Respuesta :
the answer to this question is a 3 half lives have passed
Answer: 3 half lives
Solution :
Formula used :
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives = initial amount - amount of daughter isotope= (1.28-1.12)= 0.16 g
[tex]a_o[/tex] = Initial amount of the reactant = 1.28 g
n = number of half lives = ?
Putting values in above equation, we get:
[tex]0.16=\frac{1.28}{2^n}[/tex]
[tex]2^n=8[/tex]
[tex]2^n=2^3[/tex]
[tex]n=3[/tex]
Therefore, 3 half-lives have passed since the sample originally formed.
