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¯ÓͺÍ______£®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+3O2Êýѧ¹«Ê½ 2N2+______£®
£¨3£©µ±Ç°ÎÒÊв¿·ÖÖÐѧÍÆÐС°Ñô¹âʳÌá±¹¤³Ì£®Ï±íΪijУʳÌÃijÌìÎç²Í²¿·ÖʳÆ×
Ö÷ʳ»ç²ËËزË
Ã×·¹ºìÉÕÅ£Èâ³´ºúÂܲ·¡¢³´»Æ¹Ï
ʳÆ×Öи»º¬µ°°×ÖʵÄÊÇ______£¬¸»º¬Î¬ÉúËصÄÊÇ______£¨ÌîÉϱíÖеÄÒ»ÖÖÖ÷ʳ»ò²ËÃû£©£®Ê³Ìó£Ê¹ÓÃÌúÇ¿»¯½´ÓÍ£¬½´ÓÍÖмÓÌúÇ¿»¯¼ÁÊÇΪÁË______£®

½â£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
·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£®
µãÆÀ£º±¾Ì⿼²éÁËÇâÑõ»¯ÄƵÄÐÔÖÊ£¬Íê³É´ËÌ⣬¿ÉÒÔÒÀ¾ÝÒÑÓеÄ֪ʶ½áºÏÖÊÁ¿Êغ㶨ÂɵÄʵÖÊÍê³É£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?Ðí²ýһģ£©ÇëÓÃËùѧ»¯Ñ§ÖªÊ¶½âÊÍÏÂÁÐÔ­Àí£®
£¨1£©Ï´µÓ¼ÁÄܹ»³ýÈ¥ÓÍÎÛµÄÔ­ÒòÊÇʲô£¿
£¨2£©»ðÔÖÏÖ³¡£¬Ïû·ÀսʿʹÓøßѹˮǹÃð»ðµÄÔ­ÀíÊÇʲô£¿
£¨3£©Éúʯ»ÒΪʲô¿ÉÓÃ×÷ʳƷµÄ¸ÉÔï¼Á£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?¶õÖÝ£©ÇëÓÃËùѧ»¯Ñ§ÖªÊ¶½âÊÍÏÂÁÐÔ­Àí£®
£¨1£©ÓûîÐÔÌ¿³ýÈ¥±ùÏäÀïµÄÒì棬ÊÇÀûÓÃÆä
Îü¸½
Îü¸½
ÐÔ£®
£¨2£©ÊÔÓÃÎÄ×Ö½âÊÍΪʲôÂÁÖÆÆ·¾ßÓÐÁ¼ºÃµÄ¿¹¸¯Ê´ÐÔÄÜ£º
ÂÁÓë¿ÕÆøÖеÄÑõÆøÐγÉÖÂÃܵÄÑõ»¯ÂÁ±¡Ä¤£¬×èÖ¹ÂÁ½øÒ»²½±»Ñõ»¯
ÂÁÓë¿ÕÆøÖеÄÑõÆøÐγÉÖÂÃܵÄÑõ»¯ÂÁ±¡Ä¤£¬×èÖ¹ÂÁ½øÒ»²½±»Ñõ»¯
£®
£¨3£©¹¤ÒµÉÏ¿ÉÓÃCl2ÓëNAOHÈÜÒº×÷ÓÃÖÆƯ°×·Û£ºCl2+2NaOH¨TNaCl+NaClO+H2O£®Óë´ËÀàËÆ£¬¿ÉÓÃCl2Óëʯ»ÒÈ飨Ö÷Òª³É·ÖCa£¨OH£©2£©ÖÆƯ°×·Û£¬ÊÔд³ö´Ë·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
2Cl2+2Ca£¨OH£©2=CaCl2+Ca£¨ClO£©2+2H2O
2Cl2+2Ca£¨OH£©2=CaCl2+Ca£¨ClO£©2+2H2O
£®
£¨4£©¼Ù»Æ½ð£¨Cu£¬ZnºÏ½ð£©Íâ¹ÛÓë»Æ½ð£¨Au£©¼°ÆäÏàËÆ£®ÊÔ¸ù¾Ý½ðÊôµÄÐÔÖÊд³ö¼ø±ðʱÓëÈýÀ಻ͬÎïÖÊ·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¨Ã¿Àà¸÷дһ¸ö»¯Ñ§·½³Ìʽ£©£º
¢Ù
2Cu+O2
  ¡÷  
.
 
2CuO£¨»ò2Zn+O2
  ¡÷  
.
 
2ZnO£©
2Cu+O2
  ¡÷  
.
 
2CuO£¨»ò2Zn+O2
  ¡÷  
.
 
2ZnO£©
£»
¢Ú
Zn+2HCl=ZnCl2+H2¡ü
Zn+2HCl=ZnCl2+H2¡ü
£»
¢Û
Cu+2AgNO3=Cu£¨NO3£©2+2Ag
Cu+2AgNO3=Cu£¨NO3£©2+2Ag
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÇëÓÃËùѧ»¯Ñ§ÖªÊ¶½âÊÍÏÂÁÐÔ­Àí£®
£¨1£©6000LÑõÆøÔÚ¼ÓѹµÄÇé¿öÏ¿ÉÒÔ×°Èë40LµÄ¸ÖÆ¿ÖУ®
£¨2£©´Ó΢¹Û½Ç¶È½âÊÍCO2ºÍCOÔªËØ×é³ÉÏàͬ£¬µ«¶þÕßÐÔÖÊÈ´²îÒìºÜ´ó£®
£¨3£©Óû¯Ñ§·½³Ìʽ±íʾ¹¤ÒµÉÏÓý¹Ì¿ÓëÌú¿óʯ£¨Ö÷Òª³É·ÖFe2O3£©·´Ó¦Á¶Ìú£®
£¨4£©×öľ̿ÔÚÑõÆøÀïȼÉÕʵÑéʱ£¬ºìÈȵÄľ̿ΪʲôҪÓÉÆ¿¿ÚÏòÏ»ºÂý²åÈ룮

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÇëÓÃËùѧ»¯Ñ§ÖªÊ¶½âÊÍÏÂÁÐÔ­Àí£®
£¨1£©¸øú¯Éú»ðʱ£¬ÓÃÓÚÒý»ðµÄľ²ñÓ¦¼Ü¿Õ¶ø²»Ò˽ôÃܶÑÔÚÒ»Æð£¬ÆäÄ¿µÄÊÇʲô£¿
£¨2£©ÈÕ³£Éú»îÖÐʹÓÃӲˮ»á´øÀ´Ðí¶àÂé·³£¬¼ÒÍ¥Éú»îÖг£ÓÃÀ´½µµÍˮӲ¶ÈµÄ·½·¨ÊÇ
Öó·Ð
Öó·Ð
£®
£¨3£©Óû¯Ñ§·½³Ìʽ±íʾΪʲô²»ÄÜÓÃÌúÖÆÈÝÆ÷Ê¢·ÅÁòËáÍ­£®
CuSO4+Fe=FeSO4+Cu
CuSO4+Fe=FeSO4+Cu

£¨4£©ÍâÆÅÔÚ¸øСǿ¼åºÉ°üµ°Ê±£¬·¢ÏÖ·ÅÈë³ø¹ñÄڵļ¦µ°Óв¿·Ö±ä³ôÁË£®Ð¡Ç¿¸ù¾ÝËùѧ»¯Ñ§ÖªÊ¶½¨ÒéÍâÆÅ£º¡°¿É½«¼¦µ°½þÓÚʯ»ÒË®ºóÀ̳ö±£´æ£®¡±ÕâÑù¾Í²»Òױ仵£¬ÄãÄܽâÊÍΪʲôÂð£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÇëÓÃËùѧ»¯Ñ§ÖªÊ¶½âÊÍÏÂÁÐÊÂʵ£º
£¨1£©Ï´µÓ¼ÁÄܳýÈ¥ÓÍÎÛ£¬ÊÇÒòΪËü¾ßÓÐ
 
¹¦ÄÜ£®
£¨2£©½ðÇïʱ½Ú£¬¹Å³Ç¿ª·â´¦´¦¶¼ÄÜÎŵ½âùÈ˵ľջ¨ÏãÆø£®
£¨3£©Ã¾ÌõÔÚ¿ÕÆøÖÐȼÉÕºó£¬Éú³ÉÎïµÄÖÊÁ¿±ÈÔ­À´Ã¾ÌõµÄÖÊÁ¿Ôö´ó£®
£¨4£©Å£ÄÌÔÚÏÄÌì±È¶¬Ìì¸üÈÝÒ×±äËᣮ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸