The following table needs to be completed with the particle produced in reaction. Initial nuclide Product nuclide Particle produced 94 239 P u 92 235 U 82 214 P b 83 214 B i 27 60 C o 99 60 N i 43 99 T c 44 99 R u 93 239 N p 94 239 P u Concept introduction: The fusion reaction or nuclear reaction can be defined as the reaction which follows the combination of two or more atomic nuclei and this result in the formation of one or more different atomic nuclei and subatomic particle. There is difference of the mass between the reactant and product which results the release or absorption of energy.

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Answer 1

Question

Chapter 20, Problem 71AE

Interpretation Introduction

Interpretation:The following table needs to be completed with the particle produced in reaction.

Initial nuclide Product nuclide Particle produced
94239Pu 92235U
82214Pb 83214Bi
2760Co 9960Ni
4399Tc 4499Ru
93239Np 94239Pu

Concept introduction:

The fusion reaction or nuclear reaction can be defined as the reaction which follows the combination of two or more atomic nuclei and this result in the formation of one or more different atomic nuclei and subatomic particle. There is difference of the mass between the reactant and product which results the release or absorption of energy.

Expert Solution & Answer

Answer to Problem 71AE

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Explanation of Solution

The unstable radioactive nucleus will undergo disintegration and will also emit particle such as alpha particle α or 24He , beta particle β or −10e and gamma particle γ for the stable configuration.

The conservation of mass and atomic number of any nuclear reaction can be given by the summation of summation of mass number of reactants which is equal to the summation of the mass number of products.

The nuclear reaction of Pu can be given as −

94239Pu→92235U+24He

Now, from the above reaction, the particle produced = 24He

The nuclear reaction of Pb can be given as −

82214Pb→83214Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Co can be given as −

2760Co→2868Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

4399Tc→4499Ru+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

93239Np→94239Pu+−10e

Now, from the above reaction, the particle produced = −10e

Now, the table can be given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Conclusion

It can concluded that the complete table of the particle produced in reaction is given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

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1) a) Give the dominant Intermolecular Force (IMF) in a sample of each of the following compounds. Please show your work. (8) SF2, CH,OH, C₂H₂ b) Based on your answers given above, list the compounds in order of their Boiling Point from low to high. (8)

19.78 Write the products of the following sequences of reactions. Refer to your reaction road- maps to see how the combined reactions allow you to "navigate" between the different functional groups. Note that you will need your old Chapters 6-11 and Chapters 15-18 roadmaps along with your new Chapter 19 roadmap for these. (a) 1. BHS 2. H₂O₂ 3. H₂CrO4 4. SOCI₂ (b) 1. Cl₂/hv 2. KOLBU 3. H₂O, catalytic H₂SO4 4. H₂CrO4 Reaction Roadmap An alkene 5. EtOH 6.0.5 Equiv. NaOEt/EtOH 7. Mild H₂O An alkane 1.0 2. (CH3)₂S 3. H₂CrO (d) (c) 4. Excess EtOH, catalytic H₂SO OH 4. Mild H₂O* 5.0.5 Equiv. NaOEt/EtOH An alkene 6. Mild H₂O* A carboxylic acid 7. Mild H₂O* 1. SOC₁₂ 2. EtOH 3.0.5 Equiv. NaOEt/E:OH 5.1.0 Equiv. NaOEt 6. NH₂ (e) 1. 0.5 Equiv. NaOEt/EtOH 2. Mild H₂O* Br (f) i H An aldehyde 1. Catalytic NaOE/EtOH 2. H₂O*, heat 3. (CH,CH₂)₂Culi 4. Mild H₂O* 5.1.0 Equiv. LDA Br An ester 4. NaOH, H₂O 5. Mild H₂O* 6. Heat 7. MgBr 8. Mild H₂O* 7. Mild H₂O+

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Answer 2

Question

Chapter 20, Problem 71AE

Interpretation Introduction

Interpretation:The following table needs to be completed with the particle produced in reaction.

Initial nuclide Product nuclide Particle produced
94239Pu 92235U
82214Pb 83214Bi
2760Co 9960Ni
4399Tc 4499Ru
93239Np 94239Pu

Concept introduction:

The fusion reaction or nuclear reaction can be defined as the reaction which follows the combination of two or more atomic nuclei and this result in the formation of one or more different atomic nuclei and subatomic particle. There is difference of the mass between the reactant and product which results the release or absorption of energy.

Expert Solution & Answer

Answer to Problem 71AE

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Explanation of Solution

The unstable radioactive nucleus will undergo disintegration and will also emit particle such as alpha particle α or 24He , beta particle β or −10e and gamma particle γ for the stable configuration.

The conservation of mass and atomic number of any nuclear reaction can be given by the summation of summation of mass number of reactants which is equal to the summation of the mass number of products.

The nuclear reaction of Pu can be given as −

94239Pu→92235U+24He

Now, from the above reaction, the particle produced = 24He

The nuclear reaction of Pb can be given as −

82214Pb→83214Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Co can be given as −

2760Co→2868Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

4399Tc→4499Ru+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

93239Np→94239Pu+−10e

Now, from the above reaction, the particle produced = −10e

Now, the table can be given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Conclusion

It can concluded that the complete table of the particle produced in reaction is given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

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Answer 3

Question

Chapter 20, Problem 71AE

Interpretation Introduction

Interpretation:The following table needs to be completed with the particle produced in reaction.

Initial nuclide Product nuclide Particle produced
94239Pu 92235U
82214Pb 83214Bi
2760Co 9960Ni
4399Tc 4499Ru
93239Np 94239Pu

Concept introduction:

The fusion reaction or nuclear reaction can be defined as the reaction which follows the combination of two or more atomic nuclei and this result in the formation of one or more different atomic nuclei and subatomic particle. There is difference of the mass between the reactant and product which results the release or absorption of energy.

Expert Solution & Answer

Answer to Problem 71AE

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Explanation of Solution

The unstable radioactive nucleus will undergo disintegration and will also emit particle such as alpha particle α or 24He , beta particle β or −10e and gamma particle γ for the stable configuration.

The conservation of mass and atomic number of any nuclear reaction can be given by the summation of summation of mass number of reactants which is equal to the summation of the mass number of products.

The nuclear reaction of Pu can be given as −

94239Pu→92235U+24He

Now, from the above reaction, the particle produced = 24He

The nuclear reaction of Pb can be given as −

82214Pb→83214Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Co can be given as −

2760Co→2868Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

4399Tc→4499Ru+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

93239Np→94239Pu+−10e

Now, from the above reaction, the particle produced = −10e

Now, the table can be given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Conclusion

It can concluded that the complete table of the particle produced in reaction is given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

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Answer 4

Question

Chapter 20, Problem 71AE

Interpretation Introduction

Interpretation:The following table needs to be completed with the particle produced in reaction.

Initial nuclide Product nuclide Particle produced
94239Pu 92235U
82214Pb 83214Bi
2760Co 9960Ni
4399Tc 4499Ru
93239Np 94239Pu

Concept introduction:

The fusion reaction or nuclear reaction can be defined as the reaction which follows the combination of two or more atomic nuclei and this result in the formation of one or more different atomic nuclei and subatomic particle. There is difference of the mass between the reactant and product which results the release or absorption of energy.

Expert Solution & Answer

Answer to Problem 71AE

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Explanation of Solution

The unstable radioactive nucleus will undergo disintegration and will also emit particle such as alpha particle α or 24He , beta particle β or −10e and gamma particle γ for the stable configuration.

The conservation of mass and atomic number of any nuclear reaction can be given by the summation of summation of mass number of reactants which is equal to the summation of the mass number of products.

The nuclear reaction of Pu can be given as −

94239Pu→92235U+24He

Now, from the above reaction, the particle produced = 24He

The nuclear reaction of Pb can be given as −

82214Pb→83214Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Co can be given as −

2760Co→2868Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

4399Tc→4499Ru+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

93239Np→94239Pu+−10e

Now, from the above reaction, the particle produced = −10e

Now, the table can be given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Conclusion

It can concluded that the complete table of the particle produced in reaction is given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

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Answer 5

Chapter 20, Problem 71AE

Interpretation Introduction

Interpretation:The following table needs to be completed with the particle produced in reaction.

Initial nuclide Product nuclide Particle produced
94239Pu 92235U
82214Pb 83214Bi
2760Co 9960Ni
4399Tc 4499Ru
93239Np 94239Pu

Concept introduction:

The fusion reaction or nuclear reaction can be defined as the reaction which follows the combination of two or more atomic nuclei and this result in the formation of one or more different atomic nuclei and subatomic particle. There is difference of the mass between the reactant and product which results the release or absorption of energy.

Expert Solution & Answer

Answer to Problem 71AE

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Explanation of Solution

The unstable radioactive nucleus will undergo disintegration and will also emit particle such as alpha particle α or 24He , beta particle β or −10e and gamma particle γ for the stable configuration.

The conservation of mass and atomic number of any nuclear reaction can be given by the summation of summation of mass number of reactants which is equal to the summation of the mass number of products.

The nuclear reaction of Pu can be given as −

94239Pu→92235U+24He

Now, from the above reaction, the particle produced = 24He

The nuclear reaction of Pb can be given as −

82214Pb→83214Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Co can be given as −

2760Co→2868Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

4399Tc→4499Ru+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

93239Np→94239Pu+−10e

Now, from the above reaction, the particle produced = −10e

Now, the table can be given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Conclusion

It can concluded that the complete table of the particle produced in reaction is given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Answer 6

Chapter 20, Problem 71AE

Interpretation Introduction

Interpretation:The following table needs to be completed with the particle produced in reaction.

Initial nuclide Product nuclide Particle produced
94239Pu 92235U
82214Pb 83214Bi
2760Co 9960Ni
4399Tc 4499Ru
93239Np 94239Pu

Concept introduction:

The fusion reaction or nuclear reaction can be defined as the reaction which follows the combination of two or more atomic nuclei and this result in the formation of one or more different atomic nuclei and subatomic particle. There is difference of the mass between the reactant and product which results the release or absorption of energy.

Expert Solution & Answer

Answer to Problem 71AE

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Explanation of Solution

The unstable radioactive nucleus will undergo disintegration and will also emit particle such as alpha particle α or 24He , beta particle β or −10e and gamma particle γ for the stable configuration.

The conservation of mass and atomic number of any nuclear reaction can be given by the summation of summation of mass number of reactants which is equal to the summation of the mass number of products.

The nuclear reaction of Pu can be given as −

94239Pu→92235U+24He

Now, from the above reaction, the particle produced = 24He

The nuclear reaction of Pb can be given as −

82214Pb→83214Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Co can be given as −

2760Co→2868Bi+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

4399Tc→4499Ru+−10e

Now, from the above reaction, the particle produced = −10e

The nuclear reaction of Tc can be given as −

93239Np→94239Pu+−10e

Now, from the above reaction, the particle produced = −10e

Now, the table can be given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Conclusion

It can concluded that the complete table of the particle produced in reaction is given as −

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Answer 7

Chapter 20, Problem 71AE

Interpretation Introduction

Interpretation:The following table needs to be completed with the particle produced in reaction.

Initial nuclide Product nuclide Particle produced
94239Pu 92235U
82214Pb 83214Bi
2760Co 9960Ni
4399Tc 4499Ru
93239Np 94239Pu

Concept introduction:

The fusion reaction or nuclear reaction can be defined as the reaction which follows the combination of two or more atomic nuclei and this result in the formation of one or more different atomic nuclei and subatomic particle. There is difference of the mass between the reactant and product which results the release or absorption of energy.

Answer 8

Expert Solution & Answer

Answer to Problem 71AE

Initial nuclide	Product nuclide	Particle produced
94239Pu	92235U	24He
82214Pb	83214Bi	−10e
2760Co	9960Ni	−10e
4399Tc	4499Ru	−10e
93239Np	94239Pu	−10e

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The unstable radioactive nucleus will undergo disintegration and will also emit particle such as alpha particle α or 24He , beta particle β or −10e and gamma particle γ for the stable configuration.

The conservation of mass and atomic number of any nuclear reaction can be given by the summation of summation of mass number of reactants which is equal to the summation of the mass number of products.

The nuclear reaction of Pu can be given as −

94239Pu→92235U+24He