Question

£¨13·Ö£©ÏÖÓÐA¡¢B¡¢C¡¢D¡¢EÎåÖÖÇ¿µç½âÖÊ£¬ËüÃÇÔÚË®ÖпɵçÀë²úÉúÏÂÁÐÀë×Ó£¨¸÷ÖÖÀë×Ó²»Öظ´£©¡£

ÑôÀë×Ó	H+¡¢Na+¡¢A13+¡¢Ag+¡¢Ba2+
ÒõÀë×Ó	OH¡ª¡¢C1¡ª¡¢CO32¡ª¡¢NO3¡ª¡¢SO42¡ª

ÒÑÖª£º

¢ÙA¡¢BÁ½ÈÜÒº³Ê¼îÐÔ£»C¡¢D¡¢EÈÜÒº³ÊËáÐÔ¡£

¢ÚAÈÜÒºÓëEÈÜÒº·´Ó¦¼ÈÓÐÆøÌåÓÖÓгÁµí²úÉú£»AÈÜÒºÓëCÈÜÒº·´Ó¦Ö»ÓÐÆøÌå²úÉú£¨³Áµí°üÀ¨Î¢ÈÜÎÏÂͬ£©¡£

¢ÛDÈÜÒºÓëÁíÍâËÄÖÖÈÜÒº·´Ó¦¶¼ÄܲúÉú³Áµí£»CÖ»ÄÜÓëD·´Ó¦²úÉú³Áµí¡£

ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©·Ö±ðд³öA¡¢B¡¢C¡¢D¡¢EµÄ»¯Ñ§Ê½£ºA          ¡¢B          ¡¢C        ¡¢

D             ¡¢E              £»

£¨2£©Ð´³öA¡¢E·´Ó¦µÄÀë×Ó·½³Ìʽ£º                                     £»

£¨3£©½«CÈÜÒºÖðµÎ¼ÓÈëµÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄAÈÜÒºÖУ¬·´Ó¦ºóÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ£º                                  ¡£

£¨4£©ÒÑÖª£ºNaOH£¨aq£©+HNO₃£¨aq£©=NaNO₃£¨aq£©+H₂O(1);¡÷H =£­Q kJ¡¤mol^-1¡£

д³öBÓëCÏ¡ÈÜÒº·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                                      ¡£

 £¨5£©ÔÚ100 mL 0.1 mol¡¤L^-1 EÈÜÒºÖУ¬ÖðµÎ¼ÓÈë40 mL 1.6 mol¡¤L^-1 NaOHÈÜÒº£¬×îÖյõ½³ÁµíÎïÖʵÄÁ¿Îª                    mol¡£

 

Answer 1

£¨1£©A£®Na₂CO₃¡¢B. Ba£¨OH£©₂ C.HClD.AgNO₃ E.Al₂(SO₄)₃(¸÷1·Ö)

£¨2£©3CO₃^2-+2Al^3£«£«3H₂O £½2Al(OH)₃¡ý+ 3CO₂¡ü(2·Ö)

£¨3£©c£¨Na⁺£©>c£¨C1^£­£©>c£¨HCO₃^£­£©>c£¨OH^£­£©>c£¨H⁺£©> c£¨CO₃^2£­£©

»òc£¨Na⁺£©>c£¨C1^£­£©>c£¨HCO₃^£­£©>c£¨OH^£­£©>c£¨H⁺£©£¨2·Ö£©

£¨4£©OH^£­£¨aq£©+H⁺£¨aq£©=H₂O£¨1£© ¡÷H =£­Q kJ/mol

»ò£¨a£©Ba£¨OH£©₂£¨aq£©+2HC1£¨aq£©=BaC1₂£¨aq£©+H₂O£¨1£© ¡÷H =£­Q kJ/mol

»òBa£¨OH£©₂£¨aq£©+2HC1£¨aq£©=BaC1₂£¨aq£©+2H₂O£¨1£© ¡÷H =£­2Q kJ/mol£¨2·Ö£©

£¨5£©0.016 mol£¨2·Ö£©

½âÎö:ÂÔ