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:
The concentration of the CaBr2 solution is 96 µmol/L
Explanation:
<u>Step 1:</u> Data given
Moles of Calciumbromide (CaBr2) = 4.81 µmol
Volume of the flask = 50.0 mL = 0.05 L
<u>Step 2:</u> Calculate the concentration of Calciumbromide
Concentration CaBr2 = moles CaBr2 / volume
Concentration CaBr2 = 4.81 µmol / 0.05 L
Concentration CaBr2 = 96.2 µmol /L = 96.2 µM
The concentration of the CaBr2 solution is 96 µmol/L
Answer: 1709.4 Joules
Explanation:
The quantity of Heat Energy (Q) released on cooling a heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since Q = ?
M = 18.5 grams
Recall that the specific heat capacity of copper C = 0.385 J/g.C
Φ = 285°C - 45°C = 240°C
Then, Q = MCΦ
Q = 18.5grams x 0.385 J/g.C x 240°C
Q = 1709.4 Joules
Thus, 1709.4 Joules is released when copper is cooled.
The mass of the product is <em>98.78 g.</em>
The word equation is
aluminum + chlorine → product
20.00 g + 98.78 g → <em>x</em> g
If each reactant is completely consumed, the <em>Law of conservation of Mass </em>tells us the mass of the product must be 98.78 g.