ÌâÄ¿ÄÚÈÝ
16£®ÓÐA¡¢B¡¢GÈýÖÖÓлúÎËüÃÇÔÚÒ»¶¨Ìõ¼þÏÂÓëNaOHÈÜÒº·´Ó¦£®ÆäÉú³ÉÎﻹÄÜ·¢ÉúÈçͼËùʾµÄ±ä»¯£¨²¿·Ö²úÎïÂÔÈ¥£©ÒÑÖª£º
¢ÙEÓë¼îʯ»Ò¹²ÈÈ£¬¿ÉÖƵÃÃܶÈΪ0.717 g/LµÄÆø̬Ìþ£¨±ê׼״̬ϲⶨµÄÃܶȣ©£»
¢ÚCÓëÐÂÖƵÄCu£¨OH£©2Ðü×ÇÒº¹²ÈÈÉú³ÉשºìÉ«³Áµí£»
¢ÛD¼«Ò×ÈÜÓÚË®£¬Í¨³£Çé¿öÏÂΪÎÞÉ«µ«Óд̼¤ÐÔÆøζµÄÆøÌ壬ÆäË®ÈÜÒºÖðµÎµÎÈëAgNO3ÈÜÒºÖÐ×î³õ±ä»ë×Ç£¬ÔÙ¼ÌÐøµÎÈëDÈÜÒººóÓÖ±äΪ³ÎÇåÈÜÒº£¬ËùµÃ³ÎÇåÈÜÒºÔÚÊÔ¹ÜÖÐÓëC»ìºÏ¹²ÈÈ£¬¿ÉÉú³ÉÒø¾µ£»
¢ÜGÄÜʹÐÂÖƵÄCu£¨OH£©2Ðü×ÇÒº±ä³ÎÇ壮
Çë»Ø´ð£º
£¨1£©Ð´³öÏÂÁи÷ÎïÖʵĽṹ¼òʽ£ºACH3COONH4¡¢BCH3COOCH2CH3¡¢CCH3CHO£®
£¨2£©Ð´³öÏÂÁи÷·´Ó¦µÄ»¯Ñ§·½³Ìʽ£®
¢ÙA¡úE£ºCH3COONH4+NaOH$\stackrel{¡÷}{¡ú}$CH3COONa+NH3¡ü+H2O£»
¢ÚCÓëÐÂÖƵÄCu£¨OH£©2·´Ó¦£ºCH3CHO+2Cu£¨OH£©2$\stackrel{¡÷}{¡ú}$CH3COOH+Cu2O¡ý+2H2O£®
£¨3£©Ð´³öFͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£ºCH3OCH3£®
·ÖÎö ±ê׼״̬ÏÂÃܶÈΪ0.717 g/LµÄÆø̬Ìþ£¬ÆäĦ¶ûÖÊÁ¿Îª0.717g/L¡Á22.4L/mol¡Ö16g/mol£¬ÎªCH4£¬ÒÒ´¼ÄÆÓë¼îʯ»Ò¹²ÈÈ·¢ÉúÍÑôÈ·´Ó¦Éú³É¼×Í飬ËùÒÔEΪCH3COONa£¬FÑõ»¯Éú³ÉC£¬CÑõ»¯Éú³ÉG£¬EÄܹ»Éú³ÉG£¬GÄÜʹÐÂÖƵÄCu£¨OH£©2Ðü×ÇÒº±ä³ÎÇ壬˵Ã÷GΪCH3COOH£¬FΪCH3CH2OH£¬CΪCH3CHO£¬ÔòBΪCH3COOCH2CH3£¬D¼«Ò×ÈÜÓÚË®£¬Í¨³£Çé¿öÏÂΪÎÞÉ«µ«Óд̼¤ÐÔÆøζµÄÆøÌ壬ÆäË®ÈÜÒºÖðµÎµÎÈëAgNO3ÈÜÒºÖÐ×î³õ±ä»ë×Ç£¬ÔÙ¼ÌÐøµÎÈëDÈÜÒººóÓÖ±äΪ³ÎÇåÈÜÒº£¬ËùµÃ³ÎÇåÈÜÒºÔÚÊÔ¹ÜÖÐÓëC»ìºÏ¹²ÈÈ£¬¿ÉÉú³ÉÒø¾µ£¬ËµÃ÷DΪNH3£¬AΪCH3COONH4£¬¾Ý´Ë·ÖÎö£®
½â´ð ½â£º±ê׼״̬ÏÂÃܶÈΪ0.717 g/LµÄÆø̬Ìþ£¬ÆäĦ¶ûÖÊÁ¿Îª0.717g/L¡Á22.4L/mol¡Ö16g/mol£¬ÎªCH4£¬ÒÒ´¼ÄÆÓë¼îʯ»Ò¹²ÈÈ·¢ÉúÍÑôÈ·´Ó¦Éú³É¼×Í飬ËùÒÔEΪCH3COONa£¬FÑõ»¯Éú³ÉC£¬CÑõ»¯Éú³ÉG£¬EÄܹ»Éú³ÉG£¬GÄÜʹÐÂÖƵÄCu£¨OH£©2Ðü×ÇÒº±ä³ÎÇ壬˵Ã÷˵Ã÷GΪCH3COOH£¬FΪCH3CH2OH£¬CΪCH3CHO£¬ÔòBΪCH3COOCH2CH3£¬D¼«Ò×ÈÜÓÚË®£¬Í¨³£Çé¿öÏÂΪÎÞÉ«µ«Óд̼¤ÐÔÆøζµÄÆøÌ壬ÆäË®ÈÜÒºÖðµÎµÎÈëAgNO3ÈÜÒºÖÐ×î³õ±ä»ë×Ç£¬ÔÙ¼ÌÐøµÎÈëDÈÜÒººóÓÖ±äΪ³ÎÇåÈÜÒº£¬ËùµÃ³ÎÇåÈÜÒºÔÚÊÔ¹ÜÖÐÓëC»ìºÏ¹²ÈÈ£¬¿ÉÉú³ÉÒø¾µ£¬ËµÃ÷DΪNH3£¬AΪCH3COONH4£¬
£¨1£©AΪCH3COONH4£¬BΪCH3COOCH2CH3£¬CΪCH3CHO£¬¹Ê´ð°¸Îª£ºCH3COONH4£»CH3COOCH2CH3£»CH3CHO£»
£¨2£©AΪCH3COONH4£¬AÓëNaOHÈÜÒº·´Ó¦Éú³ÉÒÒËáï§Óë°±Æø£¬»¯Ñ§·½³ÌʽΪCH3COONH4+NaOH$\stackrel{¡÷}{¡ú}$CH3COONa+NH3¡ü+H2O£¬CΪCH3CHO£¬CÓëÐÂÖƵÄCu£¨OH£©2·´Ó¦µÄ»¯Ñ§·½³ÌʽCH3CHO+2Cu£¨OH£©2$\stackrel{¡÷}{¡ú}$CH3COOH+Cu2O¡ý+2H2O£¬¹Ê´ð°¸Îª£ºCH3COONH4+NaOH$\stackrel{¡÷}{¡ú}$CH3COONa+NH3¡ü+H2O£»CH3CHO+2Cu£¨OH£©2$\stackrel{¡÷}{¡ú}$CH3COOH+Cu2O¡ý+2H2O£»
£¨3£©FΪCH3CH2OH£¬Æäͬ·ÖÒì¹¹ÌåÖ»Óм×ÃÑCH3OCH3£¬¹Ê´ð°¸Îª£ºCH3OCH3£®
µãÆÀ ±¾Ìâͨ¹ýÓлú¿òͼÍƶϿ¼²éÁËõ¥ÀàË®½â¡¢´¼µÄÑõ»¯¡¢ÒÔ¼°Òø¾µ·´Ó¦µÈµÈ֪ʶµã£¬×¢Òâ¾Ý¹ÙÄÜÍŵÄÐÔÖÊÍƶϣ¬ÌâÄ¿ÄѶȲ»´ó£®
¢ÙʵÑéÊÒÄÜÓþƾ«µÆ¼ÓÈȵÄÓÐÕô·¢Ãó¡¢Ô²µ×ÉÕÆ¿¡¢ÉÕ±
¢ÚÈôÕôÁó100mlʯÓͿɽ«ÆäÖÃÓÚ250mlÕôÁóÉÕÆ¿ÖÐ
¢ÛÐÂÖƵÄÂÈË®±£´æÔÚ×ØÉ«¹ã¿ÚÆ¿ÖУ¬²¢·ÅÔÚÒõÁ¹´¦
¢Ü²¨¶û¶àÒº£¨CuSO4Óëʯ»ÒË®°´Ò»¶¨±ÈÀý»ìºÏ£©Ê¢·ÅÔÚÌúÖÆÈÝÆ÷ÖÐ
¢ÝÈôµÎ¶¨¹ÜÖеÄÒºÃæÈçͼËùʾ£¬ÔòÆä¶ÁÊýӦΪ23.65mL£®
A£® | ¢Ù¢Ú | B£® | ¢Ù¢Û¢Ü | C£® | ¢Û¢Ü¢Ý | D£® | ¢Ú¢Û¢Ý |
A£® | ÔÚÑô¼«ÊÒ£¬Í¨µçºóÈÜÒºÖð½¥ÓɳÈÉ«±äΪ»ÆÉ« | |
B£® | µç·ÖÐÓÐ0.2molµç×Óͨ¹ýʱ£¬Ñô¼«ÓëÒõ¼«ÈÜÒº¼õÉÙµÄÖÊÁ¿²îΪ1.4g | |
C£® | Èô²âµÃÓÒÊÒÖÐKÓëCrµÄÎïÖʵÄÁ¿Ö®±ÈΪ3£º2£¬Ôò´Ë¹ý³Ìµç·Öй²×ªÒƵç×ÓÊýΪ0.1NA | |
D£® | Èô²â¶¨Ñô¼«ÒºÖÐKÓëCrµÄÎïÖʵÄÁ¿Ö®±ÈΪd£¬Ôò´Ëʱ¸õËá¼ØµÄת»¯ÂÊΪ2-d |
A£® | ÈôͼÖз´Ó¦¾ùΪ·ÇÑõ»¯»¹Ô·´Ó¦£¬µ±WΪһԪǿ¼îʱ£¬ÔòX¿ÉÄÜÊÇNaAlO2 | |
B£® | ÈôͼÖз´Ó¦¾ùΪ·ÇÑõ»¯»¹Ô·´Ó¦£¬µ±WΪһԪǿËáʱ£¬ÔòX¿ÉÄÜÊÇNH3 | |
C£® | ÈôͼÖз´Ó¦¾ùΪÑõ»¯»¹Ô·´Ó¦£¬µ±WΪ·Ç½ðÊôµ¥ÖÊʱ£¬ÔòZ¿ÉÄÜÊÇCO2 | |
D£® | ÈôͼÖз´Ó¦¾ùΪÑõ»¯»¹Ô·´Ó¦£¬µ±WΪ½ðÊôµ¥ÖÊʱ£¬ÔòZ¿ÉÄÜÊÇFeCl3 |
A£® | µÈÎïÖʵÄÁ¿µÄOH-ÓëôÇ»ù£¨-OH£©Ëùº¬µç×ÓÊýÏàµÈ | |
B£® | ³£Î³£Ñ¹Ï£¬44 g CO2ÆøÌ庬ÓÐÑõÔ×ӵĸöÊýΪ2 NA | |
C£® | 1 L 0.5 mol•L-1NaHCO3ÈÜÒºÖк¬ÓÐHCO3-µÄ¸öÊýΪ0.5 NA | |
D£® | 11.2 gÌú·ÛÓëÏ¡ÏõËᷴӦתÒƵç×ÓÊýÒ»¶¨Îª0.6 NA |
A£® | ·Åµçʱ£¬µç³ØÕý¼«·´Ó¦Ê½Îª£ºO2+H++4e-=2H2O | |
B£® | Ôö´ó¿ÕÆø½øÆø¿ÚÖ±¾¶£¬¿ÉÔö´óµç³ØÊä³öÄÜÁ¦²¢ÑÓ³¤µç³ØʹÓÃÊÙÃü | |
C£® | ·Åµçʱ£¬Ã¿Í¨Èë2.24L¿ÕÆø£¨±ê×¼×´¿ö£©£¬ÀíÂÛÉϸº¼«ÐèÒªÏûºÄ13gZn | |
D£® | µç³ØÓÃÍêºó£¬Ö»Ðè¸ü»»·â×°ºÃµÄп·Û»òµç³Øп°å¼´¿É |