Answer:
Solution A that will form a precipitate with Ksp = 2.3 x 10−4
Explanation:
Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
3S S
Where S = Solubility(mole/lit) and Ksp = Solubility product
⇒ Ksp = (3S)³ x (S)
⇒ 27S⁴ = 2.3x10−4
⇒ S = 0.05 mol/lit
Concentration of Li₃PO₄ precipitate = 0.05
<u>Solution A </u>
0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole
0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole
3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄
(Mole/Stoichiometry) 
= 0.05 = 0.24
Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.
So concentration of Li₃PO₄ is equal to 0.05.
<span> </span>
Answer
is: volume is 20 mL.<span>
c</span>₁(CH₃COOH) = 2,5 M.<span>
c</span>₂(CH₃COOH) = 0,5 M.<span>
V</span>₂(CH₃COOH) = 100 mL.<span>
V</span>₁(CH₃COOH) = ?<span>
c</span>₁(CH₃COOH) · V₁(CH₃COOH)
= c₂(CH₃COOH) · V₂(CH₃COOH).<span>
2,5 M · V</span>₁(CH₃COOH)
= 0,5 M · 100 mL.<span>
V</span>₁(CH₃COOH) = 0,5 M · 100 mL ÷ 2,5 M.<span>
V</span>₁(CH₃COOH) = 20 mL ÷ 1000 mL/L =0,02 L.
A serial dilution is the stepwise dilution of a substance in solution. Usually the dilution factor at each step is constant, resulting in a geometric progression of the concentration in a logarithmic fashion.
Explanation:
When the covalent bonds in a molecule are polarized so that one portion of the molecule experiences a positive charge and the other portion of the molecule experiences a negative charge. This separation of opposite charges creates an electric dipole.
