The enthalpy of combustion of 1 mole of benzene is 3169 kJ/mol .
The first step in answering this question is to obtain the balanced thermochemical equation of the reaction. The thermochemical equation shows the amount of heat lost or gained.
The thermochemical equation for the combustion of benzene is;
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔrH° = -3169 kJ/mol
We can see that 1 mole of benzene releases about 3169 kJ/mol of heat.
Learn more: brainly.com/question/13164491
Answer:
41.45 mL
Explanation:
Applying the general gas equation,
PV/T = P'V'/T'............... Equation 1
Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.
make V' the subject of the equation
V' = PVT'/TP'................ Equation 2
Given: P = 718 torr = (718×133.322) N/m² = 95725.196 N/m², V = 47.9 mL = 0.0479 dm³, T = 26 °C = (26+273) = 299 K, T' = 273 K, P' = 101000 N/m²
Substitute these values into equation 2
V' = ( 95725.196×0.0479×273)/(299×101000)
V' = 0.04145 dm³
V' = 41.45 mL
Explanation:
Ne] 3s¹ is the answer your welcome
a heater turns electrical energy into thermal energy
the sun turns gas into thermal energy
annd i cant think of anymore but i hope this helps you
