Question 2633
 Let Q = Qo e^(-kt),
 when t =0, Q = 40, so Qo = 40,
 So, Q = 40 e^(-kt)
 when t=710, Q = 1/2 Q0 = 20,
 solve 20 = 40 e^(-710k), or 1/2 = e^(-710k)
 we get k = ln2/710 = 0.000976264 = 9.76 *10^(-4)

 hence,Q = 40 e^(- 9.76 *10^(-4)t)
 when t = 200 yrs, Q = 40 e^(- 9.76 *10^(-4)*200)
                     = 32.91 grams

 Kenny