text:
How to predict amine donor in biochemical reactions (whether glutamine or aspartate)?
In biochemical reactions amine $(\ce{-NH2})$ transfer is done in multiple pathways (e.g. urea cycle, purine biosynthesis etc.).
In many of the situations glutamine donates amine group (e.g. xanthosine monophosphate → guanosine monophosphate). In few instances, however, aspartate donates amine $(\ce{-NH2})$ group (e.g. inosine monophosphate → adenosine monophosphate).
This seems to be very random to me. Is there any pattern in such amine transfer reactions? Can we generalise it in some way?

While that's on topic here, you may have better chances with similar questions on Biology.SE.

The answer to this question might be "whichever residue is at the appropriate location in the enzyme's active site".

The side chain or aspartate cannot DONATE an amine (-NH2) as it does not have one.. Only the side chains of asparagine (N) and glutamine (Q) in addition to Lysine (K), and argentine (R) have amine groups.. (and not all of them are capable of donating due to resonance).. Unless aspartate is the substrate (i.e. freely bound)

Go through purine de novo biosynthesis. You will see where Aspartate reacts to transfer an amine to the substrate. And I never saw a biosynthetic pathway where Asparagine donates amine group.

@A.N.Ψ, I can see that I misinterpreted the question a bit, as I was automatically thinking of amino-acid side chains in protein. However, if the aspartate is on its own (as a substate) then I can see how this is the case. I.e. the -NH3 group may be donated.. I think I should update my answer a bit in this regard.

Yes, that's what I wanted to point out.

There are several very general concepts that you have to know about in order to see a pattern in biochemical reactions driven by enzymes. A quick answer is given at the very end. Amongst many concepts to understand, two main concepts are as following:
Enzyme are classified with a specific enzyme commission (EC) number, which is a type of ID number based on catalytic function. The first number groups the enzyme into an overall group. In example, enzymes that can transfer functional groups are placed in EC.2. Within EC.2 you can then subdivide into more specific functions, e.g. transaminases that transfer nitrogen containing groups have the the EC number of 2.6, and within this group again you can find enzyme that specifically transfer amine groups. Those enzymes will then have the EC number of 2.6.1 (aminotransferases), and you can then further specify into families, e.g. aspartate aminotransferases EC.2.6.1.1.
Enzymes folds within specific functional families is evolutionary conserved and highly specific. The three-dimensional shape of enzymes within a specific family, e.g. aspartate transaminase; EC 2.6.1.1, is very conserved (the amino acids sequence identity will also be very high)
and enzymes that are grouped into the same family will have highly similar overall structure and thus very similar mechanisms of catalysing reactions. In contrast, enzymes in completely different families, with different foldes and different amino-acid sequence will have completely different catalytic mechanisms.
To predict how some enzyme will catalyse a reaction, you need to find what enzyme it is related to. Here it will also be important to remember that structure is more conserved than sequence. A good starting point is to do a BLAST search with the protein sequence that you are investigating, note the you could also do a blast search against the PDB database to find related published protein structures.
Continuing with the exsample of amino-transferases, you can even subdivide the enzymes into where the catalytic reaction will take place. In example, you have alpha-aminotransferases or omega-aminotransferases, depending on the location on the substrate that the amino will we translocated to and from.
To answer your question:
Now that we covered some very general concepts about enzymes, it is important to know that their structure and sequence are specifically designed to accommodate a particular substrate. In other words, some enzymes will take aspartate as substrate and donate its amine group - while other type of enzymes will take glutamine as substate - and donate its amine group.

As a side-note. In proteins, the amine-group in aspartate would be part of the protein backbone (amide bond / peptide bond), and the side-chain (no amine group) would therefore be the only accessible part of this amino acid. It is therefore a good idea to specify if amino acids are substrates or catalytic residues (i.e. part of the enzyme) when asking about biochemical reactions.


text:
Why isn't chlorine reduced or nitrogen oxidised in the reaction between ammonium perchlorate and aluminium?

A space shuttle’s upward thrust came from the following reaction between aluminium and ammonium perchlorate.
$$\ce{10 Al + 6 NH4ClO4 → 4 Al2O3 + 2 AlCl3 + 12 H2O + 3 N2}$$
Which statements about this reaction are correct?

Aluminium is oxidised.
Chlorine is reduced.
Nitrogen is oxidised.

I understand why aluminium is oxidized, but why are 2 and 3 wrong?

They aren't. Isn't there an "all of the above" answer?

No there isn't.

Well the question as written does not require just one of them to be correct

Could it be that the question was translated and originally authors were asking about chemical substances, not elements? I'm not sure though which language would allow for such distinction.

Notice there is "which statements ...are", not "which statement .. is". The question implies more than one possibility.

@Ian Bush alright

@andselisk this is straight from Cambridge so I don't think so

@Poutnik I never noticed. This would actually make more sense. Thanks for clarifying

The English is fine - but for a question I think it could be phrased better

@Ian Bush Some questions like to play with the proper reading, what is sometimes the key to the proper answer. It trains people not to miss details.

literally everything in the a-level could be phrased better - they just don't want many of us to understand

@Jana Fair enough, I also agree that choosing all three options would be correct.

@Poutnik you're very correct in this

Since you have multiple redox reactions going on, balancing the equation by the usual redox methods is a bit tricky.  See this answer for one way to do it.

Here,

The oxidation number of chlorine in the perchlorate radical is +7 whereas, in the product side, chlorine has an oxidation number of  minus 1 in AlCl3. As the oxidation number is reduced by 8 here (gaining 8 electrons), chlorine is, indeed, reduced.

Nitrogen has an oxidation state of minus 3 in NH4ClO4 whereas, it has an oxidation state of 0 in N2 molecule in the product side. Here, the oxidation number of nitrogen has increased by 3 via losing 3 electrons. So, Nitrogen here is oxidised.

Hence, all the options are correct.

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text:
Common Ion Effect - Ionic Equilibrium
Question

In which of the aqueous solutions of the following, dissociation of $\ce{NH4OH}$ will be minimum?
A) $\ce{NaOH}$
B) $\ce{H2O}$
C) $\ce{NH4Cl}$
D) $\ce{NaCl}$

My Thoughts
My book says that the answer is option C '$\ce{NH4Cl}$' giving the reason as common ion effect.
But I think that option A '$\ce{NaOH}$' also has a common ion as $\ce{OH-}$.
What should be the right answer and how do we compare which will cause more suppression by common ion effect?
I get why $\ce{H2O}$ will not suppress dissociation of $\ce{NH4OH}$ that much since its equilibrium constant is very low (of the order of $10^{-14}$). But then why not $\ce{NaOH}$ (which is a very strong base and thus will almost completely dissociate into constituent ions)? Which one out of $\ce{NaOH}$ and $\ce{NH4Cl}$ is a more stronger electrolyte?
Is it because $\ce{NH4Cl}$ will form a buffer with $\ce{NH4OH}$?
I am feeling really confused about this. Any help will be highly appreciated!
Final Question
I now understand that both $\ce{NaOH}$ and $\ce{NH4Cl}$ will cause decrease in dissociation of $\ce{NH4OH}$. Hence my final question is this: Which will cause more decrease in dissociation and why?

Existance of NH4OH in water is a myth. There is no dissociation like $\ce{NH4OH(aq) <=> NH4+(aq) + OH-(aq)}$ but the equilibrium $\ce{NH3(aq) + H2O <=> NH4+(aq) + OH-(aq)}$, or the true dissociation of the conjugated acid:  $\ce{NH4+(aq) + H2O <=> NH3(aq) + H3O+(aq)}$

@Poutnik True, but I don't think that's relevant here.

How can someone dissociate NH4OH in non-aqeous NaCl, NH4Cl or NaOH? Are you trying to dissociate it in their liquid phases?

@Ivan Neretin Well, partly true, as it is relavant to discourage its use at any occurance.

@Habib First you would need to have NH4OH.

@Habib There is stated "in the aqueous solutions of the following"

Oops, missed it accidentally. Thanks.

Why did you remove the formatting? How does it affect your question in a negative manner?

@Safdar Sorry about removing the formatting. Actually, I accidentally deleted this post and then posted the question again here... https://chemistry.stackexchange.com/q/138884/97862... But then I found out we can also undelete a deleted post.

Let's consider an aqueous solution, the concentration of which is $\pu{1 M}$ in $\ce{NH3}$ and $\pu{1 M}$ in $\ce{NaOH}$. Thus, following equilibrium would be taken place:
$$\ce{NH3 (aq) + H2O <=> NH4+ (aq) + OH- (aq)}\tag1$$
$$\ce{NaOH (aq) -> Na+ (aq) + OH- (aq)}\tag2$$
The $\mathrm{p}K_\mathrm{b}$ of equilibrium $(1)$ is 4.75, thus $K_\mathrm{b} = \pu{1.78E{-5}}$. In pure ammonia solution, from equation $(1)$:
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}\tag3$$
If ionized amount at equilibrium is $\alpha$, then
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{\alpha \times \alpha}{1-\alpha} = \alpha^2 \ \Rightarrow \ \therefore \ \alpha = \sqrt{\pu{1.78E{-5}}} = \pu{4.22E{-3}}$$
Assumptions: $\alpha \lt\lt 1$, and thus $1-\alpha \approx 1$, and $\alpha  \gt \gt \pu{1.00E{-7}}$ and autoionization of water can be ignored. At the end, since $[\ce{NH4+}] = [\ce{OH-}] = \alpha = \pu{4.22E{-3}}$, both of these assoumtions are correct.
Now consider, if you have $\ce{NaOH}$ in your solution. Since it is a strong base, it completely dissociate according to equation $(2)$. Thus, there is a common ion in this solution: $[\ce{OH-}] = \pu{1 M}$. Hence, from the equation $(3)$:
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{\beta \times (1+\beta)}{1-\beta} = \beta \ \Rightarrow \ \therefore \ \beta = \pu{1.78E{-5}} $$
Hence ($\alpha \gt \beta$), the ionization amount of $\ce{NH3}$ in presence of the common ion $\ce{OH-}$ is less than that of the solution when no common ions are present.
In similar way, you can prove the ionization amount of $\ce{NH3}$ is larger in the presence of the common ion $\ce{NH4+}$ Using following equilibria:
$$\ce{NH3 (aq) + H2O <=> NH4+ (aq) + OH- (aq)}\tag1$$
$$\ce{NH4Cl (aq) -> NH4+ (aq) + Cl- (aq)}\tag4$$
$$\ce{NH4+ (aq) + H2O <=> H3O+ (aq) + NH3 (aq)}\tag5$$

Thanks for spending time on answering my question. I get that addition of NaOH will cause the dissociation of NH4OH to decrease due to common ion 'OH-'. But why did you say that addition of NH4+ ions from NH4Cl will cause ionisation amount of NH3 to increase?

Wouldn't addition of NH4+ ions cause decrease of OH- ions and hence reduce the ionisation amount?

Final Question
I now understand that both NaOH and NH4Cl will cause decrease in dissociation of NH4OH. Hence my final question is this: Which will cause more decrease in dissociation and why???

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text:
How to neutralize the stink of decomposed fish?
Recently I discovered for myself an interesting way to easily produce home made plant fertilizer that is quite effective (tested myself on the plants). It's fish emulsion (so can be used also in a hydroponic system) - 1 part fish wastes(skin/bowels) to 2 parts water placed in an airtight container in a sunny spot for ca. 1-2 weeks. The problem - it stinks unbearably. Both - the gas that is produced during the decomposition and the liquid(/fats).

Is it possible to neutralize the stench somehow without loosing the fertilizing qualities of the emulsion? Especially of phosphorus.
1.1 Ideally of both - the gas and the liquid. But at least of the liquid - as that is what is used actually and then continues to spread the smell after being added to the plants.
1.2 Ideally it should be done using some usual household substances/wastes - maybe adding of lemon/orange peels along with the fish wastes? Or maybe ash/charcoals?

Ideally the resulting emulsion should be acidic as many plants like the pH range of 5 - 6.5 .

You can try to boil acetic acid/vinegar in the space that you have been working. Mixing it with the products will probably be harmful to your plants.

The recipe looks mighty like garum. Wonder how the Romans countered the smell, or maybe they did not mind it at all.

Plants like fully decomposed (composted) organic materials just like human prefer cooked food. The solution which you are preparing, is going to hurt the plants very badly. For example, cow dung is a good organic fertilizer but fresh manure will burn the plants. Why don't you bury this fish waste in ground, mix it with old leaves and let it compost? No smell because soil bacteria will take care of it.

The difference between fish hydrolysate and synthetic NPK is the world of plant beneficial microbes you are culturing and adding to the soil. Have those microbes work for you to reduce odour during the ferment.
Make a lactic acid bacteria (LAB) serum and add this along with molasses and biochar to the brew. The bacteria and molasses will kickstart fermentation to enzymatically break down the fish and it will inhibit the smell. You could also add commercially available plant probiotics and/or mychorrizal fungi to the brew to bolster the microbial diversity.
Biochar is an effective way to reduce anaerobic odours such as hydrogen sulphide and methane. Put a lid on the bucket and install a water airlock so air can escape in a very slow and controlled way.
The fertiliser could take 12 months or more to ferment so that the fish product has completely liquified. By the end of fermentation, the smell should be mild. When you apply the fertiliser you need to dilute it to at least 10 parts water, further reducing any odour.
Because it will be heavily diluted, the pH shouldn’t matter too much unless it is extreme. If you need to correct the pH to become more neutral, add ascorbic acid to lower the pH, or dolomite lime to increase it, but only enough to neutralise the pH and no more. Do not add vinegar or lemon juice during brewing as it will kill microbes. Be sure to keep the product out of the sun as UV will also kill beneficial microbes.

thank you very much! (1) How do I get lactic acid bacteria (LAB) serum at home? (2) What molasses would you recommend? (3) What should be the approx. proportions of water, fish wastes, LAB serum, molasses and biochar in that mixture?

Plants like fully decomposed (composted) organic materials just like humans prefer cooked food. The solution which you are preparing, is going to hurt the plants rather badly. For example, cow dung is a good organic fertilizer but fresh manure will burn the plants. Why don't you bury this fish waste in the ground or soil, mix it with old leaves and let it compost for 1-2 months? No smell should arise because soil bacteria and perhaps worms/insects will take care of it. Above all everything will be well covered with soil.
Do not add vinegar or lemon juice to your liquid (which should go to waste anyway). Vinegar has salt and plants don't grow well with salty water.

(1) I use it in a hydroponic system, i.e. in water solution without ground. (2) it decomposes ca. 2 weeks on the sun. (3) I have tested it and results were quite amazing - just over night melons started produce small fruits... (4) I actually do add acidic stuff/vinegar to the hydroponic solution, otherwise the pH is too high - ca. 8.

Okay, good to know that your hydroponic system is working. You may add this part in the main question.

Actually, I discovered a way of eliminating fish odor that was apparently generated from an old can of salmon, which was in the process of bursting and leaking. The can was hidden on a lower shelf in my kitchen.
I originally believed there was an issue with something on the floor and kept on treating the floor until I produced results.
The exact process I employed, however, remains the subject of a future patent.
But, I will disclose that the apparent active agent is Singlet Oxygen (or, O(1D)). The latter could interact with oxygen in air further creating reactive oxygen species (see, for example, Equation (4) presented here). Apparently, O(3P) may be formed from severe collision quenching of O(1D) atom with oxygen and acts as the major oxidant, for example, in the work by Huan Yue and colleagues ‘Exploring the working mechanism of graphene patterning by magnetic-assisted UV ozonation’. Note, O(3P) is also known as highly reactive ground-state 3P oxygen and a form of atomic oxygen.
Of practical importance, singlet oxygen has a half-life of around 45 minutes and should be considered toxic, so treated areas should be evacuated prior to treatment. On contact with water, it forms hydrogen peroxide.
Note, unlike other de-odorizing agents that disguise smells, this path actually results in a chemical breakdown with no toxic or carcinogenic residual products.
Also, an associated 2008 patent: "Use of photosensitizer for cleaning air, water or contaminated surfaces and in filters, where the photosensitizer is covalently linked and/or linked over physical interaction to surface and activated by visible light to form singlet oxygen" discussed here.


text:
Mechanism for the addition of hydrogen iodide to 3,3‐dimethylbut‐1‐yne
The following reaction mechanism was given as a solution to a solved problem in my textbook1 for the addition of hydrogen iodide to 3,3‐dimethylbut‐1‐yne:

It can be seen that that 2,2-diiodo-3,3-dimethylbutane (a geminal dihalide) is the product according to this mechanism. I arrived at a different product 2,3-diiodo-2,3-dimethylbutane (a vicinal dihalide) by using the following mechanism:

The difference is due to the methanide shift I did on the second step, converting 3,3-dimethylbut-1-en-2-ylium cation to 2,3‐dimethylbut‐3‐en‐2‐ylium cation. I consider this to be a reasonable carbocation rearrangement because after the methanide shift, the 2,3‐dimethylbut‐3‐en‐2‐ylium cation is tertiary as well as in conjugation with the double bond. Moreover, the initial carbocation has a positive charge on a more electronegative $\mathrm{sp^2}$ hybridized carbon atom. Due to this, I believe, the methanide shift produces a more stable carbocation. But, why did the author proceed without making this rearrangement?
When I discussed this in chat, there was a consensus that the product of this reaction must be a geminal dihalide as obtained by the author. The only way, I could think of, by which I can obtain the geminal dihalide even after the rearrangement discussed earlier is to do another rearrangement during the addition of second molar equivalent of hydrogen iodide as given below:

The problem with this mechanism is the carbon bearing the postitive charge after the methanide shift also has an iodine atom attached to it. Earlier, I learnt that chlorine atom is the only halogen for which the positive mesomeric effect is stronger than the negative inductive effect thereby stabilizing the positive charge. But here, due to the presence of iodine, I think the carbocation is not stable after rearrangement.
Even after neglecting this fact, there seems to be a major difference betweeen the author's mechanism and the modified mechanism, which I've emphasized using a carbon-12 labelled reactant as given below:

It can be seen that even though we obtain geminal dihalides through either of the mechanism, the products formed aren't exactly the same. One has the iodine attoms attached to the normal carbon whereas in the other they are attached to the carbon-12 atom.
In short, what happens when hydrogen iodide is added to 3,3‐dimethylbut‐1‐yne?
Reference

Solomons, et al. Organic Chemistry for JEE (Main & Advanced). Edited by MS Chouhan, Third Edition; Wiley India Private Limited. ISBN 978-81-265-6065-3

There seems to be an issue with both mechanisms on the front that an SN1 reaction would not take place since the carbocation formed is a vinylic carbocation which is highly unstable.
The actual reaction follows a termolecular mechanism where the rate of reaction is given to be:
$$\text{Rate}=[\ce{HX}]^2[\text{alkyne}]$$
Now, according to Advanced Organic Chemistry by Francis A. Carey, the reaction mechanism would be as follows:$^1$

Step $1$:
A concerted termolecular reaction...
This involves an acid/base reaction, protonation of the alkyne developing positive charge on the more substituted carbon. The π electrons act pairs as a Lewis base.
The other part is attack of the nucleophilic bromide ion on the more electrophilic carbocation creates the alkenyl bromide.

Step $2$:
In the presence of excess reagent, a second protonation occurs to generate the more stable carbocation.$^2$

Step $3$:
Attack of the nucleophilic bromide ion on the electrophilic carbocation creates the geminal dibromide.

[$1$]: The reaction mechanism stated above uses $\ce{HBr}$ and not $\ce{HI}$ instead. However this reaction takes place for $\ce{HX}$.
[$2$]: The same reaction using $\ce{HI}$ would have a comparatively lower yield of the geminal product since geminal diiodides are unstable due to the steric hinderance posed by the large size of the iodine atoms.


text:
What is an alternate way of preparing a sodium hydrogen phosphate/sodium dihydrogen phosphate buffer through titrating an acid with a strong base?
So my friend and I are stumped. This is in the context of achieving the sodium hydrogen phosphate/sodium
dihydrogen phosphate buffer by titrating the acid with a strong base.
My friend says the answer would be H3PO4 + Na3PO4. In my case, I'd think that sodium hydrogen phosphate would be the acid (but I genuinely am not sure) that we'd titrate in that scenario and all we'd need then is a strong base (although I honestly don't know how to figure out what strong base in particular you must use.)
What's correct? Or what would be the correct way?

Start with writing the 3 equilibrium chemical reactions between 4 forms of phosphate. You will see what is an acid and when.

Any process treating any mixture of $\ce{Na_{n}H_{3-n}PO4}$ ( n=0..3) by any of $\ce{Na_{n}H_{3-n}PO4}$ or $\ce{NaOH}$ to produce solution of $\ce{NaH2PO4 + Na2HPO4}$ with the desired $\mathrm{pH}$ or the molar ratio/concentration is a correct way. But most handy may be titration of $\ce{H3PO4}$ or $\ce{NaH2PO4}$ by $\ce{NaOH}$.

Both methods are correct. You may mix equal amounts of $\ce{H3PO4}$ and $\ce{Na3PO4}$. Or you may mix n moles of $\ce{NaH2PO4}$ and n / $2$ moles of NaOH.

What is your target ph?

Also you have to consider ionic strength of your buffer in addition to pH.

This picture says more than a thousand words:

You want a mixture of $\ce{HPO4^2-}$ and $\ce{H2PO4^-}$. You can start with any species (in a solution with a certain pH) and add another species whose pH is "on the other side" of the desired pH. If the pH of your starting solution is lower than the desired pH, you can also titrate with NaOH.
What you actually do (of the six options) is determined by solubility of the solution and by availability. If you titrate two phosphate solutions of the same concentration, there will be no dilution effect. If you titrate with NaOH, you would used fairly concentrated NaOH to avoid dilution, but would have to add water to a defined volume (or measure the volume after reaching your desired pH) to figure out the concentration of your buffer.
You should not titrate with HCl because then you end up with extra sodium choride in your buffer.


text:
Why is copper(II) used a catalyst here instead of as an intermediate?

Photochromic glass, used for sunglasses darkens when exposed to bright light and becomes more transparent again when the light is less bright. The depth of color of the glass is related to the concentration of silver atoms.
The following reactions are involved:
reaction $1$: $\ce{Ag+ + Cl- <=> Ag + Cl}$
reaction $2$: $\ce{Cu+ + Cl → Cu^2+ + Cl-}$
reaction $3$: $\ce{Cu^2+ + Ag → Cu+ + Ag+}$
Which statement about these reactions is correct?
A) $\ce{Cu+}$ and $\ce{Cu^2+}$ ions act as catalysts.
B) $\ce{Cu+}$ ions act as an oxidising agent in reaction 2.
C) Reaction $2$ is the one in which light is absorbed.
D) $\ce{Ag+}$ ions are oxidised in reaction $1$.

I understand why B, C, D are incorrect. I also understand why $\ce{Cu+}$ is a catalyst; however, I do not understand why $\ce{Cu^2+}$ is a catalyst and not an intermediate. It appears as a product first and then as a reactant later, unlike $\ce{Cu+}$, so shouldn't it be an intermediate instead of a catalyst?
The question is question $11$ in the October November $2009$ A-level chemistry Cambridge variant $11$ paper.

Rather Cu is the catalyst, switching between 2 oxidation states, catalysing the clearing up the glass in low light.


text:
Acidity of thiamine - how comes hydroxide anion with pK about 15 ionizes thiamine with pK about 18?
Textbooks say removing proton from thiamine's C atom will produce a resonance stabilized carbanion which is hence bioactive. That's OK, but how comes that 1) we abstract the acidic proton with the OH base whose pK is lower that for thiamine, and 2) the remaining anion acts - in bio context - in aqua medium (?), hence will immediately deprotonate water and lose its negative charge.

Actually this reactive intermediate is a persistent carbene.

In the ionized structure shown, which looks like an ylid, the carbanion is formed on an $\mathrm{sp^2}$ orbital and no resonance stabilization of this charge is possible.
Regarding your question about the ionization, remember that biochemical reactions don't always happen in water, they may happen in the active site of enzymes, which have a completely different environment: the dielectric constant is different, the $\mathrm{p}K_\mathrm{a}$ of substances change and the proximity of certain groups stabilize reactive intermediates that would not be formed in water.

Your second point is good, but first is incorrect - formal charge on carbon is only in mesomeric structure with dipolar bond, there's another one - carbenic.

I see that a carbene can be formed, but he specifically said "will produce a resonance stabilized carbanion", which does not seem to be the case

For a reaction to happen, it is sufficient if 1 in a 1000 thiamine structures are deprotonated. For a species with a $\mathrm{p}K_\mathrm{a}$ of 18 to be 50% deprotonated, you would need an aqueous solution with a pH of 18, which is not possible as far as I know.
So even if thiamine is present in aqueous solution and not (as mentioned in Rafael L's answer) in the active site of an enzyme, it still could be sufficiently deprotonated to show reactions explained by the deprotonated intermediate.


text:
Does magnesium sulfate in aqueous solution form magnesium (hydrogen-)carbonate when hydrogen carbonat is present?
I'm using an additive of magnesium sulfate to complement my fertilizer. The additive lists the following ingredients:

I know that it does NOT contain magnesium oxide, because due to historical / commercial reasons ingredients of phosphorus, potassium or magnesium are indicated as mass percentages of the oxides with the element in question. Also the merchant confirmed that the source of magnesium is magnesium sulfate.
I noticed the precipitation of a white powder at the bottom of my reservoir. I suspect it to be magnesium carbonate. The  mains water contains the following chemicals. The dose of magnesium sulfate is within the range recommended. The precipitation occurs even if I do NOT use any fertilizer, so I think It is a reaction with chemicals in mains water supply.

Is the precipitate magnesium carbonate?
How can I prevent this precipitation? What are factors influencing it besides the availability of hydrogen carbonate?
EDIT:
User AJKOER suggested that the precipitate is calcium sulfate. I don't think that is the case because:

The solubility of calcium sulfate is given at 2.1g / liter @ 20° C  (https://en.wikipedia.org/wiki/Calcium_sulfate). This equates to about 2.1 * 0.71 ~ 1.48g sulfates per liter. (mass of SO4 /  mass of CaSO4 = 0.71)
I'm adding only about 92.1 mg / liter sulfates from magnesium sulfates
The mains water contains only 13.3mg / liter sulfates
92.1 mg / liter + 13.3mg / liter <<<< 1480 mg / liter

Can downvoters elaborate how to improve the question

There are plenty of unusual things here going on, it is more likely that we are on the wrong path due to insufficient data on the mentioned product. First the pH looks suspicious, since Mg is not that basic compared to alkali metal bases, so the pH should be somewhere below 7. I guess there is more Mg than sulfuric acid stoichiometrically speaking, so that is  probably reacting with the (bi)carbonates in the water or atmospheric carbon dioxide.

@AndrewKovács But you would generally agree that a reaction with carbonates is more likely than a reaction with sulfates? Because If I add magnesium sulfate and in consequence reduce the calcium levels this would be an exercise in futility because I'm already low on calcium..

In this case of scenario since the pH is high YES it is very possible.

I will run an experiment where I add only potassium sulfate to the same tap water. If no precipitation occurs I can conclude that the precipitation which occurs after adding magnesium sulfate is indeed magnesium carbonate..Or some other form including magnesium

You might try some phosphoric acid to remove precipitates, this way you wont lose any calcium and it is actually a good addition to the fertilizer.

I'm already using phosphoric acid to set my PH to 6. Precipitation still occurs.

The white precipitate is more likely insoluble white calcium sulfate.
Per a reference on the making of fertilizer, to quote:

Calcium, magnesium, and sulfur are also important materials in plant growth. They are only included in fertilizers in small amounts...

Net ionic reaction:
$\ce{ Ca^2+ (aq) + SO4^2- (aq) -> CaSO4 (s) }$

The origin of calcium in your net ionic reaction is not yet obvious because it is not explicitly mentioned in the OP's question.  I speculate it is by your reasonable reasoning that tab water contains a varying concentration of $\ce{Ca^2+}$ ... The trigger to edit your answer however was the notation of $\ce{Ca(2+)}$ which I brought into a more mhchem-like syntax with the circumflex sign ^. Maybe this collection of examples in ChemSE meta is interesting for you: https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here?noredirect=1&lq=1

I don't think the precipitate is calcium sulfate because the calcium sulfate levels are too low. See the edit to my question.Am I missing something?

AJKOER suggested that the precipitate is calcium sulfate.
However this seems unlikely due to the following reasons:

The solubility of calcium sulfate is given at 2.1g / liter @ 20° C (https://en.wikipedia.org/wiki/Calcium_sulfate). This equates to about 2.1 * 0.71 ~ 1.48g sulfates per liter. (mass of SO4 / mass of CaSO4 = 0.71)

I'm adding only about 92.1 mg / liter sulfates from magnesium sulfates

The mains water contains only 13.3mg / liter sulfates

92.1 mg / liter + 13.3mg / liter <<<< 1480 mg / liter

It seems more likely that the precipitate is indeed magnesium carbonate.

The water contains 181mg / liter hydrogencarbonate, which is the highest concentration of a molecule in the mains water.
The solubility of magensium carbonate is much lower than the solubility of calcium sulfate. (0.14g / liter (MgCO3) vs. 2.1g / liter (CaSO4))
I'm adding about 26.9mg / liter Mg from MgSO4
The mains water contains 7.5mg / liter Mg
The solubility of magnesium carbonate is 0.14g / liter. This equates to 0.14 * 0.29g = 0.04g / liter magnesium.
26.9 mg / liter + 7.5mg / liter = 34.4mg / liter mg
34.4mg / liter Mg < 40mg / liter (solubility of magnesium carbonate with respect to magnesium)
There is 181mg / liter hydrogencarbonat which equates to about 181 * 0.93 = 178 mg / liter carbonate (CO3 2-)
Speculation: The high availability of hydrogencarbonate facilitates precipitation of magnesium carbonate.

Note: Not a chemist here so maybe I made some gross mistake in my reasoning, please point it out!


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Non-foaming surfactant to add to water
I'm looking for a surfactant that I can add to water to decrease the surface tension by 20 to 40%.  I'm working on a new kind of mixing device that is so designed such that it mixes relatively large volumes of two liquids and cannot be made small enough for laboratory scale.  So flow on the order of 1.5 to 2.5 liters/min.  I would like to be able to run my experiments for a number of minutes, so the estimate is 10 gallons for each hypothetical liquid (total amount would be 20 gallons for an experiment that would give me about 20 minutes to adjust parameters, take photographs, etc.).  Adding something like isopropyl alcohol can get expensive very fast (10% by weight in water lowers the surface tension to about 41 where pure water is 72).
Very important:  The surfactant needs to be non-foaming as release of gas will be problematic for these tests.
Thanks for any thoughts.

Conceptual thoughts about mixing two immiscible liquids (not on surface tension):

A reflux condenser used in the ochem lab equally has a circulation of about 2 L/min water (reference from an advertisement, which doesn't look too much off-range)

After the two liquids are mixed, what application will follow?  Do you target cleaning an engine, or deployment for food processing?  What are intended storage conditions once the two are mixed (e.g., time of storage / transport, temperature of storage / use)?  Altogether with the other liquid you want to mix with water, this may narrow your choice of the surfactant.

Mixing the two continuously may be achieved with a tube reactor with two inlets, one outlet, and a screw-shaped inner geometry

(reference, section «Liquid-Liquid & Liquid-Gas Mixers»)
Depending on parameters of your liquids (e.g., density, viscosity, flow rate, key word Reynolds number) you may enter the regime of turbulent flow until you divide the liquid into tiny droplets which increases the surface accessible for mixing.  The addition of surfactant may help to achieve the intended mixing, too.

(loc. cit.)

The formation of foam however is the result of an entrapment of a gas.  Whipping cream with a whisk for example intentionally aerates the cream.  Liquids like water may boil (i.e., generation of gas) because of sudden pressure changes while passing a propeller blade at a flow rate too high (cavitation), too.
Thus, mixing water and your other liquid in a tube reactor like the one mentioned above, in absence of air and at the right hydrodynamic state may prevent the formation of foam.  (You mau fight foam by ultrasound (example), though.)

Thanks, but I was hoping for a recommendation for a surfactant.

The question reads «it mixes relatively large volumes of two liquids», I understood the second liquid to be neither water, nor isopropanol.  The later because isopropanol and water mix without a miscibility gap. Thus «liquid #2» could be a hydrocarbon, ether, ester, ... Thus no specific, experiment based, recommendation.

Thanks, the purpose of the experiment is to test the effect of surface tension on the production of turbulence because of the unique geometry employed.  The liquids are 100% miscible in one another.  I just want to take some measurements at different surface tensions; not use a surfactant to disperse one liquid in another.


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How to determine the number of electron in a shell
My textbook has been mentioned that the maximum number of electron in a shell is 2n² and the octet rule. It has also said that period number signifies the number of electron shells of an element and group number signifies number of valence electrons in a shell. When I tried to distribute electrons on basis of above mentioned laws,it was not true for all elements. For eg. Rubidium-37 the electron is distributed as 2,8,18,8,1 here, there are 5 shells and valence electron 1 which satisfies the statement about period and group number but when I tried to do the same with Caesium-55 this is what it's electrons distribution look like 2,8,18,8,8,8,3 which neglects the statement given in my book.
Please tell what has gone wrong here. Please explain in simple words as I am a high school student.

Does your textbook mention Aufbau's rule?

Caesium is in the sixth period. How many shells does it have?

@Peter Do you know about s, p, d and f orbitals and blocks. Because if you do it will be easier to explain.

@Habib "Aufbau" is not a person. It is a term derived from the German word for building up. So it is Aufbau rule rather than Aufbau's rule.

I'll try to keep it simple. The actual details of why this is so has to do with quantum physics.
Each shell has several "subshells". Each "subshell" in turn holds a certain number of "orbitals". Each orbital can hold up two electrons. Rules of thumb (again, I'm not explaining why because it's probably way out of your understanding)

The $\mathrm {n^{th}}$ shell has n subshells, labelled from $0$ to $\mathrm n-1$. E.g., the $2$nd shell has two subshells, $0$ and $1$. For reference, we often call these subshells by letters, $0$ is $\mathrm s$, $1$ is p, $2$ is d and $3$ is f. The periodic table can be broken up into four blocks depending on which of these subshells the most loosely bound (valence) electron in the atom is in. (s-block, p-block, etc)

The $\mathrm{k^{th}}$ subshell can hold $2\mathrm k+1$ "orbitals". An orbital can hold up to two electrons. So, the $0$th (s) subshell can hold $1$ orbital, and thus two electrons. The $1$st subshell (p) can hold $3$ orbitals, or $6$ electrons. This is why we have the $2\mathrm n^2$ rule. There are as many orbitals in the $\mathrm n$ shell as the sum of the first mathrm n odd numbers: $\mathrm n^2$, and two electrons in each ($2\mathrm n^2$).

An important idea in physical chemistry is the Aufbau principle. Orbitals are filled according to the increasing order of their (orbitals') energies. What are the orbitals' energies? It's quite simple for the first few subshells:

$1\mathrm s, 2\mathrm s, 2\mathrm p, 3\mathrm s, 3\mathrm p$ (order)
In a hydrogen atom, these orbitals are straightforward: all subshells in the same shell have identical energies ($1 \mathrm s, 2\mathrm s=2\mathrm p, 3\mathrm s=3\mathrm p=3\mathrm d,$ etc). In other atoms, though, everything gets tangled up. Here is the order, something you will need to memorise:
$1\mathrm s, 2\mathrm s, 2\mathrm p, 3\mathrm s, 3\mathrm p, 4\mathrm s, 3\mathrm d, 4\mathrm p, 5\mathrm s, 4\mathrm d, 5\mathrm p, 6\mathrm s, 4\mathrm f, 5\mathrm d, 6\mathrm p, 7\mathrm s, [...]$ (this will do for most elements, but there are some exceptions, which you will have to memorise as well)
So the electrons don't actually fill the entire $2\mathrm n^2$ in a shell before going to the next. E.g., in iron, we have the configuration $1\mathrm s^2 2\mathrm s^2 2\mathrm p^6 3\mathrm s^2 3\mathrm p^6 4\mathrm s^2 3\mathrm d^6$. Note that the $1$ shell has $2$ electrons; the $2^{\mathrm {nd}}$ shell, $8 (2\mathrm s+2\mathrm p$); the $3^{\mathrm{rd}}$ shell, $14 (3\mathrm s+3\mathrm p+3\mathrm d)$; the $4^{\mathrm {th}}$ shell, $2$. Your textbook might call this configuration (2, 8, 14, 2)$.
Caesium has 55 electrons, they fill up like so:
$1\mathrm s^2 2\mathrm s^2 2\mathrm p^6 3\mathrm s^2 3\mathrm p^6 4\mathrm s^2 3\mathrm d^{10} 4\mathrm p^6 5\mathrm s^2 4\mathrm d^{10} 5\mathrm p^6 6\mathrm s^1$.
Why don't you add the number of electrons in each shell up and see if it matches what your textbook says.
P.S. The octet rule doesn't actually say the highest shell should have eight electrons. It says that the atom achieves stability by gaining the electronic configuration of the nearest noble gas. Noble gases have the general configuration $\mathrm s^2 \mathrm p^6$ ($8$ electrons?), but as demonstrated above, these $8$ electrons may not be in the highest shell. For example, in iron, two of the valence electrons are in the fourth shell and the other six in the third shell.


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Distribution of negative charges in iodine dioxide difluoride anion
I was drawing the structure for $\ce{[IO2F2]-}.$ I'm stuck on whether the negative charge should be given to oxygen or iodine.

Have you tried both? What looks more feasible to you?

The compound for which the V.S.E.P.R. diagram is asked for is $\ce{[IO2F2]-}$
As OP has stated, there are two possibilities for the negative charge.

One, we add the negative charge to an oxygen atom. In such a scenario, we would have three different types of connected groups: $\ce{O-}$, $\ce{O}$, $\ce{F}$.
Second, we give the negative charge to the iodine (central atom), In such a scenario, we would have two types of connected groups: $\ce{O}$, $\ce{F}$ . We also have $8$ electrons in the central atom whereas in the first case, we have seven.

Oxygen gets the electron
Here, the electrons of iodine would be distributed in the following way to complete the octet of the connected groups: $1$ electron to $\ce{O-}$, $2$ electrons($1$ for a sigma bond and one for the pi bond) to $\ce{O}$ and $1$ electron to each fluorine. This leaves the iodine with one lone pair.
Therefore, according to VSEPR, we have $4$ sigma bonds, $1$ lone pair. Which is equivalent to a see-saw shape with the fluorines at the axial position.
Iodine gets the electron
Here, the electrons of iodine would be distributed in the following way to complete the octet of the connected groups: $2$ electrons each(one for a sigma bond and one for the pi bond) to both oxygen atoms and $1$ electron to each fluorine. This leaves the iodine with one lone pair.
Therefore, according to VSEPR, we have $4$ sigma bonds, $1$ lone pair. Which is equivalent to a see-saw shape with the fluorines at the axial position.

As you can see both cases have identical geometries and no change in positions of atoms. This is because both of them are resonance structures of each other. Both are correct.

I got the core answer but u said that the structures are resonance structures of each other plz confirm that its bcuz the oxygen gives a coordinate bond to the d subshell in case if oxygen gets the electron


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Heats of combustion of pentadiene isomers
This is a problem from the book Organic Chemistry by Carey and Giuliano (10th edition), the problem is:

Match the heats of combustion with the compounds:
The compounds are 1,3-pentadiene, 1,2-pentadiene and 1,4-pentadiene and the heats of combustion are given as 761.6 kJ/mol, 768.9 kJ/mol and 777.1 kJ/mol.

I know from my knowledge of organic chemistry that 1,3-pentadiene is the most stable of the three compounds (since its π bonds are in conjugation), therefore its heat of combustion corresponds to the minimum of the given heats of combustion, ie. 761.6 kJ/mol, however I am confused about how will I compare heats of combustion of 1,2-pentadiene and 1,4-pentadiene.

(1) Please don't use $\text{...}$ for no reason; you really don't need to use, and shouldn't be using, MathJax unless you're writing equations. (2) Chemical names are not capitalised unless at the beginning of a sentence.

I will take care of what you said the next time when I use chemistry SE

Your question isn't exactly the same as https://chemistry.stackexchange.com/q/110857/16683 (yours is better - that one is poorly formulated), but the answers there address your exact case.


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Finding the reduced pressure to distill a solution containing 0.4 g of naphthalene in 25 g of chloroform at 50 °C

Chloroform boils at $\pu{62 ^\circ C}$. For a solution containing $\pu{0.4 g}$ of naphthalene in $\pu{25 g}$ of chloroform, the boiling point is elevated by $\pu{0.45 K}$, to avoid any decomposition during distillation chloroform has to be distilled at $\pu{50 ^\circ C}$, what maximum pressure can be maintained in the flask?

My approach:
I found the molality of the solution to be $0.125m$, equating it with the formula $$\Delta T_\mathrm b=K_\mathrm b\cdot m$$ I found $K_\mathrm b$ to be $3.6\ \mathrm{K\ kg/mol}$, I am not able to proceed after this step, please provide an approach to solve the problem from here.
I also tried using the formula $$\Delta K_\mathrm b=\dfrac{R {T_\mathrm b}^2}{1000l_\mathrm v}$$ but finding out the latent heat of vaporisation did not prove to be of any help.

Apply the Clausius-Clapeyron integral equation, as you know vapour tension at 2 temperatures(on indirectly ). Then you can calculate the tension at the 3rd temperature.

Can you provide your approach in detail in the answer, I am not able to apply the Clausius-Clapeyron equeation

I can,but I will not. Try harder. If I consider the time delay of your comment and time to write and post it, you could hardly have a good look at it. Consider also the Raoult law to get the 2nd vapour tension.

@Poutnik Is your answer 509.6 torr, I am asking this just to be sure?

Particular answer is of no importance, the principles and procedures are what matters. Otherwise it qualifies well for the closure as a homework-like question.

@Poutnik you can close my question if you want to, but please let me know at least that my work on this question is correct or not, I have solved it using Clausius-Clapeyron equation and obtained the answer, I have to know to confirm that I understand the equation well

Present your work first, putting it to the question. If you are not sure,comment why you think it may be wrong. Be aware SE chemistry site is not a homework solver nor homework proof reading service. It helps to understand principles and to solve non trivial tasks.

Technical note: Question titles should be like book titles. They distinguish books from each other, but give just a hint about the content, not being as much particular.

Let us continue this discussion in chat.

Identify sources where the Clausius-Clapeyron equation is applied and work these examples multiple times (e.g., https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Phase_Transitions/Clausius-Clapeyron_Equation), (yet your course notes may contain similar examples, too) until you feel good enough to explain them to someone else. Have a break. Solve them without seeing their answers. Pause. With a fresh mind, return to the problem of your question here.


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Can aluminium form blue-colored solution in liquid ammonia?
I've read that group 1 elements and group 2 elements with  exception of Be form blue-colored solutions in liquid ammonia (Mg requires an electrolytic process). What are the conditions for a metal to form a solution in liquid ammonia, and which metals (other than what I have mentioned) form blue-colored solution in liquid ammonia? Can aluminium also form such solution?

Blue is the color of the solvated electron. Only the most active metals do that.

Can Aluminium do that?

I'd say it is unlikely, but I don't know for sure.

The textbook blue solvated e- is already difficult to achieve with alkali metals... The answer might practically be yes or no. Written after seeing that is possible according to an answer below and its citation.

Consistency between question and accepted answer:  If we are allowing an electrolytic process for aluminum, then one should be allowed for magnesium and that works too.  Question edited accordingly.

The SE answers quoted in the comments provide a good broad answer but with respect to the question of aluminium – aluminium will form a blue solution in liquid ammonia following electrolytic reduction of $\ce{AlI3}$ in liquid ammonia according to J. Am. Chem. Soc. 1950, 72 (11), 5178–5180.
Europium and ytterbium have also been reported to dissolve in liquid ammonia in J. Phys. Chem. 1956, 60 (11), 1590–1591.

Yes, aluminium forms a blue colored solution in liquid ammonia. Liquid ammonia will dissolve group 1 (alkali) metals and other electropositive metals such as calcium, strontium, barium, magnesium, aluminum, europium, and ytterbium. At low concentrations (ca. $\pu{0.06 mol/L}$), deep blue solutions are formed: these contain metal cations and solvated electrons. The solvated electrons are stable in liquid ammonia and form a complex: $\ce{[e^-(NH3)6]}$.
For further details see Chemistry of the main group elements by Andrew R. Barron on openstax cnx.org.


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Where can I find a cannabinoids profile of hemp strains?
For my web article, I need to find out the content of individual cannabinoids in hemp strains, specifically for Kompolti, Carmagnola and Purple Haze. I mean the content of all the components, not just THC and CBD.
Do you know of any website that provides this information?
Thank you for answer

I’m voting to close this question because BiologySE (https://biology.stackexchange.com/) may represent a better audience than ChemSE.  The names you mention appear more like (possibly local) brand names (cf. https://industrialhempfarms.com/hemp-strains/), than systematic ones globally accepted (e.g., instead of the commoner's «ivy» the then more specific «Hedera helix L.»).  In addition, biologists (rather than chemists) were able e.g. to attenuate THC to practically zero and published this.


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Synthesis Problem?

What I tried to do is this: Opening the epoxide using $\ce{^-OCH2CH2CH3}$.
But then I have two alcohols and I get stuck. It's possible in one way but takes more than 4 steps...

One of the most helpful things to do in these types of problems is mapping the carbon skeleton of the starting material onto the product. So, I see the the propyl ether is formed from simple alkylation, and the rest of the carbon skeleton is already in place. The only issue is I need to break the terminal C-O bond of the epoxide, which we can do with a nucleophilic hydride source. That will give an alcohol that you can oxidize. I'll leave you to fill in the gaps.

Step 1 - rearrange the terminal epoxide to the methyl ketone using Pd(OAc)2, PBu3 in refluxing toluene as desribed in this JOC paper here.  It is not necessary to protect the alcohol as the paper describes the transformation of a very similar hdyroxy epoxide.
Step 2 - deprotonate the alcohol using NaH/THF at low temperature and quench with n-propyl iodide to give the required product.

Thank you so much btw :) Any simpler ways since this is an O-chem 1/2 problem.

@Waylander: May have a problem with enolate formation although pKa's are in your favor. Depends on the relative rates of alkylation.

@user55119 alternatively you could mesylate the alcohol (MsCl/Et3N) then react with the anion of 1-propanol.

@SharbelDamouni  alternatively you could react the epoxide with DABCO to do a Meinwald rearrangement to the methyl ketone - see https://pubs.acs.org/doi/10.1021/acs.joc.8b03171

@Waylander: The mesylate also has the problem of enolate formation leading to cyclopropylmethyl ketone or cyclopentanone whan the the mesylate is exposed to CH3CH2CH2O-.

@user55119 With a pKa difference of approx 3.5 units between the -OH and ketone alpha protons,  I think the first method of adding 1eq NaH portionwise to a mixture of the keto-alcohol and n-propyl iodide will give a decent yield

Perhaps so. I found a Russian paper that opens cyclopropyl methyl ketone with C1-C3 alcohols in the presence of Cu(II) and Pd(II) salts. I think all of these methods involve reactions too sophisticated for intro organic chemistry. I suspect a ketalization was expected at some point.


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How to determine the van't Hoff factor for different salts in water experimentally?

How do I experimentally determine the van't Hoff factor for different salts in water?

I'm going to dissolve each of sodium chloride, potassium chloride, and lithium chloride in $\pu{20 mL}$ of water to form $0.1$ molality standard solutions. I want to determine the actual van't Hoff factor for these ionic compounds. Theoretically, it is 2, but how do I find the actual value? I know that the formula for van't Hoff is actual number of ions/theoretical number of ions, but I don't know how to find the actual number of ions. Can someone please help?
PS: My experiment is about ionic association. I am trying to see, how it affects van't Hoff and therefore freezing point depression. I was initially going to find the depression and then use the formula to find van't Hoff, but this is too tedious.

Measuring the freezing point depression may be tedious. But it is the only way of solving this problem.

Osmotic pressure measurement might be easier @Maurice


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Adding acid to hard water to prevent limescale deposition
Acid is commonly used to remove deposited limescale. If I wanted to prevent build-up of limescale in appliances, would it make sense to add acid to the water in the first place if that doesn't harm the appliance?
As far as I understand, limescale gets produced when calcium bicarbonate turns into calcium carbonate when some of the dissolved $\ce{CO2}$ goes airborne, making the water less acidic and thereby shifting the equilibrium from bicarbonate to the less soluble carbonate.
If I now add acetic acid, according to my understanding that should keep the equilibrium on the bicarbonate side. What acid concentration do I need for different water hardnesses if this trick actually works?
I'm thinking of using it in my cleaning robot. The small valves are getting clogged very quickly. A percent of vinegar in the water shouldn't harm the tank. But prevent the clog ups.

What appliances would you be treating? Coffee maker? Hot water heater?

The amount of acid is proportional to the concentration of calcium bicarbonate. If this concentration is low, a small amount of acid will do.

@maurice Can you make it more precise? Is it equal concentration roughly? And just dependent on concentration of bicarbonate? Some actual numbers would be nice :)

One degree water hardness corresponds to around 0.3mM, so for standard hardness 10 you need 3mM acid, or 180mg acetic acid per L. At 5% weight concentration in vinegar, that's around 1mL vinegar per L of water. Does that sound right?

@ Cornelius. Your calculations seems correct.


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What is the influence of glass type on silver mirror making?
I'm considering the reaction wherein ionic silver (from the dissolved nitrate) is reduced by glucose in the presence of NaOH and forms a metalic silver layer on glass.
Reading through several topics here on stackexchange it seems that it still hasn't been clearly elucidated what exactly is happening on the glass surface for reduced silver to deposit onto / bind to it (instead of just precipitating out as a powder in solution).
I'd like to know if/how glass type might affect the process:
Can I get as perfect of a mirror on a borosilicate surface as on pure SiO2 glass?
Is the silver layer in either case uniform or will is be porous at the molecular level?
I've read that a slow reduction is key. Is it so that a silver mirror is qualitatively better produced at lower temperatures (or with highly diluted reagents) over a long time than by gently heating the solution?
Could a large temperature gradient between the solution and the glass surface be advantageous for the reaction to proceed much faster on the surface than in solution (e.g. mixing a solution at 10C into a glass container whose surface is brought to 80C)?
How can the process be optimized to produce a perfectly insulating silver layer at the molecular level?

My experience has been that all precipitated silver mirrors are slightly porous (or at least have numerous defects), are quickly damaged by exposure (e.g. to sulfides in air) and that adhesion is problematic. The process is of historical interest, and an interesting lab experiment, but for most purposes, a vacuum-evaporated, overcoated mirror is far more useful and cost efffective.

@DrMoishe_Pippik Thanks for sharing your experience. Do you know if the thickness could easily be grown to uniformity by additional exposition, or perhaps electrolysis? What steps are involved in the vacuum evaporation: Does it imply rinsing and then vacuum drying the glass surface or vacuum evaporating the entire solution? By what mechanisms does it help?

The coating is tenuous, but theoretically, could be made thicker by electroplating - though only the glass-covered side would be usable (i.e. a second surface mirror, unsuited for most optics). Vacuum vapor deposition is a thermal, or ion-mediated, process; see https://en.wikipedia.org/wiki/Silvering .

@DrMoishe Pippik: Do you know which are the mechanisms involved in the silver deposing on / adhering to the glass surface rather than just precipitating in solution? Would a thermal gradient improve it (glass heated from the dry side while the to-be-coated side is in contact with a cold solution)?

Consider that after the first monolayer on the glass, it is no longer silver adhering to glass, but silver to silver. Since this is precipitating from the solution, rather than being moderated by a force as in electroplating, it is a statistical process and irregularities would be expected to increase with the square root of thickness.

Perhaps this link will be of interest: https://chemistry.stackexchange.com/q/134278/79678.

@EdV: I've read through that discussion prior to posting my question but I didn't find an answer as to which forces are involved in the silver depositing/adhering to the glass vs just precipitating in solution.

@DrMoishe Pippik: I guess surface irregularities are not too much of a problem to me. I need to coat glassware with silver such that there will be 0 contact between my solutions and glass at the molecular level. Would it be suitable to start with this Ag deposition and electroplate from there? (I liked the chemical route as it appears easier to make an even coating this way on glass surfaces of all shapes and forms)

It is not so much the glass, it is the contact between the substrate and the reflecting layer.
Some of the problems with the silver/glucose approach, like their optical imperfections is related to the bad adhesion of elemental silver to glass.  The «normal glass mirror» in the bathroom thus uses a tin layer between the cleansed glass, and the subsequently added silver (mostly) responsible for the reflection.  Discovery channel once  aired brief documentary about this; the more interesting part may be still seen on youtube here.
Modern mirrors for telescopes and optical instruments in general do not use silver, but aluminum; the reflective surface is now not behind the glass, but on top (first surface mirrors).  One of the reasons to use aluminum instead of silver is that silver darkens over time (think about the formation silver sulfide, already on cutlery), while aluminum oxide forming on top of aluminum still is transparent enough (for visible light) that these mirrors are aluminized about once every two years only.  Willi Koorts shows the procedure in a about 20 min video here on small scale: washing the mirror, remove of Al by KOH, washing, deposit of Al in high (real high) vacuum.  (This shows the operations at a larger, $\pu{8 m}$ scale.)


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Reaction between Polyvinylpyrrolidone (PVP) and NaOH
I am using polyvinylpyrrolidone (PVP) as a dispersant in a reaction. The reaction contains concentrated NaOH (pH of the  NaOH solution is 12 to which PVP is added). When I add the PVP, the solution turns yellow. Should this happen?
I was thinking that because of this color change, maybe there is a side reaction that is occurring between PVP and NaOH. I found this article that says PVP reacts with sodium hydroxide to form a gel. But I could not find any other sources that mentioned anything like that article
The temperature of the solution is 25 °C and under agitation to help dissolve the PVP.

The German version of wikipedia (https://de.wikipedia.org/wiki/Polyvinylpyrrolidon) states in section «Eigenschaften» (properties), without literature reference, the following: «Unter alkalischen Bedingungen und erhöhter Temperatur vernetzen die Ketten zu wasserunlöslichen Produkten.».  Google translate translates this into «Under alkaline conditions and at elevated temperatures, the chains crosslink to form water-insoluble products.»

I check the alkaline condition but the temperature is room temperature.


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Polymerisation of a UV-curing resin
I have a photosensitive resin MSDS. It lists a mixture of 1,6-Hexanediol diacrylate ($\ce{C_{12}H_{18}O_4}$) with its proper CAS number and Polyurethane acrylate together with the CAS number of "Butyl acrylate/methyl methacrylate/methacrylic acid copolymer" with the formula (or rather, formulas) of $\ce{(C7H_{12}O2.C5H8O2.C4H6O2)_x}$. There's also the Photoinitiator 1-Hydroxycyclohexyl phenyl ketone listed. The additives of color and reaction inhibitor are most likely irrelevant for the actual question, so I leave them out.
My organic chemistry is somewhat rusty, but the first part of the compound I can identify as having the setup $\ce{H2C=CH-CO-O-C6H12-O-CO-CH=CH2}$. This appears to be a quite nice molecule that reacts evenly on both sides under polymerization. This material is also listed as photosensitive, which I interpret that it sheds something from both ends to activate the ends.
The other, of course is a mixture of three monomers: Butyl acrylate ($\ce{H2C=CH-CO-O-C4H9}$), methyl methacrylate ($\ce{H2C=C(CH3)-CO-O-CH3}$) and methacrylic acid ($\ce{H2C=C(CH3)-COOH}$). The first two have ester groups in them, the last has its precursor, the $\ce{-COOH}$ group (carboxylic acid). All these only seem to have one active side to me.
Assuming the four main chemicals are mixed somewhat evenly and in even numbers and what might be an ample supply of the starter Hydroxycyclohexyl-phenyl-ketone, they surely will react and polymerize under UV light of about 400 nm (+- 5), as this is how the mixture behaves in its native application, an SLA printer.
However I have a problem with these parts:

What kind of reaction does happen between the compounds? I remember vaguely that some polymerizations create waste in the shape of water (was that condensation?), but I don't really see that happening here for most of the combinations and, as a result, I fail to see what the parts of the polymer might look like in their bonded state.
What type of bonds does the resulting polymer have? This is pretty much the direct result of failing to see the reaction between the compounds. I thought I had seen a possibility to form an ester ($\ce{R_1-CO-O-R_2}$) between the methacrylic acid and the hexanediol diacrylate, but then realized the later misses an alcohol group ($\ce{R-OH}$) to attack and create water from to form the bond. So I am pretty sure that these don't bond as esters to one another, but what is the correct name for these bonds that elude me?!
What would this polymer be classed as? I remember that some polymers are classed based on their monomers, manufacturing process, the chain length, bond type, and the macro-structure, putting each into several groups. These would allowing to estimate some vague properties about them. It's most certainly not a Homogenous one like pure PolyLacticAcid, as it is basically made up of 4 different monomers. It also should not form an elastic bond due to what I know from the MSDS and that the processed resin is a somewhat brittle product. But what groups it can be put into from the polymerization process?

You need a photoinitiator to do this. Related article: Songklanakarin J. Sci. Technol. 2011, 33(2), 201-207 (http://www.thaiscience.info/Journals/Article/SONG/10761807.pdf).

that would be, according to the MSDS, CAS 947-19-3, so... 1-Hydroxycyclohexyl phenyl ketone

That'd work. Check with photo-curing coating. Similar reactanets and polymerization.

in this case, it's an SLA resin, I think the chemistry might be quite similar.

I don't get it. It's just an acrylic resin and you get polyacrylate!

@Mithoron I beg your pardon, but the last time I had had my fingers on working out any polymerization was 13 years ago.

:/ While most of your post is nice and reasoning OK, the questions in the end are not stated well. You need to write them much more precisely.

@Mithoron I hope by adding more to explain how and where I fail to see the process might help to give it at least some of the precision you ask for.

Yeah, that's better.

Polymerizations expelling water, equally known as polycondensation (PET is a typical product), are a subset of polymerizations. The radical polymerization here is more likely to proceed differently with three reaction types running concurrently: initiation, propagation, and termination.

During the initiation, light splits the phenylketone into two fragments with one unpaired electron each. This is symbolized by the dot next to the oxygen atoms, denoting that these chemicals are radicals:

These radicals may react with one electron of the double bond of a neutral acrylate in the propagation reaction. Symbolized by blue is the newly formed single bond between the two reagents (first line).  Simultaneously, the other electron of the former double bond represents a new site where again a neutral acrylate may react (second line).  This growth of the polymer chain may continue again and again, or propagate (third line).

As soon as there are radicals formed, there may be a number of reactions where these carbon-centered radicals are consumed.  This cancels the further growth of the polymer chains.  Among these termination reactions, hydrogen-transfer / redox reactions as well as dimerizations are the most prominent ones.  Some which seem likely to occur are depicted below:

Structure
Your starting material consists of a starter, and a blend of bifunctional acrylates.  Thus the curation will yield polymer chains that are cross-linked with each other.  The illustration below (for a different material) symbolized this by the blue zig-zag lines:

(Konuray et al., Polymers 2018, 10, 178 doi 10.3390/polym10020178, open access)
Because you started with multiple different monomers, it would be an example of branched copolymers.


text:
Predict the species formed by the reaction of mercury(II) chloride and iodine

$\ce{HgCl2}$ and $\ce{I2}$ both when dissolved in water containing $\ce{I-}$ ions the pair of species formed is

A. $\ce{HgI2, I3-}$
B. $\ce{HgI2, I-}$
C. $\ce{HgI4^2-, I3-}$
D. $\ce{Hg2I2, I-}$

Answer: option (c)
In this question, I was able to predict the formation of $\ce{I3-}$. So, I could eliminate option (b) and (d).  But, between option (a) and (d), I could not distinguish. Is there any specific way to guess answers for these type of questions in exam?

There should be a redox reaction taken place here: $$\ce{I3- + 2 e <=> 3I-}$$ and $$\ce{Hg^2+ + 2 e <=> Hg2^2+}$$

You are right about the formation of $\ce{I3-}$ ion (from the reaction of $\ce{I2}$ and solution containing iodide ion, let's say $\ce{KI}$ solution). Reaction between $\ce{HgCl2}$ and $\ce{KI}$ solution at $\pu{70 ^\circ C}$ produces mercury(II) iodide ($\ce{HgI2}$):
$$\ce{HgCl2 + 2 KI → HgI2 + KCl}$$
But the story doesn't end here. This mercury(II) iodide reacts with $\ce{KI}$ solution to form potassium tetraiodomercurate(II)($\ce{K2HgI4}$):

Crystallizing from a concentrated aqueous solution of mercuric iodide
with potassium iodide is the monohydrate $\ce{KHgI3.H2O}$, which is pale
orange. In aqueous solution, this triodido complex adds iodide to
give the tetrahedral tetraiodo dianion.

So, the answer is  C. but there is a typo. It should be $\ce{[HgI4]^2-}$. The overall reaction is:
$$\ce{HgCl2 + 5KI + I2 -> K2HgI4 + KI3 + 2KCl}$$

Yes there is a typo . The option is HgI4 2-.

Is there any way to guess such questions. The info is not given in ncert too.

@Hydroquinone You don't guess chemical equations. You solve it logically.

Some behaviour patterns can be derived from basic or general principles, some just guessed as theory/empirical knowledge interpolation, but the rest is to be learned as the fact.


text:
Does a minus charge represent two electrons?
In OH- 's lewis structure, we put 8 dots around O and then the H, thus, it now has 10 electrons, instead of 9 (as 8 electrons for O and 1 for H), and thus we say that it has a charge of minus one.
Am I correct?
But in organic mechanisms, many books put two dots over OH and the bring them to a bond, or vice versa, thus, I assume that the dash over OH or the minus sign means two electrons, this confuses me.

The minus sign means a charge of minus one, there is no doubt about that. As for the two dots, these represent an electron pair, which has nothing to do with the overall charge. Many neutral molecules have them as well.

This forced me to remember a nuclear physicist that made her way thinking there is an anion called ammonia.

No, you are mistaken.
Oxygen has 6 electrons (dots) on its outer shell in the Bohr model and two on the inner shell. The inner shell is, under lewis rules, irrelevant. So the outer shell, 6 out of 8, needs to be put on the 4 sides of the O: 2 doubles, 2 singles. Ignore the color note for the moment, that's for physicists view below.

Then comes H with one dot, connecting to one singleton via the bond. Now comes the negative charge -, which adds another electron (dot) to the last singleton, giving us a full 8 dots: 7 of these are held by the Oxygen, one belongs to the Hydrogen, but shared to the Oxygen:

I am a physicist. We go further and talk about orbitals that each can have 2 electrons. Hydrogen has only one, which means it has a partially filled 1s-orbital. As we go up in the periodic table we start to fill the orbitals: first each orbital gets one in order of which one has the lowest energy needed to stock them there, then they start to get their second electron, as soon as that position becomes the most energetically cheap position to put it. Starting with Selene (and again Lanthan!), that can become a mess, but we are at Oxygen and Hydrogen, no danger here.
Oxygen has 8 electrons total: First we filled 1s-orbital with two. Then we put one to the 2s-orbitals, then one on the next energetic stable ones: 2px, 2py and 2pz orbitals - there are no 1p orbitals. This means we have 2 left. So we fill up 2s fully and  one (commonly 2px) of the p-orbitals fully and are left with two partially filled 2p orbitals (2py & 2pz), telling us Oxygen can do two single-p-bonds or one double-p-bond using a smeared orbital in the yz-plane.
One of these p-orbitals now connects with the hydrogen's 1s orbital, creating the bond to the oxygen's 2py. The negative charge ends on the last 2pz-orbital. Why we do that? Because it's pretty and we claim that the electrons are smeared all over the shell of these orbitals. Electrons are not positional, they have a volume where they might be, and that's what each orbital stands for.

It's bit unusual to use a shell model and then label the occupying electrons with s, p, d, f...

@Zhe I know, but it's how you go from Bohr to Orbitals the easy way. the hard way is to solve a lot of formulae with all the quantum numbers to get which one is the next one that's energetically best.

By convention, Lewis formula often represent two electrons by a dash instead of dots.  This dash might appear like a minus sign to represent a negative charge, especially in absence of visual comparison.  The two concepts however are independent of each other.
Though the equilibrium of the autodissociation of propanol likely favors the non-dissociated state on the left hand side, one can describe the underlying the reaction with the upper reaction equation equation.  This one omits explicit non-bonded valence electrons altogether:

At a lower level, in course of the formation of the alcoholate, both electrons  of the $\ce{O-H}$ bond are transferred to oxygen, the more electronegative atom.  To discern non-bonded valence electrons of Lewis formulae and charges, some sketchers offer special signs enclosed by a circle (lower reaction equation).*
Further, if one would remove one electron from the alcoholate oxygen, the molecule would be neutral - and because of an unpaired valence electron, a radical as shown for hydroxyde ion and hydroxyl radical:

(image credit: English wikipedia).
On occasion, authors indeed use this special notation to discern charges and Lewis electrons:

Both photos display equations in Warren's retrosynthesis book, p. 111.
* To highlight the complete electron transfer, the $\ce{O-H}$ bond in the lower equation was extended beyond typical proportion.
references:
Warren, S.; Wyatt, P.  Organic Synthesis: The Disconnection Approach. Wiley,  2008, ISBN 978-0-470-71237-5 (landing page of the book at the publishers).

Non-bonded pairs of electrons (lone pairs) are shown in Lewis structures either as two dots, or as a line (like bonded pairs).
If you are using a line for lone pairs and are showing negative charges in the same figure, it is good practice to circle the negative (and positive) charges so that there is no confusion.

Does a minus charge represent two electrons?

It does not. It does not represent electrons at all. It represents an excess of negative charges (typically from electrons) compared to positive charges (typically from protons). Because you don't show inner electrons in Lewis structures, and don't show protons at all, figuring out the charge of a species (or the formal charge of an atom) has to be done as a separate calculation whose result you would add to your  Lewis structure or other representation.


text:
when using glass to glass stoppers, is it best to have them dry or use lubricant?
I have some glass flasks that have glass stoppers. It’s basically the typical glass stopper with the frosted tip which fits into a glass opening. Basically think of a typical glass volumetric flask. i attached a pic of my flasks with the stopper. they are not threaded connections. i believe they slight self lock. like i put the stopper in and push down and turn to get it tight.
So my question is, if I am travelling/moving theses flasks around in a situation where they may tip over or be upside down for some time with fluid inside them , is it better to have the glass to glass contact between the glass stopper and the glass opening fully dry? Or is it better to add some oil to the glass stopper?
If I do glass to glass and they both are dry, it’s usually pretty sealed however I believe I have always been told to use some type of oil as it helps it seal better but I tbink I am against using oil becusse I feel like the stopper has a higher chance of coming off or slightly moving causing a leak becusse of the slipperiness of the oil.
*the connection is not threaded, it is just flat like a typical glass volumetric flask”
I of course will be using parafilm/tape on top so I am just wondering which method is best so no liquid leaks out. There will be no gas build up, its just liquid like water or alcohol sometimes but i dont care if they slightly evaporate. i just dont want leaks
Basically I am looking for the best method to reduce leaks
Thanks

I think this might open up a discussion based on taste and luck of the operator. I never used dry contact. Sealing wasn't the point but removal of the apparatus. I always used silicon based lubricant in minimal amount or Teflon bands coming in sizes fitting your joint. Teflon is also available in rolls like band aid, I use this for valves of plumbing style connections and apparatus but I think it can be used for glassware as well.

Recall that a flat bottom flask shouldn't be used for a distillation. Using grease on the stopper may end up as an impurity of the material you aim to stock in these flasks, too.  The use of these flasks is restrained, though. Don't use them for alkaline liquids; once the surface between stop cock and flask is wet, lye (e.g. NaOH reacting with CO2 of air) builds a cake which may render opening the flasks difficult. If you store these flasks clean and empty for sometime, either keep the stop cocks separate from the flasks, or put a strip of paper (e.g., newspaper) between stop cock and flask.

Sometimes dry glass-to-glass sealing of the stopper leads to seizing of the stopper and can become very hard to remove. Therefore vacuum grease is often applied lightly. However, if your material in the flask can dissolve the vacuum grease, the material can become contaminated by the vacuum grease. If your material is volatile, you will need the vacuum grease to ensure an airtight seal. Another method you can use is to use teflon tape between the stopper and the ground glass joint to make a seal that will not seize. But, vacuum grease will give you the best seal.


text:
Sintered vs perforated plate Büchner funnels
I just bought a sintered-glass style Büchner funnel for some vacuum filtration tasks. There's two issues I'm having with it:

Filter material, especially activated charcoal, ends up stuck in the glass disc. I did a little bit of research trying to find out how to clean it out and the only thing I found (for the carbon, at least), was piranha solution, but I have neither the equipment nor the experience to use it safely so I'm SOL on cleaning. Fortunately the presence of carbon isn't a huge deal for my uses, though.
For paper filtering; the sintered disc goes all the way to the edge of the funnel, so unless I've got filter paper exactly the size of the funnel (which I don't), I'm worried that the 0.5-1.0mm gap I usually end up with around the edge of the paper gives fluid a path to bypass the paper. If I err on the side of larger paper it's still a poor fit with large channels around the edge.

It seems to me that the perforated-disc style of Büchner funnel would solve both of these problems; since the funnel itself can't trap filter material and also the holes don't go all the way to the edge of the disc:

(Incidentally, the pictured funnel on the left is exactly the one I own.)
My question is, given the two apparent disadvantages I've observed with the sintered-glass style funnel, what are the pros and cons of each type? More specifically, what could I do with a sintered-glass funnel that I couldn't do (as easily) with a perforated-disc funnel?
The context here is "kitchen chemistry"; I'm not a chemist I just do a few DIY projects here and there and all my knowledge comes from YouTube and Google searches. The only lab equipment I own is this funnel, a flask, and a vacuum pump. The only relevant PPE I own is a respirator, various gloves, and some chemical safety goggles.

For the sintered glass filter, add an inch or two of diatomaceous earth (Celite). Wet it under vacuum with solvent that you are using to make a pad. Now slowly filter your charcoal suspension. This will keep the sintered glass clean. Measure the inside diameter of the Buchner funnel. Use a compass to draw a circle on the filter paper and trim. Put the paper in the funnel with vacuum on. Wet paper then go through the Celite procedure. PS: The right size filter paper should be available for the Buchner funnel.

Super appreciate the pro tips I've got Celite on order anyways so it's perfect. I am using a compass I guess I just need to play around a little to get it perfect. Are there "standard" funnel sizes? Mine is 94mm but I couldn't find the Whatman papers I wanted (#1, 6, 597, and 602H) in that size, and the lack of sizes matching my funnel makes me wonder if my funnel is strangely sized. I just sent Cytiva (Whatman) a support request to ask.

Your Buchner could just be old. Back flushing the sintered funnel may help clean it but use the Celite as suggested above and give it a good stir around in the solvent to break up any lumps and cover the entire filtering surface.

Sweet thanks. I'm already using sodium bentonite as a filtering step and have some on hand so I just combined it with the charcoal step (paper -> bentonite clay -> paper -> charcoal, bottom to top) and that also seems to keep the charcoal dust out and kills two birds with one stone.

The thing on the left is not a called a Büchner funnel, but a frit. Frits are only meant to be used for stuff that can be redissolved afterwards.

The thing on the left is not usually a called a Büchner funnel, but a frit. Frits are generally meant to be used for stuff that can be redissolved afterwards, because they tend to clogg permanently, if very small particles get stuck inside.
You do not normally put a filter paper on it, because that would leak on the edge. The filter grade should be etched onto the glass somewhere. https://en.wikipedia.org/wiki/Fritted_glass Sometimes a frit can be used to hold another, powdery filtering substance, if that has a coarse enough grain size.
Advantage of the frit is the well defined pore size, you can heat it for drying, it can be mounted in a glass tube with two ground glass connectors (for evacuating), it can be used with concentrated and/or oxidising acids or other solutions, ...
Advantage of the Büchner funnel is that it does not get clogged with insoluble stuff, ease of use, cheap, available in large diameters, very mechanically solid, tolerates alkalic solutions (glass frits do too, but only briefly), ...

Hm... I'm a little confused now, because this funnel (the one I have) is described as a "Buchner filtering funnel" with "coarse frit"; and says "Compatible with StonyLab Qualitative/Quantitive Filter Paper Circles". It does give the pore size like you mentioned (80-120 microns), I didn't notice that before. The "G1" grade marking is indeed printed on the glass.

(Similar deal here)

Well, seems some people have extended the meaning. If that misnomer has reached textbooks already I don't know. But I wouldn't trust a company that sells frits that are "compatible with pape disk filters" on correct naming. ;) You have already pointed out the problem above.

I wonder if maybe it's a regional / generational / field-specific thing. Who knows. So, anyways; your answer is great and I'm going to order a... Karl-style... Buchner funnel and a vacuum adapter. Question about usage: Is it typical to use a sintered funnel with a granular/powdered solid filter material; like celite, charcoal, bentonite clay, etc.? Or do people in the lab tend to use the perforated disc style with some paper on the bottom to hold the filtering material?

Or is it just kind of a "use whatever you've got that works and that you feel like cleaning later" kind of thing?

Depends on whether you are removing unwanted solids from solutions  - then you use the sintered funnel, with or without Celite - or collecting a solid that is the material you want - then use the perforated disc (Buchner) with filter paper.

Büchner style. ;-) https://en.wikipedia.org/wiki/Ernst_B%C3%BCchner  His name also goes with that Erlenmayer-style flask with vacuum flange at the side.

It is not necessary that the sintered glass glass filters be used for something which can be dissolved later. We routinely use them for filtering synthesized silica particles or chemically modified stationary phases. The reason for using fritted funnel is have to chemical compatibility with harsh solvents (with the exception of bases, but glass frits survive a brief exposure)

@M. Farooq: How do synthesize silica particles? Can you make them very thin? Wouldn't this be better than diatomaceous earth above the fritted disk? Pure silica is less reactive, and won't leach Al/Fe/...

We don't synthesize silica particles (about 2-4 micrometers in diameter), they are made by another company. One can surface modification for chromatography. You cannot afford any impurity so modified silica has to be directly filtered over sintered glass.

I agree with user55119's suggestions (see the comment I have copy and pasted below):

For the sintered glass filter, add an inch or two of diatomaceous earth (Celite). Wet it under vacuum with solvent that you are using to make a pad. Now slowly filter your charcoal suspension. This will keep the sintered glass clean. Measure the inside diameter of the Büchner funnel. Use a compass to draw a circle on the filter paper and trim. Put the paper in the funnel with vacuum on. Wet paper then go through the Celite procedure. PS: The right size filter paper should be available for the Büchner funnel.

However, the celite bed is only good for your charcoal suspension and if you are filtering just to remove minute impurities (when your target is the filtrate). Make sure you pour your suspension slowly, slow enough not to disturb the celite bed. The filtration with celite would be slower than usual (without celite). You may use a thinner bed to have a faster filtration rate, but have to extra careful not to disturb the celite bed. Nonetheless, if you are patient enough, you can achieve your goal.
If you want to filter out your target crystals or solid, you cannot use celite. For that purpose I'd always use a Büchner funnel because it is easy and could have recovered most of materials. Also cleaning procedure is easy. If you have used a sintered glass filter in place of Büchner funnel, after use, you have to wash it thoroughly with suitable solvent to remove trapped particles (similar to charcoal particles you have experienced) before subjected to usual cleanup. That is an extra step.
One advantage of the sintered glass filter is it has a ground-glass end. Thus you can use filtering flask with ground-glass head, so that the equipment would have airtight connection for ideal vacuum filtration. For Büchner funnel, this is impossible because it needs a rubber filtering adaptor to connect with a filtering flask, which is almost always leaky.
Note:  When you buy filter papers for your Büchner funnel, always look for what size it fits. This information is provided by some vendors: For example, the filling filter paper size for $\pu{35 mL}$-Büchner funnels is those with $\pu{4.25 cm}$ in diameter. If you can't find the right filter paper, go with user55119's instructions.

Quick terminology question: Does "Büchner funnel" only refer to the perforated disc type, or does it sometimes refer to both types of funnel?

You can find enough info here: https://en.wikipedia.org/wiki/B%C3%BCchner_funnel

@JasonC That wp article is wrong. The thing on the left is not a Büchner funnel.

A Büchner funnel is a funnel with the upper section shaped as a cylinder with a a filter at the bottom of the cylinder which then tapers to a funnel shape. A glass fritted Büchner funnel has a fritted glass filter, and a porcelain Büchner funnel has perforations which a filter sets on top of. Both of them are Büchner funnels because both of them have a cylindrical upper section with a filter at the bottom. Both are able to be used under vacuum, however depending on the particulate- the porcelain one doesn’t always require a vacuum. We learn all about these things in chemistry 101in college, memorizing lab ware is one of the very first things you are required to learn. A Büchner funnel can be made of wood even, as long as it has a upper cylinder shape for the upper part with some kind of filter at the bottom, which then tapers and funnels down, that’s it.


text:
Ionisation vs Dissociation
This is a statement given in my book:

A strong electrolyte is completely ionised at all dilutions but not completely dissociated.

But the book also uses the terms "Ionisation" and "Dissociation" interchangeably in other places.
When I further tried to search about this topic I found out there is equation which governs dilution of strong electrolyte-
$\lambda_m = \lambda_o + a\sqrt C$
So is it right to say that it will only get completely dissociated at infinite dilution (which is not practically possible)?
Hence my final questions are:

Are ionisation and dissociation the same or different?

If they are not the same then why are they different?

Also the correct explanation of the above statement given in my book?

Will this change for a weak electrolyte?

Also please consider that I am in high school. So please try to explain in simple terms so that I am able to clear my confusion.
Please also tell whether I can assume both the terms to be same or not at my current high school level.

Have you learnt about solutions?

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

The parameter a in front of √C is empirical. It varies from solute to solute. And yes, the dissociation is complete at infinite dilution. Dissociation is the separation of ions when dissolving a salt in water. Ionization is what happens when an atom or a molecule looses one electron, without being modified or broken. If$\ce{NaCl}$ is dissociated, it gives two independent ions, one $\ce{Na+}$ and one $\ce{Cl-}$. If $\ce{H2O}$ is ionized, it gives an ion $\ce{H2O+}$ and one electron, but the atoms of the molecule are not separated from one another.

@Maurice I think you should make that an answer

Do you have a reference for the book in question? I suppose the book means that the ions are paired or correlated but I would not call the electrostatic interactions the equivalent of a bond in say an ionic solid.

The parameter a in front of √C is empirical. It varies from solute to solute.
And yes, the dissociation is complete at infinite dilution.
Dissociation is the separation of ions when dissolving a salt in water. If NaCl is dissociated, it yields two independent ions, one Na+ and one Cl−
Ionization is what happens when an atom or a molecule looses one electron, without being modified or broken. If $\ce{NaCl}$ is dissociated, it gives two independent ions, one $\ce{Na+}$ and one $\ce{Cl−}$
If $\ce{H2O}$ is ionized, it gives an ion $\ce{H2O}$+ and one electron, but the atoms of the molecule are not separated from one another.


text:
Baeyer-Villiger Oxidation restricted to Ketones
According to Wikipedia, Baeyer Villiger oxidation occurs at the electron deficient oxygen site, with elimination of the carboxylate ion, and migration of an R group. But it is restricted to ketones and lactones. Why are aldehydes not included in it? Moreover, since migratory amplitude of hydrogen is lower than alkyl or aryl, shouldn't aldehydes be more prone to participate in this reaction?

Check Dakin's oxidation.. seems to be a similar reaction.

@Safdar Dakin's reaction migrates phenyl, but what is the criteria for this?

https://chat.stackexchange.com/rooms/74807/conversation/phenyl-migration

The Dakin reaction is baeyer villiger oxidation for aryl aldehydes,

https://onlinelibrary.wiley.com/doi/abs/10.1002/0471264180.or043.03 This is cited in wikipedia

May be Baeyer Villiger oxidation is not for aldehydes in the past. But that has changed recently (Ref.2). The abstract of this reference states that:

A conceptually distinct, modern strategy for Baeyer−Villiger oxidation (BVO) was developed. Our novel method involves initial hydration of water to carbonyl compounds, followed by ligand exchange of hypervalent aryl-$\lambda^3$-bromane on bromane(III) with the resulting hydrate, yielding a new type of activated Criegee intermediate. The intermediate undergoes BV rearrangement and produces an ester via facile reductive elimination of an aryl-$\lambda^3$-bromanyl group, because of the hypernucleofugality. The novel strategy makes it possible to induce selectively the BV rearrangement of straight chain primary aliphatic as well as aromatic aldehydes, which is missing in the classical BVO: for instance, octanal and benzaldehyde afforded rearranged formate esters with high selectivity (> 95%) under our conditions, while the attempted classical BVO produced only carboxylic acids. This firmly establishes the powerful nature of new methodology for BVO.

The suggested mechanism and comparison it with classical Baeyer−Villiger oxidation is shown in following scheme $(\bf{A})$:

The part $\bf{B}$ of the scheme depicted computational calculation of energy diagram for acetaldehyde oxidation.
In addition, similar oxidation of acetal (an aldehyde derivative) to ester have been known for years. For example, conversion of cyclic acetals to hydroxy esters by mCPBA oxidation is published in 2002 (Ref.2).

Reference:

Masahito Ochiai, Akira Yoshimura, Kazunori Miyamoto, Satoko Hayashi, Waro Nakanishi, "Hypervalent $\lambda^3$-Bromane Strategy for Baeyer−Villiger Oxidation: Selective Transformation of Primary Aliphatic and Aromatic Aldehydes to Formates, Which is Missing in the Classical Baeyer−Villiger Oxidation," J. Am. Chem. Soc. 2010, 132(27), 9236–9239 (https://doi.org/10.1021/ja104330g).
Jin Yeon Kim, Hakjune Rhee, Misoo Kim, "Conversion of Cyclic Acetals to Hydroxy Esters by MCPBA Oxidation," Journal of the Korean Chemical Society 2002, 46(5), 479-483 (DOI: 10.5012/jkcs.2002.46.5.479).

Is the aryl group stablizing the process as in vicarious substitution?

I have attached the proposed mechanism and computational studies for your convenience.


text:
bonding in polar covalent bonds
I have recently learned that pure ionic and covalent bonds are just the extremes of a spectrum of bonds from this article from Chemguide. But I can't seem to square this with my understanding of how bonding works. In ionic compounds like $\ce{NaCl}$ for example, the difference in electronegativity is so great that it seems that electrons are transferred from $\ce{Na}$ to $\ce{Cl}$, so they both gain octet structures. Similarly, in covalent compounds like $\ce{H2}$, the difference in electronegativity is $0$, so the electrons are shared between the elements, so both gain octet structures. But in between these extremes, are the electrons shared, transferred, or a mix of both? And if it's a mix of both, how can atoms fill their octet if the electrons are "partially transferred"?

They are shared but the electrons are more likely to be found nearer to the more electronegative atom..

@Safdar So the spectrum of bonds goes from covalent, to polar, to ionic, and in covalent and polar bonds the electrons are shared between atoms, while in ionic bonds, one atom has control of both electrons in the electron pair(i.e electrons are "transferred")?

Yes.. That would be correct.

Recently, there was a similar question which was answered with electron density maps (https://chemistry.stackexchange.com/questions/138377/what-is-the-graphic-difference-between-ionic-and-covalent-bonds/138380#138380).  Perhaps this complementary perspective is helpful for you,

In a covalent bond between two atoms with an electronegativity difference, the bonding electrons are closer to the more electronegative atom in the bond. This causes a partial negative charge to form on the more electronegative atom and a partial positive charge forms on the less electronegative atom. This causes two poles to form in the compound called a dipole. Thus, there is a degree of ionic character observed in the covalent bond.
Now you need to understand charge density. In ionic bonds, cations and anions are next to one another. If the cation is small in size and has a large charge it has a high charge density. If it is bonded to an anion of large radius. The outer electrons of the anion are not firmly bonded to the nucleus and are attracted by the  strong electric field of small highly charged cation. This causes an increase in electron density between the two ions. Thus, a degree of covalent character is conferred on the ionic bond.
Now, to answer why covalent compounds are covalent and ionic compounds are ionic is answered by the following phenomenon.
Strength of dipole is measured by its dipole moment. Dipole moment is product of separated electric charge and distance between them. It has the unite Debye which has the dimensions of coulomb meter.
Now ionic compounds have quite high dipole moments like 7.0D etc. while even the most polar covalent compounds like HF have a dipole moment of 1.91D. This shows that even the most polar covalent compounds are a long way from being ionic.
So electrons are not "partially transferred". They are either transferred or shared. The sharing may become unequal or unbalanced causing bond polarity. The electron transfer may result in ions forming with different charge densities leading to attraction of electrons towards the strong electric field. The main objective is to reach a point where there is maximum stability in the compound.
Following link gives further insight into polarity and dipole moments if you find difficulty in understanding the answer.
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.03_Bond_Polarity_and_Dipole_Moments