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 20.60mL Final burette reading 20.60 m -0.40 mL 20.20ML Volume NaOH added 20.20mL moles of oxalic acid 0:001m O.1314 3 = 0,00Im moles of NaOH added 0.002m 2x 0.00lm=0.002 Molarity of NaOH O.IM Volume 2020ML

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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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 me
-0.40 mL
20.20 mL
Volume NaOH added
20.60mL
· 20.20mk
moles of oxalic acid
O.1314 g
moles
= 0.00Im
90g
つ11*メP
moles of NaOH added
Actd
0.002m
2x 0.00lm=0.002
Molarity of NaOH
of NaOH=melas
Volume
molavit
2020ML
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 me -0.40 mL 20.20 mL Volume NaOH added 20.60mL · 20.20mk moles of oxalic acid O.1314 g moles = 0.00Im 90g つ11*メP moles of NaOH added Actd 0.002m 2x 0.00lm=0.002 Molarity of NaOH of NaOH=melas Volume molavit 2020ML
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