¡¾ÌâÄ¿¡¿¾Û´×ËáÒÒÏ©õ¥ÊÇð¤ºÏ¼Á£¬Ó¦Óù㷺¡£ÏÂÃæÊǸÃÓлúÎïµÄºÏ³É·Ïߣº

Ìáʾ£º¢Ù¼×ÍéÔڵ绡µÄ×÷ÓÃÏÂÉú³ÉȲÌþ¡£

¢ÚCH3C¡ÔCHCH3CH2CHO(δÅäƽ)¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)¼×ÍéºÏ³ÉAµÄ»¯Ñ§·´Ó¦ÖÐÔ­×ÓÀûÓÃÂÊΪ________¡£

(2)BµÄ½á¹¹¼òʽΪ________¡£

(3)BÉú³ÉCµÄ·´Ó¦ÖгýÐÂÖÆCu(OH)2Ðü×ÇÒºÍ⻹ÐèÒªµÄÌõ¼þÊÇ________¡£

(4)AÓëC·´Ó¦Éú³ÉDµÄ·´Ó¦ÀàÐÍÊÇ________¡£

(5)д³öÓÉDÉú³É·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________________________¡£

(6)д³öÄÜʹ×ÏɫʯÈïÈÜÒº±äºìµÄDµÄÈýÖÖͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º________________________________¡£

¡¾´ð°¸¡¿81.25%CH3CHO¼ÓÈȼӳɷ´Ó¦CH2===CHCH2COOH¡¢CH3CH===CHCOOH¡¢

¡¾½âÎö¡¿

ÓÉÄ¿±ê²úÎï¿ÉÖªA¡¢B¡¢C¾ùÊǺ¬ÓÐ2¸ö̼ԭ×ÓµÄÓлúÎ½áºÏÌáʾÐÅÏ¢¿ÉÖªAΪÒÒȲ£¬BΪÒÒÈ©£¬CΪÒÒËá¡£¾Ý´Ë·ÖÎö¿ÉµÃ½áÂÛ¡£

(1)¼×ÍéÉú³ÉÒÒȲµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ2CH4CH¡ÔCH£«3H2£¬ËùÒԸ÷´Ó¦µÄÔ­×ÓÀûÓÃÂÊΪ26¡Â32¡Á100%£½81.25%£»£¨2£©BµÄ½á¹¹¼òʽΪ£ºCH3CHO (3)Bת»¯ÎªCʱ£¬ÓÃÐÂÖÆÇâÑõ»¯Í­Ðü×ÇÒºµÈÈõÑõ»¯¼ÁÑõ»¯Ê±ÐèÒª¼ÓÈÈ£¬¹Ê´ð°¸Îª£º¼ÓÈÈ£» (4)ÒÒȲÓëÒÒËá·´Ó¦Éú³É£¬ÕâÊÇ·¢ÉúÔÚ̼̼Èý¼üÉϵļӳɷ´Ó¦£¬¹Ê´ð°¸Îª£º¼Ó³É·´Ó¦£»(5)DÉú³ÉµÄ·´Ó¦Îª¼Ó¾Û·´Ó¦£¬·½³ÌʽΪ£º£»(6)Äܹ»Ê¹×ÏɫʯÈïÈÜÒº±äºìµÄDµÄͬ·ÖÒì¹¹ÌåÖк¬ÓÐôÈ»ù£¬¹Ê·ûºÏÌõ¼þµÄͬ·ÖÒì¹¹µÄ½á¹¹¼òʽΪ£ºCH2=CHCH2COOH¡¢CH3CH=CHCOOH¡¢¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¼×´¼ÊÇ×î»ù±¾µÄÓлú»¯¹¤Ô­ÁÏÖ®Ò»¡£¹¤ÒµÉÏ¿ÉÓöþÑõ»¯Ì¼ºÍÇâÆø·´Ó¦À´Éú²ú¼×´¼¡£

(1)ÒÑÖªÆø̬¼×´¼µÄȼÉÕÈÈΪa kJ/mol£¬2H2(g)+O2(g)= 2H2O(g) ¦¤H=-bkJ/mol£»H2O(g)=H2O(l) ¦¤H= -ckJ/mol¡£ ÔòCO2(g)+3H2(g)CH3OH(g)+H2O(g)µÄ¦¤H=_________¡£

(2)ijζÈÏ£¬ÔÚ2 LÃܱÕÈÝÆ÷ÖУ¬³äÈë2.4 mol CO2ºÍ4.4 mol H2£¬·¢ÉúºÏ³É¼×´¼µÄ·´Ó¦£¬²âµÃ¼×´¼µÄÎïÖʵÄÁ¿Ëæʱ¼äµÄ±ä»¯Í¼ÏñÈçÏÂͼÖеÄÇúÏßI£¬ÔòÇ°4·ÖÖÓ¦Í(CO2)=__________£»ÈôÔÚ1 minʱ£¬¸Ä±äijһ·´Ó¦Ìõ¼þ£¬ÇúÏßI±äΪÇúÏßII£¬Ôò¸Ä±äµÄÌõ¼þΪ___________£»¸ÃζÈÏ·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýΪ___________¡£

(3)ÔÚºãѹµÄÌõ¼þÏ£¬ÏÂÁÐÑ¡ÏîÄÜ˵Ã÷CO2+3H2CH3OH+H2O·´Ó¦ÒÑ´ïƽºâ״̬µÄÊÇ______¡£

A¡¢¦ÍÕý(H2): ¦ÍÄæ(CH3OH)=3:1

B¡¢»ìºÏÆøÌåµÄÃܶȲ»Ôٱ仯

C¡¢»ìºÏÆøµÄƽ¾ùĦ¶ûÖÊÁ¿²»Ôٱ仯

D¡¢·´Ó¦ÖÐH2OÓëCH3OHµÄÎïÖʵÄÁ¿Å¨¶ÈÖ®±ÈΪ1:1£¬ÇÒ±ÈÖµ±£³Ö²»±ä

(4)ÔÚÁíһζÈÏ·¢ÉúºÏ³É¼×´¼µÄ·´Ó¦£¬¹Ø±ÕK£¬ÏòAÈÝÆ÷ÖгäÈë1 mol CO2ºÍ4 mol H2£¬ÏòBÈÝÆ÷ÖгäÈë1.2 mol CO2ºÍ4.8 mol H2£¬Á½ÈÝÆ÷·Ö±ð·¢ÉúÉÏÊö·´Ó¦¡£ÒÑÖªÆðʼʱÈÝÆ÷AºÍBµÄÌå»ý¾ùΪa L£¬·´Ó¦´ïµ½Æ½ºâʱÈÝÆ÷BµÄÌå»ýΪ0.9a L£¬Î¬³ÖÆäËûÌõ¼þ²»±ä£¬Èô´ò¿ªKÒ»¶Îʱ¼äºóÖØдﵽƽºâ£¬ÈÝÆ÷BµÄÌå»ýΪ______L(²»¿¼ÂÇζȵı仯£¬PΪ¿É×ÔÓÉ»¬¶¯»îÈû£¬²»¿¼ÂÇ»îÈûµÄĦ²ÁÁ¦)¡£

(5)Ò»¶¨Ìõ¼þϼ״¼¿É½øÒ»²½Ñõ»¯×ª»¯Îª¼×Ëá¡£ÊÒÎÂÏ£¬½«amol/LµÄ¼×ËáÓëbmol/LµÄNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬ÌåϵÖдæÔÚc(Na+)=c(HCOO-)£¬ÊÔÓú¬aºÍbµÄ´úÊýʽ±íʾ¼×ËáµÄµçÀë³£ÊýΪ__________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³»¯Ñ§ÐËȤС×éÒªÍê³ÉÖкÍÈȵIJⶨ¡£

£¨1£©ÊµÑé×ÀÉϱ¸Óдó¡¢Ð¡Á½¸öÉÕ±­¡¢ÅÝÄ­ËÜÁÏ¡¢ÅÝÄ­ËÜÁÏ°å¡¢½ºÍ·µÎ¹Ü¡¢»·Ðβ£Á§½Á°è°ô¡¢0.5 mol¡¤L-1ÑÎËá¡¢0.55 mol¡¤L-1 NaOHÈÜÒº£¬ÊµÑéÉÐȱÉٵIJ£Á§ÓÃÆ·ÊÇ___________¡¢___________¡£

£¨2£©ÊµÑéÖÐÄÜ·ñÓû·ÐÎÍ­Ë¿½Á°è°ô´úÌæ»·Ðβ£Á§½Á°è°ô___________£¨Ìî¡°ÄÜ¡±»ò¡°·ñ¡±£©£¬Æä

Ô­ÒòÊÇ ¡£

£¨3£©ËûÃǼǼµÄʵÑéÊý¾ÝÈçÏ£º

ÒÑÖª£ºQ£½cm£¨t2£­t1£©£¬·´Ó¦ºóÈÜÒºµÄ±ÈÈÈÈÝcΪ4.18 kJ¡¤¡æ-1¡¤kg-1£¬¸÷ÎïÖʵÄÃܶȾùΪ1 g¡¤cm-3¡£

¢Ù¼ÆËãÍê³ÉÉÏ±í¦¤H£½_____________¡£

¢Ú¸ù¾ÝʵÑé½á¹ûд³öNaOHÈÜÒºÓëHClÈÜÒº·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º ¡£

£¨4£©ÈôÓÃKOH´úÌæNaOH£¬¶Ô²â¶¨½á¹û__________£¨Ìî¡°ÓС±»ò¡°ÎÞ¡±£©Ó°Ï죻ÈôÓô×Ëá´úÌæHCl×öʵÑ飬¶Ô²â¶¨½á¹û____________£¨Ìî¡°ÓС±»ò¡°ÎÞ¡±£©Ó°Ïì¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¹¤ÒµÉÏÒÔÒÒ´¼ÎªÔ­ÁϾ­Ò»ÏµÁз´Ó¦¿ÉÒԵõ½4-ôÇ»ù±âÌÒËáºÍÏ㶹ËØ-3-ôÈËᣬ¶þÕߵĺϳÉ·ÏßÈçÏÂ(²¿·Ö²úÎï¼°Ìõ¼þδÁгö)£º

ÒÑÖª£º¢ñ.RCOOR¡ä£«R¡åOHRCOOR¡å£«R¡äOH£»

¢ò. (R£¬R¡ä£¬R¡å±íʾÇâÔ­×Ó¡¢Íé»ù»ò·¼»ù)

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)·´Ó¦¢ÚÊôÓÚÈ¡´ú·´Ó¦£¬ÔòAÖйÙÄÜÍŵÄÃû³ÆÊÇ________

(2) µÄÃû³ÆÊÇ________£¬·´Ó¦¢ÝµÄ·´Ó¦ÀàÐÍΪ________¡£

(3)·´Ó¦¢ÞµÄ»¯Ñ§·½³ÌʽÊÇ________________________________________¡£

(4)ÒÑÖªG·Ö×ÓÖк¬ÓÐ2¸öÁùÔª»·¡£ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ________(Ìî±êºÅ)¡£

a£®ºË´Å¹²ÕñÒǿɲâ³öEÓÐ5ÖÖÀàÐ͵ÄÇâÔ­×Ó

b£®ÖÊÆ×Òǿɼì²âFµÄ×î´óÖʺɱȵÄֵΪ236

c£®G·Ö×ÓʽΪC12H10O4

d£®»¯ºÏÎïWÄÜ·¢Éú¼Ó¾Û·´Ó¦µÃµ½ÏßÐ͸߷Ö×Ó»¯ºÏÎï

(5)ij·¼Ï㻯ºÏÎïQÊÇ4-ôÇ»ù±âÌÒËáµÄͬ·ÖÒì¹¹Ì壬¾ßÓÐÏÂÁÐÌØÕ÷£º¢Ù±½»·ÉÏÖ»ÓÐ3¸öÈ¡´ú»ù£»¢ÚÄÜ·¢ÉúË®½â·´Ó¦ºÍÒø¾µ·´Ó¦£»¢Û1 mol Q×î¶àÄÜÏûºÄ3 mol NaOH¡£Q¹²ÓÐ________ÖÖ(²»º¬Á¢ÌåÒì¹¹)

(6)×Ðϸ¹Û²ìÓÉÒÒ´¼ºÏ³ÉÏ㶹ËØ-3-ôÈËáµÄ¹ý³Ì£¬½áºÏÏà¹ØÐÅÏ¢£¬µ±ÒÒ´¼Óë±û¶þËáµÄÎïÖʵÄÁ¿µÄ±ÈΪ________ʱ£¬Ö»Ðí3²½¼´¿ÉÍê³ÉºÏ³É·Ïß¡£Çëд³öºÏ³É·Ïß_______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿²»Ðâ¸Ö±íÃæÓÃÏõËáºÍÇâ·úËáµÄ»ìËá´¦Àíºó£¬²úÉúµÄËáÏ´·ÏÒºÖк¬ÓÐFe3+¡¢Ni2+¡¢NO3-¡¢F-ºÍ+6¼Û¸õµÄº¬ÑõËá¸ùÀë×ӵȣ®ÈçͼÊÇ×ÛºÏÀûÓøÃËáÏ´·ÏÒºµÄ¹¤ÒÕÁ÷³Ì£º

ÒÑÖª£º

¢Ù½ðÊôÀë×Ó¿ªÊ¼³ÁµíºÍ³ÁµíÍêȫʱµÄpH£º

Fe3+

Ni2+

Cr3+

¿ªÊ¼³Áµí

1.5

6.7

4.0

³ÁµíÍêÈ«

3.4

9.5

6.9

¢ÚNi2+Óë×ãÁ¿°±Ë®µÄ·´Ó¦Îª£ºNi2++6NH3[Ni£¨NH3£©6]2+

£¨1£©ÔÙÉúËáÖк¬ÓÐ______£¬²ÉÈ¡¼õѹÕôÁóµÄÔ­ÒòÊÇ______£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£®

£¨2£©ÀûÓ÷ÏÑõ»¯Ìú£¨Ö÷Òª³É·ÖΪFe2O3£©´úÌæÉÕ¼îµ÷½ÚpHµÄºÃ´¦ÊÇ______£®

£¨3£©Çëд³ö¡°×ª»¯¡±Ê±NaHSO3ÓëCr2O72-·¢Éú·´Ó¦µÄÀë×Ó·´Ó¦·½³Ìʽ£º______£®

£¨4£©ÒÑÖª[Ni£¨NH3£©6]2+ΪÄѵçÀëµÄÂçºÏÀë×Ó£¬Ôò¡°³ÁÄø¡±µÄÀë×Ó·½³ÌʽΪ£º______£®

£¨5£©ÂËÔü3µÄÖ÷Òª³É·ÖΪCa£¨OH£©2ºÍ_________________________£®

£¨6£©¾­¼ì²â£¬×îºóµÄ²ÐÒºÖÐc£¨Ca2+£©=0.004molL-1£¬Ôò²ÐÒºÖÐF-Ũ¶ÈΪ______mgL-1£¬[ÒÑÖªKsp£¨CaF2£©=4¡Á10-11mol3L-3.

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Óлú»¯Ñ§»ù´¡]MΪºÏ³É¸ß·Ö×Ó²ÄÁϵÄÖмäÌ壬ÒÔ·¼ÏãÌþAÖƱ¸MºÍ¸ß·Ö×Ó»¯ºÏÎïNµÄÒ»ÖֺϳÉ·ÏßÈçÏ£º

ÒÑÖª£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©CµÄ»¯Ñ§Ãû³ÆΪ________¡£

£¨2£©A¡úB¡¢H¡úMµÄ·´Ó¦ÀàÐÍ·Ö±ðΪ________¡¢________¡£

£¨3£©FÖÐËùº¬¹ÙÄÜÍŵÄÃû³ÆΪ________¡£GµÄ½á¹¹¼òʽΪ________¡£

£¨4£©ÊÔ¼Á1Ϊ________¡£

£¨5£©D¡úNµÄ»¯Ñ§·½³ÌʽΪ________¡£

£¨6£©QΪHµÄͬ·ÖÒì¹¹Ì壬ͬʱÂú×ãÏÂÁÐÌõ¼þµÄQµÄ½á¹¹¼òʽΪ________¡£

¢Ù±½»·ÉÏÁ¬ÓÐÁ½¸öÈ¡´ú»ù£¬³ý±½»·ÍâÎÞÆäËû»·×´½á¹¹

¢ÚÄÜÓëÂÈ»¯ÌúÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£¬1 mol Q×î¶àÏûºÄ3 molNaOH

¢ÛºË´Å¹²ÕñÇâÆ×ÓÐ5×é·å£¬·åÃæ»ýÖ®±ÈΪ6¡Ã2¡Ã2¡Ã1¡Ã1

£¨7£©²ÎÕÕÉÏÊöºÏ³É·ÏߺÍÐÅÏ¢£¬ÒÔ¼×È©ºÍÒÒȩΪÆðʼԭÁÏ£¨ÎÞ»úÊÔ¼ÁÈÎÑ¡£©£¬Éè¼ÆÖƱ¸¾Û±ûÏ©Ëᣨ£©µÄºÏ³É·Ïߣº________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Na2O2¿ÉÓÃ×÷Ư°×¼ÁºÍºôÎüÃæ¾ßÖеĹ©Ñõ¼Á¡£

£¨1£©Ä³Ñ§Ï°Ð¡×é·¢ÏÖ£ºÔÚÊ¢ÓÐNa2O2µÄÊÔ¹ÜÖмÓÈë×ãÁ¿Ë®£¬Á¢¼´²úÉú´óÁ¿ÆøÅÝ£¬µ±ÆøÅÝÏûʧºó£¬ÏòÆäÖеÎÈë1~2µÎ·Ó̪ÊÔÒº£¬ÈÜÒº±äºì£»½«ÊÔ¹ÜÇáÇáÕñµ´£¬ºìÉ«ºÜ¿ìÍÊÈ¥£»´ËʱÔÙÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿MnO2·ÛÄ©£¬ÓÖÓÐÆøÅݲúÉú¡£

¢Ùʹ·Ó̪ÊÔÒº±äºìÊÇÒòΪ____________________________________£¬ºìÉ«ÍÊÈ¥µÄ¿ÉÄÜÔ­ÒòÊÇ____________________________________________¡£

¢Ú¼ÓÈëMnO2·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________________________¡£

£¨2£©Na2O2ÓÐÇ¿Ñõ»¯ÐÔ£¬H2¾ßÓл¹Ô­ÐÔ£¬ÓÐͬѧ²ÂÏëNa2O2ÓëH2ÄÜ·´Ó¦¡£ÎªÁËÑéÖ¤´Ë²ÂÏ룬¸ÃС×éͬѧ½øÐÐÈçÏÂʵÑ飬ʵÑé²½ÖèºÍÏÖÏóÈçÏ¡£

²½Öè1£º°´Èçͼ×é×°ÒÇÆ÷£¨Í¼ÖмгÖÒÇÆ÷Ê¡ÂÔ£©£¬¼ì²éÆøÃÜÐÔ£¬×°ÈëÒ©Æ·£»

²½Öè2£º´ò¿ªK1¡¢K2£¬²úÉúµÄÇâÆøÁ÷¾­×°ÓÐNa2O2µÄÓ²Öʲ£Á§¹Ü£¬Ò»¶Îʱ¼äºó£¬Ã»ÓÐÈκÎÏÖÏó£»

²½Öè3£º¼ìÑéH2µÄ´¿¶Èºó£¬¿ªÊ¼¼ÓÈÈ£¬¹Û²ìµ½Ó²Öʲ£Á§¹ÜÄÚNa2O2¿ªÊ¼ÈÛ»¯£¬µ­»ÆÉ«µÄ·ÛÄ©±ä³ÉÁË°×É«¹ÌÌ壬¸ÉÔï¹ÜÄÚÁòËáͭδ±äÀ¶É«£»

²½Öè4£º·´Ó¦ºó³·È¥¾Æ¾«µÆ£¬´ýÓ²Öʲ£Á§¹ÜÀäÈ´ºó¹Ø±ÕK1¡£

¢ÙÊ¢×°Ï¡ÑÎËáµÄÒÇÆ÷Ãû³Æ___________£»B×°ÖõÄ×÷ÓÃÊÇ______________¡£

¢Ú±ØÐë¼ìÑéÇâÆø´¿¶ÈµÄÔ­ÒòÊÇ_________________¡£

¢ÛÉèÖÃ×°ÖÃDµÄÄ¿µÄÊÇ___________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿·Ç½ðÊôµ¥ÖÊA¾­ÈçͼËùʾµÄ¹ý³Ìת»¯Îªº¬ÑõËáD£¬ÒÑÖªDΪǿËᣬÇë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈôAÔÚ³£ÎÂÏÂΪ¹ÌÌ壬BÊÇÄÜʹƷºìÈÜÒºÍÊÉ«µÄÓд̼¤ÐÔÆøζµÄÎÞÉ«ÆøÌå¡£

¢ÙDµÄ»¯Ñ§Ê½ÊÇ_____£»

¢ÚÔÚ¹¤ÒµÉú²úÖУ¬BÆøÌåµÄ´óÁ¿Åŷű»ÓêË®ÎüÊÕºó»áÐγÉËáÓê¶øÎÛȾ»·¾³¡£Ð´³öÐγÉËáÓêµÄ»¯Ñ§·½³Ìʽ______¡£¹¤ÒµÉϳ£ÏòúÖмÓÈëʯ»Òʯ£¬¼õÉÙBÆøÌå¶Ô»·¾³µÄÎÛȾ£¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____¡£

£¨2£©ÈôAÔÚ³£ÎÂÏÂΪÆøÌ壬CÊǺì×ØÉ«µÄÆøÌå¡£

¢ÙAת»¯ÎªBµÄ»¯Ñ§·½³ÌʽÊÇ____¡£

¢ÚDµÄÏ¡ÈÜÒºÔÚ³£ÎÂÏ¿ÉÓëÍ­·´Ó¦²¢Éú³ÉBÆøÌ壬Çëд³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ_____¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¡°¹²Ïíµ¥³µ¡±£¬µÍ̼»·±££¬ÓÐÒ潡Éí£¬·½±ã³öÐС£¡°×îºóÒ»¹«À£¬ÏÂÁйØÓÚµ¥³µµÄÖÆÔì²ÄÁÏ˵·¨ÕýÈ·µÄÊÇ( )

A.ÖÆÔì³µÂÖ¸ÖȦµÄ²ÄÁÏÊǺϽðB.ÂÁºÏ½ðÖÆÔìµÄ³µ¼Ü½Ï¸ÖÖÆÔìµÄÖØ

C.ÖÆÔìÂÖÌ¥ÓõÄÏð½ºÓй̶¨µÄÈÛµãD.ÖÆ×÷·´¹â°åµÄÓлú²£Á§ÊôÓÚ¹èËáÑÎ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸