What is heat of submisation of `P_(4)O_(6)(s)?`
Given heat of sublimation of `P_(4)o_(6)` is x KJ/mol and `P-O` bond energy is y KJ/mol.

The number of P-O bonds in P_(4)O_(6) is :

The heat of atomisation of PH_(3)(g) and P_(2)H_(4)(g) are 954 kJ mol^(-1) and 1485 kJ mol^(-1) respectively. The P-P bond energy in kJ mol^(-1) is

The heat of atomisation of PH_(3)(g) and P_(2)H_(4)(g) are 954 kJ mol^(-1) and 1485 kJ mol^(-1) respectively. The P-P bond energy in kJ mol^(-1) is

The ratio of sigma bond in P_(4)O_(10) and P_(4)O_(6) is respectively :

P_(4)O_(10) is heated with water to give

The heats of atomisation of PH_(3)(g) and P_(2)H_(4)(g) are 954 kJ mol^(-1) and 1485 kJ mol^(-1) respectively. The P-P bond energy in kJ mol^(-1) is

White phosphorus is a tetra atomic solid [P_(4)(s)] at room temperature and on strong heating in absence of oxyge, it polymerizes into red phosphorus as: /_\H=-104kJ//"mol" of P_(4) The enthalpy of sublimation [P_(4)(s)toP_(4)(g)] white is 59KJ//"mol" and enthalpy of atomization is 1265 KJ//"mol" of P(g) The average P-P bond enthalpy in P_(4) molecule is:

White phosphorus is a tetra atomic solid [P_(4)(s)] at room temperature and on strong heating in absence of oxyge, it polymerizes into red phosphorus as: /_\H=-104kJ//"mol" of P_(4) The enthalpy of sublimation [P_(4)(s)toP_(4)(g)] white is 59KJ//"mol" and enthalpy of atomization is 1265 KJ//"mol" of P(g) The average P-P bond enthalpy in P_(4) molecule is:

White phosphorus is a tetra atomic solid [P_(4)(s)] at room temperature and on strong heating in absence of oxyge, it polymerizes into red phosphorus as: /_\H=-104kJ//"mol" of P_(4) The enthalpy of sublimation [P_(4)(s)toP_(4)(g)] white is 59KJ//"mol" and enthalpy of atomization is 1265 KJ//"mol" of P(g) The P-P bond enthalpy in red phosphorus joining the two tetrahedral is:

The heats of atomization of PH_(3)(g) and P_(2)H_(4)(g) are 954 kJ "mol"^(-1) and 1485kJ"mol"^(-1) respectivel. The P-P bond energy in kJ"mol"^(-1) is