ÌâÄ¿ÄÚÈÝ

I. ½ñÓÐH2¡¢Cl2»ìºÏÆø100 mL£¨±ê¿ö£©£¬ÆäÖÐÒ»ÖÖÆøÌåµÄÌå»ýΪ45mL¡£¹âÕÕʹÁ½ÖÖÆøÌå·¢Éú·´Ó¦ºó»Ö¸´µ½±ê¿ö£¬ÆøÌåÌå»ýΪ    mL¡£ÎªÁË˵Ã÷·´Ó¦ºóÆøÌåÖÐH2»òCl2ÓÐÊ£Ó࣬ʹÆøÌåͨ¹ý10 mLË®£¬²¢Ê¹Ê£ÓàÆøÌå¸ÉÔïºó»Ö¸´µ½±ê¿ö£¬
£¨1£©ÈôÈÔÊ£Óà___mL£¬Ö¤Ã÷ÓÐ___Ê£Ó࣬ÀíÓÉÊÇ_____£»
£¨2£©ÈôÈÜÒºÓÐ___ÐÔÖÊ£¬Ö¤Ã÷ÓÐ___Ê£Ó࣬ÀíÓÉÊÇ________________________¡£
II.һλͬѧÉè¼ÆÁËÒ»Ì×ÓÃŨÑÎËáºÍKMnO4¹ÌÌåÖÆÈ¡ÉÙÁ¿ÂÈÆø²¢±È½ÏÂÈÆøÓëµâµ¥ÖʵÄÑõ»¯ÐÔÇ¿ÈõµÄ΢ÐÍ×°Öã¨ÈçͼËùʾ£©¡£

£¨1£©ÏÂÁÐÈÜÒºÄÜÎüÊÕCl2µÄÊÇ________¡£

A£®±¥ºÍʳÑÎË®B£®Na2SO4ÈÜÒº
C£®NaOHÈÜÒºD£®Å¨ÁòËá
£¨2£©ÄÜ˵Ã÷Cl2µÄÑõ»¯ÐÔÇ¿ÓÚI2µÄʵÑéÏÖÏóÊÇ_______________________¡£
£¨3£©ÊµÑéÊÒÖÆÈ¡ÂÈÆøµÄÀë×Ó·½³Ìʽ__________________________¡£

¢ñ¡¢100(2·Ö) £¨1£©10£»H2£»H2²»ÈÜÓÚË®  
£¨2£©Æ¯°×£»Cl2£»Cl2ÈÜÓÚË®ºóÉú³ÉÓÐƯ°×ÐÔµÄHClO (ÿ¿Õ1·Ö)
¢ò¡¢(ÿ¿Õ2·Ö)(1)C
(2)ʪÈóµí·Û¡ªµâ»¯¼ØÊÔÖ½±äÀ¶
(3)MnO2+4H++2Cl£­Mn2++Cl2¡ü+2H2O

½âÎöÊÔÌâ·ÖÎö£º¢ñÂÈÆøÓëÇâÆøµÄ·´Ó¦ÊÇ·´Ó¦Ç°ºóÆøÌåÌå»ýÎޱ仯µÄ·´Ó¦£¬ËùÒÔ·´Ó¦ºóÆøÌåÌå»ýÈÔΪ100ml£¬
£¨1£©ÈôÓà10mlÆøÌåÔòÓàÇâÆø£¬ÒòΪÇâÆø²»ÈÜÓÚË®£»
£¨2£©ÈôÈÜÒºÓÐƯ°×ÐÔ£¬ÔòÖ¤Ã÷ÂÈÆøÓÐÊ£Ó࣬ÒòΪÂÈÆøÈÜÓÚË®ÓдÎÂÈËáÉú³É£¬´ÎÂÈËá¾ßÓÐƯ°×ÐÔ¡£
¢ò£¨1£©ÂÈÆøÈÜÓÚË®ºóÈÜÒºÏÔËáÐÔ£¬Òò´ËÓ¦ÓüîÐÔÈÜÒºÎüÊÕ£¬Ñ¡C£»
£¨2£©ÊªÈóµí·Û¡ªµâ»¯¼ØÊÔÖ½±äÀ¶ËµÃ÷ÂÈÆøÄÜÖû»³öµâ£¬´Ó¶øÄÜ˵Ã÷Cl2µÄÑõ»¯ÐÔÇ¿ÓÚI2£»
£¨3£©ÊµÑéÊÒÖÐÓÃŨÑÎËáÓë¶þÑõ»¯Ã̹²ÈÈÖÆÂÈÆø£¬ÆäÀë×Ó·½³ÌʽΪ£ºMnO2+4H++2Cl£­Mn2++Cl2¡ü+2H2O
¿¼µã£º¿¼²éÂÈÆøµÄÖÆÈ¡¡¢Ñõ»¯ÐÔµÄÅжϡ¢ÂÈË®µÄÐÔÖÊ¡¢»ìºÏÆøÌåÈÜÓÚË®µÄ·ÖÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

A¡¢B¡¢X¡¢Y¡¢Z¡¢WÁùÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬AÊǵؿÇÖк¬Á¿×î¶àµÄ½ðÊôÔªËØ£¬¶ÌÖÜÆÚÖ÷×åÔªËØÖÐBµÄÔ­×Ӱ뾶×î´ó£¬X¡¢Y¡¢Z¡¢WÔªËØÔÚÖÜÆÚ±íÖеÄÏà¶ÔλÖÃÈçÏÂͼËùʾ£¬ÆäÖÐZÔªËØÔ­×Ó×îÍâ²ãµç×ÓÊýÊǵç×Ó²ãÊýµÄ2±¶¡£Çë»Ø´ðÏÂÁÐ
ÎÊÌ⣺

£¨1£©WµÄ×î¸ß¼ÛÑõ»¯ÎﻯѧʽÊÇ           £»ZµÄÔ­×ӽṹʾÒâͼΪ      ¡£
£¨2£©A¡¢B¸÷×Ô×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ  ¡£
£¨3£©AW3¿ÉÓÃÓÚ¾»Ë®£¬ÆäÔ­ÀíÊÇ         ¡££¨ÇëÓÃÀë×Ó·½³Ìʽ±íʾ£©
£¨4£©¹¤ÒµºÏ³ÉXµÄ¼òµ¥Æø̬Ç⻯ÎïÊÇ·ÅÈÈ·´Ó¦¡£ÏÂÁдëÊ©ÖмÈÄÜÌá¸ß
·´Ó¦ËÙÂÊ£¬ÓÖÄÜÌá¸ßÔ­ÁÏת»¯ÂʵÄÊÇ                          ¡£
a£®Éý¸ßζÈ
b£®¼ÓÈë´ß»¯¼Á
c£®½«XµÄ¼òµ¥Æø̬Ç⻯ÎPʱÒÆÀë
d£®Ôö´ó·´Ó¦ÌåϵµÄѹǿ
£¨5£©±ê×¼×´¿öÏ£¬2.24L XµÄ¼òµ¥Æø̬Ç⻯Îï±»200 mL l mol L£­1XµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÈÜÒºÎüÊÕºó£¬ËùµÃÈÜÒºÖÐÀë×ÓŨ¶È´Ó´óµ½Ð¡µÄ˳ÐòÊÇ£¨ÓÃÀë×Ó·ûºÅ±íʾ£©                         ¡£
£¨6£©WY2ÔÚɱ¾úÏû¶¾µÄͬʱ£¬¿É½«¾ç¶¾Ç軯ÎïÑõ»¯³ÉÎÞ¶¾ÆøÌå¶ø³ýÈ¥£¬Ð´³öÓÃWY2£¨·Ðµã9.9¡æ£©Ñõ»¯³ýÈ¥CN£­µÄÀë×Ó·½³Ìʽ         ¡£

2013Äêµ×£¬ÉϺ£ÇàÆÖ·¢ÉúÒ»¼ÒÈý¿ÚÎóʳÑÇÏõËáÑÎÔì³ÉÁ½Äк¢ÉíÍöµÄ²Ò¾ç¡£³£¼ûµÄÑÇÏõËáÑÎÖ÷ÒªÊÇÑÇÏõËáÄÆ£¨NaNO2£©£¬ËüÊÇÒ»ÖÖ°×É«²»Í¸Ã÷¾§Ì壬ËäÈ»ÐÎ×´ºÜÏñʳÑΣ¬¶øÇÒÓÐÏÌ棬µ«Óж¾¡£ÑÇÏõËáÄƺÍÂÈ»¯ÄƵIJ¿·Ö×ÊÁÏÈçÏÂ±í£º

 
ÑÇÏõËáÄÆ£¨NaNO2£©
ÂÈ»¯ÄÆ£¨NaCl£©
Ë®ÈÜÐÔ
Ò×ÈÜ£¬ÈÜÒº³ÊÈõ¼îÐÔ
Ò×ÈÜ£¬ÈÜÒº³ÊÖÐÐÔ
ÈÛµã
271¡æ
801¡æ
·Ðµã
320¡æ»á·Ö½â
1413¡æ
¸úÏ¡ÑÎËá×÷ÓÃ
Óкì×ØÉ«µÄNO2ÆøÌå·Å³ö
ÎÞ·´Ó¦
Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©µªÔªËØ×îÍâ²ãµç×ÓÅŲ¼Ê½Îª________£¬µªÔ­×ÓºËÍâµç×Ó¹²Õ¼¾ÝÁË___¸ö¹ìµÀ¡£ÑÇÏõËáÄÆÖи÷ÔªËØÔ­×Ӱ뾶ÓÉ´óµ½Ð¡ÒÀ´ÎΪ_________£¬ÑÇÏõËáµÄµçÀë·½³ÌʽΪ£º_______________________¡£
£¨2£©ÑÇÏõËáÑÎÖж¾ÊÇÒòΪÑÇÏõËáÑοɽ«Õý³£µÄѪºìµ°°×Ñõ»¯³É¸ßÌúѪºìµ°°×£¬¼´Ñªºìµ°°×ÖеÄÌúÔªËØÓɶþ¼Û±äΪÈý¼Û£¬Ê§È¥Ð¯ÑõÄÜÁ¦£¬Ê¹×éÖ¯³öÏÖȱÑõÏÖÏó£®ÃÀÀ¶ÊÇÑÇÏõËáÑÎÖж¾ºóµÄÓÐЧ½â¶¾¼Á£®ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ(  )
A£®¸ßÌúѪºìµ°°×µÄ»¹Ô­ÐÔ±ÈÑÇÏõËáÑÎÈõ
B£®Ò©Æ·ÃÀÀ¶Ó¦¾ßÓл¹Ô­ÐÔ
C£®Öж¾Ê±ÑÇÏõËáÑη¢Éú»¹Ô­·´Ó¦
D£®½â¶¾Ê±¸ßÌúѪºìµ°°×±»»¹Ô­
£¨3£©NaNO2Óж¾£¬½«º¬¸ÃÎïÖʵķÏˮֱ½ÓÅÅ·Å»áÒýÆðË®ÌåµÄÑÏÖØÎÛȾ£¬ËùÒÔÕâÖÖ·ÏË®±ØÐë´¦Àíºó²ÅÄÜÅÅ·Å¡£´¦Àí·½·¨Ö®Ò»£ºÔÚËáÐÔÌõ¼þÏ£¬ NaNO2ÓëKIµÄÎïÖʵÄÁ¿Îª1:1ʱǡºÃÍêÈ«·´Ó¦£¬ÇÒI-±»Ñõ»¯ÎªI2£¬´Ëʱ²úÎïÖꬵªµÄÎïÖÊ(A)Ϊ________£¨Ìѧʽ£©¡£ÈôÀûÓÃÉÏÊö·½·¨ÖÆA£¬ÏÖÓÐÁ½ÖÖ²Ù×÷²½Ö裺¢ÙÏȽ«·ÏË®ËữºóÔÙ¼ÓKI£»¢ÚÏȽ«KIËữºóÔÙ¼ÓÈë·ÏË®¡£ÄÄÖÖ·½·¨½ÏºÃ£¿____£¨ÌîÐòºÅ¡£¼ÙÉè·ÏË®ÖÐÆäËüÎïÖʲ»·´Ó¦£©
£¨4£©ÈçÒª¼ø±ðÑÇÏõËáÄƺÍÂÈ»¯ÄƹÌÌ壬ÏÂÁз½·¨²»¿ÉÐеÄÊÇ
A£®¹Û²ì²¢±È½ÏËüÃÇÔÚË®ÖеÄÈܽâËÙ¶È       B£®²â¶¨ËüÃǸ÷×ÔµÄÈÛµã
C£®ÔÚËüÃǵÄË®ÈÜÒºÖеμӼ׻ù³È          D£®ÔÚËáÐÔÌõ¼þϼÓÈëKIµí·ÛÊÔÒº

µªÑõ»¯ÎïÊÇ´óÆøÎÛȾÎïÖ®Ò»£¬Ïû³ýµªÑõ»¯ÎïµÄ·½·¨ÓжàÖÖ¡£

£¨1£©ÀûÓü×Íé´ß»¯»¹Ô­µªÑõ»¯Îï¡£ÒÑÖª£º
CH4 (g)£«4NO2(g)£½4NO(g)£«CO2(g)£«2H2O(g)   ¡÷H £½£­574 kJ/mol
CH4(g)£«4NO(g) £½ 2N2(g)£«CO2(g)£«2H2O(g)  ¡÷H £½£­1160 kJ/mol
ÔòCH4 ½«NO2 »¹Ô­ÎªN2 µÄÈÈ»¯Ñ§·½³ÌʽΪ                               ¡£       
£¨2£©ÀûÓÃNH3´ß»¯»¹Ô­µªÑõ»¯ÎSCR¼¼Êõ)¡£¸Ã¼¼ÊõÊÇÄ¿Ç°Ó¦ÓÃ×î¹ã·ºµÄÑÌÆøµªÑõ»¯ÎïÍѳý¼¼Êõ¡£ ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2NH3(g)£«NO(g)£«NO2(g)   2N2(g)£«3H2O(g)¡¡¦¤H < 0
ΪÌá¸ßµªÑõ»¯ÎïµÄת»¯ÂʿɲÉÈ¡µÄ´ëÊ©ÊÇ                 £¨Ð´³ö1Ìõ¼´¿É£©¡£
£¨3£©ÀûÓÃClO2Ñõ»¯µªÑõ»¯Îï¡£Æäת»¯Á÷³ÌÈçÏ£º
NONO2N2
ÒÑÖª·´Ó¦¢ñµÄ»¯Ñ§·½³ÌʽΪ2NO+ ClO2 + H2O £½ NO2 + HNO3 + HCl£¬Ôò·´Ó¦¢òµÄ»¯Ñ§·½³ÌʽÊÇ                £»ÈôÉú³É11.2 L N2£¨±ê×¼×´¿ö£©£¬ÔòÏûºÄClO2          g ¡£
£¨4£©ÀûÓÃCO´ß»¯»¹Ô­µªÑõ»¯ÎïÒ²¿ÉÒÔ´ïµ½Ïû³ýÎÛȾµÄÄ¿µÄ¡£
ÒÑÖªÖÊÁ¿Ò»¶¨Ê±£¬Ôö´ó¹ÌÌå´ß»¯¼ÁµÄ±íÃæ»ý¿ÉÌá¸ß»¯Ñ§·´Ó¦ËÙÂÊ¡£ÈçͼÊÇ·´Ó¦2NO(g) + 2CO(g)2CO2(g)+ N2(g) ÖÐNOµÄŨ¶ÈËæζÈ(T)¡¢µÈÖÊÁ¿´ß»¯¼Á±íÃæ»ý(S)ºÍʱ¼ä(t)µÄ±ä»¯ÇúÏß¡£¾Ý´ËÅжϸ÷´Ó¦µÄ¡÷H     0 (Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°ÎÞ·¨È·¶¨¡±)£»´ß»¯¼Á±íÃæ»ýS1     S2 (Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°ÎÞ·¨È·¶¨¡±)¡£

¢ñ.ijʵÑéAС×éÉè¼ÆÉú²úÑÇÂÈËáÄÆ£¨NaClO2£©µÄÖ÷ÒªÁ÷³ÌÈçͼ£¬

ÒÑÖªNaClO2ÊÇÒ»ÖÖÇ¿Ñõ»¯ÐÔƯ°×¼Á£¬¹ã·ºÓÃÓÚ·ÄÖ¯¡¢Ó¡È¾¹¤Òµ¡£ËüÔÚ¼îÐÔ»·¾³ÖÐÎȶ¨´æÔÚ¡£
£¨1£©Ë«ÑõË®µÄµç×ÓʽΪ           £¬×°ÖâñÖз¢Éú·´Ó¦µÄ»¹Ô­¼ÁÊÇ¡¡¡¡¡¡¡¡¡¡£¨Ìѧʽ£©¡£
£¨2£©AµÄ»¯Ñ§Ê½ÊÇ¡¡¡¡ ¡¡£¬×°Öâóµç½â³ØÖÐAÔÚ¡¡¡¡¡¡¼«Çø²úÉú£¬Èô×°ÖâóÖÐÉú³ÉÆøÌåaΪ 11.2 L(±ê×¼×´¿ö)£¬ÔòÀíÂÛÉÏͨ¹ýµç½â³ØµÄµçÁ¿Îª¡¡¡¡¡¡             £¨ÒÑÖª·¨À­µÚ³£ÊýF="9.65¡Ál" 04C¡¤ mol-1)¡£
£¨3£©×°ÖâòÖз´Ó¦µÄÀë×Ó·½³ÌʽÊÇ¡¡¡¡¡¡¡¡¡¡                        ¡¡¡¡¡¡ ¡£
¢ò£®Ä³ÊµÑéBС×é²â¶¨½ðÊôÎýºÏ½ðÑùÆ·µÄ´¿¶È£¨½öº¬ÉÙÁ¿Ð¿ºÍÍ­£¬×é³É¾ùÔÈ£©£¬½«ÑùÆ·ÈÜÓÚ×ãÁ¿ÑÎËá: Sn+ 2HCl=SnCl2+H2¡ü£¬¹ýÂË£¬Ï´µÓ¡£½«ÂËÒººÍÏ´µÓÒººÏ²¢ÔÙ¼Ó¹ýÁ¿µÄFeCl3ÈÜÒº¡£×îºó¿ÉÓÃÒ»¶¨Å¨¶ÈµÄK2Cr2O7ËáÐÔÈÜÒºµÎ¶¨Éú³ÉµÄFe2+£¬´Ëʱ»¹Ô­²úÎïΪCr3+¡£ÏÖÓÐÎýºÏ½ðÊÔÑù1.23g£¬¾­ÉÏÊö·´Ó¦¡¢²Ù×÷ºó£¬¹²ÓÃÈ¥0.200mol/LµÄK2Cr2O7µÄËáÐÔÈÜÒº15.00mL¡£
£¨4£© ÁÐʽ¼ÆËãÑùÆ·ÖÐÎýµÄÖÊÁ¿·ÖÊý¡£
£¨5£©ÓÃÉÏÊöÑùÆ·Ä£Ä⹤ҵÉϵç½â¾«Á¶Îý£¬Èçͼ£º

b¼«·¢Éúµç¼«·´Ó¦Ê½      £¬µ±µÃµ½11.90g´¿Îýʱ£¬µç½âÖÊÈÜÒºÖÊÁ¿¼õÇá0.54g£¬ÔòÎýºÏ½ðÖÊÁ¿¼õÉÙ_______ g£¨½á¹û±£ÁôһλСÊý£©¡£

Ñо¿Ì¼¼°Æ仯ºÏÎïµÄ×ÛºÏÀûÓöԴٽøµÍ̼Éç»áµÄ¹¹½¨¾ßÓÐÖØÒªµÄÒâÒå¡£ÇëÔËÓÃÏà¹Ø֪ʶÑо¿Ì¼¼°Æ仯ºÏÎïµÄÐÔÖÊ¡£
£¨1£©½üÄêÀ´£¬ÎÒ¹ú´¢ÇâÄÉÃ×̼¹ÜÑо¿È¡µÃÖØ´ó½øÕ¹£¬Óõ绡·¨ºÏ³ÉµÄ̼ÄÉÃ×¹ÜÖг£°éÓдóÁ¿Ì¼ÄÉÃ׿ÅÁ££¨ÔÓÖÊ£©£¬ÕâÖÖ̼ÄÉÃ׿ÅÁ£¿ÉÓÃÑõ»¯Æø»¯·¨Ìá´¿£¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º______C£«________K2Cr2O7£«________¡ª¡ª________CO2¡ü£«________K2SO4£«________Cr2£¨SO4£©3£«________H2O¡£
¢ÙÍê³É²¢ÅäƽÉÏÊö»¯Ñ§·½³Ìʽ¡£
¢ÚÔÚÉÏÊö»¯Ñ§·½³ÌʽÉϱê³ö¸Ã·´Ó¦µç×ÓתÒƵķ½ÏòÓëÊýÄ¿¡£
£¨2£©¸ßÎÂʱ£¬ÓÃCO»¹Ô­MgSO4¿ÉÖƱ¸¸ß´¿MgO¡£
¢Ù750¡æʱ£¬²âµÃÆøÌåÖꬵÈÎïÖʵÄÁ¿SO2ºÍSO3£¬´Ëʱ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_________________________________________________________¡£
¢ÚÓÉMgO¿ÉÖƳɡ°Ã¾£­´ÎÂÈËáÑΡ±È¼Áϵç³Ø£¬Æä×°ÖÃʾÒâͼÈçͼ£¨a£©Ëùʾ£¬¸Ãµç³Ø·´Ó¦µÄÀë×Ó·½³ÌʽΪ________________________________________¡£

£¨a£©¡¡¡¡¡¡¡¡   ¡¡£¨b£©¡¡¡¡¡¡¡¡¡¡£¨c£©
£¨3£©¶þÑõ»¯Ì¼ºÏ³É¼×´¼ÊÇ̼¼õÅŵÄз½Ïò£¬½«CO2ת»¯Îª¼×´¼µÄÈÈ»¯Ñ§·½³ÌʽΪCO2£¨g£©£«3H2£¨g£© CH3OH£¨g£©£«H2O£¨g£©¡¡¦¤H¡£
¢Ù¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪK£½________¡£
¢ÚÈ¡Îå·ÝµÈÌå»ýCO2ºÍH2µÄ»ìºÏÆøÌ壨ÎïÖʵÄÁ¿Ö®±È¾ùΪ1¡Ã3£©£¬·Ö±ð¼ÓÈëζȲ»Í¬¡¢ÈÝ»ýÏàͬµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÉÏÊö·´Ó¦£¬·´Ó¦Ïàͬʱ¼äºó£¬²âµÃ¼×´¼µÄÌå»ý·ÖÊý¦Õ£¨CH3OH£©Ó뷴ӦζÈTµÄ¹ØϵÇúÏßÈçͼ£¨b£©Ëùʾ£¬ÔòÉÏÊöCO2ת»¯Îª¼×´¼·´Ó¦µÄ¦¤H________£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±£©0¡£
¢ÛÔÚÁ½ÖÖ²»Í¬Ìõ¼þÏ·¢Éú·´Ó¦£¬²âµÃCH3OHµÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯Èçͼ£¨c£©Ëùʾ£¬ÇúÏߢñ¡¢¢ò¶ÔÓ¦µÄƽºâ³£Êý´óС¹ØϵΪK¢ñ________K¢ò£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±£©¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø