The concentration of a potassium permanganate (KMNO4) solution is determined by titration with KI(aq) in the presence of excess H2SO4. A balanced chemical equation for the reaction is given below.. 2 KMNO4 + 10 KI + 8 H2SO4 6 K2SO4 + 2 MNSO4 + 5 12 +8 H20 It requires 21.4 mL of 1.62 mol L-1 KI(aq) to completley titrate 40.6 mL of KMNO4(aq). What is the molar concentration of the KMNO4 solution? 0.854 mol L-1 2.56 mol L-1 0.171 mol L-1 8.54 mol L- 0.285 mol L-1

The concentration of a potassium permanganate (KMNO4) solution is determined by titration with KI(aq) in the presence of excess H2SO4. A balanced chemical equation for the reaction is given below.. 2 KMNO4 + 10 KI + 8 H2SO4 6 K2SO4 + 2 MNSO4 + 5 12 +8 H20 It requires 21.4 mL of 1.62 mol L-1 KI(aq) to completley titrate 40.6 mL of KMNO4(aq). What is the molar concentration of the KMNO4 solution? 0.854 mol L-1 2.56 mol L-1 0.171 mol L-1 8.54 mol L- 0.285 mol L-1

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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Can you help me with the attached probelm?
A 1.250 g sample of cheese was subjected to a Kjeldahl analysis to determine the amount of protein. The sample was digested, the nitrogen is oxidized to NH₄⁺, and was then converted to NH₃ with NaOH, and distilled into a collection flask containing 50.00 mL of 0.1050 M HCl. The excess HCl is back titrated with 0.1175 M NaOH, this required 21.65 mL to reach the bromothymol blue end point. What is the reaction for the back titration for this analysis? a. NH₄⁺ + NaOH ⇌ NH₃ + H₂O + Na⁺ b. NaOH + HCl ⇌ NaCl + H₂O c. NH₃ + HCl ⇌ NH₄⁺ + Cl⁻ d. N₂ + 3 H₂ ⇌ 2 NH₃ How many moles of N is present in the cheese sample? a. 544 x 10⁻³ b. 706 x 10⁻³ c. 250 x 10⁻³ d. 794 x 10⁻³ Report the %N of the cheese sample. a. 851% b. 884% c. 033% d. 736%
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