Answer:19.6%
Explanation:
From K=0.693/t
0.693/20 =0.03465min^-1
But rate law is K =2.303/t log a/(a-x)
Substitute the value of k
0.03465=2.303/47 log a / (a-x)
Log a/ (a-x) = 0.7071
a/ (a-x) = 5.094
(a-x)/ a = 1/5.094 = 0.196
Percentage of reactants remaining after 47 minutes= 0.196× 100% = 19.6%