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How long does a current of 10 A have to flow (about the usual 'household current ') before electrons of a total mass of 1 kg are passed through the conductor cross-section?
Notes m(e–) = 9.1·10-31 kg, load 1e– = 1.602·10-19 C
Note: Identical exercise also in the subject of redox
Solution: - · 10 A = 10 C/s
- · Number of electrons per second:
- 10/1.602·10-19 = 6.24·1019
- · Total Masse(e–) per second:
- 6.24·1019·9.1E.31 = 5.68·10-11 kg
- · How long to a total of 1 kg:
- 1/5.68·10-11 = 1.76·1010 s = about 558 years
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