Ecosystem is were non living and living things live
Answer:
A. NO₃⁻ ⇒ +5
B. NO₄⁺ ⇒ +9
C. N₂ ⇒ 0
D. NH₂OH ⇒ -1
E. NO₂⁻ ⇒ +3
Explanation:
To calculate the oxidation number of an element in a compound, we have to know the oxidation number of the other elements. Then, we have to consider that the sum of the oxidation number of each atom multiplied by the subscripts is equal to the net charge of the compound.
A. NO₃⁻
This is the ion nitrate. Oxygen atoms (O) has an oxidation number of -2 because it derives from an oxide. In this case, the net charge of the ion is -1. Thus, we calculate the oxidation number of N as follows:
N + (3 x (-2)) = -1
N - 6 = -1 ⇒ N= -1 + 6 = +5
B. NO₄⁺
In this case, the sum of the oxidation numbers of O and N multiplied by the subscripts is equal to +1:
N + (4 x (-2)) = +1
N - 8 = +1 ⇒ N = +1 +8 = +9
C. N₂
The oxidation number of N is 0 because N₂ is an elemental substance.
D. NH₂OH
The oxidation number of H is +1 and -2 for O. The net charge of the molecule is 0.
N + (H x 3) + (O) = 0
N + (+1 x 3) + (-2) = 0
N + 3 -2 = 0 ⇒ N = 2 - 3 = -1
E. NO₂⁻
The net charge of the ion nitrite is -1.
N + (2 x O) = -1
N + (2 x (-2)) = -1 ⇒ N = -1 + 4 = +3
Therefore:
Highest oxidation number = B. NO₄⁺ (+9)
Lowest oxidation number = D. NH₂OH (-1)
Answer:
hydroxide ion / OH-
Explanation:
Basic solutions have a greater concentration of hydroxide ions than hydrogen (H+) ions
B. 1922 m³
F. 612 π m³
hope this helps
The answer to the question is c! If u need me to explain I will in the comments but I hope this helped ;)
