配套习题- 二氧化硅的性质和用途（2） -【青果学院】

1/5单选题

难度：

1.SiO₂属于酸性氧化物的理由主要是

A	硅是非金属元素
B	SiO₂对应的水化物是可溶性弱酸
C	SiO₂能与强碱反应生成盐和水
D	SiO₂不能与酸反应

正确答案：Qw==

5LqM5rCn5YyW56GF6Jm954S25LiN5LiO5rC05Y+R55Sf5Y+N5bqU77yM5omA5a+55bqU55qE6YW45Lmf5bCx5piv56GF6YW477yM5Lmf5bm26Z2e5Y+v5rq25oCn6YW477yM5L2G5a6D6IO95aSf5LiO5by656Kx5Y+R55Sf5Y+N5bqU77yM5omA5Lul5bGe5LqO6YW45oCn5rCn5YyW54mp44CC5bqU5b2T6YCJ5oupQ+mhueOAgg==

2/5单选题

难度：

2.有关二氧化硅的叙述不正确的是

A	与所有的酸都反应
B	能与强碱反应
C	属于酸性氧化物
D	在一定条件下能与生石灰反应

正确答案：QQ==

5LqM5rCn5YyW56GF5Y+v5Lul5LiO5rCi5rCf6YW45Y+N5bqU77yM5L2G6L+Z5piv5Liq54m55q6K55qE5Y+N5bqU77yM5bm26Z2e5LiO5omA5pyJ6YW46YO95Y+v5Lul5Y+R55Sf5Y+N5bqU77ybQemhuemUmeivr++8mzxicj7kuozmsKfljJbnoYXlj6/ku6XkuI7lpKflpJrmlbDnmoTlvLrnorHjgIHnorHmgKfmsKfljJbnianlj5HnlJ/lj43lupTvvIzlsZ7kuo7phbjmgKfmsKfljJbnianvvIznlJ/nn7PngbDnmoTkuLvopoHmiJDliIbmmK/msKfljJbpkpnvvIzlj6/ku6XkuI7kuozmsKfljJbnoYXlnKjpq5jmuKnkuIvlj5HnlJ/lj43lupTjgILlm6DmraRC44CBQ+OAgUTmraPnoa7jgII8YnI+5pys6aKY6YCJ5oupQemhueOAgg==

3/5单选题

难度：

3.下列关于CO₂和SiO₂性质的说法中，不正确的是

A	CO₂通入到水玻璃中可以得到硅酸
B	CO₂与SiO₂在物理性质上差别很大
C	SiO₂与CaCl₂溶液不反应，而CO₂通入CaCl₂溶液中可得到白色沉淀
D	CO₂和SiO₂都是酸性氧化物，都能与NaOH溶液反应

正确答案：Qw==

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

4/5单选题

难度：

4.下列说法正确的是

A	SiO₂不与HNO₃、H₂SO₄等所有酸反应
B	CO₂和SiO₂的对应水化物都是酸，其酸性碳酸比硅酸弱
C	CO₂与SiO₂均是酸酐，溶于水均得到相应的酸
D	木炭、活性炭、硅藻土、硅胶都可作吸附剂

正确答案：RA==

5LqM5rCn5YyW56GF5LiN5LiO5aSn5aSa5pWw55qE6YW45Y+N5bqU77yM5L2G5piv5Y+v5Lul5LiO5rCi5rCf6YW45Y+R55Sf5LiA5Liq54m55q6K5Y+N5bqU77yM5pWFQemhuemUmeivr++8mzxicj7kuozmsKfljJbnoYXlr7nlupTnmoTphbjkuLrnoYXphbjvvIzlhbbphbjmgKfmr5TnorPphbjlvLHvvJvkvYbkuozmsKfljJbnoYXkuI3og73kuI7msLTlj5HnlJ/lj43lupTjgII8YnI+5pWFQuOAgUPplJnor6/vvJs8YnI+57u85LiK77yM6YCJ5oupROmhueOAgg==

5/5单选题

难度：

5.将Ca(HCO₃)₂和SiO₂的混合物在高温下加热，充分反应后，所得固体的质量为原混合物质量的一半，则原混合物中Ca(HCO₃)₂与SiO₂物质的量之比及所得固体的成分分别是

A	6:5 CaO和CaSiO₃
B	5:6 CaSiO₃和SiO₂
C	6:5 CaO、SiO₂和CaSiO₃
D	无法判断

正确答案：QQ==

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