Answer :
121.5 <span>
μCi
Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.
We can use this formula;
</span>

3.8 /

=

((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5 </span>μci
Answer:
0.022 mol O
Explanation:
Mg3(Si2O5)2(OH)2
We can see that 1 mol of this substance has 3 mol of Mg.
Oxygen altogether is 5*2 (from (Si2O5)2) + 2(from(OH)2) = 10 +2 = 12
So, 1 mol of this substance has 12 mol oxygen.
So, 1 mol of this substance contains 3 mol Mg and 12 mol O, or
ratio Mg : O = 3 : 12 = 1 : 4
1 mol Mg ----- 4 mol O
0.055 mol Mg ---x mol O
x = 0.055*4/1 = 0.220 mol O