text:
Are the derivations on Gibbs free energy consistent with the interpretation of entropy?
User Buck Thorn writes here that

In chemistry we are most often interested in the effect of changes in $T$ or $P$ on $\Delta G$ for some process such as a phase change or chemical reaction which is otherwise carried out at constant $T$ and $P$.

Does this mean that the derivations using $\Delta G$ for a reaction need not be consistent with entropy interpretations?
I have been trying to prove that for a homogeneous phase at constant composition and all the processes taking place reversibly:
$V\,\mathrm{d}P - S\,\mathrm{d}T <0$ is a criterion for "favourable change" using entropy arguments only but without any success. If Gibbs energy and entropy arguments are compatible with each other can anyone showing the above only using entropy statements?

Can you explain what is not consistent?

I have been trying to prove that for a homogeneous phase at constant composition and all the processes taking place reversibly: $VdP−SdT<0$ is a criterion for "favourable change" using entropy arguments only but without any success. This makes me feel that Gibbs energy interpretation is not consistent with entropy. Can you please prove that with entropy arguments only?

Indeed is hard to see the question. Or it negates the very need for a free energy function.

@NisargBhavsar By entropy arguments, I mean entropy of the whole universe. Can you arrive at same result by showing that Entropy of the universe increases whenever VdP−SdT< 0. If no, why one cannot do so? Any condition for "favorable change" must trace its root back to entropy of universe trying to increase. Right?                            Side Note: Sorry for editing the question after being answered since that was a very fundamental error.

In a reversible process the entropy of the universe remains constant and doesn't change

@NisargBhavsar Stating it in other terms, suppose a chemical process is taking place $A \rightarrow B$,  then the absolute quantities in the equation $VdP - SdT < 0$ gets replaced by $\Delta V dP - \Delta S dT < 0$. This is a condition for equilibrium to shift towards product B$. But how can one justify this using entropy?

@orthocresol Thanks for moderating the question. I know this is not ethical to directly ping someone, but can you please see and tell me where is the problem in my understanding. Thanks in advance!

Ah, well, in this case, my thermo is a bit rusty. I can probably do straightforward questions but this is a bit more philosophical / fundamental and I don't want to make any suggestions.

$V,\mathrm{d}P - S,\mathrm{d}T <0$ for a spontaneous process, but reversible processes are not spontaneous, they're reversible and then $V,\mathrm{d}P - S,\mathrm{d}T =0$. If you are interested in the effect of p or T on a phase transition then you write $V_1,\mathrm{d}P - S_1,\mathrm{d}T =V_2,\mathrm{d}P - S_2,\mathrm{d}T$ which leads to the Clausius equation.

Gibbs free energy is an indicator of spontaneity at constant T and p, not when you change T and p. However the equation you are interested can be used to compute how G changes with T and p.

Entropy is indicator of spontaneity in isolated systems, Gibbs free energy is the mathematical equivalent in systems inside a thermal and pressure reservoir with constant external p, T.

Entropy alone cannot decide if a process is spontaneous.
Haber's process is an good example where entropy decrease but still the process on a whole is spontaneous.
The whole idea of defining free energy was to take into account both enthalpy and entropy arguments. You cannot derive the conclusion you want using only entropy interpretations. Both enthalpy and entropy need to be considered to predict the spontaneity of a process.

Coming to the result you want to prove:
We know that a process is spontaneous when $\mathrm dG<0$.
And Gibbs free energy is defined as:
$$G = H - TS$$
Now differentiating this we can get:
$$
\begin{align}
\mathrm dG &= \mathrm dH - \mathrm d(TS) \\
\mathrm dG &= \mathrm d(U+PV) - \mathrm d(TS) \\
\mathrm dG &= \mathrm dU+ P\,\mathrm dV + V\,\mathrm dP - T\,\mathrm dS - S\,\mathrm dT\\
\mathrm dG &= (\mathrm dU+ P\,\mathrm dV) + V\mathrm dP - (T\,\mathrm dS) - S\,\mathrm dT\\
\mathrm dG &= \mathrm dq + V\,\mathrm dP - \mathrm dq - S\,\mathrm dT\\
\mathrm dG &= V\,\mathrm dP - S\,\mathrm dT\\
\end{align}
$$
Therefore for a process to be spontaneous:
$$
\begin{align}
\mathrm dG&<0\\
V\,\mathrm dP - S\,\mathrm dT &<0\\
\end{align}
$$

"The whole idea of defining free energy was to take into account both enthalpy and entropy arguments. You cannot derive the conclusion you want using only entropy interpretations."  This isn't quite correct.  The idea behind the Gibbs free energy was to identify a state function that, under the common conditions of  constant T & p, could provide a criterion for spontaneity and thus for equilibrium.  This is entirely determined by the sign/value of dS_universe (0 for equilibrium; >0 for spontaneity).  However, it's not typically possible to measure this directly, [continued]

so Gibbs designed a function that, at const. T and P,  acts as a surrogate for this.  Yes, to construct such a function he needed to include the enthlapy, but the reason is not fundamentally because enthlapy plays some role in spontaneity.   It's all about the entropy, and the enthalpy happens to play a role here only to the extent it contributes to the entropy change of the universe.  See my answer here:  https://chemistry.stackexchange.com/questions/124412/what-is-wrong-in-this-argument-that-dg-must-always-be-zero/124470#124470

@theorist This entropy mentioned in the answer is the entropy of the system and not the universe. Surely the entropy of the universe is "the" deciding factor.

It's still incorrect to say that balancing entropy and enthalpy is the fundamental idea behind Gibbs free energy.  Rather, those are a consequence of the fundamental idea behind it, not the fundamental idea itself, which was to determine a state function surrogate for the entropy change of the universe.


text:
How to find the angle of emission of an X-ray during impact of accelerated cathode rays?
X rays are produced when accelerated cathode rays strike a metal target or a metal anode.
How can we find the angle at which the X-ray is emitted from the metal anode or the target when the cathode ray strikes the metal perpendicular to the surface?
Does anyone know how?

The question lacks to specify e.g, the type of X-ray radiation (energy of exciting electrons), or the nature of the target.  In parts, arguments put forward by @Amir Kalashnikov  may be misunderstood as if concepts of classical optics (e.g., for a reflection target, as if the angle between the incident ray and the surface normalwould  equate the angle between the surface normal and the reflected ray) like seen with visible light as e.g., with the photoelectric effect would always suffice.
The short answer:
Depending on the design and mode of operation of the X-ray source, you may have X-ray radiation in all directions around the target.  What does not pass the window / collimators is leakage radiation to be attenuated by shielding by the X-ray tube, or its mount to protect an operator.  The spatial distribution of this radiation around the the target may be mapped by a point detector, e.g., a counter tube, directed toward the source.

(example of an X-ray tube with rotating anode, image credit)
The longer answer:
X-ray sources with a solid target typically either deploy a high electron acceleration to harvest X-ray radiation across a thin transmission target.  Or, with much lesser electron acceleration, use a massive reflective / thick target.  These design principles affect the spatial intensity of X-ray photons around the target.
Addressing the professionals of radiology, Kahn's The physics of Radiation Therapy outlines the physics of generation of X-rays.  While the following quote focuses on bremsstrahlung, similar properties may be anticipated for the characteristic X-ray radiation, too:

"Since an electron may have one or more bremsstrahlung interactions in the material and an interaction may result in partial or complete loss of electron energy, the resulting bremsstrahlung photon may have any energy up to the initial energy of the electron. Also, the direction of emission of bremsstrahlung photons depends on the energy of the incident electrons (Fig. 3.9). At electron energies below about $\pu{100 keV}$, x-rays are emitted more or less equally in all directions. As the kinetic energy of the electrons increases, the direction of x-ray emission becomes increasingly forward. Therefore, transmission-type targets are used in megavoltage x-ray tubes (accelerators) in which the electrons bombard the target from one side and the x-ray beam is obtained on the other side. In the low-voltage x-ray tubes, it is technically advantageous to obtain the x-ray beam on the same side of the target, i.e., at 90 degrees with respect to the electron beam direction." (Kahn, p. 34)

(The emphasis was added.)
The diagram, drawn for thin transmission targets in mind, is the one below:

(Kahn, p. 34)
For reflection targets with a normal tilted to the incident electron beam, the heel effect may distort the angular distribution of the bremsstrahlung around the X-ray target. This is because the intensity of X-ray photons generated in deeper layers of the target may be more attenuated by this very material than the by X-ray photons generated closer to the target surface.

For the interested:
Aiming for a high brilliance (i.e., high number of X-ray photons per time and beam size), both target designs are subject to modelling and simulation.  This may be to identify an optimum for (electron) accelerator potential and target thickness:

(Photon generation for different W transmission targets, Nasseri 2016)
An other example -- after defining the target's design parameters -- may be about maping the space around the target:

(Omar 2018)

(Simulated spatial photon intensity distribution for a Mo reflection target, Omar 2020a)
References

Kahn, F. M.; Gibbons, J. P. Khan’s the physics of radiation therapy, 5th edition, 2014, Wolters Kluwers.

Nasseri, M. M. Determination of Tungsten Target Parameters for Transmission X-ray Tube: A Simulation Study Using Geant4 in NEK 48, 2016, 795-798; doi: 10.1016/j.net.2016.01.006, open access.

Omar, A.; Andreo, P.; Poludniowski, G.  Performance of different theories for the angular distribution of bremsstrahlung produced by keV electrons incident upon a target in Rad. Phys. Chem. 148, 2018, 73-85; doi:/10.1016/j.radphyschem.2018.02.009.

Omar, A; .Andreo, P.; Poludniowski, G. A model for the energy and angular distribution of x rays emitted from an x-ray tube. Part I. Bremsstrahlung production in Med. Phys. 47, 2020, 4763-4774; doi: 10.1002/mp.14359, open access.

Omar, A; .Andreo, P.; Poludniowski, G. A model for the energy and angular distribution of x rays emitted from an x-ray tube. Part II. Validation of x-ray spectra from 20 to 300 kV in Med. Phys. 47, 2020, 4005-4019; doi: 10.1002/mp.14360, open access.

The angle of resulting X-ray radiation, product of the impact, is related to the amount of energy or momentum of the cathode ray. Bremsstrahlung relation will amount how much energy the produced photon has.
$$hf = \frac{hc}{\lambda} = K - K'\tag{1}$$
$$\lambda_\mathrm{min} = \frac{hc}{K} = \frac{hc}{e\,\Delta V}\tag{2}$$
Back scatter the possible photon angles with the Compton effect equation,
$$\lambda' - \lambda = \frac{h}{m_\mathrm{e}c}(1 - \cos\theta)\tag{3}$$
A difference in kinetic energy is as difference in wavelength. Precise results should consider a vacuum and use imaginary numbers; also probability distribution on superposition.
For more details refer to the chapter 3 of Krane's Modern Physics [1].
Reference

Krane, K. S. Modern Physics, 3rd ed.; Wiley: Hoboken, NJ, 2012. ISBN 978-1-118-06114-5.

You can use MathJax to typeset mathematical equations in your answers. Using images for text is not appreciated. Read this, this and this to know how to typeset them.

Adding proper reference to the source is also helpful. See,  What is the recommended style of citing on chemistry.se?

This of course assumes that the bremsstrahlung x-ray is not scattered and it ignores x-ray absorption within the cathode. The discussion also ignores characteristic x-rays which are emitted in a random direction.

I thought the Compton effect were about an interaction of radiation with outer shell electrons.  On the other hand, I understood the generation of (the [element] characteristic) X-ray radiation (like Mo K $\alpha$ = 0.72 A) as a result of relaxation, after exiting inner electrons to higher levels. What do $K$ and $K'$ represent here?

@Buttonwood Energy can stand for vibrations in bond length, or shift in angular momentum.

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text:
Appropriate ways to remove aluminum from the surface of glass?
I'm looking for a way to treat the surface of borosilicate glassware such that it won't contaminate solutions of puriss reagents in conditions that pure $\ce{SiO2}$ wouldn't, yet keep the bulk physical properties (and low cost) of standard borosilicate glassware.
What are appropriate ways (i.e. without etching or contaminating the glassware) to remove aluminum from the surface of borosilicate glassware to avoid it contaminating solutions?
One (inappropriate way) would be to use an $\ce{NaOH}$ solution I guess, but that would etch the glass to some extent, and thus expose more impurities in the glass.

One standard aluminum etch is 25:5:1 phosphoric:acetic:nitric which does not attack pure SiO2 or standard glassware.

Maybe salmi (dilute NH4+/NH3), aqueous ammonia, could do the trick?
See chem.libretxts.org - on al3+ reactions with NH3

@John: Are you sure phosphoric doesn't etch glassware?

@Lars: Isn't salmi NH4Cl?

From the link NH4OH is said to form an Aluminum precipitate, so it wouldn't help solubilize the Aluminum from the glass. But maybe NH4 in an HCl solution?

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text:
What is θ in Bent's rule equation (%s - character)?
I was learning about Bent's Rule. I came across a formula
$$\cos\theta = \frac{s}{s-1}$$
I am quite confused about $\theta$. I know that it represents bond angle. But for a compound like $\ce{PCl5}$, which bonds angle has to be considered? For the equatorial bonds, $\theta = 2 \pi/2$ was used and $\theta = \pi/2$ was used for axial bonds. My doubt is that why we are not using $\theta = \pi/2$ for equatorial bonds as they also have form $\pi/2$  angle with with axial bonds.
I asked my mentor about this. He said that $\ce{PCl5}$ has non-equivalent hybridization and $\mathrm{sp^3d}$ is actually $\mathrm{sp^2 + pd}$. The equatorial bonds are $\mathrm{sp^2}$ hybridized and have bond angle as $120^\circ$. The bond angle for axial bonds is the angle formed by the axial bond and the plane formed by chlorine atoms which connected to phosphorus through equatorial bonds.
After this, I became more confused. Can anyone please tell me the meaning of $\theta$ in the formula? Is it really the bond angle? If yes, then which angle has to be considered in $\ce{PCl5}$ and why?

Your instructor is correct.

@Nisarg Bhavsar, can you please simply what he explained to me.

https://chemistry.stackexchange.com/questions/18427/why-does-f-replace-the-axial-bond-in-pcl5

TL;DR: $\theta$ is the angle between any two equivalent bonds and using  Coulson's theorem for anything beyond the second period is out of its scope.

First things first, Coulson's theorem (the generalization of said formula) cannot be applied in cases of spndm. Therefore using $\ce{PCl5}$ as a comparison is not the right idea. Also, the formula that you've stated is only when finding the bond angle between two equivalent bonds as would be proved below.
This equation is supposedly a simplification of Coulson's theorem that states,
$$1 +\lambda_1\lambda_2\cos(\theta) = 0$$
Here, $\lambda_1$ stands for the hybridization index of one bonded orbital, $\lambda_2$ denotes the hybridization index of the second bonded orbital and $\theta$ signifies the bond angle between the two orbitals. However, this rule as said above can only be applied to spn hybridisation compounds where $\sqrt{n}$ is the hybridisation index. Its formal proof is given in Wikipedia if you are curious and want to find out.
The simplified idea behind Coulson's theorem (as I understand it), is as follows.
Each set of equivalent bonds is made up of a linear combination between s and p orbitals. This is what hybridization is. Given any pair of bonds, we can find the bond angle between the two through their hybridization indexes ($\lambda_1$ and $\lambda_2$) and vice versa. Let's take the example of $\ce{CH4}$,
$\ce{CH4}$ has four equivalent $\ce{C-H}$ bonds and we know that the bond angle is $109.5^\circ$and so according to Coulson's theorem, the hybridisation of the $\ce{C=H}$ bond can be found to be,
$$1+\lambda_\ce{C-H}\cdot\lambda_\ce{C-H}\cos(\theta_\ce{H-C-H}) = 0 $$
$$1 +\lambda_\ce{C-H}^2\cos(\theta_\ce{H-C-H})  = 0 $$
\begin{align}
\therefore \lambda_\ce{C-H}^2 &= \frac{-1}{\cos(109.5^\circ)} \\
&\approx 3
\end{align}
Therefore, the $\ce{C-H}$ bonds are sp3 in nature.

Proof of $\cos \theta = \frac{s}{s-1}$
From Coulson's theorem, we have,
$$1 +\lambda_1\lambda_2\cos(\theta) = 0$$
Now, if both bonds are equivalent in nature, we get $\lambda_1 = \lambda_2 = \lambda$, therefore,
$$1 + \lambda^2\cos(\theta) = 0$$
This implies,
$$\cos \theta = \frac{-1}{\lambda^2}$$
Now for such a molecule the hybridisation would be sp$\lambda^2$ and so percentage s-character would be
$$\text{%s character} = \frac{1}{1 + \lambda^2}$$
Solving for $\lambda^2$, we get:
$$\lambda^2 = \frac{1-s}{s}$$
Substituting this in the equation for $\cos \theta$, we get:
$$\cos \theta = \frac{s}{s-1}$$
Note: An assumption that I took here is that the bonds are equivalent. If this is in fact true, then this rule can only be used to find the bond angle between any two equivalent bonds and hence in the case of $\ce{PCl5}$, only the axial-axial ($\ce{P-Cl}$) bonds or equatorial-equatorial $\ce{(P-Cl)}$ bonds can be compared. θ then is the angle between any two equivalent bonds.

Safdar Faisal is it okay if i ask something in a chat room with you ( related to chemical bonding ) ? I will delete this soon and yeah please invite if possible as i dont know how to create one


text:
What catalysts can be used to decompose nitrous oxide?
Constraints:

I'm going to need to use a lot of this stuff in various locations, so I'd prefer it not be expensive (leaving out things like platinum, palladium, and rhodium.)

While I don't believe I can get a "room-temperature" catalyst (I'm always willing to be proven wrong!) I need this to begin the decomposition reaction at as low a temperature as possible.  This will allow electric heating to get it up to heat as rapidly as possible, using as little valuable battery or fuel cell power (I haven't decided yet) as possible.

The catalyst will be mounted near (not in!) a rocket engine combustion chamber, with a flame temperature of 3,553 K.

This will be used to decompose $\ce{N2O}$ as follows:
$$\ce{2N2O -> 2N2 + O2 + heat}$$
This will (re)light LOX/LCH4 engines in-flight.  I'm thinking about a second of $\ce{N2O}$ decomposition should touch off LOX/LCH4 nicely, then cut the $\ce{N2O}$.  Landing engines will need to be lit mid-flight for a safe touchdown, and it may be necessary to relight the lifting engines for some reason, which means I'd need to have this catalyst in at least 12 locations.
8 in landing engines and  4 either in lifting engines or 1in aerospike with redundant ignition.  I'm not married to the aerospike idea yet, I'm still trying to figure out how to Regeneratively cool the center truncated spike while allowing for a "base bleed" to minimize the effects of the low pressure recirculation area at the flat end of the spike.
If I can't crack that, it will be back to four bell-nozzle lifting engines. So, minimum twelve catalyst beds per instance.


text:
Do azeotropic mixtures evaporate proportionally?
Do azeotropes evaporate, without boiling, in proportionate amounts, or does this property describe their behaviour only at boiling point?
Concretely, if 91% isopropyl alcohol/water (by volume, i.e. an azeotropic mixture) is left open at say 25 °C, will it remain in that proportion as it evaporates, or become more dilute (until just water) as the alcohol evaporates faster?
Further, above this point, would the alcohol evaporate faster 'degrading' it until 91% at which point it behaves as the answer to above? (Which would incline me to think the answer is no, if not boiling the water and alcohol would evaporate 'independently' and the alcohol concentration would fall no matter the starting point, it's not magic.)

Only at the boiling point as only here is the mole fraction of the two species the same. The boiling point has a max or min at constant $p$ and the total vpr. pressure a max or min at constant $T$.

Azeotropic composition is a function of the boiling pressure and therefore of the boiling temperature. This can be seen in looking up the composition of the constant boiling $\ce{HCl}$ azeotrope with changing atmospheric pressure. However, information on this is hard to find and the effect seems to be small in most cases. Wikipedia does discuss this a bit. Just thinking, this might be a possible research if one had access to an FTIR with a ATR trough cell so evaporation could be observed in situ.

This is a reasonable question. Basically it asked whether the pre- or below-boiling-point vapor composition can be different from the liquid composition, although they are known to be the same at the boiling point according to the property of azeotropes. We can come up with a right answer by the following careful thinking.
You are looking at a bottle of azeotrope open to air. There is always a vapor phase even below the boiling point due to the existence of a dynamic equilibrium between the vapor phase and the liquid phase. Suppose you look at the $P-x,y$ phase diagram at $T=$ 25 °C temperature, since the system is left open to the environment, the external pressure is the atmospheric pressure $P$ = 1 atm. The reason that the system is not boiling is because the vapor pressure is too low compared to the external pressure. Only if the external pressure is reduced to be equal to the vapor pressure under this temperature, the system can be brought into boiling state (note that here we are only changing pressure $P$ because temperature is kept at $T=$ 25 °C). The key question is how much the vapor pressure is for the system at $T=$ 25 °C. In fact, it can be directly read from the phase diagram at $T=$ 25 °C, which contains the full information for the vapor-liquid equilibrium. This tells us that even if the system is not boiling, the vapor pressure is still the one that can be read from the azeotropic point on the phase diagram. Accordingly, the vapor composition is determined by the azeotropic composition, meaning that it is the same as the liquid composition, regardless of whether the system is boiling or not. Finally, leaving the system open simply means that the vapor can keep evaporating to decrease the amount of the mixture, which however would not change the vapor or liquid composition. This is a fundamental reason for taking advantage of the equal vapor and liquid compositions of azeotropes to make perfume and topical mosquito repellent, both used under their boiling temperatures.
To conclude, azeotropes should evaporate, without boiling, in proportionate amounts.  However, there is a caveat about the temperature and pressure, which can fluctuate a little bit from time to time. For instance, a mixture put in a bottle at $T=$ 25 °C may have its temperature elevated to $T=$ 37 °C when rubbed or sprayed on skin, and pressure $P$ of a sunny day can be different from a rainy day. Therefore, one composition that is azeotrope under one set of $T$,$P$ may not be azeotrope under another set of $T$,$P$.


text:
Are the names for chemical elements the same in both modern Greek and classical Greek?
Have the modern Greek names for chemical elements known during antiquity (silver, copper, mercury, lead, gold, etc.) retained their Classical Greek names, or did they adopt Latin ones, or do they follow an international convention?
I'm unsure if this is the correct place to address this question as there's no Greek SE site and meta seems a bit open ended about it.

What about the SE site "History of science and mathematics? https://hsm.stackexchange.com/

Perhaps not a lot of named ‘true’ elements in Classical Greek, depending on your cutoff for when Classical ends.

I don't understand what you are asking about. Is it about which element names had greek roots? Also are names of elements different in modern Greek (from English)?

I am voting to close the question as it's unclear what exactly you are asking for.

Relevant video. In short, "molybdenum" is named after the Greek name of lead "μόλυβδος (molybdos)". Why? Because, the mineral from which molydenum was extracted was named "molybdenite(MoS2)" deriving from that name due to it being showing similar "writing properties" as lead (both lead and molydenite left a mark when rubbed against a paper).

Ther is just a few elements known to ancient Greeks or to societies where classical Greek was a scientific-like language. Name backporting does not make sense.

The question is probably limited to the old metals known in Antiquity :  iron. copper, tin, silver, gold, lead, mercury. Are these metals known in Greece the same way today as 2000 years ago ?

@Maurice Though likely not mercury under the name of mercury, because Mercury is Latin, and Hermes is Greek.  On the other hand, TLL lemma argentum vivum mentions Pliny as one reference to use hydragyri to describe the silver that naturally [without additional heating, I infer] already is liquid and sorting materials (because of bibo, and quod excidat materias hoc et liquidum quia percurrit). I'm not aware yet if there are thesauri this large about Koine and Attic Greek as TLL is about Roman Latin.

For "old"  elements (i.e., known / discovered before the advent of a body like IUPAC), there was no uniform and universal rule.  You see this heritage by comparison of the name of an element assigned in different languages; languages in neighboring countries may but need not to share their approach.  E.g., $\ce{_7N}$: nitrogen (English), nitrógeno (Spanish), nitrogen (Nowegian Norsk bokmal); azoto (Portuguese), azoto (Italian), azote (French), azot (Polish), azotas (Lithuanian), aзо́т (Russian), άζωτο (Greek); stikstof (Dutch), Stickstoff (German); dusík (Czech), etc. for a few examples you may encounter in Europe.
For newly discovered / newly synthesized elements beyond atomic number 100, IUPAC published (back in 1979) a recommendation how to name them systematically.  For this you have the roots

0 = nil, 1 = un, 2 = bi, 3 = tri, 4 = quad, 5 = pent, 6 = hex, 7 = sept, 8 = oct, 9 = enn

(Pure & Appl. Chem. 51, 1979, 381-384, open access)
Thus it was until specific agreement that unnilunium ($\ce{_{101}Unu}$) eventually was named mendelevium ($\ce{_{101}Md}$), and IUPAC continues to publish how to name the new elements (example).
Despite this report, however, the roots are neither pure Latin, nor Greek, but convention.  Because there no tenners (like decem, vīgintī), or hundreds (like centum, ducentī), etc.  Even spelling the numeri only by position, you would expect unus, duo, tres, quattuor, quinque, sex, septem, octo, novem for $1\dots9$ for the former.
For the names eventually adopted, IUPAC set the rules that these

"In keeping with tradition, elements are named after

a mythological concept or character (including an astronomical object);
a mineral, or similar substance;
a place or geographical region;
a property of the element; or
a scientist.

[...] The names of all new elements should have an ending that reflects and maintains historical and chemical consistency. This would be in general “-ium” for elements belonging to groups 1–16, “-ine” for elements of group 17 and “-on” for elements of group 18. N.B. The present recommendation is here more specific than that
written in the 2002 document."
(Pure & Appl. Chem., 88, 2016, 401-505, open access)

I am not sure if this actually answers the question at all, but I don't know what the OP is asking for. I am pretty sure though, they weren't asking for the naming conventions of superheavy elements.

That 2002 IUPAC paper doesn't appear to be providing current info.  when it says "the names of all new elements should end in “-ium”.  Yes, the placeholder names all end in -ium, but that section isn't about placeholder names, it's about actual names.  And for actual names, it's the metallic elements that end in -ium (that qualifier is seen elsewhere in that paper, not sure why they left it out there).  Halogens, however end in -ine (e.g., Tennessine, #117); and noble gases end in -on (e.g., Oganesson, #118).

@theorist As just seen, there rules by 2002 were adjusted in 2016; the answer as edited accordingly.  So far, I did not identify an additional change since then.

@ShoubhikRMaiti The second part "in keeping tradition" reads for me like a continuation of what was done in the past, equally covering the lighter elements.  E.g., tellurium, selenium, uranium about mythology, germanium / francium about geography, etc.  Well, Tc.

Worth adding that in the scandinavian languages there are also these differences, even when e.g. Norwegian and Swedish are very similar: nitrogen is "kväve" in Swedish, the equivalent of "Stickstoff" in German, but remains "nitrogen" in Danish and Norwegian.


text:
New York Times Crossword: Is "electrically balanced" in chemistry "apolar" (given), "neutral," or something else entirely?
As a lifelong New York Times crossword puzzler, I was baffled by clue 38-Across in today's puzzle (Saturday, May 22, 2021). The solution is a 6-letter word, and the clue is:

Electrically balanced, in chemistry

The word that comes to my mind is "neutral." The accepted answer is "apolar," a word I've never heard or muttered in my life. I would have gone with "non-polar" long before I hit "apolar," and I'd be in trouble due to the fine print around "non-polar" and "electrically neutral."
The Webster's Third New International Dictionary has this definition for "apolar":

having no poles - used esp. of nerve cells formerly believed to lack processes

My go-to text for historical chemistry - M. P. Crosland's Historical Studies in the Language of Chemistry - yields no answers. From the definition above, it might be something one would encounter in biology, specifically in regards to cellular biology.
Is "apolar" an accepted or commonly-used word in chemistry as a synonym for "electrically neutral," and if so, in what context has this appeared?

Apolar is certainly a valid alternative to non-polar.
Either is natural in English as it is for, e.g. "asymptomatic", or its use in chemistry for non-polar has been introduced by non native English speakers quickly translating in their works basically the same word, "apolaire" in French and "apolare" in Italian, for instance.
Let me skip this discussion as matter for another SE site.
However, a quick Google chemical oriented search showed me that apolar is less diffused than what I expected, and seems to be more recurrent in biochemistry and astrochemistry, or as suggested in a comment, computational chemistry.
For the point more relevant to Chemistry SE,
"1) Is "apolar" an accepted or commonly-used word in chemistry as 2) a synonym for "electrically neutral," and if so, in what context has this appeared?"
For 1) see above. For 2) the answer is NOT. It just means, closer to the general definition that OP has found and reported, that a chemical entity has not (electrical) poles, i.e. no excess or partial electrical charges are present over its part.
While speaking of bonds, or simple biatomic molecules, apolar and non polar means that negligible or no electrical dipole moment is present. Still, an ensemble of polar bonds can nevertheless lead, as per vectors summation, to an apolar, or non-polar if you prefere, molecule.
While apolar chemical entities are indeed neutral, the same can be true for very polar uncharged molecules.
Another use of the terms is to classify solvents. In this case partial or overall electrical dipole moments are just one of the many factors influencing the interaction between molecules of different solvents or between them and those of a solute. The interested reader can refer to the section "Other polarity scales" of this Wikipedia article https://en.m.wikipedia.org/wiki/Solvent for more details.
(In the context of the puzzle, the given word definition is somewhat acceptable, because an apolar molecule won't present much of charge whatever finely examined, unless one discerns electrons and nuclei. And the fact that - as for the title rather than the OP text - they gave a somehow vague "electrically balanced" description, makes it quite accurate for apolar, in a crossword.)


text:
How does the membrane potential change so drastically during an action potential?
I understand that the Nernst potentials
$$V = \frac{k_\mathrm{B}T}{e} \ln\left(\frac{c_\mathrm{out}}{c_\mathrm{in}}\right)$$
for all ions inside and outside a cell barely change when an action potential travels along the cell, because barely any ions transverse the membrane compared to the amount of ions present. However, the membrane potential has an enormous change of 100 mV.
Can someone provide an intuition for that behaviour?


text:
Is this ancient explosive mixture mentioned in the Apocrypha plausible?
I understand that religious discussion is not wanted here, but it's just the pretext to my question. I recently decided to read the Apocrypha. Sounded interesting. Anyway, when I got to The Book of Bel and The Dragon, there was this verse, which struck me as particularly interesting:

1:27 Then Daniel took pitch, and fat, and hair, and did seethe them together, and made lumps thereof: this he put in the dragon's mouth, and so the dragon burst in sunder and Daniel said, Lo, these are the gods ye worship.

Now, what I'm wondering (and the only reason I'm bringing this up here), is what would this do? It seems to be an extremely early reference to use of explosives, which is fascinating in a purely historical sense, but what sort of material would actually be formed by boiling a mixture of pitch, fat, and hair? Would this have some explosive tendency, as can be interpreted from the passage? I'm a chemistry major, but I haven't even gone through Organic yet, so if any of you know anything about the properties this would (or would not) have, I would greatly appreciate the knowledge.

I'm certainly not a biblical scholar but, from reading other translations, I believe the idea here wasn't that Daniel put the mixture in the dragon's mouth, upon which the mixture exploded; but rather, that the dragon burst after consuming the mixture (the implication perhaps being that the mixture reacted with the dragon's gastric juices after consumption to produce so much gas that the dragon burst).  I.e., this seems more a reference to a chemical reaction that produces gas than to one that produces an explosion.  I'd recommend checking this on https://hermeneutics.stackexchange.com

I'll add they probably were familiar with reactions that vigorously produced gas in biblical times--e.g., humans used both vinegar and sodium carbonate since at least 3500 BC.

Thank you for your answer, I will ask this on Hermeneutics. And just to clarify, is there a process that can explain what happens when the hair, fat, and pitch combined with the dragons gastric juices? You mentioned that in the dragon, it produced so much gas that the dragon burst, I am curious about that.

Well, I mentioned the idea behind it may have been that it produced so much gas the dragon burst.   I'm afraid I can't give you a chemical reaction off the top of my head. My point wasn't to say specifically that a certain chemistry was involved, but rather that the passage may have been motivated by a familiarity with gas-producing chemical reactions rather than a familiarity with explosives.

There is nothing explosive in a mixture of pitch (long chain alkanes), fat (triglcerides) and hair even in the presence of gastric acid. Deeply indigestible though. perhaps the dragon's gut burst.

Thank you both for your answer so perhaps exploding is not the right word. But, when you say the dragons gut burst, what does that mean? Like the dragons stomach ripped open as a result of the combination?

Not explosive but probably a reference to "early thermal weapons" see eg https://en.wikipedia.org/wiki/Early_thermal_weapons

Thank you because this is starting to make alot of sense. and for more clarificaiton, is there like a wesbite or something that can either explain or like a chemical formula that can explain how the hair, pitch, and fat can kill a serpent or at the very least explains how a combination like that can be toxic?


text:
Could a 'polymer' be formed by hydrogen bonding?
For example, given you had a molecule with a carboxyl functional group on either side, could it form a continuous chain of dimers? I understand this may not fit under the definition for polymer but I lack better terminology to describe such a thing. For Example:
Alternatively;

Yeah, why not? The appropriate term for this is "catemers". See the answer to these questions: https://chemistry.stackexchange.com/questions/138865/why-dont-dicarboxylic-acids-show-hydrogen-bonding-in-spite-of-the-fact-that-car ... https://chemistry.stackexchange.com/questions/28897/what-is-a-catemer/28899#28899

The two strands of DNA are largely held together by H bonds.

The water ice can be seen as such polymer.

The crystal structure of KHCO3 contains H-bond based HCO3- dimers, while for NaHCO3 there are HCO3- polymers. Or vice versa, I am not sure now.

Since a polymer is a

"molecule of high relative molecular mass, the structure of which essentially comprises the multiple repetition of units derived, actually or conceptually, from molecules of low relative molecular mass." (IUPAC Gold Book)

joining two monomers into a dimer is just a specific wording for the more general term polymer like it is for trimer, tetramer, pentamer, hexamer, etc.  While the term polymers often is used about plastics, cellulose, starch, or chitin equally are polymers, too.  Note, this definition does constraint the binding of the monomers to covalent bonds.
The depictions shown as examples by you are valid representations when describing, e.g., the hydrogen-bond pattern in organic crystals.  Here, hydrogen bond donors and hydrogen bond acceptors may contribute significantly to the overall stabilization of the supramolecule an organic crystal may be considered.  Magaret C. Etter and Joel Bernstein popularized  formalisms to identify and systematically describe these pattern.
The one described by you matches for example the situation for $p$-terephthalic acid yielding (in principle) infinite bands hold by the motif $R^2_2(8)$ (a ring of eight atoms in total, comprising 2 donor, and 2 acceptors):

(1990AccChemRes120, includes a guide how to define these graph sets)
This approach equally may be extended to describe the interactions if the monomers are chemically different:

($R^2_4(8)$ motif about the interaction of phosphine / arsine oxides and  nitroanilines in co-crystals, reviewed in 1995ACIE1555)


text:
Why are the hydrogen-carbon bonds bent in a graphical depiction of an alkene, but are straight horizontally and vertically in an alkane?
In propane, for example, the $\ce{C-H}$ bonds are depicted by straight vertical and horizontal lines. But in propene, the same bonds are bent.
Is it a rule that $\ce{C-H}$ bonds are always bent if the carbon atoms are double-bonded? The image below illustrates what I am asking.

It is a matter of taste how you project a 3D structure onto paper. Further on you will only use skeletal formulas with implicit carbon and hydrogen atoms. There is one major rule for ACS publications regarding angles in hydrocarbons: the chain angle should be 120° (which both your structures fail to comply with).

Different level of information drawn. As a rule, chemical drawing should not convey information which are not really established. But at this level all is simplified, randomly applied but still intelligible. Note that the left structure doesn't adopt the same criterion, half is "bent" half is not. It is even not a projection. But vertical bond to H at the CH2 terminal of an alkene would look even more strange and ugly

Chemicals are 3D structures so no 2D drawing represents their real structure well. Very simple drawings {like propane above) do a very bad job and many chemists would never draw them like that, preferring a 2D picture that uses ~120° angles for all the carbon and hydrogen bonds. You should infer no significance at all in the angles in the structure you show which is just an overly simple version few chemists would use.

I think it's because of hybridization. Alkanes are sp3 hybridized and thier geometry is tetrahedral, which would make drawing their 3d structures with wedged and dashed lines extremely time consuming; alkenes are sp2 hybridized and their structure is trigonal planar which means there's a 120° bond angle. The middle carbon should also have 120° bond angles as it is also sp2 hybridized.

Fundamentally, it depends on how you put a three-dimensional structure onto a 2D surface.
The Valence Shell Electron Repulsion (VSEPR) model can predict the structure of most molecules. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsion. So with the first example (propane), each bond is equally 90 degrees from each other, and in the second (propene), the bonds are 120 degrees from each other. If the bonds were placed 90 degrees away from one another, at least one gap between electron groups would be 180 degrees, thus a less stable arrangement of electrons, regardless of if it is a double bond (c=c) a or a single bond (c-c)

The first lines are correct but do not apply to the specific case. The rest is not much relevant to the discussion which should be about molecules representation and not about prediction or energy. I suggest OP to not overthink, at least not on the bases of the example s/he gave. In there, the only message is connectivity. When necessary or confusion might rise, then yes, there are rules for chemical drawing.


text:
Why does this carbocation undergo further rearrangement?

I am trying to figure out the product of this reaction. I initially thought that a carbocation would be formed at where the chlorine leaves
from:

But according to the answer stated by the textbook, it should be this:

But why does this rearrangement have to occur? Isn't the carbocation quite stable already?
Any help would be appreciated.

Can you add the entire question with the answer too? Or you just have to form the carbocation?

As comment above. However carbocation re-arrangement is a real thing. Though this example isn't even what is commonly intended by the term. It is not clear if you where introduced to the argument. I can't believe that was done by this example, though.

@RandomQuark There is no problem statement given in my book. I am just supposed to form the carbocation. And I have stated the answer as well.

I don't know if it's actually a rearrangement or more of several resonance structures of the same species.

@ShoubhikRMaiti that it what I also think and said above. Out of nothing this is a difficult exercise. It is a kind of hyoerconjugation complicated by the visual.

@Alchimista Yes I agree with you. I think OP is confusing resonance with rearrangement (which has to involve movement of whole atoms)

I find some similarity in this case and this one- https://en.wikipedia.org/wiki/Neighbouring_group_participation, see NGP of an alkene, but I'm unable to figure out the exact answer.

This is not a rearrangement, they are resonance structures of a so-called "non-classical" carbocation https://en.wikipedia.org/wiki/2-Norbornyl_cation

@orthocresol Thanks for the link. I think I found my answer.

@ShoubhikRMaiti According to the link provided by orthocresol, I believe you are right. I asked my teacher about it yesterday, who said that its called 'Wagner-Meerwein Rearrangement'. Now that I have gone through the link, I am convinced it is just a depiction of the resonance structures.

Yes it is a depiction of several resonance structures, it's definitely not Wagner-Meerwein rearrangement, because that involved movement of an H.


text:
The color of resorcinol-formaldehyde resin
At school, I did an experiment of synthesizing resorcinol-formaldehyde resin. When I added $\ce{NaOH}$ and gave some heat, the color of resorcinol-formaldehyde resin inside the test tube became reddish-brown. Can anyone tell me why?

What do you think happen if you do the same with the green tea infusion or generally phenolic compounds? And why?

Because you have made a polymer.

@Poutnik What has the green tea to do in this problem ?

@Maurice Phenolic OH groups and tendency of their conjugate bases in alkalic solutions to 1/ get intensified red/brown colour due electron delocalization 2/ to get oxidized, condensed and further deepening of colour.

@Poutnik. So the red color is due to the oxidation of resorcin, and it has nothing to do with the polymerization with methanal. Does this happen only when there is an excess of resorcin in the mixture ?

The polymerization of resorcinol creates multiple (spaghetti-like) chains containing the aromatic rings of resorcinol.  This is a simultaneous growth in multiple directions. Eventually, the chains trap each other in the solid state. In the analogy used, it is like stirring a pot of spaghetti and then freezing the content.
It may happen that an arene ring is just in proximity of an arene ring of an other chain.  Given a growing polymer chain may be flexible, it may happen that there are intramolecular "close couples", too.  If these arenes are close enough to each other, their $\pi$-electron systems may interact with each other as if they were coupled as one.  This extension of the $\pi$-electron system may move the observable absorption band toward longer wavelengths. Then,

If there are just enough of these "close encounters" per unit of volume with the same arene-arene distance, the perception of color may be intensified (e.g., pale yellow -> bright yellow).

With variation of the arene-arene distance, the intermolecular interaction of these $\pi$-electron systems equally varies. This may yield the absorption of light of different wavelengths / the subtraction of different colors from the white light by these arenes.  Thus it may happen that the color of the polymer is not uniform. If the range of light absorption is wide enough, the polymer may appear just black.


text:
Why are there exactly 5 types of two-dimensional lattices, and what distinguishes them?
In the book Chemistry: The Central Science, they introduced the following 5 types of two-dimensional lattices:
The book said that the blue square represents the unit cell, the black dots are lattice points, and the vector a and b are lattice vectors. They don't say what the solid black lines represent. I feel as though they are redundant.
According to the illustration, hexagonal lattice is just a special case of rhombic lattice where γ = 120°. So why does a rhombic lattice have 2 types of unit cells (the blue and the green) whereas a hexagonal lattice only has 1 (and why do the solid black lines for these two lattices are different?). The same question also applies to rectangular lattice. Isn't it just a special case of oblique lattice where γ = 90°?
Why does it make sense to group lattices into 5 types, and what distinguishes them?

Remember the geometry classes for kids? Why is there such thing as a square, isn't that just a special case of rectangle? Indeed it is. Then again, isn't a rectangle just a special case of parallelogram? Indeed it is. Same thing here.

Different but related in Math SE: Are planar symmetry groups and wallpaper groups the same things?

While there may be exactly 5 types of two-dimensional lattices in a book, that's just a custom, made by people who didn't think quasicrystals could exist.

First, about the black lines. I think they are helpful for the hexagonal and the rhombic lattice. For the hexagonal system, it shows you the hexagons. One hexagon contains exactly the same space as one conventional unit cell. You just have move (translate by a lattice edge) certain areas as shown below.

For the rhombic lattice, it is nice to have the black lines when you are looking at the green cell. Without it, you would not understand why you would call it rhombic.

So why does a rhombic lattice have 2 types of unit cells (the blue and the green) whereas a hexagonal lattice only has 1 (and why do the solid black lines for these two lattices are different?).

The green cell is a centered cell, which makes the math easier because it has right angles. On the other hand, you suddenly have a translation by $½a + ½b$ on top of the lattice translations that you have the keep track of. In fact, the rhombic type is usually described as a centered rectangular lattice,  e.g. https://en.wikipedia.org/wiki/Bravais_lattice#In_2_dimensions.

The same question also applies to rectangular lattice. Isn't it just a special case of oblique lattice where γ = 90°?

Compared to the oblique lattice, all the other lattices are special cases. They aren't special cases because of some rare coincident, but because of the presence of point group symmetries, i.e. mirror planes and twofold, threefold, fourfold and sixfold rotations. A mirror plane will always allow you to choose a unit cell with a right angle. A fourfold axis will always allow you to choose a unit cell that is square, and a threefold or sixfold axis will always allow you to choose a unit cell that is of the hexagonal type. As you can see, the conventional unit cell is not a hexagon but a rhombus with a 60 degree angle. This makes the math easier.
I don't know why the illustrator chose to offset the black lines and the unit cell by half a unit cell length in both $a$ and $b$ for all lattice types (except the hexagonal one). Maybe it is for better visibility. In general, any point in the unit cell will have symmetry mates in other unit cells related by lattice translations.
If you want to see some examples of lattices that contain objects, check out this link: https://www.iucr.org/education/pamphlets/21/supplement

Above is the illustration for a lattice that contains a hexagonal axis. In this case, you have six identical objects in the unit cell, related by point group symmetry. They do no fit directly into the conventional unit cell, but it has the same volume as six objects (you can try to draw the black lines showing hexogons in your mind). The object is called the asymmetric unit, and it can have various shapes as long as it maps onto the unit cell filling the space without overlap.
The animation below highlights the object in different orientations generated by the hexagonal symmetry.

+1 very nice answer! There's also a discussion of how to think about the "centered" case (in 2 and 3 dimensions) in this answer.

+1 Just for the animation. How do you do that?

@user10186832 Thanks, I traced the bird in powerpoint, made it transparent and overlayed it on the different positions in six different slides. From there, you could export slides as images and use a service like ezgif to make a gif. Instead, I used tools I have on my computer already (not-free Camtasia to record the screen, get rid of unwanted material, and turn it into an animated gif).

Here is the powerpoint

For the square, rectangular, and hexagonal lattices, the black lines show Wigner-Seitz primitive unit cells. They contain all points closer to a given lattice point than to any other lattice point. Every lattice type has an infinite number of primitive unit cells that each contain just one lattice point. The diagrams show some of those that may be helpful for recognizing the symmetries. The green cell shown for the rhombic lattice contains 2 points and is non-primitive. The different lattice types are distinguished by the symmetries they possess that are in addition to their lattice translation symmetries. For example, the rectangular lattice contains a 2-fold rotation symmetry and 2 reflection lines, while the square lattice contains a 4-fold rotation and 4 reflection lines.

No, the black lines don't show Wigner-Seitz cells. For oblique lattice these cells aren't parallelograms: they have six sides. See this animation.

Thanks for pointing out this error. I have edited my answer.


text:
Maximum non-PV work and reversibility at constant temperature and pressure
When we talk about systems at constant temperature and pressure, maximum amount of non-PV work can be extracted if process is carried reversibly and in that case it is equal to change in Gibbs energy of the system. When system does work, Gibbs energy of system decreases and vice versa.
When we have chemical reaction in the system (for example electrochemical process in galvanic cell) than Gibbs energy of the system changes because of that reaction.
Basically, what does it mean to carry chemical reaction reversibly or irreversibly? Does carrying reaction reversibly mean that system is at chemical equilibrium where reaction Gibbs energy equals zero?
If so, how can we extract any non-PV work (in our case non - PV work is electrical work since we are talking about galvanic cell) if system/reaction is already at equilibrium?
Electromotive force of cell is zero at equilibrium and such cell can't be used to power anything (electric motor for example).


text:
Major product of the dehydration of alcohols
Well I was doing some questions and I have a bit of confusion on dehydration reaction. For instance, consider following reaction:

How do I decide which adjacent hydrogen gets attracted by $\ce{HSO4-}$?

now we have 3 possibilities:

either hydrogen 1,2, or 3 can leave resulting in 3 different products. So my question is how do I identify the major product in this case?

By analysing the stability of each product and comparing them by using comparative methods, such as comparison through hyperconjugation in the final products.

Heating gives you the thermodynamically most stable product as the major product. Which is the most stable among three you are proposing?

I suspect that ring expansion is also possible.

I think it's exercise in futility - go ahead and try it in lab, or find a paper - pretty sure at least two or three products could be the major depending on how you conduct the reaction.

The mechanism of dehydration of 2° alcohol at high temperature (around 170°C-180°C) happens via $E_1$ mechanism. This means that the major product will be decided by the $\mathbf {Zaitsev}$ $\mathbf{rule}$, which says that major product will be the stables one. If you compare the three alkenes, you'll see that the middle compound you've drawn has the most hyperconjugative structures, and hence is the stablest. Therefore, that will be the major product.


text:
Mechanism for reaction between carboxylic acid and manganese(II) oxide
I am reading about redox reactions of carboxylic acids with $\ce{MnO}$ but I couldn't find any mechanism online,so can any one provide the mechanism.

@LightYagami The patent page you linked uses $\ce{MnO2}$ as oxidising agent, not MnO

Oh, and you'd like a real mechanism, or a reaction scheme? The first is most likely unknown. The second you should try to write yourself.


text:
Boiling sodium hydroxide in stainless steel cup: Solution turning to a blue color
I boiled highly concentrated sodium hydroxide in a stainless steel cup. This created a blackish layer on the bottom of the cup and turned the colour of the sodium hydroxide solution to blueish.
Am I right to assume that there was some oxidation happening at the surface of steel? Are any oxides of metals, present in common stainless steel, known to have a blue colour when dissolved in an aqueous solution?

Is your steel cup stainless steel?

Stainless steel resists to oxidation by air and water. It may be attacked by acids and bases.

Maybe try a titanium cup from a shop that sells camping or outdoors supplies?

Do you say stainless steel? Stainless steel is an alloy of $\ce{Ni, Cr, Fe}$ with other trace elements, and owes its apparent resistance to corrosion to a protective, adherent, coating of mixed chrome, nickel, and iron oxides. A large amount is probably $\ce{Cr^3+}$, which is amphoteric and will dissolve in hydroxide solution. Once the protective coating is breached chromium will react in base similarly to aluminum. The potential to $\ce{[Cr(OH)4]^-}$ is about $\pu{+1.2 V}$. Stainless steel can be much more reactive than pure iron if the protective layer is continually disrupted.

Would that solution be blue?  I thought tetrahydroxochromate (III) was green.

@Oscar Lanzi: This probably due to $\ce{Ni}$ complex.

Very interesting indeed, thank you. Now I will need to find a more etching resistant alternative to the steel... Hm

The "color" is subjective.  To discover which metal was attacked requires  actual analysis.  The reason I suggested Cr was that I have had some experience with SS corrosion and chrome was somewhat implicated and chrome is much more amphoteric than either iron or nickel.


text:
Predicting product ratio by rearrangement in Dehydration Reactions
I came across this question recently, which would have a clear major product. The OP has given satisfactory explanation for the product ratio, but wasn't sure about the major product. When I was looking for reference for better explanation, I show the following question in undergraduate textbook (Ref.1):

Dehydration of 2,2,4-trimethyl-3-pentanol with acid gives a complex mixture of the alkenes in the indicated percentages. Write a mechanism that accounts for each product.

I: 2,3,4-trimethyl-1-pentene, 29%
II: 2,4,4-trimethyl-1-pentene, 24%
III: 3,3,4-trimethyl-1-pentene, 2%
IV: 2,4,4-trimethyl-2-pentene, 24%
V: 2,3,4-trimethyl-2-pentene, 18%
VI: 2-isopropyl-3-methyl-1-butene, 3%

I thought it would be a good practice for our readers to predict the mechanism for the obtained product ratio and give explanation for the major products. Can anybody give a reasonable mechanism for this product ratio and explanation for why $\bf{I} \gt \bf{II} \ge \bf{IV} \gt \bf{V}$, and why $\bf{III}$ and $\bf{VI}$ are in such small amounts?
Late edit: It is evident that the textbook print had an error (Thanks Nisarg Bhavsar for the finding). The compound V is actually 2,3,4-trimethyl-2-pentene, not as 3,3,4-trimethyl-2-pentene in the print. I have corrected it now. Nonetheless, the question is still interesting.

References:

Robert J. Ouellette and J. David Rawn, “Chapter 9: Haloalkanes and alcohols Nucleophilic Substitution and Elimination Reactions,” In Organic Chemistry: Structure, Mechanism, Synthesis; Second Edition; Academic Press (an imprint of Elsevier): London, United Kingdom, 2019, pp. 255-298 (ISBN: 978-0-12-812838-1).

Although this question is from textbook written by a professor in well established institute, it lacks completeness. For example, the question is written such a poor way that it even didn't give any conditions or even in which solvent the reaction has performed. So, it's safe to assume that the reaction has performed in thermodynamic control. We'll say it is an acid catalyzed dehydration reaction in a refluxing condition in a protic solvent. Thus, it'd be a $\mathrm{E1}$ elimination reaction with carbocation intermediate(s). Let's see the products and their yields:

The amount of products with significant yields suggest that the reaction has gone through few relatively stable intermediates. Let's look at these possible intermediates:

The original dehydration of the substrate gives secondary carbocation, intermediate 1.
This intermediate can gives only one product, 2,4,4-trimethyl-2-pentene $(\bf{IV})$, which is Zaitsev product (there are no possibility to form a Hofmann product from this intermediate). Since $\bf{IV}$ is not the only product detected, it is fair to say rate of this product formation is slower than the carbocation (intermediate 1) rearrangement to give more stable tertiary carbocation(s). Intermediate 1 can be stabilized $(2^\circ \rightarrow 3^\circ)$ by either hydride shift (reaction path $b$) to give intermediate 2 or methide shift (reaction path $a$) to give intermediate 3. Intermediate 3 can be further rearranged by another hydride shift (reaction path $c$) to give intermediate 4, which can be less favorably $(3^\circ \rightarrow 2^\circ)$ rearranged to intermediate 5 by a methide shift (reaction path $d$). Note that this secondary carbocation, intermediate 5, can gives only one product, 3,3,4-trimethyl-1-pentene $(\bf{III})$ with the least yield percentage (2%) to justify.
Let's see how would the products would formed by these five intermediate carbocations:

Intermediates 2, 3, and 4 are tertiary carbocations with possibility to form both Zaitsev and Hofmann products. In one glance, one might think all would give favorable Zaitsev products under the conditions. However, in reality, they have given almost equal ratio of both Zaitsev and Hofmann products except for intermediate 5, which gave only 3% of the Hofmann product, 2-isopropyl-3-methyl-1-butene $(\bf{VI})$. This result is also justified by less steric hindrance on proton abstraction from either of two isopropyl groups to form the Zaitsev product, 2,3,4-trimethyl-2-pentene $(\bf{V})$, compared to other two intermediates:

In intermediate 2, a proton must be abstracted from $\ce{C}$3, which has an enormous steric hindrance created by nearby tertiary-butyl group. Thus, Hofmann product, 2,4,4-trimethyl-1-pentene $(\bf{II})$ would be the major product from this intermediate (there are two methyl groups to give this product). However, its Zaitsev product, 2,4,4-trimethyl-2-pentene $(\bf{IV})$, is also produced by intermediate 1, and hence it could be expected that $\bf{IV}$ may have significant yield as well.
In intermediate 3, a proton must be abstracted again from $\ce{C}$3, which has almost equally enormous steric hindrance created by nearby iso-propyl group in addition to methyl group on $\ce{C}$3. The two methyl groups on positively charged carbon will play a role as well. Thus, Hofmann product, 2,3,4-trimethyl-1-pentene $(\bf{I})$ would be the major product from this intermediate (there are two methyl groups to give this product). However, its Zaitsev product, 2,3,4-trimethyl-2-pentene $(\bf{V})$, is also produced by intermediate 4 as its major product, and hence it could be expected that $\bf{V}$ may have significant yield as well.

It's worth noting that according to the product ratios, it is safe to say that intermediate 3 is the major contributor during this reaction. Even though hydride shift is faster to form intermediate 2, its relatively slow rate of double bond formation to give Zaitsev product due to the steric hindrance by tert-Butyl group may have coursed the slower methide shift to dominate at the end to get intermediate 3.

The reaction pathway I would propose for the obtained products is as below:

Coming to the percentage of each product obtained, I would defer from the data provided in the question. I would believe that V would be the major product because it is both thermodynamically and kinetically the best possible product.
I would be the next best as it is formed from the best intermediate.
Than II and IV would be the next major products as they are formed from the second best intermediate. Actually I would assume that IV would have a slight higher percentage due to the fact that it is thermodynamically more stable.
III and VI are produced in very less quantities because to form the carbocation required to form these products it has to undergo a very unfavourable rearrangement which would happen only in very less proportion.
So according to me the order should be V>I>IV≥II and very minor quantities of III and VI.

You are correct about the compound V. You can do the editing to your answer now.

@MathewMahindaratne I have edited the answer in response to the edit to the question.

@ Nisarg Bhavsar: The double headed arrows are for resonance structures. In your scheme, they are not resonances but intermediate carbocations. Hydride or methide shifts did not imply resonance.

@MathewMahindaratne Opps! They were supposed to be equilibrium signs. I am editing them

@ Nisarg Bhavsar: Actually, $\bf{IV} \gt \bf{V}$ simply because hydride shift is faster than methide shift. In addition, $\bf{IV} $ can be achieved from initial secondary carbocation as well.

@MathewMahindaratne That sounds fair as well.

The idea is that the ratio of products change with change in temperature because of varied kinetic and thermodynamic factors.

@ Nisarg Bhavsar: Sure, but this question didn't give any conditions, even solvent. So, it's safe to assume this is in thermodynamic control. Say, acid catalyzed dehydration in a refluxing protic solvent.


text:
Differences between formulae for dipole–dipole interaction energy
I came across two formulae for dipole-dipole interaction energies on Chemistry LibreTexts — Dipole-Dipole Interactions:

$$V=-\frac{2\mu_1 \mu_2}{4\pi\epsilon_0r^3}\tag{3}$$

and

$$V=-\frac{2\mu^2_\ce{A}\mu^2_\ce{B}}{3{(4\pi\epsilon_0)}^2r^6}\frac{1}{k_\mathrm{B}T}\tag{7}$$

Which of the above two is the formula for the interaction energy of dipole-dipole interactions? Also, what is the difference between the two formulae?

Where did you get the equations from? The first one hides a lot of details about the orientational dependence of the dipoles and the second one has a temperature dependence which is a bit unusual.

What's unusual about that? As temperature grows, the dipoles oscillate faster and interact less - that is, if we consider their oscillations at all, which the second formula does and the first one doesn't.

In short, there is a great deal of difference between "dipoles" and "induced dipoles".

@IvanNeretin ok I mostly had seen dipole energies given in the case when they have their positions and orientations fixed in space. So the second one is an internal energy of two point dipole free particles?

@PV. Then what is your question about? Your link says it all, and in more detail than can fit in an answer.

@IvanNeretinI wanted to know more about the T term in the second equation. First is directly understood because it is just potential term but how does the second equation come from?

They are both dipole-dipole energies as from the link but their contexts are different. Eq. 3 (as numbered in the LibreTexts link),
$$V(r) = - \frac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r^{3}} \tag{3}$$
is specifically for the case when the dipoles are aligned. The more general Eq. 4
$$V(r) = - \frac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r^{3}_{12}} (\cos\theta_{12} - 3\cos\theta_{1}\cos\theta_{2}) \tag{4}$$
is preferable because it includes the dependence on the orientation of the dipoles in space. And I believe Eq. 7
$$V = - \frac{2\mu_{\mathrm{A}}^{2}\mu_{\mathrm{B}}^{2}}{3(4\pi\epsilon_{0})^{2}r^{6}} \frac{1}{k_{\mathrm{B}}T} \tag{7}$$
is the energy derived by thermally averaging the dipole–dipole energies when the dipoles are fixed in space but are free to rotate.
So Eq. 3 is the dipole–dipole energy when the dipoles are both in a specific orientation, Eq. 4 reveals the orientational dependence of the dipole-dipole energy, and Eq. 7 is the thermally averaged energy for two dipoles that have a fixed separation $r$ but are free to rotate.

What is thermal averaging of the dipole-dipole?

A thermally averaged energy is the energy you get by summing up all the energies of all the different configurations (dipole orientations) but each energy you add into your sum is weighted differently depending on the energy of that particular configuration and the temperature.

You're welcome, the thermally averaged energy is also known as the internal energy. If you want to know more about how this averaging is done exactly I would recommend looking into a topic called statistical mechanics, the subject can be quite heavy though.

I did do statistical mechanics recently, but looks like it will take some more time to fully grasp the concepts.

I take it $\theta_i$ is the angle the axis of the ith dipole makes with the displacement vector between the two dipoles, and $\theta_{12}$ is $\theta_1-\theta_2$?

The definition can be found in the link in the question its "In this formula  θ12  is the angle made by the two oppositely charged dipoles, and  r12  is the distance between the two molecules. Also,  θ1  and  θ2  are the angles formed by the two dipoles with respect to the line connecting their centers."

An equivalent and easier formula when a molecule's coordinates are known is to use vectors. The energy is then
$$V=\frac{1}{4\pi\epsilon_0}\left(\frac{\vec\mu_1\cdot\vec\mu_2}{r^3}-3\frac{(\vec\mu_1\cdot\vec R)(\vec\mu_2\cdot\vec R)}{r^5}\right)$$
where $\vec\mu_1\cdot\vec\mu_2=|\vec\mu_1||\vec\mu_2|\cos(\theta)$ with angle $\theta$ between vectors $\vec\mu_{1,2}$. The vector $\vec R$ is the vector separating the mid point of the dipoles and assumed to be large compared to the molecules dipoles themselves. The value of $r$ is the length of vector $R$ and is the separation of the two species.
The second formula you give is that averaged over angles and Boltzmann distribution of energies. It is also called the Keesom energy and should have the relative permittivity ($\epsilon$) of the medium added in the denominator. Change $\epsilon_0\to \epsilon_0\epsilon$. This equation applies when $k_\mathrm{B}T\gt \mu_1\mu_2/(4\pi\epsilon_0\epsilon r^3)$.


text:
Why doesn't the acetic anhydride moeity react with the nucleophilic carbonyl oxygen of salicylic acid?
I know that salicylic acid with acetic anhydride under acidic conditions gives acetylsalicylic acid (aspirin). But my question is, why is it that post the protonation of carbonyl oxygen of acetic anhydride as mentioned here, the nucleophilic carbonyl oxygen of the the $\ce{-COOH}$ group doesn't attack the delta postive charge on the carbonyl carbon of the protonated carbonyl group?
What makes the phenolic oxygen more nucleophilic than the carbonyl oxygen?

Well, probably no one got an idea to ask this - not without reason - making an anhydride from another anhydride is hardly probably when you can get an ester.

It may well do so, but the resulting anhydride will react with the phenolic oxygen. If the anhydride of acetyl salicylic acid does form it does not survive work up. The phenolic oxygen is more nucleophilic because the second oxygen attached to the carbonyl carbon is electron withdrawing

Even if both groups react with acetic anhydride, the mixed anhydride will hydrolyze faster than the phenolic acetate upon aqueous workup.

@Mithoron I get your point. Why the ester will be favoured and anhydride won't can be backed-up with what user55119 said. Thank you!

@Waylander Probably what you say is right as well. But regarding the nucleophilicity of the phenolic oxygen: isn't the lone pair of oxygen delocalized with the ring's pi cloud. While in the case of oxygen of carbonyl group, the lone pair won't get involved in any sort of resonance. Shouldn't this make oxygen of carbonyl carbon more nucleophilic?

Acetic anhydride actually does react with carbonyl oxygen -- in an aldehyde. The reaction is documented in Wikipedia:

Aldehydes react with acetic anhydride in the presence of an acidic catalyst to give geminal diacetates.[1] A former industrial route to vinyl acetate involved the intermediate ethylidene diacetate, the geminal diacetate obtained from acetaldehyde and acetic anhydride:[2]
$\ce{CH3CHO + (CH3CO)2O -> (CH3CO2)2CHCH3}$

A similar reaction with the carbonyl oxygen in a carboxylic acid would yield the same product as a reaction with the hydroxyl oxygen would, so the mechanisms become hard to distinguish.
Cited References

R. T. Bertz (1953). "Furfuryl Diacetate" Org. Synth., 33, 39. doi:10.15227/orgsyn.033.0039

G. Roscher "Vinyl Esters" in Ullmann's Encyclopedia of Chemical Technology (2007), John Wiley & Sons: New York. doi:10.1002/14356007.a27_419

In acidic conditions the oxygen of COOH group is not that neucleophilic as expected because it gets protonated before in acidic medium before acting as neucleophile so the best neucleophilic center in salicylic acid in acidic medium is OH which is expected to be least neucleophilic due to indulgence in resonce with ring.


text:
Reaction of acyl chloride with excess Grignard reagent
I have a reaction to which I must find the majority organic product.

I think I found the correct product but the fact that the reaction takes two moles of propylmagnesiumbromide confuses me a bit. I added an image of the product I think would be the result.

I would like to know if I found the right product, and if the two moles of propylmagnesiumbromide means that there is a second part to this reaction.

https://www.chemistrysteps.com/reaction-of-acyl-chlorides-with-grignard-and-gilman-organocuprate-reagents/ This will help you. You can Google 'acyl chloride grignard' for more explanations.

Your product would exist in the reaction medium as the MgBr alkoxide; it is unstable. You have a second equivalent of Grignard remaining.

Correct me if I'm wrong but shouldn't the grignard reagent here attack the carbonyl twice and give 4-cyclopropylHeptan-4-ol?

Cl will leave after the first grignard attack and you get a ketone again. the next eqv. reacts to give a 3 degree alcohol.

The following paragraph from Acid Halide Chemistry[1] sums up the answer to your question:

Acid chlorides react with Grignard reagents to produce tertiary alcohols. Two equivalents of the Grignard reagent are needed because the first equivalent reacts to form a ketone which then reacts with the second equivalent. Because of the high reactivity of the Grigard reagent, the reaction can NOT be stopped at the ketone.

Coming to your specific case the reaction would proceed with the following mechanism:

Reference:
(1) Acid Halide Chemistry https://chem.libretexts.org/@go/page/183124 (accessed May 24, 2021).

To prevent dehydration of this product (because it is a tertiary alcohol) keep in mind that a neutralization (during workup) with an aqueous solution of $\ce{NH4Cl}$ may be safer than the addition of $\ce{HCl}$.

@Buttonwood I absolutely agree with you on that.

Grignard reagents aren't ionic salts, they don't have a negative charge, so technically the arrow should be drawn from the C–Mg bond.


text:
Finding the major contributing tautomer of 1-(propan-2-yl)-1,2,3,5-tetrahydro-6H-pyrazolo[3,4-b]pyridin-6-one
I am having a problem generating the correct tautomer for 1-(propan-2-yl)-1,2,3,5-tetrahydro-6H-pyrazolo[3,4-b]pyridin-6-one. I am told by my collaborator that the one I am generating is wrong.
I am not able to figure out why the structure below is the wrong tautomer.

In keto-enol tautomerism, the equilibrium is mostly towards the keto form, as it should be, since $\ce{C=O}$ is a very strong and stable bond. In some cases however we find that the equilibrium sometimes shifts to enol form. This happens in cases where some additional benefit is done by the enol structure. here are two examples:
A) Removing anti-aromaticity or adding aromaticity to a ring.
B) Extended conjugation formation, providing additional stability.
C) (1,3) dicarbonyls (due to combined effect of H-Bonding and cyclization)

In the structure you have shown, the enol form is more stable because it adds aromatic character to the ring. This additional stability drives the equilibrium
towards the enol form.

Maybe the original structure contained the motif of 2-hydroxypyridine.  If this guess is correct then a 2-pyridone is one reasonable tautomer, as shown in the first line of the illustration below.  And it seems plausible to draw such an isomer for a molecule like 1H-pyrazolo[3,4-b]pyridin-6-ol, too.

upon generating tautomers using online chemaxon tool, i got this :

@srinivasanranga I wasn't yet aware about this generator you possibly refer to.  Thank you for this hint.  If so, however, pasting the first structure of 1H-pyrazolo[3,4-b]pyridin-6-ol, the program displays a preference for the pyridone motiv ($\ce{C=O}$ and $\ce{NH}$).  It is not so clear for me how the program quantifies the results down to a hundredths %, but it is really nice that it lists more than only one alternative.

I would say the enol form is favored because it establishes aromaticity of the benzene ring but the fused five-membered ring does not tautomerize as all its other atoms are sp3-hybridized (and therefore unable to do so).

This assigns much energetic stabilization to aromaticity. This indeed is, comparing cyclohexene, cyclohex-1,3-diene, and (a hypothetic) cyclo-1,3,5-triene against benzene a valuable tool. Note, comparing the equilibria for i) 2-hydroxypyridine vs. 2-pyridone in water, and ii) 2-aminopyridine vs. its imine in water suggest to equally consider hydrogen bonding. According to the Tautobase paper and its public SI (included e.g., in DataWarrior the $\log K$ is +3.5 for the former, and -6.5 for the later.


text:
Why is the silver salt ammoniated to form the Tollen's reagent?
When doing a distinguishing test between aldehydes and ketones using Tollen's reagent, is it necessary to convert $\ce{Ag+}$ ions into $\ce{Ag(NH3)2+}$ in order to oxidize the aldehyde?

Does this answer your question? Purpose of ammine in Tollens' reagent

Yes, it is important to convert your silver ion to diamminesilver(I) complex. If you do not add ammonia to silver nitrate, the silver ion is reduced so quickly that colloidal silver metal would appear and the solution would become a black, cloudy liquid and you won't get the silver mirror when it reacts with aldehyde.
It is also important that you maintain the pH of the reaction. During the preparation of Tollen reagent, you will see the use of sodium hydroxide. This is very important for maintaining the pH. During the whole reaction, the pH is between 8-10 and it goes even faster if the pH is beyond 10. The reaction also becomes more sensitive. This has been explained in details in this journal by going through the exact mechanism of the reaction.

Since a black cloudy liquid is produced, would it mean that Ag+ can still be used to distinguish between aldehydes and ketones? The only difference is that now the observation is no longer a silver mirror

@john No, it will create a false indication. The clear indication for the presence of aldehyde is the formation of mirror. If you are using only silver ions, then it will give the cloudy solution irrespective of the organic compound you are using. So, you can't distinguish aldehyde from the other organic compounds. So, presence of ammonia and basic pH is important for the formation of silver mirror.


text:
Does adding NH3 to an HCl solution facilitate the dissolution of aluminum compounds?
If we compare the two solutions hereafter, which one would better leach/dissolve aluminum metal/compounds present at/near the inner surface of the glass:
Solution A: $10\,\%\, \ce{HCl + H2O}$
Solution B: $10\,\%\,\ce{HCl + 1\% NH4OH + H2O}$
I'm asking because I'm looking for a solution to leach away aluminum (and other lesser trace metal ions) from the surface of borosilicate glass such that it won't contaminate puriss solutions in the glassware, and one suggestion was to use "salmi" (which I believe is $\ce{NH4Cl}$), so I'm now wondering if $\ce{NH3}$ in solution somehow facilitates the "grabbing" / solubility of aluminum in an $\ce{HCl}$ solution.

I don't think so.

Include the primary goal and the question background to prevent wrong assumptions and requests for clarification. The otherwise proper answer to the provided question may not lead you closer toward reaching your goal, if it is not known to readers.

I would suggest to remove "/compounds" part leaving only aluminum metal: different compounds react differently, and an answer cannot cover them all. Also, I'm not sure solution B is a thing; at least, there is a charge missing on ammonium.

@andselisk: Thank you for you suggestions. I changed NH3 to NH4OH. I wrote compound because I must say I ignore what form(s) is taken by aluminum when part of borosilicate glassware and at the surface of it (maybe a mix of Aluminisilicates / Oxide(s) / Hydroxide?).

@Poutnik: The primary goal is to "leach away aluminum (and other lesser trace metal ions) from the surface of borosilicate glass such that it won't contaminate puriss solutions in the glassware". Is this enough? If not, please tell me what additional details would be crucial to include.

If you want to eliminate other metals than aluminum, you could use a solution of ammonia, because copper, nickel, zinc and other metals make soluble complexes with ammonia.

I guess the primary goal is: not to affect the composition of those solutions by the container material. What kind of puriss solutions ? Glassware may not be the preferred material at all.

Aluminum metal is attacked by $\ce{HCl}$ solutions. But adding $\ce{NH3}$ or $\ce{NH4Cl}$ has no effect on aluminum reactions.
The same thing can be said for aluminum compounds. Adding ammonia or ammonium compounds is not recommended, because it does not help. Aluminum does not make complexes with ammonia or ammonium salts.
But I suspect this addition of ammonia or of ammonium compounds may have another effect. It may help dissolution of other metals if they are deposited on the glass together with aluminum, making complexes containing the metal and ammonia.

Thank you Maurice! So an HCl solution alone is enough to leach out any Aluminum from/near the surface of the glass? I've read online that HCl solutions dissolve aluminum but I don't understand why because I would expect an aqueous solution to turn AlCl3 into Al(OH)3 for which I found a solubility of ~2x 10^-8 g/100ml (20C).

Do you think an aqueous solution of NH4 + HCl would be better than HNO3 for some other impurities? And is it safe to assume a small amount if NH4 would NOT interfere much with the leaching of aluminum by HCl?

@Hans. First $\ce{NH4}$ does not exist. Only $\ce{NH3}$ and the ion $\ce{NH4^+}$ do exist. Second. using $\ce{HNO3}$ instead of $\ce{HCl}$ would depend of the nature of the impuritiy. Third I don't know what you mean with the question about the safety of adding ammonia.

@Hans. Concentrated $\ce{HCl}$ solutions dissolve $\ce{AlCl3}$ to form $\ce{HAlCl4}$, or the ion $\ce{AlCl4^-}$ . The hydroxide $\ce{Al(OH)3}$ is not formed in $\ce{HCl}$ solutions.


text:
Sound absorption in carbon dioxide with temperature
I am doing speed of sound measurements in $\ce{CO2}$ using ultrasonic pulses at 125 kHz. $\ce{CO2}$ is particularly difficult because it strongly absorbs ultrasound due to its multiple vibrational modes. The problem we have discovered is that absorption increases substantially as temperature rises. Is this expected, and if so is there a simple relationship between temperature and absorption coefficient?

It is not clear to me how the ultrasound translates into the excitation of vibrational modes (though apparently there is a way). That being said, yes, it is expected for more modes to "unfreeze" as the temperature rises, and no, there is no simple relationship.

Molecular vibration energy is quantized with quite big quantization steps, so it occurs significantly at elevated temperatures only. The energy needed to increase vibration is kind of composed in your case from thermal energy of molecules and the energy of sound wave. As result, higher the temperature = higher probability the sound wave will cause absorption of sound energy, causing molecule transition to higher vibration mode. It may be difficult to quantize it, more easier can be to get experiemntal data or empirical expressions. Try engineringtoolbox.

The acoustic pulses must be made up of several frequencies but all these seem very low compared to vibrational frequencies of molecules, possibly you are causing rotational levels to be excited, these are far more numerous than vibrational levels and also of much lower frequency and so their population is very sensitive temperature changes.  However, absorption even with rotational levels is in the terahertz region.

@porphyrin Even so, CO2 is a strong absorber of sound all across the acoustic spectrum, becoming extreme at frequencies above about 300kHz At 125kHz we are getting a 5x attenuation over 120mm

I don't doubt your experimental data. I'm no expert here, but as the sound wave passes does it not cause compression  and rarefaction and these cause energy loss (heating?)  if there are  intermolecular interactions in the gas caused by by van-der-waals /dispersion forces.

The acoustic wave is accompanied by a temperature oscillation. To a first approximation, the temperature oscillation is adiabatic and reversible in the bulk, but irreversible near solid boundaries. In many cases, boundary losses dominate (for example, in musical wind instruments), but as the frequency increases, and the surface-to-volume ratio decreases, bulk irreversibility may become important. Bulk losses in single-component polyatomic gasses are almost always due to loss of local thermodynamic equilibrium when the temperature oscillation describing internal degrees of freedom lags the temperature oscillation describing translational degrees of freedom.
A full discussion of these effects is given in the book "Ultrasonic Absorption" by A.B. Bhatia. His figure 5.1 shows that the attenuation for $\ce{CO2}$ peaks at about 30 kHz (STP conditions), and his figure 6.3 shows the velocity dispersion that occurs at about the same frequency. The dispersion is a large effect, with the velocity increasing by about 4% as the frequency goes from below to above the attenuation peak. The attenuation and dispersion occur when the acoustic frequency matches the rate at which the vibrational mode temperature relaxes to the translational temperature. The rotational modes relax much faster than the vibrational modes, so they do not play an important role under these conditions.
The relaxation time for the vibrational temperature will generally decrease with increasing temperature, so the attenuation peak will move to higher frequency. I am not sure what temperatures you are exploring, but if your frequency is above the attenuation peak, then you will see increasing attenuation as the temperature increases.


text:
3,4-b nomenclature for an organic compound
In this question, the compound below was named 1-(propan-2-yl)-1,2,3,5-tetrahydro-6H-pyrazolo[3,4-b]pyridin-6-one.

I assume that we can split this into three parts.

The propyl group attached to the nitrogen, namely: 1-(propan-2-yl)- (basic part of nomenclature)
The pyrazole derivative like ring, namely 1,2,3,5-tetrahydro-6H-pyrazolo (For this refer to this answer)
The pyridine ring, namely [3,4-b]pyridin-6-one.

The third part is where I am confused, what does the [3,4-b] stand for?
Another example for said nomenclature is in
Acdlabs

This leads me to assume the b stands for the fusing of the two bicyclo rings. But then, why [3,2-b] and why does the numbering not count the bridgehead carbons?

The bracketed letters gives you "which one is the face ip". When you merge the rings the second one can be fused head or cross up with respect to the other one. I hope it helps else I can better formulate it....also https://www.acdlabs.com/iupac/nomenclature/79/r79_960.htm#:~:text=3.1%20%2D%20%22Ortho%2Dfused%22,%2D1%20and%20B%2D2.

look also here https://link.springer.com/article/10.1007/s40828-016-0035-3. a, b... refers to the bond fused

And the number in bracket refers to the original numbering. The entire molecule is afterwards renumbered.

@Alchimista, according to the paper you linked "The first face described by the numbered positions 1 and 2 is lettered ‘a’, with the letters continuing alphabetically clockwise around the exterior of the parent", then how do we get the second example? shouldn't it be [3,2-d]?

given the molecule there is no need to go to d. Thieno[2.3-b]furan would be that with the heteroatoms "on the same side".


text:
Reason for three phases in total fatty matter test on soap
I am currently trying to perform the Total Fatty Matter (TFM) test on soap and for some reason I am getting three phases instead of two.
The surfactant used is SLES and other reagents include hot water, $\ce{HCl}$, petroleum ether.
I have no idea as to what could be causing the 3 phases and my knowledge in organic chemistry is very little. Any help would be greatly appreciated. Below is the procedure:
I dissolve soap in hot water and then I transfer the cooled solution to a decantation flask. I add concentrated $\ce{HCl}$ (37% stock solution) and then petroleum ether to increase extraction ratio.
I repeat this extraction two more times (and after this I am getting 3 phases) and then I purify the organic phase twice using hot water to eliminate any $\ce{HCl}$.
I then heat the organic phase to vaporize ether and calculate TFM %.
The surfactant used in the soap is sodium laureth sulfate (SLES), and I think that it may relate to why I have 3 phases but I can't tell why. What is the reason for this?

@Waylander See https://chemistry.stackexchange.com/questions/151943/three-phases-in-tfm-test-instead-of-two


text:
Titration of H+ by iodometry
Propose a titration protocol for a hydrochloric acid solution of approximate concentration 0.1M using the solutions and materials available.
Solutions:
-$V_0$ = 20mL of hydrochloric acid solution of unknown concentration.
-Solution of $\ce{KIO3}$ at $C_1$=0.1 mol/L
-solution of $\ce{KI}$ at $C_1$=0.1 mol/L
-$\ce{Na2S2O3^2-}$ solution at $C_2$ = 0.01 mol/L
-concentrated sulphuric acid
Equipment: - common glassware
-no pH meter, conductivity meter or colour indicators
Data :
-Experiment 1: a solution of $\ce{I2}$ to which a few drops of soda are added becomes colourless
-Experiment 2: a solution of iodate and iodide ions to which a few drops of sulphuric acid are added becomes brownish-orange.
-Standard redox potential = E°($\ce{S4O6^{2-}}$/$\ce{S2O3^{2-}}$)=0.08V
-E-pH diagram of iodine plotted for the species: $\ce{I2}$, $\ce{IO3^-}$, $\ce{I^-}$

My ideas: Add excess $\ce{I^-}$ and excess $\ce{IO3^-}$ to the hydrochloric acid soluton. We then have $5\ce{I^- }+ \ce{IO3^-} + \ce{6H^+} = 3\ce{I2}+ 3\ce{H2O}$. The constant for this reaction can be calculated by retrieving the associated thermodynamic constants from the diagram. The $\ce{I2}$ formed is then titrated with $\ce{S2O3^{2-}}$. The $\ce{H^+}$ concentration in the initial state can be deduced from this. However, I think that this method is not good because the first reaction is largely total and the colorimetric titration is not precise enough to determine the correct values.
Is there another method that is simpler or more accurate?

Hello, welcome; please do check out FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange? — in particular the mhchem sections: the chemical formulae and equations in your post should be typed as $\ce{Na2S2O3}$, i.e. $\ce{Na2S2O3}$ or \ce{S2O3^2-} i.e. $\ce{S2O3^2-}$.

I think it is just an academic exercise and all your teacher wanted is that you write a balanced equation showing the reaction of iodate, iodide and HCl. Iodometry is/was never meant to determine hydrogen ion concentration. It just like asking you to re-invent a triangular wheel.

@M.Farooq so the method I used is theorically right but experimentally false ?

@Julien, Yes sort of. Good for paper work but no one will do it in reality. So your proposed solution is perhaps the only solution to satisfy your instructor.

Iodate-iodide can be used to standardize STRONG acids because the reaction is very pH dependent and stops abruptly at ~pH 7.  I can't find a reference. It was in my Quant analysis Text which was lost when my company moved the lab. Simply add a slight excess of an iodate-iodide solution to a measured amount of acid and titrate the iodine with thiosulfate. No need to weigh anything if you buy standard thiosulfate but iodate can also be used as a primary standard.


text:
Three Phases in TFM test instead of two
I am currently trying to perform the Total Fatty Matter (TFM) test on soap and for some reason I am getting three phases instead of two.
The surfactant used is SLES and other reagents include hot water, HCl, petroleum ether.
I have no idea as to what could be causing the 3 phases and my knowledge in organic chemistry is very little. Any help would be greatly appreciated. Below is the procedure:
I dissolve soap in hot water and then I transfer the cooled solution to a decantation flask. I add concentrated HCl (37% stock solution) and then petroleum ether to increase extraction ratio.
I repeat this extraction two more times (and after this I am getting 3 phases) and then I purify the organic phase twice using hot water to eliminate any HCl.
I then heat the organic phase to vaporize ether and calculate TFM %.
The surfactant used in the soap is sodium laureth sulfate (SLES), and I think that it may relate to why I have 3 phases but I can't tell why. What is the reason for this?

could you explain your setup for the experiment? Right now we have nothing to go on other than the few reagents and issue you have.

@SafdarFaisal I dissolve soap in hot water and then I transfer the cooled solution to a decancatation flask. I add concentrated HCl (37% stock solution) and then petroleum ether to increase extraction ratio. I repeat this extraction two more times (and after this I am getting 3 phases) and then I purify the organic phase twice using hot water to eliminate any HCl. I then heat the organic phase to vaporize ether and calculate TFM %. The surfactant used in the soap is sodium laureth sulfate (SLES), and I think that it may relate to why I have 3 phases but I can't tell why.

@SafdarFaisal if you have any other questions or need any clarification please do ask, your help is really needed

Your message is not clear. When have you obtained three phases in your process ? Is it after adding petroleum ether to the result of hydrolysis of soap with HCl ?

Are all 3 layers clear? I think what you may have is an aqueous phase, a petroleum ether phase and a phase containing the long chain acid which is insoluble in either.

Soap is a fatty acid salt of an alkali metal, typically sodium or potassium. The fatty acid itself is always insoluble in water, and therefore acidification of the soap solution produces two layers, a lower aqueous layer with the alkali salt of the acid and an upper layer of fatty acid laced with aqueous material. This can be treated with an organic solvent to dissolve the fatty acid and expel the aqueous material, giving two immiscible layers.
I suspect that your soap is a combination of fatty acid soap plus SLES.
Sodium laureth sulfate is not a soap, it is a surfactant, and it is so constructed so as to be more soluble in water than sodium laurate (which is a soap) or sodium lauryl sulfonate (which is a surfactant) by addition of typically 3 -OCH$_2$CH$_2$O- (ethylene glycol) groups. This complicates the precipitation and extraction; another variable is the solvent: while petroleum ether is organic, it has poor solvent power, and long chain acids are not well dissolved. That's why you have to do the extraction many times.
I visualize your experiment to go this way:

When you add the concentrated HCl to the sodium laureth sulfate, you insolubilize the HLES (the acid form), but it is not easily separable. A variable here is whether there is excess acid and how much. Another variable is that the precipitation is incomplete.

You add petroleum ether to extract the HLES and reject water and excess acid, leaving, hopefully, mostly NaCl in the aqueous phase. You do this again, extracting more HLES.

However, HLES is somewhat soluble in water (because it contains 3 -OCH$_2$CH$_2$O- groups), and is quite a good emulsifier for organics, and pet ether has a lot of low mw molecules, so before the last of the HLES is extracted, you find that the HLES has attracted a lot of water and a lot of pet ether. But this combination cannot contain all the water, so it is immiscible with the lower aqueous layer; it contains too much water to be miscible with the pet ether, so it is a third phase. Excess acid in step 1 complicates things because acids increase the solubility of ethoxylated surfactants.

You get something like this:

So instead of dealing with a soluble/insoluble situation, you have a significant but incomplete precipitation of HLES, and what doesn't precipitate out manages to emulsify with some of the pet ether and create the third phase.
A possible way to improve your separation is to add NaCl to the mixture after adding the HCl to maximize the salting-out effect. Another adjustment is to minimize excess acid (add just enough acid to precipitate the HLES - measure pH with paper) or add NaOH to adjust pH to near neutral (pH 6-8), when you get the third phase. This minimizes acid solubilization of the HLES, but keep the aqueous layer saturated with NaCl by adding solid NaCl if necessary.
You could also substitute toluene for the pet ether, but there is somewhat more hazard. Diethyl ether would be the extractant of choice many years ago, but alas, we must emphasize safety.

Ethyl acetate would be a good extraction solvent