£¨10·Ö£©Ä³Ð£¿Æ¼¼½Ú»î¶¯ÖнøÁËÈçÏÂʵÑ飬Çë»Ø´ðÎÊÌâ¡£

£¨1£©Í¼AËùʾʵÑé¿É¹Û²ìµ½¡°ÌúÊ÷¡±ÉϽþÓÐÎÞÉ«·Ó̪ÊÔÒºµÄÃÞÍÅÓÉ°×É«±äΪ  É«£¬Çë½áºÏ·Ö×ӵĹ۵ã½âÊÍ£º                                         ¡£

£¨2£©Í¼BËùʾʵÑ飬½«×¢ÉäÆ÷ÖÐʯ»ÒË®×¢ÈëÆ¿ÖУ¬»á¿´µ½¼¦µ°±»¡°ÍÌ¡±ÈëÆ¿ÖУ¬¸ÃʵÑéÖÐÉæ¼°µÄ»¯Ñ§·½³ÌʽΪ                                      ¡£

£¨3£©Í¼CËùʾʵÑ飬µ±Í¨¹ýµ¼¹ÜÏòÈÈË®ÖÐͨÈëÑõÆøʱ£¬°×Á×ÔÚË®ÏÂȼÉÕ£»Ïò±ùË®ÖÐͨÈëÑõÆø£¬°×Á×ÔÚˮϲ»È¼ÉÕ¡£¸ÃʵÑé˵Ã÷ȼÉÕÐèÒªµÄÌõ¼þΪ£º¿ÉȼÎï¡¢               ºÍ                      ¡£

£¨4£©Í¼DËùʾʵÑé¹Û²ìµ½×ÏɫС»¨±äΪºìÉ«£¬Ð¡»¨±äºìµÄÔ­ÒòÊÇ

                                         £¨Óû¯Ñ§·½³Ìʽ±íʾ£©¡£

 

¡¾´ð°¸¡¿

(1)ºì£»Å¨°±Ë®³Ê¼îÐÔ£¬°±·Ö×Ó²»¶ÏÔ˶¯£¬Óë½þÓзÓ̪ÊÔÒºµÄÃÞÍŽӴ¥£¬·Ó̪Óö¼îÐÔÈÜÒº±äºìÉ«¡£  

(2) Ca (OH)2+ CO2= CaCO3¡ý+H2O

£¨3£©ÓëÑõÆø½Ó´¥£»´ïµ½×Å»ðµã£¨¸÷1·Ö£©

£¨4£©H2O+CO2 ==H2CO3

¡¾½âÎö¡¿£¨1£©Å¨°±Ë®¾ß»Ó·¢ÐÔ£¬Ò²ÊÇ°±·Ö×Ó²»¶ÏÔ˶¯µÄ½á¹û£¬°±Æø¼«Ò×ÈÜÓÚË®£¬ÆäË®ÈÜÒºÏÔ¼îÐÔ£¬¼îÒºÄÜʹÎÞÉ«µÄ·Ó̪±äºìÉ«£»

£¨2£©¶þÑõ»¯Ì¼ÆøÌåÄܺÍŨÇâÑõ»¯ÄÆÈÜҺѸËÙ·´Ó¦£¬×°ÖÃÄÚÆøѹ¼õС£¬Íâ½ç´óÆøѹ½«Ê켦µ°¡°ÍÌ¡±ÈëÆ¿ÖУ¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCa (OH)2+ CO2= CaCO3¡ý+H2O£»

£¨3£©¿ÉȼÎï×Å»ðȼÉÕµÄÈý¸öÌõ¼þÊÇ£º¢ÙÎïÖʾ߿ÉȼÐÔ£»¢Ú´ïµ½È¼ÉÕËùÐèÒªµÄ×Å»ðµã£®¢Û¿ÉȼÎïÓëÑõÆø»ò¿ÕÆø½Ó´¥£»£¨4£©Í¼DËùʾʵÑé¹Û²ìµ½×ÏɫС»¨±äΪºìÉ«£¬ÊÇÒòΪCO2+H2O=H2CO3£¬Ì¼ËáÄÜʹ×ÏÉ«µÄʯÈïÊÔÒº±äºìÉ«£®

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijУ¿Æ¼¼½Ú»î¶¯ÖнøÁËÈçÏÂʵÑ飬Çë»Ø´ðÎÊÌ⣮
£¨1£©Í¼AËùʾʵÑé¿É¹Û²ìµ½¡°ÌúÊ÷¡±ÉϽþÓÐÎÞÉ«·Ó̪ÊÔÒºµÄÃÞÍÅÓÉ°×É«±äΪ
ºì
ºì
É«£¬Çë½áºÏ·Ö×ӵĹ۵ã½âÊÍ£º
Ũ°±Ë®³Ê¼îÐÔ£¬°±·Ö×Ó²»¶ÏÔ˶¯£¬Óë½þÓзÓ̪ÊÔÒºµÄÃÞÍŽӴ¥£¬·Ó̪Óö¼îÐÔÈÜÒº±äºìÉ«£®
Ũ°±Ë®³Ê¼îÐÔ£¬°±·Ö×Ó²»¶ÏÔ˶¯£¬Óë½þÓзÓ̪ÊÔÒºµÄÃÞÍŽӴ¥£¬·Ó̪Óö¼îÐÔÈÜÒº±äºìÉ«£®
£®
£¨2£©Í¼BËùʾʵÑ飬½«×¢ÉäÆ÷ÖÐʯ»ÒË®×¢ÈëÆ¿ÖУ¬»á¿´µ½¼¦µ°±»¡°ÍÌ¡±ÈëÆ¿ÖУ¬¸ÃʵÑéÖÐÉæ¼°µÄ»¯Ñ§·½³ÌʽΪ
Ca£¨OH£©2+CO2=CaCO3¡ý+H2O
Ca£¨OH£©2+CO2=CaCO3¡ý+H2O
£®
£¨3£©Í¼CËùʾʵÑ飬µ±Í¨¹ýµ¼¹ÜÏòÈÈË®ÖÐͨÈëÑõÆøʱ£¬°×Á×ÔÚË®ÏÂȼÉÕ£»Ïò±ùË®ÖÐͨÈëÑõÆø£¬°×Á×ÔÚˮϲ»È¼ÉÕ£®¸ÃʵÑé˵Ã÷ȼÉÕÐèÒªµÄÌõ¼þΪ£º¿ÉȼÎï¡¢
ÓëÑõÆø½Ó´¥
ÓëÑõÆø½Ó´¥
ºÍ
´ïµ½×Å»ðµã
´ïµ½×Å»ðµã
£®
£¨4£©Í¼DËùʾʵÑé¹Û²ìµ½×ÏɫС»¨±äΪºìÉ«£¬Ð¡»¨±äºìµÄÔ­ÒòÊÇ
H2O+CO2¨TH2CO3
H2O+CO2¨TH2CO3
£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨10·Ö£©Ä³Ð£¿Æ¼¼½Ú»î¶¯ÖнøÁËÈçÏÂʵÑ飬Çë»Ø´ðÎÊÌâ¡£

£¨1£©Í¼AËùʾʵÑé¿É¹Û²ìµ½¡°ÌúÊ÷¡±ÉϽþÓÐÎÞÉ«·Ó̪ÊÔÒºµÄÃÞÍÅÓÉ°×É«±äΪ  É«£¬Çë½áºÏ·Ö×ӵĹ۵ã½âÊÍ£º                                        ¡£

£¨2£©Í¼BËùʾʵÑ飬½«×¢ÉäÆ÷ÖÐʯ»ÒË®×¢ÈëÆ¿ÖУ¬»á¿´µ½¼¦µ°±»¡°ÍÌ¡±ÈëÆ¿ÖУ¬¸ÃʵÑéÖÐÉæ¼°µÄ»¯Ñ§·½³ÌʽΪ                                      ¡£

£¨3£©Í¼CËùʾʵÑ飬µ±Í¨¹ýµ¼¹ÜÏòÈÈË®ÖÐͨÈëÑõÆøʱ£¬°×Á×ÔÚË®ÏÂȼÉÕ£»Ïò±ùË®ÖÐͨÈëÑõÆø£¬°×Á×ÔÚˮϲ»È¼ÉÕ¡£¸ÃʵÑé˵Ã÷ȼÉÕÐèÒªµÄÌõ¼þΪ£º¿ÉȼÎï¡¢               ºÍ                     ¡£

£¨4£©Í¼DËùʾʵÑé¹Û²ìµ½×ÏɫС»¨±äΪºìÉ«£¬Ð¡»¨±äºìµÄÔ­ÒòÊÇ

                                         £¨Óû¯Ñ§·½³Ìʽ±íʾ£©¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2012ÄêÏÃÃÅÊÐÏè°²Çø¾ÅÄ꼶ÊÊÓ¦ÐÔ¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

£¨10·Ö£©Ä³Ð£¿Æ¼¼½Ú»î¶¯ÖнøÁËÈçÏÂʵÑ飬Çë»Ø´ðÎÊÌâ¡£

£¨1£©Í¼AËùʾʵÑé¿É¹Û²ìµ½¡°ÌúÊ÷¡±ÉϽþÓÐÎÞÉ«·Ó̪ÊÔÒºµÄÃÞÍÅÓÉ°×É«±äΪ É«£¬Çë½áºÏ·Ö×ӵĹ۵ã½âÊÍ£º                                        ¡£
£¨2£©Í¼BËùʾʵÑ飬½«×¢ÉäÆ÷ÖÐʯ»ÒË®×¢ÈëÆ¿ÖУ¬»á¿´µ½¼¦µ°±»¡°ÍÌ¡±ÈëÆ¿ÖУ¬¸ÃʵÑéÖÐÉæ¼°µÄ»¯Ñ§·½³ÌʽΪ                                     ¡£
£¨3£©Í¼CËùʾʵÑ飬µ±Í¨¹ýµ¼¹ÜÏòÈÈË®ÖÐͨÈëÑõÆøʱ£¬°×Á×ÔÚË®ÏÂȼÉÕ£»Ïò±ùË®ÖÐͨÈëÑõÆø£¬°×Á×ÔÚˮϲ»È¼ÉÕ¡£¸ÃʵÑé˵Ã÷ȼÉÕÐèÒªµÄÌõ¼þΪ£º¿ÉȼÎï¡¢              ºÍ                     ¡£
£¨4£©Í¼DËùʾʵÑé¹Û²ìµ½×ÏɫС»¨±äΪºìÉ«£¬Ð¡»¨±äºìµÄÔ­ÒòÊÇ
                                        £¨Óû¯Ñ§·½³Ìʽ±íʾ£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2012Ä긣½¨Ê¡ÏÃÃÅÊÐÏè°²ÇøÖп¼ÊÊÓ¦ÐÔ¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

ijУ¿Æ¼¼½Ú»î¶¯ÖнøÁËÈçÏÂʵÑ飬Çë»Ø´ðÎÊÌ⣮
£¨1£©Í¼AËùʾʵÑé¿É¹Û²ìµ½¡°ÌúÊ÷¡±ÉϽþÓÐÎÞÉ«·Ó̪ÊÔÒºµÄÃÞÍÅÓÉ°×É«±äΪ    É«£¬Çë½áºÏ·Ö×ӵĹ۵ã½âÊÍ£º    £®
£¨2£©Í¼BËùʾʵÑ飬½«×¢ÉäÆ÷ÖÐʯ»ÒË®×¢ÈëÆ¿ÖУ¬»á¿´µ½¼¦µ°±»¡°ÍÌ¡±ÈëÆ¿ÖУ¬¸ÃʵÑéÖÐÉæ¼°µÄ»¯Ñ§·½³ÌʽΪ    £®
£¨3£©Í¼CËùʾʵÑ飬µ±Í¨¹ýµ¼¹ÜÏòÈÈË®ÖÐͨÈëÑõÆøʱ£¬°×Á×ÔÚË®ÏÂȼÉÕ£»Ïò±ùË®ÖÐͨÈëÑõÆø£¬°×Á×ÔÚˮϲ»È¼ÉÕ£®¸ÃʵÑé˵Ã÷ȼÉÕÐèÒªµÄÌõ¼þΪ£º¿ÉȼÎï¡¢    ºÍ    £®
£¨4£©Í¼DËùʾʵÑé¹Û²ìµ½×ÏɫС»¨±äΪºìÉ«£¬Ð¡»¨±äºìµÄÔ­ÒòÊÇ    £¨Óû¯Ñ§·½³Ìʽ±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸