Answer:
613 mg
Explanation:
Number of fargday's 
Here, I = 9.20 A
t = 10.5 min
= 10.5 x 60 seconds
So, 
= 0.0208 F
Here, 2e, 2F
2F = 1 mol of Ni
1 mol = 59 gm of Ni
0.0104 mol = 59 x0.0104 gm Ni
= 0.613 gm Ni
= (0.613 x 1000 ) mg of Ni
= 613 mg of Ni
Answer:
-125.4
Explanation:
Target equation is 4C(s) + 5H2(g) = C4H10
These are the data equations for enthalpy of combustion
- C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
- H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
- 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse
To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...
Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.
The reactants in the neutralization reaction are an acid and a base while the products are a salt and water.
