Over an interval of 2.0 s the average rate of change of the concentration of C was measured to be 0.0540 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 2.000 M

Explanation:

the average rate of change of the concentration =

( initial concentration - final concentration ) / time

Initial concentration = 2.000 M

the average rate of change of the concentration = .0540 M/s

time = 2 s

Putting the values

.0540 =  (2.000 - X ) / 2

0.108 = 2.000 - X

X = 1.892 M .

Final concentration = 1.892 M .