text:
How exactly does Boyle's law, Charles' law, and Avogadro's law combine to make the Ideal Gas Law?
My General Chemistry 1 lecturer derived the ideal gas law as follows:
We have $V \propto \frac1{P}$ (Boyle's law), $V \propto T$ (Charles' law), and $V \propto n$ (Avogadro's law). Combining these gives
$$V \propto \frac{nT}{P}
\\ \implies V = \frac{RnT}{P}
$$
Where $R$ is a proportionality constant.
But I'm confused as to how these proportionality expressions were combined. As far as I know, $A \propto B$ and $A \propto C$ does not imply $A \propto BC$, so how is this justified here?

Does this answer your question? How can we combine all the three gas laws into a single ideal gas equation?

@Loong Thank you, but no. I came across that question before posting this and I was still confused.

This is not how the gas laws are derived originally. However, old physics or chemistry school books taught it this way (somewhere around 1890s or early 1900s). Your teacher over simplified the derivation this way and all the key algebraic steps are missing. Old does not mean that it is false or the facts have changed, it is still very elegant if taught properly.

$A \propto B$ and $A \propto C$ does not imply $A \propto BC$, so how
is this justified here?

Yes, you are right that if $A$ is proportinal to $B$ and $A$ is also proportional to $B$, then we cannot write right away that $A \propto BC$.
However, ratio and proportion have another property, that if we say that the quantity $A$ varies jointly if it varies as a product of two variables. So you can say that the volume of a gas varies jointly with amount ($n$) and absolute temperature ($T$)., i.e., if we were to plot $V$ on the y-axis and the product nT on the x-axis, we will see a straight line at constant pressure. Note the temperature must be in Kelvins or appropriately shifted.
Also, the volume of the gas is inversely proportional to pressure P i.e., if we keep nT constant, and plot $V$ on the y-axis and P on the x-axis, we get a hyperbola (which proves inverse relationship).
Note that in ratio and proportion convention, the dependent variable is written on the left. In this case, $V$ is the dependent variable.
See the gas laws in Physics for University Students: Heat, electricity, and magnetism By Henry Smith Carhart, 1894 from Google Books (legally free to download) for a more logical derivation of the gas law using ratio and proportion properties.

Thank you for the reference! Do you know where this derivation can be found in the book (i.e. which chapter, pages)?

@Mailbox, I intentionally did not mention the page numbers so that you can download and explore the book. Search gas laws inside the book.

I see, thanks! I just wasn't exactly sure what to search for. Do you mind if I comment the page range for anyone else who might come across this post (and so you can verify that I'm looking at the right information, or suggest a different page range for further reading)?

It is pages 37-39. This is where the author uses ratios and proportions.

@Mailbox, Did you see those book pages?

Yes, those were the same pages I read.

It is much easier to do the contrary, namely to start from  $pV = nRT$, and to derive the other laws. Example :

Boyle's law. If a gas sample contains $n$ moles of gas at a constant temperature $T_0$, the pressure $p$ and the volume $V$ can vary, but the product $pV$ remains constant. So going from state $1$ to state $2$, the pressures and volumes are related by $${p_{1}*V_{1} = p_{2}*V_{2} = nRT_0 = \mathrm{constant}}$$
Charles' law. If a gas sample is maintained at constant pressure $p_o$, the volume $V$ is proportional to the temperature $T$. Using $pV = nRT$, the following expression can be written : $V/T = nR/p_o$ = constant. So that heating a gas volume $V_1$ at temperature $T_1$ to a higher temperature $T_2$, without changing $n, R$ and $p_o$, the volume becomes $V_2$, but the ratio $V/T$ remains constant and equal to $nR/p_o$ according to :$$\frac{V_{2}}{T_{2}} = \frac{V_{1}}{T_{1}} = \frac{nR}{p_o} = \mathrm{constant}$$
Avogadro's law. At fixed values of $p, V$ and $T$, the number of moles is the same for all gases. This is an immediate consequence of the law $pV = nRT$, as this law is valid for all perfect gases, and real gases are nearly perfect gases.

But is it possible to derive $pV = nRT$ without having known those three laws beforehand?

@Mailbox, Historically, Boyle's, Charles, laws were known before the pV=nRT law.

@AChem I thought so, but I didn't want to just discard this answer just because of that. If there is a way to derive $pV = nRT$ without the aforementioned laws, not only would I be interested to see that, but it would solve the potential circular reasoning issue here.

@Mailbox, Yes indeed. There is a rigorous way of deriving the ideal gas law using the kinetic theory of gases. Search this terms and you will find the derivations.

Like this?

@Mailbox, No. Kinetic theory of gases arrives independently at pV=nRT. See Wikipedia, https://en.wikipedia.org/wiki/Kinetic_theory_of_gases, See the section titled "Pressure and kinetic energy" and the following equations.

@AChem. I would be convinced if you are sure that the development of the kinetic theory of gas never uses the gas law. Are you sure of this ?

@Maurice, Why not? Kinetic theory of gases is used to derive the ideal gas law! https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107A%3A_Physical_Chemistry_for_Life_Scientists/2%3A_Properties_of_Gases/2.6%3A_Kinetic_Theory_of_Gases


text:
How to separate nylon 66 salt (hexamethylenediammonium adipate) into adipic acid and hexamethylenediamine?
After hydrolyzing nylon 66 in water (neutral hydrolysis) in a batch autoclave reactor, the resulting solution contains the organic salt hexamethylenediammonium adipate.
Is there any way to precipitate either the acid or the amine from this solution?

Ion exchange chromatography would be straightforward. Or change the pH and then separate the free base or the free acid using crystallization. Or you could use solvent extraction.

Acidify your solution and cool it, adipic acid should crystallize out. It has a solubility of 17g/L in water so you will not get all of it first time.

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Effect of pH on solubility (Lehninger's Principles)
The following problem is found in Lehninger's 8 Ed.:

I understand the reasoning in the problem statement: in a basic solution, more of the acid dissociates, and this dissolves, since the ionized form is soluble.
Applying the same reasoning to item (a), which concerns a similar (same pKa) weak acid, I conclude that it would be more soluble in aq. NaOH, however the answer key says the opposite.
The answers to the other items seem likewise "inverted".

I cannot understand how this isn't contradicting.

You have to consider whether the protonated species or the deprotonated species is the neutral form.

Please use text instead of text screenshots and cite the textbook correctly. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Theere is an error in this question. The error is it has given protonated versions of  quinoline and (2-methylthio)pyridine. That two, since they are already protonated, should dissolve in aqueous solvents. Also, as OP expected, they should also react with $\pu{0.1 M}$ $\ce{NaOH}$ since they contain acidic proton.
The question should have been asked using nutral form of structures as:

Categorize the given compounds, quinoline, m-cresol, and (2-methylthio)pyridine, whether they are more soluble in an aqueous solution of $\pu{0.1 M}$ $\ce{NaOH}$ or $\pu{0.1 M}$ $\ce{HCl}$.

Then, anybody can answer it using principles of $\mathrm{pH}$ on solubility.

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text:
Finding unknown concentration using HPLC standard curve
I calculated a standard curve equation from a given concentration and peak area, which gave me $y=1994.6x+218.4$. I made the graph with the $x$-axis μg/mL and $y$-axis peak area.
The question is, a sample of $\pu{20.00 mL}$ containing a food with the analyte had $5123$ peak area. Can I just plug in the $5123$ into the $y$ and be done, or do I divide the that answer by $\pu{20.00 mL}$?

Which volume have you injected to get the chromatograms ? Is it always the same ?

Part of your confusion is arising from the fact that you are thinking that concentration is volume dependent. It is not because it is an intensive quantity. For example, the concentration of $\ce{NaCl}$ in seawater will be the same, if you were to take a drop of seawater. The difference is the actual amount of the salt. Now that you have the equation, $y=1994.6x+218.4$, plug in the peak area of the unknown and calculate $x$. This is the concentration of your analyte in the (diluted) sample.
No further calculation can be made from the given information. Now if the mass of the food sample were given, you would now use the extra information of $\pu{20 mL}$ for further calculation.

HPLC sample peak areas are usually compared with standard and end standard peak areas to give a percent of unknown analyte concentration to Standard concentration.
You then work back through your dilutions to get, in this case, ug/ml of analyte in the original sample preparation flask.
Next you multiply by ml of that flask to get ug of analyte.  Then you divide this by sample weight and multiply by 100%.
But before doing any of this, make sure the Chromatography is producing nice sharp peaks and the software is properly integrating the standard and sample peak bases.  This will help avoid considerable error in analysis.
Instead of relying on a computer generated curve, each Chromatogram can be checked.  If Chromatography is good and standard injections are within an acceptable % RSD, then raw peak data can be entered into a spreadsheet.
These spreadsheets can be printed and presented in the analysis report, along with the stack of chromatograms.


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Group 2 hydroxide & sulphate solubility
The trend of solubility down group 2 increases for hydroxides, while it falls for sulphates. Why is that?

https://chemistry.stackexchange.com/a/97168/17368

Nobody knows how to predict solubilities, really.

Does this answer your question? Which has a greater dominance: hydration enthalpy or lattice energy?

If we consider the trend in solubility for a group II cation with a sulfate vs a hydroxide then we are looking at two very different things.
The trend for the solubility with a sulfate or carbonate is the trend for a larger dianion. If we consider the trend Be,Mg,Ca,Sr,Ba,Ra then the size of the dication is increasing.
If we have two large ion types (one cation and one anion) then these will pack better into the crystal lattice than a large and a small. As a result the energy given out when the ions are transfered from the gas phase into the solid is larger.
On the otherhand when we are considering the solubility of hydroxides and fluorides then we are dealing with small anions. Here the small cations such as magnesium two plus will pack better into the lattice with fluoride than the larger cations such as barium and radium.
The most important thing to consider is the difference in the solvation energy which is the energy given out when the gas phase ions are dispersed and dissolved into water and the lattice energy which would be given out when the gas phase ions are combined into a crystaline lattice.


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Find the w/w percentage nitrogen in an ammonium nitrate solution
MCQ Question:
What is the w/w percentage of nitrogen (N) in an ammonium nitrate solution with a density of 1.05 g/cm3 and a molar concentration of 2.1 mol dm-3?
(a) 2.8 %
(b) 11.2 %
(c) 16.0 %
(d) 5.6 %
(e) 4.2 %
My attempt:
1 dm^3 of ammonium nitrate contains 2.1 mol (from 2.1 mol dm^-3)
1 dm^3 also contains 1050 g of ammonium nitrate (from 1.05 g/cm^3)
Mass of N in NH4NO3 = 28/80 = 7/20
Using Mr = mass/moles = 1050/2.1 = 500 g/mol of NH4NO3
Therefore Mr of N = 500 x (7/20) = 175 g/mol
I'm not sure what to do after this. The answer from the mark scheme is d) 5.6%.
Is there another method I should follow?

I don't understand your calculation. For example, on the 2nd line, you state that $\pu{1 dm^3}$ contains $1050$ g ammonium nitrate. No ! It contains $1050$ g of a solution. The ratio $7/20$ is exact, and I call it later $0.35$. But I don't understand why you mean with "Mr", and the next line.
I have another approach.
$\pu{2.1 mol}$ $\ce{NH4NO3}$ weighs : $\pu{2.1 mol· 80 g/mol = 168 g}$.
The mass of $\ce{N}$ in $168$ g $\ce{NH4NO3}$ is $0.35·168 g = 58.8$ g
The weight proportion on $\ce{N}$ in the whole solution (weighing $1050$ g) is $58.8/1050 = 0.056 = 5.6$%

I took the density as the density of ammonium nitrate. Mr is molar mass.

I see where I went wrong. Thanks!


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Procedure for Determining Lead Levels in Water
I enjoyed chemistry in college and have always wanted an excuse to try it at home. A relative is currently very worried about lead from various sources. I think this is a great excuse to get back into chemistry. I was looking for a procedure that could determine the quantity of lead in a water sample. I was not able to find much on the topic but I did find this. https://chemistry-dictionary.yallascience.com/2017/05/determination-of-lead-by-edta-titration.html
My only concern with this procedure is I'm not sure what concentrations are required for a positive result, and I do not know if the test is invalidated by the presence of other metals in the solution.
Obviously "ship it off to a lab" is not the answer here, as testing for lead is only an excuse to buy some glassware. Procedures that require expensive equipment, or require a high quality fume extraction system are obviously not viable, though I can't imagine testing for lead is that complex considering you can buy test strips from the hardware store.
Edit: Mauricehas confirmed my suspicion that the included procedure will not work for the above reasons. I am still interested if anyone knows of a procedure that works in the ppm range.

The mentioned reference is valid for titrating lead solutions which don't contain other ions. If the $\ce{Pb^{2+}}$ solution contains also some other ions, like $\ce{Ca^{2+}}$ or $\ce{Mg^{2+}}$, the procedure will titrate the total amount of these ions. Furthermore, it is valid for concentrations of the order of $50$ to $1000$ mg/L $\ce{Pb}$ if the reference volume to be measured is $20$ mL. If the solutions you want to titrate contains less $\ce{Pb^{2+}}$, this amount is too small to be determined by the given method.

Thank you, I suspected as much. I will edit the question to ask directly for a new procedure.


text:
Ratio between the corresponding force constants of Cl2 and F2
MCQ Question:
Vibrational frequency (f) of a diatomic molecule is given by

where  is the force constant and  is the reduced mass. For a diatomic molecule (AX), the reduce mass is given by

where  and  are mass of atom A and atom B respectively.
If vibrational frequencies (in wavenumber terms) of Cl2 and F2 in are 915 cm^-1 and 525 cm^-1
respectively, what is the ratio between the corresponding force constants of Cl2 and F2 (Cl2 : F2)?
(a) 5.7
(b) 6.0
(c) 6.3
(d) 6.7
(e) 7.1
The answer from MCQ is A) 5.7. I tried writing two equations for Cl2 and F2 and then dividing them but I got an answer of 1.74. How can I get 5.7?

For diatomic molecules, $\mu = m/2$. This is $9.5$ for fluor and $17.75$ for chlorine.
Now $\pu{k}$ is given by :  $\pu{k = \mu* 4\pi^{2} f^2}$. If I use the index $1$ for chlorine and $2$ for fluorine, I can write : $$\pu{\frac{k_1}{k_2} = \frac{\mu_1 f_1^2}{\mu_2 f_2^2} = \frac{17.75}{9.5} * (\frac{915}{525})^2 = 5.67}$$

How did you get the reduced mass as m/2.

Use the given formula for $\mu$. Replace $\pu{m_x}$ by $\pu{m_a}$. You will obtain  :$\pu{\mu = \frac{m_a*m_a}{2m_a} = m_a/2}$


text:
Calculating percentage of Zn in a coin
Question:
An old coin found in an ancient temple is composed of zinc coated with copper. In an experiment to find the percent zinc in the coin, a student determined the weight of the coin to be 3.0 g. Then the student made several scratches in the copper coating (to expose the underlying zinc) and put the scratched coin in hydrochloric acid, where the following reaction occurred between the zinc and HCl (copper remained undissolved):
Zn(s) + 2 HCl (aq) → H2(g) + ZnCl2 (aq)
The student collected the hydrogen produced over water at 27 °C. The collected gas occupied a volume of 1 L at a total pressure of 1.02 bar. The percent zinc in the coin, the student found would be. (Assume that all the Zn in the coin dissolves.)
( a) 67.7 %
(b) 98.2 %
(c) 96.7 %
(d) 88.3 %
(e) 25.0 %
My attempt:
P = 1.02 x 10^5 Pa
V = 1 x 10^-3 m^3
T = 27 + 273 = 300K
Use PV=nRT,
(1.02 x 10^5)(1 x 10^-3)=n x .31 x 300
n = 0.04091 moles of H2
moles of Zn = 0.04091
Mass of Zn = 0.04091 x 65.39 = 2.67 g
Percentage of Zn = (2.67/3.0) x 100 = 89.2%
The mark scheme answer is D) 88.3% which is the closest answer to the value I got, but why don't I get that answer? Is it an error in my calculation, the mcq answers or something else?

I do not know how advanced the course is, are you supposed to correct for vapor pressure of water?

This is from a College Preliminary test. Apparently requires Advanced Level Chemistry knowledge only.

With considered water vapor, the true hydrogen volume, zinc mass and percentage would be lower, hopefully close to the given answer.

I'd really like to hear from an archaeologist how likely such a scenario would actually be.

If the original data gives an original weight of $3.0$ g, it means that the real weight is situated between $2.95$ g and $3.05$ g. It also means that the precision on this weight is $±0.05$ g, which correspond to ± $0.05/3.0 = 1.7$%.
The final result is also known with a precision of ±$1.7$%. And ±$1.7$% of $88.3$ is ±$1.47$ Take care of the following calculations. Don't confuse percent and percent of percent. The final value is something between $88,3 - 1.47 = 86.8$% and $88.3 + 1.47 = 89.7$%. As a consequence, your result ($89.2$%) is within these limits. It is valid.

Rules of thumb such as "3.0  g, it means that the real weight is situated between 2.95
g and 3.05 g. It also means that the precision on this weight is ±0.05 g" are not valid in real world analytical chemistry.


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Is there a reaction involving alkene formation from diazomethane?
I need help regarding the following question:

I identified some of the compounds as follows:

I think (G) should be Me₂C=CHMe because on reaction of (E) with HCl/Peroxide (Kharash effect doesn't apply) we form 2-bromo-3-chloro-2-methylbutane, which on reaction with Na/ether should be converted to Me₂C=CHMe (Wurtz reaction)...
I'm confused about the formation of (G) from (H).
Is there a way of preparing an alkene from another alkene by reacting it with diazomethane?
Or have I recognised (G) incorrectly?

Yeah, you got it wrong. Br atoms aren't next to each other.

Well, nobody would use peroxide if it didn't work. It's 1,3-dihalide and you make a cyclopropane ring, just like with diazomethane.

Is I formed from H instead of G?

@RobertDiGiovanni I too wondered the same but in the question I came across, G is formed from H and not I...

Your structure A is wrong. You won't get a alkene with molecular mass of 71 with that structure (on reaction with conc. H2SO4).

@RobertDiGiovanni May be you're right... As for the source of the question, I am not sure really!

@MathewMahindaratne Why not? Well in my opinion the alkene formed by acid catalysed dehydration would be (CH₃)₂C=C(D)CH₃... So molecular wt. = (9×1)+(2)+(5×12)=71.

Anyways thanks to all who tried helping!!

If you look at the structure, DOH leaves so MW is 70.

Your scheme is also not very clear. How did you get alkene H?

Well from what I have learnt, -OH would be protonated by the acid into -OH ₂⁺, would leave as H₂O, then rearrangement (D⁻ shift) would take place for charge stabilisation and then H⁺ would leave...

What I have finally figured out is that the problem here lies in the reaction of alkene (D) with N-Bromosuccinimide...
Here is how it should have been:
$$\ce{(CH_3)_2CH(D)CH=CH_2->[NBS] (CH_3)_2CH(Br)CH=CH_2 + (CH_3)_2C=CHCH_2Br}$$
Due to alkene stability, $\ce{(CH_3)_2C=CHCH_2Br}$ is the major product. Hence the rest of the products following (E) change:
(F): $\ce{(CH_3)_2CHCH(Br)CH_2Br}$
(I): $\ce{2-methylbut-2-ene}$
(G): $\ce{1,1-dimethylcyclopropane}$
Therefore (H) is $\ce{isobutene}$ and (J) is $\ce{2,4,4-trimethylpentene}$.

For some reason, I thought this question is going to be closed eventually. The main reason is the title asking "Is there a reaction involving alkene formation from diazomethane?" but the post of the question says something else. And, then, OP has given his/her own answer, but that did not provide an answer to the question in the title. Thus, for benefit of education, I'd try to solve so many questions within OP's post. By the way, the answer to the question in the title is yes or no. Yes, only if ketene formation is counted as an alkene formation. Otherwise, it is definitely 'no.'
Let's see OP's posted question(s) step by step:

The compound $\bf{A}$ is optically active, which have a formula of $\ce{C_nH_{2n+1}DO}$. Since, the number of unsaturation within the molecule is zero, we can conclude that the molecule is an alcohol or an ether. The positive blue color test by Victor-Mayer test tell us it is a secondary alcohol. The positive iodoform test tells it has $\ce{CH3CHOH}$- function. Thus, the structure given for is partially correct. Why partially? Because OP has ignored the next step to produce dehydroxylated product $\bf{B}$, which have molar mass of $\pu{71 g mol-1}$. Thus, the deuterium ion in $\bf{A}$ can not leave the molecule fully or partially. The structure given in $\bf{A}$ would leave the molecule at least partially if it is subjected to dihydroxylation. Therefore best structure for $\bf{A}$ would have been:

When $\bf{A}$ is subjected to dihydroxylation, it'd give two alkene compounds (cis and trans), both of which have molar mass of $\pu{71 g mol-1}$. My version of the structure $\bf{A}$ should not undergo deuterium exchange under reaction conditions so that deuterium remains 100%. The next trick is when subjected to pyrolysis after esterification of $\bf{A}$, it didn't say the (major)product would be optically active. To me, there are two different $\beta$-protons, either of which can be eliminated during the pyrolysis as shown in the scheme (Ref.1 & 2):

I assume, the "major" alkene must be optically active $\bf{D}$ because other path gave the product we received before in a different path. This optically active $\bf{D}$ would undergoes reaction to give an allyl bromide $\bf{E}$ as explain in the OP's answer. But unlike OP's structure, my product is chiral (although it gave a racemic mixture). All these and other structures are summarized in following scheme:

Only thing missing here is the Wurtz product from the chiral $\bf{F}$, which I belive dimerized to give a six-membered product due to steric hindrance of bulky $\ce{Br}$ atom.

References:

Charles D. Hurd and Ford H. Blunck, "The Pyrolysis of Esters," J. Am. Chem. Soc. 1938, 60(10), 2419–2425 (ODI: https://doi.org/10.1021/ja01277a035).
William J. Bailey and Robert Barkley, Jr., "Pyrolysis of Esters. V. Mechanism of 1,4-Elimination," J. Org. Chem. 1956, 21(3), 328–331 (ODI: https://doi.org/10.1021/jo01109a017).

Thank you for taking time to explain! As for the title, that was what my question originally was... Still could you suggest to me what could have been a better title? I am new to the platform...

Also there appears to be a problem with what you have described as (A)... It is supposed to form a salt (S) on haloform reaction, which converts into Me₂CDNH₂ on reaction with HN₃/H₂SO₄, which doesn't seem to be possible with your (A). Kindly correct me if I'm wrong...

I think it was a mistake.  The product should be Me2CDCONH2.

@MathewMahindaratne what software/tool did you use to make the diagrams on the  reactions and the arrows?

ChemDraw has done the magic.


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Why does increasing the concentration of an electrolyte decrease the mass deposited when electroplating?
I am in the process of writing a lab where the manipulated variable is the concentration of the electrolyte (zinc sulfate heptahydrate) and the responding variable is the mass of zinc deposited on a copper cathode. I electroplated each manipulation at a constant current. Hence, I found that as the concentration increases, the voltage and the mass deposited decreased.
I have researched a lot and have not found an explanation to why mass deposited decreases when current is controlled. Please help if possible!

Did you see zinc collected at the bottom of the beaker?

I did observe a few small zinc sheddings in the electrolyte solution after electroplating.
However, my procedure involved creating a concentrated 1.0 mol/L solution and diluting it for my manipulations with smaller concentrations. Does this impact the relationship?

Depositing Zinc metal requires 0.76 V.  It may be possible that too concentrated an electrolyte solution was too conductive to have a large enough voltage difference near the cathode.  So you may try various concentrations and also experiment along the axis of variable voltage.  This work should yield an excellent report and graphs even if it does not succeed.

@RobertDiGiovanni How exactly would a too conductive solution cause a lower faradaic efficiency? My guess would be these sheddings are significant or you're having a higher partial current for the HER, how exactly a high ZnSO4 causes this to happen I am not sure of.

Just a gut feeling.  "Constant current".  V = IR.  I hope it's DC.  If so, the solution (and apparatus) may be simply grounding the current.  I would increase voltage.  Using a DC transformer (like those used for heating mantles) may help.

Thank you for your replies! I will definitely consider them in my lab writeup.


text:
How did Mendeleev improve on the Newland's table?
I understand that John Newland’s law of octaves was ridiculed by the scientific community as his table failed to work past calcium. I'm trying to understand how Mendeleev’s table improved on this.
On Mendeleev’s table, do all elements in the same group even past calcium exhibit similar properties? I saw on his table that potassium and copper were in the same group. Calcium and zinc were also in the same group. It seems to me that these elements have different properties.
So why was this accepted? Don't potassium and copper exhibit different properties as well as calcium and zinc? To me, it appears that both Newlands and Mendeleev's tables had elements in the same group with very different properties.

Neither Mendeleev or Newland could properly account for the transition metals in their tables.
Newland used atomic mass as the foundation of his table, while Mendeleev (and Meyer) used other properties, such as density in addition to atomic mass.
What set Mendeleev apart is that he did not assume that all of the elements had been discovered.  Consequently, he left blank spaces on his table at atomic masses 44, 68, 72, and 100.  These spaces correspond to the elements scandium, gallium, germanium and technetium.
It's true that we credit Mendeleev with creating the modern periodic table, but his table was ordered by atomic mass, while the current table is ordered by atomic number.  This is based on work by Moseley around 1913.

All those groups of two elements are situated in columns, where the number of the column gives at least one of the oxidation numbers of the included elements. Let us look at all elements in the fourth line, from Potassium (19) to Bromine (35)
$1$) In $1$st column, Potassium(19) and Copper(29) have in common the possibility of being stable at oxidation number $+1$.
$2$) In $2$nd column, Calcium(20) and Zinc(30) have the same oxidation number $+2$.
$3$) In $3$rd column, Scandium(21) and Gallium(31) have the same oxidation number $+3$.
$4$) In $4$th column, Titanium(22) and Germanium(32) have the oxidation number $+4$.
$5$) In $5$th column, Vanadium(23) and Arsenic(33) can have the oxidation number $+5$.
$6$) In $6$th column, Chromium(24) and Selenium(34) can both have the oxidation number $+6$,
$7$) In $7$th column, Manganese(25) and Bromine(35) can both have the oxidation number $+7$.
The only trouble is the "triplet" of the $8$th column : Iron(26) - Nickel(27) - Cobalt(28), which have oxidation number +$2$ and sometimes +$3$ !! Nobody was able to explain this strange behavior and the absence of corresponding atoms after Bromine at the end of the $19$th century. Mendeleev was hoping that this strange behavior would be explained later on, in the future. A couple of years later on, the noble gas were discovered and were also put in this mysterious $8$th column.


text:
How do I assign functions for cartesian coordinates and rotations to irreducible representations of a character table?
I do not really understand the intuition behind the functions assigned to irreducible representations.
For example, in H₂O, the A₂ representation is given by
E   C₂   σv(xz)   σv(yz)
1   -1    1       -1

and the associated linear functions are x and Ry, but I am not sure where these come from.

Since 5 months have gone by with no answers or comments, I think it's fair to let you know about Matter Modeling Stack Exchange which also answers group theory questions, but make sure to delete the question from here if you're cross-posting there (or to at least put a note at the top of your question, saying that it's cross-posted).


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Permanent water marks etched onto glass
We are noticing some permanent marks etched onto various glass parts we use in lighting products that cannot be removed. These appear regardless of the type of water (tap, DI etc. Our DI water is purified to 2us/cm).
No matter the amount of rubbing with IPA or micro-fibre cloth, we cannot remove them, so it seems to be etched.
Is there some process going on here where water can somehow etch / eat the glass? Something else that we are not aware of?

Etching would be a permanent damage, but I doubt it's that.

Can you try gently rubbing a cotton ball soaked in vinegar for some time? Is it possible to remove the glass? IPA does not remove water scales and it is not meant for removing water spots.

Could be something like https://chemistry.stackexchange.com/questions/69797/e-glass-degradation-in-water but I'm sceptical

Low conductivity water is not necessarily free of non-polar contaminants. Let some of the water dry on a precleaned microscope slide and look for spots.

We tried cleaning / soaking with vinegar - no change at all. Not even a razor blade appears to affect the spots.


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How does KI catalyze the reaction with H2O2?
I've done a Google search regarding this question, and have found that there is no place online which answers this question to the degree that I want.  Unfortunately, every site online stops at the place where they say that this reaction occurs via two steps:

$\ce{H2O2 + I- -> OI- + H2O}$
$\ce{H2O2 + OI- -> H2O + O2 + I-}$

My question is, is the iodide ion abstracting the O in step 1, or is it simply bonding after the O has already been given off? Since H2O2 decomposition happens incredibly slowly, the latter seems very unlikely.  On the other hand, I can't seem to visualize why an anion would abstract the uncharged oxygen either.  Yes, O has a lone pair, but that should give it a slight negative dipole moment and make it even harder to abstract, in my mind.
Secondly, in step two, the OI- appears to catalyze the decomposition of H2O2 further.  Why?  Again, wouldn't the negative charge on the anion cause it to have no interaction with the neutral hydrogen peroxide?

Don't know the mechanism, but the oxygen in hydrogen peroxide is in an oxidized state (-1) relative to its common oxidation state (-2), so electrophilic. Maybe the iodide stabilizes the oxygen dianion and the latter is what causes the peroxide to fall apart?

For some reason you seem to think these are elementary reaction, while they obviously aren't. This is just the two reactions happening, whatever their mechanisms are is another thing.

OP stated in the question that:

I've done a Google search regarding this question (How does $\ce{KI}$ catalyze the reaction with $\ce{H2O2}$?), and have found that there is no place online which answers this question to the degree that I want.

I think OP hasn't put enough time to find out that this experiment (Decomposition of hydrogen peroxide catalyzed by iodide) is a very popular experiment amongst first year laboratory programs around most of universities in United States and around the world. For example, see Ref.1, in which it states that $\ce{H2O2}$ can act either as a oxidizing or reducing reagent:
$$\ce{H2O2 + 2H+ + 2e- <=> 2H2O} \qquad \mathrm{E^\circ} = \pu{1.776 V} \tag1$$
$$\ce{O2 + 2H+ + 2e- <=> H2O2} \qquad \mathrm{E^\circ} = \pu{0.815 V} \tag2$$
The equations $(1)$ and $(21)$ shows that $\ce{H2O2}$ can undergo spontaneous self-oxidation-reduction, and is therefore, thermodynamically unstable. Accordingly, $\ce{H2O2}$ decomposes spontaneously and exothermically, but slowly to oxygen and water at room temperature. The slow reaction rate is due to its relatively large activation energy of $\pu{76 kJ mol-1}$ (Ref.2). However, it would decompose at a much faster rate when a catalyst is introduced. For example, the large activation energy is lowered to ~$\pu{41 kJ mol-1}$ when $\ce{MnO2}$ is used as the catalyst and proceeds with a heat of reaction of $\pu{−99 kJ mol-1}$ (Ref.2). The iodide-catalyzed decomposition reaction proceeds with a heat of reaction of $\pu{−98.3 kJ mol-1}$ (Ref.3).
According to Ref.1, $\ce{I-/IO-}$ redox couple with $\mathrm{E^\circ} = \pu{0.55 V}$ and $\ce{I2/I-}$ redox couple with $\mathrm{E^\circ} = \pu{0.535 V}$ are ideal systems to catalyze $\ce{H2O2}$ decomposition. If first couple is used, the rate determining step would be:
$$\ce{H2O2 + I-  -> IO- + H2O} \tag3$$
The same the rate determining step is given in an older reference (Ref.4), so that the mechanism might be correct. The same group, Liebhafsky & Coworkers  has mentioned in later publication (Ref.5) that:

The first successful quantitative investigation in chemical kinetics, completed in 1866 by Harcourt and Esson, proved that the rate of the reaction, $$\ce{H2O2 + 2H+ + 3I- -> 2H2O + I3-} \tag4$$ is proportional to the amounts of peroxide and iodide present, and that it is “accelerated" by the addition of acid. A quarter century later Magnanini in a remarkable investigation, which we shall subsequently consider in some detail, demonstrated how this “acceleration" is related to the acid added.

The authors have measured the rate of the reaction (given in the equation $(4)$) for the first time under the simplest experimental conditions, with the result that the law, previously established (by Magnanini):
$$\frac{d[\ce{I3-}]}{dt} = k_1^\circ \frac{[\ce{I-}]}{[\ce{H2O2}]} + k_1[\ce{H+}][\ce{I-}][\ce{H2O2}] \tag5$$
As a result, they have concluded that the simplest kinetic interpretation of this rate law assumes following two rate determining steps:
$$\ce{H2O2 + I- ->[k_1^\circ] H2O + IO-} \tag6$$
$$\ce{H2O2 + H+ + I- ->[k_1]  H2O + HIO} \tag7$$
And, both steps proceed simultaneously and independently. At $\pu{25 ^\circ C}$, they have reported measured values of $k_1^\circ $ and $k_1$ as $0.69$ and  and $10.5$, respectively.

References:

John C. Hansen, "The Iodide-Catalyzed Decomposition of Hydrogen Peroxide: A Simple Computer-Interfaced Kinetics Experiment for General Chemistry," J. Chem. Educ. 1996, 73(8), 728-732 (ODI: https://doi.org/10.1021/ed073p728).
William Sweeney, James Lee, Nauman Abid, and Stephen DeMeo, "Efficient Method for the Determination of the Activation Energy of the Iodide-Catalyzed Decomposition of Hydrogen Peroxide," J. Chem. Educ. 2014, 91(8), 1216–1219 (ODI: https://doi.org/10.1021/ed500116g).
Tomoyuki Tatsuoka and Nobuyoshi Koga, "Energy Diagram for the Catalytic Decomposition of Hydrogen Peroxide," J. Chem. Educ. 2013, 90(5), 633−636 (ODI: https://doi.org/10.1021/ed400002t).
Herman A. Liebhafsky, "The Catalytic Decomposition of Hydrogen Peroxide by the Iodine-Iodide Couple. II & III. The Rate of Oxidation in Neutral, and in Acid, Solution of Hydrogen Peroxide by Iodine," J. Am. Chem. Soc. 1932, 54(9), 3499–3508 (ODI: https://doi.org/10.1021/ja01348a003).
Herman A. Liebhafsky and Ali Mohammad, "The Kinetics of the Reduction, in Acid Solution, of Hydrogen Peroxide by Iodide Ion," J. Am. Chem. Soc. 1933, 55(10), 3977–3986 (ODI: https://doi.org/10.1021/ja01337a010).


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Process of Arecoline Extraction
What would be the proper procedure to extract and isolate arecoline, an alkaloid ester, from the seed of the Areca nut, and is it possible to mass produce arecoline using that method?

Go to Google. Type "arecoline extraction". You will find at least $5$ references about how to do this extraction. But it seems difficult, because it needs a cation exchange resin, then membrane ultrafiltration, freeze dehydration and distillation under high pressure.

Here is the brief description of arecoline given by Wikipedia:

Arecoline is a base, and its conjugate acid has a $\mathrm{p}K_\mathrm{a}$ ~6.8. Arecoline is volatile in steam, miscible with most organic solvents and water, but extractable from water by ether in presence of dissolved salts. Being basic, arecoline forms salts with acids. The salts are crystalline, but usually deliquescent: the hydrochloride, arecoline•HCl, forms needles, m.p. $\pu{158 ^\circ C}$; and the hydrobromide, arecoline•HBr, forms slender prisms, mp. $177$–$\pu{179 ^\circ C}$ from hot alcohol.

Thus, you can use several different methods to isolate it from bio matrices if you have some experience in natural product chemistry. My best bet would have been the technique called steam distillation since it'd be the cheapest method to do so (Ref.1; recovery of arecoline was highest, about 96%, in the distillation method compared to other methods used). A typical procedure can be outlined as follows:

Place $\pu{2.5 g}$ 30-mesh powdered sample of arecoline nut in distillation flask. Add $\pu{5 mL}$ $\pu{0.3 N}$ $\ce{NaOH}$ solution and wet sample uniformly. Place receiver containing $\pu{10 mL}$ $\pu{0.02 N}$ $\ce{H2SO4}$ in ice bath, with stem of condenser dipping in acid. Connect apparatus and let steam bubble through material. Adjust heating so that first drop of distillate drips into receiver 2-3 min after entry of steam. Collect $\pu{150 mL}$ or more of distillate in $\pu{15 min}$. Conduct blank distillation with the reagents without sample. Titrate distillate against standard $\pu{0.022 N}$ $\ce{NaOH}$, using methyl red as indicator. From difference in titer values, deduce arecoline content. $\pu{1 mL}$ $\pu{0.02 N}$ $\ce{NaOH}$ is equivalent to $\pu{0.003108 g}$ arecoline.

The following facts are important in this distillation:

The arecoline nuts should be powdered (mesh size of 30) and then put in a alkali solution with $\mathrm{pH}$ 0f 9.0-9.1. The conditions studied for the recovery of
arecoline from areca nut in Ref.1 has showed that sodium carbonate and bicarbonate did not release the arecoline from the powdered nut. Sodium hydroxide of sufficient concentration was, however, highly effective for this purpose, as shown by data. Thus it can be seen from the data in Ref.1 that a value of 0.392% for arecoline was obtained for a 20-mesh powder, compared to 0.512% for a 30-mesh powder with the same quantity of alkali. Studies on the effect of distillation rate (Ref.1) has indicated that a rate of $\pu{10 mL min-1}$ and a distillate volume of $\pu{150 mL}$ were optimum for maximum recovery of arecoline from areca nut.

References:

E. Sankaran Nambudiri, "Estimation of Arecoline: A Rapid Distillation Method," Journal of Association of Official Analytical Chemists 1968, 51(4), 799–802 (DOI: https://doi.org/10.1093/jaoac/51.4.799).


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Nomenclature of an organic compound with multiple substituents

what is the IUPAC system of naming a compound with multiple substituents? I am not sure if the name should be something like 3-ethyne-1-hydroxy-5-yne-2-pentanone. This structure contains 2 triple bonds, 1 carbonil group and 1 hydroxy group. So which substituent I take first to name?

Follow the freely available flow chart and examples of the Brief Guide or its 4 page summary of IUPAC's Blue Book compiled in the resources page (section chemical nomenclature):


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Disposal of Aqueous Magnesium Salts
Where can I find out where (specifically) aqueous magnesium chloride and aqueous magnesium bromide can be disposed? I was trying to find out whether solutions could be diluted and washed down the drain or if more specialist waste disposal was necessary?

From the UK material safety datasheet of Magnesium Chloride here

SECTION 13: Disposal considerations
13.1 Waste treatment methods
Product
Waste material must be disposed of in accordance with the national and local regulations.
Leave chemicals in original containers. No mixing with other waste. Handle uncleaned
containers like the product itself. See www.retrologistik.com for processes regarding the
return of chemicals and containers, or contact us there if you have further questions.
Notice Directive on waste 2008/98/EC

You may wish to search the material safety datasheet (MSDS) for your own region, but it sounds like you should be using specialist waste disposal contractors if you have anything more than small amounts of it.

If it is a couple of grams, you can wash your magnesium solution down the drain. Magnesium is often present in small amounts in mineral waters. If it is a couple of kilograms, you should proceed like Waylander recommends it.


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how to neutralize stannous chloride (50ml) to make it safe for disposal
I made a small quantity of stannous chloride by dissolving tin in muriatic acid - I would like to neutralize it for safe disposal - will mixing it with soda bicarbonate dissolved in water accomplish this? and if not then what would be the best way to do this?

Please define what you mean by "safe disposal".

Yes. Adding sodium bicarbonate will produce a white precipitate of $\ce{Sn(OH)2}$ that will be slowly oxidized by air into $\ce{SnO2}$. And $\ce{SnO2}$ is totally insoluble, as it is the formula of the cassiterite, most important mineral used to extract tin on Earth.


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Does -ate suffix mean existence of -ite oxyanions?
According to online resources oxyanions are named as
Per+element+ate eg. $\ce{ClO4-}$
Element+ate eg. $\ce{ClO3-}$
Element+ite eg. $\ce{ClO2-}$
Hypo+element+ite eg. $\ce{ClO-}$
1-If I see an oxyanion should I assume that other oxyanions of same element also exists?
e.g. $\ce{BO3^{3-}}$ is borate. Do Borite, hypoborite, perborite also exist?
2- How can I write formula of an oxyanion from a given name? For example
Carbonate =$\ce{CO3^{2-}}$, while
Sulphite =$\ce{SO3^{2-}}$. Both have same  number of oxygen atoms and same oxidation state of C and S but names are different. How to know how many oxygens would be in an oxyanion?

This is about nomenclature, and when a nomenclature is established (a convened standard), this best not only considers compounds already isolated from a sample in nature, or once synthesized in a lab, but has rules powerful enough to accommodate new chemical elements and compounds.  And IUPAC's rules, because systematic names are based on the chemical composition and structure of a compound, allow to name hypothetical compounds still unknown to literature.
On the other hand, chemical nomenclature equally is subject to change, and continues to evolve gradually; all while common names based on early assumptions (on occasion today known as wrong) tend to stick.
Chemistry.se has a page resources with a section about contemporary chemical nomenclature.  IUPAC's freely available .pdf of the Red Book about inorganic chemistry is among the links,  rule IR 8-3 states:

"Table IR-8.1 also includes anions from the neutral oxoacides by successive dehydroniation.  Many of these anions also have common names that are still acceptable, in some cases despite the fact that there are now otherwise abandoned (e.g. nitrate/nitrite and perchlorate/chlorate/chlorite/hypochlorite)."

To quote (edition 2005, p. 131):

(Hill) formula
acceptable name
systematic additive name

$\ce{[ClO4]^-}$
perchlorate
tetraoxidochlorate(1-)

$\ce{[ClO3]-}$
chlorate
trioxidochlorate(1-)

$\ce{[ClO2]-}$
chlorite
dioxidochlorate(1-)

$\ce{[OCl]-}$
hypochlorite
chloridooxygenate(1-)

And similar about the analogues with bromine; tetraoxidobromate, trioxidobromate, dioxidobromate, bromidooxygenate.  This contrasts to ions of borates which, in the solid state, tend to yield are larger, interconnected units.  Hence a modern systematic nomenclature accounting for this diversity is more complex.  (For a glimpse about e.g., meta, ortho, perborates and their structures, visit e.g., Wikipedia's entry page here).

I've taken the liberty to 'convert' your MD Table so that it renders correctly. Not that big of a change. I recommend bring such things up on meta if you have questions about it. I guess there is already something about it there… I would need to look, too. Btw. I think the "LaTeX table" was alright, too. I think in general the MD Table works better though.

Thank you very much for the helpful edit.  The "LaTeX table" already followed a pattern explicitly posted as example on chemistry.meta (here), however I agree, the markup with .md is lighter (hence equally easier to read by the non-initiated) than the one with .tex.

Systematic additive names are much helpful. These should be used to avoid confusion. Thank you for valuable information.

The following rules were decided in the $19$th century.
When there is only one oxyanion containing $1$ element X and "some" Oxygen atoms, this anion is called "X-ate", whatever the number of O atoms and the charge of the anion. Example : As the only anion containing $\ce{1 C}$ and "some" O atoms is $\ce{CO3^{2-}}$, this anion is called carbonate. Important point is that there is no relation between the suffix and the number of O atoms.
If two anions are known differing only by the number of Oxygen atoms, the preceding rule is applied by the most oxygenated ion. The other anion is called X-ite. Example : In the 19th century, two anions were known containing $1$ N and "some" O atoms. They are $\ce{NO3^-}$ and $\ce{NO2^-}$. The first one is called nitrate, and the second nitrite. Here too, there is no relation between the choice of the suffix -ate or -ite and the number of O atoms. -ate means maximum, and -ite means minimum. This is the most important point of this nomenclature.
If four anions are known differing by the number of Oxygen atoms, the preceding rule is applied by the two central anions. The two extremes are called by adding a prefix "hypo" in front of X-ite for the lowest amount of O atoms. The prefix is "per" before X-ate for the largest amount of O atoms. Example : Anions containing chlorine, displayed in the original request.
If only three anions are known differing by the number of Oxygen atoms, one tries to "invent" a fourth ion, in order to apply the preceding rule.

Bromites are known and I don't see why you waste time on "questions" like this one.

Perbromate was only made in 1968...

Many people don't understand chemistry because of such doubts and quit reading. I believe every answer will be appreciated by many like me.

Not necessarily.  For example, in the case of the halo-oxyanions (like the ones you mentioned), all of the different ones exist.  This is not necessarily the case.  For example, in the nitrogen oxyanions, there are only nitrate ($\ce{NO3-}$) and nitrite ($\ce{NO2-}$).

The -ate/-ite naming system is not standardized.  The only way (that I'm aware of) to know them is to memorize them.  For example, sulfate is $\ce{SO4^{2-}}$ while nitrate is $\ce{NO3-}$.

https://en.wikipedia.org/wiki/Persulfate ...

Let me illustrate through an example...
$\ce{HClO}$ is called hypochlorous acid, the corresponding acidic radical $\ce{ClO-}$ is called hypochlorite ion. In both of these, oxidation number (O.N.) of $\ce{Cl}$ is +1.
$\ce{HClO2}$ : Chlorous acid; $\ce{ClO2-}$ : Chlorite ion;
O.N. : +3
$\ce{HClO3}$ : Chloric acid; $\ce{ClO3-}$ : Chlorate ion;
O.N. : +5
$\ce{HClO4}$ : Perchloric acid; $\ce{ClO4-}$ : Perchlorate ion;
O.N. : +7
As you can observe, "hypo-...-ite" refers to the lowest oxidation state of the element (in oxyacid form), "...-ite" is used for the next oxidation state, "-ate" for the next and "per-...-ate" for the highest oxidation state.
So basically the existence of the "-ate" and the "-ite" forms totally depends upon how many oxidation states that element can show in oxyacid (or oxyanion) form.
Also, it is to be noted that if an element exhibits two oxidation states in its oxyanions, then we generally prefer to name the ions as "-ite" and "-ate" corresponding to the lower and higher oxidation states respectively.
I hope this helps. Please feel free to correct me if I have erred anywhere.


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Why CFC naming system is so weird?
Why it can't be just as simple as just the number of C, H & F atoms respectively instead of (C - 1), (H+1) & F?

A special numbering system is to be used for fluorinated alkanes, prefixed with Freon-, R-, CFC- and HCFC-, where the rightmost value indicates the number of fluorine atoms, the next value to the left is the number of hydrogen atoms plus 1, and the next value to the left is the number of carbon atoms less one (zeroes are not stated), and the remaining atoms are chlorine.
Freon-12, for example, indicates a methane derivative (only two numbers) containing two fluorine atoms (the second 2) and no hydrogen ($1 - 1 = 0$). It is therefore $\ce{CCl2F2}$.

(English edition of Wikipedia, Chlorofluorocarbon#Numbering_system)
For that the system in part overlaps with R-numbers about media used for refrigerants, I speculate one wanted a description condensed, meaningful internationally, accessible across multiple languages, and for multiple professions. Arguably, this is a consent (e.g., Freon 112 about tetrachloro-1,2-difluoroethane, and Freon 112a about tetrachloro-1,1-difluoroethane the table compiles) with some limitations, and yet was/is powerful enough among the intended audience as an alias easy to recognize:

(image credit: English edition of Wikipedia, refrigerant)
(We chemists are not better in this regard naming a liquid "petrol ether", though there is no oxygen among the alkane molecules in question.)


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Can copper hydroxide be formed as a byproduct of water electrolysis?
I was electrolysing water with a 12v DC power supply at about 3 milliamps, using copper wire for both my anode and cathode. The solution was about 17 grams of table salt in 750 milliliters of purified water. After running the electrolysis cell for 1 minute, I left it overnight. The next day, I found a light blue powder at the bottom of the cell, which was still full of solution. Could this be copper hydroxide? If not, what could it be?
I should also note that the anode and cathode had become red, similar to what iron oxide looks like.

I just want to tell you that this experiment is pretty unsafe in general. Okay the currents are low so the amount of products formed are low, but you're forming hydrogen gas and possible chlorine gas. I would recommend you to not do this experiment for extended periods or higher currents.

Yes. The blue precipitate is $\ce{Cu(OH)2}$. Let's explain how it is produced. At the cathode, the following reaction occurs :$$\ce{2 H2O + 2e^- -> H2 + 2 OH^-}$$ So the solution becomes basic, due to the new $\ce{OH^-}$ ions being produced by the reaction. At the anode, the following reaction occurs : $$\ce{Cu -> Cu^{2+} + 2 e-}$$ which produces new $\ce{Cu^{2+}}$ ions in solutions. The new $\ce{Cu^{2+}}$ and $\ce{OH^-}$ ions are attracting one another, and when they join, they react and produce an insoluble precipitate, made of $\ce{Cu(OH)2}$ according to $$\ce{Cu^{2+} + 2 OH^- -> Cu(OH)2}$$

To note, given the reaction conditions that are given (pH 7) it is unlikely that Cu2+ is formed free in solution and more likely that the reaction going on was direct to the hydroxide. The hydroxide could have formed dendrites or other mechanically unstable structures (or just simple delamination) and just have broken off from the electrode.


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Is it possible to separate $\ce{CO2}$ from exhaled air? How?
Just for fun, I would like to see if I could make my own concentrated $\ce{CO2}$ by separating it from my own exhaled breath. I would also love to see if I could pressurize it so I could make my own carbonated water!
When you exhale, the gas that comes out of your mouth contains 78% nitrogen, 17% oxygen, 4% carbon dioxide and 1% other gases (source).
Is there any existing procedure for separating carbon dioxide from breathing air? I have done some research, but I have only found (industrial-scale) methods for separating nitrogen.

One can chemically trap one's carbon dioxide as calcium carbonate. An old school experiment to was bubble one's breath in a dilute solution of limewater. The solution would become cloudy, and further breathing into limewater would make the solution clear. The cloudiness was due to calcium carbonate. I will leave the chemistry for you to search yourself.

If you want to isolate it, you can capture CO2 in Na2CO3 solution, evaporate it and calcinate NaHCO3 to release captured CO2.

You got it all wrong. Concentrated CO2 is produced from Coca-Cola. Your exhale is just too diluted.

In my biology class they mentioned that in a spirometer there is a canister of 'soda lime' that removes CO2 from the air tank when the user exhales.

Is it correct to assume you are interested only in the specific way to obtain CO2 for the exhaled breath? I.e. not in more convenient ways e.g. from carbonated drinks.

@Poutnik Actually, the reason I wanted to obtain $\ce{CO2}$ is to make carbonated drinks. Though I'd love to hear how you could take the carbon dioxide out of a carbonated drink!

Always write clear questions with included relevant details, purpose, context or background info to receive useful answers. Otherwise the answers may be just guesses.

Have you ever opened a carbonated drink? And a warm one? Then you already know how.

As mentioned in the comments, the easiest method is to bubble your breath through an alkaline solution. Use a bubbler that produces small volume bubbles and a solution in a tall vertical column to get the best capture. The choice of alkaline salt depends on your intended recovery method. Calcium ions will lead to precipitation of most of the carbon dioxide as calcium carbonate, which can be collected as a solid and heated to release the carbon dioxide as gas (calcination). With sodium ions, the carbonate will stay in solution, and you can recover much of the carbon dioxide by acidifying the solution and collecting the released gas.
If your goal is to get relatively pure carbon dioxide directly, as opposed to a solution containing carbonates, the common method is to use a cold trap. Flow the gas over an absorbent to remove water vapor and then cool the gas. The carbon dioxide will freeze while the other gases remain gaseous. This of course requires a very cold trap. The deposition temperature is $-79.5$ $^\circ C$.

I am afraid the statement "Calcium ions will lead to precipitation of most of the carbon dioxide as calcium carbonate, which can be collected as a solid and heated to release the carbon dioxide as gas (calcination) " is not entirely correct. If we continue to bubble CO2 in limewater, the initial precipitate redissolves forming calcium bicarbonate.

@AChem - Good point.  I am assuming that initial pH is high enough to avoid bicarbonate formation, but certainly it would be possible to add enough CO2 to overcome that.


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Explanation of colours of solder plated copper pads after electrolysing a circuit board in NaCl solution
A last ditch method for identifying connections in a old multilayer circuit board that I have come up with, when you have a spare, is to put the board in salt (NaCl) solution and connect a power supply to it. One pole goes via a wire to the circuit you are trying to find (AA), the other to either the board ground (GND), or an electrode in the tank.
In the tank, you get streams of large bubbles from every pad connected to AA, and smaller fainter ones from the ground or other electrode. It is great help as it tests the entire board at once.
When doing this as an experiment, I have noticed that every ground terminal went a mid-grey colour, and that pads on the AA pole went black. I believe I was getting hydrogen at AA, and chlorine at the ground (but cannot be 100% sure on supply polarity).
This process does take the solder off the AA pads on the circuit, so I guess some copper was involved as well.
I did a quick search, but cannot find any salts that would have these colours likely to exist. The only black salt I have found is Cu(II)O, but cannot believe that is being formed in a solution- especially at both pads (more at AA).
The solder appears to be lead free (board is from 2017-2019 so is mostly tin. There may a tiny bit of silver, but it is a tiny percentage.
Would anyone like to suggest what the chemicals or mixtures of chemicals are? Electrolysing salt can give NaClO, NaOH, as well as the two gases mentioned, so there are many options.
Many thanks and I hope you enjoy this question.


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Better Way for Crushing Coconut into 500 mesh Coal for Carbonization
I am planning to use coconut as a raw material for carbonization, but I need to crush it into 500 mesh coal. Currently, I am using a mortar and pestle to grind the coconut shell into a fine powder, but the process is very time-consuming and the yield is not always consistent. Imagine half a day crushing it, and yet I only acquire 2 grams.
I am looking for a better method to crush coconut into 500 mesh coal. I have heard about using ball mills or other grinding equipment, but I am not sure which one is suitable for this purpose. For mixing with PVA for nanofiber material.
Can anyone suggest a better and more efficient method for crushing coconut into 500 mesh coal for carbonization? And what type of equipment should I use for this process? I appreciate any other suggestions or tips on how to improve the yield and consistency of the coal.

You will probably need a commercial grade mill/grinder. Please do not waste your days by using a mortar and pestle. Is this an academic project?

A hammer mill would be logical first step . There are a variety of grinding type machines ; likely depends more on what is available or how much one is willing to spend.

Do you need 500 mesh coconut shell or 500 mesh coconut coal? As the question is unclear in this. Crushing coal could be easier than crushing shells. Crushing coconut shells leads to coal only if it is intensive enough to cause carbonization.

Agree with @Poutnik. Carbonize first.  Then, if you like, reconstitute to desired size and you have ... charcoal briquettes!

@Poutnik 500 mesh of coal that was processed coconut. It is not coconut shell

The question formulation is misleading. Why not to carbonize first, crush after?

i am Sorry I gonna fix my title and create new question instead


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Why is Quinone used, while oxidising primary alcohols by Oppenauer Oxidation, instead of Acetone?
Usually, during Oppenauer Oxidation, while oxidising secondary alcohols, acetone solution is used with Aluminium Tertiary Butoxide.
This results into the oxidation of secondary alcohol into a ketone, along with isopropanol as a side product.
I was told that if primary alcohol needs to be oxidised into an aldehyde, quinone is used instead of acetone. I fail to understand why is that the case, and how that would affect the mechanism of the reaction.

Do you have specific reference for your claim?

Some authentification is needed here. What is the solvent? Certainly not quinone! If an aldehyde were produced, in many instances condensation reactions would occur in the presence of strong base, presuming that it is required.

OP has mentioned in the post that:

I was told that if primary alcohol needs to be oxidized into an aldehyde, quinone is used instead of acetone. I fail to understand why is that the case, and how that would affect the mechanism of the reaction (Oppenauer Oxidation).

I have asked OP if there specific reference for his/her claim, but didn't get a response back. Yet, I have search for the answer and found there wasn't any proof for the claim. However, I think my following answer would satisfy some clearance for OP's curiosity.
Essentially, Oppenauer Oxidation is the reverse reaction of Meerwein−Schmidt−Ponndorf−Verley (MSPV) reduction. According to Wikipedia,

MSPV reduction in organic chemistry is the reduction of ketones and aldehydes to their corresponding alcohols utilizing aluminum alkoxide catalysis in the presence of a sacrificial alcohol. The advantages of the MPV reduction lie in its high chemoselectivity, and its use of a cheap environmentally friendly metal catalyst.

The MSPV reduction was independently discovered by Albert Verley and the team of Hans Meerwein and Rudolf Schmidt in 1925. They found that a mixture of aluminium ethoxide and ethanol could reduce aldehydes to their alcohols. Ponndorf applied the reaction to ketones and upgraded the catalyst to aluminum isopropoxide in isopropanol (Ref.1-3).

The use of aluminum isopropoxide in isopropanol has led to the use of acetone in the reverse reaction, Oppenauer Oxidation introduced in 1937 (Ref.4).
Based on these facts, either ketone and aldehyde can be reduced or their corresponding alcohols can be oxidized either direction using best sacrificing ketone (acetone) or alcohol (isopropanol). Thus, need for quinone to prepare aldehyde is unheard of. For example, Ref. 5 shows a successful reduction of crotonaldehyde and cinnamaldehyde using isopropoxide, so why not the reversed reaction using acetone?

References:

Hans Meerwein and Rudolf Schmidt, "Ein neues Verfahren zur Reduktion von Aldehyden und Ketonen," Justus Liebigs Annalen der Chemie 1925, 444(1), 221-238 (ODI: https://doi.org/10.1002/jlac.19254440112).
M. Verley, Bull. Soc. Chim. Fr. 1925, 37, 871−874.
Wolfgang Ponndorf, "Der reversible Austausch der Oxydationsstufen zwischen Aldehyden oder Ketonen einerseits und primären oder sekundären Alkoholen anderseits," Angewandte Chemie 1926, 39(5), 138-143 (ODI: https://doi.org/10.1002/ange.19260390504).
R. V. Oppenauer, "Eine Methode der Dehydrierung von Sekundären Alkoholen zu Ketonen. I. Zur Herstellung von Sterinketonen und Sexualhormonen," Recueil des Travaux Chimiques des Pays-Bas 1937, 56(2), 137-144 (ODI: https://doi.org/10.1002/recl.19370560206).
William G. Young, Walter H. Hartung, and Frank S. Crossley, "Reduction of Aldehydes with Aluminum Isopropoxide," J. Am. Chem. Soc. 1936, 58(1), 100–102 (ODI: https://doi.org/10.1021/ja01292a033).


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Sacrificial Electrodes on a Car?
I know that sacrificial electrodes are used on outboard motors and in other marine applications to reduce or eliminate corrosion from salt water, etc.  Could pieces of zinc be attached to an automobile frame to prevent rusting of the steel frame?

Sacrificial electrodes needs to be dipped in an aqueous solutions to work. Out of such a liquid, they do not work.

Putting zinc in one spot will not protect another place that is not part of a complete circuit (i.e., immersed in electrolyte). A boat in water is immersed in that electrolyte.
However: the whole part is protected if covered with the sacrificial metal.
For example, automobile exhaust mufflers at one time were commonly aluminized... which worked well on the aluminized exterior, until it wore through from abrasion or attack from moisture and road salts. The exhaust system often rusted from inside, though, which was not aluminized -- had the water condensed from the exhaust, inside the muffler, been contiguous with the aluminized exterior, there might have been less rust inside.  Now, exhaust systems must meet governmental regulations to last a certain time and/or distance, and are often made from stainless steels.
Galvanized (sacrificial zinc-coated) steel is also widely used in automotive applications. Dipping in molten zinc leaves pretty spangles, which do not make for a smooth body finish, so electroplating is used, instead. If you think about it, we use electricity to deposit the zinc, then use electricity generated by oxidation of the zinc to protect the substrate.


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Why aren't the units for temperature related to kinetic energy, if temperature is defined as average kinetic energy?
Temperature, as far as I know, is a measure of the average kinetic energy of the particles in a substance. For instance, if a bar of pure gold is sitting at room temperature ($273.15 \text{ K}$), then the gold molecules within the lattice will be vibrating to some extent at some average kinetic energy in J. If the temperature of that bar of gold were somehow reduced to $0 \text{ K}$, then its particles would not be under any motion at all, and the average kinetic energy would be 0 J.
Because of this, I would imagine that temperature could be measured in units of energy, like Joules. But as far as I know, no such unit has been assigned to temperature. Why not?

Your title premise is wrong.

https://chemistry.stackexchange.com/questions/129290/what-exactly-is-temperature

Related: https://chemistry.stackexchange.com/q/121197

Imagine you would not measure water temperature since today, but the mean energy per a degree of freedom of water molecules instead.  Good luck!

@Poutnik Why is the title premise wrong?

if temperature is defined as average kinetic energy is wrong. Temperature is not defined as average kinetic energy. E.g. at the same T,  the mean kinetic energy of atom of Ar, molecules of N2, H2O and CO2 are different for each substance.

The Kinetic energies are the same the Velocities differ.

@jimchmst If the thermal energy consisted of only translation kinetic energy, your would be right. But it involves also rotation kinetic energy, vibration kinetic energy, potential energy of bond vibrations, potential energy of collisions and electron excitation energy.

Not so, the total internal energy is separated into trans. rotational and vibrational and electronic etc. The reason heat capacity increases with T is that energy absorbed is partitioned into the various modes and only part raises the temperature. Energy absorbed by molecular or electronic QM transitions does not immediately raise the temperature until the excited states collisionally relax. and distribute the energy to translation energy of the molecules. Compare photo activation to thermal activation.

In thermodynamics temperature describes how much the entropy of a system changes with a change in its internal energy, at constant volume:
$$T^{-1} = \left(\frac{\partial S}{\partial U}\right)_V$$
(well, as the equation shows the inverse of T describes that).
A change in entropy can be related to a reversible transfer of heat between system and surroundings. Temperature describes the extent to which the entropy of a system changes when non-mechanical energy (heat, energy associated with random atomic motions) is transferred to the surroundings.
Therefore temperature is a rate measured in units of energy per unit of entropy, and, as explained by other answers, is an intensive property, meaning it does not change with the size of the system if other intensive properties (such as pressure) remain the same. This is different from energy which is an extensive property (if you double the size of the system, you double the amount of energy in it).

And I thought temperature was a fundamental. That entropy was defined in terms of energy and temperature

As thermodynamic concepts both entropy and temperature are fundamental and linked. However (I think) entropy is more fundamental. Boltzmann's entropy formula $S=k_B \log \Omega$ does not include T, and energy can be defined without reference to T.

@Aurelius The second law relates a change in S to a change in energy for a given T: $dS=dq_{rev}/T = dU_V/T$. So they change hand in hand at constant V. That is the origin of the equation in my answer.

Kelvin is the SI unit for temperature. In the modern definition, it is defined as 1 Joule divided by the Boltzmann constant, or more exactly, such that the Boltzmann constant takes the value $\pu{1.3806505E−23 J K-1}$. So kinetic energy and temperature have different dimensions, but they are closely related (as are their SI units, via the set value of the Boltzmann constant).

For instance, if a bar of pure gold is sitting at room temperature (273.15 K), then the gold molecules within the lattice will be vibrating to some extent at some average kinetic energy in J.

Yes, the energy is $k_B T$, and the temperature is $T$. Related, but different quantities and different units.
To give another example where two quantities are related but have different dimensions and units, think of electromagnetic waves in a vacuum. You can characterize them by specifying either their wavelength or their frequency. That does not mean wavelength and frequency are the same, or have the same dimension. They are related, however, in this case via the speed of light.

Temperature is not energy and energy is not temperature. This problem of understanding heat, temperature and energy plagued the 19th century physicists for a long time. Indeed it is/was a tough problem with philosophical connotations. For instance can we ask what is the unit of blue color because blue color corresponds to a certain wavelength and that corresponds to certain energy in the visible spectrum. One should be able to say, how many Joule of blue color are there in a blue T-shirt? This is how temperature and energy are related. Temperature is a measurable property of matter but it is not energy. It is a measure of average kinetic energy.

The "temperature" of a particle can be expressed in eV or K. There are convenient references and calculators for that. Indeed, for plasma physics, the two units may be used interchangeably.
However, the temperature of a bulk substance is not expressed in eV, because one must take into account the energy of all particles. Is there more energy available in 1 µg of hydrogen 1,000 K hotter than the surroundings, or in 1 kg of hydrogen 1 K hotter than the surroundings?

Some properties of a material system, like temperature, are independent of the amount of matter under investigation. These properties are called intensive. Other intensive properties are color, density and flavor. However much material you have makes no difference.
In contrast, properties such as total mass, volume and energy (relative to some standard), will vary depending on how large the system is. The bigger the system, the greater will be the measure of the property. These properties are called extensive.
Energy and temperature are not the same kind of property. They are related like roses and the fragrance of roses. More roses, whether by mass, weight, or number, is a measure of an extensive property. The fragrance of roses is intensive, because it is not quantified, but identified.
There is a possible confusion if you claim that more roses equals more fragrance of roses. But, no. The fragrance is identified as the same. What you do have more of is the rose oil which is the source of the fragrance.
If you have a higher temperature, you do not have more temperature. You still have just a temperature; but it's a higher temperature - in a different system.
A good discussion can be found in many textbooks or Wikipedia (https://en.wikipedia.org/wiki/Intensive_and_extensive_properties).


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How would you use henderson equation to calculate the pH of a solution containing multiple different acids, such as carbonic acid and acetic acid?
I would like to know if it is possible to use the Henderson equation to calculate the pH of a solution containing more than one kind of weak acid.

In short, as it is, it cannot be used even for a single acid.
The Henderson-Hasselbalch equation (HHE) is nothing else but a little rearranged equation for an acid dissociation constant in the logarithmic form.
$$K_\mathrm{a}=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \implies \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{A-}]}{[\ce{HA}]}} $$
HHE is useful if the concentrations of the acid and its conjugate base (anion) are given or easily calculated, like in pH buffers, what is not our case. pH calculations are usually done without prior explicit calculation of the above concentration for HHE.

The general scenario of acid-base systems is described by set of nonlinear equations of multiple variables.
For two acids, it involves:

the equations for dissociation constants
$K_\mathrm{a,\ce{X}}=\frac{[\ce{H+}][\ce{X-}]}{[\ce{HX}]}$
$K_\mathrm{a,\ce{Y}}=\frac{[\ce{H+}][\ce{Y-}]}{[\ce{HY}]}$
$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$
the (molar) amount balances
$c_0(\ce{HX}) =[\ce{X-}] + [\ce{HX}]$
$c_0(\ce{HY}) =[\ce{Y-}] + [\ce{HY}]$
the charge balance
$[\ce{H+}] = [\ce{OH-}] + [\ce{X-}] + [\ce{Y-}]$

So we have 6 equations for 6 variables ( $[\ce{H+}]$, $[\ce{OH-}]$,$[\ce{HX}]$,$[\ce{X-}]$,$[\ce{HY}]$,$[\ce{Y-}]$ ).
The equation set can be easily modified to adapt to different number of acids, with some of them possibly being different dissociation degrees of multiprotic acids, like $\ce{H3PO4}$.
For bases like $\ce{NH3}$, we can use either
$K_\mathrm{a,\ce{NH4+}}=\frac{[\ce{H+}][\ce{NH3}]}{[\ce{NH4+}]}$
either
$K_\mathrm{b,\ce{NH3}}=\frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$
where $K_\mathrm{a,\ce{NH4+}} \cdot K_\mathrm{b,\ce{NH3}}=K_\mathrm{w}$
Solving this equation set usually leads to cubic ( or higher order) algebraic equation for $[\ce{H+}]$, where the exact solution is not convenient to use, or does not even exist. Fortunately, the process of solving can be often simplified by using assumptions of various strong inequalities ( like $[\ce{H+}] \gg [\ce{OH-}]$ or $[\ce{HX}] \gg [\ce{X-}]$). It leads to a much simpler, often quadratic solvable equation. For the resulting pH, the assumed simplifying conditions must be checked for validity, otherwise the approximation has to be rejected.
For convenient solving of the equation set, it is advised to formally replace [] variables by the single-letter symbols,
like $x$,$y$,$z$,$t$,$u$,$v$ for variables
and $a$,$b$,$c$,$d$ for given parameters.
In simple or simplified enough scenarios, one may consider in some cases just 1 acid, if acidity of both acids is very different. But even then, formulas to directly calculate the approximated pH value are derived from the original equation for acid dissociation constant.

Thanks, and if there was also a basic compound in the solution it would still be something along these lines , just adding further complexity. What might that precisely look like ?

See the updated answer.


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Dissolution of portlandite, calcite/CaCo3 phase and C-S-H due to high pH
I would like to ask about the possibility of the partial dissolution, over curing time, of portlandite, CaCo3 and C-S-H phases due to high pH (12.5 - 13) of the lime-clay-water systems (lime-stabilized clay). These phases were already newly formed phases in the lime-stabilized  bentonite formed in the short term (1 day). I noticed a decrease in their amounts over curing time.

CaCo3 is a hypothetic stoichiometric inter-metallic compound of calcium and cobalt.  // Consistent misuse of proper element symbols is frown upon very much.

A capital crime!

To clarify: after 1 day, the compounds were quantified (how?), and cured (how long?). Were new phases (new materials) identified, or were the first very crystalline products growing to amorphous clusters - lower but broader peaks?


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Do hydration (as in "solvation") and "hydration" of salts refer to the same phenomenon?
I've come across two use cases of the word "hydration", presented to me as distinct topics:

Solvation is the process of reorganizing solvent and solute molecules into solvation complexes and involves bond formation, hydrogen bonding, and van der Waals forces. Solvation of a solute by water is called hydration. And, the solute ion/molecule around which this hydration shell is formed is said to be hydrated.

(Image source)

Water of crystallization are the molecules of water which remain trapped within the lattice of a substance. The salts containing water of crystallisation are said to be hydrated.

Is it safe to assume that the two instances of the word refer to the same phenomenon?

see https://chemistry.stackexchange.com/questions/544/how-do-copper-sulphate-and-the-like-change-crystal-structure-on-adding-water-of

What is the source of the second illustration, and what actually is depicted? These are channels (the host structure, depicted with the red spheres) loaded with guest compounds as loading (blue wire models). The guest's position/location in the host, depending on the supramolecular interactions, sometimes is statistical, sometimes random. Yet the loading/discharge of guests into such a host is much more similar to adsorption/desorption (e.g. in zeolithes, MOFs), than crystallization from a solution/dissolution of a solid (e.g., formation of $\ce{CuSO4\cdot 5 H2O}$).

Water of crystallization can he inherited from the hydration sphere, especially around the cation, but it is also possible to have water molecules incorporated into other parts of the crystal structure. Hydrogen bonding from the hydrogen atoms in the water molecules to oxygen atoms in oxyanions or other water molecules is one such bonding mechanism.
Copper sulfate pentahydrate, $\ce{CuSO4 * 5 H2O}$, offers a relatively well-known example. Ruggiero et al. [1] gives an experimentally determined structure in which the copper atoms are coordinated to four of the water molecules as they would be in solution, but the fifth water molecule in the formula unit is hydrogen-bonded as shown in the illustration below (from 1; teal lines added to show the hydrogen bonds).

Reference

Ruggiero, M.; Erba, A.; Orlando, R.; Korter, T. "Origins of Contrasting Copper Coordination Geometries in Crystalline Copper Sulfate Pentahydrate." Phys. Chem. Chem. Phys. 2015, 17, 31023-31029 doi: 10.1039/C5CP05554G.


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Reduction of gingerol content in ginger using kitchen equipment
The single topic that this question focuses on is - how to remove gingerol and only gingerol from a piece of ginger in a manner that it remains safe to eat using equipment found in a well stocked domestic kitchen.
I found Ref (1) which adds information to the question suggesting that several hours above the boiling point of water is required:

After 3 h of heat treatment at 75, 100, and 125°C, the gingerol's content was reduced by 6.8, 12.6, and 42.7%, respectively, and the shogaol's content was increased by 2.95, 4.85, and 9.34 fold, respectively. After 3 h of heat treatment at 150°C, the gingerol's content was dramatically reduced by 89.5% and the increased shogaol's content was even lower than treatment at 125°C.

(See also this article)
Below I give some examples of things that I have tried.
I am trying to find out how to make preserved ginger without pungency.I assume the main thing is to break down the gingerol into safe products. Or extract it out into a fluid that can be discarded - without removing the other tastes. Experiments suggest that boiling in lemon juice has some effect. I suspect that the acidity is the operative principle. I have tried acetic acid - eg. white wine vinegar. Success is partial. Perhaps it should be heated under pressure or for a rather long time. And research suggests sodium chloride, sodium hydroxide, or potassium hydroxide.
What is the best, most effective and most safe, way to remove the pungency?
References
(1) Ho, S.-C.; Su, M.-S. Optimized Heat Treatment Enhances the Anti-Inflammatory Capacity of Ginger. International Journal of Food Properties 2016, 19 (8), 1884–1898. https://doi.org/10.1080/10942912.2015.1084633.

I rather doubt you can selectively remove gingerol and leave other flavouring compounds intact. From the structure gingerol is likely to be soluble in organics such as acetone, ethyl acetate and ether.

@Waylander given that it used to be very common to do so - I suspect that it is possible. Although I wonder that there might be multiple kinds of ginger and that this will effect the outcome. I will keep the options you suggest in mind, but they don't sound food safe. Would you say they are?

It's too broad not because of asking many questions, but because of not having a definite answer. This is even borderline off-topic. Could be considered "food chemistry", but food processing is just cooking, so ask a cook how to prepare it to be more tasty.

@PonderStibbons  Acetone is used as nail varnish remover and occurs within the body

@Mithoron So, would a question about how to remove caffeine from coffee beans be an appropriate question here? Or would this be "cooking"?

@Waylander Acetone is absolutely not applicable to food. Is even withdrawn from use in  nail varnish removers.

@BuckThorn I somehow doubt OP has access to supercritical extraction gear. There are reasons why common "chemical" solvents are not used in food processing. If you know an actual method of processing of ginger that can be used for small amounts at home, that go on and reopen.

@Mithoron  Acetone is still widely sold in the UK for nail varnish removal by major retailers, example   https://www.boots.com/boots-nail-polish-remover-200ml-10305366

@Mithoron The OP should be more clear about the scope of the problem, in other words. Hexane is still used in food processing: https://pubmed.ncbi.nlm.nih.gov/36360023/ (and no, I do not recommend use at home, even if managing to obtain some).

Here is a good start for learning about ginger: https://www.compoundchem.com/2014/11/27/

@Mithoron I am definitely tackling this as a chemistry question and I did tag it food-chemistry which is a tag that already existed. You seem to be saying it is off topic rather than unfocused - that is that a chemist here could not answer it. My reason for asking here is that it is not about making food more tasty and the members at stack exchange cookery could not answer it. It is a request for information that an industrial food chemist would have. It is the sort of question Nile Red would tackle. I believe he would be said to be a chemist.

@Mithoron I have added a statement that it is about processing using domestic equipment. However, knowledge of an industrial process would be useful to me anyway.

@Karsten Thanks. Mostly that was stuff I already know. But I was interested in this "Cooking produces zingerone, which is less pungent, and is characteristic of the ginger flavour found in gingerbread." which at least acknowledges what I am looking for. But, I have cooked ginger for an hour in various solutions - and it still did not remove the "fresh ginger" taste.

Are you using old or young ginger?

@Karsten since it is fibrous, it is likely older ginger. Of course, I have a long list of options to try. I am dubious that the answer is young ginger - as the taste I am looking for is more earthy and complex, a clove or cinnamon aromatic spice (that is not hot "spicey") taste. It is a complex and almost earthy taste of gingerbread, not the taste of ginger you get with sushi. But, I am open to suggestions. Especially since the product I know of is not fibrous. However boiling in lemon juice did reduce the fibre.

@Karsten "Cooking ginger transforms gingerol via a reverse aldol reaction into zingerone". It appears that what I am looking for is a high component of zingerone, which is produced by heat treating gingerol - but you need several hours above the boiling point of water (as atmospheric pressure) to convert enough of the gingerol to zingerone.

@mithron were you involved in the closing of the question? I just ask because I would like to be able to argue that it should be opened again. I know have enough information to answer the question myself. It is not a complicated indeterminate issue - it is a very precise and specific one. And the answer is - heat treatment above 150 degrees Celsius for several hours.

related post in meta: https://chemistry.meta.stackexchange.com/questions/5296/when-should-i-vote-to-close

Question is still open-ended and "heat treatment above 150 degrees Celsius for several hours" of food is hardly a good idea.

Maybe an enzymatic path would make more sense.  Aldolases exist, but I don’t know about them in a kitchen scenario (different from invertases, which are available in food grade).


text:
Under which conditions is Enthalpy equivalent to heat, and why can an entity (Enthalpy) be used as equal to a transfer process (heat)?
What I know: under constant pressure and assuming all work done is expansion work, enthalpy is equivalent to heat using the formula for its definition.
However, non-expansion work done during electrochemical processes alter the enthalpy value according to the equation ΔH-TΔS=-nFE, where the voltage provides the non-expansion work if I'm correct.
Under this condition, is the change in enthalpy still equivalent to the amount of heat transferred? If not, what is the physical significance of ΔH in this scenario?

Replace isobaric condition by isochoric and enthalpy by internal energy. Change of internal energy is equal to accepted heat only with zero non-volume work.  Similarly for enthalpy and isobaric process.

Yes, you are right. When there is non-pV work the equivalence fails to apply.

Heat is the transfer of thermal energy. Enthalpy is a property of a sample or system. So they are never equivalent.
The change in enthalpy is equal to the heat transfer if you have a closed system at constant pressure, and the only work done or done to the system is pressure-volume work. This follows, with a couple of steps of derivation, from the conservation of energy.

Heat content is the sum of the internal KE of the molecules+ any latent heats. It is not the process of heat transfer. The internal energy U +PV is a state function tht has been given the name enthalpy H. Its virtue is that if the only work done is PV work Q, the change in heat content, now is a state function. This allows calorimetric measurements at constant pressure to be used in calculating other state functions.


text:
Calculating equilibrium constant for a general acid base reaction
I am trying to find the equilibrium value for different acid-base reactions. I have understood that when an acid and base are in an aqueous solution, many different reactions will occur. By combining these reactions, and multiplying their equilibrium constants, we can find the equilibrium value for an acid-base reaction.
Example 1: $NH_3$ and $HCl$
To illustrate my question I will take an example: if $NH_3$ and $HCl$ are in an aqueous solution, what reactions will occur? I believe the following reactions are those that will occur:
$HCl + H_2O ⇌ Cl^- + H_3O^+$, $K_1 = K_a(HCl) = 1.3 * 10^6$
$Cl^- + H_2O ⇌ HCl + OH^-$, $K_2 = K_b(Cl^-) = K_w/K_a(HCl)$
$NH_3 + H_2O ⇌ NH_4^+ + OH^-$, $K_3 = K_b(NH_3) = 1.8 * 10^{-5}$
$NH_4^+ + H_2O ⇌ NH_3 + H_3O^+$, $K_4 = K_a(NH_4^+) = K_w/K_b(NH_3)$
$2H_2O ⇌ H_3O^+ + OH^-$, $K_5 = K_w = 10^{-14}$ (the temperature is 25 degrees Celsius)
These are all the reactions that will occur in this solution, right? I noticed that adding all these will be of no use since the combined reaction will be the autoprotolysis of water. Instead, we only add reactions 1 and 3. That gives:
$HCl + NH_3 + 2H_2O ⇌ Cl^- + NH_4^+ + H_3O^+ + OH^-$, $K_6 = K_1 * K_3 = K_a(HCl) * K_b(NH_3)$
We can subtract the autoprotolysis of water from this reaction to get:
$HCl + NH_3 ⇌ Cl^- + NH_4^+$, $K_7 = K_6 / K_w = K_a(HCl) * K_b(NH_3) / K_w = 1.3 * 10^6 * 1.8 * 10^{-5} * 10^{14} = 2.34 * 10^{15}$
So the equilibrium constant for the reaction between $HCl$ and $NH_3$ is $2.34 * 10^{15}$; in other words, it essentially goes to completion. Is this correct?
Example 2: $HCl$ and $NaOH
Another example is the neutralization reaction between $HCl$ and $NaOH$. How should one deal with $NaOH$? Here is my attempt. The reactions that occur are:
$ HCl + H_2O ⇌ Cl^- + H_3O^+ $, $K_8 = 1.3 * 10^6$
$ NaOH → Na^+ + OH^- $, $K_9$, what should $K_9$ be?
In addition to these, $Cl^- + H_2O$ and the autoprotolysis of water will occur. Similarly to the first example, these reactions will be of no use to find the equilibrium constant for the neutralization reaction.
Adding reactions 8 and 9 gives:
$HCl + NaOH + H_2O ⇌ Cl^- + OH^- + Na^+ + H_3O^+$, $K_{10} = K_8 * K_9$
Can we here subtract the autoprotolysis of water as in the first example (since there is no "2" in front of "H_2O" on the left side of the reaction)? If we can do that then the reaction becomes:
$HCl + NaOH ⇌ Cl^- + Na^+$, $K_{11} = K_8 * K_9 / K_w$
I know that reaction 11 is unbalanced. $K_{11}$ is very large since $K_a(HCl)$ is very large and $K_9$ also is very large. So the reaction will essentially go to completion. However, it feels like something is wrong because reaction 11 is not balanced. What have I done wrong?
Summary
My goal is to understand the underlying theory of acid-base reactions and equilibrium. I have here presented in these two examples what I understand. What am I misunderstanding about how to calculate the equilibrium value for general acid-base reactions?
Thank you in advance!

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

@Poutnik, thank you so much for your answer. I really appreciate it! Based on what you wrote, I should ignore that reaction between $NH_3$ and $H_2O$  (in my example nr 1) because it is negligible? Nevertheless, I want to gain a deep understanding of the topic, and thus want to consider all reactions involved and calculate the equilibrium value properly. I also understand that much bigger errors arise due to things outside of acid-base reactions, but that I am of course OK with that since it will not prevent me from understanding acid-base reactions properly.

I fully agree with that rule. Have I done a mistake in my question? If I have, although I don’t find it, I’m sorry. I will correct it immediately if you point it out. However, could you please help me with my original question. :)

Check the last 2 equations.

Thanks! I have updated reaction 10 (the next to the last one). As I wrote, the last equation is unbalanced, and that is why I feel that my approach is wrong (as I wrote in the original question).

The rule No 0: Learn to enumerate chemical reactions. The prerequisite is the knowledge of arithmetic of small natural numbers. Total charges and total atom counts must be equal on both reaction sites. No atom is created nor destroyed during any chemical reaction.
The rule No 1: Simplify everything where simplification brings negligible error, otherwise you often end with too complicated math. Considering strong acids as HCl or strong bases as NaOH as completely dissociated brings very negligible error ( with much bigger ignored error due activity versus concentration).
The rule No 2: After the calculation, verify if conditions for the simplification are met. If they are not, you have to reject simplification.
E.g., we want to derive an equation of pH of diluted solution of weak acid. The dissociation constant for a weak acid is:
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$
If we assume $\ce{H+}$ from water autodissociation is negligible, compared to  the acid dissociation ($[\ce{H+}] \gg [\ce{OH-}]$), we can simplify it using  $[\ce{H+}] \approx [\ce{A-}]$ to:
$$K_\mathrm{a} = \frac{[\ce{H+}]^2}{[\ce{HA}]}$$
If we then also assume near all acid is not dissociated ($c \approx [\ce{HA}]$), we can simplify it to:
$$K_\mathrm{a} = \frac{[\ce{H+}]^2}{c}$$
leading to
$$c \gg [\ce{H+}] \gg [\ce{OH-}] \implies \mathrm{pH} = \frac 12(\mathrm{p}K_\mathrm{a} - \log{c})$$
The rule No 3: Any respective conjugate pair acid/base (like $\ce{NH4+}$/$\ce{NH3}$) is in water solutions in first place in equilibrium with $\ce{H2O}$/$\ce{H+(aq)}$/$\ce{OH-(aq)}$ respectively. Any equilibrium with other acid/base system is indirect, dependent and redundant.
The rule No 4: Any acid-base reaction in water is generally reaction of some acid passing a proton to some base, the former becoming a base and vice versa.
$$\ce{\mathrm{acid}_1 + \mathrm{base}_2 <=>  \mathrm{base}_1 + \mathrm{acid}_2}$$

How to calculate the equilibrium constant for the reaction between $\ce{H2O}$ and $\ce{NH3}$ (in my example nr 1) and between $\ce{HCl}$ and $\ce{NaOH}$ (in my example nr 2). Could you please show me how you would do that using your method (if my method is wrong?).

Example No 1: The equilibrium constant is experimentally determined for NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq).
You have already listed it in your question as $K_\mathrm{b} = \pu{1.8E-5}$.
Example No 2: It is equilibrium $\ce{H2O <=> H+(aq) + OH-(aq)}$ with $K_\mathrm{w} = \pu{e-14}$ at $\pu{25 ^\circ C}$, as HCl and NaOH are in water (with negligible error) fully dissociated. $\ce{Na+}$ and $\ce{Cl-}$ ions are often called spectator/bystander ions that do not take part in the reaction.

Thank you very much for your response! I agree with what you say. However, my original question is how to calculate the equilibrium constant for the reaction between $H_2O$ and $NH_3$ (in my example nr 1) and between $HCl$ and $NaOH$ (in my example nr 2). Could you please show me how you would do that using your method (if my method is wrong?)

See the answer update.

Ok. Thanks for that! Sorry, I meant the reaction between $HCl$ and $NH_3$ and not between $H_20$ and $NH_3$ (my example no 1 in the question). Thank you in advance!

It is trivial,  Ka=Kw / Kb // K=[NH4+]/([NH3][H+]) = 1/Ka = Kb/Kw. // Learn to ignore bystanders.

I don't understand what you are saying. Do you mean that the $H^+$ and $Cl^-$ ions are bystanders so we should ignore them from the reaction? If that were the case, the equilibrium constant for the reaction between $HCl + NH_3$ would be the same as that between $NH_3 + H_2O$ which is $1.8 * 10^{-5}$. However, that is not true, since the reaction between $HCl + NH_3$ essentially goes to completion. Sorry for misunderstanding you!

No, I do not say that. think longer.

Sorry, but I don't get it. And I have been thinking about this for a while now. What are the bystander ions in the following reaction you are talking about: $HCl + NH_3 ⇌ Cl^- + NH_4^+$. I cannot find any.

Cl- is bystander.. As it is safe to assume there is no HCl in H2O.

It seems you need rather systematic topic study than asking on Q/A site, which just fills the gaps. /
Look at 1/ education focused site: chem.libretexts.org   2/ Chemistry cheat sheets on Hyperphysics 3/  Learning resources recommended by Chemistry SE


text:
Chemistry project on hand warmers: Salt hydrates
Currently I am working on a chemistry project about hand warmers. My goal is to find a salt hydrate which gets warmer than sodium acetate trihydrate and can supercool. I already bought magnesium nitrate hexahydrate and aluminium ammonium sulfate dodecahydrate. I have tried to supercool them but they always crystallised so far.
Do you know if it is possible to supercool them? Can I mix two of the mentioned salt hydrates? For example, sodium acetate trihydrate and magnesium nitrate hexahydrate?

Have you tried pentahydrate of sodium thiosulfate ? It was used frequently in past as the main component of fixing solution in B/W photography. In my experience, it supercooled easily and was not too sensitive until seed was introduced.

And it makes spiffy crystals when viewed under crossed polarizers! https://www.travishale.com/sodium-thiosulfate-polarized-light-microscopy/ You might even include your own photos in your project


text:
Eutectic plot - freezing point lowering
I have a doubt about the correlation between eutectic plots and freezing point lowering.
The freezing point lowering, as well as all colligative properties, is defined in the presence of a volatile solvent and a non volatile solute. In fact, by adding NaCl to water, a freezing point lowering of the resulting solution is obtained, which follows this trend:

However, a similar trend is also obtained by adding ethanol (solute) to water:

But ethanol is volatile, so it shouldn't be referring to freezing point lowering.
So, is it correct to talk about freezing point lowering in the context of these diagrams? Or do we speak about freezing point lowering only in reference to the phase diagram of a single substance, like this?

Water is volatile as well as ethanol, so I don’t understand your issue here.

@JonCuster essentially I made a distinction between the two cases:

Volatile solvent (water) + non-volatile solute (salt)
Volatile solvent (water) + volatile solute (ethanol)

Therefore, in the first plot, that trend is given by the colligative property (freezing point lowering) because we have a non-volatile solute.
Does the second graph always refer to freezing point lowering even if the solute is volatile?

or more generally, is there never a correlation between freezing point lowering and eutectic plots? So the cryoscopic lowering can be identified only in the third plot, i.e. in that of a single substance?

https://chemistry.stackexchange.com/questions/138726/do-we-have-some-rule-of-thumb-to-estimate-the-miscibility-of-metals/138742#138742 Might be of interest.

@JonCuster Thanks for the reply. I tried to take a look but it doesn't seem to answer my doubt.

btw to be more exact, its not adding salt to water but adding salt to ice at its freezing point that causes freezing point lowering.