高中化学题求解!!!高手进!!!有关有机物的,要有解析

2025-02-19 11:58:33
推荐回答(1个)
回答1:

  1. 选B

    A错误,同系物是结构相似,两分子分子式相差若干个CH2的两个物质,也就是说两个物质必须不饱和度相同,而该物质不饱和度比苯多,因此两个物质不是同系物。

    B正确,该物质分子式是C22H12,1mol C原子消耗1molO2,4mol H原子消耗1mol氧气,因此该分子1mol完全燃烧会消耗25mol氧气。

    C错误,苯环中有一个大π键,需要3个H2;其它地方有8个双键,需要8个H2,因此1 mol 该有机物最多可以和11 mol H2 发生加成反应

    D错误,该有机物关于中间对称轴对称,因此左面的氢被取代与右面的氢被取代是同一种情况,因此该有机物的一氯代物的同分异构体有6种。

  2. 答案C2H6;C2H4;60%

    设烃的分子式为CnH2n+2,烯烃的分子式为CnH2n,设烃有xL,烯烃有(2-x)L
    由化学方程式可得,xL烃完全燃烧产生nxL二氧化碳和(n+1)xL水
                    (2-x)L烯烃完全燃烧产生n(2-x)L二氧化碳和n(2-x)L水
    nx+n(2-x)=2n=4,得n=2,
    所以烃的分子式为C2H6(乙烷),烯烃的分子式C2H4(乙稀)
    (n+1)x+n(2-x)=x+2n=4.8,得x=0.8,所以A有0.8L,B有1.2L,
    所以B的体积分数为1.2/2=60%

  3. 答案:

    化合物B、C是CO2和H2O

    (1)C6H12
    (2)42g      100.8
    (3)

    (4)3种


解析:(1)0.2mol烃A含碳原子1.2mol。含氢原子2.4mol。所以1mol烃A含碳原子6mol,含氢原 .        子12mol。即烃A的分子式为:C6H12。

.      (2)生成二氧化碳和水各3mol,其中碳、氢全部来源于烃,即3mol碳、6mol氢,计算得42g;其中氧全部来源于氧气,9mol氧、4.5mol 氧气,标况下1mol气体有22.4L,所以4.5mol * 22.4mol/L =100.8L

.      (3)不能使溴水褪色,说明不含双键,但不饱和度等于1,只可能是成环造成的,且只有一个一取代物说明氢只有一种环境,再由分子式可知是环己烷

.      (4)比A少2个碳原子说明有4个C;A的烯烃的同系物也是烯烃,因此分子式是C4H8。

三种同分异构体:


如满意请采纳,不懂请追问。

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