d///////////////////////////////
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
Your reaction
.. Fe + O2 ---> FexOy
for this reaction..
.. the Fe on the left is in the 0 oxidation state
.. the Fe on the right is in the +(2y/x) oxidation state
.. the O on the left is in the 0 oxidation state
.. the O on the right is in the -2 oxidation state
meaning
.. the O is reduced... . . (it's reduced in oxidation state)
.. the Fe is oxidized.. . .(oxidation state increased)
this is a REDOX reaction
*********
AND.. it's also a synthesis reaction.. (aka combination reaction)
The molarity of (HNO₃) that was used if 2.00 L must be used to prepare 4.5 L of a 0.25M HNO₃ solution is 0.563 M
<u><em>calculation</em></u>
This is calculated usind M₁V₁=M₂V₂ formula
where,
M₁( molarity ₁) = ?
V₁( volume ₁) = 2.00 L
M₁ (molarity ₂) = 0.25M
V₂( volume₂) = 4.5 L
make M₁ the subject of the formula by diving both side of the formula by V₁
M₁ is therefore = M₂V₂/V₁
M₁ =[ (0.25 M x 4.5 L) / 2.00 L ] =0.563 M