ÌâÄ¿ÄÚÈÝ
X¡¢Y¡¢Z¡¢WΪº¬ÓÐÏàͬµç×ÓÊýµÄ·Ö×Ó»òÀë×Ó£¬¾ùÓÉÔ×ÓÐòÊýСÓÚ10µÄÔªËØ×é³É£¬XÓÐ5¸öÔ×Ӻˡ£Í¨³£×´¿öÏ£¬WΪÎÞÉ«ÒºÌå¡£ÒÑÖª:X+YZ+W
(1)YµÄµç×ÓʽÊÇ_________________________¡£
(2)Һ̬ZºÍWµÄµçÀëÏàËÆ£¬¶¼¿ÉµçÀë³öµç×ÓÊýÏàͬµÄÁ½ÖÖÀë×Ó£¬ÒºÌ¬ZµÄµçÀë·½³ÌʽÊÇ_________________________________¡£
(3)ÓÃͼʾװÖÃÖƱ¸NO²¢ÑéÖ¤Æ仹ÔÐÔ¡£ÓÐÏÂÁÐÖ÷Òª²Ù×÷£º
a.Ïò¹ã¿ÚÆ¿ÄÚ×¢Èë×ãÁ¿ÈÈNaOHÈÜÒº£¬½«Ê¢ÓÐÍƬµÄСÉÕ±·ÅÈëÆ¿ÖС£
b.¹Ø±Õֹˮ¼Ð£¬µãȼºìÁ×£¬ÉìÈëÆ¿ÖУ¬ÈûºÃ½ºÈû¡£
c.´ýºìÁ׳ä·ÖȼÉÕ£¬Ò»¶Îʱ¼äºó´ò¿ª·ÖҺ©¶·ÐýÈû£¬ÏòÉÕ±ÖеÎÈëÉÙÁ¿Ï¡ÏõËá¡£
¢Ù²½Öècºó»¹È±ÉÙµÄÒ»²½Ö÷Òª²Ù×÷ÊÇ_______________________________________¡£
¢ÚºìÁ׳ä·ÖȼÉյIJúÎïÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ ___________________________________________________________________¡£
¢Û²½ÖècµÎÈëÏ¡ÏõËáºóÉÕ±ÖеÄÏÖÏóÊÇ______________________________________
____________________________________________________________________¡£
·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ____________________________________________________¡£
(4)Ò»¶¨Î¶ÈÏ£¬½«1 mol N2O4ÖÃÓÚÃܱÕÈÝÆ÷ÖУ¬±£³Öѹǿ²»±ä£¬Éý¸ßζÈÖÁT1µÄ¹ý³ÌÖУ¬ÆøÌåÓÉÎÞÉ«Öð½¥±äΪºì×ØÉ«¡£Î¶ÈÓÉT1¼ÌÐøÉý¸ßµ½T2µÄ¹ý³ÌÖУ¬ÆøÌåÖð½¥±äΪÎÞÉ«¡£Èô±£³ÖT2£¬Ôö´óѹǿ£¬ÆøÌåÖð½¥±äΪºì×ØÉ«¡£ÆøÌåµÄÎïÖʵÄÁ¿nËæζÈT±ä»¯µÄ¹ØϵÈçͼËùʾ¡£
¢ÙζÈÔÚT1¡«T2Ö®¼ä£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_________________________¡£
¢ÚζÈÔÚT2¡«T3Ö®¼ä£¬ÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿ÊÇ£¨±£Áô1λСÊý£©______________¡£
£¨1£©
£¨2£©2NH3(I)+
(3)¢Ù´ò¿ªÖ¹Ë®¼Ð£¬Í¨ÈëÉÙÁ¿ÑõÆø
¢ÚP2O5+6OH-2+3H2O
¢ÛCuƬÖð½¥Èܽ⣬ÓÐÎÞÉ«ÆøÅݲúÉú£¬ÈÜÒºÓÉÎÞÉ«±äΪÀ¶É«
3Cu+8H++23Cu2++2NO¡ü+4H2O
(4)¢Ù2NO2 2NO+O2
¢Ú30.7
½âÎö£º¸ù¾ÝÌâÒÑÖªÌõ¼þ¿ÉÖª£¬XΪ£¬WΪH2O£¬ÓÉX+YZ+W¿ÉÖªYΪOH-£¬ZΪNH3£¬¼´
+OH-NH3+H2O
(2)Òº°±£¨NH3£©ÓëW(H2O)µçÀëÏàËÆ£¬²ÎÕÕË®µÄµçÀ룺2H2OH3O++OH-£¬Òº°±µÄµçÀë·½³ÌʽΪ£º2NH3(l)+¡£
(3)¢ÙÓÉÓÚ±¾ÊµÑéÄ¿µÄÊÇ¡°ÖƱ¸NO²¢ÑéÖ¤Æ仹ÔÐÔ¡±£¬Òò´ËÔÚ²½ÖèCÖƵÃNOºó»¹È±ÉÙµÄÒ»²½Ö÷Òª²Ù×÷ÊÇ£º´ò¿ªÖ¹Ë®¼Ð£¬·Å½øÒ»²¿·ÖÑõÆø£¬ÒÔÑéÖ¤NOµÄ»¹ÔÐÔ¡£
¢ÚºìÁ׳ä·ÖȼÉյIJúÎïΪP2O5£¬
P2O5+6NaOH2Na3PO4+3H2O£¬Àë×Ó·½³ÌʽΪ£º
P2O5+6OH-2+3H2O
¢ÛÏòÉÕ±ÖеÎÈëÏ¡HNO3ºó·¢ÉúµÄÏÖÏóΪ£ºÍƬÂýÂýÈܽ⣬ÓÐÎÞÉ«ÆøÅݲúÉú£¬ÈÜÒº±ä³ÉÀ¶É«£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
3Cu+2+8H+3Cu2++2NO¡ü+4H2O
(4)¢ÙζÈÓÉT1Éý¸ßµ½T2£¬ÆøÌåµÄÎïÖʵÄÁ¿Ôö¼Ó²¢±äΪÎÞÉ«£¬ËµÃ÷NO2·¢ÉúÁ˷ֽⷴӦ£¬Éú³ÉÎÞÉ«ÆøÌ壬½áºÏÖÊÁ¿Êغ㶨ÂÉ¿ÉÖª·´Ó¦·½³ÌʽΪ£º
2NO2(g)2NO(g)+O2(g)¡£
¢Ú¸ù¾Ý·´Ó¦¹ý³ÌÖÐÖÊÁ¿Êغ㣬ÔòT2¡ªT3ʱ»ìºÏÆøÌåÏà¶Ô·Ö×ÓÖÊÁ¿Îª£º=30.7 g¡¤mol-1£¬¼´Ïà¶Ô·Ö×ÓÖÊÁ¿Îª30.7¡£