Molar solubility is the number of moles that are dissolved in 1 L solution.
when BaF₂ dissolves it dissociates into the following ions
BaF₂ --> Ba²⁺ + 2F⁻
if the molar solubility of BaF₂ is X, then molar solubility of Ba²⁺ is X and F⁻ is 2x
then the formula for the solubility product constant -ksp is;
ksp = [Ba²⁺][F⁻]²
ksp = X * (2X)²
ksp = 4X³
since ksp = 1.7 x 10⁻⁶
4X³ = 1.7 x 10⁻⁶
X = 0.0075 M
molar solubility of BaF₂ is 0.0075 M
Answer:
Explanation:
We will need a balanced chemical equation with masses, moles, and molar masses.
1. Gather all the information in one place:
M_r: 2.016 17.03
3H₂ + N₂ ⟶ 2NH₃
m/g: 6.33 × 10⁻⁴
2. Calculate the moles of H₂
3. Calculate the moles of NH₃
The molar ratio is 2 mol NH₃/3 mol H₂.
4. Calculate the molecules of NH₃
There are 6.022 × 10²³ molecules of NH₃/1 mol NH₃.
Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.
<span>ions are solutions containing ions that react with acids or bases to minimize their effects. </span>
A shape of a 'p' orbital looks like a balloon pinched in the middle.