A stone with a mass of 2400 g is submerged in water in a rectangular tank with base dimensions 17 cm by 8 cm. The water level rises from 9 cm to 12 cm. Calculate the density of the stone in g/cm³ and

The water level rises from $9 \text{ cm}$ (from the previous problem) to $12 \text{ cm}$ when the stone is lowered. The volume of the stone is equal to the volume of the displaced water.
Change in water height ($\Delta h$) = Final height - Initial height

$$\Delta h = 12 \text{ cm} - 9 \text{ cm} = 3 \text{ cm}$$

The base dimensions of the tank are $17 \text{ cm}$ by $8 \text{ cm}$.
Volume of the stone ($V_{\text{stone}}$) = Length $\times$ Width $\times \Delta h$

$$V_{\text{stone}} = 17 \text{ cm} \times 8 \text{ cm} \times 3 \text{ cm}$$ $$V_{\text{stone}} = 408 \text{ cm}^3$$

Mass of the stone ($m$) = $2400 \text{ g}$
Density ($\rho$) = $\frac{m}{V_{\text{stone}}}$

$$\rho = \frac{2400 \text{ g}}{408 \text{ cm}^3}$$ $$\rho \approx 5.882 \text{ g/cm}^3$$

The density of the stone in $\text{gcm}^{-3}$ is $\boxed{5.882 \text{ g/cm}^3}$.

$$V_{\text{stone}} = 408 \text{ cm}^3$$

Since $1 \text{ m} = 100 \text{ cm}$, then $1 \text{ m}^3 = (100 \text{ cm})^3 = 1,000,000 \text{ cm}^3$.

$$V_{\text{stone}} = 408 \text{ cm}^3 \times \frac{1 \text{ m}^3}{1,000,000 \text{ cm}^3} = 0.000408 \text{ m}^3$$

$$\rho = \frac{m}{V_{\text{stone}}}$$ $$\rho = \frac{2.4 \text{ kg}}{0.000408 \text{ m}^3}$$ $$\rho \approx 5882.353 \text{ kg/m}^3$$

Alternatively, convert directly from $\text{g/cm}^3$:

$$1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3$$ $$\rho = 5.882 \text{ g/cm}^3 \times 1000 = 5882 \text{ kg/m}^3$$

The density of the stone in $\text{kgm}^{-3}$ is $\boxed{5882.353 \text{ kg/m}^3}$.