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What compounds smell like burning rubber or tar?
Background:
A couple of friends of mine live next door to each other in a 10-year old block of flats in London, UK. They say that in the early hours of the morning, a toxic smell appears in their flats, which smells a little bit like burning rubber or tar. It also makes their eyes burn and affects their breathing. Their attempts to inform the housing authority ended in dead ends, as the housing authority didn't seem to believe them about the smell.
In order to try to get to the bottom of this, I had some of their air sampled inside one of the flats (once when the smell was present, and once when it was not present), and sent it for Gas chromatography–mass spectrometry at a lab.
The lab only tests for about 70 specific volatile organic compounds, and none of those compounds showed much concentration change between the two samples.
Presumably, whatever is causing the smell is not one of those compounds.
Questions:

What compounds smell like burning rubber or tar?
If the lab sends me the raw data from the GC-MS test, how easy is it to identify the presence those compounds from the data?

Thiophenol is one but rubber and tar burning will produce all sorts of things

Did your friend ask someone to inspect the electrical appliances or heating elements (Dryers, water heating furnace, heaters)? https://www.quora.com/What-would-cause-my-house-to-periodically-have-a-burning-smell-usually-at-night

Check local history:

Was the land used as a dump or filled in with waste wood? Creosote is derived from tar, and it is used as a wood preservative. If creosoted wood, such as railroad-ties, was used in landfill, that could lead to a strong odor.
Coal ash or clinker can have a strong smell. Was a coal-burning power plant situated where those flats are now located? Was waste from coal-burning buried there?

Interestingly, the area was a wood yard about 15 years ago. But I doubt that's relevant, as the smell only appeared about a year ago.

Actually, the other thing that might well be relevant is that there was a fire in the building about 2 years ago (about a year before the smell appeared). The flats experiencing the smell were not affected by the fire.

Landfill pollution may take years to become apparent. Though a toxic site may be capped with soil, eventually pollutants can seep out and rise to the surface. Take soil samples, or contact a firm that does so to find cost, e.g., https://adeptus.co.uk/soil-sampling-testing-service. The fire seems a bit of a red herring, since you state, "The flats experiencing the smell were not affected by the fire."


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Corrosiveness of Deionized vs Distilled water for use in Laboratory Water Bath
I've read conflicting descriptions of the corrosiveness (or lack thereof) of very pure water. For a precision water bath, it's stated that one should fill the bath with distilled water rather than deionized of a certain purity as it would be corrosive:

5.2 Filling Bath
NOTE: DISTILLED WATER IS RECOMMENDED; IT WILL
NOT CORRODE THE BATH CHAMBER
AND IT REDUCES THE NEED FOR FREQUENT
BATH CLEANING.
DO NOT USE 18 MEG DEIONIZED WATER.
NOTE: If this is the only
source of treated water available; mix 50% with regular tap water.

Why is this? I'm struggling to find a specific mechanism by which purer water would corrode the bath.

Because there is a strong entropic driving force to suck impurities out of everywhere.

@Jon Custer When it comes to how the manual describes DI water as more corrosive than distilled, how exactly is that occurring? Doesn't there need to be a conductive medium to accelerate the process? Distilled is highly pure as well, so I'm not grasping the difference.

@Mithoron Somewhat, but not much more an answer than "the manual is just wrong" which seems to be what I'm picking up from reading similar questions here and elsewhere. I'd really like a clear mechanistic explanation to support this claim in the manual, but perhaps it is just incorrect.

What about manual? That they say not to use "18 MEG DEIONIZED WATER"? Well I don't see any suggest about it being "corrosive". I wouldn't suggest using it because it's too expensive.

You can stop struggling. Your chemical instincts are correct; the purer water is, the less corrosive.
The main factor in the corrosivity of water solutions is the ion content, the ability of the water to carry a current in galvanic reactions. Lower ions, less current, less corrosion. The second factor is the presence of dissolved ions etc. to participate in various and sundry reactions. The less of these the fewer possible reactions. It is best to use water of the highest quality available in water baths, humidifiers, drip coffee makers, CPAP devices, plating baths, even laboratory dishwashers, and definitely as a chemical reactant or solvent. Printed circuit board manufacturing uses only the highest quality DI water in any aqueous process [and they probably wish they could get rid of the pesky 10^-7M H3O+ and OH-]. [Which they do by inventing these super nonionic detergents and no residue fluxes and, once upon a time, freon rinses.]
The increased corrosiveness of pure[r] water is a myth that I think originated with the water supply companies that found that certain plumbing structures literally fell apart when high calcium water was replaced by softened water. The pipes were partially held together by buildup of CaCO3 deposits and the common ion effect prevented dissolution of the deposits. Remove Ca++ from the water the deposits dissolved and everything leaked or even fell apart.  This was probably part of the problem in Flint, MI. A change in water disturbed decades of protective coating in the pipes; the problem was deeper than removal of a coating.

You are missing the fact that pure water will happily suck atoms out of the piping. Not electrochemical corrosion, just a question of trying to come into equilibrium…

@JonCuster Oh, no, not you too. Only effect of such "sucking" is that water gradually stops qualify as distilled or deionised. And only equilibrium would be if metal was oxidised completely - saltwater is incomparably better in achieving that.

Jon, Look up the oxidation potentials of pure water compared to standard potentials to see how efficient pure water is in "sucking out ions". Water of different ion content has different effects on mineral deposits. Metals become ions by electrochemical corrosion mechanisms and an essential part of corrosion is an ionic pathway between anode and cathode no matter how close the two might be [at least down to the actual molecular level then there might be an explosion if enough reactants are present.]

@jimchmst - consider the Gibbs free energy when pure water is in contact with, well, anything. The entropic term far outweighs the enthalpy term until some impurities are introduced into the water. In a semiconductor fab, all DI plumbing is plastic for a reason - even ppb of many metals are a problem and it only takes one small section of metal plumbing to provide that. Reconsider your electrochemical argument when the system is that far from steady state much less equilibrium.

@JonCuster Do the calculation. Don't just assume the claim. In every practical way, pure water is not corrosive. Yes, contact with metal surfaces will make it less pure but the effect is vastly smaller than the effects seen with impure water where the impurities often increase the surface losses by electrochemical processes. So the surface losses to pure water are vastly lower than to dirty water.

Jon I have chemistry on my side and also practical experience I have worked in PCB manufacture, electroplating, water purification and chemical analysis and have never found that using better grade water created problems. In almost everycase there were subtle to sometimes dramatic improvements in processes when we improved water quality.  Things as simple as using RO water to make reconstituted orange juice to using 18Meg water to wash components in PCB manufacture. The only problem with superpure water is that attention must be paid to keeping it super pure.

NASA published a 23-paged reported in the 60s on the corrosion of metals in deionized water at 38 $^o$C. The report is titled "Corrosion of Metals in Deionized Water at 38° C (100° F)" By Barbara Alice Johnson, 1969.
As one can see, we cannot make sweeping statements about corrosion in deionized water. Some metals will corrode and some will not in deionized water. Long time ago, someone told a story of detecting measurable amounts molybdenum in human blood. They were using very sensitive analytical technique and it puzzled everyone as to why Mo is being detected. It turned out that the metallic syringe needle leached a few atoms of Mo in a very short period of drawing blood. Yes, deionized or distilled water will leach some metallic ions but corrosion means measurable metal loss to the extent that it causes economic damage.
Thus, corrosiveness distilled vs. deionized water is like a popular urban myth. What the manufacturer probably want is some level of slow scaling in the bath to "protect" metallic heaters.

Corrosion is corrosion If you wait for economic loss you have to wait for the bridge to collapse. Think of all the money that would have been saved by not painting the Golden Gate Bridge. A chemical engineer's job is to understand these things. Talk to Delorean  car owners for an example of poor chemical planning.

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Does the nature of the H-A bond tell us about the $\Delta$G of an acid's dissociation?
When I first learned about acid strength trends in general chemistry, I remember it was explained in terms of the character of the H-A bond. The more polar the H-A bond, the easier it was to heterolytically split it into H$^{+}$ and A$^{-}$. This explained why acid strength increased from left to right across the periodic table: as you went from left to right, electronegativity increases, so the H-A bond becomes more polar, hence more acidic. The strength of the H-A bond also contributed to the acidity; the weaker the bond, the easier to break it into A$^{-}$ and H$^{+}$, so the higher the acidity. This explained why acidity increased as you went down a group; the H-A bond became weaker, leading to higher acidity.
However, I have also seen acidity strength explained solely through the stability of the conjugate base, without mention of the nature of the H-A bond. When viewing acidity through the base stability point of view, we can explain the increase in acidity from left to right as a result of charge stabilization: in the conjugate base, a more electronegative atom is better able to stabilize the excess negative charge, therefore the conjugate base becomes more stable as the electronegativity of the atom attached to the acidic proton increases. The trend down a group can also be explained in terms of charge stabilization: as the size of an atom increases, any excess charge on the atom is spread out due to the diffuse orbitals, leading to charge stabilization. I think because I so often saw acid strength just explained in terms of base stability, I started to think that the H-A bond explanation was just a simplified version of the base stability argument.
However, I recently saw this question, which made me reassess my thinking:
Here, you can not use base stability to explain the acidity difference, because both acids have the same conjugate base. So in this case, any difference in acidity must come from the stability of the acid, not the base. In other words, the differences in the $\Delta$G's for the two dissociations must be due to a difference in the Gibbs Free Energy of the acids. This makes me think that the discussions in general chemistry about the H-A bond were really  explaining the $\Delta$G of acid dissociation, which was independent of the stability of the base. Now, I am starting to think that weak and polar H-A bonds lead to more acidity because they tend to create a more exergonic reaction (if this is the case, could someone explain why?).
This leads to my overall question: is explaining acidity through the nature of the H-A bond just another way of explaining it through conjugate base stability, or does the H-A bond tell us something about the $\Delta$G of the acid dissociation that we would not know from looking at conjugate base stability? Is it just coincidental that the nature of the H-A bond and the stability of the conjugate base contribute to the same trends in acid strength?

That is an interesting and a somewhat silly question because in aqueous solution the two compounds are the same. If Either were initially pure and added to water the eventual equilibrium would be identical giving the same [H3O+]. The deciding factor would be which of the two tautomeric forms of the acid is present in greater concentration in the solution. A Raman or IR spectrum might give the answer.

I looked up thioacetic acid it is a considerably stronger acid than acetic and exists almost entirely as the thiol tautomer; the difference from the stronger carbonyl than thionyl bond. This raises question about the acidity of the much less stable tautomer.

This is a good question, and Ill try my best!
The acidity of an acid is solely dependent on the nature of the HA bond. When learning about acidity and basicity, the concept of conjugate base stability is invariably taught to help explain the strength of various acids. But, molecules don't have brains. They arent thinking, "When I lose my proton, I'll be super stable, so I wanna be really acidic!"
The electronics of the starting acid are what determines its acidity, and as a result, things that are more acidic have relatively stable conjugate bases. The stability of the CB only helps explain its acidity (and 99% of the time, it works).
OH vs CH? Oxygen is more electronegative, pulling electron density away from H, so the electron pair done (the base) can use its electrons to grab that hydrogen more easily.
Carboxylic acid vs alcohol? Oxygen is donating electron density to the carbonyl, resulting in the oxygen of the OH bond being more electron deficient. But electronegative oxygen wants electrons, so it pulls even MORE electron density from H, making the proton even more grabbable than in an alcohol.
Thiol vs alcohol? Yeah sure, oxygen is more electronegative, but the orbital overlap between sulfur and hydrogen is garbage, that is, the energy gap between H's s orbital and sulfur sp3 orbital is much larger than that of an alcohol. Thus the S-H bond is less stable kinetically, and so bases come and snatch that hydrogen.
To answer your question completely, it is NOT coincidental that the acidity of HA correlates with the stability of the conjugate base. However, rather than thinking that HA is more acidic because the CB is more stable, think about how the two concepts go hand in hand and influence each other.
Here's another example: The iodide in tert-butyl iodide doesn't leave because it forms a more stable tertiary carbocation; the molecule doesn't know that it can form a stable tertiary carbocation. Rather, it leaves because the adjacent sigma C-H bonds donate electrons density to the C-I antibonding orbital, significantly labilizing the C-I bond. Thus, with enough energy, iodine leaves, forming the tertiary carbocation.

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What makes a compound high or low energy and why does that relate to reactivity?
After reading about thermodynamic stability, it now makes sense to me how the enthalpy of a reaction relates to the thermodynamic stability of a compound.
However, when looking for a clear way to define thermodynamic stability, a few sources mentioned how high-energy compounds are more reactive and therefore less thermodynamically stable. What actually makes a compound high, or low, energy and how does this relate to reactivity?

I suggest you read a bit more, starting e.g. here: https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation

Stronger bonds make more stable compounds both kinetically and thermodynamically. Kinetically because bonds must be broken for reactions to happen. Thermodynamically because bonds must be formed. A second factor is the reaction must be entropically favored. A third factor is the available energy thermal, photolytic or electrical must be sufficient to either not break bonds, stability, or to break bonds, reactivity.
Bonds are formed when electrons are shared between atoms, electrons are attracted to two or more nuclei. This can be a weak attraction such as intermolecular attractions; weak molecular bonds such as found in molecules such as $\ce{F2},$ $\ce{I2},$ $\ce{Li2};$ stronger bonds such as $\ce{C-H},$ $\ce{C-C},$ $\ce{C-O},$ $\ce{C-F},$ $\ce{C-Cl},$ $\ce{H-H},$ most ionic bonds and metallic bonds; and very strong bonds such as $\ce{C=O},$ $\ce{C=C},$ and the triple bonds in $\ce{N#N},$ $\ce{C#O},$ $\ce{HC#N},$ $\ce{HC#CH}.$ Double and triple bonds have unique properties because of their orbital structures.
All bonds are stable under low enough energy conditions or possibly a molecule in isolation. If energy is added: Heat, Photons, electricity, etc. Bonds will break. The energy to break sufficient bonds to cause a reaction is called the activation energy.  The bonds will either reform or form new bonds. If the reaction is entropically favored by removal of energy stronger bonds will form and the reaction will proceed. The final products are determined by the strength of the bonds in the various compounds and the distribution of the energy involved.
Molecules with weak bonds are considered to be high energy; they tend to be reactive. Molecules with strong bonds are considered to be low energy; they are more stable. Reactions happen in the direction of forming stronger bonds and increased entropy. The latter is mostly manifested in the disposition of the energy evolved in the formation of the stronger bonds.


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In bulletproof vests, why do we use ceramic plates instead of metal?
In many articles, it has been mentioned that bulletproof vests also contain ceramic plates for extra protection. Wouldn't it be better to use metal instead? Since ceramic breaks easily.

“ceramic” is not a single material, rather a wide range of them.

At the extreme deformation rates involved, none of your common sense measures of material properties hold.

Just as a starter, to get through a piece of ceramic, a bullet doesn't break the tile in two, but shatters a good portion of it into dust. Metal, otoh, just gets bent out of the way. "Cold flow" that's called, technically used for https://en.wikipedia.org/wiki/Deep_drawing

@Karl - indeed, and by shattering the material, the bullet loses lots of kinetic energy on the ceramic, rather than penetrating into the person.

I think Wikipedia provides a very good summary:

Unlike metals, ceramics are never used alone, as standalone materials, in armor systems; they are always combined with a ductile backing or support layer of metal or fiber reinforced plastic composite materials, and this ceramic-faced assembly is called ceramic armor. Ceramic materials, like glass, have high hardness and compressive strengths but low tensile strengths. Bonding a ceramic tile to a metallic or composite backing material, with high strength and good ductility, delays or mitigates tensile failure upon impact, and forces the ceramic to fail in compression.

So ceramic would never used on its own in a bulletproof vest. It will always have metal or something with similar strength.

Ceramic armor systems defeat small arms projectiles and kinetic energy penetrators by two main mechanisms: Shattering and erosion. When a hard steel or tungsten carbide projectile hits the ceramic layer of a ceramic armor system, it is momentarily arrested, in a phenomenon known as dwell. Depending on the thickness and hardness of the ceramic layer, the projectile core is then either shattered, fractured, or blunted. The projectile's remnants continue to penetrate the comminuted ceramic tile at a reduced velocity, which erodes those remnants and reduces their energy, length, and mass. The metal or fiber reinforced plastic composite layer behind the ceramic layer then arrests the projectile's fragments or its eroded remnant, and absorbs residual kinetic energy, typically via plastic deformation.

Even though ceramic breaks easily as you said, it has high compressive strength, i.e. it can resist compression to a large extent. When combined with metal backbone, this helps to reduce the kinetic energy of the bullet and thereby reduce the damage.
As an aside, I should add that this idea that bullets can only be stopped with a hard material like metal is not always true. While metals can stop bullets, they are also heavy and reduce mobility. Stopping bullets requires that its kinetic energy be absorbed and dispersed accross a larger area so that the impact is no longer felt. This can be done with materials other than metal. For example, during the Vietnam war, flak jackets (armor) without any metal plating was used. These would be made from multiple layers of ballistic nylon, along with fibreglass plates, and had much less weight than metal plated armors. The multiple layers of strong fabric and the fibreglass would spread the impact of incoming projectile across a larger area. Note, however, that these jackets were only effective against grenade fragment, shotgun pellets, and possibly small pistols. They would not work against rifles or anything stronger.
In the modern day, polymeric materials like Kevlar are used to protect against bullets. There are also many researchers working in the material science side of chemistry to find new materials that can be light but withstand impact.


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Does aqueous KOH favor SN2 or SN1 with secondary halides?
I have read that aq. KOH is equally reactive towards SN1 and SN2 with secondary halides. I'm aware that aq. KOH will participate in SN2 as well because of the lack of steric hindrance in secondary halides, but since aq. KOH is a polar protic solvent, shouldn't it favor SN1 comparatively?

I would expect a base as strong as hydroxide ion to favor SN2, as it would probably be strongly "attracted" to a carbocation while the latter is forming.


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Is vapour pressure truly a colligative property?
Colligative properties are generally defined as

A colligative property depends only on the ratio of the number of particles of solute and solvent in the solution, not the identity of the solute.

Relative lowering of vapour pressure is considered to be a colligative property.
I think, lowering of vapour pressure is not a true colligative property because in certain conditions, it does depend on the nature/identity of the solute.
Consider that we have such a solute mixed with a solvent in a solution that the solution exhibits a negative deviation from Raoult's law, i.e., the attractive forces between solute-solvent are stronger than the attractive forces between solute-solute or solvent-solvent molecules, e.g., mixture of phenol and aniline or a mixture of chloroform and acetone. The solute particles are more attracted to the solvent particles and this inhibits solvent particles from escaping easily from the solution, lowering the vapour pressure of the solvent.
In such a case, lowering of vapour pressure of solvent is clearly dependent on the nature of the solute.
Is vapour pressure truly a colligative property?

colligative properties are defined with respect to ideal solutions

@Andrew Oh, yes! As the colligative properties are based upon Raoult's law which is applicable for ideal solutions only, colligative properties should be also be defined with respect to ideal solutions. My textbook doesn't state this clearly for colligative properties.

I agree with you that generally, vapor pressure of a solution depends on, not only concentrations of its ingredients, but also on their chemical composition.
Addition of ethanol to water will decrease the boiling point of solution relative to water. While, adding acetic acid will increase the boiling point of the solution. One may think a solute that is more volatile than water(as a solvent), will always decrease the solution boiling point. This is what raoult's law predicts. But, this isn't always true and depends on the ideality of the solution. For example, consider Hydrochloric acid. The boiling point of liquid Hydrogen chloride is -85 degrees of centigrade and is more volatile than water. But after it dissolves in water, the boiling point of solution can even increases up to 108 degrees(in some concentrations) This is because the interactions between particles in Hydrochloric acid is stronger than pure water. This stronger bonding arises from dissociation of Hydrogen chloride molecules into ions. When the intramolecular forces and sizes of the components in a solution are near to each other the solution behaves more ideally. One example is a solution of n-pentane and n-hexane.
In addition, when a solution is diluted the solution becomes more ideal.
Back to your question, that wondering about vapor pressure of a solution to be a colligative property.
I think the vapor pressure in solutions that, the solute is non-volatile may only depend on solute concentration. If so, in this case the vapor pressure of the solution is a colligative property. But if the solution behaves ideally. Let me explain the reason for it.
At the surface of the solution, a number of solvent particles are exchanged by solute particles. In an ideal solution, the solute particles bond to solvent molecules, just as solvent molecules bond to each other (same intramolecular forces and molecular size). So regardless of the identity of desired solutes, the chance of solvent molecules to escape from the surface will decrease with respect to concentrations of solute. As the solute is non-volatile, the total vapor pressure is equal to solvent vapor pressure. And, that can be predicted by only the concentration of the solute(a colligative property). Conceptually, the ideality of the solution is an important factor in the answer of your question. Ideality is not an absolute property of a solution. Any solution can be considered to be non-ideal with sufficient precision in measurement.   Thanks.


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Half-life of a reversible first-order reaction of the form A ⇄ B
I am wondering if the half-life of a reversible first-order reaction of the form $\ce{A <--> B}$ is defined as

the time in which half of the amount of reactant which should react up to equilibrium, has reacted, or,

the time in which half of the total amount of reactant has reacted.

I only found this source.

Example: Species $\ce{A}$ reacts in first-order kinetics. The initial concentration is $1 \,\pu{mol  l^{-1}}$. Its equilibrium concentration is $0.3 \,\pu{mol l^{-1}}$. After $40\,\pu{s}$, its concentration is $0.65 \,\pu{mol l^{-1}}$. After $60 \,\pu{s}$. its concentration is $0.5 \,\pu{mol l^{-1}}$. Is the half-life $40 \, \pu{s}$ (because $0.35 \,\pu{mol l^{-1}}$ of total $0.7 \,\pu{mol l^-1}$ that react, have reacted), or $60 \,\pu{s}$?

I've only encountered the term 'half-life' in the context of reaction kinetics when describing elementary reactions - I've not seen anyone try to define half-life for multi-step processes (such as a reversible reaction). However, in many fields, such as medicine, half-life is used commonly to measure the time it takes for some initial condition to be reduced to half, irregardless of how exactly it would happen. So I'd guess if I had to define, I'd go with the 40 s.

So basically you would go with the assumption, that because 0.7 mol/l react, we should define the half life as the time it takes for 0.35 mol/l to react, because after 0.3 mol/l there is no net change anymore?

I misunderstood the question in my earlier comment. I'd define half-life as the time it takes for the concentration to drop to half, so from 1M to 0.5 M - which I guess is 60s in your case, right? But then again I'd prefer not to define it at all.

Related

Okay yeah 60s makes sense

The Wikipedia article on half-life states:

The term "half-life" is almost exclusively used for decay processes that are exponential (such as radioactive decay or the other examples above), or approximately exponential (such as biological half-life discussed below). In a decay process that is not even close to exponential, the half-life will change dramatically while the decay is happening. In this situation it is generally uncommon to talk about half-life in the first place, but sometimes people will describe the decay in terms of its "first half-life", "second half-life", etc., where the first half-life is defined as the time required for decay from the initial value to 50%, the second half-life is from 50% to 25%, and so on.

Concentrations of a reaction approaching equilibrium do not decay exponentially (for one, they don't approach zero as time passes).
The kinetics of a reaction approaching equilibrium can be complicated. If you start without product, kinetics are governed by the rate law of the forward reaction in the beginning. Then, as product is being made, the rate law of the reverse reaction plays into the kinetics as well. Surprisingly, as you get very close to equilibrium, the net forward rate decays exponentially (i.e. apparent first-order reaction) no matter what the rate laws for the forward and reverse reactions are. This is used in temperature-jump and other relaxation methods to study fast reactions.

[OP] I was wondering if the half-life of a reversible reaction is defined as [...]

For processes that don't follow a pure exponential decay, the half-life is ill-defined. There is no good answer to this multiple-choice question. For example, what do you do when the reaction reaches equilibrium before half of the reactant is used up?

In the reaction $A\rightleftharpoons B$ the relaxation rate constant, i.e. the rate of approach to equilibrium is exponential and given by $k=k_1+k_{-1}$. The forwards and reverse reactions have rate constants $k_1,k_{-1}$ respectively. The half-life, if you want to define it is, $\ln(2)/k$.
If the reaction is $A+B\rightleftharpoons C+D$ then the relaxation time is still exponential but now depends on the equilibrium concentrations.
You can solve the kinetics $A\rightleftharpoons B$ using $x=[A_0]-[A]=[B]$, write down the rate equation $d[A]/dt$, substitute for $x$ and then integrate $dx/dt$ from $0\to t$. Next represent the equilibrium constant as  $K_e=k_1/k_{-1}=x_e/([A_0]-x_e)$ where subscript $e$ is the equilibrium amount. You should find that
$$\displaystyle x=x_e(1-e^{-(k_1+k_{-1})t})$$
which is a rising exponential as $x$ is the amount of B at time $t$. When $t=0$ this is zero and at long times $x\to x_e$ as expected. Species A falls as B increases (from $1$ to $0.3$ in your example) and with the same rate constant $k$.
The initial amount of A is $1$ and equilibrium amount $0.3$ (so $x_e=0.7$) and the half life should be the time to reach half of the range from $1$ to $0.3$ as this is the range that A covers.


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Understanding grades of steel for cookware. 18/10 vs 18/8
I am looking for a stainless steel cookware and while researching on the best kind of stainless steel cookware, I came across this comment on youtube:

Actually 18/10 steel is better and costlier. You're talking in terms of health hazard where more nickel (10) is considered more harmful than less (8). But in reality more nickel makes a pot sturdier and shinier and less prone to chemical leaching.

So, my question is, is it true that more nickel in case of stainless steel means less leaching? I know 18/10 is better in heat distribution, is sturdier and better than 18/8, but it also has more Nickle. To me, it doesn't make sense. But then again, chemistry was not subject. Or I am worrying too much and just get either 18/8 or 18/10 steel?

What about earthen pots for cooking? Most stainless steel cookware will leach very very tiny amounts of metals into food especially if it is acidic. Also mechanical abrasion while moving the spoon also introduces metal into food. Teflon coated stuff will leach Teflon. Iron pots will leach iron. There is no way out. Aluminum cookware is also used by millions for cooking.

See this link as well: https://www.rocheindustry.com/1810-stainless-steel-vs-188-stainless-steel/#The_corrosion_resistance_of_stainless_steel_188

@AChem Earthen pots also leach depending on what soil they used. I have been using aluminium pots till now. I heard stainless steel are much better than aluminium. I have thrown every teflon coated utensil.

@AChem I saw that link while researching. It still doesn't answer my question whether more nickle in 18/10 means less leaching or more.

I strongly suspect what you put in the pan will have a far greater effect on health than the precise make up of the steel the pan is made of.

@4-K, Please keep in mind that the body has an efficient unwanted stuff eliminating system as well via excretion. If you cook with oil, as is common in South Asian food, chances of metal leaching are even lower.

@abligh we usually use toamtoes.

@AChem I am least worried about leaching. I just want to understand that does more Nickle in 18/10 means that the added nickle actually helps in less leaching as in 18/8 or not. What the guy said in comments.

@4-K, You cannot make a generic statement and no wonder you cannot find more information. Corrosion is stated with respect to certain conditions. There are hundreds of foods and cooking conditions. We cannot make a generic statement. You are looking for a wrong question. You have to specify exact conditions for a comparison. Is 18/10 or 18/8 better for cooking. Nobody can give a general answer. Both of them are food grade. Don't worry too much.

May I ask what got you worried about cooking utensils?

@AChem "more nickel makes a pot sturdier and shinier and less prone to chemical leaching"  Is the part in bold true? That's my question.

@AChem We use aluminium (heavily pitted) and non-stick cookware. So We now switched to Iron utensils. But some dishes change flavour in iron utensil so we need a stainless steel stuff too. Indians traditionally used clay and iron utensils.

@4-K, Your bolded question cannot be answered because there is NO single cooking situation. Leaching will depend on conditions and there are infinite conditions.

No stainless steel of the 18/8 family ( about 5 grades) leaches material in normal cooking service. Not to say they are immune to all possible abuse. The element that can make a difference is carbon, higher C can combine with more chrome removing it from solution where Cr provides the primary corrosion resistance. Not to worry, modern steel making generally gets C to 0.02 % max ( L grade).  I am trying to say for cookware there is no difference.

Being old , I recall Consumer Reports magazine evaluated stainless cookware over 50 years ago  One test was something like simmer sauerkraut for 100 hours.  they did find some pitting .  If that is not is recipe you use , no problem.

But what about nickle? Does more nickle means less leaching of metal in food as stated by that user in the comment? Or more nickle means more nickel leaching?

"Leeching " nickel from cookware sounds like it was written by a journalist, not a metallurgist. Closest  thing I can think of is nickel aluminum bronze may lose nickel in seawater under  rare circumstances.

Nickel is the least active metal in stainless steels so added nickel will not leach.  However, if the steel is of low quality and improperly melted etc nickel domains in the steel can be cathodic and hasten pitting corrosion of the iron and chromium. Workmanship is probably more important than 8 or10% nickel. For cooking acidic foods I use a well-seasoned cast iron saucepan [it does react a bit] or a Teflon coated saucepan. Storage is probably best in well washed polypropylene or polyethylene or soda-lime glass. [Glass leaches mostly harmless ions but breaks easily]


text:
Does chromium(III) oxide react with or dissolve in liquid sulfur dioxide?
Liquid $\ce{SO2}$ (b.p. −10 °C) is known as a versatile solvent (yet not particularly pleasant to work with) for a wide range of organic and inorganic compounds having limited solubility in conventional solvent systems. I recall that chromium(III) oxide should chemically interact with sulfur dioxide, but I cannot find a reputable source to back up this fact.
There is a related post on Reddit — Any tips for a stubborn frit filter? describing a problem with $\ce{Cr2O3}$ precipitation being stuck on the glass frit filter. Neither fusion with flux nor dissolution in alkali are suitable because they are guaranteed to permanently damage the frit. Oxide particles seem to be not fine enough to dissolve in strong acids, including oxidizing ones, to a significant degree or at a noticeable rate.
I think it would be interesting to know whether liquid $\ce{SO2}$ could potentially serve as a solvent for similar cases with “uncooperative” $\ce{Cr2O3}$ where relatively mild ambient conditions are expected.

I don't think it's soluble - SO2 is useful because it's unreactive and I don't think you could dissolve sth like Cr2O3 without reaction.

@Mithoron I fail to find either confirmation or refutation (even with a reaction). That's the issue.

Well, that would be an argument against, unless your google-fu failed you ;)

@Mithoron Now I want to research this further to prove you wrong, but I'm not sure where else to look for:( This is mildly irritating.

Well, I think it doesn't dissolve in anything without reaction and SO2 is particularly unreactive, so probably nobody even tried. Still, it would be pretty cool if it did ;)

related https://www.researchgate.net/post/Does-anyone-know-a-way-to-solve-Cr2O3

$\ce{Cr2O3}$ is one of the rare metallic oxides known to be insoluble in concentrated acids. it needs molten $\ce{NaHSO4}$ heated to a high temperature to pass into solution. This is why $\ce{Cr2O3}$ is probably insoluble in $\ce{SO2}$


text:
How to compute backward reaction rate from forward one and an equilibrium data
Suppose we have have an ideal gas mixture and a reversible elementary reaction:
$$
\ce{O + H_2 <=> H + OH}
$$
Its forward reaction rate may be computed as follows:
$$
\frac{d[\ce{H}]}{dt} = \frac{d[\ce{OH}]}{dt} = -\frac{d[\ce{O}]}{dt} = -\frac{d[\ce{H2}]}{dt} =  k^{(f)}\left[\ce{O}\right]\left[\ce{H2}\right] = A T^n e^{\left(\frac{E_a}{RT}\right)}\left[\ce{O}\right]\left[\ce{H2}\right]
$$
where $A, n$ and $E_a$ are known values. The exponents of concentrations $\left[\ce{O}\right]$ and $\left[\ce{H2}\right]$ are both equal to one due since the reaction is elementary.
Also, there is thermodynamic data for the species - heat of formation, entropy, heat capacity - for all species. For example, for H these values are:
$$
\Delta H^{298K}_f\left(\ce{H}\right) = 52.1~\text{kcal/mol}, \quad S^{300K}\left(\ce{H}\right) = 27.4~\text{cal/mol K},\quad C_p(\ce{H}) = 4.97~\text{cal/mol K}.
$$
For simplicity, heat capacity is assumed constant here.
Parameters $A, n$ and $E_a$ for backward reaction rate $\left(k^{(b)}\right)$ are unknown. According to Warnatz's et al "Combustion" book backward reaction may instead be computed from equilibrium constant, which, in turn, is determined by thermodynamic data:
$$
K_c = \frac{k^{(f)}}{k^{(b)}}= exp\left(-\Delta_R\bar{A}^0/RT\right)
$$
But how to compute $\Delta_R\bar{A}^0$?
I assume that it equals to an increment of free energy (Helmholtz function) $A = U - TS$ corresponding to consumption of one mole of each reactant, but not sure how to properly calculate it from thermo data above.

You are right that you can determine the backwards rate if you know the forward rate and the equilibrium constant. In fact, you have the right formula
$$ K_c = \frac{k^f}{k^b} $$
In general, the equilibrium constant can be calculated from the change of the Gibbs free energy during the reaction
$$ \Delta G^r = - R T \ln K_c$$
Therefore in principle you need to calculate the total Gibbs free energy of the reactants and the products and take their difference to get the equilibrium constant. How to do this depends on the nature of the data you have; it might involve quite complicated thermodynamic calculations.
Edit: I was mistaken previously, the equilibrium coefficient is expressed with the rate coefficients.

Two questions: 1. In the book eq. constant is given as the ratio of $k$'s, not $v$'s. Is it the same thing?
2. There are two constants in the book: $K_p = exp(-\Delta_RG/RT)$, where G is Gibbs energy, and $K_c = exp(-\Delta_RA/RT)$, where A is free energy. Which one do we need?

I made an edit to the post, it is indeed the ratio of the rate coefficients that you need to use. As per which equilibrium constant you need will depend on the nature of the exercise that you are trying to solve.

I forgot to mention that the ideal gas model is used. It seems that $K_p$ corresponds to partial pressures of the mixture (it is not discussed in detail in the book and thus somewhat confusing) while $K_c$ corresponds to molar concentrations. Probably, $K_c$ is what I need. But it is calculated via free energy (A), not Gibbs energy (G). Suppose we have succesfully calculated $\Delta G$, is it possible to find $\Delta A$ from it?


text:
Branch of chemistry that deals with reactions between chemicals
There are branches of chemistry like physical chemistry that hardly involve chemical reactions. Whereas, inorganic and organic chemistries pretty much definitely have a lot of chemical reactions involved.
Is there a collective name for organic and inorganic chemistries? Is there a single word for the branch of chemistry that deals with the possible products given some reactants and reaction conditions?

I think your premise is too broad. Perhaps you mean to distinguish between chemists whose work primarily involves synthesis and those who primarily deal with other aspects of chemistry?

@EdV I do not wish to distinguish chemists. It was a question about classifying topics.

Yep, as I said.

Is there a collective name for organic and inorganic chemistries? Is
there a single word for the branch of chemistry that deals with the
possible products given some reactants and reaction conditions?

As a collective term, synthetic chemistry refers to a branch of chemistry that deals with chemical reactions that produce new molecules. Please keep in mind that today these chemical branches are hard to distinguish. A synthetic chemist must know a lot of practical spectroscopy (traditional physical chemistry/analytical chemistry) in order to study his /her reactions. Also physical chemists study a lot of chemical reactions.


text:
Retrosynthesis of (1S,2R)‐1‐isopropyl‐2,4‐dimethylcyclohexane
I think that retrosynthesis of (1S,2R)‐1‐isopropyl‐2,4‐dimethylcyclohexane involves the opening of an epoxide, to control the regiochemistry in order to achieve the trans-configuration via an SN2 mechanism — which could be made from a C=C double bond. Would it be plausible to assume that the original cyclohexene is commercially available?

Are you looking for both epimers? A single enantiomer or a racemate? Your effort is rather vague. Look into monoterpenes such as limonene and pulegone.

If you are unconcerned about the undefined stereocenter and a racemic product is acceptable, you might think about a Diels-Alder reaction. You may self-answer your question. Get back to us.

Do you mean to imply a single enantiomer or the racemate including the 1R,2S enantiomer? I think you mean the latter because the configuration at C4 is undefined. In any case a mixture of diastereomers would result. If a racemic mixture of diastereomers is OK, then follow the Diels-Alder route.


text:
Would this redox titration be possible as a direct titration rather than a pseudo-back titration?
In a general chemistry lab protocol (green chemistry,  ISBN-13: 978-1482230208), students determine the content of ascorbic acid in a vitamin C tablet. There is an ascorbic acid standard in a buret, and there is also some ferric ammonium sulfate solution at known concentration, i.e. iron(III). The indicator is salicylic acid, which forms a colored complex with iron(III), but not iron(II).
The suggested protocol is a back titration where you first add a known excess amount of ferric salt to the unknown amount of ascorbic acid and add the indicator, giving a colored solution. Then, you titrate back with the ascorbic acid standard. The endpoint is the disappearance of color.
Is there are reason for the back titration in this case? In other back titrations, there is a step after adding an excess of reactant (for calcium carbonate, you add an excess of HCl and then drive out the carbon dioxide with heat, making the back titration a strong acid/strong base reaction). For the ascorbic acid reacting with ferric salt, I wonder if the endpoint is easier to see, or maybe the kinetics of complex formation motivate this (it says to let the solution sit before starting the back titration).

In my opinion, there is a mistake in the type of iron ions reacting with salicylic acid. Salicylic acid is an indicator that makes a red complex with iron(III) and not with iron(II), as stated in the first section.

You are right Maurice, the text "Green Chemistry" clearly states, iron(III)-salicylate acid complex along with its structure.

@Maurice I'll fix that.

The idea of using green chemistry to titrate vitamin C with environmentally friendly reagents is understandable, but it may not be the most appropriate method. Iodometry has traditionally been the traditional method for ascorbic acid analysis. The main reason for doing back-titration is that ferric ion is a very gentle oxidizing agent but good enough for ascorbic acid. Therefore, we wish to ensure that all of the acid in the sample is oxidized by ferric ions.
Now note that this is a pseudo-back titration, because the titrant and the analytes are the same! In other words, after adding excess iron (III) to the sample solution, a "direct" titration is being conducted between ascorbic acid and iron (III). I assume the titrant concentration is higher than ferric ammonium sulfate to speed up the reaction and ensure completion of the redox process. As an exercise, why don't you ask the students to try direct titration and back-titration and then do a statistical test to see if there is a difference. I mean half the class perform direct titration and half the class do this pseudo-back titration.

Excellent idea! Is there a preference for titrations going from colorless to color or vice versa?

No I don't think so, the color change must be rapid, that is it.

I changed the title to pseudo-back titration.

@Karsten-apprentice, Did you get different results of back vs. direct titrations?

Semester hasn't started, and this would be for the second semester of the two-semester series. Maybe I should try myself... thanks for asking though

No rush, I was just curious to see if direct titration may give the same results. It could be your J.Chem.Ed. article :-)

You are giving me a lot of J, Chem. Ed. suggestions :-). I'm happy with the one I already published, but maybe another one some day. Another comment I will have to delete, I should stop writing them.


text:
Rotten chicken smell
My boyfriend and I have a camper and he forgot to take a package of chicken out of the refrigerator before parking it. The chicken sat in the fridge which was off in the heat for a few months. We would like to pull the fridge out to dispose of it. However, many other systems run through the fridge. What chemicals will get rid of the awful smell or is it even possible? Thank you!

A good scrub with bleach is the first thing to try.

I would start more gently with soap and water and follow with wet baking soda. Many of the strongly smelling molecules produced by decay of flesh are amines, which are neutralized by the baking soda and made more water soluble.

If the entire inside of the refrigerator is stainless steel, you have a chance of killing the odor. But if there is any plastic in the interior, it will have absorbed that foul (no pun intended) smell and you will almost certainly know it every time you open the refrigerator door in the future. I hope I am wrong, but let us know what happens! Good luck! By the way, welcome and have an upvote!

BTW camping for a few months sounds like something only Mansour could do :p

Diluted vinegar is often used to wash fridges to get rid of smell.


text:
Why did neutralization of weak acid calculation differ from empirical results?
I recently stumbled upon the problem of needing to neutralize a volume of 0.33M acetic acid with 0.1M NaOH. I was curious how many mL of 0.1M NaOH I would need to bring 1mL of 0.33M Acetic Acid solution to a pH of 7.
For my calculations I used the following data:
Inputs

Ka =  0.000018
M of Acetic Acid  0.33
M of NaOH 0.1
Volume of Acetic Acid (mL):   1

Outputs:
$$
\sqrt{Ka*0.33} = 0.002437 M of H+
$$
$$
0.002437 /1000mL * 1mL = 0.000002 mol H+
$$
Volume needed to Neutralize
$$mol H+/M NaOH = 0.000024  L of NaOH$$
$$-> 0.024372   mL of NaOH$$
However, in practice this did not work. My empirical results ended up:
3.1 mL 0.1M NaOH neutralized 1mL 0.33M Acetic Acid
Why did this calculation differ from my observations? How reliable is a disassociation constant here?

Use MathJax formatting for clarity.

Since when is NaOH a weak base?

Also no idea what you even wanted to do in this calculation. Perhaps you don't get what "neutralise" means in this context... You don't need any constants, and whatnots, just equal amounts of acid and base.

I wanted to find out the volume of 0.1M NaOH would cause 1mL of 0.33M Acetic Acid solution to approach neutral ph.

So I guessed right, you just misunderstood it.

The pH is not 7 at the endpoint. Sodium acetate solution has a pH higher than 7 unless the solution is very dilute.

It looks you errorneously assume you just need to neutralize those H+ that exist in the acid solution as the equilibrium concentration. You forget that this equilibrium provides more H+ when the existing ones are being neutralised.

At pH 7, more than 99% of the acid is neutralised, so you need more than 3 mL of 0.1 M NaOH for 1 mL 0.33 M acetic acid.

No need to use dissociation constants.
$\pu{1 mL}$ of $\ce{0.33 M}$ acetic acid contains $\ce{3.3 10^{-4} mol}$ acid.
To neutralize this acid, $\ce{3.3 10^{-4} mol NaOH}$ is needed, which is included in a volume $\pu{V = n/c = \frac{3.3·10^{-4} mol}{0.1 mol/L} = 3.3 10^{-3} L = 3.3 mL}$
This amount will not go exactly to $\mathrm{p}H$ $7$, as the final $\mathrm{p}H$ will be around $9$ . But you mentioned that you only want to approach neutral $\mathrm{p}H$.  The volume of $\ce{NaOH}$ necessary to go from $\mathrm{p}H$ $7$ to $\mathrm{p}H$ $9$ is negligible in this problem.


text:
Ozone sensor with range above 5ppm?
I am attempting to find or build a sensor for controlling an ozone generator. To do this, the sensor will need to see the range of at least 0 to 5% (50000ppm). The commercially available sensors I have found so far are all in the 0 to 5ppm range; primarily used for air quality and safety. It seems like a sensor in this range should exist, but perhaps I am using the wrong vocabulary for searching.
This website mentions electrochemical ozone sensors as well but looks like they only go up to 20ppm.
Ozone obviously absorbs UV, but I have not been able to identify this being used outside of a lab environment for sensing concentration. Here is an expensive lab ozone sensor that goes up to 999ppm. Might be able to build a sensor based on this principle, but maybe I am overlooking some other common atmosphere gas that would give false readings (water vapor, VOCs, etc).
Are these the only three technologies used for measuring ozone? Other creative methods?

There is something called linear dynamic range of sensors. You are aiming for a really *insensitive" sensor which responds to 0 to 5%. With 5% ozone, it will start damaging everything.

How fancy do you want to be? Ozone absorbs infrared radiation too. Look at the infrared spectrum of ozone.

@AChem, Good idea on the infrared absorption; I will have a look!


text:
Are diffuse functions necessary for modeling unbound conjugated anions?
According to this paper, and contrary to popular belief, diffuse functions are far from being necessary for the calculation of the electron affinities of the polycyclic aromatic hydrocarbon(PAH)s, due to delocalisation of the additional charge that overweighs the Coulomb repulsion between the additional electron and the existing electrons. The paper also says that electron affinities of individual atoms do need diffuse functions for their calculation, for the fact that atoms with known positive (i.e. bound) electron affinities could be misinterpeted as having negative (i.e. unbound) electron affinities when diffuse functions are not used. All was fine- or seemed to be.
After reading the paper, I suddenly realised that some highly conjugated neutral molecules are still known to have negative electron affinities- benzene and naphthalene to name a few. Delocalisation of the additional charge would not mandate diffuse functions, while the unbound nature of the radical anion would. So one now has a dilemma- which is generally more important, delocalisation or unboundness, for deciding whether to use diffuse functions or not?

This might be better suited for Matter Modeling since there are actual basis set experts over there (Jensen, for example). The linked paper uses the old Pople basis sets, so I'm not sure how much you can take as general conclusions to modern basis sets. My personal opinion is that with a good modern basis set (e.g., def2-TZVP or better) diffuse functions are less critical on larger delocalized $\pi$-systems. I'd think a test on benzene and naphthalene would be a good idea though.


text:
Histidine-Boc deprotectection
A question on Fmoc-solid phase peptide synthesis (SPPS). Is Boc-protected Histidine too labile? Merck's website shows  Boc is labile to treatment with Piperidine. Anyone with such experience?

Comments here support this https://www.researchgate.net/post/Did-anyone-observe-spontaneous-Boc-removal-from-histidine  as does a mention here http://cem.com/media/contenttype/media/literature/ap0170v1-CEM.pdf

Also see this paper  Sieber, P.; Riniker, B. Tetrehedron Lett. 1987, 28, 6031 –6034


text:
Oxidation of toluene, ethylbenzene and styrene with potassium permanganate in acidic medium
While I was studying my organic chemistry textbook, I came across the following reactions that need more clarifications:

Oxidation of toluene (1) is the only reaction from the textbook, and I understand it.
I have a doubt about the product of ethylbenzene (2) oxidation. I have written the two products — benzoic acid (A) and 2‐phenylacetic acid (B) — which I'm not sure about. Does the ethyl substituent become a carboxylic or does it become a methyl carboxylic group?
For the oxidation of styrene (3) in addition to A and B I would propose one more product: styrene glycol (C) because I was taught that $\ce{KMnO4}$ oxidizes ethylene into diol. So, which of these three compounds will be the main product?
Lastly, are the products of reaction of ethylene with potassium permanganate in basic medium and in acidic medium different or the same? Please correct me if there are any errors.

It depends on the reaction conditions. KMnO4 is a very powerful oxidant. If you run it a high temperature with concentrated acid then phenylacetic acid loses CO2 to give toluene which is then oxidised to benzoic acid. Same with styrene, under controlled conditions you can get the diol, push harder and that oxidises further.

I do not think that the mechanism is from outside in. The benzylic position is active and ketones are isolated under less forceful conditions. Glycols do form and under acidic conditions will rapidly oxidize further, not much need to push harder.

@Waylander I did some research and I guess your explanation is the correct way to understand this.

@jimchmst Thank you too for the explanation

After asking the question, I did some research about these reactions, and I think it would be valuable if I share it with the community.
ethene becomes a diol in the presence of basic medium with KMnO4, not when there is an acidic medium. In an acidic medium, it oxidizes completely into a carboxylic acid.
Styrene however, always oxidizes completely into benzoic acid. Same happens to any kind of alkyl (alkane, alkene, alkyne) that is combined with benzene.
I think this happens because, benzene is so rich with electrons and very unsaturated, KMnO4 is really interested in oxidizing benzene, but as we all know benzene will never oxidize under these normal conditions. Instead KMnO4 tries to oxidize the Carbon atom nearest to the benzene, which is not a part of benzene but has substituted to it, and so, Benzoic acid is made.
Hope this helps someone who had the same doubts as me...

With the proper choice of oxidant and conditions alkyl etc. benzenes can be oxidized to many different compounds with varying degrees of success. This is best experienced by actually running the reactions in the lab or possibly vicariously by reading actual lab results. Hot, acidic, permanganate or dichromate are brute force oxidants that eventually will oxidize almost any carbon compound to CO2 and H2O.


text:
Why bromoacetone is more reactive towards SN2 than alkyl bromides?

Compare the rates towards SN2 reaction:

I got the point that 2-bromobutane d is a secondary halide, so the steric hinderance is highest making it least reactive among the four. Both b and c are primary halides, but c has branched isobutyl group, so because of steric effect the order should be b > c > d.
However, the answer claims that bromoacetone a is the most reactive towards SN2 despite also being primary halide. What is the exact reason for this?
My friend says if there is carbonyl on β-position, then it is highly reactive towards SN2. Is this even accurate? If yes, then what is the reason?

Does this question help?

@SafdarFaisal that is a good question but how is it related with this? As in that question the reaction proceeds by elimination i guess, but here SN2 is asked!

It was more the answer to the question. A positive charge on the alpha carbon is highly destabilized. Well here's a similar question

I think this paper gives good details on this regard: https://pubs.acs.org/doi/10.1021/ja01075a025#

As part of the paper: The favorable effect that the carbonyl and related have on sN2 reactivity probably stems in part from the absence of rate-retarding steric and field effects, coupled with a mildly rate-enhancing inductive effect. The very strong accelerating effect of these groups must, however, include some sort of an additional conjugating factor.
The conjugation responsible for carbonyl activation is caused by an overlap of the p-orbital of the carbon atom of the carbonyl group with orbitals on the entering and leaving groups, rather than a π-bond type overlap.


text:
Difference between change of enthalpy of neutralization between copper and manganese carbonate
I was doing a calorimetry experiment to calculate the change in enthalpy during neutralization reactions between metal carbonates and hydrochloric acid. I experimented with 5 metal carbonates: magnesium, zinc, copper(II), manganese(II), and calcium.
I noticed that when I reacted $\ce{MnCO3}$ with $\ce{HCl}$ and $\ce{CuCO3}$ with $\ce{HCl}$, the temperature of the solution increased unusually higher than magnesium, zinc, and calcium carbonates. After calculating the enthalpy change per mole of $\ce{MnO}$ and $\ce{CuCO3}$ when reacted with $\ce{HCl}$, it was unusually higher. For the enthalpy change in neutralization for magnesium, I obtained a value of $\pu{-45.0 kJ/mol}$ while I obtained a value of $\pu{-377 kJ/mol}$ for copper(II) carbonate and $\pu{-240 kJ/mol}$ for manganese(II) carbonate. I have double-checked my stoichiometry calculations and no mistakes were found. Could anyone please explain the large difference in change of enthalpy for copper(II) and manganese(II) carbonate?

Part of it will be differences of cation hydration ( following naked ion radius differences). Part of it will be formation of copper chlorocomplexes. Part of it may be differences in ionic lattice of the solid. But I am not sure if that explains such huge difference.

Just guessing if possibly stronger covalent character for Mn and Cu may play the role.  BTW CuCO3 is very hard to obtain, more frequent are basic forms CuCO3 . 1(2)Cu(OH)2.

@Poutnik Is there any method to calculate the theoretical value for the enthalpy of neutralization for copper and manganese carbonates?

Well there is a method to calculate theoretical value of mercury melting point. Chemists have waited like 50 years until there were available fast enough computers. Experimental determination is for both cases much easier. BTW, neutralisation enthalpy is just a part of the whole enthalpy change chain.

@Poutnik Thank you so much for your kind replies! Have a great day sir

The 3rd option would be using Hess law, considering enthalpy is a function of state. But I am not sure about the success of gathering all enthalpy changes for alternative formal process path. Like deltaH values for formal breaking solid lattice to free ions, hydration of these ions and neutralization of carbonate anion.

It is possible to calculate the enthalpy of reaction from enthalpy of formation of the reactants and products.  Tables exist with the substances in the various states.
One would expect that as ionic radius gets larger (e.g. moving down a group of metals) the strength of the ionic bond would become weaker resulting in increasing enthalpy of reaction (less energy required to break the solid lattice).  This trend is seen experimentally for metal oxides.  Metal carbonates, however, display the opposite trend - as ionic radius of the metal increases, enthalpy of reaction decreases.
Part of the reaction of a metal carbonate with acid is the liberation of CO2 gas meaning one of the carbon-oxygen bonds from the carbonate ion has to be broken.   Smaller metal ions (higher core charge) draw an oxygen and electrons from the carbonate ion towards it more strongly, allowing the remaining CO2 to be more easily liberated.  Larger metal ions pull less on the carbonate (less polarized), meaning more energy is required to break the carbon - oxygen bond to liberate a molecule of carbon dioxide.  This suggests that smaller metal ions will result in a greater (more exothermic) enthalpy of reaction.
Theoretically calculated enthalpies support this.  As ionic radius increases, enthalpy of reaction of metal carbonate with hydrochloric acid decreases (becomes more positive).  For reaction of the following metal carbonates with aqueous hydrochloric acid, calculated ΔHreaction values are:  MgCO3 : -50.1 kJ/mol; CaCO3 : -15.3 kJ/mol; BaCO3 : 1.1 kJ/mol.

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review


text:
Definition of mineralisation in context of organic chemistry
What does it mean when an organic compound, say methylene blue, is said to be mineralised? An accompanied equation shows the dye being converted into carbon dioxide and oxygen.
The only results I have found on Google are about soil chemistry.

Simply put: organic → inorganic. Mineralization is often also called TOC diminution (the term “TOC reduction” is purposely avoided due to ambiguity in chemical context). From Duffus et al. [1, p. 1263]:

mineralization
Complete conversion of organic substances to inorganic derivatives, often visible as microscopic deposits which may be associated with damage to soft tissue (e.g., in the kidney).

Related entries from [1]:

biomineralization
Complete conversion of organic substances to inorganic derivatives by living organisms, especially microorganisms.

calcification
Form of mineralization in which organic tissue becomes hardened by deposition of calcium salts within its substance.

complete mineralization
Complete breakdown of a complex organic compound to carbon dioxide, water, oxides, and oxidative inorganic products such as nitrate or sulfate.

Reference

Duffus, J. H.; Nordberg, M.; Templeton, D. M. Glossary of Terms Used in Toxicology, 2nd Edition (IUPAC Recommendations 2007). Pure Appl. Chem. 2007, 79 (7). DOI: 10.1351/pac200779071153. (Free Access)


text:
Why is it impossible to determine number of visible emission spectrum of He+
We know that atoms emit spectral lines on excitation and de excitation.
I was doing one high school question based on this theory

Number of visible lines when an electron returns from 5th orbit upto ground state in a sample of He+ ions

Answer was "Cannot be determined"
But I think it cannot be, as we know that Balmer series of hydrogen transitions are in visible side so we can calculate same wavelength emission in case of helium atom too.
Example: 3 to 2 transition in hydrogen lies in visible region... So same wavelength emission will be emitted in helium in the transition 6 to 4 using Rydberg formula. So how is the answer "cannot be determined"? If it can be then am I correct

Here you go: https://laserstars.org/spectra/ProtoHydrogen.html. Seems to be called the Pickering series. Looks like you were right. (+1)

@EdV So should the answer be 1 line? Because then only  5 to 4 transition should be considered having visible wavelength

@EdV 4-7 (5412 Å) , 4-9 (4522 Å) ,4-6 (6560 Å) and 4-8 (4859 Å) these four? but from 5 to 1 transition none of these are obtained.

@EdV My query is : how many visible transitions are there from 5th orbit to ground (1st) orbit? 1,2,3 or what?

I think the answer that was given is based on this. If the electron drop directly from 5 to 1, then 1 UV photon is emitted. But if the electron drops in cascade fashion, from 5 to 4, then 4 to 3, then 3 to 2 or 1, then you get 1 IR photon, 1 visible photon, and 1 or 2 UV photons. This looks more like a trick question than anything else. Anyway, I will comment no further.


text:
Reaction of copper(II) chloride and aluminium foil
So I did an experiment recently with copper(II) chloride and aluminum. As they reacted with each other the redox reaction made the aluminum a brown substance that is supposed to be copper and the color of the copper (II) chloride solution was supposed to be colorless, indicating that the reaction has reacted ''evenly''. Even though I made all my calculations right and put the right amount of aluminum foil in the solution, the copper(II) chloride didn't become colorless. Could there be any suggestions as to why it didn't become as expected? Im doing a lab report on this and I can't seem to figure out any possible suggestions...
The calculations:
V(CuCl2) = 50 ml = 50 cm3 = 0.05 dm3
c(CuCl2) = 0.20 mol/dm3
m(CuCl2) = ? g
n(CuCl2) = V(CuCl2) x c(CuCl2) = 0.05 dm3 x 0.20 mol/dm3 = 0.01 mol
M(CuCl2 • 2 H2O) = 170.48 g/mol
m(CuCl2) = n(CuCl2) x M(CuCl2 • 2 H2O) = 0.01 mol x 170,48 g/mol = 1.7048 ≈ 1.705 g
(The 1,705 g is the copper chloride that was used to put in the distilled water to make the solution)
The amount went into 50 ml of distilled water. For every experiment, I was supposed to only use 10-20 ml so I used 15 ml for every try.
CuIICl2 + Al0 ⇸AlIICl2 + Cu0
3CuIICl + 2Al → 2AlCl3 + 3 Cu0
V(CuCl2) = 15 ml
n(CuCl2)= 0.01 mol
m(Al) = ? g
n(CuCl2) = Al x ⅔ = 0.0066666667 mol
n(Al) = 0.0066666667  mol
m(Al) = n x M = 0,0066666667 x 27 = 0.18 g
I even showed my teacher but i have no idea what could have gone wrong :/

This may be helpful: https://chemistry.stackexchange.com/a/128099/79678. You should also show your calculations: a trivial math mistake is all it takes to ruin a calculation.

I edited them. Is there something that is wrong with the calculations?

Well one gram and 0,705 gram as well if that makes sense not 1,7 kg.

Or 1.705 grams if that makes more sense it's in Swedish so I understand if it's a bit weird but 1 gram + 0.705 gram.

Could there be unreacted Al hidden inside a shell of Cu?

So you mean that Al didn't dissolve properly in solution? Like there is a bit of coating that won't let it react fully?

That is highly probable.

@Poutnik Excuse me if I'm asking too many questions haha. So it is highly probable that the experiment wasn't done correctly or that the aluminum foil itself was the one that didn't make the solution colorless? I found a video where they used HCl as a catalyst that helped against this ''coating'' that the aluminum foil consists of.

HCl helps with Al2O3 coating, but not with Cu coating. It is not so hard to imagine Cu metal is obstacle for solution to reach Al.

Two mistakes. First : It is impossible to have a solution of $\ce{CuCl2}$ which has a concentration of $\ce{20 mol/L}$ . Look ! $\ce{20 mol CuCl2}$  weighs $\pu{2668 g}$. This cannot be dissolved in 1 liter. Second mistake : $\pu{15 mL}$ $\ce{CuCl2 0.2 M}$ does not contains $0.01$ mol $\ce{CuCl2}$.

Two other mistakes. If a sample of $\pu{15 mL}$  is taken from  $\pu{50 mL}$ of a stock $\ce{CuCl2}$ solution containing $\ce{0.01 mol CuCl2}$, the $\pu{15 mL}$ sample does not contain $\ce{0.01 mol CuCl2}$. No ! It contains $\ce{\frac{15}{50}·0.01 mol = 0.003 mol CuCl2}$. Second mistake, that I copy here : *n(CuCl2) = Al x ⅔ *. No ! It should have been written : $\ce{n(Al) = 2/3  n(CuCl2)}$. As a consequence, these $\ce{0.003 mol CuCl2}$ do react with $\ce{0.002 mol Al}$ , which weighs $\pu{0.054 g}$

Final remark. Have you weighed the sample of aluminum foil introduced into the solution? It must have been at least $\pu{0.054 g}$ ! If some copper ions were still in solution at the end, it may have been for lack of aluminum !...

@Maurice Thanks a lot for pointing the mistakes out! :) I'll make sure to fix them as much as I can.


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Why ionization energy of Thallium is higher than that of Indium?
I just normally calculated the effective nuclear charge (Zeff) for thallium and Indium from Slater's law, and I found it same for both! That is 5.(If you want calculation for answering or correcting any of my mistake please inform me)
So further moving on, My teacher told that thallium and Indium has approximately equal atomic size due to 4f  contraction.
(Note : I am in 11th standard , Inter, so I just started studying basics and inorganic chemistry for the first time , so pardon on any foolish mistake)
So I can't get "The valence electron is basically at somewhat equal distance from nucleus due to same atom size, plus those valence electron experience same effective nuclear charge (Zeff) , than according to me Ionisation potential should be same as they will expirience same protons' charge(sorry for that word) in equal distace right?  But its not! "
Please Answer it , I am extremely confused.
Some people explain it as in Thallium we are having full filled d-subshell along withy fully filled f-shell, both poor shielder and 32 proton are being added on in Thallium comparison to Indium,
But I dont find it relevant as "even after poorest of poor shielding the Overall charge an valence electron experience is same!(5) For both, so no point of saying more proton=more pull and thus higher ionisation energy.
One such answer is

I don't know whether I am correct or not, as I just started learning. Please Can you explain in easy way  and actual reason according to my calculation.
Where I am wrong with concept? too

Duplicate: https://chemistry.stackexchange.com/questions/133177/why-is-ionization-energy-of-indium-less-than-gallium

First, your Zeff calculations based on Slater's rules are correct, the values of the Zeff are 5 for both Indium's 5p and Thalium's 6p outermost electrons, as their respective shielding values are 44 and 76.
Your explnation about the lanthanide contraction makes sense, but let me take another approach. When we are talking about very heavy elements, we have to consider the so called "relativistic effects", which usually appear when the atomic number is greater than 70 aprox. So basically with these atoms that have large masses, the electrons are forced to move at higher speeds, so they are closer to the nucleus (smaller radius) and as a consequence they are more difficult to remove (higher IE).
https://www.nature.com/articles/s41467-018-06745-6
https://en.wikipedia.org/wiki/Ionization_energy#:~:text=Relativistic%20effects%3A%20Heavier%20elements%20
https://www.hindawi.com/journals/isrn/2013/689040/

Sorry but
1.) what is 'relativistic effects' ? As i told , i am a beginner. Please sir can you explain that effect?
2.) And in last para you told they are forced to move at higher speed thus lower radius but isnt  it denying the fact that both thallium and Indium has approximately equal size or radii

So,
1/ Relativistic effects are based on the theory of relativity. So with these heavy atoms is important to consider it because in these elements electrons do attain sufficient speeds for the elements to have properties that differ from what non-relativistic chemistry predicts. In other words, relativistic effects are those discrepancies between values calculated by models that consider relativity and those that do not. They can be considered to be perturbations, or small corrections, to the non-relativistic theory of chemistry, which is developed from the solutions of the Schrödinger equation. These corrections affect the electrons differently depending on the electron speed compared to the speed of light.
https://books.google.es/books?id=0xcAM5BzS-wC&pg=PA4&redir_esc=y#v=onepage&q&f=false
2/Yes, despite the fact that they have a very similar atomic radius, thalium has a few more electrons (in the same volume), so there's a much higher "electron density" (note that this term is not very accurate because is used with other mening).
"In quantum chemistry, electron density or electronic density is the measure of the probability of an electron being present at an infinitesimal element of space surrounding any given point." (From wikipedia).

If OP comments for further clarification, answerer should edit the answer, or comment, not make a new answer.

It wasn't for further clarification, it was because @Shinchan Nohara had more questions. But it seems that he hasn't seen the answer... :(

Than he should ask a new, follow-up question, this isn't a forum thread.

@jorge Bonifaz , sorry , I come to this site at long interval, but thank you for the answer , may be I am just a little immature to understand these heavy terminology until I study them completely  further more . But i know this answer is well explained for people who know all this, and help them. Thanks again

Nono, don't worry, I was just worried cause I thought you weren't answering me, I'm new at this, and I'm glad to be able to help you even a little, so if you have any other question just ask.


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Could fluorinated gases be removed from the atmosphere?
Background
Among all the discussion of 'greenhouse gas removal' (GGR) to aid climate change mitigation, almost all is concerned with removal of carbon dioxide. There are one or two mentions of chemical techniques for removing methane and nitrous oxide. I've found nothing on F-gas removal, despite destruction of HFCs being a source of carbon offsets and recent media focus on SF6 having a GWP100 of 23,500 and an atmospheric lifetime of 3,200 years; others include tetrafluoromethane (lifetime 50,000 years), dichlorodifluoromethane and chloropentafluoroethane. Presumably F-gas removal is never mentioned because concentrations even for long-lived species are unworkably low (less than one part in 109), the warming effects are far smaller than that of CO₂, it's not useful cover for the non-renewable energy industry, and production can be addressed through the Montreal Protocol and Kigali amendment (which covers HFCs but not PFCs or SF6). I am aware that F-gases represent less than 3% of GHG emissions and the main focus of mitigation has to be fossil fuels.
Question
What I would like to know is whether there is any chemical or physical method to remove F-gases from the atmosphere. How could it be done? Or is it definitively impossible?
Pointers to relevant publications would be particularly appreciated.
Initial thought: distillation
Could fractional distillation be useful? In production of liquid nitrogen, any solid impurity will be a greenhouse gas, mostly water. Apparently there's a technique to separate substances in this state. Xenon is isolated through fractional distillation of air, but occurs at around 87 parts in 109, still 104 times more than SF₆. Lackner's nitrous oxide comment linked to above mentions 'Sherwood's Rule' - 'separation costs tend to scale linearly with dilution', making this seem impossible.
Or is there any other relevant technique or chemical reaction or way of trapping fluorinated molecules at scale? Could resonance electron capture be in any way relevant?
Digression about dilution
Obviously liquefaction of the entire atmosphere might be a tad inconvenient and expensive. However, some F-gases are much denser than air, so might they pool in low, still air in drains and caves near production and waste recovery facilities? Edit: probably not - there's little sign of a CO₂ gradient in caves, so is there any research on distribution of denser substances like SF6 which can be detected in very low concentrations (measurable by electron capture detector, infra-red spectrophotometry, flame ionisation or mass spec)?  How long do they take to diffuse away?
Lackner also mentions passive absorption of carbon dioxide using natural wind.  Is there any scope for passive concentration using gravity?  A useful answer here on diffusion suggests for the heavier F-gases mentioned, you'd get a 65% ratio of concentration of still air per km.  So that implies you'd need a really deep mineshaft to collect heavier gases; about 6 km deep, Maybe tapering from a wide collector in open air down through filters could eventually achieve such a 20x concentration for SF6, and then be used in some separation process.
Sub-question: economics
Supposing a 20x concentration were located and used as the source by the compressed gas industry and it could also use a carbon price around $500/tCO2e - is there any way recovery, recycling or destruction of F-gases could be economic? Back-of-the-envelope suggests that the price of xenon (about 15 USD/L) is comparable to the carbon price of a similar sample of SF6 at the current GHG price (80 USD/tCO₂e on EU ETS), but of course the concentrations of the latter are much lower.
Sub-question and follow-on
The focus of this question remains the environmental context as above. I'm not a professional in this field and don't even know which of the 'sub-questions' I've added are relevant and what aspects have been missed. I've posted a separate question, free of environmental context, focused on SF6 at How would you remove sulfur hexafluoride from a gas sample?.

No, once mixed into the atmosphere, HFC's stay mixed and do not pool in low places (though those with low vapor pressure at Antarctic temps might accumulate there). The issue is that the gases are man-made, but widely dispersed as refrigerants, aerosol propellants, etc. Though production of some are banned, there are viloators.

Can you also search the concentration of these fluorinated compounds in the lower atmosphere? It must be in picograms/m$^3$...Imagine finding a tip of a broken needle in a haystack.

@AChem I know concentrations are measured. I don't know if it's using an electron capture detector or some other means, but presumably feasible to locate higher concentrations - indeed SF6 is used as a tracer. Those I mention range from 5-500 ppt, so nanograms per m³.

@DrMoishePippik Thanks. It does look like most HFCs are well-mixed. I was initially thinking of SF6 and the old idea that CO₂ pools in caves; there is a period when it is not well-mixed as in the Lake Nyos disaster.
Xenon is commercially extracted from air, but is at concentrations about 100 times higher - wondered if possible for F-gases. It's not only violators releasing CFCs, but that many HFCs are not completely banned.

Bulk gases denser than air can stay in depressions for a long time. In the end, in thermal equillibrium, all gase mix perfectly with each other, unless you are at very low pressure and/or very high $g$ fields. And even then they never separate, but only develop a shall concentration gradient.

In the Lake Nyla case, the CO2 was dissolved in the cold lake water at the bottom until it was disturbed. Then it came out of solution quickly like a shaken soda…

If anything concerns the whole atmosphere, the main  question must be about the method scalability, mainly in economy, resources, technology and environment aspects. Many methods for purging air of trace components, that are easy and efficient in the lab scale, are not applicable in atmospheric scale. // Environmental problems are very seldom as simple to be addressable in the scope of a single Q/A.

Short answer? No.
All gases are prefectly mixed in the atmosphere, except if you can catch them directly an the source. That's just thermodynamics.
It's already an imo crazy idea trying to remove CO2 from the ambient air. Of course it can be done, but the amount of energy needed is huge, even before you liquify or compress the gas and store it.  Anything with orders of magnitude lower concentration is just hopeless. We do not currently live on "planet free energy". ;)
If you really go and separate CO2 from air (on a really large scale, so it'd actually make a dent in the atmospheric carbon dioxide concentration), you will also capture most of the other gases you're thinking about. Except the methane, because it has a significantly lower boiling point, 80 K below CO2.
I don't know the actual numbers, but my gut feeling is CO2 makes up 80% of the global warming  problem, and 90% of the rest is, unfortunately, methane. (If you want to research the numbers, please add them here.)
So once we have excess solar or fusion energy , we can start capturing CO2, and will get some small amount of SF6 and friends with it, improving the efficiency of the climate-saving effort by a few percent. They however need special disposal, so they'd probably be more of a nuissance here.

(+1) Nice answer. If I was going to capture carbon dioxide from the atmosphere, I would grow lots of fast growing trees, harvest them, process just enough to get logs, and stash the logs in old salt mines and the like. And re-plant one for one, at least. That captures the carbon and you have wood if you need it later.

@EdV That's my favourite, too. You can make money by selling as construction wood. Turn the rest into charcoal, stash that (or sell for soil improvement), and sell the energy. Re-forest the sahara! You need large scale seawater desalination, but that's cheap in hot regions.

CO₂ removal etc is addressed in the links in the question. As percentages of the problem, it depends on timescale: see IPCC AR6 WG1 figs SPM.2 and TS.20 and Annex III - F-gases look tiny after 100 years, but I'm interested anyway. BECCS (burying tree carbon) uses huge amounts of land.

'Just thermodynamics' - statistically separation is unlikely, but not if there is something (gravity, electrical field, etc) producing a gradient. Water vapour is not well-mixed and very heterogeneous You could consider this topic like remediating heavy metal contamination.

@CedricKnight Atmospheric water freezes, melts, evaporates, and condenses all the time. It has a huge turnover by itself. None of the trace gases you are thinking about does. They are not even well water-soluble, which would make a difference. Sorry, but all I can say is "forget it". ;)

And timescale? I'm not concerned with 22nd century problems. They'll have to clean up, just like we have to for our grandparents. I know that sounds harsh, but come on. You only live once.

"statistically separation is unlikely": no, it cannot happen. The gravitational or electrical fields you mention do no exists on earth. You're orders of magnitude off.

By "statistically separation is unlikely", I meant the second law of thermodynamics is a question of probability. You could spontaneously get all the nitrogen on one side of a room and all the oxygen on the other. Unlikely is a British understatement :). Just trying to elicit some possibilities, involving energy input or not, phase changes or not. I corrected 'm' to 'km' in my question, but hope have orders of magnitude for gravity correct now.

On 'timescale' I just meant that there's a big difference between GWP10 and GWP100 as shown in that IPCC TS.20 figure. Methane emitted now has almost negligible impact in the 22nd century, but many F-gases will still be around. Some people are looking even further ahead: https://doi.org/10.1007/s10584-020-02785-4

Thanks for answers so far, and I do get the general direction this is going. 'If you want to research the numbers, please add them here.' Of 2020 GHG emissions, it looks like F-gases are 2.6%, compared to N2O at 4.9% and deforestation at 10%, with the bulk of course being coal, oil and gas.  https://wedocs.unep.org/xmlui/bitstream/handle/20.500.11822/34426/EGR20.pdf p5


text:
Literature values for the molar absorptivity of Methylene Blue
I am having trouble finding literature values for the molar absorptivity of MB dye from a scientific journal, preferably one that is recent.
I am also having trouble finding an explanation for why MB doesn't follow Beer's Law for high concentrations from a scientific journal.
I apologise if my english is hard to understand.

What is high  concentration and what is the absorbance curve like?

Wikipedia article mentions dimer formation.


text:
Where does the exchange energy released in an exchange interaction come from?
Two electrons in different orbitals can exchange with each other and if this happens they release a certain amount of exchange energy. But where does this energy come from? And wouldn't the atom collapse if exchange energy is continuously released?

The electrons are not being continually exchanged. Rather, the fact that they could be exchanged without affecting the quantum state alters their wavefunctions in a way that impacts their mutual repulsion energy. As Wikipedia explains, the distance between electrons in identical subshell and spin states is effectively increased, so they naturally have less repulsion between them.

The effect is due to the wave function of indistinguishable particles being subject to exchange symmetry, that is, either remaining unchanged (symmetric) or changing sign (antisymmetric) when two particles are exchanged. Both bosons and fermions can experience the exchange interaction. For fermions [including electrons], this interaction is sometimes called Pauli repulsion and is related to the Pauli exclusion principle. For bosons, the exchange interaction takes the form of an effective attraction that causes identical particles to be found closer together, as in Bose–Einstein condensation.
The exchange interaction alters the expectation value of the distance when the wave functions of two or more indistinguishable particles overlap. This interaction increases (for fermions) or decreases (for bosons) the expectation value of the distance between identical particles (compared to distinguishable particles).[1] Among other consequences, the exchange interaction is responsible for ferromagnetism and the volume of matter. It has no classical analogue.

The number of electron-electron repulsions that are affected in this way grows quadratically as the number of electrons in identical subshell and spin states increases: one repulsion between two electrons, three repulsions between three electrons, six between four electrons, and so on. So when we have a lot of electrons that could be unpaired with identical spins in different orbitals of a subshell, there is a growing tendency for them to favor this unpairing, leading the the effect we see in $d^5$ transition metal complexes (e.g. $\ce{Mn(II),Fe(III)}$).
Cited Reference

David J. Griffiths: Introduction to Quantum Mechanics, Second Edition, pp. 207–210.


text:
How is it possible that some celestial bodies have icy surfaces when water boils off at any temperature in absolute vacuum?
Why doesn't the ice on comets or on the moon Enceladus sublime? Enceladus doesn't have an atmosphere, so what is it that prevents Enceladus' water ice surface from becoming a gas? Did I miss something?

Ice sublimes, it does not boil. It is very slow process at low T, slowing down like exponentially with T.
Nature 2005: Atmosphere found on Enceladus

@Poutnik Thank you. There's still the question why tiny comets don't sublime around their aphelions though. How long would it take for a 1 mile (1.6 km) diameter comet to sublime (disregarding their evaporation in proximity of the Sun)?

See also https://www.engineeringtoolbox.com/ice-thermal-properties-d_576.html   for table and chart of progressive decrease of vapor pressure. Enceladus noon t  (of static surface) is reportedly near -190 °C.

And once some comes off, the partial pressure at the surface is no longer zero.

For a comet at aphelion? Much longer than the age of the universe.

Technically all solids have some equilibrium vapor pressure, but it's usually negligible.


text:
Lithium hydride from lithium fluoride
There is a method of preparation of LiH by reacting a lithium halide (chloride, bromide, or iodide) with sodium hydride in an inert organic diluent such as tetrahydrofuran. Is it possible to replace a lithium halide with a lithium fluoride?

Most of metal fluorides behave differently compared to their corresponding other halides.

One problem is the solubility of lithium fluoride. Alkali and alkaline earth fluorides are much less soluble in common solvents than the heavier halides. Lithium fluoride, for instance, is insoluble in alcohol.
Assuming you can get the reaction to go, another problem with fluoride ion in this type of reaction is it has the same charge and similar size versus hydride ion; so you tend to form solid solutions. Such solid solutions are known with various alkali and alkaline earth metal counterions, including lithium. See Ref. [1] for lithium and Ref. [2] for calcium.
Such solid solutions may not be a bad thing. The range of compositions they allow can be used to balance hydrogen storage capacity with thermal stability in thermoelectric storage applications. Examples of this application involving various combinations of sodium and magnesium fluorohydrides are given in Refs. [3-5].
References

Charles E. Messer and John Mellor (1960). "The system lithium hydride-lithium fluoride". J. Phys. Chem. 64, 4, 503–505. https://doi.org/10.1021/j100833a507.

D. Vergnat-grandjean,P. Vergnat,J.-F. Brice,R. Leveque (1979). "Infrared spectra of calcium hydridecfluorude CaF2-xHx", Physica Status Solidi (b) 96, Issue 2 p. 611-616. https://doi.org/10.1002/pssb.2220960215.

T. D. Humphries,  D. A. Sheppard,  M. R. Rowles,   M. V. Sofianos and  C. E. Buckley (2016). "Fluoride substitution in sodium hydride for thermal energy storage applications". Journal of Matheuals Chemiatry A 4, 12170-12178. https://doi.org/10.1039/C6TA03623F.

Terry D. Humphries, Jack Yang, Richard A. Mole, Mark Paskevicius, Julianne E. Bird, Matthew R. Rowles Mariana S. Tortoza M. Veronica Sofianos Dehong Yu, and Craig E. Buckley (2020). "Fluorine Substitution in Magnesium Hydride as a Tool for Thermodynamic Control". J. Phys. Chem. C 124, 17, 9109–9117. https://doi.org/10.1021/acs.jpcc.9b11211.

Sheppard, Drew & Corgnale, Claudio & Hardy, Bruce & Motyka, Ted & Zidan, Raigay & Paskevicius, Mark & Buckley, C.E.. (2014). "Hydriding characteristics of NaMgH2F with Preliminary Technical and Cost Evaluation of Magnesium-Based Metal Hydride Materials for Concentrating Solar Power Thermal Storage." RSC Advances 4, 26552 - 26562. https://doi.org/10.1039/c4ra01682c.


text:
How to remember what measure of concentration to use for colligative properties
Like the title says, how do I remember what measure of concentration(molality, molarity, mole fraction) to use for various colligative properties such as increase in boiling point when a non-volatile solute is added to a solvent or decrease in the freezing point for the same situation, osmotic pressure, decrease in vapour pressure.
I need to remember these for their use in their formulas and I wondered if someone could give a reason for the use of different measures of concentration in each of the properties, maybe I could end up remembering the logic and thus the measure of concentration.

Molality is used in cryoscopy and ebullioscopy for practical reasons.
In cryoscopy a sample of salt (or any other solute) is weighed and added to a pre-weighed amount of ice at $0°$C. The temperature change is measured, without taking care of the volume variation. It would have been much more difficult to do if you had wanted to obtain a given final volume. As you know, the masses are additives, but the volumes are not additive. Adding $\pu{V_{1}}$ mL of solute into $\pu{V_{2}}$ mL of solvent does not give a final volume equal to $\pu{V_{1} + V_{2}}$. In ebullioscopy, it is the same reasoning.

Use molality (m) for ebullioscopy and cryoscopy,as they involve temperature changes, so it would change de molarity if used (it depends on the volume).
Use molarity (M) (or preferible (c) or []) for osmotic pressure and Henry's law.
And finally, use mole fraction in Raoult's law, but because it's intrinsec to the formula.
https://en.wikipedia.org/wiki/Colligative_properties
https://en.wikipedia.org/wiki/Henry%27s_law


text:
Why do we call salts such as AgCl sparingly soluble?
Though we use the term practically insoluble for salts like $\ce{AgCl}$ in inorganic chemistry, we preferably call them "sparingly" soluble salts in physical chemistry seeing their tiny values of Solubility.
But, my thought is that this classification might have arose, because all the ionic compounds should theoretically be soluble in water (no matter how small it is) because the entropy during the dissolution process of salt increases, due to the formation of less restricted hydrated ions.
Is always the dissolution of ionic salts accompanied by entropy increase (so that the dissolution becomes feasible above particular temperature whether it is exothermic or not) and my thought correct?

If dissolution enthalpy is sufficiently positive (   lattice energy has the upper hand over ion hydration), entropy increase may not help much with decreasing Gibbs energy by dissolution.

All classifications are imperfect. The term sparingly soluble does not have a quantitative connotation.

Why "salt"? Because of lack of better name. All "salts" have both covalent and ionic component in bonding, for AgCl, more so for AgI of Ag2S, covalent component is bigger than ionic - they shouldn't be called ionic compounds.

I haven't heard of  the term  "sparingly soluble" acid, base  or covalent solutions.it seems that this term is specific/ unique for ionic salts.

No sparingly soluble is a universal term.

FTR I have asked: When was the term "Sparingly soluble" first introduced in chemistry? (cc @AChem)

Your question is a bit diffuse, but I'll try to answer.
First, yes, the dissolution of ionic salts is always accompanied by entropy increase, as far as I know, because in chemistry is often complicated to say always.
But, as @Poutnik said, it's also important to check the dissolution enthalpy and the hydrophobic efect.^[1]
On the other hand, is just a convention to say that some salts are insoluble or sparingly soluble, because even if they are all soluble but in tiny amounts, in practice you can't base much on that, the Ksp may be negligible.
[1]https://en.wikipedia.org/wiki/Hydrophobic_effect

The word "salt" comes from a rather old chemical theory, established in the $19$th century, before the "discovery" of the ions. At this time, acids were substances that turn tea yellow and cabbage red. Bases were substances that turn tea brown and cabbage blue. Other indicators were also used.
Acids formula were written $\ce{H_nX}$ and bases $\ce{M(OH)_n}$, with M being a metal, X a non-metal or a group af atoms, and $n$ an integer.
At this time, chemists discovered that each base can react with any acid according to the general equation :
Acid + Base -->  Salt + Water
This definition of "salt" (as a result of the reaction acid + base) was considered as valid even if the acid or the base was not soluble in water. For example, if an acidic solution was not acidic any more after reaction with an insoluble substance, this last substance was accepted in the category "base", even though it does not react with indicators.
If metallic oxides were also reacting with and destroying acids, they were admitted in the category "bases". So any acid can react with any base to produce a salt. A salt is by definition the result of any reaction acid + base, after loss of water.
Experiments show that $\ce{AgCl}$ can be obtained from $\ce{Ag2O}$ as a base after reaction with $\ce{HCl}$ as an acid, even though $\ce{Ag2O}$ and $\ce{AgCl}$ are sparingly soluble. So $\ce{AgCl}$ is a salt. Its poor solubility was not considered as important, in the 19th century.