The molar heat of vaporization for methane, ch4, is 8.53 kj/mol. how much energy is absorbed when 54.8 g of methane vaporizes at its boiling point?

6.42 kj
29.1 kj
137 kj
467 kj

Answer : The correct option is, 29.1 KJ

Explanation : Given,

Molar heat of vaporization for methane = 8.53 KJ/mole

Molar mass of methane = 16 g/mole

First we have to calculate the moles of methane.

Now we have to calculate the amount of energy absorbed.

As, 1 mole of methane absorbs heat of vaporization = 8.53 KJ

So, 3.425 mole of methane absorbs heat of vaporization = 3.425 × 8.53 KJ = 29.1 KJ

Therefore, the amount of energy absorbed are, 29.1 KJ

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8