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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
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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V₁ M₂ 9.) 150.0 ml of 2.0 M NaBr from 6.0 M Stock Solution (0.50 M)(100.00 mL) 2.0M (2.0 M) (150.0 mL) = (6.0 M) (V₂) 6.0 M 6.0 M V₁ M₁ M₂ 10.) 100.00 mL of 0.50 M nickel (11) Sulfate from 2.0 M Stock solution V₂ = 25 = ory Policies V₂ = 50 (2.0 M)(V₂) 2.0M M₂ V₁ 11.) 1.0 L of 0.10 M Sodium acetate from 2.5M Stock Solution = V₂ = 0.04 (1.0L) (0.10 M) = (2.5 M) (V₂) 2.5 M 2.5 M 12.) 500.0 mL of 0.80 M H₂SO4 from 12.014 stock. Solution (0.80 M) (500.0 mL) (12.0 M) (V₂) V₂ = 33.3 12.0M 12.0M
Transcribed Image Text:V₁ M₂ 9.) 150.0 ml of 2.0 M NaBr from 6.0 M Stock Solution (0.50 M)(100.00 mL) 2.0M (2.0 M) (150.0 mL) = (6.0 M) (V₂) 6.0 M 6.0 M V₁ M₁ M₂ 10.) 100.00 mL of 0.50 M nickel (11) Sulfate from 2.0 M Stock solution V₂ = 25 = ory Policies V₂ = 50 (2.0 M)(V₂) 2.0M M₂ V₁ 11.) 1.0 L of 0.10 M Sodium acetate from 2.5M Stock Solution = V₂ = 0.04 (1.0L) (0.10 M) = (2.5 M) (V₂) 2.5 M 2.5 M 12.) 500.0 mL of 0.80 M H₂SO4 from 12.014 stock. Solution (0.80 M) (500.0 mL) (12.0 M) (V₂) V₂ = 33.3 12.0M 12.0M
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