Identify each situation as an example of diffusion, osmosis,
or active transport.

is A vecause the question gives you the answe

1.11×10⁸  Electrons.

Explanation:

Applying,

Q = CV Equation 1

Where Q = Total charge on the capacitor, C = Capacitance of the capacitor, V = Voltage of the battery.

Given: C = 1.97 pF = 1.97×10⁻¹² F, V = 9.0 V

Substitute this value into equation 1

Q = 1.97×9×10⁻¹²

Q = 1.773×10⁻¹¹ C.

If the charge on one electron = 1.602×10⁻¹⁹ C.

Therefore number of electron = 1.773×10⁻¹¹ /1.602×10⁻¹⁹

Number of electron = 1.11×10⁸