0.2 mol `A_(x)B_(y)to0.5` mol `A_(2)+0.4molB_(3)`. From this data the molecular formula of AB

AB, A_(2) and B_(2) are diatomic molecules.If the bond enthalpy of A_(2,) AB from A_(2) and B_(2) od -100 kJ//"mol" , what is the bond enthalpy of A_(2)" in" kJ//mol . ?

AB,A_(2) and B_(2) are diatomic molecules. If the bond enthalpies of A_(2), AB and B_(2) are in the ratio 1:1:0.5 and the enthalpy of formation of AB from A_(2) and B_(2) is -100kJ mol^(-1) , what is the bond enthalpy of A_(2) ?

AB,A_(2) and B_(2) are diatomic molecules. If the bond enthalpies of A_(2), AB and B_(2) are in the ratio 1:1:0.5 and the enthalpy of formation of AB from A_(2) and B_(2) is -100kJ mol^(-1) , what is the bond enthalpy of A_(2) ?

AB,A_(2) and B_(2) are diatomic molecules. If the bond enthalpies of A_(2), AB and B_(2) are in the ratio 1:1:0.5 and the enthalpy of formation of AB from A_(2) and B_(2) is -100kJ mol^(-1) , what is the bond enthalpy of A_(2) ?

A and B are two element which form AB_(2) and A_(2)B_(3) . If 0.18 mol of AB_(2) weighs 10.6g and 0.18 mol of A_(2)B_(3) weighs 17.8g then -

The reaction , 2AB(g) + 2C(g) to A_(2(g)) + 2BC_((g)) proceeds according to the mechanism . I. 2AB hArr A_(2)B_(2) (fast) II. A_(2)B_(2)+C to A_(2)B + BC (slow ) III. A_(2)B+C to A_(2)+BC (fast) what will be the initial rate taking [AB] = 0.2 M and [C] = 0.5 M ? The K_(c) for the step I is 10^(2) M^(-1) and rate constant for the step II is 3.0 xx 10^(-3) mol^(-1) min^(-1)

The rate of the reaction , A_(3)Brarr(3)/(2)A_(2) +(1)/(2) B_(2), " is " x "mol.L"^(-1).s^(-1) . If the rates of formation of A_(2) and B_(2) " are " x_(1) and x_(2) "mol.L"^(-1).s^(-1) , respectively , then -