Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 19 - Nuclear Chemistry - Questions & Problems - Page 894: 19.61

Answer

We find: (a) $U_{92}^{235} + n_{0}^{1} -> Ba_{56}^{140} + 3n_{0}^{1} + Kr_{36}^{93}$ (b) $U_{92}^{235} + n_{0}^{1} -> Cs_{55}^{144} + 2n_{0}^{1} + Rb_{37}^{90}$ (c) $U_{92}^{235} + n_{0}^{1} -> Br_{35}^{87} + 3n_{0}^{1} + La_{57}^{146}$ (d) $U_{92}^{235} + n_{0}^{1} -> Sn_{62}^{160} + 4n_{0}^{1} + Zn_{30}^{72}$

Work Step by Step

We find: (a) $U_{92}^{235} + n_{0}^{1} -> Ba_{56}^{140} + 3n_{0}^{1} + Kr_{36}^{93}$ (b) $U_{92}^{235} + n_{0}^{1} -> Cs_{55}^{144} + 2n_{0}^{1} + Rb_{37}^{90}$ (c) $U_{92}^{235} + n_{0}^{1} -> Br_{35}^{87} + 3n_{0}^{1} + La_{57}^{146}$ (d) $U_{92}^{235} + n_{0}^{1} -> Sn_{62}^{160} + 4n_{0}^{1} + Zn_{30}^{72}$
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