mols Molarity of NaOH calculated from previous section 0.080m Unknown number or letter Final ml NaOH Trial 1 D Trial 2 D 13.25mL 25.5mL 0.00 mL Initial mL NaOH Total mL NaOH moles NaOH (calculated) moles HCI (calculated, and 0.00 106 makes 0.0098 moles same as???) mL HCl 25 mL 25 mL 0.0424m 0.392 (measured) M HCI (moles/L!) Trial 3 D Trial 4 38.00 m2 11.01 ml 25.5 mL 0.00 mL Ta 13.25mL 0.00106 makes 06 0098 moles 13.25 mL 12.25 mL 12.50mL 11.01 m2 0.080 x 0. 01225 = 0.00098 0.001 moles 0.000808 ma 0.001 molcs 0.000808mues 25.00 mL m 0.04 m 0.080 x 0.1325mL = 0.00106 moles Average HCI M 0.12668 m (remember, trial one may be omitted from the average if it was sloppy) Standard deviation_ Average deviation_ 3 0.080 x 0.0125 0.601 moles C 25.00 ML 4 0.03232 m x 6.ollol | 0.080x = 0.000808

mols Molarity of NaOH calculated from previous section 0.080m Unknown number or letter Final ml NaOH Trial 1 D Trial 2 D 13.25mL 25.5mL 0.00 mL Initial mL NaOH Total mL NaOH moles NaOH (calculated) moles HCI (calculated, and 0.00 106 makes 0.0098 moles same as???) mL HCl 25 mL 25 mL 0.0424m 0.392 (measured) M HCI (moles/L!) Trial 3 D Trial 4 38.00 m2 11.01 ml 25.5 mL 0.00 mL Ta 13.25mL 0.00106 makes 06 0098 moles 13.25 mL 12.25 mL 12.50mL 11.01 m2 0.080 x 0. 01225 = 0.00098 0.001 moles 0.000808 ma 0.001 molcs 0.000808mues 25.00 mL m 0.04 m 0.080 x 0.1325mL = 0.00106 moles Average HCI M 0.12668 m (remember, trial one may be omitted from the average if it was sloppy) Standard deviation_ Average deviation_ 3 0.080 x 0.0125 0.601 moles C 25.00 ML 4 0.03232 m x 6.ollol | 0.080x = 0.000808

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

Chapter4: Types Of Chemical Reactions And Solution Stoichiometry

Section: Chapter Questions

Problem 47E

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