Concept explainers
(a)
Interpretation:
The amount of each isotope present after 14 days needs to be determined.
Concept Introduction:
Half-life − It is the time required by original radioactive element to reduce to the half of its initial concentration. Thus, at half-life (
The decay of the radioactive element can be described by the following formula-
Where
N(t) − amount of reactant at time t
N0 − Initial concentration of the reactant
t1/2 − Half-life of the decaying reactant

Answer to Problem 52P
After 14.0 days,
Amount of Iodine-131 left = 62 mg
Amount of Xenon-131 formed = 62 mg
Explanation of Solution
Given Information:
N0 = 124 mg
t1/2 = 14 days
Calculation:
After 14.0 days, the initial concentration of phosphorus-32 reduces to half of its initial concentration and converts to sulfur-32.
Thus,
Hence,
Amount of phosphorus-32 left = 62 mg
Amount of sulfur-32 formed = 124 mg − 62 mg = 62 mg
(b)
Interpretation:
The amount of each isotope present after 28 days needs to be determined.
Concept Introduction:
Half-life − It is the time required by original radioactive element to reduce to the half of its initial concentration. Thus, at half-life (
The decay of the radioactive element can be described by the following formula-
Where
N(t) − amount of reactant at time t
N0 − Initial concentration of the reactant
t1/2 − Half-life of the decaying reactant

Answer to Problem 52P
After 28.0 days,
Amount of Phosphorus-32 left = 32 mg
Amount of Sulfur-32 formed = 92 mg
Explanation of Solution
Given Information:
N0 = 124 mg
t1/2 = 14 days
t = 28.0 days
Calculation:
After 28 days, amount of phosphorus-32 would be defined by N(t),where t is 28.0 days, as
Hence, the amount of phosphorus-32 decays and converts to sulfur-32. Therefore,
Amount of Phosphorus-32 left after 28.0 days = 32 mg
Amount of sulfur-32 formed after 28.0 days = 124 mg − 32 mg = 92 mg
(c)
Interpretation:
The amount of each isotope present after 42 days needs to be determined.
Concept Introduction:
Half-life − It is the time required by original radioactive element to reduce to the half of its initial concentration. Thus, at half-life (
The decay of the radioactive element can be described by the following formula-
Where
N(t) − amount of reactant at time t
N0 − Initial concentration of the reactant
t1/2 − Half-life of the decaying reactant

Answer to Problem 52P
After 42.0 days,
Amount of Phosphorus-32 left = 15.5 mg
Amount of Sulfur-32 formed = 108.5 mg
Explanation of Solution
Given Information:
N0 = 124 mg
t1/2 = 14 days
t = 42.0 days
Calculation:
After42 days, amount of phosphorus-32 would be defined by N(t),where t is 42.0 days, as
Hence, the amount of phosphorus-32 decays and converts to sulfur-32. Therefore,
Amount of Phosphorus-32 left after 42.0 days = 15.5 mg
Amount of sulfur-32 formed after 42.0 days = 124 mg − 15.5 mg = 108.5 mg
(d)
Interpretation:
The amount of each isotope present after 56 days needs to be determined.
Concept Introduction:
Half-life − It is the time required by original radioactive element to reduce to the half of its initial concentration. Thus, at half-life (
The decay of the radioactive element can be described by the following formula-
Where
N(t) − amount of reactant at time t
N0 − Initial concentration of the reactant
t1/2 − Half-life of the decaying reactant

Answer to Problem 52P
After 56.0 days,
Amount of Phosphorus-32 left = 7.75 mg
Amount of Sulfur-32 formed = 116.25 mg
Explanation of Solution
Given Information:
N0 = 124 mg
t1/2 = 14 days
t = 56.0 days
Calculation:
After 56 days, amount of phosphorus-32 would be defined by N(t),where t is 56.0 days, as
Hence, the amount of phosphorus-32 decays and converts to sulfur-32. Therefore,
Amount of Phosphorus-32 left after 56.0 days = 7.75 mg
Amount of sulfur-32 formed after 56.0 days = 124 mg − 7.75 mg =116.25 mg
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Chapter 10 Solutions
GENERAL,ORGANIC, & BIOLOGICAL CHEM-ACCES
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