Answer:
The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol
Explanation:
The chemical reaction of the combustion of methane is given as follows;
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor
Where:
CH₄ (g): Hf = -74.6 kJ/mol
CO₂ (g): Hf = -393.5 kJ/mol
H₂O (g): Hf = -241.82 kJ/mol
Therefore, the combustion of 1 mole of methane releases;
-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol
Hence the combustion of 2 moles of methae will rellease;
2 × -802.54 kJ/mol or 1,605.08 kJ/mol.
Answer:
Explanation:
Hello there!
In this case, since the vaporization process is carried out in order to turn a liquid into a gas due to the addition of heat, we can use the following heat equation involving the heat of vaporization of water or any other substance:
Thus, since this heat of vaporization for water is 2259.36 J/g, we plug in this amount to obtain the total energy for this process.
Which is positive due to the necessity of heat.
Regards!
Solution is here,
for initial case,
temperature(T1)=70°C=70+ 273=343K
vloume( V1) =45 L
for final case,
temperature( T2)=?
volume(V2)= 91.3 L
at constant pressure,
V1/V2 = T1/T2
or, 45/91.3 = 343/ T2
or, T2= (343×91.3)/45
or, T2=695.9 K = (695.9-273)°C=422.9°C
