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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 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

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
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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