A 9.00-V battery delivers a current of 1.26 A to the lightbulb filament of a pocket flashlight. (a) How much charge passes through the filament in 1.60 min?
(b) How many electrons pass through the filament?
(c) Calculate the energy delivered to the filament during that time. J
(d) Calculate the power delivered by the battery. .

What do you mean there’s no jelly beans

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

Is there any opitions? if you have the periodic table, the noble gases are 8a family, the halogen are the 7a family, and the nonmetal are  carbon, nitrogen, phosphorus, oxygen, sulfur, selenium, fluorine, chlorine, bromine, iodine and astatine.