Chapter 1, Problem 1.76P

(a)

Interpretation Introduction

Interpretation:

The mass of gold in grams is to be calculated.

Concept introduction:

The density of the substance is defined as the mass per unit volume. It is represented by $\rho$ and its unit is ${\text{kg/m}}^3$.

The formula to calculate the density of the substance is as follows:

$\text{Density of substance}\left(\rho \right)=\frac{\text{Mass of substance}\left(M\right)}{\text{Volume of substance}\left(V\right)}$        (1)

The formula to calculate the volume is as follows:

$\text{Volume}=\left(\text{Area}\right)\left(\text{depth}\right)$        (2)

The conversion factor to convert ${\text{km}}^2$ to ${\text{m}}^2$ is as follows:

${\text{1 km}}^2{\text{=10}}^6{\text{m}}^2$

(b)

Interpretation Introduction

Interpretation:

The volume of gold in cubic meters is to be calculated.

Concept introduction:

The density of the substance is defined as the mass per unit volume. It is represented by $\rho$ and its unit is ${\text{kg/m}}^3$.

The formula to calculate the density of the substance is as follows:

$\text{Density of substance}\left(\rho \right)=\frac{\text{Mass of substance}\left(M\right)}{\text{Volume of substance}\left(V\right)}$        (1)

(c)

Interpretation Introduction

Interpretation:

The value of gold in the oceans is to be determined.

Concept introduction:

The conversion factor to convert $\text{g}$ to $\text{troy}\text{oz}$ is as follows:

$\text{1 g}=\left(\frac{1}{31.1}\right)\text{troy}\text{oz}$