¸ÆÊÇά³ÖÈËÌåÕý³£¹¦ÄÜËù±ØÐèµÄÔªËØ£¬ÏÂͼËùʾΪijÖÖ²¹¸Æ¼Á¡°¸Æ¶ûÆ桱˵Ã÷ÊéµÄÒ»²¿·Ö£¬È¡1Ƭ¸Æ¶ûÆ棬·ÅÈëÊ¢ÓÐ10gÏ¡ÑÎËáµÄÉÕ±­ÖУ¬ÆäÖеÄ̼Ëá¸Æ¸úÏ¡ÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¨ÆäËü³É·ÖÓëÏ¡ÑÎËá²»·´Ó¦£©£¬ÉÕ±­ÄÚÎïÖʵÄÖÊÁ¿Îª11.34g£®ÇëÄã¼ÆË㣺
£¨1£©Ã¿Æ¬¸Æ¶ûÆæÖк¬Ì¼Ëá¸ÆµÄÖÊÁ¿£®
£¨2£©Ê¹ÓÃÕâÖÖ²¹¸Æ¼Á£¬Ã¿ÈËÿÌìÉãÈë¸ÆÔªËصÄÖÊÁ¿£®
£¨3£©ËùÓÃÏ¡ÑÎËáÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®

¡¾´ð°¸¡¿·ÖÎö£º£¨1£©ÓûÕýÈ·½â´ð±¾Ì⣬Ðë¸ù¾ÝÑÎËáÓë̼Ëá¸Æ·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¬µÃ³ö¸÷ÎïÖÊÖ®¼äµÄÖÊÁ¿±È£¬Áгö±ÈÀýʽ£¬¼´¿ÉÇó³öÿƬ¸Æ¶ûÆæÖк¬Ì¼Ëá¸ÆµÄÖÊÁ¿£»
£¨2£©ÓûÕýÈ·½â´ð±¾Ì⣬ÐëÏȸù¾ÝÔªËصÄÖÊÁ¿·ÖÊý¹«Ê½¼ÆËã³ö̼Ëá¸ÆÖиÆÔªËصÄÖÊÁ¿·ÖÊý£¬ÓÉ˵Ã÷Êé¿ÉÖª£¬Ê¹ÓÃÕâÖÖ²¹¸Æ¼Á£¬Ã¿ÈËÿÌìÐè·þÓÃ2Ƭ£¬ÔòÿÈËÿÌìÉãÈë¸ÆÔªËصÄÖÊÁ¿=ÿƬ¸Æ¶ûÆæÖÐCaCO3µÄÖÊÁ¿×2Ƭ×̼Ëá¸ÆÖиÆÔªËصÄÖÊÁ¿·ÖÊý£¬¾Ý´Ë´ðÌ⣮
£¨3£©ÓÉ£¨1£©ÒѼÆËã³ö10gÑÎËáÖÐHClµÄÖÊÁ¿£¬¸ù¾ÝÈÜÖÊÖÊÁ¿·ÖÊý=×100%¼´¿É¼ÆËã³öËùÓÃÑÎËáÖÐHClµÄÖÊÁ¿·ÖÊý£®
½â´ð£º½â£º£¨1£©ÉèÿƬ¸Æ¶ûÆæÖÐCaCO3µÄÖÊÁ¿Îªx£¬10gÑÎËáÖÐHClµÄÖÊÁ¿Îªy£®
ÍêÈ«·´Ó¦ºó²úÉúµÄ¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª£º2g+10g-11.34g=0.66g
CaCO3+2HCl=CaCl2+H2O+CO2¡ü
100     73           44
x       y           0.66g
¡à
½âÖ®µÃ£ºx=1.5g£¬y=1.095g£®
£¨2£©Ì¼Ëá¸ÆÖиÆÔªËصÄÖÊÁ¿·ÖÊýΪ£º×100%=40%
ÿÌìÉãÈëµÄ¸Æ¶ûÆæÖиÆÔªËصÄÖÊÁ¿Îª£º1.5g×2×40%=1.2g£®
£¨3£©ËùÓÃÑÎËáÖÐHClµÄÖÊÁ¿·ÖÊýΪ£º×100%=10.95%£®
´ð£ºÃ¿Æ¬¸Æ¶ûÆæÖк¬Ì¼Ëá¸ÆµÄÖÊÁ¿Îª1.5g£¬Ê¹ÓÃÕâÖÖ²¹¸Æ¼Á£¬Ã¿ÈËÿÌìÉãÈë¸ÆÔªËصÄÖÊÁ¿Îª1.2g£¬ËùÓÃÏ¡ÑÎËáÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ10.95%£®
µãÆÀ£º±¾ÌâÖ÷Òª¿¼²éѧÉúÔËÓû¯Ñ§·½³ÌʽºÍÈÜÖÊÖÊÁ¿·ÖÊý¹«Ê½½øÐмÆËãµÄÄÜÁ¦£®Ñ§ÉúÐèÈÏÕæ·ÖÎöÒÑÖªÌõ¼þÖеÄÊýÁ¿¹Øϵ£¬ÕýÈ·Êéд»¯Ñ§·½³Ìʽ£¬²ÅÄܽâ´ð£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¾«Ó¢¼Ò½ÌÍø¸ÆÊÇά³ÖÈËÌåÕý³£¹¦ÄÜËù±ØÐèµÄÔªËØ£®ÓÒͼËùʾΪijÖÖ²¹¸Æ¼Á¡°¸Æ¶ûÆ桱˵Ã÷ÊéµÄÒ»²¿·Ö£®È¡1Ƭ¸Æ¶ûÆ棬·ÅÈëÊ¢ÓÐ10gÏ¡ÑÎËáµÄÉÕ±­ÖУ¬ÆäÖÐ̼Ëá¸Æ¸úÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¨ÆäËü³É·ÖÓëÑÎËá²»·´Ó¦£©£®ÉÕ±­ÄÚÎïÖÊÖÊÁ¿±äΪ11.34g£®ÊÔ¼ÆË㣺
£¨1£©Ã¿Æ¬¸Æ¶ûÆæÖк¬Ì¼Ëá¸ÆµÄÖÊÁ¿£®
£¨2£©ËùÓÃÏ¡ÑÎËáÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¸ÆÊÇά³ÖÈËÌåÕý³£¹¦ÄÜËù±ØÐèµÄÔªËØ£®ÓÒͼËùʾΪijÖÖ²¹¸Æ¼Á¡°¸Æ¶ûÆ桱˵Ã÷ÊéµÄÒ»²¿·Ö£®ÊÔ½øÐÐÓйصļÆË㣺
£¨1£©CaCO3ÖÐCa¡¢C¡¢OµÄÖÊÁ¿±ÈΪ
10£º3£º12
10£º3£º12
£®
£¨2£©Ì¼Ëá¸ÆÖиÆÔªËصÄÖÊÁ¿·ÖÊý£®
£¨3£©Ò»°ãÀ´Ëµ£¬³ÉÄêÈËÿÌìÐè²¹¸Æ1.2g£¬ÄÇôһ¸ö³ÉÄêÈËÒ»ÌìÓ¦·þÓü¸Æ¬¡°¸Æ¶ûÆ桱

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2014?¾²°²Çøһģ£©¸ÆÊÇά³ÖÈËÌåÕý³£¹¦ÄÜËù±ØÐèµÄÔªËØ£®¸Æ¶ûÆæÊdz£ÓõIJ¹¸Æ¼Á£¬Ö÷Òª³É·ÖÊÇCaCO3£¬¸Æ¶ûÆæÿƬ2.0g£®È¡1Ƭ¸Æ¶ûÆ棬·ÅÈëÊ¢ÓÐ10gÏ¡ÑÎËáµÄÉÕ±­ÖУ¬ÆäÖÐ̼Ëá¸Æ¸úÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¨ÆäËü³É·ÖÓëÑÎËá²»·´Ó¦£©£®ÉÕ±­ÄÚÎïÖÊÖÊÁ¿±äΪ11.34g£®ÊÔ¼ÆË㣺
¢Ù·´Ó¦²úÉú¶þÑõ»¯Ì¼
0.66
0.66
g£¬ÊÇ
0.015
0.015
mol£»
¢ÚÿƬ¸Æ¶ûÆæÖк¬Ì¼Ëá¸Æ
0.015
0.015
mol£»
¢Û·þÓøƶûÆæͨ³£Ò»ÈÕ2´Î£¬Ã¿´Î1Ƭ£®ÔòÿÈËÿÌìÉãÈë¸ÆÔªËصÄÖÊÁ¿Îª
1.2
1.2
g£®
¢Ü¸ù¾Ý»¯Ñ§·½³ÌʽÁÐʽ¼ÆËãÉÏÊö·´Ó¦ËùÓÃÏ¡ÑÎËáÈÜÖʵÄÖÊÁ¿·ÖÊý£¨¾«È·µ½0.01%£©
10.95%
10.95%
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2009?ÉÛÑô£©¸ÆÊÇά³ÖÈËÌåÕý³£¹¦ÄÜËù±ØÐèµÄÔªËØ£®Ð¡·¼·þÓõÄijÖÖ¸ÆƬµÄ²¿·Ö˵Ã÷Èçͼ1£¬ËýºÜÏëÖªµÀÿÌì·þÓõĸÆƬÖÐ̼Ëá¸ÆµÄÖÊÁ¿£®ÓÚÊÇÔÚ¼ÒÖнøÐÐÁË̽¾¿£ºÈ¡2Ƭ¸ÆƬ£¬·ÅÈ˲£Á§±­ÖУ¬ÏòÆäÖмÓÈë60g °×´×£¬Á½ÕßÇ¡ºÃÍêÈ«·´Ó¦£¨¼ÙÉè¸ÆƬÖÐÆäËû³É·Ö²»Óë´×Ëá·´Ó¦£©£¬²âµÃ²£Á§±­ÖÐÎïÖʵÄÖÊÁ¿£¨m£©Ó뷴Ӧʱ¼ä£¨t£©µÄ¹ØϵÈçͼ2Ëùʾ£®·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
CaCO3+2CH3COOH¨T£¨CH3COO£©2Ca+H2O+CO2¡ü
ÇëÄã°ïС·¼¼ÆË㣺
£¨1£©·´Ó¦ºó²úÉúÆøÌåµÄÖÊÁ¿Îª
1.1
1.1
g£»
£¨2£©Á½Æ¬¸ÆƬÖÐËùº¬Ì¼Ëá¸ÆµÄÖÊÁ¿£»
£¨3£©ËùÓð״×ÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¸ÆÊÇά³ÖÈËÌåÕý³£¹¦ÄÜËù±ØÐèµÄÔªËØ£®Í¼ÎªÄ³ÖÖ²¹¸Æ¼Á¡°¸Æ¶ûÆ桱˵Ã÷ÊéµÄÒ»²¿·Ö£®È¡1Ƭ¸Æ¶ûÆ棬·ÅÈëÊ¢ÓÐ10gÏ¡ÑÎËáµÄÉÕ±­ÖУ¬ÆäÖÐ̼Ëá¸Æ¸úÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¨ÆäËü³É·ÖÓëÑÎËá²»·´Ó¦£©£®ÉÕ±­ÄÚÎïÖÊÖÊÁ¿±äΪ11.34g£®£¨·´Ó¦·½³Ìʽ£ºCaCO3+2HCl¨TCaCl2+H2O+CO2¡ü£©
ÊÔ¼ÆË㣺
£¨1£©Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿
0.66g
0.66g
£®
£¨2£©ËùÓÃÏ¡ÑÎËáÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸