at standard pressure, the total amount of heat required to completely vaporize a 100- gram sample of water at its boiling point is? no

The correct answer is - 2.26*10^5 J

Explanation:

Given:

The amount of water need to vaporized = 100 g

We know:

The heat of vaporization of water = 2260 j/g

The required heat (q)

= mass of the liquid (m) * the heat of vaporization of liquid (ΔH)

q = m*ΔH 1

So, from the equation 1,

the 100 gram sample of water being completely vaporized needs

= 2260*100

=2.26*10^5 J.

Thus, the correct answer is = 2.26*10^5 J.

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 :