Has encontrado problemas de eutrofización en Xochimilco. Los habitantes del lugar te preguntan cuál es el problema y les dices q
1 answer:
You might be interested in
Answer:
the solubility of CaCO3 is 0.015g/l 25 °C
is favored at equilibrium
Explanation:
The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?
solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)
CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)
in partial dissociation , we can say
2.25x 10^-8=
let Ca^2+=CO3^-2=S
2.25x10^-8=S*S
S^2=2.25x10^-8
S=0.00015mol/L
Converting that to g/l
the relative molecular mass of CaCO3=100g/mol
0.00015*100g/mol
0.015g/l
the solubility of CaCO3 is 0.015g/l @room temperature
is favored at equilibrium
Answer:
= 0.134;
= 0.866
The partial pressure of isopropanol = 34.04 Torr; The partial pressure of propanol = 5.26 Torr
Explanation:
For each of the solutions:
mole fraction of isopropanol () = 1 - mole fraction of propanol (
).
Given: mole fraction of propanol = 0.247. Thus, the mole fraction of isopropanol = 1 - 0.247 = 0.753.
Furthermole, the partial pressure of isopropanol = *vapor pressure of isopropanol = 0.753*45.2 Torr = 34.04 Torr
The partial pressure of propanol = *vapor pressure of propanol = 0.247*20.9 Torr = 5.16 Torr
Similarly,
In the vapor phase,
The mole fraction of propanol () =
Where, is the partial pressure of propanol and
is the partial pressure of isopropanol.
Therefore,
= 5.26/(34.04+5.16) = 0.134
= 1 - 0.134 = 0.866