ÌâÄ¿ÄÚÈÝ

¶ÌÖÜÆÚµÄËÄÖÖÔªËØX¡¢Y¡¢Z¡¢W£¬Ô­×ÓϵÊýÒÀ´ÎÔö´ó£¬ZÔ­×Ó×îÍâ²ãµç×ÓÊýÊÇX¡¢Y¡¢WÈýÖÖÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÖ®ºÍ£¬ZÓëX¡¢Y¡¢WÈýÖÖÔªËØÐγÉÔ­×Ó¸öÊýÖ®±ÈΪ1£º1µÄ»¯ºÏÎï·Ö±ðÊÇA£®B£®C£¬ÆäÖл¯ºÏÎïCÔÚ¿ÕÆøÖÐÈÝÒ×±äÖÊ£¬Çë»Ø´ð£º
£¨1£©Ð´³öZµÄÔ­×ӽṹʾÒâͼ______________________________¡£
£¨2£©Ð´³ö»¯ºÏÎïYZ2µç×Óʽ£º______________¿Õ¼ä¹¹ÐÎΪ£º_______________£¬Ð´³ö»¯ºÏÎïCµÄµç×Óʽ£º_____________»¯Ñ§¼üÀàÐÍÓУº_____________¡£
£¨3£©Ð´³ö»¯ºÏÎïCÔÚ¿ÕÆøÖбäÖʵĻ¯Ñ§·½³Ìʽ£º____________________
(1) ;
(2) £»Ö±ÏßÐÍ £» £»Àë×Ó¼üºÍ¹²¼Û¼ü
(3)2Na2O2 + 2CO2 === 2Na2CO3 + O2  £»2Na2O2 + 2H2O === 4NaOH + O2¡ü
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2012?ÉϺ£Ä£Ä⣩X¡¢Y¡¢Z¡¢WÊǶÌÖÜÆÚµÄËÄÖÖÔªËØ£¬ÓйØËüÃǵÄÐÅÏ¢ÈçϱíËùʾ£®
ÔªËØ ²¿·Ö½á¹¹ÖªÊ¶ ²¿·ÖÐÔÖÊ
X XµÄµ¥ÖÊÓÉ˫ԭ×Ó·Ö×Ó¹¹³É£¬·Ö×ÓÖÐÓÐ14¸öµç×Ó XÓжàÖÖÑõ»¯ÎÈçXO¡¢XO2¡¢X2O4µÈ£»Í¨³£Çé¿öÏÂXO2ÓëX2O4¹²´æ
Y YÔ­×ӵĴÎÍâ²ãµç×ÓÊýµÈÓÚ×îÍâ²ãµç×ÓÊýµÄÒ»°ë YÄÜÐγɶàÖÖÆøÌ¬Ç⻯Îï
Z ZÔ­×ÓµÄ×îÍâ²ãµç×ÓÊý¶àÓÚ4 ZÔªËØµÄ×î¸ßÕý»¯ºÏ¼ÛÓë×îµÍ¸º»¯ºÏ¼Û´úÊýºÍµÈÓÚ6
W WÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýµÈÓÚ2n-3£¨nΪԭ×ÓºËÍâµç×Ó²ãÊý£© »¯Ñ§·´Ó¦ÖÐWÔ­×ÓÒ×ʧȥ×îÍâ²ãµç×ÓÐγÉWn+
ÌîдÏÂÁпհףº£¨Ìáʾ£º²»ÄÜÓÃ×ÖĸX¡¢Y¡¢Z¡¢W×÷´ð£©
£¨1£©Xµ¥ÖÊ·Ö×ӵĽṹʽÊÇ
£¬ZÔªËØÔ­×Ó×îÍâ²ã¹²ÓÐ
17
17
ÖÖ²»Í¬Ô˶¯×´Ì¬µÄµç×Ó£®
£¨2£©X¡¢Y¡¢ZÈýÔªËØµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïËáÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòÊÇ
HClO4£¾HNO3£¾H2CO3
HClO4£¾HNO3£¾H2CO3
£®
£¨3£©³£ÎÂʱ£¬WµÄÁòËáÑÎÈÜÒºµÄpH
£¼
£¼
7£¨Ìî¡°=¡±¡¢¡°£¾¡±»ò¡°£¼¡±£©£¬ÀíÓÉÊÇ£º
Al3++3H2O?Al£¨OH£©3+3H+
Al3++3H2O?Al£¨OH£©3+3H+
£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£®
£¨4£©25¡æ¡¢101kPaʱ£¬32g YµÄ×îµÍ¼ÛÆøÌ¬Ç⻯ÎïÍêȫȼÉÕÉú³ÉÎȶ¨µÄÑõ»¯Îïʱ·Å³ö1780.6kJµÄÈÈÁ¿£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
CH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O£¨l£©¡÷H=-890.3KJ/mol
CH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O£¨l£©¡÷H=-890.3KJ/mol
£®

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

¾«Ó¢¼Ò½ÌÍø