A gaseous hypothetical chemical equations , is carried out in a closed vessel . The concentration of B is found to increase by `5 xx 10^(-3) mol l^(-1)` in 10 second . The rate of appearance of B isA. `5 xx 10^(-4) mol l^(-1) sec^(-1)`B. `5 xx 10^(-5) mol l^(-1) sec^(-1)`C. `6 xx 10^(-5) mol l^(-1) sec^(-1)`D. `4 xx 10^(-4) mol l^(-1) sec^(-1)`

A gaseous hypothetical chemical equations , is carried out in a closed

1.	A gaseous hypothetical chemical equations , is carried out in a closed vessel . The concentration of B is found to increase by `5 xx 10^(-3) mol l^(-1)` in 10 second . The rate of appearance of B isA. `5 xx 10^(-4) mol l^(-1) sec^(-1)`B. `5 xx 10^(-5) mol l^(-1) sec^(-1)`C. `6 xx 10^(-5) mol l^(-1) sec^(-1)`D. `4 xx 10^(-4) mol l^(-1) sec^(-1)`
Answer» Correct Answer - a Increase in concentration of `B = 5 xx 10^(-3)` mol `l^(-1)` Time = 50 sec Rate of apperance of B = `("Increase of conc. B")/("Time taken")` `= (5xx 10^(-3) mol l^(-1))/(10"sec") = 5 xx 10^(-4) mol l^(-1) sec^(-1)`.

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