Following reaction is involved in above system
HOCl(aq) ↔ H+(aq) + OCl-<span>(aq)
</span>OCl-(aq) + H2O(l) ↔ HOCl(aq) + OH-<span>(aq)
</span>
Now, if the system is obeys 1st order kinetics we have
K = [OCl-][H+<span>]/[HOCl] ............. (1)
</span>∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8<span>) ............. (2)
</span>
and now considering that system is obeying 2nd order kinetics, we have
K = [OH-][HOCl-] / [OCl-] ................. (3<span>)
</span>Subs 2 in 3 we get
K = [OH-][H+] (1 / 3.0 * 10-8<span>)
</span>we know that, [OH-][H+] = 10<span>-14
</span>∴K = 3.3 * 10<span>-7
</span>
Thus, correct answer is e i.e none of these
Calcium fluoride: CaF₂
Ca(2+) >>> Ar (argon)
F(-) >>> Ne (neon)
B. They may be gases, liquids, or solids at room temp.
Answer:
B: The sulfuric acid is not consumed or react with the reactant.
Benzaldehyde or C6H5CHO would not undergo the aldol condensation because it does not contain an alpha-hydrogen in its structure. Aldol condensation is a type of reaction that happens between an enolate and an aldehyde or ketone leading to a alkene that has a planar structure. The lack of an alpha-hydrogen would not allow for it to undergo such process since it cannot enolize. Benzaldehyde undergoes a nucleophilic reaction known as Claisen-Schmidt condensation. It has somehow same mechanism of the aldol reaction however, the nucleophilic attack on the carbonyl happens even without the alpha-hydrogen but with an enolate that is from a ketone.
