设NA为阿伏加德罗常数，下列叙述不正确的是
A．一定条件下，162g淀粉水解得到葡萄糖的分子数为n倍NA
B．1 mol C10H22分子中共价键总数为31 NA
C．室温下，4.4 g乙醛和乙酸乙酯的混合物含有的碳原子数为0.2 NA
D．88.0g干冰中含有的共用电子对对数为8NA
答案：A
A．由淀粉水解方程式：(C6H10O5)n+nH2O→nC6H12O6可知，162g淀粉水解得到葡萄糖的质量为180g，分子数为NA，不正确
B． C10H22分子中共价键总数2×10+22/2=31，故正确
C．乙醛和乙酸乙酯的最简式均为C2H4O，即44g混合物一定含有2mol碳原子，正确
D．干冰分子中含有4对电子，故88.0g干冰中含有的共用电子对对数为8NA，正确
答案为A
| 从下面方框中选择适当的词或词组填入对话，使对话意思完整正确。 | |
Student: Thanks. Teacher: How is your holiday? Student: Great! I had a 1 holiday. I traveled to Japan with my parents. Teacher: Really? What did you see there? Student: I saw a lot of 2 . Such as Mount Fuji and Sanrio Puroland. Mount Fuji is so wonderful that all of us want to climb up to the top. And Sanrio Puroland is the Hello Kitty land. We also visited the Hello Kitty's house. How lovely it is! Teacher: You must have a great time. Student: 3 . How about you? What were you doing 4 ? Teacher: I didn't 5 . But I have a good time staying at home, reading a lot of books. Student: Oh, that's interesting. It's time to have lunch. I must go. Bye. Teacher: Bye. 1._____ 2._____ 3._____ 4._____ 5._____ |
