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Determination of chloride content in aluminium chloride by titration
TLDR: Is there an as simple and direct as possible method to measure chloride content by titration?
I want to measure the content (mass fraction) of chloride in $\mathrm{AlCl_3}$ powder by titration. Purpose is to measure the purity of the aluminium chloride. For example assume the measured chloride content is 75 % then that means the $\mathrm{AlCl_3}$ is 94.0 % pure.
When I search for how to determine chloride by titration I get the following three methods, which appear to be more complex than needed: Mohr method, Volhard method, Fajans method. All of which use an indicator and require specific conditions and treatments or reagents. The Volhard method is also a back titration and seems to be the default method for some reason which I could not find.
I found the following procedure, which is basically the Volhard method but with an additional reagents (of which I do not know the purpose) and an additional dilution step.

1.) Prepare an flask with stopper with 50 ml water and chill in the freezer to form ice.
2.) Weigh about 1 g of sample and transfer it in to  the flask along with the paper and put the stopper.
3.) Allow it to come to room temperature. Add a few drops  of phenolphthalein indicator and titrate with 1N NaOH solution to the end point.
4.) Dilute the solution to 250 mL. Pipette out 20 mL in to another flask.
5.) Add 3-4 mL of conc. nitric acid , 50 mL of 0.1 N silver nitrate solution, 2 mL of nitro benzene and a few drops of ferric ammonium sulfate indicator.
6.) Titrate with 0.1 N ammonium thiocyanate solution to reddish brown end point.

No idea what the nitro benzene is for. And the additional dilution step is certainly unnecessary and lowers precision; could just weight less sample and not dilute. The NaOH step is probably to react HCl (which is formed when $\mathrm{AlCl_3}$ reacts with water and may escape and thus lower the chlorine content) to NaCl; but also not sure about that.
What speaks against just directly titrating the chloride containing sample solution with the 0.1N $\mathrm{AgNO_3}$ solution and detect the end point potentiometric?
Am I missing something? Why would people do a back titration with two solutions needed, instead of just a direct titration?

You can consult Vogel's Quantitative Analysis. There you will find all titrimetric details for chloride and explanations. Web search will only lead to confusion.

All your answers to the question are available in Vogel.

@M.Farooq - What, no love for Skoog and West?

@ToddMinehardt, I like classical books when it comes to titrimetric and gravimetric methods. Vogel's classical Quantitative Analysis from the 60s was a massive book with >1200 pages, all extensive details such as holding volumetric pipette.

Here is the textbook for the OP on Internet Archive. https://archive.org/details/textbookofquanti0006voge/page/n7/mode/2up?view=theater

@M.Farooq  Thank you; I found the fith edition to download here: https://gtu.ge/Agro-Lib/Vogels_TEXTBOOK_OF_QUANTITATIVE_CHEMICAL_ANALYSIS_5th_ed__-_G_H_Jeffery.MsuCity.pdf

There is also explained why to add nitrobenzene: "It is therefore necessary to remove the silver chloride by filtration. The filtration may be avoided by the addition of a little nitrobenzene"

However, what I still do not know is what speaks against direct titration with $\ce{AgNO3}$ and potentiometric end point detection.
Because that is the method that is currently implemented in our lab. But I could not find anything similar online.
The problem is that it produces slightly to too low assay values for the $\ce{AlCl3}$. Perhaps because of escaping HCl gas; if really that is the reason than dissolving the $\ce{AlCl3}$ cold and in a closed container should in fact be enough.

You might be searching very specific words. Search Vogel first (later editions) on Internet Archive first for potentiometric titration of chloride. How do you locate the end-point?

This problem is not as easy as it looks. It starts from the hypothesis that $\ce{AlCl3}$ is not pure in the original flask- The impurity will be supposed not to interfere later on. Only $\ce{H+}$ and $\ce{Cl-}$ will be determined by titration.
Let's suppose that the original sample of substance weighs $m_o$ and contains $n\ce{_o mol AlCl3}$. When dissolved into water, this substance reacts according to :  $$\ce{AlCl3 + 3 H2O <=> 3 H+ + 3 Cl^- + Al(OH)3} \tag{1}$$ So, if the hydrolysis is complete (and it will be during titrating by $\ce{NaOH}$), the total amount of $\ce{H+}$ and $\ce{Cl^-}$ in the solution is $3n_o$. The first titration, with $\ce{NaOH}$, determines the amount $n_1$ = $n\ce{(NaOH)}$ =  $n(\ce{H+})$ = $\ce{n({Cl^-})}$ = $3n_o$ produced by the reaction $(1)$. But, apparently, the result of this titration is not reliable enough. The author of the problem is afraid of some unwanted acidic contamination. So the actual amount $n_2$ of $\ce{H+}$ and of $\ce{Cl-}$ due to ($1$) will be obtained by another way, as will be described now.
The second titration is made to determine the amount $n_2$ of $\ce{Cl-}$ ion in solution. But it cannot be done directly, as, after the first titration by $\ce{NaOH}$, the solution contains $\ce{NaCl}$ and maybe some $\ce{NaOH}$. It is first acidified by adding some drops of $\ce{HNO3}$ to destroy this unwanted  $\ce{NaOH}$ . Then a known excess of $n_3$ mol silver nitrate is added. All the $n_2$ mol $\ce{Cl-}$ ions are precipitated as insoluble $\ce{AgCl}$. $$\ce{Ag^+ + Cl- -> AgCl(s)} \tag{2}$$ The solution contains now an excess of $n_3 - n_2$ mol  $\ce{Ag+}$ ions. Some drops of ferric ions are added (plus nitrobenzene - I don't know why). The excess of $\ce{Ag+}$ ions is titrated by gradually adding a thiocyanate solution. In the beginning, it produces the reaction :
$$\ce{Ag+ + SCN^- -> AgSCN(s)}\tag{3}$$ $\ce{AgSCN}$ is a white insoluble precipitate, which looks like $\ce{AgCl}$.
The end point can be determined with a high degree of precision. Because, when the precipitation reaction (3) is finished, the next drop of thiocyanate reacts with the ferric ions to produce a dark red color, made of the ion $\ce{Fe(SCN)^{2+}}$ which is dark red. So as soon as the last amount of $\ce{AgSCN}$ is precipitated, the solution gets red. And this point can be determined with a high precision.
If $n_4$ mol thiocyanate have been added to get to the final red color, the amount $n_2$ of $\ce{Cl-}$ ions can be determined : $n_2 = n_4 - n_3$. And the original mass of $\ce{AlCl3}$ is equal to $m = \frac{(n_4 - n_3) M(\ce{AlCl3})}{3}$, if $M(\ce{AlCl3})$ is the molar mass of $\ce{AlCl3}$. One should find that $m < m_o$

Maurice, I think HAlCl4 does not even exist.

@M. Farooq. You may be right, after all. I know that the chloraluminate ion $\ce{AlCl_4^-}$ does exist. But the existence of $\ce{HAlCl4}$ is not proven. So I will edit and modify my answer.

@Maurice  Thank you for the explanation. The $\ce{AlCl3}$ is supposed to be pure or almost pure (> 99.5 %) as it is purchased from a supplier, but we have to test the assay of it before using it in a production process.
Now, what I really do not understand is: Why is it not common to directly titrate the chloride (be it in the form of $\ce{[Al(H2O)6Cl3]}$ or $\ce{HCl}$ or $\ce{NaCl}$) with the $\ce{AgNO3}$ solution? Then the used volume of $\ce{AgNO3}$ solution directly corresponds to the amount of chloride. Why the detour via a back titration with thiocyanate?

@Wolfram. Even if the product is sold $99.5$% by the supplier, you cannot avoid that some part of $\ce{AlCl3}$ has been hydrolyzed in the flask during its storage.


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Free induction decay acquisition in NMR
Recently, there was an interesting Q&A exchange on free induction decay acquisition in NMR Is intensity in NMR spectra acquired at regular time intervals? with useful answers. I have not acquired FIDs in NMR or used the raw data but rather dealt with microwave rotational spectra free induction decay data-which is purely real with its own pecularities.
The interesting point for me in NMR-FID was the mathematical notion of a complex free induction decay. Although the FID experiment measures "real" values, since there a quadrature detection, and the two coils are orthogonal, it makes sense to treat it as a complex number.
My questions are:
(i) What is the criterion of calling one signal as real and the other imaginary from the two orthogonal coils. I mean, both coils are orthogonal to each other. I think this label must be arbitrary to call one real and the other imaginary (i.e., the labels can be exchanged). This website MRI questions says the same.
(ii) Which FID is used for displaying the NMR spectrum? Both FID from the two receivers should appear like a free induction curve since are both real experimental values. Discrete Fourier transform of each should yield the same frequency information i.e., the same NMR spectrum. There must a be phase difference but frequency information must be the same. Is this correct?

(1) What is the criterion of calling one signal as real and the other imaginary from the two orthogonal coils

Assuming that you have detection axes aligned with the $x$- and $y$- axes ($y$ being offset by $+90^\circ$ from $x$), then for a single spin with precession frequency $\omega$, the signals you get will be:
$$S_x = \cos(\omega t); \quad S_y = \sin(\omega t),$$
and then we choose to define the complex signal as
$$S := S_x + \mathrm{i}S_y = \exp(\mathrm{i}\omega t).$$
In this case $x$ is the real axis and $y$ is the imaginary axis. Complex FT of this yields a peak at frequency $\omega$, as desired. However, this isn't the only way to do it. If we wanted to swap them around, we chould choose $y$ as the real axis and $x$ as the imaginary axis. However, we would need to be careful to include a minus sign in the definition:
$$S' := S_y - \mathrm{i}S_x = -\mathrm{i}\exp(\mathrm{i}\omega t)$$
This is the same signal as before but with an additional global phase factor, which is unimportant. Notice that if we defined $S' := S_y + \mathrm{i}S_x$ we would get a peak with negative frequency. In some sense, the imaginary isn't the $x$-axis, it's the negative $x$-axis. I think the summary from this is that the imaginary axis should be offset from the real axis by $+90^\circ$ (and not $-90^\circ$).

(2) Which FID is used for displaying the NMR spectrum?

Both, actually: we don't FT only the real or imaginary part of the FID. We form a complex FID and perform a complex FT on that, i.e. $\int_{-\infty}^\infty f(t)\exp(-\mathrm{i}\omega t)\,\mathrm{d}t$ to get a complex spectrum.
Traditionally, only the real part of the spectrum is displayed to the user, though.

[...] Both FID from the two receivers should appear like a free induction curve since are both real experimental values. Discrete Fourier transform of each should yield the same frequency information i.e., the same NMR spectrum. There must a be phase difference but frequency information must be the same. Is this correct?

Yes, both real and imaginary components of the FID will contain the same frequency information. However, neither component alone is sufficient for providing quadrature detection, because one signal is
$$S_x = \cos(\omega t) = \frac{\exp(\mathrm{i}\omega t) + \exp(-\mathrm{i}\omega t)}{2}$$
and the other signal is
$$S_y = \sin(\omega t) = \frac{\exp(\mathrm{i}\omega t) - \exp(-\mathrm{i}\omega t)}{2\mathrm{i}}$$
so if each of them is individually FT'd, you get duplicate peaks at $\pm\omega$, hence the need to combine them. (If your frequencies are only positive, then of course you don't need two components; you just discard the negative frequency parts, or equivalently use a cosine transform instead, i.e. $\int_{-\infty}^\infty f(t)\cos(\omega t) \,\mathrm{d}t$. But in NMR, the frequencies are always going to have both signs.)
As for the spectra, you can reconstruct the imaginary part of the spectrum from the real part of the spectrum via a Hilbert transform. The resulting formulae relating the two parts are also known as the Kramers–Kronig relations.

Okay, very interesting. Is there a website where I can see / download raw NMR FID with both Sx and Sy. With real DFT of an FID (say in microwave spectroscopy), the FID is real. After DFT, we get a rotational spectrum which is identical for positive frequencies as well as negative frequencies. One keeps the positive portion and correct the magnitude. If I recall correctly, if the FID is complex, then this mirror image symmetry is lost. Is this what is experienced in NMR?

I'm not sure whether there are any raw NMR databases online, but if you want some data I can probably upload some of my own data somewhere. Re the rest of your comment, yes, that's precisely right.

I would be happy to see a raw output of an NMR FID and its traditional NMR spectrum too. Is time information or sampling rate available too?

@M.Farooq It is a bit too long to explain in the comment section, so I will post some information in a new chat room: https://chat.stackexchange.com/rooms/134056/discussion-on-answer-by-orthocresol-free-induction-decay-acquisition-in-nmr

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Why is there a large increase in volume when an alkali metal is dissolved in liquid ammonia?
My teacher said unexplainedly that when alkali metals dissolves in liquid ammonia, there is a large increase in the volume. I found a paper that confirms this fact [1]:

In the course of recent research it was necessary to obtain the volume change associated with the solution of lithium, sodium and potassium in liquid ammonia.
That the solution of these metals in liquid ammonia is accompanied by a relatively large increase in volume has long been known.
From the densities of sodium and potassium solutions it has been shown that these volume changes pass through a maximum at about $3N.$

I am still unaware of the reason for the increase of final volume $V_\mathrm{f}$ after reaction between metal $\ce{M}$ and liquid ammonia $\ce{NH3}$:
$$V_\mathrm{f} \gg V_0(\ce{M}) + V_0(\ce{NH3})$$
Reference

Marshall, P.; Hunt, H. Volume Changes for Metal Solutions in Liquid Ammonia. J. Am. Chem. Soc. 1955, 77 (19), 5016–5016. DOI: 10.1021/ja01624a022.

related https://chemistry.stackexchange.com/questions/72264/what-is-happening-in-this-video-of-solvated-electrons-donated-from-sodium-in-amm

Volume always changes when you dissolve something, what's the point of this question?

Rather just > than >>, as the latter implies formation of gaseous phase.

i just meant a large increase in volume i don't know if a gaseous phase is formed or not,maybe something is linked to gaseous phase for the increase.

Excellent question.  One reason I can think of is that the packing of atoms in solid alkali metals is pretty dense.  When dissolved in ammonia, this packing is undone, creating a solvated alkali metal cation and a solvated "electron".  Each of these solvated ions, consisting of shells of ammonia molecules presumably takes up a bunch of space.

@CurtF. If i consider only ammonia for an instance, then i am just adding a metal cation and a electron in between which also creates dipole-dipole attractions and is kind of filling voids present in solution so i think there should be a slight decrease or atleast not a large increase.

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Chemical Equilibrium of the Esterification Reaction Between Propanol and Ethanoic Acid
I am trying to investigate the relationship between temperature and equilibrium constant of the esterification reaction between propanol and ethanoic acid in reflux system, with conc. sulphuric acid as the catalyst. I plan to extract the ester (as I like the fruity smell) and measure the mass of the ester formed, by:

Measuring $\pu{50 cm3}$ of propanol and conc. ethanoic acid using a pipette
Add the propanol, conc. ethanoic acid and $\pu{5 drops}$ of conc. sulphuric acid into a reflux system
Put the system into a water bath set to $\pu{20 ^\circ C}$, $\pu{40 ^\circ C}$, $\pu{60 ^\circ C}$, $\pu{80 ^\circ C}$, and $\pu{100 ^\circ C}$
Wait until the system reaches equilibrium (the part I am struggling with)
Remove the two acids (and thus stopping the reaction) using excess $\ce{Na2CO3 (aq)}$
Remove alcohol using excess $\ce{CaCl2 (aq)}$
Use a separatory funnel to get rid of most of the aqueous solution
Use anhydrous $\ce{Na2SO4 (s)}$ to get rid of the remaining aqueous solution
Use distillation to obtain pure propyl ethanoate
Measure the mass of the propyl ethanoate to obtain the amount of propyl ethanoate (in $\pu{mol}$)
Use the I-C-E method to obtain the $K_\mathrm{c}$ at that temperature ($\pu{20 ^\circ C}$, $\pu{40 ^\circ C}$, $\pu{60 ^\circ C}$, $\pu{80 ^\circ C}$, and $\pu{100 ^\circ C}$)

Here comes the problem. How can I determine whether the reaction has reached chemical equilibrium or not? I plan to use the I-C-E method to obtain the $K_\mathrm{c}$ but if the system hasn't reached equilibrium yet before I quench the reaction, it will be pointless to obtain the mass of the ester. I have thought of measuring the $\mathrm{pH}$ value of the system using a data logging pH meter (since the concentration of ethanoic acid will decrease), but my teacher said it will not work as conc. sulphuric acid is present. Or maybe measure the conductivity of the mixture using a data logger (the pasco-capstone one)? Does that work?
Also, I am supposed to compare my experimental values with literature values. However, I cannot find the literature value of the $K_\mathrm{c}$ of this reaction on the internet. Does anyone know where I can find the Kc value across different temperatures of this reaction?
In addition, the boiling points of propanol ($\pu{97 ^\circ C}$), ethanoic acid ($\pu{118 ^\circ C}$), water ($\pu{100 ^\circ C}$) and propyl ethanoate ($\pu{102 ^\circ C}$) are very similar. How can I ensure that there will only be pure ester in the distillate?

You have reached equilibrium when aliquots removed from the flask give constant values for the mass of ester. You might consider gas chromatography using a non-reactive internal standard that has been calibrated for detector response to isopropyl acetate and the standard. BTW: Your experiment will probably be more successful if you refer to ethanoic acid as acetic acid as practicing chemists do. ;)

First, my school does not provide gas chromatography ;-; Second, if I remove aliquots from the flask, it will become an open system and the equilibrium cannot be measured

Open system to what? The atmosphere? Doesn't matter.

but one of the prerequisites of achieving chemical equilibrium is that it has to be a closed system

How do you think chemists monitor reactions? What do think should be done?

With my very limited Chem knowledge, i don't really know


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Titanium reaction with Sodium Percarbonate solution
I've been reading about a process for darkening Titanium using Sodium Percarbonate that consists of the following steps:

Prepare Sodium Percarbonate solution 1:5 with boiling water
Soak titanium piece in solution

I have not been able to find any technical readings on what exactly is going on with this chemical process.

These guides have also found that Titanium Grade affects the darkening of the Titanium:

An oxidation on Grade 1 surface is rather aggressive, not even and it's brown not grey... oxidation on Grade 5 is neither that aggressive nor that brown.

And that heat affects the reaction as well:

The solution is doing the job only when its warm, so I have repeated the process 3 times.

Add some hot water from time to time and stir...

Links:

https://www.rwg.bz/board/index.php?/topic/161199-how-to-force-a-patina-on-a-raw-titanium/

https://www.bladeforums.com/threads/darken-titanium-with-oxi-clean.1019026/

What exactly is going on in this reaction with the Titanium? What is the material of this new darkened layer?

The effect is similar to that of anodizing titanium (or aluminum) to build up a thin surface oxide coating. There are a few causes for the color change.

A thin, transparent, layer of oxide causes optical interference due to the reflections within the oxide layer surface and the underlying metal. This can produce a whole spectrum of colors in the titanium workpiece.
Rather than having a shiny, metallic reflective surface, a thick oxide layer is rough, at the wavelength of light, causing absorption destructive interference and therefore darkening the part.
Impurities in the metal such as silicon may cause patchy, uneven anodizing, and others, such as iron, form colored salts that fill the pores of the oxide.

Based on the first link that you gave me it appears that this sodium percarbonate bath is actually an occurrence of Type 2 Ti Anodization? I'm familiar with the Type 3 as that method seems to be well documented online, though I didn't know the name for it.


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Paracetamol infrared spectrum issues
I was studying Infrared Spectroscopy and i analyze different polymers. When i prepared paracetamol making a pill mixing 2mg of it with 200mg of $\ce{KBr}$ on a vacuum-connected press when i put it in the Infrared the equipment returned me this signal.
The equipment is infrared spectroscopy BUCK500. On the spectrum we can see starting from 1500, in 1700 the carbonyl bond $\ce{C=O}$ in 2350 some lens damage or dust and in 2900 de alkane $\ce{C-H}$ bond.
According with the Paracetamol structure must be at 3500 signal according to its -OH group and $\ce{alkene C=C}$ at 1600.
My question is if this IR is correct or is something wrong.

Yes i'll edit it. My question is if this IR spectre is correct or not if it could be something in the procedure to not appear  the -OH and the C=C group

I searched for “paracetamol IR spectrum” and found spectra. I suggest doing the same and looking at ones from authoritative sources: this is just acetaminophen (Tylenol, as one of the OTC trade names), so I would expect it to be used in an introductory IR lab experiment.

Go to NIST website NIST and check reference spectra of acetaminophen. Do not expect a 100% match in peak shapes because of instrument differences. The peak positions should match.
When you suspect that the IR instrument is not behaving well, it is a good idea to run a instrument health check with a polystyrene film. It does not require any sample preparation, the sheet is directly inserted in the light path with a window made from cardboard. You should get a very nice textbook like spectrum of polystyrene.
As per the manual of your instrument, polystyrene film is provided with the instrument. https://www.bucksci.com/pages/buck-m500-infrared-spectrophotometer
It is amazing to see that there are still scanning IRs in use! Anyway, there are no lenses, you will only find high quality super-expensive mirrors and gratings there. It is hard to damage them because it is all protected. So the artifacts appear to be of electronic nature rather than due to optics.
a) Make sure that KBr and paracetamol are "bone" dry! And warm up the instrument for at least an hour.
b) I see some abnormal artifacts. There is a sharp drop in the transmission spectrum around 700 cm$^{-1}$. This has to be addressed.
c) There is an artifact around 2350 cm$^{-1}$. This is from incorrect baseline correction. This is perhaps due to carbon dioxide. Hopefully not many students are breathing closely near the instrument or poking their nose.


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Is there a stretchable, transparent polymer like PDMS, but not so hydrophobic?
I'm looking for something like PDMS, but not as hydrophobic. I'm trying to form 2d crystals using the surface tension of an evaporating droplet to organize microspheres into an ordered array. PDMS would be perfect for this application, except it's too hydrophobic -- PDMS causes my aqueous solution to bead with a very high contact angle. I need something that allows the droplets to spread out further.
I need the material to able to be formed into sheets that can be stretched elastically, returning approximately to their original shape.
I'm looking for a material that's transparent in the visible -- doesn't have to be transparent over the entire range, but I suppose that red to near-IR is probably the part of the spectrum easiest for me to work with, so it would be best if I can transmit light through the material without much loss in that part of the spectrum.
I need the material to be available commercially or be simple to produce in a lab with limited chemistry know-how or materials -- again, PDMS would be awesome, if it weren't for the high contact angle.
I'd also be happy to try playing with the solution that the colloidal microspheres are suspended in. They're about 1um in size and formed from Fmoc-G-(Bhoc)-aeg-OH powder dissolved in water heated to 100°C. I tried previously with low concentrations of SDS, but didn't see promising results -- I have some new samples under preparation right this moment to check those earlier negative results.
Any suggestions would be greatly appreciated! Thanks!

Though not very stretchable, cellulose or cellulose-derived dialysis membrane is somewhat elastic when wet, and is hydrophilic -- so much so, that is might absorb the water, causing the opposite problem. At least it's cheap and readily available to test in your experiment. Good luck with you 2-D quasi-crystal demo!


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Do we need sulfuric acid to make TNT?
Do we actually need sulfuric acid to make trinitrotoluene? TNT's formula is $\ce{C7H5N3O6}$ so we just need toluene($\ce{C7H8}$) and nitric acid($\ce{HNO3}$). If we mix them we get TNT, and water($\ce{H2O}$)
$$\ce{C7H8 + 3HNO3 → C7H5N3O6 + 3H2O}$$
But Wikipedia states we need sulfuric acid. Why?

Please check an answer as accepted if it has answered your question.

Sulfuric acid is needed in order for a good electrophile to form. Sulfuric acid protonates nitric acid to form the nitronium ion (water molecule is lost). The nitronium ion is a very good electrophile and is open to attack. Without sulfuric acid the reaction would not occur. Hence, a nitrating mixture of concentrated nitric and sulfuric acids is used to nitrate toluene to mono-, di- and ultimately to trinitrotoluene.

(image source)


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Copper turned silver-color with crystalized soldered joints. How did this happen?
Please remove if this question if not allowed on this sub-exchange. I can post elsewhere if suggested.
I was working on a small project requiring cleaning and polishing copper pipe, but found sparkling solder joints and silver-like finish instead.
I attempted to clean copper pipe with white vinegar + salt mixture. I let the copper sit for around 30 minutes in a bucket of this solution. At this point I'm aware of the copper oxide appearing on the surface. Pictured below.
After drying off with a towel I attempted to shine the copper. I applied Copper and Brass cleaner which contains quartz, citric acid, ammonium chloride, and magnesium aluminum silicate. To my surprise, 2 changes happened to the copper/solder:

The copper has a shining silver-like appearance.
If you look closely, the joints which were soldered with lead-free silver solder are sparkling in the light (very apparent), almost as if they contained tiny crystals.

It's important to note that the sparking soldered joints only appeared after using the copper and brass cleaner. They're still dull after only removing them from the white vinegar + salt solution.
How in the world did this reaction happen? Is it possible to remove the silver finish?

I'm aware this home-depot bought copper may have small impurities of tin, iron, zinc, or nickel, but doubt that makes a difference.

Before putting the copper in a white vinegar + salt solution:

30 minutes after drying the copper pipes from the solution:

After polishing with the cleaner:

I don't think the copper pipe is pure copper. It must have other metals too. There is no way that pure copper after being exposed to copper cleaner becomes silvery.

Use a known pure copper sheet and repeat. Control experiment is very important here.

Turning pure copper into pure silver happens when copper is exposed to a soluble mercury salt. It then looks like pure silver for a couple of months.

It's rather silver in color. Not saying it's exactly silver or a silver compound.

True, but in order to get a scientific explanation, repeat the same experiment with known pure copper sheet. It won't appear silverish.

The anode here is the tin (which is subject to corrosion, aka, dissolving) and the very noble silver is the cathode (see this table of relative anodic behavior posted here). One could also argue at some point the copper becomes the cathode.
In any event, the tin goes into solution whereupon the more chemically active copper displaces it, resulting in a tin plating effect on the copper surface.
The ammonium chloride can act here in both the roles of an electrolyte, the electrochemistry part, and as a complexing agent for the tin (and copper) as part of the standard chemistry.
Note: If it is determined that the plating metal is, indeed, silver metal, then one must account for the creation of, say, aqueous $\ce{Ag(NH3)2Cl}$. This would likely involve the presence of solvated electrons (from the galvanic chemistry) interacting with oxygen exposure in the presence of the $\ce{NH4Cl}$.
Kind of an incidental, but interesting and possibly advanced example of mixed chemistry (as occurs with copper and is noted to occur with silver here).
[EDIT] At home one can test for silver metal with pure household disinfecting bleach (see the discussion here on the so-called "Bleach Test").

There is essentially no silver present in this object.

Per a source, quoting "The silver content ranges from as high as about 80 percent for very “hard” solders with high melting points, to as little as 30 percent for the industrial...".  I have once purchased silver solder and agree with this statement. So from an electrochemical viewpoint, a significant  Ag presence. Reference link: https://juxtamorph.com/silver-solder/ .

My speculation on the chemistry if plating is indeed Ag (otherwise, just Sn). First, NH4+ = H+ + NH3 where galvanic electrons e- attack H+ via H+ + e- = .H  and .H + .H = H2, along with .H + O2 = .HO2 -> H+ + superoxide radical anion ( .O2- ). Per my cited reference on the possible formation of [Ag(NH3)2]Cl, a soluble silver presence is now possible. Next, a displacement reaction either  by Sn or Cu metal in the visible form of silver plating.

Now, if a composite with 2% silver is present in a significant amount, that is still 2% of a  significant amount, which may be enough for Ag plating. So, I am not in the definite no Ag plating camp.

The copper pipe system is a large copper cathode (the copper tubing) with a small area of exposed silver solder (*) anode comprised mostly of tin. When it dissolves in the citric acid plus ammonium chloride, apparently enough silver becomes soluble to migrate over to some other copper surface and plate out - silver dendrites are a common result in concentrated silver salt solutions, but in the OP's picture, they are so tiny as to just look crystalline. Here is a picture of large silver dendrites forming on copper wire. Ref 1.

The culprit, if there can be said to be one, is the silver solder. It sounds like an upscale binder, but may have limitations. Of course, the copper and brass cleaner is part of the problem, by providing the corrosive potential.
Apparently the entire copper surface is now plated with a thin silver film that would develop dendrites if there were more silver.
Is it possible to remove the silver finish? It is probably so thin that the best way would be strictly mechanical: rubbing with abrasive powder like silica, or sandblasting with sodium bicarbonate, without chemical corrosion.

Oatey Flo 8 oz. Lead-Free Silver Solder is good for soldering copper pipe and fittings together. It is safe to use in potable water applications. This silver solder has a wide melting range of 415° to 455°F. It can be used for plumbing, HVAC, fire sprinkler and other systems.
•   95.5% tin, 3% copper, 1% bismuth and less than 2% silver alloy
•   420°F - 460°F melting range
•   Meets ASTM standard B-32 and the requirements of the safe drinking water act
•   Formulated for plumbing/potable water line applications

Ref 1. https://www.cheminst.ca/magazine/article/chemisteam-2021-silver-crystal-fern/

The Oatey solder listed in the question contains no silver, the guy writing the ad thought "silver solder" was a good sounding name. Silver solders melt above 1150 F, more accurately called silver braze.  Copper pipe, tube, and fittings are commercially pure copper[ ASTM 42 is between 99.4 and 99.9 % minimum copper per Copper Development Association data book, 1973 ( these specs do not change much with time). You are using a high tin solder containing bismuth and other elements. I can not explain how a cleaning solution could have coated the part with solder.  The silver colored structure looks like it is covered with solder as if it had been dipped in a molten bath like galvanizing or heated in a furnace. Furnace solder or braze of a smallish complicated structure as shown would not be unusual.

You are right that "less than 2% silver" could also be equivalent to 0% silver. I couldn't figure out #1. how tin could plate out on copper, and especially, #2. if the solder is being dissolved by the cleaner, why doesn't it dissolve the (slightly purer?) silvery coating? It seems like a multi-step reaction. And it is not at all a standard "silver solder", which is higher melting (hard).

It looks like a melted deposit ; but where did all the tin come from?  Not enough tin in the annular gaps  were soldered.  Tin and lead solder look the same; I have lead soldered pipes and recoated antique copper pan with pure tin. No visual indication that this is redeposited metal.


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Mechanism of alkyl benzene reaction with KMnO4/KOH to form carboxylic acids

I came across this reaction regarding formation of carboxylic acid by treating alkylbenzenes with KMnO4/KOH.
I know that alpha-(C-H) bond in this compound is easily broken down into radicals. Then it reacts to form

After this step mysteriously manganese reactions to form phenylmethanoic acid and all remaining alkyl carbon are oxidized to CO2
I want your help to understand the mechanism of this reaction and how manganese reacts further down the reaction.

this answer at least partially answers your question https://chemistry.stackexchange.com/questions/49542/mechanism-of-the-oxidation-of-alcohols-with-kmno4

i tried this exact mechanism but end product is alkylphenylketone but in above reaction which i mentioned end product is benzene carboxylic acid. I want a mechanism of this reaction to understand how reactions occurs and why certain product is favored over other.

Work through the mechanism with one of the R groups being H so that the aldehyde is formed. Remember that aldehydes readily form hydrates in aq environments.

i did the thing you are recommending and we do get a aldehyde but the original reaction has R group as alkyl so i can't just do that. But what i want is actual mechanism which explains why even the alkyl groups are converted to CO2 and how that occurs.


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Effect of pressure on the reaction rate of a reaction that produces a gas
If an irreversible reaction that produces a gas from say a solid and a liquid is in a closed system. How would the reaction rate be effected by increasing the pressure in the container?
I'm guessing that it would slow the rate of the reaction as the added gas from the reaction would further increase the pressure and that seems like it would be unfavorable. However the answer key says that the reaction rate would not be effected as there is no gas in the reactants which doesn't make sense to me.

rate of reaction will be unaffected as it is related to kinetics it will change with temperature and other factors.You can read more about that here.(https://en.wikipedia.org/wiki/Reaction_rate)
now i think in your question equilibrium is already attained and they are asking if rate of forward/backward reaction is changed or not.
so that would not change as it depends on temperature and catalyst.
note:catalyst can increase decrease rate of forward/backward reaction to equal extent hence overall equilibrium constant is still not affected.
now it is true that something is changed when pressure is increased.as you said that when pressure is increased the partial pressure of gaseous product will increase.
this would make (reaction quotient)>(equilibrium constant).
hence the reaction will then go in backward direction to re-attain equilibrium.
In case of an irreversible reaction which is going forward the rate of reaction depends only on reactants,why?
Refer this text
from the link i shared earlier
As you said it is a strictly forward reaction the products have to do nothing with the reaction as they are formed and simply remain like that.
(Note: if the reaction could go backward then it would have also depended on the products as they would have started to act like reactants for backward path)
So now the reaction rate depends only on the original reactants which are solid and aqueous form and clearly both of them are not affected by any change in pressure.
you can also see the following text for additional info.
(https://chem.libretexts.org/Courses/Bellarmine_University/BU%3A_Chem_104_(Christianson)/Phase_2%3A_Understanding_Chemical_Reactions/4%3A_Kinetics%3A_How_Fast_Reactions_Go/4.1%3A_The_Speed_of_Reactions)

Okay thank you, that makes sense to me, however in this case the reaction was not an equilibrium reaction it is strictly a forward reaction. This was the given reaction. Fe(s) + 2HCl(aq) -> H2(g) + FeCl2(aq)

please mention the whole data of question from next time

I have edited my answer, check it if that is what you wanted to ask.Also mention your research from next time while asking a question.

Thank you so much, I think I have done too many equilibrium problems and am overthinking this. But what you said makes sense.


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Hybridisation of the O atom in the hydroxyl group of ethanoic acid
Currently studying high school organic reaction mechanisms and encountered a question. It goes like this:

The C=O bond and the C–O bond of an ethanoate ion is equal at 127 pm, while the C=O bond of ethanoic acid is shorter at 121 pm than its C–O bond, which has a length of 141 pm. By drawing resonance structures, explain why this is so.

So I know why the C=O and C–O bonds in the ethanoate ion is equal because both of its resonance structures have the same significance, but I'm not so sure about the difference in the bond lengths in ethanoic acid. When I tried to draw the resonance structure for ethanoic acid, I was thinking that the O atom of the hydroxyl group is $sp^3$ hybridised (since each of the two lone pairs will occupy two separate $sp^3$ hybridised orbitals), so that means there is no resonance effect.
After scouring through the internet, the O atom of the OH group is $sp^2$ instead. Why is this so?
(p.s. high school student here, so I might not be able to understand high-level explanations just yet, sorry)

Note that ethanoic acid is the proper IUPAC systematic name, but the preferred IUPAC name is acetic acid.

This has hardly to do with any hybridisation. You're supposed to compare mesomeric structures.

Sorry, can you elaborate on what you meant by comparing mesomeric structures?

You're absolutely right in saying that in ethanoate ions, both resonance structures are equivalent - as both oxygens are the "same" in environment and position. However, in ethanoic acid, because of the presence of the hydrogen atom, the second structure is considerably less stable than the first - notice that the oxygen atom bearing a positive charge is unstable.
Why sp2 and not sp3? Well, the hybridisation of an atom is something that remains fixed in a molecule of the compound. You can't have the hydroxyl O being sp3 hybridised and yet participating in resonance - because the double bond wouldn't form without a p orbital for conjugation!
So the only sensible explanation (and experimentally verified fact) - is that the hydroxyl O is sp2 hybridised - one lone pair in an sp2 orbital of its own, the hydrogen bonded to O via another sp2 orbital, the carbon bonded via a third sp2 orbital (sigma bonding), and the second lone pair delocalised from the 2p orbital to form the pi bond in the second resonance structure. And there you have it!

I was going to post the same answer but sidsr003 got there first, well done this is the same as what I teach my undergraduates when I am teaching organic chemistry.

What is the "experimentally verified fact" you refer to?  Note that, as discussed earlier at several places in Chem SE, photoelectron spectroscopic data suggests the $\ce{O-H}$ bonds in the water molecule are roughly $\ce{sp^3}$ hybridized while the oxygen  lone pairs reside in more or less $\ce{p}$ and $\ce{sp}$ orbitals.

Thank you so much for your answer, but I am still not convinced as of why the O atom of the hydroxyl group is sp2 hybridised, since your explanation seems to assume first that there is resonance in ethanoic acid. How would you know that the O atom is sp2 hybridised, without knowing that there is in fact the resonance effect in ethanoic acid?

The experimental evidence
The bond next to the $\ce{C=O}$ double bond in acetic acid and in esters is conventionally drawn as a single $\ce{C-O}$ bond, but does not show free rotation. For example in methyl formate, the most common arrangement of atoms is this:

Rotating around the single bond (marked with red dots) does not happen readily (the conformation with the $\ce{-CH3}$ pointing at us is estimated to occur roughly a billion times less than the one shown). Similarly, the $\ce{C-N}$ bond in amides does not behave like a single bond, and amides are planar as well.
The length of the $\ce{C=O}$ double bond is about the same as in other molecules, where there is no oxygen or nitrogen single bond with the carbon. The length of the $\ce{C-O}$ double bond is a bit shorter than in, for example, methanol ($\ce{H3C-OH}$). In general, bond lengths can tell us whether a bond is a single or a double bond or something in between.
The interpretation
Somehow, the bond next to the $\ce{C=O}$ double bond has some "double bond character". The easiest explanation when writing a Lewis structure is to show a second resonance contributor. For the acetate ion, the two resonance contributors are identical (with the role of the two oxygen atoms switched). For the acetic acid molecule, one resonance contributor has formal charges (the minor one) and the other does not (the major one).
The question of the hybridization is not so important because you only have two atoms attached to the atom in question. If there are three atoms attached (as in the formamide below), you would either have a trigonal planar or a trigonal pyramidal arrangement, telling you whether it makes more sense to invoke $\mathrm{sp^2}$ or $\mathrm{sp^3}$ hybridization.

Source: https://chemistry.stackexchange.com/a/10917/72973
In this case, however, it does not matter for the major resonance contributor. The minor conributor, however, has to be $\mathrm{sp^2}$ (because you need the unhybridized p-orbital for the double bond).

The C=O bond and the C–O bond of an ethanoate ion is equal at 127 pm [...]

If the bond lengths are equal, it makes little sense to call one a single bond and the other a double bond. It would be nicer if the question read "the two bonds between carbon and oxygen in the acetate ion are of the same length, while in acetic acid, the bond between carbon and the hydroxyl group is longer than the other bond with oxygen". In this way, you don't make an assumption of the type of bond (single, double, or somewhere inbetween).

The rotation barrier between s-trans and s-cis acetic acid is a bit hard to measure because there is hardly any s-cis acetic acid present. However, it has been isolated in an argon matrix at 8 K: 10.1021/ja038341a

Thank you for the answer. Yes, I do understand that due to the resonance effect, the C=O and C–O bonds are 'partial' double bonds. But when I was drawing the resonance structure of acetic acid, I was confused as I thought it makes no sense that there is resonance in acetic acid in the first place, since I thought the O atom of the –OH group was sp3 hybridised, without an hybridised p orbital to overlap with adjacent unhybridised p orbitals. Hence, the question.


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Electrolysis in conductivity cell while measuring the conductance?
My textbook says we could measure the conductance of an electrolyte by passing an AC current through a conductivity cell and wheatstone bridge's balance.But will it not electrolysis the solution which could act as a sink of power and make us end up getting the wrong value?

It is (should be) done by low enough amplitude and high enough frequency to avoid electrolysis.

If a DC tension is applied through an electrolytic solution with a very low voltage, the ions will move towards the electrodes for probably less than one second, then this migration will stop as the solution gets polarized. The accumulation of positive charges around one pole, and of negative charges around the other pole will prevent new charges from migrating and reaching the polar regions. No electrolysis will occur. But there is a small current during this short time. Now if later on, the low voltage is suddenly reversed, the migration of the ions is reversed, and the current will also last less than one second. long enough to get the same sort of polarization of the electrodes (but with reversed signs). If the voltage is AC, the current will be maintained in the beginning of every new period. There will be no electrolysis at all. But some electric energy is spent for maintaining this comings and goings in solution.


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Why is sodium bisulfate acidic?
So, as I know, the rule says that if you get a salt from strong acid and strong base it should be neutral. In this situation with sodium bisulfate and also trisodium phosphate we get an acid salt, although bought of them are produced from strong acid and strong base. Can someone explain why they have an acidic character and perhaps also give an explanation how I can determine if a salt will have an acidic character in water?

Sodium bisulfate $\ce{NaHSO4}$ has still one acidic hydrogen, as sulphuric acid is a diprotic acid, therefore it is acidic. By other words, just a half of sulfuric acid is neutralized here. Note that the anion $\ce{HSO4-}$ lays at boundary between strong and weak acids.
OTOH, trisodium phosphate $\ce{Na3PO4}$ is strongly basic, as hydrogenphosphate anion $\ce{HPO4^2-}$ is very weak acid.

And what about Monosodium phosphate, it should be acidic too then? And that also means that strong base and weak acid doesn't always give a basic salt?

See chem.libretexts.org: Polyprotic_Acids_And_Bases

H2PO4^- is mildly acidic, as it is stronger acid than base. HPO4^2- is mildly basic, as it is stronger base than acid. PO4^3 is just strong base ( but weaker than hydroxide )


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Inhibitors reaction at home
For a school science project, my teacher asked me to find a reaction that can be slowed down with an inhibitor and film it at home/lab at school. He told not to use lemon on an apple or anything similar because it’s too common. I’ve been looking all over Internet and my kitchen but I can find anything.
Also, I can’t inhibit the reaction of catalysing hydrogen peroxide, because I’m already using this reaction for the other part of the project which is catalysts.
Is there any other inhibitor reactions that actually can be done at home?

Already well answered here: https://chemistry.stackexchange.com/a/147109/79678.

Closely related: https://chemistry.stackexchange.com/q/147094/79678.

@EdV This might be a duplicate.

Does this answer your question? Catalyst and Inhibitors home experiments

@NilayGhosh I think so. I just voted to close for that reason. Thanks!

Protease action on jello.
Fresh pineapple contains proteases including bromelain which will break down gelatin. If you add fresh pineapple to your gelatin-containing jello it will not set as the gelatin is hydrolyzed.
Canned pineapple is heat-treated, which will denature the bromelain. Thus, canned pineapple could serve as your inhibitied sample.

What is jello ? Why should it set ?  What should be the action of protease on gelatin ? Where is the inhibiton here ?

@Maurice Jell-o: https://en.m.wikipedia.org/wiki/Jell-O

The protease is “inhibited” (destroyed, really) at high temperature.

Here is a fairly simple experiment, essentially involving the replacement of $\ce{H2O2}$ with $\ce{NaOCl}$ (as commonly found in so-called Disinfecting Chlorine Bleach).
This experiment verifies the widely reported sensitivity of common chlorine bleach to sunlight, to quote a guide, for example:

Bleach rapidly degrades in the presence of light and when mixed with water.

I would proceed with two identical portions of the $\ce{NaOCl}$, where one open vessel is stored in the dark, and the other exposed to sunlight, best covered with clear plastic wrap.
Measure the respective comparative ending concentrations of the sodium hypochlorite with the addition of the same amount (but in stoichiometric excess per equation below) of hydrogen peroxide:

$\ce{NaOCl (aq) + H2O2 (aq) -> NaCl (aq) + H2O (l) + O2(g)}$

The mechanics of the photolysis, can perhaps be more likely explained from the photosensitive nature of either minor transition metal presence (from say iron or copper ions including from any incidental dust particle exposure), or just sunlight acting on hypochlorite, or even some hypochlorous acid, per the reaction:

$\ce{OCl- (aq) + H2O (aq) -> HOCl (aq) + OH-}$

based on the reported action of sunlight in this article: "Emerging investigators series: the efficacy of chlorine photolysis as an advanced oxidation process for drinking water treatment", to quote:

The photolysis of hypochlorous acid (HOCl) and hypochlorite (OCl−) produces a suite of reactive oxidants, including hydroxyl radical (˙OH), chlorine radical (Cl˙), and ozone (O3).

where such species would lead to the more rapid degradation of the bleach.
I hope suggestion is of use.


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What will be the structure of cysteine at pH=4?
As per my understanding since pKa of RSH and the amino group is >4 hence they would be in cationic form while the acid should be in anionic form. So charge on cysteine would be +1. But while looking on internet i found that nowhere the RSH group is protonated.
Please someone clarify

...hence they would be in cationic form

Protonated does not imply cationic form. Some conjugated acid/base pairs have a neutral and an anionic form (like the carboxylic acid/carboxylate pair).
The acid/base equilibrium of the thiol group is the following:
$\ce{R-SH <=> R-S- + H+}$
The left-hand side is called protonated, and the right-hand side is called deprotonated.
$\ce{R-SH2+}$ is not observed in aqueous solution.

since pKa of RSH and the amino group is > 4

The $\mathrm{p}K_\mathrm{a}$ of the ammonium group is > 4. The amino group does not act as an acid in aqueous solution, and its $\mathrm{p}K_\mathrm{a}$ is too high to be able to deprotonate an amino group in aqueous solution.

Related: https://chemistry.stackexchange.com/questions/162474/are-there-separate-quantity-symbols-for-the-pka-of-a-substance-and-the-pka-of-it


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Molecular orbital theory and Lewis structure for dioxygen
I am a bit confused about how molecular bond theory predicts pairings of unbounded electrons. For example, in $\ce{O2}$ there are two pairs of electrons on each oxygen along with the double bond. However, in the molecular orbital diagram, there are two unpaired electrons in degenerate $\pi$ antibonding orbitals. Obviously, bond order predicts the double bond. But how can one "predict" the two pairs of non-bonded electrons in $\ce{O2}$ from this?

Source: https://commons.wikimedia.org/wiki/File:Oxygen_molecule_orbitals_diagram.JPG
Pairs of non bounded electron refer to these ->  ++++++++++++ooooo++++++++++ooooo++++

Hund's first rule applies to MOs as well as AOs.

but in O$_2$ lewis structure, the electrons are $paired$, which is equivalent of two electrons in the AO being in the same orbit (ie, both red arrows are on the left).

Well, okay. So we have here two theories which contradict each other. How do we resolve this? One says that oxygen should be paramagnetic, the other says it should be diamagnetic. And it turns out that oxygen is famously paramagnetic. So one theory correctly predicts this, and the other one is wrong.

What do you mean by "the two pairs of non-bonded electrons in O2"

I provided a diagram to clarify @M.L

See Faraday and Magnetism

@AdamRadekMartinez check this out: https://chemistry.stackexchange.com/questions/73768/lone-pairs-in-mo-theory


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Derivation of the Equation $p=\rho^2 (\partial F_s/\partial \rho)_T$
I'm trying to derive a relationship between the mass-specific Helmholtz free energy,the pressure, the temperature, and the mass density. It is given by
$$p=\rho ^2\left ( \frac{\partial F_s}{\partial \rho } \right )_T$$
where $F_s$ is the mass-specific Helmholtz free energy. The steps that I carried out were:
$$dF(T,V(\rho ,T))=-SdT-pdV(\rho ,T)$$
$$=\left ( \frac{\partial F}{\partial T} \right )_VdT+\left ( \frac{\partial F}{\partial V} \right )_TdV(\rho ,T)$$
$$=\left ( \frac{\partial F}{\partial T} \right )_VdT+\left ( \frac{\partial F}{\partial V} \right )_T\left [ \left ( \frac{\partial V}{\partial \rho } \right )_Td\rho +\left ( \frac{\partial V}{\partial T} \right )_\rho dT \right ]$$
$$\Rightarrow -pdV(\rho ,T)=\left ( \frac{\partial F}{\partial \rho } \right )_Td\rho +\left ( \frac{\partial F}{\partial V} \right )_T\left ( \frac{\partial V}{\partial T} \right )_\rho dT$$
$$\left ( \frac{\partial V}{\partial T} \right )_\rho =0$$
$$\Rightarrow -pdV(\rho ,T)=\left ( \frac{\partial F}{\partial \rho } \right )_Td\rho $$
$$\Rightarrow -p=\frac{\left ( \frac{\partial F}{\partial \rho } \right )_Td\rho }{\left [ \left ( \frac{\partial V}{\partial \rho } \right )_Td\rho +\left ( \frac{\partial V}{\partial T} \right )_\rho dT \right ]
}$$
I don't see where to go from here, because carrying out the division would just yield that
$$-p=\left ( \frac{\partial F}{\partial V} \right )_T$$
which is where I started. The motivation for this is simply the fact that an expression for a mass-specific Helmholtz function given in the following pictures has been used to determine the properties of water. The first relation in part (a) in the second picture is what I'm trying to verify.

Chemistry SE site strongly recommends plain text titles for index/search reasons and due possible displaying issues in question lists.

Yup, plain text titles are always better. If you must use MathJax, please do it in inline style: $...$ rather than $$...$$ and also don't write fractions as \frac{a}{b}, just use a/b where possible.

There seems to be a dimensional inconsistency in the equation.  The LHS (p) has units of E/V.   If rho is, as you state, mass density, then the RHS has units of Erho = E(m/V).

@theorist the Helmholtz free energy considered is the mass-specific Helmholtz free energy, which has units of E/m.

Starting with dF=-SdT-PdV, you have:
$$\left(\frac{\partial F}{\partial V}\right)_T=-P$$where V is the molar volume.  The molar volume is related to the molar density $\rho$ by $$V=\frac{1}{\rho}$$Therefore, $$\left(\frac{\partial F}{\partial V}\right)_T=-\rho^2\left(\frac{\partial F}{\partial \rho}\right)_T$$
and hence
$$P=\rho^2\left(\frac{\partial F}{\partial \rho}\right)_T$$

Thanks a lot, this makes perfect sense.

$$V=\frac{m}{\rho }\Rightarrow \frac{\partial V}{\partial \rho }=-\frac{m}{\rho ^2}\Rightarrow -p\left ( \frac{\partial V}{\partial \rho } \right )_T=\left ( \frac{\partial F}{\partial \rho } \right )_T$$
$$-p\left (-\frac{m}{\rho ^2}  \right )=p\frac{m}{\rho ^2}\Rightarrow p=\frac{\rho ^2}{m}\left ( \frac{\partial F}{\partial \rho } \right )_T=\rho^2\left ( \frac{\partial F_s}{\partial \rho } \right )_T$$


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Heat capacity of solution = Solute + Solvent?
Is the heat capacity of a solution the sum of the heat capacities of the solvent and the solute?
Based on what I have learned, specific heat is determined by the interaction between molecules. This explains water's large specific heat because it has many hydrogen bonds which take a large portion of the supplied heat to break and only the rest actually increases the temperature.
When the solute dissolves into the solvent, I assume the bonds between the molecules strengthen, so the total heat capacity increases. Is this correct?
If I am correct, is the change in heat capacity negligible or significant enough to take into calculation?

Yes, it is. Heat capacity it's defined by the solvent and the solute, in proportion to its concentration. Usually, you'll find the heat capacity of a NaOH 30% solution and of a NaOH 10% solution which Is slightly different.

@AndreaPellegrini But this is obvious, as different ratios of both with the same total mass, sitting side by side, would have different heat capacity either. The question is about heat capacity of mixed versus separated.

You could keep specific heat. It is intensive property, while heat capacity is extensive. For the same total mass, their ratios for different systems are the same. // If you consider the nature of specific or molar heat capacities, it should be clear they must generally change by dissolution.

@Poutnik Thanks for commenting. So, does the heat capacity increase or decrease? + How much will it change?

Yes, it does. This much. It is case specific and rather experimental value, hard to be predicted. Aside of the direct heat capacity measurement, it can be determined by temperature dependence of dissolution enthalpy.

Very closely related: does-enthalpy-of-dissolution-change-with-temperature  // Also related Wikipedia: Enthalpy_change_of_solution

To expand on what Poutnik said, if you dissolve, say, salt in water, the physical states of the salt and water both change. The salt changes from solid to aqueous, and the water changes from only interacting with itself to interacting with both itself and the Na+ and Cl– ions. So, by definition, the heat capacity of the solution will be different from the sum of the heat capacities of the individual components. The ques., though, is how much different (and whether it's more or less) and whether that diff. is significant for your calcs. The only way to know that is to look up the actual data.


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Why is S-alanine acetylated in the synthesis of S-cathinone?
I came across a method of synthesizing enantiomerically pure S-cathinone using S-alanine which is chlorinated to the corresponding acid chloride which is used in a Friedel-Crafts acylation of benzene. In the first step, the S-alanine is N-acetylated to "protect" the amino group. I was wondering if the amino group will be replaced by chlorine by reaction with the PCl5 hence requiring protection with an acetyl group. This of course will bring about the problem of causing dehydration of the amide to a nitrile when chlorinating the carboxylic acid to an acid chloride. So why acetylate the amino group in S-alanine?

Is that last step supposed to be heat? The fact that it's equilateral makes it look more like cyclopropane. Anyway, I'd expect that the amine lone pair messes with the Lewis acid used in the FC acylation.

@Waylander The acetylated amine is an amide that is dehydrated to an imidoyl chloride which I think you refer to as "chlorimine". The imidoyl chloride upon heating decomposes to give a nitrile and HCl.

@SamuelWango To get a nitrile from an imidoyl chloride you have to heat under forcing conditions. In the absence of such treatment the imidoyl chloride readily hydolyses back to the N-acetyl compound and can be recovered unchanged.

The treatment of N-acetyl-S-alanine with PCl5 gives the acid chloride of the carboxylic acid and converts the N-acetyl group to the imidoyl chloride. This species undergoes classical F-C acylation with benzene and then on aqueous work-up the N-acetyl group is regenerated by hydrolysis.
Contrast this with the reaction of S-alanine with PCl5. This paper here shows that L-alanine forms the acyl chloride hydrochloride which will be unable to give the acylium cation on treatment with AlCl3 as it is already a charged species.


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Why adding sodium bicarbonate quenches a reductive amination using sodium triacetoxyborohydride?
I'm planning to do a reductive amination reaction between methyl pyruvate and benzylamine using sodium triacetoxyborohydride, $\ce{NaBH(OAc)3}.$ Acetic acid is used as a catalyst for protonating the iminium ion intermediate. I know that at the end of the reaction, $\ce{NaHCO3}$ is added to quench the reaction by neutralizing any unreacted acetic acid.
However, $\ce{NaHCO3}$ also reacts with any excess $\ce{NaBH(OAc)3}.$ I'm stuck on the mechanism of how this step works. This step should end up with $\ce{B(OAc)3}$ so a hydride ion is released, but I don't think hydride ions react with bicarbonate ions. References and videos I've watched used $\ce{NaHCO3}$ for quenching anyways, which confuses me. Any help?

What solvent is this done in? Can you give us more detail of the reaction conditions?

@Waylander it is done in dichloromethane, DCM. Reacting methyl pyruvate and benzylamine w/ acetic acid gives the iminium ion, and it's subsequently reduced by adding nabh(oac)3. Reaction completes ~20-25 minutes, in room temperature

Is the NaHCO3 added as solid or in solution?

It's just an equivalent of basic hydrolysis.

According to Grice et al. [1], the addition of the bicarbonate anions to the reaction mixture containing excess sodium triacetoxyborohydride and acetic acid may trigger an attack of the hydride anion on the carbon of the bicarbonate ions forming formate ions with the expulsion of the hydroxide group from the bicarbonate to give sodium hydroxide.
They describe the reaction of sodium borohydride with $\ce{CO2}$ or $\ce{HCO3-}$ and although $\ce{NaBH4}$ is a stronger reducing agent, the reaction with $\ce{NaBH(OAc)3}$ still yields the formate anions. Since this reaction occurs in addition to the neutralization reaction with acetic acid, I think it may also help in quenching the excess reducing agent.
Reference

Grice, K. A.; Groenenboom, M. C.; Manuel, J. D. A.; Sovereign, M. A.; Keith, J. A. Examining the Selectivity of Borohydride for Carbon Dioxide and Bicarbonate Reduction in Protic Conditions. Fuel 2015, 150, 139–145. DOI: 10.1016/j.fuel.2015.02.007.


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Help with mechanism: Isomerization of eugenol
I'm confused about the mechanism for the isomerization of the external alkene eugenol to the internal alkene isoeugenol. I've added an image below of their structures.

The reaction is usually carried out using a strong base (sodium or potassium hydroxide) in a polyol solvent (typically glycerol or ethylene glycol).
I have no idea why this might happen, especially given that there's no carbonyls with acidic alpha carbons in the molecule. Maybe eugenol phenoxide is responsible?

Looks like a sigmatropic rearrangement.

[1,3] sigmatropic shift?

It is thought to be a two-step reaction with cis-iso-eugenol as an intermediate as described here. The direct 1,3 sigmatropic reararrangement to the trans product has a symmetry problem.


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Old sodium hydroxide based drain cleaner become stony and brown, not hydrosoluble anymore
My drain was clogged and I used an old sodium hydroxide based drain cleaner to free it. The cleaner was quite old and inside it was brownish and monolithic, not a bunch of small white spheres anymore.
The problem is that the compound didn't dissolve in water and now I have a mass of this substance in the drain. I was wondering what made the product brown and hard and what can I do to make it soluble again.
The compound contains: sodium hydroxide, inorganic salts and excipients
Thanks!

For starters, dissolving in warmer water, stirring etc. can help, in general.

You may have been better off ditching the old stuff and buying new, but if you still have some of the material you can check if a smaller amount is still soluble, both visually and with litmus/pH paper (it probably is still alkaline). Sodium hydroxide exposed to ambient air is likely to pick up carbon dioxide and maybe water of hydration to form a carbonate, which is still soluble but much less so than the original sodium hydroxide. If this is so, then the way to get rid of it is basically enough water and patience. Not sure where the brown color came from but I suspect it's only a minor constutuent.

Thanks, eventually boiling water did the job, I was just curious of what happened from the chemical point of view.


text:
Is the work done by the system equal to negative of the work done by the surroundings on the system?
If an ideal gas expands into its surroundings through a massless piston, is the work done by the gas on the surrounds equal to the negative of the work done by surroundings on the system?

When the system does work on the surroundings, the following holds:
$$w_\mathrm{system} < 0$$ $$w_\mathrm{surrounding} > 0.$$ No matter what, this always is true:  $$w_\mathrm{system} = -w_\mathrm{surrounding}.$$ A negative value means it does work, and positive value means work is being done on it.

By Newton’s 3rd law, is the force exerted by the system on the surroundings equal in magnitude and opposite in direction to the force exerted by the surroundings on the system?  If not, why not?

As Karsten Theis said in his comment, $w_{sys} = –w_{surr}$, always. But it can be hard to think about work flow.  Thus, to help understand this result more intuitively, consider conservation of energy:
$$\Delta U_{sys} = –\Delta U_{surr}$$.
If the system is closed (no movement of matter between system and surroundings), the only way we can affect the internal energy of the system is through the flow of heat or work.  Hence:
$$\Delta U = q + w$$
And thus:
$$q_{sys} + w_{sys} =  -q_{surr} - w_{surr}$$.
Now, while it may be difficult to think about work, it should be intuitively obvious that any heat that flows into the system must have flown out of the surroundings, and visa versa.  Thus we have:
$$q_{sys} =  -q_{surr}$$
This in turn leaves us with:
$$w_{sys} =  -w_{surr}$$
From this we can see that work flow across a system boundary is just like heat flow across a system boundary: Whatever goes out of the system has to go into the surroundings, and visa versa.
This can be extended to an open system, to show that the above statement is generally true.

It can be practically tricky for a piston with friction, as w_sys = -w_surr and -w_surr -> -q_surr, effectively  w_sys = - q_surr  :-) :-)


text:
Enthalpy Departure
I'm trying to fill in the steps from my textbook on the derivation of enthalpy departure from ideal gas behavior. My textbook gives the variation of mass-specific enthalpy with temperature pressure as
$$\left ( \frac{\partial h}{\partial p} \right )_T=v-T\left ( \frac{\partial v}{\partial T} \right )_p$$
Integrating from pressure $p'$ to $p$ at fixed temperature $T$,
$$h(T,p)-h(T,p')=\int_{p'}^{p}\left [ v-T\left ( \frac{\partial v}{\partial T} \right )_p\right ]dp$$
If we use the superscript $^*$ to denote ideal gas property values, adding and subtracting $h^*(T)$ from the left hand side of the equation gives
$$\left [h(T,p)-h^*(T)  \right ]-\left [h(T,p')-h^*(T)   \right ]=\int_{p'}^{p}\left [ v-T\left ( \frac{\partial v}{\partial T} \right )_p\right ]dp$$
By the assumptions of the ideal gas model, we have
$$\lim_{p'\rightarrow 0}\left [h(T,p')-h^*(T)   \right ]=0$$
and in this limit, the following expression is obtained
$$h(T,p)-h^*(T)=\int_{0}^{p}\left [ v-T\left ( \frac{\partial v}{\partial T} \right )_p\right ]dp$$
This can be thought of as the change in specific enthalpy as the pressure is increased from zero to the given pressure isothermally. This equation can be evaluated with $pvT$ data.
The integral in the last equation can be expressed in terms of the compressibility factor $Z$ and the reduced temperature $T_R$ and reduced pressure $p_R$. Solving $Z=pv/RT$ gives
$$v=\frac{ZRT}{p}$$
On differentiation,
$$\left ( \frac{\partial v}{\partial T} \right )_p=\frac{RZ}{p}+\frac{RT}{p}\left ( \frac{\partial Z}{\partial T} \right )_p$$
$$v-T\left ( \frac{\partial v}{\partial T} \right )_p=\frac{ZRT}{p}-T\left [ \frac{RZ}{p}+\frac{RT}{p}\left ( \frac{\partial Z}{\partial T} \right )_p \right ]=-\frac{RT^2}{p}\left ( \frac{\partial Z}{\partial T} \right )_p$$
This equation can be written in terms of the reduced properties $T_R$ ad $p_R$ as
$$v-T\left ( \frac{\partial v}{\partial T} \right )_p=-\frac{RT_c}{p_c}\cdot \frac{T_R^2}{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}$$
Introducing this last equation into the equation
$$h(T,p)-h^*(T)=\int_{0}^{p}\left [ v-T\left ( \frac{\partial v}{\partial T} \right )_p\right ]dp$$
gives, on rearrangement,
$$\frac{h^*(T)-h(T,p)}{RT_c}=T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}\frac{dp_R}{p_R}=T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}d\ln p_R$$
What I'm failing to understand is how we get from the equation
$$v-T\left ( \frac{\partial v}{\partial T} \right )_p=\frac{ZRT}{p}-T\left [ \frac{RZ}{p}+\frac{RT}{p}\left ( \frac{\partial Z}{\partial T} \right )_p \right ]=-\frac{RT^2}{p}\left ( \frac{\partial Z}{\partial T} \right )_p$$
to the equation
$$v-T\left ( \frac{\partial v}{\partial T} \right )_p=-\frac{RT_c}{p_c}\cdot \frac{T_R^2}{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}$$
and then to the equation
$$\frac{h^*(T)-h(T,p)}{RT_c}=T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}\frac{dp_R}{p_R}=T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}d\ln p_R$$
In a couple books I've looked at, the algebraic and calculus manipulations are just glossed over and ignored. I'm hoping someone can help me fill in the steps.

Why is enthalpy and volume denoted by small letters?  h is usually considered as height, enthalpy is H. Similarly, v is usually velocity or speed, volume is V.

@Poutnik This is mass-specific enthalpy and mass-specific volume.

I see, you are right, it is used like that. But it is strange seeing that, I would expect rather H_s and V_s.

The first step you are missing seems trivial, just substitute $T=T_RT_c$ and same for p and remember that $T_c$ can be treated as a constant.

It seems very straightforward.  Is it the algebra that you are having trouble with?

@ChetMiller Yea, it's the algebra I'm having trouble with.

@Poutnik I see and use that notation sometimes as well.

@BuckThorn I just don't see how $$-\frac{RT_c}{p_c}\cdot \frac{T^2_R}{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}=-\frac{RT^2}{p}\left ( \frac{\partial Z}{\partial T} \right )_p$$

@BuckThorn Are these steps correct? $$v-T\left ( \frac{\partial v}{\partial T} \right )_p=-\frac{RT^2}{p}\left ( \frac{\partial Z}{\partial T} \right )_p=-\frac{RT_R^2T_c^2}{p_Rp_c}\left (  \frac{\partial Z}{\partial( T_RT_c)}\right )_p$$

It looks right. Use the following general identity for k=constant: $\left (  \frac{\partial f}{\partial( kx)}\right )_y = \frac1k \left (  \frac{\partial f}{\partial( x)}\right )_y $

@BuckThorn I can't believe I spent an entire day on this! Thanks a lot. I'm still confused as to what happens to the $p_c$ in the next equation.

First you have $dp=d(p_cp_R)=p_cdp_R$ which leads to cancelation of $p_c$. Also, use $dlnp_R = (1/p_R) dp_R$

Conclusion:
$$v-T\left ( \frac{\partial v}{\partial T} \right )_p=-\frac{RT^2}{p}\left ( \frac{\partial Z}{\partial T} \right )_p=-\frac{RT_R^2T_c^2}{p_Rp_c}\left (  \frac{\partial Z}{\partial( T_RT_c)}\right )_p$$
$$\left(  \frac{\partial Z}{\partial( T_RT_c)}\right )_p=\frac{1}{T_c}\left(  \frac{\partial Z}{\partial( T_R)}\right )_p=\frac{1}{T_c}\left(  \frac{\partial Z}{\partial T_R}\right )_{p_R}$$
$$-\frac{RT_R^2T_c^2}{p_Rp_c}\left (  \frac{\partial Z}{\partial( T_RT_c)}\right )_p=-\frac{RT_c}{p_c}\cdot \frac{T^2_R}{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}$$
$$h(T,p)-h^*(T)=\int_{0}^{p}\left [ v-T\left ( \frac{\partial v}{\partial T} \right )_p\right ]dp=-\int_{0}^{p_R}\frac{RT_c}{p_c}\cdot \frac{T_R^2}{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}d(p_Rp_c)$$
$$\Rightarrow \frac{p_c\left [h^*(T)-h(T,p)  \right ]}{RT_c}=\frac{T_R^2}{p_R}\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}p_cdp_R$$
$$\Rightarrow\frac{h^*(T)-h(T,p)}{RT_c}= T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}\frac{dp_R}{p_R}=T_R^2\int_{0}^{p_R}\left ( \frac{\partial Z}{\partial T_R} \right )_{p_R}d\ln p_R$$


text:
Can vapor pressure be defined for permanent gases?
Vapor pressure of a substance, in a given temperature, is the pressure exerted the vapor in a system where the gaseous and liquid phases of the pure substance are in equilibrium.
But, if for a temperature where a compound is obligatorily a gas and cannot be condensed into a liquid at any pressure (like a permanent gas at room temperature), is vapor pressure definable?
And if it isn't definable, does this imply that any mixture between the gas and a liquid (the gas being dissolved in the liquid) is obligatorily non-ideal?
I'm saying that because, for a system where the liquid phase is an ideal mixture and the vapor phase is an ideal gas, Raoult's Law is valid for every component for any proportion between the components, and Henry's Law volatility constant (the variant using molar fraction) for any component becomes simply the vapor pressure of that component at a given temperature.
But, if its vapor pressure isn't definable at that temperature, then it can't be equal.
Perhaps the very fact that one compound can be condensed at that temperature while the other cannot already means that their interactions in the liquid phase cannot be possibly similar?

Vapor pressure .... are in equilibrium. That is saturated vapor pressure. Vapor pressure does not have equilibrium requirement. There is water vapor pressure in atmosphere without being saturated. // We say a gaseous, not vapor phase, as a vapor is a gas.

In this case you would not say that Henry's law constant is equal to the non-existent partial pressure of the permanent gas. Take oxygen solubility in water, for instance (which is a permanent gas when T is above critical). But if the gas observes Henry's law then the solvent observes Raoult's law, and the mixture is considered ideal.

Vapor refers to the gas phase in exchange with a condensed substance, btw, so as Poutnik suggests, it is not defined for a substance which cannot be condensed.

@all Please stop perpetuating this terrible misnomer. Vapour pressure is a property of the liquid. There IS NO vapour pressure of anything in the gas phase, because what you mean is partial pressure!! The vapour pressure of the liquid is identical to the equillibrium partial pressure of it's gaseous form in the free space above the liquid.

@Karl You may be right, I may be a victim of a different naming due meteorological history, where vapor pressure and saturated vapor pressure is used in context of absolute and relative humidity as shortcut for  vapor partial pressure and saturated vapor partial pressure.  :-) So vapor pressure is partial pressure of vapor.

Even if a gas cannot be liquefied above a certain temperature, that doesn't mean it can't be directly solidified with enough pressure. For example, even hundreds of degrees above the critical temperature of water, you can directly crush the supercritical fluid into exotic ices. In principle all you need to establish a "vapor pressure" is a condensed phase, be it liquid, solid or perhaps even something more exotic.

@NicolauSakerNeto I'll bet that exotic phase will not be miscible with your original solvent. ;)

I think you are quite right, the behaviour cannot be ideal:
If you look at the solubility of a permanent gas in a liquid, you are limited to the liquid's side of Raults mixture phase diagram. It doesn't matter there's no vapour pressure on the side of the gas, because you can never get there, until you actually condense your previously permanent gas, perhaps into a solid. (Spoiler: there is probably a miscibility gap. Nonideal.)
If your gas was actually not condensable at any pressure, the curve must verge towards infinity at 100% gas, and thus obviously cannot be linear. Ergo, the mixture cannot be ideal with respect to Raults law.
qed.


text:
How to apply the Geiger–Nuttall rule to find half-life?
I'm having trouble understanding the Geiger–Nuttall rule for α-particle decay. From the textbook I'm using, the rule is defined as
$$
\log(t_{1/2}) = A + \frac{B}{\sqrt{Q_\alpha}},
$$
where the constants $A$ and $B$ have a $Z$ dependence. Here's a Geiger–Nuttall plot of the logarithm of the half-life (s) versus the square root of the $Q_α$ value (MeV):

How do we use the above equation to find the half-life? More specifically, can we use that to find the half-life of $\ce{^{226}Ra}$ nuclei?

I doubt it is used to obtain half life, unless it is extremely short or long. If you look at the logarithmic scale, is is very sensitive to values of A,B and Q.

Find alpha particle energy from radium-226 decay and apply on the the formula, getting values A and B from the line for Z=88 ( better if they are tabulated somewhere).


text:
Why do we use the ratio of Critical Pressure and Critical Temperature to calculate size of a gas molecule?
Our teacher explained that using the ratio of Critical Pressure and Critical Temperature is preferred over Critical Volume to calculate the size of a gas molecule when all these quantities are taken in terms of van der Waals constants. He told us that it was due to increased accuracy.
How exactly does using the ratio increase the accuracy?
Reference:
Critical Temperature :
$\frac{8a}{27Rb}$
Critical Pressure : $\frac{a}{27b^2}$
Critical Volume : $3b$
where a and b are van der Waals constants.

Critical temperature and pressure are easily measurable with good accuracy. OTOH $\frac{\mathrm{d}V}{\mathrm{d}p}$ or $\frac{\mathrm{d}V}{\mathrm{d}T}$ reach extremely high values near the critical point.
By other words, the ratio of the critical temperature and pressure has much smaller experimental relative error than the critical volume. Therefore obtaining $b$ from the former is more accurate.
Note that b is 4 times the true volume of 1 mol of gaseous (supposed spherical) particles in the van der Waals model. Therefore, according to the model, $V_\mathrm{molecule} = \frac{b}{4N_\mathrm{A}}=\frac{V_\mathrm{mol, c}}{12N_\mathrm{A}}$
See e.g. chem.libretexts.org: Real_Gases_and_Critical_Phenomena


text:
Why not just use the x,y,z as basis functions instead of linear/quadratic functions of these unit vectors?
I am going through Applications of Group Theory to the Physics of Solids by M. S. Dresselhaus.
In chapter 4 he states

This is with respect to the P(3) group.
I understand basis functions must transform like the given representation e.g for the identity representation $\Gamma_1$ all operators leave any unit function unchanged hence you could use any of $x,y,z,z^2$ separately as basis functions because they are left unchanged in this representation.
For a 2D representation $\Gamma_2$ often functions like $(xz,yz)$ and $(xy,x^2-y^2)$ are also stated as basis functions. Why not simply use $(x,y)$ for the 2D case?
Do we list all other basic functions in character tables for completeness? Or is it because in certain cases basis functions like $(xy,x^2-y^2)$ may be more useful than $(x,y)$
[I am thinking that the former basis function has the symmetry of a d orbital according to Harris and Bertolucci hence it may be more useful to consider in the context of looking at atoms/molecules]

In a single word, the answer is d-orbitals, as you guessed!
It's very frequently useful to determine how specific orbitals on an atom transform, because this in turn determines which orbitals it can overlap with to form bonds / molecular orbitals.
Each d subshell has five d-orbitals, and often these are written as $\mathrm{d}_{xy}$, $\mathrm{d}_{yz}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{z^2}$, and $\mathrm{d}_{x^2 - y^2}$. These transform in the same way as the respective Cartesian products in the subscripts, so for example, the $\mathrm{d}_{xy}$ orbital transforms in the same way as $xy$. (More technically, the mathematical form of the $\mathrm{d}_{xy}$ orbital is $xy$ multiplied by some spherically symmetric term, which doesn't affect the symmetry.)
In fact there are also some other uses for the quadratic terms, including Raman spectroscopy (vibrational modes that transform the same way as quadratic terms are Raman active), but I imagine that d-orbitals are the most common use, and one that most chemistry students will encounter when studying transition metal chemistry.
(The $x, y, z$ linear functions are useful for lots of things (determining symmetries of translational modes, etc.), but they're also there because of the p-orbitals, which transform in the same way as $(x, y, z)$.)


text:
How soluble is water ice in liquid H2S?
This seems to be ridiculously hard to research, as Google only wants to give me results about dissolving H2S in water (rather than the other way around)--and when I tried to get around that by looking for information on H2O/H2S phase diagrams, or eutectic mixtures, all I could get was single-substance phase diagrams for one or the other, and information about deep eutectic solvents for H2S scrubbing.
So... does anybody know how soluble water ice is in liquid H2S at -70 to -80 Celsius?

I am curious why you may need that. I assume very few chemists would be willing to measure it.

@Poutnik Because I am an SF author and it turns out that whether or not ice will dissolve is plot-relevant. (It's the sort of thing Hal Clement might've written, except much better than me.) I would've thought someone would measure on the basis that it'd be relevant to producing pure anhydrous chemicals for industrial purposes.

Now that's a responsible approach to world building.

The fact that gaseous H2S is just slightly soluble in water gives a hint about ice solubility in liquid H2S.

I think it is virtually insoluble.

I'm sure there is someone somewhere who uses cold H2S, but for typical industrial applications it is liquified by simple compression rather than chilling. If anhydrous is needed, H2O is removed from the gas phase before compression, so there is no need to know anything about the solubility of H2O in cold H2S and therefore not much motivation to measure it.

I would guess the hydrogen sulfide would rather dissolve into the ice.

@OscarLanzi Wouldn't clathrate rather form in a "cocrystallisation"? Hmm, maybe ice would turn into clathrate if pressure was high enough?

@Logan R. Kearsley What pressure range do you have in mind?

What is an industrial application for liquid H2S ? The oil industry produces mega tons of H2S which is converted to elemental sulfur and sulfuric acid, generally; never liquid.

@blacksmith37 https://www.indsci.com/en/industrial-scientific-gas-types-hydrogen-sulfide-h2s
I suppose none of the applications listed there would require chilled liquid, though; too bad for me!

@Mithoron Between one and two atmospheres; hence the low temperature range to produce liquification.

Well, under the circumstance - temp. and pressure. I still say if you simply drop some ice into liquid H2S, then nothing happens, at least not more then if you ditched it into cold diethyl ether. I think it's a bit of an X & Y problem though. Your idea is using H2S an alternative for water for life in very cold environment, perhaps?

@Mithoron Yup! Of the algae-and-bacteria variety, anyway. I got through 4 awesome experimental papers on its solvent properties and reactions with various organic molecules... and then realized, crap, none of these actually mention water!

I think liquid NH3 would be a safer bet. It's actually sometimes used as solvents in reaction. Did you see https://en.wikipedia.org/wiki/Hypothetical_types_of_biochemistry#Non-water_solvents

@Mithoron It certainly is... but everybody else thinks so, too, and I've used that conceit myself, so I'm looking for something more exotic. I have indeed seen that wikipedia page, and read most of the footnote references (and some of their references)!

Still, the more exotic, the harder - for example the clathrates issue, afaik not a problem with NH3. Heck, iirc there was a novel with FOOF as a solvent! Better go for accuracy then originality - what actual consequences of ideas would be.

Let us continue this discussion in chat.


text:
Why does Cesium have a higher Boiling point compared to Rb?
i know BP should decreese down the group as Cohesive energy decreeses down the group but i found out that cs>Rb in terms of B.P..
so what other factor is affecting this?

What's the source of your data?


text:
Confusion in finding Effective Atomic Number of a complex
In finding EAN (Effective Atomic Number) of metal in complexes, I came over one compound:
$\ce{[(CO)2RhH2Rh(CO)3 Cl]^+}$
Here the hydrogen are shown as bridging atoms and we are supposed to calculate EAN of individual rhodium atoms
The main problem is that hydrogen has a single electron, then how can it form a bridging bond?
I thought, maybe hydrogen is giving electron to one of the rhodium atoms and other rhodium is giving lone pair to hydrogen, but this doesn't seem like a good argument.

In this case, think of the hydrogen as a hydride anion and add four electrons, two from each hydride ligand, to each rhodium

Ok @Kanghun Kim . will we take O.N of each Rh same like = 4/2 =+2? But, the molecule is not symmetric

Effective_atomic_number has two different meanings: one that is the effective nuclear charge of an atom, and one that calculates the average atomic number (number of protons found in the nucleus) for a compound or mixture of materials.  Do you by chance mean rather oxidation state/number ?

H atoms are bridging atoms in the molecule $\ce{B2H6}$. Single electrons may form a bridging bond.

B2H6 is known to be electron-precise and to obey the octet rule- this can only happen if each of the bridging hydrides is counted ss a 2e- donor to each boron atom.

@Poutnik I mean number .

@Hannah Good, but what number? Or the question is just a side question without direct relation to the EAN of the complex ?

@Poutnik valence electrons actually, the heading says 18 e- rule -EAN and we are supposed to calculate EAN of Rh in the complex

I see, so the 3rd meaning. :-)


text:
What is the saturation point of dishwashing soap in water?
I am looking to measure the effect of dishwashing soap on the surface tension of water. To select appropriate values for the concentration of dishwashing soap is to find the saturation point first, but I can't find any values for it and don't know how to measure it. How can I select appropriate values for the concentration and if finding the saturation point is a way of doing that, how do I find the saturation point of dishwashing soap in water?

Dishwashing soap is a generic term. Different brands contain different mixtures of multiple surfactants and other additives in various concentrations and ratios. You would have to restrict the question to a single compound or a products, as the value is generally different for each. // You may want to search for surface tension measurement, e.g. Tensiometer

Critical Micelle Concnetration (CMC) is a commonly used parameter to measure when a solution is saturated with surfactant (soap).
CMC is the concentration where the surfactant starts to form small droplets (micelles) instead of spreading out onto the water/air surface.
One way to determine CMC is by measuring the surface tension. When the surfactant concentration is (very) low, the surface tension more or less linearly decreases with increasing concentration, but at some point the surface tension reaches a constant level, as Poutnik mentions. This concentration is called the CMC.
Kruss tensiometers: CMC and surfactant concentration

I want to point out that saturation usually refers to a situation where chemical precipitates out of solution, but it seems you are interested in soap's effect. That's very different thing.
Soap's effect is very dependent on a situation - amount of evaporation and even type of soap. But in general even few parts per million can greatly change parameters. And then even a few percents won't change much further. While actual saturation is even further, at equal parts or so.
Something like this:
1 part per million - bubbles are normal size
100 parts per million - bigger bubbles
10 000 parts per million - bubbles are large
100 000 parts per million - bubbles are as large, but you can keep dissolving soap as much as you want
Why is it so? Unlike most materials like salt and conductivity, soap's effect doesn't rely on how much soap is in the bulk of water. Soap acts only on the water's surface. As you make a bubble, soap accumulates on bubbles' surface. And because molecules are small, you have a lot of volume against small amount of surface area. So, even few parts per million in water's volume can mean almost complete fill of soap bubble's surface. And if soap bubble's surface is filled with soap molecules, nothing will happen if you keep adding soap in water's volume.
What it can affect is how much evaporation is needed for soap's film to form fully, or how thin can you extend a soap film.
If you want to make bubbles that are more than 10cm, you will need to use more tricks than just soap. Water itself evaporates too quickly at that point. You can fix it by adding glycerine, it evaporates slower. Adding just more soap won't help make bubbles bigger after this point.
Anyway, soap is very different than all other chemicals, because it acts on a surface, rather than in volume. And it is better to ask more specific question, than trying to use a term that doesn't help much with it.
If you meant what you asked, you can mix about equal amounts of liquid soap and water. But for most uses it is too much.

I suppose the OP may mean saturation in mathematical meaning, where value of measured quantity reaches a plateau on the curve of its dependence on other quantity. Like surface tension not decreasing with another soap addition any more.


text:
Most appropriate representation of 3,4,4,5-tetramethylheptane
Is there a specific correct way to draw the structure of 3,4,4,5-tetramethylheptane?

You might want to have a look at Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008).

but what is the short answer?

The second way is somewhat better. Also, all your angles in the main chain are way too acute.

Neither of the two is good. If don't know the actual angles, just go with more symbolic and less misleading representation.

@Mithoron its a rough sketch the main importance of the diagram is for the 4,4 methyl placement and nothing to do with the actual angles.

@reset You are asking about the structure, which is a set of both angles and distances. As Mithoron rightfully suggested, both do matter a lot. There are recommended/standardized values, values that tolerably represent a projection of a complex 3D object onto a 2D layer with justified adjustments, and there are incorrect ones. Given it's a trivial case and there is no objective reason to use visually incoherent bonds and odd angles, your structures are both falling into the last category.

Generally, it cannot be said that a specific representation is “right” and that all others are “wrong”. Nevertheless, the relevant guidelines given in the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008) read as follows:

GR-4.1.3 Atoms with four bonds drawn
Atoms with four bonds may be depicted in several different ways, with a preference among these ways determined by context.
When two of the adjacent atoms are each connected to no additional atoms and two of the adjacent atoms are each connected to at least one additional atom, the four bonds should be separated by one 60° angle, an opposing 120° angle, and two 90° angles. Because of the 120° angle, this style will not disrupt the orientation of a chain in which it appears. (…)

Therefore, the structure diagram of 3,4,4,5-tetramethylheptane should look like this:

Note the 120° angles of the chain, which are missing in the examples given in the question.


text:
In Lanthanides, Why is electron removed from 4f before 5p?
I've been taught that electrons are removed first from valence shell electrons with highest energy. But In Lanthanides, for example taking configuration- [Xe]4f⁴6s², If we go to +3 Oxidation State, then the configuration is [Xe]4f³.
But valence shell is 5th, so why wasn't electron removed from 5p instead of 4f?

5p electrons have significantly lower energy than 6s and 4f. 4f orbitals are being filled after 5p and 6s. Remember that with advent of d and f orbitals, the single valence shell number does not apply much.


text:
Reaction of silicon dioxide with sodium hydroxide
I recently learnt about this equation but I would like to know how NaOH reacts with the 3d structure of SiO2 and gives Na2SiO3 and H2O.

Can anyone explain with simple diagram how NaOH forms new bond in SiO2 to give the above mentioned products?Like how they replace, displace etc in the 3d structure.

Just to clarify, if you are thinking that SiO2 has a square planar structure based on that picture, it is in fact nothing like that at all.

To use simple terms, $\ce{NaOH}$ breaks the "vertical" bonds $\ce{Si−O}$ without touching the "horizontal" bonds of your picture. As a consequence, sodium silicate is not $\ce{Na2SiO3}$. It has the structure of a long polymer $\ce{Na_{2n}(SiO3)_n}$. The polymeric ion $\ce{(SiO3^{2−})_n}$ is
made of long chains of atoms $\ce{-O-Si-O-Si-O-Si-O -}$etc., where each $\ce{Si}$ atom is still attached to two negatively charged oxygen atoms $\ce{O^−}$, one "above" and one "under" the main $\ce{-O-Si-O-Si-O -}$chain. And a positive ion $\ce{Na+}$ is placed near each such oxygen atom.

ok so what happens to the OH- which comes out as water?

Also the reaction says that 1 molecule of SiO2 reacts with 2 molecules of NaOH to give 1 molecule of Na2SiO3 and 1 molecule of H20.What happens to the extra oxygen as when 2 OH-ions react one molecule of water is obtained?

Suppose all vertical Si-O-Si bonds are broken so that the only broken bond is Si-O just under Si (and never above Si). Now H2O gives one OH to the Si atom, and one H to the O atom. This produces long chains Si-O-Si-O-Si-O- where each Si atom is attached to 2 groups OH. This makes two groups $\ce{OH}$ attached to each silicium atom of the chain. Each such -OH group can react with NaOH producing  $\ce{-O^-}$ attached to the silicium atom, one $\ce{Na+}$ ion and one water molecule


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How odorless is "odorless"?
Wikipedia has entries for many odorless gasses such as Ar, CO, etc. How rigorous is odor testing? Is it possible someone with a better sense of smell could smell Ar, CO, etc?

The Reference Guide To Odor Thresholds For Hazardous Air Pollutants Listed In The Clean Air Act Amendments Of 1990 is published by the EPA (US) and is 89 pages long. You might find some of your answers there. And yes, some people have a better sense of smell than others.

Also see: https://en.wikipedia.org/wiki/Odor_detection_threshold

Smell is subjective attribute with consensual conclusion.

It's possible that many of the gases Wikipedia lists as odorless really are completely odorless, period—i.e., that they would not be detected by any human even at 100% concentration.  So I think the question is:  Do all "odorless" gases fall into that category, or are there gases that can be detected, but only by a small percentage of individuals, and only at 100% concentration, that are also designated as "odorless"?  [Also, there's a typo in your question—that should be Ar (argon), not Ag (silver).]

Odourless probably does mean odourless for good physiological reasons
The reason why common gases, such as those mentioned, are odourless is probably a consequence of they physiology of smell and its evolutionary history.
The reason why we or any other animal can smell is due to some fairly complex interactions between particular molecules in nerve cells in the nose. But those cells evolved for particular reasons. There is a big advantage, for example, to being able to detect putrefying food as it is a bad thing to eat food that isn't fresh. Putrefaction often produces a variety of gases containing nitrogen and sulfur (hydrogen sulfide and ammonia are very simple examples to which our noses are exquisitely sensitive). The same may be true for some pleasant smells for the opposite reason.
But there is absolutely no advantage in being able to smell oxygen, nitrogen, argon or carbon dioxide as they are ubiquitous in the air. Even carbon monoxide is common in the air once fire is used (though at low levels and, even though there would be an advantage in smelling it, we have not had fire and poorly ventilated rooms long enough to evolve the ability to smell it).
Even if the nerves in our noses could detect those common gases, this would be futile as the brain would edit out the constant signal–as it does with almost any repeated stimulus–and we would not notice it.
In short, smell is an adaptation to detect uncommon things that affect our survival. We can't smell ubiquitous things as there is no point at all of detecting them. For those things, odourless really does mean odourless.

There is an advantage to smelling carbon dioxide, if you're a bedbug. They identify where their prey is sleeping, or has been doing so, by olfactories that detect this slightly acidic gas.

@OscarLanzi which illustrates the point. Bedbugs don't create much CO2 but benefit from finding creatures like us who do. we exhale quite a lot so detecting it by smell would be problematic.