ÌâÄ¿ÄÚÈÝ

ij»¯Ñ§¿ÎÍâС×éÓÃÏÂͼװÖÃÖÆÈ¡äå±½¡£ÏÈÏò·ÖҺ©¶·ÖмÓÈë±½ºÍÒºä壬ÔÙ½«»ìºÏÒºÂýÂýµÎÈë·´Ó¦Æ÷A(A϶˻îÈû¹Ø±Õ)ÖС£

(1)д³öAÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ________________________________________________¡£

(2)¹Û²ìµ½AÖеÄÏÖÏóÊÇ______________________________________________________¡£

(3)ʵÑé½áÊøÊ±£¬´ò¿ªA϶˵ĻîÈû£¬È÷´Ó¦ÒºÁ÷ÈëBÖУ¬³ä·ÖÕñµ´£¬Ä¿µÄÊÇ____________________________________£¬Ð´³öÓйصĻ¯Ñ§·½³Ìʽ_____________________¡£

(4)CÖÐÊ¢·ÅCCl4µÄ×÷ÓÃÊÇ___________________¡£

(5)ÄÜÖ¤Ã÷±½ºÍÒºäå·¢ÉúµÄÊÇÈ¡´ú·´Ó¦£¬¶ø²»ÊǼӳɷ´Ó¦£¬¿ÉÏòÊÔ¹ÜDÖмÓÈëAgNO3ÈÜÒº£¬Èô²úÉúµ­»ÆÉ«³Áµí£¬ÔòÄÜÖ¤Ã÷¡£ÁíÒ»ÖÖÑéÖ¤µÄ·½·¨ÊÇÏòÊÔ¹ÜDÖмÓÈë_____________________£¬ÏÖÏóÊÇ___________________¡£

½âÎö£º±¾ÌâÖ÷Òª¿¼²éäå±½ÖÆ±¸¹ý³ÌÖеÄϸ½ÚÎÊÌâ¡£ÓÉÓÚ±½¡¢ÒºäåµÄ·Ðµã½ÏµÍ£¬ÇÒ·´Ó¦Îª·ÅÈÈ·´Ó¦£¬¹ÊAÖй۲쵽µÄÏÖÏóΪ£º·´Ó¦ÒºÎ¢·Ð£¬Óкì×ØÉ«ÆøÌå³äÂúAÈÝÆ÷¡£HBrÖлìÓеÄBr2ÔÚ¾­¹ýCCl4ʱ±»ÎüÊÕ£»ÔÚÖÆ±¸µÄäå±½Öг£»ìÓÐBr2£¬Ò»°ã¼ÓÈëNaOHÈÜÒºÖУ¬Òò·¢Éú·´Ó¦Br2+2NaOH=NaBr+NaBrO+H2O¶ø³ýÈ¥¡£Èô¸Ã·´Ó¦ÎªÈ¡´ú·´Ó¦Ôò±ØÓÐHBrÉú³É£¬ÈôΪ¼Ó³É·´Ó¦ÔòûÓÐHBrÉú³É£¬¹ÊÖ»Ðè¼ì²éDÖÐÊÇ·ñº¬ÓдóÁ¿H+»òBr-¼´¿É¡£

(2)·´Ó¦ÒºÎ¢·Ð£¬Óкì×ØÉ«ÆøÌå³äÂúAÈÝÆ÷

(3)³ýÈ¥ÈÜÓÚäå±½ÖеÄäå  Br2+2NaOHNaBr+NaBrO+H2O»ò

3Br2+6NaOH5NaBr+NaBrO3+3H2O

(4)³ýÈ¥ä廯ÇâÆøÌåÖеÄäåÕôÆø

(5)ʯÈïÊÔÒº  ÈÜÒº±äºìÉ«(ÆäËûºÏÀí´ð°¸Òà¿É)

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø