Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Mark S. Cracolice, Ed Peters
Chapter3: Measurement And Chemical Calculations
Section: Chapter Questions
Problem 67E: 0.0715 gal = _____ c m 3 2.27 x 10 4 mL = _____ g a l
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Question
Conversions: include the correct number of significant digits and follow the format provided in the image.
a. 54 mi to km
b. 1789 ft to km
![h) 75 mL to gal
mL to L to gal
75/AL 1X
1 gal
1 10^3 3.78541178/
X
X
75
10^3 x 3.78541178
gal = 0.0198129039 gal = 0.020 gal = 2.0 x 10^-2 gal](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6233d642-6c59-41d2-ba18-a053468b6458%2Fc1294d0c-7b1a-4b51-a1ec-df26cbbcf17b%2Foeyoua_processed.jpeg&w=3840&q=75)
Transcribed Image Text:h) 75 mL to gal
mL to L to gal
75/AL 1X
1 gal
1 10^3 3.78541178/
X
X
75
10^3 x 3.78541178
gal = 0.0198129039 gal = 0.020 gal = 2.0 x 10^-2 gal
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