text:
Why should I not use different battery types together?
Everyone knows about this but I'm curious about why one should never mix batteries of different types. By type I mean the same shape of the cell like AAA but with different chemical reactions and voltages inside.
Would it just mess up the voltage and the instrument that is operated with the battery? Or would it lead to unexpected reactions in the battery itself that could lead to leakage/explosion or other catastrophies?

Theoretically there must not be any issue  because only electrons are transferred through wires. You can connect different voltages of different batteries but here comes an only issue- internal resistance, for dry cell it is too high for li-po it is very low. So if you connect a li-po battery that supply huge current with a dry cell then dry cell may reach very high temperatures and may explode. I am not sure of it so added it as a comment but you can surely use different battery types with almost same values of internal resistances and voltages.

If run in parallel, one might try to charge the other one if the voltages don't quite match. I'm not aware of applications where batteries are run parallel. Usually they are in series, to get a large voltage.

They are used in parallel when you want a bigger capacity of USB power banks. They use then convertor of 1 LiIon cell voltage to 5.1V. generally, it the cell voltage difference is less then twice of their voltage raise/drop during  safe cell charging)/discharging, such connection is safe too.

Serial usage
For the serial usage, one must avoid combination of cells of different kind and health. They should be as identical as possible.
The best is using

the same chemistry
the same format
the same vendor
the same brand
the same lot
the same aging

Otherwise the cells will have the different capacity and different discharge profiles under the load.
If the serial cell chain is deeply discharged, the "weakest" cell gets discharged completely.
And what is worse, other cells start to electrolyze the unlucky cell in the opposite direction that would be its charging.
This would lead almost inevitably to heavy damage or more likely destruction of the cell, what is unwanted for rechargebles.
Physical destruction with electrolyte leakage is also possible.
For primaries, it would be rather annoyance of not fully used better cells and no easy possibility to find a matching cell.
Parallel usage
It is generally adviced not to use it, unless you know exactly, what you are doing. See also notes for non rechargeable cells below.
Dangers
If cells of different chemistry and therefore voltage range are used in parallel, they overcharge or overdischarge each other when their common voltage does not fit the overlapping of their save voltage ranges.
But even within the overlapping voltage range, cells can be overheated and thermally functionally damaged/destroyed or can even explode, depending on the cell voltage difference and cell internal resistence.
The last paragraph is also valid for cells of the same chemistry. They cannot overcharge each other, but the thermal damage from above applies to them as well.
How to do it well
Only cells of the same chemistry and performance class with about the same actual voltage (check it ) should be connected in parallel.
The voltage difference must respect expected internal resistance and acceptable resulting balancing current. Note that typical internal resistance of Modern AA NiMH cells is $\pu{30-100 m\Omega}$, so $\pu{60 mV}$ difference can cause the current $\pu{1 A}$.
Once connected and balanced, they are safe, as they autobalance themselves during charging and discharging, but consider notes below.
Cell autobalancing of already connected parallel cells
This analysis is aimed for rechargeable cells., supported by Excel simulation.
Cells in an open circurt
Balanced parallel batteries have always the same external (behind their internal resistance ) and internal voltage.
Cells under load
The initial current is in reciprocal ratio of their resistances, so high performance cells would get drained initially faster despite their lower capacity.
The cells with lower $resistance * capacity$ parameter gradually get the internal voltage lover than the others and provide therefore gradually lower current than the other cells.
This leads to the cell balancing, with the current ratio possibly even getting reversed to the initial one. The current ratio slowly converges to the capacity ratio.
Important parameters are the ratio of cell capacities and the ratio of cell internal resistances.
The best results are achieved if these ratios are reciprocal. In such an ideal case, cells are discharging in ratio of their capacities and no internal autobalancing ever occurs.
The good thing is, that the cell resistance increases with the degrading the cell capacity, as cells are aging.
Best autobalancing is achieved on low currents.  High currents lead to higher internal voltage disbalance and are more demanding for good cell matching.
If the cell resistance ratio is not optimal for the capacity ratio, autobalancing kicks on, but does not fully compensate. The internal voltages get therefore little disbalanced.
After disconnection, there is ongoing small temporary compensating current.
It is not good to combine lower capacity high performance cells with low resistance with high capacity low performance cells, especially for high loads. As it significantly raises the internal voltage disbalance and after load compensation
Cells under charging
It is generally adviced against parallel charging, as it affects charge stop detection and can cause overcharging of some cells.
That affects serial charging as well.
Non rechargable cells
Parallel connection is not generally good for different . As there can occur small balancing load after disconnection of the powered device.
OTOH, if the identical cells ( like in the serial section) are used in parallel connection all their lifetime, it is not a big deal, as these currents are minimal and short time.

And that's also why you must remove empty batteries from an appliance. If a small current can still flow, the batteries will destroy each other one by one.

Also, the longer the serie is, the bigger danger, as it is a fight  N-1 versus 1. With 2 cells only it is not a big deal. With 4 or 6, the fun may begin.

If you put different cell types in parallel, the one with highest voltage will almost definitively destroy the others. unless each of them has a dedicated overcharge protection (very unlikely).

Not necessarily, if they are of the same type and about the same charge and voltage. The big power banks uses e.g 5 3400mAH 16550 type heavy duty LiIon cells in parallel. With the open voltage getting closer, the risk of damage dramatically decreases.

You wrote above that mixed cell types in parallel may overcharge each other. I think that's nearly a cert. ;-) Same cell type and age not so problematic, but those powerbanks have a controller that monitors each cell separately. You can buy these circuits on ebay in packs of ten from China.

if they are checked before connection, there is no danger in joining. Once joined, they share the same voltage. As they share the same voltage span of operation, the connection is not problematic.They are autobalancing themselves.

You said different cell types overcharge each other but imo same cell types can also overcharge each other if one of them is less charged.

All is matter of difference. If voltage of NiMH cells can be 1-1.4V, the cell with 1.4 V cannot overcharge the cell with 1.0V. But what will happen is the charging current would be very high and would heat up both cells, depending on the cells resistance. They will both have during the charging the external voltage very roughly 1.2V, one due charging, one due discharging. I am not sure about the current value causing 0.2V drop, if it is comparable to fast 15-60 min chargers or not.

@Karl if different chemistry parallel cells have overlapping voltage range, they do not overcharge nor overdischarge each other while staying in this overlapping. But it is question of time when they leave it.

@Poutnik they may have the same voltage now, but will they retain the same voltage after you draw 50% of their combined capacity from the pair? Heck, they won't even have the same voltage the moment you put them under load. The weaker one  may not overcharge (immediately), but the stronger one will discharge faster, then experience reverse current when the load is disconnected. If it's a non-rechargeable battery, it may not like that reverse current.

@John Dvorak : Added notes about cell balancing


text:
Enzyme Kinetics - Given Km find substrate concentration at a certain velocity
From what I understand about the Michaelis Menten Model

Km defines the amount of substrate required to reach half-saturation.
1/2 Vmax corresponds to Km on the x axis
Generic formula is v = Kcat [ES] [S] / (Km + [S])

Given Km and the desired velocity in terms of reaction efficiency, can one just plug in Km into the equation and get a relational expression with [S] on one side?
Or do you manipulate the 1/2 Vmax expression?

You can solve for [S]. There are two common ways to write the equation, like you have it:
$$v_0 = k_\text{cat} [E_0] \frac{[S]}{K_\text{m} + [S]}$$
or after dividing both numerator and denominator by [S]:
$$v_0 = k_\text{cat} [E_0] \frac{1}{\frac{K_\text{m}}{[S]} + 1}$$
In this form, it is easier to see that [S] occurs once and you can solve for it.
Get [S] out of the denominator of the "big" fraction and isolate the "little" fraction:
$$\frac{K_\text{m}}{[S]} + 1 =  \frac{k_\text{cat} [E_0]}{v_0}$$
$$\frac{K_\text{m}}{[S]} = \frac{k_\text{cat} [E_0]}{v_0} - 1$$
Finally, take the reciprocal and multiply by $K_\text{m}$:
$${[S]} = \frac{K_\text{m}}{\frac{k_\text{cat} [E_0]}{v_0} - 1}$$

I'd recommend writing the ES term as $\ce{[E_0][S]}$ to explicitly indicate you are talking about the total enzyme concentration times the substrate concentration.  People might get confused by $\ce{[ES]}$, thinking you mean the concentration of the enzyme-substrate complex, as this notation is often used in e.g. derivations of the MM model.

Also, when you do that, you will find that there is an $\ce{[S]}$ on both sides of the equation, so your "explicit" solution for $\ce{[S]}$ really isn't.

My typo, now corrected.

You can also solve for $\mathrm{[S]} $ using Lineweaver-Burk plot (or double reciprocal plot). One way of writing the Michaelis Menten kinetic equation given in Wikipedia is:
$$v_{\circ} = V_\mathrm{Max} \frac {[S]}{K_\mathrm{M} + [S] } = = k_\text{cat} \mathrm{[E]_{\circ}} \frac {[S]}{K_\mathrm{M} + [S]} $$
If you take the reciprocal of both sides of the equation, you get:
$$\frac{1}{v_{\circ}} = \frac{K_\mathrm{M} + \mathrm{[S]}}{V_\mathrm{Max} \mathrm{[S]}} = \left(\frac{K_\mathrm{M}}{V_\mathrm{Max}}\right)\frac{1}{\mathrm{[S]}} + \frac{1}{V_\mathrm{Max}}$$
Thus, a plot of $\frac{1}{\mathrm{[S]}}$ vs $\frac{1}{v_{\circ}}$ is a straight line, which is called Lineweaver-Burk plot:

Now, if you know the values of $K_\mathrm{M}$ and $V_\mathrm{Max}$ for your enzyme and substrate, you can draw a plot upper-hand ($x$-$\text{intercept} = -\frac{1}{K_\mathrm{M}}$ and $y$-$\text{intercept} = \frac{1}{V_\mathrm{Max}}$) and determine your substrate concentration ($\mathrm{[S]}$) for each desired velocity ($v_{\circ}$).

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text:
What is reason of stability of metal (Sr and Ba) peroxides?
While studying inorganic chemistry I came across the following chemical equation.
$$\ce{M + Air\left(1/2O2 + N2\right) ->[Δ] MO ->[\text{Excess}~O2] MO2 (Stable) + M3N2}$$
$$\ce{M} = \ce{Sr, Ba}$$
My question is why peroxides are more stable than oxides here but in case of other alkali earth elements $(\ce{M} = \ce{Be, Mg, Ca})$ oxides are more stable.
These are the things that came into my mind:
The structure of $\ce{SrO2}$, $\ce{BaO2}$ are isomorphous to $\ce{CaC2}$. I don't think this is causing any extra stability.
The increased basic nature of oxides maybe de-stabilizing oxides when going down the group.
Or maybe the large size of $\ce{Sr^2+}$, $\ce{Br^2+}$ better fits with $\ce{O2^2-}$ in crystal lattice than $\ce{O^2-}$ which leads to increase in lattice energy of peroxides.
I am still a very beginner to inorganic chemistry so it should be helpful if someone explains in a bit more detail.

I would wager the fact the compounds gets stabilized due to similar cation and anion sizes.

@NilayGhosh It should be helpful if you can explain more about it.

Magnesium and calcium do form peroxides by other means, though not by direct combination.

@OscarLanzi True, but still oxides are more stable than peroxides in case of Mg and Ca. The question is about why peroxides are more stable in case of St and Ba not whether Ca and Mg for peroxides or not.

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text:
How were the Hoffman's and the Zaitsev's rule formulated even before the discovery of the electron?
I recently studied Zaitsev's and Hoffman's rules for deciding which product is formed via elimination and noticed that the rules for formulated much before the discovery of electrons and any conceivable atomic theory. So how did they work out such rules without even knowing how these compounds bond and behave at microscopic levels.

As Wikipedia notes, Zaitsev's rule is 'empirical' and 'subject to many exceptions'. They did a lot of chemistry and noticed trends. It didn't matter exactly how it worked, they pieced together a phenomenological theory that worked some of the time.

Yeah I read that but how did they even talk about there being double bonds after elimination if there wasn't any theory describing their existence. Also can you please elaborate on what does empirically exactly mean in this context and can I find somewhere what trends did they notice or how they concluded?

The term "double bond" stems from their time, afaik. Since the ~1930ies we know that the two bonds are not a pair of equals. They knew about the molecular geometry (or rather connectivity), and about addition and elimination.

History of chemistry can be very confusing. So hard to keep track who knew what, and when, when the old journals are hard to get, and textbooks even harder, and the nomenclature wasn't as uniform yet. Good opportunity to practice your school French and German, though. ;-)

Thanks a lot. This really clears things up. Also do you know of any standard textbook or source which has a comprehensive timeline of development of chemistry?

This might be a starting point for the history of chemistry: http://web.lemoyne.edu/~giunta/papers2.html#organic

This excerpt is from an article in J Chem Ed (J. Chem. Educ., 1961, 38 (6), p 297
DOI: 10.1021/ed038p297) describing the contents of Markovnikov's 1870 paper in Liebig’s Annalen (translated into English, I guess):

So this was before the discovery of the electron, but as you can see from the diagrams, the concept of chemical structure, atoms and valence was already in place.

Right what I wanted to add. Chemistry is first of all an experimental discipline, I always like to mention that structures like phtalocyanines or something like that were deduced solely by building/dismantling them on the bench! And just using melting points and the likes!


text:
Can sodium hypochlorite soak all the way through an unglazed ceramic cup and bleach blue paper wrapped around the outside of the cup?
My question is regarding the ionic properties of sodium hypochlorite in house hold bleach. There is a debate among potters whether bleach will kill mold growing on a porous ceramic cup. It is my understanding that sodium hypochlorite cannot penetrate a porous surface of fired clay. Someone did an experiment where they put household bleach in an unglazed cup and allowed it to soak through. There was blue construction paper wrapped around the outside of the cup and the paper turned white after a while seemingly from bleaching. So my question is this: If the ionic properties of sodium hypochlorite will not allow it to penetrate through the cup to the outside, then how did the blue paper wrapped around the outside of the cup get bleached?

Can you smell a bleach solution?

It was probably bleached by vapours.

It is not the ionic properties that control the diffusion. Diffusion is a fundamental property of molecules dissolved in a liquid. A bleach solution stored in an unglazed container can potentially diffuse to the wall. The pore size of the material will mainly control the rate of movement of the fluid (viscosity, surface tension etc.) If the pore size and the connectivity channels are very small it will take ages to wet the wall under atmospheric pressure.

Thank you for that explanation. Then why does everything I have read not recommend using household bleach to kill mold growing on porous surfaces such as concrete or drywall. Many sources ... including the Clorox Co. ... say the sodium hypochlorite molecule cannot penetrate into the porous surface to kill the mold hyphae there. It supposedly only kills mold on the surface and leaves the hyphae growing inside alone. So I’m confused by the experiment that suggested that the bleach solution was able to diffuse through the porous, unglazed ceramic cup in a matter of minutes.

You raise two points: if the unglazed pottery is very porous, it may take minutes, but if the pore size is very small it may take a very long time. I think the key point is time scale of diffusion. Secondly, bleach manufacturers will not assume that people would let their tiles, or concrete soak in bleach for several days. The typical contact time may be minutes.  That time scale may not be sufficient to eliminate the mold embedded deep inside.

Shouldn't capillarity result in a relatively fast soaking and wetting?  Broken ceramic, as I think to remember, pick up water quite easily.


text:
Efficient way to synthesize poly(alkyl)amine
I am trying to synthesize poly(ethylene-propylene)amine, i.e.:

I've read that Reductive Amination is efficient regarding the reaction yield. However, it needs aldehydes to start the reaction. To make the aldehydes I can use this way and start from the hydroxyl-terminated polymer compound. I reach to a dilemma here where my experience is not enough to decide which way to go. I can start with a aldehyde-terminated polymer at first so I don't have to make it from an OH-terminated polymer.
Which way do you suggest? A reference on how to do the one you recommend is also appreciated.

Before I offer an answer understand that I am a general synthetic chemist used to working in solution phase, and I have not worked with polymers. Absent an answer from someone who has here is how I would tackle this using OH-terminated polymer.
1 -reaction with methanesulfonyl chloride/triethylamine to activate the -OH for displacement.  This is known as mesylation more here and here
2- reaction with sodium azide in DMF or DMSO to give the alkyl azide
3- reduction of the alkyl azide to the amine with triphenyl phosphine Staudinger reduction


text:
Suitable coating material for protection of sensors
I'm building a moisture/temperature sensor for outdoor use in soil. It uses capacitive sensing for moisture and an NTC thermistor for the temperature. It's going to be used for vegetables so food safety is a concern.
I don't think the PCB is food safe, or safe from corrosion/degradation due to the soil and moisture would creep into the FR4 material in the PCB and cause my measurements to drift so I would like to coat the entire thing to protect it from the elements, keep my vegetables free from toxins, and keep the measurements from drifting.
So these are the requirements in order of importance for the coating:

Not electrically conductive
Food safe
Doesn't degrade/react in/to soil/sun light
Hydrophobic
Resistant to abrasion when insertion/removing from the soil
Affordable
Low relative permeability (electromagnetic)
Decent thermal conductivity
Not a host for bacteria

I first considered a food safe epoxy but I had difficulties finding a reputable supplier of food safe epoxies in Germany (my German sucks). I did find a supply of food safe silicone which is nice in that it has all the properties above but I could only find it in large quantities and it's quite expensive. I also considered L/HDPE which seems to have all the criteria apart from the bacterial aspect (I'm not sure about that) but it seems like melting it to have a low enough viscosity to be able to form a solid but thin coating seems hard/impossible.
My knowledge on the topic of epoxies, plastics, rubbers and chemistry in general is limited, so I'm turning to chemistry@sx for recommendations to what I can use for a coating and what process to use for coating. And maybe I've misunderstood something of the materials I've considered so far... Also I'm sorry if this is the wrong place for this question, I'm not sure where else to post.

It might be overkill, but PTFE (Teflon) looks like a great option to me.
It is temperature and chemicals-resistant. Plus, Teflon coatings are employed in cookware.
The only "bad" point is its price, but a thin coating would do the job, and not much of it would be necessary.

@The_Vinz have you worked with PTFE before? Can you offer any advice how it is applied? I know it's a thermo plastic, so I'd get a block and melt it the liquid phase and dip?

I am sorry, I didn't understand that you intended to use it in DIY. I think that working with PTFE might be hard without proper equipment, but a possible  experiment would be using some PTFE tape used for sealing threads and sold in any supermarket.
Try to tightly coat the apparatus with it, and then try to lightly melt-solder it with a  heating gun. I am not sure that it works, and probably this will require temperatures around 200°C, but maybe it's worth trying on some scrap material.

Otherwise, another good option for DIY would be using... a condom :)
Latex-free condoms are made of non-toxic polymers: polyurethane, polyisoprene, nitrile or silicone.
They are less resistant than PTFE, but maybe their shape and ease of reperibility might be an advantage.

@The_Vinz latex condoms don't have the necessary tensile strength to survive the insertion :/ (I tried)

Various DIY capacitive moisture meters use a PCB for the sensor:

While not directly related to your question I thought I'd mention that you can buy these sensors for several dollars:

Purchasing a readymade kit and relying on the discount of mass production (along with a manufacturer's warranty) is probably easier than designing your own, and etching your own PCB.
I think how you plan to make the probe plays a part in how you plan on coating it; for example you could fit the parts in a test tube, that would meet all the requirements in your list. Such a capacitive sensor would operate similar to the touchscreen of a cellphone.

I have looked at the off the shelf sensors. They all lack features that I need on a system level and are not suitable for my planned network topology. I have already manufactured PCBs and written the software. None of these sensors you show come with coating so they are not safe for use as-is with plants you intend to consume and will suffer from drift as moisture creeps into the FR4. Sticking a tube into the soil and putting the sensor in there is obviously a bad idea unless you cap the tube, as water may enter from the top it also pushes the soil further away, weakening your measured signal.

Thank you for your reply. Instead of using the above sensors exactly as-is you can order a conformal coating or if you are using them as an example then apply your own. If the test tube is a "bad idea" unless you put a stopper in it then I guess there's only one thing to do. Perhaps another answer will be forthcoming that you find more to your liking.

This whole question is about choosing a conformal coating... An answer to buy another probe and then conformal coat it isn't really helpful :/

I explained what standard practice is, and that you can do what people whom sell many of these probes do, additional coatings beyond a decent PCB material wasn't deemed necessary by many people. In the event that you wanted something simple and inexpensive a glass tube provides certain protection and food safety; unless you are prone to breaking things. As a point of interest, and stated as unrelated, you can to buy one and  choose the coated option when ordering, and get all the work done for several dollars. If your time isn't valuable then making it yourself will save a lot of money.

You have given no advice about best practice about conformal coatings, full stop. PCB manufacturers make no guarantees and pay no attention to food safety of PCB soldermask or other materials, and unless the device is RoHS compliant (you're linked probes are not) the solder on the joints or on the HASL'd pads may contain lead which is definitely poisonous. People use uncoated PCBs for their soil sensors for the same reason they use PVC piping in their hydroponics/gardens, it's cheap and they don't know that it is unsafe to use if you intend to consume the plants.

My time is valuable, but as I said in my first reply. The off the shelf products do not meet my system-level requirements.

This does not answer the question even remotely.

@WilliamR.Ebenezer If it's "not an answer" then flag it. I've earned more than 2K this past month, none of it debating in comments and very little on ELL or English.SE; where you're welcome to run your ideas by someone.

@Rob, two things: first, I was explaining my downvote, not fueling any snide-flavored debates regarding your answer. And secondly, it's best if you don't mention your reputation successes on other sites which are subject-wise unrelated. It simply doesn't make any sense. And please don't take this personally. Have a great day.


text:
How does hyperconjugation in alkenes affect bond lengths?
I understand how the diagram below results in hyperconjugation stabilising alkenes.

Essentially, the C-H sigma bonding orbital interacts with the pi bonding orbitals creating two molecular orbitals. It also interacts weakly with the pi* orbital which lowers the energy of psi 1 and 2 stabilising the system.
What I do not understand is the effect this has on bond lengths. We have done a computational investigation in the lab and have found that both the C-H and C-C bonds are weaker, while the C-C bond between the double bond and the alkyl substituent is stronger. I don't understand why.
I have received an explanation for this phenomenon using resonance structures, but I am really looking to understand how to reconcile these observations with the MO diagram above.

The MO interpretation is not really all that complicated: you are allowing the C–H σ orbital to overlap with the C=C π* orbital. This leads to a very slight population of the π* orbital, which is antibonding with respect to the doubly bonded carbons. Therefore, that C=C bond length is slightly lengthened. Likewise, the C–H σ orbital is slightly depopulated, which makes the bond length longer.
At the same time, you are forming a slightly stronger C(sp2)–C(H) bond because these new MOs have bonding interactions between the C–H σ orbital and the p orbital on C(sp2). This is not entirely made clear in the diagram above from Fleming's text, which I have addressed separately here. From the diagram, it seems that there are equal amounts of bonding character (from ψ1) and antibonding character (from ψ2) However, the main point which I wrote in that post is that the π* orbital mixes (very slightly) into both ψ1 and ψ2. This increases the C–C bonding character in ψ1 and decreases the antibonding character in ψ2, such that there is a (very slightly) shorter bond.
The actual text in the book makes this clear, but the diagram does not. I suggest reading the text carefully.

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text:
Does hydrochloric acid change in concentration over time?
I have new bottles of 1 N hydrochloric acid from Fisher. The acid comes standardized from the company:

Standardized at 25 °C to pH 8.2 against sodium hydroxide that has been standardized against NIST™ potassium acid phthalate.

Will it remain standardized after opening the bottle, or do we have to standardize it regularly?

Most likely, yes, it will require standardization at some point. If you are working at the analytical lab, there must be a schedule for testing reactants to make sure they are of analytical grade and the concentrations are correct.

Not just hydrochloric acid, concentration of any acid changes over time.

First of all, check the expiry date. In general HCl solutions are secondary standards. 1 N is just a approximation. You need to standardize it every time with a primary standard such as sodium carbonate. On the other hand, constant boiling HCl is indeed a primary standard. It is not very well known, perhaps because modern analytical chemists have become lazier.
Constant-boiling hydrochloric acid
John L. Sheldon
J. Chem. Educ., 1935, 12 (8), p 377
https://pubs.acs.org/doi/abs/10.1021/ed012p377

Boiling HCl standard is quite inconvenient to manage. The 1N one  need not to be checked every time, it is enough to do periodically, with monitoring the trend.


text:
How does HCl break down amylopectin in starch into amylose?
I'm currently researching on manufacturing bioplastics. During one of the steps, HCl is added to break down amylopectin into amylose, but I don't understand how exactly this happens. It would really help if figures were included to illustrate.

I presume you mean hydrochloric acid. Start with the mechanism of acetal hydrolysis. The topic will be covered in any reputable organic chemistry text. As a new user, be aware that you are expected to show some effort to solve the problem.  Welcome to ChemSE.


text:
Why isn't O=N-O-O the best lewis structure for NO3-?
I'm practicing lewis structures and I'm stuck on the one for NO3 -1 (nitrate ion). I understand that N has to be the central atom because it is the least electronegative, but I'm confused why all the O have to be bonded to the N atom.
With O=N-O-O, you get 3 zero formal charges and one -1 instead of one zero, two -1 and one 1, which seems better. Also, H2O2 is a straight line instead of having the both H and and one of the O around a central O atom, so I'm confused why the nitrate ion must be the opposite.
Here is the work I did:

There is nothing inherently wrong with that. It is just that we know  all three O are attached to N.

And O-O bond is quite unrealistic really. You'll get the hang of the empirical part of Chemistry on studying it a little deeper.

You win according to the rules you were given and were applying. The "correct answer" wins because that is what we observe in reality. So you discovered an exception to the rules you were given.

Peroxynitrite sure does exist. There no such thing as "best" structure. You could ask why it's much more reactive, though.


text:
Derivation of t2g eg splitting in octahedral field
I am trying to make the derivation of the d orbitals splitting under an octahedral field but I am kind of at a loss and don't know of any literature that can provide a more introductory level of this subject (if you know of any, please let me know).
I am trying to follow the book "Magnetic Oxides" by Gerald F. Dionne but I understand half of what it says in regards to this.
I know that we want to use the crystal field as a perturbation to the hamiltonian of the center ion, and for that you need to diagonalize the matrix (picture from here)

And that's it, that's where my understanding ends. The subscripts of this matrix is the quantum number m and the regular d suborbitals are almost all a linear combination of spherical harmonics with different m. So how do I diagonalize it?
The book doesn't do an explicit matrix handling and it basically just tells you from the start what the eg and t2g orbital are comprised of, I am assuming the notation means something I am unaware of,  and that they have different perturbation energies. No explicit equation that makes you understand that is shown.
I am sorry if it all seems confusing, I am confused. I know the how to derive the octahedral field (in cartesians and spherical harmonics) and how each d suborbital is written in spherical harmonics. But I don't know how to obtain the different energy splitting due to the perturbation of the octahedral field, any guidance/book or help is appreciated. Thanks!
PD: I am not a chemist, that's why I don't know of any books where I could read about this subject and why I am probably not following that well the book I was given.

Edit:
Because of all the recommendations, I have been able to understand more clearly the procedure while looking at it from a matrix perspective rather than a group theory one (completely at a loss there).
My question is that I am not getting what I should be when obtaining the element of the matrix

When trying to obtain those numbers in Mathematica I obtain values that don't make sense. I defined the d wave function like

Here you can see one matrix entry as an example

When doing the same as above for the four that appear in the image, I get that the factors before the Dq need to be

Which is not the same as the book.

Would you prefer to read the original work of Hans Bethe where he talks about splitting? I had a soft copy of the English translation (the math in his work is well above my comfort zone) https://books.google.com/books?id=5baAG1WqgYQC&pg=PA1&dq=Term+splitting+in+crystals&hl=en&sa=X&ved=0ahUKEwjQodjQ_PbhAhUGD60KHVp7B58Q6AEIMTAC#v=onepage&q=Term%20splitting%20in%20crystals&f=false

There are two chapter in C. Ballhausen's book ' Introduction to Ligand Field Theory' that describe the calculation in great detail.  It was published some time ago but you may still be able to find a copy. The first step is however to convert the spherical harmonics into Cartesian x,x,z and impose octahedral or tetrahedral symmetry on the bare ion.

You might want to consider Figgis’s “Ligand Field Theory and it’s Applications”. I skipped the math, but it does a pretty good explanation for the definition of Dq, the AOM model etc.

Thank you all for your recommendations on book, now everything is much clearer. I still have problems so I will update (and erase parts of) the old question to the pertinent content at hand now.


text:
What do we learn from an isotope peak (or M+2 peak) in a mass spectrum?
Mass spectrum of a molecule showed molecular ion peak and its isotopic
 peak in 1:1 ratio. What is the possibility of the atom present in the
 molecule?
Since it is mentioned that it has a molecular ion peak it gives the fact that it must be aromatics, olefins and not alcohols or branched hydrocarbons since the latter have less probability to show molecular ion peaks . Since they contain isotopic peaks it must mean they have chloro or bromo compounds. But I do not understand how to use ratios to identify the atom present in the molecule

The abundance of the isotopes of Chlorine 35:37 is approx 3:1 and is characteristic of chlorinated compounds.

Each element has different natural abundances for its isotopes. Two peaks with approximately 1:1 ratio indicates the presence of bromine. $\ce{^{79}Br}$ is the most common isotope, corresponding to $0.5069$ of naturally occurring bromine.  The isotopic composition for $\ce{^{81}Br}$ is $0.4931$.
What this means is that, for a molecule contaning a single bromine atom, approximately, half of the molecular ions will have $\ce{^{79}Br}$ and the other half will have $\ce{^{81}Br}$, resulting in two peaks with roughly the same intensity with $m/Z$ differing by two units.
References:
[1] https://www.nist.gov/pml/atomic-weights-and-isotopic-compositions-relative-atomic-masses
[2]https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumental_Analysis/Mass_Spectrometry/Organic_Compounds_Containing_Halogen_Atoms

In addition to the peak pattern indicating the presence of bromine, it also indicates the presence of a single bromine atom in the molecule.

@CurtF, thanks! Very good point. I'll edit that in answer.


text:
How to Scan Aromatic Dihedrals To Obtain Force Field Parameters
I am trying to generate new force field parameters based on quantum mechanical calculation. I used amber manual "parameter development" title for this and I calculated bond-angle-non bond parameters but I am having a hard time obtaning dihedral parameters for aromatics Let's consider about pyrazin.  These are gaff2 pyrazin parameters :
ca-ca-nb-ca   2    9.600       180.000          2.000
h4-ca-nb-ca   2    9.600       180.000          2.000
nb-ca-ca-nb   4   14.500       180.000          2.000
h4-ca-ca-nb   4   14.500       180.000          2.000
h4-ca-ca-h4   4   14.500       180.000          2.000

When I scanned "ca-ca-nb-ca" dihedral angels, I found that  dihedral bond is breaking at 100 degre and 111 kcal/mol energy. According to manual, if this is an ordinary dihedral ca-ca-nb-ca   should have been (111/2=)55,5 kcal/mol V/2 value. But in gaff2 it is just 9.60kcal/mol. There is a huge difference.  So, how to obtain such aromatic ring dihedral parameters by using QM calculation?
My advisors said: "you should use twist scan not rotational scan" but I am not sure even if "twist-scan" exists and I am not sure "rigid" scan is the true answer.
Thank you.
NOTE: Here is the picture.every step is 10 degree.

Did you find out how to do this?


text:
Molar flux through membrane

Consider a liquid of pure $\ce{A}$ covered by a composite membrane through which
  the vapor of $\ce{A}$ is diffusing. The membrane consists of two layers of thickness $l_1$ and $l_2$. The diffusion coefficient of $\ce{A}$ in the two layers are $D_{\ce{A},1}$ and $D_{\ce{A},2}$ respectively. On the lower side of the membrane, the concentration of $\ce{A}$ ($c^s_\ce{A})$ is controlled by the vapor pressure of $\ce{A}$. On the upper side the concentration of $\ce{A}$ is zero. The solubulity of $\ce{A}$ in the two membrane layers is the same.
(a) Derive the expression for the flux $J_\ce{A}$ through the composite membrane

The flux through a membrane is given by the equation:
$$J_\ce{A} = \frac{D_{AB}}{L}\left(c^{(m)}_{\ce{A},F} - c^{(m)}_{\ce{A},P}\right)$$
where, $c^{(m)}_{\ce{A},F}$ and $c^{(m)}_{\ce{A},P}$ are the concentrations at the edges inside the membrane.
So I would assume that $c^{(m)}_{\ce{A},F}$ = $c^s_\ce{A}$ and $c^{(m)}_{\ce{A},P}$ = $c_\ce{A}$
and because there are two membranes I would have to combine the diffusion coefficients as well as the lengths, which I do by addition to get the equation:
$$J_\ce{A} = \frac{D_{\ce{A},1} +D_{\ce{A},2}}{l_1 +l_2} (c^s_\ce{A} - c_\ce{A})$$
I am unsure if this is the correct way of deriving the expression for the flux when there are two membranes. Am I thinking correct?
Edit
I named the third concentration in the membrane $c_m$. Then I made two equations for the molar flux.
$$J_\ce{A} = \frac{D_{A,1}}{l_1}(c^s_A - c_m)$$
$$J_\ce{A} = \frac{D_{A,2}}{l_2}(c_m - c_A)$$
When rearranging and solving for $J_A$, I get:
$$J_\ce{A} = \frac{(c^s_A - c_A)}{\frac{l_1}{D_{A,1}}+\frac{l_2}{D_{A,2}}}$$
But that is the same answer as I got before

What you did is an exact equivalent of fractions addition in a "simple" way: $${1\over2}+{1\over3}={1+1\over2+3}={2\over5}.$$ Pity it never works like that.

@IvanNeretin Yeah, I thought that it couldn't be the correct way. How should it be derived?

Imagine a very thin layer of vapor between the membranes (which is not really there, but hey, that's what imagination is for). Then you know the equation for the flux to this layer, and also from this layer, and the two are equal. Can you continue?

@IvanNeretin But when making two equations, one for the flux to the layer and one from the layer, how do I use the concentrations? Will I have $(c^s_\ce{A} - c_\ce{A})$ for both fluxes?

There will be a third concentration, the one that exists between the membranes, initially unknown.

@IvanNeretin I am not sure if I understand how to solve this question. I tried to make two equations with a third concentration, but I get the same answer that I proposed in the beginning..

Now that's what I think is the correct answer. As for its relation with the one you've got before, I decisively insist that you review your opinions on the addition of fractions. Look again at the example from my first comment. What is ${1\over2}+{1\over3}$? Is it really ${2\over5}$? (Hint: it is not.)

@IvanNeretin Okay, thank you very much for all the help!


text:
Why does bromine increase the viscosity of olive oil?
Does bromine bond to the carbon in unsaturated fatty acids and break the double bonds between oxygen and carbon? Does this increase the viscosity of olive oil?
Could someone give an elaborated explanation?
This question arose from a lab experiment. Viscosity goes up because less double bondings makes the fatty acids more “packable”. But why would bromine want to break the bonds?

I briefly contemplated inquiring whether you are going to use olive oil in the lab or bromine in the kitchen, but ultimately decided against it. Yes, the addition to the double bond seems very likely. How and why would that affect viscosity is not immediately clear to me, though.

Well, its in the lab. Viscosity goes up because less double bondings makes the fatty acids more “packable”. But why would bromine want to break the bonds?

So are you interested in the rationale for increased viscosity? Or in the chemical reaction that takes place? I would think the bromine would add to the carbon-carbon double bonds rather than the carbon-oxygen double bonds.

Viscosity is one of those bulk properties where the factors that determine it can be quite difficult to figure out based on a microscopic view and first principles.

I meant the carbon.carbon bonds. Why would bromine bond there?

Olive oil is a triglyceride, and the main fatty acid is oleic acid below.

Now the bromine does not add to the double bond between oxygen and carbon; instead the bromine reacts with carbon-carbon double bond and add itself to it. If you use e.g. coconut oil which has minimal unsaturated (meaning having carbon-carbon double bond) fatty acids (and assume you don't shine light onto it, heating it or otherwise forcing the reaction to start) there should theoretically be no reaction.

But why would bromine want to break the bonds?

Simply put, bromine loves electron. The carbon-carbon double bond happens to be rich in electrons, so the bromine binds to it. While the carbon-oxygen double bond also has electrons, oxygen loves electron more than bromine and tends to keep the electron for itself, so bromine cannot add there.

Does this increase the viscosity of olive oil?

The individual molecules that make up olive oil is held together by weak forces, that arise from van der Waals' forces, which is attraction due to small, unevenness in electron distribution (which leads to very small positive and negative charges on the molecules that subsequently attract other molecules).
This is overly simplified concept though so it's not entirely accurate.
Bromine atom is very large (hence bromine, which molecules consists of only two bromine atom, exists as a liquid in room temperature). The more the electron, the more likely there are some unevenness in electron distribution, and hence stronger attraction between molecules. Also, bromine loves electron so it tends to pull electron closer to itself, creating a small but rather permanent positive and negative charge on the molecule. This also strengthen the force between brominated oil molecules, hence the viscosity goes up.

Viscosity goes up because less double bondings makes the fatty acids more “packable”

This is rather about melting point of the fat / oil, as when freezing, the molecules in liquid need to "pack" into a solid structure. In liquid however, molecules tends not wanting to maintain the "packing", or ordered structure. Adding bromine does increase the melting point but that's another story.


text:
How do we know if this following molecule is a superimposable mirror image or non-superimposable mirror image ? [See Image attached]
How do we know if this following molecule is Super Imposable mirror image or Non Super Imposable mirror image ? [See Image attached]

How do you know your left and right hands are not superimposable? (Or maybe they are?)

Yes they are super imposable ? Because if we keep the mirror image of right hand on right hand , it's the same .

Super impossable mirror images means 2 mutualy mirror images, where you can identify one such an image with the other by its translation and rotation.  Involve your imagination.

@Nityoday Tekchandani  No, left and right hands are not super impossbale. You cannot formally by a geometry way identify the left(right) hand and its mirror image right(left), implying there is a mirror symmetry.

Okay, so according to what I understood, that a real object and it's mirror image, if they look same , they are super imposable , else non super imposable ? Also, then the question's answer would be Super Imposable ?

The real task is to let you learn why it is so,, not to tell you yes or no.

@Poutnik what if we rotate our hands 180° and then superimpose them, then it will be superimposible .

@user226375 No, they will not. They do not have a plane of symmetry.

user226375 Take in your thoughts your left hand. Turn it for its thumb to point to the left as the right hand does. Now, can fingers of the left hand band downwards as easily as fingers of right hand ? Are the nails on the same side ? And fingerprint areas ? etc...

Take 2 round, soft objects like potatoes or apples.
Pin into each 4 pins of 3 different kinds ( 1 + 2 + 1 ),
pointing to the most different directions.
Call the 1st pin bromine, the 2nd one hydrogen, the 3rd one ethyl.
Do the same for the other potato to be the mirror image of the first one.
Try to rotate one of them to look like the other.
If you succeed, they are superimposable.
Now, replace one of the same pins on both objects by another -
4th kind - of pins in such a way that they are still mutual mirror images.
Try to match them the same way you tried previously.
If you succeed, they are superimposable.

The summary:

While there are 2 hydrogen atoms on a carbon atoms - or more generally, there are only 3 kinds of groups on a C atom - the groups/atoms of the same kind are interchangeble and you can rotate the molecule to match the mirror image.
If there are 4 different groups on C atoms, you cannot make such a match.
Said by other way, the former case has a plane of symmetry, the latter case does not.

Can we rotate the mirror image in x,y,z all the directions ?

Do you have in mind any reason why we cannot ? Imagine 2 pieces of LEGO. How would you determine they are identical ? You rotate them all the ways until you see they are the same - or not.

Okay, I tried this ,

Step one : https://m.imgur.com/gallery/1cLeqQd
Step two : https://m.imgur.com/gallery/Zyy2xfx
I can successfully rotate the mirror image to get it like the first one !!
Hence I've come to a conclusion that these are super Imposable.
Am I correct ?

for some reasons I cannot get the images. but if you talk about the fiirst 1+2+1 case, you are correct.

The links were wrong .. Try these :Step 1 : https://i.imgur.com/1cLeqQd.jpg  ..... Step 2 .... https://i.imgur.com/Zyy2xfx.jpg .. and I've understood the concept . Thank you

Ohkay , I just realised something, that , When we rotate the mirror image , the Wedge and Dash hydrogens replace each other, but they are with the same type of pin , hence that's okay ..? or it should be non super imposable if it's not okay to interchange wedge and dash hydrogens while rotating ..

Yes, until there are 2 hydrogens ( or more generally, there are only 3 kinds of groups on the C atoms, the groups/atoms of the same kind are interchangeble and you can rotate the molecule to match the mirror image. but if there are 4 different groups on C atoms, you cannot make such a match. Said by other way, the former case has a plane of symmetry, the latter case does not.

nice pins and table, but those carbons...  :-)

hehe those carbons are going from mental torture

About rotating to match the mirror image: https://chemistry.stackexchange.com/questions/109970/can-we-rotate-a-mirror-image-by-any-angle-to-check-if-it-is-superimposable

I will have a look, @KarstenTheis . Thank you ..

Note that  molecules can be achiral ( identical with mirror image ) even if it has 2 chiral carbon atoms ( atoms with 4 different substituents). as example see the meso-tartaric acid.

@Poutnik Alright :)


text:
Chemical Equilibria - Fraction Bound equation based on Kd, [R], [L], [RL], is actually an average?

Let's say 150 stem cells express exactly 150 copies of the same mebrane antigen X. While all differentiated cells express exactly 15 copies of X. To the cell mixture you add a large amount of 0.2 μM solution of antibody to X. The $K_D=$ 4 μM.
If you experimentally analyze the number of antibodies bound per stem cell, you will find that it is not constant.  Explain why, even for cases where the number of antigens expressed is the same, the binding follows a binomial distribution?

I found the fraction bound to be $\frac{1}{\frac{K_D}{\left[L\right]}+1\:}=f_{bound}$

I scaled the [L] contribution by the number of antigens expressed per cell, so for the differentiated case, I scaled 0.2 μM by 15, and for the normal case, I scaled 0.2 μM by 150.
the differentiated case has a lower $f_{bound}$ because it has less ligands.

To describe the binomial distribution for binding part

the antigens exist in the environment of the beaker (or wherever the reaction is taking place) so they can exchange antigens? resulting in some cells $f_{bound}$ being lower. How exactly the interaction takes place occurs stochastically.  <--- want help clarifying this part

When a reaction is at equilibrium, there will be stochastic fluctuations. These are usually not measurable (for example, if we have more than a mole of products and reactants in our reaction mixtures).
In this specific case, the fluctuations are easy to measure because the total number of antigen molecules per cell is countable, and you are separating cells (I presume) before analyzing the number of antibodies bound.
If you expect a fluctuation of $\sqrt{N}$ for N objects, that is significant for $N= 150$ but not for $N = \pu{6.022e23}$ when expressed relative to $N$.

Is it true, that the antigens produced by one cell of the same type can be used in a neighbor cell, ie can it pass into solution?

The antigen is covalently linked to the cell surface. The antibodies are shared in this experiment.


text:
For a benzene shown in a skeletal structure, what does a substituent to the center of the ring mean?

I have seen diagrams with both the substituents attached to the benzene ring by the edges and other occasions directly to the delocalized ring. What does signify?

It means "must be someplace there, dunno where exactly". Also, methanol group is not a thing at all; that's methoxy group.

Related: Structural representation of substituted benzene

Here the bond connecting the substituent with the center of the ring suggests that the substituent may be located on any of the three positions (ortho-, meta-, para-). From Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008), section GR-9.1 Small substituents [1, p. 393] (emphasis mine):

Parent structures with a substituent that is attached at one of several ring atoms may be depicted by drawing that substituent attached to a bond that extends into the center of a ring. In printed works, it is assumed that the substituent may then be attached to any of the ring atoms as is allowed by normal bonding rules. When working in electronic formats, it may be possible to specify that the substituent
  can be attached to certain of the ring atoms and not to others, depending on the capabilities of the software being used. If more than one substituent may be independently connected to various ring atoms, the bonds attached to those substituents should both extend toward the center of the ring, but should not
  themselves be connected to each other.

References

Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). Pure and Applied Chemistry 2009, 80 (2), 277–410. https://doi.org/10.1351/pac200880020277. (Free Access)

For the sake of completeness, very related: Markush Structure.


text:
Problem in terms of pC-pH plotting
I recently confronted a mission to plot a pC-pH graph('p' stands for the p-value notation, and 'C' stands for the concentration of a species), which meets the conditions below:

Plot the pC-pH graph when a monoprotic acid is added - salts and strong acids can be co-added.

Therefore, I initally assumed that the monoprotic acid to be $\ce{CH3COOH}$, strong acid to be $\ce{HNO3}$, and the salt to be $\ce{NaCl}$ in order to avoid the overlapping of the species.
But it led me to so much unknowns: so I changed the species to $\ce{CH3COOH}$, $\ce{HCl}$, $\ce{CH3COONa}$ each. However it still didn't quite work out. Is it even possible to plot this?
This is actually supposed to be an environmental engineering problem, so there might be some assumptions which won't make the circumstance way accurate but make it simpler.

Have a look at the answers https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566 and  https://chemistry.stackexchange.com/questions/100346/calculate-ph-of-a-mixture-of-a-strong-base-and-acid-knowing-only-the-ph-wt-v/100355#100355

@porphyrin So I know that those cases, especially the former one 'How to set up equation for buffer reaction?' is theoretically pointing the same notions as what I've asked. But the condition is different: in this case I have one more initial species and that's why I'm having trouble.


text:
Impact of pH? Change in Binding Affinity of protein-ligand complex based on pH and Kd?

When the $\mathrm{pH}$ was $5.0$, the temperature was $\pu{25 °C}$,the $K_\mathrm{D}$ was $\pu{5 μM}$, and [L] was equal to $K_D$, the protein(s) were half bound.

Question considers half bound condition where $[L]= _$

If $K_\mathrm{D} = \pu{20 μM}$ and $\mathrm{pH} = 6.5$, what is the impact on binding affinity?

I understand that $K_\mathrm{D}$ is the inverse of $K_\mathrm{a}$ and that $K_\mathrm{D}$ describes the degree of dissociation at equilibrium for the reaction, thus a low $K_\mathrm{D}$ corresponds to a high binding affinity complex.
I don't understand how to account for $\mathrm{pH}$ change though.

It would be nice to define what RL and $K_\mathrm{D}$ are.

Whoops. Question considers half bound condition where [L]= $K_D$

Since you are not given (or you don't provide) more information about how pH affects the protein, other than Kd, I would assume it's not really important. The point would be that the affinity goes down at the higher pH.

@NightWriter  - Yes, that was all the information I was given. But don't some proteins do better @ higher pH than lower pH?

Sure, but this is presumably just an exercise for you to reveal that you understand that lower Kd means higher affiinity and vice-versa...


text:
Is there any diprotoic acid where both pka's are less than 0?
The question states it all.
I'm working out special cases for a different problem and such an acid would create another special case.
Sulfuric Acid is the strongest that I remember. pKa's −3, 1.99

I guess any disulfonic acid with sufficiently long organic link in the middle will do.

@IvanNeretin - Super fantastic! Thank you.

@MaxW wants superacidic MHA buffer....  :-)

@Poutnik - Yes, the idea was to find an "amphoteric" salt of such an acid, eg Sodium hydrogen 1,2-Ethanedisulfonate.

Well I did look up 1,2-Ethanedisulfonic acid which exists, and the disodium salt exists. You could perhaps make the monosodium salt by reacting the acid with NaOH and drying. Does the fluorine analog even exist?

What about $$\ce{ HO-SO2-CF2-CF2-SO2-OH}$$ :-)    ( See https://en.wikipedia.org/wiki/Triflic_acid   pKa −14.7±2.0 ).

It is an intriguingly fascinating  molecule. I'd guess that either it is unstable, or that someone, somewhere, has made it. // The question with a lot of such molecules is not "Is it useful?", but rather "Can I do it?"

Ethanedisulfonic acid does indeed have $pK_{a1}=-2.06, pK_{a2}=-1.46$.
Using polycyano substutution can also work.  This paper reports the table below, note that the tetracyano hydroquinone "makes it" (in the excited state produced by irradiation).

The Wikipedia link for Methanedisulfonic acid is only for the monosulfonic acid. The disulfonic acid does exist, but I couldn't find any pKa values for it. However both pKa values would surely be less than 0.

Removed.  Misread something back then, I guess.


text:
How to calculate dew point of air-water system
I want to calculate the maximum water holding capacity of air at a given temperature (as in $\frac{max\ mass\  of\   water}{mass\  of\  dry\  air} $ vs temperature)
I looked up online and there are various models to do such calculations. However for most of them the temperature range is not more than 100 C. Are there any relations which cover a higher temperature range, upto 300-400 C?
If not, is not possible to determine saturation point above the boiling point of water?
Edit: I am considering a constant atmospheric pressure. Thus, I want to check variation only with temperature.

Above the boiling point, water becomes a genuine gas, much like air. Gases don't have saturation points. They just mix.

You can get above 100 C by increasing the pressure of course.

@IvanVeretin Water becomes a gas above the critical point, not the boiling point.

@Chet Miller No, liquid water starts becoming gas by boiling, when temperature reaches boiling point. The water phase diagram. The critical point just means above this temperatures water does not form separate phases, as their properties at given p,T just converged to the same values.

The conceptual mistake is, that the holding water vapour capacity is property of air. It is not. It is property of space. The air is just an obstacle and bystander.( with slight difference due nonideality).

@Poutnik The boiling point of water is a function of pressure, so 100 C is no magical number.  The terminology I was taught is that water in the gas phase at temperatures below the critical temperature is referred to as a "vapor," but above the critical temperature, it is referred to as a "gas."

@Chet Miller Notice that I have not mentioned 100C at all. So you say air contains CO2 vapours, as it is below its critical point, correct ? A vapour is just naming for gaseous phase of a substance, that we implicitly consider being in condensed phase. By other words, every vapour is a gas in sense of chemistry and physics. The terminology of technology and citizen life may differ.

BTW, the maximal mass of water in given mass of air at given temperature is reciprocally proportional to pressure. 0.5 atm air can contain twice as much water than 1 atm air and 4 times more than 2 atm air. So the original question does not make much sense.

To clarify, I am considering all this at atmospheric pressure. I just want to see the variation with temperature

How to calculate dew point of air-water system? I want to calculate the maximum water holding capacity of air ... However for most of them the temperature range is not more than 100 C. Are there any relations which cover a higher temperature range, up to 300-400° C? If not, is it not possible to determine saturation point above the boiling point of water?

At very high temperatures I could only find the Mollier enthalpy-entropy chart (for steam) here's one for up to 1931.67 degrees rankine (800 degrees celsius) and another for up to 2200 degrees rankine (949.072 degrees celsius):

Source: "Mollier H-S Diagram"

Source: "Mollier enthalpy-entropy chart for steam, US units"
Source: Wikipedia - Enthalpy–entropy chart

On the diagram, lines of constant pressure, constant temperature and volume are plotted, so in a two-phase region, the lines of constant pressure and temperature coincide. Thus, coordinates on the diagram represent entropy and heat.
The Mollier diagram coordinates are enthalpy $h$ and humidity ratio $x$. The enthalpy coordinate is skewed and the constant enthalpy lines are parallel and evenly spaced.

See the Wikipedia webpage for more information, or assistance reading the diagram.
For restricted ranges (of which you complained, but are useful for comparison and to view the bottom of the chart) see Wikipedia's webpage for "Psychrometric charts":

Source: "Psychrometric Chart for Sea-level pressure using SI units"
Wikipedia's webpage for "dew point temperature" shows some calculations and means to make simple estimates, but it cuts you off at 100°C.


text:
Balancing redox equation with oxygen on both sides?
The problem is to balance this equation in an acidic medium.
$$\ce{As2O3 (s) + NO3- (aq) -> H3AsO4 (aq) + N2O3 (aq) }$$
However, I am getting stuck balancing the oxygens after dividing them into half reactions.
How do you balance a redox reaction when oxygen is involved in both sides of the equation?

How do you balance it when oxygen (or any element, for that matter) is not present on either side? You simply won't be able to balance it then. That's the very idea of balancing. You must have as much oxygen (and any other element too) on the left as on the right.

Refer the following link. https://www.periodni.com/balancing_redox_equations.php?eq=As2O3%2BNO3-%3DH3AsO4%2BN2O3

It is always easy to balance redox equations by considering two half reactions. The two relevant half reactions to your redox equation are:
$$ \begin{align}
\ce{As2O3 (s) -> H3AsO4 (aq)} \\
\ce{NO3- (aq) -> N2O3 (aq)}\\
\end{align}
$$
First balance $\ce{As}$ and $\ce{N}$ in both equations. Then, balance $\ce{O}$ in both equations with water and $\ce{H}$ with $\ce{H+}$:
$$ \begin{align}
\ce{As2O3 (s) + 5 H2O (l) -> 2 H3AsO4 (aq) + 4 H+ (aq)}\\
\ce{2 NO3- (aq) + 6 H+ (aq) -> N2O3 (aq) + 3 H2O (l)}\\
\end{align}
$$
Finally, balance the charges by electrons:
$$ \begin{align}
\ce{As2O3 (s) + 5 H2O (l) -> 2 H3AsO4 (aq) + 4 H+ (aq) + 4 e-}\\
\ce{2 NO3- (aq) + 6 H+ (aq) + 4 e- -> N2O3 (aq) + 3 H2O (l)}\\
\end{align}
$$
Now add the last mass and charge balanced two equations in order to cancel electrons (this is easy, because both have 4 $\ce{e-}$s):
$$ \ce{As2O3 (s) + 2 NO3- (aq) + 2 H+ (aq) + 2 H2O (l) -> 2 H3AsO4 (aq) + N2O3 (aq)}$$
Or simply,
$$ \ce{As2O3 (s) + 2 NO3- (aq) + 2 H3O+ (aq)  -> 2 H3AsO4 (aq) + N2O3 (aq)}$$


text:
Heat and work of glucose combustion
$$\ce{C6H12O6(s) + 6 O2(g) -> 6 CO2(g) + 6 H2O(g)}$$
When looking at glucose combustion I found that the change in Gibbs free energy is greater than the change in enthalpy. If we assume the process is reversible we can set the change in Gibbs free energy to be the work available and if we assume there is no work we can set the enthalpy to be the heat. Why then would the work be less than the heat? I suspect that $\Delta G = \Delta H-\Delta (TS)$ is at work.
Cool, so can you substitute work, W and heat, Q into the equation and write W=Q−TΔS to directly show that ΔS>0? Also would you please explain how an increase in entropy would imply the discrepancy between heat and work (not mathematically).

Yes, and since you are going from a large molecule and some oxygen gas to lots of carbon dioxide and water, you can guess that the entropy increases.

Cool, so can you substitute work, $W$ and heat, $Q$ into the equation and write $W = Q - T \Delta S$ to directly show that $\Delta S >0$? Also would you please explain how an increase in entropy would imply the discrepancy between heat and work (not mathematically).

For the process to go forward, the entropy of the universe has to increase. For this specific case, the reaction is exothermic and the entropy of reaction (entropy change of the system) is positive. If all of the enthalpy is released as heat, the entropy of the universe will increase (the entropy of the surrounding increases because of the heat transfer, and the entropy of the system increases because molecules are more dispersed in the gas phase).
If there is a mix of work and heat exchange, you just have to make sure the entropy of the universe still increases. In this case, the work possible exceeds the enthalpy change. When you do maximal work, you have heat transfer into the system. The entropy decrease of the surroundings is offset by the entropy increase in the system.
So in this case $$|\text{work}| > |\Delta H|$$
For a picture, see https://chemistry.stackexchange.com/a/112958 scenario 2.

Cool, so can you substitute work, W and heat, Q into the equation and write W=Q−TΔS to directly show that ΔS>0?

No, where there is work, enthalpy change is no longer equal to heat.
Also, you have to specify which ΔS you are talking about. If it is the entropy change in the system (due to the reaction) $\Delta S_\text{sys}$, the equation below is a variation of the second law (the left-hand side is equal to the entropy increase of the universe times temperature):
$$T \Delta S_\text{sys} - q_\text{rev} > 0$$
And if you want to relate work to enthalpy, you use the first law:
$$\Delta H = q + w$$
and insert it into equation above to get:
$$T \Delta S_\text{sys} + w - \Delta H > 0$$
or for maximal work with a bit of rearrangement,
$$  w_\text{max} = \Delta H - T\Delta S_\text{sys}$$
So you can say that if the work you get out of a reaction is larger than the enthalpy change, the entropy of the reaction has to be positive.


text:
Is the molar mass of a molecules lowered by the binding energy of the chemical bonds?
When calculating the molar mass, is there any measurable difference in total mass of the molecule due to the loss of binding energy?

The difference is real, so it is measurable. What kind of balance do you have?

Einstein showed that mass and energy are interchangeable through the equation $E=mc^2$. So an exothermic reaction loses mass, an endothermic reaction gains mass. However the mass change in chemical reactions is vanishing small and is ignored. For example in chemistry the molar mass of HCl is the atomic mass of H plus the atomic mass of Cl.

@karsten what do you mean when you say "what kind of balance do you have"?

@MaxW Thanks for the explanation- I was trying to reconcile the concept of molar mass and binding energy.

@suse - She means what kind of balance do you use to weigh chemicals. There isn't a balance that can weigh with enough precision so as to measure the mass change of an explosion.

So you are saying then that the mass change isn't measurable, practically speaking?

The difference is the best observed for products of nuclear reactions, related to nucleon binding energy due the strong force.
By comparing the mass of $\pu{2 mol}$ of $\ce{^1H2}$ and $\pu{1 mol}$ of  $\ce{^4He}$, the difference is about $0.7\%$.
The chemical bonds have difference more then million times smaller.
E.g. if there is an exothermic reaction producing $\pu{500 kJ/mol}$, the mass difference per mol is $$\Delta m = \Delta E/c^2 = \pu{-5.56 ng }$$
That means, we have hardly ever scales with the needed accuracy.

Mass energy equivalence give the relevant answer and it is very small. A hydrogen molecule weighs about 4ng per mol (2g) of hydrogen less that two hydrogen atoms (using a little mental arithmetic so don't hold me to the exact number).

Sure, I do not say otherwise.

I didn't mean to imply you didn't: I was just using your calculation on a specific chemical reaction to put it in context.

Yes the binding energy of the electrons results in a loss of mass.
$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Particle} &  \text{Mass in AMU} \\
\hline
\text{Rest mass electron}     &  0.&\hspace{-1em}000548579909070(16) \\
\text{Rest mass proton}       &  1.&\hspace{-1em}007276466879(91)\\
\text{Sum rest mases}         &  1.&\hspace{-1em}007825046788\\
\hline
\text{Hydrogen atom}          &  1.&\hspace{-1em}00782503224(9)\\
\hline
\text{Hydrogen atom - Sum Rest masses} & -0.&\hspace{-1em}000000014548\\
\hline
\end{array}
$$
Difference is $1.4548\cdot10^{-8}$ amu.
1 amu is equivalent to an energy of 931.5 MeV.
$1.4548\cdot10^{-8}\times 931.5\cdot10^6 = -13.55$ ev
Guess what the binding energy of the hydrogen electron is?

A slightly better example would be to use the binding energy of two hydrogen atoms forming dihydrogen since that is a stable chemical molecule with a bond in it. A hydrogen atom forming from an ion plus an electron isn't really forming a chemical bond.


text:
Variational method: Hydrogen atom ground state in STO-3G basis expansion
I computed numerically the ground state energy of hydrogen by variational procedure, firstly expanding the wave function over the s-wave basis set STO-3G
$$\psi(r)=\sum_{i=0}^3 C_{i} e^{-\alpha_ir^2}$$
and then solving the generalized eigenvalue problem (in matrix notation)
\begin{equation}
\textbf{HC}=E\textbf{SC}
\end{equation}
As output I have all 3x3 matrices. $E$ is a diagonal 3x3 matrix and my eigenvalue is the smallest of the three on the diagonal.
My question is: what do the other two elements in the $E$ diagonal represent?

The eigenvalues you get are the energies of your orbitals. These energies don't necessarily have any physical meaning except for the fact that they can be approximately related to the true ionization energy by Koopmans' theorem.

In the case of Hydrogen I get -0.49501 H and other two positive values. These positive values are energies of which orbitals?

@Liuuuuk Orbitals are, by definition, bound states, hence have negative energy. The positive eigenvalues hardly represent anything of chemical significance. Try a larger basis set so that it can approximate both 1s and 2s orbitals, and you should get an eigenvalue close to -0.125.


text:
Derive expression for average heat flux in heat exchanger

The single-tube heat exchanger is used to increase the temperature of fluid
  I from $T_{b1}$ to $T_{b2}$. The heating oil (fluid II) used
  for this task enters at temperature $T_{02}$ and leaves at
  temperature $T_{01}$.
a) Give the expression that describes the average heat flux from fluid II to fluid I.

I don't really know how to derive the average heat flux but the equation I immedietly think of using is:
$$q = U(T_{b1} - T_{b2})$$
Still I am not sure how I should calculate the overall heat transfer coefficient and also I recall being told that when dealing with heat exchangers you should use the logarithmic temperature difference. So the expression should be:
$$q = UΔT_{\ln}$$
where the logarithmic temperature difference is
$$ΔT_{\ln} = \frac{(T_{01} - T_{b1}) - (T_{02} -T_{b2})}{\ln\frac{T_{01} -T_{b1}}{T_{02} - T_{b2}}}$$
And an attempted solution for the overall heat transfer coefficient is:
$$U = h_1 + \frac{k}{R_0\ln(R_a/R_0)} + \frac{h_2R_a}{R_0}$$
but I feel like this isn't correct since it is used when calculating the heat transfer in a heat exchanger and not the heat flux. Is it correct or is the heat flux for a heat exchanger calculated in a different way?
All help is greatly appreciated!


text:
Calculate the energy in the form of heat (in kJ) required to convert 325 grams of liquid water at 20.0 °C to steam at 115 °C

Calculate the energy in the form of heat (in kJ) required to convert 325 g of liquid water at 20.0 °C to steam at 115 °C. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities: liquid water = 4.184 J/(g K), steam = 1.92 J/(g K)

Here is what I did:
$$Q=mcT$$
$$\begin{align}Q &= 325 ( 4.184) (100 - 20)\\
&= 325 ( 4.184 ) ( 80)\\
&= 108784\ \mathrm J\end{align}$$
$$\begin{align}Q&= ml\\
Q&= 325 ( 2256)\\
&= 733200\ \mathrm J\end{align}$$
$$\begin{align}Q &= 325 (1.92) ( 115 - 100)\\
&=325 (1.92) ( 15)\\
&= 9360\ \mathrm J\end{align}$$
$$(108784+733200+9360) \mathrm{J}= 851\ \mathrm{kJ}$$
But the correct answer is $735\ \mathrm{kJ}$
Do you know the right way to do it?

Always write the values with the correct units and carry the units through the calculation. Do not omit the units while performing intermediate steps and do not just reintroduce units at the end of the calculation.

If you mess up the first part by a factor 10 and switch two digits when writing down the answer, you get 735 kJ. I would not call that correct.


text:
Why is a compound that is more stable gives out less energy upon reaction?
I am having trouble relating stability with energy released.
For example, I am looking at a statement regarding Benzene that says, since delocalised electron model of Benzene is more stable than the kekule model that has three double bonds, it gives out less energy  than the kekule model upon reaction.
I am trying to relate this stability back to my learning of electron stability on different shells. The closer to the nucleus an electron is, the more stable it is,  the more energy is released to remove it.
Why is the idea of stability completely different here?

Energy is not released when you remove an electron or break a bond.

It takes energy to break a bond or to remove an electron from an atom, not release energy. If you correct that part in your statements, the rest will be consistent.

I am sorry, I deleted the bond-related statements.

Now what is your question, really? Both examples essentially say: "has more energy = less stable".

"The closer to the nucleus an electron is, the more stable it is, the more energy is released to remove it. " This is the problem, energy is required to remove an electron from an atom. Try correcting this and see if it helps your logic.

duplicate of https://chemistry.stackexchange.com/questions/108726/why-total-heat-of-hydrogenation-of-1-3-cyclohexadiene-is-more-than-that-of-benze

I think that your question revolves around the evidence that is normally given in organic texts around the stability of aromatic compounds such as benzene.

chemguide.co.uk
The idea here is that addition reactions with benzene (such as hydrogenation) that involve bond formation that is stronger than the pi bonds that are broken are less exothermic than similar reactions with hypothetical cyclic alkenes such as cyclohexatriene. The implication is that there is some stabilization of benzene beyond what is expected from simple alkenes.
This is the main basis for the theory of aromaticity in molecules that obey the rules (planar, cyclic and having 2n + 2 pi electrons). If you wish to learn more about why benzene is more stable than expected, see this great question:
Delocalised electrons and stability of benzene
There is no conflict with the understanding of electrons in lower potential energy orbitals being more stable. It is an extension of the same idea, just using molecular orbitals rather than atomic orbitals.


text:
How long do I have to burn a carbon steel pan laced with flaxseed oil?
I've a carbon steel pan that's 26 cm in diameter, about 2 mm thick, but I doubt that it matters. I haven't been able to find a reliable resource that would tell me the smoke point of flaxseed oil, but I believe it's around 180 degrees Celsius.
I pour about 5 ml of oil on the pan, rub it in every nook and cranny, and then wipe "all" the oil off with a paper towel, and stick it in the oven set to 250.
How long should I keep the oven on? People have differing opinions on that; one says an hour, other 30 minutes, and I've even seen someone say 15 minutes.
I've done some field testing, and I nowadays burn the pan for 30 minutes, but I wonder if I could go lower. It's not just a matter of testing it, as a bad layer ruins the pan, it will flake and has to be sanded, and burnt at least 3 times after that.

You might get more traction on this on our sister site Seasoned Advice (cooking and related).

I think that site will get me biased answers. I want the science behind it.

I doubt you'll get anything else here. Quantitative scientific measures on the seasoning are hard to get. It's so thin and largely amorphous ... I never bothered about old layers, though. Does it make a difference if you remove them?

Well, I guess the most important underlying question is how long does it take for the oil to stop smoking. The layer of it is so thin that it should burn off relatively fast. The underlying layers make it better overall, it's less likely to break and it's forgiving if you scratch it. I guess it heats slightly slower with the old layers but how big of a difference can it be.

Use organic 100% flaxseed oil, unless you prefer grape seed oil.
Apply the flaxseed oil to a pan preheated to 200°F.
Remove all the flaxseed oil with a paper towel (some will remain).
Place the dry pan in oven and heat to 500°F (or 450°F if that's as high as your oven goes) and preheat the pan with the oven.
When the oven reaches temperature set a timer for one hour.
After one hour turn the oven off. Leave the pan in the oven with the door closed.
Wait two hours for the pan in the oven to cool.
Repeat the thin oil application and baking six times.
Once the pan developed a bit of a sheen it's ready.
Always use the thinnest possible coat of oil.

References:
Science:

CastIronCollector.com has a science based article: "Cast Iron Seasoning" (the polymerization of the oil and it's carbonization is what's relevant, not whether the pan is cast iron or carbon steel). They recommend grape seed oil.

Note: The cooks at ChowHound disagree with Cheryl's science but are satisfied with the result. There is also some disagreement at Cooking.SE, links below.

Cheryl's Blog - "Chemistry of Cast Iron Seasoning: A Science-Based How-To" (Same principal as CS):

The seasoning on cast iron is formed by fat polymerization, fat polymerization is maximized with a drying oil, and flaxseed oil is the only drying oil that’s edible.
...
Fat polymerization can be triggered or accelerated in a variety of ways. As best I can tell from my reading, the cast iron seasoning process is an example of “radical polymerization”. The process is initiated when something causes the release of free radicals in the oil. The free radicals then “crosslink” to form the tough, hard film you see in a well-seasoned pan.
So what is the “something” that initiates the release of free radicals in fat? Iron, for one thing. High heat, light, and oxygen, for some others. To prevent cooking oils from going rancid – i.e., breaking down and releasing free radicals – you need to store them in dark, tightly sealed containers in a cool location. To initiate or accelerate the release of free radicals, put the oil in contact with bare iron and heat it above its smoke point, which will cause even non-drying oils to release free radicals
...
Buy an organic flaxseed oil. You don’t want to burn toxic chemicals into your cookware to leach out forever more. It’s a fairly expensive oil. I paid $17 for a 17 ounce bottle of cold-pressed, unrefined, organic flaxseed oil. As it says on the bottle, shake it before you use it.
Strip your pan down to the iron using the techniques I describe in my popover post. Heat the pan in a 200°F oven to be sure it’s bone dry and to open the pores of the iron a little. Then put it on a paper towel, pour a little flaxseed oil on it (don’t forget to shake the bottle), and rub the oil all over the pan with your hands, making sure to get into every nook and cranny. Your hands and the pan will be nice and oily.
Now rub it all off. Yup – all. All. Rub it off with paper towels or a cotton cloth until it looks like there is nothing left on the surface. There actually is oil left on the surface, it’s just very thin. The pan should look dry, not glistening with oil. Put the pan upside down in a cold oven. Most instructions say to put aluminum foil under it to catch any drips, but if your oil coating is as thin as it should be, there won’t be any drips.
Turn the oven to a baking temperature of 500°F (or as high as your oven goes – mine only goes to 450°F) and let the pan preheat with the oven. When it reaches temperature, set the timer for an hour. After an hour, turn off the oven but do not open the oven door. Let it cool off with the pan inside for two hours, at which point it’s cool enough to handle.
The pan will come out of the oven a little darker, but matte in texture – not the semi-gloss you’re aiming for. It needs more coats. In fact, it needs at least six coats. So again rub on the oil, wipe it off, put it in the cold oven, let it preheat, bake for an hour, and let it cool in the oven for two hours. The picture above was taken after six coats of seasoning. At that point it starts to develop a bit of a sheen and the pan is ready for use.
If you try this, you will be tempted to use a thicker coat of oil to speed up the process. Don’t do it. It just gets you an uneven surface – or worse, baked on drips. Been there, done that. You can’t speed up the process. If you try, you’ll mess up the pan and have to start over.
The reason for the very hot oven is to be sure the temperature is above the oil’s smoke point, and to maximally accelerate the release of free radicals. Unrefined flaxseed oil actually has the lowest smoke point of any oil (see this table). But the higher the temperature the more it will smoke, and that’s good for seasoning (though bad for eating – do not let oils smoke during cooking).
I mentioned earlier there’s a myth floating around that vegetable oils leave a sticky residue. If the pan comes out of the oven sticky, the cause is one of three things:

You put the oil on too thick.
Your oven temperature was too low.
Your baking time was too short.

There is also a comment section below the article.

For cooking related advice:

Cooks Illustrated - "How to Season a Carbon-Steel Pan" (Best Method)
CarbonSteelCookware.com - "A Carbon Steel Pan Seasoning Method that Actually Works" (More Details)
The Kitchen Professor - "Best Oil for Seasoning Carbon Steel Pan" (Best oil: grape seed)
Wikipedia - "Smoke point of cooking oils":

$$\begin{array}{lcr}
\text{Fat} & \text{Quality} & & \; \text{Smoke Point} \qquad \qquad \; \\
\hline
\text{Flaxseed oil} & \text{(Unrefined)} & \text{107°C} & \text{225°F} \qquad \\
\text{Lard} & & \text{190°C} & \text{374°F} \qquad \\
\text{Grape seed oil} & & \text{216°C} & \text{421°F} \qquad \\
\text{Butter} & \text{(Clarified)} & \text{250°C} & \text{482°F} \qquad \\
\text{Sunflower oil} & \text{(Neutralized)} & \text{252-254°C} & \text{486–489°F} \qquad \\
\text{Safflower oil} & \text{(Refined)} & \text{266°C} & \text{510°F} \qquad \\
\text{Avocado oil} & \text{(Refined)} & \text{270°C} & \text{520°F} \qquad
\end{array}
$$
Answer on Cooking.SE opposing the use of high iodine oils recommends "whatever makes a nice firm bar of soap - lard, palm oil, coconut - is a good choice here".
That answers "the science" behind seasoning, offers cooking experts advice, and says what, why and how long. Admittedly there's some disagreement. Your heat and times are too short - that is both my experience and a commonality amongst the advice offered.
Video demonstration by Chef Rich with brief scientific explanation, showing the result of coating eleven times on the stovetop using flaxseed oil. Very little smoke and very shiny black finish.

The cooking SE answer that you linked. The author doesn't know what they are talking about. Recommends lard or coconut oil. My father has the same opinion; lard is best. It probably works fine for cast iron, but I tried lard once on carbon steel, and I had to sand the entire pan down because it was the worst coating I had ever seen on a pan, and it worked about as well as it looked like. And coconut oil has a smoke point so high that it doesn't even polymerize properly.

Sorry for being that guy but I've one more. That chart you got from Wikipedia. It attributes to this page, which is nowhere near credible: https://jonbarron.org/diet-and-nutrition/healthiest-cooking-oil-chart-smoke-points. It's a marketing site that seems to idolize some dude called Jon Barron. It also doesn't tell how the information was obtained, and I doubt that the author poured flaxseed oil in a glass, and heated it slowly while monitoring it's temperature.

Just a brief reminder to keep it friendly in the comments, please.  I think the comments above are okay because they are refuting the content and are not any type of personal attacks.

@jonsca I disagree, the statement about another SE user: "The author doesn't know what they are talking about." is in fact a personal attack and R/A - it is also unsupported by a link to an authoritative source. Badgering, berating and an unfriendly tone isn't the way to ask for clarification or improvement. That said, moderate as you see fit - I will ignore any rudeness and not respond. My "New Contributor" tag means that the experience should be better.

I personally find this answer quite hand-wavy as it for my taste does not contain enough science/chemistry about what is going on on the molecular level. I also would not consider any of the sources credible enough for a scientific discourse. I presume that the seasoning of kitchenware is scientifically not-so-well explored. I am sorry that the experience is not what you have expected, I hope that does not deter you from contributing more.

@Martin-マーチン To offer improvements I returned to see the cleanup and a better framed request. I prepared to spend the time/cost to make a really great answer. I'll cut my losses and help someone else with a real interest in obtaining the best answer. I retired at 20K but I expect I'll be back here sometime. As for the NC Indicator Tim's first link is very clear: "Be clear and constructive when giving feedback, and be open when receiving it.", etc.


text:
Adding an acetal to basic water
If you add an acetal to a solution containing water with a little bit of NaOH, will no reaction occur due to the stability of an acetal in the presence of a base?

Pretty much. A key step of acetal hydrolysis is protonation of one of the oxygens.

Acetals are liable in acidic conditions; they are stable in basic conditions.


text:
Why does bromine water make the viscosity of olive oil increas?

Olive oil consists of almost 100% of fat. As you can see in the table of contents below, it includes both unsaturated, monounsaturated and polyunsaturated fats in the oil.

Nutrition declaration per 100 ml (= 92 g)
Energy 3390kJ / 823 Kcal
Fat 92 g Carbohydrate 0 g
Of which saturated fat is 13 g Of which sugars are 0 g
Monounsaturated fat 68 g Protein 0 g
Polyunsaturated fat 7.2 g Salt 0 g
If you add bromine water, Br2 (aq), to the olive oil, its consistency changes significantly and it becomes more viscous. What is it that happens chemically with the olive oil when adding bromine water and why does it become more viscous?

It is a similar process to hydrogenation of plant oil on Raney nickel, producing margarines.
The saturated fats have zig-zag structure of $\ce{C-C}$ chain, what makes them good in "molecule spooning", leading to high melting point.
The addition of bromine had a similar effect.
OTOH the cus double bonds of oils are obstacles for "spooning" and the menting point is low.
The bromine is, AFAIK, used to determine the amount of unsaturated bonds in edible oils.
$$\begin{align}
\small \ce{  \\
\small -CH2-CH=CH-CH_2 - + H2 &->[Ni] \small -CH2-CH2-CH2-CH_2 -\\
\small -CH2-CH=CH-CH_2 - + Br2 &-> \small -CH2-CHBr-CHBr-CH_2 -\\
}\end{align}$$

Fatty acids are components of many types of lipids. Fatty acids are carboxylic acids with very long hydrocarbon chains, usually consists of 12-18 carbon atoms long hydrophobic chain. Olive oil is a liquid obtained from olives, which contains mixture of fatty acids. According to Wikipedia:

The composition of olive oil varies with the cultivar, altitude, time of harvest and extraction process. It consists mainly of oleic acid (up to 83%), with smaller amounts of other fatty acids including linoleic acid (up to 21%) and palmitic acid (up to 20%).

The main component of olive oil, oleic acid, is a monounsaturated fatty acid (FA; See Table) and generally average about 70% in olive oil. Another monounsaturated FA, palmitoleic acid averages in the range of 0.3-3.5%. Olive oil also contains polyunsaturated FAs such as linoleic acid (~15%) and $\alpha$-linolenic acid (~0.5%). Only about 20% saturated FAs (see Table)) are present in olive oil and most of them are palmitic acid (~13.0%) and stearic acid (~1.5%).
Structures of Common Fatty Acids (saturated), of which, palmitic (~13.0%) and stearic acid (~1.5%) are present in olive oil:
$$
\begin{array}{ccc}
\text{Name of FA} & \text{# of carbons} & \text{Structure} & \text{Melting Point} \\ \hline
\text{Lauric acid} & 12 &   \ce{CH3(CH2)10CO2H}   & \pu{44 ^{\circ}C}  \\
\text{Myristic acid} & 14 &   \ce{CH3(CH2)12CO2H}   & \pu{58 ^{\circ}C}  \\
\text{Palmitic acid } & 16 &   \ce{CH3(CH2)14CO2H}   & \pu{63 ^{\circ}C}  \\
\text{Stearic acid} & 18 &   \ce{CH3(CH2)16CO2H}   & \pu{70 ^{\circ}C}  \\
 \hline
\end{array}
$$
Structures of Common Fatty Acids (unsaturated), three of which, oleic acid (~70%, linoleic acid (~15%), and $\alpha$-linolenic acid (~0.5%) are present in olive oil:
$$
\begin{array}{ccc}
\text{Name of FA} & \text{# of carbons} & \text{Structure} & \text{Melting Point} \\ \hline
\text{ Palmitoleic acid} & 16 & \ce{CH3(CH2)5CH=CH(CH2)7CO2H}  & \pu{-1 ^{\circ}C}  \\
\text{Oleic acid} & 18 & \ce{CH3(CH2)7CH=CH(CH2)7CO2H}   & \pu{4 ^{\circ}C}  \\
\text{Linoleic acid} & 18 & \ce{CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2H} & \pu{-5 ^{\circ}C}  \\
\text{ Linolenic acid} & 18 & \ce{CH3CH2(CH=CHCH2)2CH=CH(CH2)7CO2H} & \pu{-11 ^{\circ}C}  \\
 \hline
\end{array}
$$
According to the melting points of FAs in olive oil, one can conclude that it contains a lot of double bonds (mostly mono-unsaturated). These unsaturated fatty acids contain only cis-double bonds (e.g., oleic and linoleic acid). The presence of cis-double bonds has an important lowering effect on the melting point of the fatty acid (see Table), because cis-double bonds form rigid kinks in the fatty acid chains (see the picture of oleic acid):

Keep in mind that there is no rotation around a double bond, and as a result, the unsaturated fatty acids cannot line up very well to give a regularly arranged crystal structure. On the other hand, saturated fatty acids would line up in a very regular manner, better than that of unsaturated, leading to better van der Waals-forces, thus get closer together (thicker). This is one of the reasons why they are solids in ambient conditions with high melting points (see Table).
Now, what happens when you add bromine water to olive oil? Bromine added to double bond (same reaction as clarification test for unsaturation), making the fatty acid saturated. For example, when $\ce{Br2}$ added to $\ce{C18}$ oleic acid (m.p.: $\pu{4 ^{\circ}C}$), it'd become a resemblance of $\ce{C18}$ stearic acid (m.p.: $\pu{70 ^{\circ}C}$) with two extra $\ce{Br}$ atoms replacing 2 $\ce{H}$ atoms. Now you see why olive oil getting thicker!


text:
Is there a simpler way to detect alcohol in a blood sample using IR spectroscopy?
I have seen all sorts of ways to detect alcohol using NIRS (Near InfraRed Spectrscopy) or FTIR, but they are all quite complicated in that they are using and analyzing a very wide spectrum. In addition, it is not clear as to the environment of where the alcohol is measured in. Some presentations are talking about measuring the OH-stretch of the RO-H in the region of 3200-3650 cm^-1.
Some related, but not very helpful questions are:

O-H infrared band dependence of alcohol concentration in water
These there is this book section which mentions various problems of published spectrum.

Is there a cheaper and more simple way to do this detection, using off-the-shelf components, like IR lasers or diodes?

I suppose "cheaper" would mean to use just one or very few frequencies, so that one doesn't have to analyze an entire spectrum.
What are these frequencies and how are they determined?

UPDATE:
Apparently the Blood Alcohol Concentration (BAC) is measured differently in different parts of the world. Just for reference:
0.1 g alcohol in 100 mL blood equals:

0.1% (U.S., Canada, Australia)
100 mg% (UK)
1 ‰ = 1 permille grams alcohol per 1 liter blood  (EU)
22 mmol/L (in hospitals) millimols alcohol per liter blood.

You're looking at concentrations of alcohol of approximately 0.1%. It's going to be difficult do measure that easily in an IR spectrum.

@Zhe How did you get that number?

Well, when you hit 0.5%, you're pretty close to dead if not dead, and 0.08% is the legal limit in US for driving.

The trouble with detecting alcohol (or glucose for that matter, which has been tried many times because of its utility for diabetics) in blood by IR is that blood is a very complicated mixture of things many of which also have features common to alcohol (like OH stretches). Extracting a specific signal from the noise requires vast computing power even if the signal is good enough (which is unlikely).

Interestingly, there are certain things that are significantly easier to measure. For example, a pulse oximeter reads blood O2 saturation really well, and it's just a red light that you clip to your finger.


text:
Substitution / Elimination competition
I am trying the hypothetical synthesis below, but I didn't get the product I was expecting. Instead, the reaction formed mostly styrene, in >80% yield (elimination product of the bromide)

The synthesis of an analogous product with one less carbon in the bromide (benzylic bromide) is described in the literature and I thought it would work with this one since they only differ by a carbon. Any thoughts on how I could overcome this selectivity problem?
I thought about using a milder base, but I don't think the problem here is actually the base I use, cause they will all form the enolate, the problem is the enolate abstracting the benzylic proton from the bromide and generating styrene + butyrolactone.

What was the procedure? Preform the enolate then add the bromide? Or add the base to the mixture of lactone and bromide?

Sorry, I forgot to mention... The enolate was formed first and then the bromide was added.

It is a bit too easy to form the styrene this way. Could you use phenacyl bromide instead then do a Clemmensen to get the desired product? Or react the enolate with styrene oxide and hydrogenate.

Clemmensen would not reduce the lactone, right?

Should not do so. Could always reduce it stepwise with sodium borohydride the ionic hydrogenation.


text:
Path to take when labeling R/S
I'm trying to find the R/S for the carbon marked with the asterisk. As I branched out to from the asterisk carbon to the red carbons, the methyl group has the 3rd priority (with the hidden hydrogen 4th). Here comes the confusion. In order to find the first point of difference for the two carbons, for one of the carbon, I can simply move outwards to the next green carbon (C, C, H). But for the other carbon, there are two paths I can take, one which gives (C, H, H) and the other (C, C, H). Now, which path should I take?
Assuming I take the path that gives me the (C, C, H) carbon. This gives me a tie. Again, I have two paths to take (indicated by the green arrow). Which path should I take? If I choose to go 'down', the two arrow clash at a single carbon.
I have consulted numerous textbooks but none of them seem to even surface this as an issue. In general, if there are multiply paths, which path should one take?
Extension of question: In a scenario that there are two carbon, C1 and C2 that 'ties', and it so happens that each of the two carbons has 'two paths' to take: C1 with path 1 and path 2, and C2 with path 3 and path 4. What happens if the priority of taking the different paths is not clear. Eg. Priority when taking path 1 is higher than the path taken by path 3, but path 2 has a lower priority then when taking path 4?

There is no "which". You don't choose paths; you always take them all at once.

You might be interested in this Q&A

The compound you are presenting is a natural product, compactin. The numbers of the carbon atoms are arbitrary. The digraph of C9 shows how the configuration is determined. Sphere 2 contains the atoms C8, C10, C14 and H. Clearly, hydrogen has the lowest priority. Sphere 3 has the locants C10{15,5,H}, C8{7,(7),H} and C14{H,H,H}. In a one-to-one comparison of the locants, C10 and C8 have priority over C14 because C>H. So C14 has the next lowest priority. C(7), which is a duplicate atom of atomic number 6, is attached to three atoms in sphere 4 of atomic number zero. Zero is as low as you can go. Therefore, the priorities are C10>C8>C14>H. C9 has the (S)-configuration.

You always have to compare all paths at the same level of depth.
I will use the numbering in user55119’s answer.
The first level of depth gives us $\ce{C,C,C,H}$, we can assign H-9 4th priority.
The second level of depth for the three C atoms gives us:

$\ce{C,C,H}$ for carbon 10;
$\ce{C,(C),H}$ for carbon 8; and
$\ce{H,H,H}$ for carbon 14.

Thus, carbon 14 gets 3rd priority.
The third level of depth gives us:

carbon 10

$\ce{C,C,C}$ for carbon 5
$\ce{C,H,H}$ for carbon 15

carbon 8

$\ce{C, (C), H}$ for carbon 7
nothing for ghost carbon (7)

The first point of difference is carbon 5 versus carbon 7; 5 is attached to three carbons while 7 is attached to two carbons and a hydrogen. Thus, $5>7$, thus $10>8$. This is reinforced by 15 being a real atom with bonds while (7) is a ghost atom not attached to anything. Thus, even if carbon 7 is methylated 10 still wins over 8.