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Gennadij [26K]
3 years ago
8

The molar solubility of silver bromide, AgBr in pure water is 0.0007350 mol/L. What is the

Chemistry
1 answer:
gayaneshka [121]3 years ago
7 0

Answer:

0.000000540

Explanation:

Step 1: Make an ICE chart for the solution of AgBr

"S" represents the molar solubility of AgBr

        AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)

I                           0             0

C                          +S          +S

E                           S             S

Step 2: Write the expression for the solubility product constant (Ksp)

Ksp = [Ag⁺] [Br⁻] = S × S

Ksp = S² = (0.0007350)² = 0.000000540

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3 0
3 years ago
PLEASE HELP Determine the number of neutrons for the given isotopes:
zepelin [54]
<span>Helium = 1
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3 years ago
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