Answer:
Cl⁻, Na⁺, OH⁻
Explanation:
The titration is:
CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)
In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.
Thus, ions present after the equivalence point are:<em> Cl⁻, Na⁺</em> (Don't react, spectator ions), and <em>OH⁻</em>.
The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.
A(t) = A(o) (1/2)^(t/d)
where t is the certain period of time. Substituting the known values,
A(t) = (20 mg)(1/2)^(85.80/14.30)
Solving,
A(t) = 0.3125 mg
Hence, the answer is 0.3125 mg.
Answer:
14.945798 in us fluid ounce
Explanation:
Answer:
Explanation:
When potassium chloride reacts with lead acetate in aqueous medium, it gives white precipitate of lead chloride and potassium acetate in aqueous medium as a product.
As shown in chemical reaction given below:
