If there are the same number of argon atoms in one sample and the same number of carbon atoms in another, then the value of the ratio, mass of the carbon sample/the mass of argon sample would be? a. 1.0 b. 3.3 c. 0.30 d. 12 e. 40

c. 0.30

Explanation:

One C atom has a mass of 12.0 u and one Ar atom has a mass of 40.0 u.

Let’s assume that there are n atoms of C and n atoms of Ar. Then

Mass of n C atoms = n × 12.0

Mass of n C atoms = 12.0n u

Mass of n Ar atoms = n × 40.0

Mass of n Ar atoms = 40.0n u

ratio = mass of C/mass of Ar

ratio = 12.0n/40.0n

ratio = 0.300

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

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