text:
Eutectic temperatures of multicomponent metal carbides
Some carbides do not melt at atmospheric pressure. At high enough temperature, metallic element is in vapor phase. Experimentation with these systems is complicated because of their high melting or sublimation temperatures. For this reason, not much data is available.
Is it possible to mix multiple metal carbides in proportions, which bring the melting point to significantly lower values? How low can it get, assuming the liquid phase contains at least 30 atomic percent carbon and no other non-metals?
How tight lower limit can be found for this (probably never observed) temperature by calculations?
Which metals are likely to comprise the mix in highest concentrations?
Since this is a complicated unsolved problem, partially answering it would be acceptable.

After perusing the many papers on thermodynamics of binary/ternary systems, I'm just not seeing any. One thing to note is the many different carbide phases that exist, often making it hard to actually have a mixed-metal carbide pseudobinary of a single phase in the first place (which is how I interpret your question). The few that I've found tend to have fairly small terms for the mixing of the metal components.

@JonCuster Your interpretation seems to forget the possibility that some (pseudobinary) eutectic mixtures may solidify into two phases instead of single solid phase.  I am not sure I used the term 'eutectic' correctly in my original question. I meant whatever liquid phase with metals and at least 30 atom % carbon.

Ah, OK, you are looking for any cut across a ternary (keeping $x_{C} > 30 at. %) that shows a lower-than-expected liquidus? I'll keep looking then. However, that would really require a very high enthalpy of mixing in the liquid phase to drop the liquid below the tie-line between two usually quite stable carbides (or metal + metal carbide).

@JonCuster This holds true as long as we hope ternary is enough to get an idea about the general multicomponent behavior. My intuition is, that for more different metals simultaneously present, heat capacity may be higher at temperatures comparable to the melting point of Lithium Acetylide. Am I wrong? I don't fully understand, why the enthalpy of mixing is relevant rather than entropy.

https://chemistry.stackexchange.com/questions/168754/making-a-super-eutectic/168772#168772 has some perspectives (full disclosure - my answer). From a different direction: the carbides being stable to high temperatures means they already beat out whatever entropy of mixing happens in the liquid. Now, weird things happen in associate liquid models, to be sure. But I'll keep looking around - just haven't found a clear example.

I had initially approached this from the perspective of a pseudo-binary system on a tie line between two metal carbides of the same crystal structure. One example would be across the $\ce{(Hf,Zr)C_{1-x}}$ line - that carbide show mutual solid solubility across the whole compositional range. And it shows no unexpected deviations from the melting temperature. There are several other ternary systems available in the literature that are similar. As noted in the comments, I was interpreting the question incorrectly, but at least I ruled out a number of known systems.
The other answer mentions $\ce{Li2C2}$, and the C-Li binary system looks like:

One sees the very sharp drop in melt temperature right around the 50:50 composition. The crystal structure of the $\ce{Li2C2}$ phase is a body-centered orthorhombic cell with 8 atoms, fairly unusual for most carbides. Unfortunately, there are no C-Li-Metal ternary systems listed at the ASM Alloy Phase Diagram reference site (ASM International) to see if there is any odd behavior with other carbides.
A perusal of available ternary carbide systems does not pull up much either, which is fairly frustrating. So, instead I will suggest a possible system, and explain why it could be interesting.
The first thing to note is that many carbide phases (particularly in the transition metals) have the carbon sitting on interstitial sites in a metal, and for many systems this ends up making the composition either 50:50 or with less carbon than metal. The rare earths are different, with phases like $\ce{PrC2}$ or $\ce{Pr2C3}$. There is usually a eutectic between these carbides and the rare earth metal itself, but the depth (Ce, Dy, La, Nd, Pr) is perhaps 500K at most, and is of course at lower carbon content than the carbides.
In contrast, the Be-C system looks like:

Here the eutectic is between the $\ce{Be2C}$ carbide and carbon, with a depth of some 600K or so. So now one looks at Be phase diagrams with carbide-forming metals. It turns out that zirconium is an interesting candidate. First, the Be-Zr system looks like:

Here there is a surprisingly deep eutectic between the Be-rich compounds and Zr metal, some 800K to 1000K from the melting points of either pure metal.
Then, the C-Zr diagram looks like:

Now, this has a small eutectic between the carbide and carbon of only some 500K or so, but still not bad compared with other carbide phase diagrams.
Now, looking at the carbide phases we see that the $\ce{Be2C}$ crystal structure is of the $\ce{Ca2F}$ structure, while ZrC$_{1-x}$ is the NaCl structure. It is difficult to imagine that one could go smoothly between the two to have complete solid solubility between them.
Anyway, should there not be mutual solubility in the carbide phases, and should there not be a ternary phase that is not exhibited on any of the binary diagrams, one is left with the possibility to traverse the ternary diagram from $\ce{Be2C}$ to ZrC$_{1-x}$ through liquids that, on the binary sides, have fairly steep eutectics relative to the other phases.
Sadly, the Be-C-Zr ternary system has no data that I can find. So this is only a potentially intriguing candidate for the OP.

Nice to see different approaches to this intriguing problem. As someone with very little experience with phase diagrams, I was thinking about this very differently: First, find metals that have moderately stronger affinity to carbon atom than graphite itself (Be,Mg,Al,Sc,Y). Secondly, add non-carbide-forming metals (Cu,Zn,Ga,Sn...) until solubility of carbon in the melt drops to 30 atom%. Transition metals, if any, may be present only in concentration below percolation threshold. I was hoping to see large thermal expansion and low melting point. Sadly, no data.

It may not be what you bargained for, but lithium carbide, $\ce{Li2C2}$ alone melts at only 452°C. Melting is promoted by weaker bonding than would be obtained with most metal carbides, which in turn comes fundamentally from lithium having only one valence electron to contribute to such bonds. Lithium also forms small ion pairs or clusters readily, which is also compatible with liquefication.

It is a good starting point. I was not aware of this before. I'm curious if it forms any pseudobinaries with other carbides.

@PaulKolk - Given the rather odd phase of Li2C2, I'm not sure there are other metal carbides to have a pseudobinary with.

@JonCuster What reaction prevents this typically, if we add another acetylide? Calcium acetylide for example. Or are You saying they are just immiscible?

@PaulKolk - None of the 'standard' carbides I know (say across the transition metals) have that crystal "structure" (used loosely because appartently Li2C2 is hard to crystalize in the first place). The C-Ca binary diagram at the Alloy Phase Diagram Database (https://matdata.asminternational.org/apd/index.aspx) shows only a CaC2 line compound.

Some of the issues with Li2C2 are covered in S.-M. Liang et al., Journal of Alloys and Compounds 698 (2017) 743-753 "Key experiments and challenging thermodynamic modeling of the Li-Si-C system". For example, they see no solubility between Li2C2 and SiC.


text:
Nomenclature for 5-Methylhepta-1,3,6-triene
I was practicing writing nomenclatures for several organic compounds with the help of Blue Book Essentials and I have some queries regarding the following structure:

According to the blue book, the numbering of parent chain is as follows:

Lowest locants for heteroatoms
Lowest locant(s) for indicated hydrogen
Lowest locant(s) for principal characteristic group(s)
Lowest locants for ‘ene’, ‘yne’, and hydro prefixes
Lowest locants as a set for all substituents cited by prefixes
Lowest locants for substituents in the order of citation.

According to this order, the carbon bearing the methyl group should get the lesser number, hence '3'. So the name should be 3-methylhepta-1,4,6-triene?
Please clarify this for me. Thanks!

Since CH3 isn't a principal characteristic group, I would start with the 4th rule, but to infer the side i should start from, I would first number both ways, and add up position numbers of carbon bearing the functional group/bond, and the side which gives me the highest will be the order of numbering right?

Numbering is chosen so as to give the unsaturated sites (the highest priority C atoms) the lowest numbering. It can be a 1,3,6 triene, or a 1,4,6 triene. The first one is the preferred numbering, making the methyl substituent on the 5 atom.

Thanks for clearing it up! I have another query, should I consider every other group not mentioned principal characteristic group(as per Blue Book), in terms of alphabetical order? For example, does isopropyl group get higher priority than the methyl group?

I think they're equivalent (they give the same molecules) so I forget what the IUPAC convention is there.

See this answer? https://chemistry.stackexchange.com/questions/90644/naming-tetrahalogenated-benzene?rq=1

source: Stack Exchange

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

sort: votes


text:
The active space of transition state of ch3oo+nh3
I am trying to calculate the potential energy surface of the reaction:ch3oo+nh3=ch3ooh+nh2, because of the high T1 diagnostic value, the multireference method needs to be used in Molpro. I gave the active space apace step by step, and everything seemed good until active space was (7e,6o), one active orbital refers to an o-o π orbital shows in reactant active orbital can not be included in transition states active orbital. Whatever how I rotate the orbital, it does not work, I have been struggling with this for months, so I attached an input file and ask for your advice and recommendation.
Thank you all very much!
(5e,5o):
memory,2000,m
geomtyp=xyz        ! use cartesian coordinates xmol style
geometry={
10              ! Number of atoms
CNH6O2
  C         1.572679000000      0.578668000000      0.205525000000
  H         2.316893000000      0.970482000000     -0.487615000000
  H         0.957089000000      1.394772000000      0.599585000000
  H         2.059437000000      0.052969000000      1.029229000000
  O         0.782634000000     -0.313673000000     -0.552640000000
  O        -0.136935000000     -0.901245000000      0.303994000000
  H        -1.121895000000     -0.504387000000     -0.043836000000
  N        -2.093944000000      0.366523000000     -0.112760000000
  H        -1.677239000000      1.262678000000     -0.376582000000
  H        -2.478346000000      0.505153000000      0.824559000000
}

set,spin = 1

basis=aug-cc-pvtz

   {multi;
   closed,15;
   occ,20;
   wf,35,1,1;
   rotate,15.1,16.1,0;
  ! open   define open shell orbitals
   orbprint,10;
   }
   put, molden,ch3o2h_nh2_ts1_5e5o.molden
   RS2C,SHIFT=0.25

ENCBS= energy

(7e,6o):
memory,2000,m
geomtyp=xyz        ! use cartesian coordinates xmol style
geometry={
10              ! Number of atoms
CNH6O2
  C         1.572679000000      0.578668000000      0.205525000000
  H         2.316893000000      0.970482000000     -0.487615000000
  H         0.957089000000      1.394772000000      0.599585000000
  H         2.059437000000      0.052969000000      1.029229000000
  O         0.782634000000     -0.313673000000     -0.552640000000
  O        -0.136935000000     -0.901245000000      0.303994000000
  H        -1.121895000000     -0.504387000000     -0.043836000000
  N        -2.093944000000      0.366523000000     -0.112760000000
  H        -1.677239000000      1.262678000000     -0.376582000000
  H        -2.478346000000      0.505153000000      0.824559000000
}

set,spin = 1

!basis=aug-cc-pvtz
basis=cc-pvdz

   {multi;
   closed,14;
   occ,20;
   wf,35,1,1;
   rotate,14.1,15.1;
  ! open   define open shell orbitals
   orbprint,10;
   }
   put, molden,ch3o2h_nh2_ts1_7e6o.molden
   !RS2C,SHIFT=0.2

ENCBS= energy
```

Hi, I made some changes to formatting to make the input file block look nicer. Also, i would recommend migrating this to matter modelling stack exchange - that forum is more targeted towards this type of questions.

"ch3oo+nh3=ch3ooh+nh2" seems like an ignorant and careless attempt to write a chemical equation.

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


text:
Product of 4-chlorobutan-1-ol treated with KOH in alcohol
I am looking at this problem:

I thought the KOH in alcohol would support a nucleophilic attack of hydroxide on the haloalkane, so the product I would get is B (elimination of HCl). However, this seems to be incorrect.

KOH in alcohol is a reasonably strong base. By far the most acidic proton in the 4-chlorobutan-1-ol starting material is the -OH so the first species generated is the alkoxide of the chlorobutanol

@Waylander U want to say O neg will attack at position of CL

Yes, the formation of the [5] ring is energetically favoured, Cl- is a good leaving group. This will be a fast and irreversible reaction

@Waylander thank sirr

@Karsten thanks for editing

@DUDE_WITH_J would prefer if you write questions, as much as possible, in clear language and with minimal grammatical errors. We don't demand perfection; but please maintain so that questions remain salvageable. There are various examples of good questions being asked in a bad way.

Few helpful links to help with formatting: Chem / MathJaX formatting, Upright Vs Italics. Welcome to ChemSE!

@HarikrishnanM vankam sir , I have asked a new question in clear language as you told just look at it and if you find any mistake pls tell


text:
Why can't the fructose monomer in sucrose be α-fructose?
In all academic sources, sucrose is identified as α−glucose (1-->2) β−fructose. However, I cannot find any explanation anywhere as to why the fructose monomer has to be in the β configuration. Maltose has both α and β anomers, same for lactose. Even trehalose, another non-reducing disaccharide with glycosidic linkage between two anomeric carbons, has α-α, α-β, and even β-β anomers. Why is sucrose special? And is there a disaccharide out there that has α−glucose (1-->2) α−fructose configuration?

Slides 24-28 here may be of help.


text:
Why does the 2p orbital of chlorine form a sigma bond with Be in BeCl2 and not 3p?
So I was reading my textbook when I came across, "Each of the sp hybridized orbitals overlap with the 2p orbitals of chlorine axially and form 2 Be-Cl sigma bonds."
Now this confused me because the electronic configuration of Chlorine is [Ne]3s23p5.
So, technically, shouldn't the 3p orbital of both chlorines attach to the sp orbital of beryllium? Why does the 2p orbital overlap? Also, how does it achieve this? The 2p orbitals of chlorine are all full (i.e. they all have 2 electrons). Does one 2pz electron get excited into the 3pz orbital?
I am very confused here and have tried reading online but haven't found any satisfactory answer. Any insight would be appreciated. Thanks.

Smells like a typo.

I thought so too at first, but in the very next section it says, "The three hybrid orbitals formed are oriented in a trigonal planar arrangement and overlap with 2p orbitals of chlorine to form three B-Cl bonds." This is about sp2 hybridized BCl3. A typo could've occurred once, but twice?

Would the downvoter care to explain?

Generally if you quote a textbook, it's good to provide the title, author, publisher, etc.


text:
The mechanism for the addition of HCl to propylene
Problem:
The mechanism for the addition of HCl to propylene is described below.

Complete the mechanism in step 1 with one curved arrow showing the electron movement that occurs when the double bond is broken, as well as the mechanism in step 2 with the intermediate carbocation and one curved arrow showing how $Cl^-$ is added.
Solution:
No solution was given to this problem.
What I am thinking:
Excuse how messily I've drawn everything. I just drew this fast on a notepad I specifically have to just aid me in trying to find the right solution before I write and draw everything more nicely on another notepad.

Questions:
Did I draw the curved arrows where they are supposed to be? Did I write the right carbocation? Am I thinking right?

Basically correct


text:
Is this fused ring molecule aromatic?

In my notes, I can see that it is listed as anti-aromatic because of the 16 electron ring current that would be formed on the periphery of the entire molecule were the right benzene ring drawn in the other canonical form.
But I am not sure if that would be the case, since the two benzene rings could independently be aromatic as is the case in some other compounds.


text:
Is HCl in salt form reactive with metallized aluminium plastic film?
I’m in the process of packaging an amino acid that contains HCl in salt form. The mixture is 60% base and 40% HCl salt.
Will this react to metallized aluminium plastic?

The carboxyl group is more acidic than the ammonium one.

There likely is a plastic (e.g., polyethylene) layer underneath the aluminum, to seal to the plastic bottle.
Polythene is resistant to $\ce{HCl}$, though somewhat porous. You can specify a thicker layer, as from this supplier: "20 mic./ 25mic. / 30 mic. / 40 mic.,20 mic./ 25mic. / 30 mic. / 40 mic.". N.B. Check with supplier whether thickness applies to aluminum, to polythene, or both.


text:
Reaction of Mg/Hg amalgam with acetophenone

Will Mg perform here as Grignard or will something else happen?

Note the presence of the Hg, this will not be a simple Grignard. There is a clue here http://www.orgsyn.org/demo.aspx?prep=CV1P0459

@Poutnik new to this sir. I have edited, is this fine

@DUDE_WITH_J   have a read of this   https://en.wikipedia.org/wiki/Pinacol_rearrangement

@Waylander sure sir , will read it thanks for your effort !

It helps to write in full clear sentences. I reopened but please note it is expected that you show what it is makes the solution unclear.

...and the pinacol coupling reaction and migratory aptitude.

@BuckThorn ok sirr I will remember it and Can I ask theory related Q

If it's a very closely related question it might be better to continue in chat. If your questions are repeatedly poorly received the system robot might impose a temporary ban on your ability to start more posts.


text:
Simple Calculation of Potentials for a hydrogen-oxygen Fuel cell, unexpected results from book example
I am reading "Alternative energy systems" by Hodge.
On the fuel cell section, the author has an example for calculating the potential required for the fuel cell as a function of temperature and the pressure of the products and reactants (examples 10.1 and 10.2).
The author calculates equilibrium voltage with n=2, author establishes that:
$$\Delta G^\circ _\ce{H2O}=\pu{-229kJ/mol}$$
However, when he evaluates the equilibrium potential he takes $\Delta G^\circ _\ce{H2O}=\pu{229kJ/mol}$
Later on, the author calculates the deviation of equilibrium potential when oxygen is provided at 0.21 atm pressure using Nernst equation.
$$U = U^\circ -\frac{RT}{nF}\ln\left(\frac{P_\ce{H2O}}{\sqrt{P_\ce{O2}}P_\ce{H2}}\right)$$
Keeping the sign convention where $\Delta G < 0 \rightarrow \mathrm{energy~production}$, I get negative potentials, at standard conditions I get the same potential as the author (just different sign). Keeping the negative sign in my case shows that for air at 0.21 atm the potential becomes more negative, giving me a higher absolute potential. The author on the other hand, since they removed the negative sign, when they apply Nernst eqn, showing a lower absolute potential to drive the equation.
Thank you for your help.

Why is it being downvoted? Please let me know if anything is not clear or If I should provide more details into the author's work.

Keeping the sign convention where ΔG<0→energy production, I get negative potentials. - Why, when $\Delta_\text{r} G = -nF \cdot U$ ?

You're completely right, the author writes it without the negative sign, but I have been reading a couple more electrochem books and they all stated the same formula as you. I'm guessing it is a mistake on the authors end.

Δ must be negative. If it were positive, the reaction in the fuel cell will not be spontaneous.  Δ is positive in an electrolysis, not in a fuel cell-

Agreed, I just submitted an errata form to Wiley. Thank you guys for your help.


text:
Why are bases corrosive?
If I understand correctly, an acid corrodes things because the hydronium ion attracts electrons away from whatever it is reacting with.  Do bases corrode things in a similar manner? with the metal cation operating in the same way as the hydronium ion (taking electrons)?

Note that NaCl is corrosive too.

"Corrosion is a natural process that converts a refined metal into a more chemically stable oxide. [....] In the most common use of the word, this means electrochemical oxidation of metal in reaction with an oxidant such as oxygen, hydrogen or hydroxide." from https://en.wikipedia.org/wiki/Corrosion Not exactly a perfect definition but useful. Starting from definitions is often helpful.

Check difference between acidic corrosion and alkaline corrosion: https://tft-pneumatic.com/blog/corrosion-under-acidic-alkaline-conditions/

It depends on what is being corroded.

Alkalis are caustic, because they saponify fats in skin, changing a fatty acid to soap.

Some bases are strong electrolytes, speeding galvanic corrosion.

Some metals react with the hydroxyl ion, such as zinc, forming the zincate ion, $\ce{[Zn(OH)4]^2-}$. The reaction of bases with aluminum is used to make hydrogen on demand, and to clean drains, as in this product.

Silica, $\ce{SiO2}$, is acidic, and reacts with bases to form silicates. For that reason, $\ce{NaOH}$ should not be stored in glass containers, which are attacked by dissolved $\ce{NaOH}$ (from moisture in air).

Hydroxide ion not hydroxyl radical. the oxidizer is water. The initial oxidation reaction is probably a one electron transfer to make H , Zn+ and OH- eventually OH- complexes with the Zinc ions to lower their potential and force the reaction. A radical mechanism while feasible on paper seems more complicated since OH-and H2O abound and OH radicals scarce.

@jimchmst, I stand corrected for my radical misstatement, admittedly at the root of the issue. They wouldn't be unionized (too Marxist for 'em).

@DrMoishePippik To clean drains full of aluminum? Or with hydrogen? So confusing.

@PaulKolk, NaOH+ Al is mixed in a common drain cleaner, creating effervescent hydrogen and even more heat than created by NaOH dissolving. Reference will be added above.


text:
Aqueous solutions of salts pH question
I was reading up on this "Aqueous Solutions of Salts" article on chemlibre and was doing problem 2 at the bottom:
Question: Find the pH of a solution of $\pu{0.200 M}$ $\ce{NH4NO3}$ where ($\mathrm{K_a} = \pu{1.8E-5}$).
This is the given solution:
$$\ce{NH4+ _{(aq)} + H2O_{(l)} <=> NH3 _{(aq)} + H3O+ _{(aq)}}$$
$$\frac{x^2}{0.2-x}=\frac{\pu{1E-14}}{\pu{1.8E-5}}$$
$$x=\pu{1.05E-5 M}=[\ce{H3O+}]$$
$$\mathrm{pH}=4.98$$
Why does this solution set $\frac{x^2}{0.2-x} = \mathrm{K_b}$ instead of the $
\mathrm{K_a}$? If I understand correctly, the equation in their solution is for the dissociation of an acid?
This is my work and where I got stuck:

I'm sure I'm just missing something here but any help would be appreciated!

1) We use the relation $\mathrm{K_aK_b=K_w}$, 2) obtain $\mathrm{K_b}$, and 3) use the equilibrium condition to find hydronium ion concentration. This is the method followed by the problem...

I understand the Ka and Kb relationship through Kw but am still confused as to why we use the Kb instead of the Ka in this problem, I thought since this is an acid dissociation we would use Ka.

is there a mistake in the question? $\mathrm{K_b}$ was supposed to be given, if you ask me...

Yes, there is a mistake, or a typo. The value 1.8e-5 belongs to Kb of NH3, not to Ka of NH4+. Therefore, the used Kw/Kb value correctly provides Ka of NH4+, that is needed for the approximation Ka =  [H+]^2/(c - [H+]).

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of essential parts.
// You may find useful formatting guides for texts and formulas/equations/expressions.

@Poutnik talking about canned notices it would be highly helpful if you could give the texts in Meta so that junior users like me can help the newbies out there.

@HarikrishnanM It would make them too official. Different users prefer different wording or put stress on different things. I keep mine handy on notes system at keep.google.com. and modify them often. Also having many variants, sometime combining or alternating them, some becoming obsolete, so they are dynamic. // If mods want them, I will provide them.


text:
In steady state diffusion of atoms, does dJ/dt = 0 hold true?
My textbook says that under steady state flow the flux is independent of time and remains the same at any cross-sectional plane along the diffusion direction. Further, it gives the following conditions:
$$\left(\frac{\partial c}{\partial t}\right)_x = -\left(\frac{\partial J}{\partial x}\right)_t = 0$$
According to the book shouldn't $\dfrac{\text{d}J}{\text{d}t} = 0$ hold true also, as it says that the flux is independent of time? I've looked all over the internet, but I couldn't find $\dfrac{\text{d}J}{\text{d}t} = 0$ as a condition for steady state  flow.

Chem+Math MathJax formatting reference:
Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial

The steady state corresponds to the behaviour of the solutions of the equations of conservation of a physical-chemical system when $t \to \infty$. In practice, depending on the 'inertia' of the particular system, this state will be acquired very fast (seconds) or very slowly.
For chemical species $\ce{A}$ which does not react so its rate of reaction is zero (i.e. $r_\ce{A} = 0$), the statement of conservation of mass is
\begin{equation}
  \frac{\partial C_\ce{A}}{\partial t} +
  \nabla \cdot \mathbf{J}_\ce{A} = r_\ce{A}
  \rightarrow
  \frac{\partial C_\ce{A}}{\partial t} +
  \nabla \cdot \mathbf{J}_\ce{A} = 0 \tag{1}
\end{equation}
Eq. (1) is a PDE that states the concentration of species $\ce{A}$ as  function of space and time, i.e. $C_\ce{A} = C_\ce{A}(\mathbf{r},t)$. Consequently, the flux of $\ce{A}$ will also be a function of space and time, i.e. $\mathbf{J}_\ce{A} = \mathbf{J}_\ce{A}(\mathbf{r},t)$. However, in steady state the partial derivative with respect to time is zero, and thus Eq. (1) becomes
\begin{equation}
  \nabla \cdot \mathbf{J}_\ce{A} = 0 \tag{2}
\end{equation}
Eq. (2) is also a PDE but in here the flux is a function only of space, i.e. $\mathbf{J}_\ce{A} = \mathbf{J}_\ce{A}(\mathbf{r})$. Therefore, steady-state implies not only that the flux is independent of time, but that its divergence is zero.
The condition you are trying to find is not explained in books, but perhaps because it is fairly obvious. The independence of variables with time comes when crossing out the terms that depend on time of a specific PDE.
To make an analogy to Chemistry, if I give you Schrödinger's equation in paper without the term $i\hbar \partial \Psi / \partial t$, you won't be worried about time because you don't see the variable $t$ in another part of the PDE. Thus, it just comes naturally that $\Psi = \Psi(\mathbf{r})$ (or called $\psi(\mathbf{r})$ more typically). Similarly, if I give you the statement of the conservation of mass and you don't see the variable $t$, then it is obvious that the flux won't depend on time.


text:
Is it really that simple to make dimethylmercury?
I was reading about methyl iodide and I saw a warning to not let it come into contact with sodium amalgam or it will create dimethylmercury. I'm wondering if that's really all that it takes to create such a dangerous substance. Does simply dropping sodium mercury amalgam into a container and then pouring in some methyl iodide create dimethylmercury? I'm not really sure if these chemicals would ever need to be around each other, so I don't think an accident is super likely, but it's still kind of crazy to me.

Yes it really is that easy

$\ce{Hg + 2 Na + 2 CH3I -> 2 NaI + (CH3)2Hg}$ seems quite straightforward to me. Similar reaction was used to make $\ce{(CH3CH2)4Pb}$ for gasoline, before the lead-free one.

Dimethylmercury is very toxic. Read about Karen Wetterhahn

You seem to have a wrong perspective on dangerous substances. Yes they are very easy to come by, they are all around us, they are more common than the opposite. The Nature is not friendly to us.

Yes, it is. However, it is not crazy since these precursors are not something easily obtainable directly from nature.
https://en.wikipedia.org/wiki/Sodium#Commercial_production
This illustrates the fact that sodium itself is not that simple to find, the final step in the creation of dimethylmercury is very easy, but fortunately there are some obstacles before that, which prevent accidents.

Yes, dimethylmercury is that easy. Roughly speaking, sodium amalgam is mercury endowed with excess electrons from the electropositive sodium, and methyl iodide (an electrophile) can react with the excess electrons and the mercury atoms to form dimethylmercury. The iodide ions displaced by this reduction end up precipitated from the mercury as sodium iodide.
Dimethylmercury is especially dangerous as mercury sources go because it is much more easily absorbed through the skin or inhaled than elemental mercury or inorganic mercury compounds.
The good news is there is not a lot of sodium amalgam in nature. The bad news is that there are plenty of natural processes for forming dimethylmercury; see for instance Jonsson et al[1].
Reference

Jonsson, S., Mazrui, N. & Mason, R. Dimethylmercury Formation Mediated by Inorganic and Organic Reduced Sulfur Surfaces. Sci Rep 6, 27958 (2016). https://doi.org/10.1038/srep27958

How quick is that reaction? If someone spills one on the other can it be cleaned up fast? Or should one assume its been created and treat it as worst case scenario?


text:
Stability of Nitro substituted phenylic anion
Which of the following is more stable:

What I  found on internet is 2nd structure ( meta position ) is more stable. But -NO2 shows -I (inductive effect) and -M ( metameric effect ) so ortho must be more stable. What's just different phenonmenon here stablising meta over ortho?

And here I wonder if I wanna know the site you linked, or is this another blissful ignorance scenario ;> I would say ortho, but...

@Mithoron Will here -no2 exhibit -m effect? Because my teacher says that whenever there is charge on phenyl ring negative or positive, any other substitued group will not show mesomeric effect.

One can certainly draw a structure in which o-isomer is withdrawing mesmerically. While the lone pair is not conjugated, the resulting carbenic structure still lacks charge on this C atom.

First off, the mesomeric (M) effect applies only when pi-electron conjugation is operating, and here it is not. When you deprotonate benzene, the excess electron pair left behind is in the plane of the ring where the hydrogen was; thus not conjugated with the pi system.
It is true that the inductive effect is strongest with the negative charge at the ortho position, but you have to reckon with steric effects. A nonbonding electron pair is more bulky than a bonding pair, especially when the bond is with hydrogen; so deprotonating the nitrobenzene at the ortho position invites interference and repulsion between the electron pair left on the carbon and those on the adjacent nitro group. You need to go to the meta position to keep the two sets of nonbonding electrons far enough apart for the stability you expect on charge-distribution grounds. This does not mean the more favorable steric effect with the meta isomer overcomes the stronger inductive effect with the ortho isomer, but you have to assume that might happen.

I know that loan pair of nitrogen doesn't participate in resonance but that wouldn't change the fact that -no2 is electron donating group at para and ortho position. So at the ortho and para site wouldn't a Little partial positive charge will develop due to mesomeric effect of -no2.

Sorry. -NO2 does not have a loan pair or even a lone pair at the nitrogen atom and it is pi-electron withdrawing, not donating.

Please check : https://chemistry.stackexchange.com/q/41443/129242

Still not pi donating, the ring is positively charged where a pi interaction with it occurs. The lone pairs are on oxygen, which actually takes up another lone pair when we form the pi bond to the ring.

But when I said -no2 is pi donating. It shows negative mesomeric effect ( e- withdrawing ) and creates positive charge at ring.


text:
What does the dot indicate in the molecular formula for white lead?
I came across the following molecular formula for white lead:  $2\mathrm{Pb}\mathrm{CO3}·\mathrm{Pb(OH)_2}$. What does the center dot indicate here? Is this a non-covalent interaction of some kind? https://en.wikipedia.org/wiki/White_lead

it helps - can i conclude then that white lead is a double salt based on the link you showed plus the chemical structure shown on wikipedia?

It could be as well written as $\ce{Pb3(OH)2(CO3)2}$.
The formula $\ce{2 PbCO3 . Pb(OH)2}$ is used as better visualization of composition being an intermediate between lead carbonate and lead hydroxide in ratio 2:1.
The dot has similar meaning as in formulas of salt hydrates like $\ce{CuSO4 . 5 H2O}$. Two substances incorporated into the single one in particular ratio, at level of molecular entities (general term involving molecules, atoms, ions, ionic pairs and any other kinds of the smallest building blocks), i.e. not just a mixture.
Another example are minerals malachite and azurite, containing cupric carbonate and cupric hydroxide in ratio 1:1 or 2:1, respectively.
These substances are members of a larger group called basic salts, where anions of acids are complemented with hydroxide or oxide entities, like basic chlorides of magnesium or zinc, or substances in chlorated lime.


text:
Why does boron have higher affinity for fluorine than oxygen?
$\ce{F-}$ can displace $\ce{OH}$ in $\ce{B(OH)3}$ and form $\ce{BF3}$. $\ce{BF3}$ is stable due to resonance, but donation of oxygen should be stronger, as -I power of fluorine dominates its lone pair giving power (+M).
Why is $\ce{BF3}$ more stable than B(OH)3?

Can you define exactly what your measure of stability is in this case?

The question might he better posed as "Why does fluorine have a greater affinity for just about everything (including boron) than oxygen?
The answer is first off, technically fluorine does not have a greater affinity for everything than oxygen. This answer explores counterexamples involving (early) transition metal complexes. However, in most settings the exceptionally low valence orbitals of fluorine and the availability of a vacancy in the fluorine atom to accept another electron facilitates the formation of bonding molecular orbitals with low energy. Meaning stronger, more stable bonds.

There are competing effects of resonance and steric hindrance in $\ce{B(OH)3}$ and $\ce{BF3}$. It is the reduced steric hindrance, not resonance, that dominates stability in this case. Boron has small enough ionic radius for this to happen.

Color me sceptical  - F and O have very similar covalent radii, I think. Is the question's premise even true, when BF3 hydrolyses both to $\ce{B(OH)3}$ and fluoroboric acid?

Boron forms covalent bonds, so why will we look at the ionic radii? Also steric hinderence will be close to negligible as bond angles are big(120 deg) and as said by Mithoron, O and F have nearly same radii

Both O and F are more electronegative than B. That is why ionic radius is relevant. Covalent radii of F and O vary a lot in reality, values given in tables are some kind of average of different cases. Hydrolysis involves additional hydrogen. 120 deg is not big enough for ions as small as Boron's.

How will we use ionic radii though? Boron doesn't form ionic compounds and if we take ionic radii of O and F, we are doing the opposite as they are radii of F- and O-. But due to backbonding, the partial charge on O and F will be positive(which reduces their size). Also Boron is bigger than O and F. So I don't think steric hinderance will be significant. My main doubt was that, many times in organic chemistry, we take halogens as electron withdrawing group, but O is electron giving group(with lone pair). So oxygen should make boron more stable than fluorine does.

@EagerToLearn Why not to model B3+ as an ion? It is in fact much smaller than OH- and F- . Backbonding may change their shape a bit, but doesn't make them smaller than B3+.

B3+ does not exist. Infact Boron all compounds of boron are covalent.

@EagerToLearn There are clear arguments against boron being only covalent for example in alloy systems. Coordination numbers can be totally unexplainable by covalent bonding sometimes. Even If we narrow our focus to molecular compounds, they can be modeled both as covalent or as composed of ions.

Yeah, Boron does form compounds like Mg3B2, where it is in interstitial spaces. However, in simple compounds like BF3, or boron's own crystalline structure B12, the bonds are all covalent. Its because of very very high ionisation energy. Although I would like some examples of ionic boron compounds, it is an interesting topic.


text:
Is Breslow's intermediate aromatic?
When looking at Stetter's reaction (umgepolter aldehyde + michael akzeptor -> 1,4-dicarbonyle), I was wondering wether the occuring intermediate, after the umpolung step, is aromatic. I can imagine that the resonance structure on the right exhibits aromaticity, but does the left one do so as well due to the double bond? I assume into a set of resonance structures aromaticity cannot just "dissapear", but I can not explain how the the pi-electrons behave in that structure

What is "umgepolter", "michael akzeptor", "umpolung"?

there are alot of typo

Aromaticity can indeed "disappear" between one contributing resonance structure and another when the conjugated pi-electron structure extends beyond the ring. That's because you don't have just an aromatic ring.
What happens is that with a favorable total count of pi electrons (which need not follow the 4n+2 rule when you don't have just a ring) and in some cases bond polarities, the delocalization and extra stabilization extend beyond the ring. In effect ring aromaticity is superseded by this extended delocalization. The compound in the question offers one example.
Such "extended aromatic" systems are commonplace in organic chemistry. The most often cited laboratory examples are cyclopropenone and tropone (in which a pendant oxygen is attached to a cyclopropenyl or cycloheptatrienyl ring and the polarity of the carbon-oxygen bond favors forming an aromatic-cation contributing structure); but many natural organic compounds exhibit this characteristic, including the caffeine in your coffee or tea.


text:
Deriving Nernst Equation for a hydrogen-oxygen Fuel cell
Reading across electrochemistry books (particularly the fuel cells chapter of the book "Alternative energy systems and applications" by Hodge) I came across the following formulae:
$$\Delta G=\sum_\text{Products}\Delta G-\sum_\text{Reactants}\Delta G=\Delta G^\circ+RT\cdot\ln\left(\frac{[C]^c[D]^d}{[A]^a[B]^b}\right)$$
Dividing by $-W=-nF$, we get
$$U=U^\circ-\frac{RT}{nF}\cdot\ln\left(\frac{[C]^c[D]^d}{[A]^a[B]^b}\right)=U_\text{cathode}-U_\text{anode}$$
I tried obtaining the form of Nernst equation for this fuel cell as follows:

$\ce{H2 -> 2H^+ + 2e-}$, then the potential is
$$U_{\ce{H2}/\ce{H+}}=U^\circ_{\ce{H2}/\ce{H+}}-\frac{RT}{2F}\cdot\ln\left(\frac{[\ce{H+}]^2}{[\ce{H2}]}\right)$$
$\ce{\frac12 O + 2H+ + 2e- -> H2O}$
$$U_{\ce{O2}/\ce{H2O}}=U^\circ_{\ce{O2}/\ce{H2O}}-\frac{RT}{2F}\cdot\ln\left(\frac{[\ce{H2O}]}{[\ce{H+}]^2[\ce{O2}]^{\frac12}}\right)$$

Therefore, for the overall cell potential I got is the following:
$$U_\text{cell}=U_{\ce{O2}/\ce{H2O}}-U_{\ce{H2}/\ce{H+}}$$ $$= U_{\ce{O2}/\ce{H2O}}^\circ-U_{\ce{H2}/\ce{H+}}^\circ=U^\circ_{\ce{O2}/\ce{H2O}}-\frac{RT}{2F}\cdot \left[\ln\left(\frac{[\ce{H2O}]}{[\ce{H+}]^2[\ce{O2}]^{\frac12}}\right)-\ln\left(\frac{[\ce{H+}]^2}{[\ce{H2}]}\right)\right]$$
$$ = U^\circ_{\ce{O2}/\ce{H2O}}-\frac{RT}{2F}\cdot\ln\left(\frac{[\ce{H2O}][\ce{H2}]}{[\ce{H+}]^4[\ce{O2}]^{\frac12}}\right)$$
However, the simplest form of the  Nernst Equation I usually come across for this cell is:
$$ U_\text{cell}=U^\circ-\frac{RT}{2F}\cdot\ln\left(\frac{p_{\ce{H2O}}}{p_{\ce{H2}}p_{\ce{O2}}^{\frac12}}\right)$$
I noticed I can reduce the equation I obtain with the following assumptions:

Products and reactants are in a gaseous state and behave as ideal gases.
The activities of a gas species can can approximated with relative pressures (or relative concentration): $a_s\approx[S]^s\approx\left(\frac{p_s}{1\ \mathrm{atm}}\right)^s$
Assuming hydrogen ions are in gaseous state, and $[\ce{H+}]^2\approx\frac{p_{\ce{H2}}}{1\ \mathrm{atm}}\approx[\ce{H2}]$
Substituting this assumptions into my equation yields:

$$ U_\text{cell}=U^\circ_{\ce{O2}/\ce{H2O}}-\frac{RT}{2F}\cdot\ln\left(\frac{[\ce{H2O}][\ce{H2}]}{[\ce{H+}]^4[\ce{O2}]^{\frac12}}\right)=U^\circ_{\ce{O2}/\ce{H2O}}-\frac{RT}{2F}\cdot\ln\left(\frac{[\ce{H2O}][\ce{H2}]}{[\ce{H2}]^2[\ce{O2}]^{\frac12}}\right)$$
$$=U^\circ_{\ce{O2}/\ce{H2O}}-\frac{RT}{2F}\cdot\ln\left(\frac{[\ce{H2O}]}{[\ce{H2}][\ce{O2}]^{\frac12}}\right)$$
$$=U^\circ-\frac{RT}{2F}\cdot\ln\left(\frac{p_{\ce{H2O}}}{p_{\ce{H2}} p_{\ce{O2}}^{\frac12}}\right)$$
While this assumptions "worked out" I am unsure If they are correct in any way or if my entire derivation process is flawed.

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

Rather than pointing whether the assumptions are correct or incorrect, I will show you a general method to obtain the Nernst's equation for any system. It is a more systematic and formal approach, generally found in electrochemistry books, but has the advantage that all the magnitudes pop up so you can clearly identify what each letter means.

The condition of electrochemical equilibrium for a single electrochemical reaction of $N$ species is
$$ \sum_{j = 1}^N \nu_j \tilde{\mu}_j = 0 \tag1 $$
where $\nu_j$ is the stoichiometric coefficient, and $\tilde{\mu}_j$ the electrochemical potential defined as
$$ \tilde{\mu}_j \equiv \mu_j^\circ + RT\ln a_j + z_j F \phi \tag2 $$
where $\mu_j^\circ$ is the standard chemical potential, $a_j$ is the activity of species $j$, and $\phi$ the electric potential.
The important rules are:

For electrons, we drop out its standard chemical potential and the activity, so that $\tilde{\mu}_\ce{e^-} = -F \phi^\mathrm{M}$. The letter $\mathrm{M}$ is because the electron as a species belongs to the electrode.
For hydrogen ions, we assign its standard chemical potential as zero for an aqueous solution at all temperatures, i.e. $\mu_\ce{H+}^0 \equiv 0$. However, in a fuel cell the liquid phase is not water, so we cannot eliminate it along the derivations.
For elements in their most stable state of aggregation, their standard chemical potential is zero. For our case, $\mu_\ce{H2}^0 = \mu_\ce{O2}^0 \equiv 0$.

The electrochemical reaction in the negative electrode, written as a reduction, is
$$ \ce{2H^+(aq) + 2e^-(M,n) <=> H2(g)} \tag{3} $$
and application of Eq. (1) leads to
\begin{align}
  2\mu_\ce{H+}^\circ + 2RT\ln(a_\ce{H+}) + 2F\phi^\mathrm{S,n}
  - 2F\phi^\mathrm{M,n} &= RT \ln (a_\ce{H2}) \\
  2F\phi^\mathrm{S,n} - 2F\phi^\mathrm{M,n} &=
  -2\mu_\ce{H+}^\circ - RT\ln(a_\ce{H+}^2) + RT\ln(a_\ce{H2}) \\
  -2F(\phi^\mathrm{M,n} - \phi^\mathrm{S,n}) &= -2\mu_\ce{H+}^\circ
  + RT\ln\left(\frac{a_\ce{H2}}{a_\ce{H+}^2}\right) \\
  \phi^\mathrm{M,n} - \phi^\mathrm{S,n} &= \frac{\mu_\ce{H+}^\circ}{F}
  - \frac{RT}{2F}\ln\left(\frac{a_\ce{H2}}{a_\ce{H+}^2}\right) \\
  E_\text{n} &= {E_\text{n}}^\circ -
  \frac{RT}{2F}\ln\left(\frac{a_\ce{H2}}{a_\ce{H+}^2}\right) \tag{4} \\
\end{align}
where we defined the electrode potential $E_\text{n} \equiv \phi^\mathrm{M,n} - \phi^\mathrm{S,n}$ and the standard reduction potential ${E_\text{n}}^\circ \equiv \mu_\ce{H+}^\circ /F$ for the reaction in the negative electrode.
The electrochemical reaction in the positive electrode is
$$ \ce{1/2 O2(g) + 2H^+(aq) + 2e^-(M,p) <=> H2O(g)} \tag5 $$
and application of Eq. (1) leads to
\begin{align}
  &\frac{1}{2}RT \ln (a_\ce{O2}) + 2\mu_\ce{H+}^\circ + 2RT\ln(a_\ce{H+}) +
  2F\phi^\mathrm{S,p} - 2F\phi^\mathrm{M,p}
  = \mu_\ce{H2O}^\circ + RT\ln(a_\ce{H2O}) \\
  & 2F\phi^\mathrm{S,p} - 2F\phi^\mathrm{M,p}
  = \mu_\ce{H2O}^\circ - 2\mu_\ce{H+}^\circ +
  RT\ln(a_\ce{H2O}) - RT \ln (a_\ce{O2}^{1/2})- RT\ln(a_\ce{H+}^2)  \\
  &-2F(\phi^\mathrm{M,p} - \phi^\mathrm{S,p})
  = \mu_\ce{H2O}^\circ - 2\mu_\ce{H+}^\circ +
  RT\ln\left(\frac{a_\ce{H2O}}{a_\ce{O2}^{1/2}a_\ce{H+}^2}\right) \\
  &\phi^\mathrm{M,p} - \phi^\mathrm{S,p}
  = -\frac{\mu_\ce{H2O}^\circ - 2\mu_\ce{H+}^\circ}{2F} - \frac{RT}{2F}
  \ln\left(\frac{a_\ce{H2O}}{a_\ce{O2}^{1/2}a_\ce{H+}^2}\right) \\
  &E_\mathrm{p} = {E_\mathrm{p}}^\circ - \frac{RT}{2F}
  RT\ln\left(\frac{a_\ce{H2O}}{a_\ce{O2}^{1/2}a_\ce{H+}^2}\right) \tag6 \\
\end{align}
where we defined the electrode potential $E_\text{p} \equiv \phi^\mathrm{M,p} - \phi^\mathrm{S,p}$ and the standard reaction potential ${E_\text{p}}^\circ \equiv -(\mu_\ce{H2O}^\circ - 2\mu_\ce{H+}^\circ)/2F$ for the reaction in the positive electrode.
Subtracting Eq. (6) from Eq. (5)
\begin{align}
  \require{cancel}
  E_\mathrm{p} - E_\mathrm{n} &=
  {E_\mathrm{p}}^\circ - {E_\mathrm{n}}^\circ - \frac{RT}{2F}
  \ln\left(\frac{a_\ce{H2O}}{a_\ce{O2}^{1/2}a_\ce{H+}^2}\right) +
  \frac{RT}{2F}\ln\left(\frac{a_\ce{H2}}{a_\ce{H+}^2}\right) \\
  E_\mathrm{p} - E_\mathrm{n} &=
  {E_\mathrm{p}}^\circ - {E_\mathrm{n}}^\circ -
  \frac{RT}{2F}\ln\left(\frac{a_\ce{H2O}\cancel{a_\ce{H+}^2}}
  {a_\ce{O2}^{1/2}\cancel{a_\ce{H+}^2}a_\ce{H2}}\right) \\
  \Delta E &= \Delta E^\circ -
  \frac{RT}{2F}\ln\left(\frac{a_\ce{H2O}}
  {a_\ce{O2}^{1/2}a_\ce{H2}}\right) \tag7\\
\end{align}
where we have defined the cell potential $\Delta E \equiv E_\mathrm{p} - E_\mathrm{n}$, and the standard cell potential as $\Delta E^\circ \equiv {E_\mathrm{p}}^\circ - {E_\mathrm{n}}^\circ$.
The activities of a species has a general form which is the product of the activity coefficient $\gamma_j$ and some measure of the concentration. If the gas phase behaves as an ideal gas and the ionic solution behaves as an ideal ionic solution, we can set $\gamma_j = 1$ and stick with the concentration of the species. The three species are in the gaseous phase, so we consider their measure of concentration as their partial pressures, but you can choose anyone you like. The final equation is
\begin{equation}
  \boxed{\Delta E = \Delta E^\circ -
  \frac{RT}{2F}\ln\left(\frac{p_\ce{H2O}}
                             {p_\ce{O2}^{1/2}p_\ce{H2}}\right)} \tag8
\end{equation}
Final remarks:

We crossed out the hydrogen ion concentration. This is only true if its concentration is homogeneous across the cell. In a fuel cell, $\ce{H+}$ is liberated through the oxidation in the anode, travels through the membrane, and reaches the positive electrode. Outside of equilibrium this will not hold.
As you can see, $\Delta E$ is a collection of electric potential differences, which in this case is $(\phi^\mathrm{M,p} - \phi^\mathrm{S,p}) -(\phi^\mathrm{M,n} - \phi^\mathrm{S,n})$. In conditions of equilibrium, that is called the open circuit voltage in batteries, and measuring the voltage difference at both electrodes should yield $\Delta E$. This is actually not totally true, but this needs another answer. For practical purposes, $\Delta E$ and what the voltmeter says is the same.

@MetalStorm Thank you very much for you explanation, it is very helpful.
Just to clarify, I was wondering what the $S$ stands for in $\phi^{S,n}$ and $\phi^{S,p}$. Is it the potential measured with respect to a reference electrode in the electrolyte?and $\phi^M$ is the potential at an electrode?

Hi. $\phi^\mathrm{S}$ is the electric potential of the electrolyte solution, and $\phi^\mathrm{M}$ is the electric potential of the electrode. Ions 'live' in the solution, hence the electric potential in their electrochemical potential has to be $\phi^\mathrm{S}$. Likewise, electrons 'live' in the electrodes, hence the electric potential in their electrochemical potential has to be $\phi^\mathrm{M}$. The symbols $p$ and $n$ refer to positive or negative. For example, you have two electrodes and their electric potentials are different (if they were not, electrons would not move).


text:
Why do protons in same chemical environment do not exhibit spin-spin coupling?
Reading introductory tutorials on proton NMR for the determination of chemical structure, I was told that protons can exhibit spin-spin coupling, thus creating multiplets on an NMR graph. The coupling can be explained by the fact that (referring to the diagram below) proton $H_A$ may be in the alpha or beta spin state, thus modifying the effective magnetic field "felt" by the $H_B$ proton, supposing they are in different chemical environments. A proton attached on a carbon further apart will have no influence because of the greater distance.
My question 1 is: Is this description of the spin-spin coupling accurate?

However, it seems to me that the same argument holds to justify the influence of protons in the same chemical environment on adjacent carbons or even on the same carbon. I fail to see how the same chemical environment prevents one proton from disturbing the magnetic field felt by another one. For example, in the case of methane, any given proton may "feel" 4 different configurations of neighboring proton spin states, which should be indicated by 4 distinct signals on an NMR graph. This is not what we observe.
My question 2 is: Why protons' spin states in the same chemical environment do not couple with one another?
Any insights is greatly appreciated!

Does this answer your question: https://chemistry.stackexchange.com/questions/19059/why-is-proton-coupling-through-bonds-and-how-does-this-affect-identical-protons?rq=1


text:
Does $\ce{O-CH_}3$ substituted on the meta position relative to a group on the benzene ring have a donating or withdrawing effect?
I have been confused about the behaviour of the group $\ce{O-CH_3}$ in the following two conditions:

The acidic strengths of methoxy substituted phenols: Here, according the order of acidic strengths

$$(meta>ortho>no\ substitution>para)$$
We are able to explain the highest acidic strength of meta with the fact that at meta position, the methoxy group acts as a $-I$ group.

This question in a reputable exam:

The answer given is
Here, the ethoxy group act as a donating group, activating the ring, and due to this the carbocation formed is a benzylic one, leading to the substitution of Bromine atom at the benzylic position.
I assumed that methoxy and ethoxy have almost the same behaviour.
Would this be any different if the substituent was methoxy and not ethoxy?
If not, then how can we differentiate the times when methoxy group at meta acts as donating and when it acts as withdrawing?

I assume that I have made some kind of mistake and so the downvote. Can you please clarify what way I have expressed the question incorrectly?


text:
If Internal energy can be changed by doing work also, then how come internal energy is zero in a isothermal process involving non zero work?
In the equation of First law of thermodynamics
$∆U = q + w$,
where $U$ is the internal energy, $q$ is the heat supplied and $w$ is the work done.
It is said that for an isothermal process, since the temperature remains constant, the change in internal energy is zero even if the work done is non zero.
On the other hand, thermodynamics also says that internal energy can be changed by doing work.
So supposing for a particular process the work is not zero in the above equation, then shouldn't the internal energy also be non-zero?
I don't get it why such type of consideration is taken while solving problems of the type given below:

An ideal gas undergoes isothermal compression from 5 m³ to 1 m³ against a constant external pressure of 4 N m-2. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol-¹K-¹, the temperature of Al increases by
(a) 2 K
(b) 1K
(c) 2/3 K
(d) 3/2 K

In the solution, the heat released has been equated to the work done which is
$-P(\mathrm{external})(V_2-V_1)=-4(1-5)=\pu{16 J}$ <--------I don't get this part.
What actually am I missing?

TL;DR: You can't keep the temperature of a perfect gas constant if only work is performed. Heat must be exchanged also.
Answers involving the thermodynamics of ideal gases are logically self-consistent. Start from some logical premises:
If the gas is ideal, then its internal energy depends only on T. Corollary: if the internal energy has changed, T has changed, and vice-versa.
Changes in internal energy are possible through either exchange of heat, or work, or both.
Starting from the above, you can state that:
(1) if a net amount of heat is exchanged with only one other body, but no work is done at all, then T must have changed, because the internal energy must have changed.
(2) if the internal energy has not changed, then T has not changed, and either W=-Q or W=0 and Q=0.


text:
How does graphite content affect the conductivity of pencil lead?
My peers and I started an experiment to test whether we could use the conductivity of a piece of pencil lead to determine its grade (that is, whether it is 2H, HB, 3B etc.). The paper Observational Models of Graphite Pencil Materials has a helpful list of the composition of various grade pencils, while Doğa Gürgünoğlu has made some relevant measurements of conductivities in the essay Electrical conductivities of different grade lead pencil graphite.

I had assumed conductivity would be linear in graphite composition, since the other constituents like clay and wax have negligible conductivity. However, when I graphed out the data, it seems more like an exponential growth.

Anisotropy of graphite means the conductivity changes drastically depending on its orientation, as discussed here. What else could influence the conductivity of pencil lead? Is there a theoretical explanantion of this trend?
References:

M. C. Sousa & J. W. Buchanan - Observational Models of Graphite Pencil Materials
Doğa Gürgünoğlu - Electrical conductivities of different grade lead pencil graphite

Guessing based on percolation theory, I suspect that at low graphite content there are few "good" pathways available for the current to be conducted through, and it looks relatively insulating. After you reach some critical graphite concentration, the conduction pathways start to connect en masse, and you see rapid increase in conductivity, until it saturates. This gives a roughly sigmoidal curve. A slight issue is that it's somewhat unexpected for the rapid growth to happen at such high graphite concentrations, but that could be a consequence of many real-world parameters.

It's not linear (or any specific function like exp) because it's not some true solution. If you cover some chunk of conductor in insulator, it won't conduct at all. Here's some mixing, though, so situation is intermediate.

Nicely written question and experimental data. BTW,  now you can make some hand-drawn electronic circuits, and even draw the current-limiting resistor for an LED (if your instructor appreciates humor, draw it on the lab report., with a coin cell).

@NicolauSakerNeto this sounds very interesting. Any chance you could write it into an answer with more details/additional readings? I'm a physics student, so theory is also welcome.

@DrMoishePippik by the way, the conductivity data used is not by me, but from a high school essay I found online.

I'm afraid I can't commit the time to writing an answer, sorry. Hopefully that will get you going in the right direction, though.


text:
Hydrated salts of transition metals
I am a little confused when it comes to hydrated salts and transition metals.
When it comes to non transition metals like $\ce{Mg}$, we can clearly have $\ce{MgSO4.xH2O}$. However, when it comes to an element like $\ce{Fe}$, I do not know the difference between $\ce{Fe(OH)2}$ and $\ce{[Fe(OH)2 H2O4]}$, so I am unsure if we should write $\ce{[Fe(OH)2H2O4].xH2O}$ or $\ce{Fe(OH)2.xH2O}$.

"$\ce{CuSO4\cdot5H2O}$ can also be written as $\ce{[Cu(H2O)4]SO4\cdot H2O}$." This fact might be related to your question...

$\ce{[Fe^{II}(OH)2(H2O)4]}$ is a neutral dissolved complex, formed from ferrous hydrated ion in alkalic solutions:
$$\ce{[Fe^{II}(H2O)6]^2+(aq) + 2 OH-(aq) -> [Fe^{II}(OH)2(H2O)4](aq) + 2 H2O(l)}$$
$\ce{Fe(OH)2}$  or $\ce{Fe(OH)2 . x H2O}$ is rather way to describe $\ce{Fe(OH)2(s)}$ that has rather undefined stoichiometric composition, being a result of condensation of the $\ce{[Fe^{II}(OH)2(H2O)4](aq)}$, releasing water and joining complexes with oxide or hydroxide bridging ligands.
E.g.:
$$\ce{2 [Fe^{II}(OH)2(H2O)4](aq) ->  H2O(l) + [Fe^{II}(OH)(H2O)4-O-Fe^{II}(OH)(H2O)4](aq)}$$
$$\ce{2 [Fe^{II}(OH)2(H2O)4](aq) ->  H2O(l) + [Fe^{II}(OH)(H2O)4-(OH)-Fe^{II}(OH)2(H2O)3](aq)}$$
and followed by further condensation to multicenter, polymeric substance of a typical amorphous look of transition metal hydroxides.

You said a lot of stuff that went over my head, but from what I can put together, the two iron compounds in my post are the same thing!

If the stuff went over your head then yes, they are the same thing.  :-)


text:
Silver Nitrate Solution + Flame = Mirror?
I purchased an interesting product recently, which says it contains: "Isopropyl alocohol, silver nitrate, Water-based lacquer". You apply it to a smooth surface, then use a blowtorch on it and it'll darken before turning into a mirror. I can't find any information about this reaction, the closest I can find is Tollens Test. What is this reaction and how does it work?

how to make silver adhere?

This is similar to Brashear's process. Brashear's process does use Tollen's reagent, though as described, this is a variant.
Consider that $\ce{AgNO3}$ is easily reduced to silver metal.
Tollen's reagent (diamminesilver(I) coordination complex $\ce{[Ag(NH3)2]+}$) is very easily reduced, just needing room temperature glucose or other aldehyde to create a silver mirror. In fact, the reagent is too unstable to be sold as a premixed product.
In that kit, it would seem the lacquer (dissolved in isopropanol), plus any remaining isopropanol that has not boiled off during heating, reduces the $\ce{AgNO3}$ to metallic silver when heated strongly.
BTW, you might edit your question to describe the use of the product, how well the mirror is formed, and how long it lasts -- if the lacquer burns off, silver quickly tarnishes.
Also, note that absolutely, scrupulously clean glass is needed, or the silver does not adhere.


text:
Why is Chromium hexacarbonyl diamagnetic?
So there was a question in coordination  chemistry regarding the nature of $\ce{[Cr(CO)_6]}$
The book had the question under VBT and solved it using "strong ligands cause all electrons to pair up". $$\ce{[Ar] 3d^5 4s^1\rightarrow [Ar] 3d^6}\text{ and all electrons paired}$$

Later we learnt about Crystal Field Theory and splitting in octahedral complexes.
From that the configuration of the above complex should be $$\ce{3d_{t_{2g}}^5 \ 3d_{e_g}^0\ 4s^1}$$
My answer was paramagnetic and outer orbital complex.
So why does electron from $\ce{4s}$ orbital pair up with $\ce{4d}$ electron and make the complex diamagnetic?
I only know VBT and CFT for coordinate chemistry, so how is this explained by CFT?
I didn't find much good of an explanation on the internet except that $\ce{s}$-orbitals must be emptied to take part in hybridization. But then why is it wrong to use the concept of hybridization for transition metal complexes?
My personal guess is maybe the splitting energy is so high for $\ce{4s}$ that somehow lower energy gap/mixes with $\ce{3d}$ orbital.

Due to the strong-field nature of the CO ligand, chromium hexacarbonyl is low-spin. Cr has 6 d-electrons (the d-orbitals fill first for all transition metals), which are all 6 in t2g and paired.

@Mäßige According to n+l rule 4s is filled before 3d right and ground state configuration of Cr is exceptionally $3d^5\ 4s^1$ and ligands can rearrangement of electrons in d-orbitals by splitting it can't force s orbital (non-directional) to d-orbital, right?

The n+l rule strictly applies to main group elements. The general consensus is that the transition metals’ 3d-orbitals are completely filled before the 4s orbitals in transition metal complexes. This is due to the lowered energy as per the increased effective nuclear charge, because of the electron withdrawing nature of the ligands relative to the central atom. You would be right, if the chromium center would be in the gas phase and an isolated atom without ligands.

@Mäßige Sorry to bother. But "You would be right if ........gas phase...isolated atom...." So The presence of ligands and/or water increases the energy of 4s ? So how does that happen? Also if 3d is filled before 4s, then $[Cr(H_2O)_6]$ if it exists will also have an empty s orbital?( irrespective of ligand strength)

Yes this complex would also have an empty s orbital, only the 3d orbitals would be filled. The reason is described in this question: https://chemistry.stackexchange.com/questions/33297/why-do-3d-orbitals-have-lesser-energy-than-4s-orbitals-in-transition-metals

The energy gap between 4s and 3d is very low. This, primarily is the reason for transition metals exhibiting a huge number of oxidation states. In the case of $[Cr(CO)6]$, carbonyl group acts as a strong ligand. The CFSE can be calculated by
$$CFSE = Δo[0.6 n(eg)- 0.4 n(t2g)]$$
Where n(eg) = no of electrons in eg orbital
And n(t2g) = no of electrons in t2g orbital.
In this case,
Carbonyl group creates a very strong field.
As we know for achieving stability, the energy possessed has to be the least, pairing takes place, which considerably provides a more stable compound i.e. a low spin complex.
Thus, $[Cr(CO)6]$ is a low spin where there are 5 t2g electrons.

But s orbital in non-directional, so how does ligand repulsion affect it? Strong or weak , there should be no effect of ligand repulsion on a non-directional orbital, right?

Repulsion affects s orbital before d orbital because it is in valence shell. D is in penultimate shell, so first electrons migrate from s to d then splitting takes place because the interaction of ligands is limited to valence shell.


text:
Lustre as a Chemical Property
In a text that I was reading, I found this paragraph on chemical properties:

Chemical properties describe the ability of a substance to form new substances, either by reaction with other substances or by decomposition. For example, a chemical property of the element iron is its tendency to react with oxygen from air in the presence of water to form rust (a compound composed of iron and oxygen). The conversion of rust back into iron and oxygen is a difficult chemical process. In some cases, chemical properties relate to the absence of specific changes. For example, a chemical property of the element gold is that it maintains its lustrous appearance because it resists rusting or tarnishing.

However, other sources have cited lustre as a physical property. I want to know what is the correct answer, or, if the question is wrong itself.

Lustre is a physical property, yes but the statement in the paragraph explains that the lustrous nature of Gold(which is a physical property) is preserved due to the chemical inertness of the element.
Generally, active metals lose their lustre(shine) since they react with the moisture, oxygen and other gases like hydrogen sulphide and carbon dioxide. This is a chemical property.
So, its absence in the element Gold is known as inertness towards corrosion.


text:
How exactly are mirror images of molecules rotated to determine enantiomeric pair?
I know there are many similar questions to this but none could clear my confusion over how exactly the mirrored molecules is rotated i.e on which axis it is rotated to superpose with the original molecule

Here the molecules are superimposible but I cannot figure out how the molecule is rotated to acheive this. If the molecule is rotated from y-axis (↑) then the molecule is again mirrored defeating the purpose.

If having troubles with 3D imagination, try to make yourself physical ball-stick/pin model, each stick pointing to a vertex of a thought regular tetraedr. Make 2 mutually mirror images and then try to manipulate by whatever rotation to look identical.

You can use whatever method for checking it, as long as it works ;) BTW maybe get better source than what you used for this picture.

The molecule in your example is not rotated at all. As it is drawn, VI is exactly the same as V, no rotation required.

The example is taken from Solomons by Wiley Publication

TL;DR Combined rotation and reflection bring this particular achiral molecule into a form indistinguishable from the original, in appearance.

If the molecule is rotated from y-axis (↑) then the molecule is again mirrored defeating the purpose.

No, it isn't mirrored again, it looks as if, that's the whole point. Defeat is not defeat, it is the whole purpose of the exercise, what you sought to prove. The initial reflection transformed the molecule but a rotation about a vertical axis that goes through the chiral center is enough to bring it back into an orientation indistinguishable in appearance from the original. In a chiral molecule a similar rotation cannot bring the reflected molecule into alignment no matter how hard you try. A simple exercise is to swap one of the methyl groups with a third type of substituent to see the effect of the same operations.
The question is similar to another which also deals with 2-propanol. It is worth noting that using these symmetry operations to determine whether a molecule has an enantiomer is one algorithmic test but one can define enantiomers in other ways. A chiral molecule lacks reflection (Cs) and rotoreflection symmetries (S2n). The atomic composition, connectivity, and bond order are unchanged (as are thermodynamic properties of the pure substance), but their interaction with polarized light differs. An equivalent test is to invert the sign of all dihedral angles and attempt to superimpose the molecules:

Here the molecules are shown using a Newman projection. If the blue atoms are of the same type (including isotope, etc) the two molecules are indistinguishable.
On a final note, isopropanol might not be an ideal example for such an exercise. The gas phase molecule in its ground state is classified as having C1 and not Cs symmetry, so strictly speaking its mirror images are not superimposable (that H on the hydroxyl breaks the symmetry).


text:
The use of H mass in mass defect and nuclear binding energy computations
I've noted that introductory general chemistry texts compute the mass defect (or mass difference) for radioisotopes using the proton mass $m_\ce{H}=m_{\text{p}}\approx1.00783 \ \text{u} $, instead of $m_{\text{p}}\approx1.00728 \ \text{u}$, as per the Green Book (p. 93). This affects the results of succeeding nuclear binding energy computations.
Am I missing something here conceptually? Should the mass defect computation use the actual proton mass if it is readily available instead of the mass of $\ce{H}$?

It may depend on if the context evaluates mass defects for atoms or for nuclei.

@Poutnik This is for the mass defect for atoms. Now I wonder: is the mass of H used to account for the electrons present in the isotope?

Of course it is. The value 1.0078.. is for the complete H atom, roughly 0.0005 is the electron, so about 1.0073 is the proton.


text:
About whitish layer that formed on chocolate bar
I noticed a whitish layer on the surface of a chocolate bar that I ate yesterday. I looked at the pack where it was mentioned that,

"Temperature and humidity changes may cause product to develop a whitish layer, without affecting its fitness for consumption. "
My first doubt is, is this a chemical or physical phenomenon? if it is chemical process, please elaborate.

It is solid cocoa butter, that may get "sweated" to the surface and solidified later. It can be caused by long term temperature fluctuations, or just effect if time.
So it is mainly the reversible  fat phase transitions (bordering chemical and physical as non chemical change) and matter redistribution.
If you insist, you can consider it as a physical change, as substances did not transform to something else.
Or, it could be a chemical change two, as the are several forms of solid cocoa butter and transition between them can be considered as chemical change.
All chemical phenomena are physical phenomena too, as chemistry is part of physics that physicists left for chemists. The boundary between these domains is not sharp and it sometimes does not make much sense to insist on categorization.


text:
Is it possible to calculate the theoretical rate of reaction?
I am conducting an experiment where I will be changing concentration (0.2 mol/dm3, 0.4 mol/dm3, 0.6 mol/dm3, 0.8 mol/dm3, 1.0 mol/dm3) and seeing how that affects the rate of reaction for a reaction between HCl and CaCO3.
Is it possible to mathematically calculate the theoretical rate of reaction for each variation in concentration?
Given that rate of reaction can be defined by the rate of disappearance, the experimental rate of reaction can be calculated by taking the mass loss and dividing by the time.
I don't know if the same can be done for calculating the theoretical value. I know the theoretical mass loss can be calculated by calculating the theoretical yield. However that assumes that it all reacts, and so if I divide it by the same time as before it wouldn't really be accurate right?
So I'm wondering if it's even possible to calculate the theoretical rate of reaction for each concentration of HCl. If not, is there available data online for the ROR of different concentrations of HCl?
Thank you.

As it involves dynamic turbulent heterogenous reaction, involving three different phases, it is not possible, but from prior empirical and applicable data. There are just too many unknown variables and hard-to-model processes.

I see thank you

It is always possible to postulate a kinetic rate law and test it. The chemical reaction in this case is
$$ \ce{CaCO3(s) + 2H+(aq) -> Ca^{2+}(aq) + H2O(l) + CO2(g)} \tag{1} $$
and the statement of mass conservation is
\begin{equation}
  \frac{\mathrm{d}c_\ce{H+}}{\mathrm{d}t} = -2r_\ce{H+} \tag{2}
\end{equation}
Eq. (2) assumes that there are no spatial variations of the concentration along the volume, which is ensured with a sufficiently high stirring rate. The only thing you have to do, is propose a function for $r_\ce{H^+}$.
What you can always do is apply finite differences with Eq. (2), so that
\begin{equation}
  r_\ce{H+} \approx -\frac{1}{2}
  \left(\frac{\Delta c_\ce{H+}}{\Delta t}\right) \tag{3}
\end{equation}
which is what you stated in the post that you are going to do.

Fusi et al. have studied this particular reaction. Since this is a heterogeneous system, the reaction takes place at the solid/aqueous interface between the calcium carbonate and hydrogen ions. The authors wrote the RHS of Eq. (2) in the next form
$$ r = kS^*(c^* - c_0^*)^\gamma \tag{4} $$
If you take a look at the results, they obtained an excellent agreement with experimental data. Thus, the mathematical model of the authors postulate that:

The kinetics are directly proportional to the surface available of reaction. This surface changes with time, and is a function of a particular geometry shape of the solid, which must be decided.
It is proportional to the concentration difference between the hydrogen ions $c^*$ and some threshold concentration $c_0^*$ for which the reaction stops, at a certain power $\gamma$.

As a result, the mathematical model has three unknowns: the rate constant $k^*$, the threshold concentration $c_0^*$, and the power $\gamma$.
You can do exactly the same, propose a function and see the goodness in the fit.

References
The paper can be downloaded for free in here:

Fusi, L. & Monti, Alessandro & Primicerio, Mario. (2012). Determining calcium carbonate neutralization kinetics from experimental laboratory data. Journal of Mathematical Chemistry. 50. 10.1007/s10910-012-0045-3.

S would be a complicated function of reaction progress and also of reaction rate, as evolving CO2 causes negative feedback for the reaction rate by making the surface of particles less available.

@Poutnik I had to study this topic, and nucleation/growth mechanisms for solid particles in solutions have a heavy mathematical framework based in probability. This can only be applied for models that take into account some coordinate, which was not used by the authors. In that work, a simplified approach was taken which I will add to this post later. However, I have never thought of the implications of generated gas at the surface, and the hindrance of the rate of reaction. I think it is a question that I would have trouble in answering...

In a hurry I would say that the model considers that carbon dioxide diffuses super fast from the surface and enters the air, but that won't convince anyone.


text:
Why SO2 doesn't have dative bonds?
Why doesn't the SO2 Lewis structure have, instead of two double bonds, one double bond and one dative bond? This way, the octet rule is also satisfied for sulfur.

See https://en.wikipedia.org/wiki/Sulfur_dioxide#Structure_and_bonding

It does$ $.


text:
UV-vis spectra - maximum absorption
If I have a noisy UV-vis absorption band, how do I determine the maximum absorption?  Do I look for the highest absorbance value or do I look for the most intense band and take the value from the center of the band?  The second option seems correct to me, but I don't know if those noisy portions that are not in the center of the band actually represent the maximum absorption

Do you use a dispersive instrument, or not? Is the recording of the data sequential with adjustable integration time (so and so nm/s), or all at once with a CCD array? An old rule of thumb is to record $\epsilon_\text{max} = 10^n$ at concentrations no higher than $c \approx 10^{-n}$mol/L for a 10mm cell (adjustable parameters of the sample) to stay in the linear response of the detectors. Slit width, integration time (or scan rate, nm/s), and repeating the measurements (as noise signal is random, systematic signal is not) may help on side of instrumentation/software.

You can attempt to fit the peak or part of it. That will implicitly average over the noise.

I tried to fit, but it didn't help much. The recording regime is a stopped-flow one, with a dead time of 2 ms.  I just want to know how I should evaluate the maximum

"I tried to fit, but it didn't help much. " How about showing a plot, or some sample data? How is the dead time relevant, or regime? Is your point that you can't perform averaging? Perhaps it would be more relevant to describe the UV-Vis instrument.

Do you care to monitor shifts in the spectrum maximum? Or do you merely want to maximize s/n?

Yes, I want to monitor shifts in the spectrum maximum.

Ok, I still recommend providing more info as to why a fit of some sort hasn't worked. You are left with few other options if you want to work with the data you already have (otherwise look at what Buttonwood wrote and refer to the user's manual). There is no real way to avoid noise other than filtering (which if done after the fact is equivalent to a fit). Think of noise as lack of information (or useless information).

" do I look for the most intense band and take the value from the center of the band" This is equivalent to a fit. You are making assumptions about the shape of the band. The absolutely most intense point might be due to addition of noise above the signal. If you want to report the magnitude of the signal, you'll want  to provide an estimate of noise which again involves an explicit or implicit model or fit.

If you have a noisy UV-VIS spectrum, one cannot pick the highest value in the spectrum because each point is "contaminated" with noise. I am assuming you are not measuring very large absorbance values. Since you have not shown the data, the easiest way is to do smoothing with a Savitsky-Golay filter. Wikipedia has a good explanation and it can be easily done in MATLAB. The nice thing here is that your UV signals are broad (low frequency signals) and noise is high-frequency, so, smoothing is not going to hurt the width of UV-peaks.
If you can have a look at the Fourier spectrum of your UV spectrum, you can see the nature of your noise in the stopped flow experiment. And as has been commented above, increasing slit width might help to reduce noise.

The relevant theory as well as implementations (typically Matlab, Octave, Python/Spider; spread sheet Microsoft Excel and OpenOffice Calc) ready to use can be found for instance in the superb resource A Pragmatic Introduction to Signal Processing by Tom O'Haver (University of Maryland), link.  Did I say his project is open access?

Excellent reference by Tom. Someone was selling the book at an expensive rate on Amazon. The pdf of the book is free. I have published with the author.

Thank you so much!


text:
What is the pressure of triple point and melting point for ethyl chloride?
I need the information for pressure of triple point and melting point for ethyl chloride but on the internet I found only information for temperatures which are 134,5K and 134,82K.

So what is the problem?

It would seem simple to find on the web... but when I researched the phase diagram and triple point for "chloroethane" and for "ethyl chloride", there seemed no readily availbale answer. https://chemistry.stackexchange.com/questions/53280/equation-to-calculate-the-triple-point points to a theoretical calculation, and there are paywalled references, e.g., https://link.springer.com/chapter/10.1007/978-1-4615-2522-6_231,


text:
Hydrogen–deuterium exchange in 1,4-benzoquinone
Problem

Find number of deuterium exchanges in the following compound.

Solution

 0 D-exchanges.

Question
I have been taught that hydrogen attached to the carbon which is directly or indirectly in conjugation with the ketone group will show deuterium exchange with $\ce{D2O}$ or $\ce{NaOD}$. Why does the solution claim there is no H/D exchange in p-quinone? Are the hydrogens attached to the double bonded carbons not considered in the conjugation?

Would highly appreciate if someone suggested a better tag. For now I just used a general tag.

Hydrogen attached to the carbon which will be in conjugation with electron withdrawing group after deprotonation, would be deprotonated. Is there one?

Still, I wouldn't be surprised if there was deuteration, if concentration of base was high enough.

It requires a lot of energy to remove the hydrogen atoms from the $sp^2$ hybdridised carbon atoms (involved in double bonding). So there's not really any hydrogen atom which is readily available for removal and deuterium exchange.

Okay, that makes sense. But, let's say if there were that much energy, then nothing is stopping those hydrogens from being exchanged, right? They are in conjugation with the ketone group, so I dont think there would be any problem.

Yeah well if you provide a sufficient amount of energy, they can potentially participate in deuterium exchange. It's just that usually people don't consider that as a normal answer to problems like this.


text:
pH in a buffer of Glycin and HCl
I am struggling with a task. I have the amino acid Glycin with pKa1=2,3 and pKa2=9,6.
I have 1.0 liter 0.20 mol/L solution of glycin. I add 0.1 liter 1 mol/L HCl solution. What is the pH in this buffer.
I have calculated that the isoelectric point is 5.95 for this amino acid.
Using the equation pH=pKa + log [base]/[acid], where do I go from here? Should the pKa used be 5.95?

5.95 is pI, not pKa. As pKa2 for R-NH3+ is so high, you can formally consider it just as a salt R-COO-  with pKa 2.3, resp. pKb 11.7, with the reaction R-COO- + H+ <=> R-COOH.

Are you asking about glycin or glycine? You write the former, but provide $\mathrm{p}K_\mathrm{a}$ values for the latter.

Glycine in aqueous neutral solution (i.e. add water to glycine) occurs as zwitter ion:

The $\mathrm{p}K_\mathrm{a}$ of the ammonium group is 9.6. The $\mathrm{p}K_\mathrm{a}$ of the carboxylic acid is 2.3 (quite different from acetic acid, which is around 4.7). Given that the carboxylic acid group is a stronger acid than the ammonium group, the zwitter ion forms, as shown above.
Adding HCl to the zwitter ion will protonate the carboxylate, so the $\mathrm{p}K_\mathrm{a}$ of 2.3 is relevant (the ammonium is already protonated, and remains protonated as you add strong acid). In the problem, you are adding just enough HCl to protonate half of the carboxylate. The result is a 1:1 buffer, a mixture of the zwitter ion (net neutral) and the twice protonated species (net charge of +1).
With this information, you should be able to plug in concentrations and $\mathrm{p}K_\mathrm{a}$ into the buffer equation (the Henderson-Hasselbalch equation).


text:
In electrochemisty, we write delta S= nF(dE/dT)p (at constant pressure), is (dE/dT)p cell constant?
My lecturer stated that dE/dT at constant pressure is cell constant, but isnt cell constant of a cell (l/A) the units don't match, is there any other cell constant that he might be talking about?

The lecturer means "a" cell constant, not "the" cell constant in conductometry context.
It is $\Delta_\mathrm{r} S$, not $S$. For a given temperature, it is constant only for standard conditions.
It also depends on temperature, because product and reagent heat capacities are different, so $\left(\dfrac{\partial E}{\partial T}\right)_p$ depends on $T$ too. It could be considered approximately constant for not too wide temperature range.

For the temperature dependence, let start with this equation relating the electrochemical and thermodynamic potentials:
$$-nFE^{\circ}(T) = \Delta_\mathrm{r} G^{\circ}(T) = \Delta_\mathrm{r} H^{\circ}(T)   - T \Delta_\mathrm{r} S^{\circ}(T) \tag{1}$$
$$-nF \left(\frac{ \partial E^{\circ}(T)}{\partial T}\right)_p =  \left(\frac{ \partial \Delta G_\mathrm{r}^{\circ}(T)}{\partial T}\right)_p = \\
= \left(\frac{ \partial \Delta H_\mathrm{r}^{\circ}(T)}{\partial T}\right)_p   - T \left(\frac{ \partial \Delta S_\mathrm{r}^{\circ}(T)}{\partial T}\right)_p - \Delta_\mathrm{r} S^{\circ}(T) \tag{2}$$
$$\left(\dfrac{ \partial \Delta H_\mathrm{r}^{\circ}(T)}{\partial T}\right)_p = \Delta C_p(T)\tag{3}$$
$$\left(\dfrac{ \partial \Delta S_\mathrm{r}^{\circ}(T)}{\partial T}\right)_p = \dfrac{\Delta C_p(T)}{T}\tag{4}$$
$$-nF \left(\frac{ \partial E^{\circ}(T)}{\partial T}\right)_p =  \Delta C_p   - \Delta C_p - \Delta_\mathrm{r} S^{\circ}(T) = -\Delta_\mathrm{r} S^{\circ}(T)\tag{5}$$
$$nF \left(\frac{ \partial E^{\circ}(T)}{\partial T}\right)_p = \Delta_\mathrm{r} S^{\circ}(T)\tag{5a}$$
$$\Delta_\mathrm{r} S^{\circ}(T_2)=\Delta_\mathrm{r} S^{\circ}(T_1)+ \int_{T_1}^{T_2}{ \frac{\Delta C_p(T) \mathrm{d}T}{T}}\tag{6}$$
where $\Delta C_p$ is the difference of molar heat capacities of products and reactants (including stoichiometry).
$$ \left(\frac{ \partial^2 E^{\circ}(T)}{\partial T^2}\right)_p =  \frac{\Delta C_p(T)}{nFT}\tag{7}$$

I added some $\Delta$s on Eqs. (2-5) if you don't mind.

@MetalStorm  Sure, you are right...


text:
Carbonyl and thiocarbonyl 13C NMR
According to this paper, the deshielding effect of C=S group is specifically due to the delta(11), which is the least shielded component. However, the conclusion tells me that this component is governed by two transitions: σ → π(antibonding) and π → σ(antibonding) transition.
The thing confuses me is that, as the excited energy magnitude is smaller, the paramagnetic shielding should be more negative, also means the nucleus is more deshielded. When it comes to normal thought of C=O and C=S transtion, the n → π(antibonding) is lower in energy than σ → π(antibonding) and π → σ(antibonding). So why carbon on C=S has greater chemical shift than that of C=O?


text:
Why is AgCl considered a good electrolyte? Isnt it sparingly soluble?
My textbook says AgCl has degree of dissosciation as almost 1 but its sparingly soluble. While solving to find Ksp of AgCl using kohlrauschs law we have said that molar conductivity of sparingly soluble salts in normal concentration is almost same as at infinite dilution as degree of dissosciation is 1.

"Good electrolyte" is not explicitly defined term in electrochemistry. It is more probably used in context of electrochemistry related engineering in sense of degree of its applicability in particular technical solution, giving to the solution high enough conductivity.
Electrochemistry in context of physical chemistry as a science defines

a weak electrolyte with the dissociation degree  $\ll 1$ for
$\ce{AB(solv) <=> A+(solv) + B-(solv)}$
a strong electrolyte with the dissociation degree $\approx 1$.

Note that this does not involve the electrolyte solubility.

A weak electrolyte as acetic acid is miscible with water in any ratio.
A strong electrolyte as silver chloride or baryum sulfate  have very limited solubility.

Note that being strong/weak electrolyte is rather condition dependent and usually related to typical range of not too much diluted solution.

Weak electrolytes like acetic acid converge to being strong electrolyte going to infinite dilution.
Strong electrolytes with higher ion charges like MgSO4 become not so strong at higher concentrations, forming ionic pairs.

does that mean that degree of dissociation is defined only for already solvated compounds? my understanding based on your reply is that acetic acids degree of dissosciation is low cuz CH3COOH(aq) dissoscaites less but for AgCl(s) the little part that becomes AgCl(aq) fully converts to its ions?

Yes, the degree of dissociation is defined for $\ce{AB(solv) <=> A+(solv) + B-(solv)}$. // Otherwise, you could not reasonably define it, as its value would depend on amount of present solid.

Interesting! A strong electrolyte upon reaching its saturation point would form fewer Ionic molecules before crystal formation. A careful study of supersaturation could possibly explain what is happening. The last comment indicates that since crystal formation is dependent on initiation sites, nucleation, the amount of solid present is a possible factor.

@jimchmst MgSO4 reportedly forms significant portion of ionic pairs in sea water, what is still far from being saturated solution. https://en.wikipedia.org/wiki/Ion_association   (Burgess, John (1978). Metal Ions in Solution. Chichester: Ellis Horwood. ISBN 978-0-85312-027-8.Chapter 12, Kinetics and Mechanism: Complex formation")


text:
Buildup of white powder where metal has been touching a salt stone
The other day, I opened a box for my cellar I use to store random junk and found this:

The pink stone is a salt stone from a pet store. I assume it's NaCl with impurities. The ring is some metal jewelery I found on the ground. Maybe copper or some alloy? Some white build-up has appeared where the metal thing is touching the salt stone.
The metal thing seems to be fused to the stone (I didn't pull too hard), but some of the build-up fell into the container as white power:

Silly me touched the powder to taste it. It tastes like nothing, not salty at all. But it seems the poweder did muck around with a brass (?) coin sitting below it. Especially the bottom, which was sitting against the wooden container and not touching the salt stone above it, was hit hard:

There are more of that coin in this continer and they're still as shiny as when I put them into the box.
Here are two additional photos of the thing:

The box has been sitting in my cellear for about 1.5 years. It shouldn't be above 60% rel. humidity. There's some non-stainless steel stuff and paper in the same box that don't look like they were able to pick up a lot of humidity.
So, can anyone explain what's going on here? What is the white powder composed of?

This looks like galvanic corrosion between dissimilar metals, mediated by (small amounts of) salt water.  You've basically created a battery.

maybe salt impurities

This link https://www.corrosionpedia.com/does-zinc-rust/7/7030 describes, among other air stability properties of zinc, formation conditions of powdery zinc hydroxide. The ring may be copper-zinc alloy. The conditions in Your box match the ones described in the link above: Salinity and the lack of free flowing air.
I am not sure about the black coating on Your coin, but I suspect it contains copper.
If the ring seems to be fused to salt, it definitely is because of changing air humidity over time.

It looks like maybe the white powder came from specific beads instead of the ring itself. Zinc is a reasonably common material to make beads from.


text:
Why is there anomalous trend in third ionization energy for Pr-Pm and Dy-Er?
It is not hard to observe the anomalously flat regions at Pr-Pm and Dy-Er in the trend of IE3 of lanthanides:, each described as the "first quarter" and "third quarter". Why is this the case? I used to convince myself this has something to do with the orbital and total angular momentum of the ground term symbols of the Ln(2+) and Ln(3+) ions.

Dear OP, I would advise you to take a look at the electronic configurations and see if you get any clues.


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Why does enthalpy change and not internal energy help in determination of spontaneity?
Spontaneity of a process is measured by Gibbs Free Energy
$$\Delta G= \Delta H-T\Delta S$$
$$\Delta H=\Delta U+\Delta (PV)$$
Is there an intuitive explanation for why Gibb's energy depends upon enthalpy  change rather than internal energy change.
It seems intuitive that since systems try to attain lowest energy states, negative value $\Delta U$ should increase stability and therefore spontaneity.
It also seems intuitive that since particles try to achieve equilibrium, maximum disorder/maximum microstates/maximum energy distributed and therefore positive value of entropy$\Delta S$ would increase spontaneity.
Is there a logical explanation why/where the extra $\Delta (PV)$ term helps determine the spontaneity of a process or stability of products?
I do mostly get the mathematics of the equation but not logic.

See difference between Gibbs free energy ($G=H-TS$) and Helmholtz free energy ($F=U-TS$).

And the reasoning for introducing $H = U + pV$.


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Geometrical isomerism in spiro structures
The following structure shows Geometrical isomerism (GI for short) because the -Cl and -Br are in same plane

This structure does not show GI because the -Cl and -Br are in perpendicular planes and the distances between the atoms remain same

But I have these questions which don't seem to be able to be solved by this logic
Q1)

Q2)

My immediate thinking is that since one sides has two hydrogens, it cannot show GI
Is my answer, logic correct?
I really appreciate any help you can provide.

Think again. Are really Br and Cl in same plane with C atoms?

are you talking about the first structure? If yes , then arent Br , Cl is same plane because  spiro rings are perpendicular to each other , and there are three of them .... so they will be in same plane and hence show GI

All $\mathrm{sp^3}$ hybrids are tetrahedral, which have two planes perpendicular to each other such as $\ce{CH4}$. Thus, all your four structures show in here are not planer. So, could have some mirror images not equal.

it ain't planar

The second one will definitely exhibit Geometrical Isomerism. Imagine the first ring is in the x-y plane. Then the methyl group attached to it will lie along the z axis.The second ring will be in the x-z plane. Now depending on whether the methyl group is oriented to the +z axis or -z axis, the distance between the oxygen atom of the second ring and the methyl group of the first ring will be different, thus the molecule will show GI.
As for the first problem, I don't know why your book says that it will exhibit GI. All possible configurations seem like the same to me.

but according to what I had been taught , we always needed to see the extreme ends of the spiro case and not the intermediary atoms on the ring .... So does this mean that GI in spiro is not only specific to the first 2 cases (the ones with -Cl ,-Br)?


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How to find the global minima of a supramolecular complex in computational chemistry?
So this is more of a complex problem, and my masters thesis could greatly benefit if I find an answer to this.
Essentially, I have a supramolecular inclusion complex. There is this molecule A (the host), and another molecule B (the guest). There are literature studies pointing that the inclusion complex (A+B) is much more stable in vivo than molecule B by itself.
So, there are two ways in which A and B can interact, and lets call them interaction from the "top" and from the "bottom", and my goal is to find the more stable interaction among the two using computational chemistry. The idea is to simply compare the energies of both the complex, and then compare the energy of the molecule before complexing.
The problem with this idea is, how do you model both these "top" and "bottom" complexes? I have the equilibrium geometry of both A and B separately, but if I combine them and minimise the energy, there is no guarantee that it would orient itself in the most stable state, instead, it would just form the closest stable structure. To put it better, the molecule would fall into the closest local minima, and not the global minima.
How do you mitigate this problem? There could be hundreds, or maybe thousands of local minimal to sift through, so finding that manually could be a cumbersome nightmare.
Any inputs on this would be game changing for me. Thanks!
ETA -
There are some other DFT studies on the same inclusion complex in the literature, and they have all modelled the complex only in the "top" orientation (maybe they think it is the more stable form), but they still do not claim that it is the global minima structure. Additionally, I found the crystal data for the same "top" configuration, so that could at least be a starting point, but I am still stuck on the "bottom" orientation.
Someone suggesting docking studies to me, but it is such a low level theory to depend upon for energy calculations of a complex. I do not know if I should trust molecular dynamics (or classical physics) to that degree.


text:
Covalent network phase of CO2
In a science essay in the November 1999 edition of Analog magazine by Stephen L. Gillett¹, Gillett speculated on the possibility of a compound he dubbed "diamond ether". The name derived from the fact that it would have the same structure as diamond only with an intervening oxygen atom replacing the C-C bonds with C-O-C bonds. Or to describe it another way, it would be the carbon analog of the silicate mineral cristobalite. Gillet gave as a limited example an organic  compound which consisted of a central carbon atom with four ether bonds to four organic side chains. The question is whether that structure could be repeated indefinitely. Would this be energetically impossible due to bond strains or could it at least be metastable?
¹Diamond ether, nanotechnology - and Venus
Authors: Gillett, Stephen L.
Source: Analog Science Fiction & Fact. Nov, 1999, Vol. 119 Issue 11, p38, 9 p.
Publisher Information: Dell Magazines, 1999.
Publication Year: 1999
Subject Terms: Planets -- Environmental engineering Venus (Planet) -- Atmosphere
Literature/writing
Description: Methods by which Venus can be terraformed are presented with emphasis on new ideas. The problems of achieving this environmental engineering project are discussed and the hypothetical substance of diamond ether is described.
ISSN: 1059-2113
Rights: Copyright 1999 Gale, Cengage Learning. All rights reserved.
Accession Number: edsgcl.67008930
Database: Gale General OneFile

Not that I think it very important, but most people I know would call this an inorganic compound

"Can it exist" is a different question to "can it exist under normal conditions?" The problem under normal conditions isn't likely to be bond strain; it is the existence of strong carbon-oxygen double bonds and easy pathways to generate those from carbon-oxygen single bonds.

Take a look at Santoro, M., Gorelli, F., Bini, R. et al. Amorphous silica-like carbon dioxide. Nature 441, 857–860 (2006).
Quoting from the abstract

Here we report the synthesis of an amorphous, silica-like form of
carbon dioxide, a-CO2, which we call ‘a-carbonia’. The compression of
the molecular phase III of CO2 between 40 and 48 GPa at room
temperature initiated the transformation to the non-molecular
amorphous phase.

and

Comparison with vibrational and diffraction data for a-SiO2 and
a-GeO2, as well as with the structure factor calculated for the a-CO2
sample obtained by first-principles molecular dynamics, shows that
a-CO2 is structurally homologous to the other group IV dioxide
glasses. We therefore conclude that the class of archetypal
network-forming disordered systems, including a-SiO2, a-GeO2 and
water, must be extended to include a-CO2.

So, possible but only under extreme conditions.

Yes, but not totally absurd -  48GPa is appreciably less than the maximum pressure in the Earth's mantle, we're not talking supermassive black holes here.

However, we cannot assume existence within Earth itself because the carbon dioxide is polymerized at room temperature. Under the hotter temperature in the mantle it could still be fluid instead. In the case of water we know that the high-pressure phase (Ice VII) can be included in diamonds that come to the surface and cool off, but including a-CO2 in this way appears to require more pressure than even diamond can sustain.