ÌâÄ¿ÄÚÈÝ

ÓÃ9mol/LµÄŨÁòËáÅäÖÆ³É 0.9mol/LµÄÏ¡ÁòËá 99mL£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÐèҪȡŨÁòËá
 
 mL£¬Ó¦Ñ¡ÓÃ
 
mLµÄÈÝÁ¿Æ¿£®
£¨2£©ÅäÖÆ²Ù×÷¿É·Ö½â³ÉÈçϼ¸²½£¬ÒÔÏÂÕýÈ·µÄ²Ù×÷˳ÐòÊÇ
 

A£®ÏòÈÝÁ¿Æ¿ÖÐ×¢ÈëÉÙÁ¿ÕôÁóË®£¬¼ì²éÊÇ·ñ©ˮ
B£®ÓÃÉÙÁ¿ÕôÁóˮϴµÓÉÕ±­¼°²£Á§°ô£¬½«ÈÜҺעÈëÈÝÁ¿Æ¿£¬²¢Öظ´²Ù×÷Á½´Î
C£®ÓÃÒÑÀäÈ´µÄÏ¡ÁòËá×¢ÈëÒѼì²é²»Â©Ë®µÄÈÝÁ¿Æ¿ÖÐ
D£®¸ù¾Ý¼ÆË㣬ÓÃÁ¿Í²Á¿È¡Ò»¶¨Ìå»ýµÄŨÁòËá
E£®½«Å¨ÁòËáÑØÉÕ±­±ÚÂýÂý×¢ÈëÊ¢ÓÐÕôÁóË®µÄСÉÕ±­ÖУ¬²¢²»¶ÏÓò£Á§°ô½Á°è
F£®¸ÇÉÏÈÝÁ¿Æ¿Èû×Ó£¬Õñµ´£¬Ò¡ÔÈ
G£®ÓýºÍ·µÎ¹ÜµÎ¼ÓÕôÁóË®£¬Ê¹ÈÜÒº°¼ÃæÇ¡ºÃÓë¿Ì¶ÈÏàÇÐ
H£®¼ÌÐøÍùÈÝÁ¿Æ¿ÖÐСÐĵؼÓÕôÁóË®£¬Ê¹ÒºÃæ½Ó½ü¿Ì¶ÈÏß1¡«2cm
£¨3£©ÓÉÓÚ´íÎó²Ù×÷£¬Ê¹µÃµ½µÄŨ¶ÈÊý¾Ý±ÈÕýÈ·µÄÆ«´óµÄÊÇ
 
£¨ÌîдѡÏ£®
A£®Ê¹ÓÃÈÝÁ¿Æ¿ÅäÖÆÈÜҺʱ£¬¶¨ÈÝʱ¸©ÊÓÒºÃæ
B£®Ã»ÓÐÓÃÕôÁóˮϴÉÕ±­2¡«3´Î£¬²¢½«Ï´ÒºÒÆÈëÈÝÁ¿Æ¿ÖÐ
C£®ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴ¾»£¬Ã»Óкæ¸É
D£®¶¨ÈÝʱ£¬µÎ¼ÓÕôÁóË®£¬ÏÈÊ¹ÒºÃæÂÔ¸ßÓڿ̶ÈÏߣ¬ÔÙÎü³öÉÙÁ¿Ë®Ê¹ÒºÃæ°¼ÃæÓë¿Ì¶ÈÏßÏàÇÐ
E£®°ÑÅäºÃµÄÈÜÒºµ¹ÈëÓÃÕôÁóˮϴ¾»¶øÄ©¸ÉµÄÊÔ¼ÁÆ¿Öб¸Óã®
¿¼µã£ºÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº
רÌ⣺ʵÑéÌâ
·ÖÎö£º£¨1£©ÈÜҺϡÊ͹ý³ÌÖÐÈÜÖʵÄÎïÖʵÄÁ¿²»±ä£¬ÒÀ´Î¼ÆËãÐèҪŨÁòËáµÄÌå»ý£»ÒÀ¾Ý¡°´ó¶ø½ü¡±µÄÔ­ÔòÑ¡ÔñºÏÊʵÄÁ¿Í²£»
£¨2£©ÓÃŨÈÜÒºÅäÖÃÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÒ»°ã²½ÖèΪ£ºÁ¿È¡»ò³ÆÁ¿¡¢Ï¡ÊÍ»òÕßÈܽ⡢ÀäÈ´¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔÈ¡¢×°Æ¿¡¢Ìù±êÇ©£¬¾Ý´Ë½â´ð£»
£¨3£©ÒÀ¾ÝC=
n
V
·ÖÎö£¬·²ÊÇʹnƫС»òÕßʹVÆ«´óµÄ²Ù×÷¶¼»áʹÈÜÒºµÄŨ¶ÈÆ«µÍ£¬·´Ö®Ê¹ÈÜҺŨ¶ÈÆ«¸ß£®
½â´ð£º ½â£º£¨1£©ÈÜҺϡÊ͹ý³ÌÖÐÈÜÖʵÄÎïÖʵÄÁ¿²»±ä£¬ÉèÐèҪŨÁòËáµÄÌå»ýΪV£¬ÔòV¡Á9mol/L=0.9mol/L¡Á100ml£¬
½âµÃV=10.0ml£»Ó¦Ñ¡Ôñ10mlµÄÁ¿Í²£»
¹Ê´ð°¸Îª£º10.0£»10£»
£¨2£©ÓÃŨÈÜÒºÅäÖÃÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÒ»°ã²½ÖèΪ£ºÁ¿È¡»ò³ÆÁ¿¡¢Ï¡ÊÍ»òÕßÈܽ⡢ÀäÈ´¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔÈ¡¢×°Æ¿¡¢Ìù±êÇ©£»
¹Ê´ð°¸Îª£ºADECBHGF£»
£¨3£©ÒÀ¾ÝC=
n
V
·ÖÎö£¬·²ÊÇʹnƫС»òÕßʹVÆ«´óµÄ²Ù×÷¶¼»áʹÈÜÒºµÄŨ¶ÈÆ«µÍ£¬·´Ö®Ê¹ÈÜҺŨ¶ÈÆ«¸ß£®
A£®Ê¹ÓÃÈÝÁ¿Æ¿ÅäÖÆÈÜҺʱ£¬¶¨ÈÝʱ¸©ÊÓÒºÃæ£¬µ¼ÖÂÈÜÒºµÄÌå»ýVƫС£¬ÈÜÒºµÄŨ¶ÈÆ«¸ß£»
B£®Ã»ÓÐÓÃÕôÁóˮϴÉÕ±­2¡«3´Î£¬²¢½«Ï´ÒºÒÆÈëÈÝÁ¿Æ¿ÖУ¬µ¼ÖÂÈÜÖÊnƫС£¬ÈÜÒºµÄŨ¶ÈÆ«µÍ£»
C£®ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴ¾»£¬Ã»Óкæ¸É£¬¶ÔÈÜÒºµÄÌå»ýºÍÎïÖʵÄÁ¿Ã»ÓÐÈκÎÓ°Ï죬ÈÜÒºµÄŨ¶ÈÎïÓ°Ï죻
D£®¶¨ÈÝʱ£¬µÎ¼ÓÕôÁóË®£¬ÏÈÊ¹ÒºÃæÂÔ¸ßÓڿ̶ÈÏߣ¬ÔÙÎü³öÉÙÁ¿Ë®Ê¹ÒºÃæ°¼ÃæÓë¿Ì¶ÈÏßÏàÇУ¬Îü³öµÄÈÜÒºº¬ÓÐÈÜÖÊ£¬ÈÜÖʵÄÎïÖʵÄÁ¿Æ«Ð¡£¬Å¨¶ÈÆ«µÍ£»
E£®°ÑÅäºÃµÄÈÜÒºµ¹ÈëÓÃÕôÁóˮϴ¾»¶øÄ©¸ÉµÄÊÔ¼ÁÆ¿Öб¸Óã¬Ï൱ÓÚ°ÑËùÅäÈÜҺϡÊÍ£¬Å¨¶ÈÆ«µÍ£»
¹Ê´ð°¸Îª£ºA£®
µãÆÀ£º±¾Ì⿼²éÁËÅäÖÃÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº£¬ÌâÄ¿ÄѶȲ»´ó£¬ÊìϤÈÜÒºµÄÅäÖÃÔ­Àí¡¢²Ù×÷²½ÖèÊǽâÌâ¹Ø¼ü£¬×¢ÒâÎó²î·ÖÎöµÄ·½·¨£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¹¤ÒµÉÏÔںϳÉËþÖвÉÓÃÏÂÁз´Ó¦ºÏ³É¼×´¼£ºCO£¨g£©+2H2£¨g£©?CH3OH£¨g£©¡÷H=QkJ/mol
£¨1£©ÅжϸÿÉÄæ·´Ó¦´ïµ½Æ½ºâ״̬µÄ±êÖ¾ÊÇ£¨Ìî×ÖĸÐòºÅ£©
 
£®
a¡¢Éú³ÉCH3OHµÄËÙÂÊÓëÏûºÄCOµÄËÙÂÊÏàµÈ
b¡¢»ìºÏÆøÌåµÄÃܶȲ»±ä
c¡¢»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä
d¡¢CH3OH¡¢CO¡¢H2µÄŨ¶È¶¼²»ÔÙ·¢Éú±ä»¯
£¨2£©Èç±íËùÁÐÊý¾ÝÊǸÿÉÄæ·´Ó¦ÔÚ²»Í¬Î¶ÈÏµĻ¯Ñ§Æ½ºâ³£Êý£¨K£©
ζÈ250¡æ300¡æ350¡æ
K2.0410.2700.012
¢ÙÓɱíÖÐÊý¾ÝÅжÏQ
 
0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©
¢ÚijζÈÏ£¬½«2mol COºÍ6mol H2³äÈë2LµÄÃܱÕÈÝÆ÷ÖУ¬³ä·Ö·´Ó¦ºó£¬´ïµ½Æ½ºâ²âµÃc£¨CO£©=0.2mol?L-1£¬´Ëʱ¶ÔÓ¦µÄζÈΪ
 
£»COµÄת»¯ÂÊΪ
 
£®
£¨3£©ÇëÔÚͼÖл­³öѹǿ²»Í¬£¬Æ½ºâʱ¼×´¼µÄÌå»ý·ÖÊý£¨¦Õ£©ËæÎ¶ȣ¨T£©±ä»¯µÄÁ½ÌõÇúÏߣ¨ÔÚÇúÏßÉϱê³öP1¡¢P2¡¢£¬ÇÒp1£¼P2£©
£¨4£©ÒªÌá¸ßCOµÄת»¯ÂÊ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊÇ
 
£¨Ìî×ÖĸÐòºÅ£©£®
a¡¢ÉýΠ b¡¢¼ÓÈë´ß»¯¼Á  c¡¢Ôö¼ÓCOµÄŨ¶È  d¡¢Í¨ÈëH2¼ÓѹÕË   e¡¢Í¨È˶èÐÔÆøÌå¼Óѹ   f¡¢µçÀë³ö¼×´¼
£¨5£©ÒÑÖªÒ»¶¨Ìõ¼þÏ£¬COÓëH2ÔÚ´ß»¯¼ÁµÄ×÷ÓÃÏÂÉú³É5mol CH3OHʱ£¬ÄÜÁ¿µÄ±ä»¯Îª454kJ£®ÔÚ¸ÃÌõ¼þÏ£¬ÏòÈÝ»ýÏàͬµÄ3¸öÃܱÕÈÝÆ÷ÖУ¬°´Õռס¢ÒÒ¡¢±ûÈýÖÖ²»Í¬µÄͶÁÏ·½Ê½¼ÓÈË·´Ó¦Î²âµÃ·´Ó¦´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏ£º
ÈÝÆ÷¼×ÒÒ±û
·´Ó¦ÎïͶÈëÁ¿1mol CO 2mol H21mol CH3OH2mol CH3OH
ƽ
ºâ
ʱ
Êý
¾Ý
CH3OHµÄŨ¶È£¨mol?L-1£©c1c2c3
·´Ó¦ÎüÊÕ»ò·Å³öµÄÄÜÁ¿£¨KJ£©abc
Ìåϵѹǿ£¨Pa£©P1P2P3
·´Ó¦Îïת»¯ÂʦÁ1¦Á2¦Á3
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
 
£¨Ìî×ÖĸÐòºÅ£©£®
A¡¢2c1£¾c3    B¡¢a+b£¼90.8     C¡¢2P2£¼P3      D¡¢¦Á1+¦Á3£¼1£®

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

¾«Ó¢¼Ò½ÌÍø