CuCO3 ==> CO2 + CuO
(heat is the catalyst, written above the arrow)
CuCO3 --> CO2 + CuO (gas + black powder) Further heating may result in : CuO --> 2Cu + O2 (metal copper + gas)
Describe the appearance and odor of the liquid obtained by heating copper II sulfate pentahydrate.
2Cu(NO3)2(s)+heat------->2CuO(s)+4NO2(g)+O2(g)
Upon heating copper II nitrate, the decomposition reaction is as follows: 2 Cu(NO3)2(s) --> 4 NO2(g) + O2(g) + 2 CuO(s) First the NO2 gas is released, as it is the larger molecule, then O2 gas, leaving CuO solid remaining.
It is a physical change because it has only changed state.
CuCO3 ==> CO2 + CuO (heat is the catalyst, written above the arrow)
CuCO3 --> CO2 + CuO (gas + black powder) Further heating may result in : CuO --> 2Cu + O2 (metal copper + gas)
CuSO4•5H2O + heat ---> CuSO4 + 5H2O
Describe the appearance and odor of the liquid obtained by heating copper II sulfate pentahydrate.
The thermal decomposition in this case is: CuCO3 -------------CuO + CO2
Copper has a CTE of 16.6 parts per million/degree C (16.6E-6/C)
Let the fourmula for the hydrous copper sulphate be CuSO4XH20 where X represents the number of water molecules write a balanced equation for the heating of the blue copper sulphate crystals?
2Cu(NO3)2(s)+heat------->2CuO(s)+4NO2(g)+O2(g)
The balanced equation is Cu(OH)2 (s) (heat) = CuO + H2O.
It decomposes to carbon dioxide (gas) and copper oxide (black powder) or even to metallic copper when heated up to higher temperatures (in that case also oxygen is released). You see why chemists rather put it in a reaction equation, so it can also be 'balanced': This would then take only one (=1!!) line of symbols and numbers. CuCO3 ==> CO2 + CuO and at higher temperauture: 2CuO ==> 2Cu + O2
There is a balanced equation to use for decomposition of copper II sulfate pentahydrate. It is the following: CuSO4.5H2O+heat -->CuSO4(anhydrous)+5H2O.
Cu2CO3(OH)2.H2O = 2CuO + CO2 + 2H2O