**Part A**Given: $0.010 \, M \, \text{CaCl}_2 $ ($ K_{sp} = 4.68 \times 10^{-6} $ for $ \text{Ca(OH)}_2 $)**Task:** Express your answer in molarity to two significant figures.\[ \text{[KOH]} = \, \, \, \, M \][Input box for answer][Submit] [Previous Answers] [Request Answer]**Feedback Section:**- **Incorrect; Try Again; 3 attempts remaining** - Precipitation commences when the value of the reaction quotient ($Q$) for the precipitating compound is equal to the solubility product constant ($K_{sp}$). - Calculate for $Q$ for calcium hydroxide equal to the value of $K_{sp}$, and solve it for $[\text{OH}^-]$:\[Q = [\text{Ca}^{2+}][\text{OH}^-]^2 = 4.68 \times 10^{-6}\]**Part B**[Continue to the next part]

Chemical equation:- Ca(OH)2 = Ca2+ + 2OH-

Answered: 0.010 M CaCl2 (Ksp 4.68 x 10-6 for…

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0.010 M CaCl2 (Ksp 4.68 x 10-6 for Ca(OH)2) Express your answer In molarity to two significant figures.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

**Part A**

Given: $0.010 \, M \, \text{CaCl}_2 $ ($ K_{sp} = 4.68 \times 10^{-6} $ for $ \text{Ca(OH)}_2 $)

**Task:** Express your answer in molarity to two significant figures.

\[ \text{[KOH]} = \, \, \, \, M \]

[Input box for answer]

[Submit] [Previous Answers] [Request Answer]

**Feedback Section:**

- **Incorrect; Try Again; 3 attempts remaining**
- Precipitation commences when the value of the reaction quotient ($Q$) for the precipitating compound is equal to the solubility product constant ($K_{sp}$).
- Calculate for $Q$ for calcium hydroxide equal to the value of $K_{sp}$, and solve it for $[\text{OH}^-]$:

\[Q = [\text{Ca}^{2+}][\text{OH}^-]^2 = 4.68 \times 10^{-6}\]

**Part B**

[Continue to the next part]

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