The work done when 1.0 mol of propane gas molecules react with oxygen to produce carbon dioxide and liquid water at 25 0 C has to be determined. Concept Introduction: The force F, opposing an expansion under constant external pressure p e x , then F = p e x A The work done is the force when multiplies by the distance moved, is given by W o r k d o n e , w = F o r c e × Distance w = − p e x A × h w h e r e , h × A = Δ V ∴ w = − p e x × Δ V

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Answer 1

Question

Chapter 2, Problem 2A.1ST

Interpretation Introduction

Interpretation:

The work done when 1.0 mol of propane gas molecules react with oxygen to produce carbon dioxide and liquid water at 250C has to be determined.

Concept Introduction:

The force F, opposing an expansion under constant external pressure pex, then F= pexA

The work done is the force when multiplies by the distance moved, is given by

Work done,w=Force× Distancew= −pexA× h where,h×A=ΔV∴w=− pex×ΔV

Expert Solution & Answer

Explanation of Solution

Change in volume of gas ΔV= Vf−Vi

From Ideal gas equation, ΔV=ΔnRTpex

Here, the atmosphere exerts a constant force on the gas, as the work done by the system loses energy, so it is expected to have a negative value for work done.

w= − pex×ΔV

Substituting for ΔV in the equation for work done as,

w= −pex×ΔnRTpexw= −ΔnRT

Δn is the change in number of moles of only gas molecules is given by Δn=nf−ni=3−6=−3 though number of moles cannot be negative.

w=− ΔnRTw=−((−3 mol) × 8.314 JK−1mol−1× (25+273K))w= +7432.7J=+7.43 kJ

The expansion work when dilute hydrochloric acid is poured on to some calcium carbonate is +7.43 kJ.

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Done 11:14 ⚫ worksheets.beyondlabz.com 5 (a). Using the peak information you listed in the tables for both structures, assign each peak to that portion of the structure that produces the peak in the NMR spectrum. Draw this diagram on your own sheet of paper and attach the sketch of your drawing to this question. Question 6 5 (b). Using the peak information you listed in the tables for both structures, assign each peak to that portion of the structure that produces the peak in the NMR spectrum. Draw this diagram on your own sheet of paper and attach the sketch of your drawing to this question. Question 7 6. Are there any differences between the spectra you obtained in Beyond Labz and the predicted spectra? If so, what were the differences? <

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Interpreting NMR spectra is a skill that often requires some amount of practice, which, in turn, necessitates access to a collection of NMR spectra. Beyond Labz Organic Synthesis and Organic Qualitative Analysis have spectral libraries containing over 700 1H NMR spectra. In this assignment, you will take advantage of this by first predicting the NMR spectra for two closely related compounds and then checking your predictions by looking up the actual spectra in the spectra library. After completing this assignment, you may wish to select other compounds for additional practice. 1. Write the IUPAC names for the following two structures: Question 2 Question 3 2. Predict the NMR spectra for each of these two compounds by listing, in the NMR tables below, the chemical shift, the splitting, and the number of hydrogens associated with each predicted peak. Sort the peaks from largest chemical shift to lowest. **Not all slots must be filled**

Answer 2

Question

Chapter 2, Problem 2A.1ST

Interpretation Introduction

Interpretation:

The work done when 1.0 mol of propane gas molecules react with oxygen to produce carbon dioxide and liquid water at 250C has to be determined.

Concept Introduction:

The force F, opposing an expansion under constant external pressure pex, then F= pexA

The work done is the force when multiplies by the distance moved, is given by

Work done,w=Force× Distancew= −pexA× h where,h×A=ΔV∴w=− pex×ΔV

Expert Solution & Answer

Explanation of Solution

Change in volume of gas ΔV= Vf−Vi

From Ideal gas equation, ΔV=ΔnRTpex

Here, the atmosphere exerts a constant force on the gas, as the work done by the system loses energy, so it is expected to have a negative value for work done.

w= − pex×ΔV

Substituting for ΔV in the equation for work done as,

w= −pex×ΔnRTpexw= −ΔnRT

Δn is the change in number of moles of only gas molecules is given by Δn=nf−ni=3−6=−3 though number of moles cannot be negative.

w=− ΔnRTw=−((−3 mol) × 8.314 JK−1mol−1× (25+273K))w= +7432.7J=+7.43 kJ

The expansion work when dilute hydrochloric acid is poured on to some calcium carbonate is +7.43 kJ.

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

Answer 3

Question

Chapter 2, Problem 2A.1ST

Interpretation Introduction

Interpretation:

The work done when 1.0 mol of propane gas molecules react with oxygen to produce carbon dioxide and liquid water at 250C has to be determined.

Concept Introduction:

The force F, opposing an expansion under constant external pressure pex, then F= pexA

The work done is the force when multiplies by the distance moved, is given by

Work done,w=Force× Distancew= −pexA× h where,h×A=ΔV∴w=− pex×ΔV

Expert Solution & Answer

Explanation of Solution

Change in volume of gas ΔV= Vf−Vi

From Ideal gas equation, ΔV=ΔnRTpex

Here, the atmosphere exerts a constant force on the gas, as the work done by the system loses energy, so it is expected to have a negative value for work done.

w= − pex×ΔV

Substituting for ΔV in the equation for work done as,

w= −pex×ΔnRTpexw= −ΔnRT

Δn is the change in number of moles of only gas molecules is given by Δn=nf−ni=3−6=−3 though number of moles cannot be negative.

w=− ΔnRTw=−((−3 mol) × 8.314 JK−1mol−1× (25+273K))w= +7432.7J=+7.43 kJ

The expansion work when dilute hydrochloric acid is poured on to some calcium carbonate is +7.43 kJ.

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

Answer 4

Question

Chapter 2, Problem 2A.1ST

Interpretation Introduction

Interpretation:

The work done when 1.0 mol of propane gas molecules react with oxygen to produce carbon dioxide and liquid water at 250C has to be determined.

Concept Introduction:

The force F, opposing an expansion under constant external pressure pex, then F= pexA

The work done is the force when multiplies by the distance moved, is given by

Work done,w=Force× Distancew= −pexA× h where,h×A=ΔV∴w=− pex×ΔV

Expert Solution & Answer

Explanation of Solution

Change in volume of gas ΔV= Vf−Vi

From Ideal gas equation, ΔV=ΔnRTpex

Here, the atmosphere exerts a constant force on the gas, as the work done by the system loses energy, so it is expected to have a negative value for work done.

w= − pex×ΔV

Substituting for ΔV in the equation for work done as,

w= −pex×ΔnRTpexw= −ΔnRT

Δn is the change in number of moles of only gas molecules is given by Δn=nf−ni=3−6=−3 though number of moles cannot be negative.

w=− ΔnRTw=−((−3 mol) × 8.314 JK−1mol−1× (25+273K))w= +7432.7J=+7.43 kJ

The expansion work when dilute hydrochloric acid is poured on to some calcium carbonate is +7.43 kJ.

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

Answer 5

Chapter 2, Problem 2A.1ST

Interpretation Introduction

Interpretation:

The work done when 1.0 mol of propane gas molecules react with oxygen to produce carbon dioxide and liquid water at 250C has to be determined.

Concept Introduction:

The force F, opposing an expansion under constant external pressure pex, then F= pexA

The work done is the force when multiplies by the distance moved, is given by

Work done,w=Force× Distancew= −pexA× h where,h×A=ΔV∴w=− pex×ΔV

Expert Solution & Answer

Explanation of Solution

Change in volume of gas ΔV= Vf−Vi

From Ideal gas equation, ΔV=ΔnRTpex

Here, the atmosphere exerts a constant force on the gas, as the work done by the system loses energy, so it is expected to have a negative value for work done.

w= − pex×ΔV

Substituting for ΔV in the equation for work done as,

w= −pex×ΔnRTpexw= −ΔnRT

Δn is the change in number of moles of only gas molecules is given by Δn=nf−ni=3−6=−3 though number of moles cannot be negative.

w=− ΔnRTw=−((−3 mol) × 8.314 JK−1mol−1× (25+273K))w= +7432.7J=+7.43 kJ

The expansion work when dilute hydrochloric acid is poured on to some calcium carbonate is +7.43 kJ.

Answer 6

Chapter 2, Problem 2A.1ST

Interpretation Introduction

Interpretation:

The work done when 1.0 mol of propane gas molecules react with oxygen to produce carbon dioxide and liquid water at 250C has to be determined.

Concept Introduction:

The force F, opposing an expansion under constant external pressure pex, then F= pexA

The work done is the force when multiplies by the distance moved, is given by

Work done,w=Force× Distancew= −pexA× h where,h×A=ΔV∴w=− pex×ΔV

Expert Solution & Answer

Explanation of Solution

Change in volume of gas ΔV= Vf−Vi

From Ideal gas equation, ΔV=ΔnRTpex

Here, the atmosphere exerts a constant force on the gas, as the work done by the system loses energy, so it is expected to have a negative value for work done.

w= − pex×ΔV

Substituting for ΔV in the equation for work done as,

w= −pex×ΔnRTpexw= −ΔnRT

Δn is the change in number of moles of only gas molecules is given by Δn=nf−ni=3−6=−3 though number of moles cannot be negative.

w=− ΔnRTw=−((−3 mol) × 8.314 JK−1mol−1× (25+273K))w= +7432.7J=+7.43 kJ

The expansion work when dilute hydrochloric acid is poured on to some calcium carbonate is +7.43 kJ.

Answer 7

Chapter 2, Problem 2A.1ST

Interpretation Introduction

Interpretation:

The work done when 1.0 mol of propane gas molecules react with oxygen to produce carbon dioxide and liquid water at 250C has to be determined.

Concept Introduction:

The force F, opposing an expansion under constant external pressure pex, then F= pexA

The work done is the force when multiplies by the distance moved, is given by

Work done,w=Force× Distancew= −pexA× h where,h×A=ΔV∴w=− pex×ΔV

Answer 8

Expert Solution & Answer

Explanation of Solution

Change in volume of gas ΔV= Vf−Vi

From Ideal gas equation, ΔV=ΔnRTpex

Here, the atmosphere exerts a constant force on the gas, as the work done by the system loses energy, so it is expected to have a negative value for work done.

w= − pex×ΔV

Substituting for ΔV in the equation for work done as,

w= −pex×ΔnRTpexw= −ΔnRT