ÌâÄ¿ÄÚÈÝ

¼îʯ»ÒÊÇÑõ»¯¸ÆºÍÇâÑõ»¯ÄƵĹÌÌå»ìºÏÎÊÇʵÑéÊÒ³£ÓõĸÉÔï¼Á¡£Ïà¹ØÐÅÏ¢ÈçÏÂͼËùʾ¡£Í¬Ñ§ÃÇΪȷÈÏһƿ¾ÃÖõġ°¼îʯ»Ò¡±£¨¿ÉÄÜÒѲ¿·Ö±äÖÊ»òÈ«²¿±äÖÊ£©ÑùÆ·µÄ³É·Ö£¬½øÐÐÈçÏÂ̽¾¿¡£

[Ìá³öÎÊÌâ]ʵÑéÊÒÖоÃÖõļîʯ»ÒÑùÆ·µÄ³É·ÖÊÇʲô£¿

[½øÐвÂÏë] ¾ÃÖõļîʯ»ÒÑùÆ·ÖпÉÄܺ¬ÓÐCaO¡¢NaOH¡¢Na2CO3¡¢Ca£¨OH£©2¡¢CaCO3µÈ³É·Ö¡£Óû¯Ñ§·½³Ìʽ±íʾÑùÆ·Öк¬ÓÐCa£¨OH£©2»òNa2CO3¿ÉÄܺ¬ÓеÄÔ­Òò           £¨Ð´Ò»¸ö£©

[̽¾¿¹ý³Ì]

ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©²Ù×÷¢ÙµÄÃû³ÆÊÇ       £¬ÈÜÒºCÖÐÒ»¶¨º¬       Àë×Ó¡£

£¨2£©ÈÜÒºAÖмÓÈëCaCl2ÈÜÒººó²úÉú¹ÌÌ壬˵Ã÷ÑùÆ·ÖÐÒ»¶¨º¬ÓР           £¬·´Ó¦µÄ·½³ÌʽΪ        ¡£

£¨3£©ÏòÈÜÒºAÖмÓÈëCaCl2ÈÜÒººó£¬Ö¤Ã÷CaCl2ÈÜÒº¹ýÁ¿µÄ·½·¨ÊÇ        ¡£

[ʵÑé½áÂÛ]

ÒÀ¾ÝÉÏÊöʵÑé²Ù×÷¹ý³Ì¼°ÏÖÏó£¬×ۺ϶ÔÂËÒººÍ¹ÌÌå³É·ÖµÄ̽¾¿£¬ÏÂÁжÔÑùÆ·³É·ÖµÄ·ÖÎöÕýÈ·µÄÊÇ       £¨ÌîÐòºÅ£©

¢ÙÑùÆ·ÖÐÒ»¶¨º¬NaOH        ¢ÚÑùÆ·ÖÐÒ»¶¨º¬Na2CO3       ¢ÛÑùÆ·Öк¬NaOH¡¢CaOÖеÄÒ»ÖÖ»òÁ½ÖÖ

 [ÍØÕ¹ÑÓÉì]

Ϊ²â¶¨ÊµÑéÊÒÖÐһƿ±äÖʵÄÉÕ¼îÖÐNaOHµÄº¬Á¿£¬Ä³Í¬Ñ§È¡ÊÊÁ¿µÄÉÕ¼îÑùÆ·£¬ÈÜÓÚÒ»¶¨Á¿µÄË®µÃµ½200gÈÜÒº£»ÔÙ¼ÓÈë200gÏ¡ÁòËᣨ×ãÁ¿£©³ä·Ö½Á°èµ½²»ÔٷųöÆøÅÝΪֹ£¬³ÆÁ¿ÈÜҺΪ395.6g¡£Çë·ÖÎö¼ÆË㣺

£¨1£©·´Ó¦²úÉúµÄCO2µÄÖÊÁ¿Îª        g¡£

£¨2£©ÈôËùÈ¡ÉÕ¼îÑùƷΪ50.0g£¬ÔòÑùÆ·ÖÐNaOHµÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿£¨Ð´³ö¼ÆËã¹ý³Ì£©

       

£¨3£©ÓÃͬŨ¶ÈµÄÁòËᣬ·Ö±ðÓëδ±äÖÊ¡¢²¿·Ö±äÖÊ¡¢»òÈ«²¿±äÖÊ£¨ÔÓÖʶ¼ÊÇNa2CO3£©µÄÉÕ¼îÑùÆ··´Ó¦£¬ÐèÒªÁòËáµÄÖÊÁ¿¶¼ÏàµÈ¡£´ÓÄÆÔªËØÖÊÁ¿ÊغãµÄ½Ç¶È½âÊÍÆäÔ­ÒòÊÇ£º         ¡£

¡¾½âÎö¡¿[½øÐвÂÏë]Éú³ÉÇâÑõ»¯¸ÆµÄ·½³ÌʽΪCaO+H2O¨TCa£¨OH£©Éú³É̼ËáÄƵķ½³ÌʽΪ2NaOH+CO2¨TNa2CO3+H2O

[̽¾¿¹ý³Ì]

£¨1£©½«¹ÌÒº·ÖÀëµÄ²Ù×÷Ϊ¹ýÂË£¬·Ó̪±äºìÁË£¬ËµÃ÷ÈÜÒºÏÔ¼îÐÔ£¬¹ÊÈÜÒºÖÐÒ»¶¨º¬ÓÐOH¡ª

£¨2£©ÄÜÓëÂÈ»¯¸Æ·´Ó¦µÄÊÇ̼ËáÄÆ£¬Æä·´Ó¦·½³ÌʽΪNa2CO3+CaCl2¨T2NaCl+CaCO3¡ý

£¨3£©¿ÉÒÔÈ¡ÈÜÒºCÓÚÊÔ¹ÜÖеμÓ̼ËáÄÆÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬Ôò˵Ã÷CaCl2ÈÜÒºÒѹýÁ¿

[ʵÑé½áÂÛ]ÒòΪ¼ÓË®ºóÉÕ±­±Ú±äÈÈ˵Ã÷ÑùÆ·Öк¬ÓÐÇâÑõ»¯ÄÆ»òÕßÊÇÑõ»¯¸Æ£¬»òÕßÊǶþÕ߶¼ÓУ¬¼ÓÈëÂÈ»¯¸ÆÉú³É°×É«³Áµí˵Ã÷ÑùÆ·ÖÐÒ»¶¨º¬ÓÐ̼ËáÄÆ£¬×ÛºÏÒÔÉÏ·ÖÎö£¬¹ÊÑ¡¢Ú¢Û

[ÍØÕ¹ÑÓÉì]

£¨1£©¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉ¿ÉÇó³ö·´Ó¦²úÉúµÄCO2µÄÖÊÁ¿Îª200g+200g-395.6g=4.4g

£¨2£©¸ù¾Ý¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª4.4¿Ë¾Í¿ÉÒÔ¼ÆËãÁË£¬¸ù¾Ý»¯Ñ§·´Ó¦·½³Ìʽ¼ÆË㣬¼´¿ÉµÃ³öÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿£¬¼Ì¶øµÃ³öÔ­ÑùÆ·ÖÐÇâÑõ»¯ÄƵÄÖÊÁ¿£¬¾ßÌå²½Öè¼û´ð°¸¡£

£¨3£©ÇâÑõ»¯ÄƺÍ̼ËáÄƶ¼ÊÇÿ46gÄÆÔªËØÉú³É142gÁòËáÄÆ£¬ÏûºÄ98gÁòËá

 

[½øÐвÂÏë]CaO+H2O¨TCa£¨OH£©2£¨»ò2NaOH+CO2¨TNa2CO3+H2OµÈ£¬´ð°¸ºÏÀí¼´¸ø·Ö£©

[ʵÑé¹ý³Ì]£¨1£©¹ýÂË   OH¡ª

£¨2£©Ì¼ËáÄÆ   Na2CO3+CaCl2¨T2NaCl+CaCO3¡ý

£¨3£©È¡ÈÜÒºCÓÚÊÔ¹ÜÖеμÓ̼ËáÄÆÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬ËµÃ÷CaCl2ÈÜÒºÒѹýÁ¿£¨ºÏÀí¼´¸ø·Ö£©

[ʵÑé½áÂÛ]£¨1£©¢Ú¢Û

[ÍØÕ¹ÑÓÉì]

£¨1£©4.4¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­(1·Ö)

£¨2£©½â£ºÉèËùÈ¡ÑùÆ·Öк¬ÓÐNa2CO3µÄÖÊÁ¿Îªx¡£

     Na2CO3+H2SO4===Na2 SO4+H2O+CO2¡ü¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­(1·Ö)

    106                          44

    x                            4.4g

                                          ¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­(1·Ö)

   

   ÔòÉÕ¼îÑùÆ·ÖР                                     ¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­(1·Ö)

 

´ð£ºÉÕ¼îÑùÆ·ÖÐNaOHµÄ°Ù·Öº¬Á¿Îª78.8%¡£

£¨3£©ÇâÑõ»¯ÄƺÍ̼ËáÄƶ¼ÊÇÿ46gÄÆÔªËØÉú³É142gÁòËáÄÆ£¬ÏûºÄ98gÁòËá¡­¡­¡­¡­¡­(1·Ö)

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
26¡¢Ä³»¯Ñ§ÐËȤС×éÓÃÏÂͼËùʾװÖã¨Ê¡ÂÔÁ˼гÖ×°Öã©×öÈçÏÂʵÑ飺
Ñо¿¿ÎÌ⣺̽¾¿Ê³Æ·±£ÏÊ´üµÄÔªËØ×é³É£®
²éÔÄ×ÊÁÏ£º¼îʯ»ÒÊÇÓÉÇâÑõ»¯ÄƺÍÑõ»¯¸Æ×é³ÉµÄÎüÊÕ¼Á£»ÎÞË®ÁòËáÍ­¿É×÷ΪÎüË®¼Á£¬ÇÒÎüË®ºóÏÔÀ¶É«£®
Ìá³ö²ÂÏ룺ÖƱ£ÏÊ´üµÄ²ÄÁÏ¿ÉÄÜÊÇÓÉ¡°Ì¼¡¢Çâ¡¢Ñõ¡±ÖÐÁ½ÖÖ»òÈýÖÖÔªËØ×é³É£®
ʵÑé²½Ö裺
¢Ù°´Í¼Ê¾Á¬½ÓºÃÒÇÆ÷£¬½«Ö¹Ë®¼Ð¼ÐÔÚFºóµÄÏ𽺹ܴ¦£¬BÖÐ×¢ÈëÊÊÁ¿Ë®£¬µ±ÏòAÖлº»º×¢ÈëÉÙÁ¿Ë®Ê±£¬¹Û²ìµ½×°ÖÃBÖеÄÏÖÏóΪX
G´¦ÒºÖùÉÏÉý
£¬ËµÃ÷×°ÖõÄÆøÃÜÐÔÁ¼ºÃ£®
¢Ú°´Í¼Ê¾×°ºÃÒ©Æ·ºÍ1.4g¼ôËéµÄ±£ÏÊ´üÑùÆ·£¬³ýD¡¢E¡¢F¡¢Ö¹Ë®¼ÐÍ⽫ÒÇÆ÷°´Ô­Ñù×é×°£®
¢ÛÏòAÖлºÂýעˮһ¶Îʱ¼äºó£¬½«ÒѳÆÖصÄD¡¢EÁ½¸ÉÔï¹ÜºÍδ³ÆÖØF¸ÉÔï¹Ü½Óµ½CµÄβ¶Ë£®¢Ü¸ø×°ÖÃC¼ÓÈÈ£¬Ö±ÖÁ±£ÏÊ´üËéƬÍêȫȼÉÕ£®
¢ÝÍ£Ö¹¼ÓÈȲ¢¼ÌÐøÏòAÖÐעˮһ¶Îʱ¼ä£®
¢Þ·Ö±ð¶ÔD¡¢E½øÐеڶþ´Î³ÆÖØ£®
ʵÑé·ÖÎö£º
£¨1£©¸ù¾ÝʵÑéÔ­Àí·ÖÎö£¬¸ÃÌ××°ÖÃÖÐȱÉٵIJ£Á§ÒÇÆ÷ÊÇ
¾Æ¾«µÆ
£®
£¨2£©ÊµÑé²½Öè¢ÙÖУ¬¹Û²ìµ½µÄÏÖÏóΪ
G´¦ÒºÖùÉÏÉý
£®
£¨3£©ÉèÖÃF´¦µÄ¸ÉÔï¹ÜµÄÄ¿µÄÊÇ
·ÀÖ¹×°ÖÃÍâ¿ÕÆøÖеÄCO2ºÍH2O±»EÎüÊÕ£¬Ó°ÏìʵÑé
£®B´¦µÄŨÁòËáÆðµÄ×÷ÓÃÊÇ
¸ÉÔïÑõÆø»ò³ýȥˮÕôÆø
£¬Èç¹ûûÓÐBÔò¶ÔʵÑé½á¹ûµÄÓ°ÏìÊÇ
ʹÇâÔªËØÖÊÁ¿Æ«´ó»òʹˮµÄÖÊÁ¿Æ«´ó
£®
£¨4£©ÔÚ¼ÓÈÈ×°ÖÃCÇ°£¬ÒªÏòAÖлºÂýעˮһ¶Îʱ¼äÊÇΪÁË
½«×°ÖÃÖеĿÕÆøÅž¡
£®Í£Ö¹¼ÓÈȺó¼ÌÐøÏòAÖлºÂýעˮһ¶Îʱ¼äÊÇΪÁË
½«Éú³ÉµÄ¶þÑõ»¯Ì¼ºÍË®Çý¸ÏÖÁD¡¢E×°ÖÃÖб»ÍêÈ«ÎüÊÕ
£®
£¨5£©µÚÒ»´Î¶ÔD¡¢E³ÆÖØÖÊÁ¿·Ö±ðΪ82.0g¡¢85.0g£»µÚ¶þ´Î¶ÔD¡¢E³ÆÖØÖÊÁ¿·Ö±ðΪ83.8g¡¢89.4g£»ÔòÑùÆ·ÍêȫȼÉÕºó²úÉúË®µÄÖÊÁ¿Îª
1.8
g£®
ʵÑé½áÂÛ£º
£¨6£©¸ÃÑùÆ·µÄ×é³ÉÔªËØÒ»¶¨ÊÇ
C¡¢H
£®
£¨2012?°×ÏÂÇø¶þÄ££©ÒÑÖª²ÝËᣨH2C2O4£©¼ÓÈÈʱÔÚŨÁòËáµÄ×÷ÓÃÏ»ᷢÉú·Ö½â·´Ó¦£¬Ä³»¯Ñ§ÐËȤС×é¶ÔÆä·Ö½â²úÎï½øÐÐÁËÈçÏÂ̽¾¿£®
¡¾Ìá³ö²ÂÏë¡¿
£¨1£©·Ö½â²úÎïÊÇÒ»Ñõ»¯Ì¼ºÍË®ÕôÆøµÄ»ìºÏÎ
£¨2£©·Ö½â²úÎïÊǶþÑõ»¯Ì¼ºÍË®ÕôÆøµÄ»ìºÏÎ
£¨3£©·Ö½â²úÎïÊÇÒ»Ñõ»¯Ì¼¡¢¶þÑõ»¯Ì¼ºÍË®ÕôÆøµÄ»ìºÏÎ
ÄãÈÏΪºÏÀíµÄ²ÂÏëÊÇ
£¨3£©
£¨3£©
£¨Ñ¡ÌîÐòºÅ£©£¬ÒÀ¾ÝÊÇ
ÖÊÁ¿Êغ㶨ÂÉ
ÖÊÁ¿Êغ㶨ÂÉ
£®
¡¾²éÔÄ×ÊÁÏ¡¿
£¨1£©Å¨ÁòËá¾ßÓÐÍÑË®ÐÔ£¬¿ÉÒÔ½«ÓлúÎïÖеÄÇâ¡¢ÑõÔªËØ°´Ë®µÄ×é³ÉÍÑÈ¥£®
£¨2£©°×É«µÄÎÞË®ÁòËáÍ­ÓöË®»á±äΪÀ¶É«£®
£¨3£©¼îʯ»ÒÊǹÌÌåÇâÑõ»¯ÄƺÍÑõ»¯¸ÆµÄ»ìºÏÎ
¡¾½øÐÐʵÑ顿ΪÁ˼ìÑé²ÝËáµÄ·Ö½â²úÎïµÄ³É·Ý£¬¸ÃС×éͬѧ½«²ÝËáµÄ·Ö½â²úÎïͨ¹ýÏÂͼËùʾµÄ×°Ö㨲¿·Ö¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥£©£®
£¨1£©×°ÖÃAÖÐÎÞË®ÁòËáÍ­±äÀ¶£¬×°ÖÃBÖгÎÇåʯ»ÒË®±ä»ë×Ç£¬Ö¤Ã÷²ÝËá·Ö½â²úÉúÁË
H2OºÍCO2
H2OºÍCO2
£®
£¨2£©×°ÖÃDµÄ×÷ÓÃÊÇ
¼ìÑéCO2ÊÇ·ñÒѱ»ÍêÈ«³ýÈ¥
¼ìÑéCO2ÊÇ·ñÒѱ»ÍêÈ«³ýÈ¥
£®
£¨3£©Ö¤Ã÷²ÝËá·Ö½â²úÉúÁËÒ»Ñõ»¯Ì¼µÄÏÖÏóÓ¦°üÀ¨
DÖгÎÇåʯ»Òˮδ±ä»ë×Ç£¬EÖйÌÌåÓɺìÉ«±äΪºÚÉ«£¬FÖгÎÇåʯ»ÒË®±ä»ë×Ç
DÖгÎÇåʯ»Òˮδ±ä»ë×Ç£¬EÖйÌÌåÓɺìÉ«±äΪºÚÉ«£¬FÖгÎÇåʯ»ÒË®±ä»ë×Ç
£®
£¨4£©×°ÖÃBÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Ca£¨OH£©2+CO2¨TCaCO3¡ý+H2O
Ca£¨OH£©2+CO2¨TCaCO3¡ý+H2O
£»×°ÖÃEÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Fe2O3+3CO
 ¸ßΠ
.
 
2Fe+3CO2
Fe2O3+3CO
 ¸ßΠ
.
 
2Fe+3CO2
£®
¡¾µÃ³ö½áÂÛ¡¿²ÝËá·Ö½âµÄ»¯Ñ§·½³ÌʽΪ
H2C2O4
 Å¨ÁòËá 
.
¡÷
H2O+CO2¡ü+CO¡ü
H2C2O4
 Å¨ÁòËá 
.
¡÷
H2O+CO2¡ü+CO¡ü
£®
¡¾·´Ë¼ÓëÆÀ¼Û¡¿´Ó»·±£½Ç¶È¿¼ÂÇ£¬ÉÏÊöʵÑéµÄ²»×ãÖ®´¦ÊÇ
×°ÖÃFºóȱÉÙβÆø´¦Àí×°ÖÃ
×°ÖÃFºóȱÉÙβÆø´¦Àí×°ÖÃ
£®

ÒÑÖª²ÝËᣨH2C2O4£©¼ÓÈÈʱÔÚŨÁòËáµÄ×÷ÓÃÏ»ᷢÉú·Ö½â·´Ó¦£¬Ä³»¯Ñ§ÐËȤС×é¶ÔÆä·Ö½â²úÎï½øÐÐÁËÈçÏÂ̽¾¿¡£

¡¾Ìá³ö²ÂÏë¡¿£¨1£©·Ö½â²úÎïÊÇÒ»Ñõ»¯Ì¼ºÍË®ÕôÆøµÄ»ìºÏÎï¡£

£¨2£©·Ö½â²úÎïÊǶþÑõ»¯Ì¼ºÍË®ÕôÆøµÄ»ìºÏÎï¡£

£¨3£©·Ö½â²úÎïÊÇÒ»Ñõ»¯Ì¼¡¢¶þÑõ»¯Ì¼ºÍË®ÕôÆøµÄ»ìºÏÎï¡£

ÄãÈÏΪºÏÀíµÄ²ÂÏëÊÇ     £¨Ñ¡ÌîÐòºÅ£©£¬ÒÀ¾ÝÊÇ                             ¡£

¡¾²éÔÄ×ÊÁÏ¡¿£¨1£©Å¨ÁòËá¾ßÓÐÍÑË®ÐÔ£¬¿ÉÒÔ½«ÓлúÎïÖеÄÇâ¡¢ÑõÔªËØ°´Ë®µÄ×é³ÉÍÑÈ¥¡£

£¨2£©°×É«µÄÎÞË®ÁòËáÍ­ÓöË®»á±äΪÀ¶É«¡£

£¨3£©¼îʯ»ÒÊǹÌÌåÇâÑõ»¯ÄƺÍÑõ»¯¸ÆµÄ»ìºÏÎï¡£

¡¾½øÐÐʵÑ顿ΪÁ˼ìÑé²ÝËáµÄ·Ö½â²úÎïµÄ³É·Ý£¬¸ÃС×éͬѧ½«²ÝËáµÄ·Ö½â²úÎïͨ¹ýÏÂͼËùʾµÄ×°Ö㨲¿·Ö¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥£©¡£

£¨1£©×°ÖÃAÖÐÎÞË®ÁòËáÍ­±äÀ¶£¬×°ÖÃBÖгÎÇåʯ»ÒË®±ä»ë×Ç£¬Ö¤Ã÷²ÝËá·Ö½â²úÉúÁË      ¡££¨2£©×°ÖÃDµÄ×÷ÓÃÊÇ                                     ¡£

£¨3£©Ö¤Ã÷²ÝËá·Ö½â²úÉúÁËÒ»Ñõ»¯Ì¼µÄÏÖÏóÓ¦°üÀ¨                                      ¡£                  £»×°ÖÃEÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                        ¡£

¡¾µÃ³ö½áÂÛ¡¿²ÝËá·Ö½âµÄ»¯Ñ§·½³ÌʽΪ                                    ¡£

¡¾·´Ë¼ÓëÆÀ¼Û¡¿´Ó»·±£½Ç¶È¿¼ÂÇ£¬ÉÏÊöʵÑéµÄ²»×ãÖ®´¦ÊÇ                               ¡£

ij»¯Ñ§ÐËȤС×éÓÃÏÂͼËùʾװÖã¨Ê¡ÂÔÁ˼гÖ×°Öã©×öÈçÏÂʵÑ飺

Ñо¿¿ÎÌ⣺̽¾¿Ê³Æ·±£ÏÊ´üµÄÔªËØ×é³É£®
²éÔÄ×ÊÁÏ£º¼îʯ»ÒÊÇÓÉÇâÑõ»¯ÄƺÍÑõ»¯¸Æ×é³ÉµÄÎüÊÕ¼Á£»ÎÞË®ÁòËáÍ­¿É×÷ΪÎüË®¼Á£¬ÇÒÎüË®ºóÏÔÀ¶É«£®
Ìá³ö²ÂÏ룺ÖƱ£ÏÊ´üµÄ²ÄÁÏ¿ÉÄÜÊÇÓÉ¡°Ì¼¡¢Çâ¡¢Ñõ¡±ÖÐÁ½ÖÖ»òÈýÖÖÔªËØ×é³É£®
ʵÑé²½Ö裺
¢Ù°´Í¼Ê¾Á¬½ÓºÃÒÇÆ÷£¬½«Ö¹Ë®¼Ð¼ÐÔÚFºóµÄÏ𽺹ܴ¦£¬BÖÐ×¢ÈëÊÊÁ¿Ë®£¬µ±ÏòAÖлº»º×¢ÈëÉÙÁ¿Ë®Ê±£¬¹Û²ìµ½×°ÖÃBÖеÄÏÖÏóΪX______£¬ËµÃ÷×°ÖõÄÆøÃÜÐÔÁ¼ºÃ£®
¢Ú°´Í¼Ê¾×°ºÃÒ©Æ·ºÍ1.4g¼ôËéµÄ±£ÏÊ´üÑùÆ·£¬³ýD¡¢E¡¢F¡¢Ö¹Ë®¼ÐÍ⽫ÒÇÆ÷°´Ô­Ñù×é×°£®
¢ÛÏòAÖлºÂýעˮһ¶Îʱ¼äºó£¬½«ÒѳÆÖصÄD¡¢EÁ½¸ÉÔï¹ÜºÍδ³ÆÖØF¸ÉÔï¹Ü½Óµ½CµÄβ¶Ë£®¢Ü¸ø×°ÖÃC¼ÓÈÈ£¬Ö±ÖÁ±£ÏÊ´üËéƬÍêȫȼÉÕ£®
¢ÝÍ£Ö¹¼ÓÈȲ¢¼ÌÐøÏòAÖÐעˮһ¶Îʱ¼ä£®
¢Þ·Ö±ð¶ÔD¡¢E½øÐеڶþ´Î³ÆÖØ£®
ʵÑé·ÖÎö£º
£¨1£©¸ù¾ÝʵÑéÔ­Àí·ÖÎö£¬¸ÃÌ××°ÖÃÖÐȱÉٵIJ£Á§ÒÇÆ÷ÊÇ______£®
£¨2£©ÊµÑé²½Öè¢ÙÖУ¬¹Û²ìµ½µÄÏÖÏóΪ______£®
£¨3£©ÉèÖÃF´¦µÄ¸ÉÔï¹ÜµÄÄ¿µÄÊÇ______£®B´¦µÄŨÁòËáÆðµÄ×÷ÓÃÊÇ______£¬Èç¹ûûÓÐBÔò¶ÔʵÑé½á¹ûµÄÓ°ÏìÊÇ______£®
£¨4£©ÔÚ¼ÓÈÈ×°ÖÃCÇ°£¬ÒªÏòAÖлºÂýעˮһ¶Îʱ¼äÊÇΪÁË______£®Í£Ö¹¼ÓÈȺó¼ÌÐøÏòAÖлºÂýעˮһ¶Îʱ¼äÊÇΪÁË______£®
£¨5£©µÚÒ»´Î¶ÔD¡¢E³ÆÖØÖÊÁ¿·Ö±ðΪ82.0g¡¢85.0g£»µÚ¶þ´Î¶ÔD¡¢E³ÆÖØÖÊÁ¿·Ö±ðΪ83.8g¡¢89.4g£»ÔòÑùÆ·ÍêȫȼÉÕºó²úÉúË®µÄÖÊÁ¿Îª______g£®
ʵÑé½áÂÛ£º
£¨6£©¸ÃÑùÆ·µÄ×é³ÉÔªËØÒ»¶¨ÊÇ______£®
ÒÑÖª²ÝËᣨH2C2O4£©¼ÓÈÈʱÔÚŨÁòËáµÄ×÷ÓÃÏ»ᷢÉú·Ö½â·´Ó¦£¬Ä³»¯Ñ§ÐËȤС×é¶ÔÆä·Ö½â²úÎï½øÐÐÁËÈçÏÂ̽¾¿£®
¡¾Ìá³ö²ÂÏë¡¿
£¨1£©·Ö½â²úÎïÊÇÒ»Ñõ»¯Ì¼ºÍË®ÕôÆøµÄ»ìºÏÎ
£¨2£©·Ö½â²úÎïÊǶþÑõ»¯Ì¼ºÍË®ÕôÆøµÄ»ìºÏÎ
£¨3£©·Ö½â²úÎïÊÇÒ»Ñõ»¯Ì¼¡¢¶þÑõ»¯Ì¼ºÍË®ÕôÆøµÄ»ìºÏÎ
ÄãÈÏΪºÏÀíµÄ²ÂÏëÊÇ______£¨Ñ¡ÌîÐòºÅ£©£¬ÒÀ¾ÝÊÇ______£®
¡¾²éÔÄ×ÊÁÏ¡¿
£¨1£©Å¨ÁòËá¾ßÓÐÍÑË®ÐÔ£¬¿ÉÒÔ½«ÓлúÎïÖеÄÇâ¡¢ÑõÔªËØ°´Ë®µÄ×é³ÉÍÑÈ¥£®
£¨2£©°×É«µÄÎÞË®ÁòËáÍ­ÓöË®»á±äΪÀ¶É«£®
£¨3£©¼îʯ»ÒÊǹÌÌåÇâÑõ»¯ÄƺÍÑõ»¯¸ÆµÄ»ìºÏÎ
¡¾½øÐÐʵÑ顿ΪÁ˼ìÑé²ÝËáµÄ·Ö½â²úÎïµÄ³É·Ý£¬¸ÃС×éͬѧ½«²ÝËáµÄ·Ö½â²úÎïͨ¹ýÏÂͼËùʾµÄ×°Ö㨲¿·Ö¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥£©£®
£¨1£©×°ÖÃAÖÐÎÞË®ÁòËáÍ­±äÀ¶£¬×°ÖÃBÖгÎÇåʯ»ÒË®±ä»ë×Ç£¬Ö¤Ã÷²ÝËá·Ö½â²úÉúÁË______£®
£¨2£©×°ÖÃDµÄ×÷ÓÃÊÇ______£®
£¨3£©Ö¤Ã÷²ÝËá·Ö½â²úÉúÁËÒ»Ñõ»¯Ì¼µÄÏÖÏóÓ¦°üÀ¨______£®
£¨4£©×°ÖÃBÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______£»×°ÖÃEÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______ 2Fe+3CO2

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

¾«Ó¢¼Ò½ÌÍø