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.
They dim unless more volts are added (batteries")
The pH of a liquid substance is calculated through the equation,
pH = -log[H3O+]
Substituting the given concentration of the hydronium ion to the equation above,
pH = -log[4.7x10^-4 M]
The value of pH is equal to 3.33. Thus, the pH of the solution is approximately 3.33.
Answer:D - adding a catalyst
Explanation:
the molar mass of the element is 81.36 g/mol
<u><em>calculation</em></u>
step 1 : multiply each %abundance of the isotope by its mass number
that is 79.95 x 29.9 =2391
81.95 x 70.1 = 5745
Step 2: add them together
2390.5 + 5744.7 =8136
Step 3: divide by 100
= 8136/100 = 81.36 g/mol