text:
What is the effect of NaNO2/HCl at 0-5° C on Nitro group?
In this particular reaction firstly $\ce{NaNO2/HCl}$ at $\pu{0-5°C}$ was taken and after that $\ce{H2/Ni}$ .
But what I doubt is, was there any role of $\ce{NaNO2/HCl}$ at $\pu{0-5°C}$ ?
Since only reduction occurred which can also be done by $\ce{H2/Ni}$

The only way this works is if the reduction is carried out in the presence of HNO2. Looks like whoever wrote this reversed the stages.

@Waylander this was asked in a national exam

HNO2 will have no effect on the starting material. Either the reaction scheme is incorrect, or it is poorly expressed and is meant to depict hydrogenation over Raney Nickel IN THE PRESENCE OF HNO2

@Waylander so what is effect of this reduction being carried out in presence of HNO2 ?? Will it go as usual or will have any other effect

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.

Related: https://chemistry.stackexchange.com/questions/76469/why-should-the-temperature-be-maintained-at-0-5-c-in-a-diazotisation/76550 ... https://chemistry.stackexchange.com/questions/18463/effect-of-temperature-on-benzenediazonium-during-diazotization

@MridulKumar if the reaction scheme is correct then the HNO2 has had no effect.

@MridulKumar  indeed if the diazonium species were formed I rather suspect it would be reduced by H2/Ni


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How to establish trend of carboxylic acid acidity in the presence of certain substituents?
I encountered a question in my textbook which asked the following:

At first, i thought that i can approach it in the same way i sort the effect of EDG and EWG on the acidity of phenols, that is, methoxide makes phenols a weaker acid, while halides make them stronger. But I was surprised to learn that its not the case for carboxylic acids. After conducting few research, I learnt that for methoxides, there is some sort of competition between the inductive and resonance effect, whereby resonance is what reduces the acidity by stabilising the conjugate base, and induction does the opposite. I also learnt that when both effects act at the same time, resonance always wins, that is, it's effect is stronger. I thought that in carboxylic acids, the methoxide cannot contribute to the resonance of the carboxylate ion, and its lone pair remains on the methoxide oxygen, and accordingly, the 2-methoxypropanoic acid is more acidic than propanoic acid due to the inductive effect of oxygen (high E.N.).
Sounds fine so far, right? Not until when i decided to test this hypothesis by comparing the pKa values of the following:

A website i found offered pKa values for them, which appeared to be:
Compound 1 (2-chloropropanoic acid) = 2.96
Compound 2 (2-floropropanoic acid) = 2.68
Compound 3 (2-methoxypropanoic acid) = 3.59
Compound 4 (2-aminopropanoic acid) = 2.34
And those numbers do not confirm my hypothesis at all. I expected the order of acidity to be:
2-floropropanoic acid>2-methoxypropanoic acid>2-chloropropanoic acid>2-aminopropanoic acid.
Due to their respective E.N. values.
How can i establish a trend of acidity here? Please help me understand this.

Hyperconjugative effects seem to be at play here, although it still doesn't explain the relatively high acidity of the last species.

The title hardly corresponds to the issue you get to somewhere in the end.

@Mithoron fair enough, will edit it now, thanks for pointing out

Related: Naming rules and acidity of Cl2CHCH2COOH

The source that you cited says that the given $ \mathrm{p}K_\mathrm{a}$ values are predicted.

source: Stack Exchange

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

sort: votes


text:
Electrical conductivity of a NaOH solution
We are doing a simple water electrolysis experiment but I have issues with the electrolyte solution.
Took 450ml of distilled water and slowly dissolved 50g of NaOH until heating ceased. If I'm not wrong that should be a 10% solution (in weight) and about 2.5M.
From this resource the electrical conductivity should be about 309 mS/cm, so 3.2 Ohm×cm; this seems consistent given that electrolysis uses a lot of current.
However with the highly non-scientific method of dipping the DVM probes in the solution I see more than 20 kOhm resistance, which is nowhere near what I'm expecting (sorry, no specialized equipment there). And, of course, the experiment fails (no significant current drawn)
Am I doing something wrong?
I fear that the sodium hydroxide is degraded (it's hardware store grade flakes, supposedly 98%), given that the bottle was not sealed (only a child proof cap, but not airtight). I know that it absorbs CO2 from the air but I've no idea of how much can influence it.

Note that DVM is Digital Voltmeter.
It may use too low DC voltage, what is OK for linear electronic components, but completely misleading for electrochemistry scenarios.
It may show high resistance after very short transient period, as after initial potential re-balancing, there is nothing to carry the current, if voltage is too low to perform electrolysis.
Better is to use AC several kHz with suitable amplitude.
Even with high enough DC, the I/U ratio is not equal to AC based conductance, because of electrode potential difference and possibly other boundary potential drops.

Uhm OK so the issue is that a DC current is stuck if it doesn't reach the cell potential; I'll try with an AC measurement then, thanks

Note that to get conductivity, you have to compare conductance, using identical geometry, with conductance of solution of known conductivity, typically KCl.

For posterity: it was a bad/expired batch of NaOH.
Got a fresh supply from a reputable supplier and works correctly!

Was it just the sodium hydroxide? Did you also adopt Poutnik's suggestions?

NaOH converted by air exposure to Na2CO3 would have just decreased conductivity, what itself would not cause such low measured resistance. Electrode related potential drops must be involved.

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


text:
Smarts string for querying two N substitutions at any place along bicyclic ring
I'm a newbie to SMARTs string representation and have been stuck on generating the structure that I want. I have a pattern of chemical scaffold of interest which I'd use as a query to screen a library. The pattern is a bicyclic ring (with a functional group attached) with single N substitution at every valid and possible location (the first row of attached fig) AND a bicyclic ring with two N substitutions at every valid/possible location. (the second row of attached fig)

I've been searching through the bottom of the Internet and couldn't figure out how to write this in SMARTS string. (since Smarts is a query language, I figured I'd do there first.
Alternative, is to generate SMILES string and run the substructure search independently. However, I haven't figure out a way to generate SMILES without me having to manually writing them down. If you have any pointers - please let me know!

Umm, you know two first ones are the same?

Hi @ Mithoron - Yes I'm aware of that.  The figure was just generated on the fly to depict what I was trying to do. Essentially I want to move N along the ring

You cannot search for a substructure without defining it exactly. Whether it is SMARTS or SMILES, you must define each substructure individually.


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Pressure and temperature relation of Novec 1230
I am working on a concept turbine for a school project which operates at low temperature so it's necessary for me to have a pressure temperature relation of this chemical or novec 7100 also is okay..  i looked it on the internet couldn't find the graph or any tables so if anyone is aware about this please help me out.. like what would the pressures be at 100 °C for 1 mole given a volume of 1m³

Since you are asking "what would the pressures be at 100 °C for 1 mole" (a question that doesn't make much sense), I guess you might want to read about the difference between intensive and extensive properties again.

Novec™ 1230 is basically just 1,1,1,2,2,4,5,5,5-nonafluoro-4-(trifluoromethyl)pentan-3-one. The pressure–temperature relation is given by the manufacturer as a graph:

Novec™ 7100 is a mixture of 1,1,1,2,3,3-hexafluoro-3-methoxy-2-(trifluoromethyl)propane and 1,1,1,2,2,3,3,4,4-nonafluoro-4-methoxybutane. The pressure–temperature relation is given by the manufacturer as an unfortunate value-equation:
$$\ln p=22.415-3641.9\left[1/(T+273)\right]$$
where
$p$ is pressure in pascal and
$T$ is temperature in degrees Celsius.


text:
Do diatomic isoelectronic molecules have the same bond order even when the octet rule is not followed?
It can be observed and proven that diatomic isoelectronic molecules have the same bond order when they follow the octet rule (upto 3rd period elements). But what about when they do not follow the octet rule like in transition metals? Do they still have the same bond order? If so, why?

That's not true even for CO vs N2, at least if you use computational methods.

Isoelectronic molecules have the same everything, except nucleus charge. Think of it this way: what is a bond made of? Electrons.

Do you know many diatomic molecules that do not follow the octet rule , apart from $\ce{NO}$ ?

@Mithoron Can you elaborate a little or tell me where to start searching? Do you mean that the electronic configurations of CO and N2 are different as given by molecular orbital theory?

@IvanNeretin Yeah, that's why it makes sense when they follow the octet rule, because all of them have the same number of shared electrons as the goal. But when we do not follow the octet rule, the stabilities of the specific elements are what matters right?

@Maurice I found CoO, but I haven't found any isoelectronic pairs as you said. So they don't exist? If they do, they are exceptions that don't need to follow this rule about bond order right?

@Balu. $\ce{CoO}$ is not a molecule. It is a solid made of piles of cobalt(II) and oxide ions.

@Maurice Oh, does it qualify as an ionic compound though?

There is no bond order in a pile of ions.

@Maurice Geez, could you get a bit more advanced here, not just repeat lame high school ideas? I estimate CoO has about the same covalent and ionic bonding component. Also in gas phase, especially extremely diluted, you can have, like, any molecule. In comparison with HeH+, CoO is no biggie at all.

https://en.wikipedia.org/wiki/Carbon_monoxide#Bonding_and_dipole_moment

$\ce{CoO}$ is a cobalt oxide. It has nothing to do with $\ce{CO}$. Nothing ! And $\ce{CoO}$ cannot be obtained as a gas. It is decomposed without melting or boiling when heated at 1800°C


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Why is p-dichlorobenzene the major product of chlorination of chlorobenzene?
The chlorine atom in a chlorobenzene has an ortho- and para- directing effect on any further substitution by an electrophile. Going by probability ortho- position has 2/3 chance and para- position has 1/3 chance of getting substituted by 2nd atom of chlorine. Then why substitution at para- position is preferred?

Symmetry is the key. p-Dichlorobenzene has more symmetric crystalline structure than o-dichlorobenzene, hence chlorination of chlorobenzene favors the para product. But do note that at high temperatures, the reaction favors the meta product1.

The chlorination of chlorobenzene vapour in the presence of pumice as a contact substance has been studied at 500°, 550° and 600°. The reaction proceeds violently; even when a large excess of chlorobenzene is used, carbon is formed, especially in the experiments at 550° and at 600°. Besides dichlorobenzenes a considerable quantity of higher chlorinated benzenes is formed; the yields of dichlorobenzenes are low. The mixture of dichlorobenzenes has been isolated from the reaction product and the proportions of isomerides has been determined by thermal analysis. The main product (50%-60%) of the mixture of dichlorobenzenes formed within the range of 500–600°, is meta-dichlorobenzene.
Although the chlorination of chlorobenzene at high temperatures is not suitable for an exact quantitative investigation, it is proved in this paper, that this substitution reaction belongs to the meta type at 500–600° in the presence of pumice as a contact substance.

Although the standard reaction using $\ce{FeCl3/Cl2}$ as catalysts leads to polychlorination and leaving the reaction pot for too long leads to complete chlorination to dodecachlorocyclohexane, selective catalyst is sometimes employed to yield a specific isomer as major product. Serguchev et. al. employed antimony pentafluoride-sulfur system as catalyst3.
References

Wibaut, J.P., van de Lande, L.M.F. and Wallagh, G. (1937), On the chlorination of chlorobenzene in the gaseous phase at 500-600°; The meta directing influence of the chlorine atom. Recl. Trav. Chim. Pays-Bas, 56: 65-70. DOI: 10.1002/recl.19370560109
Aver'yanov, V A. Competing chlorination of 1,1-dichloroethane and chlorobenzene. United States: N. p., 1988. Web. Link: https://www.osti.gov/biblio/6007897-competing-chlorination-dichloroethane-chlorobenzene
Serguchev, Y.A., Chernobaev, I.I. Active intermediates for the selective chlorination of chlorobenzene in the antimony pentachloride–sulfur-containing compound catalytic system. Theor Exp Chem 46, 102–106 (2010). DOI: 10.1007/s11237-010-9125-2
https://patents.google.com/patent/US3636171A/en
Why can't the electrophile attack the chlorine atom in a halobenzene?

Thanks @Nilay Ghosh. Information on further chlorination of chlorobenzene favouring meta- position AT HIGHER TEMPERATURE is a bonus to what I sought originally. Your suggestion that AT ROOM TEMPERATURE , further chlorination of chlorobenzene at para- position being due to greater stability of its CRYSTAL leads me to assume that , with BP of 323 k, para dichlorobenzene gets precipitated down in the solution, thus due to Le Chatelier's principle, as para product is precipitated, more and more para product is formed

removal of a product affects the forward reaction only if the reaction is at equilibrium and will not affect a kinetically controlled reaction. The favoring of para over ortho is probably a combo of steric effects and inductive effects in the transition state


text:
How to name this organic compound?
How would you name this compound if ethoxy ethane is disubstituted with hydroxy group on both alpha carbons? The formula of the compound is:
$\ce{H3C-CH(OH)-O-CH(OH)-CH3}$
Is this compound stable at room temperature or not?

The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.

Not a stable species

It looks like an ether of ethan-1,1-diol $\ce{CH3CH(OH)2}$ , which is the hydrated form of ethanal $\ce{CH3CHO}$. I don't think it is possible to make this ether, which could be called di(1-hydroxy)ethyl ether.

For the application of the nomenclature rules, it generally doesn't matter if the compound exists as long as it has consistent and well-understood bonding patterns and can be interpreted with a simple application of valence rules.
Using the usual straightforward nomenclature rules, you should see that the parent structure of the compound that is given in the question is ethanol. Using substitutive nomenclature, the complete name of the compound is 1-(1-hydroxyethoxy)ethan-1-ol.
In this case, however, a different rule is relevant. Since there are two identical parent structures (the two ethanol units), which are not alkanes, multiplicative nomenclature is preferred to substitutive nomenclature.

P-51.3.1 Preferred IUPAC multiplicative names
For a multiplicative name to be categorized as an IUPAC preferred name, certain restrictive conditions must be met. Multiplicative nomenclature is preferred to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical parent structures, other than alkanes when
(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical and
(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.

Therefore, the identical parent structures are expressed by the multiplicative name 1,1'-…di(ethan-1-ol).
The prefix oxy is now used for the multiplicative substituent group $\ce{{}-O-{}}$. So the complete multiplicative name is 1,1'-oxydi(ethan-1-ol).


text:
Does solid manganese reacts with atmospheric air when heated?
As discussed in my answer, manganese is not particularly reactive to air. The surface of manganese lumps oxidize to a minor extent but when it is finely divided, it becomes pyrophoric and burns in air. It tends to form the oxide $\ce{Mn3O4}$ and the nitride $\ce{Mn3N2}$. It was fine until I came across this statement from Concise Inorganic Chemistry by J.D.LEE:

The finely divided metal is pyrophoric in air, but the massive metal does not react until heated. When strongly heated, the massive metal reacts with many non-metals such as $\ce{O2, N2, Cl2}$ and $\ce{F2}$, forming $\ce{Mn3O4, Mn3N2, MnCl2}$ and a mixture of $\ce{MnF2}$ and $\ce{MnF3}$

I was under the impression that manganese is relatively inert and normally won't oxidize unless it is powdered but the book says otherwise. Nowhere, I found any source backing the statement.
Question: Does solid manganese reacts with atmospheric air when heated (if so, at what temparture)?

If something is pyrophoric when powdered, it does react when in bulk too, but may react slowly, in limited extent and just in thin surface layer. Like Al, or Fe(in dry air).

The tricky part is, what exactly do you mean by "reacts", kinetic-wise.


text:
Does manganese reacts with water at "not" standard conditions?
In my answer, I stated that manganese is said to not react with water under normal conditions although some sources say it reacts with water to liberate hydrogen gas. Does it implies that it reacts with water at harsh conditions?
WebElements and ChemLibreTexts states that "Manganese does not react with water under normal conditions" whereas Lenntech says "it reacts with water (it rusts like iron) and dissolves in dilute acids." What does rusting here mean? Does manganese rust like iron?
In some sources and Concise Inorganic Chemistry by J.D.LEE states "At room temperature very slowly reacts with water"
$$\ce{Mn + 2H2O -> Mn(OH)2 + H2}$$
and some says that it reacts and oxidizes all the way to permanganate. The results are inconsistent and sometimes contradictory.
Question: Does manganese reacts with water? If so, in what condition (standard or harsh)?

Many metals react with steam, which may be considered a reaction with water at harsh conditions. Manganese is among them.

The Lenntech phrasing "reacts with water (it rusts like iron)" is incomplete. Wikipedia states that the reaction occurs with "water containing dissolved oxygen". This would actually be like iron, which also requires dissolved oxygen to react in water. With either metal the rusting reaction involves the metal being oxidized to ions while water and oxygen are reduced to hydroxide ions, leading to the formation of $\ce{Mn(OH)2}$ or $\ce{Fe(OH)2}$; but these hydroxides are prone to additional oxidation by oxygen or air to form oxides or oxide-hydroxides with the metal in higher oxidation states.
The above mechanism requires dissolved oxygen as one of the reactants to form the hydroxide and oxide ions; neither metal appreciably displaces hydrogen from otherwise pure water at room temperature. However manganese, again like iron, will react with steam to form oxide(s) plus hydrogen.

Dissolved O2 oxidizes Fe+2 to Fe+3. It is not necessary for the reaction of Fe with water to give ferrous hydroxide. Water is reduced to H2 not to hydroxide ions. the latter have no change in oxidation state.

My interpretation is that water and oxygen are both reactants in the half equation $\ce{2 H2O + O2 +4e^- -> 4 OH^-}$. Rust formation mechanisms I have seen must include ferrous hydroxide be assembled iron(II) and hydroxide ions are formed to begin the process. See for instance https://en.wikipedia.org/wiki/Rust.

thank you for the excellent question. It was really thought provoking indeed. Here's the answer :

Yes, it (Manganese) reacts with water.
As Concise Inorganic Chemistry by J.D.LEE states it reacts with water in STANDARD conditions (if given enough time; which basically means the rate of the reaction is super slow)
As it will react in standard conditions, it might also react in harsh conditions (ie. 100°C or 50°C). "Harsh" conditions weren't properly defined in the question. So, I am unsure about what you meant there.
"Rusts like iron" means it oxidizes like iron. However, from the research done by Mingfei Zhou, Luning Zhang, Limin Shao, Wenning Wang, Kangnian Fan, Qizong Qin, it seems like Manganese Hydroxide is created upon reaction and not Manganese oxide as suggested by "rusts like iron".

Reaction of manganese with water
The interaction of manganese with water can be described by the following reaction equation: $$\ce{Mn(s) + 2H2O(l) -> Mn(OH)2 (s) + H2 (g)}$$
Most likely, the reaction is slowed down due to the poor solubility of the resulting manganese (II) hydroxide which coats the surface of the reacting manganese and prevents the metal from reacting with water. It is reasonable to say that manganese will react with warm water faster as stated in [4].
References
Hope my answer will bring a bit of clarity to the problem. It is also supplemented with some reference links concerning the interaction of manganese with water:

Ronald L. Rich. Inorganic Reactions in Water (1st ed.): "Water", p. 153.
C.E. Housecroft, A.G. Sharpe. Inorganic Chemistry: "21.8 Group 7: manganese", p. 738.
N.N. Greenwood, A. Earnshaw. Chemistry of the Elements (2nd ed.): p. 1044.
P. Patnaik. Handbook of Inorganic Chemicals: p. 542.


text:
Does E1cB take place over here?

I formed this product (reason: alkene has 6 alpha hydrogens)
---->

However, it turns out, this is the answer:

The reason why I suspect this is e1cb is because NO2 could stabilize the carboanion (conjugate base) formed after reaction with NaOCH3
I would also like to know if a substrate has two sites for E2 elimination to occur, how do I decide site which would be preferred. Is it the one with the most acidic hydrogen or the one which would lead to a more stable alkene?

Your thought carbanion formation is correct except I don't understand your point about "6 alpha hydrogens".

@MeetLalwani  It is almost always the most acidic hydrogen that decides how elimination goes.

@user55119 stability of alkene is decided by the number of alpha hydrogens right???

now I see what you are saying. For a given set of isomers the more carbon-substituted double bond increases "stability" via hyperconjugation with the attached hydrogens. That is, the heats of formation and combustion are lowered. But this is a thermodynamic issue. Your example is one of kinetics. What process occurs faster is at issue.

About the last point, it depends on the base and leaving group, for example if the base is strong, it will prefer the most acidic hydrogen. If the base is not so strong and say the leaving group is good, the stability of product would matter more.


text:
Evaporation of zinc oxide
My question involves the evaporation of silicate material (e.g., rocks, minerals, geologic material), specifically I am trying to work the number of electrons (n) that are involved in the evaporation reactions described by Sossi et al. (2019) and Sossi and Fegley (2018).
As an example, I will discuss the evaporation of zinc oxide (ZnO), which is shown in multiple sources (see footnotes) as:

Sossi et al. (2019) state that this reaction involves two electrons (n=+2), however they do not indicate how they came to this value, so I have been trying to work it out myself. I should also note that they describe these reactions as producing a metal bearing gas.
I have two lines of thinking for determining n for this reaction:
First line of reasoning
ZnO(l) has two components with charges Zn = +2 and O = -2
Zn(g) evaporates as a metal, so the charge is Zn = 0
and
1/2O2 is just oxygen gas with charge 1/2O2 = 0.5*(-2*2) = -2
This line of reasoning implies that there was a change in the charge on Zn of -2. In other words, two electrons were lost, so two electrons were involved in the reaction. It involves the production of gaseous zinc metal, which fits nicely into the description of a metal bearing gas, at least in my mind. However, this leaves the reaction unbalanced because the products have a -2 charge, which leads me to my second line of reasoning.
Second line of reasoning
ZnO(l) his a stable compound, so ZnO = 0 charge
Zn(g) evaporates as a cation, so the charge is Zn = +2
and
1/2O2 is just oxygen gas with charge 1/2O2 = 0.5*(-2*2) = -2
Now the reaction is charge balanced, but I am not clear on how two electrons were involved in this case, unless I can consider the two electrons that "went with" Zn after the evaporation of ZnO as the two electrons that are involved in the reaction. However, I don't know if Zn++(g) counts as a metal bearing gas, which the authors say is present.
I have been tying myself in knots trying to work this out, so any guidance in the right direction would be helpful.
Footnotes
Other sources that discuss the evaporation of ZnO include Anthrop and Searcy (1964) and Kodera et al. (1968)

Your following sentence is wrong (I copy it in italics without change)
1/2O2 is just oxygen gas with charge 1/2O2 = 0.5(-22) = -2
No! There is no charge $-2$ on the oxygen atom in the molecule $\ce{O2}$, and no charge at all in the molecule $\ce{O2}$ There is a charge $-2$ on the oxygen atom in the initial oxide $\ce{ZnO}$. But during the decomposition of $\ce{ZnO}$, the two electrons available on the Oxygen ion are transferred to the zinc ion $\ce{Zn^{2+}}$ which becomes a non-charged atom $\ce{Zn}$.

Thank you for catching that mistake! That means that Zn is evaporating as Zn metal (Zn+0), correct?

Boiling point of Zn is 907°C, deep below the melting point 1974°C of ZnO,  at which it is said being already decomposing.

@Poutnik I understand what you're saying, but I don't follow why that's relevant. Are you saying that my comment is incorrect?

@eyemarrs It is just a side info.

Yes, Zn is evaporating as Zn metal, mixed with neutral oxygen atoms. But when this gas is cooled down, it synthesizes back zinc oxide.


text:
Are liquid chlorine and super-shock chlorine the same?
To shock a swimming pool one uses either calcium hypochlorite ($\ce{Ca(ClO)2}$), commonly sold from at least one major brand in pouches,

or the more obscure "liquid chlorine".

I'm saying "obscure" because labels for liquid chlorine never list the compound.
But unlike the result of dissolving $\ce{Ca(ClO)2}$ in water, which results in, by definition, a bleached (white) color, liquid chlorine is, somewhat surprisingly, yellow.
Are liquid chlorine and super-shock chlorine the same?

Note that mentioned "liquid chlorine" is far from true liquid chlorine, which boils at -34°C and was used in WWI at Somma in 1915 as chemical weapon

@Poutnik Interesting (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2376985/). In 1915 a gas attack was as obscure to French soldiers as a nuclear explosion was to the Japanese in 1945, and in both cases no amount of running could save the soul of those attacked (though a bicycle or an automobile could arguably save the life of someone in the former, but perhaps not in the latter case).

@Poutnik In any case, back to the present day. The major powers have neutralized the temptation of the rogue leader of another army to use these weapons by maintaining themselves a cache. That is small consolation for the Ukrainians who have abandonned, willingly, WMDs.

@Poutnik I still digress. The important point in the present context is to highlight why chlorine gas is not sold for swimming pool maintenance. It might be highly effective. A small container would last a month (?) and could be adjusted at will to raise/lower the concentration depending on whether swimmers are expected. But the risk of misuse of chlorine gas trumps any potential benefit for civiilan use. I wonder whether this is about right.

Manipulation with elementary chlorine is not convenient and is potentially very dangerous, in case the gas escapes. Chemically bound chlorine, that is created in water in situ at reasonable concentration is much, much better way.

Liquid chlorine is probably bleach, which is a solution of sodium hypochlorite $\ce{NaOCl}$. This substance is obtained by dissolving chlorine $\ce{Cl2}$ in a solution of sodium hydroxide $\ce{NaOH}$, according to $$\ce{Cl2 +  2 NaOH -> NaOCl + NaCl + H2O}$$ Chlorine is produced by adding an acid to this solution : the acid destroys $\ce{NaOH}$ and produced free chlorine $\ce{Cl2}$ by reversing the previous equation (which should be written as an equilibrium). Even weak acids like carbonic acid (and $\ce{CO2}$) can do the job. This is why the solution deserves to be called "liquid chlorine".


text:
What are oxidation numbers of each element in CuH?
My chemistry teacher said hydrogen when bonded with metals has oxidation number (O.N.) of −1 except CuH: in CuH the hydrogen has O.N. +1 and the copper has O.N. −1.
But on the internet it is ubiquitously told that the O.N. of hydrogen is −1 in CuH too, and the compound's name is copper(I) hydride. I am confused which one is correct. If hydrogen has O.N. of +1 in CuH, why is it so?

https://chemistry.stackexchange.com/questions/118739/what-is-the-oxidation-state-of-mn-in-hmnco5

Point to note, copper hydride is non stoichiometric i.e. it is $\ce{CuH_n}$ where n = 0.95

This turned out to be a surprisingly tough cookie to give a bulletproof answer to. Thoughts based on "hey, they called it copper(I) hydride here" and products of the reaction with HCl obviously cannot be considered as real answers. We appear to lack data for appropriate methods for oxidation number determination such as XPS or BVS based on crystal structure (bond valence parameters are scarce for copper(I)). Your teacher probably were thinking about something else while providing you with the answer and likely had an image of the chart for reactivity of the metals hanging in front of them.

@andselisk I read your comment and just decided to "self-discipline" my answer. But I will note that the teacher likely did confuse the EMF series with electronegativity. In reality, hydrogen is higher in the series than its electronegativity "should" allow because of the great solvation energy of aqueous hydrogen ion.

@OscarLanzi Yes, that was my thought too. And after thinking and researching for a while, I came to the conclusion that the question is actually tricky and it could be interesting to get a real solid answer for.

@andselisk XPS may not tell us much. This compilation shows nearly the same binding energy for Cu2O as for Cu(zero). The 4s electron may be too diffuse to impact the inner shells we measure by this method.

Korzhavi and Johanssen 1 performed first-principles calculations on the electronic energy levels of several compounds of copper, hydrogen and oxygen. They find good agreement between derived and experimental properties of stable compounds such as $\ce{Cu2O}$ and correctly predict the instability of other candidate compounds such as $\ce{CuOH}$ and $\ce{CuH2}$.
For $\ce{CuH}$ the calculated density of electron states shows the hydrogen-based 1s states filled, and lower in energy than the copper-based 3d states. This is similar to the oxygen- and hydroxyl-based states in the oxide and hydroxide. The copper hydride result thereby identifies that compound as a metal hydride with positive metal oxidation state.

Reference

Korzhavyi, Pavel & Johansson, Borje. "Thermodynamic properties of copper compounds with oxygen and hydrogen from first principles." Report to Svensk Kärnbränslehantering AB, Texhnical Report TR-10-30, February 2010. Retrieved 19 June 2022.


text:
Why do we have a negative current in cyclic voltammetry curves?
Why do we have a negative current in cyclic voltammetry curves?
here are my thoughts. We have two conventions in reporting data of CV curves: the IUPAC convention and the US convention. In the IUPAC convention, the potential scan direction is from lower potential to higher and the US convention is reversed of the IUPAC convention (from higher potential to lower potential). So I think it's all about the direction of current in an electrochemical cell. because anodic current will be positive in the IUPAC convention and negative in the US convention. Therefore, maybe It does not relate to cathodic or anodic current. Can anyone explain more? what is the concept of the negative current?

In cyclic voltammetry, the applied potential cyclically alternates, going more positive, then more negative, with respect to a reference potential. The resulting currents likewise alternate between positive and negative.

Ok. That's right. I wanna ask one more question. can we have the CV that all cathodic and anodic currents of the CV are positive?

It is possible to add a constant current, either positive or negative, thereby shifting the CV up or down on the current axis. This could be deliberate, to accommodate a current-input ADC that requires unipolar input currents. Or it could be inadvertent, as a consequence of a leakage current or the like. I think this is what you mean, but correct me if I got it wrong.

Thanks for your answer. but I have another question. In electrochemical sensors (like glucose sensors) when we are adding the analyte to the electrolyte, the Cv current will shift to a positive value. (both cathodic and anodic current are positive). what do you think about it?why do we have this?

I don’t know the answer to that. Per se, adding glucose would not cause a shift of the whole CV. Unless someone here has a helpful suggestion, I think you will have a little investigation to do. This is how research typically goes.

Yeah, sure Ed. Thanks for your help. Yes, I am searching for that. but I am a little confused. By adding different concentrations of glucose the whole CV shifts up. For faster scan rate we have the same result that for faster scan rate the whole current increase and both anodic and cathodic current are positive value.


text:
Is calcium hypochlorite as effective as sodium hypochlorite in sanitizing a swimming pool from algae?
Question
From this question we know that what swimming pool stores refer to as "liquid chlorine" (sold in 10 gallon jugs) is most likely sodium hypochlorite ($\ce{NaOCl}$). In the spirit of Least Publishable Units (LPUs) I'll ask a brief sequel separate from the first question.
The worst is when microorganisms (algae) overrun the pool. In the presence of summer heat, ample sunlight, and perhaps some nutrients (just phosphates?), the green population can multiply rapidly. Is calcium hypochlorite ($\ce{Ca(ClO)2}$) (sold in crystal form) as effective as sodium hypochlorite ($\ce{NaOCl}$) in sanitizing a swimming pool from algae?
Context
A certain talented fictional chemistry teacher whose fame rests on illicit activities apparently had little difficulty maintaining good water quality in his swimming pool. Yet anyone who has tried will attest that maintaining immaculate water clarity is sometimes a nontrivial objective.
This is the second in what may be a series of questions trying to understand the basics of maintaining a swimming pool. A quick search online will show an enormous number of results—all amateurish and none technical, hence the need for learned opinion.

Yim et al. [1] studied the effectiveness of various hypochlorite salt treatments against a particular strain of bacterial cells and Speers. They found that calcium hypochlorite was much more effective than sodium hypochlorite... but it should he used with care because if its high corrosivity:

Although HTH [calcium hypochlorite] is effective for decontaminating these B. anthracis strains, it is considered extremely corrosive to metals, skin and mucous membranes, eyes, and respiratory and gastrointestinal tract [2]. Therefore, to avoid material corrosion or toxicity, alternative disinfectants with high efficacy are needed.

References

Jin-Hyeok Yim, Kwang-Young Song, Hyunsook Kim, Dongryeoul Bae, Jung-Whan Kun-Ho Seo (2021). "Effectiveness of calcium hypochlorite, quaternary ammonium compounds, and sodium hypochlorite in eliminating vegetative cells and spores of Bacillus anthracis surrogate". J Vet Sci. 2021 Jan; 22(1): e11.
Published online 2021 Jan 18. https://doi.org/10.4142/jvs.2021.22.e11
PMCID: PMC7850788
PMID: 33522163

Rogers JV, Ducatte GR, Choi YW, Early PC (2006). A preliminary assessment of Bacillus anthracis spore inactivation using an electrochemically activated solution (ECASOL) Lett Appl Microbiol. 2006; 43:482–488.


text:
How to define a chemical bond computationally?
I'm working with an amorphous system. With oxygen, my system has both covalent and ionic bond forming cations. I utilise the Wannier centre to define a covalent bond, and my theory is that if the Wannier centre is near to the line connecting the cation and anion, it is a covalent bond.
However, this method fails to identify an ionic bond as there is no region of high electronic charge density which can be localized to obtain Wannier center. How can an ionic bond be defined using a computational technique?

Matter Modeling is likely a better place to ask this.

I think you have to think carefully about what you mean by an ionic bond. For instance using the ionic model to describe MgO I would argue that, as the coulomb potential is very long ranged and can't be truncated, the oxygens sit in the potential minimum due to all the other ions in the lattice. So are all oxygens (and hence all the magnesiums) bonded to everything else? Often for ionic species a simple A bonds to B pairwise view of the world is not that useful

Have a look at the Quantum Theory of Atoms in Molecules. It might help you along.


text:
Nylon 6,6 synthesis yield
This is the nylon 6,6 reaction

Having these data available, I would need to know the theoretical quantities, in grams, of the monomers

My attempt:
Adipoyl chloride is our limiting reactant so theoretical mass is the weighted mass.
Hexamethylenediamine and adipoyl chloride are in 1:1 ratio, so amine theoretical mol are the same of adipoyl chloride: if the calculations are correct, $0.0022\;\text{mol}$.
From which, the theoretical mass of amine will be
$$ 0.0022\;\text{mol} \times 116.21\;\text{g/mol} = 0.2557\;\text{g} $$
Any advice?

No ! Your calculation gives the mass of the dimer, with $n$ = $2$. But the nylon 6,6 is a polymer with a rather high value of $n$. Its molar mass is probably higher than $100'000$. I don't know whether this value is known.

I changed the question: what I need to know are the theoretical quantities, in grams, of the monomers. I gave it a try: could you tell me if this is correct?

Your calculation is correct. $\pu{2.2 mmol}$ of the amine weighs $\pu{0.2557 g}$ and this will react with $\pu{2.2 mmol}$ adipoyl chloride, which weighs $\pu{0.4020 g}$


text:
If sulfur is directly below oxygen on the periodic table, why isn't SO the most common sulfur oxide? Instead of third, behind SO2 and SO3?
Chalcogen like oxygen and sulfur want to 'grab' (or share) two electrons to get to eight, if possible, or maybe 'give up' (or share) six, if necessary, to again get to eight... Right?
So, why is sulfur dioxide, where the sulfur shares four electrons (usually), more common than sulfur monoxide or sulfur trioxide?
Isn't sulfur dioxide a higher energy, less stable molecule?

Main group elements, from orders higher then 2nd, really don't like double bonds. Another thing is that sulfur oxidises easily. BTW I don't think I can say I like questions of "Why chemistry is as it is?" type.

Does this answer your question? Why do sulfur and oxygen form sulfur dioxide?

Sulfur monoxide is unstable and has been rarely observed in stable conditions. It has been detected in space however. Normally, it is prepared by decomposing ethylene episulfoxide but it is so unstable that it quickly dimerizes to disulfur dioxide which is also unstable and converts to stable sulfur oxides.

Nobody really knows why $\ce{SO}$ is such a rare and unstable substance. It is an orange gas at room temperature. The only way to synthesize $\ce{SO}$ is to reduce thionyl chloride $\ce{SOCl2}$ with some reagent eager to react with the chlorine atom, like sodium, silver or tin : $$\ce{SOCl2 + 2 Na -> SO + 2 NaCl}$$ The strangest property of this gas is its reaction with water. It is decomposed with the smallest amount of water vapor according to the disproportion equation : $$\ce{3SO + H2O -> H2S + 2 SO2}$$ It is an extraordinary reaction, as it requires $3$ molecules $\ce{SO}$. Another oxide is known to disproportionate with water : $\ce{NO2}$. But there is a good reason for this behavior : $\ce{NO2}$ has an uneven number of electrons. It is not the case for $\ce{SO}$.

It looks like water must be present in stoichiometric amounts for complete disproportionation. The orange color in a gas raises suspicions that it might be reactive in a redox fashion there are some loosely held electrons somewhere.


text:
Shouldn't the Cl- be placed beside the charged N+ ion when writing the structural formula of Benzenediazonium chloride?
I found the structural formula of Benzenediazonium chloride written on a website:

I believe that the Cl- ion is stabilizing the N+ ion.

So, shouldn't the structural formula be written like the above instead?

Note that the topic has two levels - a preferred formal one and structural one. For the latter, I am not aware of forming ionic pairs here.

You are right. But it is not always possible to print this formula with $\ce{Cl-}$ ion UNDER the diazo line. It depends on the programs available on your computer, specially if you replace the benzene ring by the formula $\ce{C6H5}$, and if you want to type it on one line in a text.

With respect to the point of your question, both ion placements are acceptable for the specified form of the salt's structural formula.
From section GR-7.2 Positioning of components [1, p. 383] (emphasis mine):

The various components of a salt may be positioned relative to each other in one of two ways. If one of the components consists of a single atom, it may be “paired up” with an oppositely charged atom in a larger component, with the positioning of the larger fragment determined by other considerations as discussed in GR-3. The fragments should be positioned as if there were a single bond between the cationic
and anionic centers, even if this requires that the cation be drawn on the right side of the anion. The positioning of the smaller fragment will then usually be determined according to other recommendations (that straight chains should not have arbitrary bends in them, for example).

Alternatively, the components of a salt may be depicted next to each other. This positioning is acceptable for single-atom components, and preferred for larger ones. If the components are of similar size, positively charged components should preferably be positioned to the left of negatively charged ones; if they are significantly different in size, the larger components should be positioned to the left of the smaller ones.

Notes

Negative charge should be depicted with superscripted minus $(\ce{Cl-}),$ not en dash $(\ce{Cl^\text{-}}).$
Although being acceptable, a circle is rarely used for the depiction of aromatic systems in publications beyond introductory level chemistry.
Chemical names are common nouns and inherit the capitalization rules: do not capitalize chemical names in the middle of the sentence.
An inline semistructural formula of benzenediazonium chloride could be $\ce{Ph-N2+ Cl-}.$
Even in the crystal structure ([2]; CSD Entry: BZDIZC) $\ce{Cl-}$ anion is equally distanced from nitrogen atoms in diazonium (an isosceles triangle with the legs of ~3.55 Å or ~3.23 Å):

References

Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). Pure Appl. Chem. 2009, 80 (2), 277–410. DOI: 10.1351/pac200880020277. (Free Access)
Rømming, Chr.; Karvonen, P.; Holm, A.; Nielsen, P. H.; Munch-Petersen, J. The Structure of Benzene Diazonium Chloride. Acta Chem. Scand. 1963, 17, 1444–1454. DOI: 10.3891/acta.chem.scand.17-1444. (Free Access)


text:
When a compound is soluble in a dilute mineral acid, is it necessary that it is also soluble in that concentrated acid?
For example, Barium Phosphate is soluble in dilute mineral acids. Does that also mean that it is also soluble in concentrated form of those acids?

Well. Barium phosphate $\ce{Ba3(PO4)2}$ is not soluble at all in sulfuric acid $\ce{H2SO4}$ solutions, whatever their concentrations, because it forms barium sulfate $\ce{BaSO4}$ which is an insoluble compound. On the other hand it is soluble in diluted solutions of hydrochloric acid, according to  $$\ce{Ba3(PO4)2 + 6 HCl -> 3 BaCl2 + 2 H3PO4}$$ Unfortunately it is not soluble in concentrated $\ce{HCl}$ solutions, because it produces some barium chloride $\ce{BaCl2}$ which has the strange property of being insoluble in concentrated $\ce{HCl}$ solutions, as the solubility product of $\ce{BaCl2}$ is exceeded.

Interesting facts that outline what a broad question that is...

Exceeding the solubility product means just that, it does not mean insolubility; it just informs you what the solubility is. Sodium chloride [saturated] is precipitated by adding gaseous HCl.


text:
Reaction between 1,2-bis(buta-1,3-dien-1-yl)cyclohexan-1-ol and potassium hydride in THF

Which product is formed in the following reaction?

I am quite certain that we deal with an oxy-Cope rearrangement that works at room temperature, but I don’t know how to draw the product given the symmetry.

You are correct, it is indeed an oxy-Cope rearrangement and the macroexpansion gives the 14-membered ketone. From Paquette [1, p. 13979]:

In like fashion, the macroexpansion of 38 and related compounds proceeds in a stepwise manner via enolates 39a and 39b. Since 39a does not accumulate as the reaction proceeds, the second purely Cope rearrangement must be accelerated relative to the first. Trienone 40 has served as a precursor to (+)-muscone and (−)-(3Z)-cembrene A.

The reaction was reported by Wender et al. [2, p. 3349]:

When treated with KH in THF at room temperature, both trans- and cis-10 gave a mixture of two E/Z isomers of the 15-membered tetraeneone 11 (61% and 40%, respectively). Catalytic hydrogenation of this mixture gave a 97% yield of muscone (12).

Structural assignment based on IR and NMR studies suggest the resulting product is the E,E,E-trienone [3]. The process of ring expansion via Cope rearrangement appears to be initially investigated by Cookson and Singh in late 1960s in the context of obtaining macrocyclic musk ketone [4]:

Reaction of the keto-allene (III) with vinylmagnesium bromide in tetrahydrofuran gave the vinyl alcohol (VII) in good yield.
…compound (VII) undergoes Cope rearrangement to the enol of (VIII) before it fragments. Complete rearrangement through to the ketone (VIII) before fragmentation cannot have occurred, for there
were distinct differences in the mass spectra of the alcohol (VII) and ketone (VIII)…
The compound (VII) though fairly stable, showed the appearance of a carbonyl band, evidently by rearrangement to (VIII), when left for a prolonged period.

References

Paquette, L. A. Recent Applications of Anionic Oxy-Cope Rearrangements. Tetrahedron 1997, 53 (41), 13971–14020. DOI: 10.1016/S0040-4020(97)00679-0.

Wender, P. A.; Holt, D. A.; Sieburth, S. M. Practical Method for α-Alkenyl Ketone Synthesis Based on a Facile Reductive Rearrangement of Alkynyl Halohydrins. J. Am. Chem. Soc. 1983, 105 (10), 3348–3350. DOI: 10.1021/ja00348a072.

Wender, P. A.; Sieburth, S. McN.; Petraitis, J. J.; Singh, S. K. Macroexpansion Methodology. Tetrahedron 1981, 37 (23), 3967–3975. DOI: 10.1016/S0040-4020(01)93271-5.

Cookson, R. C.; Singh, P. Expansion of the Ring of Cyclododecanone by Two and Four Carbon Atoms. J. Chem. Soc., C 1971, 1477. DOI: 10.1039/j39710001477.


text:
Which quantum numbers does the orbital angular momentum depend on?
I learned from a question that I recently solved that the orbital angular momentum depends on both the azimuthal and magnetic quantum numbers.
I did not expect this because the formula for the orbital angular momentum is $ \sqrt{l(l+1)} \frac {h}{2π} $ and so I expected it to only depend on the azimuthal quantum number.
So, why does orbital angular momentum depend on the magnetic quantum number? A detailed explanation will be appreciated.

What keeps you from searching for what the magnetic quantum number is about? // The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.

@Poutnik I can't find a detailed and understandable explanation online.

Try harder and describe what you have tried.

@Poutnik I tried "why does the orbital angular momentum depend on the magnetic quantum number" on google. That's the most I can do.

You can do much more than you say you can. See Wikipedia page for magnetic quantum number. You have nice picture there illustrating relation of orbital angular momentum and magnetic quantum number.

Note that relation of magnetic quantum number to orbital angular momentum is  the same as relation of spin quantum number to spin angular momentum.

Orbital angular momentum, like any other form of angular momentum, is a vector quantity.* Thus, it has both a magnitude as well as a direction.
The magnitude of the orbital angular momentum is given by the formula you wrote, and is thus only dependent on the quantum number $l$:
$$|L| = \hbar\sqrt{l(l+1)}.$$
However, the orientation of the angular momentum vector is partly described by the quantum number $m$. In particular, the $z$-component of the orbital angular momentum is given by
$$L_z = m\hbar.$$
It's a quirk of quantum mechanics, however, that you can't simultaneously know all three components $(L_x, L_y, L_z)$, so this is the best we can really do. Ultimately, this is because the operators $L_x, L_y, L_z$ don't commute with one another.
So, given that we know the magnitude $|L|$ as well as the $z$-component $L_z$, we can say that the orbital angular momentum lies somewhere along a cone, as the following picture from Atkins' Molecular Quantum Mechanics (4th ed.) shows. Atkins uses the general symbol $I$ to refer to some kind of angular momentum; the fact that it's orbital angular momentum doesn't change anything. Indeed, the Wikipedia page on magnetic quantum number has a very similar picture, as pointed out in the comments.

* Pseudovector, for the picky.

Pseudovector, for the picky. I was always humbled by my naive chemist approach to vectors in Usenet sci.physics and sci.physics.relativity in the past. Pseudovectors, covariant vectors, contravariant vectors, "not at all" vectors.... Oh my, the good old classic of old time volumetry. :-)

@Poutnik To be honest, I don't profess to understand it anywhere near as well as the physicists do ;) They talk about dual spaces and Hodge stars and something like that. To me, it's just called "reflect it in a mirror and it changes"!


text:
Calculating the amount of water condensed when cooling below the dew point
I've been given the following question:

Calculate the dew point temperature of a 60% humid air at $30^oC$. How much water mass wil condensate out of 1 $m^3$ of such air when the temperature is lowered to $5^oC$?

I tried to find the dew point temperature corresponding to those conditions, which is  $T_d=21.55^oC$ and then I took the density of the partial water at that temperature and at $5^oC$ from thermodynamic stables. That would be
$$
\rho=0.01943[ \frac{kg}{m^3}] ~@~ 22^oC
, \rho=0.006797[ \frac{kg}{m^3}] ~@~ 5^oC
$$
And I calculated the mass condensated by $m=(\rho_{22} - \rho_5)*[m^3]=0.01264 [kg]$
But according to our course professor, he says that all of the water vapor will condense. So
$
m=\rho_{22}*[m^3]=0.01943[kg]
$
Why is that? Does always all the water condense when cooling in constant pressure below the dew point temperature? Will it ever depend on the specific temperature I'm cooling to (in our case, $5^oC$)? Are there scenarios where my method would be correct?
Edit: It wasn't given in the question, but i think he assumed that we are cooling the air in constant pressure. First I calculated the mass of water vapor at $30^oC$ and I used the ideal gas equation with the pressure of $P(30,\nu)=RH*P_{sat}(30^oC)$, where RH stands for relative humidity.
If I look at the following $T-\nu$ diagram,

the pressure line of the dew point is $P_{sat}(22^oC)$, and if the temperature is further lowered to $5^oC$ I get
$P(5^oC,v)=Psat(22^oC)=> 5^oC$
which is far below the saturation line. I know that cooling below the dew point will condensate the excess water vapour. In that case, will all of the mass that I calculated in the start of the process condence (0.0126 kg)?

Your professor has decided to neglect the vapor pressure of water at $5°$. It is valid as a first approximation. Your result is better, if your numerical values are correct, which I have not checked. .

As a rule of thumb,maximum absolute humidity in $\pu{g m-3}$ goes approximately to 1/2 by cooling by 10 °C. So saying cooling from 22 °C to 5 °C condenses all water is very bad first approximation.

@Maurice (and also Poutnik), I've added further clarification. Is my way still the correct one?

For precise calculation, one has to involve thermal volume contraction and even condensation based contraction, as absolute humidity relates to the volume (with or without air).

The maximum vapor pressure of water is $\pu{4243 Pa}$ at $\pu{30°C}$, and $\pu{872 Pa}$ at $\pu{5°C}$. If the original humidity is $\pu{60}$%, it means that the initial vapor pressure is $\pu{0.600 · 4243 Pa = 2546 Pa}$.
The initial amount of water in $\pu{1 m3}$ humid ($\pu{60}$%) air at $\ce{30°C}$ is : $\ce{n_{H2O} = pV/RT = \frac{2546 Pa·1 m^3}{8.314 J K^{-1}mol^{-1}·303 K} = 1.0106 mol. This amount of water weighs 18.19 g}$.
If cooled down to $\pu{5°C}$, $\pu{1 m^3}$ air has a volume equal to $\pu{\frac{278}{303} 1 m^3 = 0.917 m^3}$.
At $\pu{5°C}$, the amount of water remaining in $\pu{0.917 m3}$ air (saturated in water) is :
$\ce{n'_{H2O} = pV/RT = \frac{872 Pa · 0.917 m^3}{8.314 J K^{-1} mol^{-1} 278 K} = 0.345 mol}$. This amount of water weighs $\pu{6.22 g}$
The water condensed during the cooling operation weighs : $\pu{18.19 g - 6.22 g = 11.97 g}$.
So the proportion of water condensed during the cooling to $\pu{5°C}$ of $\pu{1 m^3}$  air taken at $\pu{30°C}$ with $60$% humidity is $\pu{\frac{11.97 g}{18.19 g} = 65.80}$%. It is much smaller than $\pu{100}$%. This result is far from your teacher's statement.


text:
What is m.2m point group?
I'm using Bilbao Crystallographic Server to find 3D crystallographic point groups of the K points of anatase TiO2.
Since the space group of anatse TiO2 is $I4_1/amd$, on this page, I found the point groups of the K points (in red rectangle).

It can be seen from the table, there are point groups with dots in their names, for example $m.2m$ and $..2m$, which are not in the complete list of 3D crystallographic point groups.

What are these point groups? How to map them to the listed 3D crystallographic point groups?

These are not point groups, rather site symmetries or point symmetries, which do become
Hermann–Mauguin point group symbols if the dots are omitted.
The dots represent sets of non-contributing equivalent symmetry directions: it's a placeholder or a “dummy” for “unused” directions [1].
Wyckoff position consists of three entities: multiplicity, Wyckoff letter and site symmetry [2, p. 16]:

General and special positions: Below the point group symbol, we find a list of general and special positions (points), the latter lying on a symmetry element, and therefore having fewer ”equivalent points”. Note that the unique point at the center is always omitted. From left to right, we find:
Column 1 The multiplicity, i.e., the number of equivalent points.
Column 2 The Wyckoff letter, starting with $a$ from the bottom up. Symmetry-inequivalent points with the same symmetry (i.e., lying on symmetry elements of the same type) are assigned different letters.
Column 3 The site symmetry, i.e., the symmetry element (always a mirror line for 2D) on which the point lies. The site symmetry of a given point can also be thought as the point group leaving that point invariant. Dots are used to indicate which symmetry element in the point group symbol one refers to. For example, site $b$ of point group $4mm$ has symmetry $..m,$ i.e., lies on the second set of mirror lines, at 45° from the first set.

Simply put, the site symmetry $..m$ is a mirror reflection going through a plane perpendicular to the third symmetry direction $c$. International Tables for Crystallography Vol. A, Section 2.1.3.12. Oriented site-symmetry symbols provides definition for $m.2m$ site symmetry:

Finally, for class $4/mmm$ (full symbol $4/m\;2/m\;2/m),$ the twofold axis of $2mm$ may belong to any of the three kinds of symmetry directions and possible oriented site symmetries are $2mm.,$ $2.mm,$ $m2m.$ and $m.2m.$ In the ﬁrst two symbols, the twofold axis extends along the single primary direction and the mirror planes occupy either both secondary or both tertiary directions; in the last two cases, one mirror plane belongs to the primary direction and the second to either one secondary or one tertiary direction (the other equivalent direction in each case being occupied by the twofold axis).

References

Donnay, J. D. H.; Turrell, G. Tables of Oriented Site Symmetries in Space Groups. Chemical Physics 1974, 6 (1), 1–18. DOI:10.1016/0301-0104(74)80030-3.

Radaelli, P. G. Symmetry in Crystallography: Understanding the International Tables; IUCr texts on crystallography; Oxford University Press: Oxford; New York, 2011. ISBN 978-0-19-955065-4.

Thank you! Since site symmetry can also be thought as point groups, does it mean that if I want to find the subduction of the d atomic orbital to irreps of m.2m, I can refer to the point group mm2?


text:
Why doesn't molecularity ever exceed 3?
Recently, I learnt that molecularity of a reaction is limited to 3 and even that is quite rare. I searched online and asked my teacher about this but the explanation given was that it is difficult for more than 3 molecules to collide simultaneously.
However, I find this explanation to be superficial. I have no deep knowledge myself but I can think of a superficial counter-argument to this explanation.
Consider the following diagram:-

Here, we can observe that 8 other molecules are in contact/colliding with the inner molecules. So, going by this superifical model, a molecularity of 8 should be possible.
There, are obviously several flaws to this elementary model but it still proves the need for a better explanation than the trivial one that I found. What is it?

@Poutnik so is the reason not linked to geometry? Is it linked to probability?

Reaction molecularity has nothing to do with how many molecules can be simultaneously in contact with central one, geometry-wise. Typical 3 molecular case $\ce{3 ^4He -> ^{12}C}$ is $\ce{2 ^4He <=> ^{8}Be^{} <=>[+ ^4He]^{12}C^{} ->^{12}C + \gamma}$

Usually, only 1 key can be inserted into the lock at the same time, regardless of how many keys can simultaneously touch it. Reactions are more about properly oriented collisions than touches. even trimolecular one are usually hidden forming of intermediate, reacting with yet another one.

@Poutnik then why is a molecularity of 3 possible considering the key argument?

Some locks may be made for 2 keys. Asking after sufficient prior thinking improves thinking. Some molecules have 2 reaction centers, some  trimolecular reactions are de facto 2 fast subsequent bimolecular reactions with short-time intermediate.

"Key" can be made of two parts.

@Mithoron Would not it make the 2 parts 2 keys ? But I would not dwell too much in the key analogy. (In my experience, some people have problems with analogies not being differently pictured identities.)

Higher molecularity is possible, fourth and fifth order reactions are known. See this answer and its comments.

@NicolauSakerNeto Order versus molecularity. Complex reactions can have high order, consisting of several bimolecular reactions.

To the picture - no molecule is spherically symmetric and no molecule is eager to simultaneously react toward all directions reactants come from. Neither atoms are.

This figure is in a textbook showing unit cell types, namely, this is body-centered cubic. I.e. it is nothing to do with molecules or chemical reactions, you have just inserted a random cat photo as an "argument".

Imagine vigorously shaking a box of tiny marbles. Pairs will collide often enough, but how often will three collide simultaneously? Or four? Add in the fact others have mentioned that chemical reactions require certain orientations of the colliding molecules, and certain speeds/energies, and hopefully you can see how rare a tetramolecular or even termolecular reaction is. A termolecular reaction requires simultaneous collision of three molecules, of the right relative speeds, at the right angles to each other, in the right spatial orientations.

Often, there is no right speed, energy, orientation nor place for the 3rd molecule to react at the same time. OTOH, for big molecules, especially polymers like proteins or polysaccharides, multimolecular reactions are possible, as there is multiple potential reaction centers.

@Poutnik That's a good point, I didn't consider macromolecules. I think perhaps one could regard that sort of thing as multiple reactions occuring in different places on the large molecule, rather than a single reaction involving several molecules.

Yes, I suppose that is the case, too. Truly trimolecular reaction is hard to find. I guess this one counts  H + H + M -> H2 + M ,as H + H -> H2* -> H + H.

Energy-wise, truly bimolecular reactions of atoms or small molecules, producing a single product are rare. As they have to deal with released energy not to break the product back. It cannot be translational energy due momentum conservation. It cannot usually(unless distributed to other bonds) be vibration energy as it would break the formed bonds. It cannot be rotational energy due angular momentum conservation. It may be electron excitation energy in some cases. But the 3rd reagent is an inert reagent, with which the kinetic energy, linear and angular momentum are shared.

Usually reactions having 3rd order behaviour on further examination show that they form an intermediate complex that then reacts to form products. The chance of three species simultaneously coming into contact in less than a nanosecond being vanishingly small.
Experiments by Nobel prize winner Porter (Proc. Roy Soc. v261, p29,1961) on iodine atom recombination in the vapour phase show that while $\displaystyle \frac{d[I]}{dt}=-k[I]^2[M]$ is observed, M being an inert gas or iodine vapour  ($I_2$) itself, the mechanism is
$$\displaystyle I+M\rightleftharpoons IM, \quad I+IM \to I_2+M$$
The observed activation energy is negative which is not possible for a single step reaction.


text:
When performing electrolysis of brine, are carbon steel electrodes ok?
Are carbon steel electrodes better? Do they both eventually corrode? Is there any cheap electrode that stays inert indefinitely?

No Iron will oxidize before water and make a mess of iron oxides and carbon flakes as well a other metals in the iron.

Note that even Pt or Au anodes would be vulnerable to eventually forming $\ce{[PtCl6]^2-}$ or $\ce{[AuCl4]-}$. Possibly the best choice would be glassy carbon, but it is not (AFAIK) widely available for amateurs.

Steel electrodes for electrolysis, honestly? I think you missed a few crucial points in your textbook! Go back and read some more before the next experiment. :)

Anode materials for electrolytic production of chlorine, a corrosive gas, are not that simple. Wikipedia gives a good description of what works ... and what doesn't. The tl, dr is that in industry, titanium anodes clad with a mixture of conducting oxides such as $\ce{PtO2}$ and $\ce{IrO2}$ are used in this application (and others requiring a conductive yet chemically resistant material):

Due to the corrosive nature of chlorine production, the anode (where the chlorine is formed) must be non-reactive and has been made from materials such as platinum metal,[1] graphite (called plumbago in Faraday's time),[1] or platinized titanium.[2] A mixed metal oxide clad titanium anode (also called a dimensionally stable anode) is the industrial standard today. Historically, platinum, magnetite, lead dioxide,[3] manganese dioxide, and ferrosilicon (13-15% silicon[4]) have also been used as anodes.[5] Platinum alloyed with iridium is more resistant to corrosion from chlorine than pure platinum.[5][6] Unclad titanium cannot be used as an anode because it anodizes, forming a non-conductive oxide and passivates. Graphite will slowly disintegrate due to internal electrolytic gas production from the porous nature of the material and carbon dioxide forming due to carbon oxidation, causing fine particles of graphite to be suspended in the electrolyte that can be removed by filtration. The cathode (where hydroxide forms) can be made from unalloyed titanium, graphite, or a more easily oxidized metal such as stainless steel or nickel. [Emphasis added.]

Cited References

Faraday, Michael (1849). Experimental Researches In Electricity. Vol. 1. London: The University of London.

Landolt, D.; Ibl, N. (1972). "Anodic chlorate formation on platinized titanium". Journal of Applied Electrochemistry. Chapman and Hall Ltd. 2 (3): 201–210. https://doi.irg/10.1007/BF02354977. S2CID 95515683.

Munichandraiah, N.; Sathyanarayana, S. (1988). "Insoluble anode of α-lead dioxide coated on titanium for electrosynthesis of sodium perchlorate". Journal of Applied Electrochemistry. Chapman and Hall Ltd. 18 (2): 314–316. doi:10.1007/BF01009281. S2CID 96759724.

Dinan, Charles (1927-10-15). "The Corrosion of Durion Anodes (BSc)." Massachusetts Institute of Technology. p. 4. hdl:1721.1/87815. Retrieved 2019-09-25.

Hale, Arthur (1918). The Applications of Electrolysis in Chemical Industry. Longmans, Green, and Co. p. 13. Retrieved 2019-09-15.

Denso, P. (1902). "Untersuchungen Über die Widerstandsfähigkeit von Platiniridium‐Anoden bei der Alkalichlorid‐Elektrolyse". Zeitschrift für Elektrochemie. Wilhelm Knapp. 8 (10): 149.


text:
Continuous measurement of urea concentration in water
My colleague and I are working with fertilizers, and we are trying to continuously measure the concentration of urea in water (In situ concentration as function of time). There are two condition that we're considering, first is using distilled water, and second is with the present of some salt (i.e., KCl (0.1 mM-10mM)). Our expected concentration is within 0.1-100 mM

Is it possible to use pH meter, conductometer, or something similar? How would that even work?

Have there been anyone who've done this before?

A method can be easy, cheap or usable. Choose 2 of 3. // BTW, what concentration?

Also, what is the water quality(solute composition) and are there additional additivies? The more relevant details you provide, the more suitable answers you may get, as the optimal method choice may depend on it, if there is ever any simple/cheap enough applicable.

See Is there any method to determine urea (in water) and phosphate (KH2PO4 in water) fertilizers by UV-VIS spectrophotometer or titration method? mentioning enzyme biosensor method and spectrophotometic method.

The presence of salt would interfere with conductivity measurements. A spectroscopic or colorimetric measurement as in the link provided by Poutnik is your best bet.

Please provide more details. What do you mean by continuous?


text:
Order in Enzyme Kinetics
In an interview for post-grad level, I was asked to explain the "order" of any generic enzymatic reaction involving competitive inhibition. I assumed that Michaelis-Menten Kinetics holds good and went for "first-order". I also noted that the Lineweaver Burk plots remain linear even in competitive inhibition.
However, they asserted (at the end of the interview) that it would be "second-order" - thoughts?

It sounds like maybe you weren't including the enzyme in your order? In simple M-M kinetics, the reaction is second order at low S concentrations and only first order when the enzyme is saturated and v=kcat[E]tot. As long as the inhibitor is present at high enough concentration to be relevent, you can't saturate the enzyme with substrate. However, it's only really second order when Km(app)>>[S]

You have to distinguish overall order from order in [S]. The total order is tricky, it could be third or fourth order if you consider proton concentration and happen to have one or two catalytic groups in the active site with a pKa near the pH. In an interview, there are no right answers, only good or bad arguments to support claims. A linear Lineweaver-Burk plot does not imply first or second order. There is a switch in order depending on whether [S] is << or >> Km(app).


text:
How do I filter a solution of very fine manganese dioxide?
My synthesis requires potassium permanganate $\ce{KMnO4}$ to oxidize a side alkyl group on benzene ring. The issue is that the brown manganese dioxide $\ce{MnO2}$ is very fine so it is passing through my filter (a triple up 10 micron polypropylene filter bag).
Reading some stuff online, they say that very fine, almost nanoparticle size manganese dioxide is produced. I do not have a nanofilter, so how do I filter this solution?
I have tried making a bed of sodium chloride (grainy crystalline for kitchen use kind) to no avail. I had some satisfactory results with a very fine calcium carbonate slurry, but it interfered with the filter product, so that's no good.

Take the suspension of your product + MnO2 in an organic solvent (EtOAc is good), add water and sodium metabisulfite until all the MnO2 reduces to Mn2+ (which is water soluble) then do a liquid/liquid extraction.

One solution for filtering $\ce{MnO2}$ is to use a sintered glass funnel under vacuum. Fill the funnel ~halfway with a slurry (water or organic solvent) of Celite, then filter your reaction mixture slowly without disrupting the Celite bed. You may have to score the $\ce{MnO2}$ layer to allow liquid penetration.

Is intense centrifuging applicable ?

@user55119 Can you put your comment as the answer because it worked and I have filtered the solution. But instead of a sintered glass funnel (I don't have one), I made a wet bed of celite on the PP filter bags I have and then filtered it with buchner funnel. I even improved it so what I did was is that I put another PP filter bag over the celite layer so most of the manganese dioxide stayed on the bag itself because it couldn't pass down. Made separation from celite easy. Thanks a lot for the suggestion

Do yourself a favour and buy a ceramic (or sintered glass if that's sufficient) filter crucible. They cost ten bucks at most, with shipping. There is not point in working with unsuitable equipment.

Pulkit Sharma: Per your request. One solution for filtering $\ce{MnO2}$ is to use a sintered glass funnel under vacuum. Fill the funnel ~halfway with a slurry (water or organic solvent) of Celite, then filter your reaction mixture slowly without disrupting the Celite bed. You may have to score the $\ce{MnO2}$ layer to allow liquid penetration. I've used the technique many times for this purpose. BTW, I have found the technique useful on many occasions for breaking emulsions. Celite is diatomaceous earth, which is used in swimming pool filtration systems. PS: The Comment of @Karl is a good one. A ceramic filter crucible (Büchner) is fine with filter paper and Celite. The only downside of sintered glass funnels is that they eventually need serious cleaning. The ceramic ones are easier to maintain.


text:
Compound's reactivity in Hoffman Bromamide Reaction
I already know that the presence of electron donating groups in the soon-to-be migrating group in the Hoffman Bromamide reaction makes the molecule more reactive, but I encountered a question which asks whether p-tert butyl benzamide is more reactive (w.r.t Hoffman Bromamide) or p-methyl benzamide, and the answer is the former, but shouldn't it be the latter because t-butyl group is a better EDG than just methyl?


text:
Why does acetone give a positive test in Tollens test?
Tollens test is used to distinguish between ketones and aldehydes wherein it oxidizes aldehydes into a carboxylic acid (according to my textbook). However, another book says acetone gives a positive test so why does acetone (a ketone) give a positive test?
Thanks

What is the "another book" that says so?

The interesting thing is you have thrown out what your textbook says and have taken without doubt what the other book says as confirmed truth.  Why have not you done research of independent sources?

Acetone does NOT give a positive Tollens test.
This learning resource specifically states that Tollens may be used to distinguish between acetaldehyde and acetone. This Quora answer here confirms. This article in the Journal of Chemical Education gives a clear summary of the way the test works which precludes ketones giving a positive test.
I would advise that you discard the book that says it does.


text:
How do you calculate the internal energy of a mixture of real gases?
Is there an equation to calculate the internal energy of a gas mixture, or is it too complicated to be computed so simply? I know that there's $E_{internal}=N(degrees of freedom)\frac12kT$ and $(P+a\frac{n^2}{V^2})(V-nb)=nRT$, but are there any formulas for mixtures of real gases. I'm aware of the Wan der Waals equation, $a=\sum_{i=1}^{i=n}\sum_{j=1}^{j=n}(x_ix_j\sqrt{a_ia_i}), b=\sum_{i=1}^{i=n}\sum_{j=1}^{j=n}(x_ix_j\sqrt{b_ib_i})$, though I have no idea what it means or how to use it, so don't know whether or not it's what I'm looking for.
If this question is more appropriate for the physics stack exchange, feel free to move it.

You need to know their fugacity. Then multiply the molar ratio (or value depending on the units you need and information you have) of each real gas by its fugacity and internal energy and add those internal energies up to get the internal energy of the system.

Is the composition changing, or is it constant between the initial and final states?

@PulkitSharma How do you calculate the fugacity? The internet seems to suggest that you would need to know the fugacity coefficient (φ), but it also seems that the value of φ is dependent upon both the temperature and pressure. I feel like I'm way out of my depth with this.

@ChetMiller The variance in the composition is negligible and can be ignored.

https://www.eng.uc.edu/~beaucag/Classes/ChEThermoBeaucage/J.M.%20Smith,%20Hendrick%20Van%20Ness,%20Michael%20Abbott,%20Mark%20Swihart%20-%20Introduction%20to%20Chemical%20Engineering%20Thermodynamics-McGraw-Hill%20Education%20(2018).pdf

Read page 373 onwards. Fugacity and other reduced properties in a mixture is a real headache in itself I just don't want to get in again. I had to design a gas system once. Never again. I deal in liquids now. Even then electrolytes fuck me daily. I hate Dimethyl Acetamide.

@PulkitSharma Yeah, that's definitely way beyond me. It's giving me a headache already 'n' I'm not even close to understanding it yet. Thanks for the resource, though, I'll definitely work through it until I know what I'm doing.

For an gas mixture of constant composition, you have  $$dU=C_vdT-\left(P-T\left(\frac{\partial P}{\partial T}\right)_V\right)dV$$where U is the molar internal energy and V is the molar volume.  For a van Der Waals gas, you can show that $C_v$ is independent of volume, and is thus equal to the ideal gas heat capacity.


text:
Decomposable solvent identification in GC-FID
How can a gas chromatography with flame ionization detection (GC-FID) be used to identify a liquid solvent that decomposes at its boiling point like DMSO? In a solution with other higher boiling solvents, DMSO shows some stability at elevated temperatures but when finding its retention time for the given column, a pure sample would be injected at highly elevated temperatures, temperatures beyond its boiling and decomposition temperature so how will its retention time peak be resolved?
Full disclosure, I am separating a mixture of N-methyl-2-pyrrolidone and DMSO and I need a way to detect both of these solvents in a GC-FID.

The question is, if GC is a good method for thermally instable components.

In gas chromatography there is a common misconception that separation occurs on the basis of boiling points (it does but only on specific traditional columns). You don't have to be above the boiling point of the analyte in GC. The main condition is that you must be above the dew point of the least volatile analyte during the chromatography run. The problem is knowing the dew point. Also GC gases are strictly oxygen and water free so decomposition near the boiling point may not be an issue.
A formal method for determining inlet temperature is to inject the same standard at various inlet temperatures and plot the response. You might see a peak like response i.e., at low/high temperature inlet response is low but somewhere in the middle the response is high. If there is degradation you will lose response.
As always, best temperature and sample degradation is a trade-off. Literature survey must be done to see what temperatures people use for your analytes. DMSO is not a new compound for GC.


text:
What does it mean finely divided aluminum?
Reaction of aluminum with oxygen
Finely divided aluminum Burns with oxygen with a brilliant yellow flame at high temperature and produce white aluminum (aluminum oxide).

High surface area aluminium, think fine powder.

It's powdered aluminum.

Same as in any other context. Who are Finely and Burns?

Finely divided aluminum is a rather thin aluminum powder obtained by evaporating (boiling) liquid aluminum in a high vacuum or an argon atmosphere, and letting cool it down without touching a wall. Oxygen must be absolutely absent.


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Does the 3D structure generated with MolView violate the wedge-dash rules of "above/below the plane"?
I've been taught that in a wedge-dash diagram, the wedge represents a bond above the plane of the paper, and the dash represents a bond below the plane.
Accordingly, I expected the structure of this particular conformer of 1,2-dichloro-1,2-difluoroethane to have the fluorine groups in a trans orientation. I am aware that conformers can interconvert, but my question is about the wedge-dash representation of this particular conformer; the representation obviously cannot interconvert.

In other words, I was expecting the structure to have the following sawhorse projection:

But when I ran it through MolView, I got the following structure, in which both fluorine groups seem to be above the plane.

That is, the structure had the following sawhorse projection:

Am I missing something? Or is the 3D structure from MolView wrong? I am not referring to the R/S configurations here; my doubt is only in the wedge-dash representation.

I think you should check out how conformers work, particularly interconvert...

The molecule can rotate freely around the C-C axis. But I would interchange one $\ce{H}$ and one $\ce{F}$ atom on the left (lower) carbon atom on the picture.

@Mithoron I know that there an infinite number of possible conformers of this molecule, and that they interconvert. I wanted to know about this specific conformer, because my question is about the representation itself.

@andselisk My question is not about the R/S configurations, but the wedge-dash notation.

Get the stereochemistry right before dealing with conformation. follow the advice of @Maurice regarding H and F.

@Maurice Do you mean that the F atoms should be opposite to each other in the 3D diagram? Could you post that as an answer?

I cannot post anything as an answer, as the post has been closed. Now, yes, the $\ce{H}$ and $\ce{F}$ atoms must be interchanged in the 3D diagram, because in the original diagram, the two atoms $\ce{F}$ and the two atoms $\ce{Cl}$ are supposed to be trans.

What happens if you put the wide end of the dashed arrow on the alkyl chain? The way you drew it could have misled MolView.

@OscarLanzi I tried that and got this error: "Failed to load structure from sketcher Unknown stereocenter configuration: F, charge 0, 1 bonds (0 double)"

MolView has issues with preserving stereochemistry when converting 2D to 3D. Do not expect accurate conversion of the drawn wedge-and-dash projection into the three-dimentional model because the structure is converted to the proprietary SMILES format first, and despite SMILES having support for stereochemistry, there is no uniform open standard.
MolView sends generated SMILES to third-party services such as PubChem and CID, and numerous things may break (and they do) along the way, for example, since there is no hydrogen atoms in the SMILES generated within MolView if none explicit hydrogens were drawn, the stereochemistry interpreted by PubChem has nothing to do with the intended one because their API expects an isomeric SMILES string with explicit hydrogen atoms.
If you still want to use MolView to depict stereoisomers, use the search function with one of the following queries:

name with proper stereodescriptors: (1R,2S)-1,2-dichloro-1,2-difluoroethane

isomeric SMILES containing explicit hydrogen atoms: F[C@@H](Cl)[C@H](F)Cl

In your case you were getting wrong structure for (1S,2S)‐1,2‐dichloro‐1,2‐difluoroethane with SMILES string F[C@@H](Cl)[C@@H](F)Cl, which had nothing to do with your drawing, rather with interconversion process. Always check how the software operates and don't blindly rely on technology without understanding its limitations.


text:
How do you write fractional hydrates in words?
Apart from sequihydrate and hemihydrate (1.5 and 0.5 respectively), I can't find a way to write any other fractional hydrates.
Wikipedia states:

The notation "hydrated compound⋅nH2O", where n is the number of water molecules per formula unit of the salt, is commonly used to show that a salt is hydrated. The n is usually a low integer, though it is possible for fractional values to occur. For example, in a monohydrate n = 1, and in a hexahydrate n = 6.

Wikipedia says that fractional values can occur, but it doesn't show how to write the hydrate's name if n is fractional. Is there a simple way of writing a hydrate where n is, for example, 3.5 or 4.333...?

For n = 2.5, it could be hemipenthydrate. And hemihepthydrate for n=3.5,

No, I definitely wouldn't.

There are no IUPAC rules as to how to use multiplicative prefixes for fractional water content in crystallohydrates.
From Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), section IR-5.5 Names of (formal) addition compounds [1, p. 80] (strong emphasis mine):

For addition compounds containing water as a component, the class name ‘hydrates’ is acceptable because of well established use, even though the ending ‘ate’ might seem to indicate an anionic component. For hydrates with a simple stoichiometry, names of the classical ‘hydrate’ type are acceptable, but rules have not been formulated for non-integer stoichiometries such as that in Example 12 below. […]
Examples:
[…]

$\ce{3CdSO4·8H2O}\quad$ cadmium sulfate—water (3/8)

You need to convert the reciprocal of $n$ expressed as a decimal fraction to the simple fraction (or the ratio consisting of natural numbers) and put it in braces after the names of corresponding individual constituents of (formal) addition compounds separated by ‘em’ dash. Few examples:

$\ce{2ZnO·3B2O3·3.5H2O}\quad$ zinc(II) oxide—boron(III) oxide—water (4/6/7)
$\ce{MgCl2·4.33H2O}\quad$ magnesium(II) chloride—water (3/13)
$\ce{CHCl3·2H2S·4.5H2O}\quad$ chloroform—hydrogen sulﬁde—water (2/4/9)

Reference

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2. (PDF)


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Will ethanol precipitate lipid-saccharide compounds?
I have a cell-free solution of culture broth. It contains a wide range of compounds ranging from salts (marine broth), lipids, and proteins to polysaccharides.
My goal is to extract all type of saccharides and lipids in order to concentrate an eventual bio-surfactant produced by my cells. I'm specifically interested in those composed of saccharides and lipids linked together (e.g. glycosidic bond).
It was suggested to me to use ethanol to precipitate out these polysaccharides. I have seen this done widely in the literature.
However, I would like to understand the chemistry of this process in order to understand what is actually precipitating out (just polysaccharides, monosaccharides? proteins? lipids?). And whether it will work for both lipids bound to polysaccharides, given that lipids are soluble in ethanol.

Even solubility of well soluble sucrose (table sugar) in ethanol is just 0.6 g/100 mL.

I’m pretty sure it’s primarily due to disruption of h-bonding with water which results in h-bonding between saccharides. Obviously more of an effect the larger the molecule.

Well, you better thouroughly edit the post if you don't wanna waste the bounty. Body says very different story then title.

I don't see how the title is misleading. I seem to understand the biochemistry of ethanol precipitation. The body of my question just gives context as to why in case that helps focus the answer.

Surfactants in low concentration may be difficult to precipitate, as they are adaptive. I suppose the best would be doing separation experiments with isolated, compound class examples. What proteins do, what lipids do, what sacharides do, what liposacharides do, what their mixtures do. Because your sample may be a wild mixture with unclear interference.


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Does methyl or aldehyde group have higher priority when choosing the main chain of carbonyl compound?
As far as I know, both following names seem to be correct for the compound below. Please explain which IUPAC name is correct by which IUPAC rules:

4-formylpentanoic acid (methyl group counted in the main chain);
4-methyl-5-oxopentanoic acid (aldehyde group counted in the main chain).

Both names 4-formylpentanoic acid and 4-methyl-5-oxopentanoic acid are correct. In particular, the following rules of Section P-44 apply to both names:

The parent structure has the maximum number of substituents corresponding to the principal characteristic group, which is expressed as a suffix (-oic acid), in accordance with Rule P-44.1.1.
The principal chain has the greater number of skeletal atoms in accordance with Rule P-44.3.2.
The principal chain has the lower locant for an attached group expressed as a suffix in accordance with Rule P-44.4.1 (h).

Therefore, we have to consider the Rules of Section P-45 in order to figure out which principal chain will lead to the preferred IUPAC name.

P-45 SELECTION OF PREFERRED IUPAC NAMES
P-45.0 INTRODUCTION
Two or more names may result based on the same senior parent structure selected
according to P-44, because of different substitution patterns or multiple ccurrences of the same senior parent structure. A parent structure is defined (P -15.1) as a parent hydride, for example, benzene, a functionalized parent hydride, for example
cyclohexanol, or a functional parent compound, for example, acetic acid. Application of the criteria in this section will generate the preferred IUPAC name.

The first relevant rule of Section P-45 is about the number of substituents cited as prefixes:

P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

Therefore, the preferred IUPAC name is 4-methyl-5-oxopentanoic acid (two substituents) rather than 4-formylpentanoic acid (one substituent).

The situation would be reversed if we place more substituents on the other chain, e.g. 5,5-dichloro-4-formylpentanoic acid (three substituents) rather than 4-(dichloromethyl)-5-oxopentanoic acid (two substituents).


text:
Why do we define molar conductivity?
I've learnt about conductivity and molar conductivity in school.
I know that conductivity is the conductance of a solution of unit volume/ of unit cross-sectional area and distance between the electrodes.
The book then goes on to say "we must define a more useful quantity called Molar Conductivity."
How is Λm more useful than κ? Though Λm=κ/c, both κ and Λm vary with concentration. What is the use of defining Λm if it varies (though in an opposite manner) like κ?
And a related question: I know that Λm increases with dilution (more ionic mobility), and that κ decreases with dilution as there are less ions per unit volume. But wouldn't more ions per unit volume decrease conductivity, as it does with Λm due to interactions between the ions?
Thank you!

Molar conductivity reflects ion or ionic compound behaviour, while conductivity reflects rather their concentration. For high dilution, the former is nearly constant, the latter nearly proportional to c.
Compare molar conductivity of 2 different ionic compounds. Then compare just conductivities. The latter is not much useful until molar conductivity is calculated.
If you usefully compare conductivities of solutions of the same molar concentration(or different ones and recalculated), you implicitly compare molar conductivities.
Until very high concentrations, the conductivity increase due increased concentration is greater than its decrease due molar conductivity decreasing with concentration.
For not too high $c$ where Kohlrausch's law is still applicable:
$$\kappa = \Lambda_\mathrm{m}(c) c = (\Lambda_\mathrm{m}^{°} - K\sqrt{c})c$$
The Onsager's expression is:

$$\Lambda_m =\Lambda_m^0-(A+B\Lambda_m^0 )\sqrt{c}$$
where A and B are constants that depend only on known quantities such as temperature, the charges on the ions and the dielectric constant and viscosity of the solvent. This is known as the Debye–Hückel–Onsager equation. However, this equation only applies to very dilute solutions and has been largely superseded by other equations due to Fuoss and Onsager, 1932 and 1957 and later.

See also a scanned article at www.ncbi.nlm.nih.gov THE FUOSS-ONSAGER CONDUCTANCE EQUATION AT HIGH CONCENTRATIONS

Does anybody know the physical meaning of the parameter $K$ in front of $√c$ in Kohlrausch's law ? Is it just empirical ? Can it be justified by a reasoning ?

In Wikipedia article is written Kohlrausch suggestion of the formula was based on experimental data.

@Maurice Kohlrausch's law was emprical but theoretical explanations have been given. With Debye-Hückel theory, look into Debye-Hückel-Onsager equation (1926-27). Also Fuoss-Onsager theory (1950s or so) which extends DHO.

There is a common square root factor in them, but the relation would be more obvious if an activity coefficient were not in a logarithmic form.

@Linear Christmas. I would be pleased to get a simple answer to my question, without having to read and rediscover the long developments of the Debye-Hückel-Onsager equation, and of its extension by Fuoss-Onsager. I repeat : Can $K$in front of $√c$  be justified by a reasoning ? Yes or no ? That's all I need. Yes or no ?

@Maurice There is always some constant. I would be more curious about the square root. I suggest a dedicated question for it.

@Poutnik. Your point of view is similar to mine.


text:
Reimer-Tiemann reaction of aniline
[how reaction actually happens
I wonder is it possible for aniline to undergo Reimer-Tiemann reaction just like phenol? What is the problem in that? How to compare rates of carbylamine reaction and RTF?

Alternatively, how might dichlorocarbene react with the nitrogen of aniline?

Yes, how will dichlorocarbene react with aniline

As a recent member of ChemSE, you should be aware that you are expected to make some effort to solve the problem before you can expect to have help. Show me what your thoughts are in the next two days and then I'll post an answer. On homework questions.

https://chemistry.stackexchange.com/questions/98188/in-reimer-tiemann-reaction-why-does-phenol-attack-the-carbene-from-ortho-positio/98197#98197.      Also see this. They say that carbon is less electronegative than oxygen (and also nitrogen) thus it is more nucleophilic than them. Also consider that stable intermediate formation

Related paper: https://www.sciencedirect.com/science/article/abs/pii/S0040402001970966

I can't access it for free

The reaction of aniline with chloroform and base (Hofmann isonitrile synthesis, 1867)[1] predates the Reimer-Tiemann reaction (1876)[2] with phenol by 9 years. Strong base reacts with chloroform (1) to form dichlorocarbene (3). The sequence 3 $\rightarrow$ 9 leads to phenylisonitrile (IUPAC: isocyanobenzene). The electron pair on nitrogen in structure 6 can also, and likely preferably, be used to eliminate chloride leading to chloroimine 7.

The course of this reaction is notably different from the Reimer-Tiemann reaction. While the mechanism explains the results, it does not answer why reaction doesn't occur on the aromatic ring. Consider the reaction of phenoxide (10) with dichlorocarbene (3) using a similar mechanism to afford phenylformate (15). This ester is particularly labile to base with the irreversible formation of phenoxide. This process is a dead end; eventual reaction on the ring occurs irreversibly.
One might ask why a similar course of events doesn't occur with base and chloroimine 7 leading to N-phenylformamide (16). N-Phenylformamide is less labile toward base than phenylformate but, if N-phenylformamide is formed, it would be capable of being dehydrated by dichlorocarbene to form the protonated phenylisonitrile 8 just as reagents such as $\ce{POCl3}$ do. I leave it to the reader to write a mechanism for the reaction in the red box.
In summary, both aniline and phenoxide may react kinetically on their respective heteroatoms, except that in the case of oxygen, the reaction is non-productive providing an opportunity for an irreversible reaction on carbon.

Thanks for the explanation. I think there should not be a hydrogen bonded to the oxygen atom in product number 11

What do you mean by red box? I can't see any place to fill anything

I have completely understood your explanation but the question is why does not Ortho formyl aniline forms? Like we say that carbon is less electronegative than oxygen and nitrogen and thus more nucleophilic, then why does not it attack?

Ritil: I corrected the error in structure 11. As to your question as to why ortho-C-formylation does not occur, my Answer may not be THE answer but it is AN answer.


text:
IR spectra database for inorganic compounds
I would need a database of IR spectra of inorganic compounds, possibly free of charge. For organic compounds there is for example "AIST: Spectral Database for Organic Compounds, SDBS" (https://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi), is there something similar for inorganic compounds?

Have you checked NIST database?

SciFinder amd Reaxys will have IR spectra but it is not free.