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Earlier discussion
6 older comments kept for context.
Ismail
Mass = molar mass x I x t ÷ q x F
0.27 = 27 x I x 7200 ÷ 3 x 96500
0.27 = 19400 I ÷ 289500
cross multiply,
19400 I = 0.27 x 289500
19400 I = 78165
I = 0.4 A
Nate
1 mole of A^3+= 3×96500
= 289500
27g—>289500
0.27g—> X
C. M
27gX=0.27x289500
X =0.27×289500÷27g
X =3,126.6c
From Q=It
Making I subj of the formula
I =Q/t
Q=3126.6c
t=2hrs–>2×60×60=7200
I =3126/7200
I = 0.43...
Appr I= 0.4A
Lord Leo
Mg = 0.27g
t = 2 x60 x 60= 7200 secs
ne = 3 x 96500 = 289500
M = 27
Q= ?
mg = MQ/ne
0.27 = 27 x Q/ 289500
Cross multiply
78165 = 27 x Q
Q =78165/27
Q = It
2895 = I x 7200
2895 = 7200I
I = 2895/7200
I = 0.4A
Deviant king
What's ne??
Hassan Haruna Musa
SOLUTION
I=?
m= 0.27g
t= 2hour =2×60×60=7200s
F= 96500C
M= 1×27=27g/mil
N.E= Al+3 + 3e- =All
Therefore, N.E= 3
m/M = It/ N.E×F
0.27/27 = I×7200/ 3×96500
By cross multiplying,
0.27×289500= I×7200×27
78165= 194400×l
Divide both sides by the coefficient of letter I (current)
78165/19440 = 194400l/194400
I = 78165/19440
Therefore, I = 0.4A
Ozioma Miracle
Thanks