half life period of a first order reaction is 10 min what percentage of addiction will be completed in hundred minutes
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Answer:
99.9%
Explanation:
K= 0.693÷t 1/2
K= 2.303/t log10 (a/a-x)
= 0.693÷10
= 2.303÷100 log10 ( 100/100 - x)
x = 99.9 %
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