1/2 life question?

a)
Let n be the number of half-lives.

(1/2)ⁿ = (1/4)
(1/2)ⁿ = (1/2)²
Hence, n = 2

Time taken = (355 s) × 2 = 711 s


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b)
Let m be the number of half-lives.

(1/2)ᵐ = 15%
log(0.5ᵐ) = log(0.15)
m log(0.5) = log(0.15)
m = log(0.15)/log(0.5)

Time taken = (355 s) × [log(0.15)/log(0.5)] = 970 s

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c)
Let k be the number of half-lives.

(1/2)ᵏ = 1/9
log(0.5ᵏ) = log(1/9)
k log(0.5) = log(1/9)
k = log(1/9)/log(0.5)

Time taken = (355 s) × [log(1/9)/log(0.5)] = 1125 s

in general
.. At = Ao * (1/2)^(t / half life)
where
.. At = amount remaining after time = t has elapsed
.. Ao = initial amount
.. t = elapsed time

and you can eyeball some of this and say
.. half lives.. .. . .amount remaining
.. . .. 0.. .. .. .. .. .. . .. 1
. .. ...1. .. .. .. .. . .....1/2
.. . ...2. . .. .. ... .. ... 1/4
.. .. ..3.. .. .. .. .. . ....1/8
each time a half life has passed, the amount remaining is cut in half

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(a).. 2 half lives have passed.... t = ?
(b).. At = 0.15 * Ao -----> t = -log(0.15)/log(2) * half life = ?
(c).. (1/9) = (1/2)^(t / half life) --> t = log(9)/log(2) * half life = ?

YOU get to finish

For A 1st order reaction

rate constant, k= 0.683/t^(1/2)

given, Half-LIFE, t^(1/2)=355s

:. k= 0.693/355s = 1.95 x 10^(-3) s^(-1)

a) For A 1st order reaction 

ln[A] =ln[A]o - kt

[A]= [A]o/4

ln( [A]o/4)= ln[A]o - 1.95 x 10^(-3) s^(-1)

t=711s

b) [A]o = 100, [A]=15

t=973s

c) [A]o =1  [A]= [A]0/9

ln( [A]o/9) =ln( [A]o) - 1.95 x 10^(-3) s^(-1)

t=1126s