Answer:
The percentage of N in the compound is 0.5088
Explanation:
Mass of compound = 8.75 mg = 8.75×1000 = 8750 g
Mass of N2 = number of moles of N2 × MW of N2 = 1.59 × 28 = 44.52 g
% of N in the compound = (mass of N2/mass of compound) × 100 = (44.52/8750) × 100 = 5.088×10^-3 × 100 = 0.5088
When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2 + 2OH-
when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
so we can assume [Fe+2] = X and [OH-] = 2 X
when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17
[Fe+2]= X
[OH-] = 2X
so by substitution
4.87x10^-17 = X*(2X)^2
∴X^3 = 4.8x10^-17 / 4
∴the molar solubility X = 2.3x10^-6 M
69.9%
Explanation:
To find the mass percentage of iron in the compound in Fe₂O₃, we would go ahead to express the given molar mass of the iron to that of the compound.
Mass percentage = 
x 100
Molar mass of Fe = 55.85g/mol
Molar mass of O = 16g/mol
Molar mass of Fe₂O₃ = 2(55.85) + 3(16) = 159.7g/mol
Mass percentage = 
= 69.94% = 69.9%
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Mass percentage brainly.com/question/8170905
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The products are on the right side of the equation. For this one it would be 2AlPO4 + 3CaSO4
