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Is boiling tap water a must to destroy human mucus, proteins, fats, sugars and nucleic acids it may contain?
I understand that humans consume tap water which are chemically treated after pumped from natural or artificial reservations.
Sometimes people contaminate these reservations by spitting, pissing and shitting in them (for example, parents hold babies while swimming in a lake)
Pumped water are likely to be filtered and then chemically treated by:

ozone-based substances
fluoride-based substances
sulfate-based substances
chloride substances
Bubbling o2 oxygen allotrope into the water
Other substances
Some combination of substances
Radiation (probably artificial UV in a closed water container)

My problem
Because tap water aren't boiled by water suppliers, they contain dead saliva bacteria and other human cells, but they might also contain:

Human Mucus
Human proteins, fats, sugars and nucleic acids

My question
Is boiling tap water a must to destroy human mucus, proteins, fats, sugars and nucleic acids (and also human derived dead bacteria) it may contain?

You appear to be obsessed with the idea that boiling is necessary to destroy certain possible water contaminants. This is wrong on at least two counts (as previous answers to similar questions have explained): 1) boiling does not destroy those contaminants; and 2) other treatment methods are already completely adequate to ensure safe water.

@matt_black I am not asking about safety, but on purity from human derived materials such as saliva, urine, feces, dead bacteria derived from them, human proteins, human fats, human sugars and human nuclei --- everything the rest, everything non human, is fine with me. Do you understand my intention now?...

Please help me to ask a better question if you understand me better now.

Yes I get disgusted from the think that I might drink another human's saliva or piss in my tap water.

Boiling water is for me just a mean, not an end...

Your personal disgust reaction is irrelevant to what cities can justifiably do to provide people with clean water. Plus you seem to believe that boiling does something it doesn't achieve. Distillation might, so feel free to drink distilled water. But there are plenty of standards for what clean water supplies can  contain and current treatment methods keep the things you seem to worry about at a unmeasurably low levels.

@matt_black again, I am willing to take personal action to remove any described-human materials from my water; if boiling chemically treated water doesn't help, would would?...

I just reiterate that drinking distilled water is not healthy and should not be done. In case a new reader land here and read assertions dictated by exhaustion :)

@Alchimista your comment is in my opinion uncommunal and should be flagged as offensive.

I don't think so. I just refer to the nice comment by @matt_black in which is written "feel free to drink distilled water" assuming a kind of fatigue as ground for an assertion he would have not made. Anyway you can flag the comment and let others decide. This is not my job, just a way to spend break time learning and sometimes help others. In the specific case you got help already. We can't change answers just because you are obsessed with mucus etc. And mucus or urine in a lake even cease to be mucus and urine, to add another chemical consideration.....

@Alchimista here, by the exreme, offensive uncommunal statement just because you are obsessed with mucus, you just prove my point that was based upon assertions dictated by exhaustion :).

Boiling water will destroy living organisms like bacteria. It will not destroy chemical substances like sugars, proteins, fats and acids, which are not living species. It will probably transform their structure, coagulate the proteins, hydrolyze some sugars or fats. But this is not exactly what could be understood as a destruction.

Can I even destroy these in home kitchen?

Yes. As discussed in previous posts, boiled tap water is good and safe enough for drinking. Remember that you are drinking tap water not water from sewerage. You did not mention your country of residence. Most municipal corporations can provide you water analysis report.

I wrote I live in Thailand --- I meant to total destruction...

Do you know that there are good quality domestic water filtration system which can be attached to tap water source. Water goes through a series of filters, then activated carbon cartridge (which absorbs all organics) and eventually through a UV lamp. This is good enough to destroy/remove, yes "total destruction" of all disgusting things you can $ever$ think of. Whatever disgusting is coming to your mind, the answer is that activated carbon and UV lamp with remove or destroy it.

You can perform a very simple experiment to convince yourself that boiling will not destroy fats or sugars. This experiment is being performed countless times every day (in a slightly modified version) in kitchens all over Europe and probably other parts of the world.
Take a pot of water, pour in a shot of vegetable oil (e.g. sunflower oil) and add a spoonful of sugar. Boil the water. Let it cool down. You should be able to observe both the drop of oil (a second phase as it is immiscible with water). Furthermore, if you taste the water it will taste sweet. This is enough to show that simple oils (a type of lipids or fats) and sugars are not destroyed in solution or emulsion with water at boiling temperatures.
Your question proceeds to answer itself thenceforth.

Thank you, I have upvoted your nice answer.

You are laboring under a misunderstanding that most non-chemists have when it pertains to chemistry. You are unaware of the absolutely massive scale of tinyness that things are in.
Let's say you go to the sea and fetch a glass of water. Scoop it up. Now, you decide to have a contest with your friend. You have a magical bag that can contain anything in whatever amounts.
So you set off trying empty the water glass, one molecule at the time...
...and your friend sets off trying to empty the oceans of the world with one glass of water at the time.
You keep the same pace, the number of glass or molecule per minite.
You will lose, and with eons upon eons. You will not finish before the universe dies, though your friend migh.. In a $20$ cl glass of water there will be approx $4$ moles of water which is, if I typed it correctly, about $24920000000000000000000000$ molecules
Boiling water, purifying it, cleaning, scrubbing, leaching - none of these create pure water without contaminants. There is only a question of how many contaminants you can handle. 1 per 10000 or 1 per 10000000 makes the difference
Boiling water for 10 minutes brings the number of organic still living beings in the water down to a level where it does not harm your body. (normally - there are some parasites more durable than others - and harmful prions will be unaffected). It will still contain dead organic beings, mucus, shit, piss, toxic metals, harmful toxins - but in quantities your body can handle.
As long as tap water is measurably drinkable (potable, from french) the water works doesn't really care about the level of contaminant (of course they do, but if it is drinkable, it is drinkable, and the bar is set very low for that)
So no, boiling water is not strictly necessary as long as the level of contaminant is below whatever level of that contaminant is perceived as safe. And making the water chemically pure, as in totally free, of that contaminant requires a lot more than boiling, and is rarely done - and if you want water that clean you usually make it from it's constituent atoms or distill it repeatedly.
Don't be afraid of minute quantities of contaminants - you have been exposed to them since birth. Fun fact, if your eyes sting when swimming in a public swimming pool, it is likely ammonia (di) chloride, which is likely from urine. Enjoy!

@JohnDoea That is not to be found, no. If you, as per admission, have the understanding of the tinyness of things, this should be apparent to you. There will always be stuff that is not the right stuff in stuff. It is all about dosage. Accept it or die. (Not a threat, just a fact of life, without water you die). Also, bottled water is usually way more contaminated than tap water, unless you are in a 3rd world country. Or Flint, Michigan.

Nice answer. That is what I meant commenting on basically the same question days ago (mentioning omeophaty).


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Specific examples of all-main-group complexes with significant d-orbital contribution
It is usually said that d-orbitals are not usually used in bonding, even for so-called "hypervalent" ones, in compounds that are made purely of main group compounds.
However, after reading an article (or, strictly speaking, the preview of an article) about a calcium complex explicitly involving d-orbital contribution (see Krieck, S.; Görls, H.; Yu, L.; Reiher, M.; Westerhausen, M. J. Am. Chem. Soc. 2009), I began to wonder if there could be more d-orbital all-main-group compounds or not. The molecule depicted in the aforementioned paper is unstable in air and water, but maybe the unstability (i.e. high energy of main-group d-orbitals) could be reduced.
My guess is that, one can use the anti-bonding orbitals of a sufficiently weakly aromatic ring ("weakly" aromatic, in order to have relatively high energy) as ligands and putting many electrons in the coordination center (not necessarily metallic, and it should be reasonably electronegative, such as sulfur) so that the energies of the 4s and 3d orbitals of the coordination center reasonably match the anti-bonding orbitals of the ring and the usual Aufbau principle could be applied directly (3s2, 3p6, 4s2, 4d(1~5)).
The question is, are there any explicit examples of such complexes that are stabilised by a way similar to what I mentioned above?

Can you define "significant d-orbital contribution"? I don't know of an example with the structure you describe, but calculations on pentamethyl antimony suggest d orbital particpitation is not negligible: http://actachemscand.org/pdf/acta_vol_47_p0368-0373.pdf

Did you consider writing Dr. Krieck directly asking him?

"[...] Ca, Sr, Ba, and Ra use their inner(n−1) d-orbitals predominantly in covalent bonding contributions when sufficiently positively charged [...]. This leads to a number of peculiar structural features that bring these elements into the realm of ‘‘non-VSEPR $d^0$ systems’’ that encompass early transition elements and even lanthanides, and they have also been termed‘‘honorary d-elements’’ [4]" cited from https://media.wiley.com/product_data/excerpt/50/35273331/3527333150-343.pdf

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Emulsification and liquid liquid surface reaction

What will happen if I skip the addition of surfactant?
will the reaction happen only at the surface of the parent liquids?

If you do not add the surfactant you will not get an emulsion and the mixture will revert to a 2 phase system after stirring. Some reaction is still possible at the phase boundary but it will be orders of magnitude slower than it would be in the emulsion.

Yeah! Can you suggest some economically viable mixture components on which I may perform the above surface reaction, i.e the reaction which may occur at the 2 phase boundary! Thanks

It depends on what reaction you are attempting to do. I'm a synthetic organic chemist so my "go to" reagents for 2 phase reactions are tetra-alkyl ammonium salts such as tetrabutylammonium hydrogen sulfate

Basically I want to mimic a different system by these reactions! The only condition is that the bottom phase should be aqueous. And then the top phase maybe something which is miscible in oil (like Si oil or vegetable oil etc.).

I want to perform a surface reaction such that the new product which forms at the surface, goes on to diffuse in the 2 phases and thereby changing the thero-physical properties of these 2 phases near the reaction zone ( i.e the interface).

2 weeks back I was considering these reactions to mimic my system. You may have a look @https://chemistry.stackexchange.com/questions/125179/liquids-which-may-dissolve-polyacrylic-acid?noredirect=1#comment239534_125179

It depends a bit on what you are trying to achieve :
If you are trying to form an emulsion, for example a homopolymer emulsion resin solution, then the emulsifier is absolutely necessary.
If you are trying to extract a water soluble entity from the oil/powder phase then an emulsifier may actually interfere with the process - at least in terms of post extraction separation.
The supplied image suggests that you are trying to form a "stable" dispersion - in which case an emulsifier (or emulsifiers) of the correct type and appropriate HLB value at the correct concentration will be needed. Appropriate physical energy will also be required.

I want to just perform a surface reaction between 2 bulk phases! i.e I don't want to make emulsification droplets and just analyze the reaction film happening at the surface!

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To calculate expansion work for a process

[1 mark] To calculate the expansion work for a process,
  1) the formula $w = -P_\mathrm{ex}(V_2 - V_1)$ always works under all conditions.
  2) the formula $\mathrm dw = -P_\mathrm{ex}\,\mathrm dV$ can only be used for irreversible processes.
  3) the formula $\mathrm dw = -P_\mathrm{ex}\,\mathrm dV$ always works under all conditions.
  4) the formula $w = -nRT\ln (V_2/V_1)$ always works for any isothermal reversible process.
  5) None of the above is applicable.

I'm torn between options 1, 3 and 4. I see those 3 all as correct. Can someone point to me which one might be right and why?

3 and 4 are both correct.

Why is 1 not correct? Isn't it basically the integral of 3

A screenshot or picture of an exercise is not searchable, so I typed your question in. Please consider rewriting the text from the images in the future, so that it can be of help for future visitors. Also, it would be ideal for you to add the source of this problem (textbook/local exam).

1 is not the integral of 3 for a reversible expansion.  4 is the integral of 3 for a reversible expansion.


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Definition of t2g orbitals in CFT
My textbook defines $\mathrm{t_{2g}}$ orbitals as

$\mathrm{t_{2g}}$ stands for a set of three orbitals which are asymmetric with respect to $C_2$ axes, perpendicular to the highest $C_n$ axes, but which is symmetric in sign through the centre of inversion.

I know that “$\mathrm{t}$” is a randomly chosen letter as suggested by M. Farooq, “$\mathrm{g}$” comes from gerade (MOT), that's what  the last line says. Though “$2$” in subscript must have come from $C_2$, but I didn't get the meaning of

…which are asymmetric with respect to $C_2$ axes, perpendicular to the highest $C_n$ axes.

Please explain what does this line conveys.

I have checked the original literature, t or T, does not stand for three or triplet. It was a randomly chosen letter. You are right g is for gerade. In original works, the letter F was used alphabetically, A,B,E, F as symmetry labels.

Dos this answer your question: What are t2g and eg in CFT?

@andselisk But this post does not include the above definition or it's meaning. Wasn't the meaning of "t, e, g" explained in broader essence?

@M.Farooq what is the original literature for this? I had always assumed the symmetry labels had some German connection.

Tyberius, actually not. I was planning to write a short historical article. Except for g and u I could not authenticate any other German meaning. The rest are mostly myths including $Entartet$ for degenerate.


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Elevation in boiling point of water

$\pu{1060 g}$ of urea aqueous solution is subjected to heating. It starts boiling at $\pu{100.53 °C}.$  It is heated up to $\pu{102.12 °C}.$ Find amount of water vapourised in grams. (Pure water boiling point is $\pu{100 °C},$ $K_\mathrm{b}$ of $\ce{H2O}$ is $\pu{0.53 K kg mol-1}.)$

When water evaporates will the equivalent amount of urea also evaporate i.e. will we calculate final molality with the same amount of solute or a decreased amount?
I first calculated the initial amount of water as (53000/3) g and then taking the same moles of urea calculated the final amount which came out to be (53000/12) g. So the amount of water evaporated should be 13250 g but the given answer is 750 g.

please elaborate your doubt here, did you try the question yourself? Do you know the concept? Where are you stuck?

We have a policy which states that you should show your thoughts, effort and attempts to answer your question yourself. It'll make us certain that we aren't doing your homework for you, and that the Q/A is beneficial for broad audience.   As "homework-like questions" are considered literal homeworks, self-study questions, puzzles, worked examples etc.  Please [edit] in your full reasoning or thoughts on this. See Homework.  Otherwise, the question may get closed as "low OP effort homework-like question."

Hint: when water (only) will vaporise, molality ↑ and hence ∆Tb ↑

It's 13250g only. This time mypat answer key was completely flawed(except for maths). Most probably someone hacked the system as someone got 300 with this totally flawed key

There's 1 mole of urea and 1kg water at first, how can the mass of evaporated water be more than that of the entire solution.

@SirArthur7 Initial mass of water is not given in the question.

@user600016 I thought something like that might have happened

I am not obviously not talking about that data to be given in the question, as @Rahul Verma said, you are given 1 molal urea solution, and if you do the math you can find out that there's one mole of urea given and the rest of 1kg (1060g-(60*1)g) is water.

@Rahul Verma please recheck what you said, I agree its a 1 molal solution, but number of moles of urea is 1, because its a solution and not just urea that weighs 1060g.

@SirArthur7: yes, you're right. This was tricky! Btw, all the best for Mains

While considering colligative properties we always assume solute to be non-volatile. So, it doesn't contribute to the vapor. The amount of vapor will only be due to solvent particles.
So, in the problem we can use the following relation:
$$ \Delta T = K_\mathrm b\cdot m $$
Initially $ \Delta T  = 0.53\ \mathrm{^\circ C}$ so plugin the value of $K_\mathrm b$ and you end up with the value of initial molality of solvent as $1$. From this you can find the initial moles of solute which remain constant throughout.
Next you put $ \Delta T  = 2.12\ \mathrm{^\circ C}$ and plug in the known values of $K_\mathrm b$ and 'moles of solute' in the above formula to get the final value of solvent amount that remains in the solution.
Subtracting the final amount of solvent left from the initial amount will let you know the amount of water vaporized which is $750\ \mathrm g$.

That solution gives 13.25 Kg. Just tried it again


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Can you dissolve iodine without potassium iodide?
I don't have a sufficient amount of potassium iodide, so I wanted to know if there is other way which I can use to dissolve iodine without using potassium iodide.

It depends on the use of the solution  afterwards

Dissolving in what solvent? Are you constrained to use only water?

Aside of dissolving iodine in ethanol, suggested in the other answer, another way is to reduce a part of iodine to iodide, e.g. by the classical reaction with thiosulphate. The rest of iodine will dissolve.
The needed amount of thiosulphate is at least 1/3 of the stoichiometric amount wrt the total iodine.
$$\ce{2 S2O3^2- + I2 -> S4O6^2- + 2 I-}$$
$$\ce{I- + I2 -> I3-}$$
As 2 mols of thiosulphate reacts with 1 mol of molecular iodine, there is needed at least 2 mol of thiosulphate per 3 mols of iodine.
$$\ce{2 Na2S2O3 + 3 I2 -> Na2S4O6 + 2 NaI3}$$
Or, if the target concentration is low enough, iodine need not iodide to dissolve itself in water.
Wikipedia

Elemental iodine is slightly soluble in water, with one gram dissolving in 3450 ml at 20 °C and 1280 ml at 50 °C

Dissolve it in ethanol. Iodine I2 dissolves in ethanol to give a very well known solution sometimes called "tincture of iodine"1

I think prior to posting the answer it would be better to first ask OP in the comments whether changing the solvent would be a viable option for them. From the context it appears they want to use aqueous solution. Also, please don't forget to end the sentences with a full-stop (or other applicable punctuation marks depending on the situation).

HE SAID WITHOUT KI, doesnt that imply change?

No, it doesn't. Please don't CAPS at anyone, this isn't polite.

Okay @andselisk im sorry, just curious, what does it mean then, does he want to dissolve it in lesser KI or what?

The way I see it OP looks for an alternative to potassium iodide to promote dissolution of iodine in water. But as I said, things that are unclear or disputable should've been sorted out in the comments first.


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Chlorination of Propane
During monochlorination of propane why is chloropropane the major product formed. As far as I know using free radical mechanism 2-chloropropane is the supposed to be formed as the secondary free radical is more stable than a primary free radical. But the calculations involving reactivity order of chlorination tell otherwise. Why is this so!?

Possible duplicate: https://chemistry.stackexchange.com/questions/33199/predicting-the-products-of-the-radical-chlorination-of-propane

And what numbers do the calculations (done by whom?) tell you? And you have noticed the basic statistics of this reaction, right?

Related: Why does radical chlorination and bromination of propane occur at different positions?

The reason might be one of kinetic versus thermodynamic control.  The actual halogenation step is stage 2 in this scheme:

(reaction scheme from chem.libretexts.org)
The 2-radical might well be more stable, but that doesn't mean that it will react faster than the 1-radical.  If the reaction rate constant for the 1-radical with chlorine is much faster then the 1-chloro- product will predominate.
More likely, though is that it is that the 1-radical is formed in preference over the 2-radical in step 1. The halogen can attack at one of three positions, so statistically speaking you'd expect more 1-radical over 2-radical intermediates.  If this species then reacts immediately in step 2 without rearrangement then you'd get a predominance of the 1-product formed.
(Thanks to Karl for pointing this out).

The stability of the radical is irrelevant in the mechanism you show. It only plays a role if there is an (intra or intermolecular) equillibrium between the two radicals (which doesn´t seem to be the case). And no, "sterical hindrance" is not given here. You have a methyl:methylene ratio of 6:2, thats what makes the difference! If anything, sterics make the recombination of two methylene radicals is less probable, which would increase the yield of sec. brompropane.

Seems like you should write an answer to the question.

I wanted to leave that as an exercise for you. ;) Happy new year!

And you!  So it's a branching ratio issue?  The methyl-methylene branching ratio ratio is 6:2 in step 1?

Thats the first and simplest explanation (see also the other answer below), and it fits the bill, so why look any further. All other possible effects, where the radical stability plays a role, make the sec. brompropane more likely, but that doesn´t seem to be the case.

You can see that I've updated my answer to reflect what you've said (and given you a mention)

Let us continue this discussion in chat.

When a Cl· atom is knocking a propane molecule, it may touch it at the end of the molecule, where there are 6 H atoms. Or it may touch it by the middle, where there are are only 2 H atoms. The probability that the collision happens with the end carbon is at least 3 times higher than the other possibility. So the secondary fragment issues from this collision may be more stable than the primary fragment. It is much less frequent. Statistically speaking, more 1-choropropane than 2-chloropropane will be produced.


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Why are larger alicyclic rings more flexible?
My lecturer said that the increasing flexibility of an alicyclic ring is an inherent property. I'm not entirely sure what they mean by flexibility. Number of feasible conformations? How far the bond angles can be away from non-ideal?
Is there an actual explanation for the increasing flexibility?

It's geometry and you can get a petty good intuition by building simple models of the molecules
Saturated carbon-carbon bonds have two key properties of relevance. The angle between a pair of bonds on a carbon prefer to point to the corners of a perfect tetrahedron (~109.5 ° between each bond) if there are no spatial or geometric constraints. Also, C-C bonds can rotate freely unless something else in the molecule prevents that rotation.
In small cyclic aliphatic molecules (cyclopropane to cyclopentane) there is no easy way to satisfy the tetrahedral preference. No combination of bond rotations can give C-C-C bond angles of 109.5 °. If you build skeletal models of the compounds with tetrahedral carbon atoms this is easy to see: cyclopropane is rigid and flat; cyclobutane can twist slightly away from planarity but can't get those bond angles close to the unstrained angle, cyclopropane has slightly more freedom but is still pretty rigid.
But, from cyclohexane up there are plenty of extra freedoms. Cyclohexane can twist into a "boat" or "chair" shape and those give essentially unstrained bond angles at the carbons (the angles would be 120° in a flat hexagon, but when the bonds can also twist this does not constrain anything as enough combinations of twists are available to give an unstrained result in contrast to any of the smaller cyclic compounds). Plus, the twists cost little energy so they can interconvert easily given the amount of energy available at room temperature. Again, this is easy to see with physical models.
Higher cyclic compounds have even more freedom because there are more bonds to twist to give the ideal bond angle at carbon. So the increasing flexibility is a product of the fact that bond rotation is easy to do in the absence of geometric constraints. The more bonds you have to twist, the larger the number of conformations of the molecule available to give unstrained angles.

The answer is simple.  You have a potentially much larger number of rotatable bonds between any two atoms.  Cyclopropane is very inflexible, simply because every atom is joined to every other by a single bond.  Each extra atom added to its ring gives you an extra pivot point about which the bonds can rotate.
So large rings tend to have very many pivot points, which means that instead of behaving like a rigid ring, the molecule behaves like a chain loop.  Does this make sense?


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Oxidation number of Oxygen in peroxides: -2 or -1?
My textbook says that the oxidation number of Oxygen in peroxides is -2, but when I searched up the internet, I found it to be -1. It could probably be a misprint in my book, but I am still not sure. Not only that, the oxidation number of Hydrogen in hydrides is said to be +1 in my book, when it's supposed to be, I believe, -1? Please help me out on this.
My book says:
" The oxidation number of oxygen in most of its compounds (except peroxides, superoxides and suboxides) is -2. In peroxides (e.g., H2O2, BaO2, etc.), the oxidation number of oxygen is -2.
The oxidation number of oxygen in superoxides (e.g., KO2) and suboxides (e.g., C3O2 ) depends upon the nature of the compound. Exceptions also occur when oxygen is attatched to a more electronegative atom. For example, in the compounds like OF2 and O2F2, the oxidation numbers of oxygen are +2 and +1 respectively."

Oxygen has an oxidation state of -1 in peroxides; Hydrogen has an oxidation state of -1 in saline hydrides like $\ce{NaH}$

The textbook might be referring to the charge of the peroxo group $\ce{O2^2-}$. I suggest adding the exact quote from the textbook alongside with a complete reference and examples of chemical substances.

Ok, I will quote what my book says, just a minute..

There used to be a definition which assigned these numbers, it has changed somewhat recently, see here: Gold Book. For some more see this question: Electronegativity Considerations in Assigning Oxidation States

I have added the exact quote from my book

So, should I follow what my textbook says? I guess not?

Seems like a typo in the book to me.  One slip of my fat fingers on my cell phone, and suddenly there are 9 planets again instead of 8.

It must be a typo (the book would otherwise contradict itself). The correct oxidation number in pedoxides (and superoxides) is -1


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Why doesn't Oxygen form coordinate covalent bonds?
We know that an Oxygen atom has two lone pairs. Why doesn't oxygen then form coordinate covalent bonds using those since it has nothing to lose?

Of course it does, why anyone would think otherwise?

Take a deep breath, and think about how that oxygen reaches your brain bound to the iron atoms of heme in the hemoglobin in the red blood cells.

https://chemistry.stackexchange.com/questions/67220/how-can-oxygen-have-three-bonds?rq=1

Oxygen very much does form bonds in which both electrons come from the oxygen atom.  Examples include:
1.
The $\ce{H3O^+}$ ion at the center of the solvated proton in aqueous acids, also available as salts of some of the strongest acids such as $\ce{(H3O)(ClO4)}$
2.
Carbon monoxide, with its triple rather than double bond.
3.
Ozone, in which the oxygen atoms form a bent chain and one of the three bonds must have both electrons coming from the middle atom.  The resulting charge separation makes ozone one of the few one-element substances with a nonzero dipole moment.

The number of complex compounds that are surrounded by water molecules is quite big , a few examples of water being a ligand include [Cu(H2O)4]2+ or [Co(H2O)6]2+.
Anyway oxygen is indeed less eager to donate its electrons because its electronegativity is considerably high, nitrogen for example works better as a ligand (actually the first complex compounds to be observed and classified were with NH3 as ligands).But the examples where O donates his electrons can be found quite readily.


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Conditions for choosing quantifying and qualifying ion in LC-MS/MS
In my PhD-work, we are trying to develop LC-MS/MS methods for targeted analysis. Usually, I determine my quantifying ion/fragment based on a few different criteria:

If there is a 13C internal standard of identical structure, stick with the same fragment for both.
If not, look for the lowest noise/highest area/lowest LOD
If two fragments are similar in the previous manner, I have a look if there are different levels of contamination in the method blanks

For most compounds, that works, but for a small subset I have the case where the method blanks are similar, but for the fragment with the lowest LOD, noise is highest and/or areas are lowest between the two fragments we analyze for.
Would you go for the fragment with the highest area, highest noise (assuming a large number is more robust against small changes) and the highest LOD, or the lowest one? Is there a best practice I can refer to? There is a 1-2 order of magnitude difference between the LODs, so I really don't want to misrepresent the data.
Thanks in advance!

The limit of detection (LOD) in these kinds of assays depends on both the noise and the signal.  Usually it is defined as lowest concentration analyzed such that the signal (peak height) is three standard deviations higher than the noise.  Exactly how to define "the noise" is a critical consideration -- many different sources provide different definitions and ideas.  The best measurement of "noise" is probably in a method blank at the exact RT window you are using for quantification.  Ideally the method blank would have a sample matrix identical to your real samples, but guaranteed to be missing the target compound.
It's a little tough to know from the information provided in your question how exactly you are calculating your LODs.  If for the moment we assume you are happy with how your LODs are computed, then I would pick the method with the lowest LOD.
It's also worth mentioning that a standard practice in targeted LC-MS/MS is to monitor two separate transitions per compound.  One of those is called the "quantifier" and it is this signal that is used to compute peak areas for quantification.  The other is called the "qualifier".  The ratio of the qualifier peak area to the quantifier peak area should be monitored for all analyses, and it should be approximately the same for all samples (that contain measuremable compound).  If this ratio deviates wildly, it's a sign that the quantifier ion may not be reliable for a particular sample.


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Finding molecular formula with limited information given
I have recently stumbled upon a question I can't seem to solve :

Chlorine can also be used to synthesise a bromine-containing oxoanion that is mainly used in continuous or batch dyeing processes. When chlorine is passed through an aqueous solution containing potassium hydroxide and potassium bromide, the bromide ions are oxidised to the bromine-containing oxoanion. Careful addition of aqueous lead nitrate to the resulting solution precipitates 4.17g of $PbCl_2$. When this precipitate is filtered off and the resulting filtrate is evaporated, 0.835g of a white crystalline solid is obtained. The white solid has the following composition by mass : $K$, 23.4%; $Br$, 47.8%; $O$, 28.8%. $\mathbb \ Determine\ the\ formula\ of\ the\ white\ crystalline\ solid.$

I attempted by first finding the empirical formula of the solid using by dividing the percentage compositions by their respective elements' mass number, and ended up getting $KBrO_3$ as the empirical formula, which would make my second step to find $n$, where the molecular formula of the white crystalline solid is $K_nBr_nO_{3n}$. However, how can I go about doing this without being given how many moles of the white crystalline solid was precipitated? I can't seem to construct the equations that could potentially lead me to the mole ratio between $PbCl_2$ and $K_nBr_nO_{3n}$. If possible, can anyone tell me how to get a better grip on forming equations in scenarios such as this? I have been struggling in this topic for a while and this is likely to be one of the key reasons why. Sorry for the lengthy question!

The short answer is that you can't. Not with the information given, at least.

How many moles is 4.17g of PbCl2?

@Waylander 0.014989 moles, but how do I get the relationship between PbCl2 and the white crystalline? I am really weak in this :(

@orthocresol I feel like there might be a chance? This is all the information given in the question. Assuming (i wonder how this is justifiable) all the KCl (when chlorine is oxidised) turns into PbCl2? I'm really not sure but is it really not possible?

The equation $$
\ce{n KBr + 6n KOH + 3n Cl2 -> K_nBr_nO_{3n} + 6n KCl + 3n H2O}$$ is fully consistent with the information in the question, regardless of the value of $n$. I'm pretty sure you need something more to find the value of $n$.

An alternative (maybe easier) way of looking at it is that the equation is $$
\ce{KBr + 6 KOH + 3 Cl2 -> $\frac{1}{n}$ K_nBr_nO_{3n} + 6 KCl + 3 H2O},$$ so no matter what the value of $n$ is, the same amount of $\ce{KCl}$ will be produced, which leads to the same amount of $\ce{PbCl2}$ being precipitated. Those two equations I've written are the same, just multiplied/divided through by $n$. Also, when you react $\ce{Pb(NO3)2}$ with $\ce{KCl}$, what happens to the $\ce{KNO3}$? Has it just vanished? This problem is problematic...

@orthocresol Thank you so much! How do I construct equations like that? Also, following this equation, I think I found a solution! Since the ratio of $PbCl_2$ : $K_nBr_nO_{3n}$ is 3:1, I can first divide 4.17 by Mr of $PbCl_2$ to find number of moles of $K_nBr_nO_{3n}$ formed by dividing moles of $PbCl_2$ by 3. I can then find the molar mass of $K_nBr_nO_{3n}$ by dividing the 0.835g by the respective number of moles! The answer for $n$ which I just computed happens to be 1! Is this a valid solution? Also, how do I get better at constructing equations? What is the metheodology behind it?

Where do you get this 3:1 ratio from? Using the first equation I wrote, $6n$ moles of $\ce{KCl}$ will react to give you $3n$ moles of $\ce{Pb(NO3)2}$, so the ratio is $3n:1$ which brings us back to square one.

@orthocresol I kind of lost it, did your comment originally have 2 equations? I think I saw it there..

@orthocresol If not, maybe I got it using the ratio of $Cl_2$ : $K_nBr_nO_{3n}$, I am not sure if this is valid

I edited out the second equation, which was the followup $$\ce{3n Pb(NO3)2 + 6n KCl -> 3n PbCl2 + 6n KNO3}.$$ But that doesn't change the fact that the ratio is still $3n:1$.

@orthocresol i see, this is troubling indeed :(

On the other hand, your fundamentals seem more sound than you describe them to be. This is more of a problem with the question and not quite a problem with your chemistry skills.

@orthocresol Thanks a lot for taking your time to give me a better understanding of the question which is really wordy!

Ionic solids like NaCl, KOH, KBr, PbCl2 and KBrO3 have ratios of individual elements, but the "molecule", if you could use that term to describe it, would be the whole crystal, because all the ions are attracted to nearby ions, ad infinitum.
The white solid could be written as "K23.4% Br47.8% O28.8%", and every time this experiment is run, the same product with the same composition will be obtained. And it will behave exactly the same as the commercial product called potassium bromate with a label saying KBrO3. We prefer the simpler formula.
The numbers given for the weight of PbCl2 and the product KBrO3 do not define a specific reaction, because other reaction products and excess reactants are not specified. And to be realistic, it seems that the filtrate was not evaporated to dryness, because all these other components are not obtained. It seems that this question is posed not just to get you to divide 3 percentages by 3 atomic weights, but to decide what numbers are important to answering the question. The mole ratio between PbCl2 and KBrO3 is unimportant (it's 3) - you got the answer at KBrO3 and should stop there.
Remember this riddle: What has 4 legs and a tail, barks like a dog and concrete?
Answer: A dog. (I just threw in the concrete to make it hard.)
A common test technique of teachers is to throw in extra information to see if you know how to select the correct information to use.

We can't always simplify formulae of ionic solids; as a counterexample, consider $\ce{K2S2O8}$ which we don't write as $\ce{KSO4}$, or $\ce{Na2O2}$, or $\ce{Hg2Cl2}$. Obviously, with chemical intuition, we know that the bromate ion is indeed $\ce{BrO3-}$ and not $\ce{Br2O6^2-}$, $\ce{Br3O9^3-}$, or any other possibility, and so we write $\ce{KBrO3}$. But no information in the question allows us to deduce this.

@ orthocresol: The question was:  ℎ   ℎ ℎ  . (not the ions). There are 3 common definitions for (chemical) formula: the empirical formula, the molecular formula and structural formula. (Wikipedia) Structural formula (crystal structure) and molecular formula are not what was meant (since the compound does not consist of molecules). Wikipedia: In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound. The  only correct answer must be KBrO3.


text:
Unexpectedly low melting point of Aluminium
According to Wikipedia, the melting point of aluminium is 933.47K (660 ºC) while the melting point of magnesium is 923K (650 ºC), yet the melting point of sodium is merely 370.87K (98 ºC).
Huge difference between sodium and magnesium is expected, but small difference between magnesium and aluminium confused me.
Can anyone explain? Much appreciated

Hint: Melting points of bulk solids are mostly dependent on their Crystal structure and are not intrinsic properties of the atoms themselves.

@AniruddhaDeb would you mind further explaining how structure affects the mp?

Hint: Why do solids have a structure? What is between the atoms in e.g. solid Al, or a diamond?

@Karl I understand that Bonding (or attractive force) are existed between the atoms, like Giant covalent structure of diamond gave it the property of high melting point. However, the question here is why Al has so little difference with Mg(with merely 10K), yet so large difference between Mg and Na

Due to my insufficient knowledge, I only know these 3 are all in giant metallic structures, with different degrees of ionization which indeed show the trend of increasing mp. I tried to find the crystal structure, while Mg is simple hexagonal, and Al is face-centred cubic, which I couldn't understand how these structures help to explain this odd feature that I mentioned earlier.

Bonds is the term we are looking for. Bonds are made of ... ?

I m sorry I cannot understand, are you saying electrostatic attractive force between opposite charged particles (positive metal cation and negative electron in metallic structure)

Aluminium, Magnesium and Sodium form Metallic Bonds instead of Covalent bonds (like diamond, as discussed). The metallic bond strength depends on:

The number of electrons that become delocalized from the metal
The charge of the cation (metal).
The size of the cation.

Sodium has One delocalized electron as compared to Magnesium(2) and Aluminium(3). It's cation is also larger than Magnesium and Aluminium cations and also has a lower charge. This is why the strength of the metallic bond is lower in $\ce{Na}$ than it is in $\ce{Mg}$ and $\ce{Al}$ and hence the melting point is lower.
This also reflects in their crystal structure: $\ce{Na}$ has a body-centered cubic crystal with 68% packing efficiency whereas $\ce{Mg}$ and $\ce{Al}$ have a HCP and CCP structure respectively. Both these structures have a packing efficiency of 74%. This means that for a given unit volume, $\ce{Mg}$ and $\ce{Al}$ have more atoms in it as compared to $\ce{Na}$. This implies stronger bonds and  a higher melting point.


text:
Enthalpy of reaction between ammonia and oxygen
I have to ask this here because I asked literally everybody I know and I still disagree with them.

Using the information in the table below, calculate the heat of released from the following reaction in $\pu{kJ}$ per $\pu{g}$ of $\ce{NO}.$ (Mw. of $\ce{NO}$ is $\pu{30.01 g/mol})$
$$\ce{4 NH3(g) + 5 O2(g) -> 6 H2O(g) + 4 NO(g)}$$
$$
\begin{array}{lrrr}
\hline
  & \ce{NH3} & \ce{H2O} & \ce{NO} \\
\hline
H_\mathrm{f}^\circ~(\pu{kJ/mol}) & -46.1 & -241.8 & +90.3 \\
\hline
\end{array}
$$
a) $-13.29$
b) $-8.372$
c) $-7.541$
d) $-30.16$
e) $-5.678$

I'm convinced the answer is c.
The enthalpy of this reaction for 4 moles of NO is -905.2, to find the enthalpy for 1 mole of NO we have to divide by 4 to get to -226.3.
Then, to find it per gram of NO we have to divide by 30.01, this gives us 7.541.
Am I right ? If not, what did I do that was wrong? Where's my mistake?
Please don't put this as HW help, I asked multiple websites and went through my working 5-6 times but I still couldn't point out where I went wrong.

It seems to me that c is correct.

sorry to interrupt, do you 3 people agree its C ?

Yes, I agree with C

I also agree with C.

How did you know three people agree before the three people commented? Or were comments deleted?

The heat of reaction of a mixture is defined as the extent of reaction, $\xi$ times the enthaply of reaction, $\Delta H_{rxn}$,
$$Q_{rxn} =\xi\Delta H_{rxn} = \sum^n_{i=1}\nu_iH_{f,i}^{o} =\sum^{n,products}_{i=1}\|\nu_i\|H_{f,i}^{o} - \sum^{m,reactants}_{i=1}\|\nu_i\|H_{f,i}^{o} $$
For a reaction system,
$$\ce{4NH_3(g) + 5O_2(g) -> 6H_2O(g) + 4NO(g)}$$
In the case of the example you've provided, you are assuming this reaction is taking place at standard conditions. If we assume for a stiociometic amount of reactants, then the extend of reaction, $\xi$, should be equal to 1. Therefore,
$$Q_{rxn} = 1*\Delta H_{rxn} = 6\Delta H^o_{f,H_2O} + 4\Delta H^o_{f,NO} - 5\Delta H^o_{f,O_2} - 4\Delta H^o_{f,NH_3}$$
Since, $\Delta H^o_{f,O_2} = 0 \pu{kJ//mol}$, the equation reduces to,
$$\Delta H_{rxn} = 6\pu{mol}(-241.8 \pu{kJ//mol}) + 4\pu{mol}(90.3 \pu{kJ//mol}) - 4\pu{mol}(-46.1 \pu{kJ//mol}) $$
$$\Delta H_{rxn}= -905.2\pu{kJ}$$
Now, per gram of $\ce{NO}$, assuming the basis of 4 mol $\ce{NO}$ from the reaction equation,
$$\frac{Q_{rxn}}{m_{NO}} = \frac{\Delta H_{rxn}}{m_{NO}} = \frac{-905.2\pu{kJ}}{4_{mol, NO}(30.01\pu{g//mol})}=-7.541\pu{kJ//g_{,NO}}$$

You've used the wrong units and it confused me. $$\Delta H_{rxn}= -905.2\frac{kJ}{mol}$$ The enthaply of reaction is just regular enthalpy change with units $\pu{kJ}$. Your unit had per mole in it. I ask you, per mole of what substance? I'm 100 percent sure that the OP's friends did the exact same mistake. It is the enthaply of formation, combustion etc. which have units define for 1 mole the product. Enthalpy of reaction has units $\pu{kJ}$

Also, when you multiplied $\Delta H_f^\circ$ (which has unit $\pu{kJ/mol}$) by the respective number of moles of the substance the unit $\pu{mol}$ got cancelled. Example: $$6\Delta H^o_{f,H_2O} = 6(-241.8 \frac{kJ}{mol})$$ you should have written the unit $\pu{mol}$ with the number $6$. Which would've lead to $$\pu{mol}\times\pu{\frac{kJ}{mol}} = \pu{kJ}$$


text:
How is levulinic acid formed from sucrose?
I came across the following question:
Sucrose on treatment with conc. $\ce {HCl}$ produces?
The answer is given as levulinic acid.
So how exactly does sucrose produce levulinic acid?

Google is your friend! Search "Levulinic acid from Sucrose" and you will find.

@Waylander I found this: http://www.orgsyn.org/content/figures/CV1P0335.gif but I've always learned that $\ce {C_{12}H_{22}O_{11} + H_2O + HCl->}$ D-glucose + D-fructose. How could I've understood that the question demanded levulinic acid?

Levulinic acid can be produced from both fructose or glucose (they can isomerise into each other) https://en.wikipedia.org/wiki/Levulinic_acid,  https://pubs.rsc.org/en/content/articlehtml/2018/nj/c7nj03146g

The internet links provided in the Comments do not give much detail as to the stepwise mechanism for the transformation of fructose 1 to levulinic acid 12. I presume that you recognize that the sucrose acetal linkage is hydrolyzed in aqueous acid and that the resultant (+)-D-glucose and (-)-D-fructose are interconvertible via enolization of the carbonyl group. As a monomer, fructose 1 prefers the pyranose closed form over the furanose form 2. Facile loss of water from 2 forms cation 3 that, via the enol 4, affords aldehyde 5. Acid-catalyzed loss of two equivalents of water (steps not shown but straightforward) from 5 leads to 5-hydroxymethylfurfural (6). Cation 7 is formed  and is hydrated to provide 8. Hydration of the formyl group forms 9 followed by loss of formic acid affording 2-hydroxy-5-methylfuran 10. 2-Hydroxyfurans readily tautomerize to the keto form, in this instance to the enol lactone 11. Hydrolysis of 11 affords levulinic acid 12. I presume that the name levulinic acid arises because it is derived from (-)-fructose, aka, levulose, a name derived from the sign of its optical rotation. You may want to compare the similarities of this mechanism with the one offered by user55119 here.


text:
What is the physical property used in choosing the separation techniques for CaCO3 and KCl and NH4Cl
I've been scrolling through a chemistry past question but when I came across this I tried to solve it but it keeps confusing me

Please write your question in the Body box. Your question should be clear even without the title. (Especially, do not write your whole question in the title and then put “title says it all” in the body.)

$\ce{CaCO3}$ is not soluble in water. Both $\ce{KCl}$ and $\ce{NH4Cl}$ are soluble in water. When heated $\ce{NH4Cl}$ sublimes at 335 °C. At this temperature, $\ce{CaCO3}$ and $\ce{KCl}$ are not modified. $\ce{CaCO3}$ is decomposed in $\ce{CaO}$ and $\ce{CO2}$ at 825 °C.


text:
how to compute enthalpy of combustion for methanoic acid
My data source gives the following values for various enthalpies of formation and combustion:
formation of CO2 = -393.5
formation of HCOOH = -425
combustion of HCOOH = -254.3
Wikipedia gives liquid water's enthalpy of formation as -285.
All units are kJ/mol.
I guess the reaction is $4 \mathrm{HCOOH} + \mathrm{O}_2 \to 4\mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}$
LHS enthalpy of formation is $4(-425) = -1700$.
RHS enthalpy of formation is $4(-394) + 2(-285) = -2146$.
So, enthalpy of combusion should be $(1700-2146)/4 = -112$.
Not -254!
Can somebody please explain what I doing wrong?

By the way, methanoic acid is usually referred to as formic acid. Although the IUPAC name is methanoic acid, I have never seen anyone call it that (just like water is not referred to as dihydrogen monoxide)

@ELIJONES Although the systematic name “methanoic acid” is correct, the preferred IUPAC name according to current IUPAC recommendations is still the retained name “formic acid” (see Rule P-65.1.1.1 in the Blue Book).

$$\ce{4 HCOOH + O2 -> 4 CO2 + 2 H2O}$$

Your reaction equation is wrong. Just count the $\ce{H}$'s on both sides.
The actual equation is
$$\ce{2 CH2O2 + O2 -> 2 CO2 + 2 H2O}$$
which results in the correct enthalpy of combustion.
$$\begin{align}-2\times-393.5\ \mathrm{kJ/mol}-2\times-285.8\ \mathrm{kJ/mol}
+2\times-425.0\ \mathrm{kJ/mol}&=508.6\ \mathrm{kJ/mol}\\
&=2\times254.3\ \mathrm{kJ/mol}\end{align}$$


text:
Plausible Mechanisms for Polyketide Formation from Phloroglucinol
I am working on modifying the reaction below. I am adding the sulfonate (ignore the sulfonic acid groups shown) groups so that polysubstitution of the phloroglucinol does not occur. In terms of the reactivity of the sulfonate, there should not be any issue, as it is not acidic and will not interfere with the potassium hydroxide in the first step of the original reaction; however, I am worried about how the electron withdrawing properties of the sulfonate will affect the overall reaction. The paper does not attempt to explain the reaction mechanism of this particular step, so I am not sure how exactly this form of substitution will affect my reaction. My only guess as of now is that the keto-enol tautomerization is a stabilizing factor of the final product, rendering substitution of the -OH groups unfavorable; however, this still does not predict how this reaction would be affected by sulfonate groups?

Edit: New possible mechanism for reaction

I have really never seen any reference to a reaction mechanism that involves the dearomatization of aryl moities like this (besides the Birch reduction, which is most likely not similar to this mechanism since it involves free radicals), so I'm not sure where to start with this analysis.

The most that I could find on this reaction was that it is known as a C-alkylation reaction

It's the same mechanism as electrophilic aromatic substitution, just without the deprotonation at the end, simply because there's no proton to remove.

Why would the reaction favor the formation of the polyketide in the original reaction and not the substituted phloroglucinol? The directing effect of the -OH groups would even seem to favor this. It could be that the polyketide is just very stable, even more than its aromatic counterparts, but I'm not very sure of this. Also, why is there no Lewis-acid catalyst if this follows a similar mechanism to electrophilic aromatic substitution?

(1) the question about regioselectivity is good. I am not entirely sure myself. (2) You probably know that EAS on electron-rich rings is easy. Phenol and aniline both react with bromine, without any need for Lewis acid catalysis. This ring is at least 3 times more activated than phenol.

I have added a new possible mechanism that is based on the electrophilic aromatic substitution reaction and a keto-enol tautomerization. This would explain the choice of reagents for the reaction, sodium hydroxide followed by HCl treatment and also the long reaction time; however, this mechanism would mean that my modified reaction would not work, as there will be no hydrogen that can be removed in keto-enol tautomerization. As for the regioselectivity of the reaction, my guess is that the product is probably thermodynamically favored, not kinetically favored given the reaction time.

I may be able to modify the procedure given in the link to have a shorter reaction time, this may favor the formation of the monosubstituted product without the need for sulfonation; however, it could also mean there will be a mixture of polyketide and substituted phloroglucinol.

In your new possible mechanism with the methylated phloroglucinol, you would need base to deprotonated the OH group(s) and then most likely you obtain O-methylation. Methyl bromide is not nucleophilic enough to react without deprotonation.

I have proposed a new mechanism that is consistent with the low-nucleophilicity of alkyl bromides. This may also explain the regioselectivity of reaction towards the polyketide, as the deprotonation of the -OH group allows for the direct formation of the product in a single concerted mechanism. The product would then undergo keto-enol tautomerization to yield the most stable tautomer. This would fully explain the choice of reagents by the authors of the paper. Thank you for pointing that out!

If this mechanism is most consistent, that would mean that my original proposal of sulfonation as a type of blocking group would theoretically work, although I am concerned about a nucleophilic attack at the C-sulfonate bond instead of the C-alkyl bond, which would ruin the point of the sulfonation in the first place.

This paper involves C-alkylation of phloroglucinol with allylic and benzylic halides: https://reader.elsevier.com/reader/sd/pii/S0040402004008853?token=71D3FBC2A32DF5CDCBB98D6EBB9DEAB36539FFD37E73202A1D0B867D81B95F214BC6E8365ACB1DBCE5CCF1054115060B


text:
Is it possible to obtain the amount of air required for a combustion analysis?
I'm not very sure if the required amount of oxygen (I assume) can be found by only knowing the amounts of the produced subtances after a combustion.
To illustrate this I found this problem:

$1.14\,kg$ of octane $C_{8}H_{18}$ is burned with a certain amount of
  air. The resulting products from the combustion are known to have the
  following percentages by volume: $CO_{2} = 41.18\%$ ; $CO = 5.8%$ ;
  $H_2O_{(vapor)} = 52.94\%$. Find the weight of required air in
  kilograms. Assume the composition of the air in this combustion is
  $21\%$ of $O_{2}$ and $79\%$ of $N_{2}$ by volume and its average
  molecular weight is $28.8\,\frac{g}{mol}$.

The alternatives given are as follows:
$\begin{array}{ll}
1.&9.0\,kg\\
2.&16.4\,kg\\
3.&12.5\,kg\\
4.&45.1\,kg\\
5.&120\,kg\\
\end{array}$
What I think to approach this problem was to use the given percentages to get the number of moles using the initial moles of octane.
Using these moles I can obtain the grams of oxygen produced. Since it is given the percentage which is of oxygen in the air I could use this to find the required mass of the air to get that amount of air.
This is shown as follows: (for the sake of brevity I'm ommiting units but they are consistent.)
$\textrm{FW of octane} = 114$
$n_octane=\frac{1140}{114}=10$
Then moles of each product:
For $CO_{2} = 41.18\%$
$\frac{41.18}{100}\times 10 = 4.118$
Then in this amount of $CO_{2}$ there must be these grams of oxygen:
$4.118 \times \frac{32\,g\O}{1\,mol\,CO_{2}}= 131.776 \,g$
For $CO = 5.8\%$
$\frac{5.8}{100}\times 10 = 0.058$
$0.058 \times \frac{16\,g\O}{1\,mol\,CO}= 0.928 \,g$
For $H_2O_{(vapor)} = 52.94\%$.
$\frac{52.94}{100}\times 10 = 5.294$
$5.294 \times \frac{16\,g\O}{1\,mol\,H_2O}= 84.704 \,g$
Adding these together:
$131.776+0.928+84.704=217.408\,g\,O$
Then using the known percentage of oxygen which is $21\%$
$\frac{21}{100}x=217.408$
$x=1035.28 \,g$
Which is approximately 1.035 kg. but this answer doesn't get any close to the specified alternatives. Did I overlooked something. Or could it be that this approach is not applicable?. In this given situation is it possible to find the required mass of air?.

For the given problem there isn't much $\ce{CO}$ created. Given that the question is multiple choice an exact amount isn't needed. The problem only needs to be solve close enough to determine which choice is the answer. So assume complete combustion and then guess that the answer will be about 5% high. If octane, $\ce{C8H18}$, is completely burned with oxygen then the reaction is:
$$\ce{C8H18 + 12.5O2 -> 8CO2 + 9H2O}$$
since the problem states that 1.14 kg of octane was burned, the number of moles is:
$$\dfrac{1140}{114} = 10.0$$
Thus for complete oxidation:

125 moles of oxygen would be needed.
Air is only 21% oxygen so 125/0.21 = 595 "moles" of air are needed.
At 28.8 g/"mole" that would be 17.1 kg of air.

In reality there is about 5% less oxygen required since there is some $\ce{CO}$ produced, so $17.1*0.95 = 16.3\ \pu{ kg}$ is a good guess.

I'll leave finding the exact solution to the OP.

The reason why i did not attempted to use a complete combustion equation was due the fact that there appears other products and im referring to carbon monoxide and who knows what else as the sum of percentages given isnt $100%$. This confusion is tied to how did you obtained that there is $5%$ less oxygen required. Whats the justification for this? In the problem states carbon monoxide is $5.8%$ is it because of this? Why was i given the average molecular weight? It was not needed right?

But could be found  by mutiplying the molar fraction to their respective formula weights and adding these quantities together. But again this information was unused. So as the percentage of nytrogen.

@ChrisSteinbeckBell - Given $CO_{2} = 41.18%$ ; $CO = 5.8%$  and assuming ideal gas behavior, then normalize on the carbon atom to obtain the equivalent of 8 carbons.

@ChrisSteinbeckBell - The 5% correction was just a guess. The $\ce{CO}$ is about 10% of the $\ce{CO2}$ and contains 1/2 the oxygen. So about a 5% correction. // I used a slide rule for about a decade. You have to estimate calculations since 2.3, 23 and 230 are all on the same spot on the slide rule. Given the answers in this case just assuming complete combustion make the problem simple and gives an answer close enough to choose between the multiple choice answers.

@ChrisSteinbeckBell - $8 * (5.8/(5.8+41.18)) = 0.9877$ atoms of $\ce{CO}$ per molecule of octane. Thus there must be 7.0123 atoms of $\ce{CO2}$. I'll leave it to you to figure out how many oxygen atoms are needed.


text:
Does a halonium ion for astatine exist?
I've just begun reading about halonium ions and I noted that only the first four halogens are acknowledged to be able to form halonium ions, but hydrogen astatide can be formed.
I can't seem to find any answers on the internet, but knowing that astatine is super reactive and radioactive, I suppose it may be due to this that forming an "astatonium" molecule would be nearly impossible?
On the Wikipedia entry of hydrogen astatide, it states that there is an astatonium — listed as a conjugate acid. However, a quick Google search of the term astatonium only reveals 14 results.

Maybe it could form such ions, but the fact that the half life of astatine is 8 hours makes it very hard to handle and to source, so it is not surprising that it's not a very studied halogen, from a chemist's point of view.
Anyway, a computational chemist with some spare time could compare the energy involved in the formation of such ions to that of the other halogens, and predict if that would be thermodinamically possible.
My guess is that it could do it: less polarizable atoms like F and Cl are the big deal, but Br and I (and At?) can acquire a positive charge quite easily

Related: https://chemistry.stackexchange.com/questions/75292/is-there-a-haloalkane-for-astatine/75296?r=SearchResults#75296

And https://chemistry.stackexchange.com/questions/48354/which-is-more-electronegative-hydrogen-or-astatine/91601?r=SearchResults#91601

No astatonium ion with hydrogen is known, but it appears that a cation $\ce{S7At^+}$ is formed by replacing a sulfur atom in $\ce{S8}$ with an isoelectronic $\ce{At^+}$ moiety (the actual oxidation state of astatine, with sulfur being more electronegative, is $+3$).  The ion is proposed to explain the incorporation of astatine into the sulfur precipitate when a sodium thiosulfate solution is decomposed by acid in the presence of astatide ion.
Primary reference:
G. M. W. Visser, E. L. Diemer, Radiochimica Acta 33 145 (1983).


text:
Acetylacetonato or Acetylacetonate
I'm currently working on a project using Iron(III) acetylacetonate, which in the wikipedia page it's written Tris(acetylacetonato)iron(III), and it says it's the IUPAC name, maybe it's obvious but I'm a little confused about the nomenclature, why acetylacetonato instead of acetylacetonate, couldn't it be Tris(acetylacetonate)iron(III)? Is one name better than the other?

In IUPAC nomenclature, anionic ligands generally end in -o (-ido, -ito or -ato). See this for reference.

Both can be used. See: https://chemistry.stackexchange.com/questions/5039/where-did-the-acac-come-from-in-cracac%e2%82%83-synthesis-by-crcl%e2%82%83-reagent/5040#5040

It should be used interchangeably otherwise in Italian and probably Spanish we would miss a class. - ate already sounds - ato and I don't know what - ato ENG change into IT

Thank you, I knew it was simple, I guess it's a problem of having studied chemistry in a different language, it wasn't obvious for me. I understand then that both are correct, being the latter the normative IUPAC name, and the first one the easy, common name.


text:
Is acidic character related to reducing power?
My book says that the acidic character of halogen acids increases on going down the group because the bond strength decreases, making it easier to release the $\ce{H+}$.
But I also know that $\ce{HI}$ (considering the last element) is a strong reducing agent. Is there any way to explain this?
Won't "release of $\ce{H+}$" be equivalent to "accepting electron" (get reduced, i.e. act as oxidizing agent)? Am I wrong in saying so? Any help?

An Acidic compound donates protons (Lowry Brönsted definition)
$$\ce{HI -> H+ + I-}$$
By releasing $\ce{H+}$, you need to note that the compound as a whole releases the proton. A reducing/oxidizing agent donates/accepts electrons into it's own shells, differentiating it from an acid.
$$\ce{2HI + 2OH- -> 2H2O + I2 + 2e-}$$
Here, $\ce{I-}$ donates two electrons to form $\ce{I2}$, thereby being a reducing agent.

Why do we, then, call HI a strong reducing agent?

Because of the second reaction I mentioned: it has the ability to donate electrons rather easily and form $\ce{I2}$

I doubt that HI may react with OH- to produce I2 as written by Aniruddha Deb.Most probably HI and OH- get neutralized and yield H2O and I-.

@Arnav Das It is rather a weak reducing agent. The reduction is not because HI is acidic, but because of I- being able to be oxidized to I2.

No. Acidic character is not related to reducing power. For example, nitric acid $\ce{HNO3}$ has no reducing properties. It is a strong acid and a strong oxidant, able to carry out the oxidation of copper metal $\ce{Cu}$ to $\ce{Cu^{2+}}$ ion.

Acidity as $\mathrm{pH}$ does affect many redox potentials.
But the compound acidic character itself has no direct relation to reducing power.
E.g. superacids like $\ce{HSO3F}$  or $\ce{H2F+ + SbF6-}$ have practically no reducing power. The former probably just against free fluorine or fluorinating compounds like $\ce{ClF3}$, the latter none at all.


text:
Carcinogenicity of polynuclear hydrocarbons and benzene rings

(1933) established that polynuclear aromatic hydrocarbons (PAHs) were
carcinogenic components of pitch. ... Although less is known about
their health effects than about those of PAHs, it is clear that some
nitro-PAHs are potent mutagens and carcinogens.(1)
Benzene and polynuclear hydrocarbons containing more than two benzene
rings fused together are toxic and said to possess cancer producing
(carcinogenic) property. Such polynuclear hydrocarbons are formed on
incomplete combustion of organic materials like tobacco, coal and
petroleum.(2)

From the two sources cited above, I keep seeing that polybenzenoid compounds like 1,2-benzanthracene, 3-methylcholanthrene and 1,2- benzpyrene are mentioned to be carcinogens.

In general, why are polynuclear compounds carcinogenic? I google searched on this topic and all I could find is their effects and concentrations in various substances.
My questions are:

Why are they carcinogenic

Is there a reason for a given organic compound to be carcinogenic?

References

Potential Carcinogenic Effects of Polynuclear Aromatic Hydrocarbons and Nitroaromatics in Mobile Source Emissions, STEPHEN S.HECHT(link)
pg403, NCERT Class XI part 2(pdf link)

Search the term intercalation of DNA by polyaromatics. Not all organic compounds are carcinogens. Our whole body has thousands of organic compounds.

i meant for a given organic compound

There is no way to predict beforehand. Experiment decides it. One can make guesses from known classes of compounds by extrapolation. For example, if PAHs are bad, aromatic amines are also powerful carcinogens.

See https://chemistry.stackexchange.com/questions/125402/how-does-one-tell-or-conclude-if-a-substance-is-carcinogenic

i did not find that answer satisfactory, i do not a want an answer like experiment carried, result obtained, conclusion derived, reason unknown.

Related: https://chemistry.stackexchange.com/questions/95168/is-benzene-harmful-to-human-health/95202

https://chemistry.stackexchange.com/questions/39109/does-benzenes-resonance-structure-allow-it-to-enter-dna

and https://chemistry.stackexchange.com/questions/75590/what-component-of-the-human-dna-reacts-with-4-5-benzoapyrene-oxide-and-7-8-ben

@Haha HaHaHa Reality is often very unsatisfactory. That does not make an answer, pointing to such reality,  unsatisfactory as well.

@NilayGhosh thanks, a gem of an answer with a 130+ score of orthocresol right there, perfect.

yaa, @Poutnik thats the only sad thing about chemistry

Why are they carcinogenic?

From my point of view as a medicinal chemist, polycyclic aromatic hydrocarbons compounds tend to be metabolized by specific cytochrome P450 (CYP 450) isoforms to form highly electrophilic species, which are highly reactive and can react with the DNA of the cell and damage it. Specifically, they can react with the nucleophilic nitrogen in the DNA bases such as adenine.

Is there a reason for a given organic compound to be carcinogenic?

Mechanism for carcinogenicity can vary, in general, any compound damaging or altering the DNA can be carcinogenic. For example, highly reactive electrophiles (such as formaldehyde).
Reference

https://socratic.org/questions/how-do-polycyclic-aromatic-hydrocarbons-cause-cancer


text:
Directing effect in Intramolecular Friedel Crafts Alkylation

This reaction should cause ring closure at either the $o-$ or $p-$ position relative to the methyl substituent on the ring (as methyl is an $o,p-$directing group), giving the following products:

The question is: Which one of these is the major product? Is substitution preferred at the ortho position or para position, and why?
A few more examples of these types of questions:

This would give which of the following?

Also, I'm not too sure of the hydride shift you did in 2. won't a six-membered ring be more favoured? Just curious.

@HahaHahaha: six-membered ring formation would favorable only when it could form. I don't think that the unstable 1° carbocation would survive as intermediatre

@RahulVerma Do you have any references? I find it hard to digest the fact that 6 membered ring is not favored here.

Electron giving groups like $-OCH_3$ (by resonance) and $-CH_3$ (by hyperconjugation) are Ortho-Para directing.

But, in cases where there is no increased stability by H-bonding or other factors at ortho position, the para substituted product will give major yield.

This can simply be explained by stating that there is less steric hindernace on the para position compared to the ortho position. Which is further emphasized when you take this example:

So, the answer is para-substituted in both cases.
References
Aromatic reactivity

Reference for this?

There you go @Waylander


text:
Entropy as a state function independent of the path
When entropy is a state function and the entropy change for a reversible process is
$$\mathrm dS_\mathrm{rev} = \frac{\delta Q_\mathrm{rev}}{T} = 0,$$
how can an irreversible process be calculated by the same formula? $\delta Q_\mathrm{irr}$ is not a state function, therefore $\mathrm dS_\mathrm{irr}$ cannot be a state function.
Another description I have found was that the irreversible path should be seen as divided in a infinite amount of reversible steps. But an infinite times zero is also zero.
Can you help me please?

But the entropy change for a reversible path is 0. How can a reversible path be not zero if this is the definition?
If State A has a entropy of 5 J/K and State B has a value of 20 J/K then we would need a entropy change of 15 J/K but a reversible heat transfer would result in 0 entropy change.

Entropy change of the universe is zero for the reversible process, but the entropy changes of both the system and surroundings can be nonzero, as long as equal magnitude and opposite in sign.

I took a liberty to improve formatting a little and tweaked the appearance of the symbols (please check whether the intended meaning is preserved). Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also, regarding "an infinite times zero is also zero", I wouldn't be so sure about that.

Thank you, andselisk. I will use the codes for the next thread.

Entropy is indeed a state function, and thus depends only on the state of the system.  Hence it doesn't matter how you get from state A to state B, the entropy change will be the same.  The analogy would be that it doesn't matter which path you use to get from the base of a mountain to the summit, your elevation change will be the same.  This is because altitude is a state function: your altitude depends only on how high you are (the state of your system), not how you got there.
But, since $dS= \frac{\text{đ}q_{rev}}{T}$, one can calculate the entropy change by integrating along a path connecting the two states, but that path must be reversible.  I.e., it doesn't matter if the way the system changed from A to B was reversible or irreversible.  $\Delta S$ will be the same.  However, to calculate $\Delta S$ for that change from A to B, you need to integrate along some reversible path that connects A to B.
Also, it is possible to sum the effects of an infinite number of infinitely small steps.  That's what happens when we do an integration.  [More precisely, in an integration we calculate the sum in the limit as the step size goes to zero and the number of steps goes to infinity.]  For a graphical representation of this, take a look at the graphics in Wikipedia's entry on Reimann sums (https://en.wikipedia.org/wiki/Riemann_sum), and imagine what happens in the limit as the bar width goes to zero.
The comments by Andrew, and Chet Miller, are both useful additions.  Further expanding on Andrew's comment: The system's net change isn't affected by the path.  This means that, to distinguish between the result of making a change in a system by a reversible vs. irreversible path, we need to look at differences in effects upon the surroundings.

Perhaps worth elaborating that the irreversible paths result in different amounts of entropy change of the surroundings even if the change to the system is the same as for the reversible path.

It might be worth mentioning that there are an infinite number of reversible paths between A and B, and all of them calculate exactly the same value for the entropy change.  And if you integrate dq/T for an irreversible path, it will come out lower than for all reversible paths.

Could it also be that we look at the surroundings which has not changed but the state of the system has changed. Therefore we know that it was an irreversible process. To be sure,if an reversible isothermal expansion (against piston) increases the systems volume and therefore the entropy of the system. But because the entropy of the surroundingsa change for the exact same amount we have 0 entropy change for the universe and know that this was a reversible process.

Attachment: the entropy of the surroundings DECREASED for the exact same amount.

@AnnaDapont It sounds like you've got it right in your comments: In a reversible process, $\Delta S_{sys}$ is equal and opposite to  $\Delta S_{surr}$ , so  $\Delta S_{sys}+\Delta S_{surr}=\Delta S_{univ}=0$ . If we make the same change in the system, but instead do it irreversibly, $\Delta S_{sys}$ is still the same (because S is a state function), but $\Delta S_{surr}$ is more positive (or less negative) than in the reversible case, thus causing $\Delta S_{univ}>0$. All real processes are irreversible, and thus have $\Delta S_{univ}>0$. [NB: sys=system, surr=surroundings, univ=universe]

I appreciate all your answers. Thank you. I am pretty sure I got it now.

@AnnaDapont Feel free to accept my answer if you find it addreses you question.


text:
Electrolysis of Baking Soda separated by semi permeable membrane
I have dissolved baking soda in battery water (de-ionized), in a glass beaker and used a food handling glove (vinyl) as a semi-permeable membrane to separate a graphite electrode from a copper electrode. This is for an experiment that will use the copper electrode for the reaction and not the graphite electrode. When I apply a voltage $(\pu{12 V})$ and use a multimeter to measure the voltage, I get roughly $\pu{11 V}$ passing through the vinyl glove, however, no bubbles are produced, but if I remove the glove and have both electrodes in the beaker, then loads of bubbles are produced. So, will my experiment still work if the voltage is passing through the glove and there are no bubbles?

The primary problem is too low permeability = too high resistence of the gloves, causing high voltage drop where it should be minimal one.
The gloves are not hydrophilic enough for electrolyte soaking to get low resistance.
The remaining voltage - after subtracting the glove voltage drop - is not high enough to cause electrolysis by significant current.
You may try to use a salt bridge, as a passthru vessel with porous glass ends, or a gel formed from an electrolyte solution.
You may also want to try a compartment created from a glued filtration paper, or other hydrophilic porous material, impregnated by an electrolyte, preferably sharing one of ions with both free solutions.
For the paper, be aware  of the same leveling on both compartments, otherwise. the paper will do its major purpose filtering. Eventually, more compact paper may be better, like the thick charts for paper painting.

Ok, thanks for your help, I'll try the salt bridge and see what happens and then I'll reply on how good it was or if there are any difficulties. Your answer is much appreciated, cheers.

I used zinc with the copper and a porous plastic milk bottle as the divider and got 12 volts through, no voltage drop. But, still no fizzing, thinking that baking soda might not cause fizzing, water fizzes when electrodes are in the same container due to water splitting. Your idea of gluing filtration paper in as the divider sounds good, I'll give that a try.

A porous plastic milk bottle seems to have an internal contradiction for me. If it is a single layer plastic, it is not porous and hydrophilic. If it is porous and hydrophilic, it is a multilayer bottle with a least one layer not to be porous and hydrophilic. If it is like a foamed polystyrene cup, it is hydrophobic and gas bubbles is rather isolated then interconnected, not really porous much.

Thank You Poutnik !! The filter paper worked !! Interestingly I used 12 volts but got 6.67 volts through (I used coffee filter paper), but I got fizzing at both ends of the filter paper, so I must use hydrophillic cell dividers, makes sense, a good thing to learn.
Děkuji moc za vaši pomoc Poutnik! Měj hezký den !

You are very welcome, and thanks for the translation. :-) If the answer satisfied your question, it is supposed to mark the answer as the accepted answer. It may be helpful for future searching for  related topics, as questions with accepted answers are marked differently.


text:
Lewis acidity comparison of boron and aluminium trichlorides
Lewis acidity is a function of electron deficiency on the acceptor atom, which in this case is $\ce{B}$ and $\ce{Al}$ respectively.
Back donation is better as is the case of $\ce{B}$, due to better 2p overlap compared to 3p overlap, then the electron deficiency on the $\ce{B}$ atom should be lesser, as it's deficiency is getting filled better; but then how is $\ce{BCl3}$₃ stronger acid than $\ce{AlCl3}$, or am I mistaken?
Three things to note

This question came in the IIT- joint entrance examination in 2017, this examination rarely has factual inconsistencies. And if they do, then they get the answer key changed. For this question, it was clearly given as The lewis acidity is indeed more for $\ce{BCl3}$. So any answers which conclude it has grey zone won't be accepted unless a source is cited which states so while giving an example of the solvent dependency.
I want a clear clarification of why my reasoning is giving the incorrect answer
Please ensure the answer has citations

Related:https://chemistry.stackexchange.com/questions/80263/acidity-of-aluminium-trihalides/84484

@NilayGhosh they are two totally different questions. The one you referred to is comparing halides of boron and Al, I want to compare one specific halide between Al and B

As you are probably aware, generalizations like "$\ce{BCl3}$ is a stronger Lewis acid than $\ce{AlCl3}$" can be problematic, as the results can be dependent on the base used and the conditions (eg solvent choice).
That said, a common context for this ranking is with respect to carbonyl bases, such as in a Friedel-Crafts acylation. For these bases, $\ce{BCl3}$ is generally observed to be a stronger Lewis acid than $\ce{AlCl3}$.
To understand this observation, we need to consider what happens during the acid-base reaction. For $\ce{BCl3}$, it is likely that the dissolved molecule is a monomer and retains the trigonal planar geometry with some $\pi$ bonding occuring. In the reaction with base, the $\pi$ bonding must be broken and the geometry changed to tetrahedral. The driving force is the formation of the new $\ce{Cl3B-O=R}$ bond.
In contrast, $\ce{AlCl3}$ is most likely present in solution as the $\ce{Al2Cl6}$ dimer, with no $\pi$ bonding. Here, the oxygen of the base displaces the chloride in order to form the Al-O bond of $\ce{Cl3Al-O=R}$.
Although the B-O bond is stronger than the Al-O bond, it is difficult to predict the relative contributions of the changes in $\ce{BCl3}$ vs those in $\ce{Al2Cl6}$ because the reactions are so different, and we rely primarily on empirical results for ranking. However, some groups have attempted computational approaches. Some examples are:
Laszlo, P. and Preston, M. Determination of the Acidity of Lewis Acids. J Am Chem Soc (1990) 112:8750-8754.
Jonas, V., Frenking, G., and Reetz, M.T. Comparative Theoretical Study of Lewis Acid-Base Complexes of $\ce{BH3}$, $\ce{BF3}$, $\ce{BCl3}$, $\ce{AlCl3}$, and $\ce{SO2}$. J Am Chem Soc (1994) 116:8741-8753. DOI:10.1021/ja00098a037
UPDATE based on updated question:
For a specific counter example, there is a report that in a competition experiment in which excess chloride ion was added to a solution containing $\ce{BCl3}$ and $\ce{AlCl3}$, $\ce{AlCl4-}$ was observed to form, but no $\ce{BCl4-}$ was detected, indicating a stronger Lewis acidity of $\ce{AlCl3}$ towards the soft Lewis base chloride ion. I have seen mentions of similar trends with other soft Lewis bases, but haven't chased down the references.
Glavincevski, B. and Brownstein, S.K. Complexing and exchange of boron, aluminum, and gallium chlorides with some Lewis bases (1981) Can J Chem 59:3012-3015. DOI: 10.1139/v81-436
That paper also discusses some solvent effects, particularly acetonitrile, which as a Lewis base can form complexes with the acids.
The clear explanation of why your reasoning is incorrect is that you are not accounting for the stronger bond with B. As discussed in the references I gave, Al-O bonds are calculated to be nearly completely ionic, whereas those with B are partially covalent, so for the heterolytic cleavage reaction separating the acid from the base, the enthalpy change is greater for the boron complexes.
That explanation is also consistent with the trend reversal for soft Lewis bases, which will not form a strong covalent bond with boron.

Hey @andrew thats a good start, but could you edit your answer sccording to the guidelines you have added. Although your answer is pretty good

Well, but why shall I consider the $\ce{Al-O}$ or $\ce{B-O}$ bond strength, I want to compare lewis strength, isn't it purely a function of electron deficiency? So assuming by the medium  is not basic and the $\ce{M=O}$ bond is not formed, why is $\ce{BCl3}$ still stronger?

"Electron deficiency" is a vague qualitative concept. You need to consider both the initial state (which is what electron deficiency refers to) and the product state. In the reaction with an oxygen base, B is calculated to have a more covalent bond, so its "deficiency" is better satisfied than Al's, which is calculated to have a mostly ionic bond. With other bases, this may not be true. That's why rankings of Lewis acidity can only be done with respect to a specific base or class of bases (and a specified solvent or lack thereof).

I recommend reviewing the various approaches to quantifying Lewis acidity and basicity - eg Pearson HSAB, Drago EC, etc. They are only accurate if they include parameters from both the base and the acid, and they generally refer only to gas phase or completely unreactive solvents.

Okay that makes a lot of sense. So, in gas phase or completely unreactive solvents., is it safe to say $\ce{BCl3}$ is correct? I will accept your answer before the bounty ends unless there's a better one.

No, you seem to be missing the main point, which is that the order of acidity is dependent on the choice of base. Solvent also matters, but most rankings are assumed to refer to gas phase/unreactive solvent unless otherwise stated. Look at the papers I cited - $\ce{BCl3}$ is stronger with carbonyl bases, but $\ce{AlCl3}$ is stronger with $\ce{Cl-}$ as base.


text:
What's the purpose of an ice bath when preparing a nitration mix from KNO3?
Most preparations for nitrocellulose from $\ce{(KNO3 + H2SO4)}$ mixture require the use of an ice bath during the process of dissolving $\ce{KNO3}$ into $\ce{H2SO4}$. Last time I did this I omitted this step and the nitrocellulose turned out to be of very low quality, but there's a number of other possible causes I can think of. Therefore, for the sake of troubleshooting I have this question: What is the purpose of an ice bath during this step? Is it to just limit the fuming, or does omitting it lead to decomposition of $\ce{HNO3}$ significant enough to impact the yield? I did not see any brown $\ce{NO2}$ fumes coming out, just the typical white mist of acid.

What generally can happen if you run an (exothermic) reaction at a higher starting temperature?

It is an exothermic reaction and generates a lot of heat.

It's not very exothermic, I imagine I'd have to dump all KNO3 at once to see it runaway. So it's just an additional safety measure?

Its certainly also a safety measure. You are aware that there are three different  basic types of nitrocellulose (mono, di, trinitrate) ?


text:
Estimating protein binding and disassociation
I don't have a background in the area of drugs or pharmacokinetics/Pharmacodynamics but I am trying to understand about protein binding. I was going through this paper.
If $C_b$ is the concentration of the bound drug
$C_f$ is the concentration of the free drug
$C$ is the total drug concentration
$P$ is the concentration of protein binding sites (bound and free)
$K_d$ is the disassociation constant of the drug protein complex
For a one compartmental model the free drug can be found through:
$C_f=\dfrac{-(P+K_d-C)+\sqrt{(P+K_d-C)^2+4 K_d C}}{2}$
The assumptions used in this derivation are:

Binding occurs only to plasma proteins and follows simple
saturation kinetics. The binding process can be described by a single
macroscopic dissociation constant.
Binding equilibrium is achieved virtually instantaneously with
respect to distribution and elimination.
Aside from binding, all other processes (ie., distribution and
elimination) are linear.
Distribution and elimination processes operate only on free
drug.

My question is in how to determine the parameter values of $P$ and $K_d$.
The article has given certain properties of the drug but not a way in how these parameters are found.
The given details on a drug are:
1.The drug has a molecular weight of 150.

Drug binds only to serum albumin, whose concentration is 4.4% and whose molecular weight is 67,000.
At sufficiently low drug concentrations, the drug is 89% bound.
The distribution volume is 50 ml./kg.
Free drug has a half-life of 30 min.

Using these details is there a way to estimate the parameters of $P$ and $K_d$ and if so please explain how they can be derived.

If we call the drug L, and the binding sites S, the equilibrium is the following:
$$\ce{LS <=> L + S}$$
The equilibrium constant expression is:
$$K_d = \frac{[\ce{L}] [\ce{S}]}{[\ce{LS}]}$$
With $[\ce{L}]_\mathrm{tot}$ as the total concentration of ligand and $[\ce{S}]_\mathrm{tot}$ the total concentration of binding sites, we can solve of $[\ce{L}]$, the free ligand concentration to arrive at a quadratic expression like the one in the question.
If the ligand concentration is small compared to the concentration of available binding sites, you can estimate the ratio of free to bound ligand:
$$ \frac{[\ce{L}]}{[\ce{LS}]} \approx \frac{K_d}{[\ce{S}]_\mathrm{tot}}\tag{1}$$
Now, we can look at the statements 1 through 5 and see what information we can extract to determine $K_d$ and $[\ce{S}]_\mathrm{tot}$:
1.The drug has a molecular weight of 150
This information is not directly relevant. I'll get back to it later.
2. Drug binds only to serum albumin, whose concentration is 4.4% and whose molecular weight is 67,000
This allows us to calculate the concentration of protein capable of binding to the drug. 4.4% refers to 0.044 kg of albumin in a solution with a volume of one liter. With a molar mass of M = 67000 g/mol, we get the following concentation:
$$c(\mathrm{albumin}) = \frac{\pu{44 g / L}}{\pu{67000 g/mol}} = \pu{0.66 mM} $$
If we knew the number of binding sites for each molecule of albumin, this would also give us the concentration of sites.
3. At sufficiently low drug concentrations, the drug is 89% bound
If 89% are bound, 11% are free, and the ratio of free to bound is 0.12. "At sufficiently low drug concentration" means the drug concentration is low with respect to the concentration of binding sites, so we can use equation (1):
$$\frac{K_d}{[\ce{S}]_\mathrm{tot}} \approx \frac{[\ce{L}]}{[\ce{LS}]} = 0.12$$
So we can estimate the ratio of dissociation constant and total concentration of sites.
4. The distribution volume is 50 ml/kg
This means for every kg of body weight, the apparent volume in which the drug dissolves is 50 mL. Picture a solvent of unknown volume. You dissolve a solute and then remove some solution to test the solute solution. You should be able to calculate the unknown volume from the amount of solute added and the concentration obtained. If, however, the solute absorbs on the container wall or diffuses through it, your answer is inaccurate.
The human body has an average of 70 mL of blood for each kilogram of body mass. A substance that remains in the blood stream should have a distribution volume of 70 ml/kg if it does not bind to anything.
I'm not sure what to do with this piece of information.
5. Free drug has a half-life of 30 min.
This ensures that the binding reaction has a chance to almost attain equilibrium. If this were not the case, we could not use equilibrium calculations, and would have to resort to non-equilibrium methods such as the ones described in the paper cited by the OP (which is behind a paywall, so I just read the abstract).

Using these details is there a way to estimate the parameters of P and Kd and if so please explain how they can be derived.

The missing piece of information is the concentration of sites. If we knew the number of sites on each albumin molecule, we could calculate this concentration because we know the concentration of albumin.
Experimentally, what we are missing is data for saturating conditions. As it is, we have one equation for two unknowns, so we only know their ratio. If you assume one binding site per albumin molecule, you would be done. However, looking at the molar masses, it is entirely possible for more than one drug molecule to bind to a single albumin molecule.

Thanks a lot for the explanation. So, if there is only one binding site per albumin, is it correct to say, $[S]_{tot}=c(albumin)$

Also, is the molar mass of the drug gets used when converting mM to say mg/L? Is it the drugs molar mass that needs to be used when converting the units of K_d$ as well?

@sam_rox yes and yes


text:
Iodometric Titration: Would iodine react with sucrose as well?
I'm currently planning to do a lab experiment about the effect of sugar addition (sucrose) to orange juice on Vitamin C (ascorbic acid), also finding the degradation rate and activation energy required. To measure ascorbic acid concentration, I am going to use iodometric titration but I am wondering if the addition of sucrose would affect the results as I am not sure if iodine could also react with sucrose, not just ascorbic acid, so that the results do not show the correct concentration of ascorbic acid in the juice. If that is the case, is there any better method to measure just the ascorbic acid content? I am currently a high school student and may not be familiar with some hard concepts, so please be mindful.

Did you mean, ...orange juice or vitamin C, instead of on?

i meant the effect of sugar addition on vitamin c content in the orange juice cuz I'm going to make a homemade juice and add different amounts of sugar to see how vitamin c content in each differ

This is a fairly common undergraduate experiment. The sucrose will not do anything and it will not interfere with the ascorbic acid determination. See here for more information: https://chemistry.stackexchange.com/a/116238/79678


text:
Calculate the mass of iron in the original mixture!
I seemingly can't solve the following problem, I've spend over 30 mins on it, no progress, can any of you geniusses give a step by step solution to me? Thank you so much
Problem:
A mixture of aluminium and iron weighing 9,62g reacts with hydrogen chloride in aqueos solution according to the parallel reactions
2Al + 6HCL --> 2AlCl3 + 3H2
Fe + 2HCL --> FeCl2 + H2
A 0,0738 quantity of hydroen is evolved when the metals react completely. Calculate the mass of iron in the original mixture.
Here's the solution I found on Slader.com
1) x + y = 9.62g; Considering that x is a mass of Al, and y is the mass of Fe
2)  ; Ar (Al) = 26.98 g/mol ;Ar (Fe) = 55.85 g/mol
3)  ;;
4)  ;;
5) 0.0556x + 0.0377y = 0.366
x + y = 9.62 (multiplying with 0.0566)
0.0566x + 0.0566y = 0.535
Substracting this two equations
0.0377y = 0.169
y = 4.48 g Fe
6) x + 4.48 = 9.62
x = 5.14 g Al
-
Can someone explain step 5 to me? I really don't understand how one gets the "0.0377y = 0.169".
Thanks so much!!

Well, I understand it, but I don't like it. There's an error in line 5 of eq 5.
When you subtract the two equations, put everything in writing:
0.0556x + 0.0377y - 0.0566x -0.0566y = 0.366 - 0.535
Then  -0.0189y = -0.169, (not 0.0377y = 0.169)
Then y = 8.94g  (Fe) (answer)
(And incidentally,  x + y = 9.62g, so x = 0.68g   (Al))
I sympathize with you - but taking a little shortcut sometimes makes it difficult to detect an error. Sometimes you just have to write out everything.


text:
Emission spectra (Be and Mg)
The line says

Alkaline metals and their salts impart characteristic colour to an oxidising flame. This is because the heat from the flame excites the outermost orbital electron to higher energy level. When excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum.
Beryllium and magnesium do not impart colour to flame because electrons in beryllium and magnesium metal are too strongly bound to get excited by flame.

What if reason that Be and Mg do not impart colour to flame is not that the electrons in Be and Mg are actually bound too strongly to be excited by flame but that the radiation emitted does not fall in visible spectrum?

Provide a reference to this text please.

See: https://chemistry.stackexchange.com/questions/50467/why-lithium-gives-flame-coloration

What is "The line"? Do you mean "a paragraph" in your book?

It is disappointing when the original poster vanishes after posting a question and won't follow up those who are trying to help.

@M.Farooq Sorry I didn't see the notifications coming. The source is NCERT class 11 book, Unit 10 - The s-block elements.

@Karl yes apparently.

@NilayGhosh That was a helpful link, thanks!

+1 Had the same doubt, but you bet me to it

You are right and whoever wrote this quote "Alkaline metals and their salts impart characteristic colour to an oxidising flame. This is because the heat from the flame excites the outermost orbital electron to higher energy level. When excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum. Beryllium and magnesium do not impart colour to flame because electrons in beryllium and magnesium metal are too strongly bound to get excited by flame." is mostly wrong about Mg and Be.
The first process in flame spectroscopy is to decompose the compounds in the flame into atoms. /This requires a lot of energy. In ordinary lab flames, like that of the Bunsen burner (methane+air) the temperature is so low that alkaline earth (Ca, Ba, Sr) emit light because of their compounds in the flame not by their atoms. Recall magnesium and beryllium forms refractory oxides, so the correct explanation is that the flame temperature of ordinary oxidizing Bunsen burner is so low that it is not able to break magnesium compounds and excite beryllium in the flame.
Now if you use a proper flame, say that of acetylene-air for Mg or better acetylene-nitrous oxide flame for Be, you should see emission mostly in the UV but also very low intensity visible lines. The visible lines could be detected by a sensitive detector rather than the eyes. Nitrous oxide flame is highly colored itself (sunset pink color) and one cannot see visible emission easily without using a monochromator.

And apparently, the text comes from India's National Council for Educational Research and Training! So ironic! Thanks for clearing me up on that one, really helpful. And sorry for I could not receive the notifications earlier.

@Rew, Sorry to say but India's National Council for Educational Research and Training has to update itself. They have many outdated things. I see questions on chemical equivalents and n-factors here on these forums which were taught in the 1940s.

Yes I agree. As an ex-user of NCERT, I have regretted studying from it and all I can say is that the textbooks are full of errors and misleading information which has to updated.

@Nilay Ghosh, being outdated in syllabus was probably forgivable when internet did not exist. Today NCERT should see what is being taught around the world and compare the course contents and update them. There is a saying which is roughly translated as "If we lay the first brick of wall incorrectly, and if that wall reaches the stars, it will remain tilted." Therefore science at the school level has to improve so that students take interest in science from the beginning and better scientists can be produced at the university level.

@M.Farooq Thanks for the clarification, But I have a doubt. by "...alkaline earth (Ca, Ba, Sr) emit light because of their compounds..." you mean oxides are what causes the colouration here? Secondly, if we supplied enthalpy of atomisation would we be able to improve the issue?

@AdilMohammed, Yes you are right. If enough heat were provided one would see atomic emission not molecular compound emission of low temperature flames.


text:
Why is nitronium ion an electrophile?
I have learnt that the species which have vacant orbital(s) are called electrophiles. But in this case, the nitrogen has a complete octet. The positively charged nitrogen ion has a valency of 4, of which 2-2 are satisfied by each oxygen atoms.
Still, $\ce{NO2+}$ acts as a nucleophile. Why?

It has empty π* orbitals that are low in energy. The "low-energy" bit is important: everything has vacant orbitals, but not everything can be an electrophile.

Compare to the isoelectronic CO2.


text:
Functional Groups Identification

Identify functional groups of the following organic molecule:

A. ketone, alkene, carboxylic acid, ester
B. alkyne, ester, carboxylic acid, aldehyde
C. carboxylic acid, alkene, ketone, ester
D. ester, aldehyde, carboxylic acid, alkene

I am not sure what the answer it. I know it can't be B because there is no alkyne triple bond. However, shouldn't an alkene only exist in a straight chain? I know there is a COOH carboxylic acid, R-O-R' ether and a CHO aldehyde, and a double bond O for a ketone.
I was wondering if someone could tell the right answer.

What do you mean by "an alkene only exist in a straight chain"? https://en.wikipedia.org/wiki/Cyclohexa-1,3-diene is perfectly an alkene, in that case a conjugated diene.

Also check these https://en.wikipedia.org/wiki/Lactone esters.

The only right choice is D, because there is an ester, even though this ester is intramolecular, being made of the bridge -CO-O- in the lower part of the molecule. The aldehyde group is -CHO in the R.H.S. The carboxylic acid is in the upper part of the middle. And there are two alkene groups in the cycle at L.H.S.

Keying on ester is useless. All the choices have ester and carboxylic acid in them.

I'd call it a lactone, rather than intramolecular ester.

It can't be A&C because there is no aldehyde in both options, and also it can't be B 'cause there's no alkyne–FG in the compound, thus the only option left is D since the compound contains ester(lactone), aldehyde, carboxylic and alkene FGs as state in the option D
So I think D is the right answer.