The reaction 2N2O5(g)eys the rate law: rate = k[N2O5]. In which the specific rate constant is 0.00840 s-1. If 2.50 moles of N2O5 were placed in a 5.0-liter container at that temperature, how many moles of N2O5 would remain after 1.00 minute?

1.65 moles of N2O5

Explanation:

The reaction follows a first-order

Rate = K[N2O5] = change in concentration of N2O5/time

Let the concentration of N2O5 after 1 min be y

Initial concentration of N2O5 = number of moles of N2O5/volume of container = 2.5/5 = 0.5 M

Change in concentration = 0.5-y

k = 0.00840 s^-1

t = 1 min = 60 s

0.0084y = 0.5 - y/60

0.5 - y = 0.504y

0.504y+y = 0.5

1.504y = 0.5

y = 0.5/1.504 = 0.33 M

Number of moles = concentration × volume = 0.33×5 = 1.65 moles

the ratio of the conjugate base to the acid required to attain the desired ph.