ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Æ»¹û´×ÊÇÒ»ÖÖÓÉÆ»¹ûÖ­·¢½Í¶ø³ÉµÄËáÐÔÒûÆ·£¬¾ßÓнⶾ¡¢½µÖ¬µÈҩЧ¡£ Æ»¹û´×ÊÇÒ»ÖÖ³£¼ûµÄÓлúËᣬÆä½á¹¹¼òʽΪ£º

£¨1£©Æ»¹û´×µÄ·Ö×ÓʽΪ___________________¡£

£¨2£©1 mol Æ»¹û´×Óë×ãÁ¿½ðÊôÄÆ·´Ó¦,ÄÜÉú³É±ê×¼×´¿öϵÄÇâÆø_________________L¡£

£¨3£©Æ»¹û´×¿ÉÄÜ·¢ÉúµÄ·´Ó¦ÓÐ___________________£¨ÌîÐòºÅ£©¡£

A£®Óë NaOH ÈÜÒº·´Ó¦ B£®ÓëʯÈïÈÜÒº×÷ÓÃ

C£®ÓëÒÒËáÔÚÒ»¶¨Ìõ¼þÏÂõ¥»¯ D£®ÓëÒÒ´¼ÔÚÒ»¶¨Ìõ¼þÏÂõ¥»¯

¡¾´ð°¸¡¿

£¨1£©C4H6O5£»£¨2£©33.6£»£¨3£©ABCD

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©Æ»¹û´×µÄ·Ö×ÓʽΪC4H6O5£¬¹Ê´ð°¸Îª£ºC4H6O5£»

£¨2£©-OH¡¢-COOH¾ùÓëNa·´Ó¦Éú³ÉÇâÆø£¬Ôò1molÆ»¹û´×Óë×ãÁ¿½ðÊôÄÆ·´Ó¦£¬ÄÜÉú³É±ê×¼×´¿öϵÄÇâÆøΪ1.5mol¡Á222.4L/mol=33.6L£¬¹Ê´ð°¸Îª£º33.6£»

£¨3£©A£®º¬-COOH£¬ÓëNaOHÈÜÒº·´Ó¦£¬¹ÊAÕýÈ·£»B£®º¬-COOH£¬¾ßÓÐËáÐÔ£¬ÓëʯÈïÈÜÒº×÷Ó㬹ÊBÕýÈ·£»C£®º¬-OH£¬ÓëÒÒËáÔÚÒ»¶¨Ìõ¼þÏÂõ¥»¯£¬¹ÊCÕýÈ·£»D£®º¬-COOH£¬ÓëÒÒ´¼ÔÚÒ»¶¨Ìõ¼þÏÂõ¥»¯£¬¹ÊDÕýÈ·£»¹Ê´ð°¸Îª£ºABCD¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø