ÌâÄ¿ÄÚÈÝ

1£®ÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£®ÏÂÁÐÐðÊöÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®6g SiO2Ëùº¬·Ö×ÓÊýΪ0.1NA£¬»¯Ñ§¼ü×ÜÊýΪ0.4NA
B£®½«2mL 0.5mol/L Na2SiO3ÈÜÒºµÎÈëÏ¡ÑÎËáÖÐÖÆµÃH2SiO3½ºÌ壬Ëùº¬½ºÁ£ÊýΪ0.001NA
C£®³£Î³£Ñ¹Ï£¬32gO2ºÍO3µÄ»ìºÏÆøÌåÖк¬ÓеÄÑõÔ­×ÓÊýΪ2NA
D£®25¡æÊ±£¬PH=13µÄBa£¨OH£©2ÈÜÒºÖк¬ÓÐOH-µÄÊýĿΪ0.1NA

·ÖÎö A¡¢¶þÑõ»¯¹èΪԭ×Ó¾§Ì壻
B¡¢Ò»¸ö¹èËὺÁ£ÊǶà¸ö¹èËá·Ö×ӵľۼ¯Ì壻
C¡¢ÑõÆøºÍ³ôÑõ¾ùÓÉÑõÔ­×Ó¹¹³É£»
D¡¢ÈÜÒºÌå»ý²»Ã÷È·£®

½â´ð ½â£ºA¡¢¶þÑõ»¯¹èΪԭ×Ó¾§Ì壬¹ÊÎÞ¶þÑõ»¯¹è·Ö×Ó£¬¹ÊA´íÎó£»
B¡¢Ò»¸ö¹èËὺÁ£ÊǶà¸ö¹èËá·Ö×ӵľۼ¯Ì壬¹ÊÐγɵĹèËὺÁ£µÄ¸öÊýСÓÚ¹èËá·Ö×ӵĸöÊý£¬¼´Ð¡ÓÚ0.001NA¸ö£¬¹ÊB´íÎó£»
C¡¢ÑõÆøºÍ³ôÑõ¾ùÓÉÑõÔ­×Ó¹¹³É£¬¹Ê32g»ìºÏÎïÖк¬ÓеÄÑõÔ­×ÓµÄÎïÖʵÄÁ¿Îª2mol£¬¼´2NA¸ö£¬¹ÊCÕýÈ·£»
D¡¢ÈÜÒºÌå»ý²»Ã÷È·£¬¹ÊÈÜÒºÖеÄÇâÑõ¸ùµÄ¸öÊýÎÞ·¨¼ÆË㣬¹ÊD´íÎó£®
¹ÊÑ¡C£®

µãÆÀ ±¾Ì⿼²éÁ˰¢·üÙ¤µÂÂÞ³£ÊýµÄÓйؼÆËã£¬ÕÆÎÕÎïÖʵÄÁ¿µÄ¼ÆË㹫ʽºÍÎïÖʽṹÊǽâÌâ¹Ø¼ü£¬ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
13£®îâËáÄÆ£¨Na2MoO4£©¾ßÓй㷺µÄÓÃ;£®¿É×öÐÂÐÍË®´¦Àí¾£¡¢ÓÅÁ¼µÄ½ðÊô»ºÊ´¼Á¼°¿ÉÓÃÓÚ¾Ö²¿¹ýÈȵÄÑ­»·Ë®ÏµÍ³£»Al£¨OH£©3¹¤ÒµºÍÒ½Ò©É϶¼¾ßÓÐÖØÒªÓÃ;£®ÏÖ´Óij·Ïîâ´ß»¯¼Á£¨Ö÷Òª³É·ÖMoO3¡¢Al2O3¡¢Fe2O3µÈ£©ÖлØÊÕNa2MoO4ºÍAl£¨OH£©3£¬Æä¹¤ÒÕÈçͼ£º

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑÖªMoO3¡¢Al2O3ÓëSiO2ÏàËÆ£¬¾ùÄÜÔÚ¸ßÎÂϸúNa2CO3·¢ÉúÀàËÆ·´Ó¦£¬ÊÔд³öMoO3ÓëNa2CO3·´Ó¦µÄ»¯Ñ§·½³Ìʽ£ºMoO3+Na2CO3$\frac{\underline{\;\;¡÷\;\;}}{\;}$Na2MoO4+CO2¡ü£®
£¨2£©µÚ¢Ú²½²Ù×÷ËùºóµÄÂËÒºÖУ¬ÈÜÖÊÓÐNa2MoO4¡¢NaAlO2ºÍ¹ýÁ¿µÄNa2CO3£»¼ìÑéµÚ¢Ú²½²Ù×÷ËùµÃÂËÔüÖк¬ÓÐÈý¼ÛÌúµÄ·½·¨ÊÇÈ¡ÉÙÁ¿ÂËÔüÏ´µÓÒºÓÚÊÔ¹ÜÖеμӼ¸µÎKSCNÈÜÒº£¬ÈôÈÜÒº±äºì£¬ÔòÖ¤Ã÷º¬Èý¼ÛÌú£®
£¨3£©µÚ¢Û²½²Ù×÷H2SO4ÐèÒªÊÊÁ¿£¬Í¨³£ÊÇͨ¹ý²âÈÜÒºµÄpHÀ´µ÷¿ØH2SO4µÄÓÃÁ¿£»ÓëNa2SO4Ïà±È£¬Na2MoO4µÄÈܽâ¶ÈÊÜζȵÄÓ°Ïì±ä»¯½ÏС£¨Ìî¡°½Ï´ó¡±»ò¡°½ÏС¡±£©£®
£¨4£©ÀûÓÃÂÁÈÈ·´Ó¦¿É»ØÊÕ½ðÊôî⣮½«ËùµÃîâËáÄÆÈÜÒºÓÃËá´¦ÀíµÃµ½³Áµí£¬ÔÙ¼ÓÈȿɵÃMoO3£®Ð´³öMoO3·¢ÉúÂÁÈÈ·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º4Al+2MoO3$\frac{\underline{\;¸ßÎÂ\;}}{\;}$2Mo+2Al2O3£®
£¨5£©È¡·Ïîâ´ß»¯¼Á5.00g£¬¼ÓÈë5.30gNa2CO3£¨×ãÁ¿£©£¬¾­ÉÏÊöʵÑé²Ù×÷ºó£¬×îÖյõ½2.34g¡¡Al£¨OH£©3ºÍ6.39gNa2SO4¾§Ì壬Ôò·Ïîâ´ß»¯¼ÁÖÐAl2O3¡¢MoO3µÄÎïÖʵÄÁ¿µÄ±ÈֵΪ3£º1£®

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

¾«Ó¢¼Ò½ÌÍø