Question 7 of 15 < > 0/1 View Policies Show Attempt History Current Attempt in Progress Your answer is incorrect. How many milliliters of 0.547 M KOH are needed to react completely with 32.7 mL of 0.249M FeCl2 solution to precipitate Fe(OH)2? The net ionic equation is: Fe2*(aq) + 20H (aq) – Fe(OH)2(s) Vкон mL eTextbook and Media Hint Save for Later Attempts: 5 of 15 used Submit Answer

Question 7 of 15 < > 0/1 View Policies Show Attempt History Current Attempt in Progress Your answer is incorrect. How many milliliters of 0.547 M KOH are needed to react completely with 32.7 mL of 0.249M FeCl2 solution to precipitate Fe(OH)2? The net ionic equation is: Fe2*(aq) + 20H (aq) – Fe(OH)2(s) Vкон mL eTextbook and Media Hint Save for Later Attempts: 5 of 15 used Submit Answer

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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Mark and John, two vinegar enthusiasts, are each tasked to determine the acetic acid content of their respective vinegar concoctions by titration. First, a 1 M-labeled KOH solution was standardized against the KHP (MW = 204.22 g/mol) standard that is 99.4% pure. In the process, 0.540 g KHP was found to require 2.80 mL of the KOH solution to completely react up to the phenolphthalein endpoint. Then, Mark and John both prepared their samples by taking 10.0-mL aliquots of each vinegar and diluting them to 25.0 mL. Using the same titrant and indicator, Mark’s vinegar required 18.60 mL of the standardized titrant to reach the endpoint, while John’s vinegar required 16.50 mL of the same titrant to reach the same endpoint. Question: What is the acetic acid concentration of Mark’s and John’s vinegar in molarity given that the Concentration of KOH is 9386 M?
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A CHM 125 student performed the "Redox Titration" procedure as given in the laboratory manual. In this experiment, the titrant solution of potassium permanganate (KMNO4) was used to determine the percentage of sodium oxalate (NazC2O4) in an unknown sample. The net ionic equation for the titration carried out was: 2MnO4 (aq) + 5C20,7(aq) + 16H*(aq) → 2MN²*(aq) + 10CO2(g) + 8H2O(£) The data provided in the table below was acquired. Complete the table and determine the percent of sodium oxalate in the unknown sample based on this one titration. Molarity of KMNO, Titrant Solution: 0.0203 M Trial 1 Weight of Oxalate 0.954 g Unknown Starting Buret Reading: 0.65 mL Ending Buret Reading: 34.85 mL of permanganate sl order to use 35.00 mL of Titrant Used: mmol of MnO4 mmol of 2- Weight of NazC204 in Unknown: Percent of Na,C204 in Unknown: You wer 4-
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A 0.450 g sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is CaCO3(s) + 2HCl(aq) - CaCl2(aq) + H2O(l) + CO2(g) The excess HCl(aq) is titrated by 5.60 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO3(s) in the sample. mass percentage:
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