ÌâÄ¿ÄÚÈÝ
6£®ÉϺ£ÊÀ²©»áÓ¢¹ú¹Ý--ÖÖ×ÓÊ¥µî£¬ÓÉÁùÍò¶à¸ù͸Ã÷µÄÑÇ¿ËÁ¦[Æä·Ö×ÓʽÊÇ£¨C5H8O2£©n]¸Ë¹¹½¨¶ø³É£®Ä³Í¬Ñ§´ÓÌṩµÄÔÁÏ¿âÖÐÑ¡ÔñÒ»ÖÖÔÁÏX£¬Éè¼ÆºÏ³É¸ß·Ö×ÓÑÇ¿ËÁ¦µÄ·ÏßÈçͼËùʾ£ºÒÑÖª£º¢Ù
¢Ú
¢Û
£¨R¡¢R'¡¢R¡å±íʾÌþ»ù£©
Çë»Ø´ð£º
£¨1£©ÔÁÏXÊÇc£¨ÌîÐòºÅ×Öĸ£©£®
£¨2£©¢ÚµÄ·´Ó¦Ìõ¼þÊÇNaOHË®ÈÜÒº£¬¼ÓÈÈ£»¢àµÄ·´Ó¦ÀàÐÍÊǼӾ۷´Ó¦£»DÖйÙÄÜÍŵÄÃû³ÆôÈ»ù¡¢ôÇ»ù£®
£¨3£©¢ßµÄ»¯Ñ§·½³ÌʽÊÇ
£¨4£©CÓжàÖÖͬ·ÖÒì¹¹Ì壬ÆäÖзÖ×ÓÖк¬ÓС°
£¨5£©Ä³Í¬Ñ§ÒÔ±ûϩΪÔÁÏÉè¼ÆÁ˺ϳÉÖмäÌåDµÄ·Ïߣº±ûÏ©$\stackrel{HBr}{¡ú}$¡¡úD£¬µÃµ½DµÄͬʱҲµÃµ½ÁËÁíÒ»ÖÖÓлú¸±²úÎïM£¬ÇëÄãÔ¤²âM¿ÉÄܵĽṹ¼òʽÊÇCH3CH2CH£¨OH£©COOH£®
·ÖÎö ÓÉ·´Ó¦¢ÙÖªXÖк¬ÓÐË«¼ü£¬ÔòAΪ¶þäå´úÌþ£¬ÓÉ·´Ó¦¢Ü¡¢¢Ý¡¢¢ÞÖªBÖк¬2¸öôÇ»ù£¬CÖк¬ÓÐÒ»¸öÈ©»ùºÍÒ»¸öôÇ»ù£¬DÖк¬ÓÐÒ»¸öôÈ»ùºÍÒ»¸öôÇ»ù£¬EÖк¬ÓÐÒ»¸öË«¼üºÍÒ»¸öôÈ»ù£®½áºÏÒÑÖª2¿ÉÖªXµÄ½á¹¹Îª
£¬ÔòAΪ
£¬BΪ
£¬CΪ
£¬DΪ
£¬EΪ
£¬
FΪ
£¬ÒԴ˽â´ð£¨1£©¡«£¨3£©£»
£¨4£©CµÄ½á¹¹Îª
£¬Òì¹¹ÌåÖк¬ÓÐ
½á¹¹µÄ¿ÉÄÜÊÇËá»òõ¥£»
£¨5£©½áºÏÒÑÖª¢Û¼´È©»òͪ¿ÉÓëÇâÇèËá¼Ó³ÉºóËữ¿ÉµÃ±ÈÔÎïÖʶàÒ»¸öCµÄËᣬ±ûÏ©ÓëHBr¼Ó³É¿ÉÉú³ÉÁ½ÖÖÎïÖÊCH3CHBrCH3ºÍCH3CH2CH2Br£¬ÆäÖеÚÒ»Öֽṹ¾Ë®½â¡¢Ñõ»¯³É±ûͪºóÔÙÓëHCN¼Ó³É¡¢Ëữ¿ÉµÃDÎïÖÊ£®µÚ¶þÖֽṹ¾Ë®½â¡¢Ñõ»¯µÃµ½±ûÈ©ºóÔÙÓëHCN¼Ó³É¡¢Ëữ¿ÉµÃCH3CH2CH£¨OH£©COOH£®
½â´ð ½â£º£¨1£©XÖк¬ÓÐË«¼ü£¬ÓÉÉÏÊö·ÖÎö¿É֪Ϊ
£¬¹Ê´ð°¸Îª£ºc£»
£¨2£©¢ÚµÄ·´Ó¦Ìõ¼þÊÇNaOHË®ÈÜÒº£¬¼ÓÈÈ£»¢àµÄ·´Ó¦ÀàÐÍÊǼӾ۷´Ó¦£»DΪ
£¬¹ÙÄÜÍŵÄÃû³ÆÎªôÈ»ù¡¢ôÇ»ù£¬
¹Ê´ð°¸Îª£ºNaOHË®ÈÜÒº£¬¼ÓÈÈ£»¼Ó¾Û·´Ó¦£»ôÈ»ù¡¢ôÇ»ù£»
£¨3£©¢ßµÄ»¯Ñ§·½³ÌʽÊÇ
£¬
¹Ê´ð°¸Îª£º
£»
£¨4£©CµÄ½á¹¹Îª
£¬Òì¹¹ÌåÖк¬ÓÐ
½á¹¹µÄ¿ÉÄÜÊÇËá»òõ¥£¬ÊôÓÚôÈËáµÄΪ¶¡Ëá¡¢2-¼×»ù±ûË᣻ÊôÓÚõ¥Îª¼×Ëá±ûõ¥¡¢¼×ËáÒì±ûõ¥¡¢ÒÒËáÒÒõ¥¡¢±ûËá¼×õ¥£¬¹²6ÖÖ£¬¹Ê´ð°¸Îª£º6£»
£¨5£©±ûÏ©ÓëHBr¼Ó³É¿ÉÉú³ÉÁ½ÖÖÎïÖÊCH3CHBrCH3ºÍCH3CH2CH2Br£¬ÆäÖеÚÒ»Öֽṹ¾Ë®½â¡¢Ñõ»¯³É±ûͪºóÔÙÓëHCN¼Ó³É¡¢Ëữ¿ÉµÃDÎïÖÊ£®µÚ¶þÖֽṹ¾Ë®½â¡¢Ñõ»¯µÃµ½±ûÈ©ºóÔÙÓëHCN¼Ó³É¡¢Ëữ¿ÉµÃCH3CH2CH£¨OH£©COOH£¬ÔòM¿ÉÄܵĽṹ¼òʽÊÇCH3CH2CH£¨OH£©COOH£¬
¹Ê´ð°¸Îª£ºCH3CH2CH£¨OH£©COOH£®
µãÆÀ ±¾Ì⿼²éÓлúÎïµÄÍÆ¶Ï£¬Îª¸ßƵ¿¼µã£¬°ÑÎÕ¹ÙÄÜÍŵı仯¡¢Ì¼Á´±ä»¯¡¢Óлú·´Ó¦Îª½â´ðµÄ¹Ø¼ü£¬²àÖØ·ÖÎöÓëÍÆ¶ÏÄÜÁ¦µÄ¿¼²é£¬×¢ÒâϰÌâÖÐÐÅÏ¢¼°ÓлúÎïÐÔÖʵÄÓ¦Óã¬ÌâÄ¿ÄѶȲ»´ó£®
¢ñ£ºC£¨s£©+H2O£¨g£©?CO£¨g£©+H2£¨g£©¡÷H=+131.0kJ/mol
¢ò£ºCO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£©¡÷H=-43kJ/mol
¢ó£ºCaO£¨s£©+CO2£¨g£©?CaCO3£¨s£©¡÷H=-178.3kJ/mol
¢ô£ºC£¨s£©+2H2O£¨g£©+CaO£¨s£©?CaCO3£¨s£©+2H2£¨g£©¡÷H=akJ/mol£®
| A£® | a=+90.3 | |
| B£® | ºãκãѹÏ£¬ÔÚÒÑ´ïÆ½ºâµÄ·´Ó¦IÌåϵÖÐÔÙ³äÈëÉÙÁ¿HeʱƽºâÕýÏòÒÆ¶¯ | |
| C£® | ÆäËûÌõ¼þ²»±ä£¬¼Óѹ¶Ô·´Ó¦¢òµÄ·´Ó¦ËÙÂÊÎÞÓ°Ïì | |
| D£® | ÆäËûÌõ¼þ²»±ä£¬Éý¸ßζȿÉÌá¸ß·´Ó¦¢óÖÐCO2µÄת»¯ÂÊ |
| A£® | ³£Î³£Ñ¹Ï£¬Cu-ZnÔµç³ØÖУ¬Õý¼«²úÉú1.12LH2ʱ£¬×ªÒƵĵç×ÓÊýӦСÓÚ0.1NA | |
| B£® | 1molSO2Óë×ãÁ¿O2ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉSO2£¬¹²×ªÒÆ2NA¸öµç×Ó | |
| C£® | 2.1gDTOÖÐËùº¬ÖÐ×ÓÊýΪNA | |
| D£® | ³£Î³£Ñ¹Ï£¬28gC2H4¡¢COµÄ»ìºÏÆøÌåÖк¬ÓÐ̼Ô×ÓµÄÊýĿΪ1.5NA |
£¨1£©ÎüÁò×°ÖÃÈçͼ¶þËùʾ£®
¢Ù×°ÖÃBµÄ×÷ÓÃÊǼìÑé×°ÖÃAÖÐSO2µÄÎüÊÕЧÂÊ£¬BÖÐÊÔ¼ÁÊÇÆ·ºì¡¢äåË®»òKMnO4ÈÜÒº£¬±íÃ÷SO2ÎüÊÕЧÂʵ͵ÄʵÑéÏÖÏóÊÇBÖÐÈÜÒºÈÜÒºÑÕÉ«ºÜ¿ìÍÊÉ«£®
¢ÚΪÁËʹSO2¾¡¿ÉÄÜÎüÊÕÍêÈ«£¬ÔÚ²»¸Ä±äAÖÐÈÜҺŨ¶È¡¢Ìå»ýµÄÌõ¼þÏ£¬³ýÁ˼°Ê±½Á°è·´Ó¦ÎïÍ⣬»¹¿É²ÉÈ¡µÄºÏÀí´ëÊ©ÊÇÔö´óSO2µÄ½Ó´¥Ãæ»ý»ò¿ØÖÆSO2µÄÁ÷ËÙ¡¢Êʵ±Éý¸ßζȣ®£¨Ð´³öÁ½Ìõ£©
£¨2£©¼ÙÉ豾ʵÑéËùÓõÄNa2CO3º¬ÉÙÁ¿NaCl¡¢NaOH£¬Éè¼ÆÊµÑé·½°¸½øÐмìÑ飮£¨ÊÒÎÂʱCaCO3±¥ºÍÈÜÒºµÄpH=10.2£©
ÏÞÑ¡ÊÔ¼Á¼°ÒÇÆ÷£ºÏ¡ÏõËá¡¢AgNO3ÈÜÒº¡¢CaCl2ÈÜÒº¡¢Ca£¨NO3£©2ÈÜÒº¡¢·Ó̪ÈÜÒº¡¢ÕôÁóË®¡¢pH¼Æ¡¢ÉÕ±¡¢ÊԹܡ¢µÎ¹Ü
| ÐòºÅ | ʵÑé²Ù×÷ | Ô¤ÆÚÏÖÏó | ½áÂÛ |
| ¢Ù | È¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®£¬³ä·ÖÕñµ´Èܽ⣬µÎ¼Ó×ãÁ¿Ï¡ÏõËᣬÔٵμÓÉÙÁ¿AgNO3ÈÜÒº£¬Õñµ´£® | Óа×É«³ÁµíÉú³É | ÑùÆ·º¬NaCl |
| ¢Ú | ÁíÈ¡ÉÙÁ¿ÑùÆ·ÓÚÉÕ±ÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®£¬³ä·Ö½Á°èÈܽ⣬¼ÓÈë¹ýÁ¿CaCl2ÈÜÒº£¬½Á°è£¬¾²Öã¬ÓÃpH¼Æ²â¶¨ÉϲãÇåÒºpH£® | Óа×É«³ÁµíÉú³É£¬ÉÏ ²ãÇåÒºpH£¾10.2 | ÑùÆ·º¬NaOH |
ÒÑÖª£ºIO3-+5I-+6H+¨T3I2+3H2O 2S2O32-+I2¨TS4O62-+2I-
ijͬѧµÚÒ»²½ºÍµÚ¶þ²½µÄ²Ù×÷¶¼ºÜ¹æ·¶£¬µÚÈý²½µÎËÙÌ«Âý£¬ÕâÑù²âµÃµÄNa2S2O3Ũ¶È¿ÉÄÜÆ«¸ß£¨Ìî¡°ÎÞÓ°Ï족¡¢¡°Æ«µÍ¡±»ò¡°Æ«¸ß¡±£©£¬ÔÒòÊÇ4I-+4H++O2¨T2I2+2H2O£®
| A£® | 1LË®Öк¬ÓÐ1molH2SO4 | |
| B£® | 1LÈÜÒºÖк¬ÓÐ1molH+ | |
| C£® | ½«98gH2SO4ÈÜÓÚ1LË®ÖÐÅä³ÉÉÏÊöÈÜÒº | |
| D£® | 1LÁòËáÈÜÒºÖк¬ÓÐ96gSO42- |
| A£® | »¯Ñ§¼ü´æÔÚÓÚÔ×ÓÖ®¼ä£¬Ò²´æÔÚÓÚ·Ö×ÓÖ®¼ä | |
| B£® | Àë×Ó¼üÊÇÒõ¡¢ÑôÀë×ÓÖ®¼äµÄÎüÒýÁ¦ | |
| C£® | Àë×Ó»¯ºÏÎï¿ÉÒÔº¬¹²¼Û¼ü | |
| D£® | ·Ö×Ó¼ä×÷ÓÃÁ¦Ô½´ó£¬·Ö×ÓÔ½Îȶ¨ |