text:
Does N in NH3 ligand gains formal positive charge as N in ammonium ion?
Ammonium ion is formed when N in $\ce{NH3}$ forms a coordinate covalent bond with a proton $\ce{(H+)}$ (or amine as shown in below picture) as a result the N gains a formal positive charge and a positive charge on the proton disappears
Now my question is, in a coordinate compound when $\ce{NH3}$ act as a neutral monodentate ligand i.e. N in $\ce{NH3}$ forms coordinate covalent bond with a central metal atom. Does N gains the formal positive charge? Is there is any change in the oxidation state (or charge) of a central metal atom of the coordination complex?

Formally, yes. But don't read too much into the formal charges. They don't necessarily represent any real charge. There's no change in oxidation state; you already know that from $\ce{NH4+}$. Also, all that $\mathrm{d^2sp^3}$ "inner orbital" stuff is very outdated, and not correct. Learn it if you must, but don't propagate it.

@orthocresol. What do you have against $\ce{d^2sp^3}$ hybridization ? Are you opposed to all hybridizations or only to this one ?

Orthocresol advice of not reading too much into formal charge is sage advice.
Rephrasing your question a bit: When ammonia is coordinated to say Co as in your example, does N of the NH3 gains the formal positive charge?
From the perspective and conventional thinking of inorganic chemistry, the central Co atom is a 3+ and the NH3 ligands are considered neutral.  Formal charge is that, a formalism which tries to describe electron distribution in a molecule.  Nature does not restrict electron distribution in whole numbers [1,2, etc.] associated with the 1- charge of electrons.


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Nomenclature question about diols
I'm struggling to understand why the following organic compound:

is named systemically as 2-ethylhexane-1,3-diol (I think its common name is etohexadiol), if the longest carbon chain in the molecule is actually 7 carbons long, as labelled here:
Wouldn't the compound have a hydroxymethyl group on the 3rd carbon and a hydroxy group on the 4th? Or do I just have some big misconception about naming alcohols (very possible).
Thanks for any help anyone can give!

Welcome to ChemSE. The chain bearing the hydroxyls takes precedent over longest chain.

@user55119 Hi! Thanks for the response. Just to clarify - you name alcohols so that the parent chain is the chain with the most hydroxyl groups on it? And does this sort of rule apply for any other organic compounds? Thank you!

See this: https://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/

I'll be using the brief guide to the nomenclature of Organic Chemistry as a reference, because I could not gain access to the blue book. This source suffices for this nomenclature example.

Seniority of parent compounds
The systematic name is based on the name of the senior parent compound, which is chosen by applying the following criteria in the order described below and shown in Fig. 1, until a decision is reached. For a complete set of criteria see Ref. [8].

In the given example, both the chains contain the senior most group, which is the hydroxyl (alcohol) group. The smaller chain, however, contains 2 hydroxyl groups as compared to the larger chain, which contains one hydroxyl and one alkylhydroxyl group (which is a lower priority). Since this is the first point of difference, The smaller chain would be the parent compound.
After identifying the parent chain, it is easy to name the locants such that the seniormost groups get lower locants; As you have written, the name would finally be 2-ethylhexane-1,3-diol.
This image is possibly more detailed than this question requires, but the image and the linked reference are good resources, should you need more information on naming tricky compounds.

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Mechanism for the Formation of DMC
I have recently reviewed the synthesis of the dehydrating agent DMC (7), but I could find no reported mechanisms for its formation. I have proposed the mechanism drawn below, but I am unsure about a few steps.

To start, It is clear that this reaction is similar to the Vilsmier-Haack reaction, as the product is identical to what would be obtained under Vilsmier-Haack conditions; however, it is obvious that a different mechanism must be followed (the lone pair of the imidazolidinone (1) will not be attacking anything after all). I proposed that traces of HCl in the reaction will result in the formation of intermediate 2. The enol form of this can then attack the phthaloyl chloride (3) as expected to form intermediate 4. Of course, this will collapse to form intermediate 5. With no better nucleophiles in the reaction, the expelled chloride will attack intermediate 5. Nitrogen's lone pair could then kick out a carboxylate as a leaving group (I am not very confident about this step). This would form the desired DMC (7).
Now, I would ask why it was necessary to use phthaloyl chloride as opposed to a simple monoacid chloride. I propose this is because the phthaloyl chloride quickly destroys the nucleophilic carboxylate, which might otherwise react with the DMC to form unwanted byproducts. The resulting phthalic anhydride is not nucleophilic, and in the absence of any traces of water, it will not react with the DMC.

Now, I have 2 questions.

Does my mechanism for the formation seem correct? Only traces of HCl will be present in the reaction, so I am not sure if the concentration of the enol form will be enough to react at a reasonable rate. Also, the loss of carboxylate as a leaving group is a little bit worrisome, as always.

Because this reaction is so similar to the Vilsmier-Haack reaction, would Vilsmier-Haack conditions be a viable way to form the DMC?

Nice job, Eli! You really don't need HCl to effect enolization. 1 and 3 will do the acylation on oxygen. Basically a V-H reaction

@user55119 So, essentially, it could just be one concerted step from intermediate 1 to 4 (i.e. the phthaloyl chloride acts as a proton in a sense). As you said, in this case, no protonation or prior enolization would be required.

Exactly! Of course, phthaloyl chloride probably contains a trace of HCl.

I would say that the enolization is not only unnecessary, but the way it's written here it would impede the reaction. See the formation of the Vilsmeier reagent on Wikipedia link. The amide is nucleophilic because it's neutral and somewhat electron-rich. I might be wrong, but IMO postulating that a protonated, positively charged species attacks a neutral molecule as a nucleophile is a bit counterintuitive.

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Carbocation stability order?
so basically it's between primary benzylic carbocation which is known to be stable (benzlic) and secondary carbocation which is allylic... which is also stablized by resonance, i know that usually benzylic is more stable than allylic but when they are same degree.. here its different...
any help?

The tropylium carbocation (the one on the right) is more stable than the primary benzylic carbocation because it is not only resonance stabilised but also quasi-aromatic (which means that the charge is a part of the aromaticity) which greatly increases the stability of the cation. In contrast, the benzylic carbocation is aromatic but the positive charge is not a part of the aromaticity so it is not stabilised that much.
Hope it helps!


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Volume of compressed helium gas needed for FE test on ball valves
I'm currently doing a case study to find out the exact volume of helium gas required for FE tests on ball valves  regards to the valves' volume and pressure class and the number of helium tanks needed for it (assuming the test is being done at room temperature)
The specifications of helium gas that is being supplied to the test bench is as follow :
Helium 5.5
Purity: 99.9995
Size: X50S
Pressure: 200B

The gas capacity (with regards to the Helium Gas) 8.57NM3
The water capacity. 50L
The dimension of the cylinder tank .1,51M height
The weight of the cylinder tank (with and without the Helium Gas)  Net fill weight 1.51KG. Gross weight vary.
The maximum pressure in which the tank can be operated. Max working pressure 200 Bar
The valve outlet. BS3 Brass

Therefore my question is can the gas law be used to find the volume of helium gas needed for the tests and what are the variables that is needed in finding the volume?
ex: P1 = Valve Maximum Pressure, P2 = Helium Cylinder Tank Pressure, V1 = Valve Internal Volume (has been calculated), V2 = Volume of Helium Gas Needed (required)
Standards for the FE test are:

ISO 15848 Part 1
Shell SPE 77/312

Thank you
Edit:
In simple term, I would like to know what is the volume of helium gas needed to pressurize the valve to reach a certain test pressure?
Given that, the helium cylinder tank has a max working pressure of 200 Bar and a volume of 8.57Nm3. So how much helium gas is needed from this cylinder to pressurize a valve with an internal volume (ex : 0.91594m3) to a test pressure of 19.6 Bar

The volume that the helium gas occupies will be equal to the valve internal volume. Depending on the testing pressures, you will probably need to include the compressibility factor for helium in the ideal gas law calculations. The amount of gas used per fugitive emission test is better tracked by knowing how much mass is needed to get the valve internal volume upto testing pressure.

We can assume helium as an ideal gas, therefore the compressibility factor can be neglected right? 'by knowing how much mass is needed to get the valve internal volume upto testing pressure" can I get more explanation on this.

What is the test pressure for the FE test? You can look up the real density from a source like the NIST Webbook and confirm if Z is approx. to 1 or not at the test pressure. After you have that Z value, you can use PV=nRT to determine the number of moles of helium (and therefore mass of helium) needed to take the valve volume (V) up to the test pressure (P) at the testing temperature.

In simple term, I would like to know what is the volume of helium gas needed to pressurize the valve to reach a certain test pressure?
Here is an example :
Given that, the helium cylinder tank has a max working pressure of 200 Bar and a volume of 8.57Nm3. So how much helium gas is needed from this cylinder to pressurize a valve with an internal volume (ex : 0.91594m3) to a test pressure of 19.6 Bar (z of helium = 1.005)

My second comment tells you what calculation you need to do, with one correction to the equation - PV = ZnRT. But since you're saying the Z value for He at 19.6 barg is 1.005 it won't really matter.

This would be: PV = ZnRT : P = 19.6 barg, V = 0.91549m3, Z=1.005, R=8.31447e-5 m3 bar K−1 mol−1, T = 313K – therefore, n would be 689.83mol (based from the calculation) - from that we could find the mass by multiplying the n with 4g/mol (molar mass of helium). So, we don't really have to take into account the working pressure and the volume of helium tank ?

Correct. If you need to track the volume, you can use the mass you calculated and use the real density of the helium in the state you want to track.


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How to calculate core electron SCF Fock Matrix Elements?
I have searched everywhere but there are no worked examples of how to calculate the $\langle\chi_i |\hat{h}| \chi_j \rangle $ values for the Fock Operator using the SCF HF Roothaan method. I am given a wavefunction that is expanded as a set of basis functions, but I don't know how to apply the second derivative to something like
$$
\phi_0(x) = \sum_{\nu=1}^{4} c_{\nu}e^{-kx^2}
$$
Where we assume this is a system with two (spin-1/2, anti-aligned) particles under a harmonic oscillator potential, giving us a total Hamiltonian
$$
\hat{H} = -\frac{1}{2} \frac{\partial{}}{\partial{x_1^2}} + \frac{1}{2}kx_1^2 -\frac{1}{2} \frac{\partial{}}{\partial{x_2^2}} + \frac{1}{2}kx_2^2
$$
I am capable of calculating the overlap matrix elements for something like this, where I would get a matrix
$$
S =
\begin{bmatrix}
S_{11} & S_{12} & S_{13} & S_{14}\\
S_{21} & S_{22} & S_{23} & S_{24} \\
S_{31} & S_{32} & S_{33} & S_{34} \\
S_{41} & S_{42} & S_{43} & S_{44}
\end{bmatrix}
$$
Where each matrix element is
$$
\langle\chi_i|\chi_j\rangle
$$
Unfortunately, I am stuck when it comes to $\hat{H}|\phi_0\rangle$. I am not sure where the $x_1$ and $x_2$ terms come from in the Hamiltonian. I understand that $\phi_0$ represents the orbital that the electrons occupy, but when expanding the wavefunction into the basis set, how do we get the different $x$ terms?
Sorry if none of this makes sense, this is the product of many online learning deficiencies.

Why is your potential a quadratic?  I think the HF-SCF potential is all coulombic.

@Bertram This is altered st $V(x_1,x_2)$ is in a Harmonic potential

Are you trying to calculate elements of the fock matrix for an electronic structure calculation?  The only time I know of when the SCF method is used with a harmonic potential is for vibrational self-consistent problems.


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Decreasing order of electron density in benzene ring among the halobenzenes
What should be the order of electron density in monosubstituted halogen derivatives of benzene?
I am aware that fluorobenzene is more ring activating as compared to chlorobenzene due to the more effective +M effect of F over Cl, but I would like to know the full order among all the monohalogenated benzene compounds.
Is there any significant +M effect shown by Br, I with the benzene ring?
If yes, then how strong is it as compared to their –I effect?
Please let me know the correct final order of electron density in the benzene ring among Ph—F, Ph—Cl, Ph—Br, Ph—I, as well.

In the case of electrophilic nitration of the halobenzenes the differences between the 4 compounds is less than a factor of 10; if we include benzene in the series, then the differences are up to a factor of 50 - still small.  See this earlier answer for the correct order.  At least in the case of nitration @Adiboy's ordering doesn't match the experimental result.

Firstly, I would like to talk about the order of withdrawing effect among halogens. Since all the four halogens are electron-rich, they tend to donate electrons (+M effect). However, the high electronegativity of these elements hinders the donation by applying an even greater electron-withdrawing effect (-I effect) that overcomes the +M (resonance) effect.
In the case of halogens, the -I effect varies as -$\ce{F}$ > -$\ce{Cl}$ > -$\ce{Br}$ > -$\ce{I}$ due to the decreasing electronegative character as we move from fluorine to iodine.
The +M effect also varies as -$\ce{F}$ > -$\ce{Cl}$ > -$\ce{Br}$ > -$\ce{I}$ due to the decreasing overlap between the 2p orbitals of carbon and the 2p, 3p, 4p and 5p orbitals of the halogens respectively. Since the -I effect is stronger than the +M effect as mentioned earlier, the final order of electron-withdrawing capability is
-$\ce{F}$ > -$\ce{Cl}$ > -$\ce{Br}$ > -$\ce{I}$
Equipped with this knowledge, we come back to the question of comparing the electron densities of mono halogen benzene derivatives. From the above-mentioned trend, we can infer that the order of electron density is
$\ce{Ph-F}$ < $\ce{Ph-Cl}$ < $\ce{Ph-Br}$ < $\ce{Ph-I}$
This trend is validated by the order of reactivity of these compounds towards electrophilic substitution reactions.
Hope it helps!

Interesting. But, do you mean "electrophilic substitution reactions", instead of "nucleophilic", in the last line? Because, I think, an increase in electron density would mean an increase in electrophilic substitution reactions.

Also, due to the absence of an EWG group, nucleophilic aromatic substitution seems unlikely to me.

Yes, you are correct. I have edited my answer. Thanks for pointing out the mistake.

No problem, brother. Also, if possible, could you provide the reference for the order of reactivity of the halobenzenes towards electrophilic aromatic substitution?

Will definitely let you know when I find one


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How to determine what covalent compound a substance is?
I have been given a substance which is unknown in my class. The substance is white in colour and is in a powder form. I have conducted a few tests and have found out that the substance is a covalent compound.
Properties of the substance

Has a low melting point. When the item was melting it left a yellow/orange colour liquid, later it hardened and retained the same yellow/orange colour.
Soluble in water
Does not conduct electricity in solid state
Does not conduct electricity when dissolved in water
pH of 3-4
Soluble in Ethanol
Is not a sugar, used bennedict's solution and did not turn red instead gave of vapour and started bubbling.

From these steps I have concluded that the substance is a covalent compound. However, I need to find exactly what type of elements is in the covalent compound. If you could suggest a test to conduct to figure out what type of elements there are, it would be very helpful.
Thanks.

Is the aqueous solution of it acidic, alkaline or neutral?

I have not tested the PH level yet but if that helps, I will try it out. If the aqueous is acidic then the substance will be acidic, am I correct?

Is  it soluble in ethanol ? in an alcane like gasoline, cyclohexane, petroleum ?

I have only tested the solubility in Water, if ethanol helps, I will try that out too.

Also how would ethanol help in this situation?

I am sure that if this is a class exercise there must be the way to figure out what the sample is. But, as formulated, this question can be answered by a full course in analytical chemistry. Surely not in few lines. Perhaps someone here can guess it (a common exercise, luck, etc) but this indeed requires at least an introduction to organic analysis.Moreover, note that title and text ask for different things.

What is your teacher doing? Before asking questions, or giving out such complicated tasks? Sounds like he wasn't teaching anything.

@Karl Depending on the hidden background, it may be the opposite. The teacher can make advanced students to solve non trivial questions what need a lot of effort to solve it, and the solution may not be even reached. Such students can learn a lot from it.

@Poutnik Yea. Doing that with not so advanced students is however known to be completely counterproductive. Which seems to be the case here, imo.

The problem is huge, because there exist many millions of different compounds which are white, soluble in water, and do not conduct electricity. Here are a couple of preliminary tests.
You might first determine if it is organic, by heating it in a test tube in a Bunsen burner. It will darken and be progressively transformed into charcoal if it is organic.
Is  it soluble in ethanol ? Is it soluble in an alcane like gasoline, cyclohexane, petroleum ?
Does it react with a diluted solution of CuSO4 ? If it produces a deeper blue color, it is probably some amine like RNH2, or a complexant agent like tartaric acid. If it produces a colored precipitate (blue or green), it may be a basic substance or a carboxylic salt.
Does it discolor Bromine water ? If yes, it probably contains a double bond C=C.
Does it react with Fehling's reactant (solution of CuSO4 plus much sodium tartrate and a few NaOH) ? If it does make an orange or brown precipitate at boiling temperature, it is a reducing sugar.
Does it react with diluted solution of permanganate KMnO4 ? Try in acidic and in basic solution. If so there is a reducing agent, and further study must be done.
Does it react with Tollens' reagent (solution of silver nitrate plus ammonia) producing a silver mirror at boiling temperature ? If yes, it is an aldehyde.
Knowing if the substance contains nitrogen is a little bit more difficult, as you should do Lassaigne test : heat it with a little bit metallic sodium in a test tube to high temperature, then dissolving into water with iron compounds. This a bit more dangerous : It requires some skill
There are plenty of other tests that could be made. Solving this problem in a general way would take hours. It is nearly impossible to solve it now. Try to find a treatise of organic qualitative analysis.

My substance is soluble in ethanol, does this mean its organic?

Also, I have burnt the compound in a test tube in a bunsen burner yet the substance is not darkening and it dissolves in Ethanol. When it was burning it also released gas whihc could be smoke or vapor.

Does it react with solutions of CuSO4, or with AgNO3, or with KMnO4 ? Try ! Does it color the flame ?


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Number of stereoisomers of pentane‐2,3,4‐triol

What will be the number of stereoisomers of $\ce{CH3CH(OH)CH(OH)CH(OH)CH3}?$

By formula $2^n - 1 + 2^n + 1/2$ for $n$ being the odd number of chiral centres, it's $6;$ four optically active and two optically inactive.
Optically active  when RRS, SSR, RR, SS, but I am not getting how we get two optically inactive.

Also, this resource and this question may help you out.

The remaining two are optically inactive because each of them is a mirror image of itself; what's unclear about that?

What will be their RS configuration and IUPAC name..?

What will be iupac name of the achiral stereoisomers..?

If you read the Answers in my link, the answers will be there.

So, can we say that it is optically inactive when C1-R,C2-R and C1-S,C2-S?

Same question here

Wrap this one up!

Does this answer your question? Number of stereoisomers and symmetry of pentane-2,3,4-triol

Just ,the question left unanswered is that what will be iupac name of two stereoisomers in which middle carbon is achiral..

Rover: Read the first example here. http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/RSrs.html


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Software for determining symmetry elements and point group of drawn molecules
The title represents my overall question. I would like to have a software that able to draw molecules (especially coordination complexes) and determine the symmetry elements and point group of the molecules. I have a PC with Windows 10 system and an iPad Pro.
Thank you.

Coordination compounds / complexes may be a bit tricky in terms of «what do you define as a bond» (covalent / dative bond) and the convention what a stroke between atoms expresses (e.g. in the representation in compounds like ferrocene).  But if you already have a .mol or .pdb file in available, you could try Jmol:

(credit, includes some details how to use the program this way)
While the example is about methane as a simple organic molecule, Jmol is well capable to interact with crystal structure models, and displaying symmetries of the unit cell, too:

(credit, Hanson, J. Appl. Cryst. 43, 2010, 1250-1260 [open access])
including interactive galleries like Symmetry@Otterbein both about molecular and crystal symmetries.


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Will washing a conical flask with deionised water to remove splashes affect the titration result?
When performing a titration, once the acid/base in the burette is slowly released into the conical flask, it may splash onto the sides of the conical flask. It is then advised that we wash the sides of the conical flask with deionised water.
Will this not affect the concentration of the acid/base solution by decreasing its concentration and therefore it will take less volume from the burette to change the pH indicator?

You have already measured your acid or base into the flask. The measured amount (in moles) is what matters. Adding more water doesn't change it. Well, when you adding some volume other reagent from the burette still diluting the solution in the conical flask, any way!

With a magnetic stirrer, it should be possible to avoid any splashing, if I remember correctly. It's been a long time ... ;)

You equate amounts, not concentration. Is the same reason that in partitioning the sample you don't need exactly, say, 200 ml each time, but you must be sure to take an exact fraction of the solution and that be always the same.

In short: a small quantity of additional water will not alter the quantity (moles) of $\ce{OH-}$ of your lye, or $\ce{H3O+}$ of your acid to be characterized.  But it will dilute the intensity of the colour of your indicator used; this then hampering the visual inspection of your analysis.
True, the addition of additional water to rinse off the drop from the wall increases the total volume of your analyte, and this will lower the concentration of the acid / base to characterize.  Thus, aim to titrate without splashing the reagent solution to the beaker's wall.  It takes some practice to swirl the Erlenmeyer flask such that analyte and reagent solution are mixed with minimal splashing.
The deionized water you already used to dilute your analyte however is neutral; i.e., it neither should be acidic, nor basic.  Overall its use will add as much  hydroxy $\ce{OH-}$ as hydronium $\ce{H3O+}$ ions.  Thus, the consumption of reagent solution is not affected.


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Are the units of mole of oxygen molecules the same with the units of mole of nitrogen molecules?
Suppose we have in a box $3 \,{\rm mol}$ $\ce{O}$ and $3 \,{\rm mol}$ $\ce{N}$. Is the addition of these two quantities meaningfull? We do it all the time when we want to find the molar fraction. I can't understand if it dimensionally correct to add these two things. If a mole is like a dozen then a mole of eggs and a mole buildings can't add up.
I think a crucial step that is omitted when we want to find the molar fraction is that we should convert these moles of arbitrary entities (eggs, buildings, molecules) to same entities. I can't find anything else to justify such a process (addition of different kind of moles). Would be the following way a proper way to find the mole fraction?
$$n_{_{T}}= \frac{\,{\rm mol} \, \text{of molecules}}{\,{\rm mol} \, \text{of O molecules}}\cdot n_{_{O}} + \frac{\,{\rm mol} \, \text{of molecules}}{\,{\rm mol} \, \text{of N molecules}}\cdot n_{_{N}} $$
So $n_{_{T}}= 6\,{\rm mol}$. What about the molar fraction of oxygen? $$\chi_{_{O}}=\frac{n_{_{O}}}{n_{_T}}= \frac{\text{3 mol of O molecules}}{\text{6 mol of  molecules}}=0.5 \text{ of what?}$$
We know the ratio must be dimensionless but why? We deviding moles of $\ce{O}$ molecules by moles of molecules. Are these units different or not?

Are integers expressing  number of apples the same as integers expressing number of pears ? Are metres expressing length of roadd the same as metres expressing height of trees ?

It's a deep question, and perhaps one could say that the "of molecules" and "of O" are merely labels, and not algebraic objects like the actual unit "mol". But in any case, it works. In science, sometimes we must settle for something inexplicably working ("shut up an calculate").

Adding on to what @Poutnik said, when you’re finding mole fraction of oxygen, you’re finding the fraction of apples in a box of apples and pears. For this, you divide the number of apples by the number of apples and pears.

If the box contains 3 mol of O2 and 3 mol of N2, there is 6 mol of diatomic molecules. The oxygen molar fraction is then x_O2 = n_O2 / ( n_O2 + n_N2 ). BTW mol is the symbol of mole, as kg is the symbol of kilogram. The former is often confused. 0.5 of nothing. It is unitless. It is the same like ( 3 kg of apples ) / ( 3 kg of apples + 3 kg pear pears ) gives 0.5 as the mass fraction of apples.

You can add 1 mole eggs and 1 mole buildings and this gives you 2 moles objects. Back to Chemistry, think about perfect gases. As far P V T are concerned, it is perfectly sensible. But even with no ideality at all, you can always count total number of entities and basically define the relative composition. It is just that I don't see anything useful on adding eggs to buildings. But if you find a property of the world that depends on the molar ratio of eggs in city centre, why not?

@Alchimista But first we should convert them in units of objects, right? We can't add two different units. We can't add apple and oranges we can add fruits and fruits. For example suppose we have 5 apples from USA and 3 green apples. We may don't have 8 apples if there is no 1-1 correspondence. I mean maybe the 3 green apples were from USA.

@Poutnik "The former is often confused. 0.5 of nothing. It is unitless." Can you please elaborate a little more in this sentence?

Apples and oranges are both fruits. 3 apples and 4 oranges are 7 pieces of fruit. the molar fraction of apples is 3/7. 1 mol of O2 and 4 mol  of N2 are 5 mol of diatomic molecules. Molar fraction of oxygen is 1/5.

@Anton If there is 20 m high tree and 40 m long road, the tree is 0.5 times shorter than is the road long, 2 such trees would be needed to fit the road length. 0.5 is  unitless number. Quite obvious, is not it ? There is no special "tree metre" and "road metre".

@Anton Others have answered before me. Moreover it might be senseless* adding apple to building, but atoms to atoms why not. *but just because it is good for nothing. No that you really can't. To this I just add that mol is the unit for amount of substance, of course to know more about that substance you then refer to mole fraction of a specified component.

It is a good question which has already been somewhat addressed in the 1880s. This "field" was called quantity calculus. Calculus here is not the integration / differentiation, but rather the Latin calculus implying a method of calculation.
There is a very nice article "Quantity Calculus: Unambiguous Designation of Units in Graphs and Tables" by Mary Anne White in the Journal of Chemical Education. Please read this if you are seriously interested. Search on Google Scholar and it is free to download from there.
In quantity calculus Each physical quantity as the product of a numerical value and a unit:
physical quantity = numerical value × unit
This approach was introduced by British scientists and many leading physicists used it. Now there is there is nothing less or nothing more. Therefore your ambiguity arises from introducing another factor such as "oxygen" or "nitrogen". The unit mol does not know whether it belongs to oxygen or nitrogen.
As explained in the comments, suppose we write L symbolizing the height of a tree, then I can only write, L = 10 m. For mathematical purposes, I will not introduce "tree" anywhere in this equation. The tree is already incorporated in L (in your mind) but not in the mathematical equation. One can also write L/m =10. Now you have a pure number on both sides.
For fun: Think about taking the derivatives or log of quantities with units. What is the unit then?

Another fun is with $\ln{\frac{L_1}{L_2}}=\ln{L_1}  - \ln{L_2}$

Yes, indeed. That proves that L's must be dimensionless numbers as L/m.

@M.Farooq What troubles me is that we can use mole to measure anything. Trees, eggs, buildings etc. But we use mass to measure a specific property. It doesn't matter if it is kilograms from eggs or buildings is still mass. That is what confuses me.

No you shouldn't use moles to measure or count anything. It is good for atomic and sub-atomic scaled objects. We cannot have a mole of trees or eggs on this Earth (big big number)!! Now mol as a unit is fine just like kg. Can we not buy  1 kg of carrots? 1 kg of iron, 1 kg of paper?


text:
Regioselectivity of electrophilic addition of Halogen and Water (Halohydrin) on Alkenes?
I am currently studying for an organic chemistry exam and I just couldn't come up with a solution to the following question:
Consider the reaction of 1-Propene + Br2 with water as solvent.
Why will the nucleophilic water molecule add on C-2 and rather than on C1 of the Bromonium-Ion?
The answer seems to be, because in the transition state the bonding between Br and C will dissociate quicker than the bonding between O and C will form, which leads to a positive charge on that C-Atom in the transition state and therefore water will add on the more stable "carbokation-like" carbon.
My question is: why does the Br-C bonding dissociate quicker than the O-C bonding forms? (I read that in the pearson organic chemistry textbook). Does it have anything to do with the Hammond-postulate? Or are there any other thermodynamic or kinetics reasons? So I am basically searching for an explanation why one bond would dissociate quicker than a different bond would associate, because I guess that's the reason why there is a partial positive charge that would then be more stable on the higher substituted carbon...
Unfortunately I haven't found an answer on the internet yet and also not in other textbooks.
Maybe you can help :) Thanks!

I learnt that halohydrin addition proceeds through a classical carbocation rather than a non-classical one (like this mechanism) and that the carbocation on the disubstituted C atom is more stable than the carbocation on the monosusbtituted C atom, which is why you get a secondary alcohol.

What @AniruddhaDeb said.

Why assume that the bromonium ion is symmetrical with all of the positive charge on the bromine? Why not consider it unsymmetrical with part of the positive charge residing on the secondary carbon.?

@user55119 Yeah, some explain it by electronegativity (inductive effect), but as I mentioned, the textbook explained it differently - that the partial positive charge in the transition state is due to different speed in bonding association and dissociation... which of course leads to different energy levels of the transition state (bc positive charge is more stable on the higher substituted carbon) and therefore the reaction speed is higher.

@AniruddhaDeb well, that would explain where the positive charge comes from... do you have any references (articles, textbook, etc.)?

This is not a 100% answer, but it was too detailed to fit in the comment section.
Following up from the comment, here's a reference from Inorganic Chemistry by Morrison and Boyd:

There is evidence, of a kind we are not prepared to go into here, that these compounds are not formed by addition of preformed hypohalous acid, $\ce{HOX}$, but by reaction of the alkene with, successively, halogen and water.

Halogen adds (Step 1) to form the halonium ion; this then reacts, in part, not with the bromide ion, but with water (Step 2) to yield the protonated alcohol.
Propylene gives the chlorohydrin in which chlorine is attached to the terminal carbon. This is typical behaviour for an unsymmetrical alkene; orientation follows markovnikov's rule, with positive halogen going to the same carbon that the hydrogen of a protic reagent would. Now, this orientation would be perfectly understandable if the intermediate were an open carbocation: the initial addition of halogen yields the more stable carbocation - secondary, in the case of propylene. But the stereochemistry indicates that the intermediate is not an open carbocation, but a cyclic halonium ion.

And that's all. Quite anticlimatic, as I was expecting a reason for the regioselectivity. Most other references and papers I have seen online also do not have a concrete explanation for the regioselectivity, and they resort to hand-wavy explanations involving resonance for the same. This reaction mechanism is also quite similar to Oxymercuration-demercuration, whose regioselectivity is again explained by the "three resonating structures". I could not find diagrams of these resonating structures, and while this question asks the same, the resonance structures or the intermediate are not properly explained in the answer.
My best suggestion for you would be to accept the explanation in your textbook as that is adequate for now. Until someone more knowledgable posts a better answer with references, that should suffice.

Well, thank you for your detailed comment. I really like studying theories and mechanisms in much detail, until I really understand what is going on and why. Unfortunately nobody seems to go into depth why the regioselectivity is the way it is here. Maybe the answer is found in quantum physics and thus not easy enough to be in a basic organic chemistry book, but let's see - maybe someone can come up with another explanation, although I doubt that, since I haven't even found studies/articles about it in chemistry databases...


text:
Significance of Law of reciprocal proportions
The laws of conservation of mass, definite proportions multiple proportions makes sense and were definitely useful for chemists in the early days of modern chemistry.
The law of reciprocal proportions states that

If element A combines with element B and also with C, then, if B and C combine together, the proportion by weight in which they do so will be simply related to the weights of B and C which separately combine with a constant weight of A.

I don't understand how the law of reciprocal proportions is useful. It seems too unnecessarily complex for someone to intuitively come up with.
Is there any intuition/specific reason that Jeremias Richter came up with it?

This law has a historical importance. It was stated before the discovery of the universal notions of valence and of mole. It was important for the scientists around Dalton and Berzelius to realize that the amount of oxygen and nitrogen in various oxides were exactly in the ratio 1.000 to 2, or 1.000 to 1, or 2.000 to 1, and never 1.007 to 2, or 0.982 to 1, or 2.17 to 1.  Today it seems obvious. It was not obvious in the beginning of the 19th century. It was even surprising. But this law is no more of any use today.

I think you are talking about the law of multiple proportions, my question is about a different law

It would help to prevent wrong assumptions and interpretations if you explicitly formulated the law in the question. Omitted question info is usually expensively paid for by much longer clarification chain. Well elaborated questions prevent this.

As an example, 1 gram of sodium (Na = A) is observed to combine with either 1.54 grams of chlorine (Cl = B) or 5.52 grams of iodine (I = C). (These ratios correspond to the modern formulas NaCl and NaI). The ratio of these two weights is 5.52/1.54 = 3.58. It is also observed that 1 gram of chlorine reacts with 1.19 g of iodine. This ratio of 1.19 obeys the law because it is a simple fraction (1/3) of 3.58. (This is because it corresponds to the formula ICl3, which is one known compound of iodine and chlorine.) Similarly, hydrogen, carbon, and oxygen follow the law of reciprocal proportions.

These laws were originally empirical rules based on observations, as the discrete atomic matter theory and molar masses were to come yet. Berzelius: When two substances, A and B have an affinity for two others, C and D, the ratio of the quantities C and D which saturate the same amount of A is the same as that between the quantities C and D which saturate the same amount of B.

OK. I understand. I will try to rewrite Richter's law in the case of A = H, B = Oxygen and C = Carbon. It gives :  $4 g$ $\ce{H}$ reacts with $12 g$ $\ce{C}$ to produce methane. The same amount $4 g$ H reacts with $2·16 = 32 g$ Oxygen to produce water. So Richter's law states that if $\ce{C}$ reacts with $\ce{O}$, this reaction will involve $12 g$$\ce{ C}$ with $32 g \ce{O}$, or a multiple of $32 g$ $\ce{O}$. Apparently this law was not obvious in the 19th century.

Oh, so the law would help in predicting how 2 elements combine if we know how they both combine with a 3rd element

After defining the law of definite proportions and law of multiple proportions, we could successfully understand how two elements combine together.
The question arose: What about if we took more than two elements?. So, J. Richter came up with the law of reciprocal proportions, which is infact intuitive, I guess you are understanding it unintuitively yourself. It states

If two elements combine with a third element, then the ratio of masses of the two elements for unit mass of third element, is a simple ratio of whole numbers.

It is not that complicated; it is just like the law of multiple proportions i.e.,

If two elements combine in more than one way, the the ratio of mass of either element for unit mass of other element, is a simple ratio of whole numbers.

Now, you see the law of reciprocal proportions is nothing but kinda extended version of law of multiple proportions for three elements.
Hope this is clear :)

The law states that the ratio of the 2 elements in the compound which they make is a equal or a simple multiple of the ratio with which they combine with the unit 3rd element, but I think you have made me understand how it works

Yeah, I realised your statement was unnecessarily complicated, and I got what was unclear to you. So, posted this. :)

The comments certainly clarify the law of reciprocal proportions and are an excellent tutorial on how to write a good question but sheds little light on your question “Is there any intuition/specific reason that Jeremias Richter came up with it?”
From reading a number of short bibliographic sketches about Richter it appears his primary interest was mathematics.  The title of his doctoral thesis was, roughly translated, The use of Mathematics in Chemistry  [https://www.encyclopedia.com/science/dictionaries-thesauruses-pictures-and-press-releases/richter-jeremias-benjamin].  His study of quantitative relationships in the neutralization of acids by bases lead to the law of equivalents. [https://encyclopedia2.thefreedictionary.com/Jeremias+Benjamin+Richter ]
Based on these and similar biographical notes one might argue that he arrived at the law of reciprocal proportions not through “intuition” at all but rather through an empirically based and numbers driven approach.  When a mathematician sees a pattern or consistency in “the numbers” they then might be seek to understand the natural forces giving rise to their observations.
A modern and well written example of “math driving science” can be found here. https://www.quantamagazine.org/geometry-reveals-how-the-world-is-assembled-from-cubes-20201119/ “Scientists Uncover the Universal Geometry of Geology”

just to confirm, you mean 'law of reciprocal proportions right? not multiple proportions .

Thank you for your careful reading and catching this, yes I was focusing on the question of reciprocal proportions and will edit my answer to make that clear.


text:
Deriving an expression from Navier-Stokes under creeping flow
So I'm trying to derive
$$\mu\nabla^2\vec{v}=\nabla P-\rho g$$
from Navier-Stokes equation
$$\begin{align}\rho \left(\frac{\partial\vec{v}}{\partial t}+\vec{v}\cdot(\nabla\vec{v})\right)&=\rho\vec{g}-\nabla P+\mu\nabla^2\vec{v}\\
\rho \left(\frac{\partial\vec{v}}{\partial t}+\vec{v}\cdot(\nabla\vec{v})\right)-\mu\nabla^2\vec{v}&=\rho\vec{g}-\nabla P
\end{align}$$
under creeping flow.
I began with looking at the left-hand side of Navier-Stokes equation, which I expanded to:
$$\rho \left(\frac{\partial\vec{v}}{\partial t}+v_x\frac{\partial\vec{v}}{\partial x}+v_y\frac{\partial\vec{v}}{\partial y}+v_z\frac{\partial\vec{v}}{\partial z}\right)-\mu\left(\frac{\partial^2\vec{v}}{\partial x^2}+\frac{\partial^2\vec{v}}{\partial y^2}+\frac{\partial^2\vec{v}}{\partial z^2} \right)$$
Because of the creeping flow I new that $v_x\approx v_y\approx v_z\approx0$. Because of this I can therefore write the left-hand side as:
$$\rho \frac{\partial\vec{v}}{\partial t}-\mu\left(\frac{\partial^2\vec{v}}{\partial x^2}+\frac{\partial^2\vec{v}}{\partial y^2}+\frac{\partial^2\vec{v}}{\partial z^2} \right)$$
But from here I am stuck. I asked a friend who said that since $v_x\approx v_y\approx v_z\approx0$ we can also remove the $\frac{\partial\vec{v}}{\partial t}$-term but I find this mathematically incorrect since the derivate of a function isn't necessarily small just because the function is small. As an example the function
$$\vec{v}(t)=((t+1000)^2-10^6)\vec{e_x}+((t+100)^3-10^6)\vec{e_y}+((t+50)^2-2500)\vec{e_z}$$ is small when $t=0$ but that isn't true for the derivate in $t=0$.
Is it something I'm missing when it comes to creeping flow? Is it that $v_x$, $v_y$ and $v_z$ are constantly small and therefore we can se them as constant functions and therefore the derivate is also neglectable?

Wow. Rheology etc. are usually fine subjects for chem.SE, but this one might be better in physics or engineering.  Anyway under creep flow, all velocities should be constant in time. Re is small, and everything should be in steady state.


text:
Why is the amino acid cysteine classified as polar?
Cysteine amino acid has an embedded sulfur group in its side chain. Looking at the electronegativity difference of hydrogen and sulfur, it can be considered a non-polar side chain because the electronegativity difference is less than 0.5. This makes it non-polar but still we put it in polar category, why?

https://chemistry.stackexchange.com/questions/121570/comparing-polarity-of-amino-acids

[...] we still put it in polar category

If you search for amino acid classification, you will find that there is no agreement on how to classify cysteine.
What you should know is that cysteine has different roles in proteins. In cysteine proteases, it acts as a nucleophile (and its surrounding often makes it more prone to deprotonation compared to a netural aqueous environment). Cysteines that form disulfide bonds often are at or near the surface of the protein; once the disulfide is formed, you have a quite hydrophobic group. Protonated cysteine is incapable of making conventional hydrogen bonds, and the electronegativity of carbon and sulfur are quite similar. This explains why methionine, the other sulfur-containing amino acid, is classified as hydrophobic.

[...] , why?

The classification of amino acids is sometimes used as proxy of the tendency of amino acid side chains to be located on the surface or in the interior of a protein. Similarly confusing, proline is sometimes classified as polar amino acid; however, its side chain is made exclusively of carbon and hydrogen atoms. If you study where proline is mostly located with a folded protein, you find it is on the surface. This is not because proline is polar but because it is a helix and strand "breaker", so it is often found in turns, which are mostly on the surface of the protein.
Also, you should know that there is no all-or-nothing classification of every amino acid side chain. Tyrosine, threonine, even lysine have hydrophobic parts while being capable of making hydrogen bonds. Often, the environment is just right, with hydrophobic parts interacting with other hydrophobic parts, and all hydrogen bond potential satisfied, either by interactions with other side chains, with main chain or with solvent.
Why is cysteine not classified in a more consistent manner?
Here is an excellent starting point to answer this question, starting with the abstract:

Cysteine (Cys) is an enigmatic amino acid residue. Although one of the least abundant, it often occurs in functional sites of proteins. Whereas free Cys is a polar amino acid, Cys in proteins is often buried and its classification on the hydrophobicity scale is ambiguous. We hypothesized that deviation of Cys residues from the properties of a free amino acid is due to their reactivity and addressed this possibility by examining Cys in large protein structure datasets. Compared to other amino acids, Cys was characterized by the most extreme conservation pattern, with the majority of Cys being either highly conserved or poorly conserved. In addition, clustering of Cys with another Cys residue was associated with high conservation, whereas exposure of Cys on protein surface with low conservation. Moreover, although clustered Cys behaved as polar residues, isolated Cys was the most buried residue of all, in disagreement with known physico-chemical properties of Cys. Thus, anomalous hydrophobic behavior and conservation pattern of Cys can be explained by elimination, during evolution, of isolated Cys from protein surface and clustering of other Cys residues. These findings indicate that Cys abundance is governed by Cys function in protein rather than by the sheer chemical and physical properties of the free amino acid, and suggest that high tendency of Cys to be functionally active can considerably limit its abundance on protein surface.

Sir Karsten Theis, first I say thanks to you for your answer.. but if we consider cysteine polar just because its side chain ionize in aqueous medium than another question arises and that is.. The S-H bond in cysteine is non polar because the electronegativity difference between H and S is less than 0.5 which make the bond non polar so how this group ionizes in water which is a polar solvent?.. why hydrogen gives its own electron to sulfur which is not as powerful as oxygen or fluorine?

See https://chemistry.stackexchange.com/q/128046

And https://chemistry.stackexchange.com/a/27815


text:
Oxygen generator, heat driven, from air
What chemical can be used to absorb $\ce{O2}$ from the air at normal temperature and pressure, then release the $\ce{O2}$ at normal pressure and elevated temperature, below $\pu{900 ^\circ C}$ ($\pu{1650 ^\circ F}$), that can be cycled many times?
Closest solutions I've found are silver, but its quite expensive, and absorbs oxygen slowly at normal temperature. And possibly phosphorous, but it is toxic and needs a solution.
Same question about $\ce{N2}$. I'm interested in separating $\ce{N2}$ and $\ce{O2}$ without the use of single-use chemicals or pressure pumps. Using cycled heat as a source to drive the reaction.
I expect there is some oxide that can switch oxidation levels back and forth with temperature, like hematite and magnetite.
Idea solution will be something like $\ce{CaCO3}$ is for $\ce{CO2}$. A cheap material, that quickly reacts with $\ce{CO2}$ at normal pressure and temperature. Quickly releases the $\ce{CO2}$ at elevated temperature and normal pressure.
I did check the search, it seems people usually offer to use electrolysis for this task. But water is way too costly to electrolyze in terms of power. Is there something that can be electrolyzed with $10$ times less energy, to get oxygen, that can be safely oxidized again later? Hydrogen is rather dangerous too.

Hydrogen is valueable, oxygen not so much. Separating air via pressure-swing absorption/desorption is simple and cheap.

Search for molecular sieve oxygen separation

The only substance that absorbs $\ce{O2}$ to form an oxide at low temperature and release it at higher temperature is mercury. This experience was done by Lavoisier, Scheele and Priestley in the years around $1780$. They were able to produce pure oxygen $\ce{O2}$ by first heating pure mercury in air at about $300$°C to slowly produce mercury oxide $\ce{HgO}$, and then overheating this oxide $\ce{HgO}$ to about $400$°C. At this temperature, $\ce{HgO}$ is decomposed into mercury and pure oxygen $\ce{O2}$. This was the way oxygen was first discovered and produced.
Unfortunately, mercury ¡s unique in this respect, and cannot be replaced by another element. No metals have been found later on to have a similar property. It is a pity : presently this experiment cannot be repeated, because of the toxicity of mercury and its compounds.
This why your dream will remain at the state of dream.

Thank you, at least I can stop the search and look for other options

Who says it has to be an element?

Perhaps employing iodine in the place of mercury may work.
Some background on the proposed chemistry starting with Wikipedia on Iodic acid, to quote:

Iodic acid, HIO3. It is a white water-soluble solid...Iodic acid can be produced by oxidizing iodine I2 with strong oxidizers such as...or hydrogen peroxide H2O2,[1]...When heated, samples dehydrate to give iodine pentoxide. On further heating, the iodine pentoxide further decomposes, giving a mix of iodine, oxygen and lower oxides of iodine.

Now, the action of $\ce{I2}$ on water slowly forms $\ce{HI}$, $\ce{HOI}$ and $\ce{HIO3}$:
$\ce{I2 (s) + H2O (l) -> HI (aq) + HOI (aq) }$
$\ce{3 HOI (aq) -> 2 HI (aq) + HIO3 (aq) }$
However, apparently more rapidly, per Wikipedia on Iodine to quote:

Iodic acid is most easily made by oxidation of an aqueous iodine suspension by electrolysis or fuming nitric acid.

So to summarize the proposed chemistry per above,  the presence of oxygen from the electrolysis of water acting on a suspension of iodine forms, in part,  Iodic acid. The created $\ce{HIO3}$ can be isolated or converted into iodate followed by its thermal decomposed to liberate oxygen, iodine and iodine oxide in a partial recycling.
For those interested in the particulars of the underlying chemistry, one possible explanation is the presence of electricity (a source of solvated electrons) and oxygen could result in the formation of the superoxide radical anion per the reactions:
$\ce{e- + nH2O -> e- (aq)}$
$\ce{O2 (d) + e- (aq) -> .O2- (aq)  }$
A fast reaction of the superoxide with the $\ce{HOI}$ intermediate may result in Singlet oxygen (and the hydroxyl radical).  This is speculated from the reported reaction of hypochlorous acid with superoxide, namely:
$\ce{ .O2- + HOCl -> ^1O_2 + .OH + Cl-  }$
Singlet oxygen directly or indirectly, via its known reaction with water forming $\ce{H2O2}$ (which was also noted above to act on an iodine suspension), could explain the conversion of an aqueous suspension of iodine to $\ce{HIO3}$.
Note: If my depiction of the chemistry is correct, no need for an external electric source to introduce the needed solvated electrons. Just construct an electrochemical cell (similar to the so-called Bleach Battery) composed here of an aqueous iodine suspension with an Aluminum anode and say a copper cathode. Pass air into the cell to convert the iodine suspension to iodate, thereby capturing the oxygen for future use.
Caution: if Aluminum Iodate Hexahydrate (referred to as AIH) is possibly created in this electrochemical cell synthesis, it is considered a strong oxidizer (even as a hydrate), so mixed with the wrong compounds (say carbon from employing a graphite cathode) and heated may display notably energetics.
[EDIT] Interesting related article: High power rechargeable magnesium/iodine battery chemistry.
Also, a related galvanic cell synthesis, albeit, with advanced electrode design in this work: Energy-saving synthesis of potassium iodate via electrolysis of potassium iodine and O2 in a membraneless cell.

The Q scenario is the input are air oxygen and heat. The electricity may be consider as the heat equivalent, but the air oxygen stays. I'm interested in separating N2 and O2 without the use of single-use chemicals or pressure pumps. Using cycled heat as a source to drive the reaction.

Poutnik: In my since suggested expanded embodiment (galvanic cell with I2 suspension) air in and iodate out. Subsequent heating liberating the captured O2, and also I2 and iodine oxides (both added to water reform a suiting starting solution acting to act on the pure metal Aluminum rode functioning as the anode). Practically, a thick Al rode will last. Heating is actually mild to liberate O2 from the iodate (perhaps solar-based).

Is not alkaline pH needed to shift equilibrium to iodide and iodate ? Depending on application, there may be issue with smell of iodine vapor traces. The whole system seems to me quite complicated  Molecular sieves are widely used already.

Poutnik: As I am assuming parallels to the Bleach battery, there are two versions, one based on NaOCl and the other HOCl. I have experimented with both and my observations suggest both work well.

what core reaction is used for liberating oxygen? If we use decomposition of I2O5 into I2 and O2 at 200*C, we will loose iodine as vapor, its hard to keep it, similar to phosphorous. If we use water solutions of HIO3, we cant get oxygen out without spending quite a lot of energy for electrolysis, or a lot of metal for a battery cell. Assuming we want 1k cycles, thats many kg of aluminium, if aluminium is spent every cycle.

Solid I2 on heating sublimes and in immediate contact with a cool surface deposits solid iodine. Now, technically, heating HIO3>I2O5>I2...is within the limitations imposed per the question, however, somewhat acceptable to others may be the room temperature reaction of CO on I2O5 releasing elemental iodine! See https://en.wikipedia.org/wiki/Iodine_pentoxide. Generating CO (via several paths, including incomplete combustion) in the same vessel as the iodine salt can likely be performed safely. Again, the top of the vessel is cooled to collect the pure Iodine on warming.


text:
Why does B12H12 Borane not exist?
Whenever I searched "$\ce{B12H12}$ Borane", I can't seem to find anything, and get met with only the dianion of $\ce{B12H12}$. Why doesn't $\ce{B12H12}$ exist on its own?

The general formulae of boranes are $\ce{B_nH_{n+4}}$ and $\ce{B_nH_{n+6}}$.

Apparently polymeric BH exists - perhaps  B12H12 could be an intermediate that further polymerises. It's just too unsaturated.

Hypercloso boranes are generally unstable. However, its derivates are known. For mre details see this: https://onlinelibrary.wiley.com/doi/10.1002/1521-3773(20010504)40:9%3C1664::AID-ANIE16640%3E3.0.CO;2-O

Also see this: https://chemistry.stackexchange.com/questions/39647/why-do-closo-boranes-have-a-2%e2%88%92-charge


text:
Material Properties Dealing with "Wetting"
I am designing a rain device, and I am interested in a set of common metrics / standards for determining how something gets "wet".
(1) Is there a "water absorbance" metric for materials (ie will the water be deflected or actually penetrate the surface)
(2) Ability for the material to dry (I know this is measured in time, but is there a standard for the drying conditions)
If you know of any other standards that may be of interest please let me know.

Search contact angle measurements for surfaces.

The contact angle deals with hydrophily/hydrophoby. This is a sort of skin properties. But from the text It might be that you also look for something actual, like the weight of water absorbed for a given surface/volume.


text:
Why diethyl ether does not absorb in UV region?

Explain why diethyl ether does not have a UV spectrum, even though  it has lone-pair electrons.

Can we say that π* molecular orbital does not exist because there are no π-bonded electrons in diethyl ether?

Its uv spectrum is most likely due to non bonding lone pair to $\sigma^$ or $sigma\to \sigma^$ orbitals which may be too far into the uv (<215 nm for ether) to be measured properly on a typical spectrophotometer.

@porphyrin Ad hoc find on pages 61 and 62 of this presentation for cooled solid forms of diethyl ether: http://home.strw.leidenuniv.nl/~ewine/photo/workshops/2015-02/collected_presentations/VUV%20Leiden.pdf

@Buttonwood, this should have been a v. interesting talk. The effects of the solid state vs. gas phase are understandable in general terms but being specific for any individual case must be hard.

@porphyrin The setup of the substrate (close to Knudsen cell like deposit of a substrate on a sample holder cooled well below the melting point of substrate) and subsequent (cycles of) annealing by controlled heating reminded me on impedance / broadband dielectric spectroscopy in soft matter research about relaxation processes in solid matter / supercooled liquids.  Maybe the VUV spectroscopists deposit on comb electrodes anyway to record complementary data on the other side of the electromagnetic spectrum, too ...

We can measure the UV-Vis spectra for π → π* or n → π* electronic transitions (other transitions are forbidden according to selection rule) [1, pp. 577–578]. But in diethyl ether, being a saturated compound, it has no π-bonds. This successfully dismisses the idea of having π* orbital. And, as there is no π* orbital, we can not have a transition to the π* orbital resulting in no UV-Vis absorption spectra.
Reference

Pavia, D. L.; Lampman, G. M.; Kriz, G. S.; Vyvyan, J. R. Introduction to Spectroscopy, 5th ed.; Cengage Learning: Stamford, CT, 2015. ISBN 978-1-285-46012-3. (PDF)


text:
Order of Lewis acid character among sulfur hexafluoride, sulfur tetrafluoride and tetrachloromethane
Problem

Compare the order of Lewis acid character of $\ce{SF6},$ $\ce{SF4}$ and $\ce{CCl4}.$

My attempt
Oxidation state of $\ce{S}$ in $\ce{SF6}$ is $+6,$ so it is very electrophilic.
$\ce{S}$ has a lone pair on it in $\ce{SF4},$ but can expand its octet if attacked by a donor.
$\ce{C}$ cannot accept lone pair from donor.
So, the order should be $\ce{SF6} > \ce{SF4} > \ce{CCl4},$ but that is a wrong answer.

S wrapped by 6 F has not much  sterical chance to reach for some external e pair donor. It is almost inert.

S is in its highest oxidation state (+6) it has no electrophilicity hence not a Lewis acid.
Also, CCl4 can't have more than 8 electrons in its outermost orbit and it can not also increase it's coordination number also so it can't behave as a Lewis acid.
SF4 can accept electrons as S isn't in its highest oxidation state.
Hence order of Lewis acid character is : SF4 > CCl4 = SF6


text:
Why hydrogen gas tube produce hydrogen atomic spectrum
To measure the atomic hydrogen spectrum people often uses hydrogen gas tubes as light source.
Since the gas in the tube is the molecule $H_2$ , why we obtain the spectrum of atomic hydrogen?
My guess is that because the voltage is so high, so that the molecules are totally dissociated.
If you could give me some reference I would be grateful.

Think  thoroughly about  possible answers to your questions and search for them before asking.
Provide in the question elaboration what you already know, what you have thought about,
what you tried to search for in textbooks and online resources,
what you found, understood or not, where you failed.
Elaborated questions attract elaborated answers and vice versa.
Include in  the question all eventually relevant circumstances to prevent wrong assumptions and requests for clarifications.
Questions without  explicit own effort  are not very welcome, and may be closed.

Consider the effect of accelerated ions and relative intensity of atomic and molecular spectrum.

Of course, $\ce{H2}$ is dissociated into $2$ $\ce{ H}$ atoms, but this dissociation is not total. It is not necessary. Please think before asking !

why is not necessary?

@amiltonmoreira, if you have 10^9 molecules and 1 ion, does that prevent the ion from emitting (or absorbing) radiation??

@DrMoishePippik, I guess you mean atoms not ions (with respect to hydrogen discharge tube).

@DrMoishePippik if there is more molecules of hydrogen there is more probability to get the hydrogen molecule spectrum

The answer is what I posted here: https://physics.stackexchange.com/a/768678/313612. The high voltage results in high energy electrons colliding with the hydrogen molecules, resulting in some of the molecules dissociating into hydrogen atoms. So the discharge has hydrogen atoms, both excited and ground state, plus hydrogen molecules, both excited and ground state, plus various ions and, of course, electrons. Anything excited might emit its characteristic light as it de-excites. De-excitation might just result in heat. Please upvote my answer if it helps you! Thanks for your consideration!

One more thing: in my linked answer, the Fulcher alpha bands are from the hydrogen molecules. So you definitely get both atomic emission, from the excited hydrogen atoms, and molecular emission, from the excited hydrogen molecules.


text:
FFC Cambridge Process and impure reactants
The FFC Cambridge process with platinium electrodes allows the reduction of almost any metal oxide to a pure metal, with an oxygen byproduct.
So far, so good.  What I'm trying to understand is how it would handle mixtures of of materials.  For example, if it was a mixture of iron oide and titanium oxide, what would it do?  Which oxide would be reduced preferentially?  (I think in reverse order of reactivity?) If it was contaminated with calcium carbonate, would the chloride ions switch places with the carbonate?
Ultimately this is a question about how electroylsis works with complex solutions - so what would it do?

Where do you see chlorine ions ?

FFC process > next step is that the pellets are placed into a crucible or reduction vessel that's filled with very pure molten salt normally this will be calcium chloride.

Correct, should be chloride, fixed

Research Papers
The FFC Cambridge process and its relevance to valorisation of ilmenite and titanium-rich slag
G. Z. Chen
Pages 96-105 | Received 05 Mar 2014, Accepted 08 Aug 2014, Published online: 19 Nov 2014


text:
What's the reason that alkanolamines change their color at storage?
What's the reason that alkanolamines (e.g. monoethanolamine, diethanolamine, triethanolamine) of technical purity change their color (from colorless over yellowish to red-brown) at storage?
Is it because of oxidation, or UV-light?

Oxidative degradation by oxygen certainly occurs and has been studied 1, 2. I can't access the papers so I have no information on possible mechanisms. I've seen no mention of UV degradation, but that doesn't mean it does not happen.


text:
Equivalence point of titration of Sodium carbonate
The pH at the equivalence point of titration of $\ce{Na2CO3}$ solution with $\ce{HCl}$ is around 3.7, as shown in this titration curve:

At this point, the following reaction is completed:
$$\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}$$
But considering the fact that $\ce{NaCl}$ is a neutral salt, shouldn't the pH be closer to 7? The $\ce{H2CO3}$ formed also decomposes to form $\ce{H2O + CO2}$, and shouldn't affect the pH either. Why then is the equivalence point at such a low pH value?

Contrary to what you say, CO2 in water does affect the pH. You need a little bit more HCl to obtain the equivalence point.

@Maurice, Wouldn't the equivalence point be the point where the solution has been completely neutralized, and thus would only contain $\ce{NaCl + H2CO3}$? Also doesn't the $\ce{CO2}$ formed escape, and not cause a significant decrease in pH?

The equivalence point is not the point where the solution contains $\ce{NaCl}$ and that's all. No. The equivalent point is when the solution contains $\ce{NaCl}$ plus the small amount of $\ce{CO2}$ which remains dissolved. And this amount makes the solution a bit acidic. So the pH at the equivalence point is not $7$, but a bit less.

@Maurice, You are mixing the concept of "end-point" and "equivalence point". At the end-point there is slight excess of the acid or base. At the equivalence pt, stoichiometric amounts have reacted.

@M. Farooq. OK. You are right. I should have rewritten my text. But this does not change too much my message.

When the $\ce{Na2CO3}$ is completely neutralized by the $\ce{HCl}$, the solution will be saturated with $\ce{CO2}$, so the pH will be lower than 7 - think of carbonated beverages. Although the $\ce{CO2}$ bubbles out, not all of it bubbles out - not even if stirred. If you had stopped the titration at pH = 7, the pH reduction from 11 would have been partly from $\ce{HCl}$ and partly from $\ce{H2CO3}$.

Okay, so the low pH at the equivalence point can be attributed to just the dissolved $\ce{CO2}$, correct?

Yes. You might expect that CO2 is all expelled, but some just loves to stay in solution.

If you have a 0.1M NaHCO3 solution with Bromophenol blue indicator (pH range 3.1-4.6) titrated with (0.1M HCl) you will get an end point (the point in a titration at which the colour of the solution shows a pronounced change) of c.80-85% of the theoretical titer or equivalence point (the point in a titration at which the amount of titrant added is chemically equivalent to the amount of sample to be titrated),  ie the blue akaline solution will turn yellow more quickly than anticipated because the carbonic acid has conributed to an increase of H3O+ ions.Leaving the solution for 1-2 hours, will make it turn from yellow to green ie more alkaline as the CO2 leaves the solution under partial pressure and henrys Law.This can be shown quickly by using a water aspirator and a bung to create a vacuum in the titration vessel. Remember carbonated water is always more acidic,than distilled water which has been left to absorbed atmospheric CO2.The early end point will also be determind by how much CO2 is inside your water to start which is used for your titration solutions although this is far less than the carbonic acid contribution. The CO2 and hence the the HCO3- conc will be higher ....it is acting as a buffer and until all the excess CO2 (aq) and CO2(g) has equilibrated out of the solution into the air, the solution will remain more acidic.


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Meaning of the Term "Heavy Metals" in CofA?
I've always wondered, what is the meaning of the following terms on a certificate of analysis:

"Heavy Metals"
"Heavy Metals (as Pb)"

What exactly is comprised in "heavy metals", "heavy metals (as Pb)" and what is excluded from each? What is the difference between the two variants? I tried to find a clear standard describing this to no avail.

To someone who studied chemistry 60 years ago this is not terribly ambiguous -- "heavy metals" are metals that are heavy.  Lead, mercury, bismuth, and cadmium are prime examples, and a number of radioactive elements (such as uranium) are also considered "heavy".  While some of the (lighter) heavy metals -- iron, cobolt, and zinc, eg -- are important nutrients, many (like lead) are toxic to humans.  But note that the precise criteria vary, depending on the reason for the classification, so the term has apparently lost favor.

Related: https://chemistry.stackexchange.com/questions/65495/how-heavy-should-an-element-be-to-be-a-heavy-metal?r=SearchResults

You are right heavy metals (as Pb) is a vague term. It is being discouraged and individual analysis should be reported. When someone says "I tried to find a clear standard describing this to no avail", it is a good illustration that not all information is on the webpages. It is worthwhile to search books :-)
I used to think this was an old-school analytical terminology and it turns out to be correct because it is based on classical methods. From the Food Chemicals Codex (ISBN 0-309-08866-6), which is a 1000 paged book, has the following information. Later they describe the tests as well.
Heavy Metals (as Pb)

This test is designed to limit the content of common metallic
impurities colored by sulfide ion (Ag, As, Bi, Cd, Cu, Hg, Pb, Sb, Sn)
by comparing the color with a standard containing lead
(Pb) ion under the specified test conditions. It demonstrates that the
test substance is not grossly contaminated by such heavy metals, and
within the precision of the test, that it does not exceed the Heavy
Metals limit given as determined by concomitant visual comparison with
a control solution. In the specified pH range, the optimum
concentration of lead (Pb) ion for matching purposes by this method is
20 micrograms in 50 mL of solution. The most common limitation of the
Heavy Metals Test is that the color the sulfide ion produces in the
Sample Solution depends on the metals present and may not match the
color in the Lead Solution used for matching purposes. Lead sulfide is
brown, as are Ag, Bi, Cu, Hg, and Sn sulfides. While it is possible
that ions not mentioned here may also yield nonmatching colors, among
the nine common metallic impurities listed above, the sulfides with
different colors are those of As and Cd, which are yellow, and that of
Sb, which is orange.
If a yellow or orange color is observed, the following action is
indicated: Because this monograph does not include an arsenic
requirement, As should be determined. Any As found should not exceed 3
mg/kg. If these criteria are met, Cd may be a contributor to the
yellow color, so the Cd content should be determined. If an orange
color is observed, the Sb content should be determined. These
additional tests are in accord with the section on Trace Impurities in
the General Provisions of this book, as follows: ‘‘if other possible
impurities may be present, additional tests may be required, and
should be applied, as necessary, by the manufacturer, vendor, or user
to demonstrate that the substance is suitable for its intended
application.’’

This is good information regarding the "Heavy Metals (as Pb)" variant. Did you find a standard for the more simple "Heavy Metals" variant? Thanks a lot for your contribution!

Heavy metals means heavy metal (as used in typical sense) and it is individualized if it is written on a CoA. The manufacturer has to specify which heavy metal they are talking about. Heavy metals as Pb has a specific meaning as stated above and this term is being discouraged.

By experience, except for a very few, manufacturers don't spontaneously specify on a Certificate of Analysis which "heavy metals" they are talking about. Most would answer if one took contact with them for that purpose but it would be useful to know which are these heavy metals in a typical sense.

Does your book say which metals are supposed to be considered heavy metals in a typical sense?

"The Committee on Food
Chemicals Codex notes the importance of providing limits
for individual heavy metals as required by the source and
composition of individual food additives. Thus, it has decided
to remove from most monographs the general heavy metals
(as lead) limits and tests and, based on the current level and
availability of scientific information and on the policy stated
below, to replace them with limits and tests for specific heavy
metals such as lead, cadmium, and mercury as may be relevant
to each substance"

If I understand well, good manufacturers should provide limits for individual heavy metals and avoid using the term "heavy metal" alltogether?

Yes. They should adopt this recommended practice.

In addition to @M.Farooq answer on the food sector, if you read heavy metals as Pb in a pharmaceutical environment, it stems from a method of the United States Pharmacopoeia, active until 2018 (same for the European, Japanese and British Pharmacopoeia as well as others).
Up until then, it was normal to analyse heavy metals as a sum parameter for many drug products. The analysis was a limit test against a lead nitrate standard, where you compare the colour.
From the USP chapter 231 of  USP 39-NF 34 valid until 1. January 2018 :

This test is provided to demonstrate that the content of metallic impurities that are coloured by sulphide ion, under the specified test conditions, does not exceed the Heavy metals limit specified in the individual monograph in percentage (by weight) of lead in the test substance, as determined by concomitant visual comparison with a control prepared from a Standard Lead Solution. [NOTE—Substances that typically will respond to this test are lead, mercury, bismuth, arsenic antimony, tin, cadmium, silver, copper, and molybdenum.]

In addition for a lot of pharmaceutical products, special analysis for metals used in the manufacturing process were described in their monographs, such as limits of Arsenic, Iron, Lead and others.
Here is an article and a graph of the sensitivy of this method, which was quite bad.
It was a worldwide act and discussion, until finally in 2018 the International Council for Harmonisation (ICH) guidance for industry Q3D Elemental Impurities(ICH Q3D) was set in place  and adopted by the regional pharmacopoeias, see FDA USP .
For the European part, here  is a short "historic overview" from the European Agency.

Elemental impurities in the Ph. Eur.
The Stone age : from genesis to 2008

use of non specific wet chemical test for « heavy metals »
limit to 10 or 20 ppm > (with reference to lead)

The Bronze age : from 2008 to 2013

EMA GL on the specification limits for residues of metal catalysts or metal reagents

The Golden age : starting 2013

CHMP decision to delay the application of the EMA guideline to existing marketed products
discussions on the implementation of ICH Q3D guideline

Then the chapter <231>  was deleted form the USP and the chapters  <232> Elemental Impurities—Limits and <233> Elemental Impurities—Procedures were released.
Now, elemental impurities are analysed by ICP or AAS on an risk based approach, mostly by studying, which elements are used in the manufacturing process, as part of the packaging, colouring etc and can leach into the drug, and then analysing the content.
So, in the pharmaceutical industry, heavy metals as Pb was a specific method for the analysis of heavy metals via colour comparison test. Until 2018, pharmaceutical products could be released with a Heav metal test based on a 1800 or 1900 visual colour comaprison test which excluded a lot of metals.  Heavy metals "without Pb" did exist informal as a short version, for metals screenings and to talk about outdated methods.

Thanks for providing the USP perspective. Could you clarify what is "anium hydroxide" in the first quoted paragraph. Is it ammonium hydroxide?

@M.Farooq Copy past error. I only found a pdf online, with 2 columns and it copied both in a mix up. It`s deleted.


text:
Representing conformers using SMILES and Graph representations
I was wondering if either of SMILES or graph representations (in their vanilla form) can distinguish between two conformers of the same molecule.

Can this be enhanced by adding additional metadata on Graph nodes?
what info are necessary then?
Can this be enhanced for SMILES representations?

For the SMILES strings as defined by Daylight, you can expect these representations tell about the molecular constitution plus information about ($E$/$Z$)- and CIP's $(R)$/$(S)$-configuration. But nothing about the conformation like the boat / twist form of cyclohexane which generally is dynamic, or a faithful preservation of information like s-cis to describe the conformation of butadiene favorable for the Diels-Alder cycloaddition reaction.
Further, I speculate SMILES strings are not a faithful representation of locked conformations as in the axial $P/M$ configuration of 2,2',6,6'-substituted 1,1'-biphenyls. However, perhaps someone else may clarify this point.


text:
Why are hydrogen bonds in an antiparallel beta sheet stronger than those in parallel beta sheets?

Beta sheets are illustrated as such in most diagrams, where:

In an antiparallel β-sheet, the polypeptide strands are arranged such that a $\ce{C=O}$ and an $\ce{NH}$ from adjacent strands face each other, and the $\ce{H}$ forms an H-bond with the $\ce{O}$, with $\ce{C=O**HN}$ all lying on the same plane.
In a parallel β-sheet, the strands are literally parallel, and so the H-bond formed has to form a sort of angle, where the $\ce{C=O}$ and $\ce{NH}$ no longer lie on the same plane (when viewed from the perspective of the diagram).

Most explanations explain the stronger bond strength by stating the obvious from the diagram: that the "hydrogen bonds are aligned directly opposite each other, making for stronger and more stable bonds." (taken from chem libretexts, https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Protein_Structure/Secondary_Structure%3A_-Pleated_Sheet)
This seems intuitive, but my issue arises when considering some orbital geometry: in a peptide bond, the $\ce{C}$, $\ce{N}$ and $\ce{O}$ have orbitals with ${sp^2}$ hybridization, meaning that bond pairs/lone pairs are organized in a trigonal planar orientation around each of those atoms. As a consequence of this, one can also observe that all the atoms illustrated in the diagram should lie on the same plane, making things rather easy to visualize.
However, if one considers that the 2 lone pairs and one bond pair in the $\ce{O}$ atoms are orientated in a trigonal planar fashion, then shouldn't the ideal, strongest, H-bond is orientated at an angle (specifically $\pu{{60}^\circ}$) relative to the $\ce{C=O}$ bond? Now, if I observe the diagram again, it seems this $\pu{{60}^\circ}$ H-bond angle is better fulfilled in the parallel strand than in the antiparallel one. So shouldn't the H-bond in the parallel strand, theoretically, be stronger?
A possible explanation would be that the bond length in parallel strands would be longer, thereby weakening it. But that's not how most textbooks explain it, so what am I missing out here?

in a peptide bond, the C, N and O are all sp2 hybridized, That is not correct. A simplistic view propagated by organic chemistry that reverses the actual cause. Hybridisation is a mathematical description. The only thing we can really observe is that C, N, O are coplanar (not a surprise though). Atoms are never hybridised, only orbitals may be described this way. Terminal atoms are usually best described with sp hybrid orbitals. Linear coordination of hydrogen in hydrogen bonds is the strongest overlap.

Why are hydrogen bonds in an antiparallel beta sheet stronger than those in parallel beta sheets?

They are probably not. The difference is small, and depends on sequence context. Also, the diagrams do not reflect the typical conformation of the backbone in beta sheets. To complicate matters, most sheets are mixed rather than purely parallel or anti-parallel, and many are irregular (different length of strands, bulges, one strand part of two sheets etc).

"hydrogen bonds are aligned directly opposite each other, making for stronger and more stable bonds."

No, the geometry around the hydrogen bond is fairly similar. Here is a depiction of structures from an ab initio study showing the geometry:

[OP ..] shouldn't the ideal, strongest, H-bond is orientated at an angle (specifically 60∘) relative to the C=O bond?

Based on analyzing known structures with a single hydrogen bond for a given acceptor, the hydrogen lies in the plane of the carbonyl group but has a fairly smooth distribution with many examples having a 180 degree angle (carbonyl carbon, carbonyl oxygen, hydrogen).

A possible explanation would be that the bond length in parallel strands would be longer, thereby weakening it. But that's not how most textbooks explain it, so what am I missing out here?

The ab initio study talks about the primary effect (having a hydrogen bond), secondary effects (repulsion between partial charges of oxygen atoms, repulsion between partial charge of hydrogen atoms, C-H...O hydrogen bond in the parallel sheet) and tertiary effects (induced dipoles due to formation of hydrogen bonds). You have to consider all of these; as the differences in hydrogen bonding geometry are subtle, the secondary and tertiary effects will become significant.
Response to comments

[OP in comments] Is there some possibility that this concept is an oversimplification to explain some other biochemistry phenomena?

The question whether parallel or antiparallel sheets are more stable is complex and ill-defined. The two possibilities require very different connections between the strands, so it is impossible to separate questions of secondary structure from questions of super-secondard and tertiary structure.
In a paper from 2011, the authors Tsutsumi and Otaki state:

We found that anti-parallel sheets occupied 61.0% of the β-sheet samples, and parallel sheets occupied only 14.9%. The mixed sheets were
24.2% of the β-sheet samples. Using a Boltzman-kind of argument interpreting this small difference (4-fold preference) in terms of an energy, you would calculate an energy difference of 3 kJ/mol or so for the entire sheet. This preference, however, might easily be due to questions of folding or the stability of the parts of the structure connecting the sheets, or some other factor. Proteins do not evolve to be of highest stability, anyway, so it is not clear how function plays into the selected structures.

One system where you avoid some of the complexity is the aggregation of particles with beta strands exposed to solvent. For example, there is a study on amyloid aggregation which shows that changing the side chains can drive either parallel or antiparallel arrangement. In another system far way from sheets in globular proteins, cyclic peptides aggregated in a parallel or antiparallel fashion.
So the problem is ill-defined and conclusions might depend on the specific system under investigation. The preponderance of some sheet topologies over others in globular proteins of known structure noted by Jane Richardson in 1977 might be a result not of differences in stability but rather aspects of super-secondary and tertiary structure. So it is not clear whether a general statement about distinct stabilities has merit; the simple explanation given in textbooks certainly does not.

Thanks for the insight. A slightly more empirical study (https://pubmed.ncbi.nlm.nih.gov/9263854/) seems to agree that there is not much difference. At the same time, I wonder why the contrary is often taught in textbooks. Are there other sources of primary evidence that instead support the opposite? If not, why is it "conventional knowledge" that H bond strength is higher in antiparallel beta sheets? Thanks

Some statements seem to make a lot of sense even if there is no evidence to support them; once they make their way into a textbook with a nice "just-so" figure, they spread. I have contacted libretext to inquire about the cited section.@Heat

Thanks @Karsten Theis. While I do not dispute that the H-bond strengths of ABB and PBBs are probably very similar (I have just accepted your helpful answer), I now wonder if it is as simple as a case where a nice "just-so" figure has somehow propagated. I ask because this seems to be a very widespread concept and if I recall I also saw it in the very reputable text "Molecular Biology of the Gene", which would make for a very egregious error. Is there some possibility that this concept is an oversimplification to explain some other biochemistry phenomena? Would appreciate any insight, thanks.

***Important note: I no longer have the text. Nevertheless, I do believe I am correct in saying it is a very widely taught concept

@Heat There is a Jane Richardson paper on how common strand arragements are in proteins, and a Chou and Sheraga paper claiming a difference in stability using modelling. Overall, the literature acknowledges that sheets are twisted, and does not make arguments based purely on hydrogen bonding geometry. The issue is much more complex than asking e.g. whether a cis or trans conformation is more stable because there is no good system to compare the two states directly.

@Heat I edited my answer to reflect how subtle the question is, and how you can use different evidence to argue either way.

Backup to pdf I think this would also be very interesting to be looked at with QTAIM, or generally in the context of NCI.


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Correlating specific heat (Cp) of a substance with other substance on the basis of similar physio-chemical properties
I am modelling $\ce{H2O2}$ decomposition in ANSYS Chemkin-Pro and I have difficulty finding $C_p$ (specific heat at constant pressure) for liquid $\ce{H2O2}$ for various temperature ranges.
My ultimate aim is to find the 14 NASA polynomial coefficients so that I can input them in Chemkin-Pro. For that I need a range of $C_p$ vs $T$ data, so that I can fit the data and find the coefficients.
So I was wondering whether we can correlate specific heat of $\ce{H2O2(l)}$ with some other species on basis of same physio-chemical properties?
If yes, then I can try finding $C_p$ vs $T$ data for those other species. I know that this will cause some error but that is better than having no data at all.
I don't have a background in physical chemistry, so if anybody can help, it will be really helpful for me.


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Interplanar distance given Miller indice of the planes
Say we've a cubic crystal of unity edge length.
A set of planes in such a crystal is specified by their miller indice as $(3 2 0)$. One of these planes then has intercepts on the axes as $(2 ,3 ,\infty)$.
It's well known that the interplanar distance between two adjacent planes in a cubic lattice is given
$d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$,in our case this gives us
$d_{320} = 1/\sqrt13$.
Now if I look at the planes which form the Miller indices of $ 3 2 0$ I find that apart from a plane having intercepts as $(2 ,3,0)$ an adjacent parallel plane to this plane having the same indices will pass via origin $0,0,0$ and the distance between them is
$d_{320} = 6/\sqrt13$
Which contradicts the above result!
Can anyone please point out what's wrong in my reasoning?

Your adjacent plane isn't.

Worth a thousand words, they say.


text:
Lost track of the hydrogen in step 3 of the Krebs cycle
I'm trying to follow the Krebs cycle step by step, accounting for the number of hydrogens in
each molecule before and after reactions.
I've run into a problem starting from step 3.

Isocitrate has five hydrogens.
It undergoes oxidisation to Alpha-Ketoglutarate, with NAD+ being reduced to NADH/H+.
I would say this process entails the loss of two hydrogens from Isocitrate, and the gain of those hydrogens in NADH/H+.
What I don't understand (and this lack of understanding cascades through steps 4 and 5) is that Alpha Ketoglutarate has four hydrogens.
My arithmetic says that there should be only three.
Where does the fourth hydrogen come from?
Note:
If I start from  Step 6, my accounting method works, all the way round to Step 3.
Step 6: Succinate (four hydrogens) is oxidised to Fumarate (two hydrogens), reducing FAD to FADH2.
Step 7: Fumarate (two hydrogens) is reduced to Malate (four hydrogens), by the addition of H2O, gaining
two hydrogens.
Step 8: Malate (four hydrogens) is oxidised to Oxaloacetate (two hydrogens), reducing NAD+ to NADH/H.
Etc.
I came across a similar question at https://biology.stackexchange.com/questions/93775/what-exactly-happens-to-hydrogen-atoms-in-step-4-of-citric-acid-cycle but as this was about step 4, and for me the problem occurs one step earlier.

If you draw isocitrate with the carboxylate protonated, no more problem. The issue is that the carboxylic acid is more acidic than the CH2 group on aKG

I think you're saying that I should add one H+ to one of the COO- groups. Could you confirm this? If so, I find it confusing. Why wouldn't the diagram, which is used on many Kreb's cycle tutorials, not show this extra H+?

In biochemistry, many reactions are written without balancing them. The same can happen for ATP hydrolysis. If you write down the citric acid cycle reaction as is, not only is there a "missing" H, but the charges of reactants and products are different as well. These reactions happen in aqueous solution, and you can always add an H+ to either side, or a water to balance.

Thank you, now it's a whole lot clearer.

Your problem stems from the statement
"I would say this process entails the loss of two hydrogens from Isocitrate, and the gain of those hydrogens in NADH/H+."
This statement is incorrect. In isocitrate, the carboxylic acid functional groups are deprotonated, as would be the case at neutral pH. The oxidation reaction thus involves loss of only $\ce{H-}$ to NAD. There is no loss of $\ce{H+}$, because that has already been removed.
If you start with isocitric acid, which is isocitrate in its neutral form with $\ce{-CO2H}$ instead of $\ce{-CO2-}$, then you will need to remove $\ce{H-}$ and $\ce{H+}$ to accomplish the reaction and your above statement will be correct.
The textbook image you showed takes a confusing approach of implicitly associating an $\ce{H+}$ with isocitrate even though it is not actually on the molecule, so that "NADH/H+" is indicated as a product even though only NADH has actually been produced in the reaction shown.


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How does the cation radius affect the rate of increase of the electrical conductivity of a salt in aqueous solution?
Potassium chloride has a greater ionic radius than sodium chloride. How does this affect the  fall in rate of increase in conductivity as concentration concentrations? I know the following factors affect the plateauing of the curve:
lesser dissociation, the equilibrium is disturbed as a result of lack of water molecules to fully solvate the ions, ionic pair association and the increased visocity of the solution
In my experiment I found that the decraase of the rate of increase was greater for KCl, which I found weird.

Ionic radius is not the same as the effective hydrated ionic radius. Their orders for Li Na K are mutually reversed.

It rather affect ion mobility and therefore molar conductivity. Smaller ions have stronger field and therefore thicker effective AVG layer of attached  water molecules bound to it and following its electromigration.

Molar conductivity is better parameter than conductivity. $\Lambda = \kappa/c$ BTW I was addressing differences between ions, not concentration effect .

Let us continue this discussion in chat.

@Strawhat It would be nice if you could stop vandalizing your own posts.


text:
Pressurized Line and Rigid Tank Question
Using a pressurized line, a fluid is charged in a rigid tank with volume
$2m^3$. Derive the expression and calculate:

The internal energy $u_t$ of the fluid in that tank supposing the initial mass is
, and specific enthalpy $h_i=1670kJ/kg

Assuming a fluid is an ideal gas, derive and calculate the tank temperature using the specific heat ratio of k = 1.67 and the initial temperature of $T_i=400 ^{\circ}C$

Assuming a fluid obeys ideal gas law and a flow is choked derive and calculate the tank pressure as a function of time using $\dot{m}=0.5kg/s,t=2s,k=1.67, T_i=400^{\circ}C$.

Assume that charging occurs over a short period of time.
Starting with the control volume energy rate balance,
$$\frac{dE_{cv}}{dt}=\dot{Q}_{cv}-\dot{W}_{cv}+\sum_{i}\dot{m}_i(h_i+\frac{V_i^2}{2}+gz_i)-\sum_{e}\dot{m}_e(h_e+\frac{V_e^2}{2}+gz_e)$$
we have, after ignoring kinetic energy and potential energy effects, taking into account that no mass is exiting the tank, and also assuming the tank is rigid and insulated, we have
$$\frac{dU_{cv}}{dt}=\dot{Q}_{cv}-\dot{W}_{cv}+\sum_{i}\dot{m}_ih_i-\sum_{e}\dot{m}_eh_e$$
Upon integration and solving numerically, we have
$$\int_{0}^{t}\frac{dU_{cv}}{dt}dt=h_i\int_{0}^{t}\frac{dm}{dt}dt\Rightarrow U_{cv}=mu=mh_i\Rightarrow h_i=u=1670\frac{kJ}{kg}$$
For part 2, using the ideal gas model we have
$$u(T)-u(T_{i})=c_v(T-T_i)=RT_i$$
$$h(T)-h(T_i)=c_p(T-T_i)=RT_i+R(T-T_i)$$
I'm not sure where to go from here.