¡¾ÌâÄ¿¡¿A¡«IÊdz£¼ûÓлúÎAÊÇÌþ£¬EµÄ·Ö×ÓʽΪC4H8O2£¬HΪÓÐÏãζµÄÓÍ×´ÎïÖÊ¡£

ÒÑÖª£ºCH3CH2Br£«NaOH

£¨1£©0.2mol AÍêȫȼÉÕÉú³É17.6 g CO2£¬7.2g H2O£¬ÔòAµÄ½á¹¹¼òʽΪ____________¡£

£¨2£©D·Ö×ÓÖк¬ÓйÙÄÜÍŵÄÃû³ÆΪ_____________£»

£¨3£©¢ÙµÄ·´Ó¦ÀàÐÍΪ____________

£¨4£©G¿ÉÄܾßÓеÄÐÔÖÊΪ__________¡£

a.ÓëÄÆ·´Ó¦ b.ÓëNaOHÈÜÒº·´Ó¦ c.Ò×ÈÜÓÚË®

£¨5£©Çëд³ö¢ÚºÍ¢ßµÄ»¯Ñ§·½³Ìʽ£º

·´Ó¦¢Ú_________________£»

·´Ó¦¢ß_________________£»

£¨6£©JÊÇÓлúÎïBµÄͬϵÎÇÒ±ÈB¶à3¸ö̼ԭ×Ó£¬J¿ÉÄܵĽṹÓÐ___ÖÖ£¬Ð´³öÆäÖк¬3¸ö¼×»ù¿ÉÄܵĽṹ¼òʽ________________________¡£

¡¾´ð°¸¡¿ CH2£½CH2 ôÈ»ù ¼Ó³É ac 2CH3CH2OH+O2 2CH3CHO+2H2O CH2OHCH2OH+2CH3COOHCH3COOCH2CH2OOCCH3+2H2O 8 (CH3)3CCH2OH¡¢(CH3)2COHCH2CH3¡¢(CH3)2CHCHOHCH3

¡¾½âÎö¡¿¸ù¾Ý17.6g CO2µÄÎïÖʵÄÁ¿Îª£ºn(CO2)==0.4mol£¬Ë®µÄÎïÖʵÄÁ¿Îª=0.4mol£¬AÊÇÒ»ÖÖÌþ£¬¸ù¾ÝÔªËØÊغã¿ÉÖª0.2mol AÖÐn(C)=0.4mol£¬n(H)=0.4mol¡Á2=0.8mol£¬Ôò1molAÖÐn(C)=2mol£¬n(H)=4mol£¬AΪCH2=CH2£¬AÓëË®¼Ó³ÉÉú³ÉB£¬BΪCH3CH2OH£¬ÔòCΪÒÒÈ©£¬DΪÒÒËᣬÒÒ´¼ºÍÒÒËá·¢Éúõ¥»¯·´Ó¦Éú³ÉE£¬EΪÒÒËáÒÒõ¥£»ÒÒÏ©ÓëÇâÆø¼Ó³ÉÉú³ÉI£¬IΪÒÒÍ飻ÒÒÏ©Óëäå¼Ó³ÉÉú³ÉF£¬FΪ1£¬2-¶þäåÒÒÍ飬 1£¬2-¶þäåÒÒÍé·¢ÉúË®½âÉú³ÉG£¬GΪÒÒ¶þ´¼£¬ÒÒ¶þ´¼ºÍÒÒËá·´Ó¦Éú³ÉÒÒËáÒÒ¶þõ¥£¬¾Ý´Ë´ðÌâ¡£

(1)¸ù¾ÝÒÔÉÏ·ÖÎö¿ÉÖªAΪÒÒÏ©£¬AµÄ½á¹¹¼òʽΪCH2=CH2£¬¹Ê´ð°¸Îª£ºCH2=CH2£»

(2)DΪÒÒËẬÓÐôÈ»ù£¬¹Ê´ð°¸Îª£ºôÈ»ù£»

(3)·´Ó¦¢ÙΪCH2=CH2ÓëË®ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó³É·´Ó¦Éú³ÉCH3CH2OH£¬·´Ó¦·½³ÌʽΪCH2=CH2+H2OCH3CH2OH£¬¹Ê´ð°¸Îª£º¼Ó³É·´Ó¦£»

(4)GΪCH2OHCH2OH¡£a£®Óë½ðÊôÄÆ·´Ó¦CH2OHCH2OH+2Na¡úCH2ONaCH2ONa+H2¡ü£¬¹ÊaÕýÈ·£»b£®´¼²»ÄÜÓëNaOHÈÜÒº·´Ó¦£¬¹Êb´íÎó£»c£®ÒÒ¶þ´¼Ò×ÈÜÓÚË®£¬¹ÊcÕýÈ·£¬¹Ê´ð°¸Îª£ºac£»

(5)·´Ó¦¢Ú£ºÒÒ´¼ÔÚCu»òAg×÷´ß»¯¼ÁÌõ¼þÏ·¢ÉúÑõ»¯·´Ó¦£¬·´Ó¦Îª£º2CH3CH2OH+O22CH3CHO+2H2O£¬·´Ó¦¢ß£ºCH2OHCH2OH+2CH3COOHCH3COOCH2CH2OOCCH3+2H2O£¬¹Ê´ð°¸Îª£º2CH3CH2OH+O22CH3CHO+2H2O£» CH2OHCH2OH+2CH3COOHCH3COOCH2CH2OOCCH3+2H2O£»

(6)BΪCH3CH2OH£¬JÊÇÓлúÎïBµÄͬϵÎÇÒ±ÈB¶à3¸ö̼ԭ×Ó£¬ÔòJΪÎì´¼£¬ÓÉÓÚC5H12ÓÐ3Öֽṹ£ºCH3CH2CH2CH2CH3¡¢¡¢£¬ÆäÖÐCH3CH2CH2CH2CH3ÖÐÓÐ3ÖÖÇâÔ­×Ó£¬¼´ÓÐ3ÖÖ´¼£¬ÖÐÓÐ4ÖÖÇâÔ­×Ó£¬¼´ÓÐ4ÖÖ´¼£¬ÖÐÓÐ1ÖÖÇâÔ­×Ó£¬¼´ÓÐ1ÖÖ´¼£¬Òò´ËJ¿ÉÄܵĽṹ¹²ÓÐ8ÖÖ£¬ÆäÖк¬3¸ö¼×»ùµÄ½á¹¹¼òʽΪ(CH3)3CCH2OH¡¢(CH3)2COHCH2CH3¡¢(CH3)2CHCHOHCH3£¬¹Ê´ð°¸Îª£º8£»(CH3)3CCH2OH¡¢(CH3)2COHCH2CH3¡¢(CH3)2CHCHOHCH3¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÊôÓÚµç½âÖʵÄÊÇ

A.ÆÏÌÑÌÇB.ʳÑÎË®C.ÁòËáÍ­D.Í­Ë¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¸ß¯Á¶ÌúÖз¢ÉúµÄ»ù±¾·´Ó¦Ö®Ò»:FeO(s)+CO(g) Fe(s)+CO2(g)(Õý·´Ó¦ÎüÈÈ),Æäƽºâ³£Êý¿É±íʾΪ , ÒÑÖª1100¡æ,K=0.263

(1)ζÈÉý¸ß,»¯Ñ§Æ½ºâÒƶ¯ºó´ïµ½ÐµÄƽºâ,¸ß¯ÄÚCO2ºÍCOµÄÌå»ý±ÈÖµ__________(¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±),ƽºâ³£ÊýKÖµ__________(¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)¡£

(2)1100 ¡æʱ²âµÃ¸ß¯ÖÐc(CO2)=0.025 mol¡¤L-1,c(CO)=0.1 mol¡¤L-1,ÔÚÕâÖÖÇé¿öÏÂ,¸Ã·´Ó¦ÊÇ·ñ´¦ÓÚƽºâ״̬__________(Ìî¡°ÊÇ¡±»ò¡°·ñ¡±),´Ëʱ,»¯Ñ§·´Ó¦ËÙÂÊvÕý__________vÄæ(Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±),ÆäÔ­ÒòÊÇ_______________________________________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÀûÓù¤ÒµÒ±Á¶ÁòËáÍ­(º¬ÓÐFe2+¡¢AsO2-¡¢Ca2+µÈÔÓÖÊ)Ìá´¿ÖƱ¸µç¶ÆÁòËáÍ­µÄÉú²úÁ÷³ÌÈçÏ£º

ÒÑÖª£º¢ÙFe3+¡¢Cu2+¿ªÊ¼³ÁµíµÄpH·Ö±ð2.7¡¢5.4£¬ÍêÈ«³ÁµíµÄpH·Ö±ðΪ3.7¡¢6.4¡£

¢ÚKsp[Cu(OH)2]=2¡Á10-20

¢ÛAsO2-+H2O2+H+=H3AsO4£¬ H3AsO4+Fe3+=FeAsO4¡ý+3H+

(1)Èܽâ²Ù×÷ÖÐÐèÒªÅäÖƺ¬Í­32 g¡¤L-1µÄÁòËáÍ­ÈÜÒº1.0 L£¬ÐèÒª³ÆÁ¿Ò±Á¶¼¶ÁòËáÍ­µÄÖÊÁ¿ÖÁÉÙΪ___________g¡£

(2)²â¶¨ÈܽâÒºÖеÄFe2+µÄŨ¶È£¬¿ÉÓÃKMnO4±ê×¼ÈÜÒºµÎ¶¨£¬È¡ÓÃKMnO4ÈÜҺӦʹÓÃ

________(¡°Ëáʽ¡±»ò¡°¼îʽ¡±)µÎ¶¨¹Ü£¬ÆäÖз´Ó¦Àë×Ó·½³ÌʽΪ£º______________________¡£ÈôÒª¼ìÑéµ÷½ÚpHºóÈÜÒºÖеÄFe3+Òѳý¾¡µÄ·½·¨ÊÇ___________________________¡£

(3)Ñõ»¯ºóÐèÒª½«ÈÜÒº½øÐÐÏ¡Êͼ°µ÷½ÚÈÜÒºµÄpH=5£¬ÔòÏ¡ÊͺóµÄÈÜÒºÖÐÍ­Àë×ÓŨ¶È×î´ó²»Äܳ¬¹ý____________mol¡¤L-1¡£

(4)¹ÌÌå¢ñµÄÖ÷Òª³É·Ö³ý FeAsO4 ¡¢Fe(OH)3Í⻹ÓÐ__________________£¬ÓÉÈÜÒº¢ñ»ñµÃCuSO4¡¤H2O,ÐèÒª¾­¹ý________ ¡¢____________¡¢¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï²Ù×÷¡£

(5)ÀûÓÃÒÔÉϵç¶Æ¼¶ÁòËáÍ­×÷Ϊµç½âÖÊÈÜÒº£¬µç½â´ÖÍ­(º¬Ð¿¡¢Òø¡¢²¬ÔÓÖÊ)ÖƱ¸´¿Í­£¬Ð´³öÑô¼«·¢ÉúµÄµç¼«·´Ó¦Ê½£º________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ð¡Ã÷Ìå¼ìµÄѪҺ»¯Ñéµ¥ÖУ¬ÆÏÌÑÌÇΪ5.9mmol/L£®±íʾ¸ÃÌå¼ìÖ¸±êµÄÎïÀíÁ¿ÊÇ£¨ £©
A.Èܽâ¶È£¨s£©
B.ÎïÖʵÄÁ¿Å¨¶È£¨c£©
C.ÖÊÁ¿·ÖÊý£¨w£©
D.Ħ¶ûÖÊÁ¿£¨M£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿½üÄêÀ´Îíö²ÌìÆø¾­³£ËÁÅ°±±¾©¡¢Ìì½ò¡¢ºÓ±±µÈµØÇø£¬ÆäÖÐÆû³µÎ²ÆøºÍȼúÊÇÔì³É¿ÕÆøÎÛȾµÄÔ­ÒòÖ®Ò»¡£Òò´ËÑо¿µªÑõ»¯ÎïµÄ·´Ó¦»úÀí£¬¶ÔÏû³ýºÍ·ÀÖλ·¾³ÎÛȾÓÐÖØÒªÒâÒå¡£

(1) ¶ÔÓÚ2NO(g) + 2H2(g) = N2(g) + 2H2O(g) ¡÷H = 665 kJ/mol µÄ·´Ó¦·ÖÈý²½Íê³É£º

¢Ù 2NO(g) = N2O2(g) (¿ì)

¢Ú N2O2(g) + H2(g) = N2O(g) + H2O(g) (Âý)

¢Û ______________________________ (¿ì)£¬ÇëÍê³ÉµÚ¢Û²½µÄ»¯Ñ§·½³Ìʽ¡£Òò´Ë¾ö¶¨´Ë×Ü·´Ó¦ËÙÂʵÄÊǵÚ_____²½µÄ·´Ó¦¡£(ÌîÐòºÅ)

(2) ÒÑÖª£ºH2(g) + CO2(g) = H2O(g) + CO(g) ¡÷H = + 41 kJ/molÆû³µÎ²ÆøµÄ¾»»¯Ô­ÀíÖ÷ÒªÊÇÓô߻¯¼Á°ÑNOÓëCO·´Ó¦×ª»¯ÎªÁ½ÖÖ¶Ô´óÆøÎÞÎÛȾµÄÆøÌ壬ÊÔд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º______________¡£

¸Ã·´Ó¦ÔÚÒ»¶¨Ìõ¼þÏ´ﵽƽºâºó£¬ÎªÁËÄܼӿ췴ӦËÙÂʲ¢È÷´Ó¦ÏòÕý·½ÏòÒƶ¯£¬¿É²ÉÈ¡µÄ´ëÊ©ÓУº£¨__________£©

A£®Êʵ±Éý¸ßÎÂ¶È B£®Êʵ±½µµÍÎÂ¶È C£®Ñ¹ËõÌå»ýÔö´óѹǿ D£®Ê¹ÓÃÕý´ß»¯¼Á

¸Ã·´Ó¦²ÉÈ¡ÉÏÊö´ëÊ©ÖØдﵽƽºâºóKÖµ½«______(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±ºÍ¡°²»±ä¡±)

(3) Éý¸ßζÈÄÜÈþø´ó¶àÊýµÄ»¯Ñ§·´Ó¦¼Ó¿ì·´Ó¦ËÙÂÊ£¬µ«ÊÇÑо¿·¢ÏÖ2NO(g) + O2(g) = 2NO2(g) ¡÷H < 0 ´æÔÚһЩÌØÊâÏÖÏó¡£ÏÖÓÐij»¯Ñ§Ð¡×éͨ¹ýʵÑé²âµÃ²»Í¬Î¶Èϸ÷´Ó¦ËÙÂʳ£Êýk (´ú±í·´Ó¦ËÙÂʵÄÒ»¸ö³£Êý)µÄÊýÖµÈçÏÂ±í£º

T(K)

k

T(K)

k

T(K)

k

143

1.48 ¡Á 105

273

1.04 ¡Á 104

514

3.00 ¡Á 103

195

2.58 ¡Á 104

333

5.50 ¡Á 103

613

2.80 ¡Á 103

254

1.30 ¡Á 104

414

4.00 ¡Á 103

663

2.50 ¡Á 103

ÓÉʵÑéÊý¾Ý²âµ½v(Õý)~c(O2)µÄ¹ØϵÈçͼËùʾ£¬µ±xµãÉý¸ßµ½Ä³Ò»Î¶Èʱ£¬·´Ó¦ÖØдﵽƽºâ£¬Ôò±äΪÏàÓ¦µÄµãΪ_____µã(Ìî×Öĸ)£¬²¢½âÎöÔ­Òò£º¢Ù__________£¬¢Ú__________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿µâµÄÈÛ¡¢·ÐµãµÍ£¬ÆäÔ­ÒòÊÇ(¡¡¡¡)

A. µâµÄ·Ç½ðÊôÐÔ½ÏÈõ B. I¡ªI¼üµÄ¼üÄܽÏС

C. µâ¾§ÌåÊôÓÚ·Ö×Ó¾§Ìå D. I¡ªI¹²¼Û¼üµÄ¼ü³¤½Ï³¤

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÆÕͨˮÄàÔڹ̻¯¹ý³ÌÖÐÆä×ÔÓÉË®·Ö×Ó¼õÉÙ²¢ÐγɼîÐÔÈÜÒº¡£¸ù¾ÝÕâÒ»ÎïÀí»¯Ñ§Ìص㣬¿Æѧ¼Ò·¢Ã÷Á˵綯ÊÆ·¨²âË®ÄàµÄ³õÄýʱ¼ä¡£´Ë·¨µÄÔ­ÀíÈçͼËùʾ£¬·´Ó¦µÄ×Ü·½³ÌʽΪ2Cu£«Ag2O===Cu2O£«2Ag£¬ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ (¡¡¡¡)

A. 2 mol CuÓë1 mol Ag2OµÄ×ÜÄÜÁ¿µÍÓÚ1 mol Cu2OÓë2 mol Ag¾ßÓеÄ×ÜÄÜÁ¿

B. ¸º¼«µÄµç¼«·´Ó¦Ê½Îª2Cu£«2OH£­£­2e£­===Cu2O£«H2O

C. ²âÁ¿Ô­ÀíʾÒâͼÖУ¬µçÁ÷·½Ïò´ÓCu¡úAg2O

D. µç³Ø¹¤×÷ʱ£¬OH£­ÏòÕý¼«Òƶ¯

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÈçͼΪÏ໥´®ÁªµÄ¼×¡¢ÒÒÁ½µç½â³Ø£¬ÆäÖÐ̼°ôÉÏÓлÆÂÌÉ«ÆøÌå²úÉú¡£

£¨1£©¼×³ØÈôΪÓõç½âÔ­Àí¾«Á¶Í­µÄ×°Öã¬Ôò£ºAÊÇ________¼«£¬µç¼«·´Ó¦Îª__________________£¬µ±Ò»¼«ÓÐ1mol´¿Í­Îö³öʱ£¬ÁíÒ»¼«ÈܽâµÄÍ­___________1mol£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±¡¢¡°µÈÓÚ¡±£©¡£

£¨2£©ÒÒ³ØÖÐÈôµÎÈëÉÙÁ¿·Ó̪ÊÔÒº£¬µç½âÒ»¶Îʱ¼äºóFeµç¼«¸½½ü³Ê________É«£¬µç¼«·´Ó¦Ê½Îª_________¡£

£¨3£©Èô¼×³ØÖеç½âÖÊÈÜҺΪCuSO4ÈÜÒº£¬µç½â¹ý³ÌÖÐÒõ¼«ÔöÖØ12£®8g£¬ÔòÒÒ³ØÖÐÑô¼«·Å³öµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ_______L,Èô´ËʱÒÒ³ØÊ£ÓàÒºÌåΪ400 mL£¬Ôòµç½âºóÈÜÒºµÄpHΪ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸