ÈËÃǵÄÈÕ³£Éú»îÀë²»¿ª½ðÊô£¬¸ß¿Æ¼¼Ð²ÄÁϵĿª·¢ºÍÓ¦ÓÃÒ²ÐèÒª½ðÊô£®
£¨1£©Á¶ÌúµÄÔ­ÁÏÓÐÌúʯ¿ó¡¢
½¹Ì¿
½¹Ì¿
¡¢Ê¯»ÒʯµÈ£¬Ð´³öÓôÅÌú¿óÁ¶ÌúµÄ»¯Ñ§·½³Ìʽ
Fe3O4+4CO
 ¸ßΠ
.
 
3Fe+4CO2
Fe3O4+4CO
 ¸ßΠ
.
 
3Fe+4CO2
£®
£¨2£©ÏÖ´úÉç»áʹÓôóÁ¿µÄ¸ÖÌú£¬¸ÖÌúÓë
Ë®ºÍÑõÆø
Ë®ºÍÑõÆø
½Ó´¥ÈÝÒ×ÉúÐâÔì³ÉËðʧ£¬ÔÚ¸ÖÌú±íÃæ²ÉÓÃ
Í¿Æá
Í¿Æá
¡¢
¶Æ½ðÊô
¶Æ½ðÊô
µÈ·½·¨¿ÉÒÔ·ÀÖ¹¸ÖÌúÉúÐ⣮
£¨3£©ÉúÌúºÍ¸Ö¶¼ÊÇ
Ìú
Ìú
µÄºÏ½ð£¬ÉúÌúµÄº¬Ì¼Á¿
´óÓÚ
´óÓÚ
£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©¸ÖµÄº¬Ì¼Á¿£¬ÉúÌú±È´¿Ìú
Ó²
Ó²
£¨Ìî¡°Ó²¡±»ò¡°Èí¡±£©£®
£¨4£©îѺϽðÊÇ21ÊÀ¼ÍµÄÖØÒª²ÄÁÏ£¬¾ßÓÐÈÛµã¸ß¡¢¿ÉËÜÐԺᢿ¹¸¯Ê´ÐÔÇ¿¡¢ÓëÈËÌåÓкܺõġ°ÏàÈÝÐÔ¡±µÈÓÅÁ¼ÐÔÄÜ£®½ðÊôµÄÏÂÁÐÓÃ;£º¢ÙÓÃÀ´×÷±£ÏÕË¿£»¢ÚÓÃÀ´ÖÆÈËÔì¹Ç£»¢ÛÓÃÓÚÖÆÔì´¬²°£»¢ÜÓÃÓÚÖÆÔ캽Ìì·É»ú£®ÆäÖÐÓëîѺϽðÐÔÄÜ·ûºÏµÄÊÇ
¢Ú¢Û¢Ü
¢Ú¢Û¢Ü
£¨ÌîÐòºÅ£©£®
·ÖÎö£º£¨1£©½¹Ì¿¾ßÓл¹Ô­ÐÔ£¬ÊÇÒ±Á¶½ðÊôʱ¾­³£Óõ½µÄ»¹Ô­¼Á£¬¸ù¾Ý·´Ó¦Îï¡¢Éú³ÉÎï¡¢·´Ó¦Ìõ¼þ¼°ÆäÖÊÁ¿Êغ㶨ÂÉ¿ÉÒÔÊéд»¯Ñ§·½³Ìʽ£»
£¨2£©ÌúÓëÑõÆøºÍË®³ä·Ö½Ó´¥Ê±ÈÝÒ×ÉúÐ⣬ˮºÍÑõÆøͬʱ´æÔÚÊÇÌúÉúÐâµÄ±ØÒªÌõ¼þ£»
£¨3£©ÉúÌúµÄº¬Ì¼Á¿Îª2%¡«4.3%£¬¸ÖµÄº¬Ì¼Á¿Îª0.03%¡«2%£¬´¿Ìú½ÏÈí£»
£¨4£©îѺϽð¾ßÓÐÓÅÁ¼ÐÔÄÜ£¬¾ßÓйãÀ«µÄ·¢Õ¹Ç°¾°£®
½â´ð£º½â£º£¨1£©½¹Ì¿ÊÇÁ¶ÌúµÄÖ÷ÒªÔ­ÁÏ£¬ÔÚ·´Ó¦¹ý³ÌÖУ¬ÄܺÍÑõÆø·´Ó¦Éú³É¶þÑõ»¯Ì¼£¬ÔÚ¸ßÎÂÌõ¼þÏ£¬¶þÑõ»¯Ì¼Äܱ»½¹Ì¿»¹Ô­³ÉÒ»Ñõ»¯Ì¼£¬Ò»Ñõ»¯Ì¼ÔÙ°Ñ´ÅÌú¿óÖеÄÌú»¹Ô­³öÀ´£¬´ÅÌú¿óµÄÖ÷Òª³É·ÖÊÇËÄÑõ»¯ÈýÌú£¬ºÍÒ»Ñõ»¯Ì¼·´Ó¦ÄÜÉú³ÉÌúºÍ¶þÑõ»¯Ì¼£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
Fe3O4+4CO
 ¸ßΠ
.
 
3Fe+4CO2 £®
¹ÊÌ½¹Ì¿£»Fe3O4+4CO
 ¸ßΠ
.
 
3Fe+4CO2 £®
£¨2£©¸ÖÌúÓëË®ºÍÑõÆøͬʱ½Ó´¥Ê±ÈÝÒ×ÉúÐ⣬ÔÚ¸ÖÌú±íÃæÍ¿Æá¡¢¶ÆÒ»²ã½ðÊôµÈ·½·¨¿ÉÒÔ·ÀÖ¹¸ÖÌúÉúÐ⣮
¹ÊÌˮºÍÑõÆø£»Í¿Æ᣻¶Æ½ðÊô£®
£¨3£©ÉúÌúºÍ¸Ö¶¼ÊÇÌúµÄºÏ½ð£¬ÉúÌúµÄº¬Ì¼Á¿´óÓڸֵĺ¬Ì¼Á¿£¬ÉúÌú±È´¿ÌúÓ²¶È´ó£®
¹ÊÌÌú£»´óÓÚ£»Ó²£®
£¨4£©îѺϽð¿ÉÒÔÓÃÀ´ÖÆÈËÔì¹Ç¡¢ÖÆÔì´¬²°¡¢ÖÆÔ캽Ìì·É»ú£®
¹ÊÌ¢Ú¢Û¢Ü£®
µãÆÀ£ºÌúÓëÑõÆøºÍË®³ä·Ö½Ó´¥Ê±ÈÝÒ×ÉúÐ⣬ˮºÍÑõÆøͬʱ´æÔÚÊÇÌúÉúÐâµÄ±ØÒªÌõ¼þ£»Èç¹ûȱÉÙË®»òȱÉÙÑõÆø£¬»òÕßȱÉÙÑõÆøºÍË®£¬Ìú¾Í²»ÈÝÒ×ÉúÐ⣮
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

16¡¢ÈËÃǵÄÈÕ³£Éú»îÀë²»¿ª½ðÊô£¬¸ß¿Æ¼¼Ð²ÄÁϵĿª·¢ºÍÓ¦ÓÃÒ²ÐèÒª½ðÊô£®ÏÂͼÊǽðÊôÔÚʵ¼ÊÉú»îÖеÄÓ¦Óã®

£¨1£©µØ¿ÇÖк¬Á¿×î¸ßµÄ½ðÊôÔªËØÊÇ
ÂÁ
£®
£¨2£©¸ù¾ÝÉÏͼÍƶϣ¬½ðÊô¾ßÓеÄÎïÀíÐÔÖÊÓÐ
¾ßÓе¼µçÐÔ£¨»òµ¼ÈÈÐÔ¡¢ÑÓÐԵȣ©
£¨Ð´Ò»µã£©£®
£¨3£©¸ÖÌúÈÝÒ×ÉúÐâÔì³ÉËðʧ£¬ÇëÄãд³öÒ»Ìõ·ÀÖ¹¸ÖÌúÉúÐâµÄ´ëÊ©
ÔÚÌúÖÆÆ·±íÃæÅçÆáµÈ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?ÔæÑôÊÐÄ£Ä⣩ÈËÃǵÄÈÕ³£Éú»îÀë²»¿ª½ðÊô£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©°ÂÔ˳¡¹Ý¡°Äñ³²¡±Ê¹ÓÃÁË´óÁ¿µÄ¸ÖÌú£®¸ÖÌúÓë
ÑõÆø¡¢Ë®
ÑõÆø¡¢Ë®
Ö±½Ó½Ó´¥ÈÝÒ×ÉúÐâÔì³ÉËðʧ£¬ÔÚ¸ÖÌú±íÃæÍ¿ÓÍ¡¢Ë¢ÆáµÈ£¬¶¼ÄÜ·ÀÖ¹¸ÖÌúÉúÐ⣮
£¨2£©ÊµÑéÊÒÓÃͼʾʵÑéÄ£ÄâÁ¶Ìú·´Ó¦µÄÔ­Àí£®
¢Ùa´¦¿É¹Û²ìµÄÏÖÏóÊÇ
ºìÉ«·ÛÄ©±äΪºÚÉ«
ºìÉ«·ÛÄ©±äΪºÚÉ«
£»
¢Úb´¦µãȼ¾Æ¾«µÆµÄÄ¿µÄÊÇ
·ÀÖ¹COÅŷŵ½¿ÕÆøÖУ¬ÎÛȾ¿ÕÆø
·ÀÖ¹COÅŷŵ½¿ÕÆøÖУ¬ÎÛȾ¿ÕÆø
£»
¢ÛΪȷ±£ÊµÑ鰲ȫ£¬ÔÚͨÈëÒ»Ñõ»¯Ì¼Ö®Ç°Ó¦¸Ã
¼ìÑéCOµÄ´¿¶È
¼ìÑéCOµÄ´¿¶È
£®
£¨3£©Öû»·´Ó¦ÊÇ»¯Ñ§·´Ó¦µÄ»ù±¾ÀàÐÍÖ®Ò»£®·Ç½ðÊôµ¥ÖÊÒ²¾ßÓÐÀàËƽðÊôÓëÑÎÈÜÒºÖ®¼äµÄÖû»·´Ó¦¹æÂÉ£¬¼´»î¶¯ÐÔ½ÏÇ¿µÄ·Ç½ðÊô¿É°Ñ»î¶¯ÐÔ½ÏÈõµÄ·Ç½ðÊô´ÓÆäÑÎÈÜÒºÖÐÖû»³öÀ´£¬ÈçÔÚÈÜÒºÖпɷ¢ÉúÏÂÁз´Ó¦£ºC12+2NaBr=2NaCl+Br2£»I2+Na2S=2NaI+S¡ý£»Br2+2KI=2KBr+I2 ÓÉ´Ë¿ÉÅжϣº
¢ÙS¡¢C12¡¢I2¡¢Br2»î¶¯ÐÔÓÉÇ¿µ½Èõ˳ÐòÊÇ
Cl2£¾Br2£¾I2£¾S
Cl2£¾Br2£¾I2£¾S
£®
¢ÚÏÂÁл¯Ñ§·½³ÌʽÊéд´íÎóµÄÊÇ
B
B
£®
A£®C12+2NaI=2NaCl+I2      B£®I2+2KBr=2KI+Br2
C£®Br2+Na2S=2NaBr+S¡ý     D£®C12+K2S¨T2KCl+S¡ý
Ö÷Òª³É·Ö NH4Cl
ʹÓ÷½·¨ ²»ÄܺͲÝľ»Ò»ìºÏʹÓÃ
º¬µªÁ¿ ¡Ý25%

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÈËÃǵÄÈÕ³£Éú»îÀë²»¿ª½ðÊô£¬¸ß¿Æ¼¼Ð²ÄÁϵĿª·¢ºÍÀûÓÃÒ²ÐèÒª½ðÊô£®
£¨1£©¸ù¾ÝÈçͼËùʾµÄ½ðÊôÓ¦ÓÃʵÀýÍƶϣ¬½ðÊô¾ßÓеÄÎïÀíÐÔÖÊÓÐ
µ¼µçÐÔ¡¢µ¼ÈÈÐÔ
µ¼µçÐÔ¡¢µ¼ÈÈÐÔ
£®
£¨2£©ÈÕ³£Ê¹ÓõĽðÊô²ÄÁ϶àÊýÊôÓںϽð£¬ÏÖÓÐһЩºÏ½ð¼°Æä×é·ÖµÄÈÛµãÈçÏÂͼËùʾ£º´ÓͼÖÐÊý¾Ý¿ÉÒԵóöµÄ½áÂÛÊÇ£º
ºÏ½ðµÄÈÛµãµÍÓÚÆä×é·ÖµÄÈÛµã
ºÏ½ðµÄÈÛµãµÍÓÚÆä×é·ÖµÄÈÛµã
£®

£¨3£©¸ÖÌúÊÇÓÃÁ¿×î´óµÄºÏ½ð£®¹¤Òµ¸ß¯Á¶ÌúµÄÖ÷ÒªÔ­ÀíÊÇÓÃÒ»Ñõ»¯Ì¼½«Ìú´ÓÌú¿óʯÀﻹԭ³öÀ´£®ÈôÑ¡ÓõÄÌú¿óʯÊdzàÌú¿ó£¨Ö÷Òª³É·ÖÊÇÑõ»¯Ìú£©£¬Çëд³öÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³Ìʽ
3CO+Fe2O3
 ¸ßΠ
.
 
3CO2+2Fe
3CO+Fe2O3
 ¸ßΠ
.
 
3CO2+2Fe
£®
£¨4£©µØ¿ÇÖÐÂÁ¡¢ÌúµÄº¬Á¿Ô¶¶àÓÚÍ­£¬µ«Í­È´ÊÇÈËÀà×îÔçÒ±Á¶³öµÄ½ðÊô£¬ÒÑÓÐ6000¶àÄêµÄÀúÊ·£¬¶øÒ±Á¶Ìú±ÈÍ­ÍíÁË3000¶àÄ꣬ÖÁÓÚÒ±Á¶ÂÁÔòÖ»Óжþ°Ù¶àÄêµÄÀúÊ·£®¸ù¾Ý½ðÊôµÄ»¯Ñ§ÐÔÖÊ£¬¶ÔÕâÒ»ÏÖÏóµÄ½âÊÍΪ
½ðÊôµÄ»î¶¯ÐÔ²»Í¬£®¸ù¾Ý½ðÊôµÄ»î¶¯ÐÔ˳Ðò£¬Al¡¢Fe¡¢CuµÄ½ðÊô»î¶¯ÐÔΪAl£¾Fe£¾Cu£¬½ðÊôÔ½»îÆã¬Ô½ÄÑ´ÓÆ仯ºÏÎïÖл¹Ô­³öÀ´
½ðÊôµÄ»î¶¯ÐÔ²»Í¬£®¸ù¾Ý½ðÊôµÄ»î¶¯ÐÔ˳Ðò£¬Al¡¢Fe¡¢CuµÄ½ðÊô»î¶¯ÐÔΪAl£¾Fe£¾Cu£¬½ðÊôÔ½»îÆã¬Ô½ÄÑ´ÓÆ仯ºÏÎïÖл¹Ô­³öÀ´
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2008?±±¾©Ò»Ä££©ÈËÃǵÄÈÕ³£Éú»îÀë²»¿ª½ðÊô£¬¸ß¿Æ¼¼Ð²ÄÁϵĿª·¢ºÍÓ¦ÓÃÒ²ÐèÒª½ðÊô£®
£¨1£©µØ¿ÇÖк¬Á¿×î¸ßµÄ½ðÊôÔªËØÊÇ£¨ÌîÔªËØ·ûºÅ£©
Al
Al
£®
£¨2£©2008Äê°ÂÔË»áµÄ¸÷¸öÔ˶¯³¡Ê¹ÓÃÁË´óÁ¿µÄ¸ÖÌú£®Çëд³ö·ÀÖ¹¸ÖÌúÉúÐâÁ½Ìõ³£ÓõĴëÊ©
±£³Ö±íÃæ¸ÉÔ
±£³Ö±íÃæ¸ÉÔ
¡¢
Í¿ÓÍÆᣮ
Í¿ÓÍÆᣮ
£®
£¨3£©ÔÚAgNO3ºÍCu£¨NO3£©2 µÄ»ìºÏÈÜÒºÖмÓÈëÒ»¶¨Á¿µÄÌú·Û£¬³ä·Ö·´Ó¦ºó¹ýÂË£®Èô½«ÂËÖ½ÉϵĹÌÌå·ÅÈëÏ¡ÑÎËáÖУ¬Éú³ÉÎÞÉ«¡¢ÎÞζµÄ¿ÉȼÐÔÆøÌ壮ÔòÂËÖ½ÉÏÒ»¶¨ÓÐ
Fe¡¢Ag¡¢Cu
Fe¡¢Ag¡¢Cu
£» ÂËÒºÖÐÓÐ
Fe£¨NO3£©2
Fe£¨NO3£©2
£¨ÌîÈÜÖʵĻ¯Ñ§Ê½£©£®Çëд³öÌúÓëAgNO3ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ
Fe+2AgNO3=2Ag+Fe£¨NO3£©2
Fe+2AgNO3=2Ag+Fe£¨NO3£©2
£®
£¨4£©ÌúÓÐÈýÖÖÑõ»¯Îï¢ÙFeO ¢ÚFe2O3 ¢ÛFe3O4£®ËüÃǺ¬ÌúÔªËصÄÖÊÁ¿·ÖÊý´ÓµÍµ½¸ßµÄ˳ÐòΪ
¢Ú¢Û¢Ù
¢Ú¢Û¢Ù
£¨ÌîÐòºÅ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2011?ºþ±±£©ÈËÃǵÄÈÕ³£Éú»îÀë²»¿ª½ðÊô£®

£¨1£©¸ù¾ÝÓÒͼµÄÓ¦ÓÃʵÀý£¬ÊÔ˵³ö½ðÊô¾ßÓеÄÁ½µãÎïÀíÐÔÖÊ£®
£¨2£©ÂÁµÄ»¯Ñ§ÐÔÖʱÈÌú»îÆ㬵«ÎªÊ²Ã´ÈÕ³£Éú»îÖÐÂÁÖÆÆ·²»Ò×ÐâÊ´£¿

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸