̼ԪËØÐγɵÄÎïÖʵÄÖÖÀàÊÇ×î¶àµÄ£¬¹Ê¶Ô̼¼°Æ仯ºÏÎïµÄÑо¿ÊǺÜÓмÛÖµµÄ¡£
(1)̼ԭ×ӵĽṹʾÒâͼΪ£¬Ì¼Ô­×ÓµÄ×îÍâ²ãµç×ÓÊýΪ______£¬Ð´³öÒ»ÖÖÄãÖªµÀµÄ̼µÄµ¥ÖʵÄÃû³Æ________¡£
(2)¾Æ¾«(C2H5OH)ÊÇÒ»ÖÖº¬Ì¼ÔªËصĻ¯ºÏÎ½«100mL¾Æ¾«ºÍ100mLË®»ìºÏʱ£¬ËùµÃÌå»ýСÓÚ200mL£¬Ô­ÒòÊÇ________________¡£
(3)ÔÚ̽¾¿¶þÑõ»¯Ì¼µÄÐÔÖÊʱ£¬ÎÒÃÇÔøÓÃËĶäʯÈïȾ³É×ÏÉ«µÄ¸ÉÔïС»¨£¬½øÐÐÁËÏÂͼµÄʵÑ飬ÈÃÎÒÃÇÌåÑéÁ˶Աȡ¢·ÖÎö¡¢¹éÄÉ¡¢¸ÅÀ¨µÈ¿Æѧ·½·¨ºÍÊֶεÄÔËÓá£
±¾ÊµÑéµÃ³öµÄ½áÂÛÊÇ________¡£
(4)COÔÚÒ±½ð¹¤ÒµÖÐÓÐ×ÅÖØÒªµÄ×÷Óá£ÈôCOÖлìÓÐÉÙÁ¿µÄCO2£¬Òª³ýÈ¥ÉÙÁ¿CO2£¬ÄãµÄ·½·¨ÊÇ______________£¨Óû¯Ñ§·½³Ìʽ±íʾ£©¡£
(1)4£»½ð¸Õʯ£¨»òʯī£©
(2)·Ö×Ó¼äÓмä¸ô
(3)¶þÑõ»¯Ì¼ÄܺÍË®·´Ó¦Éú³É̼Ëá
(4)CO2+ 2NaOH==Na2CO3+H2O
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2008?¹óÑô£©Ì¼ÔªËØÐγɵÄÎïÖʵÄÖÖÀàÊÇ×î¶àµÄ£¬¹Ê¶Ô̼¼°Æ仯ºÏÎïµÄÑо¿ÊǺÜÓмÛÖµµÄ£®
£¨1£©Ì¼ÔªËصĽṹʾÒâͼΪ£¬Ì¼Ô­×ÓµÄ×îÍâ²ãµç×ÓÊýΪ
4
4
£¬Ð´³öÒ»ÖÖÄãÖªµÀµÄ̼µÄµ¥ÖʵÄÃû³Æ
½ð¸Õʯ
½ð¸Õʯ
£®
£¨2£©¾Æ¾«£¨C2H5OH£©ÊÇÒ»ÖÖº¬Ì¼ÔªËصĻ¯ºÏÎ½«100mL¾Æ¾«ºÍ100mLË®»ìºÏʱ£¬ËùµÃÌå»ýСÓÚ200mL£¬Ô­ÒòÊÇ
·Ö¼äÓмä϶
·Ö¼äÓмä϶
£®
£¨3£©ÔÚ̽¾¿¶þÑõ»¯Ì¼µÄÐÔÖÊʱ£¬ÎÒÃÇÔøÓÃËĶäʯÈïȾ³É×ÏÉ«µÄ¸ÉÔïС»¨£¬½øÐÐÁËÈçͼµÄʵÑ飬ÈÃÎÒÃÇÌåÑéÁ˶Աȡ¢·ÖÎö¡¢¹éÄÉ¡¢¸ÅÀ¨µÈ¿Æѧ·½·¨ºÍÊֶεÄÔËÓã®±¾ÊµÑéµÃ³öµÄ½áÂÛÊÇ
¶þÑõ»¯Ì¼ÄܺÍË®·´Ó¦Éú³É̼Ëá
¶þÑõ»¯Ì¼ÄܺÍË®·´Ó¦Éú³É̼Ëá
£®

£¨4£©COÔÚÒ±½ð¹¤ÒµÖÐÓÐ×ÅÖØÒªµÄ×÷Óã®ÈôCOÖлìÓÐÉÙÁ¿µÄCO2£¬Òª³ýÈ¥ÉÙÁ¿CO2£¬ÄãµÄ·½·¨ÊÇ
CO2+2NaOH=Na2CO3+H2O
CO2+2NaOH=Na2CO3+H2O
£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê°²»ÕÊ¡¸·ÑôÊоÅÄ꼶£¨ÉÏ£©ÆÚÄ©»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

̼ԪËØÐγɵÄÎïÖʵÄÖÖÀàÊÇ×î¶àµÄ£¬¹Ê¶Ô̼¼°Æ仯ºÏÎïµÄÑо¿ÊǺÜÓмÛÖµµÄ£®
£¨1£©Ì¼ÔªËصĽṹʾÒâͼΪ£¬Ì¼Ô­×ÓµÄ×îÍâ²ãµç×ÓÊýΪ    £¬Ð´³öÒ»ÖÖÄãÖªµÀµÄ̼µÄµ¥ÖʵÄÃû³Æ    £®
£¨2£©¾Æ¾«£¨C2H5OH£©ÊÇÒ»ÖÖº¬Ì¼ÔªËصĻ¯ºÏÎ½«100mL¾Æ¾«ºÍ100mLË®»ìºÏʱ£¬ËùµÃÌå»ýСÓÚ200mL£¬Ô­ÒòÊÇ    £®
£¨3£©ÔÚ̽¾¿¶þÑõ»¯Ì¼µÄÐÔÖÊʱ£¬ÎÒÃÇÔøÓÃËĶäʯÈïȾ³É×ÏÉ«µÄ¸ÉÔïС»¨£¬½øÐÐÁËÈçͼµÄʵÑ飬ÈÃÎÒÃÇÌåÑéÁ˶Աȡ¢·ÖÎö¡¢¹éÄÉ¡¢¸ÅÀ¨µÈ¿Æѧ·½·¨ºÍÊֶεÄÔËÓã®±¾ÊµÑéµÃ³öµÄ½áÂÛÊÇ    £®

£¨4£©COÔÚÒ±½ð¹¤ÒµÖÐÓÐ×ÅÖØÒªµÄ×÷Óã®ÈôCOÖлìÓÐÉÙÁ¿µÄCO2£¬Òª³ýÈ¥ÉÙÁ¿CO2£¬ÄãµÄ·½·¨ÊÇ    £¨Óû¯Ñ§·½³Ìʽ±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê½­ËÕÊ¡ÄϾ©ÊÐÐþÎäÇø¾ÅÄ꼶£¨ÉÏ£©ÆÚÄ©»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

̼ԪËØÐγɵÄÎïÖʵÄÖÖÀàÊÇ×î¶àµÄ£¬¹Ê¶Ô̼¼°Æ仯ºÏÎïµÄÑо¿ÊǺÜÓмÛÖµµÄ£®
£¨1£©Ì¼ÔªËصĽṹʾÒâͼΪ£¬Ì¼Ô­×ÓµÄ×îÍâ²ãµç×ÓÊýΪ    £¬Ð´³öÒ»ÖÖÄãÖªµÀµÄ̼µÄµ¥ÖʵÄÃû³Æ    £®
£¨2£©¾Æ¾«£¨C2H5OH£©ÊÇÒ»ÖÖº¬Ì¼ÔªËصĻ¯ºÏÎ½«100mL¾Æ¾«ºÍ100mLË®»ìºÏʱ£¬ËùµÃÌå»ýСÓÚ200mL£¬Ô­ÒòÊÇ    £®
£¨3£©ÔÚ̽¾¿¶þÑõ»¯Ì¼µÄÐÔÖÊʱ£¬ÎÒÃÇÔøÓÃËĶäʯÈïȾ³É×ÏÉ«µÄ¸ÉÔïС»¨£¬½øÐÐÁËÈçͼµÄʵÑ飬ÈÃÎÒÃÇÌåÑéÁ˶Աȡ¢·ÖÎö¡¢¹éÄÉ¡¢¸ÅÀ¨µÈ¿Æѧ·½·¨ºÍÊֶεÄÔËÓã®±¾ÊµÑéµÃ³öµÄ½áÂÛÊÇ    £®

£¨4£©COÔÚÒ±½ð¹¤ÒµÖÐÓÐ×ÅÖØÒªµÄ×÷Óã®ÈôCOÖлìÓÐÉÙÁ¿µÄCO2£¬Òª³ýÈ¥ÉÙÁ¿CO2£¬ÄãµÄ·½·¨ÊÇ    £¨Óû¯Ñ§·½³Ìʽ±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2008Äê¹óÖÝÊ¡¹óÑôÊÐÖп¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

£¨2008?¹óÑô£©Ì¼ÔªËØÐγɵÄÎïÖʵÄÖÖÀàÊÇ×î¶àµÄ£¬¹Ê¶Ô̼¼°Æ仯ºÏÎïµÄÑо¿ÊǺÜÓмÛÖµµÄ£®
£¨1£©Ì¼ÔªËصĽṹʾÒâͼΪ£¬Ì¼Ô­×ÓµÄ×îÍâ²ãµç×ÓÊýΪ£¬Ð´³öÒ»ÖÖÄãÖªµÀµÄ̼µÄµ¥ÖʵÄÃû³Æ£®
£¨2£©¾Æ¾«£¨C2H5OH£©ÊÇÒ»ÖÖº¬Ì¼ÔªËصĻ¯ºÏÎ½«100mL¾Æ¾«ºÍ100mLË®»ìºÏʱ£¬ËùµÃÌå»ýСÓÚ200mL£¬Ô­ÒòÊÇ£®
£¨3£©ÔÚ̽¾¿¶þÑõ»¯Ì¼µÄÐÔÖÊʱ£¬ÎÒÃÇÔøÓÃËĶäʯÈïȾ³É×ÏÉ«µÄ¸ÉÔïС»¨£¬½øÐÐÁËÈçͼµÄʵÑ飬ÈÃÎÒÃÇÌåÑéÁ˶Աȡ¢·ÖÎö¡¢¹éÄÉ¡¢¸ÅÀ¨µÈ¿Æѧ·½·¨ºÍÊֶεÄÔËÓã®±¾ÊµÑéµÃ³öµÄ½áÂÛÊÇ£®

£¨4£©COÔÚÒ±½ð¹¤ÒµÖÐÓÐ×ÅÖØÒªµÄ×÷Óã®ÈôCOÖлìÓÐÉÙÁ¿µÄCO2£¬Òª³ýÈ¥ÉÙÁ¿CO2£¬ÄãµÄ·½·¨ÊÇ£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸