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In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared
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the Class 11 exam syllabus. Information about In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam.
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Here you can find the meaning of In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer?, a detailed solution for In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? has been provided alongside types of In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? tests, examples and also practice Class 11 tests.