Pleas help me with my homework

(2) (a) you have a simple circuit that consists of only a battery (δvbat=1.5v) and two resistors with resistance r1=10ω and r2=5ω, connected in series with each other. what is the current running through the battery? (b) you re-arrange your circuit so now r2 is attached in parallel to r1 rather than in series. what is the current running through the battery? (c) you add an additional resistor r3=7ω on the same branch as r2. what is the current running through the battery? (d) what is the power dissipated in r3?

Nacidified solution was electrolyzed using copper electrodes. a constant current of 1.18 a caused the anode to lose 0.584 g after 1.52 ✕ 103 s. given that the charge of an electron is 1.6022 ✕ 10−19 c, calculate avogadro's number. assume that copper is oxidized to cu2+ ions.

It is reasonable to assume that the bulk modulus of blood is about the same as that of water (2.2 gpa). as one goes deeper and deeper in the ocean, the pressure increases by 10000 pa for every meter below the surface. if a diver goes down 80.0 m in the ocean, by how much does each cubic centimeter of her blood change in volume? give the answer in cubic centimeters (actually one cubic centimeter equals one milliliter).