text:
How much KMnO4 is required for the complete oxidation of ferric/ferrous oxalate in acidic medium?

Find out the amount of $\ce{KMnO4}$ required for the complete oxidation of $\pu{5 mol}$ of an equimolar mixture of ferric oxalate and ferrous oxalate.

I understand that $\ce{MnO4-}$ will be oxidised to $\ce{Mn^{2+}}$ and ferrous ions to ferric ions. I wrote the following reaction:
$$\ce{FeC2O4 + Fe2(C2O4)3 + MnO4^- + 8H+ + e- -> 2Fe^3+ + 4CO2 + Mn^2+ + 4H2O}$$
Then, I applied equivalence concept but could not get a correct answer. Any ideas?

$\ce {Fe^{ +2}} $converts to $\ce {Fe^{ +3}} $ and and carbon from +3 to +4 , calculate required equivalents for them and then equate then with equivalents of $\ce {KMnO4} $

Can you show more effort on how you applied the equivalence concept ?

I'd like the question to be tackled in two parts. In each part, we're going to deal with one of ferric oxalate and ferrous oxalate.
Since the mixture ($\pu{5 mol}$) given to us is equimolar, we have $\pu{2.5 mol}$ of ferric oxalate and $\pu{2.5 mol}$ of ferrous oxalate. I second your understanding that $\ce{MnO4-}$ will be oxidised to $\ce{Mn^{2+}}$, because the medium is acidic, and ferrous ions to ferric ions.
Ferric oxalate doesn't contain any $\ce{Fe^3+}$ ions, so our main focus here is the conversion of oxalate ($\ce{C2O4^2-}$) into carbon dioxide. The oxidation state of carbon in oxalate is $+3$, while it's $+4$ in carbon dioxide. So there's a net change of $+1$ per carbon atom. Oxalate contains two carbon atoms, so that gives us a valency factor of $2$ per mole of oxalate ion. However, for $1$ mole of $\ce{Fe2(C2O4)3}$, you'll have $3$ moles of oxalate ions that are undergoing the aforementioned change. This gives us an overall valency factor for $\ce{Fe2(C2O4)3}$ as  $2 \times 3 = 6$. The valency factor for the $\ce{MnO4- -> Mn^2+}$ change is $5$. Now, we're going to equate the equivalents as per the Law of Equivalence.
$$ \begin{align*}
n_{\ce{KMnO_4}} \times V_{f_1} &= n_{\ce{Fe2(C2O4)3}} \times V_{f_1'}\\
x \times 5 &= 2.5 \times 6 \\
x &= 3 \end{align*}
$$
For the part of the mixture that contains $\ce{FeC2O4}$, we need to change our valency factor. This is because here we're taking into account the conversion of ferrous ions into ferric ions; this conversion has a valency factor of $1$. So the overall valency factor, in this case, would be $2+1=3$. Now, since this has been done, we proceed to apply the same equation as in the previous case.
$$ \begin{align*}
n_{\ce{KMnO_4}} \times V_{f_2} &= n_{\ce{FeC2O4}} \times V_{f_2'}\\
x \times 5 &= 2.5 \times 3 \\
x &= 1.5 \end{align*}
$$
Thus, overall, it can be concluded that you'll need $\pu{3+1.5=4.5 mol}$ of $\ce{KMnO4}$ for the complete oxidation. The beauty of this? I didn't use any reaction-equilibrium equations.
This is the reason we use equivalent concept, to save ourselves from the hassle of writing (and perhaps balancing) chemical reactions.


text:
Crystallisation and crystal surface
It is my understanding that if you put table salt (as an example of polar compound) in water, the Na+ and Cl- ions will freely move through the solution. They will make the solution conductive and will independently travel through it to carry electric current.
Now the thing is, if you dry the solution, how come every Na+ finds it's Cl- to recombine into NaCl? How come there are no lonely Na+ and Cl- pairs who couldn't find their mate in time, before the solution dried completely? Can someone give me a better understanding of how do polar solutions work?

$\ce{NaCl}$ doesn't exist as molecules, it exists as unit cells with particular crystal arrangements.

@PrittBalagopal Thanks for the input, but the question  then still stands: How come there are not leftover ions that didn't fit in the crystal grid?

Crystal surface is different than interior. There is no unit cell there.

@TomášZato Have you consider the fact that both ions have a charge and they are strongly attracting each other? If the solution dries up it mea the concentration goes up, so it is even easier to find each other when they are packed in a small place.

The crystal structure of NaCl forms at the growing surface of the crystals, where there is plenty of time for the ions to sort themselves into a regular lattice. Further, each ionic species is very evenly distributed also in the solution, because of the repellent electrostatic forces between the members of each. Whenever there is a local excess of positive or negative charges, it will level itself out very quickly.
If you have something larger than sodium and chloride ions and try to crystallise very fast, you will indeed get an increasing amount of errors in the structure. Freeze-drying also tends to leave behind rather unordered structures. In both cases, this is, of course, more likely with noncharged or even nonpolar substances.

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text:
Explaining boiling using internal energy
Internal energy is summation of potential and kinetic energies of a substance
Using this interpretation of internal energy, how can I explain boiling?
Kinetic remains constant.
Potential should decrease because of an increase in average spacing.
But internal energy should increase.
Can somebody explain what flaw I am making?

You forgot about entropy...

Well I am not taught how to explain changes in internal with respect to entropy. I am just taught that if I can denote changes in kinetic and potential, I can work out changes in internal. I know this can also be explained by thermodynamic law, but I am not interested in that.

Well, this process is dominated by entropy, so you not going to understand it without considering entropy. Enthalpy is only half the picture.

Can you please then write an answer which includes entropy(I have studied the concepts of entropy so I will be able to understand the answer)? And can you please also relate why internal energy of an ICE cube increases even though volume is decreasing?(maybe this confusion is arising due to the fact that I am taught potential energy in terms of average separation instead of strength of electrostatic attraction)

You can do this in terms of enthalpy, too. How many hydrogen bonds do you need to break to vaporize water?

Okay I get your idea. So my basic assumption that increase in potential energy is always due to increase in separation is wrong?

Let us continue this discussion in chat.

Have you read this? https://en.m.wikipedia.org/wiki/Internal_energy

You are wrong when you say the potential energy of particles decrease when they move  apart.
The forces between the particles are attractive not repulsive.
Recall the definition of potential energy. It is the work done by external force on a charged particle under influence of electric field.
Therefore, the potential energy for particles, which attract each other, increases when they move apart.
So when to particles get close their potential energy decreases, hence their stability increases. This also happens to be one of the reasons why they come closer.

Can you please also relate why internal energy of an ICE cube increases even though volume is decreasing?(maybe this confusion is arising due to the fact that I am taught potential energy in terms of average separation instead of strength of electrostatic attraction)

Ice is less dense than water. Ice has quite some space between its molecules and when it melts the space between the molecules decrease. Perhaps this could be the answer. Water molecules attract each other by H-bonding.

Yes my question is exactly this. You stated increase in potential is explained by an increase in average separation. Here however, the average spacing is decreasing yet potential is increasing. Why is this happening?

There are other factors at work. Potential energy cannot be held accountable for every observation we make. This is more related to thermodynamics and out of the scope of the main question. You should consider asking another question.

So you're saying that Internal energy is not equal to only potential and kinetic energy?

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node15.html

I am aware of first law of thermodynamics. And I am able to explain the phenomenon using Heat and Work done. However, I cannot explain it using change in potential and kinetic.

I shared this link to help you clarify your concepts of internal energy. And yes, the internal energy consists the KE and PE of the particles of the system and not of the system as a whole.

I strongly suggest you to read everything that is on this site.

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text:
What organizational methods pre-date the Periodic Table?
Dmitri Mendeleev noticed patterns in elements which allowed him to design the periodic table which ultimately led to the modern periodic table.
How were elements organized before this? Was there any method that served as a crude standard?

It just so happens that I'm reading a book that provides an awesome overview of the efforts to classify the elements. An answer summarizing this will not be quick to assemble, but I'll put it together when I can. (Unless someone gets Ihde's book and beats me to it, of course.)

Aaron J. Ihde's The Development of Modern Chemistry (DMC) has an entire chapter entitled "Classification of the Elements," which includes a very nice overview of the attempts before Mendeleev to bring order to the elements.  All quotations below are from Chapter 9 (pp236-243, specifically) of the 1984 Dover edition of DMC, itself "an unabridged, slightly corrected republication of the third printing (1970) of the work originally published by Harper & Row, Publishers, Inc., in 1964" (front matter). The section and sub-section headers are generally those of DMC, with exceptions marked with an asterisk.
It is important to bear in mind when considering some of these systems that the best values of the atomic weights of many elements at the time were incorrect, either (a) due to simple experimental error, or (b) due to systematic misinterpretation of the available data, due to the limited state of knowledge at the time.  For example, the best-regarded atomic weights of many elements were one-half the accepted values of today because the values were derived from measurements on compounds whose formulas were erroneous $($e.g., the atomic weight of oxygen was calculated based on experiments with water, which was thought to have the formula $\ce{HO})$.
For interest, a short article on Aaron J. Ihde can be found on the German language Wikipedia.

Early Attempts at Classification
Döbereiner's Triads
As already noted by Buttonwood in his answer, Johann Wolfgang Döbereiner appears to have been the first to attempt to systematize the elements, after:

he observed that the atomic weight of strontium appeard to be $50$, the mean of the then accepted values for calcium $(27.5)$ and barium $(72.5)$.
...
His triads included elements with similar properties, the atomic weight of the central member being the mean of the other two atomic weights.

The Döbereiner triads (some which he was unable to complete at the time) included in DMC (using modern atomic symbols) are:
$$
\require{begingroup}\begingroup
\begin{array}{cccccccc}
\hline
\ce{Li} & \ce{Ca} & \ce{Cl} & \ce{S} & \ce{Mn} & \ce{B} & \ce{Be} & \ce{Y} \\
\ce{Na} & \ce{Sr} & \ce{Br} & \ce{Se} & \ce{Cr} & \ce{?} & \ce{?} & \ce{?} \\
\ce{K} & \ce{Ba} & \ce{I} & \ce{Te} & \ce{Fe} & \ce{Si} & \ce{Al} & \ce{Cs} \\
\hline
\end{array}
$$
DMC notes further:

Magnesium he considered as an isolated element, not part of a triad. Fluorine, as yet undiscovered but its existence clearly known, he did not include in the halogen triad. Although the atomic weight of nitrogen fell exactly between that of carbon and oxygen, Döbereiner regarded the three elements as isolated non-metals rather than as members of a triad. He also treated hydrogen as an isolated element.

Other Numerical Classification Systems

Similarity defined by an atomic weight difference of $8$:

P. Kremers [no further identification given] observed that the atomic weights of certain [elements he regarded as] non-metals in which he saw similarity differed by $8$; i.e., $\ce{O} = 8$, $\ce{S} = 16$, $\ce{Ti} = 24.12$, $\ce{P} = 32$, $\ce{Se} = 39.62$.

Atomic weights as odd vs. even multiples of four:

At the same time, [Kremers observed that] the atomic weights of certain metals fell between those of successive non-metals; i.e., $\ce{Mg} = 12$, $\ce{Ca} = 20$, $\ce{Fe} = 28$. When divided by $4$, the atomic weights of the non-metals were an even number, those of the metals an odd number.

Groupings of elements suggested by John Hall Gladstone based on (a) similar atomic weights, and atomic weights in (b) geometric and (c) arithmetic progression:

(a) $\ce{Cr/Mn/Fe/Co/Ni}$ all near $28$; $\ce{Pd/Rh/Ru}$ all near $52$
(b) $\ce{Pd}$ group $(52)$, $\ce{Pt}$ group $(99)$, and $\ce{Au}$ $(197)$; separated by multiplicative factors of approximately two
(c) $\ce{Li}$ $(7)$, $\ce{Na}$ $(23)$, and $\ce{K}$ $(39)$; separated by an additive factor of $16$

The first type [Gladstone] compared with allotropy; the second, with polymerism in organic chemistry; the third, with homologous series.

Classifications based on arbitrary mathematical progressions:

Josiah Parsons Cooke (1827-1894) of Harvard developed a classification based upon six series of elements, each series derived from the atomic weight by a mathematical formula. For example, the sixth series, composed of hydrogen and the alkali metals, was based on the formula $1+(n\times 3)$. Thus: $$\begin{array}{ccc}
\ce{H}~(1) = 1 + (0 \times 3) & ~~~ & \ce{Na}~(23) = 1 + (7 \times 3) \\
\ce{Li}~(7) = 1 + (2 \times 3) & ~~~ & ~\ce{K}~(39) = 1 + (13 \times 3) \end{array}$$

Jean Baptiste André Dumas (1800-1884) posited several arithmetic progressions in series of elements, roughly analogous to homologous series in organic compounds, in the vein of J.H. Gladstone's third type of grouping above (series names are my descriptors):

$$\def\dumhc#1#2{\ce{#1} &= 1 + (#2 \times 14)}
\textbf{Hydrocarbon Radicals} \\
\begin{align}
\dumhc{H}{0} \\
\dumhc{CH3}{1} \\
\dumhc{C2H5}{2} \\
\dumhc{C3H7}{3} \\
\dumhc{C4H9}{4} \\
\textit{general}\dumhc{}{n}
\end{align}
$$
  $$
\textbf{Halogens} \\
\begin{align}
\ce{F} &= 19 \phantom{+2~~\times 16.5)+(2~~\times 28) + 19} = \,19 \\
\ce{Cl} &= 19 + \phantom{(2~~\times} 16.5 \phantom{)+(2~~\times 28)+19} = \,35.5 \\
\ce{Br} &= 19 + (2 \times 16.5) + \phantom{(2~~\times} 28\phantom{)~+19} = \, 80 \\
\ce{I} &= 19 + (2\times 16.5) + (2\times 28) + 19 = 127
\end{align}
$$
  $$
\textbf{Chalcogenides} \\
\begin{align}
\ce{O} &= 8 \phantom{+(4~~\times 8)} = \,8 \\
\ce{S} &= 8 + \phantom{(4~~\times} 8\phantom{)} = 16 \\
\ce{Se} &= 8 + (4\times 8) = 40 \\
\ce{Te} &= 8 + (7\times 8) = 64
\end{align}
$$
  $$
\textbf{Pnictogens} \\
\begin{align}
\ce{N} &= 14 \phantom{+17+(2~~\times 44)} = ~14 \\
\ce{P} &= 14 + 17 \phantom{+(2~~\times 44)} = ~31 \\
\ce{As} &= 14+17+ \phantom{(2~~\times} 44 \phantom{)} = ~75 \\
\ce{Sb} &= 14+17+(2\times 44) = 119 \\
\ce{Bi} &= 14+17+(4\times 44) = 207
\end{align}
$$
  $$
\textbf{Alkaline Earths}\textit{ (mostly)} \\
\begin{align}
\ce{Mg} &= 12 \phantom{+(10~~\times 8)} = \,12 \\
\ce{Ca} &= 12 + \phantom{(10~~\times} 8\phantom{)} = \,20 \\
\ce{Sr} &= 12 + \phantom{1}(4\times 8) = \, 44 \\
\ce{Ba} &= 12 + \phantom{1}(7\times 8) = \, 68 \\
\ce{Pb} &= 24 + (10\times 8) = 104
\end{align}
$$

Interestingly, Dumas interpreted these "trends" as being potential evidence (obviously not borne out by subsequent investigation) for the feasibility of transmutation among specific groups of related elements.

Immediate Precursors of the Periodic Table
De Chancourtois' Telluric Helix
As noted by Buttonwood, Alexandre-Émile Béguyer de Chancourtois (1820-1886):

... conceived the idea of plotting the elements according to atomic weights on the surface of a cylinder. The circumference of the cylinder was divided into sixteen sections since the atomic weight of oxygen was $16$. The elements were plotted on a line or helix that descended at an angle of $45^\circ$ with the top of a cylinder. There was a striking resemblance in elements that were on the same vertical line.
... [H]e observed that atomic weights followed the formula $n+16n'$, the value of $n$ frequently being $7$ or $16$ [noting that none of the noble gases had yet been discovered]. ... Gaps in the helix were considered as indicating not unknown elements but different varieties of known elements. He believed, for example, that there was a form of carbon whose atomic weight was $44$.

In particular, the above Wikipedia article on de Chancourtois emphasizes that he "was the first scientist to see the periodicity of elements when they were arranged in order of their atomic weights."  A public domain image of his "Telluric Helix" is also available at the Wikipedia page, and reproduced below (click to enlarge):

Newlands' Law of Octaves
Again as noted by Buttonwood, John Alexander Reina Newlands (1837-1898):

... was the other major precursor of Mendeleev and Lothar Meyer. ... Newlands' first publication dealing with the classification of the elements appeared in 1863. It was essentially a reworking of Dumas' ideas, for Newlands was looking for numerical relationships between elements. A year later he adopted the atomic weights recommended by Cannizzaro and published a table of 37 elements subdivided into ten families. A crude repetitive character was evident, and blank spaces were left for undiscovered elements. In a table in a subsequent paper he assigned numbers to the elements [precursors to the modern atomic numbers], which were listed in order of increasing atomic weights. He used numbers for the elements in all his work thereafter. In August, 1865, he published an eight-column table listing 62 elements in order of increasing atomic weights; they were subdivided into seven horizontal families. It was in this table that he saw an analogy to the octave in music. The eighth element resembled the first element, the fifteenth resembled the first and the eighth; in other words, an interval of seven elements separated similar elements. Soon afterwards he suggested that the numerical relationships observed by earlier chemists were due to this law of octaves.
...
Newlands must be credited for taking a pioneering step toward the discovery of the periodic law, for he detected the repetition of properties when elements are arranged according to increasing atomic weights. He noted that this relation was evident only if Cannizzaro's atomic weights were used. he used blank spaces for unknown elements but failed to do this consistently, holding that perhaps the interval between repetitions was eight or nine and that this could be dealt with as new elements were discovered.

As with de Chancourtois, the Wikipedia article on Newlands includes a public domain image of his periodic chart, which is reproduced here for reference:

$\endgroup$

One attempt to order chemical elements was Döbereiner's system of triades, published in Annalen der Physik und Chemie, back in 1829 (doi 10.1002/andp.18290910217 with Wiley); or (open access with Gallica).
Although an actual view into his paper permits the speculation he was not using the atom masses we know today; rather than using specific weight and molecular weight as known / defined at his time of oxides of Ca, Ba, Sr, for example:

(The dots on top of Ca, Ba, Sr seem to adhere a nomenclature, as on top of S, Se, Te three of them; Cl, Br, I are topped by five.)

This is really interesting! Are there any other methods that you know of?

Beside the triades (groups of three), there was the attempt to order elements in groups of eight (de Chancourtois, https://en.wikipedia.org/wiki/Alexandre-%C3%89mile_B%C3%A9guyer_de_Chancourtois) and Newlands (https://en.wikipedia.org/wiki/John_Newlands_(chemist)).  To know (and retain) about it was a bit by chance, as once a professor engaged in a dedicated subsection of a chemical society offered a weekly lecture "history of Chemistry" and a pinch of own curiosity "How did the ancestors before modern times?" similar to the one by ACS (http://www.scs.illinois.edu/~mainzv/HIST/Logo/logo.php).

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text:
How pure is this chloroform?
hopefully this question is in the right place.
I'm a biologist not a chemist (so sorry if this is a stupid question!)- I'm doing some RNA extractions from brain tissue which requires the use of Chloroform. In the manual, it specifies:
Chloroform must not contain isoamyl alcohol or other additives and must be pure.
This is what it says on the label:

Does this count as 'pure'? Obviously pure is quite a subjective term, but it doesn't to my eye contain any isoamyl alcohol, so I suspect it should be OK - but I thought it best to check.

Note that this grade of chloroform is stabilised with amylene. Can you get HPLC grade? That would be purer.

Yes, I imagine that I would be able to get it - I'm just not sure whether I would need it.

Amylene is a pentene. I would guess that it is not going to react with RNA or other biological materials

To what do phosgene and amylene react? Alcohols give the respective carboxylate diesters.

Depends on amylene content which may only be a few hundred ppm. I'd use it


text:
Why is the boat form of paraldehyde not theoretically possible?
Wikipedia says the following about paraldehyde:

Theoretically, four stereoisomeric structures are possible. The
  structures (1) and (2) are known as cis- and trans-paraldehyde. The
  structures (3) (a conformer of (2)) and (4) (a conformer of (1)) don't
  exist for steric reasons.

If this is correct, it seems that the boat conformer is not theoretically possible. I would expect the boat conformer to be possible but be much less common than the chair conformer (as is the case with glucose and cyclohexane for example), but that does not seem to be the case.
Wikipedia says that conformers 3 and 4 don't exist for steric reasons, however they are still theoretically possible. To me this sounds like the nonexixtence of boat conformation is not just because of steric reasons, but there is a more fundamental reason for that.
So, my questions are:

Is it true that the boat conformer is theoretically impossible?
If yes, why is it impossible and what makes paraldehyde different from glucose and cyclohexane in this regard?

It's possible, they just ignored it because it's so minor. Every molecule has an infinite number of conformers, so if you have limited space you can't really talk about all of them. Most books on cyclohexane hardly talk about boat conformation either, except to note in passing that it exists.

The boat conformer is a transition state in any case. See for example the interconversion for cyclohexane on Wikipedia

A $\ce{CH_3}$ group is quite big. Steric hindrance is the reason.
Its not theoretically impossible, but it is energetically a lot less favourable than (1) and (2).
You wouldn't find it in nature so to speak.
to 2: A $\ce{OH}$ group is sterically a lot smaller than a $\ce{CH_3}$ group.

Thank you for your reply. I just edited my questions for further clarification. I steric hinderance is the main reason, why does Wikipedia list conformers 3 and 4 as theoretically possible but nonexixtent due to steric reasons, but does not say that the boat form as theoretically possible?

It is theoretically possible, it just is energetically so high, it couldn't exist without external help

Is it energetically much higher than structures (3) and (4)? To me it seems that if sturcutre (2) were in boat form, there were no $\ce{CH_3}$ groups in close contact with each other. In (3) the $\ce{CH_3}$ groups seem to be much closer.

Free electron pairs (on the oxiygens) have a quite substantial steric part. A boat conformer is always higher than its "neighboring" conformer and will change to it, it is, as Martin said, a transition state.


text:
Can -NH2 groups get nitrated?
In the chapter of amines, and I see reaction occuring between amines and nitrous acid. However I am thinking why I don't see reactions of amines also react $\ce{HNO3}$ in presence of stronger acid like $\ce{H2SO4}$, that can protonate $\ce{HNO3}$, causes $\ce{H2O}$ to leave and then finally have amines to react with $\ce{NO2}$, making nitroamines?

https://chemistry.stackexchange.com/questions/7/why-does-nitration-of-n-n-dimethylaniline-occur-at-the-meta-position

Yeah, but in the link, I don't see amine reacting together with HNO3 per say like the way amine react with nitrous acid.

You mean if amine group can be nitrated?

Yes, like how you create N-nitrosamine.

My first comment there

ahh... okay. So why is this nitration of amine group not taught in orgo books? Is it because of hazardous issues?

Let us continue this discussion in chat.

Yes, you can form a nitroamine (nitramine) from a secondary amine and nitric/sulfuric acids. Some explosives have a nitramine structure (e.g. tetryl, RDX, HMX, HNIW).


text:
Democritus vs John Dalton atomic theory
What is the difference between atomic theory given by Democritus and John Dalton? Because I have read in many books and figured out that both theories explain the same i.e atom is an indivisible particle and all elements differ in shape and weight of their atoms. So basically what is the difference in both the theories?

Demokrit was a philosopher, while Dalton was arguing on the basis of (somewhat limited, but still) solid, quantitative evidence. I'm not knocking Demokrit (!), but his was not a scientific statement, let alone a theory. You cannot compare the two, or at least you should not try to compare them from the view of natural science.
Both came to a similar conclusion, but they were reasoning on a completely different basis. Of course, Dalton had the advantage that the idea was already there and known to him.
In Dalton's time, the modern concept of elements was already known and widely accepted. Iron, oxygen, etc., not fire, water and stone. You could say that he knew what he was talking about, regarding atoms. And Dalton and his contemporaries had the numbers they measured.
Demokrit, on the other hand, had no idea about quantitative elemental analysis. Which ultimately was the downfall of his idea: He had no numbers that others could have proved and extended, or disproved. Only when the chemists at Dalton's time were able to make quantitative measurements the atomic theory could become useful. Demokrit had probably theorised that atoms of the same kind had the same weight (we don't know exactly, because we have very little original records from him), but Dalton now had solid evidence, that only made sense if that was indeed the case.

This may be, but one can still compare the two theories. If you think the question is too non-scientific for Chem.SE, then vote to close with a recommendation it be posted to, e.g., History of Science & Math SE.

I don't find this question off-topic, it's just that the answer is that they do not compare very well, if you look from a natural-science point of view.

Could they be usefully contrasted?

The major differences are simpler than you might think. Democritus hypothesised that everything was made of indivisible particles (essentially the atomic theory of matter) but Dalton knew the number of varieties those particles came in. Neither understood enough about why they differed (that wasn't completely clear until the neutron was discovered in the 1930s.)

you don't show their difference very well

@anna Can you be a bit more constructive? What is missing?

Well, you don't get into more detailed stuff like early atomists somewhat weird ideas about shapes of atoms etc.

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text:
How will a NPK 15-0-15 fertilizer affect this soil?
If the soil we have has pH 7,2 and we want to add this fertilizer which has 15-0-15 (NPK). What should the fertilizer contain in order for the pH to get reduced?
I believe that $\ce{K+}$ ions should derive from $\ce{KCl}$ because it won't affect the pH.If it comes from $\ce{K2SO4}$, the soil will be more alkaline.
I'm confused about the nitrogen. Should it come from urea or NH4NO3? I believe that if the ions come from urea it will be more alkaline. Or should it come from $\ce{(NH4)2SO4}$?

This Wikipedia article has tables of different fertilizers and their composition with respect to their NPK content. For example, the fertilizer sylvite is a naturally occurring mineral of KCl that is 0-0-63. Most of the fertilizers are of course synthetic mixtures. Also note that the value for P is reported as $\ce{P2O5}$ and K as $\ce{K2O}$. K and P are not actually present as these species, this is just a convention for reporting their values. This doesn't answer your pH question directly, but I'd still check out the above link.

@airhuff Thank you! My question is on a theoritical level. I just ran into this question and I was curious and confused. The only reason  I mentioned those compounds was because they were as choices.

OK, I didn't quite get the context ;) What you've said about K and urea is correct, and the chemistry for the ammonium compounds is a bit more complex. However, if you read the Wikipedia pages for the ammonium compounds, I think it will be clear which one will be likely to acidify the soil.


text:
How to calculate the height of an hcp lattice?
A hexagonal closed packing (hcp) unit cell has an ABAB type of packing. For calculating the packing fraction we require the volume of the unit cell.
Volume of hcp lattice = (Base area) $\cdot$ (Height of unit cell)
Each hexagon has a side = $2\cdot r$
Base area = $6$ (Area of small equilateral triangles making up the hexagon)
$$=6 \cdot \frac{\sqrt{3}}{4}\times(2r)^2$$
$$=6 \cdot \sqrt{3} \cdot r^2$$
Hence, volume $= 6 \cdot \sqrt{3} \cdot r^2 $ (Height of unit cell)
This is the point where I am stuck. How do I find out the height of the unit cell?
I searched in textbooks and found out that height $= 4r \cdot \sqrt{\frac{2}{3}}$. Can you please explain why is this so?

We shall try it using the similarities between hcp and ccp. Here, we know that $hcp$ and $ccp$ have similar lattice except the fact that $hcp$ is ABAB type whereas $ccp $ is ABCABC type. Hence we also know that their packing fraction $(\phi) $ is same and $$\phi = \frac{\pi} {3\sqrt{2}}$$
Now as you mentioned Volume of hcp lattice $= 6\sqrt{3} r^2h$. There are 6 atoms in total in hcp. Hence $$\frac{6\left(\frac{4}{3}\right) \pi r^3}{6\sqrt{3} r^2 h} = \frac{\pi} {3\sqrt{2}}$$ Simplifying this we obtain the height of hcp lattice $$h=4r\left(\sqrt{\frac{2}{3}}\right)$$

We get that their packing fraction is equal after evaluating volume from height,etc. Your answer is working backwards.

To calculate the height of a unit cell, consider a tetrahedral void in an hexagonal closed packing arrangement. It can be imagined as a 3 solid spheres touching each other and at the center-point, you have another sphere stacked over them. An interactive version can be viewed on this site. The situation looks like this:

If you join the centers of these four spheres, you'll get a tetrahedron. That's basically a pyramid with a triangular base. I'm assuming each edge of our tetrahedron to be equal to $a$.
Now, you have a pyramid ($ABCD$), with an equilateral base ($\Delta BCD$), I would like you to drop a perpendicular from the highest point ($A$) to the center ($G$) triangular base. If you're following me correctly, you'll have a figure like this:

All we have to do now is to calculate the length $AG$. For this, simply use the Pythagorean theorem in $\Delta AGD$.
$$ \begin{align*} AD^2 &= AG^2 + GD^2 \tag{1} \end{align*} $$
Although we know that $AD=a$, the side $GD$ remains unknown. But that's easy to calculate. The point $G$ is the centroid of $\Delta BCD$. Thus, the length $GD$ equals $a/\sqrt{3}$. Pluggging in the values in our first equation, we get $AG=a \sqrt{\frac{2}{3}}$. But note, this is half the height of our unit cell. Thus, the required height is $2a \sqrt{\frac{2}{3}}$.

In the hexagonal closest-packed structure, $a = b = 2r$ and $c = 4 \sqrt{\frac23 }r$, where $r$ is the atomic radius of the atom. The sides of the unit cell are perpendicular to the base, thus $\alpha = \beta = 90^\circ$.
For a closest-packed structure, the atoms at the corners of base of the unit cell are in contact, thus $a = b = 2 r$. The height ($c$) of the unit cell, which is more challenging to calculate, is $c = 2a \sqrt{\frac23} r = 4 \sqrt{\frac23} r$.

Let the edge of hexagonal base equal $a$
And the height of hexagon equal $h$
And radius of sphere equal $r$
The centre sphere of the first layer lies exactly over the void of 2nd layer B.
The centre sphere and the spheres of 2nd layer B are in touch
So, In $\Delta PQR$ (an equilateral triangle):
$\overline{PR} = 2r$, Draw $QS$ tangent at points
$$∴ \text{In } \Delta QRS\text{: } \angle QRS = 30^\circ, \overline{SR} = r$$
$$\cos30^\circ = \frac{\overline{SR}}{\overline{QR}}$$
$$\overline{QR} = \frac{r}{\frac{\sqrt{3}}{2}} = \frac{2r}{\sqrt 3}$$
$$∴ \overline{PQ} = \sqrt{\overline{PR}^2 - \overline{QR}^2} = \sqrt{4r^2 - \frac{4r^2}{3}}$$
$$h_1 = \sqrt{\frac{8r^2}{3}} = 2 \sqrt\frac{2}{3} r$$
$$∴ h = 2h_1 = 4 \sqrt{\frac23} r$$
Hence, in the calculation of packing efficiency of hcp arrangement,the height of the unit cell is taken as $4r\sqrt{\frac{2}{3}}$.
FROM

What does the triangle of dots mean?

How come angle QRS is 30 deg ?


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Do acyl halides react similarly to alkyl halides with nitriles with TBAB/NaOH?
I am referencing page pg. 665 in Clayden where it is shown that a nitrile can be alkylated using an alkyl halide with tetraalkylammonium chloride and NaOH. Would acyl halides behave similarly resulting in substitution at the carbon adjacent to the nitrile group?

You mean on alpha carbon? No, you'd think so?

Think they would hydrolyse with the tetraalkylammonium hydroxide as they are way more sensitive to hydrolysis than nitriles. Plus if they do deprotonate the halide can leave to give you the ketene.

I would be more worried about the alkyl halide hydrolyzing than the nitrile, but according to the text the alkylation still occurs. The procedures for hydrolyzing nitriles that I have read involve acidic conditions at reflux.


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Reaction of coumarin with phenylmagnesium bromide
I came across this problem..

I thought that the Grignard reagent, PhMgBr, would cleave the lactone and then subsequently add to the ketone formed, giving an allylic tertiary alcohol after aqueous workup. Treatment with acid would then lead to intramolecular etherification, giving 2,2-diphenyl-2H-1-benzopyran:

But then our chemistry professor gave out a different mechanism for this reaction as follows.

What is the correct product?

I immediately call BS on this. You can argue for the conjugate addition step, but you shouldn't argue for a $\mathrm{S_N2}$ at a $\mathrm{sp^2}$ carbon except under very special circumstances (which this reaction does not fit)...

Unfortunately there aren't any examples of coumarins being treated with aryl Grignards, so this is again a made up question. I much prefer your mechanism, though.

In terms of regioselectivity, at least for the first addition of the Grignard reagent, I suggest to have a look on the cumarin, that may be seen as an α,β-unsaturated carbonyl compound. Hard nucleophiles (in terms of HSAB principle) like $\ce{MeLi}$ would almost exclusively react with the carbonyl carbon, as you drew in your reaction equation.  Soft nucleophiles, such as cuprates, would prefer the 1,4-addition.
Going from $\ce{MeMgBr}$, to $\ce{EtMgBr}$, $\ce{i-PrMgBr}$, and $\ce{t-BuMgBr}$ gradually lowers the polarity of the bond between the metal (magnesium) and the directly attached carbon atom.  Hence, along with this line (in other words, as a trend), the preference to react in 1,4-position  (more like Michael reaction) increases on expense of the reaction of this nucleophile on the carbonyl carbon.  I speculate the bond in the Grignard reagent, between $\ce{Mg}$ and the phenyl group in $\ce{PhMgBr}$, is even less polarised, than in the instance of $\ce{t-BuMgBr}$, and hence conclude at least a significant portion of the product of the first addition formed will be the one of the 1,4-addition (second drawing).
As @Zhe, I however still have a difficulty to understand the subsequently suggested second addition of $\ce{PhMgBr}$ in the lower drawing, because the activation of the double bond once in conjugation with the carbonyl group is gone.

OK... So this has to do something with the HSAB theory. I shall go through. Thanks for the response.

But can you please say me why hard nucleophiles prefer attacking carbonyl carbon and soft nucleophiles prefer 1,4 addition?

@SurajS There are several factors which will be explained in detail in textbooks about organometallic chemistry.  Among the more important contributions is the ratio of charge/polarisation at a site of a reagent and size/volume over which the charge is distributed.  J. Chem. Educ., 1978, 55, 355-360 (DOI 10.1021/ed055p355), or https://retrosynthetix.com/2017/03/01/acid-base-xi/ may provide some insight.

It's drawn like a pseudo SN2,  but given that you can delocalise across the oxygen and double, coupled with the nature of the.Grignard,  it's not an SN2, but the outcome can be the same

In the second mechanism the second addition of PhMgBr may be OK if one first eliminates to a ketene and then adds PhMgBr.

A search of Chemical Abstracts for the OP's reaction of coumarin 5 with phenyl magnesium bromide forming 2,2-diphenyl-2H-chromene had 216 substructure hits but none from coumarin itself. Two typical examples are shown below.1 Suraj S is vindicated!

Although the OP's professor's proposed product 10 of the reaction is well-known, a search did not produce its formation from coumarin. There were no substructure hits either. If the reaction mixture were to contain a catalytic amount of cuprous ion and if coumarin behaves in the same fashion as an α, β-unsaturated ketone, conjugate addition would occur to form enolate 6. The enolate may well undergo elimination to afford ketene 7, which, in turn, reacts with phenyl magnesium bromide to provide ketone enolate 8. The professor's proposed displacement of phenoxide by the Grignard reagent at an sp2 center is unwarranted. Workup with water would provide ketone 9 which has been converted to 10 upon acid catalysis.
4
1) J. N. Moothry, P. Venkatakrishnan, S. Sengupta and M. Baidya, Organic Letters, 2006, 8, 4891.


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Copper (II) oxide during CuSO4 electrolysis
I was performing copper (II) sulfate electrolysis (graphite electrodes; 10V; 0,01A) today and I observed black precipitate falling from around cathode. I suppose it's copper(II) oxide. I couldn't find information about this reaction. Have you ever read somewhere about this reaction or do you know it?

How do you know this is not graphite, to begin with?

I thought about it but I used the same electrode in the electrolysis of a few solutions and precipitate occurred only when copper was reduced. I used it as a cathode in the electrolysis of dilute hydrochloric acid when it could be mechanically damaged by hydrogen bubbles (the same voltage) and nothing happened. I am aware that graphite electrodes decay but in this case, it seems to fast and as I mentioned occurred only in this solution.

When done without proper procedures, electrochemical reduction  of copper salts produce copper as a fine dust. Under water and in presence of air this dust easily oxidizes, forming black copper oxide.


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Compounds that Liberate Hydrogen from Water, Reversibly
I'm working on a paper which provides an overview of all the current methods for storing hydrogen gas as part of a renewable energy system, and I understand compressed gas and metal hydride storage well enough. I am struggling, however, to find a table, paper, database, etc.  which compares all the compounds that could be used to liberate hydrogen gas from water.
I’m looking for a compound(s) that can be stored indefinitely at ~STP and would yield hydrogen gas in a reversible reaction with water? By reversible, I mean that with the addition of electrical or heat energy, it could be transformed back to the state it was in before it reacted with the water (and thus used to generate more hydrogen gas). Ideally, all the reagents would be low cost and abundant, but this is less important provided that the reaction is reversible.
I am aware that this process is likely to be wildly inefficient, and that the round trip efficiency of any commercially available battery chemistry would make it a superior option for storing electrical energy. In the context of a broader application, however, this system could afford to have huge energetic losses in exchange for providing what is in effect a "stockpile" of hydrogen with virtually no storage cost.
If this information does not exist in an aggregated form(s), how would I go about hiring an individual or institution to compile it? Very grateful for any thoughts the community has on this subject.


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How was the potassium battery invented?
It is said that mobile phones using potassium battery can be charged within a few minutes. Was it an accidental discovery or the companies made a new design for faster charging?

https://en.wikipedia.org/wiki/Potassium-ion_battery

The search for a suitable replacement for lithium ion batteries has been ongoing pretty much since the invention of lithium ion batteries. However, the motivation for this research has primarily been cost, and shorter charge times (along with other advantages) is simply a fortuitous property of the cheaper potassium ion batteries.
Part of the reason that potassium ion batteries are much cheaper that their lithium counterparts is the relative prevalence of potassium in the Earth's crust, potassium being present at levels about 900 times that of lithium.
According to this Wikipedia article on the development of the potassium ion battery:

The prototype device used a potassium compound Prussian blue as the cathode material for its high electrochemical stability. The prototype was successfully used for more than 500 cycles. A recent review showed currently that several pragmatic materials have been successfully used as the anode and cathode for the new generations of potassium-ion batteries.
  [...]
  The potassium battery has certain advantages over similar lithium batteries (e.g., lithium-ion batteries): the cell design is simple and both the material and the fabrication procedure are cheaper. The key advantage is the abundance and low cost of potassium in comparison with lithium, which makes potassium batteries a promising candidate for large scale batteries such as household energy storage and electric vehicles. Another advantage of potassium-ion battery over lithium-ion battery is the possibility for charging faster. This means that the next generation of mobile phones based on potassium-ion batteries can be charged within a few minutes only.

So, while faster charging times are certainly a desirable characteristic, this was not the driving force for the development of potassium ion batteries.


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What do the subscripts 1/2 and 3/2 for the p-orbitals refer to?
What does $p_{1/2}$ and $p_{3/2}$ mean when referring to electron configuration?
I've seen sub shells and I'm assuming it has something to do with spin. I am asking this in regard to the  $8p_{1/2}$ and  $8p_{3/2}$ shells and the way they get filled at different energy levels?

See this earlier Q & A, Shape of the P1/2 Orbital

We prefer to not use MathJax in the title field, see here for details.

Essentially, spin-orbit coupling leads to the degeneracy of the 8p orbitals being lifted. ron's question has all the details.

Essentially, this derives from Dirac’s relativistic quantum mechanics.
In nonrelativistic quantum mechanics—the Schrödinger equation that you are probably highly familiar with—p orbitals are triply degenerate while d orbitals are 5-way degenerate. However, nonrelativistic quantum mechanics does not include the electronic property known as spin unless it is introduced in a separate step.
Dirac’s relativistic quantum mechanics solves this problem by introducing $c$ as a finite constant. Almost magically, his equations directly result in a quantum number for the spin, $\vec s$. However, the result is that the azimuthal quantum number no longer behaves like a true quantum number (i.e. it no longer describes stationary states) but thankfully the sum $\vec j = \vec l + \vec s$ does. Therefore, instead of considering just one $\vec l$ as a descriptor to decide between s, p and d orbitals, we are stuck with $\left |\vec l - \vec s\right |$ and $\left | \vec l + \vec s\right |$. For $\vec l = 1$ (a p orbital) this gives us the values of $\vec j = \frac12$ and $\vec j = \frac32$. There are two p orbitals of the $\mathrm p_\frac32$ type and one of the $\mathrm p_\frac12$ type. These are no longer degenerate, i.e. $\mathrm p_\frac12$ has a lower energy than the $\mathrm p_\frac32$ ones.
The orbital of the $\mathrm p_\frac12$ type happens to be spherical in nature. For a more thorough background, I recommend Nicolau’s extensive answer which I mostly paraphrased in the above paragraphs and which contains links to further materials.

Why $p_{3/2}$ has higher energy than $p_{1/2}$? Thanks in advance.


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Degree Of freedom Of A Critical point
What's the degree of freedom of a critical point of a single component system?
In the critical point the difference of density between the liquid phase and the vapour phase is so small the are indistinguishable, hence considered as single phase. So according the gibbs phase rule (. F=C - P +2) the degree of freedom should be 2.
But critical point is a point. And the degree of freedom of a point in the phase diagram should be zero( for example, the triple point).
So what's the correct degree of freedom   of the critical point?

There are 3 phases at the critical point. Sub-critical vapour, Sub-critical liquid and critical fluid. 3 phases -> DoF = 0
That is one way to see it - and for convenience that is how I see it. It might be a bit wobbly to regard that as rigorous analyse though.
You can also say, more rigorously, that there is 1 phase and 2 constraints:
$$\left(\frac{\partial\rho}{\partial V}\right)_T = 0 $$
$$\left(\frac{\partial^2\rho}{\partial V^2}\right)_T = 0 $$
Subtract the number of constraints from the DoF and you get 0 again.
This defines a single point on the P/T curve.


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How I can use Zaitsev's rule backwards?
In the reaction
$$\ce{H2O +CH3CH2CH=CHCH3->} ?$$
Which of the doubly bonded carbon atoms should I choose with the $\ce{OH-}$ ?

Zaitsev's rule in reverse is basically just Markonikoff's rule.

I don't understand a backwards Zaitsev's rule.
But one thing that I am sure about, is that you can decide that in which carbon which would $\ce{OH-}$ get attached by using Markovnikov's rule. This rule states that hydrogen would get attached to the double bonded carbon which has the greater number of hydrogen atoms attached to it, or conversely $\ce{OH-}$ or any other negative part would get attached to the double bonded carbon which has the lesser number of hydrogen atoms attached to it.


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How do I determine the Kc of the reaction between Fe3+(aq) + SCN-(aq)-->Fe(SCN)2+(aq) at different temperatures using spectophotometry?
Beer's law states that A=elc where A is the molar absorptivity, e is the molar absorption coefficient, l is the pathlength of light and c is the concentration. However, I am not sure how to obtain values for this reaction.
Should I use different concentrations of $\ce{Fe(SCN)++}$ and plot a graph to calculate it or is there a better way? At different temperatures will e change? Thank you.

The Beer Lambert law is $I_{trans} =I_0 \exp(-\epsilon [C]l)$ where $I_{trans}$ is the transmitted and $I_0$ the incident light intensity. Usually this is used as $I_{trans}=I_010^{(-\epsilon[C]l)}$ in which case $\epsilon$ is the decadic extinction coefficient. The term $\epsilon[C]l$ is called the optical density. The extinction coefficient depends on wavelength and is obtained by plotting  optical density vs. concentration then the slope gives the extinction coefficient at each wavelength used.
Generally the extinction coefficient does not change greatly with temperature. Any changes are due to vibrational level's populations changing as the temperature changes and this is hard to predict exactly; usually bands broaden a little and become less intense but  some vibrational bands may increase and others decrease so it would be best simply to measure at each temperature.
As far as you reaction goes you need to know the extinction coefficient from a separate experiment with known concentrations before trying to get the equilibrium constant.


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Monochlorination of 2-chloropentane and possible enantiomer products
I found a question to determine the number of possible chiral products of monochlorination of 2-chloropentane.So, first the products are (with possible number of enantiomers in the parentheses:
2,2-dichloropentane(no chiral product),
1,2-dichloropentane(4 or 2, I am not sure is there a plane of symmetry here)
2,3-dichloropentane(2 or 4)
2,4-dichloropentane(obviously none for a plane of symmetry)
2,5-dichloropentane(2,4?)
But my surprisingly the answer says it's only 5. That's even lower than my lower estimate. I am not sure how there are 5 enantiomers.

Chlorination at 2°carbon is more probable to happen here if we consider the thermodynamics of the reaction but still I am confused about having just 5 products.

I am not. Narendra Kumar answer is right.

2-chloropentane is optically active compound hence in some cases we will get only diastereomers. For example 1,2-dichloropentane is single product because it is formed by attack at carbon no. 1 hence no change in chirality of carbon no. 2.  So product formed by attack at carbon 1 (one chiral product) carbon3 (two chiral product i.e. diastereomers) carbon 4 (one chiral product, i.e. one chiral and one meso) carbon5 (one chiral product)

The answer is 5.
 See-
1) 1,2 dichloropentane --- 1 chiral
2) 2,3 dichloropentane --- 2 chiral
3) 2,4 dichloropentane --- not chiral as there is plane of symmetry
4) 2,5 dichloropentane -- 2 chiral
Therefore 2+1+2 =5 Chiral Products

why 2 chiral products when there are 2 stereocenters in 2,3-dichloropentane?

It would be better if you explain it better.

Please don't use all caps.

@pritt Balagopal  Are u sure this answer is right? and why 2 chiral products when there are 2 stereocenters in 2,3-dichloropentane?

I think there won't be 2 enantiomers for 2,5-dichloropropane. only 1. and 2,4-dichloropropane would have 1 enantiomer


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Limiting radius ratio
I recently came across this question while studying Solid State.
The coordination no. of [Ni(CN)4]2- is 4  as Ni is bonded to 4 CNs and  is planar in structure.Hence its Limiting Radius Ratio must be between 0.225 and 0.414 . But the answer given in the book is 0.414 to 0.732 (which is the limiting radius ratio for coordination no. 6) .
How is this true?Some sites agree with my answer .
Is the answer given in the book wrong or have I made any mistake in the reasoning?

I've always calculated limiting radius ratio for ionic crystals. I'm not sure if such a thing exists for coordination compounds.

I have a hunch that the charge 2- on the ion has some role to play. I am not sure. And yeah you are right . Even I calculated only for ionic crystals not for coordination compounds

If you insist, a cursory search tells me that the crystal radius of $\ce{CN-}$ is $\pu{1.77 Å}$, and the ionic radius for $\ce{Ni^2+}$ ion is $\pu{0.69 Å}$.

Then the radius ratio will be 0.3898 which agrees with my answer.

Here, limiting radius ratio is for geometry of this complex is square planar because $\ce{CN-}$ is a strong field ligand and causes pairing of unpaired electrons. So we can relate that $\ce{Ni^2+}$  occupies a square planar void of the $\ce{CN-}$ lattice. The square planar void is a cross section of the octahedral void.


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Adding hydrogen ion concentration while calculating pH
I have just started learning how to calculate pH and I have noticed that adding [$\ce{H+}$] makes a negligible difference in few cases whereas in other cases it makes a major change in the final value. Why is that?

Can you provide an example of what you mean?

pH is calculated by a logarithmic formula given by:
$$\text{pH}=-\log{\ce{[H+]}}$$
Lets say, the initial hydrogen ion concentration is $\ce{[H+]_0}$, and change in concentration is $d\ce{[H+]}$. The change in pH would be:
$$d\text{pH}=-\log{(\ce{[H+]_0}+d\ce{[H+]})}$$
This can otherwise be written as:
$$d\text{pH}=-\log{\ce{[H+]_0}}×\frac{d\ce{[H+]}}{\ce{[H+]_0}}$$
You can see that the change in pH depends on the initial hydrogen ion concentration. It's in fact inversely proportional to the initial concentration. Adding a drop of concentrated acid to a base would give a much bigger change in pH compared to adding a drop to pure water, which in turn is bigger compared to adding a drop to an acid.

So do I need to include the hydrogen ion concentration while calculating pH ?

@Justuraveragemathsstudent Of course! pH itself is a measure of hydrogen ion concentration.

sorry..I meant hydrogen ion concentration of water

@Justuraveragemathsstudent Well, it depends on the strength of the acid, and how much of it is there. If the acid produces much more ions than water does, ignore it, or else take it into consideration.

I don't follow the algebra here the change in pH is  $\delta pH=-\log(H_0)+\log(H_0+\delta H)$, rearranging gives $\delta pH=-\log(H_0)+\log(H_0[1+\delta H/H_0])$ which becomes after expanding  and then expanding the log with $\delta H/H_0 \lt 1$ and taking just the first term $\delta pH=\delta H/H_0$.


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What is the complete reaction for Cyanide ions reacting with Thiosulfate?
I was recently studying about the properties of Sodium Thiosulfate, where I came across the fact that $\ce{S2O3^2-}$ reacts with $\ce{CN-}$ ions to produce $\ce{SCN-}$ ions. This was quite useful in treating cyanide poisoning.
I have tried to find the complete reaction, but with no success. I believe this must be a redox reaction, since $\ce{S2O3^2-}$ has an average oxidation state of $+2$, while $\ce{SCN-}$ has a state of $-2$ (correct me if this is wrong). Can someone give me the complete reaction?

Redox reactions can be very complicated. In this case there's probably no complete scheme.

Source 1(If you consider $\ce{Na+}$ as counterions)

$$\ce{Na2S2O3 + NaCN -> Na2SO3 + NaSCN}$$

Sodium thiosulfate reacts with sodium cyanide to produce sodium
  sulfite and sodium thiocyanate.

Source 2(If you consider $\ce{Ag+}$ and $\ce{Na+}$ as counterions for cyanide and thiosulfate ion respectively)

$$\ce{AgCN + 2Na2S2O3 → Na3[Ag(S2O3)2] + NaCN}$$

Silver(I) cyanide react with sodium thiosulfate to produce sodium
  dithiosulfatoargentate(I) and sodium cyanide.

In this case, thiocyanate does not form, rather a complex forms.
Explanation of this reaction can be found by observing the related reactions given in the Wikipedia article of sodium thiosulfate.

Silver halides, e.g., AgBr are typical components of photographic
  emulsions, dissolve upon treatment with aqueous thiosulfate:
$$\ce{2S2O3^{2−} + AgBr → [Ag(S2O3)2]^{3−} + Br^{−}}$$

Since, $\ce{CN-}$  and $\ce{Br-}$ hold same charge,  you can easily replace it with silver bromide with silver cyanide  to get  cyanide ions as products.
Furthermore:

Sodium thiosulfate is a component of an
  alternative lixiviant to cyanide for extraction of gold to form a
  strong soluble complex $\ce{[Au(S2O3)2]^{3−}}$. The advantage of this
  approach is that thiosulfate is essentially not toxic and that ore
  types that are refractory to gold
  cyanidation (e.g. carbonaceous or Carlin-type ores) can be leached by
  thiosulfate. Some problems  include the high consumption of
  thiosulfate, and the lack of a suitable recovery technique, since
  $\ce{[Au(S2O3)2]^{3−}}$ does not adsorb to activated carbon, which is
  the standard technique used in gold cyanidation to separate the gold
  complex from the ore slurry.

This process is also suitable for silver extraction using sodium thiosulfate and silver cyanide to form corresponding silver complex, $\ce{[Ag(S2O3)2]^{3-}}$. Just replace gold cyanide with silver cyanide and it will work pretty much well.

Thanks for your answer. Can you please guide me to sources that explain the reactions of thiosulfate, particularly silver thiosulfate?

@PrittBalagopal Thanks for the feedback. I edited the answer to make it clear


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I'm trying to verify the integrity of my cheap "24k gold" leaf
In what acid would a combination of gold and copper only dissolve the copper, but leave the gold?
I've been googling around, and so far they follow very close to the same solubility rules - as in, it would react with both the copper and gold, and both were semi soluble.  If the compound is brightly colored that would help too.

Most acids don't react with gold, quite a few react with copper. Use concentrated H2SO4 to form copper sulfate, water and sulphur dioxide. The gold can then be filtered out

Well, then it wouldn't be 24k ;) Also as Archimedes found out there are other methods to find out such thing.

"Cheap" and "24k" should not be used together...

EDIT: as pointed out by Stian Yttervik, my previous answer (below) is probably too dangerous for someone without prior experience with handling hazardous chemicals.
I would therefore recommend that you use nitric acid: simply add a few drops of clear fuming nitric acid to a test tube containing your gold leaf and stir with a glass rod.
The solution will then go green/blue if it contains copper. Only do this with very small quantities of gold leaf and when outside as one of the products of the reaction with copper is nitrogen dioxide - an extremely toxic brown gas.

I would recommend concentrated $\ce{H2SO4}$, as it is easily available and produces a brightly coloured salt when reacted with copper, thus making it easy to tell whether your gold leaf is 24K or not without any expensive weighing equipment.
Make sure you do this outside or in a very well-ventilated area, as it will give off small amounts of $\ce{SO2}$, a toxic gas. Wear thick gloves and eye protection and make sure you have a water supply nearby in case of an accident.
Pour 10 ml of concentrated sulfuric acid (98%) into a boiling tube (or any other narrow Pyrex container you might have) and heat the liquid with a Bunsen burner until it starts to boil. Avoid the fumes as these are corrosive. Without removing the boiling tube from the flame, add a few small sheets of gold leaf and stir the solution with a glass rod for a few minutes.
If you find that the solution is completely colourless, there certainly was no copper in the gold leaf. However, even the slightest tinge of blue/green would indicate copper sulfate - formed by reacting copper with sulfuric acid.
When you're done, make sure to dispose of the sulfuric acid by emptying it into a litre or so of water (acid in water, not the other way around) and then neutralising this with bicarbonate of soda

While your proposal does answer the question, I am skeptic towards handing out advice to use a vicious chemical like boiling hot conc sulphuric acid. Boiling over a bunsen burner. The number things that can go wrong here, in a seriously harmful way, is mind boggling. The gold that can be recovered from this is not worth neither the price of the sufficient protective equipment you need nor the risk to your life should you forgo the equipment.


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Homemade copper etching solution
I'm trying to make an etching solution for copper (PCB) with easy to access materials. I'm using hydrochloric acid (9%), peroxide (common "oxygene" that contains many more ingredients), table vinegar, table sea salt, and bicarbonate soda. Now the recipe mentioned only hydrochloric acid and the peroxide, but it mentioned a hydrochloric acid with high concetrantion (31%). My solution with (the easily accessible) 9% hydrochloric acid didn't worked so I added the other elements and it worked fine. The problem is that I don't know what I'm doing, and what elements causes the reaction. I start with the hydrochloric acid, I put some peroxide inside. Alone it does nothing to the copper. I then add some vinegar, salt, and soda (I suppose soda helps to get oxygen inside the solution) and after a while the reaction starts and it works really quick. (the 5x5 cm copper piece gets dissolved as I like in around 5 minutes. (and whatever I have painted on it with my permanent marker remains).
So can someone explain me what reactions take place, and perhaps which is the crucial material and which I may omit?
Also when I work with that stuff my nose gets irritated, how harmful is this solution and how I may dispose it safely?

The addition of soda probably has a negative effect. I would try without it.


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Longest chain IUPAC rule
I wonder, what are the exact IUPAC organic chemistry nomenclature rules that give the preference to the chain length over number of unsaturated bonds, i.e. that justify, or prefer following names:

4-ethenyl-4-ethynylheptane over
3,3-dipropylpent-1-en-4-yne

3,5-dimethylideneheptane over
2,4-diethylpenta-1,4-diene
?

afterwards I have found this related question: What does the newest IUPAC nomenclature have against unsaturation?

In Chapter P-4 "Rules for Name Construction", section P-44 "Seniority Order for Parent Structures", of Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, it is stated that (p 490):

P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

The Blue Book lists several helpful examples, one of which is that the principal chain "octane" is senior to "hept-1-ene". Therefore, the following compound

is named 3-methylideneoctane instead of 2-ethylhept-1-ene.
The authors mention that this preference is in contrast to previous recommendations (e.g. 1979):

In acyclic parent structures the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed. Thus, the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second.

Thanks. I wonder if the reasons leading to this change are mentioned somewhere..


text:
Last electron of a transition element
The electronic configuration of nickel is
Ni: [Ar] $3d^84s^2$
Here, while writing the configuration, we fill the 3d after 4s.Hence the 28th electron enters the d orbital.
But
$\ce{Ni+}$: [Ar] $3d^84s^1$
Here we take out the last electron (28th) from 4s.
Are there any rules or conventions regarding this or am I conceptually misguided.

Yup, the transition metals do some weird things. For Ni, it prefers to keep the 3d shell intact. The point to remember is that real atoms have multiple electrons, so the nice neat picture of orbitals you learn based on hydrogen just don't hold. What is more amazing is that the simple picture holds up pretty well.

When you ionize an element, all the orbitals go down in energy but orbitals with higher angular momentum drop more than those with lower angular momentum. When you start ionizing Ni, the $d$-orbitals drop much more than the $s$-orbitals and so, in the ions, the highest energy orbitals are the $4s$.
Relatedly, I believe (and I just ran a quick calculation to confirm) your electron configuration for $\ce{Ni+}$ is incorrect and should be [Ar]$3d^9$ for the same reason.


text:
Could chromatography induce a conformational change?
We'd been studying chromatography in our instrumental analysis course and I was wondering if chromatography could induce a conformational change.
For example, you have a straight chain with a hydrophobic side and hydrophilic side; when relaxed, its conformation is such that it has a hydrophilic outer surface and a hydrophobic core.
If you run it through a normal phase column, could the stationary phase cause a change in conformation to, let's say, a hydrophilic core and hydrophobic outer surface?
Also, if this could happen, then it would initially go through the column at a rate similar to the mobile phase and then slow way down, yes? If so, what would be the implication? Would the peaks be broadened?

Compounds react and break down in the column (especially in GC). Such reactions take more energy. Thus, it doesn't seem unlikely that conformational changes couldn't be induced by the higher temperatures of a GC oven. If you are talking about HPLC: I am not sure if this can be done with only with the stationary phase. If I recall correctly, there was a study done in which scientists tried to switch the hydrophobic core with the hydrophilic surface by using different solvents. It didn't work.

There is a technique called 'molecular combing' that is used to straighten DNA's and AFM's are used to stretch proteins so it is possible that forcing molecules down columns could effect a conformational change. However, it may be that the forces applied are too small, you would have to compare with what is used in these other techniques. In particular how swiftly these forces are applied is important. The faster they are applied the stiffer molecules become.

While I do not have numbers at hand to support it, I think it is possible, chromatography on silica indeed may influence, may restrain the conformational flexibilty of molecules in the mobile phase.
Normal phase chromatography depends on adsorption.  A complete rotation around a $\sigma$-bond, even in absence of other molecules around, needs some energy to pass from one stagggered to an other staggered conformtion.

(figure source)
If there were now -- instead of methyl groups separated by three bonds, as depicted in the diagram -- two groups with a high affinity to the stationary phase, this said energetical barrier would increase.  Assuming lowered temperature and affinity of these groups "just intense enough" (e.g., deportontated sulfonic acids, not per-alkylated ammonium groups), the propability to find one molecule in such $\ce{s-cis}$ conformation will increase.  I just see a potentially relevent publication Langmuir, 2004, 20, 10639–10647 about proteins decoiling on silica.  I imagine, amphiphlic structures like reverse micelles may be equally influenced and will "swap their inner to the outer" as if they were (again) exposed to aqueous systems when facing silica.
Yet I am still hesitant to state if this may by become a constraint, leading to a situation where once conformation is looked, as in the instance of atropisomers.


text:
Reason of dissociation of water molecule in electrolysis
$\ce{H2O}$ molecule is dissociated into $\ce{OH-}$ and $\ce{H+}$ ions on reduction at the cathode in electrolysis. But what happens at the atomic level to $\ce{H2O}$ that proton and hydroxide ions get separated after gaining electrons from cathode? What really happens when water molecule gains electrons from cathode?
$$\ce{2H2O(l) + 2 e- ->  H2(g) + 2 OH-(aq)}$$

The dissociation of water to $\ce{OH-}$ and $\ce{H+}$ happens first before any interaction with the cathode. $\ce{H+}$ ions then gain an electron at the cathode to produce hydrogen. At the anode, water is oxidized (loses electrons) to produce oxygen and more $\ce{H+}$ ions.
In pure water there is a very small concentration of $\ce{H+}$ and $\ce{OH-}$ which is why the reaction is very slow and not thermodynamically favorable. It is also the reason why pure water is such a good insulator. But when you add an electrolyte the process becomes much more favourable kinetically and thermodynamically. Further proof that $\ce{H+}$ and $\ce{OH-}$ are generated before any interaction with the cathode.

A simple way of understanding what Answer 3 is saying is that one of the electrons from a hydrogen atom in a water molecule is so strongly attracted to the Anode, that it binds to the anode rather than staying with its proton. The proton (aka H+) is then repelled away from the anode (both are positively charged) and at the same time attracted to the Cathode.

"$\ce{H+}$ ions then gain an electron at the cathode to produce hydrogen".I think that the hydrogen produced at the cathode is the reduction of water molecules and not the reduction of
$\ce{H+}$ ions because the amount of $\ce{H+}$ ions is very low


text:
Trends in Slater's constant
I am aware of the various Slater rules to calculate the effective nuclear charge, $Z_\mathrm{eff}$
However, how can I decide the order in which the orbitals 2s, 3s, 3d, 3p, 4d, and 4f stand when compared regarding the effective nuclear charge on them?
Is calculating the $Z_\mathrm{eff}$ on their individual electrons the efficient way or can some trend or method be established ?


text:
Would adding water to a fish tank with sand lower the level of sand?
Let's say I have a typical 10 gallon tank with dimensions 10" x 20" x 12", and it's filled halfway (5" in height) with dry sand for hermit crabs.
If I fill the tank up with water, will the level of sand decrease?
I'm thinking that it doesn't because of this answer -- sand is insoluble.  So the water would just fill the gaps of air first.  If that's not the case though, what are some things to take in to consideration?
Not quite sure where this question belongs, so if it's in the wrong place, let me know -- or mods, feel free to move it.  Thanks!

If the submerged sand is vibrated then it will settle more than it would if dry.

If your sand is packed in a optimal sphere packing, like a hexagonal packing for example,

nothing would happen to the sand level, because it can't be packed more dense.
If the packing of the dry sand wasn't optimal (which it probably isn't 100%), then the water may help settle it through motion and friction and it would lower a bit. In any case it wouldn't lower so much as that you could easily see it with your eyes.


text:
Why do carbonyl compounds undergo nucleophilic and not electrophilic addition?
Why aldehydes and ketones undergo nucleophilic addition reactions and not electrophilic addition?
Alkenes show electrophilic addition because of pi bond, which is also present in aldehydes and ketones. But there is nucleophilic addition only. Why?

They sure do. Except that the oxygen has lone pairs, which are more nucleophilic than the $\pi$ bond. Once those electrons attack an electrophile, it further deactivates the $\pi$ bond as a nucleophile.

General Explanation
According to Feltre and Setsuo (Organic Chemistry, vol.4), the π bond is attracted to the oxygen (through mesomeric effect), leaving a partial positive charge on the adjacent carbon. Then the nucleophile attacks the carbon, leaving for the electrophile the mission to bond to the (now negative) oxygen.
To further out the question:
What I seem to miss is why the nucleophilic attack happens first (thus naming the reaction), since the exposed O- seems a lot more vulnerable to electrostatic interaction.

Aldehydes and ketones prefer nucleophilic addition reactions Instead of electrophilic addition reactions this is because in case of nucleophilic addition reaction a stable intermediate is formed as compared to the electrophilic addition reactions where the intermediate formed is carbocation which is highly unstable.

stable intermediate? $\mathrm{S_N2}$ has no intermediates, only a transition state..


text:
What is parachemistry? What does a parachemist do?
I was reading the Washington Post article An atomic town revels in its nuclear past as tunnel collapse raises contamination concerns and came across a quote from someone who was said to have been a "parachemist" at the Hanford nuclear facility in the US state of Washington.
Is parachemistry a recognized subfield of chemistry? If not, what else could this mean?

Here it is used in a rather bizarre context as a kind of alchemy. It doesn't make good sense that the Hanford Nuclear Facility would have been hiring alchemists though, at least I hope not ;)

I wonder if 'parachemist' might be a word formed by analogy with 'paramedic', so someone trained for emergency response at the facility.

@AndyW I like that image much better than the alchemist interpretation.

Another stab is possible derivation from paralytic or blind drunk.
Thus the term para-chemist.

@AndyW Or "paralegal", i.e., someone who does the menial work lawyers do not want to do.

https://en.wiktionary.org/wiki/para-#English may be hintful, except from the chemistry related subsection, paradoxically. However this word is generally never used, in present times.

A parachemist is always put directly opposite the chemist

Now I'm wondering how parachute fits in here.

Joined just to upvote Stian's comment

The word-forming element para- has many a meaning. Here it outlines change, or transformation. From the Online Etymology Dictionary,
para- "...beyond; altered; contrary..."
Parachemistry is a subfield of industrial chemistry. It is mainly concerned with chemical reactions, transformation of one chemical to the next.

parachimie [type of industrial chemistry], "Ensemble des industries dont l'activité est fondée sur la transformation des produits chimiques" (industries having to do with processing of chemicals) $-$ The Government of Canada’s terminology and linguistic data bank

Unfortunately, this definition is pretty vague and general; may I say it is akin to using the tag reaction for your questions :-). You will find parachemistry is most used in French-speaking countries. Dictionnaire Français specifies additionally

"Branche de l'industrie qui transforme les produits chimiques en biens directement utilisables par le consommateur." (A branch that deals with final processing of chemicals for consumers [including specific industries, not only mall shoppers $-$ personal guess].)

I advise against using this word.

That sounds suspiciously like chemical engineering to me.  I wonder if the two terms are synonyms.

I'll try to present the best I could find. The titles "para", "meta" and "ortho" were perhaps given by Hermann Kolbe (1818-1884) way back in the nineteenth century. For one thing, Kolbe had the reputation of being a well recognized teacher and a skilled and productive chemist. He is also well known today (partly) as the author of a critique of the foundation of stereo-chemistry, with was established by the Dutch chemist, Dr. J.H. vant Hoff. He had directed bitter remarks not only towards vant Hoff for his hypotheses (which was perhaps the boldest), but at nearly the entire realm of contemporary organic chemistry.
Background

$\ldots$ Once he gave his amanuensis the empirical formulas for three novel compounds for which their discoverer had just assigned structural formulas. The man reported back in a half hour with several more candidate structures, some of which looked more probable to Kolbe. Kolbe concluded that structure theory is a dangerous toy, especially for inexperienced chemists, and that structures are often assigned "in one's sleep," with little or no empirical warrant. $\ldots$

Privately, Frankland "entirely dissent[ed]" from this judgment and Volhard also objected privately; he joined Frankland in expressing regret over the personal character of many of Kolbe's remarks. Kolbe's response was pretty straightforward. As he remarked he must criticize and contest the chemical ideas that he considered false and worthless, he would be happy if his views were contested. He later wrote to Varrentrapp that he would rather be considered sharp-tongued than cowardly.
The crusade

In Kolbe's third and fourth retrospects, those for 1873 and 1874, he went after his favorite example of structuralist excess: Kekulé's benzene theory. He absolved Kekulé himself of much of the blame, for he was convinced that Kekulé regarded the theory merely as an intriguing and useful hypothesis. However, most chemists by this time viewed the hexagon as "infallible dogma," as "the Pope is for Catholics." They were true fanatics, Kolbe wrote, and viewed him (Kolbe) as a rank beginner, of weak understanding. They were right, he thought, for he could not understand arguments built "in the air" or "on loose shifting sand."

The end of their sand castles was not distant. Kolbe continued,

The modern chemist, who knows exactly what a chemical compound looks like in its middle and its end, how the six carbon atoms of benzene are symmetrically linked together in a plane, who then further purports . . . to have a clear conception of the spatial arrangement of the atoms, of their ortho, meta, and para positions, who determines the positions of all of the atoms in the compound, has long since abandoned the solid ground of exact science; the scientist has become a metaphysician.

He then named true physicists as "orthophysicists" as opposed to "paraphysicists" who construct hypotheses on the basis of experimental work of others and "metaphysicists" who believed that they can dispense entirely with empirical evidence.
In his own view, he was an "orthochemist" and Kekulé was a "parachemist. But as the  years went by, he increasingly believed that even the term "parachemist" was too kind; his opponents were "metachemists", to whom "chemical theories are as cheap as blackberries".

References:

Kolbe Versus the “Transcendental Chemists”: the Emergence of Classical Organic Chemistry, A. J. Rocke
The Quiet Revolution Hermann Kolbe and the Science of Organic Chemistry, Alan J. Rocke
Image and Reality: Kekulé, Kopp, and the Scientific Imagination, A.J. Rocke

Very nice! Read this through a few times, enjoyed it more each time. Perhaps some day each SE site will have it's own ortho- and para- branch to go with it's meta-?

Interesting answer, I must say.

@uhoh "para.stackexchange.com"? That would be ridiculous.

@PrittBalagopal in technical SE sites there are always those occasional questions that can't easily be closed, but are... "different". Physics SE is an example with perhaps the widest range. I'm not 100% serious, but perhaps meta-serious ;)

While interesting, I doubt this is the meaning that is intended in the Washington Post article that is linked in the question.

A parachemist, in this sense, is one who is working on issues that are related to, but not identical to classical chemistry.
Classical chemistry, as we know, involves the interactions of chemical elements, based on the properties of such elements.
Para-chemistry involves both the interaction of chemical elements, and other significant properties, such as in this case also their nuclear/radioactive properties, which may affect their chemical and physical interactions.
For example, regarding one use of the terms ortho/para chemistry see: aanda.org/articles/aa/full/2006/14/aa4420-05/… in which it states: " For each isomer, two equivalent H nuclei, of spin 1/2, couple to generate ortho (nuclear spin = 1) and para (nuclear spin = 0) species with spin statistical weights of 3 and 1, respectively. " as an example in which para chemistry is tied to the nuclear properties of the interaction.
Another example of para chemistry referring to a non-classical chemistry understanding: arxiv.org/pdf/1612.07845.pdf refers to the idea that deuterium can have 3 spin states, and thus the parachemist may need to understand such spin states, as well as the classical chemistry, in order to analyze a chemical reaction.
" D 3 + has three spin states i.e. ortho (I=1,2 and g I =16), para (I=0, g I =1) and meta (I=1,3 and g I =10). "
As another use of the term tied to its connection with properties other than purely chemistry concepts, this paper [ingentaconnect.com/content/jswt/jswt/2011/00000037/00000001/… states:
"Para chemistry covers the manufacture of products for specific uses such as, for example, agro-chemical products (insecticides, herbicides, etc.), explosive products, maintenance products, painting products, etc."
Here, I believe from my talks with industrial chemists, that the term refers to the notion that for industrial chemists, non classical chemistry concepts such as price, availability, ease of use etc. all play an important role, and not solely the classical chemistry.
Papers also exist that reference biological toxicity as a factor considered in the para-chemistry of a chemical reaction.
In an analogous way, the term ortho-chemistry, as another example, may refer to both the interaction of chemical elements, and their spatial configuration and properties, which may, in turn, affect their chemical and physical interactions.
The term ortho-para-chemistry can refer to an understanding that involves three different types of properties -- the chemical properties of the elements, properties of their spatial configuration, and their nuclear properties.
Terms such as meta chemistry also exist, and, as with the term para-chemistry, the idea expressed is that concepts and properties that are important to an understanding and analysis of a chemical reaction or sample may involve other properties outside of chemistry, but that are highly important to such analysis and understanding.
Such properties may involve the nuclear or radioactive properties, the spatial properties, the toxicity or biological properties, or others, including industrially important properties, such as scarcity or technical difficulty.

This is an interesting answer. Is there any way to verify this? Are you just best-guessing or do you know this to be so? A link or reference would be greatly appreciated.

I do know so, and as a link, see:
https://www.aanda.org/articles/aa/full/2006/14/aa4420-05/aa4420-05.right.html?display=full
in which it states:
" For each isomer, two equivalent H nuclei, of spin 1/2, couple to generate ${\it ortho}$ (nuclear spin = 1) and para (nuclear spin = 0) species with spin statistical weights of 3 and 1, respectively. "
as an example in which para chemistry is tied to the nuclear properties of the interaction.
Para-chemistry as a term can also refer to other non-classical chemistry properties of matter that affect chemical analysis.

Here is another example of para chemistry referring to a non-classical chemistry understanding: https://arxiv.org/pdf/1612.07845.pdf
in which deuterium can have 3 spin states, and thus the parachemist may need to understand such spin states, as well as the classical chemistry, in order to analyze the transaction.  " D
3
+
has
three spin states i.e. ortho (I=1,2 and g
I
=16), para (I=0, g
I
=1)
and  meta  (I=1,3  and  g
I
=10). "

This paper [http://www.ingentaconnect.com/content/jswt/jswt/2011/00000037/00000001/art00008] defines:
"Para chemistry covers the manufacture of products for specific uses such as, for example, agro-chemical products (insecticides, herbicides, etc.), explosive products, maintenance products, painting products, etc."
Here, I am merely guessing from my talks with industrial chemists, that the term refers to the notion that for industrial chemists, non classical chemistry concepts such as price, availability, ease of use etc. play an important role, and not solely the classical chemistry.

Great! it would be best if you could also include those links directly within the text of your answer! I wasn't challenging the validity of your statements, if you take a moment and check out the stackexchange tour you'll see that high quality answers should contain links or references to external sources to help verify statements of fact. It's just the way stackexchange is set up. Also, comments should be considered temporary, so anything that's important to an answer should be posted within the answer to assure it stays there permanently. Thanks!

Voting and answer acceptance are also based on the body of the answer, and in principle not on material only appearing in comments. It's for your benefit, as well as to future readers to keep linked material within the body of the answer.

"Parachemist" isn't obviously a role in a sub-field of chemistry.
A search for parachemist or para-chemist doesn't bring up much, it's mostly Spanish text. But a search for "para-chemist" hanford yields three results (other than this Q&A) – one Facebook page and two obituaries for people who had the role of a para-chemist at the Hanford site. So this may have been a fairly specific role defined at the Hanford site, or by the operators.
I proposed in a comment that it might be an analogue of paramedic – someone trained for an emergency response at the facility.
Rodrigo De Azevedo also suggested paralegal as a similar analogue. Although I think both are plausible, there's no evidence I can find to underpin either suggestion.
Regardless, I don't believe that parachemist can be Kolbe's "imperfect theorist" referred to in Berry Holmes' enjoyable answer, as it's clearly a job title and not an insult.

Can't argue with data! While this is not a complete answer, it seems to be on its way. Thank you for tracking this down.

As far as the insult, if I'm reading that correctly, it was meant to deride people who turned out to actually know what they were talking about, like "round Earthers" for example.

If it's Hanford-specific, does anyone know anyone working at Hanford, who might be asked about what "parachemist" means there?

@uhoh - Thanks. Perhaps 'insult' is the wrong term in my answer... 'ethos' or 'attitude' might be better. But Kolbe didn't seem to mean it as a very positive attribute, in any case.

Oh, I think he meant it as an insult :)

It is not an accepted sub-discipline and I'd guess that it is almost certainly a "corporate culture" specific term, if it's even actually ever used.
It would be a wild speculation to claim that the term is based on what Kolbe wrote in [when? 1850?].
My guess would be it means "near" or "similar to". The fact that the OP included it in the article is a sad indication of a clueless reporter. My best guess would be someone who has been a chemical technician but has been put in a role which normally requires a degree holding chemist (who knows whether that means a PhD, MS, BS, BA, or a 2-3 year technical degree) either as a supervisor, as some sort of inspector or in a QC role.

Wild speculation, sad, and clueless. Just sayin'


text:
Does acetamide respond positively to iodoform test?
This answer explains why acetic acid does not give a positive iodoform test.
Does acetamide respond positively to the test?

From wikipedia article of haloform reaction:

Acetyl chloride and acetamide don't give this (iodoform or in general haloform) test.

Why?

Acetamide is not a ketone which is substantially more acidic than the
  amide.  It's about a million times too weak an acid at the alpha
  protons. Hydroxide is too weak a base. (Quora)

Acetamide does not give  positive Iodoform test. The reason is that if we recall the mechanism of reaction , after incorporation of three halogen atoms on the $\alpha$ carbon of carbonyl carbon, there is a step where the OH- attacks at the carbonyl carbon and $\alpha$ CX3 - leaves. But in acetamide there is a resonance of NH2 group with Carbonyl carbon. So the electrophilicity of Carbonyl carbon decreases and it becomes difficult to attack the less electrophilic carbon by external nucleophile because the overlap of Carbonyl carbon with N of NH2 is very strong (2p $\pi$ - 2p$\pi$ overlap) .so the iodoform reaction is very difficult to occur for acetamide.

The lack of a haloform reaction does not preclude acetamide from reacting with iodine/aqueous base to give a Hofmann rearrangement.

But the question was asked about positive response in haloform test.

But the question was asked about positive response in haloform test.


text:
Treatment of allylic alcohol with thionyl chloride: why is the product rearranged?
I have read that $\mathrm{S_Ni}$ reactions involving the attack of $\ce{SOCl2}$ on alcohols proceed without the formation of discrete carbocations, and hence there is no rearrangement involved. But that doesn't explain this reaction:

Here, is the allylic carbocation formed, or is there a different explanation for this reaction?

It's not that cations don't form, it's that the formation of cations might be so fleeting that there's no time for loss of stereochemistry. In your particular example, there's no reason why we can't have a concerted $S_{\mathrm{N}}2'$ reaction though.

@Zhe The thing that I don't get is why there will be the need for Sn2'? Why not simply Sn2?

It can't be $S_{\mathrm{N}}2$. That would imply that the nucleophile is on the same carbon as where the leaving group was.

As a disclaimer, it is worth noting that the reaction of an alcohol with thionyl chloride can proceed via multiple different mechanistic pathways (Wikipedia has a pretty good overview), depending on the conditions.
In the $\mathrm{S_Ni}$ mechanism a free carbocation is not formed; instead the key intermediate is an intimate ion pair $\ce{R+...^-OSOCl}$. This is not the same as a typical $\mathrm{S_N1}$ mechanism, where the leaving group is fully dissociated.
Following the formation of the intimate ion pair, the anionic component breaks down to form $\ce{SO2}$ and chloride ion, the latter of which attacks the cationic component $\ce{R+}$ to form $\ce{R-Cl}$. The reaction proceeds with retention of configuration, because the nucleophilic chloride ion originates from the leaving group $\ce{-OSOCl}$, and hence must attack from the same face as the leaving group left from.1

If the carbocation $\ce{R+}$ is allylic, then it is electrophilic at both the α-position (i.e. where the hydroxyl group started) and γ-position (i.e. two carbons away from the hydroxyl group). Therefore, following breakdown of the $\ce{-OSOCl}$ leaving group, the chloride can choose to attack either position. The attack at the γ-position is sometimes called the $\mathrm{S_Ni'}$ mechanism.

While the example above proceeds with extremely high selectivity for the rearranged product, this is not always the case. The above reaction was reported by Young in 1962,2 and carried out in dilute ether solution, which strongly favours the $\mathrm{S_Ni'}$ pathway. According to Gu and Zakarian in Comprehensive Organic Synthesis II (Chapter 6.16),3

Thionyl chloride has been used traditionally as the reagent of choice to ensure a high degree of selectivity in favor of allylic transposition, however, the outcome is often less predictable than desired. [...] The reaction, however, is often more complicated and dependent on reaction conditions and, to a greater extent, the structure of substrate. This is illustrated in the total synthesis of codeine reported in 2011. When allopseudocodeine was treated with thionyl chloride, a nearly equimolar mixture of regioisomeric products was produced.

References

Smith, M. B. March's Advanced Organic Chemistry: Reactions, Mechanisms, and Structure, 7th ed.; Wiley: Hoboken, NJ, 2013, pp 408–409.

Young, W. G. Unexpected rearrangement and lack of rearrangement in allylic systems. J. Chem. Educ. 1962, 39 (9), 455–460. DOI: 10.1021/ed039p455.

Gu, Z.; Zakarian, A. Functional Group Transformation via Allyl Rearrangement. In Comprehensive Organic Synthesis II; Knochel, P., Molander, G. A. Eds.; Elsevier, 2014; Vol. 6, pp 636–754. DOI: 10.1016/B978-0-08-097742-3.00624-8.


text:
Is half life for first order is depend on the stochiometric coefficient of reactant or product?

I took k=6.93x10^(-2) as given in ques and use the half life formula for 1st order kinetics and solve and finally get 10 min as a ans but right ans is 5 min and i m not able to get that..... Plse tell where i m doing mistake??

Please 1) Type out your question instead of posting a picture, 2) Include your thoughts and the entire method you used to tackle the problem... we aren't here to do your homework for you, and 3) Watch your sentence construction ;)

If the reaction has the form $aA \rightarrow B + C $ then the rate at which B and C are produced is $1/a$ of the rate at which A disappears, thus
$$ -\frac{1}{a}\frac{dA}{dt}= \frac{dB}{dt}= \frac{dC}{dt}=kA$$
On this basis the half-life should be $5$ minutes in your question.

Ok got it i try to find it by the rate of apperance of O2 as given in question.... Now this help me lot to tackle next prblm... Thnx


text:
Is there a consensus on whether amber can seal in air bubbles for millions of years?
I read about deriving the oxygen content in the atmosphere in the Cretaceous by analyzing the air trapped in bubbles in amber from that period, but I have also found papers disputing the study's methods and showing that air isn't actually trapped in the bubbles because it's capable of diffusing through the amber (the amber isn't a perfect sealant), and thus the air found in that amber isn't necessarily as ancient as the amber itself.
My problem is is that almost all the relevant papers are from the late 80's; all those I've found are pro- or con-, I don't have the background to tell who is right, I haven't found a source that describes an actual consensus the field would have come to since then, and I see the claim about the original study showing what oxygen levels were in the Cretaceous cited in various places, but I find it hard to tell if it's because the objections were dismissed or if it's because it's a zombie claim.
Some sources:

What seems like the first paper to make the claim:
Gas bubbles in fossil amber as possible indicators of the major gas composition of ancient air
This source seems to be more recent and find the same conclusions but it's hard to tell (looks like at least one person is the same as the 1988 source):
Air Bubbles, Amber, and Dinosaurs - USGS
The main disagreement:
Is the air in amber ancient?
Some back-and-forth:

Google Scholar: "is the air in amber ancient"
The relative contribution of adsorption to the overall sorption and transport of small molecules in amber

An unrelated recent paper that comes to a different conclusion on Cretaceous air content (based on amber as well but carbon isotope analysis thereof):
Stable carbon isotopes of C3 plant resins and ambers record changes in atmospheric oxygen since the Triassic

Does anyone know if a consensus was arrived at on this question, and/or if one side is clearly more right than the other?

There's another paper here - http://droyer.web.wesleyan.edu/Royer_2014_Treatise.pdf - the telling thing is that it dosen't mention amber at all in a recent review of the topic.

Berner and Landis wrote another interesting paper that refers to their work from 1988 in May 2018 that can be found here http://www.ajsonline.org/content/318/5/590.abstract DOI:10.2475/05.2018.06

@jojostack  Thank you very much ! On the one hand it's frustrating that again it's the same original authors addressing the question, but on the other hand the fact that they're still working to support their hypothesis in 2018, and (according to the abstract) came to a more tentative conclusion this time, suggests that the claim is at least not consensual in the field.

There is at least a fair degree of consensus (there will always be naysayers) that the results that were performed back in the 80s and 90s by Berner and Landis were false positives that could not be confirmed by further investigations. A good historical account of this can be found for instance in Nick Lane's book "Oxygen" (first published in 2002). Quoting from this book:

The overall consensus, which has now persisted for more than a decade,
  was that the air trapped in amber could not have been ancient [...]

That "could not have been" has an air of finality about it, and you can probably take this book's word as accurate. However, all of the amber in the world has not been examined, so I would say chances may be slim but not none that some will be found containing traces of an ancient atmosphere. Probably part on the answer depends on how old the amber is, and how it was preserved. A very different question is whether gases in the amber can be regarded as representative of the atmosphere of origin, or useful. Perhaps reactions with the amber (oxidation) during burial have altered it beyond recognition.


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Explain the selectivity in bromination of this compound with two substituted benzene rings

As I know, when $\ce{Br2/FeBr3}$ is added to acetophenone ($\ce{C6H5-CO-CH3}$), Br will substitute a H from the meta position which leads to 3-bromoacetophenone. So why doesn't a Br substitute a H from the right side benzene (connected to the carbonyl group)?

The right side has got a carbonyl as a substituent, so the ring is deactivated towards electrophilic substitution.

Because that is not the most electron rich aromatic. The ring at other end of the molecule is, because it is an aniline. The first, and probably the second, bromination will go there. If you keep throwing in Bromine you will eventually get bromination of the Acetophenone ring.

Aniline is one of the strongest activating groups . It makes the ring more stable by +I and +R effects which are predominant on the Ortho and Para positions. Bromine is slightly deactivating and Ortho-Para directing . So bromine on the right benzene ring will slightly deactivate the benzene ring . Moreover, the electronegative oxygen of CO might exert a deactivating effect on the ring as well. Hence, taking the highly activating nature of aniline into account and other factors I mentioned , bromine will attach at the Para position on the left benzene ( it's Para not Meta according to the picture).


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Is (2R,3S)-2,3-dichloropentane optically inactive?
Two confusing thing about (2R,3S)-2,3-dichloropentane:

The 2 stereocenters have opposite absolute configuration.
But there is no plane of symmetry.

So, is (2R,3S)-2,3-dichloropentane optically inactive?
(This is actually a follow-up question of:
Monochlorination of 2-chloropentane and possible enantiomer products)

ad 1: the two stereo centers are different also constitutionally - structurally. So what should be the reason for optical inactivity?

You're misinterpreting the answers for the other question. Both of those compounds are chiral and count towards the 5 mentioned in the other answer.

@zhe Ok, now I understand. I had misinconception I had the notion that initial chirality of 2-carbon could change in the cholrination.

To have an optically inactive meso compound with two chiral centers:
1)  the chiral centers must have opposite configuration
2)  the chiral centers must have the same set of fragments bonded to them
Among other things the 2 carbon has a methyl group bonded to it versus the 3 carbon having an ethyl group instead.  You do not satisfy (2), so you do not have the symmetry required for a meso compound, and the molecule is still chiral and optically active.