Question

complete each nuclear fission reaction.\\(\ce{_{92}^{235}u + _{0}^{1}n \
ightarrow _{36}^{90}kr + _{56}^{a}ba + 3_{0}^{1}n}\\)\\(\ce{_{94}^{239}pu + _{0}^{1}n \
ightarrow _{c}^{b}ba + _{38}^{91}sr + 3_{0}^{1}n}\\)a: \boxed{}b: \boxed{}c: \boxed{}

Explanation:

Step1: Solve for mass number A

Sum reactant mass numbers: $235 + 1 = 236$
Set equal to product sum: $236 = 90 + A + 3(1)$
Solve for A: $A = 236 - 90 - 3 = 143$

Step2: Solve for mass number B

Sum reactant mass numbers: $239 + 1 = 240$
Set equal to product sum: $240 = B + 91 + 3(1)$
Solve for B: $B = 240 - 91 - 3 = 146$

Step3: Solve for atomic number C

Sum reactant atomic numbers: $94 + 0 = 94$
Set equal to product sum: $94 = C + 38 + 3(0)$
Solve for C: $C = 94 - 38 = 56$

Answer:

A: 143
B: 146
C: 56