ÌâÄ¿ÄÚÈÝ

ÓÉÆÏÌÑÌÇ·¢½Í¿ÉµÃÈéËᣬËáÅ£ÄÌÖÐÒ²ÄÜÌáÈ¡ÈéËᣬ´¿¾»ÈéËáΪÎÞÉ«ð¤³íÒºÌå

£¬Ò×ÈÜÓÚË®¡£ÎªÁËÑо¿ÈéËáµÄ·Ö×Ó×é³ÉºÍ½á¹¹£¬½øÐÐÏÂÊöʵÑ飺

(1)³ÆÈ¡ÈéËá0.90 g£¬ÔÚijÖÖ×´¿öÏÂʹÆäÍêÈ«Æû»¯£¬ÒÑÖªÏàͬ״¿öÏÂͬÌå»ýÇâÆøΪ0.02 g£¬ÔòÈéËáµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª_________________¡£

(2)Èô½«ÉÏÊöÈéËáÕôÆøÔÚO2ÖÐȼÉÕÖ»Éú³ÉCO2ºÍH2O(g)£¬µ±È«²¿±»¼îʯ»ÒÎüÊÕʱ£¬¼îʯ»ÒÔöÖØ1.86 g¡£Èô½«´ËÆøÌåͨ¹ý×ãÁ¿Ê¯»ÒË®ºó£¬Ôòʯ»ÒË®»á²úÉú3.00 g °×É«³Áµí£¬ÔòÈéËáµÄ·Ö×ÓʽΪ_________________¡£

(3)ÈéËá·Ö×ÓÄÜ·¢Éú×ÔÉíõ¥»¯·´Ó¦£¬Æä´ß»¯Ñõ»¯Îï²»ÄÜ·¢ÉúÒø¾µ·´Ó¦£¬ÆÏÌÑÌÇ·¢½ÍÖ»Éú³ÉÈéËᣬÊÔд³öÆÏÌÑÌÇÉú³ÉÈéËáµÄ»¯Ñ§·½³Ìʽ£º________________________________________¡£

(4)д³öÈéËáÔÚ´ß»¯¼Á×÷ÓÃÏ·¢Éú·´Ó¦£¬Éú³É·Ö×ÓʽΪC6H8O4µÄ»·×´õ¥µÄ½á¹¹¼òʽ£º

__________________________________¡£

(1)90

(2)C3H6O3

(3)CH2OH(CHOH)4CHO¡ú2CH3CH(OH)COOH

(4)

½âÎö£º(1)ÈéËáµÄÏà¶Ô·Ö×ÓÖÊÁ¿ÓÉ°¢·ü¼ÓµÂÂÞ¶¨ÂÉ¿ÉÍƳö£º

 

(2)0.01  molÈéËáÍêȫȼÉÕÉú³ÉCO2 0.03  mol£¬Éú³ÉH2O£º

ËùÒÔÒ»¸öÈéËá·Ö×ÓÖÐÓÐ3¸ö̼ԭ×ÓºÍ6¸öHÔ­×Ó£¬ÓÖÒòÆäÏà¶Ô·Ö×ÓÖÊÁ¿Îª90£¬¹Êº¬ÑõÔ­×ÓÊýΪ=3£¬ËùÒÔÈéËáµÄ·Ö×ÓʽΪC3H6O3¡£

(3)ÓÉÈéËáµÄ·Ö×ÓʽºÍÌâ¸øÐÅÏ¢Íƶϣ¬ÈéËáµÄ½á¹¹¼òʽΪ£¬¹ÊÆÏÌÑÌÇ·¢½ÍÉú³ÉÈéËáµÄ»¯Ñ§·½³ÌʽΪ£ºCH2OH(CHOH)4CHO¡ú2CH3CH(OH)COOH

(4)  

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø