text:
Why is the equilibrium constant unitless?
I read that the equilibrium constant is unitless because the molar activities of each of the species are used in the equilibrium expression, not the actual concentrations themselves. I understand that.
But it is taught that
$$K_c = \frac{k_\mathrm{fwd}}{k_\mathrm{rev}}$$
I’m having some difficulty understanding this equation. Consider the chemical equation
$$\ce{A + B <=>[$k_\mathrm{fwd}$][$k_\mathrm{rev}$] C}$$
The forward rate constant is a second order reaction and has units $\pu{M-1 s-1},$ and the reverse rate constant is a first order reaction and has units $\pu{s-1}.$ Therefore
$$\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}$$
has units $\pu{M-1},$ but it’s supposed to be unitless. Where am I going wrong?

What they don't tell you is these are not just molar activities of the species. It is activities divided by the activities at standard conditions. The latter are numerically 1, but they eat up the dimension.

@Ivan Surely you meant something else by "molar activity", as activity as such is already dimensionless (or with dimension 1) from the start.

Also strongly related (among many others): https://chemistry.stackexchange.com/questions/85529/what-is-the-correct-unit-for-the-equilibrium-constant/85532#85532, and https://chemistry.stackexchange.com/questions/5811/unit-of-the-equilibrium-constant-contradiction-of-bridgmans-theorem/104004

There are a few separate issues here to keep in mind:

$K_c$ (the equilibrium constant in terms of concentrations) is defined as
$$K_c = \prod {c_i}^{\nu_i} \tag{1}$$
Agreement between $K_c$ and the product of forward/back rate constants ($k_{\pu{fwd}}/k_{\pu{rev}}$ in the OP example) is expected only if the mechanism is correct, assuming some proportional rate law, and at equilibrium.
When dealing with gases, $K_c$ cannot generally be equated with $K_p$ (the equilibrium constant in terms of partial pressures). They may be numerically equal under some circumstances. For instance, they are numerically equal in reactions involving only ideal gases, provided the number of gaseous molecular entities remains constant during the reaction.
The (standard) equilibrium constant $K$, which simplifies to $K_a$ (under standard fugacity and some other conditions, agreements) which is defined as $$K_a = \prod a_i^{\nu_i}, \tag{2a}$$ cannot generally be equated with $K_c$. They may be numerically equal under some circumstances. $K$ itself is actually defined as follows:  $$\Delta_\pu{r} G^\circ=-RT\log(K) \tag{3}$$ whereas, strictly speaking, $K_c$ can't be used in this relation (but see below). The equilibrium constant $K$ cannot have units or this equation will not make sense.  In terms of concentrations,$$K_a = \prod \left(\frac{\gamma_ic_i}{c_i^\circ}\right)^{\nu_i} \tag{2b}$$ (To be nitpicky, one should use molality-based activities instead of amount concentration-based $c$ when saying that $K$ reduces to $K_a$).  When activity coefficients are all unity (ideal conditions), it might be that $$K_a \approx \frac{K_c}{K_c^\circ}= \prod \left(\frac{c_i}{c_i^\circ}\right)^{\nu_i} \tag{2c}$$ Strictly speaking, the equilibrium constant $K$ should remain unchanged with a change in the units used to describe concentrations. Units have no place in the constant because they cancel when ratios to reference standard states are computed (see Eq. (2)). $K_c$ and $K$ differ by the product  $$K_c^\circ = \prod ({c_i^\circ})^{\nu_i}\tag{2d}$$ (ignoring, once again, activity coefficients; standard fugacity). This difference is the source of the dangling units leading to the question in the OP.

You should therefore be careful equating $K_c$ with the ratio $k_{\pu{fwd}}/k_{\pu{rev}}$. And, assuming they are equal, the concentration equilibrium constant cannot generally be equated with $K$. However you can divide $K_c$ by $K_c^\circ$, and, assuming activity coefficients are unity, the result might be the thermodynamically correct unitless equilibrium constant (if standard fugacity), equivalent to that which appears in usual thermodynamic derivations.
Sloppy or not, associating units with an equilibrium constant is common practice. Attaching units to the equilibrium constant during reporting is a useful mnemonic when performing some computations, but it should be kept in mind that such a constant may not fulfill the requirements for computation of an associated standard free energy change, unless first modified by a ratio product of the standard concentrations.
Note that some of this is also explained in an answer to this post. IUPAC likes to use the convention standard equilibrium constant when referring to the thermodynamic equilibrium constant.

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This is a complicated question among chemical educationists- with tons of arguments over arguments. Just search do equilibrium constant have units? I would say the "units" of K are in a grey area (just like in any real world science.). You should read the nice section on Dimensionality of the Equilibrium Constant, and I will just give a background:
Now, after ages, I feel that there are mathematical benefits of sticking to the formalism of unit-less equilibrium constants. For example, you need to take logarithms of K. Quantities in transcendental functions (log, ln, e, sin, cos) cannot have units, just like pH cannot have units. It is then beneficial to keep these quantities dimensionless.
Yes, one simple argument is that we use the concept of activities and that do not have units. However, most of the calculations involving equilibria use concentrations. So the idea is to make the concentration of each species dimensionless by dividing it with unit concentration. Now, whatever the exponents be in the equilibrium constant expression, you don't worry about the units.
Regarding your query regarding rate constants, one may try to use the same argument of using dimensionless concentrations in the rate expressions. There we have logs and exponential functions too. Otherwise your dimensional analysis is correct. In general we can never write the rate expression or determine the order without an experimental support until and unless the reaction is a so-called elementary reaction.

If I use dimensionless concentrations in the rate expressions, then the units of the rate constant will always be molarity per second, which is not correct.

Your dimensional analysis is correct in the original post. If someone insists that we have to make K unitless then you make a formalism. You should make the whole rate equation dimensionless. When we raise e to the power t, where t is time, the time has to be made dimensionless. These type of functions, such as e^(-t^2) appear in making Gaussian peaks. Can you take natural log of concentration, where [C] has units of concentration? Nope. ln [C] also appears in kinetic equations. It cannot have units.


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Space groups: Herman-Maugin notation, diagrams and Wyckoff positions
I am currently taking an Introduction to Crystallography course. After studying and understanding the punctual symmetry groups (2mm, 4/m 2/m 2/m, 32,...), the flat groups and the diverse elements of symmetry that exist: helical axes, sliding planes, symmetry planes, among others.
I have doubts when building the diagram of a certain spatial group. For example, in an exercise, I am asked to construct the diagram of the spatial group Pmab.
Pmab implies the existence of:
 an ordinary plane of symmetry perpendicular to the crystallographic axis a,
 a sliding plane of type a perpendicular to the crystallographic axis b,
a sliding plane of type b perpendicular to the crystallographic axis c.
Also, Pmab is an abbreviated notation, because, in reality, there are also two helical binary axes parallel to the axes "a" and "b", and a binary axis parallel to the axis "c".
On the other hand, Pmab belongs to the orthorhombic (rhombic) system and derives from the point symmetry group 2/m 2/m 2/m 2/m.
This is all the information I get from the Herman-Maugin notation. If there is something wrong that I have understood, I hope you will tell me.
Once here I don't know how to start building the diagram, which would look like this:

As far as Wyckoff positions are concerned, I am often asked for them, but I am not quite sure what they are.

How do you know that the two binary axes along "a" and "b" are helical?

I know, because in the orthorhombic system, simplified notation tends to obviate them. But I don't understand why

That would explain how do you know there are axes in the first place. That's right, but that's not what I asked. How do you know they are helical?

I don't know, I said they're helical, because that's how it appears on the diagram.

Well, they are helical all right, but you need to know how to deduce that from "Pmab" alone.

Of course, it's one of the things I'd like to learn. It is fundamental in all of this.

Let us continue this discussion in chat.

@IvanNeretin Do you know the reason?

Well, knowing crystallography implies being able to deduce the picture above solely from the group's name $\rm Pmab$, and that's how it is done.
First you write down the symmetry elements mentioned explicitly in the group's name (that is, $\rm m,\;a$, and $\rm b$ in our case; in other examples there could have been axes among them, or even worse) and orient those accordingly, which you already know how to do. Then you combine these elements in all possible ways to find out what else is hidden beneath.
Essentially, any symmetry element is but a linear transformation that takes any vector $\bf\vec x$ to another vector $\bf A\vec x+\vec b$, where $\bf A$ is the $3\times3$ matrix of our transformation (rotation or reflection or otherwise), and $\bf\vec b$ is the shift, that is, a half-translation along the axis for a screw axis, or a half-translation in whatever direction for a glide plane, or $\bf\vec0$ in all other cases. The typical matrices are:
$$
\begin{array}{cc}
\begin{pmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix} &
\begin{pmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{pmatrix}&
\begin{pmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{pmatrix}\\
\text{Reflection $\perp$ X} & \text{Rotation about X}& \text{Center}
\end{array}
$$
Given two symmetry elements, you simply combine the two transformations by doing some matrix multiplication, which I sincerely recommend, if only to know how it feels. Alternatively, you may rely upon the wisdom of the predecessors, which says:

Two perpendicular planes give a twofold axis parallel to both.
A plane and a parallel twofold axis give another plane, perpendicular to the plane and parallel to the axis.
A plane and a perpendicular twofold axis give a center.
Two perpendicular twofold axes give a third axis, perpendicular to both.
All of the above remains true if you change

every "plane" to "plane or a glide plane" and
every "axis" to "axis or a screw axis".

When some of the combined elements have shift in them, the following happens:

If the shift is parallel to the resulting element, it becomes a part of the said element (that is, the result is going to be a screw axis or a glide plane).
If the shift is perpendicular to the resulting element, it moves the said element by half the value of the shift (that is, by $1\over4$ of a translation).

If you end up with a center of symmetry someplace other than at the origin, you relocate your origin there.

Now try to apply the rules 1-7 and see where this gets you.

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Decomposition of alkene in acidified potassium permanganate

A low boiling point liquid A, molecular formula $\ce{C5H10},$ reacts with chlorine to form B, molecular formula $\ce{C5H10Cl2},$ and with $\ce{HCl}$ (gaseous) to form substance C, molecular formula $\ce{C5H11Cl}.$ Compound A decomposes in acidifed potassium permanganate solution to form two acids D (molecular formula $\ce{C4H8O2}$) and E (molecular formula $\ce{CH2O2}$).
Find the systematic names of A, B, C, D, E.

I'd like to verify that my answers are correct and I would also like a good explanation of why the answers are what they are. My answers are:
A - pent-1-ene
B - 1,2-dichloropentane
C - 1-chloropentane (Minor), 2-chloropentane (Major)
D - butanoic acid
E - methanoic/formic acid
Currently, I doubt my answer because of C, where I got two possible compounds from applying Markovnikov's rule. Any help would be great.

gaseous hydrochloric acid or anhydrous,
we have an alkene where each side is attached to the same number of hydrogens —> both “equally substituted”.
Products in this case, 3-chloropentane and 2-chloropentane.

It wasn't specified in the question but since we're not going through anhydrous compounds, I'd say it's gaseous $HCl$. Why does your second statement have to hold true?

The alkene CANNOT be equally substituted if permanganate cleavage give formic acid. The OP's answers are correct

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text:
Are portable pen pH-meters accurate?
Are pen $\mathrm{pH}$-meters accurate? I know there are large table-top $\mathrm{pH}$-meters, but I keep seeing these small $\mathrm{pH}$-meters online that can read up to 0.1 units. My question is if these $\mathrm{pH}$-meters (look pretty cheap) are accurate or should I get the large table-top $\mathrm{pH}$-meter for my experiment (will need accuracy minimum of 1 decimal place).
(Edit: Is there also a minimum volume of sample I need before I can test it with a $\mathrm{pH}$-meter?)

I would say that it's too broad. Could you ask about a specific pH meter? Quality could vary across region, price and make..

I'd say they are as cheap as they look. But then again, a table-top pH-meter can also be wildly off unless you maintain it properly.

I've decided to get portable pH meter for 'Hanna Instruments' (https://hannainst.in/waterproof-portable-ph-meter-hi991001p.html) . It does seem like a reputable company with a quality product, compared to all the other pH meters I've seen online in the market.

Several well known instrumentation/measurement companies offer small hand-held pH meters with specified pH resolution of 0.01 - I see no reason to doubt that they fulfill those specifications.

I see. Thank you all for your inputs.

Is there also a minimum volume of sample I need before I can test it with a pH meter

Read the manual :-)

A pH meter is as accurate as it's pH electrode regular calibration by standard pH solutions. If it is not done, do not expect miracles.


text:
Would this experiment be electrolysis experiment be viable?
I plan to do a 'research project' or an experiment involving the electrolysis of dilute sulfuric acid.
The idea or 'question' is this:

Would there be a variation in the $\mathrm{pH}$ of the electrolyte at different
  points at different displacements from each electrode.

So I was wondering if $\mathrm{pH}$ in specific areas, such as around the cathode would differ from the $\mathrm{pH}$ at the exact center between the two electrodes.
One's initial assumption would be that the charged ions would be would be equally dispersed in the circuit path, but there is a reduction/oxidation of ions near the electrodes, in this case hydrogen and oxygen ions, which may mean that there would be some variation of their concentration in the cell.
So is my experiment worth doing, even if it might end up in failure, because I will need to buy a $\mathrm{pH}$ meter which my school doesn't have.
Thank you for your answers.

Litmus paper will be quite enough.

Thank you for your comment. Using litmus paper may be fine for preliminary results, but for this project, I plan to 'mathematically' interpret the results, to calculate how far a displacement from a cathode would give me x.x pH . So a pH meter might be useful in that sense.

You are aiming to measure something that can't be measured reliably. Liquids mix all the time. The displacement of such-and-such pH from cathode will fluctuate wildly, depending on the weather, any slightest vibration, and how long ago the solution was prepared. No, this is not viable.

Mathematical treatment in electrochemistry may be far more complicated than we can imagine. You would need diffusion equations, Fick's laws, and I believe what not. It is not a trivial mathematical problem.

Absolutely a good experiment!
Diffusion of ions, propelled by the electric field (i.e., current) will occur at a given, calculable rate. Deposition of those ions (and conversion to $\ce{O2 + H2}$) will occur and the $\mathrm{pH}$ will rise at the cathode (negative electrode) and fall at the anode (positive).
Then back diffusion will occur as the $\ce{OH-}$ and $\ce{H+}$ ions produced at the electrodes even out. The $\mathrm{pH}$ of the starting sulfuric acid solution should be ~3 (i.e., $\pu{1.0e-3 M}$), and the current should be ramped up high by using a higher than usual voltage (maybe 4-5 volts).
This will allow the cathode $\mathrm{pH}$ to shift by a whole $\mathrm{pH}$ or so, and the anode perhaps more (because it is closer to neutrality). The anode could even go alkaline! The center of the solution should have a fairly stable $\mathrm{pH}$.
The evolution of gas will cause considerable mixing at both electrodes, so it might help to surge the current for a few seconds, then shut it off to measure the $\mathrm{pH}$.
The problem with using an indicator is that you need to avoid having the whole solution being darker than one or the other electrodes, so you can't differentiate the color change. However, if you select an indicator which is colorless at the $\mathrm{pH}$ of the original solution, you might find one which turns color at the anode or cathode shift.


text:
Safe & Convenient method to hydrate CaO into Ca(OH)2?
I'd like to prepare a solution of $\ce{Ca(OH)2}$ from $\ce{CaO}$.
Ideally I guess I'd go for slowly adding the powder to a water-cooled round flask containing pure water but I have no mechanism to control the slow release of powder nor am I sure that the reaction wouldn't be fast enough to send the powder right back out of the flask.
Is the reaction slow enough for it to be safe and convenient to set pure water to slowly drip into the flask containing the powder (bathing in a large volume of water to dissipate the heat)? This would allow me to have my hands & mind free and control the flow to allow the heat to  slowly dissipate at the desired rate, if necessary I could change the water in the cooling bath once it reaches a certain temperature. This solution would appear more convenient though I have some concerns (if the reaction it too fast steam would pop $\ce{CaO}$ around in the flask, creating some airborne $\ce{CaO}$ coming out of it, I could do a closed system to work that around but if steam is produced I feel an open system would be safer). I'll wear protective equipment to set up everything, though I'd still rather not take unnecessary risks.
In the worse case I could hydrate a portion of $\ce{CaO}$ with steam in a cooled down space but that setup might be unnecessary.
Also, how can I calculate the amount of energy that will be released per quantity or weight from $\ce{CaO}$? (to know how much energy I'll need to dissipate)

I think you are worrying more than you should: CaO is not aggressive enough to keep you from experimenting a bit. Of course the concentration of the solution makes the difference, speaking of release of heat, but my suggestion is trying the operation on a small scale (with about 0.5 g of CaO and water scaled accordingly, for instance) to see how fast your mixing can be without releasing too much heat. If you find that heat is not a concern, or that it can be tamed by adding the base slowly, then you scale up

Remember that builders are ( or at least were ) used to perform this reaction in the scale of tens of kilograms of $\ce{CaO}$. General level of precaution for a lab scale is more than enough.


text:
Cyclic polyynes
Cyclic polyynes have been proposed as theoretical compounds to illustrate the “particle on a ring” model discussed in many quantum chemistry courses. I've looked it up on Wikipedia but there it's not clear whether one has ever been synthesised, or not.
So: have cyclic polyynes been synthesised? If not, what is the difficulty in achieving this?

This book Polyynes: Synthesis, Properties and Applications, F. Cataldo ed. has a chapter on cyclic polyynes (chapter 6 by Y. Tobe and T. Wakabayashi). This suggests that they have been prepared.


text:
What is the equation that describes the variation in concentration of an inhibitor in an heterogeneous catalysis?
I find myself in quite a pinch: I have to sustain a Chemical Physics exam, and I know for a fact that the professor often asks this particular question:

Suppose that we have both a reacting species and an inhibitor in a homogeneous solution, and we insert a solid catalyst in it. What is the equation that describes the variation of the inhibitor's concentration in this heterogeneous catalysis?

I know for a fact that four different types of flows are included (or at least the professor explained it that way): two "scalar" flows, identified by the consumption of reagent and the consumption of inhibitor, and two "vectorial" flows, identified by the creation of a gradient in concentration of the two species.
However, I can't seem to find or even build up a satisfying equation that describes it. Can someone help?

Maybe you mean Fick's laws?

Yes, that's for sure a component of the full equation, but my professor also stated that there's the change in the Michaelis-Menten equation with the (alpha) term in it. So I thought that I had to combine both the M-M and Fick, but I find myself a little lost on how to do so. Can you help me, please?

I suggest you write a set of reaction equations to guide the readers as to what the process in question is. How is the inhibitor involved, why should its concentration change (the free concentration?)...


text:
Reverse washing soda buildup on stainless steel
In a home brewing project, I use washing soda (Na2CO3) to remove labels and glue from previously used bottles. In doing this, my flip-top bottles (which have a stainless steel flip top arm) often get unruly buildup of residual soda, which tarnishes and blemishes the steel. Is there a chemical mixture I can use to reverse this process and return the steel to a near-new appearance?
Here are a couple of photos of the bottles and buildup: the left shows a typical amount of residue; the right shows my attempt to wipe it clean with white vinegar (which partially works, but is still blemished).

Sodium carbonate per se should be soluble in water. You could try washing with abundant hot water. If it doesn't work, the residue could be some trace amount of calcium carbonate. Anyway, diluted vinegar or lemon juice are food-safe acidic chemicals which help dissolving the carbonates (by protonating them to the more soluble bicarbonate)

Can’t figure out what the downvoter of the question found bad about it. Perfectly fine question, well written, chemistry related, OP even added some pictures to illustrate it, go figure. I know this SE has lots of bad, usually homework, questions, but when a good question or answer is written we should kind of encourage it right?

Eh, sometimes people downvote just because they don’t like it. Not the way SE is supposed to work, but ♂

If you use tap-water together with Sodiumcarbonate, chances are that the residue is Calciumcarbonate (unless your tap-water is very soft).
Calciumcarbonate is significantly less soluble in water, but can be dissolved in acid. I would try to put a bottle/arm in a hot 5% solution of acid (acetic acid is fine, personally I would prefer citric acid) to see if any (small) gas-bubbles are formed at the residues. If it does, it is carbonate and can be removed in an acid bath.
If not, the residues could be glue from the labels, and getting rid of this is probably more complicated.

I’ve tried with acetic acid, but still doesn’t work very well. I’ll try citric and see if that works (I’ll have to get some). Thanks!

I don't know the strength of your acetic acid solution, but given time it should easily be able to dissolve carbonates. Citric acid has some advantages: it is a stronger acid and has some complexing and emulsifying propertis that help in dissolving residues other than pure carbonate. But be aware: citric acid (powder) often contains spores from Aspergillus that may cause mould in your product. The last step in the cleaning process should always be a rinse with pure water.


text:
Proust's law/Laws of definite proportion
As per laws of constant proportion/Proust's law,Ratio of elements by mass is always constant.
But in case of
H2+O2 →H2O
Here when the chemists might have measured H2 and O2 individually they might have get a total mass of 2g+32g=34g
And on measuring the product that is H2O they might get a mass of 2g+16g=18g
I know its not balance but still as you know whenever chemist will measure individually they will get the same mass as above.Hence,how the chemists mainly Proust might have defined this law?

You have silently dropped one oxygen atom in your chemical formula. That cannot work.

But Proust might have measured individually and might have got oxygen as 32g

Yes of course! One mol of oxygen is 25 l at STP is 32 g. Two mol hydrogen is 50 l and 4 g. Together they become 2mol, 34 g of water, which take up 50l as a gas.

2 mol of water are of course 36 g.

If you take the wrong amounts of each, you get not a pure product but some mixture with the educt!
The law of constant proportions says exactly that: Whenever you do that reaction with hydrogen and oxygen properly, you always get a volume ratio of (hydrogen:oxygen:water gas) 2:1:2, and a mass ratio of 1:8:9.

Your chemical equation is not balanced, you just dropped one oxygen atom! First balance it! It should be:
$\ce{2 H2 + O2 -> 2 H2O}$
Then, you will get the correct masses.


text:
Chilling water in copper vessel

Does chilling drinking water in copper bottle result in any chemical reaction that turns it dangerous for consumption in any way?

Few days back, in summer, I carried kokum juice in copper bottle. I had a sip of it after some time when I reached office and I felt super dizzy. I felt that traveling might have an effect on me, so I stopped. I had another sip of juice after some hours I had a strong dizzy feeling. At first, the juice tasted ok but then it started tasting very very weird. When I reached home and poured the juice back out of the bottle, to my surprise, its green! The normal color of kokum juice is pinkish-reddish. So somehow copper reacted with the juice and possible resulted in some acid, even though it was at normal room temperature (or in fact cold one as the juice was prepared from cold water). I believe there shouldn't be an issue with cooling water in a copper bottle in the fridge, I just want to know if there is any slight drawback, maybe from Ayurveda point of view? Also, I hope you get the chemistry angle to this question.

Kokam is sour (probably acidic; something to do with the reaction). Water is neutral.

Are you asking about storing water in a copper vessel or storing a (likely weakly acidic) juice in a copper vessel?  The two are not equivalent.

The title ( water ) does not match the post ( juice ).

Title earlier was (mistakenly) "juice" (since I had juice experience in my mind right from the starting of writing the question as wanted to tell you that). Now I have already changed it to "water". Since beginning, the post was always: "will it do any harm with water? I know its harmful with juice at normal temperature, but want to know what happens with water at lower temperature." Anyways, responses here are super enlightening. Thanks for all the knowledge. Dont know which answer to accept.

Then the proper way would be, at the beginning, to remove minor parts with water , putting them to a different question.

Do not do it !! ( putting acidic, or rather any juice to copper bottles )
You are in danger of copper poisoning.
Generally, by food processing laws, copper is not allowed to be in direct contact with food, as there is danger of copper contamination.
Especially acidic liquids, like vinegar or citrus juices, directly slowly dissolve copper in presence of oxygen.
$$\ce{4 RCOOH + 2 Cu + O2 -> 2 (RCOO)2Cu + 2 H2O}$$
Note that cold liquids contain usually more oxygen than warm ones.
For similar reasons, even if there is not copper, it is very bad idea to keep the content of conserved fruit cans in the already  opened cans. You can otherwise see the juice corroding the can inner coating in air presence.
Soluble copper salts are usually blueish/greenish what with yellow/orange colour of juice gives green colour.
Long term usage of copper for keeping water is on the edge. It can cause chronical low dose sneaky poisoning, especially if water is treated by chlorine.
$$\ce{Cu + Cl2 -> Cu^2+ + 2 Cl-}$$
Oxidation can lead to partially soluble basic copper carbonates, soluble by traces of organic acids, related to minerals malachite and azurite.
$$\ce{2 Cu + CO2 + O2 + H2O -> CuCO3 \cdot Cu(OH)2}$$
$$\ce{3 Cu + 2 CO2 + 3/2 O2 + H2O -> 2 CuCO3 \cdot Cu(OH)2}$$
Plus, there is danger some people would want to store there other liquids than water as well, as the OP.

Biochemistry discourages copper for long term use.

Yesss that link also discourages copper for long term use and asks to give a pause of a month after three months of usage.

Long term usage can cause chronical low dose sneaky poisoning, but if Ayurveda recommends it....

There are other many benefits of copper water as suggested here, apart from destroying bacteria (like stimulates brain function, promotes digestion, improves fertility etc. I will definitely love to know the detailed analysis of those benefits and also hazards.) However, even that article asks to not to overdo it. Seems that we were over doing it for decades, drinking mostly from copper vessel. Now I may replace my plastic bottles, with some steel and some copper bottles.

Also water for keeping in copper should be free from chlorine or other chlorinating compounds, as these stimulate copper dissolution.

Are you saying that the copper pipes in my water service are "sneaky poisoning" me?

@Mahesha99 The more unrelated benefits that something is claimed to have, the more skeptical you should be about the claims. Anything claimed to "stimulate brain function, promote digestion, improve fertility" is almost certainly snake oil and, in this, case probably actively bad for you.

This is in reply to the following comment.

but if Ayurveda recommends it.... – Poutnik

The following abstract is from Journal Health Popul Nutr. 2012 Mar; 30(1): 17–21."Storing Drinking-water in Copper pots Kills Contaminating Diarrhoeagenic Bacteria" by V.B. Preethi Sudha, Sheeba Ganesan,G.P. Pazhani, T. Ramamurthy, G.B. Nair and Padma Venkatasubramanian.

Microbially-unsafe water is still a major concern in most developing countries. Although many water-purification methods exist, these are expensive and beyond the reach of many people, especially in rural areas. Ayurveda recommends the use of copper for storing drinking-water. Therefore, the objective of this study was to evaluate the effect of copper pot on microbially-contaminated drinking-water**. The antibacterial effect of copper pot against important diarrhoeagenic bacteria, including Vibrio cholerae O1, Shigella flexneri 2a, enterotoxigenic Escherichia coli, enteropathogenic E. coli, Salmonella enterica Typhi, and Salmonella Paratyphi is reported. When drinking-water (pH 7.83±0.4; source: ground) was contaminated with 500 CFU/mL of the above bacteria and stored in copper pots for 16 hours at room temperature, no bacteria could be recovered on the culture medium. Recovery failed even after resuscitation in enrichment broth, followed by plating on selective media, indicating loss of culturability. This is the first report on the effect of copper on S. flexneri 2a, enteropathogenic E. coli, and Salmonella Paratyphi. After 16 hours, there was a slight increase in the pH of water from 7.83 to 7.93 in the copper pots while the other physicochemical parameters remained unchanged. Copper content (177±16 ppb) in water stored in copper pots was well within the permissible limits of the World Health Organization. Copper holds promise as a point-of-use solution for microbial purification of drinking-water, especially in developing countries.

The paper has tabulated results of Antibacterial activity of copper pot and also Physical and chemical parameters in a copper pot as shown below.

Antibacterial activity of copper pot on drinking-water inoculated with enteric pathogens.
The level of copper that had leached into the test samples was 177±16 ppb which was well within the WHO limit of 2000 ppb in the test samples (copper pot) after incubation for 16 hours .

Discussion as per the paper.

None of the test pathogens was recovered from drinking-water stored in copper pots even after enrichment culture.
The level of copper leached in the former is far less (177±16 ppb) than that in distilled water (~420 ppb).
Safety of leached copper does not appear to be an issue since studies have shown that the current WHO guideline of 2 mg Cu/L is safe , and the levels leached in the study were 1/20th of the permissible limits.
It has been observed in the present study that the other physicochemical parameters of drinking-water remain unchanged after copper intervention, which makes them amenable for public use.

Since ancient times Ayurveda has advocated the benefits of drinking water from a copper vessel.
Scientifically speaking$^1$,

when water is stored in a copper vessel for over eight hours, very small quantities of copper get dissolved in this water. This process is called “oligodynamic effect” and has the ability to destroy a wide range of harmful microbes, molds, fungi etc. due to the toxic effect it has on living cells.

The oligodynamic effect $^2$,

(from Greek oligos "few", and dynamis "force") is a biocidal effect of metals, especially heavy metals, that occurs even in low concentrations. The health effect was known in India for more than 2700 years as their ancient texts prescribe.

In conclusion,Ayurvedic system$^4$ of medicine has great antiquity, dating back to about 5000 years B.C. Its Materia Medica contain resources in the form of drugs derived from plant, animal, metal and mineral sources.Ayurveda system   prescribes dose in limits and not in excess.
References
$^1$https://saiayurvediccollege.com/ayurveda-benefits-of-drinking-from-a-copper-vessel/
$^2$https://en.wikipedia.org/wiki/Oligodynamic_effect
$^3$https://saiayurvediccollege.com/ayurveda-benefits-of-drinking-from-a-copper-vessel/
$^4$https://saiayurvediccollege.com/ayurveda-benefits-of-drinking-from-a-copper-vessel/

Not harmful even if I drink always from copper vessel? We were doing this for more than a decade. Now that our copper vessel is broken, we have stopped it. I am guessing what damage it might have caused us. Any history of any harm that copper might have caused over prolonged use?

I have not objected against copper antimicrobial activity. I am also well aware copper is essential element in small doses.

Note that medical advices are explicitly off topic on Chemistry SE.

The above answer is in reply  to effect of Cu content in water by AYURVEDA  wich talks about moderation as presented in a paper.That said ,doctors advice is the best.

This was a good answer until I got to the "Ayurvedic system" discussion. Copper deficiency is rare, so most people are unlikely to "reap the benefits". Any copper supplementation should probably be taken under consultation with a reputable doctor.

@Water Molecule AGREED COPPER Supplimentation must be under DOCTOR SUPERVISION.

"the FDA suggests that our body needs about 12mg of copper a day. This means that you can drink about two to three glasses of water from this copper vessel" - I do not think that body can ingest pure copper. It needs to be included in organic matter by plants.

@Vashu I'm not sure what you mean by pure copper, but the body can directly take up aqueous copper ions, which are produced when copper metal is exposed to acids in the presence of chloride ions. One of the proteins involved in CTR1 (https://en.wikipedia.org/wiki/High_affinity_copper_uptake_protein_1).  Also, simple salts of copper are highly bioavailable. "Copper salts, including chloride, acetate, sulfate, and carbonate,  are  highly  bioavailable  in  animal  nutrition  when  added  to feed." https://academic.oup.com/ajcn/article-abstract/67/5/1054S/4666227

@WaterMolecule thanks, looks like I was misinformed.

There are two parts of your question. We should not mix historical medical practices with modern scientific truths. Also, this does not mean that all old medical knowledge is bad or useless. In older medicine, gold foil arsenic compounds, silver foil, crushed pearls and mercury oxides  were added in medicines,.  It does not mean that if these practices were done in ages ago, they are safe by today's standards.

Is it okay to store kokum juice in copper vessels?

For western readers, kokum juice is a berry juice (Garcinia indica), which is rich in hydroxycitric acid.
Acids in Garcinia indica. The berry is quite sour. A layman can do a quick test. Take a copper vessel and store this berry juice for half an hour. Immediately you will see discoloration and a "brand" new shiny copper surface. I have seen that on a copper cup. This implies that a significant amount of copper has dissolved. We should not store acidic juices in copper vessels.
Secondly, in older times water was not chlorinated. Chloride ions corrode copper so quickly. Add some lemon juice, salt, to a copper cup and it will corrode soon to a green color.
However, it is very much true that in older South Asian culture, copper and tin lined pots and pans were used. The coating would wash away with cooking and people would re-coat them. Aluminum took their place. Finally stainless steel came in.
Regarding storing natural water in a copper vessel, it should be fine for a shorter time (day or so). Silver vessel is a better as an anti-bacterial surface and far more inert than copper. However, nobody is rich enough to use silver (or to be born with a silver spoon as they say). Yes silverware is good for the kings and queens.
Copper is a long term liver toxin. It is also a trace essential element.
You know the Goldilock principle: Everything in moderation. Too little copper in the body, bad. Too much copper-liver damage. You just need the right amount of copper.
.


text:
Waht is the reason the electrons cannot occupy p-orbitals of the K shell?
What is the ultimate reason, that the electrons cannot occupy the p-orbitals of the K shell?
Why for example, is there no element having electron configuration 1s2 1p3? And is it possible to excite the electron(s) of hydrogen or helium to the 1p-orbital(s)?

The reason is simple: 1p orbital is not a thing at all. Why there are sport exercises where you use the strength of one of your hands (arm wrestling), or two of your hands (pull-ups), but never three? Why, really? Same thing here.

Electrons with the energy quantum number $n$ can have orbital angular momentum quantum number $\ell = 0, \dots, n-1$ only.
Therefore, with $\ell = 0$, there is just the orbital 1s and no orbital 1p with $\ell = 1$.
For the same reason, there are no orbitals 2d nor 3f.


text:
What are glues/adhesives for cardboard boxes/crates made of?
Open cardboard boxes/crates, often for foodstuffs, are often glued to hold their shape. What is the glue/adhesive made of?
I ask to know if it is biodegradable, such that it is suitable for composting/gardening.
My preliminary research seems to indicate polyvinyl acetate (PVA), which seems to biodegrade fast enough for my application.
See example box (from CTHOE at Wikimedia):

Sorry if this is the wrong Exchange, it seemed the most suitable for my question.

I found this report on adhesives for packaging in the European Union. It contains information about the different adhesives used, including natural ones. I believe you will be seeking an adhesive based on natural polymers, such as vegetable polymers, as they are biodegradable. Good luck! Link to report: http://ilsi.eu/wp-content/uploads/sites/3/2018/12/2018-Packaging-materials_10_Interactif.pdf

Thanks @Liam! I'm looking for what is in general use, not for my own use. The report you link to seem to answer just that question. It seems to indicate that PVAc, PU, and acrylic polymers are used and also hotmelt adhesives. On my example picture glue running down on the interior wall in the back. I suppose that is a hotmelt adhesive?

I would also assume that it is a hot melt adhesive from looking at the photo.

The best example for misunderstanding of biodegradability. PVAc, PU, and acrylic polymers degrade slow. The only thing because these glues are used, if they degrade they produce less harmful substances and that is why they are used in the food industry.

Because the glue is not in direct contact with food you can use basically any glue. And cardboard is easy to glue so there is no limitation. Packaging should be cheap and therefore I would not expect a biodegradable glue.


text:
Developing Computational Chemistry Software
I want to learn to develop novel methods in computational chemistry (e.g. Grimme's empirical dispersion, DFTB, new DFT functionals, etc.) What is the ideal way to learn this by oneself?
I have read books on the Hartree-Fock theory and I understand it to some degree. However, there seems to always be a large amount of details which are left unsaid and brushed under the carpet. I am looking for quite practical ressources which would teach me how to actually translate the theory into code. Is there a way to learn this other than just reading books until it sinks in?

At risk of being labelled as rude, I'll say that I think you are not likely to make much headway by yourself, unless you are some kind of once-in-a-generation genius. The obvious answer to your question is to do a PhD on it. That is not typically something you can do by yourself.

I understand your point. I have not mentioned it, but I am studying chemistry at the graduate level and use computational chemistry on a daily basis. What lacks in my education is an understanding of the code I use and how I can improve/extend it.

Developing methods and developing code are two different things, sometimes related. For the former you will have to have a firm grasp of theoretical chemistry, quantum mechanics, and related subjects. That means reading a lot of books, and when you are done with that reading plenty of papers. I guess the best way to learn these things is to join a group that actually does development (like Grimme's group). Developing code is then translating the methods for a computer to execute, it needs training and interest in writing code. It helps having a course on it.

If you want to improve on your coding, there are plenty of open-source projects available on GitHub (and similar platforms), many of them offer the possibility to contribute. Psi4 is one of them! This is probably the best way to get into the process extending/improving.

In addition to @Martin-マーチン's excellent points, I would separate numerical methods from 'coding' and all the chemistry stuff. Developing efficient and effective algorithms is different than implementing them in some language.

How do I get into method development?
As was said in the comments, one would join a group that works in this field and learn from them. From up close, there is some derivation from previous results and some trial and error involved. And yes, a PhD-sized investment is typically required.
How do I get into method implementation?
It depends on how deep you wish to go. There are open-source projects to look at and contribute to. There are publicly available exercises on implementing HF and beyond. I would not advise to implement DFT from scratch because the numerical integration grid is both critical and difficult. Same is true for the integrals, use a library for them.
How do I learn to use Program X?
It is helpful to learn one program suite well and learn others later. The reason is that many concepts are similar between the programs. Learning one program comes down to locating good worked examples, reading the manual, attending a workshop if possible. One may for instance attempt to reproduce data from a DFT benchmark, since the supporting material usually has a detailed listing, possibly down to absolute energies.

The Psi4 project recently published a set of Python notebooks and tutorials on a wide variety of methods, called Psi4NumPy- J. Chem. Theory Comput. 2018 14(7) 3504-3511 -
All the code is available as open source on Psi4NumPy GitHub:

Hartree Fock including RHF vs. UHF, DIIS, etc.
Density Functional - including grids
Properties
Geometry Optimization

Generally these are well-written and give significant extra details. Moreover, they're live code with Psi4, so you can test out the actual implementation.
Beyond the tutorials, there are scripts indicating implementations of coupled cluster, MP2, response theory, etc.
I think anyone starting out or who wants more details with a readable introduction should take a look.


text:
Which is the correct definition for metamers (or metamerism)?
I have heard two definitions of metamers.

Compounds having the same molecular formula but different number of carbon atoms on either side of the functional group.
Compounds having the same molecular formula but different alkyl groups on either side of the functional group.

Which one of these is correct?
Are pentan-2-one and 3-methylbutan-2-one metamers?

How are they different, to begin with?

@IvanNeretin Different alkyl groups might mean a more branched alkyl group, but with same amount of carbon atoms.

@IvanNeretin I asked a question in the second part of the question. I thought the answer will change according to which definition is correct.

All right, so they are different, after all. Your example fits the second definition but not the first one. Well, hopefully someone will bring in the correct definition from IUPAC.

I am pretty sure that it is the second definition,but I can't find a source for that right now. Lemme come back with a reliable source :)

I would answer but surprisingly got two downvotes. Basically your definitions 1 and 2 are equivalent, at least according to what I can remember. See the down voted A to https://chemistry.stackexchange.com/questions/78783/what-is-metamerism. There is not IUPAC definition because the idea of metamerism does not bring anything useful or practical which isn't already covered by isomerism. It is isomerism confined to the same class of compounds, eg ketones, ethers and polyethers, alkanes, ...

There is no IUPAC definition for metamers or metamerism in the gold book (not even one that makes it obsolete), and Wikipedia doesn't even have a real article about it, see Metamerism.

In chemistry, the chemical property of having the same proportion of atomic components in different arrangements (obsolete, replaced with isomer). In organic chemistry, compounds having the same molecular formula but different number of carbon atoms (alkyl groups) on either side of functional group ( i.e., -O-,-S-, -NH-, -C(=O)-) are called metamers and the phenomenon is called metamerism.

I cannot find any authoritative source for this, but there is a question discussing it on our platform: What is metamerism? Also related: Are methyl n-propyl ether and methyl iso-propyl ether metamers?
Any of the two definitions cited in the question match the general definition of isomers.

isomer (DOI: 10.1351/goldbook.I03289)
  One of several species (or molecular entities) that have the same atomic composition (molecular formula) but different line formulae or different stereochemical formulae and hence different physical and/or chemical properties.

More particularly, they match the definition of constitutional isomers.

constitutional isomerism (DOI: 10.1351/goldbook.C01285)
  Isomerism between structures differing in constitution and described by different line formulae e.g. $\ce{CH3OCH3}$ and $\ce{CH3CH2OH}$.

Metamers/ metamerism is an archaic, ambiguous, and deprecated term in chemistry. Don't use it.


text:
Does something like a fluorescent foil or a fluorescent film exist?
During my experiments I want the emitted laser beams to come back with a different wavelength than the original wavelength using a fluorescent foil or fluorescent film.
The laser I'm currently using has a wavelength of 632.8 nm. The color of the emitted beams doesn't matter (it can have every color).
Therefore I want to ask if something like this is principally possible and if a fluorescent foil/ film does exist?

Depending on type of laser (pulsed/power etc you do not define this) you may be able to change wavelength by frequency doubling (or sum frequency mixing in general), use an optical parametric oscillator, make a super-continuum, or by exciting another laser. Fluorescence films will not work.

You have a He-Ne laser, so you need a fluorescent substance that emits at longer wavelength, which is a bit of a challenge since the nominal (but not actual) infrared starts at 700 nm. You would be far better of with a green laser pointer (532 nm) or even a 405 nm laser pointer (I have one, so I know they exist). These laser pointers are cheap and readily available. The nice thing about a green laser pointer is you can just shine it on things that you think might be suitable, e.g., red plastic wrap, and see if you get fluorescence. Good luck!

@Ed V, The OP states that  'I want the emitted  laser beam to come back' not just fluorescence, which makes the thing impossible with some arbitrary fluorescent material unless it is in a laser cavity etc. In any case HeNe's and laser pointers are very feeble, and as you point out will produce a little fluorescence, but no laser beam :)

@porphyrin Ah, I assumed the OP just wanted a little fluorescence, maybe to see where the He-Ne beam was going. We usually used unsolicited business cards or uranium glass for that. So your first comment pretty much says it all. Thanks for the catch!

@Ed V, yes I've used cards the same way, often v good to detect uv light.


text:
what does the mixture between urine and dishwater cause?
I have built an rv from a short school bus.  I ran a hose from the "toilet" to a tank in the back to drain the toilet. We wash our dishes in a basin on the counter and dump the dishwater into the toilet to also drain into the tank. I am getting a tannish/white build up in the hose that is plugging the line and not letting it run into the tank. It is the consistency of toothpaste when its wet and when it dries, it is like chalk. Any ideas on what the mixture of urine and dishwater is making? Also, we use vinegar to rinse dishes, so there is a small amount of vinegar going into that hose also.

Check the composition of your water supply.


text:
Are pentan-3-one and 3-methylbutan-2-one positional isomers?
Question

Give isomeric relations between these compounds.

My answer

A and B show metamerism and positional isomerism.
A and C show chain isomerism and metamerism.
B and C show chain isomerism and metamerism.

Given answer

A and B show metamerism and positional isomerism.
A and C show chain isomerism, metamerism and positional isomerism.
B and C show metamerism and positional isomerism.

My doubt
How does A & C and B & C show positional isomerism?
Does positional isomerism even makes sense when the carbon chain skeleton is not the same? Because in definition of positional isomerism, it states that structure of carbon chain remains same.
And why does B and C does not chain isomerism?

Related, might help: https://chemistry.stackexchange.com/questions/74094/structural-isomerism-in-iso-butyl-and-tertiary-butyl-alcohol

https://chemistry.stackexchange.com/questions/117674/which-is-the-correct-definition-for-metamers-or-metamerism

In organic chemistry, compounds having the same molecular formula but different number of carbon atoms (alkyl groups, etc.) on either side of functional group (i.e., $\ce{-O-,-S-, -NH-, -C(=O) -,}$ etc.) are called metamers and the phenomenon is called metamerism (Wikipedia). Thus, you agree that all three compounds are metamers.
Positional isomers are constitutional isomers that have the same carbon skeleton and the same functional groups (and/or substituents). In Positional Isomerism, the prime carbon chain remains the same in all aspects while the position of functional groups on and/or other substituents change position on a parent structure.
In chain isomerism or skeletal isomerism, components of the (usually carbon) skeleton are distinctly re-ordered to create different structures (Wikipedia).

Accordingly, A and B show metamerism and positional isomerism (position of $\ce{C=O}$ group change on $\ce{C-C-C-C-C}$ carbon skeleton).
A and C show metamerism, chain isomerism ($\ce{C-C-C-C-C}$ carbon skeleton changes significantly to $\ce{C-C-C(-C)-C}$ skeleton), and positional isomerism (position of $\ce{CH3}$ group changes on $\ce{C-C(=O)-C-C}$ carbon skeleton from 1-to-3 position).
Similarly, B and C also show metamerism, chain isomerism ($\ce{C-C(=O)-C-C-C}$ carbon skeleton changes significantly to $\ce{C-C(=O)-C(-C)-C}$ skeleton), and positional isomerism (position of $\ce{CH3}$ group changes on $\ce{C-C(=O)-C-C}$ carbon skeleton from 4-to-3 position).

Therefore, I also agree with you that B and C should show chain isomerism. If not, then n-pentane and 2-methylbutane would not have chain isomerism! :-)
Late Addition: Example for positional isomerism with substitution:

Isn't positional isomerism related to position of the functional group? Why is CH3 considered as functional group?  If that's true, then n-pentane and 2-methylbutane must be positional isomers, since position of CH3 is changed.

Not necessarily. As long as configuration of substitution didn't change except the position, thet are considered positional isomers. See my addition to the answer.

According to the present recommendations of IUPAC, ketones are not considered to show metamerism. There must be a heteroatom in between the carbon skeleton.


text:
How hot will a carbon electrode get in an electrolysis setup?
For the electrolysis of sulfuric acid, I plan to use carbon graphite electrodes at voltages up to 6 V. I would like to know if my carbon electrodes would get considerably hot or warm (45 °C or higher).

Supposing there is a powerful DC source, then quite hot, unless small enough electrode current density is managed.
6 V is quite high voltage for electrolysis. But it is possible the source will not be able to provide the current for the voltage 6 V, unless the electrode area is small.
Consider also the graphite electrodes may get deteriorated by the released gas, releasing small carbon particles to the solution (remembering my own first electrolysis experiments with carbon electrodes from old Leclanché batteries, when I was 12).

Would it be hot enough to melt or damage dialysis tubing ?

Well, depending on geometry and electric conditions, it can have huge range of temperature, so it cannot be told a priori. You need to experiment what current the tubing can withstand. Generally, the smaller current density in mA/cm2, the better. Plus, the total electric power per solution surface will determine the solution temperature.


text:
What purpose does mercury dichloride have in fireworks?
My local newspaper ran an article covering the local rules about fireworks.  One of those rules specifically calls out "snakes and fountains containing mercury bichloride" as forbidden.
Neither Wikipedia's article on mercury(II) chloride nor its article on mercury lists any fireworks-related use.  What purpose does the mercury dichloride serve?  Is it just a way to color the sparks, or does it have some more fundamental purpose?

Most likely, mercury was used as a colorant, specifically red.  This list shows mercury providing a red color in flame tests, and the same would presumably be seen in fireworks.
Several other, less dangerous elements also give red or reddish colors.  This site identifies lithium as providing the color in red fireworks, and also mentions lithium carbonate specifically.
Be safe this 4th, handle fireworks with care if you do so at all!

In "Fireworks, the Art, Science, and Technique", Takeo Shimizu gives a very good explanation of color production. For blue, the color-producing firework chemical is thought to be copper monochloride. If there is no free chlorine in the flame, there can be no blue color.The discovery of colors produced by chlorine along with other elements belongs to Chertier, who described the phenomena in 1836.In years past, mercurous chloride, also known as calomel, was used. Calomel decomposes at ordinary flame temperatures into chlorine and mercury metal vapor. Along with potassium chlorate ,mercurous chloride(calomel) was used as the oxidizer in older blue firework star compositions, calomel was often included to insure that plenty of chlorine was available.
Today we know that mercury is a very bad actor in the environment, witnessed by the birth defects in Japan resulting from mercury poisoning, not to mention the deaths!
Making Pharaoh's Snakes

Tomasz Szymborski, Creative Commons License
Ignite a small pile of mercury(II) thiocyanate, $\ce{Hg(SCN)2}$.Mercury thiocyanate is an insoluble white solid which can be purchased as a reagent or can be obtained as a precipitate by reacting mercury(II) chloride or mercury(II) nitrate with potassium thiocyanate (All mercury compounds are toxic and probably banned for this reason).
Pharaoh's Snakes Chemical Reaction

Igniting mercury(II) thiocyanate causes it to decompose into an insoluble brown mass that is primarily carbon nitride, $\ce{C3N4}$. Mercury(II) sulfide and carbon disulfide are also produced.

$$\ce{2Hg(SCN)2 → 2HgS + CS2 + C3N4}$$

Flammable carbon disulfide combusts to carbon(IV) oxide and sulfur(IV) oxide:

$$\ce{CS2 + 3O2 → CO2 + 2SO2}$$

The heated $\ce{C3N4}$ partially breaks down to form nitrogen gas and cyanogen:
$$\ce{C3N4  → 3(CN)2 + N2}$$
Mercury(II) sulfide reacts with oxygen to form mercury vapor and sulfur dioxide. If the reaction is performed inside a container, you will be able to observe a gray mercury film coating its interior surface.

$$\ce{HgS + O2 → Hg + SO2}$$


text:
does aluminum oxide float on molten aluminum and by what mechanism?
Not really a chemistry question since no reactions happening, but couldn't find a more appropriate stack exchange site.
Considering recycling aluminum cans for scrap metal using induction heating. I've seen people do it with very unsophisticated equipment, but what I don't get is Al2O3 is denser (3.95g/mL) than molten aluminum (2.38 g/mL).
I suppose surface tension can stop any oxide that forms at the surface of the melt from sinking, but how strong is that? Is it resistant to mild stirring (e.g. when you skim off the dross)? Are there other examples of denser materials floating?

A chunk of $\ce{Al2O3}$ would sink in molten aluminum. However, $\ce{Al2O3}$ usually appears in smaller pieces (down to microns), with irregular surfaces, probably filled with air ($\ce{N2}$) from when it was cool. A mix of metallic aluminum and $\ce{Al2O3}$ could allow aluminum to sink thru the solid particles after melting, allowing the particles to rise above the surface of the liquid. $\ce{Al2O3}$ has a very high melting point.
The mass of solid particles would have some very weak structure, and some buoyancy by trapped air. The surface tension of aluminum is very high (~900 dyne/cm) , so it might very well account for the covering of the metal by a heavier substance and the non-wetting of the oxide.
This is reminiscent of floating a needle on water.
In the electrolytic production of aluminum from $\ce{Al2O3}$:

The electrolyte is a molten bath of cryolite ($\ce{Na3AlF6}$) and
  dissolved $\ce{Al2O3}$. Cryolite is a good solvent for $\ce{Al2O3}$
  with low melting point, satisfactory viscosity, low vapour pressure,
  and density lower than that of liquid aluminum (2 vs 2.3 g/cm3), which
  allows natural separation of the product from the salt at the bottom
  of the cell.

References

https://link.springer.com/article/10.1007%2Fs11663-999-0108-4
https://en.wikipedia.org/wiki/Aluminium_smelting

actually, I realized it doesn't matter if the aluminum oxide sinks or not, only that it separates when pouring. Watching some videos it seems the slag mixture is more viscous, while the molten Al can be easily poured out


text:
How do I calculate the correct sample concentration for my measurements?
I will measure absorbance for measurement of protein concentration.
The sample solutions are prepare in serial dilutions to the volume of 100 uL to obtain a certain concentration of the protein in the buffer, lets say 2 mg/mL, then 20 uL is added to the cuvette with 1 mL Bradford reagent.
Now my question is, is the protein concentration still 2 mg/mL, I mean, I know it is not in the cuvette with the reagent added, but do I have to take this into account when measuring the absorbance? Or should I recalculate the dilutions to get the concentration 2 mg/mL in total with the reagent added?
This might be a dumb question but the hour is late and my brain is scrambled eggs.
Thank you in advance

The absorbance value you are getting is due to the protein content in the cuvette. So, your calculation is based on how you plot your calibration curve. Then, you have to back calculate according to your dilution.

should I recalculate the dilutions to get the concentration 2 mg/mL in total with the reagent added?

Either way works as long as you are consistent.
Plot dilute concentration
You can use the final concentration in the cuvette and plot this against the measured absorbance. When you use this standard curve, it will also give you concentration of your unknown in the cuvette. To get the undiluted concentration of the unknown, you have to do one extra step, multiplying the concentration in the cuvette by the dilution factor.
Plot undiluted concentration
You can use the undiluted standard concentrations and plot this against the measured absorbance. When you use this standard curve, it will give you the undiluted concentration of your unknown. This only works if you always dilute by the same factor.


text:
Does water in air occur as liquid or as gas?
I've learned that in a gas, the solute has to be a gas a well. Is that always the case? How about water in air, is that liquid in a gas, or a gas in a gas?

Being a gas or a liquid is a property of solution as a whole, and not of individual components.

@IvanNeretin mm yes but we might consider the solution inside the air, for instance, some times we have aerosols each particle might be considered as a solution itself containing different compounds. I think that the question might be good if the user expands it a little bit.

Air with aerosols is not a solution.

Smoke is (mostly) a suspension of solid particles in a gas.

If there's water in your air, it's either in the form of water vapor (which is a gas), or water droplets (which are a liquid suspended in a gas).
Liquid water in air isn't a solution, it's a suspension.

or rather an aerosol, with droplets disperged in air. Dispersion of liquid in liquids is suspension. Dispersion of liquid in air is aerosol.

@Poutnik I think I'd rather stick with the more general term. All aerosols are suspensions, but not all suspensions are aerosols -- rain droplets in an updraft are too large to be considered an aerosol, but they're definitely suspended (just not stably), right?

Rather, all aerosols and suspensions are dispersions.


text:
Effective nuclear charge of oxygen atom (O) vs oxygen anion (O2-)
The size of an anion is greater compared to its parent atom because former's effective nuclear charge is lesser than that of latter.
I found on wikipedia that the effective nuclear charge can be calculated by the formula:
Zeffective = No. of protons in the nucleus - No. of non-valence electrons
For oxygen atom, the electronic config is: 1s2 2s2 2p4
For oxygen anion, the electronic config would be: 1s2 2s2 2p6
Zeffective for oxygen atom = 8 - 2 = 6
Zeffective for oxygen anion = 8 - 2 = 6
If the effective nuclear charge is staying the same, then how it can be said that the size of an anion is greater than that of its parent atom?

There are 2 phenomenas going each against the other: The effective nuclei charge, attracting Valence electrons and The electron mutual repulsion. With the same the former, more electrons = more repulsion = bigger size.

Note also computation of effective charge is not so easy. Electron shielding effectivity decreases between orbitals , s > p > d > f. Electrons in 4f, 5f orbitals do not shield well. Therefore, within lanthanoid and actinoid series, the effective charge is not constant, but increases. It is major cause of the lanthanoid contraction.

The size of an anion is greater compared to its parent atom because former's effective nuclear charge is lesser than that of latter.

This is not the best explanation. The size of the anion is larger because, as Poutnik wrote in the comment, the effective nuclear charge is almost the same (if you account for nuclear charge and inner electrons only), and there is more repulsion among the valence electrons.
The size of fluorine atoms is smaller than the size of sodium atoms because the effective nuclear charge is larger in fluorine  than in sodium.
This is the typical explanation for the biggest jump in atomic sizes. It makes sense that as you add electrons to a given nucleus, the particle gets larger. It makes sense that going down a group, atoms get larger. What needs an explanation is the dip in size going within one period.
So comparing F and Na (or any other group 17 element and the following group 1 element), we go from a high effective nuclear charge (nucleus and inner electrons combined have a +7 charge) to a low effective nuclear charge (nucleus and inner electrons combined have a +1 charge). On top of that, the outer electrons are also in the next higher shell in Na (3rd period) compared to F (2nd period).
Why are we switching explanations?
Depending on which two species you compare in size, some things change and some things stay the same. Comparing $\ce{O, O- and O^2-}$, the nucleus and inner electrons don't change, so you focus on the number of electrons in the outer shell. Comparing isoelectronic species such as $\ce{O^2-, F-, Ne, Na+, Mg^2+}$, the number of electrons stays the same, but the nuclear charge (and the effective charge) change, so you would focus on that. Comparing atoms in a group, e.g. $\ce{Li, Na, K, Rb}$, the effective nuclear charge is similar (corresponding to similar chemistry of the valence electrons) and you would focus on the valence electrons occupying higher and higher shells.

Comparing neutral atoms in order of atomic number, e.g. $\ce{O, F, Ne, Na, Mg}$, everything changes. Within a period the increasing effective nuclear charge dominates (making the atoms smaller and smaller). Jumping from one to the next period, the drastic decrease in effective nuclear is the most pronounced effect (in addition to using the next shell), resulting in a larger radius.


text:
How to account for solvation of a molecule explicitly in quantum calculation?
I'm looking for a way to account for hydrogen bonds formed with my molecule in aqueous solution explicitly in DFT calculation. I'll do it in order to compare stability of two structures in water. As far as I understand, it can be done by molecular dynamics calculation of the structure surrounded by molecules of water or by multiple random choice of H2O molecules around the structure with subsequent optimization.
Any suggestions, references to articles or programs are appreciated.

There are several approaches, with different computational cost and different scope. Can you tell more about your project? For example, which molecules (or types of molecules) are involved? Knowing this information could guide you towards an appropriate force field, if needed. It also helps to understand if your problem is a complex "conformational" problem, or if it can be solved "easily" with QM or QM/MM methods. Also, what do you want to calculate? Is it a problem about energy, geometry, or what else?

Thank you for reply! It's a problem about energy. It involves equilibrium between ketone and organic peroxide generated by nucleophilic addition of H2O2 dissolved in water to this ketone. The molecule of ketone has only C=O groups and corresponding peroxide has only C-O-C and C-O-O-C groups. I need to figure out the direction in which this equilibrium is shifted.

Do you need explicit water? If the molecule doesn't have intramolecular hydrogen bonding (eg. in sugars), implicit solvent models are quite good for calculating geometries and energies in solution. The typical process involves calculation in the gas phase, and later refinement with any implicit model (COSMO, SMD, PCM, etc...).

I believe I do. If I use only implicit model SMD the energy difference is not realistic. If I add in the system one H2O molecule on every lone pair of oxygen to simulate hydrogen bonds the results start to make sense. So I want to use something more reliable to make sure that the result I get are right.
By the way, my ketone is enolizable so this might be the reason why only implicit SMD don't perform too well.

A paper I referred to in the past is this: https://doi.org/10.1002/jlac.1995199511272 If you really wanted a precise result, you could set up a molecular dynamics simulation if the species are supported by a force field. By analyzing the geometries, if hydrogen bonding matters you will find water molecules in "specific" positions. Then you could find the average or the most representative geometry, and from that point manually include the right water molecules and go on with QM calculations (sometimes, combining implicit models and explicit water works wonderfully here)

I don't think adding explicit water molecules is a good idea, though, because it poses several questions that is hard to answer, for example how many molecules do you need? One, two, three...? I remember reading (sorry I don't have a reference for it, though :/ ) that you need at least 18 water molecules to have a realistic account. I would go back to only using an implicit solvation model, mostly if you are not an expert. What kind of functional/basis set are you using? It is possible that the "unrealistic energies" that you are talking about depend on the level of theory.


text:
To how much significant figures do I reduce my absolute uncertainty if it has more significant figures than my measurement?
They say that the absolute uncertainty in my measurement should always be reduced to 1 significant figure but I don't understand why.

There is no such rule. It is wrong, whoever is insisting on that. In real world, all that matters is standard deviation and the actual measuring instrument. If you balance can weigh up to four decimal places, and you weigh 100 times, the standard deviation can be +/- 0.0010 g [I am assuming a very bad balance]. In reality it could be +/- 0.0003 g. Perhaps this is the origin of this rule of thumb to keep 1 significant figure.

Notice the convention of providing absolute uncertainty of physical constants in Wikipedia. E.g  1.2345(67) = 1.2345 +/- 0.0067

The key point is estimation of standard deviation of estimation of measurement standard deviation on Stats SE.
The exact formula is complicated with extensive Gamma function involvement.
The essential info is the relative uncertainty of $s$ decreases very slowly with number of measurements.
It can be approximated by involving Stirling approximation for factorial of large numbers.
$$SD(s)=s\cdot \sqrt{\mathrm{e}\cdot \left(1-\frac1n\right)^{(n-1)}-1 }$$
where
SD is Standard deviation
s is estimation of standard deviation $\sigma$
e is the natural logarithm basis
n is the sample number
The estimation of relative error of $s$ is

about 33% for 5 samples
about 20% for 13 samples
about 10% for 50 samples.

So 1 significant digit for $s$ is a reasonable rule, unless there is a large enough measurement sample set.


text:
How could I prepare Iron(III) arsenate (FeAsO4)?
I have no idea in which conditions this compound is formed and would like to find out any / different ways it can be prepared (any easy way not requiring expensive equipment would be great).

Iron(III) arsenate or more commonly Ferric orthoarsenate is normally found in nature in the form of a mineral called Scorodite.
It can also be prepared chemically. From here:

The monohydrate, $\ce{FeAsO4.H2O}$, is precipitated from solution. The
  salt, when dried in air, is a dull white insoluble substance,
  containing one molecule of water. When heated at 100° C. it becomes
  anhydrous. On treatment with sodium hydrogen carbonate solution, both
  the monohydrate and the anhydrous salt cause effervescence to take
  place, a soluble double arsenate being produced. It would appear,
  therefore, that the hydrated salt is acidic, and determinations of its
  basicity indicate that its formula is $\ce{FeO.AsO2(OH)2}$.
Colloidal ferric arsenate is prepared by the action of ammonium
  hydroxide on the insoluble salt.


text:
How was this fake red mercury video made?
I Found on web video make me wondering what the hack is this , how its even possible ,so any one know what is happen can explain to me please

youtube video B_olJLjgcHc

There is no red mercury. Some mercury compounds, e.g., HgO and HgS, are red or have red varieties. Cinnabar is a naturally occurring HgS and beautiful crystals exist. Hope this helps.

Everything about that Youtube video is a cheap fake.

The video probably shows some other liquid metal (molten, i.e.). Mercury would vaporise before it goes red hot.

I'm voting to close this question because there is no intrinsic or other value in it or to what it refers.

To spare everyone watching the movie (and giving revenue to someone showing fake chemistry), here is a summary:
A container labeled "27/27 | 99%999 | 1/Kg | 001200300 | RED MERCURY | W. GERMANY | WARNING - DANGEROUS | WARNING - OPEN IN LAP" containing a red liquid is first inverted, then a metal wire (safety pin) is placed on it. The metal wire ignites.
Here is the google-translated comment of one of the viewers ("alfa wolf")

There is a battery inside the yellow metal, if you look carefully there is a gray metal compartment around the mercury, and there are two parts, one of them plus, one minus, when the wire touches the wire, and the wire burns when the high voltage passes. Mercury has no effect.

Here is a screen shot of the relevant bit. You can see that the wire starts glowing where is contacts the metal rim (smallest area of conducting, just like welding):

Do not open any shipment of chemical substances that says "open in lap".


text:
Product of Diels-Alder reaction + Consecutive Reaction with base and heat
there is a problem regarding a Diels-Alder reaction that I do not understand (see image).

The Diels-Alder reaction of the diene and the dienophile should lead to the product A. I hope I did it correctly, because the molecular formular shall be C9H6O2N2.
Now, the base triethylamine is added and heat is applied. Because it is a very strong base I would deprotonate the proton next to the ester-group -COOR, since it has a -M effect. I think applying heat leads to decarboxylation (B = carbon dioxide). But I don't know what the product C looks like, since there is another reaction to product D.
Can you please help me?
Thank you!

The pKa of the protons alpha to the nitriles is very similar to that of the proton alpha to the ester.

So if I take the proton alpha to the nitrile, then the negative charge can be stabilized by the nitrile. Can I then make a new double bond in the ring, so that a carboxylate forms? The carboxylate can then form carbon dioxide... So in the end another double bond in the ring forms and a CN- leaves. But does that even make sense?

Why does CN- leave if CO2 leaves? I think you will get a dicyanocyclohexadiene (C) which will spontaneously aromatise to give 1,2 dicyanobenzene

I am sorry, I still do not completely understand it. If CO2 leaves, then there is still a negative charge in the molecule. So what happens so that dicyanocyclohexadiene is formed?

There is also Et3NH+ present which can quench the anion. The diene will spontaneously lose H2 as the driving force for aromatisation is very powerful

@Waylander: Would you consider loss of HCN via Et3N in the last step to give benzonitrile.? Unfortunately, there is not sufficient information in the problem.

Yes, that is a possible pathway.

You have nothing to drive aromatisation without elimination, so HCN elimination is more likely in this case unless a weird ene reaction is occurring. Loss of H2  requires a catalyst in the majority of cases, which is not indicated here


text:
Are these chemical reactions with amino acids possible, and if so, what are they called?
In high school, I am taught that the synthesis of two amino acids of the form $\ce{H2NR1COOH}$ and $\ce{H2NR2COOH}$ is a condensation reaction that produces a polypeptide $$\ce{H2NR1COOH}+\ce{H2NR2COOH}\to\ce{H2NR1CONHR2COOH}+\ce{H2O}$$ where $\ce{R1}$ and $\ce{R2}$ are arbitrary components within each acid. Then I started experimenting with different positions that yield isomers of the polypeptide.
Examples

Water is formed though the pairing of the two $-\ce{OH}$ groups $$\ce{H2NR1COOH}+\ce{HOOCR2NH2}\to\ce{H2NR1COOCOR2NH2}+\ce{H2O}$$ which forms something that resembles an ester?

Water is formed through the pairing of the $-\ce{OH}$ and $\ce{H2N}-$ groups but the oxygen atom is taken instead from $\ce{C=O}$ $$\ce{H2NR1COOH}+\ce{H2NR2COOH}\to\ce{H2NR1C(OH)NR2COOH}+\ce{H2O}$$ where essentially the hydrogen atom from $-\ce{CONH}$ has shifted one place.

Are these reactions directly possible? If so, are there special names for
a) the reaction and
b) the type of product formed?

I am at an elementary level in chemistry so more detailed explanations would be appreciated.

Reaction 1
The product is called carboxylic acid anhydride. If you make sure the amino groups don't react (protecting and deprotecting) and get rid of water to prevent back-reactions, this reaction is possible.
Reaction 2
The product could be called a keto-enol tautomer (here, the enol form). The keto form is much more stable, so this product - even if made as an intermediate - would isomerize to the standard peptide bond.


text:
Calculating partial pressure equilibrium constant Kₚ given initial pressure and equilibrium pressure

A reaction
$$\ce{A(g) <=> B(g) + C(g)}$$
happens in constant volume and constant temperature. The reaction starts only with gas $\ce{A}$ (no $\ce{B}$ or $\ce{C}$) with given pressure $P_1 = \pu{6 atm}$, in equilibrium the pressure of all three gases is $P_2 = \pu{10 atm}$. Calculate $K_p.$

It seems to me like a very simple question, however it seems that I don't understand a basic concept regarding gas equilibrium. As for my understanding it is suppose to be
$$K_p = \frac{P_2 \cdot P_2}{P_2} = P_2 = \pu{10 atm}$$
But the given solution is $\pu{8 atm}$ and I really don't get what miss.

Make an I.C.E, table first.

I don't see how it make sense, since the initial amount of $A$ is $6\cdot \frac{V}{RT}$ and the final amount is $10\cdot \frac{V}{RT}$ (V,T are constants), but the amount of material can't grow.

The problem is that you are using wrong pressures.
By definition for the reaction at equilibrium partial pressures can be expressed via initial partial pressure $P_1$ and conversion factor $α$
$$
\begin{array}{ccc}
\ce{&A(g) &<=> &B(g) &+ &C(g)}\\
&(1 - α)P_1& & αP_1& & αP_1
\end{array}
$$
equilibrium constant $K_p$ is to be found as
$$K_p = \frac{P(\ce{B})\cdot P(\ce{C})}{P(\ce{A})} = \frac{α^2P_1^2}{(1 - α)P_1} = \frac{α^2P_1}{1 - α}$$
Unknown $α$ can be found by equating total pressure at equilibrium $P_2$ to the sum of partial pressures of all gaseous components at equilibrium:
$$
\begin{align}
P_2 &= P(A) + P(B) + P(C) \\
  &= (1 - α)P_1 + αP_1 + αP_1 \\
  &= (1 + α)P_1
\end{align}
\quad\implies\quad
α = \frac{P_2}{P_1} - 1 = \frac{\pu{10 atm}}{\pu{6 atm}} - 1 = \frac{2}{3}
$$
Finally, all the values can be plugged into the expression for $K_p$:
$$K_p = \frac{\left(\frac{2}{3}\right)^2\cdot\pu{6 atm}}{\left(1 - \frac{2}{3}\right)} = \pu{8 atm}$$

Thanks, may you please explain why the statement in the question that $P_2(A)=P_2(B)=P_2(C)=10atm$ in equilibrium it translated to $P_2 = P(A)+P(B)+P(C)?$ I mean, how could I deduce it from the question ?

@user5721565 I added corresponding passage to the answer.


text:
Why is hypochlorite a stronger oxidizing agent than other oxychlorides?
The reduction potentials of perchlorate, chlorate, chlorous acid and chlorine dioxide in acidic and basic solutions are listed in the table below:
$$
\begin{array}{lcc}
\hline
\text{Oxychloride} & E^\circ_\text{acidic}/\pu{V} & E^\circ_\text{basic}/\pu{V} \\
\hline
\ce{ClO4-} & 1.19 & 0.56 \\
\ce{ClO3-} & 1.21 & 0.63 \\
\ce{HClO2} & 1.65 & 0.78 \\
\ce{ClO2}  & 1.63 & 0.89 \\
\hline
\ce{}
\end{array}
$$
Why is $\ce{ClO-}$ such a strong oxidizing agent compared to other chloride oxyanions, despite its low oxidation number? Is oxidation number representative of oxidizing strength?

The TLDR here is that you provided data out of context, but the context matters. Reduction potentials apply in the context of a half reaction, not a single oxidant. In the series you have provided, there's no way to tell if this series has any kind of homology because in essence, you can't tell how much each species is reduced.

Besides the the thermodynamic aspects discussed by Matthew, consider the kinetic aspects.  Oxidation by a chlorine oxyanion involves displacement of oxygen from its bond with the chlorine.  Such a displacement, in a protic solvent such as water, must involve protonation of the oxygen:

Oxygen without the proton would have to be displaced as oxide ion, which would be a high-energy intermediate in any protic solvent.

In this article from the University of Utah[1], molecular orbitals for some phosphorous-oxygen species are discussed.  These show that oxygen can back-donate its $p$ electrons to phosphorous, strengthening the bond.  A similar back-donation to chlorine would make the oxygen harder to displace in a redox reaction, but protonation would cut down on this back-donation.

Given these considerations, oxidation by a chlorine oxyanion is kinetically favored if the oxygen can act as a base and thus form a species that is more easily displaced than a bare oxide ion.  Hypochlorite ion does that relatively well, whereas in higher oxyanions the needed basic character weakens rapidly unless we force the issue with a strongly acidic solution.
Reference
1.
Benjamin Gamoke, Diane Neff, and Jack Simons, "Nature of PO Bonds in Phosphates", J. Phys. Chem. A 2009, 113, 5677–5684. http://simons.hec.utah.edu/papers/317.pdf

In practice, kinetics trumps thermodynamics. All the tables of half-reactions are useless if you can't isolate a certain half-reaction; and you can't. The tabulated values for chlorine oxides are calculated from heats of reaction, and not from actual voltages.

First, hypochlorite ($\ce{ClO−}$) is not a strong oxidizing agent compared to other chloride oxyanions, at least in acidic medium. Under standard conditions in acidic medium, chlorous acid ($\ce{HClO2}$) is the best oxidizing agent among them due to the largest positive number for standard electrode potential:
$$\begin{align}
\ce{HClO2 + 2H+ + 2 e- &<=> HClO + H2O} &\quad E^\circ &= \pu{1.645 V} \label{rxn:1}\tag{1}\\
\ce{HClO2 + 3 H+ + 3 e- &<=> 1/2 Cl2 + 2 H2O} &\quad E^\circ &= \pu{1.628 V} \label{rxn:2}\tag{2}\\
\ce{HClO2 + 3 H+ + 4 e- &<=>  Cl– + 2 H2O} &\quad E^\circ &= \pu{1.570 V} \label{rxn:3}\tag{3}\\
\end{align}$$
Keep in mind that in the reduction half-reaction \eqref{rxn:1}, $\ce{HClO2}$ reduced to $\ce{HClO}$ with its largest positive standard electrode potential $(E^\circ = \pu{1.645 V})$ among chlorine oxoacids. For comparison, we should look at half-reactions of which each of these two reagents reducing to the same product. Thus, look at half-reactions \eqref{rxn:2} versus \eqref{rxn:4}, and \eqref{rxn:3} versus \eqref{rxn:5} where $\ce{HClO2}$ shows its superiority over $\ce{HClO}$:
$$\begin{align}
\ce{HClO + H+ + e- &<=> 1/2 Cl2 + H2O} &\quad E^\circ &= \pu{1.611 V} \label{rxn:4}\tag{4}\\
\ce{HClO + H+ + 2 e- &<=>  Cl– + H2O} &\quad E^\circ &= \pu{1.482 V} \label{rxn:5}\tag{5}\\
\end{align}$$
Perchlorate ($\ce{ClO4-}$) and chlorate ($\ce{ClO3-}$) ions, on the other hand, are also good oxidizing agents, but not as strong as $\ce{HClO2}$ or $\ce{HClO}$, yet closed. This fact is clear when compared to the half-reactions where they directly reduce to $\ce{Cl2}$ or $\ce{Cl-}$ similar to $\ce{HClO2}$ and $\ce{HClO}$:
$$\begin{align}
\ce{ClO4- + 8H+ + 7e- &<=> 1/2 Cl2 + 4H2O} &\quad E^\circ &= \pu{1.39 V} \label{rxn:6}\tag{6}\\
\ce{ClO4- + 8H+ + 8e- &<=> Cl– + 4H2O} &\quad E^\circ &= \pu{1.389 V} \label{rxn:7}\tag{7}\\
\ce{ClO3- + 6H+ + 5e- &<=> 1/2 Cl2 + 3H2O} &\quad E^\circ &= \pu{1.47 V} \label{rxn:8}\tag{8}\\
\ce{ClO3- + 6H+ + 6e- &<=> Cl– + 3H2O} &\quad E^\circ &= \pu{1.451 V} \label{rxn:9}\tag{9}\\
\end{align}$$
However, the standard reduction potentials OP is mentioning is for different reduction half-reactions $\ce{ClO3-}$ and $\ce{ClO4-}$ involved (\eqref{rxn:12} and \eqref{rxn:10}, respectively):
$$\begin{align}
\ce{ClO4- + 2H+ + 2e- &<=> HClO3 + H2O} &\quad E^\circ &= \pu{1.189 V} \label{rxn:10}\tag{10}\\
\ce{ClO3- + 2H+ + e- &<=> ClO2 + H2O} &\quad E^\circ &= \pu{1.152 V} \label{rxn:11}\tag{11}\\
\ce{ClO3- + 3H+ + 2e- &<=>  HClO2 + H2O} &\quad E^\circ &= \pu{1.214 V} \label{rxn:12}\tag{12}\\
\end{align}$$
Accordingly, when consider Latimer diagrams $(\ce{ClO4- -> ClO3- -> HClO2 -> HClO -> Cl2 -> Cl-}),$ you'd see, $\ce{HClO2}$ cannot oxidizes $\ce{ClO3-},$ although it is the strongest oxidizing agent. However, $\ce{HClO2}$ can oxidize another species, which $\ce{ClO3-}$ cannot oxidize.
Note that all reduction potential values are from: http://depa.fquim.unam.mx/amyd/archivero/TablasdepotencialesREDOX_26700.pdf (Petr Vanẏsek), which cited the following references:

G. Milazzo, S. Caroli, V. K. Sharma, Tables of Standard Electrode Potentials; Wiley, Chichester, 1978.
A. J. Bard, R. Parsons, and J. Jordan, Standard Potentials in Aqueous Solutions; Marcel Dekker, New York, 1985 (Series: Monographs in electroanalytical chemistry and electrochemistry).
S. G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at $\pu{298.15 K}$,” J. Phys. Chem. Ref. Data 1989, 18(1), 1-21 (https://doi.org/10.1063/1.555839).


text:
Club soda manufacturing process and problem in quality
How to increase $\ce{CO2}$ gas in club soda? I am planning to start a PET bottled club soda industry. My process involves chilling the water up to about $\pu{4 °C}$ and filling under pressure up to $\pu{4.5 bar}$. Of course, once the bottle is filled, pressure will be released to facilitate capping. That is the standard procedure to fill club soda in PET bottles. With this I am unable  to produce a good club soda.
Here the problem as per customers is once the cap is opened and only half of the soda is used, capped again, refrigerated, but there is no $\ce{CO2}$ gas found for the second time use. However, in other brand soda, they say, there is still gas in the soda for the second, third time use......
If I increase the pressure to $\pu{5/ 5.5 bar}$, the PET bottles exploded....
How to overcome the difficulty? How to ensure my club soda will have gas in it for the second, third time use...
What minerals can be added to water for good flavor and taste of club soda with dosage?

Perhaps trivial question, but do you allow enough time to establish gas dissolution equilibrium at the high pressure ?

I'd recommend you buy yourself a regular club soda machine and see how it works. Although why you'd want to reinvent a product that is already on the market in millions by several producers, I cannot imagine.

@Poutnik I was going to ask something similar.  Simply applying pressure doesn't guarantee dissolution. Effective mixing also helps by increasing surface area

Your polyethylene terephthalate (PET) plastic bottles must be too thin. The thickness of the plastic and the shape of the bottle both contribute to its resistance to bursting (pressure tolerance). According to SeattlePi:

Most generally used two-liter PET bottles for soda bottling begin to fail at pressures around $\pu{10.34 bars}$ (1034 kilopascals), or $\pu{150 psi}$ (pounds per square inch). This pressure, at which the bottle will burst, is more than it would normally encounter during packaging or normal use.

This pressure is also much higher than you needed to have in order to make a good soda (~$\pu{6 bars}$). So, find the provider who can make the bottle with the thickness of usual 2-L plastic bottles.
About choice of flavoring, it is upto your customers. Make few different flavored soda and see which one sell the most. I personally like pineapple burst! :-)


text:
Making oxygen from iron oxide (aka rust) with sodium chloride?
I like videogames, and I've been playing a lot of "Oxygen Not Included" recently. It's not a perfect physics simulation, but it does appear to be realistic-ish (both Specific Heat and Thermal Conductivity make up non-trivial parts of the gameplay; attempting to respect the tyranny of the rocket equation; etc.)
The latest update added in a new reaction though, that has me scratching my head a bit:

Rust (Iron Oxide) + Salt (NaCl) -> O2 + Cl

Other reactions implemented in the game make sense based on my (limited) understanding of chemistry, e.g.  Iron + Carbon + Lime = Steel, or H2O+ Electrolysis = H2 and O2, so I'm wondering if there's a rational basis for the Rust -> O2 reaction as well, or if this is the devs handwaving some chemistry for gameplay purposes. (Such as, for instance, the apparent annihilation of the Sodium into heat, above)

No this is not a feasible reaction

Chemistry in computer games always was and is still sloppy. Remember that you had you use fermium to research monkey brain in System Shock 2.

@Waylander Because of heat/pressure/reactivity concerns? Can you go into more detail about the why? Is there any basis for this in reality?

Both chlorine and oxygen are where they want to be; chlorine is most stable as Cl- and oxygen as O2-. Iron and sodium are equally in their most comfortable place. Without a lot of energy input they are not going to change.

This reaction isn't feasible and won't likely occur.
Why not?
Rust and Salt are very stable, and happy with their current state. You can see this better by looking at each product.
Rust forms from Iron and Oxygen

4Fe + 3O2 → 2Fe2O3

Salt is formed from Sodium and Chlorine

Na + Cl → NaCl

The formation of both of these products is very favorable because both molecules want to bond together to complete their valence shells.

They do not want to break their bonds, and doing so is possible but requires a lot of energy making it unfavorable. This is why we don't see salt or rust breaking down into its elements without a high amount of energy. This all applies for the overall reaction. All the reactants are already happy and don't want to break down making this reaction, possible, but not feasible due to the high amount of energy that it takes.

Strike "won't likely" and replace by "will never".


text:
Is Gibbs Free Energy of formation equal to Gibbs Free Energy of transition state?
Is activation energy equivalent to Gibbs Free Energy of transition state as related by Eyring equation?
$$E_a=\Delta ^\ddagger G \, \, ?$$
Is Gibbs Free Energy of transition state defined by the Gibbs Free Energy of formation?
$$\Delta ^\ddagger G :=\Delta _\mathrm{f} G \, \, ?$$
If not, which are the conditions for which such an approximation (second equation) would apply?

The Arrhenius equation
$$k=A\exp\left(-\frac{E_a}{RT}\right)$$
places all of the $T$-dependence in the exponential factor. The pre-exponential factor is not assumed to be temperature-dependent. By contrast, the Eyring equation
$$k=\frac{\kappa k_\mathrm{B}T}{h}\exp\left(-\frac{\Delta ^\ddagger G^⦵}{RT}\right)$$
which is justified on the basis of statistical-mechanical principles, displays a a temperature-dependence in the pre-exponential factor. So the exponential terms are not expected to be equal, at least theoretically, unless the pre-exponential factor in the Eyring equation becomes T-independent by some coincidental cancellation of terms.
As regards the second equation, I would define the free energy of the transition state more accurately as
$$\Delta ^\ddagger G :=\Delta _\mathrm{f} G(\mathrm{ts})-\Delta _\mathrm{f} G(\mathrm{ini}) $$

Tech detail: please note that using a circled minus symbol (⊖, \ominus) for the standard state isn't correct. IUPAC suggests either a superscript circle, or a plimsoll symbol. Since it looks like you intended to use plimsoll, you want to stick with a Unicode symbol ⦵ as getting it right with MathJax (or LaTeX, for that matter) is rather tricky.

@andselisk That's not the standard state, that's the transition state, and I guess I wanted plimsoll but picked the nearest thing based on detexify, but technically I guess you're right, I didn't check the with IUPAC. I remember the post not so long ago about plimsoll, that was a good one.

Either way, a transition state must have a reference point, which is normally the standard state, so there is no contradiction.  $Δ^\ddagger G^⦵$ is perfectly fine. Some books and even a Wikipedia article on transition state are using a circled minus — probably due to typographic difficulties in getting the plimsoll symbol right. Here we don't have this problem, hence my edit to comply with IUPAC rules.


text:
What happens when two galvanic cells are placed in series
I understand how a single galvanic cell works, and the relation of equilibrium and the Nernst equation.

But this appears to contradict the higher voltage in a series configuration because it basically tells you there is a limit of how much excess charges can the plates build up before the reactions halts, So how does placing two cells in series increases the potential difference?
Someone answered about the potential difference being a relative thing and I accept this idea but I don't get how does this apply to this situation.

Try these 3 things: 1/Imagine the very left electrode is grounded. 2/ Imagine there is inserted the standard H electrode to each of the 2 cells as the reference. 3/ Write down for each of 6 electrodes their electrostatic potential wrt ground and their electrochemical potential.

It is very analogical scenario as connecting capacitors or electronic DC sources ( with floating potentials ) in a serie.
With 2 cells in a serie, the same provided charge spends the doubled amount of electrode material. And vice versa for charging.
Even without looking into electrochemistry  details, energy conservation law requires doubled voltage for 2 identical cells in a serie.
In fact, such voltage doubling has nothing to do with electrochemistry.
Remember that electrolyte in cells has it electrostatic potential, and that thede potentials differ between cells by the cell voltage.
Potentials in cells, in the electrochemistry context, are floating potentials,  related to potential of the standard hydrogen electrode, would be put into the cell.
In context of physics, the same electrode in cells in a serie,  including the hydrogen one, has different electrostatic potential wrt (physical) infinity, what is conventional zero potential reference.
Electrochemical potentials of electrodes are electrostatic potentials related not to physical infinity as we are used to, but to (would be there ) the electrostatic potential of the standard hydrogen electrode in the same cell.
If cells are connected to a serie, electrostatic potentials of the same respective electrodes across the serie differ by the multiple of the cell voltage.

I get your last paragraph but i can't figure out how to apply it in the context of galvanic cells in series i think that there will be a short circuit between the inner plates and the voltage will stay the same between the outer plates

Imagine 2 transformers with floating ends of their secondary circuit. Join 1 end from 1 transformer with 1 end of the other transformer. Is there any short circuit ? Remember,joining cells is not about electrochemistry at all.

I have updated the answer.

Transformer are out of my reach for now but i know capacitors well enough , what is this standard hydrogen electrode ?

Thanks for the effort in your awnser but i didn't really understand this sentence "the same provided charge spends the doubled amount of electrode material.' what do you mean by 'spends' in this context ,

Well, Chemistry SE site is not intended to be an alternative to a serie of chemistry lessons. In fact, many of your questions are at the edge to be off topic as the homework-like question class, with not sufficient own research effort. Any question with easy answer by simple search should be off topic.

Google "standard hydrogen electrode". On of the first hits is the Wikipedia article.

Ad spends, Where do you think the energy of cells comes from ? It comes from energy rich material of electrodes.

E.g in the Daniell cell, the provided charge 1 C converts 1/2F moles of zinc to its salt., where F is the Faraday constant about 96490 C/mol. It Is de facto the product of the elementary charge and the Avogadro constant.

Let us continue this discussion in chat.

Voltage and (thermodynamic) electrode potential are slightly different things. Electrode potential refers to the maximum amount of work that can be extracted from a given single cell under very specific conditions and at a given concentration of metal ions.
Now forget about the cells shown in chemistry textbooks. First think of a hydraulic analogy. Think of two identical pumps connected in series in a closed loop, what will happen to the pressure (voltage) in the pipe? The amount of water (=current) entering pump 1 is the same as the amount of water that enters pump 2, however the pressure sensor will show increase in the pressure by a factor of 2.
Basically, voltage adds up in series for two cells.

But this appears to contradict the higher voltage in a series
  configuration because it basiclly tells you there is a limit of how
  much excess charges can the plates build up before the reactions halts

This is a misconception. There is no gradual build up of charge.