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What is the direction of the net electric dipole moment in hydronium H3O+?
Assuming electric dipole moment points towards the negative charge, what would its direction be in hydronium?
Although the bonding electron density is distorted towards the more electronegative oxygen, it already has a formal charge of +1. Assuming the partial charge on oxygen due to distorted shared e.density is less than O.5, wouldn’t the actual electric dipole moment points away from the oxygen(opposite to distortion)? I can’t find any source to check this

This is why describing molecules with Lewis structures isn't adequate.  The formal charge overestimate the sharing of electrons. In addition, you can even see the oxygen pulling even more right because its formal charge. The formal charge does not reflect the overall electron density, and the latter is more near the oxygen, as you probably already know.

@Alchimista I know the distortion of shared electron density isnt really affected by formal charges. However, I  thought the formal charge on each atom  in a perfectly covalent case supplemented with the distortion of shared e density provided by electronegativity difference would provide an accurate picture of the actual charges on each atom, no?

If you know that, that your second logical passage is not on ground. Note that the example you use can not exist. Basically formal charges are not poles of whatever component or sum dipole moment.

In aqueous medium, the proton ($\ce{H+}$) is thought to exist as hydronium, $\ce{H3O+}$. However, in reality, probably bigger complexes of water may also be associated with one proton. In liquid water, everything is interconnected with strong hydrogen bonds anyways, and it is difficult to distinguish the free species. All of this means that it would be difficult to measure the dipole moment of $\ce{H3O+}$ experimentally. I haven't found any reference that did this, and I suspect it is impossible.
So we have to turn to computational chemistry to calculate the dipole moment of $\ce{H3O+}$. For one isolated $\ce{H3O+}$ (i.e. in gas phase):

The z-axis is set along the $C_3$ axis.
$\text{M06-2X/aug-pcseg-2}$ gives a dipole moment of $\text{1.515 D}$ in z-direction. In x and y direction, the dipole values are negligible:
         DX          DY          DZ         /D/  (DEBYE)
    -0.007834    0.000072    1.515546    1.515566

This means the dipole points along +z-axis in the picture. Partial charges(ESP) fitted to the dipole moment are:
 NET CHARGES:
 -------------------------------------
 ATOM                CHARGE    E.S.D.
 -------------------------------------
 O                  -0.3668    0.0000
 H                   0.4561   -0.0000
 H                   0.4553   -0.0000
 H                   0.4554    0.0000
 -------------------------------------

As you can see, the oxygen atoms has a partial negative charge and hydrogen atoms have partial positive charges. So, the formal charges that we assign to the Lewis structure does not have much significance in this case.
(IUPAC convention of dipole moment is that the arrow points from the negative charge towards the direction of the positive charge. This is exactly opposite of what you wrote in the question)
Another method, $\text{MP2/aug-pcseg-2}$ gives $\text{1.507 D}$ along z-axis, which is close to what we got from the DFT calculation.
(Note that I have not enforced symmetry for the calculation, which is why the partial charges are slightly different for each hydrogen. This is a numerical error, because in reality all the hydrogens are equivalent)

Careful! The calculation of the dipole moment in ions is dependent on the choice of coordinate system. Please check that the center of mass is also the center of the coordinate system. The direction should be the same as in ammonia.

@Martin-マーチン I have symmetrized the molecule to ensure the C3 axis falls along z. Does the centre of mass actually matter? All the dipole moment is along the z-axis anyways, so moving it up and down along z would not change anything.

If you shift the whole molecule along an axis, you will shift a positive charge away from the origin. I would not be certain whether the program actually does account for that, corrects that, is even able to do these kinds of translations. Also, even if you have symmetrised the molecule, it is not a given that the program will use this kind of orientation. You would have to have insight into the algorithms to know about that. It should be relatively easy to check though, but it is important to do that. || Another note: Please use MathJax only for Maths or Chemistry expressions.

@Martin-マーチン I am using GAMESS, and as far as I know, it calculates dipole moments directly from the wavefunction, not from partial charges, so I think moving along axis would not change it. I have used cartesian coordinates, and it does not change the coordinate system inside, so I assumed that the same orientation would be maintained. The symmetrization is done in the GUI input builder, not by GAMESS itself.

Abstractly speaking, the whole molecule is a point charge. If you calculate all vectors from the origin, those vectors will change depending on where you put the molecule. I have not used GAMESS in a long time, but I would be surprised if it didn't do coordinate transformation internally. Your output shows that you have no symmetry, your output shows that the dipole moment is not aligned with the z-axis. While there still are numerical inaccuracies, this is not one of those. You can easily check: run a single point on a shifted coordinate system. (Add 10 to all x/y/z coordinates.)

@Martin-マーチン I have checked, adding a constant to any coordinate does not change the value of the dipole moments.

I guess that's good news then. The topic recently came up at CCL: http://www.ccl.net/cgi-bin/ccl/message-new?2021+03+12+002


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Titration curve equation/function
How can I create a graph of a titration curve (acid-base, strong or weak, not something specific), with an equation or a function. In other words, what is the equation that describes such a curve that could be used in graphing software to recreate the curve without actually needing any experimental values? Also, if they exist, what are the parameters of the function (Molarity? Volume? Etc.)

https://doi.org/10.1007/s00897000426a

This has essentially been done in a number of good answers at this site. Among other ways I do this, I use the equations in the spreadsheet screenshots in this answer: https://chemistry.stackexchange.com/a/136203/79678.

Or try Tom O’Haver’s free Excel or OpenOffice titration curve software: https://terpconnect.umd.edu/~toh/models/TitrationDemo.html. There is a lot of stuff at his website, so maybe look around there. And, of course, there are books, other free pH calculation software, etc.

@EdV Thanks, I've visited the answer you mentioned, but there's not a pH variable in any of the equations there. Do I just solve for it with a negative log?

@EdV This is for excel, I'll try it out but I prefer a mathematical function rather than a spreadseet, so that I can use it anywhere. Except if there's a way to extract the function from excel?

The equations are right in the spreadsheet screenshots! They come directly from the equilibrium constant expressions, mass balance, etc. So there is no need to use a spreadsheet. And pH is just negative log of hydrogen ion concentration, ignoring activity coefficients. So you go back and forth with those.

@EdV Alright. So for monoprotic acids I'll be fine with that. I'll try it out and if I find any problems I'll write back. Thanks!

So for monoprotic acids, only the first acid dissociation constant exists. Therefore, just take the equation in the second spreadsheet screenshot, i.e., carbonic acid titrated with NaOH, and let the second and third dissociation constants equal zero, like it says there. This greatly simplifies the equation.

I tried solving for [H+], it's a mess, haven't even tried to take the negative log of it. Any suggestions?

https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.494.1535&rep=rep1&type=pdf

Go here and scroll down past the photos of sapphire tubing. There is no problem solving for [H+], but it is very easy to just take the easy route, as explained in the link.

@EdV I didn't understand how I can graph it for -log([H+]) when it is just a concentration. The formula solved for Va that you described contained the concentration of H+. How do I solve for pH? (This is more of a maths question rather than a chemistry one)

Let us assume, arbitrarily, that you use the easy way. A spreadsheet program is convenient. So, first make a column of pH values you are interested in, e.g., values such as 1.00, 1.10, ..., 13.00. Now in the next column, put in the corresponding [H+] values. These are [H+] = 10^(-pH). So now you have two columns. Now use the Vb equation to compute Vb for each [H+] in the second column. Put these in a third column. This is very easy using an actual spreadsheet program and also easy using any computer language. Then plot pH as the vertical axis and Vb as the horizontal axis. All done!

I updated what was at the link here to show the whole thing: a spreadsheet for titration of acetic acid with MOH. Study the spreadsheet until it makes good sense, OK?

@EdV Sir thank you very much for your work. I will study the spreadsheet and I will let you know.

Good! Please copy what I posted at the link: it is only temporary and I will delete it in a three days.

I updated the spreadsheet screen shot: the volumes Va and Vb are in L, not mL. It does not really matter except to be more realistic: 0.025 L = 25 mL, which is typical in a titration.

If $\pu{1 l}$ of $\pu{1 M}$ $\ce{HCl}$ is gradually neutralized by adding $x\,\pu{mol}$ $\ce{NaOH}$ without change in volume, the $\mathrm{pH}$ of the obtained solution is given by
$$\mathrm{pH} = -\log\left(\frac{1-x}{2} + \frac{1}{2}\sqrt{(1 - x)^2 + 4\times10^{-14}}\right).$$

I want the volume to be a variable here. Suppose we have an unkown molarity weak acid we want to titrate with a strong base of known molarity. We gradually add the base solution to the acid solution. The volume is variable.

@MariosA    And that is already done in the equation I referenced in other comments: just zero out the second and third acid dissociation constants and the equation simplifies greatly. Then either use pH as the independent variable, computing base volumes as a function of pH (this is the very easy trick) or use base volume as the independent variable and compute pH as a function of base volume (a little harder, but we have computers now).

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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text:
How to dissolve hardened aluminum silicate beads back into solution?
TLDR: We use the adsorber beads (which is essentially alumino-silicate beads) to adsorb some  stuff from our process stream. The spent beads are dissolved in NaOH and sent to our wastewater treatment facility for precipitation and separation through filter presses.
Anyway, due to unavoidable circumstances, the dissolved beads in caustic solution was held up overnight before it could be pumped to wastewater treatment. The dissolved beads solidified in the vat, and would not re-dissolve on adding excess caustic. To be clear, it does dissolve only at the top of the vat, but cannot penetrate the interiors.
Can someone explain the chemistry of what might have formed (polymerization, silica gel, etc?) and how do I go about dissolving it? Should I just blast some holes in the solid with pressurized water, and then put caustic into it for increased penetration?
Much appreciated.

The chemistry of alkaline silicate is not yet perfectly understood. In the industry, sodium silicate is produced by fusing silica and $\ce{NaOH}$ or $\ce{Na2CO3}$. In both case the obtained sodium silicate is nearly insoluble in water. But it can be dissolved by using hot pressurized water. That is the way "liquid glass" is obtained. Why don't you try  using pressurized water ? Start with cold water, and try hot water in case of failure.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Why are functional groups a part of hydrocarbons?
So, I was asked a very innocent question by my friend that all hydrocarbons are defined as compounds made up entirely of hydrogen and carbon(which can be confirmed by going to any educational site like Wikipedia, Brilliant, Khan academy etc.). But, everyone almost immediately sort of contradicts it by putting functional groups as a part of hydrocarbons.
So a few questions-

why are functional groups part of hydrocarbons, despite being different from hydrogen and carbon?
if 1. is answered successfully, then is the definition wrong for hydrocarbons?

P.S. me and my friends are 10 graders, so a language simple enough for us would be much appreciated.
Thanking in advance

"putting functional groups as a part of hydrocarbons" Who is the "everyone" who does this?

My coursebook, even khan academy groups hydrocarbons and functional groups

Can you provide links or photos of your book, etc.?

https://ncert.nic.in/ncerts/l/jesc104.pdf

As far as I saw, that book does not make this claim. Even NCERT gets that one right.

That is a classic "even NCERT gets that one right"!

Hydrocarbon is an umbrella term for all compounds which contain carbon or hydrogen; benzene, methane, acetylene all are hydrocarbons. Yet you can attach words as sulfonated before the hydrocarbon(s) as a class of compounds such as a benzene sulfonic acid.
Hydrocarbons may have "hydrocarbon" functional groups such as the iso- before the common name, isobutane, isoctane etc. This is isooctane or in modern IUPAC nomenclature it is
2,2,4-Trimethylpentane. So you do have methyl groups in this hydrocarbon.


text:
Can you make R-Candy without heating the propellant mix?
I'm wondering if there's any way you can make "Rocket Candy" (or R-candy) fuel without cooking it. I'm building a model rocket that I'm hoping will get about 3 km on my first build (It's not going to be super big), and I thought R-candy was a good idea. I am just getting into this stuff, but I'm doing a ton of research and trying to be safe.
In every recipe I've seen, it says that you need to "cook it" on a stove or use some other heat source. I'm wondering if you can avoid using a stove, oven, etc.
If there's no way to make R-candy without cooking it, is there some other fuel I can make from household materials?
P.S. The ammonium–aluminum fuel isn't an option for me.

First of all, when it comes to something that may impact your future well-being in a bad way, do a thorough research on the topic. Wikipedia is a good starting point as it highlights the role of homogenization in the fuel–oxidizer mix preparation:

There are many different methods for preparation of a sugar-based rocket propellant. Dry compression does not require heating, only the grinding of the components and then packing into the motor. However, this method is not recommended for serious experimenting.[2] Dry heating does not actually melt the $\ce{KNO3},$ but it melts the sugar and then the $\ce{KNO3}$ grains become suspended in the sugar.

Heating assures homogenization and packing of the fuel–oxidizer mix to create constant burn rate. If you decide to heat the mix, it is imperative that you look up the properties of all components, especially melting points. For example, in the most common nitrate–sorbitol mix sorbitol (m.p. 95 °C) would act as a flux for potassium nitrate (m.p. 334 °C). Make sure you are using a well-functioning hot plate with a temperature controller and not open flames, otherwise the mix will ignite.
Make sure the powders are dry before you mix them. Otherwise trapped water inside molten sorbitol (remember, m.p. 95 °C and both sorbitol and nitrates are hygroscopic) will create steam pockets which may cause local explosion with the stuff dispersed around and potentially catching fire from the hot plate. Needless to say, the mix must be carefully stirred and slowly heated until fully homogenized; only then it can be poured into the firm template or the rocket body. The mix is going to be at about 150 °C and sticky, so wear heat-resisting protective gloves.
If a hot plate is not an option, you may opt out for the worse alternative of grinding the powders and pressing into shape. First, you have to produce fine powders of the fuel and oxidizer separately. Normally, separated sorbitol and potassium nitrates should not ignite because of friction, but work with small batches at a time regardless. Second, they must be thoroughly mixed and pressed to assure good contact between particles. Work with dry components in small quantities without producing too much dust flying around to prevent dust explosion.
General notes:

Work with small portions of materials at a time in a well-ventilated area using PPE.
Check your equipment thoroughly and make sure there are no sources of sparks and open flames nearby. Molten mix will ignite when overheated, and the dust is explosive.
Remaining water will quench burning due to its tremendous heat capacity and may cause unstable engine operation with a non-zero chance of explosion, so I'd avoid adding water at all regardless what preparation method you pick.

Thanks so much! I am curious, though. After looking at the dust explosion Wikipedia article, I found that flour is extremely combustible. Would you be able to use packed (or non-packed) flour as rocket fuel?

Probably yes (sugars and starch, the main component of flour, have comparable heat of combustion $\Delta_\mathrm{c}H^\circ_{298}\approx\pu{17 kJ g-1}$), but I suspect flour, being a mixture on its own, will cause a slower burning rate.

And aside from the chemistry, get all your paperwork lined up properly to do such a launch.

OoT Just curious about regulation. Can one shoot something 3 km high without some kind of " cleaning" before?

@Alchimista Take a look at this link: united states - Do you need approval from the FAA for ametuer rocket launches?


text:
Why are nanoliposome encapsulated antibiotics still so unavailable despite being so effective?
I wasnt sure how the phrase the headline question really nor if this is the right place to ask it, but i'm primarily interested in the chemical synthesis aspect and the experience of chemists regarding this subject.
In the antimicrobial literature specific liposome or nano particle encapsulations of antibiotics have been shown over and over again to show immense potential for quite a long time by now, enhancing efficacy, pharmacokinetics while at the same time reducing toxicities. My question basically is, why is this not being used really? Advanced liposomes can be found in some vaccines for example and the likes of it. But we continue to pump e.g. Aminogylcosides like Gentamicin into people which while very effective against certain hard to treat organisms can fail due to 1. pharmacokinetic limitations (low intracellular penetration etc.) and 2. toxicity and damage.
Im not interested in the discussion about efficacy for treatment in people, as that wouldnt be on the subject of chemistry and that has already been shown, but rather what is it that makes it so hard about this being available? I found literature ranging from 20 years ago showing specific PH sensitive liposome encapsulations of Gentamicin vastly increased efficacy against gram negative intracellular infections in vitro and in vivo and immensely reduced toxicity towards human cells - yet its just nowhere to be seen.
Is it that hard to synthesize/make these compounds? Does it cost a castle made of gold to produce a single vial of it? I would like to hear chemists views on the complexity/cost ratio of the production if possible. I do no have an advanced background in chemistry but it is just hard for me to imagine it would take half the world to produce some specifically charged or ph sensitive liposomes - especially if the science on what and how has already been done.

Notice that @MathewMahindaratne gave an excellent and very well informed answer to your question: he is one of the authors of the second and third references he gave! So you got an answer from an expert! Nice when that happens!

Liposomes are the first nano-drug delivery systems that have been successfully translated into real-time clinical applications. So far, all liposomal drugs are designed for intravenous applications with control release. Most of them are designed to certain targets such as cancer cells. That's why most of liposomal drugs are used in chemotherapy (Ref.1):

The main fact is those liposomal drugs are used mostly in hospitals with control of appropriate medical personals. For example, all seven liposomal drugs so far in chemotherapy are Doxil®, Depocyt®, DaunoXome®, Mepact®, Marqibo®, Myocet®, and Onivyde® as listed in Ref.1 used in cancer patients (in- or out-patients, but under the supervision of their medical doctors). Morphine sulfate extended-release liposome injection, DepoDur™, and non-opioid, bupivacaine liposome (trade name Exparel®) are analgesics used in pain management, specially after postsurgical procedures (e.g., bupivacaine liposome is a postsurgical analgesic used in the management of postsurgical pain and It can also numb an area of the body to relieve pain during surgery or medical procedures). In addition, VYXEOS® is recently FDA approved  (First approved August 3, 2017) liposomal drug for the treatment of adults with newly-diagnosed therapy-related acute myeloid leukemia (t-AML) or AML with myelodysplasia-related changes (AML-MRC). VYXEOS®, a liposomal daunorubicin/Cytarabine, is a fixed-dose combination of liposomal bound cytarabine and daunorubicin that delivers the two medications in a 5:1 molar ratio. Further, it is the ONLY treatment recommended by National Comprehensive Cancer Network® (NCCN®) for induction in sAML patients $\ge \pu{60 years}$. Also, in addition to its use in chemotherapy for leukemia, DaunoXome®, a liposomal daunorubicin, is also FDA approved to treat AIDS related Kaposi's sarcoma.
Based on these facts, therefore, it is safe to say that liposomal drugs are in use without aware of general public. Regardless to say that those treatments are very expensive as you have guessed.
I think main drawback to not available liposomal antibiotics is their high vulnerability to acidic conditions, thus no oral applications available (almost all liposome broke open in the stomach). So far, only literature available for acid stable liposome is Ref.2 and 3. However, we have to stop the development in that direction due to the lack of funding. Therefore, do not believe what you are hearing. Liposomal applications and their oral applications are coming pretty soon (someone else may follow our direction). Nonetheless, now we have a vaccine for COVID-19 within a year, don't we?
References:

Upendra Bulbake, Sindhu Doppalapudi, Nagavendra Kommineni, Wahid Khan, "Liposomal Formulations in Clinical Use: An Updated Review," Pharmaceutics 2017, 9(2), 12 (33 pages) (doi: 10.3390/pharmaceutics9020012)(PDF).
Adelphe M. Mfuh, Mathew P. D. Mahindaratne, Maritza V. Quintero, Frederick J. Lakner, Ande Bao, Beth A. Goins, William T. Phillips, George R. Negrete, "Novel asparagine-derived lipid enhances distearoylphosphatidylcholine bilayer resistance to acidic conditions," Langmuir 2011, 27(8), 4447-4455 (DOI: https://doi.org/10.1021/la105085k).
George R. Negrete, Mathew P. D. Mahindaratne, Adelph M. Mfuh, Maritza V. Quintero, "Compositions and Methods Related to Acid Stable Lipid Nanospheres," US20110268653A1 (https://patents.google.com/patent/US20110268653A1/en) (PDF).

You can't beat including your own references :)

@Buck Thorn: I don't like to do that but it is an important work among liposomes.

No worries, I was not being facetious. I think it's very cool, and referencing your own work should not be discouraged.

Great answer, thank you. I was aware those things were used in more "popular" subjects, but the lack of it specifically in the treatment of persistent infection/infection requiring advanced pharmacokinetics was what irritated me. I guess im being impatient since i know people in a very bad place who could use those drugs as a last hope really bad. I hope you are right with this stuff coming soon! Btw, as you pointed out the cost aspect - i know this is probably a question you cannot answer but what are we talking about here ballpark wise? I guess it depends also on the drug/formulation...

The FDA and other regulatory agencies require that formulations are subjected to a variety of tests. Clinical trials are laborious and expensive. Studies demonstrating efficacy in vitro, in small animal studies, or in human pre-clinical trials still have to be translated into human clinical trials. This can be a stumbling block. Quoting a reference posted by Mathew Mahindaratne:

Several liposomes were successfully translated into the clinic and other liposomal formulations are in different phases of clinical investigation. Although many of these products have been proven to be beneficial in preclinical trials, only formulations that show efficacy in clinical trials will make their way into the clinic.

Efficacy and toxicity are only part of the testing. Stability is also important. In addition you have to find a way to prepare the formulations on an appropriate scale, in a cost-effective way. Ref 1. (a little dated and from a non-mainstream journal but open access) summarizes some of the potential disadvantages with liposome formulations (distribution in the body, proneness  to  oxidation  and/or  hydrolysis, and scale-up of production).
References

Sharma Vijay K, Mishra D N, Sharma A K,Srivastava. Liposomes: Present Prospective and Future Challenges. International Journal of Current Pharmaceutical Review and Research Volume 1, Issue 2, August - October 2010
http://impactfactor.org/PDF/IJCPR/1/IJCPR,Vol1,Issue2,Article2.pdf

Thank you for your answer, i appreciate it. Its true im aware that a lot of clinical testing has to happen for drugs to be approved on a wide spectrum, i guess it sounded like they had some of those things already figured out a long time ago. But money is always the biggest factor in the end in research and in medicine.

I upvoted your excellent answer earlier! Glad to see both answers getting some upvotes. It sometimes seems upvotes are tossed around like manhole covers. Either that, or upvotes are carried in low pockets and many arms are short ....


text:
Sulfur iodide to produce alkyl iodides
Phosphorus halides react with alcohols to give alkyl halides. I think that a similar reaction would be possible with sulfur halides. In particular, sulfur iodides. While these are not stable at room temperature, they could hypothetically exist as intermediates, and quickly react with the alcohol solvent they are in. Simply combining sulfur and iodine in an alcoholic solution, with stirring and a hot water bath might be able to form this intermediate. Do you believe that such a method would work? I attached a reaction mechanism below.

I think you should better ask why SI2 isn't stable.

Carbocations aren't stable, but those are often intermediates in reactions. A lot of intermediates cannot be isolated, but are still formed in the reaction. Just because SI2 isn't stable as a pure compound does not mean that it is not a viable intermediate.

The fault of your logic is while $\ce{P}$ in $\ce{PX3}$ is a Lewis acid while $\ce{S}$ in $\ce{SX2}$ (e.g., $\ce{SCl2}$) is a Lewis base.


text:
Could you use All-Purpose Flour as a model rocket fuel?
I'm building a model rocket, and I'm trying to find a good fuel. I've done a lot of research and I know about r-candy and the aluminum-ammonium fuel, but I was wondering, before I made any of these, if I could use All-Purpose Flour instead. From what I've seen, it is highly combustible, and I think it would do a good job. However, it appears to burn slower than r-candy. Would it still burn fast enough to get a rocket launched? I also noticed that it doesn't pack easily, and I'm looking to make a solid rocket motor. Could I use water to pack it into the motor tube, then leave it to dry for a few days? Or could I pack it some other way?

Dust explosions, whether of flour, lycopodium spores or the stuff in hyperbaric explosives, can be very powerful. But bulk flour burns poorly: make a small pile and put a match to it. Not much happens. So you would need continual dust explosions to do the job and the developmental work would likely be substantial, at best.

With a nuclear explosion, maybe a steel manhole cover could be blasted into space: https://en.m.wikipedia.org/wiki/Operation_Plumbbob. Or maybe not! ;-)

This is a great book, by John D. Clark, for anyone interested in rocket fuels: Ignition! An Informal History of Liquid Rocket Propellants, Rutgers University Press Classics, New Brunswick, NJ, ©1972 and 2017.

Make some tests. It should be the exciting part. That packing seems complicated as you will end with something more solid then pressed grains and removal of water will be hard to achieve.

The problem is that once it’s packed, you don’t have an oxidant. Solid rocket fuels contain fuel and oxidant, for example a nitrate, in order to get rapid combustion.

@Andrew If I were to mix a very finely ground chlorine with it, would that be a suitable oxidant? Would it be too powerful? Or, instead, what if I poured some liquid (10%) benzoyl-peroxide over the flour?

Chlorine is a gas, so you're going to have trouble grinding it. . . You need to find something kinetically stable enough that it won't oxidize the flour until it's ignited. I think benzoyl peroxide would start reacting right away. There's a lot of info out there on good solid fuel combinations that have already been worked out. I'd suggest sticking with those rather than trying to reinvent them.

I strongly caution against grinding fuels and oxidizers together! Even just mixing finely ground fuels and oxidizers can be very dangerous. Long ago, I was grinding sulfur powder with an oxidizer that I will not name. I was taking what I thought were reasonable safety precautions and the grinding was going well. Until the ignition! I got molten sulfur on my thumb, raced out to the creek in back and submerged my hand. Turns out the sulfur and that “chlorine” oxidizer spontaneously ignite after a few hours, so I am very glad I did not make up a batch and store it. Safety first!

I don't think this question should be closed. I also think all comments by @EdV are already close to an answer. (I'm encouraging you to write one.)

@Martin-マーチン I will think about it, but I am not an explosives expert, unlike, say, matt_black (my inference based on his generally excellent answers). I have had my share of explosions when I was much younger (basement lab; lots of curiosity; minimal parental supervision), but aside from strongly cautioning against grinding fuels and oxidizers together, I was just providing a little food for thought.

@EdV I respect your decision. I (and I guess others, too) value your contributions in whatever form they come. I'm hopeful the OP gained some insight. I also do admit that I've used your commentary as a point (maybe even excuse) to justify not closing this question.


text:
INCHI-1 software input format(s)?
I am looking at the latest INCHI-1 software which supports molecules with 32 thousand atoms. I work with macro-molecules and want to use that to generate humongous InChIs. I've never used that INCHI-1 software before. I thought I could just throw in some SMILES and get InChI but I learned now that it only accepts MOLFILEs. I find this strange as the atom coordinates shouldn't matter when generating an InChI. I am inclined to write myself a SMILES to MOLFILE converter which just leaves the atom positions at zero, but then, I come to the problem of MOLFILE v2000 having a 999 atom limit. What is the simplest way to get atoms and bonds of 32 thousand atoms into the INCHI-1 system? Can I load an incorrect INCHI perhaps? Like I could just convert SMILES to a non-canonical INCHI. What is the way?
UPDATE: I have started trying this out. I generated a MOLFILE with all 0.0000 atom positions and INCHI-1 would correctly reproduce the atom and connection layers. My data source is actually my own InChI parser which generates an atom list and a link structure.
The problem with this approach is two-fold

How do I reproduce the bond degree in the MOLFILE when all I have is
atom symbol, connections, and hydrogenization from the InChI?
The MOLFILE format just cannot put more than 999 atoms in because of the
fixed column format having only 3 places for the atom numbers in the
bond line.

If my input was SMILES I would have the bond degree at least. But the 999 atom limit of the MOLFILE seems hard.
I wonder if I should use the AuxInfo format as input? And when I tried doing that with all but the coordination "layer" of this AuxInfo format I get the error

Error 40 (no InChI; Missing atom coordinates data) inp structure #1.

Finally, I did find that there is an InChI2InChI mode and I tried that with a non-canonical connection path:
Starting with the canonical test InChI:
InChI=1S/C6H12N2O4S2/c7-3(5(9)10)1-13-14-2-4(8)6(11)12/h3-4H,1-2,7-8H2,(H,9,10)(H,11,12)

I walked the connections backwards:
InChI=1S/C6H12N2O4S2/c12-6(11)4(8)2-14-13-1-3(5(9)10)7/h3-4H,1-2,7-8H2,(H,9,10)(H,11,12)

and that created the same result as the canonical starting version (which I show directly below the output)
InChI=1S/C6H12N2O4S2/c7-3(5(9)10)1-13-14-2-4(8)6(11)12/h3-4H,1-2,7-8H2,(H,9,10)(H,11,12)
InChI=1S/C6H12N2O4S2/c7-3(5(9)10)1-13-14-2-4(8)6(11)12/h3-4H,1-2,7-8H2,(H,9,10)(H,11,12)

so this is good, I might just generate a mega-pseudo-InChI instead of a MOLFILE and use that as the input format.
Now I do another test to see if the stereo layer(s) are invariant given my backwards walk of the connections, after all, in SMILES your walking direction matters for stereo. In the following I put (1) original, (2) inverted connection table, (3) output, and (4) again original to check against the output all in one block:
InChI=1S/C6H12N2O4S2/c7-3(5(9)10)1-13-14-2-4(8)6(11)12/h3-4H,1-2,7-8H2,(H,9,10)(H,11,12)/t3-,4-/m0/s1
InChI=1S/C6H12N2O4S2/c12-6(11)4(8)2-14-13-1-3(5(9)10)7/h3-4H,1-2,7-8H2,(H,9,10)(H,11,12)/t3-,4-/m0/s1
InChI=1S/C6H12N2O4S2/c7-3(5(9)10)1-13-14-2-4(8)6(11)12/h3-4H,1-2,7-8H2,(H,9,10)(H,11,12)/t3-,4-/m0/s1
InChI=1S/C6H12N2O4S2/c7-3(5(9)10)1-13-14-2-4(8)6(11)12/h3-4H,1-2,7-8H2,(H,9,10)(H,11,12)/t3-,4-/m0/s1

and lo and behold, the stereo information is conserved. This is good.
So I might be answering my own question here, but would like the authoritative answer:

MOLFILE v2000 has a 999 atom limit, syntactically there is no way to create larger MOLFILEs for input into INCHI-1
the InChI2InChI mode can be used with non-canonical connection layer and it will be canonicalized. This would be the way to deal with humongous molecules.

UPDATE:
Some people were asking to see an InChI that goes toward the 32 thousand atom length. I had shown here (in a previous edit) a freshly from my InChI peptide chain condensation algorithm on the human insulin B-chain, the INCHI-1 software would take it and return another InChI which seemed to have been re-routed and normalized, but it was incorrect! There is a problem with INCHI-1 in the /InChI2InChI mode, I ran as:
inchi-1 /InChI2InChI /STDIO`:

InChI version 1, Software v. 1.06 (inchi-1 executable)
Windows 64-bit Build (MS VS 2015) of Dec 18 2020 20:44:49

[MORE UPDATE]I thought I had found out that I cannot feed it molecules with more than 1024 atoms in that mode. If I do I get a "syntax error". But that was an error on my part. There is even a flag /LargeMolecules but even with 1400 atoms now I do not trigger any syntax errors any more.[/UPDATE]
But it gets worse!
I found the INCHI-1 software does not properly re-organize incorrect InChIs. This is quite disturbing. I make a long story short (details deleted from earlier edits to this post), even if I get all the hydrogenation right, subtly and correctly sharing the remaining H in the peptide bond between the -C(=O) and the -NC(R)-, as shown in line 1 below, still  INCHI-1 does not re-route the connection path correctly and gives me line 2 below while it should be line 3!
 1 InChI=1S/C5H10N2O3/c1(6)2(8)7-4(3)5(9)10/h4H,1,6H2,3H3,(H,7,8)(H,9,10)
 2 InChI=1S/C5H10N2O3/c3-4(5(9)10)7-2(8)1-6/h4H,1,6H2,3H3,(H,7,8)(H,9,10)
 3 InChI=1S/C5H10N2O3/c1-3(5(9)10)7-4(8)2-6/h3H,2,6H2,1H3,(H,7,8)(H,9,10)

So I'm back to square-1, meaning, I cannot use the INCHI-1 software feeding it any non-MOLFILE input hoping to get out a correctly normalized InChI connection layer, even if I have all the hydrogenation exactly correct.
So, my question remains: how can I use the INCHI-1 software with > 999 atoms getting it to produce a correct InChI?
I know there is this API, but it too wants atom positions. I do not have atom positions.

May you share / link to one of your 32k+ atoms input file? Converting many small molecules' format into an other might differ to work with few, but large ones this large. Still, why is OpenBabel (documentation) offering multiple input formats (including SMILES) to yield InChi/truncated InChi/Inchikey (entry) and APIs (e.g., to Python, C) not suitable? Or the complementary RDKit?

@Buttonwood, I will not say that OpenBabel or RDKit is not suitable, only that I abhor dependencies on libraries I don't use much or are way besides my main scope. I am a bit of a purist with a touch of NIH syndrome. But more objectively in this case, I am tracking Igor's innovations of INCHI quite closely, and so my focus is on his code, until such time that there is a real InChI specification (not just a user guide) which then would get me to implement my own InChI walker. Incidentally I am writing a pseudo-InChI serializer just now. 32k atoms is for creating full protein models.

PS: I'm working on a presentation for next week: https://cactus.nci.nih.gov/presentations/NIHInChI_2021-03/NIHInChI.html where I want to show how I "parse" a massive molecule finding the various protein chains and modifications. :)

This comment changes the context of your question quite a bit.  And it is understood, you don't want to state a superficial «generally speaking, program x works good enough for me», but rely on validated benchmarks including knowledge of the «the anatomy» (and limitations) of the program.

You do know that for proteins things like .pdb and FASTA are used? Using smiles doesn't strike me a reasonable idea.

@GuntherSchadow Some additional thoughts in the chat here.

@Mithoron what do you mean "FASTA are used?" FASTA is just a records format for essentially letter sequences. Fine for chains, not fine for mature proteins. Certainly not to create a full model of the entire finished product with all attachments.

Technicalities: (1) Please do not use update/edit statements. Rework the post as needed to be as concise as possible to the reader. A timely dependence is achieved with the edit history if someone is interested. (2) While cheminformatics is well on topic here, this is turning a lot more into the informatics theory than the actual chemistry elements. I suppose better guidance and support can be gotten at [mattermodeling.se]. || Chemistry: Is InChI able to reproduce secondary structure? Afaik InChI is only able to store connectivity; i'd assume this to be almost useless for proteins.

@Buttonwood Your link only refers to the periodic table chat room. If you want to refer back to a conversation you had, you should bookmark it there and refer to that link here. Or at least pick a persistent link at the start of the conversation. This will open the transcript instead. In a few days or weeks your link will be meaningless.

How to do this
First, I'll address the how of this problem. Although the MOL format is limited to 999 atoms in V2000, it seems like the limit for V3000 is 999999 (for all objects)? See also this link about the MOLfile format. Anyway, I can use openbabel on 1aho :
obabel -ipdb top500H/1ahoH -omol -Otmp.mol

then with:
inchi-1 ~/Data/tmp.mol -Key -LargeMolecules

I get a ridiculously long InChI, with InChIKey=SALBUZGTRUOYKX-NQUMBTFPSA-N. Note that my copy of 1aho has hydrogen atoms, so goes over the 999 limit.
Why do this
Obviously using a line notation like SMILES or InChI for macromolecules is possible, although stretching a system designed for small molecules. I take your point that post-translational modification, non-standard amino acids and so on could make it useful.
There might be more suitable formats for macromolecules, such as "HELM: A Hierarchical Notation Language for Complex Biomolecule Structure Representation"
Canonicalisation
Of course, to be a unique identifier for a modified polypeptide, a line notation would have to be canonical. InChI are by default, of course, but will be affected by tautomers.
Another possibility would be to ask Roger and Noel, who have done work on this : https://nextmovesoftware.com/blog/2014/11/26/introducing-new-formats-for-handling-macromolecules-smiles-and-inchi/


text:
Why is titanium not found in its pure state in nature?
I am doing a research project on titanium and I know it is found in ores and compounds rather than in a pure state, but I am unsure why? How does this relate to its reactivity?

$\ce{Ti}$, similarly as $\ce{Al}$, is more or less stable as metal not because it is not reactive, but because it is protected by passivation oxide layer.
Both metals are very reactive at high temperature. $\ce{Ti}$ burns in air before reaching melting point and is attacked by acids.

Titanium is one of the few elements that burns in pure nitrogen gas, reacting at 800 °C (1,470 °F) to form titanium nitride.  ... Because of its high reactivity with oxygen, nitrogen, and some other gases, titanium filaments are applied in titanium sublimation pumps as scavengers for these gases. ( Wikipedia )

This predicts the fact they are not found as free metals in the Earth crust, where high temperature high pressure chemistry is ongoing.
You may find useful reviewing Wikipedia - Titanium - Chemical_properties and/or searching for the same.


text:
Biological activity of the isotopes of the same element
I have a question regarding biological activity of different isotopes of the same elements, notably $\ce{^206Pb}$ and $\ce{^208Pb}$. Lead is known to be toxic, such that it can replace calcium from bones and iron from hemoglobin for example. Would one isotope behave differently than another in such cases?
When it comes to hydrogen/deuterium chemical activities, some difference could arise as one isotope is twice as large as another. Would this be the case for heavier elements, such as lead?

Your reasoning seems ok if radionuclides are not considered, of course. For heavy water, if I remember, there is a nice Wikipedia article.

There are interesting but subtle isotope effects for many important elements common in biological systems (eg oxygen and carbon). These are widely used to determine the origin or local environmental temperature of old specimens.

This earlier Q&A (Biological Consequences of Asteroid Mining—Death by Isotope?) might be of interest.

There is quite a bit of literature about lithium-6 versus lithium-7 for psychiatric use. Here is a review article: https://pubmed.ncbi.nlm.nih.gov/3530801/ Also this: https://pubmed.ncbi.nlm.nih.gov/32084493/

I would not expect significant differences in the chemical properties (reactivity) of these lead isotopes, the differences in atomic weight being as you point out very small. There are reports  that claim an effect. Ref 1 for instance makes such a claim, but there is no plausible mechanism presented to explain the reported isotopic discrimination. Double check what peer reviewers think (see e.g. the response in Ref 2 to the results presented in Ref 1).
References

Wu J, Liu D, Xie Q, Wang J (2012) Biological Fractionation of Lead Isotopes in Sprague-Dawley Rats Lead Poisoned via the Respiratory Tract. PLOS ONE 7(12): e52462. https://doi.org/10.1371/journal.pone.0052462

Gulson, Brian and Kamenov, George D. and Manton, William and Rabinowitz, Michael (2018) Concerns about Quadrupole ICP-MS Lead Isotopic Data and Interpretations in the Environment and Health Fields. International Journal of Environmental Research and Public Health 15(4): 723. https://doi.org/10.3390/ijerph15040723

I'm assuming we can neglect differences due to radioactivity/radioisotopes in biological systems.
Hydrogen vs Deuterium
The kinetic isotope effect - the difference in reaction rates and reactivity for different isotopes of the same element - is well known for hydrogen and its two stable isotopes protium and deuterium.  For example levels of deuteration above 25% have shown adverse effects in a variety of animals. [1]
Heavy Elements
But stable isotopes of elements as heavy as carbon can have measurably different effects in biological systems.
Carbon
In photosynthesis, lighter $\ce{^12C}$ is incorporated into plants faster than $\ce{^13C}$ creating a difference in the abundance of $\ce{^13C}$ in biological sources compared to other enviromental sources.  Furthermore different methods of photosynthesis - i.e. the $\ce{C_3}$ and $\ce{C_4}$ pathways found in different groups of plants - lead to different discrepencies in $\ce{^13C}$ abundance ($\delta^{13}\ce{C}$) which can provide information about the diets and nutrient sources of the animals that eat them.


text:
Temperature Dependence of The Conductivity of Electrolytic Solutions
I have learned in my chemistry text books that the conductivity of electrolytic solutions increase with the increasing of temperature because the ions of electrolytic solution move faster by getting more thermal energy. But a question arose in my mind if there exist any mathematical equation which can describe the conductivity of electrolytic solutions as a function of temperature. I searched about it on internet but I got nothing satisfactory. Does any equation exist which can model the electrolytic conductivity as a function of temperature?

I suppose there would be just some empirical equations, applicable just for particular cases.

One useful approximate relationship that can be used to predict the temperature dependence of the conductivity can be obtained by relating the electrical mobility to the viscosity.
For a $z_+:z_-$ electrolyte the limiting molar conductivity can be expressed as
$$\Lambda^\circ _m =  F^2\left(\frac{z_+^2\nu_+}{f_+} + \frac{z_-^2\nu_-}{f_-}  \right)$$
where the $f_i$ are frictional coefficients  for each ion, z is the ionic charge, $\nu$ is the stoichiometric coefficient, and F is the Faraday constant (for instance, for $\ce{Al2O3}$, $z_+=3$,$\nu_+=2$,$z_-=2$,and $\nu_-=3$.
The frictional coefficients can be related to the viscosity, for instance using Stokes' relation $f=6\pi \eta r$ for a sphere being driven through a uniform viscous medium with a low Reynold's number. From the temperature dependence of the viscosity, and assuming the ion radii are independent of T, you can derive a relation for the T dependence of the limiting conductivity:
$$\Lambda^\circ _m = \frac{F^2}{6\pi\eta (T)}\left(\frac{z_+^2\nu_+}{r_+} + \frac{z_-^2\nu_-}{r_-} \right)$$
The equations so far apply to dilute solutions where the viscosity is independent of the concentration of electrolyte and ion-ion interactions can be neglected.
Note that even simpler theories predict a somewhat complex dependence of the molar conductivity on temperature. One can however attempt to fit an expression inspired by those above, of the sort
$$\Lambda _m = A \times \eta (T)^{-1} $$
where A is assumed independent of T. This might work over a narrow T range and could be used to assess the importance of ion-ion interactions.

Thanks for your answer. But I haven't understood what 'z' means in your answer. Does it mean valency? Please make it clear with some examples. And do you mean Faraday's constant with 'F'? And is this formula only applicable for limiting molar conductivity but not for any molar conductivity? Please make me clear about these.

@Md.HasibulHossain Please see the modifications to the answer.

What does 'z' represent in that equation? Please explain that term with some examples. Suppose, what is the value of 'z' for Al₂O₃?

Thanks for editing your answer and making it clear.

Conductor are generally two types .Electronic conductor and electrolytic conductor. In case of electronic conductor conduction occur by moving of free electron but in case of electrolytic conductor the conduction occur by moving of ions. so with increasing the temperature, in case of electrolytes the solvent jacket surrounding the ions are destroyed as a result mobility if ions increase and conductivity increase.
From Debye Huckle limiting law
K=√((F .  I)/(∈ R T))
K^(-1)= distance of and central ion and solvent jacket.
as temperature increase than distance increase so ionic interaction decrease , mobility increase , conductivity increase.


text:
How does the force constant change when a heavier atom is used in vibrational spectroscopy?

(iii) The fundamental band in the IR spectrum of $\ce{^1H^{127}I}$ is found at $\pu{2308 cm^-1}.$ Calculate the force constant for the bond. Assume the molecule can be treated as a harmonic oscillator.
(iv) Calculate how the position of the fundamental band would be expected to differ for molecules of $\ce{^2H^{127}I}.$

I have found the force constant in (iii), but I am struggling with (iv). I don’t know how to calculate the difference. Any insight would be greatly appreciated.

You can calculate the ratio of the two frequencies as you know all the masses. Usually homework questions such as this get closed unless you explain what you have done so far and where you are stuck. As a start I suggest that you look a the vibrational spectroscopy part of your textbook where you are bound to find the equations you need.

In the harmonic oscillator approximation, the wavenumber of vibration for a diatomic molecule is
$$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$$
Here, $k$ is the force constant of your bond, and $\mu$ is the reduced mass for the two atoms.
$$\mu=\frac{m_1m_2}{m_1+m_2}$$
When you change isotopes, only $\mu$ changes. The force constant depends on the strength of the bond, which depends on the chemical nature of the elements, and can be assumed to stay constant regardless of changing isotopes.
You have to calculate the $\mu$ for the second problem, and use that with the $k$ you got from the first problem to calculate the wavenumber of the fundamental band.

Can the factor $k$ be obtained by some calculation ?

@Maurice Well, usually it's the other way round. You can measure the wavenumber by spectroscopy, and then you can use it to calculate the force constant (k). You can however, use computational chemistry, which will directly calculate the force constants of the bond.

As suggested by @porpyrin, in the second calculation you're assuming the force constant $k$ is constant in both molecules so a ratio can be taken. Let $\tilde{\nu_H}$ and $\tilde{\nu_D}$ be the hydrogen and deuterated versions respectively,
$$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$$
$$\frac{\tilde{\nu_H}}{\tilde{\nu_D}} = \frac{\sqrt{k/\mu_H}}{\sqrt{k/\mu_D}} = \sqrt{\frac{\mu_D}{\mu_H}}$$
Note: the bond strength changes as the zero point energy of the bond changes upon isotopic substitution hence this is only an approximation.


text:
Determining iron content by titration
The redox reaction is
$$\ce{KMnO4-(aq) + 8 H+(aq) + 5 e- -> Mn^2+(aq) + 4 H2O(l)}.$$
Where does $\ce{H+}$ come from in this reaction? What happened to $\ce{K+}$ in potassium permanganate?

Is this a homework question or an actual research project. Add more details.

Hints: 1) Your redox half-reaction requires acidic conditions. Sulfuric acid is an acid. 2) The potassium ions are spectators, so they are not shown in the redox half-reaction.

Will the experiment work without adding the acid ?

Then what would your acid-free redox half-reaction be?

In the experiment sheet it does not mention anything about sulfuric acid. It only mentions 100ml ground water and 6.5x10-5 mol l-1 potassium permanganate. So in this case how much sulfuric acid is needed?

The balanced half-reaction requires acidic conditions. Sulfuric acid is commonly available in teaching labs and works well in this situation. I suggest following the experimental protocol as given to you and ask the laboratory teaching assistant for specifics if you remain stuck.

In order for you to carry out this reaction, you need potassium permanganate, which provides  the ion $\ce{MnO4^-}$. You also need sulfuric acid, and quite a lot of this acid, because the equation requires $\ce{8 H^+}$. And well ! $8$ is much ! So the reaction will not work without acid. And you still need $5$ electrons, which are provided by $\ce{5 Fe^{2+}}$ ion. The $\ce{K+}$ brought by the potassium permanganate remains in the solution as spectator. They don't react.

In the experiment sheet it does not mention anything about sulfuric acid. It only mentions 100ml ground water and 6.5x10-5 mol l-1 potassium permanganate. So I’m this case how much sulfuric acid is needed?

@Lulu. $\pu{100 mL}$ ground water has a meaning. But $\pu{6.5E{-5} mol L^{-1}}$ permanganate has no meaning. One should give you an information like "XXX milliliters of potassium permanganate having a concentration of YYY mole per liter was used for titrating this amount of ground water"


text:
What non-hazardous substance can be used to locally turn sky red?
I am a horror author and I have a scene coming up in my new book where I need a red-colored gas (or a gas that would somehow be turned or viewed red when viewed from the ground) to be spilled into the atmosphere. Preferably by a tanker truck.
I need this gas to kind of hang over this small community and of course, not kill anyone. Just an eerie red sky. I looked at sulfur hexafluoride, but I think that is colorless and maybe toxic? Looking for something that would make sense here.

You'll need quite a number of tanker trucks to paint the sky.

Could you accept in your world that some colorless chemical is mixed with a vibrant pigment at the point of manufacture as a safety precaution (like odorant put into gas lines)? This alleviates the need for a specific compound. I personally haven't worked with a colored gas that didn't have some way of killing you. Colored smoke bombs are the closest thing I can think of, but they are dispersing a powdered pigment.

How about reversing the blue sky? An exotic top secret non-toxic meta-material is accidentally released, in ultra-fine powder form, and it has the property of scattering red light far more than blue light. So the sky would be locally red and the setting sun blue (or blueish). Colored gases are very uncommon and all of them are toxic to varying degrees. Also, fires can put enough particulates up so the sky goes rather deep red: search red sky pics.

hmmm, ok thank you.  If ammonium nitrate tanker exploded in my book and created nitrogen dioxide, that would be lethal correct, and not make sense of course to the story?

the sky just needs to be red locally, over a small farm town

Definitely not nitrogen dioxide or bromine! Forest fires do the job, via smoke and particulate scattering.

got it - ok thank you very much for the help!

If you are writing fiction, you could imagine a high-energy, very high pressure triatomic molecule containing one atom of oxygen, one of helium and one of neon. When it is released to the atmosphere as the truck explodes, the excited atoms of noble gases phosphoresce like neon signs for hours or days. (I said it was fiction...).

I don't know if this will work, but you may look platinum hexafluoride.  It is a dark-red volatile solid that forms a red gas. At least, it is not that hazardous as compared to bromine or nitrogen dioxide.


text:
Which of these would most likely be a strong electrolyte?

An ionic compound
A covalent compound
A polar compound
A nonpolar compound

Previously, I have been taught that polar compounds can be either strong or weak electrolytes. I also learned that ionic compounds are generally strong. Therefore, my guess would be a, an ionic compound. However, I am not entirely sure of my answers. Could you please explain to me how I could confirm and get the right answer? Thank you!

Yes. You are right. Ionic compound are strong electrolytes if they are soluble in water. $\ce{CaF2, BaSO4}$ are ionic compounds, but they are not soluble in water. I would not consider them as strong electrolytes. On the other hand, ionic compounds are a part of the bigger family of the polar compounds. But polar compounds are not always strong electrolytes. $\ce{CO, H2S, HF, H2O}$ are polar, but they are not strong electrolytes.

Critically looking at it, Ionic compound(metals+non-metals) are strong electrolytes as they are able to conduct electricity when in molten form and that's a characteristic of a good and strong  electrolytes(Ability to conduct electricity) but polar solvent have the ability of being a strong electrolytes(very rare)  and weak electrolytes. Polar solvent when separated will have partial positive and partial negative charge.
Ionic compound are strong electrolytes than any other mentioned  compounds in your question.

This was really helpful, thank you so much!


text:
Potential Energy Curve from Hartree-Fock Calculations
I am currently working on some quantum chemical calculations, in particular looking at dissociation energy curves generated from Hartree-Fock, CI, CCSD and MP2 methods for diatomic molecules. Now, I was always under the impression that in order to get a potential energy surface  (in the Born-Oppenheimer approximation)  one needs the energies for the electronic problem and the nuclear-nuclear interaction:
$$E(R) + V_{\mathrm{NN}}(R) = U(R)$$
Where $E(R)$ is the energy obtained from the solution of the electronic Schrödinger equation, $V_{\mathrm{NN}}(R)$ the internuclear coulomb potential and $U(R)$ the potential energy surface. I have now seen multiple publications and books that draw potential energy curves using only $E(R)$.
So my question: Are these publications/books wrong, or do I have a misunderstanding of how the PES is computed under the BOA?
Example publication: Dutta, A.; Sherrill, C. D. Full configuration interaction potential energy curves for breaking bonds to hydrogen: An assessment of single-reference correlation methods. J. Chem. Phys. 2003, 118 (4), 1610–1619. DOI: 10.1063/1.1531658.

In many QM codes, the value that is printed as the Hartree-Fock energy includes the nucleus-nucleus repulsion (VNN) term. Maybe that is what the books and papers reported.

I've added a publication. @ShoubhikRMaiti That would make sense. In that case, would it be sensible to fit a Morse potential to the energies as well(in the diatomic case of course)?

Why do you need to fit the energies to a Morse potential?

@ShoubhikRMaiti I'm thinking that using a fit to the energies would make it easier to read off certain quantities of interest, such as eq. bond length, dissociation energy and so on, all of which are parameters of the Morse potential.

@Martin-マーチン I'm sorry, but I am not entirely sure what you are referring to with "that curve".

I guess fitting to a Morse potential could work. I don't know much about this, but parameters of molecular dynamics force fields are derived in this way, so its not unprecedented. You should look at some papers and talk to your supervisor about whether this method is right. One problem is that you can fit the curve only if the energy curve is accurate. For example, doing CCSD on H2 dissociation would give a weird bent curve that won't fit to the eqn of Morse potential.

@ShoubhikRMaiti maybe I'm completely of here, why would ccsd give a weird curve? Shouldn't that curve be almost the same as cisd, which is full ci?

@ShoubhikRMaiti For two electrons CISD is Full CI. So should be CCSD. Depending on the implementation they should be equivalent or at least very, very close The perturbative triples should not have any effect at all. I do not have access to any paper right now so I am not sure what you are referring to.

@Martin-マーチン You are right, sorry. I realize now that I was not looking at H2 dissociation curve, when I wrote the comment. I was mainly thinking about CCSD not being able to handle multireference systems, and just wrote H2 hastily, forgetting that it has two electrons

The energy calculated by a quantum chemistry program is essential always the total potential energy $U(R)$.  For example Fig 1 and Fig 4 of Dutta and Sherrill referenced in the question both plot the total sum of electronic and nuclear energies $U(R)=E(R)+V_{NN}(R)$.
For an example, have a look at a PES I have generated for $\ce{H2}$, using Restricted Hartree-Fock and the STO-3G basis set using PySCF.  The characteristic PES, with a single minimum is actually the total sum of energies $U(R)$, labelled the "Potential Energy Surface"

N.B. The zero bond-length electronic energy is calculated using a helium nucleus as the program doesn't allow superposition of atoms.

I've update the answer to reply to this comment.
A Helium atom is used for the zero-bond-lenght case to prevent singular values due to overlapping atomic orbitals.  I was also surprised at the importance of the nuclear repulsion to cause the short distance repulsion, hence why I included the whole graph.

Notebook to calculate graph
https://colab.research.google.com/drive/1r9qcz3x_sR_8HI5nmsnKlmVMw_BO6oAM?usp=sharing

Is it right to use RHF for H2 dissociation? I thought RHF caused H2 to fragment into H+ and H- when forced to dissociate.

Thanks for the clarification. Not a fan of the helium approximation, to be honest. @ShoubhikRMaiti RHF does worse things than this. But it still is a nice  showcase.


text:
Is Fluorine more toxic than Chlorine?
Fluorine is more reactive than Chlorine. But does it mean that Fluorine can cause more damage to living tissues? If so, why wasn't Fluorine used in WW1 instead of Chlorine?

I'd assume fluorine would be harder to manufacture, store, and handle for any side attempting to wield it in combat. Also, due to its greater reactivity, fluorine gas will tend to $\ce{2F2(g) + 2H2O(g) -> 4HF(g) + O2(g)}$ with atmospheric humidity -- unlike chlorine gas, which is less reactive and heavier and thus possibly better at staying closer to the ground and/or flowing down into an enemy's trenches.

Containing and transporting gaseous fluorine would also have been a problem for WW1 level technology.

Remember elemental Fluorine was first isolated only in 1886, after causing major discomfort to many of the leading experimentalists of the day when they did try to make it (https://en.wikipedia.org/wiki/Fluorine#Isolation). Its isolation was the reason Moisson won the Noble prize in 1906, for his supreme experimental technique. By 1914 industrial preparation was likely at most in its infancy. Fluorine is a pig - whatever you have heard about it, it is worse than that.

Weapons have to be practical. Poison gas overall turned out not to be very useful. But chlorine has some practicality as it is much denser than air and fills trenches and isn't so reactive it would destroy everything near its use including your own soldiers.

Fluoride ions are indeed more ‘toxic’ than chloride ions (compare the LD50 of NaF to NaCl for a rather simple example), but that’s kind of irrelevant if you just care about potential application of fluorine gas as a chemical weapon in place of chlorine gas. But then, if you want to go for impressive halogen-based chemical weapons, you should look at chlorine trifluoride instead (that stuff is so dangerous that even the Nazis decided not to use it).

@AustinHemmelgarn Decided not to use ClF3 as rocket oxidiser, I assume (I know it was tried in the US in the 50s/60s)

@ChrisH From what I understand, they actually considered using it for chemical warfare (albeit for a very short period of time).

@AustinHemmelgarn I also saw a note that they considered it as an incendiary, just after posting that, for which it would probably be rather effective. John D Clark: so rapidly hypergolic that no ignition delay has ever been measured. It is also hypergolic with such things as cloth, wood, and test engineers, not to mention asbestos, sand, and water—with which it reacts explosively. ...metal-fluorine fire. For dealing with this situation, I have always recommended a good pair of running shoes (for those who haven't seen this quote already, wikipedia)

Obligatory Chlorine Trifluoride and Dixoygen Difluoride "Things I Won't Work With" blog posts by Derek Lowe.

@DoktorJ you can't mention dioxygen difluoride without calling it by its onomatopoeic name, FOOF

There are a lot of answers about that fluorine is much harder to storage an it is much less stable for gas usage, but the HF was very popular for poisoning - there is no antidote yet and it is almost invisible and the symptoms are not immediate.

I'm guessing because chlorine won't really corrode the metal shells they were in,but fluorine would because it's a lot more reactive and it's a hazard to the people putting the gas into the weapon because if any accidents happen and you breathe it  in it creates hydrofluoric acid ,which is toxic and corrosive and steals body calcium and can also start a fire if any of it leaks out and then that's a huge problem. And then then there's chlorine trifluoride which is so dangerous it sets fire to pretty much anything,and fluorine is a far weaker version than that,and chlorine is even weaker.

Fluorine is much more reactive than chlorine and would certainly cause more damage to living tissues. You can even check out https://www.youtube.com/watch?v=vtWp45Eewtw for some fun demonstrations of its oxidizing power too!
Likewise, compared to chlorine gas, I'd assume fluorine munitions would be significantly harder to manufacture, store, and handle for any side attempting to wield them in combat. (And, as Waylander mentioned in the comments, especially for "WW1-era technology" too!)
Also, due to its great reactivity, fluorine gas will readily react even with the atmospheric humidity to form hydrogen fluoride: $$\ce{2F2(g) + 2H2O(g) -> 4HF(g) + O2(g)}$$ which, while still toxic, is much lighter than chlorine gas and so may not be as good at staying "close to the ground" and/or "flowing down" into an enemy's trenches as intended. Or even worse -- potentially thus dispersing even back to your own troops and damaging them as well!

Bomb: canister of fluorine.

@Joshua  Chlorine trifluoride would be more practical. But think 3 times what material would be used for the canister.

@Poutnik TIL about hypervalent bonding. Fluorine is so reactive it forces chlorine to take 3 more electrons, kinda. Also, while reading, I came across "The discovery of chlorine pentafluoride rendered ClF3 obsolete as a potential rocket fuel oxidizer." So there's that.

"Fluorine is much more reactive than chlorine and would certainly cause more damage to living tissues." - Just adding that physical damage is only one of a few components of toxicity. Biochemical interactions and effects are another major component. So is radioactivity. For example, oxygen is highly reactive but the toxicity limits are extremely high, and all of the radioactive elements are toxic regardless of reactivity. Fluorine has other toxic effects, too, for example it affects thyroid hormone levels in a similar way to hypothyroidism (from unknown mechanisms).

@JasonC: Another consideration is that exposure of part of the body to a substance which destroy tissues in ways that prevent absorption into the bloodstream may be more survivable than exposure to substances that are less reactive but consequently get absorbed.

@Poutnik:  I suppose you could consider dioxygen difluoride instead, but that's even harder to contain.

@Poutnik:  Also, here's the same author's writing on chlorine trifluoride, titled "Sand Won't Save You This Time."

@JasonC Fluorine should interfere with pretty much every biochemical process. The equilibrium of any set of reactions is always in Flourine's favour.

Fluorine is in the first place much more reactive than chlorine. In contrary to chlorine, it would not damage biological tissues. It would destroy them. Pure fluorine could put the body on self-ignited fire.
Its production and manipulation would be much more expensive and difficult.
It reacts with water vapor and near anything it meets.
It would have also application disadvantage as it has just slightly greater density than air and would significantly dissipate vertically.

Also, I would add that you can protect your own troops from chlorine reasonably well. Not so with fluorine.

Yes. It can be implied from the reactivity.

From:

Fluorine gas MSDS
Chlorine gas MSDS

Section 11, toxicity:

Product
Result
Species
Dose
Exposure

fluorine
LC50 Inhalation Gas.
Rat
185 ppm
1 hours

chlorine
LC50 Inhalation Gas.
Rat
293 ppm
1 hours

So we can at least see that fluorine is more toxic to rats, for inhalation, in terms of ppm required to achieve LC50 (50% of test subjects dead) in one hour. This doesn't tell the whole story but it's relevant to gas inhalation in mammals (on the topic of WW2) and it's a good high level indicator of ultimate toxicity (regardless of specific mechanism).
The same is implied by the relative exposure limit recommendations in section 8 (I just picked comparable organizations, they're all slightly different):

Product
Organization
Workplace Limit

fluorine
OSHA PEL 1989 (United States, 3/1989)
0.1 ppm 8 hours

chlorine
OSHA PEL 1989 (United States, 3/1989)
0.5 ppm 8 hours

Although reactivity isn't necessarily an indicator of toxicity, in this case fluorine wins there, as well.
As for why they didn't use it in WW2: I don't actually have a definitive answer for this, however, according to Wikipedia:

Large-scale productions of elemental fluorine began during World War II. Germany used high-temperature electrolysis to produce tons of chlorine trifluoride, a compound planned to be used as an incendiary. The Manhattan project in the United States produced even more fluorine for use in uranium separation. Gaseous uranium hexafluoride was used to separate uranium-235, an important nuclear explosive, from the heavier uranium-238 in diffusion plants. Because uranium hexafluoride releases small quantities of corrosive fluorine, the separation plants were built with special materials.

My guess from that is two-fold:

Production processes were relatively new, but more importantly:
It was used in nuclear development, and at the time, resources involved in nuclear weapons research and processing were highly prioritized for that purpose. So it could have just been that they'd rather use the fluorine for weapons research.

Don't take my word for it on that one though.

The biggest use of poison gas was WWI - so 25--30 years before large scale production of F.  During WWII chemical weapons were stockpiled but not really used on the battlefield; they were of course used during the holocaust

In addition to the excellent comments about handling issues (not to mention the general issues with poison gas as a weapon), production of elemental flourine remains an expensive proposition even now (most use of fluorine in industry never involves conversion to the element, but essentially movement of CaF2 to HF to various flourinated products).  Conversely production of Cl2 gas was and is a massive chemical industry, with caustic soda as a byproduct.  It's much easier/cheaper than making elemental fluorine.


text:
Clemmensen reduction of α-, β-keto acids
The following is from Organic Chemistry-2 for JEE Advanced by KS Verma:

α- or β-keto acids do not undergo Clemmensen reduction.

My reasoning was that the double bonds are in conjugation which makes them reluctant to participate in such reactions. However, this doesn't seem to be the case with β-keto acids which makes me suspect that my line of reasoning might be incorrect.
What is the reason for this behavior?

Consider possible tautomers of the beta-keto acid

The likely reason for this is a steric hindrance. The Clemmensen reduction is very vulnerable to hindrance. See the image below:
There was also another reaction conducted that had a 67% yield for a gamma keto acid with resonance. See the image below:

Therefore, the main factor that makes α- or β-keto acids not undergoing Clemmensen reduction is likely not because of resonance, but primarily of steric hinderance. But this also may raise another question, why doesn't the carboxylic acid group get reduced? The reason for this is that the carbon in a carboxylic acid is not quite partially positive because OH is an electron-donating group. The mechanism involves a transfer of an electron from zinc to the partially positive carbon atom, which, as mentioned, carboxylic acid doesn't have.
However, even for diketones, the Clemmensen reduction doesn't often yield what one would expect. See the image below:

The reason is because of the mechanism. In the mechanism, radicals are formed on the carbonyl carbon. However, if you have a diradical, they can come together and form a bond if they are close enough. This is just some extra information.
The source I took the images from: https://www.sciencedirect.com/topics/chemistry/clemmensen-reduction#
Extra papers I found that may be helpful:
https://d.lib.msu.edu/etd/37284/datastream/OBJ/view
https://pubs.acs.org/doi/abs/10.1021/jo01017a512

I would agree with your argument for alpha keto acids. However Im skeptical for the beta case. I dont think steric hindrance should create trouble there. What do you say?

I don't think a definite answer exists for this case @newbie105 because the mechanism of clemmensen reduction is not fully understood yet.

Yes, there is still research into the Clemmensen reduction and I could not find any paper on the reduction of beta-keto acids. It is just that the Clemmensen reduction has shown to be greatly affected by steric effects, as shown in the first image where a methyl group on an alpha carbon prevented the reduction of that ketone. And from my second picture, a gamma keto acid was still reduced with a relatively good yield. So I would say that Beta keto acids likely does not lead to good yields because of steric hindrance. However, there needs to be more research for there to be a definitive answer.


text:
If there is no destruction of mass in the nuclear fission , then how is tremendous energy is released?
In the fission of U-235 , calculate the mass of reactants , which are U-235 and 1 low energy neutrons .
hence , the mass in reactant is = 235 + 1 = 236amu (atomic mass unit).
Now , in the products , barium , krypton are formed and 3 neutrons are emitted .
Hence , mass of products is :
mass of krypton ( 139 amu) + mass of barium (94 amu) + mass of 3 neutrons ( 3 amu ) = 139 + 94 + 3 = 236 amu (atomic mass unit) .
From above , it is clear that mass of reactants = mass of products = 236 atomic mass units
Hence , no mass is destroyed in fission reaction . hence , no tremendous energy should be released .

There is no reason why a barium nucleus should be lighter than a krypton nucleus.

If you measure the mass of each particle (nuclei, neutrons etc.) very precisely (more decimal places), and do the math, you will see that there is a slight mass loss. That mass loss is converted to energy.

Your question is based on the false assumption that the mass of nucleus is equal the sum of masses of respective count of free protons and neutrons. So you erroneously identify the nucleon numbers with multiples of the atomic mass unit.
A mass of atomic nucleus is decreased, compared to the total proton and neutron mass, by the relativistic mass equivalent of binding energy of nucleons ( energy needed to break a nucleus to elementary particles).
$$m_{\mathrm{nucleus}}\  \lt Z \cdot m_\mathrm{p} + N \cdot m_\mathrm{n}$$

$Z$ is the proton number
$N$ is the neutron number
$m_\mathrm{p}, m_\mathrm{n}$ is the proton respectively neutron  mass.

As this binding energy differs for various elements and their isotopes, the average mass per a constituent proton/nucleon is nucleus is not constant.
Therefore the mass of $\ce{^{20}_{10}Ne}$ is greater then a half of the mass of $\ce{^{40}_{20}Ca}$, as calcium nucleus has higher binding energy than neon . Similarly a nucleus $\ce{^4He}$ has by 0.7% less mass than the total mass of 2 protons and 2 neutrons, from which it consists.
During $\ce{^{235}U}$, the mass of the uranium nucleus is greater than the sum of masses of 2 created nuclei and 2-3 released neutrons.
See Nuclear binding energy curve.

you are expected to make a grade 10 student to understand this . please simplify the answer.

@anoman I was not expected to do so, as you omitted to mention it. Quality of answers reflects quality and completeness of questions.

@Maurice He rather identified the nucleon numbers with multiples of the atomic mass unit, what was wrong. The Ba and Kr specifics are just a particular extra mess, based on false premise.

$235$ for the mass of $\ce{U-235}$, he should have used $235.044$. And he should have used existing isotopes. For example Krypton-$139$ and Baryum-$94$ do not exist. Probably masses have been interchanged. Any way the calculations must be made with more precise masses.

It should be rather comment to the question.


text:
Adiabatic expansion of an ideal gas
A. For certain ideal gas Cp=8.58 cal/mol-deg. What will be the final temperature and volume when two moles of the gas at 20 degrees celsius and 15 atmosphere are allowed to expand adiabatically and reversibly to 5 atm pressure? Find W, ∆U, and ∆H.
B. Suppose that the expansion takes place adiabatically against a constant pressure of five atm, what will be the final volume and temperature of the given gas?
I already solved part A but I have no clue for part B. Especially on how to get the final temperature and pressure.

How do I get the final temperature and pressure for part B?

They tell you the final pressure.  5 atm

In the following answer I am using the physics convention of work done. Where, $W= \int P\mathrm dV$ and the first law of thermodynamics is as follows: $q = \Delta U + W$. If you want answers in the chemistry conventions than just reverse the sign of the work done. Internal energy change and heat change will be same in both conventions.

As you said you have done the first part, I will continue with the second.
First of all,
$$
\begin{align}
\hline
C_V&=C_P - R \\
&= \pu{8.58- 2 cal}\\
&= \pu{6.58 cal}\\
\hline
\end{align}
$$
As the process is adiabatic,
$$
\require{color}
\begin{align}
\hline
q&=0\\
\Delta U&=-W\\
nC_V\left(T_2-T_1\right)&=-nRP_{\text{ext}} \left(\frac{T_2}{P_2} - \frac{T_1}{P_1}\right) \\
\color{green} {T_2}&\color{green}=\pu{247.67 K}\\
\hline
\end{align}
$$
Now,
$$
\begin{align}
\hline
W&= P_{\text{ext}} \Delta V\\
W&= nRP_{\text{ext}} \left(\frac{T_2}{P_2} - \frac{T_1}{P_1}\right) \\
\color{green} W&\color{green}= \pu{599.2 cal}\\
\hline
\end{align}
$$
$$
\begin{align}
\Delta U&= -W \\
\color{green}\Delta U&\color{green}= \pu{-599.2 cal}\\
\end{align}
$$
$$
\begin{align}
\hline
\Delta H&= \Delta U + \Delta \left(PV\right) \\
\Delta H&= \Delta U + \Delta \left(nRT\right)\\
\Delta H&= \Delta U + nR\Delta T\\
\Delta H&= \pu{(-599.2 -181.32) cal}\\
\color{green}\Delta H&\color{green}= \pu{-780.52 cal}\\
\hline
\end{align}
$$


text:
Why does π-backbonding in a metal-carbonyl complex lead to a decrease in C-O bond strength?
I understand that π-backbonding in this complex involves the σ-donation of carbonyl to an empty d orbital of the metal, and the π-donation of the d electrons of the metal to the π* MO of the carbonyl. But why exactly does this lead to the weakening of the C-O bond (relative to free C-O)?

Do you know what a $\pi^*$ orbital is?

In simple words, the pi* orbital is antibonding in nature. So, putting electron density there will weaken the C-O pi bond. The CO bond contains both sigma and pi components. The pi component will be weakened.


text:
Can sugars dissolve in liquid ammonia?
Can monosaccharides and disaccharides dissolve in liquid ammonia due to hydrogen bond formation?
My rationale is that these sugars may be able to form hydrogen bond with ammonia ($\ce{NH3}$).

Welcome to Chemistry SE! Adding your rationale as to what brought you to this presumption as well as literature precedents would increase the chances of a positive, descriptive answer here.

Do you mean liquid ammonia or aqueous ammonia (ammonium hydroxide solution)?

Liquid ammonia.

Can sugars dissolve in liquid ammonia?

Yes, according to Ref.1, liquid ammonia is used to extract sugars in sugar-beet chips:

5.88 kilograms of sugar-beet chips having a moisture content of 5.4 percent and a saccharose content of 68 percent (calculated on the dried substance) are treated in an autoclave of 50-litre capacity with 18 kilograms of liquid ammonia. [...] The extracted pulp, after driving off the ammonia, contains only 1.2 percent of saccharose, which constitutes 0.4 percent of the total sugar.

Accordingly, 99.6% of sugar in sugar-beet chips are isolated in one extraction using liquid ammonia as the solvent. However, it was noted that the saccharose is not present as such in the solution, but is in the form of a compound of saccharose with ammonia. These ammonia-saccharose compounds can be obtained in a well crystallized form, which are composed of $\pu{1 mol}$ of saccharose and $\pu{2 mol}$ of ammonia and even those of other compositions Ref.1). I assume this is analogous to hydration by water molecules in aqueous solutions.
The high solubility of sugar in liquid ammonia seemingly due to H-bonding capabilities similar to water. This is evident by sucrose solubility in nonaqueous solvents such as morpholine (with a $\ce{N-H}$ bond) and N-methylmorpholine (without a $\ce{N-H}$ bond), which are $39.8$ and $\pu{0.37 g}$ in $\pu{100 g}$ of solution, respectively at $\pu{85 ^\circ C}$ (Ref.2). This difference in solubility is analogous to sucrose solubility in water and methanol, which are $210$ and $\pu{1 g}$ in $\pu{100 mL}$ of solution, respectively at $\pu{25 ^\circ C}$ (Wiki, and Ref.3). Further, sucrose is insoluble in diethyl ether.
References:

Georg Hingst, Werner Emte, "Process for recovering sugar from natural products containing it," US Patent 1958, 2,829,985 (Patent Application, 1955, US2829985A) (PDF).
Oleg K. Kononenko, Karl M. Herstein, "Nonaqueous solvents for sucrose," Ind. Eng. Chem. Chem. Eng. Data Series 1956, 1(1), 87–92 (DOI: https://doi.org/10.1021/i460001a017) (PDF).
Z. Bubník, P. Kadlec, "Chapter 5: Sucrose solubility," In Sucrose: Properties and Applications; M. Mathlouthi, P. Reiser, Eds.;  Springer: Boston, MA, 1995, pp. 101-125 (ISBN: 978-1-4613-6150-3).

Regarding the first reference: Bad Segeberg is the place of residence for Georg Hingst, so you would want to strike that.

@ njuffa: Thanks for the notation. I thought it is the name of second author. I'll change it.

I admit I have never heard about this extraction on industrial level.


text:
How would we arrange these in the increasing order of stability?
Here we have three isomers of butene there structure of who I have attached:

cis-butene
trans-butene
iso-butene

I was able to figure out that the trans isomer would obviously be more stable than the cis one due to steric repulsions among the methyl groups in the cis isomer.  But how would we judge the isobutene? As they are isomers, the number of hyperconjugative structures ends up being the same.
I had an intuition that this might relate to intermediate stability in hyperconjugation but couldn't come to any conclusions.

You might want to start with experimental results like heats of formation and then rationalize the results.

@user55119 yes but that would just tell us the order and would just be a raw answer. How can we understand how iso-butene is the most stable?

If you argue for hyperconjugation, then two methyl groups on the same sp2 carbon is better than on different sp2 carbons.

@user55119 Oh so we're indeed looking at the intermediate carbocation that will be formed during formation of the hyperconjugative structure as it would be stabilised by the positive inductive effect by the methyl groups right?


text:
Reactions with a positive Delta H and a negative Delta S are said to be nonspontaneous at ALL temperatures. But what if the temperature IS low enough?
What I mean is, that if you use G = H-TS and make the temperature low enough, couldn't you make any reaction spontaneous? For example, you have a Delta H of 85 kJ and a Delta S of -66 J/K. I've been told that this reaction is nonspontaneous at ALL temperatures, but if I make the temperature low enough, like around -1290 K, Delta G is now a negative value. Doesn't that mean that the reaction is now spontaneous at this temperature?

(-1) There is nothing below 0 K! This is absolute zero!

I’m voting to close this question because it is simply terrible!

First off, you can't go below $0$ K. That's absolute zero like @EdV said
Secondly, since $0$ K is absolute zero, you can't have the value of the temperature in negatives.

I am against closing it because the question below provides a crucial information to OP and to everyone making this big mistake in future.

Your reasoning is not wrong but you are missing a critical piece of information that invalidates your conclusion. Namely, temperature (on the absolute temperature scale) cannot go below zero.
Consider that temperature is effectively a measure of motion on the molecular scale, so you can imagine that there is a point at which all molecular motion stops.
This happens at 0 K, when all atoms are perfectly still (no rotation, no vibration, and so on) and it is not physically possible to lower the temperature any further.
If you put that limitation into the Gibbs free energy formula, you will see that there is no discrepancy.


text:
How do I calculate weight percent of BaCl2
A $\pu{0.1036 g}$ sample containing only $\ce{BaCl2}$ and $\ce{NaCl}$ is dissolved in $\pu{50 mL}$ of distilled water. Titrating with $\pu{0.07916 M}$ $\ce{AgNO3}$ requires $\pu{19.46 mL}$ to reach the end point. What is the weight percent of $\ce{BaCl2}$ in the sample?
(Molar mass of $\ce{BaCl2} = \pu{208.232 g mol-1}$ and Molar mass of $\ce{NaCl} = \pu{58.4425 g mol-1}$)
The way I did it was ;
0.1036g of sample contained 266.6745 Molar mass of $\ce{BaCl2}$ and $\ce{NaCl}$
i.e. 0.1036g of sample= 266.6745 of substance
x g = 208.232 of $\ce{BaCl2}$ only
x g=(0.1036*208.232)/266.6745
=0.08089 g
Where 0.08089 g is the mass of $\ce{BaCl2}$ present in the sample.
Weight % = (0.08089/0.1036)*100%
= 78%
I don't know if I'm correct.
what is the correct solution to the assignment please?

Doesn’t this statement of yours raise a red flag: “From my own solution : 0.1036g of sample contained 266.6745mol/L of BaCl2 and NaCl”?

Hint: Using the densities and molar masses, pure NaCl has a “molarity” of 37.13 mol/L and barium chloride’s “molarity” is 18.52 mol/L. A mixture of the two would have intermediate “molarity”.

Your approach is incorrect. Set up two equations with two different data. Remember you precipitate all $\ce{Cl-}$ in the solution as $\ce{AgCl}$.

Use algebra for solving it. Choose two unknowns : $x$ for the number of moles $\ce{NaCl}$ and $y$ for the number of moles $\ce{BaCl2}$. Now $x + 2y$ is the number of moles of $\ce{Cl-}$ ions. And it is also equal to the product Volume $\ce{AgNO3}$ times concentration of $\ce{AgNO3}$. this gives you a first equation. The second equation is related to the masses. The mass of $\ce{NaCl}$ is equal to $x$ times $58.4 g$. The mass of $\ce{BaCl2}$ is equal to $y$ times $208.2 g$. This gives the $2$nd equation. Try to write them properly, and solve the system of two equations and two unknowns. Go !

Suppose $x$ and $y$ are the amount of $\ce{BaCl2}$ and $\ce{NaCl}$ in $\pu{mol}$, respectively in original mixture. Thus the mass of $\ce{BaCl2}$ in the mixture is $208.232x$ and that of $\ce{NaCl}$ is $58.4425y$, both in $\pu{g}$. Thus,
$$208.232x + 58.4425y = 0.1036 \tag1$$
Now, the the amount of $\ce{Cl-}$ ions in original mixture is $2x+y$ in $\pu{mol}$. And all of them are precipitated by $\ce{AgNO3}$ according to:
$$\ce{Cl- + AgNO3 -> AgCl + NO3-}$$
Therefore, mole ratio of $\ce{AgNO3 : Cl-}$ needed is $1:1$. Since total $\ce{Cl-}$ in the reaction is $2x+y$, The amount of $\ce{AgNO3}$ used $(\pu{0.07916 mol L-1} \times \pu{19.46 \times 10^{-3} L} = \pu{1.540 \times 10^{-3} mol})$ is equal to $2x+y$. Thus,
$$2x+y = 1.540 \times 10^{-3} \tag2$$
Now, you have two equations for two unknowns, so you can solve them easily.
First, $(1) - (2) \times 58.4425$ gives
$$208.232x - 58.4425 \times 2x = 0.1036 - 58.4425 \times 1.540 \times 10^{-3} = 0.01360$$
$$\therefore \ x = \frac{0.01360}{91.347} = 1.489 \times 10^{-4}$$
Apply this value in the equation $(2)$, and you get:
$$y = 1.540 \times 10^{-3} - 2x = 1.540 \times 10^{-3} - 2 \times 1.489 \times 10^{-4} = 1.242 \times 10^{-3} $$
Thus, the mass of $\ce{BaCl2} = 208.232x = \pu{0.0310 g}$ and the mass of $\ce{NaCl} = 58.4425y = \pu{0.0726 g}$.
Thus, weight percent of $\ce{BaCl2} = \frac{0.0310}{0.0310 + 0.0726} \times 100 = \frac{0.0310}{0.1036} \times 100 = 29.92\%$

(+1) Glad to see I got the same answer!


text:
How does the addition of NaCl to the cathode side effect the potential of a voltaic cell: Zn(s) | Zn2+(aq) 0.1M || Cu2+(aq) 0.1M | Cu(s)
By addition of $\ce{NaCl}$ to the $\ce{Zn}$ side of the cell, I observed precipitation of $\ce{ZnCl2}$.
I measured the voltage to $\pu{1.05 V}$ before adding $\ce{NaCl}$ to the cathode side, I then observed the voltage go down by addition of the salt.
According to the Nernst equation: as the $\ce{Zn^2+}$ ion concentration reduces with precipitation, there should be an increase in voltage.
$$E = 1.05 - \frac{RT}{nF} \ln Q$$
$$E= 1.05 - \frac{RT}{nF} \ln \left(\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}\right)$$
since $\ln Q \lt 0$ for $[\ce{Zn^2+}] < [\ce{Cu^2+}]$
"$-\frac{RT}{nF} \ln Q$" term would be positive. Therefore the voltage should go up. But the opposite was observed and I don't know how to explain it.

Do you consider activity coefficients and chlorocomplexes ?

There are two mistakes here. First $\ce{ZnCl2}$ will not precipitate. It is one of the best soluble compounds of zinc. Second, you say that you add NaCl to the cathode side and the cathode of the cell is made of copper. Then you say later on that you have added NaCl in the zinc solution. So please explain what you have done and observed.

@Maurice Thank you. Yes I mixed them up, apologies. I had an equal solution of ZnSO4 (on Zink side) and CuSO4 (on copper side).  I then added som NaCl to the Zink side, anode side that is. I then observed some kind of precipitation on the anode side that fell to the bottom. I also observed the voltage go down on my voltmeter. The impression given by the instructor was that the Zn2+ would be precipitated with Cl-, thus explaining the shift in potential. However that doesn't seem correct by the calculation. Please let me know again if I'm beeing unclear, english is not my native language.

@JR19 did you try adding $\ce{NaCl}$ to cathode side you can see the color change from blue too green the copper complex formation copper aqua complex is blue where as copper chloro complex is yellow mix of these colors show green but $\ce{Zn^{2+}}$ d orbitals are filled so no color effect too bad you don't see this in anode side.

This is due to the complex formation of $\ce{Zn^2+}$ with $\ce{Cl^-}$ which will result a different half cell equilibrium in this case the anode,
Original half cell reaction,
$$\ce{Zn^2+ (aq) + 2e^- <=> Zn(s)  E^o = 0.762 V (reduction)}$$
Modified half cell reaction,
$$\ce{[ZnCl_4]^2- (aq) + 2e^- <=> Zn(s) + 4Cl^- (aq)}$$
(Since you are using a voltmeter it has infinite resistance (hypotheticaly) so the cells are in equilibrium)
The precipitation must be due to $\ce{Na2SO4}$ not $\ce{ZnCl2}$ because $\ce{ZnCl2}$ is more  soluble than $\ce{Na2SO4}$ $\ce{(281 g/L vs  357
g/L (25 °C))}$ $\ce{Na2SO4}$ $\ce{ZnCl2}$.
It should dissolve after few minutes and form the complex , try stirring the flask.
You should increase the voltage up and down with decreasing or increasing the concentration of $\ce{Zn^2+}$ ions with same solvent $\ce{ZnSO4 (aq)}$ to use the Nerst equation. Addition of different chemicals will change the cell equilibrium thus don't use the Nerst equation for those cases like this scenario. Have to refer a book for the new cell equilibrium half cell value to verify the result with the new $\ce{E^{o}_{cell}}$ value.

I don't agree with $\ce{NaCl}$ precipitation. It the precipitation really happens, that should be as $\ce{Na2SO4}$ because it has the lowest solubility among all other salts in display.

@MathewMahindaratne Yeah correct I have edited that

@Avon97. No ! If you consider that some complex with $\ce{Zn}$ and $\ce{Cl-}$ is formed, this effect will decrease the concentration of the non complexed $\ce{Zn^{2+}}$ ion. And if this concentration decreases, the potential changes from $\pu{-0.76 V}$ to a more negative value. And this is the contrary of what JR 10 has found.

@JR19. If you modify the anode compartment, and if you then observe that this change produces a decrease in the cell voltage, it means that [$\ce{Zn^{2+}}$] has increased. It is probably an effect of "activity". At high concentration, concentration must be replaced by "activity". The voltage delivered by the electrodes is not really proportional to the log of the concentration, but it is proportional to the log of the so called "activity". I explain.
Let's consider the case of $\ce{HCl}$, which is best known. If you measure the pH of $0.01$ M HCl, you find pH = $2.0$. If now you add and dissolve some $\ce{NaCl}$ to this solution, the pH goes down to $1.5$ and even to $1.2$, as if the solution gets more and more concentrated. It is hard to believe, but easy to check. Do it in the lab, if you can. Here is an explanation taken from an old issue of J. Chem. Educ. (I don't remember which one). The measured "concentration" given by an electrode is the number of moles divided NOT by the volume of the solution, but divided by the volume of the "free water", namely the volume of water not engaged in adsorption around the ions. This sort of "new concentration" is called activity. By adding more $\ce{NaCl}$, the volume of free water decreases, so that the activity of $\ce{HCl}$ increases. In $1$ liter concentrated solution of $\ce{NaCl}$, there is only about $0.2$ liter of free water. The activity is much higher than the concentration. It must be the same phenomena in your cell.

@Buck Thorn. OK. I shall edit the text, and  modify this sentence

Although we have two good answers for this question, I'd like to point out some other information to explain what's going on when you add $\ce{NaCl}$ to the anode. Understandably, increasing $\ce{NaCl}$ would make $\ce{Zn^2+}$ ions to act as they are from $\ce{ZnCl2}$. According to Ref.1, aqueous $\ce{ZnCl2}$ solutions contain at least four $\ce{Zn}$ species indicated by Ramon IR spectroscopy:

Raman studies of aqueous solutions of zinc chloride indicate the presence of $\ce{Zn(H2O)6^++},$ $\ce{ZnCl_{(aq)}+},$ linear $\ce{ZnCl2_{(aq)}},$ and $\ce{ZnCl4(H2O)2^=}$ in aqueous solutions less concentrated than ten molar. Ion concentrations are reported as functions of zinc chloride molarity. The tetrachloro complex dominates for stoichiometric zinc chloride concentrations between ∼0.5 to ∼10M with significant concentrations of $\ce{ZnCl+}$ below ∼4M zinc chloride and $\ce{ZnCl2}$ above ∼4M. For stoichiometric solutions greater than 10M, the evidence suggests a polymeric species or aggregate with structural characteristics similar to those found in the crystal.

The presence of these four species were confirmed later by X-ray scattering studies (Ref.2). Thus, it is safe to say that the standard potential of zinc ion would be adjusted significantly due to the presence of these four species (Ref.3). In addition, Ref.4 reports the increment of the conductivity of aqueous $\ce{ZnCl2}$ solutions according to their concentrations.
References:

D. E. Irish, Bruce McCarroll, T. F. Young, "Raman Study of Zinc Chloride Solutions," J. Chem. Phys. 1963, 39(12), 3436-3444 (DOI: https://doi.org/10.1063/1.1734212).
M. Maeda, T. Ito, M. Hori, G. Johansson, "The Structure of Zinc Chloride Complexes in Aqueous Solution," Verlag der Zeitschrift für Naturforschung A 1996, 51(1-2), 63-70 (DOI: https://doi.org/10.1515/zna-1996-1-210) (PDF).
Lutfullah, Helen S. Dunsmore, Russell Paterson, "Re-determination of the standard electrode potential of zinc and mean molal activity coefficients for aqueous zinc chloride at 298.15 K," Journal of the Chemical Society, Faraday Transactions 1: Physical Chemistry in Condensed Phases 1976, 72, 495-503 (DOI: https://doi.org/10.1039/F19767200495).
B. K. Thomas, D. J. Fray, "The conductivity of aqueous zinc chloride solutions," Journal of Applied Electrochemistry 1982, 12(1), 1-5 (DOI: https://doi.org/10.1007/BF01112058).


text:
Making carbon powder using fireplace and weak acid
I am doing a little home experiment trying to make as pure carbon powder as I can. I plan to then used it to eat up oxygen from a container where I want to try to melt aluminum.
My process was following:

I have put some sawdust in an iron can.
I covered the opening of the can with clay, leaving only a small opening.
I put the can in a fireplace and let it heat to about $\pu{800 ^\circ C}$.
After removing, I ended up with about 1/3 volume of black powder.
I washed the powder with water and filtered the water out.
I placed the wet powder in a beaker and added more water.
I added about 12% hydrochloric acid to dissolve calcium and metal oxides and hydroxides.
The solution started bubbling slightly, so I left it for several hours with occasional mixing.
I washed the resultant powder with water twice and left it to dry.

Did I manage to purify the carbon in the powder somewhat? Did the use of acid bring any benefit?

2x yes, obviously. I don't understand what you're asking for here. Your sequence sounds like you know what you're doing.

Only that wood ash also contains carbonates. Hence the bubbling. Hydroxides don't survive at 800°C.

@Karl I don't know if I know what I'm doing. I followed what I thought would work, and wanted to check if this is the right approach to remove impurities and leave relatively pure carbon.

The result of your first pyrolysis must be charcoal plus the mineral constituants that are present in wood ashes. Now the wood ashes are usually a mixture of : $40 - 70$ % $\ce{CaCO3}$, $5 - 10$ % $\ce{Na2CO3}$,  $5 - 10$ % $\ce{K2CO3}$, $2 - 3$ % $\ce{K2SO4}$, $2 - 5$ % $\ce{Ca3(PO4)2}$, $5 - 10$ % $\ce{MgCO3}$, $2 - 10$ % $\ce{SiO2}$, < $1$ % $\ce{FePO4}$. Washing your charcoal with water and hydrochloric acid must remove all these substances, except $\ce{SiO2}$. This impurity cannot be removed.