ÌâÄ¿ÄÚÈÝ

Ë®ÊDZ¦¹óµÄ×ÔÈ»×ÊÔ´£¬ÔÚ¹¤Å©ÒµÉú²úºÍÈÕ³£Éú»îÖÐÓÐ׿«Æä¹ã·ºµÄÔËÓã®ÏÂͼΪ×ÔÀ´Ë®³§¾»Ë®¹ý³ÌʾÒâͼ£º

£¨1£©¸Ã×ÔÀ´Ë®³§Éú²ú×ÔÀ´Ë®Ê±£¬Ê¹Óõľ»Ë®·½·¨ÓÐ
A¡¢B¡¢E
A¡¢B¡¢E
£»
A£®³Áµí    B£®¹ýÂË    C£®Öó·Ð    D£®ÕôÁó    E£®Îü¸½
£¨2£©È¡Ë®ºó¼ÓÈëÐõÄý¼Á£¨Ã÷·¯£©µÄ×÷ÓÃ
ʹÐü¸¡ÔÚË®ÖеÄÔÓÖÊÄý¾Û³Á½µ
ʹÐü¸¡ÔÚË®ÖеÄÔÓÖÊÄý¾Û³Á½µ
£»
£¨3£©ÉÏͼÎü¸½³ØÄڵĻîÐÔÌ¿ÆðÎü¸½×÷Ó㬾­³Áµí¡¢¹ýÂ˵Ⱦ»»¯´¦ÀíºóËùµÃµÄË®
²»ÊÇ
²»ÊÇ
´¿Ë®£¨Ìî¡°ÊÇ¡±»ò¡°²»ÊÇ¡±£©£®
£¨4£©¼ÒÍ¥Éú»îÖпÉÒÔÓÃ
·ÊÔíË®
·ÊÔíË®
¼ìÑéijˮÑùÊÇӲˮ»¹ÊÇÈíË®£®
£¨5£©ÓÐЩ¿ÆÑ§¼ÒÔ¤ÑÔ£º¡°ÊÀ½çÉÏ×îºóÒ»µÎË®¾ÍÊÇÈËÀàµÄÑÛÀᡱ£®Õâ¾ä»°¾¯Ê¾ÎÒÃÇÓ¦Ê÷Á¢±£»¤Ë®×ÊÔ´µÄÒâʶ£ºÒ»ÊǽÚÔ¼ÓÃË®£¬¶þÊÇ·ÀֹˮÌåÎÛȾ£®ÇëÄã¾ÙÒ»Àý½ÚÔ¼ÓÃË®µÄ×ö·¨£º
ũҵ½½¹à¸Ä½þ¹àΪÅç¹à
ũҵ½½¹à¸Ä½þ¹àΪÅç¹à
 £¨ºÏÀí´ð°¸¾ù¿É£©£®
·ÖÎö£º£¨1£©¸ù¾Ý×ÔÀ´Ë®µÄ¾»»¯¹ý³Ì½øÐнâ´ð£»
£¨2£©¸ù¾ÝÃ÷·¯µÄ×÷ÓýøÐнâ´ð£»
£¨3£©·ÖÎö»îÐÔÌ¿ÔÚ×ÔÀ´Ë®Éú²ú¹ý³ÌÖеÄ×÷Ó㻲¢ÅжϾ­³Áµí¡¢¹ýÂ˺óµÄË®ÊDz»ÊÇ´¿¾»Î
£¨4£©Ñ¡ÔñÉú»îÓÃÆ·Çø·ÖӲˮºÍÈíË®£»
£¨5£©¾ÙÀý˵Ã÷½ÚÔ¼ÓÃË®µÄ·½·¨£®
½â´ð£º½â£º£¨1£©¸Ã×ÔÀ´Ë®³§Éú²ú×ÔÀ´Ë®Ê±£¬Ê¹Óõľ»Ë®·½·¨ÓгÁµí¡¢¹ýÂË¡¢Îü¸½£»
£¨2£©È¡Ë®ºó¼ÓÈëÐõÄý¼Á£¨Ã÷·¯£©µÄ×÷Óà ʹÐü¸¡ÔÚË®ÖеÄÔÓÖÊÄý¾Û³Á½µ£»
£¨2£©»îÐÔÌ¿¾ßÓжà¿×½á¹¹£¬ÄÜÎü¸½Óж¾Óк¦ÎïÖÊ£¬ÔÚ×ÔÀ´Ë®Éú²úÖÐÆðµ½Îü¸½Ë®ÖÐÓж¾Óк¦ÓÐζÎïÖʵÄ×÷Ó㻾­³Áµí¡¢¹ýÂ˵ÄˮֻÊdzýÈ¥ÁËË®Öв»ÈÜÐÔ¹ÌÌåÔÓÖÊ£¬ÈÔº¬ÓÐÐí¶à¿ÉÈÜÐÔÎïÖÊ£¬ËùÒԵõ½µÄ²»ÊÇ´¿¾»Î
£¨4£©ÈíË®ÖеÎÈë·ÊÔíË®»á²úÉú·á¸»µÄÅÝÄ­£¬¶øÓ²Ë®µÎÈë·ÊÔíË®²úÉúµÄÅÝÄ­ºÜÉÙÇÒÓи¡Ôü£¬ËùÒÔ³£Ó÷ÊÔíË®À´¼ìÑéijˮÑùÊÇӲˮ»¹ÊÇÈíË®£»
£¨5£©½ÚÔ¼ÓÃˮӦ´Óÿ¸öÈË×öÆð£¬¿É½áºÏÉú»îÖеľßÌåÊÂÀýÀ´ËµÃ÷ÈçºÎ½ÚÔ¼ÓÃË®£¬Èçũҵ½½¹à¸Ä½þ¹àΪÅç¹àµÈ£»
¹Ê´ð°¸Îª£º£¨1£©ABE£»£¨2£©Ê¹Ðü¸¡ÔÚË®ÖеÄÔÓÖÊÄý¾Û³Á½µ£»£¨3£©²»ÊÇ£»£¨4£©·ÊÔíË®£»£¨5£©Å©Òµ½½¹à¸Ä½þ¹àΪÅç¹à£®
µãÆÀ£º±¾Ì⿼²éµÄÄÚÈݲ»ÄÑ£¬µ«Éæ¼°µÄÄÚÈݽ϶࣬¾ßÓÐÒ»¶¨µÄİÉú¶È£¬Èç¹û²»ÄܺܺõÄÍØÕ¹Ó¦±ä£¬³ö´íµÄ»ú»á¾ÍºÜ´ó£¬Òò´ËҪϸÖµķÖÎö£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
11¡¢2009Äê2ÔÂ20ÈÕÉÏÎ磬ÓÉÓÚ³ÇÎ÷Ë®³§Ë®Ô´ÊÜ·ÓÀ໯ºÏÎïÎÛȾ£¬½­ËÕÑγÇÊÐÇø·¢Éú´ó·¶Î§¶ÏË®£¬ÖÁÉÙ20Íò¾ÓÃñµÄÉú»îÊܵ½Ó°Ï죬ˮÊDZ¦¹óµÄ×ÔÈ»×ÊÔ´£¬ÔÚ¹¤Å©ÒµÉú²úºÍÈÕ³£Éú»îÖÐÓÐ׿«Æä¹ã·ºµÄÔËÓã®ÏÂͼΪ×ÔÀ´Ë®³§¾»Ë®¹ý³ÌʾÒâͼ£º

£¨1£©Îª¼ì²âË®ÖÊÊÇ·ñ´ïµ½ÒûÓÃË®µÄ±ê×¼£¬¿ÉÓÃ
pHÊÔÖ½»òpH¼Æ
¼ì²âË®µÄËá¼î¶È£®
£¨2£©ÉÏͼÎü¸½³ØÄڵĻîÐÔÌ¿ÆðÎü¸½×÷Ó㬾­³Áµí¡¢¹ýÂ˵Ⱦ»»¯´¦ÀíºóËùµÃµÄË®
²»ÊÇ
´¿¾»µÄË®£¨Ìî¡°ÊÇ¡±»ò¡°²»ÊÇ¡±£©£®
£¨3£©×ÔÀ´Ë®³§³£ÓõÄÏû¶¾¼ÁÓжþÑõ»¯ÂÈ£¨ClO2£©¡¢Æ¯°×·Û¡¢¡°84Ïû¶¾Òº¡±µÈ£®¹¤ÒµÉÏÖÆÈ¡Æ¯°×·ÛµÄ»¯Ñ§·½³ÌʽΪ2C12+2Ca£¨OH£©2¨TCaCl2+Ca£¨ClO£©2+2H2O£¬ÖÆÈ¡¡°84Ïû¶¾Òº¡±Êǽ«ÂÈÆøÍ¨ÈëÉÕ¼îÈÜÒºÖеõ½£¬·´Ó¦Ô­ÀíÓëÆ¯°×·ÛµÄÖÆÈ¡ÏàËÆ£¬Çëд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
Cl2+2NaOH¨TNaCl+NaClO+H2O
£®
£¨4£©¼ÒÍ¥Éú»îÖпÉÒÔÓÃ
·ÊÔíË®
¼ìÑéijˮÑùÊÇӲˮ»¹ÊÇÈíË®£®
£¨5£©¶¬ÌìÔÚÆû³µµÄË®ÏäÖмÓÈëÉÙÁ¿ÒÒ¶þ´¼¡¢±ûÈý´¼µÈÎïÖÊ£¬¿ÉʹÈÜÒºµÄÄý¹Ìµã
½µµÍ
£®
£¨6£©ÓÐЩ¿ÆÑ§¼ÒÔ¤ÑÔ£º¡°ÊÀ½çÉÏ×îºóÒ»µÎË®¾ÍÊÇÈËÀàµÄÑÛÀᡱ£®Õâ¾ä»°¾¯Ê¾ÎÒÃÇÓ¦Ê÷Á¢±£»¤Ë®×ÊÔ´µÄÒâʶ£ºÒ»ÊǽÚÔ¼ÓÃË®£¬¶þÊÇ·ÀֹˮÌåÎÛȾ£®ÇëÄã¾ÙÒ»Àý½ÚÔ¼ÓÃË®µÄ×ö·¨£º
ÌÔÃ×Ë®½½»¨
£®

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

¾«Ó¢¼Ò½ÌÍø