ÌâÄ¿ÄÚÈÝ

½øÐл¯Ñ§ÊµÑ飬ÐèÒªÁ˽âºÍÕÆÎÕһЩ»¯Ñ§ÒÇÆ÷µÄÓйØ֪ʶ£¬ÊԻشð£º

£¨1£©Ö¸³öÉÏͼ´ø±êºÅµÄʵÑéÒÇÆ÷µÄÃû³Æ£º¢Ù
Á¿Í²
Á¿Í²
¢Ú
ÍÐÅÌÌìƽ
ÍÐÅÌÌìƽ
¢Û
ÊԹܼÐ
ÊԹܼÐ
£®
£¨2£©ÏÖÓÐÏÂÁм¸ÖÖÒÇÆ÷£¬Çë°´ÌâÒâÒªÇóÌî¿Õ£º
ÒÇÆ÷£ºÁ¿Í²¡¢ÊԹܡ¢ÉÕ±­¡¢¾Æ¾«µÆ¡¢¹ã¿ÚÆ¿¡¢µÎÆ¿
ÓÃÓÚÊ¢·ÅÉÙÁ¿µÄÊÔ¼Á½øÐл¯Ñ§·´Ó¦µÄÊÇ
ÊÔ¹Ü
ÊÔ¹Ü
£¬ÓÃÀ´Á¿È¡Ò»¶¨Ìå»ýÒºÌåÊÔ¼ÁµÄÊÇ
Á¿Í²
Á¿Í²
£¬ÓÃÓÚ¸øÎïÖʼÓÈȵÄÊÇ
¾Æ¾«µÆ
¾Æ¾«µÆ
£¬ÓÃÓÚÖü´æ¹ÌÌåÊÔ¼ÁµÄÊÇ
¹ã¿ÚÆ¿
¹ã¿ÚÆ¿
£¬Ö»ÄÜÓÃÓÚÊ¢·ÅÒºÌåÊÔ¼ÁµÄÊÇ
µÎÆ¿
µÎÆ¿
£®
£¨3£©ÏÂÁÐÊÇÓйؾƾ«µÆÔÚ´æ·Å¡¢Ê¹Óùý³ÌµÄʾÒâĸ£¬ÇëÄãÈÏÕæ·ÖÎöͼƬËùÓÐÔ̺­µÄÒâͼ£¬½«ÆäÓÃÒâÓüòÁ·µÄÓïÑÔ±íʾ³öÀ´£®

¢Ù
¾Æ¾«µÆ²»ÓÃʱӦ¸ÇÉϵÆñ
¾Æ¾«µÆ²»ÓÃʱӦ¸ÇÉϵÆñ
¢Ú
½ûÖ¹ÓÃȼ×ŵľƾ«µÆÈ¥ÒýȼÁíÒ»Õµ¾Æ¾«µÆ
½ûÖ¹ÓÃȼ×ŵľƾ«µÆÈ¥ÒýȼÁíÒ»Õµ¾Æ¾«µÆ
¢Û
¾Æ¾«µÆÓ¦Óûð²ñµãȼ
¾Æ¾«µÆÓ¦Óûð²ñµãȼ
¢Ü
ϨÃðȼ×ŵľƾ«µÆÓ¦ÓõÆñ¸ÇÃð
ϨÃðȼ×ŵľƾ«µÆÓ¦ÓõÆñ¸ÇÃð

£¨4£©ÈôµÆÄھƾ«Á¿·ûºÏÒªÇó£®µ«ÈÔÈ»µã²»×Å£¬ÆäÔ­Òò¿ÉÄÜÊÇ
ÓÃÍê¾Æ¾«µÆʱ£¬³¤ÆÚδ¸ÇµÆñ
ÓÃÍê¾Æ¾«µÆʱ£¬³¤ÆÚδ¸ÇµÆñ
£®
£¨5£©Èô¾Æ¾«µÆµÄ»ðÑæ̫С£¬Ó°Ïì¼ÓÈÈЧ¹û£¬Äã²ÉÈ¡µÄµ÷Õû´ëÊ©ÊÇ
ÓÃÛáÛöǯ¼ÐסµÆоÏòÉϰγöÉÙÐí
ÓÃÛáÛöǯ¼ÐסµÆоÏòÉϰγöÉÙÐí
£®
·ÖÎö£º¸ù¾ÝÒÑÓеÄ֪ʶ½øÐзÖÎö£¬¾Ýͼ¼´¿ÉÖªÒÇÆ÷µÄÃû³Æ£¬ÓÃÓÚÊ¢·ÅÉÙÁ¿µÄÊÔ¼Á½øÐл¯Ñ§·´Ó¦µÄÊÇÊԹܣ¬ÓÃÀ´Á¿È¡Ò»¶¨Ìå»ýÒºÌåÊÔ¼ÁµÄÊÇÁ¿Í²£¬ÓÃÓÚ¸øÎïÖʼÓÈȵÄÊǾƾ«µÆ£¬ÓÃÓÚÖü´æ¹ÌÌåÊÔ¼ÁµÄÊǹã¿ÚÆ¿£¬Ö»ÄÜÓÃÓÚÊ¢·ÅÒºÌåÊÔ¼ÁµÄÊǵÎÆ¿£¬¸ù¾ÝͼʾָÃ÷¾Æ¾«µÆµÄʹÓÃ×¢ÒâÊÂÏ¾Æ¾«¾ßÓлӷ¢ÐÔ£¬Ò×»Ó·¢¶øÁôÏÂË®·Öµ¼Ö²»Ò×µãȼ£¬»ðÑæС¿ÉÒÔʹµÆо±ä³¤Ò»µã£®
½â´ð£º½â£º£¨1£©¾Ýͼ¿ÉÖª£¬¢ÙÊÇÁ¿Í²£¬¢ÚÊÇÍÐÅÌÌìƽ£¬¢ÛÊÇÊԹܼУ¬¹ÊÌ¢ÙÁ¿Í²¢ÚÍÐÅÌÌìƽ  ¢ÛÊԹܼУ»
£¨2£©ÓÃÓÚÊ¢·ÅÉÙÁ¿µÄÊÔ¼Á½øÐл¯Ñ§·´Ó¦µÄÊÇÊԹܣ¬ÓÃÀ´Á¿È¡Ò»¶¨Ìå»ýÒºÌåÊÔ¼ÁµÄÊÇÁ¿Í²£¬ÓÃÓÚ¸øÎïÖʼÓÈȵÄÊǾƾ«µÆ£¬ÓÃÓÚÖü´æ¹ÌÌåÊÔ¼ÁµÄÊǹã¿ÚÆ¿£¬Ö»ÄÜÓÃÓÚÊ¢·ÅÒºÌåÊÔ¼ÁµÄÊǵÎÆ¿£¬¹ÊÌÊԹܣ¬Á¿Í²£¬¾Æ¾«µÆ£¬¹ã¿ÚÆ¿£¬µÎÆ¿£»
£¨3£©¢Ù±íÃ÷¾Æ¾«µÆ²»ÓÃʱ£¬Òª¸ÇÉϵÆñ£¬¢Ú±íÃ÷²»ÄÜʹÓÃȼ×ŵľƾ«µÆÈ¥µãȼÁíÒ»Õµ¾Æ¾«µÆ£¬¢Û±íÃ÷µãȼ¾Æ¾«µÆʹÓûð²ñ£¬¢Ü±íÃ÷ϨÃð¾Æ¾«µÆʱҪʹÓõÆñ¸ÇÃ𣬹ÊÌ¢Ù¾Æ¾«µÆ²»ÓÃʱӦ¸ÇÉϵÆñ£¬¢Ú½ûÖ¹ÓÃȼ×ŵľƾ«µÆÈ¥ÒýȼÁíÒ»Õµ¾Æ¾«µÆ£¬¢Û¾Æ¾«µÆÓ¦Óûð²ñµãȼ£¬¢ÜϨÃðȼ×ŵľƾ«µÆÓ¦ÓõÆñ¸ÇÃð£»
£¨4£©¾Æ¾«¾ßÓлӷ¢ÐÔ£¬Ò×»Ó·¢¶øÁôÏÂË®·Öµ¼Ö²»Ò×µãȼ£¬¹ÊÌÓÃÍê¾Æ¾«µÆʱ£¬³¤ÆÚδ¸ÇµÆñ£»
£¨5£©»ðÑæС¿ÉÒÔʹµÆо±ä³¤Ò»µã£¬¹ÊÌÓÃÛáÛöǯ¼ÐסµÆоÏòÉϰγöÉÙÐí£®
µãÆÀ£º±¾Ì⿼²éÁ˳£¼ûÒÇÆ÷µÄʹÓÃ×¢ÒâÊÂÏÍê³É´ËÌ⣬¿ÉÒÔÒÀ¾ÝÒÑÓеÄÒÇÆ÷µÄʹÓ÷½·¨½øÐУ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2013?ÄÏ°²ÊÐÖʼ죩¸ù¾ÝÈçͼ»Ø´ðÏÂÁÐÓйØÎÊÌ⣺

£¨1£©×°ÖÃBÓɵ¼¹Ü¡¢ÏðƤÈû¡¢
³¤¾±Â©¶·
³¤¾±Â©¶·
¡¢
׶ÐÎÆ¿
׶ÐÎÆ¿
µÈÒÇÆ÷×é³É£®
£¨2£©ÊµÑéÊÒÓÃÊÊÁ¿µÄË«ÑõË®ºÍ2g¶þÑõ»¯ÃÌ·´Ó¦ÖÆÈ¡ÑõÆø£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
£º2H2O2
 MnO2 
.
 
2H2O+O2¡ü
£º2H2O2
 MnO2 
.
 
2H2O+O2¡ü
£®ËùÑ¡Óõķ¢Éú×°ÖÃΪ
B
B
£¨Ìî×ÖĸÐòºÅ£©£®ÊµÑé½áÊøºóÒª»ØÊÕ¶þÑõ»¯ÃÌ£¬¾­¹ýһϵÁвÙ×÷ºóÖ»µÃµ½1.8g¶þÑõ»¯ÃÌ£¬¼ÆËã»ØÊÕ²úÂÊ=
90%
90%
£¬Ôì³ÉµÃµ½¶þÑõ»¯Ã̵IJúÂʱÈʵ¼ÊֵƫµÍ£¬¿ÉÄܵÄÔ­ÒòÊÇ
acd
acd
£¨ÌîÐòºÅ£©£®
a£®¹ýÂËʱÂËÖ½ÓÐÆÆËð                  b£®¶þÑõ»¯ÃÌ·Ö½â²úÉúÑõÆø
c£®¶þÑõ»¯Ã̲ÐÁôÔÚÂËÖ½ÉÏ£¬Ã»ÓлØÊÕ    d£®Ï´µÓ¡¢ºæ¸Éʱ£¬ÓÐÉÙÁ¿¶þÑõ»¯ÃÌÁ÷ʧÁË
£¨3£©ÊµÑéÊÒÓôóÀíʯºÍÏ¡ÑÎËáÖÆÈ¡¶þÑõ»¯Ì¼£®ÊÕ¼¯¶þÑõ»¯Ì¼Ñ¡ÓõÄÊÕ¼¯×°ÖÃÊÇ
D
D
£¨Ìî×ÖĸÐòºÅ£©£¬ÓôË×°ÖÃÊÕ¼¯£¬¼ìÑéÆøÌåÊÇ·ñÊÕ¼¯ÂúµÄ²Ù×÷ÊÇ
ÓÃȼ×ŵÄľÌõ·ÅÔÚÆ¿¿Ú£¬ÈôϨÃð£¬Ö¤Ã÷ÒÑÂú
ÓÃȼ×ŵÄľÌõ·ÅÔÚÆ¿¿Ú£¬ÈôϨÃð£¬Ö¤Ã÷ÒÑÂú
£®
£¨4£©Ä³Ð£»¯Ñ§ÊµÑéÊÒ½øÐзÖ×éʵÑéʱÐèÒªÓÃ500gÈÜÖÊÖÊÁ¿·ÖÊýΪ9%µÄÏ¡ÑÎËᣬ¿ÉÑ¡ÓÃ
125
125
 gÈÜÖÊÖÊÁ¿·ÖÊýΪ36%µÄŨÑÎËá¼ÓÈë
375
375
 gˮϡÊͶø³É£®
ʵÑéÖÐѧ½øÐл¯Ñ§ÊµÑé²Ù×÷¿¼ÊÔ£¬×¼±¸ÁËÈý¸ö¿¼Ì⣺¢ÙÑõÆøµÄÖÆÈ¡¡¢ÊÕ¼¯ºÍÑéÂú  ¢Ú·Ö×ÓÊDz»¶ÏÔ˶¯µÄ̽¾¿  ¢Û¹ýÂË£®
£¨1£©Ð¡»ªÍ¬Ñ§³éÇ©ºó±»¼à¿¼ÀÏʦÒýµ¼ÖÁ×¼±¸ÁËÏÂÁÐÒÇÆ÷ºÍÒ©Æ·µÄʵÑę́ǰ£º

Çë»Ø´ð£º
¢ÙÖ¸³öÉÏͼÖÐÒÇÆ÷aµÄÃû³Æ£º
Ìú¼Ų̈
Ìú¼Ų̈
£»
¢ÚÓÉʵÑę́ÉÏÌṩµÄÒÇÆ÷ºÍÒ©Æ·£¬ÄãÈÏΪС»ª³éµ½µÄÊǵÚ
¢Ù
¢Ù
¸ö¿¼Ì⣻
¢ÛÒÔÏÂÊÇС»ªÍê³É¸ÃʵÑéÖ÷Òª²Ù×÷¹ý³ÌµÄʾÒâͼ£®
ÔÚ²Ù×÷¢òÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
2H2O2
 MnO2 
.
 
2H2O+O2¡ü
2H2O2
 MnO2 
.
 
2H2O+O2¡ü
£»°´ÆÀ·Ö±ê×¼£¬Ã¿Ïî²Ù×÷ÕýÈ·µÃ1·Ö£¬Âú·Ö5·Ö£¬ÄãÈÏΪʵÑéÍê±ÏºóС»ªµÃÁË
3
3
·Ö£»ÇëÌôÒ»¸öʧ·Ö²Ù×÷£¬²¢Ð´³öÕýÈ·µÄ²Ù×÷·½·¨
²½Öè¢òÖÐȡҺʱƿÈûÕý·Å£¬ÕýÈ··½·¨ÊÇÆ¿ÈûÓ¦µ¹·Å£¨»ò²½Öè¢ôÖÐÑéÂúʱ½«´ø»ðÐǵÄľÌõÉìÈëÆ¿ÄÚ£¬ÕýÈ··½·¨Êǽ«´ø»ðÐǵÄľÌõ·ÅÔÚÆ¿¿Ú£©
²½Öè¢òÖÐȡҺʱƿÈûÕý·Å£¬ÕýÈ··½·¨ÊÇÆ¿ÈûÓ¦µ¹·Å£¨»ò²½Öè¢ôÖÐÑéÂúʱ½«´ø»ðÐǵÄľÌõÉìÈëÆ¿ÄÚ£¬ÕýÈ··½·¨Êǽ«´ø»ðÐǵÄľÌõ·ÅÔÚÆ¿¿Ú£©
£®
£¨2£©Ð¡¿­Í¬Ñ§³éµ½ÁËÓëС»ªÏàͬµÄ¿¼Ì⣬ʵÑę́ÉÏÖ»ÌṩÁËÒ»ÖÖ¹ÌÌå·Ûĩ״ҩƷºÍһЩÒÇÆ÷£®
¢Ùд³öС¿­Í¬Ñ§ÖÆÈ¡ÑõÆøµÄ»¯Ñ§·½³Ìʽ
2KMnO4
  ¡÷  
.
 
K2MnO4+MnO2+O2¡ü
2KMnO4
  ¡÷  
.
 
K2MnO4+MnO2+O2¡ü
£»
¢ÚÓëС»ªÍ¬Ñ§ËùÓõÄÒÇÆ÷Ïà±È£¬Ð¡¿­Í¬Ñ§Íê³ÉʵÑéÖÁÉÙ»¹ÐèÒª
¾Æ¾«µÆ
¾Æ¾«µÆ
£»£¨ÌîÒÇÆ÷Ãû³Æ£©£¬²¢°ÑÊԹܿÚÂÔÏòÏÂÇãб£¬ÆäÄ¿µÄÊÇ
·ÀÖ¹ÀäÄýË®µ¹Á÷ÈëÊԹܵײ¿ÒýÆðÊÔ¹ÜÕ¨ÁÑ
·ÀÖ¹ÀäÄýË®µ¹Á÷ÈëÊԹܵײ¿ÒýÆðÊÔ¹ÜÕ¨ÁÑ
£®

(9·Ö)ʵÑéÖÐѧ½øÐл¯Ñ§ÊµÑé²Ù×÷¿¼ÊÔ£¬×¼±¸ÁËÈý¸ö¿¼Ì⣺¢ÙÑõÆøµÄÖÆÈ¡¡¢ÊÕ¼¯ºÍÑéÂú  ¢Ú·Ö×ÓÊDz»¶ÏÔ˶¯µÄ̽¾¿  ¢Û¹ýÂË¡£

£¨1£©   С»ªÍ¬Ñ§³éÇ©ºó±»¼à¿¼ÀÏʦÒýµ¼ÖÁ×¼±¸ÁËÏÂÁÐÒÇÆ÷ºÍÒ©Æ·µÄʵÑę́ǰ£ºÇë»Ø´ð£º

 

 

¢ÙÖ¸³öÉÏͼÖÐÒÇÆ÷aµÄÃû³Æ£º         £»

¢ÚÓÉʵÑę́ÉÏÌṩµÄÒÇÆ÷ºÍÒ©Æ·£¬ÄãÈÏΪС»ª³éµ½µÄÊǵڠ     ¸ö¿¼Ì⣻

¢ÛÒÔÏÂÊÇС»ªÍê³É¸ÃʵÑéÖ÷Òª²Ù×÷¹ý³ÌµÄʾÒâͼ¡£

 

 

ÔÚ²Ù×÷¢òÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                             £»°´ÆÀ·Ö±ê×¼£¬Ã¿Ïî²Ù×÷ÕýÈ·µÃ1·Ö£¬Âú·Ö5·Ö£¬ÄãÈÏΪʵÑéÍê±ÏºóС»ªµÃÁË     ·Ö£»ÇëÌôÒ»¸öʧ·Ö²Ù×÷£¬²¢Ð´³öÕýÈ·µÄ²Ù×÷·½·¨                            ¡£

£¨2£©Ð¡¿­Í¬Ñ§³éµ½ÁËÓëС»ªÏàͬµÄ¿¼Ì⣬ʵÑę́ÉÏÖ»ÌṩÁËÒ»ÖÖ¹ÌÌå·Ûĩ״ҩƷºÍһЩÒÇÆ÷¡£

¢Ùд³öС¿­Í¬Ñ§ÖÆÈ¡ÑõÆøµÄ»¯Ñ§·½³Ìʽ                         £»

¢ÚÓëС»ªÍ¬Ñ§ËùÓõÄÒÇÆ÷Ïà±È£¬Ð¡¿­Í¬Ñ§Íê³ÉʵÑéÖÁÉÙ»¹ÐèÒª         £»£¨ÌîÒ»ÖÖÒÇÆ÷Ãû³Æ,µ¼¹Ü¿ÉÒÔ¸ü»»£©¡£

 

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø