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Questions 6, 7, 8, and 9 pertain to different parts of the same problem. They are identifying different parts of the rate law
7 2 points (You may need to refer to your answer in the previous problem to figure this one out.) What is the exponent, nb? i
8 2 points What is the numerical value of k ? | Answer :: 5.2x10^(-8) 0.0002 :5.2 0.00005 1 5.2x10^(-4) 110.0001
2 points 11 What are the units on the rate constant, k? i Answer i no units : 1/6 1/(M^25) 8 (1/(Ms) M/S
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Answer #1

(0) tin solution :- Given A + 2B CTD Rate = K Cunits ) [A] na (BjMb (6) from given data , we can wente 512 X10-4 M/S = K 60.0(8) 7 Now in eq (ij we will put value of na= 0 and nb = 2 5:2x10-4 ms = K 10.0517° [0.0 ing? K = 5.2 M+S-! K = 5.2 = (a) Uni

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