Sn1 reactions usually proceed with a. equal amounts of inversion and retention at the center undergoing substitution b. slightly more inversion than retention at the center undergoing substitution c. slightly more retention than inversion at the center undergoing substitution d. complete inversion at the center undergoing substitution e. complete retention at the center undergoing substitution

B. slightly more inversion than retention at the center undergoing substitution

Explanation:

SN1 reaction -

In this type of reaction , nucleophilic substitution reaction takes place via formation of an intermediate , carbocation , and thereby giving the substituted product .

In the very first step of the reaction , a stable carbocation is generated , which can have the possibility to give both the products , i.e. , retention , inversion .

But , there is always more of the inversion product rather than retention .

And racemisation is 80 - 95% and the rest 20 - 5% is inversion .

true

explanation:

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years