ÌâÄ¿ÄÚÈÝ

(9·Ö) ¼×¡¢ÒÒ¡¢±ûÈýλͬѧ·Ö±ðÓÃÈçÏÂÈýÌ×ʵÑé×°Öü°»¯Ñ§Ò©Æ·£¨ÆäÖмîʯ»ÒΪ¹ÌÌåÇâÑõ»¯ÄƺÍÉúʯ»ÒµÄ»ìºÏÎï£©ÖÆÈ¡°±Æø¡£ÇëÄã²ÎÓë̽¾¿£¬²¢»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈýÎ»Í¬Ñ§ÖÆÈ¡°±ÆøµÄ»¯Ñ§·½³ÌʽΪ£º _____         __________________________¡£

£¨2£©Èýλͬѧ¶¼ÓÃÏòÏÂÅÅ¿ÕÆø·¨ÊÕ¼¯°±Æø£¬ÆäÔ­ÒòÊÇ_______________________________£®

£¨3£©ÈýλͬѧÓÃÉÏÊö×°ÖÃÖÆÈ¡°±ÆøÊ±,ÆäÖÐÓÐһλͬѧûÓÐÊÕ¼¯µ½°±£¨Èç¹ûËûÃǵÄʵÑé²Ù×÷¶¼ÕýÈ·£©£¬ÄãÈÏΪûÓÐÊÕ¼¯µ½°±ÆøµÄͬѧÊÇ___________Ìî(¡°¼×¡±¡¢¡°ÒÒ¡±»ò¡°±û¡±)£¬ÊÕ¼¯²»µ½°±ÆøµÄÖ÷ÒªÔ­ÒòÊÇ_____________________________________(Óû¯Ñ§·½³Ìʽ±íʾ)¡£

£¨4£©¼ìÑé°±ÆøÊÇ·ñÊÕ¼¯ÂúµÄ·½·¨ÊÇ£¨¼òÊö²Ù×÷·½·¨¡¢ÏÖÏóºÍ½áÂÛ£©____________________

_____________________________________________________________________________¡£

£¨5£©Èýλͬѧ¶¼ÈÏΪËûÃǵÄʵÑé×°ÖÃÒ²¿ÉÓÃÓÚ¼ÓÈÈ̼ËáÇâï§¹ÌÌåÖÆÈ¡´¿¾»µÄ°±Æø£¬ÄãÅжÏÄܹ»´ïµ½ÊµÑéÄ¿µÄµÄÊÇ___________(Ìî¡°¼×¡±¡¢¡°ÒÒ¡±»ò¡°±û¡±)£¬¸Ã×°ÖÃÖеÄNH4HCO3¹ÌÌåÄÜ·ñÓÃNH4Cl¹ÌÌå´úÌæÖÆNH3£¿______________(Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)¡£

 

£¨1£©2NH4Cl+Ca(OH)2=2NH3¡ü+2H2O+CaCl22·Ö

£¨2£©°±ÆøµÄÃÜ¶È±È¿ÕÆøµÄС1·Ö¡£

£¨3£©ÒÒ1·Ö£¬2NH3+H2SO4=(NH4)2SO42·Ö

(4) ÓÃʪÈóµÄºìɫʯÈïÊÔÖ½ÖÃÓÚÊԹܿڣ¬Èç¹ûÊÔÖ½±äÀ¶Ôò°±ÆøÒÑÂú1·Ö¡£

£¨5£©±û1·Ö£¬²»ÄÜ1·Ö

½âÎö:

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø