ÊÀ½ç»·±£ÁªÃ˽¨ÒéÈ«Ãæ½ûֹʹÓÃÂÈÆøÓÃÓÚÒûÓÃË®µÄÏû¶¾£¬¶ø½¨Òé²ÉÓøßЧ¡°ÂÌÉ«¡±Ïû¶¾¼Á¶þÑõ»¯ÂÈ¡£¶þÑõ»¯ÂÈÊÇÒ»ÖÖ¼«Ò×±¬Õ¨µÄÇ¿Ñõ»¯ÐÔÆøÌ壬Ò×ÈÜÓÚË®¡¢²»Îȶ¨¡¢³Ê»ÆÂÌÉ«£¬ÔÚÉú²úºÍʹÓÃʱ±ØÐ뾡Á¿ÓÃÏ¡ÓÐÆøÌå½øÐÐÏ¡ÊÍ£¬Í¬Ê±Òª±ÜÃâ¹âÕÕ¡¢Õ𶯻ò¼ÓÈÈ¡£ÊµÑéÊÒÒÔµç½â·¨ÖƱ¸ClO2µÄÁ÷³ÌÈçÏ£ºÇë»Ø´ðÏÂÁÐÎÊÌâ¡£

£¨1£©ClO2ÖÐËùÓÐÔ­×Ó      £¨Ìî¡°ÊÇ¡±»ò¡°²»ÊÇ¡±£©¶¼Âú×ã8µç×ӽṹ¡£ÉÏͼËùʾµç½â·¨ÖƵõIJúÎïÖÐÔÓÖÊÆøÌåBÄÜʹʯÈïÊÔÒºÏÔÀ¶É«£¬³ýÈ¥ÔÓÖÊÆøÌå¿ÉÑ¡Óà     ¡£
A£®±¥ºÍʳÑÎË®     B£®¼îʯ»Ò       C£®Å¨ÁòËá       D£®ÕôÁóË®
£¨2£©Îȶ¨ÐÔ¶þÑõ»¯ÂÈÊÇΪÍƹã¶þÑõ»¯Âȶø¿ª·¢µÄÐÂÐͲúÆ·£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ     ¡£
A£®¶þÑõ»¯Âȿɹ㷺ÓÃÓÚ¹¤ÒµºÍÒûÓÃË®´¦Àí
B£®Ó¦ÓÃÔÚʳƷ¹¤ÒµÖÐÄÜÓÐЧµØÑÓ³¤Ê³Æ·Öü²ØÆÚ
C£®Îȶ¨ÐÔ¶þÑõ»¯ÂȵijöÏÖ´ó´óÔö¼ÓÁ˶þÑõ»¯ÂȵÄʹÓ÷¶Î§
D£®ÔÚ¹¤×÷ÇøºÍ³ÉÆ·´¢²ØÊÒÄÚ£¬ÒªÓÐͨ·ç×°Öúͼà²â¼°¾¯±¨×°ÖÃ
£¨3£©Å·ÖÞ¹ú¼ÒÖ÷Òª²ÉÓÃÂÈËáÄÆÑõ»¯Å¨ÑÎËáÖƱ¸£¬»¯Ñ§·´Ó¦·½³ÌʽΪ                           £¬´Ë·¨È±µãÖ÷ÒªÊDzúÂʵ͡¢²úÆ·ÄÑÒÔ·ÖÀ룬»¹¿ÉÄÜÎÛȾ»·¾³¡£
£¨4£©ÎÒ¹ú¹ã·º²ÉÓþ­¸ÉÔï¿ÕÆøÏ¡Ê͵ÄÂÈÆøÓë¹ÌÌåÑÇÂÈËáÄÆ£¨NaClO2£©·´Ó¦ÖƱ¸£¬»¯Ñ§·½³ÌʽÊÇ                         £¬´Ë·¨Ïà±ÈÅ·ÖÞ·½·¨µÄÓŵãÊÇ                         ¡£
£¨5£©¿Æѧ¼Ò×î½üÓÖÑо¿³öÁËÒ»ÖÖеÄÖƱ¸·½·¨£¬ÀûÓÃÁòËáËữµÄ²ÝËᣨH2C2O4£©ÈÜÒº»¹Ô­ÂÈËáÄÆ£¬»¯Ñ§·´Ó¦·½³ÌʽΪ                                                     £¬´Ë·¨Ìá¸ßÁËÉú²ú¼°´¢´æ¡¢ÔËÊäµÄ°²È«ÐÔ£¬Ô­ÒòÊÇ                                                   ¡£

£¨1£©²»ÊÇ     C       £¨2£©A¡¢B¡¢C¡¢D 
£¨3£©2NaClO3£«4HCl(Ũ) =2NaCl£«Cl2¡ü£«2ClO2¡ü£«2H2
£¨4£©2NaClO2 + Cl2 ="2NaCl" + 2ClO2    °²È«ÐԺã¬Ã»ÓвúÉúÓж¾¸±²úÆ· 
£¨5£©H2C2O4£«2NaClO3£«H2SO4 = Na2SO4£«2CO2¡ü£«2ClO2¡ü£«2H2O
·´Ó¦¹ý³ÌÖÐÉú³ÉµÄ¶þÑõ»¯Ì¼Æðµ½Ï¡ÊÍ×÷ÓÃ

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÔÚClO2ÖÐÿ¸öClÔ­×ÓÓë2¸öOÔ­×ÓÐγÉÁËËĶԹ²Óõç×Ó¶Ô¡£ÕâÑùOÔ­×ӵõ½ÁË8µç×ÓµÄÎȶ¨½á¹¹£¬¶øClÔ­×ÓÔòÊÇ11¸ö×îÍâ²ãµç×Ó¡£Òò´Ë²»ÊǶ¼Âú×ã8µç×ӽṹ¡£ÔÚÉÏͼËùʾµç½â·¨ÖƵõIJúÎïÖÐÔÓÖÊÆøÌåBÄÜʹʯÈïÊÔÒºÏÔÀ¶É« £¬Ôò¸ÃÆøÌåΪNH3£¬³ýÈ¥¸ÃÆøÌåʱҪÓÃËáÐÔÎüÊÕ¼ÁÀ´ÎüÊÕ¡£Òò´ËÑ¡ÏîΪC¡££¨2£©A. ¸ßЧ¡°ÂÌÉ«¡±Ïû¶¾¼Á¶þÑõ»¯ÂÈ¡£ËùÒÔ¶þÑõ»¯Âȿɹ㷺ÓÃÓÚ¹¤ÒµºÍÒûÓÃË®´¦Àí¡£ÕýÈ·¡£B.ÓÉÓÚ¶þÑõ»¯ÂÈÊÇÒ»ÖÖÇ¿Ñõ»¯ÐÔÆøÌ壬ʳƷ´¦Óڸû·¾³ÖоͿÉÒÔ·ÀÖ¹±»ÆäËü΢ÉúÎ︯ʴ£¬Òò´ËÄÜÓÐЧµØÑÓ³¤Ê³Æ·Öü²ØÆÚ¡£ÕýÈ·¡£C£®ÉÏÊö·½·¨Éú²úµÄ¶þÑõ»¯ÂȲ»Îȶ¨£¬¼«Ò×±¬Õ¨£¬Òª±ÜÃâ¹âÕÕ¡¢Õ𶯻ò¼ÓÈÈ¡£Ê¹ÆäʹÓôó´óÊܵ½ÁËÏÞÖÆ£¬¶øÎȶ¨ÐÔ¶þÑõ»¯ÂȵijöÏֱؽ«»áÔö¼Ó¶þÑõ»¯ÂȵÄʹÓ÷¶Î§£¬¸üºÃµÄ·¢»ÓÆäÇ¿µÄÑõ»¯ÐÔ¡£ÕýÈ·¡£D.¶þÑõ»¯ÂÈÊÇÒ»ÖÖ¼«Ò×±¬Õ¨µÄÇ¿Ñõ»¯ÐÔÆøÌå,²»Îȶ¨¡¢³Ê»ÆÂÌÉ«£¬ÔÚÉú²úºÍʹÓÃʱ±ØÐ뾡Á¿ÓÃÏ¡ÓÐÆøÌå½øÐÐÏ¡ÊÍ£¬Í¬Ê±Òª±ÜÃâ¹âÕÕ¡¢Õ𶯻ò¼ÓÈÈ¡£ËùÒÔÔÚ¹¤×÷ÇøºÍ³ÉÆ·´¢²ØÊÒÄÚ£¬ÒªÓÐͨ·ç×°Öúͼà²â¼°¾¯±¨×°Öá£ÕýÈ·¡££¨3£©Å·ÖÞ¹ú¼ÒÖ÷Òª²ÉÓÃÂÈËáÄÆÑõ»¯Å¨ÑÎËáÖƱ¸ClO2µÄ»¯Ñ§·´Ó¦·½³ÌʽΪ2NaClO3£«4HCl(Ũ)£½2NaCl£«Cl2¡ü£«2ClO2¡ü£«2H2O¡££¨4£©ÎÒ¹ú¹ã·º²ÉÓþ­¸ÉÔï¿ÕÆøÏ¡Ê͵ÄÂÈÆøÓë¹ÌÌåÑÇÂÈËáÄÆ£¨NaClO2£©·´Ó¦ÖƱ¸ClO2µÄ»¯Ñ§·½³ÌʽÊÇ2NaClO2 + Cl2£½2NaCl + 2ClO2¡£ÓÉ·½³Ìʽ¿ÉÒÔ¿´³ö£º´Ë·¨ÒòΪ²»»á²úÉúÓк¦ÆøÌåCl2£¬Æø̬²úÎï¸üÉÙ£¬±¬Õ¨ÐÔ¸üС£¬Òò´ËÏà±ÈÅ·ÖÞ·½·¨µÄÓŵãÊÇ°²È«ÐԺã¬Ã»ÓвúÉúÓж¾¸±²úÆ·¡££¨5£©ÀûÓÃÁòËáËữµÄ²ÝËᣨH2C2O4£©ÈÜÒº»¹Ô­ÂÈËáÄÆÖÆÈ¡ClO2µÄ»¯Ñ§·´Ó¦·½³ÌʽΪH2C2O4£«2NaClO3£«H2SO4 = Na2SO4£«2CO2¡ü£«2ClO2¡ü£«2H2O¡£´Ë·¨ÓÉÓÚÔÚ·´Ó¦¹ý³ÌÖÐÉú³ÉµÄ¶þÑõ»¯Ì¼Æðµ½Ï¡ÊÍ×÷Óã¬ËùÒÔÌá¸ßÁËÉú²ú¼°´¢´æ¡¢ÔËÊäµÄ°²È«ÐÔ¡£
¿¼µã£º¿¼²é¸ßЧ¡°ÂÌÉ«¡±Ïû¶¾¼Á¶þÑõ»¯ÂȵķÖ×ӽṹ¡¢ÐÔÖʼ°¸÷ÖÖÖÆÈ¡·½·¨ÒÔ¼°ÓÅÁӵıȽϵÄ֪ʶ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Í­µ¥Öʼ°Æ仯ºÏÎïÊÇÓ¦Óü«Æä¹ã·ºµÄÎïÖÊ¡£
(1)Í­ÊÇÇâºó½ðÊô£¬²»ÄÜÓëÑÎËá·¢ÉúÖû»·´Ó¦£¬µ«½«µ¥ÖÊÍ­ÖÃÓÚŨÇâµâËáÖУ¬»áÓпÉȼÐÔÆøÌå¼°°×É«³ÁµíÉú³É£¬ÓÖÖªÑõ»¯ÐÔ£ºCu2£«>I2£¬ÔòÍ­ÓëÇâµâËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________
(2)ÒÑÖªCu2OÄÜÈÜÓÚ´×ËáÈÜÒº»òÑÎËáÖУ¬Í¬Ê±µÃµ½À¶É«ÈÜÒººÍºìÉ«¹ÌÌ壬ÔòCu2OÓëÏ¡ÁòËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________________£»
Cu2OÓëÏ¡ÏõËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________________£»
Ö»ÓÃÏ¡ÁòËáÀ´È·¶¨Ä³ºìÉ«¹ÌÌåÊÇ Cu2OÓëCu×é³ÉµÄ»ìºÏÎïµÄ·½·¨£º³ÆÈ¡m g¸ÃºìÉ«¹ÌÌåÖÃÓÚ×ãÁ¿Ï¡ÁòËáÖУ¬³ä·Ö·´Ó¦ºó¹ýÂË£¬È»ºó___________________¡£

(3)Cu2OÊÇÒ»ÖÖ°ëµ¼Ìå²ÄÁÏ£¬»ùÓÚÂÌÉ«»¯Ñ§ÀíÄîÉè¼ÆµÄÖÆÈ¡Cu2OµÄµç½â×°ÖÃÈçͼËùʾ£¬µç½â×Ü·´Ó¦£º2Cu£«H2OCu2O£«H2¡ü£¬ÔòʯīӦÓëµçÔ´µÄ________¼«ÏàÁ¬£¬Í­µç¼«Éϵĵ缫·´Ó¦Ê½Îª________£»µç½â¹ý³ÌÖУ¬Òõ¼«ÇøÖÜΧÈÜÒºpH________(Ìî¡°±ä´ó¡±¡¢¡°±äС¡±»ò¡°²»±ä¡±)¡£
(4)ÏÖÏòCu¡¢Cu2O¡¢CuO×é³ÉµÄ»ìºÏÎïÖмÓÈë1 L 0.6 mol/L HNO3Ç¡ºÃʹ»ìºÏÎïÈܽ⣬ͬʱÊÕ¼¯µ½2 240 mL NO(±ê×¼×´¿ö)¡£Èô½«ÉÏÊö»ìºÏÎïÓÃ×ãÁ¿µÄÇâÆø»¹Ô­£¬ËùµÃ¹ÌÌåµÄÖÊÁ¿Îª________£»Èô»ìºÏÎïÖк¬ÓÐ0.1 mol Cu£¬½«¸Ã»ìºÏÎïÓëÏ¡ÁòËá³ä·Ö·´Ó¦£¬ÖÁÉÙÏûºÄÁòËáµÄÎïÖʵÄÁ¿Îª________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Áòµ¥Öʼ°Æ仯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Óá£
£¨1£©ÒÑÖª25¡æʱ£ºSO2£¨g£©£«2CO£¨g£©£½2CO2£¨g£©£«1/xSx£¨s£©  ¡÷H£½akJ/mol
2COS£¨g£©£«SO2£¨g£©£½2CO2£¨g£©£«3/xSx£¨s£©  ¡÷H£½bkJ/mol¡£
ÔòCOS£¨g£©Éú³ÉCO£¨g£©ÓëSx£¨s£©·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ                     ¡£
£¨2£©ÐÛ»Æ(As4S4)ºÍ´Æ»Æ(As2S3)ÊÇÌáÈ¡ÉéµÄÖ÷Òª¿óÎïÔ­ÁÏ¡£ÒÑÖªAs2S3ºÍHNO3ÓÐÈçÏ·´Ó¦£ºAs2S3+10H++ 10NO3?=2H3AsO4+3S+10NO2¡ü+ 2H2O£¬µ±Éú³ÉH3AsO4µÄÎïÖʵÄÁ¿Îª0.6 mol·´Ó¦ÖÐתÒƵç×ÓµÄÊýĿΪ       £¬
£¨3£©ÏòµÈÎïÖʵÄÁ¿Å¨¶ÈNa2S¡¢NaOH»ìºÏÈÜÒºÖеμÓÏ¡ÑÎËáÖÁ¹ýÁ¿¡£ÆäÖÐH2S¡¢HS?¡¢S2?µÄ·Ö²¼·ÖÊý£¨Æ½ºâʱijÎïÖÖµÄŨ¶ÈÕ¼¸÷ÎïÖÖŨ¶ÈÖ®ºÍµÄ·ÖÊý£©ÓëµÎ¼ÓÑÎËáÌå»ýµÄ¹ØϵÈçÏÂͼËùʾ£¨ºöÂԵμӹý³ÌH2SÆøÌåµÄÒݳö£©¡£

¢ÙB±íʾ               ¡£
¢ÚµÎ¼Ó¹ý³ÌÖУ¬ÈÜÒºÖÐ΢Á£Å¨¶È´óС¹ØϵÕýÈ·µÄÊÇ      (Ìî×Öĸ)¡£
a£®c(Na+)= c(H2S)+c(HS?)+2c(S2?)
b£®2c(Na+)=c(H2S)+c(HS?)+c(S2?)
c£®c(Na+)=3[c(H2S)+c(HS?)+c(S2?)]
¢ÛNaHSÈÜÒº³Ê¼îÐÔ£¬µ±µÎ¼ÓÑÎËáÖÁMµãʱ£¬ÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ                                       ¡£
£¨4£©¹¤ÒµÉÏÓÃÁòµâ¿ªÂ·Ñ­»·Áª²úÇâÆøºÍÁòËáµÄ¹¤ÒÕÁ÷³ÌÈçÏÂͼËùʾ£º

¢Ù д³ö·´Ó¦Æ÷Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                  ¡£
¢Ú µçÉøÎö×°ÖÃÈçͼËùʾ£¬Ð´³öÑô¼«µÄµç¼«·´Ó¦Ê½              ¡£¸Ã×°ÖÃÖз¢ÉúµÄ×Ü·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

äåµÄÑõ»¯ÐÔ½éÓÚÂȺ͵âÖ®¼ä£¬ÀûÓÃÕâÒ»ÐÔÖʽâ¾öÏÂÃæµÄÎÊÌâ¡£
(1)ÄãÈÏΪ£º½«º¬ÓÐÏÂÁÐÄÄÖÖ·Ö×Ó»òÀë×ÓµÄÊÔ¼Á¼ÓÈëµ½º¬ÓÐBr£­µÄÈÜÒºÖУ¬¿ÉÒÔ½«Br£­Ñõ»¯ÎªBr2__________¡£

A£®I2 B£®I C£®Cl2 D£®Cl£­
(2)Èç¹û°ÑÂÈÆø»ºÂýµØͨÈ뺬ÓÐÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄBr£­¡¢I£­µÄ»ìºÏÈÜÒºÀ__________Ïȱ»Ñõ»¯£¬Ô­ÒòÊÇ_________________________________________
(3)°ÑÂËÖ½Óõí·ÛºÍµâ»¯¼ØµÄ»ìºÏÈÜÒº½þÅÝ£¬ÁÀ¸Éºó¾ÍÊÇʵÑéÊÒ³£Óõĵí·Ûµâ
»¯¼ØÊÔÖ½¡£ÕâÖÖÊÔÖ½ÈóʪºóÓöµ½ÂÈÆø·¢ÉúµÄ±ä»¯ÊÇ_____________£¬Ô­ÒòÊÇ_________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(1)ÏòNaBrºÍKIµÄ»ìºÏÈÜÒºÖУ¬Í¨Èë×ãÁ¿µÄCl2ºó£¬½«ÈÜÒºÕô¸É²¢×ÆÉÕ£¬×îºóÕô·¢ÃóÖÐÊ£ÓàµÄÎïÖÊÊÇ________£¬Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________________________¡£
(2)ÏòKIÈÜÒºÖеÎÈëµí·ÛÈÜÒº£¬ÏÖÏóÊÇ____________________£¬ÔÙµÎÈëÂÈË®£¬ÏÖÏóÊÇ________£¬Óйط´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©½«·Ï·°´ß»¯¼Á(Ö÷Òª³É·ÖV2O5)ÓëÏ¡ÁòËá¡¢ÑÇÁòËá¼ØÈÜÒº»ìºÏ£¬³ä·Ö·´Ó¦£¬ËùµÃÈÜÒºÏÔËáÐÔ£¬º¬VO2£«¡¢K£«¡¢SO42-µÈ¡£Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ_________________________________¡£
£¨2£©ÏòÉÏÊöËùµÃÈÜÒºÖмÓÈëKClO3ÈÜÒº£¬³ä·Ö·´Ó¦ºó£¬ÈÜÒºÖÐÐÂÔö¼ÓÁËVO2+¡¢Cl£­¡£Ð´³ö²¢Åäƽ¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£¬²¢±ê³öµç×ÓתÒƵÄÊýÄ¿ºÍ·½Ïò______________________¡£
£¨3£©ÔÚ20.00 mLµÄ0.1 mol¡¤L£­1 VO2+ÈÜÒºÖУ¬¼ÓÈë0.195 gп·Û£¬Ç¡ºÃÍê³É·´Ó¦£¬Ôò»¹Ô­²úÎï¿ÉÄÜÊÇ______________________________________________________________¡£
a£®V  b£®V2£«  c£®VO2+  d£®VO2£«
£¨4£©ÒÑÖªV2O5ÄܺÍÑÎËá·´Ó¦Éú³ÉÂÈÆøºÍVO2£«¡£ÇëÔÙдһ¸öÀë×Ó·´Ó¦·½³Ìʽ£¬ËµÃ÷»¹Ô­ÐÔ£ºSO32-£¾Cl£­£¾VO2£«__________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Èõµç½âÖʵĵçÀëƽºâ¡¢ÑÎÀàµÄË®½âƽºâºÍÄÑÈÜÎïµÄÈܽâƽºâ¾ùÊôÓÚ»¯Ñ§Æ½ºâ¡£
¢ñ.ÒÑÖªH2AÔÚË®ÖдæÔÚÒÔÏÂƽºâ£ºH2A=H£«£«HA£­£¬HA£­??H£«£«A2£­¡£
£¨1£©³£ÎÂÏÂNaHAÈÜÒºµÄpH________(ÌîÐòºÅ)£¬Ô­ÒòÊÇ_________________¡£
A£®´óÓÚ7    B£®Ð¡ÓÚ7
C£®µÈÓÚ7     D£®ÎÞ·¨È·¶¨
£¨2£©Ä³Î¶ÈÏ£¬ÈôÏò0.1 mol¡¤L£­1µÄNaHAÈÜÒºÖÐÖðµÎµÎ¼Ó0.1 mol¡¤L£­1KOHÈÜÒºÖÁÈÜÒº³ÊÖÐÐÔ(ºöÂÔ»ìºÏºóÈÜÒºµÄÌå»ý±ä»¯)¡£´Ëʱ¸Ã»ìºÏÈÜÒºÖеÄÏÂÁйØϵһ¶¨ÕýÈ·µÄÊÇ________¡£
A£®c(H£«)¡¤c(OH£­)£½1.0¡Á10£­14
B£®c(Na£«)£«c(K£«)£½c(HA£­)£«2c(A2£­)
C£®c(Na£«)£¾c(K£«)
D£®c(Na£«)£«c(K£«)£½0.05 mol¡¤L£­1
£¨3£©ÒÑÖª³£ÎÂÏÂH2AµÄ¸ÆÑÎ(CaA)µÄ±¥ºÍÈÜÒºÖдæÔÚÒÔÏÂƽºâ£ºCaA(s)??Ca2£«(aq)£«A2£­(aq)¡¡¦¤H£¾0¡£ÈôҪʹ¸ÃÈÜÒºÖÐCa2£«Å¨¶È±äС£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ________¡£
A£®Éý¸ßζȠ                  B£®½µµÍζÈ
C£®¼ÓÈëNH4Cl¾§Ìå            D£®¼ÓÈëNa2A¹ÌÌå
¢ò.º¬ÓÐCr2O72-µÄ·ÏË®¶¾ÐԽϴó£¬Ä³¹¤³§·ÏË®Öк¬5.0¡Á10£­3 mol¡¤L£­1µÄCr2O72-¡£ÎªÁËʹ·ÏË®µÄÅÅ·Å´ï±ê£¬½øÐÐÈçÏ´¦Àí£º

(1)¸Ã·ÏË®ÖмÓÈëÂÌ·¯ºÍH£«£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________¡£
(2)Èô´¦ÀíºóµÄ·ÏË®ÖвÐÁôµÄc(Fe3£«)£½2.0¡Á10£­13 mol¡¤L£­1£¬Ôò²ÐÁôµÄCr3£«µÄŨ¶ÈΪ________¡£
(ÒÑÖª£ºKsp[Fe(OH)3]£½4.0¡Á10£­38£¬Ksp[Cr(OH)3]£½6.0¡Á10£­31)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÍÑÁò¼¼ÊõÄÜÓÐЧ¿ØÖÆSO2¶Ô¿ÕÆøµÄÎÛȾ¡£
(1)ÏòúÖмÓÈëʯ»Òʯ¿É¼õÉÙȼÉÕ²úÎïÖÐSO2µÄº¬Á¿£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
_______________________________¡£
(2)º£Ë®³ÊÈõ¼îÐÔ£¬Ö÷Òªº¬ÓÐNa£«¡¢K£«¡¢Ca2£«¡¢Mg2£«¡¢Cl£­¡¢SO42¡ª¡¢Br£­¡¢HCO3¡ªµÈ¡£º¬SO2µÄÑÌÆø¿ÉÀûÓú£Ë®ÍÑÁò£¬Æ乤ÒÕÁ÷³ÌÈçͼËùʾ£º

¢ÙÏòÆØÆø³ØÖÐͨÈë¿ÕÆøµÄÄ¿µÄÊÇ_____________________________________¡£
¢ÚͨÈë¿ÕÆøºóÆØÆø³ØÖк£Ë®ÓëÌìÈ»º£Ë®Ïà±È£¬Å¨¶ÈÓÐÃ÷ÏÔ²»Í¬µÄÀë×ÓÊÇ________¡£
a£®Cl£­¡¡¡¡¡¡ b£®SO42¡ª¡¡¡¡¡¡           c£®Br£­¡¡¡¡¡¡       d£®HCO3¡ª
(3)ÓÃNaOHÈÜÒºÎüÊÕÑÌÆøÖеÄSO2£¬½«ËùµÃµÄNa2SO3ÈÜÒº½øÐеç½â£¬¿ÉµÃµ½NaOH£¬Í¬Ê±µÃµ½H2SO4£¬ÆäÔ­ÀíÈçͼËùʾ(µç¼«²ÄÁÏΪʯī)¡£

¢ÙͼÖÐa¼«ÒªÁ¬½ÓµçÔ´µÄ________(Ìî¡°Õý¡±»ò¡°¸º¡±)¼«£¬C¿ÚÁ÷³öµÄÎïÖÊÊÇ________¡£
¢ÚSO32¡ª·ÅµçµÄµç¼«·´Ó¦Ê½Îª____________________________¡£
¢Ûµç½â¹ý³ÌÖÐÒõ¼«Çø¼îÐÔÃ÷ÏÔÔöÇ¿£¬ÓÃƽºâÒƶ¯µÄÔ­Àí½âÊÍÔ­Òò£º
__________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¼×¡¢ÒÒÁ½Í¬Ñ§Ñо¿Na2SO3ÈÜÒºÓëFeCl3ÈÜÒº·´Ó¦µÄÇé¿ö¡£

²½Öè
²Ù×÷
ÏÖÏó
I
Ïò2 mL 1 mol¡¤L-1FeCl3ÈÜÒºÖмÓÈëÒ»¶¨Á¿µÄNa2SO3ÈÜÒº
ÈÜÒºÓÉ×Ø»ÆÉ«±äΪºìºÖÉ«£¬
²¢ÓÐÉÙÁ¿´Ì¼¤ÐÔÆøζµÄÆøÌåÒݳö
 
£¨1£©³£ÎÂÏ£¬FeCl3ÈÜÒºµÄpH_______7£¨Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±£©¡£
£¨2£©·ÖÎöºìºÖÉ«²úÉúµÄÔ­Òò¡£
¢Ù ¼×ͬѧÈÏΪ²½ÖèIÖÐÈÜÒº³ÊºìºÖÉ«ÊÇÒòΪÉú³ÉÁËFe(OH)3£¬Óû¯Ñ§Æ½ºâÒƶ¯Ô­Àí½âÊÍÈÜÒº³ÊºìºÖÉ«µÄÔ­Òò£º________¡£
¢Ú ÒÒͬѧÈÏΪ¿ÉÄÜÊÇ·¢ÉúÁËÑõ»¯»¹Ô­·´Ó¦£¬Íê³É²¢ÅäƽÆä·´Ó¦µÄÀë×Ó·½³Ìʽ£º
Fe3+ +SO32- +     =F e2+ +       +     
ÒÒͬѧ²éÔÄ×ÊÁϵÃÖª£º
1£®Fe2+ÓëSO32-·´Ó¦Éú³ÉÄ«ÂÌÉ«µÄÐõ×´³ÁµíFeSO3£»
2£®Ä«ÂÌÉ«µÄFeSO3Óë»ÆÉ«µÄFeCl3ÈÜÒº»ìºÏºó£¬ÈÜÒº³ÊºìºÖÉ«¡£
 
£¨3£©¼×ͬѧΪÁËÈ·ÈÏÈÜÒº³ÊºìºÖÉ«µÄÔ­ÒòÊÇÉú³ÉÁËFe(OH)3£¬Éè¼Æ²¢Íê³ÉÁËÈçÏÂʵÑ飺
²½Öè
²Ù×÷
ÏÖÏó
II
Óü¤¹â±ÊÕÕÉä²½ÖèIÖеĺìºÖÉ«ÈÜÒº
³öÏÖ¡°¶¡´ï¶ûЧӦ¡±
¼×ͬѧÒò´ËµÃ³ö½áÂÛ£ºÈÜÒº³ÊºìºÖÉ«ÊÇÒòΪÉú³ÉÁËFe(OH)3¡£¶øÒÒͬѧÈÏΪ¼×ͬѧµÃ³ö½áÂÛµÄÖ¤¾ÝÈÔÈ»²»×㣬ÒÒͬѧµÄÀíÓÉÊÇ________¡£
£¨4£©Îª½øÒ»²½È·ÈÏNa2SO3ÈÜÒºÓëFeCl3ÈÜÒº·´Ó¦µÄÇé¿ö£¬ÒÒͬѧÉè¼Æ²¢Íê³ÉÁËÈçÏÂʵÑ飺
²½Öè
²Ù×÷
ÏÖÏó
III
Ïò1 mol?L-1µÄFeCl3ÈÜÒºÖÐͨÈëÒ»¶¨Á¿µÄSO2
ÈÜÒºÓÉ»ÆÉ«±äΪºìºÖÉ«
IV
Óü¤¹â±ÊÕÕÉä²½ÖèIIIÖеĺìºÖÉ«ÈÜÒº
ûÓгöÏÖ¡°¶¡´ï¶ûЧӦ¡±
 
¢Ù ¾­¼ìÑé²½ÖèIIIÖкìºÖÉ«ÈÜÒºº¬ÓÐFe2+£¬¼ìÑéFe2+Ñ¡ÓõÄÊÔ¼ÁÊÇ_________£¨Ìî×Öĸ£©¡£
a£®K3[Fe(CN)6] ÈÜÒº         b£®KSCNÈÜÒº        c£®KMnO4ÈÜÒº
¢Ú ÒÑÖªH2SO3ÊÇÈõËᣬÇë½áºÏµçÀë·½³Ìʽ˵Ã÷²½ÖèIIIÖгöÏÖºìºÖÉ«µÄÔ­Òò£º       ¡£
£¨5£©½áÂÛ£ºÓÉÉÏÊöʵÑéµÃÖª£¬¼×¡¢ÒÒÁ½Í¬Ñ§Ëù³Ö¹Ûµã¾ùÕýÈ·¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸