If 1mole ------------- is ---------------- 6.02*10²³
than 0.25mole ----- is ---------------- x
x = [0.25mole*6.02*10²³]/1mole = <u>1,505*10²³</u>
The answer is: the distance between two nuclei is 2.35×10⁻¹⁰ m.
r(Na⁺) = 1.16×10⁻¹⁰ m; radius of sodium cation.
r(F⁻) = 1.9×10⁻¹⁰ m; radius of fluoride anion.
d(NaF) = r(Na⁺) + r(F⁻).
d(NaF) = 1.16×10⁻¹⁰ m + 1.9×10⁻¹⁰ m.
d(NaF) = 2.35×10⁻¹⁰ m; distance between two nuclei.
The sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.
prince Henry is your answer (:
Answer:
Ksp = 2.74 x 10⁻⁵
Explanation:
The solubility equilibrium for Ca(OH)₂ is the following:
Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)
I 0 0
C + s + 2s
E s 2s
According to the ICE table, the expression for the solubility product constant (Kps) is:
Ksp = [Ca²⁺] x ([OH⁻])² = s x (2s)² = 4s³
Then, we calculate Ksp from the solubility value (s):
s = 0.019 M
⇒ Ksp = 4s³ = 4 x (0.019)³ = 2.74 x 10⁻⁵
