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Work done in Irreversible Isothermal Expansion

This is the graph for the Isothermal Expansion of an Ideal Gas. Sorry for the extremely basic doubt, but I just could not understand how the expansion was Isothermal if the graph was a straight line. The pressure is changed once suddenly and then remains constant.
Why don't we consider this as an Isobaric expansion instead of Isothermal if the graph is not even an isotherm? If a gas is expanding but the pressure has to be constant then why doesn't the temperature increase?

I agree: It's not an isothermal process. It merely starts and ends at the same temperature.
It consists of two steps: an isochoric (constant-V) process, in which the pressure is decreased by cooling the gas at constant V, followed by an isobaric process, in which the volume is increased by heating the gas at constant p, returning it to its original temperature (but not its original pressure or volume).
As to what Chet Miller writes:

In short, the answer given by @theorist isn't even close to being correct.

"Isn't even close to being correct" is actually a good description of his own comment.
I stand by what I said.  The OP posted a graph that showed how the pressure and volume of an ideal gas varied during a process.  I can only answer based on what was posted. Logically, for the first leg, if p is going down at constant V, the system must be cooling.  And for the second leg, if V is going up at constant p, the system must be heating.  [We're assuming the system is closed.] It is easy to accomplish this: Simply put the system in contact with a lower-T bath while keeping V constant, and then put the system in contact with a bath at the original T, keeping p constant.
For instance, we could start at 300 K, 1 L, and 2 atm.  Then we could gradually cool at constant V to 150 K, 1 L, 1 atm.  Then we could gradually heat at constant p to 300 K, 2 L, 1 atm.
As for his niggle that T and p can never be defined during an irreversible process, that's an unfortunate and absolutist position.  In practice, it certainly is possible to define and measure T and p during irreversible processes, particularly if we make the change slow enough that the system can relax close enough to an equilibrium state such that our measuring devices can't tell the difference—or that the difference is less than the precision with which we care to report our measurements. [This is not to be confused with a true reversible process, which is an idealization that can't exist is the real world.]
And the diagram, by showing the pressure of the system throughout the process, is telling us just that: That the system is sufficiently relaxing during the process to have a measurable p. I.e., yes, normally the default assumption during an irreversible process is that the intensive variables are undefined (this is sometimes indicated using a dashed line).  But here the diagram, by using an unbroken solid line to describe p throughout the process, is telling us p is measurable.
Consider the converse: If we hold to such an absolutist view, that systems must be at equilibrium for us to be able to describe their temperature and pressure, then we can never measure or discuss the temperature or pressure of any real-world system, since no real-world system is ever truly at equilibrium.  So essentially all the temperature and pressure measurements in all the world's scientific literature are invalid.
Note that this would also prevent us from talking about the temperature or pressure change in a Joule-Thomson expansion, since such an expansion is a steady-state process so it is, by design, never at equilibrium.
I see from Chet's profile that he's a retired chemical engineer.  To be logically consistent with the view he presents here, anytime his colleagues or bosses asked him about the temperature or pressure of a system, he'd always have to say:  "No real-world system is ever at equilibrium, so they have no definable temperature or pressure."  Do you think that's what he always said?
The process Chet is ascribing to the question is different from what's pictured.  He's taking the diagram to represent the V of the system, but the p of the surroundings. Based on that assumption, his description is correct.  But, while that may be the intent of the question, there's nothing in the presented diagram to indicate that's the case.  He is, as the lawyers say, "assuming facts not in evidence."
As to his other comment:

Really???? So, in stage 1 you start out with two entries, the gas and the reservoir, at the same temperaturer. Then, spontaneously, heat flows from the gas (at constant volume) to the reservoir so that, at the end of stage 1, they are at different temperatures (with the gas colder). Do you really think that this is consistent with the 2nd law of thermodynamics?

I applaud his restraint in limiting himself to four question marks. Had he used five, I would be concerned.
He's attributing to me a bizzare process that bears no relationship to anything I said, nor to what I had in mind, nor to what's pictured by the diagram.  I have no idea how or why he came up with this.  It's so strange it took me a while to figure out what he was trying to describe, and almost seems like a straw man. To adopt Chet's parlance: Really??? [I'm not as well-trained as Chet in hyperbole, so I can only go to three question marks, max.]

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

Here is a detailed analysis of irreversible isothermal expansion and compression in a thread from PhysicsForums.com:  https://www.physicsforums.com/threads/going-from-1st-state-to-2nd-through-irreversible-isothermal-expansion.990952/. Please feel free to join and weight in.

I stand by what I said about an irreversible expansion.   For a gradual reversible expansion, we are totally in agreement.  But, to me,  the diagram indicates that the expansion takes place irreversibly.  The implication of the diagram is that the (external) pressure is suddenly reduced at time zero, and the gas is allowed to expand at this constant externally applied pressure to the final volume (in contact with a constant temperature reservoir at the original gas temperature).  To me, no other interpretation is possible.

If you disagree with my interpretation of the process diagram, then I guess we will just have to agree to  disagree.

In the process you describe, it you put the gas at 300 K in contact with a constant temperature reservoir at 150 K and cool it at constant volume, it will not cool gradually and and it will not cool homogeneously.  Then if you put the gas at 150 K in contact with a constant temperature reservoir at 300 K and heat it at constant pressure, it won't heat gradually and it won't heat homogeneously.  However, the actual process you described is irreversible, and an alternate irreversible path to the one I described.  For either path, the integral of dq/T will be less than the entropy change.

Incidentally, the problem statement says that "the pressure is changed suddenly."  Your description of first cooling at constant volume and then expanding at constant pressure is inconsistent with this.

@ChetMiller There's no point in getting into your penultimate comment until we address the issues in the last one. The OP's descripton that "The pressure changed suddenly" is an imprecise and ambiguous qualitative comment that can't be relied on to tell us exactly what happened. Is there an unambiguous source of info. about the process? Yes, it's the graph the OP supplied. From that, we can see the process has two legs: From point 1 to the corner and from the corner to point 2. Take a look at the graph and tell me what happens to the volume in the first leg. Does it change? No, it's constant.

...If it were, say, 25 L at the start of the the first leg (point 1), then it would be 25 L at the end of the first leg (the corner).  What happens to the pressure during the first leg?  The graph tells us it decreases.  So what we have is a constant-V and constant-n* process for an ideal gas, in which p decreases.  There's only one way that can happen:  T decreases.  Thus we have, exactly as a I described, cooling at constant V, i.e., isochoric cooling. [*Constant-n isn't explicitly specified, but you and I are both assuming this, since a closed container is the typical default assumption.]

The externally applied pressure can decrease instantly, without changing the temperature.  The development of viscous stress in proximity with the piston face can make this happen.  I'm guessing you've never had a course in fluid mechanics beyond the Bernoulli equation.  You are aware that the ideal gas law is applicable only at thermodynamic equilibrium, right?

@ChetMiller  "The externally applied pressure can decrease instantly, without changing the temperature. The development of viscous stress in proximity with the piston face can make this happen."   You're starting from an equililbrium state (point 1).  Given this, explain to me how what you describe happens without any change in the volume, assuming constant n (without invoking bizarre mechanisms that cleary aren't intended here, like you have dual pistons where you move one in while moving the other out, in which case there would no longer be a single externally applied pressure anyways).

Go take a course in Newtonian fluid mechanics and get back with me when you know something.  Until you have had fluid mechanics, you will never properly understand how stresses (forces per unit area) develop in rapid irreversible deformations of gases and other fluids.

How, pray tell, do you deform a gas, or decrease the "externally applied pressure", when you start in an equilibrium state, have a perfectly rigid container (no change in size or shape) and don't change the temperature?  [Barring operations clearly not intended by the OP, like accelerating the container as a whole.]  You wrote "Go take a course in Newtonian fluid mechanics and get back with me when you know something".   I don't need a course in fluid mechanics.  You need a course in logical thinking.

I'm willing to start over and be more civil this time.  I hope you will agree to start over also.

I have unfrozen the chat room for extended discussion. Please use that forum for continuing the discussion.

There is some confusion about terminology. The figure shows two processes. The curved path represents a reversible expansion in which there is pressure balance between system and surroundings throughout, so that the applied pressure can be described by the ideal gas law (or Boyle's law $p=\frac{p_1V_1}{V}$). This process is clearly isothermal, since $pV=nRT$.
The second process shown as two straight paths consists of a pressure drop and and an expansion against constant external pressure and is also isothermal (in thermodynamic lingo). The fact that the initial and final points are at equal T makes it isothermal. The temperature (and other properties except presumably composition) of the system during the process is on the other hand not well-defined. This process is irreversible.

I disagree. Isothermal means constant-T, just as isobaric means constant-p.  A process in which p_initial and p_final are the same, but in which p was varying all over the place during the process, would not be isobaric.  Same thing with isothermal.  What you are describing as "thermodynamic lingo" is not -- it runs counter to standard usage in physical chemistry  textbooks.

For instance, from Castellan 3rd Edition:  "In the isothermal cycle described above, the net work produced in the irreversible cycle was negative, that is, net work was destroyed. This is a fundamental characteristic of every irreversible and therefore every real isothermal cyclic transformation. If any system is kept at a constant temperature and subjected to a cyclic transformation by irreversible processes (real processes), a net amount of work is destroyed in the sur roundings. "

And from Engel and Reid, 3rd Edition: "The simplest processes are those in which one of P, V, or T remains constant. A constant temperature process is referred to as isothermal, and the corresponding terms for constants P and V are isobaric and isochoric, respectively."

@theorist What does the irreversible step in Castellan's isothermal cycle look like?

@theorist I realize it's not an ideal source, but the Wikipedia says: "Once obtained, these formulas can be applied to an irreversible process, such as the free expansion of an ideal gas. Such an expansion is also isothermal and may have the same initial and final states as in the reversible expansion." The temperature of the system is ill-defined during a free expansion.

In the parlance of thermodynamics, an irreversible isothermal process is defined as one in which the system and constant temperature reservoir start out at the same temperature and the system is held in contact with the constant temperature reservoir throughout the process, so that, in the end, they re-equilibrate at the same temperature.
In an irreversible process, one cannot use the ideal gas law to describe the gas behavior, because the ideal gas law is valid only at thermodynamic equilibrium.  A reversible process is characterized by a continuous sequence of thermodynamic equilibrium states, so the ideal gas law can be used for such a case.
In an irreversible process, in addition to the pressure being a function of the gas volume, it is also a function of the rate of change of volume. Fluid Mechanics tells us that this is caused by viscous stresses which contribute to the force per unit area exerted by the gas on the piston.  In an irreversible expansion, these stresses lower the pressure that the gas exerts on the piston, so that, for a massless frictionless piston, the lower pressure exactly matches the externally applied reduced constant pressure.  To provide the required match, the gas has to be expanding rapidly and at the right rate so that the viscous stresses are just right.
During the irreversible expansion, the gas temperature is not spatially uniform.  The part of the gas in contact with the constant temperature reservoir is, of course, at the reservoir temperature, but the interior of the gas becomes cooler.  However, by the end of the expansion, when thermodynamic equilibrium has been re-established, the gas temperature is again uniform and equal to the reservoir temperature.
This is all a one-stage process, and, by controlling the pressure on the piston so that it suddenly drops initially and then stays constant throughout the deformation, this can be considered a constant-pressure expansion.
In short, the answer given by @theorist isn't even close to being correct. @Buck Thorn's answer is much closer to the target.

See the following thread for lots of details on this:  https://www.physicsforums.com/threads/going-from-1st-state-to-2nd-through-irreversible-isothermal-expansion.990952/

As a followup to the answer I gave previously and to my discussions with @Theorist, I have analyzed in detail the  alternate irreversible process paths (1) specified by Theorist in his answer and (2) specified in the problem statement (and referred to in my answer as well as that of @Buck Thorn). To provide a common basis for comparison, I have derived the integral of $dQ/T_R$ for each of the two alternate irreversible process paths, where $T_R$ is the reservoir temperature exchanging heat with the gas (this is the temperature that should be used in conjunction with the Clausius inequality, as pointed out in Thermodynamics by Fermi).
For the process path specified by Theorist, it is found that $$\int{\frac{dQ}{T_R}}=nR\left(1-\frac{V_1}{V_2}\right)-nC_v\left(\frac{V_2}{V_1}+\frac{V_1}{V_2}-2\right)$$
For the process path specified in the problem statement (and by Buck Thorn and myself), it is found that $$\int{\frac{dQ}{T_R}}=nR\left(1-\frac{V_1}{V_2}\right)$$
For any reversible path, the integral is equal to the entropy change, which, of course, is the same for all three paths:$$\Delta S=nR\ln{(V_2/V_1)}$$For an irreversible path, the amount of entropy generated in the process is given by $\sigma$:
$$\sigma=\Delta S-\int{\frac{dQ}{T_R}}$$So, for the path specified by Theorist, $$\sigma_{Theorist}=nR\ln{(V_2/V_1)}-nR\left(1-\frac{V_1}{V_2}\right)+nC_v\left(\frac{V_2}{V_1}+\frac{V_1}{V_2}-2\right)$$and, for the path specified in the problem statement, $$\sigma_{problem\ statement}=nR\ln{(V_2/V_1)}-nR\left(1-\frac{V_1}{V_2}\right)$$Note that the first two terms on the RHS of these entropy generated expressions are identical; however, the third term in expression for Theorist's path is positive for all values of the volume ratio.  This indicates that, in this sense, Theorist's path is more irreversible since it involves more entropy generation.  This is primarily the result of the first step in his process which involves irreversible cooling of the gas at constant volume by use of a lower temperature reservoir.  This is not a criticism of the irreversible path he chose to specify (although it is not consistent with the problem statement). It is just a feature of that particular irreversible path.


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Expression for Gibb's free energy change for a cell reaction
According to Chemistry Libretexts, the Gibbs free energy change of a cell reaction is related to the cell voltage as
$$ \Delta G = -nFE,$$
where $n$ is the number of moles of electrons passed and $F$ is the charge on a mole of electrons.
But $E$ decreases in value through the course of the cell reaction, so  shouldn't
$$\Delta G = -\int_{q_\mathrm{i}}^{q_\mathrm{f}} E\,\mathrm dq,$$
where $q_\mathrm{f} - q_\mathrm{i} = nF?$

It's strange to see a question like that from Michael Faraday.

According to Nernst's equation,
$$E = E^o - \frac{RT}{nF}\ln Q$$
As you can see, $E$ is independent of $q$, therefore your proposed integral becomes,
$$\Delta G = - \int_{q_i}^{q_f}{E\cdot dq} = -E\int_{q_i}^{q_f}{dq} = -E\Delta q = -nFE$$

Alternatively, you can prove the the relation between Gibbs energy and cell potential as,

Multiply Nernst's equation by $nF$, we get,

$$nFE = nFE^o - RT\ln Q$$

Swapping the sides, we get,

$$-nFE = -nFE^o + RT\ln Q$$

As we know,

$$\Delta G = \Delta G^o + RT\ln Q$$
$$\Rightarrow \Delta G = -nFE$$

so you are saying  that the concentration "Q" is completely unrelated to the charge?

But 'q' is the amount of charged passed from the cathode to the anode of the electrolytic cell. And for charge to pass, a species must be reduced at the cathode and a species must be oxidized at the anode. The concentration of the species in the chemical equation changes, and hence Q is related with 'q'.. And if Q is related with q, then E is also related with q (from Nernst equation)..
Correct me if I'm wrong..

Any chemical reaction follows some "nth order kinetics", so the conc comes out to be a function of time, therefore Q should also be a function of time. Now, the integral is with respect to charge passed, hence E comes out.

@Rahul Verma if Q is a function of time, then by nernst equation, E is a function of time. Since it requires finite time to pass charge, E would vary as the charge is being passed. So E cannot be taken out of the integral since it's not constant.

@MichaelFaraday: as you know, ∆G is a state fn, so I think the integral isn't required. It is because ∆H, ∆S and T are state / point fn

source: Stack Exchange

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

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Liquid storage of transition metal oxides?
Transition metal oxides (TMOs) have a lot of interesting properties, from ferroelectricity to battery applications. Often times you can change the properties with only slight changes of oxygen stoichiometry. For example, SrTiO$_3$ is an insulating paramagnet, but becomes ferroelectric and then metallic with slight changes in oxygen stoichiometry in SrTiO$_{3+\delta}$. Another example is La$_2$CuO$_4$ which is an insulator, but with increasing oxygen content La$_2$CuO$_{4+\delta}$ becomes a superconducting metal.
One downside of this functionality is that transition metal oxides are notoriously finicky, and tend to change properties upon exposure to air, moisture, etc. In many cases, bulk polycrystals of TMOs have to stored under inert gases or vacuum to maintain their properties. Needless to say, that is quite annoying with regards to transportation.
My question is: is it possible to store transition metal oxides in liquids without significant deterioration? If so, which ones are the most practical? I'm sure the answer is "it depends", but even general suggestions and publication references would be useful.

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


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What is the unit of Z effective?
Z effective is the net positive charge experienced by an electron. But what is the unit of this net charge: coulomb (C) or microcoulomb (μC), or something else?

What is the charge of a magnesium ion? Which units are you using in your answer?

This is an example of terminology which should be taken with a grain of salt.
The term "effective nuclear charge" is often casually symbolized with $Z_\mathrm{eff}.$ This is a universally accepted simplification, but you should keep in mind that the effective nuclear charge is, strictly speaking, $Z_\mathrm{eff}e,$ where $e$ is the elementary charge.
Similarly, the full nuclear charge is $Ze,$ where $Z$ is the atomic number.
So, it would be more appropriate to refer to $Z_\mathrm{eff}$ as to effective atomic number since $Z_\mathrm{eff}$ is defined as a function of $Z$ $(Z_\mathrm{eff} = f(Z_i)).$
For example, the common form provided in every introductory physical chemistry textbook is
$$Z_\mathrm{eff} = Z − σ,$$
where $σ$ is the shielding constant of the nucleus (empirical dimensionless parameter).
Since $Z$ is dimensionless, so is $Z_\mathrm{eff}.$
However, effective nuclear charge $Z_\mathrm{eff}e$ is not, and can, for instance, be expressed in coulombs (in the SI system $e = \pu{1.602176634E−19 C}).$


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Leaving Group Tendencies in Amide Hydrolysis
When we do hydrolysis of amides in basic conditions, we come across a mechanism in which the $\ce{OH-}$ ion attacks the carbonyl forming an intermediate as given in the photo:

When the carbonyl gets formed again it ejects off $\ce{H2N-}$-that’s what my teacher says. But I have a problem that why this happens, why not the $\ce{OH-}$ group leaves off when the carbonyl gets formed again? After all, $\ce{OH-}$ is a better leaving group than $\ce{H2N-}$.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/24%3A_Organonitrogen_Compounds_II-_Amides%2C_Nitriles%2C__Nitro_Compounds/24.04%3A_Hydrolysis_of_Amides#:~:text=The%20mechanism%20involves%20protonation%20of,aqueous%20sodium%20or%20potassium%20hydroxide. Does this help?

You are generally correct that NH2- is a worse leaving group the HO-.  It is, however, not so bad that it never happens, which is why basic hydrolysis of amides is not a great reaction requiring forcing conditions and long reaction times. What makes it viable is that H2N- immediately grabs a proton from solvent and the NH3 boils out of the reaction mixture (see LeChateliers Principle).

The hydroxide leaves all the time in a non-productive reaction. But eventually, you'll get the amide to leave. This is why basic hydrolysis of amides frequently requires heating to speed up the reaction.

The reversible steps are also driven to completion by the deprotonation of the carboxylic acid that is formed after the loss of -NH2.

You can imagine following situation in this hydrolysis:
$$\ce{R-C(=O)NH2 + OH- <=>[$k_1$][$k_2$] R-C(NH2)(OH)O- \\->[$k_3$] R-C(=O)OH + H2N- ->[$k_4$] R-C(=O)O- + NH3}$$
Keep in mind that:
$$\ce{R-C(=O)NH2 + OH- ->[$k_1$] R-C(NH2)(OH)O- }$$
$$\ce{R-C(NH2)(OH)O-  ->[$k_2$] R-C(=O)NH2 + OH-}$$

In this reaction, $k_1 \gt k_2$ because of LeChateliers principle since $[\ce{OH-}]$ is in excess thus forward reaction is in favor.
Also, $k_2 \gt k_3$ because of the reason given by OP ($\ce{OH-}$ is better leaving group than $\ce{H2N-}$).
Since acid-base reaction is much faster than any other reaction, we can conclude that $k_4 \gt \gt k_3$.

Therefore, as soon as $\ce{R-C(=O)OH + H2N- }$ formed they would convert to $\ce{R-C(=O)O- + NH3}$ quickly so that $\ce{R-C(=O)OH + H2N- }$ can not able to go back to $\ce{ R-C(NH2)(OH)O-}$ intermediate. That means, when boiling with extra basic solutions, amide will be hydrolyze slowly to its corresponding carboxylate.


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Can metal matrix composites be alloyed after their creation?
I've done a little research on metal matrix composites (MMCs), purely for fun and out of interest. Different types of metal powders are combined with reinforcement materials, to produce a reinforced metal.
As the structure of the reinforcing material is an important factor, would the resulting composite then be able to be alloyed with other metals? If I made a magnesium-carbon fiber composite, could I then melt it and combine it with, let's say copper? Or would the process destroy the structure of the carbon fibers, weakening the material?
Also, what would happen, assuming alloying works, if I combined two MMCs with different reinforcements, let's say, for example, one is an aluminum-glass fiber composite, and the other is a titanium-ceramic particle composite? Could they combine?

No, it cannot...

So much depends on a number of factors:  What specific metals do you want to alloy? What temperatures do you have to attain to get the metals to alloy? What is the stability [both chemical and thermodynamic] of the matrix used?  If your were more specific as to alloying metals, relative quantities, and matrix used, more focused answers may be possible.


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Side product obtained in reduction of cuprous oxide with carbon
So my book sates that in the extraction of cuprous oxide, the following reaction takes place.
$$\ce{Cu2O + C -> 2 Cu + CO}$$
This makes sense to me, if it is seen as the following two reactions coupled.
$$
\begin{align}
\ce{Cu2O &-> Cu + 1/2 O2} \\
\ce{C + 1/2 O2 &-> CO}
\end{align}
$$
But as per the Boudouard reaction, carbon monoxide is the dominant form of oxide of carbon at higher temperatures, however using carbon for reduction is also possible at lower temperatures, which can be seen from Ellingham diagram.
So, are both $\ce{CO}$ and $\ce{CO2}$ possible as side products depending on temperature at which the extraction of $\ce{Cu}$ is done?

@ Paras Khosla. Your first equation is correct, but not your second one : $\ce{Cu_2O}$ reacts with $\ce{C}$ without getting decomposed previously into $\ce{Cu + O_2}$. $\ce{Cu_2O}$ is more stable than $\ce{CuO}$ above $1000$°C
Furthermore, $\ce{Cu_2O}$ does not react easily with Carbon in the solid state. The contact between the reacting solids is not good. The contact is better if $\ce{Cu_2O}$ is liquid. But $\ce{Cu_2O}$ melts at $1026$°C. So, the reaction starts at this temperature, forming $\ce{CO}$ which is more stable than $\ce{CO_2}$ at such a high temperature.


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Why is the 1,6 linkage in Amylopectin an alpha link and not a beta link?

The alpha link as I have understood it means that the $\ce{OH}$ molecules which make up the glycosidic bond lie on the same side of(above/ below) the plane and for beta, one lies above the plane and the other below it.
For the 1,6 linkage in Amylopectin, one $\ce{OH}$ group lies below the plane and the $\ce{CH3OH}$ group lies below the plane. Then why is the 1,6 link in amylopectin not an alpha linkage?

Your question is unclear to me and in danger of being closed. Amylopectin has alpha linkages. If the linkage were beta, amylopectin would not be digestible by humans.

The 1,6 bond in amylopectin is labeled as alpha link and it looks similar to the Beta links of cellulose. (One OH below the plane and one OH above the plane. I am talking of those OH condensed to make the glycosidic link.) So i am unable to understand how the 1,6 link in amylopectin is labelled as alpha as it very much resembles the Beta nature.

"OH molecule"? Amylopentin? is 1,6 link is? I understand that articles are difficult if you don´t come from a indoeuropean language, but this is embarassing. The effort people put into their answers is usually directly proportional to the effort you put into writing up your question!

The title and body of the question contradict each other, making it difficult to take it seriously.

Actually I didnt understand the meaning of alpha link correctly previously. I now undertsand that when two alpha anomers make up a link its called alpha link ; same for beta; and if alpha and beta make up the link its called alpha - beta. Although i previously knew what alpha anomer means i was confusing it with the meaning of what alpha link means. Anyways i have got it cleared up now.

Alpha and beta refers to the stereochemistry at the anomeric carbon. The 4 and 6 positions in glucose are not anomeric, and there is only one way to link them. So you have to look at the 1 position that is part of the glycosidic bond. Sucrose has a alpha-1, beta-2 linkage (both glucose and fructose form a glycosidic bond involving their respective anomeric carbon).

You seem to be having trouble with the terms alpha and beta and how to recognize these terms in carbohydrates. Here is an easy way to recognize them when they are drawn in the conventional way shown below.. In a Fischer projection of glucopyranose,  for example, the hydroxyl group in the beta anomer is on the left; the alpha anomer is on the right. [Notice that the italicized terms have fewer letters than the bold terms!] In the Haworth projection and the chair conformation of glucopyranose the terms beta, top and up have fewer letters than their respective terms alpha, bottom and down in the other anomer. Amateur sailors use the same mnemonic to remember that the port is on the left (red light) side of the boat and that the starboard is on the right (green light).

Source: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/carbohydrates44.pdf

Actually I didnt understand the meaning of alpha link correctly previously. I now undertsand that when two alpha anomers make up a link its called alpha link ; same for beta; and if  alpha and beta make up the link its called alpha - beta. Although i previously knew what alpha anomer means i was confusing it with the meaning of what alpha link means. Thanks.

@user55119 In my opinion you should mention that the method is valid when numbering of carbon is done in clockwise direction to make method unambiguous. eg: https://chemistry.stackexchange.com/q/43546/108386


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R/S configuration of bridging carbon in bicyclic system
I am trying to figure out the R and S configuration of bridging carbons in bicyclic systems.
In the image, there are two bicyclic systems and depending on the position of the Br substituent, the R or S configuration of the bridging carbons change. How is the configuration actually determined? I don't understand how the prioritization works in this case, as two substituents are similar. Thus, the reason must be the adjacent stereocenter... But I am not able explain it properly.

If the two substituents were similar, that would be the end of the story. However, they are not. The first atoms might be similar, but then you have to go and compare the next atoms, and so on.

@IvanNeretin Yes I have compared the next atoms but at the third atom, the both substituents meet again as their third atom is the same. and from that point, I don't know how to continue.

Are you sure the bridgehead carbons are stereogenic?

@Zhe Yes, because the solution to my exercise marked the carbons with R and S tags and when drawing these structures in ChemDraw, the R and S configurations were assigned automatically, too...

You should continue in all directions at once, not just along the cycle.

To be sure, the carbons of two bis-methylene bridges are heterotopic because of their in-space relationship to the substitution, but again, does that matter? If you say yes, then you would also need to assign a configuration to the isolated methylene unit. That has two hydrogen atoms, but they are diastereotopic. But you're not claiming that that is a stereogenic center. Therefore, something is contradictory in your reasoning.

So the assignment of Chemdraw and the solution to our exercise is kind of confusing. I totally agree with you, that the two loops are identical, hence, defining R/S configurations does not make sense. Can you explain why the solution and chemdraw assign R and S? Or what exactly does the R/S at bridging carbon mean? I mean there must be a reason why it is written like this, mustn't it?

@Faded Giant  You wrote, "2,2-dimethylbicyclo[2.2.2]octane contains two pseudoasymmetric centers". I believe you are referring to the bridgehead carbons and that you are discussing 2,2-dimethyl, not 2-bromo, 2-methylbicyclo[2.2.2]octane. If so, then the bridgehead carbons are actually prochiral rather than pseudoasymmetric. Hence the lowercase "r" and "s" descriptors do not apply (pro-r or pro-s could be used in referring to the bridgehead carbons).

Several days ago I offered a solution to the R/S assignment of the bridgehead carbons of (1R,2S,4R)-2-bromo-2-methylbicyclo[2.2.2]octane 1. During the middle of the night I realized that the method was faulty because the solution would not accommodate the (1S,2R,4S) enantiomer  3 but would rather predict (1R,2R,4R) which clearly is a non-existent diastereomer. The shortcoming lay in not applying, in a heirarchical sense, CIP rule 4b (RR/SS>RS/SR) over rule 5 (R>S).
Focusing on bromide 1, the configuration of C2 is "S" with Br>C1>C3>C9. Structure 2 is structure 1 from a different perspective to which we will refer during the ensuing discussion. To determine the configurations of the bridgehead carbons, C1 and C4, a digraph is the method of choice. In the "Digraph of C1" one starts at the non-duplicate atom C1 and traces the six possible paths back to duplicate atoms C(1) that are each attached to three phantom atoms of atomic number zero. The vertical chain has the top priority while the hydrogen (not shown) at C1 has the lowest priority. The two horizontal chains are mirror images as far as the atom connectivity is concerned and they must be assigned to the second and third priorities. In these chains C2 has been predetermined as having the S-configuration. Locate C4 in the left chain. It has the R-configuration because C3>C8>C5. Locate these atoms in structure 2 and convince yourself. [I use my hands. Point your right thumb from C4 to H and your fingers will point from C3 to C8 to C5]. In the right hand chain C4 has the S-configuration because C3>C5>C8. Check these priorities in structure 2.
Now the left hand chain is designated as RS and the right hand chain as SS. Since RR/SS>RS/SR, the order of carbon atoms attached to non-duplicate C1 is C2>C6>C7. Locate these atoms in structure 2 and convince yourself that C1 does indeed have the R-configuration!
Look at "Digraph of C4". Now that you are versed in the method, the priorities around non-duplicate C4 are C3>C8>C5 because C3-chain>SS>RS>H. C4 also is of the R-configuration! <continued>

The digraphs below confirm the assignments to the enantiomer 3 (4), (1S,2R,4S)-2-bromo-2-methylbicyclo[2.2.2]octane. I'll leave the analysis to the reader.


text:
An expression for equilibrium constant of isotope reaction
Consider the isotope exchange reaction:
$$\ce{C ^{16}O + 1/2 ^{18}O {}^{18}O <=> C^{18}O + 1/2 ^{16}O^{16}O}$$
The equilibrium constant for the reaction is:
$$K=\frac{\prod (a_i)^{n_i}_{products}}{\prod(a_i)^{n_i}_{reactants}} = \frac{[a(\ce{ ^{16}O ^{16}O})]^{1/2} a(\ce{C ^{18}O})}{[a(\ce{ ^{18}O ^{18}O})]^{1/2}a(\ce{C ^{16}O})} $$
In this chapter, from Eq. (3.3) to (3.4) the following simplification is done for the equilibrium constant:
Since activity coefficients are close to 1, one can write (eq. 3.4):
$$K= \frac{(\ce{ ^{18}O/ ^{16}O})_\ce{CO}}{(\ce{^{18}O/ ^{16}O})_\ce{O2}} = \alpha_\ce{CO-O2}$$
I understand that activity coefficient of $1$ allows us to replace it by concentration, so $a(\ce{^{16}O ^{16}O})=[\ce{^{16}O ^{16}O}] \stackrel{?}{=}\frac{1}{2} [\ce{^{16} O}]_\ce{O2}$. I don't understand how the power $1/2$ gets converted to a power $1$.
For reference, here are the chemical equation and the equilibrium expression in the original form from the linked textbook:

Your starting equation is wrong - should be without 1/2.

The 1/2 is needed to balance the oxygen isotopes on the left and right side. This equation is also Eq. (3.2) in the linked chapter.

And it should be with single exchanged oxygen in O2.

That is irrelevant; you can multiply the whole equation with any number $n$ and then ask the same question about $K^{1/n}$ when $K$ is the equilibrium constant of the multiplied equation.

The equilibrium expression written in the book is for a reaction that is not the one you wrote. You should use the correct reaction. Notice that the desired equilibrium constant  depends on dioxygen that has both a labeled and an unlabeled oxygen atom. That species is not present in your equation. I believe this is what @Mithoron is trying to tell you.

@Mithoron I edited the question to include the equation and equilibrium expression given in the textbook. They decided not to look at the mixed species (don't know why).

The notation is strange, but the result is correct.
For example, if you have 100 dioxygen molecules (so 200 oxygen atoms), 10% of the atoms are of the 16-O isotope and we distribute them randomly, we get the following:

81 both 18-O
18 mixed species
1 both 16-O

To visualize this, take all the numbers from 00 to 99, and say that the digit zero represents the 16-O. Only one number has two zeros (00), and 18 numbers have exactly one zero (01-09 and, counted by tens, 10-90).
The ratio of 18-O to 16-O is 9:1. The ratio of heavy dioxygen (both O-18) to light dioxygen (both O-16) is 81:1, i.e. the square of the atom ratio. That is why the square root disappears in the last step of the derivation.


text:
Is there an error in a Wikipedia article explaining the influence of oxidation states?
Referring to the series of oxoacids of chlorine: $\ce{HClO, HClO2, HClO3},$ and $\ce{HClO4}$, tabulated in the Wikipedia article on electronegativity, the article states, in the section near the end titled "Variation of electronegativity with oxidation number":
“As the oxidation state of the central chlorine atom increases, more electron density is drawn from the oxygen atoms onto the chlorine, reducing the partial negative charge on the oxygen atoms …” and illustrates it with a table. (NB There is an ambiguity in the use of the word "reducing" which is the opposite of "oxidising", but the intended meaning in terms of increasing or decreasing electron density is clear in the Wiki statement.)
But as I understand it (and as also given by the rules for assigning oxidation numbers), insofar as oxidation state represents the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element, the table is correct, but shouldn’t the textual statement be the converse:
“As the oxidation state of the central chlorine atom increases, more electron density is drawn from the central chlorine atom onto the oxygen atoms, increasing the partial negative charge on the oxygen atoms ...”
I reason it thus And just to be clear, let's go back one step to start with the even simpler case of $\ce{HCl}$, where clearly the oxidation states must be ($\ce{H} = +1, \ce{Cl} = -1$, respectively; total 0):
When we move on to $\ce{HClO}$, then it becomes (+1, +1, -2 respectively) as the single oxygen with stronger electronegativity than chlorine is assigned the oxidation state -2, so that balancing the total means the electronegativity of the central chlorine must change from -1 to +1 (in agreement with the table on the Wiki page, but not with the article’s statement about the direction of change of electron density.
And as we move along the series to $\ce{HClO4}$, the central chlorine is having more and more electron density sucked off it by the four more electronegative oxygen atoms, resulting in the final oxidation states (+1, +7, -8), in agreement with the table in the above-cited Wiki article, but not with the article's textual explanation.
So I conclude that the text statement in the Wiki article is currently wrong (and if so I will amend the Wikipedia article) - or else I'm wrong, and would appreciate someone pointing out my own error…?

This is a bit awkward wording there, but nevertheless it's not wrong, somewhat ambiguous, perhaps.

"drawn from the central chlorine atom onto the oxygen atoms, increasing the partial positive charge on the chlorine atom" would be better then your version, as partial charge on each one oxygen does indeed decrease, not their sum though.

@Mithoron - Excellent point. So you agree that basically I'm right and the Wikipedia article should be amended, but using your refinement. Many thanks.

Well, no it shouldn't! Such statement has nothing to do with the point the article makes - that Cl +7 doesn't release its electrons as eagerly as Cl +1.

Please do not edit the Wikipedia article, as your wording is incorrect, and the existing wording, while awkward, is correct.
What the article is trying to say is that the electron density on each oxygen is less as the oxidation state of the Cl increases. We can understand this by imagining progressive oxidation starting with $\ce{HClO}$. Here, the Cl has oxidation state +1 and the O is -2. But of course the oxygen doesn't have a full negative 2 charge. Some of that charge is shared with the chlorine and the hydrogen, neither of which has a full charge of +1.
To get to $\ce{HClO2}$, we can imagine reacting $\ce{HClO}$ with a neutral oxygen atom (oxidation state 0). After the reaction, the oxygen will have oxidation state -2, and the Cl will now be +3. Again, these are not the formal charges, but give us a hint about where charge will be located. The oxygen will have a partial negative charge, which means the chlorine will have to lose some of its electron density, increasing its actual charge. The original oxygen from $\ce{HClO}$ is still there and feels a stronger "pull" on its electron density, so it gives more to the chlorine, and the oxygen's partial negative charge becomes less negative.
Completely making numbers up, we could imagine that in $\ce{HClO}$, the charge on oxygen is -1, on chlorine is +0.25 and on H is +0.75. After adding the second oxygen to make $\ce{HClO2}$, the charge on the new oxygen might be -0.5, on the already attached O might be -0.75, +0.75 on H and +0.5 on Cl. Do you see how the negative charge on each O gets smaller in magnitude, but the total negative charge on oxygen(s) gets larger in order to offset the increased positive charge of the more oxidized chlorine? The Wikipedia statement refers to the individual O's not the sum.

Many thanks for the careful and detailed explanation, resulting in the well-clarified Wiki entry inserted by KarstenTheis .

It now reads:

The effect can also be clearly seen in the dissociation constants of the oxoacids of chlorine. The effect is much larger than could be explained by the negative charge being shared among a larger number of oxygen atoms, which would lead to a difference in pKa of log10(1⁄4) = –0.6 between hypochlorous acid and perchloric acid. As the oxidation state of the central chlorine atom increases, more electron density is drawn from the oxygen atoms onto the chlorine, diminishing the partial negative charge of individual oxygen atoms. At the same time, the positive partial charge on the hydrogen increase with higher oxidation states. This explains the observed increased acidity with increasing oxidation state in the oxoacids of chlorine.

There is a nice illustration of the partial charges of the protonated and deprotonated species on the Libretexts site. The $\mathrm{p}K_\mathrm{a}$ values quoted at Wikipedia and the textbook referenced above don't match up, especially those below zero, so there is still room for consolidation and improvement.

Source: Chlorine oxoacids pKa values from J. R. Bowser, Inorganic Chemistry (Pacific Grove, CA: Brooks-Cole,1993) according to https://wisc.pb.unizin.org/chem109/chapter/day-34/

Thanks - that makes much more sense to me (and hopefully to some others) now.

Superb! That's really helpful and illustrates the power of quantum chemistry; both Linus Pauling's electronegativity formula, and his work on early quantum chemistry are brought together here. It's just a shame I can't add a second upvote.


text:
Cis and trans electrodes
I have come across use of the terms cis and trans electrodes in the context of nanopores and driving charged biomolecules through a pore using an electric field [1]:

FIG. 1. Drawing of the cis negative electrode and trans positive electrode chambers. The arrows show the direction of electric field lines. The pore diameter $d$ and length $l$ and the capture radius $r$ are also shown.

I wondered what the origin of the term is.
I know about cis and trans isomers, but I have not come across the term before in terms of electrodes. Is there a simple explanation?
I can just assume that cis/trans are just another terms for cathode and anode, but I think I am missing something. If the cis electrode is the cathode and the trans electrode is the anode, then why not just call them that? Is there any additional meaning?
Reference

Grosberg, A. Y.; Rabin, Y. DNA Capture into a Nanopore: Interplay of Diffusion and Electrohydrodynamics. The Journal of Chemical Physics 2010, 133 (16), 165102. DOI: 10.1063/1.3495481.

Sometimes authors use buzz words in their paper in order to make their work novel or fancier than it should be. However, such newly coined terms never get popular and vanish with the author into oblivion. No, cis and trans can never mean cathode or anode. It refers to the chambers in which the electrode is place.
I looked up the Oxford's Latin Dictionary:
cis has three meanings

(in time) before, within
(in space) on the near side of, on this side of
(expressing motion) to the near side of

trans

to the other side of

Now look at the cell configuration in your link. The very first lines of the paper you linked says

DNA translocation through protein or synthetic solid state nanopores
is a two stage process that involves capture of a DNA coil by the pore
including its delivery to the pore and insertion of one of its ends
into it, followed by its motion through the nanochannel. The process
may be driven by voltage $\Delta$V typically a few hundred millivolts
applied between two macroscopic electrodes positioned on both sides of
the partition separating the cis where DNA is initially placed and the
trans to where it translocates through the pore chambers, as shown
schematically in Fig. 1


text:
Tollen's Test with Salicylaldehyde
I have read that Tollens' test is given by aldehydes. The aldehyde is oxidised to a carboxylate ion and silver is deposited.
My teacher had given me notes which pointed that salicylaldehyde does not give this test. However, I cannot find any explanation for the same. I have tried to search online and through books but I did not get anything.
I think it could be related to resonance from the −OH group as something similar is mentioned in my notes, but it is not elaborated. Can anyone please explain this?

Of course, the answer is resonance.  As described by Benet et al., a basic medium (pH>10) favours the Tollen's reaction. On the other hand, if we want to dissolve phenol in water, we will need OH- to generate phenolate. It means the hydroxyl group of salicylaldehyde is sensitive to the high pH, and this hinders attacking the other hydroxides to the carbonyl group by further stabilizing it.

Okay, so what I understood from your answer is that because of resonance from the hydroxyl group the carbonyl is further stabilized which prevents other reactions. This would also mean that if there were other groups like -NR2 that can stabilize the carbonyl then also the test would not be obtained. Is that correct ?

I don't think -NR2 or -OR (R is not H) has the same effect, but a negative response, too. I couldn't find a reference to prove this but in my opinion, the stability of the final structure has a significant influence on blocking the carbonyl group.


text:
Electrophilic addition on cyclopropane using sunlight
I encountered a weird question in an exam. The statement says that instead of radical substitution, cyclopropane undergoes electrophilic addition reaction in sunlight.
I know that all saturated compounds give substitution reactions in sunlight but the answer key states that electrophilic addition takes place.
I couldn't find any source to prove this. Any help with an explanation and a verified source would be appreciated.

Related: https://chemistry.stackexchange.com/questions/10653/why-does-cyclopropane-react-with-bromine

It should say "in addition to" rather than "instead of".
Cyclopropane suffers badly from ring strain. The bond angles in the ring are 60° rather than the normal ~109.5° when carbon makes four single bonds. In the absence of light, when in contact with $Cl_2$ or $Br_2$ the ring will break and addition occur.
In sunlight, both the expected substitution reactions and the addition reaction will occur.
Can't find the reference I thought I remembered, but here's a site or two discussing it:
https://www.chemguide.co.uk/organicprops/alkanes/halogenation.html
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.07%3A_Reactions_of_Alkanes


text:
Significance of pV work in enthalpy
If we already have $pV$ work included in the internal energy $U = q + w,$ then what is the $pV$ work in the equation for enthalpy $H = E + pV?$

Who says there's work associated with the enthalpy equation?

The equation is:
$$H = E + pV\tag{1}$$
Note: here, $pV$ is product of volume and pressure and not work done.
Differentiate, you will get
$$ΔH = ΔE + ΔpV\tag{2}$$
If pressure is constant, then you will get
$$ΔH = ΔE + p\,ΔV\tag{3}$$
Here, you can write $p\,ΔV$ is work done.
So, you can use the term work done for $pV$ in a special condition only.

Please give an intuitive sense of Enthalpy.

Briefly, enthalpy can be called as the total heat content of the system. Positive change in enthaply means process is endothermic i.e it requires heat to he supplied for reaction to happen and negative change means it gives out heat.  You can look out other answers for detailed explanations.


text:
Dehydration of primary alcohols without a β-hydrogen

What happens in dehydrations of primary alcohols (in presence of acid catalyst at high temperatures) which don't have a β-hydrogen?

For example, in the case of neopentyl alcohol, will it dehydrate as it doesn't have a β-hydrogen? Obviously, E2 reaction will not be possible, but can it react via an E1 reaction, with a rearrangement in the carbocation step?

Under what dehydration conditions?

Yes.
Most dehydration reactions proceed via E1 method in (acidic conditions because that is the most common way of proceeding with it)
So the the protons in the acidic medium attack the lone pair of oxygen. And this in turn leads to the cleavage of the C-O bond, as it is weaker than C-H bond.
So this leads to the formation of a carbocation on the carbon atom.
As you said, this undergoes rearrangements before a proton is released and thus a pi bond forms.

Is ring expansion also possible in case of primary cyclo-alcohol?

@Apurvium Yes it is. Any carbocation rearrangement happens provided the stability increases.

In this case, $\mathrm{E2}$ elimination is impossible regardless of condition used, because of lack of $\beta$-hydrogens. However, in acidic conditions, it is possible to have elimination reaction. Since 2,2-dimethylpropanol is a $1^\circ$-alcohol, initial carbocation formation is difficult. However, this formation of carbocations is accompanied by a structural rearrangement, which is the driving force for the $1^\circ$-carbocation formation, which is otherwise not favored:

Such rearrangements take place by a 1,2-shift of a neighboring alkyl group or aryl group or hydrogen, and are favored when the rearranged carbocation is more stable than the initial cation. In this case, migrating group is methyl group and carbocation changed from $1^\circ$ to $3^\circ$, which is very stable compared to initial carbocation and favored. Once $1^\circ$-carbocation formed, it quickly rearranged to $3^\circ$-carbocation and as a result, the reaction make forward progress. It can also be considered a concomitant elimination of water and methide migration where methide group acting as an intramolecular nucleophile (See the mechanism shown in bottom box of the scheme). Either way, the initial step is very slow because of very high activation energy:

Source of the pictures?

@Mithoron: Thanks for reminding. I included the source. I answered this pass midnight, so probably sleeping to forget that. :-)

If the reaction is being done in acidic conditions, dehydration of neopentyl alcohol can take place with E1 Mechanism. A 1,2 Methyl shift will happen and the major product will be 2-methylbut-2-ene.

You can also get ether formation by reaction with a second molecule of alcohol

But the yield will be poor since formation of ether will take by SN2 mechanism. 2-methylbut-1-ene can be another minor product.

that rather depends on the reaction condtions. 2-Me-2-butene can also be protonated to give fairly stable cation (see the excellent answer above) which act as an electrophile for the alcohol.

Yes, it can act as an excellent electrophile but I doubt weather a 3° alcohol will be able to act as a good nucelophile.


text:
Are colligative properties applicable for volatile liquids and gases in liquid solvent? Why/Why not?
My teacher told me that the expressions for calculation of colligative properties like lowering in freezing point are only applicable if the solute is non volatile liquid or a solid. We cannot apply the formula for gases or volatile liquids. Is this true? If so then why/ why not?
I have not been able to come up with a satisfying reasoning explaining this statement. Also I have not been able to find evidence regarding this statement online. Although one of my books does say this without an explanation but I am not convinced.

Well, if somethin's gonna boil off before your solvent then it can hardly influence b.p. like salt will...

@Mithoron Solute doesn`t necessarily need to have a boiling point lower than solvent. Also what about gas in liquid mixtures.

Yes, they are, but consider that colligative properties are often described in terms of equations whose derivation require making various assumptions about the substances involved, most importantly that the solvent and its mixtures behave close to ideally (follow Raoult's law), and that the solute does not affect the chemical potential of the solvent in the second phase (solid in freezing point depression or vapor in boiling point elevation). Textbook descriptions of freezing point depression in general assume that the solute does not partition into the solid, for instance, since partitioning would imply formation of a solid solution with a depressed (altered) chemical potential for the solid solvent, a complication too formidable to tackle in an introductory text (and even more advanced ones).

Why would volatility affect the chemical potential of the solvent in the second phase? Also what about gas in liquid mixture? If we assume that the solute doesn`t affect the chemical potential, would these mixtures follow the colligative laws?

@ManitAgarwal-Elpsycongroo If the gas is ideal then different species will not affect each other's potential. The same cannot be said of a solid mixture, even if it is ideal (the solute will still affect the chemical potential of the solid solvent in that case).


text:
What is the chemical formula for the synthesis of Hydroxyapatite (Ca10(PO4)6(OH)2)?
This might be a basic question but I am a little confused.
I have some lecture notes that says calcium phosphate ($\ce{Ca3(PO4)2}$) reacts with calcium hydroxide ($\ce{Ca(OH)2}$) to form Hydroxyapatite ($\ce{Ca10(PO4)6(OH)2}$).
So, the chemical formula (even when unbalanced) should be:
$$\ce{Ca3(PO4)2 + Ca(OH)2 -> Ca10(PO4)6(OH)2}$$
I can't figure out why $\ce{Ca10(PO4)6(OH)2}$ is the product of $\ce{Ca3(PO4)2}$ and $\ce{Ca(OH)2}$ and I am having trouble balancing it (see below). On Sigma Aldrich's website, it says the linear Formula for hydroxyapatite is $\ce{3Ca3(PO4)2·Ca(OH)2}$. I copied and pasted this directly. I assume the dot between the calcium phosphate and calcium hydroxide means 'plus'.
On the Sigma Aldrich's website, the calcium phosphate is written/balanced as $\ce{3Ca3(PO4)2}$, whereas the lecture notes I have are just $\ce{Ca3(PO4)2}$. Even with the three in front as the coefficient, I get $\ce{9Ca + Ca -> Ca10}$, $\ce{(PO4)2}$ and $\ce{(OH)2}$ which is $\ce{Ca10(PO4)2(OH)2}$, instead of $\ce{Ca10(PO4)6(OH)2}$. If I use the formula in my lecture notes, I get $\ce{Ca3(PO4)2 + Ca(OH)2 -> Ca4(PO4)2(OH)2}$, which is even more off.
Should it be $\ce{3(Ca3(PO4)2)}$ instead of $\ce{3Ca3(PO4)2}$ so the coefficient 3 can be applied to the subscript 2 on the $\ce{PO4}$ to give $\ce{(PO4)6}$?
What am I missing?

Well, it seems that before studying chemistry you must know the notation which is passed around through a word of mouth. In 3Ca3(PO4)2, the "4" refers to O and nothing else, the "2" refers to PO4, the second "3" refers to Ca, but the first "3" refers to everything after it. Pretty obvious, huh?

Stephanie's last question show that she does not master the basic convention for writing chemical formulas. I repeat her question here : "Should it be "3(Ca3(PO4)2)" instead of "3Ca3(PO4)2"  ? " Well. Stephanie should absolutely be convinced that these two formulas are identical, before going ahead with apatite structures.

@IvanNeretin I know this. But in my working out, I always (to make it clear to myself) always, always put everything in as many brackets as needed to separate the items (maths or otherwise) so I thought the 3 without the brackets was only for the Ca. Got it now. Thanks

First, let me state the standard formula of hydroxyapaptite:

The term "apatite" applies to a group of compounds  with a general
formula in the form $\ce{M10(XO4)6Z2}$, where $\ce{M^2+}$ is a metal
and species $\ce{XO4^3-}$ and $\ce{Z-}$ are anions. The particular
name of each apatite depends on the elements or radicals M, X and Z.
In these terms, hydroxyapatite (HAp) has the molecular structure of
apatite, where M is calcium ($\ce{Ca^2+}$), X is phosphorus
($\ce{P^5+}$) and Z is the hydroxyl radical ($\ce{OH-}$). This is
known as stoichiometric hydroxyapatite and its atomic ratio Ca/P is
${1.67}$. Its chemical formula is $\ce{Ca10(PO4)6(OH)2}$, with $\ce{39 {%}}$ by
weight of $\ce{Ca}$, $\ce{18.5 {%} P}$ and $\ce{3.38 {%}}$ of $\ce{OH}$.

The anion radical can be replaced with fluorine/chlorine to become fluoroapatite/chloroapatite. This formula can also be written as $\ce{3[Ca3(PO4)2]· Ca(OH)2}$ which indicates three moles of calcium phosphate and one mole of calcium hydroxide which is basically the same as the Sigma-Aldrich's formula. The confusion arises since they have omitted the brackets, you didn't considered calcium phosphate as single entity for which you had a wrong formula. Brackets are generally not given but you have consider them as a single entity or you can even use brackets conveniently for your purpose. But do note that this formula doesn't prove that it is a basic salt but is actually a product of polymerisation:

According to Bassett, the solid existing in stable equilibrium at $\pu{25 °C}$., with solutions of a range from faintly acid to nearly pure
lime-water, is oxy- or hydroxyapatite, but, although its composition
may be indicated by the formula $\ce{3Ca3(PO4)2.Ca(OH)2}$, it is not to be
regarded as a basic salt, but rather as the salt of a polymeride of
phosphoric acid, $\ce{H11P3O13}$ (or $\ce{3H3PO4.H2O}$), with one hydrogen atom
un-neutralised (like orthophosphoric acid or pyrophosphoric acid). The slowness of
formation of hydroxyapatite favours the view that it is produced by
polymerisation.

Reference

Hydroxyapatite-Based Materials: Synthesis and Characterization
By Eric M. Rivera-Muñoz, 2011, DOI: 10.5772/19123
http://calcium.atomistry.com/basic_calcium_phosphates.html


text:
Is it easier to oxidise iron(II) to iron(III) in acidic or alkaline solution?
Is it easier to oxidise $\ce{Fe^2+}$ to $\ce{Fe^3+}$ in acidic or alkaline solution?
$$
\begin{align}
\ce{Fe &-> Fe^2+(aq) -> Fe^3+(aq)}\\
\ce{Fe &-> Fe(OH)2 -> Fe(OH)3}
\end{align}
$$
In my textbook it is written that as soon as $\ce{Fe^{2+}}$ changes to $\ce{Fe^{3+}}$, it acquires $\ce{OH-}$ to form $\ce{Fe(OH)3}$. This promotes oxidation of $\ce{Fe^{2+}}$ changes to $\ce{Fe^{3+}}$. I don't quite get this why precipitate formation speeds up the reaction?

You said it yourself: the reactions are different.

@user Look, your question has been edited using standardized notations and formatting. Please make sure you use it as an example in your edits.  If you have questions regarding formatting, visit this page, this page and this one.

The experience shows that a crystal or a solution containing the ion $\ce{Fe^{2+}}$ may be useful for a couple of hours without being significantly oxidized. On the contrary, a precipitate of $\ce{Fe(OH)_2}$ starts being oxidized by oxygen from the air in less than one minute.

@user Please note, it is $\ce{Fe^2+}$ and not $\ce{Fe^{+2}}$. See: https://chemistry.stackexchange.com/questions/24769/why-ca-2-and-not-ca-2

Also see: https://chemistry.stackexchange.com/questions/90008/how-does-an-acid-prevents-or-discourges-oxidation-of-iron

@Nilay Ghosh how do you search questions on C.SE?Is there something like approach0 in M.SE?

Her is a source: THE MECHANISM OF IRON CATALYSIS IN CERTAIN OXIDATIONS that confirms Maurice's reported observations:

It is well known that ferrous iron is comparatively stable in acid solution and that it is rapidly oxidized to the ferric state by the oxygen of the air in alkaline solution.

On the question of why, I quote another work: Fe2+ adsorption on iron oxide: the importance of the redox potential of the adsorption system with citing references:

Nowadays the dominant hypothesis is that adsorption of Fe(II) by ferric and non-ferric oxides has a different mechanism. Fe(II) adsorbed on ferric oxide (like Fe2O3) can be easily oxidized by the transfer of an electron from adsorbed species to the solid (Gorski and Scherer 2011; Hiemstra and van Riemsdijk 2007; Larese-Casanova et al. 2012).

Note my added link to the work of Hiemstra based on surface complexation modeling.
I hope this helps


text:
How can I properly calculate the isoelectric point (pI) of amino acids?
The following amino acid is called lysine. I was asked to calculate its isoelectric point, with the given $\mathrm pK_\mathrm a$ values.

I've searched a lot, and the most helpful post that I found was How do I calculate the isoelectric point of amino acids with more than two pKa's?
According to orthocresol's answer:

Since the $\mathrm{pI}$ is the $\mathrm{pH}$ at which the amino acid has no overall net charge, you need to average the $\mathrm pK_\mathrm a$ values relevant to the protonation/deprotonation of the form with no net charge.

Let's call the ends $e_1, e_2$ and $e_3$ (from left to right).
Approach $\#1$

deprotonate $e_3$ (i.e., carboxyl group)
deprotonate $e_1$ or $e_2$ [neutral point]

So, $\mathrm pK_\mathrm a$'s of $e_1$ and $e_2$ are relevant.
$$\Rightarrow \mathrm{pI} = \frac{10.53 + 8.95}{2} = 9.74$$
But, is there some limit to number of protonations/deprotonations or some procedures to follow?
For instance,
Approach $\#2$

deprotonate $e_1$
deprotonate $e_3$ [neutral point]
deprotonate $e_2$ and protonate $e_1$ [neutral point]

This time, $\mathrm pK_\mathrm a$'s of $e_3$ and $e_1$ are relevant. But, the calculated $\mathrm{pI}$ isn't correct.
So, how can I validate the approaches?

Problem source: FIITJEE study material
$\mathrm pK_\mathrm a$ and $\mathrm{pI}$ values table for amino acids: https://www.anaspec.com/html/pK_n_pl_Values_of_AminoAcids.html

The first approach I'd say is the correct one. It's obvious that the isoelectric point will be between e1 and e2 because lysine is a dibasic aminoacid, therefore the two amino groups will have a bigger influence on the pI than the acidic group, thus rendering the pI basic. In these cases, as far as I know, you have to find the average of 2 pKa-values that are going to have the biggest influence on the pI. In this case, since there are 3 pKa-values of which 2 are basic, the pI will be the arithmetic mean of those values.

Related: Various pI calculators for proteins give different results: ExPASy (8.75), isoelectric.org (9.04), and Prot pi (8.475).

Isoelectric point of an amino acid is the $\mathrm{pH}$ at which the molecule carries no net charge[1]. It can be calculated by the average of the relevant $\mathrm pK_\mathrm a$ values as you have mentioned.
Your confusion seems to stem from choosing the relevant $\mathrm pK_\mathrm a$ values. For this we should refer to the titration curve of the amino acid.
For a neutral amino acid[2]:

From the curve we can infer that the $\mathrm{pI}$ is simply the average of the two $\mathrm pK_\mathrm a$ values of the carboxylic acid and the amino group.
For a basic amino acid[2]:

From the curve we can infer that the $\mathrm{pI}$ is simply the average of the two $\mathrm pK_\mathrm a$ values of the two amino groups. The $\mathrm pK_\mathrm a$ of the carboxylic acid group is not relevant.
For an acidic amino acid[3]:

From the curve we can infer that the $\mathrm{pI}$ is simply the average of the two $\mathrm pK_\mathrm a$ values of the two carboxylic acid groups. The $\mathrm pK_\mathrm a$ of the amino group is not relevant.
Here are examples for all three cases:

References:

Wikipedia

Titration curves for neutral and basic amino acids

Titration curve for acidic amino acids

But ExPASy says that pI is 8.75 for lysine (K).

@PeterMortensen please refer https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/16%3A_The_Organic_Chemistry_of_Amino_Acids%2C_Peptides%2C_and_Proteins/16.04%3A_The_Isoelectric_Point and https://www.anaspec.com/html/pK_n_pl_Values_of_AminoAcids.html


text:
Why HBr has stronger attraction between molecules than CH3NH2?

Which of the following compounds exhibits dipole-dipole forces as its strongest attraction between molecules?
a) $\ce{CO2}$
b) $\ce{CH3NH2}$
c) $\ce{Kr}$
d) $\ce{H2}$
e) $\ce{HBr}$

From what I understand, $\ce{CH3NH2}$ has an H-bond and it's the strongest form of dipole-dipole. So I thought that would be the answer.
I also knew that $\ce{HBr}$ shows the greatest electronegativity difference between $\ce{H}$ and $\ce{Br}$ atom, it would be a polar covalent bond (?!).
In that case, $\ce{HBr}$ may have stronger attraction between molecules than $\ce{CH3NH2},$ but it is not dipole-dipole. Isn't that correct? Then shouldn't the answer be $\ce{CH3NH2}$ instead of $\ce{HBr}?$
I must have some concept issues here.

Mistaking bonds for intermolecular interactions is a serious "concept issue".

Wait, I thought there would be H-bonds between CH3NH2? And each HBr molecule is attracted to other HBr molecules by a mixture of permanent dipole-dipole and dispersion forces. This is intermolecular bonding. So Hbond vs dipole-dipole, why is HBr stronger than CH3NH2? I AM trying to compare intermolecular interactions, not the bonds? Please tell me where I am making a mistake. I am going in circles and very confuse. Thank you.

Well, that's what this terrible title suggests, so edit it.

Thank you for the suggestion, I edited and made it specific.

EDIT (after title edited): $\ce{HBr}$ doesn't have stronger interaction than $\ce{CH2NH2}$, but it has dipole-dipole interaction as the strongest forces between it's molecules, which is obviously weaker than H-bonding.

Hydrogen bonding is the strongest intermolecular attraction. It is a type of dipole-dipole interaction1, but it is specific to Hydrogen.
In general, dipole-dipole interactions are considered weaker than H-bonding. The relative strength of forces are as follows,

Hydrogen bonding >  Van der Waals dipole-dipole interactions > Van der Waals dispersion forces

As per your question, (a), (c) and (d) have dispersion forces (as they're non-polar), while (b) has H-bonding as it's strongest interaction (not dipole-dipole).
Finally, as only (e) has dipole-dipole as it's strongest attraction between molecules, hence it's the answer.

Notes:
1: Dipoles are created due to difference in electronegativity. Below image shows that there is sufficient electronegativity difference to create a dipole. H-bonding and dipole-dipole interactions have same origins.

(Source: Master Organic Chemistry)

So H-bond is not a dipole-dipole? But it's said to be the strongest dipole-dipole?

@Molly_K https://chemistry.stackexchange.com/questions/35488/is-hydrogen-bonding-a-type-of-dipole-dipole-interaction https://chemistry.stackexchange.com/questions/21743/why-is-hydrogen-bonding-more-significant-than-any-other-interaction-between-dipo

@Molly_K: H-bond is the strongest dipole-dipole interaction. But, dipole-dipole forces are said to be other interactions than H-bond, in general

@RahulVerma Thank you for the clarification. But why is "dipole-dipole forces are said to be other interactions than H-bond, in general" yet the textbook emphasized H-bond it the strongest dipole-dipole intermolecular force? This is so confusing!

@RahulVerma I looked up the melting point forHbr, which is -124.6°F, and is -135.4°F for methylamine. As we also learned that higher intermolecular force (IMF) will have a higher melting point, we know that HBr is having a stronger IMF.
The question is why this is making an exception from the rule we learn in the textbook. I would really appreciate how to interpret this and also learn if there's any rule that I missed from reading.

@YiKo: 1) have a look at this image (src: masterorganicchemistry) 2) H-bonding is theoretically separated from other dipole-dipole interactions. But, in essence both of them have same origin, as said in image. 3) intermolecular forces aren't significant in determining melting point, packing fraction and crystal structure are used to compare MP. 4) intermolecular forces are significant in comparison of BP, BP(hbr) = -60, BP(ch3nh2) = -6


text:
ᴅ-amphetamine, ʟ-amphetamine
The following structural formula (source: Wikimedia) is labelled ᴅ-amphetamine:

A Fischer projection of this molecule would contain, I think, the following line:
$$\ce{H2N-C-H},$$
and I would have labelled the molecule therefore as ʟ-amphetamine.
Why am I wrong? Why is it called ᴅ-amphetamine?

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

https://chemistry.stackexchange.com/questions/44260/what-is-the-difference-between-d-and-l-configuration-and-and-%e2%88%92

(S)-d-amphetamine or (S)-(+)-amphetamine but not D-amphetamine. [Without (S)- is OK.]

The L and D nomenclature is directly transform from its origin, D- and L-phenylalanine. The following scheme summarize the transformation:

The blue box in right-hand corner explains how L- and D-$\alpha$-amino acids get their names using Fischer projection (this is similar to D- and L-monosaccharides, replacing $\ce{OH}$ group by $\ce{NH2}$ group). If you reduced the carboxylate group of amino acid to a methyl group, the optical rotation of $\alpha$-carbon change from $(S)$ to $(R)$ because second highest priority of $\ce{COOH}$ group to third highest ($\ce{CH3}$ group) based on the size of the $\ce{R}$ group (see general formula of amino acid in top right corner). For example, when $\ce{R=CH2Ph}$ group, the amino acid is phenylalanine, which has $(S)$-rotation, but reduction of caboxylate group to methyl make the rotation $(R)$. However, the product $(R)$-amphetamine still carries the original L-nomenclature (see examples in bottom red box).
Note: The conversion of phenylalanine to amphetamine in several steps were achieved in beginning of 21st century (Ref.1).This is the ultimate enanthimerically pure version.
Reference:

Dominick A. Quagliato, Patrick M. Andrae, Edward M. Matelan, “Efficient Procedure for the Reduction of α-Amino Acids to Enantiomerically Pure α-Methylamines,” J. Org. Chem. 2000, 65(16), 5037–5042 (https://doi.org/10.1021/jo000242h).

That's great input. I'm grateful for the idea to derive the prefix from phenylalanine. But is this truly the source of the prefix or just a happy coincidence?
From the comments to the question itself I came to believe that the D- was unfortunately capitalized and might have originally been a lowercase d- and chosen because of the optical effects of the substance.
Now you explain the D- based on a Fischer projection, but you write the substance with the C1 atom at the bottom and the C3 atom at the top. How does somebody who wants to name the substance know that?

D and L in sugars and amino acids are nothing to do with common d and l, which are based on optical rotation. However, they starts with stereoisomers (e.g., L- and D-glyceraldehyde). You can find a good description here.


text:
Proper method to electroplate aluminum?
I'm looking to plate some aluminum with silver using a basic solution.
My first attempt was to use sodium carbonate to make silver acetate. I sped up the process using hydrogen peroxide. I used the silver acetate as the ion source.
This simply did not work to plate any aluminum.
I concluded that this is because the aluminum was oxidized, preventing a plating from sticking.
I've come to the conclusion that the main problem with plating into aluminum is that it's able to rapidly form a highly resilient oxide layer.
Are there any common ways to prevent this oxide layer from forming? More to the point, are there any common methods to plate aluminum, specifically with silver?

It is simply impossible in aqueous solution.

@Karl if I used an oil with sufficient electrolyte?

You need a blank metallic surface, and a water-free electrolyte with silver ions in it. Im sure that´s somehow possible. Affordable? No. Can be done outside a well-equipped chemical laboratory? No. Sorry, but forget it.

Apparently, aluminum can be electroless nickel-phosphorus plated, as per https://en.wikipedia.org/wiki/Electroless_nickel-phosphorus_plating. Then the nickel plating can be silver plated, which is a standard scenario. You might also search this site for silver plating questions and answers. So the trick is several stages, which is not uncommon. Best of success!

I’m so confused now. I did this when I first started experimenting with electrochemistry (when I didn’t know what I was doing) I was using anything I had in my garage as an anode and somehow I copper plated this aluminum speed square.
All that I remember is that I was using CuSO4 I made using a copper anode and copper cathode in a vase filled with H2SO4 in distilled water (thanks NurdRage), where I switched out the cathode for this tool and it came out plated, and the strike is still totally solid!
Now that I know that it is technically impossible, I’m sure I’ll never be able to recreate the “experiment.”
Anyone have any ideas what could have contributed to my accidental success?

You sure it wasn't steel?

@tuskiomi definitely not steel. Their website says “heavy gauge aerospace aluminum alloy billet” so I tried contacting them about what the alloy is, but aren’t really counting on them telling me! Do you have any idea what alloy would allow plating in a copper sulfide electrolyte?

No ! There are no method for plating aluminum. Thousands of people have tried. In vain. It is impossible to prevent the formation of a thin layer of aluminum oxide which is waterproof at the surface of the metal. The only way of preventing it is to dip a piece of aluminum in a solution of mercuric chloride. This produces a reaction $$\ce{2 Al + 3 HgCl2 ->  3 Hg + 2 AlCl_3}$$ and the mercury makes an alloy (or amalgam) with not yet oxidized aluminum, and this prevents the formation of the continuous oxide layer. But this alloy does not prevent the oxidation of Al in contact with air. The oxidation of the amalgamated aluminum proceeds with such a speed that the growth of the aluminum oxide filaments can be seen by the naked eye. In a couple of minutes, white filaments of oxide grow to become one or two centimeters long. This extraordinary phenomena stops when the mercury has been removed from the surface by the growing oxide filaments. Of course, today, it is forbidden to carry out such experiments, because of the mercury toxicity. Fifty years ago, it was allowed. I do remember.

I know this is an old answer, but if aluminum forms an oxide layer so fast, can't you move oxygen out of the environment?

How move oxygen out of the environment ?

Pure simple straightforward electroplating is a no go onto aluminum, however aluminum can be electroplated onto other metals and itself, using chloride solutions and secondary reactions can interact with the oxide layer not enough to dissolve it but enough that a sufficient supply of surfactants will allow the softened oxide layer to be pulled away
I personally messed around with some poorly translated USSR documents on the subject alongside you tube videos for electroless plating of copper on AL in the purpose of getting a decent solder joint with very limited "success" (it immediately oxidized the copper, but I found the effect kind of wood like in appearance and used it decoratively anyways)
Some things I found to help was magnesium salts, an acidic solution, dawn brand dishsoap, boiling everything I could prior to use to remove as much trapped air and oxygen as possible, and using mechanical means to remove the oxide layer while the aluminum was submerged in the solution. There even where a few batches that turned out copper colored and maintained that coloration for several days before turning brown.
This is of course my progress so far doing something entirely different, but it is on the same base metal.
There are also older procedures for electroless and electroplating on titanium, they used some very.. uh. Toxic materials. Molten salt baths are the standard and in the end you may have to use a "low" temperature salt bath to get good repeatable quality electroplating, but I see it as a minor thing, 180°c is home candy making Temps, its definitely home chemistry Temps, just not common.

This is some really cool info.. I never thought about doing a salt bath... I suppose you do such a thing outside, I can't imagine salt plating is pleasant in any form.


text:
Entropy increase of both system and surroundings
In "why chemical reactions happen" the author (James Keeler) says that "a process which leads to an increase in the entropy of the system can be ... exothermic." Could someone please give me an example of a process which is exothermic and increases the entropy of the system? It seems strange to me how a system can release energy in the form of heat while also increasing its own entropy.

Could you edit the question to specify which author said the quoted bit, and in which book or article? Could you also edit the question to tell us why you are asking this?

We don't need to go to nuclear fusion to find an example. Lots of chemical reactions meet your criteria of being both exothermic (releasing heat to the surroundings) and increasing the entropy of the system.
A reaction is spontaneous if the free energy change of the reaction is negative ($\Delta_rG <0)$.The free energy change of a reaction at constant temperature is equal to the sum of the enthalpy change ($\Delta_rH$) and the temperature multiplied by the negative entropy change ($-T\Delta_rS$):
$$\Delta_rG = \Delta_rH - T\Delta_rS$$
One way for $\Delta_rG$ to be negative is for $\Delta_rH$ to be negative and $\Delta_rS$ to be positive. In a reaction conducted at constant pressure, the enthalpy change $\Delta_rH$ is equal to the heat exchanged between the system (reactants) and the surroundings. When $\Delta_rH <0$, the reaction is exothermic.
To find examples that meen your criteria, you only need a source of standard entropy and ethalpy data so that you can calculate $\Delta_rH$ and $\Delta_rS$ for a reaction. There are many such examples of reactions that are exothermic and have a positive entropy change of the system. Below is a simple example:
$$\ce{C(s) + O2(g) -> CO2(g)}$$
Use the NIST Chemistry Webbook, here are the standard enthalpies of formation and standard entropies of formation for carbon, oxygen, and carbon dioxide:
$$\begin{array}{|l|c|c|} \hline
 & \Delta_fH^\circ\ \mathrm{(kJ/mol)} & \Delta S^\circ\ \mathrm{J\cdot K/mol} \\ \hline
\ce{C(g)} & 0 & 5.8 \\
\ce{O2(g)} & 0 & 205.15 \\
\ce{CO2(g)} & -383.52 & 213.79 \\ \hline
\Delta_r & -383.52 & 2.84 \\ \hline
\end{array}$$
At all temperatures, this reaction is exothermic and has a positive entropy change for the  system.


text:
Finding the ratio of rate constants
A group of mountain climbers set up a camp at a $\pu{3 km}$ altitude and experience a barometric pressure of $\pu{0.69 atm}$. They discover that pure water boils at $\pu{90 ^\circ C}$  and that it takes $\pu{300 min}$ to make a "three-minute" egg. What is the ratio of rate constants at $\pu{100 ^\circ C}$ and at $\pu{90 ^\circ C}$ i.e. $\dfrac{k_\pu{100 ^\circ C}}{k_\pu{90 ^\circ C}}$?
How should I approach this problem? I have no idea what to do here, though it seems like an interesting problem. The given answer is

100

:/ Hardly interesting, or even realistic - rather trivial especially knowing the answer.

This is too clever for its own good. That's not how rate constants work, assuming this is not a zero-order "reaction." The reason I say this is that "doneness" is probably a function of the progress of the reaction in terms of the starting concentration, e.g., done is 90% reactant consumed. However, if you look at how a half-life is computed for a reaction greater than zero order, it is not a linear function of the rate constant. Therefore, the logic behind this problem is deeply flawed.

Hint: 300 minutes to make a 3 minute egg. And yes, that is unrealistically long.

@DrMoishePippik I think a "three minute egg" is just a name for creating sarcasm. This problem came in my test.

The vapor pressure of water at $\pu{90 ^\circ C}$ is $\pu{525.8 torr}$, which is $\frac{\pu{1 atm}}{\pu{760 torr}} \times \pu{525.8 torr} \approx \pu{0.69 atm}$. Thus, it is true that water boils at $\pu{90 ^\circ C}$ at altitude with $\pu{0.69 atm}$ pressure. However, taking $\pu{300 min}$ to boil an egg in same altitude is another story (think of people in Bhutan where average altitude is $\pu{3.3 km}$! How long it would take them to cook their lunch?).
Since it said in the question, we don't have a choice but go for it. However, since no additional data is given, we have to assume the edd boils linearly. That means at time zero, there would be 100% row egg and 0% boiled egg in both situation and at $\pu{100 ^\circ C}$, 0% row egg and 100% boiled egg would be remaining after $\pu{3 min}$. Therefore, the average rate of an egg boiling at $\pu{100 ^\circ C}$ is $\frac{100\%}{3} \ \pu{sec-1}$. Similarly, average rate of an egg boiling at $\pu{90 ^\circ C}$ would be $\frac{100\%}{300} \ \pu{sec-1}$. Assuming linear boiling, I plotted the percent boiling versus time in both cases:

These plots show that the rate constant $k$ is same as the rate of boiling. Thus:
$$\frac{k_\pu{100 ^\circ C}}{k_\pu{90 ^\circ C}} = \frac{\frac{100}{3}}{\frac{100}{300}} = 100$$
If you want to use the slope of plots (which are equal to corresponding rate constants):
$$\frac{k_\pu{100 ^\circ C}}{k_\pu{90 ^\circ C}} = \frac{33.333}{0.333} = 100$$


text:
Shouldn't this compound be called 2-methyl-5-(1,2-dimethylpropyl)nonane?

The book says that the name should have dimethylpropyl methyl nonane in that order. Shouldn't the methyl group come first since methyl comes before methylpropyl (in the alphabetic order)? Our prof told us to ignore prefixes such as mono, di, tri when putting groups in alphabetic order.

In addition to the answer provided, there is a special section chemical nomenclature on the resources site. For organic chemistry, select the Blue Book, or/and the flow charts in the freely available short guides.

‘5-(1,2-Dimethylpropyl)-2-methylnonane’ is not the preferred IUPAC name. The preferred name according to current IUPAC recommendations is 2-methyl-5-(3-methylbutan-2-yl)nonane.
If this is a homework exercise, however, it might be based on obsolete IUPAC recommendations from 1979 or 1993.
You are partly correct; simple prefixes are arranged alphabetically disregarding any multiplicative prefixes. The multiplicative prefixes are inserted later and do not alter the alphabetical order. For example, ‘1,2-dibromo-’ is considered to begin with ‘b’.
However, the name of a compound substituent is considered to begin with the first letter of its complete name. For example, ‘1,2-dibromobutyl-’ is considered to begin with ‘d’.
Likewise, ‘1,2-dimethylpropyl-’ is considered to begin with ‘d’. Therefore, it is arranged before ‘methyl’.


text:
How do I prepare 2-(3-azido-3-oxopropoxy)ethyl methacrylate from methacrylic acid?
I am having some trouble finding a proper synthesis route for the following product.

Originally, I was thinking of reacting methacrylic acid with a large excess of ethylene glycol (along with an acid catalyst), in accordance with Fischer esterification, to obtain the following structure.

I subsequently thought of reacting a base such as sodium hydride, NaH, with the alcohol group (removing its 'H') in the hopes of connecting it to the other fragment, which I thought would contain an alkyl halide. However, I was unsure how realistic this step might be due to the following concern: couldn't the hydride ion, which is a very strong base, attack the ester group instead of the alcohol (despite the alcohol having the lower pKa of the two groups)?
To get around this, I attempted a synthesis, pictured below, starting from ethylene glycol. My main concern with this, however, is the potential for intramolecular Fischer esterification instead of, as depicted, the esterification taking place through methacrylic acid and the acid catalyst (e.g., HCl). How valid is this concern? Additionally, please identify any mistakes I may have made in my synthesis and suggest improvements, even if you have a different synthesis route in mind.

Your esterification reaction on the second line with methacrylic acid will give an intractable mixture if you don't protect the acid of the propionate group. There will be no selectivity over which group gets esterified

Also acyl azide tends to be unstable and rearranges to isocyanate on heating. So, note the temperature.

here's how you make an acyl azide http://orgsyn.org/demo.aspx?prep=CV3P0846

You're going need a protection strategy to make this work. Here is one possible route:

Start with 2-Chloro/bromo ethanol and protect the OH with a silyl
ether, $\ce{SiMe3}$ will do, but requires anhydrous workup.
React the protected haloethanol with the anion of
3-hydroxypropionate-t-butyl ester (commercially available) ($\ce{NaH,
NaI, DMF}$) then deprotect the product (mild acidic work up for $\ce{Me3SiO}$
or $\ce{F-}$ for others).
React this product with the acid chloride of methacrylic acid
(commercially available). Isolate and purify the product.
Remove the t-butyl ester with $\ce{TFA/CH2Cl2}$, concentrate to dry and
form the acyl chloride ($\ce{SOCl2}$, cat DMF).  React this with sodium
azide to give the product following the first part of this prep
(Org. Syn.).

For "Step 2" of your answer, I am guessing (correct me if I am wrong) that NaH would react with 3-hydroxypropionate-t-butyl ester to produce the anion; however, it is unclear to me what purposes NaI and DMF would serve and/or how they might come into play. Could you please clarify?

NaI does an in situ exchange with the halide of the haloethanol to make the alkylation go better, you only need a catalytic amount. DMF is the reaction solvent; as a polar aprotic it is frequently used with NaH

Could DMF also be used effectively with NaI for the halide exchange stage? I found out that acetone is often used with NaI for the Finkelstein reaction, but I would like to minimize the number of solvents used.

It can all be done in DMF. There is no separate step with NaI.


text:
How to make Lithium medication from scratch
I take lithium carbonate as a medication. I was watching an "apocalypse-style" movie recently, which made me wonder how hard it would be to make my medication from scratch if I could no longer buy it. Some thoughts:

I might be able to make it from batteries containing lithium, but in an apocalypse, those might be too valuable. Maybe there are other sources of recycled lithium.
According to wikipedia, much of the world's lithium comes from brine pools in South America. I currently live on the California coast. I assume we don't have these...brine pools? Would it even be possible to obtain what I needed locally from natural sources?
I have read that lithium carbonate breaks up in your blood, so maybe any form of lithium would work, not necessarily lithium carbonate.
How would I figure out the dosage and the purity of my product? Lithium is well-known for its therapeutic level being close to its toxicity level.
What kind of equipment would I need?
How would I get my blood lithium level checks done to make sure I wasn't overdosing?

NOTE:
I'm just curious. I'm not planning on actually abandoning my cheap store-bought meds for my amateur attempts at following a recipe I read online that could likely poison me.

Dead batteries would probably be your best bet. There isn´t really another use from where you could easily recycle it. Of course the west of US has potential natural deposits, Salt Lake City? A geologist can sure find you sth there. The extraction and purification will be a lot of work. Battery recyling can be done at home. If you know what you´re doing, and you know a lot about chemistry so you can safely and efficiently improvise in Postapocalyptica. Getting a doctorate in chemisty takes eight years or so, better start now. ;-)

Relevant https://youtu.be/U2UQ4-NOP9E


text:
What does it mean for chemical reactions to release energy?
My Physics textbook gives the following definition of energy :

The capacity of a body to do work is defined as the energy possessed by it

This was the definition that was used to derive $U_G = mgh$ and $\text{KE} = \dfrac{mv^2}{2}$
Now, consider the following statement

Exothermic reactions release energy in the form of heat

I am unable to see how the first definition and second statement relate to each other.
As I see it, an object losing energy means that it's capacity to do work decreases. What is the object that's loosing energy here? And, what does this energy get transferred to? The atmosphere in which the reaction is taking place? I seem to have some silly misconception.
Thanks!

You might be mixing two different examples together.. In your first statement, you are talking about the physical energy of a body whereas in the second case you are taking a reaction which causes a change in chemical potential.

@Safdar I'm afraid I don't know what chemical potential is, can you provide a brief definition of it that can be interpreted by a layman? Thanks.

http://www.icsm.fr/Local/icsm/files/286/JFD_Chemical-potential.pdf Does this help?

@Safdar Looks too advanced for me. I'm not yet introduced to thermodynamics and not much of calculus too and the link that you posted includes partials as well. I'm currently in 10th grade. At what level in high school education do you think this'll come up?

Then wait until next year.. In simple terms what happens is when you have a chemical reaction, you first supply some energy to break the bonds between atoms in a molecule. This creates a compound with a higher amount of energy. that is unstable. After this the compound reacts with the other reactants to give a molecule that is more stable. This is done by releasing energy. If the energy produced is more than the energy given, it is released into the surroundings that we observe as heat. Such a reaction would be exothermic in nature.

Here, chemical potential refers to the tendency to react. The higher it is, the more likely it is to participate in a chemical reaction. This difference in chemical potential energy is enthalpy.. For an exothermic reaction the initial potential energy is greater than the final potential energy

Will 11th grade Chemistry provide a better understanding of 'energy' in the context of chemistry too?

Let us continue this discussion in chat.

You are perfectly right to be confused. That first definition is a pointless generalisation, bordering on being plain wrong. Usually (=in the terminology that 99% of the physicists and chemists in the world agree on) "work" $W$ excludes thermal energy transfer "Q".

Look energy has various forms, the formulas you listed are applicable to macroscopic objects and are due to  position and the velocities they posses  (for potential energy and the kinetic energy your'e talking about) respectively.
But when we zoom into the molecules and atoms the energies they posses are there due to completely different reasons. If you know molecules tend to vibrate and/or rotate about specific axes that contributes to one kind of energy they posses, then you can have energy due to the bonds through which they are bonded because of how the orbitals overlap (If you don't know what orbitals are, they are just regions around the nucleus where electrons are found), atoms and molecules sometimes form lattice you could say these are just clusters if them arranged in a specific manner, and a lot many things the list goes on.
Now a reaction involves two things reactants and products. So many a times it happens that the product formed is way much stable due to various different reasons than the reactant by stable itself. Now imagine a scenario you have pencil, now if I ask you to make it stand on it's tip you won't be able to but lay it horizontally it will easily rest like that so you may ask why? What happens is that the horizontal position is way stable than than the vertical because in the latter position is unstable than the former why you'd ask, it is because when you try to make it stand it on it's tip you have to correctly align the center of gravity with the tip so it perfectly balances. Same applies to atoms and molecules some configuration is more stable than the other due to various different reasons including some that I talked about above. Now when you try to keep the pencil up on it's tip it takes more effort hence more energy is required to keep it that, in way same when a chemical configuration is unstable than the other it needs more energy, when it becomes more stable it discards the excess energy it required, 'it' here means the molecule so the molecule is what gives off the energy as heat. And if you don't know fire is a result of this kind of reaction so you can probably guess were the heat/energy goes.

I assume that as you are in 10th grade you know that bonding between atoms is due to forces of attraction. To separate the atoms in the bond, you need to exert a force enough to overcome the attractive force between them. Now, as distance is being covered in the direction of the force separating the bonds. Hence, work is done. As work is done by the force, it means energy equal to the work done on breaking the bond was supplied to the molecule. Thus, energy is absorbed (It can be in any form; heat, light etc.).
Now, during bond formation, (taking the formation of a diatomic molecule) two atoms come close to one another due to their attractive force. Distance covered in the direction of the force is decreasing, actually having negative values according to the law of vectors. Then, as the distance covered is opposite to direction of force separating them, work is done on the force by the atoms forming the bond. The attractive force is doing work. Thus, energy is supplied by the bonding atoms to the environment providing separation force, equal to the work done. This is energy released to environment in the form of heat.
In exothermic reactions, energy released during bond formation is greater than energy absorbed during bond breaking. Concludingly, the extra energy is released in the form of thermal energy.
I think this is sufficient for you. You will learn in AS and A levels that
chemical bonds' thermodynamics is also related to kinetic energy and potential energy.
As distance between atoms increases, potential energy increases between them, and hence, work has to be done so energy is absorbed to separate them and vice versa.
Hope this will be helpful to you at your stage.


text:
Relationship between reaction quotient and potential energy
I am trying to calculate the Gibbs free energy in an NVT simulation. Although
$$\Delta G = -RT\ln Q,$$
in my simulation I can't calculate
$$Q = \frac{[\text{Products}]}{[\text{Reactants}]}$$
because I cant determine the concentrations of reactants and products.
Hence, I am looking for a relationship between reaction quotient $Q$ and potential energy $U$ as well as $p,$ $V,$ $T,$ i.e.
$$Q(U,p,V,T) = ?$$

$G$ itself is not equal to $RT \ln Q$; you either want $$\Delta_\mathrm{r}G ^\circ = -RT\ln K,$$ or $$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT \ln Q.$$ Aside from that, I'm highly skeptical that one can find a relationship between $Q$ and $U$, assuming that by $U$ you mean "the total internal energy of the system".

Did you meant "reaction quotient" instead of reaction coordinate?

@orthocresol I am looking for U as potential energy. Forgive my lack of proficiency with the terms, if total internal energy is essentially the same as potential energy.

Terminology aside, your first equation is still concerning, to me at least.

@orthocresol yes you are right w.r.t. the equation. However, since I will be calculating the Gibb's Free energy at each frame, I can calculate the difference without the equilibrium gibbs free energy as that would be the common term in all energies and the end result I am looking for is the difference.

Even so, then you should write $\Delta \Delta G = RT\ln \left( \frac{Q_2}{Q_1} \right)$. Also, you should use the Helmholtz free energy - not Gibbs - if working in NVT.

@BuckThorn why should I use helmholtz instead of Gibbs for NVT?  Is there a resource you can cite for this? I am trying to use eq2 from this paper and it uses G for the free energy so I assumed it has to be gibbs

The equation you are thinking of (if not actually writing) assumes the pressure is constant, which it isn't in NVT.

@BuckThorn could you please mention the eqution you're talking about.

The same one orthocresol mentions in the first comment

Maybe try googling "NVT ensemble Helmholtz", I hit this for instance: https://www.researchgate.net/post/Among_NVT_and_NPT_ensemble_which_one_is_preferable_during_MD_simulation

You might also want to post your question over at the material modeling site. It might be more appropriate there.