<h3>
Answer:</h3>
A. 1.4 V
<h3>
Explanation:</h3>
We are given the half reactions;
Ni²⁺(aq) + 2e → Ni(s)
Al(s) → Al³⁺(aq) + 3e
We are required to determine the cell potential of an electrochemical cell with the above half-reactions.
E°cell = E(red) - E(ox)
From the above reaction;
Ni²⁺ underwent reduction(gain of electrons) to form Ni
Al on the other hand underwent oxidation (loss of electrons) to form Al³⁺
The E.m.f of Ni/Ni²⁺ is -0.25 V and that of Al/Al³⁺ is -1.66 V
Therefore;
E°cell = -0.25 V - (-1.66 V)
= -0.250V + 1.66 V
= + 1.41 V
= + 1.4 V
Therefore, the cell potential will be +1.4 V
Answer:
Mass of KNO3= 10g
Molar mass of KNO3 = 101.1032g/mol
Volume = 250ml = 0.25L
No of mole on of KNO3 = mass of KNO3/Molar mass of KNO3
no of mole of KNO3 = 10/101.1032
No of mole of KNO3 = 0.09891
molarity of KNO3 = no of mole of KNO3/Vol (L)
Molarity = 0.09891/0.25 = 0.3956M
Molarity of KNO3 = 0.3956M
Answer:
monoxide
dioxide
trioxide
tetroxide
pentoxide
hexoxide
heptoxide
octoxide
nonoxide
Explanation:
Just the way it is. If there is an -a or -o infront of the oxide(or another substance that starts with vowel), the a is often dropped.
Answer:
3Mg(s) +2P(s) -------> Mg3P2(s) + energy
Keq= [Mg3P2]/[Mg]^3 [P]^2
Explanation:
The equation for the formation of magnesium phosphide from its elements is;
3Mg(s) +2P(s) -------> Mg3P2(s) + energy
Hence we can see that three moles of magnesium atoms combines with two moles of phosphorus atoms to yield one mole of magnesium phosphide. The equation written above is the balanced chemical reaction equation for the formation of the magnesium phosphide.
The equilibrium expression for the reaction K(eq) will be given by;
Keq= [Mg3P2]/[Mg]^3 [P]^2
