How many chlorine atoms will be ionised CI rarr CI^(+) +e^(-1) by the energy released from the process CI +e^(-1) rarr CI^(-) for 6.02 xx 10^(23) atoms (I.P. for CI = 1250 kJ mol^(-1) and E.A. = 350 kJ "mole"^(-1))

How many chlorine atoms will be ionised CI rarr CI^(+) +e^(-1) by the

1.	How many chlorine atoms will be ionised CI rarr CI^(+) +e^(-1) by the energy released from the process CI +e^(-1) rarr CI^(-) for 6.02 xx 10^(23) atoms (I.P. for CI = 1250 kJ mol^(-1) and E.A. = 350 kJ "mole"^(-1))
Answer» Solution :`[N_(A)XX E.A = I.P xx n 6.02 xx 10^(23) xx 350 = 1250 xx n]`