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.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.