Answer:
18.0 Ampere is the size of electric current that must flow.
Explanation:
Moles of electron , n = 550 mmol = 0.550 mol
1 mmol = 0.001 mol
Number of electrons = N
[tex]N=N_A\times n[/tex]
Charge on N electrons : Q
[tex]Q = N\times 1.602\times 10^{-19} C[/tex]
Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds
1 min = 60 seconds
Size of current : I
[tex]I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}[/tex]
[tex]=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}[/tex]
[tex]I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A[/tex]
18.0 Ampere is the size of electric current that must flow.
