-2 for O
+1 for Na
+6 for S
In Na2SO4, the oxidation state of sodium (Na) is +1, the oxidation state of sulfur (S) is +6, and the oxidation state of oxygen (O) is -2. To calculate the oxidation state of the whole compound, you can use the rule that the sum of the oxidation states in a neutral compound is zero, so in this case it would be +1*2 + (-2)*4 = 0.
The oxidation number of S in Na2SO4 is +6. This is because Na has an oxidation number of +1 and O has an oxidation number of -2. By setting up and solving an equation, we find that the oxidation number of S must be +6 to balance the charges in the compound.
The oxidation number of S in Na2SO4 is +6. This is because the oxidation number of Na is +1 and of O is -2, and the overall charge of the compound is zero. Therefore, 2(+1) + x + 4(-2) = 0, where x is the oxidation number of S, which simplifies to x = +6.
In the compound Al₂O₃, aluminum has an oxidation number of +3, and oxygen has an oxidation number of -2. This gives a total charge of zero for the compound, as it should be electrically neutral.
The oxidation number of oxygen (O) in MnO4 is -2. To find the oxidation number of manganese (Mn) in MnO4, we use the fact that the sum of the oxidation numbers in a compound is zero (since MnO4 is a polyatomic ion). Therefore, Mn must have an oxidation number of +7 in this compound.
In Na2SO4, the oxidation state of sodium (Na) is +1, the oxidation state of sulfur (S) is +6, and the oxidation state of oxygen (O) is -2. To calculate the oxidation state of the whole compound, you can use the rule that the sum of the oxidation states in a neutral compound is zero, so in this case it would be +1*2 + (-2)*4 = 0.
The oxidation number of S in Na2SO4 is +6. This is because Na has an oxidation number of +1 and O has an oxidation number of -2. By setting up and solving an equation, we find that the oxidation number of S must be +6 to balance the charges in the compound.
The oxidation number of S in Na2SO4 is +6. This is because the oxidation number of Na is +1 and of O is -2, and the overall charge of the compound is zero. Therefore, 2(+1) + x + 4(-2) = 0, where x is the oxidation number of S, which simplifies to x = +6.
In the compound Al₂O₃, aluminum has an oxidation number of +3, and oxygen has an oxidation number of -2. This gives a total charge of zero for the compound, as it should be electrically neutral.
The oxidation number of oxygen (O) in MnO4 is -2. To find the oxidation number of manganese (Mn) in MnO4, we use the fact that the sum of the oxidation numbers in a compound is zero (since MnO4 is a polyatomic ion). Therefore, Mn must have an oxidation number of +7 in this compound.
The oxidation number for oxygen (O) in KO2 is -1. Since the overall charge of the compound is 0, the oxidation number of potassium (K) is +1.
+3 for Cr and -2 for O
The oxidation number of sodium (Na) is +1. The oxidation number of carbon (C) in a compound is typically +4, except in the case of CO2 where it is +4 for each oxygen (O). In oxalate (C2O4), the overall charge is -2, so the oxidation number of oxygen (O) is -2 in this compound.
In CsAsO3, cesium (Cs) is in Group 1A, which has an oxidation number of +1. Oxygen (O) typically has an oxidation number of -2. Since the compound is neutral, the oxidation number of arsenic (As) can be calculated using the sum of the oxidation numbers in the compound, which is +5 for As in this case.
In the compound NO2, nitrogen has an oxidation number of +4 and each oxygen atom has an oxidation number of -2.
The oxidation number for oxygen (O) in K2O2 is -1. In this compound, each oxygen atom has an oxidation number of -1, which is typical for peroxides like K2O2.
In Fe2O3, iron (Fe) has an oxidation number of +3, while oxygen (O) has an oxidation number of -2. This means that each Fe atom contributes +3 to the compound, and each O atom contributes -2.