ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÎªÁ˼ì²âÊìÈâÖÐNaNO2º¬Á¿£¬Ä³ÐËȤС×é´Ó1 000 g ¸ôÒ¹ÊìÈâÖÐÌáÈ¡NaNO3ºÍNaNO2ºóÅä³ÉÈÜÒº£¬ÓÃ0.005 00 mol¡¤L£­1µÄ¸ßÃÌËá¼Ø(ËáÐÔ)ÈÜÒºµÎ¶¨¡£Æ½ÐвⶨÈý´Î£¬Çó³öÿ´ÎNaNO2º¬Á¿£¬È¡Æäƽ¾ùÖµ¡£(ÒÑÖª£º2MnO£«5NO£«6H£«===2Mn2£«£«5NO£«3H2O)

£¨1£©µÎ¶¨Ç°ÅÅÆøÅÝʱ£¬Ó¦Ñ¡ÔñͼÖеÄ________(ÌîÐòºÅ)¡£µÎ¶¨Ê±£¬¸ßÃÌËá¼ØÈÜҺʢ·ÅÔÚ___________________¡£

£¨2£©µÎ¶¨ÖÕµãµÄÅжÏÒÀ¾ÝΪ__________________________________¡£

£¨3£©ÏÂÁвÙ×÷»áµ¼ÖÂÑùÆ·º¬Á¿²â¶¨ÖµÆ«¸ßµÄÊÇ__________________(ÌîÐòºÅ)¡£

a£®×¶ÐÎÆ¿ÓÃÕôÁóˮϴºóδÓôý²âÒºÈóÏ´

b£®ËáʽµÎ¶¨¹ÜÓÃÕôÁóˮϴºóδÓñê×¼ÒºÈóÏ´

c£®µÎ¶¨¹ý³ÌÖÐÕñµ´×¶ÐÎƿʱ£¬ÓÐÉÙÁ¿´ý²âÈÜÒº½¦³ö

d£®µÎ¶¨Ç°Æ½ÊÓ¶ÁÊý£¬µÎ¶¨½áÊøÑöÊÓ¶ÁÊý

£¨4£©Ä³´ÎµÎ¶¨¹ý³ÌÖУ¬ÏûºÄ¸ßÃÌËá¼ØÈÜÒºµÄÌå»ýΪ16.00 mL¡£Ôò´Ë´ÎÇóµÃµÄNaNO2µÄº¬Á¿Îª________mg¡¤kg£­1¡£

¡¾´ð°¸¡¿ ¢Ú ËáʽµÎ¶¨¹Ü µ±µÎÈë×îºóÒ»µÎ(»ò°ëµÎ)¸ßÃÌËá¼ØÈÜҺʱ£¬ÈÜÒºÑÕÉ«ÓÉÎÞÉ«±äΪ×ϺìÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊÉ«£¬¼´ÎªµÎ¶¨ÖÕµã bd 13.8

¡¾½âÎö¡¿ÊÔÌâ·ÖÎö£º£¨1£©¸ßÃÌËá¼ØÈÜÒº¸¯Ê´Ï𽺣¬Ó¦Ê¢ÔÚËáʽµÎ¶¨¹ÜÖУ¬Îª¾«È·¿ØÖÆÅųöÆøÅÝ£¬×óÊÖÎÕסËáʽµÎ¶¨¹Ü»îÈûÅÅÆø£»£¨2£©µÎÈëµÄ¸ßÃÌËá¼ØÈÜÒº²»ÔÙÍÊɫʱ´ïµ½µÎ¶¨Öյ㣻£¨3£©¸ù¾Ý ·ÖÎöÎó²î£»£¨4£©¸ù¾Ý2MnO£«5NO£«6H£«===2Mn2£«£«5NO£«3H2O¼ÆËãÏûºÄ¸ßÃÌËá¼ØÈÜÒºµÄÌå»ýΪ16.00 mLʱ£¬NaNO2µÄÖÊÁ¿¡£

½âÎö£º£¨1£©¸ßÃÌËá¼ØÈÜÒºÄܸ¯Ê´Ï𽺣¬ËùÒÔ¸ßÃÌËá¼ØÈÜҺʢÔÚËáʽµÎ¶¨¹ÜÖУ¬Îª¾«È·¿ØÖÆÅųöÆøÅÝ£¬×óÊÖÎÕסËáʽµÎ¶¨¹Ü»îÈûÅÅÆø£¬ËùÒÔӦѡÔñͼÖеĢڣ»£¨2£©µÎ¶¨ÖÕµãʱ£¬¸ßÃÌËá¼ØÈÜÒº±»ÍêÈ«ÏûºÄ£¬µ±µÎÈë×îºóÒ»µÎ(»ò°ëµÎ)¸ßÃÌËá¼ØÈÜҺʱ£¬ÈÜÒºÑÕÉ«ÓÉÎÞÉ«±äΪ×ϺìÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊÉ«£¬¼´ÎªµÎ¶¨Öյ㣻£¨3£©a£®×¶ÐÎÆ¿ÓÃÕôÁóˮϴºóδÓôý²âÒºÈóÏ´£¬²»Ó°Ïì½á¹û£¬¹Êa´íÎó£»b£®ËáʽµÎ¶¨¹ÜÓÃÕôÁóˮϴºóδÓñê×¼ÒºÈóÏ´£¬±ê׼ҺŨ¶È±äС£¬ÏûºÄ¸ßÃÌËá¼ØÈÜÒºµÄÌå»ýÆ«´ó£¬²â¶¨ÖµÆ«¸ß£¬¹ÊbÕýÈ·£»c£®µÎ¶¨¹ý³ÌÖÐÕñµ´×¶ÐÎƿʱ£¬ÓÐÉÙÁ¿´ý²âÈÜÒº½¦³ö£¬ÏûºÄ¸ßÃÌËá¼ØÈÜÒºµÄÌå»ýƫС£¬²â¶¨ÖµÆ«µÍ£¬¹Êc´íÎó£»d£®µÎ¶¨Ç°Æ½ÊÓ¶ÁÊý£¬µÎ¶¨½áÊøÑöÊÓ¶ÁÊý£¬¸ßÃÌËá¼ØÈÜÒºµÄÌå»ýÆ«´ó£¬²â¶¨ÖµÆ«¸ß£¬¹ÊdÕýÈ·£»£¨4£©ÏûºÄ¸ßÃÌËá¼ØÈÜÒºµÄÌå»ýΪ16.00 mL£¬n(KMnO4)=0.016L¡Á0.005 00 mol¡¤L£­1=0.00008mol£¬¸ù¾Ý2MnO£«5NO£«6H£«===2Mn2£«£«5NO£«3H2O£¬n(NaNO2)= 0.00008mol¡Â2¡Á5=0. 0002mol£¬ NaNO2µÄÖÊÁ¿ÊÇ0.0138g=13.8mg£¬NaNO2µÄº¬Á¿Îª13.8mg¡¤kg£­1¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿¼×´¼ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬¹¤ÒµÉÏ¿ÉÓÃCOºÍH2ºÏ³É¼×´¼£¨´ß»¯¼ÁΪCu2O/ZnO£©£ºCO(g)+2H2(g) CH3OH(g) ¦¤H = -90.8 kJ¡¤mol-1

£¨1£©Ä³Ñо¿Ð¡×飬ÔÚζȺÍÈÝÆ÷µÄÈÝ»ý±£³Ö²»±äʱ£¬Ñо¿¸Ã·´Ó¦ÊÇ·ñ´ïµ½Æ½ºâ״̬£¬ÒÔÏÂÎåλͬѧ¶Ôƽºâ״̬µÄÃèÊöÕýÈ·µÄÊÇ__________¡£

A.¼×£ºH2µÄÏûºÄËÙÂʵÈÓÚCH3OHµÄÉú³ÉËÙÂÊ

B.ÒÒ£ºÈÝÆ÷ÄÚµÄѹǿ±£³Ö²»±ä

C.±û£ºÈÝÆ÷ÄÚÆøÌåµÄÃܶȱ£³Ö²»±ä

D.¶¡£ºÈÝÆ÷ÖÐÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿±£³Ö²»±ä

£¨2£©ÈôÔÚÒ»¶¨Î¶ÈÏ¡¢ÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬Í¶Èë2 mol CO¡¢4 mol H2£¬±£³ÖºãΡ¢ºãÈÝ£¬²âµÃ·´Ó¦´ïµ½Æ½ºâʱH2ת»¯ÂÊΪ75%£¬ÇóƽºâʱCH3OHµÄŨ¶È=__________mol/L

£¨3£©ÒÑÖªÔÚ298K¡¢101kPaÏ£º¼×´¼µÄȼÉÕÈÈ»¯Ñ§·½³ÌʽΪ£º

CH3OH(l)+ 3/2 O2(g)= CO2(g)+ 2H2O(l) ¦¤H= -725.8 kJ¡¤mol-1£»

COµÄȼÉÕÈÈ»¯Ñ§·½³ÌʽΪ£ºCO(g)+ 1/2 O2(g)= CO2(g) ¦¤H= -283 kJ¡¤mol-1;

H2O (g) === H2O(l) ¦¤H= -44 kJ¡¤mol-1

Çëд³ö¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍË®ÕôÆøµÄÈÈ»¯Ñ§·½³Ìʽ__________¡£

£¨4£©ÒÔ¼×´¼ÎªÈ¼ÁÏ£¬ÑõÆøΪÑõ»¯¼Á£¬KOHÈÜҺΪµç½âÖÊÈÜÒº£¬¿ÉÖƳÉȼÁϵç³Ø£¨µç¼«²ÄÁÏΪ¶èÐԵ缫£©

¢Ù¸Ãµç³ØÖÐͨÈë¼×´¼µÄÊǵçÔ´µÄ__________£¨ÌîÕý¡¢¸º£©¼«£¬ÈôKOHÈÜÒº×ãÁ¿£¬Ôòд³öµç³Ø×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ__________¡£

¢ÚÈôµç½âÖÊÈÜÒºÖÐKOHµÄÎïÖʵÄÁ¿Îª0.5mol£¬µ±ÓÐ0.5mol¼×´¼²ÎÓ뷴Ӧʱ£¬²úÎïÇ¡ºÃΪKHCO3ʱ£¬¸Ãµç³ØµÄ¸º¼«·´Ó¦Ê½Îª__________£¬ËùµÃÈÜÒºÖи÷ÖÖÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ__________¡£

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

¾«Ó¢¼Ò½ÌÍø