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The standard enthalpies of formation of CO2(g), H2O(l) and glucose (s) at 25°C are - 400 kJ mol-1, - 300 kJ mol-1 and - 1300 kJ mol-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is[JEE Advanced 2013]a)2900 kJb)-2900kJc)-16.11kJd)+16.11kJCorrect answer is option 'C'. 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 The standard enthalpies of formation of CO2(g), H2O(l) and glucose (s) at 25°C are - 400 kJ mol-1, - 300 kJ mol-1 and - 1300 kJ mol-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is[JEE Advanced 2013]a)2900 kJb)-2900kJc)-16.11kJd)+16.11kJCorrect answer is option 'C'. 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 The standard enthalpies of formation of CO2(g), H2O(l) and glucose (s) at 25°C are - 400 kJ mol-1, - 300 kJ mol-1 and - 1300 kJ mol-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is[JEE Advanced 2013]a)2900 kJb)-2900kJc)-16.11kJd)+16.11kJCorrect answer is option 'C'. Can you explain this answer?.

Solutions for The standard enthalpies of formation of CO2(g), H2O(l) and glucose (s) at 25°C are - 400 kJ mol-1, - 300 kJ mol-1 and - 1300 kJ mol-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is[JEE Advanced 2013]a)2900 kJb)-2900kJc)-16.11kJd)+16.11kJCorrect answer is option 'C'. 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 The standard enthalpies of formation of CO2(g), H2O(l) and glucose (s) at 25°C are - 400 kJ mol-1, - 300 kJ mol-1 and - 1300 kJ mol-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is[JEE Advanced 2013]a)2900 kJb)-2900kJc)-16.11kJd)+16.11kJCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The standard enthalpies of formation of CO2(g), H2O(l) and glucose (s) at 25°C are - 400 kJ mol-1, - 300 kJ mol-1 and - 1300 kJ mol-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is[JEE Advanced 2013]a)2900 kJb)-2900kJc)-16.11kJd)+16.11kJCorrect 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 25°C are - 400 kJ mol-1, - 300 kJ mol-1 and - 1300 kJ mol-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is[JEE Advanced 2013]a)2900 kJb)-2900kJc)-16.11kJd)+16.11kJCorrect 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 25°C are - 400 kJ mol-1, - 300 kJ mol-1 and - 1300 kJ mol-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is[JEE Advanced 2013]a)2900 kJb)-2900kJc)-16.11kJd)+16.11kJCorrect 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 25°C are - 400 kJ mol-1, - 300 kJ mol-1 and - 1300 kJ mol-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is[JEE Advanced 2013]a)2900 kJb)-2900kJc)-16.11kJd)+16.11kJCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 11 tests.