-2 for sulphur and +1 for each copper.
In CuSO4, the oxidation number of copper (Cu) is +2, and the oxidation number of sulfur (S) is +6. This is because the overall charge of the sulfate ion (SO4) is -2.
In copper sulfate (CuSO4), the oxidation number of sulfur is +6. This is because the oxidation number of copper (Cu) is +2 and oxygen (O) typically has an oxidation number of -2, so the algebraic sum in the compound should be zero.
To find the oxidation number of copper (Cu) in CuO, consider that oxygen (O) usually has an oxidation number of -2. Since CuO is a neutral compound, the oxidation number of Cu can be calculated by setting up an equation where the sum of the oxidation numbers equals zero. In this case, the oxidation number of Cu in CuO is +2.
The oxidation number of Cu in Cu2S (copper sulfide) is +1. This is because the overall charge of the compound is 0 and there are two Cu atoms each with an oxidation state of +1.
CuSO4 is copper (II) sulfate. The balanced equation for CuSO4 with water is CuSO4 + H2O reacts to become Cu+2 + HSO4-2 + OH-.
In CuSO4, the oxidation number of copper (Cu) is +2, and the oxidation number of sulfur (S) is +6. This is because the overall charge of the sulfate ion (SO4) is -2.
In copper sulfate (CuSO4), the oxidation number of sulfur is +6. This is because the oxidation number of copper (Cu) is +2 and oxygen (O) typically has an oxidation number of -2, so the algebraic sum in the compound should be zero.
To find the oxidation number of copper (Cu) in CuO, consider that oxygen (O) usually has an oxidation number of -2. Since CuO is a neutral compound, the oxidation number of Cu can be calculated by setting up an equation where the sum of the oxidation numbers equals zero. In this case, the oxidation number of Cu in CuO is +2.
Cu is oxidized. The oxidation number goes from 0 in Cu to +2 in CuSO4. S is reduced. The oxidation number goes from +6 in H2SO4 to +4 in SO2. The oxidizing agent is H2SO4 since it causes Cu to be oxidized. The reducing agent is Cu since it causes S in H2SO4 to be reduced.
The oxidation number of Cu in Cu2S (copper sulfide) is +1. This is because the overall charge of the compound is 0 and there are two Cu atoms each with an oxidation state of +1.
CuSO4 is copper (II) sulfate. The balanced equation for CuSO4 with water is CuSO4 + H2O reacts to become Cu+2 + HSO4-2 + OH-.
To balance the equation CuSO4 + Na, follow these steps: Write down the unbalanced equation: CuSO4 + Na → ? Determine the products of the reaction based on the elements involved: CuSO4 + Na → Cu + Na2SO4 Now, balance the equation by ensuring that the number of atoms for each element is the same on both sides: CuSO4 + 2Na → Cu + Na2SO4
The oxidation of any element, by itself, is zero.
CuSO4 Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O
The oxidation number of Cu in CuHSO4 is +1. This is because the overall charge of the compound is 0, and the oxidation number of H is +1, S is +6, and O is -2. By calculating the total oxidation numbers and assigning x to Cu, you can solve for x to find that Cu is in the +1 oxidation state.
Zn + CuSO4 --> ZnSO4 + Cu
Oxidation number is the charge per atom in a compound. Cl2= 2- (Cl= 1-) Cu would have to be 2+ to balance the compound, because there is only one copper atom.