Answer:
3.43 %
Explanation:
We need to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.
0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2
0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce
.00107 mol Ce x 140.116 g Ce/ mol = 0.150 g Ce
0.150 g Ce/ 4.3718 g sample x 100 = 3.43 %
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Answer:
36s^5
Explanation:
We have;
M2X3 (s)------> 2M^3+(aq) + 3X^2-(aq)
If [M^3+(aq)] = [X^2-(aq)] = s
We then have;
Ksp = (2s)^2 * (3s)^3
Ksp = 4s^2 * 9s^3
Ksp = 36s^5
Note that Ksp is known as the solubility product. It is an equilibrum equation that shows the solubility of a solute in water.
