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CuO is formed as the end product along with the liberation of SO2.
1. Iron sulfate is formed. 2. The released copper is deposited on the surface of the iron piece.
The name of the compound with the formula Cu2SO4 is Copper I sulphate or Cuprous sulphate.
2 FeSO4---------------Fe2O3 + SO2 + SO3
In the laboratory, copper(I) Iodide is prepared by simply mixing an aqueous solutions of potassium iodide and a soluble copper(II) salt such copper sulphate. : :: Cu2+ + 2I− → CuI2 The CuI2 immediately decomposes to iodine and insoluble copper(I) iodide, releasing I2. : :: 2 CuI2 → 2 CuI + I2
when blue Copper sulphate is heated, it loses its water part of crystallisation and tuns into white, anhydrous copper sulphate crystal
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CuO is formed as the end product along with the liberation of SO2.
1. Iron sulfate is formed. 2. The released copper is deposited on the surface of the iron piece.
The name of the compound with the formula Cu2SO4 is Copper I sulphate or Cuprous sulphate.
2 FeSO4---------------Fe2O3 + SO2 + SO3
copper+sulphur+oxygen then theres an arrow, but i don't know how 2 do an arrow copper sulphate
In the laboratory, copper(I) Iodide is prepared by simply mixing an aqueous solutions of potassium iodide and a soluble copper(II) salt such copper sulphate. : :: Cu2+ + 2I− → CuI2 The CuI2 immediately decomposes to iodine and insoluble copper(I) iodide, releasing I2. : :: 2 CuI2 → 2 CuI + I2
It would begin crystalising and would decrease
the answer is..... 2
a big lump of mess i dont think so copper sulphate +sodium hydroxide = Na2So4 +Cu(OH)2 WHICH IS SODIUM SULPHATE AND COPPER HYDROXIDE
penta = 5 copper(II) sulphate pentahydrate = CuSO4*5 H2O CuSO4*5 H2O + heat --> CuSO4 + 5 H2O