ÌâÄ¿ÄÚÈÝ

¹ýÑõ»¯ÇâË®ÈÜÒºË׳ÆË«ÑõË®£¬·Ðµã±ÈË®¸ß£¬Óö¹â¡¢Èȼ°ÖؽðÊô»¯ºÏÎïµÈ¾ùÄÜÒýÆð·Ö½â¡£
£¨1£©Ä³ÊÔ¼Á³§ÏÈÖƵÃ7%~8%µÄË«ÑõË®£¬Óû½«ÆäŨËõ³É30%µÄÈÜÒº£¬ÊÊÒË·½·¨ÊÇ         
£¨Ìîд±àºÅ£©¡£
a£®³£Ñ¹ÕôÁó ¡¡    b£®¼õѹÕôÁó      c£®¼ÓÈëÉúʯ»Ò³£Ñ¹ÕôÁó ¡¡¡¡ d£®¼ÓѹÕôÁó
£¨2£©Èç¹ûµÃµ½µÄË«ÑõË®ÖÐÑõÔªËصĺ¬Á¿Îª90%£¬Ôò¹ýÑõ»¯ÇâµÄ´¿¶ÈΪ           ¡£ÖÚËùÖÜÖª£¬ÇâÆøÔÚ¿ÕÆøÖÐȼÉÕÉú³ÉË®¡£ÓÐÈËÌá³ö£¬ÇâÆøÔÚ¿ÕÆøÖÐȼÉÕÒ²¿ÉÄÜÉú³ÉH2O2£¬µ«ËüÒò¸ßζø·Ö½âÁË¡£ÎªÁËÑéÖ¤ÇâÆøÔÚ¿ÕÆøÖÐȼÉյIJúÎïÖÐÊÇ·ñº¬ÓÐH2O2£¬Ä³¿ÎÍâС×éͬѧÉè¼ÆµÄʵÑé×°Öüûͼ-1¡£

£¨3£©¼×ͬѧÏë´ÓÏÂͼ-2µÄ¢Ù£­¢ÜÖÐÑ¡È¡Ìæ´úͼ£­1·½¿òÖеÄ×°Ö㬿ÉÐеÄÊÇ          £¨Ìîд±àºÅ£©¡£

£¨4£©ÈôÒÒͬѧÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒº¼ì²âµ½ÁËH2O2µÄ´æÔÚ£¬Íê³É¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º
                                 ¡ú             + ¡¡  Mn2+ +     H2O
±ûͬѧ¶ÔÒҵļìÑé·½°¸Ìá³öÁËÖÊÒÉ£ºÈôпÁ£ÓëÏ¡ÁòËáµÄ·´Ó¦ÖвúÉúÁËÉÙÁ¿H2SµÈ»¹Ô­ÐÔÆøÌ壬Ҳ»áʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«¡£Çë¶ÔÒÒͬѧµÄʵÑé·½°¸Ìá³ö¸Ä½ø½¨Ò飺                                                                       ¡£
£¨5£©¹ý̼ËáÄÆ£¨2Na2CO3?3H2O2£©Ë׳ƹÌÌåË«ÑõË®£¬¼«Ò׷ֽ⣬Æä·Ö½â·´Ó¦µÄ»¯Ñ§·½³Ìʽ¿É±íʾΪ£º2 (2Na2CO3?3H2O2) ¡ú 4Na2CO3 + 6H2O + 3O2¡ü
È¡Ò»¶¨Á¿µÄ¹ý̼ËáÄÆÔÚÃܱÕÈÝÆ÷ÖÐʹËüÍêÈ«·Ö½â£¬²âµÃÉú³ÉÑõÆø12.0g¡£ÀäÈ´µ½ÊÒκó£¬ÏòËùµÃ²úÎïÖмÓË®ÅäÖƳÉ10.6% µÄNa2CO3ÈÜÒº£¬Ðè¼ÓË®             g¡£
£¨1£©b£¨2·Ö£©
£¨2£©21.25%¡¡£¨2·Ö£¬Ð´0.2125¸øÈ«·Ö£©
£¨3£©¢Ú£¨2·Ö£©
£¨4£©2MnO­4-+5H2O2+6H+¡ú2Mn2++5O2¡ü+8H2O£¨2·Ö£¬ÎïÖÊÕýȷдȫ1·Ö£¬Åäƽ1·Ö£©£»
ÏȽ«ÖƵõÄÇâÆøͨ¹ý×°Óмîʯ»ÒµÄ¸ÉÔï¹Ü£¬È»ºóµãȼ¡££¨ºÏÀí£¬¸ø·Ö£©£¨2·Ö£©
£¨5£©433.5g£¨2·Ö£©

ÊÔÌâ·ÖÎö£º£¨1£©Ë«ÑõË®ÊÜÈÈ»á·Ö½â£¬Ö»ÓÐͨ¹ý¼õѹÕôÁó½µµÍζȺóÕô³öÒÔ·ÀÆä·Ö½â£»
£¨2£©Ë«ÑõË®Öк¬ÑõµÄÎïÖÊÓУºH2O2ºÍH2O£¬ÉèÎïÖʵÄÁ¿·Ö±ðΪxºÍy£¬ÔòÓÉË«ÑõË®ÖÐÑõÔªËصĺ¬Á¿Îª90%£¬Ôò¹ýÑõ»¯ÇâµÄ´¿¶ÈΪ±íʾΪ£º16/90%="18y+34x" ,(2x+3y)*16/(34x+18y)=90%,»¯¼òµÃ£º2x+y=1,µÃx=9-8/0.9¡£ÇóµÃ34x/(34x+18y)=21.25%¡£
£¨3£©Í¼£­1·½¿òÖеÄ×°ÖÃÊÇÆôÆ×·¢ÉúÆ÷×°Öã¬Âú×ãÌõ¼þµÄÊÇ¢Ú£¬¿ÉÒÔͨ¹ýʹ¹ÌÒº·ÖÀë´Ó¶øʹ·´Ó¦Í£Ö¹·¢Éú¡£
£¨4£©¸ù¾ÝÑõ»¯»¹Ô­·´Ó¦Ô­Àí£¬¿Éд³ö2MnO­4-+5H2O2+6H+¡ú2Mn2++5O2¡ü+8H2O¡£Ï൱ÓÚÑéÖ¤ÇâÆøÖÐÊÇ·ñº¬ÓÐH2SÆøÌ壬¿ÉÒÔ°Ñ»ìºÏÆøÌåͨÈëµ½ÁòËáÍ­ÈÜÒºÖÐÈô²úÉúºÚÉ«³Áµí£¬¼´ÓÐH2SÆøÌå¡£Ò²¿ÉÒÔÏȽ«ÖƵõÄÇâÆøͨ¹ý×°Óмîʯ»ÒµÄ¸ÉÔï¹Ü£¬È»ºóµãȼ¡£
£¨5£©¸ù¾Ý·½³Ìʽ2 (2Na2CO3?3H2O2) ¡ú 4Na2CO3 + 6H2O + 3O2¡ü£¬µ±Éú³ÉÉú³ÉÑõÆø12.0gʱ£¬Éú³É̼ËáÄÆ12*4*106/(3*32)=53g£¬Éú³ÉË®µÄÖÊÁ¿Îª£º(0.5*6/4)*18=13.5g¡£ÒªÅäÖƳÉ10.6% µÄNa2CO3ÈÜÒº£¬Ðè¼ÓË®(53/0.106 )-53-13.5=433.5g¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
Ñо¿NO2¡¢SO2¡¢COµÈ´óÆøÎÛȾÆøÌåµÄ´¦Àí¾ßÓÐÖØÒªÒâÒå¡£
£¨1£©ÀûÓ÷´Ó¦£º6NO2£«8NH37N2£«12H2O¿É´¦ÀíNO2¡£µ±×ªÒÆ1.2molµç×Óʱ£¬ÏûºÄµÄNO2ÔÚ±ê×¼×´¿öϵÄÌå»ýÊÇ       L¡£
£¨2£©¢ÙÒÑÖª£º2SO2(g)+O2(g)2SO3(g)   ¦¤H£½?196.6 kJ¡¤mol¨C1
2NO(g)+O2(g)2NO2(g)   ¦¤H£½?113.0 kJ¡¤mol¨C1
Çëд³öNO2ÓëSO2·´Ó¦Éú³ÉSO3(g)ºÍNOµÄÈÈ»¯Ñ§·½³Ìʽ                        ¡£
¢ÚÒ»¶¨Ìõ¼þÏ£¬½«NO2ÓëSO2ÒÔÌå»ý±È1:2ÖÃÓÚÃܱÕÈÝÆ÷Öз¢ÉúÉÏÊö·´Ó¦£¬ÏÂÁÐÄÜ˵Ã÷·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ         ¡£
a£®Ìåϵѹǿ±£³Ö²»±ä                    b£®»ìºÏÆøÌåÑÕÉ«±£³Ö²»±ä
c£®SO3ºÍNOµÄÌå»ý±È±£³Ö²»±ä          d£®Ã¿ÏûºÄ1 mol SO2µÄͬʱÉú³É1 molNO2
¢Û²âµÃÉÏÊö·´Ó¦Æ½ºâʱNO2ÓëSO2Ìå»ý±ÈΪ1:6£¬Ôò¸Ã·´Ó¦µÄƽºâ³£ÊýK£½          ¡£
£¨3£©Æû³µÎ²ÆøÖеÄÒ»Ñõ»¯Ì¼¿Éͨ¹ýÈçÏ·´Ó¦½µµÍÆäŨ¶È£ºCO(g)+1/2O2(g)CO2(g)¡£ÒÑ֪ijζÈÏ£¬ÔÚÁ½¸öºãÈÝÃܱÕÈÝÆ÷ÖнøÐи÷´Ó¦£¬ÈÝÆ÷Öи÷ÎïÖʵÄÆðʼŨ¶È¼°Õý¡¢Äæ·´Ó¦ËÙÂʹØϵÈçϱíËùʾ¡£ÇëÓá°£¾¡±»ò¡°£¼¡±»ò¡°£½¡±Ìîд±íÖеĿոñ¡£
ÈÝÆ÷±àºÅ
c(CO)£¯mol¡¤L¨C1
c(O2)£¯mol¡¤L¨C1
c(CO2)£¯mol¡¤L¨C1
¦Ô(Õý)ºÍ¦Ô(Äæ)
´óС±È½Ï
¢Ù
2.0¡Á10¨C4
4.0¡Á10¨C4
4.0¡Á10¨C4
¦Ô(Õý)£½¦Ô(Äæ)
¢Ú
1.0¡Á10¨C3
4.0¡Á10¨C4
5.0¡Á10¨C4
¦Ô(Õý)    ¦Ô(Äæ)
 
¡°ÉñÆß¡±µÇÌì±êÖ¾×ÅÎÒ¹úµÄº½ÌìÊÂÒµ½øÈëÁËеÄƪÕ¡£
£¨1£©»ð¼ýÉý¿Õʱ£¬ÓÉÓÚÓë´óÆø²ãµÄ¾çÁÒĦ²Á£¬²úÉú¸ßΡ£ÎªÁË·ÀÖ¹»ð¼ýζȹý¸ß£¬ÔÚ»ð¼ýÒ»ÃæÍ¿ÉÏÒ»ÖÖÌØÊâµÄÍ¿ÁÏ£¬¸ÃÍ¿ÁϵÄÐÔÖÊ×î¿ÉÄܵÄÊÇ        ¡£
A£®ÔÚ¸ßÎÂϲ»ÈÚ»¯B£®ÔÚ¸ßÎÂÏ¿ɷֽâÆø»¯
C£®ÔÚ³£ÎÂϾͷֽâÆø»¯ D£®¸ÃÍ¿Áϲ»¿ÉÄÜ·¢Éú·Ö½â
£¨2£©»ð¼ýÉý¿ÕÐèÒª¸ßÄܵÄȼÁÏ£¬¾­³£ÊÇÓÃN2O4ºÍN2H4ºÍ×÷ΪȼÁÏ£¬Æä·´Ó¦µÄ·½³ÌʽÊÇ£ºN2O4+N2H4¡úN2+H2O¡£ÇëÅäƽ¸Ã·´Ó¦·½³Ìʽ£º        N2O4+       N2H4¡ú       N2+       H2O
¸Ã·´Ó¦Öб»Ñõ»¯µÄÔ­×ÓÓë±»»¹Ô­µÄÔ­×ÓµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ       ¡£Õâ¸ö·´Ó¦Ó¦ÓÃÓÚ»ð¼ýÍƽøÆ÷£¬³ýÊÍ·Å´óÁ¿µÄÈȺͿìËÙ²úÉú´óÁ¿ÆøÌåÍ⣬»¹ÓÐÒ»¸öºÜ´óµÄÓŵãÊÇ                     ¡£
£¨3£©ÈçͼÊÇij¿Õ¼äÕ¾ÄÜÁ¿×ª»¯ÏµÍ³µÄ¾Ö²¿Ê¾Òâͼ£¬ÆäÖÐȼÁϵç³Ø²ÉÓÃKOHΪµç½âÒº£¬È¼Áϵç³Ø·ÅµçʱµÄ¸º¼«·´Ó¦Îª£º     ¡£Èç¹ûij¶Îʱ¼äÄÚÇâÑõ´¢¹ÞÖй²ÊÕ¼¯µ½33.6LÆøÌ壨ÒÑÕÛËã³É±ê¿ö£©£¬Ôò¸Ã¶Îʱ¼äÄÚË®µç½âϵͳÖÐתÒƵç×ÓµÄÎïÖʵÄÁ¿Îª      mol¡£

£¨4£©ÔÚÔØÈ˺½ÌìÆ÷µÄÉú̬ϵͳÖУ¬²»½öÒªÇó·ÖÀëÈ¥³ýCO2£¬»¹ÒªÇóÌṩ³ä×ãµÄO2¡£Ä³Öֵ绯ѧװÖÿÉʵÏÖÈçÏÂת»¯£º2CO2=2CO+O2£¬CO¿ÉÓÃ×÷ȼÁÏ¡£ ÒÑÖª¸Ã·´Ó¦µÄÑô¼«·´Ó¦Îª£º4OH¡ª4e-=O2¡ü+2H2O ÔòÒõ¼«·´Ó¦Îª£º                             ¡£ÓÐÈËÌá³ö£¬¿ÉÒÔÉè¼Æ·´Ó¦2CO=2C+O2£¨¡÷H£¾0¡¢¡÷S£¼0£©À´Ïû³ýCOµÄÎÛȾ¡£ÇëÄãÅжÏÉÏÊö·´Ó¦ÊÇ·ñÄÜ·¢Éú£¿               ÀíÓÉ                               ¡£
£¨5£©.±±¾©°ÂÔ˻ᡰÏéÔÆ¡±»ð¾æȼÁÏÊDZûÍ飨C3H8£©£¬ÑÇÌØÀ¼´ó°ÂÔË»á»ð¾æȼÁÏÊDZûÏ©(C3H6)¡£±ûÍéÍÑÇâ¿ÉµÃ±ûÏ©¡£
ÒÑÖª£º¢ÙC3H8(g)CH4(g)+HC¡ÔCH(g)+H2(g)£» ¡÷H1="156.6" kJ¡¤mol-1
¢ÚCH3CH=CH2(g)CH4(g)+ HC¡ÔCH (g)£»¡÷H2="32.4" kJ¡¤mol-1
ÔòÏàͬÌõ¼þÏ£¬·´Ó¦C3H8(g)CH3CH=CH2(g)+H2(g)µÄ¡÷H=_____kJ¡¤mol-1

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

¾«Ó¢¼Ò½ÌÍø