text:
Comprehensive list of metal alloys along with Moh's hardness scale
Material property lists, such as springermaterials.com, has lists of metal alloys and their properties. Is there a similar website that lists alloys along with their Mohs scale hardness? For springermaterials I can't seem specifically find hardness values. For example, when I search "hardness" or "Moh" in the search box nothing comes up.

The name's Mohs, not Moh or Moh's.

You can find Mohs hardness values for various pure metals, but it sounds like you want something more specialized: Mols hardness values for specific alloys.
That's probably harder to find.  But there is an alternative:  One could create a formula to convert from the types of hardness values readily availble for alloys (e.g., Vickers hardness) to Mohs.  That's what I do here.
I need to add the caveat that this is just an approximation: Mohs hardness is based on a scratch test, while Vickers hardness is based on indentation, so these two scales aren't fully comparable.
To start, I made use of a table of Vickers values for the ten Mohs reference substances, found here:   https://geology.com/minerals/mohs-hardness-scale.shtml
Then I plotted Mohs hardness vs. Vickers hardness:

You can see from the above plot that, because of the highly non-linear nature of the Mohs hardness scale, it would be difficult to create a good fit that includes diamond (the point at the far right).  Since metals probably top out at a Mohs hardness of ~$9$, I decided we could eliminate that value, and replot.  This gives a curve that would be much easier to fit:

A log-log plot nearly linearizes it, indicating it follows a power law:

This tells me I should try to fit it with a power law function. After some testing, I found that a nice fit can be obtained with a function of the form $a + b x^c$:

The above fit is:
$$\text{Mohs hardness} = -0.5853 + 0.4829 V^{0.3912}$$
where V is the Vickers hardness value.
To use it, just plug and play.  For instance, the Vickers hardness of 316 stainless steel is 152.  Substituting into the above, the estimated Mohs hardness of stainless steel is 5.0 (because of the problematic nature of comparing two different types of hardness scales, I wouldn't express the predicted Mohs values to anything beyond two significant figures; I included two guard digits beyond this in the above formula, to avoid round-off error).

Should I be amazed that you managed to fit the two properties using such a simple analytical expression, that is, that it should follow a power law given a basis in a qualitative notion of hardness with Mohs scale really a ranking? Do you have an explanation for the high quality?

No idea. But I can speculate by considering the fit of the inverse function. Without  diamond, it's: $V = -1.631 + 13.95 \text{ Mohs}^{2.270}$. I.e., actual hardness goes as $\pu{Mohs}^{2.270}$. Now let's suppose the reference minerals were chosen so that (a) the scale could go from talc to corundum in eight steps; and (b) each next step was hard enough to clearly scratch the one below. It may be that, physically, this happened by choosing minerals whose hardnesses increases as a power law (which made for an easy fit with power law). Again this just speculation....

...A  materials scientist might be able to give a better answer.  Of course, if my model had four or five free parameters rather than three, I could have cited Fermi's recounting of a quote by vonNeuman:  "I remember my friend Johnny von Neumann used to say 'With four parameters I can fit an elephant, and with five I can make him wiggle his trunk.'"  :)

Your last explanation is very funny, but I don't find it plausible. This is not a case of overfitting, you don't see weird loops in the fitted function between data points or other features that would indicate you used eg a polynomial of too high order. I speculate wildly but could it be that Vickers hardness was defined so as to exhibit the tidy quadratic dependence on Mohs that you revealed? That sounds farfetched, but the fit is truly uncanny. I suppose you may be right and Mohs did not pick the materials randomly out of a hat.

The last comment wasn't an explanation, it was just a joke.  So of course it's not plausible, because it wasn't meant to be plausible.;) I wasn't implying that my three-parameter model is overfitting. As far as your speculation on Vickers goes, I wouldn't expect Vickers was defined to correspond in some way to Mohs, since Vickers is a quantitative scale based on measured resistance to plastic deformation. It wouldn't make sense to me to compromise Vickers' quantitative information by somehow rescaling it to correspond to a qualitative scale like Mohs. But I'm not an expert on such things.


text:
How to convert from dipole derivative to atomic units?
I need to convert dipole derivative output by Gaussian into atomic units. The dipole derivative is given in Gaussian in the weird units $(\mathrm{km/mol})^{1/2}$. I know that
$\pu{1 D^2/(Å^2amu)}$ = $\pu{42.2561 km/mol}$
$\pu{1 D(ebye)= 0.393456 bohr \textrm{-} e}$
$\pu{1 bohr = 0.529 Å}$
$\pu{1 amu = 1822.8818 electron masses}$
I made the conversion (see below), but the value I get is in an improbable range (I know that the answer should be roughly 0.15 or in that order of magnitude based on literature). Is there anything wrong with the unit conversion below? I realize this is a rather dumb question but I would really appreciate if someone could point out any silly mistake I may have made.

Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown. Note that symbol "amu" is deprecated since 1961.

I guess it's a particular kind of funny to call SI units weird. Indeed, it's weird in that context as Gaussian usually adheres to ancient systems of measurement.

thank you for the suggestions and comments. Gaussian units are indeed weird, if that was your meaning. but can you answer my question or not? In general, I don't get why rather than answering the question, people feel the need to comment on presentation and other superficial things. If you had time to comment, you had time to answer.

Presentation isn't merely superficial. If folks want to check your work, they want to be able to copy and paste it into whatever program they use. They don't want to have copy it by hand, which is what happens when you put formulas in screenshots. Plus screenshots aren't searchable. Also, the way you wrote the eqns. in your 1st para. makes them less readable. As to the substance of your post, I'm not sure how you get D^2/(Å^2.amu)=42.2561 km/mol. I get  D^2/(Å^2.amu) = 6.70 x 10^–13 s^2 amperes^2/kg = 2.61 x 10^25 e^2/kg, where e is an elementary charge.[Informal formatting is OK in comments.]

It seems that conversion is designed to account for the non-std units of the Gaussian output. Anyways, this should help:  https://mattermodeling.stackexchange.com/questions/5021/what-units-are-used-in-gaussian-16-for-dipole-derivatives-output According to the answer there by Jon Kragskow, "To get the true derivatives in $[\text{D Å}^{-1}\text{u}^{-1/2}]$, you need to divide the reported derivatives in $[\text{km}^{1/2}\text{ mol}^{-1/2}]$ by $\sqrt{42.2561} \text{ km}^{1/2}\text{ mol}^{-1/2} \text{u }^{1/2} Å \text{ D}^{-1}$."  So based on that, I get $ 0.862 \text{ D Å}^{-1}\text{u}^{-1/2}$

No, you're right about formatting. I will use MathJax or something better next time. I'm not very experienced with StackExchange yet. I get the same thing 0.86...It seems there is nothing wrong with my conversion. The literature values may have been obtained in a different way. Unfortunately the paper says nothing about how it was obtained.

Usually an answer takes a lot longer to write than a comment and some comments will be helpful to any readers. And no I couldn't really answer your question as I didn't really get where you are coming from and going to. For example, I don't understand why you use amu in any of your approach. And yes, I took a swing at Gaussian for using old units, but unfortunately so does a lot of the chemistry community. If only we could all agree on a standardised system, some problems would simply go away. I guess; I obviously don't have any proof of that. (And yes, my comment want that helpful.)

source: Stack Exchange

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

sort: votes


text:
How is MnS + 2HCl -> MnCl2 + H2S a redox reaction, in terms of electron transfer?
$$\ce{MnS + 2HCl ->  MnCl2 + H2S}$$
If oxidation and reduction is taken to be the loss of hydrogen and the gain of hydrogen respectively, then I can see how the aforementioned reaction is redox — $\ce{S}$ gains hydrogen and is therefore reduced while $\ce{Cl}$ loses it, and hence is oxidized. But how can this be explained in terms of electron transfer? The oxidation state of every atom seems to be unchanged, which should mean that no reduction or oxidation has taken place:
$$\ce{\overset{\color{blue}{(2+)}}{Mn}\overset{\color{red}{(2-)}}{S}
    + \overset{\color{orange}{(+)}}{2H}\overset{(-)}{Cl}
   -> \overset{\color{blue}{(2+)}}{Mn}\overset{(-)}{Cl_2}
    + \overset{\color{orange}{(+)}}{H_2}\overset{\color{red}{(2-)}}{S}}$$

It isn't. End of story.

The reason for my asking here was because a reputable online learning platform claimed it was a redox reaction when it didn't quite fully seem to be one

Please let us know the source. It looks a little less reputable now, maybe?

It's from Khan Academy, but I won't be able to post a link to the quiz containing the problem because the questions don't seem to be static — they change for each viewer.

Well, at least now you know that Khan Academy is not a "reputable online learning platform".

If oxidation and reduction is taken to be the loss of hydrogen and the gain of hydrogen respectively...

The loss of hydrogen is oxidation in cases of homolytic breaking the bond with hydrogen when the electron goes away with the proton. Heterolytic losing hydrogen ion(=hydrated proton), when the electron stays, is not oxidation, but acid dissociation.
The former case leads to change in oxidation number/state, the latter does not.
E.g. $\ce{CH4 + O2 -> C + 2H2O}$ is oxidation, as carbon is oxidized from the oxidation state -IV to 0.
Exchanging an acidic hydrogen and a ion of an electropositive metal between some acid and salt (i.e. the metal  keeps its oxidation state) in cases like above:
$$\ce{MX_n + n HA -> n HX + MA_n}$$
does not change oxidation states of elements in involved compounds.
Therefore conversion $$\ce{MnS + 2HCl ->  MnCl2 + H2S}$$
is not a redox reaction, as Mn keeps the oxidation state +II, sulphur -II hydrogen +I  and chlorine -I.

Oxidation state involves a counting of valence electrons, so you have to look at whether and how electrons are transferred between atoms, not how the atoms themselves are transferred. Follow the definitions given by the IUPAC Gold Book for all species:
Oxidation:

The complete, net removal of one or more electrons from a molecular entity (also called 'de-electronation').

An increase in the oxidation number of any atom within any substrate

Gain of oxygen and/or loss of hydrogen of an organic substrate.

(Hydrogen and ixygen are understood to be neutral species, not hydrogen or oxide ions, in the organic chemistry definition.)
Reduction:

The complete transfer of one or more electrons to a molecular entity (also called 'electronation'), and, more generally, the reverse of the processes described under oxidation (2) and (3).

It is true that the transfer of neutral hydrogen atoms (implied in definition 3 for oxidation given above) provides a way to transfer electrons because each hydrogen atom picks up an electron, or perhaps leaves one behind if it is initially bound to an electropositive element, as it gets transferred. But here that mechanism does not apply because the hydrogen is transferred as ions, and the ions do not pick up or drop off any electrons as they move from $\ce{HCl}$ (or its dissociated solvated hydrogen ions) into $\ce{H2S}$.

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


text:
Are there separate quantity symbols for the pKa of a substance and the pKa of its conjugate acid?
For polyprotic acids, we distinguish the different deprotonation equilibria and their equilibrium constants by indices, e.g. phosphoric acid dissociation is described by $K_1$, $K_2$, and $K_3$. Considering the acid/base properties of water, there are two relevant equilibria (protonating or deprotonating water once) and equilibria with at least two other species ($\ce{H4O^2+}$ and $\ce{O^2-})$. Is there a systematic nomenclature for these equilibria, maybe something like $K_1$, $K_2$, and $K_{-1}$ and $K_{-2}$?
In tables (typically in organic chemistry), molecules are often listed alongside with the $\mathrm{p}K_\mathrm{a}$ values of the conjugate acid, e.g. comparing the basicity of amines and amides. There is potential confusion with the $\mathrm{p}K_\mathrm{a}$ values of the species listed (giving a measure of how acidic they are).
Is there are good way of symbolically distinguishing the $\mathrm{p}K_\mathrm{a}$ values of the substance itself and that of its conjugate acid?

Isn't the simple answer that pKa should always refer to the subject compound, and any other usage (such as "the pKa of ammonia is 9.2") is incorrect even if the intended meaning is clear enough?

For an aqueous solution, may you point to evidence of $\ce{H4O^2+}$ and $\ce{O^2-}$?  So far e.g., with magnesia ($\ce{MgO}$), teaching I encountered so far was it consists of $\ce{O^2-}$ ions, however upon contact with water, these would rapidly (kinetics) deprotonate the solvent to yield $\ce{OH^-}$ hydroxide as the far more stable (thermodynamics) form.

@Buttonwood I don't think twice-deprotonated or twice-protonated water exists in aqueous solution. Nevertheless, folks want to attach a pKa to reactions even when they aren't observed to show how they don't occur.

@Andrew That would be nice. However, sometimes the intent is to show that a substance will not act as a base by stating a very high pKa of the corresponding conjugage acid (which does not have a common name because it does not form readily).

Clayden et al. [1, p. 174] suggest to use $\mathrm{p}K_\mathrm{aH}$ (however, they merely introduce the symbol once and never use it again in the book relying solely on $\mathrm{p}K_\mathrm{a}$):

The ‘$\mathrm{p}K_\mathrm{a}\text{s}$’ of bases
Chemists often say things like ‘the $\mathrm{p}K_\mathrm{a}$ of triethylamine is about 10.’ (It’s actually 11.0 but 10 is a good number to remember for typical amines).
This may surprise you as triethylamine has no acidic hydrogens.
What they mean is of course this: ‘the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of triethylamine is about 10.’
Another way to put this is to write ‘the $\mathrm{p}K_\mathrm{aH}$ of triethylamine is about 10.’
The subscript ‘$\mathrm{aH}$’ refers to the conjugate acid.

Note that older literature used $\mathrm{p}K_\mathrm{b}$ written for the equilibrium
$$\ce{B + H2O <=> BH+ + OH-}\quad K_\mathrm{b} = \frac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]}\tag{1}$$
instead of uniform equilibrium with conjugate acid
$$\ce{BH+ <=> B + H+}\quad K_\mathrm{a} = \frac{[\ce{B}][\ce{H+}]}{[\ce{BH+}]}\tag{2}$$
and suggesting to use a relation
$$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w} = 14.00~(\text{at}~\pu{25 °C}),\tag{3}$$
which was not only inconvenient for comparison between two scales for acidity (for example, see What is the importance of using a pKa value instead of a pKb value when describing drug chemistry?), but was a source of errors when one didn't pay attention for what temperature the value was reported (i.e. at $\pu{20 °C}$ $\mathrm{p}K_\mathrm{w}$ is $14.17,$ not $14.00).$
There is also a conceptual flaw in using $\mathrm{p}K_\mathrm{b}$ [2, pp. 195–196]:

Today the use of $K_\mathrm{b}$ is avoided because it does not touch the heart of the matter, namely: an acid produces hydrogen ions and a base receives them.
Thus both acids and bases are best related in terms of a single quantity, their affinity for the hydrogen ion.
Such a relationship requires the use of the acidic constant $(K_\mathrm{a})$ for both acids and bases.

References

Clayden, J.; Greeves, N.; Warren, S. G. Organic Chemistry, 2nd ed.; Oxford University Press: Oxford; New York, 2012. ISBN 978-0-19-927029-3.
Albert, A. The Determination of Ionization Constants: A Laboratory Manual, 3rd ed.; Springer Netherlands: Dordrecht, 1984. ISBN 978-94-009-5548-6.


text:
What organic molecule, if any, does this diagram represent in Louis CK's "Louie" Season Four, Episode Eleven, just before the six minute mark?
I'm wondering what molecule this is, if any, and whether this molecule is even physically likely or possible.
Here is a screenshot of the scene from "Louie" (a hilarious and fascinating comedy drama series by Louis "the comic's comic" CK premiered in 2014 in the United States, ):

As andselisk states in the comments, it is adenine.

[OP] I suspect it is fake.

No, it is as real as water and the stars. What makes this a little bit complicated is how to draw it, specifically on which nitrogen (N) to put the hydrogen (H). The structure shown is the 9H adenine tautomer, supposedly the most stable tautomer.

Source: ACS Omega 2021, 6, 29, 18890-18903
Some sources show the 1H tautomer (with a hydrogen bound the nitrogen atom numbered 1), e.g. ChemSpider. Pubchem shows the 7H tautomer (the second most stable form according to the above source).

[OP in comments] Why are all your diagrams left to right reversed compared to the one displayed in the OP?

This is a flat molecule, so if you rotate it (or look at it from the other side), it appears left-to-right reversed. It is the same molecule, though. As andselisk states in the comments, there is a convention for the prefered orientation if there are no other considerations.

[OP in comments] Is it fair to say the angles of some of the bonds in the diagram displayed in the OP are wrong?

The type of diagram shown on the board of the still is called a Lewis structure. For Lewis structures, angles can be chosen at will. For rings, most people try to draw regular hexagons and pentagons. However, a house-shaped pentagon is also popular (see below, upside down in this case, and a so-called skeletal structure with the carbon atoms implied at intersections of lines).

Source: Sigma Alrich
Adenine as building block
Luckily, the adenine we care most about (as part of ATP, DNA and RNA) has a glycosidic bond at nitrogen 9 and no hydrogens directly attached to ring nitrogen atoms at all, so there is more uniformity in the molecular structures of adenosine on the web and in popular culture. Below is the structure of adenosine, with the glycosidic bond shown in blue (and the carbon atoms and most hydrogen atoms implied). In RNA and DNA, this adenosine would be linked to other building blocks via phosphodiester linkages involving its two hydroxyl groups.

[OP in comments] Since you use the word "funny", this could be a good time to ask everyone here whether there is anything comedic, however subtly, indirectly, strangely, or darkly (CK's humor is like that, IMHO), about the diagram on the board or adenine.

Well, it shows a Lewis structure, not a Louis structure...

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


text:
Voltaic cell equilibrium confusion
In a voltaic cell, taking one with $\ce{Zn}$ and $\ce{Cu }$ electrodes for an example, an EMF is formed because the equilibria at the $\ce{Zn}$ electrode is shifted more towards the $\ce{Zn^2+ + 2e-}$ side than the copper electrode’s equilibrium is shifted towards the $\ce{Cu^2+ + 2e-}$ side. Electrons, I understand, then flow from the $\ce{Zn}$ electrode to the $\ce{Cu}$ electrode and redox occurs.
The problem I have is that, if electrons migrate from the $\ce{Zn}$ electrode in this way, surely this will mess up the equilibrium at the $\ce{Zn}$ electrode and, even though it will attempt to shift back to counteract the change, the overall result will be a loss of delocalised electrons to participate in the $\ce{Zn}$ equilibrium. This would imply a decreased emf towards the copper electrode and hence an increasingly more positive $\pu{E_{cell}}$ value over time (assuming the $\ce{Zn}$ electrode to be on the left of the setup).
Am I correct or have I gone wrong and, if so, where is my mistake?

Electrons, I understand, then flow from the copper electrode to the Zn electrode and redox occurs. Exactly the opposite.

Apologies, that was a typing error which I would fix if I could make edits. It is not what my question is about, though.

What do you mean by you cannot edit? Anybody who is registered can edit their posts. AFAIK, guest users would not have 129 reputation points. // See also policy-on-(Am I right?)-questions

You must have noticed in  real life  that voltage of cells under load decreases, immediately and also in time.

So my inference is correct then?

I can’t understand this website- everyone just dislikes my genuine question but nobody offers any reason why nor any answer to my query.

Does the question manifest research effort ? Is the question clear and useful? Compare the answers with the mouse hint at the downvoting icon.

Consider electrodes as electron pumps that try to maintain their equilibrium potential, according to their electrochemical state.
When $\ce{Zn}$ and $\ce{Cu}$ electrodes (of e.g. the Daniell cell) are connected by external circuit, they mutually try to force their potential to each other, causing the current flowing across this circuit.
The potential of $\ce{Zn}$ electrode (anode) becomes higher then the equilibrium one and the oxidation of $\ce{Zn}$ occurs, partially ( because of limiting reaction kinetics) acting against increased potential, releasing electrons to the circuit and positive ions to the electrolyte.
The potential of $\ce{Cu}$ electrode (cathode) becomes lower then the equilibrium one and the reduction of $\ce{Cu^2+}$ occurs, partially acting against decreased potential, withdrawing electrons from the circuit and positive ions from the electrolyte.
The cell voltage over time decreases because the cell open voltage over time decreases, because the electrode open potentials converge to each other. Another drop of cell voltage under load is due increasing of internal resistence over time, mostly due depletion of electroactive components.

The good initial step before asking is reading relevant textbooks, and study materials, googling  keywords and studying the search results.
This prevents cases of asking for answers that have been already written in many ways on many places.
The search scope can be eventually limited to the particular address domain by adding the additional term site:, where the  can be e.g.  stackexchange.com,  libretexts.org or wikipedia.org.
Review also Chemistry SE: resources-for-learning-chemistry

Thanks for taking the time to respond, Poutnik, but my question remains unanswered: does the potential difference between the electrodes decrease progressively over time because the equilibrium at the Zn doesn’t reform electrons at the same rate that they are drawn towards and used up in redox at the Cu electrode?

No, it does not. If it did, current would stop.

the cell voltage over time decreases because the cell open voltage over time decreases, because the electrode open potentials converge to each other. Another drop of cell voltage under load is due increasing of internal resistence over time, mostly due depletion of electroactive components.


text:
Rubber permeability of hydrogen to air
I am currently studying permeation of rubber film in a university practice group.
The problem is to evaluate the permeation of hydrogen gas filled in a meteorological balloon placed in the air, but how to build a mathematical model?
In particular, it is about how the penetration phenomenon is when the rubber balloon goes up and down.
The statement is as following: A weather balloon contains hydrogen gas floats in air with its buoyancy. As balloon flies higher pressure in balloon decreases and at the same time volume of balloon increases. But because of partial pressure difference between the inside balloon and the outside balloon hydrogen permeates through rubber membrane of balloon to outside air.
How to evaluate permeation model of this phenomenon?
And also when balloon descent from high altitude to ground how this model is made?

When the balloon descends, it rather performs a free fall after its rapture, which jhappens when it has the diameter >= 3m, typically at 25-30 km. So it is a question rather for aerodynamics than gas diffusion. (  meteorology background during my enlisted military service at air force airfield in 1989-1990.

Thanks for Poutnik, and my problem is on study not real project. And I mean if hydrogen filled balloon ,which its buoyancy is not enough for its payload weight,  descent, how to model hydrogen permeation? Sorry for not enough explation

Every useful word saved in the initial question formulation is paid later by several words posted by responders and the question poster during clarification. :-)

When expanding to the twice as big diameter, the rubber becomes 4 times thinner with  bigger pores to flow through and faster diffusion through rubber itself. Not easy to simulate.

High permeability to hydrogen is why helium is generally used.

@blacksmith37, Do radiosondes last long enough for permeability to matter? As Poutnik states, they reach a maximum altitude and achieve rapture (beamed down?) so as not to be an aviation hazard. Helium is preferred, I think, for safety.

Depends on "rubber film" definition. Cheap, thin balloons lose most of the hydrogen in a day.

@blacksmith37 In meteorology, hydrogen is ( or at least was ) used for baloons, as in the supposed timeframe, permeability is not an issue. You would spend the whole He cylinder in a week or so  and helium is not exactly cheap. See also that wikipedia.org: Weather_balloon#Materials_and_equipment says The balloon is usually filled with hydrogen due to lower cost, though helium can also be used.


text:
How can I find all possible isomers of a given molecule?
I attend a course on combinatorics (mathematics) which contains many examples related to chemistry. Unfortunately, I don't have enough chemistry knowledge to feel comfortable with the examples.
For example, how can I be sure that I have produced all isomers of a given molecule?
The example I am struggling with at the moment is $\ce{C_nH_{2n+1}OH}$ for $n = 3$. There are two isomers given:

How do I know these are all there are?
Why is the molecule with $\ce{OH}$ at the bottom left (bottom of left most $\ce{C}$ atom) not another isomer?

Because it's the same thing as the first one.

Indeed, it takes some knowledge of chemistry to tell which things are the same and which are different. You need to know that the real molecules are 3D, not flat, and that a carbon is tetrahedral, and that single bonds allow rotation, and possibly more than that.

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

The first and the second drawing depict different molecules, which are isomers of each other.

Are you looking for a procedure to generate the isomers? Like SMOG by Molchanova et al. (J. Chem. Inf. Comput. Sci. 1996, 36, 888–899; https://doi.org/10.1021/ci950393z), MOLGEN, MAYGEN (each of them has pros and cons)?  Or the use of generating the isomers to explore chemical space like Reymond's GDB databases?

https://chemistry.stackexchange.com/questions/16135/how-to-determine-number-of-structural-isomers

@Buttonwood Another recent contender to this is Surge : https://github.com/StructureGenerator/surge (full disclosure - I know some of the authors)

This is far from a simple task. see the related question: https://chemistry.stackexchange.com/q/16135/81 for how complex the theory enumerating the isomers of simple hydrocarbons is. There are tens of thousands when you have just 16 carbons. There is probably some graph theory that can help, but that isn't an easy area of mathematics either.

See wikipedia: List_of_straight-chain_alkanes for the isomer numbers, including stereoisomers, for alkanes $\ce{C_nH_{2n+2}}$. For the respective alcohols  $\ce{C_nH_{2n+1}OH}$, you have to multiply each such a number by number of distingushied positions of $\ce{OH}$ group.

As drawn, there are 8 positions where the OH could replace an H, but only 3 structurally unique arrangements:

Two positions: on the centre-C (above or below are a symmetric reflection).
Two positions: on the middle of either end-C (left or right are a symmetric reflection).
Four positions: above or below at either end-C (two-way symmetry).

The diagrams for these three configurations can be rotated or reflected to match all possibilities.
To answer the question, one only has to determine how many distinct locations there are.  If there were 6 C atoms (pentane), the choices would be at an end or at the 1st, 2nd or 3rd position from an end: 4 possibilities.

There's a problem with this though, one that mathematicians will be oblivious to, but which chemists will be instinctively aware of.
Anyone not familiar with organic chemistry will observe that the C atoms have four arms branching off at 90° angles within a plane, since that's how they are represented in the diagram.
A chemist though will know that C atom arms aren't all in a plane; they spread themselves out to be equally separated in 3 dimensions.
The C is at the centre of a tetrahedron with one arm extending to each of its four corners.
For instance, a methane molecule, CH4 would shape itself as a completely symmetrical tetrahedron:

In the given propane molecule, the two C atoms at either end have four arms that are symmetrically located in 3 dimensions, again in the form of a tetrahedron.
That means that the 3 attached Hs are all symmetrically equivalent.
Rather than forming a T-shape, they are actually arranged in a triangle, whose plane is perpendicular to the diagram.
Each of the thee H positions are equivalent and indistinguishable.
So, it matters not which of the 3 Hs at the end gets replaced, in terms of the actual molecule the results will be identical.
By convention though, it is the end location that is always used in the diagram, to keep the diagram as symmetric as possible.
As a further exercise, consider the C at the centre of the diagram in the question, whose four arms are attached to two Hs and two Cs.
How many distinct ways can these connections be made?
Perhaps surprisingly, the answer is that all arrangements will be identical.
Attach one of the Hs anywhere.
No matter where the second H goes, it will be adjacent to the first.
No matter where the two Cs go, each will be adjacent to both of the Hs and to the other C.

Once there are three positions where $\ce{-OH}$ could be; then, the listing mentions six positions; then you move to a molecule with six carbon atoms: all in the first paragraph. While I disagree to count the $\ce{-H}$ at the terminal $\ce{C}$ as different to those «on the upper or lower of either end C», I speculate the presentation of the first paragraph still is a bit confusing, possibly especially for a first encounter to organic chemistry.

To do so, you need to know in advance the atoms by type (carbon, hydrogen, nitrogen, etc.), their number, and their valence.  The later states how many bonds an atom may share with an adjacent atom.  In your example, you consider hydrogen (only one bond), oxygen (up to two), and carbon (up to four).
You then introduce a constraint how these atoms are connected with each other.  In the examples above, for example, you seem to use an explicit constraint; in the hydroxyl ($\ce{-OH}$) group oxygen binds to hydrogen, but this ensemble binds to carbon only via oxygen.  An implicit constraint one may infer from the two illustrations you share is that you do not allow the formation of cycles for non-hydrogen atoms.  Instead, you want to have them as chain.
Now to put this into perspective (literally), chemists convened how to draw molecular structures e.g., on paper while these are 3D objects.  This is importance because often it does not suffice to know which atom binds to the next atom (constitutional isomer), but the spatial arrangement around e.g., carbon atoms (chirality, stereoisomers).  Each of these levels is a potential pitfall for the exhaustive generation of the isomers possible; both for chemists who need training on this as well as for computer programs.  Some generator programs (implicitly/explicitly) do not consider all of these levels.
In the examples shown by you, molecular symmetry reduces the number of isomers possible.  On paper, you might assume that any of the six hydrogen atoms at the two terminal carbon atoms may be replaced by a $\ce{-OH}$ group which already would lead to six isomers.  No, this is not the case as they are all symmetry equivalent:

Assume the three carbon atoms - symbolized by dark spheres - as a plane; the hydroxyl group (oxygen symbolized by a red, hydrogen by white spheres) may be once at this level, or above, or below.  But the whole molecule (propan-1-ol) may be rotated; the rotation of the very same group binding to $\ce{-OH}$ may be again above, at level, or below of this imaginary plane.
The second structure you drew is a constitutional isomer of the first one, it is propan-2-ol.  You might think that it would matter if the hydrogen replaced by $\ce{-OH}$ is above of this level of the three carbon atoms.  Again, because of symmetry, this is not the case, the two are symmetry equivalent and you have to consider only one isomer:

Thus, if you are set to look for the isomers of alcohols with the sum formula $\ce{C3H8O}$, there are only two isomers in total.
There would be a third isomer, an ether, if one allows for the joining $\ce{CH3CH2-O-CH3}$:


text:
What are advantages and drawbacks of using different orders of Møller-Plesset perturbation theory?
I am about to do some research in computational chemistry, however, I do not understand one method: What is the difference between second order and third order, etc, Møller-Plesset perturbation theory? I understand that MP theories include electron correlation in comparison to Hartree-Fock and how CCSD(T) can model the dispersion forces between atoms even better, but I cannot get a grasp on the difference between MP2, MP3, MP4 etc.
Also, why would I choose to use MP3 over MP2? Especially when they can be divergent, so surely I could end up with huge discrepancies in my results?

I do not understand what the issue is? This is not a homework question, can someone please clarify how it needs to be changed?

Typically, questions should include some background about what you have done to try to solve the problem and what you already understand. It helps users craft more specific answers rather than, say, having to explain the concept of perturbation theory from scratch when you already know all that.

Ok, that is much appreciated; I will try to improve the question.

Chemistry.se has a resources page including some entries point to references about computational chemistry.  Jan Jensen's http://molecularmodelingbasics.blogspot.com/ may be an other entry, too.  If you have a specific computation in mind, sibling mattermodelling.se may offer an audience more suitable than chemistry.se.

These resources have been very useful, thank you! :)

This is a very basic question, which is answered in many of the computational chemistry textbooks. Understanding the derivation of these methods is key to understand when they can be applied and when they must fail. Do the math by yourself, it really helps. Here's a sneak peek: https://chemistry.stackexchange.com/q/28346/4945


text:
Why are strong oxidizing agents "incompatible" with ferrocyanide according to SDS?
As I've mentioned in this question, I am making $\ce{Na3Fe(CN)6}$ from $\ce{Na4Fe(CN)6}$ by oxidizing the ferrocyanide using sodium permanganate. The SDS of sodium ferrocyanide as well as potassium ferrocyanide mention that they are "incompatible" with strong oxidizing agents. What reaction can occur that causes them to be incompatible? Can the reaction with strong oxidizing agents release $\ce{HCN}$, as is the case with strong acids and ferrocyanide?

The reaction $$\ce{[Fe^{II}(CN)6]^4- <=> [Fe^{III}(CN)6]^3- + e-}$$ is quite well known with standard redox potential +0.37 V. ( https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)) // Additionally, the equilibrium cyanide can be easily oxidized, e.g. to cyanate.

@Poutnik I am not sure I understand your last sentence. There is no cyanide formation in the reaction you mentioned - that's just the standard ferrocyanide/ferricyanide couple and is in fact my desired reaction. That's why I'm surprised that the SDS mentions that ferrocyanide is incompatible with strong oxidizing agents - can the ferricyanide, $\ce{[Fe^{III}(CN)_6]^{3-}}$, so formed react further with strong oxidizing agents?

"//" means for me it is a separator of different comment topic. // By equilibrium is there meant complex formation equilibrium. [Fe(CN)6]^4- <<=>  [Fe(CN)5]^3- + CN-(equilibrium cyanide),  CN- + O -> OCN- // There is possible, but I have no handy reference for it(therefore comment and not answer), that strong oxidant may attack directly the bound cyanide group. // Even if the oxidation stopped at ferricyanide, it would still make it incompatible.

The earlier answers' and comments line of thought gravitates around «what may happen to $\ce{Na3Fe(CN)6}$ or/and $\ce{Na4Fe(CN)6}$ in the presence of strong oxidizers» to the effect that MSDS express a warning.
Complementary to this, I suggest one should keep an eye on the oxidizer; because if there is an oxidation by them, simultaneously, they are going to be reduced.  Hydrogen peroxide is an example of a strong oxidizer commercially available (many chemistry labs have a bottle of 30% solution in stock [e.g., to prepare piranha solution]). Thus, a search with the three keywords «H2O2», «decomposition», and «Na4Fe(CN)6» yields for example the following entry in semantic scholar:

Corpus ID: 101692556
Effect of pH on the Rate of H2O2 Decomposition over K3Fe(CN)6 Catalyst
S. Zrnčević, Emil Pajc, V. Markov | Published 1973 | Chemistry
Pracena je brzina raspada vodikovog peroksida uz katalizator K3Fe(CN)6 u intervalu pH vrijednosti od 9 do 13 pri 30 0C. U navedenom intervalu pH u odsutnosti katalizatora dobivena je linearna ovisnost konstante brzoine reakcije o pH, a uz dodatak katalizatora u istom pH podrucju nađeni su maksimumi bruzina reakcije. Diskutira se reakcijski mehanizam u prisutnosti katalizatora.

which basically is a copy of an entry at CROSBI, the Croatian scientific bibliography.  From Google translate of the abstract (Sažetak), the authors report a linear increase of the rate of decomposition of $\ce{H2O2}$ in the range of $\mathrm{pH}\ 8\dots 13$ at $\pu{30 ^\circ{}C}$.  The presence of $\ce{K3Fe(CN)6}$ accelerates this even further.
Since the decomposition of $\ce{H2O2}$ into water and oxygen is exothermic ($\pu{–2884.5 kJ/kg}$, reference), the catalyst promoting a higher rate of reaction thus may lead into a runaway.  By analogy, I speculate if $\ce{K3Fe(CN)6}$ (the publication) acts as a catalyst here, $\ce{Na3Fe(CN)6}$ (as in the original question) equally does so, too.
Reference:
Zrnčević, S.; Pajc, E.; Vasić, Đ.; Markov, V. Ovisnost brzine raspada $\ce{H2O2}$ uz katalizator $\ce{K3Fe(CN)6}$ o pH. Kemija u industriji : časopis kemičara i tehnologa Hrvatske 1973, 1, 21-24.

Ferrocyanide $\ce{[Fe^{II}{CN}6]^{4-}}$ contains iron at the oxidation number $\pu{II}$. It can easily be oxidized to the ferricyanide ion $\ce{[Fe^{III}(CN)6]^{3-}}$containing iron at the oxidation number $\pu{III}$. This oxidation can be done by strong oxidants like elemental chlorine $\ce{Cl2}$ according to  $$\ce{[Fe^{II}(CN)6]^{4-} + Cl2 -> [Fe^{III}(CN)6]^{3-} + 2 Cl-}$$
According to Cotton and Wilkinson, ferrocyanides are extremely stable and not decomposed into cyanides and $\ce{HCN}$ by acids at ordinary temperatures. So their solutions are not poisonous. High temperatures and concentrated acids are required to carry out this decomposition. Surprisingly enough, ferricyanides are quite poisonous : they have a greater tendency to be labile and to loose one $\ce{CN-}$ ion, for kinetic reasons, according to these authors. Their reaction in water is slow but proceeds according to $$\ce{[Fe^{III}(CN)6]^{3-} + H2O -> [Fe^{III}(CN)5(H2O)]^{2-}  +  CN^-}$$
See : F. A. Cotton and G. Wilkinson, Advanced Inorganic Chemistry, Interscience Pub., Wiley & Sons, 1972, p. 865.

Cyanide can be oxidized to cyanogen [(CN)2] cyanate and eventually to CO2 and nitrogen.  In solution ferro and ferricyanides are relatively stable and do not readily release cyanide ion so ferricyanide is resistant to further oxidation. I would think that a mix of solid ferricyanide and a solid oxidant or high concentration of oxidant such as 35% peroxide could potentially react violently if initiated.

In my opinion, the statement is because the reaction between ferrocyanide and permanganate may be intense enough to cause dangers, like splattering.
I can't seem to find any info regarding HCN generation, and it is likely not due to that because even if the gas is formed, it will be oxidized to nitrogen or cyanate by the permanganate.
Also, many other quite unreactive chemicals are "incompatible" with strong oxidizing agents according to their SDS, including sodium phosphate and silicon dioxide.


text:
DNA antiparallel strains and nitrogeous bases orientation
I have a question related to the orientation of DNA antiparallel strains. DNA strains are called antiparallel because one strain is orientated from 5' to 3' and the other one from 3' to 5'. Regarding the phosphate-sugar chains that would be equivalent to say that the chains are rotated 180 degrees with respect to each other. However, in textbooks, 3' to 5' nitrogeneous bases are not shown as rotated with respect to 5' to 3' nitrogenous bases.
Why is this? If it would be the case then A-T would still make two hydrogen bonds but G-C would not make three but just one hydrogen bond!
PD: if needed I could make a picture to show my point.

Welcome to Chemistry StackExchange! My advise for you would be to stare at various 3D images or interactive models of DNA structure long enough to resolve any misunderstandings you have about the actual structure :-)

Thank you! Now it makes much more sense :)

The bases are roughly perpendicular to the axis of the double helix. In two-dimensional figures, the structure is flattened to make it easier to see.
The two strands are in the same conformation, just rotated by 180 degrees around an axis perpendicular to the helix axis. You can see this nicely in a palindromic duplex such as the one below, where the two strands (shown with black and gray carbon atoms) are rotated from the double helix orientation to a superimposed orientation and back again.

To get an interactive view of the same DNA (but without the artificial rotation to superimpose the two strands), see e.g. Proteopedia. (To run the superposition yourself, click on the "load full" button, right-click on the Jmol window, open a console, and issue the command "compare {1-12} {13-24}  ATOMS{1-12}{13-24} ROTATE TRANSLATE").
It is not easy to find a 2D drawing that captures this symmetry. Here is an example, though, where the left and the right strand look like they have the same conformation:

Source: https://universe-review.ca/I01-06-DNA.jpg


text:
How does ethanol separates salts from proteins
Grad student here messing around with famed TRIzol isolation protocol and looking to optimize the protein yield at the end. There are more direct ways to get protein from a sample but I like getting RNA, DNA and protein all in one process if possible.
Specifically I am curious to understand the wash steps in the protocol for the protein pellet. At this point the RNA and DNA phases are removed and you have this white pellet of gunk which includes your protein fraction and undesirable stuff. You proceed to perform a series of wash steps with $\pu{0.3 M}$ guanidine hydrochloride in 95% EtOH. I know the guanidine  plays a role as a detergent and relaxing the pellet. What I do not fully understand is how EtOH removes salts?
I am interested in switching to urea for the detergent role. The problem is it is not very soluble at high concentrations (~8M ideally) in ethanol. I could get fair amount in 70% ethanol but I am worried what that might do to my washes. Would I be losing protein as the water concentration increases?
Interested what the community could shed light on this issue.
Thanks in advance.

What protocol are you using, exactly (Thermo Fisher's (PDF)?)? Feel free to add a link, or, better, cite essential steps for the context. Also, what is EOTH? Do you mean ethanol EtOH? Note that Guan is a country and not an abbreviation for guanidine. Unit symbols and numbers are always separated by a space (i.e. 0.3 M), and chemical names are common nouns (no need to capitalize urea).


text:
Is there a name for the law that in chemical reactions atoms are conserved?
One characteristic of chemical reactions is the trivial fact that the number of atoms of a given element is conserved. This is not conservation of mass, albeit conservation of mass is a consequence of this fundamental property.
I guess that at some point in history philosophers found this basic property, probably somewhat around the time, when the law of constant proportions was found?
Does anybody now a name for that law? Or the name of the person that first stated it?

This isn’t actually a strict law, as evidenced by radioactive decay

It is the "Chemical equations must be balanced" law. Textbooks refer to this as the conservation of mass, even though mass is not exactly conserved. This usage reflects the history of the discoveries (8 years between conservation of mass, Lavoisier, and definite proportion, Proust),

@Andrew The question is only concerned with chemical reactions. Radioactive decay is not a chemical reaction.

@DrSvanHay - “Nuclear chemistry is the sub-field of chemistry dealing with radioactivity, nuclear processes, and transformations in the nuclei of atoms, such as nuclear transmutation and nuclear properties.” - to quote Wikipedia. Glenn Seaborg’s Nobel prize was in chemistry. Enough said.

@JonCuster, but are those processes considered to be "chemical reactions"?

For the topic of history of chemistry, consider also the SE site History of science and mathematics

" the trivial fact that the number of atoms of a given element is conserved" . There was nothing "trivial" about suggesting that atoms exist. Dalton's theory or just atomic theory would probably be the name of the theory you are searching for.

I agree with Poutnik, this fits better with the history of science and mathematics SE, as it is about the evolution (history) of a theory. A theory should mesh with existing knowledge but can also change to incorporate new facts. In this case the conservation of mass during chemical reactions had been proposed, and then new facts about atomic structure cropped up and were then meshed into an updated theory of the structure of matter.

Daltons definition of the "atom" plus mass preservation.

@RayButterworth - Well, they are reactions and I think nuclear chemists think of them as reactions. You might not. I'm sure Glenn Seaborg thought of them as as chemical reactions.

@JonCuster, I would consider transformations within the nuclei of atoms as nuclear reactions, and transformations in the relationships between two or more atoms as chemical reactions.

@RayButterworth - I'm not doubting your stance (and generally agree with it), but none of us have the authority to declare what usage everyone else should use.


text:
Why Lambert's law does not hold for photoresistors?
Why don't  photoresistors made of Si obey Lambert's cosine law? Is it related to the dark current of the photoresistor? I couldn't find any solution while searching the web.

Address this rather to Physics SE (particularly solid state/phase physics). The fact chemistry deals with various compounds and their electrons does not mean anything related to that is addressed by chemistry. // Aside of that and generally, be aware that laconic, not elaborated questions  (describe well, search, think, refer, summarize) are usually closed on the StackExchange network.

I wouldn't agree this is off-topic for Chemistry.SE. Besides, Physics.SE already have unanswered Why do some materials follow Lambert's cosine law? and the info on the underlying principles indeed seems to be scarce. I do agree though that the question is too laconic. A reference, an example of such material, or elaboration on the expected chemical effect on radiometry would improve the question.

@andselisk I have commented rather in sense Physics SE may be better, but have not checked the site for this. I do agree the solid state matter is a cross-discipline between both, but considering the Q asks rather about physical aspects.

Not sure if there are Si photoresistors. They are usually metal sulphides/selenides.

@Poutnik - plenty of silicon  photoresistors out there to be bought (although photoresistors are themselves a pretty niche product anymore).

@JonCuster Hmm, interesting, never heard of Si LDR, unless the components  are photodiodes/transistors.

Why is the material being silicon important for this question? Is it claimed somewhere that silicon photoresistors are special in their reflection directivity pattern?


text:
Are there any genuine, elementary ternary reactions?
Macroscopically, reaction rates can be of varied order.  But mechanistically, most reactions are first-order or second-order/binary (e.g. SN2, many catalyst surface reactions, dimers).  Most processes that are macroscopically ternary (e.g. protein synthesis) involve the formation of an intermediate (e.g. translation complex).
Are there any true elementary ternary reactions, involving the chance collision of 3 substrates?
Note: for this purpose, intermediate dimers are not ternary.  I'm asking if there is any known reaction with a genuine ternary transition state, only stabilized by or requiring the 3 species.

You can never prove that a reaction is ternary, even if it really is. There is always a chance of some short-lived intermediate, too unstable to observe.

Like $\ce{ 2 ^4He  <<=> ^8Be}$ and $\ce{ ^4He  + ^8Be -> ^{12}C}$ what is interstellar synthesis of carbon from helium.

@IvanNeretin nothing is certain, but pchem could probably provide supporting evidence for a proposed ternary mechanism.

The transition state will likely depend on all 3 substrates, perhaps due to a favorable electronic state in 1 caused by the presence of both others.

¶Poutnik. This mechanism is often presented in treatises on nucleosynthesis. But are you sure that it makes sense ? The nuclide $\ce{^8Be}$ has a half-life of $10 ^{-16}$ s. This time is just the duration of a collision between two alphas. In order for its reaction to occur with another alpha, this third alpha should touch the two first alphas before they leave as if it would be a ternary collision. Is it not a ternary reaction ?

@Maurice I am sure.
Ternary nuclear collision is extremely unlikely. Astrophysicists were not originally sure how 12C can be ever synthetized, but then theoretical nuclear physicists predicted there is some kind of resonance state of 8Be, that lives long enough to fuse with another 4He. Something similar like when a tiny portion of diprotons transform to deuterons in proton-proton cycle before they decay back to protons.

@Maurice See triple alpha process. // And sorry, that resonance state was related to 12C.

See: https://chemistry.stackexchange.com/questions/79080/is-there-any-empirical-upper-limit-for-order-of-reaction

We cannot really distinguish a ternary collision from two binary ones, because the binary collision takes time for the interaction to run its course and therefore any system has a nonzero chance that the third body will arrive during this time.
If the binary collision pair is stabilized, then the interaction time will be longer and the probability for a ternary interaction grows. In chemistry the poster-child example of this effect is the reduction of nitric oxide by hydrogen given by the reaction and gas-phase rate law:
\begin{align}
\ce{2NO + 2H2 &-> N2 + 2H2O},& \text{rate} &= k\ce{[NO]^2[H2]}
\end{align}
The simplest explanation for this ternary rare law is that there is some stability in the $\ce{N2O2}$ dimer[1], so a collision between two nitric oxide molecules that produces this dimer "hangs around" long enough for a hydrogen molecule to arrive and react in a significant fraction of cases, thus beginning the reduction.
Reference
1.
Stephen G. Kukolich (1982). "The structure of the nitric oxide dimer". J. Am. Chem. Soc. 104, 17, 4715–4716.
https://doi.org/10.1021/ja00381a052

If an intermediate dimer forms, then I don't think this counts as a "genuine/elementary" ternary reaction as asked.  (Not that it isn't an interesting example!)

That depends on your point of view. Is the dimer (which can decompose as rapidly as it forms) a separate species or a temporary collision state? And can you really tell? Therein lies the rub.

Yes, for the purposes of this question intermediate collision states (e.g. dimers) are separate.   QM will distinguish between a genuine ternary stabilized transition state and a dimer that rapidly collides with a third species.


text:
In the complete combustion of methane, is it methane or carbon that is oxidised?
I realise this is a rather pedantic question, but I'm wondering which application of terminology is more correct here.
I know that, in the complete combustion of methane, methane is the reducing agent. But I also know the oxidation state of hydrogen doesn't change in this reaction.
If we were asked to identify what is oxidised in this reaction, could we simply say that methane is oxidised? Or would we have to be more specific and say that 'carbon atoms in methane' are oxidised? Would either be fine?

As for me, either would be fine.

Since the four hydrogen atoms combine with oxygen, I would not say only the carbon is oxidized.

@DrMoishePippik Well, formally, H in CH4 already IS oxidized. By partial oxidation $\ce{CH4 + O2 -> C + 2 H2O}$, only carbon is oxidized. Similarly, by replacing oxygen by fluorine, hydrogen is not oxidized either. // But I do agree that factually, considering homolytic cleavage and radical mechanism, rather both C and H are oxidized by full combustion.

In the $19$th century, oxidation numbers were not known. The oxidation was the reaction of a molecule  or an atom with $\ce{O2}$. Every chemist would say that, when burning, methane is oxidized into $\ce{CO2 + H2O}$. When studying this reaction, nobody was ready to consider that only one atom could be oxidized. Oxidation was the reaction of the whole molecule. There are still people today using the word "oxidation" with this definition.
Of course today the notion of oxidation has been extended. An oxidation is not necessarily a reaction of $\ce{O2}$. And oxidation numbers have been introduced. I will not develop this well-known question here. Of course when burning methane, the Carbon atom is oxidized.
But I would like to add here a personal observation. In my country, there are two sorts of high school teachers. Most are teaching that in the $\ce{C-H}$ bond, the $\ce{H}$ atom is at the oxidation number $+1$. But some are teaching that the difference in electronegativity between carbon and hydrogen is so small that the $\ce{C-H}$ bond may be considered as non-polar, so that in hydrocarbons, both $\ce{C}$ and $\ce{H}$ atoms are at the oxidation number zero. In a hydrocarbon, all atoms are at the oxidation number zero. For these teachers, both $\ce{C}$ and $\ce{H}$ atoms are oxidized in the methane combustion.
I would be interested to hear what you think of this strange situation.

Oxidation numbers can be very arbitrary and are a tool, not a be all and end all, describing a redox reaction.  The C-H bond, in alkanes especially, is particularly nonpolar compared to the resultant O-H and C=O bonds in H2O and CO2. So if you examine the equation and the products the conclusion [to me at least] is that both are oxidized. [I would prefer an ox. no. of zero for both.] For a more polar molecule such as the oxidation of aqueous HCl to Cl2 and water by permanganate the question doesn't arise; H is +1 all along.


text:
Why does cyclopentadiene react faster than cyclohexadiene in Diels Alder reactions?
How does the length of the bridge of the resulting bicycle change the rate of the Diels Alder reaction (examples below)? I can't seem to think of how one would be more electron donating than the other to result in a more electron rich diene.

What is your source for the reaction rates?

What is your experimental evidence for this, are there kinetic data for experiments running (like in the [suggestive] illustration added) where either cyclopentadiene, or 1,3-cyclohexadiene are used as dienophile for the same diene at otherwise same conditions?  Are these differences significant (e.g., beyond $\pm 3 \sigma$)? Perhaps there is some difference in reactivity for a DA among small ring dienophiles (less than or equal to six atoms in the cyclic dienophile) by ring tension, but without hard data, this is a speculation.

First off, do we even have rate constants? I see none in the question.

I don’t have any sources about the rates, I heard a colleague mentioning this the other day, but the question still stands on how you would predict which is faster

The diene portion of the 6-ring is less coplanar than the diene portion of the 5-ring.

I'd fully expect the 5-ring to react faster than the 6-ring, because it has less conformational flexibility (cf @ron's comment and Oscar's answer). Essentially, the 5-ring is more constrained to be in a conformation that is suitable for the DA reaction. But I'd probably still like to see hard data.

Assuming the claimed difference is reaction rates is real (see the comments), we may want to consider the distance between the ends of the conjugated diene chain. You need these ends to overlap with the pi orbitals of the other alkene, and making the diene end-to-end distance closer would improve the overlap thus facilitating the reaction. Thus the pentagonal ring, by creating a closer distance than the hexagonal one, is expected to lead to a faster reaction rate.


text:
Why doesn't my estimation for the relative mass of a single lithium-7 atom work?
So I tried to calculate the relative mass of a lithium-7 atom. I added up the masses of all the protons, neutrons and electrons, like this:
mass = relative mass of 3 protons + relative mass of 4 neutrons + relative mass of 3 electrons
= 3 + 4(1838.68/1836.15) + 3(1/1836.15)
= 7.007145386
However, our bound atom would have a lower mass than this.
Our chemistry book states the relative mass of a lithium-7 atom as 7.016, which is greater than our estimation. Where des this "extra" mass come from, then? I mean, other than the particles themselves.

Yes, but why is that? That is what I am asking.

The mass of atoms cannot be calculated by addition of masses of free particles. Additionally, the mass of proton is not the relative unit of atomic mass, but 1/12 of 12C is. Mass of 4He is about 0.7% less then mass of 2p + 2n + 2e. The difference is the equivalent mass of released nuclear bonding energy.

I don't quite follow your calculation either. Using CODATA values for particles' masses, I arrive at a different theoretical number for $\ce{^7Li}$: $$3\times\pu{1.0072 u} + 4\times\pu{1.0087 u} + 3\times\pu{0.0005 u} = \pu{7.0579 u}$$

The good initial step before asking is reading relevant textbooks, googling  keywords and studying the search results.
This prevents cases of asking for answers that have been already written in many ways on many places. The Google search scope can be eventually limited to the particular address domain by adding the additional term e.g. site:stackexchange.com, site:libretexts.org or site:wikipedia.org .
// Review also Chemistry SE: resources-for-learning-chemistry

Your calculation incorrectly assumes that the mass of a proton is 1 u. All other masses used are given relative to this incorrect proton mass. Basically, you have correctly calculated that sum of the individual components of the Li-7 atom is 7.0071 times heavier than a free proton, but now you have to multiply this by the actual mass of a free proton (1.0072 u), which yields 7.0576 u, in good agreement with andselisk's more precise value. Now this value is larger than the true mass of a Li-7 atom, due to release of nuclear (and electronic) binding energy when you join the particles together.

Now i get why I was wrong. I was using masses relative to the mass of a proton. But relative isotopic masses are measured in Daltons. That's why my calculation didn't match. I did it the correct way now (considering binding energy too) and now it works. Thanks.


text:
IUPAC nomenclature of extremely long alkanes
What would be the name of a linear alkane containing 10000 carbon atoms, i.e. $\ce{C10000H20002}$?
I have searched for it, but the longest alkane I could find had 9000 carbon atoms (and was apparently called nonaliane).

Chemistry.se has a dedicated resource page for study here, including a section about nomenclature. If IUPAC did not convene about systematic names with more 9000 carbon atoms in a straight chain, either a) because nobody needs this (at present) to describe his/her synthesis, or b) there are descriptors already in use (e.g., FASTA format in biochemistry) which are more efficient than using a nomenclature designed with smaller molecules in mind.

Previously disussed: https://chemistry.stackexchange.com/questions/152181/does-the-largest-known-cycloalkane-contain-288-carbon-atoms. Btw, where did you find 9000 carbon atoms? The longest I could find is 6000 atoms: https://cen.acs.org/articles/94/i15/New-record-longest-linear-carbon.html. "Nonaliane" also is not giving me sufficient results except: https://zuzentzailea.livejournal.com/4349.html

I suppose one can treat $\ce{C10000H20002}$ as polyethylene $\require{enclose}\ce{\enclose{horizontalstrike}{\;(}CH2\enclose{horizontalstrike}{)_n}}$ with the defined degree of polymerization, resulting in source-based IUPAC name polyethene (10000 DP) or polyethene (140000 $M_\mathrm{r}$) [1, pp. 315–317]:

11.2 Structure- and source-based names for common polymers
$$
\begin{array}{lll}
\hline
\text{Structure} & \text{Source-based name} & \text{Structure-based name }\\
  & \text{(preferred given first)} & \\
\hline
\require{enclose}\ce{\enclose{horizontalstrike}{\;(}CH2\enclose{horizontalstrike}{)_n}}^\textit{a} & \text{polyethene} & \text{poly(methylene)} \\
  & \text{polyethylene}^\textit{b} & \\
\vdots & & \\
\hline
\end{array}
$$
$^\textit{a}$ The formulae $\require{enclose}\ce{\enclose{horizontalstrike}{\;(}CH2CH2\enclose{horizontalstrike}{)_n}}$ and $\require{enclose}\ce{\enclose{horizontalstrike}{\;(}CF2CF2\enclose{horizontalstrike}{)_n}}$ are more often used; they are acceptable due to past usage and an attempt to retain some similarity to the CRU formulae of homopolymers derived from other ethene derivatives.
$^\textit{b}$ Traditional names.

Further from [1, p. 351]:

Rule 1.8: Specifications about mass fractions $(w),$ mole fractions $(x),$ molar masses $(M),$  relative molecular masses $(M_\mathrm{r}),$ degrees of polymerization (DP) or the average values of the latter three quantities, may be expressed by placing the corresponding values in parentheses after the formula of the macromolecule in a manner analogous to that recommended for the naming of copolymers [6].

From subsection 8 Specification with regard to mass fractions, mole fractions, molar masses and  degrees of polymerization [1, p. 381]:

Rule 8.3
The molar mass, relative molecular mass or degree of polymerization may be included in the scheme of Rules 8.1 and 8.2 by adding the corresponding figures, followed by the symbol $M,$ $M_\mathrm{r}$ or DP, respectively.
Examples:

polybutadiene-graft-polystyrene (75:25 mass %; 90 000:30 000 $M_\mathrm{r}$) (a graft copolymer consisting of 75 mass % of butadiene units with a relative molecular mass of 90 000 as the backbone, and 25 mass % of styrene units in grafted chains with a relative molecular mass of 30 000)

polybutadiene-graft-polystyrene (1700:290 DP) (a graft copolymer consisting of a
polybutadiene backbone with a degree of polymerization of 1700 to which polystyrene
with a degree of polymerization of 290 is grafted)

Reference

IUPAC. Compendium of Polymer Terminology and Nomenclature: Recommendations 2008 (the “Purple Book”); Richard G. Jones, Jaroslav Kahovec, Robert Stepto, Edward S. Wilks, Michael Hess, Tatsuki Kitayama, W. Val Metanomski, Eds.; RSC Publishing: Cambridge, 2009. ISBN 978-0-85404-491-7.

Ah well, I guess it took me six minutes to write up my answer... I was thinking about mentioning the term degree of polymerization, but decided against it because I realized that I am unsure how to use it together with structure-based naming. Source-based naming feels not entirely spot-on here to me.

@Snijderfrey It is not surprising that polythene can contain more than 10000 carbon atoms. LDPE contains 4000-40000 carbon atoms (often branched), UHMWP can have 500000 carbon atoms, so we are not bothered to name individual alkane. Even though they can be named according to IUPAC rules, they can be quite a mouthful to say. So, we can safely classify them as "polymer" or "polythene with X degree of polymerization"

@NilayGhosh, yes, I am aware of the size of macromolecules. The answer here is good, I am just not sure about the DP given in "polyethene (10000 DP)". To my understanding, the DP should be 5000, but with the structure-based name poly(methylene) this seems strange. I asked a separate question to get some insight on how to use DP with structure-based nomenclature: https://chemistry.stackexchange.com/questions/162542/degree-of-polymerization-with-structure-based-polymer-naming

As far as I can tell, the answer is actually undefined, as the 2013 Blue Book only gives a formula for constructing so-called "numerical terms" or "multiplicative prefixes" up to the number 9999.
P-14.2 of the Blue Book gives multiplicative prefixes for all values of $n$, $10n$, $100n$, and $1000n$ (where $n$ is an integer between 1 and 9), so all numbers up to 9999 can be constructed by stringing these together. However, it stops abruptly at 9000, so we can't access anything beyond 9999.
For 9999 atoms, the corresponding prefix would be 'nonanonacontanonactanonalia' (nona = 9, nonaconta = 90, nonacta = 900, nonalia = 9000); so the linear alkane $\ce{C9999H20000}$ would be named nonanonacontanonactanonaliane, as specified by P-21.2.1 of the Blue Book:

Systematic names for the higher members of this series consist of a numerical term followed by the ending 'ane' with elision of the terminal letter 'a' from the numerical term.

The choice to stop at 9999 seems to have been conscious, although perhaps arbitrary: see e.g. https://iupac.qmul.ac.uk/misc/numb.html for more discussion. This was published in 1986:

Recently a desire has been expressed for numerical terms higher than 199. This document provides the necessary guidance, based on principles already established, for generating numerical terms up to 9999 by using the infixes "cta" (for the hundreds digits) and "lia" (for the thousands digits) in a way quite analogous to the use of "conta" for the tens digits beyond twenty.

I'm aware that this is old, but the 2013 Blue Book doesn't contain any more info beyond this.

It could be argued that 9999-ane is simply the largest parent hydride in the compound; by that logic, 10000-ane would be called 1-methylnonanonacontanonactanonaliane. I think I once wrote a script to generate general linear alkane names based on that principle. It's actually pretty tricky, since it depends on the alphanumerical order of the substituents.

I suggest to think of $\ce{C10000H20002}$ as a uniform polymer (deprecated: monodisperse polymer). The correct name according to the IUPAC Recommendations on Polymer Terminology and Nomenclature (see the Purple Book) would then be poly(methylene) in the structure-based nomenclature or polyethene (preferred over polyethylene) in the source-based nomenclature.
If you want to distinguish between poly(methylene) with different chain lengths, you would usually use the molar mass.


text:
Degree of polymerization with structure-based polymer naming
Polymers may be named using either a structure-based or a source-based nomenclature, as detailed in the IUPAC Purple Book.[1] Therefore, the polymer descibed by $\require{enclose}\ce{\enclose{horizontalstrike}{\;(}CH2\enclose{horizontalstrike}{)_n}}$ may be called either poly(methylene) or polyethene (see also this recent question). Both names are perfectly fine and by themselves do not imply how the polymer was prepared, so polyethene is not necessarily the product of a polymerization reaction.
By considering a methylene group as the repeating unit within the macromolecules in poly(methylene), it was unclear to me how the degree of polymerization (DP), a rather common number used to describe a polymer, should be used in this case. The Purple Book defines the DP as:

Number of monomeric units in a macromolecule an oligomer molecule, a block, or a chain.

Now, what is a monomeric unit?

Largest constitutional unit contributed by a single monomer molecule to the structure of a
macromolecule or oligomer molecule.

Looking for the definition of constitutional unit gives:

Atom or group of atoms (with pendant atoms or groups, if any) comprising a part of the
essential structure of a macromolecule, an oligomer molecule, a block, or a chain.

...and a monomer molecule is:

Molecule which can undergo polymerization, thereby contributing constitutional units to
the essential structure of a macromolecule.

All definitions are from the Purple Book. So, as I understand it, the term degree of polymerization is directly linked to the process of polymerization, and in case of poly(methylene) the monomer molecules, i.e. the molecules which can undergo polymerization, are ethene molecules. Also, the monomeric unit in the definition of the DP is not the methylene group.
The question is: What would be the correct DP for a poly(methylene) consisting of macromolecules with 10000 methylene groups on average? From the definitions, I would say that it must be 5000, but this would be extremely counterintuitive in combination with the structure-based name, and it would be silly to have to reformulate a structure-based name into a source-based name just to be able to understand the DP. On the other hand, it would also be strange to give different DPs for poly(methylene) and polyethene with identical average molar masses just because of the naming convention used. I would be happy if you could help me to resolve this slightly pedantic issue.
Reference

Compendium of Polymer Terminology and Nomenclature, IUPAC Recommendations 2008, IUPAC Purple Book, 2nd edition, prepared for publication by Richard G. Jones, Edward S. Wilks, W. Val Metanomski, Jaroslav Kahovec, Michael Hess, Robert Stepto, Tatsuki Kitayama, RSC Publishing, 2009 [ISBN 978-0-85404-491-7]; https://doi.org/10.1039/9781847559425.

It's polyethylene not methylene - problem solved ;)

@Mithoron, I tend to agree, but my world where this was possible was just destroyed by spending too much time looking into the Purple Book.

I would like to share some of my random thoughts on your problem.
First, when we define something, sometimes we list the exceptional conventions that do not comply with the definition. In my opinion, exceptions are acceptable if they can be completely listed. For example, if the case of polyethylene is all the exceptions of the definition of degree of polymerization, that's all right for me.
Secondly, you provided a good point in your text, that "the term degree of polymerization is directly linked to the process of polymerization, and in case of poly(methylene) the monomer molecules, i.e. the molecules which can undergo polymerization, are ethene molecules. Also, the monomeric unit in the definition of the DP is not the methylene group." Then the DP can be differently counted for the same polymer due to different uses of raw monomer and polymerizing process. This is meaningful if the concept of DP is indeed designed for evaluating the efficiency of polymerizing process, but not a quantification of the molecular weight of a polymer.
Therefore, if you want to discuss another polymer, polymethylene, for which the degree of polymerization is counted by the number of methylene units, you should first report how you polymerize it from methylene.
This may not be meaningless. Poly(ethylene glycol) (PEG) and poly(ethylene oxide) (PEO) are the same polymer. The names reflect the difference in how they are synthesized. This may be one example that the monomer and polymerization process is distinguished in nomenclature even if the resultant polymer is the same.
Therefore, if there is a synthetic route to the polymer -[-CH2-CH2-]- from the polymerization of methylene, the polymer chemists might agree to name it differently as polymethylene and for which the DP should be differently counted from how it is done for polyethylene.

Thanks for reviving my question. I see that it probably makes sense to add some context to the DP if it is ambiguous like with polyethene/polymethylene. However, I must contradict that the name of a polymer implies anything reliable about how it was synthesized, neither for polyethene/polymethylene nor for PEG/PEO. The more common distinction between PEG and PEO is the molar mass, being lower for PEG, so that its properties are influenced by the end groups. PEG should in many cases be considered an oligomer.

I am not strictly holding either version of opinion, and I agree with you 100%. But for the case of PEG/PEO, what I understand is that the difference in the ranges of molecular weight between them originates from the synthetic route. PEO commonly refers to anion ring-opening polymerization from epoxy, which will mostly reach very high molecular weight. Whereas PEG commonly refers to condensation from ethylene glycol, whose molecular weight is limited by the reversible nature of the reaction. You are right about what you have said.


text:
How compressibility factor is introduced in Clapeyron equation?
I'm reading section Experiment 13 Vapor Pressure of Pure Liquid in Experiments in Physical Chemistry [1, p. 200]:

For the case of vapor–liquid equilibria in the range of vapor pressures less than $\pu{1 atm},$ one may assume that the molar volume of the liquid $\widetilde{V}_\mathrm{l}$ is negligible in comparison
with that of the gas $\widetilde{V}_\mathrm{g}$, so that $\Delta\widetilde{V} = \widetilde{V}_\mathrm{g}.$ This assumption is very good in the low-pressure region, since $\widetilde{V}_\mathrm{l}$ is usually only a few tenths of a percent of $\widetilde{V}_\mathrm{g}$. Thus we obtain
$$\frac{\mathrm dp}{\mathrm dT} = \frac{\Delta_\mathrm{v}\widetilde{H}}{T\widetilde{V}_\mathrm{g}}\label{eqn:4}\tag{4}$$
Since $\mathrm d\ln p = \mathrm dp/p$ and $\mathrm d(1/T) = -\mathrm dT/T^2,$ we can rewrite Eq. \eqref{eqn:4} in the form
$$\frac{\mathrm d\ln p}{\mathrm d(1/T)} = -\frac{\Delta_\mathrm{v}\widetilde{H}}{R}\frac{RT}{p\widetilde{V}_\mathrm{g}} = -\frac{\Delta_\mathrm{v}\widetilde{H}}{RZ}\label{eqn:5}\tag{5}$$
where we have introduced the compressibility factor $Z$ for the vapor:
$$Z = \frac{p\widetilde{V}_\mathrm{g}}{RT}\label{eqn:6}\tag{6}$$
Equation \eqref{eqn:5} is a convenient form of the Clapeyron equation. We can see that, if the vapor were a perfect gas $(Z \equiv 1)$ and $\Delta_\mathrm{v}\widetilde{H}$ were independent of temperature, then a plot of $\ln p$ versus $1/T$ would be a straight line, the slope of which would determine $\Delta_\mathrm{v}\widetilde{H}.$ Indeed, for
many liquids, $\ln p$ is almost a linear function of $1/T,$ which implies at least that $\Delta_\mathrm{v}\widetilde{H}/Z$ is almost constant.

How is it possible to rewrite equation \eqref{eqn:4} into equation \eqref{eqn:5} in order to introduce the compressibility factor?
Reference

Garland, C. W.; Nibler, J. W.; Shoemaker, D. P. Experiments in Physical Chemistry, 8th ed.; McGraw-Hill Higher Education: Boston, 2009. ISBN 978-0-07-282842-9.

@Edgardo In the future, please cite your sources, avoid posting screenshots of text, and please try to formulate the question as clearly as possible.

The quoted transformation is an application of the chain rule:
$$\begin{align}
\frac{\mathrm{d} (\ln p)}{\mathrm{d} (1/T)} &= \frac{\mathrm{d} (\ln p)}{\mathrm{d}p} \cdot \frac{\mathrm{d}T}{\mathrm{d} (1/T)} \cdot \frac{\mathrm{d}p}{\mathrm{d}T} \\[5pt]
&= \frac{\mathrm{d} (\ln p)}{\mathrm{d}p} \cdot \left(\frac{\mathrm{d}(1/T)}{\mathrm{d}T}\right)^{-1} \cdot \frac{\mathrm{d}p}{\mathrm{d}T} \\[5pt]
&= \frac{1}{p} \cdot (-T^2) \cdot \frac{\mathrm{d}p}{\mathrm{d}T}
\end{align}$$
Substituting in the expression for $\mathrm{d}p/\mathrm{d}T$ will yield the desired equation.
If you're not comfortable with the chain rule (i.e. $\mathrm{d}z/\mathrm{d}y = (\mathrm{d}z/\mathrm{d}x)(\mathrm{d}x/\mathrm{d}y)$), or with the fact that $(\mathrm{d}x/\mathrm{d}y)^{-1} = \mathrm{d}y/\mathrm{d}x$, I'd suggest picking up a book on calculus, which will explain these things.
Note that, although this looks as though we are treating derivatives as fractions and 'cancelling out top and bottom', this is not the correct approach and will fail in other cases. Again, I refer you to a calculus book for a discussion of this.


text:
Can a TDS meter have a false initial reading due to some instrument artifact?
I have total dissolved solids (TDS) meter which is built in to a water purification system and this meter always reads high, around 90-100 when the machine is turned on, but over the next 30 seconds or so while water is produced it gradually settles down to zero.
Is this likely to be some artifact of the meter or is the purification system somehow getting more contaminated when it sits out of use?
I suppose I could test this by generating two samples: a quantity of water produced by the machine at the outset, then a second quantity produced a minute later, then testing both samples with a separate, hand-held TDS meter. However, to do this test I would have to get another TDS meter.

Sounds like there's a slug volume of water that gets moved through when you start the pump.

Your hypothesis sounds plausible. TDS meters usually measure conductivity which can change in standing water if metal ions leach from the plumbing.  What you are observing could also be a form of hysteresis. Hand held meters are not terribly expensive, unless you want a high quality device. Finally, it would help if you provided more information: type of filter, water, application (lab?), even temperature etc


text:
If a = 0 for a Van der Waals gas, what does that signify?
I know that if $a = 0$ for a gas at certain temperature and pressure, it means that the molecules of gas have almost no attractive  forces acting between them.
But does that also mean that the molecules have high repulsive forces acting between them?

High repulsive forces would probably result in a negative a.

Even the tag you have used should give you the hint the name is not Vanderwall (like some textbooks say), but Johannes Diderik van der Waals

Note that such a question is hypothetical, as real gases have positive $a$ values.

What do you mean by "high repulsive forces"? The parameter "b" also implies repulsion by the way, due to excluded volume. You can regard this excluded volume as being due to a "high repulsive force" active at very short distances (on the order of the vdW radius).

The simple answer is no. An "$a$" of zero simply means the attractive forces are so low that they don't affect the pressure of the gas in a measurable way.
As Ivan Neretin mentioned in the comments, high repulsive forces would result in a negative $a$. More specifically, a positive $a$ is due to electrostatic attraction. Thus a negative $a$ would mean that, on average, the gas molecules are close enough to feel an average electrostatic repulsion. But you will never see a negative $a$, because there will never be a net electrostatic repulsion between gas particles. Here is why:
At sufficiently high densities, real gases exert more pressure than would be predicted if they behaved ideally.  This is not due to a repulsive electrostatic interaction, it is due to an excluded volume effect.  Indeed, at even the closest average interparticle spacing that atoms and molecules can have and still be gases, their average electrostatic interaction is attractive.  For them to experience a net electrostatic repulsive interaction, they would need to be so close that they are, on average, closer than the energy minimum in their interaction potential (typically modeled as a Lennard-Jones potential). And once the particles are that close together, the substance is no longer a gas.
Rather, the positive pressure deviation from ideality at high densities comes from the fact that real gas particles take up space.  At sufficiently high densities, there is consequently reduced free volume left for the gas particles to move around (compared to what they would have if they were ideal gas point particles). Hence the term "excluded volume".
This lower free volume significantly reduces the configuration space available to the gas (i.e., the number of possible microstates), and is thus entropically unfavorable. I.e., the positive pressure deviation caused by the excluded volume effect is not energetic in origin, it's entropic.
This can be most easily confirmed by looking at equations of state for the simplest possible real gases, namely noble gases like argon.  I did this and found that, at extremely high densities, it's actually energetically favorable to compress them (confirming that their average electrostatic interactions remain attractive), but entropically unfavorable:
$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$
$$\pd{E}{V}{T} > 0 \text{ (compression is energetically favorable)}$$
$$\pd{S}{V}{T} > 0\text{ (compression is entropically unfavorable)}$$

So then if we plot a graph of interatomic potential v/s interatomic distance for the molecules of a real gas with "a"=0 what would it look like?

@Vp127 A constant potential until they "hit the wall" when they hit other molecules. Then it raises infinitely steep, as the van der Waals model does not consider compresibility of molecules.


text:
What is the reaction mechanism for formation of alkene from vicinal dihalide in presence of zinc and alcohol?

I have proposed the mechanism that zinc gives its two valence electrons to the halogen so it leaves the electrons of carbon which get shifted to form π-bond and the other halogen is pushed out. Then what is the use of alcohol in this reaction?

Alcohol is merely the solvent, without it you do not get good contact between the reactants.

Then is my mechanism correct??

More likely is Zn inserting in the C-X bond, see https://chemistry.stackexchange.com/questions/66124/reduction-of-alkyl-halides-to-alkanes  The alcohol is also acting as a proton source.

Thank you sir. You made my work easy . May god bless you .

Your mechanism leads to the alkene but it implies stereospecificity, which is not the case.https://pubs.acs.org/doi/pdf/10.1021/ja01138a042

In the vicinal dihalide, the carbon develops a slight positive charge and the halogen develops a slight negative charge. Due to this slight positive charge of carbon, zinc with its valence electrons attack the carbon atom and then carbon breaks its bond with the halogen which develops -1 charge,zinc develops +2 charge and carbon develops-1 charge. The lone pair of electrons of zinc on carbon shift to form the π bond with other carbon and the halogen atom of the other carbon also gets released with -1 charge. Hence the alkene is formed and the Zinc with +2 charge and 2 halogen atoms with -1 charge form ZnX2. The alcohol in the reaction acts as a solvent for better contact between the reagents.

Do you have any references for this?

@ You only suggested me this idea and I think it is correct.

This paper suggests that a radical process is involved in the formation of the alkyl zinc https://www.tandfonline.com/doi/full/10.1080/15685551.2014.947550  beyond that I can find no established mechanism


text:
Criteria for establishing an order of basicity among several organic compounds
In an exercise from an exam several years ago, they ask to order the following three compounds from highest to lowest basicity:

I would not know how to start the exercise without going to a database, which is what I have done as indicated below.
In the CRC Handbook of Chemistry and Physics, the following pKa values are given for these compounds:
Pyridine: pKa = 5.23
Piperidine: pKa = 11.22
1-ethylpiperidine: pKa = 10.45
So, I have deduced that the order would be: II > III > I.
But, even so, this method would not be valid for me for an exam, since I cannot resort to any reference book.
So, I would be interested in what I should look for in order to be able to answer the question satisfactorily, since I can hardly find any information in basic books on the subject.

No problem; it happens. The next thing to consider is the hybridisation of the nitrogen lone pair, and what that tells you about its basicity.

The hybridization of the nitrogen atom in pyridine is sp2. For the rest of the compounds, I believe it presents an sp3 hybridization. @orthocresol

Indeed. Which has higher energy?

Some general points of comparison of basic strengths of organic compounds are the following (The order of the points is important, check for condition 1. at first and then check the others in subsequent order)

Delocalized and Localized lone pairs of electrons - In case of the presence of any N,O,S or halogen atom, we first compare whether the lone pair on the atom is a part of Resonance. Incase if the lp resonates, then delocalization of the lp occurs decreasing Basic character. A localized lp offers greater basic strength because by the Arrhenius definition of a base, a stronger base is the one which has greater tendency to donate electrons.

Hybridization - as the s character reduces, the electronegativity decreases and the hybridization gets more Basic.
sp hybridization being most electronegative is least basic. (basic nature of hybridizations is $sp < sp^2 < sp^3$ …)

Inductive Effect - A +I group or an Electron Donating Group (EDG) increases the basicity whereas a -I group or Electron Withdrawing Group (EWG) decreases Basicity.

Some other factors which may sometimes play a role in determining Basic strength
--> Cross Conjugation - Incase if the lp is delocalized between multiple rings due to Resonance, the Basic strength reduces to a great extent.
--> Amines are generally more basic than Amides and Alcohols are generally more basic than Carboxylic Acids
--> Ortho Effect - Incase if a bulky group like any alkyl group,-NO2,-SO3H,-COOH etc. is attached at ortho position of the cyclic compound, then also basicity reduces.
--> Steric Hindrance - Incase if around the atom containing the localized lps, a lot of groups are attached, it creates Steric Hindrance due to which the atom is not able to donate its lp freely and the basicity falls.
It is not necessary that these factors will always be followed. They may be present but may not play a role in determining the Basicity. There will be several exceptions and corresponding reasons because of the experimental nature of the subject. So finally, for accurate results, you must refer to the $\ce{pka}$ values
In the question that u have asked, all the 3 compounds have localized lps. So we now check hybridization. Pyridine has $sp^2$ hybridization while the other 2 compounds have $sp^3$ hybridization making Pyridine least basic.
In my opinion 1-ethylpiperidine should have been more basic than piperidine due to a +I group or EDG attached to the former. But according to the pKa values that you have furnished, Piperidine is more basic than 1-ethylpiperidine . I'm not completely certain about this but I believe that due to too many bulky groups being attached to the N atom in 1-ethylpiperidine, it experiences Steric Inhibition which brings down its basicity to some extent. This probably provides a reason for the order of basic strength to be $II > III > I$


text:
Application of Hund's third rule to rhodium
The rhodium atom has electronic configuration: $[\ce{Kr}] \,\mathrm{4d^8\,5s^1}$; the atomic term symbol is $^4F$. The possible $j$-values are $9/2$, $7/2$, $5/2$ and $3/2$. However, when using Hund's third rule, in case of spin-orbit coupling (LS-coupling), there are multiple shells to look at. Which subshell (d or s) should I look at for the more than / less than half-filled rule? Or, if Hund's rule is not applicable, what should I do then to determine the levels? The correct answer is supposedly $9/2$, but could you give more explanation?

To clarify: the problem is not determining the levels, but the ground level. For this, the sign of the spin-orbit coupling parameter should be considered.

Another related question, but without answer: https://physics.stackexchange.com/questions/421616/how-to-use-hunds-third-rule-for-excited-electronic-configurations

For these kind of systems, the coupling between the $d$ electrons is much stronger than the coupling between the individual $d$ electrons with the single $s$ electron. You consider the $d^8$ part with LS coupling ($^3F_{4}$) as a parent term that is coupled to the $s^1$ ($^2S_{1/2}$) term. I don't think Hund's rules can be easily applied to find the value of $J$ for the ground state because of the interaction of the parent terms.

Cross posted here: https://physics.stackexchange.com/q/691298/313612.

As a remark on @Paul's comment, I found the atomic term symbol by applying the vector model to the answer to both shells separately. The J-value was found by summing the separate ones (see http://www.quimica.ufpr.br/edulsa/cq115/leitura/Spectrocopic-Terms-1.pdf, last page). This only holds for LS-coupling.


text:
Confused about concentration & temperature vs voltage of a Daniell cell
I am doing a lab that looks at how the temperature of each half cell in a $\ce{ Zn_{(s)} / Zn^{2+}_{(aq)} || Cu^{2+}_{(aq)} / Cu_{(s)} }$ cell affects the potential difference across the Cu-Zn electrodes. I'm trying to look into the theory behind the experiment. (I know that a "proper" analysis requires deeper knowledge in kinetics and physical chem, but I can't use anything beyond the high school curriculum in my background information section, please understand)
Anyways, the Nernst equation suggests that
$$E_{\ce{Zn}} = E_{\ce{Zn}}^{Ø} - \frac{RT}{2F} \ln\left( \frac{1}{[\ce{Zn^{2+}}]} \right)$$
and
$$E_{\ce{Cu}} = E_{\ce{Cu}}^{Ø} - \frac{RT}{2F} \ln\left( \frac{1}{[\ce{Cu^{2+}}]} \right)$$
where $T$ is the temperature of the species's half cell.
Since $E_{\text{cell}}=E_{\ce{Cu}} -E_{\ce{Zn}}$, I can write
$$E_{\text{cell}} = \bigg( -\frac{R}{2F} \ln \left( \frac{1}{[\text{heated species}]} \right) \bigg) \cdot T + C$$.
The intercept $C$ of the slope with the ordinate will be the potential of the half-cell that isn't heated (assumed to be constant with constant temperature).
Now I'm confused whether this implies that $E_{\text{cell}}$ increases with higher temperature if the concentration of the heated species is greater than 1, the opposite for $[\text{heated species}] < 1$, and no change at all for $[\text{heated species}] = 1$. Just seems counter-intuitive. Any help is appreciated :D

Note that standard redox potentials E° change with temperature, as standard reaction Gibbs energy generally changes with temperature too.  The temperature in the Nernst equation describes rather potential change due activity change at the given temperature.

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.


text:
Why is deoxyadenosine monophosphate called "dAMP" and not "DAMP"?
Why is deoxyadenosine monophosphate called "dAMP" and not "DAMP"?
After all, deoxyribonucleic acid is called "DNA", and not "dNA" (nor "dRNA", thanks for that insight, @Karsten Theis), and so it looks like "deoxy" or "deoxyribo" is "D" and not "d" in the case of DNA.
It seems that "dAMP" is generated using a different rule. I wonder why? "DAMP" has the advantage of being easily recalled, because it is an English word.

It would be dRNA, not dNA.

@KarstenTheis mind blown

There is also ddATP, dideoxy, used as chain terminator in DNA sequencing.

While the current answers say what d stands for, they don't really explain why D wasn't chosen instead. We could, for example, equally say that "D shows that it has deoxyribose..."
I offer this without any evidence (and without knowing whether such evidence exists). If you are interested in evidence, you will likely have to find the original article / work in which these were named, and hope that the author included some rationale.
AMP, adenosine monophosphate, is already an important biomolecule. Prefixing it with a lowercase d in dAMP suggests that it is the same as AMP, but just modified by a little bit (which is true; it just has one oxygen less). This is perhaps partly stylistic, but it also makes it easy to recognise visually as the small letter does not stand out as much. There are also other parallel examples, such as cyclic AMP (written as cAMP).

Sometimes the lowercase prefix designates isomers, iPrOH or tBuOH, sometimes function, mRNA or cDNA,  and sometimes a different molecule, dTTP or cAMP.

@KarstenTheis Indeed, that is a very wide range of examples!

+1 This was in fact my very first question as to why lowercase d is used rather than uppercase. Though this was considered as standard abbreviation, they didn't explain why did they choose the lowercase instead of uppercase. In fact, we have DNA but not dNA.

See this result from Biology: The Dynamic Science. Peter J. Russell, Paul E. Hertz, and Beverly McMillan, p.65-66 (2007):

The lowercase d in the abbreviation indicates that the nucleoside contains a deoxyribose form of the sugar.

+1 The Windows 10 screen magnifier was very useful for looking at the link, and with that I must say that it is a fabulous summary IMHO of the structure of DNA. What a pleasure. So much better than Wikipedia IMHO. It shows that my question is a fairly good and original one, I think.

+1 Though I wrote a separate answer, I did wondered why was the lowercase chosen instead of uppercase. In fact, we have DNA but not dNA.


text:
Kharasch Peroxide effect with heating
From what i have learnt, kharasch Peroxide effect is only applicable for HBr. The reason for this is for the other halogens, HCl and HI the propagation steps in the radical mechanism are endothermic. Now, can't this problem be solved by heating? Theoretically, then, Peroxide effect could be shown by HCl and HI in presence of both heat and sunlight ( required for radical mechanism)

There are two propagation steps, only with $\ce{HBr}$ are they both exothermic.  Take a look at the numbers in this earlier answer.

If the reaction is endothermic, then it cannot be proceeded by just applying heat.
$\ce{HCl}$ is a very stable acid with $\ce{H-Cl}$ much stronger than $\ce{H-Br}$ bond and hence not broken asymmetrically by the free radical generated by peroxide. Moreover, chlorination is not that selective when compared to bromination. In this case, non radical $\ce{HCl}$ are forming Markovnikov products before the radicals come to form the anti-Markovnikov products.
With $\ce{HI}$, the first propagation step is endothermic, because $\ce{H-I}$ bond is weak and it undergo homolysis readily to form free iodine radical which in turn combines to form iodide rather than getting added to the ethylenic bond.