EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
7th Edition
ISBN: 8220106637203
Author: Chang
Publisher: YUZU
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Chapter 21, Problem 21.68QP

(a)

Interpretation Introduction

Interpretation:

The equation should be written with proper abbreviation.

Concept Introduction:

  • Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively.

  7N15(p,α)6C12Parentnucleus(Projectile,ejectile)Daughternucleus

  • On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal.
  • . Half-life period: The time required to the reduce to half of its original value.
  • This is denoted by:t1/2
  • This term is popularly use in nuclear chemistry, to describe the stability & unstability of atoms or radioactive decay.
  • For determination of first order rate constant K in half –life period we use

    K=0.693t1/2 Equation.

  • The half-life in first order does not depends upon the initial concentration.

Formula:

First order rate constant:K=0.693t1/2

Age calculation:

lnNtN0=-kt

(a)

Expert Solution
Check Mark

Answer to Problem 21.68QP

The balance equation is 1940K 1840 Ar +0-1β 

Explanation of Solution

To find: for the balanced equation write Suitable abbreviation.

Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses. Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively. So for any nuclear reaction, short hand notation will be in this form that is,

1940K(0-1β ) 1840 ArParentnucleus(Projectile,ejectile)Daughternucleus

For the given reaction,

Parentnucleus-1940KDaughternucleus-1840 ArEjectile-0-1β

 So the balanced equation can be written as,

1940K 1840 Ar +0-1β 

Conclusion

The balanced equation is 1940K 1840 Ar +0-1β .

(b)

Interpretation Introduction

Interpretation:

The age of the rock should be determine & the equation should be written with proper abbreviation.

Concept Introduction:

  • Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively.

  7N15(p,α)6C12Parentnucleus(Projectile,ejectile)Daughternucleus

  • On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal.
  • . Half-life period: The time required to the reduce to half of its original value.
  • This is denoted by:t1/2
  • This term is popularly use in nuclear chemistry, to describe the stability & unstability of atoms or radioactive decay.
  • For determination of first order rate constant K in half –life period we use

    K=0.693t1/2 Equation.

  • The half-life in first order does not depends upon the initial concentration.

Formula:

First order rate constant:K=0.693t1/2

Age calculation:

lnNtN0=-kt

(b)

Expert Solution
Check Mark

Answer to Problem 21.68QP

The age of the rock is t=3.0×109yr

Explanation of Solution

To find: The age of the rock determine using the half-life equation.

At first, we have to calculate the rate constant k

 k=0.693t1/2=0.6931.2×109yr =5.8 ×1010 yr1

Then calculate the age of the rock by subtracting K into the following equation (Nt=0.18N0)

lnNtN0=-kt

Putting the value in the above equation, we can get

ln 0.181.00= (5.8 ×1010 yr1)t 

The age of the rock is found to be

t=3.0×109yr

Conclusion

The age of the rock is determine t=3.0×109yr

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Chapter 21 Solutions

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO

Ch. 21 - Prob. 21.2QPCh. 21 - Prob. 21.3QPCh. 21 - Prob. 21.4QPCh. 21 - Prob. 21.5QPCh. 21 - Prob. 21.6QPCh. 21 - Prob. 21.7QPCh. 21 - Prob. 21.8QPCh. 21 - Prob. 21.9QPCh. 21 - Prob. 21.10QPCh. 21 - Prob. 21.11QPCh. 21 - Prob. 21.12QPCh. 21 - Prob. 21.13QPCh. 21 - Prob. 21.14QPCh. 21 - Prob. 21.15QPCh. 21 - Prob. 21.16QPCh. 21 - Prob. 21.17QPCh. 21 - Prob. 21.18QPCh. 21 - Prob. 21.19QPCh. 21 - Prob. 21.20QPCh. 21 - Prob. 21.21QPCh. 21 - Prob. 21.22QPCh. 21 - Prob. 21.23QPCh. 21 - Prob. 21.24QPCh. 21 - Prob. 21.25QPCh. 21 - Prob. 21.26QPCh. 21 - Prob. 21.27QPCh. 21 - Prob. 21.28QPCh. 21 - Prob. 21.29QPCh. 21 - Prob. 21.30QPCh. 21 - Prob. 21.31QPCh. 21 - Prob. 21.32QPCh. 21 - Prob. 21.33QPCh. 21 - Prob. 21.34QPCh. 21 - Prob. 21.35QPCh. 21 - Prob. 21.36QPCh. 21 - Prob. 21.37QPCh. 21 - Prob. 21.38QPCh. 21 - Prob. 21.39QPCh. 21 - Prob. 21.40QPCh. 21 - Prob. 21.41QPCh. 21 - Prob. 21.42QPCh. 21 - Prob. 21.43QPCh. 21 - Prob. 21.44QPCh. 21 - Prob. 21.45QPCh. 21 - Prob. 21.46QPCh. 21 - Prob. 21.47QPCh. 21 - Prob. 21.48QPCh. 21 - Prob. 21.49QPCh. 21 - Prob. 21.50QPCh. 21 - Prob. 21.51QPCh. 21 - Prob. 21.52QPCh. 21 - Prob. 21.53QPCh. 21 - Prob. 21.54QPCh. 21 - Prob. 21.55QPCh. 21 - Prob. 21.56QPCh. 21 - Prob. 21.57QPCh. 21 - Prob. 21.58QPCh. 21 - Prob. 21.59QPCh. 21 - Prob. 21.60QPCh. 21 - Prob. 21.61QPCh. 21 - Prob. 21.62QPCh. 21 - Prob. 21.63QPCh. 21 - Prob. 21.64QPCh. 21 - Prob. 21.65QPCh. 21 - Prob. 21.66QPCh. 21 - Prob. 21.67QPCh. 21 - Prob. 21.68QPCh. 21 - Prob. 21.69QPCh. 21 - Prob. 21.70QPCh. 21 - Prob. 21.71QPCh. 21 - Prob. 21.72QPCh. 21 - Prob. 21.73QPCh. 21 - Prob. 21.74QPCh. 21 - Prob. 21.75QPCh. 21 - Prob. 21.76QPCh. 21 - Prob. 21.77QPCh. 21 - Prob. 21.78QPCh. 21 - Prob. 21.79QPCh. 21 - Prob. 21.80QPCh. 21 - Prob. 21.81QPCh. 21 - Prob. 21.82QPCh. 21 - Prob. 21.83QPCh. 21 - Prob. 21.84QPCh. 21 - Prob. 21.85QPCh. 21 - Prob. 21.86QPCh. 21 - Prob. 21.87SPCh. 21 - Prob. 21.88SPCh. 21 - Prob. 21.89SPCh. 21 - Prob. 21.90SPCh. 21 - Prob. 21.91SPCh. 21 - Prob. 21.92SPCh. 21 - Prob. 21.93SPCh. 21 - Prob. 21.94SPCh. 21 - Prob. 21.95SPCh. 21 - Prob. 21.96SPCh. 21 - Prob. 21.97SPCh. 21 - Prob. 21.98SPCh. 21 - Prob. 21.99SP
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