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Which reaction occurs first when 5-hexen-1-ol reacts with sulfuric acid?
I am trying to solve this question:

Find the major product of the following reaction:

Now we could either do electrophilic addition on the alkene or do dehydration of alcohol. These two different reaction pathways yield different products. So how do we decide which reaction occurs first?
I thought that electrophilic addition could be favored because of the generation of a carbocation that could then undergo $\mathrm{S_Ni}$ reaction with the $\ce{-OH}$ generating a cyclic structure. Now if the cyclic structure were more stable than the open chain product formed by dehydration of alcohol then this reaction could be favored. But I don't know how this could be proven.

In general, intramolecular reactions occur faster than their intermolecular counterparts. Most commonly this is rationalized by an entropic argument: in the case of the intermolecular reaction, two molecules (substrate + water) combine to form one (diol). This is entropically disfavored relative to the intramolecular case, in which the molecule cyclizes.
However, the biggest factor at play in determining which will occur in this case is the size of the resulting ring. If the product contains a 5, 6, or 7-membered ring, these rings are considered be unstrained (bond angles ~ 109.5 degrees). If the rings are smaller than 5 atoms or larger than 8, the rule of thumb is that the thermodynamic (enthalpic) cost of the ring strain associated with cyclization overcomes the entropic driving force to cyclize, and the intermolecular reaction is preferred.


text:
How to dissolve PLA/PCL/PMMA?
I have a research about electrospun membrane synthesis where I dissolve PLA pellets in chloroform:methanol (3:1) for 3 hours and also PCL pellets in chloroform:methanol (3:1) for 3 hours too. Another thing is I also dissolve PMMA powder in acetone for 15 minutes before the texture hardened. After that, I have to blend all of those solutions together to become homogenous solution.
The question is: Actually, I've tried to mix those 3 solutions with steps I tried to mix PLA and PCL first for 3 hours and then I mix it with PMMA for another 3 hours. I am wondering, is there any other steps to mix all of them efficiently (or with shorter time)?
If you could refer your answer with journals, I'd be glad to read it too. Thank you in any advance.

Its your own job to do your literature research. And you are asking too many questions at the same time. Lastly, you dont "dissolve for fifteen minutes". You can stir the mixture for fifteen minutes in the hope that it dissolves.

noted with thanks. I already changed it with just 1 question.

This question is extremely unclear (both missing facts and unclear wording) and therefore impossible to answer. You need to give quantities and concentrations, and report your observations. And please check how the word "to dissolve" is used in English language, because you are using it wrongly. Makes no sense.

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Explaining the result of experiment---Aluminum foil heated in boiling KMnO4 solution
I've recently been thinking back to an impromptu "experiment" I performed almost exactly 3 years ago. The setup was as follows:

A concentrated solution of KMnO$_4$ was produced by dissolving 2 grams of
  Condy's crystals in 250 ml of water. This was subsequently poured into
  an Erlenmyer flask containing a strip of aluminum foil, and the entire
  system was then heated at high on a stove until the water came to a boil. As the water evaporated, the solution lost the intensity of its violet
  hue, turning a light pink-violet instead. The stove was then turned off, and the aluminum strip was removed, having turned a golden color.

My questions are two:

What, precisely, was the chemical reaction(s) that occurred?
How much of a risk did the steam from the boiling solution pose in
   terms of potential acute manganese inhalation toxicity---and what are the potential long-term effects? Unfortunately, this
   experiment was performed on an ordinary kitchen stove and not in a
   fume hood (I'm kicking myself to this day).

My guess as to the former is that MnO$_2$ was produced, hence the observed color changes of the solution and foil. I am not sure whether this occurred via heat decomposition of the MNO$_4^-$ (which seems to occur at temperatures of >240$^\circ$C, and thus should not have been able to take place in boiling water) or oxidization of the aluminum (having somehow penetrated the surface layer of aluminum oxide). As for the latter, welp.
Thanks in advance!

Maybe solid KMnO4 decomposes at $>240^\circ\rm C$. A solution, however, is quite a different story.

Potassium permanganate solution tends to decompose with time, especially those which are not acidified. Just let an aqueous solution sit in a glass bottle and it will leave a brown film of tenacious manganese dioxide on the surface at room temperature. If water has any organic material, it will become brown and tap water is not pure water. It decomposes on the skin as well leaving a brown stain. You Al surface is no different, with a thin deposit of manganese dioxide, it may appear yellowish-golden.
Long time ago potassium permanganate was added to water wells to disinfect water. Of course it would decompose with all the organics. If you are well after three years, what is the problem?

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Assigning Cahn-Ingold-Prelog priority numbering rules for 2 groups
I was asked to draw the perspective formula of

$$\ce{(3R,4R)CH2=CH-CH(CH3)-CH(CH3)-CH2-CH3}$$

Then I had to assign priority between the two groups:

$$\ce{-CH=CH2}$$
  and
  $$\ce{-CH(CH3)-CH2-CH3}$$

Actually, they are bonded to the same atoms until the end of the first group, but one of them should have priority over the other.
Can please someone help?

When you've reached the end of your first group (after resolving the double bond), you're next comparing no atom (on the ghost C connected to C2) to a hydrogen atom (on the 4-methyl group), and any atom has a higher priority than no atom, so your second group gets the higher priority.


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Will Proving or Disproving of any of the following have effects on Chemistry in general?
I am working on a project relating to-
https://en.wikipedia.org/wiki/Millennium_Prize_Problems
I wanted to list the effects of them being proved or disproved in different aspects of science and maths. It was fairly easy to find such results on physics and maths as the questions primarily come up from them however I couldn't find any reference to effects it would have on our understanding of chemistry.
Is it because there will be no effect at all and that they are the least concern to chemists (in terms of applicability) or if there are then what are they?

To understand just how absurdly far-fetched this may come across, ask yourself the same question: what effect will it have on your life? Say, you get up, brush your teeth, have your breakfast, then BANG! $P\ne NP.$ So what?

I suspect that two problems (p-np and Navier–Stokes existence and smoothness) would have repercussions, in defining some problems as either bounded or unbounded. Guaranteeing that you can find a solution to a certain level of accuracy in a given amount of time is akin to guaranteeing that you will come out winning in a lottery if you buy enough tickets. However I admit to speculating - I am not directly familiar with the problems.

@IvanNeretin I know it wouldn't be of any importance to me in my general life but i wanted to know it's impact on a discipline or a field of study such as Chemistry

@IvanNeretin I was reading the Code of Conduct for this website. Maybe you might find the Friendly and Unfriendly part helpful :-) https://chemistry.stackexchange.com/conduct

All right, let's sum it up.

P vs NP, if solved the way that pretty much everybody expects (that is, $\rm P\ne NP$), will have no far-reaching consequences, because that's what we were thinking for quite a while now. If it would miraculously happen to be otherwise (that is, $\rm P=NP$), that would be quite a shock to many fields, especially to computer science, and by extension, to computational chemistry. But I don't believe in miracles.
Navier–Stokes (no matter how it will turn out) will have consequences in fluid dynamics, and by extension, maybe in some relatively narrow areas of chemical technology.
Yang–Mills will cause repercussions in the areas of physics which are the most distant from chemistry.
The rest of the problems, as far as I understand, have no bearing whatsoever on our material world.

"The rest of the problems, as far as I understand, have no bearing whatsoever on our material world." - Not true. The Hodge conjecture, at the very least, would impact string theory (https://mathoverflow.net/questions/221892/what-are-the-ramifications-of-the-hodge-conjecture-to-mathematical-physics), and the Birch-Swinnerton-Dyer conjecture and Riemann hypothesis have cryptographics ramifications, which would affect money, which is material (or at least is convertible into something material). The Poincare conjecture is also relevant for certain GR spacetimes.

String theory has no bearing on our world either, and won't have for a while. Poincare... well, if our universe is closed (which we are not sure about), then it is applicable and may have some consequences, though I don't see any. As for the rest, well, point taken: Riemann and probably BSD will affect the world, if only through computer algorithms. But then again, it is much like P vs NP: if what we think true is proven true, then some tentative estimates become exact, and things continue running as usual. And if it is proven not true, that would be a miracle.

... and possibly the worst version of $P = NP$, a completely nonconstructive proof of that equality.  So then we know that many problems are "easy", but we're ignorant of the easy way.

@IvanNeretin If string theory has no bearing on our world, then Yang-Mills surely doesn't either. Likewise, if P vs. NP affects our world, then Riemann and BSD do too. I'm just confused about the distinction between the things you say do have some bearing on "our material world" and those that you say have "absolutely no bearing on our material world."

That's why I said "point taken". Admittedly, Riemann and BSD do have some bearing on our world, just not on chemistry.

@EricTowers Actually, there is an algorithm that is guaranteed to be a P-time solution to an NP-problem if and only if P=NP.  (It is, however, exponential in the size of the algorithm if implemented naively, and it does not tell you if it found the algorithm, but it does tell you if it found the solution to a given case)


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why is there a net current at halfwave potential?
Im wondering why there is a current measured at a halfwave potential? According to my understanding at the halfwave potential of a diffusion controlled reduction or oxiation at an electrode, there is half of the current due to anodic and the other half of the current due to cathodic current. I assume that therefore the net current is zero, since the same amount of electrons/charge leaves and enters the electrode...

You have it all backwards. Not half, but all current is due to anodic half-reaction. Also, all of it is due to cathodic half-reaction. Each single electron participates in both.

I guess I do not really understand...
I expect that we are just observing the working electrode, right?
and here a reduction or oxidation can occur, resulting in cathodic or anodic current, respectively. is this correct?
So how is it possible that all current is anodic and cathodic at same moment in time?


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Application of Nernst Equation in Potentiometric titration of Fe2+ with Ce4+
It is well Known that to find $E_{cell}$ during titration of  $Fe^{2+}$ with $Ce^{4+}$ we consider 3 domains:
Before Equivalance point ,  $E_+ = E_{Fe^{3+}/Fe^{2+}}^o+\frac{RT}{nF}log\frac{[Fe^{3+}]}{[Fe^{2+}]} \tag{1}$
At Equivalance point , $E_+=\frac{E_{Fe^{3+}/Fe^{2+}}+E_{Ce^{4+}/Ce^{3+}}}{2} \tag{2}$
After  Equivalance point ,$E_+ = E_{Ce^{4+}/Ce^{3+}}^o+\frac{RT}{nF}log\frac{[Ce^{4+}]}{[Ce^{3+}]} \tag{3}$
$E_+$ is the electrode potential of the positive electrode where the negative electrode is a fixed reference electrode.
In the book Quantitative Chemical Analysis by Daniel C Harris, Chapter 15 says that both equation 1 and 3 can be used either of the two domains and the reason why one is used instead of the other is due to the fact that the ratio of the concentration of a particular metal ion in different oxidation state is much easily found in that domain while the other ratio usually requires a complicted equilibrium equation to be solved to obtain the solution.
my questions:

Does this mean that if by some means I knew the ratio of concentrations of both metals I could substitute it in eqn 1 and eqn 3 and end up getting the same value for $E_+$.
If not does it only hold near the Equivalance point?
If one is True , The only explanation I have is that both $Fe^{2+}$ and $Ce ^{4+}$ are in equilibrium with the same electrode. But this still doesn't make sense as usually when 2 reactions are occurring at the same container the $\Delta G_{net}=\Delta G_{1}+\Delta G_{2} $ So in some sense shouldn't $ E_+ =E_{due-to-Fe}+E_{due-to-Ce}$ be the equation we require?

If I'm missing some essential concept it would be Great if someone could point it out and probably provide some links for me to look up.

The two closing votes say this is off topic because it is not related to chemistry. This is quite absurd and strange. Potentiometry is a standard course material in modern analytical chemistry curricula.

It is an interesting question. Daniel Harris is revising his book with my former mentor. Hope he clarifies this section in the revised version. Your point number 1 is misleading. The reason is that before the titration, theoretically there is no Fe(III). So Nernst equation should not be used- electrode potential is infinite (log 0 is undefined). When you start the titration (Ce (IV) in the burette), the moment you add Ce(IV), it is immediately reduced by Fe(II) to Ce(III). You do not have the Ce(IV/III) couple in the solution. Before the endpoint, Fe(III/II) couple controls the potential of the indicator electrode.
Do you remember what that indicator electrode is made of? It is just a platinum wire and the reference electrode is usually silver-silver chloride electrode, all enclosed in a single unit. The platinum wire is oblivious to iron or cerium. All it "senses" is the charge. Imagine a single Pt electrode dipped in an iron solution and forget about the presence of cerium. Imagine that a metal is a source (sea) of electrons.
The moment you dip the electrode, instantaneous exchange of electrons takes place between the metal and the reducible or oxidizable ions. Electrons can be released by Fe(II) to form Fe(III) and similarly electrons can be gained by Fe(III) to form Fe(II). This exchange rate is dependent on the concentration ratios of Fe(II) and Fe(III).
As a result of this exchange, the Pt metal develops a charge. It is very surprising that you only need to remove a very very very small number of electrons to develop a measurable surface charge. Hence, there is no perceptible change in the concentration of Fe ions. If platinum electrode develops a charge, the interface (solution electrode boundary) develops an opposite charge until equilibrium is achieved.
Now given the fact that Pt electrode is charged i.e., if we connect Pt to another electrode, we can sense the potential difference between the Pt electrode and another electrode, which can be any other electrode. Now that you wish to have a stable. For that purpose you need an electrode whose potential is fixed i.e. a reference electrode.


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Activity of NaCl aqueous solution at high ionic strengths: Common-ion effect
Is there any theoretical way to estimate the activity of $\ce{NaCl}$ at high ionic strengths in water and that in $\pu{7 M}$ $\ce{NaCl}$ solution? This question is based on whether  solubility of $\ce{NaCl}$ in $\pu{7 M}$ $\ce{LiCl}$ solution should decrease. Solubility of $\ce{Na+}$ in $\ce{LiCl}$ solution can be approximated to
$$
[\ce{Na+}]=\frac{(K_\mathrm{sp})_{\ce{NaCl}}}{m_{\ce{LiCl}}}
$$
We have, however, assumed that both $\ce{NaCl}$ (and $\ce{LiCl}$) are in their unit activities, which is not true. How to get conservative estimates for $a_{\ce{LiCl}}$ and $a_{\ce{NaCl}}$ at high ionic strengths, say $\pu{7 M}$?

Mathematical modeling has progressed greatly since I was a student, but I believe that such activities would rely on experimental measurements rather than theory.


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Using molecular orbital for determining leaving group trend and nucleophilicity
Here is the question.

Which one has better leaving group, cyclohexanol or cyclohexane that has $\ce{OH_{2}^{+}}$ substituent? Explain.

I know the answer is the latter because if both compounds are subjected in substitution reaction, the latter will yield a weak conjugate base ($\ce{H2O}$). There are some interesting facts from there which wants me to prove the answer with molecular orbital approach.
Here, I cite the part from it.

The leaving group itself does play a slight role here in determining the energy of the $\unicode{x3c3}^*$ orbital and thus whether it is energetically favourable for the nucleophile to attack.

Based on the cited part, looks the orbital plays the role in nucleophilicity although the iodide example gives counterexample of nucleophilicity vs leaving group.

Does HOMO orbital play role in determining nucleophilicity? Based on molecular orbital of $\ce{OH-}$ and $\ce{H2O}$, the HOMO of $\ce{H2O}$ is higher than the former, thus is it correct to interpret $\ce{H2O}$ is more nucleophilic than $\ce{OH2+}$?
If nucleophilicity criteria is solely based on energy of HOMO orbital, is there a bound where the energy of HOMO orbital becomes electrophilic (or the LUMO orbital becomes nucleophilic)?
If leaving group is a thermodynamic factor (as noted by first source), would molecular orbital alone (including frontier molecular orbital) suffice to predict the leaving group order?
Can molecular orbital describes the iodide case (good leaving group but good nucleophile)?

The leaving group trend is supposed to be for similar weak bases (halides, $\ce{OTs}$, and $\ce{OMs}$).

MO theory can certainly be used to explain substitution reactions from a mechanistic perspective, and to help rationalize trends of relative reactivity. However, I think that the best way to build a mental model that gives you practical predictive power is to also incorporate the simpler ideas that mostly reduce to Lewis acid/base theory that most introductory textbooks teach.
Remember that "nucleophile" and "leaving group" are roles that molecules take when they participate in a specific type of reaction. They are not intrinsic molecular properties. MO theory can fill in part of the picture, but other factors (such as steric hindrance in the nucleophile or electrophile), solvent selection, heat, concentration, etc. can all affect reaction rates and how well a given compound takes up the role of being a nucleophile or electrophile in a given reaction context.
So, to answer your questions:
1) The energy of the HOMO does play a role in determining nucleophilicity since, from a mechanistic perspective, a substitution reaction occurs when the HOMO of a nucleophile overlaps with (and populates electrons into) the LUMO of the electrophile. However, HOMO energy levels are difficult to predict off-hand. A better explanation for the hypothetical difference in nucleophilicity between H2O and H2O+ is that H2O+ is strongly electron-deficient, with the oxygen lacking an octet. Therefore, it is extremely unlikely that the oxygen will act as a nucleophile. This is just Lewis acid/base theory.
2) The LUMO is, by definition, unoccupied. There are no electrons in the LUMO that could act in a nucleophilic manner towards another orbital. On the flip side, if the HOMO is occupied then that orbital is "full" and there is no way that it could act as an electrophile by accepting more electrons.
3) No - see the second paragraph of my answer. The energy of the electrophile's LUMO in question -- the sigma-antibonding orbital -- is certainly relevant, but many other factors are at play: sterics, the bond length between the leaving group and the electrophilic carbon, solvent effects, etc. The most useful, generalizable approach to predicting LG ability is Lewis basicity; the weaker the base, the better the leaving group. Thus, H2O is a better leaving group than OH-, Br- is better than Cl-, etc.
4) From a Lewis/acid base standpoint, one would predict I- to be a good leaving group since it is a very weak base. MO theory predicts that I- will be a better nucleophile than the other halides because its valence electrons reside in 5p orbitals, which are much higher in energy and more close in energy to the sigma-antibonding LUMO of the electrophile.


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Why is absorption spectrum of beta-carotene different from Zeaxanthin?
Here are the molecular structures of beta-carotene and zeaxanthin

These molecules only differ in the OH groups and the conjugated backbone length is exactly the same. I read that the absorption spectrum of a molecule is decided by the conjugation length. How is it then, that the absorption spectrum of the two molecules are different as shown below:

These spectra have vibronic structure, the differences might be related to different vibrations. Consider that the lineshape is an effect of unresolved vibronic transitions, local environment, and solvent effects (OH groups have a role in this), thus the small differences could be explained like this.

Do they all in same solvent or different solvents?

The spectra are very similar in energy terms. The bands/peaks are due to vibrational transitions in the first  excited state  (as mentioned by Anon). The OH groups change the electronic energy level of the conjugated bonds very slightly, in energy terms $\approx 420$ wavenumbers in $\approx 20800$ wavenumbers. A v. small perturbation to the energy consistent with the position of the OH.

One more note: solvents effect can be merely dielectric, ie for rigid molecules, or modify substantially the energy levels via conformational effect on the solute. This is particularly true in extended conjugated system. A linear chain that coils have a reduced effective conjugation length. I want to point out that one can have very drastic solvatochromic effects and they also can be modulated by more or less little differences.

@all, thank you for giving me the keywords to do further study on this


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Constant Pressure and Temperature during Phase Change
One of the solution guides to a question I was working on said that pressure and temperature is constant for a phase change. I understand why temperature is always constant for a phase change, but don't understand why pressure is. If the heat coming in and out of the system and the volume is changing, shouldn't the pressure change too, or does the solution approximate that pressure is constant because it changes by a marginal amount?
The question is:

1.00 mol of steam is reversibly compressed to form water(l) at
100 degrees celcius ($\Delta H_\text{vap}(\text{water}) = \pu{2.259 kJ/g}$). Calculate the entropy of the phase change.

I attached the solutions on the answer key:

Constant pressure and isothermal. Only a change of state occurs. As steam condenses, heat is lost from the system.
$\begin{align}&\Delta H_{\text{condensation}} = \pu{-2.259 kJ/g}=q_{\text{rev}}\\
&\Delta S_{\text{condensation}} =q_{\text{rev}}/T = q_p/T = \Delta H_{\text{condensation}}/T_{\text{condensation}} \\ &m_{\text{steam}}=\pu{1.00 mol}\times \pu{18.0 g/mol} = \pu{18.0 g} \\ &\Delta H_{\text{condensation}} = q_p=(\pu{-2259 J/g})(\pu{18.0 g})=\pu{-40700 J}\\&\Delta S_{\text{condensation}} = q_p/T=\pu{-40700 J}/\pu{373 K}=\pu{-109 J/K}\end{align}$

Is system constrained by a closed container, or, does surrounding pressure change ? If not, pressure is constant. Note that $\Delta H$ means change of internal energy at constant pressure.

@Poutnik I think what you meant to say is that it is the change in enthalpy at constant pressure.

This has nothing to do with any process you apply to a substance.  You're looking at the wrong side of the equation.  It has to do with the thermodynamic equilibrium states of a substance.

@Chet Miller. Yes, sure. :-)

sorry the question asked to calculate the entropy of the phase change. The problem didn't say whether it was in a closed container @Poutnik

Then I guess you should edit your question.

At the normal boiling point of water, the vapor pressure of water is 1 atm. If the system contains water and nothing else (no inert gas), the pressure should be 1 atm whenever there is a phase equilibrium and the temperature does not change.  The Gibbs energy change should be zero, and it should be straightforward to determine the change in entropy (when all of the steam turns into water).

You can't compress a gas unless it is in a closed container (well, if you are considering all of earth's atmosphere and have some way of cranking up the gravitational constant or the density of earth, then you might not need a container).

@KarstenTheis I'm not sure whether the container matters, but I attached the answer provided by the answer key.

For a pure substance, the phase rule implies that at coexistence of two phases only one intensive variable (pressure or temperature) can be altered independently, the other being dependent on the choice of the independent variable. So the answer is yes and no, you can choose to alter one of the two intensive variables and still retain two-phase equilibrium, but the choice of one intensive variable determines the value of the other. This fundamental principle derives from the conditions of thermodynamic equilibrium necessary for coexistence of the two phases. The set of $(p,T)$ points on the coexistence line can be described in terms of the Clapeyron equation $dp/dT = \Delta S/\Delta V$ and a single reference point $(p_{ref},T_{ref})$ on the line.
Consider a pure substance in a diathermal container whose volume we can alter at will, and kept at a fixed temperature. Assume we begin with the container completely filled with a condensed phase (say liquid) of the substance, that is, the volume is too small (or alternately, the pressure too high) to allow vapor to form. We have however chosen a temperature that corresponds to a point on the coexistence line at a lower pressure (or greater volume) than that currently in the vessel. You can bring the substance to an equilibrium between two phases by changing the volume until the pressure in the container equals the equilibrium vapour pressure - a fixed pressure at that temperature. You can then continue to change the volume (all the while keeping the temperature constant). The proportion of the two phases (their volumes) will readjust to accommodate the new total volume, with heat exchanging with the surroundings to keep the temperature constant, and the pressure returning to the value at equilibrium - the same vapour pressure - provided the volume change is not too big. Naturally if you increase or decrease the volume too much you will end up with a single phase and a different pressure below or above the vapour pressure.
Remember: you don't have to have two phases. But if you want to observe coexistence at a chosen temperature, you have no choice wrt the pressure that will ensure coexistence. What's cool is that the proportions of the two phases can be altered within a certain limited range of total volume so that p and T remain constant. I recommend consulting a p-V-T phase diagram and inspecting it carefully to see this graphically. The following is a slice in the p-V dimensions which I borrowed from this site:

Sorry, I'm still a bit confused. Does pv=nrt still apply here or do we have to use the van der waals equation? If P and T are constant then how could v change?

The equation of state for the gas will only apply to the gas component. The total volume will be a sum of the volume of gas and volume of liquid (which, being usually poorly compressible, will remain approx constant). Maybe this answer https://chemistry.stackexchange.com/questions/119024/why-does-a-nox-container-get-cold-when-nox-is-released-and-what-is-the-relation/121321#121321 to a related post will help.

v can change because the proportions of liquid and gas change.

Does this happen for every reversible phase change?

for a pure substance, yes. The phase rule dictates that p/T are mutually dependent, so if you pick T, P is also fixed.

n(total) is constant, but n(gas) changes with V(gas), so ideal gas law is not broken. @TimothyLiu


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Strange Sticky Substance on Digital Camera
I purchased a used Canon Powershot A1100 IS camera online. When I received it in the mail it had a very strange sticky substance that looked like a thin film.
It resisted attempts at cleaning. Water, water + soap, baking soda + vinegar, baking soda only, vinegar only and rubbing alcohol all did not resolve the issue. Only vigorous scrubbing kind of helped.
In addition I could not get it off my hands at all even after about 10 hand washing attempts and soaking my hands in soap and water. After my hands dried the areas where my hands touched the camera were still sticky. The only thing that helps was after it dried rubbing my hands with a paper towel. When wet it was not sticky then when it dried it came back.
The battery terminal was completely corroded with a blue color and it seems like there was some but possibly less of the substance internally.

This is part of the battery door. It is used to complete the circuit between the two AA batteries that the camera takes.
Scraping away the blue crystal only results in more of what you can already see in the picture. Metal that looks grey in some places and brown/red in others.
The closest thing I can think of is petroleum jelly but I can't be sure that, that is it. I don't have any and don't work with it much so I can't compare what it looks like.
Based on my description do you have any possible ideas on what it is?
If you do have something that matches these symptoms would you happen to know what its toxicity is and proper cleanup procedure?
Also in general how should one handle touching something they don't have knowledge of but that could be dangerous and when hand-washing is not working/removing it.
Thank you in advance for your questions and answers.

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

After the lengthy back and forth perhaps an official answer is due: the blue salt deposit on the camera battery holder door is most likely copper(II) hydroxide by reaction of leaked alkaline battery electrolyte (likely $\ce{KOH}$) with copper metal in the battery door contact. It could possibly also consist of copper(II) carbonate hydroxide formed by reaction of the copper(II) hydroxide with carbon dioxide, but the deep blue rather than green color in the picture suggests it is the previous. Copper hydroxide is regarded as moderately hazardous, but you are unlikely to be at risk unless you ingest it. Feel free to consult a MSDS for this compound (such as here) for more information.
The sticky substance coating the camera is possibly a degradation product of plastic components in the camera housing. It is not likely to be toxic unless ingested or if you are in contact with it persistently.
Without more information about how the camera was maintained it is difficult to say much more.

I agree with your final conclusion. This logically explains everything I noticed. Also the plastic explanation makes more than rubber which I thought it was previously.

I agree with @BuckThorn on the blue corrosion but I'd like to address the sticky substance. If it is in fact a degradation product of the plastic or some other substance being exuded by the plastic, I've had good luck removing these with oil-based cleaning agents. None of the cleaning products you mentioned will help very much in removing a nonpolar organic substance.
Try rubbing a small area of stickiness with a cloth moistened in WD-40, Goof Off, Goo Gone or any similar oil based cleaning product. These should work on sticky fingers too. If one of these removes the stickiness on the camera without damaging the finish, you can carefully clean the rest of the camera.

Mineral spirits (conventional paint thinner) might be a better choice than the other products mentioned, which are more likely to damage plastic.


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Bohr's postulate of quantisation of angular momentum
The Bohr's postulate of quantisation of angular momentum can be written in a formula as
$mvr = nh/(2π)$
where $m$ is mass of electron
$v$ is its velocity
$r$ is the radius of that shell
$h$ is Plank's constant
So, I was wondering that if, I want to find where approximately the thing with $n=2$ exists, or in other words, what is the velocity of electron and radius of its circular path in the second shell, I could rewrite it as
$vr = nh/(2mπ)$
or
$vr = constant$
So if I was to just increase the radius by a factor of say $z$ and velocity by a factor of $1/z$ it would give me the exact same value for $vr$ every time. Doesn't that mean I could indefinitely increase the radius and keep reducing the velocity and still stay in second shell forever?

There are (were) additional constraints on the velocity because of classical mechanics: https://en.wikipedia.org/wiki/Bohr_model#Electron_energy_levels Anyway, I'd really discourage reading too much into the Bohr model. It is severely outdated; perhaps useful to know because it provides some historical perspective for the development of quantum mechanics, but not useful to analyse in a serious fashion beyond the fact that it didn't work.

If the parameter $n$ is chosen, the velocity $v$ and the radius $r$ are defined and cannot be changed

What you're missing is that the electron (or rather, Bohr's version thereof) moves in a curved path because of the centripetal force provided by electrostatic attraction to the proton.  You therefore have
$mvr=nh/2\pi$, quantized angular momentum
and
$mv^2/r=\epsilon_0e^2/r$ centripetal force from electrostatic attraction
For each quantum number $n$ we have just one shell radius $r$ and one speed $v$ meeting both requirements.


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Is there a heatformed vinyl-like plastic that is hard as shellac when cold?
I recently started collecting shellac records, and got the idea of releasing my own music on 78rpm shellacs.
But the exact formulas and techniques for making shellac records seems pretty much lost to history, and it seems it was a very hard material to use anyway. You can use vinyl, but then it would not be playable on the old hand-cranked acoustic type 78rpm record players, as those needles have too high pressure and would destroy the softer vinyl.
So this is probably only doable if there is a plastic that is soft and possible to press to a record shape when heated, like vinyl, but that becomes a lot harder than vinyl, like shellac, when cold, and also doesn't distort when cooling. Anyone knows of something like that?

How about Bakelite? It is not thermoplastic, but its wikipedia article indicates it has some possibly useful attributes, including hardness. Not my area and I have no clue how hard a plastic would need to be to survive an old hand-crank acoustic record player. Anyway, best of success with your project!

Thanks! I think bakelite needs to be baked under pressure, or it will bubble, so that probably doesn't work. But a material that is soft under room temperature and gets hard with baking actually might work as well, I didn't consider that. The question is how to get it to an oven without destroying the grooves in that case.

@DrMoishePippik, I suggest you read up on how records are made, and then read the question again. You don't CUT into the record, it's pressed. Direct cuts exist, usually called "acetates" or "dubplates", but they are SOFTER than vinyl, I need something HARDER than vinyl.

@DrMoishePippik Yes, and this question is about the PRESSING part, not the CUTTING part. And if it's not lost to history, then please provide me with the exact composition and procedure to make the material used in a Shellac record. Because nobody else seems to know.
Again, I'm not talking about Vinyl records, which continue to be made today, I'm talking about shellac.
In general, you would do better if you stop assuming I don't know what I'm talking about.

Would working with shellac, assuming you could recover the required instructions, be a satisfactory solution to your problem?

@BuckThorn Probably, rumour has it that it was very hard to make and/or work with, but since I don't have instructions I don't actually know. It would certainly serve as a satisfactory answer to the question, even if it turns out to not be a practical solution for some reason.

Further diving into available info and videos show that a "steaming table" is enough to heat the shellac into the right temperature, so it's probably not hard to handle. So the problem is still the manufacturing, which requires the ingredients (which I don't know exactly what they are) to be ground to extremely fine powders, and then mixed in a special heating mixer, and the resulting paste is hot rolled. This will be hard to replicate. https://www.youtube.com/watch?v=2zL53iEHf_0

I found a web article, "The Origin & Many Uses of Shellac" by R.J. Wakeman, to be particularly illuminating and well-written source of information on the composition and manufacture of shellac records, for instance it explains:

Emile Berliner’s first disc records were pressed in celluloid (1894) followed by the use of vulcanized rubber (1895).(36) In 1896 Duranoid shellac records were developed and became the standard medium.(11) The formula used for making shellac records varied from firm to firm and from decade to decade; it was usually a company secret. The shellac component represented approximately five-eighths of the material used for the better quality records.(33) Essential components would include shellac, ground rock, carbon black, and cotton floc.

The article lists a few patents among the references that you could quickly search for additional preparation methods, although I found them to be somewhat evasive with regard to details, preferring to amplify the scope of the application by making general claims. The number of patents describing the manufacture of records with shellac and competing resins with a variety of additives is considerable. There is therefore no universal recipe for which to search. What common properties do the formulations share? The key is that the shellac serves as a binder, and that strength and hardness, to protect against wear of the record by the steel needle, to actively wear the needle into shape, or to prevent shattering, is largely conferred by the ground mineral filler and other additives such as abrasives and plasticizers. Apparently the fineness of the ground rock and the quality of the shellac are important.

Abrasives were frequently added to the record matrix to help wear the point of the needle to conform to the shape of the record grooves

and

When a needle enters the groove of a 78 rpm shellac record the needle point should quickly abrade to fit the record groove.

Regarding the preparation of shellac records, Wakeman writes:

To mold shellac records, powdered shellac, ground stone, carbon black, and cotton floc were measured by a formula and mixed in a revolving drum or Banbury mixer.(16) The shellac needed to be finely ground for thorough mixing with and coverage of the other components. High quality orange shellac with low wax content and few impurities was the standard.(16) One report states the shellac was ground to 80 mesh, the fillers to 200 mesh, and the coloring matter to less than 0.4 micron in size. The mixture was steam heated, causing the shellac to melt and form a thick dough which was then passed onto heated rollers and formed into a long roll. The roll was cut into sections (called “biscuits” in some factories) with adequate material to form a 10 or 12-inch record.(8,31)

So in answer to the question, "what alternatives exist to shellac with similar properties?" you would like to replace shellac with other resins, while retaining the other essential additives, such as abrasives. You would not be the first to attempt to do so. As Wakeman explains:

After the war shellac was combined with other resins to form softer and less brittle 78 rpm records.(16) Shellac was reduced in the matrix to form 15 to 30% of the mix.(16) Vulcanite (also called ebonite), a hard rubber, was sometimes included in the mix, as were Vinsol (a pine tree resin) and Manila gum (16).

Another important property is time to harden. I looked up Durium, a relatively simple shellac competitor. Durium was patented as a fast-hardening resorcinol-formaldehyde resin. Production of Durium records was faster since, unlike shellac, Durium hardened quickly and records were fabricated by pressing rather than moulding. While it has properties that represent a significant advantage over shellac (price, hardening speed, method of impression), unfortunately it also requires working with a variety of unpleasant chemicals.
For an early vinyl patent, developed during shellac shortages in WWII, see US patent 2335986. For a more recent patent providing a number of formulations alternative to shellac, see US patent 2787602.
Regarding characterization of shellac composites, this is a difficult topic to address, since the properties can be expected to depend strongly on the specific additives. I did find an old reference quoting a value for the modulus of elasticity of processed shellac as on the order of $\pu{10 \times 10^3 kg/cm2}$. You have many choices regarding a resin to match the required strength and softening temperature.  You may want to contact a company such as Ensinger plastics for guidance on selection of a thermoplastic composite.

If we get closer to actually doing this, talking to a plastics company is probably the way to go. The patent was useful, it also mentioned some harder types of plastic, like urea-formaldehyde.

@LennartRegebro Perhaps the oddest thing (imho) about the shellac disks is the addition of solids and particularly abrasives to wear down the needle, the "battle" between disk and needle. Whether this is still relevant depends on whether the playback mechanism has changed or whether the phonograph you use operates in a different way, that is, does the needle tip still need to be worn down?

Old style phonographs still have soft needles that you replace often, yes. It is indeed a very odd thing, reasonably the worn metal must end up in the grooves, which means they stop wearing down new needles, so... Uhm. Yeah. I have no idea. Maybe this is why most 78 records available today are from the 50's when most people had switched to electrical record players.
So yeah, I'm hoping it will work without abrasives. But only testing will tell.

I don't know about other plastics, nor do I know about the scarcity of shellac records, but I do know shellac is a thermoplastic. Could you take an old record, melt it, and use a die to make your own record from that? This way, you could avoid having to deal with the recipe.
By the way, the following was found on Wikipedia:
"Formulas for the mixture varied by manufacturer over time, but it was typically about one-third shellac and two-thirds mineral filler (finely pulverized slate or limestone), with cotton fibers to add tensile strength, carbon black for color (without which it tended to be an unattractive "dirty" gray or brown color), and a very small amount of a lubricant to facilitate release from the manufacturing press."
All of these ingredients, including shellac, are easy to find and the recipe is quite simple. I can't speak from the production process, but I'm sure you could figure this out.
Source: https://en.wikipedia.org/wiki/Phonograph_record#78_rpm_materials

Repurposing old broken shellac records might be doable for extremely small runs like 10 records or so, which to be honest is probably more than my music would sell. :-)


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Can classification techniques used for identifying solids be used to identify precipitates?
I am preparing for an assessed practical experiment where an unknown precipitate must be identified. I know that precipitates are insoluble, and my textbook says they are considered to be solid with reference to the aqueous solution.
I want to test for carbonate, halide, sulfate, and nitrate anions in the precipitate and understand that if it were a normal solid I would be able to test for carbonates by using a nitric acid solution, and test for halides by dissolving in nitric acid and testing with silver nitrate, and test for sulfates by dissolving the unknown solid in nitric acid and adding barium sulfate solution. I also know that I would normally be able to test for nitrates by heating the unknown solid with copper and then treating with H2SO4.
Are these classification techniques applicable to precipates as well?

You should first get some practical experience with these tests before you apply them on unknown samples.

"solid with reference to the aqueous solution"? What's that supposed to mean? Solids are solids that's it! Fresh precipitates tend to be easier to dissolve then compounds from shop, though. So slightly less aggressive treatment may be preferred.


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Bromination of an alkene in presence of water
Bromination of an alkene in presence of water results in its mechanism in a cyclic bromonium ion and "afterwards $\ce{H2O}$ binds to the carbon atom surrounded by more electron donating groups."
This is what we learned in class and I couldn't understand why. Shouldn't an electron donating group decrease the "strength" of an electrophile?
Also to my surprise, in an exercise, $\ce{H2O}$ was added to the formerly $sp2$ carbon bonded to $\ce{-OCH3}$ which donates electrons by resonace in an alkene.
But $\ce{H2O}$ was added to the cyclic bromonium and not to the alkene directly! So why to care about  the resonance effect of $\ce{-OCH3}$?

Electron donation stabilises the carbocation so that it hangs around long enough to react with a nucleophile

But carbocations are not formed in halogenation mechanism of alkenes

they are formed as an equilibrium form of the cyclic bromonium ion, particularly if stabilised

While there is no formal carbocation intermediate, the cyclic bromonium intermediate has a few different resonance forms (reference):

As you can see on the right, in the structure representing the resonance hybrid, partial positive charges build up on the electrophilic carbons connected to the bromine. The carbon that is more substituted with electron donating groups has a greater partial positive charge (and therefore more electrophilic character) because of the stabilizing effect of the substituents. That's why the nucleophile attacks at the more substituted carbon.


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Breast milk with reverse osmosis
I would like to make life lasting breast milk jewelry. I would need at least 5ml to make it. I've been doing a lot of research to get the best results, so far this questions are the result of more than a year trying to get the right answer. Would reverse osmosis remove each bacteria from it, allowing me to work with it in resin without using harsh chemicals and not molding inside the resin? How much ml of breast milk would I need to have at least 5ml after the process is done? I would need to make it into a powder after the process, which I think it could be achieved either with silica gel, or dry by the sun. What would you suggest? Thanks!

If you want to remove all the moisture content from a small sample on the order of a few mL I suggest using a lyophilizer. It should leave the solutes found in milk largely chemically unaltered. The high sugar and low water content should also serve to preserve the dry product (similarly to powdered milk). The dry product can be resuspended as needed by vortexing in an appropriate volume of distilled water. Typical solids concentration in human breast milk is ~10%.
An alternative is to isolate the milk curds (the protein fraction) by acidifying the milk, by adding for instance vinegar. You can then make casein plastic from the curds (there are many suggestions on how to do this online, and this was one source of plastics before synthetics became the norm). The acid treatment should serve as a germicidal step. You should be aware that protein content in human breast milk is approximately 3 times lower than in bovine milk (~1% versus 3%, respectively). This means that you will need roughly 3 times the volume of human milk, assuming the casein fraction is the same in both, to make the same amount of casein plastic.
Depending on whether you intend to use all of the solids or only the protein fraction of the milk, you will need (ignoring differences in density) different volumes of milk to obtain 5 mL of solids: about 50 mL of milk to make 5 mL of complete solids, and ten times more, ~500 mL of milk, to make 5 mL of protein extract. I recommend testing with bovine milk first and then using the relative solid or protein concentrations described in the references to determine the required volume of human milk.

There are tutorials about home-made galalith, like this one (based on animal milk) https://www.wikihow.com/Make-"Plastic"-out-of-Milk Eventually this would require some downscaling, of course.


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Is it possible to make micelles to pick up Microplastics from water?
I'd like to know if I could get micelles to pick up microplastics in water. I read that microplastics are sometimes charged, so can we make the lyophobic end to pick up microplastics, so that they can then be collected on the water surface and be easily disposed?
This is the idea I'm planning to pitch for a science fair, is this actually feasible?
I read that detergent micelles work well even in hard water, so can we use such detergent micelles to pick up microplastics?
Thanks!

You may want to search information on water treatment in general (for instance flocculation). While some of what you describe sounds reasonable, the mechanism you describe is not entirely clear.

Detergents are very unhealthy for practically all water organisms.

Lyophobic means solvent-hating, and here the solvent is water, so the lyophobic ends (presumably the tails, if we are discussing a normal detergent) will associate to form the hydrophobic core of the micelles. The headgroups meanwhile will preferentially interact with water, forming an interface between solvent and core.
If you add detergent to a solution containing charged plastic particles, the detergent tail is unlikely to interact with the particle unless the particle also displays amphiphilic character. More importantly, headgroups with the same charge as the particle surface will discourage association with the particle. So your proposal lacks a viable aggregation mechanism for at least one reason (charged and hydrophobic regions are not likely to associate) and possibly for a second (oppositely charged regions of detergent and particle will repel). So you need to research carefully:

the expected charge of the particles
the choice of detergent, ideally with a headgroup with charge opposite that of the particle, and which resists precipitation due to the hard water but will encourage flocculation of the particles
scalability, which is a question regarding the volume of water you intend to treat

There is always the possibility that the solution contains ions resulting in charge inversion of the charged particle, another factor to consider.
You'd be well served to look further into the steps used to treat wastewater.


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How to determine if a compound is a strong, weak, or non-electrolyte?
I'm stuck on questions asking whether a given chemical formula is a strong or weak electrolyte in an aqueous solution. I did some research online, but could not get a specific solution that I could readily comprehend.
My conception right now is that ionic compounds are strong electrolytes? But I have no idea when it comes to organic compounds or covalent bonds.
Can somebody explain?

It's best to start with basic definitions: electrolytes are compounds that allow a solvent to conduct an electric current. The mechanism is generally formation of ionic species, that is, solvated atoms or molecules carrying a net charge. As charge carriers, ions are the most important contributors to the electrical conductance of solutions.
Salts represent a reliable source of ions in solution. They tend to dissociate to some extent into their component ions in solution. It is usually ok to assume that in the presence of solvent the ions in a salt can exist either as part of the solid lattice or as free ions in solution. Only the free ions contribute significantly to conductance. Therefore solubility and conductance are directly related. A salt that dissociates to a large extent is referred to as a strong electrolyte. One that does not dissociate much is called a weak electrolyte. Generally speaking the degree of dissociation depends on the lattice free energy and the free energy of solvation of the salt in the particular solvent under given conditions of temperature and pressure. Degrees of dissociation for different salts or conditions are generally tabulated as solubilities and may be described qualitatively or quantitatively (say as solubility product $K_{sp}$). Qualitative trends are usually memorized by chemists and are taught at an early level to students. While detailed explanations can usually be found for differences in solubility, they are not usually the subject of introductory courses. Memorization (and mnemonics) tend to be the rule.
Some compounds that would not generally be regarded as salts (although are sometimes prepared in salt form) contain ionizable groups that can dissociate to form oppositely charged ions. Good examples are acidic groups, such as carboxylic groups that generate an conjugate anionic (negatively charged) molecule and hydronium (in water) upon dissociation, and basic groups, such as amines, that can accept a proton from a donor compound (such as water) to form a cation (positively charged ion). Generally speaking molecular organic compounds are regarded as weak electrolytes, since the degree of dissociation of the ionizable groups may be incomplete to a large extent. This is not universally true. Some groups readily dissociate, for instance sulfates. The degree of ionization depends inasmuch on the nature of the group in question as on the structure of the compound and the solvent conditions, including in particular identity of the solvent and temperature.
Compounds that are insoluble can obviously not contribute to the conductance and are therefore not electrolytes. Some compounds readily solubilize in the solvent (for instance nonionic polar compounds in water) but do not dissociate into ionic forms. Such compounds are not electrolytes.


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Will distillation do the job of purifying the water?
I asked about my sophisticated water purification system for a generation ship on Worldbuilding Stack Exchange, and somebody mentioned this:

If you're gonna go to all the trouble of distilling it, why not just
  have that be your only step?

And it is a good question. But, from my chemistry knowledge, this just won't work. Here are the problems I see with distillation being the only step:

Proteins in blood and urine are going to froth up from the heat. If these get into the water, the water is going to smell like urine and nobody is going to want to drink it.

This is the easiest issue to fix. Just make the apparatus tall enough and no urine froth will make it to the distallate. That or adding a trap to react with the urine froth and solidify it in between the 2 ends of the apparatus would work.

Any hydrochloric acid that is present from vomit is still going to be there because it changes the boiling point of water. You simply can't get rid of the acid via distillation alone, not even repeated distillation. You need either microbes that can break down hydrochloric acid or an acid base reaction to get rid of the acid.

Hydrochloric acid forms an azeotrope with water. In fact it forms several depending on the acid concentration. So at some point, strongly acidic water will come over. Worst part is that a mixture of hydrochloric acid and water is colorless and odorless, so the only way you could tell is by density. Start to see those lines that occur when you mix liquids of different densities and you know strongly acidic water is coming over. Also, checking for pH would help but that would mean either an acid resistant pH sensor in the container with the distillate or stopping the distillation periodically and using pH strips. Checking for a change in density is much easier.

If acid vapors escape the system, as I think some will, then everybody on the ship will get irritated lungs, and some might get "Acid Pneumonia" and die from both the irritation due to the acid and the infection that the hot acid predisposes them to. There would be some infectious agents on the generation ship to keep their immune system in their prime and to help prevent autoimmune disorders from spreading like wildfire. But if this is combined with acid vapors, it is basically a death sentence for whoever gets infected. This is not even considering the fact that acid in your eyes can make you go blind

No distillation system can 100% prevent vapors from escaping. Not even a vacuum distillation with a trap will do it.
So in my opinion, distillation is necessary but can't be the only step, otherwise WCS, you get water that smells like urine that nobody wants to drink and hot acid vapors spread all over the ship causing lung irritation and blindness and everybody dies from a combination of the acid vapors and dehydration related both to heat and infection very quickly and you just have an empty hull in space. BCS, everybody gets heartburn from drinking the acidic water.
So, is there a way to get rid of the hydrochloric acid when distilling the water? Or would the people distilling the water have to resort to an acid base reaction followed by a second distillation to get rid of the acid and any excess base?

Neutralise, then distill.

and use activated carbon to remove some of the smelly stuff....

The old school method of generating ultra-pure water by distillation was to add a small amount of potassium permanganate in a slightly alkaline solution. All the organics were oxidized by the permanganate ion in boiling water. A small of $\ce{MnO2}$ so formed during boiling would take care of the acid(s). This water was good enough for electrochemical experiments. HCl only forms an azeotrope at 20.2% composition and it boils around 110 Celsius.
Today ultrapure water is produced by using quite expensive deionizers with UV lamps. UV destroys all the organics, and then an activated carbon filter absorbs them. Water produced from such deionizer has a specific resistance of $18.2  \rm{M\Omega.cm}$.

18 MOhm is not a conductivity.

The edit is incorrect as well. It should be 18 MOhm.cm not per m.

Still wrong ;). That value is a resistivity, not conductivity. And it is 18 MOhmmeters, not centimeters. The respective conductivity is 5.5 µS/m, as given in various online sources. https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity#Resistivity_and_conductivity_of_various_materials

Third iteration, right it is specific resistance or resistivity but it is still not 18.2 MOhm meters. It is 18.2 MOhms.cm. Centimeters is a natural unit in electrochemistry because of the conductivity cell dimensions. Wikipedia is still not that trustworthy source of numbers

Ah, I messed up the reciprocal, as does wikipedia. 5.5µS/m are 180kOhmmeter, or 18000kOhmcentimeters. My old and trusted inorg chemistry textbook gives 4 µS/m, so were in the right region. I strangely didnt find a value in the CRC handbook.

We should correct the Wikipedia entry because it must be misguiding so many people.

Definitely, but I hate to correct things in wp without a proper reference.

Here is the fundamental paper: The Fundamental Conductivity and Resistivity of Water, http://esl.ecsdl.org/content/8/1/E16.short


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Thermal decomposition of K2CO3
$\ce{K2CO3}$ doesn't boil. It thermally decomposes at $\pu{1200 ^\circ C}$. However, I find conflicting information online about this. Roughly what I have found say three inconsistent things:

It doesn't decompose
At $\pu{1200 ^\circ C}$, it decomposes into $\ce{K + O + CO}$
(Chemiday)
At $\pu{1200 ^\circ C}$ it decomposes into $\ce{K2O + CO2}$ (allreaction)

What actually happens when $\ce{K2CO3}$ thermally decomposes? Most sources agree with (3). But how can that be? Wouldn't it just decompose some more? $\ce{K2O}$ decomposes at $\pu{300 ^\circ C}$ into ($\ce{K2O2 + K}$) and $\ce{K2O2}$ decomposes back into ($\ce{K2O + O2}$) at $\pu{500 ^\circ C}$.
What is going on here? Is $\ce{K2O2}$ really decomposing back again to $\ce{K2O}$? If so, then at $\pu{500 ^\circ C}$ do both of these decompositions occur, cyclically pumping out $\ce{O2}$ and $\ce{K}$?
I read elsewhere that at temperatures over $\pu{500 ^\circ C}$, $\ce{K2O2}$ decomposes into ($\ce{2K + O2}$). This would resolve the conflict. But I find the source is unreliable. What is right?

There's only mention "it starts decomposing" around 300 when heated, but that's probably just that it's not favourable in low temperatures, not high.

I suppose it's possible. But I have doubts that it becomes stable again at higher temperatures. There's no mention of its boiling point anywhere. Do any other compounds begin thermally decomposing, then reach a plateau of stability at higher temperatures?

This question serves as a prime example of how much fake and scientifically incorrect information, or more famously, fake news is there on the web. This is why ordinary web should not be anyone's first priority to search about scientific topics. There is a famous saying that "paper never refused an ink" which applies to the web as well in the sense that anyone can write anything. Use Google Scholar, which only searches research level articles (relatively more authentic).
With this type of search one can arrive at Thermal stability of potassium carbonate near its melting point. This paper rules out the possibilities of (1) and (2). The authors note that there is significant vapor pressure of carbon dioxide near the melting point, and suggest $\ce{K2O}$ formation. They also advise the readers to consult the following as the most comprehensive study on alkali carbonates. I will let you try Google Scholar for this one.
References

A. Reisman, Reactions of the Group VB pentoxides with alkali oxides and carbonates. IX. A DTA study of alkali metal carbonates, J. Am. Chem. Soc. 80 (1958) 3558-3561.


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What reactions involving hydrazine, dinitrogen tetroxide and possibly a contaminant can produce a salt?
Hydrazine $\ce{N2H4}$ and dinitrogen tetroxide $\ce{N2O4}$ form a hypergolic fuel and oxidizer pair widely used in spacecraft including Akatsuki (あかつき, 暁, "Dawn") also known as the Venus Climate Orbiter (VCO) and Planet-C.
Wikipedia states:

Hydrazine can be monoprotonated to form various solid salts of the hydrazinium cation ($\ce{N2H5+}$) by treatment with mineral acids. A common salt is hydrazinium sulfate, $\ce{[N2H5]HSO4}$, also called hydrazine sulfate. Hydrazine sulfate was investigated as a treatment of cancer-induced cachexia, but proved ineffective.
Double protonation gives the hydrazinium dication ($\ce{H3NNH3^2+}$), of which various salts are known

A comment below the Space SE question How did salt prevent Akatsuki's insertion burn? says:

Every article I see seems to parrot the same thing. I'm guessing this is yet another translation issue (Japanese to English) and the deposits are a hydrazine reaction product... which you could call a "salt". It's an interesting failure case that I hadn't heard of, thanks for the question.

Question: Are there possible reaction products from hydrazine and dinitrogen tetroxide alone that could form an ionic assembly of cations and anions and considered a salt, or would this require some additional contaminant, perhaps water or some organic material to be involved?

How about with the reaction chamber and nozzle? Of what are they made?

@DrMoishePippik I've annotated the image from this answer here: https://i.stack.imgur.com/wMImf.png I've mentioned "water, or some organic material contamination" in the question because these are common contaminations that can come from a wide variety of sources, and I think you are right that can include the reaction chamber and nozzle. Water could be (and probably is) a reaction product and could be plentiful, but my knowledge of chemistry is low and I don't know if that's enough to make something that might be considered a salt.

@DrMoishePippik The valve could potentially be pretty far away from them in terms of tubing, but I don't know much about rocket engines or what keeps the high pressure from the combustion chamber from pushing reaction products upstream in this case.

BTW, you might enjoy reading *Ignition! An  Informal  History  of Liquid  Rocket  Propellants," by John  D.  Clark, https://library.sciencemadness.org/library/books/ignition.pdf


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What will be produced when sucrose and potassium bitartrate react with each other?
What would happen if sugar and cream of tartar react with each other? Would it create an ester?
$$\ce{KC4H5O6 + C12H22O11 -> KC16H21O14 + 3H2O}$$
Is this correct? If not, why?

I don't know about potassium bitartrate but tartaric acid splits sucrose into its components(glucose + fructose). It acts as catalyst. See: http://madsci.org/posts/archives/2008-12/1228754105.Ch.r.html

Rather than "cream of tarter" I'm going to stick to the pure chemical potassium bitartrate and tartaric acid, and I'm going to speak from the point of view of a chemist about the difference between potassium bitartrate and tartaric acid (rather than a cook or winemaker).
First a simple answer to your questions and then the details. Simplistically speaking, when you add potassium bitartrate to an aqueous solution of sucrose you will get some hydrolysis (producing glucose and fructose) and some ester formation. The key word here, however, is some -- and the amount depends strongly on the reaction conditions. At room temperature, for example, you're likely to get essentially no reaction within, say a day. More details below.
Tartaric acid is a dicarboxylic acid (a weak acid). Each of its carboxyl groups (-COOH) can donate a proton (H+ ion), and when this happens it's "acting as an acid". This is important to your question because it is the H+ ion that acts as a catalyst to sucrose hydrolysis and not the tartaric acid itself. For this situation, chemists call it a specific acid catalyzed reaction. By contrast, if a specific molecule such as tartaric acid were involved in the catalytic step chemists would call that "general acid catalysis".
Acidity of molecules are gauged by assigning a pKa value. If there are 2 acidic groups in the molecule (as with Tartaric acid) then it has 2 pKa values. Adding KOH to a solution of Tartaric acid will give you potassium bitartrate. At 25C, the pKa for removing Tartaric acid's first H+ is about 3.0 and the pKa for removing potassium bitartrate's H+ is 4.3. This makes both of them what chemist's call "weak acids" because they don't release anywhere near as much H+ into a solution as a strong acid (like HCl).
The other reaction, ester formation, is also specific acid catalyzed and so would also occur as I mentioned, but to a small extent. Esters are formed in acidic solutions whenever carboxyl (-COOH) and alcohol (-OH) groups are present, as is the case here. Note that there are many -OH groups here, on both molecules.
To understand the extent of each of these reactions it's important to consider the role of water. Water is a reactant in the hydrolysis reaction and this tends to drive the reaction forward. By contrast, water is a product of the ester formation reaction. This is especially important for the ester reaction since it is highly reversible. In other words, whenever molecules collide to form an ester, it's much more likely that that ester will collide with the far more abundant water molecules and revert back to the starting materials, resulting in no net reaction.
Usually one thinks of these reactions happening somewhere near room temperature. Under these conditions neither of the above reactions happens to a great extent; they are both generally very slow. In order to push them forward you have to do things like increasing the temperature, adding more concentrated reaction materials or catalysts, or removing reaction products. Of course, in this case increasing the concentration of the reaction materials also increase the amount of catalyst (H+), but since they are such weak acids, adding HCl (or another strong acid) would be much more effective.


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styrene confusion
I am studying NMR and while looking at styrene, I can't figure out why the two hydrogens on the terminal double bond are not equivalent. Is it because one is closer to the ring in position than the other and there is not free axis motion? If so, why are the hydrogens in the ring directly left and right of the branch still equivalent?

Did you realize that one terminal hydrogen on the double bond is cis and the other  terminal hydrogen is trans  to the third hydrogen on the double bond? That means those two terminal hydrogens are not chemically and magnetically equivalent.


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Making Lyophilized Cake Lookalike using household ingredients
I'm working on a machine learning model to identify flaws in vaccines in lyophilized cake form. To train the model, I need a number of samples that look something like this:

I have vials, but I'm having trouble making a suitable cake – I need something that will stick to itself when dried...
What I've tried so far:

Salt dissolved in water/isopropyl alcohol
Baking soda dissolved in water/isopropyl alcohol

Both of these turned back into powder (instead of caking) when dry.
Next, I'm considering using powdered detergent, adding water, then letting it dry...
How would you recommend making this using common household ingredients?

Starch? Like corn starch, potato starch... Although it's not going to dissolve, so if you need to start from a solution it might be tricky.

Meringue? https://www.finecooking.com/article/how-fragile-egg-white-foams-are-transformed-into-firm-airy-meringues

Do you intend to develop this further into a commercial application or is this more proof of concept work? I have some experience in vaccine manufacturing and automated visual inspection and I'd be interested in knowing your path forward.

You may want to consider whey. Looks like Karen Smith, Dairy Processing Technologist at the Wisconsin Center for Dairy Research, already did some of the work for you. The result depends on the specific type of whey (a high score of 4 or 5 on the caking test means the material cakes readily, forming a gummy crust):

Whey  – (Scored 2-5) –  whey exhibited a wide range of caking scores. How the whey is processed has a very large effect on the tendency of the resulting powder to cake as evident in this result. Clearly, two of the samples had large amounts of amorphous lactose and without the presence of significant amounts of protein the samples readily caked.

Conclusion: choose a whey with less protein and more amorphous sugar (such as lactose). In general you are looking for a sticky substance. The amorphous sugar is sticky (it does not form a crystalline powder) and glues the material together. Bodybuilding whey may contain too much protein. Alternately milk powder (which contains plenty of lactose) may do the trick. I'm going to the market now to get supplies and run a test.
Update: I have run a home kitchen caking test on dried skim milk. The dry solids consist of 56% sugar, 40% protein, and ~1 g fat. The milk contains 11% water.
The test was similar to that described in the linked document and consists of placing ~20 g of milk powder in a small (~3 inch diameter) aluminum baking dish and allowing it to float on hot water (~$\pu{50^\circ C}$) in a closed container for about an hour.
The caking test is of course the opposite in a sense of lyophilization, for the solid is exposed to a humid atmosphere to encourage coalescence. The following image illustrates the result:

This confirms the observation in the linked document that NFDM (lower left corner of the image array) "cakes upon storage due to presence of large amounts of amorphous lactose. The protein present limits the hardness of the cake of powder."
This verifies that milk powder cakes. However I do not have a lyophilizer at home so unfortunately cannot verify that the dried milk will remain caked once freeze dried. I might yet put it in the oven to see how this affects the structure.

@Buck did you try the experiment? did it work?

@WilliamR.Ebenezer I haven't had a chance despite my comment. :-\  Maybe tomorrow....


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Does CO2 form carbonic acid in the atmosphere?
Since when $\ce{CO2}$ goes into water it makes carbonic acid, then what happens when it rains? Some of the $\ce{CO2}$ in the air joins the rain drop as it falls and while the rain drop falls and changes shape, it releases then regains $\ce{CO2}$? Is there a critical point or point of saturation within the raindrop for $\ce{CO2}$? It’s an interesting dynamic considering the complexity of storms and weather systems.

Partly related: Why is it considered acid rain with pH < 5.6?

Raindrops are just ordinary water.

Short answer
It does form carbonic acid, as rain drops are water and carbon dioxide is soluble in water.
But the content of carbonic acid is minimal as you can guess from $\mathrm{pH=5.6}$
Longer answer
Majority of dissolved $\ce{CO2}$ does not form carbonic acid, 99.83% stays as hydrated oxide.
The equilibrium $\ce{CO2}$ concentration in water follows the Henry's law and is proportional to the $\ce{CO2(g)}$ partial pressure.
Therefore, as a drop falls, the partial pressure of $\ce{CO2(g)}$ grows together with the atmospheric pressure and  $\ce{CO2}$ solubility in water would increase as well, if temperature did not raise with decreasing altitude, typically $\mathrm{ 0.0065 K/m}$.
The total effect is general decreasing $\ce{CO2}$ solubility when a drop is falling.
Sure, some drops may evaporate during their fall ( seen as precipitation patterns without raining ) and the dissolved $\ce{CO2}$ becomes gas again.
Natural rain has $\mathrm{pH=5.6}$  because of dissolved $\ce{CO2}$ and equilibrium concentration of $\ce{H2CO3}$.
If all  CO2 converted to H2CO3, the natural rain and soda water would be much more acidic, because carbonic acid is quite strong acid with pKa=3.6, instead of formal 6.3 if 100% conversion is supposed.
In carst areas, there is ongoing reversible reaction of dissolving calcium carbonate by surface rain water going down the soil and precipitating it again in caves :
$$\ce{CaCO3(s) + H2O + CO2(aq) <=> Ca(HCO3)2}(aq)$$
The calcium bicarbonate comes also to surface water and makes frequently the majority of the calcium content.( and similarly for magnesium l

Just saw on "A strange rock" that rain is one important mechanism for CO2 regulation. That wasn't very clear but I think that is how CO2 enters the soil besides long time sequestration. This is just to say that in spite of the little 0.17, it can be an important path. Plus one.

Surely it is, I have not say it is not. If all  CO2 converted to H2CO3, the natural rain and soda water would be much more acidic, because carbonic acid is quite strong acid with pKa=3.6, instead of formal 6.3 if 100% conversion is supposed.

I didn't say all should dissolved, neither. Just point out that it seems to be an important mechanism in addition to those related to carbon burial on longer time scale.

Sure. I have updated the answer.

I think that by "and so does CO2 solubility in water" you mean "and so does the amount of CO2 solubilized in water." The solubility is constant at given T right?

@Buck Thorn Solubility is constant at given temperature and partial gas pressure. The pressure at 5500m is about the half of that at sea level. But I forgot to mention that during drop falling, solubility increase due increasing pressure is generally smaller than its decrease due increasing temperature. I will update the answer.

That means, I  take gas solubility as the maximum amount of gas soluble in a unit volume at given conditions, in whatever equivalent units.

Ok, I guess I was thinking of the Henry's law constant as "the solubility". Never mind...

@Buck Thorn You would be generally right, but in context of atmospherical physics with its degrees of freedom,, a single number is not useful.

Wouldn't a falling raindrop behave approximately adiabatically, in which case its temperature would be constant, calling for use of a single Henry's constant?

I would not say so, otherwise all but small drops would be hails, as near all rain drops, especially big and fast ones, have ice origin.

Plus, the absolute solubility depends on both temperature and pressure, and temperature vertical gradient varies from -0.01 K/m to +0.0x K/m.


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When alcohol is used to sterilize, what about it changes that it can't be re-used?
I read online that when you use regular household rubbing alcohol to soak something for sterilization purposes, you're then supposed to toss it out. I'm just curious why.
Like, as long as there's a lot of liquid still left, why is it not re-usable? Is it because there's bacteria or other stuff that transferred from whatever you were sterilizing into the alcohol, therefore reducing its effective % of alcohol, and that's what's important to sterilize, so once you're below a certain % concentration it's worthless. Or is it because the alcohol has actually evaporated off and what's left is just liquid (water)?

When dealing with sterilization, I guess it's not worth leaving things to chance, I suppose. That's the whole point, right, to avoid accidental infection? It's like asking why we don't drink water from puddles on the street. Surely some contain clean water? Anyway, this has a strong whiff of being a medical question, and is too broad, since we don't know what or how you are sterilizing.

I'm sure somebody's taken up your question in a scientific study. However that would probably be more microbiology than chemistry.

Likely the process is not 100% effective so that bacteria are let around. Though it should be a problem for the soaked thing rather than the alcohol itself. The latter should be able of multiple disinfection as for bacteria are very very small. I would certainly reuse it if confined in a place and having lot of things to disinfect. I have seen hospital staff carrying around the same little alcohol and the same thermometer when measuring T for an entire lane. Besides common sense and practice, it is true that the antiseptic properties, ie igroscopicity of alcohol depend on water content.

I'm voting to close this question as off-topic because it seems better suited for MedicalSciencesSE, or at least BiologySE.


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Metal nail reaction in wall filler

Please help explain what is happening to this nail. It has been in this wall filler for a good few months.
The substance being excreted was rising up forming these stalagmites, with a liquid forming on them.
Removing the nail, there were also excretions at the part that was sunken.

What exactly is "wall filler"?

Sorry there is no chemical composition on the products label. Its used to fill up cracks in the wall... so a type of cement that hardens once applied to a surface.

Do you have a name for the product so we can look up its composition?

This is the product: http://durapaints.co.za/product/prolong-supafill/ might have to contact the company for chemical breakdown.

That's probably regular gypsum (drywall) then.

It is such a beautiful picture, perhaps good for a chemistry book cover. It is a similar phenomenon as what happens in a so-called chemical garden. Water seems to be creeping up to the nail head, and helping in oxidation of iron salts to insoluble hydroxides. Finally the gelatinous membranes act like semi-permeable membranc in the osmosis like experiment.
The figure shows several salt crystals soaked in a slightly alkaline medium.
Ref: From Chemical Gardens to Chemobrionics,Chemical Reviews, 2015, 115(16)
DOI: 10.1021/acs.chemrev.5b00014

Thanks for the comments - don't know much about chemistry but your answer does sound good. There does still seem to be some unknowns, but I guess that is why it is called a phenomenon.


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Vinegar salt reaction to produce acid; does it matter what type of vinegar I use
I am trying to get get hydrogen chloride gas by using the reaction sodium chloride + vinegar = HCl(g) + sodium acetate, does it matter what type of vinegar I use eg apple cider vinegar, malt vinegar; if it does matter where could i find some pure vinegar at a reasonable price

No it doesn't matter. The result won't change.

Since the reaction won't work for any type of vinegar, the type won't matter.

There are two show stoppers :

$\ce{HCl}$ is strong acid, $\ce{CH3COOH}$ is weak acid. The reaction is practically completely shifted toward acetic acid and sodium chloride. Especially if acetic acid is diluted.
$\ce{HCl}$ forms azeotrope with water with the maximal boiling point near 20% $\ce{HCl}$. If you heat the mixture, that would boil water and acetic acid with minimal traces of $\ce{HCl} $ that would dissolve back in the condensate.


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Is carbon dioxide soluble in petroleum ether, and why?
I googled it, and I'm finding some forum posts that say no, but no authoritative reference.
CO2 is somewhat soluble in water (1.7 g/L at 20°C) and I understand that it's mostly NOT because it forms carbonic acid: it's just dissolved CO2 molecules. Why wouldn't the same thing happen in petroleum ether?
Has anyone tried putting petroleum ether in a SodaStream device?

Carbon dioxide is a dipole so it would made perfect sense to be soluble in water.Petroleum ether is again a polar molecule so why not?

mm petroleum ether is not polar is made of aliphatic compounds and carbon dioxide has dipole 0

While most of CO2 absorbed by water stays this way, the reaction is still driving the dissolution equilibrium forward, especially in high pH.

Carbone dioxide has zero dipole, but it's 2 bonds do not. They are polar and attracted to water. Dissociation of carbonic acid plays very minor role here.

@Mithoron ..but not for low pH of water with  dissolved CO2. 1 atm CO2 leads to pH cca 3.9, the true pKa1 of H2CO3 is 3.6, so bicarbonates makes just few  0.x% of total CO2 content.

Carbon dioxide is, in fact, more soluble in nonpolar solvents than in water.
From the Wikipedia page on carbon dioxide: Solubility in water
1.45 g/L at 25 °C (77 °F), 100 kPa
(Alternative source: CRC handbook. Solubility is about twice as much at 0°C.)
From IUPAC, Solubility Data Series: Carbon Dioxide in Non-Aqueous Solvents at Pressures Less Than 200 kPa, chart on page 3: mole fraction of CO2 in hexane at 101.3 kPa, 300 K = 0.012, which corresponds to 4 g/L.
(Solubility is only slightly (10-20%) higher at 0°C.)
So the forum posts I read were wrong.
As to why, I suppose it's sufficient to argue that CO2 is a nonpolar molecule, and like dissolves like.
(It's true because it's Latin: Similia similibus solvuntur! )

Nitrogen is non polar molecule as well, and that is the reason it's solubility in water is 2 orders lower than for CO2  :-)

It is rather the CO2 bond polarity does not manifest in non polar solvents, where the dominance has zero dipole moment. Similis similibus solvuntur as Well. :-)

While $\ce{CO2}$  as a symmetric molecule has  zero dipole moment, its 2 $\ce{C=O}$ bands do have dipole moment and are polar.
Polar water molecules interact with the polar bonds what decreases   $\ce{CO2(aq)}$ chemical potential and raises solubility of $\ce{CO2}$ some 2 orders above solubility of $\ce{O2}$ or $\ce{N2}$.
That does not happen in non polar solvents likes alkanes of petroether.
BTW, why would anybody put petroether in a soda stream ??

I'd like you to put petroleum ether in a Sodastream so you could tell me if you get fizzy petroleum ether of course! (It goes without saying that I won't do it myself.)

On a more serious note, do you have any actual data on the solubility of carbon dioxide in petroleum ether? Or maybe hexane?

This is behind a paywall  https://pubs.acs.org/doi/pdf/10.1021/je0602972

That's cool but the lowest pressure they consider is 17 bar - I'd like to know about atmospheric pressure.

@Martin   ..proportionality.... 1/17 of 17 bar

It's not that simple. Dissolved CO2 vs. pressure is not linear at those pressures.


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Confusion about comparing the charge densities of benzene and fluorobenzene
I understand that the functional group $\ce{-F}$ is an electron-withdrawing group. It is expected that the difficulties of an electrophile to attack the carbon atoms are $$\ce{PhH}<\ce{2,4of PhF}<\ce{3 of PhF}\tag{1}$$Therefore, I want to demonstrate it quantitatively by computing the charge densities of carbon atoms.
Using command # opt freq hf/3-21g pop=nbo geom=connectivity
and the molecule models in Gaussian 09,
I obtained the charge densities of $\ce{PhH}$ and $\ce{PhF}$. To my surprise, the result is:
$$\begin{array}c\text{molecule}&\text{ortho}&\text{meta}&\text{para}\\\ce{PhH}&-0.239&-0.239&-0.239\\\ce{PhF}&-0.306&-0.222&-0.261\end{array}$$
By observing this result, the fact that the electrophilic substitution of $\ce{PhF}$ is likely to occur in o- and p- is verified, or the right half of $(1)$ is verified.

But why would the charge density of position 2,4 of $\ce{PhF}$ is more negative than that of benzene, given the fact that $\ce{-F}$ decreases rate of the electrophilic substitution reactions?

My speculation
i) $(1)$ may be false.
ii) The NBO charge densities between two different molecules cannot be compared.
iii) The charge density cannot indicate the difficulty of the attacking of electrophile.
Which one, or something else, can explain this strange observation? Can another method to order electron-withdrawing ability of some functional groups computationally be suggested?

This is the way of directing the electrophiles to the ortho and para places  of the benzene cycle. As you see the ortho and para places are more negatively charged.That's true. It is due to the ability to donate its electron pair to the benzene cycle.
But an electronegative element can attract the electron cloud to it. Fluorine is the most electonegative element, so it has a greater ability to attract the electron cloud of the benzene cycle. So that makes it relatively  resistant to an incoming electrophile.
This does not say that it is not a  ortho-para director. It's a ortho-para director but it deactivates the ortho and para places of the benzene cycle.

I agree with your last sentence. But I still not understand the fact that fluorine atom deactivate the ortho and para places despite that they are more negatively charged. "it has a more ability to attract the electron cloud of the benzene cycle. So that makes relatively resistant for coming for a electrophile." But it does not agree with the charge density calculation result and it contradicts with your first paragraph.

As there is two main factors to decide the charge density of the places,- the ability of donating electrons, and the ability of attract the electron cloud. Diactivating of PhF does not means that the ability of reaching electrophiles has decreased than the PhH. The ability of reaching electrophiles is still more than the PhH but it is slightly low than the other ortho para directors such as fenoxide anion, toluene, etc. For more understand see the highlighted paragraph of https://pubs.acs.org/doi/10.1021/ed080p679

Abstractly, This is usually common to helobenzenes, and basing on your question I have to say that the affect of donating electrons in to the ring  more affects the attracting of the electron cloud by the atom. So, there is still more net minus charge in the ortho and para places than the benzene. But it is slightly low than the other ortho para directors. (these two factors are known as mesomeric effect and the inducing effect.)


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electrolysis in aqueous solution. Why is hydroxide discharged at the anode when no halide ions are present
Why are hydroxide Ions discharged when halide ions aren’t present in aqueous solution during electrolysis

Please see this link: https://chemistry.stackexchange.com/a/19871/79678


text:
Can the plane of symmetry for a compound be the plane of the paper?
1,4-Dichloro-2,5-dimethylbenzene is symmetric about the plane of paper.
I have never seen a plane of symmetry positioned like this. Would it be correct to call it a plane of symmetry?

The "plane of the paper" depends on how you draw the molecule; the molecule cannot know what plane that is, there is nothing special about it.

This question seems a bit weird. Any halogenated benzene does not change if you use the paper as a mirror plane. Also the unstubstituted benzene.

The short answer is "yes" - a planar molecule like this can have the plane of the paper as a symmetric mirror plane. (See the answer below for more context about methyl groups, though.

In short: The presence of the $\sigma_h$ plane you see depends on the conformation of the methyl groups.
If your example were either benzene or 1,4-dichlorobenzene, then you were be right, the molecule would contain a plane of symmetry identical to the plane defined by the benzene ring.  This $\sigma_h$ plane then could be described as in the following example by the yellow disk:

(credit to symmetry@otterbein, example symmetry $C_{6h}$, hexakis(Me2N)benzene).
Your example however is about a methyl-substituted benzene. While the carbon atoms of the methyl groups are in the plane of the symmetry, you no longer can apply the same operation (reflection on this mirror plane) on the hydrogens of these methyl groups to yield corresponding, symmetry equivalent hydrogens on the other side of the mirror.  This is only possible if both the methyl groups share a special conformation: for each exactly one hydrogen must reside in this $\sigma_h$ mirror plane; then, the other two hydrogens of the methyl group in question are symmetry equivalent.
In all other conformations, the three hydrogens of one methyl group were no longer symmetry equivalent in respect to this $\sigma_h$ plane.  And it would require only one methyl group to be «off» to erase the $\sigma_h$ plane for the whole molecule.
I would suggest to venture out the interactive tutorials around symmetry@otterbein.  All you need is an internet browser with javascript enabled.  Later, if interested in applying the concepts of symmetry into crystallography (including the transition of Schoenflies notation used in IR-spectroscopy by the  Hermann-Maguin about point groups), you may find its sibling crystals@otterbein a valuable complement.

+1 Generally since the methyl rotation is fast at room temperature, a methyl group is often considered a sphere for the case of symmetry perception. (On exercises I give, I usually clarify this point.)

@GeoffHutchison I agree that a methyl (or trifluoromethyl) group may rotate almost at «no additional energetic cost» for molecules in solution / gas phase at room temperature and above, and by consequence, this perspective.  Of course not so for the restraints and constraints of packed, crystalline state.  Speaking of such, I assume your exercises equally discern between ferrorcene in either symmetry $D_{5d}$ or $D_{5h}$, too.

Yes - those are good thought exercises for the difference between those two point group categories - "staggered ferrocene" vs. "eclipsed ferrocene" - even though at room temperature, there's interconversion.


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Acidic strength order of mineral acids in acetic acid

Explain this acidic strength order $\ce{H2SO4} > \ce{HCl} > \ce{HNO3}$ in acetic acid.

My Attempt
Stability of conjugate base is proportional to the acidic strength of acids in acetic acid.

$\ce{SO4^2-}$ has two resonance structures.
$\ce{NO3-}$ has three resonance structures.
$\ce{Cl-}$ is stable due to octet formation.

As for electronegativity order, Cl is on par with N (almost equal).
According to stability, $\ce{HNO3}$ must me more acidic than $\ce{HCl}.$ Also, electronegativity of N & Cl is almost the same.
So, according to me, the order must be $\ce{HNO3} > \ce{H2SO4} > \ce{HCl}.$
Where am I wrong? Please explain.

The anions are so different that "comparing" them for a trend is meaningless.

The strength of the acid is not only dependent on the stability of the conjugate base but also on the proton-accepting ability of the solvent (For protic acids). HCl for example, is a strong acid in water and weak acid in pure acetic acid for this reason.

@MaxW but such questions are asked in exams.if there is no way to compare them then how should i approach such question in exams.

@ShoubhikRajMaiti i want to know more about "Effects of different solution on acid strength ". Can you please add me in a separate chat so that i can ask little bit more. please.

Note that you should consider rather $\ce{HSO4- } $ instead of $\ce{SO4^2-}$


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acid strength: Periodic trends vs delocalisation, which one is more important
In 1st year chemistry questions, the concept of acid strength is often introduced to students including the 3 factors that provide a heuristic explanation for most of the common acids and conjugate bases: Electronegativity of the atom directly attached to the protic hydrogen, resonance stabilisation and inductive effects. It is also often taught that resonance stabilisation dominate inductive effects.
Now it seems there is the following curious case: $\ce{HNO3}$ and $\ce{HCl}$. Both are considered as strong acids with negative pKa and experimental measurements showed that the pKa of $\ce{HCl}$ (and all halide acids except HF) are lower than $\ce{HNO3}$ and hence $\ce{HCl}$ is stronger.
The common explanation is often given as because $\ce{Cl}$ is more electronegative than $\ce{N}$. However such explanation is quite unsatisfactory as it ignores the fact that the $\ce{NO_3^-}$ is not only larger in size than $\ce{Cl^-}$, but has its negative charge being more delocalised among the oxygens than $\ce{Cl^-}$ (argued from its 3 resonance contributors where each oxygen has a formal charge of -2/3 and also that in MO theory, none of the antibonding orbitals of $\ce{NO_3^-}$ are occupied. Thus based on these heuristic, it seems there is every reason that the electron delocalisation of $\ce{NO_3^-}$ could have dominate the electronegativity of $\ce{Cl^-}$ as their values are only -0.12 apart (meanwhile $\ce{O}$ is roughly 0.24 apart from $\ce{Cl^-}$)
It is also hinted in this comment, that $\ce{Cl^-}$ is more solvated than $\ce{NO_3^-}$ and hence may play a role in the easier dissociation of $\ce{HCl}$. However, such concepts will not be covered in great detail for first years, and hence such a complication can be potentially confusing to students.

How can we check whether delocalisation or electronegativity effect dominates in the stability of a conjugate base, and how should we rationalise to students that electronegativity often beats delocalisation?

But in nitric acid, hydrogen is bonded to O, not N. And H-X bond strength is also an important consideration, as far as I’m aware.

Well, that adds to the confusion, O is 0.24 more electronegative than Cl, and yet Cl wins, so it means the usual 3 heuristics have not been able to give a satisfactory explanation, and yet such an answer is often given as an explanation for this HCl vs HNO3 case. Bond strengths are sometimes discussed, but often in the context of haloacids.