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Improving accuracy of water lead test
Home water tests for lead have an accuracy of about 15 ppb. If I first do a gentle/slow boil for one hour (hot plate under a glass beaker of water) to reduce 2000 mL to 20 mL, will I then effectively be able to detect 0.15 ppb?
My goal is really just 1 ppb.
My main concern now is that any lead might adhere to the glass
(though I think these home tests have a reagent which helps
dissolve lead in case of this) so, if this is truly a problem,
is there a better beaker material or boiling method to use?
Of course, if improvement were this easy, I would expect the
test kits to already be recommending it, so I am probably making
some mistake.  Please explain it.

I mean the test for non VOC salts is to literally burn the filter paper at 600C so I don't think you are going wrong concentrating the water. I do think your method should work

You might have another approach, which involves using household acid (lemon juice) to enhance recovery. The relevant publication is here. The article is: Challenges of Detecting Lead in Drinking Water Using at-Home Test Kits
Rebecca Kriss, Kelsey J. Pieper, Jeffrey Parks, and Marc A. Edwards
Environmental Science & Technology 2021 55 (3), 1964-1972,
DOI: 10.1021/acs.est.0c07614

Of course I do not know the origin of the water to be analyzed. But reducing  a water volume from $2000$ mL to $20$ ml is not so easy as it looks, because boiling tap water does usually produce a deposit of calcium carbonate, which may adsorb the lead ions $\ce{Pb^{2+}}$ during its precipitation.

@ToddMinehardt The lemon juice is basically the "reagent which helps dissolve lead" that I mentioned, so yes, of course that additionally should be optimized, but my question here is focused instead on getting rid of molecules that are not lead.

@Maurice I think your point about the calcium carbonate is the key problem since there are typically a million times more calcium atoms than lead atoms (plus we have "hard water" here!), but hoping for a more official answer.  Maybe there is even a trick that someone knows to dissolve all the mineral deposits for an accurate "color strip" test...

@bobuhito. If the solution is totally evaporated, a powder will be obtained containing all $\ce{CaCO3}$ plus a rather tiny amount of lead. Now redissolving this powder in the minimum amount of diluted nitric acid (or citric acid) can give you a solution containing calcium and lead, where the test for lead can be  carried out.

Acid should bring the metals into solution.  Find out if other metals found in water interfere with lead analysis.


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Looking For Resources on Separating Iron, Silicon Dioxide, and Sodium Chloride
I am working on an experiment and lab report for my high school chemistry course where I am given a vial with a mixture of Ferrum, silicon dioxide, and sodium chloride. I need to figure out how to separate these three without causing any chemical change to occur to any of them, and then find the percent of each in the original mixture. I think that I could figure out the percentages pretty easily with some math, but I have no idea where to start with the actual separation process. I've looked online for resources to cite, but all of them go way over my head. For context, this is the second week of chemistry, and almost no one in the class has taken chemistry before. We just finished learning the difference between pure substances and mixtures. If you have any suggestions for resources that I will actually be able to understand or any tips for research/experimentation, I would be really happy.
Thank you, and apologies if this is out of the scope of this site.

"I have no idea where to start" unfortunately this just the thing you should have and write in your question.

@Meerkat, What is the approximate size of silicon dioxide? Is it sand by chance?

This answer will assume that all the 3 substances are in solid state. First, we know that Iron, $\ce{Fe}$ as a pure metal can be attracted by a magnet, while $\ce{SiO2}$ and $\ce{NaCl}$ can't. A magnet can then separate the $\ce{Fe}$ from the other two.
Then, we also know that $\ce{SiO2}$ is insoluble in water while $\ce{NaCl}$ is soluble in water. So we can pour water into the mixture of these two substances, mix, and filter. You can obtain the filtrate as dissolved $\ce{NaCl}$ in aq form and $\ce{SiO2}$ as the residue in solid form.

This is one of those silly experiments trying to teach something. Let's play the Game. Ferrum! To separate materials, compounds, elements one must learn as much as possible about their pertinent properties. the rule is that you must not chemically change anything. That means that you must already know what a chemical reaction is, very advanced! and the difference between chemical and physical properties and that you probably have to work with the physical properties. The first thing is How much do you have and what physical properties can you use to measure how much you have. What is the definition of matter? What in that definition can you use to determine how much matter you have then use the appropriate property or perhaps more than one. Examine the sample carefully. A point here must you use the entire sample or a representative sample? Is the sample mixed? If not the job can be easier. If they are not mixed or there are few of each visual separation may be possible [Pasteur's method read about him and optical isomers] size shape and color are physical properties a microscope and time will help here. [If you are lucky there is only one particle of each.] That probably won't work but it might give you a limited sample of each to test. Now you must investigate other properties. Since you have identities, look up these materials and find out whether they are elements, compounds, metal, molecular or ionic etc. and learn about their properties. You must look for differences that you can exploit with materials and equipment on hand or reasonably obtainable. Be prepared to defend that your method is physical. If possible work with a small representative sample after determining the total amount of material, then measure a precise sample for each and prorate the totals. Good Luck When done you will know more about this than your teacher.

you have written an indigestible text worm :/

Just how is someone, who has no idea how to start, to accomplish a task without doing some reading and research. Oh yes  someone should them instead of giving them a chance to learn how to look up something themselves. The goal is to learn chemistry not learn a recipe.

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Why does aldolase not act upon glucose-6-phosphate?
I was recently asked to explain why (fructose-bisphosphate) aldolase does not act on glucose 6-phosphate in glycolysis and waits until the formation of fructose 1,6-bisphosphate. I did not know much about the enzyme structure but tried to put an unconvincing explanation on such grounds emphasizing the specificity of the substrate, which was rejected.
The instructor's answer was that such a reaction would have led to the formation of glyoxal and, in turn, toxic advanced glycation end products (AGEs). Thus, such a reaction is precluded. I am not remotely convinced, notwithstanding selection pressure, and look forward to alternative explanations or constructive critiques of the very premise of the question.
P.S.: I did some digging in Google Scholar and came across this old article, reporting on a cycloaldolase that acts upon glucose 6-phosphate to produce myo-inositol.
P.P.S.: Here's a ready-reckoner flowsheet for the glycolysis pathway.

In the aldolase reaction, the original carbonyl remains on one product, and an additional is formed on the other product. Glyoxal has two carbonyl groups on the same molecule, so I would not expect it to form even if the enzyme catalyzed a reverse aldol reaction on glucose 6-phosphate. However, this paper claims there is a direct path from glucose to glyoxal.

Preamble
Like @Karsten, I am very wary of “why?” questions in biochemistry, which often come from individuals who see themselves as some sort of evolutionary God. And their answers tend to reflect their own scientific mindset: “to the man with a hammer, everything looks like a nail”.
Nevertheless, it is not unreasonable for a chemist to look at the reactions in metabolic pathways from a chemical point of view and try to identify an overall ‘strategy’, as long as one resists the temptation to conclude that this is the only, or even the best, strategy. It may be the first one that worked and became frozen as metabolism elaborated around it, or it may just take advantage of other pathways that had already developed when it evolved. Glyceraldehyde 3-phosphate is the product of the light reactions of photosynthesis and of gluconeogenesis. Dihydroxyacetone phosphate is the immediate precursor of glycerol 3-phosphate, needed for triglyceride synthesis. Fructose 6-phosphate is a product of the pentose phosphate shunt.
In addition, one may be able to rationalize the rejection of an alternative pathway on several grounds, so who is to say that one is the ‘reason’ for rejection?
What is the question?
The problem here is that the poster’s instructor (I assume the context is student–instructor) meant one thing by “why” and the poster assumed, quite reasonably, that a mechanistic/molecular answer was required and gave a correct one. The substrate specificity of the glycolytic enzyme aldolase (strictly fructose 1,6-bisphosphate aldolase) is such that it does not ‘act’ on glucose 6-phosphate in glycolysis.
What did the instructor actually mean? The “waits until the formation of fructose 1,6-bisphosphate” suggests to me “What strategic advantage can you see in an aldolase acting on the bisphosphate form of the hexose rather than the 6-phosphate form?”, for which I provide a rationale below. But it would seem from his answer in terms of (methyl?) glyoxal formation that his concern was the reactivity of the aldehyde group in one of the triose products that would be produced instead of dihydroxyacetone phosphate.
The obvious objection to this and answer to my interpretation of the question is that the strategy of the first part of glycolysis is to convert a hexose to two molecules of a triose phosphate and to do this a hexose bis-phosphate is formed (see diagram). Not the only way, but one that works and, in the case of fructose 1,6-bisphosphate, allows easy and rapid isomerization between the two trioses. (Glyceraldehyde 3-phosphate is the form that is further metabolized to pyruvate.)

Unfair to the Instructor?
Bringing in the monophosphate, glucose 6-phosphate seems rather a pretext to introduce a story about glyoxal. But perhaps I am being unfair. Let us presume he is well aware of the need for two molecules of triose phosphate and is taking it for granted that the triose resulting from the “top” half of the hexose (C1 to C3) will be phosphorylated after an appropriate aldolase reaction. I am not going to say this would be unviable or inferior (although that is possible) but if such a strategy had been developed, surely the triose phosphate formed would no longer be suitable for conversion to glyoxal.
Don’t ask questions in answers
But just this once, the focus on glucose 6-phosphate raises another question, why have an extra step to produce fructose 1,6-bisphosphate from glucose 1,6-bisphosphate? I presume that the aldolase reaction on the latter would result in triose molecules that were chemically  ‘unsuitable’ for the subsequent steps, but my chemistry is too rusty to be sure. Is this so? And would the aldolase products of fructose 6-phosphate convert to glyoxal? (Answers provided in the comments will be used to improve this answer.)
Observations on the toxicity of Glyoxal
Although the toxicity of glyoxal is not in question, the ability of the cell to deal with this needs to be considered.
First I would like to ask whether the supposition is actually correct: that delaying the aldolase reaction until the fructose 1,6-bisphosphate stage per se prevents the formation of toxic glyoxal?
In Berg et al. Biochemistry I read the following:
“…second TIM (triose phosphate isomerase) suppresses an undesirable side reaction, the decomposition of the enediol intermediate into methyl glyoxal and orthophosphate…This labile intermediate is trapped in the active site by the movement of a loop of ten residues…”
Second I would point out that certain bacteria and even human cells synthesize glyoxal, for reasons that in some cases are not entirely clear. (As far as the chemistry goes, they use hydroxyacetone phosphate as the precursor.) However, at least in bacteria, detoxification enzymes have developed (Arch Microbiol (1998) 170 : 209–219).

If this answer, with its reference to biochemical metabolism, seems out of place here, I suggest the question be migrated to SE Biology.

As a biochemist, I feel that metabolism is definitely part of biochemistry, and biochemistry question are a good fit for either SE Chemistry and SE Biology (i.e. the OP should choose which site they prefer). Also, everything in biochemistry is connected to evolution and genetics, so I think it is fine to discuss these connections on either site.

@Karsten — Thanks. As mentioned in my revision, I could do with chemical input on the actual chemical conversions following reversal aldol condensation on various hexose phosphates.

This is the type of critique, I was looking for! You read my mind :3 I will expand on the chemical aspects ~

This is a starting point for retro-aldol reactions on glucose. You can also look into transaldolases and transketolases in the phosphate pentose pathway. It does not involve any two-carbon species, though.

@Karsten — Thanks, have downloaded it and will read and try to amplify my answer. May take a while as I'm off for a week's holiday shortly and need to clear my desk. (Perhaps I should sign up for biochemistry questions here as well as SE Biology, which has been a bit thin lately in this area.)

@David I checked the "biochemistry" tag on both sites. Biology has about twice the number of questions with that tag. The ones on chemistry tend to be more in the direction of mechanism, intermolecular interactions, quantitative - as you would expect.

When the question is "why", you need a lot of context to answer it.
It is fairly easy to answer why we think that this enzyme does not catalyze the reaction of glucose 6-phosphate. We probably base that on an experiment that did not show any reaction.
The question "why not" could go in multiple directions.
The OP interpreted it as asking for a rationale why the enzyme, as it currently exists, does not catalyze the reaction. The reasons always have the same structure - either it does not bind to the possible alternate substrate, or when bound, it does not catalyze the reaction.
The teacher apparently wanted to see a rationale for the enzyme having evolved that way. This is a slippery slope, but it is fun to imagine the what if. If it did catalyze the reaction, the teacher argues, the product would be toxic. You might also think of one of the products not having a phosphate group, so it would not be able to support substrate-level phosphorylation. Also, the reactant might diffuse out of the cell more easily without the charge of the phosphate group.
Of course, other features of the cell could have co-evolved to fix these problems. There are many reactive metabolites the cell can handle through special mechanisms. Superoxide and free radicals are an example, as is vitamin B12.

[OP] I am not remotely convinced — notwithstanding selection pressure —, and look forward to alternative explanations or constructive critiques of the very premises of the question.

All one can do is to make a convincing argument about hypotheticals, unless you have the money and time to try to make a set of engineered cells containing enzymes of altered substrate specificity (see e.g. https://www.nature.com/articles/d41586-022-02256-z). In practical terms, you have to consider the instructional context in which the question was asked. If you offer more than one way of thinking about it, even if you don't go in the direction your instructor anticipated, you might impress them nonetheless.

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Can I use a mechanical pipette to transfer strong acid?
I am an undergraduate student majoring in physics. I am looking for the best method to accurately measure the volume of strong acids like hydrobromic $(w(\ce{HBr(aq)}) = 48\,\%)$ and hydroiodic $(w(\ce{HI(aq)}) = 57\,\%)$ acids. I need milliliter volumes, and since I'm going to pour or dispense a specific volume of acid in a vial, I mean "to deliver". The only measuring device I could think of is a mechanical pipette, but I am afraid the acids or their corrosive vapors could damage the pipette's mechanism.
I don't have any friends or colleagues majored in chemistry, so I am having a hard time trying to decide which pipette to buy. I really have no clues on how actual chemists who use strong acids accurately measure them, so I'll be pleased with any kind of help.

Concentrated acids are seldom measured precisely, as their concentration is not precise either.

I think there are Hamilton syringes with glass/PTFE inserts/fittings which could resist these strong acids. But it's probably preferable to just weigh the acids instead if you can.

Maybe your advisor has friends who are chemists. It would be nice to have access to a chemist for considerations of safety and hazardous waste treatment.

Metallic pieces should be avoided, because they react with acids. But glass pipettes and rubber bulbs are resistant to strong acids like $\ce{HCl, HBr, HI}$ in solution or in a gaseous state.

PE/PP Hamilton syringe inserts should work too, unless it is for perchloric acid.

What accuracy is this supposed to have? How often do you need to do it?

If you want good, well reproducible accuracy, this there https://de.wikipedia.org/wiki/B%C3%BCrette#/media/Datei:Buerette2.jpg might be what you're going to need. Be prepared for a bill in the region of 1k for three of them. And you still need to calibrate the titre factor every few days.

Of course you can also dilute the acid, which will increase accuracy in pipetting or allow use of graduated cylinders.

Another possible solution is positive displacement pipettes, though I don't know much about them


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chemical equilibrium and activity
Here is a graph of equilibrium coeff in various salt concentrations in water.

Lets take the example $\ce{BaSO4}$.
Now as we are increasing the salt concentration in water, the following reaction is moving towards right as eq. coeff is increasing.
$$\ce{BaSO4 -> Ba^2+ + SO4^2-}$$
Now I am really confused on why as I am increasing more salt to water, the above reaction is creating more ions. So basically the more ionic the aqueous environment is, greater the generation of ions of Barium Sulphate. But isn't it bit anti-intuitive. Why the system is creating more ions (i.e. increasing K) when water already has increasing number of ions from table salt?

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Keep CH SE Q titles in plain text.

Search for activity coefficient and Debye-Hückel equation, e.g. on Wikipedia.

It's called "salting in".  Remember, pure water ionizes very little.

The $\ce{Ba^{2+}}$ ions will surround by anions
The $\ce{SO_4^{2-}}$ will be surrounded by cations

So they stabilize a lot more actually. Remember that this may take long, but thermodynamically that is probably a stable situation (i.e system lowers the energy).
But it does reach a limit and plateau at some point.
This is quite general as you can see.

You intuition is telling you that ions should suppress more ions from coming into solution and that is probably correct and were it the only thing happening it should be observed. In fact it is. Almost all ionic compounds have solubility limits. This is usually attributed to equal rates of dissolution and crystallization, but your intuitive explanation is probably just as good. Lets take a closer look at what is going on. The crystal has a lattice energy that must be supplied to dissolve it; the solvent has intermolecular attractions that must be disrupted, again requiring energy. [This is ignoring the entropy changes disrupting and rearranging order in both, in the end free energy changes are pertinent.] The energy requirements are supplied by the solvation of the ions; solvation releases energy. Ions entering solution can be solvated by solvent molecules, but would it not be nicer to have an ion of the opposite charge available to greet you? This is what is happening. This is sometimes attributed to an increase in dielectric constant but in the end it all comes down to the interaction of molecules and ions. Notice that the effect of ionic strength increase is much greater for the divalent ions of BaSO4 than the monovalent ions. I would not be surprised if the effect of ionic strength were not related to the crystal lattice energy. I have just given you a project. Report back.

This quickly got 3 downvotes Are there specific reasons? Please help me if I am violating some thing or if you think the chemistry is incorrect.

3 downvotes means that 3 individuals came to the conclusion that "This answer is not useful". If you are "violating some thing", the answer would probably be deleted. "Very low quality" is also a reason for deletion.

I did not down-vote, but I do have some comments. You could break up the text into paragraphs, one for each separate point or idea. You could focus on the effect of the presence of other ions. As you get to the core of the question, you switch from technical language to colloquial, "... nicer to have an ion of the opposite charge available to greet you?". The last three sentences don't seem to be part of an answer.

And the gibberish of the previous answer met any qualifications? "This is quite general as you can see"! I tried to answer in the language of the questioner and the last sentences are pertinent. The common approach is the effect of dielectric constant. the difference in the charge on the ions is paramount and the questioner should be encouraged to think more about the Chemistry. It's  called "salting  in"! that is used to make ice cream in the summer[after a hailstorm]

I was not comparing answers. I was giving some suggestions how you might improve this answer because your comment said "please help me". Usually, if I feel none of the answers fully address the questions, and if I find the question interesting, and if I have some time, I try to write a better answer myself rather than criticizing the other answers.

Um, you add halite to water and ice to decrease the freezing point of the water. It's not what "salting in" means.

Salting in  a highly technical term. It is usually used to decrease solubility of organic salts or organic compounds in water. that is discussed in my first three sentences. I was joking about the ice cream but if given the opportunity you should try it. I think my answer fully addressed the question. the final statements were to help the questioner to think more deeply about it, then maybe they might read about Debye -Huckel treatment.


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Is lead iodide a strong electrolyte?
So, lead iodide is insoluble. I see conflicting answers online. Some say it's a weak electrolyte because it is insoluble, others say it is a strong electrolyte because it is an ionic compound and any small amount that dissolves does so in the form of ions.
Which one is the correct answer?

Solubility and being strong/weak electrolyte are 2 independent properties.

Thank you for your answer Poutnik. So, is it a strong or weak electrolyte then?

https://en.wikipedia.org/wiki/Strong_electrolyte

From that Wikipedia article:
Strong acids, strong bases and soluble ionic salts that are not weak acids or weak bases are strong electrolytes.
It says soluble ionic salts..lead iodide is insoluble. I did my research, I just asked because I couldn't find a definitive answer!

A strong electrolyte is a solution/solute that completely, or almost completely, ionizes or dissociates in a solution. It says nothing about solubility.

Actually, since it mentions that soluble ionic salts are strong electrolytes it does refer to solubility.

Saying European cars are good does not say that American cars are not good. // Insoluble salts are not usually considered as electrolytes at all, but if they are, they usually are strong electrolytes. In fact, I cannot remember a case that would not be. It would mean both ions would hydrolyze to neutral weak acid and base.

First it has to be strongly soluble. I would argue it isn't (in water).

The definition uses the word "almost".  Educational treatments of insoluble salts assume ions are formed when we think they are tempered by different degrees of solvation or even complexation. Take the alkaline earth hydroxides barium is considered strong because it is soluble, no one cares about strontium, calcium is considered insoluble but strong because lime is used to dissolve bodies in murder mysteries, and we eat milk of magnesia to relieve dyspepsia. Teachers must stop making such definite statements without proof or data or at least rationales. Help the students think not memorize.

@OscarLanzi Then being strong/weak electrolyte would be an attribute of a solution, not of a compound.  Consider 1 mM PbI2 vs 1 mM H2SO4 vs 1 M H2SO4.

Lead(II) iodide is sparingly soluble in water at room temperature $(\pu{0.76 g/L}$ at $\pu{20 °C})$ and a bit more soluble in hot water $(\pu{4.1 g/L}$ at $\pu{100 °C}).$ In solution, all of the lead iodide is dissociated into ions. At high concentrations of iodide, the lead ions form two complexes, $\ce{PbI3-}$ and $\ce{PbI4^2-}$ [1]:

The increase of solubility of lead iodide caused by the presence of iodide ion in concentration greater than 0.1 molal may be explained by the formation of the complex ions $\ce{PbI3-}$ and $\ce{PbI4^=}.$

In any case, all of the dissolved lead iodide is present as ions (there is no neutral $\ce{PbI2(aq)}$ species), so it is a strong electrolyte. This is different from weak electrolytes such as, say, acetic acid, where there is a neutral species $\ce{CH3COOH(aq)}$ in solution which dissociates into the charged $\ce{CH3COO-(aq)}$ and $\ce{H+(aq)}$.
As an aside, the difference in color (colorless in solution, yellow as a solid) and the temperature-dependence of the solubility leads to a beautiful effect that has been called toxic golden rain, see e.g. video YouTube — Golden Rain - Growing crystals of lead iodide. Chemical reaction.
Reference

Lanford, O. E.; Kiehl, S. J. The Solubility of Lead Iodide in Solutions of Potassium Iodide-Complex Lead Iodide Ions. J. Am. Chem. Soc. 1941, 63 (3), 667–669. DOI: 10.1021/ja01848a010.

A devil advocate may object how there can be $\ce{Pb^2+(aq)}$, $\ce{[PbI3]-(aq)}$ and $\ce{[PbI4]^2-(aq)}$ without $\ce{PbI2}$.

@Poutnik You may ponder the opposite: Why do the complex ions not precipitate with a counter ion? The cite paper mentions double salts such as $\ce{RI.PbI2.H2O}$, and methylammonium lead iodide is an example.


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Database of atomic crystal structure data or images of rocks and minerals?
I am looking for crystal structure images so I can redraw them in a nice form. However, I don't seem to find much searching Google.

Perhaps this is a slightly better example (searching for Quartz instead of Obsidian):

Is there a database or place which lists all rock / mineral crystal structures? I don't know what they are supposed to look like (I did organic chemistry but not advanced inorganic chemistry or geology), so knowing that the database/place has the expected crystal structures would be good to know. Also I'd be curious to know if they have any non-perfect crystal structure images, like things with impurities or crystal structure images of realistic rock compositions.
If not, wondering what major resource I should learn to figure out how to draw them on my own at scale (meaning, what major textbook or resource lists at least the crystal structures in text, if not having images or data which can be used to render images, and so the resource has hundreds or thousands of such pieces of data for efficiently drawing many of them).

I suggest AMCSD for finding structures, and VESTA for drawing. VESTA is free, available for all major desktop systems and comes with nicely written extensive manual. Frankly, currently the question is a bit too broad and not entirely clear: "nice form" leaves too much to imagination and is subjective.

Thank you. I am not asking about the "nice form" part, that is what I am going to do, I am just asking about where to find any images or data. Will take a look at these.

Here are a couple in 3D allowing for interactive exploration: https://virtual-museum.soils.wisc.edu/displays/


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Reversible endothermic reaction
Is there an endothermic reaction that is reversible simple by adding heat?  For instance, can I put chemicals in an enclosed tube that will react to release heat, then I can add heat and reverse the reaction so that it will begin again?

Endothermic reactions are reversed by REMOVING heat. You may want to fix your question.

So you basically want a heat battery. Usually it's best to make use of phase changes (a physical process), but another chemical option which can be accessed in simple conditions is the equilibrium between nitrogen dioxide and dinitrogen tetroxide. I suppose the best possible heat battery will be represented by a process with a very high $\Delta S$, such that even a very large $\Delta H$ will still result in a process with $\Delta G$ close to zero in reasonable conditions (which, once again, sounds like a phase change near ambient conditions)

Very much not by coincidence, you are running into the idea of refrigeration (or heat pump) cycles. It's possible there are some niche applications of chemical reactions in refrigeration/heat pump cycles, so you could find an example there.

"put chemicals in a tube to release heat: that's exothermic!

Would crystallization qualify? Try a $\ce{NaCH3COO·3H2O}$ (sodium acetate) hand-warmer. Easily done, and comparatively safe as a demonstration in classroom, or even at home.
$\ce{Ca(OH)2 <=> CaO + H2O}$. This might be difficult to use as a demo, because high temperatures are involved. Though I could describe a borosilicate apparatus with two connected flasks, one containing the $\ce{Ca(OH)2}$ to be heated, the other at room temperature to capture evaporated water to be mixed back on cooling, construction of one robust enough to survive the heat involved might be tricky.
Easiest to demonstrate would be an electrochemical reaction, i.e., a secondary (storage) cell such as lead-acid or NiCd cells. Though lead is cheap, and produces enough voltage from a single cell to light a red LED, cleanup of the toxic metal and compounds formed are a downside.

Any reaction that reaches equilibrium is endothermic in one direction and exothermic in the other. This means that at equilibrium a change in heat content, with [change in equilibrium constant] or without [latent heat in a phase change] a change in temperature, will cause the reaction to establish a new equilibrium position.
Another approach is to isolate the oxidation and reduction halves of a reaction and make a cell. Now by applying an appropriate potential and adding energy the concentrations can be adjusted that the cell can supply electrical energy. This is a much more efficient method than changing the temperature. This approach is used daily by most of us to keep our cell phones operating.
The same effect should be possible by raising the temperature; unfortunately in raising the T the speed of side reactions is also increased; the phone will melt, the transistors will stop working, who knows what the lithium will do.


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Why do electrochemical sensors need reference electrodes
Here is my (limited) understanding of my question:
A pH probe consists of two half cells. One is a reference cell, constructed such that the EMF across the cell is stable and well defined.  In the case of an Ag/AgCl reference cell the silver wire is connected electrically to the negative terminal of a voltmeter, allowing the movement of electrons through the wire and eventually into the buffer solution in the reference half cell. The reference cell allows free movement of ions (current) without excess diffusion of buffer solution (commonly KCl).  Those electrons are part of a chemical reaction which turns AgCl into Ag ions and solute Cl (and vice versa). Therefore, on the other half of the liquid junction (our solution with unknown pH) there is some voltage present. This voltage is equal to the EMF of the reference half cell, which for Ag/AgCl I think is around 220mV.
This reference voltage is added to the EMF across the measurement probe (a measurement glass that is sensitive to pH according to the Nernst equation) and is approximately +/-420mV.

If this is the case, why is the measured output of the pH sensor not something like -200mV to +640mV?

If the pH glass creates it's own EMF why can't we just submerge a copper (or otherwise conductive) wire into the solution as our reference (ground)?

A practical example:
I have a commercial pH probe. I submerge it in some liquid with pH of 7 and I observe that the voltage across the cell is about 0mV as expected.  Now I place a copper wire which is electrically connected to the negative terminal of my electrometer and find that an offset voltage appears of approximately 220mV (perhaps by coincidence or as a ground loop - not yet convinced by the second).

Is it still possible to use the pH probe as normal by subtracting or otherwise calibrating out this offset?

Consider the case where you wish to remember a friend's address with a reference to a famous restaurant, but the restaurant magically keeps shifting its position without warning day by day! This would certainly be a disappointment. Such an unpredictable and constantly shifting "reference" may be useless for addressing purposes. The same holds true for copper electrodes or any electrode that isn't a reference electrode. They produce unpredictable EMF! What will happen when these metallic wires are dipped in solutions without the corresponding ions? As you may recall, Nernst's equation requires the ions to be of the same metal in contact with each other. If you want to use a copper electode, copper ions must be present in solution. For these reasons, you want a stable electrode, whose potential does not change from solution to solution, day to day and minute by minute.
Copper is quite reactive electrochemically and tends to form an oxide layer. Also note that pH measurement belongs the branch of electrochemistry called potentiometry. In potentially, ideally zero current is drawn during measurement.
BTW, nobody knows how the glass electrode works with full certainty. This is a very fancy system, if you think about it carefully. There is a thin glass membrane. The inside contains a known concentration of acid, and the outside is your test solution. In what way does the glass develop its potential difference? This has baffled electrochemists.

Yes - this makes things clearer.  I am still a little unclear on what potential develops across the reference cell.  As I understand, some potential comes from each Galvanic interface (silver wire to silver chloride, silver chloride to reference solution, etc). Why aren't those voltages superimposed - in other words: is the EMF from silver wire to unknown liquid (across liquid junction) zero?

And in a similar vein: my experimental setup shows large DC offsets when a grounded copper wire is placed. Are these offsets electrochemical or electrostatic?

One principle of electrochemistry is that no one knows the potential of a single electrode.   So we can never measure individual potentials developing across interfaces.

@EdV so in this case the copper wire (meant to provide a drain for static charge) should be galvanically isolated from the pH probe?

Yes but in practice (say in an industrial setting) the solution vessel cannot be floating for safety reasons.  I understand that copper is galvanically undesirable - but is it a problem if the grounding rod be made from stainless steel for example?

In other words, you propose replacing the reference electrode with its own solution with a logistically simpler "copper vs test solution" electrode.

Your copper electrode potential will depend on copper ions concentration in the solution. You know it, right?

Should there be no much of copper ions, you will contaminate the test solution with copper ions. Well, in some cases this is not of much importance. But the potential will vary a lot depending on the degree of contamination.

Your copper electrode potential will depend on both the pH AND the redox properties of the solution, in a complex interdependent way. Good luck knowing it with an acceptable accuracy.


text:
Reaction of hydrogen peroxide with acetic acid and potassium permanganate
I conducted an experiment where I mixed hydrogen peroxide $\ce{H2O2}$, acetic acid $\ce{CH3COOH}$ and potassium permanganate $\ce{KMnO4}$ under vigorous stirring. The reaction hasn't stopped after 15 min and the colour is cycling between pink and colourless whenever I add more potassium permanganate.
What reaction is going on? I think despite a medium created with weak acid, acidified potassium permanganate is a much stronger oxidising agent and it should react with a lot of peroxide rather than the opposite.
But hydrogen peroxide and acetic acid react to form peracetic acid $\ce{CH3CO3H}$ (PAA). I don’t know if PAA reacts with potassium permanganate or something entirely different occurs.
I am concerned about the reaction and products. The mix smells strongly of acetic acid and most of the permanganate is still in there. Could it be the manganese oxide formed in the reaction being oxidised? Well first of all no bubbles the limiting reagent of the reaction was hydrogen peroxide and I used aqueous potassium permanganate. And by most of the permanganate was still there I meant I added half a teaspoon of permanganate and I used less than 1% acetic acid and 6.5 percent peroxide. I was trying an acid catalysed reaction between and oxidiser permanganate and and reductant peroxide. When I drained the solution (by excess of water)a most of the amount added of permanganate was in the solution with manganese oxide indication it reacted with the peroxide.

Try to better explain what you have done. In which order have you mixed the three components ? And how much of each ? How many times ? Did you use solid permanganate or its solution ? In this case which concentration ? The same question for hydrogen peroxide : have you used $30$% or $3$% hydrogen peroxide ? Did you use pure acetic acid, or an aqueous solution ? Which concentration ? What is the meaning of "most of the permanganate was still in there" ? Please explain thoroughly all what you have done
!

Please be careful, peroxide and permanganate are strong oxidizers.  Acetic acid is potential fuel.  But what you may be doing is oxidizing the peroxide (with permanganate).  Did you see bubbles?

The limiting reagent was not defined. Acetic acid is a suitable solvent for permanganate or chromate oxidations at low temperatures but still should be used very carefully. This idea of mixing peroxides with a strong oxidzer and a possible reductant sounds like potential bomb making. This person is playing with chemicals not "conducting an experiment"

The Kinetics of the Acid-Catalyzed Formation of Peracetic Acid from Acetic Acid and Hydrogen Peroxide has been studied in details as early as 1965.

Well first of all no bubbles the limiting reagent of the reaction was hydrogen peroxide and I used aqueous potassium permanganate. And by most of the permanganate was still there I meant I added half a teaspoon of permanganate and I used less than 1% acetic acid and 6.5 percent peroxide. I was trying an acid catalysed reaction between and oxidiser permanganate and and reductant peroxide. When I drained the solution (by excess of water)a most of the  amount added of permanganate was in the solution with manganese oxide indication it reacted with the peroxide.


text:
Why is it not possible for water to directly attach itself to the alkene to form an alcohol in oxymercuration? What is the need for mercuric acetate?
When going through the preparation of alcohols, I came across the method of Oxymercuration-Demercuration.

Now, after going over the reaction, I didn't understand the purpose of the mercuric acetate in this reaction. According to me, H2O should be able to dissociate into H+ and OH- and then break the double bond and attach itself to the alkane. But, this does not happen because of some reason that I'm not aware of.
Image source: https://en.wikipedia.org/wiki/Oxymercuration_reaction

Your alkene would hardly dissolve in water, much less get protonated. One needs strong acid or other catalyst, like Hg(II).

It does not happen because you need a very much stronger acid. This is easily searched for using your favourite web-search tool

Alkene is a nucleophile, while water also is a nucleophile so they don't interact with each other directly, so we need something to alter alkene's property. Mercuric acetate forms a cyclic intermediate with 2 carbons of alkene where Hg has a positive charge, so it really needs that bonded electrons back. This makes the carbon site partially positive, thus it now welcomes water and oxygen donates its lone pair to the carbon while the mercury atom withdraws the electron, here the latter process is faster. Thus the process is SN2 where the transition state has a carbocation character, so now we get the intermediate that is shown in the diagram.


text:
Which Group 16 elements form stable dihalides?
My textbook says that "except oxygen, all elements of group 16 form dichlorides and dibromides", while other sources quote that only Oxygen and Sulphur form dihalides and dibromides.
Whereas, +2 Oxidation state being less stable for elements down the group, they should not be forming dihalides. So, I am unable to find concrete reason to support any of these claims, and want to ascertain which all chalcogens can actually form dihalides?

Why not to review chemistry of particular chalkogens? You would then see if a particular element forms them or not.

except oxygen, all elements of group 16 form dichlorides and dibromides

That's true because $\ce{Cl2O}$ and $\ce{Br2O}$ are not halides but rather oxides of chlorine and bromine respectively. Oxygen is more electronegative than chlorine and bromine so the compounds are oxide rather than halides.
All other group 16 elements form dichlorides and dibromides as expected.

It's not so simple, for example TeCl2 disproportionates https://en.wikipedia.org/wiki/Tritellurium_dichloride

@Mithoron If it disproportions, it must exist in the first place, doesn't it? ;)


text:
Confusion regarding classification of hydrides
According to my book,hydrides are classified as Ionic/Saltlike/Saline Hydrides, Covalent/Molecular Hydrides and Metallic/Non-stochiometric/Interstitial Hydrides.
The explanation of my book is not very clear to me  I am confused that if an S-block element & Hydrogen form a covalent compound, it will be Saline or Molecular Hydride? Same question for a P-block element which forms ionic bond with Hydrogen.
I will be grateful if you would help.

Hydrides like $\ce{NaH, CaH2}$ are ionic. Compounds or ions made by hydrogen and non-metals are covalent : $\ce{B2H6, BH4^-, CH4, SiH4, NH3, PH3, AsH3, H2O, OH-, H2S, H2Te, HF, HF2^-, HCl, HBr, HI}$. Even coordinate bonds in $\ce{AlH4^- }$ are covalent. Interstitial hydrides are relatively rare, like $\ce{H2}$  dissolved into palladium.

@Maurice - those of us coming from the materials side are more familiar with interstitial hydrides, since that is how most metal hydrides (solid phases) form. Just a matter of perspective...

Hydrides of alkaline/alkaline earth elements are almost* always ionic because the cations are very electropositive forming a strong ionic bond. Hence, they are called ionic/salt like hydrides
The hydrides of p-block elements are not very strong and are non-ionic, hence they are considered covalent hydrides.
*The key word almost is because beryllium hydride is the only s-block hydride which is covalent. $\ce{BeH2}$ is also the only known s-block hydride which can't be made by direction reaction of beryllium and hydrogen  (unlike other s-block hydrides which are made by direct metal and hydrogen interaction)
Previous discussion: Is lithium hydride a salt?

But in S block, BeH2 is a covalent compound and in P block, I think GaH3 is an ionic compound. So BeH2 & GaH3 will be Saline/Saltlike or Molecular Hydrides?

@NipunKulshreshtha I was about to update on BeH2. I will update the answer. But GaH3 is unfortunately covalent. It has been detected as a transient species in the gas phase at very low temperature. However, it is stable in its dimeric form of formula $\ce{H2Ga(H2)GaH2}$ which points towards its covalency and polymerism.

@NilayGhosh aww come on, what is so unfortunate about gallium hydride being covalent?

@OscarLanzi OP assumed gallium hydride was ionic/salt like :)


text:
What is the sample density used for a buoyancy mass correction?
For a lab exercise to calibrate volumetric glassware I was given the follow equation to correct weighings for the density of air:
$$m_\mathrm{c} = m_\mathrm{s}\,\frac{1 - \rho_\mathrm{a}/\rho_\mathrm{c}}{1 - \rho_\mathrm{a}/\rho_\mathrm{s}},$$
where:

$m_\mathrm{c}$ is the corrected mass of the sample;
$m_\mathrm{s}$ is the measured mass of the sample tared for the container;
$\rho_\mathrm{a}$ is the density of air;
$\rho_\mathrm{c}$ is the density of weights used to calibrate the balance, assumed brass;
$\rho_\mathrm{s}$ is the density of the sample.

Does $\rho_\mathrm{s}$ include the container for the sample? How is $\rho_\mathrm{s}$ determined without knowing $m_\mathrm{c}$?
If the tared container does not affect the calculation, $\rho_\mathrm{s}$ for water is straightforward.  What about other material?

You can determine mass and volume of the sample without using the buoyancy correction. There will be a secondary error (small error in density affecting the magnitude of the buoyancy correction), but it will be much smaller than the primary error (neglecting buoyancy difference between calibration standards and sample). If in doubt, you can recalculate the density with the correction and apply to the correction an iterative manner. It should converge in a single step, I bet.

If the container is tared, it does not figure in the measured mass. So, tare the container, and use the same one for the brass weights and your volumetric sample.

@Karsten also to consider is the volume of the standard weight vs sample volume as this is the difference of air mass displaced.  Brass would be around 8x the density of water.

I wonder how all this compares to the %RSD of meniscus reading errors.


text:
Why is silver considered a "type 1" metal although it has multiple oxidation states?
I noticed that the metals I learned were "type 1" back in early chemistry class as they have one oxidation state including silver, aluminum, zinc, and cadmium. However, those metals weren't reported as "type 1" when I looked online. I noticed that although silver has three oxidative states, it is still listed as "type 1" metal. Does anyone know why we still categorize it as "type 1" or have an insight into what is still considered a "type 1" metal?

I just looked it up: Type 1 has a single oxidation state, type 2 has multiple. If you look hard enough, many metals have more oxidation states than you learn in intro.

Furthermore, the main oxidation number of silver is +1. Compounds with higher oxidation number are rare, not very stable, and difficult to obtain.

What is this classification? I haven't heard of this. Can you provide a brief explanation of this classification basis?

I haven't heard of this classification of metals before. Anyways, the below image is from a nomenclature handout where silver has been considered "type 2" metal:

It's true that silver has oxidation states other than +1 but those are rarely encountered. So, it can get confusing and most of time, we may tend to consider that silver has only one oxidation state leading to it falling into "type 1" metal.

I am upvoting, since you took the time to post a valid answer, but I am now going to deliberately forget about ‘type 1’ and ‘type 2’ because they serve no needful purpose now, if they ever did at all.

@EdV I found the source: Zumdahl's Introductory Chemistry: A Foundation , Fifth Edition. Here is a web version of one of the tables.


text:
Smell through Plastic
Can smell go through plastic bags/gloves?
If it can, does the things in the bag ACTUALLY touch you?
Like cat poop, if it's in the bag my hands starts smelling but I don't feel it on my hands or see it on my hands. So is the cat poop on my hands or its just the smell of cat poop?

Smell can go through, especially after prolonged time. The poop does not touch you, unless there are holes. The smell on your hands can be just imagination as you know what is inside.


text:
Aluminum nitride hydrolysis
I wanted to ask some reference about aluminum nitride hydrolysis in water.
Specifically I would like to know if there is the formation of radicals or of some other reactive species.

Aluminum nitride $\ce{AlN}$ is made of one ion $\ce{Al^{3+}}$ and one nitride ion $\ce{N^{3-}}$. Both react with $\ce{H2O}$. The reaction is $\ce{AlN + 3 H2O -> Al(OH)3 + NH3}$. What sort of radicals do you want to introduce here ?

To avoid duplicities, it would help responders if you wrote what you had been already searching for and where you had been already looking for it.

@Maurice I have learned to take nothing for granted when it comes to chemistry. Many phenomena can be more complex than they appear to be.

I am quoting from this study[1]:

$\ce{AlN}$ powders hydrolyze in moist air at room temperature, resulting in degradation of the powders. The initial hydrolysis product is amorphous $\ce{AlOOH}$, which is further converted to a mixture of polymorphs of $\ce{Al(OH)3}$ (bayerite, nordstrandite, and gibbsite), forming agglomerates around the unreacted $\ce{AlN}$ core. In the hydrolysis each powder shows an induction period, which is attributed to slow hydrolysis of the surface oxide/oxyhydroxide layer. The powders produced by the carbothermal process show the longest induction periods.
$$
\begin{align}
\ce{AlN + 2H2O → AlOOH_{amorph} + NH3}   & \tag{R1}\\
\ce{AlOOH_{amorph} + H2O → Al(OH)3} & \tag{R2}\\
\\\hline
\ce{AlN + 3H2O → Al(OH)3 + NH3}\\
\end{align}
$$

You can find a detail explanation of the reaction in that study.
Reference:

Jinwang Li, Masaru Nakamura, Takashi Shirai, Koji Matsumaru, Chanel Ishizaki and Kozo Ishizaki, Hydrolysis of Aluminum Nitride Powders in Moist Air, 2005, DOI:  10.2240/azojomo0111


text:
What does "bulk concentration" mean?
In biogeochemistry/ecosystem models, the concentration of phytoplankton in sea water is often termed "bulk concentration" with units mol P/m^3 (moles of phosphorus per cubic meter, assuming that we know how the stoichiometry of the organic composition of phytoplankton, which is often the case in models). The unit makes sense for "normal" concentrations. What does "bulk" mean in this case?

That depends on the context which is known within biogeochemistry/ecosystem models. The unit makes sense for any concentration. By the explicit BULK they may mean in the whole water volume.

So in the model output, there are ion concentrations also in mol X per cubic meter (e.g. carbonate concentration, phosphate concentration, nitrate concentrations...) - and bulk phytoplankton concentration.

@ouranos, You can share the original text or a portion of so that readers can understand the context. Bulk concentration has many connotations. In electrochemistry it has a different meaning.

At a guess, they mean average concentration throughout the whole sample. The critters in the water would be specks of high-concentration phosphorus, and the liquid lower (unless it came from farm effluent).


text:
Why does CaSO4 break up when adding HCl but not BaSO4?
I observed in a solution with CaSO4 that the white precipitate dissolved when I added concentrated hydro chloric acid (37%). I assume CaSO4 had a divorce and Ca decided to hook up with two Cl- ions, forming CaCl2.
Having access to the big sister of Ca, Barium, I tried to stress test the marriage of BaSO4 in a similar way by enticing Ba to hook up with Cl- ions. But it seems that Barium is a very committed partner, and the precipitate could not be dissolved.
My question is: why is this difference observed, and what do I need to do to obtain the same result with Ba?
I saw a similar question on this forum with Ca (CO3), HCl and H2SO4 that were answered by mentioning an equilibrium and shifting the reaction to one side.
So I assume the general answer to my question could be that it is easier to reverse the reaction with Ca than with Ba. If that is the case, would vigorous shaking + high temperature + high concentration of H+ ions help to convince Ba to consider Cl- ions instead?
Or is BaSO4 one of those "game-over" reactions, there is no going back once you get them.
Thank you very much for your time

Sulphates of alkali earth metals dissolve like this:
$$\ce{MSO4(s) <=>[H2O] M^2+(aq) + SO4^2-(aq)}\tag{1}$$
with the solubility product constant (kept original eq. numbering):
$$K_\mathrm{sp} = a(\ce{M^2+})a(\ce{SO4^2-}),\tag{3}$$
where $a(\ce{X})$ is thermodynamic activity of the ion $\ce{X}$. For diluted solutions is can be approximated by molar concentrations (denoted as []) as
$$K_\mathrm{sp} = [\ce{M^2+}][\ce{SO4^2-}]\tag{4}.$$

$K_\mathrm{sp} \gt [\ce{M^2+}][\ce{SO4^2-}] \implies \text{net dissolution}$
$K_\mathrm{sp} \lt [\ce{M^2+}][\ce{SO4^2-}] \implies \text{net precipitation}$

As a hydrogensulphate anion is not strong acid ($\mathrm{p}K_\mathrm{a}=1.99$) , concentrated hydrochloric acid protonizes in large extent sulphate to hydrogensulphate.
$$\ce{SO4^2-(aq) + H+(aq)<=>>[conc. HCl] HSO4-(aq)}\tag{2}$$
Decreasing of sulphate concentration disbalances dissolution equilibrium and there is ongoing net dissolution until the equilibrium is reached again.
$\ce{CaSO4}$ is slightly soluble ($\pu{0.26 g/100 ml}$ at $\pu{25 ^{\circ}C}$ (dihydrate)), therefore well soluble in $\ce{HCl}$.

Solubility of $\ce{BaSO4}$ is much less, ($\pu{0.2448 mg/100 mL}$ at $\pu{20 ^{\circ}C}$ ) so it needs (hot) concentrated $\ce{H2SO4}$.
Concentrated $\ce{H2SO4}$ acts here as the solvent. $\pu{2-4\%}$ of $\ce{H2O}$ is practically all converted to $\ce{H3O+(solv)}$. Additionally, $\ce{H2SO4}$ partially autodissociates.
\begin{align}
\ce{H2SO4(l) + H2O &-> HSO4-(solv) + H3O+(solv)}\tag{5}\\
\ce{2 H2SO4(l) &<=> HSO4-(solv) + H3SO4+(solv)}\tag{6}
\end{align}
This causes extremely high activity of solvated $\ce{H+}$, extremely low activity of $\ce{SO4^2-(solv)}$,
\begin{align}
\ce{SO4^2-(solv) + H3O+(solv) &-> HSO4-(solv) + H2O(solv)}\tag{7}\\
\ce{SO4^2-(solv) + H2SO4(l) &-> 2 HSO4-(solv)}\tag{8}\\
\ce{SO4^2-(solv) + H3SO4+(solv) &-> HSO4-(solv) + H2SO4(l)}\tag{9}
\end{align}
leading to relatively high $\ce{BaSO4(s)}$ solubility.
It can be also described as
$$\ce{BaSO4(s) + H2SO4(l) <=>[H2SO4] Ba(HSO4)2(solv)},$$
in a way analogical to:
$$\ce{CaCO3(s) + H2O(l) + CO2(aq)) <=> Ca(HCO3)2(aq)}$$

Thank you Dear Poutnik for the answer! I think I understand your response, apart from the very last, regarding conc. H2SO4. I interpret Eq(1) that by adding water I can "help" to break up M from SO4-2 ions, which works for those MSO4 species that are not so fond of each other in the first place. I interpret Eq(2) that a strong Acid could help to "push" SO4 to break up from M, and bind to a hydrogen instead. Do you mean that BaSO4 does not form larger crystals (and precip.) in conc H2SO4? But how to break them up in the first place? Adding more H2SO4 seems to worsen the problem (accord. to Eq1)

I have updated the answer. I hope it would help.


text:
Suitable solvents to dissolve Aluminum Chloride (anhydrous) without generating Hydrochloric Acid (aqueous or gas)
Hypothetically, I have a chemical tank that was brought out of service and the solid deposits at the bottom of the tank primarily contain aluminum chloride. The whole system never sees water, and the aluminum chloride is generated upstream in the process as a by-product (anhydrous under tank conditions).
It's time to clean this solid out, and water flushing seems to be an easy solution. However, previous experience tells us, extensive heat and HCl (aq) and HCl (g) generation during the flooding process, corrodes the tank material and the welded joints, inner pressure rings, etc.
I want to use an alternative solvent to dissolve the aluminum chloride without generating HCl. Ideally, I would like to clean out the tank afterward with water, to get rid of the solvent and remaining solids. My first initial screen takes me toward polar aprotic solvents with high vapor pressure.
What does this community think?
Edit: After some comments, I think I should clarify. There are some other factors, such as solvent toxicity, carcinogenicity, $$ cost (tank is quite large), safe disposal of solvent at that scale, getting rid of the solvent completely from the tank, etc. might reduce the chances for some candidates.

Carbon tetrachloride $\ce{CCl4}$ and also (to a smaller degree) toluene $\ce{C6H5CH3}$ are solvents of $\ce{AlCl3}$. Toluene can be used to produce metallic aluminum by electrolysis, although the yield is low.

Hypothetically, the OP should not use $\ce{CCl4}$ because it is a carcinogen, and it sounds like the hypothetical tank is quite big.

Yes, that is correct. Both CCl4 and Toluene are carcinogens. Toluene also has low electrical conductivity, so develops static charge. This provides potential ignition sources for flammable residual chemicals in the AlCl3 solids.

Indeed the tank is quite large, and the cost of solvent, disposal, removing all solvents from the tank, etc. becomes limiting.

Several common organic solvents will dissolve this compound which has a lot of covalent character.
From Wikipedia:

Soluble in hydrogen chloride, ethanol, chloroform, carbon tetrachloride
Slightly soluble in benzene

Comments identify carbon tetrachloride as a carcinogen, ethanolhas nucleophilic oxygen and thus may react, and benzene is also carcinogenic. Chloroform might work.

Of course, chloroform also is toxic, so perhaps ethanol, despite chance of reaction, might be best. Perhaps a first cleaning with 99% C2H5OH, cycling the ethanol through a filter filled with an alkali, such as CaOH, to remove any evolved acid? Final rinse with water/ethanol (or cheap wine ;-) to remove remainder?

@DrMoishePippik This paper might be relevant – https://doi.org/10.1016/S0141-3910(96)00201-7 – which details the reaction of Al(Cl)3 with EtOH, and reports the formation of AlCl(OEt)2 and similar compounds but also (toxic) ethyl chloride, predominantly so if the temperature rises. There are lots of bad examples in chemical engineering of bad things happening when the temperature rises. I would suggest getting some of the suggested solvents and running a series of lab tests and in particular paying attention to the enthalpy changes going on...

If there is an industrial application, costs including disposal of solvent will be important.
A possible solution, if the material cannot be scraped or shoveled out, might be to dissolve it with slower reacting alcohol, rather than water.
A strong nitrogen purge into the tank and out through through an acid neutralizing trap should take care of the HCl vapor and reduce flammability hazards from solvent fumes.
Treatment of the material once out of the tank could include recovery of the alcohol through vacuum distillation.
It would be recommended to try this out on a small scale first.  IPA might be slower reacting (better) solvent than ethanol.

In some circles your last sentence suggests using Indian pale ale .


text:
Polymer purification by reprecipitation troubleshooting
I am synthesizing new polymer that has not been reported before. The structure of the polymer is based on indacenodithienothiophene unit and thiophene. I used Stille polymerization. As monomers, I used distannyl-indacenodithienothiophene unit and 2,5-dibromothiophene. 2,5-dibromothiophene has bulky tertiary amine group on 3-position. Basically I use toluene:DMF (4:1) as solvent and set temperature to 120 ℃.
I have tried:

4 mol% of $\ce{Pd[P(tBu)_3]_2}$, 72 h
4 mol% of $\ce{Pd_2(dba)_3}$, 16 mol% of $\ce{P(o-tol)_3}$, 72 h
10 mol% of $\ce{Pd_2(dba)_3}$, 40 mol% of $\ce{P(o-tol)_3}$, 72h
10 mol% of $\ce{Pd_2(dba)_3}$, 40 mol% of tBuXPhos, 72h

After the reaction is complete, first I dilute the reaction mixture with dichloromethane, and filter it through Celite pad to remove decomposed Pd residue. And then I concentrate it by rotovap (in this procedure, of course DMF is not evaporated well). Add minimal amount of dichloromethane, and add drop by drop to the 200~300 mL of methanol. Then methanol becomes cloudy and most of the grains are sticked to the wall of the flask.

When I search for this problem, I found that some people say that the cloudy methanol means insufficient polymerization, so there are many oligomers or low-weight compounds. Do the sticking to the wall occur by same reason? If I try to filter the cloudy methanol by reduced pressure filtration, then the filter paper is clogged like this:

Should I try longer reaction time, like 96 h or 120 h? Also, our lab doesn't have the Schlenk line, so it cannot afford complete inert atmosphere. Then, if I consider the decomposition of the Pd catalyst by air, does the insertion of additional Pd catalyst (and ligand) during the reaction help?

You should collect analytical data on your product in order to decide how to proceed.

@Snijderfrey What data? Like Mn or Mw of the polymer? But I can't get that value because I have trouble in recovery of the polymer.

Is this your PhD you are working on ? While it feels like I am doing your homework I cannot resist chiming in.
I worry that if you have oxygen contamination of the mixture which causes the palladium chemistry to be terminated then the chemical reaction which terminates the palladium could then form some organic product other than the one you want.
You need to consider which point in the catalytic cycle the palladium is most susceptible to poisoning with oxygen. I think that the Pd(0) complex which is a POSH version of [Pd(PPh3)4] will be the resting state of the catalyst, if this decomposes either to a palladium mirror or to a palladium(II) complex like [PdBr2(PPh3)2] then this will just result in loss of activity of the catalyst. If the oxygen does something at a different stage in the cycle it could be worse, then the oxygen could cause the formation of a different product.
I know from Monsanto acetic acid process chemistry that the RDS for that reaction is the reaction of methyl iodide with the Rh(I) complex, while with BP’s Ir based process the oxidative addition is not the RDS. There the more heavy Ir as it has a stronger C-M bond has different chemistry. I worry that a Pt catalyst would be a lot more expensive and it would also be more sluggish.
I suspect that adding more of the palladium catalyst will not do any harm as the poisoning of the system by oxygen or the formation of the palladium mirror which I used to have sometimes on glass flasks will not do too much harm.
I note that the colour is similar to many palladium compounds, I would be inclined to use thiol-SAMMS (a mesoporous silica with a monolayer of mercapto propyl groups) to remove the palladium which is still in solution from the solution of the polymer.
I think that you need to rethink the polymerisation reaction, I would want to keep the polymer in solution as long as possible during the polymerisation to stop it coming out of solution before it has fully grown. I would want to try to do the polymerisation in pure DMF in the sincere hope that the product would be more soluble in the DMF.
I would suggest that you take great care with DMF, I hate the stuff. It is viscous and toxic. I say it is an equal opportunities monster. I know that it is linked to testicular cancer in men and I am sure it does horrible things to women. I know it is a teratogen so keep away from it if you are pregnant. Some chemical companies will not allow any women to work with it. This is not a totally unreasonable thing, I know that some nuclear companies keep women under an even stricter set of rules for radiation than they do for XY chromosome owners like me.
I think that you should try to get a nitrogen / vacuum line to do this work, the problem is that you are making a polymer. If you are trying to make a long chain polymer then if a small fraction of the repeat groups are changed then you will have a very different chain length and the properties will be very different.
I would suggest sparging the mixture with argon before you add the catalyst to get the oxygen content down as low as possible before you heat the mixture up. One option would be to bubble the thing with argon at room temperature for an hour or so, add the catalyst, carry on bubbling with the argon and then start to heat up.
After the reaction time, I would allow the mixture to cool and then I would tip it into water. The sincere hope is that the majority of the DMF will remain in the water while the polymer will not. I would then redissolve in dichloromethane and then use methanol as an antisolvent.
You might also want to consider using electrochemistry as an alternative to the Pd chemistry, I think it might be possible to make a polythiophene by gentle oxidation on the anode. I suggest you read about the work of Karl Ryder, he has got into deep eutectic solvents now but when he was in Aberdeen around 2000 his group there were making things like polythiophenes.


text:
What do you think is happening to aluminium in salt water with rusted copper?
I put some aluminium foil and rusted copper(namely a Japanese 10 yen coin) in saturated NaCl solution.
After some minutes tiny amount of small bubbles were observed on the surface of the foil.
After several hours there appeared white cloudy thing at the bottom of the container.
I think this cloudy thing is aluminium hydroxide, but how could this possible?
In textbook it is written that aluminium does not react in neutral environment.

Does the book begin and end with this statement, or does it say something else about Al?

Also, look up electrochemistry. BTW, "rust" is usually used for the red oxide of iron exclusively, while "corrosion" is a more general term, applied to other metals, such as copper, as well as to iron. "Tarnish" is a surface coating on Ag, which can also be removed with the above reaction.

@DrMoishePippik Thank you for letting me know definition of rust ,corrosion, and tarnish. I didn't know these words well.  For electrochemistry, I know Al is more likely to be ionized than Cu thanks to the fact that standard electrode potential is lower in Al. But in this book it is written that Al is coated with Al2O3 and this barrier is strong and prevents further corrosion.

Ah! @satorukurita, that does show understanding of the question, and belongs in it (i.e., edit the question). And you (and the text) are correct, as long as the Al2O3 coating remains intact. If it is physically damaged, or Na+ ions react with it (forming a non-adherent aluminate), then the Al can react.

@DrMoishePippik Thank you for explaining why Al2O3 might not be a perfect coating of Al.I thought  both theory are plausible in this case.

The reaction is with water and the H3O+ and OH- ions present to attack the very thin oxide coating not Na+. Ions just hasten the process by enabling local galvanic cells Contact with copper will help things along. The bubbles on the Al are a bit problematical.  Either air is coming out of solution or H2 is being formed locally and the contact with the copper is poor. The oxide coating on Al from exposure to air and water is not very protective because it is hydrated and not uniform. Al will corrode rapidly in salt water especially when in contact with a more noble metal.


text:
Comparing melting point of neopentane and butanol
Why does something like 2,2-dimethylpropane have a much higher melting point than butan-1-ol?
I know that butan-1-ol has a hydroxyl group, and is capable of hydrogen bonding, and I know that 2,2-dimethylpropane is a compact symmetrical molecule, but I don't really see how shape and compatibility have such a huge factor on melting point compared to hydrogen bonding; or is this a special case?

Number of carbons is important too

Boiling points are subject to simple trends, but melting points are tricky.

https://chemistry.stackexchange.com/questions/65933/why-is-the-melting-point-of-tert-butyl-alcohol-140-c-higher-than-that-of-sec-bu https://chemistry.stackexchange.com/questions/92144/why-does-neopentane-have-a-higher-melting-point-than-n-pentane

You are simply comparing the wrong compounds to see the effect of a hydroxyl group. Compare tertiary butyl alcohol to neopentane. While you are at it compare melting and  boiling points of the isomeric pentanes and butanols to get a general idea of the effects of molecular shapes.


text:
Production of hydroxide ions in reactions occuring in alkaline medium
Question similar to this one but with a more precise point. For this reaction in neutral or faintly alkaline medium: (source page 20) $$\ce{2MnO4- + H2O + I- -> 2MnO2 +2OH- +IO3-}$$
If the reaction is occuring in a basic medium,  new $\ce{OH-}$ ions are produced in the reaction, then by the Le Chatelier's principle the reaction will shift backwards and would be less favourable. Also, this is in contrast to reactions in acidic mediums involving consumption of $\ce{H+}$ ions present in bulk.
So,

Why are $\ce{OH-}$ ions are being produced in this reaction?

$\ce{MnO2}$ as the product of $\ce{MnO4-}$ reduction is generaly formed in mildly acidic, neutral or mildly alkalic pH. Strongly alkalic pH prefers $\ce{MnO4^2-}$.

This is not an equilibrium reaction. It just goes to quantitative completion. In other words, the equilibrium constant of this reaction is exceedingly large. The reaction as shown below simply shows
$$\ce{2MnO4- + H2O + I- -> 2MnO2 +2OH- +IO3-}$$
that potassium permanganate can oxidize iodide ion in water as a solvent without pH adjustment. As a result, hydroxide ions are generated. If hydroxide ions were being consumed, they would be present on the left hand side. So this reaction has not been carried out in basic medium.

Since MnO2 is formed rather than reduction to Manganous ion it implies the reaction was run in neutral to basic conditions. H2O is still acidic in mild basic solutions.  The oxidation to Iodic acid is more facile in basic solution by almost one volt and the difference from iodate to periodate is about the same about -0.6v in both basic and acidic solutions. Perhaps the basic conditions help stop the reaction at iodate.

"H2O is still acidic in mild basic solutions." What does this mean?

It means that water is an acid in basic solutions. Peroxide reductions require a source of hydrogen ions, that source is water or hydrogen peroxide.


text:
Is resonance possible with consecutive double bonds?
Resonance is delocalisation of pi electrons among parallel unhybridized p orbitals,  so if we consider a molecules with 4 carbon chain with all bonds between carbon being doubly bonded then not all the unhybridized p orbitals would be parallel to each other ,
So in such a case is it possible for the molecule to have resonance,  like in Butatriene  I think there would be no resonance ? Is it right or am I studying something very very off.

Questions that consist merely just of questions without a priori elaboration are usually closed on most sites of the StackExchange network.

You'll find your answer by reading about allene.


text:
Charge acquired by sols after their preparation

This table from my textbook (p. 20), shows nature of charge on particles of listed sols, in their original or natural form. What reasons can be attributed to oxides (sols) being positively charged? Any general conclusions that can be made for their preparation or occurence?

 (Supplementary) Reasons I could find out by myself to some of these:
Positively charged sols

Hydrated metallic oxides are formed by hydrolysing metal halides, the oxides so formed preferentially adsorb metal ions present in the solution making the sol positively charged, for example $$\ce{FeCl3 + 3H2O ->[Hydrolysis] Fe(OH)3(sol) + 3HCl}$$
Here, $\ce{Fe^3+}$ ions present in the solution are preferentially adsorbed by the colloidal particles.
Basic dyes are cationic dyes, so their particles are positively charged.
Oxides - ?

Negatively charged sols

Colloidal sols of metals such as gold, silver, platinum etc. can be prepared by Bredig's arc method. The electrons disperesed in the medium are captured by the sol particles making the sol negatively charged.
Metallic sulfides are prepared by treating Metallic oxides with hydrogen sulfide, for example $$\ce{As2O3 + 3H2S ->[Double decomposition] As2S3(sol) + 3H2O}$$
The $\ce{S^2-}$ ions present in the solution are preferentially adsorbed by the sol particles making it negatively charged.
Acidic dyes are anionic dyes, so their particles are negatively charged.

(Additionally)
It will be appreciated if anyone can give idea on the case of gum, clay and charcoal too.

The net charge of sol particles depends on ion being chemically released or captured( like for weak organic acids or bases) or ions selectively adsorbed due chemical affinity. E.g. sol  particles of AgCl would be positively charged with Ag+ excess and negatively charged with Cl- excess.

@Poutnik yes I am aware of that. But how could we here judge what kind of charge a sol will acquire on its natural occurence or production? For metal sols I know they are made from Bredig arc method, hence I can conclude from there (due to electron capture) that they are negatively charged sols. But I could not shape such a general idea for other kind of sols. For example, how would one justify that naturally occuring oxides are positively charged sols? (Please suggest how I can make the question more clear if it's missing the point.)

It may be a too broad and complex question, with proper answer in a scope of a long article or a short book. Consider also removal of some cases, not representing a single or well defined sol particle type, like blood.

"The surface of red cells carry a negative charge due to the ionization of the carboxyl group of NeuNac (N-acetyl neuraminic acid), also called NANA or sialic acid. In saline, red cells will attract positively charged Na+, and an ionic cloud will form around each cell." in page on zeta potential.

More on the zeta potential, its origin and consequences.

The original, and fairly limited question, is:
"What reasons can be attributed to oxides (sols) being positively charged?"
Disregarding the method of preparation, one answer is that these metal oxides are more basic than H2O, and therefore will adsorb H$^+$ ions from a neutral (pH =7) H2O medium. The pH will rise, but it doesn't take many protons to saturate a surface, so the rise may be only about 1 or possibly 2 pH units.
The example given in the question:
FeCl3 + 3H2O −→ Fe(OH)3(sol) + 3HCl
could well result in adsorption of metal ions, but surely the production of hydrogen ions will result in a great excess of H$^+$ and adsorption of H$^+$also. FeCl3 solution is quite acidic.
On the other hand, metals, metallic sulfides, clays, etc. could be more Lewis acidic, i.e., attractive toward the oxygen atom in water. Once adsorbed, the water molecule would be more likely to release a hydrogen ion than water would, resulting in a negative charge on the solid and an increase in acidity - again, perhaps by only 1 or 2 pH units.
The bottom line is that the solvent (water) and its initial and final pH are a serious factor in determining the charge and its magnitude on suspended particles. This is not to minimize the effect of adsorption of metal ions on suspended particles when the metal ions are in significant concentrations, but these ions also have the much more plentiful H2O molecules to interact with.


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Why aluminum does not form ammonia complex ion?
I learned in my textbook that zinc forms more stable complex ion with NH3 than OH- because NH3 has only one lone pair and that makes repulsion between d orbital electron in zinc and lone pair electron in ligand smaller in NH3 complex ion than in OH- complex ion.
But I also learned that Al forms only OH- complex ion. I know Al doesn't have d orbital electron but why it can't make NH3 complex ion? Does this mean more than 1 lone pairs are involved in ligand bonding in Al(OH)4- complex ion?

Aluminium ions have strong hydrolysing tendency and aluminium has generally affinity to oxygen.

You can check: https://chemistry.stackexchange.com/questions/138924/can-aluminium-form-blue-colored-solution-in-liquid-ammonia or this sciencemadness link: http://www.sciencemadness.org/talk/viewthread.php?tid=10977

@Poutnik If it is not bothering I would appreciate it if you could explain why aluminium ions have strong hydrolysing tendency.

@NilayGhosh The paper you mentioned is interesting. I didn't know the existence of free e- (NH3)n  and the fact that it is blue in some kind of solution.

@satorukurita Generally, hydrolyzing tendency progressively grows with the ion charge. All $\ce{M^3+}$ and $\ce{M^4+}$ have strong hydrolysing tendency. $\ce{Ce^4+}$ is known to exist as a dimer with oxide or hydroxide  bridges.

related https://chemistry.stackexchange.com/questions/167995/aluminum-nitride-hydrolysis

@Mithoron thank you for explaining about hydrolyzing tendency.  So, is it okay to say that since Al3+ makes strong bond between OH⁻, it is difficult to  make complex ion with NH3? But I wonder how they can make strong bond with OH- not with NH3.

@satorukurita With high  intensity of electrostatic field around Al^3+, it is a strong advantage for OH- to have negative charge, compared to neutral NH3. Furthermore, there is possibility to form hydroxide and oxide bridges. Similar for Fe^3+.

@Poutnik Your explanation is clear. So can I say ,coulomb force is greater between OH- and Al3+ than that of NH3 and Al3+ because O is more negatively charged. And this makes it impossible for Al3+ to make NH3 complex ion?

There could be also said that coordinated H2O molecules around Al^3+ and Fe^3+ have due the strong field high tendency to release H+, becoming OH-, what starts to happen in mildly acidic pH, where is practically no free NH3.

The whole problem comes from the high charge of $\ce{Al^{3+}}$ ion : $+3$. As such it attracts strongly $\ce{H2O}$ molecules from the solution. But the central Oxygen atom of these $\ce{H2O}$ molecules is more strongly attracted than the $\ce{H}$ atoms, due to the high electronegativity of $\ce{O}$ atoms.  Hydrogen atoms from water are not so strongly attracted and they are even repelled by $\ce{Al^{3+}}$, due to the repulsion of the $+3$ charge on $\ce{Al^{3+}}$  and the partial positive charge on $\ce{H}$ atom. As a consequence, the bond $\ce{O-H}$ which exists in water and in the structure $\ce{Al···O-H}$ will be broken between $\ce{O}$ and $\ce{H}$  when water is approaching $\ce{Al^{3+}}$. And this rupture is leaving a new bond $\ce{Al^{3+}- O^{-}}$ which is charged $2+$ and repels the $\ce{H^+}$ ion into the solution. Of course the triply charged $\ce{Al^{3+}}$ is stronger than $\ce{H^+}$ to hold the intermediate Oxygen atom. Matter of fact, all solutions of triply charged ions like $\ce{Al^{3+}}$ (or $\ce{Fe^{3+}}$, etc.) are hydrolyzed and produce a strongly acidic aqueous solution. As a consequence, if ammonia is introduced into such an acidic solution, it will immediately be neutralized and transformed into ammonium ions $\ce{NH4^+}$ which cannot make a complex, for lack of available doublet. This is the reason why $\ce{Al^{3+}}$ (and $\ce{Fe^{3+}}$) ions cannot make complexes with ammonia.


text:
Why is the Gibbs free energy of titanium(IV) chloride reduction to titanium metal positive?
I use the Ellingham diagram to find the Gibbs free energy at the equilibrium state to find the partial pressure of chlorine. I have to use the Van 't Hoff reaction isotherm and the partial pressure of chlorine to determine whether it is thermodynamically feasible to reduce the titanium contained in titanium tetrachloride to form titanium metal at $\pu{800 °C}$ in an atmosphere with a chlorine partial pressure of $\pu{1E-20 atm}$ (assume an otherwise inert atmosphere).
I solved
$$\Delta_\mathrm{r}G = -RT\ln K + RT\ln Q$$
by finding $K$ at equilibrium condition and $Q$ (using new partial pressure). If $\Delta_\mathrm{r}G < 1,$ the reaction is feasible.
My answer is positive, which means the reaction is not feasible. Why the free energy is positive here? It should be negative as the reaction can occur below this partial pressure, right?

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Keep CH SE Q titles in plain text.

Might be helpful to show your calculations

How can I add picture here?

It should be $\Delta_r G \lt 0$ , not $\Delta_r G \lt 1$.

@Learnphy consider following up on all these helpful comments if you are interested in an answer here, otherwise your question may get closed.


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Why would a reaction be nonspontaneous at higher temperatures?
Typically we think of a higher temperature speeding up the reaction rate and/or supplying the activation energy of a reaction. So why is it the case that some reactions are only spontaneous at lower temperatures?
Using the gibbs free energy equation $ \Delta G = \Delta H - T \Delta S $, If I have a reaction where $\Delta H$ is negative (exothermic?) and $ \Delta S$ is negative it makes $\Delta G$ positive at higher temperatures which means the reaction is nonspontaneous at higher temperatures. Why would this be the case?

LeChatelier's principle says a reaction that creates heat get's slower (and ultimately goes backwards) at higher temperature. And btw. it's totally arbitrary which one is the "forward" direction of any reaction.

@Karl Not slower, the net reaction ultimately is backwards. LeChatelier talks about equilibrium thermodynamics, not kinetics.

@Karsten I agree, but Karl isn't completely wrong. If the equilibrium constant shifts, then the forward and backward rate constants do too (as $K = k_f/k_r$). A smaller equilibrium constant does indicate a slower forward reaction (and/or faster backward reaction).

You are mixing the thermodynamics and kinetics of a reaction, Thermodynamics just predicts if the reaction will happen or not, Kinetics predicts how fast will it happen, If the reaction is feasible then we look at its kinetics

You expect both the forward and the reverse reaction to proceed faster at higher temperature. If the reverse reaction speeds up by a higher factor, this will affect the equilibrium.
"Spontaneous" is a technical term that does not reflect our day-to-day use of spontaneous. If you substitute the longer "goes forward or would have to go forward, starting from standard state, to attain equilibrium", it might make more sense.
If I start at standard state at a given temperature, either the forward reaction will be faster than the reverse reaction, or vice versa. If I repeat the experiment at higher temperature, I will see an increased forward rate and an increased reverse rate. These increases are typically not of the same magnitude, however.
So the direction that the reaction takes to reach equilibrium, starting from standard conditions, can be different depending on temperature. With respect to the direction of the reaction as written, it can change from spontaneous to non-spontaneous, or from non-spontaneous to spontaneous. If you write the reaction in the other direction, as Karl mentions, you come to the opposite conclusion.
In conclusion, there is always one net direction of the reaction that becomes "more favored", and the other net direction that becomes "less favored" as you increase the temperature and individual rates go up.

Thank you for your answer. So basically what it means is that its opposite reaction would be favorable at the high temperature instead.

Yes, that is always the case. If a reaction is non-spontaneous, the reverse reaction is spontaneous. It just says that equilibrium is either reached in the forward or the reverse direction (unless we are already at equilibrium, in which case we don't see any net reaction).

Thermodynamics studies systems at equilibrium so changing temperature moves the equilibrium from one position to another. For a given change we wish to know if the change of state will occur spontaneously or whether some amount of work is needed to make the change occur. In the gas phase this work usually called 'PV' work such as a change in volume. From the second law there must be overall entropy production so that the total change in entropy of the 'system' i.e. the reaction, plus that of the surroundings is always positive but which can be zero if the conditions are reversible. It can be shown using the First and Second Laws that the Gibbs free energy must decrease if the total entropy production is to be positive, i.e. the Gibbs free energy change must be negative for a spontaneous reaction.
When the temperature changes we can use the Van't Hoff Iscohore to predict what happens;
$$\frac{d\ln(K_p)}{dT}=\frac{\Delta H^{\text{o}}}{RT^2}$$
which when integrated (assuming that $\Delta H^{\text{o}}$ is independent of temperature over a small temp range which is often a good approximation) produces a gradient of $-\Delta H^{\text{o}}/R$ when $\ln(K)$ is plotted vs $1/T$.
(An aside.  Notice that the entropy is not involved, the species involved are the same but in different proportions and there is always entropy of mixing which compensates. You can see this also from $G=H-TS$ where $\displaystyle \left( \frac{\partial G}{\partial T} \right)_p = -S$ is a constant as temperature changes.)
Thus for an exothermic reaction such as the formation of ammonia from hydrogen and nitrogen since $\Delta H<0$ the equilibrium constant $K_p$ must decrease as the temperature increases and less ammonia is in the equilibrium mixture as the temperature is increased, as is experimentally verified. The opposite is true for an endothermic reaction, i.e. dissociation of $\ce{N_2O_4}$.
If you have measured the forward and reverse rate constants, defining the equilibrium constant as the ratio of rate constants and using the Arrhenius equation produces
$$K_e= \frac{k_f}{k_b}=\frac{k_f^0}{k_b^0}e^{-(E_f-E_b)/RT}$$
where $k_f, k_b$ are the rate constants and $E_f, E_b$ the activation energies. An exothermic reaction has $E_b > E_f$ and the equilibrium constant decreases on increasing the temperature just as predicted by the thermodynamic argument.
Finally the Gibbs energy is related to the equilibrium constant as
$$\displaystyle \Delta G^{\text{o}}=-RT\ln(K_p)$$
so the sign of $\Delta G^{\text{o}}$, positive or negative, depends only on whether the equilibrium constant is less than or greater than one.


text:
Why is Hg2Cl2 less soluble in water than HgCl2?
I learned that $\ce{Hg2Cl2}$ is almost insoluble to hot water and $\ce{NH3}$ water in my textbook, but $\ce{HgCl2}$ dissolves well in water.
I wanted to know the reason and searched for it. It's written on some website that thanks to low shielding effect of f-orbitals and the fact that Hg has full f-orbitals, Hg has a high effective nuclear charge. And this makes it possible for the $\ce{Hg+}$ ion to form covalent metal–metal bonds. I searched further and found some information that bond between $\ce{[Hg-Hg]^2+}$ and $\ce{Cl-}$ is more ionic than the bond between $\ce{Hg^2+}$ and $\ce{Cl-}$, according to Fajans' rules.
I thought if the $\ce{[Hg-Hg]^2+}$ and $\ce{Cl-}$ bond is more ionic than the bond between $\ce{Hg^2+}$ and $\ce{Cl-}$, then $\ce{Hg2Cl2}$ would be more likely to dissolve in water because water is a polar solvent. Why is this not the case?
Is the $\ce{HgCl2}$ covalent bond so weak that it will dissociate in water? Why?
I came across the theory of negative and positive hydration [1]. Does anyone think this theory could explain the difference of solubility of these two salts?
Reference

Rodnikova, M. N. Negative Hydration of Ions. Russian Journal of Electrochemistry 2003, 39 (2), 192–197. DOI: 10.1023/A:1022317227140.

Ionic nature might not always help in checking solubility. A compound is soluble if its hydration enthalpy overcomes the lattice enthalpy. The linear structure of Hg2Cl2 and tetragonal packing might have effects on its lattice enthalpy. Did you check the enthalpies?

Nobody is able to calculate the solubility of a chemical substance in water. It is one of great mysteries of the present science. Why is potassium perchlorate the only practically insoluble potassium compound (apart from silicates) ? Why is calcium fluoride insoluble in water, when calcium chloride, bromide and iodide are  soluble in less than their volume of water ? Why ?

@Maurice Learn about lattice and hydration energy or whatever, but stop posting this annoying comment all the time. Software can nowadays even design new alloys having desired properties and you keep talking about "mysteries".

@Mithoron. I know what are lattice and hydration energies. I will stop posting this "annoying" comment if and when you are able to explain the mentioned "mysteries". Are you able ?

@Maurice This is a q & a site. If you have questions then ask them, but first check if they aren't here already.

like https://chemistry.stackexchange.com/questions/4344/solubility-of-calcium-fluoride-in-water https://chemistry.stackexchange.com/questions/494/why-is-calcium-fluoride-insoluble-in-water

Thanks everyone. It seems I've got to learn thermodynamics.

According to Fajan's rules, due to more polarising power of small cation $\ce{Hg^2+}$, the $\ce{HgCl2}$ bonds will be more covalent than those in $\ce{Hg2Cl2}$. But more ionic character of a bond does not imply more solubility.
When you put a compound, say $\ce{NaCl}$ in water (a polar solvent), the ions get solvated. Now as $\ce{Na+}$ ion is small, it has high charge density and more of water molecules surround it, increasing the stability (due to decrease in potential energy), hence higher will be the hydration enthalpy, which breaks the ionic lattice of $\ce{NaCl}$. As ionic compounds arrange themselves into lattice structure, also taking into account the coulombic force of attraction, generally ionic bond is stronger than covalent bond, especially when looking into pair of atoms forming different kinds of compounds.
Similarly, $\ce{Hg^2+}$ has higher charge density so it will realtively be more hydrated with respect to $\ce{Hg2^2+}$, also due to more covalent character of $\ce{Hg-Cl}$ bond in $\ce{HgCl2}$, its lattice would somewhat be weaker, lacking the polarity ionic compounds have. So it can qualitatively be assumed for $\ce{HgCl2}$ to be more soluble in water than $\ce{Hg2Cl2}$.
Also, as pointed by Maurice, calculating, or even comparing solubility remains much of a mystery, but many of times, in common compounds, general trends and classical theories can help explain the solubilities.

Thank you for explaining why ionic character doesn't mean more solubility.  I don't think covalent character always explains less lattice enthalpy because , for example diamond which is covalent crystal, has high lattice enthalpy (713kJ/mol). How could we tell that Hg-Cl covalent bond is weaker than ionic bond between Hg2+ and Cl-? Or am I getting something wrong?

You are correct, my saying that covalent bond is weaker is not correct. It should be, for compounds formed by same atoms, generally covalent bond is weaker than ionic bond, we compare taking the atoms same.

https://chemistry.stackexchange.com/q/11048/124959

I read through your answer again and came to understand what you meant. Thank you so much for taking time to explain your thought.


text:
What is the point of the up arrow (↑) which shows "gas evolved" or "gas liberated", if you could just write the state symbol (g)?
What is the point of the up arrow (↑) which shows "gas evolved" or "gas liberated", if you could just write the state symbol (g)? And would you write the up arrows (↑) and down arrows (↓) if the state symbols are already included?

They are pretty redundant, imo.

Probably these arrows should've been included in Arrows used in chemical reactions.

You can use one or the other. If you use the arrow, you don't need to write the state and vice versa.

@NilayGhosh Arrows should not be used at all to denote states of aggregation, it's an obsolete and non-standardized notation.

Sometimes, the up arrow is used to imply that the product is removed in the gas state, so the reaction never reaches equilibrium. If you just write (g), it might mean a closed system. However, I don't think there is consensus for interpreting it this way. For solids, you would not distinguish removing the solid from it just sitting there because continuous removal would be difficult to implement.

@Karsten Interesting, I have never noticed the open/closed system interpretation.


text:
Why do we say "octadiene" and at the same time name another compound "octanediol"?
Why do we say "octadiene" and at the same time name another compound "octanediol"? We also say for instance "pentanedione", but on the other hand we say "butadiyne". I don't see any logical pattern for when to include the infix -ne- and when to exclude it.

There's nothing "irregular" in here. You got your "ne" in butadiyne just right.

In octadiene, the double bonds are an integral part of the parent chain. So if you named it 'octanediene', it would seem to exclude the double bonds from the parent chain and making the parent chain completely saturated. So to make the double bonds inclusive, it is termed as 'octadiene'.
But in case of octanediol, the hydroxy group is not a part of the parent chain, being a functional group. So it is named 'octanediol'.

If that's the case, could one not say that all compounds beginning with "metan-", "etan-", "propan-" etc. are saturated compounds? But that does not hold true if we consider propanone.

Wherever is '-ane', it means the compound is saturated. If there was unsaturation, we would have included that in parent chain according to IUPAC rules and '-ene' would have appeared. As with propanone, it has only 3 C atoms and at the other end of double bond there is O. So there can't be 'propenone' or 'propynone'.

Nomenclature follows strict rules. The argument you present here might be a well enough for of thumb, but as a real answer to the question it doesn't hold up. I'm sure there are sources, which can be quoted to support this claim. This should probably include the current edition of the Blue Book.