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The rate law for a reaction between `A` and `B` is given by rate `= k[A]^(n)[B]^(m)`. On doubling the concentration of `A` and halving the concentration of `B`, the ratio of the new rate to the earlier rate of the reaction becomesA. `(1)/(2^(m+n))`B. m+nC. m-nD. `2^(n-m)` |
Answer» Correct Answer - D | |