ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÒÔµí·ÛΪÖ÷ÒªÔ­ÁϺϳÉÒ»ÖÖ¾ßÓйûÏãζµÄÎïÖÊCºÍ»¯ºÏÎïDµÄºÏ³É·ÏßÈçÏÂͼËùʾ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©AµÄ½á¹¹¼òʽΪ £¬·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽΪ ¡£

£¨2£©·´Ó¦¢ßÖÐÎïÖÊXµÄ·Ö×ÓʽΪ £¬·´Ó¦¢àµÄÀàÐÍΪ ¡£

£¨3£©·´Ó¦¢ÝµÄ»¯Ñ§·½³ÌʽΪ ¡£·´Ó¦¢ÞÓÃÓÚʵÑéÊÒÖÆÒÒÏ©£¬Îª³ýÈ¥ÆäÖпÉÄÜ»ìÓеÄSO2ӦѡÓõÄÊÔ¼ÁÊÇ ¡£

£¨4£©ÒÑÖªDµÄÏà¶Ô·Ö×ÓÁ¿Îª118£¬ÆäÖÐ̼¡¢ÇâÁ½ÔªËصÄÖÊÁ¿·ÖÊý·Ö±ðΪ40.68%¡¢5.08%£¬ÆäÓàΪÑõÔªËØ£¬ÔòDµÄ·Ö×ÓʽΪ ¡£

£¨5£©Çë²¹³äÍêÕûÖ¤Ã÷·´Ó¦¢ÙÊÇ·ñ·¢ÉúµÄʵÑé·½°¸£ºÈ¡·´Ó¦¢ÙµÄÈÜÒº2 mLÓÚÊÔ¹ÜÖÐ ¡££¨ÊµÑéÖпɹ©Ñ¡ÔñµÄÊÔ¼Á£º10%µÄNaOHÈÜÒº¡¢5%µÄCuSO4ÈÜÒº¡¢µâË®£©

¡¾´ð°¸¡¿£¨1£©CH3CHO

£¨2£©Br2 È¡´ú·´Ó¦

£¨3£©CH3COOH + CH3CH2OH CH3COOCH2CH3 + H2O

NaOHÈÜÒº

£¨4£©C4H6O4

£¨5£©ÓÃ10%µÄNaOHÈÜÒºµ÷½ÚÈÜÒºÖÁÖÐÐÔ£¬ÔÙÏòÆäÖмÓÈë2 mL 10%µÄNaOHÈÜÒº£¬ÔÙ¼ÓÈë4~5µÎ5%µÄCuSO4ÈÜÒº£¬¼ÓÈÈÒ»¶Îʱ¼ä¡£ÈôÓÐשºìÉ«³Áµí£¬ÔòÖ¤Ã÷·´Ó¦¢ÙÒÑ·¢Éú

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£ºÓÉÁ÷³Ìͼ¿ÉÖª£¬ÒÒ´¼Ñõ»¯ÎªAÒÒÈ©£¬ÒÒÈ©Ñõ»¯ÎªBÒÒËᣬÒÒ´¼ÓëÒÒËáÉú³ÉCÒÒËáÒÒõ¥£»ÒÒÏ©ÓëXÉú³ÉC2H4Br2£¬XÊÇä壬C2H4Br2µÄ½á¹¹¼òʽÊÇBrCH2CH2Br£¬BrCH2CH2BrÓëNaCN·´Ó¦Éú³ÉC2H4(CN)2£¬C2H4(CN)2ÊÇNCCH2CH2CN£¬NCCH2CH2CNË®½âΪHOOCCH2CH2COOH£»

£¨1£©ÒÒÈ©µÄ½á¹¹¼òʽΪCH3CHO£¬ÆÏÌÑÌÇת»¯ÎªÒÒ´¼µÄ»¯Ñ§·½³ÌʽΪ¡£

£¨2£©·´Ó¦¢ßÖÐÎïÖÊXµÄ·Ö×ÓʽΪBr2£¬BrCH2CH2BrÉú³ÉNCCH2CH2CNµÄÀàÐÍΪȡ´ú·´Ó¦¡£

£¨3£©·´Ó¦¢ÝÊÇÒÒ´¼ÓëÒÒËáÉú³ÉÒÒËáÒÒõ¥£¬»¯Ñ§·½³ÌʽΪCH3COOH + CH3CH2OH CH3COOCH2CH3 + H2O¡£¶þÑõ»¯ÁòÊÇËáÐÔÑõ»¯ÎΪ³ýÈ¥ÒÒÏ©ÖпÉÄÜ»ìÓеÄSO2ӦѡÓõÄÊÔ¼ÁÊÇNaOHÈÜÒº¡£

£¨4£©ÒÑÖªDµÄÏà¶Ô·Ö×ÓÁ¿Îª118£¬ÆäÖÐ̼¡¢ÇâÁ½ÔªËصÄÖÊÁ¿·ÖÊý·Ö±ðΪ40.68%¡¢5.08%£¬ÆäÓàΪÑõÔªËØ£¬118¡Á40.68%¡Â12=4£¬118¡Á5.08%¡Â1=6£¬(118-48-6) ¡Â16=4£¬ÔòDµÄ·Ö×ÓʽΪC4H6O4¡£

£¨5£©µí·ÛÔÚËáÐÔÌõ¼þÏÂÉú³ÉÆÏÌÑÌÇ£¬Ö¤Ã÷·´Ó¦¢Ù·¢ÉúÓÃ10%µÄNaOHÈÜÒºµ÷½ÚÈÜÒºÖÁÖÐÐÔ£¬ÔÙÏòÆäÖмÓÈë2 mL 10%µÄNaOHÈÜÒº£¬ÔÙ¼ÓÈë4~5µÎ5%µÄCuSO4ÈÜÒº£¬¼ÓÈÈÒ»¶Îʱ¼ä¡£ÈôÓÐשºìÉ«³Áµí£¬ÔòÖ¤Ã÷·´Ó¦¢ÙÒÑ·¢Éú

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø