text:
Evaporation Rate of Hydrochloric acid
I'm currently working in a Chemical Plant as an intern, and I was given a task to improve the efficiency of the blower, which is already in place right now, by designing a new blower for the appropriate flow rates.
We have a filling station in the plant for HCl liquid, which fills the HCl (30-32% conc. at 20 m$^3$/hr) into tankers for transportation purposes. The problem is when HCl is filled up in a tank, it releases fumes, which are harmful for the people present near the filling station.
We have to recommend a blower, so that most of the generated fumes can be treated at an appropriate flow rate. But we don't know how to proceed with this, as we wouldn't know about the evaporation rate of HCl while it is flowing inside the tank.
Some technical information:

Temperature, Pressure can be assumed as 30 $^{\circ}$C, 1 atm.
Flow rate of HCl inside tank: 20 m$^3$/hr.
Capacity of Tank: 25 Metric tonnes

Kudos to you for considering the health of workers! The simplest and most economical way to avoid HCl vapors is to vent (i.e., waste) at a high altitude, like flaring CH4 from petroleum drilling. However, the vented HCl is a raw material if you capture it in H2O - you should be able to get up to about 20% before the vapor pressure of HCl is evident. Run the vented vapors thru a wet column and sell the diluted HCl as a by-product (cheap?). Up to about 20% HCl has no significant odor of HCl.

I don't think these would work. We have to remove fumes from a tanker, which would be filled up and transported to another location. While being filled, the HCl will release fumes, and we need to find the amount of fumes generated, so that we can suck those out with the help of a blower, and transport them back to the HCl storage. I just need the amount of fumes generated, so that I can recommend a blower with an appropriate flow rate, and appropriate pipe diameter which can handle the pressure drop.

When HCl is put into a tank, it releases fumes... And this is not an equilibrium situation (just like pouring a carbonated beverage into a glass can be done rapidly or slowly, releasing various amounts of CO2).
You may be able to get some numbers useful for calculations (if you also specify a range) from the two references below (1 and 2). No. 1 has a nomograph that shows the concentration of HCl in the liquid and vapor, and No.2, which is a fairly large handbook of safe handling for HCl, has a different way of expressing the information:

Knowing your volumes and tank flow rates and other estimated required flow rates, along with the amount of HCl in the vapor could help decide what kind of pump you need. And don't forget to inquire of your HCl supplier, who might have significant, relatively private, but not actually confidential, information about how other customers succeed (or fail) with their methods of handling this task. Sometimes calculations don't take everything into account, but a successful solution is worthwhile, even if if is "supported" by calculations.
Ref 1: https://pubs.acs.org/doi/10.1021/i460001a002
Ref 2: https://www.jsia.gr.jp/data/handling_02e.pdf

This is a job for a competent chemical engineer not an intern; find one and suggest hiring a consultant. I am dismayed that HCl gas is released to the atmosphere and think that OSHA should get more involved with your company. Since the air volumes displaced in the delivery tank and the receiving tank are the same [and should be readily determined] a vapor recovery system should be imperative in handling such materials. Look into it. Learn all you can about the overall process. If that fails, get a better internship.
Some things to consider: 30 degrees Celsius seems a very high temperature to be handling concentrated hydrochloric acid; what is going on here? Also work with the same units; convert the masses to volumes; learn about head spaces etc. in the shipping.

Thanks for your valuable suggestions, I guess a consultant would give a better answer to this. Answering your question about the temperature, the company is situated in Pondicherry, India. The environment here is quite humid and hot, so I wrote 30 degrees Celsius as the surrounding temperature (average). I assumed this would be the temperature inside the empty tanker as well, not sure if that's the case.


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Electron affinities of the alkali and alkali earth metals
According to my book, the former has a higher EA than the latter group of elements because alkali metals will attain $ns^2$ configuration whereas an alkaline earth metal in the same period will attain $ns^2np^1$. Fully and half filled configurations are more stable but according to this article, there should be unfavorable interactions due to pairing of electrons in same orbital in alkali metals. Please explain what should be correct order and why?

https://chemistry.stackexchange.com/questions/51400/why-do-alkali-metals-have-an-exothermic-electron-affinity https://chemistry.stackexchange.com/questions/16845/why-does-k-have-a-greater-electron-affinity-than-ca

A look at any electron affinity table reveals that in factvalkali metals have higher electron affinities than the alkaline earth metals that respectively follow them. We can say the same for nonmetallic hydrogen versus helium, these also being $s$-block elements. In fact helium, beryllium and magnesium seem to have no electron affinity at all! While the effect of pairing the valence electrons in a Group 1 element is unfavorable, having to add an electron to a higher subshell as the Group 2 elements have to do is worse.
This effect of subshell occupation shows up not only with electron affinities but also with actual compounds of the metals. Most alkali metals can form compounds as anions, while alkaline earth metals are not known to have this possibility.

Well, the alkaline earth metals can be made as individual negative ions (used in tandem ion accelerators), but that is not particularly a chemistry question.

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What would be the product(s) when 2-iodobutane is heated with tertiary butoxide?
The following question from Black Book Organic Chemistry IIT JEE Advanced Level Papers [1]:

Choose the correct option(s) among the following about [P]:

A) Two C−H bonds in [P] are involved in hyperconjugation.
B) [P] cannot show stereoisomerism.
C) Hydrogenation of [P] gives mainly n-butane.
D) Monochlorination of [P] gives 3

What is the product when 2-iodobutane is heated with tertiary butoxide? Is it 1-butene, 2-butene, or both?
Bulky bases generally favor the Hoffmann product, but my teacher told that in this specific case having much less hindrance Zaitsev product dominates. I am not able to completely digest it how this happened and is this correct or not.
Answer key: C, D.
Reference

Ashish Mishra; Dr Ramu Petakamsetty. Black Book Organic Chemistry IIT JEE Advanced Level Papers; Blue Rose Publishers, 2022. ISBN 978-93-5611-183-7.

Did you attempt to search the scientific literature? This paper from 1966 reports ~2:1 preference for 2-butene over 1-butene as product (see table II): https://pubs.acs.org/doi/abs/10.1021/ja00959a018

For more on the effect of bulky bases on E2-elimination see: https://chemistry.stackexchange.com/questions/144168/how-to-determine-if-a-base-is-bulky/144169#144169

As described in most introductory organic chemistry textbooks (my usual reference is Loudon), E2 elimination to form an alkene typically produces the more substituted alkene as the major product. This outcome is the Saytzeff/Zaitsev product (roman spelling varies depending on source).
The primary exception is if a sterically hindered ("bulky") base (the standard example is t-butoxide) is used, in which case the most accessible protons are removed, leading to a terminal alkene. Similarly, if the leaving group is quite large (trialkyl amines are typical examples), the terminal alkene is favored even for small bases. This outcome is the Hofmann product.
An exception to the exception is the case of large halide leaving groups. It has been observed that the percentage of 2-alkene (ie more substituted) product in eliminations using t-butoxide as base increase when one changes the leaving group from chloride to bromide and from bromide to iodide. For the 2-halobutanes specifically, the ratio of 1-butene to 2-butene (cis or trans) products is approximately 2:1 for 2-chlorobutane, 1:1 for 2-bromobutane and 1:2 for 2-iodobutane[1]. Competing explanations were put forth for these observations (sometimes referred to as the "Ingold-Brown controversy"), and I'm not aware of a consensus even today.
[1] Brown, HC and Klimisch, RL (1966) J Amer Chem Soc 88: 1425

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Labs without equipment - virtual chem lab?
I am an "international online" high school student with very spotty access to chemistry lab materials needed for some chemistry assignments I am being given.  I would like to know if there exists online virtual chemistry labs, appropriate for a high school student, that will interactively/graphically simulate an actual lab experiment.  For example, I'd like to be able to use an online interface to declare which chemicals, in what quantities, I am placing into a specified beaker, and perhaps heated at a certain temperature for X duration.  This hypothetical chemistry simulation tool might show me the color changing of my solution, which I can then use to answer my homework questions.
I googled and found this example virtual chem lab, but it seems to only offer pre-determined labs that I can't adapt to my specific homework assignments.  I guess I need something more flexible?
Below is a sample real lab-homework question I am currently expected to perform, even though I have no chemicals or equipment:

Label one of the test tubes Fe2+. Add 4 milliliters of iron(II) sulfate to the test tube.
Label the other test tube Fe3+. Add 4 milliliters of iron(III) nitrate to the test tube.
Add 4 milliliters of potassium thiocyanate to each test tube.
Observe the contents of the test tubes, noting any evidence of a chemical reaction. Record your observations in the table. If there is no evidence of a reaction, write “no reaction.”

Likewise, here is another example homework question:

Reuse the same test tubes from part A, labeled Fe2+ and Fe3+. Be sure they’re clean.
Add 4 milliliters of iron(II) sulfate to the test tube labeled Fe2+.
Add 4 milliliters of iron(III) nitrate to the test tube labeled Fe3+.
Add 4 milliliters of potassium iodide to each test tube.
Add 1 milliliter of the prepared starch solution to each test tube.
Record your observations, noting any evidence of a chemical reaction. If there is no evidence of a reaction, write “no reaction.”

If this hypothetical chem-sim tool doesn't exist, suitable for my needs, is there some other tactic I can use to find answers in the absence of the correct materials?

How about just reading a textbook?

Virtual or real lab experiments help you predict the outcome from what you already seen. Textbooks help you to predict event outcomes you have never seen.

Make yourself a lab journal and describe what you would do and what you learned from your textbooks should come out of the experiments.


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Question regarding AIBN in toluene

Hi guys, I am trying to perform bromination of this compound using AIBN as a radical initiator. However, it seems like AIBN is only sold commercially as 0.2 M in toluene.
Is it possible to use AIBN directly as it (0.2 M in toluene) is or should I try to remove the toluene first before performing this reaction? (Not sure if toluene will be brominated instead of thymidine)
Thank you!

If AIBN solid is not available, can you get the cyclohexyl analogue 1,1′-Azobis(cyclohexanecarbonitrile) (AHCN) which will do the same job?

It is always good to explain less common acronyms like AIBN aka azobisisobutyronitrile. Even if you assumed readers able to help would know it, it frequently happens the help comes from those who are not familiar with it.

AIBN is sold by Sigma-Aldrich as a powder (98%). https://www.sigmaaldrich.com/US/en/product/aldrich/441090

Is this your other account? https://chemistry.stackexchange.com/users/125687/catmystery987?tab=profile

Before AIBN was developed as a radical initiator, the world used benzoyl peroxide example here. If none of the AIBN family of initiators are available then you can use benzoyl peroxide under similar conditions.

Hi Waylander, thank you for the reply. May I ask, if I am able to find Azobis(cyclohexanecarbonitrile) will it be better to use this instead of benzoyl peroxide?
And also is benzoyl peroxide compatible for use with benzene as the solvent?

If you can get one of the AIBN family, I would use that. Benzoyl peroxide is compatible with benzene.


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Why will a lone pair from H2O not attack a C+ atom in elimination reaction?
In step three of reaction between H2SO4 and alcohol, why will the H2O attack the H atom instead of the C+ atom? they both have positive charge and in nucleophilic substitution reactions the c+ atom is attacked.
(Image taken from chem guide )

If $\ce{H2O}$ attacks the $\ce{C+}$ atom, it will produce the original alcohol back. No chemical effect.


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Priority of Alphabetic order vs Lowest locant
I am confused about usage priority of lowest locant and alphabetical order rule(s)
For example :
3-ethyl 4,4 - dimethyl hexane
4-ethyl 3,3- dimethyl hexane
Which is considered correct (or preferred) IUPAC name?

Lowest locant has the priority over alphabetical order while numbering. But when writing name, we write according to alphabetical order. So 4-ethyl-3,3-dimethylhexane is the correct answer. One more thing, we don't consider the prefix 'di' on 'dimethyl' while deciding alphabetical order of the name.


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What is it about bleach that keeps Pharaoh ants away even after drying?
I posted this question here and one of the commentors thought that this Chemistry SE might be able to offer more sure answers.
My apartment high-rise has been afflicted with ants in the past few years.  The pest control technician says that it has become a common problem due to warming weather.  Various things have been tried, and I am leaving this in their expert hands (or rather, the apartment management is).  This question is not asking for further countermeasures.
When I find a place or pathway with many ants, I find that simply wiping with a damp cloth or vinegar doesn't keep them away for long.  Wiping is supposed to disrupt their pheromone trails, but I only find somewhat lasting effects if I wipe with undiluted household bleach.  I am pleasantly surprised that they stay away even after the bleach has dried.  After some internet research, I found that bleach leaves behind a salt residue when dried.  Is it the salt residue that keeps the ants away when wiping otherwise doesn't have lasting effect?
I've had another corroborative experience that leads me to suspect that salt residue has a repulsive effect on pharaoh ants.  I keep a large beer glass of very salty water on hand because of the dental benefits of salt water rinse.  I probably use too strong a solution, which likely has drawbacks (something I have yet to research).  It is kept on top of the fridge.  Around the glass is salt water staining.  I never see ants around it, even though ants like water.  I admit that this corroboration is not very strong, as I generally do not see ants on top of the fridge anyway.

This question is admittedly cross-disciplinary (chemical ecology/biology), but probably better suited for biology SE since they can discuss issues such as sensorial thresholds and range of chemical perception and interference, issues that have less to do with basic chemistry and more to do with neurophysiology/ecology/animal behavior.

However, you might want to explain what kind of chemical answer you are looking for (and I don't think "Is it the salt residue that keeps the ants away when wiping otherwise doesn't have lasting effect?" is a chemical question per se.

Try an experiment. Leave some of your strong salt solution to dry in an area occupied by the ants and see if they avoid it when it has dried.

Personally I use butter to convince the ants to stay away. It works. They hate butter.

@Maurice: Can you please share how you discovered this?  My online surfing indicates that they like fat, and at least one mention of butter.

@Waylander: I'll keep the dried salt solution in mind.  However, the ants show up in a food preparation area (kitchen counter, often near the sink), so I'm not keen on deviating what seems to work (bleach) for any amount of time.  The sample sizes per day are small (3 to 5 sightings per day, sometimes with application of bleach, often applied at night), so it may take a many-day controlled experiment to be sure.  And the will to avoid wiping down with bleach when I see the ants.

@ Buck Thorn: OK, thanks for your assessment of the question.  I just didn't know enough to mark the answer as correct in the Biology SE.  Would you have an opinion on that?  I feel that leaving it unmarked is irresponsible, as it could be the right answer.  I just don't know enough about the chemistry to judge.

I have not discovered the effect of butter. I just got it a couple of years ago in a museum where a real living ants hill was installed by specialists. There was plenty of people  around the nest. And some were asking questions. I remember a visitor asked the specialists how to repell ants from entering his home. The specialist mentionned the butter as an old receipt. I was an auditor, but I remembered it. Much later, I had the opportunity to test the trick at home, and I cut the ants way with a line of butter. It worked. The ants refuse to cross the line, even if it is thin.

Further web searching reveals that it depends on the ant. If what I read is to be believed, grease ants will love butter, but general ants won't because the the fat makes it hard to digest with their acid. The ant in question here is the Pharaoh ant, which has more of a sweet tooth. This page, however, implies that they like butter, as does this page. I'll play it safe for now and keep the butter idea for later.

Household bleach acts as a broad oxidizing agent of organic compounds, and this activity is rapid and usually enough to suppress the original activity of a pheromone. So my assessment would be that the answer at bio SE is correct. The residue is probably table salt. The bio question is, do the ants mind the salt, why or why not? See also: https://www.clorox.com/learn/does-bleach-leave-a-residue-dried-bleach-crystals/

In retrospect this post seems to contain plenty of chemistry. Feel free to request migration to chem SE at biology SE.

I did, days ago.  No one responded.  That's why I ended up cross-posting.  Thanks for the vote of confidence in the answer at the Biology SE.  I'll mark it as "the answer".

My bleach solution seems ill-advised, both because of the fumes and the adverse effects on laminate countertops.  It looks as if I will have to get comfortable again with ants on my food preparation surface.


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Is the entropy change positive or negative in this reaction?
Suppose I have a reversible reaction
\begin{align}
\ce{A(g) + B(g) <=> C(g)}\\
\end{align}
with equilibrium constant K.
Its ΔΗ is positive throughout.
Now, suppose I start with 1 mole each of gases A and B, and zero (negligible) moles of C. The temperature is constant.
According to $\Delta G = RT\ln(Q/K)$ , $\Delta G$ will be highly negative at my initial conditions and will increase (become less negative) as reaction proceeds forward.
Also, using the same expression, $$\Delta S = \frac{\Delta H}{T} - R \ln\left(\frac{Q}{K}\right)$$
Hence, $\Delta S$ is highly positive initially and decreases as the reaction proceeds forward. When $\Delta S = \Delta H/T$ (at $Q=K$) , no net reaction takes place further and equilibrium is maintained.
I understand it.
But how is $\Delta S$ positive? How can it be explained theoretically?
Theoretically, shouldn't $\Delta S$ be negative as number of gaseous moles decrease on going forward?
Or is this type of reaction not possible?
Source:https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.10%3A_Entropy_Changes_in_Gaseous_Reactions

If A +B = C the reaction Must be exothermic even if the only energy loss is a temporary change of kinetic to potential in an entirely elastic collision So delta H is negative and delta S is also negative. This has always bothered me if a reaction is to reach equilibrium meaning delta G = 0, delta H and delta S must have the same sign and the common penalty box with 4 options is bogus. There is really only one situation if standard conditions are ignored and a reversible reaction at equilibrium is perturbed in either direction delta G becomes negative and the reaction returns to equilibrium.

But how is dS positive? How can it be explained theoretically?

The reaction entropy rules usually taught first are about the physical state of the pure substances. If you go from a condensed state (solid or liquid) to a gas, the molecules are less constrained. Similarly, if you open up a ring, the molecule itself has more conformational freedom, and this is also reflected in an increase in entropy.
There is a second, concentration-dependent contribution to the entropy of reaction. This is the entropy of mixing. A mixture has more microstates than the separate two pure substances. This is behind the experience that ideal mixtures (when the intra- and intermolecular interactions have the same strength) don't unmix. Entropy increases whem you mix something. This strangely is the case when you go from pure reactant to a mixture of reactant and product as the reaction proceed. Similarly, there is some "unmixing" when all the reactant turns into product.
This entropy of mixing is where the $R \ln(Q)$ (or $R T \ln(Q)$) comes in. This is what makes the reaction entropy and the reaction Gibbs energy concentration-dependent.

Theoretically, shouldn't dS be negative as number of gaseous moles decrease on going forward?

Yes, going from pure reactants to pure products, the entropy change is negative. However, if reactants and products are in the same phase, you have a very high positive change in entropy as the first couple of reactant molecules turn into product. If you want, you can explain this with kinetics as well, with no back reaction (no product yet) and a robust forward reaction.

[OP in comments: What if they are in different phases? Like gases combining to form a liquid?]

One example where all "reactants" and "products" are in separate, pure phases is ice melting. If you are above the melting point, all ice will melt and the system will not reach a dynamic equilibrium. This is different for water evaporating. In a water bottle that is half-full and capped, after a while you reach equilibrium, and the same amount of water evaporates and condenses, so there is no longer a net change. The entropy of water in the gas phase is dependent on the partial pressure (the "concentration"). If you write the equilibrium expression for these processes, you will see the difference ("leaving out", or setting to one, pure solids and liquids, but not gases).
Conversely, if the last few reactant molecules turn into product and there is no more reactant left, the entropy change is negative (this is hypothetical because it is past the equilibrium, i.e. in a direction away from equilibrium). Again, using kinetics, the forward reaction rate has almost "dried up", the the reverse reaction rate is robust.

Or is this type of reaction not possible?

This type of reaction is possible. The standard Gibbs reaction energy will have a high positive value, though, so the equilibrium constant will be very low. Equilibrium will be reached quickly if you start out with pure reactants.

[OP in comments] How do you say that the Standard Gibbs Energy will be highly positive for these type of reactions?

The standard enthalpy of reaction is positive (given) and the standard entropy is negative (less particles on product side). Adding them up according to $$\Delta G = \Delta H - T \Delta S$$ gives a positive value at all temperatures, at least as high as $\Delta H$
Nomenclature
You have to distinguish $$\Delta S^\circ$$ and $$\Delta_r S = \frac{\partial \Delta S}{\partial \xi}$$ The former is what you can calculate from tabulated data. The latter is a more formal way of writing your dS, signifying the infinitesimal small change in entropy as the reaction moves forward an infinitesimal amount, i.e. when some reactant turns into some product according to the balanced chemical equation.
Textbooks, especially introductory once, are inconsistent in their nomenclature (there is an entire article on this topic for Gibbs energy of reaction). For the nomenclature and symbols for Gibbs energy, see the first slide in this lecture.

[OP in comments] In your formal notation of dS , what is the physical quantity you took partial differential with?

It is the extent of reaction $\xi$ (greek lowercase xi). It measures the amount that has reacted ($\xi$ = 1 mol means one mole has reacted if the coefficient is one, two when the coefficient is two and so on, with reactants having negative values and products positive values).

@BuckThorn I commented on the other post that editing and reopening is the way it is supposed to work. I think I understand the question though: Why does the rule (less gas particles on product side) say entropy should be negative, yet the change in (system) entropy as the reaction reaches equilibrium is positive?

I have started to understand things now.... What do you mean by "last few reactants" in fifth paragraph? The reactants are about to finish? Or equilibrium is about to be attained?

Also, in fourth paragraph, you wrote there is a high change in entropy if reactants and products are in same phase.... What if they are in different phases? Like gases combining to form a liquid?

Can you also suggest me some standard book for understanding the concepts of Entropy(and Entropy of Mixing) , Enthaply and Gibbs Energy correctly and deeply....Thanks a lot

@KarstenTheis One question more plz... How do you say that the Standard Gibbs Energy will be highly positive for these type of reactions?

For understanding these concepts correctly and deeply, I recommend taking physical chemistry. Working through a physical chemistry textbook independently is tough, but maybe with a study partner of similar interests and doing a lot of problems it would be beneficial too.

@KarstenTheis Can you suggest some particular writer's book?

No, I never had a Pchem textbook, just my lecture notes.

When the last few reactants would be turning into product, Q will tend to infinity. So entropy change must be negative....What am I missing?

According to you, standard entropy change directly depends upon the change in no of gaseous mole? It will be negative for a reaction where gaseous moles decrease

@HalfInfinity "Q will tend to infinity". Yes, good argument, my mistake. I fixed the typo.

"change in # of gaseous mole": That is a rule of thumb. I am not sure if there are any exceptions. If the number stays the same, you have to look for more subtle effects (and there is a greater risk of guessing incorrectly). In the end, it is best to rely on experimental data (i.e. thermodynamic data tables).

Let us continue this discussion in chat.


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Will my method for catching lead dust work?
I am a hobbyist and a crafter working with cast models. These modles are up to 80% lead, with the rest being tin or similar soft metals. They are made in a mould, so they often have mould lines that need to be scraped off with a knife. But doing this can create lead dust which will pollute the immediate working environment. I had the idea of doing any scraping while the model was completely submerged in water. The theory is that small particles will be caught in suspension in the water, and the water can then be filtered or otherwise safely disposed of. (In case this needs illustrating, I have made a video of my method: link). My question is essentially, "will this method successfully control the spread of lead dust into my working environment, or am I missing something important?"

Certainly this must be an estabished art. Have you consulted your peers in this field as to how they deal with potential metallic dust generation issues. With this scraping process, do you think you generate micron sized particles that can fly and remain suspended in air? Romans used to drink wine in lead glasses and all their water plumbing (lead symbol: Pb from plumbum) was based on lead.  For a century, tetraethyl lead was added to cars. Lead is everywhere.

You might consider switching to bismuth. It is low melting, dense, far less toxic than lead and relatively inexpensive.

AChem: TEL was a chemical success and environmental disaster; not something to be proud of. Wonder why the Roman Empire declined?

@jimchmst: I agree about TEL. Empires fall because of human follies. The lead story is a hypothesis, see this Science report https://www.science.org/content/article/scienceshot-did-lead-poisoning-bring-down-ancient-rome. The conclusion was "While the lead contamination was measureable, the team says the levels were unlikely high enough to be harmful, ruling out tap water as a major culprit in Rome's demise. "

@AChem there is very little discussion of this in my hobby to be honest. Most people seem to ignore it as a risk, or trust to a dust mask (which might protect them but not their environment). This is why I was moved to ask here. My fear is precisely that I am creating airborne microns by filing and scraping, and my theory is that by doing it with the model submerged in water I will catch all such particles in suspension in the water. I am just wondering if it's a good theory or not.

@TooBusyForMinis, The density of lead is pretty high, and probably if add a few drops of non-foaming detergent, water will be able to wet very small metal particles and they will sink easily-if any.

Your method might well work, but, if you care about safety, a better solution would be to use one of the many alternative non-toxic alloys to start with.

Lead-free solder (based on indium) is a possible alternative. Widely available.

Another option is to buy a small table-top soldering booth. They make acrylic booths with HEPA filtration. You'd put your workpiece in the booth and reach in from the front with your hands. Air is pulled inward and exhausted out the back. For cleanup, I'd suggest something like D-Lead wipes - they contain EDTA or something equivalent that will help to trap and retain lead and other metals in solution. You should also get a colorimetric lead test kit (I've seen them on Amazon) so you can do surface swab sampling in your work area. Color change on the swabs will allow you to visualize the effectiveness of your control measures. Lastly, tell your doctor you are working with lead, and they can arrange a blood lead level test during your annual physical.

It's risky. The water being filtered off is not good enough to reuse the water. This is because some lead dust might be small enough to fall with the liquid. It is also because if there is any metal higher than lead in the electrochemical series, some lead would fall into solution. Either way, you'd have to dispose of the water safely since it would likely be contaminated.
Would it catch the dust? Yeah it would. You'd also need to thoroughly wash the knife.  Or maybe it wouldn't. If your lead dust is super small, some amount could be flung and fly. If your dust is more like sand, well it'll probably stay at the bottom of the water.
I'm no physicist, but I would think the particles would be larger when cut underwater. So you're probably okay.
I would advise placing other precautions such as wearing a mask with a very fine filter.  For example a dust mask. It would also be a good idea to wear gloves and long sleeved clothing. But I'm not sure how you would safely clean off your clothing after use.
At that point, even if you get some lead on you, it's not gonna be super bad since there would be very little and since it is metallic. Do note that aqueous and organometallic heavy metals are very, very toxic even compared to the regular metals.
Also, the deeper your lead is, the less likely it is to escape. And whatever tool you're using, you need to make sure it has no lead dust before it leaves the water.


text:
SMARTS: distinguish between alcohol and carboxylic acid
I'm new to SMARTS and have to identify substructures with the Python package rdkit, so far so good. Let's say I have a SMILES with an carboxylic acid and a alcohol such as
OC(C1=CC(CO)=CC=C1)=O

and I want to get only the alcohol, then I would try something like this:
    mol = Chem.MolFromSmiles("OC(C1=CC(CO)=CC=C1)=O")
    toSearch = Chem.MolFromSmarts("[#6][OX2H]")
    bool = mol.HasSubstructMatch(toSearch)
    subStruct = mol.GetSubstructMatches(toSearch)
    print(bool)
    print(subStruct)

which would return:

True
((1, 0), (5, 6))

However, the substructure search also identifies the OH of the acid which I don't want. How can I exclude that? My idea was to use some kind of recursive smarts for the alcohol to exclude all those "alcohols" where the carbon is bonded to another oxygen like "=O".

Hello! You may also be interested in the Matter Modeling 'sister site'. (though please don't post questions twice)

The simplest change to make for your SMARTS pattern would be specifying the total degree of the attached carbon is 4 with "[#6X4][OX2H]". This is enough to discriminate against the carboxylic acid.
You could use the alcohol SMARTS from here, which has the comment

nonspecific definition, no acetals, aminals, and the like

The SMARTS pattern is "[OX2H][CX4;!$(C([OX2H])[O,S,#7,#15])]", which stipulates that the carbon attached to the oxygen not also be connected to O, S, N, or P.
I made a little interactive widget that takes a molecule name or SMILES in the first field and a SMARTS pattern in the second.

Awesome!!!!
Thanks for the answer and the two great links! I have one more question. This smart doesn't hit if the alcohol is on an aromatic carbon. In this Case this smarts  "[OX2H][CX4,c;!$(C([OX2H])[O,S,#7,#15])]" should work, shouldn't it? thx!

@dideldideldum - that does seem to work:  https://i.stack.imgur.com/VIfOp.png

The smarts.plus website is also useful for visualizing / debugging SMARTS patterns.


text:
Mathematical models of oscillatory chemical reactions
I am a mathematician working on real-life models in ordinary differential equations. I want to know if there are any models of oscillatory chemical reactions that consist of three ordinary differential equations where one variable is much slower than the other two. In particular I am interested in bursting behavior, which consists of alternating trains of fast oscillations with periods of rest.
I have been searching and I couldn't find what I am looking for. I found about the Belousov-Zhabotinsky reaction, but in this case we have one fast variable and two slow variables, while I'm looking for two fast variables and one slow variable.

Somewhat related

Related: https://mattermodeling.stackexchange.com/q/8812/5

related? https://wrap.warwick.ac.uk/60495/1/WRAP_computation-02-00047.pdf


text:
The difference between the freezing point of methanol and ethanol at different concentrations
I would like to ask a chemistry question:
The freezing point of methanol aqueous solution between 0% and 90% is lower than that of ethanol aqueous solution; but why is the freezing point of ethanol aqueous solution between 90% and 100% lower than that of methanol aqueous solution?
Thanks in advance!

You have not clarrified what percentages you are talking about. Is it by weight or volume?

The OP's claim that "The freezing point of methanol aqueous solution between 0% and 90% is lower than that of ethanol aqueous solution" is correct as data given in Engineering Tool Box (methanol/water) and Engineering Tool Box (ethanol/water):
$$
\begin{array}{c|c|c|c}
\text{% MeOH (v/v) or (w/w)} & \text{The freezing point } (\pu{^\circ C}) & \text{% EtOH (v/v)} & \text{The freezing point } (\pu{^\circ C}) \\
\hline
0  & 0 & 0 & 0 \\
13 \text{ or }10 & -7 & 10 &-4 \\
24 \text{ or }20 & -18 & 20 & -9\\
35 \text{ or }30 & -26 & 30 & -15\\
46\text{ or }40 & -40 & 40 & -23\\
56\text{ or }50 & -54 & 50 & -32\\
66\text{ or }60 & -71 & 60 & -37\\
75\text{ or }70 & -82 & 70 & -48\\
83\text{ or }80 & -87 & 80 & -59\\
92\text{ or }90 & -90 & 90 & -73\\
100 & -98 & 100 & -115\\
\hline
\end{array}
$$
However the two last data points are simply the comparison of the freeing point of $100\%$ methanol and that of $100\%$ ethanol. Evidently, the freeing point of $100\%$ methanol is greater than the freeing point of $100\%$ ethanol (See Table of Melting points of alcohols). Thus, there is not an unusual reason for this change in trend.

Methanol water has a eutectic 157 K ~0.88methanol; Ethanol water 150 K ~0.80ethanol. Just to show fp depression works both ways.

Would you give the references for these values?

I googled it; look it up. You should have been suspicious that the freezing point tables you listed showed no eutectic and researched a bit further. When a source seems to violate what one thinks he knows about chemistry it is time to look further; one or the other or perhaps both needs some refining.


text:
What Is This Small Peak That Always Occurs Within Ketone Infra-Red Analysis?
After looking at a bunch of IR spectrums for ketones, I have noticed this tiny peak that I circled on the diagram below. Does anyone know what it actually is and if I can always use it to distinguish a ketone vs aldehyde?

It's an overtone peak, double the frequency of the carbonyl. Look at this YouTube.

Not quite double it is from 1st to 3rd vibrational level. It would be nice to have the frequency.


text:
Azulene Conjugation?
I'm trying to figure out how azulene is an aromatic compound; I understand it is cyclic, flat, and follows Huckel's rule, but I don't understand how it is conjugated.
I have circled what I believe to be an unconjugated area in yellow. Can someone help me figure out how this molecule is conjugated? Thank you.

Azulene is a pretty special hydrocarbon. In particular, the exact linkage you highlight is what makes it special, from a graph-theoretic standpoint. I'm not sure whether Hückel's rule or simple arrow-pushing should be expected to describe it well.

Thank you for responding. I think Huckel's rule explains it well because 4(2) + 2 = 10, but what would be a good way to determine that azulene is conjugated if arrow-pushing can't describe it well?

The $\sigma$-bond circled in yellow keeps the molecule planar. If this bond were replaced with two hydrogens to form a cyclodecapentaene, the molecule would not be planar nor aromatic.

Hückel's rule likely gives the right answer for the wrong reasons - this is a recurring theme in high-school and undergraduate chemistry. A satisfactory explanation other than "it is experimentally observed to be aromatic" may be quite involved.

What definition of "conjugated" are you using?  I consider doubly (or triply) bonded carbon atoms connected by a single (lone) single (sigma) bonds to be conjugated.  For example, I'd view all the carbon atoms in this molecule as part of a chain of conjugated bonds, even though the two chains are connected by a single bond: C(C=CC(=CC=CC)C(C=CC=C)=CC=CC)=C  So, the "internal" bond of azulene might not be conjugated, but both carbon atoms that are part of that internal bond are conjugated for other reasons...

If you insist on neutral atoms then the internal linkage is not conjugated.
BUT ...
The notable thing about azulene is that the seven-and five-membered rings can be aromatic when the former is positively charged and the latter negatively charged. That puts six electrons into each ring, and we all know about the $4n+2$ rule (even if technically it should be applied to single-ring systems).
If you allow a positive charge on the seven-membered ring and a negative charge on the five-membered ring, you can render contributing structures with a double bond in the internal linkage.

http://www.chemspider.com/Chemical-Structure.8876.html
According to Wikipedia, azulene is actually known with a dipole moment having the positive end in the seven-membered ring, and its chemical reactivity (electrophilic in the seven-membered ring, nucleophilic in the five-membered ring) is also consistent with this polar contributing structure.

Hi, thank you so much for the response. I understand how a resonant structure of azulene can form a cation on the seven-membered ring and an anion on the five-membered ring; however, I still do not understand how it is conjugated. Can you provide me with a diagram showing how to conjugate it with arrow pushing? If that is not possible, how can I determine that it can be conjugated and is therefore aromatic?

The monocyclic is the first word in the definition. It really is important that this rule is not applied to polycyclic compounds. Therefore Hückel's rules do not apply to azulene. https://doi.org/10.1351/goldbook.H02867

Yes I know that, but the OP was asking for contributing structures with an internal pi bond. So I had to go along. The existence of a good contributing structure with this feature differentiates azulene from a 10-pi mono-cycle. With that, however, Hückel calculations indicate the internal pi bond is weaker in azulene than in naphthalene.


text:
How to increase basis set while running cluster calculations
I am running (opt+freq)DFT calculations using 6-311G(d) basis set on polymers with different number of monomer units. Upto 2 monomer units optimization is converged but for 4 units max displacement criteria is not converged. Which starting point do I need to consider while increasing the basis set?

What do you mean by increasing the basis set? Do you mean using a larger basis set? For the 4 unit polymer does the optimization run out of steps?

Sometimes for large molecules, it can take a long time for optimizations to converge  because the torsions are low-energy and take many steps to settle.


text:
PFR reactor: -1/r vs conversion plot
A generic reagent A is considered. The behavior equations of a CSTR reactor is the following:
$$
\tau = c_\mathrm{A,0} \intop_0^{X_\mathrm{A,final}}
\dfrac{1}{-r_\mathrm{A}} dX_\mathrm{A}
$$
where $\tau$ where it is the filling time, understood as the time required to make a fluid flow rate react whose volume is equal to the reactor volume
$$ \tau = \dfrac{V_\mathrm{reactor}}{\dot{V_\mathrm{A}}} $$
$c_\mathrm{A,0}$ is the initial concentration of A, and $X_\mathrm{A}$ and it is the conversion of A that we want to obtain
From Octave Levenspiel, Chemical Engineering Reaction, John Wiley & Sons, Third Edition, page 103, you can see how the graph $-\dfrac{1}{r_\mathrm{A}} = f(X_\mathrm{A})$ has the shape of a crescent curve

I tried to reproduce this graph, using the formula
$$ -\dfrac{1}{r_\mathrm{A}} = \dfrac{\tau}{c_\mathrm{A,0} X_\mathrm{A}} $$
and what I get is a decreasing curve

My question is: to be able to reproduce the graphic in the text, which formula should I use?

The equation seems to  show that $X_A/r_A$ is constant in time where $r_A=dC/dt$ which depends on time as does $X_A$, but this is not included in your graph.  The book's graph apparently shows what happens generally when these time dependences are added to $r$ and $X$. You should check this as I'm not an expert in chem. eng. reactors.

From http://home.ku.edu.tr/~okeskin/ChBI502/chbi502-Chapter_2.pdf
An irreversible reaction is considered.
$$\mathrm{A} \xrightarrow{k} \text{Products} $$
To find a relationship between $r$ and $X$, we start from the kinetic law
$$-r_\mathrm{A} = k\ c^n_\mathrm{A}$$
where $n$ is the order reaction. The conversion is given by the formula
$$ X_\mathrm{A} = \dfrac{c_\mathrm{A,0} - c_\mathrm{A}}{c_\mathrm{A,0}} $$
With a few simple algebraic steps, we get
$$ c_\mathrm{A} = c_\mathrm{A,0} (1 - X_\mathrm{A}) $$
Substituting in the kinetic law, we obtain
$$-r_\mathrm{A} = k\ c_\mathrm{A,0}^n (1 - X_\mathrm{A})^n$$
Multiplying the first and second members by $-1$, the desired equation is finally obtained
$$\dfrac{1}{-r_\mathrm{A}} = \dfrac{1}{k\ c_\mathrm{A,0}^n}
\dfrac{1}{(1 - X_\mathrm{A})^n}$$

This curve has been plotted for $n = 1$, but analogous curves can be obtained for $n \neq 1$. The only difference is the curve slope, which is obvious since as $n$ changes, the rate dependence on the concentration changes (and therefore from $X_\mathrm{A}$)


text:
Question about aluminium and complex ion formation in water vs in acid
My textbook states, that aluminium reacts with water after dissolving the outer layer with HgCl2, and forms Al(OH)3, an insoluble compound:
2Al + 6H2O = 2 AI(OH)3 + 3H2
In non-oxidizing acids, it forms a salt and hydrogen gas:
2Al + 6HCI = 2AlCI3 + 3H2
But, aluminium is prone to form complexes, so the next reaction states, that in acidic environment, an aquacomplex is formed:
2Al + 6H3O+ + 6H2O = 2[Al(H2O)6]3+ + 3H2
I have tried to research a chemical equation for this complex formation on the internet, and I only found aluminium and water as reactants. No acid present.
My question:

How does the aluminium aquacomplex form, if there is no acid present and if that's not possible, why does it need an acid for the reaction to happen? What role does the acid have?

Thank you

Metallic aluminum is not charged. It has to loose three electrons in order to make the ion $\ce{Al^{3+}}$, which can be complexed later on. An acidic solution is a good way for helping $\ce{Al}$ to loose its electrons, because it produces $\ce{H2}$ which is a gas and leaves the solution, allowing the reaction to proceed with efficiency.


text:
Two solutions for balancing a chemical equation?
$$\ce{6H2O2 + 2N2H4 → 2N2 + 10H2O + O2}$$
$$\ce{4H2O2 + N2H4 → N2 +6H2O + O2}$$
Which equation from above is correctly balanced? They all seem to be balanced but which one is correct?

@jimchmst The first equation was taken from the official Worked Solutions Chemistry HL - Catrin Brown and Mike Ford - Second Edition - Pearson 2014 Chapter 1 Exercise 3 (d). Do you think this was a mistake or was there an underlying implication? After reading Maurice's response, I thought this equation should not be dealt in an holistic manner as one can just simply multiply one of the separate equation by a random constant and get lots of solutions to balance this equation. Plus, H2O2 decomposes on its own right?

They are independent reactions with no stoichiometric relation and it is incorrect to add the equations as if they were. If both reactions are concurrent the products must be analysed to determine the extent of each. Possible exceptions to this are consecutive where a product reacts further.  Text books are rife with errors misconceptions and just opinions. This may be just my opinion that a chemical equation must be stoichiometrically correct, I do have the Law of conservation of mass and work of many chemists to back me up.

All your equations are correctly written. But look !
The simplest equation would be $\ce{N2H4 + 2 H2O2 -> N2 + 4 H2O}$. But plenty of other acceptable equations can be obtained by adding an arbitrary number of times the following decomposition of $\ce{H2O2}$ : $\ce{2 H2O2 -> 2 H2O + O2}$.
For example, your first equation has been obtained in doubling my simplest equation and adding once the decomposition of $\ce{H2O2}$
Your second equation is obtained by adding my simplest equation plus once the decomposition of $\ce{H2O2}$

Neither equation is correct They do not describe a possible chemical reaction!

I took too long to edit! neither equation is correct. They do not describe a possible or reasonable chemical reaction! You do point out that the equations are the result of adding two equations that are not related stoichiometrically but must also carry thru and state that this is not a legal procedure even tho it is encountered frequently, and usually improperly, when dealing with consecutive or concurrent reactions.

@jimchmst  agreed with your comment of neither equation is correct. I also received $\ce{N2H4 + 2H2O2 -> N2 + 4H2O}$ as the correct answer based on tabulated Standard Reduction Potentials of $\ce{N2H4}$ and $\ce{H2O2}$ at $\pu{25 °C}$.


text:
For cannizarro reaction is it necessary that carbonyl compounds don't have alpha hydrogen?
Till date, I have encountered carbonyl compounds without alpha hydrogens that can undergo Cannizzaro reaction. But recently I came across the following molecule and I can't understand the reason behind why it shows Cannizzaro reaction.

Welcome to ChemSE. Do you have a reference or some more details on this reaction?

In addition to user55119's request in the comments, can you give aexample with a reference how ketones without $\alpha$-hydrogen under goes Cannizarro reaction without having a hydrogen to transfer to the second ketone?

Cannizzaro reaction requires an aldehyde in the first place. Is this an aldehyde? Yes and no.


text:
Where does the energy in coal bonds come from?

I have been told that when the plants intake carbon dioxide they break carbon and oxygen apart and the carbon is utilised in making the plant's body. This process is driven by the sun's energy so when they decompose and form coal the bonds of coal have the sun's energy stored in it.

The above statement makes absolutely no sense to me.
When the sun's energy was utilised to separate carbon and oxygen how can the coal bonds store it?
If the coal bonds don't store the sun's energy where does the energy released on burning them come from?

Ever heard of photosynthesis?

@Waylander Know that pretty well.

@Waylander Do you mean that the energy in coal comes from the glucose formed in photosynthesis?

Eventually, yes, that's the ultimate source.

Yes. Photosynthesis is the engine that drives plant metabolism and growth.

What's the source of the opening statement?

Without dioxygen, there is no energy to be had from coal. So this is a terrestrial modern perspective. We call something a fuel if it releases energy during combustion. The same is true of almost any molecule you call "high energy" like ATP or hydrazine. TNT or nitroglycerine are exceptions because they don't need a reaction partner to explode.

@BuckThorn Here is a reputable agency that makes the "energy stored in bonds" claim: http://needtoknow.nas.edu/energy/energy-sources/the-sun/ . It is a bad short-hand expression you have to unlearn when you hear the complete story.

See also a similar question : https://chemistry.stackexchange.com/questions/76463/how-is-chemical-energy-stored-in-carbon

Indeed it's in some way counterintuitive to say that the sun's energy is 'stored' in the coal bonds. Energy never comes from breaking bonds, it is released when forming bonds.
In the case at hand, the sun's energy is used to break carbon-oxygen bonds through photosynthesis. You get that energy back when carbon-oxygen bonds are formed by $\ce{C}$ and $\ce{O2}$ reacting to $\ce{CO2}$: that's where the energy of burning coal comes from.
The details can be quite delicate, since in most chemical reactions new bonds are formed at the same time as old bonds are broken. It then boils down to the relative energies of those bonds. In the case of burning coal, it costs less energy to break the $\ce{C-C}$ bonds of coal than what the formation of $\ce{C=O}$ bonds yield, so the reaction is exothermic.

As has been pointed out the energy comes in the form of visible light from the sun.  The reaction written in summary or overview is
$$\displaystyle \ce{2H2O \overset{light}\to O2 + 4H^+ + 4e^-}$$
$$\displaystyle \ce{CO2 +4H^+ + 4e^-\to (CH2O) + H2O}$$
where in the first step, the oxidation of water, the light does not react directly with water. In the second step, $\ce{CH2O}$ represents carbohydrate and the electrons are carried by proteins.
The overall reactions occur in proteins in the cell membrane of chloroplasts (in green plants) and the first step is absorption of a red photon typically of $700$ nm. This causes an electron to be transferred from a chlorophyll dimer, (the special pair) down an electron transfer chain involving cytochromes and other proteins that eventually reduces the carbon dioxide, the second reaction above.
As the special pair is now short of an electron it recovers this via a metalloprotein grabbing an electron from water. Using X-ray diffraction all these protein structure are now known and ultrafast spectroscopy has measured most if not all the rates of these primary processes.
In energy terms, the couple: CO2-glucose has a redox potential $\Delta V=-0.43$ eV and for $\ce{H2O - O2}$ is $+0.82$ eV so the total redox span is $1.25$ eV and as $4$ electrons are involved this is makes $5$ eV in total. A visible photon at $700$ nm has an energy of $1.77$ eV and as one electron is released /photon absorbed then the total energy supplied is $7.1$ eV, which is more than enough for this reaction.
However, there are losses in the process, and the efficiency rarely exceeds $0.36$ which makes the useable energy available $0.63$ eV/photon which is not enough thus we need $5/0.63\to 8$ photons in total. This suggest that two photosystems (see Z-scheme) are needed in green plants and this is indeed the case.

Great analysis. And for completeness, what is coal, and how much energy is released upon its combustion?

Photosynthesis is a little like the reverse of burning
The idea that chemicals or their bonds store energy needs some clarification. The storage is always relative to some other reachable arrangement of bonds that can be achieved by some actual chemical reaction. So, if we say a specific compound "stores" energy we need to specify relative to what.
If we say coal stores energy what we usually mean is that we can burn coal (crudely, impure carbon) in air in a reaction that creates carbon dioxide and a large amount of heat. This is a simple widely observed and widely used reaction that we know releases energy.
But why? The energy is released because the total amount of energy tied up in the bonds of carbon dioxide is less than the energy tied up in the oxygen molecules and the solid carbon of the coal. When we burn coal that energy appears as light and heat we can use to warm our homes. We could, under different circumstances, react coal with other chemicals and get different outcomes (i'm sure coal will also burn in fluorine). But since the air around us has ~20% oxygen, burning coal is a very convenient way for us to produce heat.
The reason we say coal "stores" solar energy is because photosynthesis allows plants to drive the reaction in the opposite direction. Crudely (photosynthesis involves some very complex details) plants use the energy from sunlight to allow them to make a range of complicated carbohydrates by taking carbon dioxide from the air and water from the ground. (note photosynthesis also releases oxygen which is the main reason there is any in the atmosphere). The bonds in those more complicated molecules have higher net energy compared to the carbon dioxide and water they are made from and also compared to the products they would create if we burned them in air (wood burns too).
Coal is basically old plants that have been transformed by geological processes to impure carbon. So the ultimate reason we can burn it and release energy is because ancient plants captured energy from the sun to make more complicated molecules necessary for their growth. Burning is like the opposite of photosynthesis.
Coal "stores" energy because the arrangement of bonds in carbon and oxygen have higher energy than the arrangement in carbon dioxide and, in an atmosphere containing oxygen, we can release that difference as heat by burning the coal. But the ultimate reason the coal is the way it is is because plants made other complicated, higher energy, molecules by capturing solar energy to build them. Hence it is not unreasonable to say that coal "stores" energy from the sun (though it is oversimplifying a lot).

Electrons around the nuclei of atoms exist in various energy states due to quantum mechanical principles. Due to these principles, they are confined to certain configurations (shells and orbitals). Chemical bonds occur when an electron of one atom can simultaneously take up a vacant position of another atom.
For example, a neutral atom of carbon has six electrons which balance the charge of its six protons; two in a complete inner shell and four in its outer shell. The outer shell of carbon can hold up to eight electrons. A neutral atom of oxygen has eight protons and eight electrons; two in the inner shell and six in the outer shell (which can also hold up to eight electrons). So, an atom of carbon can bond with two atoms of oxygen as follows: two of the carbon atom's outer electrons are shared with one oxygen atom, allowing the oxygen atom to "complete" its outer shell, and the other two of the carbon atom's outer electrons are similarly shared with another oxygen atom.
When a carbon atom and two oxygen atoms are bonded like this, the electrons are in a lower energy state than if each of the atoms existed in an independent, neutral state. Different configurations, such as two oxygen atoms bonded to each other, or a  cluster of carbon atoms bonded to each other also each have different energy states (of their outer electrons). Plants have complex arrangements of special molecules which allow them to use energy received from sunlight to break the bonds of carbon dioxide and water molecules to build sugars and release molecular oxygen. Breaking the carbon dioxide and water bonds is all about elevating electrons in those atoms to higher energy states. This is how/where the energy is stored. When carbon and oxygen combine to form carbon dioxide, the electrons drop to lower energy states, which means energy is released.
As plants decay into substances like coal, bonds are reformed and some energy is released - that's what drives the decay process, but the bonds between carbon atoms in coal and in molecular oxygen are still at a higher energy state than in carbon dioxide.
There is one other notion here about activation potential. The resulting oxygen and sugar molecules are stable (they don't immediately react and go back to carbon dioxide and water) because it takes some amount of energy to initiate the transition. It's like a hill with a small dip at the top, you can roll a ball up the hill over a lip and into the dip and it won't roll back down again unless you first lift it back over the lip.

When the sun's energy was utilised to separate carbon and oxygen how can the coal bonds store it?

You use the word 'utilised' as if this means the energy is used; it no longer exists. But energy never disappears, it just changes form. When you take a rubber band and pull the ends apart, you consume (or better: move) energy, which is stored in the rubber band. If you now release one end of the rubber band it will snap together, the energy is released. Imagine there is a rubber band between the oxygen and the carbon atoms*, by separating the two, you add energy to the system. By combining the two, you release that energy again.
*The analogy does not hold for very long, moving the atoms further apart will not increase the energy stored in the system.

Here's a nicely simplified description of the structure of coal:

V.D Macromolecular Nature of Coals
Coals are believed to be three-dimensionally cross-linked macromolecular
networks containing dissolved organic material that can be removed by extraction.
This model offers the most detailed and complete explanation of the chemical and
mechanical behavior of coals. It is a relatively recent model and is somewhat
controversial at this writing. The insoluble portion of the coal comprises the
cross-linked network, one extraordinarily large molecule linked in a three-
dimensional array. This network is held together by covalent bonds and hydrogen
bonds, the weak interactions that play such a large role in the association of
biological molecules. The extractable portion of the coal is simply dissolved in
this solid, insoluble framework.

That was taken from a 2003 article in an encyclopedia about coal structure.
The plant matter the coal came from was somewhat similar in nature in that it also contained a lot of cross-linked organic molecules. The above can be read to imply that each lump of coal is a single solid molecule plus dissolved liquids and gasses but that's probably taking it a bit too far.
Lots of chemical reactions happen under heat and pressure including increased cross-linking of organic molecules. The heat and pressure can be viewed as shifting bonds as well as adding them. In the former case, that's still at least somewhat representative of the original sun-energy donation to the cause.
Note that the dissolved organics like methane (natural gas) are probably a mix of decomposed plant material and compounds produced under that same heat and pressure. If you want to get a sense of what those are, read up on making coke from coal for use in steel making.


text:
Why is is that tellurium(VI) fluoride is completely hydrolysed but iodine(III) fluoride isn't, even in hot water?
Tellurium(VI) fluoride is completely hydrolysed into tellurium(VI) oxide and hydrofluoric acid, but iodine(III) fluoride isn't, stopping at the iodine(III) fluoride dihydroxide/hydrofluoric acid level, even after extensive warming. Why is it that way, and how do I fix it?

Is $\ce{HIO2}$ even known? My understanding is it isn't.

Watch proper writing. Repeated confusing 0 and O  in chemical formulas is highly frown upon. Unless you have dyslexia/-graphia conditions, it is usually taken as ignorance and carelessness.

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.

Why should tellurium behave like iodine ?

And why should VI behave like III?

Why not? Tellurium hexafluoride also goes through an intermediate:

The hydrolysis of tellurium hexafluoride produces the fluoroorthotelluric acids, $\ce{TeF_n(OH)_{6–n}}$, n= 1–4, which undergo complete hydrolysis to orthotelluric acid only over a long period of time.

Ref.: Fraser, G.W., & Meikle, G.D. (1974). Hydrolysis of tellurium hexafluoride. Journal of The Chemical Society, Chemical Communications, 624-625, DOI: 10.1039/C39740000624


text:
How is the entropy of mixing related to chemical reactions in solution?
When calculating the Gibbs energy of reaction for reaction mixtures at arbitrary concentrations, you use the following expression:
$$\Delta_\mathrm r G = \Delta_\mathrm r G^\circ + RT\ln Q$$
The term $RT\ln Q$ has to do with the entropy of mixing, we are told.
How does that work for reactions that are continuously mixed? If I have a reaction like
$$\ce{A(aq) + B(aq) <=> C(aq) + D(aq)}$$
or
$$\ce{A(g) + B(g) <=> C(g) + D(g)},$$
the reactants themselves are already a mixture. In fact, when there is a solvent, even a single reactant is part of a solution, i.e. already in a mixture. If (hypothetically) the reaction goes to completion, you still have a mixture. So why do we have to consider the entropy of mixing if the reaction mixture is already mixed?

Do you mean that A+B are continuously added or that an equilibrium already exists as shown in your equation ? The calculation for entropy of mixing  is given in most textbooks as well as many times on this site.

Will you accept an answer based on a reaction of ideal gases?

@porphyrin One scenario is mixing A and B, and calculating $\Delta_r G$ at various times while the reaction reaches equilibrium.

@ChetMiller Yes, ideal gases or ideal solutions would be fine. The simpler, the better, I would think.

I'm a little unsure exactly what you are looking for but I try to show the connection between the Gibbs energy and mixing energy as reaction proceeds.  We can start with the chemical potential $\mu$ and calculate the Gibbs function. Thus,
$$\displaystyle \mu_i=\mu_i^0+RT\ln(P/P^0)+RT\ln(x_i)$$
where $P$ is the total pressure, ($P^0$ is unit pressure to make the log dimensionless and is assumed to be included from now on), $x_i$ the mole fraction of species $i$ and $\mu_i^0$ is a function of temperature only, i.e independent of composition so is unchanged as the mole fraction changes $x_i =0\to 1$.
In the reaction $A+B=C+D$ the Gibbs function is
$$\displaystyle G=n_a\mu_a+n_b\mu_b+n_c\mu_c+n_d\mu_d$$
and if we assume a perfect gas mixture, using the first equation produces
$$\displaystyle G=[n_a\mu_a^0+n_b\mu_b^0+n_c\mu_c^0+n_d\mu_d^0+RT(n_a+n_b+n_c+n_d)\ln(P)]+RT\big(n_a\ln(x_a)+n_b\ln(x_b)+n_c\ln(x_c)+n_d\ln(x_d)\big)$$
The first two terms (in square brackets) is the total Gibbs function for the four gases as if each is in a separate container at pressure $P$, the term on the right is therefore the free energy of mixing, $RT\sum_i n_i\ln(x_i)$ and if divided by temperature is minus the entropy of mixing.
If the system initially consists of only one mole each of A and B then
$$\displaystyle n_b=n_a,\quad n_c=n_d=1-n_a,\quad x_a=x_b=n_a/2,\quad x_c=x_d=(1-n_a)/2$$
and now $G$ only depends on $n_a$ which can vary between $0\to 1$
$$\displaystyle G= n_a(\mu_a^0+\mu_b^0)+(1-n_a)(\mu_c^0+\mu_d^0)+2RT\ln(P)+2RT(n_a\ln(n_a/2)+(1-n_a)\ln((1-n_a)/2)$$
and without loss of generality choose the pressure to be unity, then with some rearranging,
$$\displaystyle G= n_a(\mu_a^0+\mu_b^0)+2RTn_a\ln(n_a/2)+(1-n_a)(\mu_c^0+\mu_d^0)+2RT(1-n_a)\ln((1-n_a)/2)$$
When $n_a=1$ so that only reactants are present we find that $G=\mu_a^0+\mu_b^0+2RT\ln(1/2)$ where  $2RT\ln(1/2)$ has to be the free energy of mixing before any reaction has occurred and similarly when $n_a=0$ the same energy of mixing is present for the products there being only two species present in either case. Elsewhere both terms contribute to the mixing free energy.
The  shape of the plot of $G$ vs. $n_a$ is due primarily to the log terms, if this term were absent the plot would follow the line W-V, see figure. The lower part of $G$ shows that the free energy at equilibrium becomes even smaller than that if there were complete conversion to products, because here there are four species in various proportions not just two. Entropy of mixing depends on the mole fractions is maximal at equilibrium.
The position of the minimum, and so the equilibrium, is determined by $\mu_a^0+\mu_b^0-\mu_c^0-\mu_d^0$ which is the difference in free energies between reactants and product in their pure states. This is the case because the mixing terms are symmetrical in $n_a$ but the sloping energy terms biases this. The minimum can be calculated by differentiating $(\partial G/\partial n_a)_{T,P}=0$ or more easily and just as generally via
$$\displaystyle dG=-SdT+VdP+(\mu_a+\mu_b-\mu_c-\mu_d)dn_a$$
where the stoichiometry of the reaction was used ($-dn_a = -dn_b=dn_c=dn_d$) and at constant $T$ and $P$ when $\partial G/\partial n_a=0$, we find that
$$\displaystyle \mu_a+\mu_b=\mu_c+\mu_d$$
and substituting for $\mu_i$ and mole fractions leads to the solution for $n_a$ at equilibrium just as if $G$ had been differentiated.
What this shows is that the same relationship exists at equilibrium as in the chemical equation $A+B=C+D$ from which the equilibrium constant is normally obtained and shows the connection to the entropy of mixing as the reaction proceeds. The reaction is therefore seen as a competition between attractive forces causing the reactants to form products on the one hand and increasing entropy forming a mixture on the other.

Gibbs function of a reacting system

So, I am gonna use the most basic definition of entropy - the statistic one
S = kB ln $\Omega$
$\Omega$ is the total number of microstates possible.
In your case, different microstates can be formed by arranging momenta , energies and also by arranging POSITIONS as any molecule can be found at any possible "position" in a solution.
I will use basic principles of Permutations and Combinations to find $\Omega$.
Suppose I have 'a' and 'b' moles of A and B initially , while no C and D initially present. After a time 't', x moles of A and B have reacted to form x moles of C and D.
\begin{align}
\ce{A(aq) + B(aq) <=> C(aq) + D(aq)}\\
\end{align}
\begin{align}
\ce{t=0..... . a. . . . . . b. . .  ....0...  . .  0 ....... . . . . }\\
\end{align}
\begin{align}
\ce{t=t..... (a-x). . . . .(b-x). . ... x... . . x ...... . . . . . . }\\
\end{align}
The total number of molecules remain same in your example , and is (a+b+w) N0 , where N0 is Avogadro's number , and w is the number of moles of water.
Let me simplify the model by assuming there are (a+b+w) N0 positions available. Any molecule can "sit" at any position and can have any of the available value of momentum and energy. Changing either of position or momentum of any one molecule will lead to a new microstate.
But there's a catch. Suppose molecule 1 (m1) sits at position r1 and has energy E1 , momentum p1. Another identical molecule (i.e. the same compound) m2 has position r2 , energy E2 , momentum p2. If I exchange all - their positions, energies and momenta - I will not get a new microstate.
But if m2 was not identical, it would surely have resulted in a new microstate.
So, assume I haven't put any molecule at any position. First I assign every position a value of energy and momentum , and write down each possible arrangement in my notepad (my notepad has infinite pages...so don't worry , it won't run out ; ) ). Then I count them. There turn out to be Z possible arrangements of energy and momentum.
Now , I start placing the molecules randomly ( there are just A , B and water present for now). How many unique arrangements can I have only due to virtue of positioning? That I can easily calculate. It will be
\begin{equation}
V_{1} = \frac{[ (a+b+w) N_{0} ]!}{( aN_{0} )! * ( bN_{0} )! * ( wN_{0} )!}
\end{equation}
So, total number of microstates ($\Omega$1) = V1 * Z  Hence, S1 = kB ln( V1 Z )
Now, after time t , I again do the same.  The number of arrangements of energies and momenta would still be same , i.e. , Z .
But, the number of arrangements due to positioning will now be ,
\begin{equation}
V_{2} = \frac{[(a+b+w) N_{0}]!}  {[(a-x) N_{0}]! * [(b-x) N_{0}]! * (xN_{0})! * (xN_{0})! * (wN_{0})! }
\end{equation}
$\Omega$2 = V2 * Z  Hence, S2 = kB ln(V2 Z)
So, Change in Entropy $\Delta$S = S2 - S1 = kB ln (V2 / V1 )
Dividing V2 by V1 , one can easily see it is equal to
\begin{equation}
\frac{V_{2}}{V_{1}} = \binom{aN_{0}}{xN_{0}} * \binom{bN_{0}}{xN_{0}}
\end{equation}
\begin{equation}
Hence, \Delta S = k_{B} * \ln\left[ \binom{aN_{0}}{xN_{0}} * \binom{bN_{0}}{xN_{0}} \right]
\end{equation}
where\begin{equation}
\binom{n}{r}\end{equation} represents binomial term nCr
This is the increase in entropy due to "mixing".
I even plotted a graph to verify. Here's the graph.
I didn't know how to approximate its derivative (as the graph is broken when viewed closer) So I instead compared the graph with another one made by integrating Rln(K/Q). Both come out to be same at some value of K. This way, we have also calculated the Equilibrium Constant :)
NOTE: In the graph, I have also considered change in "Internal Entropies" apart from Entropy of Mixing. This "internal" entropy arise due to arrangements possible within a molecule. It is characteristic of a given compound.
The Black graph is the one made by integrating Rln(K/Q) and the Red haze visible behind it is the graph made by the formula we just derived.
I took a and b = 1  SA , SB , SC , SD (the molar "internal entropies" of A , B , C and D respectively) equal to 1 , 1.5 , 2.3 , 2 kJ/mol*K (not to scale....just example)  And K came out to be nearly 1.2

Sorry. I don't understand the meaning of $\pu{ \frac{V_2}{V_1}}$ with your strange parenthesis.

It is binomial term... (aNo C xNo)

I took No (Avogadro's No.)  equal to 500 in graphing calculator because its system broke down at higher values. But it doesn't matter as the values reach a limit on increasing No....So, graph is same for No = 500 and No = 6*10^23

minor formatting suggestions: using "\ln" will keep ln from being italicized, and "\left[" and "\right]" instead of just "[" and "]" will cause the size (ie height) to adjust properly to match the contents within the brackets.

What a tour de force! But you showed nicely that for Q = 1, the microstates are maximal. To calculate K, you would have to know the standard entropy and enthalpy of reaction. If you just consider the entropy of mixing, your are setting these other quantities to zero, in which case you would expect K = 1 indeed.

For large n, you can use the simplest of Stirling's approximations for ln N!: $$\ln N! = N \ln N - N$$ This will allow you to evaluate the entropy analytically rather than computationally.

@KarstenTheis In my graph, I also considered "internal" entropy. That gave deviation of K from 1 (It came out to be 1.2)..... There sure may be some more factors, but I don't think Enthalpy of Reaction is one of them. The expression for TOTAL entropy (system + surroundings) in terms of extent of reaction will NOT depend upon enthalpy. So, its partial derivative w.r.t. extent of reaction will also be independent of enthalpy. Putting this derivative zero,  we can find at what extent would total entropy be zero. This extent will give us K

@KarstenTheis Stirling's Approximation!  That's cool. That's what I needed

For a reaction of ideal gases, such as, say, aA+bB-->cC+dD, the $\Delta G_r$ in your equation is the change in Gibbs free energy in going at constant temperature T from a moles of pure A and b moles of pure B at arbitrary pressures $p_A$ and $p_B$, respectively, to c moles of pure C and d moles of pure D at arbitrary pressures $p_C$ and $p_D$, respectively.  If the pressures were not arbitrary, but, instead happened to coincide with those for an equilibrium reaction mixture of A, B, C, and D, then $\Delta G_r$ would be equal to zero, and Q would be equal to the equilibrium constant for the reaction.  Under these circumstances, the pure gas pressures would be equal to the partial pressures of the corresponding components in the equilibrium reaction mixture, and could be placed in equilibrium with the components in the mixture through 4 semipermeable membranes which allowed passage of only one of the corresponding components.
The challenge in all this is to figure out how to interpret all this in terms of a reversible process to go from the initial state of pure A and B to the final state of pure C and D so that $\Delta G_r$ can be expressed in terms of the initial and final pressures. This typically is facilitated by the use of a so-called van't Hopf equilibrium box.

I did not know about the van't Hoff equilibrium box, https://chemistry.stackexchange.com/a/127915

So for ideal gases, the entropy gain of mixing is the same as the entropy gain of expansion? It does not matter whether it is a pure gas or a mixture, the entropy depends only on the (partial) pressure.

Basically, yes.  The partial molar entropy of a component in an ideal gas mixture is the same as the molar entropy of the pure component at the same temperature and at a pressure equal to the partial pressure of the component in the mixture.  This follows from Gibbs Theorem,


text:
How can I transform Thiamine sulfate to Thiamine hydrochloride?
I have got thiamine sulfate and want to have thiamine hydro chloride. Can I obtain it by dissolving thiamine sulfate in water and adding calcium chloride water solution as long as calcium sulfate will percipitate out. After it filter the solution and boil off the water? If it's not a good method, can someone tell me a better one?

If you have ion exchange resin, that’s an easier method. Also would recommend boiling off the water under reduced pressure (eg rotavap)

Calcium sulfate is relatively soluble in water. If you want a pure sample of thiamine chloride, it is better to replace calcium chloride $\ce{CaCl2}$ by barium chloride $\ce{BaCl2}$, because it produces barium sulfate which is much more insoluble than calcium sulfate. After filtration, you obtain a pure solution of thiamine chloride (Vitamin B1). But be cautious ! Barium compounds are toxic.


text:
Solving two-electron integrals
I have been trying to teach myself Quantum Chemistry through a book, and I came across this problem and I am not sure how they solve a certain problem.
Essentially, they applied first-order perturbation theory to helium to get a correction term of $\frac{5}{8}Z$ by solving the equation given below in the question which I believe is the Coulomb integral between the two electrons in the $\mathrm{1s}$ orbital of Helium.

The next part of the question is what confuses me. They convert to spherical coordinates and the integral becomes much different.

At this point, I can solve the rest of the problem, but I am stuck on how the conversion worked, and I was wondering if anyone could help me.
Another question I have is that when solving for the electron-electron repulsion energy, would the conversion between the first integral expression to the second be the same if the electrons were in a separate orbital (i.e. For lithium between $\mathrm{1s}$ and $\mathrm{2s}$ electrons)?

I feel like this has been asked and answered here not very long ago.

@user1271772 can you point to a more direct source?

To introduce a $\theta$-dependency into the expression for $r_{12}$, use that $\vec{a}\cdot\vec{b}=ab \cos\theta$, where $\theta$ is the angle between the two vectors:
$$r_{12}=|\vec{r}_{12}|=|\vec{r}_1-\vec{r}_2|=\sqrt{(\vec{r}_1-\vec{r}_2)^2}=\sqrt{\vec{r}_1^2-\vec{r}_2^2-2\vec{r}_1\cdot \vec{r}_2}=\sqrt{\vec{r}_1^2-\vec{r}_2^2-2r_1 r_2\cos \theta}$$
with $\theta=\theta_1=\theta_2$.
Now switch to a spherical coordinate system. The integral over the full 3d-space changes as follows:
$$\int\limits_\mathbb{R^3} \text{d}\vec{r}_2 = \int\limits_0^\infty \mathrm{d}r_2\int\limits_0^{\pi} \mathrm{d}\theta_2\int\limits_0^{2\pi} \mathrm{d}\phi_2\,r_2^2 \sin\theta_2$$
where $r^2 \sin\theta$ is the value of the Jacobian determinant of the transformation between spherical coordinates and Cartesian coordinates.

It's the law of cosines. 1/norm(r1-r2) where both are vectors is the square root of (r1-r2)^2 = r1^2+r2^2-2(r1 dot r2)
The last term is the product of the norms times the cosine of the angle between them.
P.s. sorry I don't know how to format code. First time answering

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

I understood it was the rule of cosines that was used. I was more confused about how some of the added terms arose with the conversion to spherical coordinates like the integral with phi and also the $sin\theta d\theta$ term. I think I figured it out a few months back but I sort of forgot how it worked.


text:
Recovery of precipitated polymer
I have recently newly synthesized the polymer. Based on its structure I tried to precipitate its concentrated solution in dichloromethane by dropping it into methanol. It seemed to work. The methanol solution was cloudy, some gooey things that sticked to the wall of the flask were observed. Initially I tried to recover the precipitated polymer by filtering through membrane filter applying reduced pressure (Chmlab, MTF/H Grade, PTFE hydrophobic membrane filter, pore size 0.2 μm, diameter 47 mm). The solution is cloudy orange color, and the filtrate was yellow color. The filtration seemed successful, but shortly after the filtration hardly continued. I am assuming that the precipitated polymer blocked all of the pores of the membrane filter. When I stopped the filtration and checked the filter, there was no powder-like thing on the top surface of the filter, but something seemed to be adsorbed on the surface. I have tried larger pore size, 1~2 μm, but the filtrate has cloudy orange color same to the solution, and shortly after its pores were also blocked. In this situation, do I have to try centrifuging? After centrifuging, can I recover the polymer simply by pouring it onto the membrane filter and washing with methanol?

Centrifuging seems the obvious way to go. I might be inclined to put the centrifuge tube in a vac oven to dry it out after centrifuging in the hope that the dried polymer will be easier to handle.

@Waylander The polymer is sticked to the bottom of the centrifuge tube... Would change of the precipitant to ethanol or IPA help?

Can you redissolve the polymer in the tube in a little DCM, remove it to a storage vial and let the solvent evaporate to give you your sample? It may be partially soluble in EtOH or iPrOH.


text:
What drives double displacement reactions?
I have two questions:

I understand that in a displacement reaction the more reactive element displace the less reactive element. But why? In the reaction with Zinc and Copper Sulfate, we form Copper and Zinc Sulfate. But both Copper Sulfate and Zinc Sulfate are soluble in water so how does Zinc Sulfate stay together?

What drives double displacement reactions? They are not redox reactions so the reactivity series doesn't apply, so how can we explain why the anions and cations switch places. Is this energetically favorable?

Dissolved salts exist as separated, hydrated ions, moving and reacting independently. (*)
Like for $\ce{Zn}$ and $\ce{CuSO4}$:

The reaction is $\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$.
The ion $\ce{SO4^2-(aq)}$ is called a spectator/bystanding ion in the reaction context, not participating on the reaction.
The reaction can be formally written using full salt formulas, which should not be confused with molecules, as they express just ion stoichiometric ratios: $\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}$.
The driving force is here different ion affinity to electrons and related redox potential and reaction Gibbs energy.

The example for "double displacement" reaction - precipitation of $\ce{AgCl}$ from $\ce{AgNO3}$ and $\ce{KCl}$:

The reaction is $\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$.
The spectator/bystanding ions are $\ce{NO3-(aq)}$ and $\ce{K+(aq)}$.
The reaction can be again formally written using full salt formulas: $\ce{AgNO3(aq) + KCl(aq) -> KNO3(aq) + AgCl(s)}$
The driving force is here the difference in mutual ion affinity for various cation + anion combinations, leading to precipitation of the least soluble salt - $\ce{AgCl}$ in this case.
It is closely related to energy released by ion hydration versus energy released by ions forming solid ionic lattice, hold by coulombic forces.
It again depends on the reaction Gibbs energy.
The same case is when all 4 ion combinations form more or less well soluble salts.

(*) At some conditions, hydrated ions may not move independently, but they can have various degree of mutual connection of their hydration envelope, forming ion pairs.

"The driving force is here the difference in mutual ion affinity for various cation + anion combinations, leading to precipitation of the least soluble salt - AgCl in this case." I don't quite understand what this means, could you explain it in layman's terms?

There is Alice, Karen, Nigel and Carl. Alice and Carl feel strong mutual attraction, forming a couple, ending in a bad. Karen and Nigel remain single, just watching what is happening, walking aimlessly.