text:
Element Names in English
Na - Sodium - Natrium
K - Potassium - Kalium
W - Tungsten - Wolfram
Sb - Antimony - Stibium
and so forth.
[English only] Why do we not use the names that match the symbols?

Because they were named in other languages first.

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

@JonCuster All the more reason the original names should be used

There was the era when English was not the major language of science. There is interesting web site http://episodictable.com with dramatised text, audio and podcasts with stories about each element. Some include also the story of their double naming.

https://chemistry.stackexchange.com/questions/31653/did-berzelius-unify-the-names-of-chemical-elements-using-the-latin-language

You can look at this and the links therein: https://www.loc.gov/everyday-mysteries/chemistry/item/chemical-elements/

Also if you want to learn about mercury history: https://hsm.stackexchange.com/questions/13686/when-was-mercury-given-its-name

Surprisingly the history of science and math SE site does not address this question particularly, although there are a few similar questions.

Anyway, I much prefer the Latin names

I’m voting to close this question because this is better suited to SE History of Science and Mathematics

In the $18$th and $19$th century, there was a strong rivalry between chemists of France, Germany and England, specially about giving a symbol to newly found elements. If a French chemist had discovered and given a name to a new element, the German and English had to accept it, but they often proposed another name, and another symbol, just not to obey to "the enemy". If you read old French publications from the $19$th century, you may find formula like SoCl for sodium chloride. Beryllium Be was called Glucinium Gl in France up to late in the middle of the $20$th century. This stupid rivalry was finished after WWI, in $1919$, when IUPAC was created, where delegates from the world over met just to define a unique symbol for each element. Unfortunately they failed to give a common name. But they succeeded in defining a common symbol.

Perhaps the most infamous case comes not from chemistry but from astronomy.

@Maurice, Is there a reference which shows SoCl for sodium chloride? I could not find any example from Google Books. As early as 1838, NaCl was used in French works. Sodium was an English name. Why would the French embrace the symbol "So" if it ever existed.

Wikipedia reports that German chemist Ludwig Wilhelm Gilbert proposed "natronium" in 1809, the year after that element was isolated; and recorded usage of the symbol Na dates to 1814. Apart form  there is little room historically for So.

I just remember having seen once upon a time that some French chemists had used So for sodium in the 19th century. It is just a souvenir. Maybe my memory is failing.

@Maurice, I agree. I don't see any evidence of NaCl being written as SoCl. I checked early 1800s. NaCl is used there in French.

There are several reasons for this, one of which is the pigeon-hole effect. The English language consists of only 26 letters, while nature contains more than 90 elements. The letters would become repetitive if only English names were used. Secondly, English is a latecomer as a global language of science. It is a recent phenomenon dating back to the 1900s. Early chemists were comfortable with Latin/Greek perhaps due to religious norms and education of the day. Secondly, other European languages like German and French dominated chemical knowledge and chemists had to learn at least one of them. So all these symbols were "natural" for the older generation.
Sodium and potassium were discovered by Davy, an English speaker, and he wrote "Potassium and Sodium are the names by which I have ventured to call the two new substances" in H. Davy, Philosophical Transactions of the Royal Society 98,
32 (1807). However, "Natrium" compounds were well known in French, particularly Natron (sodium carbonate), at least two centuries earlier. What is natural then? Continue with the natural names or make up a new English symbol. What symbol can one use for sodium, S, but like the pigeon-hole effect it will overlap with sulfur, "So" is an English conjunction, perhaps it would cause more confusion. The same holds true for potassium, (Kalium and eventually from the Arabic Kali), its compounds were well known. German chemists had no issue with Natrium and Kalium (Na and K) because the names are still used. Wolfram is also natural to Germans, as is its symbol W, due to the fact that the mineral is well-known.

Doesn't follow. There are multiple elements with two letter abbreviations. Sodium and Sulfur could simply be distinguished by So and Su. See Chlorine and Carbon

Why do you insist that people stick to English symbols at an internationallevel? Sodium is sodium only in English. As stated earlier, English as a language of science is quite recent.

I don't. I want Latin

Natrium and Kalium are both Latin :-)

Yes, but we don't use those names

Germans do and so do many many Europeans, who is "we"? Chemistry was dominated by the German language for a very very long time.

"We" as in the title. English speaking people

And even where other languages use the English word, or maybe it is English using a word coined by the other language, the spellings may nit be the same: iodine -> Jod.

@OscarLanzi. Have you noticed that not a single element starts with J!

@Achem in English.

@OscarLanzi, Sorry, I meant officially and internationally accepted element symbol. Of course, Iodine is Jod in German and in older literature one may see J as a symbol in periodic tables.


text:
Why does bismuth-212 decay to thallium-208?
Why does $\ce{_{83}^{212}Bi}$ decay to $\ce{_{81}^{208}Tl}?$
I know that in general the $n/p$ ratio must be less than 1.5 for the nucleus to be stable, but in this case the ratio goes from 1.55 to 1.56, which is more unstable.

So what? There is no strict dependence between the n/p ratio and stability. Then again, decay into a less stable nucleus is hardly a rare thing. $\rm^{238}U$ does just that.

oh,OK. Thanks! I was told if the n/p ration becomes more than 1.5, it's automatically unstable

You mean Thallium, right? Not tellurium.  Note that 208Tl rapidly decays to 208Pb since it is not stable either. There are only so many ways to decay, and 212Bi can't get to 208Pb directly.

@JonCuster,yes. My bad

You should seriously refresh your knowledge of periodic table (Te is far away from Bi ) and element symbols (Ti is titanium, even more far away.). Thallium has the symbol Tl. //All heavy and neutron rich isotopes , that undergo alpha decay, lead to even more neutron rich isotopes, which often subsequently undergo beta decay.

Why is this question about Bi-212? Practically anything that alpha-decays increases its $n/p$ ratio as it does so, and in many cases such as U-238 that ration also starts >1.5.

That's the specific example I was taught in class

It has the direct everyday life analogy. Imagine a wrongly built, multistand, vertical holder for flower pots. Upper holders have more potential energy, but lower stand may ( or not ) be less stable, with higher risk of the pot falling down.  Similarly, the isotope, being formed by alpha particle expulsion with energy release, can be less stable than the original one.

Fundamentally, there are only a few ways that a nucleus can decay. The primary ones are fission, alpha decay, and beta (plus or minus) and electron capture (I will ignore some of the more exotic paths). This limits the options that an unstable nucleus has to become more stable.
Fission results in two smaller nuclei, with some prompt neutrons and some delayed 'evaporation' neutrons, leaving two lighter nuclei with smaller $n/p$ ratios (mostly). This does not imply that all fission products are stable themselves mind you.
An alpha decay reduces the number of protons and the number of neutrons by two each, so mathematically $n/p$ must increase. Even so the ($\alpha$ + daughter nucleus) can be a lower energy state, partly because the $\alpha$ is so strongly bound.
The $\beta+$, $\beta-$ and EC (electron capture) processes act to turn a single proton into a neutron or vice versa, whichever results in a more stable nucleus.
Looking at $^{212}$Bi (say at Wikipedia) one sees that it decays through either $\alpha$ (~36% of the time) to $^{208}$Tl or $\beta-$ (~64% of the time) to $^{212}$Po (interstingly, the $^{212m1}$Bi excited state flips those percentages). Neither daughter nucleus is stable, each undergoing the other decay to land at the stable $^{208}$Pb nucleus. Note that the "rule" that nuclei can be stable only with $n/p<1.5$ fails, as this ratio is $\approx1.54$ in the stable $^{208}$Pb product.
Broadly, looking at the various U or Th decay chains will show a variety of branching paths through different daughter nuclei until finally landing on a stable nucleus.
For more info on decays, consider one or both of:
The AME 2020 atomic mass evaluation - this is the tables with masses, binding energies, etc.
Evaluated Nuclear Structure Data Files - decays, Q values, branching ratios, etc.

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text:
Titanium-free catalysts to decrease decomposition temperature of magnesium hydride
Magnesium hydride $(\ce{MgH2})$ is a metal hydride with a relatively stable structure, at least compared to other metal hydrides. Magnesium hydride splits at around 300 °C into magnesium and hydrogen. The decomposition temperature can be halved with a catalyst.
Unfortunately, these catalysts are often titanium-based (example A and example B), which does not make this a realistic way to reduce the temperature required, at least not if made for wide-spread production, as titanium is a rare and expensive metal.
What catalysts are proven and effective at reducing the temperature required to break magnesium hydride, that don't include titanium?

I would consider titanium neither rare nor expensive. Titanium is ninth most abundant element on Earth, and is much cheaper than other transition metals used as catalysts. My friend used to ride a bicycle with titanium frame (aquired about 20 years ago) before switching to carbon fiber, and he is no Rockefeller.

The only aspect somewhat affecting price is technologically demanding production of bulk metallic  titanium, due its high affinity of oxygen and high melting point.

From Shikin et al.[1], a recommendation to press the magnesium hydride with nickel powder:

We have studied the activation of magnesium hydride decomposition by means of its pressing with a catalyst. It is established that pressing leads to the formation of metal nuclei, which favor a decrease in the temperature threshold of magnesium hydride decomposition. The introduction of catalytic additives also reduces the temperature of dehydrogenation. The most effective in this respect was found to be the addition of nickel powder.

Reference*

Shikin, I.V., Elets, D.I., Voyt, A.P. et al.  (2017). "Activation of magnesium hydride by pressing with catalytic additives." Tech. Phys. Lett. 43, 190–193. https://doi.org/10.1134/S1063785017020262.

It's seems like the cheapest catalyst to ease this decomposition is $\ce{MgO}$ (Ref.1):

Other metal catalysts, which effect this decomposition are as follows:

Evidently, $\ce{MgO}$ seemingly better metaloxide than titanium oxides as well as the best metal catalyst so far, $\ce{Nb2O5}$.
References:

José-Ramón Ares-Fernández, and Kondo-Francois Aguey-Zinsou, "Superior $\ce{MgH2}$ Kinetics with $\ce{MgO}$ Addition: A Tribological Effect," Catalysts 2012, 2(3), 330-343 (DOI: https://doi.org/10.3390/catal2030330).

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text:
Mechanisms: What is the difference between the top chemical and bottom chemical on an arrow for a mechanism?
I’ve heard the top chemical that is on top of the reaction arrow represents the “first step”, and bottom is “second step” when referring to a mechanism. My question is will the bottom chemical always act as a solution to help aid in the mechanism process after the main nucleophile/base on top attacks first.
Example: 2-bromo-3-methylbutane —> (NaOCH3 on top and CH3OH on bottom)

To the best of my knowledge, there are no strict rules, and the "bottom" chemical might be the solvent, the second step, another reactant that didn't fit in the first row, or something totally different altogether, such as a comment with reaction conditions. Related: What is meant by the numbers on reagents above and below arrows?

I write reaction equations with reagents on top of the arrow, solvents, temperatures, atmosphere and time below. If multiple steps are used then they are numbered to indicate different reactions.

In your specific example, I would read that as methanol being solvent and sodium methoxide as reagent.

There isn’t any hard and fast rule about what is supposed to go above or below the arrow, although reagents tend to go above and solvents tend to go below. Sometimes transformations are performed by subjecting the reactant to a sequence of reaction conditions. In such a case, the conditions are listed numerically above and/or below the reaction arrow.


text:
Does the ‘e’ from ‘-yne’ get dropped in compounds of higher functional groups?
What is the way of representing a triple bond in a compound of a higher functional group, like alcohol? Does the ‘e’ from ‘-yne’ get dropped like in the case of the ‘-ene’ suffix?
For example, would the following compound be named 3-methylpent-3-en-2-yn-1,4-diol or 3-methylpent-3-en-2-yne-1,4-diol according to IUPAC nomenclature?

You made an error in counting, otherwise seems OK, so I dunno what you were unsure of...

I would call it 3-methylpent-3-en-1-yne-1,4-diol

I was unsure if it should yne or yn since ene becomes en. @Mithoron

The compound also possesses cis–trans isomerism, so you might want to include E–Z notation in the name, too.

Substitutive names are formed by adding a suffix such as ‘al’, ‘ol’, ‘yl’, ‘carbaldehyde’, ‘carboxylic acid’, etc., to the name of a parent hydride, with elision of the final letter ‘e’ of the parent hydride, if any, before ‘a’, ‘i’, ‘o’, ‘u’, and ‘y’.
The corresponding rule in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows.

P-16.7.1 Vowels are systematically elided as follows:
(a) the terminal letter ‘e’ in names of parent hydrides or endings ‘ene’ and ‘yne’
when followed by a suffix or ‘en’ ending beginning with ‘a’, ‘e’, ‘i’, ‘o’, ‘u’,
or ‘y’;

Accordingly, the correct name for the compound that is given in the question is (3Z)-3-methylpent-3-en-1-yne-1,4-diol since ‘yne’ is not followed by ‘a’, ‘e’, ‘i’, ‘o’, ‘u’,
or ‘y’.
Similar examples that can be found in the Blue Book are 2-methylpent-1-en-4-yn-3-ol
and undeca-2,4,7-trien-9-yne-1,11-diol.


text:
Why does the same electron transition release photons of different frequencies for some elements?
Question: For elements in the same period with different numbers of valance electrons, why does the same electron transition release photons of different frequencies?
Example: For valance electrons moving from the 4d sub-level to the 4p sub-level, $\ce{Ca^2+}$ emits violet light, while $\ce{Cu^2+}$ emits green light.

Example case referring to JavaLab: Flame Test, which in turn claims to refer to NIST Basic Atomic Spectroscopic Data.

The energy levels are different, as are the differences between a given pair of energy levels.

@EdV Does that mean that energy levels, like the 4d sub-level, have different energies for different atoms?

@Bryan351018, Could you add more details. Calcium I and Copper I (I = unionized atom) atomic spectra are horribly complex and dense (too many lines). What figures are you referring to? You must have a reference which shows violet vs. green emission. I did not down vote your question, rather upvoted it. Don't worry about them.

@AChem I have updated the question for more details and better formatting; hopefully it's a bit better now.

You should ask Why should they have the same frequencies, considering different nucleus charges and different electron configuration? Orbital energy levels are not like fixed heights of book shelves.

@EdV Shouldn't this be posted as an answer? The current answer of "Flame tests don't actually correspond 1-to-1 with atomic emission spectra." seems to be a tangent unrelated to the original question of "Why do 'equivalent' level transitions between different atomic species emit photons of different frequencies?"

@AmateurDotCounter Actually, I deleted comments about this. For example, the simplified sodium Grotrian diagram, shown as fig. 2 here, shows the famous 3p to 3s sodium D line emissions at 589.0 and 589.6 nm. And the hyperphysics site also shows, e.g., lithium’s Grotrian diagram. The 3p to 3s transitions are quite different in energy compared with sodium.

I am glad that you updated the question because it highlights a very common misconception. First of all the JavaLab Flame Test is completely wrong for both copper, calcium and many more salts especially the Group II and transition metal salts. Consequently, the NIST reference does not apply (see below). Secondly, it is a cruelty to science if a cartoon flame test is being shown instead of the real flame test. Search Youtube for real flame tests.
Basically, the brick red color and the green color of calcium and copper respectively is not from atomic emission of Ca or Cu atoms nor it comes $\ce{Ca^{2+}}$ or $\ce{Cu^{2+}}$ ions. This color comes from small molecular species of those elements in the flame. If you had a pocket spectroscope (it is not difficult to make one at home from a DVD/CD), copper green color will appear as a range of wavelengths, which is an indication of molecular spectrum.
When we talk about atomic spectrum, we need very high temperature flames, which usually ~ 3000 K. Ordinary Bunsen burner produces a cool flame by that standards. Cooler flames cannot break molecular compounds to free atoms. Let us think of this way: You introduce $\ce{CaCl2}$ into the flame in the form of a solution. The solution will dry and produce solid salt, which will begin to vaporize and decompose in the flame. High temperature flames are electron rich, and then free calcium ion is reduced to calcium atom. In that case, one can talk about atomic emission and its corresponding electronic transitions. In those cases, one can refer to NIST tables.
For the sake of completion only the alkali metals produce an incomplete atomic emission in a Bunsen burner.

follow-up question: Difference  between a "cartoon" flame test and a "real" flame test? How do chemists do flame tests correctly?


text:
How to find the amount of ions from the dissociation of a weak acid?
Someone asked me this question,"How many ions are formed during the dissociation of 500 molecules of carbonic acid, if it dissociates in the first degree by 20%, and in the second degree by 1%?" and I don't understand this question. First degree? Second?

The word "degree" is not well chosen, when not attached to the words "of dissociation". The "first degree" means that $20$% of the $\ce{H2CO3}$ molecules are dissociated into $\ce{H+ +  HCO3^-}$ ions. The "second degree" means that $1$% are dissociated into $\ce{2 H+ + CO3^{2-}}$ ions. Now use this information to calculate the amount of $\ce{H^+, HCO3^-}$ and $\ce{CO3^{2-}}$ ions in this solution.

Degree of dissociation is the correct term with well-established definition, don't mislead the OP. There is no "probably" or need for arbitrary number of molecules whatsoever. Also, before answering the question, it's a good idea to check for answered duplicates.


text:
Difference between a "cartoon" flame test and a "real" flame test? How do chemists do flame tests correctly?
This well-received answer to Why does the same electron transition release photons of different frequencies for some elements? begins with the following intriguing passage:

I am glad that you updated the question because it highlights a very common misconception. First of all the JavaLab Flame Test is completely wrong for both copper, calcium and many more salts especially the Group II and transition metal salts. Consequently, the NIST reference does not apply (see below). Secondly, it is a cruelty to science if a cartoon flame test is being shown instead of the real flame test. Search Youtube for real flame tests.

Rather than each of us search YouTube for it, I think it would be great if a short explanation of what chemists would consider a "real flame test" were captured forever in a Stack Exchange answer. So to that end I am asking:
Question: What is the difference  between a "cartoon" flame test and a "real" flame test? How do chemists do flame tests correctly?
The next paragraph mentions a "pocket spectrometer" that one can actually make from a DVD (this is fun, I've tried it on various neon lights but never a flame) and there are discussions of spectral databases in the question and answer(s) there. Are these necessary ingredients of what a chemist would consider a "real" test?

Your interest is greatly appreciated. As a first step, let's clarify that flame tests as an analytical tool are obsolete. No professional chemist will use them to identify a given unknown sample. However, flame absorption and emission spectroscopy continue to be used routinely today at a more sophisticated level. There were so many elements discovered through flame emission that its demonstration is still worthwhile today. As for the cartoon flame test, I was simply referring to the Java app that depicted a fictitious flame test. It had some inaccurate atomic emission depictions as well as misleading electronic transitions "backed up" by the NIST database.
Flame tests of Group I, Group II, and some other selected elements should be performed using a platinum wire that has been dipped clean several times in HCl soln. Introduce the wire into the flame and heat it until no color appears in the flame. We all know that sodium is everywhere so this process ensures that the Pt wire is clean. Moreover, it is ensured that we use an oxidizing flame, that is, the flame should be very light blue to near-invisible with a clear blue cone inside. This is adjusted by checking the air flow to the Bunsen burner. Then one would dip the clean Pt wire in a concentrated solution of a chloride of Group I, II and some selected transition metals. Chlorides salts decompose and volatilize easily in a relatively low temperature flame of a Bunsen burner (methane and air) and immediately color the flame.
There are different levels of flame tests:
(a) At an elementary level, the flame is viewed with the naked eye and the flame color is noted.
(b) Next level is the use of cobalt blue glass to block sodium emission and sometimes iron contamination
(c) The third and perhaps the most effective level is to use a prism/grating spectroscope to watch the spectrum on a calibrated scale. However in that case, there must be a continuous supply of the sample to the flame. There are various methods to achieve that.


text:
Equilibrium constant for isothermal double equilibrium “shift”
I do not understand how the solutions to the following high school chemistry question are valid:

A $\pu{1 L}$ vessel containing $\pu{0.300 mol}$ of dinitrogen tetraoxide gas is allowed to come to equilibrium.
$\pu{0.200 mol}$ of nitrogen dioxide then added and the system allowed to reach equilibrium again.
The final concentration of nitrogen dioxide in the vessel is $\pu{0.600 mol L^-1}.$
The temperature is constant throughout these experiments.
Calculate the equilibrium constant $K_c$ for this reaction.

My approach

Draw up an ICE table where we account for the first equilibrium set up:

$$
\begin{array}{lccc}
\ce{&N2O4(g) &<=> & 2NO2(g)} \\
\text{I} & 0.3 && 0 \\
\text{C} & -x && +2x \\
\text{E} &  0.3-x && 2x \\
\end{array}
$$

Draw a second ICE table to account for the second shift, resulting in the difficult to solve equation:

$$
\begin{array}{lccc}
\ce{&N2O4(g) &<=> & 2NO2(g)} \\
\text{I} & 0.3-x && 2x+0.2 \\
\text{C} & +y && -2y \\
\text{E} &  0.3 - x + y && 0.6 \\
\end{array}
$$
Provided solution
$$
\begin{array}{lccc}
\ce{&N2O4(g) &<=> & 2NO2(g)} \\
\text{I} & 0.3 && 0.2\\
\text{C} & && \\
\text{E} &   && 0.6\\
\end{array}
$$
How can they just put the $0.2$ as an initial concentration? It was not there initially. Conceptually, how is this equivalent to what was done in the question?
How can I be sure/convinced that adding the $\pu{0.2 mol}$ later in the process is actually going to produce the same final equilibrium as adding it to an intermediate equilibrium?

Buck Thorn's answer already addresses the conceptual ideas, and Sam202's how to solve it brute force. In this answer, I will first show an alternate way of approaching this using the conservation of mass. Second, I will show an easier way of filling the ICE table, and use the ICE table to illustrate why a set of related starting conditions (the dollars/pounds analogy in Buck Thorn's answer) lead to the same equilibrium.
Conservation of mass
We calculate the amount of nitrogen atoms in the system. We start out with 0.6 mole (from 0.3 mole of dinitrogen tetroxide) and add another 0.2 mole (from 0.2 mole of nitrogen dioxide) for a total of 0.8 mole. If at equilibrium, 0.6 mole are in the form of nitrogen dioxide, the remaining 0.2 mole nitrogen atoms have to be in the form of dinitrogen tetroxide, showing that there are 0.1 mole of dinitrogen tetroxide at equilibrium.
ICE table
First, the ICE table correctly given by the OP:

$$
\begin{array}{lccc}
\ce{&N2O4(g) &<=> & 2NO2(g)} \\
\text{I} & 0.3-x && 2x+0.2\\
\text{C} & +y && -2y \\
\text{E} &  0.3 - x + y && 0.6\\
\end{array}
$$
And this equation is super hard to solve.

We can replace the $-2y$ with an expression containing $x$ because the value in rows "I" and "C" have to add up to the value in row "E".
$$
\begin{array}{lccc}
\ce{&N2O4(g) &<=> & 2NO2(g)} \\
\text{I} & 0.3-x && 2x+0.2\\
\text{C} & +y && 0.4 - 2x \\
\text{E} &  0.3 - x + y && 0.6\\
\end{array}
$$
Then, we can replace $+y$ (via the stoichiometry) and $0.3 - x + y$ (via addition in the column). Or you solve $-2y = 0.4 - 2x$ for $y$, and substitute into the expressions still containing $y$:
$$
\begin{array}{lccc}
\ce{&N2O4(g) &<=> & 2NO2(g)} \\
\text{I} & 0.3-x && 2x+0.2\\
\text{C} & x - 0.2 && 0.4 - 2x \\
\text{E} &  0.1 && 0.6\\
\end{array}
$$
As you can see, the $x$ is still in the table, but canceled out for the equilibrium concentrations, so we don't need to know it to calculate the equilibrium constant.

[from OP's comments to Buck Thorn's answer] How can I be sure that adding the 0.2 moles of NO2 later is going to produce the exact same result as adding it in initially?

There is an entire set of starting conditions that gives the same equilibrium. In equilibrium problems, there is no "order of addition" effect. You can add one substance first, then the other, or back and forth. As for the ICE table, I can take the equilibrium concentrations and show you the entire set of starting conditions that yield the same equilibrium (this ICE table is unusual in that we are calculating initial conditions from known equilibrium conditions):
$$
\begin{array}{lccc}
\ce{&N2O4(g) &<=> & 2NO2(g)} \\
\text{I} & 0.1 + z && 0.6 - 2 z\\
\text{C} & -z && 2z \\
\text{E} &  0.1 && 0.6\\
\end{array}
$$
As long as I don't get negative values for the initial concentrations, I get the same equilibrium. So $z$ can go from 0.3 to -0.1. Here are a couple of examples:
\begin{array}{cccc}
z &\ce{N2O4(g) &<=> & 2NO2(g)} \\
0.3 & 0.4 && 0\\
0.2 & 0.3 && 0.2 \\
0.1 &  0.2 && 0.4\\
0 & 0.1 && 0.6\\
-0.1 &  0 && 0.8\\
\end{array}
I can take the initial conditions in any row of the table, and will get the same equilibrium (one is already at equilibrium, with $z = 0$).
I should say that the initial conditions are a thought experiment anyway because as soon as you add all the reactants or all of the products, the reaction starts, so the "initial" concentrations are never present in reality.

oohhh I think I get it now. No matter what I put in initially, I am still going to get an equilibrium of NO2 and N2O4. And thus the equilibrium constant must be the same - no matter what I start off with. Frankly, it wouldn't have mattered if I put 2000 moles of NO2 into it. The Keq will always be equal

One more question then. I agree the Keq for the 2 cases is the same. But, how do I know that the actual concentration of NO2 is the same using my method and the answer's? How can I be certain that it's not actually a different set of final concentrations that still makes the Keq fraction work?

@JohnHon Equilibrium is when $Q = K$. As the reaction goes in the forward direction, $Q$ will steadily increase (monotonically), and decrease for the reverse reaction. There is only one set of concentrations (for a given total mass of reactants and products) where $Q$ is equal to $K$, so there is a single solution. If you start with a different total mass, you will also reach equilibrium, but with different concentrations of species (and still, $Q = K$).

The order in which you add the reagents doesn't matter. The amount of dinitrogen tetraoxide reacted (x) is given by $\pu{0.600 mol}=\pu{0.200 mol}+2x$. The remaining amount is $\pu{0.300 mol}−x$. From these you can compute the equilibrium constant ($x=\pu{0.200 mol}$....).
Think of an analogy: I give you x dollars and y pounds. Whether you convert some of the dollars into pounds before or after I give you the pounds doesn't matter at the end (after the final equilibrium is established). You know how many pounds you had in the beginning and in the end, therefore you know the total dollars that were converted.
You tried to find an intermediate equilibrium, but lacked the information to do so (until computing the equilibrium constant from the final concentrations), and did not need to.

Your analogy is interesting. Yes I agree that I can figure out how many dollars/pounds were converted given the initial and final. But how can we be sure it is the same if you give me the pounds later? How can I be sure that adding the 0.2 moles of NO2 later is going to produce the exact same result as adding it in initially?

Yes, the analogy works if you assume that you can predict where the equilibrium will lie in the end (you convert just the right amount of dollars before I give you the pounds). The catch is that the people who wrote the problem tell you what the final equilibrium result is (almost, they tell you half).

How you get to the final equilibrium is not important, provided you perform appropriate book-keeping. In the intermediate step you can convert an amount of N2O4 into an amount of NO2 that will result in a final equilibrium when you then add the NO2, that is, if in the intermediate step you convert (0.6-0.2)/2 moles of N2O4 into NO2. In reality it doesn't matter when the conversion happens, provided that the end result is the same.

There is a way to solve this problem with the two-step equilibrium system you proposed, although it's a bit more complicated than the solution you posted.
Let:
$A$ represent $N_2O_4$
$C$ represent $NO_2$
Then the reaction becomes:
$$\ce{A <=> 2C}$$
From your second ICE table, we can see that:
$$2x+0.2-2y=0.6$$
Setting $y$ in terms of $x$:
$$y=x-0.2$$
Since Temperature is constant, $Kc$ will remain constant, so we have:
$$Kc=\frac{C_{C2}^2}{C_{A2}}=\frac{C_{C1}^2}{C_{A1}}$$
Rearranging:
$$\left(\frac{C_{C2}}{C_{C1}}\right)^2=\frac{C_{A2}}{C_{A1}}$$
Now, we substitute all variables in terms of $x$ and $y$:
$$\left(\frac{0.6}{2x}\right)^2=\frac{0.3-x+y}{0.3-x}$$
Substituting the expression for $y$ we derived earlier:
$$\left(\frac{0.6}{2x}\right)^2=\frac{0.3-x+x-0.2}{0.3-x}$$
$$\left(\frac{0.6}{2x}\right)^2=\frac{0.1}{0.3-x}$$
We can solve this quadratic formula and keep the positive value:
$$x=0.2374$$
Solving for $y$:
$$y=0.0374$$
Substituting into $Kc$ on your second ICE table:
$$Kc=\frac{C_{C2}^2}{C_{A2}}=\frac{0.6^2}{0.3-x+y}=\frac{0.6^2}{0.3-0.2374+0.0374}=\frac{0.6^2}{0.1}=3.6$$
Substituting into $Kc$ on your first ICE table:
$$Kc=\frac{C_{C1}^2}{C_{A1}}=\frac{(2x)^2}{0.3-x}=\frac{[2(0.2374)]^2}{0.3-0.2374}=\frac{0.2254}{0.0626}=3.6$$
Either way, we get the same answer:
$$Kc=3.6$$


text:
Find the required water volume to add so that the pH of two solutions will be the same
Question.

We have two flasks: one contains 50 mL of 0.001 M HCl and the other 50 mL of 1M acetic acid.
a) How much water should be added to the more acidic solution so that the pH of the two solutions
be the same?

Attempt. So after finding both pH before adding the volume, we find that the first flask has a $\mathrm{p}H_1=-\log(10^{-3})=3$ and the second one can be found by considering $K_a(\text{acetic acid}) = \pu{1.8e-5}=\frac{x^2}{1-x}\implies x=\pu{4.25e-3 M}$. Hence $\mathrm{p}H_2=2.37$
Now here it comes the issues, following that both pH should be the same and we add a certain $V$ to the most acid one, which is the $2.37$ one: $$\mathrm{p}H_1=pH_2\implies 3=-\log(\frac{4.25\cdot 10^{-3}\cdot 0.05}{0.05+V})$$
Which if you solve this, you end up with $V\approx -0.0502L$ which doesn't make sense. And I'd rather follow a method similar to this where it starts from the equation $\mathrm{p}H_1=\mathrm{p}H_2$ since that would be how I would have reasoned through this in an exam

Adrian: The ionization of a weak acid changes with concentration so you cannot simply dilute and change the pH in a linear manner.  You know the total amount of acid in the 50mL; calculate the concentration needed to give a pH of 3 using Ka. then figure the amount of water for the amount of acid.

@jimchmst okay thanks for telling me why it is wrong. Now for the new method $3=pH_2$, you said we know the total amount of acid on 50 mL which I'm guessing you are talking about 1M? And how do I find the concentration needed using Ka? You mean to solve for y in the equation $Ka=\frac{\frac{y^2}{(0.05+V)^2}}{\frac{1-y}{0.05+V}}$? but we still have 2 variables

$$K_a=\frac{\pu{E-6}}{c - \pu{e-3}}$$ and the volume to add: $$V=50(\frac{1}{c}-1)\pu{mL}$$

In the flask that requires diluting, we have:
$$\ce{AcOH <=> H+ + AcO^-}$$
Let:
$A$ represent $\ce{AcOH}$
$C$ represent $\ce{H+}$
$D$ represent $\ce{AcO^-}$
So that the reaction becomes:
$$\ce{A <=> C + D}$$
Let:
The subscript (1) represent "before diluting with water"
We have:
$$N_{A1}=N_{Ao}-x=0.05-x$$
$$N_{C1}=N_{Co}+x=x$$
$$N_{D1}=N_{Do}+x=x$$
The relationship between $K_a$ and $K_{n1}$ is:
$$K_a=K_{n1}\left(\frac{1}{V_1}\right)^{\Delta n}$$
Considering that for this reaction $\Delta n=1$, and solving for $K_{n1}$:
$$K_{n1}=K_a \times V_1=(\pu{1.8e-5})(0.05)=\pu{9e-7}$$
Using the expression for $K_{n1}$ and the equilibrium number of moles, we can solve for $x$:
$$K_{n1}=\frac{N_{C1}N_{D1}}{N_{A1}}=\frac{x^2}{0.05-x}=\pu{9e-7}$$
$$x=\pu{2.117e-4}$$
Now, we dilute the mixture by adding water, so the equilibrium will shift again.
Let:
The subscript (2) represent "after diluting with water"
We have:
$$N_{A2}=N_{A1}-y=0.05-x-y$$
$$N_{C2}=N_{C1}+y=x+y$$
$$N_{D2}=N_{D1}+y=x+y$$
Temperature is constant, so $K_a$ does not change, and we have:
$$K_a=K_{n2}\left(\frac{1}{V_2}\right)^{\Delta n}$$
Solving for $K_{n2}$:
$$K_{n2}=K_a \times V_2$$
Substituting in the expression for $K_{n2}$:
$$K_{n2}=K_a \times V_2=\frac{N_{C2}N_{D2}}{N_{A2}}=\frac{(x+y)^2}{0.05-x-y}$$
Since we're required to dilute the acetic acid until its pH reaches 3, that means that at equilibrium, the total moles of $H^+$ divided by the resulting total volume $V_2$ must be equal to $10^{-3}$:
$$\frac{x+y}{V_2}=10^{-3}$$
Solving for $V_2$:
$$V_2=\frac{x+y}{\pu{e-3}}$$
Substituting above:
$$K_a \times \frac{x+y}{10^{-3}}=\frac{(x+y)^2}{0.05-x-y}$$
We already know $x$, so we can solve this quadratic formula for $y$, and we keep the positive value:
$$y=\pu{6.724e-4}$$
We can then solve for $V_2$:
$$V_2=\frac{x+y}{10^{-3}}=\frac{\pu{8.841e-4}}{10^{-3}}=\pu{0.8841 L}=\pu{884.1 mL}$$
Finally, we calculate the volume of water we have to add:
$$\Delta V=V_2-V_1=884.1-50=\pu{834.1 mL}$$
In conclusion, diluting the acetic acid by adding approximately 834.1 mL of pure water will increase its pH to 3.

Is not it rather complicated?

Perhaps long, but the hardest part is solving the two quadratic equations, which I wouldn't say is complicated at the level this material is usually taught.

I mean, compared to 2 simple equations in the other comment.

If what matters most to the original poster is simply answering the question, then of course. However, I don't think it's very clear where you derived those equations from and (I say this respectfully) simply posting them and just having him plug in values doesn't really provide much insight into the problem. My attempt shows a detailed analysis of how equilibrium shifts when adding water, so I think it can help understand what's happening.

That is why I have posted it as just a hinting comment (having limited resources). No derivation needed. The first one was used by the OP himself and the second is the application of the trivial reciprocity c versus V. No need for quadratic equations.

To a surprise to all of you, this expanded post does help more. Thank you Sam, it is now crystal clear

@Poutnik
Your solution is definitely more practical, but the average high-school or undergraduate student (considering this question was asked in relation to an upcoming exam) won't see it as "trivially" as you do.
I personally think members should not be discouraged from posting correct solutions out of fear of being considered "too complicated" by others, especially since that can be very subjective.

I did not want to discourage you. I just think even elaborated answer could be simpler and more compact. Just opinion. My comment was just idea to follow by OP or responder elaboration.

@Poutnik Sure, I don't disagree with that. I just think that depending on the context of the exam format, one solution can be more helpful than the other. If the format is simply multiple choice, then there's no doubt yours is more helpful. However, if the student is required to elaborate and show ICE tables and/or intermediate equilibria taking place, I think mine can be more helpful. Either way, both solutions are correct, and that's what really matters.

I admit I have never ever used ICE tables or anything similar in my life, not even as a student. Too much formal work.


text:
Using a vintage lab balance scale with a chain-dial
I have obtained an an antique Philip Harris balance scale in good, but condition aside from a lot of corrosion on the brass. I can't find a model number on it and am not sure of it's actual age, but I am guessing 1950's. It has a chain-dial on it, which is something that I'm not familiar with. I've not been able to figure out how to actually use the chain-dial or indeed any information on them. In cleaning up the scale I found that the guide for the plumb line was not correctly attached so it's possible that the chain-dial isn't attached right either.
Can anyone explain in detail how the chain-dial is used or direct me to some documentation or textbook resource on using a scale of this nature.

This is a case where including an image might help.

This user manual for a 1913 balance may or may not be helpful: https://www2.humboldt.edu/scimus/Instruments/UseCareBal/UseCareBal.htm as well as details of some different "chainomatic" balances from Becker: https://www2.humboldt.edu/scimus/Instruments/ChainBal-BKH/ChainBal.htm

Sorry I tried to include an image but stack exchange gave a server error and was unable to upload it.


text:
Why is the fugacity of a pure component not equal to the pressure calculated by an equation of state for that pure component?
I don't get why fugacity coefficients, $\phi = f/p$, of pure components are usually calculated via integrating an eos over a pressure or volume range. For example, when using a pressure explicit eos (such as the Virial-Eos for example), one can write:
$$
RT \ln \phi = \int_0^p(v-\frac{RT}{p})dp
$$
I was wondering, when we already know the real gas parameters such as the Virial-Coefficients, for example, why not calculate the actual pressure straight away?
Isn't the fugacity some kind of "real pressure", with $\phi$ serving as a conversion factor, $f = \phi p$. And isn't the pressure calculated from eos (PR, VdW, Virial etc.) also some kind of "real pressure" aswell. But why are they not equal?

No. Fugacity is related to the Gibbs free energy variation with pressure.$$dG=VdP=\frac{zRT}{P}dP=RTd\ln{f}$$So, $$RTd\ln{(f/P)}=\frac{(z-1)RT}{P}dP$$


text:
If you have an exothermic reaction in an isolated system, is enthalpy change 0?
Exothermic reaction has negative enthalpy change because energy is flowing out of the system. Isolated system means no energy is flowing in and out of the system. If you have an exothermic reaction in an isolated system, is enthalpy change 0?
Also, assume
Substance 1:  initial enthalpy of isolated system (i.e. reactants) is H1.
Substance 2:  Enthalpy of isolated system after reaction (i.e products) is H2.
Substance 3: Enthalpy of products taken out of isolated system, and left to be cooled (i.e. products cooled down to room temperature) is H3
Is enthalpy change of reaction H2-H1? H3-H1? Or neither?
Correct me if I am wrong: If you have a sealed container with gas at 500K,
the gas loses heat energy to its surroundings, then there is an enthalpy change. Since substance 3 is at a lower temp than substance 2, H3 is not equal to H2.

Guys, If there something wrong with the question/my understanding, could you please comment. I'm asking this question precisely because I don't understand.

Not downvoted by me, but if downvoted without a comment, a good guess can be the default downvote reasons displayed at mouse over the downvote triangle.

An exothermic reaction is one in which you need to remove heat from the reaction mixture in order to hold the temperature constant.  The amount of heat you need to remove to hold the temperature constant is minus the the enthalpy change for the reaction.
If you have an exothermic reaction of an ideal gas mixture in an isolated system, the internal energy change is zero, and the enthalpy change is $V\Delta P$, where V is the (constant) volume of the system and $\Delta P$ is the pressure change.
For the scenario you described, assuming no change in the number of moles in going from reactants to products, H2=H1 (really U2 = U1), and H3 < H1.

Hm, even if the molar amount does not change, temperature and therefore also pressure raise. So even if U is constant, H is not.

Enthalpy is defined as $$H = U + pV$$
Isolated systems are implicitly of constant volume, otherwise they can exchange energy by volume mechanical work.
Isolated systems have therefore constant $U$ and $V$.
$H$ is then constant if and only if $p$ is constant.
Manipulation with content of isolated systems severely break their thermodynamic context.
Neither $H_2 - H_1$, neither $H_3 - H_1$. In contrary to a general enthalpy change, the reaction enthalpy is defined as thermal exchange at isothermal and isobaric conditions.

In totally isolated system the only thing that can happen is eventual equilibrium. this means everything at the same T, P, V, chemical potential energy, nuclear, gravitational and at maximum entropy. I cannot grasp what happens to PV work since return to an original state cannot happen without unisolating the system. Instinct says it is conserved but that is not a good reason.


text:
Determining the concentration of a very small sample of sulfuric acid
How can we determine the concentration of sulfuric acid in a very small sample?
We are working on a “mechanical” process that transports a very small amount of sulfuric acid to an instrument and we need to provide the concentration of the acid on the output from our system tests (this does not need to be built into the system for operation, just bench testing validations). Our process involves levels of medium and high vacuum, I am considering the materials inside non-reactive (almost completely stainless steel), and the flow in the system is quick. I am assuming the only change to the acid is some amount of the water in the solution potentially boiling off from the vacuum, increasing the acid concentration. We know the acid solution at the input, and for safety and ease of testing at this point using a lower concentration, 0.12N solution. The output volumes are much lower than a drop of water, and I am guessing are in the 10s of microliters. We have access high quality balance and basic lab tools / glassware.
I had considered adding a drop of water to the test sample and then placing that in a refractometer to measure the diluted acid % w/w, specific gravity, and density. Then run the dilution calculations (how much water to add to concentrated acid to make low concentration acid) backwards to find the sample concentration before I added the drop of water. It seems too many things about the “concentrated acid” (the test sample) are unknowns to perform that calculation. I am also concerned that the concentration of the mixture would be too diluted to accurately measure with a refractometer.

How precise and accurate must this be? How frequently? For a quick test, could you calibrate conductivity between two point (insulated up to tip) Pt electrodes in a droplet? Test the effect of droplet size, temperature and other variations, as well as check if there are contaminants that might affect conductivity. See http://www.sulphuric-acid.com/TechManual/Properties/properties_acid_cond.htm  N.B. The curve may have more than one concentration for a conductivity value.

If you have access to a micropipette, you could do titration with a known concentration of a strong base using very small volumes. Micropipettes come in different volume ranges, but some can go as low as 0.1 microliters.

"10s of microliters" isn't much lower than a drop of water. IIRC a normal drop can be about 50 microliters. You could even don't dilute, just use some small electrodes.

Maybe turbidimetry or nephelometry? Add the small sulfuric acid droplet to 1 mL of barium chloride solution with standardized concentration. Then barium sulfate is produced, yielding a turbid solution. Calibration would be required. Good luck with whatever you decide to do!

@DrMoishePippik indeed! Is "Sulfuric acid... makes a poor electrolyte... very little of it will dissociate into ions" really true? What does that actually mean? and Details of what actually happens in cold sulfuric acid between 80 and 90%; what molecular changes cause the viscosity to skyrocket?

Spike the acid with some [Fe(CN)2(2,2'-bipy)] and rig up a mini UV/vis machine to record the colour of the droplet.
If you take a iron(II) salt and then add 2,2'bipyridine to it you will get a blood red solution. In a fumehood add sodium cyanide to this and you will get a a dark solid. Collect the solid by filtration and then wash it with water to remove any excess sodium cyanide and 2,2'bipy. Next dissolve the solid in concentrated sulfuric acid to give you a bright yellow solution. Filter this through glass wool (do not use paper or cotton wool as the sulfuric acid will char the cellulose). Now add the yellow solution to distillated water, you will now get a slurry of a deep red solid in a pale red solution.
The compound I have told you how to make is solvatochromic the more able a solvent is to accept lone pairs (act as a a Lewis acid) the more yellow the solution will be. Make a dilute solution of this iron compound in the dilute sulfuric acid and then do your experiment.
See
METAL-COMPLEXES AS COLOR INDICATORS FOR SOLVENT PARAMETERS
R.W. SOUKUP and R. SCHMID,
Journal of Chemical Education
Volume62
Issue6
Page459-462
DOI10.1021/ed062p459
Published1985

Micro pH meters are available for samples as small as 0.1 drop. The smaller they get, the more they cost ($1132, Ref 1, Ref 2). The OP mentions a fairly low concentration of acid: "We know the acid solution at the input, and for safety and ease of testing at this point using a lower concentration, 0.12N solution."

Ref 1. https://www.thomassci.com/Instruments/Electrodes-pH/_/MICRO-COMBINATION-pH-ELECTRODES?q=Micro%20Ph%20Meter
Ref 2. https://www.horiba.com/int/water-quality/support/electrochemistry/measurement-of-ph/measurement-method-for-single-micro-liter-sample-using-compact-ph-meter/

For strong sulfuric acid solutions it is not reasonable to use the pH scale

Determining the concentration of a very small sample of sulfuric acid

From Details of what actually happens in cold sulfuric acid between 80 and 90%; what molecular changes cause the viscosity to skyrocket? we can see that up to about 80% concentration the viscosity of sulfuric acid increases monotonically.
Since you (seem to be) working with pressure differentials and capillary forces, you are likely already very sensitive to viscosity and should be able to start measuring it if you have apertures or microfluidic/capillary passages of dimensions that are constant (don't vary with time).
Make a little gizmo that's sensitive to viscosity, e.g. flow rate or movement speed with a known pressure differential, and calibrate it with known dilutions. Then you've got your "concentratometer" suitable for your in-lab measurements.
Be very careful to control for the temperature - better calibrate at many closely-spaced temperatures and be sure to include a temperature reading as part of your concentration determinations.
From Details of what actually happens in cold sulfuric acid between 80 and 90%; what molecular changes cause the viscosity to skyrocket?

AChem's answer to Why does the graph of the electrical conductivity of sulfuric acid/water solutions have this knee in the ~85%-~92% range? includes this plot from Horace E. Darling in "Conductivity of sulfuric acid solutions" (Journal of Chemical & Engineering Data 9.3 (1964): 421-426.) and mentions:

There is a sharp increase in viscosity at 85%, which indicates there is a major structural change in sulfuric acid solution in the range 85-92%. Sulfuric acid forms a hydrate in this range. When the viscosity is high, the conductance goes down, there is a depression in the curve. This viscosity jump is causing the double hump. Once we are past the high viscosity range, conductance goes up again.
It is amazing how simple molecules do not stop from surprising us!

Das et al. (1997) Electrical Conductance and Viscosity of Concentrated H2SO4/H2O Binary Systems at Low Temperatures: Correlation with Phase Transitions (J. Phys. Chem. B 1997, 101, 4166-4170) do a thorough analysis and mention phase transitions and hydrate formation, but 25 years later are there more detailed molecular models of what is happening that might address the exact form of the hydrate(s) formed or any long-range correlations or ordering behind the peak in viscosity at a certain high concentration?

I think that the easiest analyses might be turbidity with BaCl2 solution or conductivity. The limiting factor would be precise measurement of the sample size. Of course 18 Meg water for the conductivity. It might be possible to make the sample size a constant systematic error


text:
What ideal gas behavior is observed by a real gas that observes Boyle's law (at Boyle's temperature)?
Boyle's temperature is a constant for a particular gas. At this temperature the real gas behaves ideally.
So, does the real gas only follow Boyle's law at this temperature, or are all the gas laws associated with ideal gas behavior applicable (or just Boyle's law as the temp is constant)?

Boyle temperature just means $pV=nRT_B$. It does not mean the gas behaves ideally. It is still approximation for a pressure range. The gas is not infinitely compressible.

Close to ideal I meant .

Being ideal has multiple aspects, fitting the ideal state equation is just one of them.

Such a gas behaves ideally in a particular context, if various aspects of its deviations from ideal behaviour mutually cancel, like in case of the Boyle law at not too high pressure.

I think I understood the gist of your unclear post, so I tried to edit it for clarity. I hope that's ok. You can revert to the original if you prefer.


text:
If entropy is a state function, how can it be calculated for both reversible and irreversible processes?

$$\Delta S = nR\ln( V_2/V_1)$$
The value of ΔS of the system is independent of the path between the
initial and final states, so this expression applies whether the change of
state occurs reversibly or irreversibly.

This is what my textbook says. However, when an ideal gas expands against constant external pressure, the process is irreversible according to the definition of reversibility (mechanical equilibrium, I Suppose)
In an isothermal expansion $U=0=q+w$, and more work is done if the process is reversible. So $dW_{rev} \ne dW$, $dQ_{rev} \ne dQ$.
How could the equation $\Delta S = nR\ln(V_2/V_1)$ apply for both reversible and irreversible processes when the amount of heat flow is different? Isn't entropy defined as heat flow divided by temperature?

"My textbook" is meaningless. The blockquote appears to be taken from Atkins' Physical chemistry. Cite the source appropriately (ACS style is preferred) and use the correct formatting (text labels and math operators should be upright) from the source. As for the question, there should be a preceding subsection in your textbook discussing entropy as state function, which should answer the question.

This may be obvious, but in case not, remember that the $\Delta S$ here is only for the system, which excludes the surroundings. The different heat flow and work in the reversible vs irreversible paths will result in different entropy changes to the surroundings, but not to the system.

Entropy change is not, in general, defined as heat flow divided by temperature.  It is determined by heat flow divided by temperature only for a reversible path between the initial and final thermodynamic equilibrium states of the system.  Fortunately, it is typically possible to identify a reversible path between the same two end states as for the irreversible path.  However, in some cases, it is a challenge.
The first step in determining the entropy change for a system undergoing an irreversible process is to establish the final state (based on the 1st law of thermodynamics). The next step is to devise a reversible path between these same two states.

The fact that entropy is a state function is what renders it path-independent. You could regard path independence as the defining property of a state function. Entropy is a state function because it depends only on the arrangement (as a statistical average) of the constituents of the system at the given values of the thermodynamic variables (T,P,V etc), not how the arrangement was obtained.
Since the entropy is a state variable, the difference in entropy must also be constant for two specific states. Since the change only depends on the states of the end points, if you find some way of determining the entropy difference, say by performing a transformation in a particular special way, then that difference must be the same for all possible processes that take you between the two states.
The rather remarkable thing is that there is a special (and ideal) way of transforming one system into another that will provide the magnitude of the entropy change for any transformation, and that is to sum the ratio of the heat produced to the temperature at each of infinitely many infinitely small steps of a reversible path, one performed infinitely slowly: $$\Delta S = \int_1^2 \frac{dq_{rev}}{T}$$
Although this is presented as the definition of the entropy (according to one formulation of the 2nd law), it is a thermodynamic definition and can lead one to miss the forest for the trees. The entropy is a property of the system, not of the path. That the above relation between heat, T, and entropy exists is remarkable (and has other implications, for instance wrt the definition of heat and T), but it is essential to remember that in general entropy is a property of the states, not of the path.
There are a lot of other posts on this site worth consulting for more on similar questions.

"You could regard path independence as the defining property of a state function" - I was taught many decades ago path independence is the defining property of a state function

@IanBush well I'm glad I got that right. There is also the fact that state functions are properties of equilibrium states, but yes, I might have introduced some unnecessary redundancy.

It's the theoretician in me getting pedantic ...


text:
Does acrylic resin in acetone react with the same acrylic resin in ethyl acetate?
I am not at all familiar with organic chemistry. I am working in museum conservation where an acrylic resin (such as Paraloid B-72) is mixed acetone and sometimes ethyl acetate to produce a resin for marking objects.
Would there be any potentially dangerous byproducts produced if a 20% solution of Paraloid B-72 in acetone was mixed with a 15% solution of Paraloid B-72 in ethyl acetate? Do these react on their own? Is it possible that the Paraloid could act as some kind of catalyst?

If you mix them you just have solution of Paraloid in mixed solvent, I don't see any reason to think it would be a problem.

Paraloid® B72 as an adhesive for archaeological ceramics and other museum objects has been well established (Ref.1) as stated in the abstract:

The proper preparation and application of Paraloid B-72 make it an excellent adhesive for archaeological ceramics and other materials. An understanding of the principles of adhesion is essential to the successful application of solvent adhesives.

Acetone is the most suitable solvent for the resin since its fast evaporation. Paraloid® B72 has replaced the cellulose nitrate lacquer used for many years in this purposes due to its long time stability (Ref.2). Other than acetone, Paraloid B-72  can be dissoved in other solvents such as toluene, xylene and ethanol, etc. The solvent effect (including that of ethyl acetate) on Paraloid B-72 has been studied and has not reported any adversary reactive nature (Ref.3).
Thus, the answer to OP's question, "Would there be any potentially dangerous byproducts produced if a 20% solution of Paraloid B-72 in acetone was mixed with a 15% solution of Paraloid B-72 in ethyl acetate?": Not necessarily as experimentally evidenced by these authors.

References:

Stephen P. Koob, "The use of Paraloid B-72 as an adhesive: its application for archaeological ceramics and other materials," Studies in Conservation 1986, 31(1), 7-14 (DOI: https://doi.org/10.1179/sic.1986.31.1.7).
Brigid Sullivan and Donald R. Cumberland, Jr., "Use Of Acryloid B-72 Lacquer For Labeling Museum Objects," Conserve O Gram 1993, No.1/4, National Park Service Publication (PDF).
Armand Vinçotte, Emmie Beauvoit, Nicolas Boyard, and Elodie Guilminot, "Effect of solvent on PARALOID® B72 and B44 acrylic resins used as adhesives in conservation," Heritage Science 2019, 7(42), 9 pages (DOI: https://doi.org/10.1186/s40494-019-0283-9).


text:
Why middle elements of transition series show more number of oxidation states
Why do the elements in the middle of the transition series show more number of common oxidation states than others?

Because they have more valence electrons to play with. Earlier elements have too few outside the core, and later ones tend to have their $d$ electrons sinking into said core. I will add that gold is known up to +5.

You say they show more number of oxidation states, and I say you got lame data. That's only most common ox. states.

Yep I wanted to mean common oxidation states.

Unfortunately what's common is opinion based ;>

related https://chemistry.stackexchange.com/questions/67510/why-do-heavier-transition-metals-show-higher-oxidation-states

https://chemistry.stackexchange.com/questions/61353/why-do-osmium-and-iridium-have-the-most-oxidation-states-of-all-the-elements


text:
Is it always necessary to boil off H2S(g) to test for group III cations in qualitative analysis?
When a sample contains $\ce{Al^3+}$ ions (group-III), to test for its presence, before adding NH4Cl/NH3 solution do we need to boil off acidic H2S added earlier to check for the presence of group-ii cations? It can be deduced that $\ce{Al^3+}$ cations would be precipitated and persist as "gelatinous hydroxide" even in the presence of sulfides since Ksp of aluminum hydroxide is far higher than that of aluminium sulfide. So it seems that removal of H2S is not necessary in this case.
But, I know that if the sample to be analysed contains $\ce{Fe^3+}$, the acidic H2S could reduce it to $\ce{Fe^2+}$ ions which would eventually precipitate as FeS even in the presence of OH- barring the detection of $\ce{Fe^3+}$ in the form of Fe(OH)3. is it exactly the reason as to why H2S should be driven off to proceed further? A detailed answer will be appreciated.

Aluminum sulfide does not exist. It is totally hydrolyzed into $\ce{Al(OH)3 + H2S}$. So it has no $\ce{K_{sp}}$, despite the information in $4$th line.

Can you give the reference for your procedure for qualitative analysis?

Well... I was instructed to do so during my labs.


text:
Calculating the pH of an exteremely dilute solution (3.6×10^-9 M) of weak acid
I've got a question about analytical chemistry that I'm having trouble solving:

For a $\pu{3.6 \times 10^{-9} M}$ solution of 1-naphthoic acid, calculate the $\mathrm{pH}$ $\left(K_\mathrm{a} = 2.1 \times 10^{-4} \right).$

I attempted to solve using an ICE table and then solving for $x$. Then solving for $\mathrm{pH}$ from there. I was able to solve the first part of the question that way (concentration was $\pu{0.048 M}$ and the $\mathrm{pH}$ was 2.51).
However, I keep getting a $\mathrm{pH}$ of 8.44 and this is incorrect.
Any help would be greatly appreciated.

You have forgotten to include autoionization of pure water.

Could you clearly state your calculations? I know you probably got 8.44 from $\mathrm{-log(3.6\cdot10^{-9})}$ and 2.51 from $\ce{[H+]^2 = (0.048)(2.1 x 10^{-4})}$, but they don't really make sense with this context as such a small concentration of a weak acid won't change the pH by anything really

It should be clear pH must be just slightly below 7 and that we can consider near all acid to be dissociated. Then you can consider it as the case of very diluted strong acid. Kw=[H+][OH-]=H+.

Have a look at this answer  https://chemistry.stackexchange.com/questions/100346/calculate-ph-of-a-mixture-of-a-strong-base-and-acid-knowing-only-the-ph-wt-v/100355#100355 and this one https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566#95566 for general method to solve these kinds of problems.


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Are there any methods to dry a wet chemical sample within roughly half an hour?
I am planning an experiment for a school project. I have roughly 50 minutes (not including equipment preparations) to perform my experiment, which aims to experimentally verify how Ksp changes with temperature. Traditional methods such as oven drying would not work because it would take way too long for the sample to be dried. I don't have a specific compound in mind that I want use, but my thought process is as follows:

Add a "poorly" soluble (Ksp on the order of 10^-2) compound to distilled water
Stir either with flask or magnetic stir bar
Vacuum filter the remaining solid
Dry the sample within roughly half an hour, if possible
Record the mass that was not dissolved, to get the mass that was dissolved and calculate Ksp from there

I don't have much experience in the lab, so I am looking suggestions of methods for  quickly drying a sample. Would it be more effective to get the Ksp using another method, such as measuring the electrical conductivity of the solution, instead of getting the mass?

What is the overall goal of the project? Is determing Ksp the only choice given to you?

Is applicable washing the solid sample by volatile solvent soluble in water (e.g. acetone) and then fast evaporation of the solvent?

@ Poutnik: I agree about acetone hoping the substance is inorganic.

You can use an inorganic stable salt and then just put that and the filter paper in a muffle funrace.

Is this a demonstration in front of people, or something you have to do from start to finish in 50 minutes? If the former, could you pre-dry a sample to move on to after showing how to do the wet part?

There is a more easy way of doing the project. This is an experiment which I did at the age of about 16 in a GCSE chemistry lesson.
Take a flask or large test tube, weigh it empty and dry. Now add to it either some potassium chlorate, potassium perchlorate or if you want a safer substance use benzoic acid. Add to the test tube a small known volume of water from a burrett. Now heat the test tube by placing it in a beaker of boiling water. Put a thermometer into the tube.
Let the mixture heat up, if the solid fully dissolves then take the tube out of the hot water beaker and put it in a test tube rack. Allow it to cool while watching it. Note the temperture at which the solid reappears. If the solid fails to dissolve then add another 1 ml of water and put it back in the hot water bath.
Each time you have determined the temperture at which the solid reappears note it in a note book and then add 1 ml more of water to the tube and keep repeating the heating up and the cooling down (with careful watching).
Now divide the mass of the solute by the formula mass to get the number of moles. Next for each entry in your notebook calculate the number of moles per litre of the solute.
Now plot a graph of the concentrations on the y axis against the temperture on the x axis. You should now have a graph of the solubility of the solute against temperture. You can use this to compute the solubility product at different tempertures.


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Can helium hydride ions eat through teflon?
Searching on the internet for a really powerful acid, I came across a species called helium hydride ion which is estimated to have a $\mathrm{p}K_\mathrm{a}$ of $-63,$ making it orders of magnitude more acidic than fluoroantimonic acid, which has a $\mathrm{p}K_\mathrm{a}$ of $-31.$
I heard that teflon is the only way to store fluoroantimonic acid. Due to its strong carbon-fluorine bonds, teflon is capable of tolerating the acidity of this superacid.
Now, teflon may be strong enough to store fluoroantimonic acid, but is teflon capable of storing the helium hydride ion, given that it is capable of protonating $\ce{O_2, H_2O, CO_2, NH_3, SO_2}$ and so on? And how do you determine or predict that (via which process)?.

You cannot bottle up helium hydride ions, no matter what the bottle would be made out of.

@Martin-マーチン teflon lost the match$?$

@EveryonesaysFU No, helium hydride ion would destroy itself first. Attempting to form a salt to store it results in proton transfer from the $\ce{HeH^+}$ to the proposed anion, so all you have is helium gas plus the acid from the anion.

@OscarLanzi: On processing the question I quickly came to conclusion it's not a question of being able to withstand the chemical corrosiveness of the ion; its rather since the ion cannot form a salt it's can you possibly make a container that can withstand the electrostatic force of any great amount of it, for which you will find the answer to be no.

The question is entirely hypothetical as there is no way to create a significant amount of helium hydride in bulk.

Helium hydride has been recently discovered by mass spectrometry at extremely low pressure and rather low temperature. It does not exist in usual conditions. It was also discovered in a planetary nebulae. So this ion cannot "stored" as you believe. Whatever their nature, positive ions alone cannot be stored in any container without the same amount of negative ions. And whatever the wall, for example teflon, $\ce{HeH+}$ ions are immediately decomposed into $\ce{He}$ and a proton when touching it.

This would be better suited as a comment, rather than an answer. And while this address a key problem with the question, it isn't actually an answer to it.

somewhat related to that: Why was helium hydride (HeH+) the universe's first molecule?

Assume that hypothetically we're able to produce $1$ liter of helium hydride$?$ Can teflon withstand it$?$ what do you exactly mean by cannot be stored $?$

@EveryonesaysFU to store macroscopic amounts of any ion we must respect electroneutrality, which means you need a counterion to form a salt. $\ce{HeH^+}$ would react with the countetion instead (see my comment to the q), and while unchanged $\ce{HeH}$ would not need the counterion this alternative would not be bound in the ground state.

If we consider the idea of helium as a conjugate base, it has an electron pair so in theory it could act as a Lewis base if a sufficently strong Lewis acid was provided. It is important to undertsand that it is rather easy to form compounds of the heavy noble gases, the lone pairs on radon can bind to silver(I) ions to form complexes. This is the reason why a siliver exchanged zeolite is so good at adsorbing radon from air.
The problem is that the smaller size of helium than radon / xenon together with the higher electronegativity of helium will make it such a weak Lewis base.
We can regard Brosted acids such as acetic acid, hydrogen flouride and sulfuric acid to be Lewis acid/base complexes. The Lewis acid in question is the proton while the conjugate base is the Lewis base.
The great problem I see with cationic helium hydride is that we will be dealing with such a one sided covalent bond that it will not be very stable. Also it will tend to break down with great ease. If we construct the orbitial diagram for a hydrogen molecule we can see that HeH+ is isoelectronic with a hydrogen molecule. I predict that the difference in the energies of the atomic orbitials will be large and the bond energy for the He-H bond will be very low.
I think that while HeH+ is formally a Bronsted acid the synthesis of this acid will be difficult to say the least. I think that while this super acid might be "the super acid of super acids" it will be impossible to make in any large amount.
I think that almost any other atom will be able to displace the helium atom in something similar to a SN2 reaction. I think it is likely to be very unstable in condensed media.
There is a paper about the HeH cation, see Zhuan Liu and Paul B. Davies, J. Chem. Phys., 1997, 107, page 337. I will adding some more to the answer.


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Air purifier: What is the reaction mechanism between HCHO and negative oxygen ion
Multiple government and commercial sources claim that the negative oxygen ion generated by air purifier can reduce HCHO concentration.
However, on google scholar, I find neither the reaction formula between them nor the reaction speed/efficiency under normal concentration and conditions.
What are some reliable source/common sense on this topic?
To my knowledge HCHO is a reductant so it makes sense that $O_3$ can effectively reduce its concentration. But the negative oxygen ion seems to be a reductant with short half-life and low concentration, casting doubts on its "purifying ability".


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What is the angle of LP-O-LP of H2O?
What is the angle of LP-O-LP of H2O, if H-O-H is 105 degree (or hybrid of sp3.86)?

In my opinion, there is no angle LP - O - LP, because these lone pairs are not bound to other atoms.

https://chemistry.stackexchange.com/questions/50906/are-the-lone-pairs-in-water-equivalent https://chemistry.stackexchange.com/questions/38597/non-bonded-orbitals-in-water

The structure of ice suggests that there is a geometry to lone pairs that does not require a partner. That there are 4 sp3 orbitals implies so

No such angle exists. The lone pairs on a water molecule do not form two lobes with no atom at the other end as they are often (mis)drawn in chemistry texts.
The molecular orbital structure of the water molecule is shown below1:

There are ten electrons in the water molecule, so the five lowest orbitals shown are the occupied ones. Going from the bottom up, we see:

a compact orbital huddled around the red oxygen nucleus, thus is the core $1s$ orbital on the oxygen.

a pair of orbitals that give bonding overlap between oxygen (red) and hydrogen (white), one shaped more or less as an atomic $s$ orbital might be and the other looking like a $p$ orbital. These constitute the two bonds in the water molecule.

the next two orbitals, and highest occu-pied, are the "$p$-like" nonbonding pairs. The fourth orbital is fairly symmetric lying in the plane of the molecule, with the node passing through the oxygen; the fifth orbital is a pure $p$ orbital straddling the entire plane, unaltered form an atomic oxygen $2p$ orbital.

With this structure, we could say the nodal planes are oriented $90°$ from each other, but directly defining an angle between the electron clouds does not really make sense.
Reference

Wikimedia Commons contributors. File:Molecular Orbitals for Water.png [Internet]. Wikimedia Commons, the free media repository; 2022 May 30, 20:15 UTC [cited 2022 Oct 16]. Available from: https://commons.wikimedia.org/w/index.php?title=File:Molecular_Orbitals_for_Water.png&oldid=660146300.


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How are the number of neutrons changed in an atom?
I'm studying isotopes in high school and I don't understand how it works.
From my understanding, when neutrons are added or removed, an isotope is created.
To calculate the number of neutrons, the atomic number is subtracted from the atomic mass and then rounded to the nearest number.
What is the reason why neutrons would be removed? What could cause neutrons to be removed from the nucleus of an atom?

Unless the atom is not stable, the number of neutrons in that atom is fixed. Changing the mass number for an isotope (make it a different isotope) is frequently not that easy.

Atomic mass and atomic number have almost nothing to do with each other. Don't get them confused.

Number of neutrons is calculated by subtracting atomic number from mass number, which is the total sum of neutrons and protons that the atom has.
Research radioactive decay to understand more.

Are you asking why various elements can have different numbers of neutrons, or how to actually change the number of neutrons in a given nucleus, or ???

All nuclei are isotopes. Adding or subtracting neutrons creates a different isotope. There are a variety of reasons why nuclei change: sometimes nuclear chemists want a different isotope; many nuclei are unstable and change spontaneously by a variety of radioactive processes.

Many elements are found in nature with differing number of neutrons. For example, hydrogen may be found with no neutron (1H, AKA protium, where the atomic mass is ~1, just a proton and a lightweight electron), with one neutron (2H, AKA deuterium, where the atomic mass is ~2, a proton, a neutron and a lightweight electron), and so forth.
All of them are called isotopes. Isotope just means that the number of neutrons is specified in the atomic mass, as in 1H and 2H. As naturally found in nature, though, most elements contain a mix of isotopes. For example, on Earth, hydrogen in the oceans contains about one atom of 2H for every 6420 atoms of 1H. However, that is just an average, and the isotopes of hydrogen can be separated comparatively easily (because of the comparatively large mass differences of its isotopes) by natural processes, such as evaporation, or by the Girdler sulfide process.
There are some monoisotopic elements, such as fluorine, that have only one stable isotope. $\ce{UF6}$ is used for separating isotopes of uranium, because of this, avoiding the hassle of also having to separate the isotopes of the element with which uranium is combined.
As for changing the number of neutrons, that is not done through most chemical processes (though one could argue that a chemical explosion can trigger a thermonuclear explosion directly). The number of neutrons may change if the element is intrinsically radioactive, as when 226Ra emits an alpha particle (2 neutrons and 2 protons), changing it to 222Rn, or when a a substance is bombarded by neutrons, as in a star or nuclear reactor.

The chemistry of an atom is determined by the behaviour of the electrons, which in turn is dictated by the number of protons in nuclei. The so the thing which defines the element is the number of the protons.
Unless we consider kinetic effects then the number of neutrons has no effect on the chemistry of an element. If I was to do an experiment with Cs133, Cs134, Cs135, Cs136 or Cs137 then the cesium will behave in the same way in all the experiments. The only stable isotope of cesium is Cs133 but the others will behave the same way.
The number of neutrons in nuceli can be changed by either radioactive decay or by reactions induced by things which strike the nuclei.
Alpha decay causes the loss of two neutrons and at the same time two protons.
Beta - decay causes a neutron to turn into a proton
Electron capture or beta + decay causes a proton to turn into a neutron.
If nuclei are struck by thermal neutrons it is possible for them to capture neutrons and thus increase their neutron number by one for each neutron which hits them. Consider for a moment if I was to put Cs133 in a thermal nuclear reactor, some of the neutrons would be captured by the Cs133 nuclei to form Cs134.
If you use faster neutrons then other reactions start to become important such as reactions where a neutron is caputred and a proton is ejected from the nucelous. For example C-14 is formed from N-14 by this reaction.
With faster and faster neutrons more bits can be spallated out of the nuceli, it is possible to have reactions where nuclei are hit by a neutron and then things like helium nuclei, tritons, deutrons and other stuff comes out. What happens is that the energy of the incoming neutron causes the product nuclei to have an energy above the ground state.
For example Pu239 if it captures a neutron it forms Pu240, the higher the energy of the Pu240 product the more likely it is to split (fission) to form two nuclei. With lower neutron energies the fraction of Pu240 nuclei that manage to relax down into the ground state by emitting gamma rays increases.
Another option is photonuclear reactions, beryllium has a very low threshold energy for such reactions. If you expose beryllium to gamma photons which are a few MeV in energy then the Be9 nuclei eject a neutron and form Be8. Be8 is very unstable and quickly undergoes an alpha decay to two helium atoms.

You are being confused by linguistics; an attempt to explain, that has failed. Elements are defined by the number of protons in the nucleus. In the neutral atom there are an equal number of electrons in their appropriate energy levels [orbitals]. The electron arrangement is responsible for Chemical reactions and varies with the element. You will learn all about that in chemistry, college and graduate school so be patient.
The nucleus of hydrogen, Protium, H, has one [1] proton. There is also another stable isotope [nuclide] of hydrogen with one proton and one neutron, Deuterium [D]. There is another "natural" isotope of hydrogen, Tritium [T] that contains, again, one proton, and 2 neutrons This isotope [or nuclide] happens to be radioactive or unstable. Other isotopes of H have been made in nuclear reactors.  This is true of most elements: there are several stable isotopes, sometimes natural radioactive isotopes, and many others have been made in nuclear reactors. The artificial isotopes are usually of little concern in most chemistry unless inserted specifically into molecules as tracers. In Chemistry we are usually talking about the natural mix of isotopes unless specifically stated or sometimes strongly implied.
Isotopes were formed in the nuclear reactors of stars, and we get to work with those that are stable enough to form into planets. An elemental symbol in chemistry almost always refers to the natural mix of isotopes the only ambiguous one is H. H can stand for the natural mix or for protium, the H isotope with one proton and No neutron. Isotopes of an element have very similar but NOT identical chemical and physical properties. This means that chemical reactions carried to completion or equilibrium will maintain the isotopic balance. Other reactions will show isotope effects. The differences are most pronounced for hydrogen H, D, and T.
Tritium, T, gives us a hint to what is going on in nuclei. It has an extra neutron and is unstable. Neutrons are needed to prevent the positive protons from repelling each other but too many neutrons are unstable in themselves. There are optimal neutron numbers for each element, usually close to the number of protons in each direction. Too many or too few neutrons and the nuclide is unstable, radioactive.  It is fascinating to lookup a table of nuclides to see which are stable and the various nuclear reactions of unstable nuclei to become stable. A nuclide with too many neutrons loses negative charge; a nuclide with too few neutrons loses positive charge; there are several ways of doing each. This is nuclear chemistry!
The mass of a proton and a neutron are almost identical; the mass of an electron is much smaller so the mass of a proton+electron is again almost the same as that of a neutron. Each nuclide [isotope] consists of a number of proton-electron couples and a number of neutrons. These can be counted, and the count is the Atomic Mass Number. The total number is measured by comparing the mass of the nuclide to the mass of a defined nuclide and relating it to Carbon-12. The atomic mass unit, amu, is 1/12 the mass of a carbon-12 atom. [this is not strictly true because the definition is slightly different based on a defined value of Avogadro's number]. The number of proton-electrons and neutrons is measured by the relative mass; the number of protons is measured from Xray spectra; the difference is the number of neutrons. in the isotope. Tha actual relative mass in atomic mass units must be measured and is very close to a whole number because the mass of a proton-electron and neutron are almost the same and the binding energy holding the nucleus together, while huge, requires a small change in mass.
Isotopes exist and chemists usually use them as they are in the natural mix. There are specific applications where specific isotopes are involved.  Chemists are not commonly involved in manufacturing isotopes. That is done by nuclear chemists and physicists using nuclear reactors, colliders etc. It is going on in the Sun and stars, a little bit in our upper atmosphere from cosmic ray interaction with matter, and the decay of radioactive nuclides that are still around, as we watch.


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Emission Spectra Light Wavelength Deviations for Mercury
The Problem: Inconsistencies between the measured and true values of mercury's spectra light wavelengths emitted following a laboratory experiment.
What I Know: The experiment conducted was viewing the colors and distances of emitted lights, through a grating, from a sample of mercury in a transformer. Purple, green, and orange were determined correctly to be the spectral light colors of mercury and the wavelengths for each must be found in centimeters. The formula $\lambda = d  \sin(\arctan((x / y))$ is required, where $d = 1/5.3 \times 10^3$ with regards to the constant distance between the grating grooves, $x = (x1 + x2) / 2$ with regards to the average x-axis length from the transformer, and $y = 99.9$ with regards to the constant y-axis length from the transformer.
Failed Calculation: The x-axis length averages were determined to be 9.25, 11.63, and 12.75 centimeters for purple, green, and orange correspondingly. After plugging the values into the formula, it resulted in wavelengths of $\pu{1.74e-5}$, $\pu{2.18e-5}$, and $\pu{2.39e-5}$ centimeters. However, the true wavelengths are $\pu{4.35e-5}$, $\pu{5.46e-5}$, and $\pu{5.77e-5}$ centimeters.
Attempted Fixes: I have already checked the data collected alongside formula input, and the formulas are what must be used when solving for the wavelengths.
With a percent error of about 60%, I do not understand how I am finding mercury's wavelengths wrong with the given formulas or what I am missing. Any help would be much appreciated.
Experiment Diagram:

@Ed V The diagram for the experiment has been added, and that is exactly how we conducted it. I just do not understand how the wavelengths are correctly deduced from the spectra light data. Thank you for your time.

The grating groove distance was initially in mm, but was converted to cm and is a constant $ 1 / 5.30 * 10^3$. Your diagram is basically how we conducted the experiment as well, and the formula is the same. I am doing something wrong with the calculations here though, I just don't know what. Again, thank you @Ed V for your time.

Those values though are very close to the true values. There may be an error in the grating distance, but with regards to an assumption of 200 lines/mm, for purple I tried to reproduce the plugging in you used as so: $λ = 200 * sin(tan^-1(9.25* 10^8 / 9.99 * 10^8))$. The result was 18.1 nm, different from 461 nm. How did you plug in your values in the equation, as my mistakes in that might also contribute to the percent error?

That conversion worked. That is all I guess, as assuming the distance as 5000 nm seems to get close to the true values and it might just be a grating length error. Have a nice day!

@Matt if you have found the answer to your question it is always okay to post answer to it. This 1) helps future readers know the question has been answered (many people don't read comments, they don't usually end up in web searches) and 2) you may pick up some additional reputation points. Once you reach 50 you can start leaving comments under other people's posts for example; privileges continue to open up as your reputation points increase. Welcome to Stack Exchange!


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Is there a molecular orbital equivalent of rehybridization?
I generally have seen the pyramidal inversion of NH3 explained in terms of rehybridization. The sp3 hybridized NH3 changes to sp2, with the lone pair in the p orbital, and then reverts to sp3 in the opposite configuration. On the other hand, something like PH3 has a higher barrier to inversion, because the poor hybridization leads to the lone pair being placed in an orbital with higher s character, and moving this lone pair into a p orbital would require more energy than in NH3.
However, hybridization is an approximation for MO theory, so I am curious as to whether there is a molecular orbital justification for why we can assume molecules rehybridize. Is there an analogous idea in molecular orbital theory that can explain the pyramidal inversion of NH3, as well as why PH3 has a higher barrier to inversion?
Thank you.


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Temperature dependence of order of reaction
In my textbook, it is given that the order of a reaction can change with temperature and pressure, but examples are not given. One possible explanation for this is that the rates of complex reactions are dependent on the rates of the elementary reactions in its mechanism, and at different temperatures, there are different rate determining steps, so the order can change. Is this explanation correct? Are there any examples for the change of order with temperature?

For gas phase reactions, concentrations are highly sensitive to temperature and pressure, so C vs t data for the same reaction will vary if performed under different P and T conditions. For liquid phase reactions, pressure changes don't matter as much as temperature changes.

Here is a contrived example. The mechanism of an enzyme-catalyzed reaction changes at high temperatures because those break the enzyme. As a consequence, the uncatalyzed reaction now is faster than the catalyzed reaction.

Your explanation seems to be correct. The inactivation or altered mechanism of an enzyme or catalyst is an example. I think the reason why examples aren't covered in a typical chemistry curriculum is that complex mechanisms are beyond the scope of introductory general chemistry. By the time students learn about those mechanisms, the fact that the reaction order of the overall reaction changes with temperature is not important compared to specific details about the mechanism itself. Whether two reactions are "the same" when they have different intermediates seems like a matter of semantics.


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Nomenclature of DIPEA
Why is Hünig's base called "N,N-Diisopropylethylamine" on Wikipedia, in the name DIPEA, as well as vendors like Sigma-Aldrich, rather than ""N,N-Ethyldiisopropylamine" (EDIPA)?
I thought these groups were named alphabetically?

'Ethyl' here is not a substituent; rather, 'ethylamine' is the name of the parent compound $\ce{CH3CH2NH2}$ (much like 'ethanol' is $\ce{CH3CH2OH}$). Notice that it isn't assigned a locant N in the same way that the isopropyl groups are.
So, there is no alphabetisation to speak of: there is only one substituent prefix, which is 'diisopropyl'.
A related question would be why 'ethylamine' is the parent compound, as opposed to 'isopropylamine', '2-propylamine', or to be precise, 'propan-2-amine'. In fact the preferred IUPAC name does use 'propan-2-amine' as the parent compound: it is 'N-ethyl-N-(propan-2-yl)propan-2-amine'.
Obviously, this is a mouthful, and in common usage the 'incorrect' but 'symmetrical' name N,N-diisopropylethylamine has prevailed.


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Does liquid mercury solubilize gold and other solid metals at room temperature?
Is it correct to say liquid mercury solubilizes gold and other solid metals at room temperature?
I know that mercury forms an amalgam (alloy) with some metals and I also know it is correct to say most alloys are a metallic solution. That's the why of the question.

"Solubilize" is not quite the right word; I'd better say "dissolve". Now to the point: yes, it does.

It does dissolves some metals but it also causes liquid metal embrittlement.  Essentially it dissolves grain boundaries and the metal will fail by brittle fracture although there is no damage visually apparent.

When I was in the army one of the guys was playing with some mercury [soldiers love danger] and got it on his wedding ring. Before he could get the ring off it broke into several pieces. He came running to the lab to get it fixed worried how to explain it to his wife. I put it in a plastic bottle as a keepsake.  Mercury dissolves gold very quickly.


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How long does acetone stay in ABS?
I know a larger part of this question is "it depends", since scales matter here. Let's bound ourselves to desktop 3D-printed objects, which will typically be no more than a few cm thick.
After an hour or so of acetone vapor smoothing, some people have noticed that ABS parts can bubble a week or two later. Below is an example of where an ABS part deformed when exposed to constant heat.

It makes sense to me that once the acetone is subsurface that it will just as easily continue to work its way deeper into the part as head toward the surface and evaporate.
On a long enough time line, though, all the acetone will eventually work its way to the surface and liberate itself from the plastic.
How can I determine that timeline? I would have thought it was hours, but empirical evidence seems to suggest it's at least weeks, if not longer.
(I'm making an assumption here and that's that acetone is a harmless solvent and will not permanently modify the plastic. It's unclear if that assumption is completely correct.)

If you're going to downvote, esp. if you're downvoting a new contributor, please take the time to explain why you think this question is not appropriate for SE.Chemistry.

Since acetone turns a solid object into goop, why do you consider it “harmless”? Do printed objects not exposed to acetone behave the same way?

Since, as you state, the effect varies, you'd need to experiment with you're printer and feed stock, to eliminate some of the many variables such as feed composition, printing temperature, heating time, thickness, temperature of piece and acetone vapor in smoothing, temperature and airflow for removing absorbed vapors, etc.

@JonCusterr "harmless" in reference to the material chemistry. I.e., water will dissolve table salt, but when the water evaporates the original salt remains chemically unchanged. I don't know if this is true about acetone and ABS.

@DrMoishePippik Those aren't my parts, but I would still like to know more about the first principles involved. I am hoping for a more rigorous approach than trial and error. Happy to rephrase the question if that would be appropriate.

@KennSebesta, rigorously, the solvent never completely leaves the workpiece; it's depletion curve is logarithmic, and, though concentration decreases over time, consider that it halves, and halves again, and halves yet again. It also would not be uniform with depth. If you could measure the rate, and establish what would be an allowable remainder of solvent, you could find this yourself.

@DrMoishePippik I agree with that, but I feel like it describes every physical process. They all asymptotically converge toward some stability point, whether it's chemistry, electricity, mechanics, etc... The difference is, as you rightly point out, the rates and the acceptable levels. I ask the question in hopes they are already well understood, and those answers can be provided here.


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1,1-Diphenylethylene subproduct/impurity
I'm working with 1,1-diphenylethylene and I've observed that, if left at room temperature in the air atmosphere, a white insoluble powder forms. I'm unable so dissolve this powder in every solvent that I've tried (acetonitrile, acetone, ethyl acetate, hexane, water, etc), so I can't get an NMR spectrum. Moreover, I can't find any paper or information online about this subproduct/impurity.
Edit: The melting point of this powder is around 120-130 ºC.
1,1-Diphenylethylene has to be stored between 2-8 ºC, but the safety data sheet doesn't say that it has to be stored under an iner atmosphere. However, we have it stored under argon in the refrigerator and the impurity only appears when leftovers of 1,1-diphenylethylene are left outside in a vial, for example. Futhermore, I observed that 1,1-diphenylethylene is also oxidized to benzophenone when left outside.

Hmmm, a polymer?

Polymerisation seems very likely

Or possibly peroxide formation, An ATR FTir will give a spectrum that should help you. More information on purity, preparation method, storage details etc. would help us.  Does the compound arrive stored under N2?

@jimchmst I've added an edit with more information.

This may be of help.


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How does the distance between the electrodes of a Galvanic cell affect the voltage between the electrodes?
I am trying to figure out how the distance between the electrodes of a galvanic cell affects the voltage between said electrons if everything else is kept the same. I have spent a few hours looking on the internet but I could not find anything.
Citations would be very much appreciated

Open or closed circuit voltage?

Hint: In an electrolyte solution, what factors affect resistance, R? Another keyword is internal cell resistance. https://en.wikipedia.org/wiki/Internal_resistance

Also remember, as Poutnik pointed out, when we talk about the voltage of Galvanic cells, we are not drawing any current. In older books, it was called null-point potentiometry.

The circuit is an open circuit. The set up lookes something like this one on the left
https://ars.els-cdn.com/content/image/1-s2.0-S0013468618313689-fx1.jpg

According to this website the resistance of an electrolyte solution is $R=\rho\frac{l}{a}$
https://www.gamry.com/Framework%20Help/HTML5%20-%20Tripane%20-%20Audience%20A/Content/EIS/Theory/Physical%20Electrochemistry%20and%20Circuit%20Elements/Electrolyte%20Resistance.htm
And isn't the resistance only half the story? Since the voltage will remain the same if the resistance changes but the current changes inversely
I do not understand what you mean in the second comment by "we are not drawing any current"

Apparently User$128138$ has not really understood the reasons and the differences for using inner and outer resistances.
Let's start from the very beginning. A galvanic cell has a certain theoretical voltage $E$ which is described by Nernst law. If the cell is connected to an outer resistance $R$, it produces a current $I$, which is not directly given by the elementary formula $I = E/R$. No ! One must take into account the inner resistance $r =\pu{\rho \frac{l}{a}}$. So that the current $I$ is linked to $E$ by the following formula $(1)$ $${I = \frac{E}{R + r}...,,,,....({1})}$$ In this case, the measured voltage on the electrodes is $E'$ < $E$ :
${E' = E - rI = E ( 1 - \frac{r}{R+r})}$, and $I ≠ 0$ , so that $E' < E$
If $R$ is infinite, the situation is like the sentence that User$128138$ does not understand, namely "we are not drawing any current". Here $I = 0$ and   $E' = E$.
If $R$ is finite, the cell produces a current $I$, which is related to the nominal voltage $E$ by the formula ($1$)
Hopefully the OP understands my explanation.