Ok it’s 38 so I hope that helps ok
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!!!
The correct answer is suspensions only. The suspension is a heterogeneous mixture that contains solid particles that are largely enough to undergo sedimentation. Usually, these particles are about one micrometer which makes these solute to be very easy to be free from their solvent and be filtered.
Answer:
2. Co(NO3)2 + H2
Explanation:
Hello,
In this case, we are evidencing a simple displacement reaction wherein the cobalt is able displace the hydrogen to produce cobalt (II) nitrate and gaseous hydrogen as a result of cobalt's higher activity:
Therefore, answer is 2. Co(NO3)2 + H2.
Best regards.
