ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÏÂÁÐÕýÈ·µÄ˵·¨Óм¸¸ö£¨ £©

¢ÙζÈÒ»¶¨£¬Ñ¹ËõÈÝÆ÷µÄÈÝ»ýÔö´óѹǿ£¬¿ÉÌá¸ßµ¥Î»Ìå»ýÄڵĻ·Ö×ÓÊýÄ¿ºÍ»î»¯·Ö×Ó°Ù·ÖÊý¡£

¢Ú·²ÊÇÄÜ×Ô·¢½øÐеķ´Ó¦¾ÍÒ»¶¨ÈÝÒ×½øÐУ¬·Ç×Ô·¢·´Ó¦¾Í²»ÄܽøÐС£

¢ÛÄÜ·¢ÉúÓÐЧÅöײµÄ·Ö×ÓÒ»¶¨Êǻ·Ö×Ó¡£

¢Ü½ðÊôµÄ¸¯Ê´·ÖΪ»¯Ñ§¸¯Ê´ºÍµç»¯Ñ§¸¯Ê´£¬ÒøÆ÷ÔÚ¿ÕÆøÖоÃÖñäºÚÊôÓڵ绯ѧ¸¯Ê´¡£

¢ÝÌúµÄµç»¯Ñ§¸¯Ê´·ÖΪÎöÇⸯʴºÍÎüÑõ¸¯Ê´£¬ÆäÖÐÎüÑõ¸¯Ê´¸üΪÆÕ±é¡£

¢Þ¶ÆÐ¿ÌúºÍ¶ÆÎýÌú¶Æ²ãÆÆ»µºó£¬¶ÔÄÚ²¿½ðÊôÈÔÈ»¾ùÓб£»¤×÷Óá£

¢ß¼×ÍéµÄ±ê׼ȼÉÕÈÈΪ890.3kJ¡¤mol£­1£¬Ôò¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ¿É±íʾΪCH4(g)£«2O2(g)=CO2(g)£«2H2O(g)¦¤H£½£­890.3kJ¡¤mol£­1

¢à±íʾÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ£ºNaOH£«HCl=NaCl£«H2O¦¤H£½£­57.3kJ¡¤mol£­1

A.1¸öB.2¸öC.3¸öD.4¸ö

¡¾´ð°¸¡¿B

¡¾½âÎö¡¿

¢ÙζÈÒ»¶¨£¬»î»¯·Ö×Ó°Ù·ÖÊý²»±ä£¬¹Ê´íÎó£»

¢ÚÄÜ×Ô·¢½øÐеķ´Ó¦²»Ò»¶¨ÈÝÒ×½øÐУ¬ÓеÄÒ²ÐèÒªÒ»¶¨µÄÌõ¼þ£¬¹Ê´íÎó£»

¢ÛÄÜ·¢ÉúÓÐЧÅöײµÄ·Ö×ÓÒ»¶¨Êǻ·Ö×ÓÊÇÕýÈ·µÄ£»

¢ÜÒøÆ÷ÔÚ¿ÕÆøÖоÃÖñäºÚÊôÓÚ»¯Ñ§¸¯Ê´£¬¹Ê´íÎó£»

¢ÝÌúµÄµç»¯Ñ§¸¯Ê´·ÖΪÎöÇⸯʴºÍÎüÑõ¸¯Ê´£¬ÆäÖÐÎüÑõ¸¯Ê´¸üΪÆÕ±é£¬ÕýÈ·£»

¢Þ¶ÆÐ¿ÌúºÍ¶ÆÎýÌú¶Æ²ãÆÆ»µºó£¬¶ÆÎýÌú¶ÔÄÚ²¿½ðÊôûÓб£»¤×÷Ó㬹ʴíÎó£»

¢ß±íʾȼÉÕÈȵĻ¯Ñ§·´Ó¦±ØÐëÉú³ÉÎȶ¨µÄÑõ»¯ÎˮµÄ״̬Ӧ¸ÃÊÇÒºÌ壬¹Ê´íÎó£»

¢àÈÈ»¯Ñ§·½³ÌʽÐè×¢Ã÷·´Ó¦ÎïÉú³ÉÎïµÄ״̬£¬¹Ê´íÎó£»

¹ÊÕýÈ·µÄÓТۢݣ¬¹ÊÑ¡B;

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø