ÌâÄ¿ÄÚÈÝ

ç²ÝËáÊǺϳÉÖÎÁÆÇÝÁ÷¸ÐµÄÒ©Î¡ª´ï·Æ£¨Tamiflu£©µÄÔ­ÁÏÖ®Ò»¡£Ã§²ÝËáÊÇAµÄÒ»ÖÖÒì¹¹Ìå¡£AµÄ½á¹¹¼òʽÈçÏ£º

£¨Ìáʾ£º»·¶¡Íé¿É¼òд³É£©

£¨1£©AµÄ·Ö×ÓʽÊÇ_________________________________________________¡£

£¨2£©AÓëäåµÄËÄÂÈ»¯Ì¼ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¨ÓлúÎïÓýṹ¼òʽ±íʾ£©ÊÇ___________£»

£¨3£©AÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¨ÓлúÎïÓýṹ¼òʽ±íʾ£©ÊÇ_______________£»

£¨4£©17.4¿ËAÓë×ãÁ¿Ì¼ËáÇâÄÆÈÜÒº·´Ó¦£¬ÔòÉú³É¶þÑõ»¯Ì¼µÄÌå»ý£¨±ê×¼×´¿ö£©Îª______L¡£

£¨5£©AÔÚŨÁòËá×÷ÓÃϼÓÈȿɵõ½B£¨BµÄ½á¹¹¼òʽΪ£©£¬Æä·´Ó¦ÀàÐÍÊÇ______________¡£

£¨6£©BµÄͬ·ÖÒì¹¹ÌåÖмȺ¬ÓзÓôÇ»ùÓÖº¬ÓÐõ¥»ùµÄ¹²ÓÐ________ÖÖ£¬Ð´³öÆäÖÐÒ»ÖÖͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ_________________¡£

£¨1£©C7H10O5

£¨2£©

£¨3£©

£¨4£©¡Á22.4 L¡¤mol-1=2.24 L

£¨5£©ÏûÈ¥·´Ó¦

£¨6£©3

 »ò  »ò

½âÎö£º£¨1£©¸ù¾ÝÆðµãºÍ¹ÕµãÓÐCÔ­×Ó¼°Ì¼ËļÛÔ­Ôò£¬¿É½«AµÄÏß¼üʽ½á¹¹¼òʽ±ä»»Îª£¬¹ÊAµÄ·Ö×ÓʽΪC7H10O5¡££¨2£©Aº¬C=CÓëBr2·¢Éú¼Ó³É·´Ó¦£¬»¯Ñ§·½³ÌʽΪ¡££¨3£©AËùº¬C=CºÍ3¸ö´¼¡ªOH¾ù²»ÓëNaOH·´Ó¦£¬ÆäËùº¬¡ªCOOHÄÜÓëNaOH·´Ó¦£¬»¯Ñ§·½³ÌʽΪ¡££¨4£©n£¨A£©==0.1 mol£¬¹ØÏµÊ½A¡ªCO2¡ü£¬n£¨CO2£©=n£¨A£©=0.1 mol£¬V£¨CO2£©=0.1 mol¡Á22.4 L¡¤mol-1=2.24 L¡££¨5£©¶Ô±ÈA¡¢BµÄ½á¹¹£¬¿ÉÖª·¢ÉúÁË´¼¡ªOHµÄÏûÈ¥·´Ó¦¡££¨6£©BµÄͬ·ÖÒì¹¹Ìå½öº¬1¸ö¡¢Ò»¸ö·ÓôÇ»ùºÍ1¸ö±½»·£¬ºÍ·ÓôÇ»ùÔÚ±½»·ÉÏÓÐ3ÖÖÏà¶ÔλÖ㬹ʹ²ÓÐÈçÏÂ3ÖÖͬ·ÖÒì¹¹Ì壺»ò»ò¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø