The combustion reaction is as expressed,
CxHy + O2 --> CO2 + H2O
The mass fraction of carbon in CO2 is 3/11. Hence,
mass of C in CO2 = (3.14 g)(3/11) = 0.86 g C.
Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g.
moles of C = 0.86 g C / 12 g = 0.0713
moles of H = 0.14 g H / 1 g = 0.14
The empirical formula for the hydrocarbon is therefore, CH₂.
Comets are usually formed of ice and other suave debris, while asteroids typically contain metals.
Answer:
Heat transfer = 3564 Jolues
The same value
Explanation:
The heat of combustion is the heat released per 1 mole of substunce experimenting the combustion at standard conditions of pressure and temperature ( 1 atm, 298 K):
Qtransfer = - mol x ΔHºc Qtransfer
So look up in appropiate reference table ΔHºc and solve the problem:
ΔHºc = - 891 kJ/mol
Qtransfer = - (4 x 10³ mol x -891 kJ/mol ) = 3564 J
if the combustion were achieved with 100 % excess air, the result will still be the same. As long as the standard conditions are maintained, the heat of combustion remains constant. In fact in many cases the combustion is performed under excess oxygen to ensure complete combustion.
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