(a) the number that goes into the elevated box (superscript) in the given nuclear reaction. (b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Question

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Answer 1

Question

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

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Students have asked these similar questions

1. An ideal gas is taken through a four process cycle abcda. State a has a pressure of 498,840 Pa. Complete the tables and plot/label all states and processes on the PV graph. Complete the states and process diagrams on the last page. Also, provide proper units for each column/row heading in the tables. Pressure (Pa) 500,000 450,000 400,000 350,000 300,000 250,000 200,000 150,000 100,000 Process ab bc cd da States P( ) V( ) 50,000 0 0.000 T = 500 K T= 200 K 0.001 0.002 0.003 0.004 0.005 Volume (m^3) Nature of Process isothermal expansion to Vb = 0.005 m³ (T = 500 K) isometric isothermal compression to V₁ = 0.003 m³ (T = 200 K) adiabatic compression to VA = 0.001 m³ b C a T() U ( ) Processes a-b Q( ) +802.852 W() AU ( ) b-c c→d +101.928 da Cycle

Plz no chatgpt I

A = 45 kN a = 60° B = 20 kN ẞ = 30° Problem:M1.1 You and your friends are on an archaeological adventure and are trying to disarm an ancient trap to do so you need to pull a log straight out of a hole in a wall. You have 1 rope that you can attach to the log and there are currently 2 other ropes and weights attached to the end of the log. You know the force and direction of the ropes currently attached are arranged as shown below what is the magnitude and direction 'e' of the minimum force you need to apply to the third rope for the force on the log to be in direction of line 'a'? What is the resultant force in direction 'a'? a ////// //////

Answer 2

Question

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

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Answer 3

Question

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Answer 6

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Answer 7

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Answer 8

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Answer 12

Ch. 43 - Prob. 1Q Ch. 43 - Prob. 2Q Ch. 43 - Prob. 3Q Ch. 43 - Prob. 4Q Ch. 43 - Prob. 5Q Ch. 43 - Prob. 6Q Ch. 43 - Prob. 7Q Ch. 43 - Which of these elements is not cooked up by...Ch. 43 - Prob. 9Q Ch. 43 - Prob. 10Q

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction. (b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Concept explainers

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Answer to Problem 1Q

Explanation of Solution

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Chapter 43 Solutions

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction. (b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Concept explainers

To find: (a) the number that goes into the elevated box (superscript) in the given nuclear reaction. (b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Answer to Problem 1Q

Explanation of Solution

Want to see more full solutions like this?

Chapter 43 Solutions

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.