Question

Questions 7, 8, and 9 refers to the H:02 decomposition experiment Remember ViM V.M Use the table below to answer the following questions mL of o.15 M KI Trial 10 ml Trial Trial mL of 0.090 M H20 20 mL mL of H2O 20 mL 30 mL Rate (ml O2lsec) 20 mL/sec 10 mL 10 mL 10 mL/sec 20 mL 10 mL 20 mL 20 mL/sec 7 For trial 1 the initial concentration of KI (after everything is added) is closest to: A 0.15M B 0.015M C 0.3M D0.03M E 0.5M FO.05M G0.005 8 Based on Trial 1, 2, and 3, the values for n and m In the expression Rate k[H202]"[r]" are: 9 Based on the rate expression and the values n and m, what would be the 02 generation rate (in units of mL 02/sec) for the concentrations below? Vol. 0.15 M KI Vol. 0.090 M H202 Vol. HO Rate (ml O:/sec) 20 mL 10 mL ??? mL/sed Trial 420 mL A 10 B 20 C 30 D 40 E50 F60 G 70 10 Which of the following is equal to one one thousandth of a liter? A 1 ml B 1 cm3 C 1cubic centimeter D All of the above E none of the above

Answer 1

[KI] can be find out by using equation M_1 V_1 = M_2V_2 On substituting values we get 0.15 m times 10 ml = M_2 times 50ml M_2 0.03 rate = K[H_2 O_2]^n [I^-]^m for finding value of m consider Trial R_3 write rate equation and solve first find the concentration of both H_2 O_2 RI trial 2 [H_2 O_2] = 0.09 times 10/50 = 0.018M [I^-] = 0.15 times 10/50 = 0.03M Trial^3 [H_2 O_2] = 0.15 times 10/50 = 0.06M value for we ? On solve ling rate eqn trial 2 R3 write rate equation trial 20 ml /sec/10ml sec K[0.018]^n [0.06]^m/K[0.018]^n[0.03]^m 2ml /sec = 2^m therefore m = 1 similarly consider Trial to find n = ? Rate eqn trial 1 % Rate eqn Trial 2 20ml sec/10ml/sec = k(0.036)^n (0.03)^n/k(0.0.18)^n (0.03)^m 2ml/sec = 2^n therefore n = 1 consider rate equation of Trial and find value of K 20ml/sec = K (0.036)(0.03)' k = 20/0.036 times 0.03 = 18518.518 Rate = 18518 times (0.036) times (0.06) = 40ml/sec 1 line = 100ml times 1/1000 = 1ml

1litre=1000cm^3

1/1000 of 1 litre=1cm^3

Answer=option D