某元素A最外层有3个电子,其氧化物中含氧52.9%,则A的相对原子质量

来源:百度知道 编辑:UC知道 时间:2024/07/05 03:38:48
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最外层有三个电子,则其氧化物为A2O3
则有48/(48+2A) = 52.9%
解得A = 21.37
答案就是如此

不过题目要是改成A在其氧化物中含量为52.9%的话
则有2A/(48+2A) = 52.9%
A = 27
为铝

你自己看着办吧 呵呵

21.4
设这种金属的符号为M,它的原子量为X
最外层有三个电子,所以与与氧化合时其化合价为+3价,故可设这种氧化物为M2O3
其氧化物中含氧52.9%,于是有(16×3)/(2X+16×3)=0.529,解得X=21.4
数据好像有点问题
应该是金属的含量为52.9%,那样原子量为27,是Al

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