11£®BͬѧÃÇÔÚʵÑéÊÒÓ÷ÏÌúм£¨º¬ÉÙÁ¿Í­£©ÖÆÈ¡ÐÂÐ;»Ë®¼ÁNa2FeO4µÄÁ÷³ÌÈçͼ£®

Çë»Ø´ðÏÂÁÐÎÊÌ⣮
£¨1£©Ïò¹ÌÌåAÖмÓÈëÏ¡ÑÎËáºó·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇFe+2H+=Fe2++H2¡ü£®·´Ó¦ºóµÃµ½µÄÈÜÒºÈôÐè½Ï³¤Ê±¼ä±£´æ£¬Ó¦²ÉÈ¡µÄ´ëÊ©ÊÇÏòÈÜÒºÖмÓÈëÉÙÁ¿Ìú·Û£®
£¨2£©ÈÜÒºBÖеÄÑôÀë×ÓÊÇFe2+¡¢H+£¬ÏòÈÜÒºBÖмÓÈëNaOHÈÜÒº£¬Ö®ºóͨÈë¿ÕÆø£¬½Á°è£¬¹Û²ìµ½µÄÏÖÏóÊÇÏȲúÉú°×É«³Áµí£¬°×É«³ÁµíѸËÙ±äΪ»ÒÂÌÉ«£¬×îºó±äΪºìºÖÉ«£¬Í¨Èë¿ÕÆøºó·¢Éú»¯Ñ§·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ4Fe£¨OH£©2+O2+2H2O¨T4Fe£¨OH£©3£®
£¨3£©Ïò¹ÌÌåCÖмÓÈëNaOHÈÜÒº²¢Í¨ÈëCl2¿ÉÖÆÈ¡Na2FeO4£¬Óйط´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ2Fe£¨OH£©3+10NaOH+3Cl2=2Na2FeO4+6NaCl+8H2O£®
£¨4£©Na2FeO4ÖÐFeµÄ»¯ºÏ¼ÛÊÇ+6£¬´ÓÑõ»¯»¹Ô­½Ç¶È·ÖÎö£¬Na2FeO4¾ßÓÐÑõ»¯ÐÔ£¬¹ÊÓÃNa2FeO4¾»Ë®Ê±£¬³ýÁ˿ɳýȥˮÖÐÐü¸¡ÓàÖÊ£¬»¹¿ÉÒÔɱ¾úÏû¶¾£®

·ÖÎö Ó÷ÏÌúм£¨º¬ÉÙÁ¿Í­£©ÖÆÈ¡ÐÂÐ;»Ë®¼ÁNa2FeO4µÄÁ÷³Ì£ºÇåÏ´·ÏÌúм±í±íÃæµÄÓÍÎÛ£¬¼ÓÈë¹ýÁ¿µÄÏ¡ÑÎËᣬÌúÈÜÓÚÏ¡ÑÎËᣬͭ²»ÈÜ£¬¹ýÂ˵õ½ÂÈ»¯ÑÇÌúºÍHClµÄ»ìºÏÈÜÒºB£¬ÏòÆäÖмÓÈëÇâÑõ»¯ÄÆÈÜÒº£¬ÖкÍÑÎËᣬ²¢Éú³ÉÇâÑõ»¯ÑÇÌú£¬Í¨Èë¿ÕÆø£¬±»Ñõ»¯ÎªÇâÑõ»¯Ìú£¬¹ýÂ˵õ½µÄ¹ÌÌåCΪÇâÑõ»¯Ìú£¬¼ÓÈëÇâÑõ»¯ÄƺÍÂÈÆøµÃµ½²úÆ·Na2FeO4£¬¾Ý´Ë·ÖÎö£®

½â´ð ½â£º£¨1£©¹ÌÌåAÖмÓÈëÏ¡ÑÎËáºó·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ£ºFe+2H+=Fe2++H2¡ü£»ÑÇÌúÀë×ÓÒ×±»Ñõ»¯£¬ÈôÐ賤ʱ¼ä±£´æ£¬¿É¼ÓÈ뻹ԭ¼ÁÌú·Û£»
¹Ê´ð°¸Îª£ºFe+2H+=Fe2++H2¡ü£»ÏòÈÜÒºÖмÓÈëÉÙÁ¿Ìú·Û£»
£¨2£©ÈÜÒºBΪÂÈ»¯ÑÇÌúºÍ¶àÓàµÄÑÎËᣬÑôÀë×ÓΪ£ºFe2+¡¢H+£»ÏòBÈÜÒºÖмÓÈëÇâÑõ»¯ÄÆ£¬ÏÈÉú³É°×É«³ÁµíÇâÑõ»¯ÑÇÌú£¬ÇâÑõ»¯ÑÇÌú±»Ñõ»¯£¬·¢Éú·´Ó¦4Fe£¨OH£©2+O2+2H2O¨T4Fe£¨OH£©3£¬Éú³ÉºìºÖÉ«µÄÇâÑõ»¯Ìú³Áµí£¬¹ÊÏÖÏóΪ£ºÏȲúÉú°×É«³Áµí£¬°×É«³ÁµíѸËÙ±äΪ»ÒÂÌÉ«£¬×îºó±äΪºìºÖÉ«£»4Fe£¨OH£©2+O2+2H2O¨T4Fe£¨OH£©3£»
¹Ê´ð°¸Îª£ºFe2+¡¢H+£»ÏȲúÉú°×É«³Áµí£¬°×É«³ÁµíѸËÙ±äΪ»ÒÂÌÉ«£¬×îºó±äΪºìºÖÉ«£»4Fe£¨OH£©2+O2+2H2O¨T4Fe£¨OH£©3£»
£¨3£©ÏòÇâÑõ»¯ÌúÖмÓÈëNaOHÈÜÒº²¢Í¨ÈëCl2¿ÉÖÆÈ¡Na2FeO4£¬·´Ó¦Îª£º2 Fe£¨OH£©3+10NaOH+3Cl2=2Na2FeO4+6NaCl+8H2O£»
¹Ê´ð°¸Îª£º2 Fe£¨OH£©3+10NaOH+3Cl2=2Na2FeO4+6NaCl+8H2O£»
£¨4£©Na2FeO4ÖÐÄÆÔªËØ+1¼Û£¬ÑõÔªËØ-2¼Û£¬»¯ºÏ¼Û´úÊýºÍΪ0£¬ÔòFeµÄ»¯ºÏ¼ÛÊÇ+6¼Û£¬ÌúÔªËØ+6¼Û£¬Ò׵õç×Ó»¯ºÏ¼Û½µµÍ£¬¾ßÓÐÑõ»¯ÐÔ£¬¿Éɱ¾úÏû¶¾£»
¹Ê´ð°¸Îª£º+6£»Ñõ»¯£»É±¾úÏû¶¾£®

µãÆÀ ±¾Ì⿼²éÁËÎïÖʵÄÖƱ¸£¬Î§ÈÆÌúÕ¹¿ª£¬Éæ¼°Ñõ»¯»¹Ô­·´Ó¦·½³ÌʽµÄÊéд£¬Á÷³ÌµÄ·ÖÎöµÈ£¬ÕÆÎÕÎïÖÊÐÔÖÊÊǹؼü£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

1£®ÏÂÁÐʵÑé×°ÖÃͼËùʾµÄʵÑé²Ù×÷£¬²»ÄÜ´ïµ½ÏàÓ¦µÄʵÑéÄ¿µÄÊÇ£¨¡¡¡¡£©
A£®
ÎÅÆøÌåµÄÆøζ
B£®
ÓÃŨÁòËá¸ÉÔïCO2
C£®
ÏòÈÝÁ¿Æ¿ÖÐתÒÆÒºÌå
D£®
Óú£Ë®ÌáÈ¡µ­Ë®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

2£®ÏÂÁйý³ÌÊôÓÚ»¯Ñ§±ä»¯µÄÊÇ£¨¡¡¡¡£©
¢Ù»îÐÔÌ¿Îü¸½ÓÐÉ«ÎïÖÊ                                                   ¢ÚÂÈˮƯ°×ÓÐÉ«²¼Ìõ
¢Û¹ýÑõ»¯ÄƶÖÃÔÚ¿ÕÆøÖР                                              ¢Ü½«ÂÈÆøͨÈëË®ÖУ¬ÈÜÒº³Êdz»ÆÂÌÉ«
¢Ý¹ýÁ¿¹ýÑõ»¯ÄƼÓÈ뺬·Ó̪µÄÈÜÒº£¬ÈÜÒºÏȱäºìºóÍÊÉ«  ¢ÞÀûÓÃÑæÉ«·´Ó¦¼ø±ðNaClºÍKCl£®
A£®¢Ù¢Ú¢Û¢ÝB£®¢Ù¢Ú¢Ü¢ÞC£®¢Ú¢Û¢Ü¢ÝD£®¢Û¢Ü¢Ý¢Þ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®¼ºÍé´Æ·ÓµÄÒ»ÖֺϳÉ·ÏßÈçͼ£º

ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÔÚNaOH Ë®ÈÜÒºÖмÓÈÈ£¬»¯ºÏÎïX ¿É·¢ÉúÏûÈ¥·´Ó¦
B£®ÔÚÒ»¶¨Ìõ¼þÏ£¬»¯ºÏÎïY¿ÉÓëŨäåË®·¢ÉúÈ¡´ú·´Ó¦
C£®ÓÃFeCl3ÈÜÒº²»Äܼø±ð»¯ºÏÎïXºÍY
D£®»¯ºÏÎïYÖв»º¬ÓÐÊÖÐÔ̼ԭ×Ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

6£®ÔÚ50mL 0.1mol/L Na2S ÈÜÒºÖÐÖðµÎ¼ÓÈë50mL 0.1mol/L KHSO4ÈÜÒº£¬ËùµÃÈÜÒºÖÐÁ£×ÓŨ¶È¹ØϵÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®c£¨SO42-£©=c£¨HS-£©=c£¨K+£©£¾c£¨OH-£©=c£¨H+£©
B£®c£¨Na+£©£¾c£¨K+£©£¾c£¨S2-£©£¾c£¨H+£©£¾c£¨OH-£©
C£®c£¨Na+£©=c£¨S2-£©+c£¨HS-£©+c£¨H2S£©+c£¨SO42-£©
D£®c£¨K+£©+c£¨Na+£©+c£¨H+£©=c£¨SO42-£©+c£¨S2-£©+c£¨HS-£©+c£¨OH-£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

16£®Ô­×ÓÐòÊýСÓÚ36µÄËÄÖÖÔªËØ£¬·Ö±ðλÓÚ²»Í¬ÖÜÆÚ£¬ÆäÖÐAÔ­×ÓºËÊÇÒ»¸öÖÊ×Ó£¬BÔ­×ÓºËÍâµç×ÓÓÐ6ÖÖ²»Í¬µÄÔ˶¯×´Ì¬£¬BÓëC¿ÉÐγÉÕýËÄÃæÌåÐÍ·Ö×Ó£¬DÔ­×ÓÍâΧµç×ÓÅŲ¼Îª3d104s1£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®A¡¢BÔªËØÐγɵÄһϵÁл¯ºÏÎïÖУ¬ÆäÖÐAÔªËØÖÊÁ¿·ÖÊýµÄ×î´óֵΪ25%
B£®ËÄÖÖÔªËØÖе縺ÐÔ×î´óµÄÊÇB
C£®CËùÐγɵÄÆø̬Ç⻯ÎÔÚÆäͬÖ÷×åÔªËصÄÆø̬Ç⻯ÎïÖзеã×îµÍ
D£®ËÄÖÖÔªËØÖеÚÒ»µçÀëÄÜ×îСµÄÊÇD

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

2£®Ä³¹¤Òµ·ÏÒºÖк¬ÓÐCu2+¡¢Mg2+¡¢Zn2+µÈÀë×Ó£¬Îª½«Æä»ØÊÕÀûÓã¬ÔÙ²ÉÓÃÁËÈçͼϹ¤ÒÕ

ÒÑÖªZn£¨OH£©2µÄÐÔÖÊÓëAl£¨OH£©3ÏàËÆËùÓÃÊÔ¼ÁÔÚÏÂÁÐÊÔ¼ÁÖÐÑ¡Ôñ
¢ÙÌú·Û ¢Úп·Û ¢ÛÏ¡HNO3 ¢ÜÏ¡H2SO4 ¢ÝÏ¡HCl ¢ÞÏ¡°±Ë® ¢ßNaOHÈÜÒº ¢àʯ»ÒË®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÊÔ¼Á1¡¢ÊÔ¼Á2¡¢ÊÔ¼Á3·Ö±ð¿ÉÒÔÊÇ¢Ú¡¢¢Ü¢Ý¡¢¢ß£¨Ìî±àºÅ£©
£¨2£©²Ù×÷2ÊÇÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢¹ýÂË£»
£¨3£©²Ù×÷3ÊÇÔÚHClÆøÁ÷ÖмÓÈÈ£»
£¨4£©¼Ó¹ýÁ¿CO2ʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽZnO22-+2CO2+2H2O=Zn£¨OH£©2¡ý+2HCO3-£»
£¨5£©ÓÃÀë×Ó·½³Ìʽ½âÊÍÊÔ¼Á7ÄÜ·ÖÀë³öMg£¨OH£©2µÄÔ­Àí£ºMg2++2OH-=Mg£¨OH£©2¡ý¡¢Zn2++4OH-=ZnO22-+2H2O£»
£¨6£©ÔÚ½ðÊôÒ±Á¶·½·¨Öз½·¨1Êǵç½â·¨¡¢·½·¨2ÊÇ»¹Ô­¼Á·¨£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®ÏÂÁбíÊöÖУ¬ºÏÀíµÄÊÇ£¨¡¡¡¡£©
A£®×¼È·Á¿È¡20.00mL¸ßÃÌËá¼ØÈÜÒº£¬¿ÉÑ¡ÓÃ25mL¼îʽµÎ¶¨¹Ü
B£®½«Ë®¼ÓÈÈ£¬KwÔö´ó£¬pH²»±ä
C£®ÓöèÐԵ缫µç½â1LŨ¶È¾ùΪ2mol/LµÄAgNO3ÓëCu£¨NO3£©2µÄ»ìºÏÈÜÒº£¬µ±ÓÐ0.2 mol µç×ÓתÒÆʱ£¬Òõ¼«Îö³ö6.4g½ðÊô
D£®NaAlO2µÄË®ÈÜÒº¾­¼ÓÈÈŨËõ¡¢Õô¸É×ÆÉÕºóÄܵõ½NaAlO2¹ÌÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

20£®ÏÂÁÐÎïÖÊÖмÈÄܸúÏ¡H2SO4·´Ó¦£¬ÓÖÄܸúÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÊÇ£¨¡¡¡¡£©
¢ÙNaHCO3  ¢ÚAl2O3  ¢ÛAl£¨OH£©3 ¢ÜAl  ¢ÝAlCl3  ¢ÞNaAlO2£®
A£®¢Û¢Ü¢ÝB£®¢Ú¢Û¢ÜC£®¢Ù¢Û¢Ü¢ÞD£®¢Ù¢Ú¢Û¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸