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The rate law for a reaction between the substances 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 will be as:A. `(1)/(2^(m+n))`B. `(m=n)`C. `(n-m)`D. `2^((n-m))` |
Answer» Correct Answer - D |
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