How are standard bond enthalpy values experimentally produced?

As defined by the IUPAC Gold Book, mean bond energy (standard bond enthalpy) is defined as:

The average value of the gas-phase bond dissociation energies (usually at a temperature of $298\ \mathrm{K})$ for all bonds of the same type within the same chemical species.
Tabulated bond energies are generally values of bond energies averaged over a number of selected typical chemical species containing that type of bond.

These energies are calculated by studying the homolytic cleavage of a bond into its respective radicals, or the colligation of two radicals to form a particular bond. Though measuring the heat from the colligation of radicals might be the most direct method to determine these values, it is also the most difficult to perform with any accuracy, and is not the preferred method.
There are a variety of ways to measure bond dissociation energies through bond cleavage—by irradiating the molecule with electromagnetic waves, bombarding it with a beam of electrons, measuring the equilibrium constant of the reaction $\ce{A-B <=>A. + B \cdot}$, or by measuring the rate by which a radical reaction proceeds$^{[1]}$.

$[1]$ Szwarc, M. Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences 1951, 207 (1088), 5–13.


Reaction of sodium hydroxide with ethanol

The hydroxyl group in ethanol has a pKa of about 16 (see wiki), which means you will not be able to deprotonate it using sodium hydroxide. Thus, the result of your titration should not be altered by the amount of alcohol in the solution.

Depending on the amount of ethanol I could envision that you are not determing solubility in pure water but in the sepcific mixture with ethanol, but that seems to be assumed in this kind of experiment, right ?


A quick search yields this.

  [Electrolytic manganese dioxide] is produced in a similar fashion as
  electrolytic tough pitch (ETP) copper: The manganese dioxide is
  dissolved in sulfuric acid (sometimes mixed with manganese sulfate)
  and subjected to a current between two electrodes. The $\ce{MnO2}$ dissolves,
  enters solution as the sulfate, and is deposited on the anode.

Unfortunately, there is no source for the information.

Some further searching yields this article, with the following information:

  EMD ($\gamma$-$\ce{MnO2}$) can be prepared from the direct electrolysis
  of an aqueous bath of manganese sulfate and sulphuric acid. The
  electrochemical deposition of $\ce{MnO2}$ on an inert electrode from
  an electrolyte containing $\ce{Mn^2+}$ occurs as per the reaction
  noted below:

  $\begin{alignat*}{3}&\textrm{anode: }\quad&&\ce{Mn^2+ + 2H2O} &&\ce{->
MnO2 + 4H+ + 2e-} \\ &\textrm{cathode: } \quad&&\ce{2H+ + 2e-}
&&\ce{->H2} \end{alignat*}$

Note that the linked article is a review, with a lot of information, both about the actual structure of $\gamma$-$\ce{MnO2}$ as well as the reactions that lead to the production of the correct crystalline structure in the presence of acid. There are a number of references that you may find useful as well.


Why are the halogens good disinfectants?

The halogens, particularly in their diatomic free states and within various oxoacids, are strong oxidizing agents by virtue of their high electronegativities, electron affinities, and reduction potentials. The polarizability of the heavier halogens also makes them almost uniquely versatile as both good leaving groups and strong nucleophiles, depending on conditions. Additionally, the relatively low bond dissociation energies of the halogens contribute to their aptitude to react by free-radical mechanisms (coupled with the exothermicity that usually accompanies the propagation of those reactions).

In their diatomic forms, they readily add across the $\pi$-bonds of alkenes and alkynes (which are ubiquitous and essential structural features of myriad types of biomolecules), and they can oxidize numerous other functional groups under the right conditions and in combination with other substances. Various oxoacids that incorporate halogens are often extremely strong oxidizers as well (household bleach, for example, is an aqueous solution of sodium hypochlorite, which is mild by comparison to certain other halogen-containing oxoacids). In those reactions, they typically serve to add oxygen to molecules in place of, e.g., hydrogen, while themselves serving as leaving groups (or parts of leaving groups).

The antiseptic properties of halogens (and halogen-containing oxoacids) are directly consequent to those chemical properties. At least one source suggests that elemental iodine, for example, reacts by mechanisms similar to the ones described above to disrupt key structures of bacterial enzymes, proteins, and lipid membranes: it oxidizes the thiol groups of cysteine or methionine to disulfides, interfering with protein synthesis; it reacts with unsaturated lipids to alter their structure; etc. These are biological equivalents of the very same reactions that are routinely carried out in organic chemistry labs. If you want more details/examples, they should be easy to find with a web search.

In short, the very same awesome chemical reactivity that makes them so indispensable in organic synthesis, also means that they are highly reactive towards (and hence, usually harmful to) all manner of organisms.


Cathode gained mass in zinc electrolysis

If the differences between the two electrode masses is small its probably due to excess electrolyte left on it or a measuring error. The white substance is  dried up zinc nitrate.

Bubbles should have been observed at both electrodes since that would have been the hydrogen, oxygen and nitrous oxide dissociating form the diluted solution of zinc nitrate.

ie: $$\ce{3e¯ + 4H+ + NO3- -> NO + 2H2O}$$ $$\ce{O2(g) + 4H+(aq) + 4e ->  2 H2O}$$

So I'm pretty sure the bubbles didn't have anything to do with the additional mass gain. $%edit$


Does sodium ethoxide (sodium ethanolate) have a boiling point, or does it decompose?

Yes, sodium ethoxide starts to decompose on heating and thus its boiling point cannot be determined. According to [1]:

  The decomposition of sodium ethoxide is observed to start above 573
  K agreeing well with that reported by Blanchard et al. The gaseous
  products formed on decomposition were mainly methane and ethylene with
  minor quantities of ethane, propylene and butylene. Solid residue was
  found to be a mixture of sodium carbonate, sodium hydroxide and
  amorphous carbon. However, in an earlier study on decomposition of
  sodium ethoxide, by collecting and condensing the evolved gases in
  liquid nitrogen and analysis by gas chromatography, the authors have
  reported the gaseous products to be mainly of ethylene and minor
  quantities of propane, heptane, ethanol and water. The authors have
  also stated that the non-condensable gas at liquid nitrogen
  temperature could be hydrogen or methane. The TG trace for
  decomposition of sodium ethoxide in their study indicated a slow and
  steady decomposition right from 273 K with a sharp weight change above
  573 K which implied that the sodium ethoxide could contain some
  amount of ethanol and moisture that evaporated on heating.[...]

Reference:

Chandran, K.; Kamruddin, M.; Ajikumar, P.; Gopalan, A.; Ganesan, V. Kinetics of thermal decomposition of sodium methoxide and ethoxide. J. Nucl. Mater. 2006, 358 (2-3), 111–128. DOI: 10.1016/j.jnucmat.2006.07.003.


In the generality the question cannot be answered with a definititve statement. You have to look at the cases. For example in the solid state structure  of CuSO$_4\cdot$5 H$_2$O, which is actually more like [CuSO$_4$(H$_2$O)$_4$]$\cdot$H$_2$O, the single crystal water (non-coordinating water) in deed forms hydrogen bonds to oxygen molecules of other coordinating water molecules. In general one could maybe predict, that coordinating water molecules for steric reasons are less likely forming hydrogen bonds to other coordinating water molecules: The oxygen points to the metal and the protons will point to outside so the likelihood of protons from coordinating water molecules  meeting oxygens of coordinating water molecules is small.


1 g of sodium bicarbonate in 400 mL of water will give an approximately 30 mM solution with a pH of ~8.4. The interesting thing about the pH of solutions of amphiprotic compounds is that so long as the concentration of the salt is significantly larger than its Ka and Kb, the solution pH only depends on the dissociation constants. Basically any amount of bicarbonate above 10 mM or so will have the same pH in solution, so pH is not a good indicator of how much acid you can neutralize with it.

I'm not really sure what good this information will do you. How much this solution will be able to affect the paper is going to depend on how much paper you use for a given volume of solution and how acidic the paper is. I don't know much about paper—it could be that it doesn't really matter if you go overboard on the bicarbonate. It might be as simple as making a concentrated solution, testing periodically that your bath hasn't become acidic, similar to photographic stop bath, which has a pH indicator that changes colour when the bath is not acidic anymore. If residual bicarbonate is a problem, rinsing with water afterwards is probably advisable.


Can a boat conformer of cyclohexane be most stable?

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


Why is 2-methyl-2,3-pentadiene achiral (allene)?

I'll start by linking to Symmetry @ Otterbein, as it's a very useful tool for looking at various structures and seeing planes an axes of symmetry.  If you're having issues visualising rotations and such, then this website should certainly help.

I suppose the first step would be to make sure that you're always, always, always visualising the structure in 3D. I'm not saying that you haven't been, but I just want to make sure.

A general explanation for testing for chirality would be to take your two structures and check to see if they are superposable. If they are, then they aren't chiral, but if they're not superposable, then chirality isn't ruled out.

If you were to take models of (R)- and (S)-bromochlorofluoromethane, you'd find that under no conditions are they superposable. They are indeed chiral.

Now consider your structure:

There is a mirror plane going right through the middle of the molecule (cutting carbons 2, 3, 4, and 5 in half, as well as one of the hydrogens on carbon 5).

Let's view it from above, with the mirror plane going through the centre of the image:

Now let's consider one of the core concepts of symmetry. If you were to only see what's on one half of the mirror plane, could you draw the entire molecule, knowing with complete confidence that you didn't draw any kind of isomer?

The answer should be yes, because that's what a mirror plane implies: it's the same on both sides of the plane, so if you know what's on one side, then you know what's on the other.

If you were to take the mirror image of this molecule through that same plane, you'd end up getting the identical molecule. If you mirror this molecule through a perpendicular axis, you'd end up getting the same thing, but with the methyl and hydrogen on carbon 5 switched around.  With the symmetry of the molecule, though, you could just rotate the molecule 180° to get the original one back.  That would make them superposable, and therefore not chiral.

When you recognise any symmetry, you'll know (or eventually intuitively see without thinking) what you need to do to generate the superposable conformation.  A reasonable analogy would be that if you write a message on white paper using black permanent marker and hold it up in the mirror, you'll see a mirrored image; BUT, realising that the paper itself becomes a mirror plane when the ink soaks through to the other side, you'll know that all you need to do is flip the paper over, showing the mirrored writing to the mirror, and see forward writing in the mirror.  This is very related to what you'd do with molecules.

I think that your best bet is to play with as many 3D models (physical or computational) as possible, doing rotations and mirror operations whenever possible.  At some point, it should click and be completely intuitive when just thinking about the molecule.  Try it on all sorts of things.  Try making a pizza with C6v symmetry.  Think about what your pizza would look like if it had D6h symmetry instead (note: very messy to make and to bake).


Major product for aldol condensation between acetophenone and acetaldehyde

I would expect the aldehyde to be the major electrophile in this reaction. The reason is that the steric demands for a nuelophile coming close to the carbonyl carbon are lower than for the same nucelophile coming close to a ketone.

I would expect that the enolate formed from the reaction of the methyl phenyl ketone (acetopheone) to be slightly more stable than the one formed from ethanal. This is because the enolate can be delocalised into the benzene ring.

If we were to form the enolate from ethanal then if this was to attack the carbonyl carbon of acetophenone then we would get a tertary alcohol as the product. The major aldo product in this reaction will be a secondary alcohol formed by the action of the enolate from acetopheone on the ethanal.


The baking soda grain size might be bigger in this new brand. When you sift, you break up individual grains that have clumped together due to moisture or being pressed together tightly but there is a fundamental particle size from how the baking soda was manufactured that would require further processing to reduce (e.g. milling). The larger grain size would adversely affect the solubility. It could also be that the baking soda you buy now is either missing some additives or has some additives that are detrimental to your final product.


Which stabilizes a carbocation more : Back Bonding or Conjugation?

I am inclined to consider the second one as a "bona-fide" carbocation, if that makes sense. The first one resembles a protonated aldehyde rather than a carbocation, so relatively speaking, the first one should be more "stable".


Would it be correct to say that ozone is an element?

It would be more correct to describe ozone as an allotrope of oxygen. Dioxygen (about 1 in five of the molecules making up air) is another allotrope.

The periodic table describes the atoms that make up elements as this is the most fundamental property. Many other elements, though, exist in multiple allotropes. Carbon, for example, has several: graphite, diamond, buckminsterfullerene and quite a few others.

All allotropes consist of only one type of atom but the bulk physical properties depend on how the atoms are joined up (flat plates of carbon hexagons with multiple bonding in graphite, a three-dimensional network of singly-bonded tetrahedral carbons in diamond).

We don't normally add too much about allotropes in a periodic table as it would make the table a little big, although some include limited information about how many there are. And, usually, only one is thermodynamically stable under normal conditions (so the physical properties will be about that one).


Stability of transition metals in aqueous solution

Both elements are stable under certain conditions in various oxidation states. In chromium they range from -2 to +6 and in manganese from 0 to +7.

However, some of them are more common than others, especially in aqueous solution. In chromium the most common are +3 and +6. As a "free" ion $\ce{Cr^{3+}}$ is most stable due to a filled s orbital and only one electron in the d shell. If you dissolve chromium(III)chloride in water you will get a complex formation.
$$\ce{CrCl3 + 6H2O <=> [Cr(H2O)6]^{3+} + 3Cl-}$$
Assuming an octahedral ligand alignment in crystal field theory, you are correct in assuming, that the $t_{2g}$ orbitals will be half filled, resulting in a more stable configuration.

For manganese the most common oxidation states are +2, +4 and +7. If you dissolve manganese(III)chloride, then you will also form a aqueous complex, which will slowly transform to manganese(III)hydroxide and precipitate.
$$\ce{MnCl3 + 6H2O <=> [Mn(H2O)6]^{3+} + 3Cl- <=> Mn(OH)3 v + 3H3+O + 3Cl-}$$
The resulting hydroxide may be oxidised by air to manganese(IV)oxide hydroxide.
$$\ce{2\overset{\color{\green}{+III}}{Mn}(OH)3 + \overset{\color{\green}{\pm0}}{O2} -> 2\overset{\color{\green}{+IV}}{Mn}O(OH)2 + 2{}^{-}OH}$$
Over time this will convert into manganese dioxide.
$$\ce{MnO(OH)2 -> MnO2 + H2O}$$


Can electrolyzing ice cause an explosion?

Bulk electrolysis of ice would likely not get very far. Electrolysis works in liquids because the liquid is able to transport an ion to complete the circuit. In the case of water electrolysis, water is needed to transport protons (or hydroxide, but we'll stick to just one thing).

The half reactions for electrolysis of water are

Oxidation at anode:

$$\ce{2 H2O (l) -> O2(g) + 2H+(aq) + 2e-}$$

The electrons travel through the wires and the protons through solution to the cathode where the following happens:

Reduction at cathode:

$$\ce{2H+(aq) + 2e- -> H2(g)}$$

Electrolysis requires ion mobility. In solid water this mobility is severely decreased. All water molecules are locked in a rigid hydrogen bond network. A proton can propagate slowly through this network by proton exchange, but it will be slow. Thus electrolysis will be slowed.

Also, electrolysis only occurs at the surface of the electrodes, which must provide a means for the gas to escape (whether they are on the surface of the ice or driven into the ice, gaps will exist for gas to escape).


The Boiling or Evaporation of Diethyl Ether

The issue here, as eluded to in your discussion, is that near the boiling point the rate of heat being removed from the ether (via rapid evaporation) was greater than the rate of heat being transferred into the ether, so that the temperature of the ether never reached it's boiling point. In fact, the ether was likely several degrees below it's boiling point due to the evaporative cooling.

The rate of heat transfer into the solution via conduction from the surrounding air is a function of the difference in temperature between the ether and the air, which in this case was very small. If the ether had been placed in an oven of a sufficient temperature, the ether would have boiled.

The mass loss that was observed is not at all unexpected, as the temperature of the ether was just below it's boiling point. Imagine how quickly water would evaporate from a beaker if the surrounding air temperature was $\pu{99^oC}$ (and assuming we're at sea level so the water doesn't boil). The vapor pressure of water, and ether, is only slightly less at a few degrees below it's boiling point than it is right at it's boiling point. (Of course the physical action of boiling also allows for more rapid evaporation, so long as sufficient heat is applied to keep up with the evaporative cooling).

Summary, TL;DR:
As the ether was close to it's boiling point, it was evaporating rapidly, resulting in rapid mass loss. The evaporative cooling near the boiling point happens at a rate to greater than the conductive heat transfer from the surrounding air, so that it never quite reaches it's boiling point, and actually cools to several degrees below it's boiling point.


Resolution depends on three factors (i) efficiency N (ii) alpha or selectivity $\alpha$ and (iii) retention factor k, which the same as the capacity factor in older texts.

Guess which factor is dependent on the column length? It is only the efficiency. If you recall the resolution equation:

$$R_s=\frac{\sqrt{N_2}}{4}\left[\frac{\alpha-1}{\alpha}\right]\left[\frac{k_2}{k_2+1}\right]$$

you need the desired resolution, say 1.5, you already know $\alpha$ and ks from the experimental retention times. All you need is to solve N. You would also know that N scales linearly with the column length. With these hints you can estimate the necessary column length.

Other ways to increase resolution, if it were a research problem: You would use a smaller thermal gradient, change temperatures, and lastly change the column length.

In the real world, people would use brute force, i.e. they would change the column, because it's the selectivity which affects the resolution the most. For example, pharmaceutical companies test about 8-16 columns at once (HPLC). The column which separates a given sample with their fixed mobile phases is chosen for further optimization.


Separating gold and silver from gold foil using hydrochloric acid and hydrogen peroxide

Hot oxidizing acids, such as nitric acid od sulfuric acid, are the more canonical methods to separate gold from less rare metals.

    My solution I assume contains a large amount of silver.

If so, shouldn't you have noticed a precipitate of silver chloride ($\ce{AgCl}$) under these conditions?


Is there are any universal methods or software for reaction rate modelling?

Most modern process simulators, including DWSIM, feature reactor models which can calculate and display reaction rates. For PFR and CSTR models you must also consider the reactor volume to estimate the reaction time.


Why does Benzoin give positive Tollen's Test?

For benzoin, a similar mechanism of tautomerisation (of alpha hydroxy ketones tautomerising to aldehydes) happens , except for one step, where instead of the hydride ion (H-) migrating to the sp2 hydridized carbon, it is the Benzene carbanion (C6H5- or Ph-) that migrates to the sp2 hybridized carbon, since it is more stable than the hydride ion (an example of the stability of benzene carbanion over the hydride ion can be found in the pinacol-pinacalone rearrangement). Other than that, the rest of the mechanism is exactly the same.

Hope this cleared your doubt.


Why do all inter-halogen compounds contain an even number of atoms?

Yes neutral $\ce{IF4}$ would be unstable because it would have one lone pair and one unpaired electron.  The extra lone pair is not a problem because Iodine is large enough to exceed an octet, but the unpaired electron makes it a free radical.


What is the effect of oxygen in welding?

According to https://www.bernardwelds.com/mig-welding-shielding-gas-basics-p152080 the purpose of shield gas in steel welding is:

  ...to prevent exposure of the molten weld pool to oxygen, nitrogen and hydrogen contained in the air atmosphere. The reaction of these elements with the weld pool can create a variety of problems, including porosity (holes within the weld bead) and excessive spatter.

There is also a picture of a weld that either didn't use a shield gas or didn't have enough. It's pretty obvious that such a weld isn't going to be very strong.

Edit: From this Wikipedia article explaining the purpose of inert gas in welding:

  Their purpose is to protect the weld area from oxygen, and water vapour.

And in what appears to be a bit of a contradiction:

  Oxygen is used in small amounts as an addition to other gases; typically as 2–5% addition to argon. It enhances arc stability and reduces the surface tension of the molten metal, increasing wetting of the solid metal. It is used for spray transfer welding of mild carbon steels, low alloy and stainless steels... Oxygen causes oxidation of the weld, so it is not suitable for welding aluminium, magnesium, copper, and some exotic metals.

When you consider that atmospheric oxygen is almost 21%, the two to five percent used in inert gas welding seems pretty small. However, though I can find no definitive reference, I'd have to say that the prevention of oxides of the metal(s) being joined is a consideration only when extremely reactive metals are being welded. Instead, it seem that the prevention of gaseous materials (probably mostly water vapor) within the weld seem to be of the most importance. Also, considering the temperatures involved in welding, the creation of oxides of nitrogen from atmospheric nitrogen and oxygen would have to be a concern.


Organic reaction of dry ice and Grignard reagent

You are correct. Grignard reagents $\ce{RMgX}$ (where $\ce{R}$ stands for some hydrocarbon group and $\ce{X}$ is a halogen, usually $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$) react with carbon dioxide to produce carboxylic acids after acidic workup:

$$\ce{RMgX ->[1)\ \ce{ CO2}][2)\ \ce{ H3O+}] RCO2H}$$

$$\ce{RMgX + CO2 -> RCO2- MgX+}$$
$$\ce{RCO2- + H3O+ -> RCO2H + H2O}$$

  In general your reasoning when approaching this problem is a good one. If you are reacting two 1-carbon reagents, then you probably have a 2-carbon product.


Check the electronegativity of hydrogen, silver and noble metals. In addition, though, consider that the product, $\ce{Ag2S}$, is a solid. The reaction is driven in one direction by precipitation, continually removing sulfur from the air.


In Crystal Field Theory what does Δo=10Dq mean?

Though I don't think I can provide a full answer, I did see this in a lecture series. $\Delta_{\mathrm{O}}$ is the ligand field splitting and it is a measure for the strong or weak field that the ligands of a complex create. Depending on the ligands the energy levels of, for example octahedral iron complexes (below), are closer to each other in a low field and farther apart for a higher field. This is dependant on $\Delta_{\mathrm{O}}$. This is also an energy difference and the units are in joules or electron volts. See the picture below. You might also want to look up Tanabe-Sugano Diagrams.

I was told (sorry, no proper reference) that the expression $\Delta_{\mathrm{O}}= \pu{10Dq}$ comes from quantum mechanics. The reason that this term is simply stated instead of defined, lies in the origin of CFT. The theory was largely empirical and  thus $\Delta_{\mathrm{O}}$ was used. This difference in energy levels could also be described in quantum systems and is called $\pu{Dq}$ after its quantum number. The difference between the levels appeared to be $\pu{10Dq}$.

I hope this is not a gross oversimplification, but a rather helpful one!


Molar mass and molar volume of monochloroethylene

You don't need to know the volume to determine the molar volume $V_\mathrm{m}$ of a substance. In your notations:

$$V_\mathrm{m} = \frac{M}{d} = \frac{\pu{62.37 g mol-1}}{\pu{2.56 g L-1}} = \pu{24.36 L mol-1}$$

Alternatively, you could've just used ideal gas law:

$$V_\mathrm{m} = \frac{V}{n} = \frac{RT}{P} = \frac{\pu{0.0821 atm L mol-1 K-1}\cdot\pu{295.8 K}}{\pu{0.9968 atm}} = \pu{24.36 L mol-1}$$

P. S. You determined molecular mass correctly, but, as MaxW pointed out, the units are wrong: you probably made a typo and put $\pu{g L-1}$ instead of $\pu{g mol-1}$.


Reaction of nitro compounds with strong alkali?

In order for a compound to act as an acid, it must have a proton which can be abstracted. As you correctly pointed out, tertiary nitro compounds have no $\alpha$ hydrogens and therefore they cannot be deprotonated at the $\alpha$ position.

Of course, they could be deprotonated elsewhere, but if the other groups are just alkyl groups then this isn't really feasible as they have such a high $\mathrm{p}K_\mathrm{a}$.


Does Amoxicillin (as trihydrate) degrade in stomach acid?

Because amoxicillin is available as oral tablets, it can be ascertained that it is not significantly degraded to a useless form in the stomach.  In fact, it has an oral bioavailability of 95% (meaning 95% of the orally ingested material goes into the blood stream rather than being destroyed in the stomach). The term for an oral medication being destroyed before it enters the blood stream is "first pass metabolism". So oral amoxicillin can be said to undergo minimal first pass metabolism. Oral potassium clavulanate also has a high oral bioavailability.


One approach would be to use various solvents. You would require solvents in which your various stains were soluble in. Alcohols / water would be suitable to dissolve the soda. Whereas an organic solvent would be suitable for washing away the gasoline. Though gasoline is typically colourless.

Another approach would be to expose the paper to copious UV irradiation, materials are coloured due to absorptions in the visible spectrum which trigger electronic excitation. These absorptions bias the transmission and reflection of otherwise white light and give the materials their colour. UV radiation is of higher energy than visible light radiation, and therefore the chemical excitations involved can prompt reactions which result in the destruction of the materials themselves.

A similar approach to invoke this chemical degradation of the coloured materials could be achieved through using a strong acid or base. This would be akin to the effect of bleach on household surfaces. Again, exposure to acids and bases can prompt reactions which result in the degradation of coloured materials.

It is worth noting that all of these methods will have some affect on the paper. Continued washing / drying of the paper with solvents will likely cause the paper to become brittle and affect the absorption of the paper, which may affect your ability to write / print on it. A similar effect may be noticed under UV irradiation, though I expect it wont be to the same extent.

A sufficiently strong acid or base may be able to completely degrade your paper as well as your coloured contaminants.


How can potassium disappear after cooking?

As Eric has mentioned in his answer, the potassium present in foods will leach out into water when you cook them.

Potassium is present in food in the form of potassium ions $(\ce{K+})$. These $\ce{K+}$ ions happen to be extremely soluble in water.

When food is boiled, the cells break down and release their contents more easily which makes the $\ce{K+}$ ions free to move into the water. The $\ce{K+}$ ions remain in the water since they are so highly soluble in it.

This happens to all water-soluble nutrients like the B vitamins, vitamin C, sodium, etc. Some ways to avoid the loss of all these nutrients would be to steam such foods instead of boiling them since the food wouldn’t come into contact with water.

Side note: The same problem occurs when you deep fry foods containing fat-soluble nutrients like vitamins A and E. These leach out easily into oil and so all those nutrients would also be lost in the process.


Equilibrium constant for a multi-step reaction

To give a specific example, let's take this set of reactions:

$$\tag{1}\ce{A + B <=> I_1}$$

$$\tag{2}\ce{I_1 + C <=> I_2}$$

$$\tag{3}\ce{I_2 <=> I_3 + D}$$

$$\tag{4}\ce{I_3 <=> E + F}$$

To get the net reaction, we add up the equations and cancel the intermediates:

$$\ce{A + B + C <=> D + E + F}$$

When the system is a equilibrium, all the reactions are at equilibrium. You can write the equilibrium constant expression based on the net reaction:

$$K_\text{net} = \frac{[\ce{D}][\ce{E}][\ce{F}]}{[\ce{A}][\ce{B}][\ce{C}]}$$

You would arrive at the same expression if you take the product of the individual equilibrium constants (and cancel):

$$K_\text{net} = K_1 K_2 K_3 K_4 =\frac{[\ce{I_1}]}{[\ce{A}][\ce{B}]} \times \frac{[\ce{I_2}]}{[\ce{I_1}][\ce{C}]} \times \frac{[\ce{I_3}][\ce{D}]}{[\ce{I_2}]} \times \frac{[\ce{E}][\ce{F}]}{[\ce{I_3}]}$$

The concentrations of the intermediates will not be zero because they are in equilibrium with the other species. They are intermediates in the sense that we care a bit less about them since they don't appear in the net equation of interest. However, they are part of the mass balance and may well be among the major species.


How does calcium carbide artificially ripen fruits?

Technically, there is no difference between naturally ripened fruit and artificially ripened fruit... they're both ripened fruit. I suppose you were looking for some sort of simple chemical test to distinguish between them; but like I said, there's no point in it since in both cases you end up with ripe fruit.

Fruits normally emit ethene (ethylene) which is essential to the ripening process. Which is why fruits picked before they're ripe are sometimes wrapped in a bag that seals in whatever little ethene they emit, thereby accelerating the ripening process.

As you would know, calcium carbide does not ripen fruit per se. Rather, it is the reaction of calcium carbide with water/moisture to release ethyne (acetylene) that's important here. Though I do not know of their mode(s) of action, apparently ethyne is almost a effective as ethylene in ripening fruits (I still need to find a reliable reference for this; though it is supposed to be a fairly well known fact).

If you let a fruit ripen naturally, or if you use ethyne instead... there's really no chemical difference at the end (apart from the usual arsenic contaminants that's associated with commercial-grade calcium-carbide).

The actual mechanism by which ethene (ethylene) and ethyne (acetylene) initiate, and thereafter, accelerate fruit-ripening is not really understood. Pectinases play a major role in the ripening process (they break down a structural component of plant cell walls called "pectin", which is why ripe fruit tend to be soft), however the scientific community has not yet arrived at a consensus as to how ethene and ethyne are linked to the activation/production of these enzymes.

However, there are a couple of subtle (non-chemical) difference between naturally and artificially ripened fruit:

Fruit that has ripened prior to plucking tend to be "fuller" than artificially ripened ones. This is because the fruit still gets an influx of nutrients while it's still attached to the tree...in other words; in the time it takes to ripen, the fruit grows as well. Artificially ripened fruit however, tend to be smaller, as they pretty much stop growing as they ripen (off the tree). ["Small", "big", "fuller" are crudely defined terms...you need some "experience" with fruits if you want to put this tip to good use].

Additionally, if you find the fruit being sold off-season (i.e- At a time of the year when it doesn't usually grow/ripen),  for a normal or somewhat cheaper price...then in all likelihood you have an artificially ripened fruit in your hand.

I reiterate:

It doesn't matter how it was ripened. It's ripe, that's all there is to it (from a consumer's PoV). Don't forget to wash and peel it before taking a bite :-)


Which molecule has maximum bond angle amongst BF₃, BCl₃ and BBr₃?

With VSEPR theory you can predict the structures of $\ce{BX3}$. Since boron only has three valence electrons, all of which are used for σ-bonding to the halogen atoms, they arrange in a trigonal planar fashion. This means that all molecules $\ce{BF3}$, $\ce{BCl3}$, and $\ce{BBr3}$ have the same point group $D_\mathrm{3h}$. From this symmetry you can deduce that the bond angle is $\angle(\ce{X-B-X}) = 120^\circ$ for all of them.

Bent's rule won't have any effect on that.

There is also considerable π-backdonation in these molecules, but that only has an effect on bond lengths, not bond angles.


Let's compare ammonium cation NH4+ and the neutral molecule, NH4 with other metals.

As Ed V pointed out, with a mercury cathode and low concentration of H+, NH4+ can be reduced to NH4 electrolytically (also with sodium amalgam), which forms an amalgam that is stable for a short period of time: the mercury bubbles up, then collapses, evolving NH3 and H2. Ammonium amalgam was discovered in 1808 by Humphry Davy and Jöns Jakob Berzelius.
https://en.wikipedia.org/wiki/Amalgam_(chemistry)

A recent (2016) paper http://scripts.iucr.org/cgi-bin/paper?xk5030 gives the crystal ionic radius of ammonium as falling between 140 and 154 pm, depending on the anion; this compares nicely with the atomic radius of mercury which is 150 - 155 pm, so an alloy is not inconceivable.

Metallic ammonium (NH4) has been considered for some off-world chemistry in a 1954 paper. The phase transition of mixed crystals of hydrogen and ammonia to metallic ammonium was found (by calculation) to take place at pressures almost certainly less than 250 kilobars, suggesting that Neptune and Uranus contain large deposits of metallic ammonium.

http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?db_key=AST&bibcode=1954MNRAS.114..172B&letter=0&classic=YES&defaultprint=YES&whole_paper=YES&page=172&epage=172&send=Send+PDF&filetype=.pdf

But on our world, the reaction of a metal like iron with a high concentration of NH4+ cation would produce H2, not NH4. The reaction could be examined on the basis of the pH, or concentration of H+ ions. As it turns out, the potentials are simply related to the concentration of H+ ion, which depends on the anion.

A 0.1M ammonia solution (anion = OH-) has a pH of 11.122 (http://www.chem.science.unideb.hu/Solutionsmonobasic.pdf). A solution of NH4Cl (anion = Cl-) gives a pH of 5.13 (https://socratic.org/questions/calculate-the-ph-of-the-following-aqueous-solutions).

When we displace from a solution of ammonium salts, we wind up displacing hydrogen. You could imagine that adding iron or zinc to a solution of NH4+Cl- could evolve neutral NH4 which immediately decomposes to NH3 plus a hydrogen radical H., which then dimerizes to H2, and the NH3 immediately redissolves in the water. Far better to think of the NH4+ dissociating in the aqueous solution before reacting with the iron or zinc, in other words, the reaction with the metal is a reaction of H+ (or, actually, hydrated H+, or hydronium ion) after dissociation of NH4+ in the presence of H2O.


How to determine the molecular formula based on molecular weight, amount of substance, and partial masses of some, but not all contained elements?

If you have $0.2\ \mathrm{mol}$ of a compound with $M_\mathrm{r}=82$, then you have $16.4\ \mathrm{g}$ of material. If there is only carbon, hydrogen and nitrogen present, and the carbon and hydrogen total up to $10.8\ \mathrm{g}$, then the amount of nitrogen must be $16.4\ \mathrm{g}-10.8\ \mathrm{g}=5.6\ \mathrm{g}$.


Maxted[1] reported in 1926 that the reaction of tartaric acid with silver produces oxalic acid, formic acid, CO2 and water as products and that the stoichiometry is 6 moles of silver reduced per tartartic acid. He proposed the following possible net reactions:

$\ce{C4H6O6 + 6 Ag+ + 2H2O -> 2(C2H2O4) + 6 Ag + 6 H+}$

$\ce{C4H6O6 + 6 Ag+ + 2H2O -> C2H2O4 + CO2 + HCO2H + 6 Ag + 6 H+}$

$\ce{C4H6O6 + 6 Ag+ + 2H2O -> 2(HCO2H) + 2CO2 + 6 Ag + 6 H+}$

For the mechanism, he concluded that initial decarboxylation to 2-hydroxy-3-oxo-propionate or splitting into two glyoxylic acid molecules were both possibilities. The glyoxylate could be oxidized to oxalic acid or oxidatively decarboxylated to formic acid, while the 2-hydroxy-3-oxo-propionate could split into glyoxylic acid and formic acid.

I'm sure someone must have done more recent mechanistic work on this to pin it down further, but that gives a pretty good idea of the possibilities.

[1] Maxted, D.R. (1926) CCLXXXVIII-The Oxidation of Tartartic Acid by Solutions of Silver Salts; J. Chem Soc. 129:2178.


Why is nylon formed using carboxylic acids, while aramids are formed with acyl halides?

Aromatic amines are less nucleophilic than aliphatic amines because the aromatic amines have the non-bonding electron pair "distracted" in the resonance with the aromatic ring. Therefore, they need a more reactive counterpart (acyl halide instead of the acid).


In the following answer I am using the Physics convention of work done. Where, $W= \int PdV$ and the first law of thermodynamics is as follows: $q = \Delta U + W$. If you want answers in the chemistry conventions than just reverse the sign of the work done. Internal energy change and heat change will be same in both conventions.
As you said you have done the first part, I will continue with the second.
First of all, $C_V=C_P - R = 8.58- 2 = 6.58$
As the process is adiabatic,
$$q=0\\
\Delta U=-W\\
nC_V(T_2-T_1)=-nRP_{ext} (\frac{T_2}{P_2} - \frac{T_1}{P_1}) \\
T_2= 247.67 K = -25. 53 ^\circ C\\\\$$
$\\Now, \\\\$
$$W= P_{ext} \Delta V\\
W= nRP_{ext} (\frac{T_2}{P_2} - \frac{T_1}{P_1}) \\
W= 599.2 cal\\\\$$
$$\Delta U= -W \\
\Delta U= -599. 2 cal\\\\$$
$$\Delta H= \Delta U + \Delta (PV) \\
\Delta H= \Delta U + \Delta (nRT)\\
\Delta H= \Delta U + nR\Delta T\\
\Delta H= (-599. 2 -181. 32) cal\\
\Delta H= -780. 52 cal\\\\$$


How did the first amino acids appear on Earth?

The universe cooled after the Big Bang. There was certainly an epoch when the universe’s overall temperature decreased from 200 °C to −40 °C throughout a very large volume. There is your test tube (more like a Florence flask) for abiogenesis and then the panspermia reservoir.

Before you suspect the preceding is clever, how and when (what ambient vacuum temperature) did elements heavier than boron arise?


How to achieve regioselective (base-catalyzed) enolization of a carbonyl group in a diketone?

As @Zhe and @Zenix have mentioned, you have to hope that there is a difference in the rate of enolate formation to differentiate the two ketones.

To do this cool a rigourously dried solution of the diketone and a trapping reagent ($\ce{Me3SiCl}$ if you're doing an anhydrous workup otherwise TIPS-Cl or MeI) to $\mathrm{-75 \, ^\circ C}$ and slowly add 1 eq. of LDA keeping the reaction mixture cold. This will trap the first formed enolate (kinetic enolate). Allow to warm, dilute with hexanes, filter (or wash with aq if you used TIPS or MeI), concentrate and analyse.

If this gives a mixture then you try making the more stable (thermodynamic enolate) and trap that. This means forming the enolate without the trapping compound ($\ce{Me3SiCl}$, MeI or TIPS-Cl), giving it time to equilibrate then adding your trapping compound. This might be done with NaH in dry DMF, left to stir at $\mathrm{-30 \, ^\circ C}$ for an hour then the trap is added, worked up and analysed.

Even then with this example I think you'll get a mixture, but you don't know until you try. If you do get a mixture then you have to rearrange your synthetic pathway so that you don't have this problem - you'll probably have a different one but that's the  way it goes!


Could you charge a recharable battery with a voltaic cell?

Any source of electricity of the required voltage and current could be used to charge a rechargeable ("secondary") cell. A home-made "primary" cell would work (though you might need a few in series to provide the needed voltage).

For that matter, you can generate DC to charge the cell many other ways:

Using a thermopile, e.g. from copper and iron wire
Magnetohydrodynamically, e.g. with flowing salt water and a magnet
Using a photovoltaic panel...


Can hydride shift and methyl shift happen one after another in a carbocation?

Multiple rearrangements can happen if the resulting species is stabilized by it. In general hydride shift is more favourable over methyl shift because of its smaller size.

The example proposed by James was not known to me but its truly an amazing example.

  James:
  It’s possible for multiple hydride/alkyl shifts to occur. One amazing example is in the biosynthesis of lanosterol.

  Multiple shifts are certainly possible, and they could happen, but generally will only happen if each shift generates a successively more stable carbocation.
  For example, you probably wouldn’t see a shift if it involved turning a tertiary carbocation into a secondary carbocation.

Source

From Wikipedia:


If the electron of a hydrogen atom transitions from energy level n=5 to n=3, which level corresponds to n1 and which level corresponds to n2


Is there any experimental evidence for OClF5?

Yes there has been experimental evidence published.  Reaction of CIF5,  with OF2: Preparation and Properties of Chlorine Oxide Trifluoride and the Existence of Chlorine Oxide Pentafluoride, Angewandte Chemie, Internation Edition, vol. 11, pages 1094-5, purports to show a series on F-19 NMR spectra of the compound, at temperatures ranging from 34 to -76 degrees C.

The series shows a relatively sharp peak at 34 degrees C, which progressively broadens and shifts upon cooling to until about -40, upon which further cooling sharpens to a peak with much different chemical shift than the room temperature peak.  The authors conclude that at room temperature there is fast exchange among the 5 F atoms, and at low temperature 4 equivalent equatorial F atoms are observed, while the peak for the axial F atom is too broad to be observed.

However, the above claim is considered refuted by "Chlorine Oxyfluorides" Advances in Inorganic Chemistry and Radiochemistry vol. 18, pages 319-98.

See On the XeF+/H2O System: Synthesis and Characterization of the Xenon(II) Oxide Fluoride Cation, FXeOXeFXeF+ J. Am. Chem. Soc., vol. 131, pages 13474–89 for further discussion and references:

  Chlorine oxide pentafluoride, ClOF5, would be the highest performing earth-storable, liquid rocket-propellant oxidizer. Although the synthesis of ClOF5 had been claimed in 1972,(1) this claim was subsequently refuted.(2) Continued efforts to prepare this compound by oxidative fluorination of substrates such as ClOF3(3-5) or ClOF4−(6, 7) have failed because the oxygen ligand is more easily oxidized than the central chlorine atom. This failure prompted a search for alternate synthetic approaches, such as the reaction of ClF5 with an oxidative oxygenator. Potential candidates for oxidative oxygenators include hypofluorous acid, HOF, a well-known unstable compound,(8, 9) and its protonated form, the H2OF+ cation,(10, 11) if indeed it should exist.


Nucleophilicity of bonding vs lone pairs

I don't know the answer to your question, but we can speculate.  A lone pair is more sterically exposed than a bonding pair of electrons and therefor more accessible.  Also a lone pair is generally more nucleophilic because the electrons are concentrated in a smaller region since they are not being shared.  These two arguments would favor the second mechanism.  Finally, when we consider the SN2 reaction where, for example, water is the attacking nucleophile; I've never seen the mechanism drawn where the attacking electrons are from the O-H bond, but rather they always seem to show involvement of the lone pair electrons.  All of these arguments favor the second mechanism.


Resources to learn second quantization formalism and how to implement it in code for quantum/computational chemistry research

You could check out any code set that implements configuration interaction wave function methods.  The Journal of Chemical Physics published an open-access article reviewing the state of ab-initio software, and it lists code sets that use CI.
 https://aip.scitation.org/doi/pdf/10.1063/5.0023185
Notably, NWChem uses the second quantization approach and is described in this open resource paper:  https://arxiv.org/pdf/1904.01131.pdf
Finding algorithms to rapidly create valid CI expansions was a big research area in the late 1970's, and the Graphical Unitary Group Approach was created.  There are several open source packages that use the GUGA, including GAMESS. https://www.msg.chem.iastate.edu/gamess/download.html   You can get GAMESS by requesting access and registering.  You could then review the implementation assuming a basic knowledge of FORTRAN.


Order of priority for including functional groups in the parent chain compared to order of priority for numbering the parent chain

There are indeed several analogous criteria for the numbering of locants and for the choice of a principal chain; however, there are also significant differences.
The most important simplified criteria for the numbering of simple acyclic compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The most important simplified criteria for the choice of a principal chain are:

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

You can find the corresponding actual wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) below. However, the IUPAC recommendations have more than 1500 pages. Hence, you should be careful when applying any simplified or abridged version of the nomenclature rules.
Also note that there is no order of priority for substituents cited as prefixes (e.g. ‘bromo’, ‘chloro’, ‘ethyl’, ‘methyl’, or ‘nitro’). All such prefixes are considered together as a set. However, you need the seniority order of compound classes in order to choose the senior class that is expressed as a suffix.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)


I have had the same question recently. And there is no simple answer. Tables (like ones from International critical tables of numerical data, physics, chemistry and technology or this) are still the 1st hand source. Formulas used are just approximating this data. Check formula here on page 8 (to get the temperature-corrected ABV you have to invert it by solving non-linear function). Alternatively check this recent study. To convert between ABV (alcohol by volume, %) and density (specific gravity SG, g/cm3) you may use this approximation:

$$\mathrm{SG} = -0.002\cdot\mathrm{ABV} + 1.0012$$

PS. There is an unconfirmed formula at vinolab:

  The hydrometer temperature correction for SG is performed with this expression:

  $$\text{corrected-reading} = r \times \frac{(1.00130346 - (0.000134722124 \cdot t) + (0.00000204052596 \cdot t^2)\ - (0.00000000232820948 \cdot t^3))} {(1.00130346 - (0.000134722124 \cdot c) + (0.00000204052596 \cdot c^2)\ - (0.00000000232820948 \cdot c^3))}$$

  where: $r$ = reading and $c$ = calibration temperature. This expression is based on °F, so the temperatures are first converted.

With some sample values it gave me results close to the table' ones (I have converted ABV to SG, amended as per temperature and formula above and finally converted back to ABV).


Which resonance structure would be favorably contributed to the actual structure?

(a) Positive charge on $\ce{C}$ versus positive charge on $\ce{O}$: Even though $\ce{O}$ is more electronegative than $\ce{C}$, its octet is fulfilled while $\ce{C}$'s is not (only 6 $e^-$s on positively charged $\ce{C}$). Therefore, second structure is more favored over the first one.
(b) Negative charge on $\ce{O}$ versus negative charge on $\ce{C}$: Here, both negatively charged $\ce{O}$ and $\ce{C}$ have their octet fulfilled. Yet, it is always stable when the negative charge is on more electronegative atom ($\ce{O}$) than on comparatively less negative atom ($\ce{C}$). Therefore, first structure is more favored over the second one.
(c) Positive charge on $\ce{C}$ versus positive charge on $\ce{O}$: Here, same argument of (a) applied here. Therefore, second structure is more favored over the first one.
(d) Charge separation versus neutral: it is always stable when the compound has no charge (neutral) than on has charge separation (with net charge is zero). Therefore, second structure is more favored over the first one.
(e) and (f) Kekulé structures of naphthalene and benzene, respectively: Of each pair, first and second structures have the same energy. Therefore, both structures of each pair have same stability equally favored.
(g) Positive charge on $2^\circ$-$\ce{C}$ in both structures: Since positive charge is on a $2^\circ$-$\ce{C}$ of each compound, the relative energy due to carbocation is same on both structures.  Yet, the first structure has its double bond conjugated to the carbonyl carbon while the second structure has isolated double bond to the carbonyl carbon (cross-conjugated). Therefore, first structure is more favored over the second one.


Can compressed CO2 sprayed onto a hot mass/liquid be an effective means to cool it?

Celebrity chef Alton Brown produces ice cream using a device called the Jet Cream.  He sprays $\ce{CO2}$ from a fire extinguisher at a stream of chocolate milk, instantly producing ice cream. The only difference in your scenario and his is the starting temperature of the liquid.

To say that $\ce{CO2}$ absorbs no heat is absurd.  If it is colder than what it touches, it should absorb heat.


Inductive effect comparison of benzylic and allyl group

Phenyl has a tendency to withdraw electrons.
see the link for the reasoning:
Inductive effect of phenyl ring
Also the allylic group would be donating via inductive effect, thus making the carboxylate ion unstable( after acid donates H+ ions)
But an sp2 carbon is more electronegative than an sp3 carbon, hence it will donate less.(more electronegative element tries to retain electrons)
So 1 has an electron withdrawing group,2 has an electron donating group which is not as good at donating as 3.This gives us the option  A

An important point to note here is that electron withdrawing groups increase H+ concentration. Think why?

Edit: Now that the Question has been edited, an edit for the solution is required.

Benzylic group has -I inductive effect as in the link. For the second compound in the question, with the allylic group, since less number of sp2 carbon are present(2 instead of 6). So we can infer that -I effect will be less. Infact it should show +I effect, especially since there is an sp2 Oxygen attached 2 carbons away, which will pull electrons as it is highly electronegative.


How does the oxidation of NaI by sodium hypochlorite to give I+ work?

I think the reaction should be done in basic solution, not acidic solution.  "One equivalent each of sodium iodide (2.77 g) and sodium hydroxide (0.74 g) was added, and the solution was cooled to 0 C. Aqueous sodium hypochlorite (34.50 g, 4.0% NaOCl) was
added dropwise over 75 min at 0-3 C" J. Org. Chem. 1990,55, 5287-5291.

So perhaps:

$$\ce{ ClO- (aq) + I- (aq) + H2O (l)  -> I+ (aq) + Cl- (aq) + 2OH- (aq)}$$

Edit: after looking at the cited article more closely, the hydroxide was only being added because of the acidity of the iodophenol product, so you could write

$$\ce{ ClO- (aq) + I- (aq) + 2H+(aq)  -> I+ (aq) + Cl- (aq) + H2O (l)}$$

also.


Can compounds (or only elements) be referred to as oxidising and reducing agents?

My textbook refers to $\ce{CuO}$ as the oxidising agent. Is this because the $\ce{Cu}$ in $\ce{CuO}$ is oxidised while $\ce{O}$ is neither oxidised or reduced, so overall it is an oxidising agent?

No. $\ce{CuO}$ is the oxidizing agent (the thing that causes something else to be oxidized) because the $\ce{Cu}$ is reduced (gains electrons, going from oxidation state $+2$ to $0$) in the course of oxidizing the nitrogen of $\ce{NH3}$, which loses electrons and goes from oxidation state $-3$ to $0$.

Conversely, $\ce{NH3}$ is the reducing agent here, because it causes something else to be reduced (here, the $\ce{Cu}$) and loses electrons in the process.

  What happens for a compound $\ce{AB}$ where $\ce A$ gets oxidised and $\ce B$ gets reduced?

If only the species $\ce{AB}$ is involved, or in the case where two of the same species $\ce A$ react to form both oxidized and reduced products:

$$
\ce{2A -> Ox + Red}
$$

the reaction is called disproportionation.

  Do I call the compound $\ce{AB}$ a reducing agent if the net change in oxidation number (between the two elements $\ce A$ and $\ce B$) is negative?

Strictly in the case of a $1\!:\!1$ compound of $\ce{AB}$ on a mole basis, if the net change in oxidation number is negative, then you would call $\ce{AB}$ an oxidizing agent, because it pulled electrons away from something else (it oxidized that other thing) in order to achieve the net negative change in oxidation number.

For $\ce{AB}$ compounds not in a $1\!:\!1$ molar ratio, a universal principle can't be stated $-$ it depends on the particulars of the mole ratio and the individual changes in oxidation number.


Mercury does not react with this compounds. However, I strongly encourage you to

1) use monolithic vial for mercury. 3d-printed pieces are very fragile in my experience. Glass vial should be fine, however I recommend double casing (keep sealed glass vial with mercury in a bigger plastic vial with sulfur dust.)

2) perform all operations with mercury over steel of plastic pan.

3) keep a piece of copper foil/plate and nitric acid at hand. Mercury drops stick to copper, freshly washed in nitric acid, which is used to collect mercury.

4) in case mercury is somewhere you don't have access to, drown it in concentrated iron (III) chloride solution. It will transform metallic mercury into mercury chloride, that does not give vapors.

5) Note, that mercury salts may react with some organic compounds, including aromatics (toluene). While risks are low, organomercury compounds accumulates in living tissues, so caution is required.


Background

One can draw conformations of n-butane with the carbon-carbon bonds oriented in certain directions and the methyl hydrogens pointing in certain directions that are chiral.  However, since rotation about single bonds is typically fast at room temperature, these chiral conformers of n-butane would only be resolvable at extremely low temperatures.  Therefore we do not typically consider n-butane to be a chiral molecule.

Take-away lesson: If a chiral conformation can readily interconvert with an achiral conformation, then the molecule cannot be resolved into enantiomers and will be optically inactive.

However, there are molecules where rotation about single bonds is restricted (for example due to steric interference) and the rotational barrier is high enough that the molecules can be resolved into their respective enantiomers at room temperature.

Molecules where the rotational barrier about a single bond is high enough to allow different conformations to be isolated at room temperature are called atropisomers.

The biphenyl pictured below (in a chiral conformation) is an atropisomer due to restricted rotation about the central single bond connecting the two benzene rings.  The steric interference that would develop between the 4 ortho groups if the molecule were to approach a planar (achiral) conformation prevents the chiral conformer from equilibrating with the achiral (planar) conformer.  Hence this biphenyl has been resolved into its two enantiomers at ambient conditions.

Trans-cyclooctene

Trans-cyclooctene is also capable of being resolved into its two enantiomers due to restricted rotation about single bonds - it also belongs to the class of atropisomers.

In the figure below trans-cyclooctene conformation (a) is planar, therefore this conformation is achiral. The two conformers pictured in (b) are chiral - these conformations are enantiomers of each other. Trans-cyclooctene will be achiral if the two enantiomers pictured in (b) can readily interconvert through achiral (planar) conformer (a).

Build a model or draw trans-cyclooctene and notice the two hydrogens on the trans double bond; one of them is "inside" the ring in the planar conformation and would cause severe steric problems as the methylene chain tries to flip through the planar conformation to the other face of the double bond as pictured below. This steric interference makes it difficult for the molecule to achieve the planar conformation.

(Image source: John D. Roberts, Marjorie C. Caserio, Basic Principles of Organic Chemistry. Excerpt available at http://www.organicchemistry.com/cycloalkenes-and-cycloalkynes/.)

Because of this destabilizing transannular interaction, trans-cycloalkenes with up to around 11 carbons do not exist as planar molecules. They exist in a twisted conformation as pictured above for trans-cyclooctene.  In your model of trans-cyclooctene, notice how you have to pull and stretch the methylene chain in order to convert trans-cyclooctene into its mirror image.  This rotational movement is highly restricted due to the internal hydrogen.  It is this barrier to rotation about singles bonds that allows trans-cyclooctene to be resolved into its two mirror image forms at room temperature.

The barrier to racemize trans-cyclooctene (e.g. pass through the planar conformation (a)) is 36 kcal/mole.  As you might imagine, as we make the chain longer it should get easier and easier to loop the methylene chain around the internal (protruding) hydrogen on the double bond.  Indeed, the barrier to interconvert the two enantiomers of trans-cyclononene is only 20 kcal/mole. Optically resolved trans-cyclononene must be kept below 0°C to prevent racemization.  The barrier to racemization in trans-cyclodecene is even lower at 10 kcal/mole.

  Are all higher trans-cycloalkenes chiral as trans-cyclooctene is?

First off, these rings can have chiral conformations, just like butane.  I think you are really asking, "Are all higher trans-cycloalkenes optically active as trans-cyclooctene is?".  That is, do these larger rings also have difficulty reaching the planar, achiral geometry?

So the answer to your original question is - it depends.  It depends on the temperature you specify.  At room temperature, trans-cyclooctene is the only trans-cycloalkene where the two enantiomeric conformations are stable enough (e.g. the barrier to reach the planar conformation is high enough) to be isolated.  If we lower the temperature to 0°C, then we can isolate the enatiomers of trans-cyclononene.  If we go low enough in temperature, we could isolate enantiomers of n-butane.


Calcium reactivity vs electronic configuration

The electronic configuration of Ca is $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2}$

It has a total 4 Energy levels. The outermost energy level has 2 electrons called valance electrons. As, Calcium is metal and it can easy donate two electrons and becomes cation (Ca+2). Now, this cation is stable.

Calcium is reactive because it can easy donate electrons by providing energy to it called ionisation energy.


This is not just some vapor pressure. This is the equilibrium vapor pressure. Thermodynamics is all about equilibrium, you know. And equilibrium, roughly speaking, is what takes place in a closed container after a billion years.

Immiscible as they are, the liquids still have some solubility in each other (maybe extremely low, but anyway). Over the course of centuries, molecules of the lower liquid will slowly creep one by one through the upper layer and vaporize. Eventually they will saturate the gas phase and create the same vapor pressure as if the upper liquid were not there.

Stirring the whole mixture, as Klaus suggested, is quite likely to speed things up a bit.


Your chemistry teacher is making a few simplifications there that make the statement false on a black-and-white true-and-false scale. Protons would repel each other electrostaticly due to their same charges. Neutrons interact with protons by the so-termed strong interaction (because it is stronger than the weak interaction; props to physicists for inventing names) which lead to stability of the nucleus, preventing it from falling apart due to the repulsion.

Hydrogen, specifically the $\ce{^1H}$ isotope, does not need neutrons to stabilise the nucleus because one proton cannot repel itself.

Note that ‘electrons crashing into the nucleus’ is a very strong simplification that you should eliminate from your language. Electrons are waves for all that matters so they are all around the nucleus at the same time. In fact, the highest probability of detecting them is right there, in the nucleus. (On a simpler scale, Bohr once invented ‘orbits’ that you can rationalise much like the orbit of a satellite, which also does not crash onto Earth. But don’t tell a quantum chemist or nuclear physicist that I said that!)

Finally, it is wrong to say ‘at least as many neutrons as protons’. $\ce{^3He}$ is a stable isotope with less neutrons than protons in the nucleus.


How to make by dilution a certain ion concentration

When you use the number of moles of calcium ions, you are actually using the number of moles of calcium nitrate. You have done the 1:2 conversion already in your question. You found the amount of nitrate ions needed to be 0.05. But then instead of converting that amount to moles of calcium ions, you were actually converting it to a number of moles of calcium nitrate, since the ratio of moles calcium ($\ce{Ca}$) to moles calcium nitrate ($\ce{Ca(NO3)2}$) is 1:1.


As propylene glycol can dissolve plastic...

This claim is absurd. Propylene glycol cannot dissolve poly(olefin) polymers especially polypropylene or polyethylene. It is possible for glycols to cause paneling of containers made to such, but not dissolve; and certainly wouldn't happen in the amount of time it takes to perform an injection.


How do hydrogen bonds break the ionic bonds of a crystal lattice?

Your presumption that it is hydrogen bonding that breaks the ionic bonds is wrong. Forget H-bonds. It is true that they exist in both pure water and in a sodium chloride solution. Water molecules surround both the Na+ and Cl- ions. The interaction there is dipole-ion. Na-OH2 and Cl-HOH. As is obvious, only one of those involve an H atom. (note that the "-" I used is not intended to suggest either a covalent or an ionic bond, rather it is intended to show the electronic interaction between the two oppositely charged species (the Cl & Na ions have full integer charge, the H & O atoms have only partial opposite charge, mostly induced.) This question can be answered just by hand-waving about the thermodynamics: the energy of the solvated Na and Cl ions is less than that of those two ions in the crystal. So, on average, the ions will leave the crystal and enter the liquid (until their concentration is so high that equilibrium is established between the two phases). So, while it is true that HOH...Cl bond can be called a H-bond, thinking about it this way has no obvious advantage as far as I can see. The dipole-ion interactions that are occurring are indiscriminate. Each ion is surrounded by lots of (oriented) water molecules. To learn more see https://en.wikipedia.org/wiki/Solvation_shell and especially https://en.wikipedia.org/wiki/Solvation


What are the products of the thermal decomposition of potassium cyanate with carbon at elevated temperature?

[...] some of the nitrogen is lost and forms something [...]

Carbon is a good and cheap reductant, often used in the processing of oxidic ores - except when metal carbides are formed. Carbon dioxide a reasonable product in these reactions.

$\ce{2KOCN + C ->[\Delta] 2 KCN + CO2}$

Let's agree that potassium cyanate is reduced to potassium cyanide, while carbon dioxide is released.


What is the length of a unit cell of CuCl assuming that it is fcc?

Well, yes, depending on the detail you want to go into. You can find a representation of the crystal structure of CuCl at, for example, http://www.webelements.com/compounds/copper/copper_chloride.html - you will see that the Cu atoms are indeed a simple fcc lattice.  The Cl atoms are on sites that are an interpenetrating fcc lattice.  The end result is the zinc blende (ZnS) lattice prototype, very familiar to semiconductor folks as the crystal structure for GaAs an associated III-V compounds. If all sites were occupied by the same atom, this is known as diamond cubic, commonly seen in diamond, Si, and Ge (amongst others).

But, yes, one way to consider the crystal (particularly for your purposes here) is an fcc crystal with a Cu-Cl basis. Just don't expect the x-ray diffraction to correspond to fcc.


Change of electron spin in Carbon after Hybridization?

Short answer: No it didn’t. But hybridisation also never happened in the way it is drawn.

The typical hybridisation picture from textbooks that you have shown us, is not ‘the truth’, it’s not ‘an approximation of the truth’ and not even ‘somewhere near to the truth’. It is no more and no less than a set of drawings trying to use one concept to explain an experiment.

The experiment in this case is the structure of methane, which is found to be tetrahedral, while the concept is that of orbitals as standing waves and solutions of the Schrödinger equation. The latter says that in a carbon atom floating around in sheer nothingness, there should be five standing waves produced by the electrons, that were given the name 1s, 2s, 2px, 2py and 2pz. But this is an entirely hypothetical construct from our understanding — not many people have been to outer space and nowhere have people observed and experimented with single, unbound carbon atoms.

Carbon is only found in compounds (on Earth and in the near space, that is). But once it’s in a compound, you need to write down a whole new Schrödinger equation, that not only includes the carbon but also all the other atoms in the molecule. If you solve that equation, you will again get a set of standing waves, but waves, that are spread out across the entire molecule. You need to perfom quite a few mathematical tricks to reduce those waves back to the orbitals you calculated for a single carbon atom — but by doing so, you have lost the predictive power of where the electrons are to go. There is no longer a way to say if any electron is spin $\alpha$ or spin $\beta$.

For methane, calculations that there are in fact four bonding orbitals at two (!) different energy levels. If hybridisation occurred, one would expect only one energy level, because all carbon orbitals would have been made equal. The orbitals generally represent a carbon ground-state orbital added to the corresponding hydrogen orbitals of the same phase (carbon’s p-orbitals oriented so that the stab in between two hydrogens). The lower energy state of course corresponds to carbon’s 2s plus the four hydrogens. (See pictures in this German lecture script, part A, section 4.8)

Take a different molecule, e. g. formaldehyde. Drawing hybridisation schemes, you would expect the oxygen to be something like $\mathrm{sp}^2$ and have a ‘rabbit ear’-like pair of free electrons. But solving the Schrödinger equation does not give you any free electrons. Instead you find a multitude of orbitals, none of which exclusively surrounds the oxygen — excepting the one that represents oxygen’s 1s in simplified diagrams. Again, reducing this to any type of structure that would tell you anything about the ‘shape’ of the free oxygen electrons means that you would lose predictive power. (Luckily in that case, though, all electrons are spin-paired, so they will remain spin-paired.)


Confusion in understanding temperature change in exothermic reactions

I think this comes down to what constitutes the system. If we think of the system as the reactants and products in a container, releasing heat will increase the temperature of the system.


Why does H₂Se have a bigger Ka2 than H₂Te

The following data are from the Rubber Book (CRC Handbook of Chemistry and Physics), 89th edition, 2009, p 8-40.

\begin{array}{lll}
\mathrm{acid} & \mathrm{pK_{a1}} & \mathrm{pK_{a2}} \\
\hline
\ce{H2S} & 7.05 & 19\\
\ce{H2Se} & 3.89& 11\\
\ce{H2Te} & 2.6 & 11\\
\end{array}

Note that $\mathrm{pK_{a1}}$ of $\ce{H2Te}$ was determined at 18 °C, all other data at 25 °C


Different masses for the same element and isotope?

As noted by Ivan in his comment, the difference is likely due to using the helium nucleus $\ce{^4He^2+}$ and not a helium atom. Shame on the Khan Academy video for not making this clear.

However, there is a good reason to only consider the nucleus. The mass defect and binding energy are consequences of the strong nuclear fore, and thus properties of the nucleus.

The choice of helium-4 is convenient: the helium-4 nucleus is the alpha particle - a common product of radioactive decay and thus well-characterized. Thanks to mass spectrometry, we can determine the mass of an alpha particle very precisely to be $4.00150608\ \text{u}$. Here is the source of that mysterious number. Again, shame on Khan Academy for not being explicit.

The electron has a mass of $5.48579909\times 10^{-4}\ \text{u}$, so a helium-4 atom would be expected to have a mass of one alpha particle and two electrons:

$$4.00150608\ \text{u} + 2\left(5.4858\times 10^{-4}\ \text{u}\right) = 4.00260324\ \text{u}$$


Why benzene takes in more delocalized electrons from its substituents even though it has six of them

Using atomic orbitals to construct molecular orbitals involves some shifting and redrawing because of overlaps. Benzene is symmetrical, so whether you draw six carbon atomic p-orbitals or squash and spread them out, you get the idea. Now, when you look at phenol, the C-O bond is unsymmetrical, so if you want to be more precise, you say that oxygen draws some electron density from the carbon thru the sigma system (because oxygen is more electronegative).

The oxygen becomes richly negative at carbon's expense, but this isn't fair(!), so it gives back a little electron density thru the pi system of molecular orbitals that it doesn't care about so much (because the pi lobes are farther from the oxygen nucleus). If you draw molecular orbitals, draw the sigma bond with more electron density toward oxygen, and draw the pi system with electron density a wee bit toward carbon.

The energy diagram of the molecular orbitals of benzene will be shifted by putting a hydroxyl on the ring. One one hand you can consider the six orbitals of the aromatic system merely being shifted in the space above the carbon ring, or, on the other hand, you could consider the three filled pi orbitals to remain essentially the same while oxygen donates some electron density to the unfilled (anti-bonding) orbitals (ignoring the sigma system, which just gets shoved around a little).

So the bottom line is that you are not putting whole electrons into someplace new, but swelling up some electron balloons while squashing some others.


How to predict precipitate upon evaporation from calcium ions and either chloride, carbonate, or nitrate counterions?

This question is testing your knowledge of general solubility trends, which largely need to be memorized. Here is a fairly comprehensive list, and there are many others scattered through the internet.

You are correct that nearly all nitrates are soluble, so that would be a bad guess.

The halides are generally pretty soluble. There are more exceptions than for nitrates, but as a rule halide salts tend to be soluble.

Carbonates however are frequently insoluble. The Group I cations tend to be an exception, but not the Group II cations, like $\ce{Ca^2+}$.

So, the general trend in solubility for the three anions in your question is $\ce{NO3- > Cl- > CO3^2-}$, so you would expect $\ce{CaCO3}$ to precipitate first.

We can verify this against the known solubilities of each of the salts at room temperature:

$\begin{array}{rr}
\pu{\ce{Ca(NO3)2}}: & \pu{1212 g/L }\\
\pu{\ce{CaCl2}}: & \pu{75g/L}\\
\pu{\ce{CaCO3}}: & \pu{0.01g/L}
\end{array}$

So, the quantitative solubility data confirms the original prediction and $\ce{CaCO3}$ will clearly precipitate first.


What is the geometry of a tri-substituted methyl free radical?

A priori, I would expect substituted methyl radicals to adopt a planar structure due to Bent’s rule.
However, I consider it possible for methyl radicals with electron-withdrawing substituents to adopt a trigonal-pyramidal structure too. In this case, Bent’s rule would imply that the electron-withdrawing substituents are ‘more electron-withdrawing’ than the radical is electron-poor. In such a case, it might be beneficial to distort the planar structure, causing rehybridisation and thus resulting in $\mathrm{sp}^n$ orbitals across the board so that all orbitals together add up to $\mathrm{sp^3}$. This structure may not be perfectly pseudotetrahedral but close enough to have a significant s-contribution to the singly-occupied orbital.
Note that $\mathrm{sp^3}$-type radicals are not inherently unstable (unlike $\mathrm{sp^3}$-type cations) as can be seen in bicyclo[2.2.2]octa-1-yl radicals (or bridgehead radicals in general). This points to $\mathrm{sp^3}$ radicals being accessible and thus the finer electronic environments mattering.

Addendum: Ron has shared links to two other questions he answered (1, 2) which argue the case that methyl radicals would be (slightly) pyramidal rather than planar in their ground state. So my a priori expectation might be wrong. I believe it might be very hard to prove the structure conclusively, as it is expected for a pyramidal radical to rapidly inverse as nitrogen does in ammonia and its derivatives so that any non-short timescale measurements would result in an averaged structure (which would be planar).


Determine reaction order and reaction rate

The rate equation is $$\ln[\ce{Fe^{3+}}] = \ln[\ce{Fe^{3+}}]_0 − kt$$ So, $a-x=[A]$. Don't subtract the  next value. Put it as it is. That's just for your understanding, however. Don't try to use the integrated rate form to plot the graph. $[\ce{Fe^{3+}}]=238 \text{ where } t=10 \text{ and } [\ce{Fe^{3+}}]=227 \text{ where } t=20$. Using the above mentioned equation: $$\ln[A] = \ln[A]_0 − kt$$ Calculate the value of $\mathrm{\ln[\ce{Fe^{3+}}]_0}$ by eliminating k. Thereafter, you can graph the above mentioned equation $\ln[\ce{Fe^{3+}}] \text{ vs time t}$ with $\ln[\ce{Fe^{3+}}]_0$ as the y-intercept. The slope will yield the value of the rate constant.

[...]that would be the rate of the reaction?

For some reason unbeknownst to me, a lot of people tend to confuse between rate and rate constant. The slope will yield the rate constant. Next,
$$\textrm{rate}=k[\textrm A]^{a} \quad \text{}$$ where $a$ is the order of the reaction with respect to reactant $\ce{A}$.
As a side-note, since this is a pseudo-first order reaction, the rate law equation, initially this: $$\text{Rate} = k^{'}[\ce{Fe^{3+}}][\ce{H2O}]  $$
now appears as this:$$\text{Rate}=k[\ce{Fe^{3+}}]$$
the variable $'k'$ in the second equation is the slope that you obtained from the graph. Here $k=k^{'}[\ce{H2O}]_o$
For more about pseudo-first order reactions you can check one of my previous answers.


Applying Avagadro's law: the relationship between moles and volume of a gas

You are correct in your approach and answer. Assuming the gas to be ideal, one can apply Avogadro's Law which states that for an ideal gas $V \propto n$.

This implies $V_1/n_1 = V_2/n_2$.

Using the given values one does obtain the volume as $568.75\ \mathrm{mL}$


How can I understand the Lewis acid-base interaction of iron(III) and water?

So, effectively, the question is: How are electrons from the covalent bond between ligand and central metal atom or ion placed with regard to the electron configuration?

$\ce{Fe^{3+}}$ has the condensed electron configuration of [Ar]$3d^5$. When the ligands approach the iron ion from the X, Y, and Z, the ion forms six bonds. It would be incorrect to state that the remaining 3d electrons are involved in the bonding with the ligands. Instead, the iron ion uses 6 orbitals from the 4s, 4p, and 4d orbitals to accept the lone pairs of electrons from the ligands. However, before they are used, the orbitals are hybridized in order to create six orbitals of equal energy.

If you want to dig deeper into the subject, I highly recommend the lecture notes from this webpage here.


Changes in physical properties in homologous series : Solubility

My question is regarding the last line, what kind of gradation in solubiltiy is observed? Increase or decrease?

Both

Solubility is not one single property, but a collection of related properties. Solubility in water is different from solubility in ethanol, which is different from solubility in benzene.

The cause of the gradation of (all of these) properties is a systematic change in intermolecular forces. The number and strength of these forces increase as molecular mass increases, which increases the boiling points and melting points. For solubility, we care about the interplay between the solute and the solvent, so we also need to know the type of intermolecular force (polar vs. nonpolar) in addition to the total strength.

For the homologous series of alcohols below, I have listed their water solubilities (mined from UC Davis's ChemWiki and from Wikipedia articles). Notice that as the number of $\ce{CH2}$ units increases, the water solubility decreases. This decrease occurs because, while the strength of nonpolar intermolecular forces (London-Dispersion) increases with each added $\ce{CH2}$ until the nonpolar interactions overpower the polar interactions (dipole-dipole and hydrogen-bonding) of the alcohol functional group. Water is a polar solvent, and with each added $\ce{CH2}$, the alcohol molecules begin to have stronger attraction to themselves than to water, and the solubility decreases.

Solubility of the homologous series of linear n-alkanols in water

Methanol $\ce{CH3OH}$ - miscible (infinite)
Ethanol $\ce{CH3CH2OH}$ - miscible (infinite)
n-propanol $\ce{CH3CH2CH2OH}$ - miscible (infinite)
n-butanol $\ce{CH3CH2CH2CH2OH}$ - 80 g / L
n-pentanol $\ce{CH3CH2CH2CH2CH2OH}$ - 22 g / L
n-hexanol$\ce{CH3CH2CH2CH2CH2CH2OH}$ - 5.9 g / L

At the same time, the solubility of this homologous series of alcohols increases in nonpolar solvents like hexane and benzene. Those data are not readily available, but a quick experiment will verify it. The increased amount of nonpolar interactions in the alcohols increases their affinity with nonpolar solvents.


What is the chemistry behind amalgamation?

An amalgam is simply an alloy of some metal with mercury.

Alloys are simply solid solutions with metals. Since the mixture of multiple elements disrupts the metal bonding, the properties are often quite different than the components. (In some cases, alloys form so-called intermetallic compounds with defined stoichiometry, but more often, the ratios are very flexible.)

I wouldn't describe the mixture as a reaction, in much the same way that I wouldn't really call dissolving sodium chloride in water as a reaction. You're forming a mixture. Water disrupts the ionic bonding in NaCl, and mercury and other metals can also accept other components or impurities.

Gallium does alloy with a lot of metals, and indium-gallium eutectic alloy is a liquid at room temperature, as is "galinstan".

As to your question about mercury dissolving the metal atoms, consider that mercury is also a metal, so it also participates in similar metal bonding.