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標題:
分子熱運動 能量守恆 問題
發問:
銅的原子量為63.54,密度 P= 8900kg/cm*3,求 1cm*3銅的分子數,估算銅分子的直徑
最佳解答:
Since by Avogadro's hypothesis, 1 mole of any substance contains 6 x 10^23 atoms (Avogadro's number) Hence, 63.54 g of copper contains 6x10^23 copper atoms. Consider 1 cm3 of copper, the mass of which is 8.9 g. Thus by simple proportion, the number of atoms = (8.9/63.54) x 6 x 10^23 = 8.4x10^22 [Note: the density of copper is 8.9 g/cm3, NOT 8900 g/cm3] Since there are 8.4x10^22 atoms packed in 1 cm3 Volume of an atom = 1/(8.4x10^22) cm3 Assume each atom is packed in a cube, hence length of one side of cube equals the diameter of the atom, Diameter of an atom = cube-root[ 1/(8.4x10^22)] cm
分子熱運動 能量守恆 問題
發問:
銅的原子量為63.54,密度 P= 8900kg/cm*3,求 1cm*3銅的分子數,估算銅分子的直徑
最佳解答:
Since by Avogadro's hypothesis, 1 mole of any substance contains 6 x 10^23 atoms (Avogadro's number) Hence, 63.54 g of copper contains 6x10^23 copper atoms. Consider 1 cm3 of copper, the mass of which is 8.9 g. Thus by simple proportion, the number of atoms = (8.9/63.54) x 6 x 10^23 = 8.4x10^22 [Note: the density of copper is 8.9 g/cm3, NOT 8900 g/cm3] Since there are 8.4x10^22 atoms packed in 1 cm3 Volume of an atom = 1/(8.4x10^22) cm3 Assume each atom is packed in a cube, hence length of one side of cube equals the diameter of the atom, Diameter of an atom = cube-root[ 1/(8.4x10^22)] cm
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