¡¾20Ì×¾«Ñ¡ÊÔ¾íºÏ¼¯¡¿ºâË®ÖÐѧ2019-2020ѧÄê¸ß¿¼»¯Ñ§Ä£ÄâÊÔ¾íº¬´ð°¸ ÏÂÔر¾ÎÄ

¢Û_______________¡£

(4)·´Ó¦¢ÛµÄv-tͼÏñÈçͼ1Ëùʾ£¬ÈôÆäËûÌõ¼þ²»±ä£¬Ö»ÊÇÔÚ·´Ó¦Ç°Ôö´óÈÝÆ÷Ìå»ýʹѹǿ¼õС£¬ÔòÆäv-tͼÏñÈçͼ2Ëùʾ¡£

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

¢Ùa1>a2 ¢Úa1b2 ¢Üb1t2 ¢Þt1=t2 ¢ß t1

(5)Óɼ״¼ÒºÏàÍÑË®·¨Ò²¿ÉÖƶþ¼×ÃÑ£¬Ê×ÏȽ«¼×´¼ÓëŨÁòËá·´Ó¦Éú³ÉÁòËáÇâ¼×õ¥

(CH3OSO3H)CH3OH +H2SO4¡úCH3OSO3H+H2O£»Éú³ÉµÄÁòËáÇâ¼×õ¥Ôٺͼ״¼·´Ó¦Éú³É¶þ¼×ÃÑ£¬µÚ¶þ²½µÄ·´Ó¦·½³ÌʽΪ____________¡£ÓëCO2ºÍH2·´Ó¦ÖƱ¸¶þ¼×ÃѱȽϣ¬¸Ã¹¤ÒÕµÄÓŵãÊÇ·´Ó¦Î¶ȵͣ¬×ª»¯Âʸߣ¬ÆäȱµãÊÇ______________¡£ 35.¡¾»¯Ñ§¡ªÑ¡ÐÞ3ÎïÖʽṹÓëÐÔÖÊ¡¿(15·Ö)

(1)µÚËÄÖÜÆÚµÄijÖ÷×åÔªËØ£¬ÆäµÚÒ»ÖÁÎåµçÀëÄÜÊý¾ÝÈçÏÂͼ1Ëùʾ£¬Ôò¸ÃÔª½ô¶ÔÓ¦Ô­×ÓµÄM²ãµç×ÓÅŲ¼Ê½Îª________________¡£

(2)ÈçÏÂͼ2 Ëùʾ£¬Ã¿ÌõÕÛÏß±íʾÖÜÆÚ±íIVA-VIIAÖеÄijһ×åÔªËØÇ⻯ÎïµÄ·Ðµã±ä»¯¡£Ã¿¸öСºÚµã´ú±íÒ»ÖÖÇ⻯ÎÆäÖÐaµã´ú±íµÄÊÇ____________¡£

(3)»¯ºÏÎï(CH3)3NÓëÑÎËá·´Ó¦Éú³É[(CH3)3NH]+£¬¸Ã¹ý³ÌÐÂÉú³ÉµÄ»¯Ñ§¼üΪ__________(ÌîÐòºÅ)¡£ a.Àë×Ó¼ü b.Åäλ¼ü c.Çâ¼ü d.·Ç¼«ÐÔ¹²¼Û¼ü Èô»¯ºÏÎï(CH3)3NÄÜÈÜÓÚË®£¬ÊÔ½âÎöÆäÔ­Òò__________¡£

(4)CO2ÔÚ¸ßθßѹÏÂËùÐγɵľ§ÌåÆäÆ·°ûÈçÏÂͼ3Ëùʾ¡£¸Ã¾§ÌåµÄÈÛµã±ÈSiO2¾§Ìå______(Ñ¡Ìî¡°¸ß¡±»ò¡°µÍ¡±)£¬¸Ã¾§ÌåÖÐ̼ԭ×Ó¹ìµÀµÄÔÓ»¯ÀàÐÍΪ____________¡£

(5)ÈçͼΪ20¸ö̼ԭ×Ó×é³ÉµÄ¿ÕÐÄÁý×´·Ö×ÓC20£¬¸ÃÁý×´½á¹¹ÊÇÓÉÐí¶àÕýÎå±ßÐι¹³ÉÈçͼ¡£C20·Ö×ÓÖÐÿ¸ö̼ԭ×ÓÖ»¸úÏàÁÚµÄ3¸ö̼ԭ×ÓÐγɻ¯Ñ§¼ü£»

Ôò C20·Ö×Ó¹²ÓÐ____¸öÕýÎå±ßÐΣ¬¹²ÓÐ______ÌõËó±ß¡£

(6)Cu2+µÈ¹ý¶ÉÔªËØË®ºÏÀë×ÓÊÇ·ñÓÐÑÕÉ«ÓëÔ­×ӽṹÓйأ¬ÇÒ´æÔÚÒ»¶¨µÄ¹æÂÉ¡£ ÊÔÍƶÏNi2+µÄË®ºÏÀë×ÓΪ______(Ìî¡°ÓС±»ò¡°ÎÞ¡±)É«Àë×Ó£¬ÒÀ¾ÝÊÇ___________¡£ Àë×Ó Sc3+ Ti3+ Fe2+ Cu2+ n2+ ÑÕÉ« ÎÞÉ« ×ϺìÉ« dzÂÌÉ« À¶É« ÎÞÉ« (7)¾§Ìå¾ßÓйæÔòµÄ¼¸ºÎÍâÐΣ¬¾§ÌåÖÐ×î»ù±¾µÄÖظ´µ¥Ôª³ÆΪ¾§°û¡£ÒÑÖªFexO¾§Ì徧°û½á¹¹ÎªNaClÐÍ£¬ÓÉÓÚ¾§ÌåȱÏÝ£¬xֵСÓÚ1¡£²âÖªFexO¾§ÌåÃܶÈΪp=5.71g/cm3£¬¾§°û±ß³¤Îª4.28¡Á10-10m£¬FexOÖÐxÖµ(¾«È·ÖÁ0.01)Ϊ_______¡£ 36.¡¾»¯Ñ§-Óлú»¯Ñ§»ù´¡¡¿(15·Ö)

ÓлúÎïAº¬Ì¼72.0%¡¢º¬ÇâΪ6.67%,ÆäÓàΪÑõÓÃÖÊÆ×·¨·ÖÎöµÃÖªAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª150£¬AÔÚ¹âÕÕÌõ¼þÏÂÉú³ÉÒ»äå´úÎïB£¬B·Ö×ӵĺ˴Ź²ÕñÇâÆ×ÖÐÓÐ4×é·å£¬·åÃæ»ý±ÈΪ2223£¬B¿É·¢ÉúÈçÏÂת»¯¹Øϵ(ÎÞ»ú²úÎïÂÔ)

ÒÑÖª¢ÙÓëÑõ»¯ÌúÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£¬ÇÒ»·ÉϵÄһԪȡ´úÎïÖ»ÓÐÁ½Öֽṹ ¢Úµ±ôÇ»ùÓëË«¼ü̼ԭ×ÓÏàÁ¬Ê±£¬»á·¢ÉúÈçÏÂת»¯RCH=CHOH¡úRCH2CHO Çë»Ø´ðÏÂÁÐÎÊÌâ

(1)BÖйÙÄÜÍŵÄÃû³ÆÊÇ___________£¬·´Ó¦¢ÜÊôÓÚ______·´Ó¦ÀàÐÍ¡£ (2)AµÄ½á¹¹¼òʽ£º___________¡£

(3)FÓëÒø°±ÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________¡£ (4)д³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ

·´Ó¦¢Ú£º________________¡£·´Ó¦¢Ü_______________________¡£

(5)NÊDZȶà1¸ö̼ԭ×ÓµÄͬϵÎͬʱ·ûºÏÏÂÁÐÒªÇóµÄNµÄͬ·ÖÒì¹¹ÌåÓÐ_____ÖÖ¡£ I.±½»·ÉÏÓÐÁ½¸öÈ¡´ú»ù£»II.ÄÜ·¢ÉúË®½â·´Ó¦£»III.ÓëÂÈ»¯ÌúÈÜÒº·¢ÉúÏÔÉ«·´Ó¦ ÆäÖк˴Ź²ÕñÇâÆ×Ϊ5×é·åµÄΪ_________(д½á¹¹¼òʽ)¡£

²Î¿¼´ð°¸

Ò»¡¢Ñ¡ÔñÌ⣺

7£®¡¾½âÎö¡¿±¾Ì⿼²éSTSECÄÚÈÝ£¬¼´¹Ø×¢¿Æѧ¼¼Êõ¡¢Éç»á¡¢»·±£¡¢ÎÄ»¯·½ÃæµÄÎÊÌ⣬ÊǸ߿¼µÄ·½ÏòÖ®Ò»¡£±¾ÌâÉæ¼°µ½µÄж«Î÷±È½Ï¶à£¬¿ÉÒÔÓÃÅųý·¨½âÌ⣬ÒòΪֲÎïÓÍÊÇС·Ö×Ó£¬ËùÒÔDÑ¡Ïî´íÎ󣬲»ÕýÈ·¡£´ð°¸¾ÍÊÇD¡£

8.¡¾½âÎö¡¿±¾Ì⿼²é»¯Ñ§¼ÆÁ¿£¬ÊǸ߿¼¿¼ÊÔ·½ÏòÖ®Ò»¡£ CuOÏà¶Ô·Ö×ÓÖÊÁ¿ÊÇ80£¬Cu2SÏà¶Ô·Ö×ÓÖÊÁ¿ÊÇ160£¬ AÏîÕýÈ·¡£NaBH4ÖÐÇâÊÇ-1¼Û£¬NaBH4+2H2O¡úNaBO2+4H2¡üÊÇÓëË®ÖеÄ+1¼ÛÇâµÄ¹éÖз´Ó¦£¬BÏîÕýÈ·¡£CÑ¡ÏîÖÐûÓиæËßÈÜÒºÌå»ý2L£¬CÏî´íÎó¡£ÖµµÃÒ»ÌáµÄÊÇDÏҪעÒâÊÇÉú³ÉÁË1molPCl5£¬²»ÊÇ1molPCl3Óë1molCl2·´Ó¦Éú³É1 mol PCl5¡£ËùÒÔ´ð°¸ÊÇC¡£

9.¡¾½âÎö¡¿±¾ÌâÊÇ¿¼²ì±ØÐÞÄÚÈÝÖеÄÓлú»¯Ñ§£¬ÊǸ߿¼±Ø¿¼ÄÚÈÝ¡£1¡«5¶ÔÓ¦µÄ½á¹¹ÖеÄÒ»ÂÈÈ¡´úÎïÖ»ÓÐ1ÖÖµÄÓÐ1ºÍ4£¬ÊÇ2¸ö£¬AÏî´íÎó¡£1¡«5¶ÔÓ¦µÄ½á¹¹ÄÜʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«µÄÊÇ2¡¢3¡¢5£¬BÏî´íÎó¡£1¡«5¶ÔÓ¦µÄ½á¹¹ÖÐËùÓÐÔ­×Ó¾ù¿ÉÄÜ´¦ÓÚͬһƽÃæµÄÓб½1¸ö£¬CÏîÕýÈ·¡£1¡«5¶ÔÓ¦µÄ½á¹¹ÖÐ2¡¢3¡¢5ÄÜʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£¬DÏî´íÎó¡£´ð°¸ÊÇC¡£±¾ÌâģʽÓÐЩÀàËÆ2017пαê1µÄÓлú»¯Ñ§ÌâÐÍ¡£ 10.¡¾½âÎö¡¿±¾Ì⿼²éÊDZØÐÞ2µÚÒ»Õ¡¶Ô­×ӽṹ¡¢ÔªËØÖÜÆÚÂÉ¡·ÄÚÈÝ£¬ÊǸ߿¼±Ø¿¼ÄÚÈÝ£¬ÉÓ´øÒ»µã¶ùÔªËØ»¯ºÏÎï֪ʶ¡£ÌâÐÍÀàËÆ2016пαê1µÄ¸ÃÄÚÈÝÌâ¡£±¾Ìâ¸ÉÈÅÐÅÏ¢±È½Ï¶à£ºWY¡¢LM¡¢¼×ÒÒ±ûµÈµÈ£¬²»Ò×ÀíÇå¡£µ«ÓɱûµÄŨÈÜÒºÓë¼×µ¥ÖʼÓÈÈ·´Ó¦£¬²úÉúÎÞÉ«ÎÛȾÐÔÆøÌå,ÒÔ¼°0. 05 mol/L±ûÈÜÒºµÄpHΪ1£¬¿ÉÍƲâ±ûÊÇŨÁòËᣬ¼×ÊÇ̼µ¥ÖÊ¡£µ«Òª´òÆƶ¨ÊÆ˼ά£ºSO2+Cl2+2H2O==H2SO4+2HCl,¶øÒªÓд´ÐÂ˼ά£º2SO2+O2+2H2O==2H2SO4¡£»¹Òª·ÖÎöµ½ÓÐÕâÑùµÄ·´Ó¦£ºSO2+ H2O2==H2SO4¡£ÍÆÀí³ö£ºWYÒÀ´ÎÊÇHCOS£¬´ð°¸ÊÇB¡£

11.¡¾½âÎö¡¿±¾ÌâÊǵ绯ѧÄÚÈÝ£¬ÊǸ߿¼±Ø¿¼ÄÚÈÝ¡£¿¼²ìÁËÔ­µç³ØµÄÓ¦Óã¨ÎÛË®´¦Àí£©¡¢µç¶ÆµÈ¡£ÓÐÒ»¶¨µÄ×ÛºÏÐÔ¡£A¼«µÄµç¼«·´Ó¦Ê½Îª

-

+H+2e=Cl+

+

+£­£­

£¬BÏî´íÎó¡£Óɸõ缫·´Ó¦¿ÉÖª£ºµ±Íâµç

·ÖÐÓÐ0.2moleתÒÆʱ£¬A¼«ÇøÔö¼ÓµÄHµÄ¸öÊýΪ0.1NA£¬AÏî´íÎó¡£·´Ó¦¹ý³ÌÖм×ÖÐÓÒ±ßÇøÓòµÄµç¼«·´Ó¦Ê½£ºCH3COO¡ª8e+4H2O==2HCO3+9H£¬pHÖð½¥½µµÍ£¬DÏî´íÎó¡£´ð°¸C¡£

12. ¡¾½âÎö¡¿±¾Ì⿼²éʵÑéÌ⣬ÊǸ߿¼±Ø¿¼ÌâÐÍ¡£±¾ÌâÐÍÓÐЩÀàËÆ2017±±¾©µÄ¿¼ÊÔÌ⡣ϡÏõËáÓëÍ­·´Ó¦Éú³ÉÎÞÉ«Ò»Ñõ»¯µªÆøÌ壬AÏî´íÎó¡£Ì¼ÓëÈȵÄŨÏõËá·´Ó¦Éú³É¶þÑõ»¯Ì¼¡¢¶þÑõ»¯µªÆø£¬BÏîÕýÈ·¡£Í­ÓëÈý¼ÛÌú·´Ó¦£ºCu+2Fe3+==Cu2++2Fe2+£¬CÏî´íÎó¡£ÌúË¿ÓëŨÁòËá·¢Éú¶Û»¯·´Ó¦£¬DÏî´íÎó¡£´ð°¸B¡£ 13.¡¾½âÎö¡¿13Ì⴫ͳÉÏ¿¼ÈÜÒºÖеġ°Èý¸öƽºâ¡±£¬±¾ÌâÊÇÈܽâƽºâ£¬ÊǸ߿¼·½ÏòÖ®Ò»¡£ÓÉͼAµã¿ÉÖª£ºBaSO4µÄsp=1.0x10-10£¬Mµã¿ÉÖªBaCO3µÄsp=2.5x10-9£¬AÏî´íÎó¡£ÓÉÓÚͬÀë×ÓЧӦºÍspÖ»Ëæζȱä¶ø±ä£¬BÏî´íÎó¡£Èôʹ0.05molBaSO4È«²¿×ª»¯ÎªBaCO3£¬ÖÁÉÙÒª¼ÓÈë1.30molNa2CO3£¬CÏî´íÎó¡£´ð°¸D¡£ 26. £¨15·Ö£©

£¨1£©ÑùÆ·ÖÐûÓÐÈý¼ÛÌú£¨»òÑùƷûÓбäÖÊ£©£¨1 ·Ö£©£¬¶þ¼ÛÌúÒ×±»ÑõÆøÑõ»¯ÎªÈý¼ÛÌú£¨1 ·Ö£©

--+

£¨2£©¢Ùecfdba£¨2 ·Ö£©¢Ú£¨3£©H2SO4 £¨1 ·Ö£©

£¨2 ·Ö£©Æ«Ð¡£¨1 ·Ö£©

£¨4£©¢Ù¾Æ¾«ÅçµÆ£¨1 ·Ö£©D¡úE¡úB¡úC £¨2 ·Ö£©

¢Ú2FeSO4¡¤xH2O

Fe2O3+SO3¡ü+SO2¡ü+2xH2O¡ü £¨2·Ö£©

¢Û²úÉúSO3ºÍŨÁòËá»áµ¼Öµ¼¹Ü»òD ÖÃÖÐUÐ͹ܶÂÈû£¨2·Ö£© 27. £¨14·Ö£©

£¨1£©MnO2+2Fe2++4H+=2Fe3++Mn2++2H2O £¨2·Ö£© £¨2£©Mn2+¡¢Mg2+¡¢Al3+£¨2·Ö£© £¨3£©d£¨2·Ö£© b£¨2·Ö£©

£¨4£©2Fe(OH)3+3ClO-+4OH-=2FeO42-+3Cl-+5H2O£¨2·Ö£©

£¨5£©¸º£¨2·Ö£© 2.24L£¨2·Ö£© 28.£¨14·Ö£©

£¨1£©£­122.5 £¨2·Ö£© £¨2£©45%£¨2·Ö£©£¨3£©¢ÚÉý¸ßζȣ¬CO2µÄת»¯ÂʽµµÍ£¨2·Ö£© ¢ÛζÈÔ½µÍ£¬Ôö´óͶÁϱÈʹCO2µÄת»¯ÂÊÔö´óµÄÔ½ÏÔÖø £¨2·Ö£©£¨4£©D£¨2·Ö£© £¨5£©CH3OSO3H £«CH3OH¡úCH3OCH3£«H2SO4£¨2·Ö£©

ŨH2SO4¸¯Ê´É豸»òÓÐÁòËá·ÏÒº²úÉú£¨2·Ö£© 35£®[»¯Ñ§¡ª¡ªÑ¡ÐÞ3£ºÎïÖʽṹÓëÐÔÖÊ]£¨15·Ö£© £¨1£©3s23p6£¨1·Ö£© £¨2£©SiH4£»£¨1·Ö£©

£¨3£©b £¨1·Ö£© »¯ºÏÎï(CH3)3NΪ¼«ÐÔ·Ö×ÓÇÒ¿ÉÓëË®·Ö×Ó¼äÐγÉÇâ¼ü £¨2·Ö£© £¨4£©¸ß£»£¨1·Ö£© sp3ÔÓ»¯£¨1·Ö£© £¨5£©12 (1·Ö) 30 (2·Ö)

£¨6£©ÓУ¨1·Ö£© Ni2+µÄ3d¹ìµÀÉÏÓÐδ³É¶Ôµç×Ó£¨2·Ö£© £¨7£©0.92 £¨2·Ö£©36£®[»¯Ñ§-Ñ¡ÐÞ5£ºÓлú»¯Ñ§»ù´¡]£¨15·Ö)

£¨1£© äåÔ­×Ó¡¢ õ¥»ù£¨2·Ö£© Ëõ¾Û·´Ó¦ £¨1·Ö£©

£¨2£© £¨2·Ö£©

£¨3£©CH3CHO+2Ag(NH3)2++2OH- ˮԡ¼ÓÈÈCH3COO-+NH4+ +2Ag¡ý+3NH3+H2O £¨4£©£¨2·Ö£©

£¨5£©9 £¨2·Ö£©£¨2·Ö£©

2·Ö£©

£¨2·Ö£© £¨