29.2 kJ of heat is required to convert an 50.0 g sample of
acetaldehyde from the liquid to gas phase. The molecular
weight of acetaldehyde is 44.05 g/mol. What is the heat of
vaporization of acetaldehyde in kJ/mol?
kJ/mol

The heat of  vaporization of acetaldehyde is 25.72 kJ/mol.

Explanation:

Mass of acetaldehyde = 50.0 g

Molar mass of acetaldehyde = 44.05 g/mol

Moles of acetaldehyde =

Amount of energy required to vaporize 50.0 grams of acetaldehyde = Q

Q = 29.2 kJ

The heat of  vaporization of acetaldehyde :

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.