earths surface area is $5.13\times10^{8} km^{2}$, and its crust has a mean thickness of $35.0 km$ and mean…

Explanation:

Step1: Calculate the volume of the Earth's crust

First, convert the surface - area and thickness to SI units. The surface area $A = 5.13\times10^{8}\ km^{2}=5.13\times10^{8}\times(10^{3})^{2}\ m^{2}=5.13\times10^{14}\ m^{2}$, and the thickness $h = 35.0\ km = 35.0\times10^{3}\ m$. The volume of the crust $V=A\times h$. $V=(5.13\times 10^{14}\ m^{2})\times(35.0\times10^{3}\ m)=1.7955\times 10^{18}\ m^{3}$. Then convert the volume to $cm^{3}$: $V = 1.7955\times10^{18}\times(10^{2})^{3}\ cm^{3}=1.7955\times10^{24}\ cm^{3}$.

Step2: Calculate the mass of the Earth's crust

Use the density formula $\rho=\frac{m}{V}$, where $\rho = 2.82\ g/cm^{3}$. Then $m=\rho V$. $m=(2.82\ g/cm^{3})\times(1.7955\times 10^{24}\ cm^{3}) = 5.06331\times10^{24}\ g$. Also convert the mass to tons: $m=\frac{5.06331\times10^{24}\ g}{1000\ g/kg}\times\frac{1\ t}{1000\ kg}=5.06331\times10^{18}\ t$.

Step3: Calculate the mass of each element

For oxygen:

The abundance of oxygen is $4.57\times10^{5}\ g/t$. The mass of oxygen $m_O=(4.57\times10^{5}\ g/t)\times(5.06331\times10^{18}\ t)=2.31393267\times10^{24}\ g\approx2.31\times10^{24}\ g$.

For silicon:

The abundance of silicon is $2.70\times10^{5}\ g/t$. The mass of silicon $m_{Si}=(2.70\times10^{5}\ g/t)\times(5.06331\times10^{18}\ t)=1.3670937\times10^{24}\ g\approx1.37\times10^{24}\ g$.

For ruthenium or rhodium:

The abundance of ruthenium and rhodium is $1\times10^{-4}\ g/t$. The mass of ruthenium or rhodium $m_{Ru/Rh}=(1\times10^{-4}\ g/t)\times(5.06331\times10^{18}\ t)=5.06331\times10^{14}\ g\approx5.06\times10^{14}\ g$.

Answer:

Part 1 of 3: $2.31\times 10^{24}$ Part 2 of 3: $1.37\times 10^{24}$ Part 3 of 3: $5.06\times 10^{14}$