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A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.Q.Heat of neutralisation per mole of each reactant isa)-2.812 kJb)-28.12 kJc)-56.24 kJd)+56.24 kJCorrect 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 A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.Q.Heat of neutralisation per mole of each reactant isa)-2.812 kJb)-28.12 kJc)-56.24 kJd)+56.24 kJCorrect 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 A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.Q.Heat of neutralisation per mole of each reactant isa)-2.812 kJb)-28.12 kJc)-56.24 kJd)+56.24 kJCorrect answer is option 'C'. Can you explain this answer?.

Solutions for A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.Q.Heat of neutralisation per mole of each reactant isa)-2.812 kJb)-28.12 kJc)-56.24 kJd)+56.24 kJCorrect 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 A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.Q.Heat of neutralisation per mole of each reactant isa)-2.812 kJb)-28.12 kJc)-56.24 kJd)+56.24 kJCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.Q.Heat of neutralisation per mole of each reactant isa)-2.812 kJb)-28.12 kJc)-56.24 kJd)+56.24 kJCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.Q.Heat of neutralisation per mole of each reactant isa)-2.812 kJb)-28.12 kJc)-56.24 kJd)+56.24 kJCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.Q.Heat of neutralisation per mole of each reactant isa)-2.812 kJb)-28.12 kJc)-56.24 kJd)+56.24 kJCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.Q.Heat of neutralisation per mole of each reactant isa)-2.812 kJb)-28.12 kJc)-56.24 kJd)+56.24 kJCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 11 tests.