23592U+10n→14456Ba+9036Kr+2X 92 235 U + 0 1 n → 56 144 B a + 36 90 K r + 2 X In the reaction above, X is

${}_{92}^{235}U{+}_0^{1}n{\to }_{56}^{144}Ba{+}_{36}^{90}Kr+2X$

In the reaction above, X is

Answer Details

235U + 1n → 144Ba + 90Kr + 2x9205636
In the reaction above, balancing the left with the right, show that X is neutron