
(a)
Interpretation:
If half-life of a radionuclide is 5.0 min, then how much fraction of the radionuclide will be present undecayed after 20 min has to be calculated.
Concept Introduction:
Radioactive nuclides undergo disintegration by emission of radiation. All the radioactive nuclide do not undergo the decay at a same rate. Some decay rapidly and others decay very slowly. The nuclear stability can be quantitatively expressed by using the half-life.
The time required for half quantity of the radioactive substance to undergo decay is known as half-life. It is represented as
The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,
(a)

Answer to Problem 11.26EP
Fraction of radionuclide that will remain after 20 min is 1/16.
Explanation of Solution
Half-life of the radionuclide is given as 5.0 min. The number of half-lives can be calculated as shown below,
The fraction of nuclide that remains after 20 min is calculated as shown below,
The fraction of sample that remains after 20 min is calculated as 1/16.
The fraction of the radionuclide sample that remains after 20 min is calculated.
(b)
Interpretation:
If half-life of a radionuclide is 5.0 min, then how much fraction of the radionuclide will be present undecayed after 30 min has to be calculated.
Concept Introduction:
Radioactive nuclides undergo disintegration by emission of radiation. All the radioactive nuclide do not undergo the decay at a same rate. Some decay rapidly and others decay very slowly. The nuclear stability can be quantitatively expressed by using the half-life.
The time required for half quantity of the radioactive substance to undergo decay is known as half-life. It is represented as
The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,
(b)

Answer to Problem 11.26EP
Fraction of radionuclide that will remain after 30 min is 1/64.
Explanation of Solution
Half-life of the radionuclide is given as 5.0 min. The number of half-lives can be calculated as shown below,
The fraction of nuclide that remains after 30 min is calculated as shown below,
The fraction of sample that remains after 30 min is calculated as 1/64.
The fraction of the radionuclide sample that remains after 30 min is calculated.
(c)
Interpretation:
If half-life of a radionuclide is 5.0 min, then how much fraction of the radionuclide will be present undecayed after 3 half-lives has to be calculated.
Concept Introduction:
Radioactive nuclides undergo disintegration by emission of radiation. All the radioactive nuclide do not undergo the decay at a same rate. Some decay rapidly and others decay very slowly. The nuclear stability can be quantitatively expressed by using the half-life.
The time required for half quantity of the radioactive substance to undergo decay is known as half-life. It is represented as
The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,
(c)

Answer to Problem 11.26EP
Fraction of radionuclide that will remain after 3 half-lives is 1/8.
Explanation of Solution
Given number of half-lives is 3 half-lives.
The fraction of nuclide that remains after 3 half-lives is calculated as shown below,
The fraction of sample that remains after 3 half-lives is calculated as 1/8.
The fraction of the radionuclide sample that remains after 3 half-lives is calculated.
(d)
Interpretation:
If half-life of a radionuclide is 5.0 min, then how much fraction of the radionuclide will be present undecayed after 8 half-lives has to be calculated.
Concept Introduction:
Radioactive nuclides undergo disintegration by emission of radiation. All the radioactive nuclide do not undergo the decay at a same rate. Some decay rapidly and others decay very slowly. The nuclear stability can be quantitatively expressed by using the half-life.
The time required for half quantity of the radioactive substance to undergo decay is known as half-life. It is represented as
The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,
(d)

Answer to Problem 11.26EP
Fraction of radionuclide that will remain after 8 half-lives is 1/256.
Explanation of Solution
Given number of half-lives is 8 half-lives.
The fraction of nuclide that remains after 8 half-lives is calculated as shown below,
The fraction of sample that remains after 8 half-lives is calculated as 1/256.
The fraction of the radionuclide sample that remains after 8 half-lives is calculated.
Want to see more full solutions like this?
Chapter 11 Solutions
Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th
- Could you please solve the first problem in this way and present it similarly but color-coded or step by step so I can understand it better? Thank you!arrow_forwardCould you please solve the first problem in this way and present it similarly but color-coded or step by step so I can understand it better? Thank you!arrow_forwardCould you please solve the first problem in this way and present it similarly but (color-coded) and step by step so I can understand it better? Thank you! I want to see what they are doingarrow_forward
- Can you please help mne with this problem. Im a visual person, so can you redraw it, potentislly color code and then as well explain it. I know im given CO2 use that to explain to me, as well as maybe give me a second example just to clarify even more with drawings (visuals) and explanations.arrow_forwardPart 1. Aqueous 0.010M AgNO 3 is slowly added to a 50-ml solution containing both carbonate [co32-] = 0.105 M and sulfate [soy] = 0.164 M anions. Given the ksp of Ag2CO3 and Ag₂ soy below. Answer the ff: Ag₂ CO3 = 2 Ag+ caq) + co} (aq) ksp = 8.10 × 10-12 Ag₂SO4 = 2Ag+(aq) + soy² (aq) ksp = 1.20 × 10-5 a) which salt will precipitate first? (b) What % of the first anion precipitated will remain in the solution. by the time the second anion starts to precipitate? (c) What is the effect of low pH (more acidic) condition on the separate of the carbonate and sulfate anions via silver precipitation? What is the effect of high pH (more basic)? Provide appropriate explanation per answerarrow_forwardPart 4. Butanoic acid (ka= 1.52× 10-5) has a partition coefficient of 3.0 (favors benzene) when distributed bet. water and benzene. What is the formal concentration of butanoic acid in each phase when 0.10M aqueous butanoic acid is extracted w❘ 25 mL of benzene 100 mL of a) at pit 5.00 b) at pH 9.00arrow_forward
- Calculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 Group of answer choices 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 choices: 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0arrow_forward
- General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage LearningChemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage LearningChemistry by OpenStax (2015-05-04)ChemistryISBN:9781938168390Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark BlaserPublisher:OpenStax
- Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage LearningWorld of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage Learning




