Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 1 - Exercises - Page 40: 115

Answer

(a)\[\underline{1.9\times \text{1}{{\text{0}}^{4}}\text{ g and 3}\text{.0}\times \text{1}{{\text{0}}^{3}}\text{ g }}\] (b) Yes

Work Step by Step

(a) The volume of a cylinder is as follows: \[\begin{align} & V=\pi {{r}^{2}}h \\ & =\left( \frac{22}{7} \right){{\left( 3.8\text{ cm} \right)}^{2}}\left( 22\text{ cm} \right) \\ & =998.4\text{ c}{{\text{m}}^{3}} \end{align}\] Calculate mass as follows: \[m=dV\] Mass of gold cylinder will be as follows: \[\begin{align} & m=\left( 19.3\text{ g/c}{{\text{m}}^{3}} \right)\left( \text{998}\text{.4 c}{{\text{m}}^{3}} \right) \\ & =1.9\times {{10}^{4}}\text{ g} \end{align}\] Mass of sand cylinder will be as follows: \[\begin{align} & m=\left( \text{3}\text{.0 g/c}{{\text{m}}^{3}} \right)\left( \text{998}\text{.4 c}{{\text{m}}^{3}} \right) \\ & =3.0\times {{10}^{3}}\text{ g} \end{align}\] Mass of gold cylinder and that of sand cylinder are \[\underline{1.9\times \text{1}{{\text{0}}^{4}}\text{ g and 3}\text{.0}\times \text{1}{{\text{0}}^{3}}\text{ g }}\]. (b) The mass of sand is \[\text{3}\text{.0}\times \text{1}{{\text{0}}^{3}}\text{ g}\] and that of gold cylinder is \[1.9\times \text{1}{{\text{0}}^{4}}\text{ g}\]. Yes, the thief sets off the alarm
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