Answer:
ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]
Explanation:
Chemical equation:
CO + O₂ → CO₂
Balanced chemical equation:
2CO + O₂ → 2CO₂
The standard enthalpy for the formation of CO = -110.5 kj/mol
The standard enthalpy for the formation of O₂ = 0 kj/mol
The standard enthalpy for the formation of CO₂ = -393.5 kj/mol
Now we will put the values in equation:
ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]
ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]
ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]
ΔH0reaction = -283 kj/mol
<span>(P1/T1) = (P2/T2)
T must be in kelvin first!</span>
The nitrogen has been reduced or has undergone reduction and it has gained one electron
Answer : The new pressure if the volume changes to 560.0 mL is, 280 mmHg
Explanation :
According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.
or,
where,
= initial pressure = 560.00 mmHg
= final pressure = ?
= initial volume = 280 mL
= final volume = 560.0 mL
Now put all the given values in the above formula, we get:
Therefore, the new pressure if the volume changes to 560.0 mL is, 280 mmHg
