text:
What is the standard industrial method for measuring caffeine content in food and drinks?
There are a lot of questions here looking for simple methods to measure caffeine content (or extract it), including one of mine here and on the Coffee site:

Home science optical absorption test for approximate caffeine quantity in coffee?
Are there any ways that coffee consumers can measure the caffeine content of what they are drinking by themselves?

But here I'd like to ask something different:
Question: Is there a standard industrial method for measuring caffeine content in food and drinks?
For example, if there were government requirements to post the amount of caffeine in a product there may be some standard way that government advised that it be done (i.e., "this is the way we're going to do it to verify what you report."). While I don't know of any such requirements, that doesn't mean there aren't any.

The answer is fortunately rather simple. Use a C18 reversed phase column and a UV detector in an HPLC system. One would make a calibration curve by injecting various concentrations into the column. Now, if an unknown sample of a drink is injected, and one sees a peak at the same retention time in the sample chromatogram, you know it is caffeine. By interpolating the area, we know its concentration. When an HPLC method is developed, it is ensured that no other component elutes at the same time. This is achieved by using gradients of mobile phase composition.
If there is a legal issue and one has to provide a 100% foolproof test, then one has to use a mass spectrometer along with a UV detector to confirm the identity of the peak at various locations, i.e., one ensures that the peak is pure.
See the examples in Agilent's note "Determination of Caffeine in Coffee Products According to DIN 20481 by Edgar Naegele from where this figure is taken:

I could see series caffeine aficionados laying out $50k for an HPLC. How else do you tune in the espresso machine for that perfect cup?

Determination of Caffeine in Coffee Products According to DIN 20481 (also here) compare figures 4 & 5. Nice!

"DIN"="Deutsches Institut für Normung" - the German ISO member. DIN norms are pretty much the requested "standard industrial method".

@JonCuster or maybe a group of them can start a local neighborhood HPLC co-op?

@uhoh, Mini portable and complete HPLCs of the size of a brief case are available now. I see no problem in running caffeine stds and samples.

@AChem that's great! I'm afraid that the compactness in volume won't translate to compactness in cost - buying one may require a brief case full of Starbucks gift cards or equivalent. So the cost sharing motivation for the co-op idea can be translated to a traveling "coffee inspector" who makes their rounds on a weekly basis on a motorcycle using an Uber Eats or Food Panda on-demand model. I like it! I'm gonna be rich!

@uhoh We used to keep a 5 pound bottle of caffeine citrate next to the always on lab break room coffee pot. The sign on it, due to me, was “Two scoops per pot, to combat the Santa menace.” So you could just start with decaf and add caf, to desired level of wiredness. (We took the big bottle of caffeine citrate away after a few weeks, because someone might have not understood and actually added the two scoops!)

@EdV Not a chemist can you explain lol

@AzorAhai-him- My comment had a typo: “Santa” should be “Sanka”, a brand of decaf coffee. In a large research group, coffee “powers” progress and decaf would be pretty much useless if you want to facilitate productivity. Our group, run by a famous chemist (not me!), typically had around 25 grad students, around 8 post-docs/sabbatical leave folks, and we had undergrads doing short research projects, though we never made a list of them because they came and went semester by semester. So the caffeine citrate was just kind of a joke, with a little point: get stuff done!

@EdV Why can't you add the two scoops?

@AzorAhai-him- Two scoops would be way too much and could have very bad medical consequences!

@EdV Ah, lol. .

for non-chemists like myself this video and lab writeup titled "Lab 7: Caffeine Quantification by HPLC" are quite helpful.


text:
What is the difference between alloying and doping?
When compared to doping, how is alloying different? What modifications to the structure are brought about by both of these?

See also https://www.researchgate.net/post/What_is_the_difference_between_alloying_and_doping_which_process_alloying_or_doping_changes_lattice_parameter

Always try to find answers on your own in offline and online resources, before asking.  It will give you more than just waiting to receive the answer. Your question will not then look like if you have not done so, what is frown upon on StackExchange site network.

Simple Google search will lead you to the research gate article Poutnik mentioned.

From an ASM International publication Alloying - Understanding the Basics one finds

Alloying may be defined as “the process of adding one or more elements
or compounds to interact with a base metal in order to obtain beneficial
changes in its mechanical, physical, or chemical properties or
manufacturing/processing characteristics.”

The section goes on:

However, elements or compounds that lead to a preferential
microstructure and subsequent improved properties are covered.
Examples of these are grain refiners (grain refining results in better forming
or higher strength), inoculants added to molten cast irons to produce
changes in graphite distribution and improvements in mechanical properties,
magnesium-containing nodulizing (or spheroidizing) additions in
ductile irons for high strength and ductility (up to 18% elongation), and
the addition of certain elements, such as calcium, sodium, strontium, and
antimony, to refine the structure of aluminum-silicon casting alloys as
well as improve their tensile properties and ductility. Also included are
discussions of some powder metallurgy (P/M) materials that technically
may fall outside the definition of alloying given above. An example is copper-
base dispersion strengthened materials. Copper can be strengthened
by using fine dispersed particles of aluminum oxide. Because this oxide is
not immiscible in liquid copper (i.e., it does not “interact”), dispersion strengthened
copper cannot be made by conventional ingot metallurgy and
alloying techniques; P/M techniques must be used. Dispersion-strengthened
superalloys made by “mechanical alloying” are also described.

This is a very broad definition, does not restrict alloying to altering (or not) the crystal structure or microstructure. Note that even dispersion hardening by separate phase particles is considered 'alloying'.
Doping is not so clearly defined by ASM. It is much more common for using trace amounts of an element to alter the electrical (think semiconductors) or optical (laser gain media) properties (at least in my experience).

Doping is adding a tiny proportion of an impurity in a nearly pure metallic sample without changing the crystalline structure of the metallic element.
Alloying does modify this structure.

Hm, so if atoms of 2 metals (neighbour lanthanoids?) can replace each other in the structure in any ratio, is it doping or alloying ?

Indeed, I think this is more restrictive than the broad range of usage seen out there. I’d more say alloying is solely a composition change, doping is an electrical or optical change. But still not right.

@Poutnik. I had not thought of the case of alloys of neighbouring lanthanoids. I should modify my definition to take car of this problem.

I think the difference is mainly in quantity and context.

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text:
Principle of Catalysts
How exactly do Catalysts work? In the cracking of Alkanes, Aluminum Oxide is used as a catalyst. However how does Aluminum Oxide decrease the conditions required to undergo cracking. At the molecular level, what is actually happening? How can a catalyst increase the rate of a reaction without being used itself?

Who says catalysts are not used? They are, but they are being regenerated. E.g  A + B -> P may be catalyzed, e.g. as A + C -> AC,   AC + B -> P + C.  Many redox reactions are catalyzed by ions of transition metals with variable oxidation state, like Fe or Mn.. The point of catalysis is that the reaction path involving a catalyst have lower activation energy(ies). Like going on a bicycle either across high hill versus 2 or more small ones.

Here is how a metallic surface catalyzes the reaction of $\ce{H2}$ + $\ce{O2}$. First $\ce{H2}$ reacts with the metal $\ce{M}$, making weak $\ce{M-H}$ bonds at its surface. As a consequence, the covalent bond between the two $\ce{H}$ atoms is extended, and weakened. Seen from outside, it looks as if the surface of the metal $\ce{M}$ is covered by nearly free $\ce{H}$ atoms. If $\ce{O2}$ molecules are touching one of these $\ce{H}$ atoms, it can make a O-H hydrogen bond in  O-O-H-M. If this hydrogen bond is weaker than the $\ce{H-M}$, the reactive molecule $\ce{HO2}$ will quit the metallic surface and reacts later with other atoms, molecules or radical to produce $\ce{H2O}$ at the end.
The catalytic effect of $\ce{Al2O3}$ on alkanes may be similar to this mechanism.

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text:
Do we know how Tylenol works?
I saw this xkcd comic and one of the unsolved entries was "How does Tylenol work?".
So I googled and found a lot of explanations for what acetaminophen does. A bunch of articles says it regulates some brain thing to reduce swelling or something. Seems like we know how it works.
What did Randall mean by this? Do we know how Tylenol works? Seemed like we do.

Sure, it does something to something to produce an effect. But just how does it regulate ’some brain thing’ in detail? Perhaps better on Bio SE.

It must be Tylenol (paraacetamol - pharmacology/acetoaminophen)

He meant what he said. You could find numerous papers about that, but if it was really known only one would suffice. Non-steroid painkillers may be more common, but are stumping scientists more than general anesthesia.

I’m voting to close this question because you'll have to ask Randall.

Just because we know one thing it does doesn't mean we know all the things it does.
And just because one of the things it does is inhibit cyclooxygenase, and another thing it does is relieve pain, doesn't mean that inhibiting cyclooxygenase always relieves pain.
And just because acetominophen does inhibit cyclooxygenase and relieve pain in most people doesn't mean it does that in all people.
And since the placebo effect is a thing, and most people don't take acetominophen while "blinded" as part of a study, and since the placebo effect is particularly strong at pain relief, a lot of the effects of acetominophen itself will often be confounded with placebo effects.
So, we know some ways that acetominophen works.  But we're pretty far away from a complete theory of how acetominophen works, because (a) biology is messy, (b) humans even moreso, and (c) we don't have a complete all encompassing theory of general pharmacology to unambiguously map particular molecules to their effects on mammalian bodies anyway.


text:
How does Hess' Law work exactly?
So I am talking about complicated compounds and not simple reactions where you subtract heats of formation. I was trying to find the heat of reaction for an aryl substituted methyl aryl sulfide being oxidised to sulfone group. Now, I calculated BDE for S=O bond and subtracted O-O energy to find heat of reaction for oxidation to sulfoxide and then repeated the step for going to sulfone. To calculate that heat experimentally, I insulated my reaction vessel and bought it to reflux temperature of the solvent. I then added hydrogen peroxide and collected the evaporated solvent and measured its volume. I used its heat of vaporization as a proxy to calculate the heat of reaction and it came out to only 20% of the Heat of formation calculated from Hess's law
I would love to know where I went wrong and if my knowledge of Hess's law is wrong.

Try to summarize your procedure mathematically. Verbose description is hard to follow.

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I assume BDE = Bond dissociation Energy.

It might help if you explain how you did this in more detail: " I insulated my reaction vessel and bought it to reflux temperature of the solvent". How did you heat the vessel to the reflux point exactly? And then you added a reagent that caused a rapid reaction (adiabatic conditions) releasing heat into the vessel and causing boiling. You trapped the boiled solvent and from its volume determined the total heat released. Looks like a lot of possible heat losses may have occurred along the way?

The Hess' law is the particular application of the fundamental law of energy conservation, applied to the total enthalpy.
It says the enthalpy change depends only on initial and final state, but not on the path/way how the final state was reached.
Particularly if we have a complicated process $$\ce{A -> D},$$
which can be in reality or formally divided to steps in 2 alternative ways, like:
$$\ce{A -> B -> C -> D},$$
$$\ce{A -> E -> D},$$
then
$$\Delta_{\ce{A->D}} H = \Delta_{\ce{A->B}} H + \Delta_{\ce{B->C}} H + \Delta_{\ce{C->D}} H\\
= \Delta_{\ce{A->E}} H + \Delta_{\ce{E->D}} H$$
If it was not so, you could generate energy from nothing, by cyclical switching between 2 states along different ways, 1 way releasing more energy than the other would consume.
E.g.: $$\ce{A -> B -> C -> D -> E -> A}$$

If we consider your specific case
$$\ce{R_1-S-R_2(l?) + 2 H2O2(aq) -> R_1-(S=O)-R_2(l?) + 2 H2O(l)}$$
divided to formal reaction steps as:
\begin{align}
\ce{2 H2O2(aq) &-> 4 ^{.}OH(aq)}\\
\ce{2 ^{.}OH(aq) &-> 2 ^{.}O^{.}(aq) + 2 ^{.}H(aq)}\\
\ce{2 ^{.}OH(aq) + 2 ^{.}H(aq) &-> 2 H2O(l)}\\
\ce{R_1-S-R_2(l?) + ^{.}O^{.}(aq) &-> R_1-(S=O)-R_2(l?)}\\
\ce{R_1-(S=O)-R_2(l?) + ^{.}O^{.}(aq) &-> R_1-(O=S=O)-R_2(l?)}
\end{align}
then the reaction enthalpy would be — by the Hess law — the sum of reaction enthalpies of particular steps. It does not matter if the steps are formal or real, as the path does not matter.
$$\Delta_\mathrm{r} H = \sum_i{\Delta_{\mathrm{r},i} H}$$

I think you make sense and lemme just calculate the deltaH with your method, confirm on monday and I'll mark my question solved.


text:
How fast does potassium hydroxide turn into potassium carbonate, in the case of leaked AA or AAA batteries?
When batteries start to leak, the highly toxic potassium hydroxide comes out, but it reacts with the carbon dioxide in the air and turn into the less toxic potassium carbonate salt. I'm wondering how fast this reaction takes place. Is it on the order of seconds, minutes, days, or weeks? Because I don't know how extra careful I should be in handling leaked batteries in toys.

KOH is not highly toxic, it is just caustic. If one get contaminated, washing with enough water, optionally with some vinegar, citric acid or citrone juice is enough. Potassium carbonate is not toxic, just mildly caustic, like sodium carbonate ( washing soda). Potassium carbonate is widely used in food industry to process most of cocoa from which chocolate is made. (turning light reddish-brown cocoa powder to dark brown one).

Despite not being "toxic" how @Poutnik pointed out, this mainly depends on how fast the carbon dioxide from the air goes into the solution that leaked off of the battery.
The actual reaction within the solution is quite fast (<milliseconds) but if you ask how long it takes to transform all the KOH to carbonate, it could be minutes to weeks in my opinion, depending on size of the leakage, amount of leaked substance, layer thickness of the leaked material, temperature etc.


text:
Removing Copper from drinking water by Electrolysis? What to keep in mind?
The drinking water in my flat contains a lot of copper. Although the amount of copper is around the allowed limit 2mg/l I would prefer to reduce it.
I know that it is at this value from a paid analysis in a dedicated laboratory.
Edit: The copper was introduced into the water when passing through long copper pipes inside the house. But because it is not my house, I cannot do anything about this aspect.
In my childhood I played around a lot with Electrolysis in general, producing funny gas, plating things with copper etc. So why not remove the copper from drinking water this way? (I know filters with activated carbon can help, but please lets think about Electrolysis here).
Lets assume a glass pot with 1l of drinking water containing 2mg of copper.
The task is to reduce copper content by an electrolytic procedure without increasing the content of other adversary substances thus without decreasing the "quality" of the treated water.
Lets take these data as a working example:

If it is possible at all, what electrode materials and arrangements, voltage, current, duration might bring me towards the defined target?

Hydrolysis isn't electrolysis... And it would be much easier to reduce it, for ex. with vitamin C.

Thanks, in the title I used the wrong word, Electrolysis was meant.

@Mithoron: What is the outcome after adding vitamin C? In what form would the copper exist afterwards? Nanoparticles? How to get it out of the reservoir?

Well, yeah, maybe this wasn't the best idea, but neither is electrolysis. Simple boiling removes quite a bit of stuff from water. If purification is really needed, it's rather reversed osmosis in use.

Is it not possible to have the water pass through steel wool. Copper ions would react with metallic iron according to : $\ce{Cu^{2+} + Fe ->  Cu  +  Fe^{2+}}$. This would replace copper ions by iron ions, which are not toxic at such concentrations.

Hm, ok, could be enforced with a few volts, a graphite-anode and the iron as cathode?
Hoped to have some monitoring (current measurement?) to see when finished....

If the given data apply, copper is mostly present as species $\ce{Cu(OH)2}$; it's probably actually saturated.


text:
Reaction of sulphuric acid with ammonium hydrogen phosphate
How will the reaction of $\ce{H2SO4}$ with ($\ce{NH4)2HPO4}$ [diammonium phosphate, commonly used DAP fertiliser in agriculture] proceed? Will phosphoric acid be produced, or a complex phosphate-sulphate salt would be the result?

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Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.

Depending on the molar ratio,
$$\ce{HPO4^2- ->[H+]H2PO4- ->[H+]H3PO4}.$$
Also remember there are no $\ce{H2SO4}$ nor $\ce{(NH4)2HPO4}$ in water solution, but $\ce{,}$ $\ce{HSO4-(aq)}$, $\ce{SO4^2-(aq)}$ , $\ce{NH4+(aq)}$ , $\ce{HPO4^2-(aq)}$,  $\ce{H2PO4^-(aq)}$ and $\ce{H3PO4(aq)}$, depending on the ratio and $\mathrm{pH}$.
For direct reaction without water, then
$$\ce{2 H2SO4 + (NH4)2HPO4 -> 2 NH4HSO4 + H3PO4},$$
with some $\ce{(NH4)2SO4}$ too.

Thank you, that’s very helpful :). From your explanation, I conclude that if the objective is to get P04 3- (soluble form of phosphate ion in acidic PH for the crop) from DAP, then a higher concentration of H2S04 should be used than an equal moles of both, and the reaction should be carried out in water.

PO4^3- is present only at very alkalic solutions. (pK_a3= 12.37)


text:
What is the reason you could use synthetic detergent in hard water？
I learned that because synthetic detergent doesn't make strong dative bond with $\ce{Ca^2+}$ or $\ce{Mg^2+}$, you could use it in hard water. But I couldn't understand why soap makes strong dative bond and synthetic detergent doesn't. For example, why $\ce{-SO3-}$ doesn't make strong enough dative bond to make scum with $\ce{Ca^2+}$?

It's not that synthetic detergents do not make any bonds with the calcium and magnesium ions present in hard water. It does make, however the salt that is formed with this bond is water soluble (note that the synthetic detergents are composed of long straight chain hydrocarbons which is hydrophobic and the ionic sulfonate group which is hydrophilic - the salt formed from hard water is thus water soluble) which gets drained away unlike salts that forms when soap is used which is insoluble and forms scum like precipitate. Hence you will see detergents making more lather than soaps.
See more details here: Why sulfonate of detergent is less likely to bind to ions present in hard water than carboxylate?

Thank you for editing and answering my question！　Could you explain why the salt is water soluble？　I want to explain it to my student.

I have edited the answer @satorukurita

Thank you for taking time to answer my question. It seems the nature of sulphonate group is the key.


text:
Mechanism of decomposition of N2O5
In my book mechanism of decomposition of $\ce{N2O5}$ was given like
\begin{align}
\mathrm{Reaction:}\\
\\
\ce{2 N2O5 &-> 4 NO2 + O2}\\
\\
\mathrm{Mechanism:}\\
\\
\ce{N2O5 &<=>[k_1][k_2]NO2 + NO3}\\
\ce{NO3 + NO2 &->[k_4]NO + NO2 + O2}\\
\ce{NO3 + NO &->[k_3]2 NO2}
\end{align}
But on stack exchange somebody gave mechanism of decomposition  like this at Steady State Approximation.
Pls tell which one is correct and why it is correct?

Which one do you think is correct?

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Now I  see I have copied the img error too, as there are 2 different $k_2$. Turning one of $k_2$ to $k_4$.

The sum of all the elementary steps in the mechanism must result in the overall balanced equation:
$$\ce{2N2O5(g)->4NO2(g) +O2(g)}\tag{R}$$
If we add up each one in your book's proposed mechanism:
$$
\require{cancel}
\begin{align}
\ce{\ce{N_2O5(g)} &<=> \cancel{\ce{NO2(g)}} +\cancel{\ce{NO3(g)}}} \tag{R1} \\
\ce{\cancel{\ce{NO3(g)}} +\cancel{\ce{NO2(g)}} &-> \cancel{\ce{NO(g)}} + NO2(g) +O2(g)} \tag{R2} \\
\ce{NO3(g) + \cancel{\ce{NO(g)}} &->2NO2(g)} \tag{R3} \\
\hline
\ce{N2O5(g) +NO3(g) &-> 3NO2(g) + O2(g)} \tag{R'}
\end{align}
$$
We can see that R' is not the same as R, so this mechanism is incorrect.
Conversely, doing the same with the other proposed mechanism:
$$
\require{cancel}
\begin{align}
\ce{\ce{N_2O5(g)} &<=> \cancel{\ce{NO2(g)}} +\cancel{\ce{NO3(g)}}} \tag{R1} \\
\ce{\cancel{\ce{NO3(g)}} +\cancel{\ce{NO2(g)}} &-> \cancel{\ce{NO(g)}} + NO2(g) +O2(g)} \tag{R2} \\
\ce{N2O5(g) +\cancel{\ce{NO(g)}} &->3NO2(g)} \tag{R3} \\
\hline
\ce{2N2O5(g)&-> 4NO2(g) + O2(g)} \tag{R''}
\end{align}
$$
Where we can see that R'' is the same as R.
We cannot conclude solely based on this that the second mechanism is necessarily correct, but we can conclude that the first mechanism is necessarily incorrect.

But on internet it is given the given same mechanism as my book mentioned https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.12%3A_Steady-State_Approximation see this.

@ArijitSingh $\ce{NO3}$ is an intermediate in this case, so it does not make sense for it to stay (not be readily consumed) and appear as a reactant for an overall reaction that does not include it in the first place.

This reaction has been studied and the mechanism has been given as exactly as one given in OP's text book.

@MathewMahindaratne I do not have full access to the paper you cited, but from the first page at the bottom, reaction (g) is not included in OP's mechanism.

For the book's mechanism, just multiply the first step (R1) by two to get the correct stoichiometry. The question is whether $\ce{N2O5}$ reacts directly with NO, or only after it dissociates. Otherwise, the two mechanisms are identical.


text:
Does the polymerization process include cooling?
The process of seasoning a cast iron pan includes covering the cast iron with a small amount of fat or oil and heating the cast iron to the oil's smoke point.  The oil recombines into polymers and bonds to the cast iron, createing a non-stick surface and protecting the cast iron from rust.
The instructions on my cast iron pans don't mention multiple layers. Other material suggest letting the pan cool between layers, however the instructions for my cast iron crepe makers state to apply ten coats of oil without cooling, but simply waiting for the oil to stop smoking before applyin the next coat.
I'm not a chemist, but I do like to understand what I'm doing and why. This question seems to be beyond the pale of internet research.
My questions are:
Does the cast iron need to cool to complete the polymerization and bonding process?
If I apply a layer of oil and wait for the oil to stop smoking, then apply another layer, have a skipped a step in the  polymerizatio or bonding process (cooling step)?
Will the bond of the seasoned layers be as strong as when the cast iron is allowed to cool between coats?
Thanks to all you who sacrifed time and brain cells to understand chemistry, and thanks to those chemists who are willing to help me understand this.

I guess that if you cooled, the layers would be more discrete, but is this better, or worse...

If you want to cure a cast iron pan simply add some cooking oil and saute some onions gently and make a fried onion sandwich. Then wash the pan very gently between uses and repeat if metal shows up. Always dry the inside completely after washing. several cookings will give a good surface. You really do not want a surface of burnt oil. This is my experience as a cook and a chemist


text:
How do concentrations of different solutes across a cell membrane affect whether water leaves or enters the cell?
I know from class that water travels to where the solute concentration is higher. I have no problem visualizing it if it's just "salt." What if there are multiple types of solutes? For example, if solute A is more concentrated in the cell and solute B is more concentrated outside, which way would the water move? Is there something that takes into account multiple solute concentrations?

The effects of multiple solutes are often considered additive, although things may get complicated at higher concentrations. In general one considers the effect of these solutes on the "activity" of water on both sides of the membrane. The difference in water activities results in an osmotic potential gradient.
There are two concepts that are useful to describe what you are interested in. One is osmolarity (or osmolality), which is the effective total concentration of solutes potentially resulting in osmotic pressure buildup across a membrane. The membrane may in this case be permeable to some of these substances however. A second concept is tonicity, which is essentially the same but considers only substances which are not freely transported across the relevant membrane. For an introduction to these topics you can read Ref. 1.
Related concepts are the total osmotic pressure and the colloid osmotic pressure$^2$. The first represents the osmotic pressure expected to result from all solutes (salts and colloids), both assumed unable to freely diffuse through the membrane. In the second case only the colloid is assumed unable to traverse the membrane.
Reference
(1) Vujovic, P.; Chirillo, M.; Silverthorn, D. U. Learning (by) Osmosis: An Approach to Teaching Osmolarity and Tonicity. Advances in Physiology Education 2018, 42 (4), 626–635. https://doi.org/10.1152/advan.00094.2018.
(2) Silbey, R. J.; Alberty, R. A.; Bawendi, M. G. Physical Chemistry, 4th ed.; Wiley: Hoboken, NJ, 2005.


text:
Why is 5-chloro-2-hexene numbered as such when it could be named 2-chloro-4-hexene and have smaller Locants?
We're learning IUPAC nomenclature of organic compounds and I came across this problem:
And it's IUPAC name is:

My question: Why did the numbering start from the left end when starting from the right end would give smaller Locants for the double bond(4) and Cl(2).
Why isn't it called 2-Chloro-4-hexene ?

An alkene-substituted halide is not the same as a halide-substituted alkene. Double-bonds take priority over halogens.

@Sam202 so we assign the smallest possible number for the double bond regardless of the position of the sidechain??

You first identify there are 2 competing functional groups: halide and alkene. Then you determine which has the higher priority (alkene in this case), and number the chain accordingly to give the highest priority group the lowest number possible.

Please note that, for almost 30 years, the IUPAC name is 5-chlorohex-2-ene and no longer 5-chloro-2-hexene.

As  Sam202 pointed out, double bond has higher priority than halogens. For example, check the IUPAC name for allyl chloride. It is 3-chloroprop-1-ene. Its never be 1-chloro-prop-2-ene.

The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound that is given in the question doesn’t have any principal characteristic group that is expressed as a suffix. Thus, Rule (c) is not relevant in this case.
Next, a low locant is assigned to the double bond according to Rule (e). Therefore, the name of the structure without further substituents is hex-2-ene rather than hex-4-ene since the locant 2 for the double bond is lower than 4.
Finally, low locants are given to any substituents that are expressed as prefixes (e.g. ‘chloro’ or ‘methyl’) according to Rule (f). Note that all prefixes are considered together; i.e., there isn’t any order of functional group priorities at this point. In this case, however, there is only one substituent that is expressed as a prefix.
Thus, the correct name is 5-chlorohex-2-ene (ignoring E and Z isomers).

Is there a particular reason why you chose to capitalize rule?

@Martin-マーチン https://ell.stackexchange.com/q/64705 ; see also "Capitalization" in the ACS Style Guide.

… it's also done like this in the Blue Book itself (with a few exceptions).


text:
How to calculate thermodynamic parameters from quadratic form of the Van't Hoff Equation?
I have drug binding data giving me Ka values for a set of associated temperatures. This allows me to plot ln(Ka) vs. 1/T, which I can fit to a second degree polynomial i.e. y = ax^2 + bx + c. From this, I would like to calculate thermodynamic parameters deltaS and deltaH.
To do this, my understanding is I would have to use a nonlinear form of the Van't Hoff equation, e.g. equation 3 from this paper:
https://www.mdpi.com/1424-8247/13/6/134/htm
Now, I have my values for a, b, and c. But I am having trouble figuring out how to use this information to get deltaS and deltaH values. In the supporting information for the above paper (https://www.mdpi.com/1424-8247/13/6/134/s1), they appear to do this, but when I plug in their values for a, b, and c, I do not get their reported values for dH and dS.
Sorry for the formatting and if what I am asking is unclear.

The quadratic fit does not come directly from eq 3. Rather it results from a Taylor expansion whose derivation is described here: https://analyticalsciencejournals.onlinelibrary.wiley.com/doi/10.1002/jssc.201100029

Described even more clearly here: https://pubs.acs.org/doi/10.1021/acsomega.9b02689


text:
Why does temperature remain constant during phase changes?
In a phase change the temperature remains constant but I can't understand why. The answers I have read state that during phase changes energy is only used to break bonds between molecules rather than increasing kinetic energy. However why is this the case? Why couldn't the supplied energy be used solely to increase the kinetic energy of particles?

If you need to overcome potential energy barrier to reach a different state, it is obvious that all energy cannot be used to increase kinetic energy when that state is reached. Like at ionization of atoms or at escaping gravity of a star or a planet.

"Temperature" is essentially movement of molecules.  So you are on the right track thinking kinetic energy.  Phase change is an energy sink moving from solid, to liquid to gas, so more energy input is required.  In other words, temperature cannot rise because the molecules cannot move faster (increase KE) until phase change is completed.

There are other phase changes like iron going from ferrite ( body centered cubic) to austenite ( face centered cubic). Thermocouples in a commercial heat-treat furnace would show a temperature drop while heating a load of steel as the phase change absorbed heat. Impressive to witness.

It's easier to see in a specific example, e.g., boiling water. At 100°C and 1 atmosphere pressure, as more heat energy is added, the pressure of the water at the spot where heat is applied exceeds the pressure pushing on it, and bubbles form and rapidly expand, pulling away that added energy. The water vapor above the boiling water can be hotter than 100°C,  but the liquid cannot... except for superheating.
Your presumption is not wrong: fairly clean water, without air bubbles, can be superheated well above 100°C, if left undisturbed -- a metastable state. The slightest disturbance, such as vibration, can cause the water to suddenly boil violently, "exploding" from the container. The remaining water is now below 100°C.
A similar phenomena is supercooling of water, in which heat energy has been transferred out of the system, yet the water does not freeze until it's disturbed (or is sufficiently cold, about −48 °C, below the freezing point of mercury). If you want to repeat the supercooling demonstration (much safer than superheating!), use a clean, smooth, plastic container  and water that is clean, and has air bubbles removed by first boiling and allowing it to cool in a container with a lid (to keep out dust) before pouring it into the bottle.
However, those answers were also correct... once equilibrium has been reached. Eventually, superheated water will flash to steam. Supercooled water will eventually freeze. And even glass can crystallize. Don't let this dichotomy faze you, though.

This is one of the over simplifications that no one thinks about after being told. For a one component system the phase rule, F = C-P+2, states for two phases in equilibrium there is one degree of freedom. When evaporating a liquid, maintaining a constant P[vapor pressure] fixes the temperature of evaporation. A variation in P causes a variation in Temperature. The effects in melting are the same. To maintain a constant steady state Pressure an external energy source is required. If the energy source is the phase itself [or the heat source is insufficient] the temperature will change until equilibrium is reached; then the constant pressure means an invariant temperature.
In normal boiling, exposed to the atmosphere when the vapor pressure reaches atmospheric pressure P cannot increase since increasing evaporation simply pushes back the atmosphere. The Vapor pressure becomes constant and the temperature becomes constant. This is assuming sufficient energy input and not excessive heating and reasonable mixing and good exposure to the atmosphere.

So you can cause a phase change through compression or expansion, during which the system will stay on the coexistence line. But what do you mean by this: "then the constant pressure means an invariant temperature" ?

Just that! when the vapor pressure equals the constant atmospheric pressure the last freedom is removed and the temperature is constant.

If the vapor pressure is changed adiabatically the vapor will condense or liquid evaporate absorbing or releasing the heat of vaporization and the equilibrium point will change temperature. If the heat is supplied or removed externally the amount of the phases will change and P will remain constant.


text:
How to determine the cell potential of chromium(III) — permanganate cell?

Determine the cell potential of the following cell:
$$\ce{Pt | Cr^3+ (\pu{0.125 M}), Cr2O7^2- (\pu{0.200 M}), H+ (\pu{0.600 M}) || MnO4- (\pu{0.150 M}), Mn^2+ (\pu{0.400 M}), H+ (\pu{0.250 M}) | Pt}$$

As far as I can tell, I've balanced the reactions properly:
$$
\begin{align}
\ce{MnO4- + 8 H+ 5 e- &-> Mn^2+ + 4 H2O} &\quad E^\circ &= \pu{+1.507 V} \tag{R1} \\
\ce{Cr2O7^2- + 14 H+ + 6 e- &-> 2 Cr^3+ + 7 H2O} &\quad E^\circ &= \pu{+1.360 V} \tag{R2}
\end{align}
$$
I don't know how to proceed. I know that when I have my voltage, I can use the Nernst equation to use the concentrations here, but the issue is that I'm not sure what comes before that.
Do I balance them by multiplying one by $6$ and the other one by $5$? which one is being reduced and which oxidized, and which one needs to be flipped around as consequence?

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You have chosen a nightmare equation. Why not find a simple reaction that you know the chemistry. Workout the standard cell and make some concentration changes that you can predict the results. Then, after you are familiar with the procedure tackle this one.

You have asked two questions. Regarding "which one is being reduced" and "which one is being oxidized", you can clarify that by using the Nernst equation
$$\Delta E=\Delta E^\circ-\dfrac{RT}{nF}\ln Q$$
In your example, you know that the reduction of $\ce{MnO4-}$ to $\ce{Mn^2+}$ (oxidation state of +7 to +2) and the oxidation of $\ce{Cr^3+}$ to $\ce{Cr2O7^2-}$ (oxidation state of +3 to +6) will take place at standard state, because $\Delta E^\circ=0.147\ \mathrm V$, and therefore $\Delta G_\mathrm r^\circ=-nF\Delta E^\circ\lt0$. Therefore, this is the redox couple that will spontaneously function as a galvanic cell at standard state.
In general, this will also be true for most exercises, since the logarithm part will not deviate much of $\Delta E^\circ$. But, in the general case you can't know in advance, so just do the following:

Arbitrarily choose the reduction and oxidation reactions.
Use the Nernst equation and calculate $\Delta E$.
If $\Delta E\gt0$, then you have guessed them right, and those electrochemical reactions will spontaneously take place in a galvanic cell. If $\Delta E\lt0$, then the answer is still correct, but you have to clarify that those reactions will not spontaneously take place. You would be right if you state that this is the external voltage that you, as surroundings, have to apply to the terminals of your cell in order for the reactions in that direction to take place.

As you said at the end, you multiply those electrochemical equations by 5 and 6 to eliminate the electrons, so you know the powers of your reactants to correctly calculate $Q$.

You can figure out the reaction from the cell notation. The general cell notation is $\text{Anode || Cathode}$, and by convention oxidation occurs at anode. Hence, here $\ce{Cr^{3+}}$ is undergoing oxidation and $\ce{MnO4-}$ is getting reduced.
Once you have determined the reaction, remember that $E^{\circ}_{\text{oxd}}=-E^{\circ}$ and voltage of the cell is calculated as $E^{\circ}_{\text{cell}}=E^{\circ}_{\text{oxd}}+E^{\circ}_{\text{redn}}$.


text:
Why are chromium (III) ions stable in aqueous medium even though they do not fulfil the 18-electron rule?
With regards to this question, I have been unable to find a convincing answer on the Internet as it seems that not many people consider exceptions to the 18-electron rule as the 18-electron rule is commonly viewed as a shorthand for determination of structural stability instead of an actual tool.
I have therefore asked my chemistry teachers, but they have given me varying responses which may even be contradictory.
The basis of my question is this: the 18-electron rule has widely been used as the rule-of-thumb to determine stability of transition metal complexes. Hexaaquachromium (III) only has 15 electrons (2x6 from all aqua ligands and 3 from the 3d3 configuration of Cr3+ ion), so from this line of argument, shouldn't Cr3+ be a decently strong oxidizing agent in aqueous medium as it attempts to fill up its t2g nonbonding orbitals with d-electrons?
One of my teachers have attempted to explain the stability of Cr3+ complexes by referencing the Frost diagram and arguing that Cr3+ ions, at least in the acidic medium, is the most redox-stable out of all possible oxidation states due to its highest free energy of formation from Cr metal, such that it is unlikely for the Cr3+ ion to change its oxidation state in order to fulfil the 18-electron rule. He also further explained that another way for Cr3+ to fill up its t2g bonding orbitals is for more ligands to bind to it, which distorts the octahedral shape and thus destabilizing the complex instead.
However, I do not find the above explanation convincing as it does not explain why Cr3+ ions in water do not spontaneously form complexes with Cr=O bonds without a change in oxidation state in order to fulfil the 18-electron rule. Another of my chemistry teacher has even mentioned that Cr3+ octahedral complex should exhibit redox instability due to its inability to fulfil the 18-electron rule unless bound to a radical ligand, which seemingly contradicts the empirical observations.
My friend has proposed the suggestion that, given the t2g nonbonding orbitals of the ML6 complex contains 3 electrons, it cannot accommodate any more ligand group orbitals as donation of ligand group orbitals will come in pairs of electrons with opposite spin which means that
a) For each ligand, one of the electrons donated must change its spin in order to fill up the t2g nonbonding orbitals without violating Pauli's exclusion principle
b) For the case of a high-spin complex, to fill up the eg antibonding orbitals but a change in spin is also required without violating Hund's rule of spin multiplicity
Therefore, it is pretty much impossible for Cr3+ aqueous ions to fulfil the 18-electron rule
However, I still do not find the ligand field approach convincing as it seems to be conceptually false: it does not make sense that electrons from ligands fill up the t2g nonbonding orbitals of the chromium center as they are the 3d orbitals which are supposed to be left untouched.
On this note, may I ask if any of the aforementioned explanations are correct in explaining the relative stability hexaaquachromium (III) despite its violation of the 18-electron rule? If not, may I ask if there is an accurate explanation, or any papers/reviews published discussing this class of exceptions?

+1. I've wondered this as well and I'm pretty sure it can only be explained quantitatively; i.e. with explicit bond strengths, potentials etc.


text:
Which is faster? Electron close to the atom's nucleus or the one far from it and why?
As we know when the distance between the atom's nucleus and the electron increase the energy of it increase but what about the speed, as physics say that the smaller the radius is the faster the body is, so does this apply to electrons or does the energy play a role in electron's speed?

What had you found before you asked,  while searching for the answer? What the law of energy conservation say ?

When you deal with electrons bound in potentials with a particular mathematical form (inverse distance potentials, here for a Coulombic potential having the form $r^{-1}$) you can invoke the virial theorem,
$$2T=−U$$
as explained for instance here.
Accordingly electrons more weakly bound to an atom (and therefore more shielded from the nuclear attraction, and further away on average) will have a higher potential energy (closer to zero, less negative) and the translational energy will therefore by smaller, too ($T>0$, $U<0$ for bound states). At the edge of the bound state, as the principal quantum number heads to infinity, the potential energy goes to zero, and so does the kinetic energy. It makes sense, since the electron would not remain bound if it had enough kinetic energy to overcome the attraction.

see also: https://chemistry.stackexchange.com/questions/119467/potential-energy-of-electron-in-excited-hydrogen-atom/119468#119468

and https://chemistry.stackexchange.com/questions/80922/what-kind-of-energies-are-present-in-an-electron/80938#80938


text:
What is the average collision duration?
So I've been trying to figure this out: What is the average collision duration of a gas in terms of macroscopic variables?
What I have figured out. Whenever two molecules collide they must reach a potential barrier (before behaving like a turning point) of
$$V_{barrier} = \frac{1}{2} \mu v_{rel}^2 \tag1$$
where, $V_{barrier}$ is the height of the potential before behaving as a turning point, $\mu$ is the reduced mass and $v_{rel}$ is the relative velocity. Thus, on average the gas molecule's collisions goes upto a potential height
$$ \langle V_{barrier} \rangle \propto T $$
Any help will be appreciated.

There are basically 2 possible approaches:
Guessing approach, based on typical molecule speed and the supposed size of molecule deformation, that could provide the order of collision lasting interval.
More precise quantitative evaluation would need the potential energy curve like E_pot=f(r), what may be available for special cases, but not generally.

There is no clear cut radius of atom and therefore no duration.

Before you start calculating, you need to define what exactly a collision is, when it starts and ends.


text:
How to determine whether a reaction is possible by using a phase diagram?
Given the following phase diagram for a binary mixture of $\ce{MoO3}$ and $\ce{Bi2O3}$:

I am asked to determine whether from $\ce{Bi2O3}$ and $\ce{Bi6Mo2O15}$, can $\ce{Bi10Mo3O24}$ and $\ce{Bi14MoO24}$ be synthesized? And why?
I have checked the given answer and it says the 1st reaction is possible, and the second one isn't. Any ideas?

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E.g. write $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$  to get $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$ (all eventually with double dollars in the display mode like $$\ce{H2SO4}$$.

Given an appropriate mixture of the two, either can be made with some effort, since those compositions fall between your two given phases. We’re there other restrictions?


text:
Assigning structure based on EPR spectroscopy
When given two copper(II) complexes - bis(N-3-chlorophenyl-salicydenaminato)copper(II) and bis(N-3-methylphenyl-salicydenaminato)copper(II) - how can you predict their EPR spectrum and hence their g average values based on their electronic properties? I am aware that bis(salicylaldiminato)copper(II) complexes have have a preference for square planar configuration, but is there any way to predict if the compound would assume a tetrahedral or square planar geometry based on its EPR spectrum?
bis(N-3-chlorophenyl-salicydenaminato) has an absorption peak in UV spectrum at 659 nm and bis(N-3-methylphenyl-salicydenaminato)copper(II) has it at 665 nm.

Yes. You can use computational chemistry to predict the electronic structure of the different possible conformers and then use that to get a bunch of EPR parameters.
Specifically, you can use ORCA and EasySpin to predict EPR spectra of your complexes and compare to an experimental spectrum.
You'll also get a bunch of other useful information about your complex, especially their relative stabilites.


text:
Chemistry behind adding methyl salicylate to iodine dissolved in petroleum jelly
We did a project at my school where we were asked to create a pain-relieving balm, inspired by Iodex. We first took 50gms of petroleum jelly and heated it in a water bath until it melted and became a clear liquid. Then we put 1.5 g of Iodine(the crystals) in it. We allowed it cool at room temperature and then added about 5 ml of methyl salicylate to get solidified Iodex, the paste with the minty smell.
I'm curious as to what goes behind the scenes and the reactions that are taking place here.

By "gms", you must mean "g" as SI symbol for the gram unit, right? Or? // Note also there is written a space between values and their units.

yes, I made the changes. Thanks for the suggestions!

I don't think there are any chemical reactions taking place here. All you are doing is creating a mixture of a non-steroidal anti-inflammatory (methyl salicylate) and an antiseptic (iodine) in a spreadable medium (petroleum jelly)

@SolitarySolus if this answer helped, make sure to accept it by clicking the checkmark.


text:
Where is the aromatic ring in the pentacene molecule?
I have the following problem: I am given the coordinates of the atoms of a pentacene molecule ($\ce{C22H14}$), and I would like to find the ring where the aromatic bond is located. Is this possible? Or is the aromatic bond kind of simultaneously located across all rings?

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The whole system is aromatic

Maybe ixaixim is looking for the position of the only benzene ring which can be drawn (with three double bonds) among the five cycles of the pentacene molecule. Is it the point ?

@Maurice Then this is any ring.

@ixaixim Possible if you can actually read the said coordinates, impossible otherwise. Can you? While we're at it, can you find the bond lengths?

@IvanNeretin yes, I can read the molecule coordinates. Yes, I can also find the interatomic distances.

Good. Then compare the C-C bond lengths in each ring. A single bond is 1.54 angstrom, a double bond is 1.34, an aromatic bond is 1.39, and you will probably observe something in between, leading to a not-very-satisfying answer "partially aromatic".

@IvanNeretin yes indeed I do! This oddity was in fact motivating my question. Thanks for explaining it the other way around, that is "the C-C bond lengths have all a mixed length, since the aromatic bond is in any ring", right?

My point about "any ring" was referring to Maurice's approach, which is based on drawing structures by hand and not on actual bond lengths. Since you have the latter, you don't need the former. If the bond lengths in some ring were exactly equal, that ring could be called fully aromatic, but I guess that is not quite so, hence we may say that the rings are partially aromatic, and maybe that some rings are more aromatic than the others.

@IvanNeretin we can indeed have some rings more stabilized than others. And put this selectivity to good use.

https://en.wikipedia.org/wiki/Clar's_rule

When you have a chain of cumulated conjugated rings either the entire chain is aromatic or it isn't. In this case it is. There is no single aromatic ring.
However, that does not mean all rings are equally stabilized. In general, with three or more rings in the chain we generally see the end rings most resistant to addition reactions, so they show greater stability than the more addition-prone middle ring(s). Such is the case with not only with pentacene, with the third ring in the chain especially prone to breaking the conjugation, but also the simpler compound anthracene.
Both ends against the middle
The case of anthracene is instructive. Although this compound is aromatic, the middle ring is relatively easily saturated at the two central carbon atoms, like a conjugated alkene undergoing 1,4-addition. The reaction with magnesium in THF is an example:

The magnesium anthracene, with its middle ring saturated, is used as an intermediate for production of reduced anthracene compounds and also magnesium hydride. In the latter case the anthracene is released in unreduced form, which can then be combined with more magnesium.
2+2 becomes greater than 5
Pentacene shows an extra form of middle-ring de-aromatization. From the Wikipedia article, we see that when an alkyl group such as methyl is substituted onto the middle ring, it spontaneously converts to a methylene group with the remaining aromatic rings now all "end" rings in naphthalene systems (reaction drawn by v8rik at English Wikipedia source):


text:
Why does base aspirin hydrolysis work better than acid aspirin hydrolysis?
Why does base aspirin hydrolysis work better than acid aspirin hydrolysis?
I conducted an experiment and found that there was greater Salicylic Acid yield for the one I added base compared to the one I added acid. Why is this so?

In what solvent system was this run?

I believe ethanol and water

Possibly aspirin is less soluble in acidic EtOH/water than in basic EtOH/water.

Does this answer your question?

Also, the mechanism given in this article explains why base is better catalysis.

Rush to close.  Unfortunately, no one checked the pKa of salicylic acid:  right around 3.  This indicates the 6 membered intramolecular carboxylate intermediate is indeed a viable weak acid hydrolysis mechanism.

Base (OH-) is the perfect nucleophile for ester hydrolysis of acetylsalicylic acid.
Acid pronation of the ester carboxyl group, which is further conjugated with benzene ring, and the far weaker OH2 nucleophile, will not work as well.
Base (and heat) hydrolysis is preferred, leaving carboxylate and OH groups.

Simple Le Chatlier's rule. Since the hydrolysis produces acid, base catalysts will remove the products (the acid), shifting the reaction.


text:
Does formal charge or the octet rule take precedence over one another?
In my AP chem test, I had a question that asked me to find the bond order of the ion $\ce{SO3^2-}$.
To draw the lewis structure, I determined that there are a total of $6+3\cdot 6+2=26$ electrons, and the sulfur had $\frac{26-8\cdot3}{2}=1$ lone pairs.

When we measure formal charges, every atom has one, and I have been told that this means the given lewis structure is inherently unstable and will attempt to form additional bonds. I use a pair on the oxygen to form a double bond, which eliminates the formal charges on the oxygen and sulfur, giving me the three (what i presume to be more stable) resonance structures

Hence the bond order is $\frac43$.
However, my teacher told me this was wrong. Apprently, due to the octet rule, the first lewis structure is the correct one where there are just three single bonds, and it's actually more stable since it satisfies the octet rule.
This makes me confused though. Which is the actual correct lewis structure (well, correct in a sense that the majority of the resonance structures exist in that form. I know that at least some aka a minor minor amount of $\ce{SO3^2-}$ would probably exist in the single bond form) that you would find bond order with?
I think this in the end comes down to whether or not formal charge or octet rule takes precedence, and based on everything I can find online, it seems formal charge does. So why does my chemistry teacher insist that the octet rule does? Am I misunderstanding something?

Apparently, your teacher is not aware that sulfur can expand its octet. Your reasoning and answer seem correct, although strictly speaking, if overall bond order was asked, the result would be 4 and not 4/3.

@Sam202 Hm not really (even though this theory is disproved/debated or w/e) he just claimed that an octet is more "stable", so sulfur must have only 3 single bonds. He ignored my argument of formal charges, and never mentioned anything about expanded octets. Also i thought bond order was number of bonds / bonding sites which would be 4/3?

Bond order of the sulfur-oxygen bond in $\ce{SO3^{2-}}$ is 4/3, but overall bond order of $\ce{SO3^{2-}}$ is 4. The way you stated the question suggests the latter was being asked.

In this case, the octet rule predominates.

First, why the downvote to whoever that did? Anyhow, @KanghunKim how so? Is there a guideline that specifies which rule dominates over the other in a given situation? BTW i checked numerous sources including several chemistry textbooks which all have the double bond, so would they all be incorrect? :( i am so confused lol

The problem is that electrons are never taught rules, so they figure it out themselves…

@Sam202 - or perhaps the teacher is aware that the idea of octet expansion in sulfur oxides has been considered outdated for decades.

@Andrew can you point to a source or paper that claims the bond order of the sulfur-oxygen bond in $\ce{SO3^{2-}}$ is 1 and not 4/3?

@Sam202 - this recent paper explains it pretty clearly I think: https://pubs.acs.org/doi/10.1021/acs.jchemed.0c00368

I believe the confusion stems from the false assumption that conventional Lewis structures are actually representative of the electronic structure of a molecule. They are not intended to be so. Rather, a Lewis structure is intended to provide enough information that a competent chemist can understand the structure of a molecule. As a trivial example, consider the Lewis structure of benzene, which uses alternating double and single bonds. Or the fact that Lewis structures only use integer charges.
So, if the question were to "draw the Lewis structure of $\ce{SO3^2-}$", then following the conventions of Lewis structures with regard to charge vs octet would make sense. Those conventions are that in first row elements (eg N, C, O), the octet rule takes precedence over reducing formal charge. For example, it is correct to have a positive charge on N in $\ce{HNO3}$ rather than having five N-O bonds. However, in second row elements like sulfur, the priority reverses, and the convention is to break the octet rule in order to minimize formal charge. For example, sulfuric acid is drawn with six S-O bonds.
But all of that is moot, since the question was about the bond order, not about the Lewis structure. For such a question about the actual electronic structure, you should use the structure that is closest to our best understanding of the actual molecule. In the case of $\ce{SO3^2-}$, it is now understood (and has been for decades) that $d$ orbitals participate negligibly in bonding, so the structure with formal charge that obeys the octet rule is a more accurate representation, and thus the bond order deduced from it is more accurate than a bond order deduced from the conventional Lewis structure.


text:
Making Sense Out of The IUPAC Exo/Endo Definition

Am I missing something or is the IUPAC endo/exo definition completely illogical? The prefixes exo- and endo- in general (also in chemistry) mean away (outwards) and towards (into), respectively. The chemical names under the shown structures are correct (e.g. it is 2-exo-bromo because the Br group is pointing away from the highest numbered bridge which is "y" based on the general structure on the left). Did IUPAC mess up the bridge numbering and what followed usage of towards/away in respect to exo- and endo-?

It's not clear what you're asking. The names are correct and so are the descriptions. What's the problem? How is the bridge numbering 'messed up'?

In the definition they write: "If the group is oriented towards the highest numbered bridge ('z' bridge, e.g. C-7 in example below) it is given the description 'exo';" First, the 'z' bridge is the lowest numbered bridge. The C-7 group is on the 'x' bridge so we are comparing it relative to the 'y' bridge which is higher numbered than 'z'.  Finally, an exo group is oriented away not towards the highest numbered bridge. I'm really confused ;/

I think OP confused about numbering scheme. I hope following diagram would explain what is meant by the highest and lowest numbered bridges:

Also, the example given in Gold Book do not do the justice because two bridges are same numbered (2-corners). If it was bicyclo[3.2.1]octane instead of bicyclo[2.2.1]heptane, then lowest numbered bridge would have been 3-corner bridge with numbers 2,3,4 (bridge heads are 1 and 5) and highest numbered bridge would have been 1-corner bridge (8).

thanks for your input. I thought the bridge numbering was based on the length and not on atom numbering. This would make a bit more sense now. However, I still believe IUPACs word combination of 'highest/lowest numbered bridge' with 'towards/away' and 'exo/endo' is misfortunate. In my opinion, exo- should be associated with away/outward and endo- with towards/inward.  This, however, would be incompatible with the highest/lowest bridge based on carbon numbering rather then the number of atoms that make up the bridge.


text:
Are solubility product (Ksp) and dissociation constant (Kd) unrelated?
Dissociation constant can be calculated from the Gibb's energy released by the dissociation, but as best I can tell, there's no way to determine solubility from dissociation constant, as things can dissolve without dissociating. Is this right?
A specific example that is perplexing me is disassociation of calcium chloride. I calculate from CRC Handbook values for aqueous $\ce{CaCl2}$ that at 300K, the pKd is 4.75, so the pCl is 1.48 and pCa is  1.78, if the pCaCl2 is 0. $\ce{CaCl2}$ is very soluble in water, so does this just mean that 1.7% of the $\ce{CaCl2}$ that is dissolved in water dissociates? I thought that salts generally dissociate almost fully, is calcium chloride just an exception?
Is it really the case that salts can only partially dissociate like this even when highly soluble?

How did you get pKd ? And how it comes pCa is so much bigger than pCl? That 1% should turn on a red light in your train of thoughts.

It does turn a red light, that's why I'm here asking this. And the pCa is bigger because there's half as much Ca as Cl.

But what about pKd?

Please do not confuse first time readers by striking things and including edit statements. If you change your question because you have made a mistake, simply do it. Then flag all comments relating to that change for deletion.

@Martin-マーチン I apologize, I wasn't aware that was possible, and I'm not sure how to do so.

No need to apologise. Just edit the question keeping the first time readers in mind. After that is done, you can flag the comments. You should be able to find a little flag icon next to the comment (when you hover over it). It'll open a menu where you can choose a reason. No longer necessary should do just fine.

Suppose we have a partially soluble ionic salt in water of the form:
$$\pu{\ce{A_aB_b(s)}}$$
With the condition: $(a≠b)$ or $(a=1$ and $b=1)$
When this salt is added to water, the following equilibrium takes place first:
$$\pu{\ce{A_aB_b(s)<=>A_aB_b(aq)}}$$
Where the equilibrium constant associated with this process would be the following, since we can approximate negligible contribution from the solid phase:
$$\pu{K_s=\ce{[A_aB_b(aq)]}}$$
The next equilibrium system that takes place is:
$$\pu{\ce{A_aB_b(aq)<=>aA^b+(aq) +bB^a-(aq)}}$$
With the equilibrium expression:
$$\pu{K_d=\ce{\frac{[A^{b+}]^a\;[B^{a-}]^b}{[A_aB_b(aq)]}}}$$
When we multiply both equilibrium constants we get:
$\require{cancel}$
$$\pu{K_{sp}=K_s\;K_d=\ce{\cancel{\ce{[A_aB_b(aq)]}}}\times \ce{\frac{\ce{[A^{b+}]^a\;[B^{a-}]^b}}{\cancel{\ce{[A_aB_b(aq)]}}}}=\ce{[A^{b+}]^a\;[B^{a-}]^b}}$$
Which corresponds to the overall process:
$$
\require{cancel}
\begin{align}
\ce{A_aB_b(s)&<=>\cancel{\ce{A_aB_b(aq)}}} \tag{$K_s$} \\
\ce{\cancel{\ce{A_aB_b(aq)}}&<=>aA^b+(aq) +bB^a-(aq)}
 \tag{$K_d$} \\
\hline
\ce{\ce{A_aB_b(s)}&<=>aA^b+(aq) +bB^a-(aq)}
 \tag{$K_{sp}$} \\
\end{align}
$$
If $K_s$ is significantly higher than $K_d$, $K_{sp}$ would also be significantly higher, even though $K_d$ is small:
$$K_s>>K_d\implies K_{sp}>>K_{d}$$
The solubility $S$ of this salt could be calculated from the following expression, assuming no initial amounts of ions are present:
$$K_{sp}=(aS)^a\;(bS)^b$$
In conclusion, both $K_s$ and $K_d$ along with their respective equilibrium systems contribute to the solubility of the salt, and not just $K_d$ independently as one could wrongly assume.

Ok, this is what I was thinking. So do highly soluble salts actually sometimes have low dissociation like I found for CaCl2? Also, there are four red "\cancel" in your equations

I don't know if your $K_d$ values for $\ce{CaCl2}$ are correct or incorrect, but as an example, at 25°C, the $K_s$ of $\ce{AgCl}$ is $3.6\times 10^{-7}$, while its $K_d$ is $5\times 10^{-4}$, giving a $K_{sp}$ of approximately $1.8\times 10^{-10}$. Also, I'm not sure why you're seeing those red terms, since both my computer and cellphone open the page normally. In short, the concentrations and text for the aqueous salt are cancelled out when multiplying $K_s$ and $K_d$ and adding both equilibrium systems, respectively.

It's showing correctly for me now. Also I checked my math and fixed it. I'm rusty with my chemistry, so I think I mixed up the solvation and dissociation because most of the experiments I did in high school were with strong acids, which by definition dissociate near fully.

The solubility of $\ce{CaCl2(s)}$ at 20°C is 745 g/L, which gives a $K_{sp}$ of approximately 1209. Assuming your $pK_d$ value is correct, this should give a $K_s$ of about $6.8\times 10^{7}$. For weak acids, $K_a$ (which is analogous to $K_d$) is used instead of $K_{sp}$.

Note that concept of Ksp is not applicable on ion molar concentrations of highly soluble salts, as ion TD activities are very different to their concentrations.

I'm not going to correct it, because it's readable enough, but for future reference: It's $K_\mathrm{s}$ and neither $K_s$ or $\mathrm{K_s}$ are correct. I know, it is way more work...

Furthermore \pu{1 unit} is only for physical units and \ce{formula/equation} is for chemical equations. They don't escape within each other and they cannot be chained, and they should only be used for what they are intended.

Dissociation constant can be calculated from the Gibb's energy released by the dissociation, but as best I can tell, there's no way to determine solubility from dissociation constant, as things can dissolve without dissociating. Is this right?

That is right. A more straightforward example would be acetylsalicylic acid (Aspirin), a weak acid. It has a certain solubility (quite low), and then dissociates at appropriate pH. More aspirin goes into solution because some of it is in the (charged) deprotonated state. The solubility of the neutral species and its acid dissociation constant are not linked directly - there are weak acids with a range of solubilities in water, and a range of acid strengths, and the two are not correlated.

I thought that salts generally dissociate almost fully, is calcium chloride just an exception?

At low concentrations, you would expect a salt to dissociate fully in water. At higher concentrations of one or both ions, ion pairs may form. They come in different flavors, contact ion pairs and solvent separated ion pairs. For calcium chloride, you could write them like this:
$$\ce{[Cl- Ca^2+ Cl-], [Cl- H2O Ca^2+ H2O Cl-] and [Cl- (H2O)_x Ca^2+ (H2O)_x Cl-]}$$
Also, you might have ion pairs with different stoichiometry, such as positively charged $\ce{[Ca^2+ Cl-]}$ or negatively charged $\ce{[Ca^2+ . 3Cl-]}$. The neutral ion pairs will not contribute to the conductivity. This is one fairly simple way to detect them.
How an ion speciates can be studied with advanced techniques such as solution state X-ray methods (EXAFS, e.g. this paper) or measuring the activity of the ions (e.g. through precipitation equilibria with a third ion). For calcium chloride, you have to have highly concentrated solutions at high temperature to see the effect of ion pair formation. An example of an ionic compound forming appreciable amounts of ion pairs is magnesium sulfate in the ocean.
There are other cases where different species co-existing in solution complicate things. Two common examples are carbon dioxide in water (either as carbon dioxide or as carbonic acid) and protons in water (either as hydronium ion or as higher-order complex with multiple waters). The pragmatic way of treating these complications is to talk about the total concentration, and use this in discussions of concentrations, reaction quotients and equilibrium constant expressions.

Is it really the case that salts can only partially dissociate like this even when highly soluble?

Most salts dissociate completely at low concentrations (behave like strong electrolytes). If they are soluble to high concentration (or you can add one or the other ion to high concentration), you increase the tendency to form coordination complexes (for transition metals) or ion pairs (for any cation).

Is this tendency to form complexes related to activity?


text:
Why does the Gallium-Aluminum Reaction Happen?
There a plenty of videos online of a small gallium sample being placed on a large aluminum sample, resulting in the former being absorbed by the latter and forming an extremely brittle alloy. What I cannot figure out is why this happens. Usually the explanation is something like the "gallium atoms invade the aluminum", but I find that very hand-wavy. It does not address why gallium is able to enter the aluminum crystalline structure so easily or how it diffuses throughout the aluminum sample so fast.
My only guess is that is has something to do with both metals being in group 13 on the periodic table has something to do with it. However, copper and silver are both in the same group and, as far as I know, placing a silver sample on a copper ingot does not result in the ingot absorbing the silver.
Just to be clear, my question is not about why a gallium-aluminum alloy is weak. My question is why is that alloy so easy to form.

Well, gallium is usually added to the aluminum as a liquid.  Your example of placing a silver sample on a copper ingot may turn out differently if the silver were molten.

I considered that gallium is a liquid being important to the reaction, but I don't see how. You could say that gallium can enter the aluminum sample easily because the interatomic forces between the gallium atoms are very weak [i.e. the sample is liquid], but that does not explain everything. Once the gallium atoms enter the aluminum they are no longer a liquid, yet they still diffuse through the entire aluminum sample.

Gallium is soluble along grain boundaries in the (polycrystalline) aluminum metal. Many elements are more soluble on grain boundaries than in the bulk. Grain boundaries also can dramatically enhance diffusivity. Finally, Ga saturating the boundary weakens it, allowing the material to fall apart along those weakened grain boundaries.

Also it helps to look at it the other way round: Al is soluble in Gallium! Whereever dissolved Al leaves a hole in the bulk (firstly and primarily at the grain boundaries), Ga can take its place. And so on. The alloy is still very close to its melting point (if it is already solid at all, that's hard to discern in such microscopic structures), so the mobility of the constituent atoms is pretty high.

BTW, a drop of mercury on gold immediately alloys, leaving a "silvery" looking stain on the gold. In a matter of hours, the mercury diffuses throughout the whole gold object, and the stain effectively disappears. This amalgam is not extremely weak, and does not destroy the object. However, "Gallium does not crystallize in any of the simple crystal structures. " (https://en.wikipedia.org/wiki/Gallium) and this makes its alloys quite weak. As for metals allowing other material to flow through, consider Pd and H2: https://www.peakscientific.com/discover/articles/hydrogen-purification-methods/


text:
Can we always use the planar representation of cycloalkanes instead of their true conformations to study their stereochemistry and chirality?
When I have drawn stereoisomers of substituted cyclohexanes and checked if they are chiral, I typically represent the ring as a planar hexagon and look for symmetry elements. For example, I would use this representation of cis-1,2-dimethylcyclohexane, and I would say that it is achiral due to reflectional symmetry:

However, I was recently reading about how some disubstituted cyclohexanes such as cis-1,2-dimethylcyclohexane are achiral not due to the symmetry in their 2D representation, but due to the fact that they exist as rapidly equilibrating chiral conformers that cancel out each other's optical activities. In this case, both the planar representation and the conformational analysis lead to the same conclusion about the chirality of cis-1,2-dimethylcyclohexane (though for different reasons). Will the planar representation and conformational analysis always lead to the same conclusion about chirality?
I think they will: I think that if a molecule can exist in an achiral conformation (the planar hexagon conformation), and you could rotate bonds to turn the achiral conformation into a chiral conformation (one of the chair conformations), then you must be able to obtain the enantiomer of that chiral conformation (here that enantiomer is the flipped chair) by performing mirrored bond rotations on the original achiral conformation (since the achiral molecule has reflectional/center of inversion/improper rotation symmetry, I think that guarantees that the mirror operations must be possible, though I am not certain). Since enantiomers have the same physical and chemical properties, they have the same thermodynamic stability, so at equilibrium, they will exist in equal amounts (the equilibrium constant for their interconversion is 1), so a sample of the disubstituted cyclohexane must exist as a racemic mixture. Is this thinking correct? Could you extend this argument to other molecules that we represent as achiral, but can exist as chiral conformers (such as butane, monosubstituted cyclopentanes, etc.)?
I also was thinking about whether it was correct to find all of the stereoisomers of the cyclohexanes by analyzing only the planar hexagon conformation. For example, to count the stereoisomers of inositol, I would use the following diagram, moving certain OH groups up or down to find the eight relative configurations, with one being chiral, which gives a total of nine stereoisomers:

I am wondering whether this always leads to the correct result, despite the majority of the inositol in a sample likely existing in a chair conformation, not a planar conformation. The nine stereoisomers drawn must be possible stereoisomers of inositol, so we know that there are at least nine stereoisomers. This leaves us with the question of whether the nine stereoisomers we have drawn represent all possible stereoisomers of inositol, or whether there are some stereoisomers that can not achieve a planar conformation, and are therefore not included in the nine we drew. I am not sure how to ensure that we have drawn all possible stereoisomers, besides saying that we can generally correspond each 3D structure of a cyclic molecule to a planar representation, and therefore all possible stereoisomers of inositol would have to be represented by one of the 9 planar representations from above.
Is it correct to analyze the planar depiction of substituted cyclohexanes for chirality and stereoisomerism, or do we need to look at the true 3D conformations?

I'd say yes. All molecules are somewhat dynamic. None of the (necessarily static) representations we are able to make can possibly be realistic. However, chirality is a geometric concept. So when we study chirality we need to use an 'ideal', rigid representation, where the molecule is 'as symmetric as possible'. That is your 2D depiction. Another way to look at it could be: is the average of all possible conformations symmetric? Take the amine example in the post you linked: the average is a planar species that does not exist, but justifies that the molecule behaves as achiral.

The only way the correct result for chirality using a planar structure might be wrong would be if there were very bulky groups as substituents that prevented easy interconversion of conformers (which otherwise interconvert very rapidly). This is certainly known in other classes of molecule like biphenyls with bulky groups or in hexahelicene where the entire chirality is a product of long distance spatial interactions preventing the possibility of a "flat" structure.

@matt_black : indeed, I had thought of biphenyls; hence the wording "average of all possible conformations". Biphenyls with bulky substituents cannot have planar conformations, thus the average will still be asymmetric, and they are chiral (only) for that reason. Still, it's true that conformations that are not possible in standard conditions can become possible when more energy is available, e.g. by heating, so this is indeed a thorny issue. To keep it simple, however, the answer to the OP is yes for the most common cases.

Is it correct to analyze the planar depiction of substituted cyclohexanes for chirality and stereoisomerism, or do we need to look at the true 3D conformations?

Yes, you can do that. As the six membered ring transitions from chair to boat to twisted to planar (with the latter having an unrealistically higher energy, so it would not be populated in reality), the configuration around the carbon atoms does not change. So if you construct the mirror image of a substituted cyclohexane, you make no mistake if you first "flatten it out" for easier drawing.
Many of the conformations of substituted cyclohexanes are chiral if you don't allow for conformational change. To take a simpler example than the cyclohexan one, if butane were frozen in the gauche conformation, it would be chiral. Because it is not frozen in that conformation, as we know, it is not chiral.


text:
How can I test for impurities in my steel wool?
I am trying to purify ferrous sulfate (for some electroplating experiments) and I came up with the following process:

Dissolve 200g of iron fertilizer in 300ml of distilled water heated to 54 °C (because solubility is highest at that temperature).
Add a splash of 15% sulfuric acid and some steel wool to prevent oxidation of iron(II) to iron(III).
Filter the solution while it's hot, cool it to 0 °C and scrape out the ferrous sulfate crystals.

Now I am wondering about impurities, because I think it smelled like hydrogen sulfide when I add the steel wool. Manganese would create hydrogen sulfide while dissolving it in sulfuric acid and is a common alloying element in steel wool (besides carbon and silicon, which would both not dissolve, right?). Also I think my ferrous sulfate looks a bit blue-ish so that made me think that there might be some chromium and nickel in the solution.
Is there a way to test for chromium or nickel with the usual household chemicals? If there is nothing in there, can I pour the leftovers of my solutions down the drain after neutralizing? There should only be iron(II/III) sulfate and small amounts of manganese sulfate in there, right?

While there are many ways to test for contaminants, your process includes crystallization, which is fairly effective for removing some unwanted impurities (though nickel salts might co-precipitate with iron salts, it should not impact use for electroplating). As for home tests, look at those used for checking drinking water, e.g., https://www.chemetrics.com/product-category/test-kits/manganese/

Iron metal vs. sulfur
Your steel wool is of course mostly iron(0), i.e. metallic iron.  This has a standard reduction potential of -0.44 V, or -440 mV.
$$\ce{Fe^2+ + 2 e−  ⇌   Fe(s)}\;\;\;\;\;\;E_0=-440\;\textrm{mV}$$
The standard reduction potential of the sulfate / sulfide system is supposedly -220 mV.
$$\ce{SO4^2- + 8H+ + 8 e−   ⇌   S^2- + 4H2O(aq)}\;\;\;\;\;\;E_0=-220\;\textrm{mV}$$
Thus, it is thermodynamically possible for the metal to reduce sulfate to sulfide.  (In actual fact, redox equilibria with sulfur are very complicated, and varying degrees of sulfite, thiosulfate, dithionate, and sulfur(0) species may also be formed.)  I'm not 100% sure if these reactions are kinetically hindered or not at room temperature in acid on the metal surface, but they are at least thermodynamically possible.  It could potentially explain the whiff of sulfide you smelled.  Since our noses are very sensitive to hydrogen sulfide, only minute amounts of this reaction would need to occur to explain the odor.
The blue color
Acidic solutions of iron(II) are blue-ish.  Depending on the concentration, they have a very similar color to solid iron(II) sulfate.  The blue color is not necessarily indicative of an impurity.
Crystallization
I agree with the comments though that crystallization should remove the vast majority of any such impurities.

Nickel, chromium and manganese can be detected by the following tests.
Chromium. Some ammonia solution ($\ce{NH3}$) is added to the acidic solution containing $\ce{Fe^{2+}}$ ions and maybe $\ce{Cr^{3+}}$ ions. The obtained precipitate is filtrated, and introduced in a test tube with $3$ mL $0.2$same  M $\ce{NaOH}$ solution. A tiny amount of $\ce{PbO2}$ is added. The test tube is heated to boiling for one minute : $\ce{Cr{OH}3¶ is oxidized into chromate ion. The obtained mixture is diluted with water and filtrated. The filtrate is neutralized by $1$M $\ce{H2SO4}$. 1 mL ether is added, then $3$ drops $3$% $\ce{H2O2}$. If the blue color (due to perchromic acid) appears in the ether phase, it is a proof of some Chromium in solution.
Nickel can be proved by the dimethylglyoxime reaction. But this is not an usual reagent.
Manganese.
A small amount of the same precipitate as obtained with $\ce{NH3}$ in previous paragraph, is introduced in a test tube. $0.5$ g $\ce{PbO2}$ + $3$ mL concentrated $\ce{HNO3}$ are added. The test tube is then heated to boiling for $30$ seconds. The obtained mixture is dropped in another test tube containing $10$ mL water. Wait for decantation. The solution takes a purple or violet color (due to permanganate ion) if manganese is present.


text:
Why is the reaction rate of tert-butyl chloride solvolysis faster in 80% water/20% acetone, than in 70% water/30% acetone?
The solvent is polar protic, so the mechanism is SN1. However, I don't understand how the composition of water and acetone in the solvent affects the reaction rate.

Water is more polar than acetone, so 80/20 water/acetone is a more polar solvent than 70/30 water/acetone.  Since the SN1 reaction has a positively charged carbocation intermediate, more polar solvents are better able to solvate and stabilize this intermediate.  That is why more water leads to faster reaction.
The relationship between solvent polarity and t-butyl chloride hydrolysis rates was investigated in 1948 by Grunwald and Winstein.  They looked at acetone/water as a solvent system, and also ethanol/water.  Here's a small portion of their data:

They were interested in describing the solvent dependence through the equation $$\log{\frac{k}{k_0}}=m Y$$, where:

$Y$ is a solvent-dependent, substrate-independent parameter, with 80/20 ethanol/water chosen as the "reference" solvent for which $Y=0$.  This parameter describes the "ionizing power" of the solvent.
$m$ is a solvent-independent, substrate-dependent parameter, with t-butyl chloride chosen as a "reference" substrate where $m=1$.
$k_0$ is the reaction rate of the reference substrate (which is t-butyl chloride, conveniently for the purposes of your question) in the references solvent (80% ethanol in water).
$k$ is the reaction rate in an arbitrary solvent with a $Y$ parameter of $Y$ of an arbitrary substrate with an $m$ parameter of $m$.

You can see from their data that the table I've excerpted that their data is qualitatively consistent with their equation.  Additional parts of their paper show a good quantitative fit too, for many (but not all) substrates and solvent systems.
It's worthwhile noting how strong the effect is.  Having 10% water improves rates by more than 10-fold relative to absolute ethanol.  And going up to 30% water improves rates by almost another 10-fold.  The bottom two entries are for acetone as the solvent.  There is almost a 10,000-fold difference in rates between 5% acetone and 80% acetone.
Later in 2007, Peters exhaustively reviewed solvent participation in SN1 reactions and mentioned there has been pushback against the simple Winstein/Grunwald mode, saying:

As of 2007, it is clear that there is no general consensus
as to the process by which the solvent participates in the
solvolysis of tert-butyl chloride. To borrow a recent statement
from Richard, “...the impression that studies of solvolysis at
tertiary carbon have resulted in a morass of experimental
data and which, when interpreted individually, provide
support for conflicting mechanistic conclusions” captures the
general sentiment of the field.

So "why" the reaction is faster with more water around is a hard question to answer precisely, but I don't think there is any doubt that more polar solvents will result in faster rates.  Unless you are reaching for world-class expertise and exacting quantitative accuracy, the simple Winstein/Grunwald model that more polar solvents have more "ionizing power" to stabilize the carbocation intermediates is probably a good place to start.


text:
Input file details for rigid PES scan
I have a planar molecule and a hydrogen atom. I wish to perform single point energy calculation for different configurations of this system where I would place the hydrogen at a certain distance away from the center of mass of the planar molecule.
I have set up the planar molecule such that it is in the XY plane (Z co-ordine zero(approximately)) with its COM at (0,0,0). Basically I want to put the hydrogen atom at different (0,0,z) positions and find the electronic energy. This is the input file I am using:
*# scan ub3lyp/6-311++g(d,p) int=ultrafine scf=xqc*

*Title Card Required*

*1 2*
 C                  0.13813800    1.25878700   -0.00021500

 H                  0.30168400    2.36487400    0.00005700

 C                  1.20739400    0.19371100    0.00012000

 H                  2.26666100    0.44051600    0.00017700

 C                  0.62471500   -1.03441700   -0.00003100

 H                  1.11284800   -2.00793900   -0.00004900

 C                 -1.13810000    0.45325300    0.00012300

 H                 -2.11780000    0.92562900    0.00023500

 C                 -0.83813500   -0.87256400   -0.00005300

 H                 -1.52746300   -1.71570300   -0.00008800

 H                  0.00000000    0.00000000    3.10000000

X 11 S 20 -0.100000

What more do I need and where do I need it? I am new to these kind of calculations, any help would be greatly appreciated.

Did you run this input file (and as shown in the post)? I recommend you try a trial run on something small and simple and troubleshoot that first. If you still have problems report those in detail.

@BuckThorn I ran this file but it says no variables in Z-matrix error.

You need to define the coordinate to step by using z-matrix coordinates; see the manual: https://gaussian.com/scan/ For examples: https://www.cup.uni-muenchen.de/ch/compchem/geom/scan.html

The input file can't work. Get rid of the asterisks. Get rid of the black lines in the structure section. Define a proper z matrix with the (or a) variable to scan through. The last line looks like input to a modredundant keyword, which is a relaxed scan. The scan keyword performs a rigid scan. I'm working from memory here as I can't try and fix it for you, but you probably want to use GIC. Here's some old stuff of mine, try to run these and adjust your setup. https://chemistry.stackexchange.com/q/49483/4945


text:
If I don’t touch my plasma/lightning globe will it still produce ozone gas
I noticed when I touched my plasma/lightning globe that my fingertips smelled like what reminded me of ozone gas.
I read somewhere that ozone gas could be harmful to the respiratory system So I came to your site to find out information about this.
I quickly discovered that people are saying that they (plasma globes) do indeed produce ozone gas.
Then somebody wrote that only occurs when touching them since that causes the arcs of electricity to intensify.
My question is if I don’t touch my plasma globe can I leave it on for extended periods of time without creating ozone gas or such tiny amounts that there’s nothing hazardous to my health to be concerned about?

Poisonous as it is, ozone is more smelly than poisonous. If you don't smell it, there is no harm.

@BuckThorn, The stated answer in the duplicate Q provides no supporting references whatsoever and it is rather speculative (duplicated cut-paste answers circulating on the web).

@AChem Maybe someone can write a better answer to that post?

@BuckThorn, Hopefully so. Hard to find reliable and specific information on this topic. Google Scholar would be a good start if someone is interested. There is a J.Chem.Ed article on plasma globes which makes a generic comment and gives reference to a high-voltage engineering book, First, any high-voltage source can produce toxic ozone gas; thus, prolonged operation of this device without adequate ventilation is not recommended


text:
Why is it wrong to name 2-methyl-1-propanol (the preferred IUPAC name of isobutanol)1-methylpropanol?
Given the compound 2-methyl-1-propanol (isobutanol), the preferred IUPAC name is the aforementioned. But why is it wrong to name it 1-methylpropanol? Given that the methyl can't go to any other of the carbons, methylpropanol has a given structure with only the hydroxyl group changing. Can you point to literature backing up the claim that the 1-methylpropanol might be confused with 1-methyl-propanol? (meaning that the number denotes the carbon on which the methyl is)

1-methyl anything is an abomination.

Why so? Methylpropanol can only have two structures. I don't need to name the methyl because it can only go to the second carbon, right?

So you are suggesting that 1-methylpropanol defaults to 2-methyl-1-propanol because methylpropanol requires that the methyl be bound at the 2 position (as it otherwise would be butanol)? I would counterargue that it is better not to leave things to interpretation. Computers have enough to worry about.

@BuckThorn Yes, exactly. I suppose that this is the case. But is it really wrong?

Wrong in the sense that it runs against the conventions of chemical nomenclature, and requires an additional layer of interpretation to resolve the ambiguity associated with the meaning of "methylpropanol". If you go one more carbon (1-methylbutanol) you run into further ambiguity. I would invoke Occam's razor.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), locants are placed
immediately in front of the part of the name to which they refer.

P-14.3.2 Position of locants
Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are
placed at the front of their names.

Therefore, the preferred IUPAC name of “2-methyl-1-propanol” actually is 2-methylpropan-1-ol.
For the same reason, the locant “1-” in the hypothetical name “1-methylpropanol” can only refer to the methyl group.
Furthermore, “1-methyl-propanol” is just a wrong spelling of the hypothetical name “1-methylpropanol”.

I don't need to name the methyl because it can only go to the second carbon, right?

IUPAC nomenclature doesn’t work like that. Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information or hidden logic involving other structures.
The parent compound propan-1-ol has several different hydrogen atoms that could be substituted. Therefore, the locant cannot be omitted in substituted propan-1-ol, e.g. 1-chloropropan-1-ol, 2-chloropropan-1-ol or 3-chloropropan-1-ol.
This principle applies to any substituent. The special situation that replacing the chloro substituent in 1-chloropropan-1-ol with a methyl substituent would actually lead to the name butan-2-ol instead of 1-methylpropan-1-ol and that replacing the chloro substituent in 3-chloropropan-1-ol with a methyl substituent would actually lead to the name butan-1-ol instead of 3-methylpropan-1-ol does not change the preferred name for 2-methylpropan-1-ol.


text:
What role does atomic radius play in the strength of a covalent bond?
How exactly do atomic radii influence the strength of a covalent bond? For instance, if I had a carbon atom covalently bonding to a fictitious nonmetal atom $\ce{X}$ and another nonmetal atom $\ce{Y}$ with equivalent electronegativity but a smaller radius, which molecule is going to have a higher bond enthalpy? $\ce{CX}$ or $\ce{CY}$, and why would that be so?

If $X$ has a higher atomic radius than $Y$, $X^-$ will be more stable than $Y^-$, since it has more surface area to efficiently distribute its negative charge, and $C-X$ would have a lower bond enthalpy than $C-Y$

@Sam202 Correct, although the situation is reversed if the C is replaced by (say) Sn. Simple HSAB theory- carbocations do not sit well with larger anions, stannocations the opposite.


text:
Are all resonance forms of pyracyclene equivalent?
This is pyracyclene:

If I’m correct, pyracyclene has four resonance forms (double carbon bonds shown in red):

Now my problem is estimating the dominance of each form. Are they equal or is there one which is more dominant?
This is what I’ve got so far:
Rules for estimating stability of resonance structures

The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets.
The structure with the least number of formal charges is more stable.
The structure with the least separation of formal charge is more stable.
A structure with a negative charge on the more electronegative atom will be more stable.
Positive charges on the least electronegative atom (most electropositive) is more stable.
Resonance forms that are equivalent have no difference in stability and contribute equally (eg. Benzene).

Pyracyclene has the same carbon atoms, the same number of covalent bonds and a formal charge of zero. I don’t see any difference in stability either. So it seems that all forms are equivalent.
Is this correct?

Good question! You've picked a molecule for which the answer is not entirely clear, from what I've just gleaned from recent and earlier publications. Welcome to chem.SE!

Which means that treating it as a generic homework, like OP did, isn't a good idea.

BTW when structures are equivalent, then there no difference at all. It doesn't have anything to do with some lame rule of thumb.

The top right and bottom left are the same by 180 rotation, the other two are related in the same way as the double bonds are in naphthalene (or benzene) by a shift single to double bond.  Whether these two types are the same is hard to assess. Have you looked for any experimental evidence, gas phase spectroscopy, or  xray of single crystals for example?

These rules are at least very sloppy worded. There cannot be a most stable resonance structure as these are all just hypothetical configurations; they simply do not exist on their own. So what you want to know or what these rules refer to are the relative contribution to the overall bonding description. Because of overall symmetry, I'd expect different contributions. A calculation might actually offer some insight into the structure as well as the bonding structure. But I unfortunately don't have the tools for that anymore.

What kind of calculation should be done? Do we expect large or small differences in contributions?


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Reaction mechanism validity
Hey I was solving some questions i found this solution to a question

I'm confused, can alpha-hydroxy acids undergo dehydration??

Two different things are  happening here. Firstly the hydroxy group is the first site of protonation, then water leaves resulting in formation of the double bond, a standard dehydration. Second the nitrile is protonating and undergoing hydrolysis (mechanism here) to the carboxylic acid via
the amide. This usually takes forcing condtions, but with hot 95% sulfuric acid that is what you've got.

Thanks for the explanation, i thought that it was alpha hydroxy elimination

@AwadheshPrajapati  if you are happy with the answer please accept it


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How can I safely recreate the "oxiclean effect" for a video production?
I am producing a parody infomercial, specifically copying the style of the early 90's OxiClean/Billy Mays spots:
https://www.youtube.com/watch?v=ZTpXh33Mbeg
As it happens in the original commercial, we need to show a solution that begins as a reddish-brown color, and then when the powder is added, it becomes clear.
With some initial research, I came across the idea of possibly using a combination of phenolphthalein and thymolphthalein in a basic solution, and then adding something acidic enough to make them both go clear.
The main questions are:

Is there anything inherently bad/dangerous with this approach?
If we would like our actor to be able to stir this solution with their bare hands, what would be a safe base solution to begin with, as well as a safe acid to add for the final solution?

Thank you for any insights or guidance.

I don't think it would be reddish-brown. More like purple maybe.

Why can't you actually use Oxiclean with a fruit juice? I know that bleaches pomegranate juice color in seconds.

Very dilute black tea might serve as the brown coloring + reddish fruit juices