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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
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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M= 78 mols Molarity of NaOH calculated from previous section_ D Trial 1 D Unknown number or letter Final ml NaOH Initial mL NaOH Total mL NaOH moles NaOH (calculated) moles HC1 Trial 2 D 13.25mL 25.5mL same as???) mL HCl (measured) M HCI (moles/L!) 0.080m 0.00 mL 13.25 mL 13.25 mL 12.25 mL 0.00106 moles 0009% moles 25 mL 25 mL 0.0424m 0.392 Trial 3 D Trial 4 38.00 m2 11.01 mL oil to noreni 25.00 mL m/ 0.04 1920X1 25.5 mL 12.50mL 11.0.1 m2 0.080 x 0.01225 = 0.00098 0.001 moles 0.000808 moles SELA (calculated, and 0.00 106 mcles 0.0098 moles 0.001 moles 0.000808mues 080 x 0.0125 0.001 moles Average HCI M 0.12668 m (remember, trial one may be omitted from the average if it was sloppy) Standard deviation m 0.032 32 TI 0.080 x 0.1325mL = 0.00 166 moles 0.00 mL Ta best m 25.00 ML 4 3 Average deviation Show HCI molarity calculation for trial 3 (starting with mL NaOH): 0.080 m NaOH x 0.01250L NaOH = 0.001 mois NACH 0.001 mois NaOH 1 mol HCl = 0.001 MOIS HCl I mol NaOH 0.080 x 6.0llol = 0.000808 Question: 0.001 mois HCl 0.0264 = 0.04 m HC1 If you needed 200 L of 6M HCl for your pool, what volume of your unknown HCl would you need to add instead (if it is NOT 6 M)? SHOW WORK! You may dispose of your NaOH solution down the drain (since NaOH is the major component of Drano®), and a stronger concentration will be required for next week. Also, turn in discussion- see end of lab. TA Initials
Transcribed Image Text:M= 78 mols Molarity of NaOH calculated from previous section_ D Trial 1 D Unknown number or letter Final ml NaOH Initial mL NaOH Total mL NaOH moles NaOH (calculated) moles HC1 Trial 2 D 13.25mL 25.5mL same as???) mL HCl (measured) M HCI (moles/L!) 0.080m 0.00 mL 13.25 mL 13.25 mL 12.25 mL 0.00106 moles 0009% moles 25 mL 25 mL 0.0424m 0.392 Trial 3 D Trial 4 38.00 m2 11.01 mL oil to noreni 25.00 mL m/ 0.04 1920X1 25.5 mL 12.50mL 11.0.1 m2 0.080 x 0.01225 = 0.00098 0.001 moles 0.000808 moles SELA (calculated, and 0.00 106 mcles 0.0098 moles 0.001 moles 0.000808mues 080 x 0.0125 0.001 moles Average HCI M 0.12668 m (remember, trial one may be omitted from the average if it was sloppy) Standard deviation m 0.032 32 TI 0.080 x 0.1325mL = 0.00 166 moles 0.00 mL Ta best m 25.00 ML 4 3 Average deviation Show HCI molarity calculation for trial 3 (starting with mL NaOH): 0.080 m NaOH x 0.01250L NaOH = 0.001 mois NACH 0.001 mois NaOH 1 mol HCl = 0.001 MOIS HCl I mol NaOH 0.080 x 6.0llol = 0.000808 Question: 0.001 mois HCl 0.0264 = 0.04 m HC1 If you needed 200 L of 6M HCl for your pool, what volume of your unknown HCl would you need to add instead (if it is NOT 6 M)? SHOW WORK! You may dispose of your NaOH solution down the drain (since NaOH is the major component of Drano®), and a stronger concentration will be required for next week. Also, turn in discussion- see end of lab. TA Initials
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