Answer:
a) [ Ca2+ ] = 3.347 E-4 mol/L
b) [ Ca2+ ] = 1.5 E-8 mol/L
Explanation:
S S 2S......in the equilibrium
⇒ Ksp = 1.5 E-10 = [ Ca2+ ] * [ F- ]² = S * ( 2S )² = 4S³
⇒ S = ∛ ( 1.5 E-10 / 4 )
⇒ S = ∛ 3.75 E-11
⇒ S = 3.347 E-4 mol/L
⇒ [ Ca2+ ] = S = 3.347 E-4 mol/L
b) NaF ↔ Na+ + F-
0.10 M 0.10 0.10
S S 2S + 0.10
⇒ Ksp = 1.5 E-10 = [ Ca2+ ] * [ F- ]² = S * ( 2S + 0.10 )²
∴∴ the Concentration: 0.10 M >>>> Ksp ( 1.5 E-10 ), son we can despise S as adding.
⇒ 1.5 E-10 = S * ( 0.10 )² = 0.01 S
⇒ S = 1.5 E-10 / 0-01
⇒ S = 1.5 E-8 mol/L
⇒ [ Ca2+ ] = S = 1.5 E-8 mol/L
Answer:
The quantity of heat lost by the surroundings is 258,5J
Explanation:
The dissolution of salt XY is endothermic because the water temperature decreased.
The total heat consumed by the dissolution process is:
4,184 J/g°C × (75,0 + 1,50 g) × 0,93°C = 297,7 J
This heat is consumed by the calorimeter and by the surroundings.
The heat consumed by the calorimeter is:
42,2 J/°C × (0,93°C) = 39,2 J
That means that the quantity of heat lost by the surroundings is:
297,7J - 39,2J = <em>258,5 J</em>
I hope it helps!
Answer:
sorry I'm a dumb small brain
Explanation:
my genes
Answer:
Biggest Radii V²⁺ > V³⁺ > V⁴⁺ > V⁵⁺ Smallest Radii
General Formulas and Concepts:
- Periodic Trends: Atomic/Ionic Radii
- Coulomb's Law
Explanation:
The Periodic Trend for Atomic Radii is down and to the left. Therefore, the element with the largest radius would be in the bottom left corner of the Periodic Table.
Anions will always have a bigger radii than the parent radii. When we add e⁻ to the element, we are increasing the e⁻/e⁻ repulsions. This will cause e⁻ to repel themselves more and thus create more space, increasing the radii size.
Cations will always have smaller radii than the parent radii. When we remove e⁻ from the element, we are decreasing e⁻/e⁻ repulsions. Since there are less e⁻, there is no need for more space and thus decreases the radii size.
Since Cations are smaller than the parent radii, the more e⁻ we remove, the smaller it will become.
Therefore, the least removed e⁻ Vanadium would be the largest and the most removed e⁻ Vanadium would be the smallest.
