hi can anyone help me with this please?
2017-10-30 11:43 am
If 0.135g of CuCl2 are dissolved in 25.00 ml, and titrated with 35.45 ml of solution ( of unknown concentration) to reach the end point, what mass of copper atoms is present in the sample at the end of the titration?
回答 (1)
2017-10-30 12:45 pm
Mass of 1 mole of CuCl₂ = (63.5 + 35.5×2) g/mol = 134.5 g
Mass of Cu in 1 mole of CuCl₂ = 63.5 g
Mass fraction of Cu in CuCl₂ = 63.5/134.5
Initial mass of Cu atoms = (0.135 g) × (63.5/134.5) = 0.0637 g
According to the law of conservation of mass, mass of Cu atoms at the end of titration = 0.0637 g
Mass of Cu in 1 mole of CuCl₂ = 63.5 g
Mass fraction of Cu in CuCl₂ = 63.5/134.5
Initial mass of Cu atoms = (0.135 g) × (63.5/134.5) = 0.0637 g
According to the law of conservation of mass, mass of Cu atoms at the end of titration = 0.0637 g
收錄日期: 2021-04-18 17:53:41
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