text:
Effect of magnetism on dissolved Na+ and Cl- ions in water
Suppose we have put some salt in water. My state of knowledge is, that a hydration envelope is formed from the oxygen side of the water molecules on the Na+-Ions and one another from the hydrogen side of the water molecules on the (Cl-)-Ions to drag these out of the crystal.
My questions now are:

When the hydration envelope is formed on the ions, does it (partially) change the charge of the ions?

(Now the main question) If I would put two magnets with opposite poles on both sides of the container, in which the salt solution is, would the ions be pulled to the side of the opposite pole, (if yes or no, then why)?

I made a little sketch an let the hydration envelope out for the sake of the little place.
And one another thing. There is no reason to downvote my question, because I am not playing around here, but the question just jumped in my head

I am not certain, but I think charges themselves are not affected by a magnetic field. And Na+ and Cl- both are closed shell i.e. no unpaired electrons, so they are diamagnetic. So overall your ions won't be attracted by magnets...

... because remember, charges are affected by a magnetic field only if they are moving. If your electrolytic solution flows in an orderly fashion, and you apply a magnetic field, only then you will see movement of the ions. This is called Hall effect (https://en.wikipedia.org/wiki/Hall_effect)

Do you mean charges rather than magnets? That’s basically the principle of electrophoresis

There are a couple of issues: first, a uniform magnetic field doesn't exert a force on stationary charges and secondly, when the charges are moving the force is at right angles to the magnetic field. There are a few ways to get the charges moving. You could have an electric current flow from the bottom of your figure to the top. The $\text{Na}^+$ ions would move with the current, up the page and since the magnetic field points to the left in your figure, the $\text{Na}^+$ ions would be pushed towards the reader by the right hand rule resulting in higher electric potential near the reader. This is called a positive Hall effect. The $\text{Cl}^-$ ions would move against the current, down the page and paradoxically would be pushed towards the reader again by the right hand rule, this time resulting in a lower electric potential near the reader: a negative Hall effect.
Another way is to have the electric current directed out of the page. Positive $\text{Na}^+$ ions move towards the reader and then are pushed down the page by the magnetic field according to the right hand rule. Negative $\text{Cl}^-$ ions move away from the reader and are also pushed down the page by the magnetic field via the right hand rule. This is the principle of magnetohydrodynamic propulsion.
Another way is to have the brine flowing up the page which forces $\text{Na}^+$ charges towards the reader and $\text{Cl}^-$ charges away from the reader by the right hand rule, both resulting in higher electrostatic potential on the near side of the stream. This would be magnetohydrodynamic propulsion in reverse operating as a generator.

+1 for a nice answer, I was about write about the Lorentz force but you already addressed it. If the velocity is zero, then the force is also zero.


text:
Is light a reactant in photochemical reactions?
According to IUPAC a photochemical reaction is a reaction caused by absorption of light. I still can't understand how we should consider light as a part of a reaction. Is a catalyst or a reactant?
In some reactions there is the $hν$ notation above the reaction arrow. For example we may have the following reaction:
$$\ce{A ->[$hν$] B}$$
I want to understand how light affect reactions and chemical equilibriums. I read somewhere about photoisomerization and that favors the less thermodynamic stable product and that make me think of how is it possible for the value of equilbrium constant to change (considering interoconversion as a "reaction").
Light in photoabsorption
When a molecule absorbs a photon it moves to an excited state. This process is associated with a probability factor, lets call it $P$. We can write such a transition as:
$$\ce{A + $hν$ -> A^*}$$
Suppose we have a box with molecules $\ce{A}$ that initially is a closed system. That is photons can enter the box. Now we use a laser and pump photons of specific wavelength into the box and after that we isolate the system, that is no energy can go in or out of the system. Does the above arrow $\ce{->}$ implies a reaction? I was thinking that if $P = 0.2$ then after the light enters the box we will have $[\ce{A}^*] = 0.2$ and $[\ce{A}] = 0.8,$ so the equilibrium constant will be $\displaystyle K = \frac{0.2}{0.8} = 0.25$ (I have omitted concentration units for simplicity).
But this doesn't make sense because first a new thermal equilibrium will be reached and the relative concentrations (populations) will be in accordance with Boltzmann distribution. Secondly if it is indeed a reaction then we must include light also. Also, the concentration $[\ce{A^*}]$ should depend on the incident light intensity.
So is photon absorption just a physical process and therefore the chemical equilibrium concept doesn't apply? I thought that if photoabsorption reaches an equilibrium such that:
$$\ce{A <=> A^*}$$
then for interconversion from cis to trans of a compound $\ce{A}:$
$$\ce{A_\textit{cis} <=> A_\textit{trans}}$$
by populating the excited state of the one isomer the equilibrium position should change. But still this doesn't make sense because Gibbs free energy change of the reaction take into account both ground and excited states of both products and reactants.
Light in chemical reactions
First the notation $hν$ can't be thought to be a catalyst as it doesn't make sense because isn't regenerated. For example, the chlorination of methane to yield chloromethane
$$\ce{CH4 + Cl2 ->[$hν$] CH3Cl +  HCl}$$
uses light to initiate the reaction. But it isn't regenerated in any other step so it shouldn't be a catalyst. Then how one should think about light in a chemical reaction? It is a reactant? In other words could we write the chlorination of methane in the following way?
$$\ce{CH4 + Cl2 + $hν$ -> CH3Cl +  HCl}$$
If that is the case can we find an equilibrium constant that includes the concentration of photons? Because for every reaction there must be a corresponding equilibrium constant according to thermodynamics.
I am asking the above because as I said I read about photoisomerization and I couldn't understand how is it possible to favor a thermodynamically less stable product via radiation.

Comments are not for extended discussion; this conversation has been moved to chat. (May I suggest to any others who are tempted to comment here: if you feel your comment should be seen by others -- and many comments should -- then please post an answer, or comment on whichever answer is relevant to your point. If it isn't, then post it in chat, or don't post it.)

It is neither reactant nor catalyst, and equilibrium concepts do not apply
There are processes other than photochemistry that to go in a direction away from equilibrium. They require mechanical or electrical work, and there is no established way to incorporate these into a chemical equation.
If we write a conceptual equation for charging a battery, it could look like this:
$$\text{empty battery} \ce{->} \text{charged battery}$$
The external power source is neither a reactant nor a catalyst; it is work done on the system, which serves to move the system away from chemical equilibrium.
If we write a conceptual equation for a refrigerator (or more generally, a heat pump), it could look like this:
$$\text{warm body + warm body} \ce{->} \text{cold body + hot body}$$
Again, the mechanical work done by the compressor is neither a reactant nor a catalyst, and it serves to move the system away from thermal equilibrium.
For both processes, you could not say that in the presence of work, an equilibrium is reached. It is just the opposite - we are moving away from the equilibrium.

[OP] So is photon absorption just a physical process and therefore the chemical equilibrium concept doesn't apply?

Photochemical reactions can move a reaction away from equilibrium (as in the example of cis/trans isomerization the OP mentions). This makes it a non-equilibrium process, so equilibrium concepts need to be expanded. If you want to describe the situation under certain irradiation conditions, you might use the term photostationary cis:trans ratio (see section on stilbene in https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/photchem.htm).

[OP] I read somewhere about photoisomerization and that favors the less thermodynamic stable product and that make me think of how is it possible for the value of equilbrium constant to change (considering interoconversion as a "reaction").

The equilibrium constant does not change. The reaction just does not go to equilibrium (or in fact moves away from equilibrium).

[OP] First the notation hν can't be thought to be a catalyst as it doesn't make sense because isn't regenerated.

Things written on top of the reaction arrow are not necessarily catalysts. In general, that spot is used for reaction conditions, such as solvent, high temperature, or "reflux".

[OP] Then how one should think about light in a chemical reaction? It is a reactant?

Some textbooks write "heat" as reactant or product, which already does not make sense (but is used in combination with Le Chatelier principle to memorize how equilibrium constant change with temperature). For photochemical reactions, however, there is a stoichiometric aspect to the role of photons; for every reaction, one photon has to be captured.
A comprehensive published treatment
The abstract of this paper (Perspective – life and death of a photon: an intuitive non-equilibrium thermodynamic distinction between photochemistry and thermochemistry) addresses some of the misconceptions for photochemical reactions. It is a bit long, so I am breaking it up into smaller parts.
First, it states that photons are not chemical reactants in photochemical reactions:

Neither the thermodynamically determined probability isotherm nor its kinetically manifest rate isotherm can be applied to photo-absorptive reactions such that the participants, including photons, may be treated as if they were chemical reactants. Photons and chemical reactants differ from each other fundamentally: firstly, a photon's energy is absolute and, in all instances of practical relevance to the present paper, independent of its surrounding electrochemical field, while the energy of a chemical reactant is relative and defined by its surrounding field; secondly, while both photons and chemical reactants can and do engage in entropy creation, only chemical reactants can engage in entropy exchange.

It then goes on to address the mistake of treating photochemical processes using equilibrium concepts:

Clarification of these differences requires identification and abandonment of fundamental historical errors in photochemical thought deriving from inappropriate overreach of analogies drawn between light and ideal gases, and including: treatment of photo-absorption as a reversible chemical reaction; attribution to light of thermal potential, or temperature (as distinct from the idealised abstraction of a ‘temperature signature’); attribution to light of exchangeable entropy content.

Then, it addresses how entropy plays a role in these processes:

We begin by addressing widespread misapprehensions concerning the perennially misunderstood concept of entropy and the frequently overlooked distinction between entropy creation and entropy exchange. Armed with these clarifications, we arrive at a useful perspective for understanding energy absorption and transfer in photosynthetic processes which, through the chemical ‘kidnapping’ of metastable excited states within structured metabolic pathways, achieves outcomes which the Second Law denies to thermal chemical reactions

"For both processes, you could not say that in the presence of work, an equilibrium is reached. It is just the opposite - we are moving away from the equilibrium". Could we say that we have an equilibrium with a constraint like when a box is partitioned by a fixed adiabatic wall to two smaller boxes with different temperature at the two sides? I mean if the refrigerator stops then the temperatures will again be equal like if we substitute the adiabatic wall with one that allows heat transfer.

@Anton I would prefer the technical term photostationary to a made up own like "equilibrium with a constraint". I edited the answer and added a link to a page on photochemical reactions.

So light just excites the reactants and reduces the activation energy?

@Anton - still thinking about this nine months later, I'm impressed. Light specifically excites the reactants but not the products. This raises the rate of the forward reaction whereas the reverse reaction keeps going at the same rate (very low, if you want to make some product). The outcome is a ratio of product to reactant that is higher than that at equilibrium. In that sense, it reduces the activation energy.

Reactants and products of chemical reactions (as well as catalysts, solvents and other chemicals that might take part in a reaction) are types of matter. Matter can be thought of as something that intrinsically has both a mass and a volume, although the linked Wikipedia article will go into greater details and varying attempts at defining matter.
Regardless of how matter is usually defined, photons (and energy in general) are not matter. They are considered to have zero mass and it does not occupy volume. Thus, they have no real place in chemical equations which deal primarily with matter.
Instead, photons in photochemical reactions are best thought of as an energy source that will transfer a discrete amount of energy onto a certain molecule. While terms such as photocatalysed are thrown around in the chemical literature a lot, these are best understood as an analogy rather than an accurate description.

For the record, even though catalysts are often written on top of reaction arrows it is always possible to spell out a detailed step-by-step mechanistic cycle starting with the original catalyst, taking various steps in which the catalyst and reagents are modified and ending with a reaction that regenerates the catalyst. These reactions are necessarily balanced and, as written in the first paragraph, entirely involve matter.

So you would agree, I guess, that including "heat" as a reactant or product is silly as well...

@KarstenTheis Absolutely!

I think you are on thin ice to claim that photons are not matter, they have energy hence mass (but zero rest mass)  and are influenced by gravity and act like particles and like waves. Molecules/atoms  also can behave like waves, as in double split experiments. Molecules also impart energy to cause reactions by collisions, e.g Lindemann, RRKM etc models of unimolecular reactions where inert molecules/atoms play this part.

Photolytic dissociation of molecules like $\ce{Cl2}$ is basically a first order reaction in the concentration of the dissociating molecule, with the rate constant equal to the photon flux $\phi$ multiplied by the absorption cross section $A_x$ of the dissociating molecule:  $$-\frac{\mathrm d[\ce{Cl2}]}{\mathrm dt}=+2\frac{\mathrm d(\ce{Cl^.})}{\mathrm dt}=k[\ce{Cl2}]$$with $$k=\phi A_x$$After that, the odd chlorine atoms can react with methane in a second reaction:  $$\ce{CH4 + Cl^. -> CH3^. + HCl}$$and the $\ce{CH3^.}$ radical can participate in subsequent reactions.
Of course the equation for the rate constant would also have to be integrated over wavelength.

In short, the chlorination is photo-induced, but the reaction itself is a plain old radical chain reaction.

Don't worry too much about wordings of whether light is a reactant or not. It essentially boils down to semantics. What is a photon anyway? A packet of energy? then what is a packet? Fenyman (Nobel laureate, Physics) wrote a story somewhere that as a newly minted PhD, his father asked him what a spontaneous emission (of photon) is.
He said: “How do you . . . think of a particle photon coming out [of the atom]
without it having been there in the excited state?”
I thought a few minutes and I said: “I’m sorry. I don’t know. I can’t explain it to
you.”
Taken from a 500 paged book Our Changing Views of Photons: A Tutorial by Bruce W. Shore by Oxford University Press.
You can realize the complexity!
All you need to worry is the mathematics and kinetics of photochemical reactions. Is light a catalyst, is it a reactant, is it a product? There are several reactions which emit light. Should I call that as one of the "products". These are all filler words. Yes, the number of photons are important and their energy is important in a reaction which is affected by light. See Einstein's laws in photochemistry.

I voted down because you add some interesting points to the discussion but dismiss the point of the OPs question.

@BuckThorn He's right, though. There are photoinduced reactions, photocatalysed reactions, luminescense, phosphoresence, chemical lasers, etc. Light needs to be treated differently in all those, and in none of it light can be treated just like a regular chemical reactant.

Exactly, there is no point in discussing whether light is a "reactant" or "product". It is like the old age argument "How many angels can land on the head of the pin"? There is no answer.

I agree with the point that the OPs question is too broad. There is no single answer, and there is room in different contexts to interpret energy from light as a reactant, product or catalyst, even if these are not universally useful or appropriate descriptions. But the first paragraph dismisses the question as having no answer ("semantics") and goes into some tangential discussion about the nature of photons. Agree the exact nature of the interaction with light is of importance, but why not say that instead? Otherwise very insightful as usual.

Indeed there is no single answer, I upvoted at least three.

I would consider photons to act like chemicals in reactions. The difference is that although they have to obey conservation of energy and linear and angular momentum, their number is not conserved so we can't balance a chemical equation by counting photons. It should be obvious that they can be absorbed by molecules and emitted as in a light stick.
There is still an effective chemical equilibrium for photons, think about blackbody radiation. As catalysts -- well, in an ordinary chemical reaction the (homogeneous) catalyst gets turned into another species at one stage of the reaction, with higher or lower energy, then gets regenerated at a later stage. The regeneration is inevitable because of the number conservation law the chemical catalyst obeys. If you think about the analogous situation with photons, chlorophyll fluoresces strongly in the infrared which could be considered as the first stage of catalysis of photosynthesis: a visible photon is absorbed and an infrared photon is emitted with the energy going towards a useful chemical reaction. The photon just leaves the reaction area never to be seen again rather than waiting around for something to bump its energy back up to a useful level. So even though we got photon in, photon out, we wouldn't consider it catalysis because the outgoing photon can't and never will be able to cause photosynthesis again (blueshifted mirrors excepted:).
But a laser could be considered catalytic because a photon stimulates the emission so it is regenerated along with an indistinguishable buddy after the reaction. If you read the above link you can see that consideration of chemical equilibrium of photons leads to prediction of lasers.

Sorry, this is reasoning by analogy, or in short, nonsense.

Wouldn't it be more accurate to think of light as a catalyst rather than a chemical? After all post-absorption the energy of the photon might be dissipated in a myriad of ways whereas (as you note) conventional chemical products conserve the identity of the individual atoms in the reagents.

While conventional definitions such as provided by the Wikipedia state that catalysts are regenerated if you weaken that condition (which is not central to their ultimate function of accelerating a particular reaction) then light can certainly fit the bill. Catalysts do not contribute to the composition of the product(s) of interest.

As impressive as your rant may be :-) I don't think it is on topic or useful for the question at hand. I have removed it, but if you really want it back, there's always the rollback button. As for the downvotes, it doesn't seem all that bad: yes, people seem to disagree with you, but the sheer number of competing answers and comments already shows that it is quite a controversial topic. You shouldn't be ashamed of your contribution to the discussion, which was made in good faith.

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Amalgamation involving an alloy
Just curious, would mercury form an amalgam with gold that is already part of an alloy, i.e a gold platinum alloy?

related: https://chemistry.stackexchange.com/q/67387/102629

Yes, this is a traditional method of gold purification from platinum group metals.

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text:
Isn't water a Lewis base in the ammonia–water reaction?

Exactly one of these statements is false. Which one?
(a) When ammonia dissolves in water, the latter behaves as a Lewis base.
(b) The conjugate acid of $\ce{H-}$ is $\ce{H2}.$
(c) Boric acid is a monobasic acid.
(d) Ammonia behaves as an acid when it reacts with sodium metal to give sodamide and hydrogen.

The key says (d) is the false statements. How? I think (a) is the one.
(d) $\ce{2NH3 + 2Na -> 2NaNH2 + H2}$
The ammonia here loses a proton, which is characteristic for acids. In addition, it attaches to the electropositive sodium.
(a) On the other hand, when ammonia dissolves in water:
$\ce{NH3 + H2O <=> NH4+ + OH-}$
Ammonia gains a proton, which is characteristic for bases.

Your reasoning is correct. Perhaps the answer key is mistaken.
For part (a), in the reaction
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
water is donating a proton $(\ce{H+}$ ion) and hence is behaving as a Brønsted acid. Since all Brønsted  acids are Lewis acids, water is behaving as a Lewis acid.
For part (d) as well, the half reactions are:
$$
\begin{align}
\ce{2 NH3 &-> 2 NH2- + 2 H+} \\
\ce{2 Na &-> 2 Na+ + 2 e-}
\end{align}
$$
It's clear that $\ce{NH3}$ is acting as an acid and $\ce{Na}$ is acting as a Lewis base by donating electrons. The $\ce{NH2-}$ ion pairs up with the $\ce{Na+}$ ion to form $\ce{NaNH2}$ and hydrogen accepts the electrons and is liberated as $\ce{H2}.$
EDIT: As pointed out by Mithoron in the comments, the reduction of ammonia is not so simple. Sodium dissolves in ammonia to give a deep-blue solution containing ammoniated electrons (reference here):
$$\ce{Na + \text{(x+y)}NH3 <=> Na(NH3)_x^+ + e(NH3)_y^-}$$
The $\ce{e(NH3)_y^-}$ eventually loses a hydrogen atom to give $\ce{NH2-}$, which then combines with the $\ce{Na+}$ giving the product.

It's not so simple. Mechanism for ammonia reduction is more like: $\ce{Na + NH3 -> Na+ + NH3-}$ then $\ce{NH3- -> NH2- + H}$.

The question then is even wronger! There are two false answers: a and d.

@Mithoron Good point, I forgot about ammoniated electrons. Updated the answer now

Yes, unfortunately that means that the statement is indeed wrong. Moreso the statement about water can be considered true as H2O acts as also as acceptor of hydrogen bond, partially donating its lone pair - such coordination is pretty weak, but for example cationic complexes of noble gases also have rather weak bonds, while being clearly Lewis adducts.


text:
Question regarding $Z$ (Compressibility factor)
Oh! This problem has been bugging me for a long time. According to Wikipedia, the Compressibility factor $Z$ is defined as the ratio of the volume occupied by a real gas to the volume occupied by an ideal gas at the same temperature and pressure.
$$Z=\frac{V_{real}}{V_{ideal}} \text{              ........(1)} $$
Now the Ideal Gas Equation is
$$P_{ideal}V_{ideal}=nRT   \text{        ....(2)}$$
Now from here we can substitute the value of $V_{ideal}$ into the above expression $(1 )$ to get
$Z=\frac{PV_m}{RT}$
Where $V_m$ is the molar volume of the gas and $P$ is the value of $P_{ideal}$ that was in the ideal gas equation $(2)$.
OK, So far so good!
Deriving the compressibility factor value from the van der Waals' gas equation:
We know for real gases the pressure exerted by them is always less than the pressure exerted by an ideal gas so we correct the pressure by the term $a\frac{n^{2}}{V^{2}}$, we get
$$P_{ideal}=P_{real}+a\frac{n^{2}}{V^{2}}$$
And the volume occupied by the gas molecules can be written as
$$V_{ideal}=V_{real}−nb$$
On substituting the above values of $P_{ideal}$ and $V_{ideal}$ in the ideal gas equation, we get our van der Waals' Gas Equation
$$(P+a\frac{n^{2}}{V^{2}})(V−nb)=nRT$$
Please note that in the above equation $P$ is $P_{real}$ and not $P_{ideal}$. Now I'm going a step further to calculate the value of Z from this equation by shifting the terms and I got this
$$Z=\frac{PV_m}{RT}=\frac{V_m}{V_m−b}−\frac{a}{RTV_m}$$
Now here's the problem that I'm facing: In the above expression the value of $P$ is not the $P_{ideal}$ but it is the value of $P_{real}$ that we had in the van der Waals' gas equation.
Doesn't that contradict the whole definition of Compressibility factor that we had learned in the first place? How this expression for $Z$ can be correct?
Thanks for reading.

related: https://chemistry.stackexchange.com/q/57909/102629

You may use \tag{N} for numbering equations, it would write right aligned  (N).

Forget about the ideal gas.  The definition of the compressibility factor is just $$Z=\frac{PV_m}{RT}$$


text:
Why does entropy increase in reactions that make more molecules?
I used to understand this (more molecules ⇒ more degrees of freedom), but do no longer, because the total number of atoms is conserved, so translational d.o.f. should just be turned into internal d.o.f.
(at least considering only those reactions in which that is the case, no particle creation or annihilation).
Example:
Two separate H atoms have 3 translational d.o.f. each (x,y,z), for 6 d.o.f. total. I bring them together to form $\ce H_2$ and now have 3 net translational d.o.f. for the whole molecule, as well as 2 rotations, and a vibration.
In fact, the vibration has 2 d.o.f. for kinetic and potential energy giving heat capacity $\frac{7}{2}RT$ for a diatomic molecule.
So why does entropy decrease in the type of reaction with a form something like
$$\ce{ nA + mB -> A_nB_m }$$
Is it perhaps related to certain internal degrees of freedom only being “unlocked” at higher temperatures because of high minimum quanta of energy per d.o.f.?  If so, that would mean that at higher temperatures, entropy no longer increases for reactions that produce more molecules.

Your interpretation is correct. As a consequence, no molecules do exist at very high temperatures. They are all transformed into atoms.

And in the interim temperature region? Between unlocking vibrations and destroying the molecule, is the entropy change per increase in number of molecules, $\frac{dS}{dn}$, a function of temperature, inversely correlated with some power of $T$, where, because of the extra d.o.f. For vibrations, for some “hot” temperature, $\frac{dS}{dn}$ is actually < 0?

Be careful ! In the reaction $\ce{nA + mB -> A_nB_m}$ the entropy does not increase. It decreases

Right, thank you.  I’ll edit it to say decrease, but my point is still Why?  Because if this is true, then at, say 100°C, it shouldn’t. This seems too different from what we were taught.  I can understand relativity and QM as modifications that correspond in the classical limit, but “entropy increases with number of molecules” only at temperatures below which rotations and vibrations aren’t unlocked (~0°-100, maybe 150°C), and in fact begins to decrease as vibrations are unlocked... seems too different. It agrees over too small a regime and. It well enough.

An atom has 3 dof.  So two atoms have 6 dof.  A diatomic molecule has 3N = 6 dof if all dof's are accessible.  However, vibrational dof's are not typically accessible at room temperature, leaving the diatomic with only 5 dof's at that temp.

Right, so “$\frac{dS}{dn} > 0$ only near room temperature” sounds much too loose, and not a very good “rule of thumb”, at least not as applicable as I was led to believe.  I’ve often boiled things in a flask from 150°C - 200°C, and I know that many processes, in particular industrial ones happen above that.
Also, do vibrations not have 2 degrees of freedom, for potential and kinetic energy, leading to heat capacity 7/2 RT? This means that not only does this “$\frac{dS}{dn} > 0$“ fail to be applicable, but in fact the opposite holds true above room temperature.

I think you are focused on the energy storage mechanism as the only way to generate entropy. But there are other ways to generate entropy. Entropy is added when molecules break apart due to the removal of constraints on the positions of atoms with respect to other atoms. That is an inherent entropy increase.


text:
Dehydrohalogenation Reactions
Why do dehydrohalogenation reactions mostly follow the E2 mechanism? Doesn't the base play a role in deciding whether a particular reaction will follow E1 or E2? For eg, a strong base would favor e2 and a weak one would favor E1. In fact, if dehydrohalogenation reactions follow the E1 mech, a carbocation would be formed which can undergo rearrangement to the carbon directly attached to the halogen, and can undergo resonance with it. Why doesn't that happen?

Also, in this question, for example, E1 reaction seems to be major. Why is that so, and what conditions have to be satisfied in dehydrohalogenation reactions for them to be E1?

I did not really understand what your actual concern is here?
"Doesn't the base play a role in deciding..." YES, the base is one of the main conditions that decides what kind of reaction is carried out.
This might seem a little weird but let me 'personify' the reagents and explain how my teacher once explained this to me:
A strong base won't wait for the carbocation to form, as a result the reaction proceeds in a concerted single step.
A weak base won't attack until a stable carbocation is formed.
The other conditions like presence of a protic solvent is also sigificant.

To answer the last question - "Why doesn't that happen?"
It is quite possible that kinetically stable products are favoured over thermodynamically stable ones. It just depends on whatever is available to satisfy the conditions.

Yes, the point about the base makes sense. But, I came across a question recently, where they asked for the dehydrohalogenated product (without mentioning the base), and the options had both E1 and E2 products. Since most dehydrohalo rxns are E2, i went for that, but the given answer was the E1 product. So, my doubt is: how do we decide which pathway it will follow? And that also led me to wonder why most dehydrohalo rxns are E2? If you google for dehydrohalogenation reactions, almost all websites mention that they are E2 reactions. Why?

http://home.iitk.ac.in/~madhavr/CHM102/Lec13.pdf

It gets very complicated to answer it in general. There should be a loop hole to figure out if the base/reagent wasn't mentioned.

Look for cues like the type of carbocation formed. For example, I've encountered questions without a base mentioned for Sn1 & Sn2. The stability of the intermediates will be relative here as ter-carbocation is stable in one type of nucleophilic but not preferable for the other type. Similar way, check the RDS outcomes to decide for now. Or maybe post the question that you were talking about.

@Desai kinetic and thermodynamic stability depends solely on temperature, right? if the temperature is low, kinetic control is favored, and thermo otherwise, if im not mistaken. So, i dont really see how that is relevant here...

@Desai I have edited my question. Also, in nucleophilic sub rxns, sn1 favors tert and sn2 favors primary, so that can be used to distinguish among them. But, for elimination, both e1 and e2 favor teriary...

@Buraian in that ppt too, they have just mentioned that the most common pathway is E2, for dehydrohalo rxns. My doubt is why? And "most common" doesnt mean always, so when will E1 product be favored?

https://chemistry.stackexchange.com/q/27380/102629

@cngzz1 that doesnt really answer my question.. any help?


text:
What does orbital mean, exactly?
My teacher told me that orbital is the probability distribution data of the electron around nucleus which is amplitude data in a way.
An example of how my teacher actually told what it means involves that you know three possibilities for a coin flip to get head and tail. Therefore, this table is what my teacher says is orbital:
Whereas in my book it says that it is the space occupied by the electron around the nucleus of an atom where the probability of finding an electron is the maximum.
Also, it says in my book that as the distance of the electron or orbital from the nucleus increase, principal quantum number increases. How is electron or orbital same thing?
Can you please show a photo that how does it also look exactly?
Both the things mean different. What does it mean exactly then?

https://chemistry.stackexchange.com/questions/103859/is-the-notion-of-orbitals-different-in-theoretical-chemistry

"Look exactly" is one thing it doesn't.

@IvanNeretin One tells that orbital is like a probability distribution table of amplitude of electron .Other 2 tells totally different thing

"Sir" is only used to directly address a person. If you write about your teacher, call him "my teacher".

Have your classmates understood anything?

@Karl, Your logic does not apply in South Asia :-) So, billions use Sir as a synonym for a male teacher and Miss or Madam for a female teacher. My sir or my madam= My teacher. Strange but true. I have heard that in Egypt every university teacher is a Doctor ;-)

@M.Farooq And in France, every teacher is a "professeur". In French. ;)

@Karl, Good to know that. What is the typical German way of addressing teachers. I have seen Chef being used for PhD supervisors.

@M.Farooq We're getting off topic here, but the usual, day-to-day address is "Herr Müller" / "Frau Meier" between adults. You can swap "Herr" and "Frau" for their title, i.e. "Doktor" or "Professor", for formal uses (in front of an audience, or when you first introduce yourself). You can alternatively swap the name for the persons' function, e.g. "Herr Professor", "Frau Bundeskanzler", also quite formal.

It sounds to me like your confusion is why these two definitions are describing the same thing.  I will attempt to clarify that.  My interpretations/slight rephrasings of these definitions are

A probability distribution of an electron around a nucleus.
A space occupied by the electron, where the probability is a maximum. (This is a bit different that what you've stated, I'll address this at the end with point 3.)

Then afterwards, point 3. Principle quantum number, $n$, increases  as distance from the nucleus increases.
The orbitals around an atom are like "slots" that you can put electrons in.
In a simplified model:   You can think of them like regions of space where the electrons can be.  If I put an electron at r = 1, then the next electron cannot be at the same location, and must go in the next lowest energy slot, r = 2.
A bit more accurately, the orbitals are probabilities associated with regions in space, ie the probability P = f(x,y,z) is function of the spatial coordinates.  That means that all the orbitals are overlapping, having a value at every (x,y,z) coordinate in space around an atom.  The individual orbitals are described by different functions though.
The previous example might turn into the first electron having a high probability (say 80%) of being at r = 1, and a low probability (the other 20%) of being outside that. This is the distribution with the lowest energy, because the electron spends most of its time close to the positive nucleus.  The next electron would go into the next lowest energy slot (higher than the first).  This distribution may have a highest probability peak of 80% at r = 2, and 10% at r = 1, and 10% spread over everywhere outside r = 2.  The location at r = 1 is therefore split 80% into orbital 1, 10% into orbital 2, and the remaining 10% over all the other orbitals such that the sum of the probabilities of all the orbitals add up to 100%.
Conclusion
The orbitals are thus a function P(x,y,z) which is a probability distribution (Def. 1) over space (Def. 2), and while the functions overlap, though with different values at each region, they have characteristic maxima.

Can you please show a photo

Here are two websites that both have lots of good visualisations
http://csi.chemie.tu-darmstadt.de/ak/immel/tutorials/orbitals/index.html
^ For atomic (s,p,d,f) orbitals go to Hydrogenic : visualisations, and click on the individual pictures.
https://winter.group.shef.ac.uk/orbitron/
^Click on the coloured "1S..." tabs on the left.
Remember that the orbitals are functions of All space, and therefore infinitely large, and cannot be shown in a finitely-sized picture.  However, much like a Gaussian distribution, you can draw a boundary and say that the electron has 99% probability of being within this region.  These probability contours are what is usually shown, and what you are looking at on these websites.
Point 3:  Principle quantum number, $n$, increases  as distance from the nucleus increases.
Electrons settle into the lowest energy state. The orbitals that have the peaks of their distributions at larger radii are higher in energy. A word of caution about definition 2:

Whereas in my book it says that it the space occupied by the electron around the nucleus of an atom where the probability of finding an electron is the maximum.

The orbital is not the region where the probability distribution is a maximum, it is the entire distribution.  However, the orbitals can be characterised by the location of their maxima.  This is simplest for the s-orbitals, which are spherical:  Higher energy orbitals have maxima at larger radii, and thus we can can give a first attempt at drawing orbitals (and a first attempt at thinking about them) as circles around the nucleus with radii equal to the maximum probability radius for a given distribution.  This says that the 1s orbital has a maximum probability at r = 1, the 2s has a probability maximum at r = 2, ... and so on.  However, the difference is, that the orbitals are not just at the specific spatial location r = n, but are a distribution across all space, with characteristic maxima at these locations.
I hope this helps.  "What is an orbital?" is a good question, that I think every chemistry student asks as some point, sometimes more than once.

The notation using $r=n$ is not standard, and, frankly, going to confuse people.

n = 1 is lowest energy state than n = 2 right.

I didn’t understand your point of probability as spatial coordinates. Therefore , I siding understand what did you mean by overlapping sir

Also m thank you very much for your effort

The location at r = 1 is therefore split 80% into orbital 1 .... point of yours you have written

That you are only talking about one electron right

Since you wrote that for the 2nd electron , at n = 2 is 80%

Is orbital like K , L , M , N shells

Sloppily written statements about probabilities are just as bad as wrong ones. -1

Orbitals are a quantum mechanical concept. Shells, orbits, and the like are semi-classical treatments that served as a stepping stone to quantum mechanics.

Good point about the potential confusion with r = n.  I was trying to treat a simplified model.  Let me clarify.  Here I am using $n \in \mathbb{Z}^+$ to be some positive integer, meaning that, in this simplified model, the electrons occupy circular orbits that have integer spacing in some units.  In the slightly more complex model, these are the radii at which the distributions have their maximum. The principle quantum number n can then be thought of as related to these radii.

I realise that the situation is a bit more nuanced, but I think this is appropriate for the level of question being asked and will not lead OP astray at this stage.

#Aridhan "n = 1 is lowest energy state than n = 2 right".  Yes.  Smaller average radii and smaller principle quantum number n both mean lower energy.  K, L, M, N shells refer to, and are merely a different notation for, the principle quantum number n.  There are (can be) several types of orbitals (s, p, d, f) associated with each shell.  I think for your question "what is an orbital?" it is sufficient to talk about s orbitals, and discussion of other shapes of orbitals might just add confusion until you understand how s orbitals act.  In general they are spherical harmonics.

Regarding "The location at r = 1 is therefore split 80% into orbital 1" and what I "mean by overlapping", One can ask two questions. 1. Given an electron is in an orbital, what is the probability that it will be at a particular location in space?  The answer to this is given by the probability distribution which is described by that orbital.  The sum of the probabilities over all space must = 1 = 100%  2. Given a particular point in space, what is the probability that an electron will be in a particular orbital?  The sum over all orbitals must = 1.

"That you are only talking about one electron right"  In the sense that each orbital is a "slot" for an electron, you have the right idea, however each orbital can contain 2 electrons, one spin up and one spin down.  It is convenient in this way to think of an orbital therefore as a 2-electron wavefunction. The shapes of the orbitals are identical for these two electrons.


text:
Reaction between barium peroxide and hypochlorous acid
When $\ce{BaO2}$ reacts with $\ce{HClO},$ the question tells us that that oxygen, barium chloride and one other product is formed.
The equation I came up with is
$$\ce{BaO2 + 2 HClO -> BaCl2 + 2 O2 + 2 H2}\tag{R1}$$
The equation given, however, is
$$\ce{BaO2 + 2 HClO -> BaCl2 + 1 1/2 O2 + H2O}\tag{R2}$$
Why is $\ce{H2O}$ produced and not $\ce{H2}?$ More importantly, what clues/rules are there that would have helped me to predict these products?

Chloric(I) acid is a strong oxidizer, even stronger than elemental chlorine. Why would it support reduction of $\ce{\overset{+1}{H}^+}$ to $\ce{\overset{0}{H}_2}?$

@andselisk Oh right, I see - so this is just a reaction to remember then, rather than to learn any 'rules' from?

Not really, common sense works here all right. One can assume that $\ce{\overset{+1}{Cl}O-}$ is a strong oxidizer even without looking up any reduction potentials simply by noting "unconventional" oxidation number of chlorine.

TLDR How on Earth would you get a reducing agent from two oxidising ones?

Because peroxide is also a reducing agent, forming $\ce{O2}$.  Study the half-reactions set up in my answer.

R1 should have one $\ce{H2}$, not two.

You may have confused barium peroxide with barium metal.  The metal can indeed displace hydrogen from hypochlorous acid, not to mention (for a metal as reactive as barium) from the water in which the acid is dissolved; but metal peroxides will produce oxygen and water from the acid.  So your missing product is water, not hydrogen.
Tricky half-reactions
If we want to balance this with half-reactions we need to take some care, because we need to simultaneously cancel protons and electrons with a weak acid like hypochlorous acid.  Given the reduction half-reaction based on chlorine being reduced from the acid:
$\ce{HClO + H^+ + 2 e^- -> Cl^- + H2O}$  Eq. 1
We must set up the peroxide oxidation to oxygen so that the electron/hydrogen ion ratio is also $2:1$.  Thus a proposed reaction
$\ce{BaO2 -> Ba^{2+} + O2 + 2 e ^-}$  Eq. 2
would be balanced but you don't have the proper ratio of electrons to protons matching the weak hypochlorous acid reduction.  Add water to get the necessary hydrogen ions, and note that this means you change the oxygen production:
$\ce{BaO2 + H2O -> Ba^{2+} + 2 H^+ + (3/2) O2 + 4 e ^-}$  Eq.3
In other words, water is also oxidized to generate the hydrogen ions necessary to effect the reduction of the weak hypochlorous acid, thus producing that extra half-mole of oxygen per mole of barium peroxide.
Four electrons to two protons in Eq. 3 matches the ratio of two electrons to one proton in Eq. 1.  So now we can double Eq. 1 and add to Eq. 3 getting the reaction that avoids both unbalanced protons and unbalanced electrons.  As we saw above, one mole of barium peroxide gives 1.5 moles of oxygen in the overall reaction:
$\ce{BaO2 + 2 HClO -> Ba^{2+} + 2Cl^- + (3/2) O2 + H2O}$.


text:
Why is aluminium oxide solution amphoteric?
Is Aluminium oxide solution amphoteric due to aluminium's electronegativity being in between 1.5 and 2.1.
If so why?
I believe it is because a AL-O-H molecule is made and it can either disassociate into a aluminium cation and a hydroxide, or a aluminium oxide anion and a proton. In this case the Al-O-H disassociates into both giving amphoteric properties, but how does electronegativity affect this?

Why is Al2O3 amphoteric? - because it can react as 'both' acid & base as amphoterism depends on the oxidation states of the oxide. (Housecroft, C. E.; Sharpe, A. G. (2004). Inorganic Chemistry (2nd ed.). Prentice Hall. pp. 173–4. ISBN 978-0-13-039913-7)
Amphoterism related - https://en.wikipedia.org/wiki/Amphoterism#:~:text=Metal%20oxides%20which%20react%20with,example%20of%20an%20amphoteric%20oxide.

Electronegativity is a relative term so I would not comment on if 'that' is the reason for amphoterism.


text:
How to find the moles of a salt added to a weak base, given ph
I would really appreciate some help with how to approach this:
I'm given that:
Methylamine (CH3NH2) is a weak base used in the manufacture of pesticides. Taking into account that the degree of dissociation of said compound is 21%, calculate:
Kb=6.1*10^-4  (This is likely an arbitrary number assigned to each student, rather than being the correct Kb, as is the degree of dissociation).
a) The initial concentration of methylamine necessary to prepare 175 mL of solution.   -  This I have managed to solve, my answer is 0.01093M, the autocorrect mechanism says it's right
Edit  -  Here is what I've done:
CH3NH2(aq) |  +     |   H2O    |  <>     |   OH-   |   +   |   CH3NH3+(aq)
 1      |              -   |                 0   |       |         0     |
 1-α    |              -   |                 +α  |       |         +α    |
Co(1-α) |              -   |                 Coα |       |       Coα     |

Kb= 6x10^-4 = (Co*.21)^2/Co(1-.21)
Solving this quadratic equation yields Co = 0.01093M;  Which is correct according to the autocorrect mechanism.
b)How many moles of methylamine chloride (CH3NH3Cl) must be added to the previous solution to obtain 217 mL of final solution with a pH = 11.16?      -   This is what I cannot fathom, I've tried making ICE charts and the Henderson-Hasselbalch equation, though I'm not having any luck, it's likely that I am doing it wrong though
Edit 2
This I've tried many times to do and feel like I'm going round in circles, hence this post.
I need to create from the previous solution above 217ml of a new buffer solution with Ph 11.16.
Therefore:
POH = 2.84   so;
10^-2.84  =  0.001445439
Which should be the OH- in the new solution.
POH  =  Pkb  + log [conjugate acid / weak base]
Kb =  6.1x10^-4
PKb = -logKb  =  3.21467
-0.37  =  Log[CA] - Log [0.01093]
10^-2.34 =  0.004612644 M
Since I need 217ml, I multiply;  0.004612644  by;   .217   giving; 0.00100943 moles of conjugate acid.  This is wrong according to the autocorrect.
I have tried doing an ICE chart again , which I won't rewrite here, and I obtained a similar figure.   That said, I have also obtained other figures which are also wrong.  So I think there must be some flaw in my reasoning here, which I haven't been able to identify myself.
Edit 3   -  I won't comment on this part since, it flows from the previous part.   Any comments will be hugely appreciated!
If 2.71 mL of 0.351 M HNO3 are added to the solution in part b), what will the pH of the final solution be?      This I feel more confident about, but cannot calculate it without first getting b)
Thanks in advance!
Edit 4  -  Base at equilibrium:

****CH3NH2(aq)
+
H2O
<>
OH-
+
CH3NH3+(aq)****

0.01093

-

0

0

-.0022953

-

0.01093 x.21

0.01093 x.21

0.0086347

-

0.0022953

0.0022953

Kb=6.1x10^-4
Desired OH- =  0.001445439
Kb = [OH-][BH+]/[CH3NH2]
BH+ = 6.1x10^-4(0.0086347/0.001445439)   =  0.003643991M
0.003643991M   x   .217  = .000790746 moles needed (wrong according to autocorrect)
Likewise
2.84  =  Pkb  + Log(conjugate acid/weak base)
[CA] =  (0.0086347/10^.374670165) = 0.003643989M
0.003643989M  x.217  = 0.000790745 moles (basically the same number)
There must be something else I am getting wrong, could it be to do with the volume, changing from .175L or original solution to .217L of new solution
Edit 5
I finally got it, the problem was I was using concentrations for the calculations, and since the volume changes, I should have been using moles.

Practical advice: Approach your task by algebraic solution first and use literal values in the end, when all the formulas are ready.  It has several benefits.

You have started from the wrong end. Start with calculation of the final solution concentration of the base. The from pKb, pOH and [B] you compute [BH+] and from it the molar amount.

Thanks, do I need to calculate the degree of dissociation again for the new solution also? Or should I use the 21% given?

@21% is a particular value for the very particular case. Start with equations of mass and charge conservation and the equlibrium equations. Then involve affordable simplifications.

Thanks, I hope I understood you correctly, I have tried to do as you say, though I could easily have made a mistake.    Unfortunately I'm still unable to achieve the correct figure, I have added it above.


text:
How can a nonlinear field gradient be generated in NMR?
I am working on the topic of optimal control in NMR, where spins undergo non-unitary evolution caused by gradient field pulses at different time intervals. I am studying the effects of this on the spin dynamics.
For my project I need a particular kind of gradient field as given by the equation $-z\exp(-z^2)$, where $z$ is the the spatial position of the region from center of the sample. All the books I had referred to (Levitt's Spin Dynamics and Keeler's Understanding NMR Spectroscopy) talk about constant gradients. So I have no idea if it is possible to create even an approximation of a spatially varying gradient as given here.
In particular, I need my field due to the gradient to look like a delta function in space. So I thought of using a derivative of a Gaussian with a small standard deviation as an approximation, which is the equation given above.

Btw., do you need a pulsed gradient, or can it be static? How many T/m, and how many T in $B_0$?

Do I understand correctly that you want B(z) to be two plateaus of different amplitude, connected by a linear slope?

Yes ,the approximation to what I am asking would be two plateaus connected by a linear slope

Regular off-the-shelve pulsed field gradient (PFG) systems only give linear gradients, obviously. However you can produce higher-order gradients with your shim coils, normally to compensate similar opposing unwanted gradients in the field of the main magnet.
So, by manipulating the shim set, you can create any odd gradient field that you can sufficiently approximate with the terms that your shim system supplies (usually up to 5th order in $z$). To find out what exactly the terms are (coordinates and absolute amplitudes), you can image them using the regular PFG system.
There are two hitches: You cannot switch the shim system quite so fast as the dedicated PFG set, and the "room temperature" shim system isn't nearly as powerful as a PFG, so you are limited in terms of gradient strength. (The "cryoshims" are stronger, but you cannot switch them, as they are superconducting.)
So, if you're OK with a static gradient of low amplitude, you can go straight ahead. If you need it to be pulsed, give it a try. It's been done before, switching shim currents during a pulse sequence. Very good chance to really get to know the insides of your spectrometer system.

If you're secretly an engineer also, check my other answer. ;)

Ah, I now remember seeing an example of this in the wild! Loening, N. M.; Keeler, J.; Morris, G. A. One-Dimensional DOSY. J. Magn. Reson. 2001, 153 (1), 103–112. DOI: 10.1006/jmre.2001.2423. (Or free version accessible here.)

Thanks for your reply, the paper you have attached seems to have employed something similar to what I am asking.It will be of great help

@orthocresol Tnx! It is to be hoped that OP can get away without hardware manipulations today. ;-)

As an alternative to (ab)using the shim coils:
For a powerful arbitrary PFG in $z$ direction (in a cryomagnet), you could just go and wind your own gradient coil. In typical high-field gradient probes, the coil is a user exchangeable part, and specifically z-coils are not hard to design and build. (Coils with x & y components are definitely harder both in math and in fabrication.)
The literature is full of accounts of people building their own probes, and if you have a competent mechanical workshop available in your institute, I say you can pull this off in a few months at low cost.


text:
Sound of ringing gels
I'm searching for an audio file of the ringing of a ringing gel - does anyone know a source?
(I already searched on different video- and sound-platforms and looked up in the literature)

From this question on Physics SE, there is a linked audio track in Soundcloud here, which is what you are looking for. The OP references a jar of hair-stying gel which, upon being tapped with a mallet, produces a "reverberating" sound instead an otherwise expected "short 'tok' sound."
Note the the ringing gels are common ingredients in pomades, shampoos, and other products where rheological properties are engineered for various reasons. See this patent application for a ringing gel, wherein it is stated:

The phenomenon of ringing is a result of the manner in which the dispersed droplets are entrapped within the network structure of a gel. It has been suggested that the network within a gel consists of a rod-like or lamellar structure. Such a network could allow dispersed particles to vibrate within a gel much like a ball bounces off the walls of a box.

The practical application of the gels appears to be mainly in the cosmetics industry.

Fascinating. I did not know this is a thing.


text:
Ring formation via oxymercuration-demercuration
I'm solving past problems from a 2014 Organic Chemistry II course (from a publicly available Harvard summer school Canvas page), when I came across this question:

The answer (which is the reactant) looks like this:

Now I understand how the alkyne would turn into an alkene, but I don't get how a furan would form from what looks like an oxymercuration-demercuration reaction. Any help would be much apprecited.

Observe that there is no nucleophile give in the mercuration step and recall that the three-membered mercury ring intermediate is very reactive. Hence, this is suggestive that the oxygen would attack the intermediate with its lone pair on the more substituted position as follows:

Edit: Oops! Missed a carbon on the last step as correctly pointed out by @Waylander. I have updated the answer with a new image that corrects the mistake.

For reference see  Addition reactions of Alkene's in David Kleins organic chemistry (around page-410 ) and for a quick reference see wiki

You are missing a carbon in the final structure - otherwise this looks correct


text:
How is o-Fluorophenol more acidic than phenol even after having Hydrogen bonding?

Source: Concepts of Organic Chemistry by Dr OP Tandon, Himanshu Pandey, Dr AK Virmani
Page no: 241

Perhaps the words in the picture are explained at greater length in the text, or perhaps actual numbers are given for the dissociation constants (https://www.quora.com/Why-is-o-flurophenol-is-more-acidic-than-p-flurophenol).
As it turns out, the dissociation constants are 8.7, 9.3. 9.9 and 10.0 for ortho, meta, para and unsubstituted phenol. Ortho-fluorophenol is the most acidic - perhaps in spite of the possible hydrogen bonding. The inductive effect of the fluoro group is surely greatest at the ortho position.
In the series of nitrophenols (Acidity order of nitrophenols), the para (p$K_a$ = 7.16) is slightly more acidic than the ortho (p$K_a$ = 7.2), so it could be said that the ortho substituted compound, which would be expected to be more acidic (if you assume that the resonance effects at ortho and para positions would be similar), but isn't, we have to look for some other explanation, and hydrogen bonding could stabilize the neutral molecule.
Well, even if the actual effect is stronger than 0.04 p$K_a$ units, the hydrogen bonding effect seems to be rather small, and is overcome in the fluorophenols by the $-I$ effect of fluorine. Just scratch out the words "H-bond decreases acidity". (Lightly.)

Interesting, this would suggest the magnitude of inductive effect depends on proxomity of groups


text:
Stereochemistry + isomer check
We start with compound M:

[Please imagine green bonds as dotted bonds going into the plane]
From compound M, we are told that the 3 membered ring is opened u using sodium azide yielding a mix of isomers N and O.
I drew the following:

and

[Again, please imagine the green bonds as dotted bonds and the blue
bonds as wedge bonds]
Given are the following two molecules:

Are my solutions the same as the ones given? If they aren't, where did I go wrong?
Many thanks!

Your solutions are not the same. You have not considered atttack by azide on the epoxide next to the ether.

First of all, for both products, you have not given the O atom in the carbonyl position. In organic chemistry, we assume that there is a C atom if no atom is mentioned specifically. Now for this reaction, the carbonyl group will be there as it is.
Now we can consider the reaction. The epoxide is away from the plane, i.e. below the plane. For the steric reason, it is more feasible for azide to attack from the upper side of the plane. Now again, azide can attack both C centre of the 3-membered epoxide ring. In that case, we'll be having two different products. But in both cases, we will have the azide group on the above of the plane and O atom in the below of the page.
Now if I assume that you have mistakenly skipped the O atom in your both products, then your first molecule matches with the first answer given below. But  the second answer given by you is not correct and doesn't match with the second answer. You can have the green and blue coloured bond in the same place and then you have to alternate the position of OH group and azide group. Then that will match the answer given below and the answer will be correct.

Thanks @aritaroy24 - you are right - I missed out the carbonyl oxygen by mistake

@aritaroy24 - I also now understand that the azide will attack from the upper side and so the N3 part will always be above the plane. Is there any where I can find out more about this type of attack for olympiad purposes? (I am a high school student in the UK - A level, if that helps)

Google is your friend! Search "nucleophilic opening of epoxides" and you will get hits like this: https://www.khanacademy.org/science/organic-chemistry/alcohols-ethers-epoxides-sulfides/ring-opening-reactions-epoxides/v/ring-opening-reactions-of-epoxides-strong-nucleophiles

@Waylander is correct. It's just a basic type of reaction. You can find it on google by just typing the words mentioned above.


text:
What is the phase of matter that has flexible volume, but constant shape?
In this YouTube video it argued that:

Solids have constant volume, and constant shape
Liquids have constant volume, but flexible shape
Gases have flexible volume, and flexible shape (in fact they don't have volume and shape at all)

So, this table came to my mind:

Do we have a phase of matter that fits in the red cell? In other words, do we have a phase of matter that keeps its shape, and only changes its volume? If so, what is the name of that phase?

Start by defining what these terms of mean, shape and volume...

Shape is quite a straightforward term but it is not a property of matter per sé. Replace it with the strength of the intermolecular forces responsible of shape and you will see that the red cell has not reason to be.

Your definitions are problematic except as very broad approximations. Consider a rubber band: it is clearly a solid, but also clearly not a substance with a constant shape; less clearly, but still true, it doesn't have constant volume either.

Doesn't the shape change with volume?

You can't really retain shape and change volume in a significant way (certainly not in a comparable way to a gas). In any case, the properties in question are akin to compressibility and Young's modulus.

There's got to be a more descriptive and general word than a balloon.

I guess you could imagine a fluid with very low dilational viscosity and infinitely high shear viscosity...No such materials are known to exist...

Gases may not have a definite volume but they certainly occupy a volume of space.

Note that none of these characteristics is absolute—with enough pressure change, all phases of matter have measurably varying volumes.  And with enough force, all phases of matter have varying shape.
Rather, these are general characteristics that provide guidelines to distinguish solids, liquids, and gases. And these general characteristics are best understood by seeing how they result from the microscopic differences between solids, liquids and gases.
The following is a simplification, but it's an appropriate simplification for a beginning student of chemistry:
Atoms/molecules in solids and liquids are very close together, so they can't be readily compressed.  The atoms/molecules in gases are far enough apart that they can be readily compressed—all you're doing is taking up the empty space. Atoms/molecules in solids are rigidly connected.  Atoms/molecules in liquids and gases aren't.  Thus the shape of solids is fixed, while that of liquids and gases is not.
Now we can ask: What microscopic properties would give you constant shape, but varying volume?  Well, in order to have varying volume, you'd need the atoms/molecules to be far apart (like in a gas).  This means they're disconnected from each other, and moving approximately independently. If they're disconected from each other, they can't hold a structure, and thus can't have have a fixed shape.
A gas contained within a perfectly uniform spherical balloon could have fixed shape but varying volume as you change the external pressure, but that's a property imposed by the balloon and the surroundings; it's not a property of the gas itself.
Granted, a perfectly uniform solid (a solid whose properties are identical in all directions; the word for this is "isotropic") could have constant shape and varying volume if the pressure around it were increased uniformly, and sufficiently (e.g., let it sink to a sufficient depth within the atmosphere of Jupiter).  But then it wouldn't fit into your table, which assumes solids have fixed shapes and volumes.

There is a mathematical argumentative approach called reductio ad absurdum...i.e., bring your own logical reasoning to the point where the result is an absurd statement if you accepted the original statement as true. In this case you can accept that there is a phase which can change its volume but keeps it geometrical shape intacr. You can try disproving your own statement using this method for solids, liquids, and gases.

Do we have a phase of matter that fits in the red cell? In other words, do we have a phase of matter that keeps its shape, and only changes its volume? If so, what is the name of that phase?

a) When solids experience stress, there is strain = deformation, change in volume.
b) Liquids are almost incompressible and they have no shape, but under extreme extreme pressures their volume does change, typically, > 100,000 psi they can become solids. The solid will have the shape of container in which it became a solid.
c) Gases acquire the shape of the container (actually a bad statement, i.e., gases occupy the space provide by the container), and their volume can be easily changed by stress.
d) The fancier state of matter, plasma, it is also an ionized gas, so it does not have a geometrical shape. However, it can be confined by using magnetic fields so that it does not touch the walls of the container, and perhaps its "volume" and "shape" can be manipulated independently. A true physicist can better answer it.


text:
Entropy change during irreversible process
Entropy change is defined as $\Delta S = \int \frac{\delta Q_{rev}}{T} $, where $Q_{rev}$ is heat transferred through a reversible process.  I’m a little confused how this works, so I have a couple of related questions which I hope will clarify what’s going on.

If a reversible process is defined as $\Delta S = 0$ then how can this integral be non-zero for a reversible process?  Or is it that just $\Delta S_{Univ} = 0 = \Delta S_{sys} + \Delta S_{surr}$ and therefore $\Delta S_{sys} = -\Delta S_{surr} \iff T_{sys} = T_{surr} $, in which case

How would one calculate it for an irreversible process, in particular, some heat, Q, transferred across a finite temperature difference?
I would try (considering isothermal transfer between infinite temperature reservoirs)
$$ \Delta S_{Univ} = Q \big( \frac{1}{T_{sys}} - \frac{1}{T_{surr}} \big) $$
But this is no longer reversible, so I need to do something else?

Since entropy is a state function, is the entropy change through any process that is irreversible, but between equilibrium states, equivalent to $\int \frac{\delta Q}{T}$ through a reversible process?

Is this equivalent to pretending I’m doing 2 reversible transfers to some fictitious bodies, the first taking Q reversibly from the system to the fictitious body, both at $T_{sys}$, and then magically dropping the fictitious body to $T_{surr}$  and transferring the heat Q to the surroundings?
If so, why use $\delta Q$ rather than $dQ$?  My understanding was that $\delta Q$ was for path-dependent processes, but there is only one fixed (adiabatic) path for reversible transfer?

Sorry for yet another entropy question, thanks in advance!

The answers to your first and third questions are "yes."  Here is a tutorial on how to calculate the change in entropy of the system or the surroundings for an irreversible process:  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/. The reference also provides worked examples.  Note that, in considering entropy changes for both the system and the surroundings from an irreversible process, one needs to first separate the system from the surroundings and then subject each of them separately to a reversible path.  These separate reversible paths do not have to match.
It simplifies things a little to think about how entropy of a system can change.  There are only two ways:  entropy transfer across the boundary of the system by heat flow in and out at the boundary temperature, and entropy generation (due to irreversibility) within the system.  In a reversible process, only the first mechanism is present.  In both cases, the transfer across the boundary is $\int{dq/T_B}$ where $T_B$ is the boundary temperature.
Regarding questions 2 and 4 (involving ideal constant temperature reservoirs), these are somewhat limiting cases.  Here, if you imagine a very thin conductive slab between the two reservoirs (with high temperature at one face of the slab and low temperature at the other face), then for the irreversible process involving the transfer of heat between the reservoirs, all the entropy generation would take place within the intervening slab, while the reservoirs would each separately experience only entropy transfer with the slab.  In the limit where the slab becomes infinitesimally thin, we would obtain the system you have envisioned, but all the entropy would still have been generated at the interface, and each reservoir would only experience entropy exchange with the interface.

Thank you for your reply. The linked tutorial is indeed helpful. Let me assemble my questions and thoughts... I think you are saying that 4. is correct, this is equivalent to doing 2 reversible transfers to fictitious bodies?  This seems similar to what you describe in the Example 1: Thermal equilibration, where the hot object has $dS = \frac{MCdT}{T}$ going from $T_h$ to $T_f$, without caring what the temperature of object 2 is, and similarly for object 2 from $T_c$ to $T_f$.

In the case of the thin barrier, do we not need to account for thermal equilibration, as heat entering from one side must travel across to the other side, with the temperature gradient getting steeper as the barrier becomes thinner? I would imagine that this internal shuffling of energy would increase entropy, as all the heat energy being piled up on one side is not in internal equilibrium.

In example 2, may I ask why you are using the constant-volume heat capacity during a non-constant-volume (expansion) process? Also is the denominator $C$ in eq. 6, $C_v$?  I think so, but you labelled it in the numerator, so just making sure. Right before Step 3, has the $T_f$ in state 2 lost the “1-...” it had in eq. 5? Not trying to be finicky, just making sure I didn’t miss something.

Are Ex. 2+3 saying that the $\Delta S = \int\frac{\delta q_{rev}}{T}$ is the same for any reversible path between two equilibrium states, and therefore irreversible adiabatic processes may have $\Delta S > 0$, but a reversible adiabatic process ($q_{rev} = 0$) will always have $\Delta S = 0$, no matter the reversible path constructed?

This sounds like the definition of a potential function (good!) in terms of closed-loop integrals being 0... which brings me back to my question:  If this is an exact differential, why do I keep being told $\delta q$ and not $dq$ or is $q_{rev}$ a special case with the normal differential is allowed?

Answer to your first question:  The thin barrier is assumed to no thermal inertia MC so its entropy does not change.  Certainly in the limit where it becomes infinitely thin, this is the case.

Answer to your second question:  For an ideal gas, the Cv is independent of volume, and internal energy depends only on temperature.  In Eqn. 6, that should be a Cv in the denominator.  I don't know what happened to the v.   And, yes, the 1 got lost.  I don't now what happened there either.  These things were there originally.  I'm going to go back and make the corrections you pointed out.

Answer to your 3rd question:  $\Delta S$ is zero for an adiabatic reversible path and for any non-adiabatic reversible path between the same two end states.

Answer to 4th question:  It is not an exact differential because there are an infinite number of reversible paths between the same two end states that give the same value for the entropy change, but different values for the heat and work.

And if the reversible paths give different amounts of heat, then they must be transferring at different temperatures, ex. Less total $Q_{rev}$ at lower $T$ such that the integral $\Delta S = \int \frac{\delta Q_{rev}}{T}$ remains the same?

Sure.  Whatever the original path is, you can always tack on a Carnot cycle or two anywhere you want.


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Is sodium aquo complex a Brönsted-Lowry acid?
I know that according to Brönsted-Lowry theory, a base is a proton acceptor while an acid is a proton donor. Also, the acid-base pair that differs by a single proton is a conjugate pair.
If this is true, then the sodium ion is a conjugate acid of sodium hydroxide. But if we look at $\ce{Na^+}$ itself, we find that it cannot donate a proton. Someone in the comments under this answer suggested that in its hydrated state, $\ce{Na^+}$ can act as a Brönsted-Lowry acid. How?

Not a duplicate of this question which is the one about which I seek clarification.

" sodium ion is a conjugate acid of sodium hydroxide" - it is not. Protonate NaOH molecule and you'll find it.

@Mithoron I think I get it. As I understand it, and as someone else wrote on this SE, writing NaOH (aq) is really writing Na+ (aq) + OH- (aq). Protonating OH- gives H20, which is the conjugate acid, not Na+. In sum, is sodium ion not a Brönsted-Lowry acid?

It is an aqua complex and has been so for quite a while now.


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Why can't we say Dimethylbut-2-ene?
Considering the IUPAC rules state that you cannot extend the main chain and that carbon can only have 4 bonds, there is only one way to have two methyl groups on a but-2-ene. Therefore, why do we still keep the '2,3'?

Well, you just wrote that and I'm also pretty sure you can say it aloud ;)

Locants are only permitted to be omitted under very specific conditions, it isn't to do with logic or ambiguity: https://chemistry.stackexchange.com/a/88150/16683

I'm glad that was useful. Hmm, I think we (the Chem.SE community) could make a more canonical Q&A for duplicates, though...


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The algebraic problem in the equilibrium of weak acids and bases
I have had some trouble verifying the systematic treatment in a weak acid base weak balance. The great equation I must arrive at is this:

and I just can't see clearly how to get that equation from these:

doublecheck your expression for [A-] at the bottom. I think you have a Ka where you should have an [H+]. . .Also don't forget to change [OH-] to Kw/[H+] in your charge balance eq.

have a look at https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566 for an alternative and simpler derivation you might find interesting.

It is a set of 4 nonlinear equations with 4 variables. Use a substitution method to eliminate 3 equations and 3 variables but [H+]. Use such a substitution where a variable is eliminated by an expression depending on still remaining variables. But as @porphyrin suggests, seriously consider simplified approach assuming some strong inequalities.

This sort of derivation is a just painful, but straightforward. As Poutnik mentioned it is just a series of equations where you eliminate variables by substitution.
So start with the basics.

$\ce{HA <=> H+ + A-}\tag{A}$
$\ce{BOH <=> B+ + OH-}\tag{B}$
$\ce{H2O <=> H+ + OH-}\tag{C}$
Given:

$\mathrm{C_A}$ is the concentration of the HA reagent
$\mathrm{V_A}$ is the volume of the HA reagent added to the solution
$\mathrm{C_B}$ is the concentration of the BOH reagent
$\mathrm{V_A}$ is the volume of the BOH reagent added to the solution

So there are six unknowns: $\ce{[HA], [A-], [BOH], [B+], [H+]}$ and $\ce{[OH-]}$
Thus we need 6 independent equations to solve the problem.
Step 1
$\mathrm{K_a} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}\tag{1-1}$
$\mathrm{K_b} = \dfrac{\ce{[B+][OH-]}}{\ce{[BOH]}}\tag{2-1}$
$\mathrm{K_w} = \ce{[H+][OH-]}\tag{3-1}$
$\ce{[H+] + [B+] = [OH-] + [A-]}\tag{4-1}$
$\ce{([HA] + [A-])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_A\cdot C_A}}\tag{5-1}$
$\ce{([BOH] + [B+])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_B\cdot C_B}}\tag{6-1}$
So now for the 6 equations just eliminate equations one at a time.
Step 2
So for example we want to know the final equation to only have the unknown $\ce{[H+]}$.  Rearrange equation (3-1) to:
$\ce{[OH-]} = \dfrac{\mathrm{K_w}}{\ce{[H+]}}\tag{3-2}$
Now in equations (1-1), (2-1), (4-1), (5-1), and (6-1) substitute for $\ce{[OH-]}$ and equation (3-1) is eliminated.  So now you're left with 5 equations and 5 unknowns.
$\mathrm{K_a} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}\tag{1-2}$
$\mathrm{K_b} = \dfrac{\ce{[B+]\cdot \mathrm{K_w}}}{\ce{[BOH][H+]}}\tag{2-2}$
$\ce{[H+] + [B+] = \dfrac{\mathrm{K_w}}{\ce{[H+]}} + [A-]}\tag{4-2}$
$\ce{([HA] + [A-])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_A\cdot C_A}}\tag{5-2}$
$\ce{([BOH] + [B+])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_B\cdot C_B}}\tag{6-2}$
Obviously as you continue it gets more and more tedious, but the process is straightforward.

The point is that you have 6 unknowns and 7 equations below. So all of the equations can't be independent. With a problem that is going to be this tedious I like to write out everything explicitly so that:
(1) I don't get confused
(2) So that I can check my work step by step.


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Who named the QM7 and QM9 datasets?
On the surface, this looks like a silly question, but I honestly can't find an answer for it.
Here's what I've found:

The dataset was introduced in Ramakrishnan R., Dral P. et al., “Quantum chemistry structures and properties of 134 kilo molecules”. Scientific Data 1 (2014). doi: 10.1038/sdata.2014.22. This paper doesn't contain the word QM9, and the dataset isn't given a name.
The aforementioned paper gives this link to download the dataset. Again, QM9 is nowhere to be found. No other name is given.
There's a readme file in the dataset folder, copied here. The first line does say Data set dsgdb9nsd, so that's my closest lead to the dataset name.
One of the authors of the paper (Pavlo Dral) has a post regarding this dataset. He says,

The data set is also called QM13 in the “Quantum Machine” website and augments previously used and smaller QM7 and QM7b data sets.

so he seemingly let someone else name it, but even then he got the number wrong (QM13 v. QM9).
On the other hand, people other than the dataset's authors name it QM9:

The quantum-machine.org page
This paper (none of the authors participated in Ramakrishnan's paper)
This master's thesis (sec 2.3.1)
This other paper (again, authors are unrelated)

It's not that I dislike the name QM9. I just want to know if there's a "proper name" for the dataset and also who chose the name QM9.

Is this a chemistry question? Does the origin of the name actually matter?

@JonCuster I don't know where else to ask it, and I've seen some QM9 questions here. The name and its origin matter to me since I'd like to refer to this dataset with a name.

I think it's fair game. IIRC it's because it's a dataset from quantum mechanics (QM) and has molecules with up to 9 "heavy atoms." The various QM7 sets have up to 7 heavy atoms, QM8 set has 8, etc.

@GeoffHutchison So, seemingly, the authors left the dataset unnamed and someone else decided to use the same scheme as the GDB-n databases?

AFAIK, that's correct - and since it came from GDB, the name makes sense (e.g. the filenames). Naming a dataset isn't usually the first thought of authors.

@GeoffHutchison Fair explanation given the circumstances, even though this feels like having a child and letting his school friends name him. Huh.

@Imp54 - well, usually school children do often name each other, the origin of many nicknames for better or worse...


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Is it better to use a smaller, more accurate measuring cylinder several times or a larger, less accurate one for the same volume?
I need to measure out 70 ml of liquid.
I have a 10 ml measuring cylinder with a maximum uncertainty of 0.1 ml and a 100 ml measuring cylinder with a maximum uncertainty of 1 ml.
I've calculated that using the 10 ml cylinder 7 times would give a 0.7 maximum uncertainty, or a 0.7/70=1% percentage uncertainty. Meanwhile, using the 100 ml cylinder, I get a 0.1 ml uncertainty, or a 1/70=1.4% percentage uncertainty. Clearly, the smaller cylinder approach is more accurate (1% < 1.4%)
Meanwhile, the mark scheme says that one must use the 100 ml cylinder, for the 10 ml cylinder would give a total 7*0.1/10=7% percentage uncertainty, compared to the 1.4% from the 100 ml cylinder. (1.4% < 7%)
Isn't adding percentages in this way fallacious? There's no way to accumulate a 4.9 ml error using a 10 ml cylinder 7 times, right? I'd appreciate if someone could clarify which approach is correct and why.

"Maximum uncertainty" is a weird concept. See propagation of error and the treatment of combining variances from multiple sources. You're case is with errors strongly correlated with one another.

@Zhe maximum uncertainty is calculated as half a division on a cylinder. The small one has divisions of 0.2 ml, while the large one has divisions of 2 ml. Hence, their maximum uncertainties are as above.
I'm fairly certain that we are meant to assume that each error has no effect on the next one, so I don't think it has to get this complicated.

Unlike problems on homework assignments, life is complicated. The standard deviation of your measurement is going to increase as you make the same measurement a bunch of times (assuming that your errors are strongly correlated).

@Zhe I appreciate the extra insight you have provided, however my only question is how to calculate maximum uncertainty in a more or less logical way, assuming that the errors are not correlated. That is, which one of the approaches described above better suits our, perhaps crude, model.

The problem is that your crude model really isn't a model at all. What is the underlying probabilistic model you are using for error? Statisticians and mathematicians have worked out how to deal with combinations of random variables, but I am not aware of formulations that deal with "maximum error." They generally deal with the concept of variance or standard deviation as a measure of that error. After all, there's no limit to how badly I can screw up a measurement...

One thing is cylinder resolution, other thing is cylinder precission ( repeatability ) and yet another cylinder accuracy ( correctness of mean value )

Of course you can accumulate a 0.7ml error by using the 10 ml cylinder 7 times. The "10 ml" mark might be 0.1 ml wrong because of a manufacturing error in the cylinder, and however carefully you measure "10 ml" you are actually getting 10.1 ml of liquid every time.

You have three types of errors:

The errors in the accuracy of your measurement.  Assuming you are reading the water levels consistently (at eye level at the bottom of the meniscus), these should be random.
To add random, uncorrelated errors, the standard method is to sum their squares and take the square root.  So, for the 0.1 mL average error for the 10 mL cylinder:
$$\text{average error for seven combined measurements} = \sqrt{7  \times (0.1 \text{ mL})^2} = 0.26 \text{ mL}$$
As you correctly intuited, it wouldn't make sense to simply multiply the 0.1 mL by 7, since this would be saying you have either a +0.1 mL error every time, or a –0.1 mL error every time, but that's not how random errors work—they are just as likely to be positive as negative. In addition, they don't all have a magnitude of 0.1 mL; rather, their magnitudes follow a distribution. Thus essentially what you have is a random walk, with a random distribution of step sizes, where the average step size is 0.1. The square-root-of-the-sum-of-the-squares method gives you the average distance you'd end up from your starting point with 7 of these steps

Error in the calibration of the cylinder.   This would be a systematic error, and it would add to the error of each measurement.  Higher qualities of cylinders (e.g., Class A glassware) have lower errors.

Error in the consistency with which the cylinder delivers the measured amount.  These cylinders are probably calibrated as "TD", which means "to deliver".  That is, it is assumed not all the water will drain out, so the markings on the cylinder are calibrated to account for that.  However, there will be some random error in how closely the amount that drains out corresponds to the assumed amount that drains out.  The only way to determine this error would be to contact the manufacturer (or to test it yourself using a scale).  In addition, if your cylinder is not clean, more water may adhere to the walls, which could cause a systematic error in the amount of water delivered.

Note also the comment by Andrew Morton that there's an additional important practical consideration, which is that if you have to use seven measurements, you could miscount and thus end up being off by  $\pm$10 mL! And this error is much easier to make when measuring with a graduated cylinder than, say, a tablespoon, because each measurement with the cylinder takes focus to get the total volume to 10.0 mL, and that focus can cause you to lose count.  There are of course ways to address this, e.g., making a mark on a piece of paper for each 10.0 mL added.

Thank you for your answer. If I understand correctly, 0.026 is the most likely uncertainty for seven measurements. However, wouldn't a maximum uncertainty specifically involve looking at the extremities, i.e. -0.1 or +0.1 every time? Although 70±0.026 ml is the most likely volume, aren't 69.3 ml and 70.7 the, though unlikely, physical boundaries to how inaccurate the volume ultimately is?
For the sake of simplicity, the question disregards errors 2 and 3. But thank you for pointing those out.

Note that, in the above analysis 0.1 mL is the average error, not the maximum error.   If 0.1 mL were the maximum error, and you made the maximum error each time, and you made it in the same direction each time then, yes, of course you'd get a maximum 0.7 mL error. But how likely is that?  You'd need to take the probability of getting the max error once, and raise it to the 7th power to get the probability of getting the max error 7 straight times.

I agree, it's quite unlikely. The question does say that 0.1 is, in fact, the maximum error. I guess I have ended up with the maximum, however unlikely, yet possible error, while you have provided the maximum most likely error.

This is a great answer.  In 3 it misses the point that if you are measuring a fluid other than pure water, the volume that sticks to the cylinder may be different.  This is another source of systematic error because the density may be different and the adhesion between the solution and the glass may be different.  Systematic errors are hard to find because you just have to think carefully about what can go wrong unless you have an accurate check.  If you have an accurate check it is often a better way to measure.

Also, there is the possibility of an off-by-one error if the counting to seven is not accurate. It might seem to be a simple task, but a distraction could lead to an error.

With the small cylinder, it's really easy to be off by 10 ml as you've miscounted the number of transfers.

@AndrewMorton Agreed, that's an important practical consideration.   I've added your point to my answer.

Keep in mind that a graduated cylinder is never meant for accurate volume transfer. You would probably use a Class-A 100 mL buret to transfer 70 mL to a vessel if accuracy were that critical say, 70.0 mL was desired.  All these general chemistry rules of thumbs are not useful for real analytical work. Experimental numbers matter the most.
Let us assume you don't have a 100 mL burette. This is how a practical scientist would do it. You would like know whether 7 transfers from a 10 mL cylinder are better or 1 transfer from a 100 mL cylinder is better? You calibrate your cylinders. You know the density of the liquid. You would transfer 10 mL, 7 times and weigh it on an analytical balance. Note the mass and from the mass-density-volume relationship, determine the volume.
Then repeat the same with a 100 mL cylinder, and transfer 70 mL in one go. Measure the mass and hence volume. Whatever delivers the right volume, is the way to go.
Practically, 7 transfers will generate more errors because you do not know the drainage time of the cylinder, i.e., how long should you wait till all the film of liquid is out of the cylinder as the last drop of liquid. Also see the rules for the propagation of error.
By error propagation rules, uncertainties cannot be added like you have shown. It is an incorrect approach. The cylinder with 0.1 mL error will not add up to 0.7 mL by any means. if you make 7 additions.

A practical scientist would know that graduated cylinders are to contain, not to deliver. So if you have seven of the 10 mL cylinders, you could fill them pretty accurately with 70 mL total. It would just be difficult to remove the entire volume, and you would have to do a lot of dishes.

And if you have a balance, you would use the 100 mL graduated cylinder to figure out the density, and then use the balance to measure out the 70 mL into the container you need it in (unless it is too heavy, in which case you would empty the 100 mL graduated cylinder, add 70 mL via the balance, and then empty it again to deliver roughly 70 mL).

Karsten, There are TD (to deliver) and TC (to contain) graduated cylinders. It would be foolish to use a TC cylinder for transferring volumes.

Thank you for the detailed answer. Perhaps I should have clarified that my question makes several unrealistic assumptions for the sake of simplicity. It is assumed that liquid is transferred entirely (disregard drainage time and the such) and that each error/uncertainty has no effect on the next one. (Frankly, I fail to see how these measurements even interfere with each other, but ok)

In essence, the problem comes down to us simply being unable to know what precise volume of liquid we get. If we want to get 10 ml, we know that every time we will get something between 9.9 and 10.1 ml. Therefore, our most inaccurate possible measurements are 710.1=70.7 ml and 79.9=69.3 ml, which are our desired volume (70 ml) plus or minus 0.7 ml (or 7*0.1 ml). Any other volume obtained will be between those two numbers. Since we are looking for maximum uncertainty, it is 0.7 ml.
Given the assumptions, I fail to see the flaw in this approach. I would appreciate if you could clarify this.

@JohnJohnson, Uncertainties do not add up linearly. It follows the rules of error propagation. You can see the maximum uncertainty as shown in the next answer is 0.026 mL by using a 10 mL cylinder 10 times not 0.7 mL. By doing a linear addition, you are eliminating the random character of the error.

@M.Farooq I did not know about TD graduated cylinders. I would imaging they are for a very limited range of liquids (i.e. not 20% glycerol or pure mineral oil).

@Karsten, Yes, but high viscosity issue applies to all TD volumetric glassware. This is why I recommended weighing to actually see how much is transferred. See reverse pipetting technique https://en.wikipedia.org/wiki/Reverse_pipetting

Meanwhile, the mark scheme says that one must use the 100 ml cylinder, for the 10 ml cylinder would give a total 7*0.1/10=7% percentage uncertainty, compared to the 1.4% from the 100 ml cylinder. (1.4% < 7%)

Let's simplify the problem and say the 10 mL cylinder has a +1% systematic error (and there is no random error), i.e. delivers 10.1 mL when filled to the 10 mL line. So filling it up to 10.1 mL seven times will deliver 70.7 mL instead of 70.0 mL. That is a 1% systematic error.
So the relative systematic error does not change comparing delivering a single volume to delivering the sum of multiple measurements. The absolute error, of course, increases (i.e. the difference between the intended and the actual volume).
In the calculation of the marking scheme, the denominator is 10 (mL), but it should be 70 (mL). If you look at the other answers, which are excellent, that is not the only thing the marking scheme got wrong. The bigger problem is that the question does not make sense in face of the reality of measuring volumes in the lab.


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Flux a specific concentration of a tracer constituent of a gas
I need to flux 250 ppm of CO2 at 1000 ml/min.
I have a gas system with a tank containing CO2 (CO2 at 5000 ppmvol, Oxygen 21%vol and Nitrogen the rest), connected with a mass flow controller (MFC1, flux max at 150 ml/min) and I have a tank of dry synthetic air (O and N), connected with a second mass flow controller (MFC2) that can be fluxed at a max of 1000 ml/min. Since I don’t have the possibility to measure the ppm of the CO2 after the dilution, I need to know a right way to dilute the CO2 in my tank (that already contains a mix of CO2 and air).
I know that I can digit the values of 50 ml/min for MFC1 and 950 ml/min for MFC2, obtaining a final flux of 1000 ml/min. But, in this way, I don’t understand the final concentration of CO2.
I don’t know if applying this formula 5000ppm*(MFC1/tot flux) [that means 5000ppm*(50ml/min : 1000 ml/min)] I am double diluting my gas (because I am considering two times the air, once mixed with the CO2 tank and the other ones in the other air tank). Or, I don’t know if this formula [5000ppm*(50ml/min : 950ml/min] is the correct one, since apparently I consider air in the CO2 tank and air in the other tank.
I don’t even know if my approach to this problem is wrong and both formulas are not correct (e.g. I should consider other parameters; the carbon dioxide is a pollutant and so it follows particular rules/formula/conversion).
Thanks.

Recalculate both flows to flows of pure CO2. Both values must match.

If I consider to flux 1000 ml/min of dry air, I don’t have carbon dioxide; if I flux 1000ml/min only from the carbon dioxide tank, I have 5000ppm of CO2 + air.  If I consider to have a tank of pure carbon dioxide at 5000ppm, I can hypothesis to flux 50ml/min from the carbon dioxide tank and mix it with 950ml/min of dry air, obtaining a tot flux of 1000m/min. This way, I should obtain 250ppm of CO2 (I applied 5000*(50/1000)). But I have carbon dioxide+air and I add other air, so, since I cannot experimentally verifying the final ppm I am fluxing, I would like to be sure on what I am doing.

@Poutnik  looking at my previous comment, to me the calculations do not match so, I would like to know which calculation is correct or where and why it is wrong.

1000 mL/min of 5000 ppmv CO2 is equivalent to 5 mL/min of pure CO2. To have 250 ppmv CO2, you need the total flow 5 1000000/250=54000=20000 mL/min, So 20 times more. In fact, it is very trivial high school calculation. Do you need 20 times diluted solution? You need 20 times bigger volume. That simple.20 times higher volume is achieved by 20 times higher flow.


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BPS & BPA treatment with 80% Ethanol
Background: BPA and BPS (Bisphenol A/S) are artificial estrogens and therefore toxic chemicals for us. They are used in the coating of thermal paper. Thermal paper is used to print in-store receipts. There is a growing demand for thermal paper that is free of BPA and BPS for the health and safety of the public.
Question: how do we test it? I heard from someone that both BPS and BPA dissolve in ethanol. If we apply 80% Ethanol solution (commonly found in hand sanitizer) on the coating side of the thermal paper then it reacts (and dissolves) with BPS and BPA; as a result, you notice black and greenish-black spots start to appear on the paper. If the paper is free of BPS/A then no spot or marks will appear on coat-sided paper.
Is this method of testing correct? Does it at least give some accurate idea? What could you suggest for testing?

Bad idea. If you have ethanol on your hand and touch the paper, you invite the bisphenol to diffuse into your skin. Buy from reputable sources, ask for test reports by reputable laboratories.

No that testing method is meaningless. The darkening of the thermal paper is, as expected, from heat or friction of your fingers. Ask the manufacturer.

The problem with any test for BPA or related compounds is that it has to be sensitive to very small amounts and to be specific for those compounds. Simple test are unlikely to be reliable or robust. Plus the description of BPA as "toxic" is disputed, especially given that it is often claimed without the important quantification for any toxin of "what amount  creates a notable effect".

The only reliable method I can come with is LC-MS (liquid chromatography–mass spectrometry) of extract of the paper in some organic solvent. The method, when properly tuned, is sufficiently fast, precise and accurate, but requires special equipment.

I agree with M. Farooq, a test that might work is cutting small pieces of thermal paper and placing them in a tube with grain alcohol or a facsimile, and allowing them to soak for a few hours, then send the sample to a lab and ask them to test for bisphenol A/S. If you can, it's just better to ask the company.
Edit: you can just soak them in grain alcohol. It's hard to do a specific test for bisphenols, but I found one for phenols in general.
Procedure:

Prepare a dilute solution of ferric chloride in grain alcohol.
Using an eyedropper, add a few drops of the solution to a teaspoon of your solution to test.
If a color change is observed, phenols are present.

This is not ideal, but it's the best you can do.

thanks everyone. sending for a lab test is expensive. I want some makeshift testing procedure, though I understand it is not completely safe. I tried on several receipts. I found no spot appearing on BPS/A free thermal paper and spots do appear on BPA free thermal paper.

thanks everyone. sending for a lab test is expensive. I want some makeshift testing procedure, though I understand it is not completely safe. I tried this idea on several receipts. I found no spot appearing on BPS/A free thermal paper, and spots do appear on BPA free thermal paper. can you please just confirm how do BPS/A react with 80% Ethanol? Should this leave any colored residue in chemical reaction?

Check my answer, I updated it.


text:
Calculating the solubility for chromium(III) hydroxide using the complex ion constant
Problem

$$
\begin{align}
\ce{Cr(OH)3(s) &<=> Cr^3+(aq) + 3 OH-(aq)} &\quad K_\mathrm{sp} &= \pu{6.7E-31} \\
\ce{Cr^3+(aq) + 4 OH-(aq) &<=> Cr(OH)4-(aq)} &\quad K_\mathrm{f} &= \pu{8E29}
\end{align}
$$
The solubility of $\ce{Cr(OH)3}$ in pure water is $\pu{1.3E-8 M}.$
The solubility of $\ce{Cr(OH)3}$ in $\pu{0.50 M}$ $\ce{NaOH}$ when the complex ion $\ce{Cr(OH)4-}$ is formed is:
a) $\pu{0.32 M}$
b) $\pu{0.59 M}$
c) $\pu{0.27 M}$
d) $\pu{0.18 M}$

Answer

 d) $\pu{0.18 M}$

My approach
$$\ce{Cr(OH)3(s) + OH-(aq) <=> Cr(OH)4-(aq)} \quad K = K_\mathrm{sp}\cdot K_\mathrm{f}$$
$$\pu{6.7E-31}\times\pu{8E29} = \frac{s}{\pu{0.50 M}}$$
$$s = \pu{0.27 M}$$
As you can see I got c), which is not the correct answer. Could anyone point out what I did wrong?