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Simply put, the products are a different color from the reactants. For example, when copper reacts with a silver nitrate solution it turns the solution from colorless to blue. This is because while the silver ions (Ag+) do not affect the solution while the copper ions (Cu2+) released by the reaction are blue in the presence of water.
29 protons and 27 electrons.
29 protons and 27electrons are present in Cu2+ ion.
Cu2(C2H3O2)4 + 2Na2S --> 2CuS + 4Na(C2H3O2) 170 mL Cu2(C2H3O2)4 X 1 L/ 1OOO mL X .3 M/ 1 Mol X 2 Mol CuS/ 1 Mol Cu2(C2H3O2)4 X 159.157 g/ 1 Mol CuS = 16.2 g
The electron configuration of Cu2+ is [Ar]3d94s0.
Most likely copper.
copper(II) phosphate can be made by mixing an aqueous copper(II) solution (i.e. CuCl2 or CuBr2) with an aqueous phosphate solution (i.e. monobasic or dibasic phosphate). A fluffy, light blue precipitate forms immediately upon addition of the copper(II) solution to the phosphate solution.
Copper exists in aqueous solutions as turquoise or blue colors, in both its 1+ and 2+ oxidation states (Cu+ and Cu2+}
Because in solution the ions Cu2+ and (SO4)2- are formed. In copper (metal) electrons can move free.
Cu2+ + I- --> Cu2I The compound created is Copper(I) Iodide
Type your answer here... it is because water is amphoteric and ammonia is basic
Cu(s) | Cu2+(aq) Au+(aq) | Au(s)
Copper two
Cu2+ + 4NH3 ----> [Cu(NH3)4]2-
Simply put, the products are a different color from the reactants. For example, when copper reacts with a silver nitrate solution it turns the solution from colorless to blue. This is because while the silver ions (Ag+) do not affect the solution while the copper ions (Cu2+) released by the reaction are blue in the presence of water.
Anhydrous copper(II) sulphate is white. When added to water, it forms a solution of CuSO4(aq) which is blue because of the Cu2+ ion, which is itself a transition metal ion.
Cu + Mg2 --------> Cu2 + Mg Cu --------------> Cu2 + 2e Mg2 + 2e --------> Mg Cu --------------> Cu2 + 2e (E = +0.35) Mg2 + 2e --------> Mg (E = -2.36V) +0.35 + (-2.36) = -2.01V --------------------------------------… Mg + Cu2 --------> Mg2 + Cu Mg --------------> Mg2 + 2e Cu2 + 2e --------> Cu Mg --------------> Mg2 + 2e (E = +2.36V) Cu2 + 2e --------> Mg (E = -0.35V) +2.36 + (-0.35) = +2.01V