A certain person had a brain that weighed 1.25 kg and contained \( 2.49 \times 10^{15} \) cella. Part 1 of 2 Assuming that each cell was completely Filied whth water (density \( =1.00 \frac{\mathrm{~g}}{\mathrm{~mL}} \)., calculate the liength of one side of sueh a cell if it were a cube. Round your answer to 3 significant digits. \( \square \) 0.00360 cm Part: 1 / 2 Part 2 of 2 If the cells were spread out into a thin layer that was a single cell thick, what would be the total surface area (in square meters) for the top side of the cell layer? Round your answer to 3 significant digits. \( \square \) \( m^{2} \) \( \square \)

To find the total surface area for the top side of the cell layer, we first need to calculate the volume of a single cell. Given that the total number of cells is \( 2.49 \times 10^{15} \), we can use the total volume filled with water: 1. Total volume = mass/density = \( 1250 \, \text{g} / 1.00 \, \text{g/mL} = 1250 \, \text{mL} = 1250 \, \text{cm}^3 \). 2. Volume of a single cell = total volume / number of cells = \( 1250 \, \text{cm}^3 / 2.49 \times 10^{15} \approx 5.02 \times 10^{-13} \, \text{cm}^3 \). Now, since the cells are cubes, the volume \( V = a^3 \) where \( a \) is the side length of one cube. Solving for \( a \): \[ a = (5.02 \times 10^{-13})^{1/3} \approx 8.00 \times 10^{-5} \, \text{cm} = 0.0000800 \, \text{cm} \] Now we need to find the total surface area when the cells are spread into a single layer. If each cell occupies \( a^2 \) on the surface, the total surface area for all cells would be: \[ \text{Total surface area} = a^2 \times \text{number of cells} = (8.00 \times 10^{-5} \, \text{cm})^2 \times 2.49 \times 10^{15} \] Calculating this gives: \[ \text{Total surface area} = 6.40 \times 10^{-9} \, \text{cm}^2 \times 2.49 \times 10^{15} \approx 159.36 \, \text{m}^2 \] Thus, the total surface area for the top side of the cell layer is approximately: \( \square \) \( 159 \, m^2 \) \( \square \)