earths surface area is (5.08\times10^{8} km^{2}), and its crust has a mean thickness of (35.2 km) and mean…

Explanation:

Step1: Calculate the volume of Earth's crust

First, convert the surface - area and thickness to SI units. The surface area $A = 5.08\times10^{8}\ km^{2}=5.08\times 10^{14}\ m^{2}$, and the thickness $h = 35.2\ km = 3.52\times10^{4}\ m$. The volume $V$ of the crust is $V=A\times h$. $V=(5.08\times 10^{14}\ m^{2})\times(3.52\times 10^{4}\ m)=1.78816\times 10^{19}\ m^{3}$

Step2: Convert the density to SI units

The density of the crust $\rho = 2.81\ g/cm^{3}=2.81\times10^{3}\ kg/m^{3}$.

Step3: Calculate the mass of the crust

Using the formula $m=\rho V$, we have $m=(2.81\times 10^{3}\ kg/m^{3})\times(1.78816\times 10^{19}\ m^{3}) = 5.0247296\times10^{22}\ kg=5.0247296\times 10^{25}\ g$

Step4: Calculate the mass of oxygen

The abundance of oxygen is $4.61\times 10^{5}\ g/t$, and since $1\ t = 1000\ kg=10^{6}\ g$, the fraction of oxygen by mass $f_O=\frac{4.61\times 10^{5}}{10^{6}} = 0.461$. The mass of oxygen $m_O=f_O\times m$. $m_O = 0.461\times5.0247296\times 10^{25}\ g\approx2.32\times 10^{25}\ g$

Step5: Calculate the mass of silicon

The abundance of silicon is $2.72\times 10^{5}\ g/t$, so the fraction of silicon by mass $f_{Si}=\frac{2.72\times 10^{5}}{10^{6}}=0.272$. The mass of silicon $m_{Si}=f_{Si}\times m$. $m_{Si}=0.272\times5.0247296\times 10^{25}\ g\approx1.37\times 10^{25}\ g$

Step6: Calculate the mass of ruthenium or rhodium

The abundance of ruthenium and rhodium is $1\times 10^{-4}\ g/t$, so the fraction of ruthenium or rhodium by mass $f_{Ru/Rh}=\frac{1\times 10^{-4}}{10^{6}}=1\times 10^{-10}$. The mass of ruthenium or rhodium $m_{Ru/Rh}=f_{Ru/Rh}\times m$. $m_{Ru/Rh}=(1\times 10^{-10})\times5.0247296\times 10^{25}\ g\approx5.02\times 10^{15}\ g$

Answer:

Part 1 of 3: $2.32\times 10^{25}$ Part 2 of 3: $1.37\times 10^{25}$ Part 3 of 3: $5.02\times 10^{15}$