(9·ÝÊÔ¾í»ã×Ü)2019-2020ѧÄê¹ã¶«Ê¡Ã·ÖÝÊл¯Ñ§¸ßÒ»(ÉÏ)ÆÚĩͳ¿¼Ä£ÄâÊÔÌâ ÏÂÔØ±¾ÎÄ

C.ÔÚÍ­ÓëŨÏõËáµÄ·´Ó¦ÖУ¬ÏõËáÖ»ÌåÏÖÑõ»¯ÐÔ£¬Ã»ÓÐÌåÏÖËáÐÔ D.¿ÉÓÃÂÁ»òÌúÖÆÈÝÆ÷Ê¢×°Ï¡ÏõËá

16£®Ä³»¯Ñ§ÐËȤС×éÏë´Óº£Ë®ÖÐÌáÈ¡µ­Ë®¡¢NaClºÍBr2£¬Éè¼ÆÁËÈçÏÂʵÑ飺

ʵÑéÖеIJÙ×÷1¡¢²Ù×÷2¡¢²Ù×÷3¡¢²Ù×÷4·Ö±ðΪ( ) A£®ÕôÁó¡¢¹ýÂË¡¢ÝÍÈ¡¡¢·ÖÒº C£®¹ýÂË¡¢Õô·¢¡¢ÕôÁó¡¢·ÖÒº

17£®¹ú¼Êµ¥Î»ÖÆÖбíʾÎïÖʵÄÁ¿µÄµ¥Î»ÊÇ A£®¸ö B£®kg C£®g¡¤mol-1 D£®mol 18£®ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ( )

A£®ÈÛÈÚ״̬Ï»òÔÚË®ÈÜÒºÖÐÄÜ×ÔÉíµçÀë³ö×ÔÓÉÒÆ¶¯µÄÀë×ӵϝºÏÎïÊǵç½âÖÊ B£®·²ÊÇÔÚË®ÈÜÒºÀïºÍÈÛ»¯×´Ì¬Ï¶¼²»Äܵ¼µçµÄÎïÖʽзǵç½âÖÊ C£®Äܵ¼µçµÄÎïÖÊÒ»¶¨Êǵç½âÖÊ

D£®Ä³ÎïÖÊÈô²»Êǵç½âÖÊ£¬¾ÍÒ»¶¨ÊǷǵç½âÖÊ 19£®ÏÂÁÐÊÂʵµÄ½áÂÛ»ò½âÊÍ´íÎóµÄÊÇ£¨ £© A£®·ÖÀëҺ̬¿ÕÆøÖÆÑõÆø¡ª¡ªËµÃ÷·Ö×Ó¿ÉÒÔÔÙ·Ö B£®¹ð»¨¿ª·ÅʱÂúÔ°Æ®Ï㡪¡ªËµÃ÷·Ö×ÓÔÚ²»¶ÏÔ˶¯ C£®ÆøÌå¿ÉѹËõ´¢´æÓÚ¸ÖÆ¿ÖСª¡ªËµÃ÷·Ö×ÓÖ®¼äÓмä¸ô

D£®Ëá¼îÖкͷ´Ó¦¶¼ÄÜÉú³ÉË®¡ª¡ªÊµÖÊÊÇH+ÓëOH¡ª½áºÏÉú²úÁËH2O 20£®ÔÚÈÜÒºÖУ¬ÄÜ´óÁ¿¹²´æµÄÀë×Ó×éÊÇ A£®Na¡¢H¡¢HCO3¡¢NO3 B£®Mg2£«¡¢OH-¡¢SO42-¡¢NO3- C£®K£«¡¢Fe3£«¡¢SO42-¡¢NO3- D£®Na¡¢H¡¢Cl¡¢OH

21£®ÏÂÁгýÔÓËùÑ¡ÓõÄÊÔ¼Á¼°²Ù×÷·½·¨¾ùÕýÈ·µÄÊÇ£¨ £© Ñ¡Ïî A B C D A£®A ´ýÌá´¿µÄÎïÖÊ NaOH(Na2CO3£© ¾Æ¾«(Ë®) CO2(HCl) Fe(Al) B£®B Ñ¡ÓõÄÊÔ¼Á ÑÎËá ±¥ºÍNa2CO3ÈÜÒº ±¥ºÍNaHCO3ÈÜÒº Ï¡ÁòËá C£®C D£®D ²Ù×÷·½·¨ Õô·¢ ÝÍÈ¡·ÖÒº Ï´Æø ¹ýÂË +

+

--£«

£«

--

B£®Õô·¢¡¢¹ýÂË¡¢ÝÍÈ¡¡¢·ÖÒº D£®·ÖÒº¡¢¹ýÂË¡¢ÕôÁó¡¢ÝÍÈ¡

22£®ÏÂÁи÷×éÀë×ÓÖУ¬ÔÚË®ÈÜÒºÖÐÄÜ´óÁ¿¹²´æ£¬ÇÒÎÞɫ͸Ã÷µÄÊÇ A£®Na+¡¢Ba2+¡¢NO3£­¡¢CO32£­ B£®Cu2+¡¢HCO3£­¡¢C1-¡¢K+ C£®Mg2+¡¢K+¡¢HCO3£­¡¢SO42£­ D£®OH- ¡¢Cl- ¡¢Na+ ¡¢Fe3+

23£®ÒªÊ¹ÂÈ»¯ÂÁÈÜÒºÖеÄAlÍêȫת»¯ÎªAl(OH)3³Áµí£¬Ó¦Ñ¡ÓõÄ×î¼ÑÊÔ¼ÁÊÇ A.NaOHÈÜÒº

B.Ï¡ÑÎËá

C.°±Ë®

D.AgNO3ÈÜÒº

24£®Éú»îÖÐÀë²»¿ª»¯Ñ§¡£´Ó»¯Ñ§µÄ½Ç¶È·ÖÎö£¬ÏÂÁÐ×ö·¨²»µ±µÄÊÇ A£®²Í×ÀÉϾƾ«Ê§»ð£¬Á¢¼´ÓÃʪĨ²¼¸ÇÃð B£®¿¾Ì¿»ðʱÔÚ»ð¯ÅÔ·ÅÒ»±­Ë®·ÀÖ¹COÖж¾ C£®×ÔÀ´Ë®³§ÓÃÂÈË®¶ÔÔ­Ë®½øÐÐɱ¾úÏû¶¾

D£®Óá°84¡±Ïû¶¾Òº¶Ô²è¾ß¡¢²Í¾ßºÍ½à¾ß½øÐÐÇåÏ´Ïû¶¾ 25£®ÏÂÁб仯¹ý³ÌÖУ¬Ã»Óз¢Éú»¯Ñ§±ä»¯µÄÊÇ A£®¹ýÑõ»¯ÄÆ·ÅÖÃÔÚ¿ÕÆøÖÐ B£®µªµÄ¹Ì¶¨ C£®ÂÈÆø¼Óѹ±ä³ÉÒºÂÈ D£®NO2 ÈÜÓÚË® ¶þ¡¢Ìî¿ÕÌâ

26£®ÏÖÓÐÏÂÁÐÁùÖÖÎïÖÊ:¢ÙÂÁ ¢ÚNa2CO3 ¢ÛCO2 ¢ÜH2SO4 ¢ÝNaOH ¢ÞºìºÖÉ«µÄÇâÑõ»¯Ìú½ºÌå (1)ÉÏÊöÎïÖÊÖÐÊôÓÚµç½âÖʵÄÓÐ_____________ (ÌîÐòºÅ)¡£ (2)д³ö¢ÚÔÚË®ÖеĵçÀë·½³Ìʽ£º__________________

(3)ÉÏÊöÎïÖÊÖÐÓÐÁ½ÖÖÎïÖÊÔÚË®ÈÜÒºÖз¢Éú·´Ó¦£¬ÆäÀë×Ó·½³ÌʽΪH£«£«OH£­=H2O, Ôò¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________________________________________¡£

(4)Ïò¢ÞµÄÈÜÒºÖÐÖ𽥵μӢܵÄÈÜÒº£¬¿´µ½µÄÏÖÏóÊÇ_____________________________¡£ Èý¡¢ÍƶÏÌâ

27£®X¡¢Y¡¢Z¡¢W¾ùΪÖÐѧ»¯Ñ§µÄ³£¼ûÎïÖÊ£¬Ò»¶¨Ìõ¼þÏÂËüÃÇÖ®¼äÓÐÈçÏÂת»¯¹ØÏµ£º

£¨1£©ÈôWΪһÖÖһԪǿ¼î£¬Y¾ßÓÐÁ½ÐÔ£¬Xµ½ZµÄÀë×Ó·½³ÌʽΪ_____________¡£ £¨2£©ÈôXÊÇÒ»ÖÖ»ÆÂÌÉ«µÄÆøÌ壬YµÄ±¥ºÍÈÜÒºµÎÈë·ÐË®ÖÐÄÜÉú³ÉºìºÖÉ«½ºÌå¡£ ¢Ù¼ìÑéÈÜÒºZÊÇ·ñ±äÖÊ×îÁéÃôµÄÒ»ÖÖÊÔ¼ÁÊÇ____________________£¨Ãû³Æ£©£» ¢ÚYÓëWת»¯ÎªZµÄÀë×Ó·´Ó¦·½³Ìʽ_________________________________£» ¢ÛÒ»¶¨Ìõ¼þÏÂWÓëË®·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________________¡£

£¨3£©X³£ÎÂÏÂÎªÆøÌ廯ºÏÎZÊÇÒ»ÖÖºì×ØÉ«ÆøÌ壬XÆøÌåµÄ¼ìÑé·½·¨____________________£¬Xת»¯ÎªYµÄ»¯Ñ§·½³ÌʽΪ______________________£»ZÓëË®·´Ó¦µÄ¹ý³ÌÖУ¬Ñõ»¯¼ÁÓ뻹ԭ¼ÁµÄÖÊÁ¿±ÈΪ__________________¡£

£¨4£©XΪһԪǿ¼îÈÜÒº£¬WÊÇÐγÉËáÓêµÄÖ÷񻮿Ìå¡£

¢ÙÔòYµ½ZµÄÀë×Ó·½³ÌʽΪ________________________________________£»

¢ÚÇë»­³öÏòº¬0.01molXºÍ0.01molYµÄÈÜÒºÖÐÖðµÎ¼ÓÈë0.1mol/LÏ¡ÑÎËáµÄÌå»ýºÍÉú³ÉÆøÌåµÄÎïÖʵÄÁ¿µÄ¹ØÏµµÄͼÏñ___________¡£

3+

ËÄ¡¢×ÛºÏÌâ

28£®I£®½üÄêÀ´£¬ÎÒ¹ú´¢ÑõÄÉÃ×̼¹ÜÑо¿»ñÖØ´ó½øÕ¹£¬µç»¡·¨ºÏ³ÉµÄ̼ÄÉÃ׹ܣ¬³£°éÓдóÁ¿ÎïÖÊ¡ª¡ªÌ¼ÄÉ

Ã׿ÅÁ££®ÕâÖÖ̼ÄÉÃ׿ÅÁ£¿ÉÓÃÑõ»¯Æø»¯·¨Ìá´¿£®Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º3C+2K2Cr2O7+8H2SO4£¨Ï¡£©=3CO2+2K2SO4+2Cr2(SO4)3 +4H2O

£¨1£©±ê³öÒÔÉÏ»¯Ñ§·½³ÌÊ½×ªÒÆµç×ӵķ½ÏòºÍÊýÄ¿___________¡£ £¨2£©´Ë·´Ó¦µÄÑõ»¯¼ÁÊÇ__________£¬Ñõ»¯²úÎïÊÇ____________£»

£¨3£©ÉÏÊö·´Ó¦ÖÐÈô²úÉú22gÆøÌåÎïÖÊ£¬Ôò×ªÒÆµç×ÓµÄÊýĿΪ___________¡£

II£®ÓÃÖÊÁ¿·ÖÊýΪ36.5%µÄŨÑÎËá(ÃܶÈΪ1.20 g¡¤cm)ÅäÖÆ³É0.5mol¡¤LµÄÏ¡ÑÎËá¡£ÏÖʵÑéÊÒ½öÐèÒªÕâÖÖÑÎËá250mL£¬ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©ÅäÖÆÏ¡ÑÎËáʱ£¬ËùÐèµÄÒÇÆ÷³ýÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜÍ⻹ÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐ___________£» £¨2£©´ËŨÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ_____________£» £¨3£©¾­¼ÆËãÐèÒª_______mLŨÑÎË᣻

£¨4£©Ä³Ñ§ÉúÔÚÉÏÊöÅäÖÆ¹ý³ÌÖУ¬ÏÂÁвÙ×÷»áÒýÆðŨ¶ÈÆ«µÍµÄÊÇ______________¡£ A£®ÈôδÓÃÕôÁóˮϴµÓÉÕ±­ÄڱںͲ£Á§°ô»òδ½«Ï´µÓҺעÈëÈÝÁ¿Æ¿¡£ B£®¶¨ÈÝʱ£¬¸©ÊÓÈÝÁ¿Æ¿µÄ¿Ì¶ÈÏß

C£®ÈÝÁ¿Æ¿ÓÃˮϴµÓºóûÓиÉÔï¾ÍÖ±½Ó½øÐÐÅäÖÆ

D£®¶¨ÈÝʱ²»Ð¡ÐļÓË®³¬¹ý¿Ì¶ÈÏߣ¬ÓýºÍ·µÎ¹ÜÈ¡³ö¶àÓàµÄË®£¬Ê¹ÒºÃæºÃµ½´ï¿Ì¶ÈÏß¡£ Î塢ʵÑéÌâ

29£®Ä³Í¬Ñ§ÓÃÈçÏÂʵÑé̽¾¿Fe¡¢FeµÄÐÔÖÊ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)·Ö±ðȡһ¶¨Á¿ÂÈ»¯Ìú¡¢ÂÈ»¯ÑÇÌú¹ÌÌ壬¾ùÅäÖÆ³É0.1mol/LµÄÈÜÒº¡£ÔÚFeCl2ÈÜÒºÖÐÐè¼ÓÈëÉÙÁ¿Ìúм£¬ÆäÄ¿µÄÊÇ_________________________________¡£

(2)¼×ͬѧȡ2mLFeCl2ÈÜÒº£¬¼ÓÈ뼸µÎÂÈË®£¬ÔÙ¼ÓÈë1µÎKSCNÈÜÒº£¬ÈÜÒº±äºì£¬ËµÃ÷Cl2¿É½«Fe2+Ñõ»¯¡£FeCl2ÈÜÒºÓëÂÈË®·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________¡£

(3)ÒÒͬѧÈÏΪ¼×ͬѧµÄʵÑé²»¹»ÑϽ÷£¬¸ÃͬѧÔÚ2mLFeCl2ÈÜÒºÖÐÏȼÓÈë0.5mLúÓÍ£¬ÔÙÓÚÒºÃæÏÂÒÀ´Î¼ÓÈë1µÎKSCNÈÜÒººÍ¼¸µÎÂÈË®£¬ÈÜÒº±äºì£¬ÃºÓ͵Ä×÷ÓÃÊÇ__________¡£

(4)Á½Í¬Ñ§È¡10ml 0.1mol/LKIÈÜÒº£¬¼ÓÈë6mL0.1mol/LFeCl3ÈÜÒº»ìºÏ¡£È¡2mL´ËÈÜÒºÓÚÊÔ¹ÜÖмÓÈë1mLCCl4³ä·ÖÕñµ´¡¢¾²Öã¬CCl4²ãÏÔ×ÏÉ«£»Ð´³ö·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________¡£

(5)¶¡Í¬Ñ§ÏòÊ¢ÓÐH2O2ÈÜÒºµÄÊÔ¹ÜÖмÓÈ뼸µÎËữµÄFeCl2ÈÜÒº£¬ÈÜÒº±ä³Éר»ÆÉ«£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________£»Ò»¶Îʱ¼äºó£¬ÈÜÒºÖÐÓÐÆøÅݳöÏÖ£¬²¢·ÅÈÈ£¬ËæºóÓкìºÖÉ«³ÁµíÉú³É£¬²úÉúÆøÅݵÄÔ­ÒòÊÇ______________£¬Éú³É³ÁµíµÄ³É·ÖÊÇ________________¡£

30£®Ä³ÊµÑéС×éµÄͬѧΪ̽¾¿ºÍ±È½ÏSO2ºÍÂÈË®µÄƯ°×ÐÔ£¬Éè¼ÆÁËÈçϵÄʵÑé×°Öá£

2+

3+

£­3

£­1

£¨1£©ÊµÑéÊÒÓÃ×°ÖÃEÖÆ±¸Cl2£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________£»ÈôÓÐ8molµÄHCl²Î¼Ó·´Ó¦£¬Ôò×ªÒÆµÄµç×Ó×ÜÊýΪ________¡£

£¨2£©¸Ã×°ÖÃÖÐÇâÑõ»¯ÄÆÈÜÒºµÄ×÷ÓÃÊÇ___________________¡£

£¨3£©Í¨ÆøºóB¡¢DÁ½¸öÊÔ¹ÜÖеÄÏÖÏó______________¡£Í£Ö¹Í¨Æøºó£¬ÔÙ¸øB¡¢DÁ½¸öÊԹֱܷð¼ÓÈÈ£¬Á½¸öÊÔ¹ÜÖеÄÏÖÏó·Ö±ðΪB£º________£¬D£º________¡£

£¨4£©ÁíÒ»¸öʵÑéС×éµÄͬѧÈÏΪSO2ºÍÂÈË®¶¼ÓÐÆ¯°×ÐÔ£¬¶þÕß»ìºÏºóµÄƯ°×ÐԿ϶¨»á¸üÇ¿¡£ËûÃǽ«ÖƵõÄSO2ºÍCl2°´1£º1ͬʱͨÈ뵽ƷºìÈÜÒºÖУ¬½á¹û·¢ÏÖÍÊɫЧ¹û²¢²»ÏñÏëÏóµÄÄÇÑù¡£ÇëÄã·ÖÎö¸ÃÏÖÏóµÄ

Ô­Òò£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©____________________¡£ ¡¾²Î¿¼´ð°¸¡¿*** Ò»¡¢µ¥Ñ¡Ìâ ÌâºÅ ´ð°¸ ÌâºÅ ´ð°¸ 1 C 19 A 2 D 20 C 3 A 21 C 4 A 22 C 5 D 23 C 6 D 24 B 7 B 25 C 8 C 9 D 10 D 11 B 12 A 13 B 14 C 15 A 16 A 17 D 18 A ¶þ¡¢Ìî¿ÕÌâ 26£®¢Ú¢Ü¢Ý Na2CO3£½2Na+£«CO2£­ H2SO4£«2NaOH£½Na2SO4+2H2O ¿ªÊ¼²úÉúºìºÖÉ«³Áµí£¬ºóÀ´³ÁµíÈܽâÏûʧ£¬ÈÜÒº×îÖÕ±äΪ»ÆÉ« Èý¡¢ÍƶÏÌâ

27£®Al3+ +4OH¡ª= AlO2¡ª+2H2O ÁòÇ軯¼ØÈÜÒº 2Fe3+ + Fe === 3Fe2+ 3Fe + 4H2O(g) === Fe3O4 + 4H2 ʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¨»òÓöÕºÓÐŨÑÎËáµÄ²£Á§°ô²úÉú°×ÑÌ£© 4NH3£«5O2

4NO£«

6H2O 1:2 SO32-+SO2+H2O=2HSO3-

ËÄ¡¢×ÛºÏÌâ 28£®A¡¢D Î塢ʵÑéÌâ

29£®·ÀÖ¹Fe2+±»Ñõ»¯ Cl2+2Fe2+£½2Fe3++2Cl- ¸ô¾ø¿ÕÆø(ÅųýÑõÆø¶ÔʵÑéµÄÓ°Ïì) 2Fe3++2I-£½2Fe+I2 H2O2+2Fe+2H£½2Fe+2H2O (Fe´ß»¯)H2O2·Ö½â²úÉúO2 Fe(OH)3 30£®MnO2+4H+2Cl

+

£­

2+

2+

+

3+

3+

K2Cr2O7 CO2 2NA 250mLÈÝÁ¿Æ¿¡¢Á¿Í² 12 mol¡¤L 10.4

£­1

Mn+2H2O+Cl2¡ü 4NA»ò2.408¡Á10 Î²Æø´¦Àí£¬ÎüÊÕ¹ýÁ¿µÄSO2ºÍ Cl2 ,ÒÔ

2+24

·ÀÖ¹ÎÛȾ»·¾³ Æ·ºìÍÊÉ« ÈÜÒº±äΪºìÉ« ÈÜÒºÎޱ仯 SO2+Cl2+2H2O£½4H++2Cl£­+SO42-