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!!!
A + B → AB: ✔ synthesis
AB → A + B: ✔ decomposition
Hydrocarbon + O2 → CO2 + H2O: <span>✔ combustion</span>
AB + CD → AD + CB: <span>✔ replacement</span>
The molecule with began to shake harder and become smaller so where they move together a lot faster
For a reaction rate that increases by a factor of 655 in the presence of a catalyst at 37°C, the activation energy of the new pathway is mathematically given as
Ea2=89.28JKoule/mol
<h3>What is the activation energy of the new pathway, all other factors being equal?</h3>
Generally, the equation for the rate constant is mathematically given as
K=Ae^{Ea1/Rt}
Therefore
Ea2=89286.083Joule/mol
In conclusion, the activation energy
Ea2=89.28JKoule/mol
Read more about Energy
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Calcium is a really soft type of metal.
~Deceptiøn
