ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿µ³µÄÊ®¾Å´ó±¨¸æÌá³ö¡°¼Ó¿ìÉú̬ÎÄÃ÷ÌåÖƸĸ½¨ÉèÃÀÀöÖйú¡±£¬·¢Õ¹Çå½àÄÜÔ´¶Ô½¨ÉèÃÀÀöÖйú¾ßÓÐÖØÒªÒâÒå¡£ÇâÄÜÊÇÒ»ÖÖ¸ßЧÇå½à¡¢¼«¾ß·¢Õ¹Ç±Á¦µÄÄÜÔ´¡£

£¨1£©ÒÔÌ«ÑôÄÜΪÈÈÔ´£¬ÈÈ»¯Ñ§ÁòµâÑ­»··Ö½âË®ÊÇÒ»ÖÖ¸ßЧ¡¢ÎÞÎÛȾµÄÖÆÇâ·½·¨¡£Æä·´Ó¦¹ý³ÌÈçͼ1Ëùʾ£º

¢Ù·´Ó¦¢ñµÄÀë×Ó·½³ÌʽÊÇ______________________________________________£»·´Ó¦¢ñµÃµ½µÄ²úÎïÓÃI2½øÐзÖÀ룬¸Ã²úÎïµÄÈÜÒºÔÚ¹ýÁ¿I2µÄ´æÔÚÏ»á·Ö³ÉÁ½²ã£ºº¬µÍŨ¶ÈI2µÄH2SO4²ãºÍº¬¸ßŨ¶ÈI2µÄHI²ã¡£¾­Àë×ÓŨ¶È¼ì²â£¬H2SO4ÈÜÒº²ãÖÐc£¨H+£©£ºc£¨SO42-£©=2.06£º1£¬Æä±ÈÖµ´óÓÚ2µÄÔ­ÒòÊÇ _______________________ ¡£

¢Ú·´Ó¦¢ò£º2H2SO4£¨l£©=2SO2£¨g£©+O2£¨g£©+ 2H2O£¨g£©¡÷H = +550kJmo1-1£¬ËüÓÉÁ½²½·´Ó¦×é³É£º¢¡£®H2SO4£¨l£©=SO3£¨g£©+ H2O£¨g£©¡÷H = +177kJmo1-1£¬¢¢£®SO3£¨g£©·Ö½â£¬Ð´³öSO3£¨g£©·Ö½âµÄÈÈ»¯Ñ§·½³Ìʽ ______________________________¡£

£¨2£©¹¤ÒµÉÏÀûÓ÷´Ó¦C£¨s£©+2H2O£¨g£© CO2£¨g£©+2H2£¨g£© ¡÷H£¾0 Ò²¿ÉÖƱ¸ÇâÆø¡£Ò»¶¨Ìõ¼þÏ£¬½«C£¨s£©ºÍH2O£¨g£©·Ö±ð¼ÓÈë¼×¡¢ÒÒÁ½¸öÃܱÕÈÝÆ÷·¢Éú·´Ó¦£¬ÆäÏà¹ØÊý¾ÝÈç±íËùʾ£º

ÈÝÆ÷

ÈÝ»ý/L

ζÈ/¡æ

ÆðʼÁ¿/mol

ƽºâÁ¿/mol

C£¨s£©

H2O£¨g£©

H2£¨g£©

¼×

2

T1

2

4

3.2

ÒÒ

1

T2

1

2

1.2

¢ÙT1¡æʱ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK= ______ £»T1 ______ T2£¨Ìî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±£©£»

¢ÚÈôÒÒÈÝÆ÷ÖдﵽƽºâËùÐèʱ¼äΪ3min£¬Ôòµ±·´Ó¦½øÐе½1.5minʱ£¬H2O£¨g£©µÄÎïÖʵÄÁ¿Å¨¶È ______£¨ÌîÑ¡Ïî×Öĸ£©¡£

A£®=1.4mol/L B£®£¼1.4mol/L C£®£¾1.4mol/L

£¨3£©¹¤ÒµÉÏ»¹¿É²ÉÓõ绯ѧ·¨ÀûÓÃH2S·ÏÆøÖÆÈ¡ÇâÆø£¬¸Ã·¨ÖÆÇâ¹ý³ÌµÄʾÒâͼÈçͼËùʾ£¬

¢Ù·´Ó¦³ØÖз´Ó¦ÎïµÄÁ÷Ïò²ÉÓÃÆø¡¢ÒºÄæÁ÷·½Ê½£¬ÆäÄ¿µÄÊÇ____________________£»

¢Ú·´Ó¦³ØÖз¢Éú·´Ó¦ºóµÄÈÜÒº½øÈëµç½â³Ø£¬µç½â×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________¡£

¡¾´ð°¸¡¿ SO2+2H2O+I2=SO42-+2I-+4H+ ÁòËá²ãÖк¬ÉÙÁ¿µÄHI£¬ÇÒHIµçÀë³öÇâÀë×Ó 2SO3£¨g£© 2SO2£¨g£©+O2£¨g£©¡÷H = +196kJmol-1 12.8 £¾ B ²ÉÓÃÆø¡¢ÒºÄæÁ÷·½Ê½µÄÄ¿µÄÊÇÔö´ó·´Ó¦Îï½Ó´¥Ãæ»ý£¬Ê¹·´Ó¦¸ü³ä·Ö 2Fe2£«£«2H£«2Fe3£«£«H2¡ü

¡¾½âÎö¡¿£¨1£©¢Ù·´Ó¦¢ñÖжþÑõ»¯ÁòºÍµâË®·´Ó¦Éú³ÉÁòËáºÍÇâµâËᣬ¸Ã·´Ó¦µÄµÄÀë×Ó·½³ÌʽÊÇSO2+2H2O+I2=SO42-+2I-+4H+£»·´Ó¦¢ñµÃµ½µÄ²úÎïÓÃI2½øÐзÖÀ룬¸Ã²úÎïµÄÈÜÒºÔÚ¹ýÁ¿I2µÄ´æÔÚÏ»á·Ö³ÉÁ½²ã£ºº¬µÍŨ¶ÈI2µÄH2SO4²ãºÍº¬¸ßŨ¶ÈI2µÄHI²ã¡£¾­Àë×ÓŨ¶È¼ì²â£¬H2SO4ÈÜÒº²ãÖÐc£¨H+£©£ºc£¨SO42-£©=2.06£º1£¬Æä±ÈÖµ´óÓÚ2µÄÔ­ÒòÊÇ£ºÁòËá²ãÖк¬ÉÙÁ¿µÄHI£¬ÇÒHIµçÀë³öÇâÀë×Ó¡£

¢Ú·´Ó¦¢ò£º2H2SO4£¨l£©= 2SO2£¨g£©+O2£¨g£©+ 2H2O£¨g£©¡÷H = +550kJmo1-1£¬ËüÓÉÁ½²½·´Ó¦×é³É£º¢¡£®H2SO4£¨l£©= SO3£¨g£©+ H2O£¨g£©¡÷H = +177kJmo1-1£¬¢¢£®SO3£¨g£©·Ö½â¡£¸ù¾Ý¸Ç˹¶¨ÂÉ£¬ÓÉ£¨II-¢¡£©¿ÉµÃSO3£¨g£©·Ö½âµÄÈÈ»¯Ñ§·½³Ìʽ2SO3£¨g£© 2SO2£¨g£©+O2£¨g£©¡÷H = +196kJmol-1 ¡£

£¨2£©¢ÙT1¡æʱ£¬¼×ÈÝÆ÷ÖУ¬ÇâÆøµÄ±ä»¯Á¿Îª3.2mol£¬ÔòË®ºÍ¶þÑõ»¯Ì¼µÄ±ä»¯Á¿Îª3.2molºÍ1.6mol£¬H2O£¨g£©¡¢ CO2£¨g£©¡¢H2£¨g£©µÄƽºâÁ¿·Ö±ðΪ0.8mol¡¢1.6mol¡¢3.2mol£¬H2O£¨g£©¡¢ CO2£¨g£©¡¢H2£¨g£©µÄƽºâŨ¶È·Ö±ðΪ0.4mol/L¡¢0.8mol/L¡¢1.6mol/L£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK=£»T2¡æʱ£¬ÒÒÈÝÆ÷ÖУ¬H2O£¨g£©¡¢ CO2£¨g£©¡¢H2£¨g£©µÄƽºâŨ¶È·Ö±ðΪ0.8mol/L¡¢0.6mol/L¡¢1.2mol/L£¬Æ½ºâ³£ÊýK=£¬ÓÉÓڸ÷´Ó¦ÎªÎüÈÈ·´Ó¦£¬Î¶ÈÔ½¸ßƽºâ³£ÊýÔ½´ó£¬ËùÒÔT1 £¾T2£»

¢ÚÈôÒÒÈÝÆ÷ÖдﵽƽºâËùÐèʱ¼äΪ3min£¬Æ½ºâʱ£¬H2O£¨g£©µÄ±ä»¯Á¿µÈÓÚ1.2mol£¬ÒòΪŨ¶ÈÔ½´ó»¯Ñ§·´Ó¦ËÙÂÊÔ½¿ì£¬ËùÒÔµ±·´Ó¦½øÐе½1.5minʱ£¬H2O£¨g£©µÄ±ä»¯Á¿´óÓÚ0.6mol£¬ËùÒÔ´ËʱH2O£¨g£©µÄÎïÖʵÄÁ¿Å¨¶ÈСÓÚ1.4mol/L£¬Ñ¡B¡£

£¨3£©¢Ù·´Ó¦³ØÖз´Ó¦ÎïµÄÁ÷Ïò²ÉÓÃÆø¡¢ÒºÄæÁ÷·½Ê½£¬ÆäÄ¿µÄÊDzÉÓÃÆø¡¢ÒºÄæÁ÷·½Ê½µÄÄ¿µÄÊÇÔö´ó·´Ó¦Îï½Ó´¥Ãæ»ý£¬Ê¹·´Ó¦¸ü³ä·Ö£»

¢ÚÓÉͼÖÐÐÅÏ¢¿ÉÖª£¬·´Ó¦³ØÖз¢Éú2Fe3£«£«H2S=S¡ý+2Fe2£«£«2H£«£¬·´Ó¦ºóµÄÈÜÒº½øÈëµç½â³Ø£¬Ñô¼«ÉÏFe2£«±»Ñõ»¯ÎªFe3£«£¬Òõ¼«ÉÏH£«±»»¹Ô­ÎªH2£¬µç½â×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ2Fe2£«£«2H£«2Fe3£«£«H2¡ü¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÆßÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ¢Ù~¢ß£¬ÆäÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬¢ÚÔªËØÊǵؿÇÖк¬Á¿×î¶àµÄ£¬¢ÝÔªËØΪÁ½ÐÔÔªËØ£¬¢Ü¢ßÁ½ÔªËØ×é³ÉµÄ»¯ºÏÎïÊÇÎÒÃÇÈÕ³£Éú»î±ØÐëµÄµ÷ζƷ£¬¢ÚºÍ¢ÞÔªËصÄÔ­×ÓÐòÊýÖ®ºÍÊǢٺ͢ÜÁ½ÔªËØÔ­×ÓÐòÊýÖ®ºÍµÄÁ½±¶¡£ÇëÓû¯Ñ§ÓÃÓï»Ø´ðÏÂÁÐÎÊÌ⣺

(1)¢Û¡¢¢Ý¡¢¢ÞµÄ¼òµ¥Àë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ_______________________¡£

(2)¢ÞºÍ¢ßµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïµÄËáÐÔÇ¿ÈõΪ___________£¾___________¡£

(3)д³ö¶þÑõ»¯¹èÓ뺬ÉÏÊöijÖÖÔªËصÄËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ_________________¡£

(4)ÓɢݺͿÕÆø¡¢º£Ë®¹¹³ÉµÄÔ­µç³ØÖУ¬ÆäÕý¼«·´Ó¦Ê½Îª_____________________¡£

(5)ÓÉÉÏÊöÔªËØÐγɵÄÎïÖÊ¿É·¢ÉúÏÂͼÖеķ´Ó¦£¬ÆäÖÐB¡¢C¡¢G Êǵ¥ÖÊ£¬BΪ»ÆÂÌÉ«ÆøÌå¡£

¢Ùд³öDÈÜÒºÓëG·´Ó¦µÄ»¯Ñ§·½³Ìʽ________________________________¡£

¢Ú»ìºÏÎïXÖеÄijÎïÖʲ»ÈÜÓÚË®£¬µ«¼ÈÄÜÈÜÓÚËáÓÖÄÜÈÜÓڼд³öÄܽâÊÍËüÔÚ¿ÁÐÔÄÆÈÜÒºÖз¢Éú·´Ó¦µÄÔ­ÒòµÄµçÀë·½³Ìʽ_________________________________¡£

¢Ûд³öµç½âAÈÜÒºµÄÀë×Ó·½³Ìʽ_______________________________¡£

¡¾ÌâÄ¿¡¿¸õÌú¿óµÄÖ÷Òª³É·Ö¿É±íʾΪFeO¡¤Cr2O3£¬»¹º¬ÓÐMgO¡¢Al2O3¡¢Fe2O3µÈÔÓÖÊ£¬ÒÔÏÂÊÇÒÔ¸õÌú¿óΪԭÁÏÖƱ¸ÖظõËá¼Ø£¨K2Cr2O7£©µÄÁ÷³Ìͼ£º

ÒÑÖª£ºNa2CO3+Al2O32NaAlO2+CO2¡ü£»

4FeO¡¤Cr2O3+8Na2CO3+7O28Na2CrO4+2Fe2O3+8CO2¡ü¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¹ÌÌåXÖÐÖ÷Òªº¬ÓÐ________________£¨Ìîд»¯Ñ§Ê½£©¡£

£¨2£©¹ÌÌåYÖÐÖ÷Òªº¬ÓÐÇâÑõ»¯ÂÁ£¬Çëд³öµ÷½ÚÈÜÒºµÄpH£½7¡«8ʱÉú³ÉÇâÑõ»¯ÂÁµÄÀë×Ó·½³Ìʽ£º____________________________________________________________¡£

£¨3£©ËữµÄÄ¿µÄÊÇʹCrO42ת»¯ÎªCr2O72£¬Èô½«´×Ëá¸ÄÓÃÏ¡ÁòËᣬд³ö¸Ãת»¯µÄÀë×Ó·½³Ìʽ£º___________________________________________________________¡£

£¨4£©²Ù×÷¢óÓжಽ×é³É£¬»ñµÃK2Cr2O7¾§ÌåµÄ²Ù×÷ÒÀ´ÎÊÇ£º¼ÓÈëKCl¹ÌÌå¡¢Õô·¢Å¨Ëõ¡¢__________¡¢¹ýÂË¡¢_______¡¢¸ÉÔï¡£

£¨5£©ËáÐÔÈÜÒºÖйýÑõ»¯ÇâÄÜʹCr2O72Éú³ÉÀ¶É«µÄ¹ýÑõ»¯¸õ£¨CrO5·Ö×ӽṹΪ£©£¬¸Ã·´Ó¦¿ÉÓÃÀ´¼ìÑéCr2O72µÄ´æÔÚ¡£Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ£º___________________________£¬¸Ã·´Ó¦_________(Ìî¡°ÊôÓÚ¡±»ò¡°²»ÊôÓÚ¡±)Ñõ»¯»¹Ô­·´Ó¦¡£

£¨6£©ÔÚ»¯Ñ§·ÖÎöÖвÉÓÃK2CrO4Ϊָʾ¼Á£¬ÒÔAgNO3±ê×¼ÈÜÒºµÎ¶¨ÈÜÒºÖеÄCl£¬ÀûÓÃAg+ÓëCrO42Éú³ÉשºìÉ«³Áµí£¬Ö¸Ê¾µ½´ïµÎ¶¨Öյ㡣µ±ÈÜÒºÖÐClÇ¡ºÃÍêÈ«³Áµí£¨Å¨¶ÈµÈÓÚ1.0¡Á105 mol¡¤L1£©Ê±£¬ÈÜÒºÖÐc(Ag+)Ϊ_______mol¡¤L1£¬´ËʱÈÜÒºÖÐc(CrO42)µÈÓÚ_________mol¡¤L1¡££¨ÒÑÖªAg2CrO4¡¢AgClµÄKsp·Ö±ðΪ2.0¡Á1012ºÍ2.0¡Á1010£©¡£

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

¾«Ó¢¼Ò½ÌÍø