text:
Why are entries missing on a solubility data chart for ionic compounds?
There is a solubility chart in my college chemistry class text for "Solubility of Ionic compounds in water". I have copied and pasted image of chart below.  I was surprised to find a dozen ionic compounds marked "--" meaning "no data".

Among the entries marked "no data" are

iron(III) iodide
iron(III) sulfide
iron(III) carbonate
iron(II) chromate
copper(II) iodide
magnesium chromate
silver hydroxide
and carbonate, chromate, and nitrate for Sn2+

It's hard to believe there is no solubility data on said ionic compounds. They should be either soluble, insoluble, slightly soluble, or an impossible compound.  I asked my professor, and he said most of said compounds are too unstable to get reliable solubility data.  In other words, the said cations and anions would not form a stable ionic compound long enough to test solubility.  He also said it might not be the case for all the "--" no data compounds. Which of the said "no data" compounds are impossible compounds as distinguished from possible stable ionic compounds for which there is really no solubility data?

Perhaps nobody has measured them yet. Or the authors did not find a readily available source.

All of iron(III) iodide, iron(III) sulfide, iron(III) carbonate, iron(II) chromate, copper(II) iodide either don't exist or are extremely unstable (guess for IronII chromate, but Fe2+ and )CrO4)2- will almost certainly undergo mutual redox). In fact of those listed it's only Magnesium Chromate that mildly surprises me, all the rest I suspect have at best very limited stability

I would ratrher say the data are incomplete. you may want to consult wikipedia - Solubility_table

In the case of magnesium chromate, good solubility in water has been known since 1931.  From Hill et al. [1], citing [2]:

The only figure on the solubility of the salt [as of 1940] appears to be that of Kohlraush[2], who reported that at 18°[C] 100 ml of saturated solution contained 60 g of magnesium chromate.

Most of the other compounds listed, as described in comments to the question, do not exist or are highly unstable because the ions are incompatible in the proposed solid. Iodide and sulfide ions are oxidized by iron(III), iodide is also oxidized by copper(II), chromate ions are reduced by iron(II) and tin(II), and tin(II) also reduces nitrate.  Carbonate ion is not redox-active, but fails to form a stable salt with some weakly-basic or amphotetic oxide forming metal ions like tin(II) or iron(III); a similar failure occurs with proposed silver hydroxide.
References
1.
Arthur E. Hill, Glenn C. Soth, and John E. Ricci, "The Systems Magnesium Chromate—Water and Ammonium Chromate—Water from 0 to 75°", J. Am. Chem. Soc. 1940, 62, 8, 2131–2134. https://doi.org/10.1021/ja01865a059
2.
Joseph Mellor, Treatise on Inorganic and Theoretical Chemistry, Vol. II (London: Longmanns, 1931), p. 275.

Besides missing the reference on magnesium chromate, the chart has some errors associated with oxides and sulfides.
Magnesium oxide is listed as insoluble in water, which it is if it is calcined at a high temperature, but otherwise it becomes (kinetically) reactive forming the hydroxide; see this answer. Barium oxide is not "soluble", but reacts in the same way as calcium, strontium, and low-temperature calcined magnesium oxides.
Among sulfides, the $\ce{S^{2-}}$ ion is a relatively strong base, so the alkali and alkaline earth metal sulfides that "dissolve" in water are in fact all prone to hydrolysis.  Calcium and magnesium sulfides do undergo more complete hydrolysis, as the corresponding hydroxides have limited solubility and their precipitation limits the basicity of the solution. Sulfide ion is also too strongly basic to coexist with ammonium ion, so ammonium sulfide as $\ce{(NH4)_2S}$ properly does not exist.

Most of the $10$ compounds from your list simply do not exist. This is why they are not reported in your table. For example iron(III) iodide does not exist. When trying to produce it by mixing $\ce{Fe^{3+}}$ and $\ce{I-}$ ions, a redox chemical reaction occurs spontaneously according to :
$$\ce{2 Fe^{3+} + 2 I^- -> 2 Fe^{2+} + I2}$$ For the other compounds, similar reactions happen spontaneously which prevent the compound to exist :
$$\ce{2 Fe^{3+} + 3 S^{2-} ->  2 FeS + S}$$
$$\ce{2 Fe^{3+} + 3 CO3^{2-} + 3 H2O -> 2 Fe(OH)3 + 3 CO2}$$
$$\ce{3 Fe^{2+} + CrO4^{2-} + 4 H2O -> Cr(OH)3 + 5 OH- + 3 Fe^{3+}}$$
$$\ce{2 Cu^{2+} + 4 I^- -> 2 CuI + I2}$$
Magnesium chromate does exist, as mentioned by Oscar Lanzi. The other compounds of the list do not exist. For example $\ce{AgOH}$ and tin compounds, according to :
$$\ce{2AgOH -> Ag2O + H2O}$$
Sn compounds, like tin(II) chromate and tin(II) carbonate are spontaneously decomposed exactly like similar iron compounds. See above. Tin nitrate is spontaneously oxidized according to :
$$\ce{Sn^{2+} + 2 NO3^- -> SnO2 + 2 NO2}$$

According to Housecroft and Sharpe FeI3 has been isolated in "inert conditions" via a photolytic reaction: 2Fe(CO)4I2 + I2 -> FeI3 + 8CO.  They do say, though, that it "readily decomposes"

"Magnesium chromate does exist".  And is used industrially, and has been known to be soluble since 1931 (see my answer).  -1 for the people who made that chart for missing that.

@Ian Bush. You are right. But I presume this table was done for students trying to mix solutions of different cations and anions. And not in inert conditions like you state.


text:
Calculate Michaelis-Menten constant of enzyme catalyzed reaction
I have:
$$\begin{array}{c|c}
\dfrac{v }{ [S]}/\pu{s^{-1}} & v \cdot 10^2/\pu{mol dm^-3 s^-1} \\ \hline
0.257\ & 5.15\\
0.895 & 4.48\\
2.00\ & 3.35\\
3.59\ & 1.8\\
4.82\ & 0.48\\
\hline
\end{array}$$
And I want to calculate $K_M$ graphically.
I know that I should use the equation:
$$\frac{1}{v}=\frac{1}{v_\mathrm{max}}+\frac{K_M}{v_\mathrm{max}[S]},$$
and that I should plot $1/v$ against $1/[S]$, and then the slope will equal $K_M$/$v_\mathrm{max}$.
I began with converting the data given from $v/[S]$ to $1/[S]$ by dividing each data point with $v$. This gave:
$$\begin{array}{c|c}
\dfrac{1}{[S]}/\pu{dm^3 mol^{-1}} & v \cdot 10^2/\pu{mol dm^-3 s^-1} \\ \hline
4.99*10^{-4}\ & 5.15\\
0.0019 & 4.48\\
0.0059\ & 3.35\\
0.0199\ & 1.8\\
0.1004\ & 0.48\\
\hline
\end{array}$$
Plotting $1/v$ against $1/[S]$ gives the graph:

But when I want to calculate $K_M$ through:
$$\text{slope} \times V_\mathrm{max} = \text{slope} \times \frac{1}{\text{intercept}}$$
I don't get the answer which is in our answer key. Unfortunately, only the answer for $K_M$ is given, and not how to get the answer. The answer is $K_M = \pu{0.0102 mol dm-3}$. Could someone explain where I am going wrong?

Your calculations are wrong on $\frac{1}{\ce{[S]}}$.

You somehow reversed the order of the data points, too. That's why there is a hyperbole and not a straight line. Because of the format of the data, it is easier to do an Eadie Hofstee fit instead of the Lineweaver-Burk you attempted.

When I am editing your question, I realized you have plot $\frac{1}{[S]}$ vs $v$ instead of $\frac{1}{[S]} $ vs $\frac{1}{v} $. That's why you got a hyperbole instead of straight-line as  Karsten Theis suggested.

@MathewMahindaratne I think the values plotted represent 1/v (off by a factor 100), but are in the wrong order, i.e. the first substrate concentration is plotted against the last rate etc.

What you have given is:
$$\begin{array}{c|c}
\hline
\dfrac{v}{[S]}, \ \pu{s^{-1}} & v, \ \pu{mol dm^-3 s^-1} \\ \hline
0.257  & 5.15 \times 10^{-2}\\
0.895 & 4.48 \times 10^{-2}\\
2.00  & 3.35 \times 10^{-2}\\
3.59  & 1.8 \times 10^{-2}\\
4.82  & 0.48 \times 10^{-2}\\
\hline
\end{array}$$
Michaelis-Menten equation for enzyme kinetics is:
$$v = \frac{V_\mathrm{max} \cdot [S]}{K_M + [S]} \tag1$$
When rearrange this equation (reciprocal) to get Lineweaver-Burk relationship:
$$\frac{1}{v} = \frac{1}{V_\mathrm{max}} + \frac{K_M}{V_\mathrm{max}[S]} \tag2$$
When multiply the equation $(2)$ by $v \times V_\mathrm{max}$, you get:
$$V_\mathrm{max} = v + \frac{vK_M}{[S]} \tag3$$
When rearrange the equation $(3)$, you get $y = mx + c$ type equation (straight-line equation):
$$v  = - K_M \cdot \frac{v}{[S]} + V_\mathrm{max} \tag4$$
The data you have is $\frac{v}{[S]}$ and $v$, which fix the relationship given in the equation $(4)$. Thus, if you can plot them against each other ($\frac{v}{[S]}$ versus $v$) you'd get a straight line with negative lope. The numerical value of the slope is $K_M$ so you can calculate K$_M$ graphically. Also, the $y$-intercept of the graph is equal to $V_\mathrm{max}$:

From the equation of the graph: $\text{|The slope|} = 0.0102$ and $y\text{-Intercept} = 0.054$
Thus, $K_M = \pu{0.0102 mol L-1}$ and $V_\mathrm{max} = \pu{0.054 mol L-1 s-1}$
Once I found OP's mistake of plotting, I was curious to see how this set of data would show the Lineweaver-Burk relationship. Thus, I make the table for $\frac{1}{[S]}$ and $\frac{1}{v}$:
$$\begin{array}{c|c}
\hline
\dfrac{1}{[S]}, \ \pu{L mol-1} & \dfrac{1}{v}, \ \pu{L s mol-1} \\ \hline
4.99  & 19.42 \times 10^{-2}\\
19.98 & 22.32 \times 10^{-2}\\
59.70  & 29.85 \times 10^{-2}\\
199.44  & 55.56 \times 10^{-2}\\
1004.17  & 208.33 \times 10^{-2}\\
\hline
\end{array}$$
As predicted by  Lineweaver-Burk, the plot of $\frac{1}{[S]}$ versus $\frac{1}{v}$ is a straight-line with a positive slope:

From the equation $(2)$, which is the equation of the graph: $\text{The slope} = 0.1891 = \frac{K_M}{V_\mathrm{max}}$ and $y\text{-Intercept} = 18.379 = \frac{1}{V_\mathrm{max}}$
Thus, $V_\mathrm{max} = \frac{1}{18.379} = \pu{0.0544 mol L-1 s-1}$ and $K_M = 0.1891 \times V_\mathrm{max} = 0.1891 \times 0.0544 = \pu{0.0103 mol L-1}$
Both graph give the same values for $K_M$ and $V_\mathrm{max}$.

Very nice answer indeed: went the extra mile! Glad I was the first upvoter! It should get the green checkmark ASAP!

Thank you very much! Truly appreciate it!

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text:
Why is the IUPAC definition of adsorption so different from other definitions?
NB: This question has been edited. Originally, some of the confusion was caused by me not knowing that condensed phase could refer to solid phases as well. This has been removed, but the confusion caused by the other elements still remains and even the first answerer to this post was unable to clear that up.
The Gold Book's definition of adsorption

An increase in the concentration of a dissolved substance at the interface of a condensed and a liquid phase due to the operation of surface forces. Adsorption can also occur at the interface of a condensed and a gaseous phase.

What I've learned is that adsorption is when a fluid, or the solute(s) within a liquid, adhere to a surface. The IUPAC definition however, only recognizes the case when the solute(s) within a substance adhere to a surface. A more limited, and to my knowledge, less useful definition. Furthermore, the formulation is very ambiguous. It does not state if it is the condensed phase or the fluid phase that contains the solute(s), or if either can contain the solute(s). The middle option would necessarily be in contradiction to the IUPAC's definition of a solution:
The Gold Book's definition of solution (shortened):

A liquid or solid phase containing more than one substance, ...

This definition does not count gaseous phases as solvents. The fluid phase can be a gas, but if the dissolved substance is always within the fluid, then in cases where the fluid phase is a gas, it would be a solvent gas. This is assuming that dissolved substance is the same as solute.
Now, for part two of this mess:
Furthermore, the IUPAC definition of absorption encroaches onto the semantic territory of adsorption, at least the territory that I've seen and been taught.
The Gold Book's definition of absorption

The process of one material (absorbate) being retained by another (absorbent); this may be the physical solution of a gas, liquid, or solid in a liquid, attachment of molecules of a gas, vapour, liquid, or dissolved substance to a solid surface by physical forces, etc.

The last part there is the, or a part of the, definition of adsorption. It described surface attraction, which I thought was clearly within the domain of adsorption. The first part of the definition says that the absorbate and absorbent create a physical solution. I cannot find any definition of that, but I assume they are referring to a homogeneous mixture where the solute isn't disassociated, but still kept within a solution complex. However, an absorbate isn't (necessarily) homogeneously distributed within the absorbent. This means they're violating their own (and countless other) defintion(s) of solution.
As mentioned, a solution is a single phase containing more than one substance, and a phase, as defined by the IUPAC, is homogeneous.
The Gold Book's definition of a phase:

An entity of a material system which is uniform in chemical composition and physical state.

Are these IUPAC definitions outdated? If so, what is the most authoritative definition of adsorption and absorption. I like to use words consistently, but given how inconsistent and/or ambiguous scientific literature can be, I want to pick the best definition in order to influence the literature, if however minutely, towards more consistency and away from ambiguity.

That's a lot of writing to criticise what is perfectly reasonable description.

@Mithoron I think the same and I've also upvoted the answer below. Yet, I am now confused by the overlap between Adsorption and Absorption, as the definition of the latter can include the former. Perhaps the key -  as I was insisting few days ago commenting a related question - is that absorption does not results in gradients. It is merely "volumic".

@Mithoron The definition is perfectly reasonable in one regard; the fact that the surface adhered to is that of a condensed substance. I did not know that "condensed" could refer to both liquid and solid phases, and not knowing that caused a part of the confusion. However, as user theorist said in his answer, he, nor me, are aware of why the definition necessitates that the adhering substance be a dissolved substance. Most cases of adsorption I've heard of involves a fluid adhering to the surface, not solutes within them (or dissolved substances within them, it appears there's a difference).

@Mithoron: solved. Indeed as I said before, in Absorption, there is no any potential for a concentration gradient per sé. It is like a fluid trapped in a sponge, which (let aside eventual Adsorption) is identical to the liquid outside. Now there is not even this point to justify so much words, indeed.

"Adsorption can also occur at the interface of a condensed and a gaseous phase". Why does not fixes your trouble but opposite causes one?

In a short email correspondence with the IUPAC, where I raised these issues, I was told that efforts are being made to revise the Gold Book. Jim McQuillan, COB of the Commission on Physiochemical Symbols, Terminology and Units, wrote this in an email:

We looked at the terms Adsorption, Absorption, and Sorption and found issues with all of the Gold book entries which underlined the urgent need to have these terms revised.

Exactly why these issues exist hasn't been made clear, and that isn't really the main issue (although my question does make it seem a bit like that). It does seem however, that this may be the result of outdatedness and/or a lack of consistency (perhaps due to poor coordination between the authors of the Gold Book, or maybe a result of poor coordination between authors within chemistry literature in general).
However, the important thing here is that the question of whether or not these definitions are problematic, has now been resolved; the IUPAC has directly confirmed the definitions' need for revision, and that the definitions are misaligned with IUPAC recommendations (see project details).

Excellent job. IUPAC glossary is not a heavenly book. All these definitions, symbolism are decided by human beings. Ten or fifteen years ago, the definition of extensive and intensive variables were messed up. I wrote to them, recieved no response but it was corrected.

The IUPAC defintion isn't more limited than the ones you've read. Instead, it is more general.
Your confusion results from not being familiar with the term "condensed phase". You think a condensed phase is a 'more condensed liquid', and consequently think IUPAC is referring to a surface formed between a liquid and a 'more condensed liquid':  "where there is an interface of two liquids, one containing a solute, and one being more condensed".
No.  A condensed phase means a liquid or a solid.  Thus, when IUPAC says the following, what they mean is that adsorption can occur at liquid-liquid, liquid-solid, gas-liquid, and gas-solid interfaces.

An increase in the concentration of a dissolved substance at the
interface of a condensed and a liquid phase due to the operation of
surface forces. Adsorption can also occur at the interface of a
condensed and a gaseous phase.

Is the IUPAC definition completely general?  No.  It doesn't account for adsorption between supercritical fluids and solids. But it at least covers all the routine phase combinations.
As to the issue of whether IUPAC's defintion of solutions (which limits the solvent to be a condensed phase) contradicts its defintion of adsorption: While it could have been written more explicitly, a close read indicates there's not necessarily a contradiction.  Let's take each sentence in turn:
The first sentence concerns a liquid phase in contact with a solid or liquid surface.  Here the adsorbent is referred to as a "dissolved substance", which would make it the solute within a liquid solution:

An increase in the concentration of a dissolved substance at the
interface of a condensed and a liquid phase due to the operation of
surface forces.

Note that the solute can be (prior to its dissolution) a solid, liqud, or gas.
The second sentence concerns a gaseous phase in contact with a solid or liqid surface.  Here no explicit mention is made of a dissolved substance:

Adsorption can also occur at the interface of a condensed and a
gaseous phase.

Thus we could parse IUPAC's language to mean that adsorption involves either the solute within a liquid solution that is in contact with a liquid or solid surface, or a gas that is in contact with a solid or liquid surface (where that gas could be pure, or part of a mixture).

I see, I did suspect that perhaps "condensed phase" meant anything more condensed than the gaseous phase. However, why does the definition only involve the increase in concentration of dissolved substance at the interface. Can't it also be the liquid or gas that adheres to the surface, and not just solutes within them? According to the IUPAC definition, a gas cannot be a solvent, so according to the IUPAC, what is adhering to the surface when a gas is the adsorbate? Also, why does the absorption definition involve adsorption: ... attatchment of ... to a solid surface.

According to the IUPAC definition of a solution, the solvent must be a liquid or solid. However, the IUPAC definition of adsorption is not restricted to solutions. Thus you could have adsorption of, e.g., gaseous acetone dissolved in gaseous nitrogen, or gaseous acetone dissolved in liquid water, onto a graphite surface.  As to why IUPAC's definition of adsorption is limited to dissolved substances, I can't say. Perhaps someone else here can. Pure acetone at a pressure of p*, and dissolved acetone at a partial pressure of p* in an approximately inert gas, should adsorb approx. equivalently.

I thought that if a substance A was dissolved in substance B, substance A is a solute and substance B is a solvent. But dissolution can happen outside of the context of solutions? If something is dissolved in something else, aren't we necessarily talking about a solution?

@user110391 What is solvent and what is solute is based on the chemist explaining the system. It is entirely* semantically. (*Well, there are cases where one makes more sense than the other, but it is still a choice)

@StianYttervik First of all, by mentioning the idiolectic differences in people's definition of solutions, are you then confirming that a substance that has a dissolved substance within it is necessarily a solution? If so, the chemist describing the system here is the IUPAC, and they're self-contradictory, since the absorbent can be a gas, and the absorbent (or absorbate, they're not clear on that so I assume both) is always a solvent according to the IUPAC. Thus, the IUPAC says that a gas cannot be a solvent but sometimes has to (not exactly though, depending on the absorb(ent/ate) thing).

I believe I should edit my question to more aptly fit the current lack of understanding and ambiguity. The core question remains the same, but the sources of confusion focused on will change, as well as one source of confusion being removed (the thing about condensed phases not necessarily being liquid but also solid).

@user110391 Well, IUPAC takes care to not describe a system, in this case. When I, for instance, refer to a 40% solution of ethyl alcohol in water, I make the distinction what is solute and solvent. You could just as easily say a solution of 60% water in ethyl alcohol. We'd be contrasting in what we emphasized, but we are in the end describing the same thing. Which is why the IUPAC must be tongue-twistingly precise, to the point of obfuscation.

@StianYttervik Your response seems a bit irrelevant. You didn't answer the whole of my comment, nor are you really answering the core of it in a relevant way. I am not talking about how to define which substances are solutes and which are solvents, as I've already been made aware of the lack of absolute rules there. What I'm talking about, more clearly shown in my revised post, is the ambiguity of the adsorption definition allowing for an interpretation where a gaseous phase is a solvent, which is in contradiction to the IUPAC definition of solutions.

@user110391  See the update I added to the end of my post.

The first sentence defines adsorption as the increase in concentration of a solute at the interface of a liquid and condensed phase. The second sentence clarifies that this interface can also be between a gas and condensed phase, yet it does not say that the increase in concentration of a solute does not apply there as well. As such, it is logical to assume that it does apply there, as the word is initially defined as such. If one doesn't use this logic, then the second sentence is basically just saying that "something" happens at the interface between a gaseous phase and condensed phase .

To follow up on my last comment: I get that a close read does not necessarily mean that adsorption between a gas and condensed phase involves a dissolved substance, but if it doesn't, then adsorption within the sentence Adsorption can also occur at the interface of a gaseous and condensed phase. actually means nothing. If adsorption is the increase in concentration of a solute only at the liquid/condensed phase interface, then that tells us nothing about what adsorption is at the gaseous/condensed phase interface.

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text:
Energy per atom in a parallel radioactivity equation
I was reading an example in my textbook and the question goes like this:

In a parallel radioactive decay, find energy liberated per atom of 'A' in MeV.

The solution is proceeded like this:
$$
 \lambda _{1}/\lambda _{2}= \frac{(T_{1/2})_2}{(T_{1/2})_1}=\frac{20}{60}=\frac{1}{3}
$$
Energy per atom of A is $$= 20\times\frac{1}{4}
+40\times\frac{3}{4}=35 MeV
$$
I have understood how we have got the first step, but I'm unable to understand the second step. Why are 1/4 and 3/4 specifically being multiplied to 20 and 40, respectively? Also, where are we using the ratio we got in the first step, in the second step? It has got me all confused. In short, how is energy per atom of A calculated?

1 and 3 in the ratio are the weights in weighted average...

A decay energy of 20 MeV or even 40 MeV (!) with a long half-life of 20 s or 60 s? What kind of radioactive decay is that supposed to be?

@Loong Imaginary ;> It's rather an arithmetic exercise and IMO such stuff should be banned altogether.

1:3 is used as 1/1+3 and 3/1+3.

The yield of C is $k_{AC}/(k_{AC}+k_{AB})$ where $k=1/T$ and is 3/4 thus 1/4 goes to B which gives 35 MeV as the answer.

You may state that, in $60$ s, one A atom has liberated $20$ MeV by the first reaction. In the same time, three A atoms have liberated $3·40$ MeV by the $2$nd reaction. The total energy liberated in $60$ seconds, is $120 + 20 = 140 $ MeV.- As a consequence, the average energy liberated by one A atom is $140$ Mev$  /4 = 35$ MeV


text:
Can you create a temporary bond in between an atom and a filter by electrically charging the filter?
I want to be able to turn a filter on and off by electrically charging the filter so that it creates a temporary bond that blocks up the filter then, by discharging the filter, releasing the temporary bonds and allowing particles through again. Is this possible?

This is what an N-95 mask does, but is always turned on.

Yes, it's possible: https://en.wikipedia.org/wiki/Electrostatic_precipitator

You about 100 years late in applying for a patent ( gas phase). Cottrell precipitator is the first name that comes to mind.

In a sense, you can increase and decrease flow through an osmotic filter with electricity. The process is called electro-osmosis. This can even be used to create a pressure differential in the electroosmotic pump.


text:
The pH of tap water changes with time
My following questions are related to $\mathrm{pH}$ values in irrigation water in plants.
I got a $\mathrm{pH}$ meter and I am measuring the tap water and I get an value about 8.
My first question: Why it is that high?
To get the $\mathrm{pH}$ down, I am using a $\mathrm{pH}$ regulator of phosphoric acid.
My second question: I now have some tap water with $\mathrm{pH}$ of about 6.5 but when this water sits for about 1-2 days, the $\mathrm{pH}$ increases back to 8. Why is it increasing?
Edit: the water with $\mathrm{pH}$ 6.5  tap water adjusted with small amounts of phosphoric acid. This then increases back to 8 when it sits for some time.

It's that high because your water supplier is (1) giving you water from a confined carbonate-hosted aquifer and a pH of around 8.3 is expected; and it bounces around because (2) it's sounding more and more like you're getting water from a well, and 6.5 is the pH of rainwater, so I'm guessing it rained near you or your water source recently, but the buffering capacity exceeds acid contributions from ambient carbon dioxide after a few days. More info, please.

What happens if you don't expose the water to air?

Pure water maintained in contact with air gets a pH value changing between $5.2$ and $5.8$ due to the $\ce{CO2}$ amount in the air, which changes from place to place. This is the consequence of the reaction : $\ce{CO2 + H2O -> H^+ + HCO3^-}$ The solution is slightly acidic because of these $\ce{H+}$ ions. Pure water in contact with air and with chalk or calcium carbonate rocks gets a pH value between $8.2$ and $8.5$, This is due to the fact that the $\ce{H+}$ions  produced by the previous rection have reacted with $\ce{CaCO3}$ according to $\ce{CaCO3 + H^+ -> Ca^{2+} + 2 HCO3^-}$


text:
Can color-changing leuco dyes be used to indicate when a certain molecule binds to another in a textile fabric?
Thermochromic leuco dyes change color when the temperature changes. I want to use leuco dyes to make a color-changing fabric. However, instead of using temperature to induce color changes, I want the color to change when a molecule woven into the fabric binds to another molecule, sort of like a lateral flow/pregnancy test.
Can leuco dyes be used to signal when a molecule binds to another and cause a color change in textiles/fabric?

Okay thank you. What I meant was I want to use the technology of the color changing fabric from leuco dye. However, instead of using temperatures, I wanted to see if a color change could occur if a molecule woven into the fabric binds to another. Sort of like a lateral flow/pregnancy test.

Ok, I reopened the question. Why limit yourself to leuco dyes though? What have you researched regarding using such dyes as indicators? Would the leuco dye respond directly or indirectly to binding of the molecule?

I just thought the dye could work like the thermachromic. However, I suppose it could be another type of dye. I haven’t been able to bind any dye that will react when binding takes place in textiles.

Thermochromic means it responds to temperature. You seem to be looking for something else, something that responds to binding of a particular type of molecule.

Yes that’s what I’m trying to find.


text:
Odd/Even swap rule for chiral compounds in stereochemistry
I had this question asked in my exam, where I was supposed to find if this pair is a diastereomer:

Now, I have learnt this rule where: Making odd number of swaps in a molecule means they both are enantiomers, and making even number of swaps means they are identical.
So according to this rule, I made even no. of swaps in the molecule to the right, which means it is identical to the inital molecule that was on the right. Now this newly formed molecule superimposes the molecule to the left perfectly. Hence, in my opinion the two molecules in the picture above should be identical. But the answer given in my answer key is that they are diastereomers, and not identical.
Is there any flaw in my justification?

The trick you are using is not always correct. Follow the standard definitions rather than some rules which cease to be followed outside of some specific cases.

These are optically active optical isomers which are not identical and thus diastereomers.

Out of curiosity, which are some of these specific cases where this trick fails to follow and why? Are these cases just arbitrary exceptions, or is there any logic as to why this trick doesn't apply?

You should be asking where does this trick apply because I don't see many cases there this is absolutely correct.

@Aashita All CC(Chiral centre)-> odd swaps-> Enantiomers and All CC-> even swaps-> Meso and In rest cases diastereomers.

While this rule is certainly not infallible, it is often used in organic compounds, especially for Fischer projections.

The odd/even rule implies that no change of configuration occurs if an even number of swaps are made at a single chiral center and change of configuration occurs if an odd number of swaps are made at a single chiral center.

Here, you have made 3 swaps at both ends which means you changed the configuration of 2 out of 4 chiral centers. Hence, you can make the inference that 2 chiral centers (C-2 and C-3) are the same, while another 2 (C-1 and C-4) are different. These are hence diastereomers.

I am assuming that you thought you made 3 + 3 = 6 swaps, so the compound should be identical. But the odd/even swap rule is only applicable to a single chiral center, hence the error.

Finally as pointed out already, this rule doesn't apply anywhere where the central atom doesn't have a tetrahedral geometry. As counterexamples, coordination complexes often have non-tetrahedral geometries:

octahedral: $\ce{[Co(NH3)6]^3+}$
square planar: $\ce{[PtCl4]^2-}$
dodecahedral: $\ce{[Ce(NO3)6]^2-}$

Note that these specific examples are optically inactive but by changing the ligands for different ones, it is possible to create optically active versions with the given geometries.
However, since organic chemistry mostly involves tetrahedral carbon as the chiral center, the rules certainly do apply there.

Hello, please refrain from using unnecessary abbreviations like "no."; it's preferable to type the word "number" instead. Please also take a look at FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange? In particular, chemical formulae should be typeset as $\ce{H2O}$, not just $H_2O$.

@Ex_Machina Thanks a lot for your answer, it helped me. I was going wrong at the part: even number of swaps are made at a single chiral center only.


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Different statements of second law of thermodynamics
This may sound trivial but I am having a hard time linking different statements of second law. What I get from second law is that heat cannot be completely converted into work and hence efficiency of heat engine is always less than one. Please explain what it has to do with entropy. Also from second law, entropy of any isolated system never decreases. How are the two statements similar? Please use simple layman terms, I am simply trying to grasp the idea.

It is not to circumvent your question but Wikipedia has an article going through the different formulations providing also an historical overview. It seems sufficiently on layman terms and would be somehow redundant to re-do it here. Still you could get a nice summary from a willing user. https://en.m.wikipedia.org/wiki/Second_law_of_thermodynamics

Please use simple layman terms, I am simply trying to grasp the idea.

Thermodynamics is a rigorous field, which is one of the things that makes it so powerful. Explaining something in simple layman terms will make it less powerful and more mysterious. One way around this is to pick and choose some parts that are necessary to answer the question and ask you to suspend your disbelief while following the argument. In this case, we need to know that in a ideal heat engine, the change in entropy of the reservoirs is:
$$ \Delta S_\mathrm{reservoir} = \frac{Q_\mathrm{rev}}{T}$$
Also, the overall change in entropy will never be negative (and for an ideal heat engine, it will be zero).

What I get from second law is that heat cannot be completely converted into work and hence efficiency of heat engine is always less than one. Please explain what it has to do with entropy.

You can figure that out because if you extract heat from a reservoir, the entropy of it increases. Your process has to increase the entropy of something else by at least the same amount, so you can't just turn thermal energy into work.

Also from second law, entropy of any isolated system never decreases. How are the two statements similar?

If it never decreases, it is zero at best. You can apply this general statement to the specific case of an ideal heat engine. In this process, you have to heat up the cold reservoir to match the entropy decrease of taking heat out of the hot reservoir. Because we are dividing by $T$, and it is different for the two reservoirs, we can extract some of the energy in form of work without breaking the second law. The higher the temperature difference, the more work we can extract for a given amount of heat transfer.

$\Delta S_\mathrm{reservoir} = \frac{Q}{\mathrm{T}}$; the reversible restriction is unnecessary because, from the (infinite) reservoir's perspective, all heat flow is reversible. But you know that, so I'm curious what was your reason for including it?

∆S = q/T, does this equation hold for irreversible?  @theorist

@pravakarwang  Only for the reservoir.  The reservoir is assumed to be so large that heat flow from the system only has an infinitesimal effect on the reservoir, and thus is effectively reversible.

@theorist got it. I also wanted to ask why that equation holds only for reversible cases in general? Will be great if u can clarify without creating another post. Thanks

@pravakarwang.  The answer to that is an entire chapter on thermodynamics.  I'd suggest searching the site and elsewhere.  This should get you started:  https://chemistry.stackexchange.com/questions/1273/calculating-entropy-why-consider-a-reversible-path  and https://chemistry.stackexchange.com/questions/112088/calculating-entropy-change-reversible-vs-irreversible-process?rq=1
and https://chemistry.stackexchange.com/questions/24518/formula-of-entropy-change?rq=1

@theorist Not everyone knows the implicit assumption that reservoir means infinite well-stirred reservoir. In a real heat engine, we would also talk of reservoirs, but these are neither infinite nor well-stirred. Also, it is hard in a simple equation to make the link that the equality holds only if we are talking about a reservoir, so folks jump to incorrect stuff like $$\Delta H_{rx} / T = \Delta S_{rx} $$


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Why neglecting Bradford Solution Volume in the assay?
I would like to know why do we not need to take account of Bradford agent solution in dilution calculation?
For example, $\pu{2.5 \mu g mL-1}$ BSA will calculate from $\pu{0.4 \mu L}$ of $\pu{1 mg mL-1} \ \text{BSA} + \pu{159.6 \mu L}$ of distill water. The additional Bradford reagent ($\pu{40 \mu L}$ in my experiment) does not count in the dilution calculation. This seems mysterious to me since the total volume is $\pu{200 \mu L}$. If the volume of Bradford reagent counted, the concentration should have been $\pu{2 \mu g mL-1}$.
I try to search for publications regarding to the protocol and most of them said just ignore the Bradford volume with no explanation. Could someone kindly explain the mechanism behind this?
Edited: I aim to determine protein concentration via Bradford assay. The standard curve is prepared by serial dilution of BSA + distill water + Bradford
serial diluion was as below:

I understand completely how to calculate the unknown concentration. However, I just want to know that why 'Bradford solution' volume can be ignored during the calculation.
Regards

How did you prepare your calibration curve? Is BSA your standard? If so, describe how you prepare standards to make the calibration curve. So, we know what you are doing.

I've edited the question, thanks.

It doesn't matter you included Bradford reagent or not in your calculation as long as you treated the same for all reading including unknowns as shown in the data. That is because, when you find the concentration of unknown sample from calibration curve, you back calculate to get the initial concentration of undiluted solution. This means you are going to substract the volume of Bradford reagent at the end.


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Why is Retention factor for Iodide greater than Bromide?
I did a separation of Bromide and Iodide ions using paper chromatography technique for 0.3% solutions of Bromide and Iodide of sodium. I got the  Retention factor equal to ~0.71 for Iodide and ~0.47 for Bromide.
So, I want to ask if ion size is playing any role in deciding this retention factor? Or, is there some another perspective that I am missing here? Specifically, what is the reason for this difference?

Asking this way is not how this site works. Answers on CH SE site are figuratively paid by the user's own effort.  Before asking, you are supposed to think, search, read, think, think more. If you still fail to find the answer, then ask, but together with explicit compact summary of ideas a/o results you got until then.

@Poutnik, I have researched on the web and most of the articles say that solvent plays a role. Here, I took the same solvent. So, I was wondering if ion size plays a role or not as I wasn't able to find a convincing source stating the same. If you get one, do let me know. And saying a user that he hasn't done any research without knowing what he has done, prior to asking... is rude.

I have not said so, so I reject accusation to be rude. Effort not shown is effectively effort not done. I could say I have been searching libraries and internet for weeks and still do nothing. Readers expect some evidence of your effort. Even if you found nothing, you are still expected to present your ideas about it.

Generally, ion radius plays a role in most processes where ions are involved. Why should Rf be exception ? Have you considered electrostatic interations of ions and cellulose polar OH bonds ?

That's what I was wondering, @Poutnik. I guessed that ion radius must play some role in this case, but couldn't find anything regarding that. No, I didn't consider the interaction of ions and cellulose polar OH bonds yet. I will do it right away, and will let you know.

Evaluate which bonding is stronger,  O(-)-H(+).....I-  OR  O(-)-H(+).....Br-, if Br- is smaller.

Obviously the latter, I guess. Okay, that means the OH will form stronger hydrogen bond with Br and won't allow those ions to move further up the paper; while it is not so much of a case for the former as it is less stronger bond. Am I right, @Poutnik?

Let us continue this discussion in chat.

@Sanu_012, Please mention your mobile phase.

The question has been answered via chat. Thank you all.


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Changing the level of theory for organometallic compounds
I'm finishing my Ph.D. in Inorganic Chemistry and having a debate with my supervisor. The work in question have two organometallic compounds A and B. The structures are based on X-Ray Diffraction, so I already have a good molecular frame and just need to optimize the hydrogens. For that I'm using PBEh-3c that usually work well for organometallics.
In the next steps, using the now pre-optimized structure, I have changed the level of theory for each task, selecting the (supposedly) best level for each, based on benchmarks:

Broken-symmetry (magnetic coupling): TPSS/Def2-TZVP
Bond Orders and spin distribution: SCS-MP2/Def2-TZVP
Energy Decomposition: DLPNO-CCSD(T)/Def2-TZVP
TDDFT: wB2PLYP/Def2-TZVP

My Supervisor is not comfortable with the mix of levels of theory, but I believe that it is OK if I'm sure that I have the correct ground-state geometry and I'm not comparing the same property for A and B with different methods (for example TDDFT using PBE for compound A and wB2PLYP for compound B).
So, the question is: is it OK to change the level of theory of my calculations to select the best setup for the property I'm trying to calculate if I know that I have a good ground-state geometry and that my calculation is not displacing the geometry?

Hi, @Buttonwood. I have not added H, just optimized the H positions using DFT because, as you said, their positions are idealized. I have listed the properties I'm interested in: broken-symmetry, absorption using TDDFT, bond orders...

I am slightly curious as to why you are not using DLPNO-CCSD(T) for all of the methods if you can afford it. I might be wrong, but I thought that DLPNO-CCSD(T) would be the most accurate of all of the methods.

@ShoubhikRMaiti I'm using it to get the Local Energy Decomposition calculated by the ORCA software. It is based on DLPNO-CCSD(T), that gives an almost full CCSD(T) result while being less expensive. And, actually, I can only afford a single point.

The way you are asking this question it appears that you are doing full structure optimisations for each property which is probably not the case. There should be some level of calibration of the ground state structure. I'm not sure that PBEh-3c is a good enough guess on the structure, so that should be made crystal clear. After that, if you qualify your results well enough, I see absolutely no issue with using the best tool for the job. I'm sure you've done your research and found benchmarks (or similar systems computed similarly); quote that and you should be fine.

I find this a very useful chemistry question and totally on topic here. (And I prefer to keep it here.) However, it would be totally unfair to not tell you about [mattermodeling.se]. It might have been asked there already, so it's probably worth checking out.

I mean I hate to suggest it, but the only way to settle this argument for real is to just do all of the calculations with all four methods and compare the results. I don't know that much about metallic compounds but I'd be pretty surprised if DLPNO-CCSD(T) wouldn't be the best for all of the properties if you can afford to do that. Even if it's not, that wouldn't give me a lot of confidence that DFT is getting the right answer for the right reason...


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Can physical absorption happen despite the formation of products?
As far as I know, the absorption of $\text{SO}_2$ in water goes under physical absorption, meaning the sulfur dioxide bubbles are held in place by Van der Waals forces, in this case Keesom forces, since both water and sulfur dioxide molecules are dipoles. This table classifies the absorption of sulfur dioxide in water as physical absorption, taken from this site.
However, sulfur dioxide does create products (within an equilibrium) with water.
$$\ce{SO2 + H2O ⇌ HSO3^- + H3O+}$$
and
$$\ce{SO2 + OH- ⇌ HSO3^-}$$
Now, with chemical absorption, the absorbate is held within the absorbent due to interfacial chemical bonding. Could it be that these products don't aid with the absorption at all? Or, could it be that they aid with the absorption in conjunction with, but perhaps to a lesser degree than, the Keesom forces, thus making the absorption more physical than chemical? If so, is this perhaps the norm?
Or, is this absorption actually a chemical one?

When you consider the dissolution of a gas like $\ce{SO3}$ , $\ce{CO2}$, $\ce{SO2}$ etc. there are three equilibria existing at the same time:
$$\ce{SO2(g)\overset{H2O}{<=>}SO2(aq)\overset{H2O}{<=>}H2SO3(aq)<=>H+(aq) +HSO3-(aq)}$$
The first equilibrium is the dissolution equilibrium of the gas, the second is the hydration equilibrium of the dissolved gas, and the third is the dissociation equilibrium of the acid.
The pKa associated with the first deprotonation of $\ce{H2SO3}$ is 1.81 from Wikipedia, so you can pretty much consider the last equilibrium as completely dissociated in dilute solutions.
You can consider $\ce{SO2(aq)}$ as the state where the $\ce{SO2}$ molecules are physically absorbed in the solution, the rest exist as $\ce{H+(aq)}$ and $\ce{HSO3-(aq)}$.
How much is physcially absorbed and how much is chemically dissociated would depend on the particular compound. Unfortunately, I cannot find any data on $\ce{SO2}$. However, in case of dissolving $\ce{CO2}$ in water, 99% exist as $\ce{CO2(aq)}$ i.e. the physically absorbed gas, and 1% as $\ce{H2CO3(aq)}$, $\ce{H+(aq)}$ and $\ce{CO3^2-(aq)}$. (reference)
So, the absorption of $\ce{SO2}$ is partly chemical and partly physical, but note that all of these reactions are reversible. Also keep in mind that the distinction between physical and chemical change is not clear-cut, and gets less and less useful the more you delve into chemistry.
The website you mention in the question gets a lot of other things wrong, for example, dissolution of $\ce{SO3}$ in water gets you sulfuric acid, which is quite stable and hygroscopic, so you would be hard pressed to reverse it, but the table lists it as physical absorption. Whereas chlorine does not dissolve that well in water, and you can drive it off by heating, but it's listed as a reversible chemical change. It doesn't make much sense to me, so I would advise you to get a more reliable source of information.


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What does the wave function of an electron look like?
In quantum mechanics, in the 'particle in a box' topic, I studied that by solving the Schrödinger equation, we can actually find out what the wave function of the particle in a box looks like. Corresponding to different energy levels, the particle can have different wave functions. For e.g., in it's ground state (lowest energy level), the wave function looks like a standing wave (with nodes at the edges of the box) if we just look at the real part of the wave function. But in reality if we look at the entire wave function (including the imaginary part), it's wave function is actually a wave that is rotating through the real and the imaginary planes. There are plenty of animations in YouTube regarding this.
Similarly, what does the wave function of an electron look like? In my book, no description has been given as to what the wave function of an electron looks like. I mean what does the plot of $Ψ(r,θ,\phi,t)$ (since we use spherical polar coordinates while solving the Schrödinger equation for an atom; $t$ is time) look like?
I don't want to just see the time-independent components of the wave function. I want to see what the wavefunction look like if we let time run (for e.g., for a particle in a box, the wave function just looks like a static wave but when we let time run, the wave function is actually a wave rotating through the real and imaginary planes). Can someone please show me a plot or animation of $Ψ(r,θ,\phi,t)$ so that I can visualise what the wave function of an electron looks like? For instance, what does the wave function of the electron in, say, a hydrogen $1s$ orbital look like? Similarly, what does the wave function of electrons in other orbitals look like?

The wave function of the 1s electron if we let time run looks exactly like the wave function of the 1s electron. It is stationary. It is the very thing you don't want to see. What is it that you want to see, then?

There was actually a recent MinutePhysics video that attempted to provide a better visual of orbitals in motion.

I want to see how the wave function of an electron in , say,  $2p_{x}$ orbital evolves with time. I want to see an animation of some sorts just like how the wave function of a particle in a box rotates through the real and imaginary planes as time passes.

@Tyberius, in that video they show the atomic orbitals, not the wave function of the electron. They show the probability density of the electron.

$2p_x$ does not evolve with time.

Do you mean an electron in an atom, in which case the wavefunction is that of an electron in a potential ,say 1S, 2P etc., or a free electron?

I think that this question touches elements not properly covered in basic and even specialised courses and does not deserve downvotes unless OP is a somewhat graduated or a teacher. Opposite it lead to nice and useful comments and answer.

On android you might look at https://play.google.com/store/apps/details?id=com.gputreats.orbitalexplorer (animation is there though two colours as in Buck Thorn answer should have been used, or a static one https://play.google.com/store/apps/details?id=com.vlvolad.hydrogenatom. Disclaimer: I don't like the way the first app explanation is worded.

@Alchemista, those apps show the atomic orbital, that is, the probability density of the electron and not the wave function of the electron. I want to see how the wave function of the electron evolves with time.

To discern the case by plots is hard for me. By in all cases the information you are referring to can be just mapped by colour changes on both meaning of the orbital term. Is the core of the answer below.

Stationary wavefunctions are solutions of the time-dependent Schrödinger equation
$$\hat{H}\psi=i\hbar\frac{\partial \psi}{\partial t} $$
for which the energy $E$ is constant (it being - like the Hamiltonian - otherwise generally time-dependent) so that
$$\begin{align}\hat{H}\psi&=E\psi\\&=i\hbar\frac{\partial \psi}{\partial t}\end{align} $$
which has solutions $\psi(t)=\psi(0)\exp(-i\phi(t))$. Here $\phi(t)=\omega t$ is a time-dependent phase factor, and $E=\hbar \omega $ is the familiar Planck-Einstein relation between frequency and energy.
The phase $\phi(t)$ is a funny thing because, although you can relate the relative phase of one wavefunction at two points in time or of the wavefunctions of two particles, you can't afaik picture the phase in some way that makes sense in a macroscopic (classical physics) sense. In other words, the phase might change, but what that is that is changing is not something that can be explained outside of mathematics. There is no coordinate system that will give meaning to the QM phase (unlike the amplitude of a classical wave, which can in general be related to spatial coordinates, or some physically measurable amplitude).
If you want to visualize then how the wavefunction (or its phase) evolves in time, in particular for a $\mathrm{1s}$ orbital, imagine a sphere representing an isosurface of the radial component of the wavefunction, where the surface color oscillates between values representing different phases (from 0 to 360 degrees), something like this:

(red, yellow, green and blue represent sequential 90 degree phase rotations)
The time-dependent phase of any orbital representing a stationary solution of the Schrödinger equation can be similarly represented as a regular oscillation between a series of colors, with regions of the orbital with the same phase being uniformly colored.
Note that the spatial contribution to a wavefunction can also be complex valued, so that different parts of an orbital can differ in both phase and amplitude. For instance, the lobes in a p-orbital represent a surface of uniform time-independent amplitude, but the phase changes sign across a node. Neither the spatial phase factor nor the time-dependent phase factor alter the amplitude of the wavefunction.

It might have been easier to show the behavior of a 2 p orbital with its famous dumb-bell shape. Here the first half-dumb-bell is green when the second one is red. And after a while, without moving, the first half-dumb-bell turns red when the second turns green, and so on.

@Buck Thorn, can you give the animations of wave functions of electrons in other orbitals also?

You might says that your animation is slowed down. Well done.

@Alchimista Yes, maybe that's not the 1s orbital, based on the frequency you can estimate what is the principal Q# :)

@BuckThorn I thought it may be worth stressing that :) also when you speak of uniform colour, well that is not very clear. See Maurice comment. I voted up but that passage can be improved.

@Alchimista thanks for your comment. Yes, this is tricky because I am showing only how the time-dependent part of the wavefunction evolves, and of course the phase is a product of time-dependent and independent parts. I'll add that to my answer when I get a chance. Plus I realized that I should have used two spheres, not one, so I need to edit anyway.

@BuckThorn I am curious now why you need two spheres . You are showing an isosurface of phi or phiw2... I will see when you'll have time.

@Alchimista turns out two spheres was unnecessary, but maybe would be less confusing (one sphere for the imaginary component, another for the real)

@BuckThorn I took it as you would necessarily need a p orbital. Now I understand what you have meant by two spheres.


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Stereoisomers of 1,3,4,6-tetramethylcyclohex-1,4-diene
How many stereoisomers are possible for following compound (1,3,4,6-tetramethylcyclohex-1,4-diene)?

My Approach
the 1,4 methyl groups lie in the plane, whereas 3,6 are possible chiral centers. However it doesn't seem like enantiomers are possible due to centre of symmetry, one structure can be meso. Apart from this I don't see any possibility of Geometric isomers as well. However the answer given is:

3

What am I missing in this?

The isomers are RR, SS, and the third is RS which is also SR

@Maurice so RR,SS are unique because there is no plane of symmetry and RS,SR are the same because of centre of symmetry. Is that the case?

I think I might have taken RR, SS to be the same structure in my working which is what went wrong

@Ashish yes, RR and SS have no centre of symmetry and RS and SR are same due to centre of symmetry.

@NisargBhavsar As seen with pen and paper, you are right.  Thanks.

For $n$ stereogenic centres, there are up to $2^n$ stereoisomers. As shown e.g., with meso-compounds, not all of these realizations mathematically possible are chemically unique.

Your compound, 1,3,4,6-tetramethylcyclohex-1,4-diene, does not have line symmetry (plane of symmetry), and hence $\ce{C}$3 and $\ce{C}$6 are chiral centers. Thus, maximum possible stereoisomers are $2^2 = 4$. However, there is a symmetry element in the structure, which is order 2 ($180^\circ$) rotational symmetry (also called point symmetry). Therefore, one stereoisomer, $(3S,6R)$ would be superimposable with it's mirror image $(3R,6S)$, thus it is a meso-isomer:

To understand this, I put "$\bullet$" symbol to indicate relevant hydrogen is above the plane of paper. And, "$\bf{.}$" symbol to indicate relevant hydrogen is below the plane of paper. Thus, first two structures have both 3,6-dimethyl groups below the plane of the paper while the third structure has 3-methyl group below and 6-methyl group above the plane of the paper. Accordingly, all chiral centers have CIP nomenclature as indicated under each structure:

The first structure is $(3S,6S)$ and the second structure is $(3R,6R)$, thus they are non-superimposable mirror images as indicated by the mirror. Therefore, they are enantiomers. To make all the groups to coincide, you may need to rotate the second structure $180^\circ$ horizontally along the axis going through $\ce{C}$3 and $\ce{C}$6. However, after the rotation, the two methyl groups at $\ce{C}$3 and $\ce{C}$6 will be now above the plane. Therefore, these two structures cannot be superimposable.
On the other hand, the third structure is $(3S,6R)$ ($\ce{C}$3 methyl is below the plane and $\ce{C}$6 methyl is above the plane). To the mirror image of this molecule, if you follow the same operations as previous steep and do another $180^\circ$ rotation on the plane of the paper (or vise versa), you'd get the same compound.  That means, the third compound is not optically active and called meso-isomer. This can be found easily by looking at the two chiral centers: They are mirror images of each other, $(3S)$ and $(6R)$ or $(3R)$ and $(6S)$.

Therefore, your compound, 1,3,4,6-tetramethylcyclohex-1,4-diene, has three stereoisomers, but only two of them are optically active.


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What product will be formed from the side chain oxidation of 4‐ethyl‐1,1‐dimethyl‐3,4‐dihydro‐2H‐naphthalene?

Benzylic hydrogen is required for the side chain oxidation of benzene. But for this compound the top branch connected to the benzene ring does not have a benzylic hydrogen. So, will the top branch be broken off in the process of oxidation?
The reagents and conditions for the side chain oxidation are $\ce{KMnO4}$ with $\ce{H2SO4}$ and heat.

I am not 100% sure but the products should be ester combinations of phthalic acid, succinic acid and ethanoic acid.

I think the benzylic dimethyl group will be resistant to oxidation and you'll end up with a dimethylphenylacetic acid there. The other site will be oxidised up to  the benzoic acid.

This is best settled in the lab.  My guess is ketones then benzoate at the tertiary carbon then destruction of the benzene ring. An interesting reaction might be with peroxide to give a phenol and ketone. Why destroy such an intriguing compound?


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The definition of n-factor, what was it when it was introduced?
I realize that n-factor is an outdated concept however I am interested in its formal definition when it was introduced.
We have many definitions now on how to calculate n-factor depending on the type of reaction and compounds involved.

For acids, n-factor is defined as the number of H+ ions replaced by 1 mole of acid in a reaction.
For bases, n-factor is defined as the number of OH– ions replaced by 1 mole of base in a reaction.
The n-factor for such salts is defined as the total moles of catioinic/anionic charge replaced in 1 mole of the salt
The n-factor of such salts is defined as the number of moles of electrons exchanged (lost or gained) by one mole of the salt.

However, none of these definitions explain what n-factor actually is. Is there a formal definition for n-factor?

n-factor is nothing but saying that if n moles of a substance react in the simplified chemical equation than n moles of that substance is called an equivalent and n-facror=1/n. It's an number used to convert between moles and equivalents.

I have only seen n-factor in South Asian textbooks. This is predominantly a South Asian college term for calculating chemical equivalents in current curricula. Here n stands for normality. You will not find this usage in British/US books.
The n-factor is simply a coversion factor for molarity to normality concentration. Since the definition of an equivalent is dependent on the type of reaction we need to be aware of several scenarios which you have listed viz. acid-base, salts, redox reactions.

You still find the term normalité for normality when describing a titration (known e.g., as analyse volumétrique). Example Uni Rennes, or other French speaking countries like CEAEQ's method to determine acidity in waste waters MA315AlcAc10 by 2016, etc. But the advantages of molarity are recognized (e.g., virtual UniSciel).

The "n-factor" is like "ICE tables", a strange concept that is used in schools in some regions in the world but that no chemist uses in real life and most chemists have never heard of.

Instead of asking what n-factor really was, a better question would have been what an equivalent was (But that would have been a duplicate)
According to the above question, the formal definition of an equivalent is as follows:

An equivalent (symbol: officially equiv; unofficially but often Eq) is the amount of a substance that reacts with (or is equivalent to) an arbitrary amount of another substance in a given chemical reaction.
In a more formal definition, the equivalent is the amount of a substance needed to do one of the following:

react with or supply one mole of hydrogen ions ($\ce{H+}$) in an acid-base reaction.
react with or supply one mole of electrons in a redox reaction.

Therefore given any chemical reaction, one equivalent of each reactant reacts to form one equivalent of each product, irrespective of the stoichiometric coefficients
$$\ce{aA + bB -> cC + dD}$$
In all such scenarios, the n-factor of the compound is defined as,
$$\textit{n}\text{-factor} = \frac{M_\mathrm{molar}}{M_\mathrm{equiv}}$$
However, why are there so many different definitions for calculating the same thing. This is because an equivalent doesn't have one fixed definition, rather depending upon the situation, the definition of an equivalent changes and hence the n-factor definition changes as well.
Therefore there is no formal definition per se for the n-factor, simply because it was never formally introduced. As M.Farooq pointed out in the comments,

I have only seen n-factor in South Asian textbooks. This is predominantly a South Asian college term for calculating chemical equivalents in current curricula. Here n stands for normality. You will not find this usage in British/US books. The n-factor is simply a coversion factor for molarity to normality concentration. Since the definition of an equivalent is dependent on the type of reaction we need to be aware of several scenarios which you have listed viz. acid-base, salts, redox reactions.


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Can oxygen form six bonds
Oxygen is a small size atom known to make only two bonds (sigma bonds), it may also form coordination bonds {oxygen ion (2-) uses its outer shell in coordination (only s & p orbitals) that theoretically form a maximum of four bonds}.
In the decavanadate ion cluster (V10O28)6-, there are oxygen atoms with one bond, two bonds, three bonds and six bonds. Can anyone explain how oxygen can bridges 3 and 6 atoms in this structure

There were numerous theoretical investigations on this polyoxometalate, and most of the point to these six bonds having ionic nature. Once this is proved, there is no element of surprise anymore.

I would suggest to draw an image of the cluster using any CIF and post V−O bond lengths to better illustrate your question.

The V-O distance varies in literature, almost 2.1077 - 2.3382 Å

It is also drawn as a solid line in all crystal structures, I believe no solid line between the atoms should be drawn if it is ionic

+1 for element of surprise. Perhaps a stable transuranic?

related https://chemistry.stackexchange.com/questions/121196/why-cant-o-form-of6


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Why is the polarity of the B-N bonds in borazine the reason for its reduced delocalisation of π electrons?
My professor mentioned that borazine is aromatic, but not as much as, say, benzene. The reason he gave for this was that the B-N bonds in borazine, being polar, inhibit resonance. He also mentioned that the same inhibition of resonance occurs in h-BN. Why is this so? What difference does bond polarity make?

Let us understand and try to answer your question with a few simple points.
First it is important to understand why exactly is borazine aromatic. Boron has an empty p-orbital and Nitrogen has a lone pair of electrons.
According to requirements for pi-backbonding, there needs to be a sigma bond between both the atoms and requirement of one empty orbital for allowing the electrons of the side atom to delocalize into it needs to be satisfied. Along with this, the lone pair which needs to be delocalized should be in kept in a pure p-orbital. Boron and Nitrogen both facilitate these requirements and thus the lone pair delocalization takes place.
Thus in a way, we could say that the aromaticity in the borazine moeity is due the pi-backdonation of unpaired electrons.

Understanding that the main reason is sideways pi-backdonation for the cyclic delocalization of electrons, one also needs to keep in mind that Boron and Nitrogen have a significant difference in their electronegativities. The polarity occurs due to the pi-backdonation but keeping in the mind the electronegativity of nitrogen is higher than boron, the delocalisation isn't as pronounced as in the case of benzene where the bonds are non polar. Thus leading to comparatively lesser delocalisation in broazine than in benzene. You could similarly compare h-BN and grpahite; in the former case the delocalisation is less pronounced due to difference in electronegativity than in the later case where a very well pronounced delocalisation is observed.

"The polarity occurs due to the pi-backdonation but keeping in the mind the electronegativity of nitrogen is higher than boron, the delocalisation isn't as pronounced as in the case of benzene where the bonds are non polar. Thus leading to comparatively lesser delocalisation in broazine than in benzene."
I was asking for the explanation of this in the question. Could you help me out with that?

The diffference in electronegativity seemed to me the prominent point in not letting the electrons fully delocalise

The point which I made, @Draculin has mentioned the same in a brief and precise manner.  So if you didn't get what I meant, refer his answer and the research paper as well.

You can think of it as: polarity causes slight concentration of pi electrons on nitrogen atoms in contrast to boron in the resonance hybrid due to which cyclic delocalization does not occur "completely"; hence the aromatic character is hindered
Further reference: https://www.researchgate.net/publication/225792945_Borazine_To_be_or_not_to_be_aromatic


text:
How are 3 center-4 electron bonds any different from 2 center-2 electron bonds?

How are the bridge bonds in $\ce{Al2Cl6}$ different from the terminal bonds?
We could have just mentioned the bridge bond formed by $\ce{Cl}$ as a coordinate bond. Why introduce this extra concept?
I can understand why 3 center-2 electron bond would have been introduced but not this one.
My guess is that during the formation of the molecule vacant orbitals take part in hybridisation which is different from usual but after formation of the bond, there is no difference.

It's showcasing ionic component of bonds and their symmetry.

Sir, I did not understand  your statement much, would you mind elaborating it. Like why does it have ionic character, if it has ionic character then why not just call it ionic bond(My guess:as Cl gets distributed between two Al, thus it has less distortion so it has less covalent/more ionic character by Fazan's rule). How does 3 centre 4 electron signify symmetry?

There are no 3c4e bonds in Al2Cl6 https://chemistry.stackexchange.com/questions/34842/why-is-bcl3-a-monomer-whereas-alcl3-exists-as-a-dimer


text:
Buffer problem in polyprotic acids
I read a problem which was as follows:

Equal volumes of $\pu{1M} \ \ce{H3PO4}$ and $\pu{1M} \ \ce{Na3PO4}$ are mixed. The $\mathrm{pH}$ of the resultant mixture is (along with this $K_\mathrm{a1}, K_\mathrm{a2}, K_\mathrm{a3}$ of the acid were mentioned)

The way I approached this problem was as follows:
$K_\mathrm{a1} = \frac{[\ce{H2PO4^-}][\ce{H^+}]}{[\ce{H3PO4}]}$, similarly $K_\mathrm{a2} = \frac{[\ce{HPO4^{2-}}][\ce{H^+}]}{[\ce{H2PO4^-}]}$ and so on...
Which gives me $K_\mathrm{a1}\cdot K_\mathrm{a2}\cdot K_\mathrm{a3} = \frac{[\ce{PO4^{3-}}][\ce{H^+}]^3}{[\ce{H3PO4}]} = [\ce{H^+}]^3$ (since $\ce{[H3PO4] = [Na3PO4]}$). Using this, I calculate the $[\ce{H^+}]$ and so the ph which comes out to be $-\log{\sqrt[3]{K_\mathrm{a1}\cdot K_\mathrm{a2}\cdot K_\mathrm{a3}}}$ which is $\frac{\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2} + \mathrm{p}K_\mathrm{a3}}{3}$.
However, it seems from the answer that the system is forming a buffer and $\mathrm{pH}$ is $\mathrm{p}K_\mathrm{a2}$
What am I doing wrong here and how does the $\mathrm{pH}$ come out to be $\mathrm{p}K_\mathrm{a2}$. Is there any such general result or theory for polyprotic acids?
EDIT: I understand now why my solution is wrong. Any help as to how to proceed further will be appreciated.

Your assumption that the phosphate ion will not hydrolyze makes your attempt incorrect. Final concentration of phosphate will not be equal to it's initial concentration.

@MathewMahindaratne Doesn't seem to me like they are the same questions because it's hydrolysis is also to be considered in this case since the concentration of Na3PO4 is also given which is not taken in the linked question. Rather the linked question would answer my question if only the concentration of H3PO4 was provided to me

@NisargBhavsar Oh, that makes sense. I understand why it's wrong now

It would be 1 M NaH2PO4 + 1 M Na2HPO4 ( hydrolysis neglected), so pH=pKa2. ( simplification assuming activity coefficient is 1, what is expected in this task context, otherwise not really justified ). It is high concentration, buffers are usually much more diluted solutions, typically 0.05-0.1 M.

Correction 0.5 M + 0.5 M - simple error.

@Ashish This helps https://chemistry.stackexchange.com/questions/40054/phosphate-buffer-system?rq=1   ?

Ok I read the question above and it starts with $\ce{H_2PO_4^-}$ and $\ce{HPO_4^{2-}}$ . I could use that in here but my doubt is that and also with comment made by @Poutnik The $Ka1$ of the acid is given as $10^{-5}$ so how can I assume that $\ce{H_3PO_4}$ is getting 100% dissociated into $\ce{H_2PO4^-}$

@Ashish You do not expect strongly acidic H3PO4 and strongly alkalic PO4^3- can survive aside each other, do you ?

Acid Base reaction! My bad, I was considering it's dissociation as a WA in solution

I closed it under the wrong question by mistake since that doen't really answer it rather if I were to choose one I would say that it was the more similar to this https://chemistry.stackexchange.com/questions/40054/phosphate-buffer-system?rq=1 please change it If it can be


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Confusion regarding DFT basics, i.e. the connection between Hohenberg–Kohn theorems, external potentials and the Kohn-Sham method
I have a question on the basics of DFT, which bases on the Hohenberg-Kohn theorem: there exist an unique mapping between the external potential, the electronic wave function and thus also the electronic density n(r). In other words, from wikipedia, where ${\displaystyle v(\mathbf {r} )}$ is defined as an external potential,
"If two systems of electrons, one trapped in a potential ${\displaystyle v_{1}(\mathbf {r} )}$ and the other in ${\displaystyle v_{2}(\mathbf {r} )}$, have the same ground-state density ${\displaystyle n(\mathbf {r} )}$, then ${\displaystyle v_{1}(\mathbf {r} )-v_{2}(\mathbf {r} )}$ is necessarily a constant."
My confusion comes from the Kohn-Sham equations, i.e.
${\displaystyle v_{\text{eff}}(\mathbf {r} )=v_{\text{ext}}(\mathbf {r} )+e^{2}\int {\frac {\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\,d\mathbf {r} '+{\frac {\delta E_{\text{xc}}[\rho ]}{\delta \rho (\mathbf {r} )}},}$
${\displaystyle \left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+v_{\text{eff}}(\mathbf {r} )\right)\varphi _{i}(\mathbf {r} )=\varepsilon _{i}\varphi _{i}(\mathbf {r} ).}$
which was derived by decomposing the energy functional of the interacting-electron system into solvable energy functionals of a non-interacting electron system affected by an alternative external potential ${\displaystyle v_\mathrm{eff}(\mathbf {r} )}$, plus an exchange-correlation functional that attempts to account for the difference between the two systems. The optimal orbitals are obtained through this iterative scheme.
My question is, since the HK theorem states the unique mapping between between an external potential and the electron density / wavefunction, why would we obtain a common electronic state when ${\displaystyle v_\mathrm{eff}(\mathbf {r} )}$ - ${\displaystyle v_\mathrm{ext}(\mathbf {r} )}$ is not necessarily a constant?
${\displaystyle v_\mathrm{eff}(\mathbf {r} )}$ in this case is the external potential experienced by a single non-interacting electron defined by the first Kohn-Sham equation, while ${\displaystyle v_\mathrm{ext}(\mathbf {r} )}$ is the external potential exerted by the stationary nuclei to the electrons, in the electron-interacting picture.

I guess more specifically, I assumed that the KS method meant that we can solve a simpler system of non-interacting electrons affected by a specific ${\displaystyle v_\mathrm{eff}(\mathbf {r} )}$ that would give the same electron density of the same system but with interacting electrons.
However, the "external potential" governed by the interacting and non-interacting systems are different, so how come they will yield the same ground state, when we minimize their respective energy expressions?

You may want to look at matter modelling SE, which is not to say in any way that it is off-topic here.

Its a tricky question but anyway when you go through the HK theorem you'll see that there is a part where you have $v_{\mathrm{ext}} - v_{\mathrm{ext}}' = H - H'$ as the electron-electron and other operators are the same for both systems and will therefore cancel. The only important part is the external (nuclear-electron) potential and you get that mapping.
Now with KS essentially what you do is take the DFT functional and switch out the kinetic energy functional to the kinetic energy operator from QM and move some other stuff around. This functional now depends on orbitals and we now want to optimise these for the lowest energies. Using some mathematical methods you can get a set of equations to do this known as the KS equations.
With the KS equations you get a set of orbital equations which you solve collectively. If you run the HK theorems with one of the KS equations the mapping between that orbital densities would be with $v_{\mathrm{eff}}$. All this is now saying is that the total electron density which maps with $v_{\mathrm{ext}}$ will be different to your orbital densities which maps to $v_{\mathrm{eff}}$ which is reasonable. You only get the total electron density when you sum all the orbital densities up.

Okay, I understand that the orbtital densities of $v_\mathrm{eff}$ that we obtain is different than $v_\mathrm{ext}$.
But how does this relate to the fact that if we know the exchange-correlation functional $E_\mathrm{EX}$ exactly, then the correct orbital density is still obtained from $v_\mathrm{eff}$, which still can be different than $v_\mathrm{ext}$? Yet as far as I know $v_\mathrm{ext}$ is the important potential in the HK theorem that defines an unique ground state electron density, while $v_\mathrm{eff}$ is an arbitrary potential defined to solve the DFT problem the KS way.

$v_{\mathrm{eff}}$ is not arbitrary it is obtained in a very specific way by a functional derivative of the relevant functionals.  There is a HK theorem that states that there is a variational principle in DFT, so to get the ground state electron density we need to obtain the electron density with the lowest energy. Therefore we need to find the electron density with the lowest energy.
If we used the KS approach and our DFT functional are formed from orbitals then we need to figure out what the best oribtals will need to be to give us our lowest energy electron density.

There is a mathematical technique to do this called the method of Lagrange multipliers this gives us the KS equations which we solve to find those orbitals.

If the KS approach is exact then using the method of Lagrange multipliers and therefore the KS equations to find the KS orbitals would give us the exact ground state density. We get $v_{\mathrm{eff}}$ in the KS equation from the method of Lagrange multipliers and it just gives an effective potential that each orbital needs to be under to form the ground state density overall.

Thank you very much for the run-down. I (think I) understood most of the DFT derivation beforehand, i.e. the two first paragraphs of your original answer and Lagrange multipliers (I meant "arbitrary" as in mathematically defined to fit a need, i.e. the first formula in my original question, rather than physically defined, i.e. $v_{ext}$).

What I had trouble understanding was the difference between $v_{ext}$ and $v_{eff}$, which you anwered in your answer and reply, i.e. $v_{ext}$ is an "effective" or "local" (as defined in some DFT resources) potential, making the two potentials not directly comparable. I.e. as you said, the total electron density will be different to any of the orbital densities, unless you add them up. Thanks a lot for the help!


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What is the name for the relationship between two elements in the same group?
I'm looking for a specific term which can be used to describe two elements in the same group of the Periodic Table. This would be useful since elements in the same group tend to have similar chemical/physical properties.
For example, is there a word that could fill in the blank: "X is a _______ of Y", where X and Y are in the same group? It has to be something that is generalisable to all groups, not just one particular group, so terms like "alkali metals" or "halogens" are too specific.

The correct (but infrequently used) term is 'congener', https://en.wikipedia.org/wiki/Congener_(chemistry)#Other

The term you are looking for is "congener". Quoting the Wikipedia:

Congeners can refer to other elements in the same group in the periodic table. For example, congeners of the Group 11 element copper are silver and gold, sometimes found together in the same ores (porphyry copper deposit) due to their chemical similarity.

or as described in the IUPAC Gold Book:

One of two or more substances related to each other by origin, structure, or function.

Note that congener doesn't only mean "two elements in the same group".  It also refers to minor chemical components in a mixture that give it distinctive characteristics (e.g., https://en.wikipedia.org/wiki/Congener_(beverages) ), and is used in biology to refer to species in the same genus (https://en.wikipedia.org/wiki/Congener).

The correct term is congener. Congeners refer to elements in the same group in periodic table.
For example, congeners of the Group 1 elements are Li , Na and K etc.
Therefore , X is a congener of Y is the right statement.


text:
Designing a mathematics course for 1st year chemistry students
Background:
I am a mathematician with very little knowledge in chemistry (I do know some
physics). I will probably teach a course "mathematics for (1st year) chemistry
students" in the future (1st semester 3h + 1h exercises per week, 2nd semester
2+1h). This will be the only (mandatory) mathematics course in the chemistry Bachelor curriculum (and even the subsequent Master curriculum, as far as I know). The students should have high school mathematics background, such as onedimensional derivates and integration (at least of polynomials), but probably no complex numbers. I will have a lot of freedom in the selection and presentation of topics.
Questions:

Which topics are commonly covered in a "mathematics for chemistry
students" courses?
Closely related: Which textbook are most commonly used?

In more detail:

Can I safely assume that mathematical proofs and even rigorous definitions can/should mostly be skipped?

How black-boxy is Quantum physics used in the typical chemistry curriculum?
Even the most basic Quantum physics will need, as minimum, some linear algebra (Hermitian, normal and unitary operators, exponention of operators, etc.) as well as Fourier transformation and a little bit of differential equations. Should these topics be coverd, or is it enough to say "one can show that in an H atom, orbitals look like this", and just talk a bit about complex valued functions in R^3?
Do you ever do calculations with spins? (Mostly spin 1/2, I would assume? That might be a motivation to show some calculations with complex 2x2 Matrices?)

Which computer mathematics software are most commonly used in a chemistry curriculum? (Mathematica? Maple?) I assume it would make sense to do some calculations and exercises using such software?

Update answering some comments:

I am in Austria (sorry, do not know how to add an according tag), but I would be interested in how these things are handled in other countries as well.
This course has been held before at our department (not by me), but I would like to get "unbiased" input from people from chemistry, and therefore did not list the traditional topics / lecture plan.
Of course I will speak to people in the Chemistry department, but I would be happy to get some ideas here first (I do not know the Chemistry department people, and it could be that a randomly selected person who happens to be willing to indulge me has some unorthodox views).

I guess the rule number 1 is to remember they are future chemists, not mathematians. I would focus on omitting proofs and using applied math levels. From QM, those things directly related to basic quantum chemistry ( H atom, quantum rotator and oscilator ) and the very basics leading to them. Very complicated or complex math including spinors may be over their head. Try to get feed back from students who passed similar courses, or from chemistry teachers,  what they think may be usefule and what may be too much.

In QM context - Perhaps some basic introduction to Lagrange and Hamilton mechanics, as I remember seeing it applied with  Schroedinger equation for the first time ever  was not intuitive.

You probably are not the first one (historically speaking) in your school teaching mathematics for chemists.  What was the must haves which was / is agreed with the lecturers / professors / dean of study within the chemistry program? Your chemical society may have division aiming at this (ex ACS) Missing a bit is applied statistics, both descriptive as well as (given time constraints) some entry into the wide field of statistical design of experiments.

I am not really talented on maths so my comment isn't that of a haughty person. What it is exactly that chemistry curriculum? In other words put at least a country tag. To get my degree in chemistry, at least formally, I went through a curriculum which can be defined advanced calculus/analysis with geometry elements. Formally pretty much that of engineers, the latter in my country pretty tough. Now it cannot be the case as you would need more hours or semesters. (Something to be addressed, even as an appendix or mere tool, is FT. That was misteriously missing, at my time).

Also a good coverage of matrix algebra at least to the level that they can talk about eigenvalues and eigenvectors and sound as though they mean it. At least Fourier series with a mention of Fourier transforms, ideally Fourier transforms - a criminal omission from my own course given the amount of FT spectroscopy that got rammed down us.

@Poutnik For the tasks you mentioned, on one hand, and GNU Octave as tool on the other, you hopefully encountered Tom O'Haver's project on signal processing.

@Buttonwood thanks for the reference, I did not. There is also very handy related project of Octave online

@Poutnik That's a nice exchange, I didn't know yet about yours.

There is a plethora of textbooks ready to use: Francis' Mathematics for chemists (1984);  Barrante's Applied mathematics for physical chemistry (1998); Póta's Mathematical problems for chemistry students (2006); Cockett&Doggett's Maths for chemists (2012); Mortimer's Mathematics for physical chemistry (2013); Kerber's Mathematical Chemistry and Chemoinformatics (2013) and many more. No need to reinvent the wheel, really. As for the software, I strongly advice against using proprietary products such as MathLab or Wolfram Mathematica in introductory courses. Teach them R/Python.

First of all, I would consult with your fellow teachers (ie mostly the chemists) about what they need. Communication always helps. You may not agree with them, but at least you know what they assume as basics taught by you. May eg they prefer to teach themselves symmetries, statistics, and such and ask you to teach more calculus or linear algebra.

I am sorry, but this is off-topic for Chemistry.SE. Yes, your target audience is chemists, and this place has the most chemists; but Chemistry.SE is for questions about chemistry, not how to teach chemistry, or how to teach maths to chemists. Have you considered Mathematics Educators Stack Exchange? I think I can migrate the question over there; and I think it would be the best to have it migrated, and to put a link at the top of your post pointing people there so that they can contribute if interested. You are also welcome to post in [chat].

Also, there used to be a Science Educators site, but I think that one died, unfortunately. It would probably have been more suitable than Maths Educators.

@orthocresol Thank you; I would appreciate the migration to Mathematics Educators Stack Exchange. However, I doubt that the users there will have much to say about this question; so I would also appreciate an according link.

@orthocresol I disagree. This question is really about what level of maths is required to study chemistry, so this isn't off-topic for chemistry. Maybe it's more suited for other sites, but it should not be off-topic here

Despite providing an answer, I have to agree that the people whose input is most important is the chem dept professors. What do they want students to know for their classes so they don't have to waste time teaching math instead of chemistry? That you don't know them is troubling, as this class should be integrated closely with their department curriculum.

In terms of German textbooks, we used the "Zachmann - Mathematik für Chemiker", which may give you insight into the rigorousness.

As I remember, by the time I took physical chemistry I had a full year of calculus, a quarter course on differential equations, and a quarter course of statistics.  (My college was on the quarter system at the time.) This was all dictated by the necessary prerequisites. If you look at the course catalog of other colleges I'm sure that you'll find similar requirements.

One thing I noticed 40 odd years ago when I was a teaching assistant was that some students struggled with integer addition, subtraction, division and multiplication. Knowing from 1:1 to 9:9 tables cold helps immensely when balancing chemical equations.

@MaxW Even just a few years ago, I had a first year university chemistry class student, not a chemistry major, who could not multiply 6 times 7. They multiplied 3 times 7 to get 21, then added 21 to that to get 42. They did this in front of the whole large class, whereupon I dumped the use of a projector-linked tablet modality (for “guide on the side” teaching) and reverted to giving lectures. This eliminated inadvertently embarrassing a student.

The first thing you need to determine is: Does the Chem Dept. only want this course to meet the requirements for freshman chemistry ["mathematics for (1st year) chemistry students"] (I assume it will be taught concurrently), or to cover more advanced topics as well ["This will be the only (mandatory) mathematics course in the chemistry Bachelor curriculum "]?  It could be the former, since they may have separate math requirements for upper division courses.

"I do not know the Chemistry department people, and it could be that a randomly selected person who happens to be willing to indulge me has some unorthodox views".  You could ask to meet with a small group of the Chemistry faculty.  And if this is just for freshman chemistry, that's often a big course that is team-taught (multiple sections), in which case it would be great if you could meet with that team.

I’m voting to close this question because it can't be answered without local knowledge, and an answer will not benefit anyone but the OP. The discussion, which might or might not be helpful to the OP, has run its course, and there is no longterm benefit in keeping the question open.

At least in the US, most undergraduate chemistry curricula do not go deep enough into the math of QM that you need to spend a great deal of time on it. Students that do take more advanced QM classes can learn the math at that time. I also am not sure it's safe to assume that all incoming students at that level have a good understanding of 1D calculus, though most will have had some exposure.
With that in mind, here are topics that I'd include (not in any specific order), structured around the relevant applications:

basic calculus review of integration and differentiation in 1D - main intro level application is chemical kinetics. Also applies to intro quantum.

setting up and solving systems of equations, especially including iterative solving methods - intro level application is solving systems of multiple equilibria by mass and charge balance

a module on statistics with a focus on dealing with experimental error and error propagation. Include methods of curve fitting and associated pitfalls. - applies to lab component of curriculum. Probably one of the more commonly overlooked aspects of math in chemistry

Fourier transform and other techniques involved in spectroscopy

Basics of symmetry operations and group theory

This shows that on topic or not at least OP should have told what is that curriculum in chemistry. I am in Academia since decades and met scientists from all over the world but I have no pale idea of what truly a bachelor is.

I would add eigenvectors/eigenvalues to the list. We also got to learn some solving of differential equations.

Somewhat surprised multivariate calculus is not there. Year 1, Day 1, lecture 1 for me (UK, mumble, mumble years ago) was classical thermodynamics, and we were getting into exact differentials within next to no time

@IanBush - Here in the US, thermo is typically presented in first year undergraduate general chem classes with very little calculus and what calculus there is is univariate. For example, see the Principles of Modern Chemistry textbook by Oxtoby and Nachtrieb. That may be because so many students enter college without having taken calculus at secondary level. They may be required to take it during the first year, but that doesn't help them for first-year chem classes.

@Andrew Thermodynamics without calculus? To what is the world coming? Our course was not dissimilar  to the classical thermo covered in (my very old) Atkins, stat thermo was second year. But the vast majority of students had studied at least univariate calculus to A level (taken at 18 in the UK).

@IanBush - To clarify, this is thermo as presented in first-year general chem classes not a more rigorous physical chem specific class which they might take later. It was taught this way when I took general chem more than 25 years ago, so it's not new. Remember that US secondary education is one year shorter than UK and without specialization (nothing similar to A level) and not known for being particularly high quality. Only the more advanced students get decent calculus classes before entering university.

The list of topics will be infinite (if that is allowed in mathematics) if you keep on thinking what chemistry students must know. This should not be the goal of this course to superficially glance over all the possible advanced and superadvanced topics which a chemist may encounter in his MSc/PhD.
We should keep in mind that these are first year students and many students would take separate mathematics / physics courses, and of course, many will not even study chemistry after their bachelors. So, the best thing is to avoid repeating the same dry topics which they will encounter again and again. It is good that you have a lot of flexibility it would be if you make this course highly application oriented. Complementing Andrew's answer...
The most important thing to teach to chemists is "mathematical common sense" in the beginning, which means developing an intuition when they get wrong answers.
(a) Rigorous concept of dimensional analysis and it's requirements. This paper would be a great introduction for first two three classes "Can One Take the Logarithm or the Sine of a Dimensioned Quantity or a Unit? Dimensional Analysis Involving Transcendental Functions", Journal of Chemical Education, 2011, 88, 1, 67–70 (link).
(b) Concepts of linear and non-linear curve fitting, types of graphing (ideally manual on a graph paper and then on computer). Reading the axes on log-log, semi-log, linear axes. Concept of extrapolation and interpolation, different coordinate systems (spherical, polar, Cartesian etc.)
(c) Differential equations relevent to second and third year quantum chemistry (simple harmonic oscillator, H- atom) and rate laws.
(d) Matrix methods to solve a system of linear equations in 3 or more variables.
(e) Obtaining approximate solutions to higher order equations (quadratic, cubic, quartic). They will need them in solving chemical equilibrium. Iterative approximations
(f) Continuous and discrete Fourier transform relevant to chemistry. This would be handy in their third or fourth year spectroscopy.
(g) Numerical differentiation and integration on discrete data sets
(h) Excel based exercises to remove noise from the data which could include moving averages, weighted moving averages, Savitsky-Golay, windowing operations with discrete Fourier transforms.
Given that you are in Austria worth checking the following German classic books.

"Rechnen in der Chemie" von Dr. techn. Ing. Walter Wittenberger: First and Second Parts. It is a nice two volume set. (link)

"Physikalisch-chemisches Rechnen mit einer Einführung in die höhere Mathematik" von Walter Wittenberger und Werner Fritz (link)

For b) and f) I would add O'Haver's project for that he shows the background and application in a true programming language, as well as (where plausible) in a spreadsheet program.  But work in the the later often is more difficult to replicate for errors easy to commit (ex 1, ex 2) and sometimes not due to the user (e.g., McCullough's papers about Excel vs. gnumeric ex 3).

Yes. O'Haver is a good friend. He has written a book on Signal Processing which he has kindly made available for free as a pdf. Amazon is selling that book for > $600 which is a rip off. It is not him.

@Buttonwood Ah, M. Farooq is too modest: he does not mention that he has co-authored several papers with Tom O’Haver! ;-)

Good luck! Give no formal proofs. Teach only with chemistry examples, say chemical kinetics and simple quantum for differential equations, spend 1 hour teaching and 2 hours student working in groups solving problems. (I have done this and it is the only way that works). Remember that chemists are not mathematicians and only want to know how to solve problems. As these students do lots of practical work they need to know how to use software to do least squares fitting and error analysis. They also need to understand fourier transforms as these are the basis of x-ray diffraction and nuclear magnetic resonance methods, both fundamental knowledge for any chemist. These can be taught largely but not exclusively pictorially.  As I said Good Luck!


text:
What Does Electrical Work Represent in Context of Electrochemical Cells?
If cell operates reversibly than change in Gibbs energy of system equals electrical work of the cell. As far as I know electrical work is work done by electric field when charge moves between certain potential difference which in case of electrochemical cells is potential difference between two electrodes: $$ W = -zF\Delta E $$
Here work is written per mole of reactant consumed or product formed in electrochemical reaction. It is said when cell is short circuited that cell does no electrical work. How is this possible if charge moves through certain potential difference? What does the fact that there is no load connected to the cell A.K.A cell is short circuited have to do with electrical work since charge will still move between two electrodes or certain potential difference?

In the case of the ideal short circuit by zero resistance, all spent G energy is converted to thermal energy within the system = cell, as ohmic lose on the internal resistance.

We are here talking about cell working reversibly which means no loses. Also, it didn't really ask my question.

@DarioMirić So are you short circuiting the cell with a superconductor? You could also run an electrical heater, using the work to produce heat. The charge moves, but initial and final state are the same (in terms of the wire; of course reactants turn into products and the heat given off).

@Karsten Theis Here it is important to note that we look at the cell working reversibly, so using superconductors is unneccesary as cell works without losses regardless of conductor used or without joule heating needed in electric heater. Electrical work in electrochemical cell is something different.


text:
Why is the A-value of methoxy group lower than that of the hydroxyl group?
I am studying conformational analysis of monosubstituted cyclohexane. "A-value" refers to the energy difference between conformers with substitutent in equatorial (more favored) and axial positions. A higher A-value means that the axial conformer will more readily convert to equatorial.
According to Wikipedia, methoxy has A-value of $0.6 \text{ kcal/mol}$ while that of hydroxyl is $0.87 \text{ kcal/mol}$. Why does methoxy, a bulkier group, have a lower A-value than hydroxyl?
I think that in both cases the group attached to the oxygen faces away from the axial hydrogens, so the size of those groups shouldn't matter very much. That is to say, the A-value of methoxy may be the same as or at the most only slightly larger than that of hydroxyl. This explanation does not suffice as its A-value is in fact smaller.
A possible reason could be that the $\text{C-O}$ bond is shorter in methoxy than in hydroxyl. However, I don't understand why this should be true. Hints would be appreciated.

I'm hazarding an educated guess here and going to say that the O-C bond is longer than the O-H bond and that there are more interaction between the axial substituents of the H from the hydroxyl group compared to the O-C-H of the longer distant from the other substituent methoxy group.