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

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
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lish632 1年前 已收到2个回答 举报
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锦衣夜心 幼苗

共回答了14个问题采纳率：100% 举报

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

1年前

7

吹儿 幼苗

共回答了47个问题 举报

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...

1年前

1
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