阿伏加德罗常数约为6.02×1023mol-1,下列叙述中正确的是 A.1.4 g乙

阿伏加德罗常数约为6.02×1023mol1,下列叙述中正确的是

A.1.4 g乙烯和丙烯的混合物中含碳原子数为0.l5×6.02×1023

B.16.2 g淀粉和纤维素混合物中共含碳原子数为0.1×6.02×1023

C.1.5g C2H6中含有共价键的数目为7×3.01×1022

D.标准状况下,2.24L CCl4中含有的分子数为0.1×6.02×1023

答案

C

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