EBK GENERAL, ORGANIC, & BIOLOGICAL CHEM
EBK GENERAL, ORGANIC, & BIOLOGICAL CHEM
3rd Edition
ISBN: 9781259298424
Author: SMITH
Publisher: VST
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Chapter 10, Problem 10.50P
Interpretation Introduction

(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 ( t1/2 )

  Concentration of reactant= Initial concentration2

The decay of the radioactive element can be described by the following formula-

  N(t)=N0(12)tt 1/2

Where

N(t) − amount of reactant at time t

N0 − Initial concentration of the reactant

t1/2 − Half-life of the decaying reactant

Expert Solution
Check Mark

Answer to Problem 10.50P

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,

  N(t=14.0 days)=N02=124 mg2N(t=14.0 days)=62 mg

Hence,

Amount of phosphorus-32 left = 62 mg

Amount of sulfur-32 formed = 124 mg − 62 mg = 62 mg

Interpretation Introduction

(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 ( t1/2 )

  Concentration of reactant= Initial concentration2

The decay of the radioactive element can be described by the following formula-

  N(t)=N0(12)tt 1/2

Where

N(t) − amount of reactant at time t

N0 − Initial concentration of the reactant

t1/2 − Half-life of the decaying reactant

Expert Solution
Check Mark

Answer to Problem 10.50P

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

  N(t)=N0(12)t t 1/2 N(t)=(124 mg) (12) 28.0 days 14.0 daysN(t)=(124 mg) (12)2N(t)=(124 mg) (14)N(t)=32 mg

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

Interpretation Introduction

(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 ( t1/2 )

  Concentration of reactant= Initial concentration2

The decay of the radioactive element can be described by the following formula- N(t)=N0(12)tt 1/2

Where

N(t) − amount of reactant at time t

N0 − Initial concentration of the reactant

t1/2 − Half-life of the decaying reactant

Expert Solution
Check Mark

Answer to Problem 10.50P

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

   N(t)= N 0 ( 1 2 ) t t 1/2

   N(t)=(124 mg)  ( 1 2 ) 42.0 days 14.0 days

   N(t)=(124 mg)  ( 1 2 ) 3

   N(t)=(124 mg) ( 1 8 )

   N(t)=15.5 mg

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

Interpretation Introduction

(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 ( t1/2 )

  Concentration of reactant= Initial concentration2

The decay of the radioactive element can be described by the following formula-

  N(t)=N0(12)tt 1/2

Where

N(t) − amount of reactant at time t

N0 − Initial concentration of the reactant

t1/2 − Half-life of the decaying reactant

Expert Solution
Check Mark

Answer to Problem 10.50P

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

   N(t)= N 0 ( 1 2 ) t t 1/2

   N(t)=(124 mg)  ( 1 2 ) 56.0 days 14.0 days

   N(t)=(124 mg)  ( 1 2 ) 4

   N(t)=(124 mg) ( 1 16 )

   N(t)=7.75 mg

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

EBK GENERAL, ORGANIC, & BIOLOGICAL CHEM

Ch. 10.2 - Prob. 10.11PCh. 10.2 - Prob. 10.12PCh. 10.3 - Prob. 10.13PCh. 10.3 - Prob. 10.14PCh. 10.3 - Prob. 10.15PCh. 10.3 - Prob. 10.16PCh. 10.4 - Prob. 10.17PCh. 10.4 - Prob. 10.18PCh. 10.4 - Prob. 10.19PCh. 10.5 - Prob. 10.20PCh. 10.5 - Prob. 10.21PCh. 10.5 - Prob. 10.22PCh. 10.6 - Prob. 10.23PCh. 10.6 - Prob. 10.24PCh. 10 - Compare fluorine-18 and fluorine-19 with regard to...Ch. 10 - Prob. 10.26PCh. 10 - Prob. 10.27PCh. 10 - Prob. 10.28PCh. 10 - Prob. 10.29PCh. 10 - Prob. 10.30PCh. 10 - Prob. 10.31PCh. 10 - Prob. 10.32PCh. 10 - Prob. 10.33PCh. 10 - Prob. 10.34PCh. 10 - Prob. 10.35PCh. 10 - Prob. 10.36PCh. 10 - Prob. 10.37PCh. 10 - Prob. 10.38PCh. 10 - Prob. 10.39PCh. 10 - Prob. 10.40PCh. 10 - Prob. 10.41PCh. 10 - Prob. 10.42PCh. 10 - Prob. 10.43PCh. 10 - Prob. 10.44PCh. 10 - Prob. 10.45PCh. 10 - Prob. 10.46PCh. 10 - Prob. 10.47PCh. 10 - Prob. 10.48PCh. 10 - Prob. 10.49PCh. 10 - Prob. 10.50PCh. 10 - Prob. 10.51PCh. 10 - Prob. 10.52PCh. 10 - Prob. 10.53PCh. 10 - Prob. 10.54PCh. 10 - Prob. 10.55PCh. 10 - Prob. 10.56PCh. 10 - Prob. 10.57PCh. 10 - Prob. 10.58PCh. 10 - Prob. 10.59PCh. 10 - Prob. 10.60PCh. 10 - Prob. 10.61PCh. 10 - Prob. 10.62PCh. 10 - Prob. 10.63PCh. 10 - Prob. 10.64PCh. 10 - Prob. 10.65PCh. 10 - Prob. 10.66PCh. 10 - Prob. 10.67PCh. 10 - Prob. 10.68PCh. 10 - Prob. 10.69PCh. 10 - Prob. 10.70PCh. 10 - Prob. 10.71PCh. 10 - Prob. 10.72PCh. 10 - Prob. 10.73PCh. 10 - Prob. 10.74PCh. 10 - Prob. 10.75PCh. 10 - Prob. 10.76PCh. 10 - Prob. 10.77PCh. 10 - Prob. 10.78PCh. 10 - Prob. 10.79PCh. 10 - Prob. 10.80PCh. 10 - Prob. 10.81PCh. 10 - Prob. 10.82PCh. 10 - Prob. 10.83PCh. 10 - Prob. 10.84PCh. 10 - Prob. 10.85PCh. 10 - Prob. 10.86PCh. 10 - Prob. 10.87PCh. 10 - Prob. 10.88PCh. 10 - Prob. 10.89PCh. 10 - Prob. 10.90PCh. 10 - Prob. 10.91PCh. 10 - Prob. 10.92PCh. 10 - Prob. 10.93CPCh. 10 - Prob. 10.94CP
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