³£ÎÂÏ£¬ÓÃ0.1000mol?L-1 NaOHÈÜÒºµÎ¶¨20.00mL 0.1000mol?L-1´×ËáÈÜÒºËùµÃµÎ¶¨ÇúÏßÈçͼ£®·ÖÎöͼÖеĢ١¢¢Ú¡¢¢Ûµã£¬Íê³ÉÏÂÁд³Ì⣺
£¨1£©µã¢ÙËùʾÈÜÒºÏÔ
 
ÐÔ£¬Ô­ÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©
 
£¬µã¢ÛËùʾÈÜÒºÖУºc£¨CH3C00-£©+c£¨CH3COOH£©
 
c£¨Na+£©£¨Ìî¡°£¾¡¢£¼»ò=¡±£©£®
£¨2£©Í¼Öеĵã¢ÚËùʾÈÜÒºÖУºc£¨CH3C00-£©
 
c£¨Na+£©£¨Ìî¡°£¼¡±¡¢¡°=¡±»ò¡°£¾¡±£©£®
£¨3£©Í¼Öеĵã¢ÙËùʾÈÜÒºÖи÷Àë×ÓŨ¶È¹ØϵÕýÈ·µÄÊÇ
 
£¨¶àÑ¡Ì⣩£®
A£®c£¨Na+£©+c£¨OH-£©=c£¨CH3C00-£©+c£¨H+£©
B£®c£¨Na+£©+c£¨H+£©=c£¨CH3C00-£©+c£¨OH-£©
C£®c£¨CH3C00-£©+c£¨CH3COOH£©=2c£¨Na+£©
D£®c£¨CH3C00-£©+c£¨CH3COOH£©=c£¨Na+£©
E£®c£¨CH3C00-£©£¾c£¨Na+£©£¾c£¨CH3COOH£©£¾c£¨H+£©£¾c£¨OH-£©
F£®c£¨CH3COOH£©£¾c£¨Na+£©£¾c£¨CH3C00-£©£¾c£¨H+£©£¾c£¨OH-£©
¿¼µã£ºËá¼î»ìºÏʱµÄ¶¨ÐÔÅжϼ°ÓйØphµÄ¼ÆËã
רÌ⣺µçÀëƽºâÓëÈÜÒºµÄpHרÌâ
·ÖÎö£º£¨1£©µã¢ÙËùʾÈÜÒºÖУ¬ÎªµÈÁ¿µÄ´×ËáºÍ´×ËáÄÆÈÜÒº£¬µã¢ÛËùʾÈÜÒº£¬Ç¡ºÃ·´Ó¦Éú³É´×ËáÄÆ£¬ÒÀ¾ÝÎïÁÏÊغã·ÖÎö£»
£¨2£©µã¢ÚËùʾÈÜÒºÖУ¬NaOHµÄÌå»ýСÓÚ20mL£¬Îª´×ËáºÍ´×ËáÄÆÈÜÒº£¬ÀûÓõçºÉÊغã·ÖÎö£»
£¨3£©µã¢ÙËùʾÈÜÒºÖУ¬ÎªµÈÁ¿µÄ´×ËáºÍ´×ËáÄÆÈÜÒº£¬Í¼Ïó·ÖÎö¿ÉÖª»ìºÏÈÜÒº³ÉËáÐÔ£¬´×ËáµçÀë³Ì¶È´óÓÚ´×Ëá¸ùÀë×ÓµÄË®½â³Ì¶È£¬ÒÀ¾ÝÈÜÒºÖеçºÉÊغ㡢ÎïÁÏÊغã·ÖÎöÅжϣ»
½â´ð£º ½â£º£¨1£©µã¢ÙËùʾÈÜÒºÖУ¬ÎªµÈÁ¿µÄ´×ËáºÍ´×ËáÄÆÈÜÒº£¬Í¼Ïó·ÖÎö¿ÉÖª»ìºÏÈÜÒº³ÉËáÐÔ£¬´×ËáµçÀë³Ì¶È´óÓÚ´×Ëá¸ùÀë×ÓµÄË®½â³Ì¶È£¬ÏÔËáÐÔµÄÔ­ÒòCH3COOH?CH3COO-+H+£»µã¢ÛËùʾÈÜÒº£¬Ç¡ºÃ·´Ó¦Éú³É´×ËáÄÆ£¬Ë®½âÏÔ¼îÐÔ£¬ÒÀ¾ÝÎïÁÏÊغã·ÖÎöc£¨CH3C00-£©+c£¨CH3COOH£©=c£¨Na+£©£»
¹Ê´ð°¸Îª£ºË᣻CH3COOH?CH3COO-+H+£»=£»
£¨2£©µã¢ÚËùʾÈÜÒº³ÊÖÐÐÔ£¬NaOHµÄÌå»ýСÓÚ20mL£¬Îª´×ËáºÍ´×ËáÄÆÈÜÒº£¬µçºÉÊغãΪc£¨OH-£©+c£¨CH3COO-£©=c£¨Na+£©+c£¨H+£©£¬c£¨OH-£©=c£¨H+£©£¬c£¨Na+£©=c£¨CH3COO-£©£»
¹Ê´ð°¸Îª£º=£»
£¨3£©µã¢ÙËùʾÈÜÒºÖУ¬ÎªµÈÁ¿µÄ´×ËáºÍ´×ËáÄÆÈÜÒº£¬Í¼Ïó·ÖÎö¿ÉÖª»ìºÏÈÜÒº³ÉËáÐÔ£¬´×ËáµçÀë³Ì¶È´óÓÚ´×Ëá¸ùÀë×ÓµÄË®½â³Ì¶È£¬
A£®µÈÁ¿µÄ´×ËáºÍ´×ËáÄÆÈÜÒºÖдæÔÚµçºÉÊغ㣺c£¨OH-£©+c£¨CH3COO-£©=c£¨Na+£©+c£¨H+£©£¬¹ÊA´íÎó£»
B£®µçºÉÊغãΪc£¨OH-£©+c£¨CH3COO-£©=c£¨Na+£©+c£¨H+£©£¬¹ÊBÕýÈ·£»
C£®ÈÜÒºÖдæÔÚÎïÁÏÊغãΪ2c£¨Na+£©=c£¨CH3COO-£©+c£¨CH3COOH£©£¬¹ÊCÕýÈ·£»
D£®ÈÜÒºÖдæÔÚÎïÁÏÊغãΪ2c£¨Na+£©=c£¨CH3COO-£©+c£¨CH3COOH£©£¬¹ÊD´íÎó£»
E£®µÈÁ¿µÄ´×ËáºÍ´×ËáÄÆÈÜÒº£¬Í¼Ïó·ÖÎö¿ÉÖª»ìºÏÈÜÒº³ÉËáÐÔ£¬´×ËáµçÀë³Ì¶È´óÓÚ´×Ëá¸ùÀë×ÓµÄË®½â³Ì¶È£¬ÈÜÒºÖÐÀë×ÓŨ¶È´óСΪ£ºc£¨CH3C00-£©£¾c£¨Na+£©£¾c£¨CH3COOH£©£¾c£¨H+£©£¾c£¨OH-£©£¬¹ÊEÕýÈ·£»
F£®ÒÀ¾ÝE·ÖÎö¿ÉÖªF´íÎ󣬹ÊF´íÎó£»
¹Ê´ð°¸Îª£ºBCE£®
µãÆÀ£º±¾Ì⿼²éËá¼î»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ¹Øϵ£¬Ã÷È·»ìºÏºóÈÜÒºÖеÄÈÜÖÊÊǽâ´ðµÄ¹Ø¼ü£¬×¢ÒâµçºÉÊغãºÍÎïÁÏÊغãµÄÓ¦Óã¬ÌâÄ¿ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

T0¡æʱ£¬ÔÚ2LµÄÃܱÕÈÝÆ÷ÖУ¬¸÷ÎïÖÊ£¨X£¨g£©¡¢Y£¨g£©¡¢Z£¨g£©£©µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯µÄ¹ØϵÈçͼ1Ëùʾ£®ÆäËûÌõ¼þÏàͬ£¬Î¶ȷֱðΪT1¡æ¡¢T2¡æʱ·¢Éú·´Ó¦£¬YµÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯µÄ¹ØϵÈçͼ2Ëùʾ£®ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢¸Ã¿ÉÄæ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪX£¨g£©+Y£¨g£©?2Z£¨g£©
B¡¢¸Ã¿ÉÄæ·´Ó¦µÄ¡÷H£¾0
C¡¢Í¼1Öз´Ó¦´ïµ½Æ½ºâʱ£¬XµÄת»¯ÂÊΪ16.7%
D¡¢T1¡æʱ£¬Èô¸Ã·´Ó¦µÄƽºâ³£ÊýK=50£¬ÔòT1£¾T0

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÀûÓ÷´Ó¦£º2NO£¨g£©+2CO£¨g£©?2C02£¨g£©+N2£¨g£©¡÷H=-746.8kJ?mol-1£¬¿É¾»»¯Æû³µÎ²Æø£¬Èç¹ûҪͬʱÌá¸ß·´Ó¦µÄËÙÂʺÍN0µÄת»¯ÂÊ£¬²ÉÈ¡µÄ´ëÊ©ÊÇ£¨¡¡¡¡£©
A¡¢½µµÍζÈ
B¡¢Ôö´óѹǿ
C¡¢Éý¸ßζÈͬʱ³äÈëN2
D¡¢¼°Ê±½«C02ºÍN2´Ó·´Ó¦ÌåϵÖÐÒÆ×ß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

3.01¡Á1023¸öOH-µÄÎïÖʵÄÁ¿Îª
 
£¬ÖÊÁ¿Îª
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚUÐιÜÖУ¬ÓöèÐԵ缫µç½â±¥ºÍNaClÈÜÒº£¬ÊµÑ鿪ʼʱ£¬Í¬Ê±ÔÚÁ½±ß¸÷µÎÈ뼸µÎ·Ó̪ÊÔÒº£¬Ôò
£¨1£©µç½â³ØÖÐÓëÍâµçÔ´¸º¼«ÏàÁ¬µÄµç¼«·´Ó¦Ê½Îª
 
£®Ôڸü«¸½½ü¹Û²ìµ½µÄʵÑéÏÖÏóÊÇ
 
£®
£¨2£©Ñô¼«Éϵĵ缫·´Ó¦Ê½Îª
 
£®¼ìÑé¸Ãµç¼«·´Ó¦²úÎïµÄ·½·¨ÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖª£º¢ÙAÊÇʯÓÍÁѽâÆøµÄÖ÷Òª³É·Ö£¬AµÄ²úÁ¿Í¨³£ºâÁ¿Ò»¸ö¹ú¼ÒµÄʯÓÍ»¯¹¤Ë®Æ½£»
¢Ú2CH3CHO+O2
´ß»¯¼Á
¡÷
2CH3COOH
ÏÖÒÔAΪÖ÷ÒªÔ­ÁϺϳÉÒÒËáÒÒõ¥£¬ÆäºÏ³É·ÏßÈçͼËùʾ£®

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÎïÖÊB¿ÉÒÔ±»Ö±½ÓÑõ»¯ÎªD£¬ÐèÒª¼ÓÈëµÄÊÔ¼ÁÊÇ
 
£®
£¨2£©B¡¢D·Ö×ÓÖеĹÙÄÜÍÅÃû³Æ·Ö±ðÊÇ
 
¡¢
 
£®
£¨3£©Ð´³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³ÌʽºÍ·´Ó¦ÀàÐÍ£º
¢Ú
 
£®·´Ó¦ÀàÐÍ£º
 
£®
¢Ü
 
£®·´Ó¦ÀàÐÍ£º
 
£®
¢Ý
 
£®·´Ó¦ÀàÐÍ£º
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÑÎA£¨º¬½á¾§Ë®£©¿É·¢ÉúÈçϵÄת»¯¹Øϵ£¨²¿·ÖÉú³ÉÎïÂÔÈ¥£©£®ÆäÖÐB¡¢D¡¢EΪÎÞÉ«ÆøÌ壬CΪ°×É«³Áµí£¬HΪºìºÖÉ«³Áµí£®ÔÚ»ìºÏÒºXÖмÓÈëBaCl2¿ÉÉú³É²»ÈÜÓÚÏ¡HNO3µÄ°×É«³Áµí£®

£¨1£©Ð´³öÓÉB¡úIµÄ¹¤ÒµÒâÒå
 

£¨2£©Ð´³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ
G¡úI
 

C¡úH
 

£¨3£©µç½â»ìºÏÒºXµÄÑô¼«µç¼«·´Ó¦Ê½Îª
 

£¨4£©ÈôÑÎA 1.960gºÍ¹ýÁ¿Å¨NaOHÈÜÒº¼ÓÈȳä·Ö·´Ó¦£¬ÊÕ¼¯µ½B 224ml£¨±ê×¼×´¿ö£©£¬½«HÏ´µÓ¡¢ºæ¸É¡¢×ÆÉÕ¡¢³ÆÁ¿µÃ0.400g£¬Ïò»ìºÏÒºÖмÓÈëÑÎËáËữ£¬ÔÙ¼ÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬½«³Áµí¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É¡¢³ÆÁ¿µÃ2.33g£¬ÔòAµÄ»¯Ñ§Ê½Îª
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

±£³ÖÓªÑø¾ùºâ£¬ºÏÀíʹÓÃÒ©ÎïÊDZ£Ö¤ÉíÐĽ¡¿µ¡¢Ìá¸ßÉú»îÖÊÁ¿µÄÓÐЧÊֶΣ¬Ò²ÊÇÇàÉÙÄêѧÉú½¡¿µ³É³¤µÄÖØÒª±£Ö¤£®
£¨1£©Î¬ÉúËØCÄÜÔöÇ¿ÈËÌå¶Ô¼²²¡µÄµÖ¿¹ÄÜÁ¦£¬´Ù½øÈËÌåÉú³¤·¢Óý£¬ÖÐѧÉúÿÌìÒª²¹³ä60mgµÄάÉúËØC£®ÏÂÁÐÎïÖʺ¬ÓзḻάÉúËØCµÄÊÇ
 
£¨Ìî×Öĸ£©£®
A£®Å£Èâ         B£®À±½·        C£®¼¦µ°
£¨2£©ÇàÉÙÄê¼°³ÉÈËȱ·¦Ä³ÖÖ΢Á¿ÔªËؽ«µ¼Ö¼××´ÏÙÖ״󣬶øÇÒ»áÔì³ÉÖÇÁ¦Ë𺦣¬¸Ã΢Á¿ÔªËØÊÇ
 
£¨ Ìî×Öĸ£©
A£®µâ           B£®Ìú          C£®¸Æ
£¨3£©ÈËÀàµÄÉúÃü»î¶¯ÐèÒªÌÇÀà¡¢
 
¡¢
 
¡¢Î¬ÉúËØ¡¢Ë®¡¢ºÍÎÞ»úÑΣ¨»ò¿óÎïÖÊ£©µÈÁù´óËØÓªÑøÎïÖÊ£®
£¨4£©°¢Ë¾Æ¥ÁÖ¾ßÓÐ
 
 ×÷Ó㮳¤ÆÚ´óÁ¿·þÓð¢Ë¾Æ¥ÁÖ£¬ÆäË®½â²úÎïË®ÑîËᠿɵ¼Ö»¼Õß³öÏÖÍ·Í´¡¢¶ñÐĵÈÖ¢×´£¬Ðè¾²Âö×¢ÉäСËÕ´ò£¨NaHCO3£©ÈÜÒº£¬ÀûÓÃСËÕ´òÓëË®ÑîËá·Ö×ÓÖеÄôÈ»ù·´Ó¦Éú³ÉË®ÑîËáÄÆ£¬Ê¹Ö¢×´»º½â£®Ð´³öË®ÑîËáÓëСËÕ´ò·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

2011Äê3ÔÂ11ÈÕÏÂÎç1µã46·Ö£¬ÈÕ±¾·¢ÉúÁË´óº£Ð¥£®ÎªÈ·±£´óÔÖÖ®ºóÎÞ´óÒߣ¬ÈÕ±¾·ÀÒß²¿ÃÅʹÓÃÁË´óÁ¿º¬ÂÈÀàÏû¶¾¼Á£¬ÆäÖÐClO2ÊÇɱ¾úЧÂʸߵÄɱ¶¾¼Á£¬¾ßÓÐÇ¿Ñõ»¯ÐÔ£®¹¤ÒµÉÏÖƱ¸ClO2»¯Ñ§·´Ó¦µÄ·½³ÌʽΪ£º2NaClO3+Na2SO3+H2SO4=2ClO2+2Na2SO4+2H2O£¬ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢ÔÚ·´Ó¦ÖÐNaClO3ÊÇ»¹Ô­¼Á£¬Na2SO3ÊÇÑõ»¯¼Á
B¡¢Óз´Ó¦¿ÉµÃClO2µÄÑõ»¯ÐÔ´óÓÚNaClO3µÄÑõ»¯ÐÔ
C¡¢¸Ã·´Ó¦ÖУ¬H2SO4Öи÷ÔªËصĻ¯ºÏ¼ÛûÓз¢Éú±ä»¯
D¡¢1mol NaClO3²Î¼Ó·´Ó¦£¬ÔòÓÐ2molµç×ÓתÒÆ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸