Determina el grado de pureza de un marmol (CaCO3), si al descomponerse 125 g del mismo se desprenden 20 litros de dióxido de carbono medidos a 15ºC y 1 atm.
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Ответ:
67.8%
Explanation:
La reacción de descomposición del CaCO₃ es:
CaCO₃ → CO₂ + CaO
Donde 1 mol de CaCO₃ al descomponerse produce 1 mol de CO₂ y 1 mol de CaO.
Usando la ley general de los gases, las moles de dioxido de carbono son:
PV = nRT.
Donde P es presión (1atm), V es volumen (20L), n son moles de gas, R es la constante de los gases (0.082atmL/molK) y T es temperatura absoluta (15 + 273.15 = 288.15K). Reemplazando los valores en la ecuación:
PV / RT = n
1atmₓ20L / 0.082atmL/molKₓ288.15K = 0.846 moles
Como 1 mol de CO₂ es producido desde 1 mol de CaCO₃, las moles iniciales de CaCO₃ son 0.846moles.
La masa molar de CaCO₃ es 100.087g/mol. Así, la masa de 0.846moles de CaCO₃ es:
0.846moles ₓ (100.087g / mol) = 84.7g de CaCO₃
Así, la pureza del marmol es:
(84.7g de CaCO₃ / 125g) ₓ 100 =
67.8%Ответ:
in the first balloon when the balloon get warm the molecules begin to distribute to the wall of the balloon and the balloon will brust because of the tension caused.
In the second case when the bLlon get cold the molecules Will atract each other then the balloon will contract because of the attraction of the molecules in the balloon.