1.

How is the rate constant ratio kx/kD related to initiator efficiency?

A. k<sub>x</sub>/k<sub>D</sub> = (1-f)/f
B. k<sub>x</sub>/k<sub>D</sub> = f/(1-f)
C. k<sub>x</sub>/k<sub>D</sub> = 1/(1-f)
D. k<sub>x</sub>/k<sub>D</sub> = (1-f)/2f
Answer» E.


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