ÌâÄ¿ÄÚÈÝ
ijÈÜÒºÖк¬ÓÐK+¡¢¢ÙÈ¡ÉÙÁ¿ÈÜÒºÖðµÎ¼ÓÈëÑÎËᣬÏÈÎÞÃ÷ÏÔÏÖÏ󣬵±ÑÎËá¼ÓÈëµ½Ò»¶¨Ìå»ýºó¿ªÊ¼²úÉú³Áµí²¢Öð½¥Ôö¶à£¬ÔÚ³ÁµíÁ¿²»ÔÙÔö¼ÓºóÓÖ²úÉúÒ»ÖÖÎÞÉ«ÆøÌ壬ÔÚÆøÌå²»ÔÙ²úÉúºó³Áµí¿ªÊ¼Èܽ⣬ֱÖÁ×îºóÍêÈ«Ïûʧ£»¢ÚÈ¡¢ÙËùµÃµÄÈÜÒº£¬¼ÓÈëBa(OH)2ÈÜÒº£¬ÎÞÃ÷ÏÔÏÖÏó¡£
£¨1£©ÔÈÜÒºÖп϶¨´æÔÚµÄÀë×ÓÊÇ________£»¿Ï¶¨²»´æÔÚµÄÀë×ÓÊÇ________£»
£¨2£©ÒÑÖªÓÃÒ»¶¨Ìå»ýµÄÔÈÜÒº½øÐÐʵÑé¢Ùʱ£¬ÐèÒªÏûºÄ0.2mol/LµÄÑÎËá5mL£¬ÕâʱÏòËùµÃÈÜÒºÖмÓÈë×ãÁ¿µÄÏõËáÒøÈÜÒº¿ÉµÃ³Áµí0.187g£¬ÔòÔÈÜÒºÖÐÊÇ·ñº¬ÓÐCl-£¿_____________¡£
´ð°¸£º
½âÎö£º
Ìáʾ£º
½âÎö£º
| £¨1£©¿Ï¶¨´æÔÚµÄÀë×ÓÊÇOH-¡¢ £¨2£©ÓÐCl-
|
Ìáʾ£º
| ¼ÓÈëÑÎËᣬÏÈÎÞÃ÷ÏÔÏÖÏó£¬ËµÃ÷ÈÜÒºÖк¬ÓÐOH-£¬ÔòÈÜÒºÖÐÒ»¶¨²»º¬ ÓÉÑÎËáÖеÄÂÈÀë×Ó·´Ó¦Éú³ÉµÄÂÈ»¯Òø³ÁµíµÄÖÊÁ¿ÊÇ
|
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ijÈÜÒºÖк¬ÓÐK+¡¢NH4+¡¢F-¡¢I-¡¢HSO3-£¬ÏòÆäÖмÓÈëÉÙÁ¿Cl2ºó£¬²»¿ÉÄܳöÏֵı仯ÊÇ£¨¡¡¡¡£©
| A¡¢I-ÊýÄ¿²»±ä | B¡¢F-ÊýÄ¿¼õÉÙ | C¡¢NH4+ÊýÄ¿Ôö´ó | D¡¢ÓÐSÎö³ö |