The specific heat of gold is 0.031 calories/gram°c. if 10.0 grams of gold were heated and the temperature of the sample changed by 20.0°c, how many calories of heat energy were absorbed by the sample?
6.2 calories
0.016 calories
6,451 calories
0.062 calories

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 :