HEAT OF VAPORIZATION Q=mx H,
If the heatyaporization of isopropyl alcohol is measured at 702 joules/gram and it requires 596,700 joules of energy to vaporize a sample, how many grams of isopropyl alcohol were in the sample? ​

b is the correct

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 :

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )