Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 6, Problem 6.98P

(a)

Interpretation Introduction

Interpretation:

The joules are in 1.00 Btu is to be calculated.

Concept introduction:

The calorie (4.184J) is defined as the amount of energy required to increase the temperature of 1.00 g of liquid water by 1.00°C. The British thermal unit is defined as the amount of energy required to increase the temperature of 1.00 lb of liquid water by 1.00°F.

The conversion factor to convert lb into g is:

  1lb=453.6g

The conversion factor to convert °C into °F is:

  1°C=1.8°F

(a)

Expert Solution
Check Mark

Answer to Problem 6.98P

1.1×103J/Btu are present in 1.00 Btu.

Explanation of Solution

1.00 Btu energy in joules is calculated as follows:

  Energy=(1.00 lb °F1.00 Btu)(1.0°C1.8 °F)(453.6 g1 lb)(4.184 J1cal)(1calg°C)=1054.368J/Btu1.1×103J/Btu.

Conclusion

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.

(b)

Interpretation Introduction

Interpretation:

The joules in 1.00 therm is to be calculated.

Concept introduction:

The calorie (4.184J) is defined as the amount of energy required to increase the temperature of 1.00 g of liquid water by 1.00°C. The British thermal unit is defined as the amount of energy required to increase the temperature of 1.00 lb of liquid water by 1.00°F.

The conversion factor to convert therm into btu is:

  1therm=100,000btu

The conversion factor to convert btu into J is:

  1btu=1054.368J

(b)

Expert Solution
Check Mark

Answer to Problem 6.98P

1.1×108J are present in 1.00 therm.

Explanation of Solution

1.00 therm energy in joules is calculated as follows:

  Energy=1.00 therm(100,000btu1therm)(1054.368J1btu)=1.054368×108J1.1×108J.

Conclusion

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.

(c)

Interpretation Introduction

Interpretation:

The mole of methane that must be burned to give 1.00 therm of energy is to be calculated.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of reactant at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔHrxn°) is as follows:

  ΔHrxn°=mΔHf (products)°nΔHf (reactants)°

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

(c)

Expert Solution
Check Mark

Answer to Problem 6.98P

1.3×102mol of methane must be burned to give 1.00 therm of energy.

Explanation of Solution

The balanced chemical equation for the reaction of CH4 and O2 is as follows:

  CH4(g)+2O2(g)CO2(g)+2H2O(g)

The formula to calculate the standard enthalpy of a given reaction (ΔHrxn°) is as follows:

  ΔHrxn°=[{1ΔHf°[CO2(g)]+2ΔHf°[H2O(l)]}{1ΔHf°[CH4(g)]+2ΔHf°[O2(g)]}]        (1)

Substitute 393.5kJ/mol for ΔHf°[CO2(g)], 285.840kJ/mol for ΔHf°[H2O(l)], 74.87kJ/mol for ΔHf°[CH4(g)] and 0 for ΔHf°[O2(g)] in the equation (1).

  ΔHrxn°=[{(1mol)(393.5kJ/mol)+(2mol)(285.840kJ/mol)}{(1mol)(74.87kJ/mol)+0}]=802.282 kJ/mol

The number of moles that must be burned to give 1.00 therm of energy is calculated as follows:

  Number of moles=1.00 therm802.282 kJ/mol(1kJ1000J)(1.054368×108J1therm)=131.4211 mol1.3×102mol.

Conclusion

A combustion reaction is a reaction in which reactant is reacted with molecular oxygen to form the product.

(d)

Interpretation Introduction

Interpretation:

The cost per mole of methane is to be calculated.

Concept introduction:

The calorie (4.184J) is defined as the amount of energy required to increase the temperature of 1.00 g of liquid water by 1.00°C. The British thermal unit is defined as the amount of energy required to increase the temperature of 1.00 lb of liquid water by 1.00°F.

(d)

Expert Solution
Check Mark

Answer to Problem 6.98P

$0.0050/mol is the cost per mole of methane.

Explanation of Solution

The number of moles in 1.00 therm is 131.4211 mol.

The cost per mole of methane is calculated as follows:

  Cost=($0.66therm)(1therm131.4211 mol)=$0.005022/mol$0.0050/mol.

Conclusion

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.

(e)

Interpretation Introduction

Interpretation:

The cost to warm 318 gal of water in a hot tub from 15.0°C to 42.0°C is to be calculated.

Concept introduction:

Specific heat capacity (c) of a substance is the amount of heat needed to raise the temperature of 1g of a substance by 1K. The formula to calculate heat required is as follows:

  q=(mass)(c)(T2T1)        (2)

Here,

T2 is the final temperature.

T1 is the initial temperature.

q is the heat released or absorbed.

c is the specific heat capacity of the substance.

Density is defined as mass per unit volume. Mass and volume are physical quantities and the units of mass and volume are fundamental units. Density is the ratio of mass to the volume. The unit of volume is derived from the units of mass and volume. The SI unit of density is kg/m3. The formula to calculate density is,

  Density=MassVolume        (3)

The conversion factor to convert gal to qt is as follows:

  1gal=4qt

The conversion factor to convert qt to L is as follows:

  1L=1.057qt

(e)

Expert Solution
Check Mark

Answer to Problem 6.98P

The cost to warm 318 gal of water is $0.85.

Explanation of Solution

Rearrange the equation (3) to calculate mass of water.

  Massof water=(Density)(Volume)        (4)

Substitute 1g/mL for density and 318gal for volume in the equation (4).

  Massof water=(1g/mL)(1000mL1L)(318gal)(4qt1gal)(1 L1.057qt)=1.203406×106g

Substitute 1.203406×106g for mass, 4.184J/g°C for cwater, 42°C for T2 and 15°C for T1 in the equation (3).

  q=(1.203406×106g)(4.184J/g°C)(42°C15°C)=1.359464×108J

The cost to warm 318 gal of water is calculated as follows:

  Cost=1.359464×108J(1therm1.054368×108J)($0.66therm)=$0.850999$0.85.

Conclusion

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.

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Chapter 6 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 6.3 - When 25.0 mL of 2.00 M HNO3 and 50.0 mL of 1.00 M...Ch. 6.3 - Prob. 6.6BFPCh. 6.3 - Prob. 6.7AFPCh. 6.3 - Prob. 6.7BFPCh. 6.4 - Prob. 6.8AFPCh. 6.4 - Prob. 6.8BFPCh. 6.5 - Prob. 6.9AFPCh. 6.5 - Prob. 6.9BFPCh. 6.6 - Prob. 6.10AFPCh. 6.6 - Prob. 6.10BFPCh. 6.6 - Prob. 6.11AFPCh. 6.6 - Prob. 6.11BFPCh. 6.6 - Prob. B6.1PCh. 6.6 - Prob. B6.2PCh. 6 - Prob. 6.1PCh. 6 - Prob. 6.2PCh. 6 - Prob. 6.3PCh. 6 - Prob. 6.4PCh. 6 - Prob. 6.5PCh. 6 - Prob. 6.6PCh. 6 - Prob. 6.7PCh. 6 - Prob. 6.8PCh. 6 - Prob. 6.9PCh. 6 - A system releases 255 cal of heat to the...Ch. 6 - What is the change in internal energy (in J) of a...Ch. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Thermal decomposition of 5.0 metric tons of...Ch. 6 - Prob. 6.15PCh. 6 - The external pressure on a gas sample is 2660...Ch. 6 - The nutritional calorie (Calorie) is equivalent to...Ch. 6 - If an athlete expends 1950 kJ/h, how long does it...Ch. 6 - Prob. 6.19PCh. 6 - Hot packs used by skiers produce heat via the...Ch. 6 - Prob. 6.21PCh. 6 - Prob. 6.22PCh. 6 - For each process, state whether ΔH is less than...Ch. 6 - Prob. 6.24PCh. 6 - Prob. 6.25PCh. 6 - Prob. 6.26PCh. 6 - Prob. 6.27PCh. 6 - Prob. 6.28PCh. 6 - Prob. 6.29PCh. 6 - Prob. 6.30PCh. 6 - Prob. 6.31PCh. 6 - Prob. 6.32PCh. 6 - What data do you need to determine the specific...Ch. 6 - Is the specific heat capacity of a substance an...Ch. 6 - Prob. 6.35PCh. 6 - Both a coffee-cup calorimeter and a bomb...Ch. 6 - Find q when 22.0 g of water is heated from 25.0°C...Ch. 6 - Calculate q when 0.10 g of ice is cooled from...Ch. 6 - A 295-g aluminum engine part at an initial...Ch. 6 - Prob. 6.40PCh. 6 - Two iron bolts of equal mass—one at 100.°C, the...Ch. 6 - Prob. 6.42PCh. 6 - Prob. 6.43PCh. 6 - Prob. 6.44PCh. 6 - Prob. 6.45PCh. 6 - A 30.5-g sample of an alloy at 93.0°C is placed...Ch. 6 - When 25.0 mL of 0.500 M H2SO4 is added to 25.0 mL...Ch. 6 - Prob. 6.48PCh. 6 - Prob. 6.49PCh. 6 - A chemist places 1.750 g of ethanol, C2H6O, in a...Ch. 6 - High-purity benzoic acid (C6H5COOH; ΔH for...Ch. 6 - Two aircraft rivets, one iron and the other...Ch. 6 - A chemical engineer burned 1.520 g of a...Ch. 6 - Prob. 6.54PCh. 6 - Prob. 6.55PCh. 6 - Prob. 6.56PCh. 6 - Consider the following balanced thermochemical...Ch. 6 - Prob. 6.58PCh. 6 - Prob. 6.59PCh. 6 - When 1 mol of KBr(s) decomposes to its elements,...Ch. 6 - Prob. 6.61PCh. 6 - Compounds of boron and hydrogen are remarkable for...Ch. 6 - Prob. 6.63PCh. 6 - Prob. 6.64PCh. 6 - Prob. 6.65PCh. 6 - Prob. 6.66PCh. 6 - Prob. 6.67PCh. 6 - Prob. 6.68PCh. 6 - Prob. 6.69PCh. 6 - Prob. 6.70PCh. 6 - Prob. 6.71PCh. 6 - Write the balanced overall equation (equation 3)...Ch. 6 - Prob. 6.73PCh. 6 - Prob. 6.74PCh. 6 - Prob. 6.75PCh. 6 - Prob. 6.76PCh. 6 - Prob. 6.77PCh. 6 - Prob. 6.78PCh. 6 - Prob. 6.79PCh. 6 - Prob. 6.80PCh. 6 - Prob. 6.81PCh. 6 - Prob. 6.82PCh. 6 - Calculatefor each of the following: SiO2(s) +...Ch. 6 - Prob. 6.84PCh. 6 - Prob. 6.85PCh. 6 - The common lead-acid car battery produces a large...Ch. 6 - Prob. 6.87PCh. 6 - Prob. 6.88PCh. 6 - Prob. 6.89PCh. 6 - Prob. 6.90PCh. 6 - Prob. 6.91PCh. 6 - Prob. 6.92PCh. 6 - The following scenes represent a gaseous reaction...Ch. 6 - Prob. 6.94PCh. 6 - Prob. 6.95PCh. 6 - Prob. 6.96PCh. 6 - Prob. 6.97PCh. 6 - Prob. 6.98PCh. 6 - Prob. 6.99PCh. 6 - Prob. 6.100PCh. 6 - Prob. 6.101PCh. 6 - Prob. 6.102PCh. 6 - Prob. 6.103PCh. 6 - Prob. 6.104PCh. 6 - Prob. 6.105PCh. 6 - Prob. 6.106PCh. 6 - Liquid methanol (CH3OH) canbe used as an...Ch. 6 - Prob. 6.108P
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