Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
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Chapter U1.14, Problem 14E
Interpretation Introduction

Interpretation:

The correct option is to be selected.

Concept introduction:

Many elements are found in their isotopic forms. They differ in their mass numbers. Some isotopes are stable while some are unstable. To study the stability of nuclei of isotopes, an isotope graph is plotted for first 95 elements.

Expert Solution & Answer
Check Mark

Answer to Problem 14E

Correct answer: Option B. have more isotopes than those with an odd atomic number.

Explanation of Solution

Reason for correct option:

The isotope graph shows isotopes of all first 95 elements found in nature and in periodic table. It shows a diagonal line and many squares marked on it. Each square represents a different isotope.

There are some inferences made by the study of isotope graph.

The diagonal line plotted on the graph represents the points with equal number of protons and neutrons. It helps to find out the stability of an isotope. The squares on the diagonal line are isotopes with equal number of protons and neutrons.

The elements with even number of protons have more isotopes than elements with odd number of protons. Since number of protons is atomic number of an element. Hence one can say that elements with even atomic number have more isotopes than elements with odd atomic number.

So the correct option is B. have more isotopes than those with an odd atomic number.

Conclusion

Reasons for incorrect options:

    A. Have only one isotope - The isotope graph is plotted with number of protons versus number of neutrons. From the graph atoms with even atomic number may have more than one isotope. Carbon has even atomic number 6. It has more than one isotope; carbon - 12, carbon-13 and carbon - 14. Hence this option is not correct.

    C. Have the same number of protons as neutrons - The isotope graph is plotted with number of protons versus number of neutrons. From the graph atoms with even atomic number does not always have equal number of protons and neutrons. Carbon - 13 has even atomic number with 6 protons but 7 neutrons. Hence this option is not correct.

    D. Are halogens- Halogens are elements which have atomic numbers 9, 17, 35, 53, 85 and 117. These are odd atomic numbers; so this option is not correct.

Chapter U1 Solutions

Living By Chemistry: First Edition Textbook

Ch. U1.2 - Prob. 1ECh. U1.2 - Prob. 2ECh. U1.2 - Prob. 5ECh. U1.3 - Prob. 1TAICh. U1.3 - Prob. 1ECh. U1.3 - Prob. 2ECh. U1.3 - Prob. 3ECh. U1.3 - Prob. 4ECh. U1.3 - Prob. 5ECh. U1.3 - Prob. 6ECh. U1.4 - Prob. 1TAICh. U1.4 - Prob. 1ECh. U1.4 - Prob. 2ECh. U1.4 - Prob. 3ECh. U1.4 - Prob. 4ECh. U1.4 - Prob. 5ECh. U1.4 - Prob. 6ECh. U1.4 - Prob. 7ECh. U1.4 - Prob. 8ECh. U1.4 - Prob. 9ECh. U1.4 - Prob. 10ECh. U1.4 - Prob. 11ECh. U1.5 - Prob. 1TAICh. U1.5 - Prob. 1ECh. U1.5 - Prob. 2ECh. U1.5 - Prob. 3ECh. U1.5 - Prob. 4ECh. U1.5 - Prob. 5ECh. U1.5 - Prob. 6ECh. U1.5 - Prob. 7ECh. U1.6 - Prob. 1TAICh. U1.6 - Prob. 1ECh. U1.6 - Prob. 2ECh. U1.6 - Prob. 3ECh. U1.6 - Prob. 4ECh. U1.6 - Prob. 5ECh. U1.6 - Prob. 6ECh. U1.7 - Prob. 1TAICh. U1.7 - Prob. 1ECh. U1.7 - Prob. 2ECh. U1.7 - Prob. 3ECh. U1.7 - Prob. 4ECh. U1.7 - Prob. 5ECh. U1.8 - Prob. 1TAICh. U1.8 - Prob. 1ECh. U1.8 - Prob. 2ECh. U1.8 - Prob. 4ECh. U1.8 - Prob. 5ECh. U1.8 - Prob. 6ECh. U1.8 - Prob. 7ECh. U1.9 - Prob. 1TAICh. U1.9 - Prob. 1ECh. U1.9 - Prob. 2ECh. U1.9 - Prob. 5ECh. U1.9 - Prob. 7ECh. U1.10 - Prob. 1TAICh. U1.10 - Prob. 1ECh. U1.10 - Prob. 2ECh. U1.10 - Prob. 3ECh. U1.10 - Prob. 4ECh. U1.10 - Prob. 5ECh. U1.10 - Prob. 6ECh. U1.10 - Prob. 7ECh. U1.10 - Prob. 8ECh. U1.11 - Prob. 1TAICh. U1.11 - Prob. 1ECh. U1.11 - Prob. 2ECh. U1.11 - Prob. 3ECh. U1.11 - Prob. 4ECh. U1.11 - Prob. 5ECh. U1.11 - Prob. 6ECh. U1.11 - Prob. 7ECh. U1.11 - Prob. 9ECh. U1.11 - Prob. 11ECh. U1.11 - Prob. 12ECh. U1.12 - Prob. 1TAICh. U1.12 - Prob. 1ECh. U1.12 - Prob. 2ECh. U1.12 - Prob. 3ECh. U1.12 - Prob. 4ECh. U1.12 - Prob. 5ECh. U1.12 - Prob. 6ECh. U1.12 - Prob. 7ECh. U1.12 - Prob. 8ECh. U1.13 - Prob. 1TAICh. U1.13 - Prob. 1ECh. U1.13 - Prob. 2ECh. U1.13 - Prob. 3ECh. U1.13 - Prob. 4ECh. U1.13 - Prob. 5ECh. U1.13 - Prob. 6ECh. U1.13 - Prob. 7ECh. U1.13 - Prob. 8ECh. U1.13 - Prob. 9ECh. U1.14 - Prob. 1TAICh. U1.14 - Prob. 1ECh. U1.14 - Prob. 2ECh. U1.14 - Prob. 3ECh. U1.14 - Prob. 4ECh. U1.14 - Prob. 5ECh. U1.14 - Prob. 6ECh. U1.14 - Prob. 7ECh. U1.14 - Prob. 8ECh. U1.14 - Prob. 9ECh. U1.14 - Prob. 10ECh. U1.14 - Prob. 11ECh. U1.14 - Prob. 12ECh. U1.14 - Prob. 13ECh. U1.14 - Prob. 14ECh. U1.15 - Prob. 1TAICh. U1.15 - Prob. 1ECh. U1.15 - Prob. 2ECh. U1.15 - Prob. 3ECh. U1.15 - Prob. 4ECh. U1.15 - Prob. 5ECh. U1.15 - Prob. 6ECh. U1.15 - Prob. 7ECh. U1.15 - Prob. 8ECh. U1.15 - Prob. 9ECh. U1.15 - Prob. 10ECh. U1.15 - Prob. 11ECh. U1.15 - Prob. 12ECh. U1.16 - Prob. 1TAICh. U1.16 - Prob. 1ECh. U1.16 - Prob. 2ECh. U1.16 - Prob. 3ECh. U1.16 - Prob. 4ECh. U1.16 - Prob. 5ECh. U1.16 - Prob. 6ECh. U1.17 - Prob. 1TAICh. U1.17 - Prob. 1ECh. U1.17 - Prob. 2ECh. U1.17 - Prob. 3ECh. U1.17 - Prob. 4ECh. U1.17 - Prob. 5ECh. U1.17 - Prob. 6ECh. U1.17 - Prob. 7ECh. U1.17 - Prob. 8ECh. U1.17 - Prob. 9ECh. U1.17 - Prob. 10ECh. U1.17 - Prob. 11ECh. U1.18 - Prob. 1TAICh. U1.18 - Prob. 1ECh. U1.18 - Prob. 2ECh. U1.18 - Prob. 3ECh. U1.18 - Prob. 4ECh. U1.18 - Prob. 5ECh. U1.18 - Prob. 6ECh. U1.18 - Prob. 7ECh. U1.18 - Prob. 8ECh. U1.18 - Prob. 9ECh. U1.18 - Prob. 10ECh. U1.19 - Prob. 1TAICh. U1.19 - Prob. 1ECh. U1.19 - Prob. 2ECh. U1.19 - Prob. 3ECh. U1.19 - Prob. 4ECh. U1.19 - Prob. 5ECh. U1.19 - Prob. 6ECh. U1.19 - Prob. 7ECh. U1.19 - Prob. 8ECh. U1.19 - Prob. 9ECh. U1.19 - Prob. 10ECh. U1.19 - Prob. 11ECh. U1.19 - Prob. 12ECh. U1.19 - Prob. 13ECh. U1.19 - Prob. 14ECh. U1.19 - Prob. 15ECh. U1.19 - Prob. 16ECh. U1.20 - Prob. 1TAICh. U1.20 - Prob. 1ECh. U1.20 - Prob. 2ECh. U1.20 - Prob. 3ECh. U1.20 - Prob. 4ECh. U1.20 - Prob. 5ECh. U1.20 - Prob. 6ECh. U1.20 - Prob. 7ECh. U1.21 - Prob. 1TAICh. U1.21 - Prob. 1ECh. U1.21 - Prob. 2ECh. U1.21 - Prob. 3ECh. U1.21 - Prob. 4ECh. U1.21 - Prob. 5ECh. U1.21 - Prob. 6ECh. U1.21 - Prob. 7ECh. U1.21 - Prob. 8ECh. U1.22 - Prob. 1TAICh. U1.22 - Prob. 1ECh. U1.22 - Prob. 2ECh. U1.22 - Prob. 3ECh. U1.22 - Prob. 4ECh. U1.22 - Prob. 5ECh. U1.22 - Prob. 6ECh. U1.22 - Prob. 7ECh. U1.23 - Prob. 1TAICh. U1.23 - Prob. 1ECh. U1.23 - Prob. 2ECh. U1.23 - Prob. 3ECh. U1.23 - Prob. 4ECh. U1.23 - Prob. 5ECh. U1.24 - Prob. 1TAICh. U1.24 - Prob. 1ECh. U1.24 - Prob. 2ECh. U1.24 - Prob. 3ECh. U1.24 - Prob. 4ECh. U1.24 - Prob. 5ECh. U1.24 - Prob. 6ECh. U1.24 - Prob. 7ECh. U1.24 - Prob. 8ECh. U1.24 - Prob. 9ECh. U1.24 - Prob. 10ECh. U1.24 - Prob. 11ECh. U1.24 - Prob. 12ECh. U1.24 - Prob. 13ECh. U1.25 - Prob. 1TAICh. U1.25 - Prob. 1ECh. U1.25 - Prob. 2ECh. U1.25 - Prob. 3ECh. U1.25 - Prob. 4ECh. U1.25 - Prob. 5ECh. U1.25 - Prob. 6ECh. U1.26 - Prob. 1TAICh. U1.26 - Prob. 1ECh. U1.26 - Prob. 2ECh. U1.26 - Prob. 3ECh. U1.26 - Prob. 4ECh. U1.26 - Prob. 5ECh. U1.26 - Prob. 6ECh. U1.26 - Prob. 7ECh. U1.26 - Prob. 8ECh. U1.26 - Prob. 9ECh. U1.26 - Prob. 10ECh. U1.27 - Prob. 1TAICh. U1.27 - Prob. 1ECh. U1.27 - Prob. 2ECh. U1.27 - Prob. 3ECh. U1.27 - Prob. 4ECh. U1.27 - Prob. 5ECh. U1.27 - Prob. 6ECh. U1.27 - Prob. 7ECh. U1 - Prob. SI1RECh. U1 - Prob. SI2RECh. U1 - Prob. SI3RECh. U1 - Prob. SI4RECh. U1 - Prob. SI5RECh. U1 - Prob. SI6RECh. U1 - Prob. SII1RQCh. U1 - Prob. SII2RQCh. U1 - Prob. SII3RQCh. U1 - Prob. SII4RQCh. U1 - Prob. SIII1RQCh. U1 - Prob. SIII2RQCh. U1 - Prob. SIII3RQCh. U1 - Prob. SIII4RQCh. U1 - Prob. SIV1RQCh. U1 - Prob. SIV2RQCh. U1 - Prob. SIV3RQCh. U1 - Prob. SIV4RQCh. U1 - Prob. SIV5RQCh. U1 - Prob. SIV6RQCh. U1 - Prob. SV1RQCh. U1 - Prob. SV2RQCh. U1 - Prob. SV3RQCh. U1 - Prob. SV4RQCh. U1 - Prob. 1RECh. U1 - Prob. 2RECh. U1 - Prob. 3RECh. U1 - Prob. 4RECh. U1 - Prob. 5RECh. U1 - Prob. 6RECh. U1 - Prob. 7RECh. U1 - Prob. 8RECh. U1 - Prob. 9RECh. U1 - Prob. 10RECh. U1 - Prob. 11RECh. U1 - Prob. 12RE

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