text:
What happens when I divide two molar flow rates in a chemical equation?
I don't quite know how summarize my question in the title. I have a combustion process given by this equation:
$$\ce{C10H22 + 15.5 O2 + 58.25 N2 -> 11 H2O + 10 CO2 + 58.25 N2}$$
This process occurs inside of a cylindrical combustion chamber of area A, but I think this is irrelevant for the purposes of my question. The combustion process is stationary, and the gas mixtures are ideal.
I wanted to know if, in general, this is true:
$$\frac{\text{molar flow rate of }\ce{N2}}{\text{molar flow rate of }\ce{C10H22}} = \frac{\text{amount of substance of }\ce{N2}}{\text{amount of substance of }\ce{C10H22}}=\frac{58.25}{1}=58.25$$
With molar flow rate I want to mean the amount of substance that leave/enter the combustion chamber in 1 second.
I'm a beginner in chemistry, but I'm not sure if I can convert a ratio between molar flows like I did, using directly the stoichiometric coefficients in the balanced equation.
The idea behind my reasoning is that:
$$\text{molar flow rate of }\ce{N2} = \frac{n_{\ce{N2}}(t+\Delta t) - n_{\ce{N2}}(t)}{\Delta t}$$
and:
$$\text{molar flow rate of }\ce{C10H22} = \frac{n_{\ce{C10H22}}(t+\Delta t)-n_{\ce{C10H22}}(t)}{\Delta t}$$
Therefore,
$$\frac{\text{molar flow rate of }\ce{N2}}{\text{molar flow rate of }\ce{C10H22}} = \frac{n_{\ce{N2}}(t+\Delta t) - n_{\ce{N2}}(t)}{n_{\ce{C10H22}}(t+\Delta t) -
 n_{\ce{C10H22}}(t)}$$
And, intuitively, this last equality should be proportional to $n_{\ce{N2}}/n_{\ce{C10H22}}=58.25/1=58.25$.

Welcome to Chem SE! If you haven't yet, take the [tour]
[https://chemistry.stackexchange.com/tour] and if you have any other questions stop by the [help center] [https://chemistry.stackexchange.com/help]. Could you clarify in what context you have seen an integrated heat capacity? It is not a term I'm familiar with.

Hi, thank you! I'm studying the combustion chamber of a civil jet, and need to study the thermodynamics that happen in there. I'm currently doing an energy balance, which includes mass flow rates. But mass flow rates can be converted to molar flow rates, which in turn lead me to this question.

Looks like solid reasoning to me.

Look, the molar flow rate is a production or consommation of a number of mole per unity of time. Your process is stationary so there is no variation of concentration with the time. What it says is if you do your reaction from 9:00 am to 10:00 am or from 9:10 am to 10:10 am your reaction will spend and produce exactly the same amount of each molecules in both cases. So there is no limitation to do what you did.


text:
What is the chemical formula for snake venom?
I haven't been able to get this question answered through the internet so far, so I was looking for a more professional angle in which this question can be answered.

Snake venom is almost entirely made up of proteins...thousand of different proteins, which means thousands of different chemical formulae.

In cobras its a protein unimaginatively its called cobratoxin. (You can see x-ray structures on web). The active site has been studied and shown (by time resolved fluorescence; it has a singe trp) to be rather rigid and by being so mimics the geometry of curare type molecules.

There are several kinds of snake venom, and as far as I know they are all complex enzymes: it means that they are complex macromolecules, made by amminoacids (these, "small molecules") bound with peptide bonds.
For enzymes and proteins, the chemical nomenclature is quite impratical: it's way easier to define the enzyme by giving the names of the amminoacid chain it's made of.
For instance, for crotamine the amminoacid sequence is: YKQCHKKGGHCFPKEKICLPPSSDFGKMDCRWRWKCCKKGS—G

The wikipedia article of snake venom clearly defines the composition of snake venom. To prevent being a wall of text, I arranged them in points and bolded the chemicals.

Proteins constitute 90-95% of venom's dry weight and they are responsible for almost all of its biological effects. Proteins are
  toxins, neurotoxins in particular, as well as nontoxic proteins (which
  also have pharmacological properties),
Enzymes (molecular weight 13-150 KDa) make-up 80-90% of viperid and 25-70% of elapid venoms: digestive hydrolases, L-amino acid
  oxidase, phospholipases, thrombin-like pro-coagulant, and
  kallikrein-like serine proteases
  and metalloproteinases (hemorrhagins), which damage vascular
  endothelium.
Polypeptide toxins (molecular weight 5-10 KDa) include cytotoxins, cardiotoxins, and postsynaptic neurotoxins (such
  as α-bungarotoxin and α-Cobratoxin), which bind to acetylcholine
  receptors at neuromuscular junctions.
Compounds with low molecular weight (up to 1.5 KDa) include metals, peptides, lipids, nucleosides, carbohydrates, amines, and oligopeptides, which inhibit angiotensin converting enzyme (ACE) and
  potentiate bradykinin (BPP). Inter- and intra-species variation in
  venom chemical composition is geographical and ontogenic.
Phosphodiesterases interfere with the prey's cardiac system, mainly to lower the blood pressure.
 Phospholipase A2 causes hemolysis by lysing the phospholipid cell membranes of red blood cells.
Amino acid oxidases and proteases are used for digestion. They also triggers some other enzymes and is responsible for the yellow
  colour of the venom of some species.
Hyaluronidase increases tissue permeability to accelerate absorption of other enzymes into tissues.
Some snake venoms carry fasciculins, like the mambas (Dendroaspis), which inhibit cholinesterase to make the
  prey lose muscle control.

While I appreciate this answer, the list format that was necessarily adopted, as well as the inevitable brevity of each entry, make me doubt the usefulness of the question.

Snake Venom is a mixture of chemical compounds. It certainly doesn't have "a chemical formula". Your question is similar to one asking "What is the color of a car?" Just like cars come in a wide range of colors, venoms have a wide variety of recipes (mixtures) and even snakes of the same species may have different compositions of their venoms.
More details are easily available, and easily found on-line. For example: https://en.wikipedia.org/wiki/Snake_venom and https://en.wikipedia.org/wiki/Venom#Snakes. For more specific information, you should A. Determine the formal taxonomic name of the species you are interested in and then B. Use Google Scholar or some other search engine which includes the academic literature. Many of these articles won't be satisfactory, I'd guess, because they generally address specific ingredients of the "witch's brew". But you'll most likely find some older articles which are broader. (Many of which will be behind pay-walls, unfortunately. You may have to visit the nearest public university for "free" access.

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text:
PCC with substituted pyridines
I was wondering if substituted pyridines can work the way pyridine does in order to oxidize alcohols to aldehydes.
Of course, I'm referring to pyridines without any functional group which can undergo oxidation reactions (eg: I was thinking methylpyridine, or trimethylpyridine, with the nitrogen untouched.)
I guess that the reactivity is the same, but is there any reference to the use of this kind of reagent?

I think the main role of pyridine is to abstract the chloride ion from the intermediate formed leading to an elimination reaction. So yeah, the substituted ones should also work. However, its efficiency might be reduced as the +I effect of the substituents will decrease the net positive charge on nitrogen in pyridine.

Collidinium chlorochromate and gamma-picolinium chlorochromate are known oxidising agents. e.g. Synth Comm. 31 2001 1253
I see no reason why other methyl pyridines should not also work.

There may be steric issues with bulky primary alcohols.

Triggered by the two chemical names mentioned, I found  it interesting to browse before and after this page google books provides: https://books.google.de/books?id=z7KZFQvfM3gC&pg=PA12&lpg=PA12&dq=picolinium+chlorochromate&source=bl&ots=T8NSAgErlb&sig=zP3B5i1hWC75odrHJXx8_-AcC2o&hl=de&sa=X&ved=0ahUKEwjrks6htKfTAhVCCSwKHc8QBewQ6AEISzAF#v=onepage&q&f=false

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text:
Are alkylidene cycloalkanes planar molecules?

Above is an Alkylidene Cycloalkane. Its structure closely resembles that of allenes.

According to my knowledge, allenes are non-planar molecules due to the two $\pi$-bonds being formed on almost perpendicular planes.
So due to the similarity in structures and optical activity, are Alkylidene Cycloalkanes also non-planar?

Yes , for two reasons. One as you suppose due to the (ideally) perpendicular angle of the H and CH3 on the cyclohexane ring relative to the H and COOH  and also due to the cyclohexane ring itself being puckered.

Related: Why is 1-ethylidene-4-methylcyclohexane chiral?

The alkylidene cycloalkane contains only two C($sp^2$) hybridised atoms, ajacent to each other, as in the instance of a "normal" double bond.  Hence I expect that in the first sphere / in $\alpha$ position to this double bond, all four substitutents adjacent to this double bond may be found almost in the same plane including the axis along the $\ce{C=C}$ bond, too.  The difference to a "normal" case of a single / isolated double bond more frequently seen in textbooks is that two of the substitutents of one side of the double bond are tethered together.  On the level of the double bond, they remain planar.
In the allene example, there are three consecutive C($sp^2$) hybridised atoms, with two adjacent $\ce{C=C}$ bonds. This differs from the instance above, and here the allene protons of one side and their carbon atom they are bond to describe -- as you correctly referred to -- a plane  almost perpendicular to the plane defined by the other two allene protons and carbon atom of the other side of this molecule.  At the level of the double bonds, they are non-planar.
Regarding the methyl group across the cyclohexane, I expect this (again) is preferentially oriented equatorial, rather than in axial alignment towards the main plane of the cyclohexane ring.


text:
Question regarding the mechanism of oxymercuration-demercuration of alkenes

Organic Chemistry by David R. Klein
The text in between says that the more substituted carbon bears a partial positive charge. Why not the other one? I am asking this considering inductive effect, although I am myself not too convinced by my inductive effect explanation.

This reaction, in my opinion is an extension of Markovnikov’s rule
To understand why Markovnikov’s rule works, consider the
structure and stability of carbocations and about the general nature of reactions
and transition states.
The first point to explore involves structure:

A great deal of experimental evidence has shown that carbocations are
  planar. The trivalent carbon is sp2-hybridized, and the three
  substituents are oriented toward the corners of an equilateral
  triangle.

Because there are only six valence electrons on carbon and all six are used in the three s bonds, the p orbital extending above and below the plane is
unoccupied.
The second point to explore involves carbocation stability. Thermodynamic measurements show that,
indeed, the stability of carbocations increases with increasing substitution so that
the stability order is tertiary > secondary > primary >methyl.

Now here comes the question…
Why are more highly substituted carbocations more stable than less highly substituted
ones?
There are at least two reasons.
Part of the answer has to do with inductive effects (as you noted in question) , and part has to do with hyperconjugation.

Inductive effects, result from the shifting of electrons in a s bond
  in response to the electronegativity of nearby atoms. (electrons from
  a relatively larger and more polarizable alkyl group can shift toward
  a neighboring positive charge more easily than the electron from a
  hydrogen

. Thus, the more alkyl groups there are attached to the positively charged carbon, the more electron density shifts toward the charge and the more inductive stabilization of the cation occurs)
Hyperconjugation, is the stabilizing interaction between a p orbital and properly oriented C- H sigma bonds on neighboring carbons that are roughly parallel to the p orbital. The more alkyl groups there are on the carbocation, the more possibilities there are for hyperconjugation and the more stable the carbocation


text:
What makes scorpions glow under UV light?
What substance (or class of substances) is responsible for the neon blue-colored fluorescence we observe when we shine UV light on scorpions?

Do note, I want to know what substance makes them glow and not why they glow... the latter being more suited for the Bio.SE.

See: http://www.kidsdiscover.com/quick-reads/makes-scorpions-glow-ultraviolet-light/

Interesting link here http://www.sciencedirect.com/science/article/pii/S1074552199800854 "The fluorescence of scorpions and cataractogenesis"

Ironically, I would have said that identifying the substance is more Bio.SE and why it fluoresces is more Chem.SE.

ever-so-slightly helpful info in video: "a nitrogenous substance in its cuticles"  More helpful, but WIRED is not a scholarly reference: Two compounds are involved in scorpion UV fluorescence: beta-carboline and 4-methyl, 7-hydroxycoumarin..

@Zhe I spent quite a bit of time deliberating over the wording I ought to use. Being a Bio. student myself, "why it fluoresces" would possibly twist the meaning of this question. For example, asking how fireflies glow is different from asking why they glow: the former occurs due to the breakdown of luciferin (this lies in the domain of Chemistry), whereas the latter is (probably) a method of locating mates at night (this is Bio. turf). Now re-read the question again ;)

Re-reading doesn't help. If you asking which specific substance makes them fluoresce, that is definitely a bio question. If you're asking about the principles of fluorescence, then it would be a chemistry question. Your wording suggests the former.

I'm voting to close this question as off-topic because identifying the specific substance (quite possibly a complex macromolecule) is more suited for Bio.SE.

I think that it is a chemistry  question, the chromophore is not unlike an indole /trp molecule in a protein, and properties of proteins is to me is part of chemistry. If as chemists we do not embrace new areas  our subject will fade away.

@Zhe How is asking which specific substance fluoresces not chemistry?  I strongly disagree with your idea that it isn't.

@matt_black You're not talking about spectroscopy or the process of identification. You're simply asking what it is (in terms of its biological function). This is not what chemistry is for. Otherwise, computer hardware engineering is chemistry because there are substances that make up the machine, and we care what they are. Alternatively, everything is just baryons and leptons interacting with photons. Why shouldn't the physicists answer this question?

@porphyrin We're not asking about properties. The property--fluorescence--is already known, and OP has specifically indicated that "why they glow" is not the point of this question. We're not talking about analyzing a sample for substance identification. We're asking about a specific substance in a biological context. Sure, it's a chemical but nowhere did we need to use chemical principles in a pure or applied manner.

@Zhe Biological function isn't the issue here as the questioner clearly stated. Identifying which molecules are responsible for something is, arguably, the core of chemistry. Compare: "which molecule cause the smell of rosemary" (clearly chemistry) with "why does rosemary smell" (probably biology).

@Zhe, not sure I understand your comment. How can you know what a substance is without analysis? I did indicate what these molecules might be, and I repeat, this is absolutely part of chemistry, not biology. And what is a 'biological context' at the molecular level if not chemistry? I repeat if we ignore new parts of our subject it will fade away and become irrelevant and we shall find that this science will be done by others and often not so well as we as chemists could do it.

@porphyrin matt_black I don't think I'm going to be able to convince either of you that I'm right, so I'm just going to drop it. Feel free to ping if you wish to discuss further, but I'm not sure we'll get anywhere.

Update: Veritasium posted a video discussing on this topic:  https://www.youtube.com/watch?v=f-Nr2z5X7Rs

As mentioned in this comment, which links to this paper, and this comment, which links to this Wired article, there are at least two molecules:
$\beta$-carboline:

4-methyl-7-hydroxycoumarin:

$\tiny\text{Yes I poached those comments, they could have been answers.}$

I think the brevity of this answer (which fully answers the question as stated) shows precisely why the question hardly contains any chemistry. Not that there's any biology in it, either. I think it is still chemistry, but all in all, not a very useful question. Now, if it asked about why these molecules fluoresce, that would be different.

@orthocresol I agree that the answer would be better if it described some of the background, like how we know these are the culprits. What do their fluorescent spectra look like, for example. Fleshed out, I think it would make a good answer.

@matt_black There is a figure from the paper I can add, and I can also give a crude comparison against some calculations. Would that be interesting?

@pentavalentcarbon Well, I think so.

@matt_black Yet, the question was not asking about how they were identified. The question only asks about what they are, and the answer to that is "they are X and Y".

@orthocresol I agree in principle with your comments, but it would be interesting to know if these molecules are just in the protein as is, or if they are chemically linked to an amino acid for example, which would be my guess. So there is much more to this yet.  It would also be interesting to know  why these compounds are there, are they photo-protectors for example, or just a byproduct of metabolism or what? Nature is generally not wasteful.


text:
Why energy of some intermediates of glycolysis increases?
It is possible that I do not understand the concept of energy, but according to the diagram in my book, the energy between steps 4 and 5 and between 6 and 8 increases. How is this possible? Also, as ATP is invested in steps 1 and 3, shouldn't the energy increase there?

Step 1:
Phosphorylation activates glucose for the following reactions in the pathway. However, the formation
of such a phosphoester is thermodynamically unfavourable and requires
energy input to operate in the forward direction. The energy comes
from ATP, a requirement that at first seems counterproductive.

$$\ce{-D-Glucose + ATP^4- → -D-glucose-6-phosphate^2- + ADP3  H+}$$
  $\Delta G$ = -16.7 kJ/mol

and energy is released.
Step 3:
Once again, the substrate that provides the phosphoryl group is ATP. Like the hexokinase/
glucokinase reaction, the phosphorylation of fructose-6-phosphate is a priming
reaction and is endergonic(energy absorbed):

Fructose-6-P + Pi +fructose-1,6-bisphosphate   $\Delta G$ = 16.3
  kJ/mol

When coupled (by phosphofructokinase) with the hydrolysis of ATP, the overall reaction
becomes exergonic:

Fructose-6-P + ATP → fructose-1,6-bisphosphate + ADP
$\Delta G$ = -14.2 kJ/mol

Step 4:
Fructose bisphosphate aldolase cleaves fructose-1,6-bisphosphate between the
C-3 and C-4 carbons to yield two triose phosphates. The products are dihydroxyacetone
phosphate (DHAP) and glyceraldehyde-3-phosphate. The reaction has an
equilibrium constant of approximately 10^-4 M, and a corresponding $\Delta G$  of +23.9 kJ/mol.
(the equilibrium is thus greatly influenced by concentration.)
Step 6:
In the first glycolytic reaction to involve oxidation–reduction, glyceraldehyde-3-
phosphate is oxidized to 1,3-bisphosphoglycerate by glyceraldehyde-3-phosphate dehydrogenase.
Although the oxidation of an aldehyde to a carboxylic acid is a highly
exergonic reaction, the overall reaction involves both formation of a carboxylic–
phosphoric anhydride and the reduction of NAD to NADH and is therefore slightly
endergonic at standard state, with a $\Delta G$ of + 6.30 kJ/mol.
(The free energy that might otherwise be released as heat in this reaction is directed into the formation of
a high-energy phosphate compound, 1,3-bisphosphoglycerate, and the reduction of
NAD.)
Step 7:
Here the glycolytic pathway breaks even in terms of ATPs consumed and produced
with this reaction.The enzyme phosphoglycerate kinase transfers a phosphoryl
group from 1,3-bisphosphoglycerate to ADP to form an ATP.
The phosphoglycerate kinase reaction is sufficiently exergonic at
standard state to pull the G-3-P dehydrogenase reaction along. (In fact, the
aldolase and triose phosphate isomerase are also pulled forward by phosphoglycerate
kinase.) The net result of these coupled reactions is

Glyceraldehyde-3-phosphate + ADP + Pi + NAD → 3-phosphoglycerate + ATP
  + NADH + H+ $\Delta G$ = -12.6 kJ/mol

(Important: Another reflection of the coupling between these reactions lies in their values
of $\Delta G$ under cellular conditions, despite its strongly negative  the phosphoglycerate kinase reaction operates at equilibrium in the erythrocyte ($\Delta G$ =0.1 kJ/mol).

So in this case you'ill notice factors that influence energy expenditure and gain. Step 1 and 3 (exegonic reactions have net -G hence ) will see energy being released (and accroding to your diagram
there is a decrease. Conversely in Step in steps 4-5 (reaction not favoured but due to concentration proceeds and is aided by energy and 6-7( energy is absorbed to form a high energy compound
1,3-bisphosphoglycerate) and 7-8 is mainly determined by reaction environment and relative concentrations of substrates and is slightly endergonic in above case.
Hope this helps
Reference
Biochemistry 4th Ed (Grisham)

The steps in question are under equilibrium. You can access all of them quickly, but the system will eventually find the low energy pyruvate at the end of the reaction. By Le Châtelier's principle, as you make more pyruvate and deplete the intermediate immediately before it, the system will be able to readjust and continually move towards making more pyruvate.


text:
How does one classify aqueous solutions of ions as acidic, basic and neutral?
The following is a homework question:

Classify the aqueous solutions of the following ions as acidic, basic or neutral:
          $\ce{C6H5NH3+, Cl- , Fe^3+ , Mg^2+ , HSO4- , NO3- , NH4+ , Al^3+ , C5H5NH+, H2PO4-, CH3CO2-, Cu^2+, Na+, Ag+, ClO4-, PO4^3-, CN-, SH-, Ba^2+}$

I am looking for an approach to solve it.
My thought process so far:

If any of the radical is highly reactive with $\ce{H+}$, then the solution will be basic as $\ce{OH-}$ is left. Similarly, if it reacts with $\ce{OH-}$, the solution will be acidic.
However, this is where I am confused. Take $\ce{Fe^3+}$ for instance. It is known that it reacts to form $\ce{Fe(OH)3}$, which is a strong base. But, according to my logic above, the solution will be acidic.

Can someone help me with this and possibly give me hints at solving the rest?

$\ce{Fe(OH)3}$ is not a soluble compound. It is actually the hydrate of iron(iii) oxide hydroxide, which is formed when $\ce{Fe^3+}$ reacts with hydroxide anions. Since it pulls hydroxide out of solution, your intuition is correct.

Your approach thus far is correct. $\ce{Fe^3+}$ is acidic; $\ce{Fe(OH)3}$ is not a strong base. Indeed, if $\ce{Fe(OH)3}$ were a strong base, it would dissociate and form $\ce{Fe^3+}$, but, as you've mentioned, the opposite process happens.
Your confusion may arise from the fact that $\ce{Fe^3+}$ is a Lewis acid, and produces protons in a "different" way than Arrhenius acids:
\begin{align*} \ce{HCl &<=> H+ + Cl-}\\ \ce{Fe^3+ + H2O &<=> Fe(OH)^2+ + H+ }\end{align*}
Either way, your thought process (1.) will work for both cases.

Will Na+ solution also be acidic then?

@Tinkidinki, no. $\ce{Na+}$ does not react with $\ce{OH-}$.


text:
Nucleophilic attack on cyclic intermediates in dibromination and oxymercuration reactions of alkenes
Why does the nucleophile attack the most substituted carbon after the formation of a cyclic intermediate (three membered ring) in the cases of halogenation and oxymercuration of alkenes, as if there were a real positive charge. Shouldn't the most electrophilic carbon be attacked by the nucleophile?

Just to clear your doubts these are electrophilic addition reactions
Case 1: oxymercuation-demercuration
The reaction is initiated by electrophilic addition of $\ce{Hg^2+}$(mercuric) ion to the alkene to give an intermediate mercurinium ion, whose structure resembles that of a bromonium ion. Nucleophilic addition of water as in halohydrin formation, followed by loss of a proton, then yields a stable organomercury product.
The final step, demercuration of the organomercury compound by reaction with sodium borohydride, is complex and involves radicals.
The regiochemistry of the reaction corresponds to Markovnikov
addition of water; that is, the $\ce{-OH}$ group attaches to the more highly substituted carbon atom, and the  $\ce{-H}$ attaches to the less highly substituted carbon.

When the π electrons of the alkene attack the mercury cation, a
  three-membered intermediate forms in which both carbons share a single
  bond with the mercury and a partial positive charge. It turns out that
  rearrangements are unlikely to occur from this intermediate (as
  opposed to the discreet carbocation intermediate in the acidic
  hydration reaction).

A water molecule then attacks the more substituted carbon, which bears a greater share of the positive charge. Cleavage of the carbon-Hg bond is accomplished with a reducing agent called sodium borohydride $\ce{NaBH4}$.
See images:

So where is the carbocation present?

The end result of an oxymercuration-reduction reaction is the Markovnikov addition of water to an alkene without rearrangement
The regioselectivity on this reaction is partly based on inductive effects resulting from the shifting of electrons in a s bond in response to the electronegativity of nearby atoms. I believe the other reason is based on hydride shift which converts the carbonation into a tertiary one and this is thermodynamically favourable. (Not entirely sure on second reason though)
Case 2: Dibromination
Halogenation of Alkenes: Addition of $\ce{Br2}$
Follows the reaction sequence (proposed by George Kimball and Irving Roberts)

that the reaction intermediate is not a carbocation but is instead a
  bromonium ion, $\ce{R2Br+}$, formed by electrophilic addition of $\ce{Br+}$ to the
  alkene.

The bromonium ion is formed in a single step by interaction of the alkene with $\ce{Br2}$ and simultaneous
loss of $\ce{Br-}$.
See image:

I have updated to include an even more detailed mechanism:

@jerepierre added extended reactions after bromination and explained possible causes for regioselection in part 1

@ReeshabhRanjan added details on where a carbocation is observed

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text:
Product of Williamson Synthesis

I read that, and was wondering that in the first reaction shown they have used a tertiary butoxide where ether is formed because there is a primary alkyl halide but in the second reaction, they have used a methoxide. The Alkene formed is due to the tertiary alkyl halide my doubt here, is that is there any reason for taking a tertiary butoxide in one and methoxide in the other? Does the product also depend on the attacking species carrying the O-?
Update:

What I am trying to say, is that in the book, they first gave 4th reaction and if they were to follow the similar reaction then they should have mentioned something like 2nd reaction but they rather changed the tert-butoxide to a methoxide. Is there a reason behind it? And what would be the products for the other two reactions? Would it be a mixed product with no component in a majority?

Yes, Because of varying basic nature. Tertiary alkyl halides will undergo elimination reaction instead of substitution.

@UddeshyaSingh OP is asking about steric bulk of nucleophile as well, I think.

The size of the alkoxide does matter. If you ran this reaction with i-propoxide versus t-butoxide, the former would react faster.

@UddeshyaSingh tert-butoxide is more basic than methoxide. The tertiary alkyl halide is made to react with the methoxide(less basic) and as far as I know, elimination is favored by more basic reactant. So I am not sure what you're saying there. Did I misunderstand you?

@Zhe I am not sure if I understand what you are trying to explain with the iso propoxide. If we change the tertiary butoxide to that, then the charge on Oxygen atom would also reduce right? And isn't steric hindrance mainly because of the species on which the attack is happening and not so much the attacking species (I may be wrong here)

@Daksh Shah +1 for presenting a question  – especially the update – showing even a handwritten, then photographed reaction schemes may be kept crisp and legible.

Both sides matter. If the electrophile is bulkier, tert-butoxide will not react, even if methoxide will. This may make a difference with something like 1-bromo-2-methylpropane.

In Williamson ether synthesis, the reactive species is an alkoxide, which can act as a strong nucleophile or base. When reacted with an alkyl halide, if the alkoxide acts as a nucleophile in an SN2 reaction, the ether product is formed. If the alkoxide acts as a base in an E2 reaction, then an alkene is formed.
So when considering a the reactivity of alkoxides, there is a competition between SN2 and E2. Which pathway will dominate is determined by several factors, and the steric nature of the nucleophile and electrophile are critical.
With respect to the alkyl halide, we know that SN2 reaction is favored by sterically unhindered electrophiles. So the trend in rates is methyl > primary > secondary >> tertiary (to the point where this is generally impossible). Steric hindrance of the nucleophile will also slow the SN2 reaction. The E2 reaction is relatively insensitive to sterics, so if there is a competition between SN2 and E2, a bulky nucleophile/base will favor E2. That is why often potassium tert-butoxide is used to perform an E2 reaction.
Let's analyze each of the reactions presented.
1) A primary alkoxide (sterically unhindered) is reacted with a tertiary bromide (very hindered). SN2 reaction is extremely slow for tertiary halides, and nine beta-hydrogens are available to deprotonate from the halide, so E2 reaction dominates, and we would expect the alkene is formed.
2) A tertiary alkoxide (very sterically hindered) is reacted with a tertiary bromide (very hindered). Again, SN2 reaction is extremely slow for tertiary halides, and nine beta-hydrogens are available to deprotonate from the halide, so E2 reaction dominates, we would expect that the alkene is formed.
3) A primary alkoxide (sterically unhindered) is reacted with methyl bromide. Not only is methyl bromide very unhindered, but there are no beta-hydrogens available to participate in an E2 reaction. SN2 is the only viable pathway, so the ether will be the only product formed.
4) A tertiary alkoxide (very sterically hindered) is reacted with methyl bromide. Again, not only is methyl bromide very unhindered, but there are no beta-hydrogens available to participate in an E2 reaction. SN2 is the only viable pathway, so the ether will be the only product formed.


text:
What is the effect of dilution on the degree of dissociation of a weak acid or base?
Please ponder on an equilibrium of a very weak base, say $\ce{MOH}$. This base undergoes a very low ionization, even in its pure form (pure form implies that there isn't the presence of a solvent). The concentration of  $\ce{OH-}$ ions, that our base generates, can be assumed to be $ \le 10^{-6} $. Now, let us add some water to $\ce{MOH}$ so as to dilute it. Ostwald's dilution law states that if the degree of dissociation ($ \alpha $) is very less as compared to unity, which in this case, is, it can be calculated by the equation $ \alpha = \sqrt{\frac{K_a}{C}} $. This leads to the conclusion that as I make the solution less concentrated (by dilution), $ \alpha $ increases.
But if I think of it in a different way, I get confused. For instance, please consider the simultaneous equilibria:
$$ \ce{MOH <=> M+ + OH-} \tag{1}$$
$$ \ce{H_{2}O <=> H+ + OH-} \tag{2}$$
Since the concentration of $ \ce{OH-} $ ions our base generated in its pure form was comparable to the concentration of $ \ce{OH-} $ ions generated by water, they may cause the first equilibrium to be shifted in the backward direction, thus leading to a decrease in $ \alpha $ upon dilution. Is there any point I'm missing which makes me conclude the opposite of what one concludes from Ostwald's dilution law?

Your argument is valid, though I believe the scenario you describe is not within the range of applicability of Ostwald's law. Allow me to recast your example in a more obvious form.
You have a solution of weak base. You add a stronger base. The increase in hydroxide concentration leads, by Le Chatelier's principle, to a lower extent of dissociation.*
This is beyond the scope of Ostwald's law, which tacitly assumes the lack of what amounts to a common-ion effect. Ostwald's law holds if the solvent which you're using to dilute your solution does not interfere (or interferes only weakly) with your electrolyte's solubility; the added solvent must serve to decrease the concentration of your electrolyte. Evidently, then, adding a stronger base will interfere with the solubility of your weak base, and is not governed by Ostwald's law.

*There is perhaps some subtlety here. As an example, consider the salt $\ce{MX}$, which dissociates weakly into $\ce{M+}$ and $\ce{X-}$. The intuition behind the greater degree of dissociation is that (1) upon dilution, the equilibrium constant for dissociation doesn't change, whereas (2) the concentration of each species decreases. Because dissociation produces more ions than was originally present, this always leads to $Q < K$, and more dissociation will occur.
If we  add a salt $\ce{MY}$, then, we are introducing two effects: (1) decreased concentration of $\ce{X-}$ and (2) increased concentration of $\ce{M+}$. Because these effects pull the equilibrium in opposite directions, we can't conclusively determine what exactly will happen. Going back to your example, if we add a slightly stronger base, then we get exactly these two contrasting effects, and the change in the degree of dissociation is indeterminate. In contrast, if we add a much stronger base, then certainly (2) dominates and the degree of dissociation is reduced.


text:
One volume of something
I am following some protocols for a reaction and after diluting my substrate in a solution I need to precipitate it in methanol.
For this I need to pour my solution into "one volume" of methanol-
What I do not understand is what is meant by 'one volume'. Is a measurement of quantity.
The protocol can be found under the Sulfation of cellulose part of the methods section in:
R. G. Schweiger, “Polysaccharide sulfates. I. cellulose sulfate with a high degree of substitution,” Carbohydr. Res., vol. 21, 219–228, 1972 (DOI: 10.1016/S0008-6215(00)82148-5).

Welcome to Chem SE! If you have any questions about the site, visit the [help]. It might help people to answer this and make the answer more applicable to others if you include a little more detail about the procedure you are following. Best of luck in getting an answer!

I'm pretty sure it means to use a volume of methanol equal to the volume of your solution, such that you would roughly (volumes aren't additive) double the total volume of your solution. Unfortunately, I don't have full free access to the article you cited so I don't have the full context of the section in question.

If the reaction till "then" was a solution of, say 500 mL, than the addition of the next liquid – from your writing, it reads like it were methanol – should be equally just 500 mL. (And 500 divided by 500 were ... just 1).
Maybe adding more than one part of methanol would no longer lead to (this much / this clean) precipitation intended, as methanol alone is quite a good solvent of your precipitate*), too.
*) Intentionally leaving open, if the precipitate is an impurity to be removed, or really refers to the product to be isolated.  You have the literature reference.


text:
what is meant by stability and potential energy of molecules in cases of formation of bonds?
molecules form chemical bonds in order to gain stability. it is said that forming bonds leads to lower potential energy of molecules and makes them stable. what does stability mean?
how formation of bond leads to lower potential energy? please  explain

The electronic energy is lowered when bonding occurs, therefore the energy levels become more stable. Stable just means they have a lower energy level.
So if we have two 1s orbitals they can combine to form a single sigma bond which has a lower energy than the components separately.


text:
Comparing energies of P−O, S−O, and Cl−O bonds
My text says that

Bond energies of P−O, S−O, and Cl−O follow the trend P−O < S−O < Cl−O.

But I think it should be a reverse of what's given because more the electronegativity difference, more will be the bond energy and lesser the bond length. This what I believed so far before seeing P−O, S−O, Cl−O trend.
Moreover, after searching the web for the values of the same I was unable to find for P−O, S−O, Cl−O, but I found values for C−O and N−O and bond energies (B.E.) for it follows C−O > N−O (which supports my reasoning).
Then why the order is different for P−O, S−O, and Cl−O? Is it an exception, or could it be explained?

I'm not sure but I think there's something to do with the bond lengths. For instance, the $\ce{N-N}$ bond in $\ce{N2}$ is shorter than the $\ce{P-P}$ bond in $\ce{P2}$. This is because of the lone-pairs on N atoms repelling each other and the large size of P makes the distribution of charges possible. Similar reasons might be at play in your case working antagonistically to the difference in electronegativity.

Your text doesn't seem to be in agreement with experimental data, and your suggested order is apparently the correct one.
I collected a set of bond dissociation energies (BDEs) $D^\circ_{298}$ as well as bond distances for free molecules in gas phase $r$, all determined experimentally except $r$ for phosphorus dioxide radical $\ce{PO2}$:
$$
\begin{array}{l|ll|ll}
\hline
\ce{X} & D^\circ_{298}(\ce{X-O})/\pu{kJ mol^-1} & r(\ce{X-O})/\pu{Å} & D^\circ_{298}(\ce{X(O)-O})/\pu{kJ mol^-1} & r(\ce{X(O)-O})/\pu{Å} \\
\hline
\ce{P} & 589^\text{[1, p. 484]} & 1.4759^\text{[2, p. 9-27]} & 540^\text{[1, p. 484]} & 1.4665^{[3]} \\
\ce{S} & 517.90±0.05^\text{[1, p. 360]} & 1.484^\text{[2, p. 9-28]} & 551.1^\text{[1, p. 360]} & 1.4308^\text{[2, p. 9-28]} \\
\ce{Cl} & 267.47±0.08^\text{[1, p. 366]} & 1.5696 ^\text{[2, p. 9-22]} & 256.3±1.3^\text{[1, p. 366]} & 1.470^\text{[2, p. 9-22]} \\
\hline
\end{array}
$$
For diatomic molecules $\ce{X-O}$ there is a trend of decreasing BDEs
$$D^\circ_{298}(\ce{P-O}) > D^\circ_{298}(\ce{S-O}) > D^\circ_{298}(\ce{Cl-O}),$$
also backed up by the increase in bond distances:
$$r(\ce{P-O}) < r(\ce{S-O}) < r(\ce{Cl-O}).$$
To tell the complete truth, comparing bond energies like that isn't entirely fair, as we are comparing bonds of various (fractional) order, e.g. in case of sulfur, it really should be a double bond $\ce{S=O}$ we are looking at.
For the sake of comparison and to demonstrate how introduction of a polyatomic system alters bonding, there is also data for $\ce{XO2}$ molecules where phosphorus and sulfur swap places and the BDEs order changes to
$$D^\circ_{298}(\ce{S(O)-O}) > D^\circ_{298}(\ce{P(O)-O}) > D^\circ_{298}(\ce{Cl(O)-O}),$$
and the bond distances order changes accordingly:
$$r(\ce{S(O)-O}) < r(\ce{P(O)-O}) < r(\ce{Cl(O)-O}).$$
References

Luo, Y.-R. Comprehensive Handbook of Chemical Bond Energies, 1st ed.; CRC press, 2007. ISBN 978-0-8493-7366-4.
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.
Computational Chemistry Comparison and Benchmark DataBase, Release 21 (August 2020) — Experimental data for PO2 (Phosphorus dioxide), Table "Atom - Atom Distances". (accessed 2022-02-19)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why doesn't the oxidation of cellulose oxidise all the alcohol groups, and how does it cause discolouration?
In cellulose, there are several alcohol groups - the two on the ring being secondary alcohols, and the one on the branch being a primary alcohol. I would expect all of these to be oxidised in the presence of any oxidising agent, such as heating under reflux with acidified potassium dichromate - this would give an aldehyde, and then a carboxyl group on the branch, whilst the two on the ring would be oxidised to ketones. However, as you can see in the diagram below (ignoring the sulfonation step which is irrelevant to the question), this is not the case with cellulose.

Why is the branched alcohol not also oxidised? Is it to do with it being a primary alcohol?
Further to this, the oxidation of these groups in paper causes discolouration. Is this because the double bonded oxygen atom then joins the delocalised system, increasing the size of the chromophore and hence bringing the frequency at which the material absorbs into the visible region? Or is there a different reason?

This depends on what reagent you use to oxidise it. As you suggested, ordinary old Cr(VI) or Mn(VII) will oxidise the primary alcohol and leave the tertiary alcohols untouched (note: they cannot be oxidised to ketones). Without specifying exactly what reagent it is, we cannot give you a definitive answer, we can only make an educated guess. Here, it seems to be something like NaIO4, which reacts with vicinal diols (two hydroxyl groups on adjacent carbons). As for why, you will need to look at a mechanism: http://www.masterorganicchemistry.com/2011/10/21/reagent-friday-sodium-periodate/

Well Noah,
Primary alcohols, in general can be oxidised more easily, if we are using suitable reagents like acidified dichromate/permanganate etc.
By looking at the reaction it looks like there has been some reagent used like lead teraacetate (LTA) or NaIO4, which attack syn, vicinal diols/(alpha)-hydroxy aldehydes/glyoxal to cleave the C-C bond and produce the oxidized product:

It has been mentioned elsewhere, that cellulose can be oxidised by NaIO4 into some carbonyl derivatives:
This reaction has been responsible for yellowing [at times referred to as discolouration] of paper:
These carbonyl groups arranged in some definite fashion give rise to yellow colour, sometimes called chromophore, which are any group responsible for proving colour to an otherwise colourless compound.

Very helpful answer - now I better understand why newspapers yellow differently (i.e. faster) than books! As yet unanswered lignin oxidation/chromophore production question...


text:
Why is the common magnesium ion Mg(II) and not Mg(I) when the second ionization energy is higher than the first ionization energy?
The first ionization energy for magnesium is given as $\pu{737.7 kJ/mol}$, and the second ionization energy is $\pu{1450.7 kJ/mol}$.
Given this information, doesn't it take $\pu{737.7 kJ/mol}$ to form $\ce{Mg^1+}$, while $\pu{737.7kJ/mol} + \pu{1450.7kJ/mol} = \pu{2188.4 kJ/mol}$ is required to form $\ce{Mg^2+}$?
Why is it then, that the $\ce{Mg^2+}$ ion is more common than the $\ce{Mg^1+}$ ion, while more energy is required to form the $\ce{Mg^2+}$ ion?

That depends on what else is accepting electrons. Just because it is less favorable did not mean that the sum when you account for the oxidant is not overall favorable. Also, you're looking at gas phase ionization energy? In solution, it's a different game.

As mentioned by Zhe, we have to look at the entire process by which an ionic compound is formed, not just the energy for a single part.
For example, consider the formation reaction of $\ce{MgO}$
$$\ce{Mg(s) +1/2O2(g)->MgO}$$ To find the enthalpy of this process, we could follow a Born-Haber cycle:
This demonstrates that while the ionization of $\ce{Mg}$ to $\ce{Mg^{2+}}$ might be unfavorable, the overall process to form the compound is not. So the ionization of $\ce{Mg}$ is driven by the favorability of the overall process.2
It is also worth noting why we do not instead form $\ce{Mg2O}$. If we were to construct a Born-Haber cycle for this compound, it would take less energy to form $\ce{2Mg^{+}(g) +O^{2-}(g)}$ than it would to form $\ce{Mg^{2+} (g) + O^{2-}(g)}$. However, the lattice energy of $\ce{Mg2O}$ would be sufficiently smaller than the lattice energy of $\ce{MgO}$ that $\ce{MgO}$ is still more favorable to form. Lattice energies are heavily dependent on the charge of the ions involved and having a $+2$ ion will lead to twice as large a lattice energy as an ion with a $+1$ charge all else being equal.
Even in this simple case, all else might not be equal. To make a more direct comparison between we can utilize the Born-Lande or Born-Mayer to obtain a calculated value of the lattice energy.

Also, in solution, the hydration enthalpy of $\ce{Mg^2+}$ is sufficiently large to compensate for the 2nd IE. The same cannot be said for, for example, Na, which has a huge 2nd IE.

@orthocresol based on this data table from the RSC, the enthalpy of hydration of $\ce{Mg^{2+}}$ would be insufficient to overcome the two ionization energies.

Well, my statement was rather simplified, there are certainly more factors that go into a proper energy cycle (hydration of the counterion for example) however the larger hydration enthalpy of $\ce{Mg^2+}$ compared to that of $\ce{Mg+}$ was the factor I thought would tilt the balance in favour of Mg(II).

@orthocresol I would agree. I just realized I sidestepped around the OPs actual question.


text:
What is the mechanism of reaction between alcohol and PCl5
Is this one correct? If not why and what is the correct one?

Strongly Related: https://chemistry.stackexchange.com/questions/33675/reaction-of-alcohols-with-pcl5-and-pcl3

Welcome to Chemistry! This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

@Sneha Srikanth Visit Wikipedia: https://en.m.wikipedia.org/wiki/Phosphorus_pentachloride and head to the Reactions subheading. The mechanism is available there.


text:
Speed distribution of lighter vs heavier gases
I cannot understand why the range of molecular speed is not always wider for a lighter gas as compared to a heavier gas .
If the same energy is supplied to both gases then wouldn't the molecular speed of the lighter gas be more?

Do you have an example of such an exception? I don't doubt that there are exceptions since the Maxwell Boltzmann distribution requires a system to be at thermal equilibrium and assumes ideal gases. However do you have a specific example of a situation where a lower molecular weight gas has a thinner distribution?

@Tyberius If i have the example of exception then why would I have asked the question

Let me rephrase my question. If you don't have a specific example, how do you know it happens? What source told you that this could be the case?

Knowing what led to this question might help to provide an explanation that addresses your specific concern. But as I mentioned in the comments to my answer, probably the main cause of deviations from Maxwell Boltzmann would be real gas behavior (attraction and repulsion) and multiatomic molecules, which have internal degrees of freedom where they can store the thermal energy.

To my understanding, the speed distribution is wider for lighter molecules. The Wikipedia page for the Maxwell Boltzmann distribution has two images at the top that convey this very well.

In the first plot, $a=\sqrt{k_bT/M}$ where $M$ is the molar mass and $k_b$ Boltzmann's constant. For a Maxwell-Boltzmann distribution, the variance is given by $$\sigma^2= \frac{a^2\cdot(3\pi-8)}{\pi}=\frac{k_b\cdot T\cdot(3\pi-8)}{M\cdot\pi}$$ So, this shows that the speed distribution will be wider for higher temperatures and smaller molar mass.

There might be some exception

@Tyberius With  infinite temperature (equivalent to an infinitely small mass) every state (velocity) has same probability density.

@santimirandarp right, but that still wouldn't lead to smaller speed distribution for lighter molecules. Equal within that limit, but not smaller. I would guess certain effects of real gases, namely attractive forces, could lead to deviations from the Maxwell Boltzmann distribution.

@santimirandarp but would it be smaller than the distribution of a large molecular weight gas at the same very high temperature. Even a large molecule would have a broad speed distribution at large temperature.

I think range is same for both gases . As the graph is extending till infinite .


text:
Why is the melting point of magnesium oxide higher than aluminium oxide?
There's a graph of the melting points of period three oxides. The melting point of magnesium oxide is several hundred Kelvin higher than aluminiumoxide. I can't find any explanations for this on the Internet or in any of my A-level textbooks — perhaps this is something more advanced I haven't come across.
The reason I am questioning the value of its melting point is I'd expect aluminium oxide to be higher. Aluminium is a 3+ ion whereas magnesium is only 2+. Aluminium is also a smaller ion so the effective electrostatic attraction on the valence electrons should be greater. I'm also pretty sure aluminum oxide has greater covalent charterer than magnesium oxide. All of these would lead me to believe it has a higher melting point, so what is the explanation for it being less?

You could apply Kapustinskii's Equation to obtain the lattice energy and see if that provides an explanation.

For what it's worth, based on your plot of melting points for these 3rd row oxides, MgO appears to fall right in line and $\ce{Na2O}$ seems to be the odd-ball.

Two ideas to consider:

Alumina is not fully ionic.  Neither is magnesia, but magnesia has more ionic character and we may see the greatest amount of ionic attraction in magnesia versus alumina which is less ionic, or soda which has only singly charged cations.

When metal oxides are melted they do not necessarily produce free metal and oxide ions.  As described by Shi et al. [1], molten alumina retains mostly four-and five-coordination of oxygen to aluminum, rather than forming "free" aluminum and oxide ions and also contrasting with the six-coordination of aluminum in the solid conundrum phase.  Presumably magnesia would behave similarly to alumina when melted, but the greater ionic bonding character in magnesia makes reduced coordination less favorable and thus enhances retaining the fully octahedrally-coordinated solid phase.

Reference
1.
Caijuan Shi, Oliver L. G. Alderman, Diana Berman, Jincheng Du, Joerg Neuefeind, Anthony Tamalonis, J. K. Richard Weber, Jinglin You and Chris J. Benmore,  "The Structure of Amorphous and Deeply Supercooled Liquid Alumina",
Front. Mater. 6:38 (19 March 2019), https://doi.org/10.3389/fmats.2019.00038.

If magnesium oxide has more ionic character shouldn't that make it's melting point lower. I would expect a greater covalent charecter to increase the melting point.
Also your idea about polyatomic species - would enough of these be present in molten aluminium oxide before it's reached it's quoted melting point?

You have to break some ionic bonds.  Solid magnesia has six of each atom/ion surrounding an opposite counterpart.  Melted magnesia will have less coordination.

Here is a simple minded answer.  High melting point compounds tend to follow the octet rule - the number of valence electrons sums to 8.  The Group IVB carbides (TiC, ZrC, HfC) have an 8 electron sum (4+4) and have very high melting points - for HfC it is approximately 7100F.  The Group IIIB nitrides and phosphides have an 8 electron sum (3+5) and are the highest melting point compounds among the rare earth metals. The Group IIA metals also follow this pattern with MgO (2+6) as the prime example.  While these compounds all have the rock salt structure, the same valence count pattern extends to other crystal structures, such as with the monoborides of the Group VB metals - TaB (5+3) does not have the NaCl structure (atom sizes fall outside of the Haag rule) yet has the highest melting point in the Ta-B system.  Following this trend, Al nitride and phosphide (3+5 electrons) also follow this pattern - they are the highest melting Al compounds.  Why does this pattern exist?  I don't know except to say that at this magic 8 valence electron sum, the electronic structure is optimized.  In density of states calculations, that valence electron sum fully fills the "bonding states" and leaves the "anti-bonding states" unpopulated, and the Fermi level occurs at a deep well.  I am not a solid state physics expert, so this is about as far as my explanation goes.


text:
At home Rochelle's salt synthesis
I tried to make Rochelle's salt. It didn't work!

7 oz cream of tartar
1 cup distilled water
heat to 180 F and keep there
add 1/2 tsp sodium carbonate at a time until there is no fizzing
filter and let cool

The solution ended with a strong yellow/brown tint. It took at least an hour of adding sodium carbonate. While cooling it rapidly formed a crystallized honey look and that was it. I reheated and added more distilled water to lighten it up a bit thinking it needed to evaporate slower. This time when it cooled nothing happened.
The sodium carbonate was 100%, and the cream of tartar was McCormick. It's possible that came with some impurities hence the brown/yellow grossness.
Is there as way to save this batch? Should I just try again? Is there a better recipe?
My first instinct is that it didn't work because you don't science with Imperial units.

Hello and welcome to Chemistry.SE. If you have any questions about how the site works you can take the short [tour] or visit the [help]. Just one comment about your question: sometimes activated carbon is added to decolorize the solution prior to filtering. I don't know if this would make a difference in your situation though.

How are you judging "success" of this process?

Tried again with 365 brand (whole foods) Cream of Tartar. Worked much better.

39g Cream of Tartar
50ml distilled water
Sodium Carbonate added slowly until stirring produces clear liquid


text:
Stability of diboron(B2) and dilithium(Li2)
Why is $\ce{B2}$ more stable than $\ce{Li2}$?
Both $\ce{B2}$ and $\ce{Li2}$ have bond order equal to 1. Moreover, $\ce{Li2}$ has only 2 electrons in the antibonding molecular orbital while $\ce{B2}$ has 4 electrons in the antibonding molecular orbital. Thus, $\ce{Li2}$ should have been more stable than $\ce{B2}$ if we consider the number of electrons in the antibonding molecular orbital as the parameter of stability. Does the fact that $\ce{B2}$ is smaller than $\ce{Li2}$ makes it more stable?

Firstly, the use of molecular orbital diagrams allows you to predict the electronic configuration, if a molecule would exist or not (as you say, evaluating the bond order), and some magnetic properties. Moreover, if you have "the same molecule" as O$_2$ and its ion  O$_2^-$, etc, you can say the higher bond order, the bigger stability.
But if the bond order is the same (your case), you can not talk about stability using those diagrams.
Coming back to the question, looking the dissociation energy (the energy relative to separated atoms) for B$_2$, you will see it is bigger in module.
So it will be more stable.
I do not have a qualitative way to estimate which of those molecules is more stable.
The way to know it theoretically is using electronic structure programmes.

I want to know how would you justify it theoretically. Because, we can't remember all the experimental data.

You can run the calculation in some electronic structure programme (for example Gaussian)

Sorry, but I am still a student. And how would I determine if this question comes in an exam.

Sorry for this question, but are you sure it was not about Be$_2$ ?

Yes, I am sure.


text:
What is pure water
Can someone please explain to me what does "pure" water mean? The definition vary so much everywhere I look. But I just want to see it from a chemistry/scientific perspective.

Interestingly, there is a classification system of what constitutes various degrees of "pure" water.

https://chemistry.stackexchange.com/questions/2870/what-is-a-pure-substance

https://chemistry.stackexchange.com/questions/575/what-do-the-different-grades-of-chemicals-mean

Hello and welcome to Chemistry.SE. If you have any questions about how this site works, you can take the short [tour] or visit the [help]. Regarding your question, I think you'll find that it's all a matter of degrees. There is no such thing as perfectly pure water (or most any other bulk substance for that matter) if you look closely enough. If you had the ability to see each individual molecule in a container of water, you would find billions of billions of non-water molecules even in the highest purity water. Hopefully someone will give you a good answer, though this is pretty broad.

@Tyberius Your comment (+1) should be the answer, as it highlights there is not "pure water" per se, rather than different thresholds and criteria of purity, suitable (or not) for certain applications in mind. As in the comment by airhuff, too.

As suggested in the comments, there isn't, strictly speaking such a thing as pure water. Pure water (pure anything) would require that there isn't a single molecule of anything else, but water in the sample. As Buttonwood mentions above, if we even could separate water pure water somehow, water will form $\ce{H+}\text{ and } \ce{OH-}$ in attempt to reestablish equilibrium.
If we are a little less stringent about what "pure" means, we can define purity via various classifications that rely on our ability to experimentally determine certain properties of water. One factor that is commonly used to judge the purity of water is its resistivity. While we often think of water as being a good conductor, this is actually due to the prevalence of charged impurities dissolved in the water. The American Society for Testing and Materials International (ATSM) includes a resistivity of $\pu{18.2 M\Omega\cdot cm}$ ($\mathrm{M\Omega}$ means Mega-Ohms) as part of the definition for water of the highest purity standard (Type I).
But as mentioned in the link, even being in the highest category might not make the water pure enough for certain applications, such as trace metal analysis. So, even still, purity really depends on the context in which it is being used and the precision of experimental methods which are a proxy for purity.


text:
Electron configuration from vanadium to chromium?
The electron configurations for vanadium is: $\ce{1s^2 2s^2 2p^2 3s^2 3p^6 3d^3 4s^2}$, and the electron configuration for chromium is: $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1}$.
In looking at the electron configurations of two consecutive elements, I noticed that an electron in the $\ce{4s}$ orbital "jumped" to the higher energy $\ce{3d}$ orbital. Why is this so?


text:
Photoelectric effect and kinetic energy
According to my textbook, electrons which are emitted due to the photoelectric effect have kinetic energy given by:
$$KE_{electron}=h\nu-h\nu_0$$
Where $\nu_0$ is the activation frequency of the metal.
My question is: when electrons are emitted in this way, is all of the photon's energy really converted to kinetic energy in the electron? Is there no loss of energy in this process?
If there is energy lost (which I suspect must be true), is it a sufficiently small amount so that we can say it is negligible for most purposes?

In your formula KE should actually be labelled KE(max). Electrons will be ejected with a range of kinetic energies from 0 to KE(max).

What exactly happens to the EMR when the photoelectric effect occurs? I would guess that the frequency of the radiation is reduced so that the energy gained by the electron is equal to the loss in photon energy, but what causes a different amount of energy transfer in different cases?

$h\,\nu$ is photon's energy; $h\,\nu _0$ is the potential energy of the electron. All energy excess is converted in kinetic energy of the eletron - that's the left side of the equation.


text:
Is it possible to make an anticatalyst?
I'm wondering if it is possible, theoretically, to create compounds which perform the opposite function of a catalyst (thus an anticatalyst). That is to say, could a compound be made which raises the activation barrier of a reaction rather than lowers it?
I know enzyme inhibitors exist, but I would argue these merely disable or slow the catalyzed reaction rather than slowing the reaction down relative to its baseline speed. (If I'm mistaken and there are enzyme inhibitors that actually lead to a slower reaction than if the enzyme weren't there at all, I would also consider that an answer).
I would think an anticatalyst could be made by having an enzyme which binds a molecule in a way that reduces the favorability of the transition state (or alternatively, stabilizes the reactants, as mentioned by Nicolau Saker Neto below). The binding to the anticatalyst enzyme would also have to be favorable enough where the reactants would not just react and ignore the slower anticatalyzed pathway (one could also assume, for the sake of argument, that enough of the anticatalyst is added where the majority of the reactant molecules are bound).

Interesting question. If you want to raise the activation energy, then either the transition state must be destabilised, or the initial state must be stabilised. The former case seems rather more unusual, because in general if an entity is added to a mixture which is capable of destabilising the transition state, the molecules reacting can simply neglect the anticatalyst. The concept of an anticatalyst, in its purest form, would have to not interact with any molecules except for when they're arranged in the transition state complex, where it would then interact very strongly. Quite odd.

Stabilisation of the initial state is much simpler. Would something like molecular sieves added to a hydrolysis reaction in non-aqueous conditions count as an anticatalyst?

@Nicolau Saker Neto It might be a somewhat contrived a scenario, but we could look at the scenario where the enzyme binds a reactant tightly enough where it won't leave to react on its own, but loosely enough where if it does manage to react(which will be slower due to being in a less reactive configuration of the transition state) it will be released.

@ Nicolau Saker Neto I can't say I'm familiar with that type of reaction process, but I would say something that stabilizes the initial state would also count.

Related article, found due to a meta post by paracetamol: http://onlinelibrary.wiley.com/store/10.1002/anie.199003551/asset/199003551_ftp.pdf?v=1&t=j4e9n9kh&s=14fd83d496bdc1b813b04620e603e291b5e0f8d9&systemMessage=Wiley+Online+Library+will+be+unavailable+on+Saturday+01st+July+from+03.00-09.00+EDT+and+on+Sunday+2nd+July+03.00-06.00+EDT+for+essential+maintenance.++Apologies+for+the+inconvenience.

Some substances do exist that slow down reactions and they're different from catalytic inhibitors/poisons. Such substances are called negative catalysts. Here is one example I can think of right now:
$$\ce{2H2O2 \xrightarrow{\large\mathrm{glycerol}} 2H2O + O2}$$
Glycerol, in this case behaves as a negative catalyst and slows down the reaction.
Another notable example you might have heard is tetraethyllead (TEL) which was used in petrol to prevent knocking.
It may be tempting to think that a negative catalyst will raise the energy of activation for a given reaction, however this isn't true. It takes away the internal energy of reactants and deactivates them to reduce the rate of a reaction.

This is very interesting. I should have looked for more synonyms to my concept initially. I think this old ACS letter by JA Young (which I believe is free to view) gets at the idea that the name negative catalyst would technically imply, by a particular definition of catalyst, something that raises the activation energy. He notes that "...defined in this way, negative catalysts do not exist, effectively" (emphasis mine). So I still wonder if hypothetically one could be made to fit this definition.

As an aside, I still prefer anticatalyst to negative catalyst. Positive/negative catalyst seems to imply a directionality to the term catalyst(it makes it seem as though a catalyst is just something that changes the speed of a reaction and positive or negative tells you whether it is increasing or decreasing. But in science and general usage, to catalyze is always to accelerate something. So in either case, I think it should be an anticatalyst to imply it does the opposite of a catalyst.

Sshhh! Don't tell anyone about TEL! We have to stop lead pollution!

@Pritt Balagopal it might be better if more people know about TEL, where it is still used, and what harm it can cause. That way people might be more motivated to find alternatives.

I think TEL breaks chain of reaction - acts as inhibitor not antycatalyst.


text:
Dipole moment and polarity of borazine / borazole / inorganic benzene
What is the dipole moment of borazine? I tried googling it but can't find a reliable source. I searched in JD Lee too. It might have it but I could not find.

This NIST publication lists the dipole moment as zero, while drugfuture.com (I've never heard of this site) reports a value of 0.50 as measured in benzene at $\pu{25^oC}$.

The simplest, but not complete, way to tell if the dipole is zero or not is look for a centre of inversion, which borazine would have if it were flat like benzene.
To be absolutely certain  the molecule's point group has to be found. To have a dipole the molecule must belong to $C_1, C_{nv}, (n>1) $ and $C_s$ point groups.
As borazine is not planar (according to wikipedia page) and seems to belong to $C_{3v}$ so it should have a dipole. However, looking at the structure the dipole has to be small due to the way the atoms are arranges around the 3 fold axis and the small displacement of the plane containing the B to that of the N atoms. If you can find B-N, N-H, B-H bond dipole  values you could calculate the dipole, at least get an approximate value.

If I remember correct, borazine is planar $D_\mathrm{3h}$?

@orthocresol yes $D_{3h} $if planar and so no dipole.

Warning: This is just an effort not complete answer. I also request higher members to give answer / put bounty as this question is not having complete answers.
Borazine is planar. But the borazine ring does not form a perfect hexagon. The bond angle is $117.1°$ at the boron atoms and $122.9°$ at the nitrogen atoms, giving the molecule distinct symmetry.
Dipole moments are forming exact $120°$ with each other resulting in net non zero dipole moment and polar molecule.With support of comments I calculated

Borazine, also known as borazole, is a polar inorganic compound. 3D model : https://chemapps.stolaf.edu/jmol/jmol.php?model=%5BBH-%5D1-%5BNH%2B%5D%3D%5BBH-%5D-%5BNH%2B%5D%3D%5BBH-%5D-%5BNH%2B%5D%3D1
All details are from Wikipedia.

According to your structure Borazine has point group D3h. Hence it has no dipole moment. See https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Symmetry_(Vallance)/01%3A_Chapters/1.04%3A_Symmetry_and_Physical_Properties

Also it would be good to reference where you got the above structure from (and picture if you didn't draw it yourself), I'm guessing https://en.wikipedia.org/wiki/Borazine

The issue with Borazine is whether it is flat. If it is, it can't have a net dipole. If it isn't flat (and you can't tell that from the picture above) and the borons are on a different plane to the nitrogens (ie the molecule is puckered) then it inevitably has a dipole but it will be perpendicular to the ring.

@matt_black I agree with you but source tell what i mentioned.

@Jay Your answer clearly doesn't agree. The problem is your assumption that the issue is whether the bonds in borazine need to be exactly 120°. Even if they are not, the molecule would not have a net dipole since it still has enough symmetry for the net bon dipoles to cancel. Only if it is non-planar is this not true.

@matt_black I understood that but this efforts will help in further answers i think. So, i don't think it is appropriate to delete till you insist. What shall I do further delete or not?

@matt_black Note the interior angles sum to 720, as in any planar, not necessarily regular, hexagon. (e.g. https://www.coolmath.com/reference/polygons-06-hexagons). When I first wrote my first comment I thought this was enough to state that this means the given structure is planar, though on rethinking the logic that's not true - it does not necessarily follow from "the sum of angles in all planar hexagons is 720" that " if the sum of angles in a hexagon is 720 it is planar". But I will be amazed if it is not so ...


text:
How do I compare the C=C bond lengths in this cumulene?
How do I arrange these chemical bonds in an increasing/decreasing order based upon their lengths?

I understand that the bond (a) will be longer than (b) because (a) is a bond between sp² hybrid and sp hybrid carbon atoms, as opposed to (b) which is a bond between two sp hybrid carbon atoms. But how do we compare bond (c)?

I know this is only meant to be a theoretical exercise, but I ran a search and this molecule does not exist (as far as I can tell). Therefore, all explanations should only be taken at face value - i.e. they are merely a prediction of the relative bond lengths, as opposed to a rationalisation of observed bond lengths.

Label the double bonds (a), (b), and (c), in correspondence with the image provided in the question. Label the carbons in the cyclopropane ring $\ce{A}$, $\ce{B}$, and $\ce{C}$, with $\ce{A}$ being the carbon with the double bond (c).
At $\ce{A}$, the usual $\ce{sp^2}$ hybridization would lead to bond angles of $120^\circ$, whereas a lack of hybridization (i.e. simple $\ce{p}$ orbitals) would lead to bond angles of $90^\circ$. The latter is not usually favored, but the high ring strain in cyclopropane leads to carbon-carbon bonds $\ce{A-B}$ and $\ce{A-C}$ having greater $\ce{p}$ character than is usual. The double bond (c) at $\ce{A}$ thus receives more $\ce{s}$ character, and is stronger than a usual $\ce{sp^2{-}sp}$ bond would be (i.e. (a)).
We conclude, therefore, that the bond strengths are (b) > (c) > (a), and hence that the bond lengths are (a) > (c) > (b).

A stronger bond does not necessarily mean it is a shorter bond. It works in this case, but it's not a correlation-causation consistent notion. One should be very, very careful using this.

@Martin-マーチン, thank you for the comment; I didn't know this. Would correcting my argument by saying that bonds with greater $s$ character are shorter be equally (in)valid?

In this case this works mainly because you are comparing carbon-carbon bonds, which is the same reason why the strength argument works. (I have been sloppy about that in my own answer, which I need to revise.) Higher s character indeed means shorter bond here since the optimal overlap (balancing Pauli repulsion, etc.) is closer to the nuclei. And this is assuming that the pi bonds are essentially the same in all bonds (they are not). But for a rudimentary analysis, this gives the correct result.

I have removed the references to my answer since I deleted it as it is not correct. I currently don't have the time to revise it thoroughly and I think your analysis is good enough from an intuitive point of view.

You can draw Hyper conjugating Structures to compare bond length. More the α Hydrogen of alkene, more are the Hyper conjugating structures, more is the delocalization of π-bond and more is the bond length (since double bond character decreases, bond length increases).
Thus bond length C would be greater than A.
As there are 4 α hydrogens around C, whereas '0' α hydrogens around A.
Therefore the order of bond length would be C>A>B


text:
Have the STRUCTURES of synthesized anthropomorphic molecules (nanoputians) been verified experimentally?
Have the molecular structures of any synthesized anthropomorphic molecules (nanoputians) been verified experimentally?
This is a different question than this question as I am asking here about the structure, not the actual shape of the molecule. The importance of the distinction was pointed out in this helpful comment. Was there some property that was measured that confirmed that the correct, desired molecular structure had been in fact correctly synthesized?

What would be satisfactory evidence to you? Surely they were characterized by the common spectroscopic techniques.

I don't currently have access to the original article, but when I looked while reading your previous question, they seemed to have done some NMR analysis of various nanoputians.

@Tyberius I don't have a copy of the paper right now either. If NMR can confirm or at least pretty-much confirm the final structure, that would be the basis of an excellent answer here. I'm assuming molecular weight alone would not be.

@jerepierre I'll defer to the authors, or to the SE community here on that. I don't have the paper now, so I can't see what they've done. And stop calling me Shirley :)

To publish a paper in the first place, you need lots of "proof" of the structure. You can't publish a paper saying "I think I made this" without NMR, MS, IR, elemental analysis, ... to back it up. Regarding your assumption that MW alone is not proof, it is true that NMR is slightly better, but it is still not 100% validation. They can be misinterpreted and there have been many cases in history of "I think I made this; here's the evidence" and then realising that they actually made something different that has qualitatively the same NMR spectrum. So, even NMR is not foolproof.

Funnily enough, the best "proof" is probably crystal structures - but that goes back to your other question. The molecular shape and the atomic positions (the other question) would be very strong proof of the connectivity (this question).

@orthocresol thanks for your perspective, I think it could represent a substantial fraction of an answer, all that's missing are the essential details in this case. I won't have access to the paper for a few days, and even then may not be able to correctly and conclusively interpret what's there.

The first article wikipedia's entry refers to (here) is not only a publication behind a paywall, but with freely available supporting information, too.  This in principle allows to replicate the work.
Browsing through the latter provides all / most / many details of how building blocks were prepared, how they were joined; including both numerical and graphical display of 1H and (broadband decoupled) 13C NMR spectral data.  Like the following, found on page 22 of the SI:

and 13C NMR (same page, below)

Perhaps (i.e. the following is a speculation) the "main" article in the journal includes additional spectroscopic data (UV, IR; MS) or characterisation (melting point, combustion analysis, etc.) as the article stretches from page 8750 to 8766.

To have a more rounded perception of this work done, it would be interesting to look at the echo by others in the community.  Which was the context this publication was cited?  Did the original researchers continue to work on these compounds, and which context?  In which ways other researchers than the authors echoed this work?  Because ACS's page indicates the publication was noticed in freely accessible media, it were worth to look up about this article in SciFinder, too.

I never thought to look for supporting information. Ok we're definitely on the right track!

I don't know if this type of analysis can confirm complete molecules, or if a mixture of incomplete molecules and/or broken molecules can produce similar spectra at this resolution.

@uhoh For some moment, I offer you an exchange here: http://chat.stackexchange.com/rooms/57191/nanoputains

I will probably edit this to add more as requested, but I have access to the original article.
As an example of how they characterized these molecules, here is the data they present for the "NanoChef":

Mp: 117-124 °C. IR (KBr): 2966, 2929, 2900,
  2867, 2225, 1580, 1481, 1350, 1230, 1022 cm-1. 1H NMR (400
  MHz, CDCl3): δ 7.63 (s, 1 H), 7.60 (s, 1 H), 7.47 (d, J ) 1.5
  Hz, 2 H), 7.40 (t, J ) 1.5 Hz, 1 H), 7.22 (s, 1 H), 6.89 (m, 4 H),
  2.39 (t, J ) 7.0 Hz, 4 H), 1.62 (sext, J ) 7.2 Hz, 4 H), 1.35 (s,
  9 H), 1.29 (s, 9 H), 1.05 (t, J ) 7.2 Hz, 6 H). 13C NMR (100
  MHz, CDCl3): δ 147.6, 136.2, 135.7, 134.4, 133.6, 129.8, 126.8,
  125.9, 124.6, 123.3, 122.2, 121.7, 108.6, 108.0, 106.0, 104.9,
  92.7, 91.4, 88.1, 79.4, 75.0, 30.9, 30.7, 28.3, 22.1, 21.3, 13.5.
  HRMS: calcd for C43H42O2 590.3185, found 590.3178.

So it appears they looked at melting point, IR, $\ce{H}$ NMR, and $\ce{^13C}$ NMR, as well as determining High Resolution Mass Spectrometry.

Thanks! I've just got a copy yesterday as well, but since it's the first synthesis paper I've ever tried to sit and read, any help is greatly appreciated.


text:
How is the acidic dissociation of acetic acid thermodynamically feasible?
For a reaction between an acid and a base
$$\ce{A1 + B1 <=> A2 + B2}$$
to be spontaneous $\ce{A1}$ must be a stronger acid than $\ce{A2}$.
With this in mind, how can this reaction  be understood in terms of spontaneity
    $$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
Also sources on the internet say that $\ce{pK_a}$ of $\ce{H3O+}$ is $-1.74$ while that of acetic acid is $4.74$.

Note that the reaction should be written as an equilibrium, and, importantly, that the equilibrium lies well to the left (i.e. most of the acetic acid is not dissociated).

Then why do those few molecules dissociate anyway? Theoretically (considering the thermodynamics of the reaction) they shouldn't right?

@Azulene You now enter the world of entropy. Suppose you have $10^{23}$ molecules of acid. Will they all behave and not dissociate? That's unlikely. So you end up with a dynamic equilibrium with some molecules recombining, other dissociating etc.

@TAR86 So do you mean to say that every molecule has different amounts of entropy?

I was thinking of the collection of molecules. In terms of pure energy ($U$), they should not dissociate. But there are so many of them - so some do, because it is not likely that they all stay associated. This is quantified by entropy, an expression of probability, which results in free enthalpy ($G$).

All chemical reactions are a balancing act between enthalpy and entropy. Strictly from  an enthalpy standpoint, going to one side of the reaction completely should be favored. However, entropy is maximized by evenly mixing components. Combining these two factors, the equilibrium position is found by maximizing the free energy which has an enthalpic and entropic component. Due to this entropic contribution, all reactions are technically equilibria; they just might be shifted so far to one side that the reaction appears to have gone to completion.

@Tyberius , I think your comment is pretty close to being a good answer...FWIW.

To expand on my comment above, I will reiterate that all chemical reactions are equilibria and this is due to the interplay of enthalpy and entropy. More specifically, it is due to enthalpy and entropy of the individual compounds of the reaction as well as an additional entropy component due to mixing.
Consider the general reaction $$\ce{A<=>B}$$ and let $\chi_\ce{i}$ denote the mole fraction of compound $\ce{i}$ in the solution or gas mixture. The enthalpy and the entropy of the compounds, combined together to form $\Delta G_{compounds}=\Delta H_{compounds}-T\Delta S_{compounds}$ or alternatively $$G=(\chi_\ce{B})\cdot H_{\ce{B}}+(1-\chi_\ce{B})\cdot H_{\ce{A}}-(\chi_\ce{B})\cdot T\Delta S_{\ce{B}}-(1-\chi_\ce{B})\cdot T\Delta S_{\ce{A}}$$ has a particular side of the reaction it favors. This component of the free energy leads to a complete preference; if this were the only factor, the solution would be all $\ce{A}$$(\chi_\ce{A}=1)$ or all $\ce{B}(\chi_\ce{B}=1)$. If we were to plot $G$ as a function of $\chi_\ce{B}$, it would either be linear increasing (reactants favored), linear decreasing (products favored), or flat (reactants and products equally favored).
However, there is an additional entropy component due to mixing, which leads to an additional free energy component. We can write the entropy of mixing as $$\Delta S_{mixing}=-\mathrm{R}[\chi_\ce{A}\ln(\chi_\ce{A})+\chi_\ce{B}\ln(\chi_\ce{B})]=-\mathrm{R}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$
where we have used the fact that $\chi_\ce{B}+\chi_\ce{A}=1$. We can then write the free energy of mixing as $$\Delta G_{mixing}=\mathrm{RT}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$ We can see graphically what these two mixing equations would look like
Combining all this together, we can write $$G_{total}=(\chi_\ce{B})\cdot H_{\ce{B}}+(1-\chi_\ce{B})\cdot H_{\ce{A}}-(\chi_\ce{B})\cdot T\Delta S_{\ce{B}}-(1-\chi_\ce{B})\cdot T\Delta S_{\ce{A}}+\mathrm{RT}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$ We can take the derivative of this function with respect to $\chi_\ce{B}$ and set it equal to zero to determine where the miniumum free energy is $$\frac{\delta G_{total}}{\delta\chi_\ce{B}}=0= H_{\ce{B}} - H_{\ce{A}}- T\Delta S_{\ce{B}}- T\Delta S_{\ce{A}}+\mathrm{RT}[-\ln(1-\chi_\ce{B})+\ln(\chi_\ce{B})]$$ $$H_{\ce{B}} - H_{\ce{A}}- T\Delta S_{\ce{B}}- T\Delta S_{\ce{A}}=-\mathrm{RT}[-\ln(1-\chi_\ce{B})+\ln(\chi_\ce{B})]$$
This second equation is the crux of all this. While we cannot solve this for an exact value without knowing the enthalpy and entropy of the compounds, we can definitively say that $\chi_\ce{B}\ne1$ and $\chi_\ce{B}\ne0$. Why? Plugging either of the values into the equation will result in the right side being $\infty$, meaning the differences in enthalpy and entropy of the components would also have to be $\infty$ for the equation to be satisfied. Thus, the minimum free energy will never be attained by letting the reaction go to completion; it will always be more favorable to allow at least some mixing even if one side of the reaction has a much lower enthalpy and higher entropy of its compounds.
Thus all reactions are technically equilibria. I say technically because the mixing free energy might be negligible compared to the overall free energy and so the reaction could appear to have gone to completion to the extent that we can observe experimentally.
So for your reaction, the reason it doesn't all remain as $\ce{CH3COOH}$ and water is that the entropy of mixing makes it favorable for some of the $\ce{CH3COOH}$ to dissociate.
(For a brief version of this argument and where I got the graph, see http://rkt.chem.ox.ac.uk/tutorials/equilibrium/entropy_mixGas.html)

Wow! Thanks a lot. This really made sense, especially that last equation.

@Azulene Not a problem. I had just received a good recap on this in my Statistical Mechanics course so and this question seemed like a convenient place to put that to use. Consider accepting the answer unless your holding out for something a little different.


text:
Is ethanol an acid or a base or amphoteric?
Is ethanol considered to be an acid or a base? Or does it act as an amphoteric species? Is acidity and basicity a relative concept? Also, can someone please explain this on the basis of the Brønsted–Lowry acids/bases theory which is based on the protonic concept; will it donate or accept $\ce{H+}$ ions? I'm unsure which acid base theory should I follow in this case, any help will be appreciated.

Hello and welcome to Chemistry.SE. If you have any question about the site you can take the short [tour] or visit the [help]. Also, as this is a homework type of question, you can read about the homework policy here. Basically you just need to show your thoughts and any work toward solving the problem yourself so that we are not just doing all of it for you. Best of luck with your problem!

If you assume the question is about Brønsted–Lowry acids/bases, does ethanol contain a proton that can be donated? Does it contain a site that can accept a proton?

Because we are talking about Brønsted–Lowry acids and bases, we are only concerned with whether or not a molecule has (or the degree to which it has) the ability to donate a proton (acid) or the ability to accept a proton (base) or to do either (amphoteric).
The case of ethanol is rather simple to evaluate. Firstly, the hydroxyl proton could be donated to a strong base, although this only happens to a minimal degree in an aqueous solution as water is a much stronger acid than ethanol. Still, ethanol has the ability to act as an acid because of the ability to donate it's hydroxyl proton.
However, aqueous solutions of ethanol are slightly basic. This is is because the oxygen in ethanol has lone electron pairs capable of accepting protons, and thus ethanol can act as a weak base.
Summary - TL/DR:
As ethanol has the ability to both donate and accept protons, it should be considered an amphoteric compound with respect to Brønsted–Lowry acid-base theory.

Acidity and basicity is a relative concept. A species can behave as an acid or a base but then that would depend on the reaction which we're talking about. Consider, for instance, acetic acid $(\ce{CH3COOH})$. It can be amusing that it ofttimes doesn't behave as an acid, contrary to its name! As an example you might want to look the reaction of its conjugate base, acetate ion, with $\ce{HCl}$.
$$\ce{HCl + CH3COONa -> NaCl + CH3COOH}$$
In this case, it can be said that acetate ion was forced to accept $\ce{H+}$ from $\ce{HCl}$, thus it was behaving as a base.
But things like can be downright confusing and thus we usually refer to species being an acid or a base in comparison to water. Alcohols (in general) are considered to be basic in nature, with the notable exception of phenol. The decreasing acidic strength order would then be summed up as:
$$\ce{Phenol \gt H2O \gt ROH}$$
Thus ethanol can be termed as a base.
Such comparisons can be easily made if you're ready to look into the $\ce{pK_a}$ values (you might want to look it up if you're unsure what it means). Water has a $\ce{pK_a}$ of $\ce{15.7}$. Alcohols generally have a $\ce{pK_a}$ in the range of $\ce{16-19}$. Thus they are weaker acids than water. A controversial case, however, is of methanol $\ce{(CH3OH)}$. It has a $\ce{pK_a=15.54}$ which makes it a little acidic than water.

Are you sure about that pKa of water? Shouldn't it be 14? Also, you shouldn't use the \ce command on pKa. For one it isn't a chemical and it also seems to have turned your decimal points into multiplication.

It's the Ionic product ($K_w$) of water which is $10^{-14}$ at $25^{\circ}C$. For calculating pKa of water, we first calculate the equilibrium constant of $\ce{H2O <=> H+ + OH-}$ which is $K_{eq} = \frac{[H^+][OH^-]}{[H_2O]}$. At room temp., $[H_2O]=55.56 M$. Cross multiplying, $K_{eq} \cdot [H_{2}O]=K_w$. The term $K_{eq} \cdot [H_{2}O]$ is known as $K_a$ and is $\approx 1.8 \times 10^{-16}$. Thus, $pK_a = 16 - \log 1.8 \approx 15.74$.

@Tyberius I'm sorry about using \ce, I'm new to this site and I'm way too obsessed with MathJax. I'll take care not to use it on my future contributions. Thanks for letting me know :)

https://chem.libretexts.org/Core/Organic_Chemistry/Fundamentals/What_is_the_pKa_of_water%3F That value of pKa doesn't work when we take into account that equilibrium constants are really defined in terms of  activities. I think the maker of \mchem had you in mind when they made it, because you can use \pu instead for what you were doing and it should format nicer.

That's true! If you take into account the activities, the value of 15.7 wouldn't be correct. I've been taught to consider the concentration of water to be constant and club it with the equilibrium constant to define a new constant. We've also been told to consider the activities of pure solids and liquids to be unity. I probably have to learn a lot about formatting on here, Quora's Latex has much less to offer.

@berryholmes You're contradicting yourself in your first comment...

@Mithoron I'm sorry but I'm not sure how am I doing that. Could you please explain...

@berryholmes You say that both $K_a=K_{eq}\cdot \ce{H2O}$ and that $K_w=K_{eq}\cdot \ce{H2O}$. This would of course imply (correctly, but not necessarily derived correctly) that $K_a=K_w=1\cdot 10^{-14}$.

@Tyberius My apologies, I made a blunder in my comment, $K_a = K_{eq}$ so when we take $\ce{[H+] = [OH-] = 10^{-7}}$, and evaluate $\frac{10^{-7} \times 10^{-7}}{55.56}$, it comes out to be $ \approx 1.8 \times 10^{-16}$ which is the value we were looking for. So $ K_a  = K_{eq} = \frac{K_w}{[H_2O]}$. I want to correct my mistake but i'm unable to edit my previous comment :(

The term $K_\mathrm{eq}$ is known as $K_\mathrm{a}$


text:
Why chlorine acts as a strong field ligand in tetrachloridoplatinate(II)?
In most of the cases I have learnted, halogens act as a weak field ligand. However, I recently learned somewhere on the web that the strength of a ligand also depends upon the size of the atom. A classic example is tetrachloridoplatinate(II): $\ce{[Pt(Cl)4]^2-}$.
Why is it so, and how can I predict the hybridization correctly in such exceptions, if I come across them for the first time?

Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral?

@orthocresol Thank you. I really tried to find it but couldn't find this answer before asking.

No problem, it is difficult to find stuff sometimes. (I didn't think it is 100% a duplicate, and therefore didn't close.)

I would like to add an answer stating a few exceptions which would come handy to you in future:

$\ce{Ni^{2+}}$ has a $3d^8$ configuration. In this case strong field ligands ($\ce{C}$ or $\ce{N}$ donors) will make the complex square planar. In all other cases, the complex will be tetrahedral.
In the case of $4d^8$ and $5d^8$ configuration and any ligand the hybridisation will be that of a square planar complex.
With $\ce{Co^{3+}}$, nearly all ligands behave as strong field ligands ($dsp^2$ hybridisation). Notable exceptions to this rule being $\ce{[CoF6]^{3-}}$ and $\ce{[Co(H2O)3F3]}$.
In case of bulky ligands (for instance $\ce{PPh3}$) present with $3d^8$ configuration, the ligands cannot get adjusted in the same plane leading to a tetrahedral shape ($sp^3$ configuration).

The last one is logical. Rest needs to be memorised probably. Thanks a lot.

This answer simplifies too much in the wrong places. Most notably, not all N donors are automatically strong-field ligands and not all strong-field ligands automatically make a nickel(II) complex square planar.

@Jan Agreed, I should've added that it's more or less general, and if it's an exam (perhaps at a high-school level) asking OP this , I'd suggest to follow the simplified rules.

I’m very much in favour of not teaching simplified rules even at high school levels if they break down somewhere. Either teach the pitfall as well or leave out the simplified rules altogether.

@Jan To be honest, I myself wasn't aware of the pitfall until you pointed out in your comment; the points I've posted in my answer are the ones they taught me in high school, I didn't take chem for higher studies.

Halogens still act as weak-field ligands even in the case of square planar $\ce{[PtCl4]^2-}$ complexes. Weak field does not automatically mean high spin and neither does strong field automatically mean low spin.
Whether a complex adopts a high or a low spin state depends on the central metal’s oxidation state, its position in the periodic table, the ligands, whether they are π-acidic, π-basic or π-neutral, whether the bond is predominantly covalent or dative, the distance between the atoms and sometimes even more subtle factors. Indeed, for some metal-ligand combinations both a high and a low spin complex are known and characterised.
One of the very few rather simple rules when it comes to high or low spin is the fact that any transition metal that is not a $\mathrm{3d}$ metal will usually have a low spin state. In very simple terms this is because the orbitals of higher transition metals are more diffuse and therefore able to form bonds which are more covalent; this in turn leads to greater energetic stabilisations and destabilisations. Therefore, ignore the chlorido ligands in your analysis; look solely at platinum and immediately assume low spin.