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Arrange the following in increasing order of acidity: water, ammonia, ethyne and ethane
I know that ethyne is more acidic than ethane because of more s character. For water and ammonia, I used the fact that oxygen is more electronegative than nitrogen and will be able to stabilize the charge better. So water will be more acidic than ammonia.Now how do I compare ammonia and ethyne?

sodamide is used to deprotonate alkynes

The order of acidic nature would be: water > ethyne > ammonia > ethane.
Since oxygen has highest electronegativity amongst the atoms of the compounds, so its hydrogen will be most acidic and hence will be best acid among them.
Ethyne contains $\mathrm{sp}$-carbon and as a result the electronegativity of the carbon in ethyne is more than $\mathrm{sp^3}$ due to increased $\mathrm{s}$-character (50%) in the bonds between the carbon atoms.
The electronegativity of the carbon increases to a great extent when  and turns out to be 3.2, which is greater than the electronegativity of nitrogen which is 3.04. Hence the hydrogen of ethyne is more acidic than that of $\ce{NH3}.$
Ethane does not possess any acidic hydrogen hence it is least acidic.

If it is ethene in place of ethane then $NH_3$ will be a better acid than ethene as in ethene the electronegativity of $sp^2$ carbon is 2.74 (33%s  character)which is less than that of nitrogen


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Nanocar racing competition and the race tracks used
In the Nanocar race competition the molecules are raced on a gold surface.
Is there a specific property of gold that it forms a good "racetrack"? On what criteria does it compare with other metal surfaces?

While adsorption certainly is needed, the gold might be also there for contingent reason, as imaging via AFM and sample preparation. For instance , if the substrate must be conductive, than something rather inert would be preferable to iron.  Let us wait for a more informed user.  I just point to (also) not intrinsic race related reason.

While I can't find specific justification for the surface, the first competition involved both gold and silver "racetracks."

Drivers gear up for world’s first nanocar race
How to build and race a fast nanocar

The competition involves propulsion and imaging using STM so the substrate must be conductive. Gold (and silver to a lesser degree) is particularly useful, since it won't easily oxidize and is easy to purify.
I've seen several comments about having 'tracks' in the gold surface, which suggests a well-defined crystal face with reconstruction, e.g. Au(100):

(Image from Wikipedia)

I know a few people interested in "theoretical" nano car racing simulations, and will revise my answer as I hear back about specific reasons.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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How to determine omitted units in a publication
I have found a 1954 paper in J. Am. Chem. Soc. which gives various temperatures in degrees but does not specify units.
A snippet of the introduction is:

Titanium reacts appreciably with fluorine above 150°…

If I were feeling charitable, I might excuse the "units are for wimps" attitude as reflecting a commonly understood standard for that publication at that time. Perhaps my searching was poor, but I couldn't find an ACS standard or style guide for the 1950s era.
Given the location and time, I would guess that the temperatures are in degrees Fahrenheit, but I can't be sure they're not Celsius. Is there any reasonable way to determine which units the authors were actually using?
I don't have access to the full paper, just the first page at the link below.
Haendler et al., J. Am. Chem. Soc., 1954, 76 (8), pp 2177–2178
DOI 10.1021/ja01637a038

These are degrees Celsius.
To prove that, you could search JACS for the publications of the same year/issue for the temperature-related physical constants such as melting point:

Douglas et al. [1] lists m.p. of $\ce{LiF}$ as $\pu{848 ± 1 ^\circ}$, whereas currently Wikipedia lists $\pu{845 °C}$.
Yankwich et al. [2] lists m.p. of malonic acid as $\pu{140 ^\circ}$ whereas currently Wikipedia lists $\pu{135 to 137 °C}$.

You may find further examples, but even at this point it's fairly obvious what those temperature units are.
References

Douglas, T. B.; Dever, J. L. Lithium Fluoride: Heat Content from 0 to 900°, the Melting Point and Heat of Fusion. J. Am. Chem. Soc. 1954, 76 (19), 4826–4829. https://doi.org/10.1021/ja01648a016.
Yankwich, P. E.; Promislow, A. L. Intramolecular Carbon Isotope Effect in the Decarboxylation of Liquid Malonic Acid Near the Melting Point. J. Am. Chem. Soc. 1954, 76 (18), 4648–4651. https://doi.org/10.1021/ja01647a050.

I don't see an easy way to determine this, without e.g. cross-referencing the data to some other source. However, maybe noteworthy is the fact that the next article in the journal (J. Am. Chem. Soc. 1954, 76 (8), 2178–2179) is written by the same first author, H. M. Haendler. In that article, there is a table giving temperatures in degrees Celsius. It stands to reason that the author would have consistently used the same units in consecutive publications.
As a side note, a tip to access old ACS papers which are two pages long: You can get the first page free from the link you have, and the second page free from the link to the article immediately after it. (The first page of the next article being, of course, the second page of the previous article.)

@orthocresol: Good point. Yet, I'd say those 2-page papers are either notes or communications, to make it technically correct. :-)

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Aldehyde and ketone reactions

Reagents to carry out above conversion, P, Q, R respectively are:
(a) $\ce{H2C=CH-CH2-Br}$, $(\ce{HO-})$, $[\ce{HO-}, Δ]$, Wacker process.
(b) $\ce{H2C=CH-CH2-Br(HO^-)}$, Wacker process, $\ce{HO-}$, $Δ$.
(c) Wacker process, $\ce{H2C=CH-CH2-Br(HO^-)}$, $\ce{HO-} (Δ)$.
(d) Wacker process, $\ce{HO-} (Δ)$, $\ce{H2C=CH-CH2-Br(HO^-)}$.

Could someone please explain the R part of this reaction? I can't understand the five-membered cyclization step. Is it aldol condensation or there's something else?

Step P is std base-driven alpha alkylation of ketone. Step Q Wacker details here takes the terminal alkene to methyl ketone then step R is base-catalysed internal aldol mechanism here gives the product shown. So the answer is (b).


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How can I save and read the CI vector during CASSCF calculations in Gaussian?
I have searched through the GAUSSIAN 16 manual and have not been able to solve these two related issues:
(1) I want to be able to save the CI vector and load it as an initial guess for a different calculation. Or at least to be able to manually set the initial CI vector.
(2) During a geometry optimization, the initial guess for the CI vector changes on each geometric iteration. This does not make much sense to me, as the final CI vector is almost always very similar to the one for the previous geometry! I have noticed that this can lead to huge orbital rotations (like starting with a CI[1] = 1 vector and swapping the HOMO for a core orbital) and then this leads to convergence issues. How do I force it to use the final CI vector from the previous iteration as the starting guess?

I have not used CASSCF before, but it seems like what you want for the first part of your question is the StateGuess keyword. http://gaussian.com/cas/


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Partial oxidation of toluene
While reading the preparation of phenol from cumene by reaction with $\ce{H2O2}$ or partial oxidation, I was curious to find out why toluene wasn't used.
I checked the literature and found that benzaldehyde would be formed in the case of toluene [1, 2].
Why does benzaldehyde form in the case of toluene?
Does the peroxide formed after initial oxidation become benzaldehyde due to its instability?
Why doesn't acetophenone form from the peroxide formed in cumene?
References

Sun, H.; Blatter, F.; Frei, H. Selective Oxidation of Toluene to Benzaldehyde by $\ce{O2}$ with Visible Light in Barium(2+)- and Calcium(2+)-Exchanged Zeolite Y. Journal of the American Chemical Society 1994, 116 (17), 7951–7952. https://doi.org/10.1021/ja00096a084.
Wang, X.; Wu, J.; Zhao, M.; Lv, Y.; Li, G.; Hu, C. Partial Oxidation of Toluene in $\ce{CH3COOH}$ by $\ce{H2O2}$ in the Presence of $\ce{VO(Acac)2}$ Catalyst. The Journal of Physical Chemistry C 2009, 113 (32), 14270–14278. https://doi.org/10.1021/jp9028062.

You can justify the formation of Benzaldehyde from Toluene by oxidising with peroxide just by looking at the mechanism of the reaction and think of migratory aptitude of Phenyl group , hydride and methyl group.
Firstly, if we delve into the mechanism of the oxidation of Cumene, we'll find that there is a rearrangement step in the mechanism. The rearrangement occurs when one of the oxygen in hydroperoxide gets protonated and leaves as water. If you now think carefully, there are both the possibilities of migration of Phenyl and methyl group to the oxygen. But Phenyl has certainly more migratory aptitude than methyl, and also the carbocation formed after the rearrangement is stable, as it has two methyl group on the $\alpha$-position to stabilise by electron release and one oxygen atom to stabilise it via donation of lone pair. Thus the migratory aptitude and stability of the carbocation formed both drive the rearrangement to prefer the phenyl migration. Therfore acetophenone is not formed due to unfavoured methyl migration.

On the other hand, the situation with toluene is slightly different. Here you have both the possibilities of hydride as well as phenyl migration. But hydride has almost similar (slightly more) migratory aptitude as Phenyl group, and what is more interesting is that the migration of hydride will lead to more stable benzylic carbocation, which is not possible by phenyl migration. These two effects cooperatively facilitate the hydride migration and therefore leading to the product Benzaldehyde.

Thus the migratory aptitude and stability of the carbocation are the two effects which create the differences in the product formation in case of oxidation of Cumene and Toluene.
(Note : For clarification, see the migratory aptitude order here : https://en.wikipedia.org/wiki/Migratory_aptitude )

Thank you for answering. I still have a doubt. In the case of cumene, wouldn't methyl shift lead to a more stable benzylic carbocation as well? Would the migratory aptitude relative rates outway the stability of intermediate always?

@thewitness Yes it can but phenyl has much much higher migratory aptitude than methyl group,so it doesn't allow the possibility of carbocation stabilization factor to dominate at all. But still the carbocation in this case is also any way tertiary and that's not very unstable. So, here the extremely high migratory aptitude of phenyl group rather than methyl group dominates here.


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method to certify the number of chiral carbons in fischer and haworth projection of glucose
While reading some notes about Glucose and its various projections:

Fischer Projection: 4 chiral centres

Haworth Projection: 5 chiral centres

My doubt is that is there any (specific) method (practical) to ascertain the number of chiral centres in both the forms.
I know that more than 99.5% of glucose exists as Haworth Projection - Closed Ring Structure and hence Linear - Fischer Projection practically does not exist.
But it (linear ie open chain glucose) could be always isolated though difficult.
Thank you!

While alpha-D-glucose does primarily exist in the cyclic form,  it does not exist as a Haworth Projection since the ring is not planar.  The chair conformation is a more accurate description. https://www.google.com/search?q=alpha+glucose+structure&client=firefox-b-1-d&tbm=isch&source=iu&ictx=1&fir=BsiLDcxZkenRdM%253A%252CI7n8CaCE0-at0M%252C_&vet=1&usg=AI4_-kQE6Wh8iXlK1MCwLjfv7RBUZiDfdw&sa=X&ved=2ahUKEwjG-9f3_MvhAhXRnuAKHdecC4IQ9QEwDHoECAoQBg#imgrc=BsiLDcxZkenRdM:

@user55119 : What you say is absolutely correct. But what my point is whether you consider simple closed ring or haworth projection or chair conformer of glucose it would be immaterial while considering the number of chiral centres, which is 5, in comparison to the chiral centre (4) in a linear representation of glucose. chem.libretexts.org ..  bottom of this page...

It is just a matter of which representation you are talking about. There is no absolute answer. Fischer used the linear representation to distinguish amongst stereoisomers. It was a simplification of Van't Hoff's more complex representation.


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Equation for reaction between chlorine fluoride and potassium bromide
The equation given in the answer key was
$$\ce{ClF + 2KBr → KCl + KF + Br2}$$
However won't the $\ce{KCl}$ also react with $\ce{ClF}$ to form $\ce{KF}$ and $\ce{Cl2}$, resulting in the net reaction being
$$\ce{2ClF + 2KBr → 2KF + Cl2 + Br2}~?$$

It may depend on molar ratio of reagents. Both equations may be correct at some conditions.

Yea poutnik I was thinking the same thing, problem was the question didn't specify. Thanks though, wanted to confirm the second equation could work.

Perhaps check ΔG on both? Just suggesting as I haven't checked anything

Let's subtract the two reactions from each other to simplify things a bit:
$$\ce{2ClF + 2KBr -> 2KF + Cl2 + Br2}\ \ \ \ \ \ \ \ \ \ [1]$$
$$\ce{ClF + 2KBr -> KCl + KF + Br2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [2]$$
$$\ce{ClF + KCl -> KF + Cl2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ [3] = [1]-[2]]$$
Now we can discuss [2] and [3] separately by splitting them up into half reactions.
Reaction 2
$$\ce{ClF + 2KBr -> KCl + KF + Br2}$$
Potassium is a spectator ion. The reduction half reaction is:
$$\ce{ClF + 2e- -> Cl- + F-}$$
The oxidation half reaction is:
$$\ce{2Br- -> Br2 + 2e-}$$
Reaction 3
$$\ce{ClF + KCl -> KF + Cl2}$$
Potassium is a spectator ion. Fluorine does not change oxidation state. Chlorine undergoes comproportionation (oxidation states +1 and -1 to oxidation state of 0). The reduction half reaction is:
$$\ce{ClF + e- -> 1/2 Cl2 + F-}$$
The oxidation half reaction is:
$$\ce{Cl- -> 1/2 Cl2 + e-}$$
Which reaction happens?
That depends on the reduction potentials, the kinetics of the two reactions and the concentration of species present. I'm not sure how you would figure it out without experimental (or computational) data.
Comparing the oxidation half reactions, you could make an argument about electronegativities of bromine vs chlorine, but I have no insight to compare the two reducation half reactions, which are very different.
The Wikipedia article on chlorine monofluoride states:

Many of its properties are intermediate between its parent halogens, $\ce{Cl2}$ and $\ce{F2}$.

If you substitute $\ce{Cl2}$ for $\ce{ClF}$ in reactions 2 and 3, they both look reasonable (reaction 3 becomes a self-exchange, i.e. equilibrium constant = 1, whereas reaction 2 reduces chlorine and oxidizes bromine, products are favored).


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Thermal conductivity and enthalphy data for nitrogen gas
I am struggling to find the data for thermal conductivity and ethalphy of nitrogen gas at reduced pressure (500 mbar). Most of the information is for nitrogen gas at 1 bar or above. Does anyone know where I can find this information? Specifically, I need to know the thermal conductivity and enthalphy of nitrogen at 295.15 K (22 degrees) and pressure of 0.5 bar (500mbar)

Transport Phenomena, Bird, Stewart, and Lightfoot.  Certainly, under these conditions, N2 can be treated as an ideal gas.  The data at 1 bar are adequate, and the values are not significantly different.

By way of comparison, using NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, I have got a specific enthalpy of $h=306.26\ \mathrm{kJ/kg}$ and thermal conductivity of $\lambda=25.601\ \mathrm{mW/(K\ m)}$ for nitrogen at a temperature of $T=295.15\ \mathrm K$ and a pressure of $p=0.5\ \mathrm{bar}$.
Note that the absolute value of enthalpy at a single state point is meaningless; it is only the difference between two different state points that matters.

Thanks, I need to find the difference between the enthalpy of nitrogen at two different set-points [(T = 295.15, P = 1 bar) and (T = 308.46, P = 0.5 bar)]. I do not have access to the NIST database. So I tried to look up the values from Perry's Engineering Handbook. I have the table for the enthalpy values but it only gives me values at certain conditions. I obtained the enthalpy for (T = 295.15, P = 1 bar), which is 306.168 kJ/kg. It would be really helpful if you could check the enthalpy at (T = 308.46, P = 0.5 bar)

The thermal conductivity is largely independent of pressure. Values for nitrogen gas at 1 atm. are (in units of $10^{-2}$ W/m/K ) 1.59 at 173.2 K, 2.40 at 273.2 and 3.09 at 373.2 K.
The values are independent of pressure because conductivity is, from kinetic theory of gases, $\kappa = n\bar v s k_B \lambda_m/6$ where $n$ is the number density of molecules, $s$ the number of translational degrees of freedom, 3, $\bar v$ the mean speed of the gas and $\lambda_m$ the mean free path. (This equation can be re-written as $\displaystyle \kappa=\frac{\bar v sk_B}{6\sqrt{2}\pi r^2}$ where $r$ is the molecular radius which shows that $\kappa$ has no pressure dependence )
The conductivity will be greatest when there are more molecules to carry the energy (large $n$) and when they can travel unrestricted through the gas, large $\lambda_m$.  However, these two terms are in inverse relationship and exactly cancel one another and hence thermal conductivity is largely independent of pressure. The range turns out to be from about $10^3 \to 10^6 $ Pa.


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Why is Lithium Chloride (LiCl) soluble in alcohols?
According do J.D. Lee's Concise Inorganic Chemistry, LiCl is soluble in alcohols because it has some covalent nature, althought it is mainly polar.
Why would LiCl have covalent character? Why doesn't this happen to NaCl?
Also, is this explanation true? A friend of mine said this solubility was because of the polar nature of small alcohols. Which of them is right?

https://en.m.wikipedia.org/wiki/Fajans'_rules is a good place to start

Small sized cations have high electronegativity due to strong electrostatic attraction between the cation to the valence electron orbital. Cl and Li have quite similar electronegativities, so a (largely) covalent bond results. Sodium has a much larger size than Li, thus has low electronegativity. The solubility of LiCl in alcohols is due to covalent nature of bonding in both of them (like-dissolves-like). The polarity of alcohols is less than that of water; you'd expect LiCl to be more soluble in water than alcohols then.

Related:  https://chemistry.stackexchange.com/questions/115112/why-do-we-use-licl-as-a-source-of-chloride-ions-as-nucleophile-but-not-other-alk/115149?r=SearchResults#115149.

The covalent character of any compound can be predicted on the basis of Fajans' Rule.
Fajans' Rule, simply stated, says that a smaller cation can polarize a given anion better than a larger cation and that larger anions are more polarizable than smaller anions.
$\ce{Li+}$ being a small cation polarizes $\ce{Cl-}$ to a good extent and since any amount of polarization increases the covalent character of a compound, $\ce{LiCl}$ can be said to have some covalent nature.
$\ce{LiCl}$ being somewhat covalent dissolves in organic solvents(Remember, Like dissolves Like) such as ethanol, acetone, ethyl acetate and pyridine.
Further, since $\ce{Na+}$ is unable to polarize $\ce{Cl-}$ to a good degree it is regarded mainly as an ionic compound.


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Electronic configuration of the actinides
I was reading about actinides and I need some help in understanding this paragraph from Concise Inorganic Chemistry by JD Lee -

The electronic structures of the actinides do not follow the simple pattern found in the lanthanides.
  Immediately after La the 4f orbitals become appreciably lower in energy than the 5d orbitals. Thus in the lanthanides the electrons fill the 4f orbitals in a regular way (apart from minor differences where it is possible to attain a half filled shell). It might have been expected that after Ac the 5f orbitals would become lower in energy than the 6d orbitals. However, for the first
  four actinide elements Th, Pa, U and Np the difference in energy between 5f and 6d orbitals is small. Thus in these elements (and their ions) electrons may occupy the 5f or the 6d levels, or sometimes both. Later in the actinide series the 5f orbitals do become appreciably lower in energy. Thus from Pt
  onwards the 5f shell fills in a regular way, and the elements become very similar.

How do we know when the energy of the orbitals of a subshell become lower or higher than the orbitals of another subshell? Can this just be found experimentally or is there some theoretical explanation too?

It is difficult to calculate the energy levels of the orbitals beyond hydrogen. "The first experimental evidence for the filling of the 5f shell in actinides was obtained by McMillan and Abelson in 1940," from Wikipedia
So the answer, as you suspected, is that experimental evidence from spectroscopy helped determine the electron configuration of the lanthanides and actinides, in particular.

Thank you for answering! I wonder, since energies of orbitals of elements beyond hydrogen can't be found (as mentioned in the comments of your first link), are the comments in the book on the increasing and decreasing energies of 5f orbitals compared to 6d just conjectures based on the electronic configuration determined by spectroscopy?

I'd say, not conjectures, but experimentally determined by spectroscopy and other means.


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Safe and responsible disposal of aluminum chloride?
This answer affirms that reacting aluminum with HCl is a (seemingly convenient) way to produce hydrogen for use in other experiments, but the byproduct, AlCl3, is rather nasty. What's the best way to dispose of it? My naive guess is that reaction with baking soda would produce carbon dioxide and aluminum hydroxide, which would be fairly harmless. Is this correct?

Sure, you can react the $\ce{AlCl3}$ with $\ce{NaHCO3}$ solution, and you'd also get $\ce{NaCl}$, in addition to $\ce{CO2}$.
Though $\ce{AlCl3}$ is toxic, it reacts readily with water, forming the less toxic chlorohydrate (used in antiperspirants!), and the baking soda should leave an innocuous sludge, safe to dispose of in a sink.
N.B.: The reaction is exothermic and bubbly, so use care when mixing in the $\ce{AlCl3}$ to avoids spatters.

If it reacts readily with water, why does the initial reaction of Al with HCl produce it rather than yielding the chlorohydrate immediately? Or does it?

@R.. It obviously doesn't, really. You get a solution of aluminium chlorohydrate, which you shouldn't drink.

You should begin by consulting the safety data sheet (SDS) provided by the manufacturer and Prudent Practices in the Laboratory to understand the hazards.
The laws of your jurisdiction and the policies of your organization will determine what amount (if any) and by what method neutralization is permitted. Environmental health and safety (EH&S) groups in some organizations deal with authorities who view quenching any regent as illegal unlicensed hazardous waste processing, justifying significant fines. EH&S might insist that you have a special pickup from your hazardous waste provider.
It is possible to quench AlCl3 cautiously in stirred ice water while keeping the pH to within the permissible range to bulk with your aqueous waste stream. Keep in mind the quench is highly exothermic and can get out of control quickly. From a practical standpoint, the volume of aqueous waste generated might make this approach more costly than having a LabPack pickup of the solid aluminum chloride waste.

According to the other answer and Karl's comment on it, there is no solid aluminum chloride waste unless you go out of your way to make it, so perhaps my question is now somewhat mismatched with my intent.

@R.. Jup. Unless you use dry HCl gas instead of dilute hydrochloric acid. ;-)

On the question "Safe and responsible disposal of aluminum chloride?", I would exercise more care (as to why, see chemistry below) as even neutralizing AlCl3 with NaHCO3 actually, in addition to CO2 and NaCl, may form insoluble aluminum carbonate albeit its apparently breaks down into aluminium hydroxide, if even formed. The latter, however, if improperly disposed of, for example, coming into contact with an organic acid, could potentially reform some soluble $\ce{Al+++}$ ions. Note: Aluminum ions are not found in nature, so living entities really do not known what to do with them. As such, do follow the recommended procedures for the more toxic chemicals.
Does anyone believe there is a problem with aluminum ion exposure?  Actually products producing said ions are currently dose limited in Europe where scientists seem to be a little more concerned about possible health effects at higher exposure levels (stating that "Aluminium is a known systemic toxicant at high doses") than say the United States.
But what are the aspects of the chemistry that actually support this cautionary behavior?  While the chemistry I present appears to be a little obtuse, it's consequences can be actually rather simply understood.  Aluminum ions can capture superoxide radicals and form a weak unstable adjunct compound therewith that, in time, breaks down releasing, unchanged, the superoxide (or in more acidic conditions with a pH under than 4.88, the more powerful $\ce{.HO2}$ radical) to engage in transition metal REDOX reactions.  So what actually is the new problem? Well, if superoxide is formed in your lungs because of exposure to transition metals dust and oxygen, for example,  it is a transient radical with a limited lifespan. However, upon jumping on board an aluminum ion, it may now have a far greater reach, especially in distant parts of your human body to engage in so-called oxidative stress reactions. Luckily, the superoxide does not live up to its name and it's actually a rather weak radical.  However, exposing parts of your brain, for example, to higher doses of any radical  activity based oxidative stress damage is not likely healthy. So perhaps not surprisingly, there is a claimed link of aluminum presence to brain disease.
Here is a chemistry based upon a review of two articles noting the pro-oxidant property of the aluminum ion carrying superoxide, which can move transition metals (like Fe, Cu, Mn,..) to lower valence states in Fenton and Fenton like reactions, recycling said reactions, and thereby contributing to so-called oxidative stress related diseases (including Alzheimer's disease,...).
The first alluded to article is “Oxidative Stress Gated by Fenton and Haber Weiss Reactions and Its Association with Alzheimer’s Disease” by Tushar Kanti Das, et al, published in Archives of Neuroscience, July 2014.
Importantly, the work cite, in Figure 4, “Formation of Aluminum Superoxide Semi reduced Radical Ion and Aluminum Superoxide Complex (43)”, with described reactions proceeding as follows (also adopting notation and water complexing from the second article):

$\ce{[Al(H2O)4](3+) + O2•− <-> [Al(O2•−)(H2O)4](2+) }$

$\ce{[Al(O2•−)(H2O)4](2+) + Fe(3+) -> O2 + [Al(H2O)4](3+) + Fe(2+) }$

Note, the above is the usual action (namely, $\ce{O2•− + Fe(3+) -> O2 + Fe(2+) }$ on superoxide recycling ferric to ferrous to promote a Fenton oxidative stress reaction, so, de facto, the superoxide is transported and released courtesy of the superoxide semi-reduced radical ion.
And,  at pH < 4.88 :
$\ce{[Al(H2O)4](3+) + O2•− + H+ <-> [Al(O2•−)(H+)H2O)4](3+) }$
$\ce{[Al(O2•−)(H+)H2O)4](3+) + [Al(O2•−)(H+)H2O)4](3+) -> 2 [Al(H2O)4](3+) + H2O2 + O2 }$
Another work: "Pro-oxidant Activity of Aluminum: Stabilization of the Aluminum Superoxide Radical Ion" by J. I. Mujika, F. Ruiperez, I. Infante, J. M. Ugalde, C. Exley, and X. Lopez in J., published in Phys. Chem. A, 2011, 115, 6717–6723, American Chemical Society.
In the Mujika article to quote “In addition, the presence of LMM ligands such as citrate could also have an indirect effect in the oxidation capacity of aluminum by augmenting the bioavailability of Al3+ species, shifting the formation of Al(OH)4- to higher pH’s. However, one should also take into account the effect of citrate chelation itself in the thermodynamic equilibrium of [AlO2•]2+ formation.”
I hope this helps understanding potential issues with Aluminum ions.


text:
Difference in properties of water as ligand vs 'in lattice'
I'm trying to understand the difference in properties between water in crystals in different positions in the lattice. In this wikipedia article It quotes:

Both notations can be combined as for example in copper(II) sulfate: $\ce{[Cu(H_2O)_4]SO_4 · H_2O}$

referring to water which is coordinately bonded to the copper and water which is in the crystal structure left over.
Firstly, is this the same as:
$\ce{CuSO_4 · 5H_2O}$ ?
And secondly, would there be a difference in how the water acted upon heating? Would it only be the one water in the lattice structure that left or all 5 waters?
i.e.
$$\ce{[Cu(H_2O)_4]SO_4 · H_2O ->[heat] [Cu(H_2O)_4]SO_4 + H_2O}$$
or
$$\ce{[Cu(H_2O)_4]SO_4 · H_2O ->[heat] CuSO_4 + 5H_2O}$$

Yes, both are the same thing. As for the second part, there is no simple rule what water molecules are eliminated first upon heating and an experiment may show unintuitive results, for instance, for copper(II) sulfate looses four water molecules first, not one (see Hydrate decomposition equation steps).


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Can the nitrogen of an amide displace a primary chloride in a SN2 reaction in the presence of a strong base?
I was looking at the following reaction and am confused.

It appears that $\ce{N}$ is attacking the $\ce{C-Cl}$ and $\ce{Cl}$ is leaving in a $\mathrm{S_N2}$ reaction. I know that $\ce{Cl}$ is a good leaving group, but $\ce{N}$ isn't a particularly soft nucleophile so I wouldn't expect it to be great at attacking a saturated carbon.
Additionaly there is $\ce{MeO-}$ in solution, even with intramolecular cyclisation being quick wouldn't there be a realitive amount of attack by $\ce{MeO-}$.
So am I missing something? Is this not as simple an $\mathrm{S_N2}$ as I have made out?
Why is sodium methoxide ($\ce{NaOMe}$) a good choice of base?

This is not an amine, it is an amide anion. Methoxide is a strong enough base to deprotonate an amide, this will happen preferentially, and an intramolecular cyclisation is strongly favoured over intermolecular. Sodium hydride would be a better choice of base

I suspect the reason NaOMe was chosen is because of the free -OH elsewhere in the molecule. There will be some deprotonation of this, but the cyclisation is strongly disfavoured.

@Waylander The cyclisation where OH attacks? Why is that?

OH attack would form a 9-membered ring, strongly disfavoured vs 6-membered.

The other issue is whether the nucleophile is the N or the O of the amide anion.

@Waylander Your comments would make a nice answer, care to write it up?

Well it is difficult to compare an intra-molecular reaction with an extra-molecular one because it really depends on the conditions. However, unless you have a very high concentration of one reactant, the intramolecular way will always be favored.
In your question, you consider that "N" is not a very good "attacking group". Well this is wrong. "N" is always happy to attack any nucleophilic group, even if it becomes "N+". However, a "CON-" is much more nucleophilic than just "CONH".
Sodium methoxide here just serves at deprotonating CONH to activate it, but MeONa (sodium methoxide) is not nucleophilic enough to interfere.

By "attack any nucleophilic group" do you mean "be a nucleophilic attacking group" or "attack an electrophilic group" or something?  In this reaction the amide anion is the nucleophile.

In accordance with what @SteffX said, it really depends on the conditions and every reaction has its own story. However a couple of factors favour the formation of the given product above others. There include the fact the MeOH might help to generate the "activated Nucleophile" anion at amide, being a polar protic solvent, it provides higher solvation to the other groups like OH, permitting the amide to take the reaction head on. Also, the greater favourability of a six membered intramolecular attack is a contributing factor. Amide anions are an unconventional nucleophile but in absence of stronger competitors, it can do the job. Hope that helps!!


text:
Does there exist a dye that will primarily bind with water in a water/oil emulsion?
I'm wondering if there exists a type of dye that will bind primarily with water when introduced to a water/oil emulsion. That way if the water is centrifuged to the bottom of the vial, it will be dyed showing better visibility for the meniscus line.
The main application of this would be for centrifugation of clear oil that has suspended water. If the oil is clear and spun out, the water may separate but seeing the distinct meniscus can be difficult. If a dye in introduced and binds to the small water content within this emulsion, once it's spun out, ideally you'd see the dyed water at the bottom. The content of water is probably 0.5 - 1.0%
I'm also somewhat aware of some water by distillation methods using dyes to detect small amounts of distilled water, but I'm unfamiliar with these and I can't find any helpful resources online about it.
I suppose it would need to be a hydrophilic dye that also resists binding to hydrocarbons?

Why, many. Pretty much any strongly colored inorganic complex ion will do.

Most food coloring uses hydrophilic dyes.

You are asking for water-soluble dyes. Typical applications are dyeing methods in microscopy and food coloring.
Water-soluble dyes for microscopy: e.g. Eosin Y, Methylene Blue, Ethyl Green.
See also this table: Wikipedia in German: Lebensmittelfarbstoff - Deutschland.

Most washable markers (for kids) contain such water-soluble dyes as well.


text:
Metal that does not deposit silver (Ag)
Hi Chemistry enthusiasts!
I am working on my scholarship exam practice but I am considered myself as a newbie so I would like to hear your opinion :)

From (1)-(4) below choose the metal that does not deposit silver (Ag)
  on the surface when immersed in aqueous silver nitrate. (1) Cu (2) Fe
  (3) Pt (4) Zn

The answer is Pt but I am not sure how I can solve this problem without any other given information such as electrode potential etc. Please advise.

You just have to know that Ag and Pt are the only noble metals over all 5 elements.  Thus Pt is the only logical choice. (Think what metals "good" jewelry is made from...)

You do need the electromotive series or reactivity series to answer that question. If you will not have access to it during such exams, you'll need to memorize the order of some common elements, and to understand the arrangement of the Periodic Table.
The most reactive metals are at the bottom left of the table, low in Group I. The "noble" or coinage metals are clustered near the center, with reactivity generally decreasing within a column. E.G., Au is less reactive than Ag, which is less reactive than Cu.
Later, you may learn about the electron configuration of atoms and how that determines reactivity, but at this point, knowing the reactivity series would get you though the test.

Thank you for pointing it out so clearly! Now I know where I should be focusing on next :)


text:
What emits the least radiation when hot?
What element with a high vapor point the emits to least light or radiation when heated to its vapor pressure point in a vacuum?
Ideally I would like to know in 0 gravity and 0 pressure which elements can be get the hottest around the sun before the vapor pressure point is reached and/or emit the least black body radiation when hot.
https://space.stackexchange.com/questions/35486/shielding-capability-of-liquid-metals

The black body radiation from all elements and compounds is the same.

i suppose an isolated proton wouldn't

As for A.K. comment: your question should be what ensemble of matter does not emit EM radiation (or Vis, depending on your exact meaning) when hot? Then the answer is in the comment by Karl. (Dark matter not considered as for no chemistry can be done ;)

@Alchimista is this better?

Say substance, so you get rid of the fact that atoms/molecules can't have a T and a vapour pressure. Hot atoms are at best moving faster. But you got an answer so.... Referring to a limit made the Q interesting.

Emissions of radiations depends on elementary particles, so NOT on molecular structure. As Karl said, it depends on "black body radiation". So, please, stop up-voting that inept answer!

Technically, a smooth or molten, unoxidized metal surface (regardless of chemistry) makes a good candidate.  This thesis from Auburn University reports the emissivities of both oxidized and unoxidized metals at temperatures typically encountered in industrial processes; note the low values for all the unoxidized metals studied in the IR range where most radiation is given off.

Yes but mecury has a low evaporation point.

@Muze — But the state does not matter in emitting electromagnetic waves!

@Steffi can you please explain more in an answer for a Layman like myself can understand.

Will anyone please tell me what is wrong with this answer?

I thought maybe @SteffX would have went more into detail with an answer but he didn't. I think your answer is good. Tks

@Muze — To my knowledge, any fundamental particle emits light when accelerated BUT acceleration in its broadest sense can mean either increasing or decreasing the speed OR changing direction (this property is used to generate X-rays). So any atom will radiate as much as its fundamental particles (black body radiation as Karl said). And there is no point about asking about "hot" compounds (how hot?) with a "high vapor pressure" in deep space.

@SteffX At what point does iron start to emit xrays before reaching it vapor point? I should have said element and not substance.


text:
Which substances have the highest/lowest enthalpy of formation per gram?
Which substances have the most extreme enthalpies of formation per gram at standard temperature and pressure?

Beryllium oxide −23.96 kJ/g
Lithium fluoride −23.75 kJ/g
Beryllium fluoride −21.84 kJ/g
Beryllium hydroxide −21.01 kJ/g
Lithium hydroxide −20.24 kJ/g
Lithium oxide −19.93 kJ/g
Boron trioxide −18.29 kJ/g
Magnesium fluoride −18.04 kJ/g
Aluminum fluoride −17.98 kJ/g
Boron trifluoride −16.76 kJ/g
Aluminum oxide −16.43 kJ/g
Water −15.86 kJ/g
Magnesium hydroxide −15.85 kJ/g
Magnesium oxide −14.92 kJ/g
Hydrogen fluoride −13.66 kJ/g
Lithium hydride −11.4 kJ/g
Boron nitride −10.24 kJ/g
Carbon dioxide −8.941 kJ/g
Lithium nitride −4.73 kJ/g
Methane −4.6 kJ/g
Ammonia −2.71 kJ/g
Diborane 1.31 kJ/g
Cyclopropane 1.33 kJ/g
Cubane 5.73 kJ/g
Dicyanoacetylene 6.579 kJ/g
Various unstable nitrogen allotropes >10 kJ/g
Free hydrogen (not bound in diatomic molecules) 216.34 kJ/g

This is interesting, but if the answer is nothing but a list of values, sources are rather important.

I calculated all these using data from either wikipedia or webbook.nist.gov

Jonathan Ray has indicated he has used data from either Wikipedia or NIST Webbook in his comment. Actually, NIST Webbook is an excellent choice because it has listed wide variety of thermochemical data. However, there were few important references and points he did not mention. For example, in that tabulated data set, the state of compound(s) wasn't clear. If you look at the data reported for calcium fluoride (as liquid and solid), you may understand my point (see below table). I'm here trying to correct those facts and some other errors in above table given by my colleague. One such erroneous data is the value given for cyclopropene (g). Actually, value given above is more closed to that of cyclopropane than for cyclopropene (see below table). For readers' benefits, I also want to emphasize that Ref.1 has tabulated a lot of data, which are NIST Webbook incorporated (in case if you don't trust NIST Webbook).
$$
\begin{array}{ccc}
\hline
\text{Compound} &\text{Formula} & M\text{, g/mol} & \Delta_\mathrm fH^\circ\text{, kJ/mol} & \Delta_\mathrm fH^\circ\text{, kJ/g} & \text{Ref.} \\\hline
\text{Beryllium oxide (s)} & \ce{BeO} & \pu{25.01} & \pu{−609.43} & \pu{-24.57} & \pu{1} \\
\beta\!-\!\text{Beryllium hydroxide (aq)} & \ce{Be(OH)2} & \pu{43.03} & \pu{−905.84} & \pu{-21.05} & \pu{2} \\
\alpha\!-\!\text{Beryllium hydroxide (aq)} & \ce{Be(OH)2} & \pu{43.03} & \pu{−902.70} & \pu{-20.98} & \pu{2} \\
\text{Lithium Hydroxide (s)} & \ce{LiOH} & \pu{23.95} & \pu{-484.93} & \pu{-20.25} & \pu{1} \\
\alpha\!-\!\text{Aluminium oxide (s)} & \ce{Al2O3} & \pu{101.96} & \pu{-1675.69} & \pu{-16.43} & \pu{1,3}\\
\text{Calcium fluoride (s)} & \ce{CaF2} & \pu{78.07} & \pu{-1225.91} & \pu{-15.70} & \pu{1} \\
\text{Calcium fluoride (l)} & \ce{CaF2} & \pu{78.07} & \pu{-1186.07} & \pu{-15.19} & \pu{1} \\
\text{Hydrogen fluoride (g)} & \ce{HF} & \pu{20.01} & \pu{-272.55} & \pu{-13.62} & \pu{1} \\
\text{Cyclopropane (g)} & \ce{C3H6} & \pu{42.08} & \pu{53.30} & \pu{1.27} & \pu{4}\\
\text{Nitric oxide (g)} & \ce{NO} & \pu{30.01} & \pu{90.29} & \pu{3.01} & \pu{1}\\
\text{Cubane (g)} & \ce{C8H8} & \pu{104.15} & \pu{622.2} & \pu{5.97} & \pu{5}\\
\text{Cubane (g)} & \ce{C8H8} & \pu{104.15} & \pu{597.1} & \pu{5.73} & \pu{6}\\
\text{Cyclopropene (g)} & \ce{C3H4} & \pu{40.06} & \pu{277} & \pu{6.91} & \pu{7}\\
\text{2-Butynedinitrile (g)} & \ce{NCCCCN} & \pu{76.06} & \pu{533.46} & \pu{7.01} & \pu{1}\\
\text{Acetylene (g)} & \ce{HCCH} & \pu{26.04} & \pu{226.73} & \pu{8.71} & \pu{1}\\
\hline
\end{array}
$$
References:

M. W. Chase, Jr., NIST-JANAF Themochemical Tables, Fourth Edition, J. Phys. Chem. Ref. Data, Monograph 9, 1998, 1–1951.
I. J. Bear, A. G. Turnbull, “The Heats of Formation of Beryllium Compounds. I. Beryllium Hydroxides,” J. Phys. Chem. 1965, 69(9), 2828–2833 (DOI: 10.1021/j100893a004).
R. C. King, G. T. Armstrong, “Heat of Combustion and Heat of Formation of Aluminum Carbide,” Journal of Research of the National Bureau of Standards–A. Physics and Chemistry 1964, 68A(6), 661–668 (http://dx.doi.org/10.6028/jres.068A.066).
J. W. Knowlton, F. D. Rossini, “Heats of combustion and formation of cyclopropane,” Journal of Research of the National Bureau of Standards 1949, 43(2), 113–115 (http://dx.doi.org/10.6028/jres.043.013).
B. D. Kybett, S. Carroll, P. Natalis, D. W. Bonnell, J. L. Margrave, J. L. Franklin, “Thermodynamic Properties of Cubane,” J. Am. Chem. Soc. 1966, 88(3), 626–626 (DOI: 10.1021/ja00955a056).
A. Bashir-Hashemi, J. S. Chickos, W. Hanshaw, H. Zhao, B. S. Farivar, J. F. Liebman, “The enthalpy of sublimation of cubane,” Thermochimica Acta 2004, 424(1-2), 91–97 (https://doi.org/10.1016/j.tca.2004.05.022).
K. B. Wiberg, W. J. Bartley, F. P. Lossing, “On the strain energy in cyclopropene and the heat of formation of the $\ce{C3H4+}$ ion,” J. Am. Chem. Soc. 1962, 84(20), 3980–3981 (DOI: 10.1021/ja00879a049).

Does spin-stabilized atomic hydrogen at $248.8$ kJ/g count?

This is a fun question.  Given that you are looking for kJ/g rather than kJ/mol, it seemed most fruitful to explore the possible compounds that can be formed from the lightest-weight elements.
So: I was unable to find any compounds to beat those in the lists provided by Jonathan and Matthew for exothermic heats of formation.  But, considering compounds with endothermic heats of formation, gaseous beryllium monohydride, $\ce {BeH_{(g)}}$, has a standard enthlapy of formation (at 298 K) of 321 kJ/mol = 32 kJ/g.
Source: Chase, M.W., Jr., NIST-JANAF Themochemical Tables, Fourth Edition, J. Phys. Chem. Ref. Data, Monograph 9, 1998, 1-1951.
[Secondary source:  https://webbook.nist.gov/cgi/cbook.cgi?ID=C13597972&Units=SI&Mask=FFFF ]
And considering polyatomic ions, gaseous helium hydride, $\ce {HeH_{(g)}^+}$, has  a standard enthlapy of formation (at 298 K) of 1348.3 kJ/mol = 269.1 kJ/g.
Source: https://atct.anl.gov/Thermochemical%20Data/version%201.122g/species/?species_number=1213
Of course, the substance with the highest enthlapy of formation is probably the omega molecule:
https://memory-alpha.fandom.com/wiki/Omega_molecule


text:
Oxidation of metals/halogens by oxygen gas in acidic aqueous solution
I had some confusion about the correct answer to the following question:

Which substance can be oxidized by $\ce{O2(g)}$ in acidic aqueous solution?

Two of the answer choices were metal cations (usually not oxidized further). The remaining three choices were:

$\ce{Br2(l)}$, $\ce{Ag(s)}$ and $\ce{Br-(aq)}$.

The standard reduction potentials were given:

$$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\
\ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V}
\end{align}
$$

The correct answer given was $\ce{Br2(l)}$, however, this did not make sense to me as the oxidation of bromine would result in a positively charged bromine ion. Am I reasoning correctly that $\ce{O2(g)}$ can oxidize both $\ce{Ag}$ and $\ce{Br-}$ in acidic solution?
Since the value of $E^\circ$ for $\ce{O2(g)}$ is greater than both of the other values, more energy is released when $\ce{O2}$ gains the electrons, and this will result in the oxidation of both $\ce{Ag}$ and $\ce{Br-}$, correct?

You do realize that the reactions are reversible. You just need to reverse sign of EMF too. $$\ce{2Br^-(aq) -> Br_2(l) + 2e^- }\quad\text{(E° = -1.065V)}$$ // Silver has enough EMF to go, but doesn't. Hopefully someone else can explain why.

@MaxW I do not suppose bromides are such strong reduction agents. Fluorides would be even much stronger. Both forward and reverse reactions share the same standard redox potential without sign flipping, as the potential includes both reactions in dynamic equilibrium.

@Poutnik - ? - for redox reaction the tables are always shown as one way redox reactions with some EMF. To get the oxidation reaction you flip reactants and products and you also flip the sign on the EMF value. (I realize that this simple approach doesn't take into account the Nernst Equation...)

@MaxW I am afraid you are mistaken. They are expressed for equilibrium 2 way reactions. The reaction visual orientation is rather formal.

By other words, particular electrodes have only one value of their redox potential, not two ones with flipped signs.

@Poutnik - Sorry. I was not trying to imply that there are two different reduction reactions. Rather I was trying to point out that you get the oxidation reaction from the reduction reaction by reversing products and reactants and that you must also reverse the sign for the EMF. I then gave the example for the bromine oxidation reaction.

I see. I understand the flipping EMF sign makes sense in context of being a measure of Gibbs energy change of the reaction. But it does not make sense in redox potentials, as they are for both way equilibrium.

The sign of the electrode reduction potential is invariant. If reflects the sign of the electrostatic charge of the electrode with respect to the hydrogen electrode.* Also remember that all electrode potentials are written as reduction these days. This is a convention set by all electrochemists all over the world along with the other conventions. The following points are needed to approach this problem. I would suggest to solve this problem conceptually along with your own reasoning. It will give you a better understanding in future.
a) Ecell=Ecathode - Eanode > 0 implies a spontaneous reaction. (Eq. 1)
b) Don't change any signs of from tabulated values for Eq. 1. Eq. 1 already takes care of that. In the tables, carefully note that all reduced species appear on the left side and all the oxidized forms appear on the right. Reduction always occurs at the cathode, oxidation at the anode.
c) Ecell < 0, it implies a non-spontaneous reaction. However the reaction is spontaneous in the reverse direction.
Coming to your question. They are asking which substance can be oxidized by oxygen? They have also given you the starting materials/ elements.
Since oxygen is supposed to oxidize Br- and Ag, so it must reduce itself. Note that only reduced forms or the species on the right side of the equations of electrode potentials can be oxidized. Br- and Ag are reduced forms, that is why they were chosen for testing via Eq 1. This implies that we choose Ecathode as the oxygen half cell and test the oxidizable materials.
For Ag+/Ag half cell, and we treat Ag+/Ag as an anode (where silver metal will be oxidized)
Ecell = E(cathode)-Eanode = 1.23 - (+0.799) > 0, this is spontaneous, at least in theory. This number does not tell you how fast the reaction will occur or will it occur or not.
For example, ask the same question, will O2 oxidize elemental hydrogen? Ecell is positive, however if you combine O2 and H2 and seal them in a flask, and wait until the age of retirement, nothing will happen.
For Br2/Br- half cell, which we are considering as anode where Br- will be oxidized
Ecell = 1.23 - (+1.065) > 0, this is also spontaneous.
In short the question is good, but your textbook answer is wrong. O2 in acidic medium, at least thermodynamically, can oxidize Ag to Ag+ and Br- to Br2.

*I don't know if you like history or not, long time ago in the 1950s-60s showing the electrostatic sign of a cell by means of a specially designed electroscope was a standard experiment in physics.
Imagine if we say H2O (l) --> H2O (g) at 100 oC
Does this mean reversing the reaction
H2O(g) --> H2O (l) will be at -100 oC?
One can see logical fallacy in reversing the sign of electrode potentials. Allan J. Bard and other leading electrochemists propose to treat the sign of the electrode potentials as constant. It was Latimer who purported the "oxidation potentials" in his very famous book, however, this school of thought is no longer appreciated by electrochemists.

RE: "at least thermodynamically" is the key phrase. I wonder if there is some kinetic or surface effect that prevents the silver from reacting.

This website clearly demonstrates that electrochemically cleaned silver is immediately oxidized by O2. https://chem.umn.edu/temperature-dependence-silver-oxide-formation "Silver metal will oxidize spontaneously upon exposure to free oxygen. This process is commonly referred to as "tarnishing". The chemical reaction describing this proces is shown below.
4 Ag (s) + O2 (g) --> 2 Ag2O (s)" We should check some older treatise's on inorganic chemistry.

Forming $\ce{Ag2O}$ on the surface and thus passivating it would explain why the "bulk" of the silver wouldn't react. In general you wouldn't detect a small current pulse as the surface passivates. // I'd quibble about silver oxide tarnishing. Silver is very shiny. So the oxide coating must be very thin and it isn't visibly apparent. So I'd only call something tarnish if I could see it.

Seems like the Al case, when it does not dissolve in conc. HNO3 by passivation.

As noted in your question, the standard reduction potentials are given as such in tables.

$$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\
\ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V}
\end{align}
$$

To get the the standard oxidation potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:

$$
\begin{align}
\ce{Ag(s) &→ Ag+ + e-  } &\quad E^\circ &= \pu{-0.799 V} \\
\ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V}
\end{align}
$$

Now there are two possible overall reactions (ignoring the Nernst equation):

$$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{Ag(s) &→ Ag+ + e-  } &\quad E^\circ &= \pu{-0.799 V} \\
\ce{O2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l)} &\quad E_\text{Total}^\circ &= \pu{+1.23 - 0.799 = +0.43 V}
\end{align}
$$

and the other overall reaction is

$$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V} \\
\ce{O2(g) + 4 H+(aq) + 4 Br-(aq) &→ 2Br2(l) + 2 H2O(l)} &\quad E_\text{Total}^\circ &= \pu{+1.23 - 1.065 = +0.16 V}
\end{align}
$$

Since the EMF for both overall reactions is positive, the reactions should be spontaneous. If the overall EMF was negative then the reaction would proceed in the opposite direction (in other words, swap the products and reactants.)
Notes:
(1) I don't know why the silver reaction doesn't happen. The EMF for the cell indicates that silver should oxidize in these conditions.
(2) The EMF's calculated ignore the Nernst equation.
(3) The oxygen half cell reaction is a bit weird. Obviously the oxygen must be dissolved in the aqueous solution to react. I'm assuming that the notation is trying to point out that the dissolved oxygen is in equilibrium with the atmospheric oxygen which has a constant concentration which is inexhaustible.

Thanks for the response. So it is clear that Br-, not Br₂, will be oxidized? Also, how did you know that Ag would not be oxidized? Is it just something that is known in chemistry? Or does it have something to do with solubility


text:
How to derive the compressibility factor for a van der Waals gas?
So as far as I know,
$$Z=\frac{p_\mathrm{r}V_\mathrm{r}}{nRT},$$
where $\mathrm{r}$ denotes the subscript for real.
Solving it for a van der Waals gas gave me the following result,
$$Z=1+\frac{p_\mathrm{r} b}{RT}-\frac{an}{VRT}+\frac{an^2b}{V^2RT},$$
where $V$ is the volume of the container.
This doesn't quite tally with the form I see in the textbooks.
Where am I going wrong?

What the textbooks say ? BTW, why the new question ? Better would be to elaborate the original one.

Provide the procedure how you have achieved the result.

See Compressibility Factor Of Gases

I have trouble typing it out in that mathy way. As you can guess, I don't know latex and i do not plan to learn it anytime soon

I did not know it a month ago either. Basics are very easy. But the point is not in the formatting itself, but in elaborated effort presented in the question.

Ah, nevermind i found the correct answer  in that answer which you had answered. Thank you :)


text:
Sunglasses that darken in sunlight
I have seen several of my (rich) friends and teachers having glasses that are colorless indoors, but take up a darker shade when exposed to sunlight.

Upon searching for the chemistry of this, I did find out that silver was being reduced in the sunlight, and reverted back to silver ions indoors.
If silver was getting reduced, what's getting oxidized? And how is it that indoor lights (fluoroscent tubes etc.) do not affect the lens?

It's a bit more complex than just silver redoxing back and forth. There is an entire industry for both glass and polymer lenses relying not only on chemical, but also physical principles (Wikipedia: Photochromism; potosensitive glass).

@andselisk Like temperature?  Isn't that just an affecting factor for the reversible reactiom?

Probably, I cannot provide a decent answer: I only  heard about basics and numerous methods of manufacturing chameleon glasses, so I'm going to wait for an answer (review?) together with you:)

As the other comments. Photochromism is surely broader than this. Still it might be that for whatever reason (lifetime of the item, feasibility, ecc.) just, or primary, this silver redox is applied in sunglasses. The page https://en.m.wikipedia.org/wiki/Photochromic_lens mentions two types of compounds as reversible redox counterparts.

As mentioned in the comments, this topic can become exceedingly complex if all the aspects of current-gen photochromic lenses are brought to the table (and I'm sure a lot of it is locked away in trade secrets). My goal here is to provide some insight into the chemical/physical mechanisms at work.
Photochromic sunglasses are a smart new application of very old knowledge. Silver salts have been isolated and manipulated for centuries, so it's surely been known they have a tendency to darken in light for a long time. This simple fact (plus a lot of auxiliary processes) was the basis of the main forms of photography all throughout the 19th and 20th century.
Because of this, the classic examples of photosensitive compounds are the silver halides ($\ce{AgX}$), particularly $\ce{AgCl}$, $\ce{AgBr}$ and $\ce{AgI}$. These compounds are colourless to slightly yellow when pure. Upon exposure to photons of appropriate frequency, the idealised sequence of events is:
$$1) \quad \ce{2 X- \xrightarrow{h\nu} X_2 + 2 e-}$$
$$2)\quad \ce{2 Ag+ + 2e^- -> 2 Ag^0}$$
So essentially, illumination of the silver halides triggers the production of elemental (i.e. metallic) silver. This is a photoredox process; we say the halide anions undergo photooxidation, and the silver cations undergo photoreduction. An interesting way to think of it is that the halides normally are not strong enough reductants to trigger the reduction of $\ce{Ag+}$ to $\ce{Ag^0}$. However, this is only true in the electronic ground state of these species - the halide ion $\ce{X-}$ in an electronic excited state has different chemical properties during its short life, and in particular it becomes a much stronger reductant, and now can reduce silver directly. This leads to another description, which can be more accurate in other contexts
$$1) \quad \ce{2 X- \xrightarrow{h\nu} 2X^{-*}}$$
$$2)\quad \ce{2 Ag+ + 2X^{-*} -> 2 Ag^0 + X_2}$$
Back to the silver. Instead of forming a shiny reflective film, the silver is formed as a multitude of tiny disconnected particles. As the photons bounce around chaotically in the maze of silver clusters, a large fraction of the incident light ends up getting absorbed (and turned into heat). On the macroscopic scale, this makes the material appear dark.
It's interesting to point out here that the element silver alone is not responsible for the darkening effect. The process starts with the absorption of light to generate a strongly reducing species, from which the silver ions then steal an electron to become silver metal. In principle, therefore, it is possible to tune this effect depending on what else is mixed with the silver ions. Indeed, as the halides get heavier, ionisation becomes easier (or alternatively, it becomes easier to reach the strongly reducing photoexcited state), and the silver halides become more photosensitive. However, it's possible to mix in something else entirely, such as electron-rich organic compounds, which could then surrender their electron to the silver ions when exposed to light. It's also not fundamental to use silver at all - all metals are dark when dispersed into very tiny clusters.

So we now put this knowledge together into one of the simplest examples of photochromic lenses. This technology may have been used in the past, but is now likely supplanted.
Like before, we start with a silver halide. For example, during the fabrication of glass, a small amount of $\ce{AgCl}$ can be added. Though it would look like otherwise ordinary glass, when struck with light, it can potentially trigger the formation of $\ce{Ag^0}$ and darken the glass.
The problem now is to make this process reversible. In the ideal reactions above, the formation of $\ce{Ag^0}$ would also require the formation of $\ce{Cl_2}$. These two compounds can in principle react to regenerate $\ce{AgCl}$ (after all, $\ce{AgCl}$ was the thermodynamic minimum, it had to be hit with a photon to turn into something else!). However, $\ce{Cl_2}$ could potentially trigger a number of other reaction, or move around in the glass (it's a gas in standard conditions), which makes the process unlikely to be reproducibly reversible.
To solve this problem, it's possible to add another pair of electrochemically active species. In particular, it turns out $\ce{CuCl}$ is a nice choice. Why? $\ce{AgCl}$ and $\ce{CuCl}$ can be mixed into glass without reacting, once again generating a colourless material. However, upon exposure to light, the following (idealised) reactions happen:
$$1) \quad \ce{ Cl- \xrightarrow{h\nu} Cl^{0} + e^-}$$
$$2) \quad \ce{Ag+ + e^- -> Ag^0 }$$
$$3) \quad \ce{Cu+ + Cl^0 -> Cu^2+ + Cl-}$$
Now the chloride is essentially a spectator species. The process could be rewritten as:
$$\text{Net ionic equation}: \quad \ce{Cu+ + Ag+ \xrightarrow{h\nu} Cu^2+ + Ag^0}$$
Over time, the tendency is for the $\ce{Cu^2+}$ to reoxidise the silver metal back completely to $\ce{Ag+}$, thus making the glass transparent again. This process is comparatively slow, however, and therefore while there is light shining on the glass, there is a buildup of $\ce{Ag^0}$, making it dark. All in all, very convenient, indeed.
Lastly, I'll briefly address why this process happens in sunlight, but not indoors. Unsurprisingly, this is an entirely intentional property. Good eye-wear glasses block invisible and damaging UV photons present in sunlight, but shouldn't block visible photons. Meanwhile, a good visible light source should create visible photons, and avoid wasting energy on harmful UV photons. The design of the materials used in both photochromic glasses and fluorescent lamps has therefore naturally lead to the situation where photochromic lenses should generally not darken with indoor light sources.
For a more hands-on approach to this whole topic, I highly recommend this excellent video.

You have certainly enlightened me and many others. Thank you for such a detailed answer.

One tiny doubt: I quote from your answer "Over time, the tendency is for the Cu2+ to reoxidise the silver metal back completely to Ag+" from which I infer the reaction must be spontaneous. But the potential for reduction of Cu2+ to Cu+ and oxidation of Ag to Ag+ is coming out to be -0.640 V. What am I missing here?

I figured that would come up. The table of standard reduction potentials is very far from applicable in this situation. In fact, copper appears to behave anomalously in water; in aqueous conditions $\ce{Cu+}$ will disproportionate into $\ce{Cu^2+}$ and $\ce{Cu^0}$, but in most other conditions it's the other way around. In short, $\ce{Cu^2+}$ is a stronger oxidiser in non-aqueous conditions.


text:
Why is the Van der Waals constant of water lesser than benzene?
Van Der Waals constant of water and benzene ($a$) are $5.537$ and $18.82$ respectively.
I have read that the value of Van Der Waals constant indicates the extent of attractive forces.
By this, shouldn't it be greater for water instead because water has hydrogen bonding?
Also, what are the factors(with priority) on which the Van Der Waals constant depends on ?

The parameter $a$ from the van der Waals equation
$$\left(p+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$$
is not to be taken only as the measure of strength of intermolecular bonding.
Rather, it is overall measure of significance and frequency of intermolecular interaction.
It may be based on purely elastic electrostatic interactions, where only the weakest van der Waals force may ( or even may not ) be involved.
The probability 2 benzene molecules interact is bigger than for 2 water molecules.
Notes also that for n-hexane is $\pu{a = 24.71 L^2\cdot bar mol^{2-}}$. It has bigger value than benzene, in spite of the weaker interaction due the lack of $\pi$ bonding. But its linear molecule is stratched with all the hydrogen atoms in many directions, so collisions are more probable.


text:
White organic solid that forms CO2 with NaHCO3, forms a diazo dye withe beta-naphthol, and reacts with bromine water
A white crystalline solid "X" gives the following chemical tests:

It liberates $\ce{CO2}$ with $\ce{NaHCO3}$
It forms a coloured dye in diazotisation and coupling with $\beta$-naphthol
With $\ce{Br2}$/$\ce{H2O}$, it forms a white precipitate of 2,4,6-tribomoaniline.

"X" can be identified as:

I thought that (d) could be the answer as COOH releases $\ce{CO2}$ from $\ce{NaHCO3}$, but can sulphonyl group release $\ce{CO2}$ from $\ce{NaHCO3}$?

Aryl sulfonic acids are certainly acidic enough to release CO2 from bicarbonates.

The title of the question is too broad. Please note that the title shall be as descriptive as possible. Generally, the wording of the title shall be established with great care; while being as concise as possible, it shall indicate, without ambiguity, the subject matter of the question in such a way as to distinguish it from that of other questions, without going into unnecessary detail.

(i) indicates presence of 'strong' acidic group: -(SO3H) and -(CO2H) since both are strongly acidic to protonate NaHCO3 and yield CO2.
(ii) indicates presence of Primary Amine which can fully undergo diazotization reaction and then with beta-naphthol form azo dye (coloured dye).
(iii) excess of bromine water forming s-tribromoaniline directly indicates that Option 2(b) is correct.
An article on point (iii): journal_iisc


text:
Influence of pH on precipitate crystallite sizes?
It's normally assumed that a higher pH of solution would slow nucleation rates and hence lead to larger crystal formation. I've found literature that it is due to solid-liquid inter-facial tension. Which is a concept that I do not fully understand.
In my experiments, higher pH solutions led to the formation of smaller crystallites. I can't make sense of why its contradicting the literature. Here are the images from my experiments:

My samples are precipitates produced from mixing aqueous $\ce{CaCl2}$ +$\ce{MgCl2}$ and $\ce{Na2CO3}$ solutions. The elevated pH images were produced through the addition $\ce{NaOH}$ to the previously mentioned mixture.
If you have any comments or ideas, it would be greatly appreciated. I am a engineering student who is trying his hand at inorganic chemistry.
Thanks in advance.

Pure interfaces reason do not take into consideration possible common ions effects, where common means whenever there is an equilibrium that links precipitation and pH.

At low pHs there is little "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ cations. Most of the carbonate species are dissolved $\ce{CO2}$, $\ce{H2CO3}$ and $\ce{HCO3-}$. Thus the precipitate forms slowly and you get relatively large crystals.
At high pHs there is a lot "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ cations. Thus the precipitate forms relatively rapidly and you get relatively small crystals.
I have no idea what literature would lead you to believe otherwise.

Aside of influence on the crystallization rate, you have also consider the direct chemical influence.
The sodium carbonate hydrolyzes in not enough alkaline solutions.
$$\ce{Na2CO3 + H2O <=> NaHCO3 + NaOH}$$
Precipitation of carbonate is interfered by solubility of Bicarbonates.
$$\ce{CaCO3 v + H2O + HCO3- <=>  Ca(HCO3)2 + OH-} $$
Addition of the hydroxide affects both processes. It shifts the equilibrium towards precipitation and also speeds up the precipitation.
The same for magnesium.
For magnesium, there is possibility of a side reaction of precipitation the insoluble hydroxide. But lower carbonate solubility will probably overrule this.
$$\ce{MgCl2 + 2 NaOH ->  Mg(OH)2 v + 2 NaCl}$$


text:
Decomposition of Sugar Temperature
I am perplexed by information on the internet about the sugar raffinose.
The melting point of raffinose is 244.4°F
The boiling point of raffinose is 1,623°F
I can't find the temperature at which it will break apart into it's constituent sugars.
The boiling point seems too high to me tbh, because it should be on fire at that point.
Is it that raffinose decomposes at it's meting point?

The boiling point is nonsense, of course. It will break apart long before that.

Do you know what temperature it will break at Ivan?

No. Or at least not precisely, so to put it. Someplace between 100 °C and 300 °C, I guess.


text:
Synthesis of a weinreb amide from an acid
The following shows the conversion of an acid to a weinreb amide:

It was taken from the IChO 2015 preparatory problem set problem 23. In this two-step synthesis, I am slighly puzzled by the first step. The rationale behind first forming the ester is that we would like to synthesise an acid derivative which can react with the amine to form the amide since direct conversion of an acid to an amide is rather difficult. However, I wonder why is the t-butyl ester formed instead of say, a methyl ester, since the former sterically-hinders the attack of the amine in the subsequent step. In fact, conversion to an acid chloride first (e.g. using thionyl chloride), instead of an ester would seem to be more optimal than conversion to an ester first since acid chlorides would be more reactive with the amine.
In summary, I would like to know why ester formation is preferred over acid chloride formation in the first step and also why a rather bulky ester is preferred over a much less bulky one. Regarding the former, I believe it could be that the formation of $\ce {HCl}$ by-product may hinder the 2nd reaction by protonating the amines. Regarding the latter, I believe there may be some electronic factors that need to be considered?

Your question is easily answered by the fact that the conditions of Me3COCl + starting acid give a mixed anhydride not an ester. The tButyl group is chosen as the hindrance it brings makes reaction at the desired C=O carbon more likely. Further reading about mixed anhydride coupling is here

Oh oops... Yes it is an anhydride... Very careless of me.

Is my reasoning regarding why an acid chloride is not generated instead in the first step correct? Reaction of the acid chloride with the amine in the 2nd step would generate HCl which would protonate the amine

I think the question has omitted the tertiary base used in both steps.


text:
Enthalpy change of gas expansion against constant external pressure

A certain mass of gas is expanded from $\pu{1 L}$ at $\pu{10 atm}$ to $\pu{4L}$ at $\pu{5 atm}$ against a constant external pressure of $\pu{1 atm}$. If initial temperature of gas is $\pu{300 K}$ and heat capacity at constant volume of preocess is $\pu{50 J °C-1}$, find enthalpy change of process.

Here are my thoughts:
$$∆H = nC_p\,\Delta T = nC_V\,\Delta T + p\,\Delta V$$
where $p$ is external pressure.
$$\Delta U = c\,\Delta T$$
with $T_1 = \pu{300 K}$ and $T_2 = \pu{600 K}$ (which I got from gas equation), so
$$\Delta U = \pu{50 J K-1} \times \pu{300 K} = \pu{1.5 kJ}$$
Still I am not getting the correct answer. I believe my methodology is right.

Look at the units of the heat capacity - it is not per mole. So your equation needs adjustment, I think.

No I did it with 50 itself taking it has heat capacity

So why do you have the "n" in your equation?

I did like this  for ∆u=c∆T T1=300K and T2 =600K(which I got from gas equation) ∆u=50 x 300=1.5kJ

So now you know ∆U for going from initial state to 600 K at the same volume. You still have to go to the larger volume and get ∆(PV) to find ∆H. The outside pressure should have no effect on your result, as both ∆H and ∆U are state functions.

??? Why, we say work is done by ext force right and how is ∆pv even work. p∆v is work

There is no need to consider work, see my answer.

In my judgment, the units on the heat capacity are incorrect.  It should be 50 J/mole-C.

@KarstenTheis you should always consider work in the initial energy balance, and then cross it off if it isn't necessary. Skipping right to the $\Delta$U or $\Delta$ H gets a person in trouble. Often the sign is backwards. There is certainly boundary work here

@CharlieCrown I didn't understand what you meant.

@CharlieCrown I should have said: While the gas does work as it is expanding against 1 atm external pressure, it turns out that the enthalpy change is independent of the external pressure (it's a state function, so it only depends on the state variables).

@AbinAntony I don't have enough time to give a real answer, but here is a somewhat related answer of mine that goes over the energy balance from start to finish with all the logic involved energy balance

Assume an ideal gas
I think we have to assume an ideal gas, otherwise it is not clear how to get the final temperature of the gas. In the OP's comments, the final temperature is determined as 600 K.
The difference between change in energy and change in enthalpy
Enthalpy is defined as:
$$H = U + PV$$
If both P and V change from state one to state two:
$$H_1 = U_1 + P_1 V_1$$
$$H_2 = U_2 + P_2 V_2$$
we can relate the change in enthalpy to the change in energy by subtracting the two equations:
$$\Delta H = \Delta U + P_2 V_2 - P_1 V_1$$
We could write $\Delta(PV)$ for $P_2 V_2 - P_1 V_1$, but maybe that is confusing.
Change in energy of the gas
In the question, the temperature, the volume and the pressure of the gas all change as we are going from initial to final state. Because energy and enthalpy are state functions, we can decide in which order these changes happen. It makes sense to use steps from initial to final that are easy to calculate. I will change temperature first, then volume (the pressure will change as well, and we will see how that plays into the calculation).
Initial state: $V = \pu{1 L}, T = \pu{300 K}$
Intermediate state: $V = \pu{1 L}, T = \pu{600 K}$
Final state: $V = \pu{4 L}, T = \pu{600 K}$
The change in energy from the initial state to the intermediate state is given by the OP:

$$\Delta U = C_V \Delta T = \pu{1.5 kJ}$$

Taking that intermediate state (V = 1L) to the final state (V = 5L) does not involve any change in internal energy if the gas is ideal, so $\Delta U = 0$ for this step. We could expand into a vacuum, or against pressure. In the latter case, the temperature would drop and we would have to transfer heat into the system to the same amount as the work it did on the surrounding.
So for the entire process, $\Delta U = \pu{1.5 kJ}$.
Change in enthalpy of the gas
$$\Delta H = \Delta U + P_2 V_2 - P_1 V_1 = $$
$$ = \pu{1.5 kJ + 5 atm 4 L - 1 atm 10 L} = $$
$$ = \pu{1.5 kJ + 10 atm L = 1.5 kJ - 0.101 kJ = 1.4 kJ}$$
What about the external pressure?
It does not matter for the calculation because the internal energy of a gas (assuming ideality) does not depend on it. In other problems, where the volume of the gas is not constrained, it might be important, but here it is not.
You could solve this problem going along another path, and keeping track of volume work and heat transferred to get at the changes in energy (using the first law of thermodynamics). This would be more complicated, though.

Since you have gotten ΔU, you can proceed most simply by converting ΔU into ΔH by method mentioned by Karsten. I believe you mistook the question asking for ΔU but not for ΔH.

Since you have assumed gas to be perfect, you can also convert ΔU into ΔH using

ΔH = ΔU + nRΔT

but you then have to find n which is not a big deal.

I will prefer just find ΔT, then take

ΔH = CpΔT, where

Cp = CV + nR, if both heat capacities are not molar, which is true based on the units you have given (i.e. they are total heat capacities of system).


text:
Nucleophilic nature of ions

Consider the following species:
$$
\underset{(\text{A})}{\ce{\overset{⊖}{O}H}}
\quad
\underset{(\text{B})}{\ce{CH3-\overset{⊖}{O}}}
\quad
\underset{(\text{C})}{\ce{\overset{⊖}{C}H3}}
\quad
\underset{(\text{D})}{\ce{\overset{⊖}{N}H2}}
\quad
$$
Arrange these species in their decreasing order of nucleophilicity.
(a) C > D > A > B
  (b) B > A > C > D
  (c) A > B > C > D
  (d) C > A > B > D

My teacher taught in polar protic solvent nucleophilicity is inversely proportional to basic character. I assumed solvent to be $\ce{H2O}$ as solvent was not mentioned in the question.
As acidity order is
$$\ce{H2O} > \ce{CH3OH} > \ce{NH3} > \ce{CH4},$$
so basicity order should be
$$\ce{CH3-} > \ce{NH2-} > \ce{CH3O-} > \ce{OH-},$$
therefore using above proportionality I arrived at the nucleophilicity order
$$\ce{OH-} > \ce{CH3O-} > \ce{NH2-} > \ce{CH3-},$$
but you can see the option is not even given. Why is it so?

−1 The question doesn't make sense to me. $\ce{NH2-}$ is hardly capable of existing in water, let alone $\ce{CH3-}$. One might as well assume that the question is about gas-phase nucleophilicities.

The problem is your assumption of H2O as solvent. The pKa of H2O is often reported as 14, vs 15.5 for methanol, but that is an artifact of H2O being the solvent for standard pKa tables. In other solvents (eg propanol), H2O is slightly less acidic than methanol, leading to (b) being the correct answer. So really this is a problem with original question not specifying the conditions.

What you probably missed was that the order you mention is for the SAME GROUP elements. For example, down the group basicity of halide ions is-
F->Cl->Br->I-
But the the nucleophilicity order in polar protic solvents is-
I->Br->Cl->F-
This reverse in order is due to higher charge density of smaller halide ions and thus better interaction with polar protic solvents.
But along a period the charge density is similar and thus the interaction between solvent and nucleophile is almost similar. Thus, along a period the order follows basicity order.

It is dangerous to relate nucleophilicity and basicity like that

To say that nucleophilicity follows basicity across a row means that,
  as basicity increases from right to left on the periodic table,
  nucleophilicity also increases. As basicity decreases from left to
  right on the periodic table, nucleophilicity also decreases. When it
  comes to nucleophilicity, do not assign this same rule when making
  comparisons between the halogens located in a column. In this case of
  moving up and down a column, nucleophilicity does not always follow
  basicity. It depends on the type of solvent you are using.
  (Nucleophile)

In general here are the rules for nucleophilicity along a group and across a period in a protic solvent

Along a group nucleophilicity and basicity for the same order, reason for this is that charge density along a group is similar for the anions as there is not much size difference. So the better base is the better nucleophile. As the best nucleophiles are the good bases that want to share their electrons with an an electrophilic center.
However when it comes to down a group nuceophilicty and basicity follow the exact opposite trend, reason for this is that charge is more dispersed in the larger anions and the extent of solvation is much lesser. Put in simpler words, the $\ce{H2O}$ molecules are in a surrounding the smaller ions and preventing them from acting as nucleophiles efficiently.

This will help you relate to what I mean
References

Nuclephile (Chemistry LibreTexts)
Wikipedia


text:
Calculating Air Density and Effect of Air Composition
I have to calculate density of atmosphere, assuming that it is:
79% : N2
21% : O2
Does the percentage of gases affect the answer, and how can we find the density with such little information?

The tricky part for newbies is, that the StackExchange site network expects questions to be more than just questions. There is expected explicitly provided elaboration effort of analysis and thoughts, explanation attempts and basic resource research.

The density of air depends on which components you consider. Water vapor has some significant effects; components with similar (and low) densities often don't. Also, the density of air cannot be calculated using ONLY the percentages of various gases (assuming you're going without the ideal gas equation, volumes of gases, and other data).

He can imply conditions according the standard atmosphere. 101325 Pa, 15 Deg C, -0.65 Deg C / 100m till the tropopause.

@farhhhhhhhhmod If you're still confused, see if these help: https://www.omnicalculator.com/physics/air-density#how-to-calculate-the-air-density; https://en.m.wikipedia.org/wiki/Density_of_air#Composition.


text:
Electrolysis - What is the basis for the discharge rules?
At the cathode, the one that is discharged is the less reactive one (compared to hydrogen) so why at the anode is the one that’s discharged more reactive (less stable) ?

What is your criterion for less reactive - the reduction potential (or qualitatively - electrochemical series)? There is an oxidation process at the anode, so the half reactions and the order of activity are switched compared to the reduction half reactions.

The reaction potential.

The reactivity is mutually reversed on an anode and a cathode. On a cathode, preferred reduction is for the redox system with the highest potential. On an anode, preferred oxidation is for the redox system with the lowest  potential.

The redox potential is a measure of tendency of  redox system to exist rather in oxidized form ( negative potential values) or reduced form ( positive values ).
Therefore the higher the standard redox potential is, the stronger oxidation effect and weaker reduction effect have the respective oxidized/ reduced forms of the redox system.
And vice versa.
Particularly, on the opposite sites of the standard redox potential table are
$$\begin{align}\ce{
Li &<=>> Li+ + e- \\
F2 + 2e- &<=>> 2 F-  \\}
\end{align}$$
The former is very strong reducting agent, while latter is very strong oxidating agent.
I will illustrate it on anode case, with the cathode case it is the same, but with the reversed potential order.
When the anode potential is externally forced to raise up, if first start to oxidate the reducted form of the redox system with lowest redox potential.
If the current is already saturated by diffusion rate of redox reagents, futher increase of potential does not increase the current until it reaches potential region of next easiest to oxidize redox system.
If the redox systems have similar values, the next system start to react on the electrode before the system
The result is, the curve current versus anode potential has a typical look of the  rounded stepwise  voltammetry curve.
See also Voltammetry,  Polarography and example of the cathodic polarographic curve.


text:
Is [Co(NH₃)₄Cl₂]Cl paramagnetic or diamagnetic?
$\ce{NH3}$ is known to be a strong field ligand, while $\ce{Cl}$ is known to be a weak field ligand.
Is $\ce{[Co(NH3)4Cl2]Cl}$ a high spin complex or a low spin complex?
I assumed this to be a high spin complex. Oxidation state of $\ce{Co}$ is $+3$.
And electronic configuration of $\ce{Co(III)}$ is $\mathrm{[Ar]~3d^6}$.
Since I assumed this to be a high spin complex — pairing of electrons of $\mathrm{3d}$ orbitals will not happen. The $6$ pairs of electrons from the ligands must be included in the $\mathrm{4s}$, $\mathrm{4p}$, $\mathrm{4d}$ orbitals — which leaves electrons unpaired in the $\mathrm{3d}$ orbital, making this a paramagnetic complex, with hybridisation $\mathrm{sp^3d^2}$.
Answer given to this question says otherwise. Hints are welcome!
This question is an excerpt of a question that appeared in the JEE(A) - 2016

Cl acts as strong field ligand with Co

@drake01 Can you link a source? Explanation?

I don't have any source rn, our teacher told this.

@drake01 Well then so is the case with this compound: https://chemistry.stackexchange.com/questions/72685/why-chlorine-acts-as-a-strong-field-ligand-in-tetrachloridoplatinateii

rule of thumb: complexes of Co+3 are low-spin, unless we are talking about hexafluorocobaltat(III) anion. Even the aquacation is low-spin.

The point is not really whether chloride or ammonia is a strong or weak field ligand, the point is $\ce{Co^3+}$ is $\mathrm{d^6}$, and virtually all "octahedral" $\mathrm{d^6}$ complexes are low spin - essentially some complexes of $\ce{Fe^2+}$  and a very small number of fluoro complexes of $\ce{Co^3+}$ are the only exceptions to the rule that all $\mathrm{d^6}$ octahedral complexes are low spin.
Why is this? Well let's use a bit of Crystal Field Theory (much of below stolen from https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Stabilization_Energy). One can consider the crystal field stabilization energy which is how stabilized an ion is compared to the free ion when in a given electronic state electronic and geometry. Here we are interested in high and low spin, and octahedral geometry. Taking a table of these values from the above link:

In the above $\Delta_\circ$ is the splitting energy of the orbitals. We can see that low spin $\mathrm{d^6}$ has the largest possible stabilization energy of any electronic configuration compared to the high spin case ($2\Delta_\circ$-P). Thus unless the splitting is very small octahedral $\mathrm{d^6}$ prefers low spin. So how do we get small splitting to see high spin $\mathrm{d^6}$? Well small splitting is favoured by

Low charge first row transition metals - for instance $\ce{Fe^{2+}}$
Very weak field ligands

And what were our exceptions above?


text:
Preparation of Copper (II) Sulfate Pentahydrate Solution
how would you make a solution of Copper (II) Sulfate Pentahydrate to achieve 150, 200, 250 and 300 mgL−1 concentration of copper? I do not have the anhydrous version of the solute. Also, when I dissolve it in water, does it become acidic? If so, how would I maintain it at pH 7 without messing up the concentration? I am going to use the solution for adsorption studies if anyone is asking. thanks for the replies!

(1) Mass of anhydrous $\to$ moles of anhydrous $\to$ moles of pentahydrate $\to$ mass of pentahydrate. (2) Yes it will become acidic, and you don't want to maintain it at pH=7.

I dont get what you mean by on number one, can you please elaborate? And about the pH level, is it highly acidic or just beteween 4-6 pH?

(1) You want to make a certain amount of solution with certain concentration; can you find out how many grams of anhydrous CuSO4 (if it were available, that is) would you need for that? (2) Slightly acidic.


text:
When amino acids react with each other, how do you know which one reacts using it's carboxylic group and which one with it's amine group?
For example the dipeptide val-ser, between the amino acids valine and serine, does valine's amino group react or does it's carboxylic group with serine's respective group?

Proteins are named from N-terminus to C-terminus, so if the same rules apply for shorter peptide chains, then it is the carboxylic acid of valine that binds to the amino group of serine.

Under physiological conditions (aqueous solution, pH near neutral), amino acids don't react with each other directly, and the equilibrium lies on the side of peptide bond hydrolysis, not formation.

When amino acids react with each other, how do you know which one reacts using it's carboxylic group and which one with it's amine group?

In the cell, amino acids first react with ATP, then form an ester with RNA, and then their amino group forms a peptide bond with the growing amino acid chain. The last step is catalyzed by ribosomes that select the next amino acid to be added using messenger RNA as a template.

For example the dipeptide val-ser, between the amino acids valine and serine, does valine's amino group react or does it's carboxylic group with serine's respective group?

In your example, valine and serine could form either valyl-serine or seryl-valine. As Tyberius writes in the comments, peptides are written so that in each peptide link, the amino acid on the left of the link provides the carboxylic acid group and the one of the right of the link provides the amino group. The very first amino acid (the one at the amino terminus) carries an amino groups that has not undergone peptide bond formation, and the last amino acid (the one at the carboxy terminus) carries a carboxylate group that has not (yet) undergone peptide bond formation.

Amino acids are organic compounds that have at least one amine group and one carboxylic group. They also contain a side chain specific to each amino acid. It is called an $\alpha$-amino acid if both amino and carboxylic groups of the amino acid are attached to the same carbon atom of its backbone (See the picture below for 21 proteinogenic $\alpha$-amino acids found in eukaryotes). If the amino and carboxylic groups are attached to separate adjacent carbon atoms of its backbone, then it is called a $\beta$-amino acid, and so on.

Naturally, basic amino group and acid group can react to form amide bond. The formation of peptides is nothing more than the application of the amide synthesis reaction. By convention, the amide bond in the peptides should be made in the order that the amino acids are written. The amine end (N terminal) of an amino acid is always on the left, while the acid end (C terminal) is on the right. For example, consider the dipeptide, $\alpha$-Asp-Leu: This is a dipeptide that is the N-(L-$\alpha$-aspartyl) derivative of L-leucine (See LHS of following picture):

Although L-aspartic acid has carboxylic group in its side chain, the free $\alpha$-amino group justify it's been a N-terminal. Based on the same reason, aspartic acid can be considered either $\alpha$- or $\beta$-amino acid. Therefore, if the peptide formation is involved with side chain carboxylic group instead (between same two aamino acids), then resultant dipeptide is called $\beta$-Asp-Leu (See RHS of the picture).


text:
Still taught to reverse oxidation half cells in electrochemistry?
In a question, Oxidation of metals/halogens by oxygen gas in acidic aqueous solution, there was a point made that reduction half-cells should not be reversed.
I was taught ($\approx 1970$) in the following manner. Given the reduction half cells:

$$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\
\ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V}
\end{align}
$$

To get the the standard oxidation potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:

$$
\begin{align}
\ce{Ag(s) &→ Ag+ + e-  } &\quad E^\circ &= \pu{-0.799 V} \\
\ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V}
\end{align}
$$

Now you can write a balanced chemical reaction as such:

$$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{4Ag(s) &→ 4Ag+ + 4e-  } &\quad E^\circ &= \pu{-0.799 V} \\
\ce{O2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l)} &\quad E_\text{Total}^\circ &= \pu{+1.23 - 0.799 = +0.43 V}
\end{align}
$$

and to get the cell potential the two half cells are added, one being positive and the other negative. If the EMF is positive the forward reaction is spontaneous, if negative the reverse reaction is spontaneous.
Is flipping the reduction reaction not the standard method taught these days to balance redox reactions?

It looks like silver-silver cell would be even better than silver zinc cell, with the added value of perpetuum mobile. :-)

For the standard silver cell vs a standard silver cell the EMF would be 0 and thus no current flow. Because of the Nernst equation the potential of a non-standard silver cell would vary if the $\ce{A_{Ag+} \ne 1.000 }$.

I know it well. But it does not make sense to write the silver  electrode potential like if silver had been better reduction agent than zinc.

@Poutnik - Huh? I wrote the Ag reaction with an EMF as -0.799V as an oxidation not a reduction so that I could balance the equation.

The point is, electrode potential is for 2-way redox reaction equilibrium, holding this potential with respect to the standard hydrogen electrode. For $\ce{Ag/Ag+}$, it is positive, not negative. With the open voltage, the oxidation and reduction is ongoing with the same rate.

I have done a significant amount of research over the past ten years to trace to origins of these electrochemical conventions and luckily got a chance to discuss these with some top electrochemists. I have been planning to write an article on this issue since it is a perpetual confusion. Basically, the origin of these "signs" issues originated in Germany and USA. The Ostwald school of thought, wrote the electrode potentials in accordance with the electrostatic sign of the electrode with reference to the standard hydrogen electrode. This was the so-called European convention. Keep in mind that electrostatic signs of the electrodes are invariant. It does not matter how you write them. A silver electrode in Ag+/Ag half-cell is always positively charged with respect to the hydrogen electrode under standard conditions.
The American school of thought, with its own influence such as Gibbs, Lewis and Randall, Ligane, Latimer, had a thermodynamic view of the signs relating electrode potentials to the Gibbs free energy. Latimer wrote a very famous book "Oxidation Potentials" which is available from Internet archives. Yes, under those conditions we can flip the signs back and forth while keeping Gibbs free energy in mind. The tug of war between European convention vs. American convention went on until the 50s-70s. Textbooks are usually 20 years behind current research...these issues lingered on.
Finally, electrochemists decided, including Allan J. Bard (from US who wrote a very influential electrochemistry book taught all over the world), that let us go back to the original notion- keep the electrostatic signs associated with the electrode potentials which are invariant with respect to the way we write them. This has "almost" terminated the sign convention issue of US vs. European signs of electrode potentials. The plus and minuses took almost a century to resolve. It is amazing to see the insight of those early scientists such as Ostwald, that world finally accepted their views.

So how do you balance the chemical equation if you don't reverse the silver half cell? The whole point in flipping the silver reaction was to get electrons on both sides of the chemical reaction so that the electrons are the same on both sides.

Certainly we need to write the silver half equation as Ag -> Ag+ e, but I would not write electrode potentials next to them and flip signs. This confuses students because in Ecell= Ecathode-Eanode, they may wonder whether they should change Eanode sign or not. The negative sign already accounts for the oxidation.

I still teach switching the sign. I find it easier to remember adding up reduction potential and oxidation potential. The half reactions are written as reduction in a table of reduction potentials, so it makes sense that you have to treat the oxidation half reaction differently.
If the cell potential is calculated from reduction potential of the cathode half reaction minus that of the anode half reaction, there are more points of possible errors, including trouble with arithmetic.
You might argue that students learn more and gain deeper insight when given more opportunity to make mistakes, so maybe my approach is inferior. For example, students can use the approach I am using without knowing which electrode is the anode and which is the cathode.

Just curious - Does the text book you use also employ that technique for balancing redox reactions?

I use multiple textbooks, and my students might use other sources entirely. Of the textbooks I use, D. M. Hanson's Foundations of Chemistry uses $E_\text{red} + E_\text{ox}$ while OpenStax Chemistry uses $E_\text{cathode} - E_\text{anode}$.


text:
Is Geologic Mineral Trapping of Chlorine Plausible?
One strategy for sequestration of $\ce{CO2}$ is mineral trapping arising from geologic sequestration.  Another strategy is mineral trapping in concrete via the Mineralization via Aqueous Precipitation or MAP process.  In particular, Alkalinity Based on Low Energy or ABLE process offers an energy efficient MAP process.  However the ABLE process produces $\ce{Cl2}$ (or $\ce{HCl}$), which, at scale, produces its own environmental impact.
One way of dealing with $\ce{Cl2}$ (or $\ce{HCl}$) might be mineral trapping via geologic sequestration.
Is this plausible?

There is a substantial market for chlorine. Why trap it?

One of the problems of geological sequestration of chlorine is that most chloride minerals are very soluble.
The number of non-soluble (or very little) minerals is limited, but there are several options. The obvious ones are sodalite ($\ce{Na8(Al6Si6O24)Cl2}$) and chlorapatite ($\ce{Ca5(PO4)3Cl}$). There are other options though.
Webmineral lists a lot of chlorine-bearing minerals, and most of those are soluble. However, there could be several minerals with high percentages of chlorine that are insoluble enough under the conditions of your reservoir. But in general, your best bet are anhydrous silicates and phosphates, because those are the most likely to be stable over a long time.
That said, this stuff could be dumped in the ocean. The $\ce{HCl}$ could be neutralised with something like feldspars, turning it into clay and a solution of harmless sodium chloride.
Alternatively, you can oxidise the HCl and make chlorine gas, a useful product.

The $\ce{Cl2}$ volumes projected for $\ce{CO2}$ sequestration via ABLE would totally saturate any market.  Dumping in the ocean is unacceptable at these levels as that merely exchanges $\ce{CO3^{2-}}$ acidification for $\ce{Cl-}$ acidification.
A possible oceanic geologic reservoir could be the pillow basalts that underlie pelagic sediments.  As I understand it pillow basalts have low compressive strength due to their rapid cooling.

@JamesBowery that’s why I said neutralise it first. Chloride anions are harmless.