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Equilibrium constant for the given reaction is C = 1020 at temperature 300 KA(s) +2B(aq)2C (s) + D(aq.)K = 1020The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K isa)-4 x 10-11b)-2 x 10-10c)-2 x 10-11d)-10-10Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Equilibrium constant for the given reaction is C = 1020 at temperature 300 KA(s) +2B(aq)2C (s) + D(aq.)K = 1020The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K isa)-4 x 10-11b)-2 x 10-10c)-2 x 10-11d)-10-10Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Equilibrium constant for the given reaction is C = 1020 at temperature 300 KA(s) +2B(aq)2C (s) + D(aq.)K = 1020The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K isa)-4 x 10-11b)-2 x 10-10c)-2 x 10-11d)-10-10Correct answer is option 'A'. Can you explain this answer?.

Solutions for Equilibrium constant for the given reaction is C = 1020 at temperature 300 KA(s) +2B(aq)2C (s) + D(aq.)K = 1020The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K isa)-4 x 10-11b)-2 x 10-10c)-2 x 10-11d)-10-10Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11. Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.

Here you can find the meaning of Equilibrium constant for the given reaction is C = 1020 at temperature 300 KA(s) +2B(aq)2C (s) + D(aq.)K = 1020The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K isa)-4 x 10-11b)-2 x 10-10c)-2 x 10-11d)-10-10Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Equilibrium constant for the given reaction is C = 1020 at temperature 300 KA(s) +2B(aq)2C (s) + D(aq.)K = 1020The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K isa)-4 x 10-11b)-2 x 10-10c)-2 x 10-11d)-10-10Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Equilibrium constant for the given reaction is C = 1020 at temperature 300 KA(s) +2B(aq)2C (s) + D(aq.)K = 1020The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K isa)-4 x 10-11b)-2 x 10-10c)-2 x 10-11d)-10-10Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Equilibrium constant for the given reaction is C = 1020 at temperature 300 KA(s) +2B(aq)2C (s) + D(aq.)K = 1020The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K isa)-4 x 10-11b)-2 x 10-10c)-2 x 10-11d)-10-10Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Equilibrium constant for the given reaction is C = 1020 at temperature 300 KA(s) +2B(aq)2C (s) + D(aq.)K = 1020The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K isa)-4 x 10-11b)-2 x 10-10c)-2 x 10-11d)-10-10Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Class 11 tests.