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Answer:
The specific rotation of D is 11.60° mL/g dm
Explanation:
Given that:
The path length (l) = 1 dm
Observed rotation (∝) = + 0.27°
Molarity = 0.175 M
Molar mass = 133.0 g/mol
Concentration in (g/mL) = 0.175 mol/L × 133.0 g/mol
Concentration in (g/mL) = 23.275 g/L
Since 1 L = 1000 mL
Concentration in (g/mL) = 0.023275 g/mL
The specific rotation [∝] = ∝/(1×c)
= 0.27°/( 1 dm × 0.023275 g/mL
)
= 11.60° mL/g dm
Thus, the specific rotation of D is 11.60° mL/g dm
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
