The rate of Formation of Carbocation mainly depends on two factors'
1) Stability of Carbocation: The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.
2) Ease of detaching of Leaving Group: The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,
R-I > R-Cl > R-F
B > C > A
If I am correct I would believe that it would be a muscle cell.
<span>Answer:
.01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L.
so .01/.015 to .005/.015 ~ 67% D to 33% L.
And thus, the enantiomer excess will be 34%.</span>
Answer:
0.192 mol.
Explanation:
- To calculate the no. of moles of a substance (n), we use the relation:
<em>n = mass / molar mass.</em>
mass of AsH₃ = 15.0 g.
molar mass of AsH₃ = 77.95 g/mol.
∴ The number of moles in 15.0 g AsH₃ = mass / molar mass = (15.0 g) / (77.95 g/mol) = 0.192 mol.