£¨14·Ö£©ÒÑÖª£º³£ÎÂÏ£¬AËáµÄÈÜÒºpH£½a£¬B¼îµÄÈÜÒºpH£½b¡£
£¨1£©ÈôAΪÑÎËᣬBΪÇâÑõ»¯ÄÆ£¬ÇÒa£«b£½14£¬¶þÕßµÈÌå»ý»ìºÏ£¬ÔòÈÜÒºµÄpH£½   ¡£ÈôËá¼î°´Ìå»ý±ÈΪ1£º10»ìºÏºóÈÜÒºÏÔÖÐÐÔ£¬Ôòa£«b£½            ¡£
£¨2£©ÈôAΪ´×ËᣬBΪÇâÑõ»¯±µ£¬ÇÒa£½4£¬b£½12£¬ÄÇôAÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ         mol¡¤L£­1£¬BÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ         ¡£mol¡¤L£­1
£¨3£©ÈôAΪ´×ËᣬBΪÇâÑõ»¯ÄÆ£¬ÇÒa£«b£½14£¬½«Ìå»ýΪVAµÄ´×ËáºÍÌå»ýΪVBµÄÇâÑõ»¯ÄÆÈÜÒº»ìºÏºó£¬ÈÜÒºÏÔÖÐÐÔ£¬ÔòÆäÌå»ý¹ØϵVA      VB£¬»ìºÏºóÈÜÒºÖеÄÀë×ÓŨ¶È¹ØϵΪc(Na£«)        c(CH3COO£­) £¨Ìî¡°£¼¡±¡°£¾¡±»ò¡°£½¡±£©¡£
£¨4£©ÈôAµÄ»¯Ñ§Ê½ÎªHR£¬BµÄ»¯Ñ§Ê½ÎªMOH£¬ÇÒa£«b£½14£¬Á½ÕßµÈÌå»ý»ìºÏºóÈÜÒºÏÔ¼îÐÔ¡£Ôò»ìºÏÈÜÒºÖбض¨ÓÐÒ»ÖÖÀë×ÓÄÜ·¢ÉúË®½â£¬ÆäË®½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ               ¡£

£¨1£©7   13   £¨2£©10¨D10     10¨D12    £¨3£©£¼   £½

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖª£º³£ÎÂÏ£¬AËáµÄÈÜÒºpH=a£¬B¼îµÄÈÜÒºpH=b
£¨1£©ÈôAΪÑÎËᣬBΪÇâÑõ»¯±µ£¬ÇÒa=3£¬b=11£¬Á½ÕßµÈÌå»ý»ìºÏ£¬ÈÜÒºµÄpHΪ
 
£®
a  ´óÓÚ7      b  µÈÓÚ7       c  Ð¡ÓÚ7
£¨2£©ÈôAΪ´×ËᣬBΪÇâÑõ»¯ÄÆ£¬ÇÒa=4£¬b=12£¬ÄÇôAÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ
 
mol/L£¬BÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ
 
mol/L£®
£¨3£©ÈôAµÄ»¯Ñ§Ê½ÎªHR£¬BµÄ»¯Ñ§Ê½ÎªMOH£¬ÇÒa+b=14£¬Á½ÕßµÈÌå»ý»ìºÏºóÈÜÒºÏÔ¼îÐÔ£®Ôò»ìºÏÈÜÒºÖбض¨ÓÐÒ»ÖÖÀë×ÓÄÜ·¢ÉúË®½â£¬¸ÃË®½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨15·Ö£©¢ñ.ÔÚÒ»¸öÈÝ»ý¹Ì¶¨Îª2LµÄÃܱÕÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£ºaA (g) + bB(g) pC(g)  ¡÷H=?·´Ó¦Çé¿ö¼Ç¼ÈçÏÂ±í£º

ʱ¼ä/(min)

n(A)/( mol)

n(B)/( mol)

n(C)/( mol)

0

1

3

0

µÚ2 min

0.8

2.6

0.4

µÚ4 min

0.4

1.8

1.2

µÚ6 min

0.4

1.8

1.2

µÚ8 min

0.1

2.0

1.8

µÚ9 min

0.05

1.9

0.3

Çë¸ù¾Ý±íÖÐÊý¾Ý×Ðϸ·ÖÎö£¬»Ø´ðÏÂÁÐÎÊÌ⣺

(1)µÚ2minµ½µÚ4minÄÚAµÄƽ¾ù·´Ó¦ËÙÂÊV(A)=                mol•L£­1• min£­1

(2)ÓɱíÖÐÊý¾Ý¿ÉÖª·´Ó¦ÔÚµÚ4minµ½µÚ6minʱ´¦ÓÚƽºâ״̬£¬ÈôÔÚµÚ2min¡¢µÚ6min¡¢µÚ8 minʱ·Ö±ð¸Ä±äÁËijһ¸ö·´Ó¦Ìõ¼þ£¬Ôò¸Ä±äµÄÌõ¼þ·Ö±ð¿ÉÄÜÊÇ£º

¢ÙµÚ2min                            »ò                                  £»

¢ÚµÚ6min                                      £»

¢ÛµÚ8 min                                     ¡£

(3)Èô´Ó¿ªÊ¼µ½µÚ4 min½¨Á¢Æ½ºâʱ·´Ó¦·Å³öµÄÈÈÁ¿Îª235.92kJÔò¸Ã·´Ó¦µÄ¡÷H=        ¡£

(4)·´Ó¦ÔÚµÚ4 min½¨Á¢Æ½ºâ£¬´ËζÈϸ÷´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýK=              .

¢ò.ÒÑÖª£º³£ÎÂÏ£¬AËáµÄÈÜÒºpH£½a, B¼îµÄÈÜÒºpH£½b

(1)ÈôAΪÑÎËᣬBΪÇâÑõ»¯±µ£¬ÇÒa=3,b=11,Á½ÕßµÈÌå»ý»ìºÏ£¬ÈÜÒºµÄpHΪ              ¡£

A£®´óÓÚ7             B.µÈÓÚ7             C.СÓÚ7

(2)ÈôAΪ´×ËᣬBΪÇâÑõ»¯ÄÆ£¬ÇÒa=4,b=12£¬ÄÇôAÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ

            mol•L£­1£¬BÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ             mol•L£­1¡£

(3)ÈôAµÄ»¯Ñ§Ê½ÎªHR£¬BµÄ»¯Ñ§Ê½ÎªMOH£¬ÇÒa+b=14,Á½ÕßµÈÌå»ý»ìºÏºóÈÜÒºÏÔ¼îÐÔ¡£Ôò»ìºÏÈÜÒºÖбض¨ÓÐÒ»ÖÖÀë×ÓÄÜ·¢ÉúË®½â£¬¸ÃË®½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º

                                                             ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìºÚÁú½­Ê¡¹þ¶û±õÊеÚÁùÖÐѧ¸ßÈýÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨15·Ö£©¢ñ.ÔÚÒ»¸öÈÝ»ý¹Ì¶¨Îª2LµÄÃܱÕÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£ºaA (g) + bB(g) pC(g) ¡÷H=?·´Ó¦Çé¿ö¼Ç¼ÈçÏÂ±í£º

ʱ¼ä/(min)
n(A)/( mol)
n(B)/( mol)
n(C)/( mol)
0
1
3
0
µÚ2 min
0.8
2.6
0.4
µÚ4 min
0.4
1.8
1.2
µÚ6 min
0.4
1.8
1.2
µÚ8 min
0.1
2.0
1.8
µÚ9 min
0.05
1.9
0.3
Çë¸ù¾Ý±íÖÐÊý¾Ý×Ðϸ·ÖÎö£¬»Ø´ðÏÂÁÐÎÊÌ⣺
(1)µÚ2minµ½µÚ4minÄÚAµÄƽ¾ù·´Ó¦ËÙÂÊV(A)=                mol?L£­1? min£­1
(2)ÓɱíÖÐÊý¾Ý¿ÉÖª·´Ó¦ÔÚµÚ4minµ½µÚ6minʱ´¦ÓÚƽºâ״̬£¬ÈôÔÚµÚ2min¡¢µÚ6min¡¢µÚ8 minʱ·Ö±ð¸Ä±äÁËijһ¸ö·´Ó¦Ìõ¼þ£¬Ôò¸Ä±äµÄÌõ¼þ·Ö±ð¿ÉÄÜÊÇ£º
¢ÙµÚ2min                            »ò                                  £»
¢ÚµÚ6min                                      £»
¢ÛµÚ8 min                                     ¡£
(3)Èô´Ó¿ªÊ¼µ½µÚ4 min½¨Á¢Æ½ºâʱ·´Ó¦·Å³öµÄÈÈÁ¿Îª235.92kJÔò¸Ã·´Ó¦µÄ¡÷H=       ¡£
(4)·´Ó¦ÔÚµÚ4 min½¨Á¢Æ½ºâ£¬´ËζÈϸ÷´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýK=              .
¢ò.ÒÑÖª£º³£ÎÂÏ£¬AËáµÄÈÜÒºpH£½a, B¼îµÄÈÜÒºpH£½b
(1)ÈôAΪÑÎËᣬBΪÇâÑõ»¯±µ£¬ÇÒa=3,b=11,Á½ÕßµÈÌå»ý»ìºÏ£¬ÈÜÒºµÄpHΪ              ¡£
A£®´óÓÚ7              B.µÈÓÚ7            C. СÓÚ7
(2)ÈôAΪ´×ËᣬBΪÇâÑõ»¯ÄÆ£¬ÇÒa=4,b=12£¬ÄÇôAÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ
            mol?L£­1£¬BÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ             mol?L£­1¡£
(3)ÈôAµÄ»¯Ñ§Ê½ÎªHR£¬BµÄ»¯Ñ§Ê½ÎªMOH£¬ÇÒa+b=14,Á½ÕßµÈÌå»ý»ìºÏºóÈÜÒºÏÔ¼îÐÔ¡£Ôò»ìºÏÈÜÒºÖбض¨ÓÐÒ»ÖÖÀë×ÓÄÜ·¢ÉúË®½â£¬¸ÃË®½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
                                                             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìºÚÁú½­¹þ¶û±õÊиßÈýÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨15·Ö£©¢ñ.ÔÚÒ»¸öÈÝ»ý¹Ì¶¨Îª2LµÄÃܱÕÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£ºaA (g) + bB(g) pC(g)  ¡÷H=?·´Ó¦Çé¿ö¼Ç¼ÈçÏÂ±í£º

ʱ¼ä/(min)

n(A)/( mol)

n(B)/( mol)

n(C)/( mol)

0

1

3

0

µÚ2 min

0.8

2.6

0.4

µÚ4 min

0.4

1.8

1.2

µÚ6 min

0.4

1.8

1.2

µÚ8 min

0.1

2.0

1.8

µÚ9 min

0.05

1.9

0.3

Çë¸ù¾Ý±íÖÐÊý¾Ý×Ðϸ·ÖÎö£¬»Ø´ðÏÂÁÐÎÊÌ⣺

(1)µÚ2minµ½µÚ4minÄÚAµÄƽ¾ù·´Ó¦ËÙÂÊV(A)=                 mol•L£­1• min£­1

(2)ÓɱíÖÐÊý¾Ý¿ÉÖª·´Ó¦ÔÚµÚ4minµ½µÚ6minʱ´¦ÓÚƽºâ״̬£¬ÈôÔÚµÚ2min¡¢µÚ6min¡¢µÚ8 minʱ·Ö±ð¸Ä±äÁËijһ¸ö·´Ó¦Ìõ¼þ£¬Ôò¸Ä±äµÄÌõ¼þ·Ö±ð¿ÉÄÜÊÇ£º

¢ÙµÚ2min                             »ò                                   £»

¢ÚµÚ6min                                       £»

¢ÛµÚ8 min                                      ¡£

(3)Èô´Ó¿ªÊ¼µ½µÚ4 min½¨Á¢Æ½ºâʱ·´Ó¦·Å³öµÄÈÈÁ¿Îª235.92kJÔò¸Ã·´Ó¦µÄ¡÷H=        ¡£

(4)·´Ó¦ÔÚµÚ4 min½¨Á¢Æ½ºâ£¬´ËζÈϸ÷´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýK=               .

¢ò.ÒÑÖª£º³£ÎÂÏ£¬AËáµÄÈÜÒºpH£½a, B¼îµÄÈÜÒºpH£½b

(1)ÈôAΪÑÎËᣬBΪÇâÑõ»¯±µ£¬ÇÒa=3,b=11,Á½ÕßµÈÌå»ý»ìºÏ£¬ÈÜÒºµÄpHΪ               ¡£

A£®´óÓÚ7              B.µÈÓÚ7             C. СÓÚ7

(2)ÈôAΪ´×ËᣬBΪÇâÑõ»¯ÄÆ£¬ÇÒa=4,b=12£¬ÄÇôAÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ

             mol•L£­1£¬BÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ              mol•L£­1¡£

(3)ÈôAµÄ»¯Ñ§Ê½ÎªHR£¬BµÄ»¯Ñ§Ê½ÎªMOH£¬ÇÒa+b=14,Á½ÕßµÈÌå»ý»ìºÏºóÈÜÒºÏÔ¼îÐÔ¡£Ôò»ìºÏÈÜÒºÖбض¨ÓÐÒ»ÖÖÀë×ÓÄÜ·¢ÉúË®½â£¬¸ÃË®½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º

                                                              ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸