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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.
11 g hydrogen are needed.
Copper(II) oxide: CuO
Both oxidation and reduction
Please mention the reaction.
No.
no reaction
6,26 g CuO are obtained.
The reaction is;CuSO4 = CuO + SO3
Copper(II) Oxide: CuO reaction with Nitric Acid: CuO + 2 HNO3 => Cu(NO3)2 + H2O Copper(I) Oxide: Cu2O reaction with Ntric Acid: Cu2O + 2HNO3 => CuNO3 + H2O
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.
1 mole CuO = 79.5454g CuO 3.2g CuO x 1mol CuO/79.5454g CuO = 0.040 mole CuO
CuO + H2SO4 --> CuSO4 + H2O
The chemical decomposition is:CuSO4------ CuO + SO3
CuO + CH3OH --> HCHO + Cu + H2O
The chemical formula for copper (ii) oxide is CuO The chemical formula for Sulphuric acid is H2SO4. So a chemical reaction between them would look like this: CuO + H2SO4 -> CuSO4 + H2O