ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÔÚºê¹Û¡¢Î¢¹ÛÓë·ûºÅÖ®¼ä½¨Á¢ÁªÏµ£¬ÊÇ»¯Ñ§Ñ§¿ÆµÄÌص㡣Ç廪´óѧÑо¿ÈËÔ±³É¹¦ÑÐÖƳöÒ»ÖÖÄÉÃ×ÏËά´ß»¯¼Á£¬¿É½«¶þÑõ»¯Ì¼×ª»¯³ÉÒºÌåȼÁϼ״¼£¬Æä΢¹ÛʾÒâͼÈçͼ(ͼÖеÄ΢Á£Ç¡ºÃÍêÈ«·´Ó¦)Ëùʾ¡£Çë¸ù¾Ý΢¹ÛʾÒâͼ»Ø´ðÒÔÏÂÎÊÌâ¡£

(1)¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____¡£

(2)ÇëÔÚ·½¿òÄÚÓÃ΢¹ÛʾÒâͼ±íʾ³ö¶¡ÎïÖÊ______¡£

(3)ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_____(Ìîд×ÖĸÐòºÅ)¡£

A ·´Ó¦Ç°ºó·Ö×ÓÊýÄ¿·¢Éú¸Ä±ä B ¸Ã·´Ó¦ÌåÏÖÁËÎÞ»úÎïÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔת»¯ÎªÓлúÎï

C ¼×Êǵ¥ÖÊ£¬ÒÒ¡¢±û¡¢¶¡¾ùΪ»¯ºÏÎï D ¸Ã·´Ó¦ÊôÓÚ¸´·Ö½â·´Ó¦

¡¾´ð°¸¡¿3H2+CO2CH3OH+H2O ABC

¡¾½âÎö¡¿

(1)ÓÉ·´Ó¦µÄ΢¹ÛʾÒâͼºÍÖÊÁ¿Êغ㶨ÂÉ¿ÉÖª£¬¶¡ÎïÖÊÊÇË®£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ

3H2+CO2CH3OH+H2O¡£

(2)ÓÉÉÏÊö·ÖÎö¿ÉÖª£¬¶¡ÎïÖÊÊÇË®£¬Î¢¹ÛʾÒâͼΪ£º

¡£

(3)A¡¢ÓÉ·½³Ìʽ¿ÉÖª£¬Ã¿4¸ö·Ö×ӱ仯³ÉÁË2¸ö·Ö×Ó£¬·´Ó¦Ç°ºó·Ö×ÓÊýÄ¿·¢Éú¸Ä±ä£¬¹ÊAÕýÈ·£»

B¡¢ÓÉÎïÖʵı仯¿ÉÖª£¬¸Ã·´Ó¦ÌåÏÖÁËÎÞ»úÎïÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔת»¯ÎªÓлúÎ¹ÊBÕýÈ·£»

C¡¢ÓÉÎïÖʵÄ×é³É¿ÉÖª£¬¼×Êǵ¥ÖÊ£¬ÒÒ¡¢±û¡¢¶¡¾ùΪ»¯ºÏÎ¹ÊCÕýÈ·£»

D¡¢¸Ã·´Ó¦Óе¥Öʲμӷ´Ó¦£¬²»ÊôÓÚ¸´·Ö½â·´Ó¦¡£¹ÊD²»ÕýÈ·¡£¹ÊÑ¡ABC¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø