Answer:
Explanation:
Hello,
In this case, since the VdW equation is:
Since the moles are 10.0 moles, the temperature in K is 300.15 K and the volume is liters is also 4.860 L (1 dm³= 1L), the pressure exerted by the ethane is:
Thus the compression factor turns out:
Regards.
80.1 g* (1 cm^(3)/ 2.70 g)= 29.666666... cm^(3).
The calculated volume should have 3 significant figures, so the final answer is 2.97*10^(1) cm^(3)
Hope this helps~
The volume decreases when the pressure is increased or when the temperature is decreased.
According to Charles's law, the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. This means that, if the temperature of the given mass of gas is decreased, its volume decreases accordingly and vice versa.
According to Boyle's law, the volume of a given mass of gas is inversely proportional to its pressure. Hence, when the pressure of a gas is increased, its volume decreases accordingly.
Summarily, the volume decreases when the pressure is increased or when the temperature is decreased.
Learn more: brainly.com/question/6590381
Answer:
Explanation:
Hello.
In this case, since the first-order reaction is said to be linearly related to the rate of reaction:
Whereas [A] is the concentration of hydrogen peroxide, when writing it as a differential equation we have:
Which integrated is:
And we can calculate the initial concentration of the hydrogen peroxide as follows:
Thus, for the given data, we obtain:
Best regards!
Respuesta:
1.50 moles
Explicación:
Paso 1: Escribir la ecuación química ajustada para la descomposición del carbonato de calcio
CaCO₃ ⇒ CaO + CO₂
Paso 2: Calcular los moles correspondientes a 150 g de CaCO₃
La masa molar de CaCO₃ es 100.09 g/mol.
150 g × 1 mol/100.09 g = 1.50 mol
Paso 3: Calcular los moles de CO₂ producidos a partir de 1.50 moles de CaCO₃
La relación molar de CaCO₃ a CO₂ es 1:1.
1.50 mol CaCO₃ × 1 mol CO₂/1 mol CaCO₃ = 1.50 mol CO₂