<span>Balanced half-reaction: Au</span>⁺<span>(aq) </span>→ <span> Au</span>³⁺(aq) + 2e⁻.
Oxidation reaction is increasing of oxidation number of element, because element or ion lost electrons in chemical reaction.
Gold lost two electrons and change oxidation number from +1 to +3.
Reduction is lowering oxidation number because element gain electrons.
A non-chlorine chemical such as iodine may be used as a
sanitizing solution. To use iodine as a sanitizing solution, it should be
around 12.5-25 ppm in water that is at least 75° F. Utensils and equipment must
be immersed for only 30 seconds because it may lose its effectivity if the pH
level gets too high due to high temperature. Discoloration may also result if
the utensils and equipment are in prolonged contact with the solution.
So C12 is the limiting reactant and P4 is the excess
Mass P4 consumed= 0.31 mol X 123.9 g/mol =38.41 g P4 consumed.
Hope i helped
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>.
