text:
Two ways to calculate dissolved oxygen
To calculate the dissolved oxygen in water at saturation pressure at 25 °C, we do:
$$
D_\ce{O2} = K_\mathrm{H} \times M \times p_\mathrm{g}
$$
$$
D_\ce{O2}~[\pu{g/L}]= 1.26\times10^{-3}~[\pu{mol/L/atm}] \times 32~[\pu{g/mol}] \times 0.21~[\pu{atm}]
$$
I used $\pu{1.26E-3 mol/L/atm}$ for the Henry's constant for $\ce{O2}$ in water at 25C which is from Warneck and Williams (2012), which was in turn taken from Battino (1981). This returns a dissolved oxygen of $\pu{8.47 mg/L}$.
Benson (1980) reports dissolved oxygen for "a body of pure water is in thermodynamic equilibrium with an atmosphere of standard composition saturated with water vapor at a total pressure $P$" as 8.26 mg/L which is a difference of 2.5 %. The Benson paper suggests an error of around 0.02 %. Why is there a difference in these two methods? Is it due to different assumptions being made? Or perhaps rounding errors in how data is presented?

To me, 8.26 vs 8.47 is no difference at all, and bold claims about 0.02% should be taken cum grano salis.

The 21% figure is for dry air. Benson assumes air is saturated with water vapor.

In either case it is a case of the grain of salt when the numbers used have a 1e-3 resolution and the precision is reported as "around" 2e-4. I'd call BS right then and there (unless OP has truncated)

The 21.0% of oxygen by volume in air is from dry air. Data for dry air from webpage of Columbia University's Department of Earth and Environmental Sciences.
Note that the "% by volume"/100 for dry air also corresponds to the partial pressure in atmospheres. For air saturated with water at 25 C, the vapor pressure of water, according to Wikipedia, is 0.0313 atm. The second column is calculated by multiplying all dry air values by (1-0.0313)=0.9687.
$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Gas} &\hspace{-5em}\text{dry air}&\ce{H2O}\text{ sat. air} \\
 & \text{% by volume} & \text{% by volume} \\
\hline
\ce{N2} &  78.084    & 75.640 \\
\ce{O2}  & 20.947    & 20.291 \\
\ce{H2O} & 0.0      &  3.13 \\
\ce{Ar}  &  0.934    &  0.905 \\
\ce{CO2} & 0.035    &  0.034 \\
\ce{Ne}  &  0.001818 &  0.001761 \\
\ce{He}  &  0.000524 & 0.000508 \\
\ce{CH4} &  0.00017  &  0.00016 \\
\ce{Kr}  &  0.000114 &  0.000110 \\
\ce{H2}  &  0.000053 & 0.000053 \\
\ce{N2O} & 0.000031 &  0.000030 \\
\ce{Xe}  &  0.0000087&  0.0000084 \\
\hline
\text{TOTAL}& 100.0027187 & 100.0026304 \\
\hline
\end{array}
$$
Now correcting the data from Warneck and Williams (evidently Warneck,  P.  and  Williams,  J.:  The  Atmospheric  Chemist’s  Companion:  Numerical  Data  for  Use  in  the  Atmospheric  Sciences, Springer Verlag, 2012) for the value of water saturated air
$\ce{O2} = \dfrac{20.291}{21}\times 8.47 = 8.18$ mg/L
This is much better agreement with the Benson (1980) data.
The OP's value of 1.26e-3 Mol/L/atm for the Henry's constant for O2 in water at 25C was evidently take from Table 8.23 of Warneck and Williams. The value also has a list error bracket of +/- 0.02.
$$\dfrac{0.02}{1.26}\times 100\text{%} = 1.6 \text{ %}$$
For the two dissolved oxygen values:
$$\dfrac{8.26-8.18}{8.26}\times100\text{ %} = -0.97 \text{ %}$$
so the values are within the error limits for Henry's constant.


text:
Self tan mousse cannot contain UV, right?
So, I bought this self tanner specificly to avoid UV radiation. Back home I noticed that the bottle says it contains ultra violet: "The inclusion of Ultra Violet fluorescence in the base colour" . I guess they meant it contains something fluorescent but what could it be and is it dangerous? (sorry about my bad English)
The product is Sunkissed Ultra Dark Selfie Ready Tan.

That's right, it can't contain UV, because UV light is not a thing you can put in a bottle. The rest I'd rather dismiss as some marketing gibberish akin to this:

(sold back in the days when radium was all the hype, this thing is probably less awful than it seems, because it could not realistically have contained more than a few atoms of radium.)

Then again, chances are that the gibberish bears some resemblance to actual facts; so what? Fluorescence is just the ability to absorb and re-emit light. Whatever its composition may be, this thing would not shine UV on you all by itself.
As to the possible dangers, you either trust the regulatory body acting in your jurisdiction, or you don't. At least these guys supposedly did test the product, which I didn't.

I looked for the whole product description and it is highly misleading.
It also states:

...and the inclusion of Ultra Violet works by reducing the bounce back of the camera lights so that your tan looks darker in the photos.

Either there is a chemical in there that's called "Ultra Violet" or it just makes no sense. I suppose the latter. UV is a radiation that cannot be "stored" and even if it would never reduce the reflecion or "bounce back" of light.
For the facts:
Flourescense would mean that there is a chemical in the tanner that absorbs light and radiates it again (just like the self glowing clock faces on watches). This would mean that you would glow in the darkness with this stuff. But this effect would not protect you from UV radiation and you can certainly not "store" it.
In general self tanners are glorified coloring agents in a fatty matrix, in the best case with a sun protection agent. The product description is utter nonsense. Whoever made it has not even a tiny bit of chemical knowledge.
For the product: I can't say if it's dangerous. I would use a different product just because of the misleading description and use a tanner from a local drug store.

Yeah, this very much sounds like it was written by someone who doesn't know what UV or fluorescence means.
I contacted the company via their website and asked for clarification:

Dear Sunkissed,
  I read about your product, "Sunkissed Ultra Dark Selfie Ready Tan," and I read that it "contains Ultra Violet fluorescence in the base colour." A customer of yours was worried that UV might be dangerous. I am a chemistry teacher, and I could assure them that there was no "UV" in the cream. I am very curious what it is that your cream contains though. You say,
  "The inclusion of Ultra Violet fluorescence in the base colour helps to intensify your colour under camera lights," and "the inclusion of Ultra Violet works by reducing the bounce back of the camera lights so your tan appears darker in photos." I am finding that hard to parse. If there is a fluorescent dye in the cream, the person using it should appear brighter in UV light (like a disco, or sunlight), but camera lights don't contain much UV. "Reducing the bounce back of the camera lights" sounds like the cream absorbs visible light, i.e. it isn't white or transparent, but that has nothing to do with fluorescence.
  I would greatly appreciate it if you could provide further information on what it is that this cream contains and how it is meant to work.
  Thank you, and have a great day,
  Martin

I'll let you know if I get a response.
Feel free to not buy this stuff again, and to leave a comment with the company or with the person hawking this cream on Youtube. Trying to make money by confusing people with $20 words is wrong.
Also: if your goal is to "avoid UV radiation," the product you want to get is sunscreen. (Or an umbrella, or a hat, or a shirt.)

No response from Sunkissed.

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text:
Double bonds in polyunsaturated fatty acids
$\omega - 3$ fatty acids are very frequently discussed.  $\omega - 6$ is also mentioned.  However, I don't recall mention of $\omega - 1$, $\omega - 2$, or $\omega - 4$.  Also, in polyunsaturated fatty acids, the double bonds are frequently all multiples of $3$ from the $\omega$ end.
Is it just some quirk of evolution that has led to this pattern or is there a chemical explanation?  For example, are double bonds at these positions more favourable than other positions?  If an $\omega - 2$ or $\omega - 4$ fatty acid was synthesised might it isomerise to $\omega - 3$?
In the cases with multiple double bonds 3 apart, is it really: pure single, pure single, pure double repeating?  Or is there some degree of delocalisation and the "single" bonds are actually a bit more than single and the "double" bond a bit less than double?  I guess that it does not go to the extreme of benzene in which all the bonds are equal or we would have developed more accurate names and notations to represent this.

There is no delocalisation here. However, if the bonds were 2 apart, then there would be some, and that would change pretty much everything. You'd have to develop a whole new set of enzymes to work with those.

@IvanNeretin Thanks, so the standard notation is accurate and not like the old alternating single and double bond diagram for benzene.  If delocalisation cannot occur at this level of separation then I guess that it partially answers my question.  The double bonds could not be closer without a very significant change in properties.

That's exactly right.

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text:
Cobalt Chloride in various solvent/water mixtures - tested
In a recent post the solvatochromic behaviour of cobalt chloride in acetone was discussed. There were some ideas how the acetone might influence the ratio between the blue $\ce{[CoCl4]^2-}$ and the pink $\ce{[Co(H2O)6]^2+}$.
An interesting thought was the change in the dielectricity constant, which would influence the dissociation of $\ce{CoCl2}$ and therefore, if lowered, cause the $\ce{Co^2+}$ and $\ce{Cl^-}$ ions to form the blue $\ce{[CoCl4]^2-}$ complex again. Atleast this is how I understood the explanation.
Using this idea any solvent, that is miscible with water and does not form a differently colored complex with $\ce{CoCl2}$ should have a similar effect if the dielectricity constant is close or lower than the one for acetone.
So I gave it a test and used the first solvents I could find in our lab, dmf, acetone, isopropanol and thf (in descending order of permittivity).
The solutions were prepared by dissolving a spatula of $\ce{CoCl2 * 6 H2O}$ in a few ml of solvent so it was fully saturated. The liquid layer was transfered to a test tube and diluted with water until the color had changed to pink.
Solvent was then added until a blue hue was visible again. To limit the amount of solvent used it was not added until a strong blue color appeared.
For the acetone you can see the best effect. It quickly turned blue again and upon adding more acetone even became quite intense.
Dmf had a much lighter solution to begin with, the blue complex was formed quite quickly but it was not as intense as for acetone.
Isopropanol gave only a poor result. A few drops of water were enough to completely discolor the solution (still more drops were added to have about equal volumes in all tubes). Even after adding a large amount of isopropanol no intense blue color showed.
THF gave good results when the layers were not allowed to mix but once the solution was stirred it quickly discolored again.
As you can see all solvents gave quite intense blue colors with the pure $\ce{CoCl2}$ but even though THF has a much lower dielectricity constant it only gave poor results. I don't really know if this can be explained by the given theory but perhaps someone can find some clues through this experiment.

What is that actual question?

Have you fully mixed your solutions? If not the final colour may be misleading. Also you seem to have added different amounts of solid hexachloride to solution and various amounts of solvent. So as not to be misled it would be best  if you used controlled amounts and then compared results, preferably by recording absorption spectra if a spectrophotometer is available of at a few wavelengths by a colorimeter if not.

It is a pity the solvents have not always been really mixed. The final mixtures are too often pink in the lower part and bluish in the upper part. It would have been advisable to transfer the final mixtures in another tube, so that the liquid becomes homogenous.


text:
Why are the sulfide ores generally converted to an oxide and then reduced?
In the elementary extraction of metals from their concentrated ores, why is the ore generally converted to an oxide, and then reduced? Why is the reduction of these metal oxides easier than the other corresponding compounds, say sulphides or carbonates.
Here are some of the elementary metals that are converted to their oxide before being reduced:

Zinc blende (sphalerite) is roasted from $\ce{ZnS}$ to $\ce{ZnO}$
Galena $\ce{(PbS)}$ is converted to $\ce{PbO}$ (Wikipedia)

Some other ores originally in their oxides remain as they were, like bauxite $\ce{(Al2O3.2H2O)}$ and hematite $\ce{(Fe2O3)}$.
An exception happens to be copper pyrite which is roasted to form copper sulphide instead of copper oxide (Wikipedia).
So, for the metals $\ce{Al, Zn, Fe}$ and $\ce{Pb}$, why is extracting the metal from the oxide favorable? And why is copper an exception?

In my book, NCERT (12) India, it is given that:

The concentrated ore must be converted into a form which is suitable
  for reduction. Usually the sulphide ore is converted to oxide before
  reduction. Oxides are easier to reduce. Thus
  isolation of metals from concentrated ore involves two major steps viz.,
  (a) conversion to oxide, and (b) reduction of the oxide to metal.

The statement in your question oversimplifies metals production from ores. For example ; Ni is mostly converted from sulfide , and has been converted to a gas ( Ni C O  - Mond process) , don't know if this is still done. Some ores are converted to oxides to aid in flotation separation , not to permit reduction with carbon ( as the primary  purpose).

Why, it's simple: you heat a metal oxide with carbon, and you get the metal you were after, plus a byproduct of $\ce{CO}$ or $\ce{CO2}$ which easily flies away. Coal is still abundant on our planet.
Try to pull that with a sulfide! I guess the reaction will not go in the first place, since $\ce{CS2}$ is a great deal less exothermic(*) than $\ce{CO2}$. In the unlikely case if it will, you'll have to deal with the byproduct. $\ce{CO2}$ can be simply dumped in the air (though this is no longer free). Dump $\ce{CS2}$, and all your neighbors will pay you an extremely unfriendly visit, possibly armed with pitchforks.

(*) So much so as to be endothermic, in fact.

Copper is less active, so instead of going via oxide, we have a chance to go straight to the metal in one step. Would be stupid not to use it.
Then there is aluminum which is not reduced like that, but then again, we don't convert it from sulfide to oxide either, primarily because we don't have much sulfide around. Should it be otherwise, it isn't immediately clear would we choose to convert it or not. Arguably, $\ce{Al}$ has no form which is suitable for reduction.
So it goes.

what about the heat of formation of sulfides and oxides? Oxides have more negative heat of formation than sulfides.

Yes, that too.$,$


text:
Is there a formula for density of any gaseous substance just based on its molar mass?

$\pu{112 ml}$ of hydrogen combines with $\pu{56 mL}$ of oxygen to form water. When $\pu{224 mL}$ of hydrogen is passed over heated cupric oxide, the cupric oxide loses $\pu{0.160 g}$ of its weight. All volumes are measured at STP. Show that the result agrees with the law of constant composition.

Although I solved the question, I came across a very interesting step in the procedure of doing so:

Weight of $\pu{112 mL}$ of $\ce{H2}$ at STP is calculated as $\frac{112 \times 2}{22400}$

Does this mean density of a gaseous substance at STP can be given as
$$\text{density} =  \dfrac{\text{Mass}}{\text{Volume}} = \dfrac{\dfrac{\text{Volume(mL)}\times\text{Molar Mass}}{22400}}{\text{Volume(mL)}} = \dfrac{\text{Molar Mass}}{22400} $$
Can we say that this holds true?

There have been different values for STP in the past. Currently at STP the molar volume is 22.7 liters.

Yes...Just knowing T, P and its identity, you can derive a density formula. [for an ideal gas]

This is true for ideal gases at standard temperature and pressure. The Volume of an ideal gas is 22.4 L/mol at 298.15 K and 1 bar.
You can check this yourself by rearranging the ideal gas equation and solving for the molar volume at STP.
Edit: Make sure you always use correct units, preferably SI in your equations. And use them in all equations.


text:
Find alloy composition by density
I have a square piece of metal from a processor heat sink that looks a lot like copper. I'm trying to find the alloy composition but I'm not sure how.
I happen to have both a digital caliper and a milligram-precision scale. The piece has a rectangular shape so volume wasn't hard to find. The volume is 5.378 cm³ and the mass is 52.395 g, that gives a density of 9,27 g/cm³. The density of copper is 8.96, nickel is 8.91, zinc is 7.14, lead is 11.34.
If I use this formula (which may be wrong):
$$p = a \cdot p_a + b \cdot p_b$$
$a$ being the volume fraction and $p$ the density. The density is higher than that of copper, this means there has to be a metal with a density higher than copper in the alloy, maybe lead? Using the formula I get 87 % copper and 13 % lead by mass, looks a bit too much.
Is this a good way to find the alloy or an alloy doesn't have a density or volume proportional to that of its components?

(1) Densities aren't additive like that (i.e. following ideal behavior) except in small ranges or unusual cases. (2) You have to consider the error in your measurement which was crude. (3) Think the most common compositions for electronic  heatsinks would be copper or aluminum alloys.

@MaxW Alright that makes sense. Would an alloy made of denser metals (eg: with lead) always be denser than one made of a less dense one (eg: aluminum) if mixed in the same volumes?

You'd have to check the phase diagram for the alloys in question. Superlatives like always and never make me wonder. There always seems to be an exception.

Besides, not every pair of metals forms an alloy. Cu/Pb, for one, does not.

No, it is not a good way to find the alloy. The best you can do is dissolve a piece of the metal and analyze the cations. It is also not a good way to find the density. You should use the method of Archimedes: submerge the piece in liquid and measure the volume increase. The density is probably wrong. Most alloys including SS are below 9. The closest I have found is Hastelloy with 9.245, but which is obviously not your material: does not look like copper, has a low heat conductivity (10 W/m.K against 400 of Cu and 200 of Al) and is very expensive.


text:
What happens to the methanol formed by the hydrolysis of methyl 2-hydroxybenzoate (oil of wintergreen) when synthesising salicylic acid?
Here are the reaction equations for reference (you may recognise this as a preliminary step for the synthesis of aspirin).
Steps for synthesis of salicylic acid:
1) methyl 2-hydroxybenzoate + sodium hydroxide -> sodium 2-hydroxybenzoate + methanol
2) sodium 2-hydroxybenzoate + hydrochloric acid -> salicylic acid + sodium chloride
Does the methanol play an underlying role in the reaction somewhere? Is the methanol filtered off? I do not know.
Thanks in advance.

The methanol plays no underlying role in the reaction
Without more detail of the reaction procedure I cannot give a complete answer, but my expectation is that after the HCl addition the reaction will be quenched by addition of water to either precipitate the salicylic acid or then be extracted into an immiscible organic solvent. The methanol will stay in the aqueous phase in either case.

Reaction mixture is heated under reflux and then allowed to cool; then the mixture is filtered under reduced pressure. Your answer supports this, thank you.

I presume that the HCl is added after the refluxed mixture is allowed to cool. If you are happy with the answer, please accept it


text:
Formation of oximes by ketones and aldehydes
Why do symmetric ketones form a single oxime and asymmetric aldehydes and ketones can form two isomeric oximes?

Because geometric isomerism.

An oxime is formed by the reaction of hydroxylamine (NH2OH) with either an aldehyde or ketone. The oxime nitrogen has a double bond (sigma and pi) to the oxime carbon, a lone pair and a single (sigma) bond to the hydroxyl oxygen. If we use valence shell electron pair repulsion (VSEPR) theory to deal with the problem.
Then the oxime nitrogen will be trigonal planar.
If the two groups on a ketone are identical then the molecule will be the same which ever way around we put the lone pair and the hydroxyl. But if the two carbon groups are different then we can have two isomers.
Consider for a moment the 2-hydroxyoximes used as LIX extraction agents for copper. There is a paper from the 1970s on the separation of the isomers of LIX 63 (T.T.Tammi, Hydrometallurgy, 1977, Volume 2, Issue 4 (October), Pages 371-380) which you migh want to read.
To make it more clear we have diagrams here of the two isomers of LIX63.

Good answer and +1 from me.  Might be worth noting that hydroxylamine_s_ react as a class -- the reaction happens w/ more than just $\ce{NH2OH}$.  O-methyl hydroxylamine gives O-substituted oximes, etc.


text:
reaction between semicarbazide and a mixture of cyclohexanone and benzaldehyde
Why is cyclohexane-semicarbazide precipitated, but after a while the precipitate is benzaldehyde-semicarbazide?

In what solvent is the reaction conducted?

Probably kinetic and thermodynamic control


text:
Why carbonyl groups are strong benzene deactivating group for electrophillic aromatic substitution?

In carbonyl group there is a resonance avaibility and also due to inductive effect it can also withdraw electron from benzene ring . As, resonance(e- donating chracter) dominates than Inductive effect , it must be a Benzene activating group.
but accorting to  L.G Wade carbony groups are strong benzene deactivating group for Electrophillic Aromatic substitution
Where am I wrong ? Explain.

Activating for nucleophilic substitution and deactivating for electrophilic aromatic substitution.

@Archer I think you got it backwards. Activating is electron-donating, and deactivating is electron-withdrawing.

A carbonyl group such as an aldehyde is a deactivating group as it withdraws electron density by both the inductive effect (through the sigma bonds) and the resonance effect which involves pi systems.
Here are three resonance forms which show how the aldehyde group takes electron density from the benzene ring in benzaldehyde. There are more resonance forms and some different but related ones account for the reduction in the stability of the cationic intermidate in an aromatic substitution reaction.


text:
Why is bromobenzene more stable than benzene?
Benzene is 45 kJ/mol more energetic than bromobenzene. I understand that due to resonance in bromine to benzene in bromobenzene increase the stability but such an enormous amount of energy can't only be dealt by resonance. There must be another reason. What is it?

Could you state the source of information?

@TanYoungBoon stated in L.G Wade"bromination  of benzene gives the product(bromobenzene) that is 45kl/mol more stable than benzene "

The corresponding section in Wade, L. G. Organic Chemistry, 8th Edition; Pearson Education: Glenview, IL, 2013; p 758. reads

Bromination of Benzene Bromination follows the general mechanism for electrophilic aromatic substitution. (…)
Formation of the sigma complex is rate-limiting, and the transition state leading to it occupies the highest-energy point on the energy diagram (Figure 17-1). This step is strongly endothermic because it forms a nonaromatic carbocation. The second step is exothermic because aromaticity is regained and a molecule of HBr is evolved. The overall reaction is exothermic by 45 kJ/mol (10.8 kcal mol).

Thus, the given value of 45 kJ/mol refers to the reaction enthalpy of the total reaction
$$\ce{Br2 + C6H6 -> C6H5Br + HBr}$$
The reaction is mainly so exothermic because of the formation of HBr. This can be shown using values for the standard molar enthalpy of formation taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.
$$\begin{align}
&\Delta_\mathrm f H^\circ(\ce{Br2(l)})&&=0\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{Br2(g)})&&=30.9\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{C6H6(l)})&&=49.1\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{C6H5Br(l)})&&=60.9\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{HBr(g)})&&=-36.3\ \mathrm{kJ/mol}\\
\end{align}$$
although the exact value of 45 kJ/mol cannot be reproduced using these values.


text:
Why AlCl3 more acidic than FeBr3?
In compounds with lone pairs or with charge it is easy to predict acidity of compounds but how to describe acidic nature of  such ($\ce{FeBr3}$, $\ce{AlCl3}$) transition element compounds ?

Well, Al is smaller than Fe, and the formal charge is the same.

https://chemistry.stackexchange.com/questions/80375/alcl3-h2o-reaction

https://chemistry.stackexchange.com/questions/89378/which-is-more-acidic-bf3-or-alcl3


text:
Why on loosing aromatic chracter still product produced is 90% in the following reaction?
In Birch reduction reactant loses it aromatic chracter  but still the product found is to be 90%.
By loosing aromaticity product must be less stable so less amount of product must be formed.
 Why still on loosing product found is to be 90%?

Because you're using sodium...

@Zhe What do you mean ?

Sodium / ammonia is very forcing conditions. Sure you lost aromaticity, but the sodium got to lose an electron.


text:
How to predict and measure excess values of reactivity ratios for copolymerization with three monomers?
I ask this question because I haven't found interesting literature on the subject and I would like to know if my way to think is good or pure bullshit. Also the question is quite long because I have to expose my whole problem and I had some trouble to make it perfeclty clear in a short text.

I'm currently in internship in a R&D emulsion polymerization group and I started to read the chapter about copolymerization in the book "Principe ofpolymerization, 3rd Ed" by George Odian. So I won't say that I'm very familiar with that topic at the moment. I will also develop in my question what I will call "excess values" related to what I already learnt in thermodynamics but I am only using these as it is, I think, better for the understanding; but not for thermodynamics purpose.
Also I haven't found a lot of questions about copolymerization here so I will first introduce a bit the topic in relation to my question.

Introduction
I mention first that I am only speaking about countinuous feed reactors i.e. the molar ratio or my monomers reamain the same during all the reaction (except the end and the beginning).
In the system of copolymerization, we have two monomers, called $\ce{M1}$ and $\ce{M2}$. We want to know how the final molar ratio of each monomers in the polymer depending on the initial molar ratio of these monomers. During the polymerization, the growing chains may either end with a $\ce{M1^*}$ or a $\ce{M2^*}$ (can be ions or radical) (these previous notations denote the whole growing chains). Then we have the four following possible reactions:
$$\ce{M_1^* + M_1 \xrightarrow{k_{11}} M_1M_1^*}$$
$$\ce{M_1^* + M_2 \xrightarrow{k_{12}} M_1M_2^*}$$
$$\ce{M_2^* + M_1 \xrightarrow{k_{21}} M_2M_1^*}$$
$$\ce{M_2^* + M_2 \xrightarrow{k_{22}} M_2M_2^*}$$
where $k_{ij}$ are the kinetic constants of the reactions.
We can then define the reactivity ratios by, $$r_1 = \frac{k_{11}}{k_{12}} \text{ and } r_2 = \frac{k_{22}}{k_{21}}$$
and then doing calculation of kenitics playing with the rate of consumption of both monomers and a steay-state approximation for both of them, one can find the "copolymerization equation" as called by George Odian":
$$\frac {d\left [M_1 \right]}{d\left [M_2\right]}=\frac{\left [M_1\right]\left (r_1\left[M_1\right]+\left [M_2\right]\right)}{\left [M_2\right]\left (\left [M_1\right]+r_2\left [M_2\right]\right)}$$
where $[X]$ denote the concentration of $X$.
As this equation isn't very practical we can introduce the notations $$f_1=\frac{\left [M_1 \right]}{\left [M_1 \right]+\left [M_2 \right]} \text{ and } F_1=\frac{d\left [M_1 \right]}{d\left [M_1 \right]+d\left [M_2 \right]} $$
And then finally get,
$$F_1=\frac{r_1 f_1^2 + f_1 f_2}{r_1 f_1^2 + 2f_1 f_2+ r_2 f_2^2}$$
Now, knowing those the values of $r_1$ and $r_2$, it's possible to predict the compostion of the final polymer knowing the compostion of the feed. I've done the graphic below on Excel using $r_1 = 0.4$ and $r_2 = 0.12$ as an example. The blue curve represents $F_1$.

The problematics
So now let called what I discussed before an ideal system, as it is "easy" to determine the reactivity ratios for only two monomers in presence. I will call "real" the case in which we "perturbe" the ideal system with a third monomer $\ce{M_3}$ in a small amount compare to the total of the two others. I would like to know how this new system of monomer behaves as I would like to be able to predict some propterties of my final polymers without making too much experiments and I also would like to generalize this method for more monomers in the mixtures.
In the following ideal will be abbreviate by $id$ and real by $\mathbb{R}$.
I need the ratio of the molar ratios of my first two monomers to remain constant after the add of the third one, i.e.$$\frac{f_1}{f_2} = \text{cste} = \tau$$.
Then after the adding of $\ce{M3}$ and with the $\tau$-condition, we have,
$$f_1^\mathbb{R} = \frac{\tau}{1+\tau}(1-f_3)$$
$$f_2^\mathbb{R} = \frac{1}{1+\tau}(1-f_3)$$

The question
Is it possible to determine with accuracy the new value of $r_1^\mathbb{R}$ and $r_2^\mathbb{R}$ in reference of their values $r_1^{id}$ and $r_2^{id}$ in the ideal system and corrected by an excess value for each of them?
It would be something like,
$$r_1^\mathbb{R}=r_1^{id}+r_1^E$$
$$r_2^\mathbb{R}=r_2^{id}+r_2^E$$
My idea is to use for example an FTIR to determine $F_1^\mathbb{R} = F_1^{experimental}$ for different values of $f_3$ so that I could define the excess values in function of this third monomer and then predict the reactivity of the whole system knowing the reactivity of the ideal system.
Using the copolymerization equation in a situation where $r_1 > r_2$ and $f_2 << f_1$ so that I can get rid of $r_2 f_2^2$ and then determine $r_1^\mathbb{R}$ from my experimental value.

For the moment I feel like I found a mine of gold but I would like to know the thought of other people with better knowledge than me on that question. I don't know that much about FTIR though but I know this is not always super accurate in complex systems. Do you think this should work (I can't do the experiment as this is not the purpose of my internship) or is this more "bullshit"?

Textbooks tend to use examples which work out to some neat mathematical equation. Evaluating numerically just isn't something that fits nicely in a book. // I didn't really analyze your question. My guess is that if you have three monomers then you'll end up with a very messy mathematical equation which doesn't nicely simply. You can evaluate it easily numerically and have various plots calculated. I'd also expect it would be unusual to find three monomers where the addition to the chain is essentially random.

@MaxW well there qre multiple questions and thought behind this one but I can't ask them all. If I change the value of $\tau$ I'd like to be able to predict the properties of my final polymer using my excess value for the third monomer. My goal is just to avoid harder calculations that may involve way more approximations to compare the reactivities of all of the monomers with the radicals/ions. I'm also looking for an easy way to do it and to confirm experimental values.


text:
On what factors does amine inversion depend
What factors promote amine inversion? Like bulky groups around the nitrogen atom or less sterically crowding groups?
Are there any other factors that affect amine inversion? Does the electronegativity of the atoms bonded to nitrogen affect amine inversion?

related https://chemistry.stackexchange.com/questions/47328/what-causes-rapid-nitrogen-inversion


text:
Why is this redox reaction possible?
I have the redox reaction $\ce{N_2H_4 {(g)} + N_2O_4 {(g)} -> N_2 {(g)} + H_2O {(g)}}$.
In $\ce{N_2O_4 {(g)}}$, the oxidation state of nitrogen is $+4$. In $\ce{N_2 {(g)}}$, the oxidation state of $\ce{N}$ is $0$. Thus, $\ce{N_2O_4 {(g)}}$ is reduced.
In $\ce{N_2H_4 {(g)}}$, the oxidation state of nitrogen is $+2$, because the oxidation state of hydrogen when bonded to nonmetals is $-1$. In $\ce{N_2 {(g)}}$, the oxidation state of $\ce{N}$ is $0$. Thus, $\ce{N_2H_4 {(g)}}$ is also reduced.
Clearly, both $\ce{N_2O_4 {(g)}}$ and $\ce{N_2H_4 {(g)}}$ cannot both be reduced. What is wrong with this?

Well first you need to balance the equation. $\ce{2N2H4_{(g)} + N2O4_{(g)} -> 3N2{(g)} + 4H2O{(g)}}$. Then what are the half cell reactions?

I'm not sure, because in both reactants, the nitrogen atom has a positive oxidation state. They can't both be reduced.

Again stop thinking about "oxidation states" and think half cells. Add both $\ce{H+ \text{and} e-}$ to one side to create a balanced equation. Start with hydrazine. $\ce{N2H4 -> \text{?}}$

I'll give you part of the solution... $$\ce{N2H4 -> N2 + 4H+ +4e-}$$. So what is the other half cell?

On an unrelated note, what oxidation state would you assign to F in HF?

The oxidation state of hydrogen is $+1$ when bonded to non-metals and not $-1$.


text:
How to make copper salicylate with electrolysis?
I am trying the following and it looks like it may be working, but I am not certain what the end product would be.
Basic electrolysis setup: two cups distilled water, with 3 g of copper wire on anode, and large graphite cathode, connected to 9 V, 1.5 A power supply.
I then added the following to the water:

45 g of pure aspirin (acetylsalicylic acid).
30 g of baking soda (sodium bicarbonate).

It has been running now for about three hours, the solution is a bright sky blue, the copper wire is about 40% dissolved. I do not know chemistry but assume that I am making several reactions form in this solution: copper carbonate, copper hydroxide, and sodium salicylate. I am not clear if the electrolysis process of these will produce a final product where all of the available copper has bonded to all the salicylate.
I would greatly appreciate an answer on what it is you think the end solution will be once the copper has fully dissolved.

Why are you using this methodology?
Copper salycilate is an insoluble salt in water, so if you put copper sulphate in water and mix this solution with salycilic acid the copper salycilate will precipitate.
If you do not have some copper sulphate it's easier to make copper sulphate using copper wires, sulphuric acid and electrolisys (see Nurdrage video https://www.youtube.com/watch?v=arlYPz3EP7A).
Finally, in your experiment, if the copper is reduced it's possible that the salicylic acid is being oxidized and therefore you do not get the desired product.

I was inspired by this: http://www.edu.utsunomiya-u.ac.jp/chem/v11n2/PerezBenitez/PEREZ_BODY.html#Anchor1718818

These were the only ingredients I had available,  that is why I used the above methodology. Except I used distilled water. When I started with just aspirin water and copper the reaction was moving very slowly, about  2 hours in I saw only a few blue crystals and the copper wire had very few blue specks on it - still looked like a new copper wire, I thought perhaps it would react faster with some baking soda. I added baking soda and within minutes I had a nice deep blue reaction... Maybe because of distilled water the reaction was slow, they used tap water in the paper I linked...

I think that what I did worked, however I think it makes more sense to do the baking soda and copper reaction first with electrolysis, and then mix the aspirin in secondly without any electrolysis. I think I used way more baking soda and aspirin than was needed because of trying to do all reactions all at once.

The baking soda (NaHCO3) is not only an electrolyte but it is also required to remove (saponify) the acetyl group from acetylsalicylic acid (aspirin).

You don't get a precipitate if you add coppper(II) sulfate to hot salicylic acid. It becomes a clear dark green liquid. I assume the low pH because of the addition of the sulfate ion causes the formation of an acidic complex. I added some bicarb and the copper salicylate formed a light green powder. I added enough till it stopped effervescing as I don't have a problem of carbonate contamination. I have never got crystals of copper salicylate; always powder.


text:
Making sulfate from elemental sulfur
I have $\approx\pu{2 g}$ of elemental sulfur that I need to convert to dissolved sulfate in DI water. What's the easiest lab chemistry that will get me there?

If you convert the whole sample, there is no fractionation. And for doing that, perhaps https://en.wikipedia.org/wiki/Sulfuric_acid#Manufacture ?

You burn the sulfur and lead the fumes through a solution of $\ce{H2O2}$. Afterwards you boil that solution for a moment to remove the rest of the peroxide.
Use a vertical tube reactor or something so you can lead a soft stream of oxygen gas (or pressurised air, in a pinch) over the sulphur.


text:
Reactivity of fluorine compound
I have tried to make sense of the article on fluorine on Wikipedia but I am still very perplexed.
I would like to know generally how reactive the fluorine compound β-(4-Fluorophenyl)-γ-aminobutyric acid; β-(4-Fluorophenyl)-GABA or "Fluorophenibut" would be as it passes through your body. Is fluorine a generally reactive element irrespective of what it is bound to or is it generally stable since it is bound to a carbon atom on an indole ring?

There is no such thing as generally reactive element.

I don't mean generally in all cases.. I mean, how likely it would dissasociate from the carbon and react with other compounds in the body?

Quite totally unlikely.

@c73c Differentiate flourine, inorganic flourides and https://en.wikipedia.org/wiki/Organofluorine_chemistry

I agree with @ivan-neretin, it is very unlikely the fluorine will be lost from this molecule

Fluorine the element is very reactive, but that doesn't mean its compounds are
Fluorine the gas is a very reactive element because it loves to take electrons from other compounds. But this is irrelevant to whether compounds containing fluorine are reactive. Consider the more common chlorine. Chlorine gas is less reactive than fluorine, but is still very reactive. It is a dangerous and corrosive compound that has been (and probably is still) used as a military weapon. But you encounter very unreactive compounds of chlorine every day. Common salt is a chlorine-containing ionic compound that makes up a significant portion of the electrolytes in your body; PVC is a fairly inert plastic made from a polymer containing chlorine carbon bonds.
The very reactivity of fluorine makes some of its compounds very unreactive. F-C bonds, for example, are very strong and often very unreactive even compared to other carbon-halogen bonds. PTFE (a long carbon chain with all the hydrogens replaced by fluorine) is a very inert polymer that is safe enough to use at the high temperatures achieved in frying pans, for example.
Some fluorine compounds are, however, very biologically active but not because their fluorines are "reactive". Fluoroacetic acid and its salts, for example, are dangerous (and naturally occurring) poisons. But their mechanism of action relies on their stability not their reactivity: they are similar enough to acetic acid that they get involved in the standard citric acid cycle but end up blocking one of the key enzymes to further reactions because the product is too stable.
Whether or not fluorophenibut is "reactive" in the body in the sense you mean is unlikely. It is a neurodepressant similar to gabapentin and pregabalin, though little is known about its safety.
Complicated molecules can't be judged as dangerous, safe or reactive just because of the elements they contain. You need to know the detailed and specific ways they interact with bodily processes. Prozac, for example, is a widely used anti-depressant that contains fluorine. Just because it does doesn't make it reactive or dangerous.

The carbon-fluorine bond is very strong. The fluorine in a drug molecule is not going to get loose and attack other molecules in your body.
What it will do is park a very strong negative charge at one spot on the drug molecule. That is extremely handy for making a molecule bind to a target. As I understand it, that's why so many drugs include one or more fluorine atoms.


text:
Why is polyethylene not called polymethylene?
I understand that the monomer that polymerises to give polyethylene is an ethylene, but after polymerisation, the smallest repeating monomer in polyethylene seems to be a methyl and not an ethyl. So why don’t we call it polymethylene instead?

Because methylene is not a compound in its own right. Who said we must name the polymer after the smallest repeating unit?

Well that’s what my polymers professor said, and that seems to be the trend for all other polymers (as far as I have learned).

For all other polymers, the smallest repeating unit is the same as the molecule that was polymerized. For polyethylene it is different.

@IvanNeretin it is, actually, a compound in its own right, albeit a radical and thus hard to handle.

@leftaroundabout No it isn't, and I don't care if Wikipedia says otherwise. A particle, yes, but not a compound.

@IvanNeretin "Who said we must name the polymer after the smallest repeating unit?" – The IUPAC Recommendations 2008 said that.

@Loong Then the common practice seems to be disagreeing with IUPAC, and I'd better step aside.

According to the Compendium of Polymer Terminology and Nomenclature (IUPAC Recommendations 2008, i.e. the ‘Purple Book’), three different types of names can be used for polymers:

traditional names,
structure-based names, and
source-based names.

When traditional names fit into the general pattern of systematic nomenclature, they are retained, in this case: polyethylene (PE)
The systematic name for a polymer requires the naming of a preferred constitutional repeating unit (CRU). If necessary, this basic name is then modified by prefixes. The systematic structure-based name for polyethylene according to the Purple Book is indeed poly(methylene).
Polymers can also be named as being derived from a monomer (or precursors), which is named according to IUPAC rules. Such names are referred to as source-based names. The systematic source-based name for polyethylene according to the Purple Book is polyethene.

Polyethylene is prepared by polymerizing ethylene. Ethylene has a double bond, thanks to which the polymerization goes on. You could not do that with a methylene group. So the name comes after the smallest unit which is actually used to prepare the polymer.

Although this certainly is the reason, I have to remark that it's a bit weird to name a compound after how it's synthesized in practice. Who's to say that a better process isn't developed in the future that will start from, say, hex-1-ene? Surely the product shouldn't suddendly be renamed to polyhexylene, if it has the same chemical structure as polyethylene. And although I reckon it will never be feasible to polymerise this from $\mathrm{CH_2}$, it's certainly conceivable.

@leftaroundabout: I think it's a bit like the fact that "sodium bicarbonate" is so called because the ratio of carbonate to sodium is double that of sodium carbonate.  In many cases, substances get named before they're fully understood, and the names stick whether or not they accurately "describe" the substance in question.


text:
Can you discriminate a monooxygenase from other enzymatic mechanisms by the requirement of NADPH?
I am working on a project about CYPs 450. It seems that most of them function as monooxygenases but there are other categories as well.
Assuming that the question makes sense, is it correct to say that CYP 450 monooxygenases always require the presence of a regenerating system (NADPH) as opposed to other CYPs 450 which are NOT functioning as monooxygenases?

I would say this is not a correct statement.  The main reason being that just because an enzyme is not a monooxygenase doesn't mean it doesn't require a regenerating system for NADPH.
Examples would be many dehydrogenase, hydroxylase and synthase enzymes. They also require NADPH regenerating systems to operate efficiently.  So, one of the CYP 450 enzymes may fall into any of these categories requiring NADPH/NADP yet not be a monooxygenase.

Thank you for your response. What about the other way round? If it requires NADPH/NAD is it a monooxygenase?

No! As stated in my previous response, many other enzymes that are NOT monooxygenases DO require NADPH/NADP.  If you mean to ask "If it is a monooxygenase, does it mean it requires NADPH/NADP?", then I believe the answer would be "yes". There may be exceptions, but none that I'm aware of.

OP's concern: Monooxygenases always require the presence of a regenerating system ($\mathrm{NADPH}$).

It is true that most Monooxygenases are nicotinamide adenine dinucleotide ($\mathrm{NADP^+}$ and its reduced form, $\mathrm{NADPH}$) dependent (Wikipedia). Monooxygenases catalyzes the incorporation of a hydroxyl group into a variety of substrates by catalyzing the reduction of $\ce{O2}$ to $\ce{H2O}$ while oxidizing $\mathrm{NADPH}$.
However, there are some exceptions such as Peptidylglycine $\alpha$-hydroxylating monooxygenase (PHM) and dopamine beta-monooxygenase (DBM). For example, PHM is a copper-, $\ce{O2}$-, and ascorbate-dependent enzyme responsible for the oxidative cleavage of C-terminal glycine-extended precursor peptides to the active $\alpha$-amidated peptide and glyoxylate (Ref.1,2):

There has also been a significant interest in describing the interactions of copper-containing enzymes with $\ce{O2/H2O2}$-derived species. The short-lived intermediates resulting from the activation of dioxygen are the key players in the mechanistic cycles in many of these metalloenzymes. For instance, In the enzyme PHM, various reduced $\ce{Cu/O2}$ species have been proposed to act as catalytically competent intermediates, yet their exact nature and their role in the enzymatic reaction is still unknown. Schismatic diagrams of recent computational mechanistic studies of PHM is shown as snapshot in three stages (Ref.3):

DBM is also a $\ce{Cu}$ containing mettaloenzyme, following similar mechanism of PHM. The three substrates of DBM are dopamine, ascorbic acid (Vitamin C), and $\ce{O2}$ and products are norepinephrine (from dopamine and $\ce{O2}$), dehydroascorbate (from Vitamin C), and $\ce{H2O}$ (from $\ce{O2}$).
References:

Betty A. Eipper, Richard E. Mains, Christopher C. Glembotski, "Identification in pituitary tissue of a peptide $\alpha$-amidation activity that acts on glycine-extended peptides and requires molecular oxygen, copper, and ascorbic acid," Proc. Natl. Acad. Sci. USA 1983, 80(16), 5144–5148 (https://doi.org/10.1073/pnas.80.16.5144).
A. F. Bradbury, M. D. A. Finnie, D. G. Smyth, "Mechanism of C-terminal amide formation by pituitary enzymes," Nature 1982, 298, 686–688 (https://doi.org/10.1038/298686a0).
Enrique Abad, Judith B. Rommel, Johannes Kästner, "Reaction Mechanism of the Bicopper Enzyme Peptidylglycine $\alpha$-Hydroxylating Monooxygenase," Journal of Biological Chemistry 2014, 289(20), 13726–13738 (doi: 10.1074/jbc.M114.558494).


text:
Explain optical activity in Biphenyl compounds
How is optical activity explained in biphenyl compounds using atropisomerism.
NOTE: I was taught that if vertical symmetry exists then optical activity is not shown by any substituted biphenyl compounds
And I was also taught that 6,6'-dimethyl[1,1'-biphenyl]-2,2'-diamine shows optical activity.
How is this possible because there exists a plane of symmetry right?

No, there is no plane. Related: https://chemistry.stackexchange.com/questions/87869/how-to-find-the-plane-of-symmetry-in-biphenyls-and-substituted-cyclohexanes/87871#87871 (part 3).

In general, there's no reason why a simple substituted biphenyl is flat either. The key is that if the barrier to rotation is low enough, then the different stereoisomers can interconvert, so you end up with a racemic mixture.
For more substituted biphenyls, strain present in the planar form is high enough that it impedes free rotation around the bond that connects the two phenyl rings. This means that the axially chiral conformations cannot easily interconvert, and thus, you can persist any enantiomeric excess that you create.

In biphenyls, the two ortho substituents in each aromatic ring must be different and bulky (e.g. COOH and NO2).
Also the two aromatic rings must not be co-planar (i.e. must not be in the same plane) but perpendicular.


text:
Where can I find example MS, IR, NMR spectra for teaching molecular spectroscopy?
I am teaching a course on analytical chemistry, and I would like to give my students MS, IR and NMR spectra to practice with.
It would be ideal if they could analyze spectra in a more "realistic" way than looking at photocopies of spectra from SDBS. Ideally we'd get some newish (not 30 years old) spectra in an interface where students could, for instance, integrate NMR peaks and zoom into multiplets, automatically get a list of wavenumbers in IR spectra, maybe even deconvolute mass spectra.
I talked to MestReNova, but we can't afford it.
BioRad's SpectraBase might be the best resource I found so far, but it's not as interactive as I'd like it to be.
There is a nice collection at UCLA specifically for teaching but it's 20 years old, mostly static, and pixelated.
I'd appreciate any pointers to suitable collections of molecular spectra, or to free (or cheap) software for spectral analysis.

Topspin is free for academic users. I don’t know if your students qualify (as an undergrad in UK, I did).

I have Topspin, yes! I wasn't sure if I should dump a fully grown-up software package on people who just learned what a chemical shift is - there's a bit of a learning curve - but maybe it's not wrong to show them the tools that they'll use in the real world.

Oh, yeah. I certainly agree. It’s hardly the... easiest piece of software to use. To be honest most people I know use Mnova. I’m the only person in my group who uses topspin... I just brought it up because it was free, of course.

TopSpin has a full set of example data (I think about 10 examples of structure elucidation) with 1H, 13C, COSY, HMBC, HSQC etc. It might not be as easy to use as MNova, but it does actually force your students to understand whats going on

Thanks guys!
Topspin example data: my course won't go beyond 1D. My academic Topspin came with a 1H 1D of cyclosporin - not a beginners' molecule. Is there more?
Any suggestions for IR, MS?

An elder way would be the via book examples, like «Organic Structures from Spectra» by Field, Sternhell, Kalman, with dozens and dozens of combined exercises in a format like the following:

and currently available in the 5th edition by Wiley (e.g., here). There is a  related project of «Interactive Spectroscopy Problems» by the University of Calgary (here), too.
German VCH published years ago Peter Bigler's training course «NMR-Spectroscopy: Processing Strategies» (e.g., here) including a CD full of 1D and 2D NMR data, which comes close to your aim of interactive sessions to process fids till the plot.  Bruker's WIN-NMR included is old (think Windows 95 era, could be still 16bit), but the experiment-based NMR data should still be processed by current software.
Almost forgotton:  NIST webbook chemistry has for many compounds which may qualify as «easy» IR, UV (condensed phase) and mass spectra, too.  While not aiming to be a teaching set, there are options to download the spectra not only as picture, but equally in the JCAMP-DX format to process, e.g., benzaldehyde.


text:
Reactivity comparison of phenol and toluene towards electrophilic substitution

Which aromatic compound is more reactive towards electrophilic substitution?

benzene
chlorobenzene
phenol
toluene

Correct answer: phenol

But, according to me, the answer should be toluene.
My reason: Toluene is an ortho-para directing group as well as an electron-donating group which increases the electron density at ortho-para position and activates the aromatic ring towards electrophilic substitution.
Where I am wrong?

And -OH is what? This description fits even more.

You have to make a comparison between phenol and toluene. You have identified what toluene does, but you haven't described what phenol does.
The hydroxyl group in phenol increases electron density in the ring by resonance (+M effect). Note that the methyl group in toluene increases electron density only by hyperconjugation and inductive effect.
Resonance effects are generally far more superior to hyperconjugation. Hence, without doubt, phenol is more activated towards EAS as compared to toluene.

"Resonance effects are generally far more superior to hyperconjugation" - Why?

@Rahul Resonance involves direct conjugation of an electron pair on an atom into the conjugated system. However, for hyperconjugation, you have to first break the C-H bond, and only then the lone pair on the carbon atom formed can resonate (Hyperconjugation is called "no bond resonance" if you recall) Hence, resonance activation of ring is better.


text:
Is it possible to represent Aluminum 6061 (AlMgSiCuCr) as a skeletal structure?
Is it possible to represent Aluminum 6061 (AlMgSiCuCr) as a skeletal formula / structure?
Linear Formula: Al-97.9% Mg-1% Si-0.6% Fe-0.4% Cu-0.28%
(possibly helpful)
Here's an example of what I mean...

Any help would be greatly appreciated, thanks.

The ratio between the different elements is a 100% fixed ratio of small integers in an organic molecule like the one you show.

In this case, probably no.  When you see copper in aluminum (as in the 6xxx series), it is probably there in small amounts for purposes of precipitation hardening.  In this case, there are tiny discrete precipitates of copper alloys or intermetallic compounds finely dispersed throughout the aluminum/silicon matrix.

No, it is not possible. There is no repeating crystal unit.

And perhaps even more to the point, 6061 aluminum it is not even a single phase, which is very common for engineering alloys. The (usually) small amounts of second phases are what gives the alloy the desired mechanical properties.

Is it one of those icosahedral phases I've heard about?

What you need to do is to check the phase diagram for the system, it may be difficult (or impossible) to get a phase diagram where Al, Si, Cu, Mg and Fe can all be varried independently of each other.
But as a start you should look at the Al / Si, Al / Cu, Al / Mg and Al / Fe phase diagrams.
Under some conditions you might have a solid solution of the dopant atoms in aluminium metal as the "solvent", under these conditions you will have a disordered solid in which some aluminium atoms are randomly replaced with things like copper.
I worry that it is not clear if the % of the elements is given in mass (weight) % or atomic %. So I think it makes the question a bit harder to answer. This book chapter may help to make things a bit more clear.


text:
Reactivity of carbon-centered radical
I've read this site for a long time but I had never asked a question, so this is my first time :).
So, i'm currently working on a computational chemistry project on radical cyclization reactions and I'm having trouble trying to understand the reactivity of the radical. Here I show one of the reactions i'm studying:

As you can see, the radical is centered on a carbon atom which is bonded to two chlorine atoms and also to an amide group. As far as I know, chlorine atoms attached to a carbon-centered radical stabilize it due to inductive effect. Also, chlorine atoms can be considered EWG due to their electronegativity. This alone would lower the energy of SOMO orbital, wouldn't it?
In addition to that, amide groups are also EWG, which again would lower the energy of SOMO orbital. So my questions are:
1) Am I correct to say that the more stable a radical is, the less reactive it is?
2) Am I ignoring some other effects which can affect reactivity?
P.S. Below I show the optimized structures of the reactant and transition state calculated at the B3LYP/6-31++G(d,p) level of theory.
Thanks for your attention.
EDIT: Well, I think I wasn't very clear about what's bothering me. I have the energies of the reactant, ts and product and they are ok, nothing wrong. It had just came to my mind that the radical is very stabilized in the reactant and I just wanted to know if my reasoning was correct. Also, in that same reasoning, could it be that a carbon-centered radical with no chlorines be more reactive? Again, this is a curiosity that came to my mind and I'm want to be sure I'm thinking it in the correct way.

What part of the radical reactivity is bothering you? I think everything you said is reasonable.

Reactivity toward cyclization with the alkene part of the molecule.

Ah so you're wondering why in an equilibrium situation, why/if the cyclized form is preferred? I think your judgement of radical stability is correct, so my first guess is that once the cyclized radical undergoes termination with H* for example, then it's stuck there, whereas a C-H bond in C(=O)CCl2H could still undergo homolysis, eventually driving reaction forward despite the difference in stability between the two radical intermediates.

I added an edit.

I question whether it is really an equilibrium. I would argue that the cyclisation occurs simply because it is faster than any other intermolecular process (e.g. H atom abstraction from something else).


text:
What type of bond would be formed if dyz orbitals overlap along the z axis?
I know that when two $\ce{p_z}$ orbitals overlap in a particular direction - along the z axis - they form a sigma bond. Now, if two $\ce{d_{yz}}$ orbitals overlap in same direction, which type of covalent bond would be formed?

Before I answer the question, you should understand that - since everything is symmetric - the answer would be the same if we took two $\ce{d_{yz}}$ orbitals along the x-axis, two $\ce{d_{xy}}$ orbitals along the y-axis, etc.
Have a look at this snippet I picked from Wikipedia Commons (by Jeremy77, public domain):

Here you can see the overlap of two $\ce{d_{xy}}$ orbitals along the y-axis constructively (in the same phase). There are four lobes overlapping, which would lead to a total of two nodal planes. Such a bond is a speciality of $\ce{d}$ orbitals and it is called the delta ($\delta$) bond.
Now, can you understand what the answer to your question would be?

See also: Which d orbitals can form Sigma, Pi, Delta bonds?

Just tell me in one word what is the answer of this question.

@RohanSachdeva I said in my answer it's the delta bond

Is delta bond resulting when $d_{x^2-y^2}$ and $p_y$ overlap along $y-axis$?


text:
How to calculate n factor for O2 and H2 in this reaction?
How do I find n factor of $\ce{O2}$ and $\ce{H2}$ if I want to use equivalence principal on a reaction like $\ce{O_2 +2 H_2 -> 2 H2O}$?

Here is the unbalanced reaction
$H_2 +O_2----> H_20$
I hope you know how to calculate oxidation number
 On product side
In $H_20$ oxidation number of $H$ is  $+1$ and $O$ is $-2$
On reactant side
oxidation number of $H$ and $O$ is  $0$
So n-factor is magnitude in change of oxidation number
So n factor of $H$ is $1-0$ i.e $1$ and $O$ is $-2-0$ i.e $2$(magnitude)
Now to balance the reaction (for 2 reactant rex) you just interchange the n factors
You get
$1O_2 +2H_2 -----> 2H_20$
Now to calculate equivalent weight formula is
$E.w=molar mass/N.F$


text:
What are the oxidation states of galium and arsenic in GaAs semiconductor?
According to the book I'm referring to ("S.M. Sze Semiconductor Devices"), gallium has an oxidation state of +1 and arsenic has that of -1. This should not be possible as each of these elements forms a covalent bond with four atoms of the other element (structure of GaAs semiconductor). For that to be possible, shouldn't both of them have one electron in $\ce{3s}$ subshell and three electrons in $\ce{3p}$ subshell?

Which book?  And how meaningful are these oxidation states anyway?

Oxidation states are meaningful because they tell about the electronic configuration that might have existed during bonding

@OscarLanzi The book I'm referring to is S.M. Sze Semiconductor Devices

In general, gallium can have +1 and +3 oxidation states, though +1 is a very rare condition. In its turn arsenic possible oxidation states are -3, +3 and +5. As it is stated in Wikipedia, gallium oxidation state in gallium arsenide is +3. Once again, arsenic has -3 oxidation state in arsenides or intermetallic compounds. Therefore, gallium arsenide oxidation states are +3 for Ga and -3 for As. This is logical because the electronegativity of As is higher then Ga. As for Simon Sze, he is an electrical engineer and physicist and not a chemist. From the mathematical point of view (-3)(+3) is the same thing as (-1)(+1).


text:
What factors contribute in making a solution more endothermic?
For example, a solution between potassium nitrate and water is more endothermic than sodium nitrate in water. Can someone explain?

We could compare the relative stabilities (which can be deduced from their enthalpy changes of formation) of the products and reactants. If the products are less stable, compared to reactants, the reaction is endothermic. Also, more quantitatively, we could also calculate the energies absorbed and released from the specific processes that occur. For example, in dissolution, we would have to look at the enthalpy changes of hydration of the ions and the lattice energies of the ionic compounds.

But what specifically makes a reaction like the solution between potassium nitrate and water have a greater enthalpy change than the solution of sodium nitrate in water.

Notice that the positive ion has changed from Na+ to K+. This would significantly change the enthalpy change of hydration of the ions, since it would then be less exothermic as a result of potassium ion's ionic radius being considerably larger than that of the sodium ion's. This may have tipped the balance between the lattice energy and the enthalpy change of hydration, resulting in the overall process becoming endothermic.

By the way, if you only concerned with the process of solution. You might want to make it clear in the question title. It seems too general currently.

So the larger the atomic radius, the weaker the bond energies which makes the reaction less exothermic/more endothermic?

https://chemistry.stackexchange.com/questions/84693/cooling-down-and-heating-of-solution-unexpectedly-during-cation-identification-t

Good info but it doesn't really answer my previous question regarding atomic radii.


text:
Electron promotion in hybridisation: A misconception
Recently, I read through a short article on Nature Chemistry, written by Michelle Francl, titled "Talking to Pauling's ghost". The article talks about how electron promotion was never something that was conceived by Pauling when he devised the theory of hybridisation. Instead, it likely arose from misinterpretations and mistranslations over the years.
It has been established very well on this site that hybridisation does not refer to a physical process and is merely a mathematical construct. This is consistent with the view presented by Francl. However, what I do not seem to understand is why the alternative view of seeing hybridisation as electron promotion and orbital "mixing" has become so prevalent, if it is so inaccurate. Clearly, there must be some truth in the alternative view. Is there?

Reference: Francl, M. Talking to Pauling’s ghost. Nature Chem 2018, 10 (7), 688–689. DOI: 10.1038/s41557-018-0099-3.

Some people believe that Earth is flat. What's surprising about that?

Sorry for being uneducated, what I have to understand under  electron promotion? With theconcept of hybridization I’m very familiar.

There's no "alternate view", just a bunch of highschoolers putting arrows in squares.

@Mithoron  I disagree. I have seen textbooks, as well as chemistry educators, advancing this view.

Is the bone of contention about something which is just useful versus what is real? The planetary model of the atom isn't real but it is a useful abstraction upon which a better model can be constructed. LCAO isn't real but it sure seems like a useful abstraction. A different abstraction is Newton's concept of gravity. Works well for 99.9% of physics that we use. There is no point in using Einstein's extension in most cases just because it is closer to reality. (Since we're still missing the grand theory of everything, I doubt that Einstein's concept of gravity is in fact "real.")


text:
What is content of chlorine in pyrene?

There is no chlorine atom in pyrene, still it have more chlorine content than gammexane.

Please explain how I can find chlorine content in pyrene.
I found these links on the Internet:

Rankerer
StudyAdda
Several other links on a Google search

Can anyone please explain this fact?

How? Just like said in the first part of your phrase. Have you seen the structural formula of pyrene? Have you seen a chlorine atom there? No? Guess why? Because it is not there. Like, really not there.

@IvanNeretin Sir, I know that pyrene is a 4 phenyl ring organic compound, but in a textbook that's widely used in India, it is mentioned that "CCl4 is used in fire extinguishers as pyrene" It isn't clear, but that's probably where the doubt arises as to why there's chlorine in pyrene

Then they must be using the word "pyrene" in a different sense.

"pyre" relates to fire. This is clearly a trade name that should not be interpreted chemically. More about the Pyrene Fire Extinguisher Compnay can be found here: https://en.wikipedia.org/wiki/The_Pyrene_Company_Limited

@IvanNeretin yaa i do agree with your points ,but i found same thing too on internet -E.G--1)   https://neetlab.com/questions/which-of-the-following-has-highest-chlorine-content/     2: https://rankerer.com/question/chemistry/H93yoNOt9rxO/which-of-the-following-has-highest-chlorine-content              3: https://www.studyadda.com/question-bank/critical-thinking/1400/103133    etc. many more links are there where i found same answer but no explaination

Question 1 asks about trade-named products. The question clearly refers to the contents of a Pyrene fire extinguisher i.e. carbon tetrachloride.

no i m not asking anything related to "fire extinguisher pyrene" i asked different question please prefer to link i provided in question

this question is correct i found so many links of this question on GOOGLE please just click on link provided above one time before downvoting my question

The answer provided by your search is not correct!! I suggest you rephrase your question or it will be closed,

@waylander did u checked link provided in question

@waylander this question is on Google how could so many website on Google show wrong question

Because tbey all copy on wrong answer perhaps. Pyrene is a 4 ring all carbon compound with no Chlorine content. I very strongly suspect that the answers you are seeing relate to a proprietary compound for fire extinguishers which is also called Pyrene.

This is a great misconception in most Indian books around here. I think it would be good if we clear it up in the answers rather than close down this question. || @RafaelNadal the links you've provided are from Indian sites only,  and you can bring thousands of such links, they'd all be wrong because they all copy from each other. The authoritative source is the Wikipedia article https://en.wikipedia.org/wiki/Pyrene

Actually i think it's a confusion because there was a company in the times of World war 2 known as Pyrene Company limited which used to manufature brass containered fire extinguishers in which carbon tetra chloride i.e. CCl4 . and hence due to this confusion some books says that it has high concentration of chlorine.

sir please prefer to link i had provided

yeah i have taken a look over that link, it's just that some people confuse pyrene ltd with actual compound pyrene and that's why the answer given to that question is pyrene, actual answer to that question should be gammaxene..


text:
Black & Green Nickel Oxide-Supplemental
A previous question (April 2017) asked about black $\ce{NiO}$ did not supply any answer. Sadly, such a referral only deters colleagues from commenting on the current question.
There is a great deal of literature on cation defects in non-stoichiometric nickel (II) oxide. There is much on charge-transfer in defective crystal structures. But, I can't find any attempt to marry the two together - which may explain how the deep colour (black) is observed.
Of course, it may simply be that I have not  discovered it yet. I did find a question on how/why copper (II) oxide is black. This one has semiconductor properties; so, $\ce{CuO}$ and non-stoichiometric $\ce{NiO}$ may be black for similar reasons?
Can anyone comment?

The non-stoichiometric form of nickel (II) oxide and copper (II) oxide are black, with the same square-planar (at metal) solid-state structure, (with crystal defects), so the metal centres are bridged by oxides. The assignment of oxidation state is +II; therefore, could consider a fragment of the structure as: -O-Cu(II)-O-Cu(II)-; but, with redox-active metals, this also:
-O-Cu(III)-O-Cu(I)- thus. Electrons can drift through the structure giving rise to intervalence electronic transitions, with strong involvement of oxygen atoms. The influence of pie-donating oxygen prevents the formation on Cu(0), the native metal.
Intervalence transitions can give rise to intense colours (e.g. black); normally, forbidden to colours arising from d-d electron transitions. (This, also seen with Prussian Blue, where Fe(II) & Fe(III) are bridged by cyanides.) A discrete Cu(II) centre would have a weak, less-intense, colour, arising from said d-d transitions.
Stabilisation of highly-polarising Cu(III) & Ni(III) is achieved by the oxide, which is a very electronegative and pi-donating ion, stimulating low activation-energy electron-flow.
Transitions between p and d orbitals (as in permanganate: metal and oxide) give rise to very intense colours. These extended oxide lattices are also susceptible to being defective, so missing ions; or, ions in the wrong location in the lattice. This also gives rise to colour. An example of this is seen in zinc (II) oxide, when heated. In this case it is not mixed valence; but, loss of oxygen from the lattice, that gives rise to defects that allow lower energy (than in "complete" ZnO) electronic transitions to occur and absorption at the UV-end of the visible region. On cooling the lattice picks up oxygen, from the air, becoming complete again, reverting to its original white from the, high-temperature, yellow colour.


text:
Hill function and stoichiometric matrix: confusion about reaction rates
When deriving the Hill function (protein $X$ and $n$ molecules of the signal $S$), my book introduces the following rates:
$$
\text{collision rate} = k_{ON} \cdot X_n \cdot S^n\\
\text{dissociation rate} = k_{OFF} \cdot [nSX],
$$
where $[nSX]$ is the complex. This translates into the following ODE:
$$
\frac{d}{dt} [nSX] = k_{ON} \cdot X_n \cdot S^n - k_{OFF} \cdot [nSX].
$$
This seems fine if I follow the $\textit{rules}$ of, for instance, this table.
On the other hand, during the class, my professor introduced the stoichiometric matrix. If we have the reaction $2A + B \leftrightarrow C$, then the corresponding system of ODEs would be:
$$
A' = -2r_1A^2B + 2r_2C\\
B' = -r_1A^2B + r_2C\\
C' = r_1A^2B - r_2C.
$$
So, my question is: why in this case we put the $2$ in front of the reaction rates while for the Hill function we do not (put the $n$ in front of it)? Is it just because the $n$ molecules of signal $S$ are not consumed nor produced?

If the reaction is $\ce{aA + bB = pP + qQ} $ then we define the rates as
$$-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{p}\frac{d[P]}{dt}=+\frac{1}{q}\frac{d[Q]}{dt}$$
As to the Hill function it seems that the complex nSX is just one species.


text:
Oligomer Terminology
As we know, dimers, trimers etc. can be produced from a certain amount monomers. In the case of a dimer being produced, two monomers dimerise.
Let's take an example of a dimer (glycine anhydride):

Two glycine monomers were dimerised for this - it's a dimer since it has 2 (poly)amide bonds.

Now, let's take an example of an ester (ethyl ethanoate):

This is where my question comes in: would it not be appropriate to call ethyl ethanoate a monomer? As we saw with glycine anhydride, it has two (poly)amide bonds, hence it's a dimer. Ethyl ethanoate has a single ester bond, so would this not render it as a monomer? Or is this incorrect logic, since ethyl ethanoate isn't an oligomer/polymer per se?

P.S. I sometimes put the word 'poly' in parenthesis since I'm not sure whether its usage is necessary in description.

According to IUPAC, a monomer molecule is a molecule that can undergo polymerization, thereby contributing constitutional units to the essential structure of a macromolecule. If you want to call ethyl acetate a monomer, what kind of polymerization do you expect ethyl acetate to do?

@Loong I don't expect it to do any polymerisation - I wasn't aware of the IUPAC definition. I was just wondering whether you could call it a monomer, based on what I mentioned about glycine anhydride.

This isn't what you were asking, but in computational chemistry, when performing multiple-molecule calculations to study intermolecular interactions, we often call single molecules monomers. By extension, there can be dimers, trimers, ..., and this has nothing to do with polymerization.

You can call it a monomer, although your audience will expect that you are about to present its oligomerization or polymerization. And according to IUPAC you should put parentheses if in the brackets you have the repeat unit of the polymer: e.g. poly(styrene).


text:
Does rearrangement occur in Friedel Crafts alkylation or not?
In the reaction of alkylation of aromatic systems with the help of Friedel Crafts alkylation - the reagent used is alkyl halide in presence of appropriate Lewis acid.
When is rearrangement observed?
If the alkyl halide is primary, the carbocation is usually not stable - as a result, a free carbocation shouldn't be formed and the attack should occur from the acid base complex itself (this is what my instructor said). However, in cases where the alkyl halide is tertiary, or where the carbocation is stable enough to be formed - rearrangement (if possible) is observed and the reaction proceeds through a free carbocation intermediate which acts as electrophile.
So in the former case, where a full fledged carbocation isn't formed, is rearrangement possible?
For example, is rearrangement possible in alkylation of benzene with n-propyl chloride in presence of aluminium chloride? Clearly, the primary carbocation is not stable - and may arrange to secondary carbocation (only if it is formed in first place)
So what really happens, and when in general and in particular can we expect rearrangements in Friedel Crafts alkylation? Also, when not to expect rearrangements? (Of course, when the carbocation isn't formed, but when is that?)
It'd be great if you could quote some credible sources along with your answer, if you wish to add certain specific reactions etc.
P.S.

Which Lewis acids allow for rearrangement to occur? As far as I know, only in presence of stronger Lewis acids does rearrangement occur - but how strong?
Does temperature also play a role for rearrangement to occur?

When you conduct the alkylation of benzene with n-propyl chloride, the product you obtain is isopropylbenzene, so a rearrangement occurred by virtue of the fact that the primary carbocation that forms will undergo a 1,2-hydride shift to form a secondary one. Most alkylation ( aromatic ring and alkyl halides ) are conducted in the presence of aluminium chloride which is one of the strongest Lewis acids. The principle is that the alkyl halide reacts with the aluminium chloride to form a salt: a carbocation and a complex ion of tetrachloroaluminate. You can have methyl chloride react in this way because even if it forms an unstable carbocation, it will readily react with the aromatic ring ( which acts as a Lewis base ). You can have methyl carbocation in solution such as in alkylation reactions, but you can’t isolate a salt containing methyl carbocation in solid state. If you had methyl chloride and aluminium chloride in solid state ( in contact with each other, they would not react to form carbocations.

Actual methyl cations in solution react to make t-butyl cations and hydrogen and are made with superacids not AlCl3, which  makes adduct with methyl chloride.


text:
Is epoxy resin plastic?
Is epoxy resin a plastic ? For clarity when I say plastic I mean of the same ilk as all the single use plastics we (should) be trying to reduce the use of.

questions seems of that same ilk as polotical agitation.

Yes. The IUPAC definition of a plastic (page 384 of document, page 18 of pdf file) is

plastic - Generic term used in the case of polymeric material that may contain other substances to improve performance and/or reduce costs.

Note 1 - The use of this term instead of polymer is a source of confusion and thus is not recommended.
Note 2 -  This term is used in polymer engineering for materials often compounded that can be processed by flow.

Before the epoxy sets it will flow, so it is a plastic.

As far as I am aware, the term plastic was in use long before the materials it is now associated with.
It was one of the terms used to describe material states/properties: elastic, plastic and fracture.
elastic = non permanent deformation, stretches and returns to original dimensions.
plastic = permanent deformation, stretches beyond the elastic stage.
fracture = the point at which the limit of plastic deformation ends by breaking.
So it may depend on how you're using the term, but I would probably use both plastic or resin for either since they both start as resins and they "can" both have a plastic deformation stage when set/cured.
Thermoplastics set, thermoset resins/plastics cure.
Anyone suggesting that epoxies should not be called plastic because they're generally incompatible with thermoplastic, is false since different formulations of thermoplastics are also generally incompatible with each other.


text:
Notation for excited states
I am reading some papers about quantum chemistry and in one I found a notation for excited states of a molecule that I don't understand at all.
In the picture below you can see the potential curves of biphenyl as a function of its torsion angle. In the description of the picture, the author speaks of states $1^1A$ and $1^1B_1$. What does he mean by these states?

They refer to the symmetry species of the states of the molecule in its particular point group.  See the answer here https://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613 for an introduction to these.

This may also be useful: https://chemistry.stackexchange.com/q/58229/194

As has already been mentioned in comments, those letters and numbers for each state correspond to their symmetry, specifically the symmetry of the (purely electronic) wavefunction. For more information regarding symmetry notation, see here, and see here for more discussion about how symmetry can be applied to molecular orbitals.
It might not be clear from the second question, but because the symmetry of the overall N-electron wavefunction results from the direct product of all occupied 1-electron wavefunctions (at least in Hartree-Fock), exciting electrons into molecular orbitals with different symmetries will lead to excited N-electron wavefunctions (excited states) with different symmetries.
To consider the specific case of biphenyl,

the $\tau = 0^{\circ}$ structure, corresponding to the leftmost values on the x-axis, will look like the 2D structure just above where both rings lie in the same plane. As an exercise left to the reader, you can figure out that the point group of the planar form of biphenyl is $D_{\mathrm{2h}}$. The point group $D_{\mathrm{2h}}$ has the following character table:
$$\begin{array}{c|cccccccc|cc} \hline
D_\mathrm{2h} & E & C_2(z) & C_2(y) & C_2(x) & i & \sigma(xy) & \sigma(xz) & \sigma(yz) & & \\ \hline
\mathrm{A_g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2,y^2,z^2 \\
\mathrm{B_{1g}} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & R_z & xy \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & R_y & xz \\
\mathrm{B_{3g}} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & R_x & yz \\
\mathrm{A_u} & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \\
\mathrm{B_{1u}} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & y & \\
\mathrm{B_{3u}} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \\ \hline
\end{array}$$
The three-dimensional symmetry of any object can always be broken down into one or more irreducible representations (irreps) belonging to the object's point group; each irrep corresponds to a row in the point group's character table. The symmetry table is the leftmost column in each irrep row. To enumerate them,

The labels from the figure are $1^1A, 1^1B_{1}, 1^1B_{2}, 1^1B_{3}, 2^1B_{3}, 2^1B_{2}, 2^1A, 2^1B_{1}$.
The labels from the character table are $\mathrm{A_g}, \mathrm{B_{1g}}, \mathrm{B_{2g}}, \mathrm{B_{3g}}, \mathrm{A_u}, \mathrm{B_{1u}}, \mathrm{B_{2u}}, \mathrm{B_{3u}}$.

You can see that these labels are almost but not quite the same as those in the figure. A minor difference is italicized versus regular/upright text; this is a matter of typographic choice and there is no physical or chemical difference between them. There are two significant differences:

The labels in the figure are prefixed with a regular number and a superscript number. The regular number helps uniquely defines the state ordering: $1^1B_{1}$ is the first state with $^1B_{1}$ symmetry, and $7^1B_{1}$ is (hypothetically) the seventh state with $^1B_{1}$ symmetry. Without the superscript number, each symmetry label only specifies the spatial symmetry, and with the superscript defined as $2S + 1$, the spin multiplicity, each label can also specify the spin symmetry. $1^1B_{1}$ is the first singlet ($S = 0$) state with $B_{1}$ symmetry, and $1^3B_{1}$ is (hypothetically) first triplet ($S = 1$) state with $B_{1}$ symmetry. See this excellent answer for more information, in particular the figure.
The labels in the figure do not have u/g in the subscript, because for most of the biphenyl structures, where $\tau \neq 0^{\circ}$, the molecular point group is no longer $D_{\mathrm{2h}}$, but one of lower symmetry. When the rings are perfectly perpendicular to each other ($\tau = 90^{\circ}$ and $\tau = 270^{\circ}$), the molecular point group will be $D_{\mathrm{2d}}$, like hydrogen peroxide, and every other intermediate value will be even lower in symmetry: $D_{\mathrm{2}}$, similar to skewed ferrocene (rather than staggered or eclipsed). To see how the irrep and its symmetry operations from a point group of higher symmetry map to those in a point group of lower symmetry, one needs a descent table. Here, most notably, all skewed structures are missing the inversion operation that would necessitate the gerade/ungerade subscript. Because most of the structures are of lower $D_{2}$ symmetry, it is easiest to represent all structures with $D_{2}$ point group labels.

For completeness,
$$\begin{array}{c|ccccc|cc} \hline
D_\mathrm{2d} & E & 2S_4 & C_2 & 2C_2' & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_2} & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_1} & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_2} & 1 & -1 & 1 & -1 & 1 & z & xy & \\
\mathrm{E} & 2 & 0 & -2 & 0 & 0 & (x,y),(R_x,R_y) & (xz,yz) \\ \hline
\end{array}$$
$$\begin{array}{c|cccc|cc} \hline
D_2 & E & C_2(z) & C_2(y) & C_2(x) & & \\ \hline
\mathrm{A} & 1 & 1 & 1 & 1 & & x^2, y^2, z^2 \\
\mathrm{B_1} & 1 & 1 & -1 & -1 & z, R_z & xy \\
\mathrm{B_2} & 1 & -1 & 1 & -1 & y, R_y & xz \\
\mathrm{B_3} & 1 & -1 & -1 & 1 & x, R_x & yz \\ \hline
\end{array}$$
See how there are fewer symmetry operations (middle set of columns) in $D_{2}$ than in $D_{\mathrm{2d}}$ than in $D_{\mathrm{2h}}$.


text:
Why are lower alcohols less flammable than their corresponding alkanes?
How to compare the general trend in amount of combustion in lower alcohols? (upto five/six carbons) Does the branching or straight chain have any effect on combustion?

Well all alcohols are not less flammable that all alkanes. The flash point of pure ethanol is 16.60 °C but the flash point of n-icosane is greater than 113 °C. Yes branching effects pretty much everything about a molecule.

Alcohols as well as alkanes can burn in air if the concentration is high enough. The flash point is the lowest temperature at which a liquid emits vapour in sufficient concentration to form an ignitable mixture with air near the surface of the liquid. Since this process depends on the evaporation of vapours from the liquid, the trend of flash point temperatures roughly follows the trend of boiling point temperatures.
$$
\textbf{} \\
\begin{array}{lll}
\hline
 & \text{Flash point in °C [1]} & \text{Boiling point in °C [2]} \\
\hline
\text{Methane}& \text{n.a.} & -161.5 \\
\text{Methanol} & \hphantom{-0}11 & \hphantom{-0}64.5 \\[3pt]
\text{Ethane} & \text{n.a.} & \hphantom{0}{-}88.6 \\
\text{Ethanol} & \hphantom{-0}13 & \hphantom{-0}78.24 \\[3pt]
\text{Propane} & -104 & \hphantom{0}{-}42.11 \\
\text{Propan-1-ol} & \hphantom{-0}23 & \hphantom{-0}97.04 \\
\text{Propan-2-ol} & \hphantom{-0}12 & \hphantom{-0}82.21 \\[3pt]
\text{Butane} & \hphantom{0}{-}60 & \hphantom{00}{-}0.5 \\
\text{Butan-1-ol} & \hphantom{-0}37 & \hphantom{-}117.6 \\
\text{Butan-2-ol} & \hphantom{-0}24 & \hphantom{-0}99.4 \\[3pt]
\text{Pentane} & \hphantom{0}{-}40 & \hphantom{-0}36.06 \\
\text{Pentan-1-ol} & \hphantom{-0}33 & \hphantom{-}137.6 \\
\text{Pentan-2-ol} & \hphantom{-0}34 & \hphantom{-}119.1 \\
\text{Pentan-3-ol} & \hphantom{-0}41 & \hphantom{-}123 \\[3pt]
\hline
\end{array}$$
Since alcohols have higher boiling points than the corresponding alkanes, also their flash points may be expected to be higher. The length and branching of the carbon chain can affect the flash point as well as the boiling point.
[1] “Flammability of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL.
[2] “Physical Constants of Organic Compounds”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


text:
What symbol should be used for the Michaelis constant?
I have found two different abbreviations for the Michaelis constant: $K_\mathrm M$ with capital subscript M, and $K_\mathrm m$ with a regular m. Which form is better for use in an article to be published?

Whatever form that the editors want.

Many journals or scientific societies have a list of recommended abbreviations.

In page 66 of the 2005 Green Book ("Quantities, Units and Symbols in Physical Chemistry"), IUPAC recommends that $K_\mathrm M$ be used as the symbol for the Michaelis constant. Note that $K$ should be italicised whereas $\mathrm{M}$ should not be italicised.
Thanks to Loong, who points out that the ACS Style Guide (Chapter 10, p 186) actually suggests $K_\mathrm m$. The Style Guide, however, also states that it is "not intended to be exclusive; alternative choices, in many cases, are acceptable".


text:
Applying salt to roads to melt ice--is it special?
Most everyone is familiar with applying salt to the roads during winter to melt ice.
I'm wondering if there's anything special about placing salt on the roads or would any other substance, sand for example, work equally well in melting the ice.

https://chemistry.stackexchange.com/questions/5023/why-calcium-chloride-is-used-to-melt-ice-over-sodium-chloride

https://chemistry.stackexchange.com/questions/10417/avoiding-snow-melting-with-salt

https://chemistry.stackexchange.com/questions/87310/melting-ice-with-salt-how-does-it-start?rq=1

https://en.wikipedia.org/wiki/Freezing-point_depression

Yes, applying salt to melt ice it is special:

Salt is a great ice melter because it causes “freezing point depression.” This means that salt helps in lowering the freezing point and, consequently, the melting point of water (the main component of snow and ice). In its pure state, water freezes at 0°C or 32°F. By using salt, that temperature can be lowered, which forces the ice to melt and prevents the water from freezing or re-freezing. This does not happen with  sand.
After melting, no additional effort is needed to clean the roads, because salt mixes with water. If you use sand, additional work will be needed to remove it from the streets afterwards.
Salt is much cheaper than other chemicals, although arguably more expensive than sand, depending on location.
Salt is recyclable, so no wastage or decompositon of salt is there in such process. Since salt becomes soluted in the melted ice, it gets washed away and absorbed by the ground. This makes salt hard to retrieve after use, and harms the environment by damaging the plants in the area.