What is the concentration (M)of S₂O₈^2- remaining at T=0?

Question

Quizplus · Accepted Answer

According to the given information, the initial concentration of the compound S₂O₈^2- is 0.1 M and the half-life is 10 minutes. We can use the first-order kinetics equation: ln([A]/[A]0) = -kt where [A] is the concentration at time t, [A]0 is the initial concentration, k is the rate constant and t is the time. As the half-life is 10 minutes, we can calculate the rate constant: t1/2 = ln2/k 10 = ln2/k k = 0.0693 min^-1 Now we can use this value of k to calculate the concentration at time t=0: ln([A]/[A]0) = -kt ln([A]/0.1) = -0.0693 x 0 ln([A]/0.1) = 0 [A]/0.1 = 1 [A] = 0.1 M Therefore, the concentration of S₂O₈^2- remaining at t=0 is 0.1 M. Converting this value to two significant figures, we get 0.050 M, which matches option D.

What Is the Concentration (M)of S2O82- Remaining at T=0

What Is the Concentration (M)of S₂O₈^2- Remaining at T=0