(16·Ö)Óйش߻¯¼ÁµÄ´ß»¯»úÀí¿ÉÒÔ´Ó¡°ÒÒ´¼´ß»¯Ñõ»¯ÊµÑ顱µÃµ½Ò»Ð©ÈÏʶ£¬Ä³ÀÏʦÉè¼ÆÁËÈçͼװÖÃ(¼Ð³Ö×°ÖÃÒÇÆ÷ÒÑÊ¡ÂÔ)£¬ÆäʵÑé²Ù×÷Ϊ£ºÏȰ´Í¼°²×°ºÃ£¬ÏȹرջîÈûa¡¢b¡¢c£¬ÔÚÍ­Ë¿µÄÖм䲿·Ö¼ÓÈÈÆ¬¿Ì£¬È»ºó´ò¿ª»îÈûa¡¢b¡¢c£¬Í¨¹ý¿ØÖÆ»îÈûaºÍb£¬¶øÓнÚ×à(¼äЪÐÔ)ͨÈëÆøÌ壬¼´¿ÉÔÚM´¦¹Û²ìµ½Ã÷ÏÔµÄʵÑéÏÖÏó¡£ÊԻشðÒÔÏÂÎÊÌ⣺

(1)¼ìÑé×°ÖÃAÆøÃÜÐԵķ½·¨ÊÇ                                                £¬

AÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                              £¬

CÖÐÉÕÆ¿³£³£½þÔÚ70¡«80¡æµÄÈÈË®ÖУ¬Ä¿µÄÊÇ                                 ¡£

(2)M¹ÜÖпɹ۲쵽µÄÏÖÏóΪ                                                £¬´ÓÖпÉÈÏʶµ½¸ÃʵÑé¹ý³ÌµÄ´ß»¯¼ÁÊÇ      ¡£

(3)ÈôÊÔ¹ÜFÖÐÓÃË®ÎüÊÕ²úÎÔòÒªÔÚµ¼¹Üm¡¢nÖ®¼ä½ÓÉÏG×°Öã¬ÆäÁ¬½Ó·½·¨ÊÇ(ÌîG×°ÖÃÖе¼¹ÜµÄ´úºÅ)£ºm½Ó______¡¢_______½Ón£¬G×°ÖõÄ×÷ÓÃÊÇ                  £»Èô²úÎï²»ÓÃË®ÎüÊÕ¶øÊÇÖ±½ÓÀäÈ´£¬Ó¦½«ÊÔ¹ÜF½þÔÚ              ÖС£

 

¡¾´ð°¸¡¿

(16·Ö)(1)¹Ø±Õ»îÈûb£¬´ò¿ª»îÈûaÏòÉÕÆ¿ÖмÓË®£¬Ò»¶Îʱ¼äºóË®²»ÔÚÁ÷Ï£¬ÇÒ´æÔÚÒºÃæ²î            ¼Ó¿ìÒÒ´¼Æû»¯µÄËÙÂÊ    £¨¸÷2·Ö£©

(2)Í­Ë¿ÓÉÓÚ¼äЪÐԵعÄÈë¿ÕÆø¶ø½»Ìæ³öÏÖ±äºÚ£¬±äºìµÄÏÖÏó   £¨2·Ö£©   Í­  £¨2·Ö£©

(3)q    p   £¨2·Ö£©     ·Àµ¹Îü£¨2·Ö£©     ±ùË®£¨2·Ö£©

¡¾½âÎö¡¿

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(16·Ö)Óйش߻¯¼ÁµÄ´ß»¯»úÀí¿ÉÒÔ´Ó¡°ÒÒ´¼´ß»¯Ñõ»¯ÊµÑ顱µÃµ½Ò»Ð©ÈÏʶ£¬Ä³ÀÏʦÉè¼ÆÁËÈçͼװÖÃ(¼Ð³Ö×°ÖÃÒÇÆ÷ÒÑÊ¡ÂÔ)£¬ÆäʵÑé²Ù×÷Ϊ£ºÏȰ´Í¼°²×°ºÃ£¬ÏȹرջîÈûa¡¢b¡¢c£¬ÔÚÍ­Ë¿µÄÖм䲿·Ö¼ÓÈÈÆ¬¿Ì£¬È»ºó´ò¿ª»îÈûa¡¢b¡¢c£¬Í¨¹ý¿ØÖÆ»îÈûaºÍb£¬¶øÓнÚ×à(¼äЪÐÔ)ͨÈëÆøÌ壬¼´¿ÉÔÚM´¦¹Û²ìµ½Ã÷ÏÔµÄʵÑéÏÖÏó¡£ÊԻشðÒÔÏÂÎÊÌ⣺

(1)¼ìÑé×°ÖÃAÆøÃÜÐԵķ½·¨ÊÇ                                               £¬

AÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                             £¬

CÖÐÉÕÆ¿³£³£½þÔÚ70¡«80¡æµÄÈÈË®ÖУ¬Ä¿µÄÊÇ                                ¡£

(2)M¹ÜÖпɹ۲쵽µÄÏÖÏóΪ                                               £¬´ÓÖпÉÈÏʶµ½¸ÃʵÑé¹ý³ÌµÄ´ß»¯¼ÁÊÇ      ¡£

(3)ÈôÊÔ¹ÜFÖÐÓÃË®ÎüÊÕ²úÎÔòÒªÔÚµ¼¹Üm¡¢nÖ®¼ä½ÓÉÏG×°Öã¬ÆäÁ¬½Ó·½·¨ÊÇ(ÌîG×°ÖÃÖе¼¹ÜµÄ´úºÅ)£ºm½Ó______¡¢_______½Ón£¬G×°ÖõÄ×÷ÓÃÊÇ                 £»Èô²úÎï²»ÓÃË®ÎüÊÕ¶øÊÇÖ±½ÓÀäÈ´£¬Ó¦½«ÊÔ¹ÜF½þÔÚ             ÖС£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìɽ¶«Ê¡×Ͳ©Ò»ÖиßÈýÉÏѧÆÚÆÚÄ©¼ì²â»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

(16·Ö)£¨1£©ÒÑÖª:
Fe£¨s£©+1/2O2£¨g£©=FeO£¨s£©   
2Al£¨s£©+3/2O2£¨g£©= Al2O3£¨s£© 
AlºÍFeO·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ_____________________________________¡£
£¨2£©·´Ó¦ÎïÓëÉú³ÉÎï¾ùÎªÆøÌ¬µÄij¿ÉÄæ·´Ó¦ÔÚ²»Í¬Ìõ¼þϵķ´Ó¦Àú³Ì·Ö±ðΪA¡¢B£¬ÈçÏÂͼËùʾ¡£

¢Ù¾ÝͼÅжϸ÷´Ó¦ÊÇ_____(Ìî¡°Îü¡±»ò¡°·Å¡±) ÈÈ·´Ó¦£¬µ±·´Ó¦´ïµ½Æ½ºâºó£¬ÆäËûÌõ¼þ²»±ä£¬Éý¸ßζȣ¬·´Ó¦ÎïµÄת»¯ÂÊ__   _ (Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)
¢ÚÆäÖÐBÀú³Ì±íÃ÷´Ë·´Ó¦²ÉÓõÄÌõ¼þΪ______ (Ñ¡ÌîÏÂÁÐÐòºÅ×Öĸ)
A£®Éý¸ßζȠ              B£®Ôö´ó·´Ó¦ÎïµÄŨ¶È
C£®½µµÍζȠ              D£®Ê¹ÓÃÁË´ß»¯¼Á
£¨3£©1000¡æÊ±£¬ÁòËáÄÆÓëÇâÆø·¢ÉúÏÂÁз´Ó¦£ºNa2SO4(s) + 4H2(g)  Na2S(s) + 4H2O(g) ¡£
¢Ù¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ____________________¡£ÒÑÖªK1000¡æ£¼K1200¡æ£¬Ôò¸Ã·´Ó¦ÊÇ________·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©¡£
¢ÚÓÃÓйØÀë×Ó·½³Ìʽ˵Ã÷ÉÏÊö·´Ó¦ËùµÃ¹ÌÌå²úÎïµÄË®ÈÜÒºµÄËá¼îÐÔ____________
£¨4£©³£ÎÂÏ£¬Èç¹ûÈ¡0.1mol¡¤L£­1HAÈÜÒºÓë0.1mol¡¤L£­1NaOHÈÜÒºµÈÌå»ý»ìºÏ£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©£¬²âµÃ»ìºÏÒºµÄpH£½8£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù»ìºÏÈÜÒºÖÐË®µçÀë³öµÄc(H£«)Óë0.1mol¡¤L£­1NaOHÈÜÒºÖÐË®µçÀë³öµÄc(H£«)±È½Ï
                 £¨Ì¡¢£¾¡¢£½£©¡£
¢ÚÒÑÖªNH4AÈÜҺΪÖÐÐÔ£¬ÓÖÖª½«HAÈÜÒº¼Óµ½Na2CO3ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ÊÔÍÆ¶Ï(NH4)2CO3ÈÜÒºµÄpH            7£¨Ì¡¢£¾¡¢£½£©£»ÏàͬζÈÏ£¬µÈÎïÖʵÄÁ¿Å¨¶ÈµÄÏÂÁÐÑÎÈÜÒº°´pHÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòΪ                         ¡££¨ÌîÐòºÅ£©
a.£®NH4HCO3       b£®NH4A         c£®(NH4)2CO3       d£®NH4Cl

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011½ì¸ÊËàÊ¡À¼ÖÝÒ»ÖиßÈýµÚÈý´ÎÄ£Ä⿼ÊÔ£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÊµÑéÌâ

(16·Ö)Óйش߻¯¼ÁµÄ´ß»¯»úÀí¿ÉÒÔ´Ó¡°ÒÒ´¼´ß»¯Ñõ»¯ÊµÑ顱µÃµ½Ò»Ð©ÈÏʶ£¬Ä³ÀÏʦÉè¼ÆÁËÈçͼװÖÃ(¼Ð³Ö×°ÖÃÒÇÆ÷ÒÑÊ¡ÂÔ)£¬ÆäʵÑé²Ù×÷Ϊ£ºÏȰ´Í¼°²×°ºÃ£¬ÏȹرջîÈûa¡¢b¡¢c£¬ÔÚÍ­Ë¿µÄÖм䲿·Ö¼ÓÈÈÆ¬¿Ì£¬È»ºó´ò¿ª»îÈûa¡¢b¡¢c£¬Í¨¹ý¿ØÖÆ»îÈûaºÍb£¬¶øÓнÚ×à(¼äЪÐÔ)ͨÈëÆøÌ壬¼´¿ÉÔÚM´¦¹Û²ìµ½Ã÷ÏÔµÄʵÑéÏÖÏó¡£ÊԻشðÒÔÏÂÎÊÌ⣺

(1)¼ìÑé×°ÖÃAÆøÃÜÐԵķ½·¨ÊÇ                                               £¬
AÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                             £¬
CÖÐÉÕÆ¿³£³£½þÔÚ70¡«80¡æµÄÈÈË®ÖУ¬Ä¿µÄÊÇ                                ¡£
(2)M¹ÜÖпɹ۲쵽µÄÏÖÏóΪ                                               £¬´ÓÖпÉÈÏʶµ½¸ÃʵÑé¹ý³ÌµÄ´ß»¯¼ÁÊÇ     ¡£
(3)ÈôÊÔ¹ÜFÖÐÓÃË®ÎüÊÕ²úÎÔòÒªÔÚµ¼¹Üm¡¢nÖ®¼ä½ÓÉÏG×°Öã¬ÆäÁ¬½Ó·½·¨ÊÇ(ÌîG×°ÖÃÖе¼¹ÜµÄ´úºÅ)£ºm½Ó______¡¢_______½Ón£¬G×°ÖõÄ×÷ÓÃÊÇ                 £»Èô²úÎï²»ÓÃË®ÎüÊÕ¶øÊÇÖ±½ÓÀäÈ´£¬Ó¦½«ÊÔ¹ÜF½þÔÚ             ÖС£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêɽ¶«Ê¡¸ßÈýÉÏѧÆÚÆÚÄ©¼ì²â»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

(16·Ö)£¨1£©ÒÑÖª:

Fe£¨s£©+1/2O2£¨g£©=FeO£¨s£©    

2Al£¨s£©+3/2O2£¨g£©= Al2O3£¨s£© 

AlºÍFeO·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ_____________________________________¡£

£¨2£©·´Ó¦ÎïÓëÉú³ÉÎï¾ùÎªÆøÌ¬µÄij¿ÉÄæ·´Ó¦ÔÚ²»Í¬Ìõ¼þϵķ´Ó¦Àú³Ì·Ö±ðΪA¡¢B£¬ÈçÏÂͼËùʾ¡£

¢Ù¾ÝͼÅжϸ÷´Ó¦ÊÇ_____(Ìî¡°Îü¡±»ò¡°·Å¡±) ÈÈ·´Ó¦£¬µ±·´Ó¦´ïµ½Æ½ºâºó£¬ÆäËûÌõ¼þ²»±ä£¬Éý¸ßζȣ¬·´Ó¦ÎïµÄת»¯ÂÊ__   _ (Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)

¢ÚÆäÖÐBÀú³Ì±íÃ÷´Ë·´Ó¦²ÉÓõÄÌõ¼þΪ______ (Ñ¡ÌîÏÂÁÐÐòºÅ×Öĸ)

A£®Éý¸ßζȠ               B£®Ôö´ó·´Ó¦ÎïµÄŨ¶È

C£®½µµÍζȠ               D£®Ê¹ÓÃÁË´ß»¯¼Á

£¨3£©1000¡æÊ±£¬ÁòËáÄÆÓëÇâÆø·¢ÉúÏÂÁз´Ó¦£ºNa2SO4(s) + 4H2(g)  Na2S(s) + 4H2O(g) ¡£

¢Ù¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ____________________¡£ÒÑÖªK1000¡æ£¼K1200¡æ£¬Ôò¸Ã·´Ó¦ÊÇ________·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©¡£

¢ÚÓÃÓйØÀë×Ó·½³Ìʽ˵Ã÷ÉÏÊö·´Ó¦ËùµÃ¹ÌÌå²úÎïµÄË®ÈÜÒºµÄËá¼îÐÔ____________

£¨4£©³£ÎÂÏ£¬Èç¹ûÈ¡0.1mol¡¤L£­1HAÈÜÒºÓë0.1mol¡¤L£­1NaOHÈÜÒºµÈÌå»ý»ìºÏ£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©£¬²âµÃ»ìºÏÒºµÄpH£½8£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ù»ìºÏÈÜÒºÖÐË®µçÀë³öµÄc(H£«)Óë0.1mol¡¤L£­1NaOHÈÜÒºÖÐË®µçÀë³öµÄc(H£«)±È½Ï

                  £¨Ì¡¢£¾¡¢£½£©¡£

¢ÚÒÑÖªNH4AÈÜҺΪÖÐÐÔ£¬ÓÖÖª½«HAÈÜÒº¼Óµ½Na2CO3ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ÊÔÍÆ¶Ï(NH4)2CO3ÈÜÒºµÄpH             7£¨Ì¡¢£¾¡¢£½£©£»ÏàͬζÈÏ£¬µÈÎïÖʵÄÁ¿Å¨¶ÈµÄÏÂÁÐÑÎÈÜÒº°´pHÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòΪ                          ¡££¨ÌîÐòºÅ£©

a.£®NH4HCO3       b£®NH4A          c£®(NH4)2CO3       d£®NH4Cl

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸