ÌâÄ¿ÄÚÈÝ

13£®»¯Ñ§ÓëÈËÃÇÉú»îÖÊÁ¿µÄÌá¸ßÒÔ¼°Éç»á·¢Õ¹ÓÐ×ÅÃÜÇеĹØÏµ£®
£¨1£©¢Ù±£»¤»·¾³ÒѾ­³ÉΪȫÈËÀàµÄ¹²Ê¶£®ÏÂÁÐÎïÖÊ»áÆÆ»µ³ôÑõ²ãµÄÊÇC£¨Ìî×Öĸ£©£®
a£®¶þÑõ»¯Ì¼    b£®¶þÑõ»¯Áò    c£®µªÑõ»¯Îï
¢Ú¹¤ÒµÉÏÖÆÔìË®Äà¡¢²£Á§ºÍ¸ß¯Á¶Ìú¶¼ÒªÓõ½µÄÔ­ÁÏÊÇb£¨Ìî×Öĸ£©£®
a£®´¿¼î    b£®Ê¯»Òʯ   c£®ð¤ÍÁ
¢Û¹âµ¼ÏËάµÄÖ÷Òª³É·ÖÊÇSiO2£¨Ìѧʽ£©£®
£¨2£©ÅÝÄ­ÂÁÊÇÒ»ÖÖÐÂÐͲÄÁÏ£¬ËüÊÇÔÚÈÛÈÚµÄÂÁºÏ½ðÖмÓÈë·¢ÅݼÁÖÆ³ÉµÄ£¬ÆäÓŵãÊÇÓ²¶È¸ß£¬ÃܶÈС£¬±Èľ²Ä»¹Çᣬ¿É¸¡ÓÚË®Ãæ£¬ÓÖÓкܴóµÄ¸ÕÐÔ£¬ÇÒ¸ôÒô¡¢±£Î£¬ÊÇÒ»ÖÖÁ¼ºÃµÄ½¨Öþ²ÄÁϺÍÇáÖʲÄÁÏ£®
¢ÙÏÂÁйØÓÚÅÝÄ­ÂÁµÄ˵·¨´íÎóµÄÊÇa£¨Ìî×Öĸ£©£®
a£®ÊÇ´¿¾»Îï    b£®¿ÉÓÃÀ´ÖÆÔì·É»ú    c£®¿ÉÓÃ×÷¸ôÈȲÄÁÏ
¢ÚÂÁÔªËØÔÚÈËÌåÖлýÀÛ¿ÉʹÈËÂýÐÔÖж¾£¬1989ÄêÊÀ½çÎÀÉú×éÖ¯Õýʽ½«ÂÁÈ·¶¨ÎªÊ³Æ·ÎÛȾԴ֮һ¶ø¼ÓÒÔ¿ØÖÆ£®ÂÁ¼°Æä»¯ºÏÎïÔÚÏÂÁг¡ºÏʹÓÃʱ±ØÐë¼ÓÒÔ¿ØÖƵÄÊÇb£¨Ìî×Öĸ£©£®
a£®ÖƵçÏßµçÀ    b£®ÓÃÃ÷·¯¾»Ë®    c£®ÖÆ·ÀÐâÓÍÆá
¢ÛÂÁÔÚ¿ÕÆøÖÐÄܱíÏÖ³öÁ¼ºÃµÄ¿¹¸¯Ê´ÐÔ£¬ÊÇÒòΪËüÓë¿ÕÆøÖеÄÑõÆø·´Ó¦Éú³ÉÖÂÃܵÄÑõ»¯Ä¤²¢Àι̵ظ²¸ÇÔÚÂÁ±íÃæ£¬×èÖ¹ÁËÄÚ²¿µÄÂÁÓë¿ÕÆø½Ó´¥£¬´Ó¶ø·ÀÖ¹ÂÁ±»½øÒ»²½Ñõ»¯£®
£¨3£©ÈçͼΪʵÑéÊÒÖÐÑÎËáÊÔ¼ÁÆ¿±êÇ©ÉϵIJ¿·ÖÄÚÈÝ£®ÊԻشðÏÂÁÐÎÊÌ⣺
¢ÙÏÂÁÐÓйØÅ¨ÑÎËáµÄ˵·¨ÖÐÕýÈ·µÄÊÇa£¨Ìî×Öĸ£©£®
a£®ÊôÓÚº¬ÑõËá    b£®ºÍÂÈ»¯ÇâµÄÐÔÖÊÏàͬ    c£®ºÜÈÝÒ×»Ó·¢
¢Ú¸ÃÑÎËáµÄc£¨HCl£©Îª11.9L mol/L£®
¢ÛÓûÓøÃÑÎËáÅäÖÆ1.19mol/LµÄÑÎËá480mL£®ÇëÍê³ÉÏÂÁÐÓйزÙ×÷ÖеĿհףº
a£®ÓÃÁ¿Í²×¼È·Á¿È¡¸ÃÑÎËá50.0 mL£¬×¢ÈëÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿µÄË®£¬»ìºÏ¾ùÔÈ£»
b£®½«²Ù×÷aËùµÃµÄÑÎËáÑØ²£Á§°ô×¢Èë500mLÈÝÁ¿Æ¿ÖУ»
c£®ÓÃÊÊÁ¿µÄˮϴµÓÉÕ±­¡¢²£Á§°ô2¡«3´Î£¬Ï´µÓÒº¾ù×¢ÈëÈÝÁ¿Æ¿ÖУ¬Õñµ´£»
d£®»º»ºµØ½«ÕôÁóË®×¢ÈëÈÝÁ¿Æ¿ÖУ¬Ö±µ½Æ¿ÖеÄÒºÃæ½Ó½üÈÝÁ¿Æ¿µÄ¿Ì¶ÈÏßl¡«2cm´¦£¬¸ÄÓýºÍ·µÎ¹Ü¼ÓÕôÁóË®ÖÁÈÜÒºµÄ°¼ÒºÃæÕýºÃÓë¿Ì¶ÈÏßÏàÇУ»
e£®½«ÈÝÁ¿Æ¿¸ÇºÃ£¬·´¸´ÉÏϵߵ¹Ò¡ÔÈ£®

·ÖÎö £¨1£©¢ÙµªÑõ»¯Îï¡¢·úÂȰº»áÆÆ»µ³ôÑõ²ã£»
¢ÚË®ÄàµÄÔ­ÁÏÊÇÕ³ÍÁºÍʯ»Òʯ£¬²£Á§µÄÔ­ÁÏÊÇ´¿¼î¡¢Ê¯»ÒʯºÍʯӢ£¬ËùÒÔÔ­ÁÏÖоùÓÐʯ»Òʯ¼´Ì¼Ëá¸Æ£»
¢Û¶þÑõ»¯¹èÊÇÖÆÔì¹âµ¼ÏËάÖ÷Òª³É·Ö£»
£¨2£©¢ÙÅÝÄ­ÂÁÖк¬ÓÐAl¡¢·¢ÅݼÁµÈ£¬ËùÒÔÊÇÒ»ÖֺϽð£¬ÒÀ¾ÝºÏ½ðÐÔÖʽáºÏÌâ¸ÉÐÅÏ¢ÅжϽâ´ð£»
¢ÚÂÁÊÇʳƷÎÛȾԴ֮һ£¬ËùÒÔÖ»ÒªÓëÈË¿ÚÇ»½Ó´¥µÄº¬ÓÐÂÁµÄÎïÖʾÍÐèÒª¼ÓÒÔ¿ØÖÆ£¬¾Ý´Ë·ÖÎö½â´ð£»
¢Û¸ù¾ÝÂÁµÄ»¯Ñ§ÐÔÖʽøÐзÖÎö£»
£¨3£©¢ÙÒÀ¾ÝÑÎËá³É·Ö¼°ÂÈ»¯Çâ×é³ÉÅжϽâ´ð£»
¢ÚÒÀ¾Ýc=$\frac{1000¦Ñ¦Ø}{M}$¼ÆË㣻
¢ÛÒÀ¾ÝÈÜҺϡÊ͹æÂɽáºÏ£¬ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÅäÖÆ·½·¨½â´ð£®

½â´ð ½â£º£¨1£©¢ÙµªÑõ»¯Îï¡¢·úÂȰº»áÆÆ»µ³ôÑõ²ã£¬¹Ê´ð°¸Îª£ºc£»
 ¢Ú¹¤ÒµÉÏÖÆÔìË®Äà¡¢²£Á§ºÍ¸ß¯Á¶Ìú¶¼ÒªÓõ½µÄÔ­ÁÏÊÇʯ»Òʯ£¬¹Ê´ð°¸Îª£ºb£»
¢Û¹âµ¼ÏËάµÄÖ÷Òª³É·ÖÊÇ£¬»¯Ñ§³É·ÖSiO2£¬¹Ê´ð°¸Îª£ºSiO2£»
£¨2£©¢ÙaÅÝÄ­ÂÁÊÇÒ»ÖֺϽð£¬ÊôÓÚ»ìºÏÎ¹Êa´íÎó£»
b£®ÒÀ¾ÝÅÝÄ­ÂÁÓŵãÊÇÓ²¶È¸ß£¬ÃܶÈС£¬¿ÉÖª¿ÉÓÃÀ´ÖÆÔì·É»ú£¬¹ÊbÕýÈ·£»
c£®ÒÀ¾ÝÌâÒâ¿ÉÖª£ºÅÝÄ­ÂÁÓ²¶È¸ß£¬ÃܶÈС£¬±Èľ²Ä»¹Çᣬ¿É¸¡ÓÚË®Ãæ£¬ÓÖÓкܴóµÄ¸ÕÐÔ£¬ÇÒ¸ôÒô¡¢±£Î£¬ÊÇÒ»ÖÖÁ¼ºÃµÄ½¨Öþ²ÄÁϺÍÇáÖʲÄÁÏ£¬¹ÊcÕýÈ·£»
¹ÊÑ¡£ºa£»
¢ÚÓÃÃ÷·¯¾»Ë®£¬ÓëÈË¿ÚÇ»½Ó´¥£¬ÐèÒª¿ØÖÆ£¬¹ÊbÕýÈ·£»
¹Ê´ð°¸Îª£ºb£»
¢Ûͨ³£Çé¿öÏÂÂÁÖÆÆ·ºÜÄ͸¯Ê´£¬ÕâÊÇÒòΪÂÁÔÚ³£ÎÂÏÂÓë¿ÕÆøÖеÄÑõÆø·¢Éú»¯Ñ§·´Ó¦£¬Æä±íÃæÉú³ÉÒ»²ãÖÂÃܵÄÑõ»¯ÂÁ±¡Ä¤£¬´Ó¶ø×èÖ¹ÂÁ½øÒ»²½Ñõ»¯£»
¹Ê´ð°¸Îª£ºÖÂÃܵÄÑõ»¯Ä¤£»
£¨3£©¢ÙaÂÈ»¯ÇâÖ»º¬Çâ¡¢ÂÈÔªËØ£¬²»º¬ÑõÔªËØ£¬¹Êa´íÎó£»
bÑÎËáÊôÓÚ»ìºÏÎïÄܹ»µçÀë²úÉúÇâÀë×Ó£¬ÂÈ»¯ÇâÊôÓÚ´¿¾»Îï²»ÄܵçÀ룬¶þÕßÐÔÖʲ»Í¬£¬¹Êb´íÎó£»
cÂÈ»¯ÇâÒ×»Ó·¢£¬¹ÊcÕýÈ·£»
¹ÊÑ¡£ºc£»
¢ÚŨÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈC=$\frac{1000¡Á1.19¡Á36.5%}{36.5}$=11.9g/mol£¬¹Ê´ð°¸Îª£º11.9£»
¢ÛaÓûÓøÃÑÎËáÅäÖÆ1.19mol/LµÄÑÎËá480mL£¬Ó¦Ñ¡Ôñ500mLÈÝÁ¿Æ¿£¬ÉèÐèҪŨÑÎËáÌå»ýΪV£¬ÔòV¡Á11.9mol/L=500mL¡Á1.19mol/L£¬½âµÃV=50.0mL£»
bÅäÖÆ480mLÈÜÒº£¬ÊµÑéÊÒûÓÐ480mLÈÝÁ¿Æ¿£¬Êµ¼ÊӦѡÔñ500mLÈÝÁ¿Æ¿£»
c¶¨ÈÝʱ£¬»º»ºµØ½«ÕôÁóË®×¢ÈëÈÝÁ¿Æ¿ÖУ¬Ö±µ½Æ¿ÖеÄÒºÃæ½Ó½üÈÝÁ¿Æ¿µÄ¿Ì¶ÈÏßl¡«2cm´¦£¬¸ÄÓýºÍ·µÎ¹Ü¼ÓÕôÁóË®ÖÁÈÜÒºµÄ°¼ÒºÃæÕýºÃÓë¿Ì¶ÈÏßÏàÇУ»
¹Ê´ð°¸Îª£º50.0£» 500£»½ºÍ·µÎ¹Ü£®

µãÆÀ ±¾Ì⿼²éÁËÉú»îÖг£¼ûµÄ»¯Ñ§ÖªÊ¶£¬Éæ¼°»·¾³ÎÛȾ¼°ÖÎÀí¡¢ÎÞ»ú·Ç½ðÊô²ÄÁÏ¡¢ºÏ½ðµÄÓÃ;ºÍÐÔÖÊ£¬Ò»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÅäÖÆ£¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
18£®Ç³ÂÌÉ«µÄÁòËáÑÇÌúï§¾§Ìå[ÓÖÃûζûÑΣ¨NH4£©2SO4•FeSO4•6H2O]±ÈÂÌ·¯£¨FeSO4•7H2O£©¸üÎȶ¨£¬³£ÓÃÓÚ¶¨Á¿·ÖÎö£®Äª¶ûÑεÄÒ»ÖÖʵÑéÊÒÖÆ·¨ÈçÏ£º
·ÏÌúм$\stackrel{ÇåÏ´}{¡ú}$$\stackrel{Ï¡ÁòËá}{¡ú}$ÈÜÒºA$¡ú_{ÊÊÁ¿Ë®}^{£¨NH_{4}£©_{2}SO_{4}¾§Ìå}$$\stackrel{²Ù×÷1}{¡ú}$$\stackrel{ÒÒ´¼ÁÜÏ´}{¡ú}$ζûÑÎ
£¨1£©Ïò·ÏÌúмÖмÓÈëÏ¡ÁòËáºó£¬²¢²»µÈÌúмÍêÈ«ÈÜ½â¶øÊÇÊ£ÓàÉÙÁ¿Ê±¾Í½øÐйýÂË£¬ÆäÄ¿µÄÊÇ·ÀÖ¹Fe2+Àë×Ó±»Ñõ»¯ÎªFe3+Àë×Ó£»Ö¤Ã÷ÈÜÒºA²»º¬Fe3+Àë×ÓµÄ×î¼ÑÊÔ¼ÁÊÇb£¨ÌîÐòºÅ×Öĸ£©£®
a£®·Ó̪ÈÜÒº      b£®KSCNÈÜÒº     c£®ÉÕ¼îÈÜÒº     d£®KMnO4ÈÜÒº
²Ù×÷IµÄ²½ÖèÊÇ£º¼ÓÈÈÕô·¢¡¢ÀäÈ´½á¾§¡¢¹ýÂË£®
£¨2£©½«Äª¶ûÑξ§Ìå·ÅÔÚÍÐÅÌÌìÆ½×óÅ̽øÐгÆÁ¿Ê±£¬ÌìÆ½Ö¸ÕëÏòÓÒÆ«×ª£¬ËµÃ÷íÀÂëÖØ£¬ÑùÆ·Çᣮ
£¨3£©ÎªÁ˲ⶨËùµÃζûÑÎÖÐFe2+µÄº¬Á¿£¬³ÆÈ¡4.0gζûÑÎÑùÆ·£¬ÈÜÓÚË®Åä³ÉÈÜÒº²¢¼ÓÈëÏ¡ÁòËᣬÓÃ0.2mol/LµÄKMnO4ÈÜÒº½øÐе樣¬µ½µÎ¶¨ÖÕµãʱ£¬ÏûºÄÁËKMnO4ÈÜÒº10.00mL£®ÔòÑùÆ·ÖÐFe2+µÄÖÊÁ¿·ÖÊýΪ14% £¨ÒÑÖª·´Ó¦ÖÐMnO4-±äΪMn2+£©£®
£¨4£©´ÓÏÂÁÐ×°ÖÃÖÐѡȡ±ØÒªµÄ×°ÖÃÖÆÈ¡£¨NH4£©2SO4ÈÜÒº£¬Á¬½ÓµÄ˳Ðò£¨ÓýӿÚÐòºÅ×Öĸ±íʾ£©ÊÇ£ºa½Ód£»e½Óf£®½«×°ÖÃCÖÐÁ½ÖÖÒºÌå·ÖÀ뿪µÄ²Ù×÷Ãû³ÆÊÇ·ÖÒº£®×°ÖÃDµÄ×÷ÓÃÊÇÎüÊÕ¶àÓàµÄNH3·ÀÖ¹ÎÛȾ¿ÕÆø£¬·ÀÖ¹µ¹Îü£®

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

¾«Ó¢¼Ò½ÌÍø