£¨2013?¼ÃÄþÄ£Ä⣩ÏÖÓк¬ÔÓÖʵÄÑõ»¯ÌúÑùÆ·£¨ÔÓÖʲ»²Î¼Ó·´Ó¦£©£¬ÎªÁ˲ⶨ¸ÃÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£¬Ä³Í¬Ñ§³ÆÈ¡ÑùÆ·10¿Ë£¬²¢ÓÃÈçͼËùʾװÖýøÐÐʵÑ飬µÃµ½ÒÔÏÂÁ½×éʵÑéÊý¾Ý£®
·´Ó¦Ç° Ñõ»¯ÌúÍêÈ«·´Ó¦ºó
·½°¸Ò» ²£Á§¹ÜºÍÑõ»¯ÌúÑùÆ·µÄÖÊÁ¿43.7g ²£Á§¹ÜºÍ¹ÌÌåÎïÖʵÄÖÊÁ¿41.3g
·½°¸¶þ ÉÕ±­ºÍ³ÎÇåʯ»ÒË®µÄÖÊÁ¿180g ÉÕ±­ºÍÉÕ±­ÖÐÎïÖʵÄÖÊÁ¿186.2g
£¨1£©¼ÆËãÁ½ÖÖ·½°¸ÖвâµÃµÄÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£º
·½°¸Ò»ÖеÄÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý
80%
80%

·½°¸¶þÖеÄÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý
75%
75%

£¨2£©±È½ÏÁ½·½°¸²âÁ¿µÄ½á¹û£¬·ÖÎöÁ½·½°¸Îó²î½Ï´óµÄ¿ÉÄÜÔ­Òò£¿
×°ÖÃÖеĶþÑõ»¯Ì¼Ã»ÓÐÈ«²¿½øÈëʯ»ÒË®ÖУ¬Ê¯»ÒË®ÖеÄÇâÑõ»¯¸Æ²»×㣬²»ÄÜÍêÈ«ÎüÊÕÉú³ÉµÄ¶þÑõ»¯Ì¼
×°ÖÃÖеĶþÑõ»¯Ì¼Ã»ÓÐÈ«²¿½øÈëʯ»ÒË®ÖУ¬Ê¯»ÒË®ÖеÄÇâÑõ»¯¸Æ²»×㣬²»ÄÜÍêÈ«ÎüÊÕÉú³ÉµÄ¶þÑõ»¯Ì¼
£®
·ÖÎö£º·½°¸Ò»£º²£Á§¹ÜÖеĹÌÌåÔÚ·´Ó¦Ç°ºóµÄÖÊÁ¿²î¾ÍÊÇÑõ»¯ÌúÖÐÑõÔªËصÄÖÊÁ¿£¬¸ù¾ÝÑõÔªËصÄÖÊÁ¿¿ÉÒÔ¼ÆËãÑõ»¯ÌúµÄÖÊÁ¿£¬½øÒ»²½¿ÉÒÔ¼ÆËãÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£»
·½°¸¶þ£ºÊ¯»ÒË®Äܹ»ÎüÊÕ¶þÑõ»¯Ì¼£¬ÉÕ±­Öз´Ó¦Ç°ºóµÄÖÊÁ¿²î¾ÍÊÇÉú³ÉµÄ¶þÑõ»¯Ì¼µÄÖÊÁ¿£¬¸ù¾Ý¶þÑõ»¯Ì¼µÄÖÊÁ¿£¬¿ÉÒÔ¼ÆËãÑõ»¯ÌúµÄÖÊÁ¿£¬½øÒ»²½¿ÉÒÔ¼ÆËãÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£®
½â´ð£º½â£º·½°¸Ò»ÖеÄÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£º
Ñõ»¯ÌúÖÐÑõÔªËصÄÖÊÁ¿Îª£º43.7g-41.3g=2.4g£¬
Ñõ»¯ÌúÖÐÑõÔªËصÄÖÊÁ¿·ÖÊýΪ£º
48
160
¡Á100%=30%£¬
Ñõ»¯ÌúµÄÖÊÁ¿Îª£º2.4g¡Â30%=8g£¬
Ñõ»¯ÌúµÄÖÊÁ¿·ÖÊýΪ£º
8g
10g
¡Á100%=80%£»
¹ÊÌ80%£®
·½°¸¶þÖеÄÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£º
½â£ºÉèÑõ»¯ÌúµÄÖÊÁ¿ÎªX£¬
Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª£º186.2g-180g=6.2g£¬
Fe2O3+3CO
 ¸ßΠ
.
 
2Fe+3CO2£¬
 160                  132
  X                   6.2g
160
132
=
X
6.2g
£¬
X=7.5g£¬
7.5g
10g
¡Á100%=75%£»
¹ÊÌ75%£®
£¨2£©Á½·½°¸Îó²î½Ï´óµÄ¿ÉÄÜÔ­ÒòÓУº×°ÖÃÖеĶþÑõ»¯Ì¼Ã»ÓÐÈ«²¿½øÈëʯ»ÒË®ÖУ¬Ê¯»ÒË®ÖеÄÇâÑõ»¯¸Æ²»×㣬²»ÄÜÍêÈ«ÎüÊÕÉú³ÉµÄ¶þÑõ»¯Ì¼£®
¹ÊÌװÖÃÖеĶþÑõ»¯Ì¼Ã»ÓÐÈ«²¿½øÈëʯ»ÒË®ÖУ¬Ê¯»ÒË®ÖеÄÇâÑõ»¯¸Æ²»×㣬²»ÄÜÍêÈ«ÎüÊÕÉú³ÉµÄ¶þÑõ»¯Ì¼£®
µãÆÀ£º»¯Ñ§ÊµÑéÏÖÏóÊÇ»¯Ñ§ÊµÑé×îÍ»³ö¡¢×îÏÊÃ÷µÄ²¿·Ö£¬Ò²ÊǽøÐзÖÎöÍÆÀíµÃ³ö½áÂÛµÄÒÀ¾Ý£¬ÕÆÎÕÎïÖʵÄÐÔÖʺÍÏ໥֮¼äµÄ·´Ó¦¹Øϵ£¬²¢ÓÐÖúÓÚÌá¸ß¹Û²ì¡¢ÊµÑéÄÜÁ¦£®ËùÒÔ£¬¶Ô»¯Ñ§ÊµÑé²»½öÒªÈÏÕæ¹Û²ì£¬»¹Ó¦ÕÆÎÕÉè¼ÆʵÑé¡¢¹Û²ìʵÑéÏÖÏóµÄ·½·¨£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?¼ÃÄþÄ£Ä⣩ÏÂÁÐͼÏóÖÐÓйØÁ¿µÄ±ä»¯Ç÷ÊÆÓë¶ÔÓ¦ÐðÊö¹ØϵÕýÈ·µÄÊÇ¡¡£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?¼ÃÄþÄ£Ä⣩¡°µÍ̼Éú»î¡±ÊÇÖ¸¼õÉÙÄÜÔ´ÏûºÄ¡¢½ÚÔ¼×ÊÔ´£¬´Ó¶ø½µµÍ¶þÑõ»¯Ì¼ÅŷŵÄÒ»ÖÖʱÉÐÉú»î·½Ê½£®ÏÂÁв»·ûºÏ¡°µÍ̼Éú»î¡±ÕâÒ»Ö÷ÌâµÄ×ö·¨ÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?¼ÃÄþÄ£Ä⣩µ¤µ¤Í¬Ñ§·¢ÏÖÂèÂè×ö°ü×Ó¡¢Âøͷʱ£¬ÏòÃæÍÅÖмÓÈë·¢½Í·Û£¬Õô³öµÄ°ü×Ó¡¢ÂøÍ·ÖÐÓÐÐí¶àС¿×£¬ËÉÈí¿É¿Ú£¬µ¤µ¤Í¬Ñ§¶Ô·¢½Í·Û²úÉúÁËÐËȤ£®
£¨1£©Ìá³öÎÊÌ⣺·¢½Í·ÛµÄÖ÷Òª³É·ÖÊÇʲôÎïÖÊ£¿
£¨2£©²éÔÄ×ÊÁÏ£ºº¬ÄÆÔªËصÄÎïÖÊÔھƾ«µÆµÄ»ðÑæÉÏ×ÆÉÕʱ£¬»á²úÉú»ÆÉ«»ðÑ森
£¨3£©ÊµÑé̽¾¿£ºÏÂÃæÊǵ¤µ¤Í¬Ñ§Éè¼ÆµÄ̽¾¿ÊµÑ飬ÇëÄã°ïËýÍê³É£®
¢Ù½«·¢½Í·ÛÈ¡Ñù£¬Ôھƾ«µÆµÄ»ðÑæÉÏ×ÆÉÕ²úÉú»ÆÉ«»ðÑ森
¢ÚÈ¡ÑùÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏ¡ÑÎËᣬ²úÉú´óÁ¿µÄÆøÅÝ£¬½«¸ÃÆøÌåͨÈë³ÎÇåʯ»ÒË®£¬³ÎÇåµÄʯ»ÒË®±ä»ë×Ç£¬Ôò·¢½Í·ÛµÄÖ÷Òª³É·ÖÖк¬ÓеÄÀë×ÓÊÇ
HCO3-»òC
O
2-
3
HCO3-»òC
O
2-
3
£¨Ð´Àë×Ó·ûºÅ£©£®
£¨4£©Ð¡½áÓë˼¿¼£º
¢Ù·¢½Í·ÛµÄÖ÷Òª³É·ÖÊÇ
NaHCO3»òNa2CO3
NaHCO3»òNa2CO3
£¨Ð´»¯Ñ§Ê½£©£®
¢Ú·¢½Í·ÛÓëÃæ·Û¡¢Ë®»ìºÏ·¢½Í²úÉúCO2ÆøÌ壬ÕâЩÆøÌå»áÔÚÃæÍÅÖÐÐγÉÐí¶àС¿×£¬Ê¹°ü×Ó¡¢ÂøÍ·ËÉÈí¿É¿Ú£®
¢ÛʵÑéÊÒÖÆÈ¡CO2¿ÉÑ¡ÔñÏÂÁÐ×°ÖÃÖеÄ
BD
BD
£¨Ìî×ÖĸÐòºÅ£©£®

¢ÜʵÑéÊÒÓÃA×°ÖÃÖÆÈ¡ÑõÆøµÄ»¯Ñ§·½³ÌʽΪ
2KMnO4
  ¡÷  
.
 
K2MnO4+MnO2 +O2¡ü»ò 2KClO3
MnO2
.
¡÷
2KCl+3O2¡ü
2KMnO4
  ¡÷  
.
 
K2MnO4+MnO2 +O2¡ü»ò 2KClO3
MnO2
.
¡÷
2KCl+3O2¡ü
£®ÓÃF×°ÖÃÊÕ¼¯ÑõÆøʱ£¬»¹¿ÉÓÃÓڲⶨÑõÆøµÄÌå»ý£¬´Ëʱ»¹ÐèÒª
Á¿Í²
Á¿Í²
£¨ÌîÒÇÆ÷Ãû³Æ£©£¬ÑõÆøÓÉ
¢Ú
¢Ú
£¨ÌîÐòºÅ£©½øÈëFÖУ®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?¼ÃÄþÄ£Ä⣩СÃ÷¼ÒÊÇÖÖÖ²Ê߲˵Äרҵ»§£¬ËûѧϰÁË»¯Ñ§ºó¸æË߸¸Ä¸£¬Ê¹ÓÃÊÊÁ¿µÄ»¯·Ê¿ÉÒÔÌá¸ß²úÁ¿£®Òò´Ë£º
£¨1£©Ëû½¨Ò鸸ĸȥÂò¸´ºÏ·Ê£¬ÏÂÁÐÊôÓÚ¸´ºÏ·ÊµÄÊÇ
CD
CD
£¨Ñ¡Ìî×Öĸ£©£®
A¡¢ÄòËØ[CO£¨NH2£©2]£»B¡¢ÁòËá¼Ø£¨K2SO4£©£»
C¡¢ÏõËá¼Ø£¨KNO3£©£»  D¡¢Á×Ëá¶þÇâ泥¨NH4H2PO4£©£®
£¨2£©Ëû»¹¸æË߸¸Ä¸£¬ÔÚÊ©ÓÃï§Ì¬µª·ÊÂÈ»¯ï§Ê±£¬
²»ÄÜ
²»ÄÜ
£¨Ñ¡Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©Í¬Ê±Ê©ÓòÝľ»ÒµÈ¼îÐÔÎïÖÊ£¬Ð´³öÂÈ»¯ï§ºÍÊìʯ»Ò·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
2NH4Cl+Ca£¨OH£©2=CaCl2+2H2O+2NH3¡ü
2NH4Cl+Ca£¨OH£©2=CaCl2+2H2O+2NH3¡ü
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?¼ÃÄþÄ£Ä⣩¹¹½¨ºÍгÉç»á£¬½¨ÉèÉç»áÖ÷ÒåÐÂÅ©´åµÄÄ¿±êÖ®Ò»ÊÇÈÃÅ©ÃñÒûÓÃÇå½àµÄ×ÔÀ´Ë®£®ClO2 ÊÇÐÂÒ»´úÒûÓÃË®µÄÏû¶¾¼Á£¬ÎÒ¹ú³É¹¦ÑÐÖƳöÖÆÈ¡ClO2 µÄз½·¨£¬Æä·´Ó¦µÄ΢¹Û¹ý³ÌͼÈçÏ£º

£¨ÆäÖР   ±íʾÂÈÔ­×Ó£¬±íʾÄÆÔ­×Ó£¬±íʾÑõÔ­×Ó£©
ÊԻشð£º
ClO2 ÖÐÂÈÔªËصĻ¯ºÏ¼ÛΪ
+4
+4
¼Û£»
¸ù¾Ý·´Ó¦µÄ΢¹Û¹ý³Ìͼд³ö·´Ó¦µÄ»¯Ñ§·½³Ì
Cl2+2NaClO2=2ClO2+2NaCl
Cl2+2NaClO2=2ClO2+2NaCl
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸