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The following article will add knowledge about “** The reaction rate constant is determined to be 0.012 M^{-1} s^{-1}. Ifafter 27 minutes the amount of A left is 0.048 M. What was the initial concentration of A? 0.049 2.53e16 0.72 19.49**“. Let’s not forget to look at the content!

## Question

The reaction rate constant is determined to be 0.012 M^{-1}s^{-1}. Ifafter 27 minutes the amount of A left is 0.048 M. What was the initial concentration of A?

0.049

2.53e16

0.72

19.49

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## Answer “The reaction rate constant is determined to be 0.012 M^{-1} s^{-1}. Ifafter 27 minutes the amount of A left is 0.048 M. What was the initial concentration of A? 0.049 2.53e16 0.72 19.49”.

We have:

K = 0,012 M^{-1}s^{-1} <- second order

1/A = 1/A_{0} + kt

1/0,048M = 1/A_{0} + (0,012 M^{-1}s^{-1})(27min x 60s/min)

=> A_{0} = 0,72M

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So, the result of this question is 0.72.

## Last words

Above is the solution of **“The reaction rate constant is determined to be 0.012 M^{-1} s^{-1}. Ifafter 27 minutes the amount of A left is 0.048 M. What was the initial concentration of A? 0.049 2.53e16 0.72 19.49“**. If you have good solutions or comments, please leave a message.

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