A 6.0 volt battery is hooked up to a light bulb, whose resistance is 3.1 ohms. How many electrons would leave the battery per second? (Hint: 1 electron = 1.6E-19 C)

To calculate the number of electrons leaving the battery per second, we need to find the current flowing through the circuit. The current can be found using Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R):

I = V / R

In this case, the voltage of the battery is 6.0 volts and the resistance of the light bulb is 3.1 ohms. Plugging these values into the equation, we get:

I = 6.0 V / 3.1 Ω
I = 1.935 A

This means that the current flowing through the circuit is 1.935 Amperes.

To calculate the number of electrons leaving the battery per second, we need to know the charge of one electron. The charge of an electron is approximately 1.6 x 10^-19 Coulombs, as given in the hint.

The total charge that passes through the circuit per second can be calculated by multiplying the current (in Amperes) by the time (in seconds):

Charge (C) = Current (A) x Time (s)

Now, we know that 1 Coulomb (C) is equal to the charge carried by 1 / (1.6 x 10^-19) electrons. So, we can calculate the number of electrons leaving the battery per second by dividing the total charge (in Coulombs) by the charge of one electron (1.6 x 10^-19 C):

Number of electrons = Total charge (C) / Charge of one electron (C)

Number of electrons = (Current x Time) / (1.6 x 10^-19 C)

Plugging in the known values:

Number of electrons = (1.935 A x 1 s) / (1.6 x 10^-19 C)

Number of electrons ≈ 1.209 x 10^19 electrons

Therefore, approximately 1.209 x 10^19 electrons would leave the battery per second.