ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿ÈËÃǵÄÉú»îºÍÉú²úÀë²»¿ªÄÜÔ´¡£
(1)Ä¿Ç°ÈËÀàʹÓõÄȼÁÏ´ó¶àÀ´×Ô»¯Ê¯È¼ÁÏ£¬Èç______¡¢Ê¯ÓÍ¡¢ÌìÈ»ÆøµÈ,ÊôÓÚ______¡¢(¡°¿ÉÔÙÉú¡±»ò¡°²»¿ÉÔÙÉú¡±)ÄÜÔ´¡£
(2)´ÓʯÓÍÖзÖÀë³öÆûÓÍ¡¢²ñÓ͵ȵĹý³Ì½ÐʯÓ͵ķÖÁó,ÔÀíÊǸù¾Ý¸÷ÎïÖʵÄ_______²»Í¬½«ÈȵÄÔÓͽøÐзÖÀë,ÊÇ_______±ä»¯;
(3)½ñÄê5ÔÂ18ÈÕ.ÄϺ£¡°¿Éȼ±ù¡±(ѧÃûÌìÈ»ÆøË®ºÏÎÖ÷Òª³É·ÖÊÇCH4)ÊԲɳɹ¦£¬±êÖ¾×ÅÖйúµÄ¿ª²ÉʵÁ¦ÒÑ´¦ÔÚÊÀ½çÇ°ÁС£¸ÃȼÁÏȼÉյĻ¯Ñ§·½³ÌʽÊÇ____________.
(4)Ç⻯þ(MgH2)¹ÌÌå¿É×÷ΪÇ⶯Á¦Æû³µµÄÄÜÔ´Ìṩ¼Á£¬ÌṩÄÜԴʱÓëË®·´Ó¦Éú³ÉÇâÑõ»¯Ã¾ºÍÒ»ÖÖ¿ÉȼÐÔÆøÌ壬·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ________,Ñ¡ÓÃÇ⻯þ×÷ΪÄÜÔ´µÄ¸öÓŵãÊÇ________.
¡¾´ð°¸¡¿ ú ²»¿ÉÔÙÉú ·Ðµã ÎïÀí CH4+202C02+H20 MgH2+2H20=Mg(0H)2+2H2 ¡ü Я´ø·½±ã£¬Ê¹Óð²È«£¬²úÎïȼÉÕÎÞÎÛȾµÈ
¡¾½âÎö¡¿±¾Ì⿼²éÁË»¯Ê¯È¼Áϼ°Æä×ÛºÏÀûÓ㬻¯Ñ§±ä»¯ºÍÎïÀí±ä»¯µÄÅб𣬻¯Ñ§·½³ÌʽµÄÊéдµÈ¡£
£¨1£©Ãº¡¢Ê¯ÓͺÍÌìÈ»ÆøÊÇÈý´ó»¯Ê¯È¼ÁÏ£¬¶¼ÊôÓÚ²»¿ÉÔÙÉúÄÜÔ´£»
£¨2£©¸ù¾ÝʯÓÍÖи÷×é·ÖµÄ·Ðµã²»Í¬£¬´ÓʯÓÍÖзÖÀë³öÆûÓÍ¡¢ÃºÓͺͲñÓ͵ȣ¬·ÖÀë¹ý³ÌÖÐûÓÐÐÂÎïÖÊÉú³É£¬ÊôÓÚÎïÀí±ä»¯£»
£¨3£©¼×ÍéȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍË®£¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCH4 + 2O2 CO2 +2 H2O
£¨4£©MgH2¿ÉÒÔºÍË®·´Ó¦Éú³ÉÇâÑõ»¯Ã¾ºÍÇâÆø£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºMgH2+2H2O¨TMg£¨OH£©2+2H2¡ü£»Ç⻯þ×÷ΪÄÜÔ´µÄÓŵãÊÇЯ´ø·½±ã£¬Ê¹Óð²È«£¬ÎÞÎÛȾ¡£
¡¾ÌâÄ¿¡¿Ê¯»ÒʯÊÇÉú²ú²£Á§¡¢ÂÈ»¯¸ÆµÈ¶àÖÖ»¯¹¤²úÆ·µÄÔÁÏ¡£Ä³Ñо¿ÐÔѧϰС×éΪÁ˲ⶨµ±µØ¿óɽʯ»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£¬È¡À´ÁË¿óʯÑùÆ·£¬²¢È¡Ï¡ÑÎËá200g£¬Æ½¾ù·Ö³É4·Ý£¬½øÐÐʵÑ飬½á¹ûÈçÏ£º
ʵ Ñé | 1 | 2 | 3 | 4 |
¼ÓÈëÑùÆ·ÖÊÁ¿/g | 5 | 10 | 15 | 20 |
Éú³ÉCO2µÄÖÊÁ¿/g | 1.76 | 3.52 | 4.4 | m |
(1)µÚ¼¸´Î·´Ó¦ÖпóʯÓÐÊ£Óࣿ_________£»
(2)±íÖÐmÖµÊÇ________£»
(3)ÊÔ¼ÆËãÕâÖÖʯ»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ_________¡£