摘要:(1)1cm3中阴.阳离子总数=()3 1cm3中Ni2+-O2-离子对数=()3× 密度= (2)解法Ⅰ:设1 mol中含Ni3+x mol, Ni2+mol. 根据电中性 3x mol +2mol=2×1 mol ∴x=0.06 Ni2+为mol=0.91 mol 离子数之比 Ni3+:Ni2+=0.06∶0.91=6∶91
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