有一个化合物[tex=1.571x1.357]46QRFrjp/bRhFtDC1AQcAw==[/tex], 分子式是 [tex=5.143x1.357]12FICuEaNNFUlY9ok4kuMn8pTVdbx7N3AqivbuUkC1dtUYs6LxdW+IF0ikl+4fa55bh0QmmOCMXBvZP+nfHpog==[/tex] 可以很快地使溴水褪色，可以与苯肼反应，[tex=1.571x1.357]46QRFrjp/bRhFtDC1AQcAw==[/tex]氧化生成一分子丙酮及另一化合物[tex=1.5x1.357]TsZ8dyhQe2G/SxZU76wR3w==[/tex]。[tex=1.5x1.357]TsZ8dyhQe2G/SxZU76wR3w==[/tex]具有酸性,同 [tex=3.071x1.0]TdpEcGrhRVEMOgzWC99o09HVjUCXwNV2mJB8x7VfzFY=[/tex] 反应则生成氯仿及一分子丁二酸。试写出[tex=1.571x1.357]46QRFrjp/bRhFtDC1AQcAw==[/tex]与[tex=1.5x1.357]TsZ8dyhQe2G/SxZU76wR3w==[/tex]可能的构造式。

试比较下列化合物的酸性大小:[tex=1.571x1.357]46QRFrjp/bRhFtDC1AQcAw==[/tex]三氯乙酸， [tex=1.5x1.357]TsZ8dyhQe2G/SxZU76wR3w==[/tex] 氯乙酸，  [tex=1.5x1.357]F9oIpvJmdoP7zb4YqQqyKg==[/tex]  乙酸，  [tex=1.571x1.357]WImqYT4z1oWRPlrECGRCRg==[/tex]羟基乙酸 

甲乙胺[tex=1.571x1.357]46QRFrjp/bRhFtDC1AQcAw==[/tex]、三甲胺[tex=1.5x1.357]TsZ8dyhQe2G/SxZU76wR3w==[/tex]和正丙胺[tex=1.5x1.357]F9oIpvJmdoP7zb4YqQqyKg==[/tex]互为异构体,试将其按沸点由高到低排列成序,并说明理由。

用化学方法区别下列各组化合物 :[tex=1.429x1.357]46QRFrjp/bRhFtDC1AQcAw==[/tex]硝基苯  [tex=1.357x1.357]TsZ8dyhQe2G/SxZU76wR3w==[/tex] 苯胺 [tex=1.357x1.357]F9oIpvJmdoP7zb4YqQqyKg==[/tex] [tex=0.857x1.0]HcQeTeQtUqN73yUJqDRZkQ==[/tex]-甲基苯胺[tex=1.429x1.357]WImqYT4z1oWRPlrECGRCRg==[/tex][tex=1.929x1.214]MshoXHUjidUXl5H1eX8/2A==[/tex] -二甲苯胺

for i in range(b.max_row): for j in range(b.max_column): print(b.cell(row=i,column=j).value)上面语句运行的结果是：__________。 A: 1 1 1 1 1 1 1 1 1 1 B: 1111111111 C: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 D: 出现异常

【单选题】CDMA通信的基站,假定基站A的码片序列是+1 +1 +1 -1 -1 +1 -1 -1,基站发射bit为101时,实际发射的信号是 A. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 B. +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 +1 +1 +1 -1 -1 +1 -1 -1 C. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 D. –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1

for i in range(1,11): for j in range(1,11): b.cell(row=i,column=j).value=1 #b是一个工作表对象for i in range(1,11): for j in range(1,11): print(b.cell(row=i,column=j).value,end=" ") print()上面程序代码运行的结果是（）。 A: 1 B: 1 1 1 1 1 1 1 1 1 1 C: 1111111111 D: 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1

下列哪个码片序列不能用于CDMA通信 A: ( -1 -1 -1 +1 +1 -1 +1 +1 ) B: ( -1 -1 +1 -1 +1 +1 +1 -1 ) C: ( -1 +1 -1 +1 +1 -1 -1 -1 ) D: ( -1 +1 -1 -1 -1 -1 +1 -1 )

请仔细观察下面行列式的计算过程，如果四个等号都理解了，请选择A，否则请选择B。 | a 1 1 1 1 | |a＋4 a＋4 a＋4 a＋4 a＋4| | 1 a 1 1 1 | | 1 a 1 1 1 | | 1 1 a 1 1 |= | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 a | | 1 1 1 1 1 | | 1 a 1 1 1 | =(a＋4) | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 1 | | 0 a–1 0 0 0 | =(a＋4) | 0 0 a–1 0 0 | | 0 0 0 a–1 0 | | 0 0 0 0 a–1 | =(a＋4)(a–1)^4

共有4个站进行码分多址通信。4个站的码片序列为：a：（-1 -1 -1 +1 +1 -1 +1 +1） b：（-1 -1 +1 -1 +1 +1 +1 -1） c：（-1 +1 -1 +1 +1 +1 -1 -1） d：（-1 +1 -1 -1 -1 -1 +1 -1） 现收到这样的码片序列：（-1 +1 -3 +1 -1 -3 +1 +1），则（ ）发送1。