Mass of KxFe(C2O4)y · zH2O : 5.60 g Mass of sample : 0.155 g Mass of FeCl3 used in preparation : 1.60 g Molarity of standard NaOH used : 0.100 V1, volume of standard NaOH required for first equivalence point : 9.000 mL V2, volume of standard NaOH required for second equivalence point : 15.90 mL Calculate the volume of standard NaOH that reacted with iron :

Mass of KxFe(C2O4)y · zH2O : 5.60 g Mass of sample : 0.155 g Mass of FeCl3 used in preparation : 1.60 g Molarity of standard NaOH used : 0.100 V1, volume of standard NaOH required for first equivalence point : 9.000 mL V2, volume of standard NaOH required for second equivalence point : 15.90 mL Calculate the volume of standard NaOH that reacted with iron :

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter4: Stoichiometry: Quantitative Information About Chemical Reactions

Section4.9: Spectrophotometry

Problem 2.1ACP: Excess KI is added to a 100.0-mL sample of a soft drink that had been contaminated with bleach,...

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The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a) BaSiF6, 0.026 g/100 mL (contains SiF62- ions) (b) Ce(IO3)4, 1.5102 g/100 mL (c) Gd2(SO4)3, 3.98 g/100 mL (d) (NH4)2PtBr6, 0.59 g/100 mL (contains PtBr62- ions)
Mass of KxFe(C2O4)y · zH2O : 4.70 g Mass of sample : 0.175 g Mass of FeCl3 used in preparation : 1.60 g Molarity of standard NaOH used : 0.100 V1, volume of standard NaOH required for first equivalence point : 7.750 mL V2, volume of standard NaOH required for second equivalence point : 19.70 mL Calculate the volume of standard NaOH that reacted with iron :
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A 1.00 g sample of NaOH which has been exposed to the air for some time is dissolve in water and water and diluted to 500.0 ml. 100.0 ml of the solution is titrated with 0.1062 N HCl (MOH indicator) and requires 38.60 ml. A second 100 .0 ml portion is titrated with excess BaCl2, diluted to 250 .0 ml and filtered. 200.0 ml of this filtrate then require 29.62 ml of the HCl ( HpH indicator) . Calculate the % NaOH and Na2CO3 in the original sample.
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