A 28.3 mL hydrochloric acid solution was standardized using 0.256 g of sodium carbonate. (Na2CO3 = 106 g/mol; HCl = 36.46 g/mol) Na2CO3 + 2 HCl → 2 NaCl + H2CO3 How many millimoles of sodium carbonate was used in the titration? 0.00242 0.00702 0.344 54.3 What is the reaction ratio of this titration? 1 mole acid to 1 mole base 1 mole acid to 2 moles base 2 moles acid to 1 mole base 1 mole acid to 3 moles base 3. How many millimoles of hydrochloric acid was used in the titration? 0.00121 0.00351 0.00483 0.0140
A 28.3 mL hydrochloric acid solution was standardized using 0.256 g of sodium carbonate. (Na2CO3 = 106 g/mol; HCl = 36.46 g/mol) Na2CO3 + 2 HCl → 2 NaCl + H2CO3 How many millimoles of sodium carbonate was used in the titration? 0.00242 0.00702 0.344 54.3 What is the reaction ratio of this titration? 1 mole acid to 1 mole base 1 mole acid to 2 moles base 2 moles acid to 1 mole base 1 mole acid to 3 moles base 3. How many millimoles of hydrochloric acid was used in the titration? 0.00121 0.00351 0.00483 0.0140
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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A 28.3 mL hydrochloric acid solution was standardized using 0.256 g of sodium carbonate. (Na2CO3 = 106 g/mol; HCl = 36.46 g/mol)
Na2CO3 + 2 HCl → 2 NaCl + H2CO3
How many millimoles of sodium carbonate was used in the titration?
-
- 0.00242
- 0.00702
- 0.344
- 54.3
What is the reaction ratio of this titration?
-
- 1 mole acid to 1 mole base
- 1 mole acid to 2 moles base
- 2 moles acid to 1 mole base
- 1 mole acid to 3 moles base
3. How many millimoles of hydrochloric acid was used in the titration?
-
- 0.00121
- 0.00351
- 0.00483
- 0.0140
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