Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
Question
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Chapter 19, Problem 19.55QP

(a)

Interpretation Introduction

Interpretation: The information regarding the elevation of Strontium 90 in cow’s milk after the Chernobyl accident is given. Various questions based on the given radioactivity are to be answered.

Concept introduction: The time taken by any species to get reduced to half of its original concentration is called half-life of that species.

To determine: The balanced equation for the decay of 90Sr .

(a)

Expert Solution
Check Mark

Answer to Problem 19.55QP

Solution

The balanced equation for the decay process is shown as,

3890Sr3990Y+10e

Explanation of Solution

Explanation

There occurs beta decay in case of 90Sr . Therefore, the final product has one more extra proton than the reactant. The balanced equation for the decay process is shown as,

3890Sr3990Y+10e

(b)

Interpretation Introduction

To determine: The number of atoms present in 200mL of glass of milk.

(b)

Expert Solution
Check Mark

Answer to Problem 19.55QP

Solution

The number of atoms present in 200mL of glass of milk is 3.125×108atom_ .

Explanation of Solution

Explanation

Given

The half life of Strontium 90 is 28.8years .

The radioactivity is 1.25Bq/L .

The volume of milk is 200mL

The conversion of years into days is done as,

1year=365days

Hence the conversion of 28.8years into days is done as,

28.8years=28.8×365days=10512days

The conversion of days into h is done as,

1day=24h

Hence the conversion of 10512days into h is done as,

10512days=10512×24h=252,288h

The conversion of h into s is done as,

1h=3600s

Hence the conversion of 252,288h into s is done as,

252,288h=252,288×3600s=9.08×108s

The half-life of first order reaction is given as,

t1/2=0.693λ (1)

Where,

  • t1/2 is the half-life period.
  • λ is the decay constant of the reaction.

Substitute the value of t1/2 in the above equation,

t1/2=0.693λλ=0.6939.08×108s=0.08×108s1

This shows about 0.08×108disintegration atom1s1 .

The relation between Becquerel and disintegration s1 is given as,

1Bq=1disintegration s1

Hence, the conversion of 0.08×108disintegration s1 to Bq is done as,

0.08×108disintegration s1=0.08×108Bq

The conversion of mL to L is done as,

1mL=0.001L

Hence, the conversion of 200mL to L is done as,

200mL=200×0.001L=0.2L

The activity in 1L of glass of milk =1.25Bq

The activity in 0.2L of glass of milk =0.2×1.25Bq=0.25Bq

The activity 0.08×108Bq is present in =1atom

The activity 1Bq is present in =10.08×108atom

The activity 0.25Bq is present in =0.25×10.08×108atom=3.125×108atom_

Therefore, the number of atoms present in 200mL of glass of milk is 3.125×108atom_ .

(c)

Interpretation Introduction

To determine: The reason behind the more concentration of Strontium 90 in milk rather than in grains, fruits and vegetables.

(c)

Expert Solution
Check Mark

Answer to Problem 19.55QP

Solution

The similarity in the chemistry of Strontium and Calcium makes Strontium more concentrated in milk rather than in grains, fruits and vegetables.

Explanation of Solution

Explanation

As Strontium is member of Group 2 alkaline earth metals and Calcium also belongs to the same group. It means that both should have the similar chemistry. As Calcium is present in milk, it ultimately leads to the more presence of Strontium 90 in milk rather than in grains, fruits and vegetables.

Conclusion

  1. a. The balanced equation for the decay process is shown as,
  2. b. 3890Sr3990Y+10e
  3. c. The number of atoms present in 200mL of glass of milk is 3.125×108atom_ .
  4. d. The similarity in the chemistry of Strontium and Calcium makes Strontium more concentrated in milk rather than in grains, fruits and vegetables

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Chapter 19 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 19 - Prob. 19.5VPCh. 19 - Prob. 19.6VPCh. 19 - Prob. 19.7VPCh. 19 - Prob. 19.8VPCh. 19 - Prob. 19.9VPCh. 19 - Prob. 19.10VPCh. 19 - Prob. 19.11QPCh. 19 - Prob. 19.12QPCh. 19 - Prob. 19.13QPCh. 19 - Prob. 19.14QPCh. 19 - Prob. 19.15QPCh. 19 - Prob. 19.16QPCh. 19 - Prob. 19.17QPCh. 19 - Prob. 19.18QPCh. 19 - Prob. 19.19QPCh. 19 - Prob. 19.20QPCh. 19 - Prob. 19.21QPCh. 19 - Prob. 19.22QPCh. 19 - Prob. 19.23QPCh. 19 - Prob. 19.24QPCh. 19 - Prob. 19.25QPCh. 19 - Prob. 19.26QPCh. 19 - Prob. 19.27QPCh. 19 - Prob. 19.28QPCh. 19 - Prob. 19.29QPCh. 19 - Prob. 19.30QPCh. 19 - Prob. 19.31QPCh. 19 - Prob. 19.32QPCh. 19 - Prob. 19.33QPCh. 19 - Prob. 19.34QPCh. 19 - Prob. 19.35QPCh. 19 - Prob. 19.36QPCh. 19 - Prob. 19.37QPCh. 19 - Prob. 19.38QPCh. 19 - Prob. 19.39QPCh. 19 - Prob. 19.40QPCh. 19 - Prob. 19.41QPCh. 19 - Prob. 19.42QPCh. 19 - Prob. 19.43QPCh. 19 - Prob. 19.44QPCh. 19 - Prob. 19.45QPCh. 19 - Prob. 19.46QPCh. 19 - Prob. 19.47QPCh. 19 - Prob. 19.48QPCh. 19 - Prob. 19.49QPCh. 19 - Prob. 19.50QPCh. 19 - Prob. 19.51QPCh. 19 - Prob. 19.52QPCh. 19 - Prob. 19.53QPCh. 19 - Prob. 19.54QPCh. 19 - Prob. 19.55QPCh. 19 - Prob. 19.56QPCh. 19 - Prob. 19.57QPCh. 19 - Prob. 19.58QPCh. 19 - Prob. 19.59QPCh. 19 - Prob. 19.60QPCh. 19 - Prob. 19.61QPCh. 19 - Prob. 19.62QPCh. 19 - Prob. 19.63QPCh. 19 - Prob. 19.64QPCh. 19 - Prob. 19.65QPCh. 19 - Prob. 19.66QPCh. 19 - Prob. 19.67QPCh. 19 - Prob. 19.68QPCh. 19 - Prob. 19.69QPCh. 19 - Prob. 19.70QPCh. 19 - Prob. 19.71QPCh. 19 - Prob. 19.72QPCh. 19 - Prob. 19.73QPCh. 19 - Prob. 19.74QPCh. 19 - Prob. 19.75QPCh. 19 - Prob. 19.76QPCh. 19 - Prob. 19.77QPCh. 19 - Prob. 19.78QPCh. 19 - Prob. 19.79QPCh. 19 - Prob. 19.80QPCh. 19 - Prob. 19.81QPCh. 19 - Prob. 19.82QPCh. 19 - Prob. 19.83QPCh. 19 - Prob. 19.84QPCh. 19 - Prob. 19.85APCh. 19 - Prob. 19.86APCh. 19 - Prob. 19.87APCh. 19 - Prob. 19.88APCh. 19 - Prob. 19.89APCh. 19 - Prob. 19.90APCh. 19 - Prob. 19.91APCh. 19 - Prob. 19.92APCh. 19 - Prob. 19.93APCh. 19 - Prob. 19.94APCh. 19 - Prob. 19.95APCh. 19 - Prob. 19.96APCh. 19 - Prob. 19.97APCh. 19 - Prob. 19.98APCh. 19 - Prob. 19.99APCh. 19 - Prob. 19.100AP
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