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For a cell reaction Pb(s) + Hg2Cl2(s) → PbCl2(s) + 2Hg(I), is 1.45 × 10–1 VK–1. Theentropy change (in J mol–1 K–1) for the reaction is [Given 1 F = 96500 C mol–1] _______.Correct answer is '27.985'. 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 For a cell reaction Pb(s) + Hg2Cl2(s) → PbCl2(s) + 2Hg(I), is 1.45 × 10–1 VK–1. Theentropy change (in J mol–1 K–1) for the reaction is [Given 1 F = 96500 C mol–1] _______.Correct answer is '27.985'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a cell reaction Pb(s) + Hg2Cl2(s) → PbCl2(s) + 2Hg(I), is 1.45 × 10–1 VK–1. Theentropy change (in J mol–1 K–1) for the reaction is [Given 1 F = 96500 C mol–1] _______.Correct answer is '27.985'. Can you explain this answer?.

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