1.

Fill in the blanks: (a) ._(92)^(235)U + ._(0)^(1)n rarr ._(55)^(142)A + ._(37)^(92) B+.... (b) ._(34)^(82)Se rarr ...+2 ._(-1)e^(0)

Answer»


Solution :`._(92)^(235)U+_(0)^(1)nrarr ._(52)^(137)A+_(40)^(97)B+X ._(0)n^(1)`
(as the reaction is balancedwith respect to NUCLEAR charge, the missing particle must be neutral i.e. NEUTRON)
Applying mass nuclear charge balancing
`34=-2+z`
`z=36" "` (This implies ELEMENT is kr)
Applying mass NUMBER balancing
`82=0+A`
`A=82`
Final balanced reaction is
`._(34)^(82)Se rarr 2 ._(-1)^(0)e + _(36)^(82) Kr`


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