¹ýÑõ»¯Çâ(H2O2)ÊÇÒ»ÖÖÎÞÉ«ð¤³íÒºÌå,ËüµÄË®ÈÜÒºË׳ÆË«ÑõË®,³ÊÈõËáÐÔ,³£ÓÃ×÷ÎÞ¹«º¦µÄÏû¶¾É±¾ú¼ÁºÍƯ°×¼ÁµÈ¡£
(1)ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ       ¡£

A£®¹ýÑõ»¯Çâ·Ö×ÓÖмÈÓм«ÐÔ¼üÓÖÓзǼ«ÐÔ¼ü
B£®H2O2ÓëH2O»¥ÎªÍ¬ËØÒìÐÎÌå
C£®34 g H2O2Öк¬ÓеÄÒõÀë×ÓÊýΪNA
D£®ÊµÑéÊÒ¿ÉÒÔÀûÓùýÑõ»¯ÇâÖÆÈ¡ÑõÆø
(2)H2O2ÊÇÒ»ÖÖ¶þÔªÈõËá,Çëд³öËüµÄµÚ¶þ²½µçÀë·½³Ìʽ:                   ¡£
(3)½«H2O2ÈÜÒº¼ÓÈëËáÐÔFeCl2ÈÜÒºÖÐ,ÈÜÒºÓÉdzÂÌÉ«±äΪ×Ø»ÆÉ«,д³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ:                                    ¡£
(4)ij³§¹¤Òµ·ÏË®Öк¬ÓÐÒ»¶¨Á¿ÂÈÆø,ΪÁ˳ýÈ¥ÂÈÆø,³£¼ÓÈëH2O2×÷ÍÑÂȼÁ,д³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ:                         ¡£

(1)AD
(2)H+H+
(3)H2O2+2Fe2++2H+=2Fe3++2H2O
(4)H2O2+Cl2=2HCl+O2¡£

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÐÛ»Æ(As4S4)ºÍ´Æ»Æ(As2S3)ÊÇÌáÈ¡ÉéµÄÖ÷Òª¿óÎïÔ­ÁÏ£¬¶þÕßÔÚ×ÔÈ»½çÖй²Éú¡£¸ù¾ÝÌâÒâÍê³ÉÏÂÁÐÌî¿Õ£º
(1)As2S3ºÍSnCl2ÔÚÑÎËáÖз´Ó¦×ª»¯ÎªAs4S4ºÍSnCl4²¢·Å³öH2SÆøÌå¡£ÈôAs2S3ºÍSnCl2ÕýºÃÍêÈ«·´Ó¦£¬As2S3ºÍSnCl2µÄÎïÖʵÄÁ¿Ö®±ÈΪ       ¡£
(2)ÉÏÊö·´Ó¦ÖеÄÑõ»¯¼ÁÊÇ       £¬·´Ó¦²úÉúµÄÆøÌå¿ÉÓà      ÎüÊÕ¡£
(3)As2S3ºÍHNO3ÓÐÈçÏ·´Ó¦£º
As2S3£«10H£«£«10NO3£­=2H3AsO4£«3S£«10NO2¡ü£«2H2OÈôÉú³É2 mol H3 AsO4£¬Ôò·´Ó¦ÖÐתÒƵç×ÓµÄÎïÖʵÄÁ¿Îª       ¡£Èô½«¸Ã·´Ó¦Éè¼Æ³ÉÒ»Ô­µç³Ø£¬ÔòNO2Ó¦¸ÃÔÚ       (Ìî¡°Õý¼«¡±»ò¡°¸º¼«¡±)¸½½üÒݳö¡£
(4)Èô·´Ó¦²úÎïNO2ºÍ11.2 L O2(±ê×¼×´¿ö)»ìºÏºóÓÃË®ÎüÊÕÈ«²¿×ª»¯³ÉŨHNO3£¬È»ºóÓë¹ýÁ¿µÄ̼·´Ó¦£¬Ëù²úÉúµÄCO2µÄÁ¿       (Ñ¡Ìî±àºÅ)¡£
a£®Ð¡ÓÚ0.5 mol        b£®µÈÓÚ0.5 mol
c£®´óÓÚ0.5 mol        d£®ÎÞ·¨È·¶¨

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÓÉ»ÆÍ­¿ó(Ö÷Òª³É·ÖÊÇCuFeS2)Á¶Öƾ«Í­µÄ¹¤ÒÕÁ÷³ÌʾÒâͼÈçÏ£º

£¨1£©ÔÚ·´Éä¯ÖУ¬°ÑÍ­¾«¿óÉ°ºÍʯӢɰ»ìºÏ¼ÓÈȵ½1000¡æ×óÓÒ£¬»ÆÍ­¿óÓë¿ÕÆø·´Ó¦Éú³ÉCuºÍFeµÄµÍ¼ÛÁò»¯ÎÁ½ÖֵͼÛÁò»¯ÎïµÄ»¯Ñ§Ê½·Ö±ðΪ________¡¢_______¡£ÔÚ·´Ó¦¹ý³ÌÖл¹ÓÐÒ»²¿·ÖFeµÄÁò»¯Îïת»¯ÎªµÍ¼ÛÑõ»¯ÎÆ仯ѧ·´Ó¦·½³ÌʽΪ____________¡£
£¨2£©±ùÍ­(Cu2SºÍFeS»¥ÏàÈۺ϶ø³É)º¬CuÁ¿½ÏµÍ¡£×ªÂ¯ÖУ¬½«±ùÍ­¼ÓÈÛ¼Á(ʯӢɰ)ÔÚ1200¡æ×óÓÒ´µÈë¿ÕÆø½øÐдµÁ¶¡£±ùÍ­ÖеÄCu2S±»Ñõ»¯³ÉCu2O£¬£¬Ã¿ÓÐ1molÑõÆø²Î¼Ó·´Ó¦£¬Éú³ÉÑõ»¯²úÎïµÄÎïÖʵÄÁ¿Îª_________¡£Éú³ÉµÄCu2OÓëCu2S·´Ó¦£¬¿ÉÖƵú¬CuÁ¿½Ï¸ßµÄ´ÖÍ­¡£
£¨3£©´ÖÍ­µÄµç½â¾«Á¶ÈçͼËùʾ¡£ÔÚ´ÖÍ­µÄµç½â¹ý³ÌÖУ¬cΪ´ÖÍ­°å£¬ Ôòa¶ËÓ¦Á¬½ÓµçÔ´µÄ_____¼«(Ìî¡°Õý¡±»ò¡°¸º¡±)£¬Èô´ÖÍ­Öк¬ÓÐAu¡¢Ag¡¢FeÔÓÖÊ£¬Ôòµç½â¹ý³ÌÖÐcµç¼«ÉÏ·¢Éú·´Ó¦µÄ·½³ÌʽÓÐ__________¡£

£¨4£©¿ÉÒÔÓ¦ÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒºµÎ¶¨·¨²â¶¨·´Ó¦ºóµç½âÒºÖÐÌúÔªËصĺ¬Á¿¡£µÎ¶¨Ê±²»ÄÜÓüîʽµÎ¶¨¹ÜÊ¢·ÅËáÐÔ¸ßÃÌËá¼ØÈÜÒºµÄÔ­ÒòÊÇ____________£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________£¬µÎ¶¨Ê±£¬×¶ÐÎÆ¿ÖеÄÈÜÒº»á½Ó´¥¿ÕÆø£¬²âµÃÌúÔªËصĺ¬Á¿»á____(Ìî¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

µç¶Æ³§¶ÆÍ­·ÏË®Öк¬ÓÐCN£­ºÍCr2O72-£¬ÐèÒª´¦Àí´ï±êºó²ÅÄÜÅÅ·Å¡£¸Ã³§ÄⶨÏÂÁÐÁ÷³Ì½øÐзÏË®´¦Àí£º

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÉÏÊö´¦Àí·ÏË®Á÷³ÌÖÐÖ÷ҪʹÓõķ½·¨ÊÇ________¡£
£¨2£©¢ÚÖз´Ó¦ºóÎÞÆøÌå·Å³ö£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________________
__________________________________¡£
£¨3£©²½Öè¢ÛÖУ¬Ã¿´¦Àí0.4 mol Cr2O72-ʱתÒƵç×Ó2.4 mol£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ________________________________________________________________________¡£
£¨4£©È¡ÉÙÁ¿´ý²âË®ÑùÓÚÊÔ¹ÜÖУ¬¼ÓÈëNaOHÈÜÒº£¬¹Û²ìµ½ÓÐÀ¶É«³ÁµíÉú³É£¬ÔÙ¼ÓNa2SÈÜÒº£¬À¶É«³Áµíת»¯³ÉºÚÉ«³Áµí£¬ÇëʹÓû¯Ñ§ÓÃÓïºÍÎÄ×Ö½âÊͲúÉú¸ÃÏÖÏóµÄÔ­Òò£º
________________________________________________________________________¡£
£¨5£©Ä¿Ç°´¦ÀíËáÐÔCr2O72-·ÏË®¶à²ÉÓÃÌúÑõ´ÅÌå·¨¡£¸Ã·¨ÊÇÏò·ÏË®ÖмÓÈëFeSO4¡¤7H2O£¬½«Cr2O72-»¹Ô­³ÉCr3£«£¬µ÷½ÚpH£¬Fe¡¢Crת»¯³ÉÏ൱ÓÚFe¢ò[FeCr]O4(ÌúÑõ´ÅÌ壬ÂÞÂíÊý×Ö±íʾԪËؼÛ̬)µÄ³Áµí¡£´¦Àí1 mol Cr2O72-£¬Ðè¼ÓÈëa mol FeSO4¡¤7H2O£¬ÏÂÁнáÂÛÕýÈ·µÄÊÇ________¡£

A£®x£½0.5£¬a£½8¡¡¡¡¡¡¡¡ B£®x£½0.5£¬a£½10
C£®x£½1.5£¬a£½8 D£®x£½1.5£¬a£½10

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(13·Ö)ÌúºÍÌúµÄ»¯ºÏÎïÔÚ¹¤ÒµÉú²úºÍÈÕ³£Éú»îÖж¼Óй㷺µÄÓÃ;¡£
£¨1£©ÔÚ¶¨Ïò±¬ÆÆÖУ¬³£ÀûÓÃÑõ»¯ÌúÓëÂÁ·´Ó¦·Å³öµÄÈÈÁ¿À´Çиî¸Ö½î£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ß£ß¡£
£¨2£©ÒÑÖª£º2Fe2O3(s)£«3C(s)£½3CO2(g)£«4Fe(s) ¡÷H£½+468.2 kJ¡¤mol-1
C(s)+O2(g)£½CO2(g) ¡÷H="-393.5" kJ¡¤mol-1¡£
ÔòFe(s)ÓëO2 (g)·´Ó¦Éú³ÉFe2 O3 (s)µÄÈÈ»¯Ñ§·½³ÌʽΪ£ß_____________________¡£
£¨3£©¿ÉÓÃKMnO4ÈÜÒºµÎ¶¨Fe2+µÄŨ¶È£¬·´Ó¦µÄÀë×Ó·½³ÌʽÈçÏ£º5Fe2£«£«MnO4£­£«8H£«£½5Fe3£«£«Mn2£«£«4H2O
¢ÙKMnO4ÈÜҺӦʢ·ÅÔڣߣߣߣߣߵζ¨¹ÜÖУ»
¢ÚÅжϴﵽµÎ¶¨ÖÕµãµÄÏÖÏóÊǣߣߣߣߣߣ»
¢ÛÓÃÁòËáËữµÄ0.020 00 mol¡¤L-1¡£KMnO4ÈÜÒºµÎ¶¨Ä³FeSO4ÈÜÒºÖÁÖյ㣬ʵÑéÊý¾Ý¼Ç¼ÈçÏÂ±í£º

Çë·ÖÎöÊý¾Ý²¢¼ÆË㣬¸ÃFeSO4ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ£ß£ß£ß£ß£ß¡£
£¨4£©ÐÂÐÍÄÉÃײÄÁÏZnFe2Ox£¬¿ÉÓÃÓÚ³ýÈ¥¹¤Òµ·ÏÆøÖеÄijЩÑõ»¯Îï¡£ÖÆȡвÄÁϺͳýÈ¥·ÏÆøµÄת»¯¹ØϵÈçÏÂͼ£º

¢ÙÒÑÖªZnFe2O4ÓëH2·´Ó¦µÄÎïÖʵÄÁ¿Ö®±ÈΪ2:1£¬ÔòZnFe2OxÖÐx=£ß£ß£ß£ß£ß£»
¢ÚÓÃZnFe2Ox³ýÈ¥SO2µÄ¹ý³ÌÖУ¬Ñõ»¯¼ÁÊǣߣߣߣߣߡ£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

³ôÑõ¿ÉÓÃÓÚ¾»»¯¿ÕÆø¡¢ÒûÓÃË®µÄÏû¶¾¡¢´¦Àí¹¤Òµ·ÏÎïºÍ×÷Ư°×¼Á¡£
£¨1£©³ôÑõ¼¸ºõ¿ÉÓë³ý²¬¡¢½ð¡¢Ò¿¡¢·úÒÔÍâµÄËùÓе¥ÖÊ·´Ó¦¡£È磺6Ag£¨s£©£«O3£¨g£©=3Ag2O£¨s£©¡¡¦¤H£½£­235.8 kJ¡¤mol£­1¡£ÒÑÖª2Ag2O£¨s£©=4Ag£¨s£©£«O2£¨g£©  ¦¤H£½£«62.2 kJ¡¤mol£­1£¬Ôò³£ÎÂÏ·´Ó¦2O3£¨g£©=3O2£¨g£©µÄ¦¤H£½________¡£
£¨2£©ÅäƽÏÂÃæ·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¨½«¸÷ÎïÖʵĻ¯Ñ§¼ÆÁ¿ÊýÌîÔÚÏàÓ¦µÄ·½¿òÄÚ£©£º
 
£¨3£©¿Æѧ¼ÒP.TatapudiµÈÈËÊ×ÏÈʹÓÃÔÚËáÐÔÌõ¼þϵç½âË®µÄ·½·¨ÖƵóôÑõ¡£³ôÑõÔÚÑô¼«ÖÜΧµÄË®ÖвúÉú£¬µç¼«·´Ó¦Ê½Îª3H2O£­6e£­=O3¡ü£«6H£«£¬Òõ¼«¸½½üÈܽâÔÚË®ÖеÄÑõÆøÉú³É¹ýÑõ»¯Ç⣬Æäµç¼«·´Ó¦Ê½Îª_______________________¡£
£¨4£©¿ÕÆøÖгôÑõµÄ¼ì²â·½·¨Êǽ«¿ÕÆøÂýÂýͨ¹ý×ãÁ¿KIµí·ÛÈÜÒº£¬ÈôÈÜÒº±äÀ¶É«£¬Ôò˵Ã÷¿ÕÆøÖк¬ÓÐO3¡£ÒÑÖªO3ÓëKIÈÜÒº·´Ó¦Éú³ÉÁ½ÖÖµ¥ÖÊ£¬Ôò¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©Ç뽫ÏÂÁÐÎåÖÖÎïÖÊ£ºKBr¡¢Br2¡¢I2¡¢KI¡¢K2SO4·Ö±ðÌîÈëÏÂÁкáÏßÉÏ£¬×é³ÉÒ»¸öδÅäƽµÄ»¯Ñ§·½³Ìʽ£º
KBrO3£«________£«H2SO4¨D¡ú________£«________£«________£«________£«H2O¡£
£¨2£©Èç¹û¸Ã»¯Ñ§·½³ÌʽÖÐI2ºÍKBrµÄ»¯Ñ§¼ÆÁ¿Êý·Ö±ðÊÇ8ºÍ1£¬Ôò
¢ÙBr2µÄ»¯Ñ§¼ÆÁ¿ÊýÊÇ________£»
¢ÚÇ뽫·´Ó¦ÎïµÄ»¯Ñ§Ê½¼°ÅäƽºóµÄ»¯Ñ§¼ÆÁ¿ÊýÌîÈëÏÂÁÐÏàÓ¦µÄλÖÃÖУº
________KBrO3£«________£«________H2SO4¨D¡ú¡­¡­£»
¢ÛÈôתÒÆ10 molµç×Ó£¬Ôò·´Ó¦ºóÉú³ÉI2µÄÎïÖʵÄÁ¿Îª________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ç⻯ÑÇÍ­(CuH)ÊÇÒ»ÖÖÄÑÈÜÎïÖÊ£¬ÓÃCuSO4ÈÜÒººÍ¡°ÁíÒ»ÎïÖÊ¡±ÔÚ40¡«50¡æʱ·´Ó¦¿ÉÉú³ÉËü¡£CuH¾ßÓеÄÐÔÖÊÓУº²»Îȶ¨£¬Ò׷ֽ⣻ÔÚÂÈÆøÖÐÄÜȼÉÕ£»ÓëÏ¡ÑÎËá·´Ó¦ÄÜÉú³ÉÆøÌ壻Cu£«ÔÚËáÐÔÌõ¼þÏ·¢ÉúµÄ·´Ó¦ÊÇ£º2Cu£«=Cu2£«£«Cu¡£
¸ù¾ÝÒÔÉÏÐÅÏ¢£¬½áºÏ×Ô¼ºËùÕÆÎյĻ¯Ñ§ÖªÊ¶£¬»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÓÃCuSO4ÈÜÒººÍ¡°ÁíÒ»ÎïÖÊ¡±ÖÆCuHµÄ·´Ó¦ÖУ¬ÓÃÑõ»¯»¹Ô­¹Ûµã·ÖÎö£¬Õâ¡°ÁíÒ»ÎïÖÊ¡±ÔÚ·´Ó¦ÖÐ×÷______(Ìî¡°Ñõ»¯¼Á¡±»ò¡°»¹Ô­¼Á¡±)¡£
(2)д³öCuHÔÚÂÈÆøÖÐȼÉյĻ¯Ñ§·´Ó¦·½³Ìʽ£º________________________¡£
(3)CuHÈܽâÔÚÏ¡ÑÎËáÖÐÉú³ÉµÄÆøÌåÊÇ______(Ìѧʽ)¡£
(4)Èç¹û°ÑCuHÈܽâÔÚ×ãÁ¿µÄÏ¡ÏõËáÖÐÉú³ÉµÄÆøÌåÖ»ÓÐNO£¬Çëд³öCuHÈܽâÔÚ×ãÁ¿Ï¡ÏõËáÖз´Ó¦µÄÀë×Ó·½³Ìʽ£º_____________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¡°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

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸