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In the following reactions,
I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4 (aq), rise intemperature = ΔT1
II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mL of 0.1 M H2SO4 (aq), rise intemperature = ΔT
II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mL of 0.1 M H2SO4 (aq), rise intemperature = ΔT
2
. Thus,- a)ΔT2 = 3ΔT1
- b)ΔT1 = ΔT2
- c)ΔT1 = 3ΔT2
- d)ΔT1 = 2ΔT2
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0....
Solution:
Given:
30 mL of 0.1 M Ba(OH)2(aq) 30 mL of 0.1 M H2SO4(aq), rise in temperature = T1
90 mL of 0.1 M Ba(OH)2(aq) 90 mL of 0.1 M H2SO4(aq), rise in temperature = T2
To determine: T2 in terms of T1
Explanation:
The given reactions are acid-base neutralization reactions that liberate heat. The heat is measured in terms of a rise in temperature.
The heat liberated in a reaction is directly proportional to the number of moles of the limiting reagent. In the given reactions, the limiting reagent is H2SO4.
Let's calculate the number of moles of H2SO4 in both reactions.
In the first reaction, the volume of H2SO4 is 30 mL, which is the same as the volume of Ba(OH)2. Therefore, the number of moles of H2SO4 is:
n(H2SO4) = Molarity × Volume = 0.1 × 0.03 = 0.003 moles
In the second reaction, the volume of H2SO4 is three times the volume in the first reaction. Therefore, the number of moles of H2SO4 is:
n(H2SO4) = Molarity × Volume = 0.1 × 0.09 = 0.009 moles
As the number of moles of H2SO4 is three times greater in the second reaction, the heat liberated will also be three times greater.
Therefore, T2 = 3T1
Hence, option A is the correct answer.
Given:
30 mL of 0.1 M Ba(OH)2(aq) 30 mL of 0.1 M H2SO4(aq), rise in temperature = T1
90 mL of 0.1 M Ba(OH)2(aq) 90 mL of 0.1 M H2SO4(aq), rise in temperature = T2
To determine: T2 in terms of T1
Explanation:
The given reactions are acid-base neutralization reactions that liberate heat. The heat is measured in terms of a rise in temperature.
The heat liberated in a reaction is directly proportional to the number of moles of the limiting reagent. In the given reactions, the limiting reagent is H2SO4.
Let's calculate the number of moles of H2SO4 in both reactions.
In the first reaction, the volume of H2SO4 is 30 mL, which is the same as the volume of Ba(OH)2. Therefore, the number of moles of H2SO4 is:
n(H2SO4) = Molarity × Volume = 0.1 × 0.03 = 0.003 moles
In the second reaction, the volume of H2SO4 is three times the volume in the first reaction. Therefore, the number of moles of H2SO4 is:
n(H2SO4) = Molarity × Volume = 0.1 × 0.09 = 0.009 moles
As the number of moles of H2SO4 is three times greater in the second reaction, the heat liberated will also be three times greater.
Therefore, T2 = 3T1
Hence, option A is the correct answer.
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In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0....
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In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4(aq), rise intemperature =ΔT1II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mLof 0.1 M H2SO4(aq), rise intemperature=ΔT2. Thus,a)ΔT2 = 3ΔT1b)ΔT1 = ΔT2c)ΔT1= 3ΔT2d)ΔT1= 2ΔT2Correct answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4(aq), rise intemperature =ΔT1II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mLof 0.1 M H2SO4(aq), rise intemperature=ΔT2. Thus,a)ΔT2 = 3ΔT1b)ΔT1 = ΔT2c)ΔT1= 3ΔT2d)ΔT1= 2ΔT2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4(aq), rise intemperature =ΔT1II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mLof 0.1 M H2SO4(aq), rise intemperature=ΔT2. Thus,a)ΔT2 = 3ΔT1b)ΔT1 = ΔT2c)ΔT1= 3ΔT2d)ΔT1= 2ΔT2Correct answer is option 'A'. Can you explain this answer?.
In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4(aq), rise intemperature =ΔT1II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mLof 0.1 M H2SO4(aq), rise intemperature=ΔT2. Thus,a)ΔT2 = 3ΔT1b)ΔT1 = ΔT2c)ΔT1= 3ΔT2d)ΔT1= 2ΔT2Correct answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4(aq), rise intemperature =ΔT1II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mLof 0.1 M H2SO4(aq), rise intemperature=ΔT2. Thus,a)ΔT2 = 3ΔT1b)ΔT1 = ΔT2c)ΔT1= 3ΔT2d)ΔT1= 2ΔT2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4(aq), rise intemperature =ΔT1II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mLof 0.1 M H2SO4(aq), rise intemperature=ΔT2. Thus,a)ΔT2 = 3ΔT1b)ΔT1 = ΔT2c)ΔT1= 3ΔT2d)ΔT1= 2ΔT2Correct answer is option 'A'. Can you explain this answer?.
Solutions for In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4(aq), rise intemperature =ΔT1II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mLof 0.1 M H2SO4(aq), rise intemperature=ΔT2. Thus,a)ΔT2 = 3ΔT1b)ΔT1 = ΔT2c)ΔT1= 3ΔT2d)ΔT1= 2ΔT2Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4(aq), rise intemperature =ΔT1II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mLof 0.1 M H2SO4(aq), rise intemperature=ΔT2. Thus,a)ΔT2 = 3ΔT1b)ΔT1 = ΔT2c)ΔT1= 3ΔT2d)ΔT1= 2ΔT2Correct answer is option 'A'. Can you explain this answer?, a detailed solution for In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4(aq), rise intemperature =ΔT1II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mLof 0.1 M H2SO4(aq), rise intemperature=ΔT2. Thus,a)ΔT2 = 3ΔT1b)ΔT1 = ΔT2c)ΔT1= 3ΔT2d)ΔT1= 2ΔT2Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4(aq), rise intemperature =ΔT1II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mLof 0.1 M H2SO4(aq), rise intemperature=ΔT2. Thus,a)ΔT2 = 3ΔT1b)ΔT1 = ΔT2c)ΔT1= 3ΔT2d)ΔT1= 2ΔT2Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In the following reactions,I. 30 mL of 0.1 M Ba(OH)2(aq) + 30 mL of 0.1 M H2SO4(aq), rise intemperature =ΔT1II. 90 mL of 0.1 M Ba(OH)2(aq) + 90 mLof 0.1 M H2SO4(aq), rise intemperature=ΔT2. Thus,a)ΔT2 = 3ΔT1b)ΔT1 = ΔT2c)ΔT1= 3ΔT2d)ΔT1= 2ΔT2Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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