a. Co in LiCoO2: The oxidation number of Li is +1 and the oxidation number of O is -2. Therefore, to find the oxidation number of Co, we can set up the equation:

(+1) + x + 2(-2) = 0

Simplifying, we get:

+1 + x - 4 = 0

Solving for x, we find that the oxidation number of Co is +3.

b. Al in NaAlH4: The oxidation number of Na is +1 and the oxidation number of H is -1. Therefore, to find the oxidation number of Al, we can set up the equation:

(+1) + x + 4(-1) = 0

Simplifying, we get:

+1 + x - 4 = 0

Solving for x, we find that the oxidation number of Al is -3.

c. C in CH3OH: Hydrogen has an oxidation number of +1 in most compounds, and oxygen has an oxidation number of -2. Therefore, to find the oxidation number of C, we can set up the equation:

x + 3(1) + (-2) = 0

Simplifying, we get:

x + 3 - 2 = 0

Solving for x, we find that the oxidation number of C is -1.

d. N in GaN: GaN is an ionic compound, meaning that the oxidation numbers must balance out to zero. The oxidation number of Ga is +3 in most compounds, so the oxidation number of N must be -3.

e. Cl in HClO2: The oxidation number of H is +1 and the oxidation number of O is -2. Therefore, to find the oxidation number of Cl, we can set up the equation:

+1 + x + 2(-2) = 0

Simplifying, we get:

+1 + x - 4 = 0

Solving for x, we find that the oxidation number of Cl is +3.

f. Cr in BaCrO4: The oxidation number of Ba is +2 and the oxidation number of O is -2. Therefore, to find the oxidation number of Cr, we can set up the equation:

+2 + x + 4(-2) = 0

Simplifying, we get:

+2 + x - 8 = 0

Solving for x, we find that the oxidation number of Cr is +6.

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