The latent heat of vaporization of water is 540 calories/gram. how many calories are required to completely vaporize 500 grams of water? 1,040 cal 40 cal 270,000 cal 1.08 cal

270,000 cal

Explanation:

In order to calculate the calories needed to completely vaporize 500 grams of water we just have to use the formula of the latent heat of vaporization, we know that it takes 540 calories to vaporize a gram of water, so we just multiply that by the number of grams:

(500 grams of water)(540 calories/gram)= 270,000 calories

It would taje 270,000 calories to vaporize 500 grams of water.

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 :