Answer:
Please review the explanation
Detailed explanation:
Based on the data provided:
Time, t (hours)__1.0__1.5__2.0__2.5__3.0
C(t) (mg/mL)__0.35_0.26_0.20_0.14_0.09
Average change of C with respect to t over the interval:
(i) [1.0, 2.0] (ii) [1.5, 2.0] (iii) [2.0, 2.5] (iv) [2.0, 3.0]
Average change is calculated as ( change in C / change in t) = C2 - C1 / t2 - t1
1) [1.0, 2.0]
C at t = 1 = 0.35; C at t = 2 = 0.20
(0.20 - 0.35) / (2 - 1) = - 0.15 / 1 = - 0.15 mg/mL hr
11) [1.5, 2.0]
(0.20 - 0.26) / (2.0 - 1.5) = - 0.06 / 0.5 = - 0.12 mg/mL hr
111) [2.0, 2.5]
(0.14 - 0.20) / (2.5 - 2.0) = - 0.06 / 0.5 = - 0.12mg/mL hr
iv) [2.0, 3.0]
(0.09 - 0.20) / (3.0 - 2.0) = - 0.11 / 1.0 = - 0.11 mg/mL hr
