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A 25.0 mL sample of 0.200 M AgNO3(aq) was allowed to react with an excess of NaCI(aq). The AgCl precipitate from the reaction was carefully dried and weighed. How many grams of precipitate should be obtained? AgNO3(aq) + NACI(aq) → AgCI(s) + NANO3(aq) О 3.13g O 0.717 g O 0.432 g O 1.70g O 1.08 g

A 25.0 mL sample of 0.200 M AgNO3(aq) was allowed to react with an excess of NaCI(aq). The AgCl precipitate from the reaction was carefully dried and weighed. How many grams of precipitate should be obtained? AgNO3(aq) + NACI(aq) → AgCI(s) + NANO3(aq) О 3.13g O 0.717 g O 0.432 g O 1.70g O 1.08 g

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
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Current Attempt in Progress A 25.0 mL sample of 0.200 M AgNO3(aq) was allowed to react with an excess of NaCI(aq). The AgCI precipitate from the reaction was carefully dried and weighed. How many grams of precipitate should be obtained? AgNO3(aq) + NACI(aq) → A£CI(s) + NaNO3(ag) O 3.13g O 0.717 g O 0.432 g O 1.70 g O 1.08 g e Textbook and Media Save for Later Attempts: 0 of 2 used Submit Answer e here to search insert & %23 3. 4 7 8. 9 R Y D pause M alt ctri B
Transcribed Image Text:Current Attempt in Progress A 25.0 mL sample of 0.200 M AgNO3(aq) was allowed to react with an excess of NaCI(aq). The AgCI precipitate from the reaction was carefully dried and weighed. How many grams of precipitate should be obtained? AgNO3(aq) + NACI(aq) → A£CI(s) + NaNO3(ag) O 3.13g O 0.717 g O 0.432 g O 1.70 g O 1.08 g e Textbook and Media Save for Later Attempts: 0 of 2 used Submit Answer e here to search insert & %23 3. 4 7 8. 9 R Y D pause M alt ctri B
Expert Solution
Step 1

Given, 

A 25.0 mL sample of 0.100 M AgNO3(aq) was allowed to react with an excess of NaCl(aq).

Volume = 25.0 mL = 0.025 L 

Molality = 0.200 M 

Molarity = number of moles/volume in L 

Number of moles = Molarity x volume in L

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