¡¾ÌâÄ¿¡¿Ä³Ö²¡Ï¸¾ú·ÖÃÚµÄÍⶾËØ£¬ÎÞÉ«£¬Ï¸Õë×´½á¾§£¬¶ÔСÊóºÍÈËÌåÓкÜÇ¿µÄ¶¾ÐÔ£¬¿ÉÒýÆðÁ÷ÏÑ¡¢Å»Í¡¢±ãѪ¡¢¾·Âεȣ¬ÒÔÖÂËÀÍö¡£¸ÃÍⶾËØΪ»·×´ëÄ£¬½á¹¹Ê½ÈçͼËùʾ£¬Çë¾Ýͼ·ÖÎö»Ø´ð£º

£¨1£©¸Ã»¯ºÏÎïÖк¬ÓÐÓÎÀëµÄ°±»ù_____________¸ö£¬ôÈ»ù________________¸ö¡£

£¨2£©¸Ã»¯ºÏÎïÊÇÓÉ_____________¸ö°±»ùËá×é³ÉµÄ£¬Çø±ðÕâЩ°±»ùËáµÄÖÖÀàÊÇÒÀ¿¿Æä½á¹¹ÖеÄ_____________¡£

£¨3£©×é³É¸Ã»¯ºÏÎïµÄ°±»ùËáÓÐ___________ÖÖ£¬ÆäÖÐÓÐ_____________¸ö°±»ùËáµÄR»ùÏàͬ£¬Õâ¸öR»ùÊÇ_______________¡£

£¨4£©¸Ã»¯ºÏÎï³ÆΪ»·×´__________ëÄ»¯ºÏÎº¬ÓÐ___________¸öëļü¡£

£¨5£©ÌîдÐéÏß¿òÄڽṹµÄÃû³Æ£ºA._________________£¬B.___________________¡£

£¨6£©¸Ã»¯ºÏÎï¾ßÓÐ8¸öµªÔ­×Ó£¬ÆäÖÐ_____________¸öλÓÚëļüÉÏ£¬____________¸öλÓÚR»ùÉÏ¡£

£¨7£©¸ÃÍⶾËØ»·ëÄÔÚÐγɹý³ÌÖÐʧȥÁË______________¸öË®·Ö×Ó¡£

¡¾´ð°¸¡¿(1)0 0 (2)7 R»ù (3)5 3 ¡ªCH3 (4)Æß 7(5)ëļü R»ù (6)7 1 (7)7

¡¾½âÎö¡¿

£¨1£©ÓÎÀëµÄ°±»ù½á¹¹Îª£­NH2£¬ÓÎÀëµÄôÈ»ù½á¹¹Îª£­COOH¡£

£¨2£©¸Ã»·ëÄÓÐ7¸öëļü£¨£­CO£­NH£­£©£¬¿ÉÍƲâÓÉ7¸ö°±»ùËá×é³É£¬ÆäÖÖÀ಻ͬÓëR»ùÓйء£

£¨3£©¸ù¾ÝR»ù¿ÉÍƲⰱ»ùËáÓÐ5ÖÖ£¬ÆäÖÐÓÐ3¸ö°±»ùµ½µÄR»ùÏàͬ£¬¶¼ÊÇ£­CH3¡£

£¨4£©¸Ã»¯ºÏÎïÓÉ7¸ö°±»ùËáÍÑË®ËõºÏÐγɣ¬³ÆΪÆßëÄ£¬º¬ÓÐ7¸öëļü¡£

£¨5£©AÊÇëļü£¬BÊÇR»ù¡£

£¨6£©Ã¿¸öëļü¶¼ÓÐ1¸öµªÔ­×Ó£¬ËùÒÔ7¸öëļüº¬ÓÐ7¸öµªÔ­×Ó£¬1¸öλÓÚR»ùÉÏ¡£

£¨7£©»·×´ÆßëÄÐγɹý³ÌÖй²ÍÑÈ¥7¸öË®·Ö×Ó¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÃÀ¹ú¡°9.11¡±¿Ö²ÀÏ®»÷ʼþÖУ¬»Ù»µµÄ½¨ÖþÎï·¢³ö´óÁ¿Ê¯ÃÞ£¬ÈËÎüÈëʯÃÞÏËάÒ×»¼·Î°©¡£Ê¯ÃÞÊǹèËáÑοóÎijÖÖʯÃ޵Ļ¯Ñ§Ê½Îª£ºCaMgxAlySi3O12£¬¸Ã»¯Ñ§Ê½ÖÐx¡¢yµÄÖµ·Ö±ðÊÇ( )

A.2¡¢2B.2¡¢3C.3¡¢2D.4¡¢3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖª¼¸ÖÖÀë×ӵĻ¹Ô­ÄÜÁ¦Ç¿Èõ˳ÐòΪI£­>Fe2£«>Br£­£¬ÏÖÓÐ200 mL»ìºÏÈÜÒºÖк¬FeI2¡¢FeBr2¸÷0.10 mol£¬ÏòÆäÖÐÖðµÎµÎÈëÂÈË®(¼Ù¶¨Cl2·Ö×ÓÖ»ÓëÈÜÖÊÀë×Ó·´Ó¦£¬²»¿¼ÂÇÆäËû·´Ó¦)

£¨1£©ÈôÂÈË®ÖÐÓÐ0.15 mol Cl2±»»¹Ô­£¬ÔòËùµÃÈÜÒºÖк¬ÓеÄÒõÀë×ÓÖ÷ÒªÊÇ________£¬Ê£ÓàFe2£«µÄÎïÖʵÄÁ¿Îª________¡£

£¨2£©ÈôÔ­ÈÜÒºÖÐBr£­ÓÐÒ»°ë±»Ñõ»¯£¬¹²ÏûºÄCl2µÄÎïÖʵÄÁ¿Îª________£¬Èô×îÖÕËùµÃÈÜҺΪ400 mL£¬ÆäÖÐÖ÷ÒªÑôÀë×Ó¼°ÆäÎïÖʵÄÁ¿Å¨¶ÈΪ________¡£

£¨3£©Í¨¹ý¶ÔÉÏÊö·´Ó¦µÄ·ÖÎö£¬ÊÔÅжÏCl2£¬I2£¬Fe3£«£¬Br2ËÄÖÖÑõ»¯¼ÁµÄÑõ»¯ÄÜÁ¦ÓÉÇ¿µ½ÈõµÄ˳ÐòÊÇ_________________________________________________¡£

£¨4£©ÉÏÊö·´Ó¦ÈôÔ­ÈÜÒºÖÐÈÜÖÊÀë×ÓÈ«²¿±»Ñõ»¯ºó£¬ÔÙµÎÈë×ãÁ¿ÂÈË®£¬ÔòI2È«²¿±»Cl2Ñõ»¯³ÉHIO3(Ç¿Ëá)¡£ÊÔд³ö´Ë·´Ó¦µÄÀë×Ó·½³Ìʽ£º_________________£»ÉÏÊöËùÓз´Ó¦¹²ÏûºÄCl2________mol¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Èçͼ±íʾ²»Í¬»¯Ñ§ÔªËØËù×é³ÉµÄ»¯ºÏÎÒÔÏÂ˵·¨²»ÕýÈ·µÄÊÇ

A. ÈôͼÖТÙΪijÖÖ¶à¾ÛÌåµÄµ¥Ì壬Ôò¢Ù×î¿ÉÄÜÊÇ°±»ùËá

B. Èô¢Ú´æÔÚÓÚƤϺÍÄÚÔàÆ÷¹ÙÖÜΧµÈ²¿Î»£¬Ôò¢ÚÊÇÖ¬·¾

C. ¢ÛÒ»¶¨ÊǺËËá»òÊÇ×é³ÉºËËáµÄµ¥ÌåºËÜÕËá

D. Èô¢ÜÊDzÎÓë¹¹³ÉÖ²Îïϸ°û±ÚµÄÒ»ÖÖ¶àÌÇ£¬Ôò¢Ü×î¿ÉÄÜÊÇÏËάËØ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÃºÌ¿È¼ÉÕ²úÉúµÄSO2¡¢CO¡¢NO2µÈÔì³ÉÁËÑÏÖصĴóÆøÎÛȾÎÊÌâ¡£

(1) CaOÄÜÆðµ½¹ÌÁò¡¢½µµÍSO2ÅÅ·ÅÁ¿µÄ×÷Óá£

ÒÑÖª£º¢ÙSO2(g)+ CaO(s)=CaSO3(s) ¡÷H=-402 kJ¡¤mol-1

¢Ú2CaSO3(s)+O2(g)=2CaSO4(s) ¡÷H=-234 kJ¡¤mol-1

¢ÛCaCO3(s)=CO2(g) +CaO(s) ¡÷H = +178 kJ¡¤mol-1

Ôò·´Ó¦2SO2(g)+O2(g)+2CaO(s)= 2CaSO4(s) ¡÷H =________ kJ¡¤mol-1

ÏòȼúÖмÓÈëCaCO3Ò²¿ÉÆðµ½¹ÌÁò×÷Óã¬Èô¹Ì¶¨2molSO2ÏàÓ¦Á¿µÄúÔÚÏàͬÌõ¼þÏ£¬È¼ÉÕʱÏò»·¾³ÊͷųöµÄÈÈÁ¿»á¼õÉÙ______ kJ¡£

(2)ÀûÓÃCOºÍH2ÔÚ´ß»¯¼Á×÷ÓÃϺϳɼ״¼£¬ÊǼõÉÙÎÛȾµÄÒ»ÖÖоٴ룬·´Ó¦Ô­ÀíΪCO(g)+2H2(g) CH3OH(g)¡÷H£¬ÔÚÌå»ý²»Í¬µÄÁ½¸öºãÈÝÃܱÕÈÝÆ÷Öзֱð³äÈë1molCOºÍ2mol H2£¬²âµÃƽºâ»ìºÏÎïÖÐCH3OHµÄÌå»ý·ÖÊýÔÚ²»Í¬Ñ¹Ç¿ÏÂËæζȵı仯Èçͼ¼×¡£

¢ÙÉÏÊöºÏ³É¼×´¼µÄ·´Ó¦ÊÇ______(Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±)·´Ó¦£¬ÅжϵÄÀíÓÉÊÇ_____¡£

¢Úͼ¼×ÖÐA¡¢B¡¢CÈýµãÖз´Ó¦ËÙÂÊ×î´óµÄÊÇ______(Ìîд¡°A¡±¡¢¡° B¡± »ò¡° C¡±)¡£

¢ÛÔÚ300¡æʱ£¬ÏòCµãƽºâÌåϵÖÐÔÙ³äÈë0.25molCO£¬0.5molH2ºÍ0.25molCH3OH£¬¸Ãƽºâ______(Ìî¡°ÏòÕý·´Ó¦·½Ïò¡±¡¢¡°ÏòÄæ·´Ó¦·½Ïò¡±»ò¡°²»¡±)Òƶ¯¡£

(3)Ò»¶¨Î¶ÈÏ£¬COµÄת»¯ÂÊÓëÆðʼͶÁϱȵı仯¹ØϵÈçͼÒÒËùʾ£¬²âµÃDµãÇâÆøµÄת»¯ÂÊΪ40%£¬Ôòx=______¡£

(4)ÀûÓÃÔ­µç³ØÔ­Àí¿É½«NO2ºÍNH3ת»¯ÎªÎÞÎÛȾÎïÖÊ£¬Æä×°ÖÃÔ­ÀíͼÈçͼ±ûËùʾ£¬Ôò¸º¼«·´Ó¦Ê½Îª______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÀûÓ÷ϱµÔü(Ö÷Òª³É·ÖΪBaS2O3£¬º¬ÉÙÁ¿SiO2)ΪԭÁÏÉú²ú¸ß´¿·ú»¯±µµÄÁ÷³ÌÈçÏ£º

ÒÑÖª:Kap(BaS2O3)=6.96¡Á10-11£¬Kap(BaF2)=1.0¡Á10-6

(1)²½Öè¢Ù³ý²úÉúSO2Í⣬»¹Óе­»ÆÉ«¹ÌÌåÉú³É£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________¡£

(2)²½Öè¢ÚµÄÄ¿µÄÊÇÖк͹ýÁ¿µÄÑÎËᣬ¼ÓÈëNaOHÈÜÒº²»Ò˹ýÁ¿£¬ÆäÔ­ÒòÊÇ__________(ÓÃÀë×Ó·½³Ìʽ±íʾ)¡£

(3)ÂËÒºµÄÖ÷Òª³É·ÖÊÇBaCl2£¬»¹º¬ÓÐÉÙÁ¿NaCl£¬Èܽâ¶ÈÊý¾ÝÈçÏÂ±í£º

ζÈ

20¡æ

40¡æ

60¡æ

80¡æ

100¡æ

NaCl

36.0g

36.6g

37.3g

39.0g

39.8g

BaCl2

35.8g

40.8g

46.2g

52.5g

59.4g

²½Öè¢ÛÒ˲ÉÓÃ_____ (Ìî¡°Õô·¢½á¾§¡±»ò¡°½µÎ½ᾧ¡±)¡£

(4)¹¤ÒµÉÏ¿ÉÓð±Ë®ÎüÊÕSO2£¬²¢Í¨Èë¿ÕÆøʹÆäת»¯Îªï§Ì¬µª·Ê¡£¸Ãת»¯ÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ__________¡£

(5)²½Öè¢ÜÉú³ÉBaF2µÄ·´Ó¦ÀàÐÍΪ____________¡£

¢ÙÈô¸Ã·´Ó¦Î¶ȹý¸ß£¬ÈÝÒ×Ôì³Éc(F-)½µµÍµÄÔ­ÒòÊÇ__________¡£

¢ÚÑо¿±íÃ÷£¬Êʵ±Ôö¼ÓNH4FµÄ±ÈÀýÓÐÀûÓÚÌá¸ßBaF2µÄ²úÂʺʹ¿¶È£¬½«Å¨¶ÈΪ0. 1mol¡¤L-1µÄBaCl2ÈÜÒººÍ0.22 mol¡¤L-1NH4FÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºÖÐc(Ba2+)=__________ mol¡¤L-1¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ a¡¢b¡¢c¡¢dÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ¡£a¡¢b¡¢dͬÖÜÆÚ£¬c¡¢dͬÖ÷×å¡£aµÄÔ­×ӽṹʾÒâͼΪ

£¬ bÓëcÐγɻ¯ºÏÎïµÄµç×ÓʽΪ¡£ÏÂÁбȽÏÖÐÕýÈ·µÄÊÇ£¨ £©

A. Ô­×Ӱ뾶£ºa£¾c£¾d£¾b B. µç¸ºÐÔa£¾b£¾d£¾c

C. Ô­×ÓÐòÊý£ºd£¾a£¾c£¾b D. ×î¸ß¼Ûº¬ÑõËáµÄËáÐÔc£¾d£¾a

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÊµÑéÊÒÓÃÂÈËá¼ØºÍ¶þÑõ»¯ÃÌÖÆÈ¡ÑõÆøµÄ·´Ó¦·½³ÌʽΪ2KClO32KCl£«3O2¡ü¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¸Ã·´Ó¦Öб»Ñõ»¯µÄÔªËصÄÃû³ÆΪ____£¬Éú³É1 mol O2ʱתÒƵç×ÓµÄÊýÄ¿ÊÇ________¡£

£¨2£©´Ó·´Ó¦ºóµÄ¹ÌÌå»ìºÏÎïÖзÖÀë³öÄÑÈÜÓÚË®µÄMnO2µÄ¾ßÌåʵÑé²Ù×÷Ãû³Æ£º________¡£

£¨3£©·ÖÀë³öµÄMnO2¿ÉÓÃÓÚʵÑéÊÒÖÆÈ¡Cl2£¬»¯Ñ§·½³ÌʽΪMnO2£«4HCl(Ũ)MnCl2£«Cl2¡ü£«2H2O£¬ÆäÀë×Ó·½³ÌʽΪ________________________________¡£

£¨4£©ÈôÁ½¸ö·´Ó¦ÖÐתÒƵĵç×ÓµÄÎïÖʵÄÁ¿Ïàͬ£¬ÔòÉú³ÉµÄO2ºÍCl2ÔÚÏàͬ״¿öϵÄÌå»ý±ÈΪ________¡£

£¨5£©ÓÃË«ÏßÇÅ·¨±êÃ÷MnO2£«4HCl(Ũ)MnCl2£«Cl2¡ü£«2H2Oµç×ÓתÒƵķ½ÏòºÍÊýÄ¿_______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ï±íÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬Çë»Ø´ðÓйØÎÊÌ⣺

¢ñA

¢òA

¢óA

¢ôA

¢õA

¢öA

¢÷A

0

2

¢Ù

¢Ú

¢Û

3

¢Ü

¢Ý

¢Þ

¢ß

¢à

4

¢á

¢â

£¨1£©±íÖл¯Ñ§ÐÔÖÊ×î²»»îÆõÄÔªËØ£¬ÆäÔ­×ӽṹʾÒâͼΪ _________________________ ¡£

£¨2£©±íÖÐÄÜÐγÉÁ½ÐÔÇâÑõ»¯ÎïµÄÔªËØÊÇ __________ £¨ÓÃÔªËØ·ûºÅ±íʾ£©£¬Ð´³ö¸ÃÔªËØÓë¢á×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï·´Ó¦µÄ»¯Ñ§·½³Ìʽ ________________________________ ¡£

£¨3£©¢ÜÔªËØÓë¢ßÔªËØÐγɻ¯ºÏÎïµÄµç×Óʽ _________________________ ¡£

£¨4£©¢Ù¡¢¢Ú¡¢¢Þ¡¢¢ßËÄÖÖÔªËصÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÖÐËáÐÔ×îÇ¿µÄÊÇ ___________£¨Ìѧʽ£©¡£

£¨5£©¢ÛÔªËØÓë¢âÔªËØÁ½Õߺ˵çºÉÊýÖ®²îÊÇ ____________ ¡£

£¨6£©Éè¼ÆʵÑé·½°¸£º±È½Ï¢ßÓë¢âµ¥ÖÊÑõ»¯ÐÔµÄÇ¿Èõ£¬Ç뽫·½°¸ÌîÈëÏÂ±í¡£

ʵÑé²½Öè

ʵÑéÏÖÏóÓë½áÂÛ

______________

________________

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸