ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿ÃºµÄ×ÛºÏÀûÓÃÊǽ«Ãº¸ô¾ø¿ÕÆø¼ÓÇ¿ÈÈ£¬Ê¹Ãº·Ö½â³ÉÐí¶àÓÐÓõÄÎïÖÊ£¬ÃºÆøÊÇÆäÖÐÒ»ÖÖ¡£ÃºÆøµÄÖ÷Òª³É·ÖÊÇʲôÄØ£¿Ä³ÐËȤС×éΪ´ËÕ¹¿ªÁË̽¾¿¡£
¡¾²éÔÄ×ÊÁÏ¡¿ £¨1£© úÆøÖпÉÄܺ¬ÓÐ CO¡¢CO2¡¢H2¡¢CH4 ÖеÄÒ»ÖÖ»ò¼¸ÖÖ¡£
£¨2£© ³£ÎÂÏ£¬ÂÈ»¯îÙ(PdCl2)ÈÜÒºÎüÊÕ CO ʱ·¢ÉúµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ£º
CO £«PdCl2£«H2O CO2£«Pd¡ý(»ÒÉ«) £«2HCl
¡¾ÊµÑé·½°¸¡¿ÐËȤС×éµÄͬѧÉè¼ÆÈçÏÂ×°ÖÃ̽¾¿ÃºÆøÖеijɷ֡£
¡¾½»Á÷ÌÖÂÛ¡¿Çë»Ø´ðʵÑéÖеÄÓйØÎÊÌ⣺
£¨1£©ÊµÑ鿪ʼºó£¬A ÖÐÎÞÃ÷ÏÔÏÖÏó£¬ËµÃ÷úÆøÖв»´æÔÚ____________________£»
£¨2£© ÈôúÆøÖÐÓÐ CO ´æÔÚ£¬Ôò BÖвúÉúµÄÏÖÏóÊÇ £»
£¨3£© д³ö C Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ___________________________£»£¨Ð´Ò»¸ö¼´¿É£©
D ÖÐŨÁòËáµÄ×÷ÓÃÊÇ_________________________¡£
£¨4£© Èô F ÖÐÎÞË®CuSO4±äÀ¶É«£¬ÔòEÖÐÒ»¶¨·¢ÉúµÄ»¯Ñ§·½³ÌʽΪ £»
£¨5£©µãȼF×°Öõ¼³öµÄÆøÌ壬ÓлðÑæ²úÉú¡£½·ãÈÏΪúÆøÖк¬ CH4£¬ÀíÓÉÊÇCH4¿ÉÒÔȼÉÕ£¬CH4 ȼÉյĻ¯Ñ§·½³ÌʽÊÇ____________________________________£¬ÀÏʦÔòÈÏΪ²»ÄÜÈ·ÈÏúÆøÖк¬ÓÐ CH4£¬ÀíÓÉÊÇ__________________________¡£ÎªÁËÈ·¶¨ÃºÆøÖÐÊÇ·ñÓÐCH4£¬ÀÏʦ½Ó×ŲÉÈ¡µÄ²Ù×÷·½·¨ÊÇ £»¹Û²ìµ½µÄÏÖÏóÊÇ___________________________________¡£ÓÉ´ËÑé֤úÆøÖÐÒ»¶¨º¬ÓÐCH4¡£
¡¾´ð°¸¡¿ ¡¾½»Á÷ÌÖÂÛ¡¿ £¨1£©CO2£¨2£©ÈÜÒºÖвúÉú»ÒÉ«³Áµí£¨3£©NaOH+ HCl NaCl +H2O »ò 2NaOH+ CO2 Na2CO3 +H2O ¸ÉÔïÆøÌå»ò³ýÈ¥ÆøÌåÖеÄË®·Ö£¨4£©H2 +CuOCu +H2O £¨5£©CH4 +2O2 CO2 +2H2O ²»ÄÜÈ·ÈÏH2ÓëE×°ÖÃÖÐµÄ CuO ÊÇ·ñÍêÈ«·´Ó¦»òFµ¼³öµÄÆøÌåÖпÉÄÜÓÐ H2 »ò»Ø´ð³öÉÏÊö 2 µã(ºÏÀí´ð°¸¼´¿ÉµÃ·Ö ) ÔÚ»ðÑæÉÏ·½ÕÖÒ»¸öÄÚ±Ú¸½ÓгÎÇåʯ»ÒË®µÄÉÕ±(ºÏÀí´ð°¸¼´¿ÉµÃ·Ö) ÉÕ±ÄÚ±Ú³ÎÇåʯ»ÒË®±ä»ë×Ç(»ò¸½ÓÐÒ»²ã°×É«ÎïÖÊ)
¡¾½âÎö¡¿
ÊÔÌâ·ÖÎö£º£¨1£©ÊµÑ鿪ʼºó£¬A ÖÐÎÞÃ÷ÏÔÏÖÏó£¬ËµÃ÷úÆøÖв»´æÔÚ¶þÑõ»¯Ì¼£»£¨2£© ÈôúÆøÖÐÓÐ CO ´æÔÚ£¬¸ù¾ÝÒ»Ñõ»¯Ì¼µÄÐÔÖʾßÓл¹ÔÐÔ¼°×ÊÁÏ¿ÉÖª£¬ BÖвúÉúµÄÏÖÏóÊÇÈÜÒºÖвúÉú»ÒÉ«³Áµí£»£¨3£© д³ö C Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽNaOH+ HCl NaCl +H2O »ò 2NaOH+ CO2 Na2CO3 +H2O £¬D ÖÐŨÁòËáµÄ×÷ÓÃÊǸÉÔïÆøÌå»ò³ýÈ¥ÆøÌåÖеÄË®·Ö £»£¨4£© Èô F ÖÐÎÞË®CuSO4±äÀ¶É«£¬ËµÃ÷ÓÐË®Éú³É£¬ÔòEÖÐÒ»¶¨·¢ÉúµÄ»¯Ñ§·½³ÌʽΪH2 +CuOCu +H2O£»£¨5£©µãȼF×°Öõ¼³öµÄÆøÌ壬ÓлðÑæ²úÉú¡£½·ãÈÏΪúÆøÖк¬ CH4£¬ÀíÓÉÊÇCH4¿ÉÒÔȼÉÕ£¬CH4 ȼÉյĻ¯Ñ§·½³ÌʽÊÇCH4 +2O2 CO2 +2H2O£»²»ÄÜÈ·ÈÏúÆøÖк¬ÓÐ CH4£¬ÀíÓÉÊÇH2ÓëE×°ÖÃÖÐµÄ CuO ÊÇ·ñÍêÈ«·´Ó¦»òFµ¼³öµÄÆøÌåÖпÉÄÜÓÐ H2ΪÁËÈ·¶¨ÃºÆøÖÐÊÇ·ñÓÐCH4£¬ÀÏʦ½Ó×ŲÉÈ¡µÄ²Ù×÷·½·¨ÔÚ»ðÑæÉÏ·½ÕÖÒ»¸öÄÚ±Ú¸½ÓгÎÇåʯ»ÒË®µÄÉÕ±£»¹Û²ìµ½µÄÏÖÏóÊÇÉÕ±ÄÚ±Ú³ÎÇåʯ»ÒË®±ä»ë×Ç£¬ÓÉ´ËÑé֤úÆøÖÐÒ»¶¨º¬ÓÐCH4¡£
¡¾ÌâÄ¿¡¿£¨2017ÇÎ÷ÄÏ£©ÏÖÓÐÁ½Æ¿ÎÞÉ«ÇÒʧȥ±êÇ©µÄ̼ËáÄÆÈÜÒººÍÇâÑõ»¯ÄÆÈÜÒº£¬ÈçºÎ¼ø±ð£¬Æä·½°¸ºÜ¶à¡£
£¨Ìá³öÎÊÌ⣩½öÌṩÇâÑõ»¯¸ÆÈÜÒººÍÏ¡ÑÎËá¡£ÄÜ·ñ½«Ì¼ËáÄÆÈÜÒººÍÇâÑõ»¯ÄÆÈÜÒºÇø±ð¿ªÀ´£¿
£¨²ÂÏë¼ÙÉ裩½«Ê§È¥±êÇ©µÄÁ½ÖÖÈÜÒº·Ö±ð±àºÅΪA¡¢B£¬¼ÙÉèAΪ̼ËáÄÆÈÜÒº£¬BΪÇâÑõ»¯ÄÆÈÜÒº¡£
£¨Éè¼Æ˼·£©ÏÖÓмס¢ÒÒÁ½Î»Í¬Ñ§·Ö±ð˼¿¼µÃ³öÈçϳõ²½¼ø±ð˼·£º
¼×£º½«Á½ÖÖÈÜҺȡÑù£¬·Ö±ðµÎÈëÏ¡ÑÎËᣬ¸ù¾ÝÊÇ·ñÓÐÆøÅݲúÉúÀ´Åжϸ÷ÊÇÄÄÖÖÈÜÒº¡£
ÒÒ£º½«Á½ÖÖÈÜҺȡÑù£¬·Ö±ðµÎÈëÇâÑõ»¯¸ÆÈÜÒº£¬Õñµ´£¬¸ù¾ÝÊÇ·ñÓа×É«³ÁµíÉú³ÉÀ´Åжϸ÷ÊÇÄÄÖÖÈÜÒº¡£
£¨1£©ÒÔÉÏÁ½ÖÖ˼·ÊÇ·ñ¶¼ÕýÈ·______£¨Ìî¡°¶¼ÕýÈ·¡±»ò¡°Ò»¸öÕýÈ·Ò»¸ö²»ÕýÈ·¡±£©¡£
£¨ÊµÑé̽¾¿£©Çë²ÎÓëÏÂÁÐʵÑé¹ý³Ì£º
£¨2£©ÇëÔÚÉÏÊöʵÑé˼·ÖÐÑ¡ÔñÒ»¸öÄãÈÏΪÕýÈ·µÄ˼·___£¨Ìî¡°¼×¡±»ò¡°ÒÒ¡±£©½øÐÐʵÑé¡£
£¨3£©Ìîд³öʵÑ鱨¸æ£º
ʵÑé²½Öè | ʵÑéÏÖÏó | ½áÂÛ£¬»¯Ñ§·´Ó¦·½³Ìʽ |
¢Ù½«A¡¢BÁ½ÈÜÒº¸÷È¡ÉÙÁ¿£¬·Ö±ð·ÅÈëÁ½Ö§ÊÔ¹ÜÖУ»¢Ú_ | AÈÜÒºÑùÆ·ÖÐÓÐ____ | ÔòAΪ___ÈÜÒº£¬BΪ__ÈÜÒº£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____¡£ |
£¨·¢É¢Ë¼Î¬£©¼ø±ð̼ËáÄÆÈÜÒººÍÇâÑõ»¯ÄÆÈÜÒº£¬»¹¿ÉÒÔ´ÓÏÂÁÐÊÔ¼ÁÖÐÑ¡È¡¡£ÇëÄãÑ¡ÔñÒ»ÖÖÄܼø±ðÕâÁ½ÖÖÈÜÒºµÄÊÔ¼Á____£¨ÇëÌîд±àºÅ£©¡£
A KOH B Na2SO4 C BaCl2 D NaCl
¡¾ÌâÄ¿¡¿ÏÂͼÊÇij¸ÆƬµÄ˵Ã÷Ê飬ÇëÔĶÁ½â´ðÏÂÁÐÎÊÌâ¡£
X X ¸Æ Ƭ Ö÷Òª³É·Ö£ºÆÏÌÑÌÇËá¸ÆC6H11O72Ca Ò©Æ·¹æ¸ñ£º2.5gÿƬº¬C6H11O72Ca 0.2g Óà Á¿£ºÃ¿ÈÕ2´Î£¬Ã¿´ÎһƬ |
¢ÙÆÏÌÑÌÇËá¸ÆÖк¬ÓÐ_________ÖÖÔªËØ¡£
¢ÚÆÏÌÑÌÇËá¸ÆÖÐ̼Óë¸ÆµÄÖÊÁ¿±ÈÊÇ_______£»
¢Û°´ËµÃ÷Êé·þÓô˸ÆƬ£¬Ã¿Ìì²¹³ä¸ÆµÄΪ£º ¡££¨Ö»Ð´Ëãʽ£¬²»Ëã½á¹û£©¡£
ÓÑÇé¸æÖª£ºC6H11O72Ca µÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª 430
£¨2£©ÏÖÓзÏÁòËá 4.9t¡£·ÏÁòËáÖÐ H2SO4 µÄÖÊÁ¿·ÖÊýΪ20£¥Óë×ãÁ¿µÄ·ÏÌúм·´Ó¦£¬¿ÉÉú²úFeSO4µÄÖÊÁ¿ÊǶàÉÙ£¿
¡¾ÌâÄ¿¡¿ÏÂͼÖУ¬ËÄÔ²¼×¡¢ÒÒ¡¢±û¡¢¶¡·Ö±ð±íʾһÖÖÈÜÒº£¬Á½Ô²µÄÏཻ²¿·ÖΪÁ½ÈÜÒº»ìºÏºó³öÏÖµÄÖ÷ҪʵÑéÏÖÏó£¬Ï±íÖв»·ûºÏͼʾ¹ØϵµÄÊÇ
¼× | ÒÒ | ±û | ¶¡ | |
A | Na2CO3 | H2SO4 | BaCl2 | ×ÏɫʯÈï |
B | (NH4)2SO4 | NaOH | Ba(NO3)2 | ÎÞÉ«·Ó̪ |
C | K2CO3 | HCl | Ba(OH)2 | ×ÏɫʯÈï |
D | HCl | Na2CO3 | AgNO3 | ÎÞÉ«·Ó̪ |