text:
Could thin layer chromatography separate enantiomers?
My textbook states the thin layer chromatography could separate enantiomers only if both/or the mobile and solid phases are chiral?
Why so?

Talking about enantiomers, we are dealing with optically active compounds. This unique physiology allows to rotate plane - polarized light in two opponent sides. Analysis of optically active compounds in complex samples is often based on chiral chromatography or capillary electrophoresis in order to separate the enantiomers. For this you require an opponent chiral reagent. The most common method  is HPLC + CD detector. Anylizing the sample youll see negative & positive peaks for the opponent enantiomers, respectively.

The reason is simple: the physical properties of chiral pairs are exactly the same.
So, for any given compound that is chiral, the melting point, the solubility, the partitioning between liquid and solid phase in a chromatography column etc. are all exactly the same for the pairs of chiral molecules. Simple physical processes can't distinguish the two molecules. The solvents and solid components in chromatography are rarely chiral, so they can't tell the enantiomers apart.
The only things than can are other chiral things. If the solid phase of a chromatography column were chiral (or the solvent, though that is a lot less common) then the column could separate the enantiomers.
Think of shaking hands. It is natural that a right handed person shakes hand with the right hand of another person but the interaction is radically different if they to shake hands with the other persons left hand. The interaction of chiral things (like hands) depends on chirality. But if you bump fists instead (fists being closer to a sphere which has no handedness) then it matters a lot less which hand your friend uses.


text:
Isopropyl alcohol still leaves evaporation residue after distillation
I just finished distilling two liters of IPA because evaporating 5 ml on a watch glass revealed it to contain quite a bit of dissolved impurities. However, it still leaves some residue, even though I used a 200 mm vigreux column.
All glassware was thoroughly cleaned beforehand. I do not get such issues with distilled acetone, for example.
What gives?

It is correct.  A long column will obviously improve separation. As the residue left by the distilled alcohol seems to be "highly non volatile" rather than acting on the distillation, a long column should prevent escape of aerosol. A 20 cm vigreaux is certainly short, especially if nested on flask of the liter scale!

How long of a vigreux would you suggest I use?

I assume the problem is, like in the watch glass the solutes after IPA evaporates, i.e. higher boiling point solutes.
Making the distillation slower and reduce the amount of aerosol to escape to distillate flask will surely diminish the amount of impurities. More time, less temperature elevation, it should solve the problem.

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text:
Why is oxidation number of Ba not +2 in this case?
According to redox rules, column 2 elements (alkali earth metals) have an oxidation number of +2, but in this reaction: $\ce{Ba + Cl2 -> BaCl2}$, the oxidation number of $\ce{Ba}$ changes from $0$ on the left to $+2$ on the right instead of starting from $0$ on the left. Why is this?

Alkaline earth metals have oxidation state of +2 when they are combined with nonmetals in compounds, such as in $\ce{BaCl2}$. In $\ce{BaCl2}$ the oxidation state of barium is +2 and the oxidation state of chlorine is -1. Since $\ce{BaCl2}$ is also a binary ionic compound, these oxidation states correspond to the charges on the ions: $\ce{Ba^2+}$ and $\ce{Cl-}$.

On the left side of the equation, barium and chlorine are present in their elemental form, and, by definition, both have oxidation state of 0.

The rules that you are referring to are for atoms in compounds, not in pure elements.

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text:
Do there exist aromatic compounds similar to azulene?
So for this question I'm considering a chemical to be "similar to azulene" if it obeys the following conditions:

It is aromatic and obeys Hückel's rule,
It is a fused bicyclic compound,
One component ring is aromatic as an anion,
The other ring is aromatic as a cation,
It is not azulene, and
It has been synthesised before and the synthesis has been confirmed.

For instance, I believe bicyclo[3.1.0]hexa-1,3,5-triene should follow all of the previous conditions except possibly the last (but I don't know that it hasn't). Do any chemicals obey all of these conditions?

Of course with focus to the aromatic core, right?

related: Can organic compounds such as hydrocarbons contain an ionic bond?

Both rings are aromatic, and charge separation isn't complete.

Also you're not asking about simple derivatives of azulene? There's a lot of them...

There is this compound which is I believe the closest to anything which you've asked for.
Except the fused part this satisfies all your conditions.
In a polar solvent the double bond breaks to form two highly stable aromatic rings(tropylium and dicyclopentene) which act as two separate poles. It also lowers the rotation barrier of this compound. Hence it even becomes soluble. Along with azulene this is as close as an organic compound gets to being ionic.

As linked by ortho, hydrocarbons can be truly ionic.

OK, so we cannot store the $\ce{C_6H_4}$ compound referred to in the question on a shelf in a lab.  But apparently it has been identified as an intermediate species, see here and, for the second page of the article, here.
In summary:  a nucleophilic substitution and elimination is carried out on a bicyclic compound, and the most plausible mechanism, backed by other experiments, is via a bicyclo[3.1.0]hexatriene intermediate.  Computations indicate that the molecule has a bond through the appropriate diagonal of the hexagon.

It seems your links don't work properly for me.

What is wrong?  What happens for me is:  The first link shows the first page of the article, the second link shows the first page of the next article which is also the second and last page of the one I targeted.  It's a sneaky way around the paywall, they need to be more careful!

I see huge list and dunno what to find there.

How strange!  It should d be on pages 1615 and 1616.  I dunno how better to explain this b/c the links work differently for me.

Link is ok here


text:
Why do stable enols give positive ferric chloride test?
It is known that phenol and all its derivatives give violet colour with neutral alkaline ferric chloride solution. However I also found out that stable enols, like 2,4 di-oxo pentane, also give this test. The reason eludes me.

The reason has to do with the complex formed. Not only phenols, but stable enols also form colored complexes with ferric chloride.

Can you be specific. Like what complex?

Ferric chloride forms an octahedral complex. For example, the phenol complex would be $\ce{[Fe(OPh)6]^{3-}}$

I tried looking for a possible complex but came up with nothing. If you could state the complex formed with enols, it would be appreciated.


text:
Which mechanism is not seen during hydrolysis of esters?

Which mechanism is not seen during the hydrolysis of esters?

AAC2
BAC2
AAL2
BAL2
AAC1

I know that hydrolysis of ester is pseudo-first order reaction but I found this question where there are completely unknown reaction mechanisms.

The reaction mechanism said actually corresponds to the following mechanisms,

$A_{AC^2} $ which means acid catalysed acyl carbon oxygen bond breaking bimolecular mechanism.
$A_{AL^2} $ which means acid catalysed alkyl carbon oxygen bond breaking bimolecular mechanism.
$B_{AC^2} $ which means base catalysed acyl carbon oxygen bond breaking bimolecular mechanism.
$B_{ALC^2} $ which means base catalysed alkyl carbon oxygen bond breaking bimolecular mechanism.

In the last one it is unimolecular, and acid catalysed and acyl carbon oxygen bond breaking mechanism.
In esters, Hydrolysis can occur via breaking of two possible bonds. If the ester is $\ce{R'-CO-O-R}$, then the $\ce{CO-O}$ bond is the acyl carbon oxygen bond, whereas the $\ce{O-R}$ bond is alkyl carbon oxygen bond. Either of these two breaks in the process of hydrolysis.
If you take usual hydrolysis of esters, they can occur in both acid and base catalysed reaction medium condition, and in that case the $\ce{R'-CO}$ part of the ester goes to the acid, and the $\ce{-O-R}$ goes to alcohol. So, this is the acyl carbon oxygen bond breaking and bimolecular as the $\ce{OH-}$(in base catalysed) and $\ce{H2O}$ (in acid actalysed) attacks at the rate determining step. So, $A_{AC^2}$ and $B_{AC^2}$ is easily seen.
In an ester, where the attack of $\ce{OH-}$ or $\ce{H2O}$ at the carbonyl carbon is highly sterically hindered (e.g. in mesitoic ester) the reaction passes through alkyl carbon oxygen bond breaking, and thus the $\ce{R'-CO-O}$ part of ester goes to acid, whereas only the R part goes to alcohol, and they are also bimolecular generally, as in the slow step, both the nucleophile attacking at the alkyl carbon remains present along with the molecule. That's why $A_{AL^2}$ and $B_{AL^2}$ are also seen but it is slightly rare.
But for showing $A_{AC^1}$, the molecule doesn't have any other chance rather than forming an acyl carbocation(as, the carbocation intermidiate follows unimolecular reactions). But if it forms acyl acrbocation in acid medium, the lone pair on the adjacent oxygen is already donated to $\ce{H+}$, so, there is no extra stabilising factor rather than any highly electron donating group, like tert butyl or phenyl attached to carbonyl carbon. But if you have those, the molecule generally prefers the alkyl bond breaking to avoid steric hindrance.So, $A_{AC^1}$ is very very rare to happen.
But $A_{AL^1}$ can be seen where R is tert butyl group. because, it can get stabilisation by forming a carbocation of its own, and the remaining portion of molecule can get stabilisation through resonance.


text:
Carbocation Stability: Tertiary versus Secondary Benzyl for Hydrobromic Acid Addition
I haven't been able to find anything concrete or consistent on this topic.
For the following reactions, which product would be favored, assuming that they both have to go through a carbocation intermediate?
What I've gathered so far:
The electron withdrawing group (-NO2) would destabilize the benzylic carbocation (through loss of electron density for stabilization), and would at least help push the reaction in the direction of the tertiary carbocation.
Thanks for any and all help.

In the first instance, bromine will attach itself on the benzylic carbon as the benzylic carbocation formed in this case would be more stable than than the tertiary carbocation due to the mesomeric effect.
However in the second case, the benzylic carbonation formed would be destabilized due to the -M effect of the nitro group, like you pointed out. So bromine would favor the the tertiary carbocation which in this case would be more stable due to hyperconjugation and inductive effects.

First off, thanks so much for the answer! As a follow-up question: as a general rule, are secondary benzylic carbocations just more stable than regular tertiary carbocations?

Unless there is an electron withdrawing group causing instability, yes. However exceptions are always there - the tertiary butly carbonation is even more stable than benzylic carbonation.


text:
Why is manganese oxide precipitated during the syn hydroxylation of alkenes with permanganate ion?
During the hydrolysis step, after the formation of the heterocyclic ring, hydrogen permanganate ion is formed but the final product is manganese oxide precipitate.
My point - why would permanganate be reduced in the first place. And if it would, then why the change to +4 oxidation state specifically, why not to some other oxidation state.

Manganese Oxide is a decay product of permanganate, if the solution is not very acidic it will be formed for sure after a redox reaction.
It might do the diol forming syn addition but some fraction seemed to reduce itself.

Manganese oxide is the product regardless of the nature of the solution.

Very low pH dissolve MnO2, you can't get it always due to that. But when you get, it is an insoluble and highly durable oxide and it will precipitate, my labcoat has them for more than 4 yours and they never got themselves clean no matter how hard we did. So you are sure reaction formed MnO2, and it is apparent that it is insoluble, so what is the question ?


text:
Why is the Diels-Alder reaction a syn addition?
It is said that the Diels-Alder reaction is a syn-addition with respect to both the dienophile and the diene. Is there a reason for that!?

https://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_17/node8.html
Of course there is a reason.

Interaction of these HOMO and LUMO is only possible as in the image, in a certain alignment which results in syn addition product.

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text:
What is the difference between ''1 C atom and 1 mole of C atoms?''

Does 1 C atom mean 1 single C atom out of Avogadro's number of C atoms = 1 single C atom out of 6.023 X 10^23 numbers of C atoms?
Does 1 mole of C atoms mean Avogadro's number of C atoms = 6.023 X 10^23 numbers of C atoms?

I am confused. Could you clarify it, please?

Clarify what? 2) is exactly what the definition says and no idea what 1) is supposed to be, 1 atom is one atom, that's it.

What is the difference between 1 grain of sand and 1 tonne of grains of sand? Does the former mean 1 grain of sand out of 10^9 numbers of grains of sand?

What is the difference between one egg and one dozen eggs?  A dozen is simply a certain count, in this case 12, that is rendered as one unit for our convenience.  Avogadro's number s just the same thing as a dozen, except we made the unit count larger than 12 in order to match it up with our conventionally used units.
Your assumptions about a "mole" are thus correct.

This sounds more like a badly-worded notation question to me.

1 C atom is just one particle of the unique element Carbon. There is virtually no way to isolate or work with a single atom of anything.
1 mole of C is a quantity, like a dozen or a gross. We use dozens for food sale, like eggs, and a gross for small items like pencils. The power of the quantity a 'mole' comes from the mass of a mole in grams. When you have 6.02 x 10^23 particles, you have the formula mass in grams.
1 dozen eggs will weigh differently than 1 dozen bicycles.
Likewise, i mole of any substance will have a mass equal to the species atomic mass, or formula mass.
So, 1 mole of Carbon will have a mass of about 12.01 g (the number written will depend on the periodic table). 1 Mole of water will have a mass of 18 g (approx)
This is why the mole is such a useful quantity.


text:
Why isn't standard Gibbs free energy always zero?
I read in my textbook that one of the standard conditions we should use when calculating the standard Gibbs free energy is having a concentration of 1 M of all the reagents and the products in the reaction. But if you use the formula $G = -RT\ln K_\mathrm e$ and you use 1 M to calculate it, you would always get zero.
So that means 1 M would be the initial concentration of reagents and products before reaching equilibrium?

See this https://en.m.wikipedia.org/wiki/Standard_Gibbs_free_energy_of_formation It says 1 mole of products only, not reactants.

$\Delta G = \Delta G^\circ - RT \ln(K)$
It isn't directly $G$, but formula for change in free energy. Also, in standard conditions $\Delta G = \Delta G^\circ$, so it isn't necessarily 0.
Relevant reading on a website

That link helped a lot!! Thank you so much ((:

@EwelinMachado If his helped a lot, you can press the V button next to his answer.


text:
Mechanism for Chromyl Chloride Formation
This is the very famous chromyl chloride test, used for detection of chloride ion in solution -
$$\ce{K2Cr2O7 + 4NaCl + 6H2SO4 -> 2CrO2Cl2 + 2KHSO4 + 4NaHSO4 +3H2O}$$
When a mixture containing chloride ion is heated with potassium dichromate and concentrated sulphuric acid, deep orange-red fumes of chromyl chloride are formed. My question is, how?
I want to know the proper mechanism for this reaction, particularly for chromyl chloride formation. I couldn't find it in several texts (and online), such as JD Lee and Atkins.
Also, if instead of chloride ion, bromide ion (or iodide) is used then 'chromyl bromide' is not formed. Instead, bromine gas is liberated, and chromium (III) sulphate is formed along with potassium (I) sulphate. Why? What is so special about the chloride ion?

Under those conditions you will have some amount of H2CrO4 in the mixture so I would start from there. Oxygen protonation and double OH2+ displacement with chloride should provide the product.

Are you sure about the mechanism? In fact, please post in the form of an answer with proper 'arrow pushing' steps. It'd be great. Thank you.

I didn't want to post an answer as I am not sure, it's my guess. Couldn't find any proof.

proper mechanism || Unlikely to happen, as the mechanisms are usually very hard to prove. But as a general idea, Cr(VI)O2X fragments are fairly stable, so as a first guess you can assume that the reaction proceeds through a series of such intermediates.

What is so special about the chloride ion || Two reasons actually. First, chloride is a lot smaller and thus better binds with ultra-small Cr(VI). Second, bromide is a lot easier to oxidize, see there electrode potentials.  CrO2F2 forms just as easy (or even easier) than CrO2Cl2

Ahh, it makes sense. Mind adding an answer?

Somewhat related: https://chemistry.stackexchange.com/questions/61493/chromyl-chloride-test-side-products

The answer may (and does) lie in the oxidation potential of the halogens. As we move down the group 17 the tendency for oxidation increases.
Refer:
$$\ce{H2SO4 + HBr -> Br2 + SO2} \tag{unbalanced}$$
but
$$\ce{H2SO4 + HCl -> X} \tag{does not occur}$$
$\ce{K2Cr2O7}$ is not able to oxidise $\ce{Cl-}$ to $\ce{Cl2}$ but it does so with $\ce{Br2}$, thus we observe reddish brown fumes of $\ce{Br2}$ as if $\ce{Br-}$ had been present, which are then passed through $\ce{NaOH}$ to confirm the presence of chloride or bromide.
Refer:
\begin{align*}
\ce{CrO2Cl2 + NaOH ->& Na2CrO4} \tag{yellow soln,unbalanced} \\
\ce{above + H+ ->& (Cr2O7)2-} \tag{orange solution} \\
\ce{Br2 + NaOH ->& NaBr + NaBrO3} \tag{light yellow soln}
\end{align*}

hey guys if this is not the answer then why doesnt any body who downvotes suggest the correct one ?.Are u guys too busy to answer thaan down vote?

perhaps you wud like to refer (E=RC-RA) E is electrode potential rc reduction potential at cathode and RA reduction potential at anode

It does not provide an answer to the main mechanism question. The selective oxidation question is quite easy.


text:
Why do chromyl bromide and chromyl iodide not exist?
It is known that chromyl chloride exists (popularly known as the product of a test for presence of chloride ion) but chromyl bromide and chromyl iodide are unknown.
Why is it so? Why don't they exist? I'd guess it has something to do with the large size of bromine and iodine compared to chlorine, but I'm not sure.
What are the possible reasons for their non existence?

Chromyl bromide  does exist. It is a deep violet needle like crystals. It is formed by reacting chromium(VI) oxide with $\ce{CF3COBr}$ and $\ce{HBr}$ in presence of phosphorus pentoxide. A less pure sample is obtained by treating chromyl chloride with excess amount of $\ce{HBr}$. The compound is hygroscopic and  is purified by vacuum filtration at 138 K. It is thermally unstable above 203 K. (From here)
Another source describing the synthesis of chromyl bromide:

In attempting to prepare chromyl bromide by the action of chromates on
  bromides in the presence of sulfuric acid, chemist obtained negative
  results. However, by the action of anhydrous hydrogen bromide in
  carbon tetrachloride on chromium(VI) oxide or dilute chromyl chloride
  in carbon tetrachloride, or of acetyl bromide on chromium(VI) oxide in
  glacial acetic acid or chromyl chloride in carbon tetrachloride, a
  deep transient purple color resembling that of permanganate is
  observed, which has been ascribed to the presence of chromyl bromide.
  This color was later used as an indicator in studying the action of
  chromyl chloride on phosphorus halides.[...] This color ascribed to
  addition compounds by Zellner, who finally isolated red-brown needles
  of chromyl bromide, contaminated with 7.5 per cent of chromyl
  chloride, by the action of a tenfold excess of liquid hydrogen
  bromide on chromyl chloride at -60°C to -80°C. Chromyl bromide
  decomposes at room temperature and could not be further purified

Chromyl iodide has not been prepared and may not exist due to the powerful oxidizing property of chromium(VI) state. From the 2nd source:

While there has been one report of the formation of chromyl iodide as
  a deep red liquid from a mixture of dichromate, iodide, and sulfuric
  acid, a majority of observers deny this observation and the existence
  of the compound. There is little likelihood that the iodide can exist,
  in view of the large free-energy change attendant on the reaction:
$$\ce{2Cr^6+ + I- -> I^5+ + 2Cr^3+}$$

What is the role of phosphorous pentoxide in this reaction?

@KanyeBest Maybe a dehydrating agent..

Chromyl Bromide does exist.  See Halogen Chemistry at page 246:

Deep violet needles of chromyl bromide are obtained by treating chromium(VI) oxide with $\ce{CF3COBr}$ and HBr in the presence of phosphorus pentoxide. A less pure sample is made by treating chromyl chloride with a large excess of hydrogen bromide. The compound, which is hygroscopic, is purified by vacuum distillation at —135° (Zellner, 1949; Krauss and Starke, 1962).

The book continues:

Chromyl iodide has not been prepared and may not exist because of the powerful oxidizing property of the chromium (VI) oxidation state.

See similarly The Chemistry Of Chromyl Compounds Chem. Rev., 1958, 58 (1), pp 1–61:

Chromyl bromide has likewise been isolated within the past few years (273,
  274). The modern work corrects earlier reports of the compound (77).
While there has been one report of the formation of chromyl iodide as a deep
  red liquid from a mixture of dichromate, iodide, and sulfuric acid (87), a majority of observers (77, 96, 128, 196, 255) deny this observation and the existence of the compound. There is little likelihood that the iodide can exist, in view of the large free-energy change attendant on the reaction:
$\ce{2Cr^{+6} + I- -> I^{+5} + 2Cr^{+3}}$

The question asked is "Why do chromyl bromide and chromyl iodide not exist?" To answer this question, one needs to go back in literature as far as 1824, when chromyl chloride was first prepared by Berzelius by distilling a mixture of a chromate, sodium chloride, and sulfuric acid (Berzelius had not believed he made chromyl chloride at the time).[1] Since then, to my knowledge, only chromyl fluoride, chromyl chloride, and chromyl nitrate of the all reported chromyl compounds have been prepared in the pure condition (note that The chromyl group has been reported to form compounds with all the halogens, and with the nitrate, cyanate, and thiocyanate groups [2]. Let's forget about chromyl nitrate, but, my question is why we don't want to know about chromyl fluoride here as well, since it is also a chromyl halide? Keep in mind that chromyl compounds are physically volatile materials, but there are wide differences in both the volatility and the liquid range of the well-characterized chromyl compounds. As liquids, all these compounds are mobile, dark red in color, and miscible with chlorinated solvents.
Since Berzelius' discovery, many works have been done on chromyl chloride. For the chromate, the choice of the compound among many workers was potassium chromate, although some used potassium dicromate. Measurements of the melting point of chromyl chloride have not been established with high accuracy, since a thermometer calibrated only in whole degrees was used at the time. The accepted value is $\pu {-96.5\pm 0.5 ^\circ C}$, which is reported for this constant in 1912.[2,3] A large number of values have been reported for the boiling point of chromyl chloride. These values, corrected to $\pu {760 Torr}$ where necessary by means of the vapor pressure curve (given in reference). The value of $\pu {116.7 ^\circ C}$ has been accepted by reference works.[3]
Although the chromyl chloride test is a qualitative test for chloride compounds, certain chlorides are reported not to yield chromyl chloride on treatment with sulfuric acid and a chromate. These are the chlorides of the noble metals (platinum, gold, silver, mercury) and antimony(III) oxychloride [2].
The preparation of chromyl fluoride has received far less attention. Most researchers interested in chromyl fluoride had used the reaction between a fluoride, a chromate, and sulfuric acid to volatilize chromyl fluoride [2]. For instance, the elusive vapors of a red chromium-fluorine compound were observed upon heating a mixture of fluorspar (the mineral form of $\ce{CaF2}$), chromates and sulfuric acid as early as 1827, yet, for a long time it was considered to be $\ce{CrF6}$.
Chromyl fluoride was prepared from $\ce{CrO3}$ and anhydrous hydrogen fluoride and isolated in pure form for the first time by Engelbrecht and Grosse in 1952.[4] The purified chromyl fluoride was reported as the violet-red crystals, which reach a pressure of $\pu{780 Torr}$  at $\pu{29.6 ^\circ C}$ and melt to an orange-red liquid at $\pu{31.6 ^\circ C}$. The vapor pressure of the compound was also reported as $\pu{885 Torr}$ at the triple point. Pure chromyl fluoride is reported as stable on storage (according to the reports, at least it makes stable, non-volatile triple addition compounds with $\ce{HF}$ and $\ce{KF}$ or $\ce{NaF}$), whereas the impure material prepared by earlier procedures by others was reported not stable and possessed variable physical and chemical properties. However, alike chromyl chloride, fluoride compound is extremely reactive, particularly with any type of oxidizable organic compound (thus, this may be one of the reasons for its instability when its impure).
Albeit the previous attempts to prepare chromyl bromide by reacting chromates with bromides in the presence of sulfuric acid had been failed, the presence of chromyl bromide was reported in 1911 when anhydrous hydrogen bromide in carbon tetrachloride was used as bromide and chromium(VI) oxide or dilute chromyl chloride in carbon tetrachloride as the chromate. Further, actions of acetyl bromide on chromium(VI) oxide in glacial acetic acid or chromyl chloride in carbon tetrachloride were also resulted a deep transient purple color resembling that of permanganate, which has been ascribed to the presence of chromyl bromide.[5] This intense permanganate-red coloration appeared and then faded rapidly to red-brown solution, further studies of which have shown that it was decomposed to $\ce{Br2}$.
The preparation of chromyl bromide (as an unstable compound) is supported by four different reactions, all of which have yielded the solutions of characteristic permanganate-red color:
$$\ce{CrO3 + 2HBr <=> CrO2Br2 + H2O}$$
The author reported that the solution has immediately assumed a permanganate-red color, which slowly faded to orange-red color.
$$\ce{CrO2Cl2 + 2HBr <=> CrO2Br2 + 2HCl}$$
The addition of hydrogen bromide solution to very dilute chromyl chloride solution (both in $\ce{CCl4}$) has produced an intense permanganate-red coloration, which has gradually faded to orange-red color.
$$\ce{(HO)2CrO2 + 2CH3COBr <=> CrO2Br2 + 2CH3CO2H}$$
Here, $\ce{(HO)2Cr2O2}$)  was prepared in situ by adding glacial acetic acid to $\ce{CrO3}$). Thus, the addition of acetyl bromide solution to $\ce{CrO3}$)  solution with trace of glacial acetic acid (all in $\ce{CCl4}$) has undergone a reaction, which may be described by above equation, producing an intense permanganate-red coloration, which has gradually faded to orange-red color.
$$\ce{CrO2Cl2 + 2CH3COBr <=> CrO2Br2 + 2CH3COCl}$$
The addition of one drop of acetyl bromide solution to extremely dilute chromyl chloride solution (both in $\ce{CCl4}$) has produced the permanganate-red coloration, which has vanished in the course of five minutes. This color was later used as an indicator (the presence of trace amount of chromyl chloride) in studying the action of chromyl chloride on phosphorus halides.[6] Forbes and Anderson have reproduced Fry's results in 1943, during which the similar colors were formed from the action of boron tribromide, silicon tetrabromide, and stannic bromide on chromyl chloride in carbon tetrachloride.[7] They suggested the decomposition of $\ce{CrO2Br2}$ is due to the reaction: $\ce{CrO2Br2 -> CrO2 + Br2}$.
The purple color was ascribed to addition compounds by Zellner, who finally isolated red-brown needles of chromyl bromide (though contaminated with $7.5\%$ of chromyl chloride), by the action of a tenfold excess of liquid hydrogen bromide on chromyl chloride at $\pu{-60 ^\circ C}$ to $\pu{-80 ^\circ C}$. Chromyl bromide decomposes at room temperature and could not be further purified.[8]
It was reported that, although there were no direct evidence to prove the results, chromyl iodide might be formed by the action of acetyl iodide on $\ce{CrO3}$ in the presence of glacial acetic acid, according to indirect evidence [5,7]. Others using various combinations of mixtures including potassium iodide and sulfuric acid, and various techniques, have obtained no evidence of the existence of the compound. It is therefore presumed not to exist.

References

J. J. Berzelius, Annalen der Physik und Chemie, 1824, 1,
1-48.
The Chemistry Of Chromyl Compounds: Winslow H. Hartford, and Marc
Darrin, Chem. Rev., 1958, 58(1), 1–61
(https://pubs.acs.org/doi/abs/10.1021/cr50019a001).
E. Moles and L. Gomez, Zeitschrift fur Physikalische Chemie,
1912, 80(5), 513-530.
Pure Chromyl Fluoride: Alfred Engelbrecht, and Aristid V. Grosse,
J. Am. Chem. Soc., 1952, 74(21), 5262–5264 (https://pubs.acs.org/doi/10.1021/ja01141a007).
The Chemistry Of Chromyl Compounds: Harry Shipley Fry, J. Am. Chem.
Soc., 1911, 33(5), 697–703
(https://pubs.acs.org/doi/10.1021/ja02218a006).
Reactions in non-aqueous solvents I. The action of Chromyl chloride
upon phosphorus trihalides: Harry Shipley Fry, and Joseph L.
Donnelly, J. Am. Chem. Soc., 1916, 38(10), 1923–1928
(https://pubs.acs.org/doi/10.1021/ja02267a002).
Cyanates and Thiocyanates of Germanium, Phosphorus, Sulfur and
Chromium: George S. Forbes, and Herbert H. Anderson, J. Am. Chem.
Soc., 1943, 65(12),  2271–2274
(https://pubs.acs.org/doi/10.1021/ja01252a003).
Versuche zur Darstellung von Chromylbromid: H. Zellner, and O.-O.
Bad Ischl, Monatshefte Fur Chemie, 1949, 80(3), 317-329
(https://link.springer.com/journal/706/80/3/page/1).

This is because of the oxidation potentials of the halogens. Br- anion is oxidised to Br2 by Cr+6 but Cl- is not. Exactly what i am saying perhaps you would like to refer (E=RC-RA) E is electrode potential RC reduction potential at cathode and RA reduction potential at anode

The std electrode potential for Cr(VI) to Cr(III) is +1.36V, same as Cl-/Cl2.  Br-/Br2 is +1.07 and I-/I2 is +0.54

@Waylander exactly what i am saying perhaps you wud like to refer
(E=RC-RA) E is electrode potential rc reduction potential at cathode and RA reduction potential at anode


text:
How to interpret detection limits
I am learning about different analytical methods such as Atomic Absorption Spectrscopy and Atomic Emission Spectroscopy and I'm having a difficult time understand how to interpret the detection limits for the different methods. For example, for AES my book says "detection limits when using an ICP are significantly better than those obtained with flame emission" follow by the table below.
I understand that the detection limit shows the lowest quantity of an element that can be detected by the instrument. However, how do I choose which is the better instrument for an experiment? And why is "ICP significantly better than flame emission"? I would assume that the larger the detection limit, the better since you have a greater limit and thus a better chance of detecting the element. Is that not the correct way to think?

That is NOT the correct way to think. The lower the detection limit of a method, the easier it is to detect the material. For example, from the data in the table, you would be able to detect 0.2 ug/ml of Ag using ICP, but only 2 ug/ml using flame emission. So, if your sample had less than 2 ug/ml and you were using flame emission, you would not be able to reliably detect the Ag in that sample. Make sense?

Oh okay, so if I didn't know the amount Ag in my sample I would choose ICP because it would allow to measure even the smallest amount of Ag in my sample but if I knew that my sample contained more than 2ug/ml then I could use flame emission.

Yes. That is the correct way to look at it.


text:
Accuracy benchmarks for atomic multipole moments
Thermochemical accuracy is regularly quoted as 1 kcal mol$^{-1}$, or 4.2 kJ mol$^{-1}$. I'm expressing some data and I want to use benchmarks like this to validate them, however, they are not all in units of energy. My analyses involve various ranks of a multipole expansion.
I have found a reference when benchmarks dipole moments$^{1}$, but I can't find anything in the literature which similarly benchmarks higher rank multipole moments. Any pointers to the literature or comments from experience would be greatly appreciated.
[1]: J. Phys. Chem. A, 2014, 118 (20), pp 3678–3687.

To the best of my knowledge, there is no definition of "chemical accuracy" for higher-order (electric) multipole moments, in part because they are difficult to measure experimentally, with relatively large uncertainties. I found one recent paper that has a small comparison, prefaced with:

There are several techniques to determine experimentally the dipole moments, [4,5] but it is still very difficult to obtain precise experimental values of higher multipole moments such as quadrupole or octupole moments, [1,6,7] independently of the experimental conditions. Theoretical calculations are therefore essential but challenging for quantum chemistry methods. The accurate calculation of these properties is highly dependent on the method employed, [8] either regarding approximate density functionals [9] or methods based on wavefunctions. [10,11] Consequently, calculating the multipole moments is a way to assess any electronic structure method.

The authors use this work as a method development showcase for the Piris natural orbital functional 6 (PNOF6), the details of which I will generally leave out, but they provide valuable data for Hartree-Fock (HF), coupled cluster with single and double excitations (CCSD), and multireference configuration interaction with single and double excitations (MR-CISD, they abbreviate as MRSD-CI after the old abbreviation SDCI). I will leave in the PNOF data, since it (to my understanding) is a natural orbital-based almost black-box correlated method when the number of occ-virt correlated pairs can be controlled. For the molecules studied (small, organic, closed-shell), CCSD in a triple-$\zeta$ basis set is most likely a reasonable choice.

Table III. $\Theta_{zz}$ component of the quadrupole moments, in atomic units, computed with the Sadlej-pVTZ basis set at the experimental equilibrium [42] geometries.
\begin{array}{lllll}
\hline
                     & \text{HF} & \text{CCSD} & \text{Expt.}             & \text{PNOF6} \\
\hline
\ce{H2}              & 0.4381    & 0.3935      & 0.39 \pm 0.01            & 0.3935 \\
\ce{HF}              & 1.7422    & 1.7156      & 1.75 \pm 0.02            & 1.6939 \\
\ce{BH}^{a}          & 2.6772    & 2.3388      & 2.3293~\text{(FCI/aVTZ)}^{b} & 2.3706 \\
\ce{HCl}             & 2.8572    & 2.7233      & 2.78 \pm 0.09            & 2.7753 \\
\ce{HCCF}            & 3.3530    & 2.9335      & 2.94 \pm 0.10            & 3.2482 \\
\ce{CO}              & 1.5366    & 1.4889      & 1.44 \pm 0.30            & 1.4562 \\
\ce{N2}              & 0.9397    & 1.1712      & 1.09 \pm 0.07            & 1.0530 \\
\ce{NH3}             & 2.1258    & 2.1661      & 2.45 \pm 0.30            & 2.1080 \\
\ce{PH3}             & 1.7217    & 1.5695      & 1.56 \pm 0.70            & 1.6507 \\
\ce{ClF}             & 0.9413    & 1.0514      & 1.14 \pm 0.05            & 1.1122 \\
\ce{CH3F}            & 0.3482    & 0.3002      & 0.30 \pm 0.02            & 0.3269 \\
\ce{C2H2}            & 5.3655    & 4.6850      & 4.71 \pm 0.14            & 5.1531 \\
\ce{C2H6}            & 0.6329    & 0.6234      & 0.59 \pm 0.07            & 0.6275 \\
\ce{C6H6}            & 5.5418    & 5.6653      & 6.30 \pm 0.27            & 6.3571 \\
\ce{CH3CCH}          & 4.2913    & 3.6939      & 3.58 \pm 0.01            & 4.1146 \\
\ce{CO2}             & 3.8087    & 3.1966      & 3.19 \pm 0.13            & 3.6012 \\
\hline
\text{MAE}           & 0.2646    & 0.0902     &                           & 0.1517 \\
\text{MAE}~(\mu_{z}) & 0.0843    & 0.0177     &                           & 0.0309 \\
\hline
\end{array}

a. Calculations performed with the aug-cc-pVTZ basis set.
b. Full CI calculation reported by Halkier [29].

I included the mean absolute error (MAE) for the dipole results from Table II to give an idea of the relative difficulty compared to calculating quadrupole moments. The quadrupole MAEs are about 3, 5, and 5 times larger than the dipole MAEs for each method.
Plotted, it looks like

which is admittedly not very useful other than to show pictorially that CCSD is generally in at least qualitative agreement with the experiment. I personally would consider HF in qualitative agreement to the level that HF usually gets geometries qualitatively correct, though bond lengths are usually too short due to lack of electron correlation. Mean-field electrostatics are the dominant contributor.
For percent errors,
\begin{array}{lrrr}
\hline
&  \delta_{\text{HF to CCSD}} &  \delta_{\text{CCSD to Expt.}} &  100 * \text{err}/\text{Expt.} \\
\hline
\ce{H2}        &                       11.33 &                           0.90 &                           2.56 \\
\ce{HF}        &                        1.55 &                           1.97 &                           1.14 \\
\ce{BH}        &                       14.47 &                           0.41 &                           0.00 \\
\ce{HCl}       &                        4.92 &                           2.04 &                           3.24 \\
\ce{HCCF}      &                       14.30 &                           0.22 &                           3.40 \\
\ce{CO}        &                        3.20 &                           3.40 &                          20.83 \\
\ce{N2}        &                       19.77 &                           7.45 &                           6.42 \\
\ce{NH3}       &                        1.86 &                          11.59 &                          12.24 \\
\ce{PH3}       &                        9.70 &                           0.61 &                          44.87 \\
\ce{ClF}       &                       10.47 &                           7.77 &                           4.39 \\
\ce{CH3F}      &                       15.99 &                           0.07 &                           6.67 \\
\ce{C2H2}      &                       14.53 &                           0.53 &                           2.97 \\
\ce{C2H6}      &                        1.52 &                           5.66 &                          11.86 \\
\ce{C6H6}      &                        2.18 &                          10.07 &                           4.29 \\
\ce{CH3CCH}    &                       16.17 &                           3.18 &                           0.28 \\
\ce{CO2}       &                       19.15 &                           0.21 &                           4.08 \\
\hline
\end{array}
this is further confirmation that CCSD is probably a good qualitative benchmark.

Table IV. $\Theta_{zz}$ and $\Theta_{xx}$ components of molecular quadrupole moments, in atomic units, computed using the Sadlej-pVTZ basis set at the experimental equilibrium [42] geometries.
\begin{array}{lrrr}
\hline
                & \text{HF}  & \text{MR-CISD} & \text{Expt.}  & \text{PNOF6} \\
\hline
\ce{H2O}~(xx)   &  1.7966    & 1.8050         & 1.86 \pm 0.02 & 1.7808 \\
\ce{H2O}~(zz)   &  0.0981    & 0.0950         & 0.10 \pm 0.02 & 0.0869 \\
\ce{H2CO}~(xx)  &  0.1019    & 0.1100         & 0.04 \pm 0.12 & 0.0516 \\
\ce{H2CO}~(zz)  &  0.0921    & 0.2230         & 0.20 \pm 0.15 & 0.1255 \\
\ce{C2H4}~(xx)  &  2.7819    & 2.3700         & 2.45 \pm 0.12 & 2.5892 \\
\ce{C2H4}~(zz)  &  1.4942    & 1.1700         & 1.49 \pm 0.11 & 1.3266 \\
\ce{O3}~(xx)    &  1.1175    & 1.2830         & 1.03 \pm 0.12 & 1.2426 \\
\ce{O3}~(zz)    & -0.2387    & 0.1680         & 0.52 \pm 0.08 & 0.3606 \\
\hline
\text{MAE}      &  0.1772    & 0.1448         &               & 0.1066 \\
\hline
\end{array}

The ozone result for Hartree-Fock most likely has the wrong sign due to its multiconfigurational ground state, with the MRCI result still in relative error of ~67%.
For octupole moments, they only present results for methane ($\ce{CH4}$), for which the first non-vanishing multipole moment is the octupole moment, due to its $T_{d}$ symmetry.

\begin{array}{ll}
\hline
                 & \Omega_{xyz}~(\text{a.u.}) \\
\hline
\text{CCSD}      & 2.0595                     \\
\text{PNOF6(14)} & 2.1142                     \\
\text{Expt.}     & 2.95 \pm 0.17              \\
\hline
\end{array}

Considering the lower bound of the experiment, the PNOF6 result is in 23% error, and considering the upper bound of the experiment, the CCSD result is in 36% error.
I won't draw any more conclusions about the data other than to say that in general, there may be large errors in the experimental data that are hard to predict (non-systematic). In that case, if you were to perform a study looking at convergence of a result with respect to some other parameter (say, cluster size), I would first look at convergence of said result with respect to method (HF, MP2, CCSD) and basis (aVDZ, aVTZ, ...) at the largest cluster size, just to check that CCSD/aVTZ is a reasonable benchmark if you're after absolute accuracy. Since experimental error is large, it may be more reasonable to just compare against higher-level, full cluster calculations.
References

0: Mitxelena, Ion; Piris, Mario. Molecular electric moments calculated by using natural orbital functional theory. J. Chem. Phys. 2016 144, 204108, DOI: 10.1063/1.4951685
1: A. D. Buckingham. Direct Method of Measuring Molecular Quadrupole Moments. J. Chem. Phys. 1959 30, 1580,  DOI: 10.1063/1.1730242
6: W. H. Flygare and R. C. Benson. The molecular Zeeman effect in diamagnetic molecules and the determination of molecular magnetic moments (g values), magnetic susceptibilities, and molecular quadrupole moments. Mol. Phys. 1971 20, 225, DOI: 10.1080/00268977100100221
7: E. R. Cohen and G. Birnbaum. Far infrared collision‐induced absorption in gaseous methane. II. Determination of the octupole and hexadecapole moments. J. Chem. Phys. 1975 62, 3807, DOI: 10.1063/1.430932
8: R. Glaser, Z. Wu, and M. Lewis. A higher level ab initio quantum-mechanical study of the quadrupole moment tensor components of carbon dioxide. J. Mol. Struct. 2000 556, 131, DOI: 10.1016/S0022-2860(00)00658-X
9: A. J. Cohen and Y. Tantirungrotechai. Molecular electric properties: an assessment of recently developed functionals. Chem. Phys. Lett. 1999 299, 465, DOI: 10.1016/S0009-2614(98)01317-7
10: P. Bundgen, F. Grein, and A. J. Thakkar. Dipole and quadrupole moments of small molecules. An ab initio study using perturbatively corrected, multi-reference, configuration interaction wave functions. J. Mol. Struct. 1994 334, 7, DOI: 10.1016/0166-1280(94)03974-P
11: J. M. Junquera-Hernández, J. Sánchez-Marín, and D. Maynau. Molecular electric quadrupole moments calculated with matrix dressed SDCI. Chem. Phys. Lett. 2002 359, 343, DOI: 10.1016/S0009-2614(02)00722-4
29: A. Halkier, H. Larsen, J. Olsen, P. Jørgensen, and J. Gauss. Full configuration interaction benchmark calculations of first-order one-electron properties of BH and HF. J. Chem. Phys. 1999 110, 734, DOI: 10.1063/1.478042


text:
Getting from pH 2 to pH 6 by mixing with water
In our book we learned that we can get an acid to a higher $\mathrm{pH}$ by mixing it with water in the relation $\frac {1}{10} $.
Now to my problem:
How much water do you need in order to get the $\mathrm{pH}$ value from $2$ to $6$. It is $\pu{250 L}$ of acid.

Try: I have $\pu{250 L}$ of acid, I will mix it with water in the relation $\frac {1}{10} $. Then I will have $\pu{2500 L}$ of this mixture, is basically means that I have needed $\pu{2250 L}$ of water to get to $\mathrm{pH}\space 3$. $\mathrm{pH}\space 4$ will have $\pu{25000 L}$ so I needed $\pu{22500 L}$ of water to get to $\mathrm{pH}\space 4$ and so on...
This would mean that I would need $2250+22500+225000+2250000=2499750 \space \mathrm{L} $ to get to $\mathrm{pH}\space 6$.
Try
I know that $c (\ce{H3O+})=\mathrm {\frac {mol}{L}}\Rightarrow 10^{-2}=\pu{\frac {mol}{\pu{250 L}}}\Rightarrow m=\pu{2.5mol}$.
Also $\mathrm{pH} (6)=c (10^{-6})$
$\frac {\pu{2.5mol}}{x\space \mathrm{L}}=10^{-6}\Rightarrow x=2500000\space \mathrm{L}$. This would mean that I need to add $\pu{2500000L}$.

What am I doing wrong, how can a fix my mistake in one of my possible solutions?

Why do you think anything is wrong in the first place?

Because I have two different solutions to my problem?

Welcome to Chemistry.SE. We’d like you to take the Tour to familiarize with the site.

In first try, everything is ok.
In second try, you find the final volume of solution, $\pu{2500000L}$, but you already have $\pu{250L}$ acid solution so in order to reach that, $\mathrm {2500000\space L-250\space L=2499750\space L}$ water is needed.


text:
Reversibility of the binding of carbon monoxide to haemoglobin
I recently learnt the basic mechanism of haemoglobin: oxygen forms coordinate bonds with the $\ce{Fe^2+}$ ion in haem groups, and this reaction is reversible to allow oxygen to be released. I also learnt that carbon monoxide does the same thing as oxygen, however, this reaction is said to be irreversible due to larger bond strengths etc. meaning that it can't be released.
What isn't clear to me is what happens to carbon monoxide if it does bind to a haem group. If the reaction is irreversible, then how is carbon monoxide removed from the body if you were to breathe it in?

That's just a matter of what you'd call irreversible in this case it's just a lot harder to unbind then O2

The reaction is not irreversible, but hemoglobin adheres to $\ce{CO}$ about 200 times more tightly than it does to $\ce{O2}$. Thus, part of the treatment for $\ce{CO}$ poisoning is to expose the patient to oxygen, which will slowly displace $\ce{CO}$ from carboxyhemoglobin.
For more on the chemistry involved, see this qestion.

Thanks for clearing that up. My exam specification says that the haem-carbon monoxide bond is irreversible, which I thought was a bit strange.

You might politely point out to the examiner the references citing reversibility, e.g. Wikipedia, and that people do recover from CO poisoning more rapidly than it would take for RBC's to be replaced (their lifetime is on the order of three months), depending on how open the examiner is to criticism.

It's not a single examiner, it's a full exam board. For me to want to mention this to them, I'd have to ask my teacher to send some form of notice to them. That of course would take time etc. haha

Sorry about that. As a teacher, I occasionally noticed mistakes in a text and would point them out to students, but an exam with an error is downright, frustratingly, unfair.

Maybe it was intentional. For the A Levels we only need to know the concept of coordination chemistry, so maybe haemoglobin is just an example for us to know; they might've said it's irreversible and left it at that. Just like we need to know that cisplatin is a chemotherapy drug but we don't need to know its full mechanism of action.


text:
How would a pH curve look like for titration of diluted weak base compared to concentrated one?
Let's say I take $\pu{80 g}$ of a weak base, dilute it with  $\pu{50 ml}$ of water and titrate it with a strong acid. I get a titration curve. Now I take again $\pu{80 g}$ of the same weak base but this time I dilute it with $\pu{200 ml}$ of water and titrate it with the strong acid. What will be the new titration curve compared to the first one?
And here is what confuses me:
On one hand, the diluted base should have lower $\mathrm{pH}$. But on the other hand, at half-equivalence point, $K_\mathrm{b}=\mathrm{pOH}$, and since it's the same base, $K_\mathrm{b}$ is the same, and thus $\mathrm{pOH}$ and $\mathrm{pH}$ at half equivalence point should be the same. I am also not sure about the endpoint.

I am first going to do the ‘flip side’ of the OP’s scenario and then basically flip it back. So we start with a $\pu{0.1 M}$ aqueous solution of benzoic acid and the initial solution volume is $\pu{25 mL}$. Benzoic acid is a monoprotic weak acid with $\mathrm{p}K_\mathrm{a} = 4.20$. The initial $\mathrm{pH}$ is $2.606$. The strong base used for the titration is $\pu{0.1 M} \ \ce{NaOH}$. At the equivalence point, the $\mathrm{pH = 8.454}$ and the solution volume is $\pu{50 mL}$ ($=\pu{25 mL}$ benzoic acid solution $+ \pu{25 mL} \ \ce{NaOH}$ solution). Thus, the $\mathrm{pH}$ is simply the $\mathrm{pH}$ of $\pu{0.05 M}$ sodium benzoate solution. After all $\pu{50 mL}$ of $\ce{NaOH}$ has been added, the $\mathrm{pH} = 12.523$ and the final solution volume is $\pu{75 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{NaOH}$ has been added and $\mathrm{pH}= 4.185$.
Now assume the initial benzoic acid solution is only $\pu{0.025 M}$, but the initial volume is $\pu{100 mL}$. The total number of moles of benzoic acid is the same as before. The initial $\mathrm{pH}$ is $2.912$: the benzoic acid solution is only 25% as concentrated. At the equivalence point, the $\mathrm{pH} = 8.254$ and the solution volume is $\pu{125 mL}$ ($=\pu{100 mL}$ benzoic acid solution $+ \pu{25 mL} \ \ce{NaOH}$ solution). The equivalence point $\mathrm{pH}$ is closer to $\mathrm{pH}=7$ because the solution is simply a $\pu{0.02 M}$ sodium benzoate solution, i.e., the number of moles of sodium benzoate is now in a volume that is 2.5 times larger, so the concentration is 40% of $\pu{0.05 M}$. After all $\pu{50 mL}$ of $\ce{NaOH}$ has been added, the $\mathrm{pH}= 12.222$ and the final solution volume is $\pu{150 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{NaOH}$ has been added and $\mathrm{pH} = 4.188$, moving slightly up toward $\mathrm{pH}=7$. See the titration curves below:

Now flipping back: the OP’s scenario involving a weak base titrated with strong acid. For reasons that will become clear, assume the hypothetical weak base has $\mathrm{p}K_\mathrm{b} = 9.80$, and that the strong acid is $\pu{0.1 M} \ \ce{HCl}$ solution. Note that the $\mathrm{p}K_\mathrm{b}$ is simply $14$ (treated as exact throughout) $- 4.20$.
As before, start by assuming $\pu{25 mL}$ of $\pu{0.1 M}$ weak base. The initial $\mathrm{pH} = 14 – 2.606 = 11.394$. At the equivalence point, the $\mathrm{pH} = 5.546$ and the solution volume is $\pu{50 mL}$ ($=\pu{25 mL}$ weak base solution $+ \pu{25 mL} \ \ce{HCl}$ solution). Thus, the $\mathrm{pH}$ is simply the $\mathrm{pH}$ of a $\pu{0.05 M}$ weak base salt solution. After all $\pu{50 mL}$ of $\ce{HCl}$ has been added, the $\mathrm{pH} = 1.477$ and the final solution volume is $\pu{75 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{HCl}$ has been added and $\mathrm{pH} = 9.815$.
Now assume the initial weak base solution is only $\pu{0.025 M}$, but the initial volume is $\pu{100 mL}$. The total number of moles of weak base is the same as before. The initial $\mathrm{pH}$ is $11.088$: the weak base solution is only 25% as concentrated. At the equivalence point, the $\mathrm{pH} = 5.746$ and the solution volume is $\pu{125 mL}$ ($=\pu{100 mL}$ weak base solution $+ \pu{25 mL} \ \ce{HCl}$ solution). The equivalence point $\mathrm{pH}$ is closer to $\mathrm{pH}= 7$ because the solution is simply $\pu{0.02 M}$ weak base salt solution, i.e., the number of moles of weak base salt is now in a volume that is 2.5 times larger, so the concentration is 40% of $\pu{0.05 M}$. After all $\pu{50 mL}$ of HCl has been added, the $\mathrm{pH} = 1.778$ and the final solution volume is $\pu{150 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{HCl}$ has been added and $\mathrm{pH} = 9.812$, moving slightly down toward $\mathrm{pH}= 7$. For the titration curves, simply imagine vertically flipping the titration curves, in the figure above, about a horizontal line at $\mathrm{pH} = 7$. Or just look at the plot:

One last thing: The $\mathrm{pH}$ values at half-equivalences were close to the respective $\mathrm{p}K$ values, as expected, but any buffer can be broken by dumping it into swimming pool full of water. So dilution may matter a great deal, or relatively little, depending on the specifics.

The equivalence is, by definition, the moment when $n_{base}=n_{acid}$.
As you titrate the same mass of base (consequently the same quantity), your equivalence will be reached at the same point.
The only difference, to me, between the two curves will be the fact that you will less see the equivalence when the solution is diluted. First curve is 80g in 50mL, second one is 80g in 200mL.

Thanks a lot! But what about the pH at the begining? is it the same? On one hand, the diluted base should have lower pH. But on the other hand, at half equivalence point, Kb=pOH, and since it's the same base, Kb is the same, and thus pOH and pH at half equivalence point should be the same.

your initial pH will change, as it will get closer of the one of water. But keep in mind that the equivalence depends only of $n_{acid}$. Hence, $n_{acid}=n_{base} \Leftrightarrow C_{acid}V_0=C_{base}V_{eq}$. So between your two titrations, pOH at equivalence and half-eq will remain the same, but volumes will change (as the quantity of base to reach the required pOH will change)

Thank you, so I understand that except the lower initial ph of the diluted base, other points should be pretty much the same (half equivalence point, volume untill equivalence point, ph at equivalence point). Did I get it straight?

That’s pretty much it, but your volumes won’t be the same, as you start from a different pH, and concentration. The quantities will be the same, but not volumes (they will change, as you don’t have the same concentrations of base). But all other values will remain unchanged.

At the beginning of the titration, the solution is a weak base. At the equivalence point, the solution is a weak acid (with some more spectator ions). The pH of these solutions changes with dilution, both in the direction of neutral pH.
In contrast, at the half-equivalence point (or midpoint) of the titration, the solution is a 1:1 buffer. As long as the buffer concentration is large compared to the concentration of hydronium and hydroxide ions, the pH will be close to the pKa (and the pOH close to the pKb), irrespective of dilution.
So if you dilute the weak base with water before the titration, nothing will change along the x-axis (i.e. half-equivalence point and equivalence point will occur at the same volume of added strong acid). Along the y-axis (where you plot the pH), the curve will "shrink" toward the pH of the buffer (i.e. the starting pH will be lower - less basic - and the pH at the equivalence point will be higher - less acidic). Past the equivalence point, the pH is controlled by the excess of strong acid. Because the strong acid is diluted into a greater volume if you add water to the weak base before starting the titration, this part of the titration curve will also show higher - less acidic - pH values.


text:
Why does the gas from dry ice flow down?
I am reading Dan Broughton's answer in replying to the question "why does the gas from dry ice flow down" here.
According to the answer, the fog is water's color and water is condensed and suspended in the CO2.
Do you agree with this answer? If so then should water have color? I see some documents saying that it is nearly colorless.

Why does the gas from dry ice flow down?
Dry ice is simply the common term used for solid carbon dioxide (CO2),
  so the gas coming off of the dry ice is CO2 gas. CO2 gas has a density
  of 1.96 kg/m^3 and air (78% N, 21% O2) is 1.29 kg/m^3. Since the CO2
  is more dense, it sinks in air.
But CO2 is colorless, so what makes it appear as a white mist? As the
  gas sublimes (change directly from solid to gas), it comes off of the
  dry ice at -78.5 C (-109.2 F).  Being well below the freezing point of
  water, any moisture in the air surrounding the dry ice condenses and
  is suspended in the CO2, creating the cloud/fog effect that we can
  see.
These two phenomena combined are what create the fog that flows along
  the ground.

Well, yes, the answer is mostly correct. Water doesn't have color, nor does the fog.

Also, being colder and denser, even a plain air fog coming off a cold object (e.g. dry ice in a plastic bag, vented away from the experimenter) will sink.

Colorless as in invisible as gas at all.

The quoted answer is correct. It just omits to mention scattering and reflection from water/ice minute particle. This is the core of your question. Why cola is dark and its foam is white is another question related to this, just in this latter example both forms are visible.

The gas flows downwards because it is more dense. It is more dense both because $\ce{CO2}$ is heavier than $\ce{N2}$ and $\ce{O2}$ and because the gas is colder than the ambient air.
The mist is water in the surrounding air condensing into microscopic droplets due to the lowered temperature.  The amount of mist depends on the humidity of the air; in perfectly dry air there will be no mist.
All of the gases involved are colourless. The mist appears white because it is in small particles and so it scatters any light falling on it.  It is the same colour as "natural" mist and for the same reason.


text:
How can I solvent exchange from water to alcohol?
Given an aqueous colloidal solution, is there a way to transfer the colloidal particles into an alcohol (assuming the colloidal particles are also stable in the alcohol)? I see in the literature that lyophilizing the aqueous solution and re-dispensing the colloidal particles is an option, but is there a method to perform this all in the wet phase?

It would help other potential answerers if you would cite the literature.

Since it's a colloid, rather than a true solution, put the water-based mixture in a dialysis membrane of appropriate pore size and immerse it in concentrated alcohol. The water will dissolve into the alcohol, while alcohol slowly seeps in to displace the water, but the colloidal particles cannot get through the pores. Depending on how little water is desired in the final product, this step may be repeated multiple times.
For some materials, inexpensive cellophane sausage casing may be suitable.


text:
What is the relation between the speed of various components of a mixture, and their respective Retardation Factors in adsorption chromatography?
Consider two components, I and II, of a mixture which is subject to adsorption column chromatography. Let I have higher adsorption than II.
I know that Retardation Factor ($R_f$) of component II will be higher than that of I. Which component will move down the column at a faster speed? Is the speed of a component in chromatography related to its $R_f$ ?

If I has really higher adsorption, it is the one retarded more, not the II.
Chromatography, in a basic terms, separates components of a mixture according to their adsorption/desorption equilibrium. If a compound does not even adsorp, it will come with solvent front and will be fastest. If a compound adsorp so much that become effectively immobile, it may not even be seen at the end.
You need to think that chemical will move when in mobile phase, when it is adsorped, it will be immobile so its change of position in unit time will decrease as much as it adsorps.

But if I is adsorbed more then it will travel less distance than II. $R_f$ is defined as ratio of distance travelled by component to distance travelled by solvent, so II will travel more distance and will have a higher $R_f$

So even though $R_f$ of II is more, will it be faster?

Yes, higher retardation factor, faster compound. I should give the first sentence as II and I not in I and II order, but nevertheless you got the point.


text:
Why does the pressure equilibrium constant exist?
Since $K_c$ can already be used for calculating the equilibrium constant for a reaction involving gases, but $K_p$ cannot be used for calculating the equilibrium constant for a reaction involving aqueous solutions, what is the use for $K_p$?
It doesn't seem like there is ever a situation where $K_c$ cannot be used and $K_p$ must be used, especially since partial pressures can be readily converted to concentration through the ideal gas law. If anything, it just seems like the existence of two equilibrium constants that more often than not don't equal each other could cause more confusion.
Thoughts?

It's easier to use $\mathrm{K_p}$ rather than $\mathrm{K_c}$ in cases pertaining to gaseous equilibrium.

https://chemistry.stackexchange.com/questions/40567/which-equilibrium-constant-is-appropriate-to-use highly descripted.

It depends on what units you use, if pressure is measured then use $K_p$, if concentration  then $K_c=K_p(RT)^{-\Delta n}$ where $\Delta n$ is the change in the number of moles of gaseous substances as the reaction goes from left to right and $p$ is the total pressure. If mole fraction $K_x=K_pp^{-\Delta n}$.

@porphyrin isn't that just derived from the ideal gas law, like I mentioned in the question?

Ah... so it actually does make a difference. Thanks @akgnit.

@KarmanyaGB Sorry, this isn't a duplicate. That question asks whether the $K$ in $\Delta G^\circ = -RT \ln K$ stands for $K_p$ or $K_c$. This question does not even once mention that equation, and is in fact, a different question.

@GaurangTandon true, but that answers why the pressure equilibrium constant exists. Its the one that is correct (under certain assumptions) for gases and Kc would not work in that equation.


text:
Is the relationship between Ka and Kb the same for all acid base pairs?
What is the relationship between $K_\mathrm{a}$ and $K_\mathrm{b}$ for acid/base reactions. I know for water $K_\mathrm{w}=K_\mathrm{a}\cdot K_\mathrm{b}$, where $K_\mathrm{w}$ is $10^{-14}$. Is this true for other acid/base pairs?

What do you mean by "other pairs"?

i meant other acids and their conjugate bases. E.g. CH3COO- and CH3COOH

First of all, let me put a statement to clarify the proper meanings of $K_\mathrm{a}$ and $K_\mathrm{b}$ for this case.
The $K_\mathrm{a}$ is the acid dissociation constant of a weak acid, and the $K_\mathrm{b}$ is the dissociation constant of its conjugate base. Alternatively, we can also say, $K_\mathrm{b}$ is the base dissociation constant of a weak base, and $K_\mathrm{a}$ is the dissociation constant of its conjugate acid.
We can deduce the same relationship, i.e. $K_\mathrm{a}\cdot K_\mathrm{b} = K_\mathrm{w}$, for these cases also. Without loss of generality, let the weak acid be $\ce{HA}$ and its conjugate base is $\ce{A^-}$ (you can also do the same [not identical] derivation considering base $\ce{B}$ and its conjugate acid $\ce{BH^+}$).
Considering acid dissociation,
\begin{align}
\ce{HA &<=> H^+ + A^-}&
K_\mathrm{a} &= \ce{\frac{[H^+][A^-]}{[HA]}},
\end{align}
and for the basic action of its conjugate base, consider,
\begin{align}
\ce{A^- + H2O &<=> HA + OH^-}&
K_\mathrm{b} &= \ce{\frac{[HA][OH^-]}{[A^-]}}.
\end{align}
Multiplying both the equations, we will have,
$$
K_\mathrm{a}\cdot K_\mathrm{b}
= \ce{\frac{[A^-][H^+]}{[HA]}}\cdot \ce{\frac{[HA][OH^-]}{[A^-]}}
= \ce{[H^+][OH^-]} = K_\mathrm{w}$$
So, it is proved that for any acid/conjugate base pair (or, base/conjugate acid pair) the identical relation, $K_\mathrm{a}\cdot K_\mathrm{b} = K_\mathrm{w}$ holds.

I have corrected the use of MathJax in your post. Please have a look here and here for more information.

@Soumik Das,the last step you did,is multiplying ka and kb for different reaction,in that case,[H+] is the concentration of ion at equillibrium of reaction 1 while,[OH-] Is concentration of [OH-] at equillibrium of second reaction,how do we know that their product must equal kw,because these are two different reactions,the result [h+][oh-]=kw is proved if both concentrations are from same aqueous solution


text:
Is there an upper limit on the coordination number of a coordination compound?
From the Gold Book, coordination number is:

In an inorganic coordination entity, the number of σ-bonds between ligands and the central atom. π-bonds are not considered in determining the coordination number.

In high school, we mostly had complexes upto coordination number six (like $\ce{[FeF6]^3-}$). But I wondered if the coordination number could go even higher.
From my search, I was able to reach this research paper (J. Am. Chem. Soc., 1979, 101 (2), pp 334–340 DOI: 10.1021/ja00496a010) where they mention lanthanide complexes of coordination number uptil nine!
So, is nine the upper limit? Or can the coordination number go higher? Also, does the upper limit exist only theoretically, or has the corresponding complex been synthesized?

Theoretically speaking, coordination number is not a thing at all, especially when we look at lanthanides with their ugly coordination polyhedra. The maximum possible coordination number is not a physical constant of any significance. Practically speaking, I vaguely remember hearing of CN up to 14 in uranium borohydrides or something.

See Can an atom bond with more than 8 other atoms?

Note: I self-marked the question as duplicate after having two CVs and then going through the linked question. The thing is - even if the titles do not match - the excellent, well-detailed answers to that question contain exactly the stuff I am looking for, and it makes little sense to report them over here.

I don't see why not. It is a matter of having

enough of an interaction between the center and the ligands, usually in the form of donor/acceptor orbital pairs, and
enough space around the coordination center.

These go hand-in-hand, as any atom that will have enough electrons to donate to ligands will also most likely have a larger atomic radius than a first-row transition metal like iron.
For example, here is a thorium complex that has CN = 15 in its crystal structure and CN = 16 in an isolated gas-phase density functional calculation:

Daly, Scott R.; Piccoli, Paula M. B.; Schultz, Arthur J.; Todorova, Tanya K.; Gagliardi, Laura; Girolami, Gregory S. Synthesis and Properties of a Fifteen-Coordinate Complex: The Thorium Aminodiboronate $\ce{[Th(H3BNMe2BH3)4]}$. Angew. Chem. Int. Ed. 2010, 49, 3379-3381, DOI: 10.1002/anie.200905797

Hmm, thanks! A small query though, "any atom that will have enough electrons to donate to ligands" isn't it in the other direction? Ligands donate electrons to metals, and not vice-versa? (pardon my basic knowledge, but that's what I was taught)

See metal-to-ligand charge transfer.


text:
Bromine radical reaction with an alkene - What factors determine whether radical addition occurs as oppose to radical abstraction?

From what I can tell, radical addition occurs when using HBr + Organic peroxides as reagents, and radical abstraction occurs when using trace amounts of bromine radicals (generated by NBS, trace HBr and UV).
However, in the end, a bromine radical is reacting with an alkene. Why / how do the preceding reagents determine the regioselectivity of the radical reaction?


text:
Do alkyl halides with an alpha nitrogen atom undergo SN1 reaction?
Today, in my organic chemistry class, I learnt about the nucleophilic substitution reaction. The notes have stated that:

Alkyl halides with $\alpha$ oxygen atom can also undergo SN1 reaction because of the heteroatom stabilised cation (oxonium ion)

An example was then given of a bromide leaving from a bromoether to give a resonance-stabilised carbocation.
I was wondering if alkyl halides with an $\alpha$ nitrogen atom can undergo SN1  reaction as well. When asked, the teacher responded by saying that with $\alpha$ nitrogen atoms, the carbocation would experience greater stabilisation and result in a faster rate of SN1 reaction for such alkyl halides.

Ask yourself if nitrogen can sustain a positive charge better than oxygen.

The reactivity of haloalkyl ethers are well documented (e.g., they were originally used to protect $\ce{-OH}$ groups during total syntheses). For instance, chloromethyl methyl ether had been used to protect alcohol and phenol $\ce{-OH}$s as $\ce{-OCH2OCH3}$ or commonly known as $\ce{-OMOM}$. They should be sealed and stored in freezer to protect from moisture because they are susceptible to decompose by reacting with water, showing their reactivity. They are also well known carcinogens. An early review on $\alpha$-haloalkyl ethers can be found in Chemical Reviews (1955)([Ref. 1]).
A recent example in Ref. 2 shows in situ preparation of $\alpha$-halo ethers to protect $\ce{R-OH}$ groups, in order to prevent researchers from exposure to carcinogenic $\alpha$-halo ethers (see Figure below):

For example for carcinogenic activity of a $\alpha$-halo ether: Human health assessment information on bis(chloromethyl)ether can be found on the IRIS website (https://www.epa.gov/iris). IRIS stands for Integrated Risk Information System, a part of EPA’s mission to protect human health and the environment.
Based on the high reactivity of  $\alpha$-halo ethers, one can expect $\alpha$-halo amine would have been even higher reactivity than them. To my knowledge, they do not exist, mainly because, they'd undergo elimination quickly to give relevant  Schiff bases (see below):
$$\ce{-> [R-CH(Cl)-NHR1] -> R-CH=NR1 + HCl} \ \mathrm {or}$$
 $$\ce{-> [R-CH(Cl)-NR1R2] -> R-CH=N+R1R2Cl-}$$
[Ref. 1]: The alpha-Haloalkyl Ethers: Lawrence Summers, Chem. Rev., 1955, 55(2), 301-353 (DOI: 10.1021/cr50002a003).
[Ref. 2]: Simple, Rapid Procedure for the Synthesis of Chloromethyl Methyl Ether and Other Chloro Alkyl Ethers: M. A. Berliner, and K. Belecki, J. Org. Chem., 2005, 70(23), 9618-9621 (DOI: 10.1021/jo051344g).

Recently, I have found some literature which does help to answer my question.
On p. 339, Clayden, Greeves, & Warren (2012) did mention that in general, $\alpha$- nitrogen, as well as sulfur, substituents can effectively provide resonance stabilisation to the carbocation, facilitating the $\ce {S_N1}$ process.

Similarly, on p. 433, Carey & Sundberg (2007) mentioned :

Adjacent atoms with one or more unshared pairs of electrons strongly stabilize
  a carbocation. Table 3.11 (p. 304) indicates the stabilization of the methyl cation by such substituents. Alkoxy and dialkylamino groups are important examples of this effect.

More interesting information
When we see such circumstances of resonance stabilisation, I believe many of us may question to what extent is this effect at play because this seemingly goes against conventional understanding as more electronegative atoms are bearing the positive charge. To this, Carey & Sundberg (2007) went on to say:

Although these structures have a positive charge on a more electronegative atom, they benefit from an additional bond that satisfies the octet requirement of the tricoordinate carbon. These “carbocations” are best represented by the doubly bonded resonance structures. One indication of the strong participation of adjacent oxygen substituents is the existence of a barrier to rotation about the C−O bonds in this type of carbocation.

If you think about it, it does make perfect sense since you are making an extra bond through this resonance effect. The exothermicity of bond formation more than compensates for the placement of the positive charge on the electronegative atom.
References
Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.


text:
Rutherford experiment
I am aware that in this experiment originally alpha particles were bombarded on gold foil. But recently I came across a question where this experiment was done with calcium and zinc instead of gold. After a bit of reading I found out that number of alpha particles deflected follows

But then I came across a question asking me to relate it with initial energy of the particle and number of  deflecting atoms ( please also explain what exactly this "number of deflecting atoms" refers to)
From my knowledge of electrostatics I know that larger the initial K.E the closer the particle can get to the atom. But I am unsure of what the exact proportion is.
For reference I am also  adding a picture of the actual question whose answer is option 1
If possible please give me an exact expression of sorts for the number of alpha particles scattered for a certain scattering angle
P.S. sorry about the poor formatting I was kind of busy. I will reformat the question as soon as possible.

The number of $\alpha$ particle scattered at a particular angle can be determined by finding out the distribution of the scattering angle ($N(\theta) \mathrm d\theta$) where $N(\theta)$ is the number of alpha paricles scattered at angle $\theta$.
First of all, if the alpha particle approach the nucleus with an impact parameter $b$, due to the repulsion the path of the alpha particle will be deflected and it will become an hyperbola, whose eccentricity can be simply calculated from a classical problem of central force motion as, $$\epsilon = \sqrt{1+ \frac{2EL^2}{\mu C^2}}$$ where $E$ is the total energy(which remains conserved and hence initial K.E), $L$ is the angular momentum ( = $\mu v_0b$), here $\mu$ is the reduced mass, and $C$ is $\frac{1}{4\pi \epsilon_0}z_{\alpha}Z^+e^2$.
Now from the knowledge of coordinate geometry, we can say that if the scattering angle is $\theta$, which is actually the supplementary angle between assymptotes of hyperbola
$$\sec\left(\frac{\pi}{2}- \frac{\theta}{2}\right)=\epsilon(\text{eccentricity})$$
and also substituting $L^2 = 2\mu Eb^2$ (this is as, $L=\mu v_0b$, and $E=\frac{\mu v_o^2}{2}$ (initial K.E), we will get
$$\sqrt{1+\left(\frac{2Eb}C\right)^2}=\frac{1}{\sin(\frac{\theta}{2})}$$
which gives
$$\frac{2Eb}C=\cot(\theta/2)$$
Now, the probablity distribution so that the alpha paticles approach at a particular
$$b=P(b)\mathrm db=\rho t2\pi b\mathrm db$$
where $\rho$  is the density of the scattering nuclei and $t$ is the thickness of the foil, but we know from just the previous result that
$$b = \frac{C\cot(\theta/2)}{2Eb}$$
now from this if you differentiate the terms to find $b\mathrm db$, and finally calculate the probablity distribution you will get
$$P(b)\mathrm db=\frac{\pi}8\rho t\frac{C^2}{E^2}\frac{\sin(\theta)\mathrm d\theta}{\sin^4(\theta/2)}=N(\theta)\mathrm d\theta$$
Now for same scattering angle, the $\theta$ parts will remain constant and for your new case
$$C'=\frac{Z_\ce{Ca}}{Z_\ce{Zn}}C=\frac23C$$
and $E'=2E$ , and also $\rho'=8\rho$ (as the density of scattering nuclei in your new case becomes $8$ times of that previous density). So, we will have by making $\theta$ parts constant,
$$\frac{N'E'^2}{\rho'C'^2}=\frac{NE^2}{\rho C^2}\to N'=\frac{NE^2\cdot8\rho\cdot\left(\frac23C\right)^2}{\left(2E\right)^2\rho C^2}=\frac89N$$
Thus if $N$ was previously $9000$, it will be now $\frac89\cdot9000=8000$.


text:
Is negative activation energy possible?
Is negative activation possible? And, in coupled reactions is there any difference?
Because I saw in the paper - Chemical Engineering Science 1996, 51 (11), 2995–2999 - the following conclusion:

Negative activation energies of reaction have been observed [...]. This unusual phenomenon can be predicted from classical theories, as a competition between two consequences of a temperature increase: greater thermal energies available to overcome the energy barrier represented by the intrinsic activation energy, and less adsorption to form reactive intermediates represented by the heat of absorption.

Does it simply mean that although the energy to overcome the activation energy barrier is there but the rate that products are formed is lower?

There are several cases like purely entropy controlled reactions (no barrier if we look at enthalpy) or non statistical affects but without context it's not possible to say what the cited paragraph is referring to.

See here for information about barrierless reactions.

When say negative activation energy, it is as if we are already talking about reactive intermediates.
https://pubs.acs.org/doi/abs/10.1021/j100476a025
Above is an example.

I have recently encountered this idea of "negative activation energy" as I was reading Chapter 11 of Elements of Physical Chemistry (5th Edition). On p. 252, the kinetics of the following reaction was studied:
$$\ce {2 NO (g) + O2 (g) -> 2 NO2 (g)}$$
A plausible and simple mechanism which was selected for this illustration, involved two steps: Firstly, the formation of the $\ce {N2O2}$ dimer from a collision of the two $\ce {NO}$ reactant molecules. It is then followed by collision of this dimer with an $\ce {O2}$ molecule. By applying the steady-state approximation (i.e. the rate of change of the dimer with respect to time approximately equals $\ce {0}$), the following overall rate constant is obtained:
$$k_\mathrm{r} = \frac{k_1 k_2}{k_1'}$$
$k_\mathrm{r}$ is the overall rate constant, $k_1$ is the forward rate constant for step $1$, $k_1'$ is the reverse rate constant for step $1$ and $k_2$ is the forward rate constant for step $2$.
When temperature increases, all of these individual rate constants for these elementary reactions increase. However, the extent of increase in their values differs between these individual rate constants. For this particular reaction, Atkins & Paula (2009) highlight that as temperature increases, the overall rate constant actually decreases. This is because the rate constant for the dissociation of the dimer increases much more as temperature increases, compared to the other rate constants in the numerator.
Mathematically, if we were to use the Arrhenius relationship for the rate constant for the overall reaction, we would say that the reaction has "negative activation energy" since that is the only way to have the rate constant decreasing in value as temperature increases (as $R, T > 0$), but it is pertinent to keep in mind that the rate of each elementary step does in fact increase as temperature increases. However, the extent to which they individually increase differs between each elementary step.
This is perhaps one way to see how a reaction can be said to have "negative activation energy" overall.
Reference
Atkins, P. W.; Paula, J. Elements of Physical Chemistry (5th ed.). W. H. Freeman & Company, 2009.

No, it's not possible to have a negative activation energy in a simple reaction such as an isomerisation because there is no possible way to draw to potential energy curves to give a negative activation energy. The best you can do is to be activationless, and is commonly observed in electron transfer reactions (see for example J. Am. Chem. Soc. 2002, 124 (19), 5518–5527).
Experimentally, however, negative activation energies can be observed when the reaction comprises several steps some of which are not identified at the time of doing the experiment. For example if each reaction follows an Arrhenius type rate constant, the first may be $k_a = A_a\exp(-E_a/RT)$ and the second $k_b=A_b\exp(-E_b/RT)$, but because there are, say, equilibria involved in the mechanism, what may be observed without realising it is that the rate is a ratio of the two individual rates:
$$\begin{align}
k_\mathrm{obs} &\propto \frac{k_a}{k_b} \\
&\propto \frac{\exp(-E_a/RT)}{\exp(-E_b/RT)} \\
&= \exp\left(\frac{-(E_a-E_b)}{RT}\right) \\
&= \exp\left(-\frac{E_\mathrm{obs}}{RT}\right)
\end{align}$$
so it is possible for $E_\mathrm{obs} = E_a - E_b$ to be positive or negative.

Why wouldn't you be able to draw potential energy curves giving a  transition complex that has lower energy than both starting material or products? See for example: 10.1002/chem.201706075
Of course, that's potential energy on the BO surface, if you factor in entropy and co you'll actually get a barrier.

@DSVA lower in energy is one thing,  a negative activation energy another.

Can you explain how this would work? Or is this just a definition thing because we have no barrier and no transition state and cannot assign an activation energy? If every point along the reaction coordinate is lower than both reactants or products like in the linked paper for example figure 2e, how do we characterize it?

@DSVA as you write in the paper you mention there is no barrier to the lowest point on the surface, so the reaction should be diffusion controlled and also will not follow Arrhenius type behaviour but determined by diffusion of the wavepacket along the reaction coordinate as common for barrierless reactions. The smallest barrier possible is zero.


text:
Living organisms are not at equilibrium with their surroundings
Why the living organisms can never be at equilibrium with their surroundings?

What is a living system? How would you define it?

Hi hafiz, the current question seems a bit unclear. Additional context,  like from which book you got this question, relevant text around it, etc.  would help. Also, please also show an attempt to understand the problem, like: what did you understand from it, and how did you think of solving it? Thanks, and welcome to Chem.SE!

I got this from the Lehninger principles of biochemistry while studying the bioenergetics

Being at equilibrium with your surroundings means that for everything (particles, chemicals, energy) that comes inside you from your surroundings an equivalent amount of it is returned to the environment, unaltered (same structure, same energy, same concentration, etc.). At the same time, and in the same manner, anything going out to the environment will be returned to you. You will not change your environment.
.
Equilibrium works both ways, so the envirionment will not change you, either.
.
If you are alive, this cannot keep up for long. You will need to use your stored energy (can't get it from your environment if you are at equilibrium with it) for your basic functions as a live being. Part of this energy will invariably become waste heat energy which you must get out of your body or you will heat up too much and die. But if you are at equilibrium with your environment, you cannot get rid of it. So...
.
But perhaps you are spore-like, and completely inactive (not even an inner timer)? Then one of two things may happen: you cannot receive anything from the environment, or you can. If you can't, you will never be able to get something to revive you. If you can, you will receive something, and then you are not in equilibrium with your environment.

It isn't just about transport into or out of the cell. Even expending "stored energy" leads to changes in concentrations of the concerned species - this, too, cannot happen in an equilibrium condition.
A cell at equilibrium is a dead cell.


text:
If both double and triple bonds are at the terminals,so prefreference would be given to?
I want to know about the preference level in order to givr IUPAC name

Possible duplicate of Why doesn't the double bond get preference for suffix in IUPAC naming of an hydrocarbon when both double bond and triple bond are present together?

The relevant rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(…)

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
(…)

If a compound contains both double and triple bonds, low locants are given first to all multiple bonds together as a set. However, if the multiple bonds are at the terminals of the chain, there still remains a choice. For example, $\ce{H2C=CH-CH2-C#CH}$ could be named “pent-1-en-4-yne” as well as “pent-4-en-1-yne” since both names correspond to the same locant set “1,4”. In this case, the lowest locant is given to the double bond. Therefore, the correct name is pent-1-en-4-yne.

Double bonds are given the preference.

Please don't spoonfeed answers to possibly homework questions.


text:
van der Waals equation for deviation of gases from ideal behavior
$(P + an^2/V^2)(V - nb) = nRT$
I understand that the term $nb$ is subtracted from the volume term as the available volume for the gas molecules to move around decreases in the case of a real gas due to the volume of the molecules themselves.
However, I don't understand why the term $an^2/V^2$ is added to the pressure term. Because of the attractive/repulsive forces, shouldn't the pressure on the walls of the container by the molecules also decrease?
(I thought about this and guessed that due to repulsive forces, the pressure must be increasing, but what about due to attractive forces? Does the sign of $a$ depend on the gas and hence affect the equation? I may be wrong here.)
Please clarify!

It's pressure under which gas is held.

The term $-an^2/V^2$ is the reduction of pressure due to attractive or cohesive forces between molecules. This is most easily seen by writing the equation as
$\displaystyle p_\text{obs}= \frac{nRT}{V-b}-\frac{an^2}{V^2}$ .
While the constants $a$ and $b$ are considered to be empirical they do have a physical basis. As you mention $b$ accounts for the finite volume of the molecules and $a$ for the attractive force between molecules.
In the centre of the gas these attractive forces are the same in every direction and so cancel out, however, as close to the boundary as it is possible to go the forces act predominately inwards towards the bulk of the gas, rather as happens in capilliary action or surface tension. If the force between adjacent molecules is resolved into components one will lie along the boundary and the other at right angles pointing inwards.  Thus we can imagine over a long enough period of time a steady force close to the boundary and acting inwards. Thus molecules on reaching the surface are deflected both by the surface and by this inwards force.
The size of the force will be proportional to the product of the number of molecules /unit area and the size of the inward force. Both of these will be be proportional to the density ($Nm/V$) or equivalently molar concentration. The reduction in pressure can thus be written as $-a(n/V)^2$ where $a$ is a positive constant that depends on the gas and is determined only by experiment.

Couldn't we also write the equation as $V = nRT/(P - an^2/V^2) - nb$, then?

yes anyway round , but why as this is not the solution to the equation as $p$ is  above, since volume its on both side of the equation.


text:
does it make any difference writing NO2 instead of O2N

can you write NO2 in all three?

You have to show connectivity, so you always have to write your molecular formulae based on what's connecting to what. In the case of the nitro group on the 6th carbon in the benzene ring, you would have to write it like that since there's a bond between carbon and nitrogen, not carbon and oxygen.

The Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), include recommendations for contracted labels in structure diagrams. In particular

GR-2.3.1 General guidelines
  Atom labels representing more than one non-hydrogen atom—also sometimes known as “contracted” labels—rely on the fact that many elements have consistent and well-understood bonding patterns. (…)

GR-2.3.3 Orientation of symbols within contracted labels
  Contracted atom labels attached to only one bond should be read outwards from that bond, usually from left to right if the bond is on the left of the label. If the bond is instead attached to the right of the label, the label will normally be read from right to left, but ambiguities can result. Accordingly, contracted labels with a bond on the right should be avoided except for simple cases, usually limited to relatively small labels containing four or fewer combined element symbols and abbreviations. (…)

GR-2.3.4 Interpretation of contracted labels
  In general, contracted labels are interpreted to fill as many valences as possible, as quickly as possible. (…)

Contracted labels for nitro groups cannot be interpreted with a simple application of valence rules, but also need some implicit charges to be added. It is equally possible to avoid entirely the issue of depiction and instead display the functional group using the $\ce{NO2}$ abbreviation. According to Table III in the Graphical Representation Standards, the left-to-right form of the label is $\ce{-NO2}$ and the corresponding right-to-left form is $\ce{O2N-{}}$.


text:
Why does the stability of hydrides decrease down the group in nitrogen family?
Why does the stability of hydrides decrease down the group in nitrogen family? Does this have to do with the fact that $\ce{NH3}$'s $\ce{N}$ is hybridised ($\ce{sp^3}$) while the rest of the group prefers unhybridised central atom for hydrogens leading to steric replusion in $\approx 90^{\circ}$ alignment of $\ce{H}$s?
Other possible explanation:
Perhaps,  the orbital overlap order is: $\ce{2sp^3-1s >  1s-3p> 1s-4p>1s-5p> 1s-6p}$ which leads to greater stability of $\ce{NH3}$ over the others.

The increasing size of the elements play a role here along with the overlapping factor.
Bismuth is humongous as compared to hydrogen so it's hydride is unstable. However nitrogen being small in size can form strong bonds with hydrogen. This also accounts for the fact that the reducing nature of the hydrides increase down the group.
As far as overlapping is concerned the bonds in the hydrides are - $\ce{2p-2p}$ up to $\ce{5p-2p}$.
Clearly the upper hydrides will be more stable as they have more effective overlapping between the orbitals.


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Is Kirchoffs law valid at all pressures?
$$\Large H_{T_\mathrm{f}, p}=H_{T_\mathrm{i},p}+\int_{T_\mathrm{f}}^{T_\mathrm{i}}c_p(T)dT$$
This can be derived by integrating: $$C_v =\left(\frac{\mathrm{d}U}{\mathrm{d}T}\right)_V$$
Applying this to both reactants and products in a reaction, and subtracting reactans from products, we can arrive to
$$\Delta H_{T_\mathrm{f},p}=\Delta H_{T_\mathrm{i},p}+\int_{T_{\mathrm{f}}}^{T_{\mathrm{i}}}\Delta c_p(T)\mathrm{d}T \tag{1}\label{a}$$
Where $$c_p(T)=\sum_i\nu_\mathrm{i}c_{p,\mathrm{i}}(T)$$
This is essentially Kirchoffs law how enthalpy varies with temperature. However, whenever I see the the  equation $\ref{a}$, it always deals with standard enthalpy changes, i.e: $$\Delta H_{T_\mathrm{f}}^{⊖}=\Delta H_{T_\mathrm{i}}^{⊖}+\int_{T_{\mathrm{f}}}^{T_{\mathrm{i}}}\Delta c_{p^{⊖}}(T)\mathrm{d}T $$
 Is this law not valid for all reactions? If so, what is the mistake in the above derivation?

Enthalpy and heat capacity are functions of both temperature and pressure.  But, at the low pressures of the standard state, it can be treated as exclusively a function of temperature.  Why do you feel that it would not be valid for all reactions?

@ChesterMiller Personally i am not convinced, but everywhere it is referenced. E.g Atkins Physical chemistry 10th edition, our lecture notes  etc. it is only stated for standard pressure. It states the top equation (how it varies for a single species) as true, but when it is applied to a reaction, it is only given for standard pressure.

It would say that you are making three separate mistakes:

The standard state pressure is any arbitrary pressure, not necessarily 1 bar, so there is no implication from the equation specifying standard state that it is only true at one particular pressure.
The important specification of being in the standard state is not the value of arbitrary pressure chosen, but that the substances are not mixed.  They are pure or in solvent as the single solute.  Also, for gases and solutions the states are fictitious.  In your derivation, you are not accounting for changes due to mixing and any changes due to being in real, rather than fictitious, states.
I don't see why you say to integrate Cv rather than Cp.

Very helpful, thank you. The non mixing thig is something that was mentioned in the very lecture we learned about this topic. So, just to be clear: The standard state(as indicated by the symbol in the quoted equation) implies that no mixing is occuring?

Also, the Cv instead of Cp was simply a typo, i copied the maths language from a related thread but forgot to edit

@Adroit yes, it means each reactant or product existing separately.

Please provide a reference for item 1. above.  In every source I have seen, the standard state pressure for gases is taken as 1 bar.

@ChesterMiller The statement is already hyperlinked to the IUPAC definition which says "a well defined but arbitrarily chosen standard pressure".   But if that is insufficient see "The standard state with respect to pressure or concentration for each state of aggregation is arbitrarily chosen as some value that can be conveniently measured" https://books.google.com/books?id=aRAoAQAAMAAJ&q=%22The+standard+state+with+respect+to+pressure%22+%22arbitrarily%22&dq=%22The+standard+state+with+respect+to+pressure%22+%22arbitrarily%22&hl=en&sa=X&ved=0ahUKEwjxjJzO28PaAhVtUN8KHZP0AXcQ6AEIJzAA

My experience has been that the state of 1 bar is virtually always used, because it is typically so close to ideal gas behavior, and thus very convenient for then determining both the temperature dependence and pressure dependence of the heat of reaction.  The important point is that the heat of reaction is pressure-dependent because the enthalpies (and thus heat capacities) of the pure reactants and products are functions of pressure.

@ChesterMiller, I agree, but I'm interpreting the subscripts of the OP as indicating that initial and final state are at the same pressure.

Yes.  Me too.  But the heat of reaction at a higher pressure (such as 10 bar) is different from the heat of reaction at a lower pressure ( such as 1 bar).

@ChesterMiller yes, I agree

If the question is, "Is the heat of reaction a function of temperature and pressure?" the answer is Yes.  If the question is, "Is the standard heat of reaction a function of temperature and pressure?," the answer is No; the standard heat of reaction applies only for reactants and products at 1 bar.  If the question is, "Are the heat capacities of the pure reactants and products functions of temperature and pressure?", the answer is Yes.  So are you asking, "How do I determine the heat of reaction at a specified temperature and an arbitrary pressure higher than 1 bar (without knowing details on how the heat capacities vary with pressure)?"


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How does the bite angle of a ligand affect the efficiency of its catalyst?
I know that optimising the bite angle of a diphosphine ligand to its metal increases catalytic efficiency, and for cross-coupling reactions larger bite angles are favoured, but how does this have an impact on the effectiveness of a catalyst?

The bite angle of a bidentate ligand often has a pronounced effect on reactivity, but it can be difficult to predict a priori whether a given reaction will be enhanced by increasing or decreasing bite angle - you can see this if you ever read a paper where new ligands / catalysts have been developed, which are often accompanied with large tables of screening in order to find an optimised set of reactions.
In general however, bite angle does correlate with the rates of oxidative addition and reductive elimination.
Reductive elimination
The rate of reductive elimination is often correlated to the bite angle of a bidentate ligand. Upon reductive elimination, the two coordinate Pd0 complex would preferentially adopt a linear geometry however with the bidentate ligand coordinating (and the chelate effect preventing it from 'falling off') this is not possible.
Larger bite angles therefore favour reductive elimination as the L-Pd-L (P-Pd-P) bond angle is closer to the desired 180o. The classic example of this was an experiment carried out by Brown in which the rate at which PhMe was eliminated from a variety of complexes was measured.
Pd(dppf)PhMe with a bite angle of 99.1o underwent rapid reductive elimination, even at low temperatures, while Pd(dppe)PhMe with a bite angle of 90.6o was stable at low temperatures, with elimination needing elevated temperatures.
Oxidative addition
The effect of bite angle on oxidative addition is less straightforward, as it depends largely on the mechanism by which the oxidative addition takes place. In general however, complexes with larger bite angles undergo faster rates of oxidation despite the fact that a smaller bite angle increases the rate of the C-X bond cleavage.


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Can an enol acetate be formed through Fischer esterification?
Since acetone is converted into its enol form under acidic conditions, could it participate in Fischer esterification to form an enol acetate? For example, would refluxing acetone, p-TSA, and acetic acid form the unsaturated ester?

My suspicion is no, although apparently you can transesterify by refluxing acetone with TsOH + allyl acetate (JOC 2012, 77, 10347)