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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 both particles A and B? 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. 210. Po and then 210. 206 Ph

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 both particles A and B? 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. 210. Po and then 210. 206 Ph

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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 both particles A and B? 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, 84Po and then 210. 84PO → B + 206 °82PB 3.852 x 10-15 kJ / particle 3.194 x 10-15 kJ / particle 3.669 x 10-15 kJ / particle 1.742 x 10-15 kJ / particle 2.764 x 10-15 kJ / particle 4.021 x 10-15 kJ / particle 2.527 x 10-15 kJ / particle 1.946 x 10-15 kJ / particle
Transcribed Image Text: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 both particles A and B? 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, 84Po and then 210. 84PO → B + 206 °82PB 3.852 x 10-15 kJ / particle 3.194 x 10-15 kJ / particle 3.669 x 10-15 kJ / particle 1.742 x 10-15 kJ / particle 2.764 x 10-15 kJ / particle 4.021 x 10-15 kJ / particle 2.527 x 10-15 kJ / particle 1.946 x 10-15 kJ / particle
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