ÌâÄ¿ÄÚÈÝ
ÓÃÈçͼËùʾװÖýøÐÐʵÑ飬½«AÖðµÎ¼ÓÈëBÖУºa£®ÈôBΪNa2CO3·ÛÄ©£¬CΪC6H5ONaÈÜÒº£¬ÊµÑéÖй۲쵽СÊÔ¹ÜÄÚÈÜÒºÓɳÎÇå±ä»ë×Ç£¬ÔòÊÔ¹ÜCÖл¯Ñ§·´Ó¦µÄÀë×Ó·½³Ìʽ £®È»ºóÍùÉÕ±ÖмÓÈë·ÐË®£¬¿É¹Û²ìµ½ÊÔ¹ÜCÖеÄÏÖÏó £®
b£®ÈôBÊÇÉúʯ»Ò£¬¹Û²ìµ½CÈÜÒºÖÐÏÈÐγɳÁµí£¬È»ºó³ÁµíÈܽ⣮µ±³ÁµíÍêÈ«Èܽ⣬ǡºÃ±ä³ÎÇåʱ£¬¹Ø±ÕE£®È»ºóÍùСÊÔ¹ÜÖмÓÈë3µÎÒÒÈ©£¬ÔÙÍùÉÕ±ÖмÓÈëÈÈË®£¬¾²ÖÃƬ¿Ì£¬¹Û²ìµ½ÊԹܱڳöÏÖ¹âÁÁµÄÒø¾µ£¬ÔòAÊÇ £¨ÌîÃû³Æ£©£¬CÊÇ £¨Ìѧʽ£©£®ÓëÒÒÈ©ÈÜÒº»ìºÏºó£¬¸ÃÈÜÒºÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ £®ÒÇÆ÷DÔÚ´ËʵÑéÖеÄ×÷ÓÃÊÇ £®
¡¾´ð°¸¡¿·ÖÎö£ºa¡¢Ì¼ËáµÄËáÐԱȱ½·ÓÇ¿£¬±½·ÓÊDz»ÈÜÓÚË®µÄ£¬µ±Î¶ȵÍÓÚ16.6¡æʱ³ÉΪÎÞÉ«¾§Ì壬µ±Î¶ȸßÓÚ¸ÃζÈʱ£¬»á±äΪÎÞÉ«ÒºÌ壻
B¡¢È©ÄܺÍÒø°±ÈÜÒºÖ®¼ä·¢ÉúÒø¾µ·´Ó¦£¬ÏòÏõËáÒøÈÜÒºÖмӰ±Ë®»òͨ°±Æø£¬µ±³Áµí¸ÕºÃÏûʧʱ£¬¿ÉÒÔ»ñµÃÒø°±ÈÜÒº£®
½â´ð£º½â£ºa¡¢CΪC6H5ONaÈÜÒº£¬ÊµÑéÖй۲쵽СÊÔ¹ÜÄÚÈÜÒºÓɳÎÇå±ä»ë×Ç£¬ËµÃ÷Óб½·ÓÉú³É£¬BΪNa2CO3·ÛÄ©£¬ËùÒÔAÊÇÒ»ÖÖÇ¿ËᣬºÍ̼ËáÄÆ·´Ó¦·Å³ö¶þÑõ»¯Ì¼£¬Ì¼ËáµÄËáÐԱȱ½·ÓÇ¿£¬½«¶þÑõ»¯Ì¼Í¨Èëµ½±½·ÓÄÆÖлáÉú³É±½·Ó³Áµí£¬¼´C6H5O-+H2O+CO2=C6H5OH+HCO3-£¬µ±Î¶ȵÍÓÚ16.6¡æʱΪÎÞÉ«¾§Ì壬ÍùÉÕ±ÖмÓÈë·ÐË®£¬±½·Ó»á±äΪÎÞÉ«ÒºÌ壬
¹Ê´ð°¸Îª£ºC6H5O-+H2O+CO2=C6H5OH+HCO3-£»»ë×DZä³ÎÇ壻
B¡¢ÍùСÊÔ¹ÜÖмÓÈë3µÎÒÒÈ©£¬ÔÙÍùÉÕ±ÖмÓÈëÈÈË®£¬¾²ÖÃƬ¿Ì£¬¹Û²ìµ½ÊԹܱڳöÏÖ¹âÁÁµÄÒø¾µ£¬ÒòΪȩÄܺÍÒø°±ÈÜÒºÖ®¼ä·¢ÉúÒø¾µ·´Ó¦£¬ÏòÏõËáÒøÈÜÒºÖмӰ±Ë®»òͨ°±Æø£¬µ±³Áµí¸ÕºÃÏûʧʱ£¬¿ÉÒÔ»ñµÃÒø°±ÈÜÒº£¬ËùÒÔAÊÇ°±Ë®£¬CÊÇÏõËáÒø£¬ÒÒÈ©ºÍÒø°±ÈÜÒº¼äµÄÒø¾µ·´Ó¦Îª£ºCH3CHO+2[Ag£¨NH3£©2]OH CH3COONH4+2Ag¡ý+3NH3+H2O£¬
¹Ê´ð°¸Îª£ºCH3CHO+2[Ag£¨NH3£©2]OH CH3COONH4+2Ag¡ý+3NH3+H2O£®
µãÆÀ£º±¾ÌâÊÇÒ»µÀ¿ª·ÅʽµÄÌâÄ¿£¬ÒªÇóѧÉú¾ßÓзÖÎöºÍ½â¾öÎÊÌâµÄÄÜÁ¦£¬ÄѶȽϴó£®
B¡¢È©ÄܺÍÒø°±ÈÜÒºÖ®¼ä·¢ÉúÒø¾µ·´Ó¦£¬ÏòÏõËáÒøÈÜÒºÖмӰ±Ë®»òͨ°±Æø£¬µ±³Áµí¸ÕºÃÏûʧʱ£¬¿ÉÒÔ»ñµÃÒø°±ÈÜÒº£®
½â´ð£º½â£ºa¡¢CΪC6H5ONaÈÜÒº£¬ÊµÑéÖй۲쵽СÊÔ¹ÜÄÚÈÜÒºÓɳÎÇå±ä»ë×Ç£¬ËµÃ÷Óб½·ÓÉú³É£¬BΪNa2CO3·ÛÄ©£¬ËùÒÔAÊÇÒ»ÖÖÇ¿ËᣬºÍ̼ËáÄÆ·´Ó¦·Å³ö¶þÑõ»¯Ì¼£¬Ì¼ËáµÄËáÐԱȱ½·ÓÇ¿£¬½«¶þÑõ»¯Ì¼Í¨Èëµ½±½·ÓÄÆÖлáÉú³É±½·Ó³Áµí£¬¼´C6H5O-+H2O+CO2=C6H5OH+HCO3-£¬µ±Î¶ȵÍÓÚ16.6¡æʱΪÎÞÉ«¾§Ì壬ÍùÉÕ±ÖмÓÈë·ÐË®£¬±½·Ó»á±äΪÎÞÉ«ÒºÌ壬
¹Ê´ð°¸Îª£ºC6H5O-+H2O+CO2=C6H5OH+HCO3-£»»ë×DZä³ÎÇ壻
B¡¢ÍùСÊÔ¹ÜÖмÓÈë3µÎÒÒÈ©£¬ÔÙÍùÉÕ±ÖмÓÈëÈÈË®£¬¾²ÖÃƬ¿Ì£¬¹Û²ìµ½ÊԹܱڳöÏÖ¹âÁÁµÄÒø¾µ£¬ÒòΪȩÄܺÍÒø°±ÈÜÒºÖ®¼ä·¢ÉúÒø¾µ·´Ó¦£¬ÏòÏõËáÒøÈÜÒºÖмӰ±Ë®»òͨ°±Æø£¬µ±³Áµí¸ÕºÃÏûʧʱ£¬¿ÉÒÔ»ñµÃÒø°±ÈÜÒº£¬ËùÒÔAÊÇ°±Ë®£¬CÊÇÏõËáÒø£¬ÒÒÈ©ºÍÒø°±ÈÜÒº¼äµÄÒø¾µ·´Ó¦Îª£ºCH3CHO+2[Ag£¨NH3£©2]OH CH3COONH4+2Ag¡ý+3NH3+H2O£¬
¹Ê´ð°¸Îª£ºCH3CHO+2[Ag£¨NH3£©2]OH CH3COONH4+2Ag¡ý+3NH3+H2O£®
µãÆÀ£º±¾ÌâÊÇÒ»µÀ¿ª·ÅʽµÄÌâÄ¿£¬ÒªÇóѧÉú¾ßÓзÖÎöºÍ½â¾öÎÊÌâµÄÄÜÁ¦£¬ÄѶȽϴó£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ijÑо¿Ð¡×éÓÃÈçͼËùʾװÖýøÐÐÍÓëŨÁòËá·´Ó¦µÄʵÑéÑо¿£®
£¨1£©Ð´³öÊÔ¹ÜBÖеÄʵÑéÏÖÏó______£®
£¨2£©Ð´³öAÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ______
£¨3£©¼ÌÐøÏòAÊÔ¹ÜÖмÓÈëH2O2£¬·¢ÏÖÍƬÈܽ⣬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º______£®
ÈôÈÔ²»²¹³äŨÁòËᣬֻҪÇóʹÍƬÈܽ⣬Ҳ¿ÉÒÔ¼ÓÈ루ÌîдÁ½ÖÖÊôÓÚ²»Í¬Àà±ðÎïÖʵĻ¯Ñ§Ê½£©______¡¢______£®
£¨4£©BÊԹܿڵÄÃÞ»¨Ó¦Õ´ÓеÄÊÔ¼ÁÊÇ______£®
£¨5£©Ð¡×é³ÉÔ±Ïò·´Ó¦ºóµÄÈÜÒºÖмÓÈë×ãÁ¿µÄÑõ»¯Í£¬Ê¹Ê£ÓàµÄÁòËáÈ«²¿×ª»¯ÎªÁòËáÍ£¬¹ýÂ˺󣬽«ÂËÒº¼ÓÈÈŨËõ£¬ÀäÈ´½á¾§ÖƵÃÁòËá;§Ì壨CuSO4?xH2O£©£®Ð¡×é³ÉÔ±²ÉÓüÓÈÈ·¨²â¶¨¸Ã¾§ÌåÀï½á¾§Ë®xµÄÖµ£®
¢ÙÔÚËûÃǵÄʵÑé²Ù×÷ÖУ¬ÖÁÉÙ³ÆÁ¿______´Î£»
¢ÚÏÂÃæÊÇÆäÖÐÒ»´ÎʵÑéµÄÊý¾Ý£º
ÛáÛöÖÊÁ¿ | ÛáÛöÓ뾧ÌåµÄ×ÜÖÊÁ¿ | ¼ÓÈȺóÛáÛöÓë¹ÌÌå×ÜÖÊÁ¿ |
11.7g | 22.7g | 18.9g |
A£®ÁòËá;§ÌåÖк¬Óв»»Ó·¢µÄÔÓÖÊ¡¡¡¡¡¡¡¡B£®ÊµÑéÇ°¾§Ìå±íÃæÓÐʪ´æË®
C£®¼ÓÈÈʱÓо§Ìå·É½¦³öÈ¥¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡D£®¼ÓÈÈʧˮºó¶ÖÃÔÚ¿ÕÆøÖÐÀäÈ´£®