How Many Electrons Pass Through a Capacitor When Discharged?

[fresh_42]

Load is capacity times voltage. Capacity is given as $20\,\mu F$ which is (see previous post) equal to $20\cdot 10^{-6}\cdot A\cdot s \cdot V^{-1}\,.$ Voltage is given as $24\,V$ so the Volt cancel in the product and we are left with $Q=20\cdot \mu \cdot F \cdot 24 \cdot V = 480\cdot \mu \cdot F\cdot V= 480 \cdot \mu \cdot A \cdot s \cdot V^{-1} \cdot V = 480\cdot 10^{-6} \cdot A \cdot s$

 

[HazyMan]

I finally managed to understand where i went wrong.
This is what i came up with. (I initially thought of taking 300x10¹³ altogether and take the 0 zeroes from the 300 and put them on the ¹³ which would give me 3x10¹⁵ but i thought it was wrong. However it realized that both methods work, but i used yours anyway.
Thanks a lot for your help!