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how do you do number 7? i attempted the problem but it isn’t right. this is a non graded practice worksheet

how do you do number 7? i attempted the problem but it isn’t right. this is a non graded practice worksheet

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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how do you do number 7? i attempted the problem but it isn’t right. this is a non graded practice worksheet
5. If 30.0 mL of 12.0 M hydrochloric acid solution is diluted to a new total volume of 500 mL, what is the molarity of the dilute acid? (ANS: 0.72 M) v₁ m₁ = √₂ m2 12M=X 72 M 36 .5 .03 X=369 6. A student pours 130 mL of water into a beaker already containing 200.0 mL of 6.0 M copper (II) sulfate solution. What is the molarity of the copper (II) sulfate solution now that it's been diluted? (ANS: 3.64 M) X=1₁2 M=3,64M 6 M = X .2 M= 112 330 ML NO 3 7. A student has 27.5 mL of a 16.0 M nitric acid stock solution. However, the student wants to create a solution with a new molarity of 1.34 M. How many mL of water would need to be added to the original solution? (ANS: 301 mL) 27.5(16) = 1.34(x) x = 328 mL 8. How would you prepare 1000. mL of a 4.50 M solution of sulfuric acid from a stock solution of 18.0 M of sulfuric acid? Be sure to calculate the amount of water needed. 201 10
Transcribed Image Text:5. If 30.0 mL of 12.0 M hydrochloric acid solution is diluted to a new total volume of 500 mL, what is the molarity of the dilute acid? (ANS: 0.72 M) v₁ m₁ = √₂ m2 12M=X 72 M 36 .5 .03 X=369 6. A student pours 130 mL of water into a beaker already containing 200.0 mL of 6.0 M copper (II) sulfate solution. What is the molarity of the copper (II) sulfate solution now that it's been diluted? (ANS: 3.64 M) X=1₁2 M=3,64M 6 M = X .2 M= 112 330 ML NO 3 7. A student has 27.5 mL of a 16.0 M nitric acid stock solution. However, the student wants to create a solution with a new molarity of 1.34 M. How many mL of water would need to be added to the original solution? (ANS: 301 mL) 27.5(16) = 1.34(x) x = 328 mL 8. How would you prepare 1000. mL of a 4.50 M solution of sulfuric acid from a stock solution of 18.0 M of sulfuric acid? Be sure to calculate the amount of water needed. 201 10
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