Hydrogenation (H2/Pd-C) of 3,4-dimethyl-3-hexene (3,4-dimethyl hex-3-ene) would be expected to form A) an enantiomeric pair for both the (E) and (Z) isomers. B) an enantiomeric pair for the (E) isomer and one meso compound for the (Z) isomer. C) one meso compound for the (E) isomer and an enantiomeric pair for the (Z) isomer. D) will not form any stereoisomers since the product is a saturated hydrocarbon. E) None of the above provide a correct response.

D) will not form any stereoisomers since the product is a saturated hydrocarbon.

Explanation:

Stereoisomerism, or spatial isomerism,refers to a type of isomerism in which  two or more molecules have the same molecular formula and atom to atom connectivity but different three-dimensional orientations of those atoms in space.

Stereoisomerism occurs in alkenes because free rotation about the C=C bond is hindered. Thus, the molecule is "locked" in a particular orientation. In a saturated alkane, there is unrestricted free rotation about the C-C bond hence straight chain alkanes do not have steroisomers.

The product formed is 3,4-dimethyl hexane which is a straight chain saturated alkane. This compound does not exhibit stereoisomerism.

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