正确答案: B

某气体能使湿润的红色石蕊试纸变蓝,该气体水溶液一定显碱性

题目:下列实验设计和结论相符的是

解析:B【解析】A项,乙醇不可以作为萃取剂,错;B项,石蕊变蓝,则肯定为碱性,正确。C项,若原溶液中含有SO32 -,生成BaSO3,再加入HCl,则与溶液的NO3- 结合,相当于HNO3,则可以氧化BaSO3至BaSO4,沉淀不溶解,故错;D项,加热时,FeCl3会水解,错。

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举一反三的答案和解析:

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  • whacver B.who


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