(9·ÝÊÔ¾í»ã×Ü)2019-2020ѧÄ꼪ÁÖÊ¡³¤´ºÊл¯Ñ§¸ßÒ»(ÉÏ)ÆÚÄ©¿¼ÊÔÄ£ÄâÊÔÌâ ÏÂÔØ±¾ÎÄ

¢Û¶¼ÊÇ˫ԭ×Ó·Ö×Ó ¢Ü¾ßÓÐÏàͬµÄ·Ö×ÓÊýÄ¿ A£®¢Ù¢Û B£®¢Ú¢Ü C£®¢Ù¢Ü D£®¢Û¢Ü

9£®ÏÂÁÐÎïÖʼÈÄܸúÁòËá·´Ó¦£¬ÓÖÄܸúÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÊÇ( ) ¢ÙNaHCO3£» ¢Ú Na2CO3£»¢Û Al(OH)3£»¢ÜNH4Cl£»¢Ý Al£»¢ÞAl2O3 A£®¢Ù¢Û¢Ý¢Þ

B£®¢Û¢Ü¢Ý¢Þ

C£®¢Ù¢Ú¢Û¢Ü

D£®¢Ú¢Û¢Ý¢Þ

10£®ÏÂÁÐÓйØÌú¼°Æä»¯ºÏÎïµÄ˵·¨ÖÐÕýÈ·µÄÊÇ£¨ £© A£®Fe3O4ÊÇÒ»ÖÖºìרɫ·ÛÄ©£¬Ë׳ÆÌúºì

B£®Fe(OH)2Ϊ°×É«¹ÌÌ壬²»ÈÜÓÚË®£¬ÄÜÔÚ¿ÕÆøÖÐÎȶ¨´æÔÚ C£®Fe2(SO4)3ÓëKSCN·´Ó¦²úÉúѪºìÉ«³Áµí

D£®³ýÈ¥FeCl3ÈÜÒºÖеÄFeCl2ÔÓÖÊ£¬¿ÉÒÔÏòÈÜÒºÖÐͨÈëÊÊÁ¿µÄÂÈÆø 11£®Ä³ÎÞÉ«ËáÐÔÈÜÒºÖУ¬Ôò¸ÃÈÜÒºÖÐÒ»¶¨Äܹ»´óÁ¿¹²´æµÄÀë×Ó×éÊÇ ( ) A£®Fe¡¢Ba¡¢NO3¡¢Cl C£®Na¡¢K¡¢SO3¡¢NO3 ÊÇ

A.¸ÃÎïÖÊÊôÓÚÁ½ÐÔÇâÑõ»¯Îï

B.¸ÃÎïÖÊÊÇAl(OH)3ºÍNa2CO3µÄ»ìºÏÎï C.1molNaAl(OH)2CO3×î¶à¿ÉÏûºÄ3molHF D.¸ÃÒ©¼Á²»ÊʺÏÓÚθÀ£Ññ»¼Õß·þÓÃ

13£®³¬µ¼²ÄÁÏΪ¾ßÓÐÁãµç×è¼°´ÅÐÔµÄÎïÖÊ£¬ÒÔY2O3¡¢BaCO3ºÍCuOΪԭÁÏ£¬¾­ÑÐÄ¥ÉÕ½á¿ÉºÏ³ÉÒ»ÖÖ¸ß㬵¼ÎïÖÊYBa2Cu3Ox¡£ÏÖÓûºÏ³É0.5 mol´Ë¸ß㬵¼ÎÒÀ»¯Ñ§¼ÆÁ¿Êý±ÈÀýÐèÈ¡Y2O3¡¢BaCO3ºÍCuOµÄÎïÖʵÄÁ¿·Ö±ðΪ£¨ £©

A£®0.50 mol¡¢0.50 mol¡¢0.50 mol C£®0.50 mol¡¢1.0 mol¡¢1.5 mol 14£®ÏÂÁÐÎïÖÊÊôÓÚµç½âÖʵÄÊÇ

A£®Í­ B£®ÂÈ»¯ÄÆ C£®Ò»Ñõ»¯µª D£®ÒÒ´¼

15£®½à²ÞÁéÓë84Ïû¶¾Òº»ìºÏ»á²úÉúÂÈÆø£º2HCl + NaClO ¡ú NaCl + Cl2¡ü + H2O£¬ÏÂÁÐ˵·¨´íÎóµÄÊÇ £®£®A.NaClO×÷Ñõ»¯¼Á C.Ñõ»¯ÐÔ£ºNaClO > Cl2

B.n(Ñõ»¯¼Á)£ºn(»¹Ô­¼Á)=1£º2 D.Cl2¼ÈÊÇÑõ»¯²úÎïÓÖÊÇ»¹Ô­²úÎï B£®0.25 mol¡¢1.0 mol¡¢1.5 mol D£®1.0 mol¡¢0.25 mol¡¢0.17 mol

+

+

2£­

£­

2+

2+

£­

£­

B£®Na¡¢NH4¡¢SO4¡¢Cl D£®Na¡¢K¡¢MnO4¡¢Br

+

+

£­

£­

++2£­£­

12£®Ë«ôÇ»ùÂÁ̼ËáÄÆÊÇÒ½ÁÆÉϳ£ÓõÄÒ»ÖÖÒÖËá¼Á£¬Æä»¯Ñ§Ê½ÊÇNaAl(OH)2CO3¡£¹ØÓÚ¸ÃÎïÖʵÄ˵·¨ÕýÈ·µÄ

16£®Cu·Û·ÅÈëÏ¡H2SO4ÖУ¬ÎÞÏÖÏ󣬵±ÔÙ¼ÓÈëÒ»ÖÖÑκó£¬Cu·ÛÖð½¥Èܽ⣬´ËÑÎÊÇ A£®NaCl B£®CuSO4 C£®KNO3 D£®Na3PO4

17£®ÔÚ¼ÓÈëÂÁ·ÛÄܷųöÇâÆøµÄÈÜÒºÖУ¬ÏÂÁи÷×éÀë×Ó¿ÉÄÜ´óÁ¿¹²´æµÄÊÇ( ) A£®Na+¡¢K+¡¢AlO2- ¡¢CO32- B£®Ag+¡¢Cu2+¡¢NO3-¡¢Cl- C£®Mg¡¢Fe¡¢I¡¢SO4 D£®NH4¡¢K¡¢SO4¡¢CO3

18£®Èç¹ûa gÄ³ÆøÌåÖÐËùº¬ÓеķÖ×ÓÊýΪb£¬Ôòc g¸ÃÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýÊÇ£¨Ê½ÖÐNAΪ°¢·ü¼ÓµÂÂÞ³£Êý£© ( )

2+

3+

-2-+

+

2-2-

22.4bcA. L

aNAA£®Ö»ÓÐFe2+

22.4baB. L

cNAB£®Ö»ÓÐFe3+

22.4acC. L

bNAC£®ÓÐFe2+ºÍCu2+

22.4bD. acNAD£®ÓÐFe3+ºÍCu2+

19£®°ÑÍ­·ÛºÍ¹ýÁ¿µÄÌú·Û¼ÓÈëµ½ÈȵÄŨÏõËáÖУ¬³ä·Ö·´Ó¦ºóÈÜÒºÖдóÁ¿´æÔڵĽðÊôÑôÀë×ÓÊÇ£¨ £© 20£®ÎªÅäÖÆÒ»¶¨Ìå»ý¡¢Ò»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÂÈ»¯ÄÆÈÜÒº£¬±ØÐëÓõ½µÄÒÇÆ÷ÊÇ

A£® B£® C£® D£®

21£®»¯Ñ§Óë»·¾³¡¢²ÄÁÏ¡¢ÐÅÏ¢¡¢ÄÜÔ´¹ØÏµÃÜÇУ¬ÏÂÁÐ˵·¨´íÎóµÄÊÇ( ) £®£®A£®µÍ̼Éú»î×¢ÖØ½ÚÄܼõÅÅ£¬¼õÉÙÎÂÊÒÆøÌåµÄÅÅ·Å

B£®ÍƹãʹÓÿɽµ½âËÜÁϼ°²¼ÖʹºÎï´ü£¬ÒÔ¼õÉÙ¡°°×É«ÎÛȾ¡±

C£®°ëµ¼ÌåÐÐÒµÖÐÓÐÒ»¾ä»°£º¡°´Óɳ̲µ½Óû§¡±£¬¼ÆËã»úоƬµÄ²ÄÁÏÊǶþÑõ»¯¹è D£®ÆôÓôóÆøÖÐϸ¿ÅÁ£ÎPM2.5£©µÄ¼à²â£¬ÒÔ×·¸ùËÝÔ´£¬²ÉÈ¡´ëÊ©£¬¸ÄÉÆ¿ÕÆøÖÊÁ¿ 22£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ A£®ËáÓêÊÇÖ¸pH СÓÚ 7 µÄ½µË® B£®Na2O2¡¢Na2O ¶¼ÊǼîÐÔÑõ»¯Îï

C£®³ÇÊÐ¿ÕÆøÖÊÁ¿±¨¸æÖаüÀ¨ PM2.5¡¢SO2¡¢NO2¡¢CO2 µÈÎÛȾÎï¡£ D£®¸ÖÊÇÓÃÁ¿×î´ó¡¢ÓÃ;×î¹ãµÄºÏ½ð 23£®ÏÂÁÐËù¸øÎïÖʵÄÐÔÖʺÍÓÃ;²»Ïà·ûµÄÊÇ

A£®Na2O2 µ­»ÆÉ«¹ÌÌå¡¢¹©Ñõ¼Á B£®Fe2O3ºìרɫ·ÛÄ©¡¢Á¶ÌúÔ­ÁÏ C£®Cl2 dzÂÌÉ«ÆøÌå¡¢×ÔÀ´Ë®Ïû¶¾ D£®SO2 ÎÞÉ«ÆøÌå¡¢¹¤ÒµÖÆÁòËá 24£®ÏÂÁÐÓйػ¯Ñ§·´Ó¦¹ý³Ì»òʵÑéÏÖÏóµÄÐðÊöÖÐÕýÈ·µÄÊÇ( ) A£®ÂÈÆøµÄË®ÈÜÒº¿ÉÒÔµ¼µç£¬ËµÃ÷ÂÈÆøÊǵç½âÖÊ

B£®Æ¯°×·ÛºÍÃ÷·¯¶¼³£ÓÃÓÚ×ÔÀ´Ë®µÄ´¦Àí£¬¶þÕßµÄ×÷ÓÃÔ­ÀíÊÇÏàͬµÄ C£®ÂÈÆø¿ÉÒÔʹÏÊ»¨ÍÊÉ«£¬ËµÃ÷ Cl2 ÓÐÆ¯°×ÐÔ

D£®ÂÈÆø¿ÉÒÔʹʪÈóµÄÓÐÉ«²¼ÌõÍÊÉ«£¬µ«Êµ¼ÊÆðƯ°××÷ÓõÄÎïÖÊÊÇ´ÎÂÈËá¶ø²»ÊÇÂÈÆø 25£®ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨ £©

A£®¹ÌÌåÂÈ»¯ÄƲ»µ¼µç£¬ËùÒÔÂÈ»¯ÄÆÊǷǵç½âÖÊ B£®Í­Ë¿Äܵ¼µç£¬ËùÒÔÍ­Êǵç½âÖÊ

C£®ÂÈ»¯ÇâË®ÈÜÒºÄܵ¼µç£¬ËùÒÔÂÈ»¯ÇâÊǵç½âÖÊ D£®CO2ÈÜÓÚË®Äܵ¼µç£¬ËùÒÔCO2Êǵç½âÖÊ ¶þ¡¢Ìî¿ÕÌâ

26£®Ä³Î¶Èʱ£¬ÔÚ2 LµÄÃܱÕÈÝÆ÷ÖУ¬X¡¢Y¡¢ZÈýÖÖÎïÖʵÄÎïÖʵÄÁ¿ËæÊ±¼äµÄ±ä»¯ÇúÏßÈçͼËùʾ¡£

£¨1£©ÆðʼʱXµÄŨ¶ÈΪc(X)£½________£»·´Ó¦´Ó¿ªÊ¼ÖÁ2·ÖÖÓÄ©£¬YµÄת»¯ÂÊΪ¦Á(Y)£½________£»ÓÃZµÄŨ¶È±ä»¯±íʾ0¡«2·ÖÖÓÄ򵀮½¾ù·´Ó¦ËÙÂÊΪv(Z)£½________¡£ £¨2£©ÓÉͼÖÐËù¸øÊý¾Ý½øÐзÖÎö£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________¡£ Èý¡¢ÍƶÏÌâ

27£®A¡¢B¡¢C¡¢D¶¼ÊÇÖÐѧ»¯Ñ§Öг£¼ûµÄÎïÖÊ£¬ÆäÖÐA¡¢B¡¢C¾ùº¬ÓÐͬһÖÖÔªËØ£¬ÔÚÒ»¶¨Ìõ¼þÏÂÏ໥ת»¯¹ØÏµÈçÏ£¨²¿·Ö·´Ó¦ÖеÄË®ÒÑÂÔÈ¥£©¡£¸ù¾ÝÌâÒâ»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈôA¡¢B¡¢CµÄÑæÉ«·´Ó¦¾ùΪ»ÆÉ«£¬CΪ·¢½Í·ÛµÄ³É·ÖÖ®Ò»£¬DµÄ¹ý¶ÈÅÅ·Å»áÔì³ÉÎÂÊÒЧӦ¡£ ¢ÙBµÄË×ÃûΪ_______________¡£

¢Ú·´Ó¦¢ñµÄÀë×Ó·½³ÌʽÊÇ_________________________________________¡£ £¨2£©ÈôA¡¢D¾ùΪµ¥ÖÊ£¬ÇÒAÎªÆøÌ壬DÔªËØµÄÒ»ÖÖºìרɫÑõ»¯Îï³£ÓÃ×÷ÑÕÁÏ¡£ ¢Ùд³öAµÄÒ»ÖÖÓÃ;_________________________¡£

¢Ú·´Ó¦¢óµÄÀë×Ó·½³ÌʽÊÇ_________________________________________¡£

¢Û¼ìÑéBµÄÈÜÒºÖÐÑôÀë×ӵķ½·¨ÊÇ________________________________________________¡£ ËÄ¡¢×ÛºÏÌâ

28£®ÊµÑéÊÒÀûÓÃÁòÌú¿óÉÕÔü(Ö÷Òªº¬ Fe2O3¡¢SiO2 µÈ)ÖÆ±¸¼îʽÁòËáÌú[Fea(OH)b (SO4)c]ÈÜÒº£¬ ²¢²â¶¨Æä×é³É¡£

(1)Ëá½þʱ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___¡£

(2)²Ù×÷ a µÄÃû³ÆÎª___£¬X Ϊ___(Ìѧʽ)¡£

(3)¼Ó CaCO3 µ÷½ÚÈÜÒºµÄ pH£¬ÆäÄ¿µÄÊÇÖкÍÈÜÒºÖеÄËᣬ²¢Ê¹ Fe2(SO4)3 ת»¯Îª Fea(OH)b (SO4)c¡£ÈôÈÜ ÒºµÄ pH Æ«¸ß£¬½«»áµ¼ÖÂÈÜÒºÖÐÌúÔªËØµÄº¬Á¿½µµÍ£¬ÆäÔ­ÒòÊÇ___(ÓÃÎÄ×Ö±íÊö)¡£

(4)Ϊ²â¶¨¼îʽÁòËáÌúµÄ×é³É£¬È¡Ò»¶¨Á¿ÑùÆ·Óë×ãÁ¿ÑÎËá·´Ó¦£¬ËùµÃÈÜҺƽ¾ù·ÖΪÁ½·Ý¡£Ò»·ÝÈÜÒºÖмÓÈë×ãÁ¿µÄ BaCl2 ÈÜÒº£¬µÃµ½°×É«³Áµí 1.7475 g¡£ÁíÒ»·ÝÈÜÒº£¬ÏȽ« Fe3+»¹Ô­Îª Fe2+£¬³ä·Ö·´Ó¦ºóÏòÈÜÒºÖÐµÎ¼Ó 0.02000 mol¡¤L K2 Cr2O7ÈÜÒº£¬ÍêÈ«·´Ó¦Ê±ÏûºÄK2 Cr2O7ÈÜÒº 50.00 mL¡£Çó¸ÃÑùÆ·µÄ»¯Ñ§Ê½(ÒÑÖª£ºCr2O72-£«6Fe2+£«14H+=2Cr3+£«6Fe3+£«7H2O)__________________¡£(Çëд³ö¼ÆËã¹ý³Ì) Î塢ʵÑéÌâ

29£®ÊµÑéÊÒÓÃNaOH¹ÌÌåÅäÖÆ250 mL 1.25 mol¡¤L£­1µÄNaOHÈÜÒº£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺ (1)ÅäÖÆÊ±±ØÐëÓõ½µÄ²£Á§ÒÇÆ÷ÓУºÉÕ±­¡¢²£Á§°ô¡¢_____________¡£

(2)ÅäÖÆÊ±£¬ÆäÕýÈ·µÄ²Ù×÷˳ÐòÊÇ(×Öĸ±íʾ£¬Ã¿¸ö×ÖĸֻÄÜÓÃÒ»´Î)________________¡£ A£®ÓÃ30 mLˮϴµÓÉÕ±­2¡«3´Î£¬Ï´µÓÒº¾ù×¢ÈëÈÝÁ¿Æ¿£¬Õñµ´

B£®ÓÃÌìÆ½×¼È·³ÆÈ¡ËùÐèµÄNaOHµÄÖÊÁ¿£¬¼ÓÈëÉÙÁ¿Ë®(Ô¼30 mL)£¬Óò£Á§°ôÂýÂý½Á¶¯£¬Ê¹Æä³ä·ÖÈܽâ C£®½«ÒÑÀäÈ´µÄNaOHÈÜ񼄯²£Á§°ô×¢Èë250 mLÈÝÁ¿Æ¿ÖÐ D£®½«ÈÝÁ¿Æ¿¸Ç½ô£¬µßµ¹Ò¡ÔÈ

E£®¸ÄÓýºÍ·µÎ¹Ü¼ÓË®£¬Ê¹ÈÜÒº°¼ÒºÃæÇ¡ºÃÓë¿Ì¶ÈÏßÏàÇÐ F£®¼ÌÐøÍùÈÝÁ¿Æ¿ÄÚСÐļÓË®£¬Ö±µ½ÒºÃæ½Ó½ü¿Ì¶È1¡«2 cm´¦ (3)ÏÂÁÐÅäÖÆµÄÈÜҺŨ¶ÈÆ«¸ßµÄÊÇ______________¡£ A£®³ÆÁ¿NaOHʱ£¬íÀÂë´í·ÅÔÚ×óÅÌ

B£®ÏòÈÝÁ¿Æ¿ÖÐ×ªÒÆÈÜҺʱ(ʵÑé²½ÖèC)²»É÷½«ÒºµÎÈ÷ÔÚÈÝÁ¿Æ¿ÍâÃæ

£­1

C£®¼ÓÕôÁóˮʱ²»É÷³¬¹ýÁ˿̶ÈÏß D£®¶¨ÈÝʱ¸©Êӿ̶ÈÏß

E£®ÅäÖÆÇ°£¬ÈÝÁ¿Æ¿ÖÐÓÐÉÙÁ¿ÕôÁóË®

30£®ÊµÑéÊÒͨ³£ÓÃÈçÏÂͼËùʾµÄ×°ÖÃÀ´ÖÆÈ¡°±Æø¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÖÆÈ¡°±ÆøÊ±·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º_________________________________________ ¡£ (2)ÊÕ¼¯°±ÆøÊ±±ØÐëʹÓøÉÔïµÄ¼¯Æø×°ÖõÄÔ­ÒòÊÇ_______________________________¡£ (3)ÏÂÁвÙ×÷²»ÄÜÓÃÓÚ¼ìÑéNH3µÄÊÇ£¨_________£© A£®ÆøÌåʹʪÈóµÄ·Ó̪ÊÔÖ½±äºì B£®ÆøÌåÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶ C£®ÆøÌåÓëÕºÓÐŨH2SO4µÄ²£Á§°ô¿¿½ü D£®ÆøÌåÓëÕºÓÐŨÑÎËáµÄ²£Á§°ô¿¿½ü

(4)³ý°±ÆøÍ⣬»¹¿ÉÒÔÓøÃ×°ÖÃÖÆÈ¡µÄ³£¼ûÆøÌåÊÇ ________________ ¡£ ¡¾²Î¿¼´ð°¸¡¿*** Ò»¡¢µ¥Ñ¡Ìâ ÌâºÅ ´ð°¸ ÌâºÅ ´ð°¸ 1 C 19 A 2 C 20 C 3 A 21 C 4 C 22 D 5 B 23 C 6 D 24 D 7 D 25 C 2Z

8 B 9 A 10 D 11 B 12 D 13 B 14 B 15 B 16 C 17 A 18 A ¶þ¡¢Ìî¿ÕÌâ 26£®5mol/L10%0.05mol/(L¡¤min)3X+YÈý¡¢ÍƶÏÌâ

27£®´¿¼î£¨»òËÕ´ò£© 2OH¨C+CO2=CO32¨C+H2O Ïû¶¾¡¢Æ¯°×¡¢ÖÆÑÎËá¡¢ÖÆÅ©Ò©¡¢ÖÆÆ¯°×Òº¡¢ÖÆÆ¯°×·ÛºÍƯ·Û¾«¡¢×ö¹¤ÒµÔ­ÁϵÈÈδðÒ»Ïî¾ù¿É 2Fe+Cl2=2Fe+2Cl È¡ÉÙÁ¿BµÄÈÜÒºÓÚÊÔ¹ÜÖУ¬ÏòÆäÖеμÓKSCNÈÜÒº£¬Èô³öÏÖºìÉ«£¬ËµÃ÷BÖеÄÑôÀë×ÓÊÇFe3+ ËÄ¡¢×ÛºÏÌâ

28£®Fe2O3£«3H2SO4£½Fe2(SO4)3£«3H2O ¹ýÂË SiO2 ÈÜÒºµÄpHÆ«¸ß£¬½«»áµ¼ÖÂÈÜÒºÖеÄÌúÔªËØÐγÉFe(OH)3³Áµí n(SO42-)=n(BaSO4)=1.7475g¡Â233g/mol=0.0075mol

ÓÉCr2O72-£«6Fe2+£«14H+=2Cr3+£«6Fe3+£«7H2O¿ÉÖª£º6Fe2+¡«Cr2O72-£¬Ôò¼îʽÁòËáÌúÖÐn(Fe)=0.02mol/L¡Á0.05Lmol¡Á6=0.006mol

¸ù¾ÝµçºÉÊØºã3n(Fe)=2n£¨SO4£©+n£¨OH£©£¬Ôòn£¨OH£©=0.006mol¡Á3-0.0075mol¡Á2=0.003mol ¼îʽÁòËáÌúÖÐn(Fe3+)£ºn£¨OH-£©£ºn(SO42-)=0.006mol£º0.003mol£º0.0075mol=4£º2£º5 ËùÒÔ¼îʽÁòËáÌúµÄ»¯Ñ§Ê½ÎªFe4(OH)2(SO4)5 Î塢ʵÑéÌâ

29£®mLÈÝÁ¿Æ¿ ½ºÍ·µÎ¹Ü BCAFED D 30£®2NH4Cl + Ca(OH)2

3+

2---3+

2+

3+

¨C

CaCl2 + 2NH3¡ü + 2H2O °±Æø¼«Ò×ÈÜÓÚË® C ÑõÆø( O2)