Question
Find the number of neutrons released by the following fission reaction:$$_{0}^{1} n+_{92}^{235} U \rightarrow_{50}^{13} S n+_{42}^{101} M o+(?) \text{neutrons}$$
Step 1
In this case, the atomic numbers are already balanced because the atomic number of a neutron is 0 and the atomic number of Uranium is 92. On the right side of the equation, the atomic number of Sn is 50 and the atomic number of Mo is 42. So, 0 + 92 = 50 + 42. Show more…
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Find the number of neutrons released by the following fission reaction: $$_{0}^{1} \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow_{50 }^{132} \mathrm{Sn}+{ }_{42}^{101} \mathrm{Mo}+(?) \text { neutrons }$$
How many neutrons are produced when $\underset{92}{235}$ U fissions in the following way: $${ }_{0}^{1} \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{50}^{132} \mathrm{Sn}+{ }_{42}^{101} \mathrm{M} \circ+\text { neutrons? }$$
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