ÌâÄ¿ÄÚÈÝ

(13·Ö)ÓÐA¡¢B¡¢C¡¢DËÄÖÖ»¯ºÏÎ·Ö±ðÓÉK+¡¢Ba2+¡¢SO42£­¡¢CO32£­¡¢OH£­ÖеÄÁ½ÖÖ×é³É£¬ËüÃǾßÓÐÏÂÁÐÐÔÖÊ£º¢ÙA²»ÈÜÓÚË®ºÍÑÎË᣻¢ÚB²»ÈÜÓÚË®£¬µ«ÈÜÓÚÑÎËᣬ²¢·Å³öÎÞÉ«Î޴̼¤ÐÔÆøζµÄÆøÌåE£»¢ÛCµÄË®ÈÜÒº³Ê¼îÐÔ£¬ÓëÁòËá·´Ó¦Éú³ÉA£»¢ÜD¿ÉÈÜÓÚË®£¬ÓëÁòËá×÷ÓÃʱ·Å³öÆøÌåE£¬E¿Éʹ³ÎÇåʯ»ÒË®±ä»ë×Ç¡£
£¨1£©ÍƶÏA¡¢B¡¢C¡¢D¡¢EµÄ»¯Ñ§Ê½¡£
A        £»B       £»C        £»D        £»E        £»
£¨2£©Ð´³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ¡£
BÓëÑÎËá·´Ó¦                   
CÓëÁòËá·´Ó¦                         
DÓëÁòËá·´Ó¦                   
EÓë³ÎÇåʯ»ÒË®·´Ó¦                   

(13·Ö)£¨1£©A£ºBaSOB£ºBaCOC: Ba(OH)2 D: K2COE: CO2(ÿ¿Õ1·Ö)
£¨2£©2H£«£«BaCO3£½CO2¡ü£«H2O£«Ba2+
Ba2+ £«2H£«£«2 OH£­£«SO42£­£½BaSO4¡ý£«2H2O
      2H£«£«CO£½CO2¡ü£«H2O
Ca2+£«2 OH£­£«CO2£½CaCO3¡ý£«H2O(ÿ¿Õ2·Ö)

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

 (12·Ö) ÓÐA¡¢B¡¢C¡¢DËÄÖÖÇ¿µç½âÖÊ£¬ËüÃÇÔÚË®ÖеçÀëʱ¿É²úÉúÏÂÁÐÀë×Ó£º(ÿÖÖÎïÖÊÖ»º¬Ò»ÖÖÑôÀë×ÓºÍÒ»ÖÖÒõÀë×ÓÇÒ»¥²»Öظ´)

ÑôÀë×Ó

Na+¡¢Ba2+¡¢NH4+¡¢K+

ÒõÀë×Ó

CH3COO£­¡¢Cl£­¡¢OH£­¡¢SO42£­

 

ÒÑÖª£º¢ÙA¡¢CÈÜÒºµÄpH¾ù´óÓÚ7£¬BÈÜÒºµÄpHСÓÚ7£¬A¡¢BÈÜÒºÖÐË®µÄµçÀë³Ì¶ÈÏàͬ£»DÈÜÒºÑæÉ«·´Ó¦ÏÔ»ÆÉ«¡£

¢ÚCÈÜÒººÍDÈÜÒºÏàÓöʱֻÉú³É°×É«³Áµí£¬BÈÜÒººÍCÈÜÒºÏàÓöʱֻÉú³ÉÓд̼¤ÐÔÆøζµÄÆøÌ壬AÈÜÒººÍDÈÜÒº»ìºÏʱÎÞÃ÷ÏÔÏÖÏó¡£

(1)AµÄÃû³ÆÊÇ____________¡£

(2)ÓÃÀë×Ó·½³Ìʽ±íʾAµÄË®ÈÜÒºÏÔ¼îÐÔµÄÔ­Òò______________________________¡£

(3)25 ¡æʱpH£½9µÄAÈÜÒººÍpH£½9µÄCÈÜÒºÖÐË®µÄµçÀë³Ì¶È½ÏСµÄÊÇ________(ÌîдA»òCµÄ»¯Ñ§Ê½)¡£

(4)25 ¡æʱÓöèÐԵ缫µç½âDµÄË®ÈÜÒº£¬Ò»¶Îʱ¼äºóÈÜÒºµÄpH________7(Ìî¡°>¡±¡¢¡°<¡±»ò¡°£½¡±)¡£

(5)½«µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄBÈÜÒººÍCÈÜÒº»ìºÏ£¬·´Ó¦ºóÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ__________________________________________________¡£

(6)ÊÒÎÂʱÔÚÒ»¶¨Ìå»ý0.2 mol¡¤L£­1µÄCÈÜÒºÖУ¬¼ÓÈëÒ»¶¨Ìå»ýµÄ0.1 mol¡¤L£­1µÄÑÎËáʱ£¬»ìºÏÈÜÒºµÄpH£½13£¬Èô·´Ó¦ºóÈÜÒºµÄÌå»ýµÈÓÚCÈÜÒºÓëÑÎËáµÄÌå»ýÖ®ºÍ£¬ÔòCÈÜÒºÓëÑÎËáµÄÌå»ý±ÈÊÇ________¡£

 

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

¾«Ó¢¼Ò½ÌÍø