text:
Problem on the wavelength of the electromagnetic radiation needed to break 1 mol of gas
Problem.
To break the bonds in 1 mol of Cl$_2$ we need E=243.4 kJ of energy. The bonds on this amount of gas can be broken by electromagnetic radiation of a certain length. Determine the wavelength.
Solution.
Firstly, there are 34 electrons in the molecule Cl$_2$. So in 1 mol of Cl$_2$, there are $n_e = 34\cdot N_A = 2.05\times 10^{25}$ electrons, where $N_A=6.022\times 10^{23}$ mol$^{-1}$ is Avogadro's constant. Now, the amount of energy needed for a single electron is (on average) $E_e = E/n_e = 1.19 \times 10^{-20}$ J.
We know that energy is related to the wavelength $\lambda$ by the relation
$$
E_e = h\frac{c}{\lambda},
$$
where $h=6.63\times 10^{-34}\, m^2 kg/s$ is the Planck constant and $c=3.00\times 10^8 \,m/s$ is the speed of light in vacuum. Thus
$$
\lambda = \frac{hc}{E_e} = 169\,nm.
$$
[Solution provided: 492 nm]
Am I doing something wrong?
UPDATE
I got the mistake thanks to @Domen (+10). I don't have to count all electrons but only those involved in the bonding. There is a single pair of electrons to be broken in a single chlorine molecule. In 1 mole of chlorine there are $N_A$ bonds, thus $n_e = N_A$. So the correct answer is $\lambda = 492$ nm.

Welcome to Chemistry StackExchange! You have a small conceptual mistake in your reasoning. Beware, there is a single covalent bond between the two chlorine atoms in $\ce{Cl_2}$. You should not calculate the number of electrons, you only need to know the number of bonds. So $n_e=1/N_A$ (where $N_A$ is the Avogadro's constant).

Oh I see, many thanks. As a side note: I haven't studied chemical bonds yet :-)

I meant $n_e = N_A$, of course.


text:
Breaking an azeotrope by using different atmospheres?
I read the threads about azeotropic mixtures and distillation, and a question arose:
Is it possible to distill an azeotropic mixtures properly (i. e., separate it into its single components) by changing the atmosphere? For example, if you distill under argon atmosphere or gaseous hydrogen chloride?
Best and stay healthy!
Wolf

I can't see the point of using something like that. Also questions aren't letters.

Sorry for not being more precise, but this question is work related:-) But I think, my question is obsolete: If you heat up the mixture to its boiling point, its vapor pressure increases to atmospheric pressure and will therefore replace the atmosphere by its own components, right? So, no effect would be visible as soon as youre mixture starts to boil.

That's also not true. Read up on things like equilibrium vapor pressure.

Okay, so, you mean, there could be a chance, that an azeotropic mixture can be separated into its single components?

If the gas reacts with one of the components like HCl you just mentioned, then sure. Otherwise, pressure works but not different gases. Sometimes, it changes compositions for non azeotropic mixture like distilling NMP in N2 vs air at atmospheric pressure (which I suspect is more due to degradation) but I've never seen azeotropes go through any change.

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text:
Trends in atomic radii across a period
I am a 12th grader. Recently, while revising the Periodic Table, I came across the statement:

As the effective nuclear charge increases across a period, the atomic radius of the elements decreases on moving from left to right in a period.

For some reason, I decided to fact-check this statement and compiled the atomic radii of the elements and graphed it. Unfortunately, I compiled the van der Waals' radii of the elements. Thus, while the graph I obtained did show the expected decreasing trend, it also showed sudden, unexpected fluctuations in the atomic radii (eg: boron, silicon).
My question is: why does the van der Waals' radius not obey the expected decreasing trend? Also, when we speak about the atomic radius, which is a better representative for this particular case: the covalent radius, or the van der Waals' radius?
I am attaching the graph I obtained for your reference:

(van der Waals' radii of the elements (in Å))

Looks like you'd like to sum up half a book in one post. Rising effective charge is just one of many effects, adding a (sub)shell is particularly important.

Related: https://chemistry.stackexchange.com/questions/124490/why-do-atomic-radii-decrease-across-a-period ... https://chemistry.stackexchange.com/questions/31640/variation-in-atomic-radii-of-elements-in-different-blocks

Atomic radii are measured in many different ways—van der Waals radii do not capture precisely the same trends as it is calculated primarily (although not exclusively) from applying the van der Waals equation to gaseous systems and approximating the molecules as spheres. Spikes when moving from the s-block to the p-block are expected (e.g. Be $\rightarrow$ B) even in standard atomic radii maps because the p-orbital is large in size and repels away from the shielding electrons. However, when you define atomic radii based instead on the intensity of the electron cloud at certain distances from the nucleus, the trend with effective nuclear charge is maintained (i.e. the silicon issue goes away even if its van der Waal radius is anomalous).

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


text:
Safely removing sodium hypochlorite crystals from hard-to-reach surfaces
I have a small bathing whirlpool which wasn’t used for over ten years. It is built into the ground and I cannot reach the disinfectant container to replace it with a new one. I can access the fill-pipe for the disinfectant which read „Sodium Hypoclorid“ at the time of the last disinfection.
When I open the 1 cm-wide diameter pipe to fill the disinfectant in, I see a layer of half-transparent salt-like material, thus my question:
Assuming that the disinfectant within the container is solid, is there some (safe) method where I can add some special solvent and then rinse the whole container so these crystals go away?

Your substance  should have been written "Sodium hypochlorite", and not "sodium hypchlorid". This substance exists only in the dissolved state. It can not be obtained at the solid state. It gets decomposed if you try. So your crystals are not sodium hypochlorite.  Next question : what are they ? Are they soluble into water ? Is it not easier to simply rub them out of the tube, instead of finding an acidic or an oxidizing solution ?

Sodium hypochlorite can be solid, but is unstable https://en.wikipedia.org/wiki/Sodium_hypochlorite

So Onelight has not obtained any crystals of sodium hypochlorite in his pool

I have only very, very little background in chemistry- so sorry for that - I am here to learn.

only one side of the fill-pipe to the disinfectant container is accessible. The inside of the disinfectant container is not accessible. So filling something in would be the best method. Sorry accidentally mixed german word „Chlorid“ and the english one.

Are the crystal pentahydrate NaOCl·5H2O ? How to solve them?! Have that one from wikipedia. Hope you guys know better.

The special solvent you want is water

yea but water doesn’t seem to do the job to make this material liquid again. I assume hot water can do the job - is it save because of the chlorine?

Thanks @Waylander. Is it enough to fill in water and then wait for 1-2 days? Will the material resolve itself in the water? Thank You!

What you are creating is bleach solution. Handle it like you would a bottle of clorox

Standard sodium hypochlorite disinfecting bleach is a mixture of $\ce{NaOCl}$, $\ce{NaCl}$, and $\ce{NaOH}$. The sodium hydroxide, also known as lye and caustic soda, is employed to capture $\ce{CO2}$ that can lower pH via carbonic acid. The latter product reacts with hypochlorite forming the more active and unstable hypochlorous acid $\ce{HOCl}$, that has an even much reduced shelf life than hypochlorite.
Now, why $\ce{NaCl}$, because the commercial preparation of chlorine bleach is from the action of chlorine on $\ce{NaOH}$, producing equal amounts of both $\ce{NaOCl}$ and $\ce{NaCl}$.
With time, the present of even dust particles with traces amount of transition metals will breakdown the $\ce{NaOCl}$ into $\ce{NaCl}$ and some problematic $\ce{NaClO3}$ (chlorate).
Since this is evidently very aged bleach, the only problematic residual compound is the chlorate. To remove the latter, the less messy dry route is to mildly heat with a small blow torch, as to quote a source:

Sodium chlorate, NaClO3, is a colorless, odorless, white solid that melts at 248°C. When heated above 300°C, it begins to lose oxygen. The ultimate products of the thermal decomposition of sodium chlorate are oxygen gas and a white solid.

where the white solid is yet more $\ce{NaCl}$.
Other paths include the use of liquid sulfuric acid, methanol,..., and there is an associated release of toxic/explosive chlorine dioxide $\ce{ClO2}$, which is a path best avoided, in my opinion.
After the heat treatment, just collect the dry salt or water hose it away.
Some references, on the decomposition of hypochlorite and also comments here.


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How can I choose the right solvent to remove pigment from cotton
I’m trying to figure out how I might be able to dissolve a pigment without destroying the substrate(cotton, very thin). I don’t know the exact pigment, so it’s a guessing game. It is violet, I suspect it’s VP23. I thought to give it a try with a non ionic surfactant, but since they got recently banned I would like to be pretty sure before trying to get some. If you have any suggestions on how I can dissolve it(or make it colourless without the use of peroxide or bleach, since they destroy the substrate)

You’re not searching a solvent, you’re searching for a surfactant. However what you are asking seems impossible

Cotton will not be destroyed by oxygen bleaches. European washing machines literally cook the cotton laundry with boiling water+detergent+oxygen bleaches and it comes out sparkling clean. Show us a picture please.

First, try soaking the cotton in transition metal rich tap water (a possible source of Fenton/Photo-Fenton generated hydroxyl radicals) and apply gentle aeration (as from a fish tank's air pump) in the presence of added citric acid (as from lemon juice) together with strong sunlight (or other especially, a UV rich light source).
This is actually a take off of historical mild sunlight assisted lemon juice bleaching, to quote a source:

Lemon juice contains citric acid, which is a natural bleaching agent. Citric acid can whiten fabrics, and it's sometimes included in skin care products to help lighten dark spots. It's this ingredient that gradually whitens hair, making it appear lighter in color.

If this is not effective enough, add a small amount of Washing Soda (Sodium Carbonate) to the procedure. This may introduce into the reaction mix, the powerful stain removing carbonate radical $\ce{.CO3-}$. It can be created from the action of the hydroxyl radical $\ce{.OH}$ on $\ce{HCO3-}$or, in basic conditions with  $\ce{CO3^2-}$.
Some supporting background, see Figueroa, “Quantification of carbonate radical formation by the bicarbonate-dependent peroxidase activity of superoxide dismutase 1 using pyrogallol red bleaching” that proposes to actually measure the carbonate radical concentration by its ability to remove a selected red dye.


text:
Are chemical equations like a cooking recipe?
Sometimes, balanced chemical equations are compared to cooking recipes. Here is one example:

Imagine if you were baking chocolate chip cookies and measured out your ingredients incorrectly. What if you added 2 tablespoons of flour instead of two cups of flour. Or if you added one cup of salt instead of one teaspoon of salt. Your cookies would definitely not be the hit of the next bake sale.
This is also true of chemistry and why stoichiometry is an important aspect of the chemical process. Balancing measures of the reactants involved to maintain the expected outcomes of the products desired.

Here are some more examples of sites using this analogy:

https://saylordotorg.github.io/text_introductory-chemistry/s09-01-stoichiometry.html
https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:chemical-reactions/x2eef969c74e0d802:stoichiometry/a/stoichiometry
https://chem.libretexts.org/Courses/Bellarmine_University/BU%3A_Chem_103_(Christianson)/Phase_2%3A_Chemical_Problem-Solving/5%3A_Reaction_Stoichiometry/5.1%3A_Chemical_Recipes

Which part of this analogy is correct, and where does the analogy fall apart?

Related: https://chemistry.stackexchange.com/q/167230

The major analogy difference is rather qualitative aspect of recipes, making mixtures to be processed, and quantitative aspects of reactions, with reactants being consumed  in some ratio.

In chemistry you don't have to eat your mess-ups

@Waylander Unless the boss would figuratively make you.

Equations aren't recipes. You have actual recipes in synthetic chemistry.

Speaking programming paradigms language, cooking recipe is imperative programming, whereas chemical equation is symbolic programming. Comparing them is like comparing apples to oranges. Tro's Chemistry: a molecular approach textbook pretty much equates stoichiometry concept (not equation per se) and cooking recipe, although I think it's an oversimplification. Take this comment with a grain of salt since I generally don't like this textbook for complete disregard for notational standards and overly dumbed-down narration.

Cooking and baking each involve a lot of chemistry themselves, and this analogy largely holds in the case of stoichiometry.
One might take liberty with a cooking or baking recipe and change it, adapt it, mess with some of the ingredients, etc. and this is not a scenario mapped to stoichiometry. The rigidity of stoichiometry comes logically from the premise that the smallest "unit" of a chemical reaction is the rearrangement of bonds between individual atoms. The law of the conservation of matter demands that all atoms at the beginning of a reaction must be present at the end of a reaction, and so these ratios of atoms are not arbitrary.
The primary way this breaks down, though, is in the end result from messing up the recipe. In baking, if your cake recipe calls for 2 cups of flour, 2 eggs, and 1 cup of sugar, using only 1 egg will get you a single, terrible cake. In chemistry, however, this will result in—to stretch the analogy—half of a great cake, 1 cup of flour, and half a cup of sugar. If you don't have enough of one thing in a chemical reaction, then the reaction still happens to some of the materials, but then it runs out and you have some "ingredients" left over.

The difference between cooking and chemistry is in the precision. To make a cake, let's admit that you need $200$ g flour, $100$ g eggs and $100$ gram sugar. But if you take $90$ g or $110$ g sugar, the result will be the same. This is not true in chemistry. To make water from hydrogen and oxygen, you need $\ce{2.02 g H2}$ and $\ce{16.00 g O2}$. If you use $\ce{2.03 g H2}$ with the same amont of oxygen, you will obtain the same amount of water, and the $\pu{0.01 g}$ excess of $\ce{H2}$ will remain at the end of the reaction.

Reversibility
All chemical reactions are reversible in the presence of sufficient energy, but you cannot unbake a cake or unfry an egg. Indeed, many reactions proceed both forwards and backwards simultaneously. But mixing flour, sugar and salt in a bowl generally does not entail simultaneous mixing and unmixing of the ingredients except perhaps at the very smallest scales.

"All chemical reactions are reversible in the presence of sufficient energy, but you cannot unbake a cake or unfry an egg." Both of those are the results of chemical reactions, though.

@nick012000 Sure, but they are not strictly chemical reactions. They are emergent phenomena that only exist at a scale far above individual molecules. For instance, if I give you a collection of 100 hundred molecules, can you identify for me, unambiguously, which ones are "fried egg molecules" vs "not fried egg molecules"? You can look at which ones are denatured vs. not, and make a guess, but a runny yolk can contain plenty of intact proteins, which makes the result ambiguous at best.

@nick012000 furthermore, the denaturing of a protein can be theoretically reversed, but in practice, the protein has so many configurations it can move through that the odds of a fully denatured protein spontaneously reforming into its original configuration are statistically zero.

@LawnmowerMan reversing chemical reactions isn't just a function of sufficient energy, it also requires sufficient INFORMATION about the system—in fact this is exactly WHY cooking is an emergent process from a series of chemical reactions. Something like for example cyclohexane combustion is also "irreversible" but it isn't cooking. Irreversibility is an entropic consideration, and this also is irrelevant to cooking as analogy

Are chemical equations like a cooking recipe?
No. A chemical equation displays the strict theoretical relations of amount of substance of the educts and resulting products.
A cooking recipe would be closer to the description of a chemical synthesis procedure. The goal of the chemical synthesis would be to achieve a chemical reaction according to a chemical equation, but would a) be very specific about how to do things (temperature, how long for the reaction to take, how to apply certain procedures, like destillation, purificiation of products etc) and b) might use different quantities from what the actual chemical equation would theoretically suggest. Some reactions can only be made work properly if a certain excess amount of educt is used, etc.
Working out unknown chemical synthesis procedures requires a lot of creativity and playing around with the amounts of substance.
Just putting together the amounts as required by the exact chemical equation will in most cases yield disappointing results.


text:
Would an effervescent tablet react with itself if ground into a nanopowder?
Effervescent tablets usually are a mixture of a weak acid, base, and binder. When dry there is no reaction. But when placed in water the acid-base neutralization releases CO2 rapidly.
If you ground a (dry) tablet into nanopowder, the acid and base grains would have enormous surface area. Would this cause them to react despite having no solvent?

Anyway, you will just obtain a surface reaction, producing the tiniest amount of gas, despite the huge surface area. Furthermore, the residual solid stuff will prevent new fresh molecules from coming into contact one another. So the rection will quickly stop.

There is possibility of water vapor adsorption and autocatalysis by produced water. Anyway, avoid sneezing toward nanopowders.

related https://chemistry.stackexchange.com/questions/34916/how-might-ammonia-be-created-in-this-mechanochemical-reaction

I guess it would react, but you'd need long time and/or temperature, not to mention water vapor in the air would influence it.


text:
How does pressure affect the color of reaction medium with equimolar amounts or reactants and products at equilibrium?
Chlorine gas is pale yellow-green, and both hydrogen and hydrogen chloride are colorless gases. According to the Le Chatelier's principle, the change in pressure will not affect the following equilibrium:
$$\ce{H2(g) + \color{green}{Cl2(g)} <=> 2HCl(g)}$$
Reduced volume will increase the partial pressures of gases. Will the color of the medium remain the same, or acquire a more saturated green tint due to increased concentration of chlorine gas?

If you evaporated a half volume of orange juice and it's color become more intense,  it would not mean additional color substance were created. It would just be more concentrated.

Doesn’t equilibrium happens in closed system?

Sure, it does, but it is not the point. It was analogy. Consider pure chlorine in a closed system, which is compressed to 1/2 volume and it's colour intensifies. Was there created extra chlorine? // You have to distinguish component mass and component concentration.

Isn't some high temperature (in some sense of high) presumed? What is the starting condition? - e.g. has a small spark been applied first...? Or what is the (potential) chemical energy? - Is some minimum energy assumed (say, after the said small spark).

By (isothermal) compression of such a system, the equilibrium is not shifted, so the chlorine mass and molar amount stays constant (with possible tiny changes due  non-ideal behaviour of gases).
But as the system volume decreased, concentration (or equivalently partial pressure) of components increased and the perceived chlorine color intensified.
If compression of the system were done by a linear way, like by transparent piston in a cylindrical vessel, the light absorption along the direction of this shortened path would be the same. As the shorter path is balanced by more intense absorption.
It means, the gas would be perceived in that direction as if colour did not change, but with more intense color in perpendicular directions.


text:
Calculating half-cell potential from two reduction potentials
It is sometimes necessary to compute E° for a given half reaction from other half reaction of known E°.
For example, the standard electrode potential for the oxidation of Titanium metal to Ti³⁺ can be obtained from the following half reaction:
Ti²⁺ + 2 e⁻ ⇌ Ti⁰ E° = -1.63 Volts ①
Ti³⁺ + e⁻ ⇌ Ti²⁺ E° = -0.37 Volts ②
Calculate E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
My answer: ② is more spontaneous reaction than ①.
Oxidation: Ti³⁺ + e⁻ ⇌ Ti²⁺ E° = -0.37 Volts ;
Reduction: Ti²⁺ + 2 e⁻ ⇌ Ti⁰ E° = -1.63 Volts
Hence,
$E^{\circ}_{rx} = E^{\circ}_{red} - E^{\circ}_{ox} = 1.63 V - 0.37 V = + 1.26 Volts $
My answer matches with author's answer. But I am doubtful about the correctness of the answer?
But, Is my logic in answering this question correct?
Would any member of chemistry stack exchange explain  me which  $E^\circ$ of half reaction be chosen as reduction $E^\circ$ and which $E^\circ$ of half reaction be chosen as oxidation $E^\circ$ in this question and why?

Answer: neither. Look at the link above for a full example of how to do this. Sure, the  numbers are slightly different, but otherwise it's entirely the same problem you're trying to solve.

@orthocresol, So $E^\circ$ for the reaction $Ti^0 \rightleftharpoons Ti^{3+} + 3 e^- = + 1.21 volts$ that means author's answer is wrong.

That’s what I’m getting, too.


text:
A method to create NH3-NH4Cl buffer solution using titration method
My aim is to create a basic buffer solution.
I know that to create a basic buffer solution, we need a strong acid and weak base. In my case I have chosen $\ce{NH3}$ and $\ce{NH4Cl}$.
Using titration, I first want to create $\ce{NH4Cl}$ shown in equation below.
$$\ce{NH3 + HCl -> NH4Cl}$$
After this step I am a bit confused.
Should I mix the $\ce{NH4Cl}$ and $\ce{NH3}$ to create the buffer?
Also I want to create a buffer solution with $\mathrm{pH} = 10$. Can anyone tell me the concentration and volumes required for each of the chemicals in this case?

I know that to create a basic buffer solution, we need a strong base and weak acid. - Exactly the opposite. This is acidic buffer (e.g. acetic acid+ sodium hydroxide).

So what type of buffer would NH3 and NH4Cl create?

basic, as it is weak base + its salt with strong acid.// acidic buffer is weak acid + its salt with strong base.

alright ill deit my original

Can you help me by answering the original question tho?

There is plenty of online info covering this topic, it would be bringing owls to Athenes. Try thorough searching. E.g. libretexts.org contains ton of study materials, so does this site, some of answers may even be mine.

I searched but couldn't get any which involved titration. Most of them just mixed powdered NH4Cl to create the buffer. But I specifically need to use titration. Any sources you could provide?

Titration is just a small detail, how to transform some substances to other ones. It can be used to set the conjugate acid/base pair ratio. The problem is, pH values cannot be predicted exactly, exact value has to be determined by measurement. Or, use published pH buffer tables.

If V1 is HCl volume titrated by V2 volume of NH3,  Then mixing V1 HCl + (n+1)V2 NH3 gives approximately  pH=9.25 + log(n) Total ammonia molarity should not be more then 0.1 M

As Poutnik correctly pointed out in the conmment section, titration is just a small detail in preparation of buffer solution. Actually it was a little confution to me. Yet, titriometric calculations can be used to set the conjugate acid/base pair ratio for a buffer with perticular $\mathrm{pH}$. Still, the $\mathrm{pH}$ value of the buffer is approximate value and cannot be predicted exactly. The exact value can be achieved by using a $\mathrm{pH}$ meter and either a dilute sulution of $\ce{NaOH}$ (if $\mathrm{pH}$ is below the expected value) or a dilute sulution of $\ce{HCl}$ (if the $\mathrm{pH}$ is above the expected value).
There are few methods to prepare buffer solutions. For example, you can simply calculate the weak base (e.g., $\ce{NH3}$) and its conjugated acid (e.g., $\ce{NH4Cl}$) using Henderson-Hasselbalch equation for the given $\mathrm{pH}$. Suppose you need to prepare $\ce{NH3/NH4Cl}$ buffer at $\mathrm{pH} = 10.0$. The relevant Henderson-Hasselbalch equation is:
$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log \left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right)$$
Since you know the  $\mathrm{pOH}$ $\left(= 14.0 -\mathrm{pH} = 14.0 - 10.0 = 4.0\right)$ and $\mathrm{p}K_\mathrm{b}$ of $\ce{NH3}$ ($4.75$), you can calculate the ratio of $\left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right)$:
$$\left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right) = 10^\left(\mathrm{pOH} - \mathrm{p}K_\mathrm{b}\right) = 10^\left(4.0 - 4.75\right) = 1.78$$
If you have $\pu{0.1 M}$ $\ce{NH3}$ solution, you need to have $\pu{0.1 \times 1.78 M}$ $\ce{NH4+}$ ($\pu{0.178 M}$) in you buffer solution. Assuming volume of the $\ce{NH3}$ solution woud not increse when certain amount of solid is dissoved, you can dissolve $0.178 \times \pu{53.5 g} = \pu{9.52 g}$ of solid $\ce{NH4Cl}$ in $\pu{1.0 L}$ of $\pu{0.1 M}$ $\ce{NH3}$ solution. That would give you the requred buffer solution with $\mathrm{pH}$ around $10$. Put a $\mathrm{pH}$ meter in the solution to measure accurrate reading. If that value is above $10$, use $\pu{3 M}$ $\ce{HCl}$ solution to readjust solution to $\mathrm{pH} = 10.0$. Similarly, if the value is below $10$, use $\pu{3 M}$ $\ce{NaOH}$ solution to readjust it to $\mathrm{pH} = 10.0$.
Now, you need to prepare a $\ce{NH3/NH4+}$ buffer at $\mathrm{pH} = 10.0$ using $\ce{NH3}$ solution and a $\ce{HCl}$ solution without using solid $\ce{NH4Cl}$.
Suppose you have conc. $\ce{HCl}$ and $\pu{500 mL}$ $\pu{0.2 M}$ $\ce{NH3}$ solution to prepare sought buffer solution. We need higher concentration of $\ce{NH3}$ solution beecause part of $\ce{NH3}$ is going to convert to $\ce{NH4+}$ during the process after strong acid/weak base reaction:
$$\ce{NH3 + HCl -> NH4+ + Cl-}$$
The conc. $\ce{HCl}$ solution should be used to minimize the increment of the volume significantly. We know by previous calculations that after adding enough conc. $\ce{HCl}$ ($\approx \pu{15 mL}$), the ratio of $\left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right)$ should still be $1.78$ if the $\mathrm{pH}$ reading of $\mathrm{pH}$ meter indicates $\approx 10$. Thus, we should put a $\mathrm{pH}$ meter in $\pu{500 mL}$ $\pu{0.2 M}$ $\ce{NH3}$ solution to measure accurrate reading when we add conc. $\ce{HCl}$ dropwise (in this way, we don't need to measure exact amount of conc. $\ce{HCl}$). When the $\mathrm{pH}$ meter reads exactly $10.0$, stop addition of conc. $\ce{HCl}$  and dilute the buffer solution to $\pu{1.0 L}$ using deionized water. At this point, the $\mathrm{pH}$ meter reading would change, but you can readjust it to $10.0$ as follows:
If the $\mathrm{pH}$ of diluted solution is above $10$, use $\pu{3 M}$ $\ce{HCl}$ solution to readjust solution to $\mathrm{pH} = 10.0$. Similarly, if the value is below $10$, use $\pu{3 M}$ $\ce{NaOH}$ solution to readjust it to $\mathrm{pH} = 10.0$.
Note that $[\ce{NH3}]$ in each solution prepared by method 1 and method 2 are not identical as well as $[\ce{NH4+}]$ in those solutions.  However, $\left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right)$ is still $1.78$. If you need accurate concentrations of each species (e.g., $[\ce{NH3}] + [\ce{NH4+}]$), you should do more accurate calculations.


text:
When does a long carbon chain become more significant to the boiling point of a substance than polarity?
I'm asked to sort the following compounds according to their boiling point (in an increasing manner): acetic    acid, sodium chloride, and pentadecane (C15H32). Obviously, sodium chloride will have the highest boiling point because it's an ionic compound. But what about the acetic acid and the pentadecane? How does one proceed when you have to choose between a long hydrocarbon chain and a polar molecule? Is there a general rule or I should just guess in a case where the chain is smaller than that of a pentadecane?

The general rules are these: Enough cents have higher value than 1 Euro. Enough Euros have higher value than many cents. Determine cents versus Euro balance.

The easiest way is to find boiling point of acetic acid and of homologic series of n-alkanes. With better experience, you may approximately remember boiling points of some compounds, or predict them from known patterns. My rough estimation is  n-octane or nhdecane may have b.p.around acetic acid b.p.

Acetic acid is liquid, higher alkanes solid, but for example HOOC-COOH is solid too. Every functional group and their number have different impact on melting/boiling point.

The question is closed, so I can't answer; but I agree tge problem was not well posed. I would have used 2-methyl-propane, which has similar MW and arrangement of its heaviest atoms compared with acetic acid, instead of pentadecane. Then the ordering intermolecular forces is clear.

Thanks for your question.
The B.P. of Acetic acid glacial is 118 C and Pentadecane is 270 C.
I think finding a general rule (or equation) for predicting the boiling point of an organic substance (its relation with factors like carbon chain length and polarity of its functional groups and...) might be difficult. But, in some cases we can guess in a number of similar substances, witch one may have a higher boiling point. Let me explain more about intermolecular forces and what may be this similarity.
The intermolecular forces are actually the electrostatic forces between positive and negative charges. These charges can appear simply by movement of electrons in a molecule. But, an instant later the new movement of electrons may decrease the charges or even naturalize them. These instant dipoles is considered to take place in every molecules and at different parts of it (with different magnitudes and directions at each instant). Although the dipoles are instant but an electrostatic force will appeared between molecules.
This type of intermolecular force is called dispersion forces(London forces). Naturally, it gets stronger by increasing the number of electrons in a molecule.
When element with high affinity to electrons (like Fluorine) are present in a molecules, permanent dipoles will appeared, among London forces. One may think that permanent dipole(dipole-dipole attraction) may be a stronger intermolecular bond than London force(for example, consider H-F and H-H). But, that's not generally correct.
Pentadecane is a compound with almost only London forces with such a high boiling point of 270 degree of centigrade. This indicates that the molecules needs high amount of kinetic energy, in order to could exit the liquid phase(to overcome the intermolecular forces and work on atmosphere pressure). The boiling point is even more than of water that has a strong hydrogen bond between its molecules.
So, generally we can't predict the trend in boiling points just by looking at types of intermolecular forces.
But, in some cases, when the substances are similar in some specifications, some assumptions can be made.
Consider the straight chain alkanes starting from methane. As more electrons results in stronger London forces, we may concluded that the boiling point must rise as the carbon chain become longer. Actually, in the experiment this is true.
Now, consider the boiling points of isooctane with eight carbon that is 99 degree and heptane with seven carbon that is 98.8 degree. Surprisingly, they are very close to each other. The branching leads to a decrease in isooctane(octane isomer) boiling point.
The intermolecular forces in this example was only London force. Adding dipole-dipole interactions (for example acetic acid in your question) makes problem more complex. You can find more information about this subject at the sites below.
https://www.chemguide.co.uk/atoms/bonding/vdw.html
https://amrita.olabs.edu.in/brch=7&cnt=1&sim=111&sub=73


text:
Why is H2N2O2, and not HNO, called hyponitrous acid?
According to IUPAC Red book page 140, hyponitrous acid, correctly said as hypodintrous acid, is $\ce{H2N2O2}$ while the compound $\ce{HNO}$, azanone also exists. So why is the dimer considered the hypo of -ous oxyacid of nitrogen over $\ce{HNO}$?

Useful links for text and formula formatting:
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@Poutnik It is the actual dimer.

In fact it HNO dimerises very fast - it's an unstable intermediate, so hyponitrous acid is the product of reaction creating HNO. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC124232/

Well, nitroxyl/azanone is a highly reactive species that spontaneously dimerizes to hyponitrous acid, which dehydrates to give nitrous oxide and water. $$\ce{HNO + HNO -> HON=NOH -> N2O + H2O}$$ with reaction rate constant of $\ce{{\it{k}} = 8 \times 10^6 M^{-1} s^{-1}}$. Hence, HNO cannot be stored or detected for long time.

I have reused the relevant portion of my answer in the related What is formal dimer?.
Comparing $\ce{HClO}$ as hypochlorous acid and $\ce{HNO}$ as eventual hyponitrous acid is misleading, as they have different structure.
$\ce{H-O-N=N-O-H}$ could be a true dimer of azanone $\ce{H-N=O}$ if the latter had structure $\ce{H-O-N}$, with the equilibrium reaction $\ce{2 H-O-N <=> HON=NOH}$.
As nitrous acid is $\ce{H-O-N=O}$, hypothetical monomeric hyponitrous acid is supposed to be $\ce{H-O-N}$ like $\ce{H-O-Cl}$, and not $\ce{H-N=O}$.
Interesting fact is that dissociation of $\ce{HNO}$ is spin-forbidden quantum event and the rate of the dissociation is very slow.

Let us assume you are discovering hyponitrous acid for the first time. You analyzed the elements and it is found to contain nitrogen, hydrogen and oxygen in a ratio of 1:1:1, and hence the empirical formula should be $\ce{HNO}$, as you wanted. The formula weight of this compound would be 1+14+16=31 (as a rough estimate).
The story of chemical formulae do not end here. People wanted to determine the molecular weight of this new compound. Chemists as early 19th century knew how to determine the molecular weight by relatively simple but elegant experiments. Only high school algebra is required.
Molecular weight measurements would tell the chemists that is 2x31 i.e., 62 not 31. This is just an example to clarify that elemental analysis and its bare formula is not enough to develop the correct formula. Molecular weight measurements are necessary.
Historically, the acid was not analyzed but its silver salts and the early discoverers assigned a formula of the salts as AgNO, implying HNO for hyponitrous acid (Ueber die untersalpetrige Säure, DOI: 10.1002/cber.18770100257).
In short, experimental measurements of the molecular weight decide that it is $\ce{H2N2O2}$ rather than $\ce{HNO}$.

Technically, per Wikipedia, hyponitrous acid ($\ce{H2N2O2}$) is an isomer of nitramide ($\ce{H2N−NO2}$) and a formal dimer of azanone ($\ce{HNO}$).
If your question is why it it is called hyponitrous acid, my suspected explanation is that the naming convention is following the early tradition established with, for example, the corresponding family of chlorine-based acids, where in moving down from chlorous acid ($\ce{HClO2}$) to hypochlorous acid ($\ce{HClO}$), there is a loss of one atom of oxygen.
So here, think of hyponitrous acid ($\ce{H2N2O2}$) as likely very early posted in the chemical literature as $\ce{HNO}$, which compared to nitrous acid $\ce{HNO2}$, is correspondingly minus one atom of oxygen from the nitrous acid.
My supposition on the $\ce{HNO}$ initial assignment is supported by the following comments from Atomistry, which is a compilation of historical chemical journal commentaries, to quote:

It was shown by Divers in 1871, that the reduction of sodium nitrate with sodium amalgam gave a salt having the empirical composition NaNO, the acid being an isomer of nitramide...
A bibliography of hyponitrous acid has been compiled by Divers, to whose researches so much of our knowledge of this interesting compound is due.

I hope this helps, and reading old chemistry texts, at times, may provide some insights.


text:
What are the "ozonated magnesium oxides" used in this product?
I recently was recommended a product for gut health/constipation that actually works quite well - the product names of the main distributors are OxyPowder and Homozon. Despite the product working really well both of the sites make a bunch of wild claims about how this product is and the history of it. The story is pretty wacky and originates from one of these old timey doctors back in the late 1800s when they were trying a bunch of weird stuff for medical conditions.
The ingredients on these products say only "Ozonated magnesium oxides" and the ones that come in capsule form come with citric acid, and powders say to also eat citric acid or something sour. This is the other tip off to me this is not just magnesium oxide, and I can tell its effect is much different than magnesium oxide.
I have seen a few similar products refer to this as "magnesium peroxide" but I don't know if that is right either. In one of the descriptions the "chemical formula" is claimed to be "2(MgO3)2" - I don't know much about chemistry to say if this is right but with the O3 in there that wouldn't be Magnesium Peroxide. I know that Magnesium Ozonide on its own is unstable.
A claim by the manufacturer "he knockoffs of Homozon don't use the special pharmaceutical grade magnesium, and they only bond oxygen, not ozone, to the magnesium. Bonding ozone to a powder (a gas to a solid) is a trade secret process, that's why no one else has the same high quality product. " They also claim they've done tests that can prove that whatever reaction is happening is releasing oxygen over 8 hours or so.
I tried to search patents but I can't find anything about this. Given that this all started in the in the old days an this is a small lab producing this it's hard for me to believe that there is some secret method used by a wacky doctor in the late 1800s.
Some of the research there lead to an older German product called Hämozon, which does make reference to "Magnesium superoxide" and MgO2, but supposedly the formula has since been expanded upon.
Any ideas?

The contents are unclear, products more so, and in the presence of Mg ions, increased oxidative stress diseases are likely. See, for example, "Pro-oxidant Activity of Aluminum: Stabilization of the Aluminum Superoxide Radical Ion" by Mujika et al, published in Phys. Chem. A, 2011, 115, 6717–6723, American Chemical Society).  Also, the article is “Oxidative Stress Gated by Fenton and Haber Weiss Reactions and Its...” by Tushar Kanti Das, et al, published in Archives of Neuroscience, July 2014.

The one well-established magnesium oxide at ambient conditions, apart from $\ce{MgO}$, is the peroxide $\ce{MgO2}$. Usually this is synthesized by deprotonating hydrogen peroxide with $\ce{MgO}$, but ozone would be a strong enough oxidizing agent to oxidize $\ce{MgO}$ to $\ce{MgO2}$. There is actually evidence of ozone even forming a superoxide $\ce{Mg(O2)2}$ [1], but most likely in the commercial product the ozone oxidizes the magnesium oxide to $\ce{MgO2}$.
Magnesium peroxide, like many metal peroxides, can release oxygen when acid is added. Thus the recommendation to add citric acid is to promote oxygen release, as opposed to any inherent difference in basicity as such.
Reference

Vol'nov, I.I., Tokareva, S.A., Belevskii, V.N. et al. "The formation of magnesium perperoxide Mg(O2)2 in the reaction of magnesium peroxide with ozone" Russ Chem Bull 19, 468–471 (1970). https://doi.org/10.1007/BF00848959.

We also do have magnesium ozonide ;) (cc @V_H)

Added ambient conditions, whic specifically favor MgO2.


text:
Can you oxidize gold using only oxygen?
I keep reading that gold does not react with oxygen, but I also see information about gold Oxides (ie: aurous oxide, Au2O and auric oxide, Au2O3)
Knowing that there is a way of getting gold oxide using multiple steps and additional chemicals (ie. decomposing aurous chloride), is there a way of getting gold oxide without any intermediate steps? Can you produce gold oxide if you have a closed system with gold and oxygen and the ability to control the amount of energy in the system?

No, it's not possible, why ask when you already heard that?

@Mithoron b/c it sounds more like what someone would say in relation to what normal people and jewlers subject gold to on a day to day basis and not what a chemist might subject gold to in a controlled experiment.

@Mithoron if this is the correct answer please post it so I can accept it.

Getting back to the post. With the downvotes this post is receiving and what I thought was a simple question, can I assume that even though chemists are not posting an answer the correct answer is that "Yes the coloquial statement that pure gold does not react with air is true regardless of the energy in the environment."

@Tolure See https://doi.org/10.1038/179212a0: gold does react with pure oxygen at elevated temperatures (~900 °C), albeit very slowly. And please refine whether you are asking about oxygen or air since the answers will be vastly different.

Thank you @andselisk for your comment. My question is specific to oxygen, I should of pharsed my comment to be "Yes the coloquial statement that pure gold does not react with oxygen is true regardless of the energy in the environment." Could you post that as an answer.

I bet I can implant oxygen ions and synthesize it, but I don’t think you mean that quite.

Well, this answer says gold does not reacts with oxygen at all. Above 160 C, any gold oxide (including intermediate oxides) will decompose.

The most common gold oxide $\ce{Au2O3.xH2O}$ is made by action of alkali on aqueous gold(III) solution. Even gold is resistant to ozone up to 100 °C.

The most stable gold oxide Au2O3(anhydrous) reportedly decomposes above 160 Deg C.

I'm sure that gold can react with oxygen even at room temperature if the oxygen pressure is high enough. As in, hundreds of thousands of atmospheres or more.

Note: There were two answers to this question, where my sulfite/oxgen/copper/chloride path was deleted as not relevant to the chemical oxidation of gold involving oxygen. Please note this August 2021 commercial gold dissolution article: "Simultaneous sulfide oxidation and gold dissolution by cyanide-free leaching from refractory and double refractory gold concentrates"
https://www.sciencedirect.com/science/article/pii/S0892687521002715 and also note that sulfite is not radicalized absence oxygen, transition metal and chloride. Does anyone know how to undelete?

@AJKOER Gold oxidation is far wider scope than gold reaction explicitly  just with oxygen.

Agreed, as to why in my cited 'answer', I argued that it was not, albeit, a simple direct oxygen path! Those curious in my cited chemistry will have to now demand an undelete.  Note: material presented therein is contained in an in process patent filing, as full disclosure, so happy if it stays deleted.

@AJKOER chill out. Sometimes good answers lose because they do not fit the question precisely enough. If you can see hidden answers, migrate to this Math SE q and observe my own misfortune.

No, need, am happy that part of my Patent's content is now secret. If anyone undeletes, read it quickly as on notification, I will delete. However, this not a matter of a 'good answer' as the ascribed rated answer cites nano gold on a support? At least, I addressed  the intent of the question (regular gold metal) and disclosed needed agents, and honestly asserted it was not an answer (implying there is no such pathway and that still is my opinion).

Actually, yes — if you cheat a little by using atomic oxygen. Ono and Cuenya[1] report that combining gold nanoparticles with atomic oxygen at 150 K results in gold oxide formation, and the oxidized nanoparticles release the oxygen as $\ce{O2}$ upon heating. The authors further note that the nano-oxide is more stable on a silica support than on a titania support, suggesting a catalytic effect of the latter material.
This finding is relevant to electrochemical oxidation of water on gold anodes, where the electrode is oxidized by the oxygen atoms from water molecules prior to evolving $\ce{O2}$ [2].
References

Luis K. Ono and Beatriz Roldan Cuenya (2008). "Formation and Thermal Stability of Au2O3 on Gold Nanoparticles:  Size and Support Effects". J. Phys. Chem. C, 112, 12, 4676–4686. https://doi.org/10.1021/jp711277u

Oscar Diaz-Morales, Federico Calle-Vallejo, Casper de Muncka  and  Marc T. M. Koper (2013). "Electrochemical water splitting by gold: evidence for an oxide decomposition mechanism".
Chem. Sci., 4, 2334-2343. https://doi.org/10.1039/C3SC50301A

Huh, does atomic count? Then ozone also would, I guess? BTW how are they making it?

@mithoron the experimental section refers to O2 plasma, so tge atomic oxygen appears to have cone from energizing O2 to make the plasma. Energy input would be acceptable since many overall spontaneous reactions need such input (e.g. heating) to be observed.


text:
What is a "formal" dimer?
I was researching for this question and I noticed the term "formal dimer":

Hyponitrous acid is a chemical compound with formula $\ce{H2N2O2}$ or $\ce{HON=NOH}$. It is an isomer of nitramide, $\ce{H2N−NO2}$; and a formal dimer of azanone, $\ce{HNO}$.

Google didn't gave any meaning of this term, rather it gave me bunch of examples:

Carbene dimerization is a type of organic reaction in which two carbene or carbenoid precursors react in a formal dimerization to an alkene.

Specifically, when simple aromatic 6n-electron systems, which are extremely sluggish in Diels-Alder additions, are incorporated into a [2n]cyclophane system, e.g. [2.2]paracyclophane which is a formal dimer of p-xylene, a dramatic increase in the rate of addition is observed in certain cases.
ORGANIC REACTIONS AT HIGH PRESSURE. DIELS-ALDER REACTIONS OF
[2. 2]PARACYCLOPHANE by Kiyoshi MATSUMOTO, CHEMISTRY LETTERS, pp. 1681-1682, 1985 (link)

What is the meaning of this term and how is it different from just dimer? Are they referring to "formal charge"?

Twins dressed in tuxedos, maybe?

Glucose $\ce{C6H12O6}$ could be considered as a formal hexamer of methanal $\ce{CH2O}$.
A formal dimer of methanal is acetic acid $\ce{CH3COOH}$ as the very contrasting example, compared to e.g. $\ce{O2N-NO2}$.
A formal trimer of methanal is lactic acid $\ce{CH3CH(OH)COOH}$.
To close the cycle, glucose can be considered also as a formal dimer of lactic acid or a formal trimer of acetic acid.
There is equilibrium $\ce{2 monomer <=> dimer}$ for "true" dimers. Formal dimers have either just atom count multiplicity, or reversible reactions are not observed and structure of the dimer is very different. That said, the boundary between true and formal dimers/oligomers is not sharp, but little fuzzy.
In some way, dimerization could be seen as reversible  glueing 2 monomers together by a chemical bond(s). The more the reaction differs from this model, the more the dimer leaves the status of a true dimer toward formal dimers. At the opposite end of the scale, there stand very formal dimers/oligomers, with no way how to get them from monomers, unless by a sophisticated synthetic path.
$\ce{H-O-N=N-O-H}$ could be a true dimer of $\ce{H-N=O}$ if the latter had structure $\ce{H-O-N}$, with the equilibrium reaction $\ce{2 H-O-N <=> HON=NOH}$.
As nitrous acid is $\ce{H-O-N=O}$, hypothetical monomeric hyponitrous acid is supposed to be $\ce{H-O-N}$ like $\ce{H-O-Cl}$, and not $\ce{H-N=O}$.
Interesting fact is that dissociation of $\ce{HNO}$ is spin-forbidden quantum event and the rate of the dissociation is very slow.


text:
Method for monitoring pyridine activation by chloroformate
I am supposed to activate a pyridine using a chloroformate, and then reduce it with hydride to form a dihydropyridine, in a similar way to the paper:
Heusler, A.; Fliege, J.; Wagener, T.; Glorius, F. Substituted Dihydropyridine Synthesis by Dearomatization of Pyridines. Angew. Chem., Int. Ed. 2021, 60 (25), 13793–13797. DOI: 10.1002/anie.202104115.
I am wondering if there was a convenient way to monitor the activation step as it seems that my substrate is much slower to activate than the ones used in the publication.

What instruments do you have access to?


text:
How do we know that rates depend on the product and not the sum of reactant concentrations
I've been struggling to understand the mathematics behind the calculation in enzyme kinetics within systems biology at an intuitive level.
Every research article and textbook says it is based on the mass-action law, and I never came across how this law was invented.
for e.g in the chemical reaction
$$
\ce{\alpha A + \beta B <=> \gamma C + \delta D}.
$$
The law of mass action gives the following kinetics
$$
V^- = {-k[C]^\gamma [D]^\delta}
$$
$$
V^+ = {k[A]^\alpha [B]^\beta}
$$
What was the reason behind taking the product, instead of taking addition.
Why is the following equation incorrect?
$$
V^- = {-k[C]^\gamma + [D]^\delta}
$$
$$
V^+ = {k[A]^\alpha + [B]^\beta}
$$
Was the law of mass-action experimentally derived or it was mathematically derived? What derivation rule was used to take the product instead of addition.
I want to understand this concept of taking the product of concentrations rather than cramming it.

https://chemistry.stackexchange.com/questions/68195/law-of-mass-action-collision-theory

https://chemistry.stackexchange.com/questions/6924/why-does-the-reaction-quotient-use-the-products-multiplications-of-reactants-a

@Dendrobium - OK, what about them (and other linked answers) not answer? I might have chosen https://chemistry.stackexchange.com/questions/14836/is-there-a-reason-for-the-mathematical-form-of-the-equilibrium-constant instead. Just saying it didn't answer you question doesn't really help the rest of us figure out what you are having difficulty with.

https://babel.hathitrust.org/cgi/pt?id=pst.000067440054&view=1up&seq=86&skin=2021
If anyone can read german here's the paper that proposed it

Note this site is not a forum for extended discussions and dialogue, although there are chat rooms available as need be. The idea is to compile "canonical" answers, the best answers that address clear questions. If there are answers that already address your problem then yours might be closed as a duplicate. OTOH if it seems unclear what you are confused about (requiring you to explain yourself in more detail - in this case justify the equation you wrote), it might also get closed.

What was the reason behind taking the product, instead of taking addition.

There are at least three reasons:

Experiment: Empirical data on reactions that were experimentally studied.
Logic: If one reactant is missing, an "addition formula" would still give a reaction even though the reaction is impossible
Theory: The combined evidence led to the concept of collision theory, where reactants collide to give a reaction. The probability of collision is given by an expression that includes the product of the concentrations of particles colliding.

The relationship between kinetics and law of mass action can be more complicated for reactions that do not consist of an elementary step, and the answers to the linked questions give some information about that topic.

Was the law of mass-action experimentally derived or it was mathematically derived? What derivation rule was used to take the product instead of addition.

The law of mass action was experimentally derived but now has a rigorous derivation as part of equilibrium thermodynamics.


text:
How to calculate the freezing point of an equimolar ethanol: water mixture using colligative properties
I am familiar with how to calculate the freezing point of a mixture with a 9:1 molar ratio of ethanol to water (see below). I can understand that ethanol, having a higher mole fraction, is considered to be the solvent. Hence in this case, we are finding the depression in freezing point assuming water to be the solute.
If I were to modify the question so that equal amounts of ethanol and water are present in the mixture, how would one proceed to find the depression in freezing point/elevation in freezing point of the mixture? What is the solvent and what is the solute in that case?

Credits:solution provided by https://www.doubtnut.com/

What is the solvent and what is the solute in that case?

To explore this, you could try both. You would get two different answers, though.

how would one proceed to find the depression in freezing point/elevation in freezing point of the mixture?

Colligative property expressions work well for low concentrations (where the solutes behave like they are "ideally diluted"). If you look at the experimental data below, you realize that for higher concentrations, the change in freezing point is no longer linear:

Source: https://commons.wikimedia.org/wiki/File:Phase_diagram_ethanol_water_s_l_en.svg
You also see that the 1:9 water: ethanol mixture might be at the edge of where the freezing point depression formula is useful (note the scale in the image is a mass ratio, not an amount ratio).


text:
Should I use DMA or tensile and compressive testing?
I have some polymer based samples (the exact composition is proprietary) that were additively manufactured, however the sample amount is severely limited so I'm limited on the amount of material testing that I can do. Therefore, would anyone recommend DMA (Dynamic Mechanical Analysis) over standard mechanical testing and why? Also, I likely don't have enough material to test all 3 orientations (x,y, and z).

What are you testing for?


text:
Mechanism for synthesis of benzofuran
Would someone be able to help me with the mechanism for the synthesis of benzofuran from salicyladehyde (taken from https://en.wikipedia.org/wiki/Benzofuran)? I have got as far as the decarboxylation step.

Which decarboxylation step? A lot of reactions there! Benzofuran 2-carboxylic acid?

I agree with @user55119, it would help if you were a bit more explicit. Consider also sharing what you've done so far -- so that people can see where you got stuck and pick up from there.

I presume there is some heating for the decarboxylation step as benzofuran-2-carboxylic acid is commercially available..


text:
How does aromaticity play out in carbon nanohoops?
Within the family of carbon phanes,one that has caught my attention is cycloparaphenylene. Considering that in practice it is just a segmented piece of a carbon nanotube, is must be aromatic in nature.
Yet it is distinct from carbon nanotubes, how does its aromatic nature differ from carbon nanotubes?
In a related topic: How would an increase in the diameter of a carbon nanohoop affects its aromaticity?
Note: Here is a cross-section of a carbon nanotube, which is in the shape of cycloparaphenylene.

Aromaticity is a tough subject. This molecule is quite special, but it reminds me more of a biphenyl than a nanotube.

You basically have a series of rings chained together. In such structures, if the individual rings can all have optimal electron counts for aromaticity then the aromatic coupling will probably be localized to the individual rings. To get aromatic coupling globally around the hoop you may need to introduce some charge, altering the electron count.
You may want to study this question and the impact of electron count on dibenzo-cyclooctatetraene in the answer.


text:
What is the reaction priority when methyl 2-(chloromethyl)benzoate reacts with potassium cyanide?
I was going through some PYQ'S of JEE Advanced where I encountered this question.

I was not able to decide whether I should go for an SN2 reaction to substitute $\ce{-Cl}$ or a nucleophilic attack on the ester.
The solution provided by a reputed coaching center says that we should prefer the nucleophilic attack on the ester but I don't understand why.
What is the reason for this?

@Waylander  I tried using that but in the end product that I am getting I have 2-hydroxy ethanoic acid instead of carboxylic acid (cooh) grp . It would be of great help if you could elaborate on what you said.Thanks

If you run the first step of the reaction with 2 or more eq of KCN both sites will react. The benzyl chloride will give the phenyl acetonitrile and the ester will give the acyl cyanide. The acid hydrolysis which follows convert the phenylacetonitrile to the carboxylic acid and the acyl cyanide will hydrolyse to the carboxylic acid. If you run reaction 1 with 1 eq of KCN then you get attack at the benzyl chloride.

@Waylander "" the acyl cyanide will hydrolyse to the carboxylic acid. "",,I am not geting this part ,I was confused as how to hydrolyze this but then I found "https://pubs.rsc.org/en/content/articlelanding/1967/j2/j29670000755#:~:text=With%20acetyl%20and%20propionyl%20cyanides,leads%20exclusively%20to%20acetic%20acid." in which it says that it would hydrolyze to acetic acid but according to the solution I shoul be getting carboxylic acid. Please look into this.

It depends on the conditions of hydrolysis. Acyl cyanides under strong acid conditions will give the alpha keto acid. Under mild basic conditions they lose cyanide to give the carboxylic acid as they are acylating agents.

@Waylander I now undersand the reaction if KCN is in excess but I fail to understand why would benzyl chloride would be attacked if KCN is 1 eq. is it beacuse of the electronic factors mentioned in the answer below or is there anything else?

The answer below is correct that the benzyl chloride is activated towards nucleophilic attack. However CN- does attack esters so you would need to choose your conditions carefully to get seelctivity

@Waylander Ohk, will keep that in mind. Thank you very much.

(cc @Waylander) Is there not a possibility of SN2 at the methyl ester itself, with the carboxylate as the leaving group? Is acyl substitution preferred over that?

@orthocresol Cyanide hydrolysis of esters is well established https://onlinelibrary.wiley.com/doi/10.1002/hlca.19740570404

@Waylander Oh no, I meant attack at the methyl group: $\ce{RCO2Me + Nu- -> RCO2- + Nu-Me}$

That appears to be the mechanism according to the paper I referenced. It appears I was wrong about acyl cyanide formation

I would prefer to go along with $\mathrm{S_N2}$ because the benzylic carbon is much more electron deficient than the carbon associated with the ester. This is because the phenyl ring is made up of $\mathrm{sp^2}$ carbons, which are more electronegative than the $\mathrm{sp^3}$ benzylic carbon. Thus it exerts a negative inductive effect which decreases the electron density on the carbon. The same effect, albeit much stronger, is also exerted on the same carbon by the chlorine.
On the other hand, the acyl carbon of the compound receives a positive mesomeric effect due to the phenyl ring. This should somewhat stabilise the electron density on the carbon atom. Here’s a diagram; check out the last three canonical forms:
Thus, the benzylic carbon, with lower electron  density , is  much more susceptible to the $\ce{CN^-}$ nucleophile and hence the reaction should favour an $\mathrm{S_N2}$ pathway. Then cyanide and ester hydrolysis give the respective acid groups.

Thanks for the answer, I understand the electronic effects that are coming into play into this question,but I have been taught that for a particulr nucleophile we first consider nucleophilic addition reactions rather than sn2 that's why I was confused.

This definitely gives an answer to my question but I am waiting so as to receive any other possible explanations to my question.


text:
Mixture is eluted too early from alumina in flash column chromatography
I have recently tried to purify my reaction mixture by column chromatography. Because my reaction mixture had poor resolution in silica-based TLC, I used neutral alumina-based TLC plate. After I determined the elution condition, I purchased the neutral alumina at Sigma Aldrich, and its particle size is 40~163 μm. Initially I tried 2 cm i.d. column and almost ~20 cm of packing with ~0.15g of crude sample. But yield was very poor.
I was advised to use smaller i.d. column and less amount of packing, so I tried using 1 cm i.d. column and 10 cm of packing with ~0.35g of crude sample. Because my major product is liquid at room temperature and I was advised that loaded sample should have ≤1 cm of height in column, I only used 200 μL of dichloromethane to dissolve the sample. I carefully loaded the sample above the top of the alumina, let it be absorbed into alumina, and added some sand. Additionally I washed the sample vial and wall of the column with a few drops of dichloromethane, and let it be absorbed into alumina too. I added fresh dichloromethane, and started elution. To carry out flash column chromatography, I applied Ar gas upon sample loading. But a few seconds later, the mixture was eluted too early without separation with very vivid UV absorption when I spotted it to the TLC plate, although the starting material moves to near the solvent front and the product hardly moves from the starting line when I checked the TLC with 100% dichloromethane. Because there was a little amount of starting material in the sample, the product was only eluted as time goes by.
But I don't understand why the mixture was eluted instantly. What was the problem? Some people suggested:

Don't apply the pressure immediately. Try to apply the pressure after
you observe the separation of the mixture.
Other people said that they don't think that applying the pressure upon the sample loading could be the problem. Instead, the smaller the i.d. of the column, the smaller should the particle size of the alumina. But '40~163 μm' of particle size is too broad distribition, so it might not hold the sample effectively. If you can't buy more smaller-size alumina, try again with 2 cm i.d. column to increase the amount of alumina while maintaining same height of packing.

Can these two be the reason of pre-elution? If not, what was the problem?

Can you dry load?

Or run as a gravity column?

Or find a different solvent or solvent mixture that is less polar than DCM.

@Waylander My product tend to show tailing behavior in TLC, so I'm afraid I can't recover all of them in gravity column. I have not tried dry loading. Today I found that the flash grade alumina (particle size 30~63 μm) is sold in some company. Is the second reason, that is, not small enough particle size the problem?

@matt_black First it is not dissolved in hexane. Is there any recommendations that is less polar than DCM?

@KrangLee There is a wide range of eluents that are less polar than DCM and you can experiment with mixtures too if you need to fine tune to get good separation.


text:
Can ions be acids or bases?
Consider a sulfate ion, its considered a weak base. Yet I thought bases and acids were compounds meaning they were always electrically neutral? Why is a sulfate ion then considered a base? And how come it suddenly becomes an acid when it bonds to hydrogen to make sulfuric acid?
Furthermore I recently learnt at school that acids release hydrogen ions in a solution, yet ammonia is somehow classified as a base when it doesn't even have any oxygen atoms present. What is going on?
How do we clearly determine whether a substance or even ion is considered an acid or a base?

https://en.wikipedia.org/wiki/Acid#Definitions_and_concepts

Both $\ce{HPO4^2-}$ and $\ce{H2PO4-}$ are acids and bases.

Many questions are already answered here.

Can ions be acids or bases?

Yes, they can. There are different definitions of acids and bases. Here, I will use the Brønsted–Lowry definition, which says that an acid is an $\ce{H+}$ donor and a base is an $\ce{H+}$ acceptor. Your example, sulfate, is a base because it can accept an $\ce{H+}$. For further discussion on sulfate as a base, see this post.

Yet I thought bases and acids were compounds meaning they were always electrically neutral?

If you want sulfate as part of a compound, you need a counter ion such as sodium. The compound would be sodium sulfate, which is a base because the sulfate ion can accept an $\ce{H+}$ while the sodium ion has no acid/base properties (e.g. in water).

And how come it suddenly becomes an acid when it bonds to hydrogen to make sulfuric acid?

Acids and bases come in conjugate pairs, for example acetic acids and acetate. As acidic acid donates its $\ce{H+}$, it has acted as acid and yields acetate, a base. Conversely, the base acetate can accept an $\ce{H+}$, acting as a base to yield an acid.
Sulfuric acid can donate two $\ce{H+}$ ions (a diprotic acid), so it is slightly more complicated. Also, it is a strong acid, so it is very difficult to accept the $\ce{H+}$ after it was donated.

ammonia is somehow classified as a base when it doesn't even have any oxygen atoms present

With the Brønsted–Lowry definition, this is easy to explain. Ammonia can accept a $\ce{H+}$ ion, yielding the ammonium cation, so it is a base. When that happens in aqueous solution, $\ce{H+}$ ions are used up, the pH increases (becomes more basic) and the $\ce{OH-}$ concentration increases.

How do we clearly determine whether a substance or even ion is considered an acid or a base?

The easiest is to dissolve some of the substance in pure water (neutral pH), and check what happens to the pH. For a cation, use e.g. the chloride salt, for an anion, use e.g. the sodium salt. So to test ammonia, you would bubble some into pure water (the pH will become basic). And to test ammonium, you would dissolve some ammonium chloride (the pH will become acidic).

A peculiar, but surprisingly accessible, case is hydroxide ion acting as both an acid and a base. In aqueous solution it would be only a base, but heating of dry hydroxides brings out the true amphoteric character. For instance, most metal hydroxides decompose into oxides plus water vapor on heating, and we can render this as an acid-base reaction between hydroxide ions:
$\ce{Mg(OH)2\overset{\Delta}{\->}MgO + H2O}$
$\ce{OH^- -> O^{2-} + H^+}$
$\ce{OH^- + H^+ -> H2O}$
This answer cites a reaction in which sodium hydroxide acts as an acid donating a proton to sodium amide:
$\ce{NaOH + NaNH2\overset{\Delta}{\->}Na2O + NH3}$
$\ce{OH^- -> O^{2-} + H^+}$
$\ce{NH2^- + H^+ -> NH3}$


text:
Will the solid formed after dehydration of a disodium hydrogen phosphate + sodium hydroxide solution be hygroscopic at STP?
Given a solution consisting of disodium hydrogen phosphate + sodium hydroxide (approaching a pH of 12), will the solid formed after dehydration of the solution be hygroscopic at STP?  Which salts are likely to precipitate?  In the event that the resulting solid is hygroscopic, how might the addition of KCl or NaCl salts to the solution, before dehydration, affect the extent to which the dehydrated solid absorbs moisture if exposed to a humid atmosphere?

pKa3 of phosporic asic is 12.67. so there would be significant initial concentration of PO4^3-. While dehydrated pH would increase and more HPO4^2- would get neutralized.

The solid is made of trisodium phosphate $\ce{Na3PO4}$. Trisodium phosphate has a pH equal to $11.9$ in a $1$% aqueous solution, according to the Merck Index, in article $8736$. If evaporation is made near room temperature, the solid is a dodecahydrate $\ce{Na3PO4·12H2O}$. It is a little hygroscopic. Caking appears after some weeks. It looses $4$ molecules water by heating.

Thanks, Maurice.  If you estimate that caking would occur after some weeks in an open atmosphere.  Would you caking could be eliminated if stored in a plastic bottle with an equivalent mass of dessicant?  How can we go beyond qualitative estimates of hygroscopicity?

Yes. Caking can be eliminated if stored in presence of a desiccant, like blue silica gel.

How does the degree of hydration (e.g. the 12 water molecules) of the resulting solid vary as a function of the temperature at which evaporation occurs?  What might be a good recommended dehydration temperature?

I see that it "Crystallizes with 8 and 12 mols of H2O"


text:
Is NH2 electron donating or electron withdrawing?
Can anyone explain why NH2 shows -I in inductive effect and +R in resonance effect? I mean NH2 is ready to take electrons in inductive effect (with its electronegativity) but in the resonance effect, it is ready to donate electrons(with its lone pair) then how are there 2 different type of qualities for NH2?

We can have two different qualities associated with a particular species ,take for instance amphiprotic species.

You did not specify it in your question, but this topic is usually taught in the context of electrophilic aromatic substitution reactions, and I will use them as an example.
Let's say you go to the lab and perform a bromination reaction using benzene and aniline as your substrates. You will see that the reaction with benzene will require a catalyst and maybe some heat to proceed. Meanwhile, the aniline will be triply brominated just by adding Br2 to it. So you can be 100% sure that NH2 is activating for electrophilic aromatic substitution reactions.
Now, you are not satisfied with an experimental result only, and want to explain it, so you test more substituents. You will notice that some of them speed the reactions, some make them slower. In fact, substituents in these two groups seem to have some properties in common.
For instance, substituents that bond to aromatic rings via an sp3 O or N seem to always speed the reaction. If the O/N is bonded to a carbonyl, the effect is less pronounced. Alkyl groups also speed the reactions, to a lesser extent.
Meanwhile, halogens, CF3, and groups with pi bonds to electronegative atoms (carbonyl, sulfonyl, nitro) make the reactions slower.
This is what we know for sure.
Now, we can try to rationalize these experimental results. Decades ago, scientists postulated that these observed effects could be separated into resonance and inductive contributions (sterics is also important). +R, -R, +I, -I are part of a theory, something we create to explain experimental results.
Molecular orbital theory is a different approach, and you could say something like "the HOMO of aniline has a different energy compared to benzene".
So what does the theory say about NH2? There is an electron pair on N that is conjugated with the aromatic ring. This conjugation increases the electron density on the ring, making it more reactive towards electrophiles (species that "want" electrons). We call this a +R effect.
What about the -I effect? N is more electronegative than C, and we know that in the absence of pi electrons, more electronegative groups (e.g. CF3) tend to slow reactions. The -I effect of N will be more clear when you study nucleophilic aromatic substitutions with benzyne, for which the effect of pi electrons is null.
Bonus question: what is the product of the nitration of aniline? Consider the effect of the acidic environment on N and how it affects its electron-donating ability.


text:
Why can't the half-life be determined in a reaction with more than one reactant?
We have a second order irreversible reaction with 2 reactants (A and B, order 1 for both) so that the initial concentrations are different. I've read that the half-life doesn't make sense in this context but I can't fully understand it.
$$\ce{A +B ->[$k$]P}$$
$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} =-\frac{\mathrm d[\ce{B}]}{\mathrm dt} = \frac{\mathrm d[\ce{P}]}{\mathrm dt} =  k[\ce{A}][\ce{B}]$$
Any explanation?

The half-life can be calculated. But it is not a constant. It depends on the initial concentrations of the reagents. This half-time is $\pu{\frac{1}{kc_0}}$ if B = A, like in the reaction  $\ce{2A -> P}$

So could it be represented in graph for the case Ao is different than Bo?

The issue is whether you wait until half of A is gone or half of B. Either could be called the half-life.

Related

Because half life is only a well defined constant for a first order reaction
Half life for simple, first order reactions is the length of time it takes for half the substance to disappear. The classic case being radioactive decay. The rate of the reaction depends only on the amount of the single substance present but the length of time for the amount of substance to half stays the same whatever the initial amount.
For more complex reactions involving more than one substance, the reactions will not be first order and the rate of the reaction depends on the amounts of two (or more) reactants. One can calculate how long it would take for half of one of the components to disappear, and you could call that a half-life, but it would be a fairly useless number. In non first order reactions that half life would depend on the specific concentrations of all the components. And it would be different for all the combinations of the starting concentrations. And it would differ over time because those concentrations change over time.
So, for first order reactions, half life is a well-defined constant but for more complex reactions it is a number which changes depending on the concentrations and changes over time. So, in those reactions, it is a totally useless number and it would rarely be helpful to calculate it or use it. That is why it is said that it doesn't make sense.

Before I actually answer the question, let's address the elephant in the room.
What is half-life?
The half life is a term that denotes the amount of time it takes for the amount of a substance to be halved. A more formal definition would be as follows.

In a chemical reaction, the half-life of a species is the time it takes for the concentration of that substance to fall to half of its initial value.

Now we've got that out of the way, let's actually talk about the question you've asked.

We have a second order irreversible reaction with 2 reactants (A and B, order 1 for both) so that the initial concentrations are different. I've read that the half-life doesn't make sense in this context.

Half-life doesn't make sense in this context. Why? It's because you don't have enough context. Go back to the "formal" definition. The half-life of a species is what is defined. This isn't the same as the half-life of a reaction (Because this term doesn't exist for anything that has more than one reactant in the mix).
However, you could ask the following question.

What is the half-life of A (or B) in said reaction?

This has more context as you've given the person enough info to actually corelate the question and definition of what they know to be half life. Can you calculate it theoretically with the data given? That's a different question, whose answer I believe is no - for the same reason, you don't have enough information.

Ok, thanks, now I understand that is specific to one species. When you say you can't calculate it is because you don't have enough information or because is not calculable anyways?

"whose answer I believe is no" - That's not (entirely) correct; if you know the reaction stoichiometry (in the question, $\ce{A + B -> P}$) then that means that there is only one unknown, because $[\ce{A}] = [\ce{A}]_0 - x$ and $[\ce{B}] = [\ce{B}]_0 - x$. The initial concentrations $[\ce{A}]_0$ and $[\ce{B}]_0$ are constants, and the only unknown is $x$. Now, I don't know whether there's a closed-form expression for the half-lives of $\ce{A}$ and $\ce{B}$ (you'd have to do some calculus, and I'm too lazy to figure that out). But you can definitely at least calculate it numerically.

@orthocresol, as it happens there is an algebraic solution, in the case $A+B\rightleftharpoons   P$ the lifetime $\tau$ is given by $1/\tau = k_1([A_e]+[B_e])+k_{-1}$ where the subscript  'e' indicates the equilibrium concentration


text:
Half-life vs. relaxation time vs. average lifetime
On page 25 of Ángel González Ureña's Cinética química (2001), appear the terms half-life time (of an intermediate) and relaxation time, but it refers to both of them with the letter $\tau$. Why is this?
Also, it defines half-life time as $\tau=\frac{c}{v}$, so when is first order $\tau=\frac{c}{kc}=\frac{1}{k}$, but this is the same value that for relaxation time (the time for the decreasing then initial concentration in $\frac1e$). So, are they essentialy the same? Can that formula be used in both cases? What I'm missing here?

Your textbook is confusing, if you define  the decay as $\sim\exp(-kt)$ then we normally define $\tau=1/k$ where  $\tau$ is called the lifetime or relaxation time. The half-life is  $t_{1/2}=\ln(2)/k$.

@porphyrin So they are the same except for a factor of $\ln{2}$, right?

They do differ by that factor, but they represent different things, $t_{1/2}$ is the time for the population to decease to $1/2$ its initial value , $\tau$ for the population to decrease to $1/e$ or $\approx 37$ %.

The confusion is between the half-life ($t_\frac{1}{2}$) and the average life time ($\tau$). These terms seem to originate from the description of the first order decay of nuclei.
For example, this source states:

Note that the radioactive half-life is not the same as the average lifetime, the half-life being 0.693 times the average lifetime.

And they illustrate this with the following set of expressions:

Why have two separate definitions? One (half-life) is the median time for a particle to decay, the other (average life time) is the average. For a symmetric distribution, these would be the same, but exponential decay is not symmetric.
Here is a juxtaposition of Spanish and English terms to the best of my knowledge:

Periodo de semirreacción: half-life of the reaction
Tiempo de relajación: relaxation time
Tiempo de vida media: average life time (of a particle, intermediate or transition state)

And finally:

Media: mean
Mediana: median

The last two seem very tricky to me because they look very similar (in both languages).

Ok, thanks for the answer. I really appreciate it. But I'm sorry, I continue with the doubt of why the relaxation time equals the average lifetime of a nucleus decay. Are they just different definitions for different concepts but they rather equal in value?

And yes, there are definitely tricky terms all around.

Maybe they only equal for first order cases, but radioactive decay only makes sense in this order, we can only change the order for relaxation time instead.


text:
Total vs. Partial derivatives in kinetics
I've recently seen partial derivatives in the velocity definition but in most cases is with total derivatives (infinitesimal change of concentration over infinitesimal change of time).
Maybe is more accurate to use partial derivatives because it shows that is only differentiation with respect time and not the other variables (which ones?).
Is this even remarkable or is just some change of notation?

The other variables can be the temperature, the pressure, the viscosity, the concentration of substances not taking part to the reaction, etc.

Why viscosity?.

Viscosity may affect the rate of molecular collisions, as there are stronger intermolecular forces.

A chemical reaction has a rate that tends to zero if the "solution" is like honey, then like glue, then like a solid.

I searched for examples of $$\frac{\partial [A]}{\partial t}$$ and could not find them. Where did you see these partial derivatives? Could you edit your question to provide the sources.

i am not familiar with this use of partial derivatives, but it wouldn't surprise me if it were either A) in a multi-order reaction and a diff eq is being described with rate dependence on other concentrations (catalyst loading, perhaps?) or B) a chemical engineering context where homogeneous mixing of solutions are not presumed and thus mechanical stresses, boundary conditions, etc. are applicable.

I've added a picture that shows an example of partial derivatives in Cinética química, Ángel González Ureña, Ed Síntesis (2005). ISBN: 978-84-773890-7-1.             Is in Spanish, this part is about consecutive reactions, and it uses a different notation as he defines [A] as (Ao-x).

Now that I see the source, I think it is just unusual notation without significance. In the example, there is only one variable used for differentiation, time. I don't think it is wrong but rather unexpected and therefor puzzling.

If we consider the ideal gas state equation as an example, in the form $V=f(n,p,T)$, then we use a partial derivative $\left(\frac{\partial V}{\partial T}\right)_{n,p}$.
But if we use it in a closed, isobaric scenario as $V=f(T)$, with $n, p$ implicitly constant, then we can use normal/total derivative $\left(\frac{\mathrm{d} V}{\mathrm{d} T}\right)$
The distinguishing of both cases is kind of formal, depending on if you formally consider the other independent variables explicitly or not.

Ok, but in this case the function was concentration, so all the variables that @Maurice said (p,T, viscosity, etc...) are what concentration depends on?

I suppose it's concentration because is in the velocity definition

And about your answer, you mean that when total derivatives in the velocity defifnition are used is because the other variables are constant, don't you?

Then apply the example on concentration or generally on any independent and dependent variables. The topic is rather mathematical than chemical.


text:
Exception to the order reaction and stoichiometric coefficient equality in elementary reactions
When a reaction is elementary,the stoichiometric coefficients always equal the order for each reactant? Is there any exception? Because at Ángel González Ureña's Cinética Química it says that this is "in general cases", but not always. I haven't found any results in google about this topic.

Elementary reactions always involve a single molecule reacting (e.g. isomerisation)  or two colliding and so react. So no exceptions. (Some texts include three molecules colliding but the chance of this is so small it never happens in practice. It is found experimentally  that an intermediate is formed, so it is a two step process.)

If the order is not equal to the stoichiometric coefficient, the reaction is not elementary.

Yes, like @Maurice, I thought this is the defining property. I guess you could have a multi-step reaction where the orders happen to be equal to the stoichiometric coefficients, so I guess the definition has to be more than that.


text:
Oscillating vs Autocatalytic reactions
Are all oscillating reactions autocatalytic? Or are all autocatalytic reactions oscillating? What's the relation between them? Am I just mixing up different things or are they basically the same?
For instance, the Belusov-Zhabotinskii is a nice example of an autocatalytic reaction, but it has an oscillating character.

Comments are not for extended discussion; this conversation has been moved to chat.

Non-oscillating autocatalytic reactions

[OP] are all autocatalytic reactions oscillating?

No, there are many examples of autocatalytic reactions that don't oscillate. On is given in the comments:

[Poutnik] Oscillators generally need some positive loopback which destabilizes them. Oscillatory reactions are not exception. E.g. Oxidation of oxalic acid by permanganate is autocatalyzed by manganese(II) ions, but is not oscillatory as there is no such a loopback.

An example that is not strictly chemistry is crystallization. Making sugar candy from a supersaturation sugar solution is faster if you add a stick with a tiny amount of solid sugar on it (a crystallization nucleus). A similar physical effect is agitating a bottle of super-cooled liquid water.
The Wikipedia article on autocatalysis lists chemical reaction examples at the end of the page. One spectacular example is the transformation of metallic tin to the non-metallic gray tin, a process called tin pest. Two examples from biochemistry are the PCR test for SARS-2 viral RNA, and the radical polymerization of acrylamide to form polyacrylamide (the product is used in gel electrophoresis of proteins).
Non-autocatalytic oscillating reactions?

Are all oscillating reactions autocatalytic?

There is an theoretical model discussed in this post, What's a minimal yet chemically-meaningful kinetic system for an oscillating reaction?. In its simplest form, it is unrealistic because there is no source of Gibbs energy to continuously drive it (i.e. to make at least one of the reactions virtually unidirectional).
I am not aware of a experimental system that oscillates yet does not contain autocatalytic elements. It is possible that there is one somewhere, but it is not among the two or three systems typically discussed as example for oscillatory reactions.

Ok, so I suppose that all oscilating reactions known are autocatalytic but there could be some exception.


text:
Is there a general equation which calculates the temperature which shifts the equilibrium mostly to the right which applies to most reactions?
Say I have the following simple reaction
$
  \ce{HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H2O(l)}
$
Does there exist a calculation which gives the ideal temperature for this reaction to occur at?
Background
I used this example as it was the easiest one I could think of from the top of my head but I mean this generally.
I tried looking this up but could only find calculations to do with how much heat is released.
Edit
Ideal meaning the point in temperature where the reaction will give the greatest yield of products. Kind of reminds me of enzymes the optimum temperature.
Like one of the commentators mentioned the haber process, the temperature at which the equilibrium shifts mostly to the right, or the minimum temperature required for the reaction to get going.
Actual question
Is there an equation which calculates the optimum/shifts equilibrium mostly to the right temperature or is it just empirical? I mean a general equation which might apply to most reactions, if there is such a thing?
Apologies for being vague.

My cat would sleep on just about any level surface in the house. Floors, windowsills, armchairs, tabletops, he doesn't seem to care. What is the ideal surface for a cat to sleep on? Same thing with your reaction.

How is 'ideal' defined?

You might want to consider using a different reaction that actually can have an optimized temperature; for example, the Haber–Bosch process (production of ammonia). Since the reaction is exothermic, the equilibrium of the reaction shifts at low temperatures to the side of the ammonia; however, high temperatures are required for an appropriate reaction rate. Therefore, there is an optimized temperature somewhere in between that leads to the maximum production rate.

Ideal for what? Fastest reaction? Most complete reaction? What about the pressure? Etc. And why do you think such an "ideal" condition exists?

You need to specify what you mean by “ideal” in this scenario. “Ideal” implies a value judgement, when here we need a measurable quantity to compare against

see edit please guys

Anytime when asking for ideal, optimal or the best, the one who asks must provide criteria, how the ideality, optimality or being the best is evaluated, as it is highly subjective criteria.
In reaction context, one of possible criteria is the maximal reaction quotient achieved at given reaction conditions, together with minimal expenses. This makes sense for slow enough reactions, what is not the case of  strong acid + base neutralisation.
The typical textbook case for optimization is catalyzed Haber-Bosch synthesis of ammonia:
$$\ce{3 H2(g) + N2(g) <=> 2 NH3(g)}$$
There are 4 major factors:

Increased temperature speeds up the reaction.
Increased pressure shifts equilibrium in favour of the product.
Increased temperature shifts the equilibrium constant in favour of the reactants.
Increased cost for higher pressure and temperature.

The reaction is performed as gases pass the solid state catalyst, ammonia is separated and remaining reactants are by loopback returned to react.
The pressure is kept as high as technologically reasonable. But there are higher time and operational expenses (device cost, energy spent) for higher pressure.
The major parameter to optimize is temperature:

If too low, reaction is limited kinetically, only small fraction would react, in spite of the equilibrium position.
If too high, reactants react fast, but just small fraction due low equilibrium constant value. And there is higher energy cost.

From evaluation of the reaction kinetics and thermodynamics, technological and cost aspects, the optimal pressure and temperature is determined.

Thank you very much. I should’ve used an equilibrium reaction… So this is looked at for each chemical equation on a case by case basis. For your example can the optimum or temperature and pressure at which the highest and fastest yield of products occurs be determined by an equation? or as an example are you just looking at the moles on the left vs right?

There is multiple empirical aspects.

I see, that’s all I needed to know, thank you.


text:
Has anyone sysnthesized perfluorocubane (AKA octafluorocubane)?
Up until now, there have been attempts at synthesizing octafluorocubane, shown below, but success has been elusive.

Has there been any success, and how might the symmetry and fluorine's electron affinity affect its properties?

[Caveat: This is not so much a question-and-answer as a desire to share Derek Lowe's article in his inimitable style. Given enough down-votes, I'll delete both Q-&-A. But enjoy the synthesis summary, cited below!]
Yes! The synthesis of perfluorocubane (or octafluorocubane, if you prefer) has been accomplished, as described by Derek Lowe, "[I]t is predicted to have the ability to hold a free electron in the middle of that cube (!) That's because of the C-F bond properties - all the electron density is being pulled towards the F atoms, so the center of the cube really has room for one to sit in there without running into a lot of interference... [T]here are eight sigma-star orbitals from those C-F bonds all pointing into the center of the cage and overlapping, and giving you effectively a stabilized vacant orbital there."
Get more details in Lowe's article!

"Quantum Mechanics: A Hand-Waving Approach" - I'd buy that book if Derek wrote it!

I'll upvote if you go into the related HN comments section and straighten-out some of those goofy comments :)

@ToddMinehardt, I'd upvote anyone who convinces Lowe to continue his series, Things I Won't Work With, or start another! They're great material for teaching, or just for personal edification and amusement.

I think the electron is trapped by the 8 positively charged carbons.

They should try dodecahedrane next!

BTW, realize that though octafluorocubane can trap an occasional electron, one could not produce a bulk sample where each molecule traps an electron due to the total Coulomb (electrostatic) repulsion. "If a pair of charges of 1 C each were 1 m apart, the force of repulsion between the two charges would be 9 billion newton's," from http://www.tecumseh.k12.oh.us/Downloads/Physics_12_23.pdf . Of course, if there's a similar molecule that can cage protons, one could, theoretically, mix the two "clathrates" to make it electrically neutral (if the charges don't unite to make hydrogen). ;-)

@DrMoishePippik one wild idea is to use an alkali metal as the electron source, complexing it with a cryptand, thus $\ce{[M(crypt)^+][C8F8^-]}$.


text:
Any cheap way to test whether a given powder has high protein content?
Protein powders are used as dietary supplements.  The photo below is for whey protein powder, which usually has about 24 grams of protein per scoop:

Image: © jirkaejc/Getty Images
Unfortunately, many forged protein powders are circulating in the market, specially as the price of genuine ones is increasing rapidly.  This stresses the necessity of testing purchased protein powders to ensure that they have high protein content.
Unfortunately, lab tests are quite expensive.  This brings my question:

Is there any simple test, or heuristic, to verify that a given powder contains high protein content?  Ideally, something that can be done at home.  I'm also open to more sophisticated tests in case they are the only way.

Note that I do not necessarily want to know the actual amount of proteins in grams.  For my purpose, it suffices to get an approximate decision that the amount of protein is generally high enough.

The answer in general will depend on your desired accuracy and the exact information you are looking for. However, in your specific case you would probably be fine using a Biuret test.
The Biuret reagent is cheap, available online, and easy to see; this test uses a complex of copper (II) sulfate with some salts and reacts with peptide bonds which link amino acids together. This should apply to whey like any other protein.
If you choose to do this test, do not consume the protein tested and make sure you keep it stored away from food or materials that may be accessed by people who don’t know what it is. It would also be wise to read the Safety Data Sheet (SDS) associated with the reagent. Proper personal protective equipment and general safety tips outlined here should be followed.
It is best examined with a ultraviolet / visible light spectrophotometer, but you don’t need to go through the trouble if you don’t need to be all that accurate. You can occasionally find these on eBay or Craigslist, but they are probably not worth your time and while they’re cheap as far as instrumentation goes, they usually aren’t worth the price either. The breaking of peptide bonds results in the originally blue liquid to become a purple-ish solution that is visible to the eye (this is called a colorimetric test).
If you have a color-perception deficiency like color-blindness, colorimetric tests can be difficult and you may prefer more dramatic color changes. There are reagents that are slightly more standard and easier to tell a change in (like the ninhydrin test) but those can get more expensive and are more applicable to a lab setting.