## Chemistry: Molecular Approach (4th Edition)

$\underline{1.99\times {{10}^{5}}\text{ kg}}$
$1\text{ }f{{t}^{3}}=28,316.8\text{ c}{{\text{m}}^{3}}$ 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 $\underline{1.99\times {{10}^{5}}\text{ kg}}$.