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Anni [7]
3 years ago
12

A 5.0 gram sample of Ag2SO4, will dissolve in 1.0 L of water. Calculate the solubility product

Chemistry
1 answer:
arlik [135]3 years ago
8 0

Answer:

Ksp = 1.64 ×10⁻⁵

Explanation:

Given data:

Mass of  Ag₂SO₄ = 5.0 g

Volume of water = 1.0 L

Solubility product = Ksp = ?

Solution:

First of all we will calculate the molarity.

Molarity = Number of moles of solute / Volume of solution in L

Number of moles = mass/ molar mass

Number of moles = 5.0 g/311.74 g/mol

Number of moles = 0.016 mol

Molarity = 0.016 mol / 1 L

Molarity = 0.016 M

Dissociation equation:

Ag₂SO₄  ⇔  2Ag⁺ + SO₄²⁻

Ksp = [Ag⁺]² [SO₄²⁻]

Ksp = [0.032]²[0.016]

Ksp = 1.64 ×10⁻⁵

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Answer:

1.0190 x 10⁻⁵ mol

Explanation:

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V KIO₃ = 10.19 mL = 10.19 mL x 1 L/1000 mL = 0.01019 L

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8 0
3 years ago
Complete the sentences to explain what a similar treatment would predict for the effective nuclear charge on boron valence elect
dsp73

Answer:

ZcH's electrons = 5-2 = +3.

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Explanation:

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4 years ago
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3 years ago
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Answer:

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