Complete each nuclear fission reaction. (}_{92)^235U+_(0)^1narrow _(30)^90Kr+_(50)^ABa+3_(0)^1n (}_{94)^239Pu+_(0)^1narrow _(c)^8Ba+_(38)^91Sr+3_(0)^1n : square B square politician C: square

To complete the nuclear fission reactions, we need to balance the equation by ensuring that the sum of the atomic numbers (the subscript) and the sum of the mass numbers (the superscript) are equal on both sides of the equation.<br /><br />Let's start with the first reaction:<br /><br />${}_{92}^{235}U+_{0}^{1}n\rightarrow _{30}^{90}Kr+_{50}^{A}Ba+3_{0}^{1}n$<br /><br />To balance the equation, we need to find the value of A for the Ba isotope. <br /><br />The sum of the atomic numbers on the left side is 92 (Uranium) + 0 (neutron) = 92.<br /><br />The sum of the atomic numbers on the right side is 30 (Krypton) + 50 (Barium) = 80.<br /><br />Since the sum of the atomic numbers must be equal on both sides, we can set up the equation:<br /><br />92 = 80 + c<br /><br />Solving for c, we get c = 12.<br /><br />Therefore, the balanced equation for the first reaction is:<br /><br />${}_{92}^{235}U+_{0}^{1}n\rightarrow _{30}^{90}Kr+_{50}^{143}Ba+3_{0}^{1}n$<br /><br />Now let's move on to the second reaction:<br /><br />${}_{94}^{239}Pu+_{0}^{1}n\rightarrow _{c}^{8}Ba+_{38}^{91}Sr+3_{0}^{1}n$<br /><br />To balance the equation, we need to find the value of c for the Ba isotope.<br /><br />The sum of the atomic numbers on the left side is 94 (Plutonium) + 0 (neutron) = 94.<br /><br />The sum of the atomic numbers on the right side is c (Barium) + 38 (Strontium) = 94.<br /><br />Since the sum of the atomic numbers must be equal on both sides, we can set up the equation:<br /><br />94 = 94<br /><br />This equation is already balanced, so there is no need to solve for c.<br /><br />Therefore, the balanced equation for the second reaction is:<br /><br />${}_{94}^{239}Pu+_{0}^{1}n\rightarrow _{56}^{141}Ba+_{38}^{91}Sr+3_{0}^{1}n$<br /><br />So, the completed nuclear fission reactions are:<br /><br />${}_{92}^{235}U+_{0}^{1}n\rightarrow _{30}^{90}Kr+_{50}^{143}Ba+3_{0}^{1}n$<br /><br />${}_{94}^{239}Pu+_{0}^{1}n\rightarrow _{56}^{141}Ba+_{38}^{91}Sr+3_{0}^{1}n$