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DIA [1.3K]
3 years ago
15

What is the solubility in pure water of ba(io3)2 in moles per liter at 25 ˚c? [ksp (25 ˚c) = 6.0  10–10]?

Chemistry
1 answer:
diamong [38]3 years ago
6 0
Ba(IO₃)₂(s) partially dissociates in water  as Ba²⁺(aq) and IO₃⁻(aq).
                Ba(IO₃)₂(s) ⇄ Ba²⁺(aq) + 2IO₃⁻(aq)
Initial          
Change        -X                +X             2X
Equilibrium                        X              2X

           Ksp = [Ba²⁺(aq)] x [IO₃⁻(aq)]²
6.0 x 10⁻¹⁰  = X * (2X)²
6.0 x 10⁻¹⁰ = 4X³
              X = 5.313 x 10⁻⁴ mol/L

Hence, the solubility of the Ba(IO₃)₂(s) is 5.313 x 10⁻⁴ mol/L

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When 0.491 grams of a protein were dissolved in 44 mL of benzene at 24.4 degrees C, the osmotic pressure was found to be 50.9 to
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4057.85 g/mol

Explanation:

Hello, the numerical procedure is shown in the attached file.

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4 0
3 years ago
If you add 25.0 mL of water to 125 mL of a 0.150 M LiOH solution, what will be the molarity of the resulting diluted solution?
Alborosie

Concentration is the number of moles of solute in a fixed volume of solution

Concentration(c) = number of moles of solute(n) / volume of solution (v)

25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.

original solution molarity - 0.150 M

number of moles of LiOH in 1 L - 0.150 mol

number of LiOH moles in 0.125 L  - 0.150 mol/ L x 0.125 L = 0.01875 mol

when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases

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therefore new molarity is

c = 0.01875 mol / 0.150 L  = 0.125 M

answer is 0.125 M

7 0
3 years ago
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