Answer:
0.68 V
Explanation:
For anode;
3Mg(s) ---->3Mg^2+(aq) + 6e
For cathode;
2Al^3+(aq) + 6e -----> 2Al(s)
Overall balanced reaction equation;
3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)
Since
E°anode = -2.356 V
E°cathode = -1.676 V
E°cell=-1.676 -(-2.356)
E°cell= 0.68 V
Answer:
Hence, 15.99 g of solid Aluminum Sulfate should be added in 250 mL of Volumetric flask.
Explanation:
To make 0.187 M of Aluminum Sulfate solution in a 250 mL (0.250 L) Volumetric flask
The molar mass of Aluminum Sulfate = 342.15 g/mol
Using the molarity formula:-
Molarity = Number of moles/Volume of solution in a liter
Number of moles = Given weight/ molar mass
Molarity = (Given weight/ molar mass)/Volume of solution in liter
0.187 M = (Given weight/342.15 g/mol)/0.250 L
Given weight = 15.99 g
Physical Change
It is being changed by the sun
Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole
There are 1.2 hr would this current have to be applied to plate out 7. 20 g of iron .
Calculation ,
Given ; Current ( I ) = 5. 68 A
In 
, the valancy of Fe is +2 .
2 moles of 
are required for the decomposition of 1 mole of Fe .
7. 20 g of Fe in moles = 7. 20 g /55.845 g/mol =0.12 mole
x moles of are required for the decomposition of 0.128 mole of Fe .
moles of are required = 0.256 moles
Charge on 1 mole of = 96500 C
Charge on 0.256 mole of = 24704 C
Current ( I )= Q/t
t =Q / I = 24704 C/5. 68 A = 4349 sec = 1.2 hr
Therefore , there are 1.2 hr would this current have to be applied to plate out 7. 20 g of iron .
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