The half-reaction are:
Cd ---> Cd(OH)₂
The oxidation number of Cd changed from 0 to +2. So, the number of mol electron transferred here is 2.
NiO(OH) --> Ni(OH)₂
The oxidation number of Cd changed from +3 to +2. So, the number of mol electron transferred here is 1.
Now, the greatest common factor would be 2. So, we use n=2 for the formula for ΔG°. F is Faraday's constant equal to 96,485 J/mol e.
ΔG° = nFE° = (2)(96,485)(1.5) =<em> 289,455 J</em>
Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
Answer is: mass of salt is 311,15 g.
V(H₂O) = 1,48 l · 1000 ml/l = 1480 ml.
m(H₂O) = 1480 g = 1,48 kg.
d(solution) = 1,00 g/ml.
ΔT(solution) = 13,4°C = 13,4 K.
Kf = 1,86 K·kg/mol; cryoscopic constant of water
i(NaCl) = 2; Van 't Hoff factor.
ΔT(solution) = Kf · b · i.
b(NaCl) = 13,4 K ÷ (1,86 K·kg/mol · 2).
b(NaCl) = 3,6 mol/kg.
n(NaCl) = 3,6 mol · 1,48 kg= 5,328 mol.
m(NaCl) = 5,328 mol · 58,4 g/mol = 311,15 g.
It is b, i learned this last year hope this helps
