ÌâÄ¿ÄÚÈÝ

ÏÂͼÊÇÔªËØÖÜÆÚ±íµÄ¿ò¼Ü

£¨1£©ÇëÔÚÉÏÃæÔªËØÖÜÆÚ±íÖл­³ö½ðÊôÔªËØÓë·Ç½ðÊôÔªËصķֽçÏß¡£

£¨2£©ÒÀ¾ÝÔªËØÖÜÆÚ±í»Ø´ðÏÂÁÐÎÊÌ⣺

A£®ÖÜÆÚ±íÖеÄÔªËآݺÍÔªËØ¢ÞµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï¼îÐÔÇ¿Èõ˳ÐòÊÇ

                            £¨Óû¯Ñ§Ê½±íʾ£©¡£

B£®ÖÜÆÚÖбíµÄÔªËآܺÍÔªËآߵÄÇ⻯ÎïµÄÈÛ¡¢·Ðµã¸ßµÍ˳ÐòÊÇ

                            £¨Óû¯Ñ§Ê½±íʾ£©¡£

C£®¢Ù¨D¢ßÔªËصÄijµ¥ÖÊÓг£ÎÂÏ»¯Ñ§ÐÔÖÊÎȶ¨£¬Í¨³£¿ÉÒÔ±£»¤ÆøµÄÊÇ

                            £¨Óû¯Ñ§Ê½±íʾ£©¡£

D£®ÔÚÉÏÃæÔªËØÖÜÆÚ±íÖÐÈ«²¿ÊǽðÊôÔªËصÄÖ÷×åÊÇ             £»È«²¿ÊǷǽðÊôÔªËصÄÖ÷×åÊÇ             £¨ÌîдÀëĸa¡¢b¡¢c¡¢d£©¡£

a£®IA×å          b£®¢òA×å        c£®¢ôA×å      d£®VIIA×å

£¨3£©ÒÑÖª¼×ÔªËØλÓÚµÚÈýÖÜÆÚ£¬ÇÒÆäÔ­×Ӱ뾶ΪͬÖÜÆÚ½ðÊôÔªËØÖеÄÔ­×Ӱ뾶×îСµÄ£¬Çëд³ö¼×µÄÑõ»¯ÎïÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ            £»ÔÚÒ»¶¨Ìõ¼þÏÂ1g¢ÙµÄµ¥ÖÊÔÚ×ãÁ¿¢ÛµÄµ¥ÖÊÖÐÍêȫȼÉÕÉú³ÉҺ̬ÎïÖÊʱ£¬·Å³öµÄÈÈÁ¿ÎªakJ£¬Çëд³ö´ËÌõ¼þϱíʾ¢ÙµÄµ¥ÖÊȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ                   £»

¢Ù¡¢¢ÛÁ½ÖÖÔªËصĵ¥ÖÊÒѱ»Ó¦ÓÃÓÚÓîÖæ·É´¬µÄȼÁϵç³ØÖУ¬ÈçͼËùʾ£¬Á½¸öµç¼«¾ùÓɶà¿×ÐÔ̼¹¹³É£¬Í¨ÈëµÄÁ½ÖÖµ¥ÖÊÓÉ¿×϶Òݳö²¢Ôڵ缫±íÃæ·Åµç¡£

Çë»Ø´ð£ºbÊǵç³ØµÄ     ¼«£»aµç¼«Éϵĵ缫·´Ó¦Ê½ÊÇ       ¡£

£¨1£©Èçͼ¡£

   £¨2£©A£®NaOH > Mg(OH)2

B£®HF£¾HCl

C£®N2

D£®b£» d

   £¨3£©Al2O3 + 2OH£­ = 2AlO2£­ + H2O

        H2(g) + O2(g) = H2O (l)£»¡÷H =£­2akJ/mol

        Õý£¬2H2+ 4OH£­£­4e£­ = 4H2O

              »òH2 + 2OH£­£­2e£­ = 2H2O

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»¯¹ØϵÈçÏÂͼËùʾ£¨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¡ÁòËáºÍNaNO2ΪԭÁÏÖƱ¸¸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