A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. (Given, `log2=0.3010,log

`k=(2.303)/(t)log((a)/(a-x))` When , a=100, x=75, t=32 `k=(2.303)/(32)log_(10)((100)/(25))=0.0433 "min"^(-1)` Time for `50%` completion `(t_(1//2))=(0.693)/(k)=(0.693)/(0.0433)~~16"min"^(-1)`

`k=(2.303)/(t)log.(a)/(a-x)` `k=(2.303)/(70)log.(100)/(100-90)` `=0.0329 "min"^(-1)`

Let `K = x^(1/log y+1/log z)xxy^(1/log z+1/log x)xxz^(1/log x+1/log y)` ` :. log K = log x[1/log y+1/log z]+log y [1/log z+1/log x]+log z [1/log x+1/log y]` Putting ` log...

`log_(3)4log_(4)5log_(5)6log_(6)7log_(7)8log_(8)9` `=(log4)/(log3)xx(log)5/(log4)xx(log6)/(log5)xx(log7)/(log6)xx(log8)/(log7)xx(log9)/(log8)` `=(log9)/(log3)=log_(3)9=2`

Correct Answer - A::B::C (1) ` log_(10)(10/2) * log_(10)(10xx2)+(log_(10)2)^(2)` ` = (1-log_(10)2)(1+log_(10)2)+(log_(10)2)^(2)=1` (2) `(log2^(2)xx3)/(log(48//4)) = 1` (3) ` -log_(5)log_(3)9^(1//10)=- log_(5)log_(3)3^(1//5)` ` = - log_(5)(1//5)=1` (4) ` 1/4 log_(sqrt3/2)(64/27) = 1/6 log_(sqrt3/2)(sqrt3/2)^(-6)=-...

Correct Answer - 2 ` N = (log_(5)250)/(log_(50)5)-(log_(5)10)/(log_(1250)5)` ` = (log_(5)250)(log_(5)50)-(log_(5)10)(log_(5)1250)` ` = (3+log_(5)2)(2+log_(5)2)-(1+log_(5)2)xx(4+log_(5)2)` ` = (log_(5)2)^(2)+5 log_(5)2+6-[(log_(5)2)^(2)+5 log_(5)2 + 4]` = 2

Correct Answer - B `"log"x/M =log k+1/n log P` `1/n=tan 45^(@) log k=0.3010` `n=1 k=2` `x/m=2xx(0.3)^(1)` `x=0.6`

Correct Answer - C `I=-e^(-x)log(e^(x)+1)+int(e^(-x)e^(x))/(e^(x)+1)dx ` `=-e^(-x)log(e^(x)+1)+int(e^(-x))/(e^(-x)+1)dx` `=-e^(-x)log(e^(x)+1)-log(e^(-x)+1)+C` `=-e^(-x)log(e^(x)+1)-log(1+e^(x))+x+C` `=-(e^(-x)+1)log(e^(x)+1)+x+C`

Correct answer is (C) 16 minutes

Correct Answer - `0.606443`