The rate constant for this second‑order reaction is 0.550 M−1⋅s−1 at 300∘C. A ⟶ productsHow long, in seconds, would it take for the concentration of A to decrease from 0.800 M to 0.380 M?

5.29 seconds

Explanation:

The reaction follows a second-order

Rate = kC^2 = change in concentration/time

k = 0.550 M^-1.s^-1

C = 0.380 M

Co = 0.800 M

Change in concentration = Co - C = 0.8 - 0.38 = 0.42 M

0.55×0.38^2 = 0.42/t

t = 0.42/0.07942 = 5.29 seconds

the millers tale is the answer.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.