ÌâÄ¿ÄÚÈÝ

£¨2008?ÉòÑô£©½ñÄê5.12ãë´¨µØÕð·¢Éúºó£¬È«¹úÉÏÏÂÖÚÖ¾³É³Ç£¬¿¹Õð¾ÈÔÖ£®ÔÚÔÖºóµÄ·ÀÒß¹¤×÷ÖУ¬³£ÓöþÑõ»¯ÂÈ£¨ClO2£©×÷Ïû¶¾¼Á£®ClO2ÊÇÒ»ÖÖ³ÈÂÌÉ«¡¢Óд̼¤ÐÔÆøζµÄÆøÌ壬11¡æÒÔÏÂΪºìÉ«ÒºÌ壬Ò×ÈÜÓÚË®£¬¼û¹âÒ׷ֽ⣮¿ÉÓÃÂÈËáÄÆ£¨NaClO3£©ÓëŨÑÎËá·´Ó¦ÖÆÈ¡ClO2£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÈçÏ£º
2 NaClO3+4HCl=2 ClO2¡ü+Cl2¡ü+2X+2NaCl£®
£¨1£©¸ù¾ÝÒÔÉÏÐÅÏ¢£¬ÇëÄã×ܽá³öClO2µÄ»¯Ñ§ÐÔÖÊ
¼û¹âÒ×·Ö½â
¼û¹âÒ×·Ö½â
£»
£¨2£©ÍƶÏÉÏÊö·´Ó¦ÖÐXµÄ»¯Ñ§Ê½
H2O
H2O
£¬ÄãÍƶϵÄÒÀ¾ÝÊÇ
ÖÊÁ¿Êغ㶨ÂÉ
ÖÊÁ¿Êغ㶨ÂÉ
£®
·ÖÎö£º£¨1£©·ÖÎöÌâÒâ¿ÉÖª£¬ÌâÄ¿Öиø³öµÄÐÅÏ¢¼ÈÓжþÑõ»¯ÂȵÄÎïÀíÐÔÖÊ£¬ÓÖÓÐÆ仯ѧÐÔÖÊ£¬¸ù¾ÝÎïÀíÐÔÖʺͻ¯Ñ§ÐÔÖʸÅÄîµÄÇø±ð£¬ÕÒ³öÊôÓÚÆ仯ѧÐÔÖʵÄÄÚÈݼ´¿É£®
£¨2£©ÍƶÏXµÄ»¯Ñ§Ê½¿ÉÒÔ¸ù¾Ý·´Ó¦µÄ»¯Ñ§·½³ÌʽºÍÖÊÁ¿Êغ㶨ÂÉ£¨ÔÚ»¯Ñ§·´Ó¦Ç°ºóÔ­×ÓµÄÖÖÀàºÍ¸öÊý²»±ä£©½øÐмÆËã¼´¿É£®
½â´ð£º½â£º£¨1£©ÒÑÖªÌõ¼þÖиø³öµÄÐÅÏ¢¼ÈÓжþÑõ»¯ÂȵÄÎïÀíÐÔÖÊ£¬ÓÖÓÐÆ仯ѧÐÔÖÊ£®ÆäÖС°¼û¹âÒ׷ֽ⡱ÊôÓÚ»¯Ñ§ÐÔÖÊ£®¹Ê´ð°¸Îª£º¼û¹âÒ׷ֽ⣮
£¨2£©¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉ¿ÉÖª£¬ÔÚ»¯Ñ§·´Ó¦Ç°ºóÔ­×ÓµÄÖÖÀàºÍ¸öÊý²»±ä£¬¹Ê¸ù¾Ý»¯Ñ§·½³Ìʽ2NaClO3+4HCl=2ClO2¡ü+Cl2¡ü+2X+2NaCl¿ÉÖª£¬·´Ó¦ÎïÖеĸ÷Ô­×ÓÓëÉú³ÉÎïÖеĸ÷Ô­×ÓÖ®²îΪ£ºNa¡¢2-2=0£»H¡¢4-0=4£»Cl¡¢2+4-2-2-2=0£»O¡¢6-4=2£®¼´2¸öXÖк¬ÓÐ4¸öHÔ­×ÓºÍ2¸öOÔ­×Ó£¬¹ÊXµÄ»¯Ñ§Ê½Îª£ºH2O£®
¹Ê´ð°¸Îª£ºH2O£¬ÖÊÁ¿Êغ㶨ÂÉ£®
µãÆÀ£º±¾ÌâÖ÷Òª¿¼²ìѧÉúÕýÈ·Çø·ÖÎïÀíÐÔÖʺͻ¯Ñ§ÐÔÖÊ£¬¸ù¾Ý»¯Ñ§·½³ÌʽºÍÁ¿Êغ㶨ÂÉÈ·¶¨ÎïÖʵĻ¯Ñ§Ê½µÄÄÜÁ¦£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø