I=nAve

3              0.6 mms^-1 or 6 x 10^-4 ms^-1

4              a)            (vol of Cu = 68.5/9000 and if this vol contains 6.0 x10²⁸delocalised electrons then 1m³ that number/that vol of Cu)

Ans = 8.5 x10²⁸

                b)            as there are 8.5 x10²⁸  delocalised electrons in 1m³ then the wire will contain that number x its volume)

Ans = 8.5 x10²²

                c)            1.4 x10⁴ C

                d)            (use Q=It)

Ans = 6.8 x10³ s

                e)            (v = s/t)

Ans = 0.15 mms^-1