Calculate the density of a mixture formed by mixing 180 cm³ of fresh water with 2800 cm³ of sea water.

Given:
Volume of fresh water $V_1 = 180 \text{ cm}^3$
Density of fresh water $\rho_1 = 1000 \text{ kg/m}^3$

First, convert the volume to cubic meters:

$$V_1 = 180 \text{ cm}^3 = 180 \times (10^{-2} \text{ m})^3 = 180 \times 10^{-6} \text{ m}^3$$

Now, calculate the mass of fresh water ($M_1 = \rho_1 V_1$):

$$M_1 = 1000 \text{ kg/m}^3 \times 180 \times 10^{-6} \text{ m}^3$$ $$M_1 = 0.180 \text{ kg}$$

Given:
Volume of sea water $V_2 = 2800 \text{ cm}^3$
Density of sea water $\rho_2 = 1025 \text{ kg/m}^3$

First, convert the volume to cubic meters:

$$V_2 = 2800 \text{ cm}^3 = 2800 \times (10^{-2} \text{ m})^3 = 2800 \times 10^{-6} \text{ m}^3$$

Now, calculate the mass of sea water ($M_2 = \rho_2 V_2$):

$$M_2 = 1025 \text{ kg/m}^3 \times 2800 \times 10^{-6} \text{ m}^3$$ $$M_2 = 2.870 \text{ kg}$$

Total mass $M_{total} = M_1 + M_2$:

$$M_{total} = 0.180 \text{ kg} + 2.870 \text{ kg}$$ $$M_{total} = 3.050 \text{ kg}$$

Total volume $V_{total} = V_1 + V_2$:

$$V_{total} = 180 \text{ cm}^3 + 2800 \text{ cm}^3$$ $$V_{total} = 2980 \text{ cm}^3$$

Convert total volume to cubic meters:

$$V_{total} = 2980 \times 10^{-6} \text{ m}^3$$

Density of mixture $\rho_{mix} = \frac{M_{total}}{V_{total}}$:

$$\rho_{mix} = \frac{3.050 \text{ kg}}{2980 \times 10^{-6} \text{ m}^3}$$ $$\rho_{mix} = \frac{3.050}{0.002980} \text{ kg/m}^3$$ $$\rho_{mix} \approx 1023.49 \text{ kg/m}^3$$

The final answer is $\boxed{1023.49 \text{ kg/m}^3}$.

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4. Explain why fish can survive under water.

Fish can survive underwater because they possess specialized respiratory organs called gills. Gills are highly efficient at extracting dissolved oxygen from the water. Water contains oxygen in a dissolved state, which fish absorb as it passes over their gills. The gills have a large surface area and a rich blood supply, facilitating the efficient exchange of oxygen from the water into the fish's bloodstream and carbon dioxide from the blood into the water.