ÌâÄ¿ÄÚÈÝ

14£®»¯Ñ§ÓëÉú»îÃÜÇÐÏà¹Ø£®
£¨1£©ÏÂÁÐÉú»îÓÃÆ·ÖУ¬ÆäÖ÷Òª²ÄÁÏÊôÓÚÌìÈ»ÏËάµÄÊÇC£¨Ìî×ÖĸÐòºÅ£©£®
A£®ËÜÁϱ£Ïʱ¡   B£®Ïð½ºÊÖÌ×   C£®ÃÞ²¼Î§È¹
£¨2£©ÎÒÃdz£ÓõÄÏ´µÓ¼ÁÇåÏ´²Í¾ßÉϵÄÓÍÎÛ£¬ÕâÊÇÒòΪϴµÓ¼Á¾ßÓÐÈ黯µÄ¹¦ÄÜ£®
£¨3£©Î¢Á¿ÔªËضÔÈËÌ彡¿µµÄÓ°ÏìºÜ´ó£¬ÈçȱÌúÔªËØ»áÒýÆðƶѪ£®
£¨4£©¶à³ÔË®¹û¡¢Ê߲˿ɲ¹³äÈËÌåÐèÒªµÄÉÙÁ¿Î¬ÉúËØ£¬´ËÀàÎïÖʿɵ÷½ÚÌåÄÚ¸÷ÖÖ»¯Ñ§·´Ó¦£®
£¨5£©ÓÃÊÊÁ¿ÑÎËá¿ÉÒÔ½«ÈÈˮƿµ¨±ÚÉϵÄË®¹¸£¨Ö÷Òª³É·ÖÊÇMg£¨OH£©2ºÍCaCO3£©³ýÈ¥£®³ýȥˮ¹¸µÄ»¯Ñ§Ô­ÀíÊÇMg£¨OH£©2+2HCl=MgCl2+2H2O£¬CaCO3+2HCl=CaCl2+H2O+CO2¡ü£®

·ÖÎö £¨1£©ÌìÈ»ÏËάÊÇ×ÔÈ»½çÔ­ÓеĻò¾­È˹¤ÅàÖ²µÄÖ²ÎïÉÏ¡¢È˹¤ËÇÑøµÄ¶¯ÎïÉÏÖ±½ÓÈ¡µÃµÄ·ÄÖ¯ÏËά£¬ÊÇ·ÄÖ¯¹¤ÒµµÄÖØÒª²ÄÁÏÀ´Ô´£»
£¨2£©Ï´µÓ¼ÁÇåÏ´²Í¾ßÉϵÄÓÍÎÛ£¬ÐγɵÄÊÇÈé×ÇÒº£»
£¨3£©¸ù¾Ý΢Á¿ÔªËصÄ×÷Óûشð£»
£¨4£©Ë®¹û¡¢Ê߲˸»º¬Î¬ÉúËØ£»
£¨5£©¸ù¾Ý³ýË®¹¸µÄÔ­ÀíÒÔ¼°»¯Ñ§·½³ÌʽµÄд·¨À´·ÖÎö£®

½â´ð ½â£º£¨1£©ËÜÁÏ¡¢Ïð½ºÊôÓÚÓлúºÏ³É²ÄÁÏ£¬ÃÞ²¼µÄ³É·ÖÊÇÏËÎ¬ËØ£¬ÊôÓÚÌìÈ»ÏËά£»
£¨2£©Ï´µÓ¼ÁÇåÏ´²Í¾ßÉϵÄÓÍÎÛ£¬ÐγɵÄÊÇÈé×ÇÒº£¬ÊôÓÚÈ黯×÷Óã»
£¨3£©È±Ìú»áÒýÆðȱÌúÐÔÆ¶Ñª£»
£¨4£©Ë®¹û¡¢Ê߲˸»º¬Î¬ÉúËØ£¬¹Ê´ð°¸Îª£ºÎ¬ÉúËØ£»
£¨5£©Ë®¹¸µÄÖ÷Òª³É·ÖÊÇMg£¨OH£©2ºÍCaCO3¶¼ÊÇÄÑÈÜÐԵ쬾ùÄÜÓëÏ¡ÑÎËá·´Ó¦£¬Éú³É¿ÉÈÜÐÔµÄÑζø³ýÈ¥£®Mg£¨OH£©2ºÍÏ¡ÑÎËá·´Ó¦Éú³ÉÂÈ»¯Ã¾ºÍË®£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºMg£¨OH£©2+2HCl=MCl2+2H2O£»Ì¼Ëá¸ÆÓëÏ¡ÑÎËá·´Ó¦Éú³ÉÂÈ»¯¸Æ¡¢Ë®¡¢¶þÑõ»¯Ì¼£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCaCO3+2HCl=CaCl2+H2O+CO2¡ü£®
¹Ê´ð°¸Îª£º£¨1£©C£»£¨2£©È黯£»£¨3£©ÌúÔªËØ£»£¨4£©Î¬ÉúËØ£»£¨5£©Mg£¨OH£©2+2HCl=MgCl2+2H2O£»CaCO3+2HCl=CaCl2+H2O+CO2¡ü£®

µãÆÀ ±¾Ì⿼²é»¯Ñ§ÓëÉú»î£¬¿¼²éÄÚÈݽ϶࣬Ã÷È·ÎïÖʵÄÐÔÖÊ¡¢·ÖÀà¡¢·´Ó¦µÄÔ­ÀíÊǽâÌâµÄ¹Ø¼ü£¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø