Answer:
hi
Explanation:
this is my first time on this app so what do i do
Answer is: A) 124 s.
c₀ = 3 mol/L.
c₁ = 0,700 mol/L.
k = 8,8·10⁻³ 1/M·s.
Integrated second order rate law is: 1/c₁ = 1/c₀ + k·t.
k·t = 1/0,700 - 1/3.
0,0088·t = 1,095.
t = 1,095 ÷ 0,0088.
t = 124 s.
c₀ - <span>initial concentration.
c</span>₁ - <span> concentration at a particular time.
k - </span><span>the rate constant.
t - time.</span>
