text:
Is it possible to increase the internal energy of an ideal gas in an isothermal process?
Ideal gas is the one in which there are no attractive or repulsive forces acting and hence its internal energy is due to kinetic energy of its molecules. We may then say, for an ideal gas internal energy is a function of temperature##.
$$E = f(T)$$
Hence for an ideal gas ∆E=0; for isothermal process.
Thus I thought it is not possible to change internal energy of ideal gas in an isothermal process
Now, I came across the following question.

Q. The internal energy of an ideal gas increases during an isothermal process when gas is
A. Expanded by adding more molecules to it
B. Expanded by adding more heat to it
C. Expanded against zero pressure.
D. Compressed by doing work.

The correct answer given is A. i.e addition of molecules.
Definitely adding more molecules would increase the energy of system since internal energy is an extensive property and related as follows
$$U = \frac{f}{2}nRT$$
My question is, would system maintain the same temperature even after addition of molecules?
My understanding is that, in an isothermal process on addition of energy
This energy must be completely used in expansion of gas or must be released into surrounding as heat.$
i.e. $$q = -W $$
If that does not happen, then net energy transfer should lead to change in temperature; i.e.
∆T ≠ 0. making it non-isothermal process.
If my above assumptions are wrong; then; I would be curious to know; where did the energy gained by the system go?
EDIT 1: Summarizing my queries here.

Is the above quoted question conveying a hypothetical scenario?
If the answer to my main question (In title) is "NO" then question proves to be hypothetical.
If question is fine; then how would we justify the increase in internal energy without changing the temperature of system?

After reading some answers and comment section; this is what I feel, I was doing wrong (I've marked what I think are the erroneous statements in my original post in italics and made bold)

The following is true only if mass of system is constant.
$$E = f(T)$$
It would be more appropriate to say; N: number of molecules.
$$E = f(N,T)$$

By adding extra molecules, the system will exchange heat to reach the same temperature and the mean thermal energy per a molecule as it had before. The extra energy is the kinetic energy of the extra molecules.

What is your own answer to this question ? An why ? What speaks for and against it ? What speaks for and against each of those 4 answers  to be the correct one ?

@Poutnik my assumption that ∆E=f(T) was wrong. Should have been ∆E= f(N,T) if mass is not constant. My other assumption again wrong was, all energy added to system on adding molecules would be transfered to surrounding to keep the process isothermal. I agree with your 1st comment. I'm just asking if my understanding of it is right. And if it is fine, you could probably add an answer.

@Poutnik now my single point against option b,c and d is since mass remaining constant ∆E= f(T). So ∆E= 0 in all those cases. We are only left with option A and that I also agree is right answer. Before reading your 1st comment I had an idea that this question portrays a hypothetical scenario.

Answer A should be qualified by adding the words "... at constant temperature and pressure."  Also, answers B and D can both cause the internal energy of an ideal gas to increase if the process is not carried out isothermally.

Thanks. Could you please help me with "how the addition of molecules doesn't affect the temperature?". Say temperature of system is 400K and we add 1 mole of molecules having 400K, would there be any temperature change in system or any heat transfer? Also if molecules are added at say 500K, would it still be isothermal?

If the addition is at the same pressure and temperature as the original gas, all that happens is that you have more volume and more moles, but the volume per mole stays the same.  If you add molecules at 500 K, then heat will be transferred between the old and the new, and the temperature of the original gas is going to rise, while the temperature of added gas is going to drop.

First law of thermodynamics is :
I have divided the answer into two parts for better understand of the concept.

For change in internal energy.

$Δ U$ or $\frac{ nfR Δ T}{2}$= $q $ or $c*m*\ Δ T$ ±$ W$ or $PdV$.
Now , for an isothermal process , $Δ T = 0 $ .
Therefore , change in internal energy and heat transfer is always = 0.
But in your answer of the Q , it states that the internal energy can be increased for an isothermal process by adding more molecules to it.

Therefore , For internal energy of a state

$U_2$ - $U_1$ = $q_2$ - $q_1$ ±$ W$ .
If we talk only about single state here. Then , that temperature is not equal = 0 . $T_i$ or $T_f$ may or may not be equal to 0.
To increase the internal energy of a system only. Yes , it can be done by adding more no of molecules. As you can notice , the heat transfer would increase.
I can see how the Q made it confusing to think that if they talk about isothermal process , the reader will mostly think of change in state. Therefore , you can only think of one state where Ti = Tf.
Yes , for each internal energy state. You can increase the internal energy of system.

Commenting upon each option :

A. Expanded by adding more molecules to it
B. Expanded by adding more heat to it
C. Expanded against zero pressure.
D. Compressed by doing work.

A is true.
B is true or similar to A.
C is not true since W = 0.
D is true.
You can make cases here and understood too.
PdV or VdP or $\delta$PV

The answer lies in the fact that when you add more molecules of the gas during the process, the mass, i.e., the number of moles of the gas undergoing the process, changes.
We know, for an ideal gas, its internal energy is a function of its temperature.
$$U=nf/2RT$$
And, change in internal energy is generally represented as,
$$∆U=nf/2R∆T$$
However, in case you add some more molecules of the gas into the system, $n$ also changes. That is,
$$∆U=f/2R∆(nT)$$
Now, coming back to the question of an isothermal process:
In an isothermal process, the temperature always remains constant. Thus, internal energy should also remain constant. Thus, by the First Law of Thermodynamics,
$$∆Q=-∆W$$
And that is fulfilled here, too.
From Thermodynamics, we know that for any process,
$$∆Q=nC∆T$$, where $C$ is the molar heat capacity of the ideal gas for the specific process.
And, $$∆W=n(C-C_v)∆T$$ because, $$∆U=nC_v∆T$$
However, as $n$ changes too in this case, and the process is isothermal, we must write,
$$∆Q=CT∆(n)$$
and,
$$∆W=(C-C_v)T∆(n)$$
and,
$$∆U=C_vT∆(n)$$
Thus, the process is still isothermal, and $$∆Q=-∆W$$ is still valid, because  the internal energy of the gas, in this case, doesn't increase due to supply of heat into the gas, or due to work done on the bas. It increases because the very quantity of the gas undergoing the isothermal process, has changed.


text:
Which is more efficient way to generate energy using hydrogen: direct hydrogen combustion or through fuel cell?
Which is more efficient way to generate energy using hydrogen: direct hydrogen combustion or through fuel cell? Please explain so that even a primary school student can understand.

Note that efficiency is just one of important parameters, when it comes to electrochemical cells. There are at least 6 other important parameters like (available power or stored energy (cycle or lifetime) per ( volume or mass or price(cycle or lifetime) )

This is mostly an engineering question. To generate heat energy, direct combustion. To generate electrical energy, perhaps a fuel cell.

If you want energy as heat, the best way to generate energy is combustion. If you want energy as electricity, the best way is a fuel cell (which also produce some heat). If you want mechanical work, using hydrogen like gas or oil in a usual combustion engine, the yield would be a maximum of $30$%. The best yield would be to produce electricity in a fuel cell, then use this electricity to run an electric motor. Here the yield is of the order of $60$%, twice as much as the combustion engine.

source: Stack Exchange

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

sort: votes


text:
How to account for dilution in the sample solution when reading calibration standards?
I have been provided the following table of calibration standards from an ICP-MS experiment.
(Some text is in German, I am operating in a location where DIN standardization is ubiquitous)

I understand that I must subtract the provided method blank from each of the provided data points before plotting them and drawing a regression line, but the sample solution I am working with (500 mg manganese slag in 10mL of a mixture of hydrogen peroxide and nitric acid) has also been diluted in a 1:10 ratio with water. How do I account for this dilution when calculating the mass concentration of manganese of the sample?

There is a very simple dilution formula:
C$_{initial}$ x V$_{initial}$=C$_{final}$ x V$_{final}$
where C and V refer to concentration and volume in consistent units.
When you analyze the sample (=Probe in your table) using the regression equation, the concentration unit of the sample will be the same as your standards ng/mL.
Lets say your diluted sample conc. from the regression equation 12 ng/mL (= C$_{final}$ see the above equation), all the variables are known to you except C$_{initial}$. You know the sample volume (=V$_{final}$), you must also know how much initial volume was used for dilution.
Once you find C$_{initial}$, one can determine the mass from the original sample volume, because analyte mass, Mn = concentration x volume.
Note: In certain methods multiple dilutions are made, then one must apply the dilution formula repeatedly for each dilution.

Thanks for pointing out that multiple dilutions could happen. In this case it was just the one, however.

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


text:
Preparation of an aqueous solution of rubidium azide from rubidium chloride
I want to start with $\ce{RbCl}$ and end up with pure rubidium azide in solution.
If I had rubidium sulfate, the following would be possible (from Wiki):
$$\ce{Rb2SO4 + Ba(N3)2 ->  2RbN3 + BaSO4}$$
The $\ce{BaSO4}$ is a precipitate and can be filtered off
Also from Wiki:
$$\ce{H2SO4 + RbCl -> RbHSO4 + HCl}$$
Would $\ce{RbHSO4}$ work with the barium azide reaction? What other reaction products would be present?
Is there any way of getting rid of residual $\ce{H2SO4}$ and $\ce{HCl}$?

Increase the concentration of RbCl and you'll get Rb2SO4.

Any azide preparation with an acid soln runs the risk of creating hydrazoic acid HN3 which is a hazardous gas.

@Waylander I agree. Azide compounds are generally unstable and prone to decomposition adding to the hazard factor. They are also toxic.

What about converting rubidium chloride to sulphate by anion exchange resin in sulphate cycle ?

I am not familiar with ion exchange ( being an electronics engineer rather than a chemist). Could you please give me some links to explain further. The reason I need to start withe rubidium chloride, is that I need to use separated isotopes(Rb87 and Rb85) and these are only available as the chloride. Application is filling cells for rubidium atomic clock


text:
Determining if the product would be cis or trans after an anti elimination on fischer projections

I find these kind of problems really confusing for some reason, but I saw a way on the web and tried doing it that way:
I considered the basic Fischer operations (Vertical: Below the plane of paper, and Horizontal: Above the plane of paper) and tried making a sawhorse representation:

But I am stuck here. Technically if I still consider the paper plane I considered while making the sawhorse projection (that is, Me and Ph are below plane of paper). During the elimination step, they are both still below the paper plane. So it seems it would give a cis-form.
But according to my book it would be trans.
If I again rotate about the sigma bond and get the other H to be anti to Br, then the Ph is on the opposite plane of the paper and would give trans.  So is it just because trans is more stable than cis so this configuration would be eliminated more than the initial one, or is it something else?
I would appreciate if someone could help me out with this and also suggest a basic approach to solve these kind of reactions when the Fischer projection is given.

Have you studied about stereochemical representations of the compound? If not, then this is required for it. If yes, then you might want to Strengthen inter converting sawhorse, Newman, wedge and dash, and Fischer into one another. Then, you will have no issue with this.

Converting to the sawhorse projection and then doing anti elimination is a perfectly correct approach, but the sawhorse projection you've drawn is incorrect. The carbon with  $CH_3$ and Br must be at the back, not at the front.

There are two possible eliminations that can happen (see the image below)

As you can see, we then have two final products depending upon which $\ce{H}$ is eliminated. From the image 1a is cis and 2a is trans. Now, how do we decide which one is the major product?
This can be based on looking at the stability of the transition state. This means we have to compare 1 and 2 to see which is more likely to be observed. As you can see 1 is a staggered configuration with the $\ce{-CH3}$ near the $\ce{-Ph}$. This means that there is some steric hinderance with the actual transition state that is observed. However, in 2, the $\ce{-CH3}$ is opposite the $\ce{-Ph}$, which leads to much lower steric hinderance. Therefore, the reaction pathway 2 is preferred over reaction pathway 1.
This means that the major product in this reaction would be 2a, which is the trans product.

As you have mentioned, the basic Fischer operations are: Vertical positions are below the plane of paper and horizontal positions are Above the plane of paper, thus you have already remember that when working with Fischer projection. What you have to remember about chiral compounds are: If you switch two groups, you get the epimer and if you switch two other group on same chiral center, you got original compound back as demonstrated below with your compound of interest (2-bromo-1-phenylpropane):

The Cahn-Ingold-Prelog configuration of the $\ce{C}$2 is $(2S)$ while that of the epimer is $(2R)$ after first switch. You get back the original $(2S)$ configuration after the second switch. I want to do this because to get $\ce{H}$ and $\ce{Br}$ in anti-positions. The $\ce{C}$1 in the molecule is not chiral. Yet we'll do the same  in case if you have to do this with a compound having two chiral centers (I'll replace one of $\ce{H}$s with a $\ce{R}$ group):

As shown, after first switch on both chiral centers, you get enantiomer (both chiral centers changed the stereochemistry). Yet, after second switch on both chiral centers should give the original compound. Make sure you change the groups as you want them to be. I put $\ce{Br}$ on the top and $\ce{H}$ on the bottom to get the best placement for the anti-situation. Yet, they are not in anti-positions since both are below the paper as shown in the second structure of the bottom line. To get into the anti-position, you must rotate the $\ce{C}$1 $180^\circ$ clockwise or anticlockwise as shown. Now $\ce{H}$ and $\ce{Br}$ in anti-positions so that the anti-elimination can occur. If that happens, the groups are in trans-positions, and the final product is Z-isomer in my case as shown. However there were two hydrogen in the $\ce{C}$1 in your case, and hence, more stable E-isomer would be the major product. Still, it would give you small amount of Z-isomer as minor product (you may see gauche-interaction between methyl and phenyl group in the anti-II conformer).
Note: If you know how to convert Fischer projections to Newman projections, it is easy to visualize and you may able to see why is the major product to be E-isomer.


text:
Why does this ring contraction take place in the following nucleophilic substitution?

While attempting this problem I thought that P1 would simply be Bromine getting substituted by OH through nucleophilic substitution.
P2 would give a product formed through ring expansion as a 4 membered ring would tend to rearrange to a 5 membered ring due to strain factors.(SN1 reaction(acidic conditions))
This is the mechanism I thought of:

This pointed towards option (2) but what's given is option (3).
Why does this happen?
It could be that the four-membered ring being strained rearranges, as Waylander suggests but if it would rearrange, why would it rearrange into a 3 membered ring?
Usually, this happens only when there is a stabilising factor and the only stabilising factor related to a 3 membered ring I can think of, is sigma bond resonance(Dancing resonance)
I would appreciate an explanation of what's happening here.

Cyclobutanes are very highly strained, more strained than cyclopropanes. If they can rearrange they will.

Even into three membered rings? I thought three membered are even more strained... Does dancing resonance(sigma resonance) operate here?

Can the 'close' voters give some feedback about the post? I have edited it to make the post more clear but I would still appreciate some constructive criticcism so that I can improve.

@ https://www.sciencedirect.com/science/article/abs/pii/S0040403900703820 This might be helpful.(Or not it is behind the paywall :/)(There is a first page preview though.)

@PrajwalTiwari If you wish you can add the mechanism which you think should be followed regarding the question in focus. :=)

@Rishi Done! Thanks for the suggestion

@PrajwalTiwari dancing resonance is not an officially accepted term for cyclopropyl methyl carbocation, stick to sigma resonance in the future.

@napstablook Alright! Sorry for that, I'm a student so don't have much idea about all that and during the lectures, my teacher usually uses these terms without worrying about 'accepted' terms. I will take care in the future :)

What software/website do you use to draw chemical structures and reactions and upload them here as an image? Would be of immense help to me. Thanks!

@TRC I have heard of chemdraw/ Chemdoodle, etc. The other more experienced members might be able to help you out with that. Here, I just used the 'preview' utility that MacOS provides since I didn't have access to any other chem structure tool.

Why did this get closed though? I provided all my approaches, thoughts, even added a mechanism that I used, what else should I have provided to make this a better post?

This reaction should be leaning towards SN1 because the approach of $\ce{H2O}$ for a back attack to the ABMO of the $\ce{C-Br}$ is heavily hindered since the neighborhood is congested due to the small ring of cyclobutane.
Once cation formation occurs the reaction will interestingly, result in a ring contraction because the carbocation formed would be very stable due to this kind of resonance.
After the ring contraction the reaction would proceed in a fairly standard nucleophilic attack by $\ce{H2O}$ resulting in the $P_1$ your answer key alludes to. The overall reaction should proceed in the manner shown below:-


text:
Why does the application of external mechanical pressure increase the performance of lithium-ion batteries?
Many electric vehicle batteries - specifically lithium-ion ones considered for this query - can be subjected to external mechanical pressure because it increases battery performance.
What components of overall battery performance are affected and why?

Generally, high external pressure can suppress forming gases due side reactions or solvent evaporations. It may also improve cooling.

The  most likely effect is to reduce the thickness of the electrolyte layer, reducing internal resistance, by squeezing the electrodes closer. There is no outgassing during discharge of most cell chemistries.

Here is a publication entitled "Analysis of the effect of applying external mechanical pressure on next generation silicon alloy lithium-ion cells." I believe your query can be addressed by the above comments and the information in the linked article (and citations within it).

@ewr3243 - I think your question is good; however, we do like to see a bit of effort from those who post here in re: getting towards an answer. Can you please edit-in your thoughts?


text:
Cleaning solution alternatives for acetate, nitrate, or polyester base motion picture film
I am in search of alternative film cleaning solutions which are safe for the film, me, and the environment.
I have used several different professional film cleaning solutions in my many years of cleaning film. The one thing they have in common is that they all require a well- ventilated work space. But I've always wondered what are their ingredients? Understandably, little to no information is available about what the actual cleaning solutions compounds are -- trade secrets, I suppose.
I would like to distinguish between two types of film cleaning methods/process I use for which I would like some suggestions on what to use or stay away from:

A re-lubricating, or conditioning, solution -- preferably a penetrating solution that helps make brittle film regain its pliable nature and yet does not dissolve the original tape adhesives from the tape that was used to hold film splices together. For this cleaning process, I gradually apply the solution to the film by running the film through a soft cotton cloth which is moistened with the solution to the point to where the whole film is eventually wet. Shortly after, I reverse the process; using solution free cotton cloths to remove the solution from the wet film. I repeat this removal process until the film is dry again. Would Paraffin Lamp Oil or Mineral Oil work at all and be safe for the film -- in the short run and long haul? If not, what might be the reasons?

A non-lubricating solution that can be used to clean film. Preferably something that will suppress static from building up and something that evaporates quickly. Hexane, petroleum ether, or cyclopentyl methyl ether comes to mind.

Doing some research, I found this interesting report https://ww3.arb.ca.gov/ei/pubs/motion_picture_film.pdf revealing some solvents used for film cleaning being Perchloroethylene, 1,1,1-Trichloroethane, other Volatile Organic Compounds, and Hydrofluoroether blend (HFE 7200).
I am aware that a solution which "helps" encourage mold growth would not be effective -- i.e. water-based. I found that quite a bit of films has first signs of mold, so Isopropyl alcohol will not work. Moreover, the solution would need to be magnetic-sound-track safe.


text:
Does ammonolysis of primary and secondary alkyl halides proceed through SN2 mechanism?
We know that the first the ammonia acting as nucleophile attack the alkyl halide forming the substituted ammonium salt, but my question is why cannot it proceed through a typical $S_\text N2$ type, mechanism for  secondary or primary alkyl halide , as ammonia is  good nucleophile, so there will be a formation $\ce{R-NH3^+}$ (halogen being a good leaving group will easily leave), then nitrogen will take the electrons from hydrogen as result of which we obtain $\ce{R-NH2}$, why do we get substituted ammonium salt first, then we have to treat it with a strong base to get amine?

Using degree convention is outdated. It's better to use primary, secondary convention.

Base is required exactly for the last step. Amine is a base so you need a stronger base to deprotonate it.


text:
Definition of dissolve and the state label (aq) in chemical equations
I am confused with the meaning of "(aq)" in chemical equations, and the meaning of "dissolved". What is the actual meaning it is supposed to convey?
For instance, in the document "Preparing Bromine Water", bromine molecules in bromine water are labelled $\ce{Br2_{(aq)}}$.

What does dissolving in water mean? Must it be that a hydration shell is formed around the molecule? See:Dissolving of non polar gases in water (liquid)

Even if no hydration shell is formed, small molecules just insert themselves into the space between solvent molecules, is it still called "dissolve"? It does seem more like being "dispersed", as in a colloid.

Must "(aq)" mean that there is a hydration shell surrounding the molecule/ion?

For non-polar molecules dissolved in water, do we label it as "(aq)" in chemical equations? Must they form hydration shells?

The abbreviation (aq) in chemical equations or symbols simply means that this material is present in water as a homogeneous phase. The state symbol has nothing to do with the actual and true "structure" in solution. In short forget about hydration shell.
From IUPAC's recommendations "Notation for states and processes, significance of the word standard in chemical thermodynamics, and remarks on commonly tabulated forms of thermodynamic functions" Pure & Appl.Chem., Vol.54, No.6, pp. 1239—1250, 1982.

the symbol for a solution in which water is the solvent (an aqueous solution) is aq; in the past this symbol has sometimes been used to denote an infinitely dilute aqueous solution, but infinite dilution should henceforward be denoted by the extra symbol $\infty$.

$\ce{CO2 (aq)}$ would imply dissolved carbon dioxide in water (i.e.,to our eyes it is a single phase, just like an unopened bottle of coke).
$\ce{Br2 (aq)}$ would imply homogeneous solution of liquid bromine in water.
$\ce{NaCl (aq)}$: Salt dissolved in water.
A colloid is a heterogeneous phase.
But above all, do not worry too much about these minor symbolism issues. A chemical equation is not meant to convey everything. It is just a shorthand notation to describe chemical reactions. A real experiment or a research paper should describe the experimental condition with sufficient details in such a way that an experienced person should be able to repeat it.
You can also use (sln) as the state symbol, when the substance is dissolved but not present in water. From the same document, I quote an example.
$$\ce{Na (s) + 5C2H5OH (l) = 0.5 H2 (g) + \{C2H5ONa + 4 C2H5OH\} (sln)}$$
Try to interpret it and see if it makes sense.

I disagree that a colloid solution is a heterogeneous phase, unless detergent or protein solutions should be considered heterogeneous.

See also the IUPAC Green Book, printed page 54. https://media.iupac.org/publications/books/gbook/IUPAC-GB3-2ndPrinting-Online-22apr2011.pdf#page=68

@Buckthorn, you are stepping in a grey area, as per the  IUPAC,  "The term (colloidal) refers to a state of subdivision, implying that the molecules or polymolecular particles dispersed in a medium have at least in one direction a dimension roughly between 1 nm and 1 μm, or that in a system discontinuities are found at distances of that order."

Detergent solution is also tricky, under what concentration, above CMC or below CMC, should we consider them as a colloid. At a molecular level every solution is heterogenous, right (if only we could see molecules).


text:
Why does CO2 diffuse out of a cylinder in the absence of a pressure difference?
Suppose I have a thermally insulated cylinder containing carbon dioxide (assuming it is ideal) at 1 atm pressure(atmospheric pressure is 1 atm). Now let's consider two cases:
CASE-1: I cut open the top of the cylinder, now the gas inside experiences a uniform 1 atm pressure and exerts the same pressure too. As there is no pressure difference so carbon dioxide should not diffuse out!
CASE-2: I pinhole the cylinder, now it should follow Graham's law of effusion and hence a net movement of air molecules inside the cylinder should take place; but how can air rush in if the pressure is the same 1 atm exerted by gas inside.
Further, my teacher told that since the partial pressure of $\ce{CO2}$ inside is much more than that of $\ce{CO2}$ outside, so $\ce{CO2}$ would diffuse out, but my point is that how can $\ce{CO2}$ differentiate whether the pressure it is subjected to, is exerted by 'X' gas or 'Y' gas. Hence the concept of partial pressure doesn't apply here. $\ce{CO2}$ would only experience that some gas is exerting 1 atm pressure, it doesn't care about the type of gas as ideal gases only care about the amount.
So, would $\ce{CO2}$ diffuse out and why?

Remember the nature of gas molecules is dynamic, not static. The same pressure of 2 gases in contact means the  number of molecules leaving each gas is compensated by molecules coming from the other gas, so the pressure difference remains constant.

https://physics.stackexchange.com/questions/565575/why-would-tennis-balls-filled-with-sulfur-hexafluoride-explode/565966#565966 has useful information.

You got answers and comment. At the root of your concern is that you ignore that the molecules in gases are in translational motion. Following the same logic you have used, you would conclude that gases are rigid or something like that.

....how can $\ce{CO2}$ differentiate whether the pressure it is subjected to, is exerted by 'X' gas or 'Y' gas. Hence the concept of partial pressure doesn't apply here. $\ce{CO2}$ would only experience that some gas is exerting 1 atm pressure, it doesn't care about the type of gas as ideal gases only care about the amount.

Let's suppose we treat these gases ideally.  In that case, you're right, the gases don't "know" about each other.  Yet differential movement can still occur in spite of this.
Let's label the inside gas X, and the the outside gas Y.  As ideal gases, each particle is essentially oblivious to the presence of the other gas particles.  They can each be considered to operate independently.
Since the temperature and pressure inside and outside are the same, the concentrations of X and Y will be identical. Further suppose, for simplicity, that gases X and Y have identical masses, and thus have the same speed distributions.
Now mark a disk of area 1 mm^2 anywhere on the wall of your container. Mark both the inside and outside surface. On the inside, the X particles are going to strike that disk at a rate that depends on the temperature, concentration, and particle mass.  Same with the Y particles striking that disk on the outside.  Since the temperature, concentration, and mass are the same for the X and Y particles, they're striking that disk at the same rate. Let's assume the rate is 100 hits per second (you could actually calculate this, but I'm not going to bother).
So, at this disk, 100 Y particles are hitting the outside wall every second, and 100 X particles are hitting the inside wall every second. Now remove the disk, turning it into a hole.
What happens?  Well, every X particle that hits it from the inside ends up exiting the container.  And every Y particle hitting it from the outside ends up entering the container. Hence initially, you have only X particles exiting, and Y particles entering.  Then as more Y particles enter the container, some Y particles exit along with the X's.  This continues until the relative concentration of particles is the same on both sides of the hole.
What does this have to do with partial pressure?  Assuming equal temperatures, partial pressure tells you the relative concentration of X and Y particles on either side of the hole. If the total pressure is 1 atm inside and out, then the initial partial pressure of X is 1 atm on the inside, and 0 atm on the outside.  And the opposite for Y.
In summary, for the same temperature and particle mass, the relative movement of X and Y particles through the hole is determined by their relative concentrations which, for equal temperatures, is expressed by their relative partial pressures.

Diffusion is a fundamental molecular phenomenon and it does not have a direction.

CASE-1 : I cut open the top of cylinder, now the gas inside
experiencees a uniform 1 atm pressure and exerts the same pressure
too. As there is no pressure difference so Carbon dioxide should not
diffuse out!

Why not? Recall if you open a bottle of perfume in a room, eventually you will smell that somebody has opened a bottle. Diffusion is not driven by pressure. Diffusion in gases is pretty fast.
In the second case, diffusion will still occur but at a much slower rate because the area is small (a pinhole). In the first case, you had simply cut open the cylinder exposing a bigger area.

Ok, I got the case-1 and case-2 but what about the partial pressure concept..is it correct to apply that in case of diffusion?

@user265825 The macroscopic net diffusion follows the opposite direction of the concentration gradient or partial pressure gradient.  It means diffusion from the region of higher partial pressure is faster than in the opposite direction. See wikipedia.org - Fick's laws of diffusion.


text:
How is most probable radius in quantum mechanical model equal to Bohr's radius in hydrogen atom
Most probable radius in 1s orbital for hydrogen electron is $\pu{0.529E-10 m}$ which is Bohr radius. But energy of electron in hydrogen atom is proportional to mean radial distance in atom. In this situation most probable radius is equal to average radius? What's going on?

The mean and most probable values of a distribution, generally speaking, are not obliged to be related in any specific way. One may be less than the other, or greater, or equal to it. That's what is going on. On a side note, the energy is not proportional to the mean radius, but that's another story.

We find the radius at which the radial distribution function of the hydrogenic 1s orbital has a maximum value by solving $\mathrm dP/\mathrm dr = 0.$
If there are several maxima, then we choose the one corresponding to the greatest amplitude (the outermost one).
The radial distribution function is given by
$$P(r) = \frac{4Z^3r^2}{a_0^3}\exp\left(\frac{-2Zr}{a_0}\right).\tag{1}$$
It follows that
$$\frac{\mathrm dP}{\mathrm dr} = \frac{4Z^3}{a_0^3}\left(2r - \frac{2Zr^2}{a_0}\right)\exp\left(\frac{-2Zr}{a_0}\right).\tag{2}$$
This function is zero where the term in parentheses is zero, which is at
$$r_\mathrm{mp} = \frac{a_0}{Z}.\tag{3}$$
For hydrogen $Z = 1,$ so $r_\mathrm{mp} = a_0.$
For more details, see Atkins' Physical Chemistry.


text:
pH meter mechanism - connecting potential differences
This question was asked here: pH probe bulb - what is happening within the glass? and How does a pH glass electrode work?.
But, the mechanism remains unanswered. How does a potential difference between the inner and outer surfaces of the glass bulb result in a measurable difference between the two electrodes?  These are two different electrical systems.  I am unable to find in the literature how the the two systems are connected.
Given:

There is an electrolytic solution connecting the electrodes through openings in the tubes housing each electrode.
A very small electric current flows between the reference electrode to the pH electrode through a high value resistor for voltage measurement.  The electrodes maintain ion balance through the chemical reactions occurring at each electrode and the movement of ions through the openings.

From a-cyclohexane-molecule answer in pH probe bulb - what is happening within the glass?:

A few side remarks.

Equilibrium is reached when the favorable binding of protons to the glass surface is balanced by the unfavorable electrostatic repulsion and chemical potential gradient that result from diffusion into the hydrated gel layer. This provides an equation relating the potential difference to the pH of the solution and allows for pH measurement.
Something has to be able to move through the inner layer of the glass membrane to conduct a current and hence allow for a measurement of the potential difference. It turns out that sodium ions can move through this inner layer, but only sluggishly---the resistance of the glass membrane is about 108Ω.

A hypothesis is:

When H+ invades the porous glass, the side with the more H+ can repulse H+ out of the opposite side.  An acidic solution being measured, ejects H+ out of the inner bulb surface and an alkaline solution allows more H+ to enter into the inner glass surface.
The positive ion concentration near the electrode next to the inner glass surface changes.
This change in positive ion concentration can not be compensated by the chemical reactions at each electrode that are maintaining a electrically neutral solution.
The change in positive ion concentration near the pH electrode tip is causing a change in the pH meter potential.
This change in the pH meter potential results in a measurable difference in voltage between the electrodes.

A glass electrode is a galvanic cell, where the membrane is a thin glass bulb, at the end of a long glass tube. The bulb is filled with some diluted HCl solution. And the bulb is made of a special glass where the protons $\ce{H+}$ are nearly free to move from one site to another one, like the electrons in a metal. I saw once that this glass contains Lanthanum $\ce{La}$. So if such a bulb (filled with a dilute HCl solution) is dipped in a more diluted acidic solution, the $\ce{H+}$ have the tendency to get out of the bulb through the membrane. But they cannot, because of the $\ce{Cl-}$ ions who attract them, and they cannot get out of the bulb.
There is a way to take care of the $\ce{Cl-}$ ions. In the center of the bulb, the end of a long silver wire (covered with insoluble $\ce{AgCl}$) is introduced to dip into the $\ce{HCl}$ solution. The unhappy $\ce{Cl-}$ ions do react with the silver wire, according to : $$\ce{Cl- + Ag -> AgCl + e-}$$ The emitted electron got through the silver wire to an external circuit containing a voltmeter and then go to another glass tube containing an electrode made of a silver wire covered by $\ce{AgCl}$ dipping in a $\ce{KCl}$ solution. The same electrode as before, but a different solution. Here, the external electrons react with $\ce{AgCl}$ producing the opposite reaction $$\ce{e- + AgCl -> Ag + Cl-}$$ Now this second glass tube has a small hole in its lower part, which is blocked by a porous ceramic. Ions can cross this barrier if there is an attraction from outside. And here there is an attraction, if this second electrode is dipped in the same weakly acidic outer solution as the first electrode (with the bulb). In this case, the $\ce{H+}$ ions having crossed the glass membrane are able to attract the $\ce{Cl-}$ ions coming out of the second electrode. This whole system of two electrodes is a galvanic cell, and the measured potential is defined by Nerns't law. The voltmeter gives a tension which is $$\pu{E = ~constant + 0.059 log[H^+] = ~ constant - 0.059 pH}$$
So the voltmeter gives directly the pH value of the outer solution.
Usually the second electrode is placed in a muff around the tube of the first electrode, taking care not to cover the thin bulb.


text:
Elimination reaction mechanism decision (order = 1 or 2)

Match the following:

\begin{array}{}
& \textbf{Column I}&&\textbf{Column II} \\
\text{(A)} & \ce{CH3-CHBr-CD3} \text{ on treatment with alc. KOH gives}& \text{(P)}& \text{E1 reaction}\\
&\ce{CH2=CH-CD3} \text{ as a major product} \\
\text{(B)} & \ce{Ph-CHBr-CH3} \text{ reacts faster than } \ce{Ph-CHBr-CD3} & \text{(Q)} & \text{E2 reaction}\\
\text{(C)} & \ce{Ph-CH2-CH2Br} \text{ on treatment with } \ce{C2H5OD/C2H5O-}& \text{(R)}& \text{E1cb}\\
&\text{gives } \ce{Ph-CD=CH2} \text{ as the major product.} && \text{reaction}\\
\text{(D)} & \ce{Ph-CH2-CH2Br} \text{ and } \ce{Ph-CD2-CH2Br} \text{ react with the }& \text{(S)} & \text{First Order}\\
&\text{same rate.} && \text{reaction}
\end{array}

The answer given in the exam solutions is

A − Q; B − Q; C − R,S; D − P,S

Question source: IIT-JEE 2006 (memory based)
My answer: A - Q ; B - Q ; C - Q ; D - Q
I am giving reasons for my answers and please correct me for the same.
B: elimination reaction not possible for weak nucleophiles so assuming strong nucleophile.
C: elimination reaction not possible (1$^0$ carbon and good leaving group) for weak nucleophiles so assuming strong nucleophiles
D: elimination reaction not possible (1$^0$ carbon and good leaving group) for weak nucleophiles so assuming strong nucleophiles. In my opinion C and D have same answers because statements doesn't have impact on answer.
EDIT: (For TRC's answer) For C: I agree that I had initially ignored the fact that an $\alpha\ce{-H}$ has been replaced by $\ce{D}$. But that doesn't imply E1cb mechanism. If E1cb occurs than also there will be no replacement of $\alpha\ce{-H}$ by $\ce{D}$ as shown below. Also $\ce{Br}$
is good leaving group than fluorine.

Lets say intermediate is formed by E1cb and reaction shown in image takes place (but it is weird and absurd thought). But than also E1cb and E2 will compete with winner being E2 according to conditions given in problem. Also if carbanion forms in equilibrium with (P), concentration of (P) will be much less and can't influence major product.

Your arrows are not standard. The arrows should be from oxygen's lone pair onto the hydrogen, and bond electrons moving down to the carbon.

I understand.  I say the end calculations must be presented in line with standard rules, otherwise , this question may confuse future readers who may not know how to correctly draw arrows. This is further compounded by the fact that most coachings do not emphasize on mechanistic study to my knowledge.

@Buraian I personally want to update all my posts but after my exam finishes. Anyway you may edit it and I may assure that I will approve it.

https://chemistry.stackexchange.com/questions/155806/e1cb-mechanism-with-replacement-of-hydrogen-step

$\ce{C-H}$ bond and $\ce{C-D}$ bonds are not the same in energy. $\ce{C-D}$ bond is stronger. So breaking $\ce{C-D}$ bond is more difficult and slower than breaking $\ce{C-H}$ bond.
I hope you know the mechanisms of $\mathrm{E1}$ and $\mathrm{E2}$ reactions. In $\mathrm{E1}$ reaction, a carbocation is formed first, which is the rate-determining step (rds). Since rds does not involve breaking of $\ce{C-H}$ (or $\ce{C-D}$) bond, hence rate is same, regardless of whether $\alpha$ hydrogen is $\ce{H}$ or $\ce{D}$. So roughly equal amounts of both products are formed.
In $\mathrm{E1}$ reaction, the rds involves breaking of $\ce{C-H}$ bond. Since $\ce{C-H}$ bond is easier to break than $\ce{C-D}$ bond, so major product is the one where $\ce{C-H}$ bond is broken during elimination.
Now let's look at the options one by one -
A) Since the product with broken $\ce{C-H}$ is major, it means it is $\mathrm{E2}$.
B) Same reasoning as above.
C) Here we see an $\ce{\alpha-H}$ being replaced by $\ce{D}$, meaning a carbanion is formed as intermediate due to very strong base ethoxide. So it is following $\mathrm{E1cB}$ mechanism.
D) They have said that rate is same for both $\ce{\alpha-H}$ and $\ce{\alpha-D}$, meaning $\mathrm{E1}$ mechanism is followed. $\mathrm{E1}$ is first order.
Disclaimer: I have been told earlier that the concept of rds is inaccurate by a very experienced user here. I agree with him. This answer is from a purely theoretical point of view, and strongly in the context of JEE only, where rds is useful and prevalent.

@Jay Out of the given options in column 2, the only mechanism with the possibility of H replaced by D is E1cb. You will agree that since the question asks for it the reaction must be following some mechanism. The H-to-D replacement rules out the possibility of E1 and E2. The only choice you have is E1cb. I'm sure you must be familiar of how frequently option-elimination is used in JEE MCQ's to get the correct answer ;-)

@Jay also, during E1cb the species with $\alpha-H$ and $\alpha-D$ are in equilibrium with the intermediate carbanion. Since C-D bond is stronger, it is favourable to form C-D bond from the intermediate carbanion and hence the concentration of species with $\alpha-D$ becomes higher, so major product is the alkene with H removed (eliminated), and D retained.

Please use $\ce{...}$ to render chemical formulae/equations. The default $...$ gives italic letters, and chemical formulae shouldn't be italic. Also, rds is very relevant in organic chemistry, it's not just in JEE. It's true that in some rare cases you can't identify a rds, but it works for most simple reactions.

@ShoubhikRMaiti Thank you for the tip on formatting - I was mainly trying to write $\alpha$ and forgot all about the italics. Also, I used to believe (and continue to believe, for I have to use it in Advanced) that r.d.s. is a relevant concept. You might see this question - https://chemistry.stackexchange.com/questions/152737/electron-density-and-reactivity-order-in-electrophilic-and-nucleophilic-addition for why I put the disclaimer. I feel electrophilic and nucleophilic additions are fairly simple reactions to work with rds.

@ShoubhikRMaiti If in the above question by me, you can't see the linked question, it's because that question got closed and deleted. In that question, I had given an answer, but Martin, a moderator here, pointed out that rds was inaccurate and downvoted my answer, without any further explanation (If you can give an answer to my question, that I have originally linked, it would be most helpful!)

I saw your question, and I think your argument with rds is completely right. There is nothing more for me to answer.

@ShoubhikRMaiti Thank you so much for the clarification!

I am so confused on why it is E1CB for option C

@Buraian - H is replaced by D. The D obviously comes from the solvent, $\ce{D2O}$. Hence it must join as $\ce{D+}$. Now to accept a positive ion, you must have a negative charge on the substrate that can bond with the positive charge. Hence in the previous step, we have a negative charge appearing on the carbon atom, which is a carbanion - hence making it an E1cb mechanism.


text:
Why doesn't the graph of radial part of wavefunction of 2p orbital change its sign?
Below I have given the graphs of radial part of the wave functions of $\mathrm{2s}$ and $\mathrm{2p}$ orbitals respectively.
The graph of the $\mathrm{2s}$ orbital makes sense. It has a radial node and about that radial node the wave function changes its sign. We can also see this in the actual diagram of the wave function of $\mathrm{2s}$ orbital.
But I am slightly confused with the graph of the radial part of wave function of the $\mathrm{2p}$ orbital. We know that $\mathrm{2p}$ orbital has two lobes. For instance, if we take $\mathrm{2p_x}$ orbital, it has two lobes, one lobe in the positive side of the $x$-axis and another lobe in the negative side of the $x$-axis. Thus the $2p$ orbital wavefunction clearly changes its sign about the origin of the $x$-axis. It changes from positive to negative sign at some point. Then why is the radial part of the wavefunction of the $\mathrm{2p}$ orbital showing that it is not changing its sign at any point?
Why is the radial part of the wave function showing us that the $\mathrm{2p}$ orbital never changes its sign? It clearly does. Can someone please explain?

First, consider what you have plotted. You plot R as a function of r. How is your x-axis supposed to differentiate between the top lobe of your 2p orbital and your bottom lobe, which both have the same values of r (but different values of $\theta$ and $\phi$? The next question you may want to consider is: the wavefunction is R (radial part) times Y (angular part). Which of these two parts is actually contributing the minus sign to the wavefunction in the other lobe? Is it R or Y?

The wave function of one of the lobes is positive and the wave function of the other lobe is negative. Then why doesn't this change of sign get reflected in the graph of the radial part of the wave function of 2p orbital

You are exactly right. But just imagine how can only the radial part differentiate between the two lobes of, let's say for consistency, the $\mathrm{p}$ orbital. Each lobe of the $\mathrm{p}$ orbital has same radial distance from the origin and thus their difference cannot be judged by the radial part of the wavefunction.
To account for this change in sign you have to look at the complete wavefunction which also takes into account $\theta$ and $\phi$.
For example look at the radial wavefunction diagram and the orbital contour diagram of $\mathrm{2p}$ orbital from Radial and Angular Parts of Atomic Orbitals[1]
Radial wavefunction(green curve):

Contour diagram:

As can be seen from the diagrams, for a given $r$ and $\phi$, $\theta$ is the deciding factor whether the sign is positive or negative.
So to account for the complete characteristics of an orbital you have to take into account the complete wavefunction and not it's part.

But now you might ask why the radial part of the $\mathrm{2s}$ orbital does describe the change in its sign.
The answer is that $\mathrm s$ orbitals are radially symmetrical and thus their wavefunctions are independent of $\theta$ and $\phi$, leaving only $r$ as the parameter which can lead to the change in sign.

Reference:
(1)   Libretexts. Radial and angular parts of atomic orbitals https://chem.libretexts.org/@go/page/1712 (accessed Jun 19, 2021).


text:
Are all chiral carbons sp3 hybridised?
I had this doubt, whether all asymmetric chiral carbons must necessarily be $\mathrm{sp^3}$ hybridized. For sure, we know that double or triple bonded carbon atoms don't shows such chirality, but is there a  possibility for carbon atom, forming 3 bonds, and having a positive charge over it? Such a carbon atom has $\mathrm{sp^2}$ hybridization, but my query is that, will such a molecule show chirality?

You know that an $sp^2$ carbon is flat, right?

Derivates of allene $\ce{H2C=C=CH2}$, like $\ce{ClCH=C=CHCH3}$ are chiral.

1,3-dimethylallene (or whatever its IUPAC name is), maybe?

Allenes don't really have 'chiral carbons' per se; the chirality element is not a point but an axis. I'm not sure whether the Gold Book has a proper entry on this, but it should.

There is a terminology problem here. Not all chirality derives from single carbon centres. For example, octahedral 6-coordinate metal centres can be chiral with bidentate ligands. Many organic molecules are chiral despite having no chiral carbons (asymmetrical allenes have been mentioned, but also look at hexahelicene).

Or look at a biphenyl with all four ortho positions substituted. The two rings occupy different planes due to steric repulsion then, and you get a chiral molecule.

There are three terms that define chirality of a molecule. They are: chirality axis, chirality plane and chirality center
According to the IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"), the definitions of these terms are as follows:

Chirality axis
An axis about which a set of ligands is held so that it results in a spatial arrangement which is not superposable on its mirror image. For example with an allene $\ce{abC=C=Ccd}$ the chiral axis is defined by the $\ce{C=C=C}$ bonds; and with an ortho-substituted biphenyl the atoms $\ce{C-1}$, $\ce{C-1'}$, $\ce{C-4}$ and $\ce{C-4'}$ lie on the chiral axis.

Chirality center
An atom holding a set of ligands in a spatial arrangement which is not superposable on its mirror image. A chirality centre is thus a generalized extension of the concept of the asymmetric carbon atom to central atoms of any element, for example $\ce{N+abcd}$, $\ce{Pabc}$ as well as $\ce{Cabcd}$.

Chirality plane
A planar unit connected to an adjacent part of the structure by a bond which results in restricted torsion so that the plane cannot lie in a symmetry plane. For example with (E)-cyclooctene the chiral plane includes the double bond carbon atoms and all four atoms attached to the double bond; with a monosubstituted paracyclophane the chiral plane includes the monosubstituted benzene ring with its three hydrogen atoms and the three other atoms linked to the ring (i.e. from the substituent and the two chains linking the two benzene rings).

According to these definitions, only an $\mathrm{sp}^3$ carbon can act as a chirality element. However, this doesn't mean that only organic compounds with $\mathrm{sp}^3$ carbons form chiral compounds. There can be a chiral axis (as in the case of allenes) or even a chiral plane (as shown in the figure below).

but is there a possibility for carbon atom, forming 3 bonds, and having a positive charge over it? Such a carbon atom has sp2 hybridization, but my query is that, will such a molecule show chirality?

You need to realize what a positive charge really is. It is a lack of electrons and so this means that there is no fourth group attached to the carbon. Its just free space which seeks electrons in order to complete its octet. So, the answer is no. such an $\mathrm{sp}^2$ carbon cannot be a chiral carbon. However, we do have a case where we can notice a chiral element in a compound that contains an $\mathrm{sp}$ carbon - asymmetric allenes. Note that here the element causing the chirality is not the carbon. but it is the axis that is aligned on the carbon.


text:
What happens to water and oxygen produced due to dissociation of nitric acid?
Nitric acid is yellow in colour due to the dissolution of nitrogen dioxide gas.
The nitrogen dioxide is obtained due to thermal dissociation of a portion of nitric acid:
$$\ce{4HNO3 -> 2H2O + 4NO2 + O2}$$
The nitrogen dioxide produced after dissociation dissolves in $\ce{HNO3}$ but what happens to the $\ce{H2O}$ and $\ce{O2}$ produced?
Shouldn't some $\ce{HNO3}$ dissolve in water as it is highly soluble? And if it does dissolve than what happens to the dissolved nitric acid?
In my book nothing is given as to what happens to the $\ce{H2O}$ and $\ce{O2}$.
So what exactly happens to $\ce{H2O}$ and $\ce{O2}$?

Having done this many years ago, using an actual retort, you get nitric acid in the chilled receiving vessel. Any oxygen produced the equation you gave simply goes into the atmosphere. Same for some water vapor; the remaining water is simply in the receiving vessel.


text:
How to dissolve brass surrounded by aluminium?
I have a broken off brass needle stuck in a caste aluminum hole. (It’s the pilot mixture jet on a carburetor).
I was wondering if there was a chemical that would dissolve the brass while not reacting to the aluminium.
Previously I’ve tried:

Using thermal expansion and contraction to loosen it.
Attempting to poke it out.
Using an air compressor to blow it out.

Concentrated nitric acid will dissolve common brasses, but not aluminum. But, even if it works well, it would be slow going for a needle in a small hole. Good luck!

Not likely. Ammonium hydroxide will attack brass, but at high concentrations and not very fast. And worse, I don't easily find data for aluminium in concentrated $\ce{NH4OH}$, probably because it has poor resistance.
I think you are stuck with mechanical options. Simplest is another carb or subsection. Drilling it out is obvious. Strangely enough, I have bits down to possible 0.010 " diameter but I don't know where you would buy tiny drill bits; likely you will need a few. They use collets rather than chucks to hold them , so you also need a collet or two. Sounds like more expense than a carburetor.


text:
Why does the graph of the electrical conductivity of sulfuric acid/water solutions have this knee in the ~85%-~92% range?
This answer to an earlier question regarding the electrical conductivity of sulfuric acid provides a graph showing the conductivity of sulfuric acid/water mixtures ranging from 0% to 100% sulfuric acid:

(Image by Horace E. Darling in "Conductivity of sulfuric acid solutions" [Journal of Chemical & Engineering Data 9.3 (1964): 421-426.], via M. Farooq here at ChemSE.)
As can be seen, the conductivity of the solution rises smoothly from 0% to a peak at approximately 30% sulfuric acid, and declines thereafter.  However, at approximately 85% sulfuric acid, conductivity reaches a local minimum, after which it actually rises slightly with increasing sulfuric-acid concentration until reaching a local maximum at approximately 92% sulfuric acid, before again dropping off, more steeply, as the concentration of sulfuric acid in the solution continues to increase to 100%.
Why does the trend of decreasing conductivity with increasing sulfuric-acid concentration temporarily reverse in the ~85%-~92% range?

interestingly the plot in this question doesn't even go out that far in concentration

Presumably the concentration in the graph is percent by weight. If so, then it's interesting to note that an equimolar mixture of $\ce{H2O}$ and $\ce{H2SO4}$ represents a 84.5 wt% solution, whereas a 1:2 molar mixture represents a 91.6 wt% solution. I'll be surprised if this is purely coincidental.

Ať 85%, the ion pairs stop being solvated by water and start being solvated by H2SO4. Solvation means ion separation and conductivity increase.

The comment by Vikki made me dig even older papers. Since conductance (not conducitivity note that Darling is using an incorrect terminology from today's standards) is inversely related to viscosity, I thought there must a sharp change in the viscosity of sulfuric acid solution as a function of concentration. This guess is not that bad. This is from 1923 paper. Rhodes, F. H., and C. B. Barbour. "The viscosities of mixtures of sulfuric acid and water." Industrial & Engineering Chemistry 15.8 (1923): 850-852.
There is a sharp increase in viscosity at 85%, which indicates there is a major structural change in sulfuric acid solution in the range 85-92%. Sulfuric acid forms a hydrate in this range. When the viscosity is high, the conductance goes down, there is a depression in the curve. This viscosity jump is causing the double hump. Once we are past the high viscosity range, conductance goes up again.
It is amazing how simple molecules do not stop from surprising us!

just asked: Details of what actually happens in cold sulfuric acid between 80 and 90%; what molecular changes cause the viscosity to skyrocket?


text:
Is hydrogen peroxide a better oxidizing agent than chlorine gas?
If we compare the standard reduction potentials of $\ce{H_2O_2}$ with that of $\ce{Cl_2}$:

$\ce{H_2O_2 + 2H^+ + 2e^- <=> 2H_2O : E^o = 1.78V}$
$\ce{Cl_2 + 2e^- <=> 2Cl^- : E^o = 1.36V}$
(Source : Wikipedia)

From the above data, as value of $\ce{E^o}$ of $\ce{H_2O_2}$ is greater, its oxidizing power is greater than that of $\ce{Cl_2}$.
But if $\ce{H_2O_2}$ is a better oxidizing agent, why does it act as a reducing agent when reacted with $\ce{Cl_2}$ (in the following reaction)?

$\ce{H_2O_2 + Cl_2 -> 2HCl + O_2}$

The above reaction is feasible, but my doubt is why does it preferentially occur even though $\ce{H_2O_2}$ is the better oxidizing agent.
Could anyone please explain this?

The first half reaction is in acidic medium. I don't think it will take place and after all these are standard state values.

Because H2O2 is a reducing agent as well, so what?

@IvanNeretin So then, how can we determine that $\ce{H_2O_2}$ is going to act as a reducing agent here, instead of an oxidizing agent?

@Pal A point to note that( though I doubt whether it is useful here) just because a reaction is thermodynamically feasible does not necessarily mean, it will be observed necessarily.

It is a problem of activation energy, not of thermodynamics..


text:
dehydration of an alcohol (3-methyl-1-phenylbutan-2-ol)
I was thinking of dehydrating this alcohol:
3-methyl-1-phenylbutan-2-ol
Its structure is:

My concern is which of the following would it form:

The problem is that the first one comes from a stable benzyl secondary carbocation while the second one comes from a secondary carbocation but forms a more substituted alkene. Which one is preferred and why? (I feel that the second one should be preferred because of Saytzeff's (Zaitzev's) rule stating that more substituted alkene is formed)
Given answer was that the first one is formed (that too trans, which is I know why). But why the first one?

The double bond in the first product is in conjugation with the aromatic system which is energetically favoured

It's better to think of the Zaitsev's rule as the more stable carbocation is preferred. The thermodynamics of this reaction show that the stability of the final product  depends on the stability of the double bond formed..

isnt zaitzev violated then ? I have learnt that hoffmann product is major only in the following cases : (i) bulky base (ii) poor leaving group (iii) gamma carbon is neo carbon (iv) thermal elimination recations

@omjoglekar Zaitev rule isn't a law of compulsion. What happens, happens. Don't get bogged down it some rule of thumb fails, rather try learning the mechanism so that you can predict products with better accuracy.

Zaitsev's rule is useful when there are no other significant factors involved. Conjugation to an adjacent aromatic system is a significant factor.

Saytzeff rule's more substituted alkene is formed when all substituents are alkyl groups, so as to have more hyperconjugative structures and stabilize the double bond. In the general case, stability by resonance > stability by hyperconjugation. So a double bond resonating with the benzene ring is more stable.

Also, if I'm not wrong, dehydration of secondary alcohols proceeds by carbocation intermediate - in that case the initial carbocation would rearrange ot give the stable benzyl cation, not the other way (again, because resonance in benzyl cation gives more stability than hyperconjugation in the others). After rearrangement, you have little room for any other product except the first one.

@Waylander Are you sure? But the t-butyl cation is more stable than benzyl cation. See : https://chemistry.stackexchange.com/a/86692/121235

@omjoglekar   The second product do not come from 2° cation but  come from stable 3° - t(butyl) cation ( after a 1,2-hydride shift) awhich is more stable than benzylic cation. I suspect that your answer key is correct.

@An_Elephant   SLIGHTLYy more stable. They will be in equilibrium which allows the more energetically favourable alkene to form.

@Waylander  Similarly like alkenes, won't "more energetically favourable" compounds form always like alkene and the stability of these cations is useless?

@An_Elephant  Not necessarily. I this case the conjugation with the aromatic system is the most important factor. In alkyl systems the cation stability will be the major factor.

Zaitsev's rules are nothing but a set of rules that were used to explain experimental results that were observed. It was not the other way around.
The real rule you have to use here is the Hammond's Postulate. This states:

The transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy.

So, in this case, the transition state is closer to the final product and so resembles the product which is a double bond. Therefore, the more thermodynamically stable product is preferred.
In essence, even Zaitsev's rule is used to explain the same thing, the reason for a more substituted alkene being favored is because of the higher number of hyperconjugative structures that are made possible due to more groups attached.
Therefore in this case, the product formed will be 3-methyl-1-phenylbut-1-ene
(E) or (Z), since the stereochemistry is unknown.

But the t-butyl carbocation formed after a 1,2-hydride shift is more stable than a benzylic carbocation  formed after 1,2-hydride shift. See this : https://chemistry.stackexchange.com/a/86692/121235


text:
Heating effect on Orthoboric Acid
I am a high school student and was studying about the effects of heating orthoboric acid, and I saw two textbooks putting different statements.
One book says,

On heating above 370K, orthoboric acid forms metaboric acid, which on further heating forms boric oxide.
$\ce{H3BO3->[∆] HBO2->[∆] B2O3}$

Another book says the following reactions:

$\ce{H3BO3->[100°C]HBO2->[160°C]H2B4O7->[red heat]B2O3}$

I searched on Wikipedia, and it says the following:

Boric acid is soluble in boiling water. When heated above 170°C, it dehydrates, forming metaboric acid ($\ce{HBO2}$):
$\ce{H3BO3 → HBO2 + H2O}$
Metaboric acid is a white, cubic crystalline solid and is only slightly soluble in water. Metaboric acid melts at about 236°C, and when heated above about 300°C further dehydrates, forming tetraboric acid, also called pyroboric acid ($\ce{H2B4O7}$):
$\ce{4HBO2 → H2B4O7 + H2O}$
The term boric acid may sometimes refer to any of these compounds. Further heating (to about 330°C) leads to boron trioxide.
$\ce{H2B4O7 → 2 B2O3 + H2O}$

I am really confused. Which one of the three is correct? Is there any finer detail?

All are correct but 2 and 3 are complete whereas 1 is incomplete.

All of them are correct. In 2nd reaction, the "red heat reaction" was meant to be the reaction occurring at 330 °C although "red heat" has its own meaning: https://chemistry.stackexchange.com/questions/108167/meaning-of-red-heat

Also in 1st reaction, "further heating" meant that temperature is increased from 160 °C to 330 °C.


text:
Why is nitrate anion substituted but bromide anion is not in Von Richter reaction?
In Von Richter reaction which is a nucleophilic aromatic substitution. As  according to this source, there is only a 37% yield of the product.I want to know what are the competing side reactions which results in such a low yield of the desired product. I searched for it on the internet but couldn't find anything related to it.
Secondly, I want to know the reason behind the substitution of nitrate anion as the major product although bromide anion is a better leaving group according to this table taken from Wikipedia

wikipedia

please mention the reason of downvote so I can edit my post accordingly and not commit the same mistake in future posts

I would expect that cyanide displacement of Br is one of the major byproducts

I expected the same thing but it would be better if someone can give a reference for it

Well, I think you see yourself how bad capitalisation, punctuation and spelling you've got there.

i have improved it to the best of my ability further edits are always welcomed

Worth reading this paper  https://pubs.acs.org/doi/abs/10.1021/jo01149a007

@Waylander this is behind paywall, is there any essential info you can give if you have read it?

I have not read it, I do not have access. It seemed to me to be likely to have information that would answer your question

@Waylander this was an ancient paper so it was available on the net, thanks it did help.

The Von-Richter reaction's accepted mechanism is given by-

As you can see, this is not a simple case of say, removal of $\ce{NO2-}$ ion, instead it involves removal of $\ce{N2}$ which provides thermodynamic stability.
the tables you use have been calculated on the basis of $\ce{pK_a}$ values which don't give the right answers in many cases due to solvent effects and the intermediates involved in specific reactions. Wikipedia itself states

It is important to note that the list given above is qualitative and describes trends. The ability of a group to leave is contextual. For example, in $S_nAr$ reactions, the rate is generally increased when the leaving group is fluoride relative to the other halogens. This effect is due to the fact that the highest energy transition state for this two step addition-elimination process occurs in the first step, where fluoride's greater electron withdrawing capability relative to the other halides stabilizes the developing negative charge on the aromatic ring. The departure of the leaving group takes place quickly from this high energy Meisenheimer complex, and since the departure is not involved in the rate limiting step, it does not affect the overall rate of the reaction

This order is a general trend and as the mechanism in this reaction suggests, is not the end-all be-all determining factor.
Coming to why this reaction gives low yield, is mainly due to the hydrolysis of $\ce{CN-}$ ion by water/alcohol to $\ce{HCOO-}$/$\ce{CH3COO-}$. This is was suggested because according to this paper$^1$

the recovery of much
unreacted starting material suggest that hydrolysis (or alcoholysis) competes
with the nitro compound for cyanide ion to the extent that hydrolytic destruction of cyanide is the yield-limiting factor; it is known that cyanide ion is rapidly
hydrolyzed in water solution at elevated temperatures

There is one thing missing from wikipedia that von-richter reactions were conducted in elevated temperatures and in sealed tubes $(\ce{150^oC-165^oC})$ both originally by von richter himself and in subsequent studies for mechanism*.
There may be other competitive side reactions but were not documented for this reaction (because it is of low yield anyway, there was not much work on it.)

*The paper by J.F.Bunnett,J.F.Cormack and Frank C.Mckay$^1$ (linked by @waylander) cites-

von Richter’s reactions were run in “alcohol” solution in sealed tubes at $\ce{180^oC-200^oC}$ or higher.

and the letter by Prof. Myron Rosenblum $^2$ while explaining the mechanism
with evidence from isotopic labelling cites-

9.5 mmoles of p-chloronitrobenzene, when treated
with 20 mmoles of potassium cyanide, and 6.2
mmoles of ammonium nitrate in a sealed tube at $\ce{160^oC}$ for
1.75 hours, gave nitrogen gas containing 0.75%
$\ce{N_2^29}$

It is normal for nucleophilic $S_nAr$ to be conducted around this temperature** so this is not surprising thing to be missed by wiki, it also means that $S_nAr$ of cyanide on bromine can probably be competitive but experiments predict a majority of unreacted products according to this table in most solvent cases:

interestingly both this table and the letter by Prof. Rosenblum (46.5% yield for the same reaction but with chlorine) suggest different yields to what wikipedia suggests.
TL;DR : Leaving group order doesn't matter here because of different mechanism involved converting $\ce{-NO2}$ to $\ce{-N2-}$ which is a very good leaving group and yield is low because of hydrolysis of $\ce{CN-}$ ion at the preferred temperature.
**Ref :. Morrison & Boyd sec 26.7 (Nucleophilic aromatic substitution: bimolecular displacement)

cited material:
$1$ : J.F.Bunnett, J.F.Cormack and Frank C.Mckay ,"Mechanism and reactivity in aromatic nucleophilic substitution reactions"J. Am. Chem. Soc. 1958 5(3), 481-490 doi 10.1021/jo01149a007
$2$ : Rosenblum, M. The Mechanism of the von Richter Reaction. J. Am. Chem. Soc. 1960, 82, 3796–3798; doi 10.1021/ja01499a090

@napstablook While aiming to bring the literature references into the style of J. Am. Chem. Soc. , the style recommended to use on chemistry.se, I noticed that reference #1 links to a publication of different year, journal and publisher than the one it (currently) states.  So, which one is reference #1?

@Buttonwood Thanks for pointing out! Apparently wikipedia had a different source cited  (both by J.F.Bunnett) I didn't think to cross-check. I will change the link

does the removal of bromine by SnAE can be a side reaction ?

@Samardeepsingh I have mentioned that while it  could have been, experimental results suggested that the reactant is largely unreacted (see the table) which means that it is mostly hydrolysis. Now for only few cases of glycol the yield of both product and the amount of unreacted reactant is low. I suspect this is because of the SnAE side reaction, however there is not much experimental observation about this reaction, since it is not very useful..


text:
Can you develop algorithms to predict synthesis of peptides?
This may be a very foolish question, but do (complex) algorithms exist to predict the steps and materials involved (or therefore different options and paths) in synthesis for the given input of for example a target molecular structure or formula? In this case for example peptides, so potentially larger molecules. But the question applies generally i guess.
If that makes sense, meaning there is a large amount of known patterns which overlap for a wide array of target at least in partial which can be applied or reverse calculated based on structure and known properties, i'm sure such things would be already available as products for the research, chemical or pharma industry.
Is there any public work for that thats mature, does it not exist to a large extend or is only available as very niche industry products?

This is a subject of considerable interest and effort in current research. There is a large article on Computer-guided retrosynthesis in the most recent issue of Chemistry in Britian. Have a look here: https://askcos.mit.edu/  and here  https://jcheminf.biomedcentral.com/articles/10.1186/s13321-020-00472-1

@Waylander Interesting, i had a feeling stuff likethis must either exist or being worked on already... i dont understand the full complexicity of chemical synthesis but i feel like it would be stupid if that wasnt the case. Thanks for providing these links!

"This may be a very foolish question" question wouldn't be bad if not for the example - peptides are pretty much strings of monomers - just a look at it should be enough to know how to make them (if you know general method) - there's nothing to "predict". Just adding one aminoacid after another.


text:
Dominant factors rendering very low solubility of large aromatic molecules
In terms of structural basis, alkanes are insoluble or immiscible in water because of such presence of C–H...O interactions that are in regime of van der Waals interaction, which tends to be regarded as dispersion contact rather than as hydrogen bond. Benzenes and other planar molecules can make stronger O–H...π interactions that render small solubility of the planar molecules.
Then, I found these statements from a lecture note (sadly I can not find the proper sources in the lecture notes and I can not find any good result with varying keywords in Google).

There are several steps of solubilization of a molecule:

Solutes must overcome solute:solute interactions in order to separate and hydrate solute molecules from one another.
Solvents must overcome solvent:solvent interactions in order to create a cavity within which a solute molecule can fit.
Both solutes and solvents must overcome solute:solvent interactions in order to reorganize its ice formation in the solvent shell.

From those above statements, it can be clearly seen that solubility of molecules is not suffice to be explained solely by looking only at single thermodynamic state of solute:solvent interaction.
Here are my questions:

To be honest, I have not found any paper mentioning crystal structure showing water orientation among larger planar molecules. Based on my intuition from above, the water molecules around planar molecules should orient to the planar surfaces because of presence of O–H...π interactions in hydrogen bonding regime. On similar reasonings, the water molecules around cyclic 3D (cyclopropanes, bicyclopropanes, etc.) should orient away from these cyclic molecules because of presence of C–H...O interactions in van der Waals regime. Are my reasonings correct?

If such solute:solvent interactions could make an ice network (or a cluster of water), does it mean thermodynamically solute:solvent interactions are trapped by themselves and can not proceed to make further solute:solute and solvent:solvent interactions? Can this particular solute:solvent interaction alone rationalize physical observation of very small solubility of larger planar molecules?

I find that this is a bit paradox for this case. Some drug discovery campaigns begin to design new chemotypes that have high sp3 characters because planar sp2 molecules have poor solubility and metabolization, although they have to put more efforts on making molecules with quaternary stereocenters. In this view, alkanes are seen to be more soluble than planar molecules although both of them should be insoluble in water. I feel that it is too confusing to say alkanes are both soluble and insoluble at same time. How should I make such words so that I do not make this confusion while keeping both chemical bonding and physical phenomenons are agree to each other.

Regarding the steps you list, I am confused because at least 1 and 2 seem to be the first steps for the process of hydration, not dehydration. Step 3 on the other hand could be argued to be a step involved in dehydration. Also keep in mind that the "iceberg" hypothesis (formation of an ice-like solvation layer) is not a universally accepted description of the structure of a hydration layer.

@BuckThorn The first step is the tendency for a planar molecule to make a stacking to other planar molecule (like π-π stacking), creating an aggregation that makes it is difficult to be solubilized although I am not too sure if colloidal aggregation fits into the context for the driving force of such solubilization of aggregates. I think the first step is a bit similar to sublimation of crystal lattice.

@BuckThorn I think so for the second step, the second step is like seeing two sides of a coin – a transition state to hydrate molecules generated from the first step and at same time, water network restructuring for the third step.

The enthalpy of dissolving in water is often approx zero or slightly negative, it is the entropy changes that cause insolubility.  Water has to bridge over the non-polar molecule which greatly restricts its H bonds.

Ok, I understand now that the third point in the top list is the last step, involving solvent rearrangement about a solute, but those three points then describe hydration (or solvation), not dehydration. It's a minor point but note that you may want to edit your post to make it clearer.

Your question is still unclear.  In particular I don't understand point 2 in the bottom. It seems to be: "can solute-water interactions be strong enough to cause a solute to be insoluble?". Where do you see an ice network forming?

@BuckThorn I have edited the first point. The first step is now mentioning hydration step explicitly.

@BuckThorn It is my hypothesis. Whether this network is due to entropy change or not, still there is no study exploring the basis of insolubility of planar molecules.


text:
Are spontaneous exothermic reactions spontaneous due to more stable products or to an increase in the entropy of the surroundings?
The lattice formation of an ionic compound from its infinitely separated gaseous ions is spontaneous under standard conditions, but I cannot figure out the 'driving force' behind this.
One way to look at this is that the charge particles try to minimize the electric potential energy of the system (as F=-dU/dx)
However, you can also look at this from the point of view of entropy. The large amount of heat given off increases the entropy of the surroundings more than the decrease in the entropy of the system, thereby increasing the total entropy of the universe.
Both of these point of views are clearly not linked as one is statistical while the other depends on the nature of the electric force.
So is the actual driving force the spontaneity of the electric force or statistical probability?

The energy could have been released in another and have the same effect on entropy.

Exothermic = you get products with lower enthalpy, which drives reaction. Entropy can hinder it or not - it's a tangential issue.

Gibbs free energy essentially gives an indication of the change in total entropy of the universe

So is the actual driving force behind the spontaniety the electric force or statistical probability?

The driving force has to be the entropy, i.e., statistical probability.
The first law (conservation of energy) provides us no way to distinguish between the favorability of two different states of a system. It only tells us energy is neither created nor destroyed—it only changes form. It doesn't tell us if one form is more favored than another.
Thus, to determine whether a change will occur, i.e., to determine if one state is favored over another, we have to look at the second law (the entropy of the universe always increases for any macroscopic process).  This does distinguish between the favorability of the intial and final states.  It's the second law that provides an arrow of time for the universe.
So we can understand why oppositely-charged ions spontaenously combine at a given temperature as follows:  At that temperature, the gain in the entropy of the surroundings, due to the transformation of potential energy into thermal energy, is greater than the loss of the entropy of the system due to the ions combining.
At higher T, for the same heat flow, the gain in entropy of the surroundings will be less:
$$\Delta S = q_{rev}/T$$
Eventually we will reach a T at which, for this process, the entropy increase of the surroundings is less than the entropy loss of the system.  At that T, the ions will remain separated.*
*The test case being considered here is a phase change so, except at the phase coexistence temperature, it's either all one phase or all the other.  With a chemical reaction it's more complicated, since you can have an additional entropy term, the entropy of mixing, and thus the relative concentrations of the reactants and products will change as T changes, rather than going from all of one to all of the other.

Two charged particles spontaneously move towards each other because of the nature of the electric force; it's not even possible to apply the concept of entropy here as there are only 2 particles. Why can't we apply the same concept to a large number of ions? Why is it necessary to invoke entropy?

Good question!  Let's turn it around and ask:  Given we can't apply entropy to two individual oppositely charged particles, how do we explain why they will spontaenously combine?  They answer is: they won;t!  Here, as you said, entropy doesn't apply, so we only have conservation of energy.  As the particles move towards each other, their PE is converted to KE.   When they collide, all that KE will be converted back to PE (deformation of the electron cloud), which will then drive them apart.  I.e., they will repeatedly collide and bounce apart....

.....This is the time-reversal symmetry we see at the level of individual particles.   Only in the presence of other particles, to which they can transfer some of that exces energy, will they remain stuck together.

Thank you. I think I found the error in my understanding. I was under the impression that all chemical phenomenon were just governed by electrostatics and forgot that the decrease in potential energy always resulting in an increase in k.e that must be dissipated as heat

@theorist two colliding charged particles might radiate light.

@BuckThorn  You're right—charged particles should radiate EM when they are accelerating towards each other, suggesting they should continuously dissipate energy as they move apart and together.   I need to think about this some more.

Both of these point of views are clearly not linked as one is statistical while the other depends on the nature of the electric force.

They are linked by conservation of energy. If the reaction is exothermic (and there is no non-PV work), it will heat up the environment, and this helps to make the product state more likely than the reactant state. Another way of describing entropy increase is a dissipation of energy or matter. For an exothermic reaction, the energy once located in the reactants is now spread out over the environment (which might be a large water bath or something of that nature).

Are spontaneous exothermic reactions spontaneous due to more stable products or to an increase in the entropy of the surroundings?

You are missing half of the necessary description. You have to consider not only how the entropy of the surrounding changes, but also how the entropy of the system (or the reacting species) changes. Otherwise, you would falsely assume that all exothermic reactions are spontaneous, which they are not.


text:
Understanding the Osmotic Pressure formula
$$\Pi = iMRT$$
Can this be rewritten as $\Pi = ORT$ where $O =$ osmolarity? The only reason I ask is because my book doesn’t talk about the relationship between osmolarity and molarity, but I came across a practice problem where I had to find the minimum pressure required to purify sea water given the osmolarity of the seawater and temperature (no compound was given).
Also, this practice problem got me thinking about the relationship between osmolarity and molarity. Without knowing what the compound is in the seawater, there’s no way to determine the molarity, correct?
A compound like glucose would have a van’t Hoff factor of 1 so it’s molarity would be equal in value to the osmolarity. However, a compound like $\ce{NaCl}$ would have a molarity equal to half the osmolarity.
Here is the word problem for context:

Reverse osmosis is a process that allows fresh water to be obtained by using pressure to force an impure water source through a semi-permeable membrane that only allows water molecules to pass. What is the minimum pressure that would be required to purify seawater at $\pu{25 ^\circ C}$ that has a total osmolarity of $\pu{1,000 mOsm L-1}$?

I don’t want an answer to this problem because I have the answer and I know how to calculate it. But if it helps to answer the general question, please feel free to use the problem as a framework for a more general answer.

What are you exactly looking for ? You have a problem. You know the solution, and you know the formula to go to this solution. What do you want us to tell you ?

I'm asking if this formula is correct: Π = ORT. I'm also asking if it is possible to find the molarity of the seawater from the given information in the problem or more generally if it is possible to find the osmolarity of a solution without knowing the solute.

Your formula is correct. You can always calculate the molarity of a solution if you know the amount of substance dissolved in a lieter of solution and also you know the formula of the solute. For instance, if you dissolve $\pu{1 mol}$ of sugar, or of alcohol, or of glycerol in water and the solution made upto $\pu{1 L}$, you obtain a solution which is $1$ molar $(\pu{1 M})$, and also $1$ osmolar $(\pu{1 Osm})$, because these substances do not dissociate in water (no ions).
If you dissolve $\pu{1 mol}$ $\ce{ NaCl}$ $(\pu{58.5 g mol-1})$, or $\pu{1 mol}$ $\ce{KNO3}$ $(\pu{101 g mol-1})$ in $\pu{1 L}$ solution, you obtain a solution which is $1$ molar, but $2$ osmolar, because each of $\ce{NaCl}$ and $\ce{KNO3}$ produce two ions in the solution. The osmolarity O is $2$ osmol/L in such solutions.
If you dissolve $\pu{1 mol}$ $\ce{CaCl2}$ $(\pu{111 g mol-1})$ or $\pu{1 mol}$ $\ce{Ba(OH)2}$ in $1$ liter water, you obtain a solution which is $1$ molar and $3$ osmolar, because each of $\ce{CaCl2}$ or $\ce{Ba(OH)2}$ produce three ions in solution. The osmolarity O in such a solution is $3$ osmol/L. This calculation is rather theoretical, because it it not possible to dissolve such a huge amount as $\pu{1 mol}$ $\ce{Ba(OH)2}$ in one liter water.

Thank you so much, this is what I thought but it’s nice to be sure. So for this problem, there is no way to find the molarity of the seawater prior to the reverse osmosis?

The sea water may contain different amounts of salt, depending on the point on earth. If the sea water contains $30$ g salt per liter, its molarity is $30/58.5 = 0.512 $ mol/L. And the osmolarity O is the double, which is $1.025$ osmol/L


text:
Does 3% hydrogen peroxide neutralize household bleach?
Wiki and lots of sources all say the same thing. "1 part of 3% hydrogen peroxide to 10 parts water". But I am not trying to neutralize water I want to neutralize laundry bleach. It could mean add 1 part hydrogen peroxide to 10 parts water and then use that to neutralize your bleach product but without knowing the  concentration of sodium hypochlorite that the mixture was intended for that would not make sense either. The directions are a complete mystery no matter how you look at them. So I will ask in the following manner to make it clear.
We have a gallon of 3% sodium hypochorite household bleach.  How much 3% hydrogen peroxide should be added to neutralize it and what will the neutralized product be that remains? The chemical reaction formula would be nice but I might be asking to too much here.

Does this answer your question? How to calculate the volume of oxygen in a reaction between hydrogen peroxide and bleach using this apparatus

Out of curiosity, is there any particular reason to not just use up laundry bleach by using it in the laundry? After all, white clothing dinginess, like rust, never sleeps.

Never mix bleach under any circumstances or for any reason. If you want to get rid of it call your county's  hazardous waste disposal department  and they can give you directions of where you can take it for safe disposal.  You can also call the manufacturer of the product and they can give you directions for disposing of it in the toilet if it is under a certain amount AND copious amounts of water are used with multiple flushes. They can give the specifics.

Never mix bleach with anything. As far as hydrogen peroxide goes the combination can explode from what I read. To dispose of it call your county hazardous waste department.

......or use it in your laundry.

Wiki and lots of sources all say the same thing. "1 part of 3% hydrogen peroxide to 10 parts water."

When talking about aqueous solutions, the concentrations can be expressed as percent by weights, $\% (w/w)$, or by volume, $\% (v/v)$, or by weight in parts of solute by weight in 100 part of solution by volume, $\% (w/v)$. The above statement in the question means if you take 100 part of the solution by weight (e.g., $\pu{100 g}$ of solution), it contains 3 part of hydrogen peroxide and 97 parts of water by weights $\left(3\% (w/w)\right)$. It also can mean that it may contain 3 part of hydrogen peroxide to 97 parts of water by volumes $\left(3\% (v/v)\right)$ if the lebel doesn't say it is whether $\% (w/w)$ or $\% (v/v)$. The third kind is 3 part of hydrogen peroxide by weight to 100 parts of solution by volume $\left(3\% (w/v)\right)$, which is common for prepared chemical reagents to use in the laboratories.
Commercial $\ce{H2O2}$ solution is usually in $\% (v/v)$. However, when prepared as a chemical reagent, it is $\% (w/v)$. That means $\pu{100 mL}$ of $3\% (w/v)$ $\ce{H2O2}$ solution has $\pu{3 g}$ or the molarity of the solution:
$$3\% (w/v) = \frac{\pu{3 g} \ \ce{H2O2}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 g} \ \ce{H2O2}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{30 g} \ \ce{H2O2}}{\pu{34.01 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{0.882 mol L-1}$$
On the other hand, since commercial $\ce{H2O2}$ solution is usually in $\% (v/v)$, and density of pure $\ce{H2O2}$ is $\pu{1.450 g mL-1}$ at $\pu{20 ^\circ C}$ (Wikipedia):
$$3\% (v/v) = \frac{\pu{3 mL} \ \ce{H2O2}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 mL}\cdot \pu{1.45 g mL-1} \ \ce{H2O2}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{43.5 g} \ \ce{H2O2}}{\pu{34.01 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{1.28 mol L-1}$$
Household bleach is aqueous $\ce{NaClO}$ $(\pu{74.44 g mol-1})$ solution, sometimes included certain percent of $\ce{NaOH}$. If you have $3\%$ solution, you can do the same calculations as above to find out the correct molarity if the solution is given whether $3\% (w/v)$ or $3\% (v/v)$ (and it is without $\ce{NaOH}$). Usually, it is always given in $\% (w/v)$. Sometimes, the default densities may be provided instead of percentages by the manufacturer, which may differ for the different manufacturer since the manufacturer may change the density by adding salt or other chemicals to the solution. Therefore, it is hard to calculate molarity using provided default density if you don't know what else other than $\ce{NaClO}$.
Let's consider the solution you have is $3\% (w/v)$ $\ce{NaClO}$:
$$3\% (w/v) = \frac{\pu{3 g} \ \ce{NaOCl}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 g} \ \ce{NaOCl}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{30 g} \ \ce{NaOCl}}{\pu{74.44 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{0.403 mol L-1}$$
The reaction is:
$$\ce{NaClO + H2O2 -> NaCl + H2O + O2}$$
Accordingly, $\pu{1.0 mol}$ of $\ce{NaOCl}$ will react with $\pu{1.0 mol}$ of $\ce{H2O2}$. Therefore, the volume $(V_\ce{H2O2})$ of $3\% (w/v)$ $\ce{H2O2}$ needed is react completely with $\pu{1.0 L}$ of $3\% (w/v)$ $\ce{NaOCl}$ can be calculated using $M_1V_1 = M_2V_2$ as follows:
$$M_\ce{H2O2}V_\ce{H2O2} = M_\ce{NaOCl}V_\ce{NaOCl} \ \Rightarrow \ V_\ce{H2O2} = \frac{M_\ce{NaOCl}V_\ce{NaOCl}}{M_\ce{H2O2}} \\ =  \frac{\pu{0.403 mol L-1}\cdot \pu{1.0 L}}{\pu{0.882 mol L-1}} = \pu{0.457 L}$$
Therefore, since you have one US gallon of bleach, which equals to $\pu{3.785 L}$, you need $0.457 \times 3.785 = \pu{1.73 L}$ of $3\% (w/v)$ $\ce{H2O2}$ solution.
Your final solution is $\frac{0.403}{1.457} = \pu{0.277 mol L-1}$ $\ce{NaCl}$ solution (of course, only if manufacturers did not add additives).

The equation is: $\ce{NaClO + H2O2 -> NaCl + H2O + O2}$
A $3\%$ sodium hypochlorite contains $\pu{30 g}$ $\ce{NaClO}$ $(\pu{0.40 mol})$ per liter. So $\pu{0.1 mol}$ $\ce{NaClO}$ is contained in $\pu{250 mL}$ bleach.
A $3\%$ hydrogen peroxide contains $\pu{30 g}$ $\ce{H2O2}$ $(\pu{0.88 mol})$ per liter. So $\pu{0.1 mol}$ $\ce{H2O2}$ is contained in $\pu{114 mL}$ of your hydrogen peroxide solution.
So mix $\pu{250 mL}$ bleach $+$ $\pu{114 mL}$ of your hydrogen peroxide solution, or any multiple or submultiple, like $1$ liter bleach and $\pu{456 mL}$ peroxide solution.
You'll obtain a dilute salt solution.


text:
Are states of matter distinguishable clearly?
Can we say a state of a matter is exactly only one of solid, liquid or gas? Can't we say something is more "liquid-like" or less "solid-like?" What is a state of slime?

My answer would be "yes". "State of matter" does not refer to only liquid, gas and solid. OTOH this question could be interpreted in other ways, for instance it is unclear whether it is meant "are states of matter always distinguishable". This is somewhat different from the Q which was deemed a duplicate, because the focus could be placed on the word clearly. It is highly unclear whether this is meant as a universal question (about theoretical definitions) or particular cases (experimental determination of the state of matter).

Every man-made classification is problematic because there are no sharp boundaries among various classes. For elementary-level discussion solid-liquid-gas classifaction is fine, but things are more complex in real life.
Slime is a famous non-Newtonian fluid, which is just a fancy way of saying that this "stuff" can flow; hence it is a fluid, but its viscosity changes with flow rate, hence the fancy term "non-Newtonian." For example, if you have ever made a soup thickener with corn starch, the spoon gets jammed in the cup if you quickly move the spoon. You feel much resistance, but if you slow down, it is liquid dispersion again. Notice we have carefully avoided liquid, or solid here, because slime is colloid (a dispersion of a solid in liquid) if you are talking about corn-starch dispersed in water.
One may ask what is the state of smoke? It is neither solid, liquid or gas. It is a colloidal dispersion of solid particulates in gas.

Definitely not. Not only do liquids and gases merge above the critical temperature and pressure, we also cannot distinguish solids from either liquids or gases by simply stating that solids do not flow. Because, under the right conditions, ultimately they do. As with many things in science, the apparent distinctions we usually see are in fact idealizations.


text:
What is 'non-expansion work' and how does ΔG represent the maximum non-expansion work that can be extracted in a reversible process?
I have just learned about Gibbs free energy and how its value indicates the change in entropy of the universe and can thus be used to find the spontaneity of a reaction. However, I found this other interpretation of Gibbs free energy on Wikipedia that I don't get at all.

The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamically closed system (one that can exchange heat and work with its surroundings, but no matter). This maximum can be attained only in a completely reversible process. When a system transforms reversibly from an initial state to a final state, the decrease in Gibbs free energy equals the work done by the system to its surroundings, minus the work of the pressure forces.

What exactly does 'non-expansion work' mean in the context of, say, a gas? How can it even do work other than by expanding or contracting?
Secondly, I thought the extractable energy during a reaction was ΔH( heat transferred during a physical or chemical reaction carried out at constant pressure). If that is actually ΔG=ΔΗ-TΔS instead, what kind of energy does 'TΔS" represent? I know that it is equal to the heat transferred into or out of the system if the process is done reversibly but I don't understand how that's relevant here.
Lastly, Wikipedia mentions that this maximum can only be extracted by a reversible process. How exactly would you extract this? For example, if a reaction has a ΔΗ=-50 and ΔG=-75, how do you get the extra 25 units of energy?

Regarding your second question you may find answers in many related Q&As on the site, for instance: https://chemistry.stackexchange.com/questions/16687/thermodynamically-reversible-processes-how-does-an-infinitesimal-step-make-a-pr?rq=1

See also: https://chemistry.stackexchange.com/questions/112900/gibbs-free-energy-and-maximum-work?rq=1

https://chemistry.stackexchange.com/questions/33816/maximum-useful-work-from-a-process?rq=1

https://chemistry.stackexchange.com/questions/29180/what-is-the-precise-definition-of-maximum-work-in-the-context-of-electrochemistr?rq=1


text:
The effect of increasing concentration of a cathode solution on the voltage in a galvanic cell
I am having trouble understanding a concept. In a galvanic cell, where electrolyte A is zinc (in a solution of zinc nitrate) and electrolyte B is copper (in a solution of copper (II) nitrate), why does voltage increase when the concentration of the copper solution increase? I know that copper is a cathode while zinc is an anode and the cathode solution is the one with the increasing concentration. I also believe that a redox reaction and equilibrium are involved in this concept but not understand how they apply?
Overall, why does voltage increase when the concentration of the cathode solution increase? what does the redox reaction have to do with this?
If it helps, the solution in Salt Bridge is 2.00 M Sodium Nitrate.

Look at the Daniell cell in my answer here. The Nernst equation, given directly under the cell schematic, gives the cell potential. So, as per the math, increasing the copper ion concentration increases the cell potential.

Saying "electrolyte X is " is wrong description.

There are a few things you can look at. First, we can write out the reaction as a whole: $\ce{Zn + Cu^2+ \rightarrow Zn^2+ + Cu}$
the nernst-equation is given by:
$\ce{E = E^\circ} - \ce{\frac{RT}{nF}lnQ}$, where $\ce{Q = \frac{[Zn^2+]}{[Cu^2+]}}$. We can see that if we increase $\ce{[Cu^2+]}$, we decreases the log value which then increases $\ce{E_{non-standard}}$.
You can also just think of it like this. Since $\ce{Cu^2+}$ is in the reactants, increasing its concentration will shift the equilibrium to the right by Le-Chatelier's principle.