»¯Ñ§Éϳ£ÓÃȼÉÕ·¨È·¶¨ÓлúÎïµÄ×é³É£®ÏÂͼװÖÃÊÇÓÃȼÉÕ·¨È·¶¨ÓлúÎﻯѧʽ³£ÓõÄ×°Öã¬ÕâÖÖ·½·¨ÊÇÔڵ篼ÓÈÈʱÓô¿ÑõÑõ»¯¹ÜÄÚÑùÆ·£®¸ù¾Ý²úÎïµÄÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É£®

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)A×°ÖÃÖзÖҺ©¶·Ê¢·ÅµÄÎïÖÊÊÇ______________£¬Ð´³öÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ

______________________________________________________________£®

(2)C×°ÖÃ(ȼÉÕ¹Ü)ÖÐCuOµÄ×÷ÓÃÊÇ______________________________________

____________________________.

(3)д³öE×°ÖÃÖÐËùÊ¢·ÅÊÔ¼ÁµÄÃû³Æ__________£¬ËüµÄ×÷ÓÃÊÇ______________£®

(4)Èô½«B×°ÖÃÈ¥µô»á¶ÔʵÑéÔì³ÉʲôӰÏ죿 __________________________.

(5)Èô׼ȷ³ÆÈ¡1.20 gÑùÆ·(Ö»º¬C¡¢H¡¢OÈýÖÖÔªËØÖеÄÁ½ÖÖ»òÈýÖÖ)£®¾­³ä·ÖȼÉÕºó£¬

E¹ÜÖÊÁ¿Ôö¼Ó1.76 g£¬D¹ÜÖÊÁ¿Ôö¼Ó0.72 g£¬Ôò¸ÃÓлúÎïµÄ×î¼òʽΪ______________£®

(6)Ҫȷ¶¨¸ÃÓлúÎïµÄ»¯Ñ§Ê½£¬»¹ÐèÒª²â¶¨________________________£®

 

¡¾´ð°¸¡¿

(1)H2O2(»òË«ÑõË®)2H2O2 2H2O£«O2¡ü(»òH2O¡¡2Na2O2£«2H2O===4NaOH£«O2¡ü)

(2)ʹÓлúÎï³ä·ÖÑõ»¯Éú³ÉCO2ºÍH2O     (3)¼îʯ»Ò»òÇâÑõ»¯ÄÆ¡¡ÎüÊÕCO2

(4)Ôì³É²âµÃÓлúÎïÖк¬ÇâÁ¿Ôö´ó

(5)CH2O ¡¡(6)²â³öÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º±¾ÊµÑéʹÓÃȼÉÕ·¨²â¶¨ÓлúÎï×é³É£¬¸ÃʵÑé×°Öð´ÕÕ¡°ÖÆÑõÆø¡ú¸ÉÔïÑõÆø¡úȼÉÕÓлúÎï¡úÎüÊÕË®¡úÎüÊÕ¶þÑõ»¯Ì¼¡±ÅÅÁС£ÊµÑé¿É²â֪ȼÉÕÉú³ÉµÄ¶þÑõ»¯Ì¼ºÍË®µÄÖÊÁ¿£¬¸ù¾Ý¶þÑõ»¯Ì¼µÄÖÊÁ¿¿ÉÇóCÔªËØµÄÖÊÁ¿£¬ÓÉË®µÄÖÊÁ¿¿ÉÇóµÃHÔªËØµÄÖÊÁ¿£¬½áºÏÓлúÎïµÄÖÊÁ¿¿ÉÇó³öOÔªËØµÄÖÊÁ¿£¬Óɴ˼´¿ÉÈ·¶¨ÓлúÎï·Ö×ÓÖÐC¡¢H¡¢O¸öÊý±È£¬Ò²¾ÍÊÇÈ·¶¨ÁËʵÑéʽ£¬ÈôÒªÔÙ½øÒ»²½È·¶¨ÓлúÎïµÄ·Ö×Óʽ£¬»¹ÐèÖªµÀ¸ÃÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿¡£

£¨1£©×°ÖÃAÊÇÖÆ±¸ÑõÆøµÄ£¬ËùÒÔ¸ù¾Ý×°ÖõÄÌØµã¿ÉÖªA×°ÖÃÖзÖҺ©¶·Ê¢·ÅµÄÎïÖÊÊÇË«ÑõË®»òË®£¬Ó¦¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ2H2O2 2H2O£«O2¡ü»ò2Na2O2£«2H2O===4NaOH£«O2¡ü¡£

£¨2£©ÓÉÓÚÓлúÎïÔÚȼÉÕ¹ý³ÌÖУ¬ÓпÉÄܲúÉúCO£¬ËùÒÔC×°ÖÃ(ȼÉÕ¹Ü)ÖÐCuOµÄ×÷ÓÃÊÇʹÓлúÎï³ä·ÖÑõ»¯Éú³ÉCO2ºÍH2O¡£

£¨3£©ÓÉÓÚÓлúÎïȼÉÕ²úÉúCO2£¬ËùÒÔE×°ÖõÄÖ÷Òª×÷ÓÃÊÇÎüÊÕÉú³ÉµÄCO2£¬Òò´ËÆäÖÐËùÊ¢·ÅÊÔ¼ÁÊǼîʯ»Ò»òÇâÑõ»¯ÄÆ¡£

£¨4£©B×°ÖõÄ×÷ÓÃÊǸÉÔïÑõÆø£¬³ýȥˮÕôÆø£¬ËùÒÔÈô½«B×°ÖÃÈ¥µô»áÔì³É²âµÃÓлúÎïÖк¬ÇâÁ¿Ôö´ó¡£

£¨5£©D¹ÜÖÊÁ¿Ôö¼Ó0.72 g£¬ÔòÉú³ÉµÄË®ÊÇ0.72g£¬ÎïÖʵÄÁ¿ÊÇ0.04mol£¬ÆäÖÐÇâÔªËØµÄÖÊÁ¿ÊÇ0.08g¡£E¹ÜÖÊÁ¿Ôö¼Ó1.76 g£¬ÔòCO2ÊÇ1.76g£¬ÎïÖʵÄÁ¿ÊÇ0.04mol£¬ÆäÖÐÌ¼ÔªËØµÄÖÊÁ¿ÊÇ0.48g£¬ÔòÔ­ÓлúÎïÖÐÑõÔªËØµÄÖÊÁ¿ÊÇ1.20g£­0.08g£­0.48g£½0.64g£¬ÎïÖʵÄÁ¿ÊÇ0.04mol£¬ËùÒÔÔ­ÓлúÎïÖÐC¡¢H¡¢OµÄÔ­×Ó¸öÊýÖ®±ÈÊÇ1:2:1£¬Ôò×î¼òʽÊÇCH2O¡£

£¨6£©ÔÙÒÑÖª×î¼òʽµÄÇé¿öÏ£¬ÒªÈ·¶¨¸Ã»¯ºÏÎïµÄ»¯Ñ§Ê½£¬Ôò»¹ÐèÒª²â³öÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿¡£

¿¼µã£º¿¼²é¸ù¾ÝÓлúÎïȼÉÕ·¨È·¶¨ÓлúÎﻯѧʽµÄ¼ÆËã¡¢ÅжÏÒÔ¼°ÊµÑé²Ù×÷

µãÆÀ£º¸ÃÌâÊÇÖеÈÄѶȵÄÊÔÌ⣬ÊÔÌâ×ÛºÏÐÔÇ¿£¬ÄÑÒ×ÊÊÖУ¬Ìù½ü¸ß¿¼¡£Ö÷ÒªÊÇ¿¼²éѧÉúÁé»îÔËÓûù´¡ÖªÊ¶½â¾öʵ¼ÊÎÊÌâµÄÄÜÁ¦£¬ÓÐÖúÓÚÅàÑøÑ§ÉúµÄÂß¼­ÍÆÀíÄÜÁ¦ºÍ·¢É¢Ë¼Î¬ÄÜÁ¦¡£¸ÃÌâѧÉúÐèÒªÃ÷È·µÄÊǸÃÀàÊÔÌâ×ÛºÏÐÔÇ¿£¬ÀíÂÛºÍʵ¼ùµÄÁªÏµ½ôÃÜ£¬ÓеϹÌṩһЩеÄÐÅÏ¢£¬Õâ¾ÍÒªÇóѧÉú±ØÐëÈÏÕæ¡¢Ï¸ÖµÄÉóÌ⣬ÁªÏµËùѧ¹ýµÄ֪ʶºÍ¼¼ÄÜ£¬½øÐÐ֪ʶµÄÀà±È¡¢Ç¨ÒÆ¡¢ÖØ×é£¬È«ÃæÏ¸ÖµÄ˼¿¼²ÅÄܵóöÕýÈ·µÄ½áÂÛ¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§Éϳ£ÓÃȼÉÕ·¨È·¶¨ÓлúÎïµÄ×é³É£®ÕâÖÖ·½·¨ÊÇÔڵ篼ÓÈÈʱÓô¿ÑõÑõ»¯¹ÜÄÚÑùÆ·£¬¸ù¾Ý²úÎïÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É£¬×°ÖÃÈçÏÂͼËùʾ£¬ÊÇÓÃȼÉÕ·¨È·¶¨ÓлúÎﻯѧʽ³£ÓõÄ×°Öã®

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²úÉúµÄÑõÆø°´´Ó×óµ½ÓÒÁ÷Ïò£¬ËùÑ¡Ôñ×°Öø÷µ¼¹ÜµÄÁ¬½Ó˳ÐòÊÇ
g f e h i c d£¨d£¬c£©ab£¨b£¬a£©
g f e h i c d£¨d£¬c£©ab£¨b£¬a£©
£®
£¨2£©C×°ÖÃÖÐŨH2SO4µÄ×÷ÓÃΪ
ÎüÊÕË®·Ö¡¢¸ÉÔïÑõÆø
ÎüÊÕË®·Ö¡¢¸ÉÔïÑõÆø
£®
£¨3£©D×°ÖÃÖÐMnO2µÄ×÷ÓÃΪ
´ß»¯¼Á¡¢¼Ó¿ì²úÉúO2µÄËÙÂÊ
´ß»¯¼Á¡¢¼Ó¿ì²úÉúO2µÄËÙÂÊ
£®
£¨4£©EÖÐCuOµÄ×÷ÓÃΪ
ʹÓлúÎï¸ü³ä·ÖÑõ»¯ÎªCO2¡¢H2O
ʹÓлúÎï¸ü³ä·ÖÑõ»¯ÎªCO2¡¢H2O
£®
£¨5£©Èô׼ȷ³ÆÈ¡0.90gÑùÆ·£¨Ö»º¬C¡¢H¡¢OÈýÖÖÔªËØÖеÄÁ½ÖÖ»òÈýÖÖ£©¾­³ä·ÖȼÉÕºó£¬A¹ÜÖÊÁ¿Ôö¼Ó1.32g£¬B¹ÜÖÊÁ¿Ôö¼Ó0.54g£¬Ôò¸ÃÓлúÎï×î¼òʽΪ
CH2O
CH2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§Éϳ£ÓÃȼÉÕ·¨È·¶¨ÓлúÎïµÄ×é³É£®ÈçͼװÖÃÊÇÓÃȼÉÕ·¨È·¶¨ÓлúÎﻯѧʽ³£ÓõÄ×°Öã¬ÕâÖÖ·½·¨ÊÇÔڵ篼ÓÈÈʱÓô¿ÑõÑõ»¯¹ÜÄÚÑùÆ·£®

¸ù¾Ý²úÎïµÄÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©A×°ÖÃÖзÖҺ©¶·Ê¢·ÅµÄÎïÖÊÊÇ
H2O2£¨»òË«ÑõË®£©
H2O2£¨»òË«ÑõË®£©
£¬Ð´³öÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ
2H2O2
 MnO2 
.
 
2H2O+O2¡ü
2H2O2
 MnO2 
.
 
2H2O+O2¡ü
£®
£¨2£©C×°Öã¨È¼Éչܣ©ÖÐCuOµÄ×÷ÓÃÊÇ
ʹÓлúÎï³ä·ÖÑõ»¯Éú³ÉCO2ºÍH2O
ʹÓлúÎï³ä·ÖÑõ»¯Éú³ÉCO2ºÍH2O

£¨3£©Ð´³öE×°ÖÃÖÐËùÊ¢·ÅÊÔ¼ÁµÄÃû³Æ
¼îʯ»Ò»òÇâÑõ»¯ÄÆ
¼îʯ»Ò»òÇâÑõ»¯ÄÆ
£¬ËüµÄ×÷ÓÃÊÇ
ÎüÊÕCO2
ÎüÊÕCO2
£®
£¨4£©Èô½«B×°ÖÃÈ¥µô»á¶ÔʵÑéÔì³ÉʲôӰÏ죿
Ôì³É²âµÃÓлúÎïÖк¬ÇâÁ¿Ôö´ó
Ôì³É²âµÃÓлúÎïÖк¬ÇâÁ¿Ôö´ó
£®
£¨5£©Èô׼ȷ³ÆÈ¡1.20gÑùÆ·£¨Ö»º¬C¡¢H¡¢OÈýÖÖÔªËØÖеÄÁ½ÖÖ»òÈýÖÖ£©£®¾­³ä·ÖȼÉÕºó£¬E¹ÜÖÊÁ¿Ôö¼Ó1.76g£¬D¹ÜÖÊÁ¿Ôö¼Ó0.72g£¬Ôò¸ÃÓлúÎïµÄ×î¼òʽΪ
CH2O
CH2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§Éϳ£ÓÃȼÉÕ·¨È·¶¨ÓлúÎïµÄ×é³É£®Èô׼ȷ³ÆÈ¡0.54gijÓлúÎïÑùÆ·£¨Ìþ»òÌþµÄº¬ÑõÑÜÉúÎ£®¾­³ä·ÖȼÉÕºó£¬ÒÀ´ÎͨÈëŨÁòËáºÍ¼îʯ»Ò£¬·Ö±ðÔöÖØ0.54g¡¢1.76g£®
£¨1£©Çó¸ÃÓлúÎïµÄʵÑéʽ£¿
£¨2£©ÒÑÖªÆäÕôÆøÃܶÈΪͬÌõ¼þÏÂH2µÄ27±¶£¬Çó¸ÃÓлúÎïµÄ»¯Ñ§Ê½£¿
£¨3£©Èô¸Ã»¯ºÏÎïÓëBr2ÒÔ1£º1¼Ó³É·´Ó¦¿É²úÉúÁ½ÖÖ»¥ÎªÍ¬·ÖÒì¹¹ÌåµÄÓлúÎÇëд³ö¸Ã»¯ºÏÎï½á¹¹¼òʽ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§Éϳ£ÓÃȼÉÕ·¨È·¶¨ÓлúÎïµÄ×é³É£®ÏÂͼװÖÃÊÇÓÃȼÉÕ·¨È·¶¨ÓлúÎﻯѧʽ³£ÓõÄ×°Öã¬ÕâÖÖ·½·¨ÊÇÔڵ篼ÓÈÈʱÓô¿ÑõÑõ»¯¹ÜÄÚÑùÆ·£®¸ù¾Ý²úÎïµÄÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É£®

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©A×°ÖÃÖзÖҺ©¶·Ê¢·ÅµÄÎïÖÊÊÇ
H2O2£¨»òË«ÑõË®£©
H2O2£¨»òË«ÑõË®£©
£¬Ð´³öÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
2H2O2
MnO2
2H2O+O2¡ü
2H2O2
MnO2
2H2O+O2¡ü
£®
£¨2£©C×°Öã¨È¼Éչܣ©ÖÐCuOµÄ×÷ÓÃÊÇ
ʹÓлúÎï³ä·ÖÑõ»¯Éú³ÉCO2ºÍH2O
ʹÓлúÎï³ä·ÖÑõ»¯Éú³ÉCO2ºÍH2O
£®
£¨3£©Ð´³öE×°ÖÃÖÐËùÊ¢·ÅÎïÖʵÄÃû³Æ
¼îʯ»Ò»òÇâÑõ»¯ÄÆ
¼îʯ»Ò»òÇâÑõ»¯ÄÆ
£¬ËüµÄ×÷ÓÃÊÇ
ÎüÊÕCO2ÆøÌå
ÎüÊÕCO2ÆøÌå
£®
£¨4£©Èô׼ȷ³ÆÈ¡1.20gÑùÆ·£¨Ö»º¬C¡¢H¡¢OÈýÖÖÔªËØÖеÄÁ½ÖÖ»òÈýÖÖ£©£®¾­³ä·ÖȼÉÕºó£¬E¹ÜÖÊÁ¿Ôö¼Ó1.76g£¬D¹ÜÖÊÁ¿Ôö¼Ó0.72g£¬Ôò¸ÃÓлúÎïµÄ×î¼òʽΪ
CH2O
CH2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§Éϳ£ÓÃȼÉÕ·¨È·¶¨ÓлúÎï×é³É£¬ÕâÖÖ·½·¨ÊÇÔڵ篼ÓÈÈʱÓô¿ÑõÆøÑõ»¯¹ÜÄÚÑùÆ·£¬¸ù¾Ý²úÎïµÄÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É£®Í¼ÖÐËùÁÐ×°ÖÃÊÇÓÃȼÉÕ·¨È·¶¨ÓлúÎï·Ö×Óʽ³£ÓõÄ×°Ö㮣¨Ã¿Ò»×éÒÇÆ÷Ö»ÄÜʹÓÃÒ»´Î£©

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²úÉúµÄÑõÆø°´´Ó×óµ½ÓÒÁ÷Ïò£¬ËùÑ¡ÔñµÄ×°Öø÷µ¼¹ÜµÄÁ¬½Ó˳ÐòÊÇ
g¡úf£¬e¡úh£¬i¡úc£¬d¡úa
g¡úf£¬e¡úh£¬i¡úc£¬d¡úa
£®
£¨2£©C×°ÖÃÖÐŨÁòËáµÄ×÷ÓÃÊÇ
ÎüÊÕË®·Ö£¬µÃµ½¸ÉÔï´¿¾»µÄO2
ÎüÊÕË®·Ö£¬µÃµ½¸ÉÔï´¿¾»µÄO2
£®
£¨3£©D×°ÖÃÖÐMnO2µÄ×÷ÓÃÊÇ
×÷´ß»¯¼Á£¬¼Ó¿ì²úÉúO2µÄËÙÂÊ
×÷´ß»¯¼Á£¬¼Ó¿ì²úÉúO2µÄËÙÂÊ
£®
£¨4£©Èô׼ȷ³ÆÈ¡7.2gÑùÆ·£¨Ö»º¬C¡¢H¡¢OÈýÖÖÔªËØÖеÄÁ½ÖÖ»òÈýÖÖ£©£¬¾­³ä·ÖȼÉÕºó£¬A¹ÜÖÊÁ¿Ôö¼Ó22g£¬B¹ÜÖÊÁ¿Ôö¼Ó10.8g£¬Ôò¸ÃÓлúÎïµÄ×î¼òʽΪ
C5H12
C5H12
£®
£¨5£©ÒªÈ·¶¨¸ÃÓлúÎïµÄ·Ö×Óʽ£¬Ôò
²»
²»
 £¨Ìî¡°ÊÇ¡±»ò¡°²»¡±£©ÐèÒª²â¶¨ÆäËûÊý¾Ý£¬ÈôÄÜÈ·¶¨Æä·Ö×Óʽ£¬ÈôÆäÒ»ÂÈ´úÎïÖ»ÓÐÒ»ÖÖ£¬ÔòÆä½á¹¹¼òʽΪ
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸