ͼ2ΪÈËÌåÖи÷ÖÖÓªÑøÎïÖʵĺ¬Á¿£¬ËûÃÇÔÚÈ˵ÄÕý³£ÉúÀí»î¶¯ÖÐÆð×ŷdz£ÖØÒªµÄ×÷Óã®
¾«Ó¢¼Ò½ÌÍø

£¨1£©ÈËÌåÉúÀí»î¶¯Ò»°ëÒÔÉϵÄÄÜÁ¿À´Ô´ÓÚ______£®
£¨2£©Î¬ÉúËØÊÇÈËÌå±ØÐèµÄÒ»ÀàÓªÑøËØ£¬Î¬ÉúËصÄÖÖÀàºÜ¶à£¬ÆäÖÐάÉúËØAÓÖ½ÐÊӻƴ¼£¬ÈËÌåȱÉÙάÉúËØA»á»¼______£¨ÌîÐòºÅ£©¢Ùҹä֢£»¢Ú»µÑª²¡£»¢ÛÈí¹Ç²¡
£¨3£©µ°°×ÖÊÊÇÉúÃüµÄÎïÖÊ»ù´¡£¬µ°°×ÖÊÊÇÓɶàÖÖ°±»ùËá¹¹³ÉµÄ£¬¸÷ÖÖ°±»ùËáµÄ½á¹¹¿ÉÓÃͼ3±íʾ£¬µ°°×ÖÊÖÐÒ»¶¨º¬ÓеÄÔªËØÓÐ______£®
£¨4£©½ðÊôÔªËØÔÚÈËÌåÖеĺ¬Á¿¹ý¸ß»òÌ«ÉÙ¶¼»áÓ°ÏìÈËÌ彡¿µ£¬½«ÏÂÁÐÔªËØÓë¸ÃÔªËز»×ã´øÀ´µÄ½¡¿µÎÊÌâÁ¬Ïß
Ca               Ó°ÏìÖÇÁ¦
Fe               ³é´¤¡¢Èí¹Ç²¡
Zn               Æ¶Ñª¡¢Éñ¾­ÐÔ¶úÁû£®
£¨1£©ÌÇÀàÊÇÈËÌåÉúÀí»î¶¯µÄÖ÷Òª¹©ÄÜÎïÖÊ£¬Ò»°ëÒÔÉϵÄÄÜÁ¿À´Ô´ÓÚ ÌÇÀ࣮
£¨2£©ÈËÌåȱÉÙάÉúËØA»á»¼Ò¹Ã¤Ö¢
£¨3£©ÓÉ°±»ùËáµÄ½á¹¹Í¼¿ÉÖª£¬µ°°×ÖÊÖÐÒ»¶¨º¬ÓеÄÔªËØÓÐ C¡¢H¡¢O¡¢N£®
£¨4£©ÈËÌåȱ¸Æ»á³é´¤¡¢Èí¹Ç²¡£»È±Ìú»áƶѪ¡¢Éñ¾­ÐÔ¶úÁû£»È±Ð¿»áÓ°ÏìÖÇÁ¦
¹Ê´ð°¸Îª£º£¨1£©ÌÇÀࣻ
£¨2£©¢Ù£»
£¨3£©C¡¢H¡¢O¡¢N£»
£¨4£©

¾«Ó¢¼Ò½ÌÍø
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2007?˳ÒåÇø¶þÄ££©Í¼2ΪÈËÌåÖи÷ÖÖÓªÑøÎïÖʵĺ¬Á¿£¬ËûÃÇÔÚÈ˵ÄÕý³£ÉúÀí»î¶¯ÖÐÆð×ŷdz£ÖØÒªµÄ×÷Óã®
£¨1£©ÈËÌåÉúÀí»î¶¯Ò»°ëÒÔÉϵÄÄÜÁ¿À´Ô´ÓÚ
ÌÇÀà
ÌÇÀà
£®
£¨2£©Î¬ÉúËØÊÇÈËÌå±ØÐèµÄÒ»ÀàÓªÑøËØ£¬Î¬ÉúËصÄÖÖÀàºÜ¶à£¬ÆäÖÐάÉúËØAÓÖ½ÐÊӻƴ¼£¬ÈËÌåȱÉÙάÉúËØA»á»¼
¢Ù
¢Ù
£¨ÌîÐòºÅ£©¢Ùҹä֢£»¢Ú»µÑª²¡£»¢ÛÈí¹Ç²¡
£¨3£©µ°°×ÖÊÊÇÉúÃüµÄÎïÖÊ»ù´¡£¬µ°°×ÖÊÊÇÓɶàÖÖ°±»ùËá¹¹³ÉµÄ£¬¸÷ÖÖ°±»ùËáµÄ½á¹¹¿ÉÓÃͼ3±íʾ£¬µ°°×ÖÊÖÐÒ»¶¨º¬ÓеÄÔªËØÓÐ
C¡¢H¡¢O¡¢N
C¡¢H¡¢O¡¢N
£®
£¨4£©½ðÊôÔªËØÔÚÈËÌåÖеĺ¬Á¿¹ý¸ß»òÌ«ÉÙ¶¼»áÓ°ÏìÈËÌ彡¿µ£¬½«ÏÂÁÐÔªËØÓë¸ÃÔªËز»×ã´øÀ´µÄ½¡¿µÎÊÌâÁ¬Ïß
Ca               Ó°ÏìÖÇÁ¦
Fe               ³é´¤¡¢Èí¹Ç²¡
Zn               Æ¶Ñª¡¢Éñ¾­ÐÔ¶úÁû£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2012?¿ªÔ¶ÊжþÄ££©ÈËÀàÉú»î¡¢Éú²úÀë²»¿ª»¯Ñ§£®
£¨1£©ÇåеĿÕÆø¡¢½à¾»µÄÒûÓÃË®¡¢¾ùºâµÄÓªÑø¶¼ÓëÈËÌ彡¿µÏ¢Ï¢Ïà¹Ø£®
¢ÙÇåеĿÕÆøÊôÓÚ
»ìºÏÎï
»ìºÏÎï
£¨Ñ¡Ìî¡°´¿¾»Î»ò¡°»ìºÏÎ£©£®
¢Ú½à¾»µÄÒûÓÃË®ÔÚ¾»»¯¹ý³ÌÖÐÏÈÓÃ
¹ýÂË
¹ýÂË
µÄ·½·¨³ýȥˮÖв»ÈÜÐÔÔÓÖÊ£¬ÔÙÓÃ
»îÐÔÌ¿
»îÐÔÌ¿

ľ̿
ľ̿
Îü¸½É«ËغÍÒìζ£®È»ºóÓÃÏû¶¾¼Á½øÐÐÏû¶¾£®XÊÇÒ»ÖÖ³£¼ûµÄ×ÔÀ´Ë®Ïû¶¾¼Á£¬¹¤ÒµÉÏÖÆÈ¡XµÄ»¯Ñ§·½³ÌʽΪ£ºCl2+2NaClO2=2NaCl+2X£¬ÔòXµÄ»¯Ñ§Ê½ÊÇ
ClO2
ClO2
£®
¢Û¾ùºâµÄÓªÑøÐèÒªºÏÀíÉÅʳ£®Ê³Æ·ÖеÄÓлúÓªÑøËØÓÐÌÇÀà¡¢Ö¬·¾¡¢µ°°×ÖÊ¡¢Î¬ÉúËØ£¬Ê³ÓÃË®¹ûÊß²ËΪÈËÌå²¹³äµÄÖ÷ÒªÓлúÓªÑøËØÊÇ
άÉúËØ
άÉúËØ
£®
£¨2£©ÒÔÏÂÊÇÎÒÃÇÈÕ³£Éú»îÖг£Óõĸ÷ÖÖÇåÏ´¼Á£¬Ñ¡ÔñÊʵ±ÓÃÆ·¿ÉÒԵõ½¸üºÃµÄÇåϴЧ¹û£®
Ãû³Æ Ï´µÓÁé ½à²ÞÁé ¯¾ßÇå½à¼Á ÎÛ×Õ±¬Õ¨ÑÎ Ïû¶¾Òº
²úÆ·
Ñùʽ
ÓÐЧ³É·Ö»ò¹¦ÄÜ ÇåÏ´ÓÍÎÛ ÑÎËá ÇâÑõ»¯ÄÆ ¹ý̼ËáÄÆ Ïû¶¾
¢ÙÎÒÃÇʹÓÃÏ´µÓÁéÇåÏ´²Í¾ßÉϵÄÓÍÎÛ£¬ÕâÊÇÒòΪËü¾ßÓÐ
È黯
È黯
¹¦ÄÜ£®
¢ÚÒÔÏÂÎïÖÊ¿ÉÒÔʹÓýà²ÞÁéÇåÏ´µÄÊÇ
AC
AC
£¨Ìî×ÖĸÐòºÅ£©£®
A£®ÌúÐâ        B£®ÓÍ×Õ      C£®Ë®¹¸
¢ÛÈô½«½à²ÞÁéÓ믾ßÇå½à¼Á»ìºÏ£¬¿ÉÒÔ·¢ÉúͼËùʾµÄ»¯Ñ§·´Ó¦£®Í¼ÖÐa΢Á£µÄ»¯Ñ§Ê½Îª
H2O
H2O
£®
¢Ü¡°ÎÛ×Õ±¬Õ¨ÑΡ±ÈÜÓÚË®ºó·Ö½âÉú³ÉNa2CO3ºÍH2O2£®Èô½«±¬Õ¨ÑÎÈÜÓÚË®ºó£¬ÔÙ¼ÓÈë×ãÁ¿µÄ½à²ÞÁ飬¹Û²ìµ½ÓÐÆøÅÝð³ö£®¼ÓÈë×ãÁ¿µÄ½à²ÞÁéºó·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
£®
¢Ý¹¤ÒµÉϽ«ÂÈÆø£¨Cl2£©Í¨ÈëÉÕ¼îÈÜÒºÖпÉÖÆÈ¡Ïû¶¾Òº£®·´Ó¦ºóÐγÉÁËNaClºÍNaClOµÄÈÜÒº£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Cl2+2NaOH=NaCl+NaClO+H2O
Cl2+2NaOH=NaCl+NaClO+H2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

ͼ2ΪÈËÌåÖи÷ÖÖÓªÑøÎïÖʵĺ¬Á¿£¬ËûÃÇÔÚÈ˵ÄÕý³£ÉúÀí»î¶¯ÖÐÆð×ŷdz£ÖØÒªµÄ×÷Óã®
£¨1£©ÈËÌåÉúÀí»î¶¯Ò»°ëÒÔÉϵÄÄÜÁ¿À´Ô´ÓÚ________£®
£¨2£©Î¬ÉúËØÊÇÈËÌå±ØÐèµÄÒ»ÀàÓªÑøËØ£¬Î¬ÉúËصÄÖÖÀàºÜ¶à£¬ÆäÖÐάÉúËØAÓÖ½ÐÊӻƴ¼£¬ÈËÌåȱÉÙάÉúËØA»á»¼________£¨ÌîÐòºÅ£©¢Ùҹä֢£»¢Ú»µÑª²¡£»¢ÛÈí¹Ç²¡
£¨3£©µ°°×ÖÊÊÇÉúÃüµÄÎïÖÊ»ù´¡£¬µ°°×ÖÊÊÇÓɶàÖÖ°±»ùËá¹¹³ÉµÄ£¬¸÷ÖÖ°±»ùËáµÄ½á¹¹¿ÉÓÃͼ3±íʾ£¬µ°°×ÖÊÖÐÒ»¶¨º¬ÓеÄÔªËØÓÐ________£®
£¨4£©½ðÊôÔªËØÔÚÈËÌåÖеĺ¬Á¿¹ý¸ß»òÌ«ÉÙ¶¼»áÓ°ÏìÈËÌ彡¿µ£¬½«ÏÂÁÐÔªËØÓë¸ÃÔªËز»×ã´øÀ´µÄ½¡¿µÎÊÌâÁ¬Ïß
Ca¡¡¡¡¡¡¡¡¡¡¡¡¡¡ Ó°ÏìÖÇÁ¦
Fe¡¡¡¡¡¡¡¡¡¡¡¡¡¡ ³é´¤¡¢Èí¹Ç²¡
Zn¡¡¡¡¡¡¡¡¡¡¡¡¡¡ ƶѪ¡¢Éñ¾­ÐÔ¶úÁû£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2007Äê±±¾©ÊÐ˳ÒåÇøÖп¼»¯Ñ§¶þÄ£ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

ͼ2ΪÈËÌåÖи÷ÖÖÓªÑøÎïÖʵĺ¬Á¿£¬ËûÃÇÔÚÈ˵ÄÕý³£ÉúÀí»î¶¯ÖÐÆð×ŷdz£ÖØÒªµÄ×÷Óã®
£¨1£©ÈËÌåÉúÀí»î¶¯Ò»°ëÒÔÉϵÄÄÜÁ¿À´Ô´ÓÚ    £®
£¨2£©Î¬ÉúËØÊÇÈËÌå±ØÐèµÄÒ»ÀàÓªÑøËØ£¬Î¬ÉúËصÄÖÖÀàºÜ¶à£¬ÆäÖÐάÉúËØAÓÖ½ÐÊӻƴ¼£¬ÈËÌåȱÉÙάÉúËØA»á»¼    £¨ÌîÐòºÅ£©¢Ùҹä֢£»¢Ú»µÑª²¡£»¢ÛÈí¹Ç²¡
£¨3£©µ°°×ÖÊÊÇÉúÃüµÄÎïÖÊ»ù´¡£¬µ°°×ÖÊÊÇÓɶàÖÖ°±»ùËá¹¹³ÉµÄ£¬¸÷ÖÖ°±»ùËáµÄ½á¹¹¿ÉÓÃͼ3±íʾ£¬µ°°×ÖÊÖÐÒ»¶¨º¬ÓеÄÔªËØÓР   £®
£¨4£©½ðÊôÔªËØÔÚÈËÌåÖеĺ¬Á¿¹ý¸ß»òÌ«ÉÙ¶¼»áÓ°ÏìÈËÌ彡¿µ£¬½«ÏÂÁÐÔªËØÓë¸ÃÔªËز»×ã´øÀ´µÄ½¡¿µÎÊÌâÁ¬Ïß
Ca               Ó°ÏìÖÇÁ¦
Fe               ³é´¤¡¢Èí¹Ç²¡
Zn               Æ¶Ñª¡¢Éñ¾­ÐÔ¶úÁû£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸