At 25°C, a saturated solution of cerium(III)
hydroxide (Ce(OH)3) contains 5.1 x 10-6 moles of
the compound in one liter of solution.
What is the reaction equation?
O Ce(OH)3(s) + Ce3+ (aq) + 3OH-(aq)
O Ce(OH)3(s) + Ce3+ (aq) + OH:(aq)
O 3Ce 3+ (aq) + OH (aq) +- Ce(OH)3(s)
O Ce(OH)3(s) + Ce 3+ (aq) + (OH-)3(aq)
DONE

Ce(OH)3(s) —> Ce3+ (aq) + 3OH-(aq)

Explanation:

In solution, Ce(OH)3 will dissociate as follow:

Ce(OH)3(s) —> Ce3+(aq) + OH-(aq)

Now, we need to balance the charge on both side to obtain the balance equation. This is illustrated below:

There is 0 charge on the left and +2 charge on the right. It can be balance by putting 3 in front of OH- as shown below:

Ce(OH)3(s) —> Ce3+(aq) + 3OH-(aq)

Now we have 0 charge on both the left and right side of the equation. Therefore, the equation is balanced.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

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