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>.
This problem is being solved using Ideal Gas Equation.
PV = nRT
Data Given:
Initial Temperature = T₁ = 27 °C = 300 K
Initial Pressure = P₁ = constant
Initial Volume = V₁ = 8 L
Final Temperature = T₂ = 78 °C = 351 K
Final Pressure = P₂ = constant
Final Volume = V₂ = ?
As,
Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
V₁ / T₁ = V₂ / T₂
Solving for V₂,
V₂ = (V₁ × T₂) ÷ T₁
Putting Values,
V₂ = (8 L × 351 K) ÷ 300 K
V₂ = 9.38 L
Two chromate ions are contained in this compound. The “di” prefix means Two
Mono- one
Di- two
Tri- three
Tetra- four
Penta- five
Hexa-six
Hepta- seven
Octa- eight
Nona- nine
Deca- ten
Sound waves are waves of growing larger and smaller, making it seem elastic through substances, such as air.
