text:
How do you find the pH of lime juice through titration?
I am doing an experiment on the effect of time on the $\mathrm{pH}$ of Lime juice.
I was wondering how you would calculate the $\mathrm{pH}$ through titrations using $\pu{1 Molar}$ $\ce{NaOH}$. I have been racking my brain on this for a few hours and searching online.
Sorry if this has already been answered. Thank you for your time.

There is no direct relation between pH and titration result, unless there is known there is negligible amount of citric acid salts, of other acids and other pH buffering systems.

To expand on @Poutnik comment, pH and titration (measuring number of moles) are two different concepts, measuring two different chemical properties. For example, 1 molar HCl is far more acid (lower pH) than 5 molar citric acid, but it would take more NaOH to titrate the citric acid.

Similarly,  a citric acid solution with lower pH can lead to both higher or lower titration volume, depending on the overall solution composition.

pH paper would be the cheapest and easiest way to estimate the pH of the lime juice. Even if titration could measure pH, it's annoying to do.

As said, titration and the concept of $\mathrm{pH}$ is very different. The process of titration is used to figure out an estimate of how much of acid $\ce{A}$ (of come concentration) will neutralize base $\ce{B}$ in an acid-base reaction (of come concentration) (also the reverse). The acid and base reaction is solely dependent on the coefficient of a chemical equation regardless of the $\mathrm{pH}$.

e.g., One mole of aqueous $\ce{HCl}$ will react always with one mole aqueous of $\ce{NaOH}$.

The $\mathrm{pH}$ in this case only indicates the "acidity" of either of the reactants based on the amount of $\ce{H3O+/OH-}$ ions that dissociated form them:
$$\mathrm{pH} = - \log[\ce{H3O+}]$$
You can of course figure out the concentration of a base/acid in titration if you know the basicity of one of the solution and the mole of acid/base in the other. You can do this by:

Let's say you want to know the concentration of a solution of ethanoic acid ($\ce{CH3COOH}$), we will have to prepare a suitable base to use as a titrant. We mix a universal indicator with the solution of $\ce{CH3COOH}$ and add it to a flask. Dissolving one mole of $\ce{NaOH}$ into x volume of water and put it in the buret. Run the buret until the resultant solution is neutral.

If the original volume of the $\ce{CH3COOH}$ solution is y and the amount of $\ce{NaOH}$ solution deposited is k.
Concentration of original acid is given by:
$$[\text{acid}] = {\frac{k}{x*\text{(original basicity of acid)}*y} }$$
as in
\begin{align}
 \\ [\text{acid}] &= {\frac{(\text{mole of}\, \ce{NaOH}\,\text{used})\cdot(\text{volume of}\, \ce{NaOH}\,\text{used})}{(\text{volume of}\, \ce{NaOH})\cdot(\text{original basicity of acid})\cdot(\text{volume of acid})} }
\end{align}
However, we still wouldn't be able to determine the $\mathrm{pH}$ of the original acid as we do not know how many of the acid in the original solution actually ionized:
$$[\text{acid}]*(\text{basicity}) \neq [\ce{H3O+}]$$
Thus we do not know the $[\ce{H3O+}]$ and the $\mathrm{pH}$ cannot be determined.

A further difficulty is the fact that the formula pH = - log [H+] is an approximation which is only valid at rather low concentrations (<< $0.1$ M). At higher concentrations, the concentration [H+] should be replaced by the activity $ a(\ce{H^+})$, which can be rather different from [H+]. For example, a solution HCl $1$ M has an activity $0.795$.


text:
Are all exothermic reactions spontaneous?

$\ce{GaAs (s)}$ is used in solar cells deployed on satellites found in the orbit of Earth. The best way to make these materials is through a technique called "molecular beam epitaxy", in which the following spontaneous chemical reaction takes place:
$$\ce{Ga(g) + As(g) -> GaAs(s)}$$
Pick the statement which best describes this process:
(A) The process is endothermic.
(B) The process is exothermic.
(C) The process leads to a decrease in the entropy of the universe.
(D) The process leads to an increase in the entropy of the system.
(E) The process leads to a decrease in the temperature of the system.

The key says the correct answer is B. I am wondering if this is because all exothermic reactions are spontaneous?

From the comments in the answers below, it seems that we lack a definition of spontaneous. Does it mean $\Delta_r G < 0$ at the current conditions or $\Delta_r G^\circ < 0$?

@KarstenTheis  Yes, that does seem to be the issue.

While is some technical thermodynamic sense, "spontaneous" might mean something different, in normal language even among chemists it means the reaction happens without an external push. Clearly many exothermic reactions are not spontaneous for the normal deifinition of the word. Burning coal is clearly exothermic, for example, but getting it to burn under normal atmospheric conditions takes significant effort and doesn't just happen.

@matt_black  Note, however, that this question, in giving options involving the entropy of the system and the universe, is specifically concerning itself with the fundamentals of thermodynamics.

No, it is not the case that all exothermic reactions are spontaneous.  More generally, it doesn't even make sense to make a broad statement that a reaction is or is not spontaneous without also specifying the concentrations of reactants and products, since for a certain relative concentration of reactants and products, the reaction might be spontaneous in the forward direction, while for another set of relative concentrations it might be spontaneous in the reverse direction. [See illustration below.]
One could, however, make the general statement that chemicals will react spontaneously if, in so doing, the entropy of the universe increases.  This, however, is not one of the options you are given.
Why, then, is "B" the correct choice?  Answer: It's because this Ga and As are forming a chemical bond, and bond formation is exothermic. In addition, there is a phase change here from a gas to a solid, and that is also exothermic.  Furthermore, since Ga and As are both monatomic gases in this reaction, no chemical bonds need to be broken to allow the Ga and As to combine. Thus, since we have chemical bonds being formed but not broken, and a phase change from gas to solid, the reaction is exothermic.
Here's a graphical illustration of the general concept articulated in the first paragraph:

I'll point out that the arrow for the reaction indicates that the reaction is irreversible. Right answer, wrong reason.

The logic of bond formation being exothermic is that otherwise, constraining the movement of the parts to be bonded would result in a decrease of entropy in the universe. For the hydrophobic effect, you gather non-polar molecules without strong interactions so that the entropy of water increases.

@MaxW.  Nope. All spontaneous reactions are thermodynamically irreversible, thus  the type of irreversibility you are speaking about it kinetic irreversibility (KI). So: (1) You can't rely on seeing single vs. dbl. arrows as a signifier of KI, since people often don't bother w/ the dbl. arrows regardless. Further I see no reason this rxn should have KI (i.e., be kinetically trapped).  (2) More to the point, the thermodynamics of the rxn are independent of the kinetics.  So, adopting your preferred linguistic style:  Wrong comment, wrong reason.

@KarstenTheis  As I explain in my comment to Buck Thorn, you can have an endothermic reaction in which the entropy of the products is less than that of the reactants, yet it can still proceed spontaneously.  Not sure how your reference to the hydrophobic effect applies here.  While there continues to be controversy about it, the standard description is that water suffers an entropy penalty (but a favorable decrease in enthalpy) in solvating non-polar solutes, because of the formation of ordered "clathrate shells" in which the water-water bonds are stronger than in bulk.

@theorist Sorry about the non-sequitur in my comment. I was wondering if my statement "any bond formation has to be exothermic" is general, and pondering whether something like aggregation of non-polar species in water would be a counterexample. In the case at hand, there is no solvent, so the entropy balance is much easier to figure out.

The entropy of the system decreases
All the reactants are mono-atomic gases, and the product is a solid. This is a strong indication that the entropy of the system decreases.
The reaction is spontaneous
That statement from the text of the question means that the Gibbs energy of reaction is negative. It is not clear if this refers to standard conditions, or to the current conditions. The question could be more specific in that regard.
Enthalpy of reaction
If the reaction entropy is negative and the reaction Gibbs energy is negative, the reaction enthalpy has to be negative as well because it is the sum of the Gibbs energy and the entropy times the (positive) temperature.
$$\Delta_r G = \Delta_r H - T \Delta_r S$$
$$\Delta_r H  = \Delta_r G + T \Delta_r S$$

Are all exothermic reactions spontaneous?

No, but all reactions that are spontaneous at standard state (interpreted to mean that at standard state, equilibrium lies in the forward direction, i.e. K > 1 and $\Delta_r G^\circ < 0$) are either exothermic ($\Delta_r H^\circ < 0$) or show a positive standard entropy of reaction ($\Delta_r S^\circ > 0$), or both. This ensures that the Gibbs energy of reaction is negative and that the entropy of the universe increases (those two are linked in the absence of non-PV work).

[theorist] The direction of spontaneity is merely a function of where the reaction mixture is relative to the equilibrium constant.

The argument I made above does not preclude finding a reaction with positive $\Delta_r H^\circ$ and negative $\Delta_r S^\circ$ that goes forward. In this case, K will be smaller 1, but reaction conditions can be chosen so that Q is smaller than 1 as well, and furthermore smaller than K, which is always possible by increasing reactant concentrations or by removing product.

[Textbook question]Pick the answer that best describes this process.

Answers A) and E) are linked. An endothermic process (A) will decrease the temperature of the system (E). So those two answers are out if there is only one correct answer. Answer C) goes against the second law of thermodynamics, so most teachers would mark it wrong. Above, I made an argument that the reaction has a negative standard entropy of reaction (less freedom for the atoms when they go from mono-atomic gas to a solid compounds); answer D) posits the opposite, so it is out.

The fact that the term "exothermic" is used in "B" properly means it is referring to the sign of the standard-state enthalpy change (see IUPAC link in comment to Buck Thorn), which means the entropy change (if used in the same expression) must, for consistency, be standard state as well.  And, as I've explained elsewhere, the fact that a rxn has a negative standard-state entropy change, and is spontaneous, places no constraints on whether it is endothermic or exothermic.  The direction of spontaneity is merely a function of where the reaction mixture is relative to the equilibrium constant.

And even if standard state wasn't intended, clearly these refer to the enthalpy and entropy changes inherent to the difference between pure reactants and pure products, not to the equilibrium mixture.

@theorist I agree with you that irrespective of the signs of reaction entropy, enthalpy, and Gibbs energy, you can always find conditions where Q < K. I will edit my answer to avoid the term "spontaneous" in my statements.

I would add that, given that the entropy change associated with the formation of the solid is almost certainly negative, the only way the reaction can proceed spontaneously at constant T and p is if it is exothermic, invoking the condition that $\Delta G = \Delta H - T \Delta S<0$ or $$\frac{\Delta H}{T} <\Delta S$$ for a spontaneous process.
That the process is exothermic can be justified using the arguments presented by theorist.

You wrote "the only way the reaction can proceed spontaneously at constant T and p is if it is exothermic (invoking the condition that $\Delta G = \Delta H - T \Delta S<0$ for a spontaneous process)".  That's actually not generally true, since one also needs to factor in the entropy of mixing.  As a consequence of the entropy of mixing, a reaction could proceed spontaneously even if $\Delta H_{rxn} > 0$ and $\Delta S_{rxn} <0$, if one were to the "left" of the equilibrium position.

@theorist entropy of mixing would contribute to the $\Delta S$ term. But sure, we can split hairs if you like.

@theorist you split hairs in that you are not reading between the lines. For the reaction as written to proceed in the forward direction (which obviously is implied) then, given that $\Delta S<0$, you must have an exothermic reaction.

It's not splitting hairs at all. You made an argument that the rxn has to be exothermic based on it being spontaenous + $\Delta S < 0$, where that $\Delta S$ referred only to the inherent entropy difference between the reactants and products ("the entropy change associated with the formation of the the solid is almost certainly negative").  That $\Delta S$  explicitly doesn't take into account the entropy of mixing.  As a consequence, your argument that the rxn must be exothermic because of this is simply incorrect...[continued]

As a simple counterexample, consider a reaction, A->B, in which $\Delta H = 10 kJ/mol$, and $\Delta S = - 10 J/(mol K)$, at 298.15K.  Then, at that temperature, $\Delta G = 13 kJ/mol => K_{eq} = 0.0053$.  Thus, if $\frac{[B]}{[A]} < 0.0053$, the reaction will proceed spontaneouslyly from reactants to products, even though $\Delta S <0$ and it is endothermic. I.e., the fact that it can proceed spontaneously from reactants to products while having a negative $\Delta S$ does not in any way constrain the reaction to be exothermic!

That's why I said it doesn't make sense to broadly label a reaction as spontaneous -- thermodynamic spontaneity is determined by where the reaction mixture is relative to the equilibrium constant, irrespective of the signs of the standard-state entropies and enthalpies.  [And when you speak about reactions being endothermic or exothermic, you are inherently speaking about standard-state changes; see:  https://goldbook.iupac.org/terms/view/E02095 and https://goldbook.iupac.org/terms/view/E02269

@theorist let me just point out that while $\Delta G ^\circ = -RT \ln K_{eq}$, it is the value of $\Delta G$ for the process that determines sponteneity. You gave a good answer to the actual question, and my answer attempted to address the source of the OPs confusion.

A exothermic reaction has the energy to proceed spontaneously. It may or may not depending on the kinetics. The reaction could be sterically hindered for example, and need a catalyst to happen at all.

Umm, it could be entropically unfavorable?

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text:
Difference between vapor-phase transport and vapor-phase deposition?
I am currently studying the textbook Physics of Photonics Devices, Second Edition, by Shun Lien Chuang. In a section discussing The Invention of Semiconductor Lasers, the author says the following:

At the Solid State Device Research Conference in July 1962, an MIT Lincoln Laboratory group and RCA Laboratories reported extremely high efficiency (85% to 100%) electroluminescence from GaAs diffused junction diodes. Semiconductor lasers were invented during September to October 1962 by four groups within 30 days [9-12] (see the review article by Dupuis in Ref. 13). They were led by Robert N. Hall of General Electric Research Development Center, Schenectady, New York; Nick Holonyak Jr. of General Electric, Syracuse, New York; Marshall I. Nathan of the IBM Research Laboratory, Yorktown Heights, New York; and Robert Rediker
  of the MIT Lincoln Laboratory, Lexington, Massachusetts. Among the four groups, only Holonyak's laser diodes and light-emitting diodes (LEDs) were created from single-crystal GaAs$_{x}$P$_{1 - x}$ alloy material grown by vapor-phase transport and were the only devices emitting in the visible region.

I had already read about vapor-phase deposition in a textbook on MEMS (microelectromechanical systems), but I was not familiar with vapor-phase transport. Reading about these two methods, they sound very similar, and I'm unsure if they're just different names for the same thing, or whether there is some difference between them. I would greatly appreciate it if someone would please take the time to explain the difference between, if any.

The paper of interest is C.M. Wolfe et al., "Growth and Dislocation Structure of Single-Crystal Ga(As$_{1-x}$P$_{x}$)", Journal of Applied Physics 36(12) 3790 (1965).
The key paragraph is:

The GaAsP of this investigation has been prepared
  by a closed-tube method utilizing a quartz ampoule
  which is cleaned and baked-out under vacuum near
  the softening point of quartz. The halide is introduced
  by subliming a metallic halide, such as PbCI2, PbI2,
  NH4CI, NH4I, etc.,ro into the reaction vessel. Typically,
  2.2 g of source GaAs and GaP, 4-8 mg of excess P,
  40-160 mg of halide, and a dopant are placed in an
  8-cm$^{3}$ ampoule as in Fig. 1. This vessel is then evacuated
  to 10$^{-6}$ Torr, sealed, and placed in a single-zone furnace
  at 950°C with a gradient of 5°-15° along the tube.
  Commercially available GaAs is used with GaP grown
  from high-purity elements by a method previously
  described.

So, yes, it is a vapor deposition technique, but uses a thermal gradient to evaporate from one place and redeposit in another place, controlled by the temperature.

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text:
Kinetic Energy Evaluation Integral Evaluation Program Ostlund
I'm reading Ostlund's Modern Quantum Chemistry. In Appendix B, the kinetic energy integral is evaluated using the Gaussian Basis functions as
T11=T11+T(A1(I),A1(J),0.0D0)*D1(I)*D1(J)
T12=T12+T(A1(1),A2(J),R2).D1(I).D2(J)
T22=T22+T(A2(1),A2(J).0.0D0).D2(1).D2(J)

I understand T11 to be the kinetic energy integral using the first atom orbitals so it will give the kinetic energy of the electron of the first atom, and this is why it is passing zero as the bond length parameter. The same is T22 but for the second atom.
What is T12 used for, aren't T11 and T22 enough to calculate the total kinetic energy?

TAR86 gives the correct mathematical answer, but given how you express yourself I thought I would try and explain it at a slightly lower level.
Your problem is in the sentence "I understand T11 to be the kinetic energy integral using the first atom orbitals so it will give the kinetic energy of the electron of the first atom"
The problem is that in a molecule the electron will be partially associated with the first atom, and partially with the second - this is almost the definition of a covalent bond! Thus yes there will be a contribution to the kinetic energy due to the first electron being on the first atom, but there will also be a contribution due to the first electron being on the second atom, and it seems to me that this is what you are missing.
Let's try and make that a bit more concrete with a little maths. As we are using an orbital approximation the first electron's behaviour is described by it's wavefunction. We don't know exactly what it looks like, but we hope it will be similar to the atomic wavefunctions on the two atoms. So let's write the wavefunction for the first electron as
$$
    \psi (1)= c_1 \phi_1 + c_2 \phi_2
$$
where $\phi_1$ is the first atomic orbital (actually contracted Gaussian basis functions) and $c_1$ is the weight of this function as determined by solving the approximation to the Schrodinger equation that we are interested in. Then if $\hat T$ is the kinetic energy operator we can obtain the Kinetic Energy of the first electron from
$$
T=<\psi(1)|\hat T|\psi(1)>=c_1^2<\phi_1|\hat T|\phi_1>+2c_1c_2<\phi_1|\hat T|\phi_2>+c_2^2<\phi_2|\hat T|\phi_2>
$$
(using Hermiticity of $\hat T$ and the fact here the wavefunction is real)
In this we can see a cross term which involves electron 1 on both atom 1 and 2, and in fact even a term which is due to electron 1 being on atom 2. So using your notation we can identify
$$
T11=<\phi_1|\hat T|\phi_1> \\
T22=<\phi_2|\hat T|\phi_2> \\
T12=<\phi_1|\hat T|\phi_2>
$$
Note more generally we should talk about basis functions rather than atoms, but here as the basis functions are associated with the atomic sites (this need not be the case) and there is only 1 basis function per site (in real calculations there will be many) we can be slightly sloppy. But also understand that the general case is not really much more complicated than the above, we simply expand the orbital for the first electron using all the basis functions, and carry on similarly.

The off-diagonal elements $T_{\mu\nu}, (\mu \ne \nu)$ are just as necessary as the diagonal elements $T_{\mu\mu}$ to represent the kinetic energy operator in the chosen AO basis set. Substituting a few of Szabo/Ostlund's equations into each other (in my edition, 3.147-3.154), one finds
$$
F_{\mu\nu} = T_{\mu\nu} + V_{\mu\nu}^{\text{nucl}} + \sum_{\lambda\kappa} P_{\lambda\kappa}\left[ \left( \mu\nu \vert \kappa\lambda  \right) - \frac{1}{2} \left( \mu\lambda \vert \kappa \nu\right) \right]
$$
where $F$ is the Fock matrix, $V^{\text{nucl}}$ represents the nuclear attraction term, $P$ is a density matrix and $\left( \mu\nu \vert \kappa\lambda  \right)$ are two-electron repulsion integrals over the AO basis functions (denoted greek indices). Of course, your Fock matrix may be diagonal already, if you have a very special case and basis set, but in general and to perform the SCF calculation, the off-diagonal parts cannot be ignored.


text:
How can they "wash away" the caffeine from Nescafe Gold to produce that "de-caf" version?
I buy and drink Nescafe Gold all the time. I also buy the smaller "caffeine-free" version of the same product. It tastes very differently, which makes me assume that the caffeine does have a lot of taste.
However, maybe it's not simply due to the lack of caffeine? Maybe the process with which they "de-caf" the original product (something about "washing it away", if I recall correctly from what I read/heard somewhere at some point) also removes other ingredients?
How can they possibly "wash away" just the caffeine from instant coffee like that?
Unless the truth is that the entire products are "built" from the ground up, separately for the caffeine-including and caffeine-less products, and the latter is not "based on" the former at all?
Even if so, they still have to "wash away" the caffeine from the raw coffee, which puts us back on square one...

Far away to know the process in details you are right for that whatever process won't specifically target just caffeine. Or better the target is caffeine but minor flavours and components will be affected too.

Decaffeination usually proceeds via extraction of caffeine from the beans using supercritical $\ce{CO2}$.

In the words of an old coffee commercial, “the natural effervescence that makes water sparkle” was used to ‘wash away’ the caffeine.

In general they use different types of solvents and methods to wash away the caffeine (or at least a great percentage of it) from the coffee beans , you can find more information here : https://www.todayifoundout.com/index.php/2011/04/how-decaffeinated-coffee-is-made/

Are you sure that DCM is still used? I know that it was employed in the 70s, but I don't think that nowadays it would be allowed in food processing

Yeah , I am not sure that it is still used today considering there are other alternatives but for example if you are doing a caffeine extraction in the lab and not on an industrial scale you can use this method and obtain quite a pure product

@DavidSterlinsky But you can't drink it to compare the taste, so it does not address the OP's question. Supercritical carbon dioxide is popular for extracting caffeine.


text:
How can you explain the solubility of BeF2?
I know that Be has a tendency to create bonds with an increase covalent character so I can tell that BeF2 is more covalent than let's say MgF2. So why is BeF2 soluble in water while MgF2 is barely soluble ? Shouldn't the ionic compound be more soluble ?
My guess is that the lattice enthalpy of BeF2 is not as great as the enthalpies of the other fluorides so it can be break down with much more ease and the energy gained from hydration outweighs the energy needed to separate the ions.

Because sucrose (sugar) is also soluble in water, maybe?  Water can dissolve plenty of covalent compounds, especially those with polar bonds or having hydrogen-bonding capability with the water.

It seems that for BeF2 the hydration enthalpy of Be2+ is really high (due to Be2+ being a small cation and having a high charge density) so the energy required to break the lattice is smaller than that gained from hydration of Be2+ while for something like MgF2 the enthalpy of the lattice is greater than the energy you would get from hydration of Mg2+. In general when dealing with compounds of the group II with F- because of the small radius of F- the lattice enthalpy is great and only the hydration energy released by the hydration of Be2+ outweigh it thus making BeF2 soluble while all the other fluorides formed with the group II elements are quite insoluble in water


text:
Is the concept of a "self cleaning" banknote plausible?
Although in 2020 there is a big movement of bankers marketers to lean the public towards cashless society, cash in the form of banknotes is still used in many if not most countries.
Cash banknotes can be very dirty in some occasions, especially old banknotes in humid countries and furthermore in humid storage areas.
Is the concept of a "self cleaning" banknote plausible?

No,  but there are currency launderers.

Something like idrophobic and self cleaning paints and coatings might be a possibility, like TiOx nanoparticles.


text:
How can subtilisin still function without its catalytic triad?
I read chapter 9 in the book Biochemistry (5th edition), by Berg, Tymoczko, and Stryer (provided in the NCBI site here). It describes the mechanism of action of the chymotrypsin enzyme. The catalysis is performed through the catalytic triad consisting of serine-195, histidine-57, and aspartate-102.
What intrigued me is the following section on the related enzyme subtilisin, which also has a catalytic triad but with different residue numbers:

9.1.5. The Catalytic Triad Has Been Dissected by Site-Directed Mutagenesis
[...] As expected, the conversion of active-site serine 221 into alanine dramatically reduced catalytic power. [...] The mutation of histidine 64 to alanine had very similar effects. [...]
These observations support the notion that the serine-histidine pair act together to generate a nucleophile of sufficient power to attack the carbonyl group of a peptide bond. The conversion of aspartate 32 into alanine had a smaller effect, although the value of $k_\mathrm{cat}$ still fell to less than 0.005% of its wild-type value.

My problem is in what it says next:

The simultaneously conversion of all three catalytic triad residues into alanine was no more deleterious than the conversion of serine or histidine alone. Despite the reduction in their catalytic power, the mutated enzymes still hydrolyze peptides a thousand times as rapidly as does buffer at pH 8.6.

A graph of the relative rates can be taken from this link:

My question is: how can it be that after replacing all the triad residues, the enzyme still possess such a great catalytic activity? What is the mechanism behind that phenomenon?

If the substrate still binds to the protein, other parts of the mechanism are still in place. Specifically, the main chain amide groups that help to stabilize the tetrahedral intermediate are still present. In the absence of serine, water will attack the carbonyl. In the presence of serine and the absence of the histidine, serine will still be the nucleophile, but a less efficient one.
Look up oxyanion hole if this part of the mechanism is unfamiliar to you.

yea, but water can attack the substrate protein even without the enzyme. why the big increase in catalysis compared to basic solution?

i'll explain my question further: water is a small molecule and it's abundant in the solution so it doesn't seems that the substrate need to be held in a particular position in order to be attacked by water. how can the oxyanion hole can perfect the catalysis if the water molecule is available in place even without it's help.  thank you!

@YtfuGjuf The oxyanion hole makes the substrate a better electrophile, just like the Asp and His in the triad make Ser (and the water in the second part of the mechanism) a better nucleophile. The substrate has to be in a particular position so that activation of the electrophile (the carbonyl) works.


text:
Are there optically active molecules in which all atoms lie in the same plane?
I've learnt that when a molecule is non-superposable on its mirror image, it exists as a pair of enantiomers and is optically active. Absence of plane of symmetry is given to be one of the properties of chiral molecules.
Are there optically active molecules in which all atoms lie in the same plane?
I think, it is not possible for a molecule to be optically active when all atoms are coplanar (lie in the same plane) because such a planar molecule would always have a plane of symmetry coinciding with the plane of atoms. But it would be helpful if you could clarify this, since there are a lot of exceptions in chemistry.

You think right.

@IvanNeretin: Thank you. However, it's quite surprising that there are no exceptions to this fact.

To be optically active, the molecule must not have a  mirror plane or inversion centre. Since all molecules that are planar have at least one mirror plane, they cannot be optically active. You have answered your question yourself.

@Magnesium There are no exceptions in geometry.

@IvanNeretin: I agree that "There are no exceptions in geometry". But I've seen a lot of exceptions in chemistry and expect something similar here :)

@Martin-マーチン: Thank you. I thought there might be some exceptions to this similar to absence of optical activity even in the presence of chirality centres and so on.

There are not too much (classes) of them.

This is not possible.
The definition of chirality ultimately relies on the superimposability of a structure on its mirror image. If they are non-superimposable, then the molecule is chiral. In terms of molecular symmetry, this is equivalent to saying:
If a molecule possesses an improper rotation axis $S_n$, then it is not chiral.
(An improper rotation is a combination of a normal rotation $C_n$ by $(360/n)^\circ$, followed by reflection in a plane perpendicular to that rotation axis.)
It turns out that the plane of symmetry and the inversion centre are the two most common forms of these (they are $S_1$ and $S_2$ respectively). Consequently, if a molecule has a plane of symmetry, it cannot be chiral. Likewise, if a molecule has an inversion centre, it cannot be chiral. There are also exotic molecules which possess neither a plane nor inversion centre, but possess a higher-order $S_n$ axis, and are therefore not chiral too.
However, the possession of a chiral centre has nothing to do with an $S_n$ axis. Consequently, there is no direct link between having a chiral centre and being chiral. Sure, there are trends that you can draw; for example, if there is only one chiral centre, then the molecule will not possess an $S_n$ axis, and thus it will be chiral. But the link is not quite as clear-cut as it is for the plane of symmetry, and consequently there are exceptions to the rule, and exceptions to the exceptions.


text:
Determining molar mass of protein from osmotic pressure

A 299 mL solution containing 21 g of a protein in toluene had an osmotic pressure of 0.053 atm at 27 degrees Celsius. What is the molar mass (in g/mol) of the protein?

We can use the formula that
$$\Pi = iMRT$$
where $\Pi$ is the osmotic pressure, $i$ the van't Hoff factor, $M$ being the molarity, $R$ being the constant (0.08206), and $T$ being temperature in Kelvin.
Using these values, we can substitute into the equation:
0.053 atm = 1 * M * 0.08206 L atm/(mol * K) * 300.15 K
When we equate this to M, we get
0.00215 = M
Now we know that there is 21 grams of toluene within the 299 mL solution. We can convert the Molarity given by multiplying it by the liters of solution there is. Therefore,
0.299 L * 0.00215 mol/L = 0.000643 mol
Finally, to get molar mass, you simply divide the mass of toluene by the mols given:
21 g / 0.000643 mol = 32639.41 g/mol
This value, however, doesn't seem to be correct, as it is absurdly large. Can anyone help tell me what I'm doing wrong?
-EDIT- I was reading through the question and realized that the answer was correct, I was just thinking about it in a different sense. Thanks to everyone for their help!

Watch this https://www.youtube.com/watch?v=tUD_d-3ZTCs. Proteins have very large molecular mass in kilodaltons. The textbook writer, or whoever developed this question, seems to have never done any practical lab work in life.

Please provide you complete attempt, including the formula with the explanation of what is in it. Otherwise it will be very difficult to help you.

Your numerical value of $R$ is correct, but the units are wrong: $0.08206$ goes with $\pu{L atm mol-1 K-1}$, not $\pu{J K-1}$. In general it will save you some pain if you make sure to include units for all quantities (such as the concentration $M$); at the very least, it prevents careless mistakes from creeping in.

By the way I sincerely doubt that 299 mL of a solution in toluene could contain 21 g of a protein. Usually the proteins are not as much soluble in toluene.


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How can sodium be reduced in the Castner–Kellner process?
In the Castner–Kellner process, why is sodium amalgam formed instead of hydrogen gas at the cathode?
If we electrolyse brine solution, then it will liberate hydrogen at cathode, not sodium, because $\ce{Na+}$ has a negative reduction potential while that of $\ce{H+}$ is (by definition) 0. What factor here causes the sodium to get reduced to form an amalgam in spite of the reduction potentials?

This is actually a nice question, for which I do not have the answer. The brine must be cold and concentrated, from what I have read. And it is even possible to electrolyze ammonium chloride, under similar conditions, to get ammonium amalgam. Upon warming, the latter yields nitrogen and hydrogen gases.

Now that I think about it, my guess is that the significant overpotential, for hydrogen gas evolution at the Hg cathode, is a big factor.

You probably want to avoid using all caps in the title, which is probably responsible for the downvotes. Capital letters makes it seem like you're SHOUTING...

@M.Farooq, you don't see many questions on MathOverflow with allcaps in the title. It's ok to say that there's a "high tendency for downvoting" (I might even agree with that, though I'm not sure), but IMO more helpful to point out possible reasons so that OP may take note and edit the current question / learn for future ones. [For what it's worth, I didn't downvote this.]

@orthocresol, I was talking about a general trend in Chem about downvoting. As I have said it again and again, the problem with downvoting is that we do not know if a high school student is downvoting it or an Einstein. In may schools, esp. in South Asia, block letters are often used by teachers in headings while writing notes or even on the blackboard.

@M.Farooq that’s a fair point; however, it’s a byproduct of the fact that this site is (or tries to be) open to both students and professionals. MO on the other hand has a much more limited scope. Re. capital letters, I don’t think it detracts from my point; the Internet has a different set of etiquette and the first time one comes on here they should be willing to learn that (and somebody should be willing to tell them, which I did). Anyway - I have my suspicions about who does all the downvoting, but that is of no use. Ultimately I cannot control who votes and how. Only my own account. :/

What I can say is that the best counter to downvotes is usually an edit. See this question: -2 before I edited, now +2, and it wasn’t even a large edit. In my experience, this works, so if you want to change the general public’s opinion towards a question, please do consider improving it yourself!

@orthocresol, I completely agree. Some people do it as mischief here. Another problem with down votes that people then avoid answering them. I agree many questions are of low quality.

I am not sure if you are familiar with the term "overpotential." With mercury electrode, as a cathode, it takes more negative potential to reduce hydrogen ions to hydrogen gas, rather it is easier for sodium ions to get reduced on a mercury surface and form an amalgam (an alloy of Hg and Na). This is a peculiarity of the mercury electrode.
Mercury cell in chlor-alkali industry

Can you explain to me a bit more about "overpotential"?

Please see https://en.wikipedia.org/wiki/Overpotential


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To get clear plasma out of a minute particle (such as fibrin), is it necessary to vortex sample before centrifugation?
I'm going to analyse plasma samples that contain a lot of fibrin. It is necessary to centrifuge the samples to be able to pipette the plasma. After the samples are centrifuged I can't vortex them because then I mix everything again and it wont be possible to pipette. Now I wonder whether it is necessary to vortex them before I centrifuge them. I don't think so because during centrifugation particles will anyway be separated by density. Or am I wrong? And in that case why? Appreciate your thoughts!

Are you talking about plasma or serum? Did you already separate either from blood sample? You need to include these specific info in your question.

I'm voting to close this question as off-topic because it doesn't seem to be about chemistry.

@Mathew Mahindaratne it is plasma and yes it is allready separated from erytrocytes. Sorry for leaving out information!

@Mithoron  It is about how chemicals separate, do you know a more suitable forum for this type of question?

Visible solid fiber particle (e.g., fibrin) will separate in higher speed centrifugation by gravity (not really by density). If your analysis is based on density gradient, you should use ultra-centrifuge (rpm is about $1.0 \times 10^5$ or higher).

@Mathew Mahindaratne, Ok! have tried with 2000 g for 10 minutes in 4 degrees. Then all other metabolites will also be separated, by gravity, density? Is there any point vortexing them before centrifugation then?


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What is maximum energy level of molecules?
I have been reading about the relation of molecular graph spectrum, that is, the set of eigenvalues of the Laplacian matrix of the molecular graph, and molecular properties. Recently, I came across a paper that finds an upper bond for maximum eigenvalue (DOI: 10.1109/ICNC.2012.6234727). To motivate their research, authors state that

The eigenvalues of adjacency matrix of the graph is used to present
  the energy level of specific electronic. Particularly, spectral radius
  of graphs is the maximum energy level of molecules.

I was wondering if anyone can elaborate on what maximum energy level of molecules is. Is it the maximum energy level among all the atoms of the molecule?

This is interesting. I don't have access to the paper, but maximum vibrational level makes some sense in that all vibrations are anharmonic and hence dissociate at some finite vibrational quantum number. I'm not sure how they define maximum energy level for the entire molecule though. This is just one way, but this would depend on the vibrational state.

Ok I got access to the paper and they don't talk about this at all in the paper. They just derive some theorems which give bounds on the spectral radius of a graph. While it may be possible there is some application for this in chemistry it's hard for me to see the connection between what they're describing and whatever type of energy of a molecule they're talking about.

Yeah, I even looked up maximum energy level of molecules in quotation and all the links in the first page are from this paper, so it seemed to me that they kinda have coined the term!

In the painting "Dream Caused by the Flight of a Bee" by Salvador Dali, among other strange items, we may see a fish flying in midair. There is also one in the drawing "Big Fish Eat Little Fish" by Bruegel the Elder. One may conclude that there is a flying fish in every picture out there. One may even publish a paper to that effect. But I believe you see the shortcomings of this way of reasoning.
Same thing here. Molecules don't have the maximum energy level. You may think of the energy required to break a  molecule, but that's not a level; rather, it is a lower bound of the continuous spectrum. You may think of the energy levels of its constituent atoms, but those are quite irrelevant to the energy levels of the molecule. Moreover, atoms don't have the maximum energy level either.
At this point you may wonder: what were the authors thinking? Were they (together with the editors) just lunatics? Not at all! Then what did they have in mind? Why, they said it loud and clear: the spectra of graphs. How are those related to the spectra of molecules? Is a molecule the same as a graph? Of course it isn't... unless we are looking at it within certain simplistic approximation where it is.
Yes, this is the old good Hückel method once again. You take a conjugated $\pi$ system, draw a graph of the molecule, solve it for eigenvalues, and get the $\pi$ energy levels of the molecule (only upside down). Naturally, there is but a finite number of those, and there inevitably is a maximum. That's what they mean when talking about the "maximum energy level of a molecule".

Now, these $\pi$ levels typically include the frontier orbitals (HOMO and LUMO), and those next to them, and answer pretty much any chemical question about the molecule. All in all, it is quite natural and easy to forget altogether about all other orbitals and energy levels of the molecule ($\sigma$ bonds and stuff). But they are still there, and what I said before is still true.
So it goes.

I can think of two reasons that a 'maximum energy' might exist for a molecule:

We can rigorously define the atomization energy for a molecule - that is, the enthalpy required to completely separate a given molecule into constituent atoms.
As mentioned above, above a certain vibrational energy, the vibration will dissociate (e.g., one can break a C-H bond in methane without atomizing the entire molecule).

I don't believe you can bound this for all possible molecules. The atomization energy will increase with the number of atoms and bonds - the more of each, the greater the enthalpy required to atomize. (Indeed, my group often uses a normalized atomization energy by dividing by the number of atoms.)
For an isolated vibration, this would depend on a particular bond strength so there may be some limit (e.g. a nitrogen-nitrogen triple bond for main group elements or a sextuple bond for metal-metal interactions). Yet, I think people would talk about this in the context of a limit to bond dissociation energies, not a maximum energy level.
The final  possibility would come in the context of electronic levels. To bind an electron, there is an energy required, which is why the standard for orbital energies is negative. Above that point, and removing an electron from an electronic state is favorable.
Yet I can't count that as a maximum 'energy level' because of course we can perform spectroscopy for higher-level electronic excited states.


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From where energy for electron promotion comes during hybridisation
I know that the short answer is that "from bond formation" but as I think electron promotion occurs before formation of bonds between atom. So it seems like a paradox.

Hybridisation occurs in our heads.

Hybridation has an advantage : it is visual and it explains perfectly the geometry of the molecules. But hybridation is not necessary to explain these structures. You can explain the geometry of the molecules without hybridation, although it is more difficult to calculate.

This is exactly the other way around. You can determine the structure, then you can propose hybridisation of orbitals.

You have to use VSEPR to determine the geometry and only then get the hybridisation; so hybridisation is really just an a posteriori justification.

I think you can find plenty of discussion of the pros and cons of hyrbidization elsewhere on this site, so I'm going keep it fairly short.
First, the premise of your question shows confusion between the paper exercise of first creating hybrid orbitals and then filling them with bonding electrons and the physical event of molecule formation, which doesn't happen in those steps. Molecules can form in all sorts of ways, and the exercise of determining electronic structure by assembling the molecule from its constituent atoms is not related to its actual formation. So the only relevant comparison is between the two representations of the final molecule - one with each bond and lone pair represented by a localized hybrid orbital and one with delocalized molecular orbitals comprised of specific $s$ and $p$ (and $d$ and $f$ as necessary) atomic orbitals. These are often called "canonical" molecular orbitals.
A key point is that electrons do not actually reside in specific orbitals. Since electrons are indisinguishable and don't have specific locations, the wavefunction of a molecule involves all of the electrons collectively occupying all of the occupied orbitals. For practical purposes, we like to think of each electron in a specific orbital because it's simpler.
With that in mind, it may be easier to understand that as long as the bonds involved in the hybridization are all identical (eg the four C-H bonds in methane or the three C-O bonds in $\ce{CO3^2-}$), the electron density described by canonical MOs is exactly identical to that described by the hybrid $sp^3$ or $sp^2$ orbital bonds. On paper, there's a difference in the space covered by each electron, but since we said before that the electrons are indistinguishable from one another, that difference isn't physically real.
Where the simple hybridization model is inaccurate is when it is applied to different bonds/lone pairs. For example, describing the O of $\ce{H2O}$ as $sp^3$ hybridized is not mathematically rigorous. The degree of $s$ and $p$ contribution (ie hybridization) to the lone pairs is different than in the O-H bonds. This creates confusion when people try to make quantitative conclusions about things like bond angles based on this qualitative representation. But working out the actual non-integer hybridization values detracts from the powers of the hybridization approach, which is its intuitive qualitative simplicity.
So the take home message is that if you have a hybridization involving identical bonds/lone pairs, it is quantitatively equivalent to canonical MOs for the purposes of calculating electron densities. If the bonds/lone pairs are not identical, integer hybridization is not quantitatively equivalent, but with more work one can calculate the appropriate non-integer hybridizations.

The important thing to note, really, is that hybridisation doesn’t really happen - what you’re learning as “Valence bond theory” is really just a dumbed-down and almost completely wrong explanation to chemical bonding that is outdated by at least 60 years.
In "proper" valence-bond theory, the many-electron wavefunction is described as a linear combination of multiple "resonance structures", with the energy variationally optimised. It can be shown that a full consideration of all possible such "structures" would be equivalent to a full configuration-interaction treatment of molecular orbital theory; however, if the Heitler-London wavefunction is the only one considered, VBT is "too covalent" while MOT is "too ionic".
In Pauling's day, it was thought that the lowest excited state of Carbon lay relatively close to the ground state, such that the promotion of one electron from the 2s to 2p orbital would be readily achievable. As that has been shown to be untrue, his model of hybridisation is hence invalid. Just keep in mind that what you're learning is both outdated and mostly wrong.
I’d suggest you look up molecular orbital theory if you’re interested in knowing more about Chemical bonding - otherwise, the book "A Chemist's Introduction to Valence Bond Theory" was pretty helpful for me to understand what modern VBT looks like.

It is important to note that valence bond theory is equal to molecular orbital theory when it is not dumbed down.

I can't just leave it because of the syllabus but can you please help me to get out of this paradoxical thing.

@Martin-マーチンthat's definitely true - I'll probably edit my answer to reflect that as soon as I have time.


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Can you create different colors of rust on iron?
I am looking to see if anyone out there can tell me if there is a way to create different colors of rust on iron. I'm specifically looking to see if I can create blue rust without heating the metal and blueing it that way. I can achieve the regular rust colors but I am looking to see if I can go beyond that. If you can help I would love the input. Thanks

Rust is a generic term for iron oxides and hydroxides, and depending on the oxidation state of the iron and the thickness of the layer, it can have many colors. Impurities in the iron also affect the color.
By controlling the amount of air and water and by changing the composition of impure iron or steel, colors from red though blue can be achieved, though the blue may be transient if it's due to interference in a thin surface film.
That said, I know of no "recipe" to provide a given color or shade; you'd need to determine that by experiment. There are chemical treatments to change the color of iron, such as using selenium dioxide to blue the metal, but the result is not rust.

As noted "rust" is a few different compounds.  Iron makes 3 oxides FeO ( black) Fe3 O4 ( black ) , Fe2 O3 ( red ). Rust is mostly hydrated Fe3O4 ( Brown ). Our lab did XRD on about a thousand samples for corrosion investigations; The tech matched patterns with the FINK Index as well as he could but there are various options : easiest to call rust "hydrated" Fe3O4.

The color of rust cannot be changed. The only thing you can do is mixing with a high amount of a blue pigment, if you want to obtain a blue color with rust.

Iron oxides and hydroxides can have different colors depending on oxidation state and thickness. "Rust" is generic and denotes multiple forms.


text:
Finding (∂U/∂V )T = n^2a/V^2 for a van der Waals gas
$\displaystyle dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV \to dU = C_v dT - C_v\left(\frac{\partial T}{\partial V}\right)_U dV$
Not sure what I do here. Thank you in advance.

For clarity  I have converted your equation to MathJax, please check that it is correct.

The title and the question are not related to one another

I guess they are related via an answer.

The more general equation you are looking for is called the thermodynamic equation of state. You can start from the differential form of U according to the 1st + 2nd laws of thermodynamics for pV work only:
$$dU = TdS-pdV$$
Taking the derivative wrt V at constant T
$$\left(\frac{\partial U}{\partial V}\right)_T= T \left(\frac{\partial S}{\partial V}\right)_T-p$$
At this point you want to use the Maxwell relation
$$\left(\frac{\partial p}{\partial T}\right)_V= \left(\frac{\partial S}{\partial V}\right)_T$$
(which follows from the Helmholtz energy being a state function) so that
$$\left(\frac{\partial U}{\partial V}\right)_T= T \left(\frac{\partial p}{\partial T}\right)_V-p \tag{tes}$$
This last expression is the thermodynamic equation of state.
The final step is to evaluate $\left(\frac{\partial p}{\partial T}\right)_V$ for the van der Waals equation of state.

The equations for du you listed in the question description can be derived by applying the definition of constant volume heat capacity and using a cyclic relation. Derivation of constant volume heat capacity can be found here : https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node18.html.
Refer to cyclic relationships which fall under the topic of Maxwell relations to understand how to convert the second term to the appropriate form.


text:
Acidic strength comparison between sulphuric acid and nitric acid
Why is $\ce{H_2SO_4}$ more acidic than $\ce{HNO_3}$. Shouldn't it be the other way around because $\ce{N}$ is more electronegative than $\ce{S}$?
My reason: $\ce{O}$ can take electrons more easily from $\ce{S}$ than from $\ce{N}$. Hence, it will be more difficult to remove $\ce{H}$ from $\ce{H2SO4}$ than from $\ce{HNO3}$. Hence, $\ce{HNO3}$ should be more acidic.

Note that by the Chemistry SE site policy, it is preferred not using MathJax formatting in titles, for indexing/searching reason.

N is not more electronegative than O !!!

I think the OP means S, as argued in the second paragraph. Should be a typo.

What about $\ce{H2SO3 and HNO2}$? If you are just arguing S vs N, they would have the same strength as $\ce{H2SO4 and HNO3}$, respectively.

@Karsten Theis By analysis of $\ce{^{17}O}$ Raman IR spectroscopy, there has been reportedly detected presence of dissolved SO2 and HSO4-, but no H2SO3.  So the equilibrium is like  $\ce{SO2 + H2O <=> HSO3- + H+}$.

The important thing about acidity is the stability of the conjugate base, which is enhanced by resonance effect and inductive effect. The more stable the conjugate base, the easier deprotonation becomes, and thus the stronger the acid.
In a $\ce{NO3^-}$ ion, the negative charge can delocalise among three O atoms, but in a $\ce{HSO4^-}$ ion, in addition to delocalisation among three O atoms, there is an additional OH group, which further stabilises the anion by inductive effect. So $\ce{HSO4^-}$ is the more stable conjugate base, and $\ce{H2SO4}$ is a stronger acid than $\ce{HNO3}$.

According to my argument, since O can easily take electrons from S, O will have a partially -ve charge. Due to this, H+ will be difficult to release and hence $\ce{H2SO4}$ should be less acidic. But I do understand the first reason so that's fine.

No that's not correct. One should not consider the attraction between the partial -ve charge on O and H+, but instead should consider the O-H bond strength. The more -ve O becomes, the less electrons it needs from H, and the weaker the O-H bond, so H actually can escape more easily.

But in that case, H- will be released not H+...

Then maybe I'm unable to rationalize your original idea. Answer edited to remove that part.

Here's my method for looking at it:

For determining acidity, we need to look which one of $\ce{H2SO4}$ or $\ce{HNO3}$ form a more stable anion after the $H+$ is extracted.

Lets condsider $\ce{H2SO4}$ first-

• There are two sites to extract a proton.
• After a proton is removed, the $O-$ is in resonance with 2 oxygen atoms attached to the sulfur atom.
Now, in $\ce{HNO3}$ -

• There is one site to extract proton
• After removal of $H+$ , there are two resonance structures but the eletrons of the oxygen that lost a proton have delocalised only once and over a shorter region as compared to $\ce{H2SO4}$
Generally, a compound with more resonating structures is more stable due to more charge distribution. Amongst other factors like electronegativity, resonance usually dominates.
Fun fact:
$\ce{H2SO4}$ is stronger acid than $\ce{HNO3}$ , this is shown by following reaction:
$\ce{H2SO4} + \ce{HNO3} ---> \ce{NO2+} + \ce{HSO4-} + \ce{H2O}$
This reaction is used in the nitration of benzene to form the electrophile.

I do believe that H2SO4 is stronger than HNO3 because also keep in mind that the  H-A bond here in H2SO4 acid is weaker. Why? Because it is one period downwards of N meaning that it has more atomic radius causing it to dissociate more than HNO3. But not much more because the difference is just one period.
Here is a link if you want to check:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwiqiuCN09_4AhUH7xoKHbOSCkYQFnoECAMQAw&url=https%3A%2F%2Fgroups.chem.ubc.ca%2Fcourseware%2FpH%2Fsection10%2Findex.html&usg=AOvVaw15gIWrcUwXxHgTZ7jzknQn


text:
Are the electrons in an atom always 'locked' with their proton?
Suppose you have a single atom, say carbon, carbon has six electrons, and six protons. Now is there anything associating the six protons locking the six electrons other than coulombic forces?
When we ionize a compound, we remove the electron from the outermost shell. Now I find this weird because, from my understanding of the Schrodinger equation, it gives orbitals which gives the probability density function of orbital as a solution. In this context, how would you explain ionization? as in supposing you ionize wouldn't the probability density functions change. What exactly are the considerations we use to describe the phenomena of ionization?
Further, the most interesting part is that the wave orbitals are distinguishable (to some extent) but not the electrons ( particle itself) why is this?

All electrons are identical. A proton would not "know" which one used to belong to it.

How do you know anything at all is a true fact?

An analogy would be a box of candy possessed by a child. If you take a candy from the box and replace it with an identical one, the child would not notice. The child can be the atom, the candy box it's orbitals and the candies themselves the electrons.

@DDD4C4U nope...

Unless some kind of entanglement :(( OP ignore this, it is a kind of joke.

[OP] Would this last electron, in any way, be related to the original atom and would the original atom 'want' to regain it's lost electron?

No. Individual electrons are indistinguishable. The experimental evidence goes along these lines (source: https://medium.com/physics-as-a-foreign-language/how-do-we-know-that-all-electrons-are-identical-part-2-7dad7d980dd1):

a pair of distinguishable particles which can be at 2 different locations has 4 possible states they could be in. Whereas a pair of fermions has only 1 possible state, and a pair of bosons has 3 possible states. This leads to very different statistical behavior for fermions and bosons, and explains why a lot of the properties of the 2 kinds of particles are so different.

All kinds of properties of our universe could not be explained with the current theories if electrons were distinguishable.

[OP]Further where do atomic orbitals arise from? are they a region in space having higher chance of electrons, created from having a collection of protons at a point? or are their existence independent of protons and dependent only on the electrons?

Orbitals are functions approximately describing the state of bound electrons. This description depends on where the protons are and on how many electrons there are. So they are dependent on both.

[OP, in the comments] there should be some sort of rational explanation as to why something is true.

The way science works, there is a theory or a model that is useful because it describes what has already been observed, and it makes predictions that are experimentally verified. In this sense, the state "individual electrons are indistinguishable" is true and has experimental backing. The question whether the explanations are rational or crazy is more a philosophical one. A science fiction writer probably would not have come up with quantum mechanics as a particularly reasonable theory, but it does a good job describing the reality we live in. So I would reformulate the statement above to say "there should be some experimental evidence for a model or theory used in science".

I asked this question way long back and with a lot more reading, I think I've come up with more specific points to answer this question.

the most interesting part is that the wave orbitals are distinguishable (to some extent) but not the electrons ( particle itself) why is this?

Every electron in a given atom is distinguishable in the sense that you will have a set of three quantum numbers which you can use to distinguish it from the other electrons inside that atom.

When we ionize a compound, we remove the electron from the outermost shell. Now I find this weird because, from my understanding of the Schrodinger equation, it gives orbitals which gives the probability density function of orbital as a solution. In this context, how would you explain ionization? as in supposing you ionize wouldn't the probability density functions change. What exactly are the considerations we use to describe the phenomena of ionization?

If the electron is given energy, it would change it quantum numbers to go up to a state of higher energy. I say 'state' instead of 'n' because there are certain cases such as the splitting of d-orbitals in certain compounds which causes the orbitals to become non degenerate. This stack post has much more detailed answers on this particular aspect.
Considering all the energy possible interactions in an atom, we can say that it must exist in some finite region of space using schrodinger equation. A contour of constant probability containing region which has 90% probability of finding electron are the shapes used to denote the orbitals.

The best explanation I found for this is from physics forums (here) by Jano L

So each particle is identical basically? There's no way to tell them apart?

Identical does not mean "no way to tell them apart"; that would be the meaning of "indistinguishable". Identical means the objects have the same intrinsic properties like mass, charge etc. but not location. Two xerox copies of a master blueprint are identical, yet they can be put in different places and you can tell them apart just by holding them apart each with different hand or placing them in two different rooms (and assigning different labels to the rooms). Different label means you can tell them apart and distinguish them, but they remain identical (same size, mass, color ...).

Another interesting thing I found is that, I am not the only person who found this crazy! Indeed one physicist John Wheeler came up with the idea that the explanation for all electrons being identical (here) and this PBS space time lecture

Also Refer Clayden page -84, page-86, has nice discussion on ionization and orbitals


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Decomposition reaction of tin(II) nitrate
What are the products formed when $\ce{Sn(NO3)2}$ is subjected to high temperatures?
I searched over the internet but I didn't get any satisfactory answer.

I'd expect SnO2.

Pure anhydrous $Sn(NO_3)_2$ does not exist. Only $Sn(NO_3)_2·20H_2O$ exists, and it is a liquid at ordinary temperature (melting point : -20°C).

No wonder you cannot find any literature on $\ce{Sn(NO3)2}$, because I don't think it exists in solid form. This is supported by the fact that I even can't find its CAS number online. This book (Ref.1) and relevant paper (Ref.2) support my suggestion, both of which state that:

Attempts to prepare a covalently bound tin(II) nitrate, by the reaction of tin(IV) tetranitrate with anhydrous nitric oxide, have produced only a white solid solid of formula $\ce{SnN2O6}$, which gives a tin(IV) Mössbauer resonance ($\delta = \pu{0.29 mm\:s-1}$, $\Delta = \pu{0.96 mm\:s-1}$)...

Nonetheless, a study on thermal decomposition of metal nitrates (Ref.3) states that:

Due to a back-donation of electronic cloud from the nitrate to an unfilled $\mathrm{d}$-orbital of transition and noble metals, their nitrates generally exhibited lower decomposition temperatures ($T_d \lt \pu{700 K}$) than those of the base metals ($\gt \pu{850 K}$).

They suggest those metal nitrates with lower decomposition temperatures decompose to corresponding oxides together with $\ce{NO2}$ and $\ce{O2}$. Thus, it is safe to assume that if $\ce{Sn(NO3)2}$ exists, it would decompose as following reaction suggest:
$$\ce{2 Sn(NO3)2 (s) + heat -> 2 SnO (s) + 4 NO2 (g) + O2 (g)}$$
However, keep in mind that this is pure speculation.
References:

R. Greatrex, “Chapter 8: Mössbauer Spectroscopy,” In Specialist Periodical Reports: Spectroscopic Properties of Inorganic and Organometallic Compounds, Volume 6; N. N. Greenwood, Ed.; The Chemical Society Burlington House: London, United Kingdom, 1973, pp. 494-622.
P. G. Harrison, M. I. Khalil, N.Logan, “Concerning anhydrous tin(II) nitrate,” Inorganic and Nuclear Chemistry Letters 1972, 8(6), 551-553 (https://doi.org/10.1016/0020-1650(72)80139-9).
Shanmugam Yuvaraj, Lin Fan-Yuan, Chang Tsong-Huei, Yeh Chuin-Tih, “Thermal Decomposition of Metal Nitrates in Air and Hydrogen Environments,” J. Phys. Chem. B 2003, 107(4), 1044-1047 (https://doi.org/10.1021/jp026961c).

$\ce{Sn(NO3)2}$ definitely does exist but only in solution and is ill-defined.. Moreover, it is written in this paper: "The dilute solution of tin(II) nitrate is reasonably stable [...] Attempts to evaporate to crystallization result in decomposition, which may occasionally be violent. The decomposition produces tin(IV) oxide, and various nitrogen compounds, principally hydroxylamine and nitrous oxide."


text:
In which order should I learn Thermodynamic
I study physical chemistry on my own and I have to decide in which order I learn the following topics
These are my topics:
Statistical Thermodynamics (especially entropy, partition function, Boltzmann factor..)/
Maxwells Relations/
Thermodynamic Potentials/
Chemical potential/
Colligative properties/
Kinetics/
Electrochemistry/
What is your opinion, your preferred order

Opinion-based questions are not a good fit for this site. Maybe you want to ask what the typical textbook order is, or some other version of the question that is fact-based rather than opinion-based.

It depends on the textbook you have. Most textbooks are written in roughly the order they are intended to be read.

I´m sure one could have different opinions on this, but I doubt that it would make much sense to stray from the tried path that basically all textbooks and profs follow.

I think it depends on what your 'end goal' , that is what you expect yourself to do by the end of studying it, based on that you can drop off of topics and save a lot of time.

Take any common PC book, and follow the order given therein.
It usually starts with ideal gas equation, enthalpy, entropy, Gibbs, chemical potential, i.e. plain thermodynamics. (that lecture was called PC1 at my alma mater)
On top of that, you can add kinetics and statistics (PC4a,b) and electrochemistry(PC5) at will, those do not depend on each other much. The basics of kinetics and electrochemistry we learned earlier however, I think in the introductory lecture for inorganic chemistry.
I´m not saying one couldn´t start with any of the latter subjects instead of plain thermodynamics, but have never heard of that being done and no idea how it would work. Would be a whole new approach to (self)teaching physical chemistry. ;)
Inbetween, you should however allow yourself a good dose of quantum mechanics (PC2&3), because that greatly improves understanding of the concepts and esp. their limitations. Some books and schools also start with QM (PC[zero]), I guess for that exact reason.
Again, I´m not saying this couldn´t be rearranged, but the above is the way most books and lectures do it. I guess that with statistics etc., you would have to stop the subject all the time to introduce another concept from basic thermodynamics, and also all the rest of chemistry remains very vague and descriptive if you don´t have a good understanding of gas laws, phase transitions, energy and temperature.

You see we had a very thorough 3 yr indoctrination in physical chemistry. ;) Followed by two more lectures on macromolecular physical chemistrys, where everything looks slightly new and different. ;)


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What is the molecularity of a reaction if it has a fractional order as per rate equation?
The rate equation of any reaction is found by using its slowest step.
My book says that molecularity of the slowest step of a reaction is equal to the order of the same reaction.
However, there are some reactions with fractional order but molecularity cannot be fractional.
Please tell whether my book is wrong or not.

"molecularity of the slowest step of a reaction is equal to the order of the same reaction" - it's generally not true.

Fractional order happens in reactions like  $$\ce{H_2  +  Br_2  <=> 2 HBr}$$ for which Bodenstein and Lind found in 1906 their results in agreement with the empirical equation :$$\frac{d[HBr]}{dt} = \frac{k[H_2][Br_2]^{1/2}}{1 + k'[HBr]/[Br_2]}$$ Thirteen years later, this kinetics law was explained by Christiansen, Herzfeld and Polanyi in terms of the following chain mechanism $$\ce{Br_2 -> 2 Br}$$$$\ce{Br + H_2 -> HBr + H}$$$$\ce{H + Br_2 -> HBr + Br}$$$$\ce{H + HBr -> H_2 + Br}$$$$\ce{2 Br->Br_2}$$Each of these equations has its own rate constant. If you want to know how to use them to establish the results from Bodenstein and Lind, it is a long job, that I will not develop here. Please read the original publications.
J.A. Christiansen, Kgl. Danske Videnskab, Selskab., Mat.-fys. Med., 1. 14 (1919) ;
K.F. Herzfeld, Ann. Physik, 59, 635 (1919) ; M. Polanyi, Z. Elektrochem., 26, 50 (1920)

Please tell whether my book is wrong or right.


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Why are colloidal dispersions stable?
Smaller particles in a colloidal dispersion are less stable than larger ones due to their higher surface to volume ratio.
Therefore Ostwald ripening occurs, a process which promotes growth of larger particles at the expense of smaller ones (alternatively aggregation of smaller particles can occur).
In principle, this process goes on until one single particle is obtained.
Then, why are colloidal dispersions with lots of small particles stable? Or rather metastable? What makes the processes promoting further growth stop (ie dissolution of smaller particles in Ostwald ripening)? And do these processes really stop or just slow down tremendously?

In colloidal dispersions all the colloidal particles possess the same charge and therefore this prevents their coagulation or aggregation to form larger particles that consequently settle down because of gravity.

Your logic is strung up in the wrong way. ;) If a colloidal dispersion is stable, then the smaller particles are not measurably less stable than the biggest ones that occur in the same dispersion. Also the reason(ing) for Ostwald ripening is thermodynamic. I does not necessarily occur on your timescale, because the kinetics can just be too slow.

Consider Brownian motion, as well as dispersants.

@Dibster7978 Stability for colloidal dispersions refer to kinetic stability. If they have the same charge how can they coagulate at first place?


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Why sigma bond has C2 symmetry not C∞?
I have seen some books indicated sigma bond has C2 symmetry. I understand pi and delta bond change their signs about C2 and C4 rotation about bond axis, but I don't  understand why sigma bond has C2 symmetry not C∞ about its bond axis.
The example of the sigma bond came from two 1s orbitals.

I have seen books about two-headed people living beyond the ocean. Somehow, this does not bother me one bit. A sigma bond is $C_\infty$-symmetric all right.

Possibly they meant C2 around an axis perpendicular to the bond, especially since 1s orbital bonding is H2, which is symmetric.

Technically speaking a $\sigma$ bond fits in the $C_{2}$ point group since $C_{2}\subset C_{\infty}$.

@Andrew the book indicates the rotation axis with the bond axis.


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Does the number of bonding MOs always equal the number of antibonding MOs
In (I think all) of the molecular orbital (MO) diagrams I have seen, I have observed that the number of antibonding molecular orbitals is the same as the number of bonding molecular orbitals.
Here are some for example:

I know that the number of atomic orbitals equals the number of molecular orbitals, but I have not heard of an explicit rule stating that the number of bonding molecular orbitals is equal to the number of antibonding molecular orbitals. Is this always true? If not, can someone please give a counter example?

Cyclopentadienyl is a counter example, as are other odd-numbered conjugated ring anions.

Have you looked at the pi orbitals of aromatic ions?

Yes, they just escaped from my memory when I was thinking about this question.


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HCP unit cell c-axis
Today I was asked to show that the perpendicular height, $H,$ of a regular tetrahedron is related to the length of a side, $L,$ of the tetrahedron, by $H = \sqrt{(2/3)} \cdot L$. I did this no problem.
I was then asked:
Explain how this is relevant in determining a relationship between the unit cell parameters, $a$ and $c$ of a hexagonal-close packed crystal structure: $c = 1.633 · a$.
Could anyone possibly help me out here?

Well, if you know what an HCP structure is, then you could have guessed that the said ratio of 1.633 was not put there on someone's whim.

No I realise it’s twice of a - I think I know what’s going on but I don’t know what they’re asking for here

Imagine the whole construction, with spheres and everything. That 1.633 is not some random decimal number. It must have been derived geometrically, you know, with square roots and stuff.

A picture is worth a thousand words (see the two tetrahedra heights making up the vertical c-axis?):

The HCP packing has layers of closest-packed spheres in an ABAB... pattern. The distance between layers is one height of a tetrahedron, and you need two layers to get a repeating structure along c, i.e. the unit cell parameter.
(picture found at https://chemistry.stackexchange.com/a/111921)


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What is the difference between quantities reported as optical rotation and circular birefringence?
Optical activity gets complicated quickly and the following question is intended to avoid threshold confusion. The linked lecture note* seems very accessible to me and Table 10.6 in particular is quite clear. On page 2 the author writes, "Optical rotation results from circular birefringence..." and then relates that idea to circular dichroism.

My question is: How does the angle we measure in a student polarimeter as optical rotation relate to the angle reported as circular birefringence? Do they merely differ by a constant or is there more to it than that?

What I think I get is that circular birefringence measures the ellipticity of light, which involves a difference in the transmission of L and R circularly polarized components of an initially plane-polarized beam. Optical rotation simply measures the change in plane of polarization of light that has passed through a sample. My guess is that the two are different, otherwise one could just translate to degrees of rotation...? But if they are both measures of optical activity it would be also be surprising if they are related in some non-linear way.
*The note is from a lecture at Washington Univ. at St. Louis and I will try to get a better cite.

This earlier answer may be helpful.

@ron: Thanks for pointing this out. I do get the connection between ORD and CD via the Kramers-Kronig relations. I think my question is different (and easier) than that one.

+1 only for "it quickly gets complicated" Yep. That's for sure. Most times I attempt to get a deeper understanding of optics, I end up nose down into a QM book, which invariably ends up with me getting into a deeper state of confusion.

It gets very complicated, there are many papers and books on the topic and this little table, from Jensen et al. 1, gives a glimpse of the anisotropic effects:

An extremely short answer is this: when linearly polarized light is incident on an optically active medium, it can be thought of as consisting of equal components of right circularly polarized (RCP) and left circularly polarized (LCP) light. The indices of refraction for these are unequal, so RCP and LCP propagate at different speeds through the medium. This is circular birefringence. Upon exiting, it is linearly polarized light that exits, not the RCP and LCP fictive components. But the differential transit speed results in a rotation of the plane of polarization of the exiting linearly polarized light. This is optical activity, so optical activity is circular birefringence. The longer the optical path, the more the rotation and so on.
This is a truly vast topic, so I recommend not jumping directly into the deep end. Shurcliff's wonderful old book 2, maybe still findable, is extremely readable and a delight: the Poincare sphere is genius from a true genius! The book by Kliger et al. 3 is a good starting point and not insanely hard to understand. The paper by Jensen et al. 1 is rewarding, but definitely not easy reading.
There are many more papers, books, animations, and so on from which to learn. But be aware that notations are all over the place and reconciling equations and terminology can be very frustrating.
References:

H.P. Jensen, J.A. Schellman, T. Troxell, “Modulation Techniques in Polarization Spectroscopy”, Applied Spectroscopy 32 (1978) 192-200 (doi.org/10.1366/000370278774331567).

W.A. Shurcliff, Polarized Light, Harvard University Press,
Cambridge, MA, 1962.  For example on archive.org.

Kliger, D. S.; Lewis, J. W.; Randall, C. E. Polarized Light in Optics
and Spectroscopy, 1st ed.; Academic Press: Boston, 1990.  To borrow for  example from archive.org.

EDIT: In response to the OP's comment, I am adding two of my tables.

Circular birefringence (optical activity)
$$
\begin{align}
[φ] &≡ \text{molar optical activity}~(\pu{deg/cm M}) \\
\mathrm{CB} &≡ 2π(n_- - n_+)l/λ ≡ 2[φ]cl × π/180
\end{align}
$$

Circular dichroism
$$
\begin{align}
[θ] &≡ \text{molar ellipticity}~(\pu{deg/cm M}) \\
\mathrm{CD} &≡ \ln 10 (ε_- - ε_+)cl/2 ≡ 2[θ]cl × π/180 \\
g &≡ \text{dissymetry ratio} ≡ Δε/ε ≡ ΔA/A \\
ε &≡ \text{mean molar absorptivity} = (ε_- + ε_+)/2
\end{align}
$$

Complex angle of rotation, $χ$
$$
\begin{align}
χ &≡ χ' + iχ'' ≡ \mathrm{CB}/2 + i\mathrm{CD}/2 = φ + iθ \\
Δε &= 4π[θ]/180\ln 10 ≅ 0.03032[θ]
\end{align}
$$

Definitions of symbols and expressions
$$
\begin{align}
\mathrm{RCP} &≡ \text{right circularly polarized light (‘+’ subscript)} \\
\mathrm{LCP} &≡ \text{left circularly polarized light (‘−’ subscript)} \\
ε_± &≡ \text{molar absorptivities for RCP and LCP}~(\pu{L/mol cm}) \\
ε &≡ (ε_+ + ε_-)/2 ≡ \text{mean molar absorptivity}~(\pu{L/mol cm}) \\
A_± &≡ \text{absorbances for RCP and LCP} \\
A &≡ (A_+ + A_-)/2 ≡ \text{mean absorbance (true)} \\
n_± &≡ \text{(real) indices of refraction for RCP and LCP} \\
c &≡ \text{concentration}~(\pu{M}) \\
l &≡ \text{path length}~(\pu{cm}) \\
λ &≡ \text{wavelength}~(\pu{cm}) = 10^{-7} × \text{wavelength}~(\pu{nm}) \\
n &≡ (n_+ + n_-)/2 ≡ \text{mean index of refraction} \\
A_l &≡ \ln 10 (A_+ + A_-)/2 ≡ \text{mean absorbance} \\
η &≡ 2πnl/λ ≡ \text{phase (radians)} \\
[θ] &≡ \text{molar ellipticity}~(\pu{degree/cm M}) \\
[φ] &≡ \text{molar optical activity}~(\pu{degree/cm M}) \\
\mathrm{CD} &≡ \ln 10 (ε_- - ε_+)cl/2 ≡ 2[θ]cl(π/180) = 2θ \\
\mathrm{CB} &≡ 2π(n_- - n_+)l/λ ≡ 2[φ]cl(π/180) = 2φ \\
a &≡ \mathrm{CD}^2 - \mathrm{CB}^2 + \mathrm{LD}^2 - \mathrm{LB}^2 + \mathrm{LD'}^2 - \mathrm{LB'}^2  \\
b &≡ 2\mathrm{CD} × \mathrm{CB} + 2\mathrm{LD} × \mathrm{LB} + 2\mathrm{LD'} × \mathrm{LB'} \\B &≡ \left[(a^2 + b^2)^{0.5} + a\right]^{0.5}/2\sqrt{2} \\
C &≡ \left[(a^2 + b^2)^{0.5} - a\right]^{0.5}/2\sqrt{2} \\
Q &≡ B + iC
\end{align}
$$

The expressions for a, b, B, C, and Q are used in the general anisotropic medium Jones optical calculus and Mueller optical calculus matrices. These two matrices are denoted by J(GAM) and M(GAM), respectively, and are given in the paper by Jensen et al.

Thank you for the larger font! It appears I am asking about the relation between CB and CD. I will do more reading and return to this. I think your answer is more relevant than the one @ron mentioned (although that is a nice answer). Thank you again!

@EdV No prob, but I'm afraid I'm going to leave Table II as is (at least for now): it gets too wide when typed in using standard font.

Ed V's answer above--which I am accepting--is correct and also allowed me to reach the conclusion on my own. I am just posting this because my confusion was not too hard to resolve once I saw the quantities involved and someone else may have the same confusion.

The short of it is: student polarimeters measure circular birefringence (CB), usually at a specific frequency, concentration, and sample length. Optical rotation is an aspect of CB.

The Wikipedia article on optical rotation conveys the following: in student polarimetry we speak of "plane polarized light" but this is best thought of as a superposition of left- and right-circularly polarized light. In an optically active solution the difference in transmission of the L and R components is ascribed to a difference in refractive indices $\Delta n$ which causes the measured rotation,
$$\Delta \theta = \frac{length\cdot \pi~ (n_L-n_R)}{\lambda}. $$
An optically active material may be described as circularly birefringent if it discriminates between L and R components of circularly polarized light according to the above relation, and so the rotation of plane polarized light through an angle $\theta$ is an example of circular birefringence. Here is a definition-oriented link.
The optical rotary power is closely connected to $\theta$ and varies (as does $\theta$) with the frequency of the light. The variance of $\theta$ with frequency is called optical rotary dispersion (ORD). Its cousin, circular dichroism (CD), is due to differential absorbance ($\Delta A$) of L and R light and also varies with wavelength. The remarkable connection of ORD and CD via the Kramers-Kronig relations (not so easy to demonstrate in practice, I guess) is described in ron's answer (linked above).


text:
What is the activation energy of a Daniell cell?
I'm having trouble finding a posted literature value for the activation energy for the Daniell cell redox reaction:
$$\ce{Cu^2+ + Zn -> Cu + Zn^2+}$$
Where might I be able to find this information?

Electron transfer is extremely fast, that finding its rate constant is not a trivial exercise.

The equation is the summary of 2 processes ongoing on 2 different places, while activation energy applies on a single elementary reaction step.

An activation energy has a sense in a homogeneous mixture. This is not the case in a Daniell cell

@Maurice, "An activation energy has a sense in a homogeneous mixture. This is not the case in a Daniell cell" is a misconception. Electrode kinetics is very vast field and people do report activation energy, see e.g., https://pubs.acs.org/doi/abs/10.1021/jp9721309


text:
Controlling Current Density in Electrowinning Cell
What are the parameters which govern the current density in an electrowinning cell? What equations relate Current Density to the applied potential difference?
Is it simply just Ohm’s Law J = σE? Where does redox at the electrodes come into play, if this is the case?

The above version of Ohm law is valid in the free solution only, and for large density range only approximately. The overall topic is complex and requires some knowledge of electrochemistry.

Do you have any recommended readings for Electrochemistry?

The system is complicated by many nonlinear processes.
As the topic is electrode focused, the primary controlled parameter should be the electrode surface current density $\sigma$, especially on the depositing electrode.
the total coltage applied on the electrolytic cell is devided on several potential differences:
$$U_\mathrm{applied} = U_{td}(\sigma) + U_\mathrm{kin}(\sigma) + U_\mathrm{diff}(\sigma) + U_\mathrm{cond}(\sigma)$$
where the particular terms are

electrode potential difference given by reaction thermodynamics
electrode potential difference given by reaction kinetics
electrode potential difference given by near electrode diffusion gradient.
free solution potential difference controlling ion migration.

For practical reasons, I advice using empiric approach :

either constant current power source, if applicable
either manual voltage control, using visual ampermeter feedback.

For further reading, I recommend:

For starters, Electrochemistry on hyperphysics
Wikipedia articles,  following internal and external links, also searching for mentioned topics.
for more advanced matters, textbooks for Physical chemistry.


text:
Reactivity in electrophillic substitution
When comparing maximum reactivity of a compound in electrophillic addition, what factors are kept in mind while comparing propene and trans-2-butene?
Why is trans-2-butene more reactive in this case than propene?

For electrophilic substitution, naturally we look for a greater electron density over the molecule. In trans-2-butene, the are two methyl groups which increase electron density over the pi bond whereas in propene there is just one methly group. In case you don't already know, alkyl groups ($R-$), increase electron density over the bonded atom. This is generally known as inductive effect.


text:
Near-IR spectroscopy measures the broad overtone and combination bands of some of the fundamental vibrations (only the higher frequency modes)
I am currently studying the textbook Infrared and Raman Spectroscopy by Larkin. In the first chapter of the textbook, the author opens (probably prematurely) with the following statement:

Near-IR spectroscopy measures the broad overtone and combination bands of some of the fundamental vibrations (only the higher frequency modes) and is an excellent technique for rapid, accurate quantitation.

This will likely be covered in later chapters, but the statement piqued my curiosity: For someone who has not yet encountered concepts such as "broad overtone", "combination bands", and "higher frequency modes", what does this all mean? Essentially, I am seeking a brief, but clearer and more elaborate, explanation of what the author has described here.
I would greatly appreciate it if someone who has knowledge of this area would please take the time to clarify.

Indeed that sentence should be taken as merely anticipating what the book will treat in following chapters.
Nevertheless, to satisfy your impelling curiosity the statement trace a combination of facts that leads to the actual spectrum of a given molecule.
Overtones IR absorption bands are those resulting from breaking of selection rules. Instead of having just transitions between adjacent vibrational levels, basically the 1<-- 0 one, transitions with a delta nu of 2 are observed as well. In other words the vibrations aren't those of an harmonic oscillator but those described by a Morse's potential.
Combination modes arise from the combination of normal modes, ie they happen when different vibrations are simultaneously excited. Normal modes which are unseen can contribute to these combination modes, as the coupling alter the symmetry.
High frequency: we are speaking about NIR, which is the high frequency range for molecular motions, and it is where vibrational overtones typically occur. The lower frequency part of the spectrum is dominated by the fundamental modes, and except for its fingerprints part and fine details, it has a strong bond related character rather than a molecular one.
If this does not help but instead brings out new terms or doubts, I would suggest that you treat the introductory part of the book for what it is, namely an introduction.
Alternatively you can refer to this link which provides a condensed explanation and includes schemes, further links, and definitions:
http://bit.ly/31Mc6lT

Any resonant frequency above the fundamental frequency is referred to as an overtone. In the IR spectrum, overtone bands are multiples of the fundamental absorption frequency.   https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones

In physics and chemistry, a selection rule, or transition rule, formally constrains the possible transitions of a system from one quantum state to another. https://en.wikipedia.org/wiki/Selection_rule

What is the "lower frequency part of the spectrum" that you are referring to? Everything at wavelengths above near-infrared?

Yes somehow the authors reiterates the the term NIR.


text:
Concentrating mixtures of volatile compounds
My interest is primarily food-related here but I think this is a fairly general chemistry question. Let's say I took some liquid like lemon juice and ran it though a rotary evaporator to get just the volatile compounds. In the case of lemon juice I'd assume I'd wind up with mostly water, and some other volatile compounds (that would smell exactly like lemon juice does). To increase the intensity of this we'd need to somehow remove water.
Similarly a common technique for making extracts is to dissolve volatile compounds of common aromatic foods into alcohol. In case this process left any non-volatile compounds you could certainly use a rotary evaporator. You wind up with a somewhat similar issue where alcohol is the primary liquid.
I'm wondering if it would be possible to concentrate such mixtures so that they contain higher ratios of smell-able volatile compounds to water/alcohol. One thing I've noticed is that most of the characteristic volatile compounds in say lemons have a much higher molecular weight than water or even alcohol. Is there someway to utilize this to like centrifuge out a more concentrated liquid? Are there good methods for either water or ethyl-alcohol specifically?

Perhaps it could interest you https://tinyurl.com/uuzpxsb   it opens a pdf

The simplest method would be fractional distillation: if you have a mixture of miscible liquids with different boiling points, as is the case in your question, then you can heat them slowly. When the boiling point of the most volatile compound in the mixture is reached, it begins to boil and the temperature does not increase until it has all boiled off. The vapour formed is cooled with a condenser and collected as a liquid. As soon as the temperature begins to increase again, all of this compound has evaporated, so it can be removed. Provided the boiling points aren't too close, you could use this to separate out all the compounds in the mixture. A diagram of the set-up (from Wikipedia) is below.

This would be easier with water than ethanol, since it has a higher boiling point so is less likely to boil at a similar temperature to your desired compound.
(The most common way to separate out mixtures in chemistry is chromatography - for example, HPLC or GC - but I'm not sure you'd want to put the results in food! HPLC, which stands for high performance liquid chromatography, involves the use of a solvent mixture which normally includes chemicals like acetonitrile or methanol, neither of which are safe to ingest.)

hmm, so something seems slightly off here. The most preventable volatile compound with an order in lemons is limonene (go figure). I know from personall experience that if you heat lemon juice up a bit (to say 160F, for 10 seconds) it loses almost all aroma. The boiling point of limonene is supposedly like 349F however.
It seems likely to me that there must be some other mechanism causing these low concentration volatile compounds to evaporate off. Any idea what's happening? I suspect whatever mechanism is occurring here would make this technique invalid.

OP: Of course it is off, it should be downvoted. Go back to the rotovapor that is basically steam distillation under reduced pressure. Except that contrary to your expectation you have to find limonene in the condensing phase, and get it off from there. Limonene is not soluble in water.

Fair enough - the OP described it as a general chemistry question so I gave a fairly general answer. I'm interested by what you say about lemon juice, because my experience is totally different! If you put lemon tea with boiling water, for example, it smells very strongly. Limonene actually has a boiling point of 176 oC, which I found surprisingly high.

It could break down at high temperature, of course - vitamin C does - but from what I could find online, I don't think it does break down at that temperature (160F is only 71 oC).


text:
are there any stable four membered chelate rings?
With reference to coordination compounds, most chelate rings formed with the metal atom and ligands are 5-member rings (including the central metal atom), such as complexes with Oxalate, ethylenediamine, dimethylglyoxime and glycinate ions as complexing agents.
Are there any compounds which contain stable 4-member chelate rings? Or any 7-member rings as well for that matter? I understand that angle strain makes these complexes hard to form but are there any exceptions to this trend?

The first thing that would come into my mind would be porphyrine based complexes. But I don't know if you were looking for something like that or if I've got your question wrong here. When I ask my students about chelate stability they often tell me about thermodynamic and kinetic stability but there is a third effect, macrocyclic stability. If the ring is already closed and stable but can coordinate different ring sizes are possible as well and square plane forms even without d8 configurations if necessary.

4-rings are more or less common with huge central atoms (like uranium).

@IvanNeretin Could you provide some examples/references please?

Four-member rings can be formed with simple oxy-anion ligands such as nitrate.  An example with iron as the central ion is given here.
Reference

G. R. A Wyllie, O. Q. Munro, C. E. Schulz, W. R. Scheidt,
"Structural and Physical Characterization of (Nitrato)iron(III)
Porphyrinates [Fe(por)(NO3)] — Variable Coordination of Nitrate",
Polyhedron, Oct 10 2007, 26(16):  4664-4672.

Yeah, acetate may also work like that if cation is small enough, like Zn.


text:
How to set up an equation to solve a rate law computationally?
I am confused on how to set up my equations to use in the solver. I have an equation
$$\ce{A + 2B -> C}$$
As a given, it is first order in A and second order in B. What I need to do is to solve for the concentrations of all three of $\ce{A}$, $\ce{B}$ and $\ce{C}$ using an ordinary differential equation (computationally in Matlab). The rate constant is given as $\pu{8.5E-4}$. I am given all initial concentrations ($\pu{1.2 mol L-1}$ for both).
To start off I have
$$r = k\ce{[A][B]2}$$
I have no idea where I should take this next, I feel completely lost here.
The second part of this is to do the same thing, but instead of being irreversible, the reaction is given as reversible.
$$\ce{A + 2B <=> C}$$
The concentrations are all given as are the forward and backward rate constants. Its first order in $\ce{A}$, 2nd in $\ce{B}$, and first in $\ce{C}$. $\ce{[A]} = \pu{1.2 mol L-1}$, $\ce{[B]} = \pu{1.8 mol L-1}$, $\ce{[C]} = \pu{0.4 mol L-1}$, $k_\ce{for} = \pu{3.4E-2}$, $k_\ce{back} = \pu{1.2E-2}$.
Again my initial thoughts are
$$r = \frac{k_\ce{for}}{k_{back}} \times \frac{\ce{[A][B]^{2}}}{\ce{[C]}}$$
But again I just don't know where to go next.
I hope I made my question understandable, if not I'll try to clarify anything. Any and all help would be appreciated, I really need it!

See, you are in the company of people who know chemistry, but not necessarily Matlab. Do you know how to solve differential equations in Matlab (say, a simple one like $y'=-y$)? If you don't, then you have come to the wrong place. And if you do, what's stopping you from solving this one?

I understand how to do the Matlab part of this (at least I think I do), but I do not understand how to get to an ODE from the reaction equation. I know how to use the correct functions in Matlab, but I don't really understand the chemistry of how to get there. In your example, I can solve y' = -y in Matlab, but what my question boils down to is how do I get to that type of equation from the reaction equation?

Wait, didn't you just write the rate expression ($r=\dots$)?

It should be set up as a series of first order ordinary differential equations like so: $$\frac{d[A]}{dt}=-k_f[A][B]^2+k_r[C]$$$$\frac{d[B]}{dt}=-2k_f[A][B]^2+2k_r[C]$$
$$\frac{d[C]}{dt}=k_f[A][B]^2-k_r[C]$$
If we substitute $[C]=[C]_0+x$, $[A]=[A]_0-x$, and $[B]=[B]_0-2x$ into these equations, they all reduce to the single equation:$$\frac{dx}{dt}=k_f([A]_0-x)([B]_0-2x)^2-k_r([C]_0+x)$$

Yes.  Thanks.  I’ll go back and change it.


text:
Determining sulfate content in Mohr's salt
I am practicing problems for my chemistry test and I am stuck on the following problem:

The value of $x$ in $\ce{Fe(NH4)2(SO4)2 ⋅ xH2O}$ can be found by determining the amount, in moles, of sulfate in the compound.
A $\pu{0.982 g}$ sample was dissolved in water and excess $\ce{BaCl2(aq)}$ was added. The precipitate of $\ce{BaSO4}$ was separated and dried and found to weigh $\pu{1.17 g}.$
Calculate the amount (in moles) of $\ce{BaSO4}$ in the $\pu{1.17 g}$ precipitate. And calculate the amount (in moles) of sulfate in the $\pu{0.982 g}$ sample of $\ce{Fe(NH4)2(SO4)2 ⋅ xH2O}.$

I was able to do the first part of the question:
$$n (\ce{BaSO4}) = \frac{\pu{1.17 g}}{\pu{233.38 g mol-1}} = \pu{0.005 mol}$$
But, I am unsure of the second part. Any help would be appreciated.

Write a reaction equation for the precipitation (include it in the post), then track the sulfate from the product to the educt. You should then be able to find the amount of the iron salt.

And please choose a descriptive title for your question, it currently values no information at all and is not helpful.

Use the number of moles sulfate precipitated to calculate the number of moles of Mohr´s Salt in the original sample. Using the molecular weight of the anhydrous salt you can calculate how much that part weighs, and the difference up to 0,982g is due to the water.

@FrankS Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

If there is $\ce{0.00501 mol}$ of $\ce{BaSO4}$ in the precipitate, this means that there was $\pu{0.00501 mol}$ of sulfate in the Mohr's salt solution and $\pu{0.00501 mol}$ of sulfate in $\pu{0.982 g}$ of Mohr's salt. And there was one half, that is $\pu{0.0025 mol}$ of $\ce{Fe(NH4)2(SO4)2.xH2O}$ in $\pu{0.982 g}$ of Mohr's salt. This allows you to calculate the molar mass of the salt, and to deduce $x$.

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