ÌâÄ¿ÄÚÈÝ

°¢·ü¼ÓµÂÂÞ³£ÊýԼΪ6.02¡Á1023 mol-1£¬ÏÂÁÐÐðÊöÖÐÕýÈ·µÄÊÇ(   )
A£®³£Î³£Ñ¹Ï£¬18.0 gÖØË®(D2O)Ëùº¬µÄµç×ÓÊýԼΪ10¡Á6.02¡Á1023
B£®ÊÒÎÂÏ£¬42.0 gÒÒÏ©ºÍ±ûÏ©µÄ»ìºÏÆøÌåÖк¬ÓеÄ̼ԭ×ÓÊýԼΪ3¡Á6.02¡Á1023
C£®±ê×¼×´¿öÏ£¬22.4 L¼×±½Ëùº¬µÄ·Ö×ÓÊýԼΪ6.02¡Á1023
D£®±ê×¼×´¿öÏ£¬a L¼×ÍéºÍÒÒÍé»ìºÏÆøÌåÖеķÖ×ÓÊýԼΪ¡Á6.02¡Á1023
BD
±¾Ìâ×ۺϿ¼²éÎïÖʵÄÁ¿ÓëÖÊÁ¿¡¢Ìå»ý¼°°¢·ü¼ÓµÂÂÞ³£ÊýÖ®¼äµÄ¹Øϵ¡£1 mol D2Oº¬ÓÐ10 mol µç×Ó£¬¹Ê18.0 g D2OËùº¬µç×ÓÊýΪ£º¡Á10¡Á6.02¡Á1023 mol-1<10¡Á6.02¡Á1023,A²»ÕýÈ·£»ÒÒÏ©¡¢±ûÏ©µÄʵÑéʽ½ÔΪCH2,¹Ê42.0 g»ìºÏÆøÌåËùº¬ÓеÄ̼ԭ×ÓÊýΪ£º¡Á6.02¡Á1023 mol-1=3¡Á6.02¡Á1023£¬BÕýÈ·£»¼×±½ÔÚ±ê×¼×´¿öÏÂΪҺÌ壬22.4 L¼×±½µÄÎïÖʵÄÁ¿Ô¶´óÓÚ1 mol£¬¹ÊC²»ÕýÈ·£»±ê×¼×´¿öÏ£¬a L CH4ºÍC2H6»ìºÏÆøÌåµÄÎïÖʵÄÁ¿Îª mol£¬Ëùº¬·Ö×ÓÊýΪ¡Á6.02¡Á1023£¬DÕýÈ·¡£¹Ê±¾Ìâ´ð°¸ÎªB¡¢D¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø