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 player ran 91.44m.
Explanation:
The problem gives you the total distance between goal line to goal line in feet, and the answer must be given in meters, so you should convert the distance the player run from ft to m, because the player run the same distance from goal line to goal line to scores the touchdown.
So, you should apply the following conversion factor:
The player ran 91.44m.
Answer: 14943.5 J
Explanation:
The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Given that,
Q = ?
Mass of water = 55.0g
C = 4.18 J/g°C
Φ = 65.0°C
Then, Q = MCΦ
Q = 55.0g x 4.18 J/g°C x 65.0°C
Q = 14943.5 J
Thus, 14943.5 joules of heat is needed to raise the temperature of water.
