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Chapter 21, Problem 16P

(a)

To determine

The amount of iron in 1.00mole of iron.

(a)

Expert Solution
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Answer to Problem 16P

1.00mole of iron consists 0.05585kg of iron.

Explanation of Solution

Given Info: The cross-sectional area of iron wire is 5.00×106m2 and the electric current is 30.0A .

Amount of iron in one mole is given as,

M=55.85g/mol

It is the standard value for iron wire.

Convert the amount of iron into kg/mol .

M=55.85g/mol×(103kg/mol1g/mol)=0.05585

Thus, 1.00mole of iron consists 0.05585kg of iron.

Conclusion:

Therefore, 1.00mole of iron consists 0.05585kg of iron.

(b)

To determine

The molar density of iron.

(b)

Expert Solution
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Answer to Problem 16P

The molar density of iron is 1.41×105mol/m3 .

Explanation of Solution

Given Info: The cross-sectional area of iron wire is 5.00×106m2 and the electric current is 30.0A .

The formula for the molar density is,

MD=ρiM

Here,

ρi is the density of iron.

M is the moles.

Substitute 7.86kg/m3 for ρi and 0.05585kg/mol for M in above equation to find the MD .

MD=7.86kg/m30.05585kg/mol=1.41×105mol/m3

Thus, the molar density of iron is 1.41×105mol/m3 .

Conclusion:

Therefore, the molar density of iron is 1.41×105mol/m3 .

(c)

To determine

The number density of iron atoms.

(c)

Expert Solution
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Answer to Problem 16P

The number density of iron atoms is 8.49×1028atoms/m3

Explanation of Solution

Given Info: The cross-sectional area of iron wire is 5.00×106m2 and the electric current is 30.0A .

The formula for the number density is,

n=AMD

Here,

A is the Avogadro number.

MD is the molar density.

Substitute 1.41×105mol/m3 for MD and 6.02×1023atoms/mol for A in above equation to find n .

n=(6.02×1023atoms/mol)(1.41×105mol/m3)=8.49×1028atoms/m3

Thus, the number density of iron atoms is 8.49×1028atoms/m3 .

Conclusion:

Therefore, the number density of iron atoms is 8.49×1028atoms/m3 .

(d)

To determine

The number density of two conduction iron atoms.

(d)

Expert Solution
Check Mark

Answer to Problem 16P

The number density of two conduction iron atoms is 1.7×1029atoms/m3

Explanation of Solution

Given Info: The cross-sectional area of iron wire is 5.00×106m2 and the electric current is 30.0A .

The formula for the number density of two conduction atoms is,

n=2AMD

Here,

A is the Avogadro number.

MD is the molar density.

Substitute 1.41×105mol/m3 for MD and 6.02×1023atoms/mol for A in above equation to find n .

n=2(6.02×1023atoms/mol)(1.41×105mol/m3)=1.7×1029atoms/m3

Thus, the number density of two conduction iron atoms is 1.7×1029atoms/m3

Conclusion:

Therefore, the number density of two conduction iron atoms is 1.7×1029atoms/m3

(e)

To determine

The drift speed of the conduction electrons.

(e)

Expert Solution
Check Mark

Answer to Problem 16P

The drift speed of the conduction electrons is. 2.2×104m/s .

Explanation of Solution

Given Info: The cross-sectional area of iron wire is 5.00×106m2 and the electric current is 30.0A .

Formula to calculate the drift speed is,

Vd=InqA

Substitute 30.0A for I , 1.6×1019C for q , 1.7×1029atoms/m3 for n and 5.00×106m2 for A in above equation to find the Vd

Vd=30.0A(1.7×1029atoms/m3)×(1.6×1019C)×(5.00×106m2)=2.2×104m/s

Conclusion:

Therefore, the total drift speed of the conduction electrons is. 2.2×104m/s .

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Chapter 21 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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