Chemistry: Molecular Approach (4th Edition)

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

Chapter 1 - Exercises - Page 40: 122


\[\underline{1.99\times {{10}^{5}}\text{ kg}}\]

Work Step by Step

\[1\text{ }f{{t}^{3}}=28,316.8\text{ c}{{\text{m}}^{3}}\] Thus, volume of the iceberg in cubic centimeters will be written as follows: \[\begin{align} & V=\left( 7655\times 28,316.8 \right)\text{ c}{{\text{m}}^{3}} \\ & =216,765,104\text{ c}{{\text{m}}^{3}} \end{align}\] Density of ice is \[0.92\text{ g/c}{{\text{m}}^{\text{3}}}\]. The relation between density\[\left( d \right)\] and mass\[\left( m \right)\] is as follows: \[m=\left( V \right)\left( d \right)\] Here, V is volume. Thus, mass of ice is written as follows: \[\begin{align} & m=\left( V \right)\left( d \right) \\ & =\left( 216,765,104\text{ c}{{\text{m}}^{3}} \right)\left( 0.92\text{ g/c}{{\text{m}}^{3}} \right) \\ & =199,423,896\text{ g} \end{align}\] Now, convert this mass of ice from grams to kilograms by the use of the following relation: \[1\text{ kg}=1000\text{ g}\] Thus, mass of ice in kilogram is written as follows: \[\begin{align} & m=199,423,896\times {{10}^{-3}}\text{ kg} \\ & =\text{1}\text{.99}\times {{10}^{5}}\text{ kg} \end{align}\] The mass of ice in kilograms is \[\underline{1.99\times {{10}^{5}}\text{ kg}}\].
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