Chemistry For Today: General, Organic, And Biochemistry, Loose-leaf Version
Chemistry For Today: General, Organic, And Biochemistry, Loose-leaf Version
9th Edition
ISBN: 9781305968707
Author: Spencer L. Seager
Publisher: Brooks Cole
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Chapter 1, Problem 1.97E

The volume of an irregularly shaped solid can be determined by immersing the solid in a liquid and measuring the volume of liquid displaced. Find the volume and density of the following:

a. An irregular piece of the mineral quartz is found to weigh 12.4 g . It is then placed into a graduated cylinder that contains some water. The quartz does not float. The water in the cylinder was at a level of 25.2 mL before the quartz was added and at 29.9 mL afterward.

b. The volume of a sample of lead shot is determined using a graduated cylinder, as in part (a). The cylinder readings are 16.3 mL before the shot is added and 21.7 mL after. The sample of shot weighs 61.0 g .

c. A sample of coarse rock salt is found to have a mass of 11.7 g . The volume of the sample is determined by the graduated-cylinder method described in (part a), but kerosene is substituted for water because the salt will not dissolve in kerosene. The cylinder readings are 20.7 mL before adding the salt and 26.1 mL after.

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The volume and density of an irregular piece of the mineral quartz weighing 12.4g and having volume change from 25.2mL to 29.9mL in a graduated cylinder are to be calculated.

Concept introduction:

The density of a substance is defined as the ratio of its mass to the volume. The expression for density is given below.

d=mV

Where,

d is the density

m is the mass

V is the volume

The volume of an irregularly shaped solid can be determined by immersing the solid in a liquid and measuring the volume of liquid displaced.

Answer to Problem 1.97E

The volume and density of an irregular piece of the mineral quartz weighing 12.4g and having volume change from 25.2mL to 29.9mL in graduated cylinder are 4.7mL and 2.64g/mL respectively.

Explanation of Solution

The volume displaced by immersing the mineral quartz in a graduated cylinder that contains water is calculated as given below.

V=VfinalVinitial

Where,

Vfinal is the volume in the graduated cylinder after immersing solid

Vinitial is the volume initially taken in the graduated cylinder

Substitute the values in the above expression as follows.

V=VfinalVinitial=29.9mL25.2mL=4.7mL

The density is calculated as given below.

d=mV

Where,

d is the density

m is the mass

V is the volume

Substitute the values in the above expression as follows.

d=mV=12.4g4.7mL=2.64g/mL

Thus, the density of mineral quartz is found to be 2.64g/mL.

Conclusion

The volume and density of an irregular piece of the mineral quartz weighing 12.4g and having volume change from 25.2mL to 29.9mL in graduated cylinder are calculated as 4.7mL and 2.64g/mL respectively.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The volume and density of a lead shot weighing 61.0g and having volume change from 16.3mL to 21.7mL in graduated cylinder are to becalculated.

Concept introduction:

The density of a substance is defined as the ratio of its mass to the volume. The expression for density is given below.

d=mV

Where,

d is the density

m is the mass

V is the volume

The volume of an irregularly shaped solid can be determined by immersing the solid in a liquid and measuring the volume of liquid displaced.

Answer to Problem 1.97E

The volume and density of a lead shot weighing 61.0g and having volume change from 16.3mL to 21.7mL in graduated cylinder are 5.4mL and 11.3g/mL respectively.

Explanation of Solution

The volume displaced by immersing the lead shot in a graduated cylinder that contains water is calculated as given below.

V=VfinalVinitial

Where,

Vfinal is the volume in the graduated cylinder after immersing solid

Vinitial is the volume initially taken in the graduated cylinder

Substitute the values in the above expression as follows.

V=VfinalVinitial=21.7mL16.3mL=5.4mL

The density is calculated as given below.

d=mV

Where,

d is the density

m is the mass

V is the volume

Substitute the values in the above expression as follows.

d=mV=61g5.4mL=11.3g/mL

Thus, the density of lead shot is found to be 11.3g/mL.

Conclusion

The volume and density of a lead shot weighing 61.0g and having volume change from 16.3mL to 21.7mL in graduated cylinder are calculated as 5.4mL and 11.3g/mL respectively.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The volume and density of rock salt weighing 11.7g and having volume change from 20.7mL to 26.1mL in graduated cylinder are to be calculated.

Concept introduction:

The density of a substance is defined as the ratio of its mass to the volume. The expression for density is given below.

d=mV

Where,

d is the density

m is the mass

V is the volume

The volume of an irregularly shaped solid can be determined by immersing the solid in a liquid and measuring the volume of liquid displaced.

Answer to Problem 1.97E

The volume and density of rock salt weighing 11.7g and having volume change from 20.7mL to 26.1mL in graduated cylinder are 5.4mL and 2.16g/mL respectively.

Explanation of Solution

The volume displaced by immersing rock salt in a graduated cylinder that contains kerosene is calculated as given below.

V=VfinalVinitial

Where,

Vfinal is the volume in the graduated cylinder after immersing solid

Vinitial is the volume initially taken in the graduated cylinder

Substitute the values in the above expression as follows.

V=VfinalVinitial=26.1mL20.7mL=5.4mL

The density is calculated as given below.

d=mV

Where,

d is the density

m is the mass

V is the volume

Substitute the values in the above expression as follows.

d=mV=11.7g5.4mL=2.16g/mL

Thus, the density of rock salt is found to be 2.16g/mL.

Conclusion

The volume and density of rock salt weighing 11.7g and having volume change from 20.7mL to 26.1mL in graduated cylinder are calculated as 5.4mL and 2.16g/mL respectively.

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Chapter 1 Solutions

Chemistry For Today: General, Organic, And Biochemistry, Loose-leaf Version

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