Answer:
a) = 0.704%
b) = 1.30%
c) = 2.60%
Explanation:
Given that:
= 
For Part A; where Concentration of A = 0.270 M
Percentage Ionization(∝) 



percentage% (∝) = 
= 0.704%
For Part B; where Concentration of B =
M



percentage% (∝) = 0.0130 × 100%
= 1.30%
For Part C; where Concentration of C= 



percentage% (∝) = 0.02608 × 100%
= 2.60%
Answer:
Nitrogen (ii) oxide
Explanation:
To know the IUPAC name for NO, we shall determine the oxidation number of N in NO.
NOTE: The oxidation number of oxygen (O) is always – 2.
Thus the oxidation number of N in NO can be obtained as follow:
N + O = 0 (ground state)
N + (– 2) = 0
N – 2 = 0
Collect like terms
N = 0 + 2
N = +2
Thus, the oxidation number of Nitrogen (N) in NO is +2.
Therefore, the IUPAC name for NO is Nitrogen (ii) oxide
The arrangement of particles that make up an ionic compound would be an ionic lattice type of crystal arrangement. An ionic lattice type of structure will be formed due to many of the ionic bonds formed between the oppositely charged ions of the metal and nonmetal.
Answer:
Empirical Formula N2O6Sr Strontium Nitrate
Explanation:
N=13.2% O=45.4% Sr=41.4%