ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿(1) A¡¢B¡¢CÈýÖÖÔªËصÄÔ­×Ó¾ßÓÐÏàͬµÄµç×Ó²ãÊý£¬¶øBµÄºËµçºÉÊý±ÈA´ó2£¬CÔ­×ӵĵç×ÓÊý±ÈBÔ­×ӵĵç×ÓÊý¶à3£» 1molAµÄµ¥ÖÊÄܸúÑÎËá·´Ó¦£¬ÔÚ±ê×¼×´¿öÏ¿ÉÖû»³ö11.2LµÄH2£¬ÕâʱAת±äΪÓëÄÊÔ­×Ó¾ßÓÐÏàͬµç×Ó²ã½á¹¹µÄÀë×Ó¡£ÊԻشð£º

¢ÙAÊÇ_____ÔªËØ£¬ CÊÇ_____ÔªËØ£»

¢ÚBµÄÀë×ӽṹʾÒâͼÊÇ_________________,ÓëBµÄÀë×Ó¾ßÓÐÏàͬµç×ÓÊýµÄ΢Á£ÖÐ,ÓÐÁ½ÖÖ·Ö×Ó¿É»¯ºÏÉú³ÉÒ»ÖÖÑÎ,¸ÃÑÎÒ׳±½â£¬ÊÜÈÈ»òÓöÈÈË®¿É·Ö½â£¬Äܸ¯Ê´²£Á§£¬Á½·Ö×Ó·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_______________________________£»

¢ÛA¡¢CÔªËØÐγɻ¯ºÏÎïA2CµÄµç×ÓʽÊÇ_________________£»

(2)ÔÚË®ÈÜÒºÖУ¬YO3n£­ºÍS2£­·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÈçÏ£º

YO3n£­ + 3S2£­ + 6H+ = Y£­+ 3S¡ý+ 3H2O

¢ÙYO3n£­ÖÐYµÄ»¯ºÏ¼ÛÊÇ_____________£»

¢ÚYÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇ_________¡£

¡¾´ð°¸¡¿ ÄÆ Áò ÂÁÀë×ӽṹʾÒâͼ£¨ÂÔ£© NH3 + HF ==NH4F Áò»¯ÄƵç×Óʽ£¨ÂÔ£© +5 7

¡¾½âÎö¡¿±¾Ìâ·ÖÎö£º±¾ÌâÖ÷Òª¿¼²éÔªËصÄÔ­×ӽṹ¡£

1molAµÄµ¥ÖÊÄܸúÑÎËá·´Ó¦£¬ÔÚ±ê×¼×´¿öÏ¿ÉÖû»³ö11.2L¼´0.5molH2£¬ËµÃ÷A³Ê+1¼Û£¬ÕâʱAת±äΪÓëÄÊÔ­×Ó¾ßÓÐÏàͬµç×Ó²ã½á¹¹µÄÀë×Ó£¬AΪÄÆ¡£A¡¢B¡¢CÈýÖÖÔªËصÄÔ­×Ó¾ßÓÐÏàͬµÄµç×Ó²ãÊý£¬¶øBµÄºËµçºÉÊý±ÈA´ó2£¬BΪÂÁ¡£CÔ­×ӵĵç×ÓÊý±ÈBÔ­×ӵĵç×ÓÊý¶à3£¬CΪÁò¡£

(1) ¢ÙAÊÇÄÆÔªËØ£¬ CÊÇÁòÔªËØ£»

¢ÚBµÄÀë×ӽṹʾÒâͼÊÇ¡£¸ÃÑÎÊÇ·ú»¯ï§£¬Á½·Ö×Ó·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇNH3 + HF ==NH4F¡£

¢Û A¡¢CÔªËØÐγɻ¯ºÏÎïNa2SµÄµç×ÓʽÊÇ¡£

(2)¢Ù¸ù¾ÝYO3n£­ + 3S2£­ + 6H+ = Y£­+ 3S¡ý+ 3H2OµÄµçºÉÊغã¿ÉµÃn=1¡£YO3n£­ÖÐYµÄ»¯ºÏ¼ÛÊÇ+5.

¢ÚY£­ÊÇ×îÍâ²ã8µç×ÓµÄÎȶ¨½á¹¹£¬ËùÒÔYÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇ7.

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø