Coal, they use to mine for it alot
Since none answered from Indian NCERT yet, let me give a simple definition from ICSE..
Latent Heat of vapourisation is the heat energy required to convert unit mass of a substance from liquid to vapour state WITHOUT ANY CHANGE IN ITS TEMPERATURE....
hope this helps!
We are given the pOH of the solution of 10.75. pOH is the property of the solution that is related to the OH ion concentration of the solution. THe formula to be followed is pOH = -log (OH); OH- = 10^-pOH. In this case, OH- = 10^-10.75 equal to B. 1.778 x 10^-11 M
Answer:

Explanation:
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 28.01 17.03
N₂ + 3H₂ ⟶ 2NH₃
m/g: 240.0
(a) Moles of NH₃

(b) Moles of N₂

(c) Mass of N₂
