¡°ÎïÖʵÄ×é³ÉÓë½á¹¹¾ö¶¨ÎïÖʵÄÐÔÖÊ¡±ÊÇ»¯Ñ§¿ÆѧÖеÄÖØÒª¹Ûµã£®°´ÕÕÕâÒ»¹ÛµãÌÖÂÛÏÂÃæµÄÎÊÌ⣮
£¨1£©ËáÈÜÒº¾ßÓÐһЩ¹²Í¬µÄ»¯Ñ§ÐÔÖÊ£¬ÊÇÒòΪËáÈÜÒºÖж¼º¬ÓР   Àë×Ó£»
£¨2£©¼îÈÜÒºÒ²¾ßÓÐһЩ¹²Í¬µÄ»¯Ñ§ÐÔÖÊ£¬ÊÇÒòΪ¼îÈÜÒºÖж¼º¬ÓР   Àë×Ó£®Òò´Ë£¬ËáÈÜÒºÓë¼îÈÜÒº·´Ó¦Ò»¶¨ÄÜÉú³É    £¬ÀýÈç    £¨Ð´»¯Ñ§·½³Ìʽ£©£®
£¨3£©¼îÓÐÇâÑõ»¯ÄÆ¡¢ÇâÑõ»¯¸ÆµÈ£¬ÇâÑõ»¯ÄÆ¿É×÷ijЩÆøÌåµÄ¸ÉÔï¼Á£¬È磺¸ÉÔï   
µÈÆøÌ壬ÇâÑõ»¯¸Æ¿ÉÓÉÉúʯ»ÒÓëË®·´Ó¦ÖƵ㬻¯Ñ§·½³ÌʽΪ£º    £®
¡¾´ð°¸¡¿·ÖÎö£º¸ù¾ÝÎïÖʵÄ×é³ÉÓë½á¹¹¾ö¶¨ÎïÖʵÄÐÔÖÊ£¬¿ÉÖª
£¨1£©ËáÔÚÈÜÒºÖÐÓÐH+£¬ÊÇËá¾ßÓÐËáµÄͨÐÔµÄÔ­Òò£»
£¨2£©¼îÔÚÈÜÒºÖÐÓÐOH-£¬ÊǼî¾ßÓмîµÄͨÐÔµÄÔ­Òò£¬ËáÓë¼î·´Ó¦ÊÇÒòΪH+ºÍOH-½áºÏÉú³ÉË®£»
£¨3£©Óüî¸ÉÔïÆøÌ壬Ö÷ÒªÀûÓÃÎüË®ÐÔ£¬²¢ÒªÇóÆøÌå²»ÄÜÓë¼î·´Ó¦£¬»îÆýðÊô¶ÔÓ¦µÄ¼î¿ÉÓɽðÊôÑõ»¯ÎïÓëË®·´Ó¦µÃµ½£®
½â´ð£º½â£º£¨1£©¸ù¾ÝËáµÄ¶¨Ò壬ËáÔÚË®ÖнâÀë²úÉúµÄÑôÀë×ÓÈ«²¿ÎªH+£¬Ôò¾ßÓÐһЩ¹²Í¬µÄ»¯Ñ§ÐÔÖÊ£¬¹Ê´ð°¸Îª£ºH+£®
£¨2£©Òò¼îÔÚË®ÖнâÀë²úÉúµÄÒõÀë×ÓÈ«²¿ÎªOH-£¬Ôò¾ßÓÐһЩ¹²Í¬µÄ»¯Ñ§ÐÔÖÊ£¬Ëá¼î·´Ó¦Éú³ÉÑκÍË®£¬
¹Ê´ð°¸Îª£ºOH-£»2NaOH+H2SO4¨TNa2SO4+2H2O£®
£¨3£©ÇâÑõ»¯ÄÆÄÜÎüË®£¬ÇâÆøºÍÑõÆø¶¼²»ÄÜÓë¼î·´Ó¦£¬Ôò¿ÉÓÃÇâÑõ»¯ÄƸÉÔïÇâÆøºÍÑõÆøµÈÆøÌ壬ÓÖÑõ»¯¸ÆºÍË®·´Ó¦Éú³ÉÇâÑõ»¯¸Æ£¬
¹Ê´ð°¸Îª£ºH2£¨»òO2£©£»CaO+H2O=Ca£¨OH£©2£®
µãÆÀ£ºÑ§»áÀûÓÃÎïÖʵĽṹ»ò¹¹³É΢Á£À´·ÖÎöÎïÖʵĻ¯Ñ§ÐÔÖÊ£¬²¢Àí½â²»Í¬µÄ·´Ó¦Óв»Í¬µÄ·´Ó¦Ìõ¼þ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

28¡¢¡°ÎïÖʵÄ×é³ÉÓë½á¹¹¾ö¶¨ÎïÖʵÄÐÔÖÊ£¬ÐÔÖʾö¶¨ÓÃ;£¬ÓÃ;ÌåÏÖÐÔÖÊ£®¡±¸ù¾ÝÕâÒ»ÀíÂۻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÉÓھƾ«ÄÜÔÚ¿ÕÆøÖÐȼÉÕ£¬ËùÒԾƾ«¿ÉÓÃ×÷
ȼÁÏ
£»
£¨2£©ÓÃÄ«Êéд»ò»æÖƵÄÓî»­Äܹ»±£´æºÜ³¤Ê±¼ä¶ø²»ÍÊÉ«£¬ÕâÒ»ÊÂʵÌåÏÖÁËÔÚ³£ÎÂÏÂ̼µÄ»¯Ñ§ÐÔÖÊ
ÔÚ³£ÎÂÏ£¬Ì¼µÄ»¯Ñ§ÐÔÖʱȽÏÎȶ¨
£»
£¨3£©ÓÉÓÚCO·Ö×ÓºÍCO2·Ö×ÓµÄ
½á¹¹
²»Í¬£¬µ¼ÖÂËüÃǵĻ¯Ñ§ÐÔÖʲ»Í¬£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

23¡¢£¨1£©»¯Ñ§ÓëÎÒÃǵÄÉú»îϢϢÏà¹Ø£®Çë´Ó£ºA£®Ì¼ËáÄÆ£¬B£®ÂÈ»¯ÄÆ£¬C£®ÇâÑõ»¯ÄÆ£¬D£®Ì¼ËáÇâÄÆ£¬E£®Ë®£¬F£®¾Æ¾«£¬G£®Ê¯ÓÍ£¬H£®ÃºµÈ°ËÖÖÎïÖÊÖÐÑ¡ÔñÊʵ±µÄ×ÖĸÐòºÅÌî¿Õ£º
¢ÙÖÎÁÆθËá¹ý¶àµÄÒ©¼ÁÊÇ
D
£¬¢Ú±»ÓþΪ¡°¹¤ÒµµÄѪҺ¡±µÄÊÇ
G

¢ÛºòµÂ°ñÁªºÏÖƼÖеġ°¼î¡±
A
£¬¢Ü×î³£¼ûµÄÈܼÁÊÇ
E

£¨2£©ÎïÖʵÄ×é³ÉÓë½á¹¹¾ö¶¨ÎïÖʵÄÐÔÖÊ£®ÊÔ·ÖÎöµ¼ÖÂÏÂÁÐÁ½×éÎïÖÊÐÔÖʲ»Í¬µÄÔ­Òò£®
¢ÙÇâÑõ»¯ÄƺÍÇâÑõ»¯±µÈÜÒº¶¼ÄÜʹÎÞÉ«·Ó̪ÊÔÒº±äºì£®Ô­ÒòÊÇ
ÈÜÒºÖж¼´æÔÚOH-
£»
¢Ú½ð¸ÕʯºÜÓ²£¬¶øʯīȴºÜÈí£®Ô­ÒòÊÇ
Ô­×ÓµÄÅÅÁз½Ê½²»Í¬
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

21¡¢¡°ÎïÖʵÄ×é³ÉÓë½á¹¹¾ö¶¨ÎïÖʵÄÐÔÖÊ¡±ÊÇ»¯Ñ§¿ÆѧÖеÄÖØÒª¹Ûµã£®°´ÕÕÕâÒ»¹ÛµãÌÖÂÛÏÂÃæµÄÎÊÌ⣮
£¨1£©ËáÈÜÒº¾ßÓÐһЩ¹²Í¬µÄ»¯Ñ§ÐÔÖÊ£¬ÊÇÒòΪËáÈÜÒºÖж¼º¬ÓÐ
H+
Àë×Ó£»
£¨2£©¼îÈÜÒºÒ²¾ßÓÐһЩ¹²Í¬µÄ»¯Ñ§ÐÔÖÊ£¬ÊÇÒòΪ¼îÈÜÒºÖж¼º¬ÓÐ
OH-
Àë×Ó£®Òò´Ë£¬ËáÈÜÒºÓë¼îÈÜÒº·´Ó¦Ò»¶¨ÄÜÉú³É
H2O
£¬ÀýÈç
2NaOH+H2SO4¨TNa2SO4+2H2O
£¨Ð´»¯Ñ§·½³Ìʽ£©£®
£¨3£©¼îÓÐÇâÑõ»¯ÄÆ¡¢ÇâÑõ»¯¸ÆµÈ£¬ÇâÑõ»¯ÄÆ¿É×÷ijЩÆøÌåµÄ¸ÉÔï¼Á£¬È磺¸ÉÔï
H2£¨»òO2£©

µÈÆøÌ壬ÇâÑõ»¯¸Æ¿ÉÓÉÉúʯ»ÒÓëË®·´Ó¦ÖƵ㬻¯Ñ§·½³ÌʽΪ£º
CaO+H2O=Ca£¨OH£©2
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

23¡¢£¨6·Ö£©ÒÑÖªXÊÇÓж¾ÇÒ²»ÈÜÓÚË®µÄÆøÌ壬YÊDz»Ö§³ÖȼÉÕµÄÆøÌ壬ZÊDz»ÈÜÓÚË®µÄ¹ÌÌ壬X¡¢Y¡¢ZÖ®¼äÓÐÈçÏÂת»¯¹Øϵ£®Çë»Ø´ðÏÂÁÐÎÊÌ⣮

£¨1£©Ð´³öXµÄ»¯Ñ§Ê½
CO
£®
£¨2£©Ð´³öYת»¯ÎªZµÄ·½³Ìʽ£º
CO2+Ca£¨OH£©2=CaCO3¡ý+H2O
£®
£¨3£©¡°ÎïÖʵÄ×é³ÉÓë½á¹¹¾ö¶¨ÎïÖʵÄÐÔÖÊ¡±ÊÇÖØÒªµÄ»¯Ñ§Ë¼Ï룮ÆøÌåX¡¢Y¶¼º¬ÓÐ
̼£¨»òC£©
ÔªËغÍ
Ñõ£¨»òO£©
ÔªËØ£¬µ«ËüÃǵÄÎïÀíÐÔÖÊ¡¢»¯Ñ§ÐÔÖʶ¼²»Í¬£®Çë¾Ù³öX¡¢YÐÔÖʲ»Í¬µÄÒ»¸öÀý×Ó
Ò»Ñõ»¯Ì¼ÓпÉȼÐÔ¶þÑõ»¯Ì¼Ã»ÓÐ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2005?ÄÏ»ãÇøһģ£©¡°ÎïÖʵÄ×é³ÉÓë½á¹¹¾ö¶¨ÎïÖʵÄÐÔÖÊ¡±ÊÇ»¯Ñ§Ñ§¿ÆÖеÄÖØÒª¹Ûµã£¬°´ÕÕÕâÒ»¹Ûµã½âÊÍÏÂÁÐÎÊÌ⣺
£¨1£©»îÐÔÌ¿µÄ±íÃæÓÐÐí¶à΢¿×£¬ÕâÖֽṹʹµÃ»îÐÔÌ¿¾ßÓÐ
Îü¸½ÐÔ
Îü¸½ÐÔ
£®
£¨2£©¹ýÑõ»¯ÇâÊÇÒ»ÖÖ³£ÓõÄɱ¾úÏû¶¾¼Á£¬ÆäÔ­ÒòÊǹýÑõ»¯Çâ·Ö×ÓÖк¬ÓÐÒ»ÖÖ½Ð×ö¡°¹ýÑõ»ù¡±£¨Èçͼ¢ÙÖÐÐéÏß¿ò±ê³öµÄ²¿·Ö£©µÄ½á¹¹£®¾Ý´ËÍƲâÏÂÁÐÎïÖÊÖУ¬¿ÉÓÃ×÷ɱ¾úÏû¶¾µÄÊÇ
¢Ü
¢Ü
£®
 ¢Ù    ¢ÚH-O-H   ¢ÛO¨TC¨TO  ¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸