ÌâÄ¿ÄÚÈÝ
¾Ý±¨µÀ£¬Ä¦ÍÐÂÞÀ¹«Ë¾¿ªÊ¼ÁËÒ»ÖÖÒÔ¼×´¼ÎªÔÁÏ£¬ÒÔKOHΪµç½âÖʵÄÓÃÓÚÊÖ»úµÄ¿É³äµçµÄ¸ßЧȼÁÏµç³Ø£¬³äÒ»´Îµç¿ÉÁ¬ÐøÊ¹ÓÃÒ»¸öÔ£®ÈçͼÊÇÒ»¸öµç»¯Ñ§¹ý³ÌµÄʾÒâͼ£®ÒÑÖª¼×³Ø×Ü·´Ó¦Ê½Îª£º2CH3OH+3O2+4KOH
2K2CO3+6H2O

ÇëÌî¿Õ£º
£¨1£©³äµçʱ£º¢ÙÔµç³ØµÄ¸º¼«ÓëµçÔ´ ¼«ÏàÁ¬£®¢ÚÑô¼«µÄµç¼«·´Ó¦Ê½Îª
£¨2£©·Åµçʱ£¬¸º¼«µÄµç¼«·´Ó¦Ê½Îª
£¨3£©Ôڴ˹ý³ÌÖÐÈôÍêÈ«·´Ó¦£¬ÒÒ³ØÖÐA¼«µÄÖÊÔö¼Ó648g£¬Ôò¼×³ØÖÐÀíÂÛÉÏÏûºÄO2 L£¨±ê×¼×´¿öÏ£©£®
| ·Åµç |
| ³äµç |
ÇëÌî¿Õ£º
£¨1£©³äµçʱ£º¢ÙÔµç³ØµÄ¸º¼«ÓëµçÔ´
£¨2£©·Åµçʱ£¬¸º¼«µÄµç¼«·´Ó¦Ê½Îª
£¨3£©Ôڴ˹ý³ÌÖÐÈôÍêÈ«·´Ó¦£¬ÒÒ³ØÖÐA¼«µÄÖÊÔö¼Ó648g£¬Ôò¼×³ØÖÐÀíÂÛÉÏÏûºÄO2
¿¼µã£ºÔµç³ØºÍµç½â³ØµÄ¹¤×÷ÔÀí
רÌ⣺
·ÖÎö£º£¨1£©³äµçʱ£¬Ôµç³Ø¸º¼«ÓëµçÔ´¸º¼«ÏàÁ¬£¬Ñô¼«ÉÏʧµç×Ó·¢ÉúÑõ»¯·´Ó¦£»
£¨2£©·Åµçʱ£¬¸º¼«Éϼ״¼Ê§µç×Ó·¢ÉúÑõ»¯·´Ó¦£»
£¨3£©ÒÒ³ØÖÐB¼«ÉÏÒøÀë×ӵõç×Ó·¢Éú»¹Ô·´Ó¦£¬¸ù¾Ý×ªÒÆµç×ÓÏàµÈ¼ÆËãÑõÆøµÄÌå»ý£®
£¨2£©·Åµçʱ£¬¸º¼«Éϼ״¼Ê§µç×Ó·¢ÉúÑõ»¯·´Ó¦£»
£¨3£©ÒÒ³ØÖÐB¼«ÉÏÒøÀë×ӵõç×Ó·¢Éú»¹Ô·´Ó¦£¬¸ù¾Ý×ªÒÆµç×ÓÏàµÈ¼ÆËãÑõÆøµÄÌå»ý£®
½â´ð£º
½â£º£¨1£©¢Ù³äµçʱ£¬Ôµç³Ø¸º¼«ÓëµçÔ´¸º¼«ÏàÁ¬£¬Ñô¼«ÉÏÇâÑõ¸ùÀë×Óʧµç×Ó·¢ÉúÑõ»¯·´Ó¦£¬µç¼«·´Ó¦Ê½Îª£º4OH--4e-¨T2H2O+O2¡ü£¬
¹Ê´ð°¸Îª£º¸º£»4OH--4e-¨T2H2O+O2¡ü£»
£¨2£©·Åµçʱ£¬¼×´¼Ê§µç×ÓºÍÇâÑõ¸ùÀë×Ó·´Ó¦Éú³É̼Ëá¸ùÀë×ÓºÍË®£¬ËùÒԵ缫·´Ó¦Ê½Îª£ºCH3OH-6e-+8OH-¨TCO32-+6H2O£¬
¹Ê´ð°¸Îª£ºCH3OH-6e-+8OH-¨TCO32-+6H2O£»
£¨3£©ÒÒ³ØÖÐB¼«ÉÏÒøÀë×ӵõç×Ó·¢Éú»¹Ô·´Ó¦£¬µ±ÒÒ³ØÖÐB¼«µÄÖÊÁ¿Éý¸ß648g£¬Ôò¼×³ØÖÐÀíÂÛÉÏÏûºÄO2Ìå»ý=
¡Á22.4L/mol=33.6L£¬
¹Ê´ð°¸Îª£º33.6£®
¹Ê´ð°¸Îª£º¸º£»4OH--4e-¨T2H2O+O2¡ü£»
£¨2£©·Åµçʱ£¬¼×´¼Ê§µç×ÓºÍÇâÑõ¸ùÀë×Ó·´Ó¦Éú³É̼Ëá¸ùÀë×ÓºÍË®£¬ËùÒԵ缫·´Ó¦Ê½Îª£ºCH3OH-6e-+8OH-¨TCO32-+6H2O£¬
¹Ê´ð°¸Îª£ºCH3OH-6e-+8OH-¨TCO32-+6H2O£»
£¨3£©ÒÒ³ØÖÐB¼«ÉÏÒøÀë×ӵõç×Ó·¢Éú»¹Ô·´Ó¦£¬µ±ÒÒ³ØÖÐB¼«µÄÖÊÁ¿Éý¸ß648g£¬Ôò¼×³ØÖÐÀíÂÛÉÏÏûºÄO2Ìå»ý=
| ||
| 4 |
¹Ê´ð°¸Îª£º33.6£®
µãÆÀ£º±¾Ì⿼²éÁËÔµç³ØºÍµç½â³ØÔÀí£¬×¢Òâµç¼«·´Ó¦Ê½µÄÊéдҪ½áºÏµç½âÖÊÈÜÒºµÄËá¼îÐÔ£¬´®Áªµç·ÖÐ×ªÒÆµç×ÓÏàµÈ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÏÂÁйØÓÚÈÝÁ¿Æ¿¼°ÆäʹÓ÷½·¨µÄÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
¢ÙÈÝÁ¿Æ¿ÊÇÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒºµÄרÓÃÒÇÆ÷
¢ÚÈÝÁ¿Æ¿Ê¹ÓÃǰҪÏȼì²éÈÝÁ¿Æ¿ÊÇ·ñ©Һ
¢ÛÈÝÁ¿Æ¿¿ÉÒÔÓÃÀ´¼ÓÈÈ
¢ÜÈÝÁ¿Æ¿²»ÄÜÓó¤ÆÚÖü´æÅäÖÆºÃµÄÈÜÒº
¢ÝÒ»¶¨ÒªÓÃ500mLÈÝÁ¿Æ¿ÅäÖÆ250mLÈÜÒº£®
¢ÙÈÝÁ¿Æ¿ÊÇÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒºµÄרÓÃÒÇÆ÷
¢ÚÈÝÁ¿Æ¿Ê¹ÓÃǰҪÏȼì²éÈÝÁ¿Æ¿ÊÇ·ñ©Һ
¢ÛÈÝÁ¿Æ¿¿ÉÒÔÓÃÀ´¼ÓÈÈ
¢ÜÈÝÁ¿Æ¿²»ÄÜÓó¤ÆÚÖü´æÅäÖÆºÃµÄÈÜÒº
¢ÝÒ»¶¨ÒªÓÃ500mLÈÝÁ¿Æ¿ÅäÖÆ250mLÈÜÒº£®
| A¡¢¢Ù¢Û | B¡¢¢Ù¢Ú¢Ü¢Ý |
| C¡¢¢Ù¢Ú¢Ü | D¡¢¢Ù¢Ú¢Û¢Ü |
¶ÔÓÚijһ´ïµ½Æ½ºâ״̬µÄ¿ÉÄæ·´Ó¦£¬Èç¹û¸Ä±äijһÌõ¼þ¶øÊ¹Éú³ÉÎïŨ¶ÈÔö´ó£¬Ôò£¨¡¡¡¡£©
| A¡¢Æ½ºâÒ»¶¨ÏòÕý·´Ó¦·½Ïò·¢ÉúÁËÒÆ¶¯ |
| B¡¢Æ½ºâÒ»¶¨ÏòÄæ·´Ó¦·½Ïò·¢ÉúÁËÒÆ¶¯ |
| C¡¢·´Ó¦ÎïŨ¶ÈÏàÓ¦¼õС |
| D¡¢Æ½ºâ¿ÉÄÜ·¢ÉúÁËÒÆ¶¯£¬Ò²¿ÉÄÜûÓз¢ÉúÒÆ¶¯ |
¶ÏÁÑÒ»¸ö¸ø¶¨µÄ¼üʱËùÏûºÄµÄÄÜÁ¿³ÆÎªÀë½âÄÜ£¬ÌṩÏÂÁÐÍéÌþµÄC-H¼üµÄÀë½âÄܸù¾ÝÌṩÐÅÏ¢ÅжÏÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

| A¡¢ÉÏÊöËĸöÀë½â¹ý³Ì¶¼ÊôÓÚÎüÈȹý³Ì |
| B¡¢Àë½âÄÜԽС£¬C-H¼üÔ½Ò×¶ÏÁÑ£¬ÇâÔ×ÓÔ½Ò×±»È¡´ú |
| C¡¢ÔÚ¹âÕÕÌõ¼þϱûÍéÓëÂÈÆø·¢ÉúÈ¡´ú·´Ó¦Ö»Éú³ÉCH3CHClCH3 |
| D¡¢2-¼×»ù±ûÍéÓëÂÈÆø·¢ÉúÈ¡´ú·´Ó¦£¬Éú³ÉµÄÒ»ÂÈ´úÎïÖ»ÓÐÁ½ÖÖ |
1.0gþÔÚÑõÆøÖÐȼÉÕºóÔöÖØ0.64g£¬µ«ÔÚ¿ÕÆøÖÐȼÉÕʱÔöÖØ²»×ã0.64g£¬ÆäÔÒò¿ÉÄÜÊÇ£¨¡¡¡¡£©
| A¡¢¿ÕÆøÖÐþȼÉտ϶¨²»ÍêÈ« |
| B¡¢¿ÕÆøÖв¿·ÖþÓëCO2·´Ó¦ |
| C¡¢¿ÕÆøÖÐþ²¿·ÖÓëN2·´Ó¦ |
| D¡¢¿ÕÆøÖÐþ²¿·ÖÓëË®ÕôÆø·´Ó¦ |