ÌâÄ¿ÄÚÈÝ
£¨7·Ö£©ÒÑÖª£º¢ÙxÊǶÌÖÜÆÚÔªËØ×é³ÉµÄµ¥ÖÊ£¬ZÊÇÒ»ÖÖ³£¼û»¯ºÏÎï¡£¢ÚA¡¢B¡¢C¡¢D¡¢E¡¢F¶¼Êdz£¼ûµÄ»¯ºÏÎ³£ÎÂÏÂCÊÇÆøÌ壬DÊÇÒºÌå¡£AÈÜÒºµÄÑæÉ«·´Ó¦Îª»ÆÉ«¡£×ÔÈ»½çÖдæÔڽ϶àµÄE¿óʯ£¬¹¤ÒµÉú²úÖг£ÓÃÕâÖÖ¿óʯÀ´ÖÆÈ¡F¡£¢Û´æÔÚÏÂÁÐת»¯¹ØÏµ£º
![]()
ÊԻشð£º
£¨1£©YµÄµç×Óʽ£º ,XÔªËØÔÚÖÜÆÚ±íÖеÄλÖá¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
¡¡¡¡C·Ö×ÓÖеĻ¯Ñ§¼üÀàÐÍ¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨2£©Yת»¯ÎªZµÄ»¯Ñ§·½³Ìʽ£º¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ ¡¡¡¡¡¡¡¡¡¡¡¡
Aת»¯ÎªEµÄÀë×Ó·½³Ìʽ£¨Z¹ýÁ¿£©:¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨7·Ö£©(1)Ca2+
2£¡¡¡¡¡¡µÚ¶þÖÜÆÚ¡¢IVA×å¡¡¡¡¹²¼Û¼ü
¡¡¡¡ (2) CaC2+2H2O£½Ca(OH)2+C2H2¡ü HCO3£+Ca2++OH££½CaCO3¡ý+H2O
¡¾½âÎö¡¿ÂÔ
£¨14·Ö£©X¡¢Y¡¢Z¡¢W¡¢QÊÇÔ×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬Ïà¹ØÐÅÏ¢ÈçÏÂ±í£º
|
Ôª ËØ |
Ïà¹ØÐÅÏ¢ |
|
X |
XÔ×ÓºËÍâ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµÄ2±¶ |
|
Y |
YµÄÆøÌ¬Ç⻯ÎïµÄË®ÈÜÒºÏÔÈõ¼îÐÔ |
|
Z |
ZÊǵؿÇÖк¬Á¿×î¶àµÄ½ðÊôÔªËØ |
|
W |
³£Î³£Ñ¹Ï£¬WµÄµ¥ÖÊÊǵ»ÆÉ«¹ÌÌå |
|
Q |
¡¡ |
¸ù¾ÝÉÏÊöÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÔªËØQÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃ______________________________£®
(2)YµÄ×î¼òµ¥ÆøÌ¬Ç⻯Îï¼×µÄË®ÈÜÒº¿ÉÓëH2O2·¢Éú·´Ó¦£¬Æä²úÎï²»ÎÛȾ»·¾³£¬»¯Ñ§·½³ÌʽΪ______________________________________£®£¨ÓÃÔªËØ·ûºÅ±íʾ£¬ÏÂͬ£©
(3)XºÍÇâÔªËØ×é³ÉµÄ»¯ºÏÎï·Ö×ÓÓÐ6¸öÔ×Ó£¬Æä½á¹¹Ê½Îª £®
(4)ÒÑÖª£ºX(s)+O2(g) =XO2(g) ¡÷H = -393.5£ëJ¡¤mol-1
2X(s)+O2(g) =2XO(g) ¡÷H = -221.0£ëJ¡¤mol-1
ÔòXOµÄȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ__________________________________________________.
(5)ÔªËØYÓëÇâÔªËØÐγɵÄÎåºËÑôÀë×ÓÒÒ£¬Ïòº¬ÒÒºÍZµÄÑôÀë×ӵĻìºÏÈÜÒºÖмÓÈë¹ÌÌåNa2O2£¬ ¼ÓÈëNa2O2µÄÎïÖʵÄÁ¿Óë²úÉú³ÁµíÓÐÈçͼËùʾ¹ØÏµ£º
![]()
д³öÓйط´Ó¦Àë×Ó·½³Ìʽ£º(ÿ¶ÎÖ»ÓÃÒ»¸öÀë×Ó·½³Ìʽ±íʾ)
o¡ªa¶Î
a¡ªb¶Î £®