ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿ÏÂͼ±íʾ¼¸ÖÖÎÞ»úÎïÖ®¼äµÄת»¯¹Øϵ£¬ÆäÖÐA¡¢B¾ùΪºÚÉ«·ÛÄ©£¬BΪ·Ç½ðÊôµ¥ÖÊ£¬CΪÎÞÉ«ÎÞ¶¾ÆøÌ壬DΪ½ðÊôµ¥ÖÊ£¬EÊǺì×ØÉ«ÆøÌ壬GÊǾßÓÐƯ°×ÐÔµÄÆøÌ壬HµÄË®ÈÜÒº³ÊÀ¶É«¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)AµÄ»¯Ñ§Ê½ÊÇ___£¬CµÄµç×ÓʽÊÇ¡¡__________¡¡£¬YµÄÃû³ÆÊÇ_____¡£
(2)·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽΪ¡¡__________________________¡£
(3)ʵÑéÊҲⶨÆøÌåEµÄÏà¶Ô·Ö×ÓÖÊÁ¿Ê±£¬µÃµ½µÄʵÑéÖµ×ÜÊDZÈÀíÂÛֵƫ´ó£¬ÆäÔÒòÊÇ(Óû¯Ñ§·½³Ìʽ±íʾ)¡¡_________________¡£
(4)19.2gµÄDÓë×ãÁ¿µÄÒ»¶¨Å¨¶ÈXµÄÈÜÒº·´Ó¦£¬½«ËùµÃµÄÆøÌåÓë_______L O2(±ê×¼×´¿öÏÂ)»ìºÏ£¬Ç¡ºÃÄܱ»Ë®ÍêÈ«ÎüÊÕ¡£
¡¾´ð°¸¡¿CuO ŨÁòËá C£«4HNO3(Ũ)=CO2¡ü£«4NO2¡ü£«2H2O 2NO2N2O4 3.36L
¡¾½âÎö¡¿
ÒÑÖª£¬A¡¢B¾ùΪºÚÉ«·ÛÄ©£¬BΪ·Ç½ðÊôµ¥ÖÊ£¬CΪÎÞÉ«ÎÞ¶¾ÆøÌ壬DΪ½ðÊôµ¥ÖÊ£¬¿ÉÍƲâAΪCuO£¬BΪC£¬CΪ¶þÑõ»¯Ì¼£¬DΪCu£»EÊǺì×ØÉ«ÆøÌåΪ¶þÑõ»¯µª£¬GÊǾßÓÐƯ°×ÐÔµÄÆøÌåΪ¶þÑõ»¯Áò£¬HµÄË®ÈÜÒº³ÊÀ¶É«ÎªÁòËáÍ¡£
(1)·ÖÎö¿ÉÖª£¬AΪÑõ»¯Í£¬»¯Ñ§Ê½ÊÇCuO£»CΪ¶þÑõ»¯Ì¼£¬µç×ÓʽΪ£º£»YΪŨÁòË᣻
(2)·´Ó¦¢ÙΪ̼ÓëŨÏõËá·´Ó¦Éú³É¶þÑõ»¯Ì¼¡¢¶þÑõ»¯ÁòºÍË®£¬·½³ÌʽΪ£ºC£«4HNO3(Ũ)=CO2¡ü£«4NO2¡ü£«2H2O£»
(3)ÑéÊҲⶨÆøÌå¶þÑõ»¯µªµÄÏà¶Ô·Ö×ÓÖÊÁ¿Ê±£¬´æÔÚ¿ÉÄæ·´Ó¦2NO2N2O4µÃµ½µÄʵÑéÖµ×ÜÊDZÈÀíÂÛֵƫ´ó£»
(4)19.2gµÄCu¼´0.3mol£¬Óë×ãÁ¿µÄÒ»¶¨Å¨¶ÈµÄÏõËáÈÜÒº·´Ó¦£¬Éú³ÉµªµÄÑõ»¯Î»¯ºÏ¼Û½µµÍÓëCuÉý¸ßµÄ»¯ºÏ¼Û×ÜÊýÏàµÈ£¬ÔòÏûºÄµÄÑõÆø»¯ºÏ¼Û½µµÍµÄ×ÜÊýÒ²ÏàµÈ£¬ÏûºÄ0.15molÑõÆø¼´±ê¿öϵÄ3.36L¡£
¡¾ÌâÄ¿¡¿ÔÚÒ»¶¨Î¶ÈÏ£¬½«ÆøÌåXºÍÆøÌåY ¸÷0.16 mol³äÈë10 L ºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢Éú·´Ó¦
X(g) + Y(g) 2Z(g) H£¼0£¬Ò»¶Îʱ¼äºó´ïµ½Æ½ºâ£¬·´Ó¦¹ý³ÌÖвⶨµÄÊý¾ÝÈçÏÂ±í¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
t / min | 2 | 4 | 7 | 9 |
n(Y)/mol | 0.12 | 0.11 | 0.10 | 0.10 |
A£®·´Ó¦Ç°2minµÄƽ¾ùËÙÂÊv(Z)=2.0¡Á10¨C5 mol/(L¡¤min)
B£®ÆäËûÌõ¼þ²»±ä£¬½µµÍζȣ¬·´Ó¦´ïµ½ÐÂƽºâÇ°v(Äæ)£¾v(Õý)
C£®¸ÃζÈÏ´˷´Ó¦µÄƽºâ³£ÊýK=1.44
D£®ÆäËûÌõ¼þ²»±ä£¬ÔÙ³äÈë0.2 mol Z£¬Æ½ºâʱXµÄÌå»ý·ÖÊýÔö´ó