Answer:
A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M
The precipitate formed is CuI
Explanation:
Step 1: Data given
The solution contains 0.021 M Cl- and 0.017 M I-.
Ksp(CuCl) = 1.0 × 10-6
Ksp(CuI) = 5.1 × 10-12.
Step 2: Calculate [Cu+]
Ksp(CuCl) = [Cu+] [Cl-]
1.0 * 10^-6 = [Cu+] [Cl-]
1.0 * 10^-6 = [Cu+] [0.021]
[Cu+] = 1.0 * 10^-6 / 0.021
[Cu+] = 4.76 *10^-5 M
Ksp(CuI) = [Cu] [I]
5.1 * 10^-12 = [Cu+] [I-]
5.1 * 10^-12 =[Cu+] [0.017]
[Cu+] = 5.1 * 10^-12 / 0.017
[Cu+] = 3.0 *10^-10 M
[Cu+]from CuI hast the lowest concentration
A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M
The precipitate formed is CuI
Answer:
c
Explanation:
total production of the three shifts/ total of prediction of each yield x 100%
Answer:
The plateaus or horizontal lines on the graph represent the transition between states of the sample. The first plateau represents the melting (or transition from solid to liquid) and the second plateau represents boiling (or transition from liquid to gas).
Explanation:
Answer:
they don't depend on the temperature
