text:
Why it the electron affinity of beryllium is greater than nitrogen?
As we know that fully filled electronic configuration is more stable than half filled electronic configuration, so in my opinion beryllium's electron affinity should be less than nitrogen's.
Is my logic correct and if so why does beryllium demonstrate such an exception?

The premise is false here -50 is clearly lower then-0,07... https://en.wikipedia.org/wiki/Electron_affinity

This is not true.  This page reveals that both electron affinities are negative, meaning the anions are unstable and, if forced to form, they will spontaneously spring the extra electron off.  Furthermore, the electron affinity is more negative for beryllium (-0.52 eV) which must accept the electron into a higher subshell, than for nitrogen (-0.0007 eV) which merely must disturb an already half-filled subshell.


text:
How does degree of freedom work
I understand that the degree of freedom is the number of ways in which a molecule can  acquire energy. I don't want you to explain the definition, but I want to see how it works. Please explain how is that a real thing, I want to see it. Try to describe the ways in which the molecule can acquire energy . I really can't see it.

Forget the energy. Degrees of freedom are the ways in which the molecule can stretch or bend.

@IvanNeretin Whoa what is that sir. How can we tell the ways in which molecules bend. Why can't they bend however they want to. This is more confusing.

Possible duplicate of What are the degrees of freedom?

Why can't you bend your arm however you want to?

@IvanNeretin I don't study biology. But it maybe due to the bone joint design. Sir maybe I get this: Nature has its own designs

OK, forget the arm. Think of a diatomic molecule. In what way(s) can it stretch or bend?

Both atoms can go upwards and go sideways and they can do it opposite directions . They can go at angles. They can bend around anywhere in the 360 degrees they got

Sure, but how many variables do you need to describe the way they are positioned relative to each other?

@IvanNeretin 3 maybe?

Well, what are those 3?

@IvanNeretin distance and angle between them ?

@IvanNeretin I first thought you asked about coordinates sir. Then I saw variables and the word "relative"

What's an angle between the two points?

@IvanNeretin that not defined. But we can draw a horizontal passing through one point and specify the other's orientation with the angle . Right sir?

No. Imagine a molecule in which the two atoms are 2Å apart and one is above the other. Now imagine a similar molecule in which the two atoms are 2Å apart and both are in the same horizontal plane. Just how are they different? They aren't.

@IvanNeretin Ok sir then I got it wrong  . I'm listening

See, there is only one variable: it is the distance between them.

@IvanNeretin ok sir

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text:
How does addition of impurities in a solid decrease the melting point?
In my textbook, it is written:
The addition of impurities in a solid decreases the melting point of the solid.
How does it do so? Why can't it increase the melting point?

A solid is a lattice held together by intermolecular forces rigidly (solids are rigid relative to a liquid). As you increase temperature, you add thermal energy to the lattice. At a certain temperature, the thermal energy overpowers the intermolecular forces (lattice energy), and the lattice breaks down. This is the melting point. The lattice energy stems from ordered structures. Impurities interrupt the crystalline structure, thereby weakening the intermolecular forces, which lowers the lattice energy and therefore the amount of thermal energy needed to melt, ie. the melting point decreases.

A liquid btw isnt a rigidly ordered structure. This flexibility allows it to incorporate impurities more efficiently into a network of maximum intermolecular forces, and so impurities can increase the boiling point. The inability to efficiently bring inpurities within an existing ordered crystal is why the mp wont increase.

Agreed. At first, I thought there's no way impurities lower mp. Like, when I add an impurity in any solvent, I'm kind of not allowing it to provide more 'space' for it to melt effectively because of it's addition. But, The kind of 'interuption' that impurities add lowers it's lattice energy and so the mp decreases. Ain't it?

one of the most prevalent (and simplest) model which explains this fact is colligative freezing point depression.
Freezing (or melting) point is temperature at which solid and liquid phase are in equilibrium, and their vapor pressures are equal. Adding (non-volatile) impurity in solid phase lowers the vapor pressure of solid phase (because you lower the "concentration" of your compound in solid phase), and phase equilibrium can be achieved at lower temperature.
Difference between melting points for pure and impure compounds can be quantified with following equation:
$$ \Delta T = i K_f b $$
where $\Delta T$ is melting point difference, $i$ is Van't Hoff factor which tells you on how many species does the impurity dissociate, $K_f$ is cryoscopic constant which tells you how great the effect is, and $b$ is molality of the impurity.
Basically, all impurities lower the melting point, if they do not interact strongly with the given compound. In some cases (i.e. urea/choline chloride) freezing point depression can be in order of 100 °C! Organic compounds in particular have pretty high cryoscopic constants, which can cause headaches in organic synthesis (in most cases, you end up with some kind of oil which has to be purified to obtain solid product).
Elevation of freezing point is possible only if pressure is decreased, but the effect is so minuscule that it is not practically useful.

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Why does 1,3-dichloropropane not show stereoisomerism?
It is a multiple choice question:

Which compound does not show stereoisomerism?
A 1,2-dichloroprop-1-ene
B 1,2-dichloropropane
C 1,3-dichloroprop-1-ene
D 1,3-dichloropropane

I can eliminate A and C as they show geometrical isomerism as there is a double bond with different priority groups on either side so these can form E and Z isomers.
Leaving B and D and by the nature being an alkane means that they cannot have geometrical isomerism as they have no double bond. Leaving us to identify which one, therefore, has optical isomerism. Which occurs from my understanding from a chiral center where a carbon is bonded to four different groups. But I cant apply this knowledge to understand why it is D as carbons in B and D are bonded to four atoms so surely they both show optical isomerism which is incorrect given the answer.

A carbon in methane is bonded to 4 atoms too. Do you understand why there is no isomerism here?

In option number D, the 3rd carbon is attached to 2 hydrogen atoms. This immediately provides your answer. For optical isomerism, case is: Carbon should be attached to 4 different atoms or group of atoms.
For B, the central(number 2) carbon is attached with Cl atom, H atom, CHCl group and CH3 group which are all different, and thus shows optical isomerism. Again implying your answer is D.

This is definition of a chiral center.

A chiral centre is an atom that has four different groups bonded to it in such a manner that it has a nonsuperimposable mirror image.

For example:

Carbon at second position in 1,2-dichloropropane is chiral, as all atoms attached to it are different.

Whereas 1,3-dichloropropane does not have any chiral center.

good, but please use PNG images, to prevent noise garbage.

@mykhal what is Noise garbage?

(There's an ugly artificial additional (thus “garbage”) noise because of lossy wavelet transformation (even though quality was set to 90%) at the edges of the graphics in the 1st image, if you ca'nt see that, depends also on monitor technology and settings.)


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Reaction of chloride of 4th A group
Why CCl4 does not react with water,while SiCl4 do react,as both central atom are surrounded by 4 chlorine atom?

Reaction of a compound with water is called hydrolysis.During hydrolysis, Si atoms of the chlorides make up of their empty d-orbitals in their valence shells to form coordinates bonds with water molecules.
However, C is in period 2 does not have d-orbitals in its valence shell to form coordinate bond with water and is not hydrolysed by water.
extra reading https://www.chemguide.co.uk/inorganic/group4/chlorides.html

I would tend to believe steric effects before $d$ orbitals.


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What is to be called as solvent?
My book says:

In a solution consisting camphor and nitrogen gas, camphor (solid) is solute whereas nitrogen gas is solvent.

Doesn't this contradict the statements that:
(1) Generally the component present in large proportion is the solvent.  I mean camphor will be in larger proportion as it is a solid. We can diffuse much nitrogen gas.
(2) Solvent determines the physical state in which solution exists. If nitrogen is solvent, then how can the solution be in gaseous state? Or if the statement is true, does that mean that the solution exists in gaseous state with camphor particles of collidial size?

Nitrogen gas can get dissolved in a condensed phase such as camphor, but not dissolve it. Unless we speak o e.g. supercritical CO2, which is not really a gas.

Regarding part 1
Camphor will readily sublime and so, in proportion to the nitrogen gas, it is the minor constituent
For part 2,

If nitrogen is solvent, then how can the solution be in gaseous state?

if the nitrogen is the solvent then of course it will be in the gaseous state as nitrogen is a gas.


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Can two lone pairs in different orbitals make a chiral centre?
Is it possible that if the two lone pairs in a molecule occupy different orbitals, then could this make the molecule chiral? What I mean is, suppose a carbon atom is attached to two different atoms and has two lone pairs in different orbitals, then could the carbon atom be considered chiral?

You could say that in methane each C-H bond is formed from a different orbital, but that doesn't make methane chiral, does it?

Chirality refers only to the positions of the nuclei. So, if we have the situation you described where one Carbon is bonded to two other atoms (e.g. CX$_2$ or CXY) it can never be chiral. (i.e. a three atom molecule can always be superimposed on its mirror image by a single rotation)
Now, why don't we consider the atomic and hybrid orbitals  populations when assigning chirality? This comes down to the fact that the atomic and hybrid orbitals are just a helpful tool for rationalizing/predicting the structure and reactivity, but they are not representations of the actual wavefunction. The true wavefunction is defined by a set of ''molecular orbitals.'' These orbitals always have the same symmetry group as the nuclear coordinates. In this case, there is a reflection through the plane defined by the three atoms (some jargon: this is called the C$_{s}$ symmetry point group). So if the nuclear positions are non-chiral, so is the wavefunction.
(More on chirality and symmetry: http://symmetry.otterbein.edu/tutorial/applications.html)

Thanku , but I never said the carbon has to be bonded to identical atoms , I said different ones

Doesn’t change the situation at all. Any three atom molecule will have at least Cs symmetry and thus cannot not be chiral. Symmetry classifications are at the heart of this.

If we have triatomic $\ce{X-Y-Z}$ and Y has 2 lone pairs and, for the sake of argument, one lone pair is in an $\ce{sp^3}$ orbital and the other lone pair is in an $\ce{sp^4}$ orbital, then isn't $\ce{X-Y-Z}$ chiral since it has no symmetry element and is non-superimposable upon its mirror image?

Ron, a three atom molecule always has at least Cs symmetry because you can always draw a mirror plane (the plane that the three define). Remember that the electronic wavefunctuon (what actually is going on) always conforms to the symmetry of the nuclear coordinates. This is a case where the hybrid orbital concept doesn’t really match well with reality.

In water the two lone pairs are in orbitals with different hybridizations (roughly p  and sp), so in this case symmetry planes remain. But if we had a case (my earlier comment) where the two orbitals had different hybridizations such that the symmetry is destroyed, the, it seems to me, the molecule is no longer superimposable upon its mirror image.

Hybridization is a simple model used to make sense of a much more complicated reality, but it does not actually reflect the positions of the electrons. So, to your earlier comment, hybridization has no meaning in this context. (In the same way that benzene is not actually alternating single and double bonds).  A classic text on this would be "Chemical Applications of Group Theory" by Cotton

@PJR Perhaps I'm missing your point, but let me rephrase one more time.  1) The hybridization, geometric location, wavefunction, etc. describing a lone pair of electrons can be responsible for why a given molecule is chiral (e.g. chiral amines). 2) When a molecule has 2 lone pairs they can exist in orbitals where the electron density has different spatial distributions (water is an example). If there are 2 other substituents attached to the atom with the 2 different lone pairs, then I would maintain that the molecule would be capable of chirality.

What you've described in the case of chiral amines is a (very cool) electronic effect (specifically, the Pseudo Jahn-Teller (pJT)) which drives the geometry away from a symmetric structure. The lone pair argument works to rationalize the resulting structure, but it does not imply that there can be asymmetric orbitals when the nuclei have a symmetry (i.e. Cs). Thus, the minimum number of atoms you need for chirality is still 4. (e.g. NHClF in shape of ammonia) With fewer than 4 atoms, pJT cannot drive the geometry of the nuclei to a symmetry without a mirror plane.

There are quite a few molecules with two lone pairs residing in different orbitals that still are achiral. A simple example would be methanethiol, $\ce{H3CSH}$. On sulphur, due to its position in the 3rd period the best description is usually to consider the atom unhybridised, which is reflected by the bond angle being close to $90°$. This puts one lone pair in the 3s and one lone pair in the 3p orbital of sulphur.
While these two lone pair orbitals are clearly different that alone would not let the molecule be chiral: an s orbital is spherical in nature and a p orbital can be approximated by the rotation of a figure of 8. The important bit is that both feature a plane of symmetry that traverses them and further contains the carbon atom and two hydrogens. Since the molecule features a plane of symmetry, it is achiral. Yes, the p orbital is actually antisymmetric with respect to this plane of symmetry but the electron phase is not exactly an absolutely measurable property.
This thought experiment could be generalised. But I will now approach the answer from the other direction and talk about how orbitals are actually derived. They are all solutions to the Schrödinger equation $\hat H\Psi = E\Psi$. Because protons and neutrons are 2000 times heavier than electrons, we apply the Born-Oppenheimer approximation in calculation that says we can consider the position of the nuclei constant, calculate an appropriate wave function, then vary the nuclei positions slightly, see if we find a lower energy, rinse and repeat until we arrive at a minimum.
Calculating the energy requires to determine the electric potential that the nuclei are causing and see how the electrons would form standing waves in said potential.
If we temporarily ignore the methyl hydrogens (we were only interested in the lone pairs on sulphur), we only have three atoms which will always form a plane. This means that the electric potential will always have a plane of symmetry. This in turn means that the electronic structure will also feature that same plane of symmetry. But a plane of symmetry means achirality; therefore two different orbitals will still obey the underlying symmetry and will not be able to cause chirality.
There’s a third argument: we cannot distinguish individual electrons. Each electron pair is identical to each other one. Thus, two electron pairs being in different orbitals will always function like two hydrogen atoms—no chirality observed.

Jan, relative to your third argument: Don't the two lone pairs have different energies, basicities and occupy different spatial locations??

I think you are mixing some concepts.
Completely agree with PJR: chirality is related to the positions of the nuclei in the molecule. If your argument was correct, water H2O will be also chiral.
You also should consider that "chirality" is not restricted to organic compounds, you can also find it in inorganic coordination complexes.

Independent of the lone pair stereochemistry, the two oxygen substituents are identical (hydrogen) in water.  Hence a plane of symmetry is maintained perpendicular to the $\ce{H-O-H}$ plane and bisecting the oxygen.  Consequently water could not be chiral.

Of course your are right, I was thinking of HOD. coming to thing about it, the two lone pair are idnetical


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Vinyl vs Acryl IUPAC Usage
I have read that the word Vinyl and Acryl, both refer to the same group.
The difference is that Acryl is used for naming when the functional group attached itself contains a carbon atom while Vinyl is used in the other case.
Then why is the name of the following compound:

Vinyl Acetate and not Acryl Acetate?

Vinyl is CH2=CH- while acryl is CH2=CH-C. So, vinyl acetate is correct; acryl acetate is not. CH2=CHCN is acrylonitrile (a nitrile is a triple-bonded nitrogen) or, in principle, vinyl cyanide. I heard a story that a chemical company tried to ship a tank car of vinyl cyanide but it was prohibited because "cyanides are toxic". It shipped without difficulty as acrylonitrile. See also, acrylic acid, acrylamide and methyl acrylate.

I saw your answer just after I commented. The point is there are no hydrogens on the bare carbon otherwise we would be using the term "allyl". All of our examples have no hydrogens on the bare carbon.

Your question is mistaken, vinyl and “acryl” are not the same (and the latter standalone name even does not mean anything in IUPAC nomenclature, it's like “acet” in ‘acetyl‘/‘acetic’). See the following structures (preferred IUPAC names/prefixes are in boldface).
$\ce{CH2=CH\bond{-}}$
vinyl
ethenyl
$\ce{CH2=CH-COOH}$
prop-2-enoic acid
acrylic acid
$\ce{CH2=CH-CO\bond{-}}$
acryloyl
prop-2-enoyl
1-oxoprop-2-en-1-yl
$\ce{CH2=CH-CO-NH2}$
acrylamide
prop-2-enamide
$\ce{CH2=CH-CO-O-CH=CH2}$
vinyl acrylate
ethenyl prop-2-enoate
(But now I understand that one might somehow deduce from some structures that the vinyl component is named acryl.. and there even might be some historic reasons, which I'm currently not aware of.)

Your answer is comprehensive and informative but inaccurate. "Acryl" or "acrylic" are alternative names for the acryloyl group and widely used in the chemical literature.


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Role of Ag2O in Wolff rearrangement
In Wolff rearrangement, what is the role of $\ce{Ag_{2}O}$ in migration of alkyl/aryl group?

There is no agreed mechanism for the Wolff. The most convincing thing to say is that it facilitates the loss of dinitrogen.

Will the reaction happen without the presence of Ag2O ?

yes, with heat or light. Some rearrangements take place at room temperature


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Why does atomic charge not balance in nuclear reactions?
In the discovery of neutron, the bombardment of beryllium with alpha particles is well known.
$$\ce{^9_4Be + ^4_2He^2+ -> ^12_6C + ^1_0n}$$
I don't understand how the electrons add up. As alpha particles do not have electrons, the total electrons should be 4 accounting for the electrons of beryllium, but after bombardment, there is a deficiency of 2 electrons in carbon atom. Shouldn't the $\ce{^12_6C}$ atom be positively charged? Like:
$$\ce{^9_4Be + ^4_2He^2+ -> ^12_6C^2+ + ^1_0n}$$

High energy alpha particles yield high energy intermediates that, yes, may be highly charged.

@Karl While I agree with your opinion that this has a physical bent to it, I nevertheless find it difficult to accept the notion that the composition of atoms would be considered off-topic for chemistry.

In nuclear reactions we leave charge balance for other processes. There are enough electrons in the universe to neutralize the C at some point.

Sure enough, carbon will be positively charged, but it will have more urgent issues to worry about at the moment.

Technically, when you write the neutron you should have 1 on top and 0 on bottom, thus $\ce{_0^1n}$.

There are two reasons why nuclear reactions don't consider (or often account for) charge.
The first is that charge is usually irrelevant to the reactions. If we are looking at changes in nuclei we can for all practical purposes ignore the electrons that would be associated with the nuclei under normal conditions. They are just irrelevant to the nuclear reaction taking place.
The second is that they do account for charge, but only in nuclear processes. Some radioactivity (isolated neutrons decay or any nucleus undergoing beta emission) can lead to the decay of neutrons into a proton and electron and some other minor particles. The electron clouds that would normally surround the nucleus are still irrelevant to these processes but the equation for the nuclear reaction should account for the emitted particles. When writing these equations only the nuclei (and the emitted particles) are accounted for and any net change of charge is ignored as a description of the nuclei involved implicitly describes the net nuclear charge (in the same way that many organic chemical structures don't explicitly draw every hydrogen as chemists can fill them in knowing that the total valence around carbon should add to 4).

Your right, the carbon will be charged, but it is not typically represented that way because it is a nuclear reaction and it is superfluous to consider charge since the charges of the atoms do not participate in the reaction (spectator particles).
$$\ce{^9_4Be + ^4_2He -> ^12_6C + ^0_1n }$$
Given that the carbon is in a beryllium target (maybe metal, maybe oxide) the carbon could just steal the electrons from the beryllium, you could think of it more like:
$$\ce{Be (s) -> Be^2+ +2e-}\\
\ce{^9_4Be + ^4_2He^2+ -> ^12_6C^2+ + ^0_1n ->[2e-] ^12_6C + ^0_1n}$$
But again charge is superfluous to the reaction and materials can have unbalanced charges (static electricity).

Another way of looking at it is as follows: the atom starts out neutral with its electrons and ultimately ends up neutral; if an electron is emitted it is eventually made up by capturing an electron. But the binding of the electron to the nucleus carries very little energy compared with the nuclear forces operating within the nucleus and driving the nuclear reaction. So the course of the electrons is of little consequence to the energy balance which is of greater interest in nuclear physics.


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Materials used in cooktop heating element, and reason for black appearance
I have been interested in the structure of heating elements/coils used in cooktops. I am referring to something such as in the following picture:

I have a cooktop at home with such elements, and at parts where there was more scraping, I can see a silver-ish colour rather than the black, suggesting the black material is a thin layer around some metal resembling stainless steel, in appearance at least.
I have read on Wikipedia that often such heating elements would have an outer layer made "of stainless steel alloys, such as Incoloy, or copper" (and similar statements elsewhere). This seems to suggest that the silver-ish layer I have seen at the scraped parts in my cooktop is indeed stainless steel.
What I am interested in is what the black layer is.
My question is why these heating elements appear black in appearance; what the black layer is.
I have a possible guess, which I would also be interested to affirm or negate with you, which is that this is due to oxidation of the metals, which perhaps is done by the manufacturer(s) in advance because they expect that the high temperature that the metal would get to during cooking would itself speed up  oxidation and the resulting discoloration, and doing the oxidation in advance (and in a controlled manner which does that uniformly, so the colour seems uniform) prevents that from happening by forming a protective layer of oxidised metal. One reason I am guessing that, is that I looked on Wikipedia at possible oxides of iron, chromium, and nickel, and all of these seem to have at least one oxide with a black appearance. Iron for example has $\ce{FeO}$ and $\ce{Fe3O4}$.

The heating element ( on the inside where you cannot see it ) is Nichrome or similar alloy ( 10 to 20 % Cr, bal. Ni ) . The surface sheath is typically Inconel 601 ( 15 Cr, 10 Fe, 0.2 Al, bal. Ni). They were Inconel 600 ( and maybe some still are 600 ) until the addition of a small amount of Al was found to increase oxidation resistance. There are several similar alloys and there are different names for the same alloy. So the black surface is oxides of Ni. Cr with some Fe. Some compositions and temperature exposure will produce a green tint in the black ( in high temperature furnace tubes of the same family of alloys). Diffraction analyses of the oxides ( from furnace tubes) often do not give specific compositions as Cr, Ni and Fe can substitute for each other.

Copper based alloys are not used for high temperature.

You are right to think that the oxide is predominantly $\ce{Fe3O4}$ this is a common coating for steels to prevent corrosion and is done intentionally through a process called bluing. Of course with alloys the oxide is impure and there is likely some chromium and Nickel doped in the material as well such as $[\ce{(Ni,Fe)^{II+}(Fe,Cr)^{III+}2O4}]$, but the predominant phase is $\ce{Fe3O4}$ with a posibly small amount of chromite ($\ce{Fe_{2-x}Cr_xO3}$).


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Magnetic couplings using DFT in a spin frustrated system
I’m working with a trimeric copper system that have the classic triangular shape and experiences spin frustration. This is my first time dealing with such system and, so far, from what I’m reading, I know that the spins resonate between ↑↑↑ ↔ ↑↓↑ ↔ ↑↑↓ and that the triangle have couplings $J1_{ab}$, $J2_{bc}$ and $J3_{ca}$.

My question is: how it is possible, using DFT, to obtain the three magnetic couplings ($J1_{ab}$, $J2_{bc}$, $J3_{ca}$)? I’m supposing that the strategy of breaking the molecule in three dimers and calculating a broken-symmetry is not going to work, right?
Thanks in advance
P.S.: I’m doing my calculations using the ORCA package.

Cross-post: http://www.ccl.net/cgi-bin/ccl/message-new?2018+10+13+001

DFT is fundamentally the wrong tool for a system like this. When you say that the spin 'resonates' between several different states what you are explicitly saying is that the wavefunction is strongly statically correlated. DFT is a single reference method, so any experimental agreement you get from it in a case like this is at worst wrong and at best right for the wrong reasons (and the range of functionals will likely cover that whole spectrum).
If we knew the exact density functional DFT would, of course, be exact. Unfortunately, real functionals are parametrized for mostly organic compounds and the occasional single metal, but certainly nothing as correlated as a spin frustrated Cu$_3$ cluster. If you want a result which is believable, you need to use a real multireference method. (It's not going to be a quick calculation, but you only need a single point so it's not too bad)


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Specific kind of plastic that won't make fabrics smell?
I store various fabrics (clothing, cloths, linens, etc.) in RubberMaid RoughNeck plastic bins.  It seems to work fine for a while, but given enough years, the contents take on a strong chemical smell.  These are not stored in a musty or damp area -- they are stored in my living room.  I assume, therefore, that the smell is from the plastic.  Before using the contents again, I need to wash them.  This means competing for machines in the laundry room of a highrise.  This entirely defeats the point of keeping the seldomly used items, which was to have them on stand-by for immediate use upon demand (as opposed to going out to re-buy them).
Are there specific types of plastic that I should look for in a bin to avoid this kind of smell from the plastic?

Hmm, that defnitely doesn't make it convenient.  Some linens are accessed more frequently than others.  But I guess it's either that or use bureaus and drawers.  The closets in my place aren't a good option because they smell funny, and research shows that it might be hidden mold (2 of the 3 closets are adjacent to the washroom, and they are the smelly ones).  It would be easier to buy my own place than to convince the landlord to rip up the walls/floors.  Did you want to post your comment as the answer?

comment moved to answer, with additional suggestions.

It is likely a plasticizer in polyethylene bins, rather than polyethylene itself, that has an odor. Storing the garments in clear polyethylene plastic bags (perhaps inside the bins), which have different plasticizers, should reduce the odor -- and some bags are designed to compress clothing with use of a vacuum cleaner, saving space.
Another option is to get an inexpensive wooden (cedar lined?) chest, or a cloth storage bag. Advantage: breathable. Disadvantage: does not protect from moisture.
BTW, some of the plasticizers include polybutylene, dioctyl  phthalate, glyceryl tribenzoate, polyethylene glycol, sunflower oil etc. Handbook of Plasticizers Ch. 11, G. Wypych. Perhaps rancid sunflower oil is the odor?

Polyethylene does not contain plasticisers. The $T_g$ of PE is far below freezing.

I'm not a chemist, but the bin is recycling code 4 and has "LDPE" stamped on it. I was prompted by the above comment to do some online searching.  According to http://sciencing.com/difference-between-polyethylene-pvc-8531363.html, LDPE is already soft.  According to http://www.foodpackagingforum.org/news/which-plastic-food-containers-are-safe-to-reuse, LDPE typically doesn't have plasticizers. DrMoishe Pippik, I apologize that I may ave prematurely asked you to post your answer.

I wonder if the residue from the fabric softener or the detergent (for the scent) may degrade over many years and start to smell like chemicals


text:
Major product of isomerization reaction
I am trying to come up with a logical mechanism for the following transformation, where 2-methylenecyclobutan-1-ol is converted to cyclopentanone under acidic conditions:

What I have tried to do was contract the ring to a methylcyclopropyl carbocation as it is exceptionally stable. I also tried to expand the ring but got an unsaturated primary alcohol instead. Can somebody suggest a proper mechanism for the reaction?

According to J. Chem. Soc., Chem. Commun. 1973, Vol. 4, 103–104: "On heating in a sealed tube at 245° for 4 h or on treatment with 5% H2SO4 at 100°
for 30 min, (3) [your starting material] is quantitatively converted into 1-methylcyclopropanecarbaldehyde." This makes a lot more sense mechanistically, and is similar to a pinacol rearrangement: https://i.stack.imgur.com/UdU5b.png

@orthocresol Yep, this is in agreement with what I know. Can you provide a more detailed explanation of this mechanism , and why it is the only correct one in the form of an answer?

https://chemistry.stackexchange.com/questions/48739/mechanism-for-ring-contraction-from-2-methylenecyclobutanol-to-1-methylcycloprop


text:
Colour of platinum
Why does platinum have silvery shine whereas platinum black (finely powdered platinum) is black in colour although both have the same constitution of pure platinum?
Due to similar composition, both of them will have electrons in similar kind of energy states. So we can't assume this due to charge transfer. Neither is there any scope of charge transfer.

https://chemistry.stackexchange.com/questions/45113/what-is-the-chemistry-behind-black-colour-of-platinum-black

In bulk metals, electrons in the conduction band are shared, creating a "sea" of reflective electrons. A nanoparticle, though, is so small that there is no conduction "band", just discrete levels enforced by the Pauli exclusion principle. Other metals in finely divided form, such as gold or silver, are also are black or darkly colored. Traditional silver emulsion photography relies on white silver compounds turning to black nanoparticles of silver when reduced.
Nanoparticles exhibit interesting color effects due to plasmons. Finely-divided gold particles produce ruby glass, used in stained glass windows, though thin bulk gold films appear green or blue in transmitted light.


text:
How do you figure out fuel specific impulse non-experimentally?
Simple question, probably very complicated answer:
Without building a test engine/rocket and finding out experimentally, that is to say, based on chemical characteristics alone, how can one compute the expected specific impulse of a given fuel?

Fuels produce energy, you can estimate how much for given mixtures from known reactions (or study those regions independently). Then you have to know how your engine transforms that into motion (or whatever else it's supposed to do). I would assume this to be an more or less everyday problem for engine engineers... I think there is another stack exchange for that...

The relevant property is the temperature of the burnt fuel when it leaves the rocket engine nozzle, i.e. when it stops exchanging impulse with the engine.
That temperature you can convert into a velocity distribution of the gas particles, multiply with their molar mass, and that's it. (somewhat less than that of course, because no all gas particles are flying in the exact same direction. depends on the nozzle type, among other influences)
The temperature you can estimate to some extent from the reaction enthalpy, but of course you have radiative and conductive losses. Which is where this becomes an engineering problem.

"Adiabatic flame temperature" is a thermodynamic state variable, at least given a fixed oxidant composition.  It probably could be used to set an upper bound on achievable impulse if you assume 100% efficiencies everywhere.  Dicyanoacetylene FTW!


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Equilibrium expression for dissolution of carbon dioxide
I was asked to give the equilibrium expression for:
$$\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)}$$
As water is a liquid and a solvent here, it shouldn't be included in the expression, hence I wrote:
$$K_c = \frac{[\ce{H2CO3}]}{[\ce{CO2}]}$$
However, the answer of the question included water in the denominator.
What is the correct answer in this situation?

In the case of heterogenous equilibria when molar concentration of pure solid or liquid involved, water in this case is constant or is independent on the amount present its concentration is omitted as per the general norm. The equilibrium constant is just modified to
$$K_c' = K_c[\ce{H2O}]$$
So the book and you are referring to different constants.


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How to derive Arrhenius equation from hard-sphere collision theory
Reading the description of the Arrhenius equation from Wikipedia:

...One example comes from the collision theory of chemical reactions, developed by Max Trautz and William Lewis in the years 1916-18. In this theory, molecules are supposed to react if they collide with a relative kinetic energy along their lines-of-center that exceeds Ea. This leads to an expression very similar to the Arrhenius equation.

I was wondering how to derive the Arrhenius equation from the mentioned hard-sphere collision theory. I tried to do so but got a totally different expression from the Arrhenius equation. The original derivation here written by Max Trautz seems to be in German, but I cannot read German. Does anyone know how to do the derivation?

Reading German physics stuff isn't rocket science, for someone who's fluent in physical English. Give it a try! ;-)


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Oxymercuration demercuration, PCC, NaCN/HCl reaction
In this reaction of the alkene, the correct answer is below but I don't understand why it goes to that. This is my thought process - reagents 1 and 2 do oxymercuration demercuration and makes a secondary alcohol, then reagent 3 turns the alcohol into a ketone, and finally reagent 4 and 5 performs nucleophilic addition and turns the alcohol into a tetrahedral with CN attached (my steps I drew below). What am I doing wrong here? Thanks

After the Cyanide formation, the last step is basically a hydrolysis step.

thanks for the reply, so does the carboxylic acid replace the CN via hydrolysis ? i am still a bit confused

Check the mechanism of the last reaction here.
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch20/ch20-3-5-1.html
It is not the displacement of nitrile by carboxylic acid(which you do not have, btw). It is the combination of water addition and the displacement of ammonia by second equivalent of water.


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What do cannabinioids have in common structurally?
I'm looking at the wikipedia page for cannabinioids, and trying to find structural similarities between them. For plant based ones it's hard enough, the only similarity I could find seemed to be 6 member rings and medium sized hydrocarbon chains. When you include endocannabinoids, it seems to get more complicated. What structural similarities do cannabinioids share?

The 1,3 oxygens off the phenyl ring and the hydrophobic alkyl chain in the 5 position, and a further hydrophobic region away from the phenyl ring attached by one or more oxygen

@Waylander Ok, thanks now that I see it I'm embarresed I didn't see it earlier, but clearly this doesn't extend to endocanabinoids like anandamide or NADA.


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Can an uncharged particle pass through a molecular bond?
Do particles ever pass through or interact with the molecular bond? Is there a physical size limit? What happens to the atoms and the particle and does this ever happen in real life?

It is quite unclear what do you mean.

Would it be possible for something to pass between two bonded atoms

A photon is an uncharged particle and it certainly interacts with electrons in a chemical bond.

like neutrinos?

The molecular bond is not a solid thing. It is merely an energetically suitable arrangement of atoms and their electrons. So another particle that happens to come close to that bond can only interact with those atomic nuclei and electrons. Whether it does interact depends.
Which particles can interact?
Considering electron-magnetic forces only:

Charged particles can interact.
Uncharged small particles can not interact.
Larger uncharged particles (e.g. other small molecules) could get polarized. This means they develop partial charges which then can interact.

How can charged particles interact with the bond?

Slow particles (with low kinetic energy) can get attracted or be repelled by the molecule, depending on the (partial) charges of the particle and molecule. If it gets attracted a chemical reaction can happen.
Fast particles (with high kinetic energy) might just pass through the molecular bond. The bond has actually much less electron density then any atom. Depending on how close it gets to an atomic nuclei, it might get deflect. Or it may even deflect the atom and break the bond. The particle could also grab or deflect an electron of the bond and thus ionize the molecule.
As a third option: there is always a slight chance of quantum tunneling.

Does this ever happen in real life?
Actually a lot: Neutrinos have no electric charge and almost no mass. As a result they can pass through matter without any consequences. They are emitted by stars and pass through earth by millions per second.
An example for charged particles would be using an electron gun to ionize molecules for mass spectrometry. The electron gun creates a beam of fast electrons, which can remove an electron from the targeted molecule upon collision (1 goes in, 2 go out).
Is there a physical size limit?
If the particle becomes larger than the molecular bond is long ... well I would not call it "passing through the bond" anymore. But the above concepts would still be the same.


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What is the driving force of this reaction?
I know that every reaction that occurs in a solution must have a driving force leading to the kicking out of some ions from the solution. These reactions include redox , double-displacements and acid/base. Now my question is if i react aqueous ammonia with aqueous hydrochloric acid what will be the driving force of this reaction?
I know that ammonium chloride is formed, but this salt is soluble in water so exists as aqueous ions. But for reactions in solutions some ions have to be removed, as a result creating a driving force for the reaction. Like in the case of Arrhenius acid-base reactions, the formation of liquid water is the driving force of the reaction.
Please help?

Why not think about reaction of gaseous NH3 and HCl - you get same thing, only not dissolved. Dissolving isn't considered a reaction, as you can get back the dissolved compound.

"for reactions in solutions some ions have to be removed"
Suppose that you have 1 mole of hydronium ions(the counteranion of ammonium cations) and you add 1 mole of hydronium ions(countercation of chloride anions). Nearly all of them will react, resulting in formation of 2 moles of water, which has negligible for the sake of this question dissociation constant $K_w=10^{-14}$. So all the mentioned ions disappear.

Pardon me, but i completely didn't understand your answer. What are you trying to say exactly?

You have here exactly the case of Arrhenius acid-base reaction, which result in the formation of liquid water.

But my question does not include water as one of the products


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Calculate the final molarity from 2 solutions
How do I calculate the final molarity when 70.0 mL of 3.0 M sodium chloride solution is added to 30.0 mL of a 1.00 M solution of sodium chloride>

Molarity (M) = moles (n) / liters (L)
70 mL of 3M NaCl = .21 mol NaCl
30 ml of 1M NaCl = .03 mol NaCl
.21 mol + .03 mol = .24 mol NaCl
70 mL + 30 mL = 100 mL = .1 L
M = n/L = .24 / .1 = 2.4 M

The hidden assumption here is that the volumes of the solutions are additive. It is not strictly true. However for the solutions involved, probably close enough. // 2 significant figures or 3? Did the original problem statement specify the first solution as 3.0 M or 3.00 M?


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Converting a thioester to an amide
What reagents can I use for the following transformation? I think this has something to do with formation of an iminium salt and reductive amination, but I'm not so sure.

No reduction is required. A solution of methylamine will do this at room temp

Waylander is right - there's no reduction involved. Thioesters and amides are at the same oxidation state. Basically any "RCOX" compound where X is a heretoatom, is the same oxidation state as carboxylic acids. This includes esters and acid chlorides.

Thioesters are particularly reactive, relative to esters and amides, so the reaction you've drawn will occur spontaneously at or close to room temperature by treatment with methylamine without other reagents or catalysts. In esters and amides, the O or N (respectively) lone pair overlaps very effectively with the carbonyl π-bond. Attack by a nucleophile would disrupt this conjugation, so esters and amides often need harsh conditions or catalysts to react with weak to moderate nucleophiles. In contrast, the S atom lone pair of a thioester does not overlap as effectively, due to longer C–S bond length and size difference between carbon 2p orbital and sulfur 3p orbital.


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Explanation for the derivative of kinetics equation
I'm working with a first rate law expression.
A + B -> C
Given: $k_1[A][B]=k_{-1}[C]$ which means   $[C]=\frac{k_1[A][B]}{k_-1}$.
The book says $\frac{d[C]}{dt}=k_1[A][B]-k_{-1}[C]$ which is where I'm lost on how to get to that step. I'm looking for an explanation into how this derivative is obtained. Thank you.

It is not a mathematically derived derivative of a different equation, but rather a consequence of the mass balance which is also valid when the system is not at equilibrium. In other words, the forward reaction increases C, the reverse reaction decreases C. The concentration change at any moment depends on the two reaction rates and the concentrations of the species at that time.

So calculus techniques are not used (like quotient rule is what I was thinking), but it's just understood the mass of reactants has to equal mass of products?

Sorry, I edited my answer while you were replying. Does the edited answer make sense?

Note that at equilibrium d[c] /dt = 0. So you want changes in C as system moves towards equilibrium.

Your equation should really be written as $\ce{A + B \rightleftharpoons C}$ to indicate that both forwards and backwards reactions can occur as indicated by $d[C]/dt = k_1[A][B]-k_{-1}[C]$. This is normally the starting point.  Only because of the reverse reaction $\mathrm{C\to A+B}$ is equilibrium is possible.  At equilibrium the rate of change of C is zero so that $k_1[A]_e[B]_e=k_{-1}[C]_e$ and then the equilibrium constant is $\displaystyle K_e=\frac{[C]_e}{[A]_e[B]_e}=\frac{k_1}{k_{-1}}$. You can see how this relates to your second equation.
Often the $e$ subscripts are not added in the expressions which can lead to confusion. It is possible to work out the integrated rate expression for this scheme in terms of the amount reacted at any time and the amounts of A and B present initially.

Stealing bits from porphyrin I'll pose...
Your equation should really be written as: $\ce{A + B <=>[k_1][k_{-1}] C}$ to indicate that both forwards and backwards reactions can occur. Now we're interested in two different situations. First what is happening before equilibrium ids reached, and secondly what happens at equilibrium.

Before equilibrium

Let's use $\ce{[A]_t}$, $\ce{[B]_t}$, and $\ce{[C]_t}$ to denote the instantaneous concentrations of species A, B, and C at some time $t$. Now at time $t$, C is increasing in concentration due to the forward reaction and that rate is given by $\ce{k_1[A]_t[B]_t}$. At time $t$ C is also decreasing in concentration due to the backward reaction and that rate is given by $\ce{k_{-1}[C]_t}$. Hence the net change in C at time $t$ is given by
  $$\dfrac{d\ce{[C]}}{d\mathrm{t}} = \ce{k_1[A]_t[B]_t - k_{-1}[C]_t}$$

At equilibrium

The rate of change of C at equilibrium is zero so that $k_1[A]_e[B]_e=k_{-1}[C]_e$ and then the equilibrium constant is $\displaystyle K_e=\frac{[C]_e}{[A]_e[B]_e}=\frac{k_1}{k_{-1}}$.


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Why can a diamond be broken using a hammer if it's the hardest natural substance known?
I've heard that diamond is the hardest natural known material but, on Google search, I found that it can easily be broken by a hammer as it's not tough.
So, what is difference between hardness and toughness? According to me, both should be the same, and if both are different than what is the toughest known substance?

To complement the excellent answers below, I'll give you an everyday example.  A glass cutting board is harder than a kitchen knife.  The cutting board will dull the knife.  However, the cutting board is MUCH easier to break (less tough) than the knife.

Hardness and toughness are not the same
Hardness and toughness are very different qualities in materials and are weakly related.
Hardness is strongly related to the more well-defined quantity of stiffness which measures how easily a compound can be deformed under stress. Glass and diamond are very stiff materials, for example. If you try to poke them with something they resist deforming to accommodate your poke. (Hardness is not perfectly aligned to stiffness because of small scale microstructures in many materials, but this is good enough for now).
Toughness is a vaguer term for materials and there isn't a simple way to measure it. This is partially because it varies depending on circumstances in a way that stiffness does not and it is a property of the overall structure and not just the materials that make up the structure. Understanding it requires some insight into why things break (and why other things don't). we need to know a lot about the small scale structure of materials not just the substances involved. For example, stainless steel (used in knives and forks) is a well-known tough material but cast iron is brittle. Both are mostly made of iron. The differences are in the crystalline structures.
One key property of tough structures is that cracks don't propagate. So a stick of glass will break easily as will a glass fibre. But a bundle of glass fibres embedded in an epoxy resin can be very tough (because the cracks in individual glass fibres are not propagated through the epoxy resin). Some tough metals can adjust the micro defects in their crystalline structures to absorb the strain that would otherwise propagate cracks. Some very soft compounds are very tough because the deform so easily that it is hard to start cracks, nylon rope for example.
Diamond is very very stiff. But it has no protection mechanism against cracks. So, like glass, once a crack has started it doesn't take a lot of energy to cause it to spread, so it may be stiff but it isn't tough.
A fuller explanation of this took material scientists and engineers a long time to work out would require a whole book to do it justice. Luckily that book has been written and is called The New Science of Strong Materials. It is well worth a read.

Definitely +1, but I preferred the full title given in the link ("The New Science of Strong Materials, Or Why You Don't Fall through the Floor") :)

Glass and diamonds are very different. Diamonds, being crystalline, have cleavage planes.

So a glass skyscraper wouldn't isn't such a good idea?

@JamieClinton No it wouldn't. And neither would a cast iron skyscraper (or railway tracks as was discovered early in the first days of railways). both rails and buildings need a good combination of stiffness and toughness. Stiff and brittle is very bad in structures the bear weight.

@matt_black lol I meant diamond, not glass.... But sounds like it has the same limitations.

Toughness is well-defined: it's a measure of the amount of energy needed to cause a crack in the material to grow.  And "hardness" isn't one measure, but (approximately) two: scratch hardness and indentation hardness.  Diamond ranks very high on both hardness measures, but other materials (such as steel or rubber) can have very different scratch and indentation hardnesses.

@mark You are right. But I was trying to simplify in a way that didn't require me to reproduce several chapters of Gordon's book on the subject. Toughness is complicated; hardness is simpler but not as simple as stiffness, but related.

Is there any known toughest substance

@Harshjain Toughness doesn't really work that way; it's mainly important in relation with how you want to use that particular material - e.g. do you want to hit other stuff with it? Hard or soft? Do you want to use it under tension? Torsion? Compression? Do you want it to hold a constant load forever, or do you get varying loads over time? At what kinds of temperatures and pressures, and what kinds of chemical environments? One reason that steel is such an awesome material (along with its low cost) is that we can make hundreds of kinds of steels for all kinds of purposes.

@Harshjain If you want the material that's toughest against being struck with a hammer against an anvil... try a pillow. If you want it to return back to its previous shape, many rubber-like materials work great - it very efficiently distributes the impact energy over a large volume. But neither of those is something you want to build skyscrapers with :)

@matt_black: I think a useful simplification may be to say that hardness relates to a material's response to levels of stress that are low enough not to cause a significant qualitative change in behavior, and can thus be used to model the interaction of materials at low stress levels, but beyond a certain point such models--if not the materials--will break down.

Diamond has cleavage planes. If you want something nearly unbreakable, try nephrite, which is a tough form of jade used by the ancient Aztecs to make axe heads! Actinolite is another related "tough as steel" mineral. These minerals are made up of interlocking strands (actinolite) or microscopic fibers (nephrite). But diamond is a regular geometric lattice, so if you hit it at the right angle, it will break. Still, you can't make a scratch on glass with nephrite -- but diamond will do the job. That's because diamond is "harder" than glass (which is mostly silicon).

True, but the detailed reason for the toughness of some minerals is complicated and related to the structure which takes a lot of background material to explain. even materials scientists took a good few years to get a grip on fracture mechanics.

So, hardness means to scratch. So can Graphene scratch diamond and how can a machine hold 1 atom thick layer of garaphene

Is there any known toughest substance

@Harshjain There's two main kinds of hardness - scratch hardness, and indentation hardness. But of course, even those are just approximations. As for holding graphene, you can hold it in your hand. There's little special about handling it - don't think that just because it's "one atom thick layer" it will just cut through anything; it handles like any other fabric, mostly. The tricky bit with scratching diamonds with graphene is that graphene is flexible - it will tend to bend under pressure, and you need pressure to scratch things. But then, have you ever cut your hand on a sheet of paper? :)

If toughness is defined as the ability of object A to damage object B, then there's no such thing as the toughest object, let alone the toughest substance. For example, a steel saw can cut a block of wood in two. The very same block of wood, used as a hammer, can break a steel saw in two.
Even if we consider similar geometries, the type of damage would depend on the kind of action involved. A block of diamond would scratch a block of iron, and diamond vise would slowly flatten an iron block, while a fast impact would rather shatter diamond.
Also, you can easily flatten or break a small piece of steel with a steel hammer.

I haven't read all the answers that have been written, but the ones I have read are very good.
This is going to be a simple one:
Yes indeed, Diamond is the hardest substance known and it refers to the resistance to be scratched. In other words, try using a metal point to scratch it and it won't be possible. While you can use diamond to scratch or indent other substances.
It does not mean that it can't be destroyed with a heavy blow, the resistance to this has another name and diamond does not possess it.
It all is reduced to the type of solid in which it crystallizes, normally a stiff and not flexible crystal lattice becomes brittle, easy to break. But as it is very dense, small distance between atoms and or ions, it does not allow penetration.


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thermochemistry problem confusion

I can understand upto where the ideal gas equation is used to calculate volume of steam. But on trying to find out $ \Delta V$ , i saw that there are no gases on the reactant side so then $P\Delta V=(\Delta n)\times R\times T$ where $\Delta n$ would be $1.39$. Then $\Delta V$ is simply the volume of the steam i.e., $42.51$. But when calculating the work a value of $42.8$ has been used for $\mathrm dV$. Why is that done? Secondly, there seems to be two different values for $\Delta$H. First they have written it to be $\pu{12470.6 cal}$. and then while calculating $\Delta U$ they have used the value of $\pu{13500 calories}$. (That value seems to have come from the given data: If we multiply $9.72\times 10^3$ (molar $\Delta H_{vap}$) by $1.39$ (number of moles) we get something around $13500$). How did those two different values appear? I also want to tell here that this is not a hw question for me, as you can see this is a solved example, so you can consider it as my attempt to understand thermochemistry better

Using units will very often reduce errors in calculations and help those you may ask for help

The $\Delta V$ is supposed to be the volume of the vapor minus the volume of the liquid.  The volume of the liquid is only 0.025 liters, which they neglect.
I don't know where they get the 12470.6 from.  I'm guessing this is an error.
(0.082)(373)(25)/18=42.48. You solve it basically the way you said.
Δ(PV)=42.48 liter−atm =1029 cal=42.48 liter-at = 1029 cal
So, ΔU=13500−1029=12471 cal

but sir how did it become 42.8 from 41.52. Even if we consider the volume of liquid, as you said we must subtract, but here they have added 0.39 . Does it have any connection with 1.39?

shouldn't we consider only the substances in gaseous phase(after you said that $\Delta$V  is volume of vapour minus volume of liquid)

I don't know where they get the 42.8 liters from. That seems to me to be an error also. If the initial volume of the liquid were taken into account, it should have been 42.48 - 0.025 =42.45 liters. As far as your second questions is concerned, the initial state is liquid and the final state is vapor. So ΔV should be the volume of the vapor minus the volume of the liquid. In their calculation, they neglect the volume of the liquid compared to the volume of the vapor.

sir , how did we get 42.48(in your comment)? so then how do i solve this problem

(0.082)(373)(25)/18=42.48.  You solve it basically they way you said.  $\Delta (PV)=42.48\ liter-atm\ =1029\ cal$.  So, $\Delta U=13500-1029=12471\ cal$


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What is the topology of a phase diagram?
This question is cross-posted from phys
Looking at various two-variable phase diagrams I was struck by that on every one I have seen so far all the phases formed simple connected regions; see, for example the phase diagrams of $\ce{H2O}$ or of $\ce{Fe-Fe_3C}$ in 1. Every phase is a connected region, no two disjoint pieces correspond to the same phase. Is this really true in general and if so why?
Going further, let us consider a three-variable phase diagram, say $p, T, y$, where $y$ maybe electric or magnetic field intensity, or some other controllable extensive/intensive variable. Is it possible that now a 3d region occupied by some phase is a kind of tube that makes a half turn so that if we fix, say, $y$ then the resulting 2d section would have two disjoint $p,T$ regions of the same phase? Can such phase "tube" exist or by some thermodynamic-geometric rule is it excluded?
A possibly more difficult question if a phase can form an annular  region, ie., homeomorphic to an annulus while surrounding another phase homeomorphic to a disk?
In short, what is the topology of the phase diagram and its sections?

There are any number of binary phase diagrams where the same crystal structure is observed in several discrete regions. For some of them, it is known that those regions can be connected using, e.g., a third element. Fe is a common case, since the fcc $\gamma$ phase sits between regions $\alpha$ and $\delta$  bcc. So, with, say, Fe-Si the bcc phase exists continually around the fcc phase, while in Fe-Ni the fcc phase extends across the entire phase diagram.

@Jon_Custer Thank you and very interesting but is it fair to say that $\alpha-Fe$ and $\delta-Fe$ are the same phase when the former is ferromagnetic and hence has domains while the latter is above the Curie temperature and is not magnetic at all?

As mentioned, in many binary diagrams the $\alpha$ and $\delta$ regions are connected, and the only question is where the Curie temperature occurs as a function of composition. But, above or below the Curie temperature the phase is considered the same.

Cross-posting the same question on two sites is not recommended.

Particularly since Fe was mentioned, I'll focus there. For pure iron, it is well known that $\alpha$-Fe, the stable phase at room temperature, is bcc. This is stable up to 1184K where it transitions to fcc $\gamma$-Fe. In turn, this transforms back to bcc $\delta$-Fe at 1667K before melting at 1810K. Yet, both $\alpha$ and $\delta$ have a bcc crystal structure.
If one looks at the Fe-Zn phase diagram:

one sees that the region of bcc phase, enclosed in dark blue, fully encompasses the green sliver of fcc $\gamma$. Using the Zn concentration as a free parameter, one can go from $\alpha$ to $\delta$ without changing phase.
In contrast, in the Fe-Pd phase diagram:

one sees a different picture. Here, Fe and Pd show complete miscibility in the fcc phase across the entire diagram. This means you can't go from $\alpha$ to $\delta$ bcc phase without going through the fcc phase. Yet, one can see that, by trading off Zn for Pd, one could manage to get around the fcc phase by some path in composition space.

You seem to be saying that the phases $\alpha$ and $\delta$ are identical because they have the same bcc crystal structure. If they are really the same phase then you have answered part of my question with a valid counter example but then why would we even bother distinguishing them? This is not just a matter of verbiage, are they really the same phase?

The $\alpha$/$\delta$ labeling is more historic than anything else, dating to early x-ray measurements of elements - as the temperature was raised the transition from bcc to fcc was seen, then from fcc to bcc and each was given their own greek letter (not quite sure what $\beta$ was mind you!). Many modern phase diagrams just call it all bcc-Fe and be done with it. If it is Fe atoms in a bcc crystal structure, it is bcc-Fe.

Is the behavior in an external magnetic field the only physical difference between the $\alpha-Fe, \beta-Fe$ and $\delta-Fe$? According to https://en.wikipedia.org/wiki/Allotropy $\beta-Fe$ is the paramagnetic bcc-Fe between 770 and 912 °C and $\alpha-Fe$ is the ferromagnetic bcc-Fe below  770°C.

Looking at modern references such as W. Xiong et al., CALPHAD 33 433-440 (2009), they call the whole bcc region $\alpha$Fe, but indicate the magnetic transition temperature across the phase. IT is just an evolution of how it is described.


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Can we convert phenol to para-nitrophenol via nitrosophenol?
Our class 12 text book says that if we treat phenol with mixture of concentrated $\ce {HNO3}$ and $\ce{H_2SO_4}$, instead of nitration, the ring is oxidized to quinone.
So, the book proposes that we should treat phenol with $\ce{NaNO2}$ and $\ce{HCl}$ to yield 4-nitrosophenol, and oxidize it to nitrophenol.
However, I could not find this method in any other standard organic chemistry textbook (eg. the book by Clayden, or by Solomons and Fryhle).
Is there any book/source where this particular reaction is illustrated in details?

The nitrosation of phenol with NaNO2 and sulfuric acid is described here: http://thepharmacistpharma.blogspot.com/2009/04/preparation-of-p-nitrosophenol.html  and the nitrosation of the related 2-Me-phenol is described here http://www.prepchem.com/synthesis-of-2-methyl-4-nitrosophenol/

Thanks. But I  am looking for some textbooks

When you use conc. HNO3 and H2SO4 for your nitration, it will lead to formation of picric acid, not quinone or any nitrophenol. Are you sure your textbook reaction is true? 2. Dilute nitric acid serves the nitration to 4-nitrophenol without needing to go through 4-nitrosophenol (although this reaction will also give 2-nitrophenol). Can you confirm why do you need the reaction to go through nitrosophenol? Do you want your end product to be pure 4-nitrophenol?

I don't know about any books but from what I know the best way to do the conversion is to make an ester(PhOCOCH3). This deactivates the ring enough so that it will not get oxidised by nitric acid.
Then treat it with conc. H2So4. This will block the para position. Otherwise, we will get almost no ortho-nitro product due to hindrance due to the ester.
The we use HNO3/H2SO4 to add a nitro group to the ortho position (only available position).
Hydrolyze the above product to get ortho-nitrophenol.

The question was about paranitrophenol, not ortho.


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Usage of platinum in flame test
Why should we use platinum wire or loop in flame test, can we use any other material ? Why can't we use glass it something?

Platinum is chemically inert ("noble") meaning it resists forming oxides or other compounds. This stability is desirable when doing a flame test because it ensures that we only see the spectrum of what we are burning. (Think how much noise we would have in our spectrum if we used a Magnesium wire to hold our sample)
Of the chemically inert metals (e.g. Au, Ag, Pd, Pt), Platinum has the highest melting point (2041º C), so it's the most useful for holding in a flame.
Glass, on the other hand, melts at a temperatures ~500º C lower than Platinum.

Of course the melting point of glass depends on the particular glass. Silica melts at 1,710°C, but soda glass melts below 600 °C.

The kind of glass you're likely to be using also has a lot of additives. You probably don't want those to contaminate your flame test.

If Mg wire was used, noise in spectrum may be the last problem...

I have done flame tests using nichrome wire.  It's cheaper (but does add a very slight yellow color to the flame.  Platinum is better in that way, but it costs a lot more).


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Etching of ZnS crystal by NaOCl
Can someone tell me the exact mechanism for etching $\ce{ZnS}$ with $\ce{NaOCl}$?
I believe it is due to $\ce{OCl-}$ ion, just need confirmation.
Many thanks.

My bet for the active oxidizing agent would be on hypochlorous acid, HOCl, which is in equilibrium with OCl(-) anion in aqueous solution. The sulfide ions on the crystal surface would displace chloride, giving an S-OH(-) species. Additional oxidation at sulfur by HOCl would lead to higher oxidation states of sulfur such as sulfite and sulfate, whose zinc salts are all more water soluble than zinc sulfide. Dissolution of this material from the crystal surface would then result in etching of the crystal.


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How to measure and transfer suspensions?
I need to measure and transfer 10 ml of a fairly dilute suspension at a time to another container.
But no matter how diligently I do it, sedimentation happens quickly after mixing and successive transfers have different strengths or concentrations of solute.
Is there a way to achieve this without increasing the viscosity of solvent?
Note: I have to start with a dilute suspension and then transfer measured quantities of it. I have no other option.

Maybe separate the suspension into its components and transfer separately?

Is it the sedimentation of the stock solution that is the problem, or the sedimentation of a diluted solution that is the problem?

Stir the dilute suspension rapidly, e.g., using a magnetic stirrer. Let a  vortex be generated whose size you can use as an estimate for vigor of stirring. With the same mixing vigor, and using the same size vessel and same original volume for the suspension, and withdrawing only a small portion (say, ~10%) for the other container(s) should enable fairly repeatable measurements.
For obsessive accuracy, you could do three withdrawals: #1 for reference, #2 for experiment, and #3 for reference. If #1 and #3 show the same concentration of solid, #2 will certainly be the same, and the concentration will be verified while sedimentation will be proven to be absent.


text:
Acetic Acid % in vinegar? Given average molarity of acetic acid
Question:

The stockroom claims the percent acetic acid in vinegar to be $2.0\%$. The density of vinegar is $\pu{1.106 g {mL}-1}$. Using average molarity given ($\pu{0.2844M}$) calculate the mass percent acetic acid in vinegar for comparison to the stockroom claim.

Not sure how to approach the problem but this is what I've done:
$$\frac{\pu{0.2844 mol}}{\pu{1L}} \times (\pu{60.05g {mol}-1})\times \left(\frac{\pu{1L}}{\pu{1000mL}}\right) = \pu{ 0.01707822 g {mL}-1}$$
if you have $\pu{1000mL}$, you'll have $\pu{0.001106 g}\text{ Acetic Acid} \div \pu{0.0000170782 g}\text{ Vinegar}$ so that will give $\%$ of acetic acid? I could use some clarification in solving this

HINT: You want: mass(vinegar)/mass(solution)

$1000$ mL of vinegar will weigh $1106$ g and contain $1000 · 0.017078$ g of acetic acid (= $17.078$ g). Then $17.078/1106 = 0.0154 = 1.54$% acetic acid.
The stockroom manager must be diluting the vinegar, perhaps for a specific experiment. The vinegar I buy is $5$% acetic acid (rarely $4$%).

According to Wikipedia, vinegar has typically 5-8%. I do not know common values found in stores now, but for some reasons I remember it had 8% in early 80s when I was a young chemist.


text:
Lattice energy of NaCl and MgCl [sic]
In an exam paper it says that the lattice energy of NaCl is more exothermic than MgCl [sic] as the Na+ cation has a smaller ionic radius than Mg+.
Here are the resources given:
What is the reasoning for for Na+ having a smaller ionic radius even though Mg+ has a higher proton number?

Mg+ is not a thing at all.

It was part of a CIE A - level chemistry exam

I'd agree with Ivan. You must be making a mistake. It should be $\ce{Mg^{2+}}$ and $\ce{MgCl2}$.

In the exam I think it was asked as hypothetical situation, so hypothetically how would you reason the ionic radius of Mg+?

We wouldn't. It makes no sense.

Ive added the resources given by the exam paper if that helps? Im sorry if this is a waste of time.

OK, then I guess you are supposed to answer that Mg+ is bigger than Na+ because of that lone 3s electron, pretty much for the same reason why Na (neutral atom) is bigger than Ne.

Oh ok! Thank you very much, that makes some sense to me and thanks for your patience :)

Then again, Mg2+ is smaller than Na+ (same electrons, more protons).

That is the exact train of thought that confused me, but the Mg2+ ion does not have the the 3s shell whereas the Mg+ ion does?

Hypothetically it is like that a real object is travelling at more than the speed of light. It's not meant anyway

When Na+ is made, the whole 3s shell is empty hence you radius is till 2p. But in Mg+ you have one electron in 3s orbital and your radius is still till 3s which makes it longer than Na+.

We actually do mean $\ce{Mg^+}$ here.  The idea is to show what factors are involved when an alkaline earth metal forms stable compounds with $\ce{M^{2+}}$ ions whereas an alkali metal favors $\ce{M^+}$.  In this case, as pointed out in some of the comments, $\ce{Mg^+}$ is bulked up by the electron that remains in the relatively diffuse, loosely held $3s$ subshell.  You need to remove that electron, making $\ce{Mg^{2+}}$, to get a compact ion that gives good lattice energies in ionic crystals.
Scandium monosulfide
Although $\ce{MgCl}$ and monatomic $\ce{Mg^+}$ are not real materials, the case of scandium monosulfude, $\ce{ScS}$ is instructive. A simple ionic model proposing $\ce{Sc^{2+}}$ and $\ce{S^{2-}}$ would require the scandium to retain a valence electron in an outer-shell orbital, apparently $3d$. In an early-group metal such orbitals are diffuse and, if occupied, would inhibit tight binding of the oppositely charged ions. Instead, according to Ref. [1] which identifies $\ce{ScS}$ as a metallic conductor, the extra electron per scandium atom is delocalized into a conduction band. This allows the sulfide ions to bind tightly with compact $\ce{Sc^{3+}}$. Thus scandium monosulfide is stable not as a conventional ionic salt, but as a compound that combines ionic and metallic characteristics.
Reference
Dismukes, J. P.; White, J. G. (1964). "The Preparation, Properties, and Crystal Structures of Some Scandium Sulfides in the Range Sc2S3-ScS". Inorg. Chem. 3 (9): 1220–1228. https://doi.org/10.1021/ic50019a004

Just out of interest Magnesium(I) compounds are known, but they are of the form [Mg-Mg]2+ rather than a bare Mg+ - so more like Mercury than Sodium. See https://pubs.rsc.org/en/content/articlelanding/2011/dt/c0dt01831g#!divAbstract , and in a way this furthers the similarity between Mg and Zn.

Not just magnesium.  Wikipedia also documents similar dimers for other alkaline earth metals.  Such dimers, of course, are easily oxidized to +2 monatomic ions.


text:
Can an EWG behave as an EDG and vice versa?
Can an Electron Withdrawing Group(EWG)(like NO2, F, Cl, OH, etc.) behave as a Electron Donating group(EDG) in a compound where a carbocation is present in order to stabilise it?
Does the position where it is present or distance from the carbocation affect it's behaviour as an EWG or EDG?
I feel that it should behave as an EDG as carbocation is electron deficient so it has a greater tendency to accept an electron.
Can we say a similar thing that an EDG behaves as an EWG in presence of carbanion as it has excess electrons?

Of course not. That's why it is called EDG or EWG. It tends to destabilize it more in the absence of other effects like resonance and hyperconjugation.

I'm afraid this may be an oversimplification; atoms care not for how we categorise them. For example, the fact that halide substituents on a phenyl ring are deactivating/electron-withdrawing yet ortho/para-directing suggests there can be simultaneous electron-withdrawing and electron-donating effects in operation, which in this case happens to have a net withdrawing effect. Now what if a bromine atom was linked to 2,4,6-trinitrobenzene? Or a pyrimidine ring? Would it still have a net electron-withdrawing effect into the ring? I have wondered this for a while myself.


text:
How to type condensed formulas in ChemDraw?
What I want is shown in the bubble 1, while what I keep getting every time I change the single bond to a double bond is shown in bubble 2. The H of the middle C in bubble 2 keeps shifting down when I want it to be in line with the C like in bubble 1. How do I do this in ChemDraw?

I have moved it from the meta site (https://chemistry.meta.stackexchange.com/ which is for discussions about the site) to the main site (https://chemistry.stackexchange.com/). I personally would say it is on topic here, but others might disagree. I do not know whether there is a user forum for ChemDraw, where this would be more suitable.

This is fairly simple to do.

Select the atom you want to change.
Right-click > Alignment > choose anything but "Automatic"

You will have to twiddle with the bond lengths to make it look ok. According to Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008),

On atoms with two attached bonds, an atom label should be oriented to minimize its overlap with any bonds in the structure. The label may be stacked vertically above or below, or may not be stacked at all depending on the orientation of the two bonds.

Thank you. Also, I don't think the fiddling with the bond lengths will be too cumbersome. The structures I made are at most four carbons long and that is when I explicitly label the atoms.

Also how can I numerically edit specific bond lengths? I can't find the option anywhere.

@SillysackButtowski Select the bond(s) and the two atoms surrounding them. Then click on the Object menu > Object Settings > Drawing and change the number under "Fixed Length".


text:
Can thiols be created using a Grignard reagent?
I'm trying to find information on the synthesis of thiols via Grignard reagents. While looking at the synthesis routes on Wikipedia, it said that a Grignard reagent will react with sulphur to form thiolates which will hydrolyse easily to the thiol; however, I'm struggling to find any references for this. Has anyone ever heard of this before or maybe tried it themselves?

Yeah that's pretty much what I expected except i'll just use ether cause for some reason nobody sells THF to individuals. Thanks for the help.

I've done this with aromatic Grignards, but I have no reference and no database access any longer.
I think when I did it, I was relying on an older colleague's instructions. It was pretty much generate your Grignard from ArBr + Mg in THF, cool in ice bath, stir vigorously and add elemental sulfur. You may need to introduce a reductive step as you can generate polysulfides.
Grignards will also add to disulfides so you could react with dibenzyl disulfide then reduce off the benzyl group.
This reference is the nearest that Google has managed to find Can. J. Chem. 70 (1992) 30

Thank you this is very helpful, definitely will be doing some tests soon.

A borohydride work-up step should minimise polysulfide formation


text:
Help understanding the equation for Spin Energy in Zero Field
I’m working with a system that experiences spin frustration and for me this is the first time dealing with such a phenomenon. To grasp the concept, I’ve decided to go to the basics and read the classic book on Molecular Magnetism by Professor Olivier Kahn.
On chapter 10.1 Kahn works to get the formula for the relative energies on Zero Field:

S' varies by an integer from 0 to $2S_{a}$ and for every S' value S varies by an integer from $|S'-S_{b}|$ to $S'+S_{b}$
My main interest is the chapter 10.1.1 Copper(II) trinuclear species where Kahn states that we have Sa = Sb = ½ and that the relative energies of the states can be deduced from equation 10.1.6 (the book says 10.6 but it was probably a typo).
$$ E(1/2,1) = 0$$
$$ E(3/2,1) = -3J/2$$
$$ E(1/2,0) = -J + J’$$
Substituting the values for S and S’ I’m failing to get the same results of 0, -3J/2 and -J + J’ and I’m getting, instead, 0.625J - 1.0J′, -0.875J - 1.0J′, −0.375J. You can see how I’m working to get these strange results here in this Jupyter Notebook.
Maybe someone with more experience could guide me on how do I get the same results as the book?
Thanks in advance

I have found the answer and I'm leaving it here in case someone needs it in the future.
Professor Kanh have arbitrarily selected E(1/2,1) as the energy zero and subtracted E(1/2,1) from all energies.


text:
Restrictions on transitions in Spectroscopy
Just for the sake of an example, let's assume a transition from one energy state to a higher energy state in atomic absorption spectrum
If the energy difference between the ground state and first excited state is E and that between ground state and second excited state is 3 E (just for say...). Now if an incident photon has an energy of 2 E so;
(Assuming the electron is initially in ground state)

Will the electron show transition to the first excited state and emit a photon of energy E

OR

Will the electron not show any transition and not accept the photon at all, since the energy of the photon does not exactly match with the energy difference between any two specific states?

The example of atomic absorption spectrum is taken up just for the sake of explaining the question, the question is rather general and pertains to any kind of spectroscopy experiment in which their occurs transition from one state to another by absorption of EM waves.

There is a another aspect to the question. A photon doesn't just have to interact with the least bound electron. A photon can interact with any electron in the atom. So the photon could excite the atom by promoting the second or third least bound electron too, if it has enough energy.

@MaxW The point I'm trying to make is that if the energy of each and every possible transition, be it from any state to any state doesn't match that of the photon(which has a energy sufficiently greater than the least energetic transition), so is it possible that the electron accepts a little amount of energy from the photon does the transistion and at the same time ejecting a photon of the remainder energy?

The gist is that all sorts of things are possible, what you need to focus on is the probability of such occurrences. The reason that atomic absorption works is that a photon of exactly the right energy for the transition has a high probability of exciting the atom. If a lower energy photon must be scattered, then the probability is much lower.

@MaxW Ok so let me conclude it this way(correct me if I'm wrong). Both the possibilities mentioned in the question can take place, but it's just a matter of probability as to which will take place(and mostly the probability of the former is greater). Also if you could tell why is the probability of the former greater.

If the energy is to be transferred from the photon to your atom one energy gap must match the energy of the photon. Thus absorption may occur from the first to second excited state in your example as the energy in both is 2E. However, there has to be some population in the first excited state for this to happen, zero population zero absorption. In fact it is a bit more subtle in that there has to be a difference in population between the two levels involved. It is possible for the photon to stimulate a transition from the upper level to the lower one, just as from the lower one to the upper one. The reason why this seems odd is that in most cases the population in the upper level is vastly smaller than in lower ones (via Boltzmann distribution) and the population in the upper level can often be set to zero without serious error. This is particularly true in electronic spectroscopy, as in the case of atoms, but far less so for pure rotational transitions in molecules which occur in the microwave region of the spectrum. You will find details in any good spectroscopy book.
Thus neither of your cases occurs but something different; transitions between the first to second excited state. How strong this transition is depends on the population in the two levels and on selection rules involving angular momentum and electron spin changes.

The transitions between quantum states follow selection rules.
https://en.wikipedia.org/wiki/Selection_rule
During any transition, energy, momentum and quantum angular momentum (spin) must be preserved. Those are the hard rules. Also the incoming EM wave has to create an electric field oscillation that overlaps with the electronic waveform oscillation that would occur during the transition (they couple). This overlap can be significant for a given transition and the EM wave frequency matching its energy difference, which makes for a percievable probability of absorption, or very low, in which case the transiton is forbidden. How strictly forbidden depends on the circumstances (neighbouring atoms, collisions, ...).
The usual transitions absorb the complete energy of an incoming photon. Then the difference in spin quantum number between the two states must be 1, the spin of a single photon. Some variable amount of energy always goes into translational energy, pushing the molecule that is hit. That's why absorption lines always have a width. The momentum of the photon is retained in that push, so the molecule has to fly off in an angle that matches excess energy and the momentum.
Raman IR transitions grab a small amount of energy off of a higher energy EM wave / photon (or add to it). The spin difference between the states must therefore be zero, and the momentum and direction of the leaving photon must also be taken into account, which dictates the energy difference.
Crazier transitions are possible. Two photons could be absorbed at the same time. Very low probability unless you have an extremely bright light source, but this allows totally different transitions, because now the same wavelenght has double the energy, and you have the spin from two photons.
So, answering your actual question, there are several possibilities to absorb a photon with an energy that does not exactly match the given transition.
(And then of course you have the intrinsic bandwidth of the transition, due to brownian motion.)


text:
Reaction of phoshorus pentasulfide with ethers
Why does phosphorous pentasulfide on reaction with ethers give dialkyl sulfides?
I searched for this reaction in my organic chemistry reference books and the internet but couldn't find it.
What is the mechanism of the reaction? I expected the oxygen as nucleophile to attack P atom but am not sure of the steps that follow.

You searched your reference books and the internet and couldn't find the reaction. How do you know it actually happens? Do you mean an ether (ROR) or some other oxygen functionality such as an amide?

I found one claim of ether(ethylene gylcol monoalkyl ether) reacting with phosphorus pentasulfide though I am not sure this is what you are looking for. See: https://apps.dtic.mil/docs/citations/AD0673686


text:
Why can't sulphate ion be discharged in the anode of electrolysis?
Given that there is an aqueous solution of sulfuric acid ($\ce{H2SO4}$), $\ce{2H+}$ is reduced in the cathode and $\ce{OH-}$ is reduced in the anode. Why is $\ce{OH-}$ preferentially discharged over $\ce{SO4^2-}$ in the anode?
One explanation I've found is that because $\ce S$ in $\ce{SO4^2-}$ has an oxidation number of $+6$ which is a maximum and thus cannot be oxidized any further. However, can't the $\ce O$ in $\ce{SO4^2-}$ be oxidised instead as it has an oxidation number of $-2$ (just like the case in $\ce{OH-}$ where $\ce{O^2-}$ is oxidised and $\ce{H+}$ is not oxidised according to the half equation $\ce{2OH- <=> 1/2 O2 (g) + H2O (l) + 2e-}$)? Therefore, the argument that $\ce S$ cannot be oxidised because it has an oxidation number of $+6$ doesn't seem to be valid.

In aqueous solution the anode will oxidize the water instead, as this reaction can be done at much lower potential than oxidizing the sulfate ion.
Sulfates can be oxidized if you take the water away.  Potassium peroxydisulfate can be prepared by electrolyzing a solution of potassium bisulfate ($\ce{KHSO4}$) in sulfuric acid solvent.


text:
How is silver chloride more soluble in brine than water?
I heard that:

Silver chloride is more soluble in very concentrated NaCl solution (brine) than in pure water.

Is it true? I mean, won't the common ion effect operate? Or is this maybe due to some kind of complex formation as the NaCl solution is highly concentrated?

I mean won't the common ion effect operate?

The common ion effect always applies for pure substances. With the added sodium chloride, you have changed the effective solvent and changed the ions.

Or may be this is due to some kind of complex formation as NaCl solution is highly concentrated?

Exactly, before your ionic equation was:
$$\ce{Ag+ (aq) + Cl- (aq) <=> AgCl (s)}$$
but with brine, the ionic equation is:
$$\ce{AgCl (s) + Cl- (aq) <=> [AgCl2]- (aq)}$$
Which has a higher solubility constant, and where as before more chloride made the $\ce{AgCl}$ less soluble, if you note that you have switched regimes and $\ce{Cl-}$ has switched from precipitating a silver ion to dissolving a silver chloride molecule, which means now with increased chloride concentration, silver chloride becomes more soluble.

The answer by A.K. is eminently reasonable, but I wondered if stability constants were available. I quote from the summary of Ref. 1 :

The solubility of silver chloride in concentrated solutions of various
chlorides was determined mainly at 25°.
The solubility is nearly doubled in going from o° to 25°, the rate of increase above and below 25° being nearly logarithmic.
The solubility is sharply proportional to integral powers of the chloride
concentration throughout considerable ranges, a fact explained by assuming the existence of the complex anions
AgCl3$^{^-2}$,  AgCl4$^{-3}$ and possibly AgCl5$^{-4}$. No evidence of the ion AgCl2$^{^-}$is found.

It seems that the simplest explanation is only the beginning of understanding the solubility increase.
References

The Solubility of silver chloride in chloride solutions and the existence of complex argentichloride ions, J. Am. Chem. Soc. 1911, 33, 12, 1937– 1946 Publ. Date: December 1, 1911 https://doi.org/10.1021/ja02225a008

Interesting, I have found earlier today AgCl2- formation constant 1.8e5 on libretexts somewhere. OTOH  [AgCl5]4- seems to me suspicious.  https://chem.libretexts.org/Ancillary_Materials/Reference/Reference_Tables/Equilibrium_Constants/E4%3A_Complex_Ion_Formation_Constants

@Poutnik: Yes, indeed, interesting on both issues! Is the 1911 account accurate about AgCl2-, and, sure, AgCl5(4-) is unlikely, but AgCl3(2-) or AgCl4(3-) don't seem so impossible. But it could be easier to just call it AgCl2- and let it go because the solubility of the complex(es) is still so small.


text:
Calculation of rate law for multi step reaction

Please help me to solve this one. I'm having a bit of a hard time grasping these concepts.

In the pre-equilibrium you treat the first reaction as an equilibrium to find $[\ce{A}]$ and substitute this into the second rate expression $\mathrm d[\ce{P}]/\mathrm dt=k_2[\ce{A}][\ce{B}]$. In the steady state approach you assume that the intermediate species $[\ce{A}]$ has a small rate of change thus let $\mathrm d[\ce{A}]/\mathrm dt = 0$ and find $[\ce{A}]$ this way.


text:
Intuition of following the lowest curvature mode to find a saddle point on a PES
In this paper A dimer method for finding saddle points on high dimensional potential surfaces using only first derivatives, it is mentioned that

The saddle point is a maximum along the lowest curvature mode.

The curvature in this paper refers to the second derivate of the PES.
I know that mathematically, a saddle point minimizes the PES in all but one direction, and it is the highest point on the minimum energy path (MEP). But this cannot help me understand why the saddle point is a maximum along the lowest curvature mode.


text:
Acid hydrolysis of ethyl acetate uses
I was wondering if the above mentioned reaction has any uses? I looked up applications and significance and couldnt find anything. I came across Saponification but that involves a base and is a completely different reaction. Then I thought the products maybe useful but i realised that ethanol and acetic acid are used to produce said ester. then what exactly is the point of this reaction?

The acid hydrolysis of ethyl acetate may not be very useful but the acid hydrolysis of esters as an alternative to saponification is important. Imagine the acetate ester of 4-iodobutan-1-ol (ICH2CH2CH2CH2OAc). Base treatment of this ester can effect saponification and substitution of iodide by hydroxide. On the other hand, acid hydrolysis of the ester group would leave the primary iodide intact.

The products of the reaction are perfectly miscible with water and are harmless to environment in low concentrations. My guess that the leftovers of ethyl acetate can be turned to ethanol and acetic acid before disposal.
For example here you can see that ethyl acetate is forbidden from sewer disposal while ethanol and acetic acid are allowed.
https://www.vumc.org/safety/waste/chemical-waste-sewer-disposal

Not sure why acetone, methanol, or ethylacetate would be disallow specifically. considering all 3 are common in consumer products.


text:
How to distinguish between iron and nickel based on the mass of precipitate formed with sodium hydroxide?

For the reactions shown below, we added $\pu{5.00 mL}$ of $\pu{0.0390 M}$ $\ce{NaOH}$ to a test tube containing one of the two cations $\ce{Ni^2+}$ or $\ce{Fe^3+}$ and recovered $\pu{0.00695 g}$ of precipitate.
  \begin{align}
\ce{Ni(NO3)2(aq) + 2 NaOH(aq) &-> Ni(OH)2(s) + 2 NaNO3(aq)}\\
\ce{Fe(NO3)3(aq) + 3 NaOH(aq) &-> Fe(OH)3(s) + 3 NaNO3(aq)}
\end{align}

How much precipitate in moles would be recovered theoretically if the ion was $\ce{Ni^2+}$?

For this answer I got $\pu{0.0000975 mol}$, but that is wrong and I don't understand why.

How much precipitate in moles would be recovered theoretically if the ion was $\ce{Fe^3+}$?
How much precipitate in grams would be recovered theoretically if the ion was $\ce{Ni^2+}$?
How much precipitate in grams would be recovered theoretically if the ion was $\ce{Fe^3+}$?

As far as I can see your calculation checks out. Maybe it's an issue with the answer checking system?
For the amount of substance of precipitate if $\ce{Fe^3+}$ is the ion, you carry out a very similar calculation as before. You find the amount of substance of $\ce{NaOH}$ in the $\pu{5 mL}$ solution and then divide that by 3 to find the amount of substance of $\ce{Fe(NO3)3}$ in this particular reaction. Pay attention to the relative reacting amounts of reactants. The product has the same amount of substance as $\ce{Fe(NO3)3}$ so that's your answer for question 2.
questions 3 and 4 are very simple. Now that you have the theoretical amount of substance of the precipitates you can convert them into masses by manipulating the
\begin{align}
\frac{\text{formula mass}}{ \text{relative molecular mass}}
&= \text{amount of substance}\\
\Longleftrightarrow \text{amount of substance} \times \text{relative molecular mass}
&= \text{mass}
\end{align}
You can get the relative molecular masses of $\ce{Ni(OH)2}$ and $\ce{Fe(OH)3}$ by using your periodic table.


text:
Instability due to neutrons
Why does an excess of neutrons leads to instability?
For example, both $\ce{^{3}H}$ and $\ce{^{4}H}$ are unstable, with respective half-lives of $3.89\cdot 10^8$ and  $1.39\cdot 10^{-22}$ seconds.
By my reasoning,  neutron-rich nuclei should be strongly bound by the strong nuclear force, and should therefore be more stable. Additionally, neutrons, unlike protons, are neutral and therefore do not contribute to repulsive electromagnetic forces.

Think of it this way: a pure neutron by itself is unstable.

The key here is that a system will tend towards the lowest energy state; in other words, for a process to be spontaneous, the final state must have lower energy than the initial state.
The mass of a neutron is slightly higher than the combined mass of the proton, electron and neutrino that result from beta decay, meaning that the decay will be energetically favoured. This might be easier to understand by considering the decay of a free neutron:
$$\text{n}^0 \rightarrow \text{p}^+ + \text{e}^-+\bar{\nu}_\text{e}$$
Whether this process will occur in a nuclear neutron is determined by the $N/Z$ ratio and the atomic number. In small neutron-rich nuclei, the process above will predominate, but as the nucleus gets larger, more neutrons are needed to counteract the electromagnetic repulsion between protons (which has a longer range than the strong nuclear force). Again, the reason is energy: the unbound nuclei collectively have higher energy than a nucleus, due to nuclear binding energy. Hence, for larger nuclei, neutron decay is not energetically favourable.