The reaction is first order in cyclopropane and has a measured rate constant of k=3.36×10−5 s−1 at 720 k. if the initial cyclopropane concentration is 0.0445 m, what will the cyclopropane concentration be after 235.0 min

0.0277 M.

Explanation:

The integral rate law of a first order reaction:

Kt = ln ([A₀]/[A]),

where, k is the rate constant of the reaction (k = 3.36 × 10⁻⁵ s⁻¹),

t is the time of the reaction (t = 235.0 min = 14100 s),

[A₀] is the initial concentration of cyclopropane ([A₀] = 0.0445 M)

∵ Kt = ln ([A₀]/[A]),

∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]

Taking the exponential of both sides:

1.6 = (0.0445 M)/[A]

∴ [A] = (0.0445 M)/1.6 = 0.0277 M.

i think its 100

i hope this !

it is b that is the answer.