ÌâÄ¿ÄÚÈÝ

1£®ÏÂÁÐÐðÊöÍêÈ«ÕýÈ·µÄÒ»×éÊÇ£¨¡¡¡¡£©
¢ÙÆøÌåĦ¶ûÌå»ýԼΪ22.4L•mol-1
¢Ú±ê×¼×´¿öÏ£¬22.4LË®ÖÐËùº¬·Ö×ÓÊýΪNA
¢Û100mL O.1mol•L-1µÄH2S03ÈÜÒºÖУ¬º¬ÓеÄÀë×ÓÊýԼΪO.03NA
¢Ü³£Î³£Ñ¹Ï£¬1.6gO3Ëùº¬µÄÔ­×ÓÊýΪNA
¢Ý9g D2OÖк¬Óеĵç×ÓÊýΪ5NA
¢Þº¤ÆøµÄĦ¶ûÖÊÁ¿Îª8g£®
A£®¢Ù¢Û¢Ü¢ÝB£®¢Ü¢Ý¢ßC£®¢ÜD£®¢Ù¢Ü

·ÖÎö ¢ÙÆøÌåĦ¶ûÌå»ýÓëËù´¦µÄ״̬Óйأ»
¢Ú±ê×¼×´¿öÏ£¬Ë®ÎªÒºÌ¬£»
¢ÛS032-µÄË®½â£¬»áʹÈÜÒºÖеÄÀë×Ó¸öÊýÔö¶à£»
¢ÜO3ÓÉÑõÔ­×Ó¹¹³É£»
¢ÝD2OµÄĦ¶ûÖÊÁ¿Îª20g/mol£»
¢ÞĦ¶ûÖÊÁ¿Îªg/mol£®

½â´ð ½â£º¢ÙÆøÌåĦ¶ûÌå»ýÓëËù´¦µÄ״̬Óйأ¬²»Ò»¶¨ÊÇ22.4L/mol£¬¹Ê´íÎó£»
¢Ú±ê×¼×´¿öÏ£¬Ë®ÎªÒºÌ¬£¬¹Ê²»Äܸù¾ÝÆøÌåĦ¶ûÌå»ýÀ´¼ÆËãÆäÎïÖʵÄÁ¿£¬¹Ê´íÎó£»
¢ÛS032-µÄË®½â£¬»áʹÈÜÒºÖеÄÀë×Ó¸öÊýÔö¶à£¬¹Êº¬ÓеÄÀë×ÓÊý¶àÓÚO.03NA£¬¹Ê´íÎó£»
¢ÜO3ÓÉÑõÔ­×Ó¹¹³É£¬¹Ê1.6g³ôÑõÖеÄÑõÔ­×ÓµÄÎïÖʵÄÁ¿n=$\frac{1.6g}{16g/mol}$=0.1mol£¬¸öÊýΪNA¸ö£¬¹ÊÕýÈ·£»
¢ÝD2OµÄĦ¶ûÖÊÁ¿Îª20g/mol£¬¹Ê9gÖØË®µÄÎïÖʵÄÁ¿n=$\frac{9g}{20g/mol}$=0.45mol£¬¶ø1molÖØË®Öк¬10molµç×Ó£¬¹Ê0.45molÖØË®Öк¬4.5molµç×Ó¼´4.5NA¸ö£¬¹Ê´íÎó£»
¢ÞĦ¶ûÖÊÁ¿Îªg/mol£¬º¤ÆøΪµ¥Ô­×Ó·Ö×Ó£¬¹Êº¤ÆøµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª4£¬¹Êº¤ÆøµÄĦ¶ûÖÊÁ¿Îª4g/mol£¬¹Ê´íÎó£®
¹ÊÑ¡C£®

µãÆÀ ±¾Ì⿼²éÁË°¢·üÙ¤µÂÂÞ³£ÊýµÄÓйؼÆË㣬ÊìÁ·ÕÆÎÕ¹«Ê½µÄʹÓúÍÎïÖʵĽṹÊǽâÌâ¹Ø¼ü£¬ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
14£®²ÝËáÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬³£ÓÃÓÚÓлúºÏ³É£¬ËüÊôÓÚ¶þÔªÈõËᣬÒ×ÈÜÓÚÒÒ´¼¡¢Ë®¡¢Î¢ÈÜÓÚÒÒÃÑ£®ÒÔ¸ÊÕáËéÔü£¨Ö÷Òª³É·ÖΪÏËάËØ£©ÎªÔ­ÁÏÖÆÈ¡²ÝËáµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º
ÒÑÖª£ºNO2+NO+2NaOH¨T2NaNO2+2H2O£¬2NO2+2NaOH¨T2NaNO2+H2O£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²½Öè¢ÙË®½âÒºÖÐËùº¬Ö÷ÒªÓлúÎïΪCH2OH£¨CHOH£©4CHO£¨Ð´½á¹¹¼òʽ£©£¬½þȡʱҪÏëÌá¸ßË®½âËÙÂʳý¿É¼õС¸ÊÕáËéÔüÁ£¶ÈÍ⣬»¹¿É²ÉÈ¡µÄ´ëÊ©ÊÇÊʵ±Éý¸ßζȣ»½þȡʱÁòËáŨ¶È²»Ò˹ý¸ß£¬ÆäÔ­ÒòÊǸÊÕáËéÔü»á±»Ì¿»¯£®
£¨2£©²½Öè¢Ú»áÉú³É²ÝËᣬÏõËáµÄ»¹Ô­²úÎïΪNO2ºÍNOÇÒn£¨NO2£©£ºn£¨NO£©=3£º1£¬Ôò·´Ó¦µÄÀë×Ó·½³ÌʽΪC6H12O6+12H++12NO3-$\frac{\underline{\;´ß»¯¼Á\;}}{\;}$3H2C2O4+9H2O+3NO¡ü+9NO2¡ü£¬Éú³ÉµÄNO2¼°NOÈô±»NaOHÈÜÒºÍêÈ«ÎüÊÕ£¬Ôò²úÎïÖеÄÁ½ÖÖÑÎn£¨NaNO2£©£ºn£¨NaNO3£©=3£º1£®
£¨3£©²½Öè¢ÛµÄÖ÷ҪĿµÄÊÇʹÆÏÌÑÌdzä·ÖÑõ»¯Îª²ÝËᣮ
£¨4£©Éè¼Æ²½Öè¢ßºÍ¢àµÄÄ¿µÄ·Ö±ðÊdzýÈ¥²ÝËá±íÃæË®·Ö¡¢³ýÈ¥ÒÒ´¼£¬Ê¹Æä¿ìËÙ¸ÉÔ
£¨5£©ÎªÁËÈ·¶¨²úÆ·ÖÐH2C2O4•2H2OµÄ´¿¶È£¬³ÆÈ¡10.5g²ÝËáÑùÆ·£¬Åä³É250mLÈÜÒº£¬Ã¿´ÎʵÑéʱ׼ȷÁ¿È¡20.00mL²ÝËáÈÜÒº£¬¼ÓÈëÊÊÁ¿µÄÏ¡ÁòËᣬÓÃ0.10mol•L-1µÄKMO4±ê×¼ÈÜÒºµÎ¶¨£®Æ½Ðеζ¨Èý´Î£¬ÏûºÄKMO4±ê×¼ÈÜÒºµÄÌå»ýƽ¾ùΪ26.00mL£®Ôò²úÆ·ÖÐH2C2O4•2H2OµÄÖÊÁ¿·ÖÊýΪ97.5%£®

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

¾«Ó¢¼Ò½ÌÍø