¡¾ÌâÄ¿¡¿Ãº·ÛÖеĵªÔªËØÔÚʹÓùý³ÌÖеÄת»¯¹ØϵÈçͼËùʾ£º
(1)¢ÚÖÐNH3²ÎÓë·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______¡£
(2)½¹Ì¿µªÖÐÓÐÒ»ÖÖ³£¼ûµÄº¬µªÓлúÎïßÁà¤()£¬Æä·Ö×ÓÖÐÏàÁÚµÄCºÍNÔ×ÓÏà±È£¬NÔ×ÓÎüÒýµç×ÓÄÜÁ¦¸ü___________(Ìî¡°Ç¿¡±»ò¡°Èõ¡±)£¬´ÓÔ×ӽṹ½Ç¶È½âÊÍÔÒò£º________¡£
(3)¹¤ÒµºÏ³É°±ÊÇÈ˹¤¹ÌµªµÄÖØÒª·½·¨¡£2007Ä껯ѧ¼Ò¸ñ¹þµÂ¡¤°£Ìضû֤ʵÁËÇâÆøÓ뵪ÆøÔÚ¹ÌÌå´ß»¯¼Á±íÃæºÏ³É°±µÄ·´Ó¦¹ý³Ì£¬Ê¾ÒâÈçͼ£º
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________(Ñ¡Ìî×Öĸ)¡£
a. ͼ¢Ù±íʾN2¡¢H2·Ö×ÓÖоùÊǵ¥¼ü
b. ͼ¢Ú¡úͼ¢ÛÐèÒªÎüÊÕÄÜÁ¿
c. ¸Ã¹ý³Ì±íʾÁË»¯Ñ§±ä»¯Öаüº¬¾É»¯Ñ§¼üµÄ¶ÏÁѺÍл¯Ñ§¼üµÄÉú³É
(4)ÒÑÖª£ºN2(g) £« O2(g) = 2NO(g) ¦¤H = a kJ¡¤mol-1
N2(g) £« 3H2(g) = 2NH3(g) ¦¤H = b kJ¡¤mol-1
2H2(g) £« O2(g) = 2H2O(l) ¦¤H = c kJ¡¤mol-1
·´Ó¦ºó»Ö¸´ÖÁ³£Î³£Ñ¹£¬¢ÙÖÐNH3²ÎÓë·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ________¡£
(5)Óüä½Óµç»¯Ñ§·¨³ýÈ¥NOµÄ¹ý³Ì£¬ÈçͼËùʾ£º
¢ÙÒÑÖªµç½â³ØµÄÒõ¼«ÊÒÖÐÈÜÒºµÄpHÔÚ4~7Ö®¼ä£¬Ð´³öÒõ¼«µÄµç¼«·´Ó¦Ê½£º________¡£
¢ÚÓÃÀë×Ó·½³Ìʽ±íʾÎüÊÕ³ØÖгýÈ¥NOµÄÔÀí£º__________¡£
¡¾´ð°¸¡¿4NH3+5O24NO+6H2O Ç¿ CºÍNÔ×ÓÔÚͬһÖÜÆÚ(»òµç×Ó²ãÊýÏàͬ)£¬NÔ×Ӻ˵çºÉÊý¸ü´ó£¬Ô×Ӱ뾶¸üС£¬Ô×Ӻ˶ÔÍâ²ãµç×ÓµÄÎüÒýÁ¦¸üÇ¿ bc 4NH3(g) + 6NO(g) = 5N2(g) + 6H2O(l) ¦¤H = (3c-3a-2b) kJ¡¤mol-1 2HSO3- + 2e- + 2H+ = S2O42- + 2H2O 2NO + 2S2O42- +2H2O = N2 + 4HSO3-
¡¾½âÎö¡¿
(1)°±ÆøÔÚ´ß»¯¼ÁÌõ¼þÏÂÓëÑõÆø·´Ó¦Éú³ÉÒ»Ñõ»¯µªºÍË®£¬ÎªÖØÒªµÄ¹¤Òµ·´Ó¦£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ4NH3+5O24NO+6H2O£»
(2)ÓÉÓÚCºÍNÔ×ÓÔÚͬһÖÜÆÚ(»òµç×Ó²ãÊýÏàͬ)£¬NÔ×Ӻ˵çºÉÊý¸ü´ó£¬Ô×Ӱ뾶¸üС£¬Ô×Ӻ˶ÔÍâ²ãµç×ÓµÄÎüÒýÁ¦¸üÇ¿£¬ËùÒÔNÔ×ÓÎüÒýµç×ÓÄÜÁ¦¸üÇ¿£»
(3)a£®µªÆøÖÐÁ½¸öµªÔ×ÓÖ®¼äΪÈý¼ü£¬¹Êa´íÎó£»
b£®·ÖÎöÌâÖÐͼ¿ÉÒÔÖªµÀ£¬Í¼¢Ú±íʾN2¡¢H2±»Îü¸½ÔÚ´ß»¯¼Á±íÃ棬¶øͼ¢Û±íʾÔÚ´ß»¯¼Á±íÃ棬N2¡¢H2Öл¯Ñ§¼ü¶ÏÁÑ£¬¶Ï¼üÎüÊÕÄÜÁ¿£¬ËùÒÔͼ¢Ú¡úͼ¢ÛÐèÒªÎüÊÕÄÜÁ¿£¬¹ÊbÕýÈ·£»
c£®ÔÚ»¯Ñ§±ä»¯ÖУ¬µª·Ö×ÓºÍÇâ·Ö×ÓÔÚ´ß»¯¼ÁµÄ×÷ÓÃ϶ÏÁѳÉÇâÔ×Ӻ͵ªÔ×Ó£¬·¢Éú»¯Ñ§¼üµÄ¶ÏÁÑ£¬È»ºóÔ×ÓÓÖÖØÐÂ×éºÏ³ÉеķÖ×Ó£¬ÐγÉеĻ¯Ñ§¼ü£¬ËùÒԸùý³Ì±íʾÁË»¯Ñ§±ä»¯Öаüº¬¾É»¯Ñ§¼üµÄ¶ÏÁѺÍл¯Ñ§¼üµÄÉú³É£¬¹ÊcÕýÈ·£»
´ð°¸Ñ¡bc¡£
(4)¢ÙÖÐNH3²ÎÓëµÄ·´Ó¦Îª£º4NH3(g) + 6NO(g) = 5N2(g) + 6H2O(l)£»
ÒÑÖª£ºN2(g) £« O2(g) = 2NO(g) ¦¤H = a kJ¡¤mol-1 i£»
N2(g) £« 3H2(g) = 2NH3(g) ¦¤H = b kJ¡¤mol-1 ii£»
2H2(g) £« O2(g) = 2H2O(l) ¦¤H = c kJ¡¤mol-1 iii£»
¸ù¾Ý¸Ç˹¶¨ÂÉiii¡Á3- i¡Á3-ii¡Á2¿ÉµÃ4NH3(g) + 6NO(g) = 5N2(g) + 6H2O(l) ¦¤H=(3c-3a-2b)kJ¡¤mol-1£»
(5)¢ÙÒõ¼«·¢Éú»¹Ô·´Ó¦£¬¾Ýͼ¿ÉÖªÑÇÁòËáÇâ¸ùÀë×ӵõç×Ó±»»¹ÔÉú³ÉS2O42-£¬µç½âÖÊÈÜÒºÏÔÈõËáÐÔ£¬ËùÒԵ缫·´Ó¦Ê½Îª£º2HSO3-+2e-+2H+=S2O42-+2H2O£»
¢Ú¾Ýͼ¿ÉÖªS2O42-ÓëÒ»Ñõ»¯µª·¢ÉúÑõ»¯»¹Ô·´Ó¦£¬Éú³ÉµªÆøºÍÑÇÁòËáÇâ¸ù£¬¸ù¾ÝµÃʧµç×ÓÊغ㡢Ô×ÓÊغãºÍµçºÉÊغ㣬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2NO+2S2O42-+2H2O=N2+4HSO3-¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ò»¶¨Î¶Èʱ£¬Ïò2.0 LºãÈÝÃܱÕÈÝÆ÷ÖгäÈë2 mol SO2ºÍ1 mol O2£¬·¢Éú·´Ó¦£º2SO2(g)£«O2(g)2SO3(g)¡£¾¹ýÒ»¶Îʱ¼äºó´ïµ½Æ½ºâ¡£·´Ó¦¹ý³ÌÖвⶨµÄ²¿·ÖÊý¾Ý¼ûÏÂ±í£º
t / s | 0 | 2 | 4 | 6 | 8 |
n(SO3) / mol | 0 | 0£®8 | 1£®4 | 1.8 | 1.8 |
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ( )
A. ·´Ó¦ÔÚÇ°2 s µÄƽ¾ùËÙÂÊv(O2) £½ 0£®4 mol¡¤L£1¡¤s£1
B. ±£³ÖÆäËûÌõ¼þ²»±ä£¬Ìå»ýѹËõµ½1.0 L£¬Æ½ºâ³£Êý½«Ôö´ó
C. ÏàͬζÈÏ£¬ÆðʼʱÏòÈÝÆ÷ÖгäÈë4 mol SO3£¬´ïµ½Æ½ºâʱ£¬SO3µÄת»¯ÂÊСÓÚ10%
D. ±£³ÖζȲ»±ä£¬Ïò¸ÃÈÝÆ÷ÖÐÔÙ³äÈë2 mol SO2¡¢1 mol O2£¬·´Ó¦´ïµ½ÐÂƽºâʱn(SO3)/n(O2)¼õС
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁнâÊÍÊÂʵµÄ·½³ÌʽÕýÈ·µÄÊÇ
A.¹¤ÒµÒ±Á¶ÂÈ»¯ÂÁÖÆÂȵ¥ÖÊ£º2AlCl3(ÈÛÈÚ)2Al+3Cl2¡ü
B.ÏòÂÈ»¯ÂÁÈÜÒºÖмÓÈë¹ýÁ¿°±Ë®£¬²úÉú°×É«³Áµí£ºAl3+ + 3OH- = Al(OH)3¡ý
C.½«Ìú·ÛÓëË®ÕôÆø¹²ÈÈ£¬²úÉúÆøÌ壺2Fe£«3H2O(g)Fe2O3£«3H2
D.µç½â±¥ºÍÂÈ»¯ÄÆÈÜÒº£¬²úÉúÆøÌ壺2NaCl+2H2O2NaOH£«H2¡ü£«Cl2¡ü
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿HI³£ÓÃ×÷Óлú·´Ó¦ÖеĻ¹Ô¼Á£¬ÊÜÈȻᷢÉú·Ö½â·´Ó¦¡£ÒÑ֪ʱ£º£¬Ïò1LÃܱÕÈÝÆ÷ÖгäÈë1molHI£¬Ê±£¬ÌåϵÖÐÓ뷴Ӧʱ¼ätµÄ¹ØϵÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÖУ¬ÕýÈ·µÄÊÇ
A.minÄÚµÄƽ¾ù·´Ó¦ËÙÂʿɱíʾΪ
B.Éý¸ßζȣ¬ÔÙ´Îƽºâʱ£¬
C.¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý¼ÆËãʽΪ
D.·´Ó¦½øÐÐ40minʱ£¬ÌåϵÎüÊÕµÄÈÈÁ¿Ô¼ÎªkJ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¶ÁÏÂÁÐÒ©Æ·±êÇ©£¬ÓйطÖÎö²»ÕýÈ·µÄÊÇ£¨¡¡ ¡¡£©
Ñ¡Ïî | A | B | C | D |
ÎïÆ·±êÇ© | ±¥ºÍÂÈË®1.01¡Á105 Pa£¬20 ¡æ | Ò©Æ·£º¡Á¡Á¡Á | ̼ËáÇâÄÆNaHCO3Ë×ÃûСËÕ´ò£¨84 g¡¤mol£1£© | ŨÁòËáH2SO4 ÃܶÈ1.84 g¡¤mL£1Ũ¶È98.0% |
·ÖÎö | ¸ÃÊÔ¼ÁӦװÔÚÏð½ºÈûµÄϸ¿ÚÆ¿ÖÐ | ¸ÃÒ©Æ·²»ÄÜÓëƤ·ôÖ±½Ó½Ó´¥ | ¸ÃÎïÖÊÊÜÈÈÒ×·Ö½â | ¸ÃÒ©Æ·±êÇ©ÉÏ»¹±êÓÐ |
A. AB. BC. CD. D
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿·ûºÏÏÂͼËùʾÌõ¼þµÄÀë×Ó×éÊÇ£¨¡¡ ¡¡£©
A. Ba2£«¡¢Mg2£«¡¢NO3-¡¢CO32-B. H£«¡¢Ba2£«¡¢Al3£«¡¢Cl£
C. K£«¡¢Ba2£«¡¢Cl£¡¢HCO3-D. NH4+¡¢Ba2£«¡¢Fe2£«¡¢Cl£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÈýÂÈ»¯Áù°±ºÏîÜ[Co(NH3)6]Cl3 ÊdzȻÆÉ«¡¢Î¢ÈÜÓÚË®µÄÅäºÏÎÊǺϳÉÆäËüһЩº¬îÜÅäºÏÎïµÄÔÁÏ¡£ÏÂͼÊÇij¿ÆÑÐС×éÒÔº¬îÜ·ÏÁÏ£¨º¬ÉÙÁ¿Fe¡¢Al µÈÔÓÖÊ£©ÖÆÈ¡[Co(NH3)6]Cl3 µÄ¹¤ÒÕÁ÷³Ì£º
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³ö¼Ó¡°ÊÊÁ¿NaClO3¡±·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ______________¡£
£¨2£©¡°¼ÓNa2CO3 µ÷pHÖÁa¡±»áÉú³ÉÁ½ÖÖ³Áµí£¬·Ö±ðΪ_______________________£¨Ìѧʽ£©¡£
£¨3£©²Ù×÷¢ñµÄ²½Öè°üÀ¨_____________________________¡¢ÀäÈ´½á¾§¡¢¼õѹ¹ýÂË¡£
£¨4£©Á÷³ÌÖÐNH4Cl³ý×÷·´Ó¦ÎïÍ⣬»¹¿É·ÀÖ¹¼Ó°±Ë®Ê±c(OH£) ¹ý´ó£¬ÆäÔÀíÊÇ_________________¡£
£¨5£©¡°Ñõ»¯¡±²½Ö裬¼×ͬѧÈÏΪӦÏȼÓÈ백ˮÔÙ¼ÓÈëH2O2£¬ÒÒͬѧÈÏΪÊÔ¼ÁÌí¼Ó˳Ðò¶Ô²úÎïÎÞÓ°Ïì¡£ÄãÈÏΪ___________£¨Ìî¡°¼×¡±»ò¡°ÒÒ¡±£©Í¬Ñ§¹ÛµãÕýÈ·£¬ÀíÓÉÊÇ_________________________________¡£Ð´³ö¸Ã²½ÖèµÄ»¯Ñ§·½³Ìʽ£º________________________________
£¨6£©Í¨¹ýµâÁ¿·¨¿É²â¶¨²úÆ·ÖÐîܵĺ¬Á¿¡£½« [Co(NH3)6]Cl3 ת»¯³ÉCo3£«ºó£¬¼ÓÈë¹ýÁ¿KI ÈÜÒº£¬ÔÙÓÃNa2S2O3±ê×¼ÒºµÎ¶¨(µí·ÛÈÜÒº×öָʾ¼Á)£¬·´Ó¦ÔÀí£º2Co3£«£«2I££½2Co2£«£«I2£¬I2£«2S2O32££½2I££«S4O62££¬ÊµÑé¹ý³ÌÖУ¬ÏÂÁвÙ×÷»áµ¼ÖÂËù²âîܺ¬Á¿Êýֵƫ¸ßµÄÊÇ_______¡£
a£®ÓþÃÖÃÓÚ¿ÕÆøÖÐµÄ KI ¹ÌÌåÅäÖÆÈÜÒº
b£®Ê¢×°Na2S2O3±ê×¼ÒºµÄ¼îʽµÎ¶¨¹ÜδÈóÏ´
c£®µÎ¶¨½áÊøºó£¬·¢Ïֵζ¨¹ÜÄÚÓÐÆøÅÝ
d£®ÈÜÒºÀ¶É«ÍËÈ¥£¬Á¢¼´¶ÁÊý
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÁòËáпÊÇÖÆÔìп±µ°×ºÍпÑεÄÖ÷ÒªÔÁÏ£¬Ò²¿ÉÓÃÓÚʪ·¨µç½âÖÆп£¬ÈçͼΪÓÉп»ÒÖÆZnSO4¡¤7H2O¾§ÌåµÄ¹¤ÒÕÁ÷³Ì¡£
ÒÑÖª£º
¢Ùп»ÒµÄÖ÷Òª³É·ÖΪZnO£¬»¹º¬ÓÐCuO¡¢PbO¡¢MnOºÍFeO£»
¢Ú¡°ÂËÔü2¡±µÄÖ÷Òª³É·ÖΪFe(OH)3ºÍMnO(OH)2¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©MnO(OH)2ÖÐMnÔªËصĻ¯ºÏ¼ÛΪ___¡£
£¨2£©ÎªÌá¸ß½þ³öЧÂÊ£¬Ð¿»ÒÔÚ¡°Ëá½þ¡±Ç°¿É²ÉÈ¡µÄ´ëÊ©ÓÐ___£»¡°Ëá½þ¡±Ê±£¬ÈôÁòËáŨ¶È¹ý¸ß£¬¿ÉÄÜ·¢Éú¸±·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___¡£
£¨3£©¡°ÂËÔü1¡±µÄÖ÷Òª³É·ÖΪ___¡£
£¨4£©¡°Ñõ»¯¡±Ê±£¬Ðè¿ØÖÆÈÜÒºµÄpH=5.1£¬Fe2£«±»Ñõ»¯µÄÀë×Ó·½³ÌʽΪ___¡£
£¨5£©²Ù×÷aΪ___¡¢¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡£
£¨6£©ZnSO4¡¤7H2O²úÆ·µÄ´¿¶È¿ÉÓÃÅäλµÎ¶¨·¨²â¶¨¡£
׼ȷ³ÆÈ¡Ò»¶¨Á¿µÄZnSO4¡¤7H2O¾§Ìå¼ÓÈë250mLµÄ׶ÐÎÆ¿ÖУ¬¼ÓˮԼ20mL£¬ÔÙ¼ÓÈë2-3µÎ5%µÄ¶þ¼×·Ó³È×÷ָʾ¼Á¡¢Ô¼5mLÁùÑǼ׻ùËÄ°·»º³åÈÜÒº£¬Ò¡ÔÈ¡£ÓÃÒѱ궨µÄ0.0160mol¡¤L-1EDTAÈÜÒºµÎ¶¨£¬µÎ¶¨ÖÁÈÜÒºÓɺì×ÏÉ«±ä³ÉÁÁ»ÆÉ«£¬¼´ÎªÖÕµã(ZnSO4¡¤7H2OÓëEDTA°´ÎïÖʵÄÁ¿Ö®±È1£º1·´Ó¦)¡£ÊµÑéÊý¾ÝÈçÏÂ±í£º
ZnSO4¡¤7H2O²úÆ·µÄ´¿¶ÈΪ___ (±£Áô2λÓÐЧÊý×Ö)¡£
£¨7£©¹¤ÒµÉϲÉÓöèÐԵ缫×÷Ñô¼«µç½âZnSO4ÈÜÒº¿ÉʵÏÖʪ·¨Á¶Ð¿£¬µç½â¹ý³ÌÖеÄÀë×Ó·½³ÌʽΪ___¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÓлúÎïAÊÇÌìÈ»Ï𽺵ĵ¥Ìå¡£ÓлúÎïFÊÇÒ»ÖÖ»·±£ÐÍËÜÁÏ£¬Ò»ÖֺϳÉ·ÏßÈçͼËùʾ£º
ÒÑÖª£º
¢Ù
¢Ú+R3-COOH£¨¡ªR1¡¢¡ªR2¡¢¡ªR3¡¢¡ªR¾ùΪÌþ»ù£©
Çë»ØÈÝÏÂÁÐÎÊÌ⣺
(1)AÓÃϵͳÃüÃû·¨ÃüÃûΪ__________________________¡£
(2)A·Ö×ÓÓëCl2·Ö×Ó·¢Éú1£º1¼Ó³Éʱ£¬Æä²úÎïÖÖÀàÓÐ____________ÖÖ(²»¿¼ÂÇÁ¢ÌåÒì¹¹)¡£
(3)BÓë×ãÁ¿H2·´Ó¦ºóµÄ²úÎïµÄ½á¹¹¼òʽΪ £¬ÔòBµÄ½á¹¹¼òʽΪ___________________£¬1¸öB·Ö×ÓÖÐÓÐ________¸öÊÖÐÔ̼Ô×Ó¡£
(4)C·Ö×ÓÖеĹÙÄÜÍÅÃû³ÆΪ_____________________¡£
(5)д³ö·´Ó¦¢ÜµÄ»¯Ñ§·½³Ìʽ£º______________________________________________________¡£
(6)GÊÇCµÄͬ·ÖÒì¹¹Ì壬GÄÜ·¢ÉúË®½â·´Ó¦ºÍÒø¾µ·´Ó¦£¬1¸öG·Ö×ÓÖк¬ÓÐ2¸ö̼ÑõË«¼ü£¬ÔòGµÄ¿ÉÄܽṹ¹²ÓÐ___________ÖÖ(²»¿¼ÂÇÁ¢ÌåÒì¹¹)¡£
(7)Éè¼ÆÓÉÓлúÎïDºÍ¼×´¼ÎªÆðʼÔÁÏÖƱ¸CH3CH2CH(CH3)CH =CHCOOCH3µÄºÏ³É·Ïß______________(ÎÞ»úÊÔ¼ÁÈÎÑ¡)¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com