Complete the reaction: n + _{92}^{235}U to _{ | vedclass

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(A) In a nuclear fission reaction, both the mass number and the atomic number must be conserved on both sides of the equation.Let the missing nucleus

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$_{36}^{89}Kr$

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(A) In a nuclear fission reaction, both the mass number and the atomic number must be conserved on both sides of the equation.Let the missing nucleus be $_{Z}^{A}X$.The reaction is: $_{0}^{1}n + _{92}^{235}U 	o _{56}^{144}Ba + _{Z}^{A}X + 3(_{0}^{1}n)$.Conservation of atomic number (charge): $0 + 92 = 56 + Z + 3(0) \implies 92 = 56 + Z \implies Z = 36$.The element with atomic number $36$ is Krypton $(Kr)$.Conservation of mass number: $1 + 235 = 144 + A + 3(1) \implies 236 = 147 + A \implies A = 89$.Therefore, the missing nucleus is $_{36}^{89}Kr$.

Complete the reaction: $n + _{92}^{235}U \to _{56}^{144}Ba + .... + 3n$

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