A 0.9550 g sample of ASA was dissolved in 40.1 mL of standard NaOH. The excess NaOH was neutralizedusing 18.1 mL of a standard HCI.Note that 1.0 mL of the standard acid is equivalent to 0.0273 g sodium carbonate and 1.0 mL of the standardbase is equivalent to 0.1005 g KHP.MW: KHP = 204.22 NazCO3 = 106.0 ASA = 180.16ASA+ NaOH+H20ÓHNaOH+ NaOH+ CH;COONNa acetateNaNaNa AcetylsalicylateNa salicylateх'ss NaOH + HCІNaCl + H,OThe Normality of the STD. Base isThe Normality of the STD. Acid isThe weight of pure ASA in grams isThe H of aspirin isThe % ASA in the sample is???

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A 0.9550 g sample of ASA was dissolved in 40.1 mL of standard NaOH. The excess NaOH was neutralized using 18.1 mL of a standard HCI. Note that 1.0 mL of the standard acid is equivalent to 0.0273 g sodium carbonate and 1.0 mL of the standard base is equivalent to 0.1005 g KHP. MW: KHP = 204.22 Na2CO3 = 106.0 ASA = 180.16 ASA +NaOH + H,O `Na HO + NaOH + CH;COONa Na acetate `Na `Na Na Acetylsalicylate Na salicylate x'ss NaOH + HCІ NaCl + H2O The Normality of the STD. Base is The Normality of the STD. Acid is The weight of pure ASA in grams is ? ?

A 0.9550 g sample of ASA was dissolved in 40.1 mL of standard NaOH. The excess NaOH was neutralized using 18.1 mL of a standard HCI. Note that 1.0 mL of the standard acid is equivalent to 0.0273 g sodium carbonate and 1.0 mL of the standard base is equivalent to 0.1005 g KHP. MW: KHP = 204.22 Na2CO3 = 106.0 ASA = 180.16 ASA +NaOH + H,O `Na HO + NaOH + CH;COONa Na acetate `Na `Na Na Acetylsalicylate Na salicylate x'ss NaOH + HCІ NaCl + H2O The Normality of the STD. Base is The Normality of the STD. Acid is The weight of pure ASA in grams is ? ?

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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