Select the correct term(s) to complete each sentence. a) The is assigned a relative intensity of 100. base peak molecular ion peak b) The is the most intense peak in the mass spectrum. molecular ion peak base peak c) The represents the original molecule that has only lost an electron. molecular ion peak base peak d) could be/represent a smaller, charged fragment of the original molecule. The base peak The molecular ion A radical cation e) is/represents a positively charged species. The base peak The parent ion The molecular ion A radical cation

a) Base Peak

b) Base Peak

c) Molecular Ion

d) Base Peak and Radical Cation

e) The Parent Ion, Molecular Ion, the Radical Cation and the Base Peak

Explanation:

a) The Base Peak is assigned a relative intensity of 100.

b) The Base Peak is the most intense peak in the mass spectrum.

c) The Molecular Ion Peak represents the original molecule that has only lost an electron.

d) Base Peak and Radical Cation could be/represent a smaller, charged fragment of the original molecule.

e) The Parent Ion, the Molecular Ion, the Radical Cation/ the Base Peak is/represents a positively charged species.

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