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How many milliliters of 3.0 m h2so4 are required to react with 0.80 g of cuo in the equation cuo h2so4 cuso4 H2O?
We need to go from grams of CuO to mL of H2SO4. Atomic weight of CuO = 63.55 g Cu + 16 g O = 79.55 g CuO (.80 g CuO) * (1 mol CuO / 79.55 g CuO) = .0100566 mol CuO (g CuO cancel) Since the moles of CuO is a 1:1 ratio to H2SO4 (see balanced equation) we know that: mol CuO = mol H2SO4 or 0.0100566 mol CuO = 0.0100566 mol H2SO4 3.0 M of H2SO4 means that there is 3 mol / 1 L. So we can divide this by the moles to get L then mL of H2SO4 (0.0100566 mol H2SO4) * (1 L H2SO4 / 3 mol H2SO4) * (1000 mL H2SO4 / 1 L H2SO4) = 3.4 mL H2SO4 (mol H2SO4 and L H2SO4 cancel) So 3.4 ml of H2SO4 is needed to react with 0.80 g of CuO.
What is the balanced equation for the reaction between copper oxide and sulfuric acid?
The balanced chemical equation for the reaction between copper(II) oxide (CuO) and sulfuric acid (H2SO4) is: CuO + H2SO4 → CuSO4 + H2O
Equation for when copper oxide reacts with sulfuric acid?
The reaction between copper oxide (CuO) and sulfuric acid (H2SO4) produces copper sulfate (CuSO4) and water (H2O). The balanced chemical equation for this reaction is: CuO + H2SO4 -> CuSO4 + H2O
What is the formula for copper oxide and hydrogen sulphate?
copper oxide- CuO hydrogen sulphate- h2SO4
What is the balanced equation for the reaction of sulphuric acid and copper oxide?
This will depend upon which version of copper oxide you have - copper I oxide or copper II oxide. For copper I oxide: H2SO4 + Cu2O --> Cu2SO4 + H2O For copper II oxide: H2SO4 + CuO --> CuSO4 + H2O
How many milliliters of 3.0 m h2so4 are required to react with 0.80 g of cuo in the equation cuo h2so4 cuso4 H2O?
We need to go from grams of CuO to mL of H2SO4. Atomic weight of CuO = 63.55 g Cu + 16 g O = 79.55 g CuO (.80 g CuO) * (1 mol CuO / 79.55 g CuO) = .0100566 mol CuO (g CuO cancel) Since the moles of CuO is a 1:1 ratio to H2SO4 (see balanced equation) we know that: mol CuO = mol H2SO4 or 0.0100566 mol CuO = 0.0100566 mol H2SO4 3.0 M of H2SO4 means that there is 3 mol / 1 L. So we can divide this by the moles to get L then mL of H2SO4 (0.0100566 mol H2SO4) * (1 L H2SO4 / 3 mol H2SO4) * (1000 mL H2SO4 / 1 L H2SO4) = 3.4 mL H2SO4 (mol H2SO4 and L H2SO4 cancel) So 3.4 ml of H2SO4 is needed to react with 0.80 g of CuO.
What is the balanced equation for the reaction between copper oxide and sulfuric acid?
The balanced chemical equation for the reaction between copper(II) oxide (CuO) and sulfuric acid (H2SO4) is: CuO + H2SO4 → CuSO4 + H2O
Equation for when copper oxide reacts with sulfuric acid?
The reaction between copper oxide (CuO) and sulfuric acid (H2SO4) produces copper sulfate (CuSO4) and water (H2O). The balanced chemical equation for this reaction is: CuO + H2SO4 -> CuSO4 + H2O
What is the formula for copper oxide and hydrogen sulphate?
copper oxide- CuO hydrogen sulphate- h2SO4
How do you chemically make copper sulphate?
Copper (II) sulphate, CuSO4 is prepared by mixing copper (II) oxide, CuO with sulfuric acid, H2SO4. CuO + H2SO4 --> CuSO4 + H2O
CuO plus HCL-CuCl2 plus H2O?
CuO + 2HCL - CuCl2 + H2O
What is the balanced equation for the reaction of sulphuric acid and copper oxide?
This will depend upon which version of copper oxide you have - copper I oxide or copper II oxide. For copper I oxide: H2SO4 + Cu2O --> Cu2SO4 + H2O For copper II oxide: H2SO4 + CuO --> CuSO4 + H2O
What is the Ionic equation of copper oxide and sulphuric acid?
The ionic equation for the reaction between copper oxide (CuO) and sulfuric acid (H2SO4) can be written as: CuO(s) + H2SO4(aq) → Cu2+(aq) + SO42-(aq) + H2O(l) This equation shows the dissociation of the reactants into their respective ions in solution.
What happens when dilute sulphuric acid is added to copper oxide?
When dilute sulphuric acid is added to copper oxide, a chemical reaction occurs, resulting in the formation of copper sulfate and water. The reaction can be represented by the equation: CuO + H2SO4 → CuSO4 + H2O.
Why does fizzing occur in a reaction between copper ii oxide and sulphuric acid?
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
Is Cu oxidized or reduced in CuO plus H2 yields Cu plus 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.
What is the symbol equation for copper carbonate plus sulphuric acid to copper sulphate plus water plus carbon dioxide?
Copper carbonate + sulphuric acid = copper sulphate + water + carbon dioxide
