For the following reaction, the equilibrium constant Kc at 298 K is 1.6 andtimes; 1017.When equal volumes of 0.06 M Fe2+(aq) and 0.2 M S2andndash;(aq) solutions are mixed, the equilibriumconcentration of Fe2+(aq) is found to be Y andtimes; 10andndash;17 M. The value of Y is andndash;andndash;andndash;andndash;andndash;andndash;andndash;andndash;Correct answer is between '8.92,8.93'. Can you explain this answer?

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For the following reaction, the equilibrium constant K_c at 298 K is 1.6 × 10¹⁷.

When equal volumes of 0.06 M Fe²⁺(aq) and 0.2 M S^2–(aq) solutions are mixed, the equilibrium concentration of Fe²⁺(aq) is found to be Y × 10^–17 M. The value of Y is ––––––––

Correct answer is between '8.92,8.93'. Can you explain this answer?

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For the following reaction, the equilibrium constant Kc at 298 K is 1....

y = 8.93

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For the following reaction, the equilibrium constant When equal volumes of 0.06 M Fe2+(aq) and 0.2 M S2–(aq) solutions are mixed, the equilibrium concentration of Fe2+(aq) is found to be Y x 10–17 M. The value of Y is ……… . Correct answer is '8.93'. Can you explain this answer?

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For the following reaction, the equilibrium constant Kc at 298 K is 1.6 × 1017.When equal volumes of 0.06 M Fe2+(aq) and 0.2 M S2–(aq) solutions are mixed, the equilibriumconcentration of Fe2+(aq) is found to be Y × 10–17 M. The value of Y is ––––––––Correct answer is between '8.92,8.93'. Can you explain this answer?

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For the following reaction, the equilibrium constant Kc at 298 K is 1.6 × 1017.When equal volumes of 0.06 M Fe2+(aq) and 0.2 M S2–(aq) solutions are mixed, the equilibriumconcentration of Fe2+(aq) is found to be Y × 10–17 M. The value of Y is ––––––––Correct answer is between '8.92,8.93'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about For the following reaction, the equilibrium constant Kc at 298 K is 1.6 × 1017.When equal volumes of 0.06 M Fe2+(aq) and 0.2 M S2–(aq) solutions are mixed, the equilibriumconcentration of Fe2+(aq) is found to be Y × 10–17 M. The value of Y is ––––––––Correct answer is between '8.92,8.93'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the following reaction, the equilibrium constant Kc at 298 K is 1.6 × 1017.When equal volumes of 0.06 M Fe2+(aq) and 0.2 M S2–(aq) solutions are mixed, the equilibriumconcentration of Fe2+(aq) is found to be Y × 10–17 M. The value of Y is ––––––––Correct answer is between '8.92,8.93'. Can you explain this answer?.

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