ÌâÄ¿ÄÚÈÝ
84Ïû¶¾Òº¡±ÄÜÓÐЧɱÃð¼×ÐÍH1N1²¡¶¾,ijͬѧ¹ºÂòÁËÒ»Æ¿Ä³Æ·ÅÆ¡°84Ïû¶¾Òº¡±,²¢²éÔÄÏà¹Ø×ÊÁϺÍÏû¶¾Òº°üװ˵Ã÷µÃµ½ÈçÏÂÐÅÏ¢:
¡°84Ïû¶¾Òº¡±:º¬25% NaClO¡¢1 000 mL¡¢ÃܶÈ1.19 g¡¤cm-3,Ï¡ÊÍ100±¶(Ìå»ý±È)ºóʹÓá£
Çë¸ù¾ÝÒÔÉÏÐÅÏ¢ºÍÏà¹ØÖªÊ¶»Ø´ðÏÂÁÐÎÊÌâ:
(1)¸Ã¡°84Ïû¶¾Òº¡±µÄÎïÖʵÄÁ¿Å¨¶ÈΪ¡¡¡¡¡¡¡¡mol¡¤L-1¡£
(2)¸Ãͬѧȡ100 mL¸ÃÆ·ÅÆ¡°84Ïû¶¾Òº¡±Ï¡ÊͺóÓÃÓÚÏû¶¾,Ï¡ÊͺóµÄÈÜÒºÖÐc(Na+)¡Ö¡¡¡¡¡¡¡¡mol¡¤L-1(¼ÙÉèÏ¡ÊͺóÈÜÒºÃܶÈΪ1.0 g¡¤cm-3)¡£
(3)¸Ãͬѧ²ÎÔÄ¸ÃÆ·ÅÆ¡°84Ïû¶¾Òº¡±µÄÅä·½,ÓûÓÃNaClO¹ÌÌåÅäÖÆ480 mLº¬25% NaClOµÄÏû¶¾Òº¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ¡¡¡¡¡¡¡¡¡£
A.Ñ¡ÓÃ480 mLµÄÈÝÁ¿Æ¿
B.ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴ¾»ºó,Ó¦ºæ¸É²ÅÄÜÓÃÓÚÈÜÒºÅäÖÆ
C.ÀûÓùºÂòµÄÉÌÆ·NaClOÀ´ÅäÖÆ¿ÉÄܵ¼Ö½á¹ûÆ«µÍ
D.ÐèÒª³ÆÁ¿µÄNaClO¹ÌÌåÖÊÁ¿Îª143 g
¡¾½âÎö¡¿(1)¸ù¾Ý![]()
![]()
(2)4.0 mol¡¤L-1µÄNaClOÈÜҺϡÊÍ100±¶ºó,ÆäŨ¶ÈΪ0.04 mol¡¤L-1,¼´c(Na+)=0.04 mol¡¤L-1¡£
(3)Ñ¡ÏîA,ʵÑéÊÒûÓÐ480 mLµÄÈÝÁ¿Æ¿,ӦѡÓÃ500 mLµÄÈÝÁ¿Æ¿;Ñ¡ÏîB,ÅäÖÆ¹ý³ÌÖÐÐèÒª¼ÓÈëË®,ËùÒԾϴµÓ¸É¾»µÄÈÝÁ¿Æ¿²»±Øºæ¸ÉºóÔÙʹÓÃ;Ñ¡ÏîC,ÓÉÓÚNaClOÒ×ÎüÊÕ¿ÕÆøÖеÄH2O¡¢CO2¶ø±äÖÊ,ËùÒÔÉÌÆ·NaClO¿ÉÄܲ¿·Ö±äÖʵ¼ÖÂNaClO¼õÉÙ,ÅäÖÆµÄÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿¼õС,½á¹ûÆ«µÍ;Ñ¡ÏîD,ѡȡ500 mLµÄÈÝÁ¿Æ¿½øÐÐÅäÖÆ,±ØÐë°´500 mLÈÜÒº¼ÆËãËùÐèÈÜÖʵÄÁ¿,ËùÒÔÐèÒªNaClOµÄÖÊÁ¿Îª
0.5 L¡Á4.0 mol¡¤L-1¡Á74.5 g¡¤mol-1=149 g¡£
´ð°¸:(1)4.0 (2)0.04 (3)C
ÒÑÖª·´Ó¦mX£¨g£©+nY£¨g£©⇌qZ£¨g£©µÄ¡÷H£¼0£¬m+n£¾q£¬ÔÚºãÈÝÃܱÕÈÝÆ÷Öз´Ó¦´ïµ½Æ½ºâʱ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
| ¡¡ | A£® | ͨÈëÏ¡ÓÐÆøÌåʹѹǿÔö´ó£¬Æ½ºâ½«ÕýÏòÒÆ¶¯ |
| ¡¡ | B£® | XµÄÕý·´Ó¦ËÙÂÊÊÇYµÄÄæ·´Ó¦ËÙÂ浀 |
| ¡¡ | C£® | ½µµÍζȣ¬»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿±äС |
| ¡¡ | D£® | Ôö¼ÓXµÄÎïÖʵÄÁ¿£¬YµÄת»¯ÂʽµµÍ |