The "normal" lead content in human blood is about 0.40 part per million (i. e., 0.40 g of lead per million grams of blood). a value of 0.80 part per million (ppm) is considered to be dangerous. how many grams of lead are contained in 6.0 x 10^3 g of blood (the amount in an average adult) if the lead content is 0.62 ppm?

3.7 × 10⁻³ g Pb

Explanation:

0.62 parts per million means that there are 0.62 g of lead per million grams of blood (10⁶ grams of blood). In order to find the grams of lead on 6.0 × 10³ grams of blood (the amount in an average adult), we will use a conversion fraction.

6.0 × 10³ g blood × (0.62 g Pb/ 10⁶ g blood) = 3.7 × 10⁻³ g Pb

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

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)