Answer:
463.0 g.
Explanation:
- We can use the following relation:
<em>n = mass/molar mass.</em>
where, n is the mass of copper(ii) fluoride (m = 4.56 mol),
mass of copper(ii) fluoride = ??? g.
molar mass of copper(ii) fluoride = 101.543 g/mol.
∴ mass of copper(ii) fluoride = (n)(molar mass) = (4.56 mol)(101.543 g/mol) = 463.0 g.
At STP
1L O₂ → 2L CO₂
xL O₂ → 5.0L CO₂
x=2.5 L
Incomplete Dominance. Hope this helps
@LightningEmerald1
How do I show work for this?
Answer:
When the concentration of F- exceeds 0.0109 M, BaF2 will precipitate.
Explanation:
Ba²⁺(aq) + 2 F⁻(aq) <----> BaF₂(s)
When BaF₂ precipitates, the Ksp relation is given by
Ksp = [Ba²⁺] [F⁻]²
[Ba²⁺] = 0.0144 M
[F⁻] = ?
Ksp = (1.7 × 10⁻⁶)
1.7 × 10⁻⁶ = (0.0144) [F⁻]²
[F⁻]² = (1.7 × 10⁻⁶)/0.0144 = 0.0001180555
[F⁻] = √0.0001180555 = 0.01086 M = 0.0109 M
Hope this Helps!!!
