ÌâÄ¿ÄÚÈÝ

£¨10·Ö£©Í¨¹ýѧϰ¼îµÄÐÔÖÊ£¬ÎÒУ»¯Ñ§ÐËȤС×éµÄͬѧ¶ÔʵÑéÊҵġ°ÇâÑõ»¯¸ÆÊÇ·ñ±äÖÊÒÔ¼°±ä³ÖµÄ³Ì¶È¡±Õ¹¿ªÁË̽¾¿»î¶¯£¬ÇëÄãһͬ²ÎÓ룺
¢ÅÌá³öÎÊÌ⣺ÇâÑõ»¯¸ÆÊÇ·ñ±äÖÊ£¿
¢Æ²ÂÏëÓë¼ÙÉ裺ͬѧÃǾ­¹ýÌÖÂÛÈÏΪÑùÆ·´æÔÚÈýÖÖ¿ÉÄÜÇé¿ö£ºÃ»ÓбäÖÊ£»²¿·Ö±äÖÊ£»ÍêÈ«±äÖÊ¡£
¢ÇʵÑé̽¾¿£º

ʵÑé²½Öè¼°²Ù×÷
ʵÑéÏÖÏó
ʵÑé½áÂÛ
È¡ÑùÓÚÊÔ¹ÜÖУ¬¼ÓÈë×ãÁ¿ÕôÁóË®Õñµ´£¬¾²ÖÃ
¢ÙÈ¡ÉϲãÇåÒº£¬µÎÈëÎÞÉ«·Ó̪ÊÔÒº
¢Úµ¹È¥ÉϲãÇåÒº£¬ÔÙÏòÊÔ¹ÜÖÐ×¢ÈëÏ¡ÑÎËá
¢ÙÎÞÉ«·Ó̪ÊÔÒº±äºì
¢Ú                 
²¿·Ö±äÖÊ
¢ÙÎÞÉ«·Ó̪ÊÔÒº²»±äºì
¢Ú                 
                          
¢Ù                 
¢ÚûÓÐÆøÅݲúÉú
                          
¢ÈС½áÓë˼¿¼£º
¢ÙСÂûͬѧ¸ù¾ÝÉÏÊöʵÑé·½°¸½øÐÐʵÑ飬ȷ¶¨¸ÃÑùÆ·ÒѲ¿·Ö±äÖÊ£¬Çëд³öʵÑéÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
                                                        
                                                         
¢ÚÇâÑõ»¯¸ÆË׳ÆÊìʯ»Ò£¬Å©ÒµÉϳ£ÓÃÀ´                      ¡£

£¨10·Ö£©
¢ÇʵÑé̽¾¿£º

ʵÑé²½Öè¼°²Ù×÷
ʵÑéÏÖÏó
ʵÑé½áÂÛ
 
¢Ù
¢ÚÓÐÆøÅݲúÉú
 
¢Ù
¢ÚÓÐÆøÅݲúÉú
È«²¿±äÖÊ
¢ÙÎÞÉ«·Ó̪ÊÔÒº±äºì
¢Ú
         Ã»ÓбäÖÊ
¢ÈС½áÓë˼¿¼£º
¢ÙCa(OH)2+CO2 = CaCO3¡ý+H2O     CaCO3+2HCl = CaCl2+H2O+CO2¡ü
¢Ú¸ÄÁ¼ËáÐÔÍÁÈÀ

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ͨ¹ý½üÒ»ÄêµÄѧϰ»¯Ñ§£¬Ê¹ÄãÄõ½ÁË´ò¿ªÎïÖÊÊÀ½ç´óÃŵÄÔ¿³×£®ÈÃÎÒ ÃÇ×ß½øÉ­ÁÖ¹«Ô°£¬ÓÃÄãѧ¹ýµÄ»¯Ñ§ÖªÊ¶À´½âÊÍÏÂÁÐÎÊÌ⣮
£¨1£©È˹¤ÏûÓê´ø¸øÉ­ÁÖ¹«Ô°ÑÞÑôÌ죬È˹¤ÏûÓêÆäʵ¾ÍÊÇÖܱߵØÇøµÄÈ˹¤ÔöÓ꣬ÈÃÓêµÎÌáǰ½µÂäÏÂÀ´£¬ÈÃÎÚÔÆÌáÔçÏûÉ¢£®Äãѧ¹ýµÄ¿ÉÓÃÓÚÈ˹¤ÔöÓêµÄÎïÖÊÊÇ
CO2
CO2
£»
£¨2£©Ô°ÄÚÏÊ»¨Ê¢¿ª£¬Ò»ÖÖ½Ð×öÔ¼¾»¨µÄºìÉ«»¨Ö­ÓöËáÈÜÒº±ä³ÉdzºìÉ«£¬Óö¼îÈÜÒº±ä³É»ÆÉ«£¬ÕâÖÖ»¨Ö­µÄÐÔÖÊÓëʵÑéÊÒÖг£ÓõÄָʾ¼Á
×ÏɫʯÈïÊÔÒº
×ÏɫʯÈïÊÔÒº
ÐÔÖÊÏàËÆ£»
£¨3£©Ê¢·ÅÉ«²ÊѤÀöµÄÏÊ»¨£¬ÓеÄÊÇÓÃÆ¯ÁÁµÄÌÕ´ÉÈÝÆ÷£¬ÌÕ´ÉÊôÓÚÏÂÁвÄÁÏÖеÄ
B
B
£¨ÌîÐòºÅ£¬ÏÂͬ£©£»
A£®½ðÊô²ÄÁÏ    B£®ÎÞ»ú·Ç½ðÊô²ÄÁÏ    C£®ºÏ³É²ÄÁÏ    D£®¸´ºÏ²ÄÁÏ
£¨4£©Ô°ÄÚÓÐÒ»¸öÅ©ÒÕÔ°£¬×¼±¸µÄ»¯·ÊÖÐÊôÓÚ¸´ºÏ·ÊµÄÊÇ
BC
BC
£»
A£®NH4NO3    B£®KNO3    C£®NH4H2PO4    D£®£¨NH4£©2SO4
£¨5£©ÔÚÔ°ÄÚÓÐÐí¶àÀ¬»øÏ䣬À¬»øµÄ´¦Àí·½·¨ÓзÖÀà»ØÊÕ£®ÌîÂñºÍ¸ßζѷʵȣ®ÏÂÁÐÔ°ÄÚÊÕ¼¯µ½µÄÀ¬»øÄÜ»ØÊÕÀûÓõÄÊÇ
AD
AD
£®
A£®¿Õ¿óȪˮƿ    B£®¿ÚÏãÌÇ      C£®¹ûƤ      D£®Ò»´ÎÐÔËÜÁÏ·¹ºÐ£®

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

¾«Ó¢¼Ò½ÌÍø