ÌâÄ¿ÄÚÈÝ
̼ÔڵؿÇÖеĺ¬Á¿²»¸ß£¬µ«ËüµÄ»¯ºÏÎïÊýÁ¿Öڶ࣬¶øÇÒ·Ö²¼¼«¹ã¡£
¸ù¾ÝÉÏͼ¼°Ëùѧ֪ʶ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÉÏͼÖÐÊôÓÚ½ð¸Õʯ½á¹¹µÄÊÇ £¨ÌîдͼÐòºÅ£©£»Í¼¢ÜÒ²ÊÇ̼µÄÒ»ÖÖµ¥ÖÊ£¬Æ仯ѧʽΪ £»Í¼¢ÙΪ̼ԪËصÄÔ×ӽṹʾÒâͼ£¬³£ÎÂÏÂ̼µÄ»¯Ñ§ÐÔÖÊ________£¨Ìî¡°»îÆá±»ò¡°²»»îÆá±£©¡£
£¨2£©ÒÑÖªXÊÇÓж¾ÇÒ²»ÈÜÓÚË®µÄÆøÌ壬YÊDz»Ö§³ÖȼÉÕµÄÆøÌ壬ZÊDz»ÈÜÓÚË®µÄ¹ÌÌ壬X¡¢Y¡¢ZÖ®¼äÓÐÈçͼת»¯¹Øϵ£º
¢ÙXµÄ»¯Ñ§Ê½Îª___________£»
¢ÚYÓëʯ»ÒË®·´Ó¦Éú³ÉZµÄ»¯Ñ§·½³ÌʽΪ_________£»
¢ÛÆøÌåX¡¢YÖÐËùº¬ÔªËØÏàͬ£¬µ«ËüÃǵĻ¯Ñ§ÐÔÖʲ»Í¬£¬ÆäÔÒòÊÇ___________.
£¨3£©½«Ì¼·ÛÓëÑõ»¯Í»ìºÏºó¼ÓÇ¿ÈÈ£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________¡£
£¨4£©ÔÚ¿Æѧ¼ÒÑÛÀCO2ÊÇ¿ÉÒÔÀûÓõÄÖØÒª×ÊÔ´¡£ÔÚÒ»¶¨Ìõ¼þÏ£¬CO2ºÍ½ðÊôÄÆ·´Ó¦¿ÉÒÔÖÆÈ¡½ð¸Õʯ£¬·´Ó¦µÄ·½³ÌʽÊÇCO2+4NaC£¨½ð¸Õʯ£©+2Na2O £¬Çë¼ÆËã92g½ðÊôÄÆÀíÂÛÉÏ¿ÉÒÔÖƵú¬Ì¼96%µÄ½ð¸ÕʯµÄÖÊÁ¿¡£
¸ù¾ÝÉÏͼ¼°Ëùѧ֪ʶ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÉÏͼÖÐÊôÓÚ½ð¸Õʯ½á¹¹µÄÊÇ £¨ÌîдͼÐòºÅ£©£»Í¼¢ÜÒ²ÊÇ̼µÄÒ»ÖÖµ¥ÖÊ£¬Æ仯ѧʽΪ £»Í¼¢ÙΪ̼ԪËصÄÔ×ӽṹʾÒâͼ£¬³£ÎÂÏÂ̼µÄ»¯Ñ§ÐÔÖÊ________£¨Ìî¡°»îÆá±»ò¡°²»»îÆá±£©¡£
£¨2£©ÒÑÖªXÊÇÓж¾ÇÒ²»ÈÜÓÚË®µÄÆøÌ壬YÊDz»Ö§³ÖȼÉÕµÄÆøÌ壬ZÊDz»ÈÜÓÚË®µÄ¹ÌÌ壬X¡¢Y¡¢ZÖ®¼äÓÐÈçͼת»¯¹Øϵ£º
¢ÙXµÄ»¯Ñ§Ê½Îª___________£»
¢ÚYÓëʯ»ÒË®·´Ó¦Éú³ÉZµÄ»¯Ñ§·½³ÌʽΪ_________£»
¢ÛÆøÌåX¡¢YÖÐËùº¬ÔªËØÏàͬ£¬µ«ËüÃǵĻ¯Ñ§ÐÔÖʲ»Í¬£¬ÆäÔÒòÊÇ___________.
£¨3£©½«Ì¼·ÛÓëÑõ»¯Í»ìºÏºó¼ÓÇ¿ÈÈ£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________¡£
£¨4£©ÔÚ¿Æѧ¼ÒÑÛÀCO2ÊÇ¿ÉÒÔÀûÓõÄÖØÒª×ÊÔ´¡£ÔÚÒ»¶¨Ìõ¼þÏ£¬CO2ºÍ½ðÊôÄÆ·´Ó¦¿ÉÒÔÖÆÈ¡½ð¸Õʯ£¬·´Ó¦µÄ·½³ÌʽÊÇCO2+4NaC£¨½ð¸Õʯ£©+2Na2O £¬Çë¼ÆËã92g½ðÊôÄÆÀíÂÛÉÏ¿ÉÒÔÖƵú¬Ì¼96%µÄ½ð¸ÕʯµÄÖÊÁ¿¡£
£¨1£©¢Ú C60 ²»»îÆÃ
£¨2£©¢Ù CO ¢ÚCO2+Ca£¨OH£©2=CaCO3¡ý+H2O ¢Û·Ö×ӵĹ¹³É²»Í¬
£¨3£© C+2CuO2Cu+ CO2¡ü
£¨4£© 12.5g
£¨2£©¢Ù CO ¢ÚCO2+Ca£¨OH£©2=CaCO3¡ý+H2O ¢Û·Ö×ӵĹ¹³É²»Í¬
£¨3£© C+2CuO2Cu+ CO2¡ü
£¨4£© 12.5g
ÊÔÌâ·ÖÎö£º£¨1£©½ð¸ÕʯµÄÓ²¶È´ó£¬¹ÊÊÇ¢Ú £¬Í¼¢Ü½á¹¹ÐÎ×´Ïñ×ãÇò£¬Ó¦ÊÇC60
£¨2£©ÓÉXÊÇÓж¾ÇÒ²»ÈÜÓÚË®µÄÆøÌ壬ÄÇôXÊÇÒ»Ñõ»¯Ì¼£¬Ò»Ñõ»¯Ì¼È¼ÉÕÉú³ÉµÄÊǶþÑõ»¯Ì¼£¬¹ÊYÊǶþÑõ»¯Ì¼£¬¶þÑõ»¯Ì¼ºÍʯ»ÒË®·´Ó¦Éú³ÉµÄÊÇ̼Ëá¸Æ¡£
£¨3£©Ì¿·Û¾ßÓл¹ÔÐÔ£¬ÄÜ»¹ÔÑõ»¯ÍÉú³ÉͺͶþÑõ»¯Ì¼£»
£¨4£© ÀûÓû¯Ñ§·½³ÌʽÖÐÄÆÓë½ð¸ÕʯµÄÖÊÁ¿±È£¬½áºÏ½ðÊôÄƵÄÖÊÁ¿£¬¿ÉÇó³ö´¿¾»µÄ½ð¸ÕʯµÄÖÊÁ¿¡£ÔÙÓýð¸ÕʯµÄÖÊÁ¿³ýÒÔ´¿¶È£¬¼´ÊǺ¬ÔÓÖʵĵĽð¸ÕʯµÄÖÊÁ¿¡£
É裺̼µÄÖÊÁ¿ÎªX
CO2+4NaC£¨½ð¸Õʯ£©+2Na2O
4¡Á23 12
92g X
92/12=92g/X
X=12g
12g/96%=12.5g
´ð£º¿ÉÒÔÖƵú¬Ì¼96%µÄ½ð¸Õʯ12.5g
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿