The unknown of this problem is the experimental percent of water in the compound in order to remove the water of hydrogen, given the following:
Mass of crucible, cover and contents before heating 23.54 g
Mass of empty crucible and cover 18.82 g
Mass of crucible, cover, and contents after heating to constant mass 20.94 g
In order to get the answer, determine the following:
Mass of hydrated salt used = 23.54 g – 18.82 g = 4.72 g
Mass of dehydrated salt after heating = 20.94 g – 18.82 g = 2.12 g
Mass of water liberated from salt = 4.72 g – 2.12 g = 2.60 g
Then solve the percent of water in the hydrated salt by:
% water = (mass of water / mass of hydrated salt) x 100
% water = 2.60 g / 4.72 g x 100
% water = 55.08 % in the compound
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Answer:
hope it helped you.
Explanation:
The properties of matter that do not depend on the size or quantity of matter in any way are referred to as an intensive property of matter. Temperatures, density, color, melting and boiling point, etc., all are intensive property as they will not change with a change in size or quantity of matter.
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
