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Decalcifying metal tubes using organic acids
I have a water that is saturated with Calcium Magnesium Carbonate $\ce{CaMg(CO3)_2}$ (Dolomite). This precipitates in particularly metal pipes and whenever the water is heated. Instead of ion exchange (Sodium for the group 2 metals), reverse osmosis or deionisation, how about reducing the pH to say $\pu{6.6 - 6.9}$ with an organic acid such as Acetic acid or Citric acid? Would that corrode the $\ce{Fe/Zn/C}$ containing surfaces whilst the more electropositive elements are still present?

Reducing the pH of tap water to $6.6$ or $6.9$ will not corrode pieces of Zinc or Iron metals. Even pH $5$ is not sufficient.

There are no more electropositive elements here.

You got it backwards: increase the pH and the hydroxide salts $\ce{Mg(OH)_2}$ and $\ce{Ca(OH)_2}$ - which are insoluble in water - precipitate. Wash away with circum-neutral pH water. This is one of the ways it's handled on a commercial or city-wide scale.

A water softener should prevent future scale deposits . It will replace Ca and Mg with Na.

@Todd Minehardt: increasing the pH will ppt the carbonates, since the soluble species in the hard water are bicarbonates. Probably a good way would be to add Mg(OH)2, which is essentially insoluble to begin with. it would convert itself (and the bicarbonates) to carbonate (which is actually more soluble!), but would than not ppt when heated. The resultant pH might be a bit bitter (~9.5-10, not quite 10.5, because of dissolved MgCO3).

@JamesGaidis We actually deal with this all the time here in central Texas. Adding any soluble salt of magnesium or calcium increases the hardness of the water, which is not acceptable. What we do is precipitate the hydroxides of magnesium and calcium by increasing caustic alkalinity with potassium hydroxide. Thus, the hydroxide salts precipitate and hardness decreases. What we're left with is high-pH drinking water with low hardness. Here is the 2021Q2 water quality report.

$6.6$ is neutral for pretty much all purposes. So is $6.0$. So is $5.0$. You need to go way lower than that. Table vinegar may have $\rm{pH}\approx2.5\dots3.0$, and that's what you want to use to hear that gentle fizz of dissolving carbonates.
Yes, this is a viable method. No, it will not corrode steel pipes to any significant extent, provided that you apply it for minutes, not days, and flush the system with water afterwards.
So it goes.

If by "metal", you mean steel: Our water guy recommended sulfamic acid for cleaning. I don't know the chemistry but this is old, standard technology. Water treatment companies like Nalco could have information and supplies. NACE ( national association of corrosion engineers) would have a variety of references ; for a price. We used citric acid and ultrasound for cleaning steel and causing no damage to the steel.


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Why is there a hydride shift in the hydration of 1-methyl-5-phenyl-1,2,3,4,5,6,7,8-octahydronaphthalene?

I was attempting a question here in which you had to find a product after the hydration of alkenes.
Can someone explain why was there a hydride shift in this reaction?
According to me, initially, the positive charge should be more stable as it is tertiary and has 4 alpha hydrogens for hyperconjugation effect whereas, after the shift, the positive charge occurs on phenyl which should be unstable.

after the shift, the positive charge occurs on phenyl which should be unstable.

Are you sure about that? Firstly, the positive charge isn't 'on' the phenyl, it's on the Carbon to which the phenyl ring is attached and hence it enables resonance.
Out of many of the factors we look for the stability of a carbocation (which include the ones you mentioned like positive inductive effect in a tertiary Carbocation, Hyperconjugation, )
Resonance takes a major chunk of the competition along with hyperconjugation of the neighbouring Carbons which is what Oscar pointed out as well.
In general, the Carbocation formed is highly stabilised due to resonance with the whole phenyl ring so there's no problem in forming such a carbocation.
When you'll try and draw the resonating structures, you'd realise that the delocalisation of the charge throughout the ring stabilises it.
Thus,

Resonance
Hyperconjugation with 3 alpha Hydrogens
Tertiary Carbocation

All these factors ensure high stability of this carbocation

Is resonance effect better than hyperconjugation in terms of stbility of carbocation?

In general, yes. Resonance is usually the major stabilising factor followed by hyperconjugation. Especially when there's resonance with a phenyl group, the number of resonating structures and a great extent of delocalisation of charge highly stabilise the carbocation.

Besides that, you don't really lose a lot of hyperconjugation. The rearranged carbocation is still tertiary as well as being benzylic.

Also could you please tell me if the positive charge was present directly on phenyl,would it be unstable?

Yes. For stability through resonance, You need conjugation of pi bonds, charges, etc. But when it's directly on the vinylic carbon, ( i.e. the double bonded carbon) there is no conjugation so resonance is ruled out. Also, the carbon here being $\ce{sp2}$ is more electronegative(since, more s character) which instead destabilises the carbocation by pulling electron density from an already electron deficient region

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Does this molecule show cis-trans isomerism?

I want to find if the above molecule shows cis-trans isomerism, per an assignment. I am assuming that such isomerism, if present, will be shown by the substituted chlorines.
It seems clear to me that the two chlorines in the picture lie in different planes. Therefore, the molecule will show optical isomerism, not cis-trans. However, the answer key says that it shows both optical and cis-trans isomerism.
I think the answer key is wrong. But if it is right, could someone please explain to me how this molecule shows cis-trans isomerism?

Please see duplicates, especially the second one. There isn't a clear answer, it's more of semantics / a technicality, rather than any real useful chemistry concept.

Just open the cyclohexane ring (similar to the pattern shown here and here) and apply the CIP rules to assign the double bond.  Not evident from the drawing (missing wedges), though, is the configuration around the stereogenic centre.

The ring is in the plane perpendicular to the double bond. Can you see now why it can show geometric isomerism?

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Why does a larger bond angle correspond to greater s-character?
In trialkylamines $\ce{NR3}$, the $\ce{R}$ alkyl group attached to nitrogen increases the electronegativity of the nitrogen atom. I also found in this question that this happens due to the repulsion between the lone pair of the nitrogen atom and the $\ce{R}$ group, which increases the bond angle and hence the s-character. (This has a direct link with Bent's rule.)
Why does the increase in bond angle increase the s-character of the compound? Please provide the orbital structures with the explanation.

This may help https://en.wikipedia.org/wiki/Isovalent_hybridization

@Rishi Does it mean the increase in bond angle shortens the bond length??

are you asking for the explanation for Bent's rule?

https://chemistry.stackexchange.com/questions/151863/what-is-%ce%b8-in-bents-rule-equation-s-character/151868#151868

@SafdarFaisal This only provides mathematical proof. What is the theoretical explanation behind it?

hybridisation is mathematics. It's the linear combination of s and p. what you are asking for is the proof for coulson's theorem..

@SafdarFaisal Can u provide a link with a precise explanation of that theorem if that explains my doubt?

https://en.wikipedia.org/wiki/Bent%27s_rule#Formal_theory

In symmetric trialkylamines, $\ce{NR3}$, the bond angle under consideration is $\ce{\angle (R'-N-R)}$, where $\ce{R' = R}$
In this answer, it has been proven using Coulson's Theorem, that for the bond angle between any two equivalent bonds, we can use the formula,
$$\cos \theta = \frac{s}{s-1}$$
A brief explanation of Coulson's Theorem and how it is used to prove this property follows.
Coulson's Theorem states that the bond angle between any two bonds can be written as
$$\cos \theta = -\frac{1}{\sqrt{\lambda_i\lambda_j}}$$
The way this is proved is as follows,
Any bond can be written to be a linear combination of $\mathrm s$ and $\mathrm p$, orbitals. Since $\mathrm s$ and $\mathrm p$ are two wavefunctions that are orthogonal to each other, we state that the hybridised orbital formed by their linear combination to be $\psi = N(\mathrm s +\sqrt{\lambda}\mathrm p)$, where $N$ is the normalization factor. In such a case, we say that the orbital in question is $\mathrm {sp}^{\lambda}$ hybridised. Therefore in our case we can say that the $\ce{N-R}$ bond and $\ce{N-R'}$ bond are $\psi_R = N_1(\mathrm s +\sqrt{\lambda_1}\mathrm p_1)$ and $\psi_{R'} = N_2(\mathrm s +\sqrt{\lambda_2}\mathrm p_2)$.
Now, we know that $\mathrm s$ has no direction and $\mathrm p$ is directional. Therefore, the angle between the $\mathrm p$-orbitals will be the bond angle $\theta$.
Also, the two bonds should not have anything in common. i.e. they should be orthogonal to each other. This implies that the inner product needs to be zero.
The inner product for two vectors a and b would be $\left \langle a | b \right \rangle = |a||b|\cos(\theta)$ which is equivalent to the dot product in this case.
Therefore, we get,
$$\left\langle N_1(\mathrm s + \sqrt{\lambda_1}\mathrm p_1)|N_2(\mathrm s + \sqrt{\lambda_2}\mathrm p_2)\right\rangle = 0$$
Simplifying and expanding, we get,
\begin{align}
\left\langle N_1(\mathrm s + \sqrt{\lambda_1}\mathrm p_1)|N_2(\mathrm s + \sqrt{\lambda_2}\mathrm p_2)\right\rangle &= \left\langle \mathrm s|\mathrm s \right\rangle + \left\langle \mathrm s|\sqrt{\lambda_1}\mathrm p_1 \right\rangle + \left\langle \mathrm s|\sqrt{\lambda_{2}}\mathrm p_2 \right\rangle + \left\langle \sqrt{\lambda_1}\mathrm p_1|\sqrt{\lambda_2}\mathrm p_2 \right\rangle \\
0&= 1 + 0+ 0+ \sqrt{\lambda_1\lambda_2}\cos(\theta)
\end{align}
This is because $\left\langle \mathrm s|\mathrm s \right \rangle$ is basically the magnitude of $\mathrm s$ which is equal to one and $\mathrm s$ is orthogonal $ \mathrm p $ making those inner products zero.
Finally, we end up with Coulson's theorem which states (as mentioned before),
$$1 + \sqrt{\lambda_1\lambda_2}\cos (\theta) = 0$$
A more detailed and expansive proof has been given by Martin - マーチン here in How is Bent's rule consistent with LCAO MO theory?
A worked out example of how Coulson's theorem can be applied to find the hybridization of the carbons in cyclopropane here by ron
Hybridization is basically the mathematics of superposition of wave functions in order to generate hybridized orbitals and hence it can be explained in the best way mathematically.
Using the first equation mentioned under the assumption that $\lambda_1$ = $\lambda_2$, we can see that as the bond angle increases, the value of s-character will also increase.
A brief explanation of equation (1) follows,

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:
$$\cos \theta = \frac{s}{s-1}$$

Sorry to be nitpicking here: bond in question is sp hybridised; that is not correct (or sloppy form). The bond orbital of the central atom is a spX hybrid orbital (if hybridisation is an applicable approximation). || write the σ-bond formed between them is also sloppy. There is not going to be a bond between orbitals at the same atom.

Also, you use $\lambda_1$ and $\lambda_2$, which explicitly states that the two hybrid orbitals compared do not have to be equal. The special case you are mentioning would be $\lambda_1=\lambda_2$.

Related: How is Bent's rule consistent with LCAO MO theory? (including full derivation of Coulson's Theorem). Why does cyclopropane react with bromine? Ron's answer has a worked example.

within hybrid theory, R groups carry non-bonding electrons (typically in s orbitals) beside the orbitals for bonding that they lend Nitrogen for use in hybridization, so now Nitrogen has more s orbitals to mix with their own p orbitals. It’s not so much this increase in s character (non-directional) rather than the decrease in p character (directional) that influence the bond angle because the p orbitals (like z, x, y axes) determine how hybrid orbitals point in space. Sure Nitrogen is still labeled sp3 hybridized in textbooks, but the ratio between s and p orbital contributions is no longer 25% as in the R=hydrogen case. It’s just a slightly higher percentage depending on how large the R group is.

Hybridisation is not a theory; it is a mathematical interpretation within the molecular orbital theory. This post can otherwise benefit from shorter sentences, some paragraphing.


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Gluconeogenesis
So I've been doing research on the production of lactic acid, however, I found that lactate will first reform into pyruvate and then go through gluconeogenesis reforming D-glucose. From what I understand, it requires 4 ATP and 2 GTP, so Question: why wouldn't the pyruvate go straight to the citric acid cycle and electron transport chain instead of reforming glucose first?
My research for context:
Carbohydrate Fermentation: The production of Lactic Acid for Anaerobic Glycolysis
Lactic acid, also known as 2-hydroxypropanoic acid, is formed primarily by muscle cells when the body requires Adenosine Triphosphate(ATP) and there isn’t enough O2 for the citric acid cycle and electron transport chain(ETC) to occur.1,2 As a result, the cell will use anaerobic glycolysis as its primary form of energy production. In glycolysis, Nicotinamide adenine dinucleotide (in its oxidized form NAD+) is used when transforming glyceraldehyde 3-phosphate into 1,3-bisphosphoglycerate. However, by the end of the reaction, the NAD+ is left in its reduced form NADH. If O2 were present, the pyruvate would undergo a link reaction on its way to the matrix of the mitochondria, where the citric acid cycle occurs, followed by the ETC in the inner membrane space (IMS). Typically, this NADH would be used as an electron donor in the ETC, where the majority of ATP is generally produced during cellular respiration (34 out of the overall 38 ATP).3 However, in order to be reused for glycolysis, it is oxidized back into NAD+.

As seen in Figure 1, D-glucose, a six-carbon carbohydrate, undergoes glycolysis in the cytoplasm forming primarily two pyruvate 3-carbon molecules, 2NADH and 2ATP. From there, NADH and H+ react with each pyruvate molecule temporarily forming a lactic acid molecule which is deprotonated in solution forming lactate. Roughly 20 mmol of lactate is formed per kilogram of tissue daily. Although anaerobic glycolysis does produce ATP, it isn’t as efficient as the aerobic ETC, nor can it operate from prolonged periods of time.4,5
The reaction can only occur under these circumstances for between one and three minutes until the lactic acid is exposed to O2. At this point, lactate is transferred to the liver, where it undergoes gluconeogenesis and is reformed into glucose for aerobic respiration. First, lactate catalyzed by lactate dehydrogenase (LDH) reforms pyruvate via C3H6O3+(NAD+) C3H3O3-+NADH, and then the pyruvate is reformed into glucose for aerobic respiration via2 C3H3O3-+6NTPC6H12O6.6,8
Until O2 levels normalize, the lactic acid will build up in the cell, known most commonly as lactic acidosis, reducing the pH in the muscle tissue. This in turn inhibits ATP production as not as much H+ would be available for the anaerobic glycolysis to occur, therefore reducing ATP produced, resulting in fatigue.9
At 15˚C and one atmosphere of pressure, lactic acid is colorless, odorless, and is an alcohol and an acid. Although it can be found in two optically active forms, L(+)-lactic acid, D(-)-lactic acid, L(+)-lactic acid is the most common in human tissue as it is the only isomer formed during muscle contraction. Lactic acid exists as a conjugate base called lactate at pH 7.4, but it does not affect the chirality (L versus D). Normal levels of L-lactic acid in human blood vary slightly between sources but are generally between 0.5 mmol/l and 1 mmol/l.2,5
Cells use anaerobic glycolysis as a short-term solution for energy production. It allows the cell to produce ATP without the need for oxygen; however, after prolonged periods of use, raised lactate levels can reduce the pH of the muscle tissue in less ATP production and, therefore, cause fatigue.4,6
References:
Garlotta, D. A Literature Review of Poly(Lactic Acid). Journal of Polymers and the Environment 9, 63–84 (2001). https://doi.org/10.1023/A:1020200822435
Komesu, A., Oliveira, J. A. R. d., Martins, L. H. d. S., Wolf Maciel, M. R., and Maciel Filho, R. (2017). "Lactic acid production to purification: A review," BioRes. 12(2). 4364-4383.
Bettelheim, F. A.; Brown, W. H.; Campbell, M. K.; Farrell, S. O.; Torres, O. J.; Madsen, S. Introduction to general, organic, and biochemistry, 11th ed.; Cengage: Boston, MA, 2020.
Lund, J., Aas, V., Tingstad, R.H. et al. Utilization of lactic acid in human myotubes and interplay with glucose and fatty acid metabolism. Sci Rep 8, 9814 (2018). https://doi.org/10.1038/s41598-018-28249-5
Miroslav Pohanka, "D-Lactic Acid as a Metabolite: Toxicology, Diagnosis, and Detection", BioMed Research International, vol. 2020, Article ID 3419034, 9 pages, 2020. https://doi.org/10.1155/2020/3419034
Kraut, J., & Madias, N. (2014). Lactic acidosis. The New England journal of medicine, 371(24), 2309-2319. Report #: 24. http://dx.doi.org/10.1056/nejmra1309483 Retrieved from https://escholarship.org/uc/item/5z25r8s8
Muscle Fatigue: Lactic Acid or Inorganic Phosphate the Major Cause?
Håkan Westerblad, David G. Allen, and Jan Lännergren
Physiology 2002 17:1, 17-21
Exton, J.; Ui, M.; Lewis, S.; Park, C. Mechanism of Glucagon Activation of Gluconeogenesis. Gluconeogenesis 1972, 21 (10), 945–990
M Nanang et al 2018 IOP Conf. Ser.: Earth Environ. Sci. 197 012049


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Chiral center or not? Similar substituents, but still two possible structures!
As far as I know, the central carbon shouldn't be a chiral center because it has two identical substituents. And yet, these two compounds should be different from each other.. Is it possible to denote the central carbon with R/S convention? They seem to be meso-compounds, but I'm unsure of their naming.

The lower case designations (r/s) refer to stereochemistry, not chirality. You may be interested in this: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/RSrs.html

Related: Stereocenter or not?

Also related: https://chemistry.stackexchange.com/questions/91539/mismatch-of-plane-of-symmetry-vs-gold-books-definition-for-a-pseudochiral-carbo/91892#91892

In this case, there are 3 stereogenic centers, they are $\ce{C-3}$, $\ce{C-4}$, $\ce{C-5}$
$\ce{C-3}$,$\ce{C-5}$ are straightforward. Using the main CIP rules (1-3), we get $\ce{C-3}$ to be $R$ and $\ce{C-5}$ to be $S$
The term 'pseudoasymmetric' is defined in Subsection P-92.1.1 (d) of the IUPAC Blue Book as follows.

Stereogenic units are called pseudoasymmetric (center, axis or plane) when they have distinguishable ligands 'a', 'b', 'c', 'd', two and only two of which are nonsuperposable mirror images of each other (enantiomorphic). (…). The 'r/s' (…) stereodescriptors describing a pseudoasymmetric stereogenic unit are invariant on reflection in a mirror (for example 'r' remains 'r', and 's' remains 's'), but are reversed by the exchange of any two ligands ('r' becomes 's', and 's' becomes 'r'). Lower case stereodescriptors are used to describe pseudoasymmetric stereogenic units

$\ce{C-4}$ is a pseudo asymmetric center since two groups attached differ only in stereochemistry. In such a scenario, we turn to CIP rule 5, which states

Atom or group descriptor 'R', 'M' or 'Seqcis' precede 'S', 'P' and 'Seqtrans', respectively.

That is, $R$ > $S$ in priorities

Now, using this, we can now name the compound using the following rules.

P-91.2.1.1 Cahn-Ingold-Prelog (CIP) stereodescriptors
Some stereodescriptors described in the Cahn-Ingold-Prelog (CIP) priority system, called 'CIP stereodescriptors', are recommended to specify the configuration of organic compounds, (…), and in the nomenclature of natural products (…). The following stereodescriptors are used as preferred stereodescriptors (…):
(a) 'R' and 'S', to designate the absolute configuration of tetracoordinate (quadriligant) chirality centers;
(b) 'r' and 's', to designate the absolute configuration of pseudoasymmetric centers;
(…)

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants
are assigned to them in the following decreasing order of seniority:
(…)
(j) When there is a choice for lower locants related to the presence of stereogenic centers or stereoisomers, the lower locant is assigned to CIP stereodescriptors Z, R, M, and r (pseudoasymmetry) that are preferred to E, S, P, and s, respectively, which are preferred to the non-CIP stereodescriptors cis, trans, or r (reference), c, and t (…).
(…)

Therefore this compound will be named (3R,4r,5S)-3,5-dichloro-4-methylheptane
Similarly for the other case, where the methyl group points away from you, we get:

As noted by Safdar, the two chiral compounds are meso-compunds.  Like for meso-tartaric acid, there a is a mirror plane in the molecule.  It is possible to name them unambiguously, too.  You may use a page like the one about e.g., ChemDraw to train yourself in this part of chemical nomenclature:

While the compound is chiral, it is better to speak about the three atoms with wedges attached to them as stereogenic centers.  As you will learn later on, there equally is e.g., axial chirality.

I usually say "chirality center" to distinguish it from whether or not the compound is chiral or not, but thanks, I'll start using stereogenic center instead.

The central $\ce{C}$ is said to be pseudochiral. This means that its chirality will depend upon the configuration of the branches attached to it. For example, in the compounds you've drawn, the central $\ce{C}$ indeed is, in fact, achiral. They are diastereomers and not enantiomers, and both are meso compounds, hence optically inactive.
However, if you had drawn one of the $\ce{Cl}$ in dash position, then both your compounds would again be diastereomers, but they would be optically active and the central $\ce{C}$ would be chiral, since the attached groups are different.


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Compressibility factor and deviation from ideality
Consider these statements :

At Boyle point, real gases show ideal behaviour.
Above  the Boyle point, real gases show positive deviation from ideality (forces of attraction between the particles are very feeble).
Below Boyle point, real gases first show decrease in Z value with increasing pressure, reaches a minimum value, increases again on further increase in pressure.

Questions:

What is the actual significance of Z ? (I mean, intuitively )
What is meant by positive deviation from ideality? If this is a measure of how Z value increases, I don't get it.

Ideal gases do not have forces of attraction between the constituent  particles.
If we consider Z as:
V(real)/V(ideal),
how can a real gas show a positive deviation from ideality? ie. How can a real gas with a small amount of intermolecular force of attraction expand more than an ideal gas with zero intermolecular force of attraction?

What is meant by negative deviation from ideality? What happens to a ideal gas when we apply pressure to it? Will we be able to compress it indefinitely  But won't there be repulsive interactions even in such gases due to decrease in stability when huge amount of pressure is applied?

Any answer would be greatly appreciated.

'Z' is a measure of deviation of a real gas from ideal behaviour.

The effect of not negligible own volume of real gas molecules goes against the effect of real gas cohesive forces. The value of Z depends on which of the effects takes the upper hand.

I think I don't understand.  Does that mean that they will be more spread out? Is there any assumption that the pressure is constant? Why is this applicable to certain gases?

Let's start thinking qualitatively.

What we call "volume" of a gas is the sum of the "free" volume where the molecules are free to move, plus the volume of the molecules themselves. The collision theory is only valid for a volume which is the free space available to the moving molecules.

What we call "pressure" of a gas is the force exerted by the particules when touching the walls of the container. But this force in a compressed gas is not simply due to its kinetic energy due to the temperature. It is diminished if there is an attraction between the particules.

In a gas like $\ce{H2}$, there is practically no attraction between individual molecules. The pressure is due to the collisions with the walls due to particules whose energy is simply proportional to $\pu{T}$. So the total volume is equal to the theoretical value $\ce{V_o = nRT/V}$, plus the volume of the molecules. It increases with the pressure. So the measured $\pu{V}$ is greater than $\pu{nRT/p}$ and  $\pu{Z = pV/nRT}$ increases with $\pu{p}$.
In a gas like $\ce{CO2}$ or $\ce{CO}$, the pressure measured on the wall is smaller than the theoretical values coming from the kinetic theory of gases, because the measured pressure is slowed down by their mutual attraction. That is why
$\pu{Z = pV/nRT}$ decreased by increasing the pressure

Is that why in van der Waals' correction, they have written that V(ideal)=V(real)-nb?

Yes. it is the origin of the van der Wals correction

What @Poutnik meant was as follows:
An ideal gas is considered to have very negligible or almost no volume of the molecules it is comprised of. It also has, as you say, zero interaction energy.
But the change in volume of the real gas depends on which of the two factors overpowers the other: (1) Real gas molecules have certain volume (2) Interaction Energy is non zero in between real gas molecules.
Not the volume of an individual molecule but the interaction energy changes with changing parameters of pressure and temperature (Implying that as we change temperature and pressure, the interaction between the molecules is affected). Thus Z is a measure which takes into consideration these factors (pressure, temperature and volume). Intuitively speaking in simple terms, it literally stands by its name: 'Compressibility factor'.
If Z<1, then for a fixed amount of pressure applied, a real gas will be better compressed than an ideal gas. Implying that if Z<1 then gas is more compressible (than what it would have been as an ideal gas). On similar lines you can understand what Z>1 means.
Now Z is formulated as
Z = PV/RT

Look at the graph above. It shows various gases showing deviation from ideality. The picture is pretty much self explanatory.
Boyles Point/Temperature is a specific temperature for each gas slightly around and at which the gas tends to behave almost ideally. Now this must tell you that the ideality of a gas is also temperature dependent.
Boyles Temperature is estimated by
T=a/Rb
where R is universal gas constant; b is the volume excluded(unoccupied) by 1 mole of the gas from the volume if the gas has been ideal; value of 'a' depends upon the attractive forces between the molecules and is directly proportional to it (Thus how the molecules behaves in terms of interacting with fellow molecules is also considered) So now you can see how all the factors play an altogether role in the gas behavior.
[P.S. I wanted to post this in a brief manner in the comment section but couldn't fit in. All required edits are open. I really tried to put it into simple words]


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Why is the trans-effect of nitrile ion (cyanide) weaker than acetonitrile (methyl cyanide) in octahedral chromium complexes?
I was reading a paper which built a series of trans-philicity (a term they coined to indicate both kinetic trans-effect and thermodynamic trans-influence) from extensive calculations. And I found that in the series, $\ce{:\!CN-}$ is shown to have a weaker trans effect than $\ce{MeC#N \!:}$ [1]  (Note: this is for octahedral complexes)
However, this trend seems odd to me, because I cannot explain it in terms of electron donating and withdrawing capacities. I had heard that the trans-effect is strongest in ligands which were either strong $\sigma$-donor or strong $\pi$-acceptor or a combination of both. This is why $\ce{CN-}$, $\ce{CO}$ etc. have very strong trans-effects.
Now, in the experimental trans-effect series that is commonly found in inorganic textbooks, we have:
$$\ce{H2O < NH3 < ... < CO < CN-}$$
$\ce{NH3}$ is near the very beginning of the series. $\ce{MeC#N}$ should be a worse $\sigma$-donor than $\ce{NH3}$ (as the lone pair is in an $\mathrm{sp}$-orbital). It also seems unlikely that $\ce{MeCN}$ will be a better $\pi$-acceptor than $\ce{CN-}$ because $\ce{MeCN\!:}$ binds via the more electronegative $\ce{N}$ while $\ce{:\!CN-}$ binds with the less electronegative $\ce{C}$ (which would mean the tendency to accept electron density in the $\pi$-orbitals should be lower for $\ce{MeCN}$).
So, I don't understand how it is possible to have the trans-effect order $\ce{CN- < MeCN}$. Is there any explanation for this?

Reference:
[1]. A. C. Tsipis, "Building trans-philicity (trans-effect/trans-influence) ladders for octahedral complexes by using an NMR probe", Dalton Trans. 2019, 48, 1814-1822 (DOI: 10.1039/C8DT04562C).

According to the reference mentioned in the question (Ref.1):

The term ‘trans-influence’, being a long-established concept of broad relevance in the realm of inorganic chemistry, was defined first in 1966 by Pidcock et al. as the ability of ligand L in a complex to weaken the metal–ligand bond trans to itself. This ground-state phenomenon should be distinguished from the kinetic phenomenon called the ‘trans-effect’, which is the effect of coordinated ligand L upon the rate of substitution reactions of the ligand in trans-position to L.
Note: Pidcock et al. 1966: Ref.2

Considering the high sensitivity of the $\ce{^{13}C}$-$\mathrm{NMR}$ isotropic shielding tensor elements to small structural/electronic changes, the authors of Ref.1 have published a reliable trans-philicity ladder for octahedral $\ce{[Cr(CO)5L]^{−/0/+}}$ complexes using $\ce{^{13}C}$-$\mathrm{NMR}$ isotropic shielding tensor elements. In  $\ce{[Cr(CO)5L]^{−/0/+}}$ complex, $\ce{L}$ represents a wide variety of ligands (50 ligands) commonly used in coordination and organometallic chemistry. Briefly, all $\ce{^{13}C}$-$\mathrm{NMR}$ isotropic shielding tensor elements and other parameters have been calculated using PBE0/Def2-TZVP(Cr)∪6-31G(d,p)(E)/PCM and PBE0/Def2-TZVP(Cr)∪6-311++G(d,p)(E)/PCM computational protocols set in dichloromethane solution where the latter protocol is more sophisticated than the former.
I think, major drawback in this publication is the lack of experimental date to support the calculations. For instance, the authors admit that to the best of their knowledge, experimental data for $\delta \ \ce{^{13}C}$-$\mathrm{NMR}$ chemical shifts of $\ce{[Cr(CO)5L]^{−/0/+}}$ complexes are available only for the $\ce{Cr(CO)6}$ complex and the “free” $\ce{CO}$ ligand, which are $212$ and $\pu{184.4 ppm}$, respectively.
When compared the calculations of $\delta \ \ce{^{13}C}$-$\mathrm{NMR}$ chemical shifts of the $\ce{Cr(CO)6}$ complex and the “free” $\ce{CO}$ ligand employing the two computational protocols, the PBE0/Def2-TZVP(Cr)∪6-31G(d,p)(E)/PCM predicted $\delta \ \ce{^{13}C}$-$\mathrm{NMR}$ chemical shifts of $210.2$ and $\pu{186.1 ppm}$, respectively for two compounds, while the protocol PBE0/Def2-TZVP(Cr)∪6-311++G(d,p)(E)/PCM) predicted $\delta \ \ce{^{13}C}$-$\mathrm{NMR}$ chemical shifts of $226.8$ and $\pu{197.6 ppm}$, respectively for the same two compounds:
$$\begin{array}{l|cc}
\hline
\text{Compound} & \ce{\delta \ ^{13}C} \text{ (calculated)}^1 & \ce{\delta \ ^{13}C} \text{ (calculated)}^2  & \ce{\delta \ ^{13}C} \text{ (experimental)} \\
\hline
\ce{Cr(CO)6} \text{ (complex)} & \pu{210.2 ppm} & \pu{226.8 ppm} & \pu{212.0 ppm} \\
\ce{CO} \text{ ('free' ligand)} & \pu{186.1 ppm} & \pu{197.65 ppm} & \pu{184.4 ppm} \\
\hline
\end{array}\\
^1 \text{From protocol PBE0/Def2-TZVP(Cr)∪6-31G(d,p)(E)/PCM;} \\ ^2\text{ From protocol PBE0/Def2-TZVP(Cr)∪6-311++G(d,p)(E)/PCM.}
$$
Evidently, the GIAO/PBE0/Def2-TZVP(Cr)∪6-31G(d,p)(E)/PCM computational protocol is a better performer in the calculation of the $\ce{^{13}C}$-$\mathrm{NMR}$ spectra of $\ce{[Cr(CO)5L]^{−/0/+}}$ complexes than that of PBE0/Def2-TZVP(Cr)∪6-311++G(d,p)(E)/PCM one. Nevertheless, the differences of the calculated $\Delta\sigma \ \ce{^{13}C}$-$\mathrm{NMR}$ descriptors of trans-philicity for the complexes using either protocol were minimal.
Yet the authors have mentioned that:

It can be seen that the $\mathrm{NMR}$ trans-philicity ladders constructed by the two computational protocols are similar with some minor local reversed orders in the trans-philicity series of similar ligands. The PBE0/Def2-TZVP(Cr)∪6-31G(d,p)(E)/PCM computational protocol predicts for the $\ce{NCR}$ ligands the order: $\ce{NCH \gt NCPh \gt NCMe}$, while the PBE0/Def2-TZVP(Cr)∪6-311++G(d,p)(E)/PCM computational protocol predicts the order: $\ce{NCMe \gt NCPh \gt NCH}$. Consideration of the $\sigma$-donor and $\pi$-acceptor abilities of the $\ce{NCR}$ ligands supports the order predicted by the PBE0/Def2-TZVP(Cr)∪6-31G(d,p)(E)/PCM computational protocol.

Thus, I could argue that the difference in two protocols make this difference than actual situation. Unless we have experimental data to support the finding, it is just speculation.

Reference:

A. C. Tsipis, "Building trans-philicity (trans-effect/trans-influence) ladders for octahedral complexes by using an NMR probe", Dalton Trans. 2019, 48, 1814-1822 (DOI: 10.1039/C8DT04562C).
A. Pidcock, R. E. Richards, and L. M. Venanzi, "$\ce{^{195}Pt–^{31}P}$ nuclear spin coupling constants and the nature of the trans-effect in platinum complexes," J. Chem. Soc. A 1966, 1707–1710 (DOI: https://doi.org/10.1039/J19660001707).

I don't understand how this answers the question? I was asking about the difference between CN- and MeCN. MeCN has a higher trans-philicity than CN- with both basis sets. Is there any experimental data you know about the trans-effect of MeCN vs CN- that disproves the order from the calculation? Otherwise your answer is just explaining what's written in the paper.

@ Shoubhik R Maiti: Before you question, did you read the paper? It was stated that two protocols give opposite results. Thus, none is conclusive. That's what I said in my answer. Please read the paper first.

Of course I read the paper, that's why I mentioned it in the question. The two protocols give different results for RCN groups, but I am not asking about the order of that. (And HCN is not the same as CN-, they are different ligands in the paper) In both protocols the order of CN- and MeCN is the same. Are you arguing that because the methods predict different order for RCN ligands, the whole series is questionable?

Did you have any other evidence that it is different than what the paper said?

I wrote in the question why I thought the order in the paper does not seem right, I was mainly looking for a orbital / inorganic based explanation.

Well, seems like I misunderstood your question. I'll see some evidence to add.


text:
Polythiophenes versus Polypyrroles
I've been searching for the differences in semiconductor properties of polythiophene (pth) and polypyrrole (pry).
Most resources just state their ability to form polarons and can be doped. However, I wanted to find out if there were differences in band gaps or any other differences due to the change of heteroatom (sorry for the vagueness)

Consider that these type of materials have properties (of all kinds) spanning broad ranges as for are highly dependent on preparation or even batch. What I can easily tell you is that PTh is the working horse for a huge number of groups, while PPy found much less interest. Moreover there is a mess in terminology. For instance you mention band gap, there could be already different ways to interpret that. Whatever difference you might find will be due to the fact that Py is less aromatic than Th. In short, to date, PPy does not seem to be of much interest at least in organic electronics.

A crucial difference is that neutral polythiophene (in particular re-P3HT) is a reality. The Py based counterpart, to my knowledge, has still to come to existence. One probably need a chapter to discuss why it is. Likely it all starts on the different reactivity of Th and Py themselves, or that of their dimers trimers... Chemistry of Py conjugated systems is anyways interesting in biological systems, from blood cells to pigments). Sorry but I can't answer in coincise form because of what these comments say.


text:
Nitrogen with shortest bond length in histidine?
The question asks what nitrogen in the ring of histidine has the shortest bond length.
I was thinking that since it has resonance, the initial thought of the nitrogen with the $\pi$ bond is incorrect. However, I imagine the resonance structure shown in the figure below makes the greatest contribution to the resonance hybrid. Therefore making my first guess correct. The question also asks for a single line of reasoning so I am confused how to present my thinking concisely.

It is correct that the $\ce{N}$ with the double bond has a localized lone pair, because of its $sp^2$ hybridization, due to which the lone pairs are in the same plane as ring, and hence not conjugated.
However, the other $\ce{N}$ is clearly in conjugation, hence, if you draw the $5$ possible resonance structures, you would notice that each bond in the ring is a double bond for exactly $2$ of them. However, since there is a positive charge on electronegative $\ce{N}$ in all of them except the one you have drawn, this structure would have maximum contribution. Thus, the shortest bond length indeed would be the one you've shown as a double bond in your figure, although in reality it would have bond order $<2$ due to resonance.


text:
Reaction between sodium carbonate and hydrochloric acid
The 2 steps to the reaction between $\ce{Na2CO3}$ and $\ce{HCl}$ are:

$\ce{Na2CO3 + HCl -> NaCl + NaHCO3}$
$\ce{NaHCO3 + HCl -> NaCl + CO2 + H2O}$

If $\ce{Na2CO3}$ is in excess and we add limited amount of $\ce{HCl}$ (e.g. 2:1), why is it that after the addition of limited $\ce{HCl}$ , we still have $\ce{NaHCO3}$ present together with $\ce{Na2CO3}$?
What I was thinking is that as $\ce{NaHCO3}$ is formed, the $\ce{HCl}$ in the solution will react with it through step 2 to form $\ce{CO2}$ and $\ce{H2O}$. So, half of the $\ce{HCl}$ will react with $\ce{Na2CO3}$ and the other half will react with $\ce{NaHCO3}$. And so, if this happens there will only be $\ce{Na2CO3}$ present at the end. (since all the $\ce{NaHCO3}$ that is formed from step 1 will be reacted with $\ce{HCl}$ upon formation)
Why wouldn't this be the case? Why must step 2 only occur after step 1 is completed?

Remember that carbonic acid is a weak acid and there are acido-basic equilibrii between the respective forms.

Also what you described isn't what normally goes under the term mechanism, or at least I think so.

You are right. When adding a small amount of $\ce{HCl}$ to a solution of $\ce{Na2CO3}$ you produce some $\ce{NaHCO3}$. When adding more $\ce{HCl}$, you cannot avoid that some of this $\ce{NaHCO3}$ does react with the newly created $\ce{NaHCO3}$ producing some $\ce{CO2}$. But this $\ce{CO2}$ will immediately react with $\ce{Na2CO3}$ which is still in excess according to :
$$\ce{CO2 + Na2CO3 + H2O -> 2 NaHCO3}$$ So it looks as if some more $\ce{NaHCO3}$ has simply been created by adding $\ce{HCl}$ to $\ce{Na2CO3}$

There are two ways to react sodium carbonate (A) with hydrochloric acid (B):

You can add A to B, or

You can add B to A.

If the ratios differ from 1:1, you can get different reaction products by adding them differently. For instance, if you dump (A to B) a solution of 0.1 mole of Na$_2$CO$_3$ into a solution of 0.1 mole of HCl at a moderate rate, with stirring, the products will be 0.05 moles of CO$_2$ fizzing off instantly, plus 0.1 moles of NaCl, plus 0.05 moles of unreacted Na$_2$CO$_3$. You won't find any NaHCO$_3$ because your reaction Step 2 occurs rapidly enough to go to completion before all the Na$_2$CO$_3$ has been added. You have Steps 1 and 2 occurring for half of the Na$_2$CO$_3$, leaving half of the Na$_2$CO$_3$ unreacted. This scheme conforms to your supposition, but is not elegant chemical technique.
On the other hand, if you add, slowly, by using a tube to add (B to A) 0.1 mole of Na$_2$CO$_3$ at the bottom of a solution of 0.1 mole of HCl, something like the picture, you should be able to get a solution of 0.1 mole of NaHCO$_3$ plus 0.1 mole of NaCl. All the CO$_2$ will have been trapped and no bubbling out will have occurred. This is not as easy as it looks, and requires a lot of patience, but is a clear demonstration of your reaction Step 1.


text:
Why is acylation giving the major product compared to alkylation in the reaction between benzene and chloroacetyl chloride?
Why are we getting this products as major and the other as minor ?

According to my knowledge, the carbocation formed on the $\alpha$-carbon of the carbonyl chloride is more stable than the carbocation formed on the doubly bonded carbon.
Shouldn't the major and minor product be reversed ?

Then your knowledge is incorrect - acyl cations are way more stabilised then alkyl.

@Mithoron how ?? the carbocation is formed on doubly bonded carbon. so how is it stable ?

Did you bother you read Safdar's answer? Not the "not an answer" you marked as accepted, which missed the point entirely...

@Mithoron i never saw the edit. now it is better and does explain the concept

This can be explained by comparing the stability of the carbocations formed in Friedel-Crafts acylation and Friedel-Crafts alkylation using chloroacetylchloride. This is because the rate determining step in both reactions is the formation of the carbocation.

As you can see, the acyl carbocation is resonance stabilized, whereas the carbocation formed in the alkylation process is very weakly resonance stabilized.[1]
According to supporting information to J. Chem. Educ. 2012, 89 (1), 147–149[2]

This o-toluidine selective attack at the acyl carbon atom of the bifunctional reagent is illustrative of the significantly greater reactivity of nucleophiles toward acylation, compared to alkylation

Therefore the acylation gives you the major product compared to the Friedel-Crafts alkylation product.
I was able to find a reference to your a similar reaction in J Flow Chem 2019, 9 (1), 35–42 with difluorobenzene instead of benzene.

A series of experiments were performed reacting a solution of 1,3-difluorobenzene (6) and aluminum trichloride ($\ce{AlCl3}$) in nitromethane [...] however increasing the temperature and time increased conversion to $22–74\%$ (entries 2–6). At $\pu{90 °C}$ (entry 5) the product was accompanied by the 2-acylated regioisomer. To further improve the reaction, the amounts of chloroacetyl chloride (7) and $\ce{AlCl3}$ were increased.These modifications
gave the highest conversion, $79\%$ conversion [...]

\begin{array}{c|ccc} \hline
\bf{Entry} & \bf{Residence\,time\,(min)}& \bf{Temp\,(^\circ C)} & \bf{Conversion\,(\%)} \\ \hline
1 & 5 & 50 & 0 \\
2 & 5 & 60 & 22 \\
3 & 10 & 60 & 45 \\
4 & 15 & 60 & 46 \\
5 & 15 & 90 & 77 \\
6 & 25 & 70 & 74 \\
7 & 25 & 80 & 77 \\
8 & 25 & 80 & 79 \\ \hline
\end{array}

Another paper that talks about the acetylation of benzene and toluene is J. Am. Chem. Soc. 1971, 93 (25), 6964–6967 which includes multiple reagents including chloroacetyl chloride.
Reference:

Creary, X. Stabilization of α-keto cations by carbonyl conjugation. J. Am. Chem. Soc. 1981, 103 (9), 2463–2465. DOI: 10.1021/ja00399a078.
Demare, P.; Regla, I. Synthesis of Two Local Anesthetics from Toluene: An Organic Multistep Synthesis in a Project-Oriented Laboratory Course. J. Chem. Educ. 2012, 89 (1), 147–149. DOI: 10.1021/ed100838a.

No the products shouldn’t be reversed as, to put simply, acylation is preferred over alkylation.
This is a conclusion based on various experimental results and there are many reasons that can be given for the same.
As said in the concluding lines of this article from Chemistry Libretexts,

Friedel-Crafts Acylations offer several synthetic advantages over Friedel-Crafts Alkylation. These advantages provide greater control over the production of reaction products

I looked up online but couldn’t find any relevant reason particular to the acylation of chloroacetyl chloride, as asked in your question. But I did find some explanations to why acylation is preferred over alkylation:

In friedel crafts aklylation there is a chance of polyalkyation. Because the alkyl group increases benzene ring electron density and ring becomes activated to another elelctrophilic substitution.... thus polyalkyation occurs….But in Fridel crafts acylation there is no chance of polyacylation because the acyl group (RC=O) decreases benzene ring electron density and henceforth the second acylation does not take place.

During acylation reaction the carbon chain of the acyl halide does not rearrange and hence there's no possibility of formation of Isomeric ketone compound and thus 100% product yield is obtained in acylation reaction.

This is an answer to a question asked on Quora.
Again, these are all just explanations given to explain the experimental results and might not be conclusive or convincing enough.

Sorry for a misunderstanding. "This answer doesn't explain why acylation takes place over alkylation but rather that the product  of acylation is preferred compared to the products generated by alkylation when each reaction takes place in isolation."No the products shouldn’t be reversed as, to put simply, acylation is preferred over alkylation", what is the proof for this statement?


text:
How are Quantum Dot molecules formed and what properties do they hold apart differently from singular quantum dots?
I am trying to understand the science and phenomena behind the function and operation of quantum dot molecules, but the only articles I run into talk about the synthesis of them. Could this be explained to me? Thank you

Did you read Wikipedia? There is nothing unique about quantum dots (QDs).  QDs are nanometer sized crystals of semiconductors with some interesting optical properties.

A) There are quite a few different ways by which  Quantum Dot molecules are formed. As M.Farooq sir has said , you can check about them on Wikipedia or even on many different sites. Let me take you step by step.
B)Quantum dots (QDs) are man-made nanoscale crystals that that can transport electrons. When UV light hits these semiconducting nanoparticles, they can emit light of various colors.
This article also talks about their function https://www.nanowerk.com/what_are_quantum_dots.php .
C) Now , you say that articles only talk about their synthesis and your Q above is how they are formed ? And inside the title , you Q is again different.
What is *Synthesis ? Synthesis means to combine a number of different pieces into a whole.
Synthesis or formation are more or less , the same thing. If you see.
Example : Synthesis or formation of molecules from atoms. General statement but mean same thing. Experiment it yourself by searching difference between formation and synthesis.
D) Now , how are they formed ?

Quantum dot molecules are formed by coherent tunneling between two individual quantum dots. The coherent tunneling leads to the formation of delocalized states that are truly molecular in nature, with bonding and antibonding orbital states.

For more details : https://sites.udel.edu/dotygroup/home/quantum-dot-molecules/
Now , you must understand. Just like molecules are formed from atoms. This is a general meaning. Similarly, there are different kinds of quantum dot molecules formed by different methods or procedure which mostly come from the same basic meaning as I have written above.
For example : How are graphene quantum dots formed ?
Pyrolysis is one of the simplest methods of synthesizing graphene quantum dots. In this method, GQDs are formed via carbonization of small organic molecules
https://www.intechopen.com/books/smart-drug-delivery-system/graphene-quantum-dots-from-emergence-to-nanotheranostic-applications. For more details.
Your question inside what you have written is completely different from your title.


text:
How change in potential energy corresponds to change in enthalpy?
So while reading about activation energy , I came across this graph :

Here, change in potential energy of reactant and product is depicted as ${ΔH}$, but at constant pressure and temperature , shouldn't that be ${ΔU}$. Because change in thermal energy is zero, So change in potential energy should be ${ΔU}$.
According to me, that gap in the graph should be equal to ${Q}$ + ${W}$ not just ${Q}$ (${=ΔH}$).
This snap from 'Wiley's Solomons, Fryhle & Snyder Organic Chemistry' also conveys that change in potential energy is ${ΔH}$ .

Why do you say change in thermal energy is 0 ? At constant p and T, heat going in / out = $\Delta H$

Constant T doesn't mean no heat flow. Thus, e.g., if a reaction evolves thermal energy, the thermal energy flows out of the system to maintain constant T.

I never said heat flow is zero . Thermal energy is proportional to Temperature , So as ΔT=0 wouldn't the change in thermal energy be zero. I think it this way as ΔU = Δ(Thermal energy) + ΔPE . (U meaning Internal Energy)

According to me, that gap in the graph should be equal to Q + W not just Q (=ΔH).

Does U depend on just T and P, or do you think that it might also depend on chemical composition?

@ChetMiller I think U contains both thermal (Kinetic) energy and potential energy of the molecules , and thermal energy is dependent on the temperature (maybe also on pressure for real gases ) while potential energy is dependent on the the chemical composition ,So that way I guess it should be dependent both on temperature and chemical composition.

Right, so the products have different U than the reactants, even at the same temperature.

Yes , no problem with that

@ChetMiller But it doesn't answers my query.

Can anybody please help me out ? If question isn't clear to you please let me know !

OK, I think I see what you are confused about. So: (1) You are correct that U contains both thermal  E (TE) (thermal energy = kinetic energy) and potential E (PE). (2) You are incorrect in saying that $\Delta T = 0$ means same TE. The products and reactants will (with rare exceptions) have diff. heat capacities, and thus contain different amounts of TE at the same T. (3)You are right to be confused by the graph. Its vertical axis is mislabeled.   Neither $\Delta H$ nor $\Delta U$ equal the change in PE, even at const. T (see #2).

(4) Is the graph from a published work or some other publicly-available source (e.g., the internet)?  If so, you need to give proper attribution (even if the graph is wrong), and indicate the source.   (5)  See my answer here, which directly addresses your question:   https://chemistry.stackexchange.com/questions/77414/is-enthalpy-the-same-thing-as-potential-energy/121235#121235

@theorist I got what you are saying , the diagram is originally from here https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.04%3A_Potential_Energy_Diagrams but I found similar graph and statements in other books also.

On Wikipedia here https://en.wikipedia.org/wiki/Activation_energy , in a diagram the difference between the energy of product and reaction is labeled as ΔH , but it isn't clear that Energy labeled on Y-axis is which energy (Potential , Total , gibbs free ..) ?


text:
What is electronic configuration of manganese in the manganate(VI) anion?
The electronic configuration of the central atom of $\ce{[MnO4]^2-}$ if given out to be $e^1\,t_2^0$
I want to know how to identify which one of the d-orbitals does this configuration utilizes because when I applied Crystal Field Theory, my answer came out to be $t_2^1\,e^0$
I already found out the oxidation number for $\ce{Mn}$ to be $+6$ and using the structure of the compound (which is tetrahedral) determined the electronic configuration as stated above but could not determine the priority order for which orbitals are to be correctly filled.
Where am I going wrong?

And in tetrahedral symmetry which of the orbitals has lowest energy? Is that the same as in an octahedral field?

In tetrahedral geometry, considering CFT, the ligands are ions that approach from positions that are not axial in nature. This means that the $\mathrm{d}$-orbitals that are aligned along the axis are at a lower energy level compared to the orbitals that are aligned between the axes. Therefore, we notice two degenerate energy levels as shown below for tetrahedral symmetry of ligands.
From Chemistry Libretexts − Non-octahedral Complexes

The splitting of the energies of the orbitals in a tetrahedral complex ($Δ_\mathrm t$) is much smaller than that for an octahedral complex ($Δ_o$), however, for two reasons: first, the d orbitals interact less strongly with the ligands in a tetrahedral arrangement; second, there are only four negatively-charged regions rather than six, which decreases the electrostatic interactions

Therefore, the orbitals will be filled in the order e and then in the t2 orbitals.
Therefore since we need to fill one electron into these orbitals, we get $e^1\,t_2^0$ and not $t_2^1\,e^0$


text:
Density and Viscosity for Solution: 80% Liquid Methane CH4 with Solution of 20% Gas Nitrogen N2 on Surface of Titan (the largest moon of Saturn)?
Problem:

Need to estimate the density and dynamic viscosity of such a mixed fluid: 80% Liquid Methane CH4 with solution of 20% Nitrogen Gas N2 with temperature of 94K and pressure of 1,467 hPa (the surface environment on Titan, the largest moon of Saturn).
Will the effect of Nitrogen be ignorable to the composition's density and viscosity?
When the chemists say composition with 80% A + 20% B, do you mean percentage by volume or by mass? sometimes it seemed assumed as a common sense in the paper, but non-chemist people may find it obscure.

Research done:
Origin/Context: Table 1 of journal paper "Cassini radar observation of Punga Mare and environs: Bathymetry and composition".
I tried wolframe alpha to give me an automatic answer, but its data base probably constrained to environment of earth.
In situ data (to this date) does not exist for the liquid methane lake's fluid property on Titan, so we have to estimate it.
This might be too basic for any chemist, but I do find it quite difficult.
Appreciate any help!

Using NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, I estimated the properties of a mixture of $80\ \%$ methane and $20\ \%$ nitrogen (by mass) at a temperature of $T=94\ \mathrm K$ and a pressure of $p=1.467\ \mathrm{bar}$.
At this temperature and pressure, the mixture is a liquid with a density of $\rho=486.24\ \mathrm{kg\ m^{-3}}$ and a viscosity of $\eta=161.13\ \mathrm{\mu Pa\ s}$.
By way of comparison, pure methane has a density of $\rho=447.14\ \mathrm{kg\ m^{-3}}$ and a viscosity of $\eta=176.28\ \mathrm{\mu Pa\ s}$.
You should check yourself if the difference is significant or if the contribution of nitrogen can be ignored.

May I ask a further question that, will the liquid methane be a Newtonian fluid or a non-Newtonian fluid at that state?


text:
Rate of a reaction when volume of the system varies with time
Background
I was solving this question from this book and I was not able to get any of the options.

The gaseous reaction $\ce{n_1 A_{(g)} -> n_2 B_{(g)}}$ is first order with respect to ‘A’. The rate constant of reaction is ‘k’. The reaction
is studied at a constant pressure and temperature. Initially, the moles of ‘A’ was ‘a’ and ‘B’ was not present.
How many moles of ‘A’ are present at time t?
A) $\mathrm{a.e^{-kt}}$
B) $\mathrm{a.e^{−n_1kt}}$
C) $\mathrm{a.e^{−n_2kt}}$
D) $\mathrm{a(1− e^{−n_1kt})}$

Answer:

 B

When I looked into the solutions, it seemed, for me that the solution is wrong.
Solution given in the book

 $$\mathrm{r_{rxn} = \frac{-1}{n_1} . \frac{dn_a}{dt} = k.n_A}$$
$$\mathrm{\implies n_A = n^o_A.e^{-n_1kt}= a.e^{−n_1kt} }$$

My approach to the question
The rate of a chemical reaction is defined as
$$\mathrm{ r= \frac{1}{\mu_i}\frac{dC}{dt}}$$ where $\mathrm{dC}$ is the infinitesimal change in concentration of reactant or product in infinitesimal interval of time $\mathrm{dt}$ , ${\mu_i}$ is the stoichiometric coefficient of reactant or product.
When volume of system, $\mathrm{V}$ is constant, $$\mathrm{\frac{dC}{dt}=\frac{1}{V}\frac{dn}{dt}}$$
But if volume of the system changes then $$\mathrm{\frac{dC}{dt}=\frac{d(n/V)}{dt}}$$
which is equivalent to $$\mathrm{\frac{dC}{dt}=\frac{1}{V}.\frac{dn}{dt} - \frac{n}{V^2}.\frac{dV}{dt}}$$ and proceeded further using the ideal gas equation.
So, is the solution provided in the book correct or my approach?

Your reasoning does make sense. But how can we guess what you call  "the solution provided by the book" ?

@Maurice look for the image of the solution. He has inserted that

But I don't get why you feel the solution is incorrect. What part did you feel wrong, please explain?

@Jayadithya This may help. https://chemistry.stackexchange.com/questions/43558/what-does-gases-at-constant-pressure-mean

@Jayadithya The first question’s solution looks wrong to me.

@Jayadithya Why not just ask about the first question, questions which generally ask about too many questions in one post might not get answers. It’s totally upto you though.

@Rishi I thought the first question is right and the answer as well. Will you please explain why it isn't right?

@Raviraj Here first we will ve getting the equation in terms of concentration and then we can transform that equation into that of moles but since $\mathrm{V_o ≠ V_t}$ thus it won't be transformed as it has been given in the solution. Let me give a short proof here only. $$\mathrm{-\frac{1}n_1 \frac{d[A]}{dt} = k[A]}$$ and then through further mathematical simplification. $$\mathrm{[A]_t=[A]_oe^{-n_1kt}}$$ then transforming into moles; $$\mathrm{n_A=(n_A+n_B)e^{-n_1kt}}$$ where this transformation occurs due to the fact that for individual gases only the temperature is constant(cont...)

The answer and the solution given in the book are wrong
I have searched and found the following in Chemical Kinetics by Laidler which talks about the volume change during reaction. It states:

....Equation (1.11) gives a general definition of the rate of reaction, but Eqs. (1.13)-(1.14) apply only if the volume remains constant during reaction. To remove the restriction of constant volume, we differentiate the relationship $$\mathrm{n_B = [B]V}$$ for a species B and obtain
$$\mathrm{dn_B = V.d[B] + [B]dV   \tag{1.15}}$$
Insertion into Eq. (1.11) gives
$$\mathrm{r = \frac{1}{\mu_B}.\frac{d[B]}{dt} + \frac{[B]}{\mu_B . V}.\frac{dV}{dt}   \tag{1.16} }$$
This becomes equivalent to Eq.(1.13) when volume remains constant. The final terms in Eq.(1.16) is a correction for the change in concentration brought by change in volume. Even if no reaction were occuring an increase in volume decreases in concentration of the substances present. The rate of this decrease is equal to $\mathrm{\frac{[B]}{V}.\frac{dV}{dt}}$ and therefore addition of the term $\mathrm{\frac{[B]}{\mu_B . V}\frac{dV}{dt}}$ is required to make $\mathrm{r=0}$ for this case of no reaction.....

Using the ideal gas equation and the first order rate law
$$n_{\mathrm{1}}\ce{A} \ce{->} n_{\mathrm{2}}\ce{B}$$
\begin{array}{|c|c|} \hline
\text{Time} &\text{Number of moles of A }& \text{Number of moles of B} \\ \hline
\text{t=0} & a & 0\\
\text{t=t} & a-n_1x& n_2x \\
\text{t= infinite} & 0 &\frac{n_2a}{n_1} \\ \hline
\end{array}
Using the ideal gas equation $pV = nRT$ at the time 't' when $n_1x$ moles of $\ce{A}$ are consumed,
$$pV_t = (a-n_1x+n_2x)RT$$
$$\implies V_t = \frac{(a-n_1x+n_2x)RT}{p}$$
Using the first order rate law,
$$\mathrm{\frac{{d(n_A/V_t)}}{dt} = k\frac{n_A}{V_t}}$$
Substituting $\mathrm{n_A} = a-n_1x$ and $V_t = \frac{(a-n_1x+n_2x)RT}{P}$ in the above rate law and solving the differential equation, the result is
$$\mathrm{\frac{akt}{V_0} = x - \frac{n_2x}{n_1} - \frac{a}{n_1}.ln(\frac{a-n_1x}{a})}$$
References

Chemical Kinetics, 3e by Keith J. Laidler, Pearson publications.

Atkins Physical Chemistry 11e Volume 3 Molecular Thermodynamics and Kinetics.

When the reactants and products are both gaseous and at same pressure and temperature you might want to prefer using volume as a unit of measurement instead of concentrations as it simplifies the equations tremendously. Using volume as
a unit of measurement is possible because the moles of each gas $\ce A$ and $\ce B$ will be proportional to their volumes.
This way instead of dealing with three variables you now are concerned with only two. And we all can agree that solving a differential equation even in two variables can sometimes be tricky lest it be in three.
Thus for the reaction,
$$\ce{n1A_{(g)}->n2B_{(g)}}$$
The rate law can be written as,
$$-\frac{1}{n_1}\frac{\mathrm dV_{\ce A}}{\mathrm dt}=kV_{\ce A}$$
Which simplifies to,
$${(V_{\ce A})}_t={(V_{\ce A})}_0e^{-n_1kt}$$
Now the same expression can also be written in the form of moles as the volume was proportional to the number of moles.
$${(n_{\ce A})}_t=ae^{-n_1kt}$$
Which is same as the answer provided by your book.

Coming to your solution:
Your solution is also technically correct. But you are overlooking the fact that the moles of a gas and it's volume are not independent variables. They are interdependent on each other and thus creating a differential equation containing two interdependent variables does no good.
Thus before differentiating the $V/n$ quantity limit it to only one variable. If you will eliminate $n$ than it will be equivalent to the approach used in this answer and if you eliminate $V$ than it will be equivalent to the solution provided by your book.

“Using volume as a unit of measurement is possible because the moles of each gas A and B will be proportional to their volumes.” Could you explain this in more detail in your answer because I am getting something else, through ideal gas equation;$$\mathrm{P_AV_A=n_ART}$$ Here note that Volume of gas A will be equal to that of the ‘volume-changing container’ say V then; $$\mathrm{\frac{P_AV}{n_A} = constant}$$ Now using partial pressure equation for gas A and simplifying; $$\mathrm{\frac{P (n_A)V}{(n_A+n_B)n_A}=constant}$$ where P is the total pressure $$\mathrm{\frac{V}{n_A+n_B}=constant}$$

@Rishi When I mention volume of A, I theoretically mean the fraction of the total volume that would be occupied by A.

I might be missing something totally obvious but won’t the gases occupy the entire volume available(assuming ideal conditions) where this entire volume changes with time.

@Rishi Yes it will occupy the complete volume. But if theoretically we consider that both gases are having independent volumes than the volume occupied is the volume being talked about.

This is an intriguing problem. Atkins 5th edn. shows a problem where azomethane (g) decomposes to ethane (g) and N2 (g) at 600 K. It gives partial pressure values for azomethane vs time, and asks to show that it is a first order reaction. The solution says "the partial pressure of a gas is proportional to its concentration", and it proceeds to show that by plotting $ln(p_{azomethane})$ vs $t$, one gets a straight line, hence the reaction is indeed first order. This would seem to support NisargBhavsar's answer, as partial volumes and partial pressures are the same.

And it is true: it is easy to show that $p_A = C_ART$. So if one works literally, from $\frac {dC_A}{dt} = - kn_1C_A$, it follows that $\frac {dp_A}{dt} = - kn_1p_A$, and one does not have to worry about changing volumes at all. Perhaps the confusion comes from the fact that, in a reaction where 2 mols of gas form from 1 mol of gas, the partial pressure is not linearly related to the 'conversion' ($1- \frac {n_A}{n_{A,0}}$), as one normally assumes in this kind of problems. Still, only one answer can be correct, unless they are both saying the same in different forms.

And in fact NisargBhavsar's answer makes sense mechanistically. Saying that azomethane decomposes by a first order mechanism means that each molecule of azomethane reacts independently from its own concentration. It does not need to 'collide' with other molecules to react. So even though its decomposition causes an increase of the total volume, that does not change the relative rate. In fact I suspect that even in a solution, if you start a 1st order reaction and then keep adding solvent, you will still get a constant $\frac 1 C \frac{dC}{dt}$. To be verified, but I am relatively confident.

@user6376297 I might be stupid here but in your first comment, “as partial volumes and partial pressures are the same” can you prove this, mathematically? I don’t see how this is true.

You're right @Rishi , it is not the same. My statement was imprecise. Apologies. https://en.wikipedia.org/wiki/Partial_pressure#Partial_volume_.28Amagat.27s_law_of_additive_volume.29 . So NisargBhavsar's answer is still correct (can anyone prove otherwise?), though maybe not formally correctly worked out. I think the fundamental element to consider here is the meaning of 'first order'. That implies by necessity that a molecule reacts 'by itself' at a constant relative rate, regardless of how large or how small the volume gets during the reaction. Otherwise Atkins would be wrong (?!?).

In fact I see now that I made quite a mishmash of this, confusing the partial pressure, partial volumes, etc. I will write an answer to clarify.

As I mentioned, I agree with Nisarg Bhavsar's conclusion.
For the justification, I think a more formal one can be obtained starting from this Wikipedia article on mass balance:
https://en.wikipedia.org/wiki/Mass_balance#Ideal_batch_reactor
In this case the system is indeed closed, so the mass (not concentration) balance is:
$r_A \cdot V = \frac {dn_A}{dt}$
The reaction rate $r_A$ by convention is moles of A produced minus transformed, per unit time, per unit volume. [That's why its product by the instant reaction volume $V$ gives indeed the change in moles per unit time].
It is stated that the reaction is first order in $A$.
This implies the following:
$r_A = -k \cdot n_1 \cdot C_A$
The stoichiometric coefficient $n_1$ must be added because, again by convention, the rate constant $k$ is given for the 'standard' case where $n_1 = 1$.
Substituting in the mass balance:
$-k \cdot n_1 \cdot C_A \cdot V = \frac {dn_A}{dt}$
As $C_A \cdot V = n_A$, this becomes:
$-k \cdot n_1 \cdot n_A = \frac {dn_A}{dt}$
which leads to the same conclusion as the one from the book and from Nisarg Bhavsar, i.e.:
$n_A(t) = n_{A,0} \cdot e^{-k \cdot n_1 \cdot t}$
The conversion or advancement is:
$y_A(t) = \frac {n_{A,0}-n_A(t)}{n_{A,0}} = 1 - e^{-k \cdot n_1 \cdot t}$
This is in line with the idea that the conversion of a first order reaction does not depend on the initial concentration.
In practice, if one wanted to follow this reaction experimentally, they could sample it, taking only a very small volume of gas out, and measure the concentration of $A$ in the sample, e.g. by gas chromatography.
Knowing what the total volume of the reaction is at the sampling time, one could calculate the total number of moles of $A$, to fit the above integrated equation.

Could one use 'partial volumes' instead of number of moles?
Yes. Given that it is stated that this reaction is occurring at constant (total) pressure and temperature, and the mixture is ideal:
$V = n \cdot \frac { R \cdot T} {P} = n \cdot constant$
$V_A = n_A \cdot constant$
So one can substitute $n_A$ by $V_A$, and the equation still holds.

Why did Atkins' example about azomethane use partial pressure?
$\ce{CH3-N=N-CH3(g) \rightarrow CH3-CH3(g) + N2(g)}$
If one starts with pure azomethane, this is clearly a reaction (in the gas phase, at 600 K) where the total number of moles of gas are doubled when the reaction is complete.
Atkins states that the reaction is first order in azomethane, and the system is closed.
So the same equation as above applies (this time with stoichiometric coefficient $1$):
$-k \cdot n_A = \frac {dn_A}{dt}$
Atkins uses the following integrated law:
$\ln {\frac p {p_0}} = - k \cdot t$
where $p$ is the partial pressure of azomethane (not sure how it was determined experimentally).
By showing that the $\ln p$ vs $t$ data can be fitted to a straight line, it concludes that the reaction was first order.
However, this requires that the reaction be at constant volume (not mentioned by Atkins, but in fact mentioned in other books, like Silbey, Alberty - Physical Chemistry).
Atkins' argument starts from the assumption that:
$\ln {\frac {[A]} {[A]_0}} = - k \cdot t$
and then it says "the partial pressure of a gas is equivalent to its concentration", thus substituting $[A]$ by $p_A$.
It is true that for an ideal gas mixture:
$p_A = x_A \cdot P = \frac {n_A} {n} \cdot P$
and, as:
$P = n \cdot \frac {R \cdot T} V$
$\implies p_A = n_A \cdot \frac {R \cdot T} V = C_A \cdot {R \cdot T}$
So, as long as the temperature is constant, it is indeed legitimate to substitute $C_A = [A]$ by $p_A$, in a ratio where the constant term ${R \cdot T}$ cancels out.
But this is missing the point.
The real question is: where did the integrated rate law come from in the first place?
Going back to the mass balance:
$-k \cdot C_A \cdot V = \frac {dn_A}{dt}$
As far as I can tell, the only way for this to become the integrated rate law used by Atkins is by assuming constant volume. Then, dividing by $V$ on both sides and taking $V$ into the differential:
$-k \cdot C_A = \frac {d \frac{n_A} V}{dt} = \frac {dC_A}{dt}$
which, integrated, is indeed:
$\ln {\frac {C_A} {C_{A,0}}} = \ln {\frac {[A]} {[A]_0}} = - k \cdot t$
If, like in the example from the original post, the total pressure, not the total volume, were constant, one could not derive this equation.
The one based on the number of moles of course would still be valid, but then:
$n_A(t) = n_{A,0} \cdot e^{-k \cdot t}$
$p_A = \frac {n_A} {n} \cdot P$
$n = n_A + n_{ethane} + n_{N_2} = n_A + 2 \cdot (n_{A,0} - n_A) = 2 \cdot n_{A,0} - n_A$
$p_A = \frac {n_A} {2 \cdot n_{A,0} - n_A} \cdot P$
$n_A = \frac {2 \cdot n_{A,0} \cdot p_A}{p_A + P}$
$\frac {2 \cdot n_{A,0} \cdot p_A(t)}{p_A(t) + P} = n_{A,0} \cdot e^{-k \cdot t}$
$p_A(t) = P \cdot \frac {e^{-k \cdot t}} {2 - e^{-k \cdot t}}$
I tested this out numerically and it seems to work, but then again, I am not a physical chemist, so if anyone sees any errors in my logic or if I am using circular arguments, please do correct me.
On the other hand, if Nisarg Bhavsar and I are correct, maybe someone should warn Jayadithya that the accepted answer is not correct (is it?).


text:
References for the Molecular Distance Geometry Problem
Note: This question has also been posted here.

The molecular distance geometry problem (MDGP) consists of two sub-problems:

Given observations of noisy distances between atoms in a molecule, estimate the values of the true distances.
Given these estimated distances, compute the locations of the atoms.

More formally, the first sub-problem can be stated as:
Given the datasets $\mathcal{D}_1,\mathcal{D}_2,\dots,\mathcal{D}_n$ of noisy distances for the atoms defined by the points $\mathcal{S} = \{x_1,x_2,\dots,x_n\}$, estimate the $n \times n$ symmetric distance matrix $\mathbf{A} = (d_{ij})$, where $d_{ij} = \lvert\lvert x_i - x_j\rvert\rvert$ and $x_i \in \mathbb{R}^K$ for $i,j \in \{1,2,...,n\}$.
The second sub-problem can then be formulated as:
Given $\hat{\mathbf{A}} = (\hat{d}_{ij})$, which is an estimate of $\mathbf{A} = (d_{ij})$, find the points $x_1,x_2,...,x_n$ such that $\lvert\lvert x_i - x_j\rvert\rvert = \hat{d}_{ij} \ \forall \ i,j$.
The second sub-problem is well-studied in the literature. If all distances $d_{ij}$ are given, and if $K=3$, then this problem can be solved using a linear order of operations [1]. Alternatively, if only a small subset of these distances are given, then it is possible to infer the rest of the unknown distances using specific geometrical constraints, such as the triangle inequality [2].
I am currently interested in the first sub-problem. More precisely, are there references that explore different noise models for the distances between atoms and references that attempt to estimate these distances?

You posted the same question on Matter Modelling SE : https://mattermodeling.stackexchange.com/questions/5088/references-for-the-molecular-distance-geometry-problem . Note that cross-posting the same question on different sites is not recommended.

Sorry, I wasn't sure which site would be more appropriate. Which one do you think is more appropriate for this question?

Personally, I would leave both questions as they are, but edit them to add the link to the question on the other site so that people know there is a cross post. I am not a moderator though, so it's just my personal opinion.


text:
Normal phase dead time marker
In reverse-phase HPLC, Thiourea, Uracil, Nitromethane or $\ce{KNO3}$ are used as dead time markers because of their polarity and therefore weak interactions with the stationary phase. However, I could not find any substances typically used as dead time markers in normal-phase-applications. These would have to have small, non-polar molecules and a high boiling point (I know that's a little contradictory since nonpolar molecules only form weak intermolecular bonds).
Thanks in advance.

Interesting and complex question for normal phase LC! Couple of years ago, it was addressed in quite detail. Most of the practitioners have accepted 1,3,5-Tri-tert-butylbenzene for normal phase LC because it is extremely non-polar. It works with most normal phase mobile phases. Boiling point has nothing to do with NPLC.
However, if you really want true "void" volume of a column, pycnometry is the way to go. It is quite simple.
Determination of void volume in normal phase liquid chromatography in the Journal of Chromatography A See paper here


text:
Will a Crooke's Tube stop working after sometime?
A Crooke's Tube is a cold-cathode tube, wherein electrons are released from the cathode when ionised gas particles strike it with speed. Following a chain reaction, other gas particles also get ionised over time and do the same. But I guess there would be a time when all the gaseous atoms would have been ionised and travelled towards the cathode. So, no more electrons should be released now...and thus no cathode rays.
Or is there something happening, like the ionised atoms becoming normal again. It arises one more question in my mind....what happens to the ions after they strike the cathode?

If only it were that easy to make ultra high vacuum. Sadly, your chamber outgasses material and you need to keep pumping on it to just maintain the vacuum.

Over time, the residual pressure in a gas discharge tube does drop. One of the main causes  for gas loss is sputtering. Ions sweep gas atoms from the tube and embed them in metal sputtered from the cathode, as does an ion vacuum pump.
For this reason, neon and fluorescent lamps show a gradual increase in voltage required to sustain the glow discharge, and the glass near the electrodes gradually darkens from sputtered metal.
However, loss of ions is not an issue - if atoms are broken apart into electrons and ions at one end of the discharge tube, they're happily reunited at the other end - though with different electrons. Of course, in the dark (discharge), all electrons look the same.

okay...thanx a lot

If the cations recombine with the electrons, then won't they release energy in form of light....as happens in glow discharge (negative glow)

@user95732, Absolutely! It's called the cathode glow, https://www.merriam-webster.com/dictionary/cathode%20glow . Furthermore, when high-energy electrons and ions smash into the electrodes, they release other wavelengths than visible light, i.e., X-rays. This is from braking, or Bremsstrahlung, radiation.  (should have an  eszett, not ss, IMHO).

Gotcha! But please tell me one thing more...when these electrons (cathode rays) strike the glass wall, some are transmitted, some absorbed and some reflected. What does absorption mean? Do the electrons loose all their energy in exiciting the electrons of glass atoms....and stop midway in the glass? Can they stop midway in the glass, between the atoms?

@user95732, yes,right again! Electrons and ions can be implanted into solids. In non-conductive materials, ionic defects (F-centers, https://en.wikipedia.org/wiki/F-center), Frrenkel defects, https://en.wikipedia.org/wiki/Frenkel_defect and hole-centers, http://www.minsocam.org/msa/collectors_corner/arc/color.htm, can cause coloration, as in quartz or diamonds exposed to ionizing radiation. Of course, if enough charge builds up, the material would arc over internally, even though it is a nonconductor. See https://www.caratsdirect2u.com/Articles.asp?ID=278

When ions strike an electrode, they are discharged and become electrically neutral. Cations, or positive ions, are attracted by the cathode (negative pole). Anions, or negative ions, are attracted by the anode (positive pole). Once discharged, they may again go into the gas phase, with or without chemical change.


text:
Why does one drain strainer rust while the other does not?
About a month back, I acquired drain strainers for my shower and kitchen sink.  The package I got included one of each, which I installed immediately.
Within two weeks, the shower strainer was showing spots of rust, and currently has several large holes where the wire mesh has disintegrated.  The kitchen strainer is perfectly fine (except for the cheese I haven't managed to clean off yet.
What could be causing the difference in rusting?  More importantly, what can I do to prevent the replacement shower strainer from rusting this quickly?

Some steel meshes are stainless steel, so they are rust resistant, and other steel meshes might be not stainless steel, to save a little money. You might try testing them with a magnet: some stainless steels are nonmagnetic or almost so, while ordinary steels are often ferromagnetic. If the magnet sticks to the shower drainer steel mesh, probably that drainer should be upgraded to a stainless steel mesh one. And take the magnet to the store to avoid non-stainless steel.

My guess would be that they are chrome plated steel . The one in the shower had thin or defective ,or damaged plating.

@blacksmith37 What you suggest is more likely than my comment, especially given that the two strainers were in the same purchased package. I guess “stainless steel” ain’t what it used to be!

Switch places, see what happens.

@IvanNeretin unfortunately, they're different sizes.


text:
When do repulsive forces in a gas actually become dominant?
We say that repulsive forces become dominant when Z>1
(i.e. Vobserved > Vtheoretical).
But when the difference between Vobserved and Vtheoretical  is less than b, shouldn't attractive forces still be dominant as the effective volume (i.e. volume in which gas molecules can move freely) of real gas is less than ideal gas?
b is the volume occupied by 1 Mole of gas molecules
Effective volume of real gas =Vobserved - b

You confuse the effect of the non-negligible own gas molecule volume with repulsive forces. The former leads to higher than expected pressure at given V,T even without repulsive forces.

It's worth emphasizing that there are no repulsive forces between non-ionized gas particles, other than those that keep them from overlapping with each other when they collide (i.e., other than the forces that cause the individual particles to take up space).  Other than that, the net force between gas particles is always attractive.  For the particles to feel a net repulsive force, they'd need to be squeezed together so closely that they would no longer be a gas.

To elaborate on theorist's comment above, there are both repulsive and attractive forces, the attractive ones have a much longer range than the repulsive ones which occur on 'contact' of the two molecules or atoms. Have a look at the shape of a Lennard-Jones potential.


text:
Why do Enantiomers have different chemical properties with optically active reagents?
As I understand Enantiomers have similiar physical properties but they do show different chemical properties with optically active reagents. This point is something which I am not able to logically comprehend. Why would a compound which being only different in rotation of direction of PPL behave differently in a chiral environment ?
I tried searching and found through the following answer
Why are optically active compounds abundant in nature?
here in comments it has been crudely mentioned that one of the enantiomers may react to some enzyme that specifically works on it....  Which makes me even more confused.
I will be glad if this could be cleared up.
Regards

I would be also cautious saying that they have the same physical properties. First, the main difference is what you are aware of. Second, at least in principle, you can imagine more. Like the friction on a chiral surface, or the viscosity in a chiral medium, etc. Again, as per Tyberius answer.

@Alchimista Thanks for your inputs :) Yes I do get the differences now..

Suppose you are introduced to someone, and you want to shake their hand.  They extend their right hand to do so.  No matter how you position your body or hands, you cannot grasp their right hand with your own left in the manner appropriate for a conventional handshake.  Your two hands have approximately the same kind of geometric relationship as a pair of opposite enantiomers, and the effect that only one of them has the right shape to complete that handshake is actually pretty analogous to the fact that sometimes only one of a pair of enantiomers has the right shape for a particular reaction.

@RishiKirti yet, as pure compounds themselves, then yes, all physical properties are the same except for the obvious one. This I add to prevent you getting troubles in the class ;)

I think a simplified version of the enzyme example is helpful here. Consider this image of binding a molecule to a receptor:

Using a model of enantiomers where the attached groups are just differently shaped blocks, it is clear to see that if one enantiomer fits, the other one can not, regardless of the way it is oriented. The different rotation of light is a consequence of this different arrangement of attached groups, which in turn has an effect on reactivity in chiral environments.
So while certain physical properties that just depend on the composition and arrangement of the groups in space will remain the same for each enantiomer, any property that depends on the relative positioning of these groups will differ. This includes optical rotation, but also chemical interactions between chiral species or more physical interactions or more exotic effects like the interaction with the electroweak field.

It is worth pointing out what exactly enantiomers are: exact mirror images of each other. Most of what enantiomers do and don't do can be understood by remembering that basic definition and applying it logically.
First, let's compare a chiral compound to an achiral one. Imagine an achiral molecule in front of a (molecular-sized) mirror. The orientation does not matter; the mirror will show a mirror image. Take a screenshot of the picture in the mirror. By definition, it will be possible to take that achiral molecule and rotate it so that it looks exactly like the screenshot you just took, except that you aren't using the mirror.
Whichever macroscopic property of the pure material you decide to measure, the resulting value is typically one of two things: either a simple number with a unit (i.e. it is a non-directional property; e.g. melting point) or a directional vector (e.g. rotation of polarised light. Each individual molecule will contribute identically to the simple-number properties; whereas the directional properties will typically be averaged out as there will always be a molecule oriented exactly opposite in a large enough sample.
Enantiomers on the other hand cannot be mapped onto their mirror image. Thus, it will be impossible to find a molecule oriented in exactly the opposite way in a pure sample of a single enatiomer. However, you are able to produce an equally pure sample of the opposite enantiomer, i.e. the exact mirror image. Now, the simple-number type of macroscopic properties will remain unaffected by which enantiomer you are measuring but those properties that are vector-based will be opposite, as they cannot be internally compensated.
Therefore it follows that unlike your first sentence enantiomers do not have similar properties but identical ones save those where direction actually matters. The most common example of an experiment where direction matters is the rotation of polarised light.
However, this difference is a consequence of the molecular structure; not the other way around. It is worth remembering that.
Now we move on to the interaction between a chiral compound and another compound, be it achiral or chiral. You can do some macroscopic experiments to investigate this, if the thought experiment isn't conclusive enough. For example, take a (small!) figure or sculpure or whatever of a human being or animal whose arms or legs are doing different things. This could be your chiral molecule. As an achiral object, you could take e.g. a mug. Play around with the two to see how they interact. You will be able to get 'better fits' one way but not the other way. For example, if your figure has one arm bent at an angle while the other is straight, this angled arm can probably easily lock around the mug's handle while they are both standing on the table; not, however, using the other arm. This is akin to a chiral molecule reacting with an achiral one in a stereospecific manner.
To understand how two chiral molecules interact, use two such asymmetric figures and see what you can do. One way will 'fit' notably 'better' than the other way. Finally, image one of the figures being a mirror image of itself. Now, the interaction will be markedly different and they may not even fit together at all. (Of course, it would be ideal if you had some kind of mirror images of figures but I'm not sure if that's all that common. Some kind of irl 3D tetris blocks perhaps?)
This is the gist behind the image in Tiberius' nice and succinct answer. The ramblings herein only serve to further illustrate the point and to allow a thought path to be re-understood.
Finally, I will briefly address that Alchemista's comment can also be understood in this context but I will argue that Alchemista's examples are not physical properties of the pure compound and thus not really applicable to the first part of this answer.

Thanks for clearing up things :) Yes the finger example does make sense.


text:
Why is milk not sold in food-grade stainless steel bottles?
I have grown up in a country where milk is stored in a steel container in the fridge at home. That’s what makes me wonder why milk is not sold that way?
Nowadays the steel container single wall is cheap anyway. Given the fact that steel is infinitely recyclable, I would think it is far cheaper than single-use disposable plastic or even reusable glass. Is there any scientific reason like bacteria formation when stored in steel? Or corrosion due to an acidic nature?

I don't see a reason it would be connected to chemistry of steel. It's just heavy and difficult to process.

There is no such thing as "food grade" stainless.  There are stainless steels traditionally used in the food industry , primarily 316 and 304.

PE-coated cartons are designed to be recyclable, and I suspect that the environmental impact of people who don't offer those for recycling is lower than the environmental impact of putting stainless steel into rubbish dumps. Bulk transport of milk is done in stainless steel tanks.

If you can solve the issues steel has, glass would still beat it. In fact, some places use it, with the niche clientele that is able to not throw the containers away.

This question probably makes more sense if you live in a country where bottle recycling is a regular feature

Your country doing it one way does not mean that it is the best way.  Even before reading the Answers, my first thoughts were, "steel is expensive and heavy compared to plastic and paper.  (Heck, in Canada, milk is sold in thin plastic bags!!!

I believe it is cost.  According to 1, food-grade stainless steel works well as a container for milk.
However, the article also says that the wholesale cost of a 0.568 L (19.2 fluid ounce) glass milk bottle in Great Britain was 0.35 £ ($0.50) in 2020.
By contrast, I searched a few wholesale sites, and the least expensive staineless steel water bottle I could find in that size (20 oz.) was $3.19 in quantities >2000[2].  [US cost.]
On the other hand, stainless steel does have a longer lifetime than glass, which reduces the effective cost. But that would only apply if customers actually returned the bottles, which many don't; and the deposit doesn't make up for the cost of the stainless steel container.
References

Tomasz Błażejewski, Stuart RJ Walker, Rukayya Ibrahim Muazu, Rachael H Rothman, Reimagining the milk supply chain: Reusable vessels for bulk delivery, Sustainable Production and Consumption,
Volume 27, 2021, DOI: 10.1016/j.spc.2021.02.030.
https://www.silkletter.com/venture-20-oz-stainless-steel-sports-bottle.html

Also, plastic is lighter and reduces transport costs.  If you ship it far enough, a plastic bottle will save more than its own cost compared to the steel one.  This is also an argument for plastic over glass which used to be common.  Finally, the recycling of steel is not free.

@badjohn  The article I linked addresses those issues in a lot of detail; though it concerns wholesale, rather than retail, delivery of milk.

Sorry, I didn't follow the links and just read your text.  For the benefit of other lazy readers, it might be nice to mention them.

There are other types of milk containers, like cardboard (with a very thin plastic (polyethylene) coating on the inside) - for instance, used for the majority of milk products where I live. It has started to be recycled here (not just incinerated). Plastic-coated paper

The other advantage of plastic or glass over steel is that it's transparent - One of the first signs the milk is starting to go bad is curdled lumps clinging to the side of the bottle.  With steel, you might not notice that until it starts to smell or the lumps are solid enough to pour out, by which point the milk has already been partly bad for a few days.  (Cardboard containers have the same problem, with the added drawback that they don't seal very well and are thus more likely to spoil quickly.)

@DarrelHoffman: Transparent containers also let in light, which causes vitamins to degrade and may impact flavor (e.g., see http://dx.doi.org/10.3168/jds.S0022-0302(02)74080-0 ).

While cost of one steel bottle is much higher than one single use disposable Plastic packaging, steel is reusable hundreds if not thousands of times. Steel is strong and durable as well to last long time. We have been using some single walled steel utensils for 4 generations now and still probably last long.

@Brian True, but you probably store your milk in the fridge most of the time, and as long as the door is closed, there's no light in there.  The bit of light it gets when you take it out to use is probably too little to have much noticeable effect.  (If you leave it out long enough for that to be a problem, you've got worse problems than light to deal with as well.)  The only issue in that case is that the fridges in the store are still lit, but hopefully you're checking the dates and not buying milk that's been on the shelf for too long...

@DarrelHoffman: The problem isn't at home.  The problem is it being in supermarket stock rooms, supermarket displays, and other lighted refrigerators.

Any steel container is going to quickly show wear and dents in it.  If you use one thin enough to be cost and weight effective, even more so.
Milk is one of those products that has to look clean and fresh on the shelf.  A dented can isn't going to work well in this situation.

That's a really good point.  The analysis I cited was for commercial deliveries, where customers are concerned more about practicality than presentation.  Retail is an entirely different ballgame.

A dented bottle is also probably not going to get deposit back. We have to be responsible with the bottle and return it in acceptable condition to get full deposit back. This way we can keep bottle in circulation for long and eliminate single use waste.

Polyethylene-coated cartons are designed to be recyclable, and I suspect that the environmental impact of people who don't offer those for recycling is lower than the environmental impact of putting stainless steel into rubbish dumps.
Glass is very recyclable, but there are additional transport costs associated with the weight and volume overheads of glass containers compared to PE-coated card, and broken glass is more dangerous than broken card.
Milk has a limited use time because there will be bacteria in it, regardless of the packaging used to contain it. Milk which has been filtered to remove (some) bacteria is available (e.g. sold as Cravendale, but even that has a limited lifetime). UHT milk is perhaps not the same product as normal milk, depending on what it's going to be used for.
Bulk transport of milk is done in stainless steel tanks which are cleaned between uses: the re-use of a milk-tanker is pretty-much guaranteed.
Overall, it is cheaper to use PE-coated cartons than any other package for small volumes of milk.