The solubility of copper sulfate pentahydrate at 100 0C is 114 g/100 g water.
To calculate the grams of copper sulfate needed to produce 29.8g of water, you need to consider the molar mass of water and copper sulfate. Then use stoichiometry to determine the amount of copper sulfate needed.
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 % of Cu by mass in CuSO4 will be greater than the % of Cu by mass in the pentahydrate (5H2O) because in the hydrate there is added mass (5 H2O = 90 g) but no added Cu.
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 + heat ---> CuSO4 + 5H2O
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
To calculate the grams of copper sulfate needed to produce 29.8g of water, you need to consider the molar mass of water and copper sulfate. Then use stoichiometry to determine the amount of copper sulfate needed.
No. The % of Cu by mass in CuSO4 will be greater than the % of Cu by mass in the pentahydrate (5H2O) because in the hydrate there is added mass (5 H2O = 90 g) but no added Cu.
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
CuSO4•5H2O + heat ---> CuSO4 + 5H2O
Copper sulfatepentahydrate is a solid at 20 degrees Celsius has a density of 2.28g/cm3 (cubed)... and its' chemical symbol is CuSO4-5H2O
CuSO4•5H2O
CuSO4 5H2O
CuSO4 * 5H2O
cuso4 - 5h2o= cuso4 + 5h20 + heat
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
The atomic mass of CuSO4 is 159.61 g/mol, and that of 5 H2O is 90.08 g/mol. To find the percentage of water, we calculate (90.08 g/mol / (90.08 g/mol + 159.61 g/mol)) * 100, which is approximately 36.1%. So, the crystal contains around 36.1% water to the nearest tenth.