A certain person had a brain that weighed 1.35 kg and contained \( 3.82 \times 10^{10} \) cells. Part 1 of 2 Assuming that each cell was completely filled with water (density \( =1.00 \frac{\mathrm{~g}}{\mathrm{~mL}} \) ), calculate the length of one side of such a cell if it were a cube. Round your answer to 3 significant digits. \( 3.27 \times 10^{-3} \) cm Part: \( 1 / 2 \) I 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. Par 2
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To find the total surface area when the cells are spread out in a single layer, we first need to establish the volume occupied by all the cells. Since the weight of the brain is given as 1.35 kg and assuming all the cells are filled with water (density = 1.00 g/mL), we can convert weight to volume: 1.35 kg = 1350 g → Volume = 1350 mL = 1350 cm³. Now, if we divide the total volume by the total number of cells \( (3.82 \times 10^{10}) \): \[ \text{Volume of one cell} = \frac{1350 \text{ cm}^3}{3.82 \times 10^{10}} = 3.53 \times 10^{-8} \text{ cm}^3. \] If we assume each cell is a cube, the length of one side \( s \) can be calculated using: \[ s^3 = 3.53 \times 10^{-8} \text{ cm}^3 \Rightarrow s = (3.53 \times 10^{-8})^{1/3} \approx 3.27 \times 10^{-3} \text{ cm}. \] For the total surface area (SA) of the cells when they are laid out in a single layer, we can find the total area by multiplying the number of cells by the area of one cell: Each cell, being a cube, has an area of: \[ A_{\text{cell}} = 6s^2 \] Since \( s \approx 3.27 \times 10^{-3} \) cm, we find: \[ A_{\text{cell}} = 6 (3.27 \times 10^{-3})^2 \approx 6.43 \times 10^{-5} \text{ cm}^2. \] Now multiplying by the total number of cells gives: \[ SA_{\text{total}} = 3.82 \times 10^{10} \times 6.43 \times 10^{-5} \text{ cm}^2 \approx 2.45 \times 10^6 \text{ cm}^2. \] Finally, converting square centimeters to square meters (1 m² = 10,000 cm²): \[ SA_{\text{total}} \approx 2.45 \times 10^6 \text{ cm}^2 \times \frac{1 \text{ m}^2}{10^4 \text{ cm}^2} \approx 245 \text{ m}^2. \] Thus, the total surface area for the top side of the cell layer is approximately **245 m²** when rounded to three significant digits.