µÚÒ»·¶ÎÄÍø - רҵÎÄÕ·¶ÀýÎĵµ×ÊÁÏ·ÖÏíÆ½Ì¨

¸ßÈý»¯Ñ§¾«Ñ¡ÊÔÌâ»ã±à£ºÎÞ»úÍÆ¶Ï110Ìâ - ͼÎÄ

À´Ô´£ºÓû§·ÖÏí ʱ¼ä£º2025/5/18 16:24:08 ±¾ÎÄÓÉloading ·ÖÏí ÏÂÔØÕâÆªÎĵµÊÖ»ú°æ
˵Ã÷£ºÎÄÕÂÄÚÈݽö¹©Ô¤ÀÀ£¬²¿·ÖÄÚÈÝ¿ÉÄܲ»È«£¬ÐèÒªÍêÕûÎĵµ»òÕßÐèÒª¸´ÖÆÄÚÈÝ£¬ÇëÏÂÔØwordºóʹÓá£ÏÂÔØwordÓÐÎÊÌâÇëÌí¼Ó΢ÐźÅ:xxxxxxx»òQQ£ºxxxxxx ´¦Àí£¨¾¡¿ÉÄܸøÄúÌṩÍêÕûÎĵµ£©£¬¸ÐлÄúµÄÖ§³ÖÓëÁ½⡣

£¨5£©ÈôB¡¢CµÄÏ¡ÈÜÒº»ìºÏºó£¨²»¼ÓÈÈ£©ÈÜÒº³ÊÖÐÐÔ£¬Ôò¸ÃÈÜÒºÖÐÀë×ÓŨ¶È´Ó´óµ½Ð¡µÄ˳ÐòÊÇ

--´ð°¸£®.(1)NH3;CO2;Al(OH)3+OH=AlO2+2H2O

2---(2)´óÓÚ;CO3+H2OHCO3+OH (3)NaOH>Na2CO3>AlCl3>NH4HSO4

++-(4)NH4+H+2OH=NH3¡ü+2H2O +2-++-(5) C(Na)>C(SO4)> C(NH4)>C(H) =C(OH)

42£®¡¾½­ËÕÊ¡ÄÏͨÖÐѧ2008ѧÄê¶ÈµÚһѧÆÚÆÚÄ©¸´Ï°¸ßÈý»¯Ñ§¡¿Ä³ÖÐѧ»¯Ñ§½Ì²ÄÉÏÓÐÈçϽéÉÜ£º¡°ÔÚÇâÑõ»¯ÄÆÅ¨ÈÜÒºÖмÓÈëÑõ»¯¸Æ£¬¼ÓÈÈ£¬ÖƵð×É«¹ÌÌå¾ÍÊǼîʯ»Ò£¬ËüÊÇË®ºÍ¶þÑõ»¯Ì¼µÄÎüÊÕ¼Á¡±¡£ÓÐÁ½¸öʵÑéС×é¾ö¶¨Í¨¹ýʵÑéÀ´Ì½¾¿¼îʯ»ÒµÄ×é³É¡£ £¨1£©µÚһС×éÉè¼ÆµÄʵÑé·½°¸ÈçÏ£º

w.w.w.k.s.5.u. c.o.m

ÏÂÁйØÓÚÈÜÒºÖÐn(OH-)¡¢n(Ca2+)¡¢n(Na+)Ö®¼äµÄ¹ØÏµÊ½ÕýÈ·µÄÊÇ________ A. n(Na+)+ 2n(Ca2+)= n(OH-) B. 2n(Na+) + 2n(Ca2+)= n(OH-) C. n(Na+)+ n(Ca2+)= 2n(OH-)

ÏÖÓÐ4.0g¼îʯ»Ò,ÉèÆäÖÐn(Na+)= xmol, n(Ca2+)= ymol,ÇëÌîдÏÂÁпոñ: ¼îʯ»ÒµÄ¿ÉÄÜ×é³É NaOH, CaO NaOH, CaO, Ca(OH)2 40x+56y©‚4.0©‚ 40x+74y NaOH, Ca(OH)2 B NaOH, Ca(OH)2 H2O C x,yÖ®¼ä¹ØÏµÊ½ (µÈʽ»ò²»µÈʽ) A

°Ñ±íÖÐA¡¢CµÄ¹ØÏµÊ½ÌîÔÚÏÂÃæ¿Õ¸ñÉÏ£º

A£º__________________________C£º______________________ ¡£

£¨2£©µÚ¶þС×é²éµÃÈçÏÂ×ÊÁÏ£ºÇâÑõ»¯¸ÆÔÚ250¡æÊ±²»·Ö½â£¬¸ßÓÚ250¡æÊ±²Å·Ö½â£»ÇâÑõ»¯ÄÆÔÚ580¡æÊ±²»·Ö½â¡£ËûÃÇÉè¼ÆÁËÈçÏ·½°¸²¢µÃ³öÏàÓ¦Ïà¹ØÊý¾Ý£ºÈ¡ÊÐÊÛ¼îʯ»Ò4.0g£¬ÔÚ250¡æÊ±¼ÓÈÈÖÁºãÖØ£¬²âµÃ¹ÌÌåÖÊÁ¿¼õÉÙÁË0.6g,Ê£Óà¹ÌÌåÔÚ580¡æÊ±¼ÌÐø¼ÓÈÈÖÁºãÖØ£¬²âµÃ¹ÌÌåÖÊÁ¿ÓÖ¼õÉÙÁË0.7g.Çëͨ¹ý¼ÆËãÈ·¶¨¸Ã¼îʯ»Ò¸÷³É·ÖµÄÖÊÁ¿·ÖÊý?£¨Òª¼ÆËã¹ý³Ì£©

´ð°¸£®£¨1£©(3·Ö)A A£º40x+56y=4.0 C£º40x+74y©‚4.0 £¨2£© (3·Ö)NaOH£¥=13£¥ Ca(OH)2£¥=72£¥ H2O£¥=15£¥

43£®¡¾½­ËÕÊ¡ÄÏͨÊÐ2009½ì¸ßÈýÆÚÖмì²â11Ô¡¿ (6·Ö)ÏÂͼ±íʾµÄ·´Ó¦¹ØÏµÖУ¬²¿·Ö²úÎï±»ÂÔÈ¥£¬ÒÑÖª1mol°×É«¹ÌÌå·ÛÄ©XÍêÈ«·Ö½âºó£¬»Ö¸´µ½ÊÒΣ¬Éú³É°×É«¹ÌÌåA£¬ÎÞɫҺÌåB£¬ÎÞÉ«ÆøÌåC¸÷1mol£¬E¡¢X¡¢GµÄÑæÉ«·´Ó¦¾ùΪ»ÆÉ«¡£ w.w.w.k.s.5.u. c.o.m

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)д³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£ºX£º G£º D£º

(2)д³öGÓëC·´Ó¦µÄ»¯Ñ§·½³Ìʽ (3)д³öX+E¡úAµÄÀë×Ó·½³Ìʽ

´ð°¸£®.(1)X£ºNaHC03 G£ºNaAIO2 D£ºAl(OH)3 (2)NaAlO2+C02+2H20=NaHCO3+Al(OH)3¡ý

-Ò»2-(3)HCO3+0H=CO3+H20

44£®¡¾½­ËÕÊ¡Æô¶«ÖÐѧ2009½ì¸ßÈý½×¶Îµ÷Ñв⡿[]£¨8·Ö£©ÏÂÁпòͼÖеÄA~HÊÇÖÐѧ»¯Ñ§Öг£¼ûµÄ°ËÖÖÎïÖÊ£¬ËùÓÐÎïÖʾùÓɶÌÖÜÆÚÔªËØ×é³É£¬ÆäÖÐB¡¢C¡¢G¶¼º¬ÓÐͬһÖÖÔªËØ¡£B¡¢E¡¢FΪµ¥ÖÊ¡£³£Î³£Ñ¹ÏÂDΪҺÌ壬E¡¢FÎªÆøÌ壬FÊÇ¿ÕÆøµÄÖ÷Òª³É·ÖÖ®Ò»¡£CÓÉÁ½ÖÖÔªËØ×é³É£¬Æä¾§ÌåÈÛµã¸ß¡¢Ó²¶È´ó£¬ÉÁ˸מ§Ó¨µÄ¹âÔó¡£A~HÓпÉÄÜ·¢ÉúÈçÏÂת»¯£¬Æä·´Ó¦¹ØÏµÈçͼËùʾ£º

E µãȼ D F A

¼ÓD C ¼ÓD G B A H

£¨1£© д³öE+F¡úDµÄ»¯Ñ§·´Ó¦·½³Ìʽ___________________________________¡£ £¨2£© BµÄ»¯Ñ§Ê½_______£¬AµÄµç×Óʽ__________¡£

£¨3£© ÈçÉÏͼËùʾ£¬AÓëCÁ½ÖÖ¹ÌÌå»ìºÏºó£¬¼ÓÈëÒºÌåD£¬¿ÉÄÜ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ

________________________________£¬_________________________________¡£

´ð°¸£º£¨1£©2H2+O2 2H2O £¨2£©Al ÂÔ£¨Na2O2µÄµç×Óʽ£©

£¨3£©2Na2O2+ 2H2O = 4NaOH + O2¡ü Al2 O3£« 2NaOH= 2NaAlO2£«H2O

45£®¡¾½­ËÕÊ¡ÉäÑôÖÐѧ2009½ì¸ßÈýµÚ¶þ´ÎÔ¿¼¡¿ (10·Ö)A¡«I·Ö±ð±íʾÖÐѧ»¯Ñ§Öг£¼ûµÄÒ»ÖÖÎïÖÊ£¬ËüÃÇÖ®¼äÏ໥¹ØÏµÈçÏÂͼËùʾ£¨²¿·Ö·´Ó¦Îï¡¢Éú³ÉÎïûÓÐÁгö£©£¬ÇÒÒÑÖªGΪÖ÷×åÔªËØµÄ¹Ì̬Ñõ»¯ÎA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖÎïÖÊÖоùº¬Í¬Ò»ÖÖÔªËØ¡£

ÇëÌîдÏÂÁпհףº

£¨1£©A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖÎïÖÊÖÐËùº¬Í¬Ò»ÖÖÔªËØµÄÔ­×ӽṹʾÒâͼΪ_________£» £¨2£©ÈôC¡úDΪ»¯ºÏ·´Ó¦£¬Ôò´ËʱµÄ»¯Ñ§·½³Ìʽ______ ___£» £¨3£©Ð´³ö·´Ó¦¢ÜµÄÀë×Ó·½³Ìʽ_____________ _£» £¨4£©Ð´³ö·´Ó¦¢ÛµÄµç¼«·½³Ìʽ£ºÑô¼«£º______________ Òõ¼«£º____________ £¨5£©´ÓÄÜÁ¿±ä»¯µÄ½Ç¶È¿´£¬¢Ù¢Ú¢Û·´Ó¦ÖÐÊôÓÚ¡÷H£¾0µÄ·´Ó¦ÊÇ_______¡££¨ÌîÐòºÅ£©

´ð°¸£º

£¨1£© (2·Ö)

£¨2£©8Al£«3Fe3O4

£­

4Al2O3£«9Fe (2·Ö)

£­

£¨3£©2Al£«2OH£«2H2O£½2AlO2£«3H2¡ü(2·Ö)

2£­£­3+£­

£¨4£©Ñô¼«£º6O£­12e£½3O2¡ü£»Òõ¼«£º4Al£«12e£½4Al (ÿ¿Õ1·Ö) £¨5£©¢Û (2·Ö)

46£®¡¾³£ÖÝÒ»ÖÐ2009½ì12Ô¸ßÈý»¯Ñ§Ô¿¼¡¿ (10·Ö)ÏÂͼÖÐÿ¸ö·½¿ò±íʾÓйصÄÒ»ÖÖ·´Ó¦Îï»òÉú³ÉÎ·½¿òÖÐ×ÖĸÊÇÎïÖÊ´úºÅ£¬ÒÑÖªB¡¢D¡¢E¡¢F¡¢X¾ùΪµ¥ÖÊ£¬ÊÔÌî¿Õ¡£

D A µç½â H2O µãȼ H AgNO3ÈÜÒº Ï¡ÏõËá °×É«³Áµí E F G H2SO4 X µãȼ I B µãȼ Na2SO4 ×ãÁ¿GÈÜÒº J C D ¢Åд³öA¡¢BµÄ»¯Ñ§£ºA___________£¬B____________¡£

¢ÆÐ´³öIºÍJ·´Ó¦µÄ»¯Ñ§·½³Ìʽ__________________________________________¡£

д³öEºÍË®·´Ó¦µÄÀë×Ó·½³Ìʽ_________________________________________¡£

¢Çд³öCÈÜÒººÍD·´Ó¦µÄÀë×Ó·½³Ìʽ_____________________________________¡£

´ð°¸£º

¢ÅA£ºNaCl B£ºS

£«£­

¢ÆNa2O2£«SO2£½Na2SO4 2Na£«2H2O£½2Na£«2OH£«H2¡ü

£­£«£­£­

¢ÇSO32£«Cl2£«H2O£½2H£«SO42£«2Cl

47£®¡¾09½­ËÕ¸ÓÓܺ£Í·ÖÐѧ¸ßÈý»¯Ñ§µÚ¶þ´ÎÔ¿¼¡¿(¹²10·Ö)ÏÖÓÐA¡¢B¡¢C¡¢D¡¢E¡¢X¡¢YºÍ

MµÈÎïÖÊ£¬ËüÃǾùΪ´¿¾»ÎÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢ÉúÏÂÁÐת»¯£º

ÆäÖУºX¡¢YΪ³£¼û˫ԭ×ÓÆøÌåµ¥ÖÊ(XΪÓÐÉ«ÆøÌå)£»BΪ³£¼û½ðÊôµ¥ÖÊ£¬ÖÊÈí£»CÈÜÓÚÇ¿ËáºÍÇ¿¼î£»EΪÓÉÁ½ÖÖ³£¼ûÔªËØ(°´Ô­×Ó¸öÊýl£º1)×é³ÉµÄÒºÌ壻AΪºÚÉ«»¯ºÏÎï¡£ÉÏÊö¸÷²½×ª»¯ÖÐֻд³öÆäÖÐÒ»ÖÖÉú³ÉÎÆäËüÉú³ÉÎïûÓÐд³ö(Ò²ÓпÉÄÜ·´Ó¦Ö»ÓÐÒ»ÖÖÉú³ÉÎï)¡£ ÊԻشð£º

(1)д³ö»¯Ñ§Ê½£ºx £¬ E ¡£ (2)д³öÀë×Ó·½³Ìʽ£º

A¡úX£º £» C¡úD£º ¡£ (3)X+E·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ ¡£ ´ð°¸£º

¡¢(1)X£ºCl2 E: H2O2(¸÷2·Ö) (2)MnO2+4H+2Cl

++

-

¡÷ 2+

Mn+Cl2¡ü+2H2O

Al2O3+6H=2Al+3H2O(¸÷2·Ö)

(3)Cl2+H2O2=2HCl+O2(¸÷2·Ö) (ÿ¿Õ2·Ö£¬¹²10·Ö)

48£®¡¾2009½­ËÕ¸ÓÓܺ£Í·ÖÐѧ¸ßÈý»¯Ñ§µÚ¶þ´ÎÔ¿¼¡¿Ã¾¼°ÆäºÏ½ðÊÇÒ»ÖÖÓÃ;ºÜ¹ãµÄ½ðÊô²ÄÁÏ£¬

ĿǰÊÀ½çÉÏ60%µÄþÊÇ´Óº£Ë®ÖÐÌáÈ¡µÄ¡£Ä³Ñ§Ð£¿ÎÍâÐËȤС×é´Óº£Ë®É¹ÑκóµÄÑα(Ö÷Òªº¬Na¡¢Mg¡¢Cl¡¢BrµÈ)ÖÐÄ£Ä⹤ҵÉú²úÀ´Ìáȡþ£¬Ö÷Òª¹ý³ÌÈçÏ£º

+

2+

£­

£­

3+

¸ßÈý»¯Ñ§¾«Ñ¡ÊÔÌâ»ã±à£ºÎÞ»úÍÆ¶Ï110Ìâ - ͼÎÄ.doc ½«±¾ÎĵÄWordÎĵµÏÂÔØµ½µçÄÔ£¬·½±ã¸´ÖÆ¡¢±à¼­¡¢ÊղغʹòÓ¡
±¾ÎÄÁ´½Ó£ºhttps://www.diyifanwen.net/c2rzan4ce1z6h1tx45d7638ccg96mxg00734_8.html£¨×ªÔØÇë×¢Ã÷ÎÄÕÂÀ´Ô´£©

Ïà¹ØÍÆ¼ö£º

ÈÈÃÅÍÆ¼ö
Copyright © 2012-2023 µÚÒ»·¶ÎÄÍø °æÈ¨ËùÓÐ ÃâÔðÉùÃ÷ | ÁªÏµÎÒÃÇ
ÉùÃ÷ :±¾ÍøÕ¾×ðÖØ²¢±£»¤ÖªÊ¶²úȨ£¬¸ù¾Ý¡¶ÐÅÏ¢ÍøÂç´«²¥È¨±£»¤ÌõÀý¡·£¬Èç¹ûÎÒÃÇ×ªÔØµÄ×÷Æ·ÇÖ·¸ÁËÄúµÄȨÀû,ÇëÔÚÒ»¸öÔÂÄÚ֪ͨÎÒÃÇ£¬ÎÒÃǻἰʱɾ³ý¡£
¿Í·þQQ£ºxxxxxx ÓÊÏ䣺xxxxxx@qq.com
ÓåICP±¸2023013149ºÅ
Top