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At 407 K the rate constant of a chemical reaction is 9.5 × 10–5s–1 and at 420 K, the rate constant is 1.9 × 10–4s–1. Calculate the frequency factor of the reaction: [rounded up to two decimal places]
    Correct answer is between '4.90,5.20'. Can you explain this answer?
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    At 407 K the rate constant of a chemical reaction is 9.5 × 10–5s–1 and at 420 K, the rate constant is 1.9 × 10–4s–1. Calculate the frequency factor of the reaction: [rounded up to two decimal places]Correct answer is between '4.90,5.20'. Can you explain this answer?
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    At 407 K the rate constant of a chemical reaction is 9.5 × 10–5s–1 and at 420 K, the rate constant is 1.9 × 10–4s–1. Calculate the frequency factor of the reaction: [rounded up to two decimal places]Correct answer is between '4.90,5.20'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about At 407 K the rate constant of a chemical reaction is 9.5 × 10–5s–1 and at 420 K, the rate constant is 1.9 × 10–4s–1. Calculate the frequency factor of the reaction: [rounded up to two decimal places]Correct answer is between '4.90,5.20'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 407 K the rate constant of a chemical reaction is 9.5 × 10–5s–1 and at 420 K, the rate constant is 1.9 × 10–4s–1. Calculate the frequency factor of the reaction: [rounded up to two decimal places]Correct answer is between '4.90,5.20'. Can you explain this answer?.
    Solutions for At 407 K the rate constant of a chemical reaction is 9.5 × 10–5s–1 and at 420 K, the rate constant is 1.9 × 10–4s–1. Calculate the frequency factor of the reaction: [rounded up to two decimal places]Correct answer is between '4.90,5.20'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry. Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free.
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