| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â
| ÇâÑõ»¯Î↑ʼ³ÁµíʱµÄpH | ÇâÑõ»¯Îï³ÁµíÍêȫʱµÄpH | |
| Fe3+ | 1.9 | 3.2 |
| Fe2+ | 7.0 | 9.0 |
| Cu2+ | 4.7 | 6.7 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêÒø´¨¶þÖи߿¼Ô¤²â£¨×ÛºÏÌ⣩»¯Ñ§¾í ÌâÐÍ£ºÊµÑéÌâ
ijÁ¶Ìú·ÏÔüÖк¬ÓдóÁ¿µÄCuS¼°ÉÙÁ¿ÌúµÄÑõ»¯Î¹¤ÒµÉÏÒԸ÷ÏÔüºÍNaClΪÔÁÏÉú²úCuCl2¡¤2H2O¾§Ì壬Æä¹¤ÒÕÁ÷³ÌÖÐÖ÷񻃾¼°±ºÉÕ¡¢Î²Æø´¦Àí¡¢Ëá½þ¡¢µ÷¿ØÈÜÒºpH¡¢¹ýÂË¡¢Õô·¢½á¾§µÈ¡£±ºÉÕ¹ý³ÌÖз¢ÉúµÄÖ÷Òª·´Ó¦Îª£º CuS+2NaCl+2O2 = CuCl2+Na2SO4
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)±ºÉÕʱ»¹ÓÐ·ÏÆø²úÉú£¬ÆäÖÐÒ»ÖÖÊÇÖ÷ÒªµÄ´óÆøÎÛȾÎï¡£ÈôÔÚʵÑéÊÒÖÐÒÔ¼îÒºÎüÊÕ´¦ÀíÖ®£¬ÏÂÁÐA¡¢B¡¢C×°ÖÃÖпÉÐеÄÊÇ________(Ìî×Öĸ)£»ÈôÑ¡Óü××°Öã¬ÔòÉÕ±ÖеÄϲãÒºÌå¿ÉÒÔÊÇ_______¡£![]()
(2)µ÷¿ØÈÜÒºpHʱ²ÉÓÃpHÊÔÖ½À´²â¶¨ÈÜÒºµÄpH£¬ÔòÕýÈ·µÄ²Ù×÷·½·¨ÊÇ___________________ ¡£
(3)Èô×îºóËùµÃµÄÂËÒºÖеÄÒõÀë×ÓÖ»ÓÐS042-ºÍCl-£¬Ôò¼ìÑéÂËÒºÖÐÕâÁ½ÖÖÀë×ÓµÄʵÑé²Ù×÷______ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ì¸£½¨Ê¡Ü¼³ÇÈýУ¸ßÈýÏÂѧÆÚµÚ¶þ´Î¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ
£¨12·Ö£©
ijÁ¶Ìú·ÏÔüÖк¬ÓдóÁ¿CuS¼°ÉÙÁ¿ÌúµÄ»¯ºÏÎÌúÔªËØµÄ»¯ºÏ¼ÛΪ+2ºÍ+3¼Û£©£¬¹¤ÒµÉÏÒԸ÷ÏÔüΪÔÁÏÉú²úCuCl2µÄ¹¤ÒÕÁ÷³ÌÈçÏ£º![]()
±ºÉÕ¹ý³ÌÖз¢ÉúµÄÖ÷Òª·´Ó¦Îª£ºCuS+2NaCl+2O2¸ßÎÂCuCl2+Na2SO4¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©±ºÉÕǰÏȽøÐзÛËéµÄÀíÓÉÊÇ_____________________________________________¡£
£¨2£©ÊÔ¼ÁAӦѡÓá¡¡¡ ¡¡£¨Ìî±àºÅ£©¢ÙNaClO¡¢¢ÚCl2¡¢¢ÛH2O2ÈÜÒº¡¢¢ÜŨÁòËá
¼ÓÈëÊÔ¼ÁAµÄÄ¿µÄÊÇ____________________________________________________________¡£
£¨3£©ÉÏÊö¹ýÂ˲Ù×÷ÖÐÐèÒªÓõ½µÄ²£Á§ÒÇÆ÷ÓÐ_____________________________________________¡£
£¨4£©´ÓCuCl2ÈÜÒºµÃµ½CuCl2¾§ÌåµÄ²Ù×÷ÊÇ £¨Ð´³ö²Ù×÷Ãû³Æ£©£¬¸Ã²Ù×÷±ØÐèÔÚHClÆøÁ÷ÖнøÐУ¬ÔÒòÊÇ_____________________________________________£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
£¨5£©ÈçºÎ¼ìÑéÂËÒºBÖÐÊÇ·ñ»¹ÓÐδ³ÁµíµÄFe3+______________________________£¬
¼ÓÊÔ¼ÁAµ÷½ÚÈÜÒºpH=4ʱ£¬c(Fe3+)Ϊ_______________¡££¨ÒÑÖªKsp[Fe(OH)3]=2.6¡Á10£39£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄ긣½¨Ê¡Ü¼³ÇÈýУ¸ßÈýÏÂѧÆÚµÚ¶þ´Î¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ
£¨12·Ö£©
ijÁ¶Ìú·ÏÔüÖк¬ÓдóÁ¿CuS¼°ÉÙÁ¿ÌúµÄ»¯ºÏÎÌúÔªËØµÄ»¯ºÏ¼ÛΪ+2ºÍ+3¼Û£©£¬¹¤ÒµÉÏÒԸ÷ÏÔüΪÔÁÏÉú²úCuCl2µÄ¹¤ÒÕÁ÷³ÌÈçÏ£º
![]()
±ºÉÕ¹ý³ÌÖз¢ÉúµÄÖ÷Òª·´Ó¦Îª£ºCuS+2NaCl+2O2¸ßÎÂCuCl2+Na2SO4¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©±ºÉÕǰÏȽøÐзÛËéµÄÀíÓÉÊÇ_____________________________________________¡£
£¨2£©ÊÔ¼ÁAӦѡÓá¡¡¡ ¡¡£¨Ìî±àºÅ£©¢ÙNaClO¡¢¢ÚCl2¡¢¢ÛH2O2ÈÜÒº¡¢¢ÜŨÁòËá
¼ÓÈëÊÔ¼ÁAµÄÄ¿µÄÊÇ____________________________________________________________¡£
£¨3£©ÉÏÊö¹ýÂ˲Ù×÷ÖÐÐèÒªÓõ½µÄ²£Á§ÒÇÆ÷ÓÐ_____________________________________________¡£
£¨4£©´ÓCuCl2ÈÜÒºµÃµ½CuCl2¾§ÌåµÄ²Ù×÷ÊÇ £¨Ð´³ö²Ù×÷Ãû³Æ£©£¬¸Ã²Ù×÷±ØÐèÔÚHClÆøÁ÷ÖнøÐУ¬ÔÒòÊÇ_____________________________________________£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
£¨5£©ÈçºÎ¼ìÑéÂËÒºBÖÐÊÇ·ñ»¹ÓÐδ³ÁµíµÄFe3+______________________________£¬
¼ÓÊÔ¼ÁAµ÷½ÚÈÜÒºpH=4ʱ£¬c(Fe3+)Ϊ_______________¡££¨ÒÑÖªKsp[Fe(OH)3]=2.6¡Á10£39£©
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - 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