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:
Explanation: n=m/M(molar mass)
n=24.3 grams/(16+2x1.008)grams/moles(molar mass of H2O)
n=24.3grams/18.016grams/moles
n=1.35moles
Answer:
1) increases
2) smaller
Explanation:
Generally, as electron- electron repulsion increases and more electrons are added to the atom while Z is held constant, the electron cloud size is increased. The size of the anion formed is usually measured as the size of this extended electron cloud. Hence the larger electron cloud means a larger anion size compared to the size of the neutral atom.
For a cation, the converse is true and the cation is found to be smaller than the neutral atom.
Answer:Label the parts of this wave.
A:
✔ crest
B:
✔ amplitude
C:
✔ trough
D:
✔ wavelength
Explanation: