Answer:
ΔH3 = -110.5 kJ.
Explanation:
Hello!
In this case, by using the Hess Law, we can manipulate the given equation to obtain the combustion of C to CO as shown below:
C(s) + 1/2O2(g) --> CO(g)
Thus, by letting the first reaction to be unchanged:
C(s) + O2(g)--> CO2 (g) ; ΔH1 = -393.5 kJ
And the second one inverted:
CO2(g) --> CO(g) + 1/2O2(g) ; ΔH2= 283.0kJ
If we add them, we obtain:
C(s) + O2(g) + CO2(g) --> CO(g) + CO2 (g) + 1/2O2(g)
Whereas CO2 can be cancelled out and O2 subtracted:
C(s) + 1/2O2(g) --> CO(g)
Therefore, the required enthalpy of reaction is:
ΔH3 = -393.5 kJ + 283.0kJ
ΔH3 = -110.5 kJ
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It’s a nonet because an octet is 8, a quartet is 4, and a duet is 2.
Answer:
the work done on the gas is 4,988.7 J.
Explanation:
Given;
number of moles of the monoatomic gas, n = 4 moles
initial temperature of the gas, T₁ = 300 K
final temperature of the gas, T₂ = 400 K
The work done on the gas is calculated as;
For monoatomic ideal gas: 
Where;
R is ideal gas constant = 8.3145 J/K.mol
Therefore, the work done on the gas is 4,988.7 J.
