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:
A family in the periodic table is a group of related elements.
Explanation:
Answer:
Conduction, Convection and Conduction
Explanation:
Answer:
ΔH°_rxn = -195.9 kJ·mol⁻¹
Explanation:
4NH₃(g) + 3O₂(g) ⟶ 2N₂(g) +6H₂O(g)
ΔH°_f/(kJ·mol⁻¹): -45.9 0 0 -241.8
The formula relating ΔH°_rxn and enthalpies of formation (ΔH°_f) is
ΔH°_rxn = ΣΔH°_f(products) – ΣΔH°_f(reactants)
ΣΔH°_f(products) = -6(241.8) = -1450.8 kJ
ΣΔH°_f(reactants) = -4(45.9) = -183.6 kJ
ΔH°_rxn = (-1450.8 + 183.6) kJ = -1267.2 kJ
Boyle Law says “the pressure of fixed amount of ideal gas which is at constant temperature is
inversely proportional to its volume".<span>
P = 1/V
<span>Where, P is pressure of the ideal gas and V is volume of the ideal gas.</span>
<span>For two situations, this law can be added as;
P</span>₁V₁ = P₂V₂<span>
</span><span>14 lb/in² x V₁ = 70 lb/in² x 500 mL</span><span>
</span><span>V₁ =
2500 mL</span><span>
Hence, the needed volume of atmospheric air = 2500
mL
<span>Here, we made two </span>assumptions. They are,
1. The
atmospheric air acts as ideal gas.
2.
Temperature is a constant.
<span>We didn't convert the units to SI units since
converting volume and pressure are products of two numbers, they will cut off. </span></span></span>