Reaction: 2HCI + Ca(OH)2 →2H20 + CaCl2 Initial Burette Volume of 0.O5M HCI Volume of saturated calcium Final Burette Volume of HCI hydroxide solution 0.019 L 0.006 L 0.015 L Analysis 1. Calculate the volume and mols of HCl required to reach the end point. Volume (0.015L) - (0.006L) = 0.009 L Molarity = 0.05 M = 0.05 mol/L Number of moles of HCI (0.05mol/L) (0.009L) = 0.00045 mol 2. Use mols of HCl to determine the mols of calcium hydroxide.
Reaction: 2HCI + Ca(OH)2 →2H20 + CaCl2 Initial Burette Volume of 0.O5M HCI Volume of saturated calcium Final Burette Volume of HCI hydroxide solution 0.019 L 0.006 L 0.015 L Analysis 1. Calculate the volume and mols of HCl required to reach the end point. Volume (0.015L) - (0.006L) = 0.009 L Molarity = 0.05 M = 0.05 mol/L Number of moles of HCI (0.05mol/L) (0.009L) = 0.00045 mol 2. Use mols of HCl to determine the mols of calcium hydroxide.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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answer number 2 plz I included the answers form number 1 to help.
![Reaction:
2HCI + Ca(OH)2 →2H20 + CaCl2
Volume of saturated calcium
Initial Burette
Final Burette Volume of HCI
hydroxide solution
Volume of 0.05M HCI
0.019 L
0.006 L
0.015 L
Analysis
1. Calculate the volume and mols of HCl required to reach the end point.
Volume
(0.015L) - (0.006L) = 0.009 L
Molarity = 0.05 M = 0.05 mol/L
Number of moles of HCl
(0.05mol/L) (0.009L) = 0.00045 mol
2. Use mols of HCl to determine the mols of calcium hydroxide.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F26703abc-b411-475c-9236-6ed9f9a36303%2F58e3cb2a-63d6-4cc2-ac42-6b365a48ee8a%2Fqvvy1ll_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Reaction:
2HCI + Ca(OH)2 →2H20 + CaCl2
Volume of saturated calcium
Initial Burette
Final Burette Volume of HCI
hydroxide solution
Volume of 0.05M HCI
0.019 L
0.006 L
0.015 L
Analysis
1. Calculate the volume and mols of HCl required to reach the end point.
Volume
(0.015L) - (0.006L) = 0.009 L
Molarity = 0.05 M = 0.05 mol/L
Number of moles of HCl
(0.05mol/L) (0.009L) = 0.00045 mol
2. Use mols of HCl to determine the mols of calcium hydroxide.
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