The mass of mercury in grams contained in each mL of lake water is to be calculated. Also, a road map for the solution is to be drawn. Concept introduction: Meter square ( m 2 ) is the SI unit of the area. The English equivalent of the area is miles square ( mi 2 ) . The conversion factor to convert the area from mi 2 to ft 2 is, ( 0.3048 ) 2 m 2 = 1 ft 2 Mass is a physical quantity and its SI unit is kilograms ( kg ) . It is the measure of the quantity of the matter present in an object. The other units of mass are grams, milligrams. 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. In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows: ( beginning unit ) ( Final unit beginning unit ) = Final unit

Question

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Answer 1

Question

Chapter 1.4, Problem 1.6BFP

Interpretation Introduction

Interpretation:

The mass of mercury in grams contained in each $mL$ of lake water is to be calculated. Also, a road map for the solution is to be drawn.

Concept introduction:

Meter square $(m2)$ is the SI unit of the area. The English equivalent of the area is miles square $(mi2)$ . The conversion factor to convert the area from $mi2$ to $ft2$ is,

$(0.3048)2 m2=1 ft2$

Mass is a physical quantity and its SI unit is kilograms $(kg)$ . It is the measure of the quantity of the matter present in an object. The other units of mass are grams, milligrams.

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.

In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows:

$(beginning unit)(Final unitbeginning unit)=Final unit$

Expert Solution & Answer

Answer to Problem 1.6BFP

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 1

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

Explanation of Solution

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 2

The mass of mercury discharged in the lake is $75,000 kg$ .

The conversion factor to convert the mass from kilograms $(kg)$ to grams $(g)$ is,

$1 kg=1000 g$

The mass of mercury in the lake in grams is,

$Mass(g)=75,000 kg(1000 g1 kg)=7.5×107 g$

The surface area of the lake is $4.5 mi2$ and the depth of the lake is $35 ft$ .

The conversion factor to convert the area from $mi2$ to $ft2$ is,

$1 mi2=(5280)2 ft2$

The surface area of the lake in $ft2$ is calculated as:

$Area(ft2)=4.5 mi2((5280)2 ft21 mi2)=1.25452800×108 ft2$

The volume of the lake is calculated as:

$Volume of lake(ft3)=(Area of lake(ft2))(depth of lake(ft))$ (1)

Substitute $1.25452800×108 ft2$ for the area of the lake and $35 ft$ for depth of lake in equation (1).

$Volume of lake(ft3)=(1.25452800×108 ft2)(35 ft)=4.3908×109 ft3$

The conversion factor to convert volume from $ft3$ to $m3$ is,

$1 ft3=0.02832 m3$

The conversion factor to convert volume from $m3$ to $cm3$ is,

$1 m3=1×106 cm3$

The conversion factor to convert volume from $cm3$ to $mL$ is,

$1 cm3=1 mL$

The volume of the lake in $mL$ is,

$Volume of lake(mL)=4.3908×109 ft3(0.02832 m31 ft3)(1×106 cm31 m3)(1 mL1 cm3)=1.24349×1014 mL$

The mass of mercury in grams present in one $mL$ of lake water is calculated as,

$Mass(g)of mercury per mL=Mass of mercury discharged in lake(g)Volume of lake(mL)$ (2)

Substitute $7.5×107 g$ for the mass of mercury discharged in the lake and $1.24349×1014 mL$ for the volume of the lake in equation (2).

$Mass(g)of mercury per mL=7.5×107 g1.24349×1014 mL=6.0314×10−7 g/mL=6.0×10−7 g/mL$

Conclusion

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

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Answer 2

Question

Chapter 1.4, Problem 1.6BFP

Interpretation Introduction

Interpretation:

The mass of mercury in grams contained in each $mL$ of lake water is to be calculated. Also, a road map for the solution is to be drawn.

Concept introduction:

Meter square $(m2)$ is the SI unit of the area. The English equivalent of the area is miles square $(mi2)$ . The conversion factor to convert the area from $mi2$ to $ft2$ is,

$(0.3048)2 m2=1 ft2$

Mass is a physical quantity and its SI unit is kilograms $(kg)$ . It is the measure of the quantity of the matter present in an object. The other units of mass are grams, milligrams.

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.

In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows:

$(beginning unit)(Final unitbeginning unit)=Final unit$

Expert Solution & Answer

Answer to Problem 1.6BFP

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 1

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

Explanation of Solution

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 2

The mass of mercury discharged in the lake is $75,000 kg$ .

The conversion factor to convert the mass from kilograms $(kg)$ to grams $(g)$ is,

$1 kg=1000 g$

The mass of mercury in the lake in grams is,

$Mass(g)=75,000 kg(1000 g1 kg)=7.5×107 g$

The surface area of the lake is $4.5 mi2$ and the depth of the lake is $35 ft$ .

The conversion factor to convert the area from $mi2$ to $ft2$ is,

$1 mi2=(5280)2 ft2$

The surface area of the lake in $ft2$ is calculated as:

$Area(ft2)=4.5 mi2((5280)2 ft21 mi2)=1.25452800×108 ft2$

The volume of the lake is calculated as:

$Volume of lake(ft3)=(Area of lake(ft2))(depth of lake(ft))$ (1)

Substitute $1.25452800×108 ft2$ for the area of the lake and $35 ft$ for depth of lake in equation (1).

$Volume of lake(ft3)=(1.25452800×108 ft2)(35 ft)=4.3908×109 ft3$

The conversion factor to convert volume from $ft3$ to $m3$ is,

$1 ft3=0.02832 m3$

The conversion factor to convert volume from $m3$ to $cm3$ is,

$1 m3=1×106 cm3$

The conversion factor to convert volume from $cm3$ to $mL$ is,

$1 cm3=1 mL$

The volume of the lake in $mL$ is,

$Volume of lake(mL)=4.3908×109 ft3(0.02832 m31 ft3)(1×106 cm31 m3)(1 mL1 cm3)=1.24349×1014 mL$

The mass of mercury in grams present in one $mL$ of lake water is calculated as,

$Mass(g)of mercury per mL=Mass of mercury discharged in lake(g)Volume of lake(mL)$ (2)

Substitute $7.5×107 g$ for the mass of mercury discharged in the lake and $1.24349×1014 mL$ for the volume of the lake in equation (2).

$Mass(g)of mercury per mL=7.5×107 g1.24349×1014 mL=6.0314×10−7 g/mL=6.0×10−7 g/mL$

Conclusion

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

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Answer 3

Question

Chapter 1.4, Problem 1.6BFP

Interpretation Introduction

Interpretation:

The mass of mercury in grams contained in each $mL$ of lake water is to be calculated. Also, a road map for the solution is to be drawn.

Concept introduction:

Meter square $(m2)$ is the SI unit of the area. The English equivalent of the area is miles square $(mi2)$ . The conversion factor to convert the area from $mi2$ to $ft2$ is,

$(0.3048)2 m2=1 ft2$

Mass is a physical quantity and its SI unit is kilograms $(kg)$ . It is the measure of the quantity of the matter present in an object. The other units of mass are grams, milligrams.

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.

In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows:

$(beginning unit)(Final unitbeginning unit)=Final unit$

Expert Solution & Answer

Answer to Problem 1.6BFP

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 1

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

Explanation of Solution

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 2

The mass of mercury discharged in the lake is $75,000 kg$ .

The conversion factor to convert the mass from kilograms $(kg)$ to grams $(g)$ is,

$1 kg=1000 g$

The mass of mercury in the lake in grams is,

$Mass(g)=75,000 kg(1000 g1 kg)=7.5×107 g$

The surface area of the lake is $4.5 mi2$ and the depth of the lake is $35 ft$ .

The conversion factor to convert the area from $mi2$ to $ft2$ is,

$1 mi2=(5280)2 ft2$

The surface area of the lake in $ft2$ is calculated as:

$Area(ft2)=4.5 mi2((5280)2 ft21 mi2)=1.25452800×108 ft2$

The volume of the lake is calculated as:

$Volume of lake(ft3)=(Area of lake(ft2))(depth of lake(ft))$ (1)

Substitute $1.25452800×108 ft2$ for the area of the lake and $35 ft$ for depth of lake in equation (1).

$Volume of lake(ft3)=(1.25452800×108 ft2)(35 ft)=4.3908×109 ft3$

The conversion factor to convert volume from $ft3$ to $m3$ is,

$1 ft3=0.02832 m3$

The conversion factor to convert volume from $m3$ to $cm3$ is,

$1 m3=1×106 cm3$

The conversion factor to convert volume from $cm3$ to $mL$ is,

$1 cm3=1 mL$

The volume of the lake in $mL$ is,

$Volume of lake(mL)=4.3908×109 ft3(0.02832 m31 ft3)(1×106 cm31 m3)(1 mL1 cm3)=1.24349×1014 mL$

The mass of mercury in grams present in one $mL$ of lake water is calculated as,

$Mass(g)of mercury per mL=Mass of mercury discharged in lake(g)Volume of lake(mL)$ (2)

Substitute $7.5×107 g$ for the mass of mercury discharged in the lake and $1.24349×1014 mL$ for the volume of the lake in equation (2).

$Mass(g)of mercury per mL=7.5×107 g1.24349×1014 mL=6.0314×10−7 g/mL=6.0×10−7 g/mL$

Conclusion

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

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Answer 4

Question

Chapter 1.4, Problem 1.6BFP

Interpretation Introduction

Interpretation:

The mass of mercury in grams contained in each $mL$ of lake water is to be calculated. Also, a road map for the solution is to be drawn.

Concept introduction:

Meter square $(m2)$ is the SI unit of the area. The English equivalent of the area is miles square $(mi2)$ . The conversion factor to convert the area from $mi2$ to $ft2$ is,

$(0.3048)2 m2=1 ft2$

Mass is a physical quantity and its SI unit is kilograms $(kg)$ . It is the measure of the quantity of the matter present in an object. The other units of mass are grams, milligrams.

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.

In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows:

$(beginning unit)(Final unitbeginning unit)=Final unit$

Expert Solution & Answer

Answer to Problem 1.6BFP

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 1

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

Explanation of Solution

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 2

The mass of mercury discharged in the lake is $75,000 kg$ .

The conversion factor to convert the mass from kilograms $(kg)$ to grams $(g)$ is,

$1 kg=1000 g$

The mass of mercury in the lake in grams is,

$Mass(g)=75,000 kg(1000 g1 kg)=7.5×107 g$

The surface area of the lake is $4.5 mi2$ and the depth of the lake is $35 ft$ .

The conversion factor to convert the area from $mi2$ to $ft2$ is,

$1 mi2=(5280)2 ft2$

The surface area of the lake in $ft2$ is calculated as:

$Area(ft2)=4.5 mi2((5280)2 ft21 mi2)=1.25452800×108 ft2$

The volume of the lake is calculated as:

$Volume of lake(ft3)=(Area of lake(ft2))(depth of lake(ft))$ (1)

Substitute $1.25452800×108 ft2$ for the area of the lake and $35 ft$ for depth of lake in equation (1).

$Volume of lake(ft3)=(1.25452800×108 ft2)(35 ft)=4.3908×109 ft3$

The conversion factor to convert volume from $ft3$ to $m3$ is,

$1 ft3=0.02832 m3$

The conversion factor to convert volume from $m3$ to $cm3$ is,

$1 m3=1×106 cm3$

The conversion factor to convert volume from $cm3$ to $mL$ is,

$1 cm3=1 mL$

The volume of the lake in $mL$ is,

$Volume of lake(mL)=4.3908×109 ft3(0.02832 m31 ft3)(1×106 cm31 m3)(1 mL1 cm3)=1.24349×1014 mL$

The mass of mercury in grams present in one $mL$ of lake water is calculated as,

$Mass(g)of mercury per mL=Mass of mercury discharged in lake(g)Volume of lake(mL)$ (2)

Substitute $7.5×107 g$ for the mass of mercury discharged in the lake and $1.24349×1014 mL$ for the volume of the lake in equation (2).

$Mass(g)of mercury per mL=7.5×107 g1.24349×1014 mL=6.0314×10−7 g/mL=6.0×10−7 g/mL$

Conclusion

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

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Answer 5

Chapter 1.4, Problem 1.6BFP

Interpretation Introduction

Interpretation:

The mass of mercury in grams contained in each $mL$ of lake water is to be calculated. Also, a road map for the solution is to be drawn.

Concept introduction:

Meter square $(m2)$ is the SI unit of the area. The English equivalent of the area is miles square $(mi2)$ . The conversion factor to convert the area from $mi2$ to $ft2$ is,

$(0.3048)2 m2=1 ft2$

Mass is a physical quantity and its SI unit is kilograms $(kg)$ . It is the measure of the quantity of the matter present in an object. The other units of mass are grams, milligrams.

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.

In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows:

$(beginning unit)(Final unitbeginning unit)=Final unit$

Expert Solution & Answer

Answer to Problem 1.6BFP

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 1

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

Explanation of Solution

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 2

The mass of mercury discharged in the lake is $75,000 kg$ .

The conversion factor to convert the mass from kilograms $(kg)$ to grams $(g)$ is,

$1 kg=1000 g$

The mass of mercury in the lake in grams is,

$Mass(g)=75,000 kg(1000 g1 kg)=7.5×107 g$

The surface area of the lake is $4.5 mi2$ and the depth of the lake is $35 ft$ .

The conversion factor to convert the area from $mi2$ to $ft2$ is,

$1 mi2=(5280)2 ft2$

The surface area of the lake in $ft2$ is calculated as:

$Area(ft2)=4.5 mi2((5280)2 ft21 mi2)=1.25452800×108 ft2$

The volume of the lake is calculated as:

$Volume of lake(ft3)=(Area of lake(ft2))(depth of lake(ft))$ (1)

Substitute $1.25452800×108 ft2$ for the area of the lake and $35 ft$ for depth of lake in equation (1).

$Volume of lake(ft3)=(1.25452800×108 ft2)(35 ft)=4.3908×109 ft3$

The conversion factor to convert volume from $ft3$ to $m3$ is,

$1 ft3=0.02832 m3$

The conversion factor to convert volume from $m3$ to $cm3$ is,

$1 m3=1×106 cm3$

The conversion factor to convert volume from $cm3$ to $mL$ is,

$1 cm3=1 mL$

The volume of the lake in $mL$ is,

$Volume of lake(mL)=4.3908×109 ft3(0.02832 m31 ft3)(1×106 cm31 m3)(1 mL1 cm3)=1.24349×1014 mL$

The mass of mercury in grams present in one $mL$ of lake water is calculated as,

$Mass(g)of mercury per mL=Mass of mercury discharged in lake(g)Volume of lake(mL)$ (2)

Substitute $7.5×107 g$ for the mass of mercury discharged in the lake and $1.24349×1014 mL$ for the volume of the lake in equation (2).

$Mass(g)of mercury per mL=7.5×107 g1.24349×1014 mL=6.0314×10−7 g/mL=6.0×10−7 g/mL$

Conclusion

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

Answer 6

Chapter 1.4, Problem 1.6BFP

Interpretation Introduction

Interpretation:

The mass of mercury in grams contained in each $mL$ of lake water is to be calculated. Also, a road map for the solution is to be drawn.

Concept introduction:

Meter square $(m2)$ is the SI unit of the area. The English equivalent of the area is miles square $(mi2)$ . The conversion factor to convert the area from $mi2$ to $ft2$ is,

$(0.3048)2 m2=1 ft2$

Mass is a physical quantity and its SI unit is kilograms $(kg)$ . It is the measure of the quantity of the matter present in an object. The other units of mass are grams, milligrams.

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.

In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows:

$(beginning unit)(Final unitbeginning unit)=Final unit$

Expert Solution & Answer

Answer to Problem 1.6BFP

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 1

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

Explanation of Solution

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 2

The mass of mercury discharged in the lake is $75,000 kg$ .

The conversion factor to convert the mass from kilograms $(kg)$ to grams $(g)$ is,

$1 kg=1000 g$

The mass of mercury in the lake in grams is,

$Mass(g)=75,000 kg(1000 g1 kg)=7.5×107 g$

The surface area of the lake is $4.5 mi2$ and the depth of the lake is $35 ft$ .

The conversion factor to convert the area from $mi2$ to $ft2$ is,

$1 mi2=(5280)2 ft2$

The surface area of the lake in $ft2$ is calculated as:

$Area(ft2)=4.5 mi2((5280)2 ft21 mi2)=1.25452800×108 ft2$

The volume of the lake is calculated as:

$Volume of lake(ft3)=(Area of lake(ft2))(depth of lake(ft))$ (1)

Substitute $1.25452800×108 ft2$ for the area of the lake and $35 ft$ for depth of lake in equation (1).

$Volume of lake(ft3)=(1.25452800×108 ft2)(35 ft)=4.3908×109 ft3$

The conversion factor to convert volume from $ft3$ to $m3$ is,

$1 ft3=0.02832 m3$

The conversion factor to convert volume from $m3$ to $cm3$ is,

$1 m3=1×106 cm3$

The conversion factor to convert volume from $cm3$ to $mL$ is,

$1 cm3=1 mL$

The volume of the lake in $mL$ is,

$Volume of lake(mL)=4.3908×109 ft3(0.02832 m31 ft3)(1×106 cm31 m3)(1 mL1 cm3)=1.24349×1014 mL$

The mass of mercury in grams present in one $mL$ of lake water is calculated as,

$Mass(g)of mercury per mL=Mass of mercury discharged in lake(g)Volume of lake(mL)$ (2)

Substitute $7.5×107 g$ for the mass of mercury discharged in the lake and $1.24349×1014 mL$ for the volume of the lake in equation (2).

$Mass(g)of mercury per mL=7.5×107 g1.24349×1014 mL=6.0314×10−7 g/mL=6.0×10−7 g/mL$

Conclusion

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

Answer 7

Chapter 1.4, Problem 1.6BFP

Interpretation Introduction

Interpretation:

The mass of mercury in grams contained in each $mL$ of lake water is to be calculated. Also, a road map for the solution is to be drawn.

Concept introduction:

Meter square $(m2)$ is the SI unit of the area. The English equivalent of the area is miles square $(mi2)$ . The conversion factor to convert the area from $mi2$ to $ft2$ is,

$(0.3048)2 m2=1 ft2$

Mass is a physical quantity and its SI unit is kilograms $(kg)$ . It is the measure of the quantity of the matter present in an object. The other units of mass are grams, milligrams.

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.

In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows:

$(beginning unit)(Final unitbeginning unit)=Final unit$

Answer 8

Expert Solution & Answer

Answer to Problem 1.6BFP

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 1

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The road map to calculate the mass of mercury in grams contained in each $mL$ of lake water is as follows:

Chemistry: The Molecular Nature of Matter and Change (Custom Package), Chapter 1.4, Problem 1.6BFP , additional homework tip 2

The mass of mercury discharged in the lake is $75,000 kg$ .

The conversion factor to convert the mass from kilograms $(kg)$ to grams $(g)$ is,

$1 kg=1000 g$

The mass of mercury in the lake in grams is,

$Mass(g)=75,000 kg(1000 g1 kg)=7.5×107 g$

The surface area of the lake is $4.5 mi2$ and the depth of the lake is $35 ft$ .

The conversion factor to convert the area from $mi2$ to $ft2$ is,

$1 mi2=(5280)2 ft2$

The surface area of the lake in $ft2$ is calculated as:

$Area(ft2)=4.5 mi2((5280)2 ft21 mi2)=1.25452800×108 ft2$

The volume of the lake is calculated as:

$Volume of lake(ft3)=(Area of lake(ft2))(depth of lake(ft))$ (1)

Substitute $1.25452800×108 ft2$ for the area of the lake and $35 ft$ for depth of lake in equation (1).

$Volume of lake(ft3)=(1.25452800×108 ft2)(35 ft)=4.3908×109 ft3$

The conversion factor to convert volume from $ft3$ to $m3$ is,

$1 ft3=0.02832 m3$

The conversion factor to convert volume from $m3$ to $cm3$ is,

$1 m3=1×106 cm3$

The conversion factor to convert volume from $cm3$ to $mL$ is,

$1 cm3=1 mL$

The volume of the lake in $mL$ is,

$Volume of lake(mL)=4.3908×109 ft3(0.02832 m31 ft3)(1×106 cm31 m3)(1 mL1 cm3)=1.24349×1014 mL$

The mass of mercury in grams present in one $mL$ of lake water is calculated as,

$Mass(g)of mercury per mL=Mass of mercury discharged in lake(g)Volume of lake(mL)$ (2)

Substitute $7.5×107 g$ for the mass of mercury discharged in the lake and $1.24349×1014 mL$ for the volume of the lake in equation (2).

$Mass(g)of mercury per mL=7.5×107 g1.24349×1014 mL=6.0314×10−7 g/mL=6.0×10−7 g/mL$

Answer 12

Conclusion

The mass of mercury in grams contained in each $mL$ of lake water is $6.0×10−7 g/mL$ .

Answer 13

Ch. 1.1 - Is the following change chemical or physical? Ch. 1.1 - Is the following change chemical or physical? Ch. 1.1 - Decide whether each of the following processes is...Ch. 1.1 - Decide whether each of the following processes is...Ch. 1.4 - A chemistry professor can walk a mile in 15...Ch. 1.4 - The rhinovirus, one cause of the common cold, has...Ch. 1.4 - Prob. 1.4AFP Ch. 1.4 - Prob. 1.4BFP Ch. 1.4 - An intravenous nutrient solution is delivered to a...Ch. 1.4 - Nutritional tables give the potassium content of a...

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Answer to Problem 1.6BFP

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