EBK CHEMISTRY: ATOMS FIRST
EBK CHEMISTRY: ATOMS FIRST
3rd Edition
ISBN: 8220103675505
Author: Burdge
Publisher: YUZU
Question
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Chapter 20, Problem 20.34QP

(a)

Interpretation Introduction

Interpretation:  For the given nuclear reaction, X should be identified and the equation should be balanced.

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.

To find: The value of X in the given all nuclear equation

(a)

Expert Solution
Check Mark

Answer to Problem 20.34QP

715N+11P126C+42α;Xis157N

Explanation of Solution

Explanation

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,

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

For the given reaction a, Short hand notation is X(p,α)6C12 . From the notation it is clear that,

Parentnucleus-XProjectile-pDaughternucleus-6C12Ejectile-α

The given chemical equation can be written as,

X+11P126C+42α

On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal. So the X will be 157X . By analyzing the X, atomic number of X is 7 and the atomic mass is 15. So it should be the isotope of nitrogen. It is found that X is 157N .

 So the balanced equation can be written as,

715N+11P126C+42α .

(b)

Interpretation Introduction

Interpretation:  For the given nuclear reaction, X should be identified and the equation should be balanced.

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.

To find: The value of X in the given all nuclear equation

(b)

Expert Solution
Check Mark

Answer to Problem 20.34QP

2713Al+12H2512Mg+42α;Xis2512Mg

Explanation of Solution

Explanation

For the given reaction b, Shorthand notation is 2713Al(d,α)X . From the notation it is clear that,

Parentnucleus-2713AlProjectile-dDaughternucleus-XEjectile-α

The given chemical equation can be written as,

2713Al+12HX+42α

On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal. So the X will be 2512X . By analyzing the X, atomic number of X is 12 and the atomic mass is 25. So it should be the isotope of Magnesium. It is found that X is 2512Mg

 So the balanced equation can be written as,

2713Al+12H2512Mg+42α .

(c)

Interpretation Introduction

Interpretation:  For the given nuclear reaction, X should be identified and the equation should be balanced.

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.

To find: The value of X in the given all nuclear equation

(c)

Expert Solution
Check Mark

Answer to Problem 20.34QP

5525Mn+10n5625Mn+γ;Xis5625Mn

Explanation of Solution

Explanation

For the given reaction c, Shorthand notation is 5525Mn(n,γ)X . From the notation it is clear that,

Parentnucleus-5625MnProjectile- nDaughternucleus-XEjectile-γ

The given chemical equation can be written as,

5525Mn+10nX+γ

On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal. So  X will be 5525X . By analyzing the X, atomic number of X is 25 and the atomic mass is 56. So it should be the isotope of Manganese. It is found that X is 5625Mn .

 So the balanced equation can be written as,

5525Mn+10n5625Mn+γ .

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Chapter 20 Solutions

EBK CHEMISTRY: ATOMS FIRST

Ch. 20.2 - Prob. 20.2.2SRCh. 20.2 - What is the change in mass (in ka) for the...Ch. 20.3 - Prob. 20.3WECh. 20.3 - Prob. 3PPACh. 20.3 - Prob. 3PPBCh. 20.3 - Prob. 20.4WECh. 20.3 - Prob. 4PPACh. 20.3 - Prob. 20.3.1SRCh. 20.3 - Prob. 20.3.2SRCh. 20.4 - Prob. 20.5WECh. 20.4 - Prob. 5PPACh. 20.4 - Prob. 5PPBCh. 20.4 - Prob. 5PPCCh. 20.4 - Prob. 20.4.1SRCh. 20.4 - Prob. 20.4.2SRCh. 20 - Prob. 20.1QPCh. 20 - Prob. 20.2QPCh. 20 - Prob. 20.3QPCh. 20 - Prob. 20.4QPCh. 20 - Prob. 20.5QPCh. 20 - Prob. 20.6QPCh. 20 - Prob. 20.7QPCh. 20 - Prob. 20.8QPCh. 20 - Prob. 20.9QPCh. 20 - Prob. 20.10QPCh. 20 - Prob. 20.11QPCh. 20 - Prob. 20.12QPCh. 20 - Prob. 20.13QPCh. 20 - Prob. 20.14QPCh. 20 - Prob. 20.15QPCh. 20 - Prob. 20.16QPCh. 20 - Prob. 20.17QPCh. 20 - Prob. 20.18QPCh. 20 - Prob. 20.19QPCh. 20 - Prob. 20.20QPCh. 20 - Prob. 20.21QPCh. 20 - Prob. 20.22QPCh. 20 - Prob. 20.23QPCh. 20 - Prob. 20.24QPCh. 20 - Prob. 20.25QPCh. 20 - Prob. 20.26QPCh. 20 - Prob. 20.27QPCh. 20 - Prob. 20.28QPCh. 20 - Prob. 20.29QPCh. 20 - Prob. 20.30QPCh. 20 - Prob. 20.31QPCh. 20 - Prob. 20.32QPCh. 20 - Prob. 20.33QPCh. 20 - Prob. 20.34QPCh. 20 - Prob. 20.35QPCh. 20 - Prob. 20.36QPCh. 20 - Prob. 20.37QPCh. 20 - Prob. 20.38QPCh. 20 - Prob. 20.39QPCh. 20 - Prob. 20.1VCCh. 20 - Prob. 20.3VCCh. 20 - Prob. 20.4VCCh. 20 - Prob. 20.40QPCh. 20 - Prob. 20.41QPCh. 20 - Prob. 20.42QPCh. 20 - Prob. 20.43QPCh. 20 - Prob. 20.44QPCh. 20 - Prob. 20.45QPCh. 20 - Prob. 20.46QPCh. 20 - Prob. 20.47QPCh. 20 - Prob. 20.48QPCh. 20 - Prob. 20.49QPCh. 20 - Prob. 20.50QPCh. 20 - Prob. 20.51QPCh. 20 - Prob. 20.52QPCh. 20 - Prob. 20.53QPCh. 20 - Prob. 20.54QPCh. 20 - Prob. 20.55QPCh. 20 - Prob. 20.56QPCh. 20 - Prob. 20.57QPCh. 20 - Prob. 20.58QPCh. 20 - Prob. 20.59QPCh. 20 - Prob. 20.60QPCh. 20 - Prob. 20.61QPCh. 20 - Prob. 20.62QPCh. 20 - Prob. 20.63QPCh. 20 - Prob. 20.64QPCh. 20 - Prob. 20.65QPCh. 20 - Prob. 20.66QPCh. 20 - Prob. 20.67QPCh. 20 - Prob. 20.68QPCh. 20 - Prob. 20.69QPCh. 20 - Prob. 20.70QPCh. 20 - Prob. 20.71QPCh. 20 - Prob. 20.72QPCh. 20 - Prob. 20.73QPCh. 20 - Prob. 20.74QPCh. 20 - Prob. 20.75QPCh. 20 - Prob. 20.76QPCh. 20 - Prob. 20.77QPCh. 20 - Prob. 20.78QPCh. 20 - Prob. 20.79QPCh. 20 - Prob. 20.80QPCh. 20 - Prob. 20.81QPCh. 20 - Prob. 20.82QPCh. 20 - Prob. 20.83QPCh. 20 - Prob. 20.84QPCh. 20 - Prob. 20.85QPCh. 20 - Prob. 20.86QPCh. 20 - Prob. 20.87QPCh. 20 - Prob. 20.88QPCh. 20 - Prob. 20.89QPCh. 20 - Prob. 20.90QPCh. 20 - Prob. 20.91QPCh. 20 - Prob. 20.92QPCh. 20 - Prob. 20.93QPCh. 20 - Prob. 20.94QPCh. 20 - Prob. 20.95QPCh. 20 - Prob. 20.96QPCh. 20 - Prob. 20.97QPCh. 20 - Prob. 20.98QPCh. 20 - Prob. 20.99QPCh. 20 - Prob. 20.100QPCh. 20 - Prob. 20.101QP
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