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

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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