ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÒÑÖª²ð¿ª1 molÇâÆøÖеĻ¯Ñ§¼üÐèÒªÏûºÄ436 kJÄÜÁ¿£¬²ð¿ª1 molÑõÆøÖеĻ¯Ñ§¼üÐèÒªÏûºÄ498 kJÄÜÁ¿£¬¸ù¾ÝÈçͼÖеÄÄÜÁ¿Í¼£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©·Ö±ðд³ö¢Ù¢ÚµÄÊýÖµ£º

¢Ù__________£»¢Ú__________¡£

£¨2£©Éú³ÉH2(Xg)ÖеÄ1 mol H¡ªO¼ü·Å³ö__________kJµÄÄÜÁ¿¡£

£¨3£©ÒÑÖª£ºH2O(l)H2O(g)¡÷H£½+44 kJ mol1£¬ÊÔд³öÇâÆøÔÚÑõÆøÖÐÍêȫȼÉÕÉú³ÉҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£º___________________________________¡£

¡¾´ð°¸¡¿£¨1£©¢Ù1370 ¢Ú1 852

£¨2£©463

£¨3£©2H2(g)+ O2(g)2H2O£¨l£© ¦¤H=482 kJ mol1

¡¾½âÎö¡¿£¨1£©¢ÙµÄÄÜÁ¿Îª¶ÏÁÑ2 mol H¡ªH¼ü¼üÄܵÄÊýÖµºÍ1 mol O=O¼ü¼üÄܵÄ×ܺͣ¬Í¬ÑùµÄµÀÀí£¬¢ÚµÄÄÜÁ¿Îª4 mol HÓë2 mol OÐγÉ2 mol H2O(g)ÊͷŵÄÄÜÁ¿£¬¼´ÐγÉ4 mol H¡ªO¼üÊͷŵÄÄÜÁ¿¡££¨2£©ÐγÉ4 mol H¡ªO¼üÊͷŵÄÄÜÁ¿Îª1 852 kJ£¬ÔòÉú³ÉH2(Xg)ÖеÄ1 mol H¡ªO¼ü·Å³ö463 kJµÄÄÜÁ¿¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÁòËáÌúï§aFe2(SO4) 3¡¤b(NH4) 2SO4¡¤cH2O]¹ã·ºÓÃÓÚ³ÇÕòÉú»îÒûÓÃË®¡¢¹¤ÒµÑ­»·Ë®µÄ¾»»¯´¦ÀíµÈ¡£Ä³»¯¹¤³§ÒÔÁòËáÑÇÌú£¨º¬ÉÙÁ¿ÏõËá¸Æ£©ºÍÁòËáï§ÎªÔ­ÁÏ£¬Éè¼ÆÁËÈçϹ¤ÒÕÁ÷³ÌÖÆÈ¡ÁòËáÌú李£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÁòËáÑÇÌúÈÜÒº¼ÓH2SO4ËữµÄÖ÷ҪĿµÄÊÇ_________________£¬ÂËÔüAµÄÖ÷Òª³É·ÖÊÇ_______________¡£

£¨2£©ÏÂÁÐÎïÖÊÖÐ×îÊʺϵÄÑõ»¯¼ÁBÊÇ________£»·´Ó¦µÄÀë×Ó·½³Ìʽ ¡£

a£®NaClO b£®H2O2 c£®KMnO4 d£®K2Cr2O7

£¨3£©²Ù×÷¼×¡¢ÒÒµÄÃû³Æ·Ö±ðÊÇ£º¼×______________£¬ÒÒ___________________¡£

£¨4£©ÉÏÊöÁ÷³ÌÖУ¬Ñõ»¯Ö®ºóºÍ¼ÓÈÈÕô·¢Ö®Ç°£¬ÐèÈ¡ÉÙÁ¿ÈÜÒº¼ìÑéFe2+ÊÇ·ñÒÑÈ«²¿±»Ñõ»¯£¬Ëù¼ÓÊÔ¼Á ÄÜ·ñÓÃËáÐÔµÄKMnO4ÈÜÒº£¿ (Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)ÀíÓÉÊÇ£º ¡££¨¿ÉÓÃÓïÑÔ»ò·½³Ìʽ˵Ã÷£©

£¨5£©¼ìÑéÁòËáÌúï§ÖÐNH4+µÄ·½·¨ÊÇ____________________________¡£

£¨6£©³ÆÈ¡14.00 gÑùÆ·£¬½«ÆäÈÜÓÚË®ÅäÖƳÉ100 mLÈÜÒº£¬·Ö³ÉÁ½µÈ·Ý£¬ÏòÆäÖÐÒ»·ÝÖмÓÈë×ãÁ¿NaOHÈÜÒº£¬¹ýÂËÏ´µÓµÃµ½2.14 g³Áµí£»ÏòÁíÒ»·ÝÈÜÒºÖмÓÈë0.05 mol Ba (NO3) 2ÈÜÒº£¬Ç¡ºÃÍêÈ«·´Ó¦¡£Ôò¸ÃÁòËáÌú淋Ļ¯Ñ§Ê½Îª______________________¡£

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

¾«Ó¢¼Ò½ÌÍø