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1. Calculate the energy released by the reaction given in sample problem 12 and compare it to the energy released by the alpha decay of californium-252. 252 Cf → 120Xe + *44! 98 54Xe + 19Ru +4 in 108 25Cf → 29Cm + He 248Cm + He Isotope Atomic Mass (u) 252Cf 98 252.081626 140y 54Xe 139.92165 108 44RU 107.91017 248 960 248.072349 He 4.001506

1. Calculate the energy released by the reaction given in sample problem 12 and compare it to the energy released by the alpha decay of californium-252. 252 Cf → 120Xe + *44! 98 54Xe + 19Ru +4 in 108 25Cf → 29Cm + He 248Cm + He Isotope Atomic Mass (u) 252Cf 98 252.081626 140y 54Xe 139.92165 108 44RU 107.91017 248 960 248.072349 He 4.001506

Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter25: Nuclear Chemistry
Section: Chapter Questions
Problem 68IL
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Please answer no. 1 and 2 in the attached image.

Note: For No. 1, Sample problem 12 is also attached in the second image.

Sample Problem 12 I The spontaneous fission of californium-252 releases xenon–140 and ruthenium- 108. How many neutrons are released in this process? > Writing the following reaction, we have 252 2Cf 140Xe + 10°Ru + ? ¿n 544 Adding the atomic numbers, we could see that the totals are balanced. For the mass numbers 252 = 140 + 108 + (x)(1) in which x is the number of neutrons. Solving for x 252 = 248 +x=x= 252 – 248 = 4 Therefore, there are four neutrons that are products of the reaction. The balanced reaction is 252, 98 140y 54Xe + 44RU +4 ¿n 108
Transcribed Image Text:Sample Problem 12 I The spontaneous fission of californium-252 releases xenon–140 and ruthenium- 108. How many neutrons are released in this process? > Writing the following reaction, we have 252 2Cf 140Xe + 10°Ru + ? ¿n 544 Adding the atomic numbers, we could see that the totals are balanced. For the mass numbers 252 = 140 + 108 + (x)(1) in which x is the number of neutrons. Solving for x 252 = 248 +x=x= 252 – 248 = 4 Therefore, there are four neutrons that are products of the reaction. The balanced reaction is 252, 98 140y 54Xe + 44RU +4 ¿n 108
1. Calculate the energy released by the reaction given in sample problem 12 and compare it to the energy released by the alpha decay of californium-252. 2Cf 140y. 252 54Xe + 9Ru +4 ¿n 25Cf → 248Cm + ŽHe Isotope Atomic Mass (u) 25Cf 252.081626 140Xe 139.92165 108 107.91017 24Cm 248.072349 He 4.001506 2. The CNO-IV cycle is a related cycle to the CNO-I cycle but is only seen on massive stars. It starts with an oxygen-18 nuclide and conducts the following steps: a hydrogen fusion with a gamma ray release, a hydrogen fusion with a release of an alpha particle, a hydrogen fusion with a gamma ray release, a positron emission, a hydrogen fusion with a gamma ray release, and a positron emission. Determine the nuclear reactions and draw a cycle that represents the CNO-IV cycle.
Transcribed Image Text:1. Calculate the energy released by the reaction given in sample problem 12 and compare it to the energy released by the alpha decay of californium-252. 2Cf 140y. 252 54Xe + 9Ru +4 ¿n 25Cf → 248Cm + ŽHe Isotope Atomic Mass (u) 25Cf 252.081626 140Xe 139.92165 108 107.91017 24Cm 248.072349 He 4.001506 2. The CNO-IV cycle is a related cycle to the CNO-I cycle but is only seen on massive stars. It starts with an oxygen-18 nuclide and conducts the following steps: a hydrogen fusion with a gamma ray release, a hydrogen fusion with a release of an alpha particle, a hydrogen fusion with a gamma ray release, a positron emission, a hydrogen fusion with a gamma ray release, and a positron emission. Determine the nuclear reactions and draw a cycle that represents the CNO-IV cycle.
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