If the dimensions of a cube-shaped cell with a volume of 1 pm were to
double, what would be the new surface area of the cell?
o a: 12 um
o b. 24 uma
o c. 6um
o d. 3 um

B

Explanation:

Assuming the units are all supposed to be the same

A cube shaped cell with a 1 um³ volume would have a dimensions of 1 x 1 x 1   (because volume = length x width x height and 1 x 1 x 1 =1)

If we double the dimensions, we have a cube with the dimensions 2 x 2 x 2.

The surface area of each face of the cube is the length x width, so 2 x 2. This means the surface area of each face of the cube is 4 um².

There are 6 faces of a cube, so the total surface area is 6 x 4 um² = 24 um².

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.