If a solution was created by mixing 250 mL of pure isopropyl alcohol with enough water to make 1.00 L of solution. What is the % (v/v) concentration? 2.5 % 20 % 25 % 70 %
If a solution was created by mixing 250 mL of pure isopropyl alcohol with enough water to make 1.00 L of solution. What is the % (v/v) concentration? 2.5 % 20 % 25 % 70 %
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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![If a solution was created by mixing 250 mL of pure isopropyl alcohol with enough
water to make 1.00 L of solution. What is the % (v/v) concentration?
O 2.5 %
20 %
25 %
O 70 %](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90acb466-191d-42a7-9183-4a1182b098d1%2F65c540a2-872b-4bb5-99e1-bae921e09263%2Fxq9eci_processed.jpeg&w=3840&q=75)
Transcribed Image Text:If a solution was created by mixing 250 mL of pure isopropyl alcohol with enough
water to make 1.00 L of solution. What is the % (v/v) concentration?
O 2.5 %
20 %
25 %
O 70 %
![Aluminum reacts with chlorine to form aluminum chloride. The equation for this
reaction is 2 AlI(s) + 3 Cl2(g) → 2 AICI3(s). What is the percentage yield for a reaction
in which 33.5 g of aluminum is reacted with excess chlorine to produce 164.5 g of
aluminum chloride? Use the molar mass values given here.
MAI = 26.9815 g/mol
%3D
MCl2 = 70.9064 g/mol
%3D
MAICI3 = 133.3411 g/mol
%3D
66.8 %
100.6
99.4 %
49.7 %](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90acb466-191d-42a7-9183-4a1182b098d1%2F65c540a2-872b-4bb5-99e1-bae921e09263%2Finnwatp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Aluminum reacts with chlorine to form aluminum chloride. The equation for this
reaction is 2 AlI(s) + 3 Cl2(g) → 2 AICI3(s). What is the percentage yield for a reaction
in which 33.5 g of aluminum is reacted with excess chlorine to produce 164.5 g of
aluminum chloride? Use the molar mass values given here.
MAI = 26.9815 g/mol
%3D
MCl2 = 70.9064 g/mol
%3D
MAICI3 = 133.3411 g/mol
%3D
66.8 %
100.6
99.4 %
49.7 %
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