Explanation The atomic mass of 1 1 H is 1 .0078 amu . The atomic mass of 1 2 H is 2 .01410 amu . The mass of neutron is 1.0087 amu . Number of protons in 1 2 H = 1 Number of neutron in 1 2 H = 1 The stated reaction is, 1 1 H + 1 1 H → 1 2 H+ 1 0 e The mass defect is calculated by the formula, Δ m = [ Mass of 1 2 H nucleus +Mass of positron ] − [ 2×Mass of 1 2 H nucleus ] Δ m = ( [ ( Atomic mass of 1 2 H nucleus − Mass of electron ) + Mass of positron ] − [ 2(Atomic mass of 1 1 H-mass of electron) ] ) Substitute the value of the atomic mass of 1 2 H , the number of protons; the mass of the 1 1 H proton and that of the neutron in the above equation. Δ m = 2 .01410 − [ 2 ( 1.00782 − 5.4858 × 10 − 4 ) ] Δ m = − 0.0044 amu / nucleus The conversion of amu / nucleus to Kg/nucleus is done as, 1 amu = 1.66 × 10 − 27 Kg Therefore, the conversion of − 0.0044 amu / nucleus into kg/nucleus is, − 0.0044 amu / nucleus = ( − 0.0044 × 1.66 × 10 − 27 ) kg/nucleus = - 7 . 3 0 × 1 0 - 3 1 K g / n u c l e u s _ Therefore, the mass defect ( Δ m ) of 1 2 H is - 7 . 3 0 × 1 0 - 3 1 K g / n u c l e u s _

Bundle: Chemistry, Loose-Leaf Version, 10th + OWLv2, 4 terms (24 months) Printed Access Card

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ISBN: 9781337537933

Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher: Cengage Learning

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Question

Chapter 19, Problem 55E

Interpretation Introduction

Interpretation:

The amount of energy released per gram of hydrogen reacted for the given reaction is to be calculated.

Concept introduction:

The sum of masses of the component nucleons and the actual mass of a nucleus is known as the mass defect and it can be used to calculate the nuclear binding energy.

Expert Solution & Answer

Chapter 19 Solutions

Bundle: Chemistry, Loose-Leaf Version, 10th + OWLv2, 4 terms (24 months) Printed Access Card

Ch. 19 - Prob. 1RQ Ch. 19 - Prob. 2RQ Ch. 19 - Prob. 3RQ Ch. 19 - Prob. 4RQ Ch. 19 - Prob. 5RQ Ch. 19 - Prob. 6RQ Ch. 19 - Prob. 7RQ Ch. 19 - Prob. 8RQ Ch. 19 - Prob. 9RQ Ch. 19 - Prob. 10RQ

Ch. 19 - Prob. 1Q Ch. 19 - Prob. 3Q Ch. 19 - Prob. 4Q Ch. 19 - Prob. 5Q Ch. 19 - Prob. 6Q Ch. 19 - Prob. 7Q Ch. 19 - Prob. 8Q Ch. 19 - Prob. 9Q Ch. 19 - Prob. 10Q Ch. 19 - Prob. 11Q Ch. 19 - Prob. 12Q Ch. 19 - Prob. 13Q Ch. 19 - Prob. 14Q Ch. 19 - Prob. 15E Ch. 19 - Prob. 16E Ch. 19 - Prob. 17E Ch. 19 - Prob. 18E Ch. 19 - Prob. 19E Ch. 19 - Prob. 20E Ch. 19 - Prob. 21E Ch. 19 - Prob. 22E Ch. 19 - Prob. 23E Ch. 19 - Prob. 24E Ch. 19 - Prob. 27E Ch. 19 - Prob. 28E Ch. 19 - Prob. 29E Ch. 19 - Prob. 30E Ch. 19 - Prob. 32E Ch. 19 - Prob. 34E Ch. 19 - Prob. 35E Ch. 19 - Prob. 36E Ch. 19 - Prob. 37E Ch. 19 - Prob. 38E Ch. 19 - Prob. 39E Ch. 19 - Prob. 40E Ch. 19 - Prob. 41E Ch. 19 - Prob. 42E Ch. 19 - Prob. 43E Ch. 19 - Prob. 44E Ch. 19 - Prob. 45E Ch. 19 - Prob. 46E Ch. 19 - Prob. 47E Ch. 19 - Prob. 48E Ch. 19 - Prob. 49E Ch. 19 - Prob. 50E Ch. 19 - Prob. 52E Ch. 19 - Prob. 53E Ch. 19 - Prob. 54E Ch. 19 - Prob. 55E Ch. 19 - Prob. 56E Ch. 19 - Prob. 57E Ch. 19 - Prob. 58E Ch. 19 - Prob. 59E Ch. 19 - Prob. 60E Ch. 19 - Prob. 61E Ch. 19 - Prob. 62E Ch. 19 - Prob. 63E Ch. 19 - Prob. 64E Ch. 19 - Prob. 65AE Ch. 19 - Prob. 66AE Ch. 19 - Prob. 67AE Ch. 19 - Prob. 68AE Ch. 19 - Prob. 69AE Ch. 19 - Prob. 70AE Ch. 19 - Prob. 71AE Ch. 19 - Prob. 72AE Ch. 19 - Prob. 73AE Ch. 19 - Prob. 74AE Ch. 19 - Prob. 75AE Ch. 19 - Prob. 76AE Ch. 19 - Prob. 77AE Ch. 19 - Prob. 78AE Ch. 19 - Prob. 79AE Ch. 19 - Prob. 80AE Ch. 19 - Prob. 81CWP Ch. 19 - Prob. 82CWP Ch. 19 - Prob. 83CWP Ch. 19 - Prob. 84CWP Ch. 19 - Prob. 85CWP Ch. 19 - Prob. 86CWP Ch. 19 - Prob. 87CP Ch. 19 - Prob. 88CP Ch. 19 - Prob. 89CP Ch. 19 - Prob. 90CP Ch. 19 - Prob. 91CP Ch. 19 - Prob. 92CP Ch. 19 - Prob. 93CP Ch. 19 - Prob. 94CP Ch. 19 - Prob. 95IP Ch. 19 - Prob. 96IP

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Francium was discovered as a minor decay product of actinium-227. Write the nuclear equation for the decay of actinium-227 by alpha emission.

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