text:
Why should we not add the H+ ion conc coming from water before calculating pH of acidic/basic solution?
For eg:
If we are calculating pH of the soln of a strong acid, we shall do this:
pH= - log [H+]
Where H+ is the concentration coming from that acid.
Why shall we not add to this concentration, the H+ ions coming from water itself (By the autoionization of water) before calculating the pH?
Does water not ionize in presence of the acid/base?
Thank you.

Unless you're working with extremely dilute solutions ($\pu{\mu mol/L}$ or lower) of a strong acid, the amount of $\ce{H+}$ contributed by water is negligible for pH calculations.

It is rather removed than not added, in cases where simplification error is much smaller than error due ignoring activity cofficients <> 1.  Presence of acids or bases suppresses water auto-ionization. If there is 1e-5 mol/L of H+ or OH-, there is 1e-9 mol/L of H+ or OH- due water ionization, i.e. 10000 times less.

Pure water is very weakly dissociated, meaning that the amount of H+ and OH- is very small. If we look at the autoprotolysis equilibrium of water
$$\ce{H2O <=> H+ + OH-}$$
and its autoprotolysis constant at 25 °C
$$\ce{K_w = [H+][OH-] = 1.01\times10^{-14}}$$
According to the equation above we know that the H+ and OH- are the same, so we can easily calculate that the concentration of H+ is around $\ce{1.005\times10^{-7}}$ mol/L
Now let's assume you add a small amount of a strong acid, such as HCl, so that the concentration of this acid solution is 0.001 mol/L. If we assume a complete dissociation of HCl according to the following equation
$$\ce{HCl -> H+ + Cl-}$$
we know that the concentrations of H+ and Cl- ions are the same, and equal to 0.001 or $\ce{1.0\times10^{-3}}$ mol/L. Thus, several orders of magnitude higher than the amount of H+ ions coming from water.
So when calculating the pH we can use only the concentration of H+ from HCl dissociation, where $\ce{pH = -log_{10}(1.0\times10^{-3}) = 3}$. Or you can sum up the concentration of H+ coming from HCl and $\ce{H2O}$, where $\ce{pH = -log_{10}(1.0\times10^{-3} + 1.005\times10^{-7}}) \approx 3$ or more specifically 2.99996. But of course, this amount of decimal numbers makes no sense.

Be aware of supression of water auto-dissociation by H+ excess. With pH=3 is pOH=11 so there is 1e-11 mol/L of H+ and OH- from water dissociation.


text:
Why does the ozone layer deplete faster in winter months over the poles?
Why is the depletion of the ozone layer greater in the winter months over polar regions?
Is it because there is more release of $\ce{NO_x}$ due to the increased use of fossil fuels in heating which depletes the $\ce{O3}$ layer?

(Polar vortex).

As mentioned in the comments, "polar vortex" is one of the prime reason ozone layer depletes faster in winter months over the poles. This is beautifully explained here [1]:

Under normal conditions, the amount of stratospheric ozone depends on the amount of sunlight reaching a certain geographic area of the atmosphere. The seasonal variation is low in low latitudes because sunlight is fairly constant year-round. In high latitudes, sunlight goes way down in the winter months. Ozone typically "builds up" to higher values over the poles during the winter and early spring in each hemisphere. Because this season is offset by 6 months in the Northern and Southern hemispheres, the effect is seen at the North and South poles roughly 6 months apart.
[...] The air masses above the poles become isolated from the rest of the atmosphere during their winter and early spring seasons due to a phenomenon known as the "polar vortex". In simplest terms, this vortex is a spinning, funnel shaped region of the atmosphere that forms in late fall and early winter over a pole, allowing chemical reactions in the enclosed air mass to be enhanced due to the lack of mixing with other, lower latitude, air masses. The effect of the pollutants we have added to the atmosphere are thus enhanced in these isolated regions of the atmosphere. The Antarctic vortex over the South Pole is more effective at isolating this region of the atmosphere during the austral winter than is the corresponding arctic vortex. A second feature of the polar stratosphere that is unique and probably aids the polar ozone depletion is polar stratospheric clouds. These very high altitude clouds are composed of ice crystals, sometimes greatly enriched in nitrogen oxide species ("$\ce{NO_x}$") that can enhance the ozone degradation reactions discussed above. These ice particles can react with various forms of Chlorine in the atmosphere and accumulate the molecule $\ce{ClONO2}$, which is a source of ozone depleting Cl radicals. Once spring time comes, this $\ce{ClONO2}$ decomposes and allows ozone degradation reactions can occur.

One of the other reason is polar stratospheric clouds (PSCs).

The very low winter temperatures in the Antarctic stratosphere cause  polar stratospheric clouds (PSCs) to form. Special reactions that occur on PSCs, combined with the relative isolation of polar stratospheric air, allow chlorine and bromine reactions to produce the ozone hole in Antarctic springtime.

References:

https://www.soest.hawaii.edu/GG/ASK/ozonehole.html
https://csl.noaa.gov/assessments/ozone/2010/twentyquestions/Q10.pdf
https://uk-air.defra.gov.uk/research/ozone-uv/moreinfo?view=antarctica-hole-explained
http://cimss.ssec.wisc.edu/wxwise/ozone/OZONE5.html
https://scied.ucar.edu/learning-zone/atmosphere/ozone-layer

source: Stack Exchange

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

sort: votes


text:
Why is silver cyanide used for electroplating
(Partially answered by Safe silver electroplating solutions)
Why is silver cyanide used as an electrolyte in silver plating? (source) There are other (cheaper, less toxic) silver salts, such as silver nitrate. The linked question says that using the nitrate results "makes a fairly rough film that can be somewhat easily removed with abrasion". How does the anion affect the adhesion of the film of silver to the cathode? Why does cyanide result in a strong stable layer of silver, but nitrate (or otherwise) does not?

Silver cyanide is used for silver electroplating because of the concentration of the free silver ions in solution which is so low that it is not far from zero. Due to the high values of its complex equilibrium formation constant, the complex $\ce{[Ag(CN)2]^-}$ is practically not dissociated in $\ce{Ag^+}$ ions. If such a complex solution is electrolyzed, only the rare free silver ions are discharged at the cathode. So at the beginning of the electrolysis, these free $\ce{Ag^+}$ ions produce a discontinuous deposit of metallic atoms on the cathode.
Let's consider the mechanism of electrolysis in the microscopic level. As the first metallic atoms produced on the cathode by the beginning of the electrolysis look like small points or bumps, these tiny bumps will attract electric field lines. As a consequence, the next positive ions will get discharged most probably on this bump, making it bigger, if the solution is concentrated enough. More and more positive ions will be attracted and discharged on this point. The metallic deposit will be localized around specific points. The surface of the cathode becomes rough and course.
But if the free ion concentration is extremely low, as with $\ce{Ag^+}$ ions in $\ce{[Ag(CN)2]^-}$ solutions, the first $\ce{Ag}$ atom is too far from the next $\ce{Ag+}$ ion. This metallic ion is not attracted by the $\ce{Ag}$ "bump". It is attracted by the whole cathode. It touches the cathode and get discharged equally on its surface, independently from the position of the previously deposited Ag atom. The Ag deposit makes a smooth layer which looks like a mirror.
To summarize, the formation of a mirror can only occur in a solution of a silver complex where the free (non-complexed) ion concentration is extremely weak, like in $\ce{[Ag(CN)2]^-}$ solutions. Solutions of other silver compounds, like silver nitrate, will only produce rough silver deposits by electrolysis. No mirror !

Note the free cyanide is highly toxic. Cyanide bound to metal ions is less toxic, often much less or even less toxic than table salt. Potassium_ferrocyanide $\ce{K4[Fe(CN)6]}$ has LD50 $\pu{6400 mg/kg}$ (oral, rat), what would be about $\pu{500 g}$ for $\ce{75 kg}$ person, assuming the same LD50. $\ce{NaCl}$ has LD50 $\pu{3000 mg/kg}$ (oral,rat).
Silver cyanide AgCN is hardly used, being insoluble. It would be soluble potassium silver cyanide K[Ag(CN)2](commercially available), with silver strongly bound to cyanide.
Silver (or metal in general) ions too easily available for deposition on cathodes, as are silver ions from silver nitrate, lead to diffusion controlled metal deposition. This prefers exposed metals surfaces which are amplified. This is taken care, aside of solution thickeners, by using metal complexes like $\ce{[Ag(CN)2]-}$ that have to be broken. Additionally, cathodes do not attract anions that migrate to cathodes only to address concentration gradient or to maintain electro-neutrality of solutions.

For curiosity, metallic silver dissolves in potassium cyanide solution, evolving hydrogen. Gold needs to be pushed by oxygen to be dissolved in cyanide.
$$\ce{2 Ag(s) + 4 CN-(aq) + 2 H2O(l) \\
-> 2 [Ag(CN)2]-(aq) + 2 OH-(aq) + H2(g)}$$
$$\ce{4 Au(s) + 8 CN-(aq) + 2 H2O(l) + O2(g) \\
-> 4 [Au(CN)2]-(aq) + 4 OH-(aq)}$$

+1. [That's why $\ce{K4Fe(CN)6}$ is used in non-toxic cyanation reaction.

@NilayGhosh My professor of chemistry on high school told us once that when he had been a young teacher, he had regularly demonstrated the lack of its toxicity by eating the salt during the lecture. He later concluded that it was too crazy and stopped.

@Poutnik It's also crazy because a student might misunderstand and ingest a different cyanide salt in attempt to copy, but with a fatal result.

@Poutnik. The first equation between $\ce{Ag}$ and $\ce{CN^-}$ is not correct. It contains $4$ H atoms on the left-hand side, and only $2$ H atoms on the right-hand side.

@Maurice Try again :-)  4 = 2 + 2.

@Poutnik. The first equation (with $\ce{Ag + CN^-}$) does not contain $\ce{H2}$ on the right-hand side. This information is missing.

@Maurice It does contain H2, you may have it "behind the corner".  I see it there viewed on phone browser.

@Maurice I have line-breaked the equations.

The problem with electroplating the more noble metals such as copper, silver and gold is that these metals will react with the substrate [cathode] with a competitive immersion plating reaction. This corrodes the surface, causes an uneven, less adherent plate and contaminates the solution. There are two common methods, usually used in conjunction, to combat this: The first is to use a thin electroless coating [or strike plating] of copper or even nickel that will not conversion plate under the conditions. This is also the method to plate a non-conductive surface. [Tollens reagent is an example of an electroless plating method]. The second method is to reduce the chemical activity of the metal ion with an appropriate complexing ion. The complexing agents cause ion transport problems and require good mixing and appropriate anodes, pure silver is best. There are proprietary baths to do this some of which use cyanide as the complexing agent. Electroplating has a high art content and you will have to contact the manufacturers for the actual proprietary mixes and recommended procedures.
The classical method to measure quantitative current flow is the silver coulometer, a cell placed in series with a silver anode, a silver nitrate solution and a platinum cathode. No danger of immersion plating and I understand the plating adhered well. Electrolytic deposition of copper on a platinum cathode can give a quantitative removal of copper [beautifully even and adherent] with appropriate control of voltage and pH.

Silver cyanide is commonly used for electroplating because it is a highly conductive and efficient electroplating solution. When used in electroplating, silver cyanide ions are deposited onto a substrate by passing an electric current through the solution. The silver ions are attracted to the cathode (negative electrode) and are reduced to metallic silver, which is then deposited onto the substrate.
There are several reasons why silver cyanide is a good choice for electroplating:
High conductivity: Silver cyanide is a highly conductive solution, which makes it an efficient choice for electroplating. This is because the higher the conductivity of the solution, the faster the electroplating process can be completed.
High silver content: Silver cyanide contains a high concentration of silver ions, which allows for a thin and uniform layer of silver to be deposited onto the substrate.
Easy to use: Silver cyanide is a stable and easy-to-use electroplating solution, making it suitable for use in a variety of electroplating applications.
Good corrosion resistance: Silver has excellent corrosion resistance, making it a good choice for electroplating applications where corrosion resistance is important.
Overall, silver cyanide is a popular choice for electroplating due to its high conductivity, high silver content, ease of use, and good corrosion resistance.

Except for silver cyanide is insoluble, with solubility product $K_\mathrm{sp}=\pu{5.97E−17}$

BTW, silver nitrate solution has high conductivity (because nitrate  is soluble), has high silver content, is easy to use. Corrosion resistance is property of silver, not of its salts. And yet, silver nitrate is not well suitable for silver electroplating.

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


text:
Why wavelength of Rubidium flame test(red violet) is more than Lithium(crimson red)
My teacher show this table from my book

Here, you can see Red violet wavelength is more than Lithium, which is against order of visible light which increase according to acronym 'VIBGYOR'
According to it violet has lesser wavelength than red(or it's shades)
what is the basic reason?

Look up the Grotrian diagrams for lithium and rubidium and see what the emission-relevant energy levels are. Emission from excited energy levels in atoms is vastly more involved than some simple rainbow color ordering. See here for lithium and sodium.

These types of questions keep coming up. I wish teachers did a better job at explaining these beautiful concepts.

Realize, also, that there are many bright emission lines, and that our perception of the color is the combination of them. See https://physics.nist.gov/PhysRefData/Handbook/Tables/rubidiumtable2.htm for a list of Rb lines, and http://www.alexpetty.com/2014/09/21/the-periodic-table-of-light/ for some spectra, including Rb.

Thank you all, but I am a mere high school (class 11) student I do not know such higher terminology, please can you explain in a easier way (my level) in form of answer, so that I can answer my teacher.

Excited atoms, of each of the elements you list, can emit multiple colors of light. It happens that rubidium emits some red light, whence how it was named, and it also emits other colors, i.e., 780 nm: this is from the 5p to 5s de-excitation. Look up “rubidium Grotrian diagram”. Then, as @DrMoishePippik said, what you perceive as the emission color is what your eyes and brain interpret the mix of emission colors to be. Even a simple neon bulb produces hundreds of emission lines, despite just looking like a reddish-orange color (to me, anyway).

Following my own advice (!), check out rubidium’s emission spectrum here and see the red and violet emission lines. Scroll down as necessary. Also, purple is a mixture of colors whereas violet denotes a range of colors, e.g., 405 nm is violet, as is 419.5 nm, but there is no single wavelength that denotes purple. Same deal with brown.


text:
Is there a difference between Hartree-Fock method and LCAO?
I have to dive in some quantum chemistry for a quantum machine learning project and I came across the so-called Hartree-Fock method.
In one of the reference I used, they considered electrons as independent so that the many-electron wave function (is it of the full molecule or only for one atomic nucleus ?) can be described as a Slater determinant.
So far no problem but then I came across the linear combination of atomic orbitals method and I wanted to know how it relates to Hartree-Fock. Because looking at some quantum chemistry textbook, it appears that for $n$ AO we can produce $n$ MO, say in the case of $\ce{H2O}$ it would a priori be 3 (for the oxygen) and 2 (for the $\ce{H2}$ fragment). But the Slater determinant accounts for all doesn't it ?
Thank you for your answer :)

Obviously, this is a topic which much can be written on. I'll try to keep it short and simple. The LCAO concept merely says that MOs are formed from AOs through linear combination, i.e.
$$\psi_n = \sum_i c_{ni}\phi_i$$
where $\phi_i$'s are AOs, $\psi_n$'s are MOs, and $c_{ni}$'s are the coefficients of the $i$-th AO in the $n$-th MO.
It offers no way of determining the coefficients $c_{ni}$. Sometimes it can be determined through symmetry (let's say $\ce{H2}$, the bonding MO obviously has equal contributions from both hydrogen 1s orbitals). And sometimes you can get qualitative results (in $\ce{HF}$, the bonding orbital is mostly fluorine 2p).
But in general, simply knowing the form of the MOs isn't enough, and you need some kind of quantitative method to determine the coefficients — which is where methods such as Hartree–Fock come in. Using the iterative method prescribed in Hartree–Fock theory you can actually obtain values for $c_{ni}$ (and from there, orbital energies, etc.)
The concept of a Slater determinant is completely separate. It is just a convenient way of representing the total electronic wavefunction as a product of one-electron wavefunctions (the $\psi_n$'s above are one-electron wavefunctions), but it still has nothing to do with finding the coefficients $c_{ni}$. The total electronic wavefunction refers to all electrons in the molecule, by the way—it would not be correct to say that an electron belongs to one specific nucleus.

Strictly in HF theory you don't need to introduce LCAO at all, or more generally a basis - the HF Hamiltonian is a well defined approximation to the exact Hamiltonian, introduction of a basis is a further step.

Thank you for your answer! So you're essentially saying that LCAO is indenpendent of Hartree-Fock theory (the fundamental idea, it is useful for computing the coefficients you've mentionned) and is for one-electron wavefunction, right ? Now, if I want to know the many-electron wavefunctions for independent electrons, whether it is for an atom or  a molecule, I may use Slater determinant to account for both the separablity and the antisymmetric principle ?

and while I am it, LCAO is merely a sort of basis change from molecular orbitals to atomic orbitals, right ? hence the fact that n AOs lead to n MOs, right ?

@QMLSorbonne Yes, that's my understanding of it. And yes, LCAO is precisely that- you can call it a basis change, but I think a choice of basis is more precise, because it is the only basis we're actually using.

There's an infinite number of basis choices, MO and AO are just two options, but because they all span the same n dimensional Hilbert space there must be n entries in whatever basis you choose. And then as you say you can apply matrices to transform from the representation of the wavefunction in one basis to that in another

OK many thanks for all your responses @orthocresol!!


text:
What is the decomposition of potassium nitrite?
I can find plenty of information on the decomposition of potassium nitrate into potassium nitrite and free oxygen, but apparently the resulting potassium nitrite likes to further decompose and produce even more free oxygen.  I cannot however find the reaction for this, and its products.  Do any of you know?

$\ce{4 KNO2 -> 2 K2O + 2 N2 + 3 O2}$
Temperature is too high for $\ce{K2O2}$ or $\ce{KO2}$

You may want to cite this article. It mentioned the exact temperature: 800 C. https://pubs.acs.org/doi/pdf/10.1021/ja01561a015


text:
How to calculate absolute charge of a protein
My understanding is that hydrogen bonds formed by proteins require the NH2 / COOH to be neutral. Hence to find out when the hydrogen bonds are the strongest will depend on the absolute charge of the protein and not the net charge. Hence how do you calculate this absolute charge or is there another way to determine when the strongest hydrogen bonds will be formed when changing the pH condition?
Note: For absolute charge I do not need the actual charge in coulombs but more of the relative charge the NH2 / COOH will bear when changing the pH conditions

Free -COOH/-NH2 groups in proteins ( like in glutamic acid or lysin) are too acidic/basic to stay neutral, so very majority of them is in their ionic form. So there would be rather ionic pair bonding  R - COO- ... NH3+ - R // For the charge of the protein, you need pH and detailed aminoacid sequence.

You might want to start by asking whether the premise of your post here is correct: "My understanding is that hydrogen bonds formed by proteins require the NH2 / COOH to be neutral. " <-- is this true? How did you come to this conclusion? What atoms in proteins commonly form H-bonds?

I recommend starting learning about proteins form beginning, maybe go back to aminoacids...

You are asking a number of questions based on a false assumption. It leads to confusing sentences like this one: "Hence how do you calculate this absolute charge or is there another way to determine when the strongest hydrogen bonds will be formed when changing the pH condition?" The question in the title can be answered, however, and there are theoretical and experimental approaches to determine the total charge of a protein.


text:
How to write formula for potassium hydrogenpyrophosphite?
I know that pyrophosphite by itself is $\ce {({P_2}{O_5})^{4-}}$. My question is about the hydrogen part- how to know how many hydrogens to add? The answer key for this question says the compound is $\ce {KH_3P_2O_5}$ implying that I should add 3 hydrogens. But why not add 1 hydrogen and have $\ce {K_3}$ instead?

Closely related CH SE: structure-of-pyrophosphorous-acid

As phosphorous acid is biprotic:
$$\ce{HO-PH(=O)-OH},$$
the respective pyrophosphorous acid is biprotic too:
$$\ce{HO-PH(=O)-O-PH(=O)-OH}$$
and pyrophosphite is:
$$\ce{^{-}O-PH(=O)-O-PH(=O)-O-}$$
Therefore, potassium hydrogen pyrophosphite would be:
$$\ce{KHP2H2O5}$$
respectively
$$\ce{KH3P2O5}$$
By other words, from 4 hydrogens of pyrophosphorous acid, only 2, bound to O, are acidic. The other 2, bound directly to P, are not.
Similarly, hypophosphorous acid is monoprotic, as only 1 of 3 hydrogen atoms is acidic:
$$\ce{HO-PH2=O}$$

Pyrophosphite is not analogous to pyrosulphite, nor phosphite to sulfite. There is no $\ce{PO3^3-}$ nor $\ce{P2O5^4-}$.
Pyrophosphorous acid is in this context similar to acetic acid. Not all their hydrogen atoms are acidic, so even if fully neutralized, some  remain. Sodium acetate is $\ce{CH3COONa}$, not $\ce{CNa3COONa}$.

BTW, such an acid salt is hard to achieve due to multiple equilibriums between all phosphites and hydrogenphosphite species. The closest we can have is potassium dihydrogenphosphite, $\ce{KH2PO3}$ which forms the addendum, $\ce{KH2PO3.H2(HPO3)}$.

@NilayGhosh I have quite intentionally avoided physical availability of the acid, considering formal aspect. I suspect the author of the task may have made it up to trick the resolvers, catching them on unusual attributes of acids of phosphorus.


text:
What is the chemical equation for enthalpy of solution of hydrated salts?
If I take the anhydrous salt $\ce {CuSO_4}$, the equation for its dissolution in water would be:
$$ \ce {CuSO_4}_{(s)} + \ce {aq}.\rightarrow \ce {CuSO_4}_{(aq)}  $$
Now for the hydrated salt $\ce {{CuSO_4}\cdot {5H_2O}}_{(s)}$, is it this:
$$ \ce {{CuSO_4}\cdot {5H_2O}}_{(s)} + \ce {aq}.\rightarrow \ce {CuSO_4}_{(aq)}  $$
or is the output the "aqueous version" of the hydrated salt ($\ce {{CuSO_4}\cdot {5H_2O}}_{(aq)}$), if that's even a thing? I think it's the first one, but I don't understand why the salt would become anhydrous when dissolved.

Note that the current convention is not to write the compound state as a subscript. E.g $\ce{H2(g)}$ as $\ce{H2(g)}$

It is the first one. The salt does not become anhydrous when dissolved; quite the contrary, it becomes, er, dissolved. "CuSO4" in "CuSO4 (aq)" is not an indication of being anhydrous. Quite the contrary.

AVS should have written : $$\ce{CuSO4  + aq. ->  Cu^{2+} + SO4^{2-}}$$

$\ce{CuSO4(aq)}$ is just a shortcut for $\ce{Cu^2+(aq) + SO4^2-(aq)}$.
$$\ce{CuSO4(s) ->[H2O]Cu^2+(aq) + SO4^2-(aq)}$$
$$\ce{CuSO4 . 5 H2O(s) ->[H2O]Cu^2+(aq) + SO4^2-(aq)}$$
As the symbol (aq) means implicitly involved, indefinite amount of ion-hydrating water, the reactions do not need to be enumerated wrt the water molecule count.
Crystal water abandons ions and becomes ordinary water, possibly but not necessarily taking first dibs in ion hydration.


text:
In combustion method for analysing molecular formula of gaseous hydrocarbon, is pressure of the gaseous mixture always constant?
I understand the we have to look at the volumes, but the fact that volume of a gas is directly proportional to the number of moles of the gas holds only at constant pressure. Also we are applying Gay-Lussacs law, so shouldnt the pressure be constant at all times?
I don't understand why the gaseous mixture is taken to be at constant pressure, the diagram of the eudiometer tube is not able to convince me that the pressure of the mixture is constant.
If the volume of the gas is changing, the height of the water column inside the tube is also changing. Since atmospheric pressure is constant then on changing height of water column shouldn't the pressure of the gaseous mixture also change?
Please explain how the pressure of the mixture is constant.

Mathematically constant is not the same as constant in context of measurement.
For the latter,

below the error threshold it is considered constant
above the threshold are applied corrections.

The volume measurement can be adjusted in such a way outer and inner water levels are aligned anth therefore volumes are measured at constant pressure(safe the trend of atmospheric pressure).
Or, corrections can be made for hydrostatic pressure.
Or, deviations can be neglected, depending on scenario demand on result accuracy.

The topic has in fact nothing to do specifically with chemistry. It is the general topic of mathematical versus scientific precision.
A measured quantity is within a given scenario considered constant if:

Deviations are below measurement resolution threshold
Deviations are observed but negligible compared to other error sources.
Deviations are not negligible wrt other error sources, but can be neglected in context of required accuracy of the result vulue.

Is pressure in the eudiometer tube always constant?  Probably not if combustion is involved.
However, once allowed to cool, pressures will be constant. Volumes change. This is done so volume (as residual oxygen) can be measured and compared with the original volume of oxygen.
Hydrocarbon analysis is based on ratio of carbon (atomic weight 12) and hydrogen (atomic weight 1).
More moles of oxygen are consumed for a given mass of hydrocarbon if the ratio of hydrogen content is higher, as in CH4 (4:1) as compared with C8H18 (2.25:1).
Alkali is used to strip CO2 from the combustion mixture, leaving only residual oxygen.
Atmospheric pressure is used to keep pressure in the reactor constant (just hope the barometer does not change much while running the experiment). Volume changes based on moles Oxygen consumed by combustion.


text:
Carbocation rearrangement in dehydration of an alcohol

In this problem, I know that initially a positive charge is formed at the carbon bearing the hydroxyl group , yielding a secondary carbocation.
But, I can't go any further with that. What kind of ring expansion could this be ?
How would we end up with 2 fused rings of different sizes and why?
Is there any alkyl shift or hydride shift preceding the ring expanison?
I don't know how to predict the direction in which the expansion proceeds after the first step.
Can someone pls help me out with this?
Thanks in Advance.

Answer should be 1.
Refer to the picture below, this may help you:


text:
What does chemically happen to rubber when it soaks in mineral spirits type 1?
Mineral spirits(mineral turpentine)type 1 is a mixture of aliphatic and alicyclic hydrocarbons with a maximum content of 25% aromatics
and less than 0.1% benzene and all kind of impurities.
I have noticed that if i soak piece of rubber(2cm^3 cube)in mineral spirits type 1,the rubber swells.I assume that a specific type of hydrocarbons diffused into the rubber and cause it to swell.
Which type of hydrocarbons are diffused into the rubber and responsible for the swelling?(the aromatic ones?)and is there a way to remove the mineral spirit(residues)from the swollen rubber or the hydrocarbons in the mineral spirits are actually bonding to the rubber molecules,so it can't be removed?

Mostly aromatics, some alicyclic.
Probably the only way is evaporation, at temperature being trade off between patience and thermal damage.
Be aware that of the rubber object may not revert exactly to original shape or size. Some residues may remain and arrangement of macromolecules may change.

@poutnic,As you said that even after evaporation,some residues may remain.Do you mean solid residues that can not be removed?and for that reason,if the rubber was food grade or suitable for potable water,those residues,after the evaporation,make the rubber not suitable to use with food or water anymore?

I have meant solvent residues. Some molecules may be bound by van der  Waals' force stronger then others.

Okay,so does that mean that the rubber will not be suitable to use with food or water anymore(assuming that they were suitable before they were soaked in mineral spirit type 1)and there is nothing to do about that?or it depends on some factors?

I have not  commented food compatibility aspect intentionally.

Okay,i understand.The problem is that i know(by reading)that the white spirits,type 1,contains a toxic materials like benzene etc.So if they are part of the solvent residues,the rubber won't be suitable anymore to use with food or water.But i know,also that everything is related to the amount of the toxic residues and this is what i was trying to understand,if the rubber can be suitable again to the same application(after i give it long time to evaporate)or it won't be safe,no matter what.But again i understand the sensitive part of this question.Thanks.


text:
Regarding colour visibility due to F-centre
Below shown image is a cut out from the NCERT book.

Referring to the highlighted sentences:
I know that an $\ce{e^-}$ absorbs and emits light of a particular wavelength when bound in an atom and corresponding to that wavelength a particular colour is visible.
But in the case of F-centres as NCERT explains.....
that an $\ce{e^-}$ in the F-centre when excites(i.e. absorbs a photon of a particular wavelength) and then de-excites, emits that photon and due to that ionic crystals possess some color.
But here comes the problem, how can an free $\ce{e^-}$ (present in the F-centre) excite?
What's the phenomenon here which is going on?
I am not able to understand that how will an $\ce{e^-}$ present in the anionic vacancies excite?
Overall how F-centres impart color?
Link to the book.

It isn’t a free electron - it is bound to the point defect.

A "sea" of free electrons looks silvery, i.e., reflective, as in solid and liquid metals, such as aluminum or gallium, and even in ionized dissolved metals, such as sodium in $\ce{NH3}$ (although the $\ce{Na+ in NH3}$ looks blue, at first, as electrons are interspersed by $\ce{NH3}$ molecules, as it becomes more saturated, it appears metallic, i.e., specularly reflective). The key thing to understand is that delocalized electrons can reflect EMR.
However, in the F-center, the electrons are not free, but are bound by the surrounding ions. These point defects not only stay in place, but can even be used for long-term data storage, as in nitrogen vacancies in diamond.

You can explain the phenomenon using both crystal field theory and band theory.
Crystal field theory: The electron occupies the atomic site of a vacant Cl-; it will occupy the position that minimizes the repulsion of the surrounding Cl- ions. When you excite it, the electron will move and different higher energy levels will be available, dependent on the symmetry of the anion's environment and the strength of the crystal field.
Band theory: In NaCl there's wide gap between valence and the conduction band, so no visible radiation is absorbed. Vacancies in the Cl-substructure (filled by electrons) forms new levels in the band-gap; hence lower-energy excitations become allowed and some radiation in the visible spectrum can be absorbed.
see K. Nassau: The origins of color in minerals. American Mineralogist, Volume 63, pages 219-229, 1978


text:
Chemistry Thermodynamics and Sign Convention
Can anyone pls explain me the sign conventions that we use in chemistry thermodynamics for heat and work...also kindly explain how to identify what will happen to heat in positive or negative work and vice-verse...

https://chemistry.stackexchange.com/questions/66088/why-do-chemistry-and-physics-have-different-sign-convention-in-thermodynamics?rq=1

There exist two ways for defining the sign of the work, depending who was defining it.
The definition proposed by the chemists implies that all energies that are added to a system are positive. $\Delta U = Q + W$
The approach of physics teachers is different. It is a mechanical approach. It implies that the system is like a steam machine which is heated to produce useful work. The heat is positive when entering the system. But the work is positive when it produces some useful effect (with this heat) outside of the machine. For physicists, the internal energy $U$  is this fraction of the heat that remains inside the system and does not produce useful work. For physicists, $\Delta U = Q - W$. So as the work produced by the system is positive, it means that if some work is entering the system, this work must be negative.
Edit : More details for AtharvZope. Imagine a syringe half filled with a volume $V_o$ of a gas (air or water vapor). If the syringe is heated by introducing a heat quantity $Q$ under constant atmospheric pressure $p$, its gas volume increases to $V_1 > V_o$. So $\Delta V$ is equal to $\Delta V = V_{fin} - V_{in} = V_1 - V_o > 0$. This sign is the same for both definitions of work. As a consequence, in such a phenomena, $p\Delta V $ is positive all over the world, for chemists or for physicists. But, if $p\Delta V $ is always positive, the work $W =|p\Delta V|$ has not the same sign everywhere. $p\Delta V$ is always positive, but $W$ is negative for chemists as the corresponding energy is lost by the system ($W_{chem} = - p\Delta V$), and the same work is positive for physicists, because the syringe has produced a work which could be used for moving a car or producing electricity, etc. ($W_{phys} = + p\Delta V$)
Whatever the choice of the sign of $W$, the change in internal energy $\Delta U$ is the same for physics and for chemistry : $\Delta U = Q + W_{chem} = Q - W_{phys} = Q - p\Delta V$

Interesting. I have not heard about it as  American versus Europian way( in contrary to redox potentials). I have heard about originally the way of chemistry - work done on a system (Q + W) and the way of physics - work done by a system ( Q - W ), later rather as way of science and way of engineering. But it has many exceptions and deviations, with good  habit to explicitly say what convention is used.

Can you pls explain the american system in a more simpler way...if you don't mind...pls

Seconding Poutnik - I do not think there is a distinction between America and Europe, but rather between scientific disciplines. I learned q-w in American physics classes and q+w in American chemistry classes.

@Andrew. These "geographic" definitions were given to me by my chemistry teachers during my studies once upon a time. It may have been wrong. So I have edited my answer to follow your informations. I have replaced the names of the continents by the names of the scientific disciplines.

Thank You Very much!

According to the international standard ISO 80000 Quantities and units – Part 5: Thermodynamics,

for a closed thermodynamic system
$$\Delta U=Q+W$$
where $Q$ is amount of heat transferred to the system and $W$ is work done on the system provided that no chemical reactions occur

The same sign convention is used in the IUPAC Green Book Quantities, Units, and Symbols in Physical Chemistry.

The given equation in integrated form is $\Delta U=Q+W$. $Q\gt0$ and $W\gt0$ indicate an increase in the energy of the system.


text:
What is the maximum number of silver(Ⅰ) ions in one litre of a 0.003 M sodium sulfide solution?

What is the maximum number of silver(Ⅰ) ions that can be present dissolved in one litre of a $\pu{0.003 M}$ $\ce{Na2S}$ solution?

According to my book, silver(I) reacts with sulfide producing $\ce{Ag2S}$ with the solubility product constant
$$K_\mathrm{sp}(\ce{Ag2S}) = [\ce{Ag+}]^2\,[\ce{S^2-}] = \pu{8E-51}.$$
Now here is what I don't understand. I have seen several questions where they ask to solve for the maximum solubility of a substance in water, in this case $\ce{Ag2S}$. The usual thing they do is set up an equation with $x$, like $K_\mathrm{sp} = (2x)^2\,x.$
But in our case, we are given an initial concentration $c_0(\ce{Na2S}) = \pu{0.003 M}$, from which we know that the concentration $c_0(\ce{S}) = \pu{0.003 M}$. We also know the $K_\mathrm{sp}$ value, and they ask specifically for the silver(I) ions, not silver(I) sulfide.
So I am not sure if I need to proceed like above, i.e. set up an equation with $x$ and solve for it, because it seems to me that would give the maximum solubility of $\ce{Ag2S}$, not the maximum number of silver ions.

Try setting up an ICE table, and be aware about the concentration of $\ce{S2-}$ in the row with Initial concentration.

@M.L You mean $\ce{Ag_2}$ instead of $\ce{S_2}^-$ right ? There is only one $\ce{S}$ for every two $\ce{Ag_2}$ or $\ce{Na_2}$ as far as I know

I meant $\ce{S^{2-}}$

If $\ce{[Ag+]}$ = x, and if $\ce{[S^{2-}] = 0.003}$, it is possible to write $\ce{K_{sp} = x^2 ·0.003 = 8 10^{-51}}$. So the solubility of silver ions is $$\ce{x = \sqrt(8 10^{-51}/0.003) = 1.63 10^{-24}}$$ and this result is nearly $x = 1/N_A$ where $N_A$ is the Avogadro number. It means that the concentration of the silver ion is about $1$ ion per liter.

@Maurice That makes sense, but why is there no "2" in front of the x in this case ? For every S in $Ag_2S$, there are two Ag, so shouldn't there be a "2" like $(2x)^2 \cdot 0.003 = 8 \cdot 10^{-51} $ ?

@M.L How would an ICE table help ? As Maurice says, we would simply need to solve $K_{sp} = [Ag]^2[S]$ for $Ag$, that is $[Ag]^2 = K_{sp} / [S] $. I don't understand why there is no "2" in front of the Ag, because for every S, there are two Ag

@wengen But when $\ce{Ag2S}$ dissociates, it dissociates into $\ce{Ag+}$ and $\ce{S^{2-}}$, so $\ce{[S^{2-}]}$ changes too, so it would be $k_{sp} = (2x)^2(0.003 + x)$

Product solubility problems with non-zero initial concentrations can be generalized for a dissociation reaction of the form:
$$\ce{A_aB_b(s)<=>aA^{b+}(aq) + bB^{a-}(aq)}$$
With the condition: ($a≠b$) or ($a=b=1)$, and $K_{sp}>Q_{sp}$
And a general equilibrium expression:
$$K_{sp}=[A^{b+}]^a\;[B^{a-}]^b$$
Since initial concentrations are non-zero:
$$[A^{b+}]=[A^{b+}]_o + ax$$
$$[B^{a-}]=[B^{a-}]_o + bx$$
The resulting expression can be solved for x:
$$K_{sp}=\left([A^{b+}]_o + ax\right)^a\;\left([B^{a-}]_o + bx\right)^b$$
In our particular case, the dissociation reaction is:
$$\ce{Ag2S(aq)<=>2Ag+(aq) + S^2-(aq)}$$
So we can define $a$ and $b$ from the stoichiometric coefficients:
$$a=2$$
$$b=1$$
Initially, no $\ce{Ag+}$ is present, but $\ce{S-}$ is, so in terms of initial concentrations we have:
$$[A^{b+}]_o=[\ce{Ag+}]_o=0$$
$$[B^{a-}]_o=[\ce{S^{2-}}]_o=\pu{0.003M}$$
So the resulting equilibrium expression would be:
$$K_{sp}=\left(2x\right)^2\;\left(0.003+x\right)=8\cdot10^{-51}$$
Solving for $x$:
$$x=\pu{8.2\cdot10^{-25}M}$$
Calculating the resulting concentration of $\ce{Ag+}$ at equilibrium:
$$[\ce{Ag+}]=2x=\pu{1.64\cdot10^{-24}M}$$
Finally, the number of silver ions can be calculated using Avogadro's constant and the volume of the solution given:
$$N_{\ce{Ag+}}=[\ce{Ag+}]\;V\;L=(\pu{1.64\cdot10^{-24}mol/L})(\pu{1L})(\pu{6.022\cdot10^{23}ions/mol})≈\pu{1 ion}$$

Thank you for your thorough answer ! I have a last question, I’m a bit confused when we need to set up an equation with x and 2x, and when we need to solve for [A+] directly like [Ag+], in other word what does the x and 2x actually represent compared to [Ag+] and [S-] ? Don’t both x and [Ag+] represent some form of concentration ? What’s the difference between them ? Are those different approaches ? At the end, you say that [Ag+] is equal to two x, because for every [S-], there are two [Ag+] ? But then why do we get the same result than if we simply did [Ag+]=sqrt( K_sp / [S-]) like @Maurice ?

You are correct in acknowledging the stoichiometric ratio between species, but that's already accounted for when including the stoichiometric coefficients ($a,b$) when calculating the equilibrium concentrations. $x$ is arbitrary, so you could associate $[\ce{Ag+}]$ with $x$ or $2x$, as long as its proportion with $\ce{S^{2-}}$ is preserved. If you choose $x$ for $[\ce{Ag+}]$, then the corresponding factor for $[\ce{S^{2-}}]$ would be $\frac{x}{2}$. Conversely, if you choose $2x$ for $[\ce{Ag+}]$, then the corresponding factor for $[\ce{S^{2-}}]$ would be $x$.

I prefer the second option, since it guarantees all factors involving $x$ will be whole numbers. Note that the reason why i had to multiply $x$ by two near the end of my answer is because I defined the concentration of $\ce{Ag+}$ at equilibrium as $\ce{[Ag+]=[Ag+]_o + ax}$, where a=2 since silver's molar coefficient in the reaction is 2. If I had left it as $x$, then that would only represent half of the equilibrium concentration of silver. Simply put, $x$ is the change in concentration per molar coefficient of a particular species.


text:
Should I give an endothermic reaction the exact energy it needs to occur or can I give it in parts?
I know this might be a silly question, but should I sustain the exact energy or more for the enthalpy of endothermic reaction to occur or can I give it in parts?
Let's say a certain reaction requires $100\ \mathrm{kJ}$ to occur, do I have to give it $100\ \mathrm{kJ}$ or more to occur or can I give it $50\ \mathrm{kJ}$, two times? Assuming that no energy is lost. If it is possible, what about real-life, where energy is in fact lost?

It's not a matter of silliness, but fundamental misunderstanding. You think you need to freeze coal to burn it? No? So why you'd think something similar about reaction that is endo- instead of exothermic?

It may depend on the phenomena. If you want to heat one or two liters water, you may provide $50$ kJ than $50$ kJ later on. But if you want to break a bond whose bond energy is $100$ kJ/mol, you must provide $100$ kJ/mol. Providing half of it is useless. This energy will be lost.

@Maurice Does that mean that the bond will absorb the 50 kJ but do nothing with them? Where will the 50 kJ go?

@BelalBahaa. The bond will simply not absorb $50$ kJ/mol

No reaction has to be given the exact activation energy to proceed
There is a deep misunderstanding here about what activation energy means. It isn't a property of the bulk reaction that is happening, but more like a property of the individual molecules undergoing reactions.
The point is that all the molecules in a bulk compound or mixture do not have the same energy. Simplistically consider a reaction where two different molecules need to collide with more than a certain energy to cause a reaction (the net result of which might absorb more energy ("endothermic") or release more energy ("exothermic"). Many of the individual molecules will bang together without enough energy and nothing will happen. Some will have enough energy and the reaction will proceed. But the individual molecules will always have a wide range of individual energies (As Geoff says, in a Boltzmann distribution). What matters for whether the reaction happens is not how much total energy there is, but what proportion of the individual molecules have more than the activation energy threshold.
That proportion is what varies with the energy you add to the system. It isn't about getting the added energy exactly  right: it is about getting enough to achieve some proportion of the molecules with the right level.
In an endothermic reaction this is still true. But if the final products absorb energy, the amount of energy remaining in the system will fall and fewer molecules will have enough energy to leap the activation energy barrier, so the reaction rate may fall (perhaps low enough to stop the reaction). In which case you can add incrementally more external energy to compensate for the internal loss and keep the reaction rate up.
But, at no point can you add the exact "right amount" because energy is always distributed among all the molecules and the issue is what proportion have enough energy for the reaction to happen.
So, in the bulk, it doesn't matter how the energy is added (partwise or all at once) as long as there is enough for some proportion of the molecules to react.

+1 certainly. But I think it's also important to emphasize that adding a little energy at a time is not necessarily going to do what you think because of distribution into other vibrational / rotational / translational modes.

In principle, you could add 50% energy to the motion that goes over the reaction barrier, which moves you up some of the vibrational energy levels .. and then another 50%. But in reality, that's going to be hard because in most molecules, there are dissipation mechanisms -- so it would be very hard to "keep" the energy in the motion (vibration) needed to overcome the reaction barrier.
Sometimes it's useful to use metaphors when talking about kinetics.
Let's consider that you need to jump over a barrier to get from a starting point to another place:

If the figure attempts to jump 50% of the height, it's not that they can jump another 50% to get over the wall.
In detailed chemical kinetics theory, molecules will have a distribution of different energies (the Boltzmann distribution) but either they have the energy to go "up and over" or they do not and stay on one side of the reaction barrier.
Now this metaphor isn't perfect because some chemical processes (e.g., heating water) can use repeated additions of energy. Also, few chemical reactions are one dimensional and some reactions occur via catalysts or quantum tunneling, so they're not really "jumping over" a barrier.
You might wonder, what happens to the first 50 kJ/mol. In the metaphor, the figure "falls back" because it does not get over the wall. There are a few mechanisms in which the molecule may dissipate energy, for example going into translational or rotational kinetics, or other low-energy vibrational modes.


text:
Can NH4Cl react with HNO3 to give NH4NO3?
I assume that the reaction would be
$\ce{HNO3(aq) + NH4Cl(aq) -> HCl(aq) + NH4NO3(aq)}$
But will this reaction really happen?

Be aware that all 4 compounds, dissolved in water, are dissociated to respective ions. E.g. $\ce{NH4Cl(aq)}$ is just a shortcut for $\ce{NH4+(aq) + Cl-(aq)}$.
So it is not combination of $\ce{HNO3}$, $\ce{NH4Cl}$, $\ce{HCl}$ and $\ce{NH4NO3}$, but of (hydrated) ions $\ce{H+}$, $\ce{NO3-}$, $\ce{NH4+}$ and $\ce{Cl-}$.
Ammonium nitrate cannot be formed even by crystallization of the less soluble salt:
$$\ce{NH4Cl(s,aq) + HNO3(aq) -> NH4NO3(s) + HCl(aq)},$$
because it is more soluble than ammonium chloride.
OTOH, if the mixture has minimum of water, there can be reaction like:
$$\ce{NH4Cl(s,aq) + HNO3(l,aq) <=> NH4NO3(s,aq) + HCl(aq,g)}$$
But, in such high concentrations, there would be significant oxidation of chlorides, forming nitrogen oxides, chlorine and nitrosyl chloride.

As a comment to Poutnik's message, it should be mentioned that the original $\ce{HNO3}$ reacts with the $\ce{HCl}$ produced in a second reaction like $\ce{HNO3 + 3 HCl -> NOCl + Cl2 + 2 H2O}$, which occurs in the solution called aqua regia.

It was already said, by other words.

Dissolving those two compounds in water would lead to four species of ions floating around (without discussing solvation and oxidation).
To get the solid compounds back again would require removing the water, e.g., with a vacuum pump, to get the anions and cations to link (crystallize) again. Then the question arises, which of the resultant compounds are least soluble, and would precipitate first?
In this way, brines are separated into (relatively) purer substances through fractional crystallization. For example, solar evaporation ponds are used commercially to separate $\ce{NaCl}$, $\ce{K2SO4}$, and $\ce{MgCl2}$ from the mix of ions in Utah's Great Salt Lake brine.

Redox reactions aside, the reaction would tend to go the other way, precipitating ammonium chloride.
Below, we note the solubilities of a few ammonium salts in water at 20°C, translated into moles of ammonium ions per kilogtlram of water. Mass solubility from which the molar solubilities are computed are given in the linked sources.
$\ce{NH4Cl}: 6.8$ (https://en.wikipedia.org/wiki/Ammonium_chloride) (interpolated between 0 and 25°C)
$\ce{NH4HSO4}: 10.0$ (https://www.scbt.com/p/ammonium-bisulfate-7803-63-6) (it is assumed that in the acid solutions described below sulfate ions would be converted to bisulfate)
$\ce{NH4NO3}: 18.7$ (https://en.wikipedia.org/wiki/Ammonium_nitrate)
Thus given sufficient salt concentrations, the addition of $\ce{HCl}$ would precipitatethe less soluble ammonium chloride from solutions of the other salts listed:
$\ce{NH4HSO4 + HCl -> NH4Cl(s) + H2SO4}$
$\ce{NH4NO3 + HCl -> NH4Cl(s) + HNO3}$
As described by Maurice in a comment, given the high solute concentrations required plus the formation of nitric acid, the reaction with ammonium nitrate would likely be accompanied by oxidation of the ammonium and chloride ions. With sulfuric acid the oxidizing action is not as strong, so the (bi)sulfate reaction would be cleaner.
The reduced solubility of ammonium chloride may be attributed to its relatively efficiently packed lattice structure. This salt has a caesium chloride type simple cubic structure, with ammonium ions oriented so that their protic hydrogen atoms are facing alternating chloride ions. With this arrangement both ionic bonding and hydrogen bonding serve to stabilize the solid. The bisulfate and nitrate, with bulkier and nonspherical anions, do not offer as good a fit in the solid lattice and thus the solid is more easily attacked by the water solvent.


text:
Which water is easier to filter? Rain water or tap water
I understand the water greatly varies depending on the region and filtration differs as well. Is it better to filter rain water or tap water from contamination like hormones, pesticides, and cloud seeding, which are hard to filter? Is there a better place/SE to ask this?

Why do you need to filter hormones out of rainwater?

Ordinary filtration cannot remove dissolved impurities.

If you fear hormones in rainwater, then you have a problem indeed, and not one that can be solved at Chem.SE.

@IvanNeretin there is a certain level of hormones in all tap water from agriculture and birth control and maybe rain water but it is not the limiting factor. Which water is easier to filter?

@Justintimeforfun, There are hundreds of other compounds in water too. It all depends on what levels are you looking at. More and more sensitive machines can detect a few molecules. I feel that you have misread or watched some pseudoscience videos. Yes, there are human hormones in waste water along with all types of junk that one can imagine, but that does not mean that tap will contain all that. The way Nature has been designed, it takes care of all those things. Bacteria, sunlight, oxygen all work for you in Nature!

Why would a hormone in rainwater? Too fresh?

"I understand the water greatly varies depending on the region and filtration differs as well. " Then maybe you'd care to be more specific?

First thing: All surface water and high ground water started out as evaporated ocean water that condensed on a nanoparticle, scrubbed the air as the droplet formed and fell, landed, extracted whatever it flowed on, eventually collected in streams, ponds, lakes, rivers temporarily, picking up ions, biomass, possibly suspended solids and exchanging gases on its way back to the Oceans. This IGNORES industrial pollution that can be pervasive [Remember how everything cleared up in the start of the pandemic].
Each phase has its own purification problems. First step is an analysis. The least complicated is rainwater collected and settled; then new surface water, Municipal sources [tap water] usually have some pretreatment and usually supply an analysis; then high-level ground water such as springs, then shallow wells, deeper wells; Finally sinks such as the Dead Sea, the Great Salt Lake, the ocean, and sewage [space station].
A rain water story: I taught high school chemistry for 2 years at a school 2 blocks from Kodak Park, Eastman Kodak's industrial complex. While in class a sudden thunderstorm caused a deluge and everyone said, "Let's check it for acid rain".  A girl sitting next to the window held out a bucket to collect water; after several seconds she screamed and pulled her arm in! There were black tendrils hanging all over it. We quickly wiped it off and got her to the girl's room to wash her arm and to the nurse to get checked out, no irritation. We never checked for acid rain or had the substance analysed.

What was the substance? Something from the roof?

You provided incomplete detail (region you live on, what the quality of your tap water is and it origins from, and if cloud seeding is really relevant on your area, or if there are other industrial sources that emit particles and aerosol nearby) to get more than a generic answer.
In Germany for instance the answer would be different as in the USA or in agricultural regions with heavy pestcide use elseshere.
Using ultra-filtration membrane combined with activated carbon filters often is good enough, and proven to improve typical tap water. If you have a problem with heavy metals a little more effort is needed.
Q: rain or tap water easier to filter?
A: it depends, see above
Q: which contaminants are more complicated to get rid of by filtering?
A: nanoparticles, nitrates, heavy metals,...

I just wanted to know overall is it easier or better to filter tap water over rain or snow even. +1 thanks for answering

BTW: why was I downvoted? I want to learn ;)


text:
If you give an electron in a subshell more energy, does it simply "jump" to the next energy level?
Suppose you have an electron in the $\ce{2s}$ subshell of an atom. If energy is given to it, does it simply jump to the next energy level (into the $\ce{3s}$ subshell), or does it move into $\ce{2p}$?

Maybe not duplicate but definitely related: https://chemistry.stackexchange.com/questions/166687/fundamental-doubts-about-energy-levels-vs-shells-vs-subshells-vs-orbitals

Transitions must happen with a change in the l quantum number to conserve momentum. the 2s to 3s transition does not happen. The lowest is 2s to 2p actually a doublet. The 2p state can be excited to 3s.

Depends on the amount of energy you're supplying.
For example, if you supply light of different wavelengths, the amount of energy each holds is different. Your electron will jump to a subshell corresponding to that energy difference.

Transitions also happen thermally. There are vibrational and rotational changes with partial excitation, the Raman effect.


text:
How does phenol show tautomerism?
I read that atleast one alpha carbon has to be sp3 wrt the functional group (and has to have atleast 1 alpha Hydrogen atom) for it to show tautomerism.
I am able to see that in the keto isomers of phenol.
However looking at phenol (enol isomer), how does one predict it can show tautomerism?
I'm wondering if some method can be used to identify that.

All enols exhibit tautomerism, at least to some extent, but not all ketones can tautomerise.
Conditions for showing tautomerism (for ketones) are listed below:

Presence of at least one alpha H at the $\ce{sp^3}$ hybridized alpha C (i.e. C just next to the carbonyl carbon)
In case of alpha-beta unsaturated ketones/aldehydes at least one alpha H
should be present at the gamma carbon (i.e. third to the carbonyl C).

Also keep in mind some basic rules such as Bredt's Rule.
For better understanding refer to this.

The absolute comment on ketones is troublesome.

@jimchmst What do you mean?

I was thinking something like benzaldehyde or benzophenone could have an ionic enol to give a hydroxy enol. but di-t-butyl ketone cannot without a methyl shift to give an alpha hydroxy alkene [not an enol].  very unstable!

@jimchmst not an ionic enol but it seems carbene enol which I don't think would be stable(why would it loose aromaticity , forming carbene). Also any of the mentioned compounds doesn't satisfy the conditions.


text:
What happens to a solute's potential energy during diffusion?
I was thinking, if you have a large amount of water, with an uneven solute concentration, diffusion will lead to an even concentration of solute throughout the solution.
Intuitively,that would mean that our system initially has a large amount of potential energy, and that it decreases to a minimum.

I feel like this raises two questions :
Firstly, from which force does this inital "osmotic" potential energy derive ?
Secondly, under what form is this potential energy realeased once concentration equilibrium is reached ?
What's more, if you consider a cell membrane with a great concentration difference between one side, and another, simple diffusion will lead solute molecules to pass through the membrane. So, in order to pass through the membrane, a solute would have to expend potential energy : the diffusion reaction would have a certain activation energy.
Under what form is this energy then dissipated ? It would first become kinetic enrgy, allowing the solute to pass through the membrane, but what form would it then take ? Simply heat energy ?
Any answer would be greatly appreciated, thanks.

It seems to be a problem of entropy, and not a problem of internal energy.

Both entropy and enthalpy of mixing are important and looks like OP hadn't looked up neither.

Energy can be recovered across a concentration gradient, as in osmotic power production.
But you ask, if I understand correctly what happens if the energy is not harvested? The same thing that would happen if you dropped a weight:

If the weight is attached so as to spin a generator, then the potential energy can be converted to electric energy.
If the weight is allowed to impact a surface, instead, then the potential energy is converted to heat.

If economics is considered the dismal science, thermodynamics is equally grim. Consider Ginsberg's Theorem

You can’t win.
You can’t break even.
You can’t get out of the game.


text:
Functional group for glycine that had being titrated (i) HCl (ii)NaOH
i am kinda confused with this question. They asked for the functional group of (i) glycine hydrochloride  (ii) glycine that titrated with NaOH. I did take a look many times especially for glycine hydrochloride. But the R-group is still the same, that is H, not the NH3+ Cl- (if I'm not mistaken, NH2 is amino group).
Oh, and when they mentioned glycine species from both of these mixtures, what did they mean by it?
Please enlighten me since I am totally lost here.

Glycine:
$$\ce{NH3^{+}-CH2-COO- + OH- -> NH2-CH2-COO- + H2O}$$
Glycine hydrochloride:
$$\ce{NH3^{+}-CH2-COOH + 2 OH- -> NH2-CH2-COO- + 2 H2O}$$


text:
Are salts (e.g. NaCl) soluble in liquid metals?
I'm curious whether any salt would at all dissolve in a liquid metal, such as gallium, mercury, or some other metal in the liquid phase?
A Google search of "solubility of NaCl in Mercury" results in papers on Mercury's solubility in molten salt. I have tried googling various combinations of salts and metals but with no success. I have little knowledge of searchable chemistry resources beyond google, and any suggestions of other places to look would be great.
Any ideas or resources that point me in the right direction would be appreciated (including better search terms).

Well, obviously they are usually insoluble, at least the ones you can approximate as "ionic". If metallic character of these "salts" is bigger, you could expect some solubility.

The AgCl-NaCl pseudo binary shows a miscibility gap at lower temperatures and complete solid solubility at higher temperatures. But you are only asking about liquid solubilities, with the salt as a unit or as separate entities? Please clarify.

Yes, to varying extent.  So just as a starter for your bibliographic research:

The solubility of salts in metals, a technical report by the US Department of Energy turned in by December 31st, 1960:

"The results show that $\ce{BiCl3}$ and $\ce{Hg2Cl2}$ are most soluble, the former being completely miscible with Bi at 778 deg C, while the latter is soluble to the extent of 7 mole% at 600 deg C. $\ce{PbCl2}$ is slightly soluble (1 mole% at 1,000 deg C) while the remaining salts are considered insoluble in their metals. There does not appear to be any single criterion for predicting the solubility of salts in their metals."

Bredig, M. A.; Johnson, J. W.; Smith, T. M. Miscibility of Liquid Metals with Salts. I. The Sodium-Sodium Halide Systems. J. Am. Chem. Soc. 1955, 77, 307–312; doi 10.1021/ja01607a016.  So far, there are 66 citations ACS identified just for this publication.

This equally is something to consider for e.g., molten-salt batteries (an entry on wikipedia, and an illustrative video).

The solubility of oxygen, in effect cuprous oxide, in molten copper is important in the smelting and processing of this metal.
Silvain et al. 1 give the copper-oxygen phase diagram below:

Thus at $\pu{1200 °C}$, roughly $7$ atomic percent oxygen may be dissolved into molten copper; even at $\pu{1100 °C}$ about $2$ atomic percent may be dissolved. Since $\ce{Cu2O}$ is stable at these temperatures, the solution is effectively a solution of cuprous oxide in the metallic liquid. Above $\pu{1223 °C}$ where cuprous oxide decomposes, the metallic phase is effectively a nonstoichionetric oxide salt dissolved in copper. Above $\pu{1345 °C}$ the solubility limit disappears completely as metal and salt merge into a single phase.
Since copper is smelted and often processed as a liquid, this oxygen/oxide solubility must be considered in order to avoid contamination of the final copper product with precipitated oxide.
Reference

J.F Silvain, J.L Bobet, J.M Heintz (2002). "Electroless deposition of copper onto alumina sub-micronic powders and sintering".
Composites Part A: Applied Science and Manufacturing, 33(10), 1387-1390. https://doi.org/10.1016/S1359-835X(02)00153-7.

Similarly, molten silver is reportedly known as "misbehaving", dissolving oxygen too, excluding it while solidifying. That causes splashes of molten metal when breaking just formed and thin meta crust.

One difference between silver and copper is that during solidification, silver has no stable oxides. So the oxygen is evolved as gas (silver spitting/splashing) rather than precipitated as a solid (contamination of copper metal).

Sure, I have not said otherwise. But the oxygen solubility in molten silver is very interesting. If I did not know it happens I would say "nonsense".

Bredig et.al. did an extensive study on the miscibility of liquid metals with salts especially alkali metal-alkali metal halide system. Later Dworkin et.al. worked together with M.A. Bredig and continued the study pertaining to alkaline-alkaline earth halide and rare earth-rare earth halides system. They together published a series of 11 papers under the title "Miscibility of Liquid Metals with Salts"

Miscibility of Liquid Metals with Salts I-V by Bredig et.al.- Paper 1, Paper 2, Paper 3, Paper 4, Paper 5
Miscibility of Liquid Metals with Salts VI-XI by Dworkin et.al. - Paper 6, Paper 7, Paper 8, Paper 9, Paper 10, Paper 11

Simultaneously, Yosim et.al. worked with heavy metal-heavy metal chloride system. You can find the reference in @Buttonwood's answer.


text:
Question about entropy generated in isothermal expansion of an ideal gas
So, this has been bugging me for a while.
Consider the isothermal expansion of an ideal gas in a container. As the gas is allowed to expand, it does work on the surroundings which causes the temperature of the gas to drop. Since it is isothermal expansion, energy must flow via heat into the gas from the surroundings to maintain the temperature. The total change in entropy is given by
$$\mathrm dS_\text{total}=\mathrm dS_\mathrm{system}+\mathrm dS_\mathrm{surroundings}$$
Now, the entropy change for the system is
$$\mathrm dS_\mathrm{system}=\mathrm dS_\mathrm{produced}+\mathrm dS_\mathrm{exchanged}$$
where, $\mathrm dS_\mathrm{produced}$ is internally produced entropy in the system and $\mathrm dS_\mathrm{exchanged}$ is entropy change due to exchange of energy via heat.
Let the amount of heat exchanged be
$$\delta q_\mathrm{system}=-\delta q_\mathrm{surroundings},$$
so $\mathrm dS_\mathrm{exchanged}$ over the temperature $T$ is $$\frac{\delta q_\mathrm{system}}T=-\frac{\delta q_\mathrm{surroundings}}T$$
Then, $$\mathrm dS_\mathrm{total}=\mathrm dS_\mathrm{system}+\mathrm dS_\mathrm{surroundings}\\
=\mathrm dS_\mathrm{produced}+\frac{\delta q_\mathrm{system}}T-\frac{\delta q_\mathrm{surroundings}}T\\
=\mathrm dS_\mathrm{produced}$$
My question is that why should we account for the energy given via heat by surroundings which is $\frac{\delta q_\mathrm{system}}T$, to the system, as it is already used by the gas to maintain the temperature. For an irreversible isothermal expansion the internally produced entropy will be greater than zero due to increase in volume of the gas. But how will heat exchanged add to total entropy when it is only used to maintain the temperature of the system.

Let $T_B$ be the temperature of the interface between the surroundings and the system.  For an ideal surroundings reservoir (typically assumed), the entropy change of the surroundings is $\Delta S_{surroundings}=\frac{-q_{system}}{T_B}$.
If the expansion is irreversible, the average temperature of the gas system during the expansion is less than the initial (and final temperature) $T=T_B$, assuming that the surroundings reservoir is always at the initial temperature of the system T. From Cauchy's relationship, it follows that $$\Delta S_{system}=\frac{q_{system}}{T_B}+\sigma=\frac{q_{system}}{T}+\sigma$$where $\sigma$ is the generated entropy due to irreversibility during the process.
So we have $$\Delta S_{total}=\frac{q_{system}}{T}+\sigma+\frac{-q_{system}}{T}=\sigma$$
To get the entropy change for the system for a reversible path, we need to obtain the reversible isothermal work (which is equal to the reversible isothermal heat $q_{rev}$), and calculate the entropy change for the system from that:  $$\Delta S_{system}=\frac{q_{rev}}{T}$$So, if we combine previous equations, we obtain: $$\sigma=\frac{q_{system,rev}-q_{system,irreversible}}{T}$$
ADDENDUM
It might be helpful to think of this in terms of a comparison between two separate processes.
Process 1 is the irreversible isothermal expansion we have already been discussing.
Process 2 is a two step process. Step 1 consists of the isothermal reversible expansion we have been discussing, to the same final volume as Process 1. In Step 2 of this process, we use a stirrer to stir the gas at constant volume (generating viscous thermal energy and entropy) while at the same time removing the mechanical energy that the stirrer put in as heat.  The amount of mechanical work that we allow the stirrer to do is $\sigma T$, where $\sigma$ is the entropy change in the irreversible Process 1. So, the heat removal in Step 2 is $\sigma T$ and the entropy change in Step 2 is 0.  So, for the overall Process 2, the heat added to the system, the work done on the system, and the entropy change are:  $$q=q_{rev}-T\sigma$$ $$w=w_{rev}-T\sigma$$ $$\Delta S=\frac{q_{rev}}{T}=\frac{q}{T}+\sigma$$  These are exactly the same as our irreversible isothermal Process 1.

Thank you so much! $ \Delta S_{system}=\frac{q_{system}}{T_B}+\sigma=\frac{q_{system}}{T}+\sigma$ is the part I'm confused about. I don't get why we have to include $\frac{q_{system}}{T_B}$ when all that part of the equation is doing is to maintain the temperature of system. Is it that as you said, that if the expansion is irreversible then average temperature of system during expansion is less than initial temperature, so entropy is less and this  part $\frac{q_{system}}{T_B}$  is then used to make up for the loss of entropy due to expansion of gas. Is it like that?

No.  In the irreversible expansion, entropy is generated within the gas as a result of viscous dissipation of mechanical energy to thermal energy; and  this then requires less heat be transferred  from the surroundings to the system to maintain the temperature.  Also, less work is done on the surroundings in the irreversible change as a result of the lower heat transfer.

See my ADDENDUM in the answer.

Thank you for replying. So basically the heat that enters the system, some part of it is reduced due to viscous thermal energy ? I'm sorry but am not getting it.

No.  Just less heat has to enter the system and less work gets done to get to the same final state as in the reversible case.

Okay, so I think what you are trying to say is that because entropy is generated within the gas as a result of viscous dissipation of mechanical energy to thermal energy, less heat is exchanged by surroundings and less work is done on the surroundings. Am I right now? But why consider the heat added in system while calculating entropy produced in the system when all it is( heat) doing is maintain the temperature?And thanks for bearing with me,  English is not my first language.

There are two ways the entropy of a system can change:  1.  By heat exchange with the surroundings at the boundary temperature and 2. by entropy generation within the system as a result of transport processes at finite rates:  viscous friction, heat conduction, molecular diffusion.

Thank you so much!


text:
Term Symbol for Carbon
Currently studying how to "compute" term symbols. My book gives the example of carbon. Carbon has the electron configuration $(1s)^2(2s)^2(2p)^2$. We can ignore full orbitals, so we only look at $(2p)^2$.
Now the possible values for $L$ and $S$ are:
$l_1 = 1, \ l_2 = 1 \quad \Rightarrow \quad L = 0, 1, 2$
$s_1 = 1/2, \ s_2 = 1/2 \quad \Rightarrow \quad S= 0, 1$
We get the possible terms: $ \sideset{^3}{}D, \ \sideset{^1}{}D, \ \sideset{^3}{}P, \ \sideset{^1}{}P, \ \sideset{^3}{}S, \ \sideset{^1}{}S $
with $(2l_1 + 1)(2l_2 + 1)(2s_1 + 1)(2s_2 + 1) = 36$ states.
Now they use the following table where they consider the above 36 states as a two electron function with the quantum number $M_L = m_l + m_l^{'}$ and $M_s = m_s + m_s^{'}$

Now they start to check which entries they can cancel and the first argument they bring is:

In the cells on the diagonal, both electrons have identical quantum numbers. That's now allowed according to the Pauli principle. We now have 30 states

They continue to cancel states but that's not important to my questions.
My main problem is: I don't understand the above quote. If I look at the diagonal, I see $(2,1), (2, -1), (0, 1), (0,-1), (-2, 1), (-2, -1)$, how are those quantum numbers the same? I first thought that they meant the diagonal from the left lower corner to the upper right one but that's not the case.
On another note, I'm also not sure I completely understand why we know that $l_1 =1, \ l_2 = 1$ and why the $s_i > 0$.
Edit: Ah I think they actually meant that for the diagonal entries, we have $m_l^{'} = m_l$ and $m_s^{'} = m_s$ and not the values in the cell (the totals).

Most of these terms are related to excited states. Only one is used for the fundamental state :$\ce{^3P}$

'In the cells on the diagonal, both electrons have identical quantum numbers. That's not allowed according to the Pauli principle.' [...] I don't understand the above quote.

If two electrons have the same quantum numbers, it means that they are in the same orbital ($m_l$) with the same spin ($m_s$). This is forbidden by the Pauli exclusion principle — you can't have two electrons with the same spin in the same orbital.

how are those quantum numbers the same

The quantum numbers of electron 1 are on the left column, those of electron 2 are on the top column. The diagonal entries correspond to (row, column) pairs where electrons 1 and 2 have the same quantum numbers. So yes, your edit is correct.

I'm also not sure I completely understand why we know that $l_1 = 1$, $l_2 = 1$ and why the $s_i > 0$.

$l$ here refers to the azimuthal quantum number of the orbitals the electrons are in. Since you are considering a $2\mathrm{p}^2$ configuration, both electrons are in p-orbitals, so they both have $l = 1$.
$s$ here is the spin quantum number, it is related to the magnitude of the spin angular momentum. Spin quantum numbers are always non-negative and for electrons is $1/2$. You are probably confusing it with $m_s$ which relates to the projection of the spin angular momentum and takes values between $-s, -s+1, \ldots, s$, i.e. $\pm 1/2$ for electrons.
Why use $l$ and $s$ instead of $m_l$ and $m_s$? It comes down to how angular momentum coupling is worked out. In this case, the objective is to couple two sources of angular momentum: that of electron 1 with that of electron 2.
For any two sources of angular momentum with quantum numbers $J_1$ and $J_2$, the resulting sum can have quantum numbers
$$J = |J_1 - J_2|, |J_1 - J_2| + 1, \ldots, J_1 + J_2$$
$J$ is just a generic symbol for some source of angular momentum. Note that here, $J_1$, $J_2$, and $J$ are the quantum numbers which relate to the magnitude of the angular momentum. It's got nothing to do with the projection quantum numbers $m_J$.
In the case of the two p-orbital electrons, we have that $J_1 = 1/2$ and $J_2 = 1/2$, so the allowed values for $J$ are
$$J = 0, 1, 2$$
which are called $S$, $P$, and $D$ states respectively (because spectroscopists have weird terminology sometimes).
It just so happens that in the case of orbital angular momentum, instead of calling the quantum number a generic $J_1$ and $J_2$, we have a special symbol for it $l_1$ and $l_2$. The resulting summed angular momentum is called $L$ instead of just $J$. And for spin angular momenta, we call it $s_1$, $s_2$, $S$.
Coupling of angular momenta is a rather large topic and can't be covered in detail here, but you can find it in pretty much any good textbook on quantum mechanics.


text:
Resolution in mass
I'm pretty confused about mass resolution. The resolution of my instrument is 1000 and I have two masses at 85.0639 and 85.0651. So, the difference is 0.0012.
I know that (M1-M2)=M/R.
But, I'm not sure if my spectrometer will be able to separate these ions.

It is not very clear what those "1000" means. As resolution is usually meant the smallest difference distinguished by device reading. E.g. the classical, old school  2-arm mechanical scales had resolution 0.0001 g.

My bad, I meant mass resolution.

From your Q it seems 1/1000 is the relative resolution of the device, so for relative mass 85, resolution is 0.085 > 0.0012 so it seems you are out of luck. // If it is wrong, that it is probably your fault due bad description of scenario.

related/helpful: Resolution in Mass Spectrometry

The mass resolution $R$ of a mass spectrometer is defined as
$$R = \frac{m}{\Delta{}m} $$
with $\Delta{}m$ the mass difference between two masses still be resolved, $\Delta{}m = m_1 - m_2$.  Thus, if your spectrometer specs detail out  $R = 1000$, then you record masses  of $m_1 = 100.0$ and $m_2 = 100.1$ as two separate signals.*
In your example, $m_1 = 85.0639$ and $m_2 = 85.0651$, or a $|\Delta{}m| = 0.0012$.  So you would need a resolution of
$$ R = \frac{85.0639}{0.0012} \approx 70887$$
to record these masses as separate signals. This number leans more toward a high-res spectrometer (e.g., to substitute a combustion analysis), than the simpler bench-tops routinely hyphenated at the exit of a HPLC.
* This omits how you define a peak in mass spectrometry.  For simplification, this back on the envelope computation assumes zero-width of the signals and no partial overlap.

+1, There is a non-ending argument about the terms spectroscopy vs. spectrometry :-) Your mass spectroscopy reminds me of my undergraduate analytical chemistry teacher held me outside the class and said that I was the only person in class who wrote mass spectroscopy in the exam. Why did you write that? It is mass spectrometry! I still see very established chemists arguing about it.

@AChem The inconsistency (twice spectroscopy, twice spectrometry) is fixed. With "The term mass spectroscope continued to be used even though the direct illumination of a phosphor screen was replaced by indirect measurements with an oscilloscope." from wikipedia, I agree, since its recordings no longer rely on emitted/captured electromagnetic radiation, MS is better off  as spectrometry. Google Ngram


text:
Energy of a molecule using MOs diagram (case of CH4)
I'm trying to find the ground state energy of CH4 in order to compare it with simulations I made. I came across this diagram of energy (not sure it gives the true energies though, if you could confirm...) and I wanted to know whether if to determine the ground state energy, we simply have to add up the energy of each MOs for each electron ? Here for instance we'd have E(ground) = (-2 x 25.7 - 6 x 14.8) eV ?
Thank you for your answer!

See https://chemistry.stackexchange.com/a/164470/5017

Note also that your MO diagram here also ignores the $1s$ electrons from the carbon atom, so you cannot easily compare with a quantum calculation that uses all electrons.


text:
Why is the van der Waals constant b of NH3 less than that of N2?
According to the actual data, van der Waals constant $b$ of $\ce{NH3}$ gas is $\pu{0.0371  L mol-1}$ and that of of $\ce{N2}$ gas is $\pu{0.0387 L mol-1}$.
But it should be,
$$b_{\ce{NH3}}>b_{\ce{N2}}$$
This is because structure of $\ce{NH3}$ is a tetrahedral while that of $\ce{N2}$ is linear, so excluded volume of $\ce{NH3}$ should be greater when compared to that of $\ce{N2}$. As not much difference in size between them, bond length of $\ce{N-H}$ in $\ce{NH3}$ is $\pu{1.008 Å}$ while that of $\ce{N#N}$ is $\pu{1.097 Å}$, why is the opposite thing happening?
Refer here for data of van der Walls constant b
Refer here for bond length(s)

For one thing nitrogen atoms are much bigger than those of H. You also have to take into consideration the distribution of electron density, which becomes localized between atoms during bonding.

But size should not as much higher because $\ce{N_2}$ has triple bond while $\ce{NH_3}$ has 3 single bond which I think should not as much higher as we think of, and also on the other hand $\ce{NH_3}$ is tetrahedral which should make more volume of sphere of exluded volume.

Have you looked up other information on the relative dimensions of these molecules? For instance bond length? Since that is instrumental to your argument, please do the geometric calculations and include them into the post.

There's difference of just approx
0.09 Å. You can see in the post now*. But the tetrahedral structure will make overall molecule slightly larger due to 3 bonds in 3D at angle ≈107°.

Be aware that van der Waals equation is not a tool to experimentally determine sizes of molecules. It is just a mathematical model. Note that there is strong correlation between values of both a and b parameters, with relation to critical p and T. So values of parameter b in context of the molecule effective size are not easily comparable between molecules of very different behaviour in context of intermolecular forces.

You might now account for the atomic radii to determine the longest end-to-end distance in the molecule. For NH add the atomic radius of H at one side, On the other side you need to consider the effect of the angle in the tetrahedral molecule. For NN add twice the atomic radius of N. (this is highly approximate of course but a starting basis for a  comparison of predicted geometries). For radii see for instance https://en.wikipedia.org/wiki/Atomic_radius#Calculated_atomic_radius

From a purely geometrical standpoint and using some simple assumptions the similarity and order of b parameters should not be all that surprising.
For ammonia the longest distance in the molecule can be taken as the sum of the length of one NH bond plus the H radius, plus the projection at an angle of $73^ \circ$ of that same distance ($r_\text{H}+r_\text{NH}$) on the opposite side of the N center. This gives  $d_\ce{NH3} = 1.9877 Å$.
For dinitrogen we can sum the bond length and twice the computed radius of a nitrogen atom, arriving at $d_\text{NN}=  2.217 Å$.
The ratio $d_\ce{NN}/ d_\ce{NH3} = 1.1154$ compares favorably with the alternative estimate $(b_\ce{NN}/ b_\ce{NH3})^{1/3} =   1.0142$ using the van der Waals parameters as estimates of (or proportional to) the molecular volume. The latter could be argued to be smaller because it represents an effective radius, an average of interactions along different angles of approach rather than only along the longest distance in the molecule.

So, what if someone didn't knew the data? What will be the factor to decide it?

@Siddharth You chose a difficult example. The b parameters (and molecular volumes) are very similar. For some cases the difference in molecular volume is more obvious and decisive. For others (eg acetonitrile vs benzene!) it's not (the difference in b is negligible but one would expect a large difference in volume). In general regard b as determined by volume but potentially also interactions (consider comment by Poutnik). It is empirically determined (experimental).


text:
Can we calculate van der Waals' constant a for the following case?

The given Z vs P curve for 1 mole of a gas at 400K starts at Z=1 & P=0.
The slope at the point when the curve again intersect Z=1 is 0.005. The critical temperature of the gas is 500K.
My approach:
From the van der Waals' equation,
$$(P+\frac{a}{V^2})(V-b)=RT$$
On expanding,
$$\frac{PV}{RT}=Z=1-\frac{a}{VRT}+\frac{Pb}{RT}+\frac{ab}{V^2RT}$$
And  $$ PV=ZRT  $$ $-(1)$
By both equations,
$$Z=1-\frac{a}{VRT}+\frac{Zb}{V}+\frac{ab}{V^2RT}$$ $-(2)$
Differentiating with respect to Z for
Equation (1)
$$ PV'+VP'=RT$$
Equation (2)
$$ 1= \frac{aV'}{V^2RT}+\frac{b}{V}-\frac{bV'}{V^2}-\frac{2abV'}{V^3RT}$$
& $$T_c=\frac{8a}{27Rb}$$
Now I am stuck, I don't even know if I am moving in the right direction because things are becoming too complex to solve.
Where,
a, b = van der Waals' gas constants
Z = Compressibility Factor
V = Volume of gas when curve again intersects Z = 1
R = Universal gas constant
T = 400K
$T_c$ = Critical Temprature

@Poutnik yes angle a pressure dependent and it is specific angle on the point when curve intersects Z=1.

I mean something else. For given pressure, angle depends on pressure unit. Tangens of angle is 1000x bigger for kPa than for Pa.

Yes, you are right. But I couldn't edit that one because it is an image from web.

You need Z=f(p), not Z=(p,V(p)). But as you are interested in region near Z=1, you can do substitution V=RT/p.

Can you please explain briefly in the answer?

Does not $dZ/dP=b/RT$ ? As you know $T_c$ and the slope then you know $a$ and $b$.

@porphyrin, I think it's not because as you see in the equation 2. $Z$ depends on $V$ too, which is a variable. It this was true The $Z$ vs $P$ graph will left only a straight line as there's no variable in your slope.

You can use substitution V = ZRT/P, but it helps just partially, getting the implicit cubic function f(Z,P)=0.

@Siddarth, you are right, silly me! Try this, substitute $PV=RT$ and solve for $Z=0$ to get $P$, next  differentiate Z to get slope whose value you know and use Tc to connect a and b.

You were going in the right direction, But I has an extra step.
You already had proved the following equations.
$$ PV'+VP'=RT-(1)$$$$ 1= \frac{aV'}{V^2RT}+\frac{b}{V}-\frac{bV'}{V^2}-\frac{2abV'}{V^3RT}-(2)$$$$T_c=\frac{8a}{27Rb}-(3)$$$$\frac{PV}{RT}=Z=1-\frac{a}{VRT}+\frac{Pb}{RT}+\frac{ab}{V^2RT}-(4)$$
Substituting values $T_c=500K$, $T=400K$, $R=0.08$ (it will make calculation easier), $P'= \frac{dP}{dZ} = \frac{1}{0.05}=200 $ (at POI) in equation $(1),(2),(3)$ gives,
$$ \frac{32V'}{V}+200V=32-(A)$$$$ 1= \frac{aV'}{32V^2}+\frac{b}{V}-\frac{bV'}{V^2}-\frac{abV'}{16V^3}-(B)$$$$\frac{a}{b}=135-(C)$$
Now, if you look at the equation $(4)$. If you you put $Z=1$
$$\frac{a}{VRT}=\frac{Pb}{RT}+\frac{ab}{V^2RT}$$
Canceling out $ \frac{1}{VRT} $ and Substituting $P=RT/V$ gives,
$$a=32b+\frac{ab}{V}⟹\frac{a}{b}=32+\frac{a}{V}⟹135=32+\frac{a}{V}$$
So,
$$\frac{a}{V}=103-(D)$$
So, we're done.
By equation $(A),(B),(C),(D)$,
$$a= \frac{46532}{2575}≈18.07$$ $$b=\frac{46532}{347625}≈0.134$$
$$V=\frac{46532}{265225}≈0.175$$
$$V'=-\frac{47648768}{2813772025}≈-0.017$$

First, there is a mistake in the fifth expression. It should be written $$PV' + P'V = RT$$
Second, it is not clear what Siddharth is looking for. If he wants a way of calculating $a$ with other van der Wals parameters, the simplest way is to draw $a$ from the final formula giving the critical temperature :  $$\ce{a = \frac{27}{8}bRT_c}$$ Here the curve $Z(P)$ is not needed.

Actually b is unknown. But this gives the relation $\frac{a}{b}≈135$. That I've used.

Can you tell what is the mistake in fifth expression, So I could correct it.

@Siddharth. You have written $PV' + V'P = ...$ If this had been correct, it could have been rewritten $PV' + V'P = 2 PV'$. But, care ! The position of the second   "prime" is wrong. You should have written $PV' + P'V = ..$

Oh yes! Sorry that was typing mistake. How could I further proceed to the question?

What are you looking for ? What is the question ?

Sorry van der Waal**

@Siddharth "van der Waals"  :-)

No, no ! Please write van der Waals ! W A A L S ! With two A letters !

Yes, "van der Waals". My keyboard has auto correct problem.


text:
By what mechanisms does Amoxicillin break down while in storage?

Assuming the following conditions, what are the primary mechanisms of breakdown?

Dry powder form (not in any pill or packaging)
Stored in a glass bottle
Not exposed to light
22C ambient temperature

For a start, I suggest to look into the documentation that comes with the medication. What does it say? Of course the active ingredient may degrade, but also other parts of the formulation. Intact packaging is obviously important.

Each is different, and even different manufacturers of the same formulation could have differing storage recommendations. The best you can do is research each online, e.g., https://medlineplus.gov/druginfo/meds/a685001.html#storage-conditions , and look for commonalities.

@DrMoishePippik These guidelines seem tailored to consumers. If that's the case they're not going to say "store at -80C" even if the drug would last indefinitely at that temperature. I'm looking for more a chemistry angle on this. They've already published a shelf life so they have no reason to explain how to keep it longer than that. Furthermore there have been some studies showing that a lot of emergency stockpile medications in the US are needlessly thrown out for exactly this reason.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7040264/

I'm not looking for recommendations, I'm looking for the mechanisms of breakdown and how they can be minimized.

Extreme temperature can cause also mechanical breakdown of tablet assembly, what can cause big impact on absorption of the drug, considering absorption rate and it's place in digestion tract.

@Poutnik Ok, but lets just consider the drug compounds themselves for now.

For compounds, there is universal advice of low T, darkness and low humidity, with need of proper temperation in both ways. ( To prevent high or condensing package humidity when cooled down too fast even with desiccant. And similarly condensing external humidity at shocking warm up.) // Particular advises are highly specific.

P.S.: So are mechanisms of degradation.

One can't generalize about conditions. For example, low temperatures may cause a medication delivered as an emulsion to separate through freezing, or crack the shell of a capsule as contents expand, contract or crystallize.

This gives some ideas what happens in the presence of water and oxygen: https://www.sciencedirect.com/science/article/abs/pii/S0045653513002245

Why does nobody speak about Drew's inquiry, namely the mechanism of breakdown ?

Karsten's reference to Gozlan et al., Chemosphere (2013) 91(7), 985-992 appears to show that the major breakdown route under the reported wet conditions was to open the beta-lactam ring hydrolytically to form the corresponding penicilloic acid.
In any other conditions, the strained 4-membered beta-lactam ring seems on many reckonings still likely to be the most reactive (~unstable) part of the molecule, together with its fused thiazolidine ring partner.
Under the storage conditions mentioned in the question (i.e. 'dry' but in an unspecified hence presumably ordinary bottle), there may be a shortage of water but probably not a complete absence. So it might be expected that some hydrolytic decomposition to the penicilloic acid would happen unless the 'dry' character has been taken even further by freeze-drying and storage in sealed ampoules.
Under freeze-dried conditions, there might still be residual room for slower ring-opening reactions between amoxicillin molecules.


text:
What will remove this screen printed logo without destroying the polycarbonate safety sunglasses lens?
I bought some safety sunglasses and didn't notice until I'd taken the tags off that there's an annoying logo painted or printed right in my field of view.  (You can feel with your fingertips or fingernails that it is actually on the surface and not, for instance, laser etched into the surface.)  What household or garage chemicals would take it off without destroying the polycarbonate lens?
I recently did some PVC pipe work and I'm confident the purple primer would take this right off.  I'm not confident it wouldn't also dissolve the polycarbonate.  PVC... polycarbonate... is the latter likely to be resistant to something that dissolves the former?
Otherwise what else should I try?  I also have a can of stuff for cleaning paint brushes.

You might experiment in a very small spot with a cotton applicator and milder solvent, e.g. 91% or stronger ethyl or isopropyl alcohol, then try stronger solvents such as naphtha and then diethyl ether, if the previous doesn't work. N.B. These are either flammable or highly flammable -- use caution.

Test also on other PC objects like old CD/DVD before the glasses.


text:
Properties of glyceral with respect to cooking/baking?
Out of curiosity I went looking for information or videos of people taste-testing monosaccharides and I came away surprised that I could find this for virtually every sugar except for the simplest (triose) monosaccharide: glyceral (glyceraldehyde). There is a lot about glycerol, which may make searching for glyceral more difficult. In any case I was curious what it tastes like and if it could be used for baking.

Well, it's also sweet, but I don't think there's much point in using it.

Why not? Depending on how sweet it is it could be used as a sugar substitute, depending on its taste it could be useful (e.g. maltose), or it could change the structure/mouthfeel of baked goods...

Only hypothetically - why use something you'd need to synthesize vs stuff that occurs in nature in large amounts?

For fun! For high-end molecular gastronomy. For industrial baking...

One challenge is that the cyclization is much less favorable than for all the other aldoses, so the aldehyde is much more reactive with amines and such that are present in most foods. Maybe useful if you want to promote Maillard chemistry


text:
Do two atoms 'participating' in a single covalent bond rotate in only one direction, without rotating the entire molecule?
A Biology textbook stated that, "single bonds allow the atoms they join to rotate freely about the bond axis".
This definition is not clear enough for me to answer the question, "do two atoms 'participating' in a single covalent bond rotate in only one direction, without rotating the entire molecule?", like this:

I think I have read that single bonds permit rotation, because the overlap between two atoms' orbitals, participating in a single bond, does not change with rotation in one direction, whereas, in double bonds, another pair of orbitals is shared between the two atoms, therefore, rotation of one atom, in any direction, independent of the other, would cause one of these orbital pairs to cease overlap.
So, I take it this is the reason why atoms participating in single bonds, rotate independently of one another, in a specific direction, whereas, atoms participating in double bonds rotate with one another, unitedly, regardless of the direction of rotation.
Moving on, let us say we have an atom single covalently bonded to two atoms, such as our good old friend H2O, let us assume it is linear (which it is not in reality):
H — O — H
If we rotate the single oxygen atom 'sideways' (in line with the hydrogen atoms), I would assume the single bonds between the hydrogen atoms and oxygen atom, would rotate with the oxygen atom to maintain the overlap between the two hydrogens' orbitals and the oxygen's orbitals.
In reality H2O looks more like this:

Therefore, I would assume that rotating the oxygen atom in any direction (up, down, left and right), would result in the orbitals of the hydrogen and oxygen atoms losing their overlap with one another and to maintain overlap, the single bonds would rotate with the oxygen atom.
Thus, why I think atoms in a single covalent bond can only rotate in one direction, without rotating the entire molecule.

Atoms are not rotating, groups are. More precisely, molecules are rotating, vibrating and oscillating, but but what happens "internally" to a single atom is irrelevant, as long as it doesn't change position.

Rather than thinking of rotation of the atoms, it makes more sense to think of a rotation of the groups that are attached to the atom - similar to the first figure you showed. This phenomenon is called internal rotation and in principle is not completely free. When you take an ethane molecule and start to rotate the two methyl groups around the connecting CC bond, you will find that the potential energy that you have to overcome is sinusoidal with maxima and minima corresponding to the eclipsed and staggered positions, respectively (see the figure below). You can calculate this profile with QM software, but qualitatively, it emerges from the interplay between Pauli repulsion and orbital overlap. For single bonds, the barrier is on the order of a few kcol/mol, much less than the thermal energy typically available to a molecule at room temperature and the two groups can more or less rotate freely. For double bonds, the barriers are so high that the molecule cannot overcome the potential hill and no rotation is observed.
Note that, in terms of degrees of freedom, internal rotation corresponds to a vibration mode rather than a rotation because the atoms of the molecule move with respect to each other.

Figure source.

Assuming that the molecule is stable and well-behaved, molecular rotations happen along single covalent bonds, in whatever direction, because that is the only thing that the molecule has enough energy to do.
Silly example:
The energy cost of turning your car wheel to the left or the right isn't too bad, but attempting to rotate it in other directions (say upwards or downwards) would require more strength than most people could muster. Also, it would wreck your car. In the same way, a molecule ripping itself apart spontaneously is certainly possible, however, the likelihood is incredibly small.
Lastly, in a void, the rotation of the entire molecule has no effect on the overall energy of the molecule and therefore we don't typically care about it too much.


text:
Are there smart materials that dissolve in cold water faster?
I know things dissolve quicker in hotter solvents but I was wondering if there existed an exception to the rule. If not water, then what about other solvents and/or non-solid solutes?

https://chemistry.stackexchange.com/questions/21855; https://physics.stackexchange.com/questions/311409/

Note that better and faster are 2 different things. A compound may dissolve at lower temperature more ( thermodynamics), but slower (kinetics).

Just one example is Poloxamer 407 which forms a hydrogel at room temperature at sufficiently high concentration. Adding solid Poloxamer 407 to water at room temperature would easily result in Poloxamer 407 powder encapsulated in gel which would strongly slow down the dissolution process. At a colder temperature, such as 4 °C, the gel does not form, so dissolution is typically achieved in a shorter time.
So yes, there are examples of compounds that dissolve faster in colder conditions. As usual when it comes to kinetics, the mechanism is highly relevant.

Kind of similar when classical starch sol/gel is to be prepared, in iodometry or culinary context. It must be suspended in cold water before mixed with hot one.