Refer to the table below. Credits to https://terpconnect.umd.edu/~wbreslyn/chemistry/naming/IonicCharge2.jpg
Cations with (+ ) charges lose electrons in order to obtain an octet (8 valence electrons) when they ionically bond with another ion. We're looking for the ions that loses electrons here. So, from the table:
Al 3+ , S 2- , O 2-, Ag + , Ne ( noble gas, no charge)
Since Al and Ag has (+) charges, they are going to lose electrons to form ionic bonds with other atoms.
The statement that describes the chemical reaction is D chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromide<span>. The symbol "Cl" represents chlorine. The symbols in the brackets show the physical state of the substance, (g) is gaseous, (s) is solid, (aq) is aqueous and (l) is liquid.</span>
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.
I'm sure that to calculate the freezing point depression <span>subtract</span> solution's freezing point and the freezing point of it's pure solvent. According to the formula.