Answer:
82.4 s
Explanation:
Find the NUMBEr of half lives...then multiply by 54.3
2.27 = 6.5 (1/2)^n
log (2.27/6.5) / log (1/2) = n = 1.52 half lives
1.52 * 54.3 = 82.4 s
Answer:
Explanation:
To calculate the cell potential we use the relation:
Eº cell = Eº oxidation + Eº reduction
Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .
So the species that is going to be oxidized is the Aluminium, and therefore:
Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V
Equally valid is to write the equation as:
Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species
These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.
Answer:
3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)
Explanation:
In solubility rules, all ammonium and nitrates ions are solubles and all sulfates are soluble except the sulfates that are produced with Ca²⁺, Sr²⁺, Ba²⁺, Ag⁺ and Pb²⁺. That means the NH4NO3 and the Al2(SO4)3 produced are both <em>soluble and no precipitate is predicted. </em>
The reaction is:
<h3>3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)</h3>