Respuesta :
Answer : The half life of 28-Mg in hours is, 6.94
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = time passed by the sample = 48.0 hr
a = initial amount of the reactant disintegrate = 53500
a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520
Now put all the given values in above equation, we get
[tex]k=\frac{2.303}{48.0}\log\frac{53500}{42520}[/tex]
[tex]k=9.98\times 10^{-2}hr^{-1}[/tex]
Now we have to calculate the half-life.
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]9.98\times 10^{-2}=\frac{0.693}{t_{1/2}}[/tex]
[tex]t_{1/2}=6.94hr[/tex]
Therefore, the half life of 28-Mg in hours is, 6.94
