ÌâÄ¿ÄÚÈÝ

ÏÂͼÊÇÔªËØÖÜÆÚ±íµÄ¿ò¼Ü£¬ÒÀ¾ÝÔªËØÖÜÆÚ±í»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÖÜÆÚ±íÖеÄÔªËØ¢ÝºÍÔªËØ¢ÞµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï¼îÐÔÇ¿Èõ˳ÐòÊÇ________(Óû¯Ñ§Ê½±íʾ)£¬ÖÜÆÚ±íÖеÄÔªËØ¢ÜºÍÔªËØ¢ßµÄÇ⻯ÎïµÄ·Ðµã¸ßµÍ˳ÐòÊÇ________(Óû¯Ñ§Ê½±íʾ)£®

(2)Çëд³ö¢ÚµÄÇ⻯ÎïµÄµç×Óʽ________£¬¢ÛµÄÇ⻯ÎïµÄ½á¹¹Ê½________£®

(3)¢Ù¡«¢ßÔªËØµÄµ¥ÖÊ£¬ÔÚ³£ÎÂÏ»¯Ñ§ÐÔÖÊÎȶ¨£¬Í¨³£¿ÉÓÃ×÷±£»¤ÆøµÄÊÇ________(Ìîд·Ö×Óʽ)£®

(4)Çëд³ö¹¤ÒµÉÏÖÆ±¸¢ßµ¥ÖʵĻ¯Ñ§·½³Ìʽ________£®

´ð°¸£º
½âÎö£º

¡¡¡¡(1)NaOH£¾Mg(OH)2£¬HF£¾HCl£®

¡¡¡¡(2)£¬H£­O£­H£®

¡¡¡¡(3)N2£®

¡¡¡¡(4)2NaCl£«2H2O£½2NaOH£«H2¡ü£«Cl2¡ü(Ìõ¼þ£ºÍ¨µç)£®


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2012?Ìì½òÄ£Ä⣩A¡¢B¡¢C¡¢D¡¢EÊÇÖÐѧ»¯Ñ§³£¼ûµÄ5ÖÖ»¯ºÏÎÆäÖÐA¡¢BÊÇÑõ»¯Îµ¥ÖÊX¡¢YÊÇÉú»îÖг£¼ûµÄ½ðÊô£¬ÊÔ¼Á1ºÍÊÔ¼Á2·Ö±ðΪ³£¼ûµÄËá»ò¼î£®Ïà¹ØÎïÖʼäµÄת»¯¹ØÏµÈçÏÂͼËùʾ£¨²¿·Ö·´Ó¦ÎïÓë²úÎïÒÑÂÔÈ¥£©£º

Çë»Ø´ð£º
£¨1£©¢Ù×é³Éµ¥ÖÊXµÄÔªËØÔ­×ӽṹʾÒâͼÊÇ
£»
¢Ú×é³Éµ¥ÖÊYµÄÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃ
µÚËÄÖÜÆÚµÚ VIII×å
µÚËÄÖÜÆÚµÚ VIII×å
£®
£¨2£©ÈôÊÔ¼Á1Ϊǿ¼îÈÜÒº£¬ÔòXÓëÊÔ¼Á1·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
2Al+2H2O+2OH-=2AlO2-+3H2¡ü
2Al+2H2O+2OH-=2AlO2-+3H2¡ü
£®
£¨3£©¢ÙÈôÊÔ¼Á1ºÍÊÔ¼Á2Ïàͬ£¬ÇÒEÈÜÒº¼ÓÈÈÕô¸É²¢×ÆÉÕºó¿ÉµÃµ½A£¬ÔòAµÄ»¯Ñ§Ê½ÊÇ
Fe2O3
Fe2O3
£®
¢Ú½«ÎïÖÊCÈÜÓÚË®£¬ÆäÈÜÒº³Ê
ËáÐÔ
ËáÐÔ
£¨Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£©£¬Ô­ÒòÓÃÀë×Ó·½³Ìʽ¿É±íʾΪ
Al3++3H2O?Al£¨OH£©3+3H+
Al3++3H2O?Al£¨OH£©3+3H+
£®
£¨4£©ÈôÊÔ¼Á2ΪϡÁòËᣬ¹¤ÒµÉÏÒÔE¡¢Ï¡ÁòËáºÍNaNO2ΪԭÁÏÖÆ±¸¸ßЧ¾»Ë®¼ÁY£¨OH£©SO4£¬ÇÒ·´Ó¦ÖÐÓÐNOÉú³É£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ
2FeSO4+2NaNO2+H2SO4=2Fe£¨OH£©SO4+Na2SO4+2NO¡ü
2FeSO4+2NaNO2+H2SO4=2Fe£¨OH£©SO4+Na2SO4+2NO¡ü
£®
£¨5£©Ç뽫D+Y¡úEµÄ¹ý³ÌÉè¼Æ³ÉÒ»¸öÄܲúÉú³ÖÐø¶øÎȶ¨µçÁ÷µÄÔ­µç³Ø×°Öã¨Ê¹ÓÃÑÎÇÅ£©£¬ÔÚ¿ò¸ñÄÚ»­³öʵÑé×°ÖÃͼ£¬²¢ÔÚͼÖбê³öµç¼«ºÍÊÔ¼ÁµÄÃû³Æ£®

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

¾«Ó¢¼Ò½ÌÍø