Franco was told that he had mycarditis. He went home and researched the disease, and he determined that he had
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Answer:
P IBr: 15.454atm
I₂: 0.923 atm
P Br₂: 0.923atm
Explanation:
Basados en la reacción:
I₂(g) + Br₂(g) ⇄ 2 IBr(g)
La constante de equilibrio, Keq, es definida como:
<em>Se cumple la relación de Keq = 280 cuando las presiones están en equilibrio</em>
Usando PV = nRT, la presión inicial de IBr es:
P = nRT / V; 0.500mol*0.082atmL/molK*423.15K / 1.00L = <em>17.3 atm</em>
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Siendo las presiones en equilibrio:
P IBr: 17.3 - 2X
P I₂: X
P Br₂: X
<em>Donde X representa el avance de reacción.</em>
Remplazando en Keq:
280 = (17.3 - 2X)² / X²
280X² = 4X² - 69.2X + 299.29
0 = -276X² - 69.2X + 299.29
<em>Resolviendo para X:</em>
X = -1.174 → Solución falsa. No existen presiones negativas
X = 0.923 → Solución real
Así, las presiones parciales en equilibrio de cada compuesto son:
P IBr: 17.3 - 2X = <em>15.454atm</em>
P I₂: X =<em> 0.923atm</em>
P Br₂: X = <em>0.923atm</em>
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