Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 17 - Radioactivity and Nuclear Chemistry - Exercises - Problems - Page 639: 83

Answer

2

Work Step by Step

U235 + n → Xe144 + Sr90 + 2n See only one neutron can cause fission in U235. So, the total number of neutrons will be 236. So, the number of neutrons released will be 236 - 234 = 2
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