Ó¦Óõ绯ѧÊéºóϰÌâ´ð°¸Ñî»Ô - ¬ÎÄÇì±à ÏÂÔØ±¾ÎÄ

2£©¹¯²Û·¨µÄÖ÷ÒªÓŵãÊÇËùµÃ¼îÒºµÄŨ¶È¸ß£¬½Ó½ü50£¥£¬¿ÉÖ±½Ó×÷ΪÉÌÆ·³öÊÛ£¬¶øÇÒ´¿¶È¸ß£¬¼¸ºõ²»º¬Cl¡£ÆäÖ±Á÷µçÄÜÏûºÄËä¸ß£¬µ«Ëü²»ÐèÒªÕô·¢Å¨Ëõ¼îÒºµÄºó´¦Àí²Ù×÷£¬¹Êÿ¶Ö¼îµÄ×ÜÄܺÄÈÔºÍÆäËû¶þ·¨Ïà·Â£¬¶øÇÒ¶ÔÑÎË®µÄ¾»»¯ÒªÇó²»Ïñ¸ôĤ²ÛÄÇÑù¸ß¡£´ÓÉú²úÄÜÁ¦ÉÏ¿´£¬¹¯²ÛµÄÓÅÔ½ÐÔÊÇËùÓõçÁ÷Ãܶȴ󣬶øÇÒ¿É´ó·ù¶ÈµØ±ä»¯£¬¿É±Ü¿ª³ÇÊÐÓõç¸ß·å£¬ËæÊ±µ÷ÕûµçÁ÷Ãܶȡ£

¹¯²ÛµÄÖ÷ҪȱµãÊÇÓй¯¶¾ÎÛȾ»·¾³£¬±ØÐëÑϸñ¿ØÖÆÅÅ·ÅÎÛË®Öй¯µÄº¬Á¿£¬°´Ê±¼ì²é²Ù×÷¹¤È˽¡¿µ×´¿ö£¬¼ÓÇ¿ÀͶ¯±£»¤´ëÊ©¡£¹¯µÄ¼Û¸ñ¹ó£¬Í¶×Ê´óÊÇËüµÄÁíһȱµã¡£

3£©Àë×ÓĤ²Ûµç½âµÄÓŵãÊÇûÓй¯ºÍʯÃ޵Ĺ«º¦£»ËùµÃNaOH²»º¬C1£¬ºÜ´¿£¬ÆäŨ¶È¿É´ï10£¥¡«40£¥£¬¹ÊÕô·¢Å¨ËõµÄºó´¦Àí·ÑÓÃÒªÉٵöࣻµçÁ÷ÃܶȿɱȸôĤ²ÛËùʹÓõĴóÒ»±¶¶øÈÔ±£³Ö3.1¡«3.8VµÄ²Ûµçѹ£¬¡°×ÜÄܺġ±(°üÀ¨µç½âÓõ硢¶¯Á¦ÓõçºÍÕôÆøÏûºÄ)Ïà¶Ô½ÏµÍ£¬Ò»°ã½Ï¸ôĤ·¨ºÍ¹¯·¨µÍ25£¥ÒÔÉÏ£¬Ã¿Éú²úÒ»¶ÖÉÕ¼î¿ÉÊ¡½Ó½ü1000kW¡¤hµç¡£³£°Ñ¶àÖ»µç½â³Ø»ã¼¯×é×°³ÉѹÂË»úʽµÄµç½â²Û£¬µ¥²Û²úÂʿɴï100¶ÖNaOH/Ä꣬¶ø²ÛÌå»ý±ÈǰÁ½·¨µÄµç½â²ÛСµÃ¶à£¬ÌرðÊÊÓÃÓÚС¹æÄ£Éú²ú£¬²úÁ¿¿É°´Êг¡ÐèÒªµ÷½Ú¡£ÆäȱµãÊÇÀë×ÓĤµÄʹÓÃÊÙÃü¼°Àë×ÓĤµÄÐÔÄÜ»¹ÐèÒª¸Ä½ø¡£ÀýÈ磬ΪÁËÑÓ³¤Àë×ÓĤµÄʹÓÃÊÙÃü£¬ÒªÇó°ÑÑÎË®¸öµÄ¸Æ¡¢Ã¾Àë×Óº¬Á¿½µµÍÁÐ10ÊýÁ¿¼¶¡£Ä¿Ç°£¬Àë×ÓĤµÄÐÔÄÜ»¹ÔÚ²»¶Ï¸Ä½ø£¬´Ó·¢Õ¹Ç÷ÊÆ¿´£¬Àë×ÓÖ¬²Û×îÓÐÉúÃüÁ¦£¬Ô¤¼ÆÔÙ¹ýÈô¸ÉÄ꽫ȫ²¿È¡´úÆäËûµÄµç½â²Û¡£

30. ijÂȼÓÃË®Òø²Ûµç½âʳÑÎË®£¬Í¨¹ýµç½â²ÛµÄµçÁ÷Ϊ100 kA¡£ÎÊÀíÂÛÉÏÿÌìÉú²ú¶àÉÙC12¡¢H2ºÍNaOH£¿ÉèÑô¼«µÄµçÁ÷ЧÂÊΪ97£¥£¬ÎÊʵ¼ÊÉÏÿÌìÉú²ú¶àÉÙC12£¿(ÒÔÿÌì24h¼Æ)

½â£º£¨1£©ÉèµçÁ÷ЧÂÊΪ100£¥£¬ÂÈÆø°´ÏÂʽ·´Ó¦Ê½¼ÆË㣺

-6

-1

-

2Cl??2e?Cl2(g)

ÿÌìµçÁ¿Q=It=100000¡Á24¡Á60¡Á60£½8640000000 C ÿÌìÉú²úµÄC12£ºN(ÂÈÆø)£½Q/£¨2¡Á96485£©£½44773.80 mol

ÿÌìÉú²úµÄNaOH£ºN(NaHgm)= Q/£¨2¡Á96485£©=89547.60 mol= N(NaOH) ÿÌìÉú²úµÄH2£¨Æø£©£ºN£¨ÇâÆø£©= N(NaHgm)/2=44773.80 mol £¨2£©ÉèÑô¼«µÄµçÁ÷ЧÂÊΪ97£¥£¬Ã¿ÌìÉú²úµÄC12£º

N(ÂÈÆø)£½¦ÇQ/£¨2¡Á96485£©£½0.97¡Á44773.80 mol£½43430.59 mol

31£®ÂȼҵÖÐÔÚÒõ¼«µÄ¸Ä½ø·½Ã棬ÈËÃÇÊÔÓÃÑõ»¹Ô­·´Ó¦´úÌæHµÄÎö³ö

+

1/2O2·´Ó¦£¬¼´ÓÃÒõ¼«·´Ó¦£º

´úÌæÎöÇâ·´Ó¦¡£

?H2O?2e£½2OH?£¬???0.410V£¬

(1)ÊÔд³ö¸Ä½øÇ°µÄµç¼«·´Ó¦ºÍ×Ü·´Ó¦£»(2)Òõ¼«·´Ó¦¸Ä½øºó£¬ÀíÂÛ·Ö½âµçѹ½µµÍ¶àÉÙ£»(3)ÀíÂÛÉÏ£¬Ã¿Éú²ú1¶ÖNaOH½«½ÚÔ¼¶àÉÙµçÄÜ?

½â£º(1)¸Ä½øÇ°£ºÑô¼«·´Ó¦Îª£º

2Cl??Cl2?2e ???1.36V

21

Òõ¼«·´Ó¦Îª£º

2H2O?2e?H2?2OH? ???-0.828V

×Ü·´Ó¦Îª£º

2NaCl?2H2O?2e?H2?2NaOH?Cl2

E??2.188V

(1)¸Ä½øºó£º Ñô¼«·´Ó¦Îª£º

2Cl??Cl2?2e ???1.36V

1O?H2O?2e?2OH? ???0.401V 21 ×Ü·´Ó¦Îª£º2NaCl?O2?H2O?2e?2NaOH?Cl2

2E??0.959 V

Òõ¼«·´Ó¦Îª£º

ÀíÂÛ·Ö½âѹ½µµÍ£½2.188-0.959£½1.229 V (3)ÀíÂÛµçÄܼÆË㹫ʽΪ£ºW?Ee?(m/M)zF

ÀíÂÛÉÏÿÉú²ú1¶ÖNaOH½«½ÚÔ¼µçÄÜ

£½£¨2.188-0.959£©*£¨1000*1000/40£©*2*26.8£½1646860 V¡¤A¡¤h £½1646860 W¡¤h=1646.860 kW¡¤h

µÚ°ËÕ µç»¯Ñ§¸¯Ê´Óë·À»¤

43£®¼ò»¯µÄ?¡ªpHͼÒÔ¿ÉÈÜÐÔÀë×ÓŨ¶È×ܺÍСÓÚ10 mol¡¤L×÷Ϊ½ðÊô

-6

-1

¸¯Ê´Óë·ñµÄ½çÏÞ£¬ºÏÀíÂð?ʲô½Ð¸¯Ê´Çø£¬¶Û»¯ÇøºÍÎȶ¨Çø? ´ð£ººÏÀí¡£

44.

?(?Cu?2?/Cu)±È

?(?H??/H2,Pt)¸ß£¬ÎªÊ²Ã´CuÔÚ³±Êª¿ÕÆøÖÐÒ²¸¯

Ê´??(Ti2?/Ti)±È??(H?/H2,Pt)µÍµÃ¶à£¬µ«ÎªÊ²Ã´TiÓÖÊÇÁ¼ºÃµÄÄ͸¯Ê´½ðÊô?

?(?Zn2?/Zn)±È?(H?/H2,Pt)µÍ£¬µ«ÓÃZnͰʢˮ²»¸¯Ê´£¬ÎªÊ²Ã´?

?(H?/H2,Pt)½â£º(1)¾¡¹Ü??(Cu2?/Cu)±È?¸ß£¬ CuÔÚ³±Êª¿ÕÆøÖлᱻ¸¯Ê´£¬

Ò²¾ÍÊÇÍ­ÔÚº¬ÓÐÈܽâÑõµÄËá»òË®½éÖÊÖÐÔò¿ÉÄܱ»¸¯Ê´£¬ÒòΪͭµÄ±ê×¼µç¼«µçλ±ÈÑõµÄÁ½¸ö·´Ó¦µÄ±ê×¼µç¼«µçλ¾ù¸ü¸ºÒ»Ð©¡£ £¨2£©¾¡¹Ü??(Ti2?/Ti)±È??(H?/H2,Pt)µÍµÃ¶à£¬ÒòΪîÑÔÚ¿ÕÆøÖÐÉú³ÉÁËÒ»²ã

Ñõ»¯Ä¤TiO2£¬Õâ²ãÑõ»¯Ä¤¸ô¾ø³±Êª¿ÕÆø£¬½«±£»¤½ðÊôîÑÃâÔⸯʴ¡£ £¨3£©¾¡¹Ü??(Zn2?/Zn)±È??(H?/H2,Pt)µÍ£¬µ«Ð¿ÔÚÔÚ³±Êª¿ÕÆøÖÐÈÝÒ×Éú³ÉÒ»

22

²ãÖ÷ÒªÓɼîʽ̼Ëáп×é³ÉµÄ±¡Ä¤£¨3Zn(OH)2?ZnCO3£©£¬Õâ²ã±¡Ä¤ÓÐÒ»¶¨µÄÄ͸¯Ê´ÄÜÁ¦£¬Òò´ËÓÃZnͰʢˮ²»¸¯Ê´¡£

45£®¾Ö²¿¸¯Ê´Ö»ÊǾֲ¿ÎÊÌ⣬Ϊʲô˵±È¾ùÔȸ¯Ê´¸üΪÑÏÖØ?ÊÔ¾ÙÀý˵Ã÷¿×Ê´£¬¾§¼ä¸¯Ê´£¬Ó¦Á¦¸¯Ê´ÆÆÁѺͳåÊ´¡£

½â£ºÕâÖÖ½ðÊô/ÈÜÒº½çÃæµÄ²»¾ùÒ»ÐÔÊDzúÉú¾Ö²¿¸¯Ê´µÄÔ­Òò¡£¾Ö²¿¸¯Ê´µÄΣº¦±È¾ùÔȸ¯Ê´ÒªÑÏÖØµÃ¶à£¬ÒòΪ½ðÊô¸¯Ê´µÄÑô¼«·´Ó¦ºÍ¹²¶óÒõ¼«·´Ó¦£¬ÓÉÓÚ½ðÊôÈÜ/Òº½çÃæµÄ²»¾ùÒ»¶ø²úÉúÁ˿ռä·ÖÀ룬Ñô¼«·´Ó¦ÍùÍùÔÚ¼«Ð¡µÄ¾Ö²¿·¶Î§ÄÚ·¢Éú£¬´Ëʱ×ܵÄÑô¼«ÈܽâËÙÂÊËäÈ»ÈԾɵÈÓÚ×ܵĹ²¶óÒõ¼«·´Ó¦ËÙÂÊ£¬µ«ÊÇÒõ¼«µçÁ÷ÃܶÈ(µ¥Î»Ãæ»ýÄڵķ´Ó¦ËÙÂÊ)È´´ó´óÔö¼ÓÁË£¬¼´¾Ö²¿µÄ¸¯Ê´Ç¿¶È´ó´ó¼Ó¾çÁË¡£ÀýÈçÒ»¸ù¾ùÔȸ¯Ê´µÄÌú¹Ü¿ÉÒÔÁ¬ÐøÊ¹Óúܳ¤Ê±¼ä¶øÎÞ´ó°­£¬µ«Èç¾Ö²¿¸¯´©¾ÍÖ»Äܱ¨·Ï¡£

¢Ù¿×Ê´ÊÇÔÚ½ðÊô²ÄÁϱíÃæÉÏÐγÉÖ±¾¶Ð¡ÓÚ1mm£¨²¢Ïò°åºñ·½Ïò·¢Õ¹µÄ£©Ð¡¿×µÄ¾Ö²¿¸¯Ê´¡£ÀýÈ磬½ðÊô²ÛÄÚ½éÖÊ·¢Éúй©£¬´ó¶àÊÇ¿×Ê´Ôì³ÉµÄ£¬¶øÇÒËüµÄ·¢Õ¹ËÙ¶ÈÒ²ÊǺܿìµÄ£¬´ó¶àΪÿÄêÊýºÁÃס£

¢Ú¾§¼ä¸¯Ê´ÊÇÓɾ§½çµÄÔÓÖÊ»ò¾§½çÇøÄ³Ò»ºÏ½ðÔªËØÔö¶à»ò¼õÉÙÒýÆðµÄ£¬ÑØ×ŽðÊô²ÄÁϵľ§½ç²úÉúµÄÑ¡ÔñÐÔ¸¯Ê´£¬¾¡¹Ü¾§Á£¼¸ºõ²»·¢Éú¸¯Ê´£¬µ«ÈÔÈ»µ¼Ö²ÄÁÏÆÆ»µ¡£ÀýÈ磬²»Ðâ¸Öƶ¸õÇø²úÉúµÄ¾§¼ä¸¯Ê´£¬ÊÇÓÉCr23C6µÈ̼»¯ÎïÔÚ¾§½çÎö³ö£¬Ê¹¾§½ç½üÅԵĸõº¬Á¿½µµ½°Ù·ÖÖ®¼¸ÒÔÏ£¬¹ÊÕⲿ·ÖÄÍÊ´ÐÔ½µµÍ¡£¢ÛÓ¦Á¦¸¯Ê´ÆÆÁÑÊÇÖ¸Ó¦Á¦ºÍÒ»ÖÖÔÚÌØ¶¨¸¯Ê´½éÖʹ²Í¬´æÔÚ¶øÒýÆðµÄÆÆÁÑ¡£ÆÆÁÑÓÐÑØ¾§(¾§½çÆÆÁÑ)ºÍ´©¾§(¾§Á£ÆÆÁÑ)Á½ÖÖ¡£ËüÃǶÔÓÚÊÜÓ¦Á¦µÄÆ÷еΣº¦×î´ó£¬Èç¸ßѹ¹øÂ¯¡¢·É»úÉϲàÃæ±¡±Ú¡¢¸ÖË÷¡¢»úÆ÷µÄÖáµÈ£¬Èç¹û·¢ÉúÕâÀฯʴ¾Í¿ÉÄÜͻȻ±ÀÁѶøÄð³Éʹʡ£

¢Ü³åÊ´ÊÇÔÚ³å»÷µÄ»úе×÷ÓÃÏ£¬²ÄÁϱíÃæ·¢ÉúÄ¥ËðµÄͬʱÓÖ¼ÓÈ븯ʴ×÷Óã¬Á½ÕßÏ໥´Ù½ø£¬²úÉúÑÏÖØµÄÇÖÊ´¡£ÆøÏàÁ÷ÌåÖеÄÒºµÎ¡¢ÒºÏàÁ÷ÌåСµÄ¹ÌÌå·Ûľ¡¢ÒºÌåÖÐÐýÎвúÉúµÄ¿ÕѨ¡¢Íä¹ÜµÈ²¿Î»·¢ÉúµÄÎÐÁ÷µÈ£¬¶¼ÄÜÆÆ»µ±íÃæÄ¤£¬¼ÓËÙ¸¯Ê´¡£±©Â¶ÔÚÔ˶¯Á÷ÌåÖеÄËùÓÐÀàÐ͵ÄÉ豸¶¼»áÔâµ½³åÊ´¡£Èç¹ÜµÀϵͳµÄÍäÍ·¡¢·§ÃÅ¡¢±ÃµÈ¡£

46.ʲô½Ð¾Ö²¿µç³ØºÍ¾Ö²¿µçÁ÷£¬ËüÃÇÓë¾Ö²¿¸¯Ê´¹ØÏµÈçºÎ£¿ÊÔÁбí¶Ô¾ùÔȸ¯Ê´ºÍ¾Ö²¿¸¯Ê´µÄÌØµã¼ÓÒÔ˵Ã÷¡£

½â£ºµ±½ðÊô±íÃæº¬ÓÐһЩÔÓÖÊʱ£¬ÓÉÓÚ½ðÊôµÄµçÊÆºÍÔÓÖʵĵçÊÆ²»¾¡Ïàͬ£¬¿É¹¹³ÉÒÔ½ðÊôºÍÔÓÖÊΪµç¼«µÄÐí¶à΢СµÄ¶Ì·µç³Ø£¬³ÆÎªÎ¢µç³Ø(»ò¾Ö²¿µç³Ø)£¬´Ó¶øÒýÆð¸¯Ê´²úÉú¾Ö²¿µçÁ÷¡£¶ÔÓÚ¾Ö²¿µç³Ø»ò¾Ö²¿¸¯Ê´¶øÑÔ£¬ÓÉÓÚ½ðÊôÈÜ/Òº½çÃæµÄ²»¾ùÒ»¶ø²úÉúÁ˿ռä·ÖÀ룬Ñô¼«·´Ó¦ÍùÍùÔÚ¼«Ð¡µÄ¾Ö²¿·¶Î§ÄÚ·¢Éú£¬´Ëʱ×ܵÄÑô¼«ÈܽâËÙÂÊËäÈ»ÈԾɵÈÓÚ×ܵĹ²¶óÒõ¼«·´Ó¦ËÙÂÊ£¬µ«ÊÇÒõ¼«µçÁ÷ÃܶÈ(µ¥Î»Ãæ»ýÄڵķ´Ó¦ËÙÂÊ)È´´ó´óÔö¼ÓÁË£¬¼´¾Ö²¿µÄ¸¯Ê´Ç¿¶È´ó´ó¼Ó¾çÁË¡£ÀýÈçÒ»¸ù¾ùÔȸ¯Ê´µÄÌú¹Ü¿ÉÒÔÁ¬ÐøÊ¹ÓúÜ

23

³¤Ê±¼ä¶øÎÞ´ó°­£¬µ«Èç¾Ö²¿¸¯´©¾ÍÖ»Äܱ¨·Ï¡£¶ø¾ùÔȸ¯Ê´¶¼ÊÇÑô¼«ºÍÒõ¼«·´Ó¦ÊÇÔÚ½ðÊô±íÃæÏàͬµÄλÖ÷¢ÉúµÄ£¬ÕâÑùÒýÆðµÄ½ðÊô¸¯Ê´ÊǾùÔȵ쬳ÆÎª¾ùÔȸ¯Ê´»òÈ«Ãæ¸¯Ê´¡£ÀýÈçÒ»¿éп»ò¸Ö½þÔÚÏ¡ÁòËáÖУ¬Í¨³£ÔÚÈ«²¿±íÃæÉÏÒÔ¾ùÔÈËÙ¶ÈÈܽ⡣ 47.»­³öÒ»ÖÖµäÐ͵ĽðÊô¶Û»¯I??ÇúÏߣ¬²¢ËµÃ÷¸÷¶ÎµÄÒâÒå¡£ÈçºÎÄÜ˵

Ã÷½ðÊô¶Û»¯µÄÔ­ÒòÊǽðÊô±íÃæÐγÉÖÂÃÜÑõ»¯ÎﱻĤµÄ½á¹û¡£

½â£º47ÌâͼÊÇÒÔFe/lmolH2SO4½ðÊôµç¼«Îª»ù´¡£¬Ãè»æÒ»¸ö·´Ó³¶Û̬½ðÊôÑô¼«¼«»¯Ò»°ãÌØÕ÷µÄ¼«»¯ÇúÏßʾÒâͼ£¬Õû¸öÇúÏ߿ɷֳÉËĸöÇø£º

47Ìâͼ£º ÌúÔÚlmolH2SO4ÈÜÒºÖеÄÎÈ̬Ñô¼«¼«»¯ÇúÏß

(1)A-KÇø: ½ðÊô»îÐÔÈܽâÇø¡£´Ó¸¯Ê´½ðÊôµÄ×Ô¸¯Ê´µçλ¦Åcµ½¶Û»¯µçλ¦Å

p

£¬½ðÊôµç¼«µÄÑô¼«µçÁ÷ÃܶÈËæ×ŵçλ¦ÅµÄÉý¸ß¶ø²»¶ÏÔö´ó¡£µ±¦Å£½¦Åpʱ

½ðÊôµç¼«µÄÑô¼«µçÁ÷Ãܶȴﵽ×î´óÖµiÁÙ¡£iÁÙ³ÆÎªÁÙ½ç¶Û»¯µçÁ÷Ãܶȡ£ÀýÈ磬ÔÚÕâ¸öµçÎ»Çø¼äÖÐ,½ðÊôµÄÌúµç¼«ÔÚlmolH2SO4Öв»¶ÏÒÔFeÐÎʽ»îÐÔÈܽ⣬ÈܽâÏÂÀ´µÄFe»ý¾ÛÔڵ缫Óëµç½âÖʽçÃæ£¬µ±ÆäŨ¶È´ïµ½FeÓëSO4

¡ª

2+

2+

2

2+

µÄŨ¶È»ýʱ£¬FeSO4¿ªÊ¼ÒÔÑÎĤÐÎʽ³Á»ýÔÚ½ðÊô±íÃæ¡£

p

(2)K-IÇø : ͨ³£³ÆÎª»î»¯Ò»¶Û»¯¹ý¶ÉÇø¡£ ´Ó¦ÅÖÁ¦Å

F

ÊÇÇúÏߵĵڶþ²¿

·Ý¡£¦ÅF½Ð×ö»î»¯µçλ¡£ÔÚÕâÒ»µçÎ»Çø¼ä£¬½ðÊô±íÃæ×´Ì¬·¢Éú¼±¾çµÄ±ä»¯£¬½ðÊô±íÃæ´¦ÓÚ²»Îȶ¨µÄ״̬£¬ÔÚÓÐЩÇé¿öÏ£¨ÈçÉÏÊöµÄFe£¯lmolH2SO4Ìåϵ£©£¬µçÁ÷ÃܶÈÓдó·ù¶ÈµÄÕñµ´¡£

(3)I-EÇø: ³ÆÖ®Îª¶Û»¯Çø»òÎȶ¨¶ÛÌ¬Çø¡£ µ±µçλ¸ßÓÚ¦Å

F

ʱ£¬½øÈëÇúÏß

µÄµÚÈý²¿·Ý¡£´Ëʱ£¬½ðÊô±íÃæÓÐijÖÖÖÂÃܵġ¢Èܽ⻺ÂýµÄ±¡Ä¤Éú³É£¬½ðÊôµÄÑô¼«µçÁ÷ÃܶȽ«ÏÔÖøÏ½µ£¬½ðÊô±»ÈÏΪ´¦ÓÚ¶Û»¯×´Ì¬¡£ÔÚÒ»¸öÏ൱¿í¹ãµÄµçλ·¶Î§ÄÚ£¬½ðÊôµÄÑô¼«ÈܽâµçÁ÷²»±ä²¢±£³ÖÔÚÒ»¸öºÜµÍµÄˮƽ¡£ÓÐЩ½ðÊôÔÚÕâÒ»µçÎ»Çø¼äµÄÑô¼«Èܽâ²úÎïÒ²²»Í¬ÓÚ»îÐÔÑô¼«ÈܽâʱµÄ²úÎï¡£ÀýÈ磬ÌúÔÚlmolH2SO4ÈÜÒºÖУ¬»îÐÔÑô¼«ÈܽâµÄ²úÎïÊǶþ¼ÛµÄÑÇÌúÀë×Ó£¬¶øÔÚÕâÒ»µçÎ»Çø¼ä£¬Ñô¼«Èܽâ²úÎïÊÇÈý¼ÛµÄÕýÌúÀë×Ó¡£ (4)E-GÇø ÕâÒ»²¿·Ý½Ð×ö¹ý¶Û»¯Çø¡£¦Å

TP

½Ð×ö¹ý¶Û»¯µçλ¡£ÕâÒ»²¿·ÝµÄ

ÇúÏßµÄÌØÕ÷ÊÇ£¬Ñô¼«µçÁ÷ÃܶÈÔÙ´ÎËæµçλµÄÉý¸ß¶øÔö´ó¡£ÕâÊÇÒòΪÔÚ¹ý¶Û»¯Çø£¬½ðÊôµç¼«ÉÏ·¢Éúеĵ缫·´Ó¦£¨»ò¶Û»¯Ä¤±»ÆÆ»µÁË£©¡£´ËʱµÄÑô

24