text:
How do we find hybridization when molecule has resonance?
Consider the following image which shows the resonance structures of $\ce{CO_3^{2-}}$ ion. What is the hybridization of $\ce{O}$? How do we proceed in such cases? In one resonance structure, the top oxygen has $\ce{sp^3}$ hybridization and in others, it has $\ce{sp^2}$ hybridization. In the hybrid structure, shown in the second image, I don't even know how to proceed.

All the oxygens should be considered to be sp² hybridised, with its resonance-contibuting p orbitals being unhybridised and acting solely as pi-bonds.

@KanghunKim Not really, sp is an accurate description for, like, all terminal atoms.

Does this answer your question? What is the hybridization of terminal fluorine atoms in molecules like boron trifluoride?

Boron trifluoride may seem different, but is isoelectronic with carbonate anion.

The only important thing to consider for the description in terms of hybridisation is the molecular structure. It is therefore irrelevant if you are using a resonance structure, all resonance structures, or the resonance hybrid, or even some kind of completely different diagram. The structure for all of these is the same.
Therefore, to determine a reasonable hybridisation scheme, you should first look at local coordination of the atom. Please keep in mind that the $x$ in $\mathrm{sp}^x$ doesn't need to be an integer value.
For this, please look up Bent's rule and Coulson's theorem. You may start with my answer here: How is Bent's rule consistent with LCAO MO theory?
Please also keep then in mind that only orbitals hybridise and any combination of hybrid orbitals my be a valid description.*
If you keep this in mind, the following definition makes quite a bit more sense, hybridization in the Gold Book (DOI: 10.1351/goldbook.H02874)

Linear combination of atomic orbitals on an atom. Hybrid orbitals are often used in organic chemistry to describe the bonding molecules containing tetrahedral (sp³), trigonal (sp²) and digonal (sp) atoms.

Considering this, here are some zero order approximations, which my come in handy:

Coordination
Examples
Approximate Hybrid Orbital

linear
carbon in acetylene or carbon dioxide, or any terminal atom**
sp

trigonal
carbon in $\ce{CO3^2-}$, boron in $\ce{BH3}$, carbon in acetone
sp²

tetrahedral
carbon in methane, sulfur in $\ce{SO4^2-}$
sp³

bent
oxygen in water
sp² - sp³

trigonal pyramidal
nitrogen in ammonia
close to sp³

trigonal bipyramidal
phosphorus in $\ce{PCl5}$
model breaks down

octahedral
sulfur in $\ce{SF6}$
model breaks down

After writing all this, I've realised that I have commented on this before: What is the hybridization of the carbonyl oxygen in a carboxylic acid?

Footnotes:
* The terminology we use for hybridisation actually is just an abbreviation:
$$\mathrm{sp}^x = \mathrm{s}^\frac{1}{x+1}\mathrm{p}^\frac{x}{x+1}$$
In theory $x$ can have any value; since it is just a unitary transformation, the representation does not change, hence
\begin{align}
 1\times\mathrm{s}, 3\times\mathrm{p}
   &\leadsto 4\times\mathrm{sp}^3 \\
   &\leadsto 3\times\mathrm{sp}^2, 1\times\mathrm{p} \\
   &\leadsto 2\times\mathrm{sp}, 2\times\mathrm{p} \\
   &\leadsto 2\times\mathrm{sp}^3, 1\times\mathrm{sp}, 1\times\mathrm{p} \\
   &\leadsto \text{etc. pp.}\\
   &\leadsto 2\times\mathrm{sp}^4, 1\times\mathrm{p}, 1\times\mathrm{sp}^{(2/3)}
\end{align}
There are virtually infinite possibilities of combination.
Or see more: Are the lone pairs in water equivalent?
** I have written about this a few times already. These should provide also data:

What is the hybridization of terminal fluorine atoms in molecules like boron trifluoride?
What is the hybridization of chlorine in vinyl chloride?
Hybridisation of terminal nitrogen in diazomethane


text:
What easily accessible material has melting point between 500 °C and 590 °C?
Is there an available at home material with the melting point between 500 °C and 590 °C?

Plastics usually melt below 400 °C.
Tin and lead also below that.
Aluminum is melting above 600 °C.

I made a deep search around the web with no success.

It's interesting that there appears to be no chemical element with m.p. in this range.

As @Poutnik says a salt mix will do it. That temperature is in a range to heat-treat and process steels so some vendor of heat-treat supplies will have a salt mix available. I never worked with them so can't be more specific.

You could try contacting one of the metals suppliers that sells aluminum alloys on Amazon, and asking them if they have one that would melt in your desired temperature range.  I'd suggest finding a supplier that sells on Amazon, as that tells you they are willing to sell to consumers, in small quantites.  You'll need to tell them if it has to melt at a specific temperature, or if it's acceptable if it melts over a  temperature range (which happens with some alloys). I can't vouch for the following company, since I've never ordered from them, but one example is https://www.onlinemetals.com

I admit, I'm a physicist and more of an engineer than a chemist, but to choose a material you need to consider what it will be used for and come into contact with.  For example molten salts can be quite corrosive.  If you're going to put some granules in a container to see if the container has reached a certain point while you couldn't otherwise monitor it, you'll also need to consider the heat flow in, and the heat capacity and latent heat, otherwise you will actually be recording whether it's exceeded the melting point for some time.

The salt mixes for steel treating are generally contained in cast iron , so minimally corrosive.

What is the application?  What is your budget?  How much do you need?  Impossible to answer without more information.

In plain-speak, what are ya working on?

Also, what facilities do you have available?  The idea of DIY alloying has come up, but do you have a furnace?

The problem with alloys is that they often do not have a clear melting point. They start melting at the solidus temperature. But only a small fraction is melting. To get all the salloy in the liquide state, you have to heat it to a much higher temperature, called liquids point.

Temperature-indicating or thermo crayons are used by welders. These melt at specific temperatures.
Not as an endorsement, but to give you an idea of what's available, the crayons from McMaster-Carr come in these temperatures in the range you are interested in:

932 °F (500 °C)
950 °F (510 °C)
1000 °F (537.8 °C)
1022 °F (550 °C)
1050 °F (565.6 °C)

Another brand name is "Tempilstik" (again, not an endorsement).
Temperature-indicating crayons are available at any welding supply.

F and C are different…

Great idea - OP wants 932-1094 degrees F and- the same link lists test crayons at 932°F 950°F  1000°F   1022°F  and 1050°F   which are all in OP's required range.

Oh good gravy, I completely missed that the OP wanted C.  I'll edit the post.  Thanks @JonCuster and Criggie.

Unusually named "tempilstik" ; any weld supply should have them.

It might not qualify as 'available at home' but MakeItFrom has a database with engineering material you can search:

MakeItFrom.com is a curated database of engineering material properties that emphasizes ease of comparison. It is not a datasheet dump: every listed material is an internationally recognized generic material. The data is sourced from published standards, academic literature, and supplier documentation.
To find information on a particular material, browse from the list below, or search from the menu. Aside from searching by material name, you can also search by property values. Once on a material's page, you can search for a second material for a side-by-side comparison.

This query lists all materials with a melting point between 500 and 590 degrees Celsius; mostly aluminium alloys.
There are some salts in this list on The Engineering Toolbox which qualify too and can be bought online, but e.g. silver iodide (melting point 558°C) is toxic and I would rather avoid that.

The problem is that this alloy needs to either be identified (which would require elemental analysis, XRF, or powder XRD), or ordered, which manufacturers really like to do in bulk amounts, which would be an overkill for home experiment. Other than that, a good source of information.

@andselisk a metal stockholder will supply fairly small quantities of a standard alloys, in forms they hold.  The Al alloys produced by that table are mostly casting types. I'm used to machining types, and my supplier doesn't carry any of those.  But in the next city there's a supplier of ingots of casting alloys that has quite a few of the listed ones.  I reckon I could jump on the bike and have an ingot in my hand this afternoon.  So "at home" should be doable.

Name
Formula
m.p./°C

Copper(II) chloride
$\ce{CuCl2}$
498

Barium perchlorate
$\ce{Ba(ClO4)2}$
505

Potassium fluoroborate*
$\ce{KBF4}$
530

Antimony trisulfide (stibnite)
$\ce{Sb2S3}$
550

Calcium nitrate
$\ce{Ca(NO3)2}$
561

Sodium cyanide
$\ce{NaCN}$
563

Cadmium bromide
$\ce{CdBr2}$
567

Strontium nitrate
$\ce{Sr(NO3)2}$
570

Potassium periodate
$\ce{KIO4}$
582

Iron(II) iodide
$\ce{FeI2}$
592

* It is not difficult to buy it.

560°C for KOH is suspiciously high, compared to NaOH. I have checked Wikipedia, it says 360.

I even have a small bottle of strontium nitrate at home, as one does! (+1)

OP is asking about home materials and not laboratory chemicals.

I don't know the specific chemistries of these,  but the general problem with melting crystalline compounds is that they often contain water of crystallization and some of them will "melt" by dissolving into their own water of crystallization at a lower temperature.  Also beware of toxic,  corrosive and even explosive risks. Perchlorates are strongly oxidizing and Sodium Cyanide is  notorious as a poison.

There ought to be an alloy with the desired melting point. It doesn't seem to be commercially available, but you could experiment by obtaining tin or zinc, plus copper powder (or scrap brass / bronze), and melting them together. Copper has a high melting point, tin just 232 °C and zinc 420 °C. Copper + tin = bronze and copper + zinc = brass.
The commercially available forms have higher melting points than you want. The forms with more tin or zinc than usual may not be very good in other metallurgical terms, but it's certain that there's a composition that melts at the temperature you want.

And here's a Cu-Zn phase diagram suggesting you'll need something like 90% Zn, 10% Cu.  Al-Zn (phase diagram alloys might be easier to make as you're nearer 50:50, on a flatter part of the curve, and Al is much easier to melt than Cu when you make the alloy.  This time commercial alloys have a melting point that's too low

Copper will dissolve in a liquid metal well below its melting point. Specifically from experience, in solder (lead-tin alloy in those days). Soldering-iron bits are iron coated copper. After any slight perforation of the iron coating, the liquid solder quite rapidly dissolves the underlying copper and the bit "craters" and becomes fairly useless. Uncoated bits fan be filed flat but the solder soon dissolves copper off the flat surface.

Good point re soldering tips.  I've done that enough (and used stiff copper wire to make consumable custom tips. The acid in the flux may play a part too. I'm not sure how well copper would dissolve in zinc though, hinting more at sticking to tin (modern soldering being Sn+Cu or Sn+Ag anyway)

As mentioned, various aluminium alloys melt in this range.  Alloys and other mixtures don't have a single melting point.  They melt over a temperature range between the solidus and liquidus temperatures.
For example alloy 4043 is readily available as welding wire. Solidus 573.9 °C, Liquidus 632 °C.  You would need to use a phase diagram to determine the liquid fraction at intermediate temperatures.

If the solidus is so far away form the liquidus, you will never be able to really know the temperature of your system. At 573°C, the first drop of liquid appears. But the alloy is not yet liquid. You will have to heat it to 632°C to get a clear liquid. So the difference between the appearance of your meal at 590°C and 620°C is not significant.

Based on a search on Matweb, BK7 glass meets your requirements and is close to the middle  of your range (559°C melting point).
It's available in small quantities on eBay for reasonable prices.
If you're looking for something with a sharp melting point (for example as a calibration point for a thermocouple or RTD) you don't want a glass though.

I found out that h-k9l(chinese version) is the equivalent to bk7 glass.Does the melting point of h-k9l the same as bk7?(557 °C)

source: Stack Exchange

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text:
Addition of HOCl to ethyne
What is the product formed in addition of HOCl to ethyne? Is it dichloroacetaldehyde, or perhaps the enol form of dichloroacetaldehyde could react further with HOCl to form chloral hydrate / chloral?

https://chemistry.stackexchange.com/questions/134154/reaction-of-acetylene-with-hypochlorous-acid

All of it is rather dubious, as aldehydes oxidise very easily.

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text:
Hypothetical symmetric carbon peroxide
Could a hypothetical carbon peroxide $\ce{C5O12}$ with the following structure (MolView — 3D model) exist, and I what its properties would be (other than decomposing, of course)?

I tried looking it up myself, but I couldn't find any information on it or anything similar.

Huh, this thing could be perhaps an interesting oxidiser, if it didn't decompose outside of argon matrix ;)

Have a read: https://en.wikipedia.org/wiki/Oxocarbon

I think the weakness here is the multiple C-O single bonds on each carbon. Those will eliminate readily in favor of C=O bonds, so I don't think this thing would even be metastable, but I could be wrong.

Complex/compact molecules are better presented with white background, fog showing depth, thinner bonds and smaller atomic radii. I used Jmol with zshadepower=1, wireframe 0.075 and spacefill 0.2 to produce a new static image to supplement the one that MolView generates. cc @Editors Please don't use lossy formats for images and don't apply compression to PNG images.

Your molecule basically consists of a bunch of 1,2-dioxolane rings fused together.
Since 1,2-dioxolane exists, I would guess that your molecule probably does too, at least under suitable conditions (which might involve an ultra-cold argon matrix in complete darkness). As noted in the comments below, though, the presence of multiple peroxide groups on each outer carbon might be enough to make it unstable even under ideal conditions, so take that with a grain of salt.
In any case, with all those fused peroxide rings packed tightly together, synthesizing it is going to be another matter. Still, enterprising chemists have managed to prepare some pretty crazy molecules, so if it's stable at all (however marginally), I wouldn't rule out the possibility that someone might some day create it.
Anyway, I expect that your molecule, like organic peroxides in general, will be rather highly reactive. If heated, it will likely thermally decompose into smaller, more stable molecules such as $\ce{CO2}$ and $\ce{O2}$ (and maybe $\ce{CO}$), as in: $$\ce{C5O12 -> 5 CO2 + O2}$$
This reaction is exothermic, and can thus self-accelerate if the sample is concentrated enough and not sufficiently well cooled, leading to runaway (and potentially even explosive) decomposition. Impurities that can react with the peroxide groups (which your molecule is entirely covered in!) are also likely to accelerate this decomposition.
In theory, if your molecule was stable enough to safely store and handle, it could make a pretty good explosive and/or a rocket propellant, either on its own or as an oxidizer combined with something less oxygen-rich as fuel in a bipropellant mix. That said, the world is full of compounds that could be great explosives or rocket fuels, if only they were stable enough — the real challenge in high-energy chemistry is not making a molecule that goes boom, it's making one that will not go boom until you want it to.

As noted in my comment above, the difference between this beast and 1,2 dioxolane is the multiple oxygens attached to each carbon, which I would expect to readily eliminate. They're not stable like acetals are because the alkyl peroxide leaving group here is much better than the alkoxide in an acetal.

Carbon is not very good at forming stable peroxide compounds (prospective chemists are early and often warned about ethers turning into explosive peroxides), so this compound with multiple peroxide linkages is going to go boom.
But if we are careful to bond oxygen just to carbon, then a surprising variety of oxocarbons are known. Far beyond just the widely known carbon monoxide and carbon dioxide, they have a wide variety of structures and properties, including linear molecules, ring-based systems, and polymers. The linked WP article gives a good overview.
Linear dioxides: an odd feature
One of the most notable types of oxocarbons are linear dioxides. Most readers are aware of $\ce{CO2}$, but there can also be $\ce{C2O2}$, $\ce{C3O2}$, etc; chains as long as 21 carbon atoms are known.
Among these the most stable are $\ce{CO2}$ of course, $\ce{C3O2}$, and $\ce{C5O2}$ (the latter is stable in solution up to room temperature), whereas $\ce{C2O2}$ is highly unstable and $\ce{C4O2}$ can be kept only in the dark in a frozen argon matrix. Moreover, longer $\ce{C_nO2}$ are reported in the WP article only for odd values. What favors the odd-numbered members?
"Linear aromaticity"? The revenge of 4n
In these molecules all the atoms are $sp$ hybridized, so each atom contributes two $p$ orbitals to the delocalized pi system and all the pi orbitals are doubly degenerate. Thus a closed-shell electronic structure with increased stability due to pi-electron delocalization is favored by a $4n$ pi-electron count, not $4n+2$ as in the usual sort of aromaticity. With each carbon atom contributing two electrons to the pi orbitals and each oxygen contributing three, what might be called the "linear aromatic" structures are those with an odd number of carbon atoms, which are indeed the more stable and predominant ones.
Peroxocarbon anions
Although neutral oxocarbons with peroxide linkages are obtained only in extreme circumstances, both peroxycarbonate and peroxydicarbonate ions are well-known and reasonably accessible.
Peroxycarbonate, $\ce{O^- - O - CO2^-}$, can be made by electrolyzing molten lithium carbonate or cold, concentrated aqueous lithium carbonate solution. It is known as such only in solution, but the protonated ion $\ce{HO - O - CO2^-}$ has been isolated as the potassium and rubidium salts [1-4]. Note the location of the protonic hydrogen; the terminal peroxy oxygen is more basic than the single oxygen atoms attached to carbon. The terminal peroxy oxygen is effectively cut off from the pi bonding in this ion, so its negative charge remains (formally) localized. In addition, the attached proton may be incorporated into a five-membered hydrogen-bonded ring (cf. this discussion of Caro's Acid).
Peroxydicarbonate, $\ce{CO2^- - O - O - CO2^-}$, is made by electrolysis of molten lithium carbonate or by electrolyzing a saturated aqueous solution of rubidium carbonate. This ion is stable enough for its potassium and rubidium salts to have been isolated (with no protonation of the ion)[5-7]. The light blue potassium salt decomposes with loss of oxygen and carbon dioxide, but only slowly below 141°C[6].
References
1.
Mimoza Gjikaj (2001), "Darstellung und strukturelle Charakterisierung neuer Alkali- bzw. Erdalkalimetallperoxide, -hydrogenperoxide, -peroxocarbonate und -peroxohydrate" Archived 2012-02-25 at the Wayback Machine. Doctoral Thesis, University of Köln. 115 pages.
2.
Adam, Arnold; Mehta, Mathias (1998). "KH(O2)CO2·H2O2—An Oxygen-Rich Salt of Monoperoxocarbonic Acid". Angewandte Chemie International Edition. 37 (10): 1387–1388. https://doi.org/10.1002/(SICI)1521-3773(19980605)37:10<1387::AID-ANIE1387>3.0.CO;2-3. ISSN 1521-3773. PMID 29710903.
3.
M. Mehta and A. Adam (1998), Z. Kristallogr., Suppl. Issue 15 p. 53. Cited by Gjikaj.
4.
M. Mehta and A. Adam (1998), Z. Kristallogr., Suppl. Issue 15 p. 46. Cited by Gjikaj.
5.
Constam, E. J.; von Hansen, A. (1896). "Elektrolytische Darstellung einer neuen Klasse oxidierender Substanzen". Zeitschrift für Elektrochemie. 7: 18 ff. https://doi.org/10.1002/bbpc.189600032 (inactive 31 October 2021).
6.
Dinnebier, Robert E.; Vensky, Sascha; Stephens, Peter W.; Jansen, Martin (2002). "Crystal Structure of K2[C2O6]". Angewandte Chemie International Edition. 41 (11): 1922–1924. https://doi.org/10.1002/1521-3773(20020603)41:11<1922::AID-ANIE1922>3.0.CO;2-T.
7.
Dinnebier, Robert E.; Vensky, Sascha; Jansen, Martin (2003). "Crystal and Molecular Structure of Rubidium Peroxodicarbonate Rb2[C2O6]". Chemistry: A European Journal. 9 (18): 4391–4395. https://doi.org/10.1002/chem.200304914. PMID 14502625.

Cool, but what about peroxides?

@mithoron there's the rub, unfortunately. The oxocarbons seen in the referenced article do not seem to include peroxides ... yet.

The best I could do was discuss a couple anions with a peroxide linkage.


text:
Interpretation of permanganate-stained TLC spots
When I stain TLC spots with $\ce{KMnO4}$, I found that some spots appear differently as time goes by. Here is the example:

The leftmost one is the plate right after staining, and the others are listed sequentially with time.
Question 1: As time goes by, some white stains appeared and widened. Does this mean some materials exist in that area, or is this just result of diffusion? Are the white stains due to original yellow spots, or indicate the existence of different materials?
Question 2: Some white stains (numbered as 1, 2, 3) appeared in the same manner, but disappeared. Then should I consider the existence of this white stain? And should I record staining with time-scale whenever I stain the TLC with $\ce{KMnO4}$?

As you have seen, potassium permanganate is very reactive. It can react with almost anything organic.
Question 1: Yes diffusion of the single component is causing the broadening of the spots.
Question 2: The explanation of spots turning white is that either you have acid in your eluent or the compound themselves are relatively acidic. When manganese dioxide reduces in the presence of acids, it is converted to manganese ions (which are very light pink in color).
Record the stain immediately. That is its correct retention.

In the rightmost plate, the second column has pale-yellow spot at the top but leftmost doesn't have. In fact this spot was observed under UV radiation. But this spot doesn't appear right after staining. I am somewhat nervous whether I use information of the rightmost plate.

If you are concerned, you will have to try two or three different solvent eluent systems just to check how many spots does a mixture generate. By chance is an HPLC available?


text:
Carbohydrates with two hydroxyl group on one carbon atom?
Every carbohydrate I keep encountering so far has only one hydroxyl (–OH) group attached to every carbon atom in the carbon chain (in Fischer projections): either on the left, or on the right of the carbon, but never both.
This makes me wonder:
Are there any carbohydrates with –OH groups on both sides?
If there aren't, why? Is there any physical or chemical rule that would prevent this?
According to this website:
http://colapret.cm.utexas.edu/courses/Nomenclature_files/alcohols.htm
it seems to be possible for alcohols to have two –OH groups on one carbon:

Many compounds have more than one hydroxyl group; two hydroxyl groups are called "diols". If the hydroxyl groups are on adjacent carbons, then it is called a "glycol." If the hydroxyl groups are on the same carbon, then it is called a "gem" (for geminal, or twin) diol.

so what about carbohydrates?

Geminal diols are generally unstable with respect to loss of water (forming an aldehyde/ketone instead). This is generally true whether it's a carbohydrate or some other alcohol. Exceptions are rare. https://en.wikipedia.org/wiki/Geminal_diol

Besides the examples  of geminal diols cited in orthocresol's link, ninhydrin also comes to mind.

Uhh....glucose?


text:
Relation between concentration and activation energy
Will increasing the concentration of the reactants in a chemical reaction increase the proportion of molecules with an energy greater than the activation energy $E_\mathrm{a}$?
It seems to me that the proportion of molecules with energy greater than $E_\mathrm{a}$ should remain the same after increasing concentration because no extra energy is supplied. $E_\mathrm{a}$ is solely dependent upon the temperature of the molecules in the system, correct?
Assume that the volume $V$ and the temperature $T$ of the system is unchanged when the concentration of the molecules is increased.

If there is twice more (in both cases randomly chosen) people in the room, is there more than twice people with higher salary than $x ?

The population of an ensemble of molecules (so the molecules that would have $E>E_\mathrm{a}$) is determine by the Boltzmann equation (below a simplify version of it)
$$
\frac{N_i}{N_\text{tot}} = g_i\exp\left(-\frac{\Delta G_i}{RT}\right),
$$
where $\Delta G$ is the difference in energy between the molecule you're considering and the so called ground state in $\pu{kJ mol^-1}$, and $g$ is the number of states with the same energy.
So, increasing the number of molecules will change the frequency of the molecules with that energy, but the ration over the whole population will remain the same.


text:
Why is there no charge in NH3
The formation of $NH_3$ is by covalent bond,i.e sharing of electrons take place. Here $N$ now has 8 electrons in it's outermost shell,$3$ more than it's original state,so it should have charge $-3$. Furthermore each hydrogen has $2$ electrons in it's outermost shell,$1$ more than usual,so total charge on hydrogen should be $3\times(-1)$. So the net charge on $NH_3$ should be $-6$. Then why is it neutral?

How many valence electrons are there in ammonia? Is it really 8 + 2 + 2 + 2?

You have counted each of the 6 electrons from 3 bonds twice.

Sorry but I didn't understand,since this is a covalent bond,electrons are shared,so $N$ has now the electrons of $H$ and $H$ has the electrons of $N$,going by not nitrogen has obtained $3$ more electrons from the hydrogens and each hydrogen has obtained one more electron on $N$,that's how I actually considered the charges.

@TanjiroO Each shared electron must be counted as just a half of an electron. Similarly as in crystallography cells, an atom shared with N cells is counted as 1/N atoms. So nitrogen atom in NH3 has 4 + 6/2=7 electrons, hydrogen atoms 2/2=1 electron. How simple, dear Dr. Watson!

Let's actually count protons and electrons.
Nitrogen has  7 protons and each hydrogen has 1 proton which makes a total of 10 protons.
Nitrogen has 10  electrons out of the  3 that are shared with the hydrogen atoms.
Each Hydrogen has 2 electrons out of the 1 is shared with the nitrogen atom.
We get a total of 10 protons and 10 electrons.But why 10 electrons?Because 3 electrons are in both sets of nitrogen and electron protons.

Instead of counting the total number of protons and electrons,shouldn't we be counting respective charges?Like nitrogen has $7$ protons and $10$ electrons now,so it has a charge of $-3$ and so on

Study set theory I recommend this video:https://www.youtube.com/watch?v=yCwnifwVjIg

I actually know about set theory,what is the $C$ that you mentioned? Also why aren't we considering charges on respective atoms? The thing you did is you derived 10 electrons and 10 protons,but I am talking about comparing proton number of an atom and the excess of electron that atom gets.

Which means I am dealing with things individually

C is the # of electrons


text:
Restoring potential of atoms in a solid
Above $\pu{0 K}$ the atoms in a solid are vibrating. However, what kind of potential restores each atom in the starting position?
Consider the crystal lattice of NaCl. The potential energy of a $\ce{Na+}–\ce{Cl-}$ pair is
$$E(r) = -\frac{1}{4\pi \varepsilon_0}\frac{e^2M}{r^2} + \frac{B}{r^m},\tag{1}$$
where $M$ is the Madelung constant, $B$ and $m$ are the material-depending constants.
When a $\ce{Na+}$ ion and a $\ce{Cl-}$ ion get very close, the second term of the equation is increased very much and it becomes the restoring potential in the vibration of the ions.
In a  crystal lattice made of $\ce{Si}$ atoms above $\pu{0 K}$ each atom oscillates at the Debye frequency from its starting position as part of a quantum harmonic oscillator:
$$\frac{-h^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x) + V(x)\Psi(x) = hf\left(\frac 1 2 + N\right)\Psi(x),\tag{2}$$
and $x=0$ will be the position of equilibrium of the atom.
What kind of expression is $V(x)?$ To which power do we raise the variable $x$?

Please edit this and typeset your equations. The GIF images are less amenable to searching.

I don't know how to do it.

Come on, you've been a member on this site for 8 months now. Read the docs. You give, you get.

Zero kelvin is not OK, it's $\pu{0 K}$. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Can I ask this?The LaTeX editor generates a string of characters can I copy paste it instead  of uploading the image?

@JunSeo-He In general, no. LaTeX and MathJax are vastly different systems that happen to share similar markup. Even if you are a LaTeX user, there is little to no reason to use code generators as usually they fail at something slightly more advanced than a couple of fractions, or produce horrible non-semantic code. And keep in mind copypasting anything without understanding how it has been produced is a bad idea.

@JunSeo-He You have just given two examples for $V(x)$. So what is exactly your question?

What is a empirical formula for the attraction/repulsion of 2 atoms covalently bonded?


text:
How is a beta particle an electron?
I've previously read that a beta particle is just a fast moving electron or positron that emerges out of a neutron and turns it into a proton.
I'm really confused by this since particle physics states that an electron is a fundamental particle and neutrons
and protons are made of another fundamental particle, quarks.

So what is the problem? https://de.wikipedia.org/wiki/Datei:Standard_Model_of_Elementary_Particles-de.svg

The OP thinks this interaction is mediated by photons but in fact it is mediated by Z bosons.

https://chemistry.stackexchange.com/questions/10453/understanding-beta-decay

I do not think that this question in any shape or form is a chemistry question. This is pure particle physics.

I’m voting to close this question because it is an off-topics (Particle Physics) question.

More a physics question, so I googled it: "quarks when neutron decays"

[wikipedia: neutron] The decay of one of the neutron's down quarks into a lighter up quark can be achieved by the emission of a W boson. By this process, the Standard Model description of beta decay, the neutron decays into a proton (which contains one down and two up quarks), an electron, and an electron antineutrino.

And here is a picture:

A proton or a neutron can become a neutron or proton by a process called neutrino capture in which due to conservation of charge a electron or a positron must be emitted.But this is not the complete story.
The complete story is this:

A neutron can capture a neutrino and with the help of the weak interaction it is transformed to a electron and a proton.
For more reading :https://en.wikipedia.org/wiki/Neutrino_detector

p+ is a proton.


text:
How to oxidize black nickel plating
Looking to force a patina on a black nickel plating over brass to mimic age.  I know black nickel can oxidize brown or yellow but I don't know how to oxidize it.  I don't see anything on the web. Can someone please guide me?

You can see the dark gray color of oxidized nickel on any electric range heating element ; they are Nichrome about 80 % Ni. I have seen many oxidized Nickel alloys, Inconels, Incoloys, high nickel stainless like HP , HK, etc ( about 32 % Ni ) ; they are all the same dark gray after heating.

You might try various selenide or sulfide chemicals used in gun bluing, but would need to experiment.


text:
Lone Pairs Hybridization
Why are lone pairs considered when determining the hybridization orbitals of an atom? If the lone pairs will lead towards a filled orbital and won't perform any bonding, why can't they be left unhybridized?

They can, and often are. better described as unhybridized.

https://chemistry.stackexchange.com/questions/80962/what-is-the-hybridization-of-terminal-fluorine-atoms-in-molecules-like-boron-tri


text:
How does 25% CO2/75% Ar welding gas exist?
What state does a 25% $\ce{CO2}$ / 75% $\ce{Ar}$ mixture exist in under high pressure? This mixture is a common welding shielding gas. I'm not sure if manufacturers are referring to this mixture by partial pressures or by mass, though I would think partial pressures. For both cases, I calculated the partial pressures of the gasses in the mix. In both cases, the $\ce{CO2}$ is in the gas state (34-35 bar) and the $\ce{Ar}$ is in the supercritical phase (around 110 bar). I don't really understand supercritical fluids, though.
So what state does this mixture exist in? Do supercritical fluids mix with gases the same way two regular gases would? How does this mixture behave inside the bottle?
For the record, here are the numbers I arrived at:
Calculation Conditions:

Condition
Value

Volume
0.1 m3

Total Pressure
2100 psia / 14479 kPa

Temperature
293 K

$\ce{CO2}$ gas constant
0.1889 kJ/kg$\cdot$K

$\ce{Ar}$ gas constant
0.2081 kJ/kg$\cdot$K

If by pressure, it's easy:

Gas
Partial Pressure

$\ce{CO2}$
3619.7 kPa

$\ce{Ar}$
10859.2 kPa

If by mass, it's a little more involved. I calculate the pressure per mass in kPa/kg using
$$
\frac{P}{m} = \frac{RT}{V}
$$

Gas
Pressure per Mass

$\ce{CO2}$
553.48 kPa/kg

$\ce{Ar}$
609.73 kPa/kg

The total pressure has to equal 14479 kPa. So the total pressure is given by
$$
P_T = \left(\frac{P}{m}_\ce{CO2}\right) m_\ce{CO2} + \left(\frac{P}{m}_\ce{Ar}\right) \left(\frac{\%Ar}{\%CO2}\right)m_\ce{CO2}
$$
Solving for $m_\ce{CO2}$ gives
$$
m_\ce{CO2} = \frac{P_T}{\left(\frac{P}{m}_\ce{CO2}\right) + \left(\frac{P}{m}_\ce{Ar}\right) \left(\frac{\%Ar}{\%CO2}\right)} = \pu{6.077 kg}
$$
Thus the Argon mass would be 18.230 kg.
The ideal gas law gives the pressures for these gas masses:

Gas
Pressure

$\ce{CO2}$
3363.4 kPa

$\ce{Ar}$
11115.6 kPa

The gas is sold by molecule percentage. Seems unlikely there is a good thermodynamic reason for the delivered gas to deviate from that percentage.

@JonCuster so that would be equivalent to partial pressure, then. Does a mixture of a supercritical fluid and a normal gas act like a gas practically?

short answer: yes it does. ;)

As long as there is no first order phase transition there is no issue. And going from a supercritical fluid to a 'normal' gas around the critical point is not a first order phase transition.


text:
How to find the wave equation in a 1D box with a 2 different V(x) values
So this is for a simple infinite well $1$D box. The bounds are defined as
$V(x)=0$, for $0<x<\frac{L}{2}$, and $V(x)=V_0$ for $\frac{L}{2}<x<L$ with everywhere in the box having a potential $0$.
Now, my first wave equation from bound I think was
$\psi(x)=A\sin(\frac{2n\pi}{L})$
but now I'm lost on what to do when evaluating the wave function from $\frac{L}{2}<x<L$
I started by evaluating from $\psi(\frac{L}{2})=A\sin(kx)+B\cos(kx)=0$, but it doesn't feel as simple as when I had to evaluate the first bounds, since I don't know which or even if $A$ or $B$ will be eliminated.
I've now added a photo below to show what the box looked like to clear confusion.
I've been told my $V_0=6E_1^0$, where $E^0_1=\frac{h^2}{8mL^2}$

Also, here is my actual questions, since I'm probably phrasing them wrongly.

Your question is contradictory. Assuming that your "box" is the region $[0, L]$, in your first sentence you state that half of the box has $V = 0$ and the other half has $V = V_0$. In your second sentence you state that the potential is $0$ throughout. Also, if the potential in the box isn't a finite constant, then it's not an infinite square well, and you should not expect the solutions to be the same: for example, why should $\psi(L/2) = 0$? This was true of the original system because the wavefunction must be zero outside of the box where $V$ is infinite, but that's not applicable here.

Made sure to update and add a photo to maybe cover what i was to clear any miscommunication.

Okay, that helps. Well, then it's not right to say that "everywhere in the box [has] a potential 0", and also I probably wouldn't call this an infinite square well.

@orthocresol well then how would i go about finding the wavefunction at $V_0$? Would it include $A$ and $B$ from the base wave equation $A\cos(kx)+B\cos(kx)$, or does one of them get eliminated? I'm more confused at what happens at $\frac{L}{2}$

@2OpForTheSystem Understanding in the first place why $sin(x something)$ is the solution (your version is incorrect, missing $x$) would help you a lot. How familiar are you with the derivation of the "particle in the box"? It would be easier to start from there instead of fixating on an incorrect solution.

Quantum Theory (David Bohm) has a good discussion on square well potentials that you might find useful. This includes transmission coefficients, which you might need for the tunneling part.

You have two differential equations to solve throughout the whole region - one in the left half, one in the right half. They have different solutions with different arbitrary constants. What do you think should happen at the boundary between these two solutions (i.e. at L/2)? At that point both are valid, what does that imply? (Yes, I know ... but let's keep it simple)

@IanBush Does it imply that both $\psi(x)$ at $V(x)=0$ and $V_0$ are going to be identical solutions at the point $\frac{L}{2}$?

Can you give a good reason why they shouldn't? If they weren't equal what would that imply? Would that be physically reasonable? Maybe a read of something like http://labman.phys.utk.edu/phys222core/modules/m10/wave_functions.html might help


text:
Lobry de Bruyn-Van Ekenstein reaction
I've recently read about this reaction called Lobry de Bruyn-Van Ekenstein reaction, where simple glucose is converted to fructose and mannose by adding a base or acid, as I've read this is supposedly called Base/Acid catalyzed isomerization but I wasn't able to find any reaction mechanism for the reaction.

A very similar reaction I've seen in synthesis of vanillin from eugenol where eugenol is converted to isoeugenol by using $\ce{KOH}$ as base and it stated that it was base catalyzed isomerization but there wasn't any mechanism mentioned.

I would like to know the mechanism for this base catalyzed isomerization reaction. Any help would be great!

That's a fancy name for what's essentially keto-enol tautomerisation. https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(McMurry)/22%3A_Carbonyl_Alpha-Substitution_Reactions/22.01%3A_Keto-Enol_Tautomerism read until the bottom as it explains the isomerisation of your alpha-hydroxy carbonyl compound. The eugenol one is not so easily explained.

Thanks for that but I'm unable to understand how eugenol to isoeugenol is a base catalysed isomerisation?

A quick search of Google Sholar shows that this named reaction is pretty common, but what a mouthful!

Have you went through Wikipedia? Also, you need to capitalize the names of the discoverers. I have done that for you.

@NilayGhosh thanks for the edit! yes I've gone through Wikipedia but I still did not manage to understand the mechanism for eugenol to isoeugenol conversion even if base catalyzed isomerization means simple keto enol tautomerism if base or acid just activates the mechanism for keto enol tautomerism, I don't understand how the second reaction occurs.

also the reaction has been carried out successfully I've seen a person on YouTube carried out the exact same synthesis but he did not mention the mechanism of the reaction and many sources just say its a base catalyzed isomerization type reaction but they did not provide any mechanism for that.

Did you Google "mechanism eugenol isoeugenol" and look at Images?


text:
Doubt regarding trends in modern periodic table
We know that atomic radius decreases along a period and increases along a group. (Same goes for metallic character)
But if we take 2 elements A & B(A is at somewhere in the top and left in periodic table and B is somewhere is below and right of B) then which among A and B will have greater atomic radius and more metallic character?

If you increase both (positive) A and B, will A/B increase or decrease? If you get answer for this, you will have answer for your question.

The question is looking very vague, to me at least.

Title is not matching the question (if I understand the question).

Such generalisations are hardly useful. Despite having a somewhat similar atomic diameter and "metallicity", A and B will have very different chemical properties.

If the period effect for B overrules the group effect, B will have the greater atomic radius than A.
If the group effect for B overrules the period effect, B will have the smaller atomic radius than A.
The same for the element metallic character.
For the p elements, you can see the metalicity boundary goes somewhat diagonally.


text:
How does the temperature of HCl affect the reaction rate of HCl & Mg reaction?
I'm Olivia. I'm experimenting with HCl and Mg. My purpose is to find the change of the reaction rate while I increase the temperature of HCi.
According to the study, when I increase the temperature of the reactant, the time of the reaction should be shorter, which means that the reaction rate should be higher.
However, when I was in the lab, I met a problem. When I increase the temperature of HCl, the time it took for Mg to dissolve in HCl became longer.
I put 0.1g of Mg into 50ml of HCl and repeat that 4 times. I heat the HCl to 20, 40, 60, and 80 degrees Celsius in 4 separate beakers. Then, I put 0.1g of Mg into it, and the time it took to dissolve completely was 22, 25, 29 and 29 seconds.
I'm pretty sure that I weigh and measure the Mg and HCl properly; the size of a piece of Mg is also the same, which means that the surface area should be the same.
I'm truly confused right now because observations are completely different from what I learned. Is this one a special case, or if there was something that I might have done incorrectly?
Thank you for your time to look at my questions and provide a valuable answer.

You have chosen a very complicated system, being an inter-phase heterogenic reaction. As the effective reaction rate is not determined solely by the reaction kinetic, but also by the contact surface. transient layer thickness and properties, rate of diffusion of reactants and products and variable local concentrations.

.... as the result, while the reaction rate at the same "reactant meeting rate" would increase with temperature, the effective reaction rate can decrease, as  the  "reactant meeting rate" may be decreased. If considering the Arrhenius equation for thje reaction rate, while the exponential factor would increase, the frequency factor may decrease even more.


text:
What is the effect of adding another solute to the solubility of the initial solute?
Let's say we have a saturated salt solution with 35 g/100 mL NaCl in water. If we add another solute up to its saturation concentration, for example 182 g/100 ml sucrose (C12H22O11), what happens?
Knowing that the solubility of NaCl is 35g/100 mL and that of sucrose is 182 g/100 ml, which one would precipitate? To clarify, are 35 g NaCl and 182 g sucrose combined soluble in 100 mL aqueous solution?

Knowing that solubility of NaCl is 35g/100 mL and sucrose 182 g/100 ml, which one would precipitate?

I found data on sucrose and glucose, shown below.

Source: https://pubs.rsc.org/en/content/articlehtml/2017/fo/c6fo01497f
At room temperature, the binary water sucrose mixture can have up to 58 volume percent sucrose, and the binary water glucose mixtures can have up to 44 volume percent glucose. If you try to dissolve both at these levels, there would be no water left.
For some supersaturated solutions of glucose and sucrose in water (e.g. 60% sucrose and 30% glucose), you can have either glucose (dashed line) or sucrose (solid line) precipitate. Which one comes out of solution would depend on the kinetics of the two crysallization processes.

What is the effect of adding another solute to the solubility of the initial solute?

For ideal solutions, there is no effect. You could explore this with two solutes of extremely low solubility (because with low concentrations, solutions approach ideal solutions). So adding silver bromide to barium sulfate should not affect its solubility. This example works because the salts have very low solubility in water whereas silver sulfate and barium bromide have moderate solubility.
You could also use two molecular solids as example, e.g. caffeine and aspirin in cold water.
The reason there is an effect in the case of sucrose is its high solubility. At the high concentration, neither sucrose nor water behaves ideal (e.g. their activity coefficients are far from 1).

Thank you so much, Not only you edited my question but also you tried to answer it. You are awesome. While your answer tackles different aspects of the problem that I didn't know, there is still a minor part about the effect of organic solutes on salt solutions that is missing.

There is a phase diagram in this paper: https://www.hindawi.com/journals/jchem/2018/4849639/


text:
What are some three+ stage cyclical reactions that only go one direction?
I was imagining the origins of life and supposed that "humans are just a bunch of cyclical chemical reactions in a coat."  Which made me wonder, what sort of cyclical chemical reactions exist in nature?
For the purpose of this question, I'm defining cyclical as "three or more stages, only ever going in one direction."
I'm also defining stages as "distinct molecular arrangements which are not in-between each-other."  Example: picking up one Oxygen, then picking up another, then dropping both and starting over wouldn't count because step 2 is just sort of "in-between."
EDIT:  A lot of the context of this question was removed for being off topic, but I believe it is important: I did attempt to research this subject before asking, but I ran into a wall where I do not have any of the formal education required to know the correct key-terms for phrasing this question on any search engine.  And I still believe it is worth apologizing for that.

Welcome to Chem.SE. I have removed extra sentences which were not part of the question. We just need your question and the research/effort you have done and references if any.

@Poutnik if you'd like to post your comment as an answer, I will accept it.  Thank you for sharing this.

Here are 2 quite known examples of cyclic reactions:

6-member ring of CNO catalytic cycle of hydrogen stellar fusion, which is used with stars more massive than the Sun, which uses proton-proton chain reaction
12-member ring Krebs/citric acid  cycle, used by mitochondria of eukaryot cells, which is central point of cell metabolism to produce ATP (adenosine triphosphate), a ready to use unit of energy.


text:
Using bond energies to calculate the released energy of a neutralization reaction
I am trying to make a simulation of a neutralization reaction that simulates the change in enthalpy as the reaction progresses. The strategy I have decided to go with is using the energy of the covalent bonds to find the difference in energy released (bond being formed) and energy consumed (bond breaking) by the system. The system I am working with is
$$\ce{HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)}$$
What is giving me trouble is that I can't find a reliable way to do this. The closest have gotten to figuring it out is this answer, but it does not finish the example in what bonds are being broken/formed and what values I should be using for the bond energies (I have searched up tables of bond energies but I am still a little confused on how to use them in this context). If anyone has a more clean cut explanation and could go more in depth showing how this works It would be much appreciated!

Be aware that both HCl and NaOH are fully dissociated before neutralization, which is just H+(aq) + OH-(aq) -> H2O(l)

Quote at least the essential relevant info from referred links for the question making sense for future readers even with  links being broken.

No bonds are broken when $\ce{HCl}$ and $\ce{NaOH}$ solutions are mixed, because the only reaction that occurs is  $\ce{H^+ + OH- -> H2O}$. Just one covalent bond is formed between $\ce{H}$ and $\ce{O}$. Covalent bonds $\ce{H-Cl}$ do not exist any more in aqueous $\ce{HCl}$ solutions


text:
How do thermodynamic variables behave in reversible and irreversible processes?
I read this definition:

reversible process is one which can be stopped at any stage and reversed so that the system and surroundings are exactly restored to their initial states. An irreversible process is one in which heat is transferred through a finite temperature.

But I just don't get it yet. How can we really differentiate between an irreversible and a reversible process with only measurements or calculations? How do U, H, T, Q and P behave? I only clearly understand that in a reversible process maximum work is done, we are taking more steps to change from one state to the other and entropy must stay constant, the order of the particles of the system can't be restored once it changes, if we had 1 mole it would be hard for $6.022 \times 10^{23}$ atoms to be in the same state as before.
But it is not as clear, for me, how the other variables are affected if we have a reversible or irreversible process.

The difference is this:  In a reversible process, when you restore the system to its original state, the surroundings are also restored to their original state (note that the entire cyclic process, including the movement away from the initial state, and the restoration to the initial state, must be reversible).  Hence all the state functions and state variables of both the system and surroundings are returned to their original values.
In an irreversible process, when you restore the system to its original state, the surroundings are not restored to their original state.  Hence, while the state functions and state variables of the system are returned to their original values, not all the state functions and state variables of the surroundings return to their original values.
I.e., what distinguishes reversible and irreversible processes is not the final state of the system, but the final state of the surroundings.  To put it another way, for the same change in state of the system, the difference between a reversible and irreversible process manifests itself through different effects upon the surroundings.
Focusing on the $2$nd law:
In a reversible process, when the system is returned to its original state, $\Delta \text{S}_{system}=0$ (because $\text{S}$ is a state function) and $\Delta \text{S}_{surroundings}=0$ (because the process is reversible).  Hence $\Delta \text{S}_{total}=0$.
In an irreversible process, when the system is returned to its original state, $\Delta \text{S}_{system}=0$ (because $\text{S}$ is a state function) but $\Delta \text{S}_{surroundings}>0$. We know that $\Delta \text{S}_{surroundings}>0$ because the $2$nd law requires that, for an irreversible process, $\Delta \text{S}_{total} = \Delta \text{S}_{system}+\Delta \text{S}_{surroundings} > 0 $ .

Nice answer. // There is yet another point of view as reversible process ( reversed by reversed external change ) versus spontaneous/irreversible process ( not reversed by reversed external change ). // And yet another POV as reversible as always in equilibrium versus irreversible as the opposite.

In thermodynamics, we usually specify the demarcation between the system and the surroundings such that all irreversibility occurs within the confines of the system, while the surroundings are taken to be operated reversibly assuming ideal constant temperature reservoirs and mechanical equipment involving conservative forces (e.g., tiny pebbles added and removed from a piston at different elevations).
Irreversibility within a system is the result of finite rate transport processes including viscous dissipation of mechanical energy to internal energy via finite velocity gradients, heat conduction at finite temperature gradients, and mass diffusion at finite concentration gradients.  Another contributor to irreversibility is chemical reaction proceeding at finite reaction rate.  All these irreversibilities result in entropy generation within the system, and make it impossible to return both the system and surroundings to their initial states.
Examples of irreversible process involving these mechanisms include rapid expansion or compression of a gas within a cylinder, heat transfer between hot and cold bodies at significantly different temperatures, and mixing of different gases or of the same gas at significantly different concentrations.
Regarding U, H, T, and P, for a reversible process, these are uniform spatially throughout the system.  For an irreversible process, if you are willing to accept the idea that, even though the system is not at equilibrium, the internal energy- and enthalpy per unit mass can be defined locally, then these vary with spatial position within the system, and integrate over differential mass to give the total U and H for the system.  Similarly, T and P vary with spatial position within the system for an irreversible process.

The following is correct:

the order of the particles of the system can't be restored once it changes, if we had 1 mole it would be hard for 6.022x10^23 atoms to be in the same state as before.

But thermodynamics doesn't really worry about this. Instead, statistical thermodynamics provides a theoretical bridge between the macroscopic and microscopic pictures. In thermodynamics you are looking at a macroscopic system and you ignore the atomic structure of matter. It does not make sense to say that a macroscopic system is ever static when viewed from an atomic perspective, but in thermodynamics this is not a concern, because the macroscopic properties such as U, H, T, and P remain constant, thanks to the statistical behavior of the very large number of particles making up a thermodynamic system. We use the value of those macroscopic properties and the composition of the system to define its state. A transition to another state involves changes in the macroscopic properties and/or the composition. A return to the original state, whether reversible or irreversible, means a return to the original values of U, H, T, P, and S.
Other answers address what it means to have a reversible versus irreversible process in terms of the total change in the entropy of the universe. It is that total entropy change that is zero for a reversible process (not the entropy of the system).


text:
Difference between SN' and elimination reactions
My professor has recently taught us about SN' reaction example in which 3-bromo-3-methyl-but-1-ene was reacted with with KCN to yield 4-methyl-pent-3-ene-nitrile as the major product.When I asked him why elimination didn't take place,my professor replied that it was based on experimental data.
Can anyone please clarify whether  the reaction is correct?If yes,then why elimination is not taking place.
Also,I was very surprised that I couldn't find any article on SN' reactions online..

Many factors contribute as to whether SN1, E1, SN2, E2 occur. By that way, it is not SN'. It is SN1

@M.L No, I am talking about SN '( prime) reactions, not $SN _1$.

What are SN1' and SN2' reactions?

@andselisk  Thank you for the related link. Can you please explain why in my case and also in the related answer why E2 elimination is not taking place to form the major product?

Honestly, I have not heard of this type of reaction before. But why can't we logically say this is the major product because SN1 leads to allylic carbocation where you can actually insert the CN in the specified position leaving the double bond in a more stable position.


text:
Why does the factor of ⅔ come in when we calculate net flux due to diffusion for an ideal gas
As per Atkins' Physical Chemistry (11E) Pg:694. We arrive at a crude estimate for the net flux through an imaginary flux plane. We find it to be $$J_{z}=-\frac{1}{2}v_{mean}\lambda(\frac{d\mathcal{N}}{dz})_{z=0}$$ Where $v_{mean}$ is the mean speed of the ideal gas and $\lambda$ is the mean free path. Now according to Atkins, there should be a correction factor due to Orientational Averaging which comes out to be $2/3$. A small demonstration was also given in form of the figure below.

I am trying to get more mathematical proof of why the factor of $2/3$ appears. If anyone can recommend to me how to approach it or suggest me any reading material where this thing is explained in detail please do help.


text:
If I heat a solution to bring a solute in solution, will it precipitate when it goes back to room temperature?
Suppose I'd like to bring a solute into solution at a concentration above its solubility at $\pu{25 ^\circ C}$, so I heat the solution and achieve a concentration above the $\pu{25 ^\circ C}$ solubility. Once the solution is back to room temperature, will the solute precipitate?

Mostly likely it will. This is called recrystallisation and is often used to purify a crystallisable material

Or it won't. This is called a supersaturated solution (honey often is supersaturated).

and can a supersaturated solution stay so for a very long time, or will it eventually precipitate? I'm interested specifically in amino acids like adenine hemisulfate that seem to have low solubility in water, but many protocols suggest preparing stock solutions with concentrations above the room temperature solubility by warming the solution, and I'm worried that they may slowly precipitate with time

Another good example is glycerol, which melts at $17.9°$C. But it can be kept liquid for years at much lower temperatures, due to its high viscosity. During my chemistry studies, I remember the professor told us once that the first crystals had been obtained in a flask of glycerin that was nearly forgotten in winter in a truck in a railway station in northern Sweden. Apparently these unique crystals have been used later on for starting the crystallization of all the glycerin crystals in the world.

The you should add this detail to the question.

Generally, your conclusion is correct However:

As mentioned in comments, supersaturated solutions can endure for a long time: consider the $\ce{NaCH3COO.3H2O}$ heating pad, which dissolves in its own water of crystallization at 58°C, and doesn't crystalize, even when greatly cooled, until disturbed (e.g., by a "clicker"). Similarly, starting crystallization can be difficult, requiring additional chemicals and seed crystals, or ultrasound, or just scratching a glass vessel.

Another possibility is a retrograde temperature solubility curve. Common calcium sulfate,$\ce{CaSO4}$ and lead halide perovskites are far less soluble in hot solutions. In that case, one would chill the solution to completely dissolve the solute, and slowly heat it to promote precipitation and crystallization! BTW, that's why precipitates from heating water cause boilers (and percolators) to need occasional "descaling".

A nice answer, but it talks rather about exceptions than rules. Generally, it will crystallize, the time would differ. //Scale deposition is mostly due Ca/Mg carbonates.  Solubility of CaSO4.2H2O is 2.55/2.05 g/L at 20/100 °C, so unless it is very mineralised mineral water, it would stay dissolved.


text:
How to identify isochoric process from the problem's wording?
Problem 19 from NEET's Solved Paper 2013:

The amount of heat energy required to raise the temperature of $\pu{1 g}$ of helium at NTP from $T_1$ to $T_2$ (in kelvin) is
\begin{align}
&\text{(a)}~\displaystyle\frac 3 8 N_\mathrm{A}k_\mathrm{B}(T_2 - T_1)
&\quad
&\text{(b)}~\displaystyle\frac 3 2 N_\mathrm{A}k_\mathrm{B}(T_2 - T_1) \\
&\text{(c)}~\displaystyle\frac 3 4 N_\mathrm{A}k_\mathrm{B}(T_2 - T_1)
&\quad
&\text{(d)}~\displaystyle\frac 3 4 N_\mathrm{A}k_\mathrm{B}\frac{T_2}{T_1}
\end{align}

My chemistry teacher said the “language of question” gives away the process is isochoric and suggested to use the following equation to solve the problem:
$$\mathrm dQ = nC_V\Delta T = \frac f 2nR\Delta T$$
yielding the correct answer $\text{(a)}~\displaystyle\frac 3 8 N_\mathrm{A}k_\mathrm{B}(T_2 - T_1).$
How can this be deduced conceptually? What exactly points to an isochoric process in this case?

Welcome to CH SE site!
Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
Suitable formatting can be done according to formatting math/chem expressions/equations.

Search for molar heat capacity of ideal gases at constant p ( isobaric process ) and constant V ( isochoric process ) // Look also at molar versus specific heat capacity. // Generally, search properly before asking. The site tries to avoid answering questions for which answers can be easily found.

1 g of He and 3/8 point there.

I guess the implication of the problem statement, especially NTP, suggests (slightly) that the process is at constant pressure.  But it certainly is not clear.

@ChetMiller c_v(1 g He) =(approx) 3/8Na.kb, cp(1 g He)=5/8Na.kb // I do agree the wording itself (NTP) speaks for isobaric process, but reportedly correct formula speaks for isochoric process.

Note that NTP cannot be assumed to be at constant p, as it does not assume constant T either. My interpretation is NTP is meant as  initial conditions.

@Ujjawal Mishra where did you get struck?

@Infinite Where it is written that the Volume is constant. I think there should be constant pressure, because in question NTP is mentioned. So, why we are using that Heat given = n Cv ΔT ?

At NTP both temperature and pressure is constant. Then what do you mean by raising temperature.

For your better understanding first forgot "at ntp" in the question. Then understand the definition of internal energy.

The heat given to a system can be utilised in increasing its internal energy or doing work against the surroundings. This is popularly known as first law of thermodynamics.  The change in the internal energy brings the change in the temperature and vise versa. So to increase the temperature one should increase the internal energy of the system. We know that, $$ \Delta U = nC_v \Delta T $$ According to first law of thermodynamics, the heat given to the system at constant volume is known as internal energy. ( $\because$ work done ($W$) = 0 at constant volume )$$ \Delta Q = \Delta U $$ $$\implies \Delta Q = nC_v\Delta T $$ I think this clarifies your doubt.

This avoids the major question why isochoric and not isobaric. :-)

In question, they had mentioned 1 gram of Helium. So, now the moles of helium is fixed, and the temperature is changing. Thus, now there are two possibilities, that the volume can change or the pressure will change. Now, how we will decide that what will change ? @Poutnik

@UjjawalMishra  Reread all what has been posted, including comments. It is there. Molar heat capacity of monoatomic ideal gas is (3/2)R for isochoric process, (5/2)R for isobaric one.

It should be pointed out that ∆U=ncv∆T is true for isochoric processes with real substances (assuming constant heat capacity), any process with an ideal gas, but not generally true for any process with any substance.

No you are misconcepted. It is true for all processes.

You can verify in Wikipedia.

@Poutnik Sir please answer this question with reliable explanation. Because I want to understand this important concept. Pardon me for inconvenience. ( I'm new in Thermodynamics lesson.

@UjjawalMishra Chemistry SE site does not support answering without explicit significant effort to understand it on your own.


text:
Book reccomendations for Pertubation Theory?
I'm sorry if this is an inappropriate place to ask this, but I'm still trying to get the hang of this site. So my book is Thomas Engel and Philip Reid Physical Chemistry $3^{rd}$  ed. I'm on the Chapter 17: "Commuting and Non-commuting Operators and the Surprising Consequences of Entanglement
However, it is missing the Chapter on Perturbation Theory, which I was told should've been where chapter 18 is. Instead of Ch. 18, we have A Quantum Mechanical Model for the
Vibration and Rotation of Molecules. Where are some good sources for Pertubation Theory that would have this?

My personal suggestion is to skip the "physical chemistry" books and go straight to the "physics" books, or the "quantum mechanics" books. Griffiths was my favourite when I started.

Cohen-Tannoudji is recommended in the duplicate, it's a standard text for QM too.


text:
Raman spectra of peptide
This is my first time measuring Raman spectra of a peptide sample. It is in solid form. I do think I see some Raman peaks but I am not sure how to assign them. I have a peptide sequence through which I can make tentative assignments but I can't pinpoint which one belongs to which amino acid. This is due to the possibility that there could be more than one contender for a given peak/band.
I would be grateful if someone could help me out or point me to some other source/reference I could lookup.
Thanks in advance.

Peak assignment in spectroscopy is not a trivial task especially for a large molecule. I am not an expert in this area but you should consult someone who can do modelling for you and try to predict its spectra.


text:
Various pathways of an alpha diketone reaction

This is a question from JEE Advanced 1997.
I have found two possible products of the reaction:

By intramolecular aldol condensation:

By benzil-benzilic acid type rearrangement:

Which reaction takes place here?

#1 is a very strained species, whereas the benzil-benzilic acid rearrangement is a well-precedented pathway. Occam's razor applies. Your choice.

@Waylander I agree! However, I wonder if there is any stabilizing effect on that...

I don't think there is enough stabilisation to make #1 a viable product.

Furthermore there is very little literature mention of 2-phenylcyclopropenone - it if could be prepared by this route then there would be multiple examples.

There is something here that will be of use.


text:
What is the mechanism of the reaction between 1(3H)-Isobenzofuranone and benzaldehyde?
I have been trying to deduce a reaction mechanism for the reaction between the two compounds to give the compound labeled 2 in the diagram but I do not think I have a satisfactory answer. My thinking is that a strong base deprotonates the carbon adjacent to the oxygen in the cyclic system which then attacks the carbonyl of the benzaldehyde in an aldol type reaction to give the final product.
Is this a correct mechanism and if not, which is? I have tried looking for the reaction conditions and reagents but Wikimedia does not cite any references.

Redraw the double bonds in the benzene ring and you will find that the reaction site is an active methylene and can attack the aldehyde in a straightforward vinylogous aldol reaction. It could probably go under acid catalysis e.g. Toluene/PTSA Dean-Stark reflux. See more here: pubs.acs.org/doi/10.1021/cr990247i

Maybe it's just me, but that looks like most obvious step in this reaction.


text:
How many elements have been identified for which there are no known spectral lines?
Background:
@ProfRob's answer to If there were undiscovered elements (119 on) in a star's spectral lines, could we tell? in Physics SE begins:

I think that would be very difficult indeed. Unfortunately there are many elements, many isotopes and different ionisation states. Literally millions of transitions. Missing data and incorrectly estimated oscillator strengths mean there are often "unexplained" weak absorption lines in high resolution stellar spectra.

and later

I am not convinced by Gopka et al.'s identification of a single absorption line of Einsteinium.

Which may refer to Gopka et al. (2008) Identification of absorption lines of short half-life actinides in the spectrum of Przybylski’s star (HD 101065). As an aside Wikipedia's Przybylski's Star says:

Przybylski's Star also contains many different short-lived actinide elements with actinium, protactinium, neptunium, plutonium, americium, curium, berkelium, californium, and einsteinium being detected. The longest-lived isotope of einsteinium has a half-life of only 472 days. Other radioactive elements identified in this star include technetium and promethium.15
15Gopka et al. (2008)

Question:
How many elements have been identified for which there are no known spectral lines?
By "spectral lines" I mostly mean UV-Vis-NIR spectroscopy where optical telescope (space and ground) equipped with spectral analysis equipment can record atomic transitions.

Please take this to chat.

How many elements have been identified for which there are no known spectral lines?

tl;dr: Eighteen, or all of them above A=100.

One place to start is a 2015 article by H. Backe et al. in Nuclear Physics A on "Prospects for laser spectroscopy, ion chemistry and mobility measurements of superheavy elements in buffer-gas traps". For the elements Np, Pu, Am, Bk, Cm, Es, and Fm they note:

In each actinide spectrum tens of thousands of spectral lines can be observed. The levels are organized into terms, some dozen of terms form a configuration, and there are a dozen or more configurations. The order of hierarchy is not evident since there is considerable overlapping of different configurations and since the terms are not pure in any coupling scheme they must be described as mixtures (configuration mixing). In most cases, the levels can be identified only by comparison with theoretical calculations.

The rest of the paper then goes on to discuss how things like single ion traps or in-beam measurements could be used in the future to investigate the superheavy elements which are made one atom at a time in accelerators (as opposed to neutron absorption in high flux reactors, which is only good up to Fermium). Two comments: first, it is not surprising that with ~100 electrons on an atom there could be 1000s of spectral lines. Second,  finding one or even a few lines in the spectrum of a star hardly seems to justify connecting those to a specific element, particularly a superheavy element.
A recent example of an in-beam measurement of Einsteinium was just published in Physics Review C, where a reactor-generated "bulk" (4 picograms of $^{255}$Es to 4 nanograms of $^{254}$Es) sample was ionized, accelerated, and laser resonance ionization spectroscopy was used. Based on Figure 1 in that paper, as of right now $^{255}$Fm is the heaviest element that has optical spectroscopy data.
Another paper of interest would be from Physica Scripta, particularly Figure 7 which shows the broad range of isotopes for which laser spectroscopy has been performed at on-line radioactive beam facilities. It indicates the $^{255}$Fm isotope, but does not contain the later publication of $^{253}$Es and $^{255}$Es in the reference above, presumably since they did not have a time machine available.
So, as of March 2022, no elements above Fermium (A=100) have any spectroscopic lines identified for either neutral or ionized species.
Comments have asked about several specific rare lighter elements. The emission spectra of Actinium was pretty well measured by William F. Meggers et al., Journal of Research of the National Bureau of Standards Vol 58(6) June 1957, identifying 500 lines between $206.2$ and $788.682$ nm. Meggers had previously worked on Promethium, published in the Journal of Research of the National Bureau of Standards Vol. 46(2), February 1951. More than 2200 lines between $220$ and $690$ nm were measured. A 2006 paper in Reports on Progress in Physics by E. Gomez et al., titled 'Spectroscopy with trapped francium: advances and perspectives for weak interaction studies' nicely covers work up to then on Francium using single ion traps. To quote the abstract:

The studies include the location of energy levels, their hyperfine splittings and their lifetime. All of these levels are close to the ground state. The results show a remarkable agreement with calculated ab initio properties to a degree that is comparable with other stable alkali atoms.

Apparently they intend (although I have not followed up on it) to perform parity non-conservation measurements to test the weak interaction, and will use precision spectroscopy of Francium to do it.

@jon you mention astatine but what about francium? I ask because WP dies not report a spectrum for this element either.

NIST lists 4 lines.


text:
Reaction of long-term stored zinc acetate dihydrate
I was investigating solubility of zinc acetate dihydrate for various organic solvents, and for this I took out zinc acetate dihydrate in my lab containor. When I opened it very big aggregation was formed and some powders looked like transparent crystal. In various papers that synthesize zinc oxide nanopowder the researchers often use zinc acetate dihydrate as precursor and 2-methoxyethanol as its solvent. So I tried to dissolve it in 2-methoxyethanol. As I dissolve it and stirred for about 1 hours at room temperature, the solution turned somewhat cloudy. Ethanol did not dissolve it, and it turned opaque white powder as I raise temperature to about 60 ℃. When I report this problem to my senior he said that is strange and let me use recently bought zinc acetate dihydrate. 2-methoxyethanol successfully dissolved it, and the solution did not turn cloudy. I think that the old zinc acetate dihydrate reacted with air or water during long-term storage, but I can't find the information for it. Anyone who knows this?

add a drop or two of acetic acid. This often helps. Zince acetate, as a salt of a weak base and a weak acid is prone to hydrolysis.

My guess is the old lot of zinc acetate dihydrate may have partially lost its crystal water (reportedly loses water at 100 °C). Or it may gain water in humid regions . This  could affect the dissolution in  ways:

Assuming it is still dihydrate, more (or less) of zinc acetate is used than thought.
Missing (or extra) crystal water = less (more) of solvent, which is less (more) polar.

Have you checked if the fresh lot acts as expected?
Theoretically, you could determine mean zinc content (e.g. AAS or EDTA complexometry) and eventual adding of extra water if it was lost.
My advice is - as the purpose is to determine solubilities in solvents - forget about using the old lot. It is too poorly defined substance, as the water content change may not be even across all crystals sizes.
For investigation of matter behavior, you need to keep experiments as much controlled as possible. Using old lots brings unknown parameters.

Another possibility coming on mind is slow hydrolysis by crystal and aerial water and evaporation of acetic acid, leaving basic zinc acetate and/or zinc oxide/hydroxide. Like:
$$\begin{align}
\ce{Zn(CH3COO)2 . 2 H2O(s) &-> Zn(OH)2(s) + 2 CH3COOH(g)}\\
\ce{Zn(CH3COO)2 . 2 H2O(s) &-> ZnO(s) + 2 CH3COOH(g) + H2O(g)}
\end{align}$$
See also Wikipedia: Zinc_acetate#Basic_zinc_acetate stating that:

Heating $\ce{Zn(CH3CO2)2}$ in a vacuum results in a loss of acetic anhydride, leaving a residue of basic zinc acetate, with the formula $\ce{Zn4O(CH3CO2)6}$.

Speculatively, long term storage may lead to similar results for previously dehydrated acetate with the summary reaction:
$$\ce{4 Zn(CH3COO)2 . 2 H2O(s) -> Zn4O(CH3COO)4(s) + 2 (CH3CO)2O(g) + 8 H2O(g)}$$


text:
Why do I measure some remaining voltage after disconnecting an electrolyte from a voltage source?
I'm not at all an expert in electrochemistry and want just understand a little what is going on:

I have a NaCl solution in water, connected most simply by plain copper conductors.
My observation is: When the switch is closed, the electrolyte is connected to 5V. After opening the switch, I measure a slight voltage of ~100mV. What could be the reason for this voltage? It seems, that the arrangement is slightly (!) charged like a battery and inherits a voltage source - could that be and if yes, what are the reactions going on?
BTW, when I switch polarity of the battery, also the remaining voltage will invert to a negative voltage of about ~100mV. So it is not just some offset voltage of the voltmeter.

Consider the electrolytic cell as a very very inferior rechargeable cell. The residual voltage is caused by differences of composition of electroactive particles at both electrodes,  created during electrolysis..

So what I observe is quite normal...?

It will serve you better if you try to answer it yourself.

I'm not a educated in chemistry. What reactions are going on on the electrodes? There are Ions Na+, Cl-, but also H+, OH- ... but anyway, i'll try

In fact, you need not any knowledge of chemistry for that, just read carefully the 1st comment.  Said by other words, a perfectly symmetric cell creates zero voltage output. If some process like electrolysis breaks this symmetry, it produces nonzero voltage.

You're not the first to notice this effect: Saltwater's Second Wave: Aquion Has Emerged From Bankruptcy Under a New Owner. That's a business-oriented article, but you can trace back from there to discover the science behind it.

@Poutnik: You mean that due to electrolysis the concentrations of ions in the solution will be different on both electrodes. Therefore a potential is established. Is this disbalance  of concentrations stable or would there be a "self discharge" because of diffusion process due to a concentration gradient? I guess, that the voltage disappears after stirring the liquid - right?

@RayButterworth The linked article says absolutely nothing about this effect. Voltage behavior in lithium-ion batteries after electrochemical discharge and its implications on the safety of recycling processes is a far better reference.

@MichaelW I encourage you to ask this question over on https://electronics.stackexchange.com/. Because we discourage cross-posting, you'll need to delete the question here. I have no doubt you will get several satisfactory explanations from the electronics/electrical engineering crowd.

@ToddMinehardt, the article isn't about lithium-ion batteries, but about an alternative therefor, salt-water batteries (which don't require recycling).

@RayButterworth The article is completely devoid of anything scientific. There is literally one sentence that uses the word "saltwater." If you want to contribute meaningfully, there are myriad other references you could have chosen. Further, the OP is asking about the effect observed, which is NOT unique to salt water.

@ToddMinehardt, I wasn't trying to answer the question.  I was pointing out that others have discovered this effect too, specifically with salt-water, and that there are now commercial products available that make use of it (potentially creating a cheap and safe way of storing solar or wind energy). The OP might find that of interest, even if it doesn't directly answer the question.

The electrolysis of $\ce{NaCl}$ solutions is not the best way to handle this problem, because of secondary reaction occurring during the electrolysis between $\ce{Cl2}$ produced at the anode, and $\ce{OH-}$ ions produced at the cathode.
Let me first describe a simple case : electrolyzing an aqueous solution of sulfuric acid $\ce{H2SO4}$. The following reactions are carried out on the electrodes : $$\mathrm{Cathode, (-) } : \ce{2H+ + 2 e- -> H2}$$ $$\mathrm {Anode, (+)}: \ce{2 H2O ->  4 H+ + O2 + 4 e-}$$ At the end, a couple of bubbles of $\ce{H2}$ remain adsorbed on the cathode and $\ce{O2}$ on the anode. Of course, much more of these gases have been produced previously, which have been lost by bubbling and getting out of the solutions around the electrodes. But the few remaining bubbles on the electrodes make up a weak galvanic cell. When the switch is off, this cell is able to produce a small current for at least a couple of seconds, according to the equations $$\mathrm{Anode, (-)} :  \ce{H2 -> 2 H+ + 2 e-}$$ $$\mathrm{Cathode, (+)}: \ce{O2 + 4 H+ + 4 e-  -> 2 H2O}$$ This behavior does not last long, because of the ridiculous small amount of $\ce{H2}$ and $\ce{O2}$ available on the electrodes.
It is worth while pointing out that $\ce{H2}$ is produced by electrolysis at the cathode (negative pole). But at the end of the electrolysis, when the two electrodes are directly connected, the system is a galvanic cell. The electrode holding the $\ce{H2}$ gas is still the negative pole, but it is the anode now. Anode is always the electrode where reduction happens.
With solutions of $\ce{NaCl}$, the behavior is similar, although the chemistry is more complicated, because of secondary reactions occurring during the electrolysis between $\ce{Cl2}$ produced at the anode, and $\ce{OH-}$ ions produced at the cathode, like : $$\ce{Cl2 + 2 OH- -> Cl- + ClO- + H2O}$$ $$\mathrm{and, when }   \mathrm{hot } :  \ce{3 ClO- -> ClO3^- + 2 Cl-}$$


text:
Notational inquiry regarding nuclear chemistry
I'm aware that the notation for an element (with a charge) is as follows:
$$\large ^{~~~\text{mass number}}_{\text{atomic number}}(\text{element symbol})^{\text{charge}}$$
However, during beta decay (of a neutron to a proton and a negatively charged electron) we write it almost as if the beta decay emitted two negatively charged electrons. Allow me to explain.
$\ce{^14_6C -> ^14_7N + ^0_{-1}e}$ is the reaction of negative beta decay of carbon-14. But, the fact that in the right side of the reaction we wrote $\ce{^14_7N}$ by itself already means that has a charge of 0. The issue here is that every proton has a positive charge of 1, and so this means there must be another negatively charged electron we haven't accounted for (since the other one that is emitted appears to not be in this consideration, because if it was already summed then we'd have $\ce{^14_7N-}$ in the right-side of the reaction.
Any clarification would be very much appreciated.

You do realize $\ce{_6^{14}C}$ and $\ce{_7^{14}N}$ are nuclei, don't you?

Yes, that's the notation of the things inside the nuclei. However, for $^{14}{7}N$ to have a neutral charge wouldn't that mean we had to emit another negative-charged electron on top of the one that split off, the ^{0}{-1} e?
For example, when I read $^{14}_{7} N$ I assume that means its charge is neutral.

You can use $\ce{^14_7N}$ instead, it comes out much better. When you write $^{14}_7 N$, there are a couple of problems: firstly the TeX / MathJax engine thinks that you're trying to apply the super/subscripts to whatever came before it. This is less problematic if there's nothing that comes before it, but in the equation in your original question, it was trying to attach them to the plus sign. Next, the alignment is kinda wrong; you want the super- and subscripts to be right-aligned. Thirdly, the N comes out italicised, whereas chemical symbols should be upright.

Appreciate it orthocresol, will apply that.

Nuclear physics folk generally ignore what all happens to the electrons. Partly they just don't care much about those light weight gnats buzzing around the nucleus, but also because an energetic charged particle departing the nucleus (beta, positron, alpha) may well cause some undetermined number of electrons to fly away as well, so there is no unique balance of electrons on the remaining nucleus.

I see Jon Custer, so you are saying that they write the resulting nucleus as if its neutral charge but completely ignore whether it actually is a neutral charge or not. That is honestly very helpful and I really appreciate it. I try to understand things to the best extent rather than just skipping them, hence this question which may seem basic but also unorthodox in a way.


text:
How much calcium hydroxide is needed to raise the pH of an aqueous sulphuric acid solution?
Lets say dissociation constants $K_{\mathrm{a}1}$ & $K_{\mathrm{a}2}$ of $\ce{H_2SO_4}$ in water are $1000$ and $0.012$ respectively. So the first dissociation step happens completely:
$$\ce{ H_2SO_4 (l) + H_2O (l) \rightleftharpoons HSO_4^- (aq) + H_3O^+(aq)}$$
Where: $\ce{[HSO_4^-] = [H_2SO_4]_0)}$, let those be $c_0$. And the second step:
$$\ce{ HSO_4^-(aq) + H_2O(l) \rightleftharpoons SO_4^{2-}(aq) + H_3O^+(aq)}$$
So the IE table for the second step would look something like this:

$\ce{[HSO_4^-]}$
$\ce{[SO_4^{2-}]}$
$\ce{[H_3O^+]}$

Initial
$c_0$
0
$c_0$

Equilibrium
$c_0(1-\alpha)$
$c_0\alpha$
$c_0(1+\alpha)$

So at equilibrium, $K_{\mathrm{a2}} = \frac{[\ce{SO_4^{2-}][H_3O^+][HSO_4^-]}} {(1 - \alpha)} = \frac{c_0\alpha(1+\alpha)}{(1-\alpha)}$. Which can be solved for $\alpha$: $c_0\alpha^2+(c_0+K_{\mathrm{a2}})\alpha-K_{\mathrm{a2}}=0$
Then, $\alpha$ can be used to obtain the $p$H of the solution at equilibrium:
$p$H = - $log\ce{[H_3O^+]}$)= - $log$ $ { \pu{c_o( 1+\alpha)}}$)
So, if this solution is to be neutralized with $\ce{Ca(OH)_2}$ to a certain pH, lets say $11$, how would I calculate the amount of lime needed to increase the $p$H by that much?
Thank for your time

Why choosing $\ce{Ca(OH)2}$ and not $\ce{NaOH}$? Neutralization sulfuric acid with $\ce{Ca(OH)2}$ produces $\ce{CaSO4}$, which is not very soluble in water ($0.0047$ M, from solubility product). And measured solubility is much higher : $0.018$ M (Journal of Chemical Education 77, 12, Dec. 2000, p. 1558). This is due to other ions like $\ce{[Ca(OH)]^+}$ and  $\ce{HSO4^-}$.

Please have a look at the mhchem extension of MathJax for correct typography.

If you wanted pH = 11 then acid would be irrelevant and the basic properties of hydroxide would be needed for calculation.


text:
Between compounds having same functional groups, why does the one with greater -I effect have higher acidic strength
Consider the two compounds $\ce{FCH2COOH}$ and $\ce{ClCH2COOH}$.
Since fluorine is more electronegative than $\ce{Cl}$ it will show a greater –I effect.
It will pull the the electrons closer towards it and give $\ce{CH2}$ a small positive charge and so on to $\ce{COO-}$.
The $\ce{COO-}$ in compound 1 will be more stable but have  a greater positive  less negative charge than compound 2.
 Since $\ce{COO-}$ in compound 2 has lesser charge it should attract $\ce{H+}$ better than compound 1
In short, why is the strength of an acid in an organic compound directly proportional to the stability of the acid's conjugate base?

I think this question is conflating a few concepts. Stability is a relative concept, so in order to compare two completely different compounds, you need to establish a common baseline first. Then it's depending on exactly that reference what behaves how. So the question you need answered is which state predominates in an acid base reaction to establish some kind of acid strength measure. For most acids and bases, water seems an excellent choice as reference.

I fixed your mixup with the charges. Still unclear?

Less (absolute, i.e. less negative) local charge on the oxygen atoms of the conjugate base means the energetic penalty for removing the $\ce{H+}$ is lessened.
Or in other words: The -I effect of the halogen atoms weakens the COO-H bond.


text:
Are the solid supports used in oligonucleotide synthesis reusable?
So in solid-support oligonucleotide synthesis, the protected nucleotides are attached at either the 3' or 5' end to a linker, which is then itself attached to a a solid support. After synthesis, the linker and oligonucleotide are cleaved from the solid support and each other. Deprotection and purification occur.
What I don't understand, is if I'm using a nucleosidic support, is the support reusable? I mean, once I cleave the linker and oligonucleotide from the support, presumably the support is no longer nucleosidic. But if I can just reuse the support and attach my own different linker and nucleoside to it, then what exactly is the advantage of universal supports?
This may be a stupid question, but I cannot for the life of me find the answer to it and I struggle to answer it intuitively myself because I have no background in biochemistry or even chemistry in general. I do not doubt I am totally misunderstanding something here.

No, you cannot reuse a solid support.
See, the solid support is usually a polymer, as for instance a silica based resin, chemically functionalized with something that will be attached to this resin, but that will also be able to react with the oligonucleotide's first monomer (last, actually since the synthesis is 3' to 5'), and have this monomer attached to it at the end of the 4 steps synthetic cycle. In the next cycle another monomer will attach to the already attached monomer and that will repeat as many times as the number of nucleotides in the desired oligo.
Traditionally, solid supports were sold with the first synthetic monomer already attached to it. So when you configured the synthesis you would have to analyze what was the last nucleotide in the oligo sequence, and use the solid support with that specific nt to initiate the synthesis from the second nt.
Latter, they developed universal supports, which would not have any nucleotides already attached, meaning you could just use the same resin to synthesize any oligos, regardless of the sequence, and of course, now you would start the synthesis by the last nt, and not by the second last (remember, the synthesis starts by the last nt, the last nt in the oligo sequence is the first to be attached, and the first nt in the sequence is attached in the last cycle).
After the synthesis you have to release the oligo from the solid support by cleaving the linker. To use the same support again, you would have to chemically attach a new linker, but only after really assuring that absolutely no trace of the oligo you synthesized in it before were left. That would be so laborious that is definitely not worthy.
To be sure you are synthesizing perfectly free from cross contamination oligos, use always fresh new solid supports, including the inert frits (if you are using a resin).

This is helpful, but you open up by stating it can't when you conclude later that it shouldn't be reused, or at least that it would be time consuming and laborious to do so (probably not worth the time and money).

Yes, you are right, it is possible, but it is not because it is possible, that you can do it. In practical terms when you are in the oligo synthesis universe you just CANNOT reuse a solid support. Not only because it is not worthy, but mainly due to quality issues - really serious quality issues. So, it is feasible, but it is not allowed, you just can't.


text:
Difference between the specific heat capacity of potassium chloride solution and sodium chloride solution
I am curious about the difference between the specific heat capacity of a potassium chloride solution (KCl + water) and a sodium chloride solution (NaCl + water) - particularly the specific heat capacity of which one is greater and by how much (a number would be nice!). You may assume that all variables for the two solutions are the same - volume, concentration, etc.
I assume that this would have to be linked to ionic bonding to understand which molecules are affected by heating the solution but I am not aware of the differences in bonding in the two solutions, and more importantly, how to represent it mathematically.
A simple one-liner such as "Solution X has a greater specific heat capacity than Solution Y by a factor of n" would be perfect (some reasoning/explanation would be great as well).

The Handbook of Chemistry and Physics gives the following heat capacities for $\ce{KCl}$ and $\ce{NaCl}$ solutions at $\pu{18°C}$ :

Mole fraction x
0.04
0.02
0.01

$\ce{KCl}$
3.46 $\pu{J g^{-1} K^{-1}}$
3.78 $\pu{J g^{-1} K^{-1}}$
3.966 $\pu{J g^{-1} K^{-1}}$

$\ce{NaCl}$
3.68 $\pu{J g^{-1} K^{-1}}$
3.895 $\pu{J g^{-1} K^{-1}}$
4.025 $\pu{J g^{-1} K^{-1}}$

Would you know how to find the (approximate) value of any mole fraction and at any temperature using this table?


text:
What will happen in the volume occupied capacity of each gas when separator is removed?
Consider a situation where two gases occupying volumes $V_1$ and $V_2$ and having partial pressures $p_1$ and $p_2$, respectively, are kept separated in a rigid container. If the separator is removed, is it true that the first gas will occupy the whole volume and, similarly, the second gas will occupy the whole volume $(V_1 + V_2)$?
If yes, why?
If no, is the reason that the molecules of each gas needs some fixed space so in general they will occupy some quantity $V^ʹ$ and $V^{ʹʹ}$ in space such that $V^ʹ + V^{ʹʹ} = V_1 + V_2$?

One handy way of visualizing the behaviour of gasses in a container is to imagine them as ping-pong balls continually bouncing around in a vacuum sealed room, without gravity or friction.
Pressure on the walls is caused by their bouncing off them, and room temperature is related to their average speed.
In the described situation, imagine that one gas is represented by red balls, and the other by blue balls.
At the instant the barrier is removed, all of one side of the room will be red and all of the other blue.
Gradually, more of the balls of one colour will move into the area occupied by the other.
Eventually, each colour will be uniformly distributed throughout the container.
In studying the behaviour or characteristics of one specific colour (type of gas), one can totally ignore the presence of any other colours.
The only effect of interaction between colours is mid-air collisions, which randomly transfer momentum and velocity between individual balls, but which make no over-all difference once a long-term stable state has been reached other than equalizing the temperatures of each of the colours.
And if different coloured balls happen to have different masses, again the only effect this has is in the transfer of momentum and velocity, and again it has no significance in the long-term.
The pressure caused by one colour is known as its partial pressure.
The sum of the partial pressures of each of the colours always adds up to the total pressure (proof: turn out the light and all balls are the same colour).