ÌâÄ¿ÄÚÈÝ
ÇëÓÃËùѧ»¯Ñ§ÖªÊ¶½âÊÍÏÂÁÐÔÀí£®
£¨1£©ÇâÑõ»¯ÄƹÌÌåÆضÔÚ¿ÕÆøÖУ¬ºÜÈÝÒ×±äÖÊ£¬Æä±ä»¯µÄ»¯Ñ§·½³Ìʽ¿É±íʾΪ£º
£¬ËùÒÔÇâÑõ»¯ÄƹÌÌå±ØÐëÃÜ·â±£´æ£®21ÊÀ¼Í½ÌÓýÍøʵÑéÊÒÊ¢·ÅÇâÑõ»¯ÄÆÈÜÒºµÄÊÔ¼ÁÆ¿²»ÄÜÓò£Á§Èû£¬ÆäÔÒòÊÇÔÚ³£ÎÂÏ£¬ÇâÑõ»¯ÄÆÓë²£Á§ÖеĶþÑõ»¯¹è»ºÂýµØ·¢Éú·´Ó¦£¬²úÎïʹƿ¿ÚÓëÆ¿ÈûÕ³ºÏÔÚÒ»Æ𣬷´Ó¦µÄ»¯Ñ§·½³ÌʽΪ £®
£¨2£©ÄÜÔ´¡¢»·¾³ÓëÈËÀàµÄÉú»îºÍÉç»á·¢Õ¹ÃÜÇÐÏà¹Ø£®Ä¿Ç°£¬ÈËÀ໯ʯȼÁÏΪÖ÷ÒªÄÜÔ´£¬³£¼ûµÄ»¯Ê¯È¼ÁÏ°üÀ¨Ãº¡¢Ê¯ÓÍºÍ £®Îª¼õÉÙÎÂÊÒÆøÌåµÄÅÅ·Å£¬ÈËÃÇ»ý¼«Ñ°ÕÒ²»º¬Ì¼ÔªËصÄȼÁÏ£®¾Ñо¿·¢ÏÖ£¬NH3ȼÉյIJúÎïûÓÐÎÛȾ£¬ÇÒÊÍ·Å´óÁ¿ÄÜÁ¿£¬ÓÐÒ»¶¨Ó¦ÓÃÇ°¾°£®½«È¼ÉÕ·´Ó¦µÄ»¯Ñ§·½³Ìʽ²¹³äÍêÕû£º4NH3+3O2
2N2+ £®
£¨3£©µ±Ç°ÎÒÊв¿·ÖÖÐѧÍÆÐС°Ñô¹âʳÌá±¹¤³Ì£®Ï±íΪijУʳÌÃijÌìÎç²Í²¿·ÖʳÆ×
ʳÆ×Öи»º¬µ°°×ÖʵÄÊÇ £¬¸»º¬Î¬ÉúËصÄÊÇ £¨ÌîÉϱíÖеÄÒ»ÖÖÖ÷ʳ»ò²ËÃû£©£®Ê³Ìó£Ê¹ÓÃÌúÇ¿»¯½´ÓÍ£¬½´ÓÍÖмÓÌúÇ¿»¯¼ÁÊÇΪÁË £®
£¨1£©ÇâÑõ»¯ÄƹÌÌåÆضÔÚ¿ÕÆøÖУ¬ºÜÈÝÒ×±äÖÊ£¬Æä±ä»¯µÄ»¯Ñ§·½³Ìʽ¿É±íʾΪ£º
£¨2£©ÄÜÔ´¡¢»·¾³ÓëÈËÀàµÄÉú»îºÍÉç»á·¢Õ¹ÃÜÇÐÏà¹Ø£®Ä¿Ç°£¬ÈËÀ໯ʯȼÁÏΪÖ÷ÒªÄÜÔ´£¬³£¼ûµÄ»¯Ê¯È¼ÁÏ°üÀ¨Ãº¡¢Ê¯ÓͺÍ
| ||
£¨3£©µ±Ç°ÎÒÊв¿·ÖÖÐѧÍÆÐС°Ñô¹âʳÌá±¹¤³Ì£®Ï±íΪijУʳÌÃijÌìÎç²Í²¿·ÖʳÆ×
Ö÷ʳ | »ç²Ë | ËØ²Ë |
Ã×·¹ | ºìÉÕÅ£Èâ | ³´ºúÂܲ·¡¢³´»Æ¹Ï |
·ÖÎö£º£¨1£©ÇâÑõ»¯ÄƹÌÌåÒ×ÎüË®³±½â£¬Ò׺ͶþÑõ»¯Ì¼·´Ó¦¶ø±äÖÊ£»ÇâÑõ»¯ÄÆÓë²£Á§ÖеĶþÑõ»¯¹è»ºÂýµØ·¢Éú·´Ó¦
£¨2£©¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉ£¬»¯Ñ§·´Ó¦Ç°ºóÔªËصÄÖÖÀàºÍÔ×ӵĸöÊý²»±ä£¬¿ÉÒÔÍÆÖªÎïÖʵĻ¯Ñ§Ê½£®
£¨3£©ÈËÀàÐèÒªµÄÓªÑøÎïÖÊÓÐÌÇÀà¡¢ÓÍÖ¬¡¢µ°°×ÖÊ¡¢Î¬ÉúËØ¡¢Ë®ºÍÎÞ»úÑΣ»ÈËÌåȱ·¦ÌúÔªËØÈÝÒ×»¼È±ÌúÐÔƶѪ£®
£¨2£©¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉ£¬»¯Ñ§·´Ó¦Ç°ºóÔªËصÄÖÖÀàºÍÔ×ӵĸöÊý²»±ä£¬¿ÉÒÔÍÆÖªÎïÖʵĻ¯Ñ§Ê½£®
£¨3£©ÈËÀàÐèÒªµÄÓªÑøÎïÖÊÓÐÌÇÀà¡¢ÓÍÖ¬¡¢µ°°×ÖÊ¡¢Î¬ÉúËØ¡¢Ë®ºÍÎÞ»úÑΣ»ÈËÌåȱ·¦ÌúÔªËØÈÝÒ×»¼È±ÌúÐÔƶѪ£®
½â´ð£º½â£º
£¨1£©ÇâÑõ»¯ÄƹÌÌåÆضÔÚ¿ÕÆøÖУ¬ºÜÈÝÒ×±äÖÊ£¬Æä±ä»¯µÄ»¯Ñ§·½³Ìʽ¿É±íʾΪ£ºCO2+2NaOH=Na2CO3+H2O£¬ËùÒÔÇâÑõ»¯ÄƹÌÌå±ØÐëÃÜ·â±£´æ£®ÊµÑéÊÒÊ¢·ÅÇâÑõ»¯ÄÆÈÜÒºµÄÊÔ¼ÁÆ¿²»ÄÜÓò£Á§Èû£¬ÆäÔÒòÊÇÔÚ³£ÎÂÏ£¬ÇâÑõ»¯ÄÆÓë²£Á§ÖеĶþÑõ»¯¹è»ºÂýµØ·¢Éú·´Ó¦£¬²úÎïʹƿ¿ÚÓëÆ¿ÈûÕ³ºÏÔÚÒ»Æ𣬷´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
SiO2+2NaOH=Na2SiO3+H2O£®
£¨2£©ÄÜÔ´¡¢»·¾³ÓëÈËÀàµÄÉú»îºÍÉç»á·¢Õ¹ÃÜÇÐÏà¹Ø£®Ä¿Ç°£¬ÈËÀ໯ʯȼÁÏΪÖ÷ÒªÄÜÔ´£¬³£¼ûµÄ»¯Ê¯È¼ÁÏ°üÀ¨Ãº¡¢Ê¯ÓÍºÍ ÌìÈ»Æø£®Îª¼õÉÙÎÂÊÒÆøÌåµÄÅÅ·Å£¬ÈËÃÇ»ý¼«Ñ°ÕÒ²»º¬Ì¼ÔªËصÄȼÁÏ£®¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉ£¬¿É½«È¼ÉÕ·´Ó¦µÄ»¯Ñ§·½³Ìʽ²¹³äÍêÕû£º4NH3+3O2
2N2+6H2O£®
£¨3£©ºìÉÕÈâÖк¬ÓзḻµÄµ°°×ÖÊ£¬ºúÂܲ·¡¢»Æ¹ÏÖк¬ÓзḻµÄάÉúËØ£®Ê¹ÓÃÌúÇ¿»¯½´ÓÍ¿ÉÒÔΪÈËÌåÌṩÌúÔªËØ£¬Äܹ»·ÀֹȱÌúÐÔƶѪ£®
¹Ê´ð°¸Îª£º£¨1£©CO2+2NaOH=Na2CO3+H2O£» SiO2+2NaOH=Na2SiO3+H2O
£¨2£©ÌìÈ»Æø£» 6H2O
£¨3£©ºìÉÕÅ£È⣻ ³´ºúÂܲ·¡¢³´»Æ¹Ï£»·ÀÖ¹³öÏÖȱÌúÐÔƶѪ
£¨1£©ÇâÑõ»¯ÄƹÌÌåÆضÔÚ¿ÕÆøÖУ¬ºÜÈÝÒ×±äÖÊ£¬Æä±ä»¯µÄ»¯Ñ§·½³Ìʽ¿É±íʾΪ£ºCO2+2NaOH=Na2CO3+H2O£¬ËùÒÔÇâÑõ»¯ÄƹÌÌå±ØÐëÃÜ·â±£´æ£®ÊµÑéÊÒÊ¢·ÅÇâÑõ»¯ÄÆÈÜÒºµÄÊÔ¼ÁÆ¿²»ÄÜÓò£Á§Èû£¬ÆäÔÒòÊÇÔÚ³£ÎÂÏ£¬ÇâÑõ»¯ÄÆÓë²£Á§ÖеĶþÑõ»¯¹è»ºÂýµØ·¢Éú·´Ó¦£¬²úÎïʹƿ¿ÚÓëÆ¿ÈûÕ³ºÏÔÚÒ»Æ𣬷´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
SiO2+2NaOH=Na2SiO3+H2O£®
£¨2£©ÄÜÔ´¡¢»·¾³ÓëÈËÀàµÄÉú»îºÍÉç»á·¢Õ¹ÃÜÇÐÏà¹Ø£®Ä¿Ç°£¬ÈËÀ໯ʯȼÁÏΪÖ÷ÒªÄÜÔ´£¬³£¼ûµÄ»¯Ê¯È¼ÁÏ°üÀ¨Ãº¡¢Ê¯ÓÍºÍ ÌìÈ»Æø£®Îª¼õÉÙÎÂÊÒÆøÌåµÄÅÅ·Å£¬ÈËÃÇ»ý¼«Ñ°ÕÒ²»º¬Ì¼ÔªËصÄȼÁÏ£®¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉ£¬¿É½«È¼ÉÕ·´Ó¦µÄ»¯Ñ§·½³Ìʽ²¹³äÍêÕû£º4NH3+3O2
| ||
£¨3£©ºìÉÕÈâÖк¬ÓзḻµÄµ°°×ÖÊ£¬ºúÂܲ·¡¢»Æ¹ÏÖк¬ÓзḻµÄάÉúËØ£®Ê¹ÓÃÌúÇ¿»¯½´ÓÍ¿ÉÒÔΪÈËÌåÌṩÌúÔªËØ£¬Äܹ»·ÀֹȱÌúÐÔƶѪ£®
¹Ê´ð°¸Îª£º£¨1£©CO2+2NaOH=Na2CO3+H2O£» SiO2+2NaOH=Na2SiO3+H2O
£¨2£©ÌìÈ»Æø£» 6H2O
£¨3£©ºìÉÕÅ£È⣻ ³´ºúÂܲ·¡¢³´»Æ¹Ï£»·ÀÖ¹³öÏÖȱÌúÐÔƶѪ
µãÆÀ£º±¾Ì⿼²éÁËÇâÑõ»¯ÄƵÄÐÔÖÊ£¬Íê³É´ËÌ⣬¿ÉÒÔÒÀ¾ÝÒÑÓеÄ֪ʶ½áºÏÖÊÁ¿Êغ㶨ÂɵÄʵÖÊÍê³É£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿