In a gas chromatography experiment, a mixture of two compounds (a and b) was separated for analysis. a chart recorder that fed paper at a constant rate of 100 mm. per minute was used to record the chromatogram, and a mark was made on the chart at the time the sample was injected. the first peak on the chromatogram corresponds to compound a, and that peak was observed at a distance of 5.19 cm from the sample injection point, and it had a height of 3.05 cm and a width at half-height of 1.16 cm. the peak corresponding to b was observed at a distance of 8.23 cm from the injection point, and it had a height of 4 cm and a width at half-height of 1.9 cm. what was the retention time for compound b (in seconds)? (enter your answer as a number only.)

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