How much heat is required to vaporize 33.8 g of water at 100 degrees celsius

Answer : Heat required will be 76.42 KJ

Explanation : Here we have the data as 33.8 g of water and temperature is 100°C

To calculate the heat required we need to use the formula as,

q= mol of water / heat of vaporization of water.

So we have convert the mass into moles,

33.8g / 18g = 1.877 moles of water.

and heat of vaporization is 40.7 KJ

Now, q = 1.877 X 40.7 KJ = 76.4 KJ

So the heat required will be 76.4 KJ

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 :

2b2t

explanation:

2b2t