A) 4400 kj of heat released into surroundings
<h3>Further explanation</h3>
Reaction
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O, the ∆H is –2200 kJ
Reaction exothermic( ∆H=-, released heat to surrounding) and for combustion of 1 mole of C3H8
So for two moles of C3H8, the enthalpy :

Answer:
the mass required to inflate a 72 L bag is 191.491 g
Explanation:
reaction:
conditions:
- V = 72.0 L
- STP: P = 1 atm ∧ T = 298 K
gas law:
- PV = RTn
- R = 0.082 atm * L / K * mol
⇒ n = PV / RT
⇒ n = ((1 atm) * ( 72.0 L)) / (0.082 atm*L / K*mol) * (298 K)
⇒ n = 2.946 mol
⇒ m = n * Mw = ( 2.946 mol ) * ( 64.99 g/mol)
⇒ m = 191.491 g
Answer:

Explanation:
From the question we are told that:
Chemical Reactions:
X=A⇌B,ΔG= 14.8 kJ/mol
Y=B⇌C,ΔG= -29.7 kJ/mol
Z=C⇌D,ΔG= 8.10 kJ/mol
Since
Hess Law
The law states that the total enthalpy change during the complete course of a chemical reaction is independent of the number of steps taken.
Therefore
Generally the equation for the Reaction is mathematically given by

Therefore the free energy, ΔG is


