What is the partial pressure of 0.50 mol ne gas combined with 1.20 mol kr gas at a final pressure of 730 torr?

it would be 215 torr for the answer for edg

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.