Answer:
2.1 kg of water
Explanation:
Step 1: Given data
- Moles of lithium bromide (solute): 4.3 moles
- Molality of the solution (m): 2.05 m (2.05 mol/kg)
- Mass of water (solvent): ?
Step 2: Calculate the mass of water required
Molality is equal to the moles of solute divided by the kilograms of solvent.
m = moles of solute/kilograms of solvent
kilograms of solvent = moles of solute/m
kilograms of solvent = 4.3 mol /(2.05 mol/kg) = 2.1 kg
It tends to in increase because the size of the atoms increase.
Answer:
Neutral solution is formed.
Explanation:
When the hydrochloric acid and sodium hydroxide which is a strong base are combined together, it produces sodium chloride which is a salt and water. This solution is known as Neutral solution because the solution do not have the characteristics or properties of either an acid or a base. If the concentration of one of the reactant is higher as compared to another reactant so the product has the characteristics of that reactant.
A) Fe⁰ ----> Fe⁺³ +3e⁻ oxidation | *2
b) <u>Cu⁺² + 2e⁻ -----> Cu⁰ reduction |*3</u>
c) 2Fe⁰ +3Cu⁺² -----> 2Fe⁺³ + 2Cu⁰
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
