text:
What should be the name of the given compound?

Following compound is named as

(1) 6-Mercaptocyclohex-4-ene-1,3-diol
(2) 1-Mercaptocyclohex-2-ene-4,6-diol
(3) 1-Mercaptocyclohex-5-ene-2,4-diol
(4) 4-Mercaptocyclohex-2-ene-1,5-diol

As per my knowledge if I start numbering the compound from the alcohol then to mercapto group, then alkene group.
So, according to me, the numbering should be from bottom alcohol as number 1 and making the way to the top alcohol as number 3. Then mercapto will come at number 4 and alkene at 5.
So as per me the name should be 4-mercaptocyclohex-5-ene-1,3-diol, but according to the answer key the answer should be (1). Is there any rule which I am missing out on?

The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

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:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

For the compound that is given in the question, the principal characteristic group is $\ce{-OH}$. It is expressed as suffix “ol”. Thus, low locants are assigned first to the $\ce{-OH}$ groups according to Rule (c), which yields “…1,3-diol”.
Next, a low locant is given to the double bond according to Rule (e), which gives “4-ene-1,3-diol”.
This step also establishes the numbering of the other atoms of the ring so that there is no longer a choice for the last substituent. Nevertheless, strictly following the rules, a low locant can be assigned to the remaining substituent according to Rule (f).
Thus, the complete name is 6-sulfanylcyclohex-4-ene-1,3-diol. Note that the prefix “mercapto” is no longer recommended.


text:
Why do the NMR peaks shift when H+ concentration and temperature are varied on the reversible hydrolysis of pyruvic acid to 2,2-dihydroxypropionate?
We performed two experiments. Each of the peaks are referring to the methyl peaks in both reactant and product of the hydrolysis reaction.

We varied $\ce{H+}$ concentration of the reaction and the peaks move ever so slightly right (up field) as acid concentration increases. Why?
As more acid is added there is more pyruvic acid present than the hydrate. But pyruvic acid is more deshielded, so I would have expended it to shift downfield because more deshielded molecules are present.

We varied temperature (45, 50, 55, 60 and 65 °C) and the peaks move slightly left (down field) as temperature is increased. Why?
The only possible explanations I can come up with are that the temperature changes conformations in a way that is more deshielded, but I don't know why specifically?

What nucleus are you monitoring? What is the solvent, pH etc

Ask yourself: generally speaking, what does it mean when a nucleus becomes deshielded? If you know what that means, you are probably in as good a position as anyone to propose a cause.

Proton NMR, My hydrogen ion concentrations are: 0.000E+00, 4.444E-04, 8.889E-04, 1.333E-03, 1.778E-03, 2.222E-03

I understand that shielding is relating to electron density, but I am still note sure why they would shift, based solely on changes in temperature and acid concentration

I would begin by looking at the inductive effect of different groups and attempting to rationalize the observations in terms of this: https://en.wikipedia.org/wiki/Inductive_effect#Inductive_effect

I recommend you add some spectra to illustrate the observations. As it stands there are dynamic effects that would not be easy to describe in words but would be quite apparent if you show a series of spectra (as function of pH or temperature). You don't need to show the entire spectra, just the properly referenced methyl region.

Like, you want the NMR spectra???

Sure, you can take screenshots of overlayed spectra, emphasizing methyl region. You don't have to do this, but I strongly think (can't be sure) that it might help to clarify what is happening.

I . dont have the spectra, they were on a computer and we just recorded the numebrs

Let us continue this discussion in chat.

source: Stack Exchange

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text:
Does the volume change when the reactions occur?
I have this reaction (all substances are gaseous):
$$
\begin{align}
\ce{A + 4.5 B &→ C + 2 D + 2 E}\tag{R1}\\
\ce{C + 7.5 B &→ 8 D + 2 E}\tag{R2}
\end{align}
$$
Does the volume change when the reactions has occurred?
Because if so, the concentration changes too, right?

The volume changes can only be predicted if at least one of the substances A, B, C, D and E is gaseous. If the substances are not gaseous, no predictions can be made.

Just edited it(all substances are gaseous)

I don't understand why sometimes questions that classify as homework (of course they might even be genuine users Q) get answered. OP poses a trivial question without attempts/discussion at all.

@Alchimista To poorly paraphrase Michael Caine, "...some men just want to watch the world learn".

A volume change from a reaction occurs if the difference in mole of gases does not equal to $0$
In order to find the answer for your question, we will have to add both reactions together, resulting in:
$$\ce{A + 4.5 B + C + 7.5 B -> C + 2 D + 2 E + 8 D + 2E}$$
We can shorten the reaction by adding up all duplicates and removing both $\ce{C}$ from each side, resulting in:
$$\ce{A + 12 B -> 10 D + 4E}$$
Now if we calculate the difference in mole gasses we get:
$$\Delta n = \pu{(10 + 4)mol} - \pu{(1 + 12)mol} = \pu{1 mol}$$
Based on the positiv difference in mole of gasses we can say that the reaction will change in volume, that is it will expand.

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text:
n-Factor of P4O6
I had a question in my exam:

Which of the following can be the $n$-factor of $\ce{P4O6}?$
A. 8
B. 9
C. 10
D. 12

I went for options A and B as in cold water $1$ mole of $\ce{P4O6}$ on hydrolysis give $4$ moles of $\ce{H3PO3}$, whereas in hot water it gives $3$ moles of $\ce{H3PO4}$ and $1$ mole of $\ce{PH3}$, thus getting $n$-factor  $8$ and $9,$ respectively.
However, the answer given is A, B and C. How can the $n$-factor be $10?$

Assuming the n-factor is a measure of dissociation, what happens when you dissolve (for instance) H3PO4 in water?

Whoever designed this question doesn't understand the equivalent weight concept. Equivalent weight is dependent on a reaction. So the key question is n-factor for which reaction?

@sjb-2812, n factor is very old concept of equivalent weights which is still taught in India. It has nothing to do with dissociation.

Anyway it is necessary that Zenix tells us what is the n-factor, for him.

As M. Farooq pointed in the comments, the concept of $n$-factor is applied to the reactions a given compound participates in, namely of the following types:

acid-base, neutralization (in a context of Arrhenius approach, i.e. the amount of transported hydronium ions);
redox (total change in oxidation state per mole of the substance);
precipitation and double decomposition reactions.

This concept is so rarely used these days that I actually had to look this up to refresh my memory.
Apparently, $\ce{P4O6}$ doesn't have any protons to transfer, so the first category is irrelevant.
I also don't think that the question counts basicity of the acids that can be formed by hydrolysis or decomposition of $\ce{P4O6}$ in water, so I doubt your reasoning is relevant to this question.
Besides, in hot water there is also elementary phosphorous among the products, and the stoichiometry is going to be quite different and won't justify $n$-factor $9.$
However, since the decomposition of $\ce{P4O6}$ in hot water is in fact a redox reaction:
$$\ce{6 \overset{+3}{P}_4O6(s) + 24 H2O(l) ->[Δ] 8 \overset{0}{P}(s, red) + 15 H3\overset{+5}{P}O4(aq) + \overset{-3}{P}H3(g)}\tag{R1}$$
$$n\text{-factor}(\ce{P4O6}) = \frac{8 × |3- 0| + 15 × |3 - 5| + 1 × |3 - (-3)|}{6} = 10\tag{1}$$
which explains answer C.
Oxidation with oxygen, another redox reaction, explains answer A:
$$\ce{\overset{+3}{P}_4O6(s) + 2 O2(g) ->[\pu{50 - 120 °C}] \overset{+5}{P}_4O10(s)}\tag{R2}$$
$$n\text{-factor}(\ce{P4O6}) = 4 × |3 - 5| = 8\tag{2}$$
Another redox reaction, for example, thermal decomposition in vacuum, could also demonstrate another $n$-factor:
$$\ce{4 \overset{+3}{P}_4O6(s) + 2 O2(g) ->[\pu{210 - 250 °C}] 3 \overset{+3}{P}_2\overset{+5}{P}_2O8 + 4 \overset{0}{P}(s,red)}\tag{R3}$$
$$n\text{-factor}(\ce{P4O6}) = \frac{3 × 2 × |3 - 5| + 4 × |3 - 0|}{4} = 6\tag{3}$$
And so on, so it looks like the question requires you to come up with the examples of the reactions that would be impossible for the listed $n$-factors.
I think this is an unjust choice for an exam as oxy-compounds of phosphorous have rich chemistry and I still cannot come up with the suitable process for the $n$-factor $9$.
Notes
The reactions are adapted from [1, pp. 167]; they can also be found in Russian and German Wikipedia articles on phosphorous(III) oxide.
References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


text:
Chromium cyanide complexes
I have rarely seen any chromium $(\ce{Cr})$ square planar complexes and I have been told that $\ce{[Cr(CN)4]-}$ is tetrahedral.
So, if the statement is actually correct, are there any other known examples of square planar complexes of chromium?

This doesn't appear to be a rare case and there are plenty $\ce{Cr^2+}$ square planar complexes.
I found 15 square planar complexes with well-determined crystal structures $(R_\mathrm{int} < 10\%)$ by applying the following geometrical restrains:

Chromium must have C.N. exactly 4 (T4).
Chromium is allowed to be coordinated by virtually any element of periodic table $\ce{El}$.
Dashed lines signify any bond type.
All four angles between neighbors should lead to nearly ideal square planar geometry: $∠(\ce{El-Cr-El}) = \pu{(90.0 ± 0.5)^\circ}$ (by allowing higher tolerance the number of structures grows in geometrical progression: for $∠(\ce{El-Cr-El}) = \pu{(90.0 ± 5.0)^\circ}$ there were nearly 100 structures).

List of compounds

CCDC ACACCS
[1]
bis(2,4-Pentanedionato)-chromium(II)
CCDC BHPZCR
[2]
bis(Dihydro-bis(1-pyrazolyl)borato) chromium(II)
CCDC CRPORT
[3]
α,β,γ,δ-Tetraphenylporphinato-chromium(II) toluene solvate
CCDC DAQDUV
[4]
hexakis(Tetrahydrofuran)-magnesium tetrakis(perfluorophenyl)-chromium(II) tetrahydrofuran solvate
CCDC DAYNEW
[5]
bis(Pyridinium) diaqua-dibromo-chromium(II) dibromide
CCDC EGINEN
[6]
bis((μ2-Chloro)-hexakis(tetrahydrofuran)-di-lithium) tetrakis(pentachlorophenyl)-chromium(II)
CCDC EQIVOP
[7]
trans-Dichloro-bis(2,6-dimethylpyridine)-chromium(II)
CCDC HARKIU
[8]
trans-bis(tris(t-Butoxy)siloxy)-bis(diethylammonio)-chromium(II)
CCDC KOMRAF
[9]
Tetrapyridine-chromium(II) bis(hexafluorophosphate) dipyridine clathrate
CCDC PITVAP
[10]
bis(Tetra-n-butylammonium) tetrakis(isothiocyanato)-chromium(II)
CCDC PUQPEY
[11]
bis(1-t-butyl-3-(di-t-butylphosphino)imidazol-2-ylidene)-dichloro-chromium
CCDC PUQPIC
[11]
dichloro-bis(1-(di-t-butylphosphino)-3-mesitylimidazol-2-ylidene)-chromium tetrahydrofuran solvate
CCDC PUQPIC
[12]
bis((Phenothiazinyl-N)-tetrahydrofuran)-chromium(II)
CCDC XEBZAF
[13]
trans-bis(4-t-Butylpyridine)-dichloro-chromium(II)
CCDC ZOYCUL
[14]
Dichloro-bis((trimethylphosphine)trimethylsilylamido)-chromium(II)

References

Cotton, F. A.; Rice, C. E.; Rice, G. W. The Crystal and Molecular Structures of Bis(2,4-Pentanedionato)Chromium. Inorganica Chimica Acta 1977, 24, 231–234. https://doi.org/10/bhjhh5.
Dapporto, P.; Mani, F.; Mealli, C. Vanadium(II) and chromium(II) Complexes with Poly(1-Pyrazolyl)Borate Ligands. Crystal and Molecular Structures of Potassium Tris[Dihydrobis(1-Pyrazolyl)Borato]Vanadium(II) Monoalcoholate and Bis[Dihydrobis(1-Pyrazolyl)Borato]chromium(II). Inorg. Chem. 1978, 17 (5), 1323–1329. https://doi.org/10/dzfdh8.
Scheidt, W. R.; Reed, C. A. Stereochemistry of the Toluene Solvate of α,β,γ,δ-Tetraphenylporphinatochromium(II). Inorg. Chem. 1978, 17 (3), 710–714. https://doi.org/10/cvqg73.
Forniés, J.; Martín, A.; Martín, L. F.; Menjón, B.; Zhen, H.; Bell, A.; Rhodes, L. F. The First Structurally Characterized Homoleptic Aryl-Manganese(III) Compound and the Corresponding Isoleptic and Isoelectronic chromium(II) Derivative. Organometallics 2005, 24 (13), 3266–3271. https://doi.org/10/bz3r8w.
Ladd, M. F. C.; Larkworthy, L. F.; Leonard, G. A.; Povey, D. C.; Tandon, S. S. Crystal and Molecular Structure of a Complex Bromide Containing Planar Diaquadibromochromium(II) Units. J. Chem. Soc., Dalton Trans. 1984, No. 10, 2351. https://doi.org/10/cj5z95.
Alonso, P. J.; Forniés, J.; García‐Monforte, M. A.; Martín, A.; Menjón, B.; Rillo, C. A New Series of Homoleptic, Paramagnetic Organochromium Derivatives: Synthesis, Characterization, and Study of Their Magnetic Properties. Chemistry – A European Journal 2002, 8 (17), 4056–4065. https://doi.org/10/cx9r27.
Gibson, V. C.; Newton, C.; Redshaw, C.; Solan, G. A.; White, A. J. P.; Williams, D. J. Low valent chromium complexes bearing N,O-chelating pyridyl-enolate ligands $\ce{[OC(Bu^t)(=2-CHN5H3Me-x)]^-}$ $(x = 3–6)$. Dalton Trans. 2003, No. 24, 4612–4617. https://doi.org/10/ffdkvr.
Terry, K. W.; Gantzel, P. K.; Tilley, T. D. chromium(II) and Chromium(III) Tri-Tert-Butoxysiloxy Complexes. Inorg. Chem. 1993, 32 (23), 5402–5404. https://doi.org/10/dfw68w.
Fochi, Giovanni.; Straehle, Joachim.; Gingl, Franz. Disproportionation of the Bis(Benzene)Chromium Cation Induced by Pyridine. Crystal and Molecular Structure of Two Tetrapyridine Complexes Containing chromium(II) and Chromium(III). Inorg. Chem. 1991, 30 (24), 4669–4671. https://doi.org/10/c3ffnj.
Larkworthy, L. F.; Leonard, G. A.; Povey, D. C.; Tandon, S. S.; Tucker, B. J.; Smith, G. W. Crystal Structures and Magnetic Behaviour of the Two Forms of Tetra-n-Butylammonium Tetraisothiocyanatochromate(II). J. Chem. Soc., Dalton Trans. 1994, No. 9, 1425. https://doi.org/10/cgtgbm.
Ai, P.; Danopoulos, A. A.; Braunstein, P. N-Phosphanyl- and N,N′-Diphosphanyl-Substituted N-Heterocyclic Carbene Chromium Complexes: Synthesis, Structures, and Catalytic Ethylene Oligomerization. Organometallics 2015, 34 (16), 4109–4116. https://doi.org/10/ggc5mx.
Edema, J. J. H.; Gambarotta, S.; Meetsma, A.; Spek, A. L.; Smeets, W. J. J.; Chiang, M. Y. chromium(II) Amides: Synthesis and Structures. J. Chem. Soc., Dalton Trans. 1993, No. 5, 789. https://doi.org/10/fcfcdg.
Cotton, F. A.; Dikarev, E. V.; Gu, J.; Herrero, S.; Modec, B. Alkylpyridine Complexes of Tungsten(II) and chromium(II). First Rotational Isomers of $\ce{W2X4L4}$ Molecules with $D_\mathrm{2h}$ and $D_2$ Symmetries. Inorg. Chem. 2000, 39 (23), 5407–5411. https://doi.org/10/c5cfr4.
Miekisch, T.; Mai, H. J.; zu Köcker, R. M.; Dehnicke, K.; Magull, J.; Goesmann, H. Silylierte Phosphanimin‐Komplexe von Chrom(II), Palladium(II) und Kupfer(II). Die Kristallstrukturen von $\ce{[CrCl2(Me3SiNPMe3)2]},$ $\ce{[PdCl2(Me3SiNPEt3)2]}$ und $\ce{[CuCl2(Me3SiNPMe3)]2}.$ Z. Anorg. Allg. Chem. 1996, 622 (3), 583–588. https://doi.org/10/bm3fq8.

Clearly the porphyrin complex is cheating as that could never be tetrahedral ;)

@Jan No (DABDA), your zinc is cheating, my chromium is adapting:D Could be worse, and sit way above the plane, like naughty lanthanoids do:)

An article by Bradley et al. [1] documents $\ce{[Cr(N(SiMe3)2)2(THF)2]}$ as a square planar complex.  However, the $\ce{[Cr(N(SiMe3)2)3NO]}$ complex is pseudotetrahedral $(C_\mathrm{3v}).$
Figures below are provided by andselisk, the are from the reference above.
References

Bradley, D. C.; Hursthouse, M. B.; Newing, C. W.; Welch, A. J. Square Planar and Tetrahedral Chromium(II) Complexes; Crystal Structure Determinations. J. Chem. Soc., Chem. Commun. 1972, No. 9, 567–568. https://doi.org/10/br8qtf.


text:
What is the use of both mol and mass representing the same thing i.e. amount of something?
A mol is a unit but also has a magnitude. How can a unit have a magnitude? And if mole is used to measure an amount of substance than what does mass measure?

I don't understand what is being asked here. Strictly following your logic, there is no difference between mass and volume, or even between any extensive properties whatsoever, which is just wrong.

Maybe the language got a bit complex. I simply want to ask why there is a need of both mol and mass to measure the amount of a substance

If you have 1 mol of O2 and 1 mol of N2, do they have the same mass? No, moles and mass measure 2 different things. At least I think this is what you are asking.

Mole itself doesn't have a magnitude, but the value of mole does, in fact it is the magnitude. Any measurement has a magnitude (ie for the mass of 5 kg the magnitude will also be 5).
I think you are confusing it with scalar measurements (ie forces) which also have a direction. For example 5 Newtons 'in that direction'.

Mole is similar to a dozen (just a bit larger). It is used to count how many atoms, molecules, atoms or anything else there are.
If you think about mass as "the amount of stuff there is in an object" you can confuse it with mols
but the actual definition of mass is:

A property of a physical body and a measure of its resistance to acceleration (a change in its state of motion) when a net force is applied.

which is different to mols (which is used to count objects).


text:
What does ethylene glycol decompose to when exposed to air for an extended period of time?
I came across this article on wikipedia on ethylene glycol stating that it decomposes fairly quickly. I was under the impression that ethylene glycol was dangerous to let leach into soil, but I think I'm mistaken as wikipedia states

it [ethylene glycol] breaks down in air in about 10 days and in water or soil in a few weeks. It enters the environment through the dispersal of ethylene glycol-containing products, especially at airports, where it is used in deicing agents for runways and aeroplanes.

What does it decompose into ($\ce{CO2 + H2O}$?) and what is the mechanism?
Does this mean that a tub of ethylene glycol exposed to air will eventually contain only water?
Is the reaction the same when ethylene glycol penetrates soil around runways?
What is the activation energy for the decomposition and at what temperature will it no longer decompose?

This image highlights the oxidation process of ethylene glycol fairly well

( Taken from Ethylene glycol: properties, synthesis, and applications )

Since it has little tendency to evaporate, unless it is kept in direct sunlight, it would probably oxidize to a mixture of the above liquid compounds (glyoxal, glyoxalic acid) and not directly to water. If it is kept in sunlight, the reaction would be photochemically catalysed and a it may degrade to carbon dioxide faster, but according to theoretical principles you would not get much water.
Ethylene glycol is relatively hydrophilic and also has a low tendency to separate from water, so the only major degradation process available to it in soil would be biodegradation by either aerobic or anaerobic process, which is NOT the same as atmospheric oxidation1
Not sure about reaction energetics regarding the same

Most of the danger regarding glycol spills (and other chemicals in the environment) comes from absorption by other organisms; since glycol is comparatively less volatile and more hydrophilic than, say ethanol, it has a greater tendency to be absorbed by organisms.
1 Taken from William R. Roy, in Handbook of Antistatics (Second Edition), 2016


text:
What happens when ferric chloride is dissolved in water?
Ferric chloride can be prepared by dissolving hematite in hydrochloric acid:
$$\ce{Fe2O3 + 6HCl -> 2FeCl3 + 3H2O }\tag{1}$$
But the resulting ferric chloride will be exposed from its formation to water.
After investigating a little, I found that anhydrous ferric chloride, when dissolved in water, hydrolysed and forms ferric hydroxide:
$$\ce{FeCl3 + 3H2O -> Fe(OH)3 + 3HCl }\tag{2}$$
But isn't it supposed that, when dissolved in water, hexahydrate is formed? What really happens, hydrolyses or hydrates?

There is a great deal of difference between pure water and acidified water. Also, I don't believe it will hydrolyze completely even in pure water.

Related: https://chemistry.stackexchange.com/questions/75762/why-would-a-solution-of-fecl2-appear-brown-yellow-sort-of-like-fecl3/75768

Iron(III) chloride indeed hydrolyses in water, or to be precise, $\ce{Fe^3+}$ does assuring acidic medium.
Hydrolysis isn't completely suppressed even in strongly acidic solutions.
In the absence of other factors, what happens exactly depends on the concentration of chloride and temperature.
In diluted cold solution a typical formation of metal aqua ions takes place:
$$
\begin{align}
\ce{FeCl3 + 4 H2O &<=> [Fe(H2O)4Cl2]+ + Cl-}\tag{R1.1}\\
\ce{[Fe(H2O)4Cl2]+ + H2O &<=> [Fe(H2O)5Cl]^2+ + Cl-} &\quad \mathrm{p}K_\mathrm{a} &= 0.30\tag{R1.2}\\
\ce{[Fe(H2O)5Cl]^2+ + H2O &<=> [Fe(H2O)6]^3+ + Cl-} &\quad \mathrm{p}K_\mathrm{a} &= 1.48\tag{R1.3}\\
\ce{[Fe(H2O)5Cl]^2+ + H2O &<=> [Fe(H2O)4(OH)Cl]+ + H3O+} &\quad \mathrm{p}K_\mathrm{a} &= 3.05\tag{R1.4}\\
\end{align}
$$
In concentrated cold solution aquacomplex is also accompanied by  tetrachloroferrate(III):
$$\ce{2 FeCl3 + 4 H2O <=> [Fe(H2O)4Cl2)]+ + [FeCl4]-}\tag{R2}$$
In boiling-hot solution iron(III) chloride decomposes to hydroxochloride
$$\ce{FeCl3 + 2 H2O ->[Δ] Fe(OH)2Cl(s) + 2 HCl}\tag{R3}$$
which is partially hydrolyzed further to colloidal iron(III) hydroxide ($\ce{Fe(OH)2Cl}$ is still more soluble than $\ce{Fe(OH)3}$).
Finally treatment with steam results in iron(III) oxide:
$$\ce{2 FeCl3 + 3 H2O ->[\pu{350 - 500 °C}] Fe2O3 + 6 HCl}\tag{R4}$$
Reactions were adopted from [1, pp. 420–421].
References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


text:
Catalyst discovery
How are catalysts for chemical reactions found?
Given a hoped for reaction, let's take the hydrogenation of carbon monoxide to methanol as an example, do chemists:

Proceed directly from some property of the reactants to an obvious (to the initiated) catalyst according to some grand unified theory of chemistry
Perform a computerized search over some set of simulations of different catalyzed reactions
Proceed by trial and error, trying different catalysts in the lab and hoping to find a good one
Proceed by trial and error but informed by some negative criteria. "Obviously, the catalyst can't contain [element] because [reason]"

And given a proposed catalyst how much can a chemist predict about the reaction. Would the optimal temperature and pressure be calculable?

Unanswered duplicate: How are catalysts discovered for a particular reaction?

Point 1 would be beautiful, but often things behave funnier in the flask then they do on paper, and the known "gtu" (the laws of quantum mechanics) exists, but is computationally quite expensive. Point 2 has the same limitations given above, but there are some examples of such applications, eg: https://doi.org/10.1021/jacs.8b10499 Point 3 and 4 would go together, and they are the "classical" way: you try some plausible things, and see if they work. You missed point 5: serendipity and observation

I hope this question gets a well-written answer!

All of the paths given by the OP can lead to discovery of catalysts. Usually, it is a combination of strategies and an accumulation of knowledge from different sources. I will give two examples, ammonia synthesis and enzymatic reactions.
Ammonia synthesis
$$\ce{N2(g) + 3H2(g) <=> 3NH3(g)}$$
The synthesis of ammonia from the elements is one of the largest scale reactions nowadays, and it provides most of the fixed nitrogen (i.e. other than elemental) for human food globally (some nitrogen is fixed by bacteria, especially in roots of legumes). In this sketch, I am using "Ammonia Synthesis Catalysts: Innovation And Practice" by Liu Huazhang as one of the sources. There is also a wikipedia article on one of the players, Alwin Mittasch, that I used.
Fritz Haber figured out that high pressure and low temperature would give the best yield for the reaction (as it is exothermic and the number of gas molecules decreases). This provided the "negative criteria", point 4 made by the OP: the catalyst has to be stable at high pressure. He developed a circulating process to enrich the product and received a patent for it in 1908. Haber and Bosch, working for the BASF company, brought this to industrial scale in the early 1900s, resulting in the Haber-Bosch process. To find a catalyst, they screened thousands of potential substances (strategy 3 given by the OP). They found an efficient osmium catalyist, but osmium was to expensive for commercial use.
The next improvement was a combination of serendipity and rational followup. Mittasch and Wolf, both working for Bosch, were key players. Wolf tried a magnetite from Gallivare, Sweden, that had been "sitting on a shelf" as a catalyst, and it showed promise. There were earlier observations that iron is a good catalyst, but pure iron did not work. Mittasch systematically followed up, figuring out how to reproduce the result by making a catalyst from $\ce{Fe3O4, K2O, CaO, Al2O3 and SiO2}$.
This catalyst, with minor modifications, is still used today. Haber and Bosch won the Nobel prize in 1918, and Ertl won the Nobel prize in 2007 for elucidating the molecular mechanism of the catalysis. Research is ongoing to find other paths to nitrogen fixation (see e.g. here), using other catalysts inspired by biological nitrogen fixation or different precursors (e.g. nitrogen and water for photocatalytical synthesis of ammonia).
Enzymatic catalysis
Enzymes have been developed by trial and error in the course of evolution. It is thought that often enzymes catalyzing on particular reaction led to related enzymes that catalyze related reactions.
A rational way to find a protein that catalyzes a given reaction is to make a transition-state analog. Then, you raise antibodies against that analog. In theory, the antibodies should act as catalysts for the desired reaction, and often they do (catalytic antibodies). This might be the best example of strategy 1 given by the OP.
In 2018, Frances Arnold was awarded a Nobel prize for directed evolution. She was able to mutate known enzymes (or more generally proteins) to catalyze reactions that have no known counterpart in biological systems (for example making carbon-silicon bonds). The process is rational (as in having a plan) but involves randomization and screening steps.

How are catalysts for chemical reactions found?

The answer is all of the above.


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What is the mass fraction of trichloroethene in the liquid phase using vapour pressure and mole fraction in gas phase?

Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene ($\ce{C2HCl3}$) and trichloromethane ($\ce{CHCl3}$) is established at $\pu{25 °C}$. The mole fraction of $\ce{CHCl3}$ in the vapour phase is $0.73$. What is the mass fraction of $\ce{C2HCl3}$ in the liquid phase?  Round your answer to two significant figures.

The vapour pressures of trichloroethene and trichloromethane at $\pu{25 °C}$ are:
$$\begin{align}P_\text{vap}(\ce{C2HCl3}) &= \pu{73.0 mmHg}\\[0.5em]
P_\text{vap}(\ce{CHCl3}) &= \pu{199.1 mmHg}\end{align}$$
So, what I did was I found mole fraction of $\ce{C2HCl3}$ and then used the two mole fractions along with the vapour pressures to find the total pressure of the solution.
$$P_\text{vap}= \frac{0.73}{199.1}+\frac{0.27}{73} = \pu{165.053mmHg}$$
Then, from Raoult's Law I know that the mole fraction in liquid phase is equal to mole fraction in vapour phase, multiplied by vapour pressure, divided by total pressure. From that, I found the mole fraction of both things in liquid phase. I use the mole fraction to find mass of both, and then did mass of $\ce{C2HCl3}$ divided by the total mass that I calculated. I got an answer of
$\frac{15.68}{120}=0.13$ but it says that it's wrong. I'm not sure where I messed up?

There is actually small error in the calculation part of total pressure and in the Raoult’s law formulae.
According to Raoult’s law:
$$ P_A= P^0 _ A \chi_A $$
i.e. partial pressure of a component in a solution($P_A$) is the product of the vapour pressure of the pure solvent A at that temperature($ P^0_A$) and it's mole fraction in liquid phase($\chi_A$).
Also, we have
$$ P_A= P_{total}\chi^‘_A$$
where:
$P_A$=partial pressure of component A
$P_{total}$=total vapour pressure of the solution
$\chi^{’}_A$=mole fraction of component A in vapour phase
Considering these equations along with Dalton’s law of partial pressure (total vapour pressure of a solution is the sum of vapour pressures of its individual components)$$P_{total}=P_A+P_B$$
we get:
$$P^0_A\chi_A=P_{total}\chi^{‘}_A$$
Substituting the respective values:
$$ (199.1){\chi_{\ce{CHCl3}}}=P_{total}(0.73)$$
and
$$ (73){\chi_{\ce{C2HCl3}}}=P_{total}(0.27)$$
Since $${\chi_{\ce{CHCl3}}}+{\chi_{\ce{C2HCl3}}}=1$$
we have two equation with two unknowns which upon solving we get ${\chi_{\ce{CHCl3}}}=0.4978$
and ${\chi_{\ce{C2HCl3}}}=0.5022$ and you shall continue in the same way to get the mass fraction.
Hope it helps.

As far as the question is concerned, you do not need the vapor pressure. In order to calculate the mass fraction of $\ce{C2HCl3}$, we first have to calculate the molar mass of $\ce{C2HCl3}$ and $\ce{CHCl3}$.
$$M(\ce{C2HCl3}) = \pu{131.38 g mol-1}$$
$$M(\ce{CHCl3}) = \pu{119.37 g mol-1}$$
With these two calculated we can calculate the absolute mass relative to $\pu{1 mol}$, due to the fact of ${\pu{0.73 mol}}/{\pu{1 mol}}$ being relative to a total of $\pu{1 mol}$ of molecules. Now if we calculate the absolute masses, we get:
$$M(\ce{C2HCl3}) = \pu{131.38 g mol-1}\times\pu{0.27\frac{mol}{mol}} = \pu{35.47 g mol-1}$$
$$M(\ce{CHCl3}) = \pu{119.37 g mol-1}\times\pu{0.73\frac{mol}{mol}} = \pu{87.14 g mol-1}$$
And the total mass in the system results in:
$$m_\text{total}=\pu{35.47 g mol-1}+\pu{87.14 g mol-1}=\pu{122.61 g mol-1}$$
And with the total relative mass given, we can calculate the mass fraction of $\ce{C2HCl3}$, as follows:
$$w(\ce{C2HCl3})=\frac{\pu{35.47 g mol-1}}{\pu{122.61 g mol-1}}=0.29$$
$$w(\ce{CHCl3})=\frac{\pu{87.14 g mol-1}}{\pu{122.61 g mol-1}}=0.71$$

Did you notice the mole fraction provided (0.73) is for the vapor phase and the mass fraction asked is for the liquid phase?


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Organic Chemistry - Effect of methyl substituent in Aromatic Substitution via Benzyne

The following reaction has been taken from my study material. This shows that the benzene ring is undergoing a nucleophilic aromatic substitution via benzyne mechanism (presence of a strong base like $\ce{NaNH2}$). My question is:
Why is substitution preferred at the ortho position?
The following link explains the mechanism quite well and gives an example of ortho-chlorotoluene undergoing a similar reaction. Since both proceed via the same benzyne intermediate, why is there a difference in the products formed by meta-chlorotoluene? Or is my study material wrong in saying that o-aminotoluene is a major product? Any help regarding this is appreciated!
Attaching the related reaction's image:

I am not sure though but maybe the -NH2 group at ortho may stabilise more the hyperconjugative structures of the product.

The Me group has no directing effect in the deprotonation so there are 2 protons (o & p) that can be removed with equal probability by the NaNH2 to give the benzyne. These give 2 possible benzynes (2 and 3 relative to Me). The nucleophile will add with equal facility to either end of either benzyne giving a mixture of 3 isomers in 1:2:1 ratio. The major isomer is the m- aminotoluene because it can arise from both of the possible benzynes.

Ok that makes a lot of sense. So that means that m-aminotoluene and not o-aminotoluene is the major product, which means that the first reaction is wrong.


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Determining heat capacity of gasses
I need a little help on answering a question which popped into my head while I was reading about gas laws:

Suppose that I have three canisters, each holding a different gas of
  unknown identity [let's call them gasses A, B, and C] . I also happen
  to have another tank of a different volume in which these gasses can
  be mixed as I please, and a pump to remove the gasses from the mixing
  tank afterwards. Each of these containers is fitted with a pressure
  and temperature gauge. Furthermore, I have a heating apparatus which
  can be used to heat or cool the gasses inside any of the containers. I
  want to determine the molar heat capacity of each of the gasses.
  Assuming that no energy is lost to heating of the containers, is it
  possible to determine the molar heat capacities of gasses A, B, and C
  using this setup? If so, how? If not, why not?

I presume that yes, it is possible to do so. I'm thinking that, for instance, gas A at room temperature could be pumped into the mixing chamber, then the amount of moles of gas A in the mixing tank could be calculated using the PV=nRT formula. Then, gas B could be heated up to a certain temperature and pumped into the tank. Based on the fact that we know the amount of moles of gas A in the tank and the change in temperature and pressure, we can calculate the amount of moles of gas B. Afterwards, the mixing tank can be purged and a different combination of gasses can be pumped in. Rinse and repeat. Afterwards,I would get a system of equations that looks something like the following:
AF=BG
BH=CJ
CN=AL
Where A, B, and C are the heat capacities of each of the gasses, and F,G,H,J,N, and L each equal the amount of moles of a particular gas multiplied by the observed change in temperature and are known. Solving for A, B, and C would yield the molar heat capacity of each gas.
Is my reasoning correct?

That is a very detailed and complex scenario that "popped into" your head. Does the heating device tell you how much heat it delivers?

@KarstenTheis No.

I think if you know the heat capacity of one gas, you could figure out the heat capacity of the others. However, I don't think your set of equations is able to get absolute values (i.e. if I find a solution and then multiply all heat capacities by 2, it would still be a solution to the equations).


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What is the difference between chemical equilibrium and dynamic equilibrium?
Is chemical equilibrium and dynamic equilibrium the same thing? Both point to the reversible reactions and no net changes in the concentrations of product and reactant.
Dynamic equilibrium from Wikipedia:

In chemistry, and in physics, a dynamic equilibrium exists once a reversible reaction occurs. Substances transition between the reactants and products at equal rates, meaning there is no net change. Reactants and products are formed at such a rate that the concentration of neither changes.

Chemical equilibrium from Wikipedia:

In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system. Usually, this state results when the forward reaction proceeds at the same rate as the reverse reaction. The reaction rates of the forward and backward reactions are generally not zero, but equal. Thus, there are no net changes in the concentrations of the reactant(s) and product(s). Such a state is known as dynamic equilibrium.

main difference between dynamic and static is that Dynamic mean motion and statics mean rest

Chemical equilibrium is a type of dynamic equilibrium, but not every dynamic equilibrium is a chemical equilibrium.
In a chemical equilibrium there is no change on the macroscopic scale. That means that if you look at the system it seems like nothing is happening, but at molecular scale there are reactions going on and the rate of forward reaction = rate of backward reaction.
There is another example of dynamic equilibrium (although not the best one, as it is not a closed system):
Imagine a bath into which water is flowing in at constant rate. Water is also flowing out of the bath at the bottom with the same rate.
This is an example of dynamic equilibrium, which has nothing to do with chemistry. If you look at the bath, the water level is constant, so on the macro level nothing is happening, but there is flow of water, so on micro level water from the tap is replacing water flowing out of the bath.

I would call the second case (the bath example)  a steady state system, see https://chemistry.stackexchange.com/questions/110794/is-there-a-difference-between-equilibrium-and-steady-state

A chemical equilibrium concerns chemical reactions. There should be at least a forward- and backward reaction between two species but more complex systems with multiple individual reactions may occur. The important observation is that there is no macroscopic change to the chemical constituents of the system, i.e. the concentrations of all reaction partners remain the same despite all the individual reactions happening. Chemical equilibria can always be written with reaction equations such as below
$$\ce{H2SO4 + 2 H2O <=> HSO4- + H3O+ + H2O <=> SO4^2- + 2 H3O+}$$
Before I get to dynamic equilibria, it is important that I briefly note equilibria in general. In equilibria, there must be some kind of balance but there need not be any movement. For example, in a tug-of-war, two groups of people are pulling the same rope in different directions. When there is no macroscopic movement, it means that both sides are pulling with the same force. This is an equilibrium. However, even if you go down to a microscopic level, you will still notice no movement (i.e. the rope would seem (mostly) stationary). Thus, this type of equilibrium is classified as a static equilibrium: there is no change to the overall state because there is no change to any individual states.
In a dynamic equilibrium, the macroscopic state remains the same but if you care to look at the individual microscopic parts you will notice them switching between states. For example, consider two sealed rooms, one filled with nitrogen and one with oxygen separated by a door. Once that door is opened, gases can flow back and forth. After a certain time, the concentrations in both rooms of both gases will have equilibrated (say to a $4:1$ ratio if that is the ratio of room sizes). Individual gas molecules will still flow back and forth through the open door, but as their concentrations are identical on both sides the flow rate for individual gases is the same and the constant back and forth will not affect the overall ratio of the rooms. This is a dynamic equilibrium because the individual molecules move back and forth (microscopic change) while the overall state remains the same. However, it cannot be written as a chemical equation and thus it is not a chemical equilibrium.

Discussing both the macroscopic and the atomic level is critical to defining a dynamic equilibrium in chemistry.


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Calculate voltage, cost and time of nickel coating

To nickel with a bath of $\ce{NiSO4}$ a cathodic current of $\pu{0.3 A/cm2}$ is applied with a square plate of side $\pu{5 cm}$ is used to nickel a certain substrate. Due to the formation of $\ce{H2}$ the current efficiency is $75\%.$ If the resistance of the solution is $\pu{0.4 \Omega}$ and the price of the current $\pu{1 c€/kWh},$ determine the voltage to be applied, the energy cost and the
  required time to nickel a total surface of $\pu{2 m2}$ with a coating thickness of $\pu{0.03 cm}.$
  Data: $M(\ce{Ni}) = \pu{58.71 g/mol}$; $\delta(\ce{Ni}) = \pu{8.9 g/cm3}.$

First I’m going to show you what I got

I’m sure this is wrong because I didn’t use given data. But I don’t know where should I use mass of nickel.
I’ve never seen this type of exercise before and I don’t where should I start from.

This is just a friendly suggestion but in such questions I always recommend writing down all known quantities before starting so that one doesn't end up jumbled in all the data

The volume of nickel plated is surface area times the thickness.
That's $2m^2 \times \pu{0.03 cm} $.
We'll use cubic cm as the units so it is: $2m \times 1m \times \pu{0.03 cm} = \pu{200 cm} \times \pu{100 cm} \times \pu{0.03 cm} = \pu{600 cm3} $
This helps to find the mass as the density is given as $\pu{8.9 g/cm3}$.
The mass = density $\times$ volume = $ \pu{8.9 g/cm3} \times \pu{600 cm3} = \pu{5.34 kg}$
Convert this to moles of Nickel: $\displaystyle \frac{5.34 \times \pu{10^3 g}}{\pu{58.71 g/mol}} = \pu{90.96 mol} $.
We'll come back to this.
Now for the current, $\pu{0.3 A/cm2}$ applied to a $5 \times \pu{5 cm}$ square plate, meaning that the total current on the plate at a given time is $\pu{0.3 A} \times \pu{25 cm2} = \pu{7.5 A}$ or $\pu{7.5 C/s}$. However, only a fraction of that actually reduces the nickel ($75\%$) so $\pu{5.625 A}$.
The voltage is then $ V = IR = 5.625 \times 0.4 $
The time calculation involves the actual nickel.
$I = \displaystyle \frac{q}{t}$
Where $q =$ total charge
We have $\pu{90.96 mol}$ of Nickel reduced.
$\ce{Ni^{2+} + 2e^- -> Ni}$
That's $\pu{2 mol}$ of electrons per mole of Nickel. So we need $\pu{181.92 mol}$ of electrons.
The charge per mol of electrons is Faraday's constant of $\pu{96485.3 C/mol}$
The total charge is just $\pu{181.92 mol} \times \pu{96485.3 C/mol}$.
Then we rearrange the formula of current for time $\displaystyle t = \frac{q}{I} = \frac{\pu{181.92 mol} \times \pu{96485.3 C/mol}}{\pu{5.625 C/sec}}$
$\text{time} = \pu{3120463 seconds}$ or $\pu{866 hours}$.
That's sort of the basic idea I'd imagine. I may have made a mistake somewhere in between. The time though while seems long is in line with this:
https://sciencing.com/calculate-electroplating-7597391.html
There, they plated just one mole of Cu and with a higher current. That took 2 hours. We have a lot more of $\ce{Ni}$ and smaller current.
Then $\text{Energy} = \text{power} \times \text{time} = VIt$ and you can calculate the cost from there.

Thank you so much for your nice explanation!!


text:
How to prevent feasibility of intermolecular attack of 1,2-diol?
I was given a problem that requires me to synthesize (1R,2R)-2-methoxycyclohexanol from cyclohexene. I tried it with syn-hydroxylation of alkene $\pi$ bond with $\ce{OsO4}$ and NMO then directly transformed the hydroxyl group to methoxy group with $\ce{NaOH}$ and $\ce{CH3I}$.
During the protonation of one hydroxyl group, I notice that it would be attacked from the other hydroxyl group and I think it would make ring closure, which is what I do not want. Is the intermolecular attack of 1,2-diol feasible on any starting material (cyclic or polycyclic)? How to prevent the intermolecular attack to get desired product?

Huh? You worry about epoxy and not about selectivity?

@Mithoron I do not think I need to concern selectivity issue because the $\ce{CH3I}$ is one equivalent thus it will not methoxylate both of hydroxyl groups.

@Mithoron Oh, protecting group (i.e. TMSCl, py)? But is it a solution for that problem?

Why not form the epoxide from the alkene and open that with methoxide or PTSA/MeOH?

@Waylander Does the ring open via SN2 mechanism so the $\ce{OMe}$ group will be inverted? Or the PTSA will make the $\ce{OMe}$ on retention?

Basic MeO- opening is SN2. Acidic MeOH opening goes via protonation and a carbocation intermediate so you will lose the stereochemical integrity that way. MeO- is a poor leaving group. I don't think you would have any great problem with your original idea, except for a bit of dimethoxylation pulling your yield down

@Waylander Ah, the product in the problem is (1R,2R)-2-methoxycyclohexanol, the opening makes the product is (1R-2S).

Then go with your original route, you will not get epoxide formation. Form the monoanion with 1eq of NaH, quench with MeI. You will get a bit of di-OMe and a bit of recovered diol and no epoxide

@Waylander I think the original route looks good. Thank you for helping me.


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Copper plating with switched anode and cathode
Please help!
I was working on a mock research project/lab for my chemistry class testing the rate of electroplating of copper vs. voltage. I had copper and nickel electrodes and .1M copper sulfate solution. I made a huge mistake, however, and plugged the positive end into my nickel electrode and the negative end into the copper. This meant that the copper in solution was attracted to the copper electrode, not the nickel. I however, didn't notice because as I ran more and more trials, an orange buildup was in fact showing up on my nickel electrode that looked just like copper plating. When I measured the mass, however, of the nickel electrodes, they consistently went down (the higher the voltage, the bigger the difference). What happened? Why did the mass of the nickel electrode go down (shouldn't it have stayed the same)? What was being plated onto the nickel?
Thank you!

Your anode was nickel, by mistake. It was being oxidized and started to dissolved. Nothing surprising rather. The color on nickel could be a very thin film of copper deposited from reduction of copper ions on the nickel surface. Nickel is above copper in electrochemical series.


text:
Entropy as an intrinsic property of matter
I would like to ask how we can see entropy as an intrinsic property of molecules, namely as something depending also on their geometry (and not only as gross disorder). Is this what is called the "shape entropy" or there's no relation between the two concepts? Which are the parameters that determine this "geometric" or conformational entropy?
EDIT:
The answer written by Buck Thorn provides an interesting mathematical insight of entropy considering the conformations or "geometries" available to molecules in thermal equilibrium.
Is there also a qualitative analysis of this phenomenon (in terms of the possible conformations playing a role in entropy - the parameters I was referring to in the question)?

Maybe it's just me but I can't tell if your question is even valid. Even if it is I doubt it would actually be about chemistry.

@Mithoron I ask you kindly to re-evaluate this; my chemistry teacher mentioned it, and I would not know which other field this could be related to

@Mithoron my teacher talked about entropy in stretching and movement of the bonds, rotations and so on... But I would like to delve deeper into this

Well, you'd need to completely rewrite it. I have a feeling you'd need to learn a bit more to properly word it, though :(

@Mithoron yeah, probably, English is not my native language. If you did sense something from my confused words, could you pls suggest an edit?

I removed your polite greeting since it's considered superfluous. Please see the  site guidelines if you have questions on this or other rules regarding posting questions.

As an example consider an idealised/simplified model of butane, which has trans and gauche configurations as shown in the figure. We shall suppose these have different energies due to interactions between the protons on carbon $1$ and $4$ as bond C2 to C3 rotates.

We take the energy of the trans state at $0^\text{o}$ to be zero, $E_0 = 0$. The energy of the other two conformations, measured relative to this, have minima at $120^\text{o}$ and $240^\text{o}$ and are $E_{120}$ and $E_{240}$.
The partition function is the sum of the statistical weights of the energy levels. The statistical weight of a state $i$ with energy $E_i$ is the Boltzmann factor $\displaystyle e^{-E_i/k_BT}$, multiplied by the degeneracy of that state $g_i$. The partition function is therefore $Z=\sum_i g_ie^{-E_i/k_BT}$.
The partition function for our model butane is therefore  $\displaystyle Z = 1 + e^{-E_{120}/ k_BT} + e^{-E_{240}/ k_BT}$. By symmetry, the energy of the $120$ and $240$ states are the same and the probability of being in either is $\displaystyle 2e^{-E_g / k_BT}/Z$ where $E_g$ is an abbreviation for $E_{120}$ and $E_{240}$. These two levels are accidentally degenerate.
Once the partition function has been found then the entropy $S$ can be calculated as can the internal energy $U$ and other thermodynamic quantities; $\displaystyle U=RT^2\left(\frac{d\ln(Z)}{dT}\right)_V;\; S=R\ln(Z) +\frac{U}{T}$.
If you calculate the conformational entropy you will find it is zero at low temperatures, because all molecules are in the lowest trans state, and rises to 3R at high temperatures.
Try not to think of entropy as disorder, it is generally more useful to think of it as the number of ways of placing the molecule in its many possible energy states.

Really thank you for the clear and detailed explanation! I've just a doubt, and I don't know if the answer might already be in the sentence "The energy of the other two conformations [... ] have minima at 120° and 240°". Well my question is: can we theoretically talk about infinitely many possible conformations, some of which are only the most probable? Is this the meaning of minima of energy? Again thanks a lot

yes you can do this if you want, and may be necessary for some problems, then of course you will have to find an infinite sum which may or may not be easy.

You can use the tools of statistical mechanics to compute the entropy of molecules in thermal equilibrium.
For molecules in an ideal gas the molar entropy is
$$S=R\ln z +RT\left(\frac{\partial z}{\partial T} \right) _V - R \ln N_A + R$$
Here $z$ is the molecular partition function
$$z=\sum_i e^{\epsilon_i/kT}$$
where the summation extends over all states available to a molecule and $\epsilon _i$ is the energy of state $i$ (states degenerate in energy are counted separately).
If you can identify all the possible conformations ("geometries") available to the molecule and their respective possible energy states, you can use the formula to compute the entropy.
Qualitatively speaking, if you increase the number of conformations you increase the entropy, since the summation in the partition function now extends over more states.
You might invoke Boltzmann's equation
$$p_i=\frac{e^{-\epsilon_i/kT}}{z}$$
to compute the probability of finding a particular molecule in state i.
Reference
McQuarrie and Simon, Statistical Thermodynamics, University Science Books (1999).


text:
What is the white powder that I have extracted from cinnamon?
In Advanced higher chemistry, I am extracting cinnamaldehyde using this procedure- https://www.flinnsci.ca/api/library/Download/ba6ab218508047ea843bb0d9e580be16
. After evaporating the hexane, instead of the cinnamaldehyde (a liquid), a solid white power is formed. We have no clue what it is and think it may be either something that is naturally extracted from the cinnamon bark or something that was added during 'production' (got it from Tesco's).

I guess the aldehyde may oxidized to solid acid, similarly as liquid  benzaldehyde easily oxidizes to solid benzoic acid.

"6 grams of cinnamon can contain from to 42 to 189 mg cinnamaldehyde." http://herbpedia.wikidot.com/cinnamaldehyde  That's not very much.
"Cinnamon consists of a variety of resinous compounds, including cinnamaldehyde, cinnamate, cinnamic acid, and numerous essential oils." https://www.hindawi.com/journals/ecam/2014/642942/
The extraction method starts with 10 grams of cinnamon and salt is used to salt out the cinnamaldehyde from the steam distillate. Could the solid be salt carried over from the separation funnel? If the solid you obtained is more than about 400 mg, it might well be contaminated with NaCl. You could add a milliliter or two of H2O to see if you get a (saturated) salt solution plus the cinnamaldehyde oil on top. If the cinnamon you started with was very old or had very little cinnamaldehyde to begin with, the amount of oil you obtain might well be overshadowed by a little NaCl contamination.


text:
Oxidation States of D Block Elements
These might be basic question but they are quite confusing to me?
If $\ce{Sc+}$ doesn't exists, why does $\ce{Ti^{2+}}$ exist?
on the similar account why does this trend increases at first( till Mn) and then decreases?
My book just concludes that the general trend is due to increase in $Z_{\mathrm{eff}}$ and decrease in atomic radius.
A little insight will be really  appreciated.

Actually an organometallic scandium(I) compound is known, from P.L. Arnold at al. (1996).  As with organometallic compounds generally, the metal is covalently bonded, and the use of $3d$ orbitals as valence orbitals (there are 14 shared valence electrons associated with each scandium atom) in this and similar compounds favors including scandium as a transition metal.
In predominantly ionic settings, however, it is true that scandium(I) would be unlikely.  Because scandium is only at the beginning of the transition metals it has relatively low ionization energies for its valence electrons, and this low energy cost tends to be overcome by increasing lattice or solvation energy associated with higher ion charges.  With ionically bonded scandium the balance is all in favor of more ionization and more electrostatic attraction, whereas with titanium the slightly higher ionization energies and the fact that a noble gas configuration requires four rather than three positive charges makes lower charge ions more accessible.


text:
Atomic emission spectrum assumption n_f = 3
It's a standard question on the atomic emission spectrum.

An atomic emission spectrum of hydrogen shows three wavelengths: 1875 nm, 1282 nm, and 1093 nm. Assign this wavelengths to transitions in the hydrogen atom.

I understand how to calculate it, but from both the answers in my book and the answer from slader.com. Both just say

since these wavelengths are larger than those of visible light, then we assume $n_\mathrm{f} = 3.$

However, I don't understand why we could assume $n_\mathrm{f} = 3.$ Is it because we just look at Paschen series (Bohr series, $n′ = 3$) and the Paschen lines all lie in the infrared band? Could anyone explain it?

I am sorry for being off topic, but @andselisk how do we create direct hyperlink?

@Zenix If you are referring to Markdown, then inline markup would be [Text](https://site.domain.com). Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Another method is using the provided button: a "link" symbol, which you can click, and then insert the https:// for your selected text.

@Molly_K True, but there is nothing to click on when you post a comment, and this is an "anchoring" method which is fine for a link or two, but when you have a couple of dozens of URLs in your post it quickly becomes tedious to keep a track on what pointer refers to which site.

As the wavelengths are in the infrared region hence the transition occurs at $n > 2.$
Now, if you may have seen the data, these values are all common values of Paschen series. But even if you haven't, you can do a simple test by finding the shortest wavelength of all the series after Balmer. You will find only the shortest wavelengths of Paschen and Brackett are suitable for having values lying in this range as the shortest wavelength for Pfund is $\pu{2279 nm}.$
Then also as the shortest wavelength for Brackett is $\pu{1458 nm},$ you can be sure that the second and third are definitely lying in the Paschen series and check for the remaining wavelength which would be a fairly easy job now.
Directly assuming that the final energy level is $n = 3$ isn't correct, I think, but some analysis can help us to draw the above-mentioned conclusions.

@andselisk thanks for editing this however i am facing this difficulty thst whenever i use mathjax it automatically adds line breaks. How do I change this?

For inline formulas, enclose the formula in $...$. For displayed formulas, use $$...$$. Probably you will find a more in-depth overview on Math.SE helpful: MathJax basic tutorial and quick reference.

No prob. Also, note that here using MathJax for the quantities here wasn't really necessary: 1458 nm looks fine on its own. \pu{…} macro comes in handy in more complex situations like $R = \pu{8.21E-5 m3 atm K-1 mol-1}$ ($R = \pu{8.21E-5 m3 atm K-1 mol-1}$). Keep in mind though that \pu{…} is currently only available here on Chemistry.SE and won't work on other SE sites such as Physics.SE where you have to enforce proper typography with bare hands and use ~, \,, \mathrm{…} etc. :)

@andeselisk got your point.

@StackUpPhysics, thanks for your response, I think it would be essential to have the Series spectrum next to the question to make the assumption. (or able to pull that information out of my brain if there's anything to pull...)

@Molly_K Yeah it would be a good to have it. However i believe if this was asked in an exam you won't have it and yes it's very difficult to pull these values out if you have brief encounters with them. Maybe someone who's been working in these spectrums or doing close analysis might see such insights but that is not what's expected of the intended audience of the question i believe.


text:
Is (2R,3S)-3,4-dimethylpentan-2-ol feasible to be dehydrated with POCl3 and pyridine?
I was given a problem of reaction between (2R,3S)-3,4-dimethylpentan-2-ol and $\ce{POCl3}$ and pyridine. I know that $\ce{POCl3}$ will be attacked with hydroxyl group and the hydrogen on hydroxyl will be cleaved with pyridine so the hydroxyl group becomes good leaving group.
The reaction mechanism proceeds on E2 mechanism no matter whether the alcohol is primary, secondary, or tertiary. The problem is E2 mechanism can proceed if the stereochemistry of antiperiplanarity is fulfilled. When I drawn the stereochemistry of the alcohol, it turned that methyl on alpha-carbon and beta-carbon made the beta-hydrogen on these carbons cis to the hydroxyl group (the methyl on alpha carbon is wedge so the bonded hydrogen and hydroxyl group are dash), thus the reaction can not proceed.
Is that correct? If it is correct, will the reaction mechanism proceed on other mechanism?

I am uncertain whether or not you are assuming that the E2 product of a "cis elimination" is the (E)-alkene. Alternatively, you were given the geometry of the alkene and were asked to provide a mechanism for its formation. In either case, let's go through the steps.
You are correct that the alcohol 1 reacts with phosphorus oxychloride to form derivative 2a in addition to pyridine hydrochloride. Recall that there is free rotation about the C2-C3 bond in 2a, which is shown as its Newman projection in 2b in the anti-periplanar conformation ready for E2-elimination. Clearly, this elimination leads to (Z)-3,4-dimethylpent-2-ene. (Z)-4. If indeed it is the stereoisomer (E)-4 that is the target, then one can invoke SN2 displacement on 2a by chloride to invert the configuation at C2 leading to chloride 3a. Now E2-elimination of 3b leads to (E)-4.

The reaction will take place via E2 mechanism as the C2-C3 $\sigma$ bond will undergo rotation, giving the product. The concentration of the minor product will also be comparatively more as compared to the (2R,3R) or (2S,3S) form, but it will not form the major product. Stereochemistry plays an important role in these reactions when the C-C bond cannot rotate about it's axis and hence no reaction occurs. An example is given below:

In the first reaction, bond rotation occurs and E2 elimination takes place, whereas since the second compound is cyclic, no bond rotation can occur and the compound undergoes an E2 elimination, giving the less substituted product as the major product.

You drew the 2S,3R enantiomer, not that it matters. E2 anti elimination of the phosphorus derivative would give the Z-alkene. Elimination in the second case probably gives more of the less substituted alkene.

Ah yes my bad. Edited with a better image now


text:
Electrons and Atomic Mass
Is the mass of electrons covered when talking about atomic mass. For example, the mass of Carbon is $12.011\space\text{amu}$. If this is including the mass of electrons (assuming the atom is neutral) the nucleus' mass is $12.011-6(\frac{1}{1840})=12.00773913$
The reason I need to know is I'm studying mass defect in relation to protons and neutron. If I inadvertently include electrons in my calculation, my binding energy will also be wrong.

Good question. The mass of electron is so small relative to neutrons/protons that it can be neglected in binding energy calculations. Electron mass in amu is 0.00054. You can try both approaches and see the differences.

Note that 12.011 amu is the weighted isotope average mass. The mass of a neutral carbon-12 atom in the ground state (electronic and nuclear) is 12.0 amu (exactly), defining the amu. So, yes, it includes the mass of the 6 electrons.

(And, it is kind of funny that this was asked by a user named 'Dalton' - you'd think they would know their own definition!).


text:
How can i convert .gjf or .chk file to .wfn file?
Please explain to me in details. I can't understand corresponding contents in manual.

You cannot convert gjf files to wfn files: gjf files contain only the input information, wfn files contain some of the output information.
As to the chk file: I guess that you intend to do QTAIM analysis on the file. Many softwares for AIM (eg: the wonderful and free Multiwfn) let you use a chk file (or a fchk file) as input. I don't know if they can save a wfn file from that, but the AIM analysis can be performed anyway.
If you specifically need the wfn file, you can generate it with Gaussian:

In a new calculation, by specifying the keyword output=wfn and, after the coordinates, filename.wfn. See Dr. Barroso's blog for nice examples, eg: https://joaquinbarroso.com/2016/02/16/quantifying-%CF%83-holes-g09-and-multiwfn/
For an already performed calculation (for which a *.chk file is available) you can do the same; adding geom=allcheck option, the calculation uses the geometry and results in the already converged *.chk, and also outputs a wfn file

Some other caveats are listed here: http://aim.tkgristmill.com/knownissues.html
(for instance, for the particular cases in which DENSITY=CURRENT, POP=NOAB, etc. are needed)
A couple of examples I wrote ("!" is seen as a comment line by Gaussian, I wrote it to put the blank line at the end of the code on stack exchange).
Optimization of HF molecule, with generation of both a *.chk file and a *.wfn file:
%chk=HF.chk
#n output=wfn B3LYP/6-31G(d) Opt

 Hydrogen fluoride

0 1
F          0.09138        0.00000        0.00000
H         -0.84046        0.00000        0.00000

HF.wfn

!

Now, let's assume that the calculation was run without generation of wfn file (you have a *.chk file and you want a *wfn file):
%chk=HF.chk
#n output=wfn B3LYP/6-31G(d) geom=allcheck guess=only

HF.wfn

!

Alternatively:
%chk=HF.chk
#n output=wfn B3LYP/6-31G(d) geom=check guess=(read,only)

 Hydrogen Fluoride

0 1

HF.wfn

!

This is quite a good answer for an absolutely terrible question! For the sake of completeness though: nowadays it is better to use wfx files instead as they support more functions. Using multiwfn, you do not need to write a wave function file; format the checkpoint file and load it. The rest is the same as in the tutorials.

Also please note that your last examples will repeat the calculation. You can suppress this by guess=only.

Edited with Martin's suggestions!


text:
how can carbon dating even work? Its at atomic level!
I am a homeschooling parent, so pardon me asking stupid questions.
Everywhere I read, it is said carbon based lifeforms eat carbon and after their death that atom starts a decay.
But what I cannot understand, we dont eat or work with food at atomic levels. Yes, life forms breaks down molecules from food and use them as they wish. But that chemical process does not effect the atom structure. I dont think any life form has capacity or energy to break down atoms.
So if one atom of C14 was having a decay already before we ate it, it keeps having it while it gets digested and goes on in the same way if life is alive or dead. If another C14 was not having a decay, then it starting a decay has nothing to with the life form is alive or dead.
As Dr. Manhattan(Watchmen 2009) said atoms dont care if you are alive or dead. They stayed there at the same place before and after.
So how can a lifeform have any effect in the C14 decay at all?
Got some good answers, so I clarifying my questions:
How can we be sure that we have eaten only fresh out of factory C14? It may be already decayed by hundred or thousand of years  when we eat it. It applies to all living beings. If there is fresh C14 in the air, there is also decaying levels of C14 as old as 50K years. So when you check my c14 count, you would have a wrong answer of when I died cause origin of C14 was even before I was my great grandfather was.

The life form doesn't have any effect on the decay at all; the point is that once the organism is dead, it cannot replenish itself with C14 any more.

Take a look at https://chemistry.stackexchange.com/questions/123691/why-does-radiocarbon-dating-only-work-in-nonliving-creatures and https://chemistry.stackexchange.com/questions/74629/how-does-radiocarbon-dating-work .  These should help you understand how carbon dating works.

As theorist points out, there is a very good comment by @MaxW here that answers your question.

You are correct in the fact carbon-14 does not care whether or not a life form is alive; however, this is irrelevant in the case of carbon dating. You may be familiar with the carbon cycle, which is, in a nutshell, the recycling of carbon through carbon dioxide. Plants are, of course, constantly using carbon dioxide from the atmosphere to generate large bio molecules that we can eat. We are also constantly recycling the carbon, ingesting it in the form of food, and exhaling it in the form of carbon dioxide, so it is clear that living organisms are constantly replacing their "old" carbon from carbon from atmospheric carbon dioxide. The key factor to carbon dating is that atmospheric carbon dioxide has a higher percentage of Carbon-14. This is caused by high energy cosmic rays which convert nitrogen-14 (the most abundant isotope of nitrogen) into carbon-14 nuclei, which in turn result in a greater percentage of our carbon content being carbon-14. When we die, we are no longer replenishing our carbon from the atmosphere (higher carbon-14 content due to cosmic rays), so the carbon-14 will gradually decay into more stable nuclei, leaving a lower percentage of carbon-14 in our remains. By using the half-life of carbon-14 nuclei, we are able to calculate the time when a certain living organism was no longer exposed to our atmosphere's high concentration of carbon-14. As a side note, carbon-14 dating is fairly accurate because the relative concentration of carbon-14 has remained constant in our atmosphere.

How can we be sure that we have eaten only fresh out of factory C14? It may be already decayed by hundred or thousand of years  when we eat it.It applies to all living beings. If there is fresh C14 in the air, there is also decaying levels of C14 as old as 50K years. So when you check my c14 count, you would have a wrong answer of when I died cause origin of C14 was even before I was my great grandfather was.

We can't and aren't. We eat (and therefore our bodies contain) a mixture of fresh-out-of-factory C14, and that produced 50 years ago, and that produced 100 years ago, and so on. Dead guys don't eat, and that's what makes the difference.

Supposed I had eaten only carrots and lots of them. All those carrots had atoms of C14 produced in the air 12000 years ago. When I die tomorrow, and you find c14, wont you see stale c14 atoms and assume I died 12000 years ago?

@thevikas Be aware the rate of 14C atom decay is independent on how old are the atoms. Living bodies maintain quite stable 14C/12C ratio, that follows the ration in air by metabolism, that recycles body content in much shorter time than is the 14C half time. This recycling stops after death and 14C content decreases by the exponential way.

By other words, radiocarbon dating does not measure,how old 14C atoms are,but calculates what time it would take to reach the measured 14C/12C ratio.

ok. I understand better. but are we assuming all different life forms, in all seasons, in all age groups and regions have same ratio? I think thats too much assumption. Diets of humans is different in just 100kms. Should it not make a massive difference in how much carbon I have or in 14C/12C ratio?

This ratio of carbon-14 to carbon-12 is very accurate among all living organisms. The diet does not matter. All that matters is that all of our food contains approximately the same ratio of carbon-14 to carbon -12. All of the carbon you ingest comes from the food we eat, which was just recently synthesized by a plant from atmospheric carbon dioxide. Do not get caught up on the age of organisms, diet, etc. All of that is irrelevant. Every living thing on the Earth will recycle a fixed ratio of carbon-12 and carbon-14 until they die, and from then on, the ratio will start to change.

Carbon-14 decays no matter what the chemistry. New carbon-14 is made in the atmosphere when nitrogen-14 is hit by cosmic radiation replenishing carbon-14.
Living organisms continuously exchange carbon with the atmosphere. Carbon in fossils and in rocks does not.
That’s why carbon dating gives the time of death rather than the time of birth but with lifespans much shorter than the age of fossils, this differences is rarely significant.


text:
How much CO2 is emitted by burning coal to produce 1 kWh of energy
I'm a little embarrassed to ask this question at this stage of the climate game but, in searching, I seem to get contradictory answers.
For example, here: https://www.eia.gov/tools/faqs/faq.php?id=73&t=11
we see 215.4 lbs CO2 giving 1 million btu heat. Since 1 million btu = 293 kWh, we end up with 0.735 lbs/kWh.
But at this site: http://shrinkthatfootprint.com/electricity-emissions-around-the-world
we see 1 kg/kWh = 2.2 lbs/kWh
New York times https://www.nytimes.com/2007/11/07/business/businessspecial3/07carbon.html
gives 1.9 lbs/kWh
These are very different numbers (too different for the problem to be type of coal, at least for eia vs NYT) and I have found other numbers at other sites.
Can anyone point me to a site that I can trust for this number and others like it (related to energy production)?
Or else point out the elementary error I am making?

You probably mean electric energy, not thermal energy.

The value from EIA is pounds of $\ce{CO2}$ produced per kWh of thermal energy produced, while those from the latter two sources are pounds of $\ce{CO2}$ produced per kWh of electrical energy produced.  Since the thermal efficiency of coal-fired plants is ~37% (~37% of the coal's thermal energy is converted to electrical energy), converting the EIA figure to pounds of $\ce{CO2}$ produced per kWh of electrical energy produced gives us:
$$ 0.735 \frac{\text{lbs }\ce{CO2}}{\text{kWh thermal energy}} x \frac{\text{1 kWh thermal energy}}{\text{0.37 kWh electrical energy }} = 1.99 \frac{\text{lbs }\ce{CO2}}{\text{kWh electrical energy}}, $$
i.e., about the same as the NY Times and shrinkthatfootprint figures.


text:
Problem from Ionic Equilibrium

This seems to be a pretty easy problem but I cannot understand why we do not consider the contribution of $\ce{H+}$ ions by water (water dissociates into $\ce{H+}$ and $\ce{OH-}$) and only that of $\ce{HCl}$ when we find the total $\ce{H+}$  ion concentration, in the same way both $\ce{NaCl}$ and $\ce{HCl}$  contribute towards the concentration of Chlorine.
Is the solution wrong or is my reasoning incorrect?

It'd be nice if you transcribed the image into text since images are not searchable.

What kind of old textbook ist that? Please add a citation. The usage of normality is deprecated for years, decades even.

@Martin A Book of physical chemistry by OP Tandon. Its Indian.

The contribution by water would be of the order $\pu{10^{-7} M}$ at Room Temperature which is negligible in comparison to that provided by $\ce{HCl}$ .
In the second case the contributions by both $\ce{NaCl}$ and $\ce{HCl}$ towards Chlorine Ions are comparable hence both need to be included.
However nevertheless one may at times need to account for contributions from Water say for instance one needs to calculate the pH of $\pu{10^{-8} M}$ $\ce{HCl}$ as here it wouldn't be negligible but instead comparable.

oops. You've already answered. I'll delete mine.

But if we consider that Water has 15 milli equivalents, its concentration will be 15/100 = 0.15M . Won't its dissociation into H+ and OH- lead to the formation of H+ with a similar concentration?

Or is its degree of dissociation just very low?

@JohnTony Yes, quite low.

@William R. Ebenezer Nice to see an IITian here! Hope to be one in the future.

Please note that element names, compound names, and things like 'room temperature' are not capitalised.

@Martin-マーチン Do you mean the full words aren't capitalized or all the first letter? In case of former it was not the case in my answer and in the latter isn't it correct to capitalize the first letter of element names?

@JohnTony All the best! Don't lose focus. Ever.


text:
Radicals in Gaussian
I want to do a PES scan of a radical structure in Gaussian. However, I'm not completely sure how to prepare an input file. I've figured out that charge and multiplicity are 0 and 2. What I don't know is how to indicate where a radical is. Will Gaussian figure it out itself by bonds and electrons count or do I have to indicate it explicitly?

Do you remember the postulate "The state of a quantum mechanical system is completely specified by its wavefunction"? Gaussian will find the wavefunction of your input system, which includes the location of the radical in that case (when dealing with wavefunctions, the spin density is the property you want to look at).
Of course, in some cases you can (and should) do something "manually". For example, the 1-propyl radical will lack a hydrogen atom on the first carbon, while the 2-propyl radical will lack a hydrogen atom on the second carbon

You are approximately solving the electronic Schrödinger equation. Using Gaussian means applying molecular orbital approximation. Radicals are not localised in this formalism. You cannot specify "where it is".

In case it's not clear yet, Gaussian will figure it out itself. You just need to input the charge and multiplicity with your coordinates and other standard fare.

Question for downvoters: what exactly is your problem?

@Anna I am not one of the down-voters, but I'd like to give some insight into why you might have attracted them. Broad computational questions which focus about a certain software are not popular with everyone. Conceptual questions (independent of the software) have a better stance. Your question also has a little also appears that you have not spent enough time with the program manual. Such questions are often seen as homework questions which lack in presenting enough own thoughts and effort.

@Martin-マーチン, thanks. However, this is a question on how to prepare a system for an experiment, so I have to be specific. Not giving the name of the software would be like asking a question about spectroscopy and not saying which one.
And judging the question as a "homework" is just ridiculous. I was asking for advice on my experiment. (Saying this to people who closed the question.)
If asking for advice is not a proper question, then what is this place? And if computational questions shouldn't be asked, why is there such a tag?

@Anna computational chemistry is much more than using a program. While there are certain things for which it is important to know the program you are using, in many cases another program can do a similar thing as the concepts are exactly the same. This platform and to gather questions and answers of a broadly applicable nature, it is not so much a forum for discussions. Most of our users expect that a question is well researched, and that the effort going into it is shown. Your question is very basic and can be answered following plenty of tutorials on the net and reading the manual.

Since you have not included your prior research and where exactly you got hung up on, some users will perceive it as a homework type question; and vote that way. While I don't necessarily agree with that verdict, the decision is made by a community. You always have the chance to [edit] your question to include more details and your thought process. However, I believe your question should be answered through the comments now.


text:
What apparatus is needed to get a balloon full of anhydrous carbon dioxide from solid carbon dioxide?
I'm currently planning a research project (new to this) and we need to bubble CO2 through our reaction mixture, however we are not able to use compressed gas so have been advised to pass CO2 from dry ice over a drying agent and syphon it into a balloon. Is there any literature on this procedure, any help much appreciated.

That`s not research, thats a lab course task. ;) How about you propose some setup, and we tell you if it would work?

Hint: Look up lab procedures of Grignard reactions with carbon dixoxyde in the reaction flask.


text:
Phenolphthalein and methyl orange end points comparison
In determining the solubility constant for Ca(OH)2, how does phenolphthalein end point compare with the methyl orange end points?

See wikipedia article on pH indicators.

The suitability of indicators is given not only by their range, but also by the direction of the transition.
Generally, the transition is always much better from the light colour to the more intense colour, as the equivalence point and it's approaching is much easier to track.
So for MO (red 3.1- yellow 4.4), it is toward acidic, for PHPH (8.2 clear -10.0 violet/magenta), it is toward alkaline.
Both indicator ranges are about equally suitable for strong acid strong base titrations.
Usual issues with  $\ce{CO2}$, needing heating and cooling, do not apply in the case of $\ce{Ca(OH)2}$, due insolubility of $\ce{CaCO3}$.
As $\ce{Ca(OH)2}$ is supposed titrated by an acid, I see MO with transition yellow to read as a better candidate than PHPH with violet to clear.


text:
Ka for reaction from tables
For the reaction below:
NH4+ + H2O -> NH3 + H30+
I have the following values from a table to use:

And in the solution manual they tell me to use the Ka for this reaction. Why is this? Why do we use the Ka for NH3 when it is NH4 acting as an acid?

As you pointed out $\ce{NH3}$ is a base and $\ce{NH4+}$ is an acid. There is really no point in listing both in a table on separate lines since $$K_a \times K_b = K_w$$

You are strictly right. Ka refers not to the acidity constant of $\ce{NH3}$ but to that of its conjugate acid, $\ce{NH4+}$

@BuckThorn states the matter closer to the 'truth." The column labeled "pKa" should really be labeled  "pKa of conjugate acid." As chemists we understand the implicit relationship.


text:
Which biphenyl is optically active?

Which biphenyl is optically active?

I know that it can never be 1. I don't think it will be 4 either as I read that it should be ortho-substituted biphenyl.
So, when I look at 2 and 3, I can't make out which one should be the correct answer.
Should it always be two ortho substituents on each phenyl group for it to be optically active?
Also, what if the two groups on the same phenyl are different, will it always be optically active in that case?

Check Atropisomerism

Biphenyl 2 is the only optically active compound here. These stereoisomers are due to the hindered rotation about the 1,1'-single bond of the compound (Ref.1). Biphenyl 3 is not optically active, because partially allowed rotation about the 1,1'-single bond of the compound (rotation is only partially restricted). To illustrate this phenomenon, I depicted the following diagram:

Note that compound 3 can rotate through two simultaneous $\ce{I}$ and $\ce{H}$ atoms allowing last $180 ^\circ$ rotation, which is well illustrated in the diagram posted by Karsten Theis.
References:

Paul Newman, Philip RutkinKurt Mislow, "The Configurational Correlation of Optically Active Biphenyls with Centrally Asymmetric Compounds. The Absolute Configuration of 6,6'-Dinitro-2,2'-diphenic Acid," J. Am. Chem. Soc. 1958, 80(2), 465-473 (https://doi.org/10.1021/ja01535a054).

So only biphenyl 3 in your diagram is optically active? Why is that so? They all look pretty much the same, 2 different atoms at the same positions in each case?

I think your third diagram has an error in the lower left corner-- it looks like you are saying the hydrogens in Biphenyl 3 will turn into bromine when you rotate it certain directions.

I think your numbering is wrong in the beginning. In your second box (for biphenyl 2) your neatly explain how the mirror images are not superposable. Therefore 2 should be optically active. The same reasoning would technically hold for biphenyl 3, but the interconversion barrier is low, i.e. supposedly accessible at room temperature. Therefore it would me optically active, but it racemises immediately. (Your third box also contains a small copy paste error in the lower left corner.)

@Martin-マーチン I edited it accordingly (swapped 2 and 3 in the first part).

(3) is not chiral, so maybe it is (2).

So should it have 2 different groups at ortho positions of each biphenyl? What if there are 2 groups on one biphenyl and different 2 groups on the other biphenyl?

@studious To restrict rotation, you need two large groups in the ortho position of one phenyl, and at least one large group in the ortho position of the second phenyl. If you have identical groups on one phenyl (2-A, 6-A) and identical groups on the other (2'-B, 6'-B) none of the conformations are chiral.

TL;DR: It's 2 (2,2'‐dibromo‐6,6'‐diiodo‐1,1'‐biphenyl) and maybe even 3 (2,2'‐diiodo‐1,1'‐biphenyl; 2,6‐dimethyl‐1,1'‐biphenyl) at higher temperatures.
Introduction
According to the IUPAC gold book chirality is defined the as the geometric property of a rigid object of being non-superposable on its mirror image. Orr in other words, such an object has neither a mirror plane, nor a centre of inversion.
For examples, please see this Q&A: What is the perfect definition for chirality?
Asymmetric substituted biphenyl derivatives may possess axial chirality. This is quite well explained in Matthew's answer (at the time of writing, I think he has the numbering wrong though). This arises (mainly) due to hindered rotation, see atropisomers in the IUPAC gold book and on Wikipedia.
Analysis
Based on this, we can immediately rule out biphenyl 4 (2,6‐dimethyl‐1,1'‐biphenyl). I have simulated the structure[1] and the equilibrium structure is close to C2v.

The remaining compounds will have chiral conformers as they are asymmetric substituted. The first compound 1 (3‐iodo‐3'‐nitro‐1,1'‐biphenyl) has no (C1) symmetry. The equilibrium structures of the second compound 2 (2,2'‐dibromo‐6,6'‐diiodo‐1,1'‐biphenyl), as well as the third compound 3 (2,2'‐diiodo‐1,1'‐biphenyl; 2,6‐dimethyl‐1,1'‐biphenyl), have C2 symmetry. These are potential candidates for atropisomerism.
The question now becomes, which one is (most) hindered. I have (very crudely) simulated the interconversion barriers.[2] As a reference, I have also included 1,1'-biphenyl as 0. The barrier is calculated in a zeroth order approximation as the difference of the conformer(s) and the highest energy image.
\begin{array}{cr}
\text{Structure} & \Delta E /\pu{kJ mol-1}\\\hline
\mathbf{0}       &  11.0   \\
\mathbf{1}       &   9.7   \\
\mathbf{2}       & 208.4   \\
\mathbf{3}       & 107.6   \\
\mathbf{4}       & \text{n.a.}   \\\hline
\end{array}
Note that I have only considered one conversion path, which actually only would make a difference for 3 anyway.
We can immediately spot that there is not much hindrance in 1, as its value is in the same ballpark as 0. The value for the latter is approximately the same as what Henry Rzepa calculated.[3,4] For the cinephiles, here is the animated minimum energy path (MEP):

We can also deduce that 3 is more hindered, but conversion is probably still doable at room temperature. Without further analysis, I'd make an educated guess that the large size of iodine is the reason for the higher barrier. In the local minima the structures are probably stabilised by non-covalent interactions, while during the transition Pauli repulsion will have a significant effect.
When you cool down the system enough, you should be able to differentiate the two forms.

The highest barrier is for 2, which is not really surprising given the bulky halogens used. During the transition the molecule has to distort significantly as the MEP will show.

Notes & References

Optimisation has been done on the semiempirical GFN2-xTB level of theory. (a) Mulliken Center for Theoretical Chemistry, xTB on Github, xTB documentation, xTB: J. Chem. Theory Comput. 2017, 13 (5), 1989–2009 & J. Chem. Theory Comput. 2019, 15 (3), 1652–1671.
Cartesian coordinates (Xmol format) of the optimised structure:

28
 energy: -37.106373072108 gnorm: 0.000617993871 xtb: 6.3.0 (preview)
C          -0.74457039092978     0.00668380904848     0.00403847586541
C          -1.43960105448658    -0.45330956062940     1.12306879090173
C          -1.43111456476918     0.48819737048977    -1.11130113065019
C          -2.82681693607605    -0.42604266071721     1.11315113205938
C          -2.81837054760201     0.50359494897240    -1.09421940563873
C          -3.51401551801468     0.04941173994657     0.01133914771101
H          -3.37205771113385    -0.78013743294762     1.97649654009736
H          -3.35684427561920     0.87533913913887    -1.95437486971859
H          -4.59411676716931     0.06680911498157     0.01451573070096
C           0.73741240721740    -0.01399221137562    -0.00121303974220
C           1.45737190356414     1.07903119590108     0.46615300935510
C           1.42452698329909    -1.12468405637333    -0.47649552340433
H           0.86686558778550    -1.97662540315982    -0.84058815377695
H           0.92547790883778     1.94454881021216     0.83667570046312
C           2.80773675401329    -1.14118822737596    -0.48485609074287
C           2.84061966747731     1.06166601900200     0.45746804395492
H           3.33358058669268    -2.00908593002598    -0.85653458093421
H           3.39191821123738     1.91650859033419     0.82242788728035
C           3.51831774061953    -0.04807951153356    -0.01828710942118
H           4.59840909035030    -0.06111117644652    -0.02569905851317
C          -0.68780013266629    -0.96774724332108     2.31428913997988
H          -1.37323200644906    -1.27392883680533     3.10034826852651
H          -0.06975635021791    -1.82092520510799     2.03621620918938
H          -0.02338556354040    -0.19893824461916     2.70781448086228
C          -0.66947839003714     0.98202574672165    -2.30497384579965
H          -0.01808999107620     1.80968104054958    -2.02509754296330
H          -0.03719586612218     0.19173089684959    -2.70896585204466
H          -1.34904789687459     1.32012824269556    -3.08395328163034

The interconversion has been calculated with the same level of theory using the Nudged Elastic Band (NEB) algorithm of ORCA 4.2.0 (Forum, WIREs Comput. Mol. Sci. 2018, 8 (1)). The path had a total of 15 images, here is the example file for 2 along with the necessary coordinate files. If you want to try this at home, on my laptop (old!) with a single core this calculation took about 2 hours and 40 minutes.

! XTB2 NEB
%neb
NEB_End_XYZFile "conformer_end.xyz"
Nimages 13
end

*xyzfile 0 1 conformer_start.xyz

Xmol conformer_start.xyz of conformer 1:
22
 energy: -44.541438926695 gnorm: 0.000273788204 xtb: 6.3.0 (preview)
C          -0.73842516239767     0.12099397705083     0.10023099347275
C          -1.44934703887101     1.30959596420843    -0.04635714487514
C          -1.46459032906802    -1.05364356156319     0.26275562729962
C          -2.83260221816166     1.33338660213393    -0.03856888366719
Br         -0.49031781314024     2.94358437512547    -0.26534599360856
C          -2.84794544579427    -1.04806937194567     0.27338021399222
I          -0.43874303211246    -2.89099212545445     0.49464849107241
C          -3.52725076437943     0.14798638986577     0.12187031074955
H          -3.35400213047419     2.27002355888684    -0.15598934111452
H          -3.39039633102590    -1.97145652424781     0.40014772926329
H          -4.60579519489111     0.15562863225057     0.12966905294877
C           0.73854134684734     0.12608798707818     0.09393840630342
C           1.44940095781461     0.26239315529407     1.28380896054353
C           1.46477927365802     0.00836392380816    -1.08597026657536
I           0.43777759512819    -0.19754357173773    -2.92573037522205
Br          0.49031834581031     0.43262730661288     2.92353161238781
C           2.84816021024059     0.01980313175890    -1.08292737332063
C           2.83266002230666     0.27532688568945     1.30522345586451
H           3.39064596265727    -0.07352515350963    -2.01027791369359
H           3.35396295523884     0.38108640783580     2.24326316270063
C           3.52740515329100     0.15308581626132     0.11533500559347
H           4.60593238792661     0.16262139848447     0.12098160378166

Xmol conformer_end.xyz of conformer 2:
22
 energy: -44.541436788329 gnorm: 0.000524902200 xtb: 6.3.0 (preview)
C          -0.73832361282032     0.11983310306066     0.10840520776248
C          -1.44801951621468     1.30797897222752    -0.04711380834348
C          -1.46589375900469    -1.05507032640145     0.26341702181708
C          -2.83136671292522     1.33298447781511    -0.04044389495310
Br         -0.48672217710205     2.94010480951284    -0.27007089450699
C          -2.84922363862597    -1.04840911658639     0.27193924049683
I          -0.44386752640863    -2.89430038039744     0.49812818908251
C          -3.52726058220375     0.14843604873707     0.12049840516296
H          -3.35201572570858     2.26977601051251    -0.16025533523296
H          -3.39258284243005    -1.97170525823195     0.39578778183960
H          -4.60581188092237     0.15713306581713     0.12812396988730
C           0.73861053874909     0.11915254649018     0.09757065282745
C           1.43899785537134    -0.05963599109061    -1.09283010773567
C           1.47533465683736     0.30394299799110     1.26237171334068
I           0.46538312669624     0.58600628876877     3.10142029802479
Br          0.46500728057852    -0.30932233235425    -2.71351652270371
C           2.85860590331322     0.30295779336713     1.24612034762903
C           2.82215760585763    -0.06398015847435    -1.12722135693164
H           3.40910925948393     0.44613646389996     2.16234539469840
H           3.33549654196190    -0.20778602755665    -2.06467923531197
C           3.52731730663299     0.11776860585707     0.04879749941162
H           4.60580925392358     0.11518160422810     0.03305180289059

On ωB97XD/6-31G(d,p) about $\pu{3 kcal mol-1}$, which is in real units $\pu{12.5 kJ mol-1}$. Henry Rzepa: Conformational analysis of biphenyls: an upside-down view
Possibly relevant dissertation by David Vonlanthen Biphenyl-Cyclophanes: The Molecular Control over the Conductivity of Single-Molecule Junctions (in German): pdf at researchgate. He cites the value for 1,1'-biphenyl as $\pu{2-3 kcal mol-1}$.


text:
How is Zn not a transition metal?
A transition metal can be defined as an element that possesses an incomplete sub-level in one or more of its oxidization states. In the textbook I'm reading, it claims that zinc is not a transition metal because it has a full $d$-sub-level in all its oxidization states.
A quick google reveals that zinc has oxidization states $-2, 0, +1$, which means that zinc(with oxidization number $+1$) has an incomplete d-sub-level and is a  transition metal.
What's going on here? Is my textbook incorrect?

Zinc does not have an incomplete $d$ subshell, it does not use any of that subshell for valence orbitals and even its heats of fusion and vaporization betray a lack of typical transition-metal properties.  It just does not fit.  But hey, we still have scandium through copper in that period.

So is zinc just a special case? I would imagine cadmium should have these properties if not?

Cadmium isn't a transition metal either.

Zinc in the +1 oxidation state is $\text{[Ar]}3d^{10}4s^1$, and even in its highest, most common known oxidation state +2 (which the quoted values above seem to have forgotten) it's still $\text{[Ar]}3d^{10}$. No known zinc species in what we normally consider the realm of chemistry breaks that complete $3d^{10}$ subshell, and we would need a major revamp of our calculations and models if any ever does.  Moreover, the thermophysical properties of zinc also betray a loss of transition-metal character.  Zinc is just not a transition element.

As a followup to my question, I want to state $\ce{Zn}$ is not a transition metal also because of the Aufbau Principle. According to the principle, electrons are removed from the subshell with the highest quantum number first. So in a case such as $\ce{Zn^2+}$, electrons are moved from the $4s$ subshells instead of the $d$-subshells. The $d$-subshell remains full.


text:
If water freezes at 32 °F (0 °C) at sea level, then what temperature does ice melt?
If water freezes at 32 °F, what happens to that newly formed ice at that temperature? Does it start melting as soon as it’s frozen, or does it require more heat to melt again? If it does require more heat to melt, then how much?
And what would happen if you were in between that melting point and the water’s freezing temperature of 32 °F?

There is no between.

Possible duplicate of How can melting point equal freezing point?

Melting points equal to freezing points, not limited to water.
Which of the processes takes place depends on if the surrounding provides or takes away thermal energy.
If there is neutral thermal balance at freezing/melting point temperature, the both opposite processes occur, at the same rate, similarly as the opposite chemical reactions at chemical equilibrium.

Poutnik's answer is spot on, but I think it's worth elaborating a bit. As pointed out, the melting and freezing temperatures are identical. At thermodynamic equilibrium, which means the entire system is at exactly the same temperature, some fraction will be ice and some fraction will be liquid. if heat is removed or added from the system (very slowly), the fraction that is solid will increase or decrease, but as long as there is still some liquid and some solid, the system is still at equilibrium.
When no heat is being added or removed from the system, the ice is constantly melting, but at exactly the same rate as some of the water is freezing, so there is no change in the fraction that is solid. But the specific bit that was frozen at one moment might be liquid the next.
In real life, it can be hard to heat or cool things so slowly that they remain close to equilibrium. If you heat the air around a block of ice, it is easy for the air temperature to get well above the melting point while the temperature within the solid remains low, so it may appear as if you have to heat it above the melting temperature in order to get it to melt.
Likewise, the air temp can get well below freezing while the liquid remains warming, again creating the impression that you need to go below 32 F in order to get freezing to occur. In both cases, overshooting helps shorten the time it takes to get to freezing/melting point, but it isn't a reflection of the equilibrium process.
One unusual case is supercooling, where you actually do get the liquid below the freezing temp without forming ice. This is also just due to the cooling happening more quickly than the rate of ice nucleation in the given system. The system is not at thermodynamic equilibrium even though the temp may be uniform. We call this a meta-stable state.

See melting and boiling …really equilibrium? https://chemistry.stackexchange.com/questions/122929/melting-and-boiling-really-equilibrium


text:
Which of the following isomers of 2,3‐dihydroxy‐4‐methoxy‐4‐oxobutanoic acid are identical?

Which of the following are identical?

A and B are identical
A and B are diastereomers
A and C are enantiomers
A and B are enantiomers

What I know is that when we rotate a Fisher projection by 180°, then we get identical compounds (I might be wrong). But how to tell if the compounds are enantiomers of each other or not?
For that how to get the same group, i.e. $\ce{-COOH}$ in this case at the top to compare the compounds?
Also, is there any trick to tell that if they are diastereomers, enantiomers or identical by the hydroxyl group positions in Fisher projections?

It is so because A and B are enantiomers. Tell me why.

Rotating a Fischer projection by 180 deg. does not always give an identical compound.

On rotating the compound B by 180°, we get the two -OH groups on the left side and the two H atoms on the right. In other words, B would then be an enantiomer of A.
Note: on rotating a Fischer projection by 180°, the compound remains the same; it's just that all the atoms have to be rotated, i.e. the atoms on C3 in this case would go to C2 and vice versa.

So it should be one -OH group or two -OH groups on the opposite sides of 2 molecules for them to be enantiomers of each other?

And what about option B and C? Are they enantiomers because of the position of 1 -OH group is opposite?

Tech note: ˚ is a Unicode symbol that means "a ring above" (U+02DA); if you want to depict a degree symbol, then use °, "a degree sign" (U+00B0). Cherry-picking a random character that looks approximately the same as the correct one is a horrible practice.

B, C and A,C form diastereomer pairs: diastereomers are isomers that are not mirror images of each other, while enantiomers are exact mirror images of each other.

@andselisk I did not know that! alt+j on my mac gives me this character, which I always assumed was a degree sign. Thanks for letting me know.

@AniruddhaDeb On macOS a degree symbol is called via Option+Shift+8, I believe.