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The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 250C are –400 kJ/mol,–300 kJ/mol and –1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at250C isa)+2900 kJb)– 2900 kJc)–16.11 kJd)+16.11 kJCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared
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the JEE exam syllabus. Information about The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 250C are –400 kJ/mol,–300 kJ/mol and –1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at250C isa)+2900 kJb)– 2900 kJc)–16.11 kJd)+16.11 kJCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam.
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The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 250C are –400 kJ/mol,–300 kJ/mol and –1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at250C isa)+2900 kJb)– 2900 kJc)–16.11 kJd)+16.11 kJCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 250C are –400 kJ/mol,–300 kJ/mol and –1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at250C isa)+2900 kJb)– 2900 kJc)–16.11 kJd)+16.11 kJCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 250C are –400 kJ/mol,–300 kJ/mol and –1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at250C isa)+2900 kJb)– 2900 kJc)–16.11 kJd)+16.11 kJCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 250C are –400 kJ/mol,–300 kJ/mol and –1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at250C isa)+2900 kJb)– 2900 kJc)–16.11 kJd)+16.11 kJCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.