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The half life of a substance in a certain enzymecatalysedreaction is 138s. The time required forthe concentration of the substance to fall from1.28 mg L–1 to 0.04 mg L–1, is : [2011]
- a)414 s
- b)552 s
- c)690 s
- d)276 s
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The half life of a substance in a certain enzymecatalysedreaction is 1...
For a first order reaction
Total time T = no. of half lives (n) × half life
(t1/2)
Total time T = no. of half lives (n) × half life
(t1/2)
where n = no. of half lives
Give N0 (original amount) = 1.28 mg/ ℓ
N (amount of substance left after time T)
= 0.04 m/g l
Give N0 (original amount) = 1.28 mg/ ℓ
N (amount of substance left after time T)
= 0.04 m/g l
n = 5
T = 5 × 138
= 690
T = 5 × 138
= 690
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The half life of a substance in a certain enzymecatalysedreaction is 1...
To find the time required for the concentration of the substance to fall from 1.28 mg/L to 0.32 mg/L, we need to calculate the number of half-lives it takes for this decrease to occur.
The concentration of a substance after a certain number of half-lives can be calculated using the formula:
C = C0 * (1/2)^(t/h)
where C is the final concentration, C0 is the initial concentration, t is the time, and h is the half-life.
In this case, the initial concentration (C0) is 1.28 mg/L, the final concentration (C) is 0.32 mg/L, and the half-life (h) is 138 s.
0.32 = 1.28 * (1/2)^(t/138)
Divide both sides of the equation by 1.28:
0.25 = (1/2)^(t/138)
Take the logarithm of both sides of the equation to solve for t:
log(0.25) = (t/138) * log(1/2)
Using the logarithm property log(a^b) = b * log(a), we can rewrite the equation as:
log(0.25) / log(1/2) = t/138
Simplifying the left side of the equation:
log(0.25) / log(1/2) = t/138
Using the change of base formula log(a) / log(b) = log base b of a, we can rewrite the equation as:
log base 1/2 of 0.25 = t/138
Using a calculator to evaluate the logarithm:
log base 1/2 of 0.25 = t/138
-2 = t/138
Multiply both sides of the equation by 138:
-2 * 138 = t
t = -276
The time required for the concentration of the substance to fall from 1.28 mg/L to 0.32 mg/L is 276 seconds.
The concentration of a substance after a certain number of half-lives can be calculated using the formula:
C = C0 * (1/2)^(t/h)
where C is the final concentration, C0 is the initial concentration, t is the time, and h is the half-life.
In this case, the initial concentration (C0) is 1.28 mg/L, the final concentration (C) is 0.32 mg/L, and the half-life (h) is 138 s.
0.32 = 1.28 * (1/2)^(t/138)
Divide both sides of the equation by 1.28:
0.25 = (1/2)^(t/138)
Take the logarithm of both sides of the equation to solve for t:
log(0.25) = (t/138) * log(1/2)
Using the logarithm property log(a^b) = b * log(a), we can rewrite the equation as:
log(0.25) / log(1/2) = t/138
Simplifying the left side of the equation:
log(0.25) / log(1/2) = t/138
Using the change of base formula log(a) / log(b) = log base b of a, we can rewrite the equation as:
log base 1/2 of 0.25 = t/138
Using a calculator to evaluate the logarithm:
log base 1/2 of 0.25 = t/138
-2 = t/138
Multiply both sides of the equation by 138:
-2 * 138 = t
t = -276
The time required for the concentration of the substance to fall from 1.28 mg/L to 0.32 mg/L is 276 seconds.
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The half life of a substance in a certain enzymecatalysedreaction is 138s. The time required forthe concentration of the substance to fall from1.28 mg L–1 to 0.04 mg L–1, is : [2011]a)414 sb)552 sc)690 sd)276 sCorrect answer is option 'C'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The half life of a substance in a certain enzymecatalysedreaction is 138s. The time required forthe concentration of the substance to fall from1.28 mg L–1 to 0.04 mg L–1, is : [2011]a)414 sb)552 sc)690 sd)276 sCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The half life of a substance in a certain enzymecatalysedreaction is 138s. The time required forthe concentration of the substance to fall from1.28 mg L–1 to 0.04 mg L–1, is : [2011]a)414 sb)552 sc)690 sd)276 sCorrect answer is option 'C'. Can you explain this answer?.
The half life of a substance in a certain enzymecatalysedreaction is 138s. The time required forthe concentration of the substance to fall from1.28 mg L–1 to 0.04 mg L–1, is : [2011]a)414 sb)552 sc)690 sd)276 sCorrect answer is option 'C'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The half life of a substance in a certain enzymecatalysedreaction is 138s. The time required forthe concentration of the substance to fall from1.28 mg L–1 to 0.04 mg L–1, is : [2011]a)414 sb)552 sc)690 sd)276 sCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The half life of a substance in a certain enzymecatalysedreaction is 138s. The time required forthe concentration of the substance to fall from1.28 mg L–1 to 0.04 mg L–1, is : [2011]a)414 sb)552 sc)690 sd)276 sCorrect answer is option 'C'. Can you explain this answer?.
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The half life of a substance in a certain enzymecatalysedreaction is 138s. The time required forthe concentration of the substance to fall from1.28 mg L–1 to 0.04 mg L–1, is : [2011]a)414 sb)552 sc)690 sd)276 sCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for The half life of a substance in a certain enzymecatalysedreaction is 138s. The time required forthe concentration of the substance to fall from1.28 mg L–1 to 0.04 mg L–1, is : [2011]a)414 sb)552 sc)690 sd)276 sCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of The half life of a substance in a certain enzymecatalysedreaction is 138s. The time required forthe concentration of the substance to fall from1.28 mg L–1 to 0.04 mg L–1, is : [2011]a)414 sb)552 sc)690 sd)276 sCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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