A9.5 v battery supplies a 2.0 ma current to a circuit for 6.0 h . you may want to review (page) . for with math skills, you may want to review: conversion between seconds and hours part a how much charge has been transferred from the negative to the positive terminal

= 43.2 Coulombs

Explanation:

Using the formula;

Q = it

where Q is the charge in coulombs

I is the current in Amperes and

t is the time is seconds

I = 2.0 mA = 0.002 A

t = 6.0 h = 6 × 3600 seconds

Therefore;

Q = 0.002 × 6 × 3600 s

   = 43.2 Coulombs

Therefore the charge that has been transferred from the negative to the positive terminal is 43.2 coulombs.

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