Answer:
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Explanation:
<u>Step 1: </u>Data given
The solution contains 0.036 M Cu2+ and 0.044 M Fe2+
Ksp (CuS) = 1.3 × 10-36
Ksp (FeS) = 6.3 × 10-18
Step 2: Calculate precipitate
CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36
FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18
Calculate the minimum of amount needed to form precipitates:
Q=Ksp
<u>For copper</u> we have: Ksp=[Cu2+]*[S2-]
Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]
[S2-]= 3.61*10^-35 M
<u>For Iron</u> we have: Ksp=[Fe2+]*[S2-]
Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]
[S2-]= 1.43*10^-16 M
CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Answer:
<u>STEP I</u>
This is the balanced equation for the given reaction:-

<u>STEP II</u>
The compounds marked with (aq) are soluble ionic compounds. They must be
broken into their respective ions.
see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).
On breaking them into their respective ions :-
- 2KOH -> 2K+ + 2OH-
- H2SO4 -> 2H+ + (SO4)2-
- K2SO4 -> 2K+ + (SO4)2-
<u>STEP III</u>
Rewriting these in the form of equation

<u>STEP </u><u>IV</u>
Canceling spectator ions, the ions that appear the same on either side of the equation
<em>(note: in the above step the ions in bold have gotten canceled.)</em>

This is the net ionic equation.
____________________________

- KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.
[Alkali metal hydroxides are the only halides soluble in water ]
Answer:
A
Explanation:
The mallet transfers kinetic energy, then that energy is transferred to the orange ball, then the purple ball, thus making the purple ball move