ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿1,2£¶þäåÒÒÍéµÄÖƱ¸ÔÀíÊÇ£ºCH3CH2OHCH2=CH2¡ü+H2O CH2=CH2+Br2¡úBrCH2¡ªCH2Br£»Ä³¿ÎÌâС×éÓÃÏÂͼËùʾµÄ×°ÖÃÖƱ¸1,2£¶þäåÒÒÍé¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)×°ÖÃBµÄ×÷ÓÃÊÇ_______¡£
(2)Èý¾±ÉÕÆ¿ÄÚ¼ÓÈëÒ»¶¨Á¿µÄÒÒ´¼£Å¨ÁòËá»ìºÏÒººÍÉÙÁ¿´ÖÉ°£¬Æä¼ÓÈë´ÖÉ°Ä¿µÄÊÇ_______£¬ÒÇÆ÷EµÄÃû³ÆÊÇ_______¡£
(3)×°ÖÃCÄÚ·¢ÉúµÄÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______¡£
(4)¼ÓÈÈÈý¾±ÉÕÆ¿Ç°£¬ÏȽ«CÓëDÁ¬½Ó´¦¶Ï¿ª£¬ÔÙ½«Èý¾±ÉÕÆ¿ÔÚʯÃÞÍøÉϼÓÈÈ£¬´ýζÈÉýµ½Ô¼120¡æʱ£¬Á¬½ÓCÓëD£¬²¢Ñ¸ËÙ½«AÄÚ·´Ó¦Î¶ÈÉýÎÂÖÁ160¡«180¡æ£¬´ÓµÎҺ©¶·ÖÐÂýÂýµÎ¼ÓÒÒ´¼£Å¨ÁòËá»ìºÏÒº£¬±£³ÖÒÒÏ©ÆøÌå¾ùÔȵØͨÈë×°ÓÐ3.20mLÒºäå(¦ÑÒºäå=3g/cm3)ºÍ3mLË®µÄDÖÐÊԹܣ¬Ö±ÖÁ·´Ó¦½áÊø¡£
¢Ù½«CÓëDÁ¬½Ó´¦¶Ï¿ªµÄÔÒòÊÇ_______
¢ÚÅжϷ´Ó¦½áÊøµÄÏÖÏóÊÇ_______
(5)½«´ÖÆ·ÒÆÈë·ÖҺ©¶·£¬·Ö±ðÓÃË®¡¢ÇâÑõ»¯ÄÆÈÜÒº¡¢Ë®Ï´µÓ£¬²úÆ·ÓÃÎÞË®ÂÈ»¯¸Æ¸ÉÔ¹ýÂ˺óÕôÁóÊÕ¼¯129¡«133¡æÁó·Ö£¬µÃµ½7.896g1,2£¶þäåÒÒÍé¡£1,2£¶þäåÒÒÍéµÄ²úÂÊΪ_______¡£
(6)ÏÂÁвÙ×÷ÖУ¬²»»áµ¼Ö²úÎï²úÂʽµµÍµÄÊÇ(ÌîÕýÈ·´ð°¸µÄ±êºÅ)_______£®
a.ÒÒϩͨÈëäåˮʱËÙÂÊÌ«¿ì b£®×°ÖÃEÖеÄNaOHÈÜÒºÓÃË®´úÌ森
c.È¥µô×°ÖÃDÉÕ±ÖеÄË® d.ʵÑéʱûÓÐC×°ÖÃ
e£®DÖеÄÊÔ¹ÜÀï²»¼ÓË®
¡¾´ð°¸¡¿°²È«Æ¿·À¶ÂÈû ·À±©·Ð ׶ÐÎÆ¿ SO2+2OH£=SO32¡ª+H2O ¼õÉÙÆøÌå¶ÔäåË®½Á¶¯£¬¼õÉÙäåÕôÆøµÄ»Ó·¢ DÖÐÊԹܵÄäåË®ÍÊÉ«(»òÊÔ¹ÜÀïµÄÒºÌå±äΪÎÞÉ«) 70% b
¡¾½âÎö¡¿
ʵÑéÊÒÖƱ¸1£¬2-¶þäåÒÒÍ飺Èý¾±ÉÕÆ¿AÖз¢Éú·´Ó¦ÊÇÒÒ´¼ÔÚŨÁòËáµÄ×÷ÓÃÏ·¢Éú·Ö×ÓÄÚÍÑË®ÖÆÈ¡ÒÒÏ©£¬ÒÒ´¼·¢ÉúÁËÏûÈ¥·´Ó¦£¬·´Ó¦·½³ÌʽΪ£ºCH3CH2OHCH2=CH2¡ü+H2O£¬Èç¹ûDÖе¼Æø¹Ü·¢Éú¶ÂÈûʹʣ¬AÖвúÉúµÄÒÒÏ©ÆøÌå»áµ¼ÖÂ×°ÖÃBÖÐѹǿÔö´ó£¬³¤µ¼¹ÜÒºÃæ»áÉÏÉý£¬ËùÒÔ×°ÖÃBÖг¤²£Á§¹Ü¿ÉÅжÏ×°ÖÃÊÇ·ñ¶ÂÈû£¬×°ÖÃBÆ𻺳å×÷Óã¬Å¨ÁòËá¾ßÓÐÍÑË®ÐÔ¡¢ÎüË®ÐÔºÍÇ¿Ñõ»¯ÐÔ£¬ÄÜÑõ»¯ÒÒ´¼£¬¿ÉÄÜÉú³ÉµÄËáÐÔÆøÌåΪ¶þÑõ»¯Áò¡¢¶þÑõ»¯Ì¼£¬×°ÖÃCÖзÅÇâÑõ»¯ÄÆÈÜÒº£¬·¢Éú·´Ó¦SO2+2NaOH=Na2SO3+H2O£¬CO2+2NaOH¨TNa2CO3+H2O£¬³ýÈ¥ÔÓÖÊÆøÌ壬ÒÒÏ©º¬Óв»±¥ºÍ¼üC=CË«¼ü£¬ÄÜÓë±Ëص¥ÖÊ·¢Éú¼Ó³É·´Ó¦£¬DÖÐÒÒÏ©ºÍäå¼Ó³ÉÉú³É1£¬2-¶þäåÒÒÍ飬·´Ó¦Îª£ºCH2=CH2+Br2¡úBrCH2¡ªCH2Br£¬ÖƵÃ1£¬2-¶þäåÒÒÍé¡£
(1)Ò»µ©ÆøÌå·¢Éú¶ÂÈû£¬×°ÖÃBÄÚµÄÒºÌå½øÈëµ½³¤²£Á§¹ÜÄÚ£¬Æðµ½°²È«Æ¿µÄ×÷Óã»
Òò´Ë£¬±¾ÌâÕýÈ·´ð°¸ÊÇ£º°²È«Æ¿·À¶ÂÈû£»
(2)¸øÒºÌå¼ÓÈÈ£¬ÒײúÉú±©·ÐÏÖÏó£¬ËùÒÔÈý¾±ÉÕÆ¿ÄÚ¼ÓÈëÒ»¶¨Á¿µÄÒÒ´¼£Å¨ÁòËá»ìºÏÒººÍÉÙÁ¿´ÖÉ°£¬Æä¼ÓÈë´ÖÉ°Ä¿µÄÊÇ·À±©·Ð£»ÒÇÆ÷EµÄÃû³ÆÊÇ׶ÐÎÆ¿£»
Òò´Ë£¬±¾ÌâÕýÈ·´ð°¸ÊÇ£º·À±©·Ð£»×¶ÐÎÆ¿£»
(3)ÒÒ´¼ºÍŨÁòËá·¢Éú·´Ó¦£¬ÖƱ¸³öµÄÒÒÏ©ÆøÌåÖк¬ÓÐÔÓÖÊÆøÌå¶þÑõ»¯Áò£¬Òò´Ë»ìºÏÆøÌåͨ¹ýÇâÑõ»¯ÄÆÈÜÒº£¬ÎüÊÕÁ˶þÑõ»¯Áò£¬Ìá´¿ÁËÒÒÏ©£»¶þÑõ»¯ÁòÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³ÉÑÇÁòËáÄƺÍË®£¬Àë×Ó·½³ÌʽΪSO2+2OH£=SO32¡ª+H2O £»
Òò´Ë£¬±¾ÌâÕýÈ·´ð°¸ÊÇ£ºSO2+2OH£=SO32¡ª+H2O£»
(4) ¢Ù¼ÓÈÈÈý¾±ÉÕÆ¿Ç°£¬ÏȽ«CÓëDÁ¬½Ó´¦¶Ï¿ª£¬ÆøÌå¾Í²»ÄܽøÈëµ½×°ÓÐäåµÄ×°ÖÃÄÚ£¬¼õÉÙäåÕôÆøµÄ»Ó·¢£»
Òò´Ë£¬±¾ÌâÕýÈ·´ð°¸ÊÇ£º¼õÉÙÆøÌå¶ÔäåË®½Á¶¯£¬¼õÉÙäåÕôÆøµÄ»Ó·¢£»
¢ÚÒÒÏ©ºÍäå·¢Éú¼Ó³É·´Ó¦Éú³ÉÎÞÉ«µÄäå´úÌþ£¬µ±DÖÐÊԹܵÄäåË®ÍÊÉ«ºó£¬¸Ã·´Ó¦½áÊø£»
Òò´Ë£¬±¾ÌâÕýÈ·´ð°¸ÊÇ£ºDÖÐÊԹܵÄäåË®ÍÊÉ«£»
(5)äåµÄÎïÖʵÄÁ¿Îª=0.06mol£¬¸ù¾Ý·´Ó¦£ºCH2=CH2+Br2¡úBr-CH2-CH2-Br¿ÉÖª£¬Éú³É1£¬2-¶þäåÒÒÍéµÄÁ¿Îª0.06mol£¬ÖÊÁ¿Îª0.06mol¡Á188g/mol=11.28g£¬
Ôò1,2-¶þäåÒÒÍéµÄ²úÂÊΪ¡Á100%=70%£»
Òò´Ë£¬±¾ÌâÕýÈ·´ð°¸ÊÇ£º70%¡£
(6) a.ͨÆøËٶȹý¿ì£¬·´Ó¦²»³ä·Ö£¬½«µ¼Ö²úÂʽµµÍ£¬a²»ºÏÌâÒ⣻
b.×°ÖÃE×÷ÓÃÊÇÎüÊÕβÆø£¬²»Ó°Ïì²úÎï²úÂÊ£¬b·ûºÏÌâÒ⣻
c.DÉÕ±ÖеÄË®ÆðÀäÈ´×÷Óã¬È¥µô½«µ¼ÖÂäå»Ó·¢¼Ó¾ç£¬´Ó¶øʹ²úÂʽµµÍ£¬c²»ºÏÌâÒ⣻
d.×°ÖÃCÖеÄNaOH¿É³ýÈ¥CO2¡¢SO2µÈÔÓÖÊ£¬ÕâЩÔÓÖÊÈô²»³ýÈ¥£¬Ôò»áÓëäåË®·´Ó¦½µµÍ²úÂÊ£¬Ë®ÎüÊÕ¶þÑõ»¯ÁòÆøÌåµÄЧ¹û²î£¬»áµ¼Ö²úÎï²úÂʽµµÍ£¬d²»ºÏÌâÒ⣻
e.DÖÐÊÔ¹ÜÀïµÄË®ÔÚÒºäåµÄÉÏ·½£¬ÊǶÔÒºäå½øÐÐÒº·â£¬¼õÉÙäåµÄ»Ó·¢£¬ÈôDÖеÄÊÔ¹ÜÀï²»¼ÓË®£¬äå»Ó·¢µÄ¶à£¬²úÎïµÄ²úÂʽµµÍ£¬e²»ºÏÌâÒ⣻
Òò´Ë£¬±¾ÌâÕýÈ·´ð°¸ÊÇ£ºb¡£