ij»¯Ñ§Ì½¾¿Ñ§Ï°Ð¡×éÉè¼ÆÈçÏÂͼװÖÃÖÆÈ¡ÏõËᣨ¼Ð³ÖºÍ¼ÓÈÈÒÇÆ÷¾ùÒÑÂÔÈ¥£©¡£ÊµÑéÖпɹ©Ê¹ÓõÄÒ©Æ·ÓУºNa2CO3¡¢NaHCO3¡¢NH4HCO3¡¢Na2O2¡¢NaOHÈÜÒººÍË®¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©×°ÖÃAÖз¢ÉúµÄ»¯Ñ§·´Ó¦·½³ÌʽÊÇ                               ¡£
£¨2£©³·È¥×°ÖÃDÖеļÓÈÈ×°Öú󣬲¬Ë¿ÈÔÈ»±£³ÖºìÈÈ£¬ÕâÊÇÒòΪDÖз¢ÉúµÄ»¯Ñ§·´Ó¦ÊÇÒ»¸ö         £¨Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±£©·´Ó¦¡£
£¨3£©×°ÖÃFÖÐÊ¢·ÅµÄÊÇ                  ÈÜÒº£¬Æä×÷ÓÃÊÇ                                          ¡£
£¨4£©ÊµÑé¹ý³ÌÖУ¬ÒªÊ¹NH4HCO3³ä·Öת»¯ÎªHNO3£¬»¹ÒªÔÚ×°ÖÃDÖÐͨÈë¹ýÁ¿µÄÑõÆø¡£¼×ͬѧÌáÒéÔÚC´¦Á¬½ÓÒ»¸öÖÆÈ¡ÑõÆøµÄ×°Öã¬ÒÒͬѧÈÏΪ¿ÉÖ±½ÓÔÚAÖÐÔÙ¼ÓÈëÉÏÊöÌṩҩƷÖеÄÒ»ÖÖÎïÖÊ£¬ÕâÖÖÒ©Æ·µÄ»¯Ñ§Ê½ÊÇ                 

£¨1£©NH4HCO3NH3£«H2O£«CO2¡ü
£¨2£©·ÅÈÈ
£¨3£©NaOH   ÎüÊÕ¿ÉÄÜδ·´Ó¦µÄµªÑõ»¯Î·ÀÖ¹ÎÛȾ¿ÕÆø
£¨4£©NaHCO3

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ÂÁþºÏ½ðÒѳÉΪÂÖ´¬ÖÆÔì¡¢»¯¹¤Éú²úµÈÐÐÒµµÄÖØÒª²ÄÁÏ£®Ñо¿ÐÔѧϰС×éµÄÈýλͬѧ£¬Îª²â¶¨Ä³ÂÁþºÏ½ð£¨Éè²»º¬ÆäËüÔªËØ£©ÖÐþµÄÖÊÁ¿·ÖÊý£¬Éè¼ÆÏÂÁÐÈýÖÖ²»Í¬ÊµÑé·½°¸½øÐÐ̽¾¿£®ÌîдÏÂÁпհף®
¡¾Ì½¾¿Ò»¡¿
ʵÑé·½°¸£ºÂÁþºÏ½ð
ÇâÑõ»¯ÄÆÈÜÒº
²â¶¨Ê£Óà¹ÌÌåÖÊÁ¿
ÎÊÌâÌÖÂÛ£º
£¨1£©³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÂÁþºÏ½ð·ÛÄ©ÑùÆ·£¬¼ÓÈë¹ýÁ¿µÄNaOHÈÜÒº£¬³ä·Ö·´Ó¦£®ÊµÑéÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
2A1+2NaOH+2H2O¨T2NaA1O2+3H2¡ü£¨»ò2A1+2NaOH+6H2O=2Na[A1£¨OH£©4]+3H2¡ü£©
2A1+2NaOH+2H2O¨T2NaA1O2+3H2¡ü£¨»ò2A1+2NaOH+6H2O=2Na[A1£¨OH£©4]+3H2¡ü£©
£®
£¨2£©¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿Ê£Óà¹ÌÌ壮ÈôδϴµÓ¹ÌÌ壬²âµÃþµÄÖÊÁ¿·ÖÊý½«
Æ«¸ß
Æ«¸ß
£¨Ìî¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±£©£®
¡¾Ì½¾¿¶þ¡¿
ʵÑé·½°¸£ºÂÁþºÏ½ð
ÑÎËá
²â¶¨Éú³ÉÆøÌåµÄÌå»ý
ʵÑé×°Öãº
ÎÊÌâÌÖÂÛ£º
£¨1£©Ä³Í¬Ñ§Ìá³ö¸ÃʵÑé×°Öò»¹»ÍêÉÆ£¬Ó¦ÔÚA¡¢BÖ®¼äÌí¼ÓÒ»¸ö×°Óмîʯ»ÒµÄ¸ÉÔï×°Öã®ÄãµÄÒâ¼ûÊÇ£º
²»ÐèÒª
²»ÐèÒª
 £¨Ìî¡°ÐèÒª¡±»ò¡°²»ÐèÒª¡±£©£®
£¨2£©Îª×¼È·²â¶¨Éú³ÉÆøÌåµÄÌå»ý£¬ÊµÑéÖÐӦעÒâµÄÎÊÌâÊÇ£¨Ö»ÒªÇóд³öÆäÖÐÒ»µã£©£º
¼ì²é×°ÖõÄÆøÃÜÐÔ£¨»òºÏ½ðÍêÈ«Èܽ⣬»ò¼ÓÈë×ãÁ¿ÑÎËᣬ»òµ÷ÕûÁ¿Æø¹ÜCµÄ¸ß¶È£¬Ê¹CÖÐÒºÃæÓëBÒºÃæÏàƽµÈºÏÀí´ð°¸£©
¼ì²é×°ÖõÄÆøÃÜÐÔ£¨»òºÏ½ðÍêÈ«Èܽ⣬»ò¼ÓÈë×ãÁ¿ÑÎËᣬ»òµ÷ÕûÁ¿Æø¹ÜCµÄ¸ß¶È£¬Ê¹CÖÐÒºÃæÓëBÒºÃæÏàƽµÈºÏÀí´ð°¸£©

¡¾Ì½¾¿Èý¡¿
ʵÑé·½°¸£º³ÆÁ¿x gÂÁþºÏ½ð·ÛÄ©£¬·ÅÔÚÈçÓÒͼËùʾװÖõĶèÐÔµçÈÈ°åÉÏ£¬Í¨µçʹÆä³ä·Ö×ÆÉÕ£®
ÎÊÌâÌÖÂÛ£º
£¨1£©Óû¼ÆËãMgµÄÖÊÁ¿·ÖÊý£¬¸ÃʵÑéÖл¹Ðè²â¶¨µÄÊý¾ÝÊÇ
×ÆÉÕºó¹ÌÌåµÄÖÊÁ¿
×ÆÉÕºó¹ÌÌåµÄÖÊÁ¿
£®
£¨2£©ÈôÓÿÕÆø´úÌæO2½øÐÐʵÑ飬¶Ô²â¶¨½á¹ûÊÇ·ñÓÐÓ°Ï죿
ÊÇ
ÊÇ
£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º¸£½¨Ê¡ÏÃÃÅÊÐ2012£­2013ѧÄê¸ßÒ»ÏÂѧÆÚÆÚÄ©¿¼ÊÔÄ£ÄâÊÔÌâ(1)£­»¯Ñ§ ÌâÐÍ£º058

ÂÁþºÏ½ðÒѳÉΪÂÖ´¬ÖÆÔì¡¢»¯¹¤Éú²úµÈÐÐÒµµÄÖØÒª²ÄÁÏ£®Ñо¿ÐÔѧϰС×éµÄÈýλͬѧ£¬Îª²â¶¨Ä³ÂÁþºÏ½ð(Éè²»º¬ÆäËüÔªËØ)ÖÐþµÄÖÊÁ¿·ÖÊý£¬Éè¼ÆÏÂÁÐÈýÖÖ²»Í¬ÊµÑé·½°¸½øÐÐ̽¾¿£®ÌîдÏÂÁпհף®

[̽¾¿Ò»]ʵÑé·½°¸£ºÂÁþºÏ½ð²â¶¨Ê£Óà¹ÌÌåÖÊÁ¿

ÎÊÌâÌÖÂÛ£º(1)³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÂÁþºÏ½ð·ÛÄ©ÑùÆ·£¬¼ÓÈë¹ýÁ¿µÄNaOHÈÜÒº£¬³ä·Ö·´Ó¦£®ÊµÑéÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ________£®

(2)¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿Ê£Óà¹ÌÌ壮ÈôδϴµÓ¹ÌÌ壬²âµÃþµÄÖÊÁ¿·ÖÊý½«________(Ìî¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±)£®

[̽¾¿¶þ]ʵÑé·½°¸£ºÂÁþºÏ½ð²â¶¨Éú³ÉÆøÌåµÄÌå»ý

ʵÑé×°Öãº

ÎÊÌâÌÖÂÛ£º(1)ijͬѧÌá³ö¸ÃʵÑé×°Öò»¹»ÍêÉÆ£¬Ó¦ÔÚA¡¢BÖ®¼äÌí¼ÓÒ»¸ö×°Óмîʯ»ÒµÄ¸ÉÔï×°Öã®ÄãµÄÒâ¼ûÊÇ£º________(Ìî¡°ÐèÒª¡±»ò¡°²»ÐèÒª¡±)£®

(2)Ϊ׼ȷ²â¶¨Éú³ÉÆøÌåµÄÌå»ý£¬ÊµÑéÖÐӦעÒâµÄÎÊÌâÊÇ(Ö»ÒªÇóд³öÆäÖÐÒ»µã)£º________

[̽¾¿Èý]

ʵÑé·½°¸£º³ÆÁ¿xgÂÁþºÏ½ð·ÛÄ©£¬·ÅÔÚÈçÏÂͼËùʾװÖõĶèÐÔµçÈÈ°åÉÏ£¬Í¨µçʹÆä³ä·Ö×ÆÉÕ£®

ÎÊÌâÌÖÂÛ£º(1)Óû¼ÆËãMgµÄÖÊÁ¿·ÖÊý£¬¸ÃʵÑéÖл¹Ðè²â¶¨µÄÊý¾ÝÊÇ________£®

(2)ÈôÓÿÕÆø´úÌæO2½øÐÐʵÑ飬¶Ô²â¶¨½á¹ûÊÇ·ñÓÐÓ°Ï죿________(Ìî¡°ÊÇ¡±»ò¡°·ñ¡±)£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

ÂÁþºÏ½ðÒѳÉΪÂÖ´¬ÖÆÔì¡¢»¯¹¤Éú²úµÈÐÐÒµµÄÖØÒª²ÄÁÏ£®Ñо¿ÐÔѧϰС×éµÄÈýλͬѧ£¬Îª²â¶¨Ä³ÂÁþºÏ½ð£¨Éè²»º¬ÆäËüÔªËØ£©ÖÐþµÄÖÊÁ¿·ÖÊý£¬Éè¼ÆÏÂÁÐÈýÖÖ²»Í¬ÊµÑé·½°¸½øÐÐ̽¾¿£®ÌîдÏÂÁпհף®
[̽¾¿Ò»]
ʵÑé·½°¸£ºÂÁþºÏ½ðÊýѧ¹«Ê½²â¶¨Ê£Óà¹ÌÌåÖÊÁ¿
ÎÊÌâÌÖÂÛ£º
£¨1£©³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÂÁþºÏ½ð·ÛÄ©ÑùÆ·£¬¼ÓÈë¹ýÁ¿µÄNaOHÈÜÒº£¬³ä·Ö·´Ó¦£®ÊµÑéÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ______£®
£¨2£©¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿Ê£Óà¹ÌÌ壮ÈôδϴµÓ¹ÌÌ壬²âµÃþµÄÖÊÁ¿·ÖÊý½«______£¨Ìî¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±£©£®
[̽¾¿¶þ]
ʵÑé·½°¸£ºÂÁþºÏ½ðÊýѧ¹«Ê½²â¶¨Éú³ÉÆøÌåµÄÌå»ý
ʵÑé×°Öãº
ÎÊÌâÌÖÂÛ£º
£¨1£©Ä³Í¬Ñ§Ìá³ö¸ÃʵÑé×°Öò»¹»ÍêÉÆ£¬Ó¦ÔÚA¡¢BÖ®¼äÌí¼ÓÒ»¸ö×°Óмîʯ»ÒµÄ¸ÉÔï×°Öã®ÄãµÄÒâ¼ûÊÇ£º______ £¨Ìî¡°ÐèÒª¡±»ò¡°²»ÐèÒª¡±£©£®
£¨2£©Îª×¼È·²â¶¨Éú³ÉÆøÌåµÄÌå»ý£¬ÊµÑéÖÐӦעÒâµÄÎÊÌâÊÇ£¨Ö»ÒªÇóд³öÆäÖÐÒ»µã£©£º______
[̽¾¿Èý]
ʵÑé·½°¸£º³ÆÁ¿x gÂÁþºÏ½ð·ÛÄ©£¬·ÅÔÚÈçÓÒͼËùʾװÖõĶèÐÔµçÈÈ°åÉÏ£¬Í¨µçʹÆä³ä·Ö×ÆÉÕ£®
ÎÊÌâÌÖÂÛ£º
£¨1£©Óû¼ÆËãMgµÄÖÊÁ¿·ÖÊý£¬¸ÃʵÑéÖл¹Ðè²â¶¨µÄÊý¾ÝÊÇ______£®
£¨2£©ÈôÓÿÕÆø´úÌæO2½øÐÐʵÑ飬¶Ô²â¶¨½á¹ûÊÇ·ñÓÐÓ°Ï죿______£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ij»¯Ñ§Ì½¾¿Ñ§Ï°Ð¡×éÄâÀûÓÃÏÂÁÐ×°ÖòⶨijºÚÉ«ÌúµÄÑõ»¯Îï¿óÑùµÄ»¯Ñ§Ê½£¨Éè¿óÑù²»º¬ÆäËüÔÓÖÊ£©¡£Ö÷Òª»¯Ñ§·´Ó¦Îª£º

    ¢ÙH2C2O4£¨ÒÒ¶þËᣩCO2¡ü+CO¡ü+H2O  ¢ÚFeO+CO=Fe+CO2

ËûÃÇÒª²â¶¨µÄÊý¾ÝÊÇ¿óÑùÖÊÁ¿ºÍ·´Ó¦¢ÚËùÉú³ÉµÄCO2µÄÖÊÁ¿[ÓÃm£¨FeO£©ºÍm£¨CO2£©±íʾ]¡£ËùÐèÒÇÆ÷£¨Á¬Í¬Ò©Æ·£©µÄʾÒâͼÈçÏ£¨±ØҪʱÓеÄÒÇÆ÷¿ÉÖظ´Ê¹Óã©

£¨1£©¼ÙÉèÏ´Æø£¨»òÎüÆø£©×°Öñû¡¢¶¡¡¢ÎìÄڵķ´Ó¦½øÐеö¼ÍêÈ«£¬°´ÆøÌå´Ó×óÖÁÓÒµÄÁ÷Ïò½«ÉÏÊöÒÇÆ÷×é×°ÆðÀ´£¨ÓýӿÚ×Öĸa¡¢b¡­¡­±íʾ£©£º

     ½Ó    £¬     ½Ó     £¬     ½Ó     £¬    ½Ó     £¬  c  ½Ó  f 

£¨2£©ÊµÑéÖÐÓÐÒ»¼þÒÇÆ÷Ç°ºóÓõ½Á½´Î£¬Ç°±ßÓÃËüµÄÄ¿µÄÊÇ                £¬ºó±ßÓÖÓÃËüµÄÄ¿µÄÊÇ                                                ¡£

£¨3£©ÎªÁ˱£»¤»·¾³£¬ÔõÑù´¦ÀíÒÇÆ÷Ä©¶Ëµ¼³öµÄCO£¨Ð´³ö¾ßÌå·½·¨£©                     

£¨4£©¸ÃС×é²âµÃm£¨FeO£©=15£®2g£¬m£¨CO2£©=11£®0g£¬Ôòx£ºyΪ           

    A£®4£º5        B£®1£º1      C£®2£º3    D£®3£º4

ÔÚ¸ÃʵÑéÖУ¬·´Ó¦Ç°FeOΪºÚÉ«£¬·´Ó¦ºóÉú³ÉµÄFeҲΪºÚÉ«£¬ÄÑÒÔÈ·¶¨FeOÊÇ·ñÍêÈ«±»»¹Ô­£¬ÈôÈÔÓÐFeOÊ£Ó࣬ÔòËù²âµÄx£ºyÖµ±Èʵ¼ÊÖµ       £¨ÌîÆ«µÍ£¬Æ«¸ß»ò²»Ó°Ï죩²Éȡʲô´ëÊ©¿ÉÒÔ±ÜÃâÉÏÊöÔ­ÒòÒýÆðµÄżȻÎó²î£º

                                                                            

                                                                            

£¨5£©ÈÔÓÃÉÏÊöÌṩµÄÒÇÆ÷ºÍÒ©Æ·£¨±ØҪʱ¿ÉÒÔÉÙÑ¡ÓÃÒ²¿ÉÒÔ°´ÐµÄ˳Ðò½øÐÐ×é×°£©£¬Í¬Ñù²â¶¨Á½ÖÖÎïÖʵÄÖÊÁ¿£¬±ã¿É¼ÆËã³öxºÍyµÄ±ÈÖµ£¬Ð´³öÒª²â¶¨µÄÁ½ÖÖÊý¾Ý¿ÉÄܵÄ×éºÏ£º

¢Ù                     ¢Ú                        ¢Û                      

£¨Èô²»¹»ÈýÖÖ×éºÏ£¬¿É²»±ØÌîÂú£»Èô¶àÓÚÈýÖÖ×éºÏ£¬¿É×ÔÐÐÔö¼Ó£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸