(18·Ö)¼×¡¢ÒÒÁ½³Øµç¼«²ÄÁ϶¼ÊÇÌú°ôÓë̼°ô£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÈôÁ½³ØÖоùΪCuSO4ÈÜÒº£¬·´Ó¦Ò»¶Îʱ¼äºó£º
¢ÙÓкìÉ«ÎïÖÊÎö³öµÄÊǼ׳ØÖеÄ________°ô£¬ÒÒ³ØÖеÄ________°ô¡£
¢ÚÒÒ³ØÖÐÑô¼«µÄµç¼«·´Ó¦Ê½ÊÇ_______________________________________________ _
(2)ÈôÁ½³ØÖоùΪ±¥ºÍNaClÈÜÒº£º
¢ÙÒÒ³ØÖÐ̼¼«Éϵ缫·´Ó¦ÊôÓÚ____________(Ìî¡°Ñõ»¯·´Ó¦¡±»ò¡°»¹Ô·´Ó¦¡±)£¬Ð´³öÒÒ³ØÖиõ缫·´Ó¦µÄÀë×Ó·½³Ìʽ_______________________________________¡£
¢Ú¼×³ØÖÐ̼¼«Éϵ缫·´Ó¦Ê½ÊÇ_____________£¬¼ìÑé¸Ã¸º¼«·´Ó¦²úÎïµÄ·½·¨ÊÇ_____________________________ÓÃÀë×Ó·½³Ìʽ±íʾ_______________________________
¢ÛÈôÒÒ³ØתÒÆ0.02 mol e£ºóֹͣʵÑ飬³ØÖÐÈÜÒºÌå»ýÊÇ200 mL£¬ÔòÈÜÒº»ìÔȺóµÄpH£½________¡£
(1)¢Ù̼¡¡Ìú
¢Ú4OH££4e£===2H2O£«O2¡ü
(2)¢Ù Ñõ»¯·´Ó¦ 2Cl££2e£=Cl2¡ü£
¢Ú2H2O£«O2£«4e£=4OH£¡¡µÎ¼Ó2µÎÌúÇ軯¼ØÈÜÒº£¬¿´µ½ÓÐÀ¶É«³Áµí³öÏÖ£¬ËµÃ÷Éú³ÉÁËFe2£« 3Fe2£« +2[Fe(CN)6]3- = Fe3[Fe(CN)6]2£¨À¶É«£©£¬Éú³ÉÎï½ÐÆÕ³ʿÀ¶
¢Û13
½âÎö(1)¢Ù̼£¨Õý¼«£©¡¡Ìú£¨Òõ¼«£©
¢ÚÑô¼«ÎªÌ¼°ô£¬Ê§µç×Ó£¬4OH££4e£===2H2O£«O2¡ü
(2)¢Ù Ñõ»¯·´Ó¦ 2Cl££2e£=Cl2¡ü£
¢Ú2H2O£«O2£«4e£=4OH£¡¡µÎ¼Ó2µÎÌúÇ軯¼ØÈÜÒº£¬¿´µ½ÓÐÀ¶É«³Áµí³öÏÖ£¬ËµÃ÷Éú³ÉÁËFe2£« 3Fe2£« +2[Fe(CN)6]3- = Fe3[Fe(CN)6]2£¨À¶É«£©£¬Éú³ÉÎï½ÐÆÕ³ʿÀ¶
¢ÛH£« Ũ¶ÈΪ£º0.02mol/0.200L=0.1mol/L,pH= 13
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(18·Ö)¼×¡¢ÒÒÁ½³Øµç¼«²ÄÁ϶¼ÊÇÌú°ôÓë̼°ô£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÈôÁ½³ØÖоùΪCuSO4ÈÜÒº£¬·´Ó¦Ò»¶Îʱ¼äºó£º
¢ÙÓкìÉ«ÎïÖÊÎö³öµÄÊǼ׳ØÖеÄ________°ô£¬ÒÒ³ØÖеÄ________°ô¡£
¢ÚÒÒ³ØÖÐÑô¼«µÄµç¼«·´Ó¦Ê½ÊÇ________________________________________________
(2)ÈôÁ½³ØÖоùΪ±¥ºÍNaClÈÜÒº£º
¢ÙÒÒ³ØÖÐ̼¼«Éϵ缫·´Ó¦ÊôÓÚ____________(Ìî¡°Ñõ»¯·´Ó¦¡±»ò¡°»¹Ô·´Ó¦¡±)£¬Ð´³öÒÒ³ØÖиõ缫·´Ó¦µÄÀë×Ó·½³Ìʽ_______________________________________¡£
¢Ú¼×³ØÖÐ̼¼«Éϵ缫·´Ó¦Ê½ÊÇ_____________£¬¼ìÑé¸Ã¸º¼«·´Ó¦²úÎïµÄ·½·¨ÊÇ_____________________________ÓÃÀë×Ó·½³Ìʽ±íʾ_______________________________
¢ÛÈôÒÒ³ØתÒÆ0.02 mole£ºóֹͣʵÑ飬³ØÖÐÈÜÒºÌå»ýÊÇ200 mL£¬ÔòÈÜÒº»ìÔȺóµÄpH£½________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìºÓÄÏÊ¡ÙÈʦÊи߶þµÚ¶þ´ÎÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
(18·Ö)¼×¡¢ÒÒÁ½³Øµç¼«²ÄÁ϶¼ÊÇÌú°ôÓë̼°ô£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÈôÁ½³ØÖоùΪCuSO4ÈÜÒº£¬·´Ó¦Ò»¶Îʱ¼äºó£º
¢ÙÓкìÉ«ÎïÖÊÎö³öµÄÊǼ׳ØÖеÄ________°ô£¬ÒÒ³ØÖеÄ________°ô¡£
¢ÚÒÒ³ØÖÐÑô¼«µÄµç¼«·´Ó¦Ê½ÊÇ_______________________________________________ _
(2)ÈôÁ½³ØÖоùΪ±¥ºÍNaClÈÜÒº£º
¢ÙÒÒ³ØÖÐ̼¼«Éϵ缫·´Ó¦ÊôÓÚ____________(Ìî¡°Ñõ»¯·´Ó¦¡±»ò¡°»¹Ô·´Ó¦¡±)£¬Ð´³öÒÒ³ØÖиõ缫·´Ó¦µÄÀë×Ó·½³Ìʽ_______________________________________¡£
¢Ú¼×³ØÖÐ̼¼«Éϵ缫·´Ó¦Ê½ÊÇ_____________£¬¼ìÑé¸Ã¸º¼«·´Ó¦²úÎïµÄ·½·¨ÊÇ_____________________________ÓÃÀë×Ó·½³Ìʽ±íʾ_______________________________
¢ÛÈôÒÒ³ØתÒÆ0.02 mol e£ºóֹͣʵÑ飬³ØÖÐÈÜÒºÌå»ýÊÇ200 mL£¬ÔòÈÜÒº»ìÔȺóµÄpH£½________¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com