point where energy transfer gets less efficient towards water and easier towards the heating mantle.

The heat transfer is limited by the flask material heat transfer rate, which is proportional to the thermal gradient within the flask. The external flask temperature will follow.
On the water side, liquid can jump out of flask in form of the thick foam, powered by the vapour. This is the major limiting factor.
On external side, at high enough power, the eventual plastic holders/features may start to melt/decompose. At higher power, the flask may start shining red. Or start to melt if metallic.

No, It is not.The former is material constant, the latter is not. If we consider for simplicity a rectangular block model, then heat_transfer_rate = thermal_conductivity * area / thickness * temperature_difference

For glass, another limitation is endurance against thermal shocks

My flask, diam 280mm & thickness 8mm) is made of borosilicate. I have tried to calculate the heat transfer rate and it seems to be around 2000W if I consider the heating mantle to be 200 °C and the extract boiling inside. So this should not be the choke point.
Does it mean that if the extract is forming foam then less energy is transfered to the extract and more to the heating mantle?
I naively thought that the more I put energy into this system the more water vapour I get...

Sorry I deleted the first comment, a bit new to the format.

I mean with too much heat, water can boil out, especially if foaming, like if overheat boiling soup.

This hasn’t happened yet, but it is better to add nucleation centers if done with new glassware and pure water, to avoid humongous bubbles forming.
I’m still trying to figure out if the water itself has a limit at which rate it can “accept“ energy

I did not have in mind overheating, rather continuous too high rate of vapour bubbles. No, water does not have such a limit,but a particular scenario often has.


text:
At what pressure is the triple point of ethane?
Depending on source there are two givens for the triple point of ethane: 0.8 Pa or 1.1 Pa. Only one of them can be correct, which one is?

Could you share your sources?

Wikipedia and NIST webbook say 1.1Pa, that has to be correct.

@Safdar The 0.8 Pa value reportedly is from the NIST, however no younger than from 1988.

Could the 0.8 value be outdated?

@Giovanni NIST reports 1.1 Pa-https://webbook.nist.gov/cgi/cbook.cgi?ID=C74840&Mask=4. Can you share the link where you found that out?

@RobinSingh One source had been the Wikipedia page "Triple point" but I changed it to the 1.1 value. However, some days later I saw the 0.8 value on another site but I dunno which one anymore. I think the 0.8 value is simply outdated.


text:
Why does lattice enthalpy decrease with increasing ionic size?
Lattice enthalpy decreases as ions get larger, but I have found two explanations:

The charge density is greater in smaller ions, so greater attraction
The ions are themselves able to get closer together, so greater attraction

Do these two explanations relate in any way? What would be the best explanation?

Think about Coulomb's inverse-square law:
$$F = k\frac{q_1q_2}{r^2},$$
where $q$ is the magnitude of the charge, $r$ is the distance, and $k$ is the Coulomb’s constant.
Mathematically speaking, as $r$ increases, the magnitude of $F$ becomes smaller. As far as the charge density, as it increases, the magnitude of $F$ increases.
So both explanations are saying the same thing from different point of views. As ionic radii increases, charges are further apart so attraction decreases. As charge density increases, the attraction increases since the charges are increasing in magnitude.
Just as $\ce{MgO}$ is stronger then $\ce{NaCl}$ since $\ce{Mg}$ and $\ce{O}$ have smaller ionic radii and they both bare a $2+$ and $2-$ charge, respectively, making the lattice enthalpy greater.


text:
Formulations for wax thermostatic element for effect in alternative temperature range
Wax thermostatic elements are used automobile cooling systems and anti-scald mixing valves to control flow rates or to adjust mixing of flow inputs. These devices operate on the principle of thermal expansion of "waxes" to convert heat energy to mechanical energy (Wikipedia article).  A review of thermostatic mixing valves for sale indicates that the common formulation for the "wax" within the element in those devices expands in the range 20 °C to 50 °C.
Is it possible to formulate a "wax" that effectively changes volume in a lower temperature range such that it could operate a mixing valve, beginning at 5 °C, for instance, or are the materials typically used in this application not effective at temperatures below 20 °C?

If you read the whole article you linked, it says the range is -15 C to 120 C, so yes, 5 C is within the possible range.

Although typical applications for wax elements are centered around 40 °C, it is possible to create formulations for wax elements that change volume in the lower range of 5 °C to 20 °C.  The example below, although not linear in the lowest temperature extreme of the range of interest, behaves roughly linearly in the majority of the range.


text:
Driving force in distillation
If the pressure is lowered in a distillation column, then the equilibrium curve will move further away from the diagonal (bulge out more) in an equilibrium diagram. This leads to a decrease in ideal step stages required to reach the separation requirement.
My question is why does the ideal step stages decrease? Is it only because of how the stages are drawn; meaning if the equilibrium curve is more bulged less horizontal and vertical stages are required. Or has it to do with the driving force increasing as pressure is lowered?
All help is appreciated. Thank you!

In an x-y diagram, the "bulging out" is demonstrating the increase in relative volatility of the components; the number of stages is a strong function of relative volatility, so an increase in relative volatility leads to less stages.


text:
What do (2S,3S) and similar notations mean?
When my textbook is discussing stereoisomers with more than one chiral centers, it's using notation like (2S,3S) being an enantiomer with (2R,3R) and so forth. I understand R and S configuration, and I'm guessing there are two items in each brackets to represent the two chiral centers (so 2S, 3S would at least mean that both chiral centers have an S configuration), but I don't understand what the numbers preceding the configuration mean. What does the 2 in 2S mean and what does the 3 in 3R mean? I was unable to find an explanation in the text and it seems that their meaning is assumed. Can I get some help?

Same thing as in any formula - number of atom. I have a feeling you have a bad day ;)

The number refers to the carbon position in the molecule. R and S indicate the orientation around of the chiral atom.

@Mithoron No sir, just a momentary frustration.

When naming a organic compound, S/R-notation comes if there is a chiral atom present. The S/R-notation always accompanies a numerical such as (2S,3R)- where the numerical refers to the chiral carbon position in the molecule according to the IUPAC numbering while R and S representate the orientation of the chiral atom according to the Cahn–Ingold–Prelog priority rules. For example, let's look at glyceraldehyde molecules where all this stereochemistry started from:

The IUPAC name of glyceraldehyde is 2,3-dihydroxypropanal. However, it has a one chiral carbon at $\ce{C}$2 and hence has two stereoisomers. The priority order of groups attached to chiral carbon is $\ce{-OH \gt -CHO \gt -CH2OH \gt -H}$. Hence you can look at the $1 \rightarrow 2 \rightarrow 3$ priority group rotation through the bond of chiral carbon and least priority group, which is $\ce{H}$ here (though $\ce{C-H}$ bond). Accordingly, the rotation for the first compound is clockwise as shown in the image, thus it carries (R)-notation. Since chiral carbon is 2 according to the IUPAC rules, the complete notation is (2R)-. Similarly, the mirror image is caring (2S)-notation since it has counterclockwise rotation as depicted in the image. Therefore complete names of these two isomewrs are (2R)-2,3-dihydroxypropanal and (2S)-2,3-dihydroxypropanal (Note that these isomers are called enantiomers).
In addition, if a $\ce{H}$ in $\ce{C}$3 is replaced by another different group, say it is $\ce{CH3}$ group, it will also become chiral and can have either (S) or (R) rotation. Suppose you replace $\ce{H}$ in $\ce{C}$3 of (2R)-2,3-dihydroxypropanal by a $\ce{CH3}$ group making it has (S) rotation. By addition of $\ce{CH3}$ group make it 2,3-dihydroxybutanal. With correct stereochemistry, its correct name is (2R,3S)-2,3-dihydroxybutanal.

You appear to have an understanding of R/S nomenclature. The numbers, 2 and 3, are simply the carbons that are stereogenic. In the compounds shown below, the carbon bearing the hydroxyl group is C2 and not C3 because the hydroxyl group has priority over the bromine. Of course, carbon numbering begins at the terminus of the chain proximate to the hydroxyl group. (2S,3R)-3-Bromobutan-2-ol and (2S,3S)-3-bromobutan-2-ol are diastereomers of one another. Each one has an enantiomer, (2R,3S)-3-bromobutan-2-ol and (2R,3R)-3-bromobutan-2-ol, respectively. In your example, (2S,3S) and (2R,3R), you are dealing with enantiomers. I hope my example is of help to you.

The 2 and 3 are just the second and third carbons in the longest carbon chain.
For instance, take the molecule 1-chloro-1-fluoroethane.
When you draw out the molecule, look for all carbons where there are four different $\ce{R}$-groups, or side chains, attached. This means the carbon they are attached to is a chiral center. If a molecule has a chiral center, that means the molecule comes in at least two kinds of isomers called stereoisomers. Two isomers have a molecular formula that is the same, but the structures and arrangements of the two molecules are different. Stereoisomers are isomers where two molecules cannot be superimposed on one another because they are mirror images at the chiral center.

In this case, one of the carbons is bonded to four different $\ce{R}$-groups: a fluorine atom, a chlorine atom, a methyl group, and an unmarked hydrogen atom. Therefore, this is a chiral center, meaning it has two different mirror images: S and R.
Since there is only one chiral center, then there are two chiral forms of this molecule: (1S)-1-chloro-1-fluoroethane and (1R)-1-chloro-1-fluoroethane. When only one chiral center exists, you can write it like this simply without parenthesis or numbers because we know which chiral center you are referring to since there is only one.
For a harder example, take another molecule: 4-fluoro-3-methylheptane.

You identify the longest chain, which is 8 carbons long. Then you look for all carbons with 4 different $\ce{R}$-groups. In this case, they are the third and fourth carbons. The rest are bonded to at least 2 hydrogens, making them achiral.
This means that there are four stereoisomers of this molecule: (3R,4R), (3R,4S), (3S,4R) and (3S,4S).
If all chiral centers have mirror images of each other, then the two molecules are considered enantiomers, meaning the ENTIRE molecule is a mirror image of the other. There are two pairs of enantiomers: (3R,4R) and (3S,4S), as well as (3R,4S) and (3S,4R). They are enantiomers of each other because EVERY carbon center has its chirality reversed.
If not all chiral centers are the same or different, then the two molecules are called diastereomers. There are 4 diastereomer combinations for this example, one of which is (3R,4R) and (3R,4S), because the chirality of carbon 2 is the same, but the chirality of carbon 3 is different.
You can also determine the number of stereoisomers by calculating $2^n$, where $n$ is the number of chiral centers. For instance if a molecule has four chiral centers, then there will be ($2^4 = 16$) sixteen different stereoisomers.
Finally, the last major trick with stereochemistry is meso-compounds. Two stereoisomers are meso-compounds if the molecule has a plane of symmetry or rotational symmetry. Meaning, if you can draw a line anywhere across the molecule in such a way where both sides are symmetrical, or rotate the molecule in such a way where you can get the same molecule without turning 360 degrees, there will be at least one set of meso-compounds present. Rotational symmetry is harder to notice and easier to forget to check for. When I took organic chemistry, we had a compound given to us for one of our exams that we had to draw out (at first with no stereochemistry in mind), identify the chiral centers, then draw out the stereoisomers, provide the parenthetic notation for each isomer, then identify a set of enantiomers, a set of diastereomers, and, if present, a set of meso-compounds. The compound our professor chose was 1,3-dicyclobutoxycyclopentane, which does have plane of symmetry, and therefore, has a set of meso-compounds.


text:
How is a photoelectron spectrum assigned?
The photoelectron spectrum of a given molecule is commonly used as experimental evidence of the energy ordering of molecular orbitals (See here:
10.4: Photoelectron Spectroscopy, and many other physical chemistry books).
How is the experimental photoelectron spectrum assigned? Depending on the molecule you can do an accurate electronic structure calculation, but can it be correctly assigned using only the information in the spectrum (and selection rules, maybe)?

To be honest, to me it feels like the proof of "orbitals" by photoelectron spectroscopy appears like a of a circular story or a snake biting its own tail. The textbook story goes like this, Xray photoelectron peaks prove that orbitals exist. The next statement is that the XPS peaks are due to electrons ejected from various orbitals. You have correctly guessed nobody ever tells us how they assigned the transitions of hydrogenic like orbitals to almost all elements of the periodic table.

For an alternative perspective, see John Ogilvie's "Do orbitals exist?" in the Journal of Chemical Education. Also see "Are Orbitals Observable? by Hyle

Forget about XPS or photoelectron spectra, I have asked various spectroscopists and physicists how did people assign electronic transitions in a simple spectrum such as that of sodium? It is a lost art. Ten years have passed and I am still looking for an answer.

@M.Farooq thanks for the comments. I wanted to know if there was an origin to this story that you described as "a snake biting its own tail". I'm a computational chemist so I'm aware of their limitations. I was really curious to know if there was a "pure" experimental path to assign these spectra. Too bad that this is a lost art. This could be a nice story to be included in textbooks.

Dear Antonio, I have used XPS in my research (for analysis) but nobody could tell me how the spectra were assigned. May be we should start with the earliest papers on XPS. As I said, it has been ten years nobody has told me how to experimentally assign the sodium spectral lines. A photoelectron spectrum is way more complicated I guess.

The snake story refers to an alleged dream by a chemist named Kekule, who was thinking about the structure of benzene. He saw a dream of a snake biting its own tail which gave him a hint of a cyclic structure. I have asked in physics forums, ResearchGate, only get hopeless answers. So far I have not been able to find experimental methodology assignment of electronic transitions in atoms (most likely it is buried in the German literature of 1920s-1930s).


text:
How can nitrogen be both inert and bond with everything in periodic table?
Aren't below statements from this link contradicting each other?

Nitrogen bonds to almost all the elements in the periodic table    except the first three noble gases, helium, neon, and argon

Nitrogen gas is mostly used as an inert atmosphere whenever the    oxygen in the air would pose a fire, explosion, or oxidising problems

How can one element be both inert and bond with everything at the same time? Aren't inertness and bonding mutually exclusive?

Are you aware that the noble gases mentioned are also inert in the first place?

...[cont] If you see the Wiki page of Argon compounds, there exist cations like $\ce{ArN2+, Ar(N2)2^+, ArN2O+, ArNH+}$ etc. which has been detected albeit unstable. There also exists polyatomic van der Waals argon-nitrogen molecules. Also, a high pressure Van der Waals solid, $\ce{(N2)6Ne7}$ has been detected.

Nitrogen as an atom can be bound to any other atom of the periodic table. But nitrogen as a molecule is something else. It is $\ce{N2}$ and the two atoms are strongly bound in the molecule. It is even extremely difficult to separate the two atoms from the molecule $\ce{N2}$, because $\ce{N2}$ contains a triple bond : there are three covalences between the two atoms. That is why $\ce{N2}$ is nearly inert. A huge amount of energy is needed to break these three bonds. And it is usually impossible to do it in a lab at ordinary temperatures. Some plants (beans, peas) are able to do it and to metabolize $\ce{N2}$ by a somewhat mysterious process.

Ah ok so the first point refers to the atom $N$, and the second point refers to the molecule $N_2$. Got this XD Thank you so much!

For chemical vapor deposition techniques one usually uses ammonia as a nitrogen source.

@Maurice Inert relates with kinetics. Why bond energy which is a thermodynamic factor relates with kinetics?


text:
What is the temperature needed for a reflux using a mix of solvents as the reaction medium?
I had a quick query regarding the setup of a reaction. The procedure that I am working from states that I need to put my reaction mixture under reflux for 3 hours. The solvents used are diethyl ether (30 mL) and THF (10 mL). How do you determine the temperature needed? Do I go via the boiling point of diethyl ether or THF?
If more details are needed regarding the reaction I am carrying out then please let me know. But my query leans more towards the determination of mixed solvent systems in general and how one would determine the reflux temperature needed.

Why do you need to know an exact number?: you can turn down the heat when reflux is achieved and that will be the right temperature. This gives a simple self-calibrating way to get the right temperature where you don't even need to measure the temperature.

Put a thermometer in your reaction when it is refluxing, and you will know.

Or do you mean the bath temperature outside your reaction vessel?

Hi Matt and Karl, I am a student and didn't know this, I'm very new to some of the practical chemistry I'm carrying out. The paper I was working from didn't give the temperature used so I just wondered. Anyway thanks for your help

You don't have to know the temperature if you supposed to reflux a reaction mixture. You add correct amounts of reactants and solvent(s), and heat it until mixture is boiling. However, make sure you have efficient cooling system for your upright condenser, specifically when you are heating diethyl ether mixtures (no flame; use a heating mantle or a hot plate).  If you really want to know the temperature for reporting purposes, follow Karl's instructions.

Generally speaking, the mixing solvents may yield azeotropic mixtures with / without a maximum / minimum temperature along the boiling points of the two components -- which does not show for $\ce{Et2O}$/THF (http://www.ddbst.com/en/EED/AZD/AZD%20Diethyl%20ether%3BTetrahydrofuran.php), but e.g., $\ce{CHCl3/MeOH}$ and $\ce{EtOH/H2O}$ do (https://en.wikipedia.org/wiki/Azeotrope).  With something added, the boiling point may increase (ebullioscopy) - so, rely on an experiment.

Trial and error, but most often the boiling point of the lower-boiling solvent.
The absolutely correct answer would be to look at a phase diagram, determine whether you have a positive or negative azeotope (or non at all), find out where on the phase diagram you are and then derive the temperature. That would tell you approximately at what temperature your mixture will boil – after you factored in boiling point increase due to dissolved compounds.
In practice, you need to set your oil bath to some 5 or 10 degrees higher than the boiling point of the lower-boiling solvent anyway. So you can just ballpark it, selecting a temperature that would keep the lower-boiling solvent refluxing. Come back after 15 minutes, if it is not refluxing, increase by 5 or 10 degrees.

And for your specific case: refluxing ether is room temperature ;)


text:
Why can't an "upside down" chiral molecule neutralize the polarized light effect?
Steve Mould did a nice video a few months back called "Why Sugar Always Twists Light To The Right" (link is timestamped at the critical part).

However his explanation was not complete.
He explains that molecules that have mirror symmetry, ultimately cancel each other out because any polarized light that gets warped by one molecule, gets de-warped by the next "mirrored" molecule.
However, he continues, chiral molecules (which are like hands, in that left and right cannot be "stacked"), don't do this because they don't have mirrored molecules, so any effect on polarized light passing through never gets "undone". It only gets "twisted" in one direction.
Here's what remains unexplained:
Even two left hands CAN be stacked in such a way that they WILL cancel each other out. Yes perhaps they will not be oriented in a "high five pose", but in a floating stew why can't they be lying on top of eachother in such a way that they completely cover eachother and undo eachother's effect?

Sugar is a generic name. What Mould probably meant was that sucrose always twists (plane polarized) light to the right. (+)-Dextrose (D-glucose) is dextrorotatory and (-)-levulose (D-fructose) is levorotatory. Both of them are sugars.

A right-handed screw, turned upside down, is still right-handed.

@Mithoron A bit but I still don't get why there aren't "upside down" molecules in the stew as well

This question is basically a duplicate of that linked in the comment by @Mithoron

Take the spring out of a ball pen, and have a good look. It's chiral, and the twist cannot be undone. A flat molecule (a flat hand, neglecting the different palm and back side) is NOT chiral.

Two left hands don't "cancel each other out" they reinforce each other. If they "completely cover each other" they both have the same effect.

I comment again because the doubt of OP seems to rises because s/he is missing the mechanism of optical rotation. Thus this discussion togheter with the above should fix the issue: https://physics.stackexchange.com/questions/15503/how-do-optically-active-compounds-rotate-plane-polarized-light

The key point is that you need the high-five orientation for the two effects to cancel each other out.


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Why doesn’t it burn?
In this video by Chatzida - 11 Fascinating Chemistry Experiments (Compilation) there’s an experiment where they light a bank note on fire, but it doesn’t burn. Why doesn’t it burn?

In order for a combustible substance to ignite, it must first attain a certain temperature called the 'autoignition point', which is the minimum temperature that the substance requires to begin combustion. In this case, the banknote didn't 'burn' because the water present absorbs the majority of the energy given off by the combustion of ethanol, which keeps the temperature of the banknote below its ignition point. As per the explanation given by the video, "water protects the paper by absorbing the energy of alcohol combustion". Water is particularly good at absorbing energy with a minimum increase in temperature due to its hydrogen bonding, which is the attractive forces between the polar ends of water molecules.

The solvent was ignited with a lighter so it isn't the "autoignition point" but rather the "ignition point".


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What does enthalpy of formation at room temperature mean?
$\ce{AlCl3}$ has a standard enthalpy of formation some high negative value at $298\ \mathrm K$ $(-705.63\ \mathrm{J/mol})$. It means it's an exothermic reaction i.e will release heat.
But my question is how can it release heat when it (aluminium) has not even started reacting at room temperature with chlorine.

Shouldn't it be -705.63 kJ/mol, not J/mol?

The value -705 doesn't matter and unit also I am asking if such a large energy is released i.e it's an exothermic reaction in 298k or room temperature,my question is how is it possible because at 298k Al and cl2 doesn't even react

That was asked like a dozen times - it has nothing to do if it reacts, or even exists in room temp. It's just a standard value.

The molar enthalpy of formation at room temperature is not dependent on the reaction effectively happening at the room temperature.
It follows the general principle of state variables, being related to the Hess's law, which refers to the fundamental law of energy conservation. The principle says, that the change of enthalpy ( or any other state variable ) does not depend on the way, how you reach the new state, but only on the initial and final states themselves. If it had not been so, we would have been able to easily construct a perpetuum mobile that would have been switching between 2 states along 2 different paths, producing energy from nothing.
Particularly applied to the $\ce{AlCl3(s)}$ formation enthalpy, it can be formally expressed as the sum of respective enthalpies of these formal steps  (see the Hess's law):
Enthalpy of $\ce{Al}$ atomization: $\ce{Al(s) -> Al(g)}$
Enthalpy of $\ce{Cl}$ atomization: $\ce{3/2 Cl2(g) -> 3 Cl(g)}$
Enthalpy of reaction : $\ce{3 Cl(g) + Al(g) -> AlCl3(g)}$ ( more exactly $\ce{Al2Cl6(g)}$ )
Enthalpy of condensation : $\ce{AlCl3(g) -> AlCl3(s)}$
Another way of dealing with measurement of formation or reaction enthalpy, that are too slow at room temperature, is measuring it at high temperature.
Then $$\Delta H_{\mathrm{r},\mathrm{T}_1}= \sum_\mathrm{reactants} n_i \cdot C_i \cdot ( T_2 - T_1)  +  \Delta H_{\mathrm{r},\mathrm{T}_2} + \sum_\mathrm{products} n_i \cdot C_i \cdot ( T_1 - T_2) $$
$n_i$ are molar amounts.
$C_i$ are molar heat capacities.
That means the reaction enthalpy at room temperature = Enthalpy of heating up the reactants + reaction enthalpy at high temperature + enthalpy of cooling down the products.
All kinds of enthalpy chnages can be combined and sumed, i.e. enthalpy of formation, reaction, phase change, heat exchange etc.

Yeah thanks. Let me sure am I right in my understanding. The enthalpy change is not measured directly by performing a particular reaction in this case Al +Cl2 but actually observing various other enthalpy changes that may be enthalpy of atomisation enthalpy of formation of Al etc . In other cases where so4²- etc involved then we can actually interpret the enthalpy of formation of so4²- in the reaction to calculate the enthalpy of formation of any other substance like FeSO4 etc.

Yes. See also the answer update. All kinds of enthalpy chnages can be combined, i.e. enthalpy of formation, reaction, phase change, heat exchange etc.

@ Debakant. Forget about sulfate. You are right in saying that the enthalpy change is not measured directly, but is obtained by actually observing various enthalpy changes and combining them. First you measure the amount of heat $\ce{Q1}$ necessary to heat aluminum metal up to the temperature where it reacts with chlorine. Second you measure the amont of heat $\ce{Q2}$ produced by the reaction. Third you measure the amount of heat $\ce{Q3}$ produced by cooling the obtained $\ce{AlCl3}$ to room temperature. These three values are then combined to get your final enthalpy change.

You heat your calorimeter with Al and chlorine (electrically, recording the energy you put in) from 298K until the reaction starts.
You record the final temperature after the reaction.
After cooling the calorimeter down to 298K, you heat it (with the AlCl3 in it) until it is at the same temperature again, and record the energy needed.
Now you subtract the two energies, and (if you put 1mol Al and 3/2 mol Cl2 in the calorimeter) the difference will be, lo and behold, 706 kJ.