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35. A 100 mL sample is titrated to the methyl orange (alkalinity color indicator) end point with 2 mL of 0.5 N H₂SO4. Based on the calculation equation of alkalinity: xeq. wt. CaCO, x (1000 mg g) mL X N Total alk= acid acid mL sample What is the total alkalinity in mg L-¹ as CaCO3 (Note the equivalent weight of CaCO3 is 50 g eq¹)? A) 5 B) 50 C) 500 D) 1000 E) 5000 36. Which of the following is NOT one way that clays can develop charge? A) Isomorphic substitution

35. A 100 mL sample is titrated to the methyl orange (alkalinity color indicator) end point with 2 mL of 0.5 N H₂SO4. Based on the calculation equation of alkalinity: xeq. wt. CaCO, x (1000 mg g) mL X N Total alk= acid acid mL sample What is the total alkalinity in mg L-¹ as CaCO3 (Note the equivalent weight of CaCO3 is 50 g eq¹)? A) 5 B) 50 C) 500 D) 1000 E) 5000 36. Which of the following is NOT one way that clays can develop charge? A) Isomorphic substitution

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.148QP
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35. A 100 mL sample is titrated to the methyl orange (alkalinity color indicator) end point with 2 mL of 0.5 N H₂SO4. Based on the calculation equation of alkalinity: xeq. wt. CaCO, x (1000 mg g) x N mL acid acid Total alk = mL What is the total alkalinity in mg L-¹ as CaCO3 (Note the equivalent weight of CaCO3 is 50 g eq-¹)? A) 5 B) 50 C) 500 D) 1000 E) 5000 36. Which of the following is NOT one way that clays can develop charge? A) Isomorphic substitution B) Imperfections in the crystal structure C) Proton reaction D) Sorption E) Hydration 37. A soil with a high clay and organic matter content typically has A) low B) high C) neutral D) no E) basic 38. What is the PZC value of Kaolinite? 0.2 Surface charge (C m-²) 0.1 A) 2 B) 3 C) 4.6 D) 8 E) >10 0 -0.1 -0.2 Montmorillonite, Feldspar 2 SIO₂ x-MnO₂ Kaolinite 39. Oxidation refers to a process of electrons. A) gaining; gaining sample 6 a-FeOOH Calcite was cosso 000000 200000 Me 8 6-AL₂0₁ 7 10 MgO 00000000 7517 12 pH CEC. (9 electrons while reduction refers to a process of 71
Transcribed Image Text:35. A 100 mL sample is titrated to the methyl orange (alkalinity color indicator) end point with 2 mL of 0.5 N H₂SO4. Based on the calculation equation of alkalinity: xeq. wt. CaCO, x (1000 mg g) x N mL acid acid Total alk = mL What is the total alkalinity in mg L-¹ as CaCO3 (Note the equivalent weight of CaCO3 is 50 g eq-¹)? A) 5 B) 50 C) 500 D) 1000 E) 5000 36. Which of the following is NOT one way that clays can develop charge? A) Isomorphic substitution B) Imperfections in the crystal structure C) Proton reaction D) Sorption E) Hydration 37. A soil with a high clay and organic matter content typically has A) low B) high C) neutral D) no E) basic 38. What is the PZC value of Kaolinite? 0.2 Surface charge (C m-²) 0.1 A) 2 B) 3 C) 4.6 D) 8 E) >10 0 -0.1 -0.2 Montmorillonite, Feldspar 2 SIO₂ x-MnO₂ Kaolinite 39. Oxidation refers to a process of electrons. A) gaining; gaining sample 6 a-FeOOH Calcite was cosso 000000 200000 Me 8 6-AL₂0₁ 7 10 MgO 00000000 7517 12 pH CEC. (9 electrons while reduction refers to a process of 71
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