The first-order decomposition of N2O5 at 338K has a rate constant of 2.70 x 10-3 s-1

The first-order decomposition of N2O5 at 338K has a rate constant of 2.70 x 10-3 s-1. If the initial concentration of N2O5 is 2.88M, what is the concentration of N2O5 after 75.0 minutes?would you please help me i totally forgot about this subject
i need a brief explanation to be ready for my finals?

The concentration-time equation for a first order reaction is

ln ([A]t / [A]o) = -kt

[A]t = concentration of A at time t
[A]o = initial concentration of A
k = rate constant
t = time (same units a s rate constant)

So in our problem,

k = 2.70 x 10^-3 s^-1 which means the time must be expressed in seconds.
t = 75.0 min x (60 s / min) = 4500 s
[N2O5]o = 2.88 M

ln ([N2O5]t / [N2O5]o) = -kt
ln ([N2O5]t / 2.88) = -(2.70 x 10^-3)(4500)
ln [N2O5]t - ln 2.88 = -12.15
ln [N2O5]t - 1.06 = -12.15
ln [N2O5]t = -11.09
[N2O5] = e^-11.09 = 1.5 x 10^-5 M