ÌâÄ¿ÄÚÈÝ
£¨8·Ö£©ÈçͼËùʾ£¬¼×¡¢ÒÒ¡¢±ûÊÇÈýÖÖ³£¼ûµ¥ÖÊ£¬X¡¢Y¡¢ZÊdz£¼û»¯ºÏÎËüÃÇÖ®¼äÓÐÈçÏÂת»¯¹ØÏµ£º![]()
¢ÅÈô¼×ÊǶÌÖÜÆÚ½ðÊôµ¥ÖÊ£¬ÒÒ¡¢±ûÊǶÌÖÜÆÚ·Ç½ðÊôµ¥ÖÊ£¬X¡¢Y¡¢ZÖÐÖ»ÓÐÒ»ÖÖÊÇÀë×Ó¾§Ì壬
ÊÔÍÆ¶Ï£º
¢ÙXµÄµç×ÓʽÊÇ______________________________¡£
¢ÚXÓë¼×·´Ó¦µÄ»¯Ñ§·½³Ìʽ____________________________________¡£
¢ÆÈô¼×ÊÇÆøÌåµ¥ÖÊ£¬±ûͨ³£ÊÇÒºÌ壬YºÍZ¾ßÓÐÏàͬµÄÑôÀë×Ó£¬XÓëZº¬ÓÐÏàͬµÄÒõÀë×Ó£¬ÊÔÍÆ¶Ï£º
¢Ùд³öZµÄ»¯Ñ§Ê½_______________________¡£
¢Úд³öXÓë×ãÁ¿µÄ¼×ÔÚÈÜÒºÖÐÍêÈ«·´Ó¦µÄÀë×Ó·½³Ìʽ£º________________________ _ ¡£
¢Å¢Ù
£¨2·Ö£© ¢Ú2Mg+CO2==2MgO+C£¨2·Ö£©
¢Æ¢ÙFeBr3£¨2·Ö£© ¢Ú3Cl2+2Fe2++4Br-==2Fe3++2Br2+6Cl-£¨2·Ö£©
½âÎö
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¢ÙC ¢ÚH2O2 ¢ÛNa ¢ÜFe ¢ÝHNO3£®
| A¡¢½ö¢Ù¢Û¢Ü | B¡¢½ö¢Ù¢Ú¢Ý | C¡¢½ö¢Ù¢Ú¢Û¢Ý | D¡¢¢Ù¢Ú¢Û¢Ü¢Ý |
| A¡¢¢Ù¢Ú¢Û | B¡¢¢Ú¢Û¢Ü | C¡¢¢Ù¢Ú¢Ü | D¡¢¢Ù¢Ú¢Û¢Ü |
¼×£ºN2£¨g£©+3H2£¨g£©?2NH3£¨g£©£»¡÷H=-92.4kJ/mol
ÒÒ£º2NO2£¨g£©?N2O4
±û£ºH2£¨g£©+I2£¨g£©?2HI£¨g£©
ÔòÏÂÁÐÓйØËµ·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
| A¡¢Èô¼×µÄÌå»ýΪ2L£¬¾¹ý10Ãëºó·´Ó¦´ïµ½Æ½ºâ״̬£¬·Å³öÈÈÁ¿Îª55.44U£¬ÔòH2µÄ·´Ó¦ËÙÂÊÊÇ0.09mol/£¨L?s£© | B¡¢Èô¼×¡¢ÒÒÖз´Ó¦´ïµ½Æ½ºâʱµÄÌå»ýÏàͬ£¬ÔòÁ½ÈÝÆ÷ÖÐËùº¬ÎïÖʵÄÁ¿¿ÉÄÜÏàͬ | C¡¢ÈôÒÒ¡¢±ûÖз´Ó¦´ïµ½Æ½ºâʱµÄÌå»ý¡¢Ñ¹Ç¿¾ùÏàͬ£¬ÔòÒÒÖÐNO2µÄת»¯ÂÊΪ50% | D¡¢Èô¼×¡¢ÒÒ¡¢±ûÖз´Ó¦¾ù´ïµ½Æ½ºâ״̬ʱ£¬Ôò¼×ÖÐÎïÖÊµÄÆ½¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä£¬ÒÒÖÐÎïÖʵÄÑÕÉ«²»±ä£¬±ûÖеÄζȲ»±ä |