# An area of "1 hectare" was filled to a depth of 0.1*m, and then filled with soil whose density was 1*g*cm^-3. What mass of soil was used?

Mar 7, 2017

$1 , 000 \cdot k g$

#### Explanation:

We must measure the volume of the plot of the land, and then multiply by the density:

$\text{1 hectare}$ $=$ $100 \cdot m \times 100 \cdot m = 10000 \cdot {m}^{2}$

We multiply this value by the depth, to give a volume:

$10000 \cdot {m}^{2} \times 0.1 \cdot m = 1000 \cdot {m}^{3}$

And we multiply this volume by the density to get the mass:

$1000 \cdot {m}^{3} \times 1 \cdot g \cdot c {m}^{-} 3 = 1000 \cdot {m}^{3} \times 1 \cdot g \cdot {\left({10}^{-} 2 \cdot m\right)}^{-} 3 =$

$1000 \cdot {m}^{3} \times 1 \cdot g \cdot {10}^{6} \cdot {m}^{-} 3 =$

$\cancel{1000} \cdot \cancel{{m}^{3}} \times 1 \cdot \cancel{g} \times \cancel{{10}^{-} 3} \cdot k g \cdot \cancel{{g}^{-} 1} \times {10}^{\cancel{6} 3} \cdot \cancel{{m}^{-} 3} =$

${10}^{3} \cdot k g$

I think this is right (all care taken, but no responsibility admitted!). It is always too easy to make an error in these sorts of calculations. We required a mass, and got a mass. Anyway, please check the calculation. Again, I suspect that the actual density of soil would be much more than $1 \cdot g \cdot c {m}^{-} 3$.