the pressure of the gas increases
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
Answer:
The standard enthalpy of formation of methanol is, -238.7 kJ/mole
Explanation:
The formation reaction of CH_3OH will be,

The intermediate balanced chemical reaction will be,
..[1]
..[2]
..[3]
Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:
We get :
..[1]
..[2]
[3]
The expression for enthalpy of formation of
will be,



The standard enthalpy of formation of methanol is, -238.7 kJ/mole
Answer:
PV=nRT where P=pressure in atm, V=volume is liters, n=numbber of moles, R=gas constant, 0.08206 L-atm/mole KL, and T=temperature in K (273 + C). So (5.67atm)(99.39L)=n(0.08206 L-atm/mol.K)(328.94K), solve for n, the number of moles, n=20.9 moles.
Explanation: