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¿

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

¾«Ó¢¼Ò½ÌÍø