Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasynumerical

Half-life period of the radioactive element A is 10 days. Amount of A left on the end of 11th day starting with 1 mole A is ${(\frac{1}{2})}^{\frac{n + 6}{10}}$ mole. What is the value of n.

Answer:

5.00

Solution:

$A_{t} = A_{0} \times {(\frac{1}{2})}^{n o o f h a l f l i f e}$
${(\frac{1}{2})}^{(\frac{n + 6}{10})} = {(\frac{1}{2})}^{(\frac{11}{10})}$
$n = 5$

Stream:NTA_ABHYASSubject:ChemistryTopic:Chemical KineticsSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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