(There is no part A) Part B. Standardization of Sodium hydroxide solution Balanced chemical equation: Trial 1 Trial 2 Trial 3 Mass of oxalic acid 0.1314g NA NA Initial burette reading 0.40mL
(There is no part A) Part B. Standardization of Sodium hydroxide solution Balanced chemical equation: Trial 1 Trial 2 Trial 3 Mass of oxalic acid 0.1314g NA NA Initial burette reading 0.40mL
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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Is data correct?

Transcribed Image Text:Experiment 11
Data to be completed and uploaded.
Only one trial!
(There is no part A)
Part B. Standardization of Sodium hydroxide solution
Balanced chemical equation:
Trial 1
Trial 2
Trial 3
Mass of oxalic acid
0.1314g
NA
NA
Initial burette reading
0.40mL
Final burette reading
20.60 mL
-0.40 me
20.20mL
Volume NaOH added
20.60mL_
· 20.20m
moles of oxalic acid
00014m
0.1314 g
90g
moles of NaOH added
molcs
= 0.00145 m
0.00292m
2x 0.00146m=0.00292
Molarity of NaOH
OF NaOH= welas
Volume
0.00292 m
20,20ML
molaviH
メ01445m

Transcribed Image Text:Part C. Determination of Unknown
Balanced chemical equation:
Hal + Na OH -o Nacl t Ha0
Unknown number _NA
Trial 1
Trial 2
Trial 3
Volume of unknown HCl
10.00mL
NA
NA
titrated
Initial burette reading
0.90mL
Final burette reading
30.90mL
Volume NaOH added
20 00mL
50.00ML XO.144 m=43.2m
4.335m
moles NaOH added
(Use average molarity from part B
and multiply it by the volume you added)
a00435 m
moles HCI
4.335m
=0.4335m
10.00 mL
molarity of HCl
4235 M
Actual Molarity (from instructor)
0.3150M
percentage error
375%-
O 4335 M -0.3150m
X100)= 37.:
0.3150M
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