Final answer:
To find how many electrons leave a 9.0-V battery per minute for a bulb with a 1.9 Ω resistance, the current is calculated using Ohm's Law and then the charge flowing per minute. Dividing this charge by the charge of an electron gives the number of electrons, which is 1.77 x 10^21 electrons.
Step-by-step explanation:
The question is asking how many electrons leave a 9.0-V battery per minute when it is connected to a bulb with a resistance of 1.9 Ω. First, we need to calculate the current flowing through the bulb using Ohm's Law, which states that current (I) equals voltage (V) divided by resistance (R). Thus, I = V/R.
For a 9.0-V battery and 1.9 Ω bulb, I = 9.0 V / 1.9 Ω = 4.74 A.
Next, we calculate the number of coulombs per minute, knowing that 1 ampere equals 1 coulomb per second. So, 4.74 A means 4.74 coulombs per second, and for a minute (60 seconds), it would be 4.74 C/s * 60 s = 284.4 C.
The charge of one electron is approximately 1.602 x 10-19 coulombs. Therefore, the number of electrons can be calculated by dividing the total charge by the charge of one electron:
Number of electrons = Total charge / Charge of one electron = 284.4 C / 1.602 x 10-19 C = 1.77 x 1021 electrons.
So, 1.77 x 1021 electrons leave the battery per minute.