ÌâÄ¿ÄÚÈÝ
£¨6·Ö£©Ä³µª·ÊNH4HCO3ÖлìÓÐÉÙÁ¿(NH4)2CO3£¬ÏÖ²ÉÓÃÏÂÁз½°¸²â¶¨¸Ãµª·ÊÖÐ(NH4)2CO3µÄÖÊÁ¿·ÖÊý£º³ÆÈ¡5.7 gÉÏÊöÑùÆ·Óë2.0 mol/L NaOHÈÜÒº»ìºÏ£¬ÍêÈ«Èܽâºó£¬µÍμÓÈÈʹÆä³ä·Ö·´Ó¦(¸ÃζÈÏÂï§Ñβ»·Ö½â)£¬²¢Ê¹Éú³ÉµÄ°±ÆøÈ«²¿±»ÁòËáÎüÊÕ£¬²âµÃ°±ÆøµÄÖÊÁ¿ÓëËùÓÃNaOHÈÜÒºÌå»ýµÄ¹ØÏµÈçͼËùʾ£º
![]()
£¨1£©AµãǰÑùÆ·ÓëNaOH·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________
£¨2£©ÎªÊ¹Éú³ÉµÄ°±Æø±»ÁòËáÎüÊÕʱ²»·¢Éúµ¹Îü£¬¿ÉÒÔÑ¡ÓÃÏÂÁÐ×°ÖÃÖеÄ________¡£
![]()
£¨3£©ÑùÆ·ÖÐ(NH4)2CO3µÄÖÊÁ¿·ÖÊýÊÇ________%(±£ÁôһλСÊý)¡£
£¨1£©HCO3¡ª£«OH£===CO32¡ª£«H2O £¨2£©ABD £¨3£©16.8
¡¾½âÎö¡¿
ÊÔÌâ·ÖÎö£ºÓÉͼÏó¿ÉÖª£¬Aµãǰ²¢ÎªÉú³ÉNH3£¬Ôò·¢Éú·´Ó¦ÎªHCO3¡ªÓëOH£Ö®¼äµÄ·´Ó¦£¬Àë×Ó·½³ÌʽΪ£ºHCO3¡ª£«OH£=== CO32¡ª£«H2O£¬ÎªÊ¹Éú³ÉµÄ°±Æø±»ÁòËáÎüÊÕʱ²»·¢Éúµ¹Îü£¬ÔòÎüÊÕ×°ÖÃΪÄܷŵ¹ÎüµÄ×°Öã¬Ñ¡ABD¼ÓÈë2.0 mol/L NaOHÈÜÒº70mlʱ£¬ÈÜÒºÈÜÖÊΪNa2CO3£¬Éè»ìºÏÎïÖÐNH4HCO3 xmol£¬(NH4)2CO3 ymol£¬¿ÉÁеÈʽ¹ØÏµ£º![]()
½â¿ÉµÃ£ºx=0.06mol y=0.01mol£¬¿ÉµÃÑùÆ·ÖÐ(NH4)2CO3µÄÖÊÁ¿·ÖÊý=![]()
=16.8%.
¿¼µã£ºÍ¼Ïñ¼ÆËãÎÊÌâ¡£
£¨12·Ö£©¸ßÃÌËá¼ØÊÇÒ»ÖÖµäÐ͵ÄÇ¿Ñõ»¯¼Á£¬ÎÞÂÛÔÚʵÑéÊÒ»¹ÊÇÔÚ»¯¹¤Éú²úÖж¼ÓÐÖØÒªµÄÓ¦Óá£ÈçͼK10?3ÊÇʵÑéÊÒÖÆ±¸ÂÈÆø²¢½øÐÐһϵÁÐÏà¹ØÊµÑéµÄ×°ÖÃ(¼Ð³ÖÉ豸ÒÑÂÔ)¡£
![]()
ͼK10?3
£¨1£©ÖƱ¸ÂÈÆøÑ¡ÓõÄҩƷΪ¸ßÃÌËá¼ØºÍŨÑÎËᣬÏàÓ¦µÄÀë×Ó·½³ÌʽΪ___________________________¡£
£¨2£©×°ÖÃBµÄ×÷ÓÃÊÇ________________________£¬ÊµÑé½øÐÐʱCÖпÉÄÜ·¢Éú¶ÂÈû£¬Çëд³ö·¢Éú¶ÂÈûʱBÖеÄÏÖÏó£º______________________________________¡£
£¨3£©×°ÖÃCµÄʵÑéÄ¿µÄÊÇÑéÖ¤ÂÈÆøÊÇ·ñ¾ßÓÐÆ¯°×ÐÔ£¬Îª´ËCÖТñ¡¢¢ò¡¢¢óÒÀ´Î·ÅÈë____(Ñ¡¡°a¡±¡°b¡±»ò¡°c¡±)¡£
a | b | c | |
I | ¸ÉÔïµÄÓÐÉ«²¼Ìõ | ʪÈóµÄÓÐÉ«²¼Ìõ | ʪÈóµÄÓÐÉ«²¼Ìõ |
¢ò | ¼îʯ»Ò | ŨÁòËá | ÎÞË®ÂÈ»¯¸Æ |
¢ó | ʪÈóµÄÓÐÉ«²¼Ìõ | ¸ÉÔïµÄÓÐÉ«²¼Ìõ | ¸ÉÔïµÄÓÐÉ«²¼Ìõ |
£¨4£©Éè¼Æ×°ÖÃD¡¢EµÄÄ¿µÄÊDZȽÏÂÈ¡¢äå¡¢µâµÄ·Ç½ðÊôÐÔ¡£µ±ÏòDÖлº»ºÍ¨Èë×ãÁ¿ÂÈÆøÊ±£¬¿ÉÒÔ¿´µ½ÎÞÉ«ÈÜÒºÖð½¥±äΪºìרɫ£¬ËµÃ÷ÂȵķǽðÊôÐÔ´óÓÚäå¡£´ò¿ª»îÈû£¬½«DÖеÄÉÙÁ¿ÈÜÒº¼ÓÈëEÖУ¬Õñµ´E£¬¹Û²ìµ½µÄÏÖÏóÊÇ________________________________¡£¸ÃÏÖÏó________(Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)˵Ã÷äåµÄ·Ç½ðÊôÐÔÇ¿Óڵ⡣
ÏÂÁÐÓйػ¯Ñ§ÓÃÓï±íʾ²»ÕýÈ·µÄÊÇ
| A£®ÑõµÄÔ×ӽṹʾÒâͼ£º |
| B£®Na2O2µÄµç×Óʽ£º |
| C£®HClOµÄ½á¹¹Ê½£ºH¡ªO¡ªCl |
| D£®ÖÐ×ÓÊýΪ16µÄÁòÀë×Ó£º |