(1)①锌比铜活泼,铜与稀硫酸不反应,形成原电池时,锌为负极,被氧化,铜为正极,正极发生还原反应生成氢气,电极方程式为2H
++2e
-=H
2↑,
故答案为:没有;锌;2H
++2e
-=H
2↑;
②正极发生2H
++2e
-=H
2↑,负极发生Zn-2e
-=Zn
2+,
n(H
2)=
=0.5mol,则转移1mol电子,
由电极方程式可知消耗n(Zn)=0.5mol,n(H
+)=1mol,
则c(H
2SO
4)=
=1mol/L,
c( ZnSO
4)=
=1mol/L,
故答案为:1;1;
(2)A.乙烯被氧化生成二氧化碳气体,引入新的杂质,应用溴水,故A错误;
B.乙醇和水混溶,不能用作萃取剂,应用苯或四氯化碳,故B错误;
C.淀粉在酸性条件下水解生成葡萄糖,葡萄糖与氢氧化铜浊液的反应应在碱性条件下进行,应在水解之后加入碱调节溶液至碱性,故C错误;
D.乙酸乙酯不溶于饱和碳酸钠溶液,可用于饱和碳酸钠溶液分离,故D正确;
故答案为:D;
(3)硬脂酸甘油酯与氢氧化钠反应生成C
17H
35COONa与甘油,反应方程式为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/faedab64034f78f0fbb848427a310a55b2191ccb?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
+3NaOH→3C
17H
35COONa+
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/024f78f0f736afc377bfedd4b019ebc4b64512cb?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
.
故答案为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/faedab64034f78f0fbb848427a310a55b2191ccb?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
+3NaOH→3C
17H
35COONa+
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/024f78f0f736afc377bfedd4b019ebc4b64512cb?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
(4)氨基酸含有氨基和羧基两种官能团,其中氨基具有碱性,羧基具有酸性,故答案为:羧基;氨基;
(5)戊烷有正戊烷、新戊烷和异戊烷三种同分异构体,
①若A的一氯代物只有一种结构,则应为新戊烷,结构简式为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/3bf33a87e950352ab324f7b95043fbf2b2118b3e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,故答案为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/3bf33a87e950352ab324f7b95043fbf2b2118b3e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
②若B无支链,应为正戊烷,结构简式为CH
3(CH
2)
3CH
3,故答案为:CH
3(CH
2)
3CH
3;
③戊烷只有3种同分异构体,则C应为异戊烷,故答案为:异戊烷.