½«µÈÎïÖʵÄÁ¿µÄA£¬B»ìºÏÓÚ2LµÄÃܱÕÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£º3A(g)+B(g)
xC(g)+2D(g)
¦¤H<0¾5minºó´ïµ½Æ½ºâ£¬Æ½ºâʱ²âµÃDµÄŨ¶ÈΪ0.5mol/L£¬C(A)£ºC(B)=1£º2£¬CµÄƽ¾ù·´Ó¦ËÙÂÊΪ0.05mol/(L¡¤min)
(l)xµÄֵΪ_______________;
(2)AµÄƽºâŨ¶È____________;
(3)ÈôºãκãÈÝ£¬ÔòÈÝÆ÷ÖÐÆ½ºâʱµÄѹǿÓëÆðʼʱµÄѹǿ±È___________;
(4)ÏÂͼÊÇijһʱ¼ä¶ÎÖи÷´Ó¦ËÙÂÊÓë·´Ó¦½ø³ÌµÄÇúÏß¹ØÏµÍ¼£¬»Ø´ðÏÂÁÐÎÊÌ⣺
![]()
¢Ù´¦ÓÚÆ½ºâ״̬µÄʱ¼ä¶ÎÊÇ____________¡£
¢Ú
¡¢
¡¢
ʱ¿ÌÌåϵÖзֱðÊÇʲôÌõ¼þ·¢ÉúÁ˱仯£¿
___________¡¢
_____________ ¡¢
_____________£¨ÌîA-E£©
A£®ÉýΠB£®½µÎ C£®¼Ó´ß»¯¼Á D£®¼Óѹ E£®¼õѹ
¢ÛÏÂÁи÷ʱ¼ä¶Îʱ£¬AµÄÌå»ý·ÖÊý×î¸ßµÄÊÇ____
A.
B£®
C£®
D£®![]()
£¨1£©1 £¨2£©0.5mol/L £¨3£©9:10 £¨4£©¢Ùt0¡ªt1 t2¡ªt3 t3¡ªt4 t5 --t6 ¢ÚA¡¢C ¡¢E ¢ÛD
¡¾½âÎö¡¿
ÊÔÌâ·ÖÎö£º£¨1£©V(C)=¦¤C/¦¤tËùÒÔ¦¤C= V(C)¡Á¦¤t=0.05mol/(L¡¤min) ¡Á5min=0.25 mol/L¡£ÔÚ·´Ó¦ÖÐC(C):C(D)=X:2 0.25: 0.5= X:2£¬ËùÒÔX=1.£¨2£©¦¤C£¨A£©:¦¤C(D)=3:2 ¦¤C£¨A£©: 0.5=3:2. ¦¤C£¨A£©=0.75¡£¼ÙÉ迪ʼʱ¼ÓÈëµÄA¡¢BµÄÎïÖʵÄÁ¿Îª2m,ÔòÆðʼʱA¡¢BµÄÎïÖʵÄÁ¿Å¨¶ÈΪm mol/L.¸ù¾Ý·´Ó¦±ä»¯µÄA Ũ¶ÈΪ0.75 mol/LºÍ·´Ó¦·½³ÌʽÖÐA¡¢BµÄϵÊý¹ØÏµ¿ÉÖª·´Ó¦±ä»¯µÄBŨ¶ÈΪ0.25 mol/L¡£ËùÒÔÆ½ºâŨ¶ÈC(A)=(m-0.75) mol/L ,C(B)= (m-0.25) mol/L. ÒòΪC(A): C(B)=1:2ËùÒÔ(m-0.75) :(m-0.25) =1:2,½âµÃm=1.25. ËùÒÔAµÄƽºâŨ¶ÈΪ£ºC(A)=(m-0.75) mol/L=(1. 25-0.75) mol/L=0. 5mol/L¡££¨3£©¿ªÊ¼Ê±A¡¢BÆøÌåµÄÎïÖʵÄÁ¿Îª2.5 mol,ÆøÌåµÄ×ÜÎïÖʵÄÁ¿Îª5 mol¡£Æ½ºâʱ¸÷ÖÖÆøÌåµÄÎïÖʵÄÁ¿Îªn(A)= 0. 5mol/L¡Á2L=1mol; n (B)= (m-0.25) mol/L¡Á2L= (1.25-0.25) mol/L¡Á2L =2mol;n(C)= 0.25 mol/L¡Á2L =0. 5mol/L ;n(D)= 0.5mol/L¡Á2L =1mol.ƽºâÊ±ÆøÌåµÄ×ÜÎïÖʵÄÁ¿Îª£¨1+2+0. 5+1£©mol=4.5molËùÒÔn(ƽºâ)£ºn(¿ªÊ¼)= 4.5£º5=9:10.¸ù¾Ý°¢·üÙ¤µÂÂÞ¶¨ÂɵÄÍÆÂÛÔÚÏàͬÎÂÏàͬÌå»ýµÄÃܱÕÈÝÆ÷ÖУ¬ÆøÌåµÄѹǿ±ÈµÈÓÚËüÃǵÄÎïÖʵÄÁ¿Ö®±È¡£ËùÒÔP(ƽºâ)£ºP(¿ªÊ¼)= n(ƽºâ)£ºn(¿ªÊ¼)= 9:10.£¨4£©¢Ù¿ÉÄæ·´Ó¦ÔÚÆ½ºâʱÕý·´Ó¦¡¢Äæ·´Ó¦µÄËÙÂÊÏàµÈ¡£ÓÐͼ¿É֪ƽºâʱ¼ä¶ÎΪ£ºt0¡ªt1 t2¡ªt3 t3¡ªt4 t5 --t6 ¢ÚÔÚt1ʱ£¬Õý·´Ó¦¡¢Äæ·´Ó¦µÄËÙÂʶ¼Ôö´ó£¬ÇÒV(Äæ)´óÓÚV(Õý)£¬ËµÃ÷·´Ó¦ËÙÂʼӿ죬ƽºâÄæÏòÒÆ¶¯£¬¿ÉÄÜÊÇÉý¸ßζȡ£Ñ¡A¡£ÔÚt3ʱÕý·´Ó¦¡¢Äæ·´Ó¦µÄËÙÂʶ¼Ôö´óÇÒV(Äæ)µÈÓÚV(Õý)£¬ËµÃ÷·´Ó¦ËÙÂʼӿ죬ƽºâ²»Òƶ¯£¬¿ÉÄÜÊǼÓÈë´ß»¯¼Á¡£Ñ¡ÏîΪ£ºC ¡£t4ʱÕý·´Ó¦¡¢Äæ·´Ó¦µÄËÙÂʶ¼¼õС£¬ÇÒV(Äæ)´óÓÚV(Õý)˵Ã÷·´Ó¦ËÙÂʼõÂý£¬Æ½ºâÄæÏòÒÆ¶¯£¬¿ÉÄܵı仯ÊǼõСѹǿ¡£Ñ¡ÏîΪE¡£¢ÛÈôʹAµÄÌå»ý·ÖÊý×î¸ß£¬ÔòƽºâÄæÏòÒÆ¶¯×î´ó²Å·ûºÏÒªÇó¡£t0¡ªt1ƽºâ״̬£» t2¡ªt3 ƽºâÄæÏòÒÆ¶¯´ïµ½ÐÂµÄÆ½ºâ£¬Aº¬Á¿±Èǰһ¶Î¸ß£» t3¡ªt4ƽºâÃ»ÒÆ¶¯£¬Óët2¡ªt3¶ÎÏàͬ£» t5 --t6ƽºâÄæÏòÒÆ¶¯£¬´ïµ½ÐÂµÄÆ½ºâ¡£¹ÊAº¬Á¿×î¸ßµÄΪt5 --t6Ñ¡ÏîΪ£ºD¡£
¿¼µã£º¿¼²éÍâ½çÌõ¼þ¶Ô»¯Ñ§Æ½ºâµÄÓ°Ïì¼°·´Ó¦·½³ÌʽÖиöÎïÖʵÄËÙÂʹØÏµµÈ֪ʶ¡£
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêÉÂÎ÷Ê¡Î÷°²Ò»ÖиßÒ»ÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£º¼ÆËãÌâ
£¨8·Ö£©½«µÈÎïÖʵÄÁ¿µÄA¡¢B»ìºÏÓÚ1 LµÄÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÈçÏ·´Ó¦
3A(g)£«B(g)
xC(g)£«2D(g)£¬¾5 minºó£¬²âµÃDµÄŨ¶ÈΪ0.5 mol/L£¬c(A)¡Ãc(B)£½3¡Ã5£¬CµÄƽ¾ù·´Ó¦ËÙÂÊΪ0.1 mol/(L¡¤min)¡£Çó£º
(1)´ËʱAµÄŨ¶Èc(A)£½________mol/L£¬·´Ó¦¿ªÊ¼Ç°ÈÝÆ÷ÖеÄA¡¢BµÄÎïÖʵÄÁ¿£ºn(A)£½n(B)£½________mol¡£
(2)BµÄƽ¾ù·´Ó¦ËÙÂÊv(B)£½________mol/(L¡¤min)¡£
(3)xµÄֵΪ________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ìÉÂÎ÷Ê¡¸ßÒ»ÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º¼ÆËãÌâ
£¨8·Ö£©½«µÈÎïÖʵÄÁ¿µÄA¡¢B»ìºÏÓÚ1 LµÄÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÈçÏ·´Ó¦
3A(g)£«B(g)
xC(g)£«2D(g)£¬¾5 minºó£¬²âµÃDµÄŨ¶ÈΪ0.5 mol/L£¬c(A)¡Ãc(B)£½3¡Ã5£¬CµÄƽ¾ù·´Ó¦ËÙÂÊΪ0.1 mol/(L¡¤min)¡£Çó£º
(1)´ËʱAµÄŨ¶Èc(A)£½________mol/L£¬·´Ó¦¿ªÊ¼Ç°ÈÝÆ÷ÖеÄA¡¢BµÄÎïÖʵÄÁ¿£ºn(A)£½n(B)£½________mol¡£
(2)BµÄƽ¾ù·´Ó¦ËÙÂÊv(B)£½________mol/(L¡¤min)¡£
(3)xµÄֵΪ________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì½Î÷Ê¡¸ßÒ»ÏÂѧÆÚµÚÁù´Î¶Î¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
(4·Ö)½«µÈÎïÖʵÄÁ¿µÄA¡¢B»ìºÏÓÚ2 LµÄÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÈçÏ·´Ó¦3A(g)£«B(g)
xC(g)£«2D(g)£¬¾5 minºó£¬²âµÃDµÄŨ¶ÈΪ0.5 mol/L£¬c(A)¡Ãc(B)£½3¡Ã5£¬CµÄƽ¾ù·´Ó¦ËÙÂÊΪ0.1 mol/(L¡¤min)£®Çó£º
(1)´ËʱAµÄŨ¶Èc(A)£½ mol/L£¬·´Ó¦¿ªÊ¼Ç°ÈÝÆ÷ÖеÄA¡¢BµÄÎïÖʵÄÁ¿£ºn(A)£½n(B)£½ mol¡£
(2)BµÄƽ¾ù·´Ó¦ËÙÂÊv(B)£½ mol/(L¡¤min)¡£ (3)xµÄֵΪ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - 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