Question

20£®Ì¼ËáÄƺÍ̼ËáÇâÄÆÊÇÉú»îÖг£¼ûµÄÎïÖÊ£®Çë»Ø´ðÏÂÁÐÎÊÌ⣮
£¨1£©Ì¼ËáÄÆË׳ƴ¿¼î»òËÕ´ò£¬µÈÎïÖʵÄÁ¿µÄ̼ËáÄƺÍ̼ËáÇâÄÆÓë×ãÁ¿ÑÎËᷴӦʱÉú³ÉCO₂µÄÁ¿£¬Ç°Õß=ºóÕߣ¨Ì¡¢£¼»ò=£©£®
£¨2£©Ì¼ËáÇâÄÆ¿ÉÖÎÁÆθËᣨ0.2%¡«0.4%µÄÑÎËᣩ¹ý¶à£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºHCO₃^-+H⁺=H₂O+CO₂¡ü£®
£¨3£©ÏÖÓÐÒÔÏÂÎïÖÊ£º¢ÙNaCl¾§Ìå ¢ÚÈÛÈÚµÄKNO₃¢Û°±Ë® ¢ÜÂÁ ¢ÝCl₂ ¢Þ¾Æ¾«£¨C₂H₅OH£©Çë»Ø´ðÏÂÁÐÎÊÌ⣨ÓÃÐòºÅ£©£ºÒÔÉÏÎïÖÊÖÐÄܵ¼µçµÄÊǢڢۢܣ»ÒÔÉÏÎïÖÊÖÐÊôÓÚÇ¿µç½âÖʵÄÊÇ¢Ù¢Ú£®
£¨4£©³ýȥ̼ËáÄÆÖлìÓеÄ̼ËáÇâÄƵĻ¯Ñ§·½³Ìʽ2NaHCO₃$\frac{\underline{\;\;¡÷\;\;}}{\;}$Na₂CO₃ +CO₂¡ü+H₂O
£¨5£©ÏÂÁÐʵÑé·½°¸ÖУ¬²»ÄÜ׼ȷ²â¶¨Na₂CO₃ºÍNaHCO₃»ìºÏÎïÖÐNa₂CO₃ÖÊÁ¿·ÖÊýµÄÊÇC
A£®È¡a¿Ë»ìºÏÎï³ä·Ö¼ÓÈÈ£¬¼õÖØb¿Ë
B£®È¡a¿Ë»ìºÏÎïÓë×ãÁ¿Ï¡ÑÎËá³ä·Ö·´Ó¦£¬¼ÓÈÈ¡¢Õô¸É¡¢×ÆÉÕ£¬µÃb¿Ë¹ÌÌå
C£®È¡a¿Ë»ìºÏÎïÓë×ãÁ¿Ï¡ÁòËá³ä·Ö·´Ó¦£¬ÒݳöÆøÌåÖ±½ÓÓüîʯ»ÒÎüÊÕ£¬ÔöÖØb¿Ë
D£®È¡a¿Ë»ìºÏÎïÓë×ãÁ¿Ba£¨OH£©₂ÈÜÒº³ä·Ö·´Ó¦£¬¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É£¬µÃb¿Ë¹ÌÌ壮

Answer 1

·ÖÎö £¨1£©Ì¼ËáÄÆË׳ƴ¿¼î»òËÕ´ò£¬¸ù¾Ý·´Ó¦µÄ·½³Ìʽ¼ÆË㣻
£¨2£©Ì¼ËáÇâÄÆÓëÑÎËá·´Ó¦Éú³ÉÂÈ»¯ÄÆ¡¢Ë®ºÍ¶þÑõ»¯Ì¼£¬ÒÔ´ËË®½âÀë×Ó·½³ÌʽΪ£»
£¨3£©µç½âÖÊÔÚÈÛÈÚ״̬»òË®ÈÜÒºÖе¼µç£¬½ðÊô¿Éµ¼µç£»³£¼ûÇ¿µç½âÖÊÓÐÇ¿Ëᡢǿ¼îºÍ´ó¶àÊýÑΣ»
£¨4£©Ì¼ËáÇâÄƼÓÈÈ·Ö½âÉú³É̼ËáÄÆ£¬¿ÉÒÔÓüÓÈȵķ½·¨³ýȥ̼ËáÄƹÌÌåÖÐ̼ËáÇâÄÆ£»
£¨5£©A¡¢Ì¼ËáÇâÄƵIJ»Îȶ¨ÐÔ£¬¼ÓÈȷֽ⣬ÀûÓùÌÌå²îÁ¿·¨¼´¿É¼ÆËã³ö¹ÌÌåÖÐ̼ËáÇâÄƵÄÖÊÁ¿£»
B£®¸ù¾ÝÄÆÊغ㣬¿ÉÁз½³Ì×éÇó½â£»
C¡¢Ó¦ÏÈ°ÑË®ÕôÆøÅųý²ÅºÏÀí£»
D¡¢¸ù¾ÝÖÊÁ¿¹Øϵ£¬¿ÉÁз½³Ì×éÇó½â£®

½â´ð ½â£º£¨1£©Ì¼ËáÄÆË׳ƴ¿¼î»òËÕ´ò£¬NaHCO₃ÓëHCl·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºNaHCO₃+HCl=NaCl+H₂O+CO₂¡ü£»Na₂CO₃ÓëHCl·´Ó¦µÄ»¯Ñ§·½³ÌʽΪNa₂CO₃+2HCl=2NaCl+H₂O+CO₂¡ü£¬ÓÉÒÔÉÏÁ½¸ö»¯Ñ§·½³Ìʽ¿ÉÒÔ¿´³ö£¬µÈÎïÖʵÄÁ¿µÄNa₂CO₃ºÍNaHCO₃·Ö±ðÓëHCl·´Ó¦£¬²úÉúCO₂µÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º1£¬
¹Ê´ð°¸Îª£º´¿¼î»òËÕ´ò£»=£»
£¨2£©ÒòNaHCO₃ÓëHCl·´Ó¦µÄ»¯Ñ§·½³ÌʽΪNaHCO₃+HCl=NaCl+H₂O+CO₂¡ü£¬Ôò·´Ó¦µÄÀë×Ó·½³ÌʽΪHCO₃^-+H⁺=H₂O+CO₂¡ü£¬
¹Ê´ð°¸Îª£ºHCO₃^-+H⁺=H₂O+CO₂¡ü£»
£¨3£©¢ÚÈÛÈÚµÄKNO₃¢Û°±Ë®Öк¬ÓÐ×ÔÓÉÒƶ¯µÄÀë×Ó£¬Äܵ¼µç£¬¢ÜÂÁΪ½ðÊô¾§Ì壬Äܵ¼µç£¬£º¢ÙNaCl¾§Ìå¡¢¢ÚÈÛÈÚµÄKNO₃Ϊǿµç½âÖÊ£¬
¹Ê´ð°¸Îª£º¢Ú¢Û¢Ü£»¢Ù¢Ú£»
£¨4£©Ì¼ËáÇâÄƼÓÈÈ·Ö½âÉú³É̼ËáÄÆ£¬¿ÉÓüÓÈȵķ½·¨À´³ýȥ̼ËáÇâÄÆ£¬·´Ó¦·½³ÌʽΪ£º2NaHCO₃$\frac{\underline{\;\;¡÷\;\;}}{\;}$Na₂CO₃ +CO₂¡ü+H₂O£¬
¹Ê´ð°¸Îª£º2NaHCO₃$\frac{\underline{\;\;¡÷\;\;}}{\;}$Na₂CO₃ +CO₂¡ü+H₂O£»
£¨5£©A¡¢NaHCO₃ÊÜÈÈÒ×·Ö½âÉú³É̼ËáÄÆ¡¢Ë®ºÍ¶þÑõ»¯Ì¼£¬ËùÒÔͨ¹ý¼ÓÈÈ·Ö½âÀûÓòîÁ¿·¨¼´¿É¼ÆËã³öNa₂CO₃ÖÊÁ¿·ÖÊý£¬¹ÊÕýÈ·£»
B£®Na₂CO₃ºÍNaHCO₃¾ù¿ÉÓëÑÎËá·´Ó¦Éú³ÉË®¡¢¶þÑõ»¯Ì¼ºÍÂÈ»¯ÄÆ£¬ËùÒÔbg¹ÌÌåÊÇÂÈ»¯ÄÆ£¬ÀûÓÃÊغ㷨¿É¼ÆËã³öNa₂CO₃ÖÊÁ¿·ÖÊý£¬¹ÊÕýÈ·£»
C¡¢»ìºÏÎïÓë×ãÁ¿Ï¡ÁòËá³ä·Ö·´Ó¦£¬Ò²»áÉú³ÉË®ºÍ¶þÑõ»¯Ì¼£¬ËùÒÔÒݳöµÄÆøÌåÊǶþÑõ»¯Ì¼£¬µ«»á»ìÓÐË®ÕôÆø£¬¼´¼îʯ»ÒÔö¼ÓµÄÖÊÁ¿²»ÊǶþÑõ»¯Ì¼µÄÖÊÁ¿£¬²»Äܲⶨº¬Á¿£¬¹Ê´íÎó£»
D¡¢Na₂CO₃ºÍNaHCO₃¶¼ÓëBa£¨OH£©₂·´Ó¦£¬·´Ó¦µÄ·½³ÌʽΪCO₃^2-+Ba²⁺=BaCO₃¡ý¡¢HCO₃^-+OH^-+Ba²⁺=H₂O+BaCO₃¡ý£¬Òò´Ë×îºóµÃµ½µÄ¹ÌÌåÊÇBaCO₃£¬ËùÒÔ¿ÉÒÔ¼ÆËã³öNa₂CO₃ÖÊÁ¿·ÖÊý£¬¹ÊÕýÈ·£»
¹ÊÑ¡C£®

µãÆÀ ±¾Ìâ×ۺϿ¼²éÁËNa₂CO₃ºÍNaHCO₃µÄÐÔÖÊ¡¢»ìºÏÎï¼ÆËã¡¢¶ÔʵÑé·½°¸ÆÀ¼ÛµÈ£¬ÌâÄ¿ÄѶÈÖеȣ¬×¢ÒâÀí½âÖÊÁ¿·ÖÊý¶¨Ò壬½áºÏÎïÖʵÄÐÔÖÊ·ÖÎöʵÑé·½°¸£®