ÌâÄ¿ÄÚÈÝ
(¹²16·Ö) ¢ÅÏÖÓÃÎïÖʵÄÁ¿Å¨¶ÈΪa mol/LµÄ±ê×¼ÑÎËáÈ¥²â¶¨V mL NaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È£¬ÇëÌîдÏÂÁпհףº
¢Ù ËáʽµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»ºó£¬»¹Ó¦¸Ã½øÐеIJÙ×÷ÊÇ___¡ø___¡£
¢Ú ÓÒͼÊÇËáʽµÎ¶¨¹ÜÖÐÒºÃæÔڵζ¨Ç°ºóµÄ¶ÁÊý£º
ÊÔÓÃÓйطûºÅ±íʾ¸Ã´ý²âNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È£ºc (NaOH) = ___¡ø___ mol/L¡£
¢Û ÈôÔڵζ¨Ç°µÎ¶¨¹Ü¼â×첿·ÖÁôÓÐÆøÅÝ£¬µÎ¶¨ºóµÎ¶¨¹Ü¼â×첿·ÖÆøÅÝÏûʧ£¬Ôò²â¶¨µÄNaOHÎïÖʵÄÁ¿Å¨¶È»áÆ«___¡ø___¡£
¢ÆÒÑÖª¿ÕÆøÖк¬ÓÐN2¡¢O2¡¢CO2¡¢H2SµÈÆøÌå¡£ÊÔÅжÏÏÂÁе樲Ù×÷µÄÖյ㡢·ÖÎö³öÏÖÏÂÁÐÏÖÏóµÄÔÒò¡££¨»òÓÃÓйØÀë×Ó·½³Ìʽ±íʾ£©
¢Ù ÒÔ·Ó̪Ϊָʾ¼Á£¬¼îÒºµÎ¶¨ËáÒº£¬µ½ ¡ø ΪÖյ㡣30sºóµ¼ÖÂÍÊÉ«µÄÔÒò£º ¡ø ¡£
¢Ú ÒÔµí·ÛΪָʾ¼Á£¬ÓÃNa2S2O3µÎ¶¨I2(2S2O32¡ª+I2 £½ S4O62¡ª+2I¡ª)µ½ ¡ø ΪÖյ㣬Լ5minºóÈÜÒºÓÖÏÔÉ«µÄÔÒò£º ¡ø ¡£
£¨¹²16·Ö£©
¢Å£¨6·Ö£©¢ÙÓÃÒÑÖªÎïÖʵÄÁ¿Å¨¶ÈµÄ±ê×¼ÑÎËáÈóÏ´2µ½3´Î£¨2·Ö£©
¢Úc (NaOH) = (V2-V1)a/V £¨2·Ö£© ¢Û¸ß£¨2·Ö£©
¢Æ£¨10·Ö£© ¢ÙÏÔÏÖdzºìÉ«ÇÒÔÚ°ë·ÖÖÓÄÚ²»ÍÊÉ«£»£¨2·Ö£© ¿ÕÆøÖдæÔÚCO2 ¡¢H2S £¬ CO2+OH¡ª==HCO3¡ª£¬ H2S+OH¡ª==HS¡ª+H2O£¨4·Ö£¬ÐðÊöºÏÀí¼´¿É£¬ÏÂͬ£©
¢ÚÀ¶É«Ç¡ºÃÍÊÈ¥£»£¨2·Ö£© ¿ÕÆøÖдæÔÚO2£¬O2+4I¡ª+4H+===2H2O+I2£¨2·Ö£©
£¨Ã¿¿Õ2·Ö£¬¹²16·Ö£©¶ÌÖÜÆÚÔªËØX¡¢Y¡¢Z¡¢W£¬Ô×ÓÐòÊýÒÀ´ÎÔö´ó¡£³£Î³£Ñ¹Ï£¬Ö»ÓÐWµÄµ¥ÖÊÎªÆøÌå¡£ËüÃǵÄ×î¸ßÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÒÀ´ÎΪ¼×¡¢ÒÒ¡¢±û¡¢¶¡¡£¼×¡¢ÒÒ¡¢±ûÊÇÖÐѧ»¯Ñ§Öеij£¼ûÎïÖÊ£¬ÆäÖÐÖ»ÓÐÒÒÄÑÈÜÓÚË®£¬ÇÒÄܺͼס¢±û·´Ó¦µÃµ½³ÎÇåÈÜÒº¡£¸ù¾ÝÒÔÉÏÐÅÏ¢ÌîдÏÂÁпհףº
¢Å»³öWµÄÔ×ӽṹʾÒâͼ_________________________________________£»
¢Æ½«ÒҺͼס¢±û·Ö±ð·´Ó¦ºóµÃµ½µÄÈÜÒº»ìºÏ£¬¹Û²ìµ½µÄÏÖÏóÊÇ___________________
___________________ £¬Á½ÈÜÒº»ìºÏʱËù·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________________________________£»
¢ÇÏÂÁÐÊÂʵÄÜÖ¤Ã÷ZºÍW·Ç½ðÊôÐÔÇ¿ÈõµÄÊÇ£¨Ñ¡ÌîÐòºÅ£©__________________________£»
| A£®µ¥ÖʵÄÈ۵㣺Z£¾W2 |
| B£®ËáÐÔ£º¶¡£¾±û |
| C£®ÔÚÈÜÒºÖУºW2£«H2Z£½2HW£«Z |
| D£®Îȶ¨ÐÔ£ºHW£¾H2Z |
F.ÈܽâÐÔ£º¶¡£¾±û
¢ÈÓÃYµ¥ÖʺÍÉú»îÖÐ×î³£ÓõĽðÊô×÷µç¼«£¬Óõ¼ÏßÁ¬½Ó²åÈë¼×µÄÈÜÒºÖй¹³ÉÔµç³Ø£¬¸ÃÔµç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½Îª_______________________________________________ £»
¢É¹¤ÒµÉÏÒÔXWΪÔÁÏ¿ÉÒÔ½øÐÐÐí¶à»¯Éú²ú£¬¼×ºÍW2¶¼ÊÇÆäÖ÷Òª²úÆ·¡£Ð´³ö¹¤ÒµÉÏÒÔXWΪÔÁÏÉú²ú¼×ºÍW2µÄ»¯Ñ§·½³Ìʽ_______________________________________________
_______________________________________________£»ÈôÒªÉú²ú80.0 kg¼×ÎïÖÊ£¬ÖÁÉÙÐèÒªXW______________kg£¬Í¬Ê±¿ÉµÃW2_____________m3£¨±ê¿ö£©¡£
£¨¹²16·Ö£©¢ñ£®£¨1£©ÁòËá;§ÌåµÄÈܽâ¶ÈËä´ó£¬µ«Èܽâ¹ý³Ì½ÏÂý£¬ÊµÑéÊÒ³£ÓÃÈÈË®ÅäÖÆÒÔ¼Ó¿ìÈܽâËÙÂÊ£¬µ«³£³£»á²úÉú»ë×Ç£¬Çë¼òҪ˵Ã÷ÔÒò______________________£¬ÈçºÎ²ÅÄÜÓÃÈÈË®ÅäÖÆ³ö³ÎÇåµÄ½ÏŨµÄCuSO4ÈÜÒº____________________£»
£¨2£©Ï¡Na2SÈÜÒºÓÐÒ»ÖÖ³ô¼¦µ°ÆøÎ¶£¬¼ÓÈëAlCl3ÈÜÒººó£¬³ô¼¦µ°ÆøÎ¶¼Ó¾ç£¬ÓÃÀë×Ó·½³Ìʽ±íÊ¾ÆøÎ¶¼Ó¾ç¹ý³ÌËù·¢ÉúµÄ»¯Ñ§·´Ó¦______________________________________________
¢ò£®Ä³ÉÕ¼îÑùÆ·Öк¬ÓÐÉÙÁ¿²»ÓëËá×÷ÓõĿÉÈÜÐÔÔÓÖÊ£¬ÎªÁ˲ⶨÆä´¿¶È£¬½øÐÐÒÔÏµζ¨²Ù×÷£º
| A£®½«ÈÜÒº×ªÒÆÖÁ250 mLÈÝÁ¿Æ¿ÖУ¬¼ÓË®ÖÁ¿Ì¶ÈÏߣ» |
| B£®ÓÃÒÆÒº¹Ü(»ò¼îʽµÎ¶¨¹Ü)Á¿È¡25.00 mLÉÕ¼îÈÜÒºÓÚ×¶ÐÎÆ¿Öв¢¼Ó¼¸µÎ¼×»ù³È×÷ָʾ¼Á£» |
| C£®ÔÚÌìÆ½ÉÏ׼ȷ³ÆÈ¡ÉÕ¼îÑùÆ·w g£¬ÔÚÉÕ±ÖмÓÕôÁóË®Èܽ⣻ |
| D£®½«ÎïÖʵÄÁ¿Å¨¶ÈΪm mol?L£1µÄ±ê×¼H2SO4ÈÜҺװÈëËáʽµÎ¶¨¹Ü£¬µ÷ÕûÒºÃæ£¬¼ÇÏ¿ªÊ¼¿Ì¶ÈÊýΪV1 mL£» |
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÕýÈ·µÄ²Ù×÷²½ÖèµÄ˳ÐòÊÇ(Ìîд×Öĸ)
________¡ú________¡ú________¡ú________¡ú________£»
(2)Öյ㵽´ïµÄÏÖÏóÊÇ________________________£»
(3)ÈôËáʽµÎ¶¨¹ÜûÓÐÓñê×¼H2SO4ÈóÏ´£¬¶Ô²â¶¨½á¹ûÓ°Ïì________£¬µÎ¶¨Ç°ÑöÊÓ¶ÁÊýµÎ¶¨ºó¸©ÊÓ¶ÁÊý¶Ô²â¶¨½á¹ûÓ°Ïì________ £»(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£¬ÆäËû²Ù×÷¾ùÕýÈ·)¡£
(4)¸ÃÉÕ¼îµÄ°Ù·Öº¬Á¿ÊÇ________¡£