The number of significant figures in the measurement 0.01305 kg is
a. 5.
b. 3.
d. 4.
c. 2

The number of significant figures in the measurement 0.01305 kg is :

d. 4

Explanation:

First know the rules of significant figure:

1. All non-Zero are counted as significant figure .

Example : In 234749 = 6 non-zero digits are present = 6 significant figure

2. The zero preceding the non-zero and decimal is considered non-significant

Example :

In 0.0345 = only three significant figures are present . The bolded Zero are non-significant

003125  only 4 significant figures are present

3. The Zeros between two non-zero are always significant

Eg: 305 = This zero is significant = there are 3 significant figure

In this measurement :

0.01305 kg are significant digits = 4

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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