The iron (55.847 g/mol) in a 700.0 mL sample of a natural water was determined by precipitation of thecation as hydrated Fe203. The precipitate was filtered, washed, and ignited in a crucible to form Fe203(MW = 159.69 g/mol). The mass of the product was 0.1168 g. Calculate the ppm iron in the water.A:В:Oc:OD:OE:F:OG:H:99.75116.71136.55159.76186.92218.70255.87299.37Submit AnswerTries 0/99

Given: Volume of sample = 700mL Mass of Fe2O3 = 0.1168g

The iron (55.847 g/mol) in a 700.0 mL sample of a natural water was determined by precipitation of the cation as hydrated Fe203. The precipitate was filtered, washed, and ignited in a crucible to form Fe203 = 159.69 g/mol). The mass of the product was 0.1168 g. Calculate the ppm iron in the water. A: OB: C: OD: O E: F: G: OH: 99.75 116.71 136.55 159.76 186.92 218.70 255.87 299.37 Submit Answer Tries 0/99

The iron (55.847 g/mol) in a 700.0 mL sample of a natural water was determined by precipitation of the cation as hydrated Fe203. The precipitate was filtered, washed, and ignited in a crucible to form Fe203 = 159.69 g/mol). The mass of the product was 0.1168 g. Calculate the ppm iron in the water. A: OB: C: OD: O E: F: G: OH: 99.75 116.71 136.55 159.76 186.92 218.70 255.87 299.37 Submit Answer Tries 0/99

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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