text:
How to conveniently separate Cd impurities from ZnO without contaminating it?
How could I separate Cd impurities from ZnO without contaminating it and without using overly expensive equipment or extremely dangerous solvents?
I don't want to contaminate it by introducing other heavy metals or being left with more than trace levels of undesired solvents.
I'd be fine losing some amount (say max 10-20%) of my ZnO in the process, if necessary.
Please ask me for any further clarity I could bring to this question.
I know nothing about this so I'd be happy with any contributions on how to easily separate any reasonable portion of $\ce{Cd}$ salts from $\ce{ZnO}$ in general (precipitating it out or solving it away from $\ce{ZnO}$ for instance). Cycling the zinc through a series of reaction and then back to ZnO would be fine also (if not leaving trace levels of anything biologically harmful).

Have you thought of using a calutron?

Purity of $\lt \pu{26 mg}/\pu{100 mL}$ of $\ce{Cd}$ salt in solution is okay?

@A.K. Hello A.K! I had actually never heard of calutron. I'll read about it! (though I'm not sure I can get access to the necessary equipment)

@MathewMahindaratne 26 mg/100 mL may be a little bit too high. My target end result should at the very minimum be under 5ppm of dry product,  preferably <0.5 ppm. Though this doesn't need to be a one step process, and I welcome any measure that can allow me to conveniently separate some amount of Cd salts from my ZnO samples (I can then look at all the separation possibilities and make up the right process based on the equipment I need and my requirements)

Think about treating it with caustic to form a zincate ... crystallise to remove Cd http://www.dwc.knaw.nl/DL/publications/PU00011954.pdf add the pure zincate to water to dissolve and adjust pH of slurry / solution to return zinc oxide.  That's how I'd start, there are most probably cleverer ways of doing it.  And if you keep your waste liquors it may be worth attempting a different purification of them.  Hope this helps.

@EdV Adding high purity Zn dust per se is no pb as long as it's cost effective. 99.999% Zn costs about 200x as much as 99,9% ZnO so the required ratio and effective impact would need to be small enough to reflect that (it's not a problem if I need to add 0.1% pure Zn, but it may become if it's much more than that). I also have to be careful not to introduce other impurities as even 99.999% Zn contained up to 10ppm of I don't know what. Could you describe the mechanism you have in mind? ZnO is not much soluble itself so I ignore how I could use filtration to remove precipitated Cd for instance?

Could just calcinating the sample and then rinsing it with very pure water be a solution? I don't understand combined solubilities well but there's a ~3:1 ratio between the individual solubility values of CdO and that of ZnO, so if I could ensure all my Zn & Cd would exist as ZnO/CdO exclusively (the point of the calcination) then I could maybe reduce Cd impurities by up to 48ppm CdO per liter of pure water (at a minimal loss of ZnO). Of course the calcination step ads  undesired cost/complexity but perhaps I could borrow a blowtorch and just invest (manufacture?) in a high purity crucible.

Then would come the question of non-contamination in that step but it would still be interesting to know if it would work in theory. Would there otherwise be a chemical method to ensure all species in the sample exist as oxides exclusively?

@EdV It's good throwing in ideas even if they don't end up being the final solution, it adds to the reflection.

Since you actually considered the use of a calutron, I assume that you are interested in separation of relatively small amounts of pure ZnO - on the order of a few grams.
Cadmium and zinc are similar in many respects, which makes me think of rare earths. One of the separation techniques used there is ion exchange chromatography. Resins with special affinity for many ions (either to catch it or reject it) are available. Some contaminants are not easy to remove by conventional ion exchange resins. In many cases, very specific resins have been developed for these contaminants. Selective resins from Rhom and Haas, www.lenntech.com, are available today for the removal of:
• Boron
• Cadmium, mercury and other heavy metals
• Chromate
• Lead
• Nickel
• Nitrate
• Perchlorate
and some other contaminants.
On the more researchy side, the stability constants of the tetra ammonium complexes of zinc and cadmium differ by a large factor, unlike their complexes with EDTA and similar chelators. For Zn(NH3)4++, the dissociation constant is 9.8 x 10^-10; for Cd(NH3)4++, it is 2.5 x 10^-7. (Reference Book of Inorganic Chemistry, Latimer and Hildebrand, 1951, 3rd ed., pp 135-6). The impure ZnO could be dissolved in NH4Cl (which will also dissolve CdO) and the solution run thru a column. Here's the research: a column of what? Perhaps a column of the impure ZnO. Since the zinc complex is more stable (more soluble), it should concentrate in the first eluate, leaving CdO behind. Solid ZnO would rip NH3 off the cadmium complex and become soluble, so the ZnO will turn into CdO. I have no idea how useful this scheme could be, but the big difference in stability constants is an eye-catcher.

Thanks a lot for this answer. There's some very nice thinking in there! There's one point I don't understand though: Does ZnO + NH4Cl yield Zn(NH3)4++ and not ZnCl2?

NH4Cl is equivalent to NH3 + HCl. ZnO dissolves with 2 NH4Cl and goes to ZnCl2 + H2O + 2 NH3. Then the 2 NH3 complex with Zn++, giving Zn(H2O)2.(NH3)2++, or maybe 50% Zn(NH3)4++, or a mix, depending on how strong the ammonia bonding is. Probably fairly strong if you can dissolve insoluble Zn compounds by reducing the concentration of Zn++/Zn(H2O)4++ in solution.

I will set up to test your idea! It has a bit of elegance :)

I forgot to say that you can add more NH3 /NH4OH to get all the zinc complexed. But since zinc complexes NH3 more strongly than cadmium does, an insufficiency of NH3 would tend to leave the cadmium uncomplexed as Cd++ (or, probably hydrated). This might magnify the difference and make separation easier or more complete.


text:
Why is the order of reflection simplified to n = 1?
My professor told us that for the sake of simplicity, we will assume that the order of reflection for Braggs' Law is always equal to 1 but failed to explain why. Is it because d(h,k,l) for materials has been determined using an order of reflection 1?

The common form of the Bragg equation is $\lambda= 2 d_{hkl}\sin(\theta)$ but in crystallography it is used as $\displaystyle \lambda =\frac{2d_{hkl}}{n}\sin(\theta)=2d_{nh,nk,nl}\sin(\theta)$ where the indices are divided through by $n$ to give $\lambda= 2 d_{hkl}\sin(\theta)$. The indices $hkl$ have to be integers (from derivation via Laue equations) but can also be multiples of integers, i.e. they can have a common factor. When the $hkl$ do have a common factor they are not strictly lattice planes but planes parallel to $(h/n, k/n,l/n)$ with a spacing $1/n$ of the lattice. Suppose a plane is (210) then each lattice plane passes through a lattice point (atom). If the indices were (420) then not all planes are equivalent as only half of the points (atoms) pass through a lattice plane so they are not equivalent. Thus sticking with $n=1$ ensures we always use for example, (210), and not (420) or (630) etc., and so always calculate the correct value for $d_{hkl}$

source: Stack Exchange

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

sort: votes


text:
Can someone explain 3‐(propan‐2‐yl)cyclohex‐1‐ene forming a carbocation?
Can someone explain this one, especially the formation of the secondary carbocation (in the solution)?

Solution:

After the electrophilic attack by $\ce{H+}$ on the π bond, I had expected the carbocation formation on the other side (the left end of the π bond) because of more hyperconjugation thus ruling out the chance of its rearrangement to the tertiary site. I've seen this happen in other questions too.

Carbocation: something with a positive charge on carbon. Carbonation: the thing that makes coke fizzy.

A more complete solution here relies on knowledge that the protonation is not productive, in the general sense that there is no capturing nucleophile. This means that the protonation is reversible. This should be clear based on the last step of losing a proton. That means it doesn't matter which end of the double bond is attacked--both are. But only the one that leads to the rearrangement matters here.

You are correct that the $\alpha$-carbon would be protonated... and then deprotonated, and back and forth, until eventually, the $\beta$-carbon is protonated, the hydride shift proceeds, and the molecule is trapped in the product state. Since protonating the $\alpha$-carbon doesn't result in some other reaction, it doesn't really matter which secondary cation is more stable. I think the important thing here is recognizing where and when a hydride shift can occur.

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


text:
Other than CO2 release, what are the disadvantages of precipitating metal carbonates using NaHCO3 instead of Na2CO3?
Be, Mg, Ba, Zn, Ni, Ca, Cu, Fe, Mn, Cd, Pb, Hg, Al carbonate salts are all insoluble.
I (perhaps naïvely) believe that I could precipitate carbonates from those mixing a soluble salts of these (e.g. CaCl2) with Na2CO3.
I (perhaps naïvely) believe that it would also work using NaHCO3, yielding CO2 and H2O as additional byproduct(s).
So I'm wondering, what are the advantages of using Na2CO3 over NaHCO3 for the purposes of precipitating carbonate salts of these metals? or, conversely, what are the disadvantages of precipitating metal carbonates using NaHCO3 instead of Na2CO3?

Note that bicarbonates of Ca and Mg, (possibly of some others as well), are soluble, they are bases of water carbonate(temporary) hardness.

@Poutnik: Thanks Poutnik, I didn't think of bicarbonates at all. So the CO2 released from NaHCO3 would induce the formation of a portion of soluble metal bicarbonates alongside their carbonate salts, which would decrease the yield for the latter? Do you have an idea of the order of the proportion between the bicarbonates and simple carbonates at room temperature (i.e. is it a significant loss?) Would performing the reaction in a near vacuum help circumvent this issue (maybe by preventing the presence of dissolved CO2, in addition to eliminating reactions at the solution's surface)?

There can be full loss. Why not to use carbonates, or Bicarbonates with hydroxide ? General rule is to maximize concentration of ions, that you expect to be in precipitate.

@Poutnik I can get bicarbonate with ~ twice lower levels of impurities than carbonate. So I'd like to go with the bicarbonate. I could either produce carbonate from it in the oven (though it would cost a lot of electricity), or use it directly in place of carbonate. Another aspect is that I think bicarbonate is a better way to store a carbonate source (as it needs to go through two steps to decompose into Na2O vs only one for Na2CO3). Having the metal carbonation reaction occur in a vacuum would cost very little energy compared to heating up the bicarbonate, if that would be of any help.

I had in mind addition of hydroxide to bicarbonate, or vice versa, to produce carbonate, before mixing with metal salts.

@Poutnik I'm not sure I understand your last comment. How can I produce hydroxide? Should I slowly hydrate CaO > then add Na2CO3 > filter out the precipitating CaCO3 > mix the NaOH solution with NaHCO3? and then add the metal salt?

I was not speaking about producing hydroxide, but carbonate.

@Poutnik But how do you add hydroxide to the solution to produce carbonate?

Hmmm, are you sure chemistry is the right thing for you ?

But because of limited solubility if sodium bicarbonate , it is better to add bicarbonate to solution of computed equivalent mass of sodium hydroxide..


text:
Reaction between 1-methylcyclopentanol and potassium permanganate
In my basic organic chemistry class, I learned that only primary and secondary alcohols undergo oxidation (using $\ce{KMnO4}$).
However, what is the outcome of reacting tertiary alcohol (1-methylcyclopentanol) with a strong oxidant like $\ce{KMnO4}$ with no heat?
P.S. I am specifically interested to know what happens with no heat, because according to my minimal knowledge I know if heat was added then it is dehydration.

There is no reaction.
t-BuOH may be used as a solvent for permanganate reactions. For example, see Abiko et al. [1].
References

Abiko, A.; Roberts, J. C.; Takemasa, T.; Masamune, S. $\ce{KMnO4}$ Revisited: Oxidation of Aldehydes to Carboxylic Acids in the Tert-Butyl Alcohol - Aqueous $\ce{NaH2PO4}$ System. Tetrahedron Letters 1986, 27 (38), 4537–4540. https://doi.org/10/c3t9xj.

When do organic compounds get oxidized to CO2 then?

When you use forcing conditions: high temp. conc reagents. The OP asked about reactions without heating.


text:
Is chlorophenylmethane the correct IUPAC name rather than chloromethylbenzene?
I thought that the correct IUPAC name for the molecule below (benzyl chloride) would be chloromethylbenzene because the ring always gets priority over the chain unless a primary functional group is present. However, it seems the correct name is chlorophenylmethane. Can someone explain why that is?

The preferred IUPAC name is, indeed, (chloromethyl)benzene, built from substitutive prefix chloromethyl. The parentheses around "chloromethyl" are required because it is a compound substituent (a simple substituent attached to another simple substituent, specifically, methyl and chloro in this case). The Blue Book [1, p. 66] has plenty of examples similar to yours:

$$
\begin{array}{lll}
\ce{C6H5-CH2-NCS} &\quad &\text{benzyl isothiocyanate} \\
  & &\text{(isothiocyanatomethyl)benzene (PIN)} \\
  & & \\
\ce{C6H5-NC} &\quad &\text{phenyl isocyanide} \\
  & &\text{isocyanobenzene (PIN)}
\end{array}
$$

Note that common names for this molecule also include: benzyl chloride and α-chlorotoluene.
References

Favre, H. A.; Powell, W. H. IUPAC “Blue Book” Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names; IUPAC Recommendations; Royal Society of Chemistry: Cambridge, England, 2013. ISBN 978-0-85404-182-4.

The remark about the use of parentheses is correct. However, the actual question was whether the preferred name is chloro(phenyl)methane or (chloromethyl)benzene.

@Loong You are right. I was (and still am) too lazy to look up the relevant rules so just decided to edit in a brief line about the parentheses; I leave it to you if you want to expand further or write another answer.

Is 1-phenyl methyl chloride possible?


text:
Why does roasting requires lower temperature than melting point
Why do processes like roasting and calcination require a temperature lower than the melting point? Since intermolecular forces are lower in a liquid won't it be easier to oxidise (roasting) / decompose (calcination) in this state than in a solid?

If you melt it ,the product is a complicated slag , rather than a low sulfur ore.

Reason 1:
For example if you have an porous ore and you melt it, the surface area will decrease dramatically. If you just roast it, the surface area will stay about the same.
A higher surface area will aid calcination proceses as more oxygen gets into contact with the surface and will eventually react.
Reason 2:
A higher temperature means more energy to heat the ore and this means it costs more money.
Reason 3:
Some oxides will volatilize when temperature is high (e.g. litharge).


text:
Why does the hexachlorocopper(II) ion not form?
I am not entirely sure if this question has been answered before, but I was wondering why when copper forms a complex ion with chlorine there are only 4 chlorine ligands. I asked my teacher and he said that it was because chlorine is a weak field ligand. I am not particularly satisfied with this answer as tetrahedral, octahedral and other geometries can be formed by weak field ligands.
So my question is essentialy why is it $\ce{[CuCl4]^{2-}}$ instead of something else like $\ce{[CuCl6]^{4-}}$

4- is a lot of charge to pack into a small volume

Is it possible to explain it using LFT or CFT at all? I would like a constructive way to determine whether a certain metal-ligand complex is feasible or not.

In general, CFT and LFT lead to "Structure preference energy" (SPE) estimations, which predict the likely preferred coordination number and structure based just on the number of d electrons and the strength of the ligand (ie whether complex is high or low spin). However, in the case of $d^9$ (like the Cu(II) you have here), the difference in energy between octahedral and square planar by SPE is 0, so it doesn't tell us which is more likely. SPE calculations also do not account for sterics or crystal packing considerations. Instead, we have to rely on observed trends or more complex math.

@Andrew correct me if I'm mistaken but $\ce{[CuCl4]^{2-}}$ is tetrahedral not square planar? Also could you point me in the direction of some websites or books I could read in order to learn more about calculating SPE estimations?

Yes and no. In solution it's a slightly distorted tetrahedron, but in crystal structures it is more flattened and sometimes completely square planar depending on the counterion.


text:
Why does Freundlich adsorption isotherm not have a term for surface area?
Why does Freundlich adsorption isotherm not have a term for surface area?
The formula given by Freundlich adsorption isotherm:

$$ \frac{x}{m}= k P ^{1/n}$$
  Where $x$ is the mass of gas adsorbed on mass $m$ of the adsorbent and pressure $P$. $k$ and $n$ are constants which depend on the nature of the absorbent and the gas at a particular temperature.

There is no term for the surface area involved for adsorption and I had read that adsorption increases with the increase in surface area so if I take a highly porous substance, it will adsorb more gas on it. What am I missing here?
Edit: Actually, what I mean is that, let's say I have a single piece of adsorbent, it has a certain value of $x$ for a given $P$. Then I break it into 10 pieces, The value given by the formula is still the same, while there will for sure be an increase in $x$, due to increased surface area.

The equation is empirical where $x/m$ is equivalent to the fractional surface coverage often labelled $\theta$ as in the Langmuir isotherm. $k$ and $n >1$ are found by fitting the data. The Freundlich isotherm is unrealistic as there is no limiting value for $\theta$ but data is often followed by this expression for $\theta $ in the range $0.2\to 0.8$

The ratio
$$\frac{x}{m}= k P ^{1/n}$$
can be re-expressed in terms of the area $A$, surface-to-volume ratio $\lambda$, and density $\rho$ (note that $m=A\rho/\lambda)$:
$$\frac{x}{A}= k' P ^{1/n}$$
where
$$k'=k\rho/\lambda$$
The advantage of using $k$ as opposed to $k'$ is that you don't need to determine $\lambda$ to know how much (as in, what mass) adsorbent to use in a particular application (see e.g. Table 4 in ref 1).
There are two types of adsorbent material you are usually interested in:

Porous material; in this case, breaking up the particles will not significantly increase $A$, and therefore not change k (since $\lambda$ will remain approx constant).
Small particles with high $A/V$ ratio and adsorption on an outer surface of the particle. Now breaking up particles into smaller ones matters a lot, but this would be reflected in a different value of $k$, since $\lambda$ increases and $k\propto\lambda$.

Parameter $k'$ is more fundamental than $k$, as it reflects how much surface area the adsorbed molecules require, keeping in mind the range of applicability of the equation, as it does not predict e.g. monolayer saturation. Note that in neither scenario would a reduction in the size of the particles be expected to change $k'$ by much - certainly not in the first case, and, in the second case, depending on the importance of changes in adsorbent curvature or edge/corner effects.
Reference

G. P. Jeppu, T. P. Clement, "A modified Langmuir-Freundlich isotherm model for simulating pH-dependent adsorption effects," Journal of Contaminant Hydrology 2012, 129–130, 46–53 (DOI: 10.1016/j.jconhyd.2011.12.001).

Actually, what I mean is that ,let's say I have a single piece of Adsorbent, it has a certain value of x for a given P. Then I break it into 10 pieces , The value given by the formula is still the same, while there will for sure be an increase is x, dur to increased surface area.

@user232243 there are two types of adsorbent material you are usually interested in. (1) Porous material; in this case, breaking up the particles will not significantly increase A, and therefore not change k (since $\lambda$ will remain approx constant). (2) Small particles with high A/V ratio. Now breaking up particles into smaller ones matters a lot, but this would be reflected in a different value of k, since $\lambda$ increases and $k\propto\lambda$.

@user232243, echoing Buck thorn here, you say in the question that $k$ depends on the "nature of the absorbent", and what this answer is saying, I think, is that part of what "nature of the absorbent" is, is, well, its specific surface area.  So if you break up the absorbent into ten pieces, then its "nature" has changed and thus so would the value of $k$.


text:
Will sodium thiosulfate react with nitrate and/or nitrite ions
I am doing nitrate and nitrite quantification in raw water and treated water using a colorimetric method. The manufacturer of the instrument states that water with chloride will interfere with the test result. I was thinking to inactivate the chloride in the water with sodium thiosulfate. However, I am not sure how is sodium thiosulfate react with the nitrate and nitrite in my sample. Thanks in advanced for any guidance provided! Cheers!

Do you mean "chloride" or "chlorine"?  I'm guessing the latter but am not sure.  Chloride ion, i.e. the same thing found in a simple solution of table salt and water, doesn't react with thiosulfate.

First, typically only high concentrations of chloride will interfere with the colorimetric test...I assume you are using NitraVer or something similar? If you aren't using seawater (i.e., $\ce{Cl-} \gt \pu{100 mg/L}$) then you will be okay. But, let's say you are using seawater. There is an easy solution, simply make your standards in an aqueous matrix containing the same amount of chloride. (I hope you are making standards or have a check standard and are not just blindly trusting the UV-Vis result...)
Second, thiosulfate is a reductant and thus could potentially reduce nitrate and nitrite. However, your most powerful answer will be your own tests - you need to do control experiments. Make some nitrate and nitrite solutions of known concentrations and then mix them with thiosulfate at concentrations you would use. Check the nitrate/nitrite concentrations after mixing and determine if thiosulfate reduced them.


text:
What is the unit of the Rydberg Constant?
In the image, the Rydberg constant seems to have two units. How is that possible?

Inverse centimeters are/were a commonly used unit for spectroscopy. http://how-it-looks.blogspot.com/2010/07/how-to-convert-to-and-from-wavenumbers.html

But why did they equate it with joules?

To convert between joules (units of energy) and inverse centimetres (conventional units of wavenumber) simply requires multiplication by a factor of $hc$. This is because
$$E = h\nu = \frac{hc}{\lambda} = hc\bar\nu$$
Of course, they are not literally equal, but they are physically equivalent.


text:
What do brightness and intensity of spectral lines depend on?
My book states that it depends upon

the number of photons of same frequency or wavelength absorbed or emitted.

From what I understand, if many hydrogen atoms within the discharge tube emit at the same wavelength, then, on passing the radiation through a prism, these waves will overlap on the photographic plate because all of them will get deflected to the same extent by the prism. This should increase the intensity. Is my understanding correct?
Secondly, how can the number of absorbed photons of the same wavelength create any difference in the intensity of spectral lines? As far as I know, absorption only results in black lines on the bright background corresponding to the absorbed wavelengths.

The body is somehow less clear than the title Q. If you ask for why the same element shows abs/emiss lines of different intensity look for oscillator strength and/or *transition moment" here or in the web.

Hint: If you want understanding in chemistry, don't generalise too much, and too early. Start with a specific example and a specific method. Otherwise you can only learn facts by heart, and never know if they apply on your current problem.

Absorption and emission lines are (mostly) two sides of the same coin. So their intensities are strongly related.

And any explanation that says the "strength" of a spectral line is related to the number of photons emitted (or absorbed) is not an explanation at all but merely a more precise definition of what you mean by "strength".

The answer "The number of photons of same frequency or wavelength ABSORBED or EMITTED" from your book is a general way of saying when spectral lines are brighter (or darker). Your understanding of this general mechanism is (I think) mostly correct.
From what I understand, the differences in intensity of different spectral lines, however, is related to two (possibly more) things:
1) If a photon hits an atom (or molecule) and that photon has the exact energy needed to excite that atom does not necessarily means that the photon WILL be absorbed. There is an absorption probability associated with each energy transition. For example, one transition might have a 80% probability of absorption, while another has 0% (because the transition is forbidden).
2) Another factor that influences intensity is related to what state the system is in. For example, if two transitions, lets call them 0->2 and 1->2 both have a 100% probability of adsorption, but all the atoms/molecules are in state 0, then the 1->2 transition will not happen. That is, it will be dark/low intensity.
I hope this is helpful.

We are not sure of the complete context. In simple words, the intensity is proportional to the number of photons impinging on the detector per unit area. This also implies that we can measure the power (in Watts) of the radiation, which is proportional to the number of photons. One can say that a laser is a high intensity source because a lot of photons impinge on a small area.
A high intensity spectral (absorption or emission) indicates that transition occurred in a larger number of atoms i.e. that electronic transition occurred in a large population of atoms.
A more rigorous definition is provided by IUPAC:
spectral radiant intensity,  I
The radiation intensity,  I at wavelength λ per unit wavelength interval. The SI unit is  W m−1 sr−1, but a commonly used unit is  W nm−1 sr−1
Note that sr symbol indicates the solid angle measured in steradians.

You are not explaining the strength of a spectral line at al or describing what it depends on: you are just defining what strength means.

Write a separate detailed answer rather than serving as an unsolicited reviewer and editor. This is rather a voluntary work.


text:
Difference between base-promoted and base-catalysed
Why is halogenation of ketones in base described as being "base-promoted" and not "base-catalysed"?
If the reaction is not "catalyzed" then how is the base used up?

For example, the base used in idoform reaction did not get regenerated. Therefore, it is needed as a reactant, not as a catalyst. Thus, it is a "base promoted" reaction, not "base catalysed."

https://en.wikipedia.org/wiki/Ketone_halogenation The second image shows you the mechanism, and why the base is not a catalyst

Does "base-promoted" simply mean that the mechanism requires base? Base-catalyzed means that the base is regenerated during the reaction. The base in the aldol condensation is catalytic. In the Claisen condensation or saponification the base needs to be stoichiometric in that it is consumed during the reaction by deprotonation. IMO both sets of conditions may be considered base-promoted.


text:
Exothermic: enthalpy or heat?
I am used to thinking of exothermic reactions as being defined as having negative $ΔH.$  However, I also am used to hearing/thinking of exothermic reactions as "releasing heat".
Since change in enthalpy $∆H = Q + VΔP,$ it seems that enthalpy and heat flow $(Q)$ are not strictly the same thing. Is it not therefore possible to have a reaction which absorbs heat $(+Q),$ but has a decrease in pressure, such that the product $-VΔP > Q,$ thus both absorbing heat and being exothermic (or if exo-/endothermicity is determined by $Q,$ then an exothermic reaction with positive $∆H)?$
As a sort of corollary, is there an example of a thermodynamic process that changes only the pressure (thus $V∆P)$ with no heat flow or $PV$ work? $(Q = 0 = PΔV)$

Exothermic reactions don't necessarily have to release heat, although they commonly do; they simply have to release excess energy. Heat doesn't always equal energy.

"Since change in enthalpy ∆H = Q + VΔP, it seems that enthalpy and heat flow (Q) are not strictly the same thing."

First, note that, when you write "∆H = Q + VΔP," you are assuming the only type of work is PV-work.  That's a reasonable assumption for these discussions.  It's just important to make that assumption explicit.
Given this, the quoted statement is correct.  Note, however, that many chemical reactions are carried out in an open container, such that the system stays at constant pressure (∆P = $0$) throughout the reaction. Thus, under this condition,  ∆H = Q + VΔP = Q.  I.e., the reason ∆H is a particularly useful thermodynamic function is that, under the common condition of ∆P = $0$, it corresponds directly to an important quantity: the amount of heat that flows into or out of the system.

"Is it not therefore possible to have a reaction which absorbs heat (+Q), but has a decrease in pressure, such that the product -VΔP > Q, thus both absorbing heat and being exothermic (or if exo-/endothermicity is determined by Q, then an exothermic reaction with positive ∆H)?"

Here it comes down to a definition of terms.  The IUPAC Gold Book defines an "exothermic reaction" as "A reaction for which the overall standard enthalpy change ΔH⚬ is negative" ( https://goldbook.iupac.org/html/E/E02269.html ), where "standard" means P = 1 bar. Thus it doesn't cover how one would define exothermic or endothermic if the process is not at constant pressure.*   Having said that you could, in principle, have a reaction that takes place, say, at constant volume, and that releases thermal energy (Q < $0$) but, because of a sufficiently large and positive VΔP term, has ΔH > $0$. But, in practice, $\mid{Q}\mid$ would have to be very small in order for the VΔP term to dominate.

"As a sort of corollary, is there an example of a thermodynamic process that changes only the pressure (thus $V∆P)$ with no heat flow or $PV$ work? (Q = $0$ = PΔV)"

Yes.  Consider, for instance, an isolated system, which is a system in which no flow of heat or work is allowed, and that is closed (no flow of matter).  The pressure could increase in an isolated system if, for instance, a chemical reaction is taking place in that system that evolves gas, and/or for which (were we to remove the adiabatic constraint) Q would be negative.
*Indeed, the IUPAC definition doesn't even seem to cover how one would define exothermic or endothermic if the reaction took place under constant P $\neq$ 1 bar, which seems quite limiting!  Further, it only defines the term for a "reaction", which IUPAC seems to take to mean a "chemical reaction" (typing "reaction" into the search field redirects you to "chemical reaction"), thus excluding important processes like phase changes.

I really have to wonder why IUPAC came up with that definition, rather than simply generalizing to endothermic meaning q>0, the latter afaik including the IUPAC definition.

Yes, and a benefit of defining exo/endothermic in terms of the sign of Q would be that such a definition follows the natural meaning of the words: giving off or taking in thermal energy. The only thing I can think of is that the same chemical reaction has Q=$0$ if it's carried out adiabatically. Hence they may have decided it was safer to define it in terms of the change in a state function under certain conditions (constant P = 1 bar) rather than the integral of a path function (Q). Of course, they could have defined it using the sign of Q under isothermal conditions.

The heat of reaction $\Delta H^\circ$ is defined as the difference in enthalpy between the pure products and pure reactants, all at 1 bar pressure and 25 °C.  If this $\Delta H^\circ$ is negative, the reaction is regarded as exothermic.
If we are dealing with all ideal gas species, where the enthalpy is independent of pressure and the heat of mixing is zero, and the reaction takes place in a process mixture maintained at constant pressure, the heat added to the reaction mixture to maintain constant temperature is equal to $\Delta H^\circ.$


text:
Does benzene react with steam to create an alcohol?
Benzene has a molecular formula of $\ce{C6H6},$ right? Surely it has some alkene-like properties because of the double bonds, so can it undergo a hydration reaction to create alcohol?

Getting the double bonds in benzene to react like alkenes is possible, but also rather non-trivial, and steam will not do it.

At what temperature? Probably well above the critical point of both.

Benzene doesn't have double bonds. Each of the six carbon-carbon bonds is identical, somewhere between a single and a double bound.

Strange as it may seem, benzene can be reacted with steam over a suitable catalyst [1].  The result is a complete breakdown to simpler molecules like methane and carbon dioxide.  Typical conditions are 0.055 atm partial pressure benzene, 0.60 atm partial pressure steam, and 370 °C temperature.
The cited reference also considers the breakdown of alkylbenzenes with steam under similar conditions.

Reference

Masahiro Saito, Yoshio Sohda, Michiaki Tokuno, Yoshiro Morita, Reactions of Benzene or Alkylbenzenes with Steam over a Silica-Supported Nickel Catalyst, Bulletin of The Japan Petroleum Institute, 1972, Volume 14, Issue 1, Pages 54-63.


text:
How to calculate the value of heat absorption/ heat release during water electrolysis?
The overall reactions with the enthalpies of water electrolysis at room temperature can be represented by the equation below:
$$\ce{2H2O + \pu{237.2 kJ mol-1} + \pu{48.6 kJ mol-1} → 2 H2 + O2}$$
Where $\pu{237.23 kJ mol-1}$ is the maximum electrical work can perform and should be calculated by:
$$
\begin{align}
\Delta H^\circ &= \pu{0 kJ mol-1} - (\pu{-285.83 kJ mol-1}) \\
  &= \pu{285.83 kJ mol-1}
\end{align}
$$
$$
\begin{align}
\Delta S^\circ &= 0.5 \times\pu{205 J mol-1 K-1} + \pu{130.6 J mol-1 K-1} - \pu{70 J mol-1 K-1} \\
  &= \pu{163.1 J mol-1 K-1}
\end{align}
$$
$$
\begin{align}
\Delta G^\circ &= \Delta H^\circ - T\Delta S^\circ \\
  &= \pu{285.83 kJ mol-1} - \pu{298 K}\times\pu{163 J mol-1 K-1} \\
  &= \pu{237.23 kJ mol-1}
\end{align}
$$
where $\pu{48.6 kJ mol-1}$ is the net value of heat absorption, and can be calculated as the following:
$$
\begin{align}
T\Delta S^\circ &= \pu{298 K}\times\pu{163.1 J mol-1 K-1} \\
  &= \pu{48.6 kJ mol-1}
\end{align}
$$
This is the overall heat absorption during water electrolysis. I still don't know the heat release/absorption for the anode and cathode electrochemical reaction separately.
For the anode electrochemical reaction:
$$\ce{2 H2O → O2 + 4 H+ + 4 e-}$$
I believe this is a heat absorption process.
For the cathode electrochemical reaction:
$$\ce{2 H+ + 2 e- → H2}$$
I believe this is a heat release process.
How to calculate the value of heat absorption at anode, and heat release at cathode separately?


text:
What is the major product of the reaction between 4-chlorobut-1-ene?

When acetic acid looses hydrogen, an acetate ion is formed which attacks 4-chlorobut-1-ene, forming methyl butenoate. How does the cyclic structure form?
Does the lone pair of oxygen attack the butene moiety internally?

In solvolysis of simple primary cyclopropylmethyl systems the rate is
  enhanced because of participation by the $\sigma $ bonds of the ring. The ion
  that forms initially is an unrearranged cyclopropylmethyl cation that is
  symmetrically stabilized, that is, both the 2,3 and 2,4 $\sigma $ bonds help
  stabilize the positive charge(page 464 ,JerryMarch ,Advanced Organic Chemistry
  Reactions, Mechanisms,and Structure,Sixth Edition).

The reaction of various cyclopropylmethyl systems are given below.

In the question 4-chlorobut-1-ene  (1) undergoes rearrangement as shown in the scheme 3 and scheme 4 to give 1a, 1b and 1c.Assuming solvolysis with acetic acid ,1a, 1b and 1c would give 2, 3 and 4.

Which would be the major product? I feel it would be 3

@thewitness , the product composition could follow the one I suggested in Scheme 1.But this reaction is with acetic acid and composition may change.Access to a journal could verify or disprove it.


text:
What is this fluorinated organic substance?
The corresponding author of the paper where this formula was published as appendix passed away. Can someone help me with identifying what is the name of this compound and why have strange zig-zag between Nitrogens

That is generally known as Selectfluor, a source of electrophilic fluorine. The zig-zag line is a 2-D representation of the third ethylene $\ce{-CH2-CH_{2} -}$ unit that links the two nitrogens. more here and wikipedia

thank you very much. do you know if there is other way to donate F?

Yes, there are several other sources of electrophilic fluorine. This Wikipedia article is a good starting point  https://en.wikipedia.org/wiki/Electrophilic_fluorination

The name of the compound is 1-chloromethyl-4-fluoro-1,4-diazoniabicyclo[2.2.2]octane bis(tetrafluoroborate) (CAS #: 140681-55-6), which is commonly known as Selectfluor, a trademark of Air Products and Chemicals (see Waylander's comment elsewhere). Different view of the compound is given below (to you to understand the zig-zag feature):

Introduced in 1992, this organic salt is used as a fluorine donor in organic synthesis (Ref.1). For example of using Selectfluor as a source of fluorine, see Ref.2:

References:

R. Eric Banks, Suad N. Mohialdin-Khaffaf, G. Sankar Lal, Iqbal Sharif, Robert G. Syvret, "1-Alkyl-4-fluoro-1,4-diazoniabicyclo[2.2.2]octane salts: a novel family of electrophilic fluorinating agents," J. the Chem. Soc., Chem. Commun. 1992, (8), 595-596 (DOI: 10.1039/C39920000595).
Timothy J. Barker, Dale L. Boger, "$\ce{Fe(III)}$/$\ce{NaBH4}$-Mediated Free Radical Hydrofluorination of Unactivated Alkenes," J. Am. Chem. Soc. 2012, 134(33), 13588-13591 (https://doi.org/10.1021/ja3063716).

is this substance toxic to cells or not environmentally friendly?

Not related to the question Which application do you use for these figures? Is there an open-source alternative (if the one you're using is proprietary)?

@SSimon yes Selectfluor is a toxic substance. PPE is very necessary when handling it.

@Waylander thank you. Is there any substance that can give F ions but that is not toxic?

@SSimon - there are many compounds that can give an "F ion" -- i.e., fluoride. For example, NaF gives F ions when dissolved in water. It is sufficiently nontoxic that it is added to toothpaste and drinking water. However, the point of Selectfluor is that is does not donate an ion; it donates a neutral F which forms a covalent bond with the target.  Any compound that can perform this function on a wide variety of organic compounds must be toxic.  Why? Because people are made out of a wide variety of organic compounds, and we do not want to be fluorinated.


text:
Change buffer in dialysis?
I have a protein that I purified in PBS buffer, pH 7. I will do dialysis to remove salt and will then further purify the protein with ion exchange chromatography.
I will need to use another buffer (Tris-HCl) with pH 8 in the chromatography, and so my question is:
Can I change the buffer system for my protein in the dialysis step? Or would you recommend me to use the Tris-HCl buffer from the very start and then just adjust the pH for the chromatography?
Thanks in advance!

Tris-HCl works from pH ~7 to ~9. Does the processes upstream of the filtration work with Tris-HCl? See also: diafiltration and buffer exchange.


text:
What makes polyimides good polymers for fibers?
Wikipedia says polyimides are known for “thermal stability, good chemical resistance, excellent mechanical properties”. I guess that's why they're used industrially as fibers.
But where do they derive there good stability (thermal, mechanical and chemical) from? How do these properties differ from, e.g., polyamides?

Chain stiffness.

Polyimides (PIs) are high-temperature engineering polymers originally developed by the DuPont™ Company. Polyimides exhibit an exceptional combination of thermal stability (>500 °C), mechanical toughness, and chemical resistance.It was based on pyromellitic dianhydride and 4,4’diaminodiphenyl ether.

image source
The presence of n-π conjugation between non-pair electron of nitrogen atom and π electrons of the carbonyl group makes them resistant to chemical agents and moisture.

image source
The linearity and stiffness of the cyclic backbone allow for molecular ordering. This phenomenon results in lower coefficients of thermal expansion (CTE) than those found for thermoplastic polymers having coiled, flexible chains.
Additionally, the morphology of long, linear ordered chains provides solvent resistance to the aromatic polyimides.
The comparatively rigid structure of polyimides provides high glass transition temperature (Tg > 300°C) and imparts good mechanical strength and high modulus.
The thermal stability of polymer can further be improved by incorporating aromatic rings on the backbone and/or side groups. In addition to such high thermal stability, the nature of the chemical structure consisting of rigid imide and aromatic rings always provides:

Excellent mechanical toughness
Excellent dielectric properties
High chemical resistance

These result in  thermoxidative stability, high mechanical strength, high modulus, excellent insulating properties, and superior chemical resistance.
Polyimide layers are suitable matrix for incorporation of metal, salts, chromophores as nanoscale particles to obtain of nanocomposite materials. It was discussed the possibility of use polyimides in materials chemistry and nanomaterials, one of these applications is the use for making biomedical implants for neurology, ophthalmology, biosensor device and chips which are a powerful tool in clinical diagnostics.
Another important trend is use in electronics and optoelectronics such as dielectric substrates and intermediate barrier layers, creating nanocomposite films with various nanosized particles such as dyes, metal, dielectric and other clusters.
Typical polyimide parts are not affected by commonly used solvents and oils — including hydrocarbons, esters, ethers, alcohols and freons. They also resist weak acids but are not recommended for use in environments that contain alkalis or inorganic acids.

This is a question about poliimides, They don't exhibit hydrogen bonding...

@Mithoron ,polyimides exhibit similar behavior.I shall update for the same ,polyimides,as soon as possible.

Thing is as OP mentioned this is rather about differences, which are quite big...

@Mithoron , i have edited to polyimides. which do not undergo hydrogen bonding and poly amides do.


text:
Products of the oxidation of 1,4-dimethylcyclohexane

I am having difficulty identifying the second product. Is it an ether, or a kind of epoxide? I am out of ideas on how to reach it.
However, I do have a general idea how to receive the first product, in the means of knowing that I have to first turn the 1,4-dimethylcyclohexane to an alkyl halide and then into an alkene and then an epoxide.
I would be grateful to receive an idea of how to start.

It's an ether, obviously.

I would imagine you can access it via 1,4-dimethylcyclohea-1,4-diene

would it be logical that after reaching 1,4-dimethylcyclohea-1,4-diene doing the following:
1.hydration (to add OH next to Me)
2.hydrohalogenation( to add Br next to the opposite Me)
3.NaOH, H2O(to reduce OH to O-, thus allowing an intramolecular Sn2 reaction to occur).
open to corrections.


text:
Oxidation state of carbon
How do we find the oxidation state of individual carbon in Mg₃C₂. Explain with the help of structure if possible. For the average oxidation number, we can exploit the fact that the molecule will be electrically neutral, but what about the individual atoms of carbon?

Hint: The (formal) oxidation state of magnesium should be +2. Deduce the formal ON of carbon now given the overall charge is 0.

@Saifuddin Please avoid edits which are rendering already provided answer (a good and a complete one, like the one you've received) irrelevant (even partially). Based on what you are asking (oxidation state of an element), changing compound's formula (Mg3C2 → Mg2C3) is a big deal. At this point it would be better to leave the originally formulated question untouched, and pay more attention when composing new questions in the future.

While in your task it was probably Mg2C3, this edit somewhat invalidated answer. Exact example isn't important here so I rollbacked it. You can make another question about sesquicarbide

I'll update my answer to include compounds like $\ce{Mg2C3}$

Method 1: Assume the compound is ionic
For binary compounds consisting of a metal (but not a transition metal) and a nonmetal, the best method of determining oxidation numbers is to assume that the compound is ionic and determine the charge on the corresponding ions. The oxidation number of a monatomic ion is equal to its charge. This a good assumption because most compounds of this type are ionic, and it will work on many compounds that might be covalent. To make this work, you need to memorize some typical charges for ions:

Alkali metals - Li, Na, K, Rb, Cs - are typically 1+
Halides - F, Cl, Br, I - are typically 1-
Alkaline earth metals - Be, Mg, Ca, Sr, Ba - are typically 2+
Chalcogenides - O, S, Se, Te - are typically 2-

These are typically sufficient to deduce charges for most binary ionic compounds.
Let's do some examples.

Definitely ionic: $\ce{Li3P}$

First, we assume that this compound is comprised of lithium ions and phosphide ions. The lithium ion is $\ce{Li^+}$, and there are three of them for a total of $3+$. Thus the charge on phosphorous must be $3-$. The oxidation number of lithium is +1, and the oxidation number of phosphorous is -3.

Definitely covalent: $\ce{BeH2}$

Beryllium hydride is covalent, but this approach still works. If we assume this compound is a beryllium cation, $\ce{Be^2+}$, and two hydride anions, we can work it out. The hydride anions must be $\ce{H^-}$ to make the compound neutral. Thus, the oxidation number of beryllium is +2, and the oxidation number of hydrogen is -1.

Method 2: Based on structure
If we know the structure, we can use an alternative method. This method works best for covalent compounds, and does not work well for ionic lattices or network compounds. For each atom, we can assign the oxidation number as follows:

Each bond to a less electronegative atom contributes -1 to the oxidation number
Each bond to a more electronegative atom contributes +1 to the oxidation number
Each bond between two atoms of the same electron contributes +0 to the oxidation number

Let's look at some examples:

$\ce{H2O2}$

The structure of $\ce{H2O2}$ is $\ce{H-O-O-H}$. Each oxygen atom has one bond to hydrogen (-1) and one bond to oxygen (+0) for an oxidation number of -1. Each hydrogen has one bond to oxygen (+1) for an oxidation number of +1.

$\ce{HCN}$

The structure of $\ce{HCN}$ is $\ce{H-C#N}$. The hydrogen atom has one bond to carbon (+1), for an oxidation number of +1. The carbon atom has one bond to hydrogen (-1) and three bonds to nitrogen (+3) for an oxidation number of +2. The nitrogen atom has three bonds to carbon (-3), for an oxidation number of -3.

$\ce{NaC2H}$

This compound is an ionic compound with one of the ions having covalent bonds. The structure is $\ce{Na^+ ^-C#C-H}$. The charge on sodium is 1+, so its oxidation state is +1. The first carbon atom has a formal charge of 1- and three bonds to another carbon atom (+0). The formal charge is added to the contributions from the bonds to get an oxidation state of -1. The other carbon atom has an oxidation state of -1 because of one bond to hydrogen (-1) and three bonds to the first carbon (+0). The oxidation state of the hydrogen atom is +1.

Your compound $\ce{Mg3C2}$
Either approach will work for $\ce{Mg3C2}$. If we treat it as ionic (a valid assumption based on results reported in this paper by Hongzhe Pan, et al.), there are three $\ce{Mg^2+}$ cations in the formula unit, and carbon needs to have a corresponding charge to make the compound neutral.
In the same paper by Pan, et al., the structure of a $\ce{Mg3C2}$ monolayer is described as a honeycomb kagome lattice. Below is a representation of part of this lattice:
.
In this lattice, each magnesium is bonded to two carbon atoms, and carbon is more electronegative than magnesium. Each carbon atom is bonded to three magnesium atoms.
The similar compound $\ce{Mg2C3}$ is a little trickier. If we use the ionic method, we can get an average oxidation state for each carbon atom of $\dfrac{+4}{3} \approx +1.333$. An atom cannot have a fractional oxidation state, so the total +4 must be distributed unequally over the three carbon atoms. If we look at the structure, we can see that one carbon atom (the one in the middle) is different than the other two.
$$\ce{Mg^2+ ^2-C=C=C^2- Mg^2+}$$
Using the formal charges on the carbon atoms, you can determine the oxidation states.


text:
Do I have to change my sample buffer before IEX chromatography?
I will use IEX chromatography to purify a protein and will then use a equilibration buffer Tris-HCl, pH 8 to set the start conditions for my column and optimize the binding of my protein. Then inject my sample and further on elute with the same buffer but with a salt gradient.
My sample will contain my protein as well as the PBS buffer (pH 7) that I used in previous purification steps.
This question might be dumb but is this applicable, i.e. can I inject my sample with the PBS buffer to the column after equilibration with another buffer, or do I have to change my sample buffer in some way before the chromatography?
Thanks!

Depends on the injection volume. If the sample volume is very small, then it should not matter. IEX peak shape is dependent on the sample solvent in some conditions. Try the experiment at a small scale.

OK, thanks! But is it standard to have the same buffer for the sample as well as the buffer for the chromatography?

Are you doing preparative HPLC or analytical? What are the column dimensions? Is it a simple glass column IEX. What is your sample volume?

Bed size is 7 × 25 mm, volume 1 mL. Sample volume, I don't know. I suppose it is preparative since my goal is to obtain fractions of my specific protein.

I would say stick to the correct protocol you column volume is very small.

OK, so I don't have to change the sample buffer you think? Still I need to know how much sample I should load, and this is decided by the column capacity? Forgive my dumb questions, this is new to me. Thanks!

No, I meant follow the original protocol as suggested by the manufacturer.

@Firas Yes. The vendor (of the pre-packed column or the resin) typically lists a dynamic binding capacity value specific to the rt (retention time).


text:
Molar conductivity of transition metal complexes solutions

Which of the following has the highest molar conductivity?
(A) Diaminedichloroplatinum(I)
(B) Tetraamminedichlorocobalt(III) chloride
(C) Potassium hexacyanoferrate(II)
(D) Hexaaquochromium(II) bromide

Since mobility depends on the size, charge and the number of particles, how can we determine the extent of molar conductivity from the given data?
The solution provided to me considered only the number of ions that each gives. But is this justified? Because they all have different sizes, charges and thus different mobility.

Charge transport depends on the presence of charge carriers, and an increase of the concentration of charge carriers may increase the electric conductivity.  This is the basis of the answer given to you which (implicitly) assumes each of these complexes dissociate, e.g.
$\ce{K4[Fe(CN)6] <=> 4 K+ + [Fe(CN)6]^{4-}}$
in your solvent completely.  In this view, charge transport happenz without moving these carriers on a macroscopic level (like  the salt bridge used to connect electrochemical cells). You may imagine a bucket brigade, instead people exchanging buckets of water, ions shuttle electrons.
On the other hand, there are examples where charge transport is coupled to the transport of matter (e.g., ions) on a macroscopic scale.  Proteins and nucleic acids, for example, may be separated by this (e.g., gel electrophoresis); their different size (and shape, and charge) quite noticeably influence how quickly they travel in an electric field.


text:
Reaction mechanism of rearrangement
I am currently studying for my organic chemistry exam, but there is one problem I do not understand. Unfortunately, I do not have any solutions.
See image for the problem.

My first thought was the Dienone-phenol-rearrangement. But in order to do so, I need to add the alkyl substituent and I need a dienone. So can I deprotonate the phenol using potassium carbonate? (Or is it too weak as a base? The pKa of bicarbonate is around 10, so maybe it is too weak) But if it is possible, then the free electron pair of the oxygen can form a ketone... I could then use the alkene with a chloride substituent, so that the aromatic ring can attack it. Then I would have the dienone. But with the Dienone-phenol-rearrangement, the alkyl substituent can only move by one carbon-atom in the ring...
Is my thought process wrong? Can you please help me?
I did not have any lectures on rearrangement reactions, I had to learn them by myself, but by now, I have not found any similar reactions.
Thank you!

The first reaction is O-alkylation of p-cresol to give a 4-methylphenyl allyl ether derivative 3. The reagent in the first box should be 1-bromo-3-methylbut-2-ene (1; see the top box in the picture), which would undergo $\mathrm{S_N2}$ reaction with phenolic anion (2) in refluxing acetone. Note that potassium carbonate is a strong enough base to complete this reaction (this is surely a strong base than potassium bicarbonate).

The product 3 from alkylation would undergo Claisen rearrangement upon heating to give final alkylated cresol 4 in solvent-free condition. Remember, Claisen found his famous rearrangement first in solvent-free conditions, while he was trying to find the melting point of newly synthesized naphthyl allyl ether! (Picture is from https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3873100/ and modified accordingly)

Potassium carbonate is a perfectly good base for the alkylation of phenol ($\mathrm{p}K_\mathrm{a} = 10$) with a good electrophile, in this case 3,3-dimethylallylbromide. The reaction you are looking for is Claisen rearrangement which proceeds by a 3,3-sigmatropic rearrangement mechanism.

Image from ref 1

The first step is nucleophilic substitution.

It is possible for nucleophile to attack directly at the allylic position, displacing the leaving group in a single step, in a process referred to as $\mathrm{S_N2'}$ substitution. This is likely in cases when the allyl compound is unhindered, and a strong nucleophile is used. The products will be similar to those seen with $\mathrm{S_N1'}$ substitution. Thus reaction of 1-chloro-2-butene with sodium hydroxide gives a mixture of 2-buten-1-ol and 3-buten-2-ol:

Step 1: Williamson Ether synthesis

para-Cresol reacts with 3,3-dimethylallylbromide to give 1 and 2 using potassium carbonate in acetone as shown below.

Step 2: 3,3-Sigmatropic rearrangement (Aromatic Claisen rearrangement)

Compounds 1 and 2 undergo 3,3-sigmatropic rearrangement (Aromatic Claisen rearrangement)${^1}$. Compound 1 gives 3 via sigmatropic shift followed by tautomerism to give 4 (as shown below).

Similarly, 2 gives 5 via sigmatropic shift followed by tautomerism to give 6 (as shown below).

Net reaction
As per the question given, the following is the scheme.

References
${^1}$:https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3873100/
${^2}$:https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/Chapter_18%3A_Ethers_and_Epoxides%3B_Thiols_and_Sulfides/18.04_Reactions_of_Ethers_-_Claisen_Rearrangement
${^3}$:J. Am. Chem. Soc.19891112511-519 Publication Date:January 1, 1989
https://doi.org/10.1021/ja00184a018


text:
Conductance of the mixture of two solutions with given resistance values

If $R_1$ and $R_2$ are the resistances of two solutions of equal volume what is the conductance of the mixture in the same conductivity cell?

In general, I grasped neither the approach, nor the meaning very well.
Do we add the resistances as now both the solution are there to resist, or do we add the conductance or conductivity of them as now both conduct?
As now the volume is doubled, do I divide by 2 or multiply? The area is doubled, but the ions per unit volumes is reduced to half (or maybe not, I'm not sure).
Also, how can we directly add and ignore the inter-particular interaction?
The answer given is

$$\frac{1}{2}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$

@EdV But what about the Electrocostatic attractions??

There is an excellent treatment of electrolytic conductivity in Wikipedia. However, the OP’s main question involved a specific equation, pertaining to the conductance of a mixture of equal volumes of two electrolyte solutions, and this was not specifically discussed in the Wikipedia article:
$$\text{conductance} = \frac{1}{2}\left(\frac{1}{R_{1}} + \frac{1}{R_{2}}\right)$$
For simplicity, assume the electrolyte concentrations in the two solutions are in the “Goldilocks” zone: not so high that activities and fancy theory are needed, but not so low that the conductivity contribution due to water’s autoionization has to be taken into account. Neither solution is concentrated, so the ions just bounce around like they would in the unmixed solutions. This would certainly be incorrect in relatively concentrated solutions, but that is not the case at hand.
In this idealized scenario, a solution’s conductivity, i.e., specific conductance, is directly proportional to the mobile charge carrier (i.e., ion) concentrations in the solution. The conductance of a solution, denoted by $G,$ is directly proportional to the solution’s conductivity. The resistance of the solution, denoted by $R,$ is simply the reciprocal of the conductance and, likewise, the resistivity, i.e., specific resistance, is the reciprocal of the conductivity.
Now consider two electrolyte solutions that can be mixed without causing any chemical reaction. In particular, no precipitate forms, no gas is evolved, no complexes form, and so on. Measurements on the first solution provide values of $G_1$ and $R_1$, while measurements on the second solution provide values of $G_2$ and $R_2$. If equal volumes of the two solutions were well mixed, and a conductivity measurement was performed on the mixed solution, the conductivity of the mixed solution, $G_\mathrm{mix}$, would be
$$G_\mathrm{mix} = \frac{G_1}{2} + \frac{G_2}{2} = \frac{1}{2}(G_1 + G_2)$$
since each solution, when mixed with an equal volume of the other solution, had its constituent ion initial concentrations halved. But the total conductance is just the sum of the two conductances, $G_1 = 1/R_1$, and $G_2 = 1/R_2$, so
$$G_\mathrm{mix} = G_\mathrm{tot} = \frac{G_1}{2} + \frac{G_2}{2} = \frac{1}{2}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$
Like all approximations, this one is only as good as the foundational assumptions upon which it is predicated.


text:
How do you purify phosphoric acid from cheap commercial (hardware-store) products?
I am looking for idea's on how to purify phosphoric acid for lab use, as the the price for reagent grade V's home-use ended with me buying a product designed as a rust-killer for steel... The percentage is actually quite high in this product (being 40%), but the unlisted agents added are obviously not going to result in any useful outcomes in lab use.. Has anyone got any ideas on safe purification method's in this case?

It may be more expensive than buying a reagent grade acid.

Depends a lot on your lab use. Also I find 83% technical grade phosphoric acid online for 10€ per liter, five liters for twenty. That's too expensive?


text:
Would the mixture of Latex Rubber and the Squid's chitin make a biodegradable plastic?
I don't know much of Chemistry but I put the chemical structures of each. Thank you!

Could you provide more background? How would you combine the two polymers?

Background would mean: What is the inspiration behind the combination? What drives you to ask this?

The wikipedia does a good job of explaining the concept of biodegradation:

In practice, almost all chemical compounds and materials are subject to biodegradation processes. The significance, however, is in the relative rates of such processes, such as days, weeks, years or centuries.

Saying that something is "biodegradable" is not very informative unless you know what standard rate is being used to define something as "biodegradable".
Also, often people convolute the following two concepts when referring to something as biodegradable:

the origin of a material
the ease (or rate of decay) with which something decays, ultimately by the action of organisms

The two materials (chitin and latex) you want to combine are natural products (well, natural rubber latex is, latex need not be) and biodegradable according to the broad definition of the concept. However they degrade at different rates, with chitin degrading more quickly compared to natural rubber (see e.g. the references below).
Perhaps your question regards what kind of biodegradation rate can be expected for a composite prepared by combining chitin with natural rubber latex. This is difficult to answer generally as much depends on the microstructure of the composite, but I would expect exposed chitin to decay rather rapidly whereas the latex would decay at a much slower rate.

References

Karsten Rose and Alexander Steinbüchel. Biodegradation of Natural Rubber and Related Compounds: Recent Insights into a Hardly Understood Catabolic Capability of Microorganisms. Appl Environ Microbiol. 2005 Jun; 71(6): 2803–2812. doi: 10.1128/AEM.71.6.2803-2812.2005
Sara Beier and Stefan Bertilsson. Bacterial chitin degradation—mechanisms and ecophysiological strategies. Front Microbiol. 2013; 4: 149. doi: 10.3389/fmicb.2013.00149


text:
What is the difference in Da of ESI analytes with multiple charges?
This is a follow-up question to Why are isotopes an issue in reading mass spectra?
A follow-up question: electrospray ionization (ESI) produces multiply charged species and charge can be >40 (for the purposes of m/z ratio), and assuming the difference in mass for an isotope is ~1 Da (with each additional neutron for an analyte at a charge of +1 in positive ionization mode) and assuming there is only one additional isotope, would the difference in weight for two isotopes each with a charge +10 be 0.1 Da?

would the difference in weight for two isotopes each with a charge +10 be 0.1 Da?

Short answer:  yes
Long answer: You said

assuming there is only one additional isotope

This is a weird assumption, as it is often not true.  Molecules containing both $\ce{H}$ atoms and $\ce{C}$ atoms will have isotopic contributions from $\ce{H}$, $\ce{D}$, $\ce{^13C}$, and $\ce{^12C}$.  Anything that has a sulfur atom will have contributions from $\ce{^32S}$, $\ce{^34S}$, $\ce{^33S}$, and $\ce{^36S}$.  However none of those things really affect the answer to your question as I understand it.  Isotopes are separated by almost 1 (or 2) Da increments.
The key is almost.  Ions with a +10 charge would have isotopologues that were almost 0.1 Da apart, but not exactly.  If the ion in question were only made of carbon, then a more exact value would be 0.10034 Da apart.  Not all instruments can resolve a mass difference of 0.1 Da from 0.10034 Da, but some can.

More realistically if you have a molecule with Br, then the distance between isotopes:

Will be 2 $m/z$ units in case of 1 charge
And 1 $m/z$ units in case of 2 charges

You can use this information to determine if there are molecules ionized twice — their isotopes appear closer than expected.
Eventually since ESI/APCI adds $\ce{H+}$ for each charge, given original EMW was 100 Da, then:

$m/z$ in case of 1 charge: $(100+1)/1=101$ and $(102+1)/1=103$
$m/z$ in case of 2 charges: $(100+2)/2=51$ and $(102+2)/2=52$

Or in case of negative ionization you'll need to subtract 1 and 2 Da, respectively.


text:
What is meaning of LAV2P basis set?
I am working on energy optimization of molecular structure using DFT with Schrödinger software and I found that job is running on LAV2P basis set. I want to know what is the meaning of this basis set in particular and how it is different from other basis sets.
Also, I want to compare which basis set corresponds to this in Gaussian software.
P.S. The only info that I found on my research on the internet is the full form of LAV2P.

If I'm right, the more common name for this basis is LANL2DZ (for Los Alamos National Lab double zeta). You can find out more about it at and the way it is entered in different programs at https://www.basissetexchange.org/

Indeed, as shown in http://glab.cchem.berkeley.edu/glab/schrodinger.old/jaguar_command_reference/jaguar_files_basis_ecp.htm It looks like the definition of the basis set is the same of LANL2DZ https://www.basissetexchange.org/basis/lanl2dz/format/gaussian94/?version=0&elements=11


text:
questions about half cell of Daniell cell
My questions are as follows:
For the Daniell cell, we assume that ZnSO4 and CuSO4 are 1M aqueous solutions. What if we use pure molten ZnSO4 and CuSO4 instead?
Assume that we prepare an isolated half-cell containing just a zinc metal bar immersed in a ZnSO4 solution, without any connection to any other half-cells (no salt bridge or wire).
Do you think the Statement below is true?
In the microscopic scale, a few number of Zn atoms of the metal bar are first oxidized to Zn2+ ions and while migrating to the aqueous solution leave their electron on the electrode. But since there is no way for the electrons of the electrode to migrate to, the accumulation of negative charge on the electrode and positive charge in the solution, force the  Zn2+ ions to reduce back to Zn on the electrode.
This process also occurs for an isolated half-cells of Cu-CuSO4. Is it true? Is it right to say that here, the first few Cu atoms oxide to Cu2+ ions and leave the electrode negatively charged.
What is the reason for the separation of the first few Zn atoms from the electrode and oxidizing as Zn2+? (What is the driving force?)
Why doesn’t just "neutral" Zn atoms separate from the electrode and migrate to the solution? Is it necessary for the Zn atoms to leave their electrons on the electrode and migrate to the solution as Zn2+ ions?
Is it possible that the Zn electrode, instead of releasing Zn2+ ions to the solution, adsorb Zn2+ of the ZnSO4 solution and accordingly the electrode get positively charged and the solution get negatively charged?
Does it occurs for Cu bar and the CuSO4 solution?

When you have melted CuSO4, let me now. :-) Many of heavy metal sulphates decompose to metal oxides before melting.

Note that metals are not formed by neutral atoms, but rather there is a lattice of metal ions with delicalized valence electrons, forming conducting  electron energy band.


text:
What caused the flashes in the video footage of Chernobyl?
There is video footage of nearby villages in Pripyat, in the first week after the Chernobyl disaster. When I watch it there are villagers leading normal lives in normal clothing. Some Russian nuclear scientists pass by in full radioactive suits. Then there these clearly visible "flashes" in the air; I have looked on youtube but haven't found the video that I watched a long time ago. My layperson understanding is that these flashes are caused by an extreme amount of radiation.
Here's footage that shows the flashes in the air. They are subtle orangish flashes that are quick. But there's one for instance just after 3 minutes.
This isn't the original one that showed the flashes better. Anyway:
Chernobyl
From Wikipedia "Ionized-air glow":

Within minutes after the steam explosion that caused the Chernobyl
  accident at 01:23 local time, a number of employees at the power
  station went outside to get a clearer view of the extent of the
  damage. One such survivor, Alexander Yuvchenko, recounts that once he
  stopped outside and looked up towards the reactor hall he saw a "very
  beautiful" LASER-like beam of light bluish light, caused by the
  ionization of air, that appeared to flood up into infinity.

My questions are:

What amount of radioactive material is needed to make those flashes in the air happen?
Is radiation alone enough?
Are other factors such as weather, temperature or other prior chemical elements in the air needed to allow this flashing to happen?

Comments are not for extended discussion; this conversation has been moved to chat.

The white flashes seen in the video are the result of radioactive contamination of the original film. Airborne radioactive particles were deposited onto the film and caused a local overexposure of the negative, quite similar to the historic experiment of Antoine Henri Becquerel. That's why the flashes only appear in individual spots in individual frames of the film. (This is also explained in the documentary The Battle of Chernobyl, which also contains these film excerpts.)

There was no corresponding visual phenomenon in the air that could be directly observed by people in Pripyat.
The ambient dose rates measured in 26 locations in Pripyat on 26 April 1986 at noon (i.e. about the time when the film was recorded) were about 0.038–1.9 R/h (i.e. about 0.38–19 mGy/h; with a geometric mean of 2.4 mGy/h). Such dose rates cannot cause any directly visible effects in air.
The flashes on the film are not related to any visual phenomenon that was caused by nuclear reactions, radiation, fire or other chemical reactions at the open reactor during the night of the accident.


text:
Synthesis of an epoxide from 1,2-diol

Why is there no formation of epoxide when the two OH-groups are on different planes? I do understand the epoxidation mechanism. However, I would be grateful if someone cleared the specific requirements for epoxidation.

Read up on under what conditions SOCl2 reacts with ROH to give RCl. One set of conditions is SN2; another "internal return", i.e., no inversion.

In the above mechanism base is seen at the second step which is not the case

Thionyl chloride first reacts with the alcohol to form an alkyl chloro sulfite,which gives various stereochemical products as shown here.

Extending this to the current question , alkyl chloro sulfite 2 is formed.${Cl^-}$ attacks in a ${S_N^2}$ mode to give 3.
Base deprotonates 3 to give 4.
Alkoxide and chloride in 4 are in correct trans configuration for a ${S_N^2}$.This results in 5(B in your scheme).

If we take C in your question , and follow the scheme above, 9  is formed.In 9 alkoxide and chloride are not in correct configuration for ${S_N^2}$.Therefore
C in your question does not undergo epoxidation.

Detailed answer but the OP didn't make much of an effort. That is why I offered hints and no answer. Respectfully, I think we should refrain from quickly offering answers when little effort is made by the OP and hints are offered in the comments. Let the OP do some work.

@user55119 ,noted,the question was intreasting for an answer.


text:
Is sodium acrylate soluble in ethanol?
I am trying to make sodium acrylate by neutralizing acrylic acid with sodium hydroxide. I further want to photo-pattern it, which requires the addition of photoinitiator to sodium acrylate.
The problem is acrylic acid and photoinitiator are soluble in ethanol, but NaOH is not. Hence I'm looking for a common solvent for sodium acrylate and the photoinitiator.

Try sodium ethoxide or sodium tert-butoxide in EtOH as your base


text:
How can electron move between two metals without the presence of electric field?
Electron can move from one location to another if it feels a force that is if there is a presence of electric field. Now suppose I have a $\ce{Zn}$ rod immersed in $\ce{CuSO4}$ solution. Electrons are transferred from $\ce{Zn}$ to $\ce{Cu}$. What results in this motion of electrons from $\ce{Zn}$ to $\ce{Cu}$. How can electron move without a field?

Still, the question is legitimate. Electrons won't move without a field, will they?

You might want to check: https://chemistry.stackexchange.com/questions/23450/what-causes-electrons-to-move-from-zinc-to-copper?rq=1

Even better: Why do Electrons leave the Zinc in a Galvanic Cell

My interpretation of this question, which I will upvote, is that @GracemarsMars wants to know exactly what happens, and WHY it happens, when a cupric ion, in aqueous solution, gets reduced by a bulk piece of zinc metal. It obviously happens, and the thermodynamics and driving force are understood, but what is the step-by-step reaction mechanism? I do not know the answer and even doubt that people like Prof. Bard know (and I do not say this lightly). It is an extremely difficult thing to study. If I am wrong in this, I would absolutely love to see a posted answer explaining the mechanism!

@EdV I agree that there is nothing trivial about the problem, particularly if you dwell into mechanisms of electron transfer etc, but I think that the OP might not be familiar with the Nernst equation. After all, that provides a quantitative description of the correspondence between differences in the chemical potential of the electrons at the two electrodes and the resulting voltage. Not quite "QED" but I think if you want a really detailed explanation maybe you need to roll up the sleeves and start digging into more advanced literature. There is bound to be lots of case-specific variation.

I see the question as "how cupric ions moves to the Zn rod?". This does not need an electrochemical answer. Was it so?

The wording of the question hides the answer.
Electrons do not move from Zn to Cu. They move from neutral zinc to cupric ion Cu++. The field arises from the mix of H2O, H+, OH-, SO4--, and Cu++ all around the Zn rod. Could electrons be forced onto Zn from OH- or SO4-- ? No.There is no gain in stability; energy would be needed. Or could electrons be attracted to H+; yes, and even more so to Cu++. And the final products (Zn++ and neutral Cu) would be more stable.

How can electron move without a field?

There is a field, the field of two nuclei (shielded by some electrons, if you will). In the figure below, the top shows the potential energy (y-axis, arbitrary units) of a free electron in a homogeneous electric field (changing along the x-axis). It will move along the gradient, like in an old-style TV with a cathode ray tube.

If you treat the situation of an electron transfer in a pre-Planck way, it will look like the lower panel (again, the y-axis is the potential in arbitrary units, and the x-axis location). There are two infinitely deep wells in the potential energy landscape, and the electron would disappear in either of them.
Instead, quantum mechanics describes that the energy of the electron is quantized, and there are bound states near each nucleus. Using a one-electron approximation, if the highest occupied orbital of one atom/molecule/ion (species) has a higher energy than the lower unoccupied orbital of the other species, the electron will transfer "when given a chance" (the latter has to do with distance and other subtleties).
From which to which species electrons will transfer is described by the electrochemical series (or the reduction potential). If the two nucleic are sufficiently close, the electron can tunnel, so it can overcome a high barrier without input of energy. Rudolph A. Marcus received the 1992 Nobel prize for work on electron transfer in chemistry; you can read the Nobel lecture here.

I think there is a misconception in your answer about what QM does or does not provide you with. Much of what you encounter in basic QM treatments is a "classical" treatment of the potential (as opposed to relativistic), what you call "pre-Planck". What is new is the dual particle/wave description of particles. What QM does is do away with the exclusively particle description of matter (it also describes the photon as a particle with zero rest mass). So those singularities remain in a "static" (eg Born-Oppenheimer) description of the potentials.

I haven't read a good explanation of why particles should not dive head first into those singularities, only that they don't because of quantization, as you point out, where quantization is a property of the standing waves that define the solutions of the time-independent SE.

But good link, will have to read!


text:
Looking for a dye which emits around 680 nm
For a specific experiment, I'm looking for a dye which would emit around 680 nm. Preferably one I could buy easily commercially. Are there any databases which might give the emission spectra of specific dyes? Or any other methods to find dyes at around $\ce{680 nm}$.

I suggest searching for “laser dyes” online. There should be lots of them and they are listed by emission wavelength. Cresyl violet 670 perchlorate emits around 670 nm and nile blue 690 perchlorate around 690 nm. Several companies sell quality laser dyes and dyes for other fluorescence applications.

Note that many laser dyes have toxicity issues, so check the MSDS (SDS).

There are two dyes commonly used in biochemistry research with $\lambda_\mathrm{Ex}$ of around $\pu{650 nm}$. They are Alexa Fluor 647 from ThermoFisher and Cyanine5 (Cy5) from Lumiprobe. Two example of these two categories are illustrated in the following figure. Either probe can be ordered according to the way you want to conjugate to your substrate (azide and alkyne groups are shown here). For example, IgG conjugate of Alexa Fluor 647 with amide linkage is shown in the insert.

I believe there is an Alexa Fluor 680 available.

@evildemonic: Yes, it is. Actually, Alexa Fluor 660 has the requested $\lambda_\mathrm{ex}$. However, I realized that after I made the slide so I thought it wouldn't heart $\pu{15 nm}$ difference by my experience. :-)


text:
Are there any gases that can not be turned into liquid, or liquids that can not be frozen solid?
My son points out that some solids can not be melted (because they combust, for example). He wonders if there are any liquids that can not be frozen, or gases that can not be liquified.

Related:   https://chemistry.stackexchange.com/questions/29729/why-cant-helium-be-solidified-at-ordinary-pressures.

Some liquids separate on freezing. For example, Sn/Pb solder, other than the eutectic alloy, form separate phases of tin-in-eutectic or lead-in-eutectic on cooling. Thus, on freezing, the alloy changes from a fully-dispersed soluton to separate components. See http://www.uefap.com/reading/exercise/ess3/alex2.htm

Related: https://chemistry.stackexchange.com/questions/37413/sublimation-of-iodine/37416#37416

All gases can be compressed into a condensed phase, assuming no chemical reaction occurs, but it might be interesting to point out that for Helium (to my knowledge uniquely) just reducing the temperature is not sufficient, pressure is required as well - https://en.wikipedia.org/wiki/Liquid_helium


text:
Question about calculating enthalpy change
$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)$$
$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)$$
Both $\Delta H$'s given by the above formulas should be equivalent.
However, I found one special case that the above 2 formulas give 2 different results.
$$\ce{2 H+ + 2 e- → H2}$$
$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)= 0-0=0$$
$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)<0$$ because there is only bond formation, no bond break.
How to explain this difference? Can I say the heat is released during this process, because $$ \Delta H $$ is smaller than 0?

Where in the calculation is $\Delta H_f$ of the electron?

I think the $$ \Delta Hf $$ of the electron should be zero

The approach to estimating the enthalpy change in a reaction by using bond dissociation energies assumes homolytic cleavage of the bonds: each atom gets one electron from the covalent bond.
The bond dissociation energy of the $\ce{H-H}$ bond is reported as 436 kJ/mol.  The relevant reaction is:
$$\ce{H2(g) -> 2H.(g)}$$
Note that this reaction is not the same as $\ce{H2 -> 2H+ + 2e-}$. The homolytic cleavage   reaction produces hydrogen atoms, not hydrogen ions. If we look up the standard enthalpy of formation for hydrogen atoms, it is +218 kJ/mol. Since the standard enthalpy of formation for $\ce{H2}$ is defined as zero, using the enthalpies of formation we also get $\Delta H = +436$ kJ/mol for this reaction.
It is generally hard to use bond dissociation energies for heterolytic reactions (reactions that produce ions from molecules).


text:
How pure gold is obtained back from dissolved state in Aqua regia?
I know Aqua regia is used for purifying gold by dissolving gold in it, but how that dissolved gold is obtained back in the pure state?
I searched everywhere, but I didn't get the answer. So, I am asking here.

Pretty much anything can reduce gold.

The Niels Bohr's story has already been mentioned several times in related questions(1, 2); and there is nothing special about reducing gold, there are numerous agents, both organic and inorganic (e.g. citric acid or tin(II) chloride) that can be utilized. The choice is also dictated by the post-refining process ("Affinage"), if any.

Gold is not really purified by dissolving it into aqua regia.  Most base metals dissolve in aqua regia, too.  The actual purification comes from choosing a reducing agent that acts on the gold but leaves the less reducible base metal ions in solution.

There is a historic anecdote about this question. De Hevesy dissolved his gold Noble prize medals in aqua regia during World war II. After the war, he found the solution undisturbed and precipitate d back his gold medals from the solution without any loss in weight. See: https://en.m.wikipedia.org/wiki/George_de_Hevesy

By electrolysis. In general electrolysis is a very common way of purifying metals if you have them dissolved, since you can selectively control the reaction by adjusting the potential current. Al and Cu are also obtained through electrolysis of their salt solutions.
Electrolysis is a redox reaction that occurs in the non spontaneous direction driven by a external potential. In electrolysis we have two electrodes (a cathode and an anode) submerged in a solution with the salt of the metal we want to purify. As it always happens in electrochemistry, the reduction reaction will occur in the cathode, while the oxidation in the anode.
Since in a gold solution obtained through Aqua Regia we have the gold forming a tetrachloride complex (AuCl4), the oxidation reaction we will have will be the production of chlorine gas from the chloride. So the global redox reaction that will be occurring will be:
$$\ce{2[AuCl4]- + 6e- -> 2Au + 8Cl-}$$
$$\ce{6Cl- -> 3Cl2 + 6e-}$$
And if you look at the standard potentials of the two reactions in this Wikipedia webpage you can see that you just need a 0.43 V external potential for the reaction to occur. The negative electrode should be made out of gold and it will grow during the process by the deposit.
Industrially this process is done by having hundreds of electrodes connected in series and submerged into the gold salt solution. In this image you can see a factory for Cupper purification.

You can see from the electrodes lifted by the crate that they have copper lamina that grow during the process. Electrolysis is normally used when you want very high purity in your product, like gold and copper destined for electric applications in which any oxygen impurities greatly decrease conductivity.
Another great thing about electrolysis is that it can be used to easily apply a thin layer of the metal over a piece, if the piece is conductive itself. This is how gold coated audio jacks are produced, by using the audio jack adapter as the cathode.

You say:

I searched everywhere, but I didn't get the answer.

When I type youtube gold recovery from chloroauric acid into Google, the first link is to a Youtube video with a visible demonstration of converting chloroauric acid to gold.
The video shows that metabisulfite can reduce chloroaurate to form tiny tiny grains of gold powder, as a reddish dust.  A chemical reaction for this might be:
$$\ce{4 HAuCl4(aq) + 3 Na2S2O5(aq) +9H2O -> 4 Au(s) + 6 NaHSO4 + 16 HCl(aq)}$$


text:
Complete Combustion of Inorganic, Hydrogen Containing Covalent Compounds in the Atmosphere
From what I've seen in alkanes, their complete combustion creates carbon dioxide, which has carbon in its highest possible oxidation state.  I've been wondering what complete combustion would look like in similar compounds, such as ammonia, phosphine, hydrogen sulfide, and silane.
After looking at through several search results of The combustion of <compound> and <compound>'s reaction with oxygen, I was surprised to only see that the search results for silane and phosphine generally showed that the combustion products attained their maximum oxidation state.
For example, the general result for ammonia seems to be nitrogen monoxide or nitrogen dioxide, which make sense as products for incomplete combustion, but I would think that a complete combustion would produce dinitrogen pentoxide.  Hydrogen sulfide's combustion also seems to form sulfur dioxide or disulfur dioxide rather than sulfur trioxide.
This really confused me, since I was expecting the main products of all these reactions to be fully oxidized nonmetallic oxides.  My question is, how can one predict what the fully combusted products of an inorganic, hydrogen containing compound look like?

Links for my ammonia query:
http://www.questionsolutions.com/ammonia-gas-reacts-molecular-oxygen-gas/
https://answers.yahoo.com/question/index?qid=20150615223643AArPdtu
https://answers.yahoo.com/question/index?qid=20080504104957AAgQrL3
Links for my hydrogen sulfide query:
https://chemiday.com/en/reaction/3-1-0-151
https://www.sciencedirect.com/science/article/pii/S0082078473800463

The "completeness" of combustion depends on the proportion of the reactants, and on the temperature cycle, not only the peak temperature, but the rate at which the products are quenched. For example, using ammonia and air as fuel in internal combustion engines produces mostly $\ce{H2O}$ and $\ce{N2}$, with less than 0.1% $\ce{NO_x}$ according to Kobayashi, et. al. and Zacharakis-Jutz.
The greatest oxidation state of a substance is often not thermodynamically stable. Perchlorates, chlorine dioxide , $\ce{H2O2}$ and similar compounds are somewhat unstable (i.e., BANG) and are not generally released in great quantities by combustion.


text:
Where does the protons and electrons go in this redox reaction?

In the above oxidation reaction, does the 2 protons and electrons released get accepted by oxygen molecules? If not, where do they go?
Appreciate all answers :)

Yes, they get accepted by oxygen molecule (or any other oxidant, for that matter). Where else could they go, really?

Even the protons? Because I learnt that only electrons are exchanged between molecules in a redox reaction.

It depends on the oxidant.

In my school's experiment, orange juice is exposed to oxygen. So oxygen?

That's right.$;$

So oxygen does accept the 2 protons?

Yes, and the electrons too. That's how it turns into H2O.

See this: https://chemistry.stackexchange.com/a/116242/79678


text:
Is there any book where specific heat capacities are given?
I want to find the value of the specific heat capacity of aluminium nitrate nonahydrate, in order to calculate mixing effects in a chemical reaction. I am looking for a source, such as a book or link to an online PDF, which would be appreciated.

Does you reaction occur in water solution, by any chance?

Yes, the entire reaction is taking place in water at 60 deg C.

Then you don't really need what you ask about, and that for two reasons: because your specific compound does not exist in water solution, and because its heat capacity is negligible compared to that of water.

There are several databases of thermochemical properties, a number of them being listed on this Wikipedia page (at the bottom). Most of them are not freely available online, but require subscription. I've searched online but couldn't find that information freely available.
Another possible way to get the information would be through a search of the published literature (scientific papers). You can, for example, find this information on magnesium nitrate in a 1944 paper in J. Am. Chem. Soc.. However, I had no such luck with Al(NO3)3 · 9 H2O .

Clarification: the answer above is for the crystalline compound aluminium nitrate nonahydrate, which as @Ivan stated, may not be what you actually need for your calculation if the reaction is in water.

As a first step, there's also th Dulong-Petit approximation that molar heat capacity of solids is ≈ 3 R which should be OK at room temperature. Of course reality doesn't behave ideally, but it should be good for getting the order of magnitude.


text:
How is Mn+7 in the Permanganate ion considered as momentarily reduced to +6 in LMCT?
I saw this passage on a book and it says that Ligand to Metal Charge Transfer (LMCT) overpowers d-d transitions thus the permanganate ion has an intense purple color. However, it was mentioned that the Mn+7 ion was momentarily reduced to Mn+6 in the charge transfer, how is that?


text:
What causes oxygen cylinders to explode when in contact with oil (oiled cloth)?
Perhaps I did not put it correctly, but from the welders I know that if you leave the oiled rag on the cylinder valve, it can explode. At least, I saw how greasy oily rags that just lay in the trash can are smoldering. It is interesting to me to understand the principle of this phenomenon: whether it depends on the amount of oil, oxygen, pressure in the cylinder. If possible, I would like to see the chemical reactions occurring.

What kind of oil are we talking about? What application for the rag?

@BuckThorn We are talking about some oil(maybe machine), and rag for cleaning.

Oxygen rated valves for gas delivery are special rated, including no oils and no plastics. Using the wrong (not oxygen rated) part in an oxygen system can and has resulted in explosions. 100% pure oxygen is much better at getting things to burn than air is.

Summary
⤑ Oils are mostly made up of Fatty Acids ("FAs")
⤑ FAs oxidize and release a lot of energy
⤑ The reaction is proportional to the O2 content of the gas in contact with FAs
⤑ The reaction is also proportional to the contact surface area between FAs & surrounding gas
⤑ Rags offer a high contact surface area between the liquid soaked into them and the surrounding atmosphere
⤑ Maybe that increasing the surface area between air and oil (via rags) also increases volatile FAs which yields an extreme increase in contact area between FAs & O2
Tentative Explanation
I think that oils are made up of mon-/di-/tri-glycerides (at least natural oils): One glycerine bound to 1, 2 or (most often) 3 fatty acid molecules. Fatty acid chains can be oxidised, using up O2 and releasing water + a lot of energy (that's part of our metabolism, and yields much more energy per gram than the metabolism of carbohydrates). I think that the less saturated a fatty acid chain is, the more readily it can be oxidised, which is why linseed oil, which has a very high Alpha-linolenic acid content (a polyunsaturated fatty acid), can undergo spontaneous combustion. A rag exposes oil to a high contact surface with the surrounding atmosphere. Now if in addition to all of that the atmosphere is high in O2, you have perfect conditions for a high-energy-yielding reaction to massively occur.
The local temperature, oil-to-air surface area, air pressure, and air O2 content are the factors which most affect the likeliness of spontaneous combustion. The amount of oil will mostly affect how much of the reaction can occur / how long it can occur (it's the fuel in the reaction). It has a minor impact on the reaction speed only proportional to the fact that there has to be some correlation between oil volume and contact surface area.


text:
Is methanol optically active?
Strictly on theoretical basis, not based on experimental results.
I can very well be wrong, but I think that methanol should be
optically inactive because it has a plane of symmetry passing through it.
My teacher however insists that it is optically active as you can create two non-superimposable mirror images out of it.
Who is correct and why?

You can create two STATIC images of methanol that are non-superimposable mirror images but that does not make methanol optically active. Every conformation has a mirror image, i.e., a racemate, or a plane of symmetry. Both conditions lead to a lack of optical activity. Consider changing classes.

The -OH group is pretty much free to rotate around. If not for this fact, your teacher would be wrong just once.

Is your teacher an engineer?

Thanks @user55119 for your answer. I however for some reason can't tick your answer because I found it helpful.

Just to sum up what has already been pointed out in the comments: Methanol can have infinite conformations in free space due to free rotation of the C-H and C-OH sigma bonds. Hence, for any particular spatial conformation you can think of for the molecule, it can always develop a perfect mirror image for the same almost instantly(in case you are wondering about some time lag between them) as conformers are in chemical equilibrium.
So what this essentially means is that, suppose some conformation A of methanol has polarized light in some direction (assuming the right-handed coordinate system, let's say the positive X direction) in some fashion(let's say circularly polarized, clockwise). Then, this particular interference pattern of light will be internally compensated by a conformation B of the molecule which will circularly polarize the light in the anticlockwise direction in the negative X direction. So, the emergent light ray will appear to be unpolarized for the observer


text:
Arrhenius equation using concentration
I've been conducting some experiments to determine the temperature dependence of a reaction. To do this, I've been conducting the experiments at different temperatures and generated a graph of $\ln{[\ce{X}]}$ (where $[\ce{X}]$ is the concentration of the product) vs $T^{-1}$. I was told in a passing comment that because the reaction is first order the gradient of this graph is equal to $E_\mathrm a/R$ in the Arrhenius equation
$$
\ln{k} = \ln{A} - \frac{E_\mathrm a}{RT},
$$
but I don't understand why I can substitute $\ln k$ for $\ln [\ce{X}]$.
Any help is much appreciated.

You should get back to the person who told you in passing, and ask for a complete explanation. You say [X] is the concentration of product - but at what time? If [X] is zero at t = 0, and you measure [X] after the same time has passed for all reactions, a larger [X] means a larger k. Whether k is proportional to [X] does depend on the rate law. Could you edit your question to clarify how you did the experiment (initial conditions) and when you measured the product concentration? Otherwise, giving a good answer is impossible.

Zero order, perhaps, or pseudo zero order?

Normally one would plot $\ln(X/X_0) $ vs time ($t$) (for a first order reaction and $X_0$ being initial amount ) and obtain $k$ at the prevailing temperature then plot $k$ vs $1/T$, so, are you sure you have'n't got $t$ and $T$ muddled?


text:
Electronic States of the (unstable) tetrahedral CrCl4
I was probing for first order Jahn-Teller instability in the coordination complex CrCl4 (2 electrons in E set). Taking the direct product yields the states A1 + A2 + E, so I took the symmetric and antisymmetric direct products too to see which states represented a singlet and triplet configurations.
To my surprise, the symmetric direct product yields 2A1 + E while the antisymmetric product yields A2 -A1. I believe that I probably made a mistake somewhere (but can’t find it), so did I make a mistake somewhere in my working, or is there an actual meaning behind a negative irreducible representation?
Thanks!

have you tried to use a calculator for this? A negative irrep shouldn't be possible. Using a calculator show whether you made a mistake in determining irreps or counting symmetry elements.

I’ve tried it for point group Td, and it really does result in -A1.

than that suggests to me that you have something wrong with how you counted symm elements. As far as I'm aware, for anything in chemistry having negative irreps is unfeasible.

Could you try and evaluate the direct product when you have the time? I feel like I probably got too rusty and made a mistake somewhere, but can’t find it.

Based on this direct product table the antisymmetric part should be A2.

There’s actually nothing there for E x E for Td point group, though.

that's my bad for missing it. I tried to send one with the table easily accessible but didn't see that it was printed poorly. The top row of table 2 is mislabeled, it should be the same order as the first column. But otherwise the table is correct. What you want is labeled B2 x E, but it is actually the correct direct product for E x E. I would have redirected you to the Atkins Direct product table where this is displayed correctly, but I always have trouble linking PDFs from mobile.