Five water molecules.
CuSO4•5H2O + heat ---> CuSO4 + 5H2O
their isn't one CuSO4 is an anhydrous salt which will absorb water so the way to find out how much is in it is to find out the difference in water befor and after addition and calculate it by finding the mols of water absorbed incomplarison with the number of mols of CuSO4 used. it is normally wrighten nH2O. CuSO4
No, the percent by mass of copper in CuSO4 5H2O will be different than in CuSO4 because CuSO4 5H2O includes water molecules in addition to the copper sulfate compound itself. Therefore, the total mass of CuSO4 5H2O will be greater, resulting in a lower percent by mass of copper in CuSO4 5H2O compared to CuSO4.
29.8g H2O = 1.66 mol H2O Molar Mass CuSO4 * 5H2O = 249.6 g mol CuSO4 * 5H2O --> 5 mol H2O 249.6 g CuSO4 * 5H2O/1 mol CuSO4 * 5H2O Times * 1mol CuSO4 * 5H2O/5mol H2O Times* 1.66 mol H2O = 82.6 g CuSO4 * 5H2O
The balanced equation for the heating of copper(II) sulfate pentahydrate (CuSO4•5H2O) is: CuSO4•5H2O(s) -> CuSO4(s) + 5H2O(g). This reaction decomposes the pentahydrate compound into anhydrous copper(II) sulfate and water vapor.
CuSO4.5H2O(s) --> CuSO4(s) + 5H2O(g)
CuSO4 * 5H2O ----> CuSO4 + 5H2O. This is true because CuSO4 * 5 H2O is a salt weakly bounded to water, that is why it is hydrous. When it decomposes, the weak bonds are broken making the products above. CuSO4*5H2O formula is [Cu(OH2)4]SO4*H2O CuSO4 + 5H2O --> [Cu(OH2)4]SO4*H2O
CuSO4 * 5H2O
CuSO4 5H2O
The chemical equation for the dissolution of copper(II) sulfate pentahydrate in water is: CuSO4 5H2O (s) Cu2 (aq) SO42- (aq) 5H2O (l)
The molecular mass of sodium thiosulfate heptahydrate is 248,18 g.
Na2B4O7.10H2O + 2HCl --> 4H3BO3 + 2NaCl + 5H2O