Chemistry for Today: General, Organic, and Biochemistry
Chemistry for Today: General, Organic, and Biochemistry
9th Edition
ISBN: 9781305960060
Author: Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher: Cengage Learning
Question
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Chapter 10, Problem 10.39E
Interpretation Introduction

(a)

Interpretation:

The given reactions are to be completed by identifying the missing nuclear symbols.

Concept introduction:

The type of radioactive decay in which an alpha particle is emitted by the nucleus of an atom such that an atom of another element is produced after decay is known as alpha decay. An alpha particle is a helium nucleus. The radioactive decay in which a positron or an electron is emitted is known as beta decay.

Expert Solution
Check Mark

Answer to Problem 10.39E

The complete reaction is,

49Be+24α612C+01n

Explanation of Solution

The given reaction is,

49Be+?612C+01n

The net mass is obtained by subtracting the mass number of the species on the left-hand side from the mass number of the species on the right-hand side. The right-hand side consists of carbon atom and a neutron and the left hand side consists of beryllium atom. Therefore, the net mass is,

Netmass=(12+1)9=4

The net charge is obtained by subtracting the charge of the species on the left-hand side from the charge of the species on the right-hand side. Therefore, the net charge is,

Netcharge=(6+0)4=2

The species that has +2 charge, that is atomic number, and a mass equal to 4 is an alpha particle, that is, 24α. Therefore, the given equation using appropriate notations and formulas is,

49Be+24α612C+01n

Conclusion

The complete reaction is,

49Be+24α612C+01n

Interpretation Introduction

(b)

Interpretation:

The given reactions are to be completed by identifying the missing nuclear symbols.

Concept introduction:

The type of radioactive decay in which an alpha particle is emitted by the nucleus of an atom such that an atom of another element is produced after decay is known as alpha decay. An alpha particle is a helium nucleus. The radioactive decay in which a positron or an electron is emitted is known as beta decay.

Expert Solution
Check Mark

Answer to Problem 10.39E

The complete reaction is,

1327Al+24α1530P+01n

Explanation of Solution

The given reaction is,

1327Al+?1530P+01n

The net mass is obtained by subtracting the mass number of the species on the left-hand side from the mass number of the species on the right-hand side. The right-hand side consists of phosphorus atom and a neutron and the left hand side consists of aluminum atom. Therefore, the net mass is,

Netmass=(30+1)27=4

The net charge is obtained by subtracting the charge of the species on the left-hand side from the charge of the species on the right-hand side. Therefore, the net charge is,

Netcharge=(15+0)13=2

The species that has +2 charge, that is atomic number, and a mass equal to 4 is an alpha particle, that is, 24α. Therefore, the given equation using appropriate notations and formulas is,

1327Al+24α1530P+01n

Conclusion

The complete reaction is,

1327Al+24α1530P+01n

Interpretation Introduction

(c)

Interpretation:

The given reactions are to be completed by identifying the missing nuclear symbols.

Concept introduction:

The type of radioactive decay in which an alpha particle is emitted by the nucleus of an atom such that an atom of another element is produced after decay is known as alpha decay. An alpha particle is a helium nucleus. The radioactive decay in which a positron or an electron is emitted is known as beta decay.

Expert Solution
Check Mark

Answer to Problem 10.39E

The complete reaction is,

1327Al+12H1225Mg+24α

Explanation of Solution

The given reaction is,

1327Al+12H?+24α

The net mass is obtained by subtracting the mass number of the species on the right-hand side from the mass number of the species on the left-hand side. The left-hand side consists of aluminum atom and a hydrogen atom and the right-hand side consists of an alpha particle. Therefore, the net mass is,

Netmass=(27+2)4=25

The net charge is obtained by subtracting the charge of the species on the right-hand side from the charge of the species on the left-hand side. Therefore, the net charge is,

Netcharge=(13+1)2=12

The species that has +12 charge, that is atomic number, and a mass equal to 25 is an isotope of magnesium, that is, 1225Mg. Therefore, the given equation using appropriate notations and formulas is,

1327Al+12H1225Mg+24α

Conclusion

The complete reaction is,

1327Al+12H1225Mg+24α

Interpretation Introduction

(d)

Interpretation:

The given reactions are to be completed by identifying the missing nuclear symbols.

Concept introduction:

The type of radioactive decay in which an alpha particle is emitted by the nucleus of an atom such that an atom of another element is produced after decay is known as alpha decay. An alpha particle is a helium nucleus. The radioactive decay in which a positron or an electron is emitted is known as beta decay.

Expert Solution
Check Mark

Answer to Problem 10.39E

The complete reaction is,

714N+24α817O+11H

Explanation of Solution

The given reaction is,

714N+24α?+11H

The net mass is obtained by subtracting the mass number of the species on the right-hand side from the mass number of the species on the left-hand side. The left-hand side consists of nitrogen atom and an alpha particle and the right-hand side consists of an atom of hydrogen. Therefore, the net mass is,

Netmass=(14+4)1=17

The net charge is obtained by subtracting the charge of the species on the right-hand side from the charge of the species on the left-hand side. Therefore, the net charge is,

Netcharge=(7+2)1=8

The species that has +8 charge, that is atomic number, and a mass equal to 17 is an isotope of oxygen, that is, 817O. Therefore, the given equation using appropriate notations and formulas is,

714N+24α817O+11H

Conclusion

The complete reaction is,

714N+24α817O+11H

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Chapter 10 Solutions

Chemistry for Today: General, Organic, and Biochemistry

Ch. 10 - Prob. 10.11ECh. 10 - Prob. 10.12ECh. 10 - Prob. 10.13ECh. 10 - Prob. 10.14ECh. 10 - Prob. 10.15ECh. 10 - Prob. 10.16ECh. 10 - Prob. 10.17ECh. 10 - Prob. 10.18ECh. 10 - Prob. 10.19ECh. 10 - Prob. 10.20ECh. 10 - Prob. 10.21ECh. 10 - Prob. 10.22ECh. 10 - Prob. 10.23ECh. 10 - Prob. 10.24ECh. 10 - Prob. 10.25ECh. 10 - Prob. 10.26ECh. 10 - Prob. 10.27ECh. 10 - Prob. 10.28ECh. 10 - Prob. 10.29ECh. 10 - Prob. 10.30ECh. 10 - Prob. 10.31ECh. 10 - Prob. 10.32ECh. 10 - Prob. 10.33ECh. 10 - Prob. 10.34ECh. 10 - Prob. 10.35ECh. 10 - Prob. 10.36ECh. 10 - Prob. 10.37ECh. 10 - Prob. 10.38ECh. 10 - Prob. 10.39ECh. 10 - Prob. 10.40ECh. 10 - Prob. 10.41ECh. 10 - Prob. 10.42ECh. 10 - Prob. 10.43ECh. 10 - Prob. 10.44ECh. 10 - Prob. 10.45ECh. 10 - Prob. 10.46ECh. 10 - Prob. 10.47ECh. 10 - Prob. 10.48ECh. 10 - Prob. 10.49ECh. 10 - Prob. 10.50ECh. 10 - Prob. 10.51ECh. 10 - Prob. 10.52ECh. 10 - Prob. 10.53ECh. 10 - Prob. 10.54ECh. 10 - Prob. 10.55ECh. 10 - Prob. 10.56ECh. 10 - Prob. 10.57ECh. 10 - Prob. 10.58ECh. 10 - Prob. 10.59ECh. 10 - Prob. 10.60ECh. 10 - Prob. 10.61ECh. 10 - Prob. 10.62ECh. 10 - Prob. 10.63ECh. 10 - Prob. 10.64ECh. 10 - Prob. 10.65ECh. 10 - Prob. 10.66ECh. 10 - Prob. 10.67ECh. 10 - Prob. 10.68ECh. 10 - Prob. 10.69ECh. 10 - Prob. 10.70ECh. 10 - Prob. 10.71ECh. 10 - Prob. 10.72ECh. 10 - Prob. 10.73ECh. 10 - Prob. 10.74ECh. 10 - Prob. 10.75ECh. 10 - Prob. 10.76ECh. 10 - Prob. 10.77ECh. 10 - Prob. 10.78ECh. 10 - Prob. 10.79ECh. 10 - Prob. 10.80ECh. 10 - Prob. 10.81ECh. 10 - Prob. 10.82ECh. 10 - Prob. 10.83ECh. 10 - Prob. 10.84ECh. 10 - Prob. 10.85ECh. 10 - Prob. 10.86E
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