ÌâÄ¿ÄÚÈÝ

±±¾©Ê±¼ä£¹ÔÂ25ÈÕ21ʱ10·Ö04Ã룬ÎÒ¹úº½ÌìÊÂÒµÓÖÓ­À´Ò»¸öÀúÊ·ÐÔʱ¿Ì£¬ÎÒ¹ú×ÔÐÐÑÐÖƵÄÉñÖÛÆߺÅÔØÈË·É´¬ÔÚ¾ÆȪÎÀÐÇ·¢ÉäÖÐÐÄ·¢ÉäÉý¿Õ£¬ÓÚ28ÈÕÏÂÎç»Øµ½ÃÀÀöµÄµØÇò¡£¡°ÉñÖÛ¡±ÆߺŻ㼯ÁË´óÁ¿×îпƼ¼£¬·¢Éä»ð¼ýµÄ³£¹æÍƽø¼ÁΪËÄÑõ»¯µªºÍÆ«¶þ¼×루C2H8N2£©£¬5.00g C2H8N2ÍêȫȼÉտɷųö212.5kJÈÈÁ¿¡£ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨    £©
A£®È¼ÉÕ²»Ò»¶¨ÓÐÑõÆø²Î¼Ó£¬Ò²²»Ò»¶¨ÊÇ·ÅÈÈ·´Ó¦
B£®»ð¼ýµã»ðºó£¬Åç³öµÄºìÉ«»ðÑæÊǽðÊôµÄÑæÉ«·´Ó¦²úÉúµÄ
C£®»ð¼ýȼÁÏȼÉÕÖ÷ÒªÊǽ«»¯Ñ§ÄÜת±äΪÈÈÄܺ͹âÄÜ£¬¿ÉÄܶԻ·¾³²úÉúÎÛȾ
D£®Æ«¶þ¼×ëÂȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽÊÇ£ºC2H8N2(g)+2N2O4(g)=2N2(g)+2CO2(g)+4H2O(g)£»¡÷H=-2550kJ/mol
C
ȼÉÕ·´Ó¦ÊÇ·¢¹â·¢ÈȵķÅÈÈ·´Ó¦£¬µ«²»Ò»¶¨ÓÐÑõÆø²Î¼Ó£¬ÈçÄÆ¡¢ÇâÆøµÈÔÚÂÈÆøÖÐȼÉÕ£¬AÏî´íÎó£»»ð¼ýµã»ðºóÅç³öµÄºìÉ«»ðÑæÊÇËÄÑõ»¯µª·Ö½âÉú³ÉµÄ¶þÑõ»¯µª£¬BÏî´í£»»ð¼ýȼÁÏȼÉÕºó°Ñ»¯Ñ§ÄÜת»¯ÎªÈÈÄܺ͹âÄÜ£¬ÓÉÓÚȼÉÕºó²úÎïÖлìÓÐÉÙÁ¿¶þÑõ»¯µª£¬»á¶Ô»·¾³²úÉúÒ»¶¨µÄÎÛȾ£¬CÏîÕýÈ·£»¾ÝÌâÒâÖªC2H8N2ӦΪҺ̬²»ÎªÆø̬£¬¼´ÈÈ»¯Ñ§·½³ÌʽÖÐӦΪC2H8N2£¨l£©£¬DÏî´íÎó¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨8·Ö£©·Ï¾ÉÓ¡Ë¢µç·°åµÄ»ØÊÕÀûÓÿÉʵÏÖ×ÊÔ´ÔÙÉú£¬²¢¼õÉÙÎÛȾ¡£·Ï¾ÉÓ¡Ë¢µç·°å¾­·ÛËé·ÖÀ룬Äܵõ½·Ç½ðÊô·ÛÄ©ºÍ½ðÊô·ÛÄ©¡£
£¨1£©ÏÂÁд¦ÀíÓ¡Ë¢µç·°å·Ç½ðÊô·ÛÄ©µÄ·½·¨ÖУ¬²»·ûºÏ»·¾³±£»¤ÀíÄîµÄÊÇ   £¨Ìî×Öĸ£©
A£®ÈÈÁѽâÐγÉȼÓÍB£®Â¶Ìì·ÙÉÕ
C£®×÷ΪÓлú¸´ºÏ½¨Öþ²ÄÁϵÄÔ­ÁÏD£®Ö±½ÓÌîÂñ
£¨2£©ÓÃH2O2ºÍH2SO4µÄ»ìºÏÈÜÒº¿ÉÈܳöÓ¡Ë¢µç·°å½ðÊô·ÛÄ©ÖеÄÍ­¡£ÒÑÖª£º
Cu(s)£«2H£«(aq)£½Cu2£«(aq)£«H2(g) ¡÷H£½64.39kJ¡¤mol£­1
2H2O2(l)£½2H2O(l)£«O2(g) ¡÷H£½£­196.46kJ¡¤mol£­1
H2(g)£«1/2O2(g)£½H2O(l)  ¡÷H£½£­285.84kJ¡¤mol£­1
ÔÚH2SO4ÈÜÒºÖÐCu ÓëH2O2·´Ó¦Éú³ÉCu2£«ºÍH2OµÄÈÈ»¯Ñ§·½³ÌʽΪ£º          ¡£
£¨3£©¿ØÖÆÆäËüÌõ¼þÏàͬ£¬Ó¡Ë¢µç·°åµÄ½ðÊô·ÛÄ©ÓÃ10%H2O2 ºÍ3.0mol¡¤L£­1
H2SO4ÈÜÒº´¦Àí£¬²âµÃ²»Í¬Î¶ÈÏÂÍ­µÄƽ¾ùÈܽâËÙÂÊ£¨¼ûÏÂ±í£©
ζȣ¨¡æ£©
20
30
40
50
60
70
80
Í­µÄƽ¾ùÈܽâËÙÂÊ
£¨¡Á10£­3mol¡¤L£­1¡¤min£­1£©
7.34
8.01
9.25
7.98
7.24
6.73
5.76
µ±Î¶ȸßÓÚ40¡æʱ£¬Í­µÄƽ¾ùÈܽâËÙÂÊËæ×Å·´Ó¦Î¶ȵÄÉý¸ß¶øϽµ£¬
ÆäÖ÷ÒªÔ­ÒòÊÇ                                          ¡£
£¨4£©ÔÚÌá´¿ºóµÄCuSO4ÈÜÒºÖмÓÈëÒ»¶¨Á¿µÄNa2SO3ºÍNaCl ÈÜÒº£¬¼ÓÈÈ£¬Éú³É
CuCl µÄÀë×Ó·½³ÌʽÊÇ                                    ¡£

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

¾«Ó¢¼Ò½ÌÍø