Answers+to+Ch5

187lmol265 lmolmin211min

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Unformatted text preview: aight line. (b) k = 0.200 (c) ln(0. 497/0. 100)= 0.200 t ; t = 8.00 h It takes 8.00 h to reduce the concentration from 0.497 M to 0.100 M and an additional 4.00 h to reduce the original concentration to 0.497M . Total time is 12.00 h. (d) rate = 0.200(0.415 M ) = 0.0830 mol/L· h 39. (a) ln(150. 0 mg/43. 2 mg)= k ×0. 750 h; k = 1.65/h (b) t 1/2=0. 693/1. 65h = 0.420 h 3 (c) ln1/(1 − 0. 95)=1. 65/h ×t ; t = 1.8 h 53. (a) k =1/[Ao ]k=1/(1. 51 min)(0. 250 mol/L)= 2.65 L/mol· min (b) t =(1/ [A] −...
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