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F 97. A 0.9300 g sample of a mixture of Nace and Mylle is treated with excess Ag (NO3) cag). The precipitate is filtered off then dried. If the the percentage by mass of Nall 2.616 g what is dry precipitate weighs the original mixture? In -AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA Molar mass N=14019 0 = 16.00 g U = 35.45 9 Ag = 107.9 Na = 22.99 گیا = 24.31 moles of AgNO3 = given, mass of mixture =0.9300 g dry precipitate = 2.676 g rxn 1: Nall+AgNO3 Agle + NaNO3 rxn 2 mg l mass of precip molar mass z - Ag No₂ 12Agel + Mg (NO₂) ₂ = 2.676 143.35 = 0.0 1.8 667 59 6 17 Molar Mass AgNO3 = 107 9 14.01 + = 160 Vo moles of Nall = 0.018667596 x = 0.01866796 169.919 molar, mass 4- = 107.9 mass of Nall= (0.01866796) (58.44) = 1.0909 55582 Percentage by mass of Nall = mass Agul + 35.45 = 143.35 l૦૧૦%5582 0.9300 3116.00) molar mass Nall = 22.99 +35.45 = 58.44 ×100

F 97. A 0.9300 g sample of a mixture of Nace and Mylle is treated with excess Ag (NO3) cag). The precipitate is filtered off then dried. If the the percentage by mass of Nall 2.616 g what is dry precipitate weighs the original mixture? In -AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA Molar mass N=14019 0 = 16.00 g U = 35.45 9 Ag = 107.9 Na = 22.99 گیا = 24.31 moles of AgNO3 = given, mass of mixture =0.9300 g dry precipitate = 2.676 g rxn 1: Nall+AgNO3 Agle + NaNO3 rxn 2 mg l mass of precip molar mass z - Ag No₂ 12Agel + Mg (NO₂) ₂ = 2.676 143.35 = 0.0 1.8 667 59 6 17 Molar Mass AgNO3 = 107 9 14.01 + = 160 Vo moles of Nall = 0.018667596 x = 0.01866796 169.919 molar, mass 4- = 107.9 mass of Nall= (0.01866796) (58.44) = 1.0909 55582 Percentage by mass of Nall = mass Agul + 35.45 = 143.35 l૦૧૦%5582 0.9300 3116.00) molar mass Nall = 22.99 +35.45 = 58.44 ×100

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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Answer should be 24%, I keep getting 117% which is wrong obviously. What am I doing wrong 

F 97. A 0.9300 g sample of a mixture of Nace and Mylle is treated with excess Ag (NO3) cag). The precipitate is filtered off then dried. If the the percentage by mass of Nall 2.616 g what is dry precipitate weighs the original mixture? In -AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA Molar mass N=14019 0 = 16.00 g U = 35.45 9 Ag = 107.9 Na = 22.99 گیا = 24.31 moles of AgNO3 = given, mass of mixture =0.9300 g dry precipitate = 2.676 g rxn 1: Nall+AgNO3 Agle + NaNO3 rxn 2 mg l mass of precip molar mass z - Ag No₂ 12Agel + Mg (NO₂) ₂ = 2.676 143.35 = 0.0 1.8 667 59 6 17 Molar Mass AgNO3 = 107 9 14.01 + = 160 Vo moles of Nall = 0.018667596 x = 0.01866796 169.919 molar, mass 4- = 107.9 mass of Nall= (0.01866796) (58.44) = 1.0909 55582 Percentage by mass of Nall = mass Agul + 35.45 = 143.35 l૦૧૦%5582 0.9300 3116.00) molar mass Nall = 22.99 +35.45 = 58.44 ×100
Transcribed Image Text:F 97. A 0.9300 g sample of a mixture of Nace and Mylle is treated with excess Ag (NO3) cag). The precipitate is filtered off then dried. If the the percentage by mass of Nall 2.616 g what is dry precipitate weighs the original mixture? In -AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA Molar mass N=14019 0 = 16.00 g U = 35.45 9 Ag = 107.9 Na = 22.99 گیا = 24.31 moles of AgNO3 = given, mass of mixture =0.9300 g dry precipitate = 2.676 g rxn 1: Nall+AgNO3 Agle + NaNO3 rxn 2 mg l mass of precip molar mass z - Ag No₂ 12Agel + Mg (NO₂) ₂ = 2.676 143.35 = 0.0 1.8 667 59 6 17 Molar Mass AgNO3 = 107 9 14.01 + = 160 Vo moles of Nall = 0.018667596 x = 0.01866796 169.919 molar, mass 4- = 107.9 mass of Nall= (0.01866796) (58.44) = 1.0909 55582 Percentage by mass of Nall = mass Agul + 35.45 = 143.35 l૦૧૦%5582 0.9300 3116.00) molar mass Nall = 22.99 +35.45 = 58.44 ×100
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