it is equal theres your answer np :)
Answer:
The standard enthalpy change for the reaction at 
is -2043.999kJ
Explanation:
Standard enthalpy change (
) for the given reaction is expressed as:
![\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
Where 
refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
![\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%28-393.509kJ%2Fmol%29%5D%2B%5B4mol%5Ctimes%20%28-241.818kJ%2Fmol%29%5D-%5B1mol%5Ctimes%20%28-103.8kJ%2Fmol%29%5D-%5B5mol%5Ctimes%20%280kJ%2Fmol%29%5D%3D-2043.999kJ)
The answer is: the mass of carbon is 420.6 grams.
m(C₈H₁₈) = 500 g; mass of octane.
M(C₈H₁₈) = 114.22 g/mol; molar mass of octane.
n(C₈H₁₈) = m(C₈H₁₈) ÷ M(C₈H₁₈).
n(C₈H₁₈) = 500 g ÷ 114.22 g/mol.
n(C₈H₁₈) = 4.38 mol; amount of octane.
In one molecule of octane, there are eight carbon atoms:
n(C) = 8 · n(C₈H₁₈).
n(C) = 8 · 4.38 mol.
n(C) = 35.02 mol; amount of carbon.
m(C) = 35.02 mol · 12.01 g/mol.
m(C) = 420.6 g; mass of carbone.
Answer:An increase in temperature commonly will increase the rate of reaction. An growth in temperature will improve the common kinetic electricity of the reactant molecules.
Explanation:
It turns liquid to gas. Hoped this helped!!
