(11·Ö)ij¿ÎÍâ»î¶¯Ð¡×éÉè¼ÆÁËÒÔÏÂʵÑé·½°¸ÑéÖ¤AgÓëŨHNO3·´Ó¦µÄ¹ý³ÌÖпÉÄܲúÉúNO¡£ÆäʵÑéÁ÷³ÌͼÈçÏ£º

(1)²â¶¨ÏõËáµÄÎïÖʵÄÁ¿
·´Ó¦½áÊøºó£¬´ÓÈçͼB×°ÖÃÖÐËùµÃ100 mLÈÜÒºÖÐÈ¡³ö25.00 mLÈÜÒº£¬ÓÃ0.1 mol¡¤L£­1µÄNaOHÈÜÒºµÎ¶¨£¬Ó÷Ó̪×÷ָʾ¼Á£¬µÎ¶¨Ç°ºóµÄµÎ¶¨¹ÜÖÐÒºÃæµÄλÖÃÈçÉÏͼËùʾ¡£ÔÚBÈÝÆ÷ÖÐÉú³ÉÏõËáµÄÎïÖʵÄÁ¿Îª________mol£¬ÔòAgÓëŨÏõËá·´Ó¦¹ý³ÌÖÐÉú³ÉµÄNO2Ìå»ýΪ________mL¡£

(2)²â¶¨NOµÄÌå»ý
¢Ù´ÓÉÏͼËùʾµÄ×°ÖÃÖУ¬ÄãÈÏΪӦѡÓÃ________×°ÖýøÐÐAgÓëŨÏõËᷴӦʵÑ飬ѡÓõÄÀíÓÉÊÇ______________________________________________________________________
________________________________________________________________________¡£
¢ÚÑ¡ÓÃÉÏͼËùʾÒÇÆ÷×éºÏÒ»Ì׿ÉÓÃÀ´²â¶¨Éú³ÉNOÌå»ýµÄ×°Öã¬ÆäºÏÀíµÄÁ¬½Ó˳ÐòÊÇ________(Ìî¸÷µ¼¹Ü¿Ú±àºÅ)¡£¢ÛÔڲⶨNOµÄÌå»ýʱ£¬ÈôÁ¿Í²ÖÐË®µÄÒºÃæ±È¼¯ÆøÆ¿µÄÒºÃæÒªµÍ£¬´ËʱӦ½«Á¿Í²µÄλÖÃ________(Ñ¡ÌϽµ¡±»ò¡°Éý¸ß¡±)£¬ÒÔ±£Ö¤Á¿Í²ÖеÄÒºÃæÓ뼯ÆøÆ¿ÖеÄÒºÃæ³Öƽ¡£
(3)ÆøÌå³É·Ö·ÖÎö
ÈôʵÑé²âµÃNOµÄÌå»ýΪ112.0 mL(ÒÑÕÛËãµ½±ê×¼×´¿ö)£¬ÔòAgÓëŨÏõËá·´Ó¦µÄ¹ý³ÌÖÐ________(Ìî¡°ÓС±»ò¡°Ã»ÓС±)NO²úÉú£¬×÷´ËÅжϵÄÒÀ¾ÝÊÇ
________________________________________________________________________
________________________________________________________________________¡£
(1)0.008   268.8  
(2)¢ÙA ÒòΪA×°ÖÿÉÒÔͨÈëN2½«×°ÖÃÖеĿÕÆøÅž¡£¬·ÀÖ¹NO±»¿ÕÆøÖÐO2Ñõ»¯  ¢Ú123547(1547¿¼ÂÇÒ²¿É)¡¡¢ÛÉý¸ß
(3)ÓРÒòΪNO2ÓëË®·´Ó¦Éú³ÉµÄNOµÄÌå»ýСÓÚÊÕ¼¯µ½µÄNOµÄÌå»ý(89.6<112.0)
(1)B×°ÖÃÖз¢ÉúµÄ·´Ó¦ÊÇ3NO2£«H2O===2HNO3£«NO£¬ÓÃNaOHÈÜÒºµÎ¶¨µÄÊÇHNO3£¬Ôòn(HNO3)£½4n(NaOH)£½4¡Á0.1 mol¡¤L£­1¡Á(20.40£­0.40)¡Á10£­3 L£½0.008 mol£¬ÔòÉú³ÉNO2µÄÌå»ýΪ£º0.008 mol¡Á¡Á22.4 L/mol¡Á103 mL/L£½268.8 mL¡£(2)¢ÙÑ¡A×°Ö㬿ÉͨÈëN2°Ñ×°ÖÃÖеĿÕÆøÅž¡£¬¢ÛÓ¦±£Ö¤Á¿Í²ÄÚµÄÒºÃæÓ뼯ÆøÆ¿ÖеÄÒºÃæÏàƽ£¬Ê¹Á¿Í²Î»ÖÃÉý¸ß¡£(3)ÓÉ(1)ÖªÉú³Én(NO)£½n(HNO3)£½0.004 mol£¬¼´Îª89.6 mL£¬¶øÊÕ¼¯µÄNOΪ112.0 mL£¬¹ÊAgÓëŨHNO3·´Ó¦¹ý³ÌÖвúÉúÁËNO¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

(18·Ö)ijУ»¯Ñ§ÊµÑéÐËȤС×éÔÚ¡°Ì½¾¿Â±Ëص¥ÖʵÄÑõ»¯ÐÔ¡±µÄϵÁÐʵÑéÖз¢ÏÖ£ºÔÚ×ãÁ¿µÄÏ¡ÂÈ»¯ÑÇÌúÈÜÒºÖУ¬¼ÓÈë1¡«2µÎäåË®£¬Õñµ´ºóÈÜÒº³Ê»ÆÉ«¡£
(1)Ìá³öÎÊÌ⣺     Fe3£«¡¢Br2ÄÄÒ»¸öµÄÑõ»¯ÐÔ¸üÇ¿£¿
(2)²ÂÏ룺
¢Ù¼×ͬѧÈÏΪÑõ»¯ÐÔ£ºFe3£«>Br2£¬¹ÊÉÏÊöʵÑéÏÖÏó²»ÊÇ·¢Éú»¯Ñ§·´Ó¦ËùÖ£¬ÔòÈÜÒº³Ê»ÆÉ«ÊǺ¬________(Ìѧʽ£¬ÏÂͬ)ËùÖ¡£
¢ÚÒÒͬѧÈÏΪÑõ»¯ÐÔ£ºBr2>Fe3£«£¬¹ÊÉÏÊöʵÑéÏÖÏóÊÇ·¢Éú»¯Ñ§·´Ó¦ËùÖ£¬ÔòÈÜÒº³Ê»ÆÉ«ÊǺ¬__________ËùÖ¡£
(3)Éè¼ÆʵÑé²¢ÑéÖ¤£º
±ûͬѧΪÑéÖ¤ÒÒͬѧµÄ¹Ûµã£¬Ñ¡ÓÃÏÂÁÐijЩÊÔ¼ÁÉè¼Æ³öÁ½ÖÖ·½°¸½øÐÐʵÑ飬²¢Í¨¹ý¹Û²ìʵÑéÏÖÏó£¬Ö¤Ã÷ÁËÒÒͬѧµÄ¹ÛµãȷʵÊÇÕýÈ·µÄ¡£¹©Ñ¡ÓõÄÊÔ¼Á£º
a£®·Ó̪ÊÔÒº  b£®CCl4   c£®ÎÞË®¾Æ¾«  d£®KSCNÈÜÒº
ÇëÄãÔÚÏÂÁбí¸ñÖÐд³ö±ûͬѧѡÓõÄÊÔ¼Á¼°ÊµÑéÖй۲쵽µÄÏÖÏó¡£(ÊÔ¼ÁÌîÐòºÅ)
 
Ñ¡ÓÃÊÔ¼Á
ʵÑéÏÖÏó
·½°¸1
 
 
·½°¸2
 
 
(4)Ó¦ÓÃÓëÍØÕ¹
¢ÙÔÚ×ãÁ¿µÄÏ¡ÂÈ»¯ÑÇÌúÈÜÒºÖмÓÈë1¡«2µÎäåË®£¬ÈÜÒº³Ê»ÆÉ«£¬Ëù·¢ÉúµÄÀë×Ó·´Ó¦·½³ÌʽΪ_____________________________________________________________¡£
¢ÚÔÚ100 mL FeBr2ÈÜÒºÖÐͨÈë2.24 L Cl2(±ê×¼×´¿ö)£¬ÈÜÒºÖÐÓÐ1/3µÄBr£­±»Ñõ»¯³Éµ¥ÖÊBr2£¬ÔòÔ­FeBr2ÈÜÒºÖÐFeBr2µÄÎïÖʵÄÁ¿Å¨¶ÈΪ_________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

£¨9·Ö£©ÓÐÈËÉè¼ÆÒ»¸öNa2O2ÓëCO2·´Ó¦ÊµÑé×°ÖÃͼÈçÏ£º
 
´ò¿ªÖ¹Ë®¼Ð,·¢ÏÖCO2ͨ¹ý¹üÓÐNa2O2µÄÍÑÖ¬ÃÞ£¬¿É¹Û²ìµ½ÍÑÖ¬ÃÞ¾çÁÒȼÉÕÆðÀ´£®
£¨1£©ÓÉʵÑéÏÖÏóËùµÃ³öµÄÓйØNa2O2ÓëCO2·´Ó¦µÄ½áÂÛÊÇ£º
a:ÓÐÑõÆøÉú³É£ºb:                                 
£¨2£©¼×¡¢ÒÒÁ½Î»Í¬Ñ§¸÷³ÆÈ¡ÖÊÁ¿Îªm gµÄ¹ýÑõ»¯ÄÆÓë¶þÑõ»¯Ì¼·´Ó¦ºóµÄÑùÆ·£¬²¢ÓÃÏÂͼËùʾÒÇÆ÷²â¶¨ÑùÆ·µÄ×é·Ö¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù¼×ͬѧͨ¹ýʵÑé²âµÃµÄÊý¾ÝÊÇÑõÆøµÄÌå»ý£¬¸Ãͬѧ¶ÁȡʵÑéÊý¾ÝʱӦעÒâÀäÈ´ÖÁÊÒΡ¢____________¡¢ÑÛ¾¦ÊÓÏßÓë°¼ÒºÃæ×îµÍ´¦ÏàÇС£
¢ÚÒÒͬѧͨ¹ýÁ¬½ÓÒÇÆ÷¢Ù¢Ú½øÐÐʵÑ飬Ëû²âµÃµÄÊý¾ÝÊÇ             ¡£°´Ëû²âµÃµÄÊý¾Ý¼ÆËã³öµÄʵÑé½á¹ûÆ«¸ß£¬ÀíÓÉÊÇ                     ¡£
¢ÛΪÁ˲âµÃ׼ȷµÄʵÑéÊý¾Ý£¬ÇëÄ㽫ÒÒͬѧµÄʵÑé×°ÖýøÐиĽø£¨Ã¿ÖÖÒÇÆ÷ֻ׼ʹÓÃÒ»´Î£©£¬Ð´³ö¸÷ÒÇÆ÷½Ó¿ÚµÄÁ¬½Ó˳Ðò                           ¡£ 
¢Ü°´¢ÛÉè¼ÆµÄʵÑé×°ÖýøÐÐʵÑ飬Èô²âµÃʵÑéÇ°ºó×°ÖâڵÄÖÊÁ¿·Ö±ðÊÇw1 gºÍw2 g£¬ÔòÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊýΪ                       ¡£
¢ÝÔڢ۵ĸĽø×°ÖÃÖÐÓÉÓÚÊܵ½ÒÇÆ÷µÄ¾ÖÏÞ£¬ÊÇ·ñÒ²Óв»×ãÖ®´¦       £¨ÈôûÓУ¬´Ë¿Õ²»±Ø»Ø´ð£»ÈôÓУ¬ÇëÒ»²¢Ëµ³ö²»×ãµÄÀíÓÉ£©                  ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÒÑÖªFeSO4ÔÚ²»Í¬Ìõ¼þÏ·ֽâµÃµ½µÄ²úÎﲻͬ£¬¿ÉÄÜÊÇFeOºÍSO3£¬Ò²¿ÉÄÜÊÇFe2O3¡¢SO3ºÍSO2¡£Ä³Ñо¿Ð¡×é̽¾¿Ôھƾ«ÅçµÆ¼ÓÈÈÌõ¼þÏÂFeSO4·Ö½âµÄÆøÌå²úÎï¡£ÒÑÖªSO3µÄÈÛµãÊÇ16.8¡æ£¬·ÐµãÊÇ44.8¡æ¡£

£¨1£©×°ÖâòµÄÊÔ¹ÜÖв»×°ÈκÎÊÔ¼Á£¬Æä×÷ÓÃÊÇ_______________________£¬ÊԹܽþÅÝÔÚ50¡æµÄˮԡÖУ¬Ä¿µÄÊÇ________________________________¡£
£¨2£©×°ÖâóºÍ×°ÖâôµÄ×÷ÓÃÊÇ̽¾¿±¾ÊµÑéÆøÌå²úÎï³É·Ö¡£ÇëÍê³ÉʵÑéÉè¼Æ£¬Ìîд¼ìÑéÊÔ¼Á¡¢Ô¤ÆÚÏÖÏóÓë½áÂÛ¡£
ÏÞÑ¡ÊÔ¼Á£º3 mol¡¤L£­1 H2SO4¡¢6 mol¡¤L£­1 NaOH¡¢0.5 mol¡¤L£­1 BaCl2¡¢0.5 mol¡¤L£­1 Ba(NO3)2¡¢0.01 mol¡¤L£­1ËáÐÔKMnO4ÈÜÒº¡¢0.01 mol¡¤L£­1äåË®¡£
¼ìÑéÊÔ¼Á
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
×°ÖâóµÄÊÔ¹ÜÖмÓÈë________________¡£
²úÉú´óÁ¿°×É«³Áµí£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO3¡£
×°ÖâôµÄÊÔ¹ÜÖмÓÈë________________¡£
______________________________
______________________________
______________________________
______________________________
£¨3£©×°ÖâõµÄ×÷ÓÃÊÇ·ÀֹβÆøÎÛȾ»·¾³£¬ÉÕ±­ÖÐÓ¦¼ÓÈëµÄÊÔ¼ÁÊÇ              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐʵÑé²Ù×÷»ò¼ìÑéÕýÈ·µÄÊÇ
A£®ÊÕ¼¯ÂÈÆøB£®×ªÒÆÈÜÒºC£®Ä£Ä⹤ҵÖƱ¸²¢¼ìÑé°±ÆøD£®Óú£Ë®ÖÆÉÙÁ¿ÕôÁóË®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐʵÑé²Ù×÷ÕýÈ·µÄÊÇ
¢ÙÓÃÏ¡ÁòËáÏ´µÓ³¤ÆÚ´æ·Åʯ»ÒË®µÄÊÔ¼ÁÆ¿£»
¢ÚÓôøÏð½ºÈûµÄ×ØÉ«ÊÔ¼ÁÆ¿´æ·ÅŨÁòË᣻
¢ÛÓÃÑÎËáËữ¹ýµÄFeCl3ÈÜÒº£¬³ýÈ¥H2»¹Ô­CuOʵÑéÁôÔÚÊÔ¹ÜÄÚµÄÍ­£»
¢Ü²â¶¨Ä³ÈÜÒºµÄpHʱ£¬ÏÈÓÃÕôÁóË®ÈóʪpHÊÔÖ½£¬ÔÙ˦½à¾»¡¢¸ÉÔïµÄ²£Á§°ôպȡ¸ÃÈÜ, ÒºµãÔÚÊÔÖ½ÉÏ£¬²¢Óë±ê×¼±ÈÉ«¿¨±È½Ï£»
¢Ý·ÖҺʱ£¬·ÖҺ©¶·ÖÐϲãÒºÌå´ÓÏ¿ڷųö£¬ÉϲãÒºÌå´ÓÉÏ¿Úµ¹³ö£»
¢ÞÕôÁóʱ£¬½«Î¶ȼƵÄË®ÒøÇò¿¿½üÕôÁóÉÕÆ¿Ö§¹Ü¿Ú¡£
A£®¢Ù¢ÛB£®¢Ú¢Û¢ÜC£®¢Ú¢ÛD£®¢Û¢Ý¢Þ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

£¨12·Ö£©Ä³Ñ§ÉúΪÁËÑéÖ¤ÇâÆø»¹Ô­Ñõ»¯Í­µÄ²úÎÉè¼ÆÁËÏÂͼËùʾµÄʵÑé×°Öá£

¢Åд³ö±àºÅÒÇÆ÷µÄÃû³Æ£º¢Ú            
¢ÆŨH2SO4µÄ×÷ÓÃÊÇ                               ¡£
¢Ç±¾ÊµÑéÐèÒª¼ÓÈȵÄ×°ÖÃΪ       £¨Ìî×Öĸ±àºÅ£©¡£
¢Èд³ö×°ÖÃC¡¢DÖпɹ۲쵽µÄÏÖÏó£ºC                  £¬D                    ¡£
¢É DÖз´Ó¦Ã¿Éú³É1¸öË®·Ö×Ó£¬×ªÒƵĵç×ÓÊýΪ             ¸ö¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

£¨8·Ö£©Ä³Ñ§ÉúÍùÒ»Ö§ÊÔ¹ÜÖа´Ò»¶¨µÄ˳Ðò·Ö±ð¼ÓÈëÏÂÁм¸ÖÖÎïÖÊ(Ò»ÖÖÎïÖÊÖ»¼ÓÒ»´Î)£º
A£®KIÈÜÒºB£®µí·ÛÈÜÒºC£®NaOHÈÜÒºD£®Ï¡H2SO4E£®ÂÈË®
·¢ÏÖÈÜÒºÑÕÉ«°´ÈçÏÂ˳Ðò±ä»¯£º¢ÙÎÞÉ«¨D¡ú¢Ú×Ø»ÆÉ«¨D¡ú¢ÛÀ¶É«¨D¡ú¢ÜÎÞÉ«¨D¡ú¢ÝÀ¶É«¡£ÒÀ¾ÝÈÜÒºÑÕÉ«µÄ±ä»¯£¬»Ø´ðÏÂÁÐÎÊÌ⣺
(1)¼ÓÈëÒÔÉÏÒ©Æ·µÄ˳ÐòÊÇ(дÐòºÅ)_______________________________________¡£
(2)¢Ù¨D¡ú¢Ú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________________________________________¡£
(3)ÈÜÒºÓÉ×Ø»ÆÉ«±äΪÀ¶É«µÄÔ­ÒòÊÇ_______________________________________¡£
(4)¢Û¨D¡ú¢Ü·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

ÓÃÈçͼËùʾװÖýøÐÐʵÑ飬½«ÒºÌåAÖðµÎ¼ÓÈëµ½¹ÌÌåBÖУ¬»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÅͼÖÐD×°ÖÃÔÚʵÑéÖеÄ×÷ÓÃÊÇ            ¡£
¢ÆÈô AΪ30%H2O2ÈÜÒº£¬BΪMnO2£¬CÊ¢ÓÐÇâÁòËᣨH2S£© ±¥ºÍÈÜÒº£¬Ðý¿ªEºó£¬CÖгöÏÖdz»ÆÉ«»ë×ǵÄÏÖÏó£¬Ð´³öCÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌΪ                        ¡£
¢ÇÈôAΪŨÑÎËᣬBΪKMnO4£¬CÖÐÊ¢ÓÐKIµí·ÛÈÜÒº£¬Ðý¿ªEºó£¬CÖеÄÏÖÊÇ                              £»¼ÌÐøͨÆøÌåÓÚCÖУ¬×ã¹»³¤µÄʱ¼äºó£¬·¢ÏÖCÖÐÈÜÒºµÄÑÕÉ«Ïûʧ£¬ÕâÊÇÒòΪÔÚÈÜÒºÖÐI2Äܱ»Cl2Ñõ»¯ÎªHIO3£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·´Ó¦·½³Ìʽ                   ¡£
¢ÈÈôAΪŨ°±Ë®£¬BΪÉúʯ»Ò£¬CÖÐÊ¢ÓÐAlCl3ÈÜÒº£¬Ðý¿ªE£¬×ã¹»³¤µÄʱ¼äºó£¬CÖеÄÏÖÏóÊÇ                  £¬CÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ             ¡£
(5)ÈôBΪ¿é×´´óÀíʯ£¬CΪÈÜÒº£¬ÊµÑéÖй۲쵽ÈÜÒº±ä»ë×Ç£¬ÔòËáA²»ÒËÓÃÏÂÁеÄ________¡£
A£®HClB£®HNO3C£®H2SO4D£®CH3COOH

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸