150.0 ml of 2.0 M NaBr from (2.0 M) (150.0 mL) = (6.0 6.0 M V₁ M₁ 100.00 mL of 0.50 M nicke OCK Solution (0.50 M)(100.00 mL) = ( 2.0M V₁ 316 OCE
150.0 ml of 2.0 M NaBr from (2.0 M) (150.0 mL) = (6.0 6.0 M V₁ M₁ 100.00 mL of 0.50 M nicke OCK Solution (0.50 M)(100.00 mL) = ( 2.0M V₁ 316 OCE
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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![**Dilution Calculations from Stock Solutions**
---
**9. Preparation of 150.0 mL of 2.0 M NaBr from a 6.0 M Stock Solution**
To find the volume of the stock solution (V₂) needed:
\[
(2.0 \, \text{M})(150.0 \, \text{mL}) = (6.0 \, \text{M})(V_2)
\]
Solving for \(V_2\):
\[
V_2 = \frac{(2.0 \, \text{M})(150.0 \, \text{mL})}{6.0 \, \text{M}} = 50 \, \text{mL}
\]
---
**10. Preparation of 100.0 mL of 0.50 M Nickel(II) Sulfate from a 2.0 M Stock Solution**
To find \(V_2\):
\[
(0.50 \, \text{M})(100.0 \, \text{mL}) = (2.0 \, \text{M})(V_2)
\]
Solving for \(V_2\):
\[
V_2 = \frac{(0.50 \, \text{M})(100.0 \, \text{mL})}{2.0 \, \text{M}} = 25 \, \text{mL}
\]
---
**11. Preparation of 1.0 L of 0.10 M Sodium Acetate from a 2.5 M Stock Solution**
To determine \(V_2\):
\[
(1.0 \, \text{L})(0.10 \, \text{M}) = (2.5 \, \text{M})(V_2)
\]
Solving for \(V_2\):
\[
V_2 = \frac{(1.0 \, \text{L})(0.10 \, \text{M})}{2.5 \, \text{M}} = 0.04 \, \text{L}
\]
---
**12. Preparation of 500.0 mL of 0.80 M H₂SO₄ from a 12.0 M Stock Solution**
To calculate \(V_2\):
\[
(0.80 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00437ad0-c521-4cbf-b1d5-8afcb4493cf2%2F8cfe1ec3-5b9d-4f35-a30f-75a7986fb9f5%2Fhytlre_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Dilution Calculations from Stock Solutions**
---
**9. Preparation of 150.0 mL of 2.0 M NaBr from a 6.0 M Stock Solution**
To find the volume of the stock solution (V₂) needed:
\[
(2.0 \, \text{M})(150.0 \, \text{mL}) = (6.0 \, \text{M})(V_2)
\]
Solving for \(V_2\):
\[
V_2 = \frac{(2.0 \, \text{M})(150.0 \, \text{mL})}{6.0 \, \text{M}} = 50 \, \text{mL}
\]
---
**10. Preparation of 100.0 mL of 0.50 M Nickel(II) Sulfate from a 2.0 M Stock Solution**
To find \(V_2\):
\[
(0.50 \, \text{M})(100.0 \, \text{mL}) = (2.0 \, \text{M})(V_2)
\]
Solving for \(V_2\):
\[
V_2 = \frac{(0.50 \, \text{M})(100.0 \, \text{mL})}{2.0 \, \text{M}} = 25 \, \text{mL}
\]
---
**11. Preparation of 1.0 L of 0.10 M Sodium Acetate from a 2.5 M Stock Solution**
To determine \(V_2\):
\[
(1.0 \, \text{L})(0.10 \, \text{M}) = (2.5 \, \text{M})(V_2)
\]
Solving for \(V_2\):
\[
V_2 = \frac{(1.0 \, \text{L})(0.10 \, \text{M})}{2.5 \, \text{M}} = 0.04 \, \text{L}
\]
---
**12. Preparation of 500.0 mL of 0.80 M H₂SO₄ from a 12.0 M Stock Solution**
To calculate \(V_2\):
\[
(0.80 \
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