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Is there any difference between getting dissolved and getting dissociated?
In the context of electrolytes only, is there any difference between electrolyte getting dissolved and electrolyte getting dissociated?
According to my professor, they are different. He says a substance may dissolve completely but it may not dissociate completely and a substance may not dissolve completely but amount that dissolves can dissociate completely and, thus, substances which dissolve (either wholly or partially), amount that gets dissolved if dissociate completely are called strong electrolytes and amount of substances which gets dissolved if dissociates sparingly are called weak electrolytes (although it may dissolve completely).
How is it possible? I mean, how can solubility be different from dissociating?
If something dissociates it means it’s dissolving and in the context of electrolytes if something dissolves in water it has to dissociate into ions and even Arrhenius’ theory of electrolytic dissociation sys that substance which are water soluble, dissociates into ions and get dispersed in water.
And if someone says they are different to them, I ask the following two questions:

If solubility and dissociation are different then what would be the criteria for getting dissolved and what will be criteria for getting dissociated (for me the criteria for getting dissolved is dissociation).
how would you arrange the following in increasing order of 1.solubility
2.dissociation

NaCl, MgCl2, FeCl2, ZnCl2

Sugar dissolves amazingly well, but never dissociates; how about that?

Two other examples that immediately come to mind are $\ce{CO2}$ and the infamous "ammonium hydroxide" solutions.

@Ivan Neretin I'm talking in the context of electrolytes only

@MaxW the question is subject to electrolytes context only

There is no clearcut distinction between electrolytes and non-electrolytes.

Both $\ce{CO2}$ and $\ce{NH3}$ dissolved in water will change the conductivity of water greatly. For that I'd surely call them electrolytes.

HgCl2 is soluble in water, but is practically non dissociated.
CaF2 is dissociated in solution, but extremely insoluble,

This is a simple logical and semantic problem, which is not a problem at all. Your professor is right and wrong- both at the same time. He is creating a classification which does not exist and which is meaningless.
Look at the word origin of electrolyte:

Etymology from OED: < electro- comb. form + ancient Greek λυτός that may be
  dissolved, soluble (see -lyte comb. form). Compare electrolysis n.,
  electrolyse v.
and the portion -lyte "< ancient Greek -λυτος, combining form (in e.g.
  εὐδιάλυτος : see eudialyte n.) of λυτός that may be dissolved, soluble
  < λύειν to loose (see lysis n.) + -τός, suffix forming verbal
  adjectives. Compare French -lyte, German -lyt.
A substance which gives rise to ions when dissolved (typically in
  water) or fused; a liquid or gel which contains ions and can be
  decomposed by electrolysis; spec. the conducting fluid used in a
  battery.

Does this solve the semantics problem?
By definition, an electrolyte is soluble in a given solvent. It is also understood among electrochemists that an electrolyte is added to a solvent to impart conductivity to that solvent. I will never call sugar an electrolyte because nobody would use sugar to increase conductivity of water. You should ask: Does a given substance produce ions in a solution? If yes, it can be called as an electrolyte.
Look at this logic: grapes is a fruit but not all fruits are grapes.
Electrolytes are soluble in the solvent, but not all soluble substances in that solvent behave as electrolytes. Anything which produces ions in that solvent, can be labelled as an electrolyte.
Q-1 If solubility and dissociation are different then what would be the criteria for getting dissolved and what will be criteria for getting dissociated (for me the criteria for getting dissolved is dissociation).
The measure of dissociation is conductivity. How much electrical resistance is offered by that dissolved (and dissociated) substance in that solvent
Q-2 how would you arrange the following in increasing order of 1.solubility 2.dissociation
NaCl, MgCl2, FeCl2, ZnCl2
There is no correlation between solubility and dissociation.  If I want to study their dissociation I will measure the conductivity of their solutions systematically.

Not meaningless, as you stated. Compounds, per my answer below, can display important and large differences in their respective abilities to be dissolved and undergo dissociation.

@Buttonwood, The etymology of -lyte must have a typo mistake in OED for German cognate $-lyt$. OED writes "-lit". Duden doesn't give the details of Elektrolyt origin https://www.dwds.de/wb/Elektrolyt#wb-1

@M.Farooq The French dictionary «Le noveau petit Robert» notes «électrolyte» as indexed since 1838 as derived from English «electrolyte» (*1834) as composite of electro (you know as derived from greek electron (amber), now related to electricity) and greek lutos (soluble).  It is both a general (non chemistry specific, even if you find e.g., «Furane» with Hill formula C4H4O) reference, as well backed by the reference, the dictionary of the l'Academie Française (https://www.dictionnaire-academie.fr/article/A9E0713, https://www.dictionnaire-academie.fr/article/A9E0675). Epoch of Faraday.

@Buttonwood, Good to know this French etymological dictionary. I wan't aware of it. It impressive that you know so many languages.

Here are four examples of substances dissolving:
$$\ce{C6H12O6(s) <=> C6H12O6(aq)}\tag{1}$$
$$\ce{CH3COOH(l)<=> CH3COOH(aq) <=> CH3COO-(aq) + H+(aq)}\tag{2}$$
$$\ce{NaCl(s) <=> Na+(aq) + Cl-(aq)}\tag{3}$$
$$\ce{SO3(g) <=> SO3(aq);\ \  SO3(aq) + H2O(l) <=> HSO4-(aq) + H+(aq)}\tag{4}$$
Because NaCl is an ionic solid, the only way it can dissolve (3) is to also dissociate. Glucose is a molecular solid, so it only dissolves (1). Glacial acetic acid is a molecular liquid, and it dissolves in water and then partially dissociates (3), whereas sulfur trioxide dissolves in water and then reacts with it (4).

$\ce{NaCl, MgCl2, FeCl2, ZnCl2}$

Some transition metals cations hang on to their counter anions, or pick up some more in solution, together with water as a ligand.
For example for zinc cations:

The highest chloro complex, [ZnCl4]2-, is tetrahedral with a Zn-Cl bond length of 2.294(4) Å. The trichloro complex, [ZnCl3]-, which coordinates one water molecule, is pyramidal with the Cl-Zn-Cl angle 111°. The Zn-Cl and the Zn-H2O bonds are 2.282(4) and 1.9 Å, respectively. The two lower complexes, [ZnCl2] and [ZnCl]+ , cannot be separated by Raman spectra. The average Zn-Cl distance in these complexes is 2.24 Å, and the average Zn-H2O distance is 1.9 Å. In [Zn(H2O)6]2+ the Zn-H2O distance is 2.15 Å.

Source: https://www.researchgate.net/publication/273506504_The_Structure_of_Zinc_Chloride_Complexes_in_Aqueous_Solution
So for these metal cations, the story is complicated and there is a complex mixtures of species, with concentrations dependent on anion concentrations, pH and temperature. A classic example is cobalt chloride, where you can follow the equilibrium easily because of color changes, see e.g. https://www.chemedx.org/blog/multi-colored-equilibrium-experiment

Dear Karsten, I think can also include the example of SO3 and even NO2 gases. When they dissolve they $react$ with water and make it highly conductive. Still we will not call SO3 and NO2 as electrolytes, because they react with water to form products which ionize very very well.

@M.Farooq Nice suggestion, I included SO3.

@M.Farooq can we call what mentioned  in the  above comment"they react with water and make it highly conductive" partial ionization not dissociation ?

No, it is a chemical reaction. The products of the chemical reaction are ionizable.

[EDIT] I start with the following definition:

Dissociation, in chemistry, the breaking up of a compound into simpler constituents that are usually capable of recombining under other conditions. In electrolytic, or ionic, dissociation, the addition of a solvent or of energy in the form of heat causes molecules or crystals of the substance to break up into ions (electrically charged particles). Most dissociating substances produce ions by chemical combination with the solvent. The idea of ionic dissociation is used to explain electrical conductivity and many other properties of electrolytic solutions.

I now present an example of two molecules, chlorine oxides ($\ce{Cl_xO_y}$) and also chlorine, all of which are gases at room temperature, to state my case that there can be a major difference between dissolve and dissociate creating distinct and important properties in a solvent of water.
For example, $\ce{ClO_2}$ is a gas that is highly soluble in water (some 8 g/L at 20 °C). However, per Wikipedia , to quote:

It does not hydrolyze when it enters water, and is usually handled as a dissolved gas in solution in water.

Although technically, there are many (including important) 'produce ions by chemical combination of the molecule with' water (see my note below).
In contrast, we have $\ce{Cl_2O}$, which per Wikipedia is described as:

At room temperature it exists as a brownish-yellow gas which is soluble in both water and organic solvents. Chemically, it is a member of the chlorine oxide family of compounds, as well as being the anhydride of hypochlorous acid.

As a result of the latter, we have:
$\ce{Cl2O + H2O <=> 2 HOCl}$
And, further aqueous hypochlorous acid can create ions (hence an electrolyte, albeit, very weak):
$\ce{HOCl <=> H+ + OCl-}$
In fact, dichlorine monoxide is very soluble due to hydrolysis in water (some 143 g of $\ce{Cl_2O}$ per 100 g water), but also can dissolve in organics (like CCl4 and CHCl3).
So in water, $\ce{ClO_2}$ mostly dissolves and $\ce{Cl_2O}$ largely dissociates, which seemingly presents significant differences. And, finally, in contrast to chlorine, it is but mildly soluble (1/10 of $\ce{ClO_2}$) and mildly dissociates, per below, which varies significantly with temperature:
$\ce{Cl2 + H2O <=> H+ + Cl- + HOCl}$
Therefore, it appears that all three gases have distinct (and important) properties connected to solubility and propensity to undergo dissociation. If a chemistry student of mine were to conflate solubility and dissociation, in regard to these compounds, I would feel bad if I contemplated assigning a final grade above C.
A technical note, with respect to $\ce{ClO_2}$, I explain the dissociation of chlorine dioxide due to its self-reaction in water as chlorine dioxide is an example of a stable free radical:
$\ce{.ClO2 + .ClO2 <=>  Cl2O4}$
$\ce{Cl2O4 + H2O <=>  HClO2 + HClO3}$
where both created acids form ions in water. [EDIT] A supporting source makes the statement:

In air, chlorine dioxide readily dissociates both thermally and photochemically and may form chlorine, oxygen, hydrogen chloride, HClO3, HClO4.ClO, chlorine peroxide, and/or chlorine trioxide, dependent on temperature and humidity. Chlorine dioxide dissociates in water into chlorite and chloride, and to a lesser extent into chlorate (Budavari et al. 1996).

I would also note that $\ce{Cl_2O}$ dissolves before undergoing dissociation, and the dichlorine monoxide reversibly, to an extent, will escape from solution (especially concentrated or on heating) and is responsible for the 'chlorine-smell' of bleach (per Watts, p. 16) and as also reported by Watts, the ability to concentrate HOCl by distilling off half of the solution (due to the partial release of the $\ce{Cl_2O}$).
[EDIT] Now, as to the answer to the 2nd question, the answer depends on a substance's ability to dissolve and undergo dissociation, to rank one must measure the electrical current produced by my adopted definition. However, I would argue that the results can vary in the presence of dissolved oxygen for acidic ferrous salt, for example. The explanation is that the ferrous can be converted into ferric per an electrochemical reaction in the presence of H+ (per hydrolysis of FeCl2) and O2 as dissolved or from air, upon exposure over time. As such, the FeCl2 with time and air exposure may seemingly present different electrical properties (due actually to dissociation). Similarly, per this work: The reaction of oxygen with magnesium chloride, again measurable changes in voltage may occur. Apparently, MgCl2/O2 is NOT exactly inert either, so again a possible departure in ranking with time in the presence of gaseous or dissolved oxygen.


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The Evasive Ether: Existence of diethynyl ether?
I tried to find online whether the compound diethynyl ether, $\ce{HC#COC#CH}$, exists in a stable state at STP, mainly out of curiosity, but also because it would be interesting to study the properties of a molecule with such interesting characteristics (eg. the very long bent shape, the highly exposed oxygen atom, the potential for multiple different polymerization reactions). However, I spent about 30 minutes scrolling through page after page of search results with only results about diethyl ether, and came up empty-handed. I'm wondering if anyone here has any info/sources/knowledge about this evasive ether. I've included some images to clarify what molecule I'm talking about, as well as to possibly help jog your memory. Any and all information/data about the molecule is appreciated.

Room-Temperature Chemical Synthesis of C2 by K Miyamotohad a footnote: (21) Diethynyl  ether was  too  unstable  to  be  isolated.  See:  Serratosa,  F.  Acetylene 243Diethers: A Logical Entry to Oxocarbons. Acc. Chem. Res. 16, 170–176 (1983)

I doubt I've ever seen any molecules with $\ce{R'-C#C-O-C#C-R''}$ group. On the other hand, there are definitely some with sulfur, $\ce{R'-C#C-S-C#C-R''}$.

And if acetylenediol. was detected, then this should be quite doable compound.

You can find a nice discussion of diethynyl ether and some of its derivatives here:
Ooi, I. H., & Smithers, R. H. (1989). Bis [(trimethylsilyl) ethynyl] ether: a moderately stable C4H2O derivative. The Journal of Organic Chemistry, 54(6), 1479-1480.
https://pubs.acs.org/doi/pdf/10.1021/jo00267a053
They write that "little or nothing is known about diethynyl ether", and they were only able to find theoretical (INDO) calculations in their literature search about the compound, as opposed to actual reports of its synthesis. They do mention that the following authors managed to synthesize a dimethyl derivative, $\ce{Me-C#C-O-C#C-Me}$, but reported that it explodes at room temperature (and I infer, from what Ooi and Smithers wrote, that diethynyl ether would be even more unstable):
Brandsma, L., & Arens, J. F. (1962). Chemistry of acetylenic ethers LVI Di‐(alk‐1‐ynyl) ethers and thioethers. Recueil des Travaux Chimiques des Pays‐Bas, 81(6), 510-516.
https://onlinelibrary.wiley.com/doi/abs/10.1002/recl.19620810610
They themselves (Ooi and Smithers) were able to synthesize a di-trimethylsilyl derivative, $\ce{Me_3Si-C#C-O-C#C-SiMe_3}$ which, while stable when dissolved in $\ce{CH_2Cl_2}$ and stored in a freezer, degrades overnight when stored neat at room temperature.
In summary, for $\ce{R-C#C-O-C#C-R}$, larger R's provide added stability:
R = H (diethynyl ether): If it were ever synthesized, it would probably have a very short half-life, except under specalized conditions.
R = Me: Has been synthesized, but explodes at room temperature
R = $\ce{Me_3Si}$: Low to moderate stability, as described above.   Ooi and Smithers write that the added  stability provided by the trimethylsilyl groups "is probably both steric and electronic."
See also:
Arens, J. F. (1960). Ethynyl ethers and thioethers as synthetic intermediates. Advances in organic chemistry, 2, 117.

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pH indicator that will fluoresce under UV light
I'm interested in a pH indicator that will strongly fluoresce under a typical black light that anyone could buy.  I want to use an acid base reaction to get an idea of how fast two liquids mix in a new type of mixing device.  Therefore the pH indicator has to have a very rapid reaction rate with either acid or base so that mixing is the rate limiting step.
I remember someone using coumaric acid decades ago for something similar. He showed me some beautiful photographs of how the fluorescence went from bright green to colorless when the pH went from basic to acidic. However, I don't remember the particular isotope and can no longer contact this person.  In my planned tests, I was thinking of using a 20% excess of either acid or base (depending upon the chosen indicator) so that after the liquids mixed and reacted, the resulting solution would be either strongly basic or strongly acidic.  Of course this depends on the indicator.  But something like the coumaric acid idea would work well in my case where I want to take photographs.  Ideally, I would like the chosen indicator to be environmentally safe.
Thanks.

https://www.google.com/url?sa=t&source=web&cd=14&ved=2ahUKEwiEn-L4k83oAhVPOs0KHfYNAkkQFjANegQICRAB&url=https%3A%2F%2Fwww.ncbi.nlm.nih.gov%2Fpmc%2Farticles%2FPMC3647017%2F&usg=AOvVaw3oEkDke0eV-pwd9dneB3LR

E.g., https://www.thermofisher.com/de/de/home/references/molecular-probes-the-handbook/ph-indicators/overview-of-ph-indicators.html mentions some of these as suitable for a determination in living cells.

I've collected fluorescent rocks. Typical lighting washes out any fluorescence so you really need to look at UV fluorescence in the dark. Also glass doesn't transmit UV well. // How about adding a strong base, say NaOH, to an acidic solution with phenolphthalein? The indicator goes from colorless to pink.

@MaxW Agreed, behind the (normal glass) window is no sun burn. The OP could mount the UV lamp from the TLC box over the beaker containing solution A and receiving reagent solution B to bypass this limit.  Yet equally there is fluorescence microscopy with samples dyed (e.g., to mark nucleic acids) with stains sometimes excited by UV, too (e.g., DAPI, https://en.wikipedia.org/wiki/DAPI).

Going the other way, you could use quinine (from ordinary tonic water or more concentrated solution) as the fluorophore and quench the fluorescence with chloride ion (from ordinary salt water). For the light source, use a blacklight or 405 nm BluRay laser diode or UV LED, etc.

Possibly this answer would help you: https://chemistry.stackexchange.com/a/111455/61578

@EdV: BODIPY dyes are very expensive so won't fit into OP's budget. :-)

@MathewMahindaratne Ah, your linked answer is your usual excellence! I have bookmarked it for the next time! As for the BODIPY, that link was what popped up in a quick search of “fluorescent pH indicators”. Personally, I would try the salt water and tonic water: safe, inexpensive, and no nasty dye mess if a leak happens. I had a CHEE colleague who studied fluid mixing, using ethylene glycol and rhodamine 6G laser dye. Cool under blacklight illumination, but it attracted lots of ants and dye ended up all over.

@EdV:  The salt water and tonic water idea sounds interesting.  However, I haven't been able to find a reference indicating how fast the quenching reaction is.  The only thing I saw was a paper where the assumption was made that the reaction was fast enough to result in a diffusion-limited process.  My mixer will mix two low-viscosity liquids in a few milliseconds or less.  So, do you know how large the reaction rate constant is? Also, the concentration of quinine in tonic water is only 82 ppm.  This will result in very low concentrations reducing the reaction rate below what might be expected

My gut feeling is that quenching will occur in under a millisecond. The quinine concentration can be increased a great deal by starting with quinine sulfate and making higher concentration solutions in 0.05 M sulfuric acid. As well, the higher the salt concentration, the better. In any event, the problem is basically the same for mixing a pH dependent fluorescent dye and acid or base. High concentrations are good!

@EdV:  My calculatons show that 82 ppm of quinine is a molar concentration of about .000253 moles/liter.  That's about 50 to 100 times lower than the concentrations I was thinking about using for an acid-base reaction using an indicator. The solubility of quinine sulfate in water is 2mg/mL or by comparison 2000 ppm.  I've seen some tonic water syrups available.  I wonder if I couldn't mix higher concentrations using syrup and water.  My understanding is that the 82 ppm amount is health imposed limitation.

Yeah, the 82 ppm is a health imposed limit. You can definitely use the syrups and I just checked ebay: quinine sulfate powder is available. It costs about 4 dollars per gram.

I'm looking for "black light" LED units on amazon.  Looks like wavelengths are 395 to 400 nm.  One manufacturer says to choose a slightly lower wavelength for better fluorescense (around 375 to 390 nm).  Any advice is appreciated.

The wavelength range you give would be somewhat better, but not by much, and the 395 to 400 nm LEDs are inexpensive, so you can run a dozen. The peak absorbance wavelength is 350 nm, but there is no reason to pay more for LEDs with that nominal wavelength. And without a spectrometer, you cannot easily tell if you are getting what you paid for. Sorry I did not see your comment before now!

It looks like I can't buy quinine sulfate since I don't have an established commercial address.  Can anyone suggest other fluorescent indicators that I could buy as an individual or another way to get quinine (maybe from a concentrated syrup or something similar).

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Does aluminium foil have an odour/gas off?
I have used some aluminium foil to cover some windows and I notice a light smell in my small flat. I could be mistaken and it could something as I've been moving things around but this leads me to wonder does aluminium foil have an odour/gas off?
I notice it sticks to things as well so might this be an added solvent or what property of the foil gives it this characteristic?

https://chemistry.stackexchange.com/questions/7916/why-can-we-smell-copper

Aluminium doesn't smell. But then again, foil isn't pure aluminium. There might be a polymer coating, to begin with...

Standard household aluminum foil has a light oil coating. You can buy lab grade foil that has been solvent cleaned. It is, of course, more expensive.

Here is related material I once posted:

Also, commentary in Bretherick, Volume 1, page 23-24, to quote:
A student mixed aluminium foil and drain cleaner in a soft drink bottle, which started emitting gas. Another student carried the bottle outside and was claimed to have been overcome by the toxic fumes [1]. Most drain cleaners are alkalis, so that aluminium will dissolve to produce hydrogen. The bleach that it is is suggested may have been present will produce no toxic fumes in alkali, and one would be surprised to find arsenic or antimony compounds present. If the collapse was not purely hysterical, the remaining, though remote, possibility would be phosphine. The soft drink the bottle had contained was one of the many perhaps best described as impure dilute phosphoric acid [2]. "
A related comment attributing the formation of gases like arsine and stibine from nascent hydrogen per this source.
Education on nascent hydrogen interaction is important so as to avoid toxic gases that may be inadvertently and unexpectedly created in significant amounts.

Now, in the context of extensive employment of Al foil in your home, there is a potential danger that an acid/base is attacking the foil. This could release active (nascent) hydrogen atoms.
The latter could interact, for example, with compounds in physical contact, like paint (containing perhaps zinc sulfide). This may, hypothetically, result in the release of odorous hydrogen sulfide. This is a problematic gas, but even more deadly gases could arise also (per source above).
As such, I would advise removing the Aluminum foil (or, at least areas in contact with paints,...).

There is no nascent hydrogen! It is an incorrect notion: see the short wiki article on “Nascent hydrogen”!

+1 More about drain cleaner: Understanding sodium hydroxide drain cleaner's exothermic dissolution of aluminum; does the aluminum need to be oxidized before it dissolves? and also Is it known for sure that bases feel slippery because of the production of soap/surfactant?

Nascent hydrogen is a historical convention due to what is now understood to be hydrogen atom radical .H (= H+ + e-), which functionally can reside on the Al surface, at least for a period of time. It chemically behaves as a  (e-,H+) pair, for example with PbS:  Pb(2+)S(2-) + 2 (e-,H+) --> Pb + H2S

Note, I also state in my answer "active (nascent) hydrogen atoms".  Believe it or not, this chemisorb H on Al has been presented as a tool in the hydrometallury processing of metal ores (I will provide an ebook link if anyone requests).

Personally, I don't mind the odor from aluminum foil, whatever little there may be. Aluminum foil (household style) contains aluminum and its hard-to-separate sesquioxide (Al2O3), but also oil from the manufacturing process used to make it. If the odor you are sensing really comes from the foil, then that might be it. (though in my opinion unlikely) If so, there are more high quality options available for purer foil. Usually, I don't sense any smell from aluminum foil, except for when it gives off a "burnt" odor from an electrical discharge during a plasma globe experiment.
So if it is a burnt type of smell and the foil is the problem, it could be from the sunlight scorching the foil to the point where it emits odors. This would most likely happen over a long period of time, which i can't clarify at the moment because i don't think I have ever left out aluminum foil that long under direct sunlight before. The sticking phenomenon could be related to the solvent like you suggested, and this also might be a side affect of long term exposure to the sunlight.
If the smell itself is not like the burnt smell from aluminum conducting current, than it could still be from the oil solvent that was left under exposure to the sun. Aluminum itself should not be responsible for the stickiness, but the smell, like I said before, could be from scorched aluminum sesquioxide, and if not that, then the oil coating.


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Does direction matter in breaking a bond?
Im not sure if this question belongs here or in another community. I was wondering if the direction that we apply a force in order to break a bond matters and how it affects strength. Say  we have one compound e.g NaCl. Compared to breaking the bond between the Na ion and Cl ion in a tensile manner, would it require a different force to "tear" it in a shearing manner.

Why, surely the tensile strength and shear strength are two different things, otherwise why would we bother to have two different terms for those?

Normally those two types of forces are mentioned for materials in bulk and they make sense in that context. I was wondering if the differences still applied down to the atomic level so I made sure to specify that I am talking about "one compound" in the question.

I think the problem is not that trivial. You are talking about cleavage anisotropy of crystals. Search this keyword. NaCl crystal is isotropic, it implies directions does not matter.

Although you ask specifically about NaCl, unfolding proteins using force (breaking H bonds) the direction in which the force is applied does matter as does the rapidity of applying that force.

Twenty kilograms of NaCl is still one compound. You probably wanted to say "one molecule" or "one formula unit".

@M.Farooq Thank you, I will look more into it but from an initial glance it seems like that the papers looking into that topic are still talking about crystal structures in bulk

@porphyrin thats very interesting. That definitely says something about directionality affecting strength at a molecular level. I will look more into it. Thanks

@IvanNeretin Yea it felt weird typing "compound" but I didn't want to say molecule because NaCl is not covalently bonded.

There is an example here 'Pulling geometry defines the mechanical resistance of a beta-sheet protein',  Brockwell, et al.. Nature Structural Biology (2003), 10, 731

Thank you for the reference. In the example they present, it seems to have required a lesser force to shear than to rupture. The article is examining somewhat of a "bulk" material with multiple hydrogen bonds but the results are still interesting.

A possible real-world example of directional force use may be apparent from the commercial application of so-called magnetizers (see discussion here).
I once experimented with a small inexpensive one and it appears to accelerate select reactions.
A commercial example for reducing the amount of chlorinating agent required in public swimming pools in England demonstrated success to reduce scale build-up, moderate pH and increase available chlorine in a pool, employing magnetic field treatments here.
Some discussion details on how magnetic fields may potentially impact chemical reactions per comments posted previously on StackExchange.
An important point concerns radical based reactions, which per chlorine-based chemistry in sunlight treated swimming pools, is likely extensive and may be potentially impacted. This could particularly alter pool chemistry especially surrounding the fate of the chlorine radical, where an alteration of spins along with a possible decrease in the reverse recombination reaction to Cl2 via the normal reactions:
$\ce{Cl2 + hv -> •Cl + •Cl}$
$\ce{ •Cl + •Cl -> Cl2}$
where the above reaction is impaired due to spin changes by magnetic fields, resulting in more chlorine radicals. The latter can slowly react with water, for example, importantly producing the powerful hydroxyl radical:
$\ce{•Cl + H2O <=> HCl + •OH}$
Source: See Table l in "Photochemistry of HCl and Other Minor Constituents in the Atmosphere of Venus" by Prinn, R.G. as an example.
More detail on mechanics is provided by this source, to quote:

Radical pairs in chemical reactions are known to respond to magnetic fields. This involves pairs of transient radicals being created simultaneously, such that the two electron spins, one on each radical, are correlated. Their chemical fate is largely dictated by weak magnetic interactions, so that you get radicals with unpaired electron spins either antiparallel (singlet state) or parallel (a triplet state).
Even though the interactions of electron spins with an applied field is very weak, there is enough time to affect the balance between singlet and triplet radical pair states and therefore to change the yields of the products formed from them,...


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Disubstituted cyclic alcohols - looking for IUPAC rule
Given a simple cyclic alcohol:

We can name it cyclopentanol and using cyclopentan-1-ol wouldn't be correct. When more than one alcohol group is added:

we must use numbers and name it as cyclopentane‐1,1‐diol. However, what about situation like below?

In my understanding such compound should be named 1‐methylcyclopentan‐1‐ol and that's how I always named them. However I got into an argument with a person claiming the name of 1‐methylcyclopentan‐1‐ol is invalid and 1‐methylcyclopentanol must be used instead.
Does anyone maybe have a better google-fu and can locate a rule which specifies how such situation should be handled according to Gold Book or Blue Book?

"...1‐methylcyclopentan‐1‐ol is invalid and 1‐methylcyclopentanol..." - what's the difference? butene can be written as but-1-ene. isn't it?

@RahulVerma I'm pretty sure he's asking which one is IUPAC official.

Please note that your use of hashed wedges is not correct.

For preferred IUPAC names, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), explicitly stipulates when locants are omitted. Most of these rules apply to the omission of the locant ‘1’.
In particular:

P-14.3.4.2 The locant ‘1’ is omitted:
(…)
(c) in monosubstituted homogeneous monocyclic rings;
(…)

One example for this rule given in the Blue Book is cyclohexanethiol. Likewise, cyclopentanol is the preferred IUPAC name for the first compound that is given in the question.

This rule, however, is not applicable to the other two compounds that are given in the question since they are not monosubstituted.
Furthermore, we can find a general rule that is relevant in this case:

P-14.3.3 Citation of locants
In preferred IUPAC names, if any locants are essential for defining the structure of the parent structure or of a unit of structure as defined by its appropriate enclosing marks, then all locants must be cited for the parent structure or that structural unit. For example, the omission of the locant ‘1’ in 2-chloroethanol, while permissible in general usage, is not allowed in preferred IUPAC names, thus the name 2-chloroethan-1-ol is the PIN. (…)

Therefore, the preferred IUPAC names are cyclopentane-1,1-diol and 1-methylcyclopentan-1-ol, respectively. The omission of the locant ‘1’ is not permissible for the preferred IUPAC name.

Nevertheless, the name ‘1-methylcyclopentanol’ describes the correct compound and may be used in general nomenclature.

Sad to see loong's work not acknowledged.


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Elimination reaction
In this question i understand that we have to break C- N bond. But which one to choose?

If the question has been sufficiently answered, please accept it!

This is an example of Cope Elimination where the base takes up hydrogen from the least sterically hindered Carbon atom to avoid crowding in the Transition state. Hence, Hoffman Alkene is obtained and ethene is formed.
The reaction is not an example of E1cb since the Leaving Group is a stable alkene whereas the basic requirement for a reaction to be categorised as E1cb is that the leaving group should be a weak base.


text:
Counting number of acidic-H in a molecule
In a recent exam, following question was asked:

Number of acidic hydrogen present in X are:

I've found 8 of them as shown:

7 of them (marked with numerals) were correct according to solution, but 8th one (pointing with red arrow) was not. Is it because of the fact that the carbon was bridge-headed or something else?

I guess the trans conformation of the decalin-like ring might not permit the transition from the p orbital to the antibonding of π C=O for the marked hydrogen

@Yusuf Hasan: How do you know that the molecule has a trans conformation (sic)? Perhaps you meant configuration. I suspect that the inventors of the problem assumed that the order of deprotonation  would be 6, 3(4), 1(2).... The enolate from 3(4) might render 8 less acidic. All in all, it is a terribly constructed question.

It depends on the criteria used. In aqueous solution, only the H form COOH is acidic when heated by NaOH. The H7 is also acidic, but not in water. All H atoms are weak acids, sometimes extremely weak.

An aldehyde or a ketone with an alpha hydrogen forms a carbanion that resonates to enolate form. This leads to two canonical structures that are in resonance. The example below is a polyphenol compound extracted from turmeric, called curcumin.

source: Biomaterials. 2010 May;31(14):4179-85. doi: 10.1016/j.biomaterials.2010.01.142. Epub 2010 Feb 23. (https://www.ncbi.nlm.nih.gov/pubmed/20181392)
The hydrogens between two keto groups are most acidic. This is due to formation of two equivalent enol structures stabilized by hydrogen bonding. The structural requirement would be planarity of atoms involved in hydrogen bonding.
Below are 3-D structures of the given compound in the question above.

Structure without hydrogens

Bridgehead carbon is 4. Keto groups are at 2 , 5 and 8 (shaded in grey). Keto groups on 5 and 8 are not in the same plane . Hence Hydrogen on 4 , 22 (see figure with hydrogens) does not form stable hydrogen bonded enol structures as discussed in the above example.
Therefore, hydrogen at 8th one (pointing with red arrow) may not be acidic.
Note: 3-D structures have been drawn on Chem3Dpro

Structure with hydrogens

Structure with hydrogens (2nd view)

Thanks for your 3D structures, it was very helpful. But, is it true that those keto groups will always be non-coplanar (when both placed adjacent to bridgehead carbon), or is it specific to this one? (I'm asking this bcoz, I'd been provided only with 2D structure)

@Rahul verma if two keto groups are in two 6 membred rings seperated by a carbon bonded to these two rings , the two keto groups may not be in the same plane.

Why is hydrogen 5 acidic then? The two carbonyls are non co planar too.

It is because of the fact that a double bond is not usually stable at the bridge head position in smaller fused rings.
Therefore the hydrogen cannot be considered as acidic.

I don't understand. Where did the double bond come from? We're just asked to find the acidic hydrogens, right?

You may relate this with hyperconjugation and look into what makes a hydrogen acidic

Are you referring to the Bredt's rule?

Yes I was indeed referring to Bredt's rule


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Odd looking carbocation stability order
I am studying introductory organic chemistry, suddenly this one started looking odd,
The carbocation stability for three compounds as given by my book is shown below,

Let me call those as compounds A>B>C as given in image. It is very well known that 'N' and 'O' are electron withdrawing group(shows -I effect) and they can also show +M effect( some texts call it as Resonance effect ) by delocalising their lone pair over the vacant p-orbital of the carbocation.
Now all i want to know is, How to decide whether +M effect or -I effect will operate in this case? If I consider only inductive effects, the order would be C>A>B   which is wrong in this case.
One part of my text book says that +M effect > -I effect (in more than 99% of cases). But then too the next question arises, why +M effect of 'N' is more than that of 'O'?
I got stuck in this one and i am now asking whether such questions are feasible without conducting experiments.
Explanation with the principles which are familiar to my level would be nice.

You already stated (correctly) mesomeric effects and inductive effects may occur simultaneously, and sometimes one of them dominates the other.  Think a moment about rules about electrophilic aromatic substitutions on benzene: For the entry of a second substitutent, you have substitutents directing preferentially either a) toward ortho- and para-substitution, or b) toward meta-substitution -- where both the mesomeric influence typically is more important. And c) the halogens where both mesomeric effects (free electron pairs) and inductive ones ($\Delta{}EN$) affect the reaction.

If hetro atom with lone pair is neighboring to carbocation, then that lone pair can be donated to empty orbital on carbocation (see figure below). These orbitals are said to be in conjugation. Conjugation leads to delocalization of electrons resulting in resonance structures.

Quoting from "Organic Chemistry" by T.W. GRAHAM SOLOMONS , CRAIG B. FRYHLE .SCOTT A.SNYDER
12edition:

Structures in which all the atoms have a complete valence shell of electrons
  (i.e., the noble gas structure) are more stable.

The third structure (your structure C) does not have conjugation. Hence it is comparatively less stable then 1st and 2nd structures.
Among 1st and 2nd structures , due to lower electronegativity of Nitrogen compared to Oxygen , Nitrogen has greater electron pair donating tendency. Therefor the order of stability is as follows.

How to decide whether +M effect or -I effect will operate in this case?

The extent of stabilizing effect follows the order: $\ce{\text{Mesomeric} > \text{Hyperconjugation} > \text{Inductive}}$.
In general, this order is based on extent of $\ce{e-}$ transfer. In mesomeric effect, $\pi$-bonds are in conjugation which completely transfers $\ce{e-}$ density to carbocation. On the other hand, in inductive effect, induction of charge takes place, which partially transfers $\ce{e-}$ density through $\sigma$-bond. Hence, it is weaker.

why +M effect of 'N' is more than that of 'O'?

As $\ce{N}$ has lower electronegativity than $\ce{O}$, hence it has greater $\ce{e-}$-donating tendency, therefore greater $\ce{+M-\text{effect}}$

According to you, when one answers theoretical questions like this, our thought process should be like , 1) +M > -I . 2) if (1) is true, COMPLETELY IGNORE INDUCTIVE EFFECTS. 3)find the strongest +M ??? Does this always work? I hope you understand that ignoring the inductive effects pesters me a bit.

@MukunthA.G: 1) It's not feasible to cover every detail of the answer, unless you've asked in the post itself. 2) You asked for "find the strongest +M?" - I would suggest you to either search via google for previously asked question on this site, or maybe ask a new question. 3) You can read ron's answer in: Competing resonance and inductive effects in a substituted benzene...

...Also, you should prevent asking multiple questions here. You may've gone through the how to ask a good question page earlier, but it would be nice to have a look at it, if you feel like it.

I'd try to give my explanation with the principles simple enough so that it would familiar to your level of education:
Rahul Verma has given simple enough explanation to answer most part of your question. I said here 'simple enough' because of your subsequent comments. However, I have to point out that mesomeric effect ($M$) almost always predominate inductive effect ($I$), even when the distance between two concerning centers are neighbors like in the case in hands where inductive effect is in its highest level.
When consider structures (A) and (B), the electron-withdrawing inductive effect ($-I$) of (B) is lager than that of (A) simply because electronegativity of $\ce{O}$ is higher than that of $\ce{N}$. However, the electron-donating mesomeric effect ($+M$) of (A) is lager than that of (B), which can be simply explained by following reason you have learned in general chemistry classes:
Consider $\ce{H2O}$ and $\ce{NH3}$ for their affinity for electron donating:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
The solution is always basic ($\mathrm{p}K_\mathrm{b}$ of ammonia is about 4.75) because ammonia donate its lone pair to grab a proton from water. It'd never be other way around like: $\ce{NH3 + H2O <=> NH2- + H3O+}$. By this phenomenon along, we can assume electron donating ability of $\ce{NH3}$ is comparatively larger than that of $\ce{H2O}$. Hence, one can consider that $(+M)_\ce{N} \gt (+M)_\ce{O}$.
Based on these two comparisons, it is evident that $(-I+M)_\ce{N} \gt\gt  (-I+M)_\ce{O}$ when compared to net positive value. Therefore, one can conclude that (A) is stable than (A).


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Unable to form the product formed via rearrangement

Ok so I first reacted one of the carbonyl groups with NH2OH to form an oxime. And then a Beckmann rearrangement takes place followed by tautomerism. Then the lone pair on nitrogen did an intramolecular attack on the other carbonyl to form a six membered ring. But I am unable to form the correct product which is option (A). Any help on how to proceed and is my approach correct?

Related: https://chemistry.stackexchange.com/questions/124166/product-of-the-reaction-between-hexane-2-5-dione-and-ammonium-carbonate/124171#124171

Amide nitrogens are not nucleophilic, the oxygen is. No need for the Beckmann.

Forget the Beckmann, the oxime is still nucleophilic through N (imagine the N-hydroxy enamine tautomer if that helps). Have it attack the second carbonyl intramolecularly (and hence favoured) to give a dihydropyridine. Dihydropyridines are unstable with respect to aromatisation unless substituted with strongly electron withdrawing groups. The N-hydroxy-dihydropyridine spontaneously will lose H2O to give the pyridine.


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Chemical equilibrium — why multiplication, rather than addition?
Why do we multiply (and not take the sum) the concentration of products and reactants when more than one entity is present in either side and rise the power to the coefficient term?
Suppose either
$$\ce{aA + bB <=> cC +dD}$$
described by
$$K_{\mathrm{eq}} = \frac{[\ce{C}]^c \cdot [\ce{D}]^d} {[\ce{A}]^a \cdot [\ce{B}]^b}$$
where we multiply the concentrations and rise the power to the coefficient. Or the dissociation of
$$\ce{A_aB_b <=> aA+ + bB-}$$
described by
$$K_{\mathrm{sp}} = [\ce{A}]^a \cdot [\ce{B}]^b$$
again based on a product, and not the sum.
This seeks a clarification of math based on chemistry. How to explain it to an 8th grader?

Please provide an explicit and specific example; otherwise, it is guess work /  asking an oracle if you refer to e.g., the equilibrium constant, or solubility product.

Please recheck the edited question.

To render chemical and mathematical equations better, ChemSE is supported by mathjax and mchem.  This works well for the questions, answers, and comments.  The commands are shown on ChemSE's meta site (https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here).  Your are very welcome to venture out the syntax and use it for future posts.

There may be a proper answer,  but the answer may not be applicable well for the target audience.

For the 8th grader, hmmm...
Let suppose this funny equilibrium equation:
$$\ce{2 S <=> H}$$
Where $\ce{S}$ are ice skaters on a giant skating ground and $\ce{H}$ are "collision heaps" of 2 skaters laying on the ice.
Now, what is the frequency of collisions leading to a heap, if there is "skater concentration" $\ce{ [S] }$ ?  It should be $k_1 \ce{[S]}$ as frequency should be proportional to $\ce{[S]}$, correct ?
But wait ! When there is more skaters, each skater would collide with  other skaters more frequently,  proportionally to $\ce{[S]}$ as well !
So collision frequency would be like $k_1 \ce{[S]}^2$
Now the heap "recovery":
It should be proportional to $\ce{[H]}$, as more heaps, more skaters would be  raising at the same time. So the heap recovery rate would be like $k_2 \ce{[H]}$.
Now imagine the equilibrium on the skating ground, where the rate of collisions is equal to the recovery rate
and number of collision heaps is constant.
Then $k_1 \ce{[S]}^2= k_2 \ce{ [H]}$
If we express the equilibrium constant $K=k_1/k_2$, then $$ K = \frac{\ce{[H]}}{\ce{[S]}^2}$$
Do you see ? No sum, but multiplication. Twice as many skaters leads to 4 times more heaps.
You can create similar system, where just for fun collide just boys and girls together.

Nice analogy! But all all the A's should be replaced with S's, to be consistent with your first equation.  And, just to make the origin of the exponent fully explicit, I'd recommend writing $K = \frac{[H]}{[S][S]} =   \frac{[H]}{[S]^2}$

Thanks, weird I have not notice. :-) Fixed.


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Energy of the Orbital versus Energy of the Electron
When introducing the hydrogenic atomic orbitals, it is said that the 2s and 2p orbitals are degenerate. I have also read that for atoms with more than one electron, like He, the 2s and 2p are NOT degenerate.
How does the addition of a second electron in the 1s orbital change the energy of the 2s and 2p orbitals?

What are orbitals, come to think of it? They are certain solutions of the Schrödinger equation which were found for H atom. Now why would they be related in any way to the solutions for any more complicated atom?

Title has very little to do with the rest, here.

Let's imagine triplet $\ce{He}$, hence one electron in the 1s. The other electron could not be in the 1s, too (it would have to have opposite spin, thus, singlet $\ce{He}$). It is in either the 2s or one of the 2p orbitals.$^{1}$ Among other things, the Coulomb interaction with the 1s orbital/electron will affect their orbital energies. Given that the 1s and 2s have the same symmetry, but each of the 2p have a symmetry different from them, it is reasonable to assume that the Coulomb interaction between 1s and 2s would be different from the one of 1s and 2p.$^{2}$ Thus, the orbital energies are different.
$^{1}$ Though I would think that the electron density would always be spherically symmetric for the isolated atom.
$^{2}$ This can also be calculated when assuming some parameters for the helium orbitals (which can be calculated exactly for the one-electron case).


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Aromaticity in resonance stabilisation
In this question;

(2) is, apparently, the correct answer, as it has the most resonating structures.
In comparing resonance- structure stability, shouldn't you also check whether it destabilizes the species by making it anti-aromatic?
For example, in option (1), resonance into the left ring would result in it having 8π-electrons, making it anti-aromatic.
Is this not considered because it's just a resonance hybrid, and Hückel's rule isn't considered for hybrids?

Of course you need to consider it. "most resonating structures" is largely irrelevant. (2) has both rings aromatic, that's the reason.

So (2) isn't it?

Yes! as @Mithoron explained.


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EDS analysing the elemental composition
Just wondering if I'm interpreting this correctly. So in SEM, and particularly the EDS mode, the beam irradiation refers to the beam irradiation from the thermionic emission gun. This paper measures the EDS spectra with varying degrees of irradiation in a 2D structure of BiOCl, noting that the structure of the compound changes under high beam irradiation to disintegrated domains of bismuth nanoparticles. I'm wondering is this a problem that is often encountered in EDS when trying to analyse the elemental composition of a sample in many other compounds?
Many Thanks

Of course the electron beam tears up the bonding on the surface. The SEM is like a 50 cal machine gun shooting at a brick wall.

Ok thanks, I'm just reading about SEM for first time today, so I don't know much about it. Would it have a big effect on most sample materials, and would the primary reason choosing a shorter live time be that a change in composition occurs?

I'm not sure what you're trying ask. Let me add that a SEM operates in a high vacuum, and the sample must be coated with a conducting film. So an actual loss of material from the surface (and SEM EDS is a surface technique...) is due to volatile molecules being formed which can dissipate into the vacuum chamber. From instance $\ce{Cl-}$ in a compound going to $\ce{Cl2}$. // Does that help?

Thanks this helps in understanding. I meant if you were to operate an SEM in EDS mode, would it be a concern that you will get an inaccurate reading of the composition of the elements in the sample, or does this generally only happen at high irradiation. I've read that you generally need to get more than 10,000 counts, so I'm wondering how much of a problem would the altering in elemental composition due to electrons bombarding the sample be at this scale.

IT DEPENDS...  How much an analysis might change can't be known without knowing ALL the experimental details. The gist is this. You analyze long enough to get a spectra. Say 30 seconds. You zap the specimen for 10 minutes more, then you analyze the same spot again for 30 seconds. Are the first and second spectra the same?

@AleksejusPacalovas, This paper is missing a fundamental information, what the electron beam energy? Read more and more papers and don't trust the spam and bologna of a single paper.

@M.Farooq: The paper notes " All imaging results were obtained with an accelerating voltage of 200 kV unless otherwise specified"


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How are molecular orbitals obtained in quantum chemistry?
When trying to solve for a molecule's molecular orbitals, we start off with the total molecular hamiltonian. However, given the difficulty of solving the Schrödinger Equation for this complicated hamiltonian we can introduce the Born-Oppenheimer approximation. By assuming that the nuclei of molecules are relatively fixed, our molecular hamiltonian can be reduced to the electronic hamiltonian. Then, we can use the electronic hamiltonian and solve for the electronic wavefunctions of our molecule.
This calculation for the wavefunctions is significantly easier. What is confusing to me is what the solution to the electronic Schrödinger equation represents.
Are these solutions considered the molecular orbitals?

I think you are close to the right idea, but let me try to clarify some points.
The full Schrodinger equation for a molecule should depend on the $3N$ nuclear coordinates, as well as the $3n$ electronic coordinates, where $N$ and $n$ are the number of nuclei and electrons respectively. This is challenging both due to the number of coordinates and, in particular, due to the coupling between electronic and nuclear coordinates. However, with the Born-Oppenheimer approximation, we assume they are decoupled and solve for the electronic part of the wavefunction at a fixed nuclear geometry.
The issue here is that we still can't solve the Schrodinger equation for an $n$ electron system with $n>1$. To simplify further, we can make the Hartree-Fock approximation, which (in one way of phrasing it) assumes that the $n$ electron wavefunction can be formed from the antisymmetrized product of $n$ one electron wavefunctions/orbitals. These orbitals can be determined by solving a nonlinear system of equations and then stitched together in the form of a Slater Determinant to make an approximate $n$ electron wavefunction.
While chemists like to frame discussions in terms of orbitals, strictly speaking orbitals are always an approximation (except for hydrogen) and the electronic structure of a molecule is determined by a $n$ electron wavefunction. Even though orbitals aren't quite real, they still can give a pretty good qualitative description of a molecule's behavior.
For a bit more about the meaning of orbitals, you might want to check out this previous question: Is it correct to talk about an empty orbital?

Just to be clear, when you say " electron wavefunction can be formed from the antisymmetrized product of  one electron wavefunctions, which we call orbitals", is the molecular orbital = n electron wave function?

The molecular orbitals are one-electron wavefunctions that are combined together to make an $n$ electron wavefunction.

Ok I think I understand your answer now. It was difficult for me to understand initially because my professor is constructing MOs via the LCAO method. Looking through my notes, it seems the "one-electron wavefunctions" that you mention, are they MOs we obtain via LCAO? Also, for a given electronic state, the overall electronic wave function involves some product of those MOs (and a spin part), and it represents the OVERALL state of our molecule.

LCAO adds another layer to it. Basically, we can't solve for the MOs analytically, but if we define a basis of atomic orbitals(AOs), we can approximately express the MOs as linear combinations of the AOs. But yes, the overall state of the molecule is particular product of the MOs, which have a spatial and a spin component.

Since the is some confusion about the many- and one-electron wave function and which one the actual solution to the Schrödinger-Equation is: I would add the Slater-Determinant to the explanation.


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L-lysine acetate synthesis
Is it possible to synthesize L-lysine acetate in the following steps? First, mix L-lysine HCl with an excess of NaOH and just enough water so that the total water in the product is enough to solve all NaCl. Decant the water to obtain L-lysine free base. Then either mix with excess (concentrated) acetic acid and dry the product in a hermetic environment with silica gel, or pour acetic acid in a pot, wrap a cheesecloth on the top, spread l-lysine over it and let the acetic acid steam gently without cooking the lysine.
Aside: According to an article from 2015, L-lysine is usually available in hydrochloride and acetate forms because the free base form is unstable. However, L-lysine acetate seems to have gone out of the market. Is there a reason for that? Is l-lysine citrate also stable and could it be made by simply mixing l-lysine free base with citric acid?

In principle yes. That is essentially what is done in this patent: https://patents.google.com/patent/CN102070477B/en


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EDS of BiOCl (Conductivity)
For EDS mode of SEM, would you need to create a conductive path to ground for a BiOCl sample material. I have researched and found that gold and carbon can be used for such purposes, but I thought that BiOCl is conductive enough for grounding to occur.
The reason I'm asking this is I was reading a research article, and I came across the following EDS spectra of BiOCl:

I believe that on the left side "Cl" in purple is mislabelled, as Chlorine only has K-series emission, and the energy of such emission is greater than oxygen, so this led me to believe that the material is has to be smaller than Oxygen, and Carbon seems to be the obvious choice, but the research paper doesn't mention using anything apart BiOCl, and the mount which consists of Al and Cu.
Many Thanks

Frankly I'd guess that low energy peak labeled Cl is just noise. In a SEM EDS setup the Si(Li) detector needs to be cooled with liquid nitrogen. To prevent an ice buildup the x-rays must pass through a Be window. In order withstand air pressure and prevent slow air/moisture leakage into the detector, the Be window has to be thick enough to seal. Hence no such low energy x-ray is going to pass into the detector. // Note $\ce{Cl K_\alpha}$ escape peak is at 2.622 - 1.74 = 0.882 Kev, about the same energy as the Cu L lines.

Why are you taking away 1.74 keV from Kalpha of chlorine?

A $\ce{Cl K_\alpha}$ x-ray enters the $\ce{Si(Li)}$ detector and excites an electron in the $\ce{Si K}$ shell. The Si excited state decays with the emission of an $\ce{Si K\alpha}$ x-ray. The xray sometimes escapes from the detector carrying off 1.74 keV of energy.

Ok is see. So just 2 more follow up questions: 1. Would this mean that the 1.74 keV electron would also be detected by the detector too (1/2*4pi chance of each such electron not leaving the silicon matrix) 2. Why does the Cu line at roughly 8 keV not have substantial noise at 8-1.74 keV?

It isn't a 1.74 keV electron, it is a 1.74 keV x-ray. The Cu escape peak would definitely be there. It is just too low in intensity to be visible because the Cu x-rays go deeper into the detector on average making it less likely for the $\ce{Si K\alpha}$ x-ray to escape.


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What is the precipitate of aqueous copper (II) and polysulfides?
So if you heat up a piece of sodium engulfed in a large excess of sulfur, what you would get (from what I've read) is sodium polysulfide $\ce{NaS_n}$. But how large can $n$ get? If you left sodium polysulfide in molten sulfur, would you increase $n$?
If you tried to dissolve the sulfur-polysulfide mixture in water and filtered it, I assume you would get a basic aqueous hydrogen polysulfide solution. If you then poured in a solution of copper sulfate, besides $\ce{Cu(OH)2}$, would $\ce{CuS_n}$ also precipitate out with an arbitrarily large $n$, or would the larger polysulfides break up into elemental sulfur and lower polysulfides first and precipitate out $\ce{CuS_n}$ with a more discrete number $n$?
And if you acidified the polysulfide solution until you have only $\ce{HS-}$, $\ce{H2S}$ and suspended sulfur in the solution, from which you could get the precipitate $\ce{CuS}$, would you be able to calculate the initial mass of sodium used from the mass of copper sulfide?

What $\ce{NaS_n}$ compounds would result from putting Na in sulfur is an interesting question. I'd guess that such a reaction would create a complicated mixture of compounds. There are probably certain configurations that would be quasi-stable, such a six-membered ring.

Depending on the preparation, see  https://pubs.acs.org/doi/abs/10.1021/ja00166a054,  the 'n' may not be too arbitrary.


text:
How come I get different Gibbs free energy for the same reduced equation?
I want to calculate the Gibbs free energy of this reaction which can be:
$\ce{H2(g) + 1/2O2(g) -> H2O(g)}$
or
$\ce{2H2(g) + O2(g) -> 2H2O(g)}$
I am using the entropy values (J/molK): H2(g) - 130.684, O2(g) - 205.138, H2O(g) - 188.7
And enthalpy values (kJ/mol): H2(g) - 0, O2(g) - 0, H2O(g) - 241.8
With the equation ΔG=ΔH−TΔS and temperature 298.15K. How come I get ΔG = -457kJ/mol for the full equation and -228kJ/mol for the reduced equation, shouldn't  I get the same Gibbs free energy since they're the same reaction (except one is reduced)?
Shouldn't both be considered "correct" or is one "more correct" than the other?

Welcome on the ChemSE! We have quite a good Latex support here. Type in $\ce{H2O}$ and you will get $\ce{H2O}$. Look how beautiful became your reaction equation now. :-)

The equilibrium constant changes, too. The only thing that does not change is the standard cell potential because that is per electron, so to speak, and not per mol of reaction. It is kind of weird that an intensive quantity (i.e. molar Gibbs energy) depends on how you choose the coefficients. At least it does not depend on the amount of substance that reacts.

The change in Gibbs free energy is written per mole. For reactions, consider it "per mole of reaction." Since the stoichiometry of the second reaction is exactly double the first, "one mole of the first reaction" is half "one mole of the second reaction." Therefore, the value of the change in free energy is doubled as expected.
For an arbitrary reaction, there is no standard.
For the standard free energy of formation of a substance, the canonical reaction should generate one mole of the substance from its constituent elements (the first reaction here).


text:
How to store small amounts of gas
I need to store small amounts of gas. I already saw devices that look similar to the image below, but I don't know how they are called. Does someone here know how these devices are called? I am also open for alternative ideas. I mainly need some kind of storing devices that works with low pressure and that is capable of pushing the gas back if pressure is removed. A ballon for example wouldnt work since it requires to much force to get the gas inside. It would be awesome if someone here know more and can help me with this.

More of an engineering problem than a chemistry problem. What gasses? How much gas do you want to store? What/how do you want to use it?

No idea, but you should have a thick glasware catalogue somewhere in your lab. It´s bound to be in there. (Btw. your nick has one letter missing. Hopfenkaltschale?)

A rubber balloon does require some force to get filled. A mylar balloon (typically filled with helium) can be purchased empty (or can be emptied) and requires almost no force to fill; then can be squashed manually to eject the gas. They are available in many sizes.
Rubber balloons also tend to go flat or leak, so their storage utility, especially for low m.w. gases, is short-term (a day or so). Aluminized mylar balloons can hold helium or hydrogen for long periods (weeks and even a month or two). For short-term, you could also use a plain polyethylene sandwich bag - after all, many objects are shipped with bags of air in the boxes to help protect them from damage. Use a needle to insert the gas and retrieve it; tape the area around the needle insertion - like an intravenous operation! Or, you can tape a plastic or glass tube into the bag.
Longer storage would require a gas sampling or collection tube, a glass container with a valve at one or both ends. The internet has many references to these.


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Addition of Methyl Grignard and use of HClO4 in this question

Okay so the first step is the use of Grignard. I know that Grignard does a 1,2 addition(generally) and the attack happens directly on the carbonyl,and I think HClO4 basically does protonation.Not sure about the role of NaOH. Also the final answer given is 3,5 dimethylphenol. The final product is devoid of the oxygen which is present in the ring. How should I approach this question?

I think you are correct that the HClO4 is merely a protonation reagent, don't know why that was chosen. The NaOH, I think, is to do an intramolecular aldol cyclisation. To get the product 3,5-dimethylphenol the Grignard cannot be adding 1,2. Try drawing tautomers of the starting material and see if Grignard addition to one of those makes a reasonable pathway.

@Waylander I tried but i dont see an intramolecular aldol possible as the compund itself is a ring,i dont see how anintramol aldol can occur(its already cylised). A little more help would be nice. Thanks a lot!

The ring has to open. The MeMgCl attacks one of the carbons bonded to the oxygen in the ring giving an unsaturated dicarbonyl. The NaOH forms the anion which does the intramolecular aldol

Yup! Got it after seeing your answer! Just one doubt. How did u know that the attack had to happen on a tautomer? Basicaaly u have done a 1,4 attack right? thx!

My feeling is that showing the tautomer makes it easier to understand why the attack is at that point. As you correctly commented, unmodified Grignards will usually attack 1,2. This is a special case.

Consider the resonance structures of the pyranone. MeMgCl attacks the structure shown second. The perchloric acid workup hydrolyses the enol ether to give the diketone shown bottom right. NaOH promotes the internal aldol cyclisation to give 3,5-dimethylphenol.

They're resonance structures, not tautomers - not that it detracts from the other steps

You are correct, I have amended my answer


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What is the best way to make a substantial amount of soot in a high school laboratory?
I need to make about 50 grams of soot (carbon black particles) in a high school laboratory. What would be the best way to do this?
The purpose of needing 50 grams of soot is because I would like to carry out an experiment that investigates how the percentage soot in oil (e.g. diesel) affects the viscosity of the mixture. To measure the viscosity, I would have to put 500 ml of oil in a measuring cylinder and do the ball-bearing method. The highest percentage I would like to test is 10% soot and that calls for around 50 g of it.

Soot and CB have a very diverse, and different, morphology. CB particles usually have an onion like structure.

There is an extensive literature on producing carbon black see for instance Donnet, Jean-Baptiste, ed. Carbon black: science and technology. CRC Press, 1993.
Instead of adapting the Roman recipe for producing lamp black explained on page 19, I would use olive oil rather than resin or turpentine (that is quite a nasty substance). For doing it on a small scale maximizing the yield maybe I would try (under the fume hood)  to heat the oil in a flask attached to a water-cooled Liebing condenser after reaching the smoke point the soot should start to accumulate on the walls of the condenser. It should be quite easy to recollect the carbon black from the wall of the condenser with nylon bristle and maybe from a little funnel with a slight offset fixed on the exit of the condenser. All this should be carried out using heat-resisting glassware in a lab under the supervision of an adult.
P.S.

The highest percentage I would like to test is 10% soot and that calls
  for around 50 g of it.

This means that you will need more than 50 g for the whole experiment so if you test 10 different proportion you will need at least 80 g performing the experiment only once. It could require a lot of oil to produce 80 grams.

It is not clear if your target is carbon black - there are commercial sources for this (e.g., here) - or soot. Ideally, composition wise speaking carbon black would be just carbon, while soot possibly is a blend of compounds, depending on what was partially burned, under which conditions (e.g., oxygen / carbon ratio, temperature, residence time in the reactor) assuming a continous flow reactor.  But what about the reproducible distribution of shape and size of the particles?
Both automotive industry as well as raffineries know that viscosities of the oils depend e.g., on temperature, pressure and rotational speed (shear forces) of moving parts in an engine, as well as the composition of the oil / lubrificants used. Thus they use standarized test engines and test protocols to probe the endurance of these liquids (example). To ease exchange of the know-how in this field, CIMAC is one adress to retain.  Beside a brief survey about oil degradation, abstracts of their database are available for non-members, too.  Thus maybe CIMAC's PR office may assist in establishing a contact with a partner already performing such tests which is close to your area.

Try burning turpentine.
It produces a very soot rich flame, per my recollection.
Another path per the article: Preparation of Carbon Nanoparticles from Candle Soot.To quote:

Carbon nanoparticles (CNPs) were prepared by simply mixing candle soot into ethanolic Nafion solution under ultrasonic radiation for 90 min. Well-defined CNPs could be obtained after centrifugation and washing.

[EDIT] My colleague mentions other choices for a school lab, citing benzene as a non-choice. The reason, let's hear from what the CDC says about it, to quote:

Immediate signs and symptoms of exposure to benzene.
People who breathe in high levels of benzene may develop the following signs and symptoms within minutes to several hours:
Drowsiness
Dizziness
Rapid or irregular heartbeat
Headaches
Tremors
Confusion
Unconsciousness
Death (at very high levels)

Another option suggested is toluene, comments from Toluene Safety Tips from MSDS experts – MSDSonline, to quote:

Toluene is a highly flammable liquid and it can cause mild damage to the skin and the eyes. However, the most-common hazard associated with this chemical is inhalation. Products containing toluene can produce dangerous fumes which can cause nausea, headaches, unconsciousness, and even death if inhaled.

Note, my final choice, for high school labs in general, is safe ordinary candle soot with a reference citing a preparation path to nanoparticles. Now, I could make other more potentially efficient and likely problematic suggestions, but I sleep well with this conservative/responsible choice.

Turpentine (pinene) has a C/H ratio of only 0.625, toluene is much better at 0.875, beaten by benzene (no good for a high school lab) and polystyrene, with a ratio of one.

Thanks for the list of what I would not use in a high school lab (which you did note).  Insurance cost and lab experiments gone wrong are likely good reasons for schools not to have chemistry labs, which can be mitigated, in part, by avoiding dangerous and toxic reagents in the hands of students.

Well, smoke/soot is per se toxic if inhaled. What you cite for toluene is a default statement for any flameable liquid. Ethanol does the same, and certainly turpentine.

Per a source: Candle soot can damage property, but has not been shown to have any adverse health effects. To keep a candle from sooting, make sure to trim the wick to ¼” before burning.  Link: https://www.keapbk.com/blogs/keap/are-candles-bad-for-you-safe-toxic-unhealthy-soot-wick-wax-fragrance-indoor

I think the problem lies in the original demand for fifty grams.  :)

Nobody should follow the unsafe and dangerous practices in this answer.

I just realized that one could take one inexpensive candle and surround it with other candles and tie them all together. Upon igniting all the candles, one can significantly change the efficiency of the candle approach (just take precautions to avoid getting burned and perform in a safe, ventilated space).


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Significance of shapes of Conical flask and Kjeldahl flask
What is the practical significance of the shapes of conical flask and Kjeldahl flask?

For context, see this: https://chemistry.stackexchange.com/questions/46828/why-are-erlenmeyer-flasks-shaped-the-way-they-are

Question: What is the practical significance of the shape of conical flask (in US: Erlenmeyer flask)?

An Erlenmeyer flask (also known as a conical flask or a titration flask) is a type of laboratory flask, which consists of a flat bottom, a conical body, and a cylindrical neck. The flask is created by the German chemist, Richard August Carl Emil Erlenmeyer (1825–1909) in 1860, and hence its name. With its wide flat bottom, it was perfect for heating solutions (large surface area). Hot vapor condenses on the upper section of slanted sides of the Erlenmeyer flask, reducing solvent loss. Also, these slanted sides and small mouth of the flask cause the hot solutions in the flask to cool slowly, keeping the empty air space hot. This feature greatly aids certain techniques in chemistry, especially appropriate for recrystallization. For example, during the hot filtration step, when the hot solution is filtered through a fluted filter paper into the receiving Erlenmeyer flask, hot vapors from the boiling solvent would keep the filter funnel warm, avoiding the premature crystallization on the funnel and its stem.
Equally, the sloping sides prevent risk of spillage of solutions in it, therefore ideal to hold content, which needs to be mixed by swirling. This nature also makes them suitable for use in titrations. Again, the flat bottom of the flask has the advantage of giving its content a uniform color across the entire flask, which makes it easier to determine sharp end points.

Question: What is the practical significance of the shape of Kjeldahl flask?

In 1883, Danish chemist Johan Kjeldahl (1849–1900) of the Carlsberg Laboratory published a method to determine nitrogen content in organic matter. This is the first accurate, simple, and speedy method to determine nitrogen content in organic matter including protein content in food, which bares discoverer's name, the Kjeldahl method. The first step of the method is the digestion of sample in concentrated sulfuric acid at $\pu{360–410 ^\circ C}$, which decomposes the sample by oxidation to liberate the reduced nitrogen as ammonium sulfate. This heating would cause splash-back, which is a treat because of conc. $\ce{H2SO4}$. Thus, along with his famous method, Kjeldahl had also developed a piece of laboratory equipment he developed in 1888, which bares his name as well, Kjeldahl's flask. The long-necked, round-bottomed (actually pear shaped) flask was ideal for avoiding splash-back when heating solutions. The long-neck also function as air condenser.

Wow you explained it so well thanks!

An additional point is that in many South Asian countries, people pronounce it as "Jel-dahl" flask or method. Just like in Knorr noodles, K is not silent, same here. This "j" sound  is somewhat close to "y". "K'ell-dahl" may be close. Listen to the actual Danish pronunciation: https://forvo.com/word/kjeldahl/


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Basic Water Dissolving Salt Questions
I have a basic question about how water dissolves salt. In the Khan Academy explanation, it says that the H and Cl atoms attract each other while the O and Na atoms attract each other.
My question: Does this weaken the bond between Na and Cl within a single NaCl molecule? Or does it weaken the bonds between Na of one molecule and Cl of another salt molecule? Or does dissolution weaken both?

NaCl doesn't exist as molecules; it's an ionic solid.

What is the difference? Also, how important is that distinction at the beginner level of chemistry I'm at?

See: https://www.chemguide.co.uk/atoms/structures/ionicstruct.html and it's really quite important. If you haven't learnt that yet, you will very soon.

There are no $\ce{NaCl}$ molecules in the solid salt nor in solution.
In solid salt, there are $\ce{Na+}$ ions, surrounded by 6 $\ce{Cl-}$ ions, while each $\ce{Cl-}$ is similarly surrounded by 6 $\ce{Na+}$ ions. Here is the Wikipedia picture of the structure.
During the dissolution, water molecules reach for a particular ion, oriented toward it by the opposite partial charge of their electric dipole.
$\ce{Na+}$ ions are torn away and wrapped by $\ce{O}$ side of $\ce{H2O}$ molecules, which have a partial negative charge.
$\ce{Cl-}$ ions are torn away and wrapped by $\ce{2H}$ side of $\ce{H2O}$ molecules, which have a partial positive charge.
Ions end hydrated in solution like $[\ce{Na(OH2)}_n]^+$ and $[\ce{Cl(H2O)}_m]^-$

Got it. Thank you! Not sure if this question belongs in the same thread or not, but how can we still call it "salt" if the Na and Cl are separated? If the ionic bonds are broken, how does it retain its salty taste? Hasn't it undergone a chemical reaction?

Taste of the solid salt does not exist. Tongues do not know the taste of salt otherwise but as the taste of dissolved  and separated ions. BTW the equimolar mixed solution of NaCl and KNO3 has the same taste as the similar solution of KCl and NaNO3, as there is no difference in the solution.

Interesting. How far separated can those ions be and still have the taste of salt? Also, why are they still called ions if they're separated? How is the Na still giving the Cl an electron?

It is limited by concentration threshold of salt taste detection. When they are separated, and even when in solid salt, no Na+ ion belongs to  a particular Cl- ion and vice versa. In solution, they diffuse independently. Why not to call them ions when they have a charge ? Why giving? Na gave it before the crystal of salt was created.


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Creating 1L of 1000 ppb stock solution of Pb2+
I’m testing on samples that require a concentration of 1000 ppb (or 1 mg/L) of Pb 2+. My first thought was to simply dissolve 1 mg of Lead nitrate in 1 L of deionized distilled water to create 1 L of stock solution at the desired 1000 ppb (1 mg/L) concentration.
Is the process as simple as that, or is 1 mg proportionally too small a mass of solute to dissolve in 1 L of distilled water? Should I create a more concentrated stock solution and dilute it to 1000 ppb by adding requisite amounts of water? What concentration stock solution would be recommended in this case?
All the best!

Making the solution in a single step is inaccurate up to impossible.

Try to find a crystal of lead nitrate with the mass close enough to $\pu{1 mg}$
Try to weight up $\pu{1 mg}$ on scales with good enough accuracy.
Try to make such $\pu{1 ppm}$ solution stable enough.

The usual thing is to create a concentrated stock solution ( e.g $\pu{1000 ppm}$ and creating the working solution in 1 (1000 times) or 2 step ( 20x50 times) dilution whenever needed.
There is need to consider stabilizing by an acid, to prevent hydrolysis and adsorption. Very diluted solutions, like for AAS, are often stored in PE instead of laboratory glass to prevent ion Vs glass surface interactions.

When you say “steps”, do you mean creating a moderately concentrated stock solution (e.g. 1000 ppm) and diluting it to create a solution of the desired concentration?

I mean, using 1 or 2 stock concentrations, if dilution in 1 step would be too big to be practical or accurate. E.g. you may dilute it 1000 times in 1 step, or 20x50 times in 2 steps.

That makes sense. Would dissolving a “1 mg crystal of lead nitrate” in 1 L of water be an accurate way to create a stock solution at 1000 ppb? Will every volume of water in that 1 L contain the same concentration of lead if the container was perhaps shaked/agitated thoroughly?

It would, IF...   But as the premise is false, it would not.

Alright. So if I understand correctly, my best bet would be to dilute 1000 ppm of solution with the necessary amount of distilled water to yield a 1000 ppb stock solution?

Be aware there is probably need to use auxiliary chemicals like nitric acid to make diluted solutions stable ( e.g. to prevent hydrolysis ). It seems you have not done your homework to get familiar with the topic.

I will definitely look into stabilizing chemicals; thank you for the heads ups!


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What's the effect of Pressure change in Chemical Equilibrium if any reactant or product is Solid or Liquid?
I am studying La-Chatelier's Principle of Chemical Equilibrium. I got a problem of Pressure change in Chemical Equilibrium.
I know that Pressure change If a reversible reaction is in equilibrium state, the effect of Pressure change depends on the number of moles of reactants & products when all are in gaseous state.
But what happens if any of them is liquid or solid?
Like a reaction:
$$\ce{2A(g) + B(s) <=> C(g) + D(l)}$$
I am pursuing my studies in 10th Grade.In the textbook,I learned that if pressure increases,The Equilibrium state shifts from left to right if the number of moles of gaseous reactants is greater than that of products.And if,
Number of moles of Reactants=Number of moles of Products,
Then there will be no effect of Pressure change.While Studying,it came to my mind what will be the change if any reactant or Product is in Solid/Liquid state?
I didn't got such answer in internet.
What will be the effect of pressure change in chemical equilibrium in the  following reaction above?
Eagerly waiting to know the answer.

Le Chatelier's principle applies regardless of the phases of the reactants and products.
You just get a bigger effect if gases are involved, since the change in volume is larger.
Specifically, let $\Delta$V be the change in volume in going from reactants to products.  If $\Delta$V > $0$, then increasing the pressure will shift the reaction to the left (towards reactants).  If $\Delta$V < $0$, then increasing the pressure will shift the reaction to the right.
Also, it's fundamentally about the change in volume, not the change in the number of moles.   The reason why you were taught to look at the number of moles is that, if all the reactants and products are in the gas phase, then the side with the greatest number of moles will have the greatest volume.
But this no longer necessarily applies if you have mixed phases.  In the latter case, you need to look more closely to determine the sign of  $\Delta$V.   For your sample reaction,
$$\ce{2A(g) + B(s) <=> C(g) + D(l)},$$
it is clear that $\Delta$V < $0$, because you have two moles of gas on the left, and one mole of gas on the right, and the difference in volume for the gases dominates over the difference in volume for the condensed phases.  Hence, for this reaction, as you increase the pressure, it will shift to the right (towards  products).

Ok I understood, But One more question, What happens if there is no gaseous reactants or products in the reaction, like                                        2A(l)+B(s) => C(l) +D(l) [reversible], What will be the effect of pressure on this following reaction in chemical equilibrium?

@Sindid  Same thing.  You need to sum the molar volumes of the reactants (multiplying the molar volume of A by two), and compare this to the sum of the molar volumes of the products.  BTW:  If you find an answer helpful, you should consider acknowledging this by upvoting it.

Whenever we talk about equilibrium equations, we deal with active masses of the reactants of the product. All the conclusions regarding the effect on equilibrium caused due to various changes in conditions, including pressure are drawn with respect to these active masses.
What is active mass?
The term active mass means the concentration of the reactants & products  expressed in moles per litre (molar concentration). Active mass is usually expressed by enclosing the symbol of the reactant in square bracket [    ].
The active mass of solids and pure liquids is a constant quantity (unity) and solvent (excess) is considered as one, because:

There is no change in activity with the change in quantity or volume of vessel.
Density of pure solids and liquids is constant and molecular mass is also constant.

However, this is not applicable to the substance in aqueous solution or gaseous state because their amount in a given volume can vary, which is exactly the case you are dealing with at your level in general.
Hence, no change in equilibrium is brought out due to change in pressure corresponding to Solid or Liquid reactants only.

This is incorrect.  As a simple example, consider the solid->liquid phase transition in water.  Increasing the pressure lowers the temperature at which this occurs (i.e., favors the liquid phase), because the molar volume of the liquid is less than the molar volume of the solid.

Similarly, increased pressure by adding oxygen to nitrous oxide before compression avoids condensation of nitrous oxide, as chemical potential of liquid increases with pressure. ( used in medicine ). It is rather special case, as the critical T is near room T.

@theorist I get your point. However, it is mentioned that the active mass is taken  as unity since "Density of pure solids and liquids is constant and molecular mass is also constant", which implies that the concept is applicable only in cases where during the reaction/process the density remains constant as opposed to the case of 'phase transition' you mentioned above.

@Poutnik There are many details to this concept of active mass and many cases like you mentioned, but since the answer was demanded by a tenth grader, I wrote it the way that would cover most of the problems at that level.

@HappyFeetMe   Still no.  There is a pressure effect on the relative amount of reactants and products anytime the reactants have a different volume from the products.  This is true even if the densities of the products and reactants are assumed to be pressure-independent.

It doesn't matter what aggregate state the reactants and products are in. What matters are the volumes. The rule of thumb about gaseous substances is because only the gaseous substances have significant volume per mole in more or less normal conditions.
Imagine you have a closed container with 1 mole of water and 1 mole of oxygen in 20 liter volume at temp 20C (they don't react, but no worry). Shrink the volume to 10 liter and you will see that only the gas (oxygen) has shrunk, water didn't.
If the conditions are not usual as in the above example and you make the liquids and solids to shrink and expand measurably (say, by applying a GPa range pressure) then all the liquids and solids will also participate in the equilibrium shift.

LeChatlier's principle is based on the difference in volume between the reactants and products, not on whether their volumes are themselves pressure-dependent.  I.e., in a reaction involving condensed phases, it doesn't operate because the condensed phases change density as you change the pressure; rather it operates based on the difference in volume between the reactants and products.  I.e., what matters is the value of $\Delta$V,  not dV/dp.

@theorist LeChatlier's principle can be expressed for volume, pressure, temperature or something else (e.g. voltage for some systems). It still holds.

You seem to have misunderstood my comment.  I wasn't saying Le Chatelier's principle is restricted to pressure and volume.  Rather, I was saying two things: (1) IF you are using it for pressure-dependence, then what matters is $\Delta$V (between products and reactants); and (2) Your last paragraph is incorrect.  The liquids and solids don't need to "shrink and expand measurably" to "participate in the equilibrium shift".  Indeed, the individual liquids and solids don't need to change density at all!  All that matters is that there is a volume difference between products and reactants.


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Is there any evidence, *any evidence at all*, that nascent hydrogen actually exists?
As far as I know, nascent hydrogen is nothing more than an especially magical unicorn entity that either makes reactions happen or at least greatly facilitates reactions happening. Many years ago, it was assumed that nascent hydrogen existed, but it was never characterized in any way because it did its work so fast as to escape detection. Now I am seeing "answers" to questions that invoke nascent hydrogen. I consider nascent hydrogen to be entirely bullcrap, pardon my language. Would someone please tell me what is going on with this?

Since I've started answering questions here, I've often repeated my mantra against d orbital hybridisation and the likes. I know this isn't particularly helpful, but prepare to type many comments and be patient. Spread the word, people have learned it that way, and because they might not know any better, they'll continue to propagate outdated theories.

This, unfortunately, remains part of the syllabus in some parts of the world.

@orthocresol it's taught in my country too, in high school,  they include all theories that once had been considered to be "successful", VBT, VSEPR, CFT, till MOT. Don't know why they don't exclude outdated theories.

@Zenix - Don't throw the baby out with the bath water. VBT and CFT won't explain everything but they do work well enough for certain applications.

Believing in nascent hydrogen is not quite as outdated as believing in phlogiston or a flat earth.

There are certain reactions that don't occur with molecular hydrogen, but do occur with hydrogen at the moment of appearance. That's a fact, and it needs an explanation. Whether you call it nascent hydrogen or otherwise is up to you.

What is the current teaching about intermediate products between [(H3O) (H2O )n]+ and H2 ?

@MaxW Don't throw out VBT at all; just throw away how it is taught. You'll get to the same solution as with MOT if you see it through. CFT has been amended to LFT, incorporating MOT. Surprisingly that has made it easier to understand, not more complex. So this one really is an outdated theory that shouldn't be taught at all.

You should read the Solution to nascent hydrogen challenge by Juris Meija & Alessandro D’Ulivo Link. The authors vehemently deny the existence of any nascent hydrogen.
You will find that the concept of nascent hydrogen and nascent oxygen is still taught in some Indian chemistry textbooks. This term has literally vanished from US & European chemistry books. I just thought to search an elite journal JACS  and this term is present in 2011 in the abstract of
Highly Selective Ammonia Synthesis from Nitrate with Photocatalytically Generated Hydrogen on CuPd/TiO2, J. Am. Chem. Soc. 2011, 133, 5, 1150-1152

"The continuous generation of nascent hydrogen atoms on the surface of
  the CuPd nanoalloy, where Cu and Pd are homogeneously mixed, led to
  the high selectivity for ammonia."

Well, this is exactly a classical example of the usage of nascent hydrogen when Zn could reduce nitrate ions in an alkaline medium. The texts of 1940s describe those experiments very well. See Mellor's Treatise of Inorganic Chemistry.
Another 2012 RSC journal Chemical reduction of an aqueous suspension of graphene oxide by nascent hydrogen


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Is it true to say that as of 2020, chemists can create any gemstone, from 1 to 10 level of Mohs' Scale strengths, artificially?
Please, I have a question here as a non chemist:
Is it true to say that as of 2020, chemists can create any gemstone, from 1 to 10 level of Mohs' Scale strengths, artificially?
If so, please detail if it causes any serious financial damages to the "old school" gemstone industry of picking and mining and polishing gemstones from natural environments and deposits respectively, or if the two industries keep thriving one alongside the other.

All the exemplars in the hardness scale are minerals, but all minerals are not gemstones.

"Gemstones" each have a characteristic fundamental "shape", you cannot force them to grow into another one.

Defining “gemstone” is a bit tricky. Roughly speaking, a gemstone is a mineral specimen, natural or synthetic, that has been fashioned by a gem cutter. There is no requirement that a finished gemstone be flawless, have a naturally occurring crystal habit, be hard on the Moh’s scale, be chemically resistant, have no easy cleavage(s), be rare as a mineral, and so on. This question could get a really long and detailed answer.

Keeping in mind that the original Mohs scale is qualitative, arguably we can also make a select few materials "above 10", in the sense that they would scratch diamond without themselves being scratched by diamond. One that comes to mind right now is aggregated diamond nanorods.

@Karl if it's okay I ask; why for example should it be different than growing a watermelon in a box to get a quadrangular shape?

Confine each growing crystal in a box? Well, that might work, if your box can stand the conditions, and you can suppress more crystals to grow from the walls. Chances are that touching the wall induces all kinds of growth defects. And it sounds a bit impractical to me.


text:
Specific conductance of KCl
Question:

A conductivity cell containing distilled water has a resistance of $\pu{454 k \Omega},$ while the same cell containing $\ce{KCl}$ solution prepared in the same sample of water shows resistance of $\pu{2150 \Omega}.$ If the plates of the conductivity cell have area $\pu{0.5 cm^2}$ each and are separated by $\pu{1 cm},$ find the cell constant and also calculate the specific conductance of $\ce{KCl}.$

My Attempt
First things first. I found out the cell constant, that is, $G^*:$
$$G^* = \frac{l}{A},$$
where $l$ is the distance between the square plates and $A$ is the surface area of the square plates — hence, $G^* = \pu{200 m^-1}.$ Now I also know that,
$$\frac{1}{R} = \frac{\kappa}{G^*},$$
where $\kappa$ is the conductivity, and $R$ is the resistance of the solution.
Plugging in $R = \pu{2150 \Omega}$ and $G^* = \pu{200 m^-1}$ yields the answer $\kappa \approx 0.093.$ [Which is incorrect.]
Note
I did not use the value of resistance of distilled water that is provided in the question. What am I doing wrong? I also feel that the sentence "…also calculate the specific conductance of $\ce{KCl}$" is a bit weird? Shouldn't be referred to as "$\ce{KCl}$ solution"?

Correct in both cases. The resistance of the cell with distilled water is of course irrelevant.

Thanks Karl for clarifying that. Also, given answer is: $0.0812 Sm^{-1}$. Question is from the InChO 2005.

It´s of course clear here that they mean the solution and not the bulk solid or molten KCl.

And your numbers check out. No idea where the error is. Maybe in that textbook.

I'm solving this book: "Indian National Chemistry Olympiad (INChO): Theory Papers with Solutions 2005 - 2007" Published by Homi Bhabha Centre for Science Education

@McSuperbX1, Your method is correct. Textbooks are written by human beings, and it is not unusual to find 1-2 errors especially in the answers.


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Reaction rate and temperature
I recall being told that for a given reaction, every 10 degree rise in temperature the reaction rate approximately doubles. Is this true?

It is a simple question, a simple answer would suffice.

Is the purpose of this forum to give answers that are for the cogniscenti or the questioners?


text:
Action of KMno4/H+ and heat in this reaction

The first step uses LAH which is a hydride donor. So H- attacks any one carbonyl carbon and forms an aldehyde and acid attached to the ring.
Then Conc. KMno4 is used. As far as i know, it cleaves c=c and oxidises them into acids.
 The third step then says heat, so it leads to the formation of the anhydride on both sides.
However, this product has nothing to do with the correct answer which is option (A) . Somehow the product formed consists of 2 seprate 5 membered rings. Any idea on where i went wrong?
edit::

The $\ce{LiAlH4}$ will reduce the cyclic anhydride to the diol.
$\ce{KMnO4}$ re-oxidises both the alcohol groups (diol formed) to carboxylic acids and also oxidises the double bond to the diacid so the product you have drawn is correct, but it is an intermediate. You then have to consider the effect of heating this (to a pretty high temperature, I would assume).
Product A Heating causes a double dehydration to produce the di-anhydride shown but the right hand pair of acids do not recombine, instead the top acids combine and the bottom acids combine.
Product B Heating dehydrates the right hand diacid to the anhydride, the left hand diacid does a Claisen-ester-type thermal condensation followed by decarboxylation to give a ketone in  the left hand ring.

Got it now! Thanks! Just one doubt. Here a claisen occurs between the 2 acids on the left(product b) . For this a carbanion must be formed right? so should the questions mention a base? I have attached an image for this. Pls tell me if my rxn is correct.

Yes, that is correct. I think if base was introduced then the salt of the di-acid is less likely to undergo the thermal condensation


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Elimination reaction with bad leaving groups
If a halogen is present, as in case of 2-fluorobutane, which is reacted upon by alc. KOH then, shouldn't the product be according to Zaitsev's rule.
Then, why does the removal of hydrogen takes place, does this happens in case of halogens only or in case of all the bad leaving groups?

Sorry, your question is unreadable. You can write perfectly good long sentences in English, but you need to have a flawless grammar to remain legible. In the meantime, split it in two or three, add a fourth.

Alkyl halides with bad-leaving groups (like $\ce{F-}$) having EWG at $\ce{\beta}$-position undergoes these reactions, also known as $\ce{E_{1CB}}$, viz. elimination unimolecular conjugate base.
Example:

aldol condensation

photo-induced decarboxylation

Reagent: bulky or strong bases or both (like $\ce{K+(^tBu)-, LDA, NaOH, KNH2}$, etc) along with some heating.
Mechanism:

$\ce{OH-}$ leaving is unlikely. It is more like $\ce{H2O}$ leaving after exchange hydrogen from available water molecules.

@MathewMahindaratne: yes, thanks for pointing out. I've added some examples from Wikipedia.


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Metalation of thiophene
So lately I've been studying the concerted metalation-deprotonation of thiophene and I came to the conclusion that the alpha positions of thiophene are the most reactive, which is why one of those alpha carbons (either at the 2 or 5 position) will be deprotonated en metalated.
But why are the alpha carbons on thiophene the most reactive ones and why is thiophene preferred during concerted metalation-deprotonation if there are other heterocyclic compounds that are more reactive?
Thanks!