初三化学题。 A,B,C,D,E是初中化学中常见的不同类别的物质。已知A是单质,C是红棕色固体,E是水溶液可...

2025-03-15 16:53:14
推荐回答(4个)
回答1:

“A是单质,C是红棕色固体”,再结合图示C→A,又“A、B、C、D、E是初中化学常见的不同类别的物质”,可推测C为氧化铁,则A为铁;结合图示可知,D能分别与A、C反应,而能同时与金属和金属氧化物反应的物质应是酸,故D为酸(如盐酸);再结合图示和酸的通性分析,由于B、E也能与D反应,故B、D应分别是碱和盐;利用复分解反应的规律和条件可知,像“碱+盐→碱+盐”这一类都符合题意,再结合上面的分析可知,E可以是碳酸钠、B可以是氢氧化钠,因为二者均能与盐酸反应,且满足E到B的转化;当然,E也可以是氢氧化钙、而B可以是碳酸钙,因为二者也均能与盐酸反应,也能满足E到B的转化。
(1)A:Fe E:Na2CO3(或K2CO3) C:Fe2O3
(2)Na2CO3+Ca(OH)2=CaCO3+2NaOH(或用K2CO3)炼铁
(3)A是铁,D为酸

回答2:

(1) :A,E,C的化学式分别为
Fe Na2CO3 Fe2O3
(2) :由E转化为B的化学方程式为Na2CO3+Ca(OH)2=CaCO3↓+2NaOH,
由C转化为A的原理在工业上常用于:炼铁。
(3) :根据题意,A,B,C,D,E是不同类别的物质,A是单质,D是:酸(如盐酸等强酸)

回答3:

A:Fe B:NaOH C:Fe2O3 D:HCl(酸就行) E:Na2CO3(强碱弱酸盐)
这样应该可以了吧。工业上那个是炼钢。都是不同类别的。酸碱盐单质氧化物都在了。

回答4:

1 Fe NaHCO3 Fe2O3
2 2NaHCO3==== Na2CO3+ H2O+CO2↑
3 酸

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