ÌâÄ¿ÄÚÈÝ
£¨17·Ö£©ÓлúÎïAÓÉ̼¡¢Çâ¡¢ÑõÈýÖÖÔªËØ×é³É£¬¿ÉÓÉÆÏÌÑÌÇ·¢½ÍµÃµ½£¬Ò²¿É´ÓËáÅ£ÄÌÖÐÌáÈ¡£¬´¿¾»µÄAΪÎÞÉ«Õ³³íÒºÌ壬Ò×ÈÜÓÚË®¡£ÎªÑо¿AµÄ×é³ÉÓë½á¹¹£¬½øÐÐÁËÈçÏÂʵÑ飺
£¨1£©³ÆÈ¡A 9.0g£¬ÉýÎÂʹÆäÆû»¯£¬²âÆäÃܶÈÊÇÏàͬÌõ¼þÏÂH2µÄ45±¶¡£ | £¨1£©ÓлúÎïAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª |
£¨2£©½«´Ë9.0gAÔÚ×ãÁ¿´¿O2³ä·ÖȼÉÕ£¬²¢Ê¹Æä²úÎïÒÀ´Îͨ¹ý¼îʯ»Ò¡¢ÎÞË®ÁòËáÍ·ÛÄ©¡¢×ãÁ¿Ê¯»ÒË®£¬·¢ÏÖ¼îʯ»ÒÔöÖØ14.2g£¬ÁòËáÍ·ÛĩûÓбäÀ¶£¬Ê¯»ÒË®ÖÐÓÐ10.0g°×É«³ÁµíÉú³É£»ÏòÔöÖصļîʯ»ÒÖмÓÈë×ãÁ¿ÑÎËáºó£¬²úÉú4.48LÎÞÉ«ÎÞζÆøÌ壨±ê×¼×´¿ö£©¡£ | £¨2£©9.0gÓлúÎïAÍêȫȼÉÕʱ£¬¾¼ÆË㣺 Éú³ÉCO2¹²Îª mol£¬ Éú³ÉµÄH2O g£¬ ÓлúÎïAµÄ·Ö×Óʽ ¡£ |
£¨3£©¾ºìÍâ¹âÆײⶨ£¬Ö¤ÊµÆäÖк¬ÓÐ-OH¼ü£¬-COOH»ùÍÅ£¬C-H¼ü£»ÆäºË´Å¹²ÕñÇâÆ×ÓÐËÄ×é·å£¬Ãæ»ý±ÈΪ1©s3©s1©s1¡£ | £¨3£©AµÄ½á¹¹¼òʽ |
£¨4£©¾ºìÍâ¹âÆײⶨ£¬AµÄÒ»ÖÖͬ·ÖÒì¹¹ÌåÖУ¬´æÔÚ-OH¼ü£¬»¹º¬ÓÐÈ©»ù£¬C-O¼ü£»ÆäºË´Å¹²ÕñÇâÆ×ÓÐÎå×é·å£¬Ãæ»ý±ÈΪ1©s2©s1©s1©s1¡£ | £¨4£©AµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ |
£¨5£©Èç¹û¾ºìÍâ¹âÆײⶨ£¬AµÄÒ»ÖÖͬ·ÖÒì¹¹ÌåÖУ¬´æÔÚ-OH¼ü£¬»¹º¬ÓÐC=O£¬C-O¼ü£» ÆäºË´Å¹²ÕñÇâÆ×ÓÐÁ½×é·å£¬Ãæ»ý±ÈΪ1©s2¡£ | £¨5£©AµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ |
£¨1£©90 £¨2£©0.3 5.4 C3H6O3
£¨3£©CH3CH (OH) COOH £¨4£©OHCCH(OH)CH2OH £¨5£©HOCH2COCH2OH
½âÎö
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿