ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÏÂÁÐÝÍÈ¡Óë·ÖÒº½áºÏ½øÐеIJÙ×÷£¨ÓÃ×÷ÝÍÈ¡¼Á£¬´ÓµâË®ÖÐÝÍÈ¡»Ç£©´íÎóµÄÊÇ£¨ £©

A.µâË®ºÍ¼ÓÈë·ÖҺ©¶·ºó£¬¸ÇºÃ²£Á§Èû£¬ÓÒÊÖѹס·ÖҺ©¶·¿Ú²¿£¬×óÊÖÎÕס»îÈû²¿·Ö£¬°Ñ·ÖҺ©¶·µ¹×ª¹ýÀ´Õñµ´

B.¾²Ö㬴ý·ÖҺ©¶·ÖÐÒºÌå·Ö²ãºó£¬ÏÈʹ·ÖҺ©¶·ÄÚÍâ¿ÕÆøÏàͨ£¨×¼±¸·Å³öÒºÌ壩

C.´ò¿ª·ÖҺ©¶·µÄ»îÈû£¬Ê¹Ï²ãÒºÌåÈ«²¿ÑØÊ¢½ÓÒºÌåµÄÉÕ±­ÄÚ±ÚÂýÂýÁ÷³ö

D.×îºó¼ÌÐø´ò¿ª»îÈû£¬ÁíÓÃÈÝÆ÷Ê¢½Ó²¢±£´æÉϲãÒºÌå

¡¾´ð°¸¡¿D

¡¾½âÎö¡¿

A£®µâ²»Ò×ÈÜÓÚË®£¬Ò×ÈÜÓÚËÄÂÈ»¯Ì¼£¬ÔÚ·ÖҺ©¶·ÖÐÝÍÈ¡¡¢·ÖÒº£¬ÔòÓÃÓÒÊÖѹס·ÖҺ©¶·²£Á§Èû£¬×óÊÖÎÕס»îÈû²¿·Ö£¬°Ñ·ÖҺ©¶·µ¹×ª¹ýÀ´Õñµ´£¬È»ºó¾²Ö㬲Ù×÷ºÏÀí£¬AÕýÈ·£»

B£®Ê¹·ÖҺ©¶·ÄÚÍâ¿ÕÆøÏàͨ£¬Æ½ºâÄÚÍâÆøѹ£¬ÈÜÒº·Ö²ãºó±£Ö¤ÒºÌå˳ÀûÁ÷³ö£¬BÕýÈ·£»

C£®·ÖÒº²Ù×÷ʱ£¬·ÖҺ©¶·ÖÐϲãÒºÌå´ÓÏ¿ڷųö£¬Ï²ãÒºÌå·ÅÍêʱÁ¢¼´¹Ø±Õ»îÈû£¬ÉϲãÒºÌå´ÓÉÏ¿Úµ¹³ö£¬ÒÔÃâÁ½ÖÖÒºÌåÏ໥ÎÛȾ£¬CÕýÈ·£»

D£®·ÖÒº²Ù×÷ʱ£¬·ÖҺ©¶·ÖÐϲãÒºÌå´ÓÏ¿ڷųö£¬Ï²ãÒºÌå·ÅÍêʱÁ¢¼´¹Ø±Õ»îÈû£¬ÉϲãÒºÌå´ÓÉÏ¿Úµ¹³ö£¬ÒÔÃâÁ½ÖÖÒºÌåÏ໥ÎÛȾ£¬D´íÎó£»

¹Ê´ð°¸Îª£ºD¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ClO2ÊÇÒ»ÖÖÓÅÁ¼µÄÏû¶¾¼Á£¬³£½«ÆäÖƳÉNaClO2¹ÌÌ壬ÒÔ±ãÔËÊäºÍÖü´æ£¬¹ýÑõ»¯Çâ·¨±¸NaClO2¹ÌÌåµÄʵÑé×°ÖÃÈçͼËùʾ¡£

ÒÑÖª£º¢Ù2NaC1O3+H2O2+H2SO4=2C1O2¡ü+O2¡ü+Na2SO4+2H2O

2ClO2+H2O2+2NaOH=2NaClO2+O2¡ü+2H2O

¢ÚClO2ÈÛµã-59¡æ¡¢·Ðµã11¡æ£¬Å¨¶È¹ý¸ßʱÒ×·¢Éú·Ö½â£»

¢ÛH2O2·Ðµã150¡æ

£¨1£©±ùˮԡÀäÈ´µÄÄ¿µÄÊÇ___¡£

£¨2£©¿ÕÆøÁ÷ËÙ¹ý¿ì»ò¹ýÂý£¬¾ù½µµÍNaClO2²úÂÊ£¬ÊÔ½âÊÍÆäÔ­Òò£¬¿ÕÆøÁ÷ËÙ¹ýÂýʱ£¬__¡£

£¨3£©Cl-´æÔÚʱ»á´ß»¯ClO2µÄÉú³É¡£·´Ó¦¿ªÊ¼Ê±ÔÚCÖмÓÈëÉÙÁ¿ÑÎËᣬClO2µÄÉú³ÉËÙÂÊ´ó´óÌá¸ß£¬²¢²úÉú΢Á¿ÂÈÆø¡£¸Ã¹ý³Ì¿ÉÄܾ­Á½²½Íê³É£¬Ç뽫Æä²¹³äÍêÕû£º

¢Ù___£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¢ÚH2O2+Cl2=2Cl-+O2+2H+

£¨4£©NaClO2´¿¶È²â¶¨£º

¢Ù׼ȷ³ÆÈ¡ËùµÃNaClO2ÑùÆ·10.0gÓÚÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®ºÍ¹ýÁ¿µÄµâ»¯¼Ø¾§Ì壬ÔÙµÎÈëÊÊÁ¿µÄÏ¡ÁòËᣬ³ä·Ö·´Ó¦£¨C1O2-µÄ²úÎïΪCl-£©£¬½«ËùµÃ»ìºÏÒºÅä³É250mL´ý²âÈÜÒº£»

¢ÚÈ¡25.00mL´ý²âÒº£¬ÓÃ2.0mol¡¤L-1Na2S2O3±ê×¼ÒºµÎ¶¨(I2+2S2O32-=2I-+S4O62-)£¬ÒÔµí·ÛÈÜÒº×öָʾ¼Á£¬´ïµ½µÎ¶¨ÖÕµãʱµÄÏÖÏóΪ__£¬Öظ´µÎ¶¨3´Î£¬²âµÃNa2S2O3±ê׼Һƽ¾ùÓÃÁ¿Îª20.00mL£¬Ôò¸ÃÑùÆ·ÖÐNaClO2µÄÖÊÁ¿·ÖÊýΪ___¡££¨M(NaClO2)=90.5g/mol£©

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

¾«Ó¢¼Ò½ÌÍø