Answer:
140 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 3 atm
- Initial temperature of the gas (T₁): 280 K
- Final pressure of the gas (P₂): 1.5 atm
- Final temperature of the gas (T₂): ?
Step 2: Calculate the final temperature of the gas
We have a gas whose pressure is reduced. If we assume an ideal behavior, we can calculate the final temperature of the gas using Gay-Lussac's law.
T₁/P₁ = T₂/P₂
T₂ = T₁ × P₂/P₁
T₂ = 280 K × 1.5 atm/3 atm = 140 K
Answer:
answer is option 2 hope it helps
Answer:
Explanation:
Hello,
Considering the chemical reaction, the enthalpy of reaction is given by:
ΔH°rxn=ΔfHCO2+ΔfHH2O-ΔfHC8H18
(ΔfHO2=0)
Taking into account that the reaction produces energy, ΔH°rxn is negative. No, solving for ΔfHC8H18:
ΔfHC8H18=-ΔH°rxn+8*ΔfHCO2+9*ΔfHH2O
ΔfHC8H18=-(-5104.1 kJ/mol)+9*(-292.74kJ/mol)+8*(-393.5 kJ/mol)
ΔfHC8H18=-678.56 kJ/mol
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