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Calculate the amount of heat evolved during the complete combustion of 100 ml liquid benzene from the following data :
18 gm of graphite on complete combustion evolve 590 kJ heat
15889 kJ heat required to dissociate all the molecules of 1 liter water into `H_(2)` and `O_(2)`
The heat of formation of liquide benxene is `50kJ//"mol"`
Density of `C_(6)H_(6)(l)=0.87gm//ml`

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Calculate the amount of heat evolved during the complete combustion of 100 ml of liquid benzene from the following data. (i) 18 g of graphite on complete combustion evolve 590 KJ heat (ii) 15889 KJ heat is required to dissociate all the molecules of 1 litre water into H_(2) and O_(2) (iii) The heat of formation of liquid benzene is 50 KJ//mol (iv) Density of C_(6)H_(6)(l)=0.87 g//ml

The heat librerated on complete combustion of 1 mole of CH_(4) gas to CO_(2)(g) and H_(2)O(l) is 890 kJ. Calculate the heat evolved by 2.4 L of CH_(4) on complete combustion.

The heat evolved in the combustion of methane is given by the following equation : CH_(4)(g)+2O_(2)(g)to CO_(2)(g)+2H_(2)O(l) , Delta H = -890.3 kJ How many grams of methane would be required to produce 445.15 kJ of heat of combustion ?

On complete combustion of 2 gm methane -26575 cals heat is generated. The heat of formation of methane will be (given heat of formation of CO_(2) and H_(2)O are - 97000 and - 68000 cais respectivvely) :

The heat of combustion of CH_(4) is -400 KJ mol^(-1) . Calculate the heat released when 40g of H_(2)O is formed upon combustion :-

The combustion of 1 mol of benzene takes place at 298 K and 1 atm . After combustion, CO_(2)(g) and H_(2)O(l) are produced and 3267.0 kJ of heat is librated. Calculate the standard entalpy of formation, Delta_(f)H^(Θ) of benzene Given: Delta_(f)H^(Θ)CO_(2)(g) = -393.5 kJ mol^(-1) Delta_(f)H^(Θ)H_(2)O(l) = -285.83 kJ mol^(-1) .

The heat liberated on complete combustion of 7.8g benzene is 327kJ . This heat has been measured at constant volume and at 27^(@)C . Calculate heat of combustion of benzene at constant pressure at 27^(@)C . (R=8.3 J mol^(-1)K^(-1))

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