text:
Does a unit cell have to contain a whole number of atoms?
I was given this problem recently to compute the number of ions in a unit cell of ferrous oxide $\ce{FeO}$. The data given was:

Side length $a = \pu{5 Å}$
Density $d = \pu{4 g/cc}$

Using the formula $\displaystyle d = \frac {ZM}{N_0 a^3}$, I obtained $Z = 4.18$. However, the answer had rounded $Z$ off to $4.$
Could a unit cell have  a fractional number of atoms? Intuitively, it doesn't make sense for a cell to have $0.18$ of an atom. If, by experimental analysis, we obtain a fractional number of atoms in a unit cell, do we round them down as is done in this problem?
P.S. Another thought would be that such odd numbers are obtained due to stoichiometric defects.

$Z \in \mathbb{N}$ (always, by the definition of the number of formula units); 2. Both $a$ and $d$ are given to the single significant figure. How did you manage to obtain the more precise answer with three significant figures using this data?

@andselisk: Would you please elaborate your comment as an answer to the question?

Crystallographic background
A unit cell cannot contain parts of atoms from a formula unit, otherwise it would not be a geometric unit of repeatability.
The simplest condition arising from translational symmetry is that the total number of atoms present in the unit cell must be either equal to or be a multiple of the number of atoms in the chemical formula, which is an equivalent to the integer number of formula units per unit cell: $Z \in \mathbb{N}.$
Note, however, that the asymmetric unit (crystallographically independent region) contains the $Z/n$th part of the atoms of the formula unit, where $n$ is the order of the crystallographic point group.
Thus, the asymmetric region may contain an integer number of formula units, one formula unit or its fractional part, meaning that some atoms located on symmetry elements, are "divided" across several asymmetric regions.
Data precision
It appears you are using the correct formula for the crystallographic density $ρ$ (I strongly encourage not to use $d$ as a symbol for density as in crystallography $d$ is reserved for spacing between planes):
$$ρ = \frac{MZ}{N_\mathrm{A}V}$$
where $M$ is molar mass, $N_\mathrm{A}$ is Avogadro's number and $V$ is the unit cell volume.
Although you can use the exact value for $N_\mathrm{A} = \pu{6.02214076E23 mol-1}$ and a precise tabulated value $M(\ce{FeO}) = \pu{71.844 g mol-1},$ note that you are given both $a$ and $d$ (experimental data) to the single significant figure, which prohibits your answer from being more precise than these parameters, and you must present a rounded number $Z = 4$ and not $Z = 4.18,$ which doesn't make either crystallographic or mathematical sense.
In real life situation there always will be a systematic difference between the measured and crystallographic densities due to the presence of defects, and the deviation of $Z$ from the integer value is often due to statistically filling of the proper point group by the atoms of a different kind.

The density presented by andselisk is the one based on the diffraction experiment and the subsequently built model (_exptl_crystal_density_diffrn in the .cif files you find in databases, https://www.iucr.org/__data/iucr/cifdic_html/1/cif_core.dic/Iexptl_crystal_density_diffrn.html).  You should at least check by an estimation, if this density is reasonable.  The same files may include an entry about _exptl_crystal_density_meas, measured e.g., by floatation of your sample in liquids (https://www.iucr.org/__data/iucr/cifdic_html/1/cif_core.dic/Iexptl_crystal_density_meas.html).


text:
Which change occurs in all exothermic reactions?

Which change occurs in all exothermic reactions?
A) Bubbles of gas are released from the mixture.
B) Light energy is produced.
C) The temperature of the mixture decreases.
D) The temperature of the mixture increases.

I'm stuck with choosing between A, B and C, but I can't tell which one is right.

What is the definition of an exothermic reaction?

exothermic reactions give out heat and raise the temperature of the surroundings, they make chemical bonds and gives out energy to the surroundings.

Well, that pretty much answers it.

I'm still confused though, i've read that a exothermic reaction releases it as heat or light energy...

That's true. So what?

@IvanNeretin I assume the temperature of an exothermic reaction mixture that produces light can theoretically decrease, if you would let the light escape.

And where does that released heat go if not into the mixture?

to the environment

Can you please recheck your initial choice? You can definitely easily rule out three of the four. If the fourth is the correct answer we can discuss some more.

I just want to know the answer man I'm getting bare confused, I think it's B

Wait, you have a big misunderstanding here. ;) To check if a reaction is exothermic, you let it run in an isolated box, where nothing, neither heat or light or particles can get out. Then you check the temperature afterwards. The box has ideally no (in reality you calibrate that out) heat capacity.

does the temperature in the mixture decrease then??

No! If a reaction is exothermic, it produces heat! The mixture is hotter afterwards! ;)

so would the answer be D?

Exothermic reactions produce heat. Some of them may additionally produce (non thermal or thermal ) light or gas.

So the answer would be that the temperature of the mixture increases?

Yes. That's precisely what the word "exothermic" means. The rest is but smoke and mirrors. Imagine a question:  what do all red-haired people have in common? (A) They are male. (B) They are over 6 feet tall. (C) They are named Donald. (D) They have red hair.

see: https://chemistry.stackexchange.com/questions/35298/what-is-enthalpy and https://chemistry.stackexchange.com/questions/117456/exothermic-enthalpy-or-heat

Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!

This is a poorly worded question (no fault of the OP). Heat implies energy lost to the surroundings (without work being performed), not increase in the temperature of the sample. An exothermic reaction is one which, when carried isobarically and isothermally (constant temperature and pressure), releases heat, the heat being then equal to the enthalpy change of the system ($q_p=\Delta H$). IUPAC simply defines an exothermic reaction as one with a negative enthalpy change, which amounts roughly to the same statement. A reaction might be exothermic and carried out isothermally, such that the temperature of the sample would remain unaltered. However if the sample is thermostated - the reaction is carried out isothermally - then during the course of the reaction you might, with a thermometer, register a deviation from the set temperature before it returns to the set point. This is what the question is alluding to and is the common experience in a practice: exothermic reactions emit heat (if given a chance to do so) and the sample might become warm (if not given the chance to emit heat, ie the reaction is carried out adiabatically, and no work is performed).

While there are lots of posts addressing the definitions of exothermic and enthalpy, I leave this answer for the time being given the imho rather confusing and possibly misleading commentary that follows the question.

I have no quibbles with Buck’s answer, which is correct (and I upvoted it), but consider another plausible scenario. Suppose the OP’s stated multiple choice question is simply an exam question from an exam given during the first semester of a typical two semester general chemistry course. Perhaps the course sequence is for science majors but not chemistry majors, per se. This will be assumed below.
Then the putative exam question would be quite reasonable because students would have learned about exothermic and endothermic reactions and would likely have seen some of these common reactions in lecture demonstrations. Four specific examples from my own experience, teaching such a course multiple times, include 1) burning magnesium in air (lots of heat and blinding white and UV light released), 2) throwing calcium metal into not much water in a big beaker (lots of heat produced in the roiling and boiling reaction mixture, the beaker gets too hot to touch, hot water vapor spews up and out of the beaker), 3) the super-saturated sodium acetate ‘hot ice’ demo (which is used in some consumer ‘cold packs’) and 4) dissolution of ammonium nitrate in water in a beaker (the solution gets quite cold and students are told that this is another way that ‘cold packs’ have been made).
Students are told that the first three reactions are exothermic, as are lots of other reactions familiar from everyday life, e.g., burning wood in a fireplace or combustion of gasoline in an automobile engine, are also exothermic and the reactions are performed specifically because heat is desired. The dissolution of ammonium nitrate is described as endothermic and the reason for this is deferred to the next semester because students have not been introduced to entropy.
And there is the issue: students are introduced to thermodynamic concepts in much the same fashion as hospital patients are introduced to intravenous drips: slowly, for the common good of them and the teacher. During the first semester, students encounter the concepts of system, surroundings, energy and energy changes, enthalpy and enthalpy changes, Hess’s law and standard enthalpies of reaction. But it is not until the second semester that students are introduced to entropy, free energy, changes in them and equilibrium.
Arguments abound about best practice in teaching the beginning material in general chemistry and whether or not it should even be taught at all: some favor just starting with organic chemistry. People go apoplectic about some of the simplifications and admitted over-simplifications, e.g., VSEPR with d orbitals, that are perpetrated in the first year course. But it is not an upper level undergraduate course or a graduate course and students would be poorly served if everything had to be meticulously and profoundly correct. This does not mean that teaching incorrect material is justified. It simply means that there is a place for zero order approximations in the pedagogy. Of course, rigorous quantum mechanics is correct in so far as we know. But, in light of the great Richard Feynman’s experience teaching physics to insanely bright freshman CalTech students, it would take amazing courage to attempt such an endeavor in first year chemistry, especially for non-majors.
Now, with regard to the OP’s question, the answer is simply “D”: heat is released in exothermic reactions. For this question and its implicit course context, it does not matter where the heat goes: if the beaker gets hot (as in the calcium and water demo), or you can get burned by touching it, then it was exothermic. So answer “C” is wrong. The first two answers are obviously incorrect: burning magnesium (or wood or paper or ...) does not involve bubbles and light is not released when calcium metal is dropped into water.

I understand your points, particularly the need to keep things relatively simple. From the OPs comments it appears that the confusion had to do with definitions of system, surroundings and heat, which is not surprising. The problem would have been averted if the student had registered that an exothermic reaction in an insulating rigid vessel increases the system's temperature.

@BuckThorn Exactly! It pops up practically every semester: a student will say that since the heat left the system and went to the surroundings, then this must mean the system is cooler than before the reaction. Understandable, and a good teachable moment, yet always a bit frustrating that the curtain cannot be pulled all the way back right away.

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text:
Is there any collection of charge density maps for small molecules? Any good data at all?
Despite wide variety of existing experimental methods producing electron/charge density maps, I failed to find a good dataset of experimental data for small (1-3 atoms) molecules.
There are datasets for integral properties such as https://cccbdb.nist.gov/ . It is fairly trivial to find atomic spectra data on NIST site and with some effort it is possible to obtain electronic spectra data for some molecules (though apparently there is no good unified database).
Regrettably, I failed to find any collection of experimental charge density data for small molecules. Furthermore, on second though I cannot remember where charge density maps in QC textbooks come from. Calculated charge densities are fairly easy to obtains, but experimental charge densities are a different matter. In fact, "experimental charge density hydrogen molecule" search gave me zero relevant references.
Any suggestions?

My background is macromolecular crystallography; while experimental data was kept locked in the research laboratories in the 1980s, they are now widely available through the international protein database. For example, you can view the electron density of crambin at http://www.rcsb.org/3d-view/1EJG.
The database for smaller structures is the Cambridge Structural Database. They have implemented depositing experimental data (structure factors) in the 2000s, but I did not find an efficient way to search for those structures that come with structure factors.
Charge density maps are often visualized as 2D-sections of the experimental density minus the density expected from isolated atoms (deformation density). One software program to calculate these is MoPro. Another program is CrystalExplorer, which shows deformation densities in 3D. In all cases, you would have to have access to the structure factors to make your own figures.

Is there any collection of charge density maps for small molecules? Any good data at all?

I don't think so.

Say I want a small (7-15 molecule) set of die hard, hi quality experimental electron density data for well-defined small molecules (3-6, definitely no more than 8 atoms including hydrogen), what would you suggest to go for?

I would contact the authors of MoPro (Christian Jelsch, Benoît Guillot, Cristian Iordache) - they must have a set of structures with experimental data. Also, maybe contact the CCSD - they have a freely-available teaching set of coordinates, and maybe they want to include a new set where structure factors are available. The classic structures are oxalic acid dihydrate and formamide. If the molecule is much smaller, it is not in the solid phase (or contains heavy metals that are difficult computationally). Of course, ionic structures like sodium carbonate would work. @permeakra

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text:
Multiple bonds solely with sp3 hybridised orbitals?
I have fond that similar question has been asked already (why don't carbons use sp3 instead of sp2 orbitals to make double bonds?), but I don't think it was formulated in a manner that was touching the concept I am interested in.
I would like to know is it reasonable to consider double and triple bonds as result of overlapping solely sp3 hybridised orbitals. I imagine that alternative view on carbon-carbon double bond in ethene could be that two pairs of sp3 hybridised orbitals from each carbon overlap - giving four molecular orbitals, two of which would be bonding and fully occupied while other two would be empty. Moreover, geometry of such constructed double bond would be correct as all four hydrogen atoms bonded by remaining four sp3 orbitals from the two carbons would be coplanar (together with these two carbon atoms as well) and length of the bond would also be shorter than for single bond. I believe that similar case could be made for carbon-carbon triple bond.
Below I am presenting you a simple figure I have made to graphically demonstrate the idea.

I do recall seeing this concept is some textbook being presented as legitimate thing, I do not remember where it was, unfortunately.
Could you please explain it to me if this concept is reasonable idea, and if not - then why not? It would also be great to know how does energetic levels in such instance compare to the system with multiple bonds created by overlaps including hybridised and nonhybridized orbitals. (The one thing that appears in my mind is that in commonly accepted version of multiple bonds there is sigma bond that has electron density directly between nuclei making this arrangement more favourable. However, there are systems like two electron three centre bonds, as in B2H6, in which three centres are hold in similar arrangement to discussed above)

The theory of hybridization is a method developed by Pauling to explain the geometry of the molecules and ions. It has an advantage : it works. But it has never been proved that hybridization exists. And there are serious chemists who have shown that using non hybridized atoms orbitals provide the same result as with hybridization. The only trouble is that it is much more difficult to follow the calculations done for minimizing the energy without hybridization. So to go back to your question, you may of course combine your sp3 orbitales to make simple, double, and triple bonds. It will work

I continue. Using sp3 orbitals to make simple, double and triple bonds does work. The distance between carbon atoms in simple, double and triple bonds is exactly what coud be calculated using 1, 2 or 3 sp3 orbitals as you did it. The only trouble is that it is difficult to use this technique to explain mesomerism and the conjugation of double bonds.

@Maurice Nobody has ever claimed that hybridisation exists. The sp, sp2, sp3 are useful mathematical approximation for orbitals to be used to form bonds via LCAO instead of plain hydrogenlike AOs. If you use them for a bond where they are not applicable, you surely won´t get correct bond lenghts etc., will you?

https://chemistry.stackexchange.com/questions/32407/why-can-there-not-be-more-than-one-sigma-bond-in-a-set-of-bonds

Dear @Maurice
thank you for your comments that are answering my question. I see a lot of value in them. Indeed, I have found some examples of using only non-hybridised s and p orbitals for explaining bonding in simple alkane molecules.

Dear @Mithoron, thank you for getting involved in this discussion. And directing me to two other threads on this portal. They are interesting and very useful.


text:
Why is liquid germanium denser than solid germanium?
Why are some elements like Ge and Si have liquids denser than their solids? In water it is explainable by H bonding, but no H bonds or such are there in Ge and Si.
Well, I want an explanation about the structure formed by Ge and Si in their solid state which makes them less dense than their liquids, and also the bonds formed

If this helps, (I have taken this from @matt_black answer here.) "There isn't really a simple single explanation of why some solids are less dense than their associated liquids. The general explanation (which doesn't really explain much) is that the solid has a structure and, sometimes, that structure takes us more space than the average structure that occurs in the liquid."

Thanks for that Zenix, but I wanted a more detailed answer, so I modified the question accordingly as well.

Si and Ge are 4-fold coordinated diamond structure semiconducting solids. Their liquids are 8 to 12-fold coordinates metallic phases. The diamond crystal structure is fairly open.

Gallium and bismuth are also more dense, as liquids near their melting points, than are their respective solids at room temperature.

Don’t forget antimony as well...


text:
Cyanotype photography with potassium ferrocyanide
Some years ago, due to a translation error that I only discovered later, I made cyanotypes using a mixture of ammonium ferric citrate and potassium ferrocyanide instead of ferricyanide. This worked, although they seemed to be less sensitive (reacted slower to sunlight) than the guide I was trying to follow, but I didn't have a way to measure.
Is it safe to make this substitution? (I have lots left over and it would be good if I could use them up)
Is the resulting blue pigment the same Prussian blue as would result from using potassium ferricyanide? (some of the pigment seemed to wash away, whereas the guide said that only the unreacted reagents would do so)
Is there really a difference in sensitivity between this mixture and the ferricyanide mixture?


text:
Indicators of Cu(I) or Cu(II)
We performed an experiment in class where we placed a piece of copper wire in a solution of silver nitrate, we were tasked with predicted the mass of copper that should react and the mass of silver that should be formed. My question is, what are indicators of Cu(I) and Cu(II) ions?
$$\ce{2 AgNO3(aq) + Cu(s) -> Cu(NO3)2(aq) + 2 Ag(s)}$$
I need to know this in order to calculate reaction yield and efficiency (whether the product is $\ce{CuNO3}$ or $\ce{Cu(NO3)2}.$

You already have the balanced net equation.

Yes, but that was under the assumption that it was Cu(II).

Copper(I) nitrate is not the kind of copper(I) complex that would exist or form in an aqueous solution. As @EdV pointed out, you already have it all.

As noted in the comments, $\ce{Cu+}$ is not stable in aqueous solution. It tends to disproportionate to $\ce{Cu^0}$ and $\ce{Cu^{2+}}$. You can therefore assume that all but a negligible amount turned into $\ce{Cu^{2+}}$.
Source:
Holleman/Wiberg: Lehrbuch der Anorganischen Chemie, de Gruyter, 101. edition, 1995.

There is very well-written chapter "The chemistry of monovalent copper in aqueous solutions" in Advances in inorganic chemistry, Volume 64 [1, pp. 220–223, DOI: 10.1016/B978-0-12-396462-5.00007-6] which extensively covers as to why copper(I) is unlikely to exist in aqueous solution and why nitrate is a poor ligand for the purpose of preserving monovalent copper:

The redox potentials of copper in acidic aqueous solutions are as follows:
$$
\begin{align}
\ce{Cu+(aq) + e- &-> Cu^0(s)}, &\quad E^\circ &= \pu{0.521 V} \tag{1}\\
\ce{Cu^2+(aq) + e- &-> Cu^+(s)}, &\quad E^\circ &= \pu{0.153 V} \tag{2}\\
\ce{Cu^2+(aq) + 2 e- &-> Cu^0(s)}, &\quad E^\circ &= \pu{0.342 V} \tag{3}\\
\end{align}
$$
From these values, it is clear that $\ce{Cu+(aq)}$ is unstable and
  disproportionates in aqueous solutions:
$$\ce{2Cu+(aq) <=> Cu^2+ + Cu^0(s)}, \quad K_{1.4} = \pu{(1 - 1.7) × 10^{6} M-1} \label{rxn:4}\tag{4}$$
However, if $\ce{Cu+(aq)}$ is formed in a clean aqueous medium, the disproportionation reaction has often a relatively long induction period as reaction \eqref{rxn:4a} is endothermic due to the $\ce{Cu^0(s)}$ lattice energy.
$$\ce{2Cu+(aq) -> Cu^2+(aq) + Cu^0(aq)}\label{rxn:4a}\tag{4a}$$
Furthermore, the concentration of $\ce{Cu^0(aq)}$ in nonacidic aqueous solutions is limited due to the low solubility of $\ce{CuOH}/\ce{Cu2O}:$
$$\ce{Cu+(aq) + OH- <=> CuOH(s)/Cu2O(s)} \quad K_\mathrm{SP} = \pu{1E-14 M-2} \label{rxn:5}\tag{5}$$
Naturally, the concentration of Cu(I) species in aqueous solutions
  can be increased by adding appropriate ligands.
[…]
…the instability of $\ce{Cu+(aq)}$ stems from the difference between the solvation energies of the two cations and the second ionization potential of copper.
  This is correct for all transition-metal ions, but as the ionization energies increase along the series faster than the solvation energies, the lower oxidation states become relatively more stable.
  In order to shift reaction \eqref{rxn:4} to the left ligands which
  preferentially bind to $\ce{Cu(I)},$ must be added to the solution.
[…]
$\ce{Cu(I)L_n}$ complexes can be stabilized in aqueous solutions via
  employing ligands with one or more of the following properties:

Ligands that have a lower coordination number with $\ce{Cu(I)}$
  complexes than with $\ce{Cu(II)}$ complexes; these $\ce{Cu(I)}$ complexes are stabilized due the entropy gain in reaction \eqref{rxn:7}:

$$\ce{Cu(II)L_m + e- -> Cu(I)L_n + $(m - n)$ L} \label{rxn:7}\tag{7}$$

Ligands that are π acids and therefore prefer $\ce{Cu(I)}$ over $\ce{Cu(II)}$ and shift the redox potential of the $\ce{Cu(II)/Cu(I)}$ couples anodically.
Hydrophobic ligands, which often also increase the radius of
  the complex, decrease the stabilization of $\ce{Cu(II)}$ by decreasing
  its solvation energy.
Ligands which enforce a tetrahedral coordination sphere, or
  one which approaches this geometry, on the central cation.
Ligands that are soft bases, for example, thiols. Thus, for example, the biological copper-trafficking proteins, the metallochaperone Atx1, and the copper-transporting P-type ATPases have a highly conserved CXXC metal binding motif which binds a single $\ce{Cu(I)}$ ion via a two-coordinate site consisting of two cysteines.

All these types of ligands stabilize $\ce{Cu(I)}$ complexes by shifting the redox potentials of the $\ce{Cu(II)/Cu^0(s)}$ and $\ce{Cu(I)/Cu^0(s)}$ couples and thus shifting the equilibrium constant of reaction \eqref{rxn:4} to the
  left.
  However, ligands might stabilize $\ce{Cu(I)}$ complexes also
  kinetically by inhibiting the disproportionation reaction by
  inhibiting the formation of $\ce{Cu^0(s)}$ via blocking the approach of two copper atoms, this is, for example, the mechanism by which
  some $\ce{Cu(I)}$ enzymes are stabilized.
Naturally, all these ligands also increase the solubility of $\ce{Cu(I)}$ by competing with reaction \eqref{rxn:5}.

References

Inorganic/Bioinorganic Reaction Mechanisms, 1st ed.; Eldik, R. van, Ivanović-Burmazović, I., Eds.; Advances in inorganic chemistry; Academic Press: London; Waltham, MA, 2012; Vol. 64. ISBN 978-0-12-396462-5.

I upvoted since as my experience with copper chemistry certainly confirms the role of pH and ligands. However, commercial copper leaching employs NH3, O2 and copper (which is, in part based on an electrochemical cell) and latter processes, added NH4+ which demonstrated improvement. Why? Perhaps because NH4+ = H+ + NH3 with the H+ removed by created solvated electrons, or OH-, may be a latent source of ammonia, or is there another unaddressed factor?


text:
Mercury bulb on spirit thermometer?
I recently broke a Quickfit distillation thermometer and the bulb had liquid metal in it. I assumed it's mercury and acted accordingly. However, the column looks like spirit filled (blue liquid). Has anyone seen those, and if so, do you know if the metal is indeed mercury or something else? The model is Quickfit MF 31/2 (not MF 31/2S which is the newer one). I have never seen mercury and spirit combined in other thermometers. Just curious that's all.

One of the thermometer's designs than would use both mercury and ethanol is a Six's thermometer. Also, a photo might help.

Added a photo. This is of another one we have. The bulb is definitely metal filled, but not the column.

Look closely at the tube at the different angles. Many thermometers these days use mercury or galinstan and the colored (painted) line along the inner side of the tube that makes the metal appear colored, blue in this case (see e.g. this image).

It may just be a blue dye deposited on the back of the tube


text:
Do any pharmaceuticals increase in potency or toxicity over time?
Pursuing a question raised in these comments:
Are there any known pharmaceuticals that increase in potency or toxicity during storage, or over periods of months to years?
If not, what if we broaden the question to include any bioactive chemical?
My hypothesis is that the answer is no due to the following considerations:

If we limit the scope to pharmaceuticals we often find that the bioactive components of a chemical are specific isomers.  In that case other isomers are relatively inert.  Over time racemization can only decrease the concentration of the bioactive isomers,* so under this mechanism potency can only decrease with time.  (I can't find any examples in which non-therapeutic isomers are more toxic than the therapeutic isomers.)
The other mechanisms of chemical change that I can think of chemically affecting a bioactive substance during extended storage are heat, light, evaporation, and atmospheric oxidation and moisture.  We might rule out all of these but heat in the case of properly stored pharmaceuticals, which are formulated and/or packaged to prevent exposure to these influences.  But even if we allow for exposure to any or all of these (keeping heat within reason – say under 60°C) I can't think of any examples by which these processes increase toxicity of any chemical.

Any specific examples or general suggestions to the contrary are welcome!

* Silas W. Smith, Chiral Toxicology: It's the Same Thing…Only Different, Toxicological Sciences, Volume 110, Issue 1, July 2009, Pages 4–30, https://doi.org/10.1093/toxsci/kfp097
provides an extensive survey of this question.  I certainly could have missed something during reading, but it does state early on that in drugs offered only as enantiomers "...the other isomer has no clinical effect or adverse effects in vivo."  Some examples show that while "adverse" effects (which I'll classify under "toxicity") can be found in non-therapeutic isomers, they do not exceed the adverse effects found in the therapeutic isomer.  So in the worst case it appears that a pharmaceutical can lose potency but not toxicity over time due to racemization, but this does not constitute an increase in "toxicity."

See https://www.nytimes.com/2014/02/11/science/do-some-drugs-become-dangerous-after-expiration.html and https://www.consumerreports.org/drug-safety/the-problem-with-expired-medication/

@DrMoishePippik the first article mentions anecdotal evidence that "old" tetracycline was associated with some adverse reactions, but then cites a 2004 review that concludes no causal link has been established (much less any hypothesis for how tetracycline might degrade into something toxic, or what the toxic chemical might be).  The second link just repeats the FDA's generic advice to "throw out expired meds because they may have reduced potency."  Neither is informative to this question.

In The Wolf of Wall Street, it's asserted that old Quaaludes (specifically Lemmon 714's) increased in potency. See this clip for the cinematic treatment.


text:
Mechanism for Nagata Alkylation
I am looking into the mechanism for the Nagata Alkylation; however, I cannot find anything more specific than the mechanism listed below.

I am only refering to the formation of the first intermediate, not the cyclization of the ortho-quinone methide. I would assume that the phenylboronic acid first forms an ester with the phenol and then the lewis-basic carbonyl attacks the boron. This might then make an electrophillic carbon that can undergo an alkylation reaction. I am not very sure about this proposal of mine, and I was wanting to ask if anyone had any suggestions as to how this mechanism might be occuring.

This reaction of ortho-alkylation of phenols was first indroduced by Nagata and coworkers in 1979 (Ref.1). Accordingly, for example, reaction of a phenol with a large excess of an aldehyde and a  slight excess of benzeneboronic acid in refluxing benzene (or toluene) in the presence of $\pu{0.1 mol}$ equivalent of propanoic acid (or trichloroacetic acid) gave about 97% of the expected product, a 2-phenyl-4 H-1,3,2-benzodioxaborin, as a crystalline product (Ref.1):

Note that the presence of propanoic acid is critical in the reaction (no propanoic acid, no expected product). The authors have suggested that mechanistically, the reaction would have taken place by coordination of an aldehyde molecule to the boron atom of a probable intermediate I (i.e. a substituted  phenyl benzeneboronate), followed by [3,3]-sigmatropic rearrangement (as indicated in II) to give the product, a substituted benzodioxaborin.
References:

Wataru Nagata, Kyo Okada, Tsutomu Aoki, “ortho-Specific $\alpha$-Hydroxyalkylation of Phenols  with Aldehydes. An Efficient Synthesis of Saligenol Derivatives,” Synthesis 1979, (5), 365-368 (DOI: 10.1055/s-1979-28683).


text:
What effect is on the rate and rate constant when concentration or temperature increases/decreases?
I have a few questions regarding rate and rate constant. I am wondering what effect certain concentrations, temperature, etc., has on rate and rate constant. Would be of great help if you can clarify this for me!
If a concentration was halved (and all other factors remained the same), does that mean rate would decrease? If instead the concentration was doubled, I would assume that it would also double.
For rate, if the temperature of the solution decreased, does that mean that it would also decrease since it means it would take less time?
For rate constant, if the concentration was halved, I think there would be no effect (because concentration effects rate only). But, what if the concentration was doubled, does that also mean that there's no effect or would the rate constant decrease?
For rate constant, let's say the ionic strength of the solution increased. Does that mean that the rate constant decreases?
Thanks for clarifying!

Assuming you are speaking about an elementary reaction, and by that I mean a reaction on the form $\ce{A -> B}$ the rate of the reaction is
$$r = \frac{\mathrm d[\ce{B}]}{\mathrm dt} = -\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]$$
Most of the time $k$ can be express following the Arrhenius law:
$$k = A\exp\left(\dfrac{-E_\mathrm{a}}{k_\mathrm{B}T}\right)$$
Where $A$ is the pre-exponential factor (the constant rate when them temperature is really high), and $E_\mathrm{a}$ is the activation energy of the reaction.
In this case the initial concentration has no affect on the rate constant, only the temperature. Looking at the first equation the initial concentration has effect on the rate, this effect will depend on the order of the reaction…
$$r = k[\ce{A}]^2$$
$$r = k[\ce{A}]^3$$
Here a second and third-order reaction. If you study a complex mechanism, it can happen that initial concentration will affect the global rate constant, it can also happen with some approximation:
$$\ce{A + B -> C} \qquad r = k[\ce{A}][\ce{B}]$$
If $\ce{A},$ for example, is in really large quantity compared to $\ce{B},$ you can admit that for every time the concentration will approximately stay the same $[\ce{A}] = [\ce{A}]_0$ and then put it in the constant
$$k' = k[\ce{A}]_0$$
I hope that helped! If not, you need to study more.

A is (collision frequency) times (orientation factor), shouldn't collision frequency increase with increase in concentration. Or is it a constant, perhaps calculated when temperature is high?


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Should I be concerned about a galvanic reaction between silver solder and stainless steel?
I would like to solder a drain fitting onto my ultrasonic cleaner's tank. The fitting and the tank are both stainless steel. During its operation, the tank is filled with a cleaning solution and heat is applied to accelerate the process.
I'm concerned the silver solder joint and or the tank itself may become compromised overtime. Could there be a galvanic reaction between the solder and the tank, especially in the presence of mildly corrosive cleaning agents and heat?

It depends on the composition of your inox. There are several sorts of inox. But usually the potential of usual inox alloys is not very different from the silver redox potential. So the corrosion looks improbable.

The composition of Silver solders, themselves, as presented in this reference shows large differences in percentages of the anodic zinc and the cathode metal, with respect to possible galvanic based corrosion:

The compositions for the most commonly used jewelry grade silver solders are as follows, in descending order of melting point, from Knuth’s “Jewelers’ Resource”. Their compositions may vary with the different manufacturers, especially in the zinc content:
IT (extremely hard) 80Ag, 16Cu, 4Zn (sometimes no zn)
Hard 75Ag, 22Cu, 3Zn
Medium 70Ag, 20Cu, 10 Zn
Easy 65 Ag, 20Cu, 15 Zn
Extra Easy 56 Ag, 22 Cu, 17 Zn, 5 Sn
Easy Flo 45 Ag, 15 Cu, 16 Zn, 24 Cd

There is at least one study, to quote from Corrosion Issues in Solder Joint Design and Service:

Corrosion is an important consideration in the design of a solder joint. It must be addressed with respect to the service environment or, as in the case of soldered conduit, as the nature of the medium being transported within piping or tubing. Galvanic-assisted conosion is of particular concern, given the fact that solder joints are comprised of different metals or alloy compositions that are in contact with one-another.....
Functionally, however, it is the loss of material, be it the filler metal or the loss of nearby substrate-material, that most sibtificantly impacts solder joint performance and reliability. Material loss degrades the joint’s capacity to support a mechanical load, provide hermetically for a container structure, or sustain continuity in an electrical circuit.....
The predominance of one or more corrosion mechanisms, be it uniform, pitting, or crevice corrosion, as well as the rates of material loss, are very sensitive to the alloy properties (composition, phase distribution, oxide layer chemistry and thickness, etc.) and the service environment.

So, to answer your question, there is likely cause for your concern, but based on the last comment, a correct assessment likely requires the precise nature of the Silver solder employed and what chemicals, and associated mixes, that could act on the solder joint over time.


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Equations for the oxidation reactions of tea teaflavins and EGCG with potassium permanganate
For my chemistry investigation, I used Stamm's potassium permanganate method to investigate tea tannins with potassium permanganate to find the tannin content of teas. The method assumes complete oxidation of tea polyphenols to carbon dioxide.
Assuming that the tea polyphenols consist of only theaflavins, $\ce{C29H24O12},$ and Epigallocatechin gallate (EGCG), $\ce{C22H18O11},$ could someone help me with finding the equation of the reaction between the listed polyphenols and potassium permanganate respectively, i.e. theaflavin reaction with potassium permamganate and EGCG reaction with potassium permanganate (preferably 2 equations for each redox, but 1 overall equation should be fine too)?
I tried looking for these everywhere, but no source seems to give me the equations I need.

I upvoted your question and provided a full answer, but you also got two downvotes on the question. And as for half-reactions, your two compounds cannot be oxidized in one fell swoop, so there are no reasonable half-reactions for the highly complicated mess of piecemeal oxidations of degraded fragments of the compounds. Same as for the decay of a dead leaf: its carbon will ultimately be in carbon dioxide. So you are all set.

The two organic compounds are ultimately being converted to carbon dioxide, $\ce{CO2}$, and water, $\ce{H2O}$, just as they would be in complete combustion with pure oxygen gas, $\ce{O2}$, in a bomb calorimeter. Thus the two balanced combustion equations would be as follows:
$$\ce{C29H24O12 + 29 O2 -> 29 CO2 + 12 H2O} \tag{1}$$
$$\ce{C22H18O11 + 21 O2 -> 22 CO2 + 9 H2O} \tag{2}$$
So now assume that potassium permanganate, $\ce{KMnO4}$, is being used as the oxidizing agent, with manganese dioxide, $\ce{MnO2}$, as the product. Then, as per Ron's answer (https://chemistry.stackexchange.com/a/85865/79678), and assumed reasonable conditions therein, we have the following reaction:
$$\ce{4 MnO4- + 4 H+ -> 4 MnO2 + 3 O2 + 2 H2O} \tag{3}$$
Showing the spectator potassium ions, and arbitrarily assuming hydrochloric acid as the source of the hydrogen ions, yields
$$\ce{4 KMnO4 + 4 HCl -> 4 MnO2 + 3 O2 + 2 H2O + 4 KCl} \tag{4}$$
Since 29 $\ce{O2}$ and 3 $\ce{O2}$ have no common factor other than unity, multiply equation 1 by 3, equation 4 by 29, add them, cancel the equal number of $\ce{O2}$ molecules on each side and the result is
$$\ce{3 C29H24O12 + 116 KMnO4 + 116 HCl -> 87 CO2 + 116 MnO2 + 94 H2O + 116 KCl} \tag{5}$$
Since 21 $\ce{O2}$ is 7 times 3 $\ce{O2}$, use equation 2 as is, multiply equation 4 by 7, add them, cancel the equal number of $\ce{O2}$ molecules on each side and the result is
$$\ce{C22H18O11 + 28 KMnO4 + 28 HCl -> 22 CO2 + 28 MnO2 + 23 H2O + 28 KCl} \tag{6}$$

It is really surprising that your molecules are transformed to $\ce{CO2}$. But if it is the case, if all your theaflavin is oxidized into $\ce{CO2}$, the half-equation is :
$$\ce{C29H24O12 + 46 H2O -> 29 CO2 + 116 H+ + 116 e-}$$
The other half-equation depends on the pH of the solution. In highly acidic solution, it produces $\ce{Mn^2+}$. In nearly neutral solution, it produces $\ce{MnO2}.$


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Solubility of alcohols in non-polar solvents
The solubility of an alcohol in a non-polar solvent (like hexane) increases with size of the alcohol, as the non-polar chain increases. However, as the chain keeps increasing, will the solubility eventually start to drop because the molecule becomes too large?

Yes, I expect so.

Think about the extreme. I wouldn't expect polyethylene to be very soluble in hexane.

@Max. The problem is related to alcohols, not to  alcanes

@Maurice - The terminal -OH group wouldn't make any significant difference in solubility in hexane between a extremely long alkane and a extremely long n-alcohol.

Actually, polyethylene is soluble in organic solvents such as toluene, xylene, and mehylene chloride at slightly elevated temperatures. You may have even noticed this if you tried to wash a weigh dish with an organic non-polar solvent. This is an interesting question, but I would infer that the length of the chain would not have a significant effect on the solubility of an alcohol of high chain length.

Edit: Weight boats are actually made of polystyrene, but the same idea applies

@EliJones Like with solubility of ions, wouldn't the solubility decrease as the chain length increases? Why do big molecules like polystyrene dissolve in non-polar solvents? Does the polymer chain break into smaller chains or something along those lines?


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For molecular orbitals, how is an energy of zero set or defined?
When labeling the energies of molecular orbitals, often some anti-bonding MOs are shown with a positive energy. For example, MolCalc has the following energies for methane (the anti-bonding MO #8 at +19.47 eV depicted on the right):

I'm trying to understand what these numbers mean. I get that electrons in MOs #2 through #5 help to keep atoms bound (having them occupied contributes to bonding), whereas electrons in MOs #6 through #9 would lead to an increase in bond lengths or the molecule falling apart. On the other hand, MO #1 does not participate in bonding, right (basically, it is the inner 1s electrons of carbon)?
So what defines the zero reference point of these orbitals? It can't be the free electron because I thought at least for the hydrogen atom, even the excited states are states where the electron is still bound (i.e. energies are negative). It can't be the gradient of increasing all bond lengths (i.e. with the extreme case being separate atoms) because MO #1 would be hardly affected by that, given that it hardly participates in bonding.
This question is not limited to one particular software or author. Here is another example (source: https://www.researchgate.net/publication/234126374_Electron_rectification_through_donor-acceptor-heterocyclics_connected_to_cumulenic_bridge_A_computational_study):

I realize that the language in this question is sloppy, reflecting my poor understanding. I encourage anyone who answers to use the correct terminology.

The reference point must be the free electron. What you thought for the hydrogen atom is right. But you aren't calculating a hydrogen atom.

I think it will be worth contacting the company to ask for details. I had similar issues with the Matlab wavelet analysis toolbox and after asking around the world, it turned out to be something very arcane and specific to the way Matlab does it. They also had consistent labeling errors too in their toolbox, which are being worked upon.

The answer technically depends on the method. In most cases, meaning HF or DFT with standard basis sets, it is infinite separation of all bare nuclei and all electrons. One exception known to me is VASP (a periodic, solid-state code), which uses pseudopotentials to avoid having to model the core singularity and core electrons. There, the zero-point is in reference to some atomic state.
Note that the MO energies are basis-set dependent, which is most pronounced and most easily understood for core electrons.

If free electrons are defined as zero, what would happen to the methane molecule when you excite an electron in the MO approximation. There is no empty orbital with an energy lower than a free electron, so would it just ionize? What about for aromatics that absorb UV light? I would think the anti-bonding orbitals should have lower energy than the free electron because otherwise, UV light would not lead to absorption and emission of light but rather - again - to ionization.

As you excite one electron, the other MOs would not stay the same (unless you are in the frozen orbital approximation, which is poor). There are methods dedicated to the description of such excited states: time-dependent HF (TDHF), CI singles (CIS), time-dependent DFT (TDDFT), see https://pubs.acs.org/doi/abs/10.1021/cr0505627

So is there any physical interpretation of assigning a positive energy to the unoccupied anti-bonding MOs?

Much like Koopman's theorem, they can be used to estimate the electron affinity (in the frozen orbital approximation). But the basis-set dependence is again very high.


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Can someone explain the article below?
This is the article: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5611670/
I'm not a Chemist, so I can't understand the article very well.
The conclusion states that the molecules tested should not be called "piezolytes". Is this a nomenclature issue only? My only question is: does the presence of such molecules in deep sea organisms help in keeping proteins in their naturally folded state, resistant to denaturation due to high pressure or not?

From the abstract of the article in question:

The results suggest that the pressure stability of a protein in solution is not directly affected by the presence of these proposed piezolytes, and so they cannot be granted this distinction.

So the answer is "no", the supposed piezolytes do not appear to have the purported function. They do stabilize proteins against thermal denaturation, but not against pressure-induced destabilization.

I see. So there's likely no "holy grail" of deep-sea adaptation, such as those piezolytes, could one say that instead the biochemical elements of deep-sea creatures are all adapted to the pressure, for example using modified versions of land-animal proteins that count on high pressure to maintain their natural state and synthesis?

Yes that's what I'd conclude in any case from the study. Similarly to hyperthermostable proteins. But perhaps the piezolytes don't act alone?


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How does gefitinib affect the EGFR(ecptor) and why is there relapse
Gefitinib has been shown to be an effective tyrosine kinase inhibitor in a fraction (~$10$%) of non-small cell lung cancer patients.
These patients are characterized as having a mutation (usually a deletion or missense) in the EGFR gene affecting the ATP-binding site.
I have two questions:
-- How does gefitinib impede the proliferation and apoptosis pathway? Does it in effect "ace out" the binding of EGF? If this is correct, how does it accomplish that? If not, I would appreciate a brief explanation as to what the successful mechanism is.
-- It seems that (similar to imatinib, another tyrosine kinase used to treat CML) successfully treated patients tend to relapse after a period of time. Why is this?

I suspect this question would fit better on the Biology or Medical Sciences SE pages.

How does gefitinib impede the proliferation and apoptosis pathway? Does it in effect "ace out" the binding of EGF? If this is correct, how does it accomplish that? If not, I would appreciate a brief explanation as to what the successful mechanism is.

No, EGF binds on the extracellular side (to the ectodomain), and gefitinib binds on the intracellular side (to the tyrosine kinase, purple in the figure). The inhibitor interferes with ATP binding (see below, middle panel - the part boxed in green has structural similarities to adenine in ATP). The ATP would otherwise phosphorylate the tyrosines of other signalling proteins. So even if EGF binds to the ectodomain, there is no downstream effect because the tyrosin kinase domain is inhibited.

Source of images: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6523254/
I recommend section 3.2 to learn about how various inhibitors work.

It seems that (similar to imatinib, another tyrosine kinase used to treat CML) successfully treated patients tend to relapse after a period of time. Why is this?

Additional mutations might directly prevent the drug from binding, or change signalling pathways in a different fashion to render the drug inefficient.


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Why aren't Boron and Aluminium assigned to group 3 of periodic table? What determines the group?
I've been curious about this 3D representation of the periodic table "Mendeleev's Flower" and was trying to study it, wondering if it reveals any regularities that are not obvious from classic periodic table.

But since it's hard to look at a 3D table on a flat computer screen, I found flat equivalents.
Theodor Benfey's periodic table (1964):

J. F. Hyde's periodic table (1975):

There I found a discrepancy: Hyde assigned Boron and Aluminium to group 3 (IIIA), whereas everywhere else B and Al are in the group 13 (IIIB).
I can totally see why Hyde split the table like this (apart from being obsessed with Silicon): such a split creates an absolutely stunning symmetry.
You can see this symmetry on the Hyde's table with colors and dashed lines that cross Silicon and connect related elements.
This can't be just a stretch/exaggeration/far-fetching/"squeezing the universe to fit into an appealing model, even though the universe resists". It's just too good!
Well, to be honest, he did do some universe-squeezing: B and Al have electron configuration of s²p¹ — just like other elements of group 13 (IIIB), whereas elements of group 3 (IIIA) have electron configuration of d¹s². Since Al and B appear in the d¹s² column, Hyde had to make a note inside B and Al cells that they actually are s²p¹.
But the symmetry is just too good. It reveals stunning periodicity that no other periodic table is capable of.
And then I remembered this short-form periodic table:

Russian Wikipedia says the short form was banned by IUPAC in 1989 for some reason, but it's still used in Russian schools because it makes figuring out valence values easier.
And this table does not have the split. Boron and Aluminium appear in the same column regardless of whether you assign them to 3 (IIIA) or 13 (IIIB). I think this is absolutely beautiful.
I believe this Al/B positioning thing has scientific significance, but I don't understand the meaning of the groups.
My questions:

What does the group actually mean? Is it simply the electron configuration and nothing else?
What determines Al and B being placed in group 3 (IIIA) and not 13 (IIIB), other than the electron configuration?
Why is the short-form periodic table considered obsolete, even though it clearly has something going on that the full-form table is not revealing?
Is the Hyde's table missing any periodical properties that the classic table has?

By "short form" you mean the old system with roman numerals and main groups and side groups, right? I think they decided to have a uniform nomenclature, and then basically flipped a coin. ;)  Btw. I agree that for highschool chemistry (9th grade or so), the old system is better.

My knowledge of Russian is too rusty to identifty on the Russian wiki a reason why the short form was banned.  On https://www.meta-synthesis.com/webbook/35_pt/pt_database.php?Button=1950-1999+Formulations, entry 1994, however is one form suggested by Fathi Habashi including a link to his argumentation in J. Chem. Educ.  To quote from his conclusion to put Al just next to Mg to the left, and B to the right: «Boron is a metalloid with a high melting point, aluminium is a typical metal with low melting point, gallium, indium, and thallium are less typical metals.»

OT: The element´s abbreviations in your table are using Latin letters, while the rest is Kyrillic. Is that standard usage in Russian language?

@Karl «By "short form" you mean the old system with roman numerals and main groups and side groups, right?» — Yes. My point is that this form reveals periodicity that the 18-column form completely obscures. Can't understand why that's desirable.

@Buttonwood «My knowledge of Russian is too rusty to identifty on the Russian wiki a reason why the short form was banned» — Russian wiki does not explain the reasons. It only says that IUPAC ruled in 1989 that this form should no longer be used.

@Buttonwood «by Fathi Habashi» — This is awesome! I understand this is not a complete answer, but I believe this info deserves to be promoted from comment to answer!

@Karl «The element´s abbreviations in your table are using Latin letters, while the rest is Kyrillic. Is that standard usage in Russian language?» — Yes, Russian chemistry uses latin abbreviations. Every Russian knows what C2H5OH stands for.

Makes sense, too. ;) It is the same with units and symbols for physical quantities, right?

One answer could be that chemists don´t care much about the periodic system. It is awesome when you are a beginner, starting to get get a feeling, when everything starts making sense, and then new things just fall into place. Goodness, I loved it.But later on, you find that there are a lot of exceptions in those beautiful trends, and then you specialise on some corner of the chemistry universe, and the periodic system becomes one of the poster on your office wall. ;)

@Karl «It is the same with units and symbols for physical quantities, right?» — Nope, units and multipliers are localized in Russian. E. g. a gigapascal is ГПа.

Back in 1933, John Drury Clark published 1 his "race track" periodic table that subsequently became famous as the cover of Life magazine in 1949:

Looks like B and Al are nicely positioned. Just need to add an outer track for elements 105 - 118, element names and symbols for elements 97 - 103 and element 41 is niobium, Nb.
1 Clark, John D. (November 1933). "A new periodic chart". Journal of Chemical Education. 10 (11): 675–677. Bibcode:1933JChEd..10..675C. doi:10.1021/ed010p675.
Clark is also the author of Ignition! An Informal History of Liquid Rocket Propellants, Rutgers University Press Classics, New Brunswick, NJ, ©1972 and 2017.

Element 41 need not be "niobium".  "Columbium" is also a name for that element (cf. tungsten/wolfram) and the symbol Cb is still used in some industrial applications such as reporting the Chemistry of heats of steel.

Yeah, I know that obsolete name still persists.


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lewis structure of O3
This is my first question, sorry if I don't know how to write a good question.
why is O3 shaped like this?:
and not shaped like each Oxygen molecule is connected to two other molecules?
(i did not find an image for that)

The (non)existence and (in)stability of molecules is related to the energy of the configuration. And it just so happens that the configuration of O3 that you show is more stable than "cyclic ozone". This might be because of the strain, which would appear in a three-membered ring (similarly to the strain in cyclopropane).
However, the molecule has been studied theoretically, or at least the wikipedia page says so. (https://en.wikipedia.org/wiki/Cyclic_ozone)


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Non-trivial cases where the frozen core approximation is bad?
In computational chemistry, it is extremely common to freeze the core electrons after an initial Hartree-Fock calculation, meaning that the core electrons are left out of any treatment of the electron correlation. Which orbitals are considered core depends on the atom, but for the second row, it usually just means the 1s electrons are not correlated.
This is an intuitive approximation, as one would expect that correlation of the valence electrons would have the greatest effect on any actual chemistry.
Now, clearly one can imagine situations where freezing the core electrons is a very bad idea. One such case is if you were interested in modelling x-ray spectroscopy. I am less interested in this type of example as this is almost just a user error, and hence I would consider this a somewhat trivial example.
So, to be concrete, are there any known examples in which treatment of correlation of the core electrons results in a qualitative change in the modelling of a process which is primarily mediated by the valence electrons?
My guess is if there is an example it occurs in the third row and beyond.

It's also very bad for NMR properties, but I assume that falls under "trivial" too. :)

@orthocresol haha ya I think so. Obviously it's good to know those things though. Maybe it is worth highlighting those cases?

Somewhat non-trivial-ish but the relative energy of DMSO and its structural isomer, Me-O-S-Me (methyl methanesulfenate), is significantly affected (and just alllmost qualitatively impacted) when considering frozen (with a standard basis set) vs. fully correlated calculations.

@LordStryker I would say that's exactly the kind of thing I'm interested in. Is that published somewhere? Or do you have numbers? I'd be interested to see more of what causes that change.

@jheindel Ask me in a couple months. Hopefully I'll have a DOI to send you. ;)

@LordStryker Haha I look forward to it.

The frozen core approximation (FCA) falls apart whenever you need orbital relaxation of the core to accurately describe your system. From a theory-based argument, there are multiple electronic structure methods which rely on this (CCSD, orbital optimized methods) to better account for correlation effects. However, given that the core orbitals do not change appreciably during most processes (ionization, binding, standard chemical reactions), the error introduced by the FCA is all but completely cancelled (you can still get sub kcal/mol accuracy with the FCA).
However, one case where it rears its head is when you're trying to calculate the single-point energy of a system. In this case, the FCA will get in the way of recovering (for example) the full CI result in the full basis set limit.


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Why are there so many different formulations for energy? When do I use each one?
I understand that energy is a fairly arbitrary concept, and that it completely depends on your definition of energy. So far in my class, we have discussed:

Entropy, $S$ (not really energy, but it seems to be closely related);
Total internal energy, $U$, for which it is impossible to find an absolute value for;
Enthalpy, $H=U+PV$;
Gibbs energy, $G=H-TS$.

My questions are (and sorry if this is kind of loaded):

Where do these formulations arise? Why is there not just one version that trumps the rest? (Total internal energy sounds like a candidate)
In which scenarios would I be using one version over another?

If you are not going to elaborate the question by your effort to answer the question yourself, you may rather want to post it to Quora site. As it is, it does not fit the site policy and will be probably voted to be closed.

Why do we use kinetic energy, potential energy, thermal energy and not just total energy ?

@Poutnik Has it exactly right: the answer to all three of your questions is simply to pay close attention to what the teacher/professor says.

We use them in context with different conditions. It's often easier to evaluate/control systems with different constraints (constant pressure etc.) using particular thermodynamic potentials (the energy functions above). These functions also describe different aspects of the "total energy" the system possesses, like Poutnik says. It's somewhat like an aircraft trying to remain level by monitoring gravitational potential energy with respect to the earth, which sounds more rational than trying to keep an eye on the total mechanical energy the aircraft possesses.

As you see, each correspondence makes it easier to monitor the system and plot responses meaningfully.
It's also easy to determine equilibrium conditions in each kind of system shown above: minimization of the respective thermodynamic potential.

Table taken from Lecture Notes of Prof. Srabani Taraphder, IIT Kharagpur.


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Steric inhibition of resonance in substituted benzoic acid
Why does attaching an $\ce{-NO2}$ group at the ortho position of benzoic acid cause $\ce{-COOH}$ to become perpendicular to the plane of the benzene ring? Why doesn't the nitro group get pushed out while $\ce{-COOH}$ stays on the plane?
Also, in the case of two $\ce{-NO2}$ groups attached on the ortho  and meta positions of benzoic acid, both the nitro groups become perpendicular to the benzene ring while $\ce{-COOH}$ remains at an angle to the plane. Why doesn't it become perpendicular this time?
Here is what I referred to for the structure of 2-nitrobenzoic acid :https://pubchem.ncbi.nlm.nih.gov/compound/2-Nitrobenzoic-acid#section=3D-Conformer&fullscreen=true
Here is what I referred to for the structure of 2,3-dinitrobenzoic acid :https://pubchem.ncbi.nlm.nih.gov/compound/2,3-diNitrobenzoic-acid#section=3D-Conformer&fullscreen=true
If you feel these structures arent correct please do give a reference to where I can find the right structure. :)

It's quite not how you say.

Then what is it like? Could you explain...? Im pretty confused

https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases

That thread is specifically about the consequences of SIR and SIP... I already know about that. It was after learning that when I looked up the 3D structure of o-nitrobenzoic acid and got the doubt... I just want to know why -COOH rotates instead of -NO2 which stays perfectly planar... I want the reason behind the phenomenon not the consequence.

Structure of o-nitrobenzoic acid. Check out the 3D conformer given in this link :https://www.google.com/url?sa=t&source=web&rct=j&url=https://pubchem.ncbi.nlm.nih.gov/compound/2-Nitrobenzoic-acid&ved=2ahUKEwi454XKoILoAhUIfH0KHV3vAEEQFjACegQIBhAB&usg=AOvVaw3JvBtRwipcskI1y9JGj49F

Better question is if you can really trust Puchem.


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Selective halogenation of an alkane
Is it possible to selectively halogenate a certain position on an alkane chain, and also get a good yield?
I understand that to form such haloalkanes, taking a different substrate other than alkane (like a substituted alkane with the substituent attached at that position) would be a good choice, but is there a favourable synthesis to convert an alkane, say, n-pentane to 3-chloropentane?
Free radical bromination works well for smaller alkanes (due to its selectivity for secondary and tertiary positions) but it would not make much good out of longer ones. One could distil out the desired product after the free radical halogenation, but that doesn't seem quite effective.
P.S. The question may sound silly but only intends to know whether such synthesis is possible or not.

A good way of making 3-Chloropentane from n-pentane - no.  As you observe, a free radical halogenation produces a mixture of products and separating out 3-Cl-pentane from 2- and 1- chloropentane is not a job anyone would undertake lightly. If you need a clean sample of 3-Cl then start from something else (e.g. pentan-3-ol) and do the functional group interconversions.

Or from pent-1-ene - allylic chlorination and then hydrogenation.

If you are lucky or good at enzyme engineering, you might find an enzyme that does regioselective catalysis, like described e.g. in this paper.

This is an outstanding problem in organic synthesis right now. As mentioned by Karsten you could probably do this with enzyme engineering or directed evolution methods. Others have used transition metal catalysts
or photoredox-mediated organic photocatalysts. However, no method that I know of achieves halogenation in high amounts of regioselectivity that you describe.


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Safety of IPA and sodium hypochlorite mixture?
Is the mixture of 97% isopropyl alcohol and 0.5% of sodium hypochlorite safe?
Do they react together and form any compounds ?
Does this mixture present more danger to humans and animals than its ingredients separately ?

The SDS that I looked at indicated that oxidizing agents are incompatible with isopropyl alcohol.

As for above if this is meant to be a disinfectant or so, mixing the two would likely reduce the activity of at least one ingredient.


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Endothermic reaction absorbs heat from environment
I read on my school book about chemistry the reaction:
$$\ce{H2O + NH4Cl}$$
with $T(\ce{H2O}) = \pu{25 °C}.$
The temperature of the final solution is $\pu{20 °C}$.
I know that it is an endothermic reaction, and the initial solution absorbs temperature from the environment.
I don't understand why the temperature of the final solution is lower than the initial solution, I think it should not be lower because  all the heat that it needs comes from the environment.

The environment is whatever is closest to the reactants, and that is principally the solvent the reaction is conducted in

@Waylander I am sorry, what do you mean? The environment is a solvent?

As I said in comment 1, the environment is whatever is around the reactants. the bulk solvent is closest so that is where the energy (as heat energy) comes from. Hence the reaction mixture gets colder.

@Waylander thanks, so it means that the reaction absorbs energy (heat) from the H2O and NH4Cl?

@Ed V so I'll do for sake of clarity

@Edoardo yes microscopicalky whatever is around the reacting entitities is the surrounding. No need to look further out, though is of course possible.

An exothermic reaction occurs when the temperature of a system increases due to the evolution of heat. This heat is released
into the surroundings, resulting in an overall negative quantity for the heat of reaction (-ΔE ).
An endothermic reaction
occurs when the temperature of an isolated system decreases while the surroundings of a non-isolated system gains heat.
Endothermic reactions result in an overall positive heat of reaction (+ΔE ).
Exothermic and endothermic reactions cause energy level differences and therefore differences in enthalpy (ΔE ), the sum of
all potential and kinetic energies.
ΔH is determined by the system, not the surrounding environment in a reaction.
A system
that releases heat to the surroundings, an exothermic reaction, has a negative ΔH by convention, because the enthalpy of the
products is lower than the enthalpy of the reactants of the system.
The enthalpies of these reactions are less than zero, and are therefore exothermic reactions.
A system of reactants that absorbs
heat from the surroundings in an endothermic reaction has a positive , because the enthalpy of the products is higher than
the enthalpy of the reactants of the system.
Because the enthalpies of these reactions are greater than zero, they are endothermic reactions.
EDIT ( See this )
I understand your confusion. See, in endothermic reaction the product will be "colder"....this might seem counterintuitive as in endothermic reaction heat is absorbed by the system form the environment. Therefore the heat absorbed must increase the temperature right?
No! See temperature shows the direction of heat flow. If the temperature of the system increased ( assuming that previously the system and environment were in thermal equilibrium ~ same temperature ) this would mean that heat should flow from system to the environment as the system has a higher temperature than environment.
The following example makes everything clear:
$$T_{(env)}= 25°$$
$$A + B _{(25°)}  = C _{(20°)}  ...... (1)$$
$$A' + B' _{(25°)}  = C' _{(30°)} ..... (2)$$
Compare the temperature of system and environment in the above rxn.
Therefore in rxn. 1 heat flows from environment to system ( from 25° to ~20°).
This heat energy is used to break bonds and stuff.( Endothermic )
And in rxn. 2 heat flows from system to environment ( from 30° to 25°).( Exothermic )
Also see this practical video:
https://youtu.be/NHOHKH3808c

Thanks for your answer Navoneel! does it mean that the system of reactants absorbs heat from environment or from the solution?

Heat is absorbed from outside the system, that is the environment.

@Edoardo See my EDITS.

I think you should make the point that such analyses depend on the system being isolated, specifically that system is adiabatic.

@MaxW I thought it would be better to use isobaric open system, which is what found in our lab experiment . Therefore we could relate to real life practical situation.

Thanks, but in in rxn. 1 if the system absorbs heat from the environment, the T of the system should increase, and the heat energy  should be bigger = T(system) + T(absorbed from environment) - T(final solution)

@Edoardo I think you are forgetting that temperature can't be absorbed! Heat is absorbed and that heat is used to break bonds

But if the system absorbs heat from the environment, the temperature of the environment decreases?

@Edoardo You just can't add and subtract temperature like this! What gets added is heat. Heat is not temperature!

@Edoardo Environment can never get it's temperature reduced. It is considered a huge heat reservoir whose heat content is never reduced or increased!.....This is just an assumption. The temperature of the immediate environment in contact with the system does get cold....... Please see the video I linked.

I saw the video, but what would have happened if there was no water under the beaker? (thanks for your patience)

@Edoardo the wood and air would have got cold, it's not about water. Water is used to show how cold it was getting!

Ok, so the system needs heat from the environment, to absorbs the heat it needs to decrease its temperature, and after the reaction the system absorbs heat from environment because the system is colder, is it right?

@Edoardo Yes after the rxn is over if the system is left for some time, it will come to thermal equilibrium with the environment (25°).

@Edoardo One thing to note, the system doesn't decrease the temperature to absorb heat, it is the other way, heat is being used in chemical changes which reduces the temperature of immediate surroundings.( Which in turn helps heat flow in).

@Edoardo I would suggest you to go through a book or article on chemical equilibrium and understand the principal of Le Chatelier's .... Then you can get a true sense of the dynamic situation!..https://courses.lumenlearning.com/boundless-chemistry/chapter/factors-that-affect-chemical-equilibrium/

I got it, Thanks again Navoneel for your patience :)

@NavoneelKarmakar What if the system (reactants) and the surroundings are at the same temperature? How then the endothermic reaction can proceed if heat flow can't happen?

Energy is needed for separating positive from negative ions in the dissolution process. This energy is taken in the surrounding water. Water is loosing energy in the dissolution process. That is why the temperature of the water decreases. There is nothing special in using $NH_4Cl$. The same phenomena happens when dissolving a salt like $NaCl$ or any other one.
It may happen that a dissolution is exothermic. This happens with $NaOH$ for example. In this case a new phenomena is happening. One of the ions separated from the solid substance attracts an important number of water molecules that get adsorbed or fixed in the outer layer around each ion. This happens with ions "derived" from the water formula, like $OH^-$ ions or  $H_3O^+$ ions, which get surrounded by a lot of oriented water molecules. This creation of new bonds is exothermic, as every bond formation.

This answer is only applicable for cases where lattice enthalpy is concerned ( ie. solution formation) , what about reaction between Ba(OH)2 and NH4OH where temperature of product is lower than reactant and the process is endothermic


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Water-Soluble Photoinitiators for Free-Radical Photopolymerizations?
Could anyone recommend me a good proven free-radical photoinitiator available commercially suitable for (meth)acrylate-based UV-light photo-polymerizations ocurring in water phase, preferably with high water solubility and moderate-to-high quantum yield of photoinitiation? Commercial suppliers would also be highly appreciated. Thank you.

Most azo-based water-soluble thermal initiators can also be excited by UV light and therefore used as a photoinitiator in radical polymerizations, for example, V-50 and V-501. They are inexpensive and available from many commercial sources but they do not usually have very high quantum yield. Depending on your specific application scenario, you can also choose phenone-based water-soluble initiators, such as 2-hydroxy-4′-(2-hydroxyethoxy)-2-methylpropiophenone, which has a higher quantum yield.


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Naming acids, oxidation numbers and charges
I have been studying chemistry for a short time and studied the nomenclature of inorganic compounds. I got to naming acids, and got really confused with several things, including the definition of oxidation number. Let me explain myself:
I started studying using the book Chemical Principles: The Quest for Insight and in the naming acids section, it is said that an acid is formed by an oxoanion and positive hydrogen. To write the formula of an acid, use the prefixes to find the oxoanion (which I don't really know hot to do, because take "nitrate". How do I know the number of oxygens needed? How do I know if it is $\ce{NO-}$, $\ce {NO_2-}$ or $\ce {NO_3-}$?) and then add as much hydrogens as negative charge there is.
Then I came across another equivalent method that uses the oxidation numbers of the elements: it says you have to memorize a set of oxidation numbers of each element, like $\ce{Cl}$ has $-1, 1, 3, 5, 7$, etc. Then a table is shown associating each oxidation number to a prefix and suffix, like +1 corresponds to hypo-, -ous for $\ce{Cl}$, etc. Then combine this element with the corresponding charge with $\ce{O^2-}$ and add water ($\ce{H2O}$) to it. Let me give an example to show the two methods:
Take sulfuric acid.
Method I) The corresponding oxoanion is Sulfate ion, $\ce{SO4^2-}$ which has charge 2-, and so we add 2 $\ce{H+}$ to it, thus getting $\ce{H2SO4}$.
Method II) The corresponding charge for sulfur according to the table for the suffix -ic is +6. Thus we have the ions $\ce{O^2-}$ and $\ce{S^6+}$. Exchanging the oxidation numbers gives the compound $\ce{S2O6}$, which simplifies to $\ce{SO3}$. Adding the water molecule (i. e. adding 1 $\ce H$ and 2 $\ce O$) gives $\ce{H2SO4}$.
So I have a few questions regarding this: why do both methods work and which one is "more correct"? I would say II is the most correct since acids are said to be molecular compounds, but I'm not sure at all. Also, in a general acid $$\ce{H_a X_b O_c}$$
which are the oxidation states of each of the elements? I thought the oxidation number of $\ce X$ was negative, the oxidation number of $\ce H$ was +1 and the oxidation number of $\ O$ was (exceptionally) 0. It turns out that's wrong, right? Or is the oxidation number not the same as the charge?
Thanks in advance and sorry for the long question

The method used for naming the acids was adopted in the beginning of the 19th century. It goes like this. We first chose all the known acids with $b = 1$ in the formula $H_aX_bO_c$.
If there is only one such acid, its name is made of the name of X plus the suffix $-ic$. Ex.: $H_3BO_3$ is the $boric$ acid. $H_2CO_3$ is the $carbonic$ acid.
If there are two acids, the most oxidized is named by the previous rule. In the less oxidized acid, the suffix is changed into $-ous$. Ex. $HNO_2$ and $HNO_3$ are known. The first is nitrous acid, the second is nitric acid.
If there are four such acids, the two acids in the middle are named after the previous rule. The name of the most oxidized is obtained by adding the prefix $per-$. The name of the least oxidized is made by adding the prefix $hypo-$.
Ex. Four acids are known with $X$ = $Cl$ : $HClO$ is $hypochlorous$ acid ; $HClO_2$ is  $chlorous$ acid ; $HClO_3$ is $chloric$ acid ; $HClO_4$ is $perchloric$ acid.
There are no known cases where more than 4 different acids are known. If there are only three known acids, one guesses(!) a fourth formula. Ex. With $X = Br$, the following acids are known :  $HBrO, HBrO_3$ and $HBrO_4$. Their names are $hypobromous$ acid, $bromic$ acid, and $perbromic$ acid.

Hello and thanks for your answer. My question wasn't quite about the naming of acids themselves, which I think I understand, but the equivalence of your method with the one I talked about, using H2O added to the molecule. Thanks!


text:
Fluorescent polycarbonate changing optical qualities after melting
So I'm trying to figure out how to melt fluorescent poly without it changing. What it's doing after it's melted, under UV it's fluorescing a sort of opaque white instead of it's original color. Keep in mind that normally fluorescent acrylics look almost the same under UV as they do in daylight, just a little more brilliant. However after I melt the stuff I get this opaque white crap. Could someone point me in a direction to look for answers as to what's happening, why it's happening, and how to avoid it?
Thanks a heap
About how fluorescent poly is supposed to look under UV:

Are you sure it is polycarbonate? And are you clear that polycarbonate is a very different material to acrylic? Also, it might help to show a before and after for the fluorescence as that might provide some clues.

Actually I'm not positive whether it's poly or acrylic.


text:
H2O2 Storage and Package
I want to keep a medical-grade 3% H2O2 solution in a glass vial as the container for medicinal applications. It was claimed that inside of the vial should be covered with a thin layer of silicone to prevent H2O2 decomposition. Is there any need for this?

I have never seen a requirement like coating the inner surface of the glass bottle with silicone before. Hydrogen peroxide doesn't react with glass, but is prone to decomposition being exposed to the light, so switching to the amber glass might be a good idea. Out of curiosity, what is your source of information?

You may find more of a difference with stabilized H2O2 (e.g. addition of phosphates or stannates) than with a silicone layer. See http://www.h2o2.com/faqs/FaqDetail.aspx?fId=11 . Many over-the-counter brands of H2O2 contain  stabilizers... which might or might noit affect your use.


text:
Is there a crystal structure of ammonia in ionic form?
Like solid $\ce{PCl5}$, does solid $\ce{NH3}$ exist as $\ce{[NH4+][NH2-]}$ (i.e. ammonium amide)?

Now that's a different question, with an answer "normally no, but..."

According to the work of Ninet et al. published in 2014 [1], ammonia has a β-phase at 150 GPa, where it transitions into ionic crystalline form, forming $\ce{[NH4+][NH2-]}.$
References

Ninet, S.; Datchi, F.; Dumas, P.; Mezouar, M.; Garbarino, G.; Mafety, A.; Pickard, C. J.; Needs, R. J.; Saitta, A. M. Experimental and Theoretical Evidence for an Ionic Crystal of Ammonia at High Pressure. Phys. Rev. B 2014, 89 (17), 174103. DOI: 10.1103/PhysRevB.89.174103.


text:
How does one determine whether a chemical requires hazardous waste disposal?
It is pretty easy to find information about the actual hazards of a chemical in the SDS, but I have been struggling to sort out which classifications require that the compound be handled differently, especially when it comes to disposing of waste. I'd like to be able to determine what the restrictions on a chemical are by reading the SDS. For reference, I am in the USA. I know it varies state-by-state, but I am just looking for the "average" protocol, or else just the federal requirements.
For example, I calibrate/check water quality sensors. Some sensors are calibrated with NaCl, which obviously I can pour down the drain. For others I use less common/experimental stuff like p-Toluenesulfonic acid monohydrate as a stable substitute for quinine sulfate. In this case, the SDS clearly says not to pour it down the drain. What I am wondering about is how to interpret the myriad of different hazard classifications shown below, and which of them sets the legal requirement that it not be poured down the drain.
Chemical waste is not common where I work, which is why I need to be able to determine before I buy something whether I'll need to find a disposal service to handle it.

The requirements are those of the organisation to whom you hand over your waste for disposal.

Karl, there are legal requirements set by the EPA (probably others) that require certain chemicals be disposed of differently.

I've seen too many SDSs that say 'dispose of in accordance to all federal, state, and local laws.' You can't expect some company to divine where everybody is and state how to dispose of stuff in every entity in the world. Work with your university/company ES&H folks.

Please give some background to this question. You are in the US. What are you doing there. What kind of institution pays you, and pays for your chemicals. Etc. You cannot dispose of your chemical waste yourself. Whoever does it for you sets the rules. If you don´t trust them to know the laws and regulations, well, find someone else.

I think my intent is being misunderstood. I do not expect the SDS to provide disposal guidelines for the reasons you listed @JonCuster. As I stated in the question, I am just looking for the federal requirements.

Sadly, federal requirements tend to be focused on transportation, and then things like RCRA (https://www.epa.gov/laws-regulations/summary-resource-conservation-and-recovery-act). Pouring down the drain is your local or state environment department (for example, no, I can't pour even salt water down the drain - no chemicals go down the drain). Find your local/regional hazardous waste disposal company and work with them.

You may find information from the US Federal government relating to this questions at the following sites:
https://www.epa.gov/hwgenerators/steps-complying-regulations-hazardous-waste
https://www.epa.gov/rcra/resource-conservation-and-recovery-act-rcra-regulations
Furthermore, your organization (theoretically) has a local waste discharge permit that it must adhere to as well. If chemical waste is not common where you work, you or the responsible party should review the permit requirements and the operations taking place.


text:
Why does hydrazine have a higher boiling point than water?
$\ce{N2H4}$ has a boiling point of 114 °C, which is higher than the boiling point of water. Why is this so?
Both $\ce{N2H4}$ and $\ce{H2O}$ have two lone pairs, so the number of hydrogen bonds per molecule seems to be the same. Also, the $\ce{O...H}$ hydrogen bond (21 kJ/mol) is stronger than the $\ce{N...H}$ hydrogen bond (13 kJ/mol). So, given this information, why is it that $\ce{N2H4}$ has a higher boiling point?

Maybe partially due to the molecular weight difference?

H2O isn't fully using it's lone pairs,  some O atoms are getting only one, not two H-bonds

The reason is in the molar masses. For N2H4 it is 32 g/mol. For H2O, it is 18 g/mol. Heavier moles must get more energy for passing in the vapor phase.

Being a larger molecule, i suspect you can excite more vibrational modes before the hydrogen bonds break.


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Electrode potentials in a potentiostat
I am not a chemist but for a university project i am dealing with simulations concerning a potentiostat using a 3-electrode setup.
For example in the book Electrochemical Methods (Bard & Faulkner, Second edition 2001) on page 642 they give this example of a (adder) potentiostat which inputs three different potentials $e_1, e_2$ and $e_3$:

While the working electrode is grounded, point $S$ is on virtual ground and by using Kirchhoff's rule they obtain
$$-i_{ref} = i_1 + i_2 + i_3.$$
By using Ohm's law they rewrite this equation into
$$-e_{ref} = e_1 \cdot \frac{R_{ref}}{R_1} + e_2 \cdot \frac{R_{ref}}{R_2} + e_3 \cdot \frac{R_{ref}}{R_3},$$
with $e_{ref}$ being the potential of the working electrode with respect to the reference electrode.
With all resistors having the same value this can be expressed as
$$-e_{ref} = e_1 + e_2 + e_3$$
They state:

Thus the circuit maintains the working electrode at a potential equal
  to the weighted sum of the inputs

As far as I understand, the electrode of interest is the working electrode which potential should be varied (with the reference electrode providing a stable reference potential) to study its properties. But from the description above the potential of the working electrode does not change and instead the potential at the reference electrode changes.

Bard and Faulkner are outstanding electrochemists, but that potentiostat figure always annoys me because it makes it seem as though current flows through the reference electrode and that the working electrode is literally grounded. In reality, the reference electrode should conduct extremely low currents (pA, really) and the current through the working electrode is what you care about, as a function of the electroactive species in th potentiostat's solution.
Therefore, consider Fig. 15.24 and surrounding text from from Horowitz and Hill 1:

In this three wire potentiostat, the bottom operational amplifier (op amp) simply works as a transimpedance amplifier, converting current at the working electrode to voltage. In this regard, it is basically a very low input impedance ammeter, which approximates an ideal ammeter. So the working electrode is held at virtual ground, which typically means less than a millivolt from true ground.
The potential at the inverting input of the upper op amp must equal the input reference voltage, $\mathrm V_{ref}$, to within less than a millivolt. Note that $\mathrm V_{ref}$ can be from a ramp generator, a function generator or from another op amp, as in the inverting summing op amp in the figure from Bard and Faulkner. Importantly, note that virually no current flows through the reference electrode, because of the extremely high input impedance of the op amp inputs, so it will function properly and not become polarized or damaged.
The potential between the working and reference electrodes is $\mathrm V_{ref}$ because the working electrode is clamped to virtual ground and the potential of the reference electrode is clamped at $\mathrm V_{ref}$. The solution in the cell has an impedance, which is generally complex, and the solution spans the volume between the counter electrode (also called the auxilliary electrode) and the working electrode. The reference electrode is physically placed as close to the working electrode as is feasible.
Now let $\mathrm Z_{c}$ denote the solution impedance between the counter and reference electrodes and let $\mathrm Z_{u}$ denote the solution impedance between the reference and working electrodes. Then we have a non-inverting voltage follower with gain, as shown is this figure:

The output voltage, $\mathrm V_{o}$, is the voltage at the counter electrode. Therefore, $$\mathrm V_{o} = [(Z_{c} + Z_{u})/Z_{u}] \times V_{ref} \tag{1}$$
But the voltage at the junction of $\mathrm Z_{c}$ and $\mathrm Z_{u}$ , i.e., in the solution where the reference electrode is located, is $\mathrm V_{ref}$. Therefore $$\mathrm V_{ref} = [Z_{u}/(Z_{c} + Z_{u})] \times V_{o} \tag{2}$$
If solution impedances change, the upper op amp will source or sink current, as necessary, by changing $\mathrm V_{o}$, thereby maintaining the solution potential, at the reference electrode, equal to $\mathrm V_{ref}$.
1 P. Horowitz, W. Hill, The Art of Electronics, 2nd Ed., Cambridge University Press, Cambridge, ©1989, pp. 1015-1016.

Bard and Faulkner show a similar extension to the the circuit as the one you presented. Since the working electrode is grounded doesn't the statement "Thus the circuit maintains the working electrode at a potential equal to the weighted sum of the inputs" is wrong, when in fact the reference electrode is subject to change. Do I understand it correctly, that when we ignore $Z_c$ and $Z_u$ (e.g. by changing $V_{ref}$ rapidly), the working electrode is always at ground potential and the potential of the reference electrode is subject to change?

The figure in your question is messed up in several ways: the reference current is negligible and would go the other way anyway. To measure the current would require, say, a known resistor between the op amp output and the counter electrode. So I see no point in beating that dead horse. In the H & H circuit, the upper op amp provides whatever current is necessary to keep the solution at the input reference potential. No actual reference electrode, e.g., Ag|AgCl, is shown, but if that were inserted in the obvious way, then the op amp would compensate as expected. So the circuit maintains the

solution in the vicinity of the working electrode at the user-specified potential.

A colleague of my reports the page reference in The Art of Electronics is typo'd: he reckons it should be pp. 1015-1016.

Thanks for pointing out my error: your colleague is right! I just corrected the mistake. I intend to buy the 3rd edition, released a whopping 29 years after the second edition.


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Software for simulating NMR spectra
I am wondering is there any open source software that predict NMR spectra by giving the chemical shift?
Something like Spinach library http://spindynamics.org/group/?page_id=12 (Matlab is not free, so...).

Actually, this is pretty much a duplicate of https://chemistry.stackexchange.com/questions/64861/modern-open-source-tools-for-simulation-of-nmr-spectra?rq=1 I will transplant my answer there...

There is one difference, in that here OP is specifically looking for a programme where you feed chemical shifts in and get a spectrum, whereas the linked question is (presumably) looking for a programme where you feed a structure in and get a spectrum. So it is not exactly the same, to be honest. But I think it is close enough that I would not feel too bad about duplicate-closing it (it is useful for future readers to consolidate info in one question).

There is still the old WINDNMR programme developed by Hans Reich: https://www.chem.wisc.edu/areas/reich/plt/windnmr.htm which apparently should still work on modern versions of Windows.
It is not quite as sophisticated as Spinach, but if all you want is a 1D spectrum with various multiplets, then it is probably good enough.
If you have the computational resources, then there is software developed by Stefan Grimme's group https://xtb-docs.readthedocs.io/en/latest/enso_doc/enso.html, but you can't specify shifts manually, you need to generate them via a quantum chemical calculation using ORCA. Technically, all of these are free to use, but obviously it is not always practical. But in any case, if you are interested, the relevant publication is Angew. Chem. Int. Ed. 2017, 56 (46), 14763–14769. It should be possible now to run all the steps using just one ORCA input file; I suggest looking in the ORCA manual for more information.

For what it's worth, if you have the programming knowledge required, it's easy enough to write a (e.g.) Python script that is capable of simulating two- or three-spin (maybe four-spin) systems. At the end of the day, an NMR experiment is "just" a bunch of unitary operators acting on density matrices, i.e. lots and lots of matrix multiplication. For example, you could take a look at some of my MATLAB code here which simulates the first FID of a sensitivity-enhanced HSQC experiment. It is fairly trivial to port this to Python, since numpy provides you with pretty much every function you might need.
[Disclaimer: the code will not necessarily stay there forever.]
The only problem is that this scales exponentially with the number of spins, so unless you do some serious optimisation of the code (like that done in Spinach), it becomes intractable very quickly. But if your molecule can be "decomposed" into several spin systems which are small, then the overall spectrum is just the sum of the spectrum of each spin system, so you might be able to get away with simpler code. I've never thought about how you might do that, although my guess is that you would want to construct a Hamiltonian matrix, then pass it to a function which would call itself recursively if the matrix can be written in block-diagonal form.


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Tools for Molecular Simulation
I am a high school student and I don't know a whole lot about molecular simulation. I wish to view different conformers of cyclohexane and the changes that occur when substituents are attached (for example cis-1,4-di-tertbutylcyclohexane attains a twist boat shape) through molecular simulation.
The only easy-to-use tool I could find was MolView, but I can't see the above molecule getting into a twist boat shape when I ran an energy minimization because it swapped the cis to a trans form (which is in a chair conformation). Couldn't find the molecule through PubChem as well, so that I could import it.
I would also like to try out ortho effect on substituted benzene rings. Basically, I want to see a few simple molecules, (not anything like proteins and other macromolecules) and run an energy minimization on them.
So, is there any easy-to-use, free molecular energy minimization simulation program that can help me achieve this?

Usually free and easy-to-use do not go well together. You could check out QwikMD: http://scc.acad.bg/ncsa/articles/library/Library2016_Supercomputers-at-Work/Molecular_Dynamics/qwikmd-tutorial.pdf

A very simple way is using molecular mechanics: it only applies to very simple molecules, but it should be enough for your purpose. Avogadro is a free software which allows you to build structures and to minimize them via molecular mechanics, through a variety of "methods" (force fields).
The software outputs an estimate of the energy of your geometry, which you can compare to that if other geometries (for instance, boat vs chair cyclohexane).
Note that these results are very approximate, the methods work properly only in the case of neutral and covalent molecules, but anyway they provide good results for a qualitative analysis.
More accurate methods, capable of working with more complex molecules (in order of accuracy: semiempirical methods, like PM3, PM6, AM1... << Ab initio methods like HF and DFT methods like B3LYP, M06, PBE...) exist in other software packages, but this is a good starting point for learning.
Molecular mechanics works by treating angles, bonds, etc. as subject to forces (Imagine them as springs) which constrain them to an "optimal" conformation, as seen in sets of typical and simple molecules.
The other methods work by solving quantum-mechanical equations, with or without some approximations, and thus can treat "atypical" geometries which don't work with molecular mechanics.
To make a example, molecular mechanics works well for whole sets of typical molecules like our amino acids, but fails to predict the geometry of strange molecules, like tetrahedrane, which shows very atypical angles.
Within these limits, you can use molecular mechanics for free and with not much computational and software hassle

There is a proposal for a Stack Exchange site for Materials Modeling which is currently in the commitment phase. It would be great if you could help building the site by committing to the proposal.