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Can a battery be built where several electrodes share the same electrolyte?
If you have two metals i.e. iron / copper and an electrolyte solution you can make a battery. Typically all of the designs have several of these connected in series to boost the output. But I have seen some instances where people claim you can connect several nails sharing the same lemon as an electrolyte, connect the metals in sequence and get that boost in output.
Example: https://www.sciencebuddies.org/stem-activities/lemon-battery
If that is the case could you do the same with several nails (iron/copper) connected in sequence but all sharing the same saltwater electrolyte and still get the boost in power output?

Can you provide a reference for the several nails sharing the same lemon?

I added a reference that uses foil and copper pennies but it does show them connected in series on the same lemon. Link has been added to the main post.

Thanks! So what you have is two series cells: lemons have membranes such that you do not have a simple contiguous fluid beneath the rind. So you could slice the lemon into pieces and make a lemon cell from each piece with its two electrodes. Then two or more gives a battery proper.

Say you had a single cell consisting of a large (plastic) drum of salt water. And on separate sides of this drum (say the circumference is 2 feet) you had the electrode/ anode pairs and then you connected them in sequence via a copper wire. Would the size of the cell potentially provide enough distance to allow for a boost or would all the power likely flow between the electrode/anodes via the electrolyte and not via the series.

I think your second scenario is the likely one. But the nice thing about your question and proposed scenario is that they are fairly easy to test. You could use a big plastic bowl, clothespins to hold electrodes to the side, wire and metal paper clips to make connections, and a cheap DMM to measure the voltage.

@EdV I had the same/similar thought that if you peel the lemon you could see where to put the nail pairs so that they occupy different lemon segments and thus utilise the electrolyte in several separated cells

Connecting 2 electrodes dropped in the same electrolyte container causes the shortcut of the cell they form. 2n electrodes do not form n cells. they form 2n(2n-1)/2 cells, n(n-1) of them with the same electrodes and n-1 of them have a shorcut.

@CaiusJard I agree. I upvoted all three answers because they are all informative and address the question and they supplement each other. The accepted answer includes the important recognition that the shared electrolyte effectively puts parallel resistances (“leakages”)  across each cell. So compartmentalizing the electrolyte, as in using the different lemon segments, is beneficial. This also shows that a balance exists: if the electrolyte is too conductive, the battery is shorted quite a lot. But if the electrolyte is not sufficiently conductive, then the battery will be very low powered.

I've only basic training in chemistry, since my competences are in Electrical Engineering, but I'd like to tackle your question from another angle.
I'm assuming your question is sort of an X-Y problem and you are not really interested in increasing the efficiency of an actual lemon-based cell.
From an engineering POV, what you want from a power source is, simply speaking, to extract the maximum amount of energy without losing much in the process.
There are 3 basic quantities involved in this: energy, power and efficiency. Voltage and/or current are of much lesser importance, since there are electronic methods that can boost one of those at the expense of the other.
For AC generators, a transformer can increase/decrease the voltage while decreasing/increasing current (with a given load) at the same time.
For DC generators such as batteries, you can add a switching DC/DC converter that can step the voltage up or down as needed with high efficiency.
So the point is: is yours an efficient method to improve either the power or the energy output of your cell?
As already pointed-out, your "multi-cell" arrangement can be modeled as a series connection of cells with every cell having a resistor in parallel. This latter models the fact that the electrolyte of the various cells is shared among them.
Those parallel resistances represent an additional load to the cells. This wastes power and energy. Are these losses negligible or not? To answer this you should consider other losses in the system, e.g. the internal resistance of the electrolyte, the contact resistance of the connection wires and the resistance of the wires themselves.
To do a meaningful comparison, you have to know the value of those parallel resistances, and this is probably tricky business.
But remember, the energy stored in a cell depends on the volume of the electrolyte (well, probably on its mass, but assuming a more or less constant density...), so you should ask if there is a better method to suck out the energy out of a given volume of electrolyte. Well this is what is already done when you need more energy: you build a bigger cell. And if you need also more power, you increase its current rating (since it's voltage can't be changed) by increasing the surface of the electrodes.
If you need higher voltage, for a given power, you simply put a DC/DC boost converter between the cell and the load. Nowadays, this is what is done in lots of products that can be powered with a single AA or AAA cell, for example.
In this way you avoid entirely those extra losses represented by those parallel resistances.
BOTTOM LINE
As already pointed out in other answers and comments, your idea could work, but it's inefficient. To increase the efficiency you should increase those parallel resistances and this means more distance between pairs of electrodes. This implies a need for a bigger volume of electrolyte.
This is also (probably) inefficient because there are already other well-tested means to extract the energy from the same amount of electrolyte and provide to the end-user (the load) the voltage level it needs to work properly.
Moreover your system is much more complicated to set-up, even if you could make the losses negligible, since the output voltage can only be set by multiples of the cell voltage, and you would need a DC/DC converter (or an DC/AC converter if your load needs AC) anyway to produce the voltage the load needs.

(+1) Excellent answer, covering a number of important issues, and welcome to the chemistry SE! One thing, though: the electrolyte is not the energy source in a simple voltaic cell, even for the lemon cell. The cohesive energy is the main source of energy. So in the present case, with zinc and iron electrodes in seawater, the zinc is the main energy source. There is little iron ion content, per se, in seawater.

Reference for the cohesive energy statement in my other comment: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95 (10) (2018) 1801-1810. The Zn and Cu Daniell cell is addressed in great detail. TL; DR Cohesive energy differences are the major factor in explaining the behavior of this famous galvanic cell.

@EdV Thanks! And also thanks for the clarification. As I said, I'm not a chemistry specialist and my electrochemistry basics dates back to ~30yrs ago. I never heard the term "cohesive energy" (unless it is called in a completely different way in Italian). I hope I can get a bit of spare time to give a quick refresher to my rusty electrochemistry notions. Thanks for the pointer.

Glad to have EEs here! I had a grad minor in EE, but having real pros around keeps thing more grounded (pardon the pun)! Klaus’s paper is nice because it is way too easy to jump to a simplistic explanation like electronegativity differences, etc. And, for the Daniell cell, zinc and copper are adjacent elements in the periodic table, so he addresses why zinc gets oxidized and copper ions get reduced.

@EdV. I appreciate puns. They help engineers keep their feet on solid ground. ;-)

Yes, you could have multiple electrodes in the same electrolyte, but to some extent, that would short-circuit the battery. For example, if you stack copper and silver coins with blotting paper (bp) between them, in the order:
Cu bp Ag Cu bp Ag ... Cu bp Ag and immerse the whole in an electrolyte, rather than just wetting each piece of blotting paper, some of the electricity will "sneak around" the edges, using up some current from the battery and reducing voltage because of internal resistance voltage drop, as well as shortening the life of the battery.
The same applies to iron and copper tacks, stuck into a lemon and connected as above. That said, if the non-connected electrodes are closely spaced, and the wired-together tacks are widely separated, not too much juice would be lost from the lemon battery for a brief demo, e.g., lighting a 3 volt blue LED.

As EdV and the OP  (who appears to be from electrical engineering side) postulate in the comments, the (-) and (+) poles will be oblivious of the other electrodes. You have to use separate lemons.

@M.Farooq DrMoishePippik appears to claim that because the lemon is not a very good conductor, the smaller resistances within each cell and greater resistance between cells should be enough to create a greater voltage, at least for a short time. S/he's not claiming it's efficient or that 98% of the available energy isn't wasted in self-heating.

I've done this many years ago with some success.  The separation ratio I recall was rather larger than shown, and we used a voltmeter so the current drawn would be tiny.  I think an experiment is in order but won't have time tonight

@ChrisH, Worth trying. The distances have to be larger, I believe.

I'd double the distance at least, but the theory behind it is correct.

What if all of those nails were connected in parallel with the other nails of their type?

Parallel is no problem in he same electrolyte; it is equivalent to a larger electrode.

Yes, you can have same the electrolyte and a pair of two different metals, but the key point is that if you wish to increase voltage difference, you need to connect them in series and use separate containers for each pair, and of course each pair must be connected. Your postulate in the comment is correct. If we use a large bucket, only the pair connected to the terminals of the battery will be responsible for the voltage. The rest of the pairs sitting in the same bucket are sort of "short circuited".
Now, there is a problem with same electrolyte batteries. The Nernst equation cannot predict the potential. This is why for theoretical purposes, it is beneficial to match the ions in solution and the metallic electrode, i.e., Zn must dip in zinc ions (zinc sulfate). Just by chance, I was reading a translation of a century old book (from German): Quantitative analysis by electrolysis. It is amazing how much people knew in the 1890s. There is a list of batteries, whose names we have never heard of including a "gravity battery". The two solutions of copper sulfate and zinc sulfate are separated not by membranes but by their densities and this battery was quite popular in telegraphic days.
Of course real battery science is more an art. This is the only branch of modern electrochemistry where fast progress is needed. If someone can beat the electron transfer kinetics he/she will deserve a Nobel Prize. You see lead acid batteries, despite all the bad things about lead, are still used worldwide, because its electron kinetics is fast and electrochemists do not have a viable alternative.
I do recall a series potato battery, which could light a small bulb here (https://www.livescience.com/62570-potato-battery-conduct-electricity.html). Replace potatoes with salt water beakers or any inert salt.

Ok extreme example. Say you have a Electrode/Anode pair that will react well with sea water. The ocean is reasonably big so its possible to have enough separation that the pairs of metal won't short each other.
Once you have achieved this minimum distance (no idea what it might be but pretty sure if they are far enough away from each other the single cell could service both reactions without them sharing in the reaction with each other) is it likely that the minimum distance will be too great to productively attach the items in serial?

I upvoted: good answer. As for the idea by @JosephU., a simple test might be to carefully open up (with needle nose pliers) a standard 9 V battery, spread the 6 individual cells apart while keeping their electrical connections, and then placing that exposed battery in a bowl of saltwater. Will it give the original (nominal) 9 V output? I may try this for fun: I have seen 9 V batteries with blue cylindrical internal cells, like wikipedia shows in its 9 V battery article.

@EdV Worth trying.


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Translating from milligrams in the pill to Ki at the receptor
Many places (e.g. Wikipedia) report the action of a drug on the various receptors, transporters, ion channels, and the like in terms of the Ki(nM). (This is for drugs that act primarily through such channels). They really ought to, but usually don't,  also tell you whether the mechanism is competitive, uncompetitive, noncompetitive, as these can affect the rate of binding by a factor of 2 (or more?). For purposes of this question, assume we are dealing with competitive binding.
Is there a way to go from drug dosage and these Ki values to the level of binding of particular receptor? For example, suppose we have 100% absorption (or we look at the pharmacokinetics for greater precision). If you know the molecular weight of a drug, and blood volume (average about 5 liters, though you should adjust for body weight) then you can calculate the concentration in nM. (I am guessing that I can neglect the volume of blood cells and leakage into the lymphatic system, though I may be wrong about this). Then I think you can just multiply by the Ki.
My goal is to draw a line between the interactions that matter and those that don't under standard dosage. More precisely, I  am trying to compare receptor and transporter bindings for several drugs in the same drug class at the usual prescribed dosage for each. I observe that the range of reported Ki(nM) is often six orders of magnitude or greater. So even if you don't know the binding mechanism, one should still be able to divide the actions into those definitely present, often at near saturation, those that are only trivially present, and those in a range that might be affected by doubling or halving the dose. The difference in magnitude caused by different binding mechanisms should only matter for those in the middle ground.
However, I am not a pharmacist, biochemist, or doctor, and I am very aware that every step in the chain of logic above might be mistaken. This is about trying to think clearly about the effects of related drugs, and the information is not going to be used to set dosage for any animal or human. So I think it is OK for me to look for, and for you to offer, pretty good information that may not have 100% scientific validation.

There are just too many unknowns in this to be able to estimate level of binding without a lot more information. You cannot assume 100% absorption. Where is the site of action? If it is brain then you have to consider transport across the blood/brain barrier. There may be preferential uptake into some tissues. You need information about the rate of uptake into the blood stream and metabolism too.

Have you read further in the literature, eg a search for "drug binding ki receptor"? I hit for instance this right away: https://pubs.acs.org/doi/pdf/10.1021/acschemneuro.7b00185 Like Waylander, I think ignoring the"D" in ADME might get you bad results. OTOH for a highly soluble drug with few side-effects ignoring other targets might not be so bad (<- opinion).

In vivo "target engagement" assays are notoriously complex, especially for non-covalent ligands.  Your target might be intracellular.  It might be partially expressed on the cell surface of an epithelial and thus exposed to the plasma, but there might be a second population of receptors/targets that are not at the cell surface, but e.g. on the mitochondrial surface.  Those two populations might be exposed to different levels of ligand.

Well, not exactly what I hoped to hear (but thanks, Buck Thorn!, This article is very helpful, & I'm working my way through it).  I'm trying to compare about 15 antidepressents, focusing on non-transporter receptors.  Antidepressants are differentially effective on various special populations, such as seniors, those with treatment-resistant depression, those with concommitant cognitive problems (which don't abate with depressed mood), to see if receptor bindings correlate to the variation. But I cann't begin to address these questions unless I know which receptors are bound.

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How the concentration of anodic half cell increases in galvanic cell?
According to my textbook

Concentration of anodic half cell increases and that of cathodic half decreases when two electrodes are joined by a wire.

I have also read that it is connected with a Salt bridge which prevents the flow of the electrolyte ions and supplies the required cation or anion (from the inert electrolyte in the salt bridge) and overall its electrically neutral . Then how they can state that concentration increases in one half cell only ,that is both cells are provide with sufficient ions ? Please correct me if I am wrong somewhere in understanding the concept.

Look at the Daniell cell here. The zinc metal anode is oxidized to zinc ions, so the zinc ion concentration in the anode reservoir increases as the cell provides electrical power. Copper ions get reduced at the cathode, so the copper ion concentration is reduced in the cathode reservior.

@Ed V okay sir , I understood . I was considering anions too , I thought that in cathodic cell there will be accumulation of anions which will increase the concentration of anions there .

Cations and anions must be always balanced. Any tiny disbalance would cause strong potential gradient leading to rebalancing.

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Are there any ways of modifying molecules to increase their probability of two photon absorption?
While checking two photon microscopy, the use of high flux femtosecond laser is required for the dual absorption of photons, due to the low probability of these two photons being absorbed. Is there any way of modifying molecular electronic structure to increase this probability?

I cannot write a nice answer because I lack the basics of the optics involved, as well as the quantum mechanical tool linking structures to the phenomenon. However, as a fact, modifications are certainly pursued and in most cases starting from well know dyes, or conjugated polymers, or even short or small conjugated systems. Substitution with electron donor and acceptor groups leading to internal charge transfer is a recurrent modification. At least, one rationale is a very high absorption coeff already for the first absorption, as whatever happens to the "first" photon influences the TPA

People have been using nanoparticles to "enhance" two photon fluorescence.  This is so-called plasmonic enhancement. If you recall Raman spectroscopy, a similar effect surface enhanced plasmon resonance increases the intensity of Raman scattering. Search these keywords.

There is an excellent article that addresses your question is great detail- the design requirements for two-photon fluorescence "Two-Photon Absorption and the Design of Two-Photon Dyes". Go to Google Scholar and search this title.

BTW, one of the simplest ways to increase two-photon absorption is decreasing temperature. See https://pubs.acs.org/doi/10.1021/jp303789s and https://pubmed.ncbi.nlm.nih.gov/17583328/ (which mentions molecular conformation change with temperature).


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Raising melting point of water at STP (using disolved impurities)
Typically, water experiences melting point depression when it is impure. The more impure it is, the lower the melting point, up to a point.
I understand you could raise the melting point by increasing pressure, as discussed here: How to RAISE the melting point of water?
However, are there any impurities known that increase the melting point of the water. Ideally, whilst still being mostly water.

I understand you could raise the melting point by increasing pressure It is rather converting ice to more compact ice structures with higher melting points. But the needed pressure is not convenient. :-) // Impurities generally decrease melting point of major substance. Many salt hydrates have high melting points, rather dissolving themselves in own crystal water, but I guess it is not what you want.

When this kind of topic comes up, I like to cite tetrabutylammonium hydroxide 30-hydrate. Consisting of 67.5% water by weight, it can be isolated as a pure solid with a melting point around 27 °C. You could interpret this as an impurity which raises the melting point of ice by almost 30 °C. This is the hydrate with the highest water content I know that is a solid at room temperature, but others may exist with even more water.

When the pressure is about 1 bar, the melting point decreases when increasing the pressure, This tendency is reversed if the pressure is extremely high, and higher thant 1000 bar, which is not easy to obtain,

The more D and T atoms in water, the higher melting point.

https://chemistry.stackexchange.com/questions/150905/effect-of-impurities-on-melting-and-boiling-points

See "How to RAISE the melting point of water?" at https://chemistry.stackexchange.com/questions/25123/how-to-raise-the-melting-point-of-water


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Effect of R/S configuration on reactions not involving the stereocentre
An exam question involved the following reaction -
 +  forms
Both reactants (2-methylbutanoic acid and 2-methylbutanamine) are optically active. The question used pure R-enantiomer of the acid, and a racemic mixture of the amine, amine in excess.
However, it also suggested that the R isomer of acid will not react equally well with both enantiomers of the amine. I could not understand why, since the reaction does not involve any stereocentre. My best (and probably silly) guess would be that one configuration would encounter greater steric hindrance during attack of the nitrogen atom on carbon, so that product will be formed less. However, I could not find any resource to support this, nor did I find this very likely as the tertiary carbon atoms are separated by 4 bonds - so steric hindrance probably won't have a large effect.
Could anyone please confirm if the configuration does indeed affect the reactivity, and if yes, how and why?

The effect might be very small. but it's definitely nonzero.

@orthocresol But what is the reason for that effect? Is it steric hindrance, as I proposed?

What are the reaction conditions?

@Waylander Only these reactants are given, nothing else is mentioned (that same exam where students are expected to know everything by telepathy), but amide was the intended product - I'm sure of that.

@Waylander Also, this is just a specific example I encountered - my question is about the general case. If R/S might affect reactivity during formation of esters or ethers or secondary amines as well, apart from amides as above.

The only reason I can think of given the paucity of information on the reaction conditions is that the reaction of one enantiomer is slightly more hindered than the other

The bottom line is that it will be different, simply because the products and transition states are diastereomeric. Thus, there’s no way it can be exactly the same. Having a model or a rationale to understand which one is faster is nice, but secondary: you must first accept that they are different.

@orthocresol I see now - two diastereomers are inherently different in other properties than optical rotation, too. I was mainly thinking about the two individual molecules. Thank you for clarifying.

If you got your answer, feel free to self-answer it :)

@NilayGhosh Done :) I still don't have a source though, nor a model/theory that can reasonably predict which one out of the two products will be favoured. If you have any resources regarding that, it'll be really helpful.

Thanks to @orthocresol for helping me via the comments. Here is a reasonable explanation for it -
The final product comprises of two different diastereomers, of configurations RR and RS. Diastereomers, unlike enantiomers, have some inherently different properties, other than optical rotation.
Consequently, the reaction that forms two different diastereomers will proceed through intermediates and transition states that will also have different properties. Since they differ in aspects other than optical rotation, their products will be formed in different amounts. Just because the products are different, the amounts formed are also bound to be different - they essentially become two different reactions. Even if the difference is very slight.


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Calculating the weight of a chemical within a liquid mixture
I want to know the weight of ethylene glycol in a mixture of Shellzone 50/50 Antifreeze. I have taken the mid-range of 40-60% from the safety data sheet(SDS) for the amount of ethylene glycol in the mixture and used the number 50 to estimate that 50% of the mixture is ethylene glycol. The specific gravity of the Shellzone 50/50 Antifreeze is 0.909, also taken from SDS, so I have calculated the solution to be 7.59 lbs/gal.
Would I be correct in saying that 1 gallon of Shellzone 50/50 Antifreeze contains 3.79 lbs of antifreeze?
EDIT:
Would I be correct in saying that 1 gallon of Shellzone 50/50 Antifreeze contains 3.79 lbs of ethylene glycol?
See below for calculations.

I believe that the amount of ethylene glycol in one gallon of Shellzone 50/50 Antifreeze would be 3.79 lbs because the concentration is a (% w/w) according to the SDS. Therefore, 40 - 60 percent of the mixture should contain 3.03 - 4.55 lbs of ethylene glycol or 3.79 lbs at its midpoint.
A coworker discussed with me that this may be incorrect because I did not use the specific gravity of ethylene glycol. I chose not to use the specific gravity of ethylene glycol because ethylene glycol and Shellzone Antifreeze 50/50 are not equivalent. Even though both solutions have a chemical in common, they are not the same solution.
I do not think it would be appropriate to use that number to calculate its mass since they are not equivalent. Similarly, if I were to apply the same logic to the other constituents, I would need to use the specific gravity for water and diethylene glycol to calculate the weight of each chemical. I think that this would be unnecessary since the relative density is already given and that the mixture is designed to be homogenous. It would seem fair to suggest that the weight of each constituent can be calculated by using the concentration and the relative density.
The only other potential problem with this calculation is the variation in the actual specific gravity due to temperature but that is not important for the purpose of this calculation.

Provide detailed calculation and reasoning why it should be correct and why it may not be correct. Start with replacing the word antifreeze with ethylene glycol, as the antifreeze is the whole mixture.

You procedure is correct. 50% w/w means 1/2 of total mass. You would need specific gravity of the antifreeze AND of ethylene glycol, if you worked with V/V%, similarly as for ethanol. As 50% V/V ethanol is 1 volume of ethanol + 1 volume of water, giving 1.96 volumes of the mixture(volumes are not additive).

@Poutnik, I am glad that my edit was able to help you understand more clearly. I appreciate you taking the time to answer my question.

Please give us values in usual units ! And not in lbs per gallon. Is lbs a pound ? And why is the word antifreeze crossed ? What is the meaning of your first sentence? I copy : "a mixture of Shellzone 50/50 Antifreeze" ? Try to be clear !

@Maurice I  wanted to ask if he means (UK wine/US) gallon OR Imperial gallon, omitting corn/dry gallon as probably not applicable to liquids. OTOH lbs means pounds I guess, which are hopefully same everywhere.

@Poutnik.Sorry ! My question was not for you. It was for Brian Smithers

@Maurice  I got it. :-) My comment was rather a comment to your comment than answering your comment.


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How many stereogenic centers exist in the atropine?

I'm pretty sure that both C bonded to N are stereogenic centers. I don't know about the others, any help? Is this correct?

I think there is only one chiral center in atropine.

@Curt F.: The ring is symmetric so all three carbon centers shown in are not chiral. The sole chiral center is the $\alpha$-carbon to carbonyl group.

Hi, actually I meant stereogenic center, I don't know if it's different from the definition of chiral center.

In this question you can look at the IUPAC definition of the chiral center. According to the definition, carbons that are attached to nitrogen are chiral centers — they have 4 different chemical groups attached to them. The $\alpha$-carbon to the ester group is also a chiral center. However, the third carbon (from top to bottom) that you circled is not a chiral center because two groups from the cycle are identical.

Yes, nevertheless, the "third carbon" is a pseudoasymmetric center (r or s).

IUPAC defines chirality centre (https://doi.org/10.1351/goldbook.C01060) as An atom holding a set of ligands in a spatial arrangement which is not superposable on its mirror image. That doesn't necessarily mean that it has to be four different groups, although that is a convenient rule of thumb.

@Loong you're right but question asks for chiral centers

@Martin-マーチン in definition that you referred to it's written Cabcd. And it's also written that this is a generalization of concept of asymmetric carbon atom which is defined as I wrote (https://doi.org/10.1351/goldbook.A00479). If I am wrong somewhere, please correct me.

Hi, actually I meant stereogenic center, I don't know if it's different from the definition of chiral center.

Well, the second part is in principle an example, where it generalizes the concept of Cabcd to other species. It is not necessary for the definition.

I assume the OP means atropine in which case the structure presented has an extra methylene group. The renderings of a single enantiomer of atropine by ChemDraw 20.1 and JSmol are shown below.
Using the terminology of Mislow and Siegel, the bridgehead carbons and the benzylic carbon are stereogenic and chirotopic, Thus, the R/S designation. The remaining carbon labeled "r" is stereogenic but achirotopic (r/s). Therefore, there are four stereogenic centers three of which are chirotopic.


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Can stainless steel be electrically insulated with (oven-dried) Sodium Silicate Solution?
Can I electrically insulate stainless steel by painting Sodium Silicate solution on it and heating it to 200C for 30min or so?

The solution should be conductive, and I've been told (to be verified) sodium silicate itself is non-conductive. Will it work from an electrical standpoint?

Will it work chemically in the long term, or will it corrode the stainless steel or something?

I don't think it would actually work in practice. Even if you somehow manage to make a layer good enough for insulation, it would probably easily wear off.

The issue at hand has little to do with the properties of sodium silicate or a baked coating of it. The question is whether any such coating would either form or be robust once formed when applied as described on untreated stainless steel. this seems very unlikely. And unnecessary as there are plenty of paints and varnishes that are already known to work.

Do you know of one that works well above 450C and conducts heat acceptably well?

The short answer is yes since1:

Amorphous  silica  is,  in  general,  a  good  electric  insulator
but  exhibits  a  finite  dc  electrical  conductivity,  especially
when  impurity  alkali  ions  are  contained

You will also find more info and specific references on that paper(ref. 1). For sure tests will be needed depending on the service conditions (it seems that porosity is also a critical aspect of the insulation properties).
It will also work well for improving the corrosion resistance of an existing coating. From the abstract of this paper2:

To improve the corrosion resistance of phosphate coatings, the
phosphated hot-dip galvanized (HDG) sheets were post-sealed with
sodium silicate (water glass) solutions. The results show that after
the silicate post-treatment the pores among zinc phosphate crystals
are sealed with the films containing Si, P, O and Zn, leading to the
formation of the continuous composite coatings on the surface of HDG
steel. The corrosion resistance of the composite coatings depending on
concentration of sodium silicate and post-sealing time is greatly
improved by the silicate post-treatment.

References

Minoru Tomozawa, Chapter 3 - Amorphous silica, Editor(s): Hari Singh Nalwa, Silicon-Based Material and Devices, Academic Press, 2001,
Pages 127-154, ISBN 9780125139090, DOI: 10.1016/B978-012513909-0/50005-2
Bi-lan Lin, Jin-tang Lu, Gang Kong, Synergistic corrosion protection for galvanized steel by phosphating and sodium silicate post-sealing,
Surface and Coatings Technology, Volume 202, Issue 9,
2008, Pages 1831-1838, ISSN 0257-8972, DOI: 10.1016/j.surfcoat.2007.08.001.

Thank you Jojostack! Will this electrically insulate the stainless steel?

yes, when all water is boiled out you're left with non conductive glass. it shouldnt be redissolved by atmospheric moisture

The paper does not show that a simple sodium silicate solution does a good job of insulating steel. It only shows that a steel already heavily treated and coated with other substances can be further protected with sodium silicate (if fact the silicate closes pores in the phosphate coating on galvanised steel). So irrelevant to the question as asked.

added It will also work well for improving the corrosion resistance of an existing coating as it was reported in the abstract

The water glass will have no affect on the stainless (304 ?). But insulation will depend on thickness of glass and the voltage, so specifics would need to be researched. I understand water glass redissolves if exposed to water.


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Why sulfonate of detergent is less likely to bind to ions present in hard water than carboxylate?
I was reading about why detergents do not form scum with hard water but soaps do, and according to wikepedia it has to do something with electrostatic interaction of sulfonates( most commonly) with ions present in hard water
These substances are usually alkylbenzene sulfonates, a family of compounds that are similar to soap but are more soluble in hard water, because the polar sulfonate (of detergents) is less likely than the polar carboxylate (of soap) to bind to calcium and other ions found in hard water.
https://en.wikipedia.org/wiki/Detergent
I am not able to find any reason for this anywhere?
According to me it has to do something with Lattice energy and size of sulfonate and carboxylate as sulfonate is a big ion relatively so it has less lattice interaction with ions like calcium and sodium, but i am not sure about it

You should really brush up on formatting...

Use slant or bold for highlighting details, not make whole paragraphs less readable, in particular.

Is lattice even involved in detergent action? Those things take place on micelles as far as I know. . .

He is asking why the sulfonate salts do not form precipitates in water not about detergent action.

If I knew nothing about soaps, sulphonates and detergents, then, based on general calcium behaviour toward acetic and sulphuric acid, I would rather expect sulphonates to cause trouble with calcium. Perhaps the difference is in calcium complexation abilities of R-COO- vs R-SO2-O-.

Well. Solution of sodium alkylsulfonates are good detergents, because they produce calcium alkylsulfonates in hard water, and these calcium derivates are soluble in water. By contrast, usual soap produce calcium alkylcarboxylates (like oleate, palmitate, etc.) in hard water, and these compounds are not soluble in water. Soap may be a good detergent : it removes stains. But it replaces the stain by a deposit of calcium carboxylate. This is not what is expected when cleaning dirty clothes.
Anyway, the problem of the solubility is not yet perfectly understood. For example, why is $\ce{CaCl2}$ extremely soluble in water ($56$ g. $\ce{CaCl2}$ in $100$ mL water), and why is $\ce{CaF2}$ extremely insoluble in water, despite the fact that chlorine and fluor should have similar chemical properties ?

Is this difference in solubility  due to lattice effect?

can you add some data to support your answer please?

Nobody really knows how to calculate the solubility. If you find something, it is the Nobel Prize for you. There are plenty of publications which "explain" the solubility of a group of substances. They take care of lattice energy, electronegativity, geometry, ligand field, entropy, hydrogen bonds, etc. But these articles have too many exceptions.

It is probably related to the micellar behavior differences of the soap and sulfonate. The carboxylate is a stronger base so the hydrophlic portion will be less ionic and less able to stabilize the micelle in the water. What data I found indicates that Ca++ and Mg++ stearates etc. are at best slightly soluble. I would think that a micelle of these could be even more insoluble.  Even detergents have a limited ability to solubilize organics and fats.


text:
Leakage during Celite filtration
I'm conducting a synthesis involving crown ether and Pd catalyst.
In many papers, prior to column chromatography, they filter crude mixture through a Celite pad.
I tried to, but leakage of Celite occurred. Referring to this paper : https://pubs.acs.org/doi/10.1021/jo005577d (see General Procedure part), I started the reaction with 10 mL of toluene as solvent, and continued work-up with 60 mL dichloromethane. In some cases the filtrate seemed to contain Celite, and the other cases, once I turn on the vacuum pump, some of Celite powder fell down to the collecting flask. What did I mistake? I used Buchner filter, and almost-fitted filter paper (55mm diameter, 1 μm pore size). Basically I referred to this video : https://youtu.be/oAzScPuZqjU. (In this video, after first filtration to make pad, he washes collecting flask and continues second filtration. Both images I uploaded are about second filtration.)

The problem is the "almost fitted" filter paper. Do you have access to sintered glass filters?

According to Vogel's practical organic chemistry, it says that a filter paper should cover the entire perforate plate, but its diameter should be slightly less than the inside diameter of funnel. so I thought, 'Is the filter paper "too fitted?"', so I tried to cut some of its circumference and filtrated again, but it didn't work. (even before I turn on the vacuum pump, Celite slurry almost fell down.) I have sintered glass filter, but I don't have appropriate-sized filter paper.

sorry for misleading word, 'almost-fitted' should be corrected to 'almost completely-fitted'.

The reason that I referred to filter paper in the case of sintered glass filter is in that video, he recommends to use second filter paper to maintain Celite pad unbroken.

@KrangLee If you are sure that paper fitting isn't an issue, then probably celite particles size is. Are you using Celite 545? Alternatively, make sure you are not rapidly applying vacuum — this might cause some particles to puncture the filter paper.

Yes, I use Celite 545. I usually follow these steps:

put filter paper.
wet the paper and turn on vacuum shortly, then turn off.
put appropriate amount of Celite.
pour enough solvent, and slightly mix.
turn on vacuum, make Celite pad.
put second filter paper.
wet paper.
start filtration.

You mean that at first, I should do filtration without vacuum? (for your information my vacuum pump can't control the power. so whenever I saw some texts saying "When you do filtration, suction gently at first", I couldn't do anything)

As I remembered just now, it seemed that after I wet the filter paper and then turn on the vacuum, the paper flap up. How do I prevent this? I completely wet the paper aforetime, but once I turn on vacuum, the solvent evaporated quickly and paper flap up.

If you have a sintered glass filter you do not need a filter paper. Make up the celite/solvent slurry in the filter then apply the suction to form the Celite pad.


text:
Ionic mobilities and conductivities of lithium and potassium cations in acetonitrile–propylene carbonate solution
Recently I calculated the ionic mobility and molar ionic conductivity values for $\ce{Li+}$ and $\ce{K+}$ cations in an acetonitrile–propylene carbonate binary mix solution (8:2 molar fraction ratio).
For AN–PC 8:2, $T = \pu{273 K},$ dynamic viscosity of the binary AN–PC solution $η = \pu{0.737*10^3 Pa s}$ I calculated the values for ionic mobility
$$μ = \frac{ez}{6πηa}\quad(\pu{m^2 V^-1 s^-1})\tag{1}$$
and molar ionic conductivity
$$λ = zμF\quad(\pu{S m^2 mol^-1}).\tag{2}$$
According to this the mobility of $\ce{K+}$ is 5 times higher than that of $\ce{Li+}$, and the molar ionic conductivity of $\ce{K+}$ is 27 times higher than that of $\ce{Li+}.$
Is it correct to calculate with these formulas?

$\eta$ is unitless magic number or dynamic viscosity with a unit ? It seems rather big.  How did you get $a$ ? Does it consider solvation ? How have you got 5 and 27 multiples ? They should be the same.

I used the ionic radius of K and Li: 1.45 Å and 0.92 Å respectively. (Johnson, O. (1973). Ionic Radii for Spherical Potential Ions. I. Inorganic Chemistry, 12(4), 780–785)

Remember that in water, hydrated lithium ion is bigger than potassium one, with lower mobility and molar conductivity, even if a naked lithium ion is smaller than a naked potassium ion. It guess it will be similar for other polar solvents. By other words, you cannot use ion radius, unless it is explicitly stated it is the (effective) hydrated ion radius.

The problem is that in context of ion mobility in solvents, the efficient ion radius is calculated from the actual ion mobility and not vice versa. As the effective ion radius is scenario dependent.

Thanks for the clarification. So after all, we could say, that K+ ions have higher molar ionic conductivity therefore mobility (lower hydrated ionic radius) in water regardless of the accompanied anion it has (e.g. LiTFSi vs KTFSi, or LiCl vs KCl)? According the Kohlrausch's law, this is what I conclude. Is it right? How to interpret this law for other electrolyte solutions: acetonitrile or propylene carbonate for example?


text:
How to sustain iodine vapor in a big container (V=200 L)?
I am lost in a quest on how to have a big container in which I can have iodine vapor at a known temperature.
We intend send iodine into an air flow, and observe fluorescence of iodine in this flow. This method is known as planar laser induced fluorescence. The challenge lies in the fact that, we need to be able to send iodine above a certain concentration for about several minutes. The idea is to have a large volume of iodine vapor ready, so that we do not run out of iodine vapor during the experiment and we do not need to wait too long between experiment sessions for iodine to sublimate.
In order to do that, I came up with the idea that we will have a big volume of container, and we will heat this container. The bigger the container, the more iodine vapor mass will be ready for us to send to the experimental chamber. Otherwise, if we had a smaller volume, iodine would deplete quicker and I would have to wait for sublimation.
So, with this purpose, I want to do be able to get a container, and I am trying to come up with a way to heat it up. What kind of a barrel, or container of some other type would be suitable for me, and how could I heat it so that it is around 80-100 C?

Your message is not clear. You speak of an air flow. Well ! Do you need a heated container with motionless air + iodine or will you produce an air jet ? On the other hand, iodine reacts with most metals, producing iodides. Have you taken this into account ?

Well, the idea is to first heat container long enough so that there's saturated iodine in the container. Then I will generate flow into and out from the container, there will be an air jet.
And about the reaction with metals, yes I thought about it, that's why I was thinking of a plastic barrel.
Have I made it clear or is something still missing?

You should describe the expt. in more detail. What sort of air pressure, what sort of I2 concentration, flow rate, expanding into vacuum or not etc. It would be better to do this dynamically, pass gas (air or O2 or N2?) over reservoir of solid iodine at  temp to get vpr pressure you need.

Why don't you pre-heat the air rather?

Usual plastics are good solvent to iodine $\ce{I2}$. So I would not recommend using  a plastic container for iodine vapor. The best container would be in glass, as metals are oxidized by iodine vapor, and plastics absorb iodine.


text:
Is there any relation between reaction quotient and rate constant in the context of order of the reaction?

How many times will the rate of the reaction
$$\ce{2 A + B -> A2B}$$
change if the concentration of substance $\ce{A}$ is doubled and that of substance $\ce{B}$ is halved?

The given solution takes
$$\mathrm{rate} = k[\ce{A}]^2[\ce{B}]\tag{1}$$
which is absurd to me since the order of the reaction is an experimental value and not necessarily stoichiometric coefficient.
Here is my attempt:
$$K_\mathrm{i} = \frac{[\ce{A_2B}]}{[\ce{A}]^2[\ce{B}]}\tag{2}$$
and doing the same for $K_\mathrm{f}$ just replacing $[\ce{A}]$ with with $2[\ce{A}]$ and $[\ce{B}]$ with $[\ce{B}]/2$ and then dividing the two equations to get $2.$
This just so happens to match up with the answer. Is there any relation and can I use reaction quotients this way?

Be aware that at equilibrium, forward and backward rates are equal, regardless of the complexity of the overall reaction and particular values of concentrations of involved substances.

@Poutnik how would affect the order of the reaction though? its an experimental value so I don't see what difference would it make.

The equilibrium constant is an experimental value as well and must be consistent with the reaction rates.

@Poutnik but the reaction isnt give at equilibrium in this question

It does not matter.  Why do you ever consider reaction quotient for the forward reaction rate?

There is no direct relation between reaction quotient or eq. constant on one side and the single way reaction rate or rate constant on the other side. A single value from the former two can be related to a wide range from the latter two and vice versa.

When you are implying reaction quotient, you should use $Q$ instead of $K$. As $Q$ is defined at any concentrations other than equilibrium.

When you are changing $[\ce{A}]$ and $[\ce{B}]$, $[\ce{A2B}]$ may change also, which you have neglected.

From the given data, the change in rate of reaction cannot be calculated for the reason you have pointed out in the beginning.

@andselisk About the edit: If A and B denote a particular atom then they can be represented upright, if A and B can be any atom then they can be in italic?

Symbols for chemical elements and symbols/abbreviations referring to them are always upright. The only exception I can come up with from the top of my head are locants depicted with italicized Roman letters, e.g. N,N-dimethylaniline.


text:
Is the lone electron pair of an amide nitrogen part of the π system when building a Hückel matrix?
Is the lone electron pair of the nitrogen atom part of the conjugated π system in the α,β-unsaturated amide pictured below?

Is there a general rule for choosing when to include a nitrogen atom in a Hückel matrix?

I have edited your post for clarity. At the risk of incurring ill-will from our German folks, I have replaced your "ue" (yes, it's correct; no, most people don't know that) in the title and added back the umlaut in the text.

Pro tip: get to know a compose key and never fear Unicode German or Greek letters again with simple and intuitive keystrokes like RAlt " u for letter ü or RAlt * p for letter π (cc @ToddMinehardt).

@ToddMinehardt So to get it right - is or is not ue acceptable EN transcription  for  DE ü ? ( I guess such transcriptions would get hard time for CZ ones ) // An alternative is to have a short list of frequent UC characters and codes, performing either copy/paste either invoking unicode insertion by the code, like CTRL+SHIFT+U<unicode>ENTER on Linux.

Actually I am German, so I have the ü right on my keyboard. I have just always read the "ue" spelling in English literature, so I thought I'd not confuse you with umlauts :D

I guess you can include the lone pair in the Huckel matrix, but it will probably become very difficult to solve by hand. If you have a computer, that will solve it. If you ignore the lone pair, your system is only a linear chain of p-orbitals, it will give that diagonal looking matrix (not a diagonal matrix!!) that you could solve with a pen and paper.

@Poutnik - I assumed Charles was German and/or was helping us optimize for search engines.

@Poutnik Writing English with occasional extras, you might consider as keyboard layout US international if characters stay Latin. Then «ü» either is alt+y, or "+u. This equally adds e.g., the µ character to your tool box, too.

@Buttonwood Thanks, it may be useful. But it may also further mess up things, if one already uses 2 or 3 layouts. Android has exceptionally small advantage here.

Setting aside the arguments over umlauts, yes you can add the nitrogen atom to the Hückel matrix; in fact, theoretically you should do so and allow the matrix eigenfunctions and eigenvalues to decide whether the nitrogen atom in this case is really conjugated. The carbonyl oxygen can also be coupled in.
But there is a catch. Since the nitrogen or
oxygen atom has different atomic orbital energy levels from the carbon atoms, and the orbital overlap with other atoms will be different, you cannot just put in the same numbers that you would with a carbon atom there. Various estimates are available for these heteroatom parameters, for instance this one given by Texas A&M University which includes common heteroatoms for the diagonal elements and their bonds with carbon for the off-diagonal elements.
Once you have decided upon the matrix elements you can solve the matrix for its eigenfunctions (molecular orbitals) and eigenvalues (energy levels) the same way you would with an all-carbon system, but the equations will probably be more complex and require numerical methods to solve efficiently.


text:
Variation of vapour pressure with application of external pressure
Many times study of fact that with pressure of a substance only depends on the temperature and nature of the substance recently I study about dew point and bubble point.
Then I came up with this question to which I didn't find any on Google :-
Suppose we have a system as shown in figure:

In the above system the piston is touching the liquid surface that is there is no vapour currently.
Now we slowly move the piston up, keeping the temperature same and the vapour starts forming in the container.

Now the question is how will this pressure applied by the vapours of the liquid so form (i.e. the vapour pressure at equilibrium) will vary with the change in external pressure at the same temperature ?
Will the vapour pressure be equal to the saturated vapour pressure of that substance at that temperature  ? or if it changes will it not contradict the statement used in almost all articles over the net teaching vapour pressure , stating that vapour pressure of a substance only two points on temperature and nature of that substance?

As the approximation, the saturated vapour pressure does not depend on the total pressure, only on the substance and temperature.  (*)
If there is nothing but the vapor in the gaseous phase under the piston then the amount of vapor and its pressure accomodate to external pressure. (**)

If the external pressure is kept greater than saturated vapor pressure for given temperature, all vapour will eventually condense and released heat will dissipate, assuming constant temperature is maintained.
If the external pressure is kept lower than saturated vapor pressure, all liquid will eventually evaporate and needed evaporation heat will be extracted from surrounding.
If the external pressure is by chance equal to the saturated vapor pressure, liquid and vapor will maintain the equilibrium with the same vapour pressure as the external one.

The piston stays in rest if and only if the net force on it is zero, therefore if actual vapour pressure ( saturated in equilibrium ) is equal to the total external pressure caused by external, non-vapor based forces acting on the piston. Otherwise it starts moving toward the lower pressure.
There is assumed the lifting of the piston is slow enough for liquid to have time to establish saturated vapour pressure at any time. Fast piston lifting can end with higher volume and less than saturated vapor pressure, which may not be possible to be topped again due lack of remaining liquid.

(*) - Finer analysis tells us that for gas+vapor mixtures, liquid saturated vapor pressure increases little with the total pressure. This is proportional to liquid molar volume and such an increase is usually negligible. Unless temperature and pressure are near the liquid critical point where it can make the difference.
(**) - Atmospheric pressure is just one of possible sources or components of external pressure on the piston. Atmospheric pressure can be

the only part of external pressure, like in idealized scenario with a free moving, massless, frictionless  piston
missing, e.g. if the system is placed in vacuum
a part of total external pressure, like if the weight caused by the piston mass cannot be neglected and/or if explicit mechanical external force is applied.

What is the relation between vapour pressure and the external pressure ? Are they equal ? Can we apply Boyle's law here ?

Does the piston move if they are not equal?  Boyle's law is for ideal gases, and as approximation for real "permanent gases. It is not applicable for vapor, if increased pressure above saturated vapor pressure causes condensation. Even before condensation starts, there is already big deviation from this law.

What if the question states to consider vapor as ideal gas will Boyle's law be valid then ? I don't think that it is necessary for piston to move when the pressure is unequal and the reason I'm saying that is because of a particular case in which we move the piston all the way up until all the liquid become vapour at that point the sum of external pressure and vapour pressure should be equal to atmospheric pressure ( maybe ) ?

Such assumption does not make sense, ideal gas never condenses, what is the direct opposite to being a vapor.

No, external pressure being sum of atmospheric pressure and eventually additional pressure would be equal to saturated vapour pressure, assuming lifting the piston is slow enough. (Fast moving would end with higher volume and less than saturated vapor pressure.)

So you say vapour pressure > atmospheric pressure

No, I say you should read properly and think independently. In the piston scenario, vapor pressure relates to external pressure. Atmospheric pressure is just one of possible sources of external pressure on the piston. It can be the only part, can be missing or part of total external pressure.


text:
Why do boron salts impart a characteristic green colour to the flame
In salts of boron it exists as a cation in the B(III) form. It is known to impart green colour in several of its salts like borates, boron triflourides, boric acid etc. If there are no electrons in its valence shell to excite (in its cationic form) and hence obtain these characteristic colors, then what is responsible for this effect?

Ions are not responsible for the color. The color you are seeing is due to "atomic emission" resulting from the atomization of ions

@NilayGhosh does this also imply that alkali metal salts would, in general, appear colorless but impart color when in flame?

@Ashish The salts don't necessarily have to be colorless but they will impart characteristic colors when in flame. In fact this can be used for qualitative analysis to check if and which alkaline metal is contained in your sample. Sodium for instance will give a bright yellow flame. See for instance https://i.ytimg.com/vi/9oYF-HxtoYg/maxresdefault.jpg

A great substance to get the green flame is boric acid methylester B(OCH3)3, readily formed from $\ce{B2O3}$ in methanol ( AFAIK with few drops of $\ce{H2SO4}$ additions ). Ethanol adds yellow significantly.

Sometimes atomic bonds are resposible. E.g. a blue color of methan flame is reportedly caused by C-H radical.

@Ashish Reply to your comment: Yes.

@Poutnik. In his book, Herzberg says that the blue color of methane flame is due to $\ce{C2}$, as its vibrational structure is typical of $\ce{C2}$

@Maurice Reviewing https://en.wikipedia.org/wiki/Flame and  http://guilford.eng.yale.edu/pdfs/CH.pdf  it may be both ways.

Don’t forget flame species such as CaOH, SrOH, etc. An earlier version of the publication below, freely available on the web, appeared in Journal of Pyrotechnics, No. 18, 2003. Color Values and Spectra of the Principal Emitters in Colored Flames,
W. Meyerriecks* and K. L. Kosanke. Boron is not discussed in the paper, but a few other common elements are discussed.

You wrote "If there are no electrons in its valence shell to excite (in its cationic form)....".  Irrespective of whether the green color is atomic or molecular in origin, you should note that $\ce{B^{3+}}$ does indeed have valence electrons:  2 of them!  When you remove all of the neutral atom's original valence electrons to create a cation (as is typical of cations of main-group elements), the electrons in the shell below become the valence electrons, giving the ion the electron configuration of the next lower noble gas—which, for B, is He.

And that such boron cations don't exist in any compound or even any flame, is no problem? One would truly extreme temperatures to triply ionise boron, or anything else.

Whenever you see low temperature colored emission in flames, it is invariably because of the molecular species, do not attribute all flame colors due to atomic emissions right away (alkali metals are exceptions). This beautiful green color of boron flames is due to a small molecules containing boron. The green band centered at 547 nm has puzzled spectroscopists for a while. It has been attributed to a radical BO$_2$ and people have excluded B$_2$O$_3$. See for example
Kaskan, W. E., and R. C. Millikan. "Source of Green Bands from Boron‐Containing Flames." The Journal of Chemical Physics 32.4 (1960): 1273-1274.
There are two requirements in order to see the green color of boron-flames.
(i) You must have a lot of "OH" entities in the flame, which means an oxygen rich is good.
(ii) Boron must be present too (obviously!)
Even elemental boron when burnt properly can generate this characteristic color which is often seen with boron and alcohol solutions.

Now an interesting question: what do we expect from boron's atomic emission? It indeed has valence electrons. The strongest emission is in the UV range which your eyes can never see! Around 200 nm. This is the proof that boron "atoms" are not emitting this green color. You can see this in Grotrian diagram which simply shows what wavelengths (Angstrom units) to expect in an atomic emission. Bor III means neutral boron atom (in German). This is from a old book by Grotrian, " Graphische Darstellung der Spektren von Atomen und Ionen mit ein, zwei und drei Valenzelektronen: "  Freely available from Google books.


text:
How to transform 'Cartesian gradient' taken from the formatted checkpoint file into the input orientation?
I would like to extract the energy gradient at a particular geometry with many digits from the formatted checkpoint file. However, Gaussian prints the gradient components related to the standard orientation. My question is how to convert this kind of gradient into its counterpart corresponding to the input orientation.

Do you have symmetry turned on? I think if you turn off symmetry Gaussian won't change the orientation of the molecule. Otherwise you need the matrix that transforms between the two orientations. By the way, you may want to check Matter Modelling SE as well.


text:
Why do different products form when alkali halide reacts with zinc in polar protic and aprotic solvents?
Let's take an alkali halide such as $\ce{CH3CH2Cl}$ the two reactions in polar protic and aprotic solvents are as follows:
$$
\begin{align}
\ce{CH3CH2Cl &->[Zn, CH3OH] CH3CH3}\tag{R1} \\
\ce{2CH3CH2Cl &->[Zn, dry ether] CH3CH2CH2CH3}\tag{R2}
\end{align}
$$
What is the reason that we obtain different products in different solvents? How does the mechanism of reaction differ? Also, please correct me if I'm wrong regarding any of the following reactions, they were mentioned in my class notes.

$\ce{CH3\color{blue}{C}H2ZnCl}$ is potentially proton-basic at the blue carbon atom.

Think of the zinc acting like magnesium, forming Grignard (they're not exactly same, of course, but a fair comparison). In the first case you have a proton source, which then will form alkane. In the second, it will undergo SN2.


text:
Major product formed when 4-Methoxybenzil is reacted with a base
What will be the major product formed in the given reaction:

I am not able to apply any of the reactions or mechanisms I know of in this case. The only thing I am able to predict is that the $\ce{OH-}$ will probably do a nucleophilic approach on the carbon atom directly attached to the phenyl group (adjacent to the starred carbon). I am not sure how to proceed after that. It will be of great help if someone could guide me with a proper mechanism as to how this proceeds or whether this is any standard reaction.

This is the Benzilic Acid rearrangement - diagram sourced from here
The Wikipedia entry is as good a place to start as any  Wikipedia

In addition to the answer by @Waylander, I would add something since his answer is about symmetrical benzil.
In case the substituents on the two phenyl rings are different, it must be noted that the first step occurring in this reaction is attack of $\ce{OH^-}$ as a nucleophile on the carbonyl $\ce{C=O}$. Nucleophilic addition is preferred at that carbon atom which is more electron deficient. In your case, methoxy group at para position increases electron density on left carbon. So the attack of $\ce{OH^-}$ occurs on the right carbon. Proceed with that as the first step, for the final product.

You are correct

Although your suggestion is reasonable based upon accumulated knowledge, how does one demonstrate the selectivity in the p-methoxy example? One might argue that although the anisyl carbonyl is less reactive toward base, the anisyl ring is favored in a bridged migration. Certainly nothing in the structure of the benzylic acid proves which carbonyl is attacked by hydroxide.

@user55119 You're right, my answer is completely based on theoretical knowledge, based on what I have read in a textbook. I would say that since the first step favours the existence of intermediate where $\ce{OH^-}$ is attached to the benzyl carbonyl (and not the anisyl one), that intermediate has greater quantity existing after the first step, and hence has a greater chance to react further. I might be completely wrong, of course. My knowledge is extremely limited - I am still yet to enter college. Perhaps someone more qualified can be of more help...

Imagine that the benzyl carbonyl is ten times more reactive than the anisyl AND that this step is reversible. If the migratory aptitude (look up the term) is 100X more likely for anisyl than phenyl and that this step is irreversible, then which step determines which group is migrating? For a point of comparison, there is more equatorial bromocyclohexane than axial but it is the axial conformation that has the greater rate constant for E2 elimination than the equatorial. You may wish to construct an energy diagram to illustrate these reactions. Read about  the Curtin-Hammett principle.

Ran out of characters.... The anisyl group may be expected to have a greater migratory aptitude than the phenyl group owing to the electron donating ability of the p-methoxy group.

@user55119 I know that the anisyl will have greater migratory aptitude - that's why I said earlier that I couldn't provide a satisfactory answer to your query. My textbook, unfortunately, is largely focussed on the exam that I'll be appearing for this year, and does not provide any references for the facts stated in it. In any case, the OP hasn't accepted either of the two answers here -  so if you can find any good resource regarding the selectivity, please do post it in a new answer. It will be of great help to me too :) Thank you for pointing out the flaw in my argument in the first place.

The is a possible experiment that might be useful.  Compare the rate of rearrangement of benzil vs. the bis-p-methoxybenzil. If benzil reacts faster, then attack at more reactive carbonyl is important. If the bis-anisyl cmpd reacts faster, migratory aptitude is where it is at.


text:
Calculating molecular formulas of a combustion reaction
The Question
In an experiment, 10 cm$^3$
of an organic compound in the gaseous state was sparked with an excess of oxygen. 20 cm$^3$
of
carbon dioxide and 5 cm$^3$
of nitrogen were obtained among
the products. All gas volumes were measured at the same
temperature and pressure. What are the possible molecular
formulas that would fit these data?
My Attempt
Since there is one volume of $\ce{N2}$ produced from 2 volumes of the organic compound, there must be one $\ce{N}$ atom in the compound. Also, 4 volumes of $\ce{CO2}$ is formed from 2 volumes of the organic compound and thus the number of carbon atoms must be 2. It follows that the compound must be $\ce{C_{2}H_{x}N}$. Therefore the compound must be $\ce{C_{2}H_{7}N}$, $\ce{C_{2}H_{5}N}$, or $\ce{C_{2}H_{3}N}$.
The answer key states that the only compounds which satisfy the conditions are $\ce{C_{2}H_{7}N}$ and $\ce{C_{2}H_{3}N}$.
Why isn't $\ce{C_{2}H_{5}N}$ a valid solution, given that the balanced chemical reaction of it reacting with oxygen satisfies all the conditions?

You are right. I don't see why $\ce{C2H5N}$ cannot be accepted.

Ohk thanks I asked cause all the answers available online also gave the same answers. Thanks!


text:
How can I make cerium(IV) chloride?
How can I make CeCl4?
I have some ideas but am worried I'm going to mess it up since this is the first time I've worked with lanthanides.

To the best of my knowledge, there are no known $\ce{LnCl4}$ species.
Although $\ce{CeCl4}$ has been mentioned in older literature, more recent publications tend to refute its existence attributing the observed reactions to $\ce{CeCl3}$ and $\ce{[CeCl6]^2-}$ species, which could be fairly easily isolated and characterized.
Among cerium(IV) tetrahalides, only $\ce{CeF4}$ is known, and solely thanks to the fluorine's ability to support high oxidation states [1, pp. 17, 21]:

Fluorine is well known to promote high oxidation states.
Factors associated with this are the high lattice energies associated with the small ﬂuoride ion, along with the very small $\ce{F–F}$ bond energy (due to non-bonding electron pair repulsions) as well as high bond energies involving ﬂuorine (not relevant in this case).
Because of the larger size of the $\ce{Cl-}$ ion, there is going to be much less difference between the lattice energies of $\ce{CeCl3}$ and $\ce{CeCl4},$ and this is probably the determining factor.
The higher $\Delta_\mathrm{at}H(\ce{Cl})$ of $\pu{121.5 kJ/mol}$ also mitigates against the formation of $\ce{CeCl4}.$

You can, however, stabilize cerium(IV) tetrachloride with neutral organic ligands [2, p. 310] such as dimethylsulfoxide, triphenylphosphine oxide, 2,2′-bipyridine-N,N′-dioxide and tris(dimethylaminido)phosphine oxide allowing to produce adducts such as $\ce{[CeCl4(dmso)3]}$ [3], $\ce{[CeCl4(tppo)2]}$ [3], $\ce{[CeCl4(bpyO2)2]}$ [4] and $\ce{[CeCl4(tdpo)2]}$ [5], respectively.
References

Cotton, S. A. Lanthanide and Actinide Chemistry; Inorganic Chemistry: A Textbook Series; Wiley: Chichester, England; Hoboken, NJ, 2006. ISBN 978-0-470-01005-1.
Handbook on the Physics and Chemistry of Rare Earths; Gschneidner, K. A., Ed.; Elsevier North Holland: Amsterdam, 2006; Vol. 36. ISBN 978-0-444-52142-2.
Březina, F. Verbindungen Des Vierwertigen Cers Mit Triphenylphosphin, Fallweise Mit Triphenylphosphinoxid. Collect. Czech. Chem. Commun. 1971, 36 (8), 2889–2894. DOI: 10.1135/cccc19712889.
Březina, F. Koordinationsverbindungen Des α,α’-Dipyridyl-N,N’-Dioxids Mit Salzen Des Vierwertigen Cers. Collect. Czech. Chem. Commun. 1974, 39 (8), 2162–2167. DOI: 10.1135/cccc19742162.
Du Preez, J. G. H.; Rohwer, H. E.; De Wet, J. F.; Caira, M. R. Preparation and Structure of Tetrachlorobis[Tris(Dimethylamido)Phosphine] Oxidecerium(IV). Inorganica Chimica Acta 1978, 26, L59–L60. DOI: 10.1016/S0020-1693(00)87177-7.

As noted by @ansdelisk, $\ce{CeCl4}$ hasn't been isolated yet as a pure compound. It exist in a solution although not for too long as it as hydrolyze/decomposes to $\ce{Ce(III)}$ and chlorine. As noted here:

Ceric chloride, $\ce{CeCl4}$, has not been isolated. It can be obtained in solution, but the solution cannot be kept without decomposing. Ceric hydroxide dissolves in cold concentrated hydrochloric acid with the production of a dark red solution; chlorine, however, is slowly evolved, and ultimately a solution of cerous chloride remains. The decomposition proceeds rapidly in hot solutions.

An old literature points out that $\ce{CeCl4}$ has been made using $\ce{CeO2}$ and a solution of $\ce{(POCl3 + SnCl4)}$:

$\ce{(POCl3 + SnCl4)}$ solutions were prepared by volume in a dry box. The $\ce{SnCl4}$ content was verified by chemical analysis for $\ce{Sn}$. This solution and $\ce{CeO2}$ were placed in sealed ampoules, heated to $\pu{120 °C}$ for 2 hours to increase the rate of solution, and then rotated in an air thermostat at $\pu{20 °C}$ for 2 hours. Without preheating, equilibrium was established after 200 hours. Preheating to $\pu{120 °C}$ lowered the
equilibration time at $\pu{20 °C}$ to 2 hours.

In general, Ce(IV) salts are metastable in aqueous solution and is a strong oxidizing agent that oxidizes hydrochloric acid to give chlorine gas. The reason is somewhat written in the Wikipedia article:

Despite the common name of cerium(IV) compounds, the Japanese spectroscopist Akio Kotani wrote "there is no genuine example of cerium(IV)". The reason for this can be seen in the structure of ceria itself, which always contains some octahedral vacancies where oxygen atoms would be expected to go and could be better considered a non-stoichiometric compound with chemical formula $\ce{CeO_{2−x}}$. Furthermore, each cerium atom in ceria does not lose all four of its valence electrons, but retains a partial hold on the last one, resulting in an oxidation state between +3 and +4. Even supposedly purely tetravalent compounds such as $\ce{CeRh3}$, $\ce{CeCo5}$, or ceria itself have X-ray photoemission and X-ray absorption spectra more characteristic of intermediate-valence compounds. The 4f electron in cerocene, $\ce{Ce(C8H8)2}$, is poised ambiguously between being localized and delocalized and this compound is also considered intermediate-valent.


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How is the equilibrium expression (law of mass action) related to the rate law?
Every chemistry textbook I've read will have a chapter on the rate law. It will say something like, given a reaction $a\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D}$, the rate law (for the forward reaction) is given by:
$$
\text{rate} = k[A]^x[B]^y
$$
and the text will usually stress that $x$ and $y$ are experimentally determined and unrelated to the stochiometric coefficients.
Then, there will be a subsequent chapter about equilibrium expressions. Using the same reaction, $a\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D}$, the equilibrium expression will be given by:
$$
K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}
$$
In this case, the stochiometric coefficients are relevant. I would be willing to blindly accept this, but the text will then go on to "derive" the equilibrium expression from the rate law by noting that, in equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. That is, $rate_{forward} = k_f[A]^a[B]^b = rate_{reverse} = k_r[C]^c[D]^d$. Rearranging this equation, we get:
$$
\frac{k_f}{k_r} = K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}
$$
How are we suddenly able to use the stochiometric coefficients in the rate law when deriving the equilibrium expression? What am I missing?

The equilibrium constant is derived from the rate law and not vice versa. For an elementary reaction, the rate law is written as $$K={[A]}^a{[B]}^b$$
But for a complex reaction, it is not dependent on the stoichiometric coefficient as the rate would be dependent on the multiple elementary reactions taking place.
The equilibrium constant which we use is for an elementary reaction and not for a complex reaction, hence the stoichiometric coefficients would determine the equilibrium constant.

That's what I thought. It is incredibly confusing when textbooks do this. The "assuming an elementary" reaction is a small blip in the equilibrium expression chapter, but in the rate law chapter they mention a million times that the powers are experimentally determined.


text:
Is powdered acid dangerous?
I was told that an acid is only dangerous because it releases $\ce{H+}$ ions when in contact with water.
Therefore, powdered acid should be completely safe on skin provided that your skin is not covered with sweat.
Am I correct, or is my logic flawed?

What is ‘powdered acid’ - many strong acids are aqueous solutions.
There are some strong organic acids that are non-aqueous, but these will cause damage with or without water.
There are some weak acids (eg citric) found in foods which do come as powders, and are not harmful in small quantities with or without water

@NotEvans. Powdered acid like Citric Acid powder.

Oleum has no water, and you wouldn't like to touch it, I assure you. Being solid makes a difference because of hindered mixing.

Citric acid is found in food and has many household applications, it isn’t really a very strong acid at all.
That said, contact with skin might cause some irritation, especially in large quantities. Even when ‘dry’ your skin still contains moisture, so the idea that keeping things dry isn’t really relevant

If you need an inorganic example, take sodium hydrogen sulfate.

The state of a compound, solid, liquid, or gaseous is a property which is unrelated to the property / strength of a Brønsted or Lewis acid of said compound.  Thus, you can not infer from its physical appearance if an acid is safe enough to be touched by skin; it may be irritant / corrosive even in absence of water (e.g., p-TsOH as a solid acid, triflic acid as a liquid acid).  Similar, solid Brønsted bases like KOH or KOtBu are not magically safe to touch once obtained as a solid instead of in solution.  The interaction of these acids and bases with the skin may be influenced by water. Your skin however includes many other chemicals which may interact and equally react as an acid, or a base, thus react with external acids and bases accordingly in absence of water, too.
In addition, there are plenty of chemicals (gaseous, or liquid, or solid) which neither are considered as an acid, nor as a base; still, for other reasons than those above these are not harmless enough to be touched over extended periods of time by human skin.  Their origin, natural, biological, processed, or entirely artificial, is not a sufficient criterion to render them safe, or potentially dangerous, either.  Equally note, skin is not uniform.

Or at least, there isn't a known relation of scientific interest. Even if there is stochastic dependence, I anticipate it to be rather weak.

There's also acids which are also poisonous, like hydrofluoric acid.


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Why is "activity" of gaseous component equal to its partial pressure but that of aqueous component its concentration?
While solving an electrochemistry problem I had to calculate the reaction quotient of this reaction:
$$\ce{2Fe(s) + 4H+(aq) + O2(g) -> 2Fe^{2+} (aq) + 2H2O (l)}$$
It turns out to be: $$\ce{Q=\dfrac{[\ce{Fe}^{2+}]^{2}}{[\ce{H+}]^4p\ce{O2}}}$$
where square brackets mean concentration and $p$ denotes partial pressure. Here's what I do know: activity of solid and liquid components is 1, and that of aqueous components is their concentration.
However, why is the activity of gaseous components their partial pressure? Is this fact experimental or can it be derived?

@Maurice I picked it straight from an online lecture, but a Google search of the exact phrasing of the question returns multiple solution sites like this one. I don't know about plausibility, but looking at the amount of results I'm sure this is from some famous problem book.

Are you sure of your equation ? Because usually oxygen $\ce{O2}$ reacts with iron in acidic solution to produce ferric ions $\ce{Fe^{3+}}$, according to : $$\ce{4 Fe  + 3 O2 + 12 H^+ ->  4 Fe^{3+} + 6 H2O}$$

Thermodynamics courses usually start by calculating amounts of heat and amounts of work enterring a container. And the work is $\pu{p\Delta V}$. No mention of concentration ! Just the pressure. Afterwards, enthalpy is introduced, then gas chemistry is developed, always using pressures. Equilibrium constants are then introduced, always with gases and pressures. And suddenly appears the necessity of defining equilibrium constants in liquid phase, where pressures have no obvious meaning. Here the authors of the course state that, thanks to Henry's law, it makes sense to replace the pressure by the concentration (or the activity) which is easier to determine than pressure. Concentration or activity are supposed to be proportional to pressure. Fortunately, this change of parameter does work well in practice. The equilibrium constants determined by using concentration allow nice predictions, and are verified for example in electrochemistry (Nernst law). But it should be known that as soon as a gas intervenes in a chemical reaction, it is safer to use its pressure.


text:
Hydrogen Bond Length and Lattice Density
The subject of the density of solid water is discussed on Physics Stack Exchange in this post.
In the post, it is said that..

the way the molecule is angled ensures that a specific (energetically
favorable) orientation leaves a relatively large amount of open space

However, how can the orientation alone be responsible for density? With a short enough bond length, can not any solid lattice be more dense than its liquid form?
Would the density of a lattice structure not depend on the bond length as well? (In the case of water, would the density of the lattice not depend on the bond length of its (dipole-dipole Hydrogen bonds).
If Hydrogen bonds are indeed stronger than other types of intermolecular bonds, why is the bond length so long?

The bond angles are $\approx 120$ meaning hexagons are formed, if water had $90$ degree bonds, squares, which would be more dense, would form thus geometry can influence density. Secondly bonds have a minimum energy at a certain distance and this influences density.

Some solids are denser than their liquids. And don't forget that liquds have bonds as well.

Here is a sketch of a layer of ice:

The hydrogen bonds are fairly short (177 pm, compared to a covalent O-H bond length of 100 pm), and the distance between hydrogen-bonded atoms is shorter than the minimal distance in the absence of the hydrogen bond (about 195 pm).
What should be apparent from the image is that there is a lot of space in the center of the hexagons (the picture does not show that every second water is a bit higher and the ones in between a bit lower). For a 3D view see http://biomodel.uah.es/en/water/p1.htm). Trying to squeeze in another water molecule would not work for two reasons. First, there is not enough space without disrupting existing hydrogen bonds. Secondly, that extra water molecule would have no hydrogen-bonding partners.
Liquid water has a less orderly structure and fewer hydrogen bonds, so the water molecules get closer on average. It does not, however, have shorter hydrogen bonds or let non-hydrogen bonded hydrogen and oxygen molecules approach closer than the non-bonded minimal distance.

The bond length, intermolecular forces, and molecular shape all determine the lattice structure, which in turn determines the density of the substance. Thus, with a small enough bond length, solids can be more dense than liquids due to a much more tight lattice structure.
Water is somewhat unique in its densities due to its bent molecular geometry. Most other substances tend to flow away from each other in their liquid form due to their kinetic energy, but when they become solid they don't have enough kinetic energy to move away from each other, making them closer to each other and thus denser in solid form. However, the attraction between water molecules causes them to use their kinetic energy while in a liquid state to get closer to each other. However, there are two lone electron pairs that belong to the oxygen atom in a water molecule, and when water becomes a solid, it no longer has enough kinetic energy to counter the lone electron pairs' repulsions, so the water molecules in ice are slightly farther away due to the repulsion, thus making ice less dense than liquid water. There are other molecules that are like water: more dense in their liquid form than solid form.
As an extra note, you said that hydrogen intermolecular forces are stronger than other intermolecular forces, but we should acknowledge that hydrogen intermolecular forces are weaker than ion-dipole and ion-ion intermolecular forces, explaining why things such as saltwater (ion-dipole) and table salt (ion-ion) have higher densities than water.


text:
Sodium hydroxide reaction with manganese dioxide
I came across a website that said it is possible for sodium hydroxide to react with manganese dioxide at temperatures near 0 degrees Celsius. The reaction goes:
2MnO2 + 3NaOH ⇄ MnO(OH) + Na3MnO4 + H2O
[
Source: https://chemiday.com/en/reaction/3-1-0-7458
Is this actually a feasible reaction and if so how would one tilt the equilibrium to the right?

Chemiday is not a reliable source. This time, however, they copypasted this reaction from the Russian textbook Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000, p. 397. How do I know? They didn't bother with proper chemical names and transcribed Russian spelling ("gipomanganate"). Don't use that website. As for the question, the answer is evident if you apply Le Chatelier's principle (hint: temperature aside, $\ce{MnO(OH)}$ is precipitate).

Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown.


text:
Reaction of cyclooctatetraene with sulfuric acid
Problem

The correct statement is
(A) P & Q are aromatic compound and Q has $\mathrm{sp^3}$-hybridized carbon atom.
(B) P is aromatic with 10 π electrons and Q is aromatic with 2 π electrons.
(C) P is aromatic and Q is anti-aromatic.
(D) P is aromatic and Q is non-aromatic.

Answer

(A) P & Q are aromatic compound and Q has sp3-hybridized carbon atom.

Question
I got P easily; the two $\ce{K}$ atoms would lose two electrons and form the cyclooctatetraene dianion, which will become planar and aromatic with 10 π electrons.
However, I could not find what Q will be. My first step was a proton getting attached to one carbon of any of the four $\ce{C=C}$, which will give $\mathrm{sp^3}$ hybridization on that carbon which accepted the proton. There will now be another adjacent carbon with a positive charge on it, which will be delocalized through the 6 π bonds.
However, the presence of $\mathrm{sp^3}$ carbon in between will prevent complete conjugation throughout the ring.
I don't think the next step can be the attack of $\ce{HSO4^-}$ as a nucleophile, because $\ce{HSO4^-}$ is a stable anion and hence a very poor nucleophile. Water isn't mentioned, so I couldn't assume attack of $\ce{OH^-}$ as a nucleophile either.
In any case, I don't see how Q can be aromatic. What am I missing?

This is a rather unusual and interesting case of aromaticity, which has been given a special name: homoaromaticity.
The Wikipedia page does a quite nice job of explaining what's going on. As you state, protonation of cyclooctatetraene generates a $\ce{C8H9^{+}}$ cation containing an $\mathrm{sp^3}$-hybridised carbon atom between all the other $\mathrm{sp^2}$-hybridised carbons. However, what's surprising is that this does not imply aromaticity must be broken. If you comfortably understand that the cyclooctatetraenide anion $\ce{C8H8^{2-}}$ is aromatic, you probably also know that the tropylium cation $\ce{C7H7^{+}}$ is also aromatic. As it turns out, this aromaticity is sufficiently strong that if you add a $\mathrm{sp^3}$-hybridised $\mathrm{CH_2}$ linkage, the cation pushes the $\mathrm{sp^3}$ carbon out of the way to allow through-space conjugation and maintain aromaticity.

The homotropylium cation is exactly what cyclooctatetraene rearranges into upon protonation. The evidence for the aromatic nature of the homotropylium cation is ironclad. NMR spectrometry is one of the best techniques to determine the presence and strength of aromaticity in a compound, due to aromatic ring currents. In the homotropylium cation, the hydrogen atoms in the $\mathrm{CH_2}$ linkage are magnetically inequivalent, and indeed, starkly so - their $\mathrm{^1H}$ chemical shifts differ by over 6 ppm.  In particular, one hydrogen atom lies above the ring and is heavily shielded by the aromatic ring current, generating a signal at -0.73 ppm which is very unusual for a carbocation.
It would be excellent to study the crystal structure of the homotropylium cation, but there do not seem to be any experimental data for examples containing the parent ion, only derivatives such as a hydroxylated homotropylium cation (in fact, I'm surprised the parent ion hasn't been crystallised as a carborane superacid salt). Nevertheless, structural evidence for aromaticity can also be found in the derivatives, in the approximate coplanarity of the $\mathrm{sp^2}$-hybridised carbons and the unusual distance between the $\mathrm{sp^2}$ carbons directly attached to the $\mathrm{sp^3}$ carbon (longer than almost any C-C single bond, but far too short for isolated carbon atoms in the absence of aggressive steric constraints).
For some more information, check out this reference: The homotropylium ion and homoaromaticity, Ronald F. Childs, Accounts of Chemical Research, 1984, 17 (10), 347-352 DOI: 10.1021/ar00106a001

Thanks a lot. The structures and Wikipedia links clarified it for me (I don't know anything about NMR except its full form, so I couldn't make sense of that paragraph. But of course more learnt users will certainly find that part useful too).

No problem. I've added a picture with less abstract representation of the structures, just to make it clearer.

I guess it's a tiny leap for extended pi-systems becoming aromatic, it's a bit funky that this description works best for cations. The wonderful $\ce{C7H8}$, which should possibly be the best representative is mentioned in the footnotes. I guess it just sounds funky enough. Apart from this, one point that should always be mentioned when written down: There are no such thing as hybridised atoms. They might be atoms which can be described with hybrid orbitals, but that is just it.


text:
Why doesn't limestone layer on house walls (now dust) flow off when it rains due to dissolved carbon dioxide?
What happens  to the limestone layer formed on house walls after whitewashing?
Why does it not flow off when it rains, as rainwater contains $\ce{CO2}$?
That will covert $\ce{CaCO3}$ into calcium bicarbonate that is soluble in water?
When  $\ce{Ca(OH)2}$ is applied on a wall for whitewashing it reacts with atmospheric  $\ce{CO2}$ to give $\ce{CaCO3}$ (and a shiny appearence), but we also learned that, when we pass  $\ce{CO2}$ from aqueous solution,  $\ce{CaCO3}$ will change into sodium hydrogen carbonate ($\ce{CaHCO3}$) that is soluble in water, so we know $\ce{CaCO3}$ was present on wall (from above discussion), then supposing it rains that will also contain dissolved $\ce{CO2}$ obviously. So why doesn't this dissolved $\ce{CO2}$ in rainwater react with $\ce{CaCO3}$ of the wall and wash off by forming $\ce{CaHCO3}$ (soluble in water)?

Please check spelling and grammar before posting. It is difficult to understand what you mean otherwise and as a  result your post is apt to get less attention.

It is not a question of if, but how fast. The answer is very slowly.

When applying a solution or an emulsion of $\ce{Ca(OH)2}$ on a wall, this substance gets deposited on the wall. But it will soon be transformed into calcium carbonate $\ce{CaCO3}$ which is insoluble in water. Now the rain may fall down against this wall. And this water contains a small amount of dissolved $\ce{CO2}$. So, of course, this $\ce{CO2}$ can react with the deposit of $\ce{CaCO3}$ according to the reaction : $$\ce{CaCO3 + CO2 + H2O -> Ca^{2+} + 2 HCO3^-}$$ producing a solution containing the ions $\ce{Ca^{2+}}$ and $\ce{HCO3^-}$ as if it was a solution of the non existant $\ce{Ca(HCO3)_2}$.
If I read correctly your message, you are surprised that this reaction does not dissolve your deposit when it's raining. This is a good question. But you should realize that rain drops contain an extremely low amount of $\ce{CO2}$, maximum $1$ gram per liter. Each rain drop contains less than $0.01$ milligram $\ce{CO2}$. And the rain does not stay a long time on the wall. Furthermore, the reaction is not immediate. So the rain does dissolve $\ce{CaCO3}$ on your wall, but at such a low rate that it will take years and even centuries for the deposit to be eliminated by rain washing.


text:
Question regarding conflicting results of checking the electron-preciseness of isoelectronic species via NBO and EHM calculations
It is well known that the tert-butyl cation's empty 2p orbital on the central carbon is stabilised by hyperconjugation from neighboring C-H bonds. This would mean that the central 2p "lone pair" is antibonding in character in the tert-butyl cation; the same would go for the empty boron 2p orbital of trimethylborane and filled nitrogen 2p orbital of trimethylamine.
However, when I did NBO calculations using the minimal STO-3G basis set on the aforementioned three species (note that the lattermost species is "two electrons too rich" of being isoelectronic with the first two) on Gaussian, I got a surprising answer- the empty "lone pair"s of both the tert-butyl cation and the trimethylborane molecule did turn out to be antibonding in character, but the filled "lone pair" of the trimethylamine molecule was not antibonding in character.
Defining electron-preciseness of a molecule being the statement, if true, that "all molecular orbitals of molecule in question, up to (and including) the highest lying orbital(s) that are not antibonding, are doubly filled and all other orbitals are empty", this would mean that both the trimethylborane molecule and the tert-butyl cation are electron-precise, while the trimethylnitrenium cation, being two electrons poor of the electron-precise trimethylamine molecule, would actually be electron-decifient. Since two isoelectronic species must be either both electron-precise, both electron-rich or both electron-deficient, this means that the NBO calculation method I used (Gaussian, STO-3G) does not accurately predict whether a "lone pair" is of antibonding character or not.
My question now proceeds; which is electron-precise (via the above EHM-based definition) of the following species, trimethylborane or trimethylamine?

You may also consider asking this in matter modelling SE, which is not to say that it is off-topic here.

Did you account for the difference in geometry between the molecules? The trimethylnitrenium cation is presumably trigonal planar, whereas the trimethylamine is trigonal pyramidal, so the molecular orbitals will be different.  Just removing two electrons from the trimethylamine orbital diagram doesn't get you the MO diagram of the cation.

I don't think that would create changes in the (anti)bondingness of orbitals, since e.g. the trigonal pyramidal trifluoromethanide ion is still 2 electrons rich of being electron-precise.

If you say to a quantum chemist that the calculation was STO-3G they'll just laugh - the results are not trustworthy. I suggest, before you leap to any complicated conclusions, that you redo the calculations with something a bit more believable. And what level of theory are you using? Hartree-Fock? DFT? Something else?

Hartree-Fock is what I used.


text:
Number of products formed in radical chlorination on a chiral center
If an organic compound with a chiral center of a specific (R) or (S) configuration, is chlorinated on that carbon center through a radical mechanism, will one product be formed or two? Basically will the compound undergo racemization or not?
For example consider the case of: $\ce{C(CH3)(CH2CH3)(Ph)(H)}$ when monochlorinated, a radical intermediate will form. Now since the radical intermediate appears $\ce{sp^3}$ to me in this case I think that the configuration of compound about the chiral center will remain conserved when Chlorine radical later combines with this intermediate form. Is this true or not? Your analysis and views are appreciated.
(Please ignore effects of steric crowding in the compound, this example is purely for theoretical analysis)

I would agree with you, for the same reason that you indicated.

@Ashish https://chemistry.stackexchange.com/questions/34976/what-is-the-geometry-of-an-alkyl-radical?noredirect=1&lq=1  (sort of duplicate...)

From the JEE point of view, the final product would almost certainly be a racemic mixture. The link @Rishi provided says precisely that, how the barrier for inversion in the intermediate radical is extremely low -similar to amine inversion - so it rapidly interchanges between R/S forms, so the final product is also roughly equal in R/S (exactly equal in the theoretical, ideal case).


text:
Intuition for relation between electrode potential and reaction quotient
Consider the following reaction in a galvanic cell: $$\ce{X(s) + Y+(aq) -> Y(s) + X+(aq)}$$
According to the Nernst equation, $$E =E_\circ - \dfrac{RT}{F}\ln\left(\dfrac{\ce{[X+]}} {\ce{[Y+] }} \right)$$
Thus, the lower the concentration of the product ion, the higher the electrode potential.
I am trying to figure out why this is true.
According to me, there are two factors at play:

Some metals are more "comfortable" being cations than others are, due to solvation effects, low effective nuclear charge leading to high electropositivity, and so on. The higher the "comfort difference" between the metals in the cell, the higher the electrode potential.

As electrons leave the metals that are comfortable being cations, they leave mutually repelling positively charged particles in their wake that increase the energy of the system. This is undesirable for stability, so the higher the product cations' concentration, the lower the electrode potential.

I believe taking the fraction of the concentrations of product and reactant ions rather than their absolute figures nicely captures this tradeoff. This is because even though increasing the product concentration will increase repulsion between cations, a proportionate increase in the reactant concentration will increase the "discomfort" mentioned in point 1. Electrons will again flow to the discomfited cations at the same rate.
Is this model correct? Is it complete?

Let's consider an example of your galvanic cell : the Daniell cell, made of a zinc anode ($\ce{Zn}$) and a copper cathode ($\ce{Cu}$). When zinc $\ce{Zn}$ is in contact with water, it "prefers" being transformed into the cation $\ce{Zn^{2+}}$ in order to be dissolved in water. In your language, you say that they are more "comfortable" in water. Of course these new ions would repel one another in solution, and prevent new zinc atoms to produce the same operation. The reaction could not take place with more than $1$ or $2$ ions in solution. But it seems to me that you forget what happens in solution. These newly created $\ce{Zn^{2+}}$ ions are attracting negative ions from where they are in excess, for example in the cathode region, where they are repelled. Matter of fact, if the cathode zone contains a solution of $\ce{CuSO4}$ where $\ce{Cu^2+}$ ions are discharged by the arriving electrons, the remaining sulfate ions $\ce{SO4^{2+}}$ are repelled by these same arriving electrons, and are attracted by the anode zone. As a result, these "unwanted" $\ce{SO4^{2+}}$ cross the membrane between the two electrodes, so as to neutralize electrically the newly created $\ce{Zn^{2+}}$ ions in the anode zone. Is it what you expected to hear ?

Ah, then the reasoning in my post is incorrect. If there is no "cost" for electrons to migrate from anode to cathode (in the form of increased repulsion between zinc cations) why does reaction quotient matter while calculating electrode potential? I know that the straightforward answer is that it is because the Nernst equation is just the Gibbs-Helmhotz in disguise, but I want to get some intuition.

@Ray Bradbury. Why do you want more than Nernst equation ?

I figured it out myself, and it was elementary. Obviously electrons will go where there are more positively charged particles (cations), and obviously they will move faster if there's a great difference between cation concentrations. If there are much more cations in the anode than in the cathode, the electrons won't ever want to leave. But thank you for your answer, you pointed out a pretty significant error in my thinking.