ijÌþA 0.2 molÔÚÑõÆøÖÐÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷1.2 mol£®ÊԻشð£º
(1)ÌþAµÄ·Ö×ÓʽΪ________£®
(2)ÈôÈ¡Ò»¶¨Á¿µÄÌþAÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷3 mol£¬ÔòÓÐ________gÌþA²Î¼ÓÁË·´Ó¦£¬È¼ÉÕʱÏûºÄ±ê×¼×´¿öϵÄÑõÆø________L£®
(3)ÈôÌþA²»ÄÜʹäåË®ÍÊÉ«£¬µ«ÔÚÒ»¶¨Ìõ¼þÏÂÄÜÓëÂÈÆø·¢ÉúÈ¡´ú·´Ó¦£¬ÆäÒ»ÂÈÈ¡´úÎïÖ»ÓÐÒ»ÖÖ£¬ÔòÌþAµÄ½á¹¹¼òʽΪ________£®
(4)ÈôÌþAÄÜʹäåË®ÍÊÉ«£¬ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬ÓëH2¼Ó³É£¬Æä¼Ó³É²úÎᄇⶨ·Ö×ÓÖк¬ÓÐ4¸ö¼×»ù£¬ÌþA¿ÉÄÜÓеĽṹ¼òʽΪ________(д³öÒ»ÖÖ¼´¿É)
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(1)ÌþAµÄ·Ö×Óʽ____________£¬B¡¢CµÄ·Ö×Óʽ·Ö±ðÊÇ_____________¡¢_____________¡£
(2)ÈôÈ¡Ò»¶¨Á¿µÄÌþAȼÉÕºóÉú³ÉB¡¢C¸÷3 mol£¬ÔòÓÐ___________gÌþA²Î¼ÓÁË·´Ó¦£¬È¼ÉÕʱÏûºÄ±ê×¼×´¿öÏÂO2____________L¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(1)
(2)ÈôAÄÜʹäåË®ÍÊÉ«£¬ÇÒÔÚ´ß»¯¼Á´æÔÚÏÂÓëH2¼Ó³ÉµÄ²úÎï·Ö×ÓÖк¬ÓÐ4¸ö¼×»ù£¬ÔòA¿ÉÄܵĽṹ¼òʽΪ(ÈÎдһÖÖ) __________________________________¡£
(3)ijÓлúÎïµÄ·Ö×ÓʽΪCxHyO2£¬ÈôxµÄÖµÓëA·Ö×ÓÖеÄ̼Ô×Ó¸öÊýÏàͬ£¬Ôò¸Ã·Ö×ÓÖÐyµÄ×î´óֵΪ__________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012Äêɽ¶«ÁijÇÝ·ÏØʵÑé¸ßÖи߶þµÚÈý´ÎÄ£¿é²âÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ
£¨Ã¿¿Õ2·Ö£¬¹²10·Ö£©Ä³ÌþA 0.2 mol ÔÚÑõÆøÖÐÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷1.2 mol¡£ÊԻشð£º
£¨1£©ÌþAµÄ·Ö×ÓʽΪ_____________¡£
£¨2£©ÈôÈ¡Ò»¶¨Á¿µÄÌþAÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷3 mol£¬ÔòÓÐ________gÌþA²Î¼ÓÁË·´Ó¦£¬È¼ÉÕʱÏûºÄ±ê×¼×´¿öϵÄÑõÆø___________L¡£
£¨3£©ÈôÌþA²»ÄÜʹäåË®ÍÊÉ«£¬µ«ÔÚÒ»¶¨Ìõ¼þÏÂÄÜÓëÂÈÆø·¢ÉúÈ¡´ú·´Ó¦£¬ÆäÒ»ÂÈÈ¡´úÎïÖ»
ÓÐÒ»ÖÖ£¬ÔòÌþAµÄ½á¹¹¼òʽΪ__________________¡£
£¨4£©ÈôÌþAÄÜʹäåË®ÍÊÉ«£¬ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬ÓëH2¼Ó³É£¬Æä¼Ó³É²úÎᄇⶨ·Ö×ÓÖк¬ÓÐ
4¸ö¼×»ù£¬ÌþA¿ÉÄÜÓеĽṹ¼òʽΪ______________£¨Ð´³öÒ»ÖÖ¼´¿É£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ìºÓÄÏÊ¡º×±ÚÊиßÒ»ÏÂѧÆÚµÚ¶þ´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ
(12·Ö)ijÌþA 0.2 molÔÚÑõÆøÖгä·ÖȼÉÕºó£¬Éú³É»¯ºÏÎïB¡¢C¸÷1.2 mol£¬ÊԻشð£º
(1)ÌþAµÄ·Ö×ÓʽÊÇ________________¡£
(2)ÈôÈ¡Ò»¶¨Á¿µÄÌþA³ä·ÖȼÉÕºó£¬Éú³ÉB¡¢C¸÷3 mol£¬ÔòÓÐ________gµÄA²Î¼ÓÁË·´Ó¦£¬È¼ÉÕʱÏûºÄ±ê×¼×´¿öϵÄÑõÆø________L¡£
(3)ÈôÌþA²»ÄÜʹäåË®ÍÊÉ«£¬µ«ÔÚÒ»¶¨Ìõ¼þÏÂÄÜÓëÂÈÆø·¢ÉúÈ¡´ú·´Ó¦£¬ÆäÒ»ÂÈ´úÎïÖ»ÓÐÒ»ÖÖ£¬ÔòÌþAµÄ½á¹¹¼òʽÊÇ________________¡£
£¨4£©ÈôÌþAÄÜʹäåË®ÍÊÉ«£¬ÔÚ´ß»¯¼Á×÷ÓÃÏÂÓëÇâÆø¼Ó³É£¬Æä¼Ó³É²úÎ¾²â¶¨·Ö×ÓÖк¬ÓÐ4¸ö¼×»ù£¬ÔòÌþAµÄ½á¹¹¼òʽ¿ÉÄÜÊÇ__________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìÕã½Ê¡¸ß¶þµÚ¶þ´ÎÔ¿¼»¯Ñ§£¨Àí£©ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
£¨14·Ö£©Ä³ÌþA 0.2 mol ÔÚÑõÆøÖÐÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷1.2 mol¡£ÊԻشð£º
£¨1£©ÌþAµÄ·Ö×ÓʽΪ_____________¡£
£¨2£©ÈôÈ¡Ò»¶¨Á¿µÄÌþAÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷3 mol£¬ÔòÓÐ________gÌþA²Î¼ÓÁË·´Ó¦£¬È¼ÉÕʱÏûºÄ±ê×¼×´¿öϵÄÑõÆø___________L¡£
£¨3£©ÈôÌþA²»ÄÜʹäåË®ÍÊÉ«£¬µ«ÔÚÒ»¶¨Ìõ¼þÏÂÄÜÓëÂÈÆø·¢ÉúÈ¡´ú·´Ó¦£¬ÆäÒ»ÂÈÈ¡´úÎïÖ»ÓÐÒ»ÖÖ£¬ÔòÌþAµÄ½á¹¹¼òʽΪ__________________¡£
£¨4£©ÈôÌþAÄÜʹäåË®ÍÊÉ«£¬ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬ÓëH2¼Ó³É£¬Æä¼Ó³É²úÎᄇⶨ·Ö×ÓÖк¬ÓÐ4¸ö¼×»ù£¬ÌþA¿ÉÄÜÓеĽṹ¼òʽΪ_______________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com