ÌâÄ¿ÄÚÈÝ
ÓûÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÇâÑõ»¯ÄÆÈÜÒº250mL£¬Íê³ÉÏÂÁв½Ö裺
¢ÙÓÃÍÐÅÌÌìÆ½³ÆÁ¿NaOH¹ÌÌåµÄÖÊÁ¿£¬Èô×óÅÌÉÏ·ÅÓÐ8gíÀÂ룬ÓÎÂëλÖÃÈçͼ£¬ÌìÆ½µÄÖ¸ÕëÔÚ±ê³ßÖм䣬ÔòËù³ÆNaOH¹ÌÌåµÄʵ¼ÊÖÊÁ¿Îª ¿Ë£®

¢Ú½«³ÆºÃµÄÇâÑõ»¯ÄƹÌÌå·ÅÈë ÖУ¬¼ÓÕôÁóË®½«ÆäÈ«²¿Èܽ⣬´ý ºó£¬½«ÈÜ񼄯 ÒÆÈë mLµÄÈÝÁ¿Æ¿ÖУ®
¢ÛÓÃÕôÁóˮϴµÓÈܽâËùÓÃÒÇÆ÷ ´Î£¬½«Ï´µÓÒºÒÆÈë ÖУ¬ÔÚ²Ù×÷¹ý³ÌÖв»ÄÜËðʧµãµÎÒºÌ壬·ñÔò»áʹÈÜÒºµÄŨ¶ÈÆ« £¨¸ß»òµÍ£©£®
¢ÜÏòÈÝÁ¿Æ¿ÄÚ¼ÓË®ÖÁ½Ó½ü¿Ì¶ÈÏß1-2cm´¦Ê±£¬¸ÄÓà СÐĵؼÓË®ÖÁÈÜÒº°¼ÒºÃæÓë¿Ì¶ÈÏßÏàÇУ¬Èô¼ÓË®³¬¹ý¿Ì¶ÈÏߣ¬»áÔì³ÉÈÜҺŨ¶ÈÆ« £¬Ó¦¸Ã £®
¢Ý×îºó¸ÇºÃÆ¿¸Ç£¬ £¬½«ÅäºÃµÄÈÜÒºÒÆÈë Öв¢ÌùºÃ±êÇ©£®
¢ÙÓÃÍÐÅÌÌìÆ½³ÆÁ¿NaOH¹ÌÌåµÄÖÊÁ¿£¬Èô×óÅÌÉÏ·ÅÓÐ8gíÀÂ룬ÓÎÂëλÖÃÈçͼ£¬ÌìÆ½µÄÖ¸ÕëÔÚ±ê³ßÖм䣬ÔòËù³ÆNaOH¹ÌÌåµÄʵ¼ÊÖÊÁ¿Îª
¢Ú½«³ÆºÃµÄÇâÑõ»¯ÄƹÌÌå·ÅÈë
¢ÛÓÃÕôÁóˮϴµÓÈܽâËùÓÃÒÇÆ÷
¢ÜÏòÈÝÁ¿Æ¿ÄÚ¼ÓË®ÖÁ½Ó½ü¿Ì¶ÈÏß1-2cm´¦Ê±£¬¸ÄÓÃ
¢Ý×îºó¸ÇºÃÆ¿¸Ç£¬
¿¼µã£ºÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº
רÌ⣺ʵÑéÌâ
·ÖÎö£º¢Ù¸ù¾ÝÍÐÅÌÌìÆ½µÄ³ÆÁ¿ÔÀí½øÐнâ´ð£»
¢Ú³ÆÁ¿ÇâÑõ»¯ÄÆÊ±£¬½«ÇâÑõ»¯ÄÆ·ÅÔÚÉÕ±ÖгÆÁ¿£¬ÇâÑõ»¯ÄÆÈܽâ¹ý³Ì·ÅÈÈ£¬±ØÐë´ýÈܽâµÄÈÜÒºÀäÈ´ºóÑØ×Ų£Á§°ô×ªÒÆµ½250mLÈÝÁ¿Æ¿ÖУ»
¢ÛÅäÖÆÒ»¶¨Å¨¶ÈµÄÈÜҺʱ£¬±ØÐëÏ´µÓÉÕ±ºÍ²£Á§°ô2-3´Î£¬½«Ï´µÓÒº×ªÒÆµ½ÈÝÁ¿Æ¿ÖУ»ÈôûÓÐÏ´µÓÉÕ±ºÍ²£Á§°ô£¬»áµ¼ÖÂÅäÖÆµÄÈÜҺŨ¶ÈÆ«µÍ£»
¢Ü¶¨ÈÝʱ×îºóÐèҪʹÓýºÍ·µÎ¹Ü½øÐж¨ÈÝ£»Èô¼ÓË®³¬¹ý¿Ì¶ÈÏߣ¬µ¼ÖÂÅäÖÆµÄÈÜҺŨ¶ÈÆ«µÍ£¬´Ë´ÎÅäÖÆÊ§°Ü£¬ÐèÒªÖØÐÂÅäÖÆ£»
¢Ý¶¨ÈݽáÊøºóÐèÒª½øÐÐÒ¡ÔȲÙ×÷£¬×îºó½«ÅäÖÆµÄÈÜÒº×ªÒÆµ½ÊÔ¼ÁÆ¿ÖУ®
¢Ú³ÆÁ¿ÇâÑõ»¯ÄÆÊ±£¬½«ÇâÑõ»¯ÄÆ·ÅÔÚÉÕ±ÖгÆÁ¿£¬ÇâÑõ»¯ÄÆÈܽâ¹ý³Ì·ÅÈÈ£¬±ØÐë´ýÈܽâµÄÈÜÒºÀäÈ´ºóÑØ×Ų£Á§°ô×ªÒÆµ½250mLÈÝÁ¿Æ¿ÖУ»
¢ÛÅäÖÆÒ»¶¨Å¨¶ÈµÄÈÜҺʱ£¬±ØÐëÏ´µÓÉÕ±ºÍ²£Á§°ô2-3´Î£¬½«Ï´µÓÒº×ªÒÆµ½ÈÝÁ¿Æ¿ÖУ»ÈôûÓÐÏ´µÓÉÕ±ºÍ²£Á§°ô£¬»áµ¼ÖÂÅäÖÆµÄÈÜҺŨ¶ÈÆ«µÍ£»
¢Ü¶¨ÈÝʱ×îºóÐèҪʹÓýºÍ·µÎ¹Ü½øÐж¨ÈÝ£»Èô¼ÓË®³¬¹ý¿Ì¶ÈÏߣ¬µ¼ÖÂÅäÖÆµÄÈÜҺŨ¶ÈÆ«µÍ£¬´Ë´ÎÅäÖÆÊ§°Ü£¬ÐèÒªÖØÐÂÅäÖÆ£»
¢Ý¶¨ÈݽáÊøºóÐèÒª½øÐÐÒ¡ÔȲÙ×÷£¬×îºó½«ÅäÖÆµÄÈÜÒº×ªÒÆµ½ÊÔ¼ÁÆ¿ÖУ®
½â´ð£º
½â£º¢ÙÍÐÅÌÌìÆ½µÄ³ÆÁ¿ÔÀíΪ£ºÒ©Æ·ÖÊÁ¿+3.2g=8g£¬ÔòÇâÑõ»¯ÄƵÄÖÊÁ¿Îª4.8g£¬
¹Ê´ð°¸Îª£º4.8£»
¢ÚÇâÑõ»¯ÄƾßÓи¯Ê´ÐÔ£¬Ó¦¸Ã·ÅÔÚÉÕ±ÖгÆÁ¿£»ÇâÑõ»¯ÄƹÌÌåÈܽâʱ·Å³öÈÈÁ¿£¬±ØÐëÀäÈ´ºóÔÙÑØ×Ų£Á§°ô×ªÒÆµ½250mLÈÝÁ¿Æ¿ÖУ¬
¹Ê´ð°¸Îª£ºÉÕ±£»ÀäÈ´£»²£Á§°ô£»250£»
¢ÛÅäÖÆ¹ý³ÌÖÐÐèҪϴµÓÉÕ±ºÍ²£Á§°ô2¡«3´Î£¬½«Ï´µÓÒº×ªÒÆµ½ÈÝÁ¿Æ¿ÖУ¬·ñÔòÅäÖÆµÄÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿Æ«Ð¡£¬ÈÜҺŨ¶ÈÆ«µÍ£¬
¹Ê´ð°¸Îª£º2¡«3£»ÈÝÁ¿Æ¿£»µÍ£»
¢ÜÏòÈÝÁ¿Æ¿ÄÚ¼ÓË®ÖÁ½Ó½ü¿Ì¶ÈÏß1-2cm´¦Ê±£¬¸ÄÓýºÍ·µÎ¹Ü¶¨ÈÝ£¬Èô¼ÓË®³¬¹ý¿Ì¶ÈÏߣ¬µ¼ÖÂÈÜÒºÌå»ýÆ«´ó£¬»áÔì³ÉÈÜҺŨ¶ÈÆ«µÍ£¬ÐèÒªÖØÐÂÅäÖÆ£¬
¹Ê´ð°¸Îª£º½ºÍ·µÎ¹Ü£»µÍ£»ÖØÐÂÅäÖÆ£»
¢Ý¶¨ÈݺóÐèÒª½«ÈÝÁ¿Æ¿ÉÏÏ·ת·´¸´Ò¡ÔÈ£¬×îºó½«ÅäÖÆµÄÈÜÒº×ªÒÆµ½ÈÝÁ¿Æ¿ÖУ¬
¹Ê´ð°¸Îª£º½«ÈÝÁ¿Æ¿ÉÏÏ·ת·´¸´Ò¡ÔÈ£»ÊÔ¼ÁÆ¿£®
¹Ê´ð°¸Îª£º4.8£»
¢ÚÇâÑõ»¯ÄƾßÓи¯Ê´ÐÔ£¬Ó¦¸Ã·ÅÔÚÉÕ±ÖгÆÁ¿£»ÇâÑõ»¯ÄƹÌÌåÈܽâʱ·Å³öÈÈÁ¿£¬±ØÐëÀäÈ´ºóÔÙÑØ×Ų£Á§°ô×ªÒÆµ½250mLÈÝÁ¿Æ¿ÖУ¬
¹Ê´ð°¸Îª£ºÉÕ±£»ÀäÈ´£»²£Á§°ô£»250£»
¢ÛÅäÖÆ¹ý³ÌÖÐÐèҪϴµÓÉÕ±ºÍ²£Á§°ô2¡«3´Î£¬½«Ï´µÓÒº×ªÒÆµ½ÈÝÁ¿Æ¿ÖУ¬·ñÔòÅäÖÆµÄÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿Æ«Ð¡£¬ÈÜҺŨ¶ÈÆ«µÍ£¬
¹Ê´ð°¸Îª£º2¡«3£»ÈÝÁ¿Æ¿£»µÍ£»
¢ÜÏòÈÝÁ¿Æ¿ÄÚ¼ÓË®ÖÁ½Ó½ü¿Ì¶ÈÏß1-2cm´¦Ê±£¬¸ÄÓýºÍ·µÎ¹Ü¶¨ÈÝ£¬Èô¼ÓË®³¬¹ý¿Ì¶ÈÏߣ¬µ¼ÖÂÈÜÒºÌå»ýÆ«´ó£¬»áÔì³ÉÈÜҺŨ¶ÈÆ«µÍ£¬ÐèÒªÖØÐÂÅäÖÆ£¬
¹Ê´ð°¸Îª£º½ºÍ·µÎ¹Ü£»µÍ£»ÖØÐÂÅäÖÆ£»
¢Ý¶¨ÈݺóÐèÒª½«ÈÝÁ¿Æ¿ÉÏÏ·ת·´¸´Ò¡ÔÈ£¬×îºó½«ÅäÖÆµÄÈÜÒº×ªÒÆµ½ÈÝÁ¿Æ¿ÖУ¬
¹Ê´ð°¸Îª£º½«ÈÝÁ¿Æ¿ÉÏÏ·ת·´¸´Ò¡ÔÈ£»ÊÔ¼ÁÆ¿£®
µãÆÀ£º±¾Ì⿼²éÁËÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº£¬ÌâÄ¿ÄѶÈÖеȣ¬×¢ÒâÕÆÎÕÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº·½·¨£¬ÊÔÌâÅàÑøÁËѧÉúÁé»îÓ¦ÓÃËùѧ֪ʶµÄÄÜÁ¦£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÏÂÁÐÀë×Ó·´Ó¦·½³ÌʽÕýÈ·µÄÊÇ£¨¡¡¡¡£©
| A¡¢ÇâÑõ»¯±µÈÜÒºÖеμÓÉÙÁ¿Ï¡ÁòË᣺Ba2++2OH-+2H++SO42-=BaSO4¡ý+2H2O |
| B¡¢¸¯Ê´·¨ÖÆ×÷Ó¡Ë¢Ïß·°å£ºFe3++Cu=Fe2++Cu2+ |
| C¡¢´×Ëá³ýË®¹¸£ºCO32-+2H+=Ca2++H2O+CO2¡ü |
| D¡¢AlCl3ÈÜÒº³ÊËáÐÔµÄÔÒò£ºAl3++3H2O=Al£¨OH£©3+3H+ |
0.02molµÄÂÁµ¥ÖÊÔÚÑõÆøÖÐÍêȫȼÉÕ£¬È¼ÉյIJúÎïÓë2.00mol/LµÄÑÎËá·´Ó¦£®ÈôÒª·´Ó¦ÍêÈ«£¬Ôò×îÉÙÐèÒª¶àÉÙºÁÉýµÄÑÎËᣨ¡¡¡¡£©
| A¡¢15cm3 |
| B¡¢20cm3 |
| C¡¢30cm3 |
| D¡¢60cm3 |