Suppose there was a release of 0.572 mole of particle A and 0.792 mole of particle B and both particles are accelerated to 5.73 MeV. What is the energy in terms of kJ/mole for particle A? Hint: 1 MeV = 1 x 106 ev and 1 eV = 1.602 X 10-19 J and 1 kJ = 1000 J, and finally in 1 mole there are Avogrado's number of particles which is 6.02214 x 1023 particles/mole. 21083Bİ → A + 210, °84P0 and then 210 °84P0 → B + 206 °82PB 2.036 x 1015 kJ / mole particle A 9.403 x 10-14 kJ / mole particle A 1.829 x 10-15 kJ / mole particle A 1.207 x 10-15 kJ / mole particle A 1.605 x 10-15 kJ / mole particle A 1.412 x 10-15 kJ / mole particle A 9.807 x 10-14 kJ / mole particle A

Suppose there was a release of 0.572 mole of particle A and 0.792 mole of particle B and both particles are accelerated to 5.73 MeV. What is the energy in terms of kJ/mole for particle A? Hint: 1 MeV = 1 x 106 ev and 1 eV = 1.602 X 10-19 J and 1 kJ = 1000 J, and finally in 1 mole there are Avogrado's number of particles which is 6.02214 x 1023 particles/mole. 21083Bİ → A + 210, °84P0 and then 210 °84P0 → B + 206 °82PB 2.036 x 1015 kJ / mole particle A 9.403 x 10-14 kJ / mole particle A 1.829 x 10-15 kJ / mole particle A 1.207 x 10-15 kJ / mole particle A 1.605 x 10-15 kJ / mole particle A 1.412 x 10-15 kJ / mole particle A 9.807 x 10-14 kJ / mole particle A

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter18: Nuclear Chemistry

Section: Chapter Questions

Problem 18.CCP

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