Econnect ter 1: HW 2 Question 9 (o value 0.00 points 1 out of 3 attempts Assis C

Question

Econnect ter 1: HW 2 Question 9 (o value 0.00 points 1 out of 3 attempts Assis Ch Vie Vie She Gui Pra Prir Be sure to answer all parts. A certain person had a brain that weighed 1.15 kg and contained 2.50 x 1010 cells. Assuming that each cell was completely filled with water (density 1.00 g/mL), calculate the length of one side of such a cell if it were a cube. cm 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 one side of the cell layer? Rep References eBook & Resources Difficulty: Medlum Multipart Answer Type here to search

Explanation / Answer

Part a

Mass of a cell of brain = (1.15 x 1000 g) / (2.5 x 10^10 cells)

= 4.6 x 10^-8 g/cell

Volume of one cell = mass of cell/density

= 4.6 x 10^-8 g/cell / 1 g/mL

= 4.6 x 10^-8 mL/cell

= 4.6 x 10^-8 cm3/cell

Volume of a cube = 4.6 x 10^-8 cm3/cell

a3 = 4.6 x 10^-8 cm3

a = (4.6 x 10^-8 cm3)1/3  

= 0.0035 cm

= 3.5 x 10-3 cm

Part b

Total volume of cells = mass of brain / density

= (1.15 x 1000 g) / (1 g/mL)

V = 1150 mL = 1150 cm3

Surface area = V/a

= (1150 cm3) / (0.0035 cm)

= 328571.42 cm2 x 1m2/10000cm2

= 32.85 m2

= 3.285 x 101 m2

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