ÓÃpHÊÔÖ½²â¶¨ÈÜÒºpHµÄÕýÈ·²Ù×÷ÊÇ£¨¡¡¡¡£©

A. ½«Ò»Ð¡¿éÊÔÖ½·ÅÔÚ±íÃæÃóÉÏ£¬Óò£Á§°ôպȡÉÙÁ¿´ý²âÒºµãÔÚÊÔÖ½ÉÏ£¬ÔÙÓë±ê×¼±ÈÉ«¿¨¶ÔÕÕ

B. ½«Ò»Ð¡¿éÊÔÖ½ÓÃÕôÁóË®Èóʪºó·ÅÔÚ±íÃæÃóÉÏ£¬Óò£Á§°ôպȡÉÙÁ¿´ý²âµãÔÚÊÔÖ½ÉÏ£¬ÔÙÓë±ê×¼±ÈÉ«¿¨¶ÔÕÕ

C. ½«Ò»Ð¡ÌõÊÔÖ½ÔÚ´ý²âÒºÖÐպһϣ¬È¡³öºó·ÅÔÚ±íÃæÃóÉÏ£¬Óë±ê×¼±ÈÉ«¿¨¶ÔÕÕ

D. ½«Ò»Ð¡ÌõÊÔÖ½ÏÈÓÃÕôÁóË®Èóʪºó£¬ÔÚ´ý²âÒºÖÐպһϣ¬È¡³öºóÓë±ê×¼±ÈÉ«¿¨¶ÔÕÕ

   

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¢ñ.ÏÂÁÐÓйØʵÑé²Ù×÷»ò½á¹ûµÄ˵·¨ÖÐÕýÈ·µÄÊÇ                  £¨Ìî×Öĸ£©

A£®µÎ¶¨Ê±£¬ÑÛ¾¦Ó¦Ê¼ÖÕ×¢Êӵζ¨¹ÜÄÚÒºÃæµÄ±ä»¯

B£®ÓüîʽµÎ¶¨¹ÜÁ¿È¡0.10 mol¡¤L£­1µÄKMnO4ÈÜÒº15.10 mL

C£®Ëá¼îÖк͵ζ¨Ö®Ç°£¬×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»¼´¿É£¬²»ÄÜÓôý²âÒºÈóÏ´

D£®ÓÃpHÊÔÖ½²âÁ¿Ä³ÈÜÒºµÄpHʱҪÏȽ«ÊÔÖ½Èóʪ

E£®µÎ¶¨¹Ü¾­ÕôÁóˮϴ¾»ºó£¬Ö±½Ó×¢Èë±ê×¼Òº£¬½«Ê¹²âµÃµÄ´ý²âҺŨ¶ÈÆ«¸ß

F£®Óù㷺pHÊÔÖ½²âÁ¿H2SO4ÈÜÒºµÄpHʱ£¬²âµÃpH=3.2

G£®²â¶¨Ëá¼îµÎ¶¨ÇúÏߣº¿ªÊ¼Ê±²âÊԺͼǼµÄ¼ä¸ô¿ÉÉÔСЩ£¬µÎ¶¨ÖÁÖյ㸽½üÔòÒª´óЩ

¢ò.£¨1£©ÓÉÇâÆøºÍÑõÆø·´Ó¦Éú³É1 molË®ÕôÆø·ÅÈÈ241.8kJ£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                       ¡£ÒÑÖªH2O(l) £½ H2O(g) ¦¤H £½£«44 kJ¡¤mol£­1 Ôò±ê×¼×´¿öÏÂ33.6 L H2 Éú³ÉҺ̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇ                    kJ¡£

£¨2£©»¯Ñ§·´Ó¦¿ÉÊÓΪ¾É¼ü¶ÏÁѺÍмüÐγɵĹý³Ì¡£»¯Ñ§¼üµÄ¼üÄÜÊÇÐγɣ¨»ò²ð¿ª£©1mol»¯Ñ§¼üʱÊÍ·Å£¨»òÎüÊÕ£©µÄÄÜÁ¿¡£ÒÑÖª°×Á×P4ºÍP4O6µÄ·Ö×ӽṹÈçÏÂͼËùʾ£º

ÏÖÌṩÒÔÏ»¯Ñ§¼üµÄ¼üÄÜ£ºP-P 198 kJ¡¤mol£­1 P-O 360 kJ¡¤mol£­1,ÑõÆø·Ö×ÓÄÚÑõÔ­×Ó¼äµÄ£¨O=O£©¼üÄÜΪ498 kJ¡¤mol£­1¡£ÔòP4+3O2 = P4O6µÄ·´Ó¦ÈȦ¤HΪ              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¢ñ.ÏÂÁÐÓйØʵÑé²Ù×÷»ò½á¹ûµÄ˵·¨ÖÐÕýÈ·µÄÊÇ                 £¨Ìî×Öĸ£©

A£®µÎ¶¨Ê±£¬ÑÛ¾¦Ó¦Ê¼ÖÕ×¢Êӵζ¨¹ÜÄÚÒºÃæµÄ±ä»¯

B£®ÓüîʽµÎ¶¨¹ÜÁ¿È¡0.10 mol¡¤L£­1µÄKMnO4ÈÜÒº15.10 mL

C£®Ëá¼îÖк͵ζ¨Ö®Ç°£¬×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»¼´¿É£¬²»ÄÜÓôý²âÒºÈóÏ´

D£®ÓÃpHÊÔÖ½²âÁ¿Ä³ÈÜÒºµÄpHʱҪÏȽ«ÊÔÖ½Èóʪ

E£®µÎ¶¨¹Ü¾­ÕôÁóˮϴ¾»ºó£¬Ö±½Ó×¢Èë±ê×¼Òº£¬½«Ê¹²âµÃµÄ´ý²âҺŨ¶ÈÆ«¸ß

F£®Óù㷺pHÊÔÖ½²âÁ¿H2SO4ÈÜÒºµÄpHʱ£¬²âµÃpH=3.2

G£®²â¶¨Ëá¼îµÎ¶¨ÇúÏߣº¿ªÊ¼Ê±²âÊԺͼǼµÄ¼ä¸ô¿ÉÉÔСЩ£¬µÎ¶¨ÖÁÖյ㸽½üÔòÒª´óЩ

¢ò.£¨1£©ÓÉÇâÆøºÍÑõÆø·´Ó¦Éú³É1 molË®ÕôÆø·ÅÈÈ241.8kJ£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                      ¡£ÒÑÖªH2O(l) £½ H2O(g) ¦¤H £½£«44 kJ¡¤mol£­1 Ôò±ê×¼×´¿öÏÂ33.6 L H2 Éú³ÉҺ̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇ                    kJ¡£

£¨2£©»¯Ñ§·´Ó¦¿ÉÊÓΪ¾É¼ü¶ÏÁѺÍмüÐγɵĹý³Ì¡£»¯Ñ§¼üµÄ¼üÄÜÊÇÐγɣ¨»ò²ð¿ª£©1mol»¯Ñ§¼üʱÊÍ·Å£¨»òÎüÊÕ£©µÄÄÜÁ¿¡£ÒÑÖª°×Á×P4ºÍP4O6µÄ·Ö×ӽṹÈçÏÂͼËùʾ£º

ÏÖÌṩÒÔÏ»¯Ñ§¼üµÄ¼üÄÜ£ºP-P 198 kJ¡¤mol£­1 P-O 360 kJ¡¤mol£­1,ÑõÆø·Ö×ÓÄÚÑõÔ­×Ó¼äµÄ£¨O=O£©¼üÄÜΪ498kJ¡¤mol£­1¡£ÔòP4+3O2= P4O6µÄ·´Ó¦ÈȦ¤HΪ             ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêºÚÁú½­Äµµ¤½­Ò»Öи߶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÊµÑéÌâ

¢ñ.ÏÂÁÐÓйØʵÑé²Ù×÷»ò½á¹ûµÄ˵·¨ÖÐÕýÈ·µÄÊÇ                 £¨Ìî×Öĸ£©

A£®µÎ¶¨Ê±£¬ÑÛ¾¦Ó¦Ê¼ÖÕ×¢Êӵζ¨¹ÜÄÚÒºÃæµÄ±ä»¯
B£®ÓüîʽµÎ¶¨¹ÜÁ¿È¡0.10 mol¡¤L£­1µÄKMnO4ÈÜÒº15.10 mL
C£®Ëá¼îÖк͵ζ¨Ö®Ç°£¬×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»¼´¿É£¬²»ÄÜÓôý²âÒºÈóÏ´
D£®ÓÃpHÊÔÖ½²âÁ¿Ä³ÈÜÒºµÄpHʱҪÏȽ«ÊÔÖ½Èóʪ
E£®µÎ¶¨¹Ü¾­ÕôÁóˮϴ¾»ºó£¬Ö±½Ó×¢Èë±ê×¼Òº£¬½«Ê¹²âµÃµÄ´ý²âҺŨ¶ÈÆ«¸ß
F£®Óù㷺pHÊÔÖ½²âÁ¿H2SO4ÈÜÒºµÄpHʱ£¬²âµÃpH=3.2
G£®²â¶¨Ëá¼îµÎ¶¨ÇúÏߣº¿ªÊ¼Ê±²âÊԺͼǼµÄ¼ä¸ô¿ÉÉÔСЩ£¬µÎ¶¨ÖÁÖյ㸽½üÔòÒª´óЩ
¢ò.£¨1£©ÓÉÇâÆøºÍÑõÆø·´Ó¦Éú³É1 molË®ÕôÆø·ÅÈÈ241.8kJ£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                      ¡£ÒÑÖªH2O(l) £½ H2O(g) ¦¤H £½£«44 kJ¡¤mol£­1Ôò±ê×¼×´¿öÏÂ33.6 L H2Éú³ÉҺ̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇ                   kJ¡£
£¨2£©»¯Ñ§·´Ó¦¿ÉÊÓΪ¾É¼ü¶ÏÁѺÍмüÐγɵĹý³Ì¡£»¯Ñ§¼üµÄ¼üÄÜÊÇÐγɣ¨»ò²ð¿ª£©1mol»¯Ñ§¼üʱÊÍ·Å£¨»òÎüÊÕ£©µÄÄÜÁ¿¡£ÒÑÖª°×Á×P4ºÍP4O6µÄ·Ö×ӽṹÈçÏÂͼËùʾ£º
ÏÖÌṩÒÔÏ»¯Ñ§¼üµÄ¼üÄÜ£ºP-P 198 kJ¡¤mol£­1 P-O 360 kJ¡¤mol£­1,ÑõÆø·Ö×ÓÄÚÑõÔ­×Ó¼äµÄ£¨O=O£©¼üÄÜΪ498 kJ¡¤mol£­1¡£ÔòP4+3O2 = P4O6µÄ·´Ó¦ÈȦ¤HΪ             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêºÚÁú½­Äµµ¤½­Ò»Öи߶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÊµÑéÌâ

¢ñ.ÏÂÁÐÓйØʵÑé²Ù×÷»ò½á¹ûµÄ˵·¨ÖÐÕýÈ·µÄÊÇ                  £¨Ìî×Öĸ£©

A£®µÎ¶¨Ê±£¬ÑÛ¾¦Ó¦Ê¼ÖÕ×¢Êӵζ¨¹ÜÄÚÒºÃæµÄ±ä»¯

B£®ÓüîʽµÎ¶¨¹ÜÁ¿È¡0.10 mol¡¤L£­1µÄKMnO4ÈÜÒº15.10 mL

C£®Ëá¼îÖк͵ζ¨Ö®Ç°£¬×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»¼´¿É£¬²»ÄÜÓôý²âÒºÈóÏ´

D£®ÓÃpHÊÔÖ½²âÁ¿Ä³ÈÜÒºµÄpHʱҪÏȽ«ÊÔÖ½Èóʪ

E£®µÎ¶¨¹Ü¾­ÕôÁóˮϴ¾»ºó£¬Ö±½Ó×¢Èë±ê×¼Òº£¬½«Ê¹²âµÃµÄ´ý²âҺŨ¶ÈÆ«¸ß

F£®Óù㷺pHÊÔÖ½²âÁ¿H2SO4ÈÜÒºµÄpHʱ£¬²âµÃpH=3.2

G£®²â¶¨Ëá¼îµÎ¶¨ÇúÏߣº¿ªÊ¼Ê±²âÊԺͼǼµÄ¼ä¸ô¿ÉÉÔСЩ£¬µÎ¶¨ÖÁÖյ㸽½üÔòÒª´óЩ

¢ò.£¨1£©ÓÉÇâÆøºÍÑõÆø·´Ó¦Éú³É1 molË®ÕôÆø·ÅÈÈ241.8kJ£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                       ¡£ÒÑÖªH2O(l) £½ H2O(g) ¦¤H £½£«44 kJ¡¤mol£­1 Ôò±ê×¼×´¿öÏÂ33.6 L H2 Éú³ÉҺ̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇ                    kJ¡£

£¨2£©»¯Ñ§·´Ó¦¿ÉÊÓΪ¾É¼ü¶ÏÁѺÍмüÐγɵĹý³Ì¡£»¯Ñ§¼üµÄ¼üÄÜÊÇÐγɣ¨»ò²ð¿ª£©1mol»¯Ñ§¼üʱÊÍ·Å£¨»òÎüÊÕ£©µÄÄÜÁ¿¡£ÒÑÖª°×Á×P4ºÍP4O6µÄ·Ö×ӽṹÈçÏÂͼËùʾ£º

ÏÖÌṩÒÔÏ»¯Ñ§¼üµÄ¼üÄÜ£ºP-P 198 kJ¡¤mol£­1 P-O 360 kJ¡¤mol£­1,ÑõÆø·Ö×ÓÄÚÑõÔ­×Ó¼äµÄ£¨O=O£©¼üÄÜΪ498 kJ¡¤mol£­1¡£ÔòP4+3O2 = P4O6µÄ·´Ó¦ÈȦ¤HΪ              ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(08ÏÃÃÅÊÐÖʼì)ÏÂÁÐʵÑé·½°¸ºÏÀíµÄÊÇ      £¨    £©

       A£®ÓÃʪÈóµÄpHÊÔÖ½²â¶¨Ä³ÑÎÈÜÒºµÄpH

       B£®ÓÃÏ¡ÁòËáÇåÏ´³¤ÆÚ´æ·ÅÂÈ»¯ÌúÈÜÒºµÄÊÔ¼ÁÆ¿ÄÚ±ÚËù¸½×ŵIJ»ÈÜÎï

       C£®ÓüÓÈÈ·¨·ÖÀëI2ºÍNH4C1µÄ»ìºÏÎï

       D£®À´Á˼ӿìµÃµ½ÇâÆø£¬¿ÉÒÔÓô¿´â´úÌæ´Öп

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸