En un matraz, disponemos de 100 g de gas oxígeno que se encuentran a 1 at de presión y 273 K de temperatura. Calcular : a) el nú
1 answer:
Answer:
Explanation:
Dado que:
masa de oxígeno gaseoso = 100 g
presión = 1 atm
temperatura = 273 K
(a)
número de moles de oxígeno contenidos en el matraz = masa de oxígeno / masa molar de oxígeno
= 100 g / 16 gmol⁻¹
= 6.25 moles
(b) El número de moléculas de oxígeno es el siguiente:
Dado que 1 mol de oxígeno gaseoso contiene 6.023 * 10²³ moléculas de oxígeno.
Entonces, 6.25 moles contendrán:
= (6.25 × 6.023 * 10²³) moléculas de oxígeno.
≅ 3.764 × 10²³ moléculas de oxígeno.
(c) El número de átomos de oxígeno es:
= 2 × 3.764 × 10²³
= 7.528 × 10²³ átomos de oxígeno
(d) Usando la ecuación de gas ideal
PV = nRT
El volumen ocupado por el oxígeno =
Volumen ocupado por oxígeno =
Volumen ocupado por oxígeno= 14185.76 m³
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Answer : The enthalpy of neutralization is, 56.012 kJ/mole
Explanation :
First we have to calculate the moles of HCl and NaOH.
The balanced chemical reaction will be,
From the balanced reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of NaOH
So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH
Thus, the number of neutralized moles = 0.08 mole
Now we have to calculate the mass of water.
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water =
Now we have to calculate the heat absorbed during the reaction.
where,
q = heat absorbed = ?
= specific heat of water =
m = mass of water = 400 g
= final temperature of water =
= initial temperature of metal =
Now put all the given values in the above formula, we get:
Thus, the heat released during the neutralization = -4480.96 J
Now we have to calculate the enthalpy of neutralization.
where,
= enthalpy of neutralization = ?
q = heat released = -4480.96 J
n = number of moles used in neutralization = 0.08 mole
The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, 56.012 kJ/mole