How much heat is required to vaporize 166g of water at it's boiling point, if water's heat of vaporization is equal to 40.7kJ/mol?
kJ

375.3KJ

Explanation:

The following data were obtained from the question:

Mass of water = 166g

Heat of Vaporisation (ΔHv) = 40.7kJ/mol

Heat (Q) =..?

Next, we shall determine the number of mole in 166g of water. This is illustrated below:

Mass of H2O = 166g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Number of mole = Mass/Molar Mass

Number of mole of H2O = 166/18

Number of mole of H2O = 9.22 moles.

Now, we can obtain the heat required to vaporise the water as shown below:

Q = n·ΔHv

Q = 9.22 mol x 40.7kJ/mol

Q = 375.3KJ

Therefore, the heat required to vaporise the water is 375.3KJ.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.