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·0225mL x.1M = 2,25 × 103 mol NaOH LLS 10 mul Acid 1125×15 D IL * ASSAS t 01125M yej + 70 D 1.129 F wular st = mols Acit + ว A A 10/125001 = (1295) ✓ ' DAS Q 68 12-1.44 A 94.044 1.125×103 = 125 molar map Molar Mass = 106 g/mol pure, ubstances mixed AST, V A A 4x Kl I 2 40096) sp C +25 2. (04) 2 + 20H x(2x2)=111 I olbot C0160+22 080 0 5.2851×15 X 00101 25 (0080+5)/45) (0.03214) = 3.00×1017 4.07252 5.78645×10.5 2.177 32104 2.732×167 (0190142) (.0020)(45)

·0225mL x.1M = 2,25 × 103 mol NaOH LLS 10 mul Acid 1125×15 D IL * ASSAS t 01125M yej + 70 D 1.129 F wular st = mols Acit + ว A A 10/125001 = (1295) ✓ ' DAS Q 68 12-1.44 A 94.044 1.125×103 = 125 molar map Molar Mass = 106 g/mol pure, ubstances mixed AST, V A A 4x Kl I 2 40096) sp C +25 2. (04) 2 + 20H x(2x2)=111 I olbot C0160+22 080 0 5.2851×15 X 00101 25 (0080+5)/45) (0.03214) = 3.00×1017 4.07252 5.78645×10.5 2.177 32104 2.732×167 (0190142) (.0020)(45)

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
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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Hello I need help clarifying how this person got a molar mass 106 g/mol. The image shows the how person got the answer, I am just still confused. And once clarified I would like to see the clarified written steps of how the answer was gotten.

·0225mL x.1M = 2,25 × 103 mol NaOH LLS 10 mul Acid 1125×15 D IL * ASSAS t 01125M yej + 70 D 1.129 F wular st = mols Acit + ว A A 10/125001 = (1295) ✓ ' DAS Q 68 12-1.44 A 94.044 1.125×103 = 125 molar map Molar Mass = 106 g/mol pure, ubstances mixed AST, V A A 4x Kl I 2 40096) sp C +25 2. (04) 2 + 20H x(2x2)=111 I olbot C0160+22 080 0 5.2851×15 X 00101 25 (0080+5)/45) (0.03214) = 3.00×1017 4.07252 5.78645×10.5 2.177 32104 2.732×167 (0190142) (.0020)(45)
Transcribed Image Text:·0225mL x.1M = 2,25 × 103 mol NaOH LLS 10 mul Acid 1125×15 D IL * ASSAS t 01125M yej + 70 D 1.129 F wular st = mols Acit + ว A A 10/125001 = (1295) ✓ ' DAS Q 68 12-1.44 A 94.044 1.125×103 = 125 molar map Molar Mass = 106 g/mol pure, ubstances mixed AST, V A A 4x Kl I 2 40096) sp C +25 2. (04) 2 + 20H x(2x2)=111 I olbot C0160+22 080 0 5.2851×15 X 00101 25 (0080+5)/45) (0.03214) = 3.00×1017 4.07252 5.78645×10.5 2.177 32104 2.732×167 (0190142) (.0020)(45)
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