Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 5, Problem 5.133AP

(a)

Interpretation Introduction

Interpretation: The unfilled rows are to be completed by multiplying each molar mass and specific heat pair. The units of the resulting values in column 4 are to be stated.

The average of the values of column 4 is to be calculated.

The missing atomic masses and their link with corresponding elements given in the table with the use of mean value from part (b) are to be stated.

The missing specific heat values with the use of average value from part (b) are to be stated.

Concept introduction: As per Dulong and Petit, the product of atomic masses (μ) and specific heat (cs) of metals are nearly constant.

μ×cs=constant

The average of the given values is calculated by the formula,

Average=Observedvalues(μ×cs)Numberofobservation(μ×cs)

The missing atomic mass from mean of μ×cs is calculated by the formula,

Missingatomicmass=Meanofμ×csSpecificheat

The missing specific heat values with the use of average value is calculated by formula,

MissingSpecificheatvalue=Meanofμ×csAtomicmass

To determine: The unfilled rows by multiplying each molar mass and specific heat pair.

(a)

Expert Solution
Check Mark

Answer to Problem 5.133AP

Solution

The solution is stated below.

Explanation of Solution

Explanation

Given

The atomic mass (μPb) of element Lead is 207.2g/mol .

The atomic mass (μAu) of element Gold is 197.0g/mol .

The atomic mass (μSn) of element Tin is 118.7g/mol .

The atomic mass (μZn) of element Zinc is 65.4g/mol .

The atomic mass (μCu) of element Copper is 63.5g/mol .

The atomic mass (μFe) of element Iron is 55.8g/mol .

The specific heat (cs,Pb) of element Lead is 0.123J/(g°C) .

The specific heat (csAu) of element Gold is 0.125J/(g°C) .

The specific heat (cs,Sn) of element Tin is 0.215J/(g°C) .

The specific heat (cs,Zn) of element Zinc is 0.388J/(g°C) .

The specific heat (cs,Cu) of element Copper is 0.397J/(g°C) .

The specific heat (cs,Fe) of element Iron is 0.460J/(g°C) .

As per Dulong and Petit, the product of atomic masses (μ) and specific heat (cs) of metals are nearly constant.

μ×cs=constant

The product of specific heat and atomic mass of metal Lead is,

=207.2g/mol×0.123J/(g°C)=25.48J/mol°C

The product of specific heat and atomic mass of metal Gold is,

=197.0g/mol×0.125J/(g°C)=24.62J/mol°C

The product of specific heat and atomic mass of metal Tin is,

=118.7g/mol×0.215J/(g°C)=25.52J/mol°C

The product of specific heat and atomic mass of metal Zinc is,

=65.4g/mol×0.388J/(g°C)=25.37J/mol°C

The product of specific heat and atomic mass of metal Copper is,

=63.5g/mol×0.397J/(g°C)=25.20J/mol°C

The product of specific heat and atomic mass of metal Iron is,

=55.8g/mol×0.460J/(g°C)=25.66J/mol°C

The product of specific heat and atomic mass of every metal has common unit J/mol°C .

All of the values of atomic mass, specific heat, and their product are summarized in Table 1 .

Element Atomic mass (μ) Specific heat (cs) (μ×cs)
Lead 207.2g/mol 0.123J/(g°C) 25.48J/mol°C
Gold 197.0g/mol 0.125J/(g°C) 24.62J/mol°C
Tin 118.7g/mol 0.215J/(g°C) 25.52J/mol°C
Zinc 65.4g/mol 0.388J/(g°C) 25.37J/mol°C
Copper 63.5g/mol 0.397J/(g°C) 25.20J/mol°C
Iron 55.8g/mol 0.460J/(g°C) 25.66J/mol°C

Table 1 .

(b)

Interpretation Introduction

To determine: The average of the values of column 4 is to be calculated.

(b)

Expert Solution
Check Mark

Answer to Problem 5.133AP

Solution

The average of the values of column 4 is 25.30J/mol°C_ .

Explanation of Solution

Explanation

The average of the given values is calculated by the formula,

Average=Observedvalues(μ×cs)Numberofobservation(μ×cs)

The number of observation of (μ×cs) are 6 .

Substitute the values of (μ×cs) present in column 4 in above expression.

Average=(25.48+24.62+25.52+25.37+25.20+25.66)J/mol°C6=151.8586=25.30J/mol°C

Hence, the average of the values of column 4 is 25.30J/mol°C_ .

(c)

Interpretation Introduction

To determine: The missing atomic masses and their link with corresponding elements given in the table with the use of mean value from part (b).

(c)

Expert Solution
Check Mark

Answer to Problem 5.133AP

Solution

The solution is stated in explanation.

Explanation of Solution

Explanation

The missing atomic mass from mean of μ×cs is calculated by the formula,

Missingatomicmass=Meanofμ×csSpecificheat (1)

The mean of μ×cs is 25.30J/mol°C . The missing atomic masses of elements are calculated below.

Bismuth element with specific heat of 0.120J/(g°C) .

Substitute the values of specific heat and the mean of μ×cs in equation (1).

Missingatomicmass=25.30J/mol°C0.120J/(g°C)=210.83g/mol

The atomic mass of Bismuth in modern periodic table is 208.98g/mol . Thus, it indicates that calculated value is close to actual value but not 100% accurate.

An unknown element with specific heat of 0.233J/(g°C) .

Substitute the values of specific heat and the mean of μ×cs in equation (1).

Missingatomicmass=25.30J/mol°C0.233J/(g°C)=108.6g/mol

The calculated atomic mass of unknown element is close to Silver (Ag) which has atomic mass of 107.86g/mol in modern periodic table. Thus, it indicates that calculated value is close to actual value, but not 100% accurate.

An unknown element with specific heat of 0.433J/(g°C) .

Substitute the values of specific heat and the mean of μ×cs in equation (1).

Missingatomicmass=25.30J/mol°C0.433J/(g°C)=58.4g/mol

The calculated atomic mass of unknown element is close to Silver (Ag) which has atomic mass of 58.69g/mol in modern periodic table. Thus, it indicates that calculated value is close to actual value, but not 100% accurate.

Hence, the Dulong and Pettit assumptions about atomic masses are not 100% accurate.

(d)

Interpretation Introduction

To determine: The missing specific heat values with the use of average value from part (b).

(d)

Expert Solution
Check Mark

Answer to Problem 5.133AP

Solution

The missing specific heat values of Platinum and Sulfur are 0.129J/(g×°C)_ and 0.788J/(g×°C)_ .

Explanation of Solution

Explanation

The missing specific heat values with the use of average value is calculated by formula,

MissingSpecificheatvalue=Meanofμ×csAtomicmass (2)

The mean of μ×cs is 25.30J/mol°C . The specific heat values are calculated below.

Platinum with atomic mass of 195.1g/mol .

Substitute the values of atomic mass and the mean of μ×cs in equation (2).

MissingSpecificheatvalue=25.30J/mol°C195.1g/mol=0.129J/(g°C)

Sulfur with atomic mass of 32.1g/mol .

Substitute the values of atomic mass and the mean of μ×cs in equation (2).

MissingSpecificheatvalue=25.30J/mol°C32.1g/mol=0.788J/(g°C)

Hence, the missing specific heat values of Platinum and Sulfur are 0.129J/(g×°C)_ and 0.788J/(g×°C)_ .

Conclusion

  1. a. The product of specific heat and atomic mass of every metal has common unit J/mol°C .
  2. b. The average of the values of column 4 is 25.30J/mol°C_ .
  3. c. The Dulong and Pettit assumptions about atomic masses are not 100% accurate.
  4. d. The missing specific heat values of Platinum and Sulfur are 0.129J/(g×°C)_ and 0.788J/(g×°C)_

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Chapter 5 Solutions

Chemistry

Ch. 5.5 - Prob. 11PECh. 5.6 - Prob. 12PECh. 5.6 - Prob. 13PECh. 5.7 - Prob. 14PECh. 5.7 - Prob. 15PECh. 5.7 - Prob. 16PECh. 5.8 - Prob. 17PECh. 5 - Prob. 5.1VPCh. 5 - Prob. 5.2VPCh. 5 - Prob. 5.3VPCh. 5 - Prob. 5.4VPCh. 5 - Prob. 5.5VPCh. 5 - Prob. 5.6VPCh. 5 - Prob. 5.7VPCh. 5 - Prob. 5.8VPCh. 5 - Prob. 5.9QPCh. 5 - Prob. 5.10QPCh. 5 - Prob. 5.11QPCh. 5 - Prob. 5.12QPCh. 5 - Prob. 5.13QPCh. 5 - Prob. 5.14QPCh. 5 - Prob. 5.15QPCh. 5 - Prob. 5.16QPCh. 5 - Prob. 5.17QPCh. 5 - Prob. 5.18QPCh. 5 - Prob. 5.19QPCh. 5 - Prob. 5.20QPCh. 5 - Prob. 5.21QPCh. 5 - Prob. 5.22QPCh. 5 - Prob. 5.23QPCh. 5 - Prob. 5.24QPCh. 5 - Prob. 5.25QPCh. 5 - Prob. 5.26QPCh. 5 - Prob. 5.27QPCh. 5 - Prob. 5.28QPCh. 5 - Prob. 5.29QPCh. 5 - Prob. 5.30QPCh. 5 - Prob. 5.31QPCh. 5 - Prob. 5.32QPCh. 5 - Prob. 5.33QPCh. 5 - Prob. 5.34QPCh. 5 - Prob. 5.35QPCh. 5 - Prob. 5.36QPCh. 5 - Prob. 5.37QPCh. 5 - Prob. 5.38QPCh. 5 - Prob. 5.39QPCh. 5 - Prob. 5.40QPCh. 5 - Prob. 5.41QPCh. 5 - Prob. 5.42QPCh. 5 - Prob. 5.43QPCh. 5 - Prob. 5.44QPCh. 5 - Prob. 5.45QPCh. 5 - Prob. 5.46QPCh. 5 - Prob. 5.47QPCh. 5 - Prob. 5.48QPCh. 5 - Prob. 5.49QPCh. 5 - Prob. 5.50QPCh. 5 - Prob. 5.51QPCh. 5 - Prob. 5.52QPCh. 5 - Prob. 5.53QPCh. 5 - Prob. 5.54QPCh. 5 - Prob. 5.55QPCh. 5 - Prob. 5.56QPCh. 5 - Prob. 5.57QPCh. 5 - Prob. 5.58QPCh. 5 - Prob. 5.59QPCh. 5 - Prob. 5.60QPCh. 5 - Prob. 5.61QPCh. 5 - Prob. 5.62QPCh. 5 - Prob. 5.65QPCh. 5 - Prob. 5.66QPCh. 5 - Prob. 5.67QPCh. 5 - Prob. 5.68QPCh. 5 - Prob. 5.69QPCh. 5 - Prob. 5.70QPCh. 5 - Prob. 5.71QPCh. 5 - Prob. 5.72QPCh. 5 - Prob. 5.73QPCh. 5 - Prob. 5.74QPCh. 5 - Prob. 5.75QPCh. 5 - Prob. 5.76QPCh. 5 - Prob. 5.77QPCh. 5 - Prob. 5.78QPCh. 5 - Prob. 5.79QPCh. 5 - Prob. 5.80QPCh. 5 - Prob. 5.81QPCh. 5 - Prob. 5.82QPCh. 5 - Prob. 5.83QPCh. 5 - Prob. 5.84QPCh. 5 - Prob. 5.85QPCh. 5 - Prob. 5.86QPCh. 5 - Prob. 5.87QPCh. 5 - Prob. 5.88QPCh. 5 - Prob. 5.89QPCh. 5 - Prob. 5.90QPCh. 5 - Prob. 5.91QPCh. 5 - Prob. 5.92QPCh. 5 - Prob. 5.93QPCh. 5 - Prob. 5.94QPCh. 5 - Prob. 5.95QPCh. 5 - Prob. 5.96QPCh. 5 - Prob. 5.97QPCh. 5 - Prob. 5.98QPCh. 5 - Prob. 5.99QPCh. 5 - Prob. 5.100QPCh. 5 - Prob. 5.101QPCh. 5 - Prob. 5.102QPCh. 5 - Prob. 5.103APCh. 5 - Prob. 5.104APCh. 5 - Prob. 5.105APCh. 5 - Prob. 5.106APCh. 5 - Prob. 5.107APCh. 5 - Prob. 5.108APCh. 5 - Prob. 5.109APCh. 5 - Prob. 5.110APCh. 5 - Prob. 5.111APCh. 5 - Prob. 5.112APCh. 5 - Prob. 5.113APCh. 5 - Prob. 5.114APCh. 5 - Prob. 5.115APCh. 5 - Prob. 5.116APCh. 5 - Prob. 5.117APCh. 5 - Prob. 5.118APCh. 5 - Prob. 5.119APCh. 5 - Prob. 5.120APCh. 5 - Prob. 5.121APCh. 5 - Prob. 5.122APCh. 5 - Prob. 5.123APCh. 5 - Prob. 5.124APCh. 5 - Prob. 5.125APCh. 5 - Prob. 5.126APCh. 5 - Prob. 5.127APCh. 5 - Prob. 5.128APCh. 5 - Prob. 5.129APCh. 5 - Prob. 5.130APCh. 5 - Prob. 5.131APCh. 5 - Prob. 5.132APCh. 5 - Prob. 5.133APCh. 5 - Prob. 5.134APCh. 5 - Prob. 5.135APCh. 5 - Prob. 5.136APCh. 5 - Prob. 5.137APCh. 5 - Prob. 5.138APCh. 5 - Prob. 5.139APCh. 5 - Prob. 5.140APCh. 5 - Prob. 5.141APCh. 5 - Prob. 5.142APCh. 5 - Prob. 5.143AP
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