Both HF and HBr experience H-bond, dipole, and dispersion forces. These forces allow the compounds to have stronger bonds with each other. HF would experience a higher amount of attraction between its molecules. This is because fluorine is more electronegative than bromine.
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<h3>Answer:</h3>
7.57 × 10⁻²² g of F
<h3>Solution:</h3>
Data Given:
Number of Molecules = 8
M.Mass of BF₃ = 67.82 g.mol⁻¹
Mass of Fluorine atoms = ?
Step 1: Calculate Moles of BF₃
Moles = Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Putting value,
Moles = 8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Moles = 1.33 × 10⁻²³ mol
Step 2: Calculate Mass of BF₃:
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting values,
Mass = 1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹
Mass = 9.0 × 10⁻²² g
Step 3: Calculate Mass of Fluorine Atoms:
As,
67.82 g BF₃ contains = 57 g of F
So,
9.0 × 10⁻²² g will contain = X g of F
Solving for X,
X = (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g
X = 7.57 × 10⁻²² g of F
Put it in a beaker. Use a smaller beaker filled half way with ice and water and place in the larger one. It should be about an inch or two above the mixture. Heat over a Bunsen burner and the naphthalene will deposit on the bottom of smaller beaker.
And in this way, nephthalene be separated from the mixture of KBR and sand.
It releases energy (what I am saying in other words that "D. Releases" would be your answer)
