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15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with hydrogen With the volume was 42 ml.of sparking the resulting mixture was occupied by27 ml.to the mixture 10 ml of oxygen was added and on explosion again the volume fell to 19 ml find the formula of oxide of oxygen that was originally admitted in eudiometry both explosion led to formation.?
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15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with...
Solution:
Given,
Volume of oxide of nitrogen taken = 15 ml
Volume of hydrogen added = 42 ml
Volume of resulting mixture after sparking = 27 ml
Volume of oxygen added = 10 ml
Volume of resulting mixture after second explosion = 19 ml
Let's assume the formula of the oxide of nitrogen to be NxOy.
Step 1: Determination of the volume of nitrogen present in the oxide of nitrogen
From the given data,
Volume of nitrogen in the oxide of nitrogen = Volume of oxide of nitrogen - Volume of oxygen added
= 15 - 10
= 5 ml
Step 2: Determination of the volume of oxygen present in the oxide of nitrogen
From the given data,
Volume of resulting mixture after sparking = Volume of nitrogen + Volume of hydrogen
= 5 ml + 42 ml
= 47 ml
Let the volume of oxygen in the oxide be x ml.
Then, the volume of nitrogen in the oxide will be (5 - x) ml.
From the equation of reaction,
NxOy + (2y)H2 → N2 + (2y)H2O
The volume of nitrogen released after the explosion = volume of nitrogen in oxide = (5 - x) ml
The volume of hydrogen consumed after the explosion = (2y) x 2 = 4y ml
The volume of water vapour formed after the explosion = (2y) x 2 = 4y ml
So, the volume of nitrogen and water vapour formed after the explosion = 5 ml - 27 ml = -22 ml
Therefore, we have the equation: (5 - x) ml + 4y ml = -22 ml
Step 3: Determination of the volume of oxygen present in the oxide of nitrogen after the second explosion
From the given data,
Volume of resulting mixture after second explosion = Volume of nitrogen + Volume of hydrogen + Volume of oxygen
= (5 - x) ml + 4y ml + 10 ml
= (5 - x + 4y + 10) ml
= (15 - x + 4y) ml
From the equation of reaction,
NxOy + (2x)O2 → 2NOx
The volume of oxygen consumed after the explosion = (2x) x 2 = 4x ml
The volume of nitrogen oxide formed after the explosion = 2(5 - x) ml = 10 - 2x ml
The volume of the mixture after the explosion = 19 ml
Therefore, we have the equation: 4x ml + (10 - 2x) ml = 19 ml
Solving the two equations, we get:
x = 3 and y = 2
Therefore, the formula of the oxide of nitrogen is N2O3.
Given,
Volume of oxide of nitrogen taken = 15 ml
Volume of hydrogen added = 42 ml
Volume of resulting mixture after sparking = 27 ml
Volume of oxygen added = 10 ml
Volume of resulting mixture after second explosion = 19 ml
Let's assume the formula of the oxide of nitrogen to be NxOy.
Step 1: Determination of the volume of nitrogen present in the oxide of nitrogen
From the given data,
Volume of nitrogen in the oxide of nitrogen = Volume of oxide of nitrogen - Volume of oxygen added
= 15 - 10
= 5 ml
Step 2: Determination of the volume of oxygen present in the oxide of nitrogen
From the given data,
Volume of resulting mixture after sparking = Volume of nitrogen + Volume of hydrogen
= 5 ml + 42 ml
= 47 ml
Let the volume of oxygen in the oxide be x ml.
Then, the volume of nitrogen in the oxide will be (5 - x) ml.
From the equation of reaction,
NxOy + (2y)H2 → N2 + (2y)H2O
The volume of nitrogen released after the explosion = volume of nitrogen in oxide = (5 - x) ml
The volume of hydrogen consumed after the explosion = (2y) x 2 = 4y ml
The volume of water vapour formed after the explosion = (2y) x 2 = 4y ml
So, the volume of nitrogen and water vapour formed after the explosion = 5 ml - 27 ml = -22 ml
Therefore, we have the equation: (5 - x) ml + 4y ml = -22 ml
Step 3: Determination of the volume of oxygen present in the oxide of nitrogen after the second explosion
From the given data,
Volume of resulting mixture after second explosion = Volume of nitrogen + Volume of hydrogen + Volume of oxygen
= (5 - x) ml + 4y ml + 10 ml
= (5 - x + 4y + 10) ml
= (15 - x + 4y) ml
From the equation of reaction,
NxOy + (2x)O2 → 2NOx
The volume of oxygen consumed after the explosion = (2x) x 2 = 4x ml
The volume of nitrogen oxide formed after the explosion = 2(5 - x) ml = 10 - 2x ml
The volume of the mixture after the explosion = 19 ml
Therefore, we have the equation: 4x ml + (10 - 2x) ml = 19 ml
Solving the two equations, we get:
x = 3 and y = 2
Therefore, the formula of the oxide of nitrogen is N2O3.
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15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with hydrogen With the volume was 42 ml.of sparking the resulting mixture was occupied by27 ml.to the mixture 10 ml of oxygen was added and on explosion again the volume fell to 19 ml find the formula of oxide of oxygen that was originally admitted in eudiometry both explosion led to formation.?
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15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with hydrogen With the volume was 42 ml.of sparking the resulting mixture was occupied by27 ml.to the mixture 10 ml of oxygen was added and on explosion again the volume fell to 19 ml find the formula of oxide of oxygen that was originally admitted in eudiometry both explosion led to formation.? 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 15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with hydrogen With the volume was 42 ml.of sparking the resulting mixture was occupied by27 ml.to the mixture 10 ml of oxygen was added and on explosion again the volume fell to 19 ml find the formula of oxide of oxygen that was originally admitted in eudiometry both explosion led to formation.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with hydrogen With the volume was 42 ml.of sparking the resulting mixture was occupied by27 ml.to the mixture 10 ml of oxygen was added and on explosion again the volume fell to 19 ml find the formula of oxide of oxygen that was originally admitted in eudiometry both explosion led to formation.?.
15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with hydrogen With the volume was 42 ml.of sparking the resulting mixture was occupied by27 ml.to the mixture 10 ml of oxygen was added and on explosion again the volume fell to 19 ml find the formula of oxide of oxygen that was originally admitted in eudiometry both explosion led to formation.? 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 15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with hydrogen With the volume was 42 ml.of sparking the resulting mixture was occupied by27 ml.to the mixture 10 ml of oxygen was added and on explosion again the volume fell to 19 ml find the formula of oxide of oxygen that was originally admitted in eudiometry both explosion led to formation.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with hydrogen With the volume was 42 ml.of sparking the resulting mixture was occupied by27 ml.to the mixture 10 ml of oxygen was added and on explosion again the volume fell to 19 ml find the formula of oxide of oxygen that was originally admitted in eudiometry both explosion led to formation.?.
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15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with hydrogen With the volume was 42 ml.of sparking the resulting mixture was occupied by27 ml.to the mixture 10 ml of oxygen was added and on explosion again the volume fell to 19 ml find the formula of oxide of oxygen that was originally admitted in eudiometry both explosion led to formation.?, a detailed solution for 15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with hydrogen With the volume was 42 ml.of sparking the resulting mixture was occupied by27 ml.to the mixture 10 ml of oxygen was added and on explosion again the volume fell to 19 ml find the formula of oxide of oxygen that was originally admitted in eudiometry both explosion led to formation.? has been provided alongside types of 15 ml of oxide of Nitrogen was taken in eudiometry tube and mixed with hydrogen With the volume was 42 ml.of sparking the resulting mixture was occupied by27 ml.to the mixture 10 ml of oxygen was added and on explosion again the volume fell to 19 ml find the formula of oxide of oxygen that was originally admitted in eudiometry both explosion led to formation.? theory, EduRev gives you an
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