Cheper 1 Homberenk Question 30 of 30 (1 pernf) I Question Attempt: 1 of 3 Marco 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 $$ =1.00 \frac{f}{\mathrm{mI}} $$, calculate the length of one side of such a cell if it were a cube. Round your answer to $$ \mathbf{3} $$ significant digits. $$ 3.27 \times 10^{-3} \mathrm{~cm} $$ 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.

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To determine the total surface area of the cell layer when all cells are spread out into a single layer, we'll follow these steps:

### **Given:**
- **Number of cells (N):** $$ 3.82 \times 10^{10} $$
- **Side length of each cubic cell (s):** $$ 3.27 \times 10^{-3} $$ cm

### **Step 1: Convert Side Length to Meters**
First, convert the side length from centimeters to meters:
$$
s = 3.27 \times 10^{-3} \, \text{cm} = 3.27 \times 10^{-5} \, \text{m}
$$

### **Step 2: Calculate the Area of One Cell**
The area of one face of a cubic cell is:
$$
\text{Area per cell} = s^2 = (3.27 \times 10^{-5} \, \text{m})^2 = 1.069 \times 10^{-9} \, \text{m}^2
$$

### **Step 3: Calculate Total Surface Area**
Since all cells are spread out in a single layer, the total surface area $$ A $$ is:
$$
A = N \times \text{Area per cell} = 3.82 \times 10^{10} \times 1.069 \times 10^{-9} \, \text{m}^2
$$
$$
A \approx 40.82 \, \text{m}^2
$$

### **Final Answer:**
$$
\boxed{40.8 \ \mathrm{m^{2}}}
$$
The total surface area of the cell layer is $$ 40.8 \, \mathrm{m}^{2} $$.

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