Electricity

[victorias]

---

Members don't see this ad.

For this question, I got the right answer but I used a different method - I didn't take charge/electron into account.

Q = It = nF
n = (2x10^20)/(6.02x10^23) = 3 x 10^-4

So, I = nF/t ~ (3x10^-4)(10^5)/5 = ~ 6A

Is my method still correct?

 

[Bruskie_77]

Yes!

 

[aldol16]

For this question, I got the right answer but I used a different method - I didn't take charge/electron into account.

You actually did take charge into account. The Faraday constant gives you the charge per mole of electrons - that's its physical meaning. Think about what you're doing before you just go apply equations!