ÌâÄ¿ÄÚÈÝ

ÒÑÖª±½µÄͬϵÎïAµÄ·Ö×ÓÖеÄ̼ԭ×ÓÊýΪÆø̬Á´ÌþB·Ö×ÓÖÐ̼ԭ×ÓÊýµÄ3±¶¡£Èô½«A¡¢BÁ½ÌþµÄÆø̬»ìºÏÎïÒÔÈÎÒâ±ÈÀý»ìºÏ£¬Ö»Òª×ÜÖÊÁ¿Ò»¶¨£¬ÍêȫȼÉÕʱµÄºÄÑõÁ¿¾ÍΪ¶¨Öµ¡£
120 ¡æʱ£¬½«Ò»¶¨Á¿µÄÉÏÊö»ìºÏÆøÌåÍêȫȼÉÕºó£¬ËùµÃÆøÌåÏȺóÒÀ´Îͨ¹ýÊ¢ÓÐ×ãÁ¿µÄCaCl2ºÍ¼îʯ»ÒµÄ¸ÉÔï¹Ü£¬²âµÃ¸ÉÔï¹ÜÒÀ´ÎÔöÖØ7.20 gºÍ26.4 g?¡£ÊÔͨ¹ý¼ÆËãºÍÍÆÀíÍê³ÉÏÂÁÐÎÊÌâ¡£
(1)ÌþBµÄ·Ö×Óʽ¡£
(2)AµÄͬ·ÖÒì¹¹ÌåÉõ¶à£¬ÆäÖб½»·ÉϲàÁ´²»³¬¹ýÁ½¸öµÄËùÓÐͬ·ÖÒì¹¹Ìå¹²ÓÐ__________ÖÖ¡£
(1)BµÄ·Ö×ÓʽΪC3H4 (2)5
(1)ÒÀÌâÒâÖª£¬A¡¢BÁ½ÌþµÄʵÑéʽÏàͬ¡£
n(H)=¡Á2="0.800" mol
n(C)=="0.600" mol
¹ÊAºÍBµÄʵÑéʽΪC3H4¡£
ÉèA¡¢B·Ö±ðΪ(C3H4)3nºÍ(C3H4)n£¬nÖ»ÄÜÈ¡1£¬·ñÔò(C3H4)3nÓë±½µÄͬϵÎïͨʽ²»·û¡£¹ÊBµÄ·Ö×ÓʽΪC3H4¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø