# See Details?

May 27, 2017

@Chan, placing the stone in the beaker clearly displaced........

#### Explanation:

@Chan, placing the stone in the beaker clearly displaced........$7 \cdot m L$ of water. You notice the same thing when you have a bath. You lower yourself into a bath, and you notice that the water is displaced. If you have filled your bath too full, you get water on the bathroom floor you klutz.

If you have the mass of the stone (or whatever object), you can calculate the density of the stone, because density is given by the quotient:

$\text{Density,}$ $\rho = \text{Mass"/"Volume}$..........

As regards the equivalence of millilitres and cubic centimetres...........it is a fact that $1 \cdot m L \equiv 1 \cdot c {m}^{-} 3$.

Now we know that $1 \cdot {m}^{3}$ (which is a HUGE volume) $\equiv 1000 \cdot L$

$1 \cdot c {m}^{3} = {\left(1 \times {10}^{-} 2 \cdot m\right)}^{3}$. Why? Because the prefix $\text{centi} \equiv {10}^{-} 2$. And thus............

$1 \cdot c {m}^{3} = {\left(1 \times {10}^{-} 2 \cdot m\right)}^{3} = 1 \times {10}^{-} 6 \cdot {m}^{3}$.

Sometimes I remember that $1 \cdot {m}^{3}$ (which is indeed a HUGE volume!), is $1000 \cdot L$, and that there are ${10}^{3} \cdot c {m}^{3}$ in a $L$. An equivalent term for the $\text{litre}$ is $\text{cubic decimeter}$, ${\mathrm{dm}}^{3}$.

And thus $1000 \cdot L = {10}^{3} \cdot L \times {10}^{-} 6 \cdot {m}^{3} = 1 \times {10}^{-} 3 \cdot {m}^{3}$.

Equivalently $1 \cdot L = 1 \cdot {\mathrm{dm}}^{3} = {\left(1 \times {10}^{-} 1 \cdot m\right)}^{3} = {10}^{-} 3 \cdot {m}^{3}$, as required.