Ñõ»¯»¹Ô­·´Ó¦ÔÚ»¯Ñ§¹¤ÒµÖÐÓ¦Óù㷺£®
¢ñ¡¢ÓÐÈçͼ1ËùʾµÄʵÑé×°Öã®
£¨1£©ÈôÓÃʯīµç¼«µç½âNaHCO3ÈÜÒº£¬Yµç¼«ÉϲúÉúµÄÆøÌåÊÇ
O2
O2
£®
£¨2£©ÈçÒªÔÚÌúÉ϶ÆÍ­£¬ÔòXµç¼«µÄ²ÄÁÏÊÇ
Ìú
Ìú
£¬Yµç¼«·´Ó¦Ê½ÊÇ
Cu-2e-=Cu2+
Cu-2e-=Cu2+
£®
¢ò¡¢Ä³µç¶ÆÍ­³§ÓÐÁ½ÖÖ·ÏË®ÐèÒª´¦Àí£¬Ò»ÖÖ·ÏË®Öк¬ÓÐCN-Àë×Ó£¬ÁíÒ»ÖÖ·ÏË®Öк¬Cr2O72-Àë×Ó£®¸Ã³§ÄⶨÈçͼ2ËùʾµÄ·ÏË®´¦ÀíÁ÷³Ì£®
»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©¢ÚÖÐʹÓõÄNaClOÈÜÒº³Ê¼îÐÔ£¬Ô­ÒòÊÇ
ClO-+H2O?HClO+OH-
ClO-+H2O?HClO+OH-
£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£®
£¨2£©¢ÛÖз´Ó¦Ê±£¬Ã¿0.4mol Cr2O72-תÒÆ2.4mole-£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ
3S2O32-+4Cr2O72-+26H+¨T6SO42-+8Cr3++13H2O
3S2O32-+4Cr2O72-+26H+¨T6SO42-+8Cr3++13H2O
£®
£¨3£©È¡ÉÙÁ¿´ý¼ìË®ÑùÓÚÊÔ¹ÜÖУ¬¼ÓÈë¹ýÁ¿NaOHÈÜÒº£¬ÓÐÀ¶É«³ÁµíÉú³É£®ÏòÉÏÊöÓÐÀ¶É«³ÁµíµÄÐü×ÇÒºÖмÓÈëNa2SÈÜÒº£¬ÓкÚÉ«³ÁµíÉú³É£¬ÇÒÀ¶É«³ÁµíÖð½¥¼õÉÙ£®ÇëÄãʹÓû¯Ñ§ÓÃÓ½áºÏ±ØÒªµÄÎÄ×Ö˵Ã÷·ÖÎöÆä¹ý³Ì
´ý¼ìË®ÑùÖл¹ÓÐCu2+£¬¼Ó¼î·¢ÉúCu2++2OH-¨TCu£¨OH£©2¡ý£¬ÔÙ¼ÓÈëNa2SÈÜÒº£¬CuS±ÈCu£¨OH£©2¸üÄÑÈÜ£¬Ôò·¢ÉúCu£¨OH£©2£¨s£©+S2-£¨aq£©¨TCuS£¨s£©+2OH-£¨aq£©£®
´ý¼ìË®ÑùÖл¹ÓÐCu2+£¬¼Ó¼î·¢ÉúCu2++2OH-¨TCu£¨OH£©2¡ý£¬ÔÙ¼ÓÈëNa2SÈÜÒº£¬CuS±ÈCu£¨OH£©2¸üÄÑÈÜ£¬Ôò·¢ÉúCu£¨OH£©2£¨s£©+S2-£¨aq£©¨TCuS£¨s£©+2OH-£¨aq£©£®
£®
¢ó¡¢¹¤Òµ¿ÉÓõç½â·¨À´´¦Àíº¬Cr2O72-·ÏË®£®
£¨1£©Fe2+ÓëËáÐÔÈÜÒºÖеÄCr2O72-·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
Cr2O72-+6Fe2++14H+¨T2Cr3++6Fe3++7H2O
Cr2O72-+6Fe2++14H+¨T2Cr3++6Fe3++7H2O
£®
£¨2£©ÊµÑéÊÒÀûÓÃÈçͼ3ËùʾģÄâ´¦Àíº¬Cr2O72-µÄ·ÏË®£¨·ÏË®ÖмÓÈëÊÊÁ¿NaClÒÔÔö¼ÓÈÜÒºµÄµ¼µçÐÔ£©£¬ÔÚÒõ¼«ÇøµÃµ½½ðÊôÇâÑõ»¯Îï³Áµí£®ÓÃƽºâÒƶ¯Ô­Àí½âÊͳÁµí³öÏÖÔÚÒõ¼«¸½½üµÄÔ­Òò
Òõ¼«·´Ó¦ÏûºÄÁËË®ÖеÄH+£¬´òÆÆÁËË®µÄµçÀëƽºâ£¬´Ù½øÁËË®µÄµçÀ룬ʹÈÜÒºÖÐOH-Ũ¶ÈÔö´ó£¬ÈÜÒºµÄ¼îÐÔÔöÇ¿
Òõ¼«·´Ó¦ÏûºÄÁËË®ÖеÄH+£¬´òÆÆÁËË®µÄµçÀëƽºâ£¬´Ù½øÁËË®µÄµçÀ룬ʹÈÜÒºÖÐOH-Ũ¶ÈÔö´ó£¬ÈÜÒºµÄ¼îÐÔÔöÇ¿
£®
£¨3£©Óõç½â·¨´¦Àí¸ÃÈÜÒºÖР0.01mol Cr2O72-ʱ£¬µÃµ½³ÁµíµÄÖÊÁ¿ÊÇ
8.48
8.48
 g£®
·ÖÎö£º¢ñ¡¢£¨1£©µç½â̼ËáÇâÄÆÈÜÒº£¬Ñô¼«ÉÏÇâÑõ¸ùÀë×ӷŵ磬Òõ¼«ÉÏÇâÀë×ӷŵ磬ʵ¼ÊÉÏÏ൱ÓÚµç½âË®£»
£¨2£©¸ù¾Ýµç¶ÆµÄ¹¤×÷Ô­Àí֪ʶÀ´»Ø´ð£»
¢ò¡¢£¨1£©ÀûÓÃÑεÄË®½âÀ´·ÖÎöÑÎÈÜÒºÏÔ¼îÐÔµÄÔ­Òò£»
£¨2£©ÀûÓÃÿ0.4molCr2O72-תÒÆ2.4molµÄµç×ÓÀ´¼ÆËã±»»¹Ô­ºóCrÔªËصĻ¯ºÏ¼Û£¬ÔÙÊéдÀë×Ó·½³Ìʽ£»
£¨3£©¸ù¾ÝÍ­Àë×ÓÓëÇâÑõ¸ùÀë×Ó·´Ó¦Éú³É³Áµí¼°³ÁµíµÄת»¯À´·ÖÎö£»
¢ó¡¢£¨1£©Fe2+Àë×ÓÓëCr2O72-Àë×Ó·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³ÉFe3+Àë×ÓºÍCr3+Àë×Ó£»
£¨2£©Ëæ×ŵç½â½øÐУ¬ÈÜÒºÖÐc£¨H+£© Öð½¥¼õÉÙ£¬c£¨OH-£©Å¨¶ÈÔö´ó£»
£¨3£©¸ù¾ÝCr2O72-+6Fe2++14H+¨T2Cr3++6Fe3++7H2O£¬Cr3++3OH-¨TCr£¨OH£©3¡ý¡¢Fe3++3OH-¨TFe£¨OH£©3¡ý½øÐмÆË㣻
½â´ð£º½â£º¢ñ¡¢£¨1£©Yµç¼«ÎªÑô¼«£¬Ñô¼«ÉÏÇâÑõ¸ùÀë×ӷŵ磺4OH--4e-=O2¡ü+2H2O£¬
¹Ê´ð°¸Îª£ºO2£»
£¨2£©µç¶Æʱ£¬¶Æ²ã½ðÊô×÷Ñô¼«£¬´ý¶Æ½ðÊô×÷Òõ¼«£¬ËùÒÔXµç¼«µÄ²ÄÁÏÊÇÌú£¬Yµç¼«µÄ²ÄÁÏÊÇÍ­£¬µç¼«·´Ó¦Ê½Îª£ºCu-2e-=Cu2+£»
¹Ê´ð°¸Îª£ºÌú£»Cu-2e-=Cu2+£»
¢ò¡¢£¨1£©NaClOÈÜÒº³Ê¼îÐÔ£¬ÊÇÒò´ÎÂÈËá¸ùÀë×ÓË®½âÉú³ÉÇâÑõ¸ùÀë×Óµ¼Öµģ¬ÔòÀë×Ó·´Ó¦Îª£ºClO-+H2O?HClO+OH-£¬
¹Ê´ð°¸Îª£ºClO-+H2O?HClO+OH-£»  
£¨2£©Ã¿0.4molCr2O72-תÒÆ2.4molµÄµç×Ó£¬É軹ԭºóCrÔªËصĻ¯ºÏ¼ÛΪx£¬Ôò0.4mol¡Á2¡Á£¨6-x£©=2.4mol£¬½âµÃx=+3£¬ÔòÀë×Ó·´Ó¦Îª3S2O32-+4Cr2O72-+26H+¨T6SO42-+8Cr3++13H2O£»
¹Ê´ð°¸Îª£º3S2O32-+4Cr2O72-+26H+¨T6SO42-+8Cr3++13H2O£»
£¨3£©ÒòÍ­Àë×ÓÓëÇâÑõ¸ùÀë×Ó·´Ó¦Éú³ÉÇâÑõ»¯Í­³Áµí£¬CuS±ÈCu£¨OH£©2¸üÄÑÈÜ£¬Ôò¼ÓÈëNa2SÈÜÒºÄÜ·¢Éú³ÁµíµÄת»¯£¬
¹Ê´ð°¸Îª£º´ý¼ìË®ÑùÖл¹ÓÐCu2+£¬¼Ó¼î·¢ÉúCu2++2OH-¨TCu£¨OH£©2¡ý£¬ÔÙ¼ÓÈëNa2SÈÜÒº£¬CuS±ÈCu£¨OH£©2¸üÄÑÈÜ£¬Ôò·¢ÉúCu£¨OH£©2£¨s£©+S2-£¨aq£©¨TCuS£¨s£©+2OH-£¨aq£©£®
¢ó¡¢£¨1£©ÑÇÌúÀë×ÓÓëCr2O72-·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬±»»¹Ô­ÎªCr3+È»ºóÉú³ÉCr£¨OH£©3³Áµí£¬ÖظõËá¸ù¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Äܽ«Éú³ÉµÄÑÇÌúÀë×ÓÑõ»¯ÎªÈý¼Û£¬¼´6Fe2++Cr2O72-+14H+=6Fe3++2Cr3++7H2O£»
¹Ê´ð°¸Îª£ºCr2O72-+6Fe2++14H+¨T2Cr3++6Fe3++7H2O£»
£¨2£©Ëæ×ŵç½â½øÐУ¬ÈÜÒºÖÐc£¨H+£© Öð½¥¼õÉÙ£¬´òÆÆÁËË®µÄµçÀëƽºâ£¬´Ù½øÁËË®µÄµçÀ룬ʹÈÜÒºÖÐOH-Ũ¶ÈÔö´ó£¬ÈÜÒºµÄ¼îÐÔÔöÇ¿£¬Éú³ÉFe£¨OH£©3ºÍCr£¨OH£©3³Áµí£¬½ðÊôÑôÀë×ÓÔÚÒõ¼«Çø¿É³ÁµíÍêÈ«£»
¹Ê´ð°¸Îª£ºÒõ¼«·´Ó¦ÏûºÄÁËË®ÖеÄH+£¬´òÆÆÁËË®µÄµçÀëƽºâ£¬´Ù½øÁËË®µÄµçÀ룬ʹÈÜÒºÖÐOH-Ũ¶ÈÔö´ó£¬ÈÜÒºµÄ¼îÐÔÔöÇ¿£»
£¨3£©¸ù¾ÝCr2O72-+6Fe2++14H+¨T2Cr3++6Fe3++7H2O£¬Cr3++3OH-¨TCr£¨OH£©3¡ý¡¢Fe3++3OH-¨TFe£¨OH£©3¡ýÖª0.01mol Cr2O72-£¬¿ÉÉú³É0.02molCr£¨OH£©3£¬0.06molFe£¨OH£©3£¬ÖÁÉٵõ½³ÁµíµÄÖÊÁ¿ÊÇ0.02mol¡Á103g/mol+0.06mol¡Á107g/mol=8.48g£¬
¹Ê´ð°¸Îª£º8.48£»
µãÆÀ£º±¾Ì⿼²éÁ˵ç½âÔ­Àí¡¢Ñõ»¯»¹Ô­·´Ó¦£¬×¢ÖØÁË»¯Ñ§Óëʵ¼ÊÉú²úµÄÁªÏµ£¬Ñ§ÉúӦѧ»áÀûÓÃÎïÖʵÄÐÔÖÊ¡¢ÔªËصĻ¯ºÏ¼Û¡¢µç×ÓÊغãµÈÀ´½â´ð£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

[»¯Ñ§¡ªÑ¡ÐÞ»¯Ñ§Óë¼¼Êõ]

H2O2µÄË®ÈÜÒºÊÇÒ»ÖÖ³£ÓõÄɱ¾ú¼Á¡£

£¨1£©H2O2¾ÉµÄ¹¤ÒµÖÆ·¨ÊÇÒÔPtΪÑô¼«¡¢Ê¯Ä«ÎªÒõ¼«µç½âÁòËáÇâï§ÈÜÒº£¬ÔÙ½«µç½â²úÎïË®½â¡£»¯Ñ§·½³ÌʽΪ£º

¡£Ð´³öµç½âÁòËáÇâï§ÈÜÒºµÄÒõ¼«ºÍÑô¼«·´Ó¦¹¤¡£

Òõ¼«£º                                                            £»

Ñô¼«£º                                                            £»

£¨2£©H2O2ÐµĹ¤ÒµÖÆ·¨ÊÇÏȽ«ÒÒ»ùÝìõ«¼ÓÇ⻹ԭ£¬ÔÙ½«Öмä²úÎïÈ¥ÇâÑõ»¯¼´µÃH2O2£¬·´Ó¦·½³ÌʽΪ

ÒÒ»ùÝìõ«Ôڴ˱仯¹ý³ÌÖеÄ×÷ÓÃÒÔ¼°Óë¾ÉµÄ¹¤ÒµÖÆ·¨Ïà±Èй¤ÒµÖÆ·¨µÄÓŵãÊÇ          

                                                                               

                                                                              ¡£

£¨3£©H2O2µÄʵÑéÊÒÖÆ·¨Ö®Ò»Êǽ«¹ýÑõ»¯±µ¼ÓÈ뵽ϡÁòËáÖУ¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ       

                                                                              

£¨4£©Ð´³öÒ»ÖÖ¶¨Á¿²â¶¨H2O2Ë®ÈÜÒºÖÐH2O2º¬Á¿µÄ»¯Ñ§·½³Ìʽ

                                                                               

£¨5£©Ð´³ö·ûºÏ(4)·´Ó¦Ô­ÀíµÄ¼òҪʵÑé²½Öè

                                                                              

                                                                               

                                                                              

                                                                              

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

[»¯Ñ§¡ªÑ¡ÐÞ»¯Ñ§Óë¼¼Êõ](15·Ö)

H2O2µÄË®ÈÜÒºÊÇÒ»ÖÖ³£ÓõÄɱ¾ú¼Á¡£

£¨1£©H2O2¾ÉµÄ¹¤ÒµÖÆ·¨ÊÇÒÔPtΪÑô¼«¡¢Ê¯Ä«ÎªÒõ¼«µç½âÁòËáÇâï§ÈÜÒº£¬ÔÙ½«µç½â²úÎïË®½â¡£»¯Ñ§·½³ÌʽΪ£º

¡£Ð´³öµç½âÁòËáÇâï§ÈÜÒºµÄÒõ¼«ºÍÑô¼«·´Ó¦¹¤¡£

Òõ¼«£º                                                            £»

Ñô¼«£º                                                            £»

£¨2£©H2O2ÐµĹ¤ÒµÖÆ·¨ÊÇÏȽ«ÒÒ»ùÝìõ«¼ÓÇ⻹ԭ£¬ÔÙ½«Öмä²úÎïÈ¥ÇâÑõ»¯¼´µÃH2O2£¬·´Ó¦·½³ÌʽΪ

ÒÒ»ùÝìõ«Ôڴ˱仯¹ý³ÌÖеÄ×÷ÓÃÒÔ¼°Óë¾ÉµÄ¹¤ÒµÖÆ·¨Ïà±Èй¤ÒµÖÆ·¨µÄÓŵãÊÇ          

                                                                              

                                                                             ¡£

£¨3£©H2O2µÄʵÑéÊÒÖÆ·¨Ö®Ò»Êǽ«¹ýÑõ»¯±µ¼ÓÈ뵽ϡÁòËáÖУ¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ       

                                                                              

£¨4£©Ð´³öÒ»ÖÖ¶¨Á¿²â¶¨H2O2Ë®ÈÜÒºÖÐH2O2º¬Á¿µÄ»¯Ñ§·½³Ìʽ

                                                                              

£¨5£©Ð´³ö·ûºÏ(4)·´Ó¦Ô­ÀíµÄ¼òҪʵÑé²½Öè

                                                                              

                                                                              

                                                                              

                                                                              

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º¹þʦ´ó¸½ÖС¢¶«±±Ê¦´ó¸½ÖС¢ÁÉÄþÊ¡2010Äê¸ßÈýµÚ¶þ´ÎÁª¿¼£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

[»¯Ñ§¡ªÑ¡ÐÞ»¯Ñ§Óë¼¼Êõ] (15·Ö)

H2O2µÄË®ÈÜÒºÊÇÒ»ÖÖ³£ÓõÄɱ¾ú¼Á¡£

£¨1£©H2O2¾ÉµÄ¹¤ÒµÖÆ·¨ÊÇÒÔPtΪÑô¼«¡¢Ê¯Ä«ÎªÒõ¼«µç½âÁòËáÇâï§ÈÜÒº£¬ÔÙ½«µç½â²úÎïË®½â¡£»¯Ñ§·½³ÌʽΪ£º

¡£Ð´³öµç½âÁòËáÇâï§ÈÜÒºµÄÒõ¼«ºÍÑô¼«·´Ó¦¹¤¡£

Òõ¼«£º                                                            £»

Ñô¼«£º                                                            £»

£¨2£©H2O2ÐµĹ¤ÒµÖÆ·¨ÊÇÏȽ«ÒÒ»ùÝìõ«¼ÓÇ⻹ԭ£¬ÔÙ½«Öмä²úÎïÈ¥ÇâÑõ»¯¼´µÃH2O2£¬·´Ó¦·½³ÌʽΪ

ÒÒ»ùÝìõ«Ôڴ˱仯¹ý³ÌÖеÄ×÷ÓÃÒÔ¼°Óë¾ÉµÄ¹¤ÒµÖÆ·¨Ïà±Èй¤ÒµÖÆ·¨µÄÓŵãÊÇ          

                                                                              

                                                                              ¡£

£¨3£©H2O2µÄʵÑéÊÒÖÆ·¨Ö®Ò»Êǽ«¹ýÑõ»¯±µ¼ÓÈ뵽ϡÁòËáÖУ¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ       

                                                                              

£¨4£©Ð´³öÒ»ÖÖ¶¨Á¿²â¶¨H2O2Ë®ÈÜÒºÖÐH2O2º¬Á¿µÄ»¯Ñ§·½³Ìʽ

                                                                               

£¨5£©Ð´³ö·ûºÏ(4)·´Ó¦Ô­ÀíµÄ¼òҪʵÑé²½Öè

                                                                              

                                                                              

                                                                               

                                                                              

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºËÄ´¨Ê¡Ä£ÄâÌâ ÌâÐÍ£ºÌî¿ÕÌâ

ºÏ³É°±¹¤Òµ¼°ÏõËṤҵ¶Ô¹¤¡¢Å©Éú²úºÍ¹ú·À¶¼ÓÐÖØÒªÒâÒå¡£Çë»Ø´ð£º
£¨1£©ÒÔN2ºÍH2ΪԭÁϺϳɰ±Æø¡£ÒÑÖª£ºN2(g)+3H2(g)2NH3(g)£»¡÷H= -92.4 kJ/mol£¬
¢ÙºÏ³É°±¹¤ÒµÖвÉÈ¡µÄÏÂÁдëÊ©¿ÉÒÔÓÃÀÕÏÄÌØÁÐÔ­Àí½âÊ͵ÄÊÇ£¨ÌîÑ¡ÏîÐòºÅ£©_______¡£  
a£®ÓÃÌú´¥Ã½¼Ó¿ì»¯Ñ§·´Ó¦ËÙÂÊ            
b£®²ÉÓýϸßѹǿ£¨20 MPa~50 MPa£©
c£®½«Ô­ÁÏÆøÖеÄÉÙÁ¿COµÈÆøÌå¾»»¯³ýÈ¥  
d£®½«Éú³ÉµÄ°±Òº»¯²¢¼°Ê±´ÓÌåϵÖзÖÀë³öÀ´
¢ÚÒ»¶¨Î¶ÈÏ£¬ÔÚÃܱÕÈÝÆ÷ÖгäÈë1 mol N2ºÍ3 mol H2²¢·¢Éú·´Ó¦¡£
a£®ÈôÈÝÆ÷ÈÝ»ýºã¶¨£¬´ïµ½Æ½ºâʱÆøÌåµÄѹǿΪԭÀ´µÄ7/8£¬N2µÄת»¯ÂÊΪ§Ñ1£¬´Ëʱ£¬·´Ó¦·ÅÈÈΪ_________kJ£»
b£®ÈôÈÝ»ýΪ4L£¬µ±½øÐе½µÚ4·ÖÖÓʱ´ïµ½Æ½ºâ£¬Éú³ÉNH3Ϊ1.0 mol£¬Ôò´Ó¿ªÊ¼µ½´ïƽºâʱNH3µÄƽ¾ùËÙÂÊv(NH3)=_________¡£
c£®ÈôÈÝÆ÷ѹǿºã¶¨£¬Ôò´ïµ½Æ½ºâʱ£¬ÈÝÆ÷ÖÐN2µÄת»¯ÂÊΪ§Ñ2£¬Ôò§Ñ2_______§Ñ1£¨Ìî¡°£¾¡¢£¼»ò£½¡±£©¡£
£¨2£©ÒÔ°±Æø¡¢¿ÕÆøΪÖ÷ÒªÔ­ÁÏÖÆÏõËá¡£
¢ÙNH3±»ÑõÆø´ß»¯Ñõ»¯Éú³ÉNOµÄ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ______________________¡£
¢ÚÏõË᳧³£ÓÃÈçÏÂ2ÖÖ·½·¨´¦ÀíβÆø¡£
a£®´ß»¯»¹Ô­·¨£º´ß»¯¼Á´æÔÚʱÓÃH2½«NO2»¹Ô­ÎªN2¡£
ÒÑÖª£º
2H2£¨g£©+O2£¨g£©=2H2O£¨g£© ¡÷H=£­483.6kJ/mol                  
N2£¨g£©+2O2£¨g£©=2NO2£¨g£© ¡÷H=+67.7kJ/mol
ÔòH2»¹Ô­NO2Éú³ÉË®ÕôÆø·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ___________________¡£
b£®¼îÒºÎüÊÕ·¨£ºÓÃNa2CO3ÈÜÒºÎüÊÕNO2Éú³ÉCO2¡£Èôÿ9.2gNO2ºÍNa2CO3ÈÜÒº·´Ó¦Ê±×ªÒƵç×ÓÊýΪ0.1mol£¬Ôò·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_______________¡£
£¨3£©Ëæ×ŶԺϳɰ±Ñо¿µÄ·¢Õ¹£¬Ï£À°¿Æѧ¼Ò²ÉÓøßÖÊ×Óµ¼µçÐÔµÄSCYÌÕ´É£¨ÄÜ´«µÝH+£©Îª½éÖÊ£¬ÓÃÎü¸½ÔÚËüÄÚÍâ±íÃæÉϵĽðÊôîٶྦྷ±¡Ä¤×öµç¼«£¬ÊµÏÖÁ˳£Ñ¹¡¢570¡æÌõ¼þϸßת»¯Âʵĵç½â·¨ºÏ³É°±£¨×°ÖÃÈçͼ£©¡£îٵ缫AÊǵç½â³ØµÄ_______¼«£¨Ìî¡°Ñô¡±»ò¡°Òõ¡±£©£¬¸Ã¼«Éϵĵ缫·´Ó¦Ê½ÊÇ__________________ ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸