Answer:
The solution will not form a precipitate.
Explanation:
The Ksp of PbI₂ is:
PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)
Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] <em>Concentrations in equilibrium</em>
When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:
[I⁻] = 0.00345M × (328mL / (328mL+703mL) =<em> 1.098x10⁻³M</em>
[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) =<em> 5.469x10⁻³M</em>
<em />
Q = [I⁻]²[Pb²⁺] <em>Concentrations not necessary in equilibrium</em>
If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.
Replacing:
Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹
As Q < Ksp, the solution is not saturated and <em>will not form a precipitate</em>.
Answer:
a) 119 g/mol
Explanation:
-We apply the formula for freezing point depression to obtain the molality of the solution:

#We use the molality above to calculate the molar mass:

Hence, the molar mass of the compound is 119 g/mol
Hello!
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You'll need to react
7,5 moles of Sodium with sulfuric acid to produce 3.75 moles of sodium sulfate
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First of all, you need to balance the reaction. The balanced reaction is shown below (ensuring that the Law of Conservation of Mass is met on both sides):
2Na + H₂SO₄ → Na₂SO₄ + H₂
Now, all that you have to do is to use molar equivalences in this reaction applying the coefficients to calculate the moles of Sodium that you'll need:
Have a nice day!
C. The number of protons in the atoms nucleus.