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A thief uses a bag of sand to replace a gold statue that

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 103P Chapter 2

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 103P

A thief uses a bag of sand to replace a gold statue that sits on a weight-sensitive, alarmed pedestal. The bag of sand and the statue have exactly the same volume, 1.75 L. (Assume that the mass of the bag is negligible.)

(a) Calculate the mass of each object, (density of gold = 19.3 g/cm3; density of sand = 3.00 g/cm3)

             (b) Did the thief set off the alarm? Explain.

Step-by-Step Solution:
Step 1 of 3

Problem 103P

A thief uses a bag of sand to replace a gold statue that sits on a weight-sensitive, alarmed pedestal. The bag of sand and the statue have exactly the same volume, 1.75 L. (Assume that the mass of the bag is negligible.)

(a) Calculate the mass of each object, (density of gold = 19.3 g/cm3; density of sand = 3.00 g/cm3)

             (b) Did the thief set off the alarm? Explain.

Solution: Here, we are going to calculate the mass of sand and gold given in the problem.

Step1:

Density of any substance may be defined as mass per unit volume. If ‘d’ be the density, ‘m’ be the mass and ‘v’ be the volume of any substance, then, numerically,

d =                 -------(1)

Step2:

Given, density of gold = 19.3 g/cm3

        Volume of gold statue = 1.75 L = 1.75 x 1000 cm3                [1 L = 1000 cm3]

                                = 1750 cm3

        Substituting the values in equation (1), we get,

                                19.3 g/cm3 =  m / 1750 cm3        

                                        m = 19.3 x 1750...

Step 2 of 3

Chapter 2, Problem 103P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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