The heat evolved in combustion of benzene is given by the following equation :
`C_(6)H_(6)+15//2O_(2)to3H_(2)O+6CO_(2)` : `DeltaH=-3264.6 kJ`
Which of the following quantities of heat will be evolved when `39g` of benzene are burnt in an open container ?

The heat evolved in the combustion of benzene is given by the equation C_(6)H_(6)(l)+7(1)/(2)O_(2)(g)to 6CO_(2)(g)+3H_(2)O(l) , Delta H =- 781 Kcal mol^(-1) Which of the following quantities of heat will be evolved when 39 g of benzene is burnt in an open container

The heat evolved in the combustion of benzene is given by the equation C_(6)H_(6)(l)+7(1)/(2)O_(2)(g)to 6CO_(2)(g)+3H_(2)O(l) , Delta H =- 781 Kcal mol^(-1) Which of the following quantities of heat will be evolved when 39 g of benzene is burnt in an open container

The heat evolved in the combustion of benzene is given by the equation: C_(6)H_(6)(g) + (15)/(2) O_(2)(g) to 6CO_(2)(g) + 3H_(2)O(l), DeltaH = - 3264.6 kJ "mol"^(-1) The heat energy changes when 39 g of C_(6)H_(6) are burnt im an open ontainer will be:

According to the reaction C_(6)H_(6)+15//2O_(2)to3H_(2)O+6CO_(2) ( DeltaH=-3264.6 kJ // mol ) the energy evolved when 3.9g of benzene is burnt in air will be

The heat evolved in the combustion of benzene is given by. the equation: 'C_6 H_6(g)+(15)/2 O_2(g) rarr 6 CO_2^(g)+3 H_2 O(l), Delta H=-3264.6 kJ mol^(-1)' The heat energy change when '39 g' of 'C_6 H_6' are burnt in an open container will be:

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 ?

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 ?