分子熱運動 能量守恆 問題

2009-04-26 12:14 am
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

回答 (1)

2009-04-26 10:06 am
✔ 最佳答案
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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