Heating is necessary to transform sugars in enediols which react with Cu(2+).
Cu+H2SO4 Gives CU2(SO4)2+H2O+S2O2
sugars are converted to enediols by benedict's reagent on boiling. these enediols reduce Cu(II) to Cu(I) which then forms CuOH (yellow in color). on heating CuOH yields Cu2O which is orange/red in color.
Cu(I), cuprous, Cu+.
The element copper (Cuprum in Latin) is symbolised by Cu
You can't convert between linear units and cubic units.
Heating is necessary to transform sugars in enediols which react with Cu(2+).
What are the high and low heating values for heptane ((Btu/cu ft)
You'll get first: blue Cu-sulfate pentahydrate crystall's and after further heating: white anhydrous Cu-sulfate powder.
Zn(s) --> Zn2+(aq) + 2e : Oxidation Cu+(aq) + 1e --> Cu(s) : Reduction
Copper is a good conductor to to it's mixed-valence Cu(II)/Cu(III) oxides which gives it zero electrical resistance.
Cu+H2SO4 Gives CU2(SO4)2+H2O+S2O2
Laboratory preparation of copper oxide: copper salt such as CuSO4 reacts with alkali (NaOH) to generate Cu(OH)2, and then Cu(OH)2 is decomposed by heating to prepare CuO.   CuSO4+2NaOH==Cu(OH)2↓+Na2SO4   Cu(OH)2=△=CuO+H2O Heat copper (Cu) to make copper react with oxygen (O2) to generate oxidation Copper (CuO).   2Cu+O2==2CuO Copper oxide can be prepared by heating metallic copper in air or decomposing copper hydroxide, copper nitrate, basic copper carbonate, etc.
The balanced equation is Cu(OH)2 (s) (heat) = CuO + H2O.
Cu(2-) does not exist, because Cu does not gain electrons at all being a metal. When it gives off two electrons Cu(2+) ions are formed.
mg+cuso4 gives mgso4+cu2+
Dissolving and hydratation reaction (this gives the blue color of Cu hydrated ions)CuSO4 + 4H2O --> Cu(H2O)42+ + SO42-followed by two differend protolysis reactions, of which the first is relatively stronger acidic than the second (wich is very weakly basic):Cu(H2O)42+ + H2O ==> Cu(OH)(H2O)3+ + H3O+SO42- + H2O --> HSO4- + OH-