ÌâÄ¿ÄÚÈÝ


Áò»¯¼î·¨Êǹ¤ÒµÉÏÖÆ±¸Na2S2O3µÄ·½·¨Ö®Ò»£¬·´Ó¦Ô­ÀíΪ£º

        2Na2S+Na2CO3+4SO2==3Na2S2O3+CO2   £¨¸Ã·´Ó¦¡÷H>0£©

ijÑо¿Ð¡×éÔÚʵÑéÊÒÓÃÁò»¯¼î·¨ÖƱ¸Na2S2O3¡¤5H2OÁ÷³ÌÈçÏ¡£

£¨1£©ÎüÁò×°ÖÃÈçͼËùʾ¡£

¢Ù×°ÖÃBµÄ×÷ÓÃÊǼìÑé×°ÖÃAÖÐSO2µÄÎüÊÕЧÂÊ£¬BÖÐÊÔ¼ÁÊÇ           £¬±íÃ÷SO2ÎüÊÕЧÂʵ͵ÄʵÑéÏÖÏóÊÇBÖÐÈÜÒº            ¡£

¢ÚΪÁËʹSO2¾¡¿ÉÄÜÎüÊÕÍêÈ«£¬ÔÚ²»¸Ä±äAÖÐÈÜҺŨ¶È¡¢Ìå»ýµÄÌõ¼þÏ£¬³ýÁ˼°Ê±½Á°è·´Ó¦ÎïÍ⣬»¹¿É²ÉÈ¡µÄºÏÀí´ëÊ©ÊÇ                 ¡¢                ¡££¨Ð´³öÁ½Ìõ£©

£¨2£©¼ÙÉ豾ʵÑéËùÓõÄNa2CO3º¬ÉÙÁ¿NaCl¡¢NaOH£¬Éè¼ÆÊµÑé·½°¸½øÐмìÑé¡££¨ÊÒÎÂʱCaCO3±¥ºÍÈÜÒºµÄpH=10.2£©

ÏÞÑ¡ÊÔ¼Á¼°ÒÇÆ÷£ºÏ¡ÏõËá¡¢AgNO3ÈÜÒº¡¢CaCl2ÈÜÒº¡¢Ca(NO3)2ÈÜÒº¡¢·Ó̪ÈÜÒº¡¢ÕôÁóË®¡¢pH¼Æ¡¢ÉÕ±­¡¢ÊԹܡ¢µÎ¹Ü

ÐòºÅ

ʵÑé²Ù×÷

Ô¤ÆÚÏÖÏó

½áÂÛ

¢Ù

È¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®£¬³ä·ÖÕñµ´Èܽ⣬___________________¡£

_______________

ÑùÆ·º¬NaCl

¢Ú

ÁíÈ¡ÉÙÁ¿ÑùÆ·ÓÚÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®£¬³ä·Ö½Á°èÈܽ⣬___________________¡£

_______________

ÑùÆ·º¬NaOH

£¨3£©Na2S2O3ÈÜÒºÊǶ¨Á¿ÊµÑéÖеij£ÓÃÊÔ¼Á£¬²â¶¨ÆäŨ¶ÈµÄ¹ý³ÌÈçÏ£º×¼È·³ÆÈ¡a g KIO3£¨»¯Ñ§Ê½Á¿£º214£©¹ÌÌåÅä³ÉÈÜÒº£¬¼ÓÈë¹ýÁ¿KI¹ÌÌåºÍH2SO4ÈÜÒº£¬µÎ¼Óָʾ¼Á£¬ÓÃNa2S2O3ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄNa2S2O3ÈÜÒºµÄÌå»ýΪV mL¡£Ôòc(Na2S2O3)£½_________mol¡¤L£­1¡££¨Ö»ÁгöËãʽ£¬²»×÷ÔËË㣩

ÒÑÖª£ºCr2O72£­£«6I£­+14H+=== 2Cr3+£«3I2£«7H2O     2S2O32£­£«I2===S4O62£­£«2I£­


£¨14·Ö£©£¨1£©¢ÙÆ·ºì¡¢äåË®»òKMnO4ÈÜÒº £» ÈÜÒºÑÕÉ«ºÜ¿ìÍÊÉ«»òÖ¸³öÏàÓ¦ÊÔ¼ÁµÄÕýÈ·ÑÕÉ«£¨ºìÉ«¡¢»ÆÉ«µÈ£©ºÜ¿ìÍÊɫҲ¿ÉµÃ·Ö¡£  

¢ÚÔö´óSO2µÄ½Ó´¥Ãæ»ý¡¢¿ØÖÆSO2µÄÁ÷ËÙ¡¢Êʵ±Éý¸ßζȣ¨´ðÆäÖжþÌõ¼´¿É£©

£¨2£©

ÐòºÅ

ʵÑé²Ù×÷

Ô¤ÆÚÏÖÏó

¢Ù

µÎ¼Ó×ãÁ¿Ï¡ÏõËᣬÔٵμÓÉÙÁ¿AgNO3ÈÜÒº£¬Õñµ´¡£

Óа×É«³ÁµíÉú³É

¢Ú

¼ÓÈë¹ýÁ¿CaCl2ÈÜÒº£¬½Á°è£¬¾²Öã¬ÓÃpH¼Æ²â¶¨ÉϲãÇåÒºpH

Óа×É«³ÁµíÉú³É£¬ÉϲãÇåÒºpH´óÓÚ10.2

£¨3£©6000a/214V »ò3000a/107V   (ÆäËüÕýÈ·´ð°¸Ò²¸ø·Ö)     


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø