text:
Why do Fullerenes have even number of carbon atoms?
My teacher started a chapter called "Carbon and its compounds", Which is just an intro to organic chemistry . In which she taught about fullerenes structure and examples and also told that fullerenes have even number of carbon atoms. I got curious and asked why but she didn't answer me. So I went and googled it but still did not get any answer. Is there any exception or any reason behind it?

I think you should think merely on geometry. Rather than why is how they can form. It will be the same for some "regular" polyhedron of that type.

A guess, fullerenes structures are ascribed to have symmetry. To be symmetric around an axis of symmetry implies a factor of 2 in structural units. I was also contemplating as to whether it was related to a charge balance, but it said that fullerenes apparently possess a charge imbalance!

And how many carbons are in double bond? 2! If number wasn't even, then it would be a radical and two of them would bind to each other.

If some people in the room shake hands with each other, then the total number of shaked hands is even. It is as simple as that.

The structure of (closed) fullerenes is governed by Euler’s polyhedron formula: vertices (carbon atoms) + edges (bonds) - faces = 2. Geometric constraints determine that at least some of the faces must be pentagons or heptagons, and that the number of atoms must be even.


text:
What is the fundamental difference between temperature and energy?
We know that temperature is proportional to the average kinetic energy of a system. We also know that temperature is not the same as energy because the temperature is intensive while the temperature is extensive. This lead me to think that temperature was some sort of energy density, like how density represents the "average" mass over a volume. However, we cannot express Kelvin in terms of Joules. So if temperature is not just an intensive version of energy, what is it really? How can we link entropy's connotations with disorder with the units J/K?

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text:
What fluid can I store metal filings in so they do not oxidize?
I am working on an art project that will involve storing filings of various metals (e.g., iron, copper, zinc) in a colorless fluid. I want to prevent the metals from oxidizing for several years, so water is out. I already have 70% isopropyl alcohol. Will this work, or will the remaining 30% water be a problem? If not, what fluid would you suggest?

I feel like mineral oil is a frequently used, but I don't know if that works for years...

Mineral oil would be my go to

And seal the container after backfilling with nitrogen.

Alcohol reacts slowly with some metals, e.g., aluminum and probably zinc. Since sodium is normally stored under pure mineral oil, as others suggest, that might be best. For lower viscosity, try a lighter alkane, e.g. dodecane, or naphtha. See https://en.wikipedia.org/wiki/Alkane N.B. flammability hazard: not only will alkanes burn, but finely divided metal is pyrophoric!

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How does Metol (4-(methylamino)phenol sulfate) react when developing black and white photos?
I know that during the development process the developing agents reduce exposed Ag ions to silver. But what happens to Metol which is one of the developing agents?

"Quick questions" without elaboration effort are not very welcome, and may be closed. Always think about and search for possible answers thoroughly before posting a question.
Chemistry SE site expects to explicitly elaborate the question by at least basic textbook/online search, writing down what was found/understood/tried/failed.

^Wrong duplicate. Correct duplicate; https://chemistry.stackexchange.com/q/14680/23561

@Poutnik the problem is that I have contacted three different chemistry teachers, searched throughout the internet, and I was not able to find how Metol (developer D-72) reacts during the developing process of resin-coated paper?

@Megija The problem with short questions is that respondents can just guess what you did or did not do and can make unnecessarily wrong assumptions. The best approach is to write elaborated question with listing of what you tried and failed and also the question background. I am not an organic chemist, so I would be guessing by guts or search as well as you.

@A.K.  It is unclear if the linked Q/A consider specifically metol oxidation reactions, unless it is deeply in the 2nd level links.

The analogous B/W developer agent p-hydroquinone is oxidised by silver ions  in alkaline environment to p-benzoquinone:
$$\ce{O^{-}-C6H4-O- + 2 Ag+  -> O=C6H4=O + 2 Ag }$$
The related molecule 4-aminophenol ( contained in the Agfa developer Rodinal ) reportedly according to pubs.rsc.org oxidates to NN′-bis-(p-hydroxyphenyl)-2-hydroxy-5-amino-1,4-benzoquinone di-imine, but the analogical structure is not possible for N-methyl-4-aminophenol as the methyl derivate.

Benzoquinone imines. Part 16. Oxidation of p-aminophenol in aqueous solution, Keith C. Brown  and  John F. Corbett

Abstract:
Oxidation of p-aminophenol gives NN′-bis-(p-hydroxyphenyl)-2-hydroxy-5-amino-1,4-benzoquinone di-imine. Data from reactions using potassium ferricyanide as oxidant indicate that this product is formed from 3 molecules of p-aminophenol and that rate-controlling steps depend on solution pH. Thus, rate-controlling steps are coupling of protonated monoimine and neutral p-aminophenol at pH <8, coupling of protonated monoimine and p-aminophenolate at pH 8–10, and above pH 10, coupling of neutral monoimine and p-aminophenolate. The resulting aminohydroxydiphenylamine is rapidly attacked at C-4 by a second molecule of monoimine to give the reduced form of the above di-imine, to which it is then oxidized by a third molecule of monoimine.

with the structural formula  from chemsrc.com.

It think it's wrong even for aminophenol AFAIK it makes azo or azoxyphenol when oxidised.

I suppose you are right. Indophenols are blue dyes.


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How to determine the unknown molecule using the infrared spectroscopy?

An unknown sample was analysed using infrared spectroscopy. What is
the compound?

So I am having difficulty in using the infrared data to determine the unknown compound. I keep on misinterpreting the information on the Infrared spectrum. According to the answer provided, it is ethanamide. But how? What does the strong and broad range tell you? Is it $2730-3100$ or $3200-3600$? Why can't it be $O-H$?
What about the range between $1500 - 2000$, does that correspond to $C$ double bond $C$ or $C$ double bond $O$?
How can I interpret the diagram correctly? The range seems to overlap it selves?

You can't identify an unknown compound by manually analyzing its IR spectra. Computers can do it by pattern matching the unknown spectra to a library of spectra, assuming that the unknown sample is pure. // You could perhaps choose the correct compound from say 4 choices.

With the information provided it is not possible to identify this moelcule as an amide as the crucial information about the amide C=O and N-H signals is not given. This will provide hte missing information  https://personal.utdallas.edu/~scortes/ochem/OChem_Lab1/recit_notes/ir_presentation.pdf

@Waylander Sorry, it is a multiple choice question. Mind taking a look again? Thanks

@MaxW Hi Max, thanks for commenting. It is a multiple choice question, see the update. Many thanks.

@Negrawh  In future please give us the whole question first time

First, eliminate ethylene (ethene) because the strong absroption from 3200 to 3600 requires $-OH$ or $-NH$.
Ethanol could be suggested by the 3200-3600, but lacks a strong absorption at 1050-1410, so eliminate ethanol.
The strong peak at 1700 suggests acid or amide $C=O$.
The lack of strong absorption between 2500-3000 eliminates carboxylic acid, so the answer must be: ethanamide!
This process is only possible because it is a multiple choice question. In the actual world, a sample would be analyzed thusly:

if it's ethylene, it will be in a gas sample tube, and has a recognizable odor.
if it is ethanol, it evaporates quickly, and has a recognizable odor.
if it is ethanamide, it evaporates slowly and has a fishy odor.
if it is ethanoic acid, it smells like vinegar.

In addition to James's answer, you can often tell primary amines and amides apart from OH groups because they will have two peaks in the 3300 - 3500 region. Your sample clearly shows this double peak, suggesting an NH2 group. Combined with the strong c=o absorbance at 1700 and you have an amide.


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Effect of mixing acids on solubility
Please help me clarify the title if you come up with a better way to put it (you certainly will).
My question is the following:
If you make a mix of equal parts of various acids (Bicarbonate, Hydrochloric, Nitric, Sulfuric, etc.) and throw some metal in there that will ionize and wait for a while until you have an equilibrium, what will that resulting equilibrium look like vs. the known solubilities of the expected salts? Where and in what form will the metal ions most likely end up? Will the metal's solubility tend towards an average of the expected solubilities? Or tend towards the solubility of the least soluble metal salt? Or of the most soluble metal salt? Or exceed the solubility of the most soluble salt?
This is assuming that no acid is / would have been limiting in reacting with the metal in the first place.
My assumption is that most metal ions would end up precipitating as the least soluble salt but I'm very much unsure of myself.
Upon Poutnik's advice, here are two examples I provided in the comments:
Case No 1: I drop 1 mole Cadmium metal in a solution with 5 moles of sulfuric acid and a great excess of water beyond the solubility limit of the product (I assume it would dissolve entirely), I then add 5 moles of phosphoric acid. Would that (a) precipitate most of the nickel (almost as if only phosphoric acid was present), or will (b) the solubility be somewhere between that of the sulfate and phosphate salts, or (c) most of the nickel remain in solution as if only the sulfate was present?
Case No 2: I drop 1 mole of lithium and 1 mole of cadmium metal into 2 moles of sulfuric acid with enough water for both to be entirely dissolved. I then add one mole of phosphoric acid, leave the solution in a sealed container with heavy stirring and come back 1000 years later. I then filter out and analyze the precipitate: Will it be all cadmium / mainly cadmium / an even mix of both cadmium and lithium phosphate salt? (knowing both cadmium and lithium form insoluble phosphate salts, but the solubility of the lithium salt is several orders of magnitude above that of the cadmium salt). Let's assume there is no double salt formed for the sake of simplicity.

I didn't know that. Do I understand well that bicarbonate is amphoteric?

I assumed the acid was reacting with the base (calcium or magnesium in the case of the corresponding carbonates), thus releasing carbonate beyond its water solubility such that water expels CO2

Note that $\ce{LiH2PO4}$ solubility is very good. solubility of $\ce{Cd(H2PO4)2}$ and $\ce{CdHPO4}$ is not available. Also $\ce{Cd3(PO4)2}$ solubility in strong acidic environment is much higher than in water.

Hmm. All of the parameters affecting solubility makes it very complicated. I'll try to understand that better.

All strong acids dissociate in water: $\ce{HX + H2O -> H3O+ + X-}$.
Metals reacts with hydronium ions, not with a particular acid, e.g. like $\ce{M(s) + 2 H3O+ -> M^2+ + H2(g) + H2O}$.
( With nitric acid, it is more complicated due its reduction to $\ce{NO}$ or $\ce{NO2}$, but it is not important now. )
Metallic cations and acid anions are not paired with a particular counter ion, until they start precipitating. They all form together a "ionic soup".
As a rule of thumb, the guidance is the salt solubility. But in case of mixed anions and/or cations and non stoichiometric cation/anion ratios, more important is solubility product
$$K_\mathrm{sp}=[\ce{M^n+}]^{m} \cdot [\ce{X^m-}]^{n}$$
The salts that would reach their solubility product would start precipitate,
but then hard-to-control kinetic factors of crystallization initiation and propagation take part.
The excess of the counter ion decreases solubility of the salt, compared to salt solubility in water. It is similar effect as gravimetric precipitation of very little soluble salts, where excess of the counter ion causes much lower ion residual concentration, compared to being just in water.
Other ions may alternate salt solubility by side reactions  ( if applies ) with the considered ions, or by affecting activity coeeficients - see below.

For the case 1, sulphates would have the minor effect of decreasing somewhat the activity conefficients that therefore somewhat increased solubility of the phosphate. But it is general aspect of ions presence.
For the case 2: It depends on if there is formed a double salt like LiCdPO4 or not.

In particular specific cases, there is possibility of forming double salts, where precipitate consists of integrated more than 1 cation or anion. E.g. crystallization from solution containing $\ce{Fe^2+, NH4+, SO4^2-}$ may lead to crystallization of "Mohr salt" $\ce{(NH4)2Fe^{II}(SO4)2 \cdot 6 H2O}$. Similarly for alums with $\ce{K}$ and $\ce{Al/Cr/Fe^{III}}$ as  $\ce{M^{I}M^{III}(SO4)2 \cdot 12 H2O}$.

During rapid crystallization there occurs also coprecipitation:

There are three main mechanisms of coprecipitation: inclusion, occlusion, and adsorption. An inclusion occurs when the impurity occupies a lattice site in the crystal structure of the carrier, resulting in a crystallographic defect; this can happen when the ionic radius and charge of the impurity are similar to those of the carrier. An adsorbate is an impurity that is weakly bound (adsorbed) to the surface of the precipitate. An occlusion occurs when an adsorbed impurity gets physically trapped inside the crystal as it grows.

In the case of rather concentrated solutions of well soluble salts, there is complication we should consider the ion activities  instead:
$$K_\mathrm{sp}=a_{\ce{M^n+}}^{m} \cdot a_{\ce{X^m-}}^{n}$$
Acticity coefficients for diluted solutions can be calculated by Debye-Huckel theory, but for concentrated solutions it is hard to theoretically predict them. We can use concentrations for rough predictions, but the numbers may not fit well.

Thanks for this elaborate answer Poutnik! As a conclusion, should one expect most of the metal to precipitate as the least soluble of the possible salts (assuming there is enough of the counter-ion)? Or its solubility to be increased by the presence of counterions for which the corresponding salt has high solubility?

I'm not sure what the phenomenon of passivation is. Until now I thought it only involved the solubility of a compound (such that upon reacting it would form an insoluble surface layer protecting the metal or alloy from further reaction and dissolution).

So for case 1, the addition of a counter-ion with which the metal ion forms a salt with very low solubility would not decrease its solubility but very slightly increase the amount of metal ion in solution? For case 2, let's assume there is no double salt formed for the sake of simplicity. (I had initially not thought about the possibility of double salts)

Please, delete eventually already obsolete comments. Generally, if you need to clarify an asnwer, use comments. If you have rather followup questions, extend the question and put in a temporary comment about Q update. You can notify the user by @username  .

I followed your recommendation about the comments. If you know to provide an answer for case no 2 when the possibility of double salts is left out (for simplicity), it would be really great to have this new tool of knowing what to expect to happen in such a case.

What is the answer for case 2 if one assumed that no double salt is formed?

Hm, you should be able to apply mentioned general information here.


text:
Can the electrochemical stability of a aqueous KOH solution be increased by adding more KOH?
My goal is to increase the efficiency of a nickel-iron battery. One of the options to reach this goal is:suppressing the hydrogen production. Due to electrolysis, the electrolyt will start discomposing at a certain potential (when charging, about 1.6 Volt). I expect that the electrochemical stability/window of the solution increases when the water activity decreases.
Recently, i read an article about molecular crowding electrolytes, a proces in which the water activity is lowered due to the presence of other molecules. I would like to know if it is possible to decrease the water activity, due to molecular crowding, by increasing the concentration of KOH in the elektrolyt.

I have some doubts about the outcome.
Optimal setting for all major commercial  cells is well optimized during the years of development. Without bringing a new technology in chemistry or materials of cells, the improvement is not much probable.
Increasing of hydroxide concentration would decrease water activity. But it may have also adverse effects, like partial dehydration and deactivation of active electrode components (hydroxides/hydrated oxides), leading to worse cell performance and lifetime, or it may affect conductivity. Highly concentrated solutions are quite viscous and affecting the proper procedure of the cell assemmly.


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Naming of perchloric acid
Why is perchloric acid called so? I am aware of the rules, but why is it not called hyperchloric acid? Or is it correct to also refer to perchloric acid as hyperchloric acid?

I would be more interested why it's even still called like that. IUPAC introduced newer nomenclature ages ago.

We have extra suffix for each of positive oxidation numbers in Czech, so we do not have such problems. Kyselina chlorná HClO, chloritá HClO2, chlorečná HClO3, chloristá HClO4.

In the mid 19th century, chemists apparently named it 'hyperchloric acid' with composition HClO7. But it was later renamed to 'perchloric acid' with a corrected composition $\ce{HClO4}$. So, in common nomenclature, acids with highest O.S is given the per- prefix. There is always IUPAC name to help you in case common name causes confusions. The compound  is actually $\ce{ClO3(OH)}$. So, the IUPAC name is " hydroxidotrioxidochlorine".

I doubt anybody would call it hydroxidotrioxidochlorine in nearby future.

@Poutnik. Indeed. That is quite a mouthful.

Thank you @NilayGhosh By any chance is there a peroxy acid of chlorine? If so, is it called perchloric acid?

Acids should contain the peroxo bond -O-O-H in order to call themselves the "peroxy acid". Unfortunately, there are no acids of chlorine which contains such bonds. So, there is no "peroxy acid" of chlorine. OTOH, there is a nitrogen oxyacid which is called "peroxynitric acid". Observe the bonds carefully.

Oh okay then. Thanks! Since there is no peroxy acid, there would be no confusion regarding perchloric acid and per[peroxy]chloric acid, which was what I wanted to ask about.

This nomenclature comes from a long time ago. In the $19$th century, it was decided that the acidic compounds containing H, O and another atom X would be named from the name of this third atom X.
If only one molecule of such an acid was known, the name would be made by adding the ending -ic. Example: Only one acid is known with a Carbon atom, and it is $\ce{H2CO3}$, which is called carbonic acid.
If two acids were known differing in the number of oxygen atoms, the most oxidized would follow the preceding rule. In the least oxidized, the ending would be changed into -ous. Example : In the $19$th century, the two following acids were known containing a nitrogen atom. Well  !$~\ce{HNO3}$ was called nitric acid and $\ce{HNO2}$ nitrous acid.
If more than two acids were known differing in the number of oxygen atoms, the chemists would copy the case of the chlorine atom, where four acids were known : $\ce{HClO, HClO2, HClO3, HClO4}$. In this case, the two acids situated in the middle of the list are named according to the previous method : $\ce{HClO2}$ is chlorous acid, and $\ce{HClO3}$ is chloric acid. For the two extreme cases, it was decided to add prefixes, hypo- for the lowest number, and per- for the highest number. As a consequence, $\ce{HClO}$ became hypochlorous acid, and $\ce{HClO4}$ was called perchloric acid.
Today, the same regulations is applied, but the criterion of the number of oxygen atoms has been replaced by the oxidation number. The suffix -ous and -ic must be used if the corresponding acids have a difference of $+2$ in their oxidation numbers.
It should also be mentioned that the previous nomenclature is valid for acids having only one X atom. If an acid have two X atoms, the prefix di- should be added, and tri- for three X atoms. For example, $\ce{H2S2O5}$ is called disulfurous acid, and $\ce{H2Cr2O7}$ is dichromic acid. There are also other prefixes like ortho- and meta- to distinguish between acids having one X and the same oxidation number, but a difference in the number of H atoms.

In my understanding, the question was not about naming nomenclature the OP is reportedly familiar with. It was rather an etymology question why per- and not hyper-.

Maybe the use of the prefix "per-" is due to the influence of the French language. For example, in French the verb "forer" means "to make a hole in the ground". And if we add the prefix "per-", we get the verb "perforer" which means "to punch, to pierce, to make a hole so as to see the light through it".

Thanks for your answer Maurice but like @Poutnik mentioned, my question was why we refer to the acid as perchloric and not hyperchloric acid

@Metaxylene I guess the answer is similar as for peroxide, where there is also  superoxide, but not hyperoxide ( at least not in English, because CZ and DE do have hyperoxide ).

Sorry, didn't quite get what you meant. Could you please clarify? And why the confusion? Wouldn't just sticking to naming the acid hyperchloric acid at the time reduce confusion?

@Metaxylene Is there any confusion ? Naming conventions as for any other language say many "because" to many "whys". They do not always follow straight logic nor optimal way. Note that per- is not a convenience for hyper- but alternative prefix. Also, the pair hypo/hyper prefixes is used for low/high state of the same phenomena, what is not directly applicable here. There is no hypochlorate nor hyper/perchlorite.  And last but not least, per- is shorter and that counts.

Understood. Thank you. You've cleared any confusion I had about your previous post.


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Comparative Solubility of Double Salts
Is there a tendency that could be used to generally forecast how a double salt might dissolve in aqueous medium compared to the simple salts of its constituents?
Example:
There is metal A, metal B and acid C.
The solubility of AC and BC are known. Could one use these as a rule of thumb to know what to expect from the solubility of say ABC?
Would it (1) Be fairly in between the solubilities of AC and BC? (2) Be fairly close to that of the least soluble among the two (3) Be fairly close to that of the most soluble among the two, or could it (4) exceed or lie beneath the solubility of the least soluble of the two?
Poutnik enlightened me on factoring in the possibility of double salts in another question which prompted this interrogation in my mind so I thought I'd make a separate question specifically for this topic.

I suppose there is no simple rule, it can be any from the 4 cases.

I doubt it can exceed the solubility of the two. Other than that, everything is possible.


text:
Number of moles and Avagadro's number
Counting # of atoms don't match in rounding?
The total number of atoms contain in 1.0L as 1000g of liquid water must equal the total number of atoms contained in a gas released by 1000g of liquid water.
A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms. Avogadro's number is 6.023×10^23 molecules/mole.
One mole of water =6.023×10^23 molecules of water.
1 molecule of water = 3 atoms of water. One mole of water =3×6.023×10^23=1.807×10^24 atoms. Now
Atomic mass of H2O = 16 + 2 = 18g/moles Number of moles = Mass Given/(Atomic Mass) or Molecular Mass
Number of moles = 1000g/18
Number of moles = 55.556
55.56 moles of water as liquid = 55.56 x 3×6.023×10^23 = 1.0039x10^26
"Total number of atoms = 1.0039x10^26 "
The mass of 1.0L as 1000g or 55.56 moles of water produces 1866 L of gas (1244L of H2 and 622L of O2).
Number of moles of hydrogen = 1244/22.4 = 55.54
Number of moles of oxygen = 622/22.4= 27.76
Atoms contain in 55.54 moles of hydrogen = 55.56 x 6.023×10^23= 3.346x10^25
Atoms contain in 2x 55.54 moles of hydrogen = 2x55.56 x 6.023×10^23= 6.69x10^25
Atoms contain in 27.76 moles of oxygen = 27.76 x 6.023×10^23=1.67x10^25
"Total number of atoms = 8.36 x10^25 "
Where did I make the error in rounding?
i didn't mention STP but it was presumed

Always try to approach the task first by the general algebraic solution and plug in the numbers in the end.  It has several benefits. 1/ It allows you better grasp on underlaying principles. 2/ It allows both of you  and readers to more easily spot a solution error or conceptual mistake. 3/ It gives the Q/A higher and more permanent value to be reused/applied for similar questions. 4/ It follows the site policy to focus on explaining principles and procedures, rather than solving of particular tasks with the value for 1 person only. 5/ You will avoid the intermediate rounding errors.

Note the purpose of the site is not proof reading. Is there any principle or concept you have trouble to apply ?

Note that moles of oxygen atoms and moles of oxygen gas are not the same thing. Same with moles of hydrogen atoms and moles of hydrogen gas. Both of those gases are diatomic.

Avogadro number is $\pu{6.02214076E23}$. (exactly by definition)

Use MathJax for eventual formatting for formulas or expressions.
E.g., instead of 0.123456x10^78, write $\pu{0.123456E78}$, what will look $\pu{0.123456E78}$.

1 molecule of water = 3 atoms of water.  / 1 molecule of water contains 3 atoms is better formulation.

The following sentence, taken from your text, is wrong :  ... the mass of 1.0L  of water produces 1866 L of gas (1244L of H2 and 622L of O2). NO !  1866 L water vapor does not "produce" H2 and O2. Water can be decomposed into H2 and O2, but this would require a huge amount of energy, and cannot be made easily in the vapor phase. It can be made by electrolysis in the liquid phase. But the volumes of H2 and O2 are not related to the volumes of the liquid phase.

This is an extended comment...
I assume the statement that you are trying to prove is:

The total number of atoms contain in 1.0L as 1000g of liquid water must equal the total number of atoms contained in a gas released by 1000g of liquid water.

It is a rather general statement. The statement doesn't say anything about temperature,  pressure, volume, or composition of the gas.  I'm guessing that you're trying to use gaseous hydrogen and oxygen since water doesn't have a vapor pressure of 1 atmosphere at STP.

Avogadro's number is $6.023 \times 10^{23}$ molecules/mole.

You must be using an old book/reference. $\mathrm{N_A} \approx 6.022 \times 10^{23}$ had been used for quite a while. But in 2019 the SI base units were redefine and the value now is exactly $6.02214076 \times 10^{23}$.

Atomic mass of H2O = 16 + 2 = 18g/moles
Number of moles = Mass Given/(Atomic Mass) or Molecular Mass
Number of moles = 1000g/18
Number of moles = 55.556

The calculation for number of moles of water has a massive mistake. You calculated the mass of water to only 2 significant figures but end up with 5 significant figures for the moles of water. Significant figures are a crude way of doing error propagation, but proper use prevents exactly this kind of mistake. You could easily lookup the molecular weight of water which is 18.015 g/mol to 5 significant figures.
You then go on to talk about "55.56 moles of water as liquid" and then "Atoms contain in 55.54 moles of hydrogen". This is just sloppy.

Number of moles of hydrogen = 1244/22.4 = 55.54

Here again you've used an old value, 22.4, for the volume of a gas at STP. Since 1982, STP is defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of exactly $10^5$ Pa (100 kPa, 1 bar). This gives a molar volume of 22.711 liters/mole.

Now in general I agree with user Poutnik's advice. Try to write out an algebraic formula if you can. However if the formula is going to be large and messy, then look at all the given values and decide how many significant figures should be in the answer. I'd use two extra significant figures in intermediate calculations to prevent rounding errors as well as possible.


text:
Halogenation by Phosphorous tribromide vs using HBr
My textbook introduced $\ce{PBr3}$ by stating that it leads to an inversion in stereochemistry since it always goes through an SN2 pathway, unlike using $\ce{HBr}$.
My question is that since $\ce{Br-}$ is a reasonably good nucleophile, so I assume it would most likely proceed through SN2 in a substrate with a secondary carbon anyway. Both pathways seem to me to be extremely similar with the exact same attacking species eventually ($\ce{Br-}$) and an excellent modified leaving group attached ($\ce{H3O+}$ and $\ce{OHPBr2}$).
What exactly is the difference and the advantage (other than being non-acidic) that make halogenation by $\ce{PBr3}$ a better method?

HBr can give rise to great amounts of alkene through elimination due to the SN1 pathway with secondary alsohols.

How would this be avoided when using PBr3?

There is no SN1 pathway with PBr3

The problem with $\ce{HBr}$ is that it can almost always cause an SN1 reaction, even if the alcohol is primary.
Consider the reaction of propanol with $\ce{HBr}$. Even though the resulting leaving group would make a primary carbocation if it left on its own, this can be avoided via a concerted rearrangement, thus making it possible for an SN1 reaction to happen. This doesn't happen with $\ce{PBr3}$ of its difference in mechanism.

If, as your mechanism shows, the reaction is concerted, then there is no primary carbocation formed. Is this true?

Yes, which is what makes the SN1 reaction viable even for primary carbocations. I'm not exactly sure why (since the LG is good in the PBr3 reaction), but Solomons & Fryhle 10th edition mentions "The reaction of an alcohol with PBr3 does not involve formation of a carbocation and usually occurs without rearrangement of the carbon skeleton" in chapter 11.9, which means that it wouldn't have issues of potential concerted rearrangements.

Your diagram shows HBr, not $\ce{PBr3}$
. I'd be surprised if a primary alcohol didn't give a primary bromide with $\ce{PBr3}$, i.e., an $\ce{SN2}$ reaction. But with HBr, you show the rearrangement as a concerted hydride shift, indicating no intermediary primary carbocation

I believe I was unclear with my wording. I am trying to convey that the water group leaving on its own does not involve the concerted rearrangement (and would thus result in a primary carbocation), but because the concerted rearrangement happens, the formation of a primary carbocation is avoided entirely. Hopefully my edit better reflects this.


text:
Neutralization reactions
I saw that most neutralization reactions produce water while some do not.
How can I know if water will be a product in a given neutralization reaction?
I initially thought that water is a product only when an acid reacts with an alkali (producing hydroxide ions) but there apparently exist some counterexamples like the reaction between ethanoic acid and copper oxide.

Your counterexample does give water on neutralization: $\ce{CuO + 2CH3COOH -> Cu(CH3COO)2 + H2O}$

yeah, so although CuO is not an alkali, it still gives water and that's why it's a counterexample of my initial thought.

$\ce{CuO}$ is basic in nature. Most metal oxides eg $\ce{CaO}$, $\ce{MgO}$ etc are basic in nature and neutralize acids in aqueous media. These are divalent and the oxygen anion accepts two protons to form $\ce{H2O}$.

Hint: Ask yourself and try to thoroughly search for an answer, what is an acid, what is a base and when their reactions produce water.

There are plenty more bases than hydroxide or carbonates.

Neutralizations of any acid with oxides, hydroxides and carbonates always produce a salt and water.
The only neutralizations that do not produce water are reactions of an acid $\ce{HX}$ with ammonia $\ce{NH3}$ or with organic derivates of ammonia, like amines. These reactions produce an ammonium salt $\ce{NH4X}$ and no water
$$\ce{NH3 + HX -> NH4X}$$ If the acid is $\ce{H2Y}$ or $\ce{H3Z}$, the neutralization reactions are $$\ce{2  NH3 + H2Y -> (NH4)2Y}$$ $$\ce{3 NH3 + H3Z -> (NH4)3Z}$$ It never produces $\ce{H2O}$. If one atom $\ce{H}$ from $\ce{NH3}$ is replaced by an organic radical $\ce{R}$, a primary amine $\ce{RNH2}$ is obtained and the neutralization reaction produces also an ammonium salt $\ce{RNH3X}$ $$\ce{RNH2 + HX -> RNH3X}$$ Similar reactions can be obtained with secondary and tertiary amines. All these reactions produce salts and no water.

Hmm, how about CaO + SiO_2 = CaSiO3? Isn't that considered a neutralization reaction despite not producing water? @Maurice

@Shane. You are right. The reaction $\ce{CaO + SiO2 -> CaSiO3}$ is considered a neutralization reaction despite not producing water, because both reactants  are reacting with water to produce a hydroxide ($\ce{Ca(OH)2}$ and an acid $\ce{H2SiO3}$. These two last substances do react together according to : $$\ce{Ca(OH)2 + H2SiO3 -> CaSiO3 + 2 H2O}$$ I should have added this comment into my message.

The General Chemistry answer is for any strong base and strong acid there will be a molecule of water produced and a salt.
BOH + HA --> H2O + BA
Giving credit where it is due, @waylander, @Poutnik, @Dongwi Min, and @Aniruddah Deb make excellent assertions regarding your question and I'd refer you to this Quora post (https://www.quora.com/What-is-formed-by-the-reaction-between-metal-oxide-and-water) that talks a bit about productions of metal hydroxides within water.


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Can we predict the IMF strength of Acetone vs Glycerol?
I was recently assigned a lab where I was to observe several properties of the chemicals Acetone and Glycerol then use my observations to predict their overall Intermolecular Force strength.
The properties I observed were:
Surface Tension: Here, I found that Acetone had a much higher surface tension than Glycerol. (I tested this by seeing how many drops of each substance would stay on a penny: Acetone had 19 and Glycerol had 13)
Viscosity: Here, I found that Glycerol had a much higher viscosity than Acetone. (I did this by dropping a penny through a 500mL graduated cylinder: Took 0.4 seconds for Acetone and 5ish seconds for Glycerol)
Boiling Point: Researching online, I found that Glycerol had a much higher boiling point than Acetone.
Glycerol also took much longer than Acetone to evaporate.
However, this seems to be conflicting evidence. I have learned that a higher surface tension is associated with overall stronger IMFs (suggesting that Acetone has stronger IMFs), but the other pieces of evidence seem to suggest that Glycerol has stronger IMFs.
My question is, can anybody tell me if Acetone's IMFs really are stronger than Glycerol, or is it the other way around? Additionally, if any of my experimental results seem wrong, can somebody please point them out?

Surface tension test seems doubtful. I would rather check, how many drops is needed for 1 mL ( or other arbitrary volume ), corrected for density..

So the intermolecular forces would be stronger in glycerol then in acetone. Glycerol can form hydrogen bonds while acetone is not very polar and can’t significantly form h bonds with itself.
The surface tension for glycerol is 63.4 $\frac{mN}{m}$ while acetone is 25.2 $ \frac{mN}{m}$ (both @ 20 C)
I supposed the error in your experiment arises from using pipette to determine the amount of material on the coin. Unless your using very precise pipettes, counting drops isn’t necessarily the best way to quantify the amount of material added. Additionally the density of glycerol is higher then acetone which needs to be accounted for in the test.


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Viewing and capturing real time values of Electronic balance measuring instrument on television set and computer monitor larger screens
https://en.wikipedia.org/wiki/Weighing_scale
Is it possible to view and capture real time values of Electronic balance measuring instrument using wired, wireless,CCTV technology on television set or computer monitor larger screens?
Scenario
Chemistry labs - Electronic balance interfaced wired/wireless with the television set or computer monitor i.e.Larger display screen display taking into account.Viewing the mass recording of a object in real time.

I’m voting to close this question because it's not really about chemistry.

At my local big box store one can readily find wireless home security camera sets, often with up to 8 cameras or more, and a regular monitor. Includes ability to select one or all cameras on the display.

Three possibilities come to mind:
(1) Purchase a scale with a digital output e.g, https://www.arlynscales.com/digital-indicator/industrial-scales-with-analog-and-digital-outputs/ , send it to your computer, and output it to a monitor (depending on the scale's output format, it may be possible to send it directly from the scale to a monitor)
(2) Point a USB webcam towards the scale's digital readout, plug that into to your computer, and then display on a monitor.
(3) Same as no. 2, but use a digital camcorder that has a digital output that can be displayed directly on a monitor, without the need for a computer.

Thanks. Same above solutions are applicable also to Television set ?

Depends on the type of video input formats your TV can recieve.  You'd need to check your TV.


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Carbocation rearrangement
I am dealing with some basic reactions in organic chemistry. One of these is the acid reflux of alcohols. Something I came across was hydride shifts. I know how and why they occur, but I don't understand why they are called 1,2 hydride or methyl shifts. What do these numbers indicate? I initially thought it would be the position of the carbon atoms involved in the shift, in the principal chain, but that was probably wrong because a shift occurring at the 3rd carbon atom was also called a 1,2 shift. Help?

1,2 means adjacent carbon atoms (not necessarily locant numbers on the principal chain)

I believe that in the examples where a 1,2 hydride shift was mentioned, the hydrogen atom would have 'shifted' between two adjacent carbon atoms. I think the 1,2 refers to the fact that the carbon atoms involved in the shift are adjacent.
A 1,3 shift would mean a the hydrogen atom shifts to the second carbon from the original carbon atom it was bonded to, if you get what I mean.
The same applies to methyl shifts.


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How to read character table?
I am having some problems reading character table. I will describe what I understand and thought to be right.
Consider the $\pi$ MOs formed from overlap of p-orbitals in benzene.
These two degenerate $\pi$ MOs below have $E_{1g}$ symmetry:

I don't understand why that is the case. Based on what I have read, if a symmetry operation $g$ maps an orbital to itself, e.g.

then $\chi'(g) = 1$.
Elif $g$ maps the orbital to negative itself (invert the sign of orbital),

then $\chi'(g) = -1$.
Else, $\chi'(g) = 0$.
For two degenerate orbitals as above, final $\chi(g)$ is the sum of $\chi'(g)$ for two orbitals, where $\chi(g)$ is character of a symmetry operation $g$, that is recorded in the character table.

Applying this, consider the $C_6$ of benzene,

Not mapped to $\pm$ itself, so $\chi'(g) = 0$.

Also not mapped to $\pm$ itself, so $\chi'(g) = 0$.
So $\chi(g) = 0 + 0 = 0$.
But if we look at the character table for $D_{6h}$, for the symmetry species $E_{1g}$, $\chi(C_6) = 1 \neq 0$.
Enlighten me, please, I must have understood something terribly wrong.

"else $\chi(g) = 0$" - this is not true, see e.g. https://chemistry.stackexchange.com/a/112717/16683. Apart from that you have the right idea, I think.


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Can a weak acid react to give a stronger acid?
I was trying to understand why $\ce{AgCl}$ does not dissolve in $\ce{HNO3}$ and the reaction I could think of was something like this:
$$\ce{AgCl + HNO3->AgNO3 + HCl}$$
I could not understand why this reaction wouldn't take place. Research on similar reactions further led me to the question: "Can a weak acid react to give a stronger acid (Although I understand that in the above reaction both acids are strong)?"

See reaction of boric acid, with sugars, glycerol or generally vicinal alkohols with OH group on neighbour atoms.

Come to think of it, not an acid at all can react to give a strong acid.

Can a weak acid react to give a stronger acid? Well, yes, if you set up the right conditions. For instance, sulfuric acid (estimated first pK = -3) can be heated with $NaCl$ to boil off $HCl$ (pK = -5.9) according to Wikipedia. $HCl$ is driven from the reaction arena as a gas.
You will claim that’s a special case, with unusual conditions (the boiling off). But in the question, $AgCl$ is tightly bound and leaves the reaction arena, leaving $H^+$ and $NO_3^-$ to bounce around freely and separately. The $AgCl$ just sits there.
To put some numbers on a scale, 1 gram of $AgCl$ in water is 99.9% bound as solid (and therefore out of the reaction arena: solubility is 0.0009 g/L). Less than 0.1% is “loose” - ionized and available for reaction with other ions. On the other hand, 1 gram of nitric acid in water is 100% ionized.
The binding in a weak acid is not as tight as in the case of $AgCl$, but is not nearly as loose as in $HNO_3$. Formic acid, a fairly strong weak acid, has a pK = 3.75. One gram (0.022 mole) in a liter of water will be approximately 10% ionized (“loose”), and 90% unionized. Now the fact that formic acid is totally dissolved and seems “loose” in one sense, and is also in a mobile equilibrium, exchanging protons with water faster than you can blink, doesn’t cancel out the fact that formate ions are mostly bound up to protons. If a stronger acid were produced, the freer protons would attach to the (few) formate ions and bind them up even more. So there is no chance for formic acid to loosen up more than its pK allows - unless you can take those protons from the reaction arena. They don’t go away to a stronger acid unless you can take that whole stronger acid molecule away from the reaction arena. Formic acid is not very stable, but I propose that a mixture of concentrated formic acid (say, 90%) plus $NaCl$ and gentle heating, below the boiling point of formic acid (100.8 C), would evolve $HCl$ at first, but as sodium formate is formed, the ionization of the acid would be depressed until it is too weak to form enough hydrogens to give $HCl$ to drive off. So this case would resemble the reaction of sulfuric acid with salt, but could not be driven to completion. (Even with $H_2SO_4$, only the first proton is available; the second is too weak to drive off $HCl$ efficiently.)

Your cited reaction proceeds, but in the reverse directions as follows:
$$\ce{HCl (aq) + AgNO3 (aq) -> AgCl (s) + HNO3 (aq)}$$
So apparently, a weak acid can react to give a 'stronger acid', albeit, in dilute form, with the creation of an insoluble precipitate.
This is relatedly evident, as one can also produce a dilute version of the acids like HI, HBr, HCl, HNO3, H2SO4, for example, from the action of Oxalic acid on the respective aqueous salts $\ce{CaI2, CaBr2, CaCl2, Ca(NO3)2}$ and finally $\ce{K2SO4}$.
In fact, it did a thread on it on another forum, with the underlying reaction given, for example, by:
$$\ce{H2C2O4 (s) + CaI2 (aq) -> Ca(C2O4)2 (s) + 2 HI (aq)}$$
which parallels the reaction cited above, but why? Because the corresponding formed oxalate salt is of very low solubility, barring a concentrated acidic medium, as is also the case of Silver chloride in dilute Nitric acid, per Wikipedia on Silver chloride, to quote:

Solubility: soluble in NH3, conc. HCl, conc. H2SO4, alkali cyanide,...; insoluble in alcohol, dilute acids.

A side note, caution apparently distilling down, in the case of created H2SO4 by Oxalic acid (in an attempt to directly create highly concentrated H2SO4), has been observed, in the presence of formed highly concentrated sulfuric acid, to then act on the Oxalic acid. This can result in a gaseous decomposition of the Oxalic acid itself, as noted by a sudden dangerous ejection of the created concentrated H2SO4 from the vessel.


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Is this statement valid for reactions of all orders?
My textbook (not an international standard) states that for the first order reactions only

The quantity of reactant remaining after $n$ half lives is
$$A_n = \frac{A_0}{2^n},$$
where $A_0$ is the initial concentration and  $A_n$ is the concentration after $n$ half lives.

Shouldn't this statement be true for all orders of reaction?
Upon calculation (for 0 order reaction), I observed that half life time reduced with progress of reaction.
Doing same calculation for First order reaction, I observed that half life time remained constant with progress of reaction.
So, is it this(inconsistency of half life time period) due to which Half life concept is not extended to other order of reactions?
I only wanted to confirm regarding concentration after n half lives,
as in a single half life of any order reaction, concentration of reactant reduces to half of its initial value.

Does this answer your question? Half-life equation for 2nd order kinetics

Reactions of all other orders do not have half-lives.

@IvanNeretin can you please give one example? I am unaware of it.

@Poutnik I understand that for a reaction of any order, other than 1st order, half life depends on the initial reactant concentrations. However, we are not talking about the expression of half life here.
Regardless of the order of a reaction, with the progress of any reaction, reactant concentration reduces, and it halves at half-life. So, unless a reaction reaches equilibrium before consumption of half of the reactants, it should have a half life time. So, if a reaction indeed have a half life or multiple half lives that expression for n-half lives is valid for any order of a reaction.

@SujayGhosh Well, if for the 2nd order reaction $\ce{A + B -> C}$ is $[A]=2 [B]$, how is defined the half concentration ? $[A]/2$ or $[B]/2$ or $([A] + [B])/2$ or $[A][B]/2$ ? Half-time is very unuseful concept for these rate orders, even if I agree we can formally define it. What if $[A]=10000 [B]$ ?

@Poutnik Thanks understood your point. So only for $A$ ⟶ $B$ reaction, half life is useful for any order. Hope I got it right?

@Poutnik but according to IUPAC, half life is the half concentration only. No mention of half of initial rate. https://goldbook.iupac.org/terms/view/H02716

@SujayGhosh   *The half life of a reaction has meaning only in special cases:
For a first-order reaction, the half life of the reactant may be called the half life of the reaction.
For a reaction involving more than one reactant, with the concentrations of the reactants in their stoichiometric
ratios, the half life of each reactant is the same, and may be called the half life of the reaction.*  As I have said previously, it may be formally defined, but it is not practically used nor called so.


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Bond dissociation energy for polar molecules
This formula relates to how Pauling determined the values for electronegativity for all the elements and how delta (if big) shows that the molecule is very polar. Delta being:
$\Delta =\Delta H\left(AB\right)-\Delta \:0.5\left(\Delta \:H\left(A2\right)+\Delta \:H\left(B2\right)\right)$
$\Delta H\left(AB\right)$ is the experimental bond dissociation energy. The difference in experimental and theoretical bond dissociation energy is caused by the electronegativity of the molecule.
The more electronegative the atom the bigger the delta value. But why is that?


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Is it possible to vacuum distill citric acid?
I'm wondering if

it is at all possible to vacuum distill citric acid without the slightest decomposition;
it is something that can easily be done in a small lab with widely available equipment.

I've looked on internet and found the following data:

~154 °C melting point (anhydrous citric acid)
~175 °C starts decomposing
~310 °C boiling point
~$10^{-8}$ mmHg vapor pressure at 25 °C

If you have enough to do a real distillation (not say Kugelrohr), you should have enough for a recrystallization, and I suspect you'll get better results.

@Zhe This is true, but they have different effects and thus could be complementary. Do you know if such a distillation is possible and easy to set up in a small lab?

Itaconic acid is formed by the distillation of citric acid. https://en.wikipedia.org/wiki/Itaconic_acid

@User55119 That's interesting. I had never heard of itaconic acid before. The wikipedia page does not explain how dry distillation of citric acid turns it into this other substance. Is it the thermal degradation product of citric acid?

Given that the citric acid has an appreciable vapor pressure at 25 °C, I'd look into purification using vacuum sublimation.

@MaxW Good point, but I don't trust that number. 10mmHg vapour pressure should give a well notiveabele acidic smell, which citric acid doesn't have.

The thermal conversion of citric acid is a decarboxylative dehydration to form the double bond of itaconic acid.

@MaxW & Karl:I don't know how to format the vapor pressure properly for clear reading but it should read as ten to the minus 8th power. It's a fairly low vapor pressure

@User 55119: Thanks for that information! Would you know at which temperature this decarboxylative dehydration starts taking place (i.e. the temperature under which it will not occur at all)?

@Hans 10^–8 mmHg  written as $\pu{e-8 mm Hg}$  is shown as $\pu{e-8 mm Hg}$ ( inline mode)

Or 10^-8 can be written as $10^{-8}$ seen as  $10^{-8}$. See MathJax for eventual formatting of formulas or expressions.

Here is the Org. Syn. prep: http://www.orgsyn.org/demo.aspx?prep=CV2P0368 The itaconic anhydride is distill at 175-190oC but the citric acid is heated in the pot with an open flame. Hydrolysis of the anhydride affords the diacid.

@Poutnik: I updated the text with the appropriate format. Thank you!

@User55119: Useful information. The distilling temp seems to correspond to the temperature at which citric acid starts to decompose so maybe itaconic acid is a / the main degradation product.


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In which of the these structures C-3 carbon is not a chiral centre?

In which of the these structures C-3 carbon is not a chiral centre?

I think that the answer to the above question should be both (b) and (c) as C-3 is attached to same groups and also they have a plane of symmetry hence making it achiral. But, the answer given to this problem is (d) with the reasoning that only in (d) option configurations of the both the carbon attached to C-3 is same while the configurations of both the carbon attached to C-3 is different in (b) and (c).
Should the configurations of C-2 and C-4 carbons even matter here?

I suggest you read this: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/RSrs.html

This question should also help you out.


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Reaction of Methyl Cyanide with nitroethane
I want to know the products of reaction of Methyl Cyanide with nitroethane in presence of (Sodium ethoxide + ethanol) and then hydrolysed in concentrated Acidic medium.
According to me ,reaction should be like shown in below picture
Can anyone confirm me if mine is correct?
If not,then please tell the correct products.Thanks!

look up the Nef reaction

@Waylander That contains the hydrolysis of nitro group part and seeing that I think mine is correct.Thanks.

Except for the carbon atom with 5 bonds. And is this really only a 'hydrolysis'?

@user6376297 that carbon atom is a typo ,remove H from there :p. And yes,it is Hydrolysis in Acidic medium(with concentrated acid)+heat

@Ashutosh yes, I imagined it was a typo, but you know, if this were a test, they would not forgive that. The small pain I'm causing to your brain by pointing this out might help you avoid this mistake when it matters :)

Concerning the hydrolysis: if you boil a carboxylic acid ester with water and a catalyst, you get an alcohol and an acid. None of the atoms involved changes oxidation state. In your reaction, would you say it's the same? See this and this. So I was pointing out the fact that there's more to it than just addition of water followed by some cleavage.

@user6376297 Yes I appreciate you correcting my mistake ,sorry if it looked like I didn't . And Yes,I saw the links you shared and conversion of nitro into ketone looks more than just addition of water. Maybe because I am high school student so my teacher didn't want to go into detail and just told us that this is hydrolysis.Though ,seeing nef reaction I guess my product looks correct? Or am I missing something?

Sodium ethyl nitronate is not a strong enough nucleophile to attack the sp carbon in acetonitrile (commonly used as a solvent because of low reactivity). The possible reaction may be the hydrolysis of acetonitrile to sodium acetate by NaOH which is present as an impurity in the preparation of NaOEt.


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Why does ink form tendrils on wet paper
When a drop of ink or watercolour paint is placed on wet paper, rather than uniformly diffusing in all directions it forms tendrils which rapidly expand out from the spot.
Why is this and what is the chemistry or physics occuring?
This video may further show the type of thing I am referring to
https://motionarray.com/stock-video/black-silver-tendrils-pack-216423
It appears that the pigment solvent forms discrete channels outwards through the water. I wondered if this might be a surface tension interaction or something else.

If you look closely at paper, you will see that the paper itself is not uniform.

If you’re talking about a gentle placement of the drop on the paper, I’d guess it’s likely due to the fibrous structure of the paper providing “easier” paths for the ink to flow. If you’re talking about an impacting drop, the drop will vibrate when it impacts the paper until the forces of surface tension are exceeded and the drop “splatters”.


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Equilibrium reaction where the number of moles of reactants and products are the same
Le Chatelier's principle says that for an equilibrium reaction where all reactants and products are gases a change in pressure will change the number of moles of the product such to counteract the change in pressure.
My textbook also states that this only affects reactions in which the number of reactants and products are different.
I'm trying to think of a gaseous equilibrium reaction where the number of moles of reactants and products are the same.
All suggestions welcome.

And the question is? Well I guess what it is, but what I infer isn't acceptable. In theory, you could get a list so long that it wouldn't fit into one post.

A simple example:
If $T > \pu{100 °C},$ and $p \leq \pu{1 atm},$ then water will be in the gas phase and:
$$\ce{CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)}$$
Note, however, that while this reaction's equilibrium will be highly insensitive to pressure changes (within the regime that all molecules remain in the gas phase), it will not be entirely insensitive to pressure.
That's because, even though there are the same number of moles of gas on the LHS and RHS, these are real gases, and thus their molar volumes will differ slightly. This is a small effect compared to the effect of actually having different numbers of moles of gases on the LHS and RHS.


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Does the non-serotonin-like half of LSD resemble anything of significance?
LSD's resemblance to serotonin explains many of its effects on the brain, but I noticed only half of it resembles serotonin. The question I have, is does the other half resemble anything, particularly anything of significance?
In this image the bottom half resembles serotonin:

In this image, the other half is shown, and I wonder if there's anything that it resembles.

[Image by By Ben Mills - Own work, Public Domain, https://commons.wikimedia.org/w/index.php?curid=14584151]

I don't see how the first structure is supposed to resemble serotonin. Anyway, as for the second fragment, it appears to be N,N-diethyl-2-methylpropanamide, CAS 33931-44-1.


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Reaction of Glucose + NaOH + PVP
I'm trying to synthesize of silver particles by using glucose as reducing agent, PVP as dispersant, and NaOH as pH control. Before adding silver nitrate solution, I've produced a solution which is combination of glucose, PVP and NaOH in distilled water.
In the polymer solution, I see that the color changes, from colorless to dark orange/brown solution. Is there any reaction that occurs in the polymer solution which contains glucose, PVP, and NaOH due to the changes of color?
I know from researching that the solution is expected to turn yellow when glucose and sodium hydroxide are mixed, which I observe, but I do not understand why the solution changes to darkish orange to brown when PVP is added?


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Geometry of the transition metal complex in solid dipotassium bis(oxalato) copper(II) tetrahydrate
I was thinking of the geometry of the transition metal complex in solid $\ce{K2[Cu(ox)2] · 4 H2O}$ solid around the $\ce{Cu(II)}$ ion.
At first I thought square planar, but is it possible or is there any chance that it could be octahedral with two water molecules since it is a hydrate?
Or do brackets always indicate what is directly complexed in this case?

Yes, square brackets are used to denote a coordination sphere, but they may sometimes be used incorrectly. Ideally you want to check the crystal structure, luckily there is one [1]:

The $\ce{Cu(II)}$ atom has square-planar geometry with two bidentate oxalate ligands. Each $\ce{K+}$ cation is coordinated by eight oxygen atoms from one bidentate and two unidentate oxalate groups and four water molecules.

Fig. 1. (a) Crystal structures of $\ce{[K2Cu(ox)2 · 4 H2O]}$ (1) and (b) $\ce{[K2Cu(ox)2 · 2 H2O]}$ (2) with atom numbering scheme. The thermal ellipsoids are drawn at 50% probability.

Reference

Fan, J.; Sun, W.-Y.; Okamura, T.; Yu, K.-B.; Ueyama, N. The X-Ray Crystal Structural Characterization of Dipotassium Bisoxalato Copper(II) Tetrahydrate, $\ce{[K2Cu(Ox)2·4 H2O]}$ $(\ce{Ox}$ = oxalate Dianion). Inorganica Chimica Acta 2001, 319 (1), 240–246. DOI: 10.1016/S0020-1693(01)00466-2.


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Carboxyllic acid Test
I'm currently in high school doing a project in chemistry. So I had to make a project where concepts from physical chemistry will be used alongside organic chemistry.
After going through my book I decided to create a test for carboxylic acids using electrochemistry. The basis of the test is:

Aldehydes, ketones and alcohols are non ionic in aqueous solution
while carboxylic acids are ionic. Therefore  carboxylic acid can be
used as an electrolyte.

So I finalized the following:
Lead, copper (my school doesn't have much chemicals and reagents and mostly available one was lead and copper so I stuck with it). The thing is, boiling the given organic compound with elemental lead, hydrogen peroxide and taking the precipitate obtained and measure the weight and dissolve it in test tube with water this acts as an electrolyte for lead half cell and make a copper half cell and connect the copper electrode and lead electrode with wires and put a suitable salt bridge. If there is reading in ammeter connected to the wires then the given organic compound contains carboxylic acid.
Is my method okay?
Edit:

1)since someone pointed out a lot of flaws ,  can I do like this:
Taking the aqueos solution of the organic compound as electrolyte and putting Zn,Cu electrodes in it and connecting the electrodes using a wire and connecting the wires to an ammeter ,(will this work like a lemon battery?)

Try to be clear. Which organic compound do you boil with lead and hydrogen peroxide ? What is the nature of the precipitate obtained ? Why do you weigh it ? Why and how is this precipitate suddenly soluble into water ? Another fundamental question : Why do you choose a carboxylate salt ? Of course carboxylates can be used as electrolyte. But carboxylates are decomposed at the anode through the Kolbe reaction, which gives CO2 and a hydrocarbon. Why use lead ? It is toxic ! Why not use Cu in CuSO4 at both electrodes ? And you weigh both electrodes before and after electrolysis.

while carboxylic acids are ionic. Therefore carboxylic acid can be used as an electrolyte.  This is not correct. Carboxylic acids CAN dissociate to ions, but there is a equilibrium and are called for the reason weak electrolytes. See Study.com  - lessons - weak-electrolyte-definition-examples E.g just about 1% of acetic acid is dissociated at pH 2.75. 6% vinegar has pH about 2.38, so just a fraction of % is present as ions. .

if the compound I took contains COOH then it boils with H2O2 and lead to form lead acetate, this precipitate is soluble in water, and it is white in colour. I didn't choose a carboxylate salt but I'm testing for it. If in the first step itself I'm not getting a precipitate then can I say compound  doesn't contain COOH? Please clarify me , I'm not clear that's why asked for help. And I'm weighing them because to record what molar aqueous solution I'm taking.

@Jagan. It is the first time you are speaking of acetic acid ! I could not guess it. Anyway electrolyzing lead acetate is not a good idea. Specially with a membrane. Lead is toxic. Acetate ions are decomposed by electrolysis. The membrane increases seriously the resistance of the bath.

I'm sorry but what exactly is a membrane here xD, I'm still at school and in my textbooks they haven't mentioned about membranes in electrochemistry

@Jagan Membranes in this context ( also  called a diaphragm ) means a separator of solutions at each of respective electrodes, that does not allow mixing both solutions, but allows migration of ions in the electrostatic field. A trivial example would be a filtration paper soaked by an electrolyte ( assuming the sale liquid levels ), or similar hydrophilic material. An alternative is a "salt bridge".

Thank you for all the responses and answers.

Let's be positive. The edited proposal has merit. You can pass a current thru a solution of a carboxylic acid and measure it with an ammeter (a more sensitive milliammeter would be better).
How about this: check for conductivity. Commercial conductivity meters use AC current and an AC ammeter. Use two inert, identical electrodes for this, and use DC if you can't arrange an AC apparatus. Do a conductivity measurement (i.e., measure the ohms resistance of the solution with the two electrodes taped to a certain distance separation (e.g., two copper wires (or stainless steel) 1 cm apart, 1-2 cm immersion, taped to a plastic ruler) so you can do the identical measurement on another solution.
Then, for a second test, examine the conductivity of that same solution diluted by a factor of 10. Now here's where you are on your own: the conductivity will be less, but not exactly 10 times less, because as the carboxylic acid is diluted, it will ionize to a greater percentage, and therefore carry more current at the same voltage than its concentration reduction predicts.
I say you are on your own, because the experimental results will not be as clear as you might wish (lots of opportunity for scatter), but this will enable you to explore the physical chemistry idea of the ionization of carboxylic acids and dilution and the question of voltage vs current, DC vs AC, electrode composition (reactivity); also bring up pH.
Above all, keep it simple! Consider two variables and plot one against another. You would be surprised how difficult that will be and how there are many more variables just waiting to hop into your experiment!
But you will have fun learning about all the variables and writing up the results and impressing your teacher with your comprehension. Good luck!


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The desirable property of plastics
My teacher once asked me that "one desirable property of plastics is that they can be made into very complicated shapes, Explain how this is done as a reaction"
I never know how to answer this question, can anyone help

It is not due to a reaction. It is due to the relatively low fusion temperature of the usual polymers. When heated a little above 100°C, polyethylene, polypropylene and similar polymers can fill any mould in the liquid state. When cooled down, the object is recovered

Is this the explanation

@Maurice This is only true for thermoplastics. But thermosetting or other plastics that cure are often liquids than can be used in complex moulds before they set.

The property of "moldability" can be based on several properties of plastics
There are broadly two types of polymer commonly used. Polymers that set and thermoplastics that can be molded when hot. Both of these types can have the desirable property of being liquids that can fill a complex mold before becoming rigid (though by far the commonest is the thermoplastic that merely requires heat to make it flow).
Thermoplastics can be supplied to manufacturers as beads of the already polymerised compound that can be melted in the feed to the moulding machine. Examples include polystyrene, polypropylene, polyethylene and many others. These form flowable liquids when hot enough so that a mold can be filled easily during manufacturing and the polymer will set into a rigid shape when cooled. This makes for very easy manufacturing.
This is how a very wide variety of complex shaped can be made. PET bottles, Lego bricks, toy soldiers, plastic shells for computers and printers, car bumpers and many, many other complex components are made this way.
But other complex components can be made from plastics that set from chemical reactions and cannot be thermally molded. These work as long as the setting reaction is slow enough and the components are liquid. There are a variety of such materials which usually involve mixing a (fairly) liquid monomer or oligomer (examples include methyl methacrylate, some silicones, epoxies and some polyesters. Usually the process involves mixing a controlled amount of the monomer (eg methyl methacrylate which was often the basis for home modelling kits) with another ingredient that triggers polymerisation (often a radical initiator). With the right mix of the two ingredients, the polymerisation speed can be controlled so the speed of polymerisation is slow enough that the liquid can fill a complex mold before starting to set. This process is more often used in small batches as it isn't always well adapted to speed because the curing process can take minutes to hours.
The thing that matters in both thermoplastics and thermosets is that the components are liquids that can fill complex molds before setting rigid. In some cases the setting mechanism is based on cooling; in others it is based on chemical reactions.


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When is ammonia and water a strong ligand?
While studying the spectrochemical series, my sir said that as ammonia and water lie in the center, their behaviour as a ligand is tough to judge. For example if we have CO as a ligand, we know that the electrons of the central atom tend to pair up, but he said that the same cannot be predicted for water and ammonia. Is this true, or is there some logic which will help to predict whether the electrons pair up or not. For example in [Co(NH)6]+3, the electrons pair up and the hybridization is d2sp3. But such pairing up is not seen at times in other compounds.

You should not be so concerned with whether the ligand is strong- or weak-field. That is only half of the story. After all, you are looking at a metal-ligand complex which can be high-spin or low-spin. Although the ligand plays an important part in determining that, the metal is equally important. See: https://chemistry.stackexchange.com/a/59171/16683

Also, "hybridisation is d2sp3" - please note that the use of hybridisation for transition metal complexes is an outdated and inconsistent theory: https://chemistry.stackexchange.com/q/76726/16683


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How many alternative chemical ways to transport oxygen in living creatures are known?
Mammals and many other groups of animals usually transport oxygen using haemoglobin and other complex proteins, the core of which is based on an iron coordinated to a porphyrin. There are plenty of subtle differences in the protein structures but are there any notable alternatives to that central iron-porphyrin structure?
Obviously there are other chemicals that can do a similar job. Polyfluorinated hydrocarbons were once considered as a viable emergency alternative to donor-blood as they are very good at transporting oxygen. But those are not something that living creatures are likely to develop.
So the question is are there either possible or known alternative chemistries for transporting oxygen in living creatures?

Mollusks and others use alternative complex using copper called hemocyanin. Reduced complex (Cu(I)) is colourless, oxidized (Cu(II)) is blue.

Some other types include copper and hem-like iron active centers: https://www.compoundchem.com/2014/10/28/coloursofblood/

I knew there was one with cobalt, but it isn't Vulcan, it's this purely artificial one: (https://en.wikipedia.org/wiki/Coboglobin

The answer, in short, is no. But there is a host of other oxygen-binding proteins https://en.wikipedia.org/wiki/Hemoglobin#Other_oxygen-binding_proteins

Yes! A variety of metalloproteins are used for oxygen transport. These include proteins that bind to porphyrins with different central metal atoms, as well as proteins that bind the metal atoms directly, without a porphyrin ligand. Fun fact- hemocyanin, which uses two copper atoms as prosthetic groups to bind oxygen, is responsible for the blue color of horseshoe crab blood!
Here are some relevant links:
https://en.wikipedia.org/wiki/Respiratory_pigment
https://en.wikipedia.org/wiki/Hemocyanin
https://en.wikipedia.org/wiki/Hemoglobin (scroll to "other oxygen-binding proteins" near the bottom)


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How come the C1 is above the ring in fructose-6-phosphate and C1 is below the ring in fructose-1,6-bisphosphate?
Does the phosphofructokinase-1 flip the position of the covalent bonds? If I've understood correctly, C2 is a chiral atom, so it would make a difference if C1 goes from above the C2 to below the C2.

A picture would help.

In the cell, both the phosphate and the bis-phosphate occur predominantly in the beta form. As hemi-ketals, both rings can open and isomerize. This changes the stereochemistry of C2, the anomeric carbon. However, enzymes might be specific for one or the other form (in the case, they would react with that form, and more of that form would be made through isomerization of the other, non-substrate anomer).
For the E. coli phophofructokinase-2 enzyme, crystal structures of both substrate (3uqd) and product (3n1c) are available. In both models, the sugars are in the beta anomeric form.
According to wikipedia, this is the beta form of fructofuranose:

Again according to wikipedia, this is the beta form of fructofuranose 6-phosphate (but comparing it to the non-phosphorylated form, it is not):

Finally, again according to wikipedia, this is the beta form of fructofuranose 1,6-bisphospate:

So there is a name/structure mismatch for the middle structure, which might have propagated through the internet.


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How can we define optical isomerism?What is the example of non enatiomeric optical isomers?
My book has very vaguely explained these concepts,I can't understand anything from there.
Through a little bit of research on the internet I came to following conclusion:

that optical isomers are either dextrorotatory or levorotatory
all enantiomers are either dextro or levo so they are optical isomers.
Some diastereomers may also be related as d- or l- (though I feel the angle of rotation will be different) and thus can be optical isomers.

I feel I am missing something or have the wrong concept entirely.
These are pics from my book.
In the heading of optical isomers there are 4 isomers of tartaric acid mentioned,why is that?So how many types of optical isomers are there?

Your book is pretty bad, I'm afraid to say. The most helpful thing for you would probably be for you to find a better book. But in the meantime, it would be better if you could (a) say which book it is, and (b) don't post screenshots, but rather quote the relevant paragraphs of text. You can [edit] your post to include this information.

There are, in the sense you are asking the question, three types of optical isomers: +, -, and optically inactive. This is detailed in pages of the textbook you have uploaded.
You should think about optical isomers in how they are determined. Imagine you have a machine which we will call and "optical isometer". To use this machine, you put in any compound in solution, and it will tell you one of three outputs:

"This compound polarises light in the clockwise direction"
"This compound polarises light in the anti-clockwise direction"
"This compound does not polarise light"

Now, our optical isometer doesn't know any chemistry, but it does know how to tell if light is being polarised by our compound in solution. If you fundamentally think of isomers as identical in their molecular formula, but differing in their spatial arrangement of atoms, you need only imagine as optical isomers being different in how the spatial arrangement of atoms polarises light.
Let's imagine we have a solution of a tartaric acid, but we don't know what isomeric form it is in. The options are:

meso-tartaric acid
(+)tartaric acid
(-)tartaric acid
A mixture of (+) and (-)tartaric acid.

Where I think you have been confused by the textbook is that option number 4 doesn't actually describe different spatial arrangements of tartaric acid, rather it describes a mixture of options 2 and 3. So when it says there are four isomeric forms, it isn't strictly correct about that.
Now, how will we determine what isomer of tartaric acid we have in solution? One thing that we can do to find this out would be to use our optical isometer. We put our solution in and it outputs: "This compound polarises light in the clockwise direction". We know that of our four possible options, only (+)tartaric acid polarises light in the clockwise direction. Likewise, we would know it to be (-)tartaric acid if the light were polarised in the anti-clockwise direction.
What if our optical isometer output, "This compound does not polarise light"? Then this means that our compound is optically inactive. We know that both the meso- form and the racemic (mixture of (+) and (-)) don't polarise light in either the clockwise or anti-clockwise direction, and so we would need further analysis to find out for sure!
There are other types of light polarisation, but not in the sense this book is talking about optical isomers. This textbook seems to be a little confusing, but these ideas are actually really well explained on places like wikipedia or khan academy, so next time you get stuck, don't worry too much about the textbook but give other resources a shot!
Hope this helped,
Edan

You're starting backwards here.
The first term to start with is chiral. A chiral object is non-superimposable on its mirror image.
A chiral molecule has two enantiomers--the two optically active mirror images. The optical rotation is opposite (levo versus dextro) for each of the pair.
Diastereomers are structures that have the same connectivity but are not enantiomers. These most commonly result from a molecule having multiple stereocenters, where we can identify pairs of such molecules with the same connectivity but the sense of chirality in the stereocenters are not all the same nor all different.
In the case of tartaric acid, there is a chiral form. The chiral form has two enantiomeric isomers that have opposite optical rotation. Then, there is a third isomer that is diastereomeric to the other two. Due the level of symmetry (and this will be hard to explain without going into more detail), there is no separate fourth isomer.

hmm..but I still dont know how to define optical isomerism.and why diastereomers are written in the heading of optical isomers?

Diastereoisomers ARE a type of optical isomer. An optical isomer, more generally known as a steteoisomer, or spatial isomer differs from other spatial isomers only in the spatial disposition of its atoms. Changes in composition and connectivity is not allowed between steteoisomers.


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Explanation for a basic decomposition of water experiment
My son is trying to write a conclusion for a basic decomposition of water experiment using a 9 V battery (with a test tube placed over each terminal) placed inside a salt water solution.  The results were: the tube over the negative terminal nearly completely filled with presumably hydrogen gas while the tube over the positive terminal only had a minute amount of presumably oxygen gas.
His textbook didn't explain these results, other than to hint at the fact that the results were due to there being twice as many hydrogen atoms in the product.  He has several questions that I cannot answer.

If the hydrogen and oxygen molecules take up the same amount of space, then why did the tube over the negative terminal have much more than twice the gas as the other one?
The atomic mass of oxygen is nearly 16 times that of hydrogen, so then even if there was twice as much hydrogen in the product, wouldn't the oxygen still take up more space?

Electrolyzing a NaCl solution does produce hydrogen at the cathode, but no oxygen at the anode. Chlorine $\ce{Cl2}$ is produced at the anode, with maybe a small proportion of oxygen as an impurity. Unfortunately this $\ce{Cl2}$ gas is relatively soluble in water. That is why you have obtained relatively few gas at the anode. The mas of the atoms has no effect on the volume of the gases produced at each electrode.

If the hydrogen and oxygen molecules take up the same amount of space, then why did the tube over the negative terminal have much more than twice the gas as the other one?

Good questions but keep in mind that you did not mention the concentration of sodium chloride. If you have too much salt in water, chlorine gas is produced (eventually bleach) instead of oxygen. This is why you did not see the expected 1:2 vol. ratios.
If possible, use the lowest possible salt concentration for the experiment or even better, can you use a small amount of Epsom salt? Then there is no issue of chlorine formation and only oxygen gas will form. You will certainly see 1:2 vol. ratio.

The atomic mass of oxygen is nearly 16 times that of hydrogen, so then even if there was twice as much hydrogen in the product, wouldn't the oxygen still take up more space?

This issue was tackled by scientists long time ago. The key point is that if you have 1 billion molecules of $\ce{H2},$ and 1 billion molecules of $\ce{O2},$ they will occupy the same volume at a given temperature and pressure. There is a named law behind it. The volume of molecules way too small as compared to the volume they occupy.
The radius of oxygen molecule is 152 picometer (symbol: $\pu{pm}$) and that of hydrogen is 120 picometer $(\pu{1 pm} = \pu{10^{-12} m}),$ so masses and volumes do not correlate exactly.

IIRC, electrolysis of water is actually electrolysis of sulphuric acid. However, it is difficult for the layman to get, apart from old lead-acid batteries (and is too dangerous for children to experiment with). @MFarooq, Epsom salts is magnesium sulphate.  Does this provide the sulphate ions in the same way as sulphuric acid does?

@CSM, Interesting point. In order to make water conductive which has a very resistance, we need ions to carry current which can be protons (addition of acid) or even hydroxide (from KOH, NaOH), or just a neutral salt like KNO3, Na2SO4, and MgSO4. All work. It is the electrolysis of the water molecule.

Oxygen is soluble somewhat well in water, while hydrogen ist hardly soluble. Depending on how long the experiment lasted, this might explain the difference.
Regarding your second question: Gases you would reasonably encounter in school experiments can well be approximated by an ideal gas, that is, they all occupy the same volume per amount of substance.
These explanations are based in my recollection of my own school education, so take them with a grain of salt.
As for M. Farooq and Maurice’s remarks regarding the formation of chlorine gas: My teacher used a mild acid (citric acid or similar) to produce the required ions to conduct electric current in the water. This is something you might try as well.

From what you describe, I think the results depend a lot on (1) the salt added to the water to enhance conductivity, and (2) the composition of the wires used.

As other answers mention, using a table salt (NaCl) electrolyte can and usually will produce chlorine gas at the positive terminal (anode). This is because chlorine is easier to produce than oxygen for reasonably high salt concentrations. You will probably notice a chlorine smell if there is chlorine being produced. If you used baking soda as the electrolyte (which I recommend and think makes for an easier-to-explain experiment) only oxygen gas should be produced at the anode. Even with only oxygen, you might not get it all in a test tube, because...
The wires may be reacting with and using up the oxygen or chlorine produced at the anode. If your solution has turned green/blue (likely copper wires) or rusty orange/brown (iron) during the experiment, then this is likely the culprit for where your missing oxygen is. You are extracting pure oxygen or chlorine right at the metal surface, so it makes sense that the metal would react! Whatever doesn't react with the electrode would float up into your collecting tube.

As previous answers have already mentioned, for gases at reasonably low pressures (less than ~5 atmospheres), all the molecules regardless of weight take up about the same space, so you can get away with using the Ideal Gas Law. So double the number of molecules means double the volume.
I'm sure you've already looked at the Electrolysis of Water Wikipedia page, but I include the link because it's probably helpful in writing up your son's results. Plus, there are a lot of links in it to more references if that's not enough for you. Good luck with the report!


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What colour ranges would a solution of Methyl Orange and Thymol Blue have
Would the solution of Thymol blue and Methyl orange tell the difference between an acid with ph 2.1 and anther acid with ph 2.2?
I am a chemistry beginner, so I don't understand the reactions on that good of a level.

It is rather a question, if your eye is able to tell the difference. Thymol blue has range 1.2(red)-2.8(yellow), methyl orange 3.1(red)-4.4(yellow). So effectively 1.2(red)-2.8(orange=yellow+red). It is possible, but I would not bet on being able to notice 0.1 pH difference, putting it side by side.

The answer of this question depends on the relative amount of Thymol blue and Methyl orange. If the acidic mixture contains roughly equal amounts of the indicators, or more Methyl orange, the only thing to be observed in an acidic solution is that its color is red up to about pH $4$, where Methyl orange turns yellowish. If the Methyl orange is present in rather small amounts, its effect of the color change would not be significant : the color change from red to yellow would be visible at about pH $2$. Whatever your choice of indicators, there will not be any visible difference between pH $2.1$ and pH $2.2$.


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Need something that is dishwasher safe that breaks down paraffin
I work for a company that has a lab where there work leaves a lot of dishes coated in paraffin. Today I learned that they wash these in a non commercial dishwasher to remove the stuff.... one issue they paraffin solidifies and gunks up the washer. They have been replacing these every few months or so. After talking with the lab staff, they expressed concern and are open to using an extra step to remove the paraffin before washing. What most of the internet has to say is obviously do not wash paraffin in a dishwasher. Feeling a little s.o.l here, does anyone know anything that could potentially help, so we can end the vicious cycle of dishwasher destruction.

Is this paraffin wax that is on the dishes? If so, then a wash with heptanes should dissolve it off. Collect the washing in a sealable bin for disposal.

In my previous lab, they used a lot of paraffin for making microscope slides. They kept the tools in an oven at 60 C to melt the paraffin and just let it drip off the tools. Everything was coated in newspapers to catch any paraffin that dripped off onto tables or the floor. So maybe you could try heating the dishes to about 60 C to melt off excess paraffin.

On edge, into a catch container, of course.

There are dishwashers made for using solvents in them. Tend to be pricy, but so it goes.

This may be doable, per a source, to quote:

Paraffin wax is mostly found as a white, odorless, tasteless, waxy solid, with a typical melting point between about 46 and 68 °C (115 and 154 °F).

So as a pre-treatment, try applying a jet of steam to facilitate the removal of the paraffin wax.

Even a heat gun as a first treatment. Or a lab oven (put a layer of aluminum foil on the bottom to catch the wax).


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General Question about transferring Solution from containers (Workplace Related)
At my workplace we will soon begin diluting a solution to specific concentrations, this is done within a specific container.
The solution to be diluted is provided by a manufacturer in the following type of bottle :

The contents inside this bottle can only be retrieved with a syringe. However, we need to transfer volumes in the ranges of $ 0.015 \,mL - 0.06 \,mL $  ,with syringes we cannot be confident (because we have to eyeball it real hard).
So we think of course to transfer from the bottle to a beaker and then using a micropipette to transfer to the desired container and dilute. The problem is that the solution to be diluted is extremely expensive , and every unnecessary transfer to a beaker will cause some of our solution to be lost , beakers need to be also washed THOROUGHLY if they are to be reused.
Question : Are there any viable instruments that have a needle to enter the bottle and enough accuracy to transfer $ 0.015 \,mL - 0.06 \,mL $ ? Can you think of any way we can get around our problem?. We hope we can do just 1 transfer from Manufacturer bottle to our desired container to be diluted. Upon researching online we have not had much luck.

Opening , and replacing the cap of Manufacturer bottle gives problems with sterile properties of solution , as well of greatly reduces the expiry date.

Transfering from manufacturer bottle to beaker and then micropipetted to desired container to be diluted , gives rise to problems of sterile , as well as time consuming due to having to thoroughly clean the beaker, causes loss of solution.

I know this question may not be appropriate for this site , but I do thank you for your help and time.

Do you have Hamilton syringes? That is the way, along with appropriate use of them.

@Alchimista these look promising! they are a bit in the expensive side but i suppose there's not much that can be done , thanks for the input! Have to check cost vs earning to see if they are worth it.

Yes, they are expensive and requires proper operation and care.


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Why does the Avogadro's number have the unit of reciprocal mole?
In my understanding,
$$1 \ mol \equiv N_A \ (atoms)$$
(Don't say it might not be atoms but molecules and also don't say what's it's not even a unit)
Then the unit should be $N_A = 1  \ mol/atoms = 1 \ mol$, not the reciprocal mole. My first question is what's wrong.
My second question is, we know that defining $atoms$ as being unit is absurd, because it's purely a number. And same applies to mole. There's not difference between them in the sense that they are dimensionless. But why is the mole a SI unit?

The unit is $\pu{1 mol} = N_\mathrm{A} \text{ particles}$ . What is strange on that ? Note that there is the unitless Avogadro number and the Avogadro constant with dimension $\pu{mol^-1}$. Then $n [mol] \cdot N_\mathrm{A} [1/mol] = N$. The molar amount with 1 mol as the unit is useful concept, as many quantities relate rather to the number of particles than to their total mass.

You may take any non-unit and call it a unit, and as long as it doesn't contradict anything, you'll be fine. That's why mole is a unit.

"we know that defining atoms as being unit is absurd, because it's purely a number." the same applies to dollars or what ever currency. It seems you are fine with more abstract units (length for instance) but have problem with countable hardware.

Does this answer your question? What is the dimension of Avogadro's constant?

I realised where my confusion lies - so I'm leaving an answer so that other people learning chemistry for the first time are able to fully understand what's actually going on about Avogadro's number.
We have to make a clear distinction between Avogadro's number and Avogadro's constant.
The former is a unitless real number, defined as $6.02214076\times 10^{23}$.
The latter is a physical (or shall I call chemical?) constant, and thus has a unit. As @Poutnik and other people have commented on my question, the fact that $mol$ does not have a dimension doesn't mean it can't be a unit (the same applies to radian as a measure of angles); and that means there should exist a constant called the Avogadro's constant with the dimension $mol^{-1}$ so that the product of the two is dimensionless, i.e.
$$n \ [mol] \ \times N_A \ [mol^{-1}] = N \ [particles]$$
Here it's trivial that [particles] is not a unit, simply because we did not feel the need to (and thus chose not to) assign any unit, dimension, and whatever onto that thing called number of particles.
Again, although $mol$ is dimensionless it should still be treated as a unit so in order to cancel that out to produce a unitless number $N$ we can conclude that we need a number with value $6.02214076\times 10^{23}$ and unit $mol^{-1}$, or namely Avogadro's constant.

dimension of the mole is mole, as of the kilogram is kilogram and  of the metre is metre. Where is the problem ?

You have solved your confusion but please see Poutnik comment.

@Poutnik Yes you are right. I agree with that, and in my answer I was just emphasising that. I was originally confused about that.


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Water solubility of different types of Iron III
For an experiment I require an aqueous solution of Iron(III). My school has the following compounds, which of them are water soluble? I believe that only Iron(III) oxide is insoluble, but I would like to verify.

Iron(III) oxide
Iron(III) oxalate
Iron(III) chloride-hexahydrate

Many thanks in advance.

Have you thoroughly consulted the topic with textbooks, Wikipedia and Google ?

Yes, according to Google, Iron(III) Oxide is insoluble, Iron(III) Oxalate and Iron(III) Chloride-Hexahydrate are soluble. Is this correct?

Ferric oxalate is reportedly rather just slighly soluble, so for practical means, consider it insoluble. Ferric chloride hexahydrate is well soluble, routinely used in dissolving copper from printed circuit boards.
You can find this quite easily.

Next time, more thorough searching.

Anyway the choice of Iron(III) oxalate is not happy, because it is a substance that can only be synthesized in the darkness. It is destroyed by the visible light, according to the equation $$\ce{Fe2(C2O4)3 + light ->  2 FeC2O4 + 2 CO2}$$ This reaction  is even the basis of the photochemical process called heliography.

Something similar happens for ammonium ferric citrate.

Iron(III) oxide is insoluble.
Iron(III) oxalate is, in practical situations, insoluble.
Iron(III) chloride-hexahydrate is soluble.

Aditionally, Iron(III) oxalate is destroyed by visible light, and I intend to complete my experiment during the day, especially considering that involves colour changes.


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Chiral centres with stereoisomeric substituents
The structure of one particular isomer of 4-chlorohepta-2,5-diene is shown here. I am confused about how two of the propenyl groups attached to the central carbon have the same atoms with different configurations.

More specifically:

Is the carbon shown in the image is a chiral carbon or pseudo-chiral carbon?

Will this compound show optical isomerism, i.e. does it have an enantiomer?

Are the two alkenyl groups considered to be different from each other, or the same?

If the attached side centers were chiral centers (e.g $\ce{-CHClCH3}$) with different configurations, instead of geometrical isomers, then would this compound show optical isomerism?

the carbon connected to 4 different groups is chiral, because it's connected to 4 different groups. 2) weather the whole molecule shows optical isomerism depends on one parameter. if the molecule has reflective symmetry or not. A molecule with no reflective symmetry is chiral. So I think this would be a chiral molecule. 3) the 2 circled groups are different. they are also stereoisomers E and Z or cis and trans. Why are they that? because you can't turn one group into another without breaking the double bonds. 4) I don't understand what you're asking here. An edit is needed.

Let's address your concerns one at a time:

The structure 1 you have drawn is chiral with $\ce{C4}$ having four different groups attached to it. $\ce{C4}$ has the (S)-configuration with the CIP priorities Cl>(Z)-propenyl>(E)-propenyl>H.
The structure has an enantiomer of the (R)-configuration. Just switch any two groups: Cl, (Z)-propenyl, (E)-propenyl or H. Each enantiomer will have the same specific rotation but opposite in sign.
The two propenyl groups are different stereoisomers, just as different as the (Z)- and (E)-2-butenes.
If the two propenyl groups were changed to -CHClCH3, say, (2R,4R)-2,3,4-trichloropentane 2a ($\ce{CH3CHClCHClCHClCH3}$), then the central carbon, $\ce{C3}$, is chirotopic but non-stereogenic. That is to say, if the chlorine and hydrogen at $\ce{C3}$ were switched as in 2b, the compound would still be chiral but not a different stereoisomer. Rotate 2a by 180o about a vertical axis and you will obtain 2b. They are identical.

If the two propenyl groups were changed to -CHClCH3, say, (2R,4R)-2,3,4-trichloropentane 2a (CH3CHClCHClCHClCH3), then the central carbon, C3, is chirotopic but non-stereogenic. I am unable to understand your this line. I want to know if these geometrical centrical were replaced by chiral centres like in 2a and 2b then the C3 carbon will remain chiral or not?

I chose the 2R,4R-isomer. If I had chosen the 2R,4S-isomer, then C3 would be achirotopic and stereogenic. Two diastereomers would arise: 2R,3r,4S and  2R,3s,4S. This link should help for both your original diene and the trichloro examples: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/RSrs.html


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Significance of enthalpy of hydration of salts
I've been researching for weeks but can't find anything on a real life process of hydration, as my experiment is determining the relationship between the enthalpy of hydration and different types of salts. When would knowing the enthalpy of hydration be vital outside the lab?

Do you have doubts about hydration  enthalpy specifically, or about enthalpy in general ?

abd.shd_247 , welcome to the exchange!
The enthalpy of hydration would be necessary to understand in the industrial context:
One of my professors of Analytical Chemistry told me that, at a plant he used to work for,  a group of engineers neglected to factor in the enthalpy of hydration when making vats of concentrated NaOH in massive drums. As such, they added water to the NaOH salt, of which there was hundreds of kilograms, and as the water began to hydrate the salt, the heat was so great that it melted the steel on the drums and flooded the bottom floor of the lab with a highly corrosive NaOH solution.
As such, the enthalpy of hydration will always be an important factor in industrial chemistry, not just in gen chem labs.


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Reaction of unsaturated lactone and alcohol

Image source
I found this equation to turn this compound into an ketoester but I am unsure of the reaction mechanism.
I postulate that the alcohol attacks the carboxyl carbon, the O becomes formally negative. It reforms the double bond and the leaving is the O in the ring which takes the electrons in the bond.
Here my thought differs from the image (gotten from one of the United States Patent Office Patents). I thought the end product would be an ester OHCCCCOOR. Alcohol then alkene and at the end an ester but the picture shows a ketone and no alcohol.

You are correct that the alcohol atacks the carbonyl then the ring oxygen leaves in a manner analagous to ester transesterification. The intermediate generated is an enol which picks up a proton to for the ketone.

image from here: 1

Oh keto-enol Tautomer. I sometimes have trouble seeing where this can happen. Is there a way to know where this would occur or do you simply have to see many examples and at a certain point you can just see it?

Most of the time the keto form prevails unless there's extended conjugation. It is worth noting that in this molecule enolisation towards the ester group is the more favoured.