ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÈçͼΪʵÑéÊÒijŨÑÎËáÊÔ¼ÁÆ¿ÉϵıêÇ©£¬ÊÔ¸ù¾ÝÓйØÊý¾Ý»Ø´ðÏÂÁÐÎÊÌ⣺

¸ÃŨÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ ______ £®

È¡ÓÃÈÎÒâÌå»ýµÄ¸ÃÑÎËáÈÜҺʱ£¬ÏÂÁÐÎïÀíÁ¿Öв»ËæËùÈ¡Ìå»ýµÄ¶àÉÙ¶ø±ä»¯µÄÊÇ ______ £®

A.ÈÜÒºÖÐHClµÄÎïÖʵÄÁ¿ÈÜÒºµÄŨ¶È

C.ÈÜÒºÖеÄÊýÄ¿ÈÜÒºµÄÃܶÈ

ijѧÉúÓûÓÃÉÏÊöŨÑÎËáºÍÕôÁóË®ÅäÖÆ450mLÎïÖʵÄÁ¿Å¨¶ÈΪϡÑÎËᣮ

¸ÃѧÉúÐèÒªÁ¿È¡ ______ mLÉÏÊöŨÑÎËá½øÐÐÅäÖÆ£®

ÅäÖÆʱ£¬ÆäÕýÈ·µÄ²Ù×÷˳ÐòÊÇÓÃ×Öĸ±íʾ£¬Ã¿¸ö×ÖĸֻÄÜÓÃÒ»´Î ______ £»

A.ÓÃ30mLˮϴµÓÉÕ±­´Î£¬Ï´µÓÒº¾ù×¢ÈëÈÝÁ¿Æ¿£¬Õñµ´

B.ÓÃÁ¿Í²×¼È·Á¿È¡ËùÐèŨÑÎËáµÄÌå»ý£¬ÂýÂýÑر­±Ú×¢ÈëÊ¢ÓÐÉÙÁ¿Ë®Ô¼µÄÉÕ±­ÖУ¬Óò£Á§°ôÂýÂý½Á¶¯£¬Ê¹Æä»ìºÏ¾ùÔÈ

C.½«ÒÑÀäÈ´µÄÑÎËáÑز£Á§°ô×¢Èë500mLµÄÈÝÁ¿Æ¿ÖÐ

D.½«ÈÝÁ¿Æ¿¸Ç½ô£¬µßµ¹Ò¡ÔÈ

E.¸ÄÓýºÍ·µÎ¹Ü¼ÓË®£¬Ê¹ÈÜÒº°¼ÃæÇ¡ºÃÓë¿Ì¶ÈÏßÏàÇÐ

F.¼ÌÐøÍùÈÝÁ¿Æ¿ÄÚСÐļÓË®£¬Ö±µ½ÒºÃæ½Ó½ü¿Ì¶ÈÏß´¦

ÔÚÅäÖƹý³ÌÖУ¬ÏÂÁÐʵÑé²Ù×÷¶ÔËùÅäÖƵÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈÓкÎÓ°Ï죿Ìî¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£®

I¡¢ÓÃÁ¿Í²Á¿È¡Å¨ÑÎËáʱ¸©Êӹ۲찼ҺÃæ ______

II¡¢ÓÃÁ¿Í²Á¿È¡Å¨ÑÎËáºó£¬Ï´µÓÁ¿Í²´Î£¬Ï´µÓҺҲתÒƵ½ÈÝÁ¿Æ¿ ______

III¡¢ÈÜҺעÈëÈÝÁ¿Æ¿Ç°Ã»Óлָ´µ½ÊÒξͽøÐж¨ÈÝ ______

ÈôÔÚ±ê×¼×´¿öÏ£¬½«VLHClÆøÌåÈÜÓÚ1LË®ÖУ¬ËùµÃÈÜÒºÃܶÈΪd£¬Ôò´ËÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ ______ Ìî×Öĸ

A. B. C. D.

¡¾´ð°¸¡¿ 12 BD BCAFED Æ«µÍ Æ«¸ß Æ«¸ß A

¡¾½âÎö¡¿£¨1£©¸ù¾ÝÎïÖʵÄÁ¿Å¨¶ÈÓëÈÜÖÊÖÊÁ¿·ÖÊýµÄ»»Ë㹫ʽ¼ÆËã¡£

£¨2£©È¡ÓÃÈÎÒâÌå»ýµÄ¸ÃÑÎËáÈÜÒº£¬ÈÜÒºµÄŨ¶È¡¢ÈÜÒºµÄÃܶȲ»ËæËùÈ¡Ìå»ýµÄ¶àÉÙ¶ø±ä»¯£¬HClÎïÖʵÄÁ¿¡¢Cl-µÄÊýÄ¿¶¼ÓëÌå»ýµÄ¶àÉÙÓйء£

£¨3£©¢Ù¸ù¾Ýc£¨Å¨ÈÜÒº£©V£¨Å¨ÈÜÒº£©=c£¨Ï¡ÈÜÒº£©V£¨Ï¡ÈÜÒº£©¼ÆËã¡£

¢ÚÅäÖÆÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄʵÑé²½ÖèΪ£º¼ÆËã¡¢³ÆÁ¿£¨»òÁ¿È¡£©¡¢Èܽ⣨»òÏ¡ÊÍ£©²¢ÀäÈ´ÖÁÊÒΡ¢×ªÒÆ¡¢Ï´µÓ¡¢³õ²½Õñµ´¡¢¶¨ÈÝ¡¢µ¹×ªÒ¡ÔÈ¡¢×°Æ¿Ìù±êÇ©¡£

¢Û¸ù¾Ý¹«Ê½cB=½øÐÐÎó²î·ÖÎö¡£

£¨4£©ÓÉHClµÄÌå»ý¼ÆËãHClÎïÖʵÄÁ¿ºÍÖÊÁ¿£¬ÓÉÈÜÒºµÄÖÊÁ¿ºÍÃܶȼÆËãÈÜÒºµÄÌå»ý£¬×îºó¸ù¾Ýc£¨HCl£©=¼ÆËãÑÎËáÎïÖʵÄÁ¿Å¨¶È¡£

£¨1£©ÎïÖʵÄÁ¿Å¨¶ÈÓëÈÜÖÊÖÊÁ¿·ÖÊýµÄ»»Ë㹫ʽΪc=£¬Ôò¸ÃŨÑÎËáÎïÖʵÄÁ¿Å¨¶ÈΪ=12mol/L¡£

£¨2£©ÈÜÒºÊǾùÒ»¡¢Îȶ¨µÄ»ìºÏÎȡÓÃÈÎÒâÌå»ýµÄ¸ÃÑÎËáÈÜÒº£¬ÈÜÒºµÄŨ¶È¡¢ÈÜÒºµÄÃܶȲ»ËæËùÈ¡Ìå»ýµÄ¶àÉÙ¶ø±ä»¯£¬HClÎïÖʵÄÁ¿¡¢Cl-µÄÊýÄ¿¶¼ÓëÌå»ýµÄ¶àÉÙÓйأ¬´ð°¸Ñ¡BD¡£

£¨3£©¢ÙÅäÖÆ450mLÈÜҺӦѡÓÃ500mLÈÝÁ¿Æ¿¡£¸ù¾Ýc£¨Å¨ÈÜÒº£©V£¨Å¨ÈÜÒº£©=c£¨Ï¡ÈÜÒº£©V£¨Ï¡ÈÜÒº£©£¬Á¿È¡µÄŨÑÎËáµÄÌå»ýΪ=12.5mL¡£

¢ÚÅäÖÆÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄʵÑé²½ÖèΪ£º¼ÆËã¡¢³ÆÁ¿£¨»òÁ¿È¡£©¡¢Èܽ⣨»òÏ¡ÊÍ£©²¢ÀäÈ´ÖÁÊÒΡ¢×ªÒÆ¡¢Ï´µÓ¡¢³õ²½Õñµ´¡¢¶¨ÈÝ¡¢µ¹×ªÒ¡ÔÈ¡¢×°Æ¿Ìù±êÇ©£»ÅäÖÆʱ£¬ÕýÈ·µÄ²Ù×÷˳ÐòΪBCAFED¡£

¢Û¸ù¾Ý¹«Ê½cB=·ÖÎö¡£

I.ÓÃÁ¿Í²Á¿È¡Å¨ÑÎËáʱ¸©Êӹ۲찼ҺÃ棬ËùÁ¿È¡µÄŨÑÎËáÌå»ýÆ«µÍ£¬n£¨HCl£©Æ«µÍ£¬ËùÅäÖƵÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈÆ«µÍ¡£

II.ÓÃÁ¿Í²Á¿È¡Å¨ÑÎËáºó£¬Ï´µÓÁ¿Í²2~3´Î£¬Ï´µÓҺҲתÒƵ½ÈÝÁ¿Æ¿ÖУ¬n£¨HCl£©Æ«¸ß£¬ËùÅäÖƵÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈÆ«¸ß¡£

III.ŨÑÎËáÏ¡ÊÍʱʱ·ÅÈÈ£¬ÈÜҺעÈëÈÝÁ¿Æ¿Ç°Ã»Óлָ´µ½ÊÒξͽøÐж¨ÈÝ£¬ËùÅäÏ¡ÈÜÒºÌå»ýÆ«µÍ£¬ËùÅäÖƵÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈÆ«¸ß¡£

£¨4£©n£¨HCl£©==mol£»m£¨ÈÜÒº£©=m£¨HCl£©+m£¨H2O£©=mol36.5g/mol+1g/mL1000mL=g£¬V£¨ÈÜÒº£©=gdg/mL=mL=10-3L£»´ËÈÜÒºÎïÖʵÄÁ¿Å¨¶ÈΪmol£¨10-3L£©=mol/L£¬´ð°¸Ñ¡A¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø