Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 2 - Atoms, Molecules, and Ions - Exercises: 57

Answer

(a) $O^{17}_8$ (b) $Cl^{37}_{17}$ (c) $Co^{60}_{27}$ (d) $Fe^{57}_{26}$ (e) $I^{131}_{53}$ (f) $Li^7_3$

Work Step by Step

(a) Given that $z$ i.e atomic number is 8 so, it is oxygen and its mass$=8+9=17$ we can represent it as $O^{17}_8$ (b) As $A=37$ i.e mass number is 37. This shows that the element is chlorine which can be represented as $Cl^{37}_{17}$ (c) As $Z=27$ and $A=60$ which shows the element is cobalt and can be represented as $Co^{60}_{27}$ (d) Number of protons is 26 and number of neutrons is 31 so mass number $=26+31=57$ and the element is $Fe$ which we can represent as$Fe^{57}_{26}$ (e) $A$ and $Z$ are the subscripts which show mass number and atomic number respectively. Given that the element is $I$ so $Z$ is 53. Now we can represent it as $I^{131}_{53}$ (f) As $Z=3$ and number of neutrons$=4$ so mass number $=3+4=7$. Thus the element is $Li$ and can be represented as $Li^7_3$
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