¡¾ÌâÄ¿¡¿ÒÑÖª£º
(1)д³öÏÂÁз´Ó¦²úÎïµÄ½á¹¹¼òʽ£º
H2C=CHCH=CHCH3+H2C=CHCHO¡ú _______________¡£
(2)ÒÔijÁ´ÌþAΪÆðʼÔÁϺϳɻ¯ºÏÎïG µÄ·¾¶ÈçÏÂ(ͼÖÐ Mr ±íʾÏà¶Ô·Ö×ÓÖÊÁ¿)
¢Ùд³ö·´Ó¦ÀàÐÍ B¡úC£º_______£¬F¡úG£º_______¡£
¢Úд³öÏÂÁÐÎïÖʵĽṹ¼òʽ£ºA£º________£¬F£º_______¡£
¢Ûд³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
B¡úC£º________________£»D¡úE£º_______________¡£
¢Üд³ö G ÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ·½³Ìʽ____________________¡£
¡¾´ð°¸¡¿
(1)»ò£»
(2)¢ÙË®½â·´Ó¦(È¡´ú·´Ó¦)£»¼Ó³É·´Ó¦£»
£»
¢ÛBrCH2-C(CH3)=CH-CH2Br+2NaOHCH2(OH)-C(CH3)=CH-CH2(OH)+2NaBr£»
HOOC-C(CH3)=CH-COOH+2CH3CH2OHH5C2OOC-C(CH3)=CH-COOC2H5+2H2O
¢Ü
¡¾½âÎö¡¿
ÊÔÌâ·ÖÎö£º(1)¸ù¾ÝÐÅÏ¢£¬H2C=CHCH=CHCH3+H2C=CHCHO¡ú»ò£¬¹Ê´ð°¸Îª£º»ò£»
(2)ijÁ´ÌþAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª68£¬Ôò·Ö×ÓÖÐ̼Ô×Ó×î´óÊýĿΪ=5¡8£¬ËùÒÔ¸ÃÌþµÄ·Ö×ÓʽΪC5H8£¬Æä²»±¥ºÍ¶È==2£¬Ôò¸ÃÌþÊÇȲÌþ»ò¶þÏ©Ìþ£»AºÍäå·¢Éú¼Ó³É·´Ó¦Éú³ÉB£¬¸ù¾ÝAºÍBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Öª£¬AºÍäåÊÇ1£º1¼Ó³É£¬ÔòAÊǶþÏ©Ìþ£¬Á½¸öË«¼üͬʱÁ¬ÔÚÒ»¸ö̼Ô×ÓÉϲ»ÊÇÎȶ¨½á¹¹£¬ÔòAµÄ½á¹¹¼òʽΪ£º£¬Bͨ¹ýË®½â·´Ó¦£¬äåÔ×Ó±»È¡´úÒýÈë2¸öôÇ»ùÉú³ÉC£®¸ù¾ÝDºÍEµÄÏà¶Ô·Ö×ÓÖÊÁ¿¿ÉÅжϣ¬DÔÚŨÁòËá¼ÓÈÈÌõ¼þÏÂÓëÒÒ´¼·¢Éúõ¥»¯·´Ó¦Éú³ÉE£¬DÖк¬ÓÐ2¸öôÈ»ù£¬¹ÊBΪBrCH2C(CH3)=CHCH2Br£¬CΪHOCH2C(CH3)=CHCH2OH£¬DΪHOOCC(CH3)=CHCOOH£¬EΪCH3CH2OOCC(CH3)=CHCOOCH2CH3£®¸ù¾ÝÒÑÖªÐÅÏ¢¿ÉÖª£¬FµÄ½á¹¹¼òʽΪ»ò£¬ÔòGµÄ½á¹¹¼òʽ»ò¡£
¢Ù¸ù¾ÝÒÔÉÏ·ÖÎö£¬BÔÚÇâÑõ»¯ÄƵÄË®ÈÜÒºÖз¢ÉúË®½â·´Ó¦Éú³ÉC£¬FÓëBr2·¢Éú¼Ó³É·´Ó¦Éú³ÉG£¬¹Ê´ð°¸Îª£ºÈ¡´ú·´Ó¦£»¼Ó³É·´Ó¦£»
¢Ú¸ù¾ÝÒÔÉÏ·ÖÎö£¬AΪ£¬FΪ»ò£¬¹Ê´ð°¸Îª£º£¬»ò£»
¢ÛBÔÚÇâÑõ»¯ÄƵÄË®ÈÜÒºÖз¢ÉúË®½â·´Ó¦Éú³ÉC£¬ÔòB¡úC·½³ÌʽΪ£ºBrCH2-C(CH3)=CH-CH2Br+2NaOHHOCH2-C(CH3)=CH-CH2OH+2NaBr£»¸ù¾ÝDºÍEµÄÏà¶Ô·Ö×ÓÖÊÁ¿¿ÉÅжϣ¬DÔÚŨÁòËá¼ÓÈÈÌõ¼þÏÂÓëÒÒ´¼·¢Éúõ¥»¯·´Ó¦Éú³ÉE£¬ÔòD¡úEµÄ·½³ÌʽΪ£ºHOOC-C(CH3)=CH-COOH+2C2H5OHC2H5OOC-C(CH3)=CH-COOC2H5+2H2O£¬¹Ê´ð°¸Îª£ºBrCH2-C(CH3)=CH-CH2Br+2NaOHHOCH2-C(CH3)=CH-CH2OH+2NaBr£»HOOC-C(CH3)=CH-COOH+2C2H5OHC2H5OOC-C(CH3)=CH-COOC2H5+2H2O£»
¢ÜG ÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ·½³ÌʽΪ£¬¹Ê´ð°¸Îª£º¡£
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¿¡¿ÒÑÖªQºÍRµÄĦ¶ûÖÊÁ¿Ö®±ÈÊÇ9£º22£¬ÔÚ·´Ó¦X+2Y=2Q+RÖУ¬µ±1£®6gXÓëYÍêÈ«·´Ó¦ºó£¬Éú³É4£®4gR£¬Ôò²ÎÓë·´Ó¦µÄYºÍÉú³ÉÎïQµÄÖÊÁ¿Ö®±ÈΪ
A£®46:9
B£®32:9
C£®23:9
D£®16:9
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¼×ÍéÖеÄ̼Ô×ÓÊÇsp3ÔÓ»¯£¬ÏÂÁÐÓÃ*±íʾ̼Ô×ÓµÄÔÓ»¯ºÍ¼×ÍéÖеÄ̼Ô×ÓÔÓ»¯×´Ì¬Ò»ÖµÄÊÇ £¨ £©
A. CH3*CH2CH3 B. *CH2£½CHCH3 C. CH2£½*CHCH2CH3 D. HC¡Ô*CCH3
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÊµÑéÊÒÀïͨ³£ÓÃMnO2ÓëŨÑÎËá·´Ó¦ÖÆÈ¡ÂÈÆø£¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºMnO2 + 4HCl(Ũ) MnCl2 + Cl2¡ü+ 2H2O
£¨1£©Óõ¥ÏßÇÅ·¨±íʾ¸Ã·´Ó¦µç×ÓתÒƵķ½ÏòºÍÊýÄ¿£º___________¡£
£¨2£©Ôڸ÷´Ó¦ÖУ¬ÈçÓÐ1 mol Cl2Éú³É£¬±»Ñõ»¯µÄHClµÄÎïÖʵÄÁ¿ÊÇ___________£¬×ªÒƵç×ÓµÄÊýÄ¿ÊÇ_____________¡£
£¨3£©Ä³Î¶ÈÏ£¬½«Cl2ͨÈëNaOHÈÜÒºÖУ¬·´Ó¦µÃµ½º¬ÓÐClO-ÓëClO3-ÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã1µÄ»ìºÏÒº£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ _________________________ ¡£
£¨4£©±¨Ö½±¨µÀÁ˶àÆðÎÀÉú¼äÇåϴʱ£¬Òò»ìºÏʹÓá°½à²ÞÁ顱£¨Ö÷Òª³É·ÖÊÇÑÎËᣩÓë¡°84Ïû¶¾Òº¡±£¨Ö÷Òª³É·ÖÊÇNaClO£©·¢ÉúÂÈÆøÖж¾µÄʼþ¡£ÊÔ¸ù¾ÝÄãµÄ»¯Ñ§ÖªÊ¶·ÖÎö£¬ÔÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©_________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÎïÖÊÓëËáµÄ·´Ó¦ÖУ¬Ö»±íÏÖ³öËáÐÔ×÷ÓõÄÊÇ£¨¡¡¡¡£©
A.Cu£«2H2SO4£¨Å¨£©===CuSO4£«SO2¡ü£«2H2O
B.C£«4HNO3£¨Å¨£©===CO2¡ü£«4NO2¡ü£«2H2O
C.3Fe£«8HNO3£¨Ï¡£©===3Fe£¨NO3£©2£«2NO¡ü£«4H2O
D.CuO£«2HNO3£¨Ï¡£©===Cu£¨NO3£©2£«H2O
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÎïÖʵķÖÀà·½·¨ÓжàÖÖ·½·¨£¬ÏÂÁжÔÎÞ»ú»¯ºÏÎï·ÖÀàÈçͼ£º
£¨1£©ÉÏͼËùʾµÄÎïÖÊ·ÖÀà·½·¨Ãû³ÆÊÇ____________________¡£
£¨2£©ÒÔK¡¢Na¡¢H¡¢O¡¢S¡¢NÖÐÈÎÁ½ÖÖ»òÈýÖÖÔªËØ×é³ÉºÏÊʵÄÎïÖÊ£¬·Ö±ðÌîÔÚϱíÖТڡ¢¢Ü¡¢¢ÞºóÃæ¡£
ÎïÖÊÀà±ð | Ëá | ¼î | ÑÎ | Ñõ»¯Îï | Ç⻯Îï |
»¯Ñ§Ê½ | ¢ÙHNO3 ¢Ú_______ | ¢ÛNaOH ¢Ü_______ | ¢ÝNa2SO4 ¢Þ_______ | ¢ßCO2 ¢àSO3 | ¢áNH3 |
£¨3£©Ð´³ö¢ßÓëÉÙÁ¿µÄ¢ÛÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ_______________________¡£
£¨4£©Ð´³öÂÁÓë¢ÛÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ______________________¡£
£¨5£©ÓÒͼÊÇijѧУʵÑéÊÒ´Ó»¯Ñ§ÊÔ¼ÁÉ̵êÂò»ØµÄŨÁòËáÊÔ¼Á±êÇ©ÉϵIJ¿·ÖÄÚÈÝ¡£ÏÖÓøÃŨÁòËáÅäÖÆ480 mL 1 mol¡¤ L£1µÄÏ¡ÁòËá¡£
¿É¹©Ñ¡ÓõÄÒÇÆ÷ÓУº¢Ù½ºÍ·µÎ¹Ü¢ÚÉÕÆ¿¢ÛÉÕ±¢Ü²£Á§°ô¢ÝÒ©³×¢ÞÁ¿Í²¢ßÍÐÅÌÌìƽ¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
a£®¸ÃÁòËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ __________ mol¡¤ L£1¡£
b£®ÅäÖÆÏ¡ÁòËáʱ£¬»¹È±ÉÙµÄÒÇÆ÷ÓÐ ______________ (дÒÇÆ÷Ãû³Æ)¡£
c£®¾¼ÆË㣬ÅäÖÆ480mL 1mol¡¤ L£1µÄÏ¡ÁòËáÐèÒªÓÃÁ¿Í²Á¿È¡ÉÏÊöŨÁòËáµÄÌå»ýΪ_________mL¡£
d£®¶ÔËùÅäÖƵÄÏ¡ÁòËá½øÐвⶨ£¬·¢ÏÖÆäŨ¶È´óÓÚ1 mol¡¤ L£1£¬ÅäÖƹý³ÌÖÐÏÂÁи÷Ïî²Ù×÷¿ÉÄÜÒýÆð¸ÃÎó²îµÄÔÒòÓÐ ___________¡£
A£®¶¨ÈÝʱ£¬¸©ÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß½øÐж¨ÈÝ ¡£
B£®½«Ï¡ÊͺóµÄÏ¡ÁòËáÁ¢¼´×ªÈëÈÝÁ¿Æ¿ºó£¬½ô½ÓמͽøÐÐÒÔºóµÄʵÑé²Ù×÷¡£
C£®×ªÒÆÈÜҺʱ£¬²»É÷ÓÐÉÙÁ¿ÈÜÒºÈ÷µ½ÈÝÁ¿Æ¿ÍâÃæ¡£
D£®ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴµÓºóδ¸ÉÔº¬ÓÐÉÙÁ¿ÕôÁóË® ¡£
E£®¶¨Èݺ󣬰ÑÈÝÁ¿Æ¿µ¹ÖÃÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬±ã²¹³ä¼¸µÎË®ÖÁ¿Ì¶È´¦¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ñо¿·¢ÏÖ£¬NOxºÍSO2ÊÇÎíö²µÄÖ÷Òª³É·Ö¡£
£¨Ò»£©NOxÖ÷ÒªÀ´Ô´ÓÚÆû³µÎ²Æø¡£
ÒÑÖª£ºN2£¨g£©+O2£¨g£©2NO£¨g£©¡÷H£½+180.50 kJ/mol
2CO£¨g£©+O2£¨g£©2CO2£¨g£©¡÷H£½-566.00kJ/mol
£¨1£©ÎªÁ˼õÇá´óÆøÎÛȾ£¬ÈËÃÇÌá³öÔÚÆû³µÎ²ÆøÅÅÆø¹Ü¿Ú²ÉÓô߻¯¼Á½«NOºÍCOת»¯³ÉÎÞÎÛȾÆøÌå²ÎÓë´óÆøÑ»·¡£Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ__________¡£
£¨2£©T¡æʱ£¬½«µÈÎïÖʵÄÁ¿µÄNOºÍCO³äÈëÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬±£³ÖζȺÍÌå»ý²»±ä£¬·´Ó¦¹ý³Ì£¨0-15min£©ÖÐNOµÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçͼËùʾ¡£
¢ÙT¡æʱ¸Ã»¯Ñ§·´Ó¦µÄƽºâ³£ÊýK= __________£»Èô±£³ÖζȲ»±ä£¬ÔÙÏòÈÝÆ÷ÖгäÈëCO¡¢N2¸÷0.8mol£¬Æ½ºâ½«____________Òƶ¯¡££¨Ìî¡°Ïò×󡱡¢¡°ÏòÓÒ¡±»ò¡°²»¡±£©
¢ÚͼÖÐa¡¢b·Ö±ð±íʾÔÚÒ»¶¨Î¶ÈÏ£¬Ê¹ÓÃÖÊÁ¿Ïàͬµ«±íÃæ»ý²»Í¬µÄ´ß»¯¼Áʱ£¬´ïµ½Æ½ºâ¹ý³ÌÖÐn£¨NO£©µÄ±ä»¯ÇúÏߣ¬ÆäÖбíʾ´ß»¯¼Á±íÃæ»ý½Ï´óµÄÇúÏßÊÇ__________¡££¨Ìî¡°a¡±»ò¡°b¡±£©
¢Û15minʱ£¬Èô¸Ä±äÍâ½ç·´Ó¦Ìõ¼þ£¬µ¼ÖÂn£¨NO£©·¢ÉúÈçͼËùʾµÄ±ä»¯£¬Ôò¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇ________¡£
£¨¶þ£©SO2Ö÷ÒªÀ´Ô´ÓÚúµÄȼÉÕ¡£È¼ÃºÑÌÆøµÄÍÑÁò¼õÅÅÊǼõÉÙ´óÆøÖк¬Áò»¯ºÏÎïÎÛȾµÄ¹Ø¼ü¡£
£¨3£©Óô¿¼îÈÜÒºÎüÊÕSO2¿É½«Æäת»¯ÎªHSO3-¡£¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_____________¡£
£¨4£©Èçͼµç½â×°Öÿɽ«Îíö²ÖеÄNO¡¢SO2·Ö±ðת»¯ÎªNH4+ºÍSO42-¡£
¢Ùд³öÎïÖÊAµÄ»¯Ñ§Ê½_____________£¬Ñô¼«µÄµç¼«·´Ó¦Ê½ÊÇ_____________
¢Ú¸Ãµç½â·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿2016Äê4ÔÂ22ÈÕÊÀ½çµØÇòÈÕµÄÖ÷ÌâΪ¡°ÕäϧµØÇò×ÊÔ´£¬×ª±ä·¢Õ¹·½Ê½¡ª¡ªÌá¸ß×ÊÔ´ÀûÓÃЧÒ桱¡£ÏÂÁÐ˵·¨²»·ûºÏ¸ÃÖ÷ÌâµÄÊÇ( )
A. ÀûÓÃÅ©×÷Îï½Õ¸ÑÖÆÈ¡ÒÒ´¼
B. »ØÊյعµÓÍ£¬ÖƱ¸ÉúÎï²ñÓÍ
C. ·ÙÉշϾÉËÜÁÏ£¬·ÀÖ¹°×É«ÎÛȾ
D. ¿ª·¢ÀûÓø÷ÖÖÐÂÄÜÔ´£¬¼õÉÙ¶Ô»¯Ê¯È¼ÁϵÄÒÀÀµ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ä³À໯ѧ·´Ó¦¿ÉÓÃÈçÏÂͨʽ±íʾ£ºA+B¡úC+D+H2O Çë°´ÒªÇó»Ø´ðÎÊÌ⣺
£¨1£©ÈôCµÄÏ¡ÈÜÒºÏÔÀ¶É«£¬DΪºì×ØÉ«ÆøÌ壬ÔòBÈÜÒºµÄÃû³ÆÊÇ______________£¬·´Ó¦ÖÐÿÉú³É1 mol H2OʱתÒƵĵç×ÓÊýĿΪ_________¡££¨ÒÔNA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£©
£¨2£©ÈôAΪµ¥ÖÊ£¬C¡¢D¶¼ÊÇÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×ǵÄÆøÌå¡£ÔòBµÄ»¯Ñ§Ê½Îª_____________£¬A¡¢BÔÚ»¯Ñ§·´Ó¦·½³ÌʽÖл¯Ñ§¼ÆÁ¿ÊýÖ®±ÈΪ__________________¡£
£¨3£©ÈôAΪÑõ»¯ÎC¡¢DÖÐÓÐÒ»ÖÖÊdz£¼ûµÄÓж¾ÆøÌåµ¥ÖÊ¡£¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________¡£
£¨4£©ÈôCÊÇÒ»ÖÖ¼îÐÔÆøÌ壬ÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬DÊÇÒ»ÖÖ¹ÌÌå¿É×÷¸ÉÔï¼Á£¨ÖÐÐÔ£©£¬ÔòʵÑéÊÒÖÆÈ¡ÆøÌåCµÄ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ ¡£
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