Respuesta :

aencabo
A(t) = A(o)(½)^t/h 
A(t) /Ao = 0.5^t/h 
Where A(t) = amount present, A(o) = initial amount, t = elapsed time, h = half-life 
If 95% has been converted then At/Ao = 0.06 t½ = 15.0 h 
log(0.06) = log(0.5)t/15 
15×log(0.05)/log(0.5) = t 
t = 60.9 h 

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