Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
I’m pretty sure it’s abode and cathode
Answer:
Repeated SN2 reactions occur leading to the formation of a racemic mixture
Explanation:
S-2-iodooctane is a chiral alkyl halide with an asymmetric carbon atom. The presence of an asymmetric carbon atom implies that it can rotate plane polarized light and thus lead to optical isomerism. The two configurations of the compound are R/S according to the Cahn-Prelong-Ingold system.
However, when S-2-iodooctane is treated with sodium iodide in acetone, repeated SN2 reactions occur since the iodide ion is both a good nucleophile and a good leaving group. Hence a racemic modification is formed in the system with time hence we end up with (±)- Iodooctane.
Explanation:
so for this u have to use this equation where
Moles = number of particle/6.02×10^23
= 3.045 × 10^24/6.02×10^23
= 5.0581
write it to 3 S.F so 5.06 moles
If there is a constant loss of 
the concentration of 
will be affected and decrease since 
is a main component of 
.
Hope it helped,
BioTeacher101
