text:
Proof for Arrhenius equation
The Arrhenius equation states that:
$$k=A\mathrm e^{\frac{-E_\mathrm a}{R T}}$$
What is the mathematical proof for this equation?

Chemistry doesn't work like that. It is not about mathematical proofs at all. When the Arrhenius equation came about, it was an empirical observation. Sure, it is compatible with (read: can be kinda explained by) certain models. Does that constitute a mathematical proof? As for me, not really.

In science, theories, hypothesis and particular mathematical models cannot be proven to be right.
It is the principal limitation. They can be proven right only formally as mathematical construct, proving consistency. In relation to reality, there can be just confirmation or refutation of their agreement with experimental data.
The Arrhenius equation is based on the model that supposes:
There is a frequency A of geometrically  favourable molecular collisions, that might lead to their reaction. As simplification, it is  implied A does not depend on T. But we know it does, as molecules move faster at higher T.
The exponential factor is based on the Boltzmann distribution, predicting the probability molecules would have enough thermal energy to pass the energy barrier - the reaction activation energy.
The confirmation can be done by measuring dependency of reaction kinetics and equilibrium constants on temperature. The latter follows  van't Hoff equation,  that is derived from Arrhenius equation, applying it to both forward and reversed reaction.

Although it is not in any way a mathematical proof, a theoretical rationale for the general form of the empirical Arrhenius relationship is provided by transition state theory.
The equation that results from transition state theory is $$k=\frac{\kappa k_BT}{h}e^{\frac{\Delta S^\ddagger}{R}}e^{\frac{-\Delta H^\ddagger}{RT}}$$
where the $^\ddagger$ symbol represents properties of the transition state of the reaction, and $\kappa$ is a constant specific to the given reaction (related to, but not the same as the Arrhenius prefactor).
I've deliberately split the two exponential terms, since the second is similar (though not exactly equal) to the exponential term in the Arrhenius equation. The activation energy is a function of the enthalpy change of the transition state, but isn't exactly the same.
The first exponential shows that the Arrhenius pre-factor is a function of the entropy change of the transition state, but the pre-exponential terms explicitly include the temperature-dependence, which is ignored in the Arrhenius equation.


text:
Does an aryl compound with any functional group attached to it, undergo benzylic oxidation?
>
If you look at this compound, only the benzylic carbon on the top has a hydrogen attached to it, so how do we oxidize only the top part without touching the lower one (not sure if this product is right), and what happens to the $\ce{-CH2}$ in the middle?
Also, does benzylic oxidation with $\ce{KMnO4}$ happen only with alkyl chains (R)? What if it had a carbonyl instead (a ketone let's say) or any other organic functional group attached to a benzene ring, what would happen to it? For example, Molecule Y here:

Why did it get oxidized to phthalic acid if the lower benzylic carbon doesn't have any hydrogen atoms attached to it?

Would you like to check this out, also a duplicate of a previously answered question. Also would you mind editing the pictures you've posted to make them more clear?

@SirArthur7 Thanks, honestly Im not that satisfied with the answers on the other questions, I updated the picture, and it still is not clear what happens if what attached to benzene group is cyclic with 1 carbon without benzylic hydrogens  like the second picture or contains a carbonyl like the picture i added first clearer

" The abstraction of a benzylic hydrogen atom is consistent with the fact that arenes with no benzylic hydrogens, such as tert-butylbenzene, do not get oxidised. "
Then what about my examples? When 1 side has no benzylic hydrogens in the cycle and the other does, logically one side will turn into COOH, what about the other? like the second picture.

Yes sure, your question isn't a duplicate and IMHO the 1st link I gave you wasn't either. The answered question is largely general and there hasn't been much of discussions targeted at specific functional groups. On a different note, can you cite your image sources? It's quite helpful for others to analyse the issue accordingly. As for the post, let's hope someone answers it well :)

@MathewMahindaratne Thank you but this link has already been given to me and it did not answer my question. (Picture 2)

source: Stack Exchange

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sort: votes


text:
Can a boat conformer of cyclohexane be most stable?
It is well-known that the "chair" conformation of cyclohexane is the most stable of its conformers. But is there any cyclohexane derivative where other notable conformers (boat, half-chair) are the most stable?
My own intuition is that if there are 4 very bulky groups in the 1, 2, 4, 5 positions then the boat conformer would be preferred.

You don't want your cyclohexane to be a part of a larger polycyclic structure, I guess?

Yes, I don't want, say, a decaline derivative.

How about something where there is a carbonyl group on the ring, forcing a planar bonding geometry at that carbon; exclude that as well?

A simple force field optimisation should anwer this question.

How about norbonane molecule?

Tempting, but I suspect the OP won't go for anything multicyclic.

I judt went through my university notes as we had a big part on six membered cycles conformation. However I did not find examples in which that happens. Discussions were always on the axial vs equatorial aspect. Sugars included.

In my experience, the boat conformation is always a transition state.

TL;DR:
It is highly unlikely that there is a non-multicyclic derivative with a stable boat conformation.
It needs a lot of convincing and quite bulky groups to destabilise the chair conformation and stable another conformation like the twist-boat.
Conformations of cyclohexane
There are quite a few imaginable conformations,[1] but for cyclohexane itself there are only two stable: the chair and the twist-boat conformation. The other ones are transition states, which are given in the comments of the linked question.

A related question is Where is bifurcation situated on cyclohexane isomerisation PES? Unfortunately it is not answered. However, more insight is given by the articles of H. Rzepa on his blog and the pdf of some lecture notes.[2]
Below is an animation of the boat conformation's imaginary frequency, which will yield the twist-boat conformation upon relaxation.

Why is this actually the case? There is a very intuitive answer: There are eclipsed conformations where they do not have to be, see the image below. This stress can be easily relieved in this twist-boat conformation, in which all bonds are staggered (and the chair conformation, too, as it is the global minimum).

You can even do a little bit more, for example looking at non-covalent interactions. I have optimised all structures on the RI-BP86/def2-SVP level of theory using ORCA 4.2.0. Then I have used Multiwfn to analyse NCI with promolecular densities.[3,4] You will see that the isosurface extends further to the hydrogens in the boat and twist boat conformations. Note that this is more a visualisation of these properties than actual evidence for their existence. It should help in understanding, but it shouldn't be used as a general argument. There are plenty of sources explaining why staggered conformations are generally lower in energy than eclipsed ones.

Locking in a boat conformation
You have probably heard of a conformational anchor. Adding groups so that they will lock into a specific conformation. Like when you attach a tert-butyl group in one of the positions, it will rise the energy of the conformation when it is in axial position.
If you do not want to consider extended ring systems to lock the conformation into position, like the earlier mentioned norbornane and derivatives, you'll probably get structures to prefer a twist-boat conformation, but probably not one that will result in an actual boat conformation. I have tried some conformational sampling with CREST on the semiempirical GFN2-xTB level of theory and multiple attempts have failed so far.[5] In my attempts the closest I have gotten was with (1R,2S,3S,4R,5S,6r)‐1,2,3,4,5,6‐hexamethylcyclohexane, where the twist boat conformation is only about $\pu{1 kJ mol-1}$ higher in energy, see the images below (hydrogen omitted for clarity).

I encourage you to get a molecular viewer and look at the structures yourself.[6] You will quickly see how crowded the ring actually is and that there is not much space to put much larger groups anyway. I believe this illustrates the problem well.
However, I was able to push it further and have found a beast that is predominantly available in the twist-boat conformation. Therefore I have added tert-butyl moieties in 1, 2, 4, and 5 position just as you have proposed, (1R,2S,4R,5S)‐1,2,4,5‐tetra‐tert‐butylcyclohexane. The CREST program found two (very, very similar) conformers within a $\approx\pu{25 kcal mol-1}$ window.[7] There was no chair conformation in this window, and the second structure is barely in the given interval, therefore I will not include it here.[8]

Notes & References:

For example, see what Geoff demonstrates for pyranose sugars in Conformations of Cyclohexane. There is also a more general article about ring puckering: https://doi.org/10.1021/ja00839a011, pdf via the Internet Archive
The pdf via the Internet Archive
Johnson, E. R.; Keinan, S.; Mori-Sánchez, P.; Contreras-García, J.; Cohen, A. J.; Yang, W. Revealing Noncovalent Interactions. J. Am. Chem. Soc. 2010, 132 (18), 6498–6506. DOI: 10.1021/ja100936w.
I have previously written about it: How to identify hydrogen bonds and other non-covalent interactions from structure considerations?
(a) Mulliken Center for Theoretical Chemistry, xTB on Github, xTB documentation, CREST: Phys. Chem. Chem. Phys. 2020, xTB: J. Chem. Theory Comput. 2017, 13 (5), 1989–2009 & J. Chem. Theory Comput. 2019, 15 (3), 1652–1671
A good open source choice to view the following Cartesian coordinates is Avogadro.

  36
 GFN2-xTB       -37.96990166 (“chair”)
 C          1.2520828623       -0.5447977556        0.4005613406
 C          2.5529154246       -1.2427641817       -0.0035234170
 H          2.5670950259       -1.4609755375       -1.0683723566
 H          2.6812224740       -2.1742430344        0.5412168974
 H          3.4037282124       -0.6057253357        0.2302177479
 H          1.1934851832       -0.5794558732        1.4947912916
 C          1.2932369007        0.9399180068       -0.0046696921
 C          1.6543517514        1.1471873131       -1.4778364887
 H          1.5052577715        2.1882681334       -1.7560128481
 H          2.7000386472        0.9041963949       -1.6490857234
 H          1.0549606283        0.5311695599       -2.1388312879
 C          0.0004268408        1.6743272530        0.3958681428
 C          0.0003369080        1.9795490247        1.8972278766
 H         -0.8803447484        2.5595415091        2.1655307399
 H          0.0002805364        1.0701573015        2.4915141491
 H          0.8810286139        2.5594814678        2.1656262412
 H          0.0007645822        2.6443828331       -0.1162274411
 C         -1.2926832357        0.9406761004       -0.0050185229
 C         -1.6531427367        1.1479070715       -1.4783512638
 H         -2.6988310362        0.9052087273       -1.6499938197
 H         -1.0536713182        0.5316482470       -2.1390476329
 H         -1.5036207617        2.1889143513       -1.7565761588
 H         -2.1018819331        1.4053817317        0.5738089641
 C         -1.2524695951       -0.5439792518        0.4004522605
 C         -2.5537665456       -1.2411162270       -0.0035423382
 H         -2.5679998358       -1.4596918660       -1.0683148885
 H         -2.6827814315       -2.1722962125        0.5415232111
 H         -3.4041581021       -0.6033840583        0.2298854735
 C         -0.0003947478       -1.2538292818       -0.1423844973
 C         -0.0008645772       -2.7329645415        0.2464953252
 H          0.8741384569       -3.2373492283       -0.1525769397
 H         -0.0008947662       -2.8427074989        1.3295318583
 H         -0.8762127488       -3.2367530896       -0.1525580475
 H         -0.0003167563       -1.1995260135       -1.2346248660
 H         -1.1939183223       -0.5785009660        1.4946918825
 H          2.1024953812        1.4040515111        0.5745201353
  36
 GFN2-xTB       -37.96945643 (“boat”)
 C          1.3009843031        0.3904694151        0.6116442813
 C          2.7523842286        0.8697430480        0.5759961739
 H          3.1370164012        0.8924092382       -0.4400705597
 H          3.3851298683        0.2141081553        1.1682244589
 H          2.8236187015        1.8700221417        0.9985166065
 H          1.0589906677        0.1993477976        1.6630540690
 C          0.3343786187        1.4770082561        0.0930304054
 C          0.8370012219        2.1686754632       -1.1755196945
 H          0.0729240573        2.8439373660       -1.5550562687
 H          1.7313073547        2.7544698863       -0.9832626635
 H          1.0594535849        1.4416329460       -1.9531205354
 C         -1.0843113664        0.9157582793       -0.1622022415
 C         -2.1249907625        1.9785105445        0.1901580387
 H         -3.1130958026        1.6979927608       -0.1639981009
 H         -2.1696260162        2.1243458743        1.2679063910
 H         -1.8631471288        2.9288081963       -0.2704747138
 H         -1.1814251431        0.6916654706       -1.2293744660
 C         -1.3009787861       -0.3905116297        0.6116386064
 C         -2.7523814313       -0.8697957935        0.5761213628
 H         -3.1371030728       -0.8924849595       -0.4399103572
 H         -3.3850864608       -0.2141683561        1.1683995711
 H         -2.8235618371       -1.8700681103        0.9986656463
 H         -1.0588901236       -0.1993916529        1.6630242750
 C         -0.3343953584       -1.4770058647        0.0929158564
 C         -0.8370714193       -2.1686218804       -1.1756271655
 H         -0.0730166306       -2.8438887577       -1.5552044908
 H         -1.7314363652       -2.7543627913       -0.9834449686
 H         -1.0594386262       -1.4414807272       -1.9531506454
 C          1.0843324761       -0.9157815011       -0.1622407002
 C          2.1250455546       -1.9784641695        0.1902373202
 H          2.1696137891       -2.1242687584        1.2679947854
 H          1.8631584859       -2.9287608420       -0.2703581429
 H          3.1131889321       -1.6980202130       -0.1638895009
 H          1.1815268495       -0.6916894600       -1.2294091391
 H         -0.2613905266       -2.2395020738        0.8774919294
 H          0.2613017165        2.2394206185        0.8776918141

The default is actually $\pu{6 kcal mol-1}$, but I refuse to give in to these medieval units.
Cartesian coordinates of the lowest structure of (1R,2S,4R,5S)‐1,2,4,5‐tetra‐tert‐butylcyclohexane:

  66
 GFN2-xTB       -69.56788868 (“twist-boat”)
 C          1.2637535413       -0.4535845271       -0.7036170298
 H          0.8234262820       -0.3514280493       -1.7011134458
 C          0.2874611950       -1.2968906916        0.1262691467
 H          0.4862260613       -1.1072837877        1.1760079950
 H          0.4711152326       -2.3509422427       -0.0576548430
 C         -1.2138745760       -1.0160032197       -0.1668918408
 H         -1.5059514690       -1.6533453308       -1.0111946700
 C         -2.0580638654       -1.5542451449        1.0386873052
 C         -3.5291500453       -1.7879146105        0.6673596086
 H         -4.0389307694       -2.2795212021        1.4949415613
 H         -3.6040044082       -2.4413551590       -0.2001570459
 H         -4.0628013225       -0.8693626960        0.4623391177
 C         -1.5380651982       -2.9543377210        1.4269474062
 H         -1.4806005503       -3.6016258516        0.5545295073
 H         -2.2263828398       -3.4107111998        2.1359237707
 H         -0.5631185572       -2.9193311608        1.9023111302
 C         -1.9997742111       -0.6828458212        2.2953624607
 H         -2.6171623926       -1.1292449936        3.0737145663
 H         -2.3680382876        0.3190177054        2.1083809564
 H         -0.9863340848       -0.6190135645        2.6836308403
 C         -1.2637584056        0.4535847074       -0.7036134097
 H         -0.8234213043        0.3514344229       -1.7011060021
 C         -0.2874690577        1.2968787699        0.1262850454
 H         -0.4862254771        1.1072466416        1.1760203260
 H         -0.4711273811        2.3509334181       -0.0576150207
 C          1.2138643519        1.0160021804       -0.1668873426
 H          1.5059281496        1.6533459288       -1.0111931129
 C          2.0580613014        1.5542510848        1.0386835495
 C          3.5291313266        1.7879648142        0.6673219595
 H          4.0628193815        0.8694121253        0.4623879257
 H          4.0388969819        2.2796720426        1.4948537541
 H          3.6039479309        2.4413286270       -0.2002563066
 C          1.5380254724        2.9543167424        1.4269934992
 H          0.5630890040        2.9192624252        1.9023748186
 H          1.4805221407        3.6016312375        0.5545978702
 H          2.2263377642        3.4106933389        2.1359728252
 C          1.9998372135        0.6828176443        2.2953360365
 H          2.3681483571       -0.3190197145        2.1083145750
 H          0.9864074015        0.6189230717        2.6836201429
 H          2.6172202830        1.1292244475        3.0736878514
 C         -2.5734381625        1.2159695783       -1.0649400350
 C         -3.4459088573        0.3435906290       -1.9774479664
 H         -3.6675713251       -0.6208273126       -1.5422795084
 H         -4.3867433184        0.8479997186       -2.1893090423
 H         -2.9342730730        0.1735856167       -2.9232393678
 C         -2.1913064757        2.4425123599       -1.9212170334
 H         -1.7500553845        3.2379982242       -1.3271846647
 H         -3.0851290926        2.8470606980       -2.3920018369
 H         -1.4898086285        2.1738538773       -2.7080018465
 C         -3.3790378509        1.7599694390        0.1164846666
 H         -3.8515396128        0.9790436919        0.6998702684
 H         -2.7421665034        2.3465041110        0.7744832021
 H         -4.1670675305        2.4137947017       -0.2548140206
 C          2.5734370335       -1.2159683519       -1.0649370663
 C          3.3790279628       -1.7599706209        0.1164927971
 H          3.8516735927       -0.9790428294        0.6997637198
 H          2.7421180248       -2.3463565373        0.7745861394
 H          4.1669484076       -2.4139382233       -0.2547885352
 C          3.4459325455       -0.3435930502       -1.9774226323
 H          4.3867849824       -0.8479910147       -2.1892312850
 H          3.6675602100        0.6208308804       -1.5422534239
 H          2.9343365617       -0.1736062497       -2.9232383093
 C          2.1913055807       -2.4425004188       -1.9212296089
 H          3.0851285923       -2.8470461925       -2.3920160549
 H          1.7500476265       -3.2379922732       -1.3272105313
 H          1.4898131110       -2.1738289364       -2.7080150254

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text:
Shielding effects and atomic size
The atomic size on going from aluminum to gallium decreases because of poor shielding effect of the $(n-1)d$ electrons, but on going from copper to zinc, the size increases due to the same shielding effect. Why ?

Aluminum and gallium are on different periods. When an additional shell is (fictitiously) added to aluminum, the size is seen to decrease. This is, as correctly mentioned, due to the poor shielding offered by the $(n-1)d$ orbitals to the incoming electrons in the ultimate $p$ subshell.
However, between copper and zinc, the matter seems to be different–their electronic configurations are different.
While copper has a $4s^13d^{10}$ arrangement, zinc has a $4s^23d^{10}$. I would presume the fully filled spherical $4s$ orbital to considerably repel (also read as: shield) the d-electrons,  and hence the effect appears as a minute increase in size.
There can be deeper discourses, but I think this should suffice for now.


text:
If I dissolve 0.1M of CaCl2.2H2O in water, do I get a 0.1M CaCl2 solution?
If I dissolve $\pu{0.1 M}$ of $\ce{CaCl2.2H2O}$ in water, do I get a $\pu{0.1 M}$ $\ce{CaCl2}$ solution?

@Pastudent - you don't neglect the water of hydration. You should dissolve 0.1 moles of the hydrate in say 750 ml of water and then add water in a volumetric flask to get "exactly" 1.000 liter of solution total.

First, try to be clear. Nobody can dissolve 0.1 M of anything in water. 0.1 M is not an amount. It is a concentration. I think you should replace "M" by "mole".
Second, you must mention the amount of water used to dissolve your product, or the final volume. I think you should rewrite your demand in such a way : "If I dissolve 0.1 mole CaCl2·2H2O in enough water to get a final volume of one liter, do I get a 0.1 M CaCl2 solution?"
And here the answer would be "Yes !"

Kudos for pointing out the flaw in how the units were specified in the problem statement. Newbies need that sort of clarification to understand how all the terminology fits together.

At least the title was correct.

NO ! The title is not correct !! It is impossible to dissolve 0.1 M of anything in water, without specifying the volume of water.  The title and the question are exactly the same series of words !!


text:
What happens to the crystal structure of ice as it gets colder and colder
Not sure if this belongs on Chemistry or Physics stackexcahnge.
As water freezes into ice, it turns into a solid, and expands as it forms a crystal structure.  But what happens to this crystal structure as it continues to cool?  For example, if you take ice and cool it down to -50C, how does the crystal structure alter?  -100C?
I would assume that it contracts, similar to any other crystal as it cools.
I'm familiar with the different phases of ice, which require different amounts of pressure and such to create.  For my example above, I'm talking about taking a cup of water, and slowly freezing it over time with roughly no change to the pressure in the room.

Not all exotic phases of ice require pressure to create.

https://en.wikipedia.org/wiki/Ice#Phases

May have an answer here

Not all crystals contract monotonically as the temperature is lowered. Some start to expand again in some temperature range. Thermal expansion coefficients are not constants.

Well, the easiest way to address this question is take a look at the phase diagram for ice. Ordinary ice as we know consists of lots of hexagonal rings of water molecules. First off, as temperature is decreased well below $0~^\circ$C, there will be a slight contraction of the lattice due to decreased thermal contribution to vibrational frequencies. This will cause the average $\ce{OH}$ bond length to decrease which will cause a subsequent contraction of the average $\ce{OO}$ distance.

It is not obvious in any way that there should be another phase transition at low temperatures and pressure for ice Ih, but it turns out there is. One can understand why there is a phase transition by observing that in the low temperature limit, entropic effects will make a diminishing contribution to the free energy of any state, and the ground state will in the low temperature, low pressure limit will simply be the global minimum of the potential energy surface.
Now, experiments demonstrate that while the oxygen atoms in ice Ih are all arranged in these hexagonal rings, the hydrogen atoms are essentially distributed randomly subject to the conditions that two hydrogen atoms are covalently bound to each oxygen atom, each oxygen receives two hydrogen bonds, and no two hydrogen atoms point towards each other. These are known as the ice rules (or sometimes Bernal-Fowler rules). This, however, is not the most stable structure that one can imagine as the protons are disordered, and hence there is not optimal cooperativity in the direction of the hydrogen bonds.
For this reason, people thought there might be a low-temperature phase of ice which is proton-ordered. That is, you have the same lattice as in regular ice Ih, but the protons all point in the same direction when forming hydrogen bonds. This is easier to see, so I have put a picture of the lattice below.

This phase of ice is called ice XI. I think they are numbered based on the order they were discovered, which is very annoying.
In practice, even though this is essentially the ground state of bulk water, it is very uncommon to find in nature, and very difficult to synthesize in a lab. This is because at such low temperatures, the rates of proton rearrangements are very slow. These rearrangements occur by proton transfer in ice, or by rotating to rearrange which oxygen the hydrogen bonds are donated to. These processes individually have high barriers, and hence have low rates even at relatively high temperatures and pressures. So, going from ice Ih to ice XI occurs much more slowly than most phase transitions.
You can read about how ice XI is prepared in practice on the wikipedia page for ice XI, but basically you introduce defects which act as catalysts for the rearrangements for the hydrogen atoms.
TL;DR:
To be really pedantic, the answer to your question depends on how long you are willing to wait as you keep cooling down the block of ice. That is, if you have an ice cube and cool to e.g. 5K, then nothing will happen besides a slight contraction of the lattice. If you keep it at that temperature for, literally something like 1,000 years (evidence from Antarctic ice cores), then your ice cube will look exactly the same, but microscopically, the protons will be organized much more neatly. How pleasant.

There are two structures of ice that are found in that temperature range, hexagonal (common on earth) and cubic (less common on earth). To read more about those, see e.g. https://www.pnas.org/content/110/29/11757.
In the absence of a phase change, the density increases with decreasing temperature (for comparison, the density of water is on the order of 1000 g/L):

Data from: https://www.engineeringtoolbox.com/ice-thermal-properties-d_576.html


text:
Reaction of Fe2+ and S2- to form FeS

Consider this reaction:
$$\ce{Fe^{2+} (aq) + S^{2-} (aq) <=> FeS (s)}$$
The equilibrium constant for the reaction is $1.6 \times 10^{17}$ at $\pu{298K}$. If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^2+}$ and $\pu{0.2 M}$ of $\ce{S^2-}$ are mixed, the concentration of $\ce{Fe^2+}$ at equilibrium is found to be $A \times 10^{-17}$. Find $A$.

Let $x$ amount of $\ce{Fe^2+}$ is converted to $\ce{FeS}$. Then
$$K = \displaystyle {\frac{1}{(0.06 - x)(0.2 - x)}}$$ Which then leads to the quadratic equation
$$ x^2 - 0.26x + 0.012 - 6.25 \times 10^{-18} $$
Solving this the $A$ I got is not the answer.
Answer is $8.93$

Related: Does the concentration of a solution change when you add another solution to it? http://chemistry.stackexchange.com/questions/127510

This is very poorly worded question, and really not just an equlibrium question. This is actually a solubility product question. Consider the given reaction:
$$\ce{Fe^{2+} (aq) + S^{2-} (aq) <=> FeS (s)} \tag{1}$$
If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^2+}$ and $\pu{0.2 M}$ of $\ce{S^2-}$ are mixed, since the equilibrium constant for the reaction is $1.6 \times 10^{17}$ at $\pu{298K}$, we can assume that all $\ce{Fe^2+}$ are reacted because it is the limiting reagent and would give $\pu{0.06 M}\times V$ amount of solid $\ce{FeS}$ (where $V$ is the volume of $\ce{Fe^2+}$).
However, the author of the question ignore that $\ce{FeS}$ would gives solubility product of $\ce{Fe^2+}$ and $\ce{S^2-}$.
Since most sulfides of transition metals are insoluble in water, following equilibrium would establish after the original reaction:
$$\ce{FeS (s)  <=> F^2+ (aq) + S^2- (aq)}\text{;} \,\,\, K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]=4.9 \times 10^{-18} \tag{2}$$
Here, $K_\mathrm{sp}$ has not be provided, so I found it in here at $\pu{298K}$ ($\pu{25 ^\circ C}$). The concentration of $\ce{Fe^2+}$ at this equilibrium is given as $A \times 10^{-17}$. We can also calculate the concentration of $\ce{S^2-}$ from equation $(1)$ that $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ (since equal volumes has made final volume dobuled and hence concentration of remaining ions are halves). Since this value is much larger than the $[\ce{S^2-}]$ generated by dissociation of $\ce{FeS}$ ($\approx \sqrt{4.9 \times 10^{-18}}$), we can assume $\pu{0.07 M$ is the same concentration of $\ce{S^2-}$ at this equilibrium. Accordingly:
$$K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]= A \times 10^{-17} \times 0.07 = 4.9 \times 10^{-18}$$
$$\therefore \; A = \frac{4.9 \times 10^{-18}}{1.0 \times 10^{-17} \times 0.07}  = 7.0 $$
Note that this final answer is highly depends on the $K_\mathrm{sp}$ value. Thus, the deviation of this answer from your given answer is based on what value should have been given at the first place.
Suppose we ignore the solubility product as author may be suggesting: Then, once the original reaction is completed, significant amount of $\ce{S^2-}$ would remain unreacted since it is in excess (compared to $[\ce{Fe^2+}]$). That amount would be $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ as indicated above. Since at this point, $[\ce{Fe^2+}]$ is $A \times 10^{-17}$, the equilibrium constant is set up as follows as OP has done:
$$K = \frac{1}{[\ce{Fe^2+}][\ce{S^2-}]}=\frac{1}{[A \times 10^{-17}][0.07]}= 1.6 \times 10^{17}$$
$$\therefore \; A = \frac{1}{1.0 \times 0.07 \times 10^{-17}\times 1.6 \times 10^{17}}= 8.93$$
Although, this setup gives the expected answer, it is not the correct way to solve this (at least to my knowledge).

The $\mathrm{K}\mathrm{sp}$ was provided in the problem statement, just in an unusual way. $$\mathrm{K}\mathrm{sp} = \dfrac{1}{1.6\cdot10^{17}} =6.25\cdot10^{-18}$$

I think this is the correct way to solve this.  Published solubility constants vary; the OP was supposed to just use the given one (provided in reciprocal form) and demonstrate the larger concepts (e.g. dilution effects) with it.

I guess I should explain more clearly. For the given chemical reaction in equation 1, the activity of the solid is defined as 1. It doesn't matter at equilibrium if there is 1 microgram of FeS(s) or 1 kg of FeS(s) in equilibrium with the aqueous solution, the amount of FeS dissolved in the solution will be the same.

Consider this reaction:
  $$\ce{Fe^{2+} (aq) + S^{2-} (aq) <=> FeS (s)}\tag{1}$$
  The equilibrium constant for the reaction is $1.6 \cdot 10^{17}$ at $\pu{298K}$.

This is nothing but a weird statement of the $\mathrm{K}_\mathrm{sp}$ for FeS. Instead of concentrations let's use activities, $a$, for the species which is formally correct.
$$\mathrm{K}_\mathrm{eq} = \dfrac{a_\mathrm{FeS}}{a_\ce{Fe^{+2}}a_\ce{S^{-2}}}\tag{2}$$
Since FeS is a solid, its activity, in an aqueous solution, is 1 by definition. But $\mathrm{K}_\mathrm{sp} = a_\ce{Fe^{+2}}a_\ce{S^{-2}}$ so:
$$\mathrm{K}_\mathrm{eq} = \dfrac{1}{\mathrm{K}_\mathrm{sp}}\tag{3a}$$
$$\mathrm{-- or --}$$
$$\mathrm{K}_\mathrm{sp} = \dfrac{1}{\mathrm{K}_\mathrm{eq}} = 6.25\cdot10^{-18}\tag{3b}$$
Now, let's make the typical substitutions:
$$a_\ce{Fe^{+2}} \approx [\ce{Fe^{+2}}]\tag{4}$$
$$a_\ce{S^{-2}} \approx [\ce{S^{-2}}]\tag{5}$$
Therefore:
$$\mathrm{K}_\mathrm{sp} =\ce{[Fe^{+2}][S^{-2}]}\tag{6}$$

If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^{+2}}$ and $\pu{0.2 M}$ of $\ce{S^{-2}}$ are mixed, the concentration of $\ce{Fe^{+2}}$ at equilibrium is found to be $A \times 10^{-17}$. Find $A$.

Since the volume of the mixed solution is twice the volume of either of the solutions of the individual ions, the concentrations will be cut in half. So nominally the final solution would have $\pu{0.03 M}$ of $\ce{Fe^{+2}}$ and $\pu{0.1 M}$ of $\ce{S^{-2}}$.
We can assume that the FeS precipitates quantitatively. Obviously the $\ce{S^{-2}}$ is $\pu{0.07 M}$ in excess, so we can assume that is the final concentration of the  $\ce{S^{-2}}$. Thus
$$\ce{[Fe^{+2}]_\mathrm{final}} = \dfrac{\mathrm{K}_\mathrm{sp}}{[S^{-2}]_\mathrm{final}}= \dfrac{6.25\cdot10^{-18}}{0.07}= 8.9\cdot10^{-17}$$

RANT ON I hate these kinds of stupid problems that use significant figures incorrectly. If the answer is supposed to be given to three significant figures then the $\mathrm{K}_\mathrm{sp}$, $\ce{[Fe^{+2}]}$ and $\ce{[S^{-2}]}$ should be given to three significant figures too. RANT OFF

When you mix the solutions, the iron that was in solution A is now diluted over both A and B.  Same is true for the sulfide coming from B.  With the A and B components being equal in volume you then have to divide your concentrations in half, thus
$K=\dfrac{1}{(0.03-x)(0.1-x)}=\dfrac{1}{(y)(0.07+y)}$
where $y=0.03-x$ is the iron concentration you want.  Conclude that
$y(y+0.07)=1/K$
and simplify by noting that $y$ will be much less than $0.07$.


text:
Reorganization during ionisation for d block elements
This is a quote from my textbook:

The irregular trend in the first ionisation enthalpy of 3d lmetals,can be accounted for by considering that the removal of one electron alters the relative energies of 4s and 3d orbitals. So the unipositive ions have dn configuration with no 4s electrons.There is thus a reorganization energy accompanying ionisation with some gains in exchange energy as number of electrons increases and from transference of s electrons into d orbitals.However the value of Cr is lower because of the absence of any change in d configuration.

What I basically understand from this is that during first ionisation, electron is lost from $\mathrm{4s}$ orbital because it have higher energy and forms uni-positive ion and the remaining one electron is transferred back into $\mathrm{3d}$ lower energy orbital as energy gap between $\mathrm{3d}$ and $\mathrm{4s}$ lowers. This transition of electrons into lower orbital gives a net decrease in ionisation enthalpy (where energy is released as reorganization energy). But I don't understand what they mentioned about $\ce{Cr}$.
https://www.chemguide.co.uk/atoms/properties/3d4sproblem.html
In this site when they explain about Vanadium electronic configuration it has been mentioned that Vanadium in its unipositive ion state have 4 electrons in $\mathrm{3d}$ rather than 3 electrons in $\mathrm{3d}$ and 1 electron in $\mathrm{4s}$ which supports the idea that electrons is transferred from $\mathrm{4s}$ to $\mathrm{3d}$.
Is it true that during ionisation there is a transference of electron from $\mathrm{4s}$ to $\mathrm{3d}$ orbital?Why the text mention reorganization and exchange enthalpy when talking about ionisation?
Text source:NCERT textbook ($\mathrm{d}$- and $\mathrm{f}$-blocks, page 215)

This doesn't answer your whole question, but I think the point with Cr is that it's configuration is already 3d5 4s1 (for reasons explained in the Chemguide article you linked), so when it's ionised, it has no electrons in its 4s orbital and there is no need to reorganise.
Compare this with, say, vanadium, which is 3d3 4s2 when neutral. If an electron is removed from the 4s orbital, there's still one electron left in the 4s which is then transferred to the 3d, and there is reorganisation.

Is there some energy release during reorganization

As your textbook notes, the relative energy of atomic orbitals is not fixed as you start ionizing your atom. As you ionize, all orbitals drop in energy. However, higher angular momentum orbitals drop more than lower ones, and specifically in your case, the $\mathrm{3d}$ orbitals drop below the $\mathrm{4s}$ orbitals. So consider vanadium, the atom has a $\mathrm{[Ar]4s^23d^3}$ configuration, but after you ionize it, the $\mathrm{3d}$ orbitals drop below the $\mathrm{4s}$ ones and you get $\mathrm{[Ar]3d^4}$ as the electron configuration of the $\ce{V+}$ ion. So your measured ionization energy includes both the energy it takes to remove an electron from $\ce{V}$ and also some reorganization energy. You know there is some reorganization energy since two electrons (both $\mathrm{4s}$ ones) have "moved" but you only removed one.
Now, what about $\ce{Cr}$? Recall that $\ce{Cr}$ is a bit special and has a $\mathrm{[Ar]3d^54s^1}$ configuration when neutral. After we remove an electron (in this case, a $\mathrm{4s}$ one), we have an $\mathrm{[Ar]3d^5}$ configuration for $\ce{Cr+}$. Note that no electrons had to move around besides the one we removed. So there is no reorganization energy in the case of $\ce{Cr}$, as your textbook notes.

I read from above mentioned site that 3d orbital have less energy than 4s orbital before ionisation,thus ionisation occur from 4s rather than 3d.And since  reorganization provides some stability(due to electron moving from higher to lower orbital)...Cr doesn't get to have this stability so shouldn't that mean a higher ionisation enthalpy for Cr.

Cr is mentioned as in Cr d5 configuration exists which is stable so no transfer of electrons from s to d orbitals occur. That's why 1st ionization enthalpy is less due to absence of reorganisation and exchange energy


text:
Why Zn has highest ionisation enthalpy in 3d series?
Zn which has the highest ionisation enthalpy in 3d series.The reason given in my textbook is:

The value of zinc is higher because it represent ionisation from 4s level.

This is not correct because for every 3d element ionisation is from 4s orbital. The explanation I found for this is from this site is:
https://www.chemguide.co.uk/atoms/properties/ies.html

Increase in nuclear charge outweigh the electron electron repulsion of 4s electrons rendering Zn an higher ionisation enthalpy than Cu.

This seems like a probable explanation but if Zn has higher ionisation enthalpy than Cu due to its increased nuclear charge.Why can't Zn have smaller atomic radius than Cu due to its increased nuclear charge which can decrease the size of the atom. But experimentally it has been found that Zn is bigger in size than Cu.
So why does Zn has the highest ionisation enthalpy in 3d series?
TEXT SOURCE:NCERT textbook (chapter:d and f block)

Where is the atomic radius of Zn rendered larger than that of Cu?  In Wikipedia they are the same and so are a few preceding elements to boot.

In my textbook it's given that Zn has a higher atomic radii than Cu

The reference for the textbook being ...

The real reason for this happening is because the orbitals self overlap in the d-block; refer to the band theory.
http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/Metalle.htm
and check out this question that I've asked
https://chemistry.stackexchange.com/questions/123802/why-does-copper-have-a-high-enthalpy-of-atomisation-when-it-has-a-fully-filled-d

I think a more suitable question would be "is atomic radii and ionization independent properties"

There is band structure formed in the metallic lattice  in the crystals of d and f block elements due to partially filled orbitals. But as you move along the periodic table to 'Zn' , you will find that the orbitals are completely filled hence there is no possibility for inter-orbital mixing. Another thing worth noting is that due to the filled d-orbital, zn and copper behaves more like s-block rather than d-block elements.
"The filled d-shells in these elements contribute little to interatomic interactions, which are dominated by the s-electrons through metallic bonds. Unlike metals with incomplete d-shells, metallic bonds in copper are lacking a covalent character and are relatively weak. "
An alternate explanation can be on the metallic lattice:
Copper has a face centered cubic structure and zinc has a hexagonal close packed structure
reference : wikipedia

why the downvote?

In the Zinc atom, the levels $4s$ and $3d$ are completely occupied. This increases the stability of the atom. And electrons need more energy to be removed.


text:
Order of Reaction differs in equation and linearisation
I have given the following equation: $A + 2B \rightarrow C$
$ v = –k\cdot [A]\cdot[B]^2$
gives me the orders 1 for A and 2 for B. That results in a total order of the reaction of 3.
So far so good. But I have given the following plot for linearization:

But this linearization with $\frac{1}{[A]}$ defines a 2nd Order for A, not 1st.
The question is what does this mean for the given equation? And in which sequence could steps of the reaction occur?

As this appears to be homework the rules here dictate that should add some thoughts about what is going on and what you have done to understand the data, otherwise the question may be closed without an answer.

This is an extremely  interesting problem. Could you tell us more about this reaction ? What are the structure of A and B ? In which concentration ? How do you measure [A] ? Have you ever measured [B] !

@Maurice Unfortunately not, that's all the Information I have. But an idea: As the plot makes it clear, which order A must be, perhaps the reaction occurs as $2A + 4B \rightarrow 2C$. That could indicate that the given equation is not an elementary reaction and can be split up in (for example) following elementary reactions: $2B + A \rightarrow B_2A $ and then $2B_2A \rightarrow 2C$ . Then we would have one reaction with order 3 and one with order 1.  stoichiometricly it would be the same as the given equation, but it couldn't happen that way. What do you think?


text:
What does it mean to 'fuse' ores and why do we use alkali metals in the process?
For the extraction of chromium, fusion of chromite ore with sodium or potassium carbonate in excess of air is done. What exactly happens in this 'fusion' process?
According to my textbook (class XII chemistry, NCERT): d- and f-block element, extraction of chromite ore
According to the dictionary: To melt or fuse (ores) in order to separate the metallic constituents.
Now,  while this makes sense and is sort of (implied) by my textbook? how exactly is this process done? I want to learn the technicalities related to it! Further I've heard we use alkali metals in fusion, why do we use them ?

Why "excess of air"? Is the iron (II) supposed to oxidise to iron(III)? You definitely need an excess of carbonate.

Chromeisenstein, the bane of first year chemistry students who still need a few credits out of the last task in their quantitative inorganic analytics lab. ;)

Check my previous comment. The first year analytics lab is where most chemists see this reaction for the first and last time. Where did it turn up for you?

it came up for me in my highschool textbook : ncert chemistry class 12 textbook. I asked the question because I could not find any answer to this directly in my textbook nor the internet from a direct search.

That´s bad if a school textbook just throws a few facts but does not explain them.

"To fuse" is another word for "to melt" (e.g. "heat of fusion"). Specifically, if you say you want to fuse two materials, you melt them in the hope that they will mix.
In this case, you melt the carbonate, and hope that the chromite will dissolve in it. Because e.g. $\ce{Cr2O3}$ has a melting point of $\pu{2435 °C}$, chromite ($\ce{Fe(II)Cr2O4}$) typically around $\pu{2140 °C}$. That is out of reach with a common Bunsen burner, and even if you did reach it, it would still be the same insoluble mess after cooling down.
When your chromite is dissolved in alkali carbonate, the mixture easily dissolves in $\ce{HCl}$, and you can properly run your wet analytics.
Soldering for example is a very similar process: Your soldering iron cannot nearly melt copper, but the molten tin/lead/etc. fuses (partly dissolves/melts) with the copper surface, and you get a continous metallic connection.

RE: "if you say you want to fuse two materials, you melt them" -- You're really only melting the alkali carbonate which then chemically reacts with the ore.

@MaxW I didn´t want to split too many hairs in the first two sentences. ;)

As a comment of Karl's message, I would like to add that chromite $\ce{FeCr_2O_4}$ is the most important Chromium mineral. When mixed with $8$ times its weight of sodium carbonate, and heated to high temperature, $\ce{Na_2CO_3}$ melts at $850°$C and reacts with chromite and air according to
$\ce{8 Na_2CO_3  + 4 FeCr_2O_4  +  7 O_2  ->  8 Na_2CrO_4  +  2 Fe_2O_3  +  8 CO_2 }$.
In this equation, both $\ce{Fe}$ and $\ce{Cr}$ atoms are oxidized. And this produces a mixture of soluble sodium chromate and insoluble iron oxide, which is separated from the chromate by dissolution into water.

I find this oxidation scheme a bit suspect. Have you got a source for it? Where would the oxygen come from?

The oxygen would be from the "excess of air" as mentioned in the OP

@Waylander How do I add an "excess of air" to a molten blob of carbonate?

Bubble compressed air through it with a needle

@Waylander And contaminate your sample with the burnt off needle tip? Not sure if I´d do that.

Only if you buy cheap needles

In the industry, they add sodium or potassium nitrate to $Na_2CO_3$  to produce the necessary Oxygen. The reaction is :  $$26 Na_2CO3 + 20 FeCr_2O_4 + 28 NaNO_3 --> 40 Na_2CrO_4 + 10 Fe_2O_3 + 26 CO_2 + 14 N_2$$

I have to read up again on industrial inorganic chemistry one of these days. One of our profs actually gave a lecture on that, but it´s a long time, and there was only so much she could cover in one hour per week in one semester.


text:
Rate determining step for parallel reaction?
I have always read that the rate determining step is the slowest step in a reaction (source: Wikipedia).

In chemical kinetics, the overall rate of a reaction is often approximately determined by the slowest step, known as the rate-determining step (RDS) or rate-limiting step.

However on this website (source), it says that the rate determining step for a parallel reaction is the fastest step. I was quite confused about this statement.
For a parallel reaction $\ce{B <- A -> C}$, where $\ce{A -> B}$ has rate constant $k_1$ and $\ce{A -> C}$ has rate constant $k_2$, the equations (from solving the first-order linear differential equations for A, B, and C, give: $A = A_oe^{-(k_1+k_2)t}$,  $B = \frac{k_1*A_o}{k_1+k_2}*(1-e^{-(k_1+k_2)*t})$, and $C = \frac{k_2*A_o}{k_1+k_2}*(1-e^{-(k_1+k_2)*t})$.
But I do not know how to proceed from there.

Let's suppose that $\ce{A -> C}$ is the fast step in the parallel reaction scheme.  This implies that $k_2 \gg k_1$.  Thus, $k_1 + k_2 \approx k_2$, it is therefore (approximately) true to write the rate of consumption of $\ce{A}$ as $A_0 e^{-k_2t}$.  This is what is meant by the fast reaction "controlling" rate for parallel reactions.


text:
Beer's law: UV-Vis absorbance values exceeding 2
In classical analytical chemistry, absorbance values in Beer's law can theoretically range from 0 to infinity.
As general rule of thumb neither high, nor very low absorbance are used for constructing calibration curves. The reason for avoiding high absorbance is that when absorbance is 2, only 1% light reaches the detector and with 3, only 0.1% light reaches the detector. Analytical chemists were taught to avoid absorbance > 1.5.  Low absorbance values are avoided because it hard to distinguish the light beam with and without the cuvet in place. The question then arises how much is the real absorbance and how much is contributed by noise.
Nowadays (= within the last decade), I am seeing UV-Vis spectrophotometers have absorbance ranges from 0 to 3, and some have even more. One day a PhD student showed an absorbance > 5 for a spectra, apparently without even thinking twice on the meaning of absorbance. Similarly, researchers showed an absorbance spectrum with the maximum at absorbance of 3.
My question is this a fundamental flaw in Beer's law that high absorbance values are erroneous or it is limitation in detector technology and the intensity of light sources? Please ignore the chemical reasons for deviations in Beer's law.
Imagine, if we had very high intensity light source, such as xenon arc lamp, instead of a typical deuterium or a tungsten lamp, would we get less error in the absorbance measurements of concentrated solutions? Absorbance is a ratio technique, but the number of photons reaching the detector will be high, even with conc. solutions, with intense light sources.
Can any spectroscopist shed some light on this issue (no pun intended) whether high intensity sources would beneficial in UV-Vis spectrophotometry of concentrated solutions?

High intensity light sources, per se, do not solve the problem. Some expensive instruments can get to 5 or 6. But Tom O'Haver has a paper (I will send it if I can find it again!) where he discusses curve fitting transmission profiles and this effectively gets you to much higher absorbances: 100 or higher! Of course, there are no photons at the center of the profile, but the width carries information. But this is NOT a panacea. If I cannot find the paper, just ask Tom: it is very nice and I have computer sims demonstrating it.

Okay, great to know. Do you recall the title of the paper. Absorbance of 100! This is amazing. He spent a significant portion of his life on xenon light source.

Of course, this is really for atomic profiles! No way UV-vis with solutions could use this idea. I will send along the paper's title ASAP!

High intensity UV/Vis light sources are terrible for your analyte, because it gets heated and/or destroyed.

@Karl, of course lasers are to be avoided, but xenon arc lamps are common in spectrofluorometers and at least commercial UV-Vis. The trick is not to irradiate the sample but have brief pulses.

Beer's law limitation can be more easily understood if the coefficient $\epsilon$ is considered as a cross-section effect. Let's start with a broad beam or a bundle of rays having a section $A$ (in $m^2$). Its measured intensity is $I_o$. Now suppose there exist plenty of tiny obstacles suspended in the bundle of rays, and that each such particle behave like a dark disk perpendicular to the bean, with a surface or a cross-section $\sigma$. $\sigma$/A is the fraction of $I_o$ that is blocked off by one such particule. If there are $n$ particules in the whole cylindrical bundle, the fraction of $I_o$ that is captured is $\Delta$$I$/$I_o$ = $n·\sigma$/$A$.
Now let l be the length of the bundle where the particules are present. The volume of this bundle filled with dark particles is $V=A·l$. So the fraction of $I_o$ absorbed by the particules is $\Delta$$I$/$I_o$ = $n·\sigma ·l/A·l$ = $(n/V)·\sigma ·l$ = $c·\sigma ·l$, if $c = n/V$ is the concentration of the particules in the volume crossed by the beam.
Going to the limit, this expression becomes : $dI/I_o$ = - $\sigma ·l·dc$.
Integration gives : $ln(I/I_o)$ = - $\sigma·c·l$
This last expression is similar to Beer's law :  $log(I/I_o) = - \epsilon ·c'·l$. I have written $c'$ because in the beginning of my text $c$ is the number of particules per volume unit. And the concentration in Beer's law is in mol/L. $c' = c/N_o$, where $N_o$ is Avogadro's number.
In the end we obtain the final formula : $\epsilon = N_o·\sigma /2.3$
Now if a particule is a disk of $0.2$ $nm$ diameter (dimension of a small molecule), the corresponding $\epsilon$ is about $10^5 L mol^{-1} cm^{-1}$. This is the order of magnitude of $\epsilon$ for the usual dye.
But if $\epsilon$ = $N_o·\sigma /2.3  = 10'400 m^2/mol$. And this is the point where I would like to go today. It means that if one mole of the dye molecules were all to be touched by the beam, they would have to cover a surface of $10'400 m^2$. This is as large as a football ground. It also means that when the concentration becomes too high, Beer's law does not apply any more. The majority of the molecules of the dye are hidden behind others.
You may even calculate that when $c > 1/ (\epsilon · cm), $ then $  : I/I_o < 0.01$, the absorbance is $> 2$, and this is the limit of application of Beer's law, because most of the molecules are hidden behind some other ones at higher values of $c$.
Sorry for the length of the demonstration.

Maurce, in a real spectrophotometer a very small portion of the solution is irradiated. The solid angle so formed encloses a relatively small number of molecules.

Maurice, Interesting perspective on cross-sections. Do you have a reference which shows this limit in terms of cross sections.

I think that explaining absorption as if molecules are discs shading others behind them is a misconception. The important property is the transition dipole, (think of a vector) and this takes various orientations in space, depending on molecular orientation, which determines the probability of absorption along with its size. This can vary greatly, i.e. the extinction coefficient varies by many orders of magnitude between different types of transitions. The experimental limit on the Beer-lambert law is only signal to noise and not for any fundamental reason.


text:
Rate of reaction involving different stoichiometric coefficients
Suppose I have a reaction (Say it is the Haber process):
$$\ce{N2 + 3H2 <=> 2NH3}$$
If we have to calculate the average rate of reaction of the above we follow the reasoning the rate of consumption of $\ce{N2}$ is one- third that of $\ce{H2}$ and half that of production of $\ce{NH3}$. We go ahead and divide the change in concentration of $\ce{N2}$ or $\ce{H2}$ or $\ce{NH3}$ by their stoichiometric coefficient. My doubt is why rate of reaction is taken as the rate of consumption of $\ce{N2}$? Why can't it be the rate of production of $\ce{NH3}$? Why can't we say that the rate of reaction is the rate of production of $\ce{NH3}$ or half the rate of consumption of $\ce{N2}$ or one and half times that of rate of consumption of $\ce{H2}$?

See wikipedia's article on reaction rate

If you added a space after each punctuation, and perhaps a line break somewhere, this would become a lot more readable. Btw. if you put chemical formulas into curly braces with backslash "ce" in front, they look a lot better.

The Haber process is reversible (corrected). The last phrase of question is "...one and half times that of rate of consumption of $\ce{H2}$" (not $\ce{NH3}$, whch is also corrected). If that's not what you have intended, you may change the corrections back.

The rate of this reaction is the same as the rate of N2 consumption, because N2 happens to have a stochiometric factor of 1.

See also https://chemistry.stackexchange.com/questions/38167/do-coefficients-matter-in-rate-law and https://chemistry.stackexchange.com/questions/32304/law-of-mass-action

My doubt is why rate of reaction is taken as the rate of consumption of N2?

It's done that way because rates have units, and most people don't pay enough attention to the units.  Usually in chemistry, rates are units of moles per unit time (per unit volume or per unit catalyst mass or per unit surface area or something).  But you need to know moles of what?
By writing the reaction as $\ce{N2 + 3 H2 <=> 2 NH3}$, you are implicitly defining that a mole of "reaction" corresponds to one mole of $\ce{N2}$, two moles of $\ce{NH3}$, and three moles of $\ce{H2}$.  If you wrote the reaction as $\ce{\frac{1}{2}N2 + \frac{3}{2}H2 <=> NH3}$, then one mole of this new reation would correspond to one mole of $\ce{NH3}$, but only half a mole of $\ce{N2}$, etc.

Why can't it be the rate of production of NH3? Why can't we say that the rate of reaction is the rate of production of NH3 or half the rate of consumption of N2 or one and half times that of rate of consumption of H2?

It could be any of those things.  The choice is up to you.  You just have to be consistent.  You might this related chem.SE question and answer helpful.
When you say "the rate", you have to specify what rate you are referring to.  And if you talk about rates "of the reaction", the agreed upon reference is that you are counting moles of species that have stoichiometric coefficients of 1.

"The choice is up to you" -1 // "The rate" has a definition just as assigning coefficients to the chemical equation does. The equation $\ce{\frac{1}{2}N2 + \frac{3}{2}H2 <=> NH3}$ has not assigned the coefficients "properly."

What I try to do when I answer questions is to explain things logically and to dispel confusion.  Your strategy might be more focused on what you feel is "proper".  Multiple strategies for answering could be effective, so I urge you to consider writing your own.

It depends how we define the rate,so we have different values of $k$


text:
Question about a line from Rosenfeld's paper on optical rotation
There are a lot of papers on optical rotation which cite Rosenfeld's (German) 1928 paper "Quantum mechanical theory of natural optical rotation..." [Quantenmechanische Theorie der naturlichen optischen
Aktivitat von Flussigkeiten und Gasen, Zeitschrift für Physik, Volume 52, Issue 3-4, pp. 161-174] as background. Here is a paid link--
Apologies that I cannot link to the paper. I have a paper copy and can't find a paywall-free version.
The symbols below are: $\lambda$--wavelength; $c$ is the speed of light; $\nu$ is frequency; $n$ is the index of refraction; $\rho$ is a quantity that does not change between equations (42) and (43).
The last two equations in that paper (42), (43), are a conclusion about an angle of rotation $\theta:$
$$\Delta n = \frac {2\pi ~\rho}{\lambda~ n} \tag{42}$$
and
$$\theta = \frac {2\pi \nu~\cdot \Delta n}{c~\cdot ~ 2} = \frac{2 \pi^2\rho}{\lambda^2} \tag{43}$$
My question is pretty dumb. If we insert the expression for $\Delta n$ into the equality in (43) we get:
$$\frac{2\pi\nu}{c}\cdot \frac{2\pi \rho}{2\lambda~ n} = \frac{2\pi^2\rho}{\lambda^2}\to \frac{\nu}{c~n}=\frac{1}{\lambda}\to\lambda =\frac{cn}{\nu} $$
We know already--and Rosenfeld mentions this a few lines up and at eq. (21)--that
$$\lambda = \frac{c}{\nu~n}$$
which forces $n$ to be $1$. In general $n \neq 1$.
Can someone tell me what, if anything, has gone awry here?
Edit: After looking at several versions of Rosenfeld's equation I see $\theta$ is proportional to the trace of the rotation matrix...the constant of prop. can take various forms and is not the focus of this approach.

In the medium, both the speed of light in vacuo and the wavelength are smaller by a factor of 1/n, but the frequency, nu, is independent of n. Does this help in your substitution?

I feel it must be a typo. It is not uncommon to find typos in old papers and of course in new papers. Can you find another reference on circular birefringence and see how delta n is defined?

@M.Farooq: Yes this could be nothing more than a typo, and for all I know it was corrected in the next edition of the journal. Am reluctant to assume it. Your instinct seems good and I will try to find a similar expression.

Of course so many landmark papers had typos and those great scientists readily accepted their mistakes. I recently wrote an article on the history of the term "shot noise" and highlighted a mistake by Schottky (1920). When people used his value the charge on the electron was completely wrong but that paper was a landmark paper. Just curious why are you reading a 1928 paper? I usually go from modern to the oldest.

@M.Farooq: Rosenfeld is credited with the insight that (still) underlies a lot of current work on optical rotation. Seemed like it might be worth the investment of time--I have only begun to look at it.

Equation 21 is correct: it shows that lambda is the wavelength in the medium, with c and nu being in vacuo. So when I crunch through, I get no contradiction: (nu/c) = (1/the vacuum lambda).

So in eqn 42, lambda is the wavelength in the medium and on the RHS of eqn 43, that lambda is the wavelength in vacuo. It equals n times lambda in the medium.

And here is what I have: https://chemistry.meta.stackexchange.com/a/4552/79678

Ed V's linked sandbox calculation is really nice. If no one else has an idea I see no reason not to accept that as correct. It makes sense of this in an elegant way.

I deleted what was in the Sandbox II while doing another answer, but I will put it back. Since I put the bounty on the question, I cannot win it! If worse comes to worse, daniel should use it, mutatis mutandis, to post a bountiful answer! No sense letting the bounty vanish away, softly and suddenly, like the hunter who spies a snark! (Apologies to Lewis Carroll.)

@EdV: Oof! You should get your bounty points back as anything I would post at this point would be a paraphrasing of your answer! I think if I award myself the bounty I can turn around and award it back to one of your meritorious earlier answers?

Assuming you do not get a better answer (and you are the judge of that) and assuming it is not unethical or in violation of any rules here, then that is fine with me! So please post the answer and I will award you the bounty! Thanks, daniel!

This answer was suggested by Ed V. It resolves the apparent problem nicely and readers with some fluency in the subject would probably not find the transition from $\lambda$ (medium) to $\lambda$ (vacuum) too troublesome.
We begin with $c = \nu \lambda_o$ in a vacuum and $\frac{c}{n} =\nu\lambda $ in a medium, in which $n$ is th refractive index.
In equation (42) of Rosenfeld's paper we have (using $\lambda n =\lambda_o$):
$$\Delta n =  \frac{2\pi\rho}{\lambda n}= \frac{2\pi\rho}{\lambda_o}.\hspace 74 mm
 ~~~(42)$$
Then in equation (43) above we get $\lambda_o^2$ in the denominator:
$$\theta = \frac{2\pi\nu}{c}\cdot \frac{\Delta n}{2}=\frac{2\pi\nu}{c}\cdot\frac{\pi\rho}{\lambda_o}=\frac{2\pi^2\nu\rho}{c\lambda_o}=\frac{2\pi^2\rho}{\lambda_o^2}\hspace 30 mm (43) $$
and finally substitution of (42) into (43) gives
$$\theta = \frac{2\pi\nu}{c}\cdot \frac{2\pi\rho}{2\lambda n}= \frac{2\pi^2\rho}{\lambda_o\lambda n}=\frac{2\pi^2\rho}{\lambda_o^2}. $$


text:
Trends in the acid strength of oxoacids
I was working on a problem regarding the relative acidity of oxoacids in the form $\ce{HYO3}$ where $\ce{Y}$ represents a group 17 element.
The question (which is from a worksheet created by a chemistry instructor at my institution) is as follows:

Which of the following is the most acidic: $\ce{HBrO3}$, $\ce{HFO3}$, $\ce{HIO3}$, and $\ce{HClO3}$?

The given answer is $\ce{HIO3}$. However, in my understanding, the acidity of oxoacids is driven by bond polarity, which is affected by the electronegativity of atom $\ce{Y}$, as this question suggests. For oxoacids of the same structure, only differing in the atom $\ce{Y}$, I believed that the acid strength increases as atom $\ce{Y}$ increases in electronegativity. Thus I believed that $\ce{HFO3}$ would have the greatest relative acid strength. As a matter of fact, I believed that $\ce{HIO3}$ would have the lowest relative acid strength as $\ce{I}$ is the least electronegative of the four $\ce{Y}$ atoms on this list. I would appreciate it if you could help me understand why this anomaly, that $\ce{HIO3}$ is stronger in acidity than $\ce{HFO3}$ contrary to the general trend regarding the electronegativity of $\ce{Y}$, applies in this case.

OK, first things first.  How in the world is there HFO3?

Oh, it did not occur to me that HFO3 might not exist. I still think, though, that if one was to go by the electronegativity-acid strength trend, then HIO3 would actually be the weakest out of all of the acids listed, as Br, Cl and F are all more electronegative than I.

There is some link between this acidity trend for oxoacids of the halogens and the acidity trend of the hydrohalides HX. The notion of charge dispersal over large atomic volumes being more important than electronegativity considerations remain relevant.

You compare electronegativity when you have similar sized atoms.


text:
Ion chromatography of anions
Using an eluent of 1.00 mM NaHCO3 and 8.00 mM Na2CO3 in Milli-Q water, chromatographs were taken of the following anions: fluoride, chloride, nitrate, sulfate And phosphate. Why were the  retention times of sulfate and phosphate longer than those of chloride, nitrate, and fluoride? The retention time of phosphate is not longer than the retention time of sulfate, why? What data do I use to find pH

In ion chromatography there are multiple retention mechanisms, the ion-exchange is the primary one. However, let us keep it simple.
You should be able to explain why monovalents retain less than divalents? Use Coulomb's law! The force of attraction between the ion and the ion-exchange on stationary phase is proportional to the product of their charges and inversely proportional to the hydrated radius. Think about it why fluoride comes out before chloride, almost universally on all IC columns.
Another complication, this explanation breaks down with large anions like iodide, thiocyanate and perchlorate. They retain extremely long on most IC columns. They are called polarisable ions. These ions are sticking because of some other reasons.
Coming to divalents and trivalents: The retention of divalents and trivalents are very complicated. It depends on the chemistry of the stationary phase and its cross-linking and the very nature of the charged site on the ion-exchanger. However you can do a ball-park at an elementary level. Hard to predict in general.
Is phosphate ion always 3(-) or it can have 2(-) or 1(-) charge? Is the ionization pH dependent? The pKa (third proton) is >12. Is the mobile phase pH above 12?


text:
Effect of intramolecular hydrogen bonding on the stability of a molecule
What will be the effects of intramolecular hydrogen bonding on the stability of a molecule? Will it increase or decrease the overall stability of a molecule? Or will it not have any impact on the stability?
Why do I think intramolecular hydrogen bonding increases the stability?
I've learn that formation of a bond lowers the energy or increases the stability of a molecule. Thus, formation of intramolecular hydrogen bond makes the molecule more stable.
Why do I think intramolecular hydrogen bonding decreases the stability?
Let us consider the following molecule:

Intramolecular hydrogen bonding exists between oxygen atom of the carbonyl group and hydrogen atom of the hydroxyl group. Due to the hydrogen bond, these oxygen and hydrogen atoms are under attractive forces. I think this causes strain in the molecule, thereby increasing the energy or decreasing the stability of the molecule.
However, I don't know which of the above two arguments is correct. It would be helpful if you could explain the effects of intramolecular hydrogen bonding on the stability and the appropriate reason behind it.

Given that an unstabilizing effect won't take place you are left with a stabilization effect. This kind of doubts come from chemistry on paper. It might be related to phrasing. Besides this note, in the example sketched the effect is that of overall stabilization. The conformer giving room to internal H bond has in fact more bonds. Note that the formed cycle is not a strained one. But

.... again in case of strained system the H bond won't take place. If it lowers stability perhaps only detailed mechanisms studies would be able to ascertain that (as for we can see the molecules with and without H bond as two different ones existing in equilibrium).


text:
Why does the concentration of the acid equals that of the base at the mid-point of titration?
Consider this graph of the titration of a weak acid and a strong base:

The buffering region is strongest at the midpoint of titration because there is equal concentrations of acetic acid and its conjugate base.

My question is why at the midpoint of titration does the concentration of the acid equal the concentration of its conjugate base? Also, why does the $\mathrm{p}K_{a}$ of the acid = the $\mathrm{pH}$ at this point?

Using as an example the monoprotic acid HA, by definition the pKa of HA  is the pH at which $\ce{[HA] = [A-]}$.

Can you explain it please, I don't know what happens. @Karl

A + B -> C, OK?  How much B do you have to add so that A is eqal to C.

At the midpoint of titration, the concentration of acid is just approximately equal to its conjugated base, as there must be considered the acid dissociation and ( in less extend ) salt hydrolysis.
If we take the definition(simplified using concentrations):
$$K_\mathrm{a}=\frac {\ce{[H+][A-]}}{\ce{[HA]}}$$
we can see that by definition:
$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log { \frac {\ce{[A-]}}{\ce{[HA]}}}$$
$\mathrm{p}K_\mathrm{a}$ is $\mathrm{pH}$ when both items of the conjugate pair have the same concentration.
$\mathrm{pH}$ near titration midpoint is approximately equal
$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log { \frac {n_\mathrm{base}}{
n_\mathrm{acid, init} - n_\mathrm{base}
}}$$
This function has minimal slope = maximal buffer capacity at $$n_\mathrm{acid, init} = 2 \cdot n_\mathrm{base}$$

But still, why is just in the midpoint that the acid is equal to the base?

@Positron12 If you pass a half of your journey, does  the passed segment have the same length as the remaining one, or different ? At the midpoint, the half of acid is neutralized, forming the salt, which is its conjugated base, being in ratio 1:1 to the acid.

Then the same should apply at the equivalence point, why doesn't pKa = pH at it then?

Why should it ? Is at the equivalence the acid/base conjugated pair at 1:1 ratio ? There is almost no acid there.


text:
How to predict the products of an inorganic dehydration reaction?
For example, this reaction is a dehydration reaction
$$\ce{ HNO3 + P4O10 -> N2O5 + HPO3} $$
If I were only given the reactants side of the equation, how would I predict the product's side of it?
What I have seen already:
How can I predict if a reaction will occur between any two (or more) substances?
I wish to know how many $ \ce{H2O}$ molecules a dehydrating agent would remove.

Related: https://chemistry.stackexchange.com/questions/50684/how-can-i-predict-if-a-reaction-will-occur-between-any-two-or-more-substances

$\ce{P4O10}$ is known to be a dehydrating agent. It removes two hydrogens and one oxygen from any substance it may encounter.
In order to withdraw $\ce{H2O}$ from nitric acid, it must find at least two $\ce{H}$ atoms. One molecule $\ce{HNO3}$ is not enough. So you have to use at least two molecules of $\ce{HNO3}.$ What remains from two $\ce{HNO3}$ after removing one $\ce{H2O}?$ By subtraction you find that it remains $\ce{N2O5}.$
What is more difficult to discover is the molecule made with $\ce{P4O10}$ reacts with water. It could be $\ce{HPO3}$ or $\ce{H3PO4}.$ By simply counting the atoms before and after the arrow, you may discover that $\ce{HPO3}$ is formed. So, the equation may be
$$\ce{4 HNO3 + P4O10 -> 2 N2O5 + 4 HPO3}.$$

how do you know if compound has dehydrating properties or not?

The only use of $P_4O_{10}$ is as a dehydrating agent. It is probably the most efficient dehydrating agent.

oh so it's like a general knowledge thing?

Chemistry contains a lot of empirical knowledge, that cannot be derived from basic principles. Some of them can be derived by quantum chemistry calculations, but it is not the path you want to follow. E.g. computation of correct mercury melting point has waited for 50 years until  recently, when powerful enough computers were available.


text:
Regarding the infinitesimal form of the First Law of Thermodynamics
First Law of Thermodynamics is expressed as[1]

The internal energy of an isolated system is constant.

If $w$ denotes the work done on a system, $q$ for the energy transferred as heat to the system, and $\Delta U$ for the resulting change in internal energy, then it follows that $$\Delta U = q + w \label{1}\tag{1}$$
$$\mathrm{d}U = \mathrm{d}q + \mathrm{d}w \label{2}\tag{2}$$
What I don't understand is how does $\eqref{2}$ follow mathematically from $\eqref{1}$. Taking the limit $\Delta U \rightarrow 0$ gives the left-hand side but how does $q$ change into $\mathrm{d}q$ and $w$ into $\mathrm{d}w$? I understand the interpretation of both the equations physically. Is this related to the fact that $U$ is a state function whereas $q$ and $w$ are not?
Reference:
[1] Atkins, P.; de Paula, J. Physical Chemistry, 10th ed; Oxford UP: Oxford, U.K., 2014.

This is the reason why some authors do not write dq and dw, but use a sort of  crossed symbol of d, that I am unable to represent here. It looks like a d with a superimposed / bar.

It's just the authors playing fast and loose (inventing their own version) with the mathematics.

Mathematically, neither dq neither dw are total differentials, but their sum is.

Some texts use $\delta$ to indicate inexact differentials. Not to be confused with $\partial$.

The first law of thermodynamics is a statement of energy conservation thus and defines the internal energy E as an extensive state function. In an infinitesimal transformation, the first then law reduces to
$$\mathrm{d}E = \mathrm{d}Q + \mathrm{d}W$$
where $\mathrm{d}E$ is a total (exact) differential for infinitesimal transformation.
However, $\mathrm{d}Q$ and $\mathrm{d}W$ are not exact ($Q$ and $W$ are not state functions); $Q$ and $W$ in a thermodynamics transformation are process-dependent. All of these are properties of functions of more than one variables


text:
Residue from biomass incineration
What gasses are typically produced when biomass is incinerated at high temperatures (600 C) in high oxygen environments and what is a typical composition of the ashes left behind?

Google is your friend. I found this in seconds  https://www.researchgate.net/figure/Chemical-composition-of-wood-ash-from-various-species-of-timber-14_tbl1_270163491

This is only the ash, not gasses released. It's also specific to wood ash, how similar is wood ash to say food waste or other biomass that's burnt?


text:
How (in)flammable is "isopropyl alcohol" really?
I have an electric device which frequently gets very oily (it's a vaporizer for herbs). It came with a few individually packaged small "tissues" drenched in "isopropyl alcohol". I've used up them all now to clean the insides.
After each cleaning session, I wait a few minutes before I put in new herbs and heat it up again (to 365-420 degrees F). Every time, it feels scary, because "isopropyl alcohol" is said to be "extremely" flammable. I feel like it's going to explode every time, but it luckily never does. Since they shipped those with the product, and didn't mention a word about this in the "manual", I assume it's safe.
But how can it be safe? Is it not hot enough? Does it evaporate so quickly as to never actually be left in there when I heat it up?
I also bought a separate, very expensive bottle of what I assume is "high-quality" isopropyl alcohol and poured it into a glass container with a firm plastic lid, which I keep in a wardrobe. Luckily, it doesn't seem to evaporate into nothing, like used to happen with other alcohol which I previously attempted to store in the same container. After a couple of weeks, half of it had disappeared! That doesn't happen with "isopropyl" alcohol, apparently, but why doesn't it?
As you can hear, I wonder both how (in)flammable it really is, and what makes it not evaporate inside the box, but does evaporate very quickly (apparently) when not restricted by a closed container.

Yes, it is very flammable. If you search on youtube "isopropyl alcohol jet" you'll hit some nice videos with demonstrations of combustion of isopropanol vapors in 20-l bottles.

Isopropyl alcohol is very similar to ordinary alcohol. I burns like ordinary alcohol, which we call ethanol, produces a flame very similar to ethanol. It boils at nearly the same temperature. It is not drinkable.

The Flash Point of a liquid is the minimum temperature at which a liquid gives off vapor within a test vessel in sufficient concentration to form an ignitable mixture with air near the surface of the liquid. The flash point is therefore dependent on the boiling point and consequent vapor pressure of the liquid (Source of Vapor Pressure):
$$
\begin{array}{l|c c c}
\text{Flammable Liquid} & \text{Boiling Point at $\pu{1 atm}$} & \text{Flash Point} & \text{Vapor pressure at $\pu{25 ^\circ C}$} \\
\hline
\text{Diethyl ether} & \pu{35 ^\circ C} & \pu{-45 ^\circ C} & \pu{532.7 mmHg} \\
n\text{-Pentane} & \pu{36 ^\circ C} & \pu{-40 ^\circ C} & \pu{434.3 mmHg} \\
\text{Hexane} & \pu{69 ^\circ C} & \pu{-26 ^\circ C} & \pu{151.2 mmHg} \\
\text{Ethyl acetate} & \pu{77.1 ^\circ C} & \pu{-4 ^\circ C} & \pu{94.7 mmHg} \\
\text{Methanol} & \pu{65 ^\circ C} & \pu{11 ^\circ C} & \pu{126.89mmHg} \\
\text{Ethanol} & \pu{78 ^\circ C} & \pu{12 ^\circ C} & \pu{58.8 mmHg} \\\text{Isopropanol} & \pu{82.2 ^\circ C} & \pu{12 ^\circ C} & \pu{43.6 mmHg} \\
\text{Water} & \pu{100 ^\circ C} & \text{N/A} & \pu{23.7 mmHg} \\
\end{array}
$$
By definition, any liquid with a flash point less than $\pu{37.8 ^\circ C}$ ($\pu{100 ^\circ F}$) is considered to be a flammable liquid. Any liquid with a flash point between $\pu{100-200 ^\circ F}$ is considered combustible. In general, the relative hazard of a flammable liquid increases as the flash point decreases.
However, OP's isopropyl wipes are soaked with about 70% isopropanol and DI water (the best combination for better sanitation). For example, see these wipes on market. As a consequence, the flash point ($\approx \pu{18 ^\circ C}$) is increase on these wipes. Also, since vaporizer has good ventilation system (I assumed), these vapors never would get to saturation point in addition to combination of water and isopropanol vapors. That would reduce the flammability of vapors even with a spark (here are heating coils, I assume). That's probably why the user manual did not warn fire hazard.
OP's question about evaporation from an open bottle: As you can see in the given table, vapor pressure of ethanol and isopropanol is quite high in room temperature ($\approx \pu{25 ^\circ C}$) compared to that of water. If bottle is open, the vapor escapes without ever getting saturated like wipes in closed box. Thus, alcohol in the bottle keep evaporate and, as a result, volume in the bottle reduces with time. On the other hand, the closed space in the box of wipes eventually get saturated with alcohol/water vapors, and evaporation stops at that point.

Let's read the information from a source that has to be right, a supplier's (Fisher Scientific) Safety Data Sheet.

SECTION 9 : Physical and chemical properties
Appearance (physical state,color):  Clear colorless liquid
Explosion limit lower:  2%
Explosion limit upper:  12.7%
...
Flash point (closed cup): 12.0°C
Auto/Self-ignition temperature:  425.0°C
...

OSHA defines "(in)flammable" as having a flash point below 93°C, which this material certainly does.  But to generate the flash you need a spark.  Without the spark you have to go to the much higher auto ignition temperature to make fire, which is the 425°C/797°F figure.  With the spark -- well, you're above that 12°C/54°F figure.  Fortunately in your case there were no sparks.
If you are heating or even working with isopropanol, keep smoking and other flame or spark sources away, because one spark ... see above.


text:
t2g and eg electronic configuration
I have been told that with the help of crystal field theory, we can decide the electronic configuration ($\mathrm{t_{{2g}}}$ $\mathrm{e_g}$ configuration) of the metal ion in a complex.
For example, in octahedral complexes, if the ligand is a strong field ligand, pairing of electron in $\mathrm{t_{{2g}}}$ will be favorable, rather than the electron occupying an $\mathrm{e_g}$ orbital. The opposite is true for a weak field ligand.
The spectrochemical series which helps us relatively identify which ligand is a strong field ligand, and which is a weak field one.
I have two questions:

I understand that the ligands towards left side are strong field, and the ligands towards right side are weak field. But let’s say we take a ligand in the middle. Is it strong field or weak field? Which ligand from the left side is the last strong field ligand? Is it even possible to determine that?
Let’s say we have the following compound:
$\ce{[CoCl2(en)2]Cl}$. Now, as $\ce{Cl}$ is a weak field ligand and en is a strong field ligand, how exactly do we decide the electronic configuration? Is there any general method which considers cases like this?

Please bear in mind that I am a high school student, and I might not be able to understand any  advanced theory involved.

Related: https://chemistry.stackexchange.com/q/64645/41556

@Tyberius: Thank you for the link. It’s very helpful.


text:
I was wondering if anyone could provide me with a method to solving as I do not know what I am doing wrong after multiple attempts?
A portland cement sample contained 20% SiO2 by weight derived
from two silicate compounds, SiO2.2CaO and SiO2.3CaO that
are present in the cement in the mole ratio 3 : 4. Determine
the percent by weight of each silicate compound in the
cement.
Answer key from book  : 24.57%, 43.43%
My answer: 36.03%, 63.09%
Assuming the first silicate compound as W1 & the second one as W2 and their overall composition contribution to the sample as 100%. I got one equation W1+W2= 357.4 and using the mole ratios given, I equated them to get W1= .564 *W2. upon resubstituting that into the first equation, I got 128.8 g for W1 and 228.51 g for W2. Calculating the percentages for each, I got 36.03% and 63.9 % respectively. –
Note: I do find the percentages from the answer key weird too as they don't quite indicate a 100% composition contribution from the constituents. Or, maybe I am just missing something.
The help is much appreciated.

You should include you solution as well.l, not just your result.

Ok. Assuming the first silicate compound as W1 & the second one as W2 and their overall composition contribution to the sample as 100%. I got one equation
W1+W2= 357.4  and using the mole ratios given, I equated them to get
W1= .564 *W2. upon resubstituting that into the first equation, I got 128.8 g for W1 and 228.51 g for W2. Calculating the percentages for each, I got 36.03% and 63.9 % respectively.

Technical note: The question improvement should be done in the question itself, not in the comments. Factual note: There is not said W1&W2 contribution is 100%. You also do not respect their ratio 3:4.

Yes, I think this is the main mistake. w1+w2 is not equal to the weight of cement.

You should give your question some meaningful title.

@Karl I honestly do not understand how your comment helps solve the problem. But sure, your concern has been duly noted.

SE is not just a forum, questions and answers should be written in a way so they can be found and used later by someone else, who gets stuck with a similar problem. Instead of trying to be ironic, you should rather do something about it.

Just to enforce the comment of M. Farooq. The exercise is fine there is no reason the two must add the 100.  And generally speaking one can as well ask for the percent content of, eg, Cao, in a batch of pure Ca(OH)2...

This problem seems to have wrong data ! Look !
The molar mass of $Ca_2SiO_4$ is $172$. The molar mass of $Ca_3SiO_5$ is $228$. If you mix $3$ moles $Ca_2SiO_4$, (= $516$ g), and $4$ moles $Ca_3SiO_5$, (= $912$ g), the total weight is : $516 + 912  = 1428$ g. This mixture contains $3+4 = 7$ moles of $SiO_2$, which weighs $420$ g. In this mixture, the ratio $SiO_2$/total mass = $420/1428$ = $29.4$%. Well. This is contrary to the initial data saying that the proportion of $SiO_2$ is $20$% !!! Where is the mistake or the misconception ?

Thanks for your reply Maurice. I have a link here that will direct you to site I retrieved  the question from. I just rechecked it and it has been presented as is.
Pg 72, problem 3.44 -
https://vdocuments.site/stoichiometry-and-process-calculations-1.html

No, I am afraid the above calculation is not correct. Why are you assuming the total weight of calcium silicates is the total weight of cement?


text:
Molten lithium carbonate conductivity
Does anyone know or could make an educated guess as to the conductivity/resistivity of molten lithium carbonate? Ideally, around 750 degrees C.

Similar to the conductivity of any other molten alkali salt.

Lithium carbonate begins to decompose at 730C

Hope this would help.

The conductivity of molten lithium carbonate has been determined for vast range of temperatures (Ref.1). The abstract of which states that:

The present communication reports the results of an investigation of the properties of surface tension, density, and electrical conductance for molten $\ce{Li2CO3}$, $\ce{Na2CO3}$, and $\ce{K2CO3}$ and some mixtures in the temperature range of $750^\circ –\pu{1000 ^\circ C}$. The surface tensions are approximately twice the values for the corresponding chlorides; the densities and electrical conductance are quite comparable to those of the chlorides. The ionic nature of the molten carbonates is examined from the physicochemical criteria based on these properties; and the mechanism of electrical transport is considered in the light of current theoretical concepts. Relative to $\ce{Na2CO3/K2CO3}$ mixtures, the surface tensions and partial molal volumes indicate but little deviation from the predictions for thermodynamically ideal mixtures.

The conductivity of molten lithium carbonate/metal carbonate mixture has also been determined for range of temperatures (Ref.2), the abstract of which states that:

The electrical conductivities of the molten binary salt mixtures $\ce{Li2CO3–X2CO3}$ ($\ce{X: Na, K, Rb,}$ and $\ce{Cs}$) and $\ce{Na2CO3–Z2CO3}$ ($\ce{Z: K, Rb,}$ and $\ce{Cs}$) were measured using an ac two-probe technique and electrochemical impedance spectroscopy. The electrical conductivities of the eutectic compositions of $\ce{(Li_{0.52}Na_{0.48})2CO3}$ and $\ce{(Li_{0.62}K_{0.38})2CO3}$ at $\pu{923K}$ were $2.06$ and $\pu{1.31Scm−1}$, respectively. The conductivity of molten $\ce{Li2CO3–X2CO3}$ ($\ce{X: Na, K, Rb,}$ and $\ce{Cs}$) and $\ce{Na2CO3–Z2CO3}$ ($\ce{Z: K, Rb,}$ and $\ce{Cs}$) decreases with an increase in the ionic radius of $\ce{X}$ or $\ce{Z}$. We have succeeded in correlating the equivalent conductivity of a binary carbonate mixture $\Lambda$, with the composition ($x_\mathrm{A}$ and $x_\mathrm{B}$), the equivalent conductivities of the pure components $\ce{A}$ and $\ce{B}$ ($\Lambda_\mathrm{A}$ and $\Lambda_\mathrm{B}$), and two parameters ($\lambda_\mathrm{A}$ and $\lambda_\mathrm{B}$) as $\Lambda = x^2_\mathrm{A}\Lambda_\mathrm{A} + x^2_\mathrm{B}\Lambda_\mathrm{B} + x_\mathrm{A}x_\mathrm{B}(\lambda_\mathrm{A}x_\mathrm{A} + \lambda_\mathrm{B}x_\mathrm{B})$. The pair of interaction parameters ($\lambda_\mathrm{A}$ and $\lambda_\mathrm{B}$) can be estimated for a binary system of molten carbonates using the relationship between the cationic radii and the equivalent conductivities of the pure carbonates.

The table 107 in page 85 of government technical report (Ref.3) gives some insight to the conductivity (equivalent conductance ($\Lambda$)  and specific conductance ($\kappa$)) of molten lithium carbonate (m.p.: $\pu{618 ^\circ C}$ or $\pu{891 K}$; eq. wt.: $\pu{36.95 g/eq.}$) in different temperatures together with density ($\rho$) and viscosity ($\eta$) at the same temperature:
$$
\begin{array}{c|ccc}
\text{Temperature in }\pu{ K}  & \Lambda & \kappa & \rho & \eta \\
\hline
1010 & 82.88 & 4.097 & 1.8260 & - \\
1020 & 84.58 & 4.172 & 1.8222 & - \\
1030 & 86.31 & 4.249 & 1.8185 & - \\
1040 & 88.06 & 4.326 & 1.8148 & - \\
1050 & 89.83 & 4.404 & 1.8111 & 4.64 \\
1060 & 91.63 & 4.483 &  1.8073 & 4.34 \\
1070 & 93.46 & 4.563 &  1.8036 & 4.01 \\
1080 & 95.31 & 4.644 & 1.7999 & 3.67 \\
1090 & 97.19 & 4.726 &  1.7961 & 3.36 \\
1100 & 99.09 & 4.808 & 1.7924 & 3.10 \\
1110 & 101.02 & 4.892 & 1.7887 & 2.91 \\
1120 & 102.98 & 4.976 & 1.7850 & 2.83 \\
\hline
\end{array}
$$
The tabulated values of specific conductance ($\kappa$) for molten $\ce{Li2CO3}$ in this table were calculated from the data of Ref.1, Using the following quadractic equation for specific conductance (the precision is $3.00 \times 10^{-4}\; (0.0066\%))$:
$$\kappa = 0.9877 - 1.3529 \times 10^{-3} T + 4.3873 \times 10^{-6} T^2 $$
The uncertainty of the specific conductance values is estimated to be about $1.5\%$. The ref.3 listed this as the best equation for temperature range $\pu{1018-1118 K}$. However, the ref.3 did not provide the relevant units for these calculations and, unfortunately, I do not have access to the Ref.1.
The Ref.3 also provided three other equations for their calculations (I mention this only for interesting viewers):
$$\rho = 2.2026 - 0.3729 \times 10^{-3} T$$
The ref.3 listed this as the best equation for temperature range $\pu{1011-1115 K}$ (The uncertainty of the values is estimated to be about $0.2\%$).
$$\eta = - 5259.12 + 14.8091 T - 1.38581 \times  10^{-2 }T^2 + 4.31294 \times  10^{-6} T^3$$
The ref.3 also listed this as the best equation for the temperature range $\pu{1046-1122 K}$ exracted from Ref.4 (The uncertainty of the values is estimated to be about $3.0\%$).
$$\Lambda = 754.5 e^{-\frac{4438}{RT}}$$
The ref.3 listed this as the best equation for equivalent conductance (The uncertainty of the values is estimated to be about $0.15\%$). Here, the units are given as $\pu{ohm-1 cm2 equiv-1}$. The equation is derived from Arrhenius type equation, $\Lambda = A_\Lambda e^{-\frac{-E_\Lambda}{RT}}$. For $\ce{Li2CO3}$, $A_\Lambda = \pu{754.5 ohm-1 cm2 equiv-1}$ and $E_\Lambda = \pu{4438 cal mol-1}$.
References:

George J. Janz, Max R. Lorenz, “Molten Carbonate Electrolytes: Physical Properties, Structure, and Mechanism of Electrical Conductance,” J. Electrochem. Soc. 1961, 108(11), 1052-1058 (doi: 10.1149/1.2427946).
Toshikatsu Kojimaa,b,z, Yoshinori Miyazakia, Katsuhiro Nomuraa,* and Kazumi Tanimoto, “Electrical Conductivity of Molten $\ce{Li2CO3–X2CO3}$ ($\ce{X: Na, K, Rb,}$ and $\ce{Cs}$) and $\ce{Na2CO3–Z2CO3}$ ($\ce{Z: K, Rb,}$ and $\ce{Cs}$),” J. Electrochem. Soc. 2007, 154(12), F222-F230 (doi: 10.1149/1.2789389).
G. J. Janz, F. W. Dampier, G. R. Lakshminarayanan, P. K. Lorenz, R. P. T. Tomkins, In Molten Salts: Volume 1, Electical Conductance, Density, and Viscosity Data; National Standard Reference Data Series: National Bureau of Standards 15, US Department of Commerce, National Bureau of Standards, US Government Printing Office: Washington DC, 1968.
George J. Janz, Fumihiko Saegusa, “Molten Carbonates as Electrolytes: Viscosity and Transport Properties,” J. Electrochem. Soc. 1963, 110(2), 452-456 (doi: 10.1149/1.2425785).


text:
why is water not compatible with lewis acids in a friedel-crafts alkylation
I know that it destroys the lewis acid character but that doesn't tell me anything.

Lewis acids are strong electron acceptors, the oxygen of water has an electron pair that is very available.

The catalyst used for a Friedel-Crafts reaction is anhydrous aluminum chloride $\ce{AlCl3}$. But this chloride reacts extremely violently with water. Try to put maybe 0.1 g $\ce{AlCl3}$ in a test tube, and add one or two drops of water. You'll soon observe an extremely violent and exothermic reaction. It produces a huge amount of gaseous $\ce{HCl}$ and the tube becomes so hot that you cannot hold it any more with your fingers. The equation is $$\ce{AlCl3 + 3H2O -> Al(OH)3 + 3 HCl}$$ If you continue adding water, the gaseous $\ce{HCl}$ is soon redissolved into this water, giving a solution containing now $\ce{H3O+}$ and $Cl^-$ ; and this solution redissolves $\ce{Al(OH)3}$ in the reaction : $$\ce{Al(OH)3 + 3 H3O+ -> [Al(H2O)6]^{3+}}$$ The surprising consequence of this operation is that evaporation of this solution produces $\ce{AlCl3·6H2O}$ or $\ce{[Al(H2O)6]Cl3}$. And this compound has no activity for a Friedel-Crafts reaction. Friedel-Crafts needs anhydrous $\ce{AlCl3}$


text:
If we use Magnesium instead of the sodium with alkyl halide in dry ether then will a reaction similar to Wurtz reaction will occur
I know that a grignard reagent will formed will be formed but if you look at his question then the answer of this question is b. so I just want to know if something like wurtz reaction will take place if there is some another explanation for this answer.

Consider that many Grignard reagents are commercially available including cyclohexyl magnesium bromide, and thus are stable in solution for months if not years.

The mix of products would depende very strongly on how the reaction was conducted and how it was worked up. And on the exact amount of magnesium used. I'd say, without that detail, it is a bad question.

In 1930, Palmer and Kipping1 reported the formation of cyclohexyl magnesium bromide in ether in 50-60% yield. They did not make an effort to detect the coupling product, "bicyclohexyl" [choice (b)] in the Grignard reagent itself. Subsequent reaction of the Grignard reagent with SiCl4 led to the isolation of bicyclohexyl and some cyclohexene. Although the authors did not speculate on the source of these products, it is likely that they are part of the 40-50% of the "missing" Grignard reagent formed by Wurtz-like coupling and elimination, respectively.

The following year, Gilman and Zoeller2 reported their results on the relative merits of forming cyclohexyl magnesium bromide and chloride. Cyclohexyl chloride gave generally higher yields than the corresponding bromide, and under their conditions, higher yields than what Palmer and Kipping had obtained.

To answer your question, it appears that both bicyclohexyl and cyclohexene may be by-products during the formation of cyclohexyl Grignard reagents.
1) K. W. Palmer and F. C. Kipping, J. Chem. Soc., 1930, 1020.
2) H. Gilman and E. A. Zoellner, J. Am. Chem. Soc., 53, 1931, 1945.


text:
Reproducibility and Standard Deviation
Would a set of chemical data with a relative standard deviation of 0.6% or 20% be more reproducible?
My reasoning is that because it's easier to make mistakes and have a wide range of data, the 20% would be more reproducible, even though it is not accurate.
A RSD of 0.6% would be extremely precise and therefore harder to reproduce.
At least this is my thoughts.  Thank you for your help!

Both accuracy and precision are important. Either results could be just as wrong as both may be nowhere near to the true value when there is a systematic error, i.e. both could have low accuracy no matter what the std dev is. Generally is it better to be accurate and have a large std deviation than a small std dev and be inaccurate. (Of course it is best to be both accurate and precise)

When the RSD is larger, it  is easier to reproduce the results. But there is a lower value of this reproducibily, as there is higher probability the reproducibily has been reached when it should not.

Both accuracy and precision are desired for proper data analysis. They are not mutually exclusive. This is heavily dependent on the person and the measurement tool as well. You must have heard of the idiom "A bad workman quarrels with his tools"... this is true in science as well. Even if you have million dollar instrument, and the operator is unskilled, the data will be garbage. The premise that it is easier to make mistakes is not valid and generally RSD of 1-5 % is routine in scientific measurements. An RSD of 20% would be considered too high.

Think in terms of a discrete probability distribution. Then, a small standard deviation implies that the probability mass functions (PMFs, aka spikes) near the center of the distribution are higher. Conversely, a large standard deviation implies smaller PMFs spread out.
A higher PMF implies a greater likelihood of matching a result near the center of the distribution. Hence, more likely reproducible.


text:
For what processes is Mayer's formula applicable?
I know that
$$C_p - C_V = R, \label{eqn:1}\tag{1}$$
but I have also seen
$$C_p - C_V = R/M \label{eqn:2}\tag{2}$$
being used in case of ideal gas.
In what cases \eqref{eqn:1} is used and in what cases \eqref{eqn:2} is used?

Numbers don't mean anything until they have a dimension. What are the dimensions of all these things?

As Poutnik says, you need an “m” subscript for the heat capacities in the first equation.

The former difference is for molar heat capacity in J/K/mol, the latter for specific heat capacity J/K/kg, or J/K/g, depending on M unit.


text:
Chemical Equilibrium - Le Chatelier's Principle, Change in Volume

Question: Consider the following reaction at equilibrium at a total pressure that we will call $P_1$.
$$\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)}$$
Suppose the volume of the system is compressed to $\frac 12$ its initial volume and then equilibrium is reestablished. The new equilibrium total pressure will be:
a. Twice $P_1$
b. Three times $P_1$
c. 3.5 times $P_1$
d. less than twice $P_1$
e. unchanged

My thought process: Since I knew the volume had decreased, that meant the pressure had increased and so the reaction will head towards the side with less moles, which would be the products. I thought that since the volume halved, that meant the pressure should increase two times.
However, the answer is D. Could someone please explain this to me?

Immediately after you reduce the volume the pressure is doubled. But the forward reaction then leads to a reduction in amount of gas, and therefore in pressure (at constant V and T).

Expanding on what Buck Thorn said in the comments and providing a more rigorous answer.
The partial pressure would change as follows:
$$\begin{array}{c|c|c|c}
& \ce{SO2} & \ce{O2} & \ce{SO3} \\
\hline
\text{Initial state} & P_A & P_B & P_C \\
\text{On halving volume} & 2P_A & 2P_B & 2P_c\\
\text{Final state} & 2P_A-2x & 2P_B-x & 2P_c + 2x\\
\end{array}$$
where $P_A + P_B + P_C = P_1$
Since initial and final states are in equilibrium,
$$ K_p = \frac{P_C^2}{P_A^2 \cdot P_B} = \frac{(2P_C+2x)^2}{(2P_A-2x)^2 \cdot (P_B+x)}$$
We do not need to solve this equation: looking at both the sides, we can see that the value of x would be positive (for a negative x, the numerator would decrease and denominator would comparatively increase due to the $(2P_A - 2x)^2$ term increasing more than the $(P_B + x)$ term due to the square). You could go ahead and solve the equation for a more accurate answer but looking at the options, we can make do with an approximation.
Adding the partial pressures at the final state, we have
$$P_2 = 2(P_A + P_B + P_C) - x = 2P_1 - x < 2P_1$$
Which corresponds to option (d)