text:
Is Hydrogen ion an acid?
I had an argument with a friend about this question. He says it is not an acid But I think it is as it itself can increase the pH value.

This is an un-question. If you argue about it or have an opinion on it, then you are wrong.

@Ivan Neretin. Why do you think it is an un-question ? This proposed question may have been a consequence of Arrhenius' theory of acids. Why not ?

Can you have a mole of hydrogen ions in a beaker? Yes and no.

Not to complicate things but the "hydrogen ion" is one of the fundamental particles - our friend the proton.

@Hal technically, since protons are composed of the fundamental particle, quarks, a proton would be considered a hadron.

@DVAL98 Your point is well taken.  The use of "fundamental particle" was not the best of terms.  Given the discipline [chemistry not physics] and the apparent level of the question, it seemed helpful to make the point that a proton is one of the three particles most often discussed by chemists.  It also helps make the point that H+ is never along, and hence emphasize  the importance of the H3O+ discussion.

@Hal I apologize if that sounded pompous. In it was just semantics which can be annoying and your point was well delivered (and important). I just wanted to clarify for OP/future readers. The hard part of communication on online forums is trying to convey a point without seeming condescending.


text:
How to find the oxidation number of Co and Cr in [Co(NH3)6] [Cr(C2O4)3]?
In the question I actually got I had to name the above compound, and I know how to name it but I don't know how to find the oxidation number of the above compound(which has to be mentioned in the IUPAC name). I actually know how to find the oxidation number in complexes but here both the ions are complex ions which makes it confusing. If someone needs the name its Hexaamminecobalt(?) trioxalatochromate(?)

So, what you propose to be the name?

$\ce{Co}$ is known to have two possible oxidation numbers in solution : +II and +III. Both make complexes with 6 $\ce{NH3}$. If $\ce{Co}$ is at +II, the charge of the complex is $2$+, as $\ce{NH3}$ is not charged.
If $\ce{Co}$ is at +III, the charge of the complex is $3$+ for the same reason.
For chromium, it is similar, as $\ce{Cr}$ may have oxidation number +II and +III. But the ligand oxalate is charged $-2$. The complex of $\ce{Cr}$ with three oxalate ions must compensate the charges of the cobalt complex (+$2$ or +$3$). So the charge of the chromium complex has to be -$2$ or -$3$. In other words, the complex ion is either $\ce{[Cr(C2O4)3]^{2-}}$ or $\ce{[Cr(C2O4)3]^{3-}}$. Let's compare the two possibilities, and the oxidation numbers of $\ce{Cr}$ in these two structures.
To get a complex $\ce{[Cr(C2O4)3]^{2-}}$, $\ce{Cr}$ oxidation number should be equal to $x$, with $x + 3(-2) = -2$. So $x$ = $+4$ or +IV. And unfortunately, this oxidation number IV does not exist for $\ce{Cr}$ in solution. This structure has to be abandoned,
On the other hand, to get a complex $\ce{[Cr(C2O4)3]^{3−}, Cr}$ must have an the oxidation number $y$ given by : $y +3(-2) = +3$. So $y$ = $+3$ = +III, and this is possible and even welcome. It corresponds to one of the requirements developed previously. As a final result, the complex ion $\ce{[Cr(C2O4)3]^{3-}}$ has a charge $3-$ that can be coupled with $\ce{Co(NH3)6^{3+}}$.
As a consequence, both Co and Cr are at oxidation number +III in the original structure $\ce{[Co(NH3)6][Cr(C2O4)3]}$

You could just say the charge of the ligands adds up to negative six, so the cations have to have a total of six charges as well.

source: Stack Exchange

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

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text:
When titrating to find the concentration of a dissolved acid, does the volume of solvent matter?
I am conducting a titration to find the mass of aspirin present in tablets. I will be dissolving the aspirin tablet into a solution of ethanol and distilled water to dissolve it. But, does the volume of ethanol or distilled water added actually matter? Or does it just need to be enough to completely dissolve the aspirin?

You want to dissolve the tablet completely. However you don't want a huge excess of water. So 1.5 to 2 times the amount of the solvent mixture to dissolve the whole tablet.

@MaxW, all right, got it. Does the volume of solvent need to be precise, i.e. do I need to measure it using a pipette or will a measuring cylinder do?

A graduated cylinder will certainly do. The notion is that you want to use about the same volume of solvent for each trial.

Concentration != mass. For concentration volume is crucial, for mass - not really.

At very great dilutions, impurities could have some effect on the end point. For example, even previously-distilled water will absorb some small amount of $\ce{CO2}$ from air, making it slightly more acid. However, if the aspirin is present in reasonable amount, that would have negligible effect. So, as stated by @MaxW, consistency, with a moderate theoretical excess of solvent, is important, but don't use greatly excessive amounts of solvent.

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


text:
Can someone explain why the electrical conductivity of magnesium chloride decreases with increasing concentration?
I was under the impression that the electrical conductivity should increase initially, then begin to plateau, then fall past the saturation point for strong electrolytes. However, from this, I see that the conductivity of magnesium chloride does not follow this ( it is decreasing with increasing concentration) , BUT the data could be wrong though. Would someone mind explaining what I should expect when I conduct the experiment myself?

You simply started your data range at too high a concentration to see the initial increase.  Semantic Scholar gives data for the conductivities of several aqueous electrolyte solutions at 25°C, including magnesium chloride.  The strongly increasing portion of the magnesium chloride curve goes up to about 10%, which corresponds to about one molar (where your data happens to start).

Not increasing at an increasing g rate.  Compare conductivities with the mass percents whose increments are non uniform (1%, 2%, 5%, 10%, etc.).  10% is just over 100 g/L including the increased density and the molecular weight of MgCl2 is 95.  Finally your trend at >1M is not very strong and its direction at a given concentration may be affected by measurement uncertainties.

If you want to go up to 1M what you choose is good.


text:
What are the consequences of mixing Ferric Chloride Solution, distilled vinegar, baking soda and water?
I was attempting to etch and blade with a ferric chloride solution. I did not have enough so I filled a glass with vinegar and water (3 parts vinegar to 1 part water) then added 2 oz. of ferric chloride solution. I added baking soda later to neutralize the ferric chloride solution, but was met with a deep red foam. I quickly added more baking soda and flushed the solution down a deep sink in my basement. I rinsed out the sink and glass with water and continued to add baking soda to neutralize any ferric chloride solution that had been spread by the red foam. What reaction occurred and is this and do I need to worry about it?

Adding baking soda to anything acidic like vinegar has this effect, releasing gaseous carbon dioxide. Ferric chloride is there for the color. Additionally, it forms gel of ferric hydroxide.

I suspect the red salt you are seeing is Iron (III) carbonate, which was likely created from, as you noted, the neutralization of aqueous Iron (III) chloride with Baking Soda (in excess?) per the reactions:
$\ce{FeCl3 (aq) + 3 NaHCO3 (aq)-> 3 NaCl (aq) + 2Fe(HCO₃)₃(aq)}$
$\ce{2Fe(HCO₃)₃ (aq) → Fe₂(CO₃)₃ (s) + 3H₂O (l) + 3CO₂ (g)}$
$\ce{2Fe(HCO₃)₃ + HAc (aq) → Fe₂(Ac)₃ (s) + 3H₂O (l) + 3CO₂ (g)}$
where the last reaction could also lead to Iron (III) acetate from the vinegar presence, which is also subject to further neutralization by the Baking Soda.


text:
When converting between a hydrated electrolyte amount in mass and in milliequivalents (meq), why are the water molecules taken into account?
A textbook I'm reading called "Ansel's Pharmaceutical Calculations, 13th edition" defines the milliequivalent (meq) thus (p. 187):

This unit of measure is related to the total number of ionic charges in solution,
and it takes note of the valence of the ions. In other words, it is a unit of measurement of the
amount of chemical activity of an electrolyte.

It then goes on to demonstrate how can one convert the amount of electrolyte by mass to the amount in meqs and vice versa, and it gives the following example (pasting as a picture to keep all the original formatting):

Now see the note below the exercise, highlighted in yellow. The text does not explain any further, and I'm a bit confused.
If the amount in milliequivalent represents the number of ionic charges in solution or "chemical activity", why do we need to use the atomic weight of the hydrated ionic compound? Do the water molecules contribute anything to the chemical activity of the solution? If one meq means essentially one charge, the amount of meqs obtained when using a molecular weight that includes the hydration molecules is not representative of the amount of ions in the solution (as given by molar amounts).
So what is the reason for including the weight of water molecules in the calculations?

If you can, try to avoid using the obsolete unit called "equivalent". it was banned in the middle of the 20th century, because it often produces inconsistencies when handling redox reactions. The notion of equivalent weight works all right for acid-base reactions, but it fails with redox chemistry. As an example, 1 mole KMnO4 is always 158.03 g. But 1 equivalent is 158.03/5 g in  acidic solution or 158.03/4 g in neutral solution. So the risk of confusion is great. So try to use moles instead of equivalent ! This risk of confusion will not appear with moles.

Thanks. Many medical references still provide electrolyte dosages in meq and therefore doctors prescribe electrolytes in meq dosages, so I have to use it too...

When you want to obtain a solution of $\ce{CaCl2}$ in water, you can do it by dissolving anhydrous compound $\ce{CaCl2}$, or  the hydrated compound $\ce{CaCl2·2H2O}$. It does not matter which one to choose, provided you add enough water to compensate the difference and obtain at the end the wanted total volume.
If you want to obtain $1$ mL of a final solution that has to be $4$ meQ/ml, you must dissolve $4$ mEq of $\ce{CaCl2}$ or $4$ mEq of $\ce{CaCl2·2H2O}$ in $1$ mL . It does not matter which one you use. The only requirement is that the number of Equivalent has to be $4$ mEq. But the Equivalent weight  $\ce{CaCl2}$  is $111/2 = 55.5 $ g, and the equivalent weight of $\ce{CaCl2·6H2O}$ is $73.5$ g
To get the final solution with $\ce{CaCl2}$ you must choose to dissolve $4$ mEq $\ce{CaCl2}$, which weighs $\ce{(4·55.5}$ mg) = $222$ mg $\ce{CaCl2}$.
To get the same final solution with $\ce{CaCl2·6H2O}$ you must choose to dissolve $4$ mEq $\ce{CaCl2·6H2O}$, which weighs $\ce{(4·73.5}$ mg) = $294$ mg $\ce{CaCl2}$,
Do you understand ?

Yes, I understand this. What I don't understand is why do we need to take the water molecules into account when we are interested in meq of the electrolyte itself (e.g., when we want to give magnesium sulfate to someone with hypomagnesemia).


text:
Is there an experimental way of measuring the critical concentration of pigment in paint?
With an assigned, generally paint-based, experimental project I've been looking up CPVC or critical pigment volume concentration in binders. CPVC defines the point of addition of pigments after which follows a severe drop in blistering and increase in rusting on the applied surface due to the pigment not while binding with the binder having voids in it. But going past the critical point can also make the paint fragment and crumble or maybe even lose just some of its pigment, in conclusion lose efficiency.
Paint manufacturers use this to make paints a certain type by knowing the CPVC of the mixture with which they can adjust to make the fitting paint for the job but the way they know is by applying the general formula for determining the point which is

$$CPVC = \frac{1}{ 1 + OA * ρ / K }$$

With the 'ρ' standing for the specific gravity of the pigment, 'K' specific gravity of the binder and 'OA' which is the oil absorbent number of the pigment depended of the exact pigment and the binder. But excluding it is there a way that a clear mixture of paint without any additions like extenders for example and just acrylic polymer binder and an inorganic pigment type can be used to calculate the near CPVC of it. Can it maybe be done by centrifuging a dry paint sample in a centrifuge tube and later sampling the super natant that will contain an ammount of pigment and binder that would dissolve in it  and fall apart from the rest of the paint and quantitatively analysing it in a high performance liquid chromatograher? By increasing temperature or the pH it could also affect the stability (the CPVC) so can that method be used effectevly? Is there a better one and not by just assuming the CPVC by observing the paint on a wall?
Any help could be really helpful sense im young and do not know the process well, only by researching the concept itself, and it's hard to find details and think of a valid project idea.

An image of the CPVC effect is shown in the diagram below. The red color is the binder, the black is the pigment and the white is air (because there is not enough binder to coat all the pigment). The lines depict the change in properties you would expect, and you already have the formula to correlate the paint properties with the raw material properties.

BTW, economics generally suggests maximizing the inexpensive pigment/filler concentration, but adding too much increases permeability to water, which is bad in the case of metals which might corrode. On the other hand, blistering can be caused by corrosion of the substrate which then disadheres some of the film if it is too rich in binder; blistering is minimized by allowing the film to break a little. Of course, then corrosion continues - you just don't see it as much until it is too far gone. The best paint will protect from corrosion with low permeability and without disadhering.
So, how to measure a property related to CPVC if you can't pick one of these? First, you need some of the paint film; I assume you can start with the plain binder and add different amounts of the pigment/filler. Mix well, do not incorporate unnecessary air bubbles when mixing. Cast a film of constant thickness on aluminum foil by using a threaded rod; let the paint dry.

Then do your experiments!

Flexibility. If you bend the painted aluminum foil over a rod of defined radius, the rich binder film will not crack, but as more filler is added more and more cracking tendency will be noticed.

Permeability. If you put a drop of water (or perhaps alcohol) on the dried film, it will wet and spread. You may be able to differentiate between different levels of added pigment. If the aluminum foil substrate does not give good results, you can apply the paint to a decent grade of paper. Then a drop of water or other solvent might wick all the way thru and appear on the other side of the paper.

Glossiness. Ideally, you should have some kind of electronic measurement, but ranking the specimens subjectively (with your own eyes) is better than nothing. In practice, flat paints have a high pigment concentration and are not durable when washed; semi-gloss paints have less pigment, more binder, and can be washed more often. (I always wondered why they didn't go all the way to "full gloss", but I guess the binder was never so good as to give much better performance.)

Washability. Apply a water stream to the painted aluminum foil (like a dripping faucet: constant height, constant volume per minute), watch the paint film disappear.

And try coming up with some more tests!


text:
Question on d block contraction
For size trend down a group, it is known that:$$Al\gt Ga$$
Due, to d block contraction of gallium.
Why isn't this the case when it comes to silicon and germanium?
Won't the poor shielding effect from the d electrons have an effect here?

The Answer on the SE for a similar question does not really answer the question.

See this: https://imgur.com/MZ9dpRe

@NilayGhosh sorry, That didn't help me much.(-_-;).Is it because it is just an observed property?

Does this help?

Please use meaningful titles.

If germanium is supposed to be bigger than aluminum, in fact, its empirical values correspond to $\pu{125pm}$ for each one respectively, but germanium being lower in the periodic table is bigger at the same time which is why we say that $\ce{Al < Ga}$. Now yes, the atomic radius of silicon is $\pu{118pm}$ while as we said before, the atomic radius of germanium is $\pu{125pm}$, in the same way the size is deduced based on its position in the periodic table, so $\ce{Si < Al < Ga}$.
This occurs because there is a relationship between the effective nucleus charge and the electronic levels of the atom, since the positive nucleus attracts the electrons, but in turn, the electrons repel each other (except those of the first level), it is by This means that the forces and counter forces end up having an effect on the valence electrons, and these determine the atomic radius of the atom. This difference of attraction by the nucleus and repulsion by the electrons increases as there are more electronic levels, that is, if I have a compound with Z = 11 and e = 11 that we will call "A", it will be greater than a compound "B" whose Z = 10 e = 10, leaving B <A since although both have the same number of electrons as protons, compound A has an electronic configuration $\ce{1s^2 2s^2 2p^6 3s^1}$ with 11 protons while compound B has an electronic configuration of $\ce{1s^2 2s^2 2p^6}$ with 10 protons, that is, compound A has a higher electronic level than compound B, which gives a larger atomic radius.
Reference
1.https://es.wikipedia.org/wiki/Anexo:Radio_at%C3%B3mico_de_los_elementos_qu%C3%ADmicos
Ah, I just realized that it was Galio and not Germanio, sorry.


text:
How do I synthesise this tertiary amide from these 5 compounds?
So basically I'm at a loss as to how to proceed. I started with the idea of converting the primary alcohol into an aldehyde and that is where I stoped. I understand that each substituent of the Nitrogen is somehow made from B, C and D and that the cynaonitrile is supposed to help you make the amide and the Gringnard Reagent is for adding a methyl group somewhere, maybe to C. Im completely at a loss and I'm just getting frustrated now :(

You seem to have got the uses of the compounds right, but could not bring that knowledge together to apply it here. As you said, you first use the Grignard to add a methyl group to D

Let's call this compound K. Then, we use $\ce{NaCN}$ to convert the alcohol into a nitrile, then to a carboxylic acid rather than an amide, as this would be easier to work with for us.

Let's call this compound L. Now we are ready to start creating the actual compound. Nitrate C and then reduce it to create the amine and reduce the ketone.

Then add L to the product, followed by K. This order is followed as adding the alcohol before the acid may create a tertiary amine rather than the amide we need.

I think you could expand on how to add K to the final intermediate, just adding the amide to the alcohol is not going to work.

Oh this is great I understand now! I think I was trying to be too complicated by trying to make an amide first and then adding substituents to it. For the final part is it all just substitution reactions?

@ChemGeek addition of L should be a simple acid-base neutralization, but as pointed out by Waylander addition of K might not be as simple as I think it is. I'll update the answer once I'm 100% sure

@AniruddhaDeb I think I would reverse the steps by forming the amide last and forming the  secondary amine by a reductive amination on p-MeO-acetophenone

Unfortunately, Chem Geek might believe that cyanide may displace hydroxide, and that water will hydrolyze a nitrile to a carboxylic acid. Does tin effect a Clemmensen reduction? How is the amide prepared? Melting the aniline and L together? The last alkylation is not detailed enough.


text:
What is the product obtained when methyl vinyl ketone is reacted with 1,2-ethanedithiol and the product is further hydrogenated with Raney Nickel?
I was solving questions from aldehyde, ketones and ethers and I came across a question that asks:

What is the product obtained when Methyl vinyl ketone is reacted with 1,2-Ethanedithiol and the product is further hydrogenated with Raney Nickel?

My thoughts:

I thought that the reaction could be similar to the one where we react a ketone with ethylene glycol, then on further hydrogenation the double bond breaks. I may be wrong. Any help will be highly appreciated

Also, do note that it is Raney nickel, not "Rancy" nickel.

The answer could be 3 or 4.
1,2-ethanedithiol will react with the carbonyl group to form a dithiane (similar to the ethylene glycol reaction). Reaction of this with Raney Nickel/H2 desulfurises it, so the whole process is effectively a reduction of the carbonyl group to CH2. It depends on the reaction conditions whether the double bond survives. Raney Nickel can reduce alkenes.
You can read more about it here 1 and here 2

In the presence of Ni(R)/H2 I would expect desulfurization  and double bond reduction, i.e., n-butane. In the presence of degassed Ni(R), the double bond may survive. However, because the desulfurization is likely radical in nature, 2-butenes in addition to 1-butene would be expected. Your thoughts?

I have modified my answer. Just stirring in EtOH with Raney Ni can desulfurize some of these. I think this is a poor example to teach this reaction  with as there are these other possibilities . I suspect this was set without thinking too hard about it, or reference to the literature.

I agree. This type of question does a disservice to the facts.


text:
Why does this formula work?
My teacher gave us a formula for determining if a M-O-H like compound will act like a base or not.without explaining much.
Base if:$$|\chi_o-\chi_M|\gt|\chi_o-\chi_H|$$
Flipped inequality if acid.
I am not unable to understand why the formula works.I think its because if the diffrence in electronegativity between Oxygen and element m is high.
It is  more likely for M-O-H to break into M and O-H
.Not certain about that either.

The bond with greater electronegativity difference has greater tendency to break with help of polar $\ce{H2O}$ molecules to hydrated ions, unless it is ionic already. Electrons then stay at the oxygen as the most electronegative atom from all the three.
So if $\ce{M}$ is less electronegative then $\ce{H}$, there will be'
$$\ce{M^{(+)}-O^{(-)}-H(aq) <=> M+(aq) + OH-(aq)}$$
but if $\ce{M}$ is more electronegative than $\ce{H}$ , then there will be:
$$\ce{M-O^{(-)}-H^{(+)}(aq) + H2O <=> MO-(aq) + H3O+(aq)}$$
( The charge in parenthesis is a partial charge.)

Thanks for the answer, you explained it very well. Can you tell me a possible reason why my post was downvoted? I don't use chem SE much and I would like to improve upon the framework of my questions.

Hard to tell. The most probable reason may be the downvoter considers it not enough elaborated by explicit own thoughts based on prior search of available resources.


text:
Combining Lead pipe with Zinc or Copper pipes in a domestic water installation
Grateful if someone assists in explaining this example that appears on a high school text book. The problem can be summarized as follows:
"Old water networks supply water to a residential area via Pb lead pipes. A user wants to fit his own  installation with pipe made out of either Zn or Cu.  The intent is to reduce the lead presence in the water. Condition: the lead pipe of the municipality will remain unchanged up to the water gauge/meter at the house input.
Account for the presence of O2 in the water and give an explanation on your choice Zinc or Copper."
The solution suggested in the book shows Cu pipes, and excludes the use of Zn.
I would chose Zn. Reason would be that O2 would preferentially oxidize Zn instead of Pb. Thus preventing the formation of Pb oxide/hydroxides and consequent leaching in the water.
Thought I overlooked something and looking for an explanation to the book' choice which I assume is correct.. thanks for your support

Zinc pipe / tubing is not made . It would require technical development to make long lengths necessary for plumbing. It is low strength like lead but does not have good ductility like lead, and embrittles with age . So a poor engineering choice. Like lead , soldering zinc requires skill  , unlike soldering copper.

Zinc is more sensitive to atmospheric oxidation than copper. So in contact with oxygenated water, zinc will be more rapidly destroyed than copper. So tubing made in zinc should be replaced pretty often. Should they last more than one year ?

Thanks Maurice. Sure does make sense from a reduction potential stand point. But how would the use of Cu prevent the Pb from being oxidized and entering in solution as a lead oxyde or hydroxide. The piping life is not the main concern.   Omar

This leaves an open question. Is the fact that the main piping is of lead just there for the narrative (never understood why text book do so instead) or is important/crucial for the right answer? On the basis of this answer the main pipe could be whatever and one shall use cupper. Is there any fine detail?

yes indeed, the lead presence is crucial because the final intent of the excercise is to show how to prevent high lead concentration in the drinking water. in the solution they show Cu being the best choice to reduce Pb concentration. it also shows the importance of Oxygen presence.

Copper will not reduce Pb concentration in water.

The focus of the analysis should be first on the anodic index (see, for example, table here), which relates to the electrochemical corrosion potential of the metal serving as the anode. Second, materially important is the current state of the old municipality Lead pipes.
Per the source, the anodic index of Pb is high (see chart) above copper, so historically it has already electrochemically corroded likely inducing a protective Lead oxide coating in the presence of dissolved oxygen.
However, Zn metal is more anodic than the Pb anode, and as such, would readily corrode and may introduce high levels of Zn ions (even levels considered problematic/toxic).
As such, copper metal, which is lower in the anodic index relative to the Pb electrode, is likely a better choice to preserve the current state of affairs, which is in apparent agreement with the reputed cited textbook 'correct' answer.


text:
How does DeepHF (a CRISPR sgRNA design tool) compute binding free energy?
I am looking at the code of DeepHF[1] that computes $\Delta G_\text{binding}$, and I see that it breaks a sequence into overlapping dimers and compute a weighted sum according to a dictionary with 16 key values:
def dG_binding(seq):
    seq = seq.lower()
    dG = {'aa': -0.2, 'tt': -1, 'at': -0.9,
          'ta': -0.6, 'ca': -1.6, 'tg': -0.9,
          'ct': -1.8, 'ag': -0.9, 'ga': -1.5,
          'tc': -1.3, 'gt': -2.1, 'ac': -1.1,
          'cg': -1.7, 'gc': -2.7, 'gg': -2.1, 'cc': -2.9}

    seq = seq.replace( 'u', 't' )
    binding_dG = 0
    dGi = 3.1
    for i in range( 0, len( seq ) - 1 ):
        key = seq[i:i + 2]
        binding_dG += dG[key]
    binding_dG += dGi
    return binding_dG

Where do the values of each dimer come from? I couldn't find where these values in dG stem from.
Reference

Wang, D.; Zhang, C.; Wang, B.; Li, B.; Wang, Q.; Liu, D.; Wang, H.; Zhou, Y.; Shi, L.; Lan, F.; Wang, Y. Optimized CRISPR Guide RNA Design for Two High-Fidelity Cas9 Variants by Deep Learning. Nat Commun 2019, 10 (1), 4284. DOI: 10.1038/s41467-019-12281-8.

The simplest model for estimating binding energies (or melting temperatures) is to consider each base pair individually, without regard to sequence context. This is the rationale for the 4 deg vs. 2 deg rule of thumb for DNA duplex, i.e. the Wallace–Ikatura relation $$T_\mathrm{m}(\pu{°C}) = \mathrm{4(G+C) + 2(A+T)},$$ where G+C refers to the number of G/C base pairs and A+T to the number of A/T basepairs in the duplex DNA. You can refine this model by considering the effect of ionic strength. More sophisticated models go beyond just considering the composition to include sequence information.
The nearest-neighbor model for binding is the simplest of these more sophisticated models, and it looks like that's what is used here. For DNA:DNA duplexes, you can find  parameters on Wikipedia. The DNA:DNA model has less parameters because of symmetry lacking for the RNA:DNA hybrids (i.e. AG/CT is the same as TC/GA for the former, but AG/CU is different from TC/GA).
The code does not indicate the units, which might be kJ/mol or kcal/mol. The parameters are usually derived from experimental data and some fitting algorithm, see e.g. https://pubs.acs.org/doi/abs/10.1021/bi00035a029
Missing in this implementation is the contribution of the first and the last base pair.

Are there some library that implemented all this simple models? biopython offers some melting temperature module https://biopython.org/docs/1.75/api/Bio.SeqUtils.MeltingTemp.html. Is there something with more options/models?

@0x90 This might be a starting point: https://biology.stackexchange.com/questions/33032/biopython-big-discrepancy-calculating-rna-melting-temperature-over-literature?rq=1

Here is a fairly new paper with an updated set of parameters. The supplementary material has a web-based calculator implementing these: https://academic.oup.com/nar/advance-article/doi/10.1093/nar/gkaa572/5871365

@KarstenTheis Does "4 deg / 2 deg rule of thumb" refer to the Wallace–Ikatura relation  for  denaturation/melting temperature $$T_\mathrm{m}(\pu{°C}) = \mathrm{4(G+C) + 2(A+T)}?$$

@andselisk Yes, I'll add the technical term to my answer.

@KarstenTheis Thank you, I was also wondering whether "4+2 rule of thumb" would make more sense — there is summation (+), not division (/). Do you know what name people in the field normally use?

I edited to say "4 deg vs. 2 deg rule". I was using the slash like in "and/or" or "his/her", neither of which indicate division. @andselisk


text:
Rate expression with water as one of the reactants
When finding an equation for d[A]/dt from reaction mechanisms and $\ce{H2O}$ is a product/reactant, does $\ce{H2O}$ have to be included or left out like for equilibrium constant expressions?
For example, in the reaction of carbon dioxide with water,
$$\ce{CO2(aq) + H2O(l) -> H2CO3(aq)}$$
which expression describes the kinetics correctly (assuming that the reaction above shows the elementary step):
$$ \mathrm{rate} = k_1 \cdot [\ce{CO2(aq)}] \tag{1}$$
or
$$ \mathrm{rate} = k_2 \cdot [\ce{CO2(aq)}] \cdot [\ce{H2O(l)}] \tag{2}$$

I think people will ask you to reformulate the question, as it's not clear and lacks relevant details. E.g. is water a solvent as well? Because then what usually happens is that you consider its concentration approximately constant (as it is most of the time much larger than the one of the other reagents), and that makes it 'disappear' from the rate law. So e.g. a reaction with order 2 where one of the reactants is water becomes 'pseudo first order', just because the concentration of water becomes part of the constant.

Ask yourself, if there are cases when you can consider [H2O] being constant, including it implicitly into the rate constant.

In the dissolution of $\ce{CO2}$ in water, the concentration of $\ce{H2O}$ may be considered as constant.


text:
Why can't cyclohexylbenzene be called benzylcyclohexane?
The following structure represents the compound cyclohexylbenzene:

Why can't the benzene ring be treated as a substituent and the compound be called benzylcyclohexane?

There is no benzyl there, just phenyl.

First of all see the comment above. Then, consider that the "why" where dealing with nomenclature is always a rule. Studying the chemistry of cyclohexane, your proposal would be appropriate and reasonable. But imagine someone searching for your work. S/he would have to look for two different options.... In other words, plenty virtual infinite examples like that exist. Nomenclature is needed to uniquely determine what is what in a general homogenous way readable by all.

And even with nomenclature on place, there exist accepted or tolerated names, either for historical reasons or because belongs to different systems of rules.

Same size rings, the one with more double bonds is the parent. But it is phenylcycloheptane where 7>6.

The relevant rule for the seniority order of parent structures in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-44.4 SENIORITY CRITERIA APPLICABLE TO RINGS, RING SYSTEMS, OR ACYCLIC CHAINS
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(…)

P-44.4.1.1 The senior ring, ring system or principal chain has the greater number of multiple bonds; for the purpose of this criterion, mancude rings or ring systems are considered as consisting of noncumulative double bonds [criterion (a) in P-44.4.1].

Therefore, the name of the compound that is given in the question is cyclohexylbenzene rather than “phenylcyclohexane” since benzene has more multiple bonds than cyclohexane.
Note that this rule is not to be confused with the rules for systems composed of rings and chains, which are less strict. For systems composed of rings and chains, the context may favor the ring or the chain; nevertheless, the ring is selected as the parent structure for the preferred IUPAC name. In case of cyclohexylbenzene, however, benzene is always the parent structure – not only for the preferred IUPAC name.


text:
Why is HgI2 a colored complex?
Why is $\ce{HgI2}$ a colored complex it has fully filled d-orbits. According to the book, one which has fully filled d-orbital has no color in transition elements. So, why is this compound colored?

It seems that you have missed quite a few key points while explaining your question. Firstly, Hg is a d-block element which is not considered to be a transition element since it has complete d-orbital. The same holds for other members of the group like Zn and Cd Also, the statement that complexes containing fully filled d-orbitals are not coloured is incorrect. Here's my explanation:
Colour of compounds and complexes containing transition elements may be due to two reasons: d-d transitions and charge transfer.
d-d transition is the point that your book seems to be talking about. But that is not the reason why HgI2 is coloured. The actual reason for the colour is charge transfer. Charge transfer occurs when the compound has a high covalent character. This occurs when the anion is highly polarisable and can distort its electron cloud easily. $I^{-}$ has very high polarisability which results in the anionic charge getting easily transferred to the $Hg^{2+}$ cation. This process releases some energy which falls in the visible spectrum. Hence we can say that compounds like KMnO4, HgI2, etc. are coloured.
Hope it helps!


text:
Are the potentials of the electrolytes equal in Galvanic cell?
My book defines Electrode Potential for a Galvanic Cell as follows:

A potential difference that develops between the electrode and the electrolyte is called Electrode Potential

$$E= V_{electrode} - V_{electrolyte}\tag1$$
It further defines cell potential as:

The potential difference between the two electrodes of a galvanic cell is called Cell Potential.

Mathematically, I interpret it as
$$E_{Cell} =  V_{electrode,anode} - V_{electrode,cathode}\tag2$$
However my book further mentions that

The cell potential is the difference between the electrode potential of the anode and cathode.

I interpret this as:
$$E_{Cell} = E_{anode} - E_{cathode}$$
From (1)
$$E_{cell}=(V_{electrode,anode} - V_{electrolyte,anode} )- (V_{electrode,cathode} - V_{electrolyte,cathode})\tag3$$
To keep the above consistent with (2),
$$V_{electrolyte,anode} = V_{electrolyte,cathode}\tag4$$

I find (4) illogical as there is no basis for the two electrolytes to be at same potential.
What is going on here?

There is a potential difference between an electrode and an electrolyte, but it is not directly measurable and the electrolyte potential is rather undefined.
We do not use electrolyte potentials as reference potentials, as reference must be well defined. By mechanical analogy, it would be like defining the relative height wrt a level that is not clearly given. By other words, the conventional electrode potential is NOT the potential difference between the electrode and electrolyte, even if this difference does exist.
By convention, the electrode potential is considered the potential difference between the electrode and the standard hydrogen electrode(SHE) $E - E^{\circ}_{\ce{H2/H+/Pt}}$.
The cell potential is called rather the cell voltage or the cell electrode potential difference. The cell voltage is difference of electrode potentials, which are difference of electrode potentials wrt SHE.
$$U_\mathrm{cell} = E_\mathrm{cathode} - E_\mathrm{anode} \\ = ( E_\mathrm{cathode,absolute} - E^{\circ}_{\ce{H2/H+/Pt}}) - (E_\mathrm{anode,absolute} -  E^{\circ}_{\ce{H2/H+/Pt}})\\
= E_\mathrm{cathode,absolute}  - E_\mathrm{anode,absolute}$$
Note that cathode is the more negative one for electrolytic cells, but more positive one for galvanic cells.
The cell voltage and the potential electrode difference wrt SHE ( = the cell voltage if the other electrode were replaced by SHE ) does not depend on the electrolyte potential.

So do we simply ignore the electrolyte potentials in my eq 3 or does my eq 3 does not make any sense?

Secondly is it something like the "electrolyte potential is very small and hence we neglect  it" or "electrolyte potential of any solution is almost equal to that of 1M HCl of SHE" or "it is taken out of the equation even though it is large and not equal to 1M HCl"?

The last one.  The cell voltage and the potential electrode difference wrt SHE ( = cell voltage if the other electrode were replaced by SHE ) does not depend on the electrolyte potential.

By other words, the conventional electrode potential is NOT the potential difference between the electrode and electrolyte, even if this difference does exist.


text:
What will be the product of Wolff–Kishner reduction in presence of hydrogen peroxide?

My textbook exercise gives this reaction.
What I observe is only the alkene part at the bottom changes to alkane.
I know the procedure of ordinary Wolff–Kishner reaction (shown in img given below), but peroxide makes it difficult.
1 and 2 represents the catalyst addition order.
Please explain why this happens?
Does this reaction remove double bonds?

Welcome to ChemSE! I suggest you explore "diimide". Wolff-Kishner is not involved. Note that the carbonyl is intact.

Thank you. Yes, I think you are right. This Wikipedia page helps: https://en.m.wikipedia.org/wiki/Reductions_with_diimide . I had no idea of it.

FYI: The chemistry was well-known prior to the advent of Wikipedia. Yes, it is right.

Ofc, it was. I used Wikipedia to understand. Your comment was helpful.

@user391330 if you got your answer, feel free to self-answer your question. Self-answering is encouraged here. It will help future readers and also helps you in earning some reps.

I understand the mechanism (https://upload.wikimedia.org/wikipedia/commons/8/81/DiimideMech1.png) now . But, the H2O2 part still worries me.


text:
Energy loss in the process for exothermic reactions
Some chemical reactions while producing the product molecules releases heat to the surroundings and such reactions are called “exothermic reactions”. Is it true that such systems must be showing energy loss in the process meaning “ΔE ” must be negative?

You are right, if you admit that the energy $E$ included in your $"\Delta E" $ correspond to what chemists call "internal energy" at constant volume and "enthalpy" at constant pressure.

No.  If means that to maintain the temperature of the products equal to the temperature of the reactants, you need to remove heat, and, form the first law, if you do this, then $\Delta E$ is negative.  However, if you don't remove the heat, the temperature of the products will be higher than those of the reactants, and $\Delta E$ of the system will then be zero (as required by the first law).

Yes , you are right to say that become from 1st law.It also depends on the Macroscopic work done on or by the system.

@Robert Patrick I assumed he was talking about a rigid container.  I removed that from my first draft to keep the answer simple.

$\delta$ E = q+W . So in your case q which is your “Heat loss” is negative but if somehow there is some W done on the bod( By compressing a piston ;one way ).Then your E can be positive.So it depends.

@user277768 At the OPs level of understanding, I made the determination that it would be confusing to him to include work.  Actually, the "heat of reaction" is defined as $\Delta H$, assuming constant pressure and temperature, in which case some work is already included.


text:
Balancing redox equation involving ethanol and dichromate ion
$$\ce{C2H6O(l) + Cr2O7^{2-}(aq) -> C2H4O2(aq) + Cr^3+(aq)}$$
Basically, I have written the half equations, but I have having trouble with converting ethanol to acetic acid ($\ce{C2H6O -> C2H4O2}$).
I do not know how to balance this. I tried adding $\ce{H2O}$ to either side but I didn't get an equal number of oxygens or hydrogens on either side.

Duplicate. Does this help? https://chemistry.stackexchange.com/questions/140462/balancing-a-reaction-where-dichromate-ion-reacts-with-ethanol-to-give-criii-an/140478#140478

@user55119 yes it is duplicate.

More usual is using structural formulas of ethanol and acetic acid, as $\ce{C2H5OH + CH3COOH}$

Of course you must add $\ce{H2O}$ on the left hand side, in order to compensate the excess of $\ce{O}$ atoms at right. Then add $\ce{4 H+}$ on the right hand side ! And $4$ electrons to compensate the charges. And that's over.
$\ce{C2H6O + H2O -> C2H4O2 + 4 H+ + 4 e-}$. Where is the problem ?

How come you added 4H+ to the RHS, shouldn't it be 2H+ because we only have 2 hydrogen in oxygen

Oh wait i think i see, because we need to balance the H+ ??


text:
Trouble understanding interplanar spacing using miller indices
We know that two parallel planes have the same Miller indices, this implies that we can have an infinite number of parallel planes close to one another all of which have the same Miller indices.
But it's claimed that

the interplanar distance between two adjacent planes is given as $$d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$$.

I can't understand what adjacent means here?
to me the next parallel plane to any plane is infinitely close to it, hence the $$d_{h k l }=0$$
Can someone please explain what exactly is interplanar spacing?
Thank you.

Take a simple cubic lattice, look along one cube-side direction, lets call it z, or (001), so h=0, k=0, and l=1. The planes of atoms are not infinitely close together. They are one cube side apart.

Thank you, but say I take a parallel plane halfway from the first plane it'll be given by (001) as well but it's distance is one half cube length. Then I can take a parallel plane $1/4$ of cube length distance away it'll be specified by (001) as well but it's distance isn $1/4$ of cube length and so on...

But there aren’t planes of atoms there.

Is it that the Miller indices of a plane are only defined only for planes passing via actual lattice points? I didn't find this addition of definition on the web or my text book.

That is the point, yes.

Oh Thank you. Could you copy paste your comment as an answer, I'd be glad to give a 10+ :).

The lattice planes are a property of the lattice based on dividing up the unit cell. The cell can be divided up indefinitely to give all the Miller indices for example $(\bar{10}\;20\;0);(10\;23\;20)$ etc. Atoms do not need to be on a lattice plane to add to the diffracted signal. The structural information is in the intensity of the diffracted spots not their position. (2 2 4) is parallel to (1 1 2) etc. as shown in the figure. In the general case it is usual to use vectors to find the perpendicular distance using formulas for finding a point at a given distance from a plane.
(Edit. I added (2 2 4) to indicate that it is a different plane but parallel to (1 1 2), in crystallography it is conventional to make all parallel planes have the same Miller indices, (h k l) (as they all scatter in the same way) by removing any common factor. )

Thank you, but why does the yellow plane have indices $(224)$ and not $(112)$ ? As far I've read I'd go on like this for the yellow plane : take  reciprocal of the intercepts and I'd have $1/2,1/2,1$ and now reduce to lowest integers by multiplying by 2 I'd get $1 1 2$ which is incorrect as you've shown the indices to be $2 2 4$.

I've used your image with credits to you, I hope you are okay with that, here https://physics.stackexchange.com/questions/596123/should-we-reduce-the-miller-indices-down-to-lowest-integers-or-not

yes, you are correct, these are conventionally given the same indices as the planes are parallel, I put 224 just to indicate the difference, then the indices are conventionally reduced to remove any common factor.

@porphyrin For looking at the geometry of the planes, removing a common factor from the Miller index makes sense. For indexing diffraction data and for Fourier transforms, you have to keep the higher-order indices intact.

I can't understand what adjacent means here?

For first order reflections, you only consider parallel planes that contain crystal lattice points. If you have a first order reflection (i.e. the Miller indices have no common divisor), the plane closest to a given origin will go through a/h, b/k and c/l. There will also be an plane through the origin, and the distance between the two will be the interplanar spacing.

In the figure, the plane through the origin is not shown. Instead, a second parallel plane (through 1, 1, 2) is shown. You might notice that the distance from the origin to the first plane is the same as the distance between the first and second plane.
For reflections of higher order, there are additional planes in between the planes of the corresponding first order reflection. The (4 4 2) reflections, for example, would have twice as many planes, with half the interplanar distance.
In a diffraction experiment, this corresponds to a larger Bragg angle. In using data from a diffraction experiment, these higher order reflections correspond to higher resolution.

[...]But it's claimed that [...]

The formula given by the OP is for cubic systems only. For other systems, you get more complicated formulas. For a simpler general treatment, you have to introduce the reciprocal lattice parameters $a^*, b^*, c^*$, and the relevant relation (so-called Laue condition) is:
$$ d^*_{hkl} = h a^* + k b^* + l c^*$$
To obtain $d_{hkl}$, take the reciprocal of the magnitude of $d^*_{hkl}$ .


text:
Why does it so long for a Lithium-ion battery to be charged from 98% to 100%?
Why does it so long for an Lithium-ion battery to be charged from 98% to 100%?
It is not even comparable to the effort I exerce to pump my bike tires from 95 psi to (the minimally usable) 100 psi. There the effort is a few times more than that needed to pump from 35 psi to 40 psi. With Apple's Lithium ion batteries, it takes at least an order of magnitude more to charge from 98-100% as it takes to charge from 8-10%.
Does the analogy hold? Is it actually chemically harder to charge? Is it detrimental to charge any faster (for heat, perhaps)? Is the internal charger simply delaying the charge to reduce the duration it will stay fully charged (because that itself is detrimental)?
The difference in rate-of-charging may be particularly noticeable this time because I was using a 20 W charger.

Well, in some sense, chemistry is involved in everything, from psychology to cosmology, but it does not mean chemists are experts for everything.

@Poutnik I'd have to dig out the reference, but chemists and other people who work in areas that are more inter-disciplinary tend to be better at understanding more things.

@Zhe I do not disagree, but people not involved in chemistry often stretch this assumed knowledge a lot, taking chemists almost as magicians who control the mystery of knowledge of everything. :-)

What does it mean (to Apple engineers) that the battery is 98% charged? How do they measure it?

@KarstenTheis The output voltage probably correlates well with the amount of charging.

@KarstenTheis These devices usually implement hardware charge counters, based on current integration. This charge is compared to the nominal or actually measured capacity.

@Zhe Dependence voltage versus charge is highly nonlinear and depends on the way of cell usage and age.

@Poutnik Yeah, I'm guessing even sampling the voltage is not an easy thing to do. I didn't know about charge counters, so thanks for pointing that out.

@Zhe I must correct myself here. Phones do have current sensors, but the passed charge is , at least at some devices, counted by a software way by current sampling. Unfortunately for charge management, Android devices seems to derive charge percentage from the voltage, so the percentage scale is nonlinear, dropping much faster near 0%, compared to near 100%.  I have no  personal experience with iPhones nor other Apple devices.

It could be more chemistry oriented if we assume that 98 and 100 refer indeed to the amount of charge actually stored in a Li ions battery.

The short answer is: The ending stage uses much smaller charging current, so it lasts longer. TCharging of any cell in the contant voltage mode, not limited to lithium cells, leads to asymptotically decreasing of the charging current and progressively slowing down charging proces.
The reason behind is the charging voltage must not cross the maximal allowed voltage to avoid detrimental or even dangerous side chemical processes. These processes  may include :

forming metallic lithium on surface of graphite-lithium intercalate $\ce{C6Li_x}$
reaction of lithium with the dialkylcarbonate solvent
irreversible collapse of $\ce{Li_xCoO2}$ structure, forming $\ce{CoO2}$ and $\ce{Li2O}$, if too many of lithium ions leave the compound.
forming gases from side reactions, like alkanes and ethers.
a rupture of the cell
ignition of the volatile flammable solvent, powered by metallic lithium.

This observed slow end of charging phenomena is not iPhone specific, but applies to all devices using Li-ion/Li-Po cell technology. It is matter of the lithium cell protection-circuit design and the device charging control software/firmware design.
The usual Li-ion/Li-Po charging schema is

Optional Trickle charge ( the link terminology ) by  low current to revive deeply  discharged cell.
Charging by a constant current until the battery maximal voltage threshold is reached. This happens typically when 70-80% of the actual cell capacity is reached, the lower value for older cells with greater internal resistancy.
Charging switches to the constant voltage charging, with the charging current asymptotically decreasing toward zero. Charging is then usually stopped when the currect drops below the chosen minimal charging current threshold. This mode provide lower charge increase per time, as the average current is lower.
Some charging algorithms use as the last  step the "topping" charging mode of staying at the maximal voltage for some time, to push into the cell the maximal charge ever possible, with expense of the cell life-time. This topping significantly prolongs the total charging time, as the average charging current is minimal here.

I have answered from my memory,but there is plenty of related online resources, like e.g. powerelectronics - charging-liion-batteries-for-maximum-run-times

That all makes sense, but what is your level of confidence in your answer? When you write "chemistry is involved in everything, from psychology to cosmology, but it does not mean chemists are experts for everything.", I get the impression that you're taking a stab at the answer, but are not really sure. Can you clarify either way?

@Sam I have not fully got your last sentence before the ending question. I have not answered from the chemistry point of view, as chemistry has little to do with that ( It has, but indirectly, via the critical cell maximum voltage). I have answered from point of view of many years of experience in lithium cell maintenance. All chargers charge them asymptotically, so it is clear that the ending amount of charge takes a lot longer than the same but starting amount of charge. –

By other words, I am quite confident in that, what is based on quite a lot of reading and experience. I think if you searched about the topic little bit before asking, you would reach the similar conclusion and would not even ask. It is quite widely spread public knowledge about CC/CV charging of lithium cells.

If the question is junk, let's not sully the site with noise. Please delete your answer, so that I can delete my question. Thanks.

No need to delete, there are much worse questions here.

@Sam you should be aware that the answerer had invested time and effort into answering your question and you shouldn't tell him to delete the answer(which means all the time and effort will go to waste). Everyone is welcome to contribute here and your question though closed will be helpful for future readers (only difference is that you will not get further answers) and as Poutnik said, there are much worse questions. If you got your answer, please accept the answer by clicking the tick sign or edit your question to clarify your doubt.

@NilayGhosh AFAIK there is another difference that only the author, moderators and senior users with closing/reopen vote privilege can see a closed question. Try to log out and find the question.


text:
When writing a paper should I describe the procedure thoroughly?
I have not written any article on a chemistry subject yet, and I want to use good practices from the start.
What I have noticed reading papers, is that a lot of times it is not specified the exact procedure that is followed in a reaction, for example by saying the exact tools that are used, or the parameters used in a technique, not even in supporting information documents.
At first I thought that some information was not really relevant, but as I read more articles I found more things like “X product is centrifugated” or “Y is sonicated”, without further detailing the exact procedure that was followed, the set parameters or even the devices that were used, which can obviously be very important for a given reaction.
My opinion would be that the more information the better and I think that by omitting that information you are wasting time and resources to any researcher that tries to replicate your experiment, as well as making it difficult to check the replicability of the procedure.
Then my question is, how far should I go in describing the procedure? Should I mention even which glassware I use for a given reaction? Maybe sometimes it is pretty obvious what should be used, but it is additional information that does not hurt to know and may be useful for obtaining the exact same results. However, it is as far as I have seen not a common practice and there must be a reason behind that.
Maybe omitting this information is preferable in articles as it uses less words and it avoids making articles excessively wrong, if that is the case, what about reports or thesis with a less restricting length limits?
Thank you for your time.

I will assume that you are a student or at least not that experienced in experimental, wet lab chemistry based on your tags. If you are writing for fellow novices, the more descriptive the better. A good source is Organic Syntheses. http://www.orgsyn.org/

Usually chemists like to think they have an understanding as to what aspects of the reaction matter for reproducibility, and what aspects aren't (and can therefore be omitted). Some reactions are very forgiving; others less so. To an extent, that is something you gain through experience with the particular reaction you are reporting. But in general, people are more likely to appreciate extra detail than the other way round. Also, sometimes people find that they are quite wrong, and that some previously overlooked bit was actually crucial for reaction success :-)

In better journals, sometimes more detail is in the supporting information. I find most academic procedures shockingly poor in their descriptive detail and I'm not always sure if it's unintentional

@orthocresol I agree, as a new chemistry researcher I feel like I should write down every little step, as I'm not all so sure that even using a given glassware over another can really affect the reaction outcome, for example heat can distribute differently depending on the shape or the size of the container, which i know can be an important factor when synthesising things like nanoparticles, on the other hand, as a new researcher I see most of the papers lack this information, and I don't want to make the mistake of putting too much irrelevant information.

There can be too much detail. I always used a lot of detail in my papers, but once I started teaching and reading lab reports on a regular basis, I realized that my students like to write too much in their methods sections, often including what labels they put on each tube. But in general, I think you should assume that anyone reading your paper knows about as much as you knew about the topic, and you just need to give them enough detail to repeat your procedure. And I definitely went into more detail in my thesis than in my papers.

"Charles, it isn't necessary to tell us the incredibly obvious.
The merely obvious will do." -- Hawkeye Pierce, "Life Time" ;-)

Use the following criterion in mind when you write a method in a paper: If you were reading this synthesis procedure from a book or a paper, could you replicate it or not? Do you feel anything is missing for proper duplication? When I review a methodology section of a papwe, I keep this criterion in mind: "Using this written information, can I reproduce his/her work?", otherwise I ask for more information.
Since you mentioned synthesis, a good sample procedure could be from Organic Synthesis http://www.orgsyn.org/
See how they write good procedures, and these methods are verified by independent person. Additionally, some journal put restrictions on word count, but still you want more details. For that, Supporting Information file exists on most good quality journals. I once saw someone even added pictures of their synthetic procedure/apparatus. Nobody will mind but it is not common.
Odd cases: Mentioning the source of glassware might be important if glassware surface were affecting the reaction. I "re-discovered" this only once. Sometimes the age of a solvents might affect the reaction. Tetrahydrofuran when used a solvent can have traces of peroxide which can show new unexpected products.
Now, there is a trend these days to intentionally hide some information so that competitor's cannot catch up with your work or write the method in a sloppy bare minimum way. This is really sinister but it happens and such practices must be curbed. Ayway scientists are not saints-they are human beings like everyone.


text:
How to break bonds of tea on a curtain?
If I throw tea on my curtain , it looks as if the curtain has absorbed tea molecules but is it true to say that since matter has spaces between them.So these tea molecules go into that space and stick with the molecules of curtain which makes them hard to be removed.
If I want to break bonds of tea molecules with curtain molecules,I rub it so that there is heat formed which break the bond of tea with the curtain.After some time , The colour does appear to go away.
Is it that it got stuck with the molecules of my hand after breaking with curtain molecules ?
Is there any other way to completely remove that liquid from it ?
I did spill tea on my curtain but found it to be really fascinating.

"Tea Molecules", "Curtain Molecules"??

Tea and curtain  are made up of large no of molecules.

When I talk about them at microscopic level,that is what I wanted to write.

Well if you just clean by friction certainly there is some transfer from the fibers to your hand. Perhaps some grease on our skin work as a kind of solvent. Stuff like that. The bonds you are referring to are not really hard chemical bonds, in the sense that every component retains its own structure and individuality. This sort of "adhesivity" involves various interactions of electrical nature or, in cases, the so called hydrogen bridge. The latter goes in the direction of a chemical bond. In all cases there must be an affinity between the stainer and the stained, as well as a specially...

... tailored one in case of detergents. Search for detergents, soap, and their working mechanism. I am sure you'll get an insight about this kind of phenomena. Some dirt can really stain and happily stay where it landed. In this case one need some more chemistry, i.e. damaging a dye will cause the stain to disappear...

@Alchimista am I correct about spaces between matter part ?.Thank you for telling til here.I will search what you have written.

Well of course there is alway space. At first between fibers. Let me put it this way: there is much more surface than that of the curtain.

As stated above tea is made up of hundreds of different substances, add milk to it and it becomes a more complex mixture. Anyway, detergent formulation chemists and textile chemists study those questions that you are trying to ask. Similarly, the cloth used in making a curtain is also a chemically complex material.
Keep in mind the molecular dimensions. Molecules are really really small, so the spacing between the threads is like the whole universe to the molecules. So the space between the threads/fibers of the curtain do not matter at all.
Many food stains are compounds which absorb visible light. Usually such molecules have benzene rings as a part of their structure. There are several possibilities. The stain is chemically bonded to the fiber (worst case) or it is simply adsorbed. Tea has plenty of polyphenols which absorb light. One can use substances which destroy the visible light absorbing part of the molecule or the chromophores. As a result you see that the stain has gone away.
If you rubbed your tea stain and you saw some lightening of the stain, it could just be an optical illusion because the way threads are arranged and they reflected light. The heat generated from rubbing cannot make or break bonds, it is a very low temperature. Tea has already been boiled in water.


text:
Why do reusable heat packs have a limited time frame before the reaction becomes irreversible?
I have two "rechargeable" heat packs which from my understanding operate on the phase change
$$ CH_3COONa_{(aq)} \rightleftharpoons CH_3COONa_{(s)} + \Delta H.$$
Triggering the crystallization releases heat, and leaving them in boiling water returns it to its supersaturated form. The packs say if they are left in solid form for more than two months, they may not be able to return to aqueous form. I left one solid for many months, and sure enough, I couldn't quite redissolve the crystals. It gets 90% dissolved, but the remaining crystals just trigger the forward reaction as the pack re-cools, even after multiple attempts. What mechanism prevents this?

Writing sodium acetate as NaCH3COO is quite unusual, even if not necesserily wrong. The usual way is CH3COONa, following rather the organic way of writing  structure formula than the inorganic ways to have the metal first. For the latter, it has the metal first, as there is the acidic hydrogen first. For CH3COOH, the acidic hydrogen is the last. It would be more correct, if the summary formula was used, instead of the structural one, like NaC2H3O2.

Ryan - Exactly how did you try to "redissolve the crystals"?

@MaxW wrapped in a cloth, boiling them in water for 15 minutes. Those are the instructions on the product

Ryan - Couple of followup questions. (1) Are you reasonable close to sea level? (ie not in mountains at 10,000 feet.) (2) You boiled them for 15 minutes after the water + pack started to boil? (ie you didn't start timing when you added cold pack to boiling water...)

@MaxW yeah I’m at sea level. I dropped them in cold, but actually, one time I left them in the pot and forgot about them for like an hour and most of the water boiled away and the one bad one still didn’t fully revert. So I think it was thoroughly heated...

Why do reusable heat packs have a limited time frame before the
reaction becomes irreversible?

This reaction should never become irreversible until and unless there is bacterial growth or loss of water for some reasons. For example, one manufacturer explains that the pads can re-used for 18 years! http://www.thermo-pad.com/frequently-asked-questions/
We normally think that plastics are impermeable to gas exchange or other small molecules. They are not just like our skin. Do remember that a balloon eventually gets deflated with time?  Moisture is no exception.
Search Moisture permeability of polymers and you will find 100s of articles.

How many times can they be used?
There is no limit as to the number of times a Thermo-Pad can be used.
There are pads still in use after eighteen years. The pad will not
operate if there is a hole in the vinyl. Thermo-Pad is manufactured
from the same vinyl as is used in water beds and comes with a one year
warranty.
What are the white crystals that form in the liquid pad?
If the Thermo-Pad has been stored in its crystallized state for an
extended period of time, a small amount of water can evaporate from
the pad and cause the solution to become more concentrated. This
higher concentration is indicated by “snowflake like” crystals in the
liquid pad. The pad will still operate but at a slightly higher
temperature. This water loss can be substantially reduced by storing
the pad in its liquid state and may be reversed by extending the
boiling times and by leaving the pad in water to cool.

The sodium acetate that shows this behavior is not anhydrous $\ce{NaCH3COO}$ as you write. It is the hydrated salt $\ce{NaCH3COO·3H2O}$. The anhydrous salt does not show the phenomena of surfusion. Only the hydrated form does it. If you keep the solid form of the hydrated salt a long time, I think it will slowly loose water by evaporation through the plastic bag. So it will be partly transformed into anhydrous acetate. And by heating the whole system, it will never melt entirely. So it may never return to its supersaturated form. It is necessarily to add a few drops water to get supersaturation again.

It is trihydrate $\ce{CH3COONa . 3 H2O}$, not dihydrate.

Are not the packs supposed to be heated in water? Would storage in a humid place preserve the pack?

@Poutnik, You are right. It is a trihydrate. I am sorry.

@Alchimista. I don't know whether storage in a humid place will preserve the pack. Maybe !

So why does it lose moisture in solid form, but no evaporation in liquidform? Is it a surface area thing?

Huh, not sure how this works out. The pack surely contains residual unbound water. Do you have a reference that the hydrate is unstable at RT in water?

It doesn't seem that the manufacturer would choose a plastic that had a high permeability for water. However planned obsolescence isn't just a baseless conspiracy theory.


text:
How can one say pressure is constant in thermodynamics?
When a gas expands by  b charles law pressure increases
$$\frac{p}{t}=k$$
thermodynamics says:
$$dQ=dU+PdV$$
so herfe one is considering constant pressure but how b charles law says it changes?So shouldn't it be:
$$dQ=dU+PdV+VdP$$
can't we apply a variable external pressure?

The volume work done by the system is dW=p.dV, not p.dV + Vdp. You may confuse it with H=U+pV -> dH=dU + d(pV) = dU + p.dV + V.dp.

Charles law is at constant volume,

Sometimes in thermodynamics we wish to constrain pressure to be constant.  [Or we may wish to do a thermodynamic analysis for a system that is at constant pressure, which is a common laboratory condition.]   However, that is a choice.  Pressure is not generally constrained to be constant in thermodynamics.
For a closed system:
$$d\text{U} = \text{đ}q + \text{đ}w$$
[Note that I'm using a different sign convention from you for work, but the results come out the same.]
If the only type of work is $pV$- work, then  $\text{đ}w = -p_{ext}dV$, where $p_{ext}$ is the external pressure applied to the system by the surroundings.  Thus we have:
$$d\text{U} = \text{đ}q - p_{ext}dV,$$
where the pressure can change in any way you please during the process.
If the process is reversible, then (in that idealized case) the system is in continous equilibrium with its surroundings. As as a consequence, $p_{sys} =p* = p_{ext}$, and we can write:
$$d\text{U} = \text{đ}q - pdV$$
[*The convention is that if a subscript is not present, the variable is assumed to apply to the system.]

Ok so if pressure changes that means work is done and the basic violation of an assumtion m that there isno exteenal work.

No, a change in pressure does not necessarily mean work is done.  For instance, if you heat up a gas at constant volume, the pressure will increase, but no pV-work is done, because dV = 0.

So if both pressure and volume change that means external work is done?

Not quite.  If the volume changes, and the pressure is non-zero, then work is done, regardless of whether the pressure changes or not.  Thus the better statement is "If the pressure is non-zero, and the volume changes, then work is done."  It's helpful to understand the physics underlying work:  Work is a force exerted over a distance.  So in order to have work, you need force (≡ pressure) to be non-zero, and distance = change in position ≡ change in volume to be non-zero.

I am not talking about work done by the system but when work is being done on the system then.

What I wrote applies regardless of whether work is being done by the system (expansion) or on the system (compression). The only difference between the two is the sign.  Note also that the work and heat flows for a process are always defined by what changed in the surroundings.  Thus the p you use in calculating work is always the external pressure, regardless of whether the work is done by the system or on the system.  I think if you have more questions about work you might want to search chem.se (this site) and physics.se, and then, if you're still unclear, pose a separate question.

Charle's Law has nothing to do with the first law of thermodynamics, it simply states that for a gas, at constant pressure, volume will be directly proportional to temperature. Whereas, the first law of thermodynamics states that $\text{d}U=\text{d}Q+\text{d}W$, where $W$ is the work done on the system, which is equal to $-P\text{d}V$.
Now, it directly comes from the definition of work done, which is $\text{d}W=\vec{F} \cdot \text{d}\vec{x}$ and so if external pressure applied is $P$ over the cross sectional area $A$, we will have $F=PA$ and so $\text{d}W=-PA \text{d}x=-P\text{d}V$. The negative sign is taken as the direction of displacement and force applied by atmosphere is in opposite direction, so to make it positive, negative sign  is multiplied.

How can you say it is a constant "Force"?


text:
Will Raman photon counts/Spectral resolution increase if laser beam becomes parallel?
Most Raman spectrometer has the laser converging into focus with the sample at center like this.

Usually how many millions of H2O atoms will be covered by one spot size?
What will happen if the laser beam would be made parallel instead of converging. Will it increase the counts of the spectrum? Will it improve the spectral resolution? Or will both become worse?
Water Raman:

Do you have any own thoughts about this? Please elaborate!

The collection optics image a small volume in the sample, you need to get as many photons into this region as possible to get strong signals. This normally means focussing the laser, but parallel beam would work if intensity is sufficient. The resolution is limited by slits and grating or CCD pixels per mm and grating, as appropriate , assuming that you laser bandwidth is narrow, as it usually is. You could try going to 2nd order to increase resolution but intensity will be lower. (You should look up Confocal parameter and Gaussian beams to understand laser focusing )

Spectral resolution is not dependent on the laser intensity. It is determined by the state of the sample ( gas, liquid or solid) and more importantly by the components of the instrument itself. For example, what is the slit width anf the number of lines on the grating? What type of detector is there? If it is charged coupled device (CCD) then the spectral resolution will also be dependent on it.

So what happens if the laser is made parallel covering more sample than just spot size? What will happen to the spectrum?

It might affect the intensity only. Some surface Raman experiments which use polarized light are angle sensitive.

I disagree with the statement that spectral resolution is independent of laser intensity. You can have power broadening (saturation effect) and Doppler broadening due to the wave front curvature for example. It is of course true that these effects are probably much smaller than the instrument effects you mention.

Right, those effects must be quite small for the Raman spectrometer this OP is using (as seen in his other posts).


text:
Energy levels of combined orbitals
I saw the MO diagram for a NH3 molecule but couldn't understand why the combined atomic orbitals were in the positions they were, and if these positions were fixed.

Why is the sigma anti bonding lower in energy than the pi anti-bonding? Is it something you have to learn off by heart for each common molecule?

This diagram is very awful. It mixes hybridisation and group orbitals, which does not make sense. In a way it is mixing MO theory and VB theory, but in the worst possible way...

@Martin-マーチン do you have a book/chapter recommendation where it is explained better?

hav not read, but looks better: https://en.wikibooks.org/wiki/Advanced_Inorganic_Chemistry/NH3_Molecular_Orbitals This looks quite correct: https://commons.wikimedia.org/wiki/File:NH3_MO.png Basic texts on quantum chemistry should cover this in detail... see Resources for learning Chemistry

In the strictest sense, the orbital energies are simply eigenvalues of the multi-electron Schrödinger equation; unfortunately, the multi-electron Hamiltonian takes a nasty form and is impossible for us to solve exactly, so we can't get any insight on the energies from that approach...
Instead we inadvertently resort to LCAO-based methods, whereby exploiting symmetry the interacting atomic (or symmetry) orbitals can form molecular orbitals of the correct form. The energy of these is another issue; for some back-of-the-envelope calculations, severe approximations such as Hückel MO theory can be useful for a rough idea, but more accurate estimates just require much more complicated computational methods (e.g. the acclaimed Hartree-Fock).
As these methods resort to a large basis of AOs (and not just the simplified 2 AOs mix in- and out-of-phase to give 2 MOs picture) and other complications such as electron correlation, I'm afraid there is no simple rule to give an accurate enough estimate unless thorough calculations are made. So, to forwardly answer your question, rather impractically we can't know a priori the ordering of the orbitals in the general case.
Certain semi-empirical rules (based on the experience from multiple common systems and "chemical intuition") exist, such as the (often useful) approximation formulated for organic molecules $\sigma < \pi < l.p. < \pi^* <\sigma^*$ though this is very rough and suitable for organic systems, usually.
In the case of ammonia (as given in the question) a particular ordering is found; it should be noted that the two types of bonding interactions are both $\sigma$-interactions (i.e. locally cylindrically symmetric; there is no nodal plane within each individual bond). Despite the double degeneracy, the $e$ set still constitutes a $\sigma$-type MO (with a node in the overall MO shape, rationalising a higher energy than the $a_1$); so the ordering in ammonia is $\sigma_{a_1} < \sigma_e < lp_{a_1} < \sigma_{a_1}^* < \sigma_e^*$ -not so counter-intuitive after all, just different types of $\sigma$ MOs possessing different energies. (As a sidenote, the doubly deg. $\sigma_e^*$ are those allowing ammonia to act as a weak $\pi$-acceptor, placing it higher than simple $\sigma$-donors in the spectrochemical series.)
Summary: MO energy calculation especially in polyatomic species is hard and done with complex computational methods. Unfortunately, then, we can't have a general rule for the energy ordering in advance. However, often we can make educated guesses in analogy to systems we have already seen.
(For a more detailed look into the methods used to obtain the energies see Atkins & Friedman Molecular Quantum Mechanics, 4th ed.; Chapters 8–9)


text:
Why does the emission of dansyl group diminish with time in this experiment?
I don't know how to make sense of this. I what is happening when an enzyme (carboxypeptidase, which contains tryptophan as its only chromophore and uses a Zn(II) ion in its active center) hydrolises a dansyl-Gly-Phe peptide. The enzyme and the peptide are mixed and the reaction is studied by fluorescence measurements: the whole reaction is excited with 280 nm light, and emissions at 310 nm (for Trp) and 550 nm (dansyl) are studied (see the two graphs below).

All I know is that this is an energy transfer process, which is known to lose effectiveness following a $1/r^{6}$ law regarding distance between emitter and absorber. Also, I know how Trp and dansyl emission spectra overlap, and I know that the enzyme is active but does not show fluorescence if Co(II) is used instead of Zn(II). My guess is that Trp emits light which is absorbed by the dansyl group due to the overlap in their emission/absorbance diagrams, but I don't see why the dansyl emission graph would go down w/ time. Isn't the number of dansyl groups constant, even if they are not bonded to the peptide in the end.

I agree it is confusing. You expect the trp fluorescence to be quenched by dipole-dipole transfer to the Dansyl so the trp fluorescence decreases and the Dansyl increases as the reaction proceeds. If the detector measures negative signal as the intensity increases the data makes more sense. Ignore the horizontal signals and initial signal 1/4 of a square  as this is probably due to scattering from the duration of the excitation source used. (btw Forster dipole transfer is  non-radiative transfer, D to A, not fluorescence and reabsorption. The spectral overlap ensures energy conservation.)


text:
Molarity of a mix between solid and solution
$\pu{50 g}$ of solid $\ce{NaOH}$ is added to $\pu{1.5 L}$ of $\pu{2.4 M}$ $\ce{NaOH}.$ Water is then added to the mix until the solution reaches $\pu{5 L}.$ I need to find molarity of the final solution.
I am a bit confused because I have different answers with my friends, my approach was as follows.
The molarity $x$ when I add solid $\ce{NaOH}$ to $\pu{1.5 L}$ $\ce{NaOH}$ assuming $n_1 = n_2:$
$$x = \frac{\pu{1.5 g} \cdot \pu{2.4 M}}{\pu{1.55 g}} = \pu{2.322 M}$$
The molarity $y$ after I add more water also assuming $n_1 = n_2:$
$$y = \frac{\pu{1.55 g} \cdot \pu{2.322 M}}{5} = \pu{0.72 M}$$
What is wrong with my approach?

Always try to approach the task first by general algebraic solution and plug in the numbers in the end.  It has several benefits.
1/ It allows you better grasp on underlying principles.
2/ It allows both of you  and readers to more easily spot a solution error or conceptual mistake.
3/ It gives to the Q/A a higher and more permanent value to be reused/applied for similar questions.
4/ It follows the site policy to focus on explaining principles and procedures, rather than solving of particular tasks with the value for 1 person only.
5/ You will avoid the intermediate rounding errors.

Additionally, always use units in expressions with literal numbers like above.

is it still wrong sir?

Try to plug in the numbers AFTER you get the final algebraic expression that provides the result. For formatting of chem/math formulas/expressions, use can use MathJax,

Why not to use e.g. $$c = \frac {n}{V_2} = \frac {\frac{m}{M} + c_1 \cdot V_1}{V_2}$$

How moles of NaOH do you have in total? Your final volume is 5L so divide (moles of NaOH) by 5 to get the molarity.

@Poutnik - You edited the question and added units to the OP's mathematical expression $1.52.4/1.55$. For the 1.52.4 part I think the OP was trying to get the moles of NaOH in the solution. I think the 1.55 in the denominator came from assuming that the solid NaOH had a density of 1 and adding the volume of the NaOH solution to get that the final volume of the solution after adding the solid NaOH was 1.55 Liters.

@MaxW AFAIK, I have not added units, intentionally, aside of changing gr  to g, unless accidentally. OP had added them after my early comments, before I applied the formatting.

@Poutnik - Sorry, you're correct. The OP added the nonsensical units in the first edit of the  post. Your very positive contributions to the site should have given me pause before I made such a stupid accusation.

@MaxW Your appreciation is very welcome. :-)

I am sorry to not make a correct format for asking questions, I have already found my mistake. Thank you for the advise sir @Poutnik, I'll write better questions then. Thank you too sir MaxW for the formula.


text:
Searching for a fireproof sponge?
(I know this is a bit off topic, but I don't know where else to ask. If you know a better StackExchange site, please tell me)
I'm working on a project, and I want to basically create a "flaming sail". For this, I'm planning to drench a cloth in methanol and then lighting it, and for this, I need a fabric that both retains a lot of moisture and is flame proof, as this should be reusable.

https://www.mscdirect.com/browse/tnpla/06641153?cid=ppc-google-New+-+Welding+%26+Soldering+-+PLA_s1SU9UK6M___164124449405_c_S&mkwid=s1SU9UK6M%7cdc&pcrid=164124449405&rd=k&product_id=06641153&gclid=CjwKCAiA-_L9BRBQEiwA-bm5fkdEzExtD89KdCgcNdt7GYiCQovemOQ0xRNz67h4L4w0lVb1y9PcHhoCVksQAvD_BwE  Asbestos fabric might work but the downside to experimenting at home with asbestos just is not worth it.  So why post this suggesting?  I hope it a) directs you to something that will work and b) keeps you away from something that is just not worth doing.

How large? Can you use a metal fabric?

Perhaps glass wool or the glass/ceramic tape used by glassblowers as a safe asbestos substitute.

Mix the methanol with water, perhaps 50/50%, and most fabrics will be cooled by evaporation and not burn. BTW, add a little salt, copper sulfate, lithium chloride or other salts with bright spectral lines to make the flames colorful.

You could try a fiberglass or carbon fiber welding blanket.  Here's an example: https://www.harborfreight.com/6-ft-x-8-ft-fiberglass-welding-blanket-95015.html  One doesn't normally think of fiberglass as being absorbent but, with the right weave, it could have enough channels to absorb methanol via capillary action.

One possibility is to imbue with Sodium bicarbonate. The action of heating the Baking Soda, $\ce{NaHCO3}$, as a powder absorbs energy while releasing $\ce{H2O}$ and $\ce{CO2}$ to assist in retarding the flame per the reaction:
$\ce{2 NaHCO3 (s) ->  Na2CO3 (s) + H2O (l) +  CO2 (g)}$
Also, lesser-known, but important per a source, to quote:

Even though the exact modes of action are not well established, the $\ce{NaOH}$ formed may promote catalytic recombination of the radical species needed for flame propagation, causing the flames to extinguish [23]

Note, it is known that heating Sodium carbonate in the presence of $\ce{H2O}$ can be a source of $\ce{NaOH}$ (see comments here) to facilitate the above cited mechanism.
So, safe inexpensive Baking Soda is my recommendation.

Fiberfrax is the standard material for high temperatures in refinery, chem. plants, etc. There are also other brands , all readily available on the net. Ads do not give detailed composition other than "alumina and silica" .  Fiberglasses may not have  high melting /softening temperatures. There are higher temperature fibers like thoria, but are expensive and not easy to find.


text:
Can we observe the shapes of MOs?
Condensed question formulation:

Is there an experimental method to directly visualise the 3D form of a MO wave function, or at least the electron density associated with it?

Full statement:
Molecular orbital theory is a crowning achievement of modern Chemistry: it is by far the most accurate description of chemical bonding we possess and accounts for the vast majority of related phenomena.
An array of evidence exists to confirm predictions from MO theory. Ab initio computational methods such as Hartree-Fock are able to produce a very accurate description of orbital coefficients and energies, which match experimental data for the energies (via photoelectron spectroscopy etc.) or predicted reaction mechanisms/FMO interactions (in kinetic runs).
But do we have any experimental evidence for the shape of MOs?
I'd presume the most feasible way of observing some measure related to the MO wave function $\Psi$ would be to follow a probabilistic approach and try imaging the electron density assumed to be given by $|\Psi|^2$.
Of course, X-ray diffraction first springs to mind; the notorious issue with that being the phase problem. In my superficial understanding: due to the loss of phase information, in going from the XRD pattern to the electron density isosurface (via ab initio methods), essentially we model some initial guess phase and try to fit it to the intensity data.
Since usually the location of nuclei is of far greater practical interest, the formalisms developed to model the phases (e.g. Hansen-Coppens) tend to focus on charge densities localised on each atom (though they are not necessarily of spherical symmetry). This is useful in actually getting the form of the molecule, but produces a picture different to the expected electronic occupation of delocalised MOs.
Is it possible to instead produce the delocalised picture if our ab initio phase guess is based on a RHF calculation? But even if possible, does this have any more legitimacy than the localised picture? I.e. it seems, as far as XRD is concerned, these are equally possible interpretations of the same intensity data where we just forced some phase information as convenient.
Sorry for the long description, this is an in depth presentation of my thoughts on the issue thus far. Any experimental technique/fundamental aspect of MO behaviour I'm missing would be appreciated. But the bottom line is, as stated in the beginning, can we experimentally visualise the shape of MOs or at least their modulus squared?

There are two problems here: 1) Molecular orbitals coming from a one-electron model of the wavefunction, i.e. they are essentially an approximation. 2) Wavefunction, just like the molecular orbitals are not observables. In your question, you said that 2) is not an issue, as you accept electron densities, too, which are observables. Still, we have a problem with 1). While MOs are not real physical objects, a possible solution to your problem is to find physical properties that well approximated in a simple orbital picture / one MO Note: HF is far from an accurate description of chemistry.

https://chemistry.stackexchange.com/questions/5376/molecular-imaging-any-surprises-to-be-had

https://chemistry.stackexchange.com/questions/57784/has-anyone-even-taken-a-picture-of-a-molecule-to-confirm-the-geometry-predicted

Thanks for the suggestions and the quite helpful sign towards the right direction! Though not exactly the same question, the second link brought STM into my attention and eventually helped me find this paper on STM for individual MO imaging -which is the sort of technique I was asking about.

Molecular orbital theory is a crowning achievement of modern Chemistry
This grand value judgement is matter of opinion, but ok.

it is by far the most accurate description of chemical bonding we possess
well, that's just false.

Ab initio computational methods such as Hartree-Fock are able to produce a very accurate description of orbital coefficients and energies, which match experimental data for the energies (via photoelectron spectroscopy etc.) or predicted reaction mechanisms/FMO interactions (in kinetic runs).

The very question you pose is addressed thoroughly in this open access work:
https://www.nature.com/articles/ncomms9287 .
The short answer to your question is that the electron density can be mapped using a technique akin to diffraction as described in their work.  You mentioned a distinction between the electron density and the orbitals, and the orbital is not itself observable; but as you suggested, the electron density can be mapped.

Thank you very much, this is outstanding! I was not familiar with the ARPES technique, which seems to provide the exact answer I was looking for.
Two quick additional questions (for anyone with better understanding in these methods than my half an hour earlier ignorant self): (a) is the angle-resolved part of ARPES which makes it not suffer from the phase problem of XRD? (b) do any further techniques as suitable as ARPES or STM (mentioned elsewhere in comments) for single MO imaging exist?


text:
Which of the following is the strongest oxidizing agent?

Which of the following is the strongest oxidizing agent?
A) $\ce{I2}$
B) $\ce{Ag^2+}$
C) $\ce{Cu^2+}$

The answer is apparently B) but I think it has to be A) because it is the only non-metal which actually wants to lose electrons. I just wanted to confirm whether it’s an incorrect solution or my own understanding is wrong.

Have you checked the electrode potential tables?

A compound wanting to lose electrons is a reducing, not oxidating agent. BTW, are you aware of any  Ag(II) salt that is used as commonly as I2 or CuSO4 ?

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Standard_Reduction_Potential


text:
Why is steam assumed to be an ideal gas here?
A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporisation at 100
o
C.
Δ
vap
H
⊝
for water at 373K =40.66 kJ mol

Because you need to assume it's an ideal gas to convert $p \Delta V$ to $\Delta n R T$  (recall that, for an ideal gas, $pV=nRT$).  The other assumption being made is that $\Delta V$ due to the loss of the liquid water is negligible.

Liquid water to gas has very negligible volume change ? @theorist

No.  You have two effects:  (1) you are decreasing the moles of liquid water; (2) you are increasing the moles of gaseous water by an equal amount.  Effect #1 causes a decrease in volume.  Effect #2 causes an increase in volume.  In doing the above calculation, you are ignoring effect #1, and only considering the change in volume due to effect #2.  I.e., you are not subtracting, from $\Delta V$, the change in the  volume of liquid water, because it is negligible compared to the gain in volume of gaseous water.  Specifically, $| \Delta V_{liquid}| \approx 1/1000 | \Delta V_{gas}|$

Using photos/screenshots of ( often even hand written ) text instead of typing text itself is highly discouraged and can lead to ignoring a question by potential helpers or even its closure. The image text content cannot be indexed/searched for nor reused in the answers ( and scripts are challenges to interpret ). That puts unnecessary extra effort on others. Consider  copy/pasting or rewriting of the essential parts.
Use MathJax for eventual formatting of formulas or expressions.

@Poutnik I have typed the question which is what needs to be searched.You can’t search the question by searching for solution.I get your point of not putting texts but if sometimes it is a big thing to type , photos can be easier.You have option for putting images as well.Images can be reused as well.Like you have written to rewrite essential parts - I wrote the question in which I had doubt and typed the doubt too.If it too big or maybe too complex sometimes to type it , then it also takes a lot of time of the person who is typing the doubt.

You have an option for putting photos.

I have also been typing my question using mathjax in mathematics stack exchange.But sometimes if it too big , I wrote the essential parts and put others as photo.

@Poutnik I would also be happy to know then how can I save my time and improve the quality of my question

Consider saving time of others in the first place. What if one wants to quote a part of your question ? Consider also the Q/A should have a permanent value for more users, not a one time value for just you. BTW it is not a long text, there are much longer ones here.

How can I easily and saving less time , convert it to mathjax.

Good is the Markdown Shortcuts for StackExchange script running within TamperMonky or similar addon in Firefox and I guess chrome too. Seach the chemistry.meta Se site for it. E.g.selection of H2SO4 and pressing CTRL+ALT+E put there $\ce{H2SO4}$, what looks like $\ce{H2SO4}$. See also https://github.com/GaurangTandon/MarkdownShortcutsForSE

I have a tablet.

Let us continue this discussion in chat.

At the equilibrium vapor pressure of water at 298 K, the PVT behavior of water vapor is very close to that of an ideal gas.

The critical temperature of water is 647 K, its critical pressure is 22100 kPa, and its equilibrium vapor pressure at 298 K is 3.169 kPa.  So the reduced temperature is 0.461 and its reduced pressure is 0.000143.  From generalized compressibility factor charts, at this reduced temperature and pressure, the compressibility factor z is virtually indistinguishable from a value of 1, characteristic of an ideal gas.   Also, from the steam tables, the specific volume of water vapor under these conditions is 43.36 m^3/kg = 0.7805 m^3/mole.  So, $$\frac{Pv}{RT}=\frac{(3160)(0.7805)}{(8.314)(298.15)}=0.995$$


text:
vacuum pack of a substance/substrates
What equipment can I put a substance or substrate under vacuum pack?
After buying this SERS silver substrates (which can rapidly decay in air). I need to immediately put them in vacuum packaging or container. How do I do that?

The most artisanal way to do it would be having a box filled with, at least, nitrogen. Or slowly flushing it continuously. A dessicator can be already around. And a vacuum pump.

I need something not dangerous. How do you wrap them with vacuum packaging? What instrument is used to create the vacuum package in the 2nd picture?

@Jtl These are vacuum pouches, which are, for example, used in food industry, too (example).  The sealers don't need to be big (example).

Flushing said box (@Alchimista's comment) with argon instead of nitrogen could be better because different to nitrogen, argon is more dense than air and thus creates a blanket above the items in the box.  It is a bit more expensive per litre, though.

@Buttonwood of course. I said at least. But anyway the box seems not to be the desired storage. OP want to pack them. S/he need a sealer or some workaround. There are even box for food which might help.

These are meant to be used all 5 pcs at once. https://www.oceaninsight.com/products/sampling-accessories/raman/sers/  but I need only one at a time. So if I repack them briefly, would the exposure to oxygen when unpacking them degrade them? By how much? see also https://www.researchgate.net/publication/215649986_SERS_Not_To_Be_Taken_for_Granted_in_the_Presence_of_Oxygen

@Jtl Maybe an other group at your uni (e.g., inorganic chemistry, catalysis) has a glovebox providing a dry and inert intermediate storage for the set of five samples at room temperature.  (There gloveboxes with a freezer, too; but less frequently seen.)  From there, they either could handout one of the slides at a time when needed, or repackage them in samples of one piece in small pouches, too.

It's for home use. I graduated from the university dozens of years ago and no access to one now. So it has to be something to be done at home. So if I repack the 5 substrates into individual vacuum sealer. Would the brief transfer exposed them to oxygen and degrade them? Or not?

What guys? please tell me. If I buy the 5 silver substrates and transfer them immediate to vacuum package using vacuum sealer. It won't degrade the substrate right? I'm learning SERS from home. Catching up with molecular science. Thanks.

@Jtl Sealing the samples under vacuum both quickly and efficiently (perhaps sample in a vacuum sealed pouch, which in turn again is inside a vacuum sealed pouch) will increase the chances your samples survive.  Perhaps recording a reference spectrum of a pristine sample allows you to recognize if subsequent samples are still good enough after months of storage.  (Possibly the chemical vendor may provide some hints, too.) This may buy you time to keep them outside a glove box (https://www.erowid.org/archive/rhodium/chemistry/equipment/glovebox.html, https://www.youtube.com/watch?v=4nJ1FKZNQVA).