¡¾ÌâÄ¿¡¿(1)±±¾©°ÂÔË»áÏéÔÆ»ð¾æ½«Öйú´«Í³ÎÄ»¯¡¢°ÂÔ˾«ÉñÒÔ¼°ÏÖ´ú¸ß¿Æ¼¼ÈÚΪһÌå¡£»ð¾æÄÚÐÜÐÜ´ó»ðÀ´Ô´ÓÚ±ûÍéµÄȼÉÕ£¬±ûÍéÊÇÒ»ÖÖÓÅÁ¼µÄȼÁÏ¡£ÊԻشðÏÂÁÐÎÊÌ⣺

¢ÙÈçͼÊÇÒ»¶¨Á¿±ûÍéÍêȫȼÉÕÉú³ÉCO2ºÍ1molH2O(l)¹ý³ÌÖеÄÄÜÁ¿±ä»¯Í¼£¬ÇëÔÚͼÖеÄÀ¨ºÅÄÚÌîÈë¡°+¡±»ò¡°¡± ___¡£

¢Úд³ö±íʾ±ûÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ£º___¡£

¢Û¶þ¼×ÃÑ(CH3OCH3)ÊÇÒ»ÖÖÐÂÐÍȼÁÏ£¬Ó¦ÓÃÇ°¾°¹ãÀ«¡£1mol¶þ¼×ÃÑÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ·Å³ö1455kJÈÈÁ¿¡£Èô1mol±ûÍéºÍ¶þ¼×ÃѵĻìºÏÆøÌåÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ¹²·Å³ö1645kJÈÈÁ¿£¬Ôò»ìºÏÆøÌåÖУ¬±ûÍéºÍ¶þ¼×ÃѵÄÎïÖʵÄÁ¿Ö®±ÈΪ___¡£

(2)¸Ç˹¶¨ÂÉÈÏΪ£º²»¹Ü»¯Ñ§¹ý³ÌÊÇÒ»²½Íê³É»ò·ÖÊý²½Íê³É£¬Õû¸ö¹ý³ÌµÄ×ÜÈÈЧӦÏàͬ¡£ÊÔÔËÓøÇ˹¶¨ÂɻشðÏÂÁÐÎÊÌ⣺

¢ÙÒÑÖª£ºH2O(g)¨TH2O(l) ¡÷H1=Q1kJ/mol

C2H5OH(g)¨TC2H5OH(l) ¡÷H2=Q2kJ/mol

C2H5OH(g)+3O2(g)¨T2CO2(g)+3H2O(g) ¡÷H3=Q3kJ/mol

Èôʹ23gҺ̬ÎÞË®¾Æ¾«ÍêȫȼÉÕ£¬²¢»Ö¸´µ½ÊÒΣ¬ÔòÕû¸ö¹ý³ÌÖзųöµÄÈÈÁ¿Îª___kJ¡£

¢Ú̼(s)ÔÚÑõÆø¹©Ó¦²»³ä·Öʱ£¬Éú³ÉCOͬʱ»¹²¿·ÖÉú³ÉCO2£¬Òò´ËÎÞ·¨Í¨¹ýʵÑéÖ±½Ó²âµÃ·´Ó¦£ºC(s)+O2(g)¨TCO(g)µÄ¡÷H.µ«¿ÉÉè¼ÆʵÑé¡¢ÀûÓøÇ˹¶¨ÂɼÆËã³ö¸Ã·´Ó¦µÄ¡÷H£¬¼ÆËãʱÐèÒª²âµÃµÄʵÑéÊý¾ÝÓÐ___¡£

¡¾´ð°¸¡¿ C3H8(g)+5O2(g)=3CO2(g)+4H2O(l)¡÷H=2215kJ/mol 1:3 1.5Q10.5Q2+0.5Q3 ̼¡¢Ò»Ñõ»¯Ì¼µÄ±ê׼ȼÉÕÈÈ

¡¾½âÎö¡¿

(1)¢ÙÓÉͼ¿ÉÖª£¬·´Ó¦ÎïµÄ×ÜÄÜÁ¿´óÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£»

¢ÚÓÉͼÏó¿ÉÖª£¬±ûÍéÍêȫȼÉÕÉú³É1molË®µÄìʱä¡÷H=553.75kJ/mol£»

¢ÛÒÀ¾ÝÈÈ»¯Ñ§·½³Ìʽ½áºÏ»ìºÏÆøÌåÎïÖʵÄÁ¿ºÍ·ÅÈÈÁÐʽ¼ÆËãµÃµ½¶þ¼×ÃѺͱûÍéÎïÖʵÄÁ¿Ö®±È£»

(2)¢ÙÒÀ¾Ý¸Ç˹¶¨ÂɼÆËã¿ÉµÃ£»

¢ÚÉè¼ÆʵÑé¡¢ÀûÓøÇ˹¶¨ÂɼÆËãC(s)+ O2(g)¨TCO(g)µÄ¡÷H£¬ÐèÒªÖªµÀ̼ºÍÒ»Ñõ»¯Ì¼µÄȼÉÕÈȲÅÄܼÆËãµÃµ½¡£

(1)¢ÙÓÉͼ¿ÉÖª£¬·´Ó¦ÎïµÄ×ÜÄÜÁ¿´óÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£¬¸Ã·´Ó¦Îª·´Ó¦·ÅÈÈ£¬¡÷HΪ¡°¡°£¬¹Ê´ð°¸Îª£º£»

¢ÚÓÉͼÏó¿ÉÖª£¬±ûÍéÍêȫȼÉÕÉú³É1molË®µÄìʱä¡÷H=553.75kJ/mol£¬Ôò·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºC3H8(g)+5O2(g)=3CO2(g)+4H2O(l)¡÷H=2215kJ/mol£¬¹Ê´ð°¸Îª£ºC3H8(g)+5O2(g)=3CO2(g)+4H2O(l)¡÷H=2215kJ/mol£»

¢Û1mol¶þ¼×ÃÑÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ·Å³ö1455kJÈÈÁ¿£¬Èô1mol±ûÍéºÍ¶þ¼×ÃѵĻìºÏÆøÌåÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ¹²·Å³ö1645kJÈÈÁ¿£¬Éè1mol»ìºÏÆøÌåÖжþ¼×ÃÑÎïÖʵÄÁ¿x£¬±ûÍéÎïÖʵÄÁ¿Îª1x£¬ÓÉÈÈ»¯Ñ§·½³Ìʽ¿ÉÖª±ûÍéȼÉÕ·ÅÈÈ2215 (1x) kJ£¬ÓÉÌâÒâ¿ÉµÃ¹Øϵʽ16451455x=2215 (1x)£¬½âµÃx=0.75£¬Ôò»ìºÏ±ûÍéÎïÖʵÄÁ¿Îª0.25mol£¬»ìºÏÆøÌåÖбûÍéºÍ¶þ¼×ÃÑÎïÖʵÄÁ¿Ö®±È=0.25:0.75=1:3£¬¹Ê´ð°¸Îª£º1:3£»

(2)¢Ù½«ÒÑÖªÈÈ»¯Ñ§·½³ÌʽÒÀ´Î±àºÅΪ¢Ù¡¢¢Ú¡¢¢Û£¬ÓɸÇ˹¶¨ÂÉ¿ÉÖª£¬¢Û¢Ú+¢Ù¡Á3µÃÈÈ»¯Ñ§·½³ÌʽC2H5OH(l)+3O2(g)=2CO2(g)+3H2O(l) £¬Ôò¡÷H1=(3Q1Q2+Q3)kJ/mol£¬Èôʹ23gҺ̬ÎÞË®¾Æ¾«ÎïÖʵÄÁ¿Îª0.5mol£¬ÍêȫȼÉÕ£¬²¢»Ö¸´µ½ÊÒΣ¬ÔòÕû¸ö¹ý³ÌÖзųöµÄÈÈÁ¿Îª(1.5Q10.5Q2+0.5Q3)kJ£¬¹Ê´ð°¸Îª£º1.5Q10.5Q2+0.5Q3£»

¢ÚÉè¼ÆʵÑé¡¢ÀûÓøÇ˹¶¨ÂɼÆËãC(s)+ O2(g)¨TCO(g)µÄ¡÷H£¬ÐèÒªÖªµÀ̼ºÍÒ»Ñõ»¯Ì¼µÄȼÉÕÈȲÅÄܼÆËãµÃµ½£¬¹Ê´ð°¸Îª£ºÌ¼¡¢Ò»Ñõ»¯Ì¼µÄ±ê׼ȼÉÕÈÈ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¢ñ.Ò˳ÇÊжõÎ÷»¯¹¤³§ÒÔ¸õÌú¿ó(Ö÷Òª³É·ÖΪ FeO ºÍ Cr2O3£¬º¬ÓÐ Al2O3¡¢SiO2 µÈÔÓÖÊ)ΪÖ÷ÒªÔ­ÁÏÉú²ú»¯¹¤Ô­ÁϺ췯ÄÆ(Ö÷Òª³É·Ö Na2Cr2O7¡¤2H2O)£¬¹¤ÒÕÁ÷³ÌÈçÏÂͼ£º

i.³£Î£¬NaBiO3²»ÈÜÓÚË®£¬ÓÐÇ¿Ñõ»¯ÐÔ£¬¼îÐÔÌõ¼þÏ£¬Äܽ«Cr3+ת»¯Îª CrO42£­¡£ii£®

½ðÊôÀë×Ó

Fe3+

Al3+

Cr3+

Fe2+

Bi3+

¿ªÊ¼³ÁµíµÄpH

2.7

3.4

5.0

7.5

0.7

³ÁµíÍêÈ«µÄpH

3.7

4.9

5.9

9.7

4.5

(1)²½Öè¢ÛÐè¼ÓÇâÑõ»¯ÄÆÈÜÒº£¬´ËʱpHÒªµ÷µ½5µÄÄ¿µÄÊÇ__________________¡£

(2)д³ö¢Ü·´Ó¦µÄÀë×Ó·½³Ìʽ____________________________________¡£

(3)½«ÈÜÒº H ¾­¹ýÕô·¢Å¨Ëõ£¬ÀäÈ´½á¾§£¬¹ýÂË£¬Ï´µÓ£¬¸ÉÔï¼´µÃºì·¯ÄÆ´Ö¾§Ì壬¾«Öƺ췯ÄÆ´Ö¾§ÌåÐèÒª²ÉÓõIJÙ×÷ÊÇ______________(Ìî²Ù×÷Ãû³Æ)¡£

¢ò.¾­¼ì²â¶õÎ÷»¯¹¤³§µÄ¹¤Òµ·ÏË®Öк¬ 5.00¡Á10-3 mol¡¤L-1 µÄ Cr2O72-£¬Æ䶾ÐԽϴ󡣸û¯¹¤³§µÄ¿ÆÑÐÈËԱΪÁ˱ä·ÏΪ±¦£¬½«·ÏË®´¦ÀíµÃµ½´ÅÐÔ²ÄÁÏ Cr0.5Fe1.5FeO4£¨Fe µÄ»¯ºÏ¼ÛÒÀ´ÎΪ+3¡¢+2£©£¬ÓÖÉè¼ÆÁËÈçϹ¤ÒÕÁ÷³Ì£º

(1)µÚ¢Ù²½·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_________________________________________¡£

(2)Óûʹ 1L ¸Ã·ÏË®ÖÐµÄ Cr2O72-Íêȫת»¯Îª Cr0.5Fe1.5FeO4¡£ÀíÂÛÉÏÐèÒª¼ÓÈëFeSO4¡¤7H2OµÄÖÊÁ¿Îª_________g (ÒÑÖª FeSO4¡¤7H2O µÄĦ¶ûÖÊÁ¿Îª 278 g/mol)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿2019ÄêÊÇÔªËØÖÜÆÚ±í·¢±í150ÖÜÄ꣬ÆÚ¼ä¿Æѧ¼ÒΪÍêÉÆÖÜÆÚ±í×ö³öÁ˲»Ð¸Å¬Á¦¡£Öйú¿ÆѧԺԺʿÕÅÇàÁ«½ÌÊÚÔøÖ÷³Ö²â¶¨ÁËî÷£¨49In£©µÈ9ÖÖÔªËØÏà¶ÔÔ­×ÓÖÊÁ¿µÄÐÂÖµ£¬±»²ÉÓÃΪ¹ú¼Êбê×¼¡£î÷Óë﨣¨37Rb£©Í¬ÖÜÆÚ¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

A. InÊǵÚÎåÖÜÆÚµÚ¢óA×åÔªËØ

B. 11549InµÄÖÐ×ÓÊýÓëµç×ÓÊýµÄ²îֵΪ17

C. Ô­×Ӱ뾶£ºIn>Al

D. ¼îÐÔ£ºIn(OH)3>RbOH

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚÒ»¶¨Î¶ÈϵÄ2 LµÄÃܱÕÈÝÆ÷ÖУ¬¼ÓÈë3 mol AºÍ1 mol B£¬·¢ÉúÈçÏ·´Ó¦£º3A(g)£«B(g)2C(g)£«3D(s)£¬5 min´ïµ½Æ½ºâʱ£¬n(B):n(C) =1:3¡£

£¨1£©0¡«5 minÄÚÓÃB±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ_______£»´ïµ½Æ½ºâʱ£¬AµÄת»¯ÂÊΪ_______¡£

£¨2£©´ïµ½Æ½ºâʱÈÝÆ÷ÄÚÆøÌåѹǿÓ뷴ӦǰÈÝÆ÷ÄÚÆøÌåѹǿ֮±È_________¡£

£¨3£©Î¬³ÖÈÝÆ÷µÄζȲ»±ä£¬ÈôËõСÈÝÆ÷µÄÌå»ý£¬Ôòƽºâ½«Ïò_____(Ìî¡°ÕýÏòÒƶ¯¡±¡°ÄæÏòÒƶ¯¡±»ò¡°²»Òƶ¯¡±)¡£

£¨4£©´ïµ½Æ½ºâºó£¬Èô±£³ÖζȲ»±ä£¬½«C´ÓÈÝÆ÷ÖзÖÀë³öÒ»²¿·Ö£¬Ôò»¯Ñ§Æ½ºâ³£Êý____(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. 1 L 0.1 mol¡¤L£­1NaClOÈÜÒºÖк¬ÓеÄClO£­ÎªNA

B. 1 mol FeÔÚ1 mol Cl2Öгä·ÖȼÉÕ£¬×ªÒƵĵç×ÓÊýΪ3NA

C. ³£Î³£Ñ¹Ï£¬32 g O2ÓëO3µÄ»ìºÏÆøÌåÖк¬ÓеķÖ×Ó×ÜÊýСÓÚNA

D. ±ê×¼×´¿öÏ£¬22.4 L HFÖк¬ÓеķúÔ­×ÓÊýĿΪNA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÖÓÐÈçͼԭµç³Ø×°Ö㬲åÈëµç½âÖÊÈÜҺǰCuºÍFeµç¼«µÄÖÊÁ¿ÏàµÈ¡£

(1)µ±µç½âÖÊÈÜҺΪϡÁòËáʱ£¬ÌúƬ×÷________¼«£¬Í­Æ¬ÉϵÄÏÖÏóÊÇ________.ͼIÖмýÍ·µÄ·½Ïò±íʾ__________(Ìî¡°µç×Ó¡±»ò¡°µçÁ÷¡±)µÄÁ÷Ïò¡£

(2)µ±µç½âÖÊÈÜҺΪijÈÜҺʱ£¬Á½¼«(ÓÃM¡¢N±íʾ)µÄÖÊÁ¿±ä»¯ÇúÏßÈçͼIIËùʾ£¬Ôò¸Ãµç½âÖÊÈÜÒº¿ÉÒÔÊÇÏÂÁÐÖеÄ________(Ìî´úºÅ)¡£

A.Ï¡ÁòËá B.CuSO4ÈÜÒº C.Ï¡ÑÎËá D.FeSO4ÈÜÒº

Èôµç½âҺΪËùÑ¡ÈÜÒº£¬Ôòµç¼«NµÄµç¼«·´Ó¦Ê½Îª________£¬ÈÜÒºÖÐÑôÀë×ÓÒƶ¯·½ÏòÊÇ________£¬m=________g¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓöþÑõ»¯Ì¼Éú²ú»¯¹¤²úÆ·£¬ÓÐÀûÓÚ¶þÑõ»¯Ì¼µÄ´óÁ¿»ØÊÕ¡£¶þÑõ»¯Ì¼ºÍÒÒ¶þ´¼ÔÚZnO»òпÑδ߻¯Ï¿ɺϳÉ̼ËáÒÒÏ©õ¥¡£

CO2++H2O

£¨1£©Ð¿»ù̬ԭ×ÓºËÍâµç×ÓÅŲ¼Ê½Îª_________£»Ð´³öÒ»ÖÖÓëCO2»¥ÎªµÈµç×ÓÌåµÄ·Ö×ӵĻ¯Ñ§Ê½£º__________¡£

£¨2£©Ð¿ÑÎË®ÈÜÒºÖÐZn2+¿ÉÓëH2OÖ®¼äÐÎ³É [Zn£¨H2O£©6]2+£¬Ìṩ¿Õ¹ìµÀµÄÊÇ_____£¨Ìî΢Á£·ûºÅ£©¡£

£¨3£©Ì¼ËáÒÒÏ©õ¥ÖÐ̼ԭ×ÓÔÓ»¯¹ìµÀÀàÐÍΪ______£»1mol̼ËáÒÒÏ©õ¥Öк¬ÓеļüÊýĿΪ______¡£

£¨4£©ÉúÎïÖÊÄÜÊÇÒ»Öֽྻ¡¢¿ÉÔÙÉúµÄÄÜÔ´¡£ÉúÎïÖÊÆø£¨Ö÷Òª³É·ÖΪCO¡¢CO2¡¢H2µÈ£©ÓëH2»ìºÏ£¬´ß»¯ºÏ³É¼×´¼ÊÇÉúÎïÖÊÄÜÀûÓõķ½·¨Ö®Ò»¡£¼×´¼´ß»¯Ñõ»¯¿ÉµÃµ½¼×È©£¬¼×È©ÓëÐÂÖÆCu£¨OH£©2µÄ¼îÐÔÈÜÒº·´Ó¦Éú³ÉCu2O³Áµí¡£

¢Ù¼×´¼µÄ·Ðµã±È¼×È©µÄ¸ß£¬ÆäÖ÷ÒªÔ­ÒòÊÇ____________£»

¢Ú¼×È©·Ö×ӵĿռ乹ÐÍÊÇ______________£¨ÓÃÎÄ×ÖÃèÊö£©£»

£¨5£©¿¹»µÑªËáµÄ·Ö×ӽṹÈçͼ 1 Ëùʾ£¬ÍƲ⿹»µÑªËáÔÚË®ÖеÄÈܽâÐÔ£º______£¨Ìî¡°ÄÑÈÜÓÚË®¡±»ò¡°Ò×ÈÜÓÚË®¡±£© £»Ò»¸öCu2O ¾§°û£¨¼ûͼ 2£©ÖУ¬Cu Ô­×ÓµÄÊýĿΪ______ ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¡°Öк͵ζ¨¡±Ô­ÀíÔÚʵ¼ÊÉú²úÉú»îÖÐÓ¦Óù㷺¡£ÓÃI2O5¿É¶¨Á¿²â¶¨COµÄº¬Á¿£¬¸Ã·´Ó¦Ô­ÀíΪ5CO+I2O55CO2+I2¡£ÆäʵÑé²½ÖèÈçÏ£º

¢ÙÈ¡250 mL£¨±ê×¼×´¿ö£©º¬ÓÐCOµÄijÆøÌåÑùƷͨ¹ýÊ¢ÓÐ×ãÁ¿I2O5µÄ¸ÉÔï¹ÜÖÐÔÚ170 ¡æϳä·Ö·´Ó¦£»

¢ÚÓÃˮһÒÒ´¼Òº³ä·ÖÈܽâ²úÎïI2£¬ÅäÖÆ100 mLÈÜÒº£»

¢ÛÁ¿È¡²½Öè¢ÚÖÐÈÜÒº25.00 mLÓÚ׶ÐÎÆ¿ÖУ¬È»ºóÓÃ0.01 mol¡¤L-1µÄNa2S2O3±ê×¼ÈÜÒºµÎ¶¨¡£ÏûºÄ±ê×¼Na2S2O3ÈÜÒºµÄÌå»ýÈç±íËùʾ¡£

µÚÒ»´Î

µÚ¶þ´Î

µÚÈý´Î

µÎ¶¨Ç°¶ÁÊý/mL

2.10

2.50

1.40

µÎ¶¨ºó¶ÁÊý/mL

22.00

22.50

21.50

£¨1£©²½Öè¢ÚÖÐÅäÖÆ100 mL´ý²âÈÜÒºÐèÒªÓõ½µÄ²£Á§ÒÇÆ÷µÄÃû³ÆÊÇÉÕ±­¡¢Á¿Í²¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜºÍ____________________¡£

£¨2£©Na2S2O3±ê׼ҺӦװÔÚ__________£¨Ìî×Öĸ£©ÖС£

£¨3£©Ö¸Ê¾¼ÁӦѡÓÃ__________£¬ÅжϴﵽµÎ¶¨ÖÕµãµÄÏÖÏóÊÇ____________________________________¡£

£¨4£©ÆøÌåÑùÆ·ÖÐCOµÄÌå»ý·ÖÊýΪ__________£¨ÒÑÖª£ºÆøÌåÑùÆ·ÖÐÆäËû³É·Ö²»ÓëI2O5·´Ó¦£º2Na2S2O3+I2=2NaI+Na2S4O6£©

£¨5£©ÏÂÁвÙ×÷»áÔì³ÉËù²âCOµÄÌå»ý·ÖÊýÆ«´óµÄÊÇ__________£¨Ìî×Öĸ£©¡£

a£®µÎ¶¨Öյ㸩ÊÓ¶ÁÊý

b£®×¶ÐÎÆ¿Óôý²âÈÜÒºÈóÏ´

c£®µÎ¶¨Ç°ÓÐÆøÅÝ£¬µÎ¶¨ºóûÓÐÆøÅÝ

d£®ÅäÖÆ100 mL´ý²âÈÜҺʱ£¬ÓÐÉÙÁ¿½¦³ö

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖª¡¢¡¢¾ùΪÖ÷×åÔªËØ£¬·ÖÎö±íÖÐÊý¾Ý£¬ÅжÏÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨£©£¨ £©

ÔªËØ

×îÍâ²ãµç×ÓÊý

b

a

Ô­×Ӱ뾶/

0.152

0.143

0.186

A.ÓëλÓÚͬһÖ÷×壬ÇÒÔÚµÄÉÏÒ»ÖÜÆÚ

B.ÓëλÓÚͬһÖ÷×壬ÇÒÔÚµÄÏÂÒ»ÖÜÆÚ

C.ÓëλÓÚͬһÖÜÆÚ£¬ÇÒµÄÔ­×ÓÐòÊýСÓÚµÄÔ­×ÓÐòÊý

D.ÓëλÓÚͬһÖ÷×壬ÇÒµÄÔ­×ÓÐòÊýСÓÚµÄÔ­×ÓÐòÊý

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸