ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ä³»¯Ñ§ÐËȤС×éÀûÓÃÏÂͼװÖýøÐÐÒÒËáÒÒõ¥ºÏ³ÉºÍ·ÖÀëµÄʵÑé̽¾¿£¬Çë»Ø´ðÒÔÏÂÎÊÌâ

£¨1£©Ð´³öºÏ³ÉÒÒËáÒÒõ¥µÄ»¯Ñ§·½³Ìʽ_______________________________¡£

£¨2£©ÒÇÆ÷bµÄÃû³Æ________£¬Í¼ÖÐÆðÀäÄý»ØÁ÷×÷ÓõÄÊÇ______£¨Ìî£á¡¢£â¡¢£ã¡¢£ä¡¢e£©¡£

£¨3£©ÎªÁËÌá¸ßÒÒËáÒÒõ¥µÄ²úÂʿɲÉÈ¡µÄ´ëÊ© _______________________________

£¨4£©¾­¹ý0.5h¼ÓÈÈ·´Ó¦ºó£¬½«·´Ó¦×°ÖÃcÖдֲúƷתÒÆÖÁdÖнøÐÐÕôÁó¡£

ÎïÖÊ

98.3%ŨÁòËá

ÒÒËáÒÒõ¥

ÒÒËá

ÒÒ´¼

ÒÒÃÑ

Ë®

·Ðµã

338¡æ£¬

77.1¡æ

118¡æ

78.5¡æ

34.6¡æ

100¡æ

¸ù¾ÝÉϱí·ÖÎö£¬ÕôÁóºóµÃµ½µÄÒÒËáÒÒõ¥ÖУ¬×îÓпÉÄܺ¬ÓÐ________________ÔÓÖÊ¡£

¡¾´ð°¸¡¿ CH3COOH£«CH3CH2OHCH3COOCH2CH3£«H2O ÀäÄý¹Ü £á ½«CH3COOCH2CH3¼°Ê±ÕôÁó·ÖÀ룻·´Ó¦Î¶Ȳ»Ò˹ý¸ß£¬¼õÉÙCH3COOH ¡¢CH3CH2OHµÄ»Ó·¢£»·´Ó¦Îï¿ØÖÆÎÞË®Ìõ¼þ£¬Å¨ÁòËáµÄÎüË®×÷ÓÃÓÐÀûÓÚƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬Ìá¸ß²úÂÊ¡££¨Ð´³öÆäÖÐÒ»Ìõ¼´¿É£© CH3CH2OH

¡¾½âÎö¡¿£¨1£©ºÏ³ÉÒÒËáÒÒõ¥µÄ»¯Ñ§·½³ÌʽΪCH3COOH£«CH3CH2OHCH3COOCH2CH3£«H2O£»£¨2£©ÒÇÆ÷bµÄÃû³ÆÀäÄý¹Ü£¬aÊÇÖ±ÐÎÀäÄý¹Ü£¬Í¼ÖÐÆðÀäÄý»ØÁ÷×÷ÓõÄÊÇa£»£¨3£©ÓÉÓÚõ¥»¯·´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬Ìá¸ßÒÒËáÒÒõ¥µÄ²úÂʿɲÉÈ¡µÄ´ëʩΪ£º½«CH3COOCH2CH3¼°Ê±ÕôÁó·ÖÀ룻·´Ó¦Î¶Ȳ»Ò˹ý¸ß£¬¼õÉÙCH3COOH CH3CH2OHµÄ»Ó·¢£»·´Ó¦Îï¿ØÖÆÎÞË®Ìõ¼þ£¬Å¨ÁòËáµÄÎüË®×÷ÓÃÓÐÀûÓÚƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬Ìá¸ß²úÂÊ£»£¨4£©Óɱí¸ñÖеÄÊý¾Ý¿ÉÖª£¬ÒÒ´¼ÓëÒÒËáÒÒõ¥µÄ·Ðµã½Ó½ü£¬ÔòÕôÁóºóµÃµ½µÄÒÒËáÒÒõ¥ÖУ¬×îÓпÉÄܺ¬ÓÐCH3CH2OH ÔÓÖÊ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÊµÑéÊÒÒªÅäÖÆ480mL 0.2 mol¡¤L£­1 NaClÈÜÒº£¬ÊµÑéÊÒÖ»Óк¬ÓÐÉÙÁ¿ÁòËáÄƵÄÂÈ»¯ÄÆÑùÆ·£¬Éè¼ÆÈçÏ·½°¸£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢ñ£®´ÖÑÎÌá´¿

£¨1£©¹ÌÌåAµÄ³É·ÖΪ________________£¨Ìѧʽ£©¡£

£¨2£©ÊÔ¼Á2µÄ»¯Ñ§Ê½Îª_____________£¬ÅжÏÊÔ¼Á2ÊÇ·ñ¹ýÁ¿µÄ·½·¨_______________£¬²Ù×÷2µÄÃû³Æ__________________¡£

¢ò£®ÅäÖÆÈÜÒº

£¨1£©ÓÃÍÐÅÌÌìƽ³ÆÈ¡ÂÈ»¯ÄÆ£¬ÆäÖÊÁ¿Îª________g¡£

£¨2£©ÅäÖƹý³ÌÖÐÐèҪʹÓõÄʵÑéÒÇÆ÷³ýÁËÍÐÅÌÌìƽ¡¢Ò©³×¡¢Á¿Í²¡¢²£Á§°ô¡¢ÉÕ±­¡¢½ºÍ·µÎ¹ÜÍ⣬»¹ÐèÒª___________¡£

£¨3£©ÏÂÁÐÖ÷Òª²Ù×÷²½ÖèµÄÕýȷ˳ÐòÊÇ¢Ù¡ª____¡ª_____¡ª ____¡ª¢Ü£¨ÌîÐòºÅ£©¡£

¢Ù³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÂÈ»¯ÄÆ£¬·ÅÈëÉÕ±­ÖУ¬ÓÃÊÊÁ¿ÕôÁóË®Èܽâ

¢Ú¼ÓË®ÖÁÒºÃæÀëÈÝÁ¿Æ¿¾±¿Ì¶ÈÏßÏÂ1~2ÀåÃ×ʱ£¬¸ÄÓýºÍ·µÎ¹ÜµÎ¼ÓÕôÁóË®ÖÁ°¼ÒºÃæÓë¿Ì¶ÈÏßÏàÇÐ

¢Û½«ÈÜҺתÒƵ½ÈÝÁ¿Æ¿ÖÐ

¢Ü¸ÇºÃÆ¿Èû£¬·´¸´ÉÏϵߵ¹£¬Ò¡ÔÈ

¢ÝÓÃÉÙÁ¿µÄÕôÁóˮϴµÓÉÕ±­ÄڱںͲ£Á§°ô2¡«3´Î£¬Ï´µÓҺתÒƵ½ÈÝÁ¿Æ¿ÖÐ

£¨4£©ÏÂÁÐÇé¿ö¶ÔËùÅäÖƵÄNaClÈÜҺŨ¶ÈÓкÎÓ°Ï죿£¨Ñ¡Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©

¢ÙÈÝÁ¿Æ¿ÓÃÕôÁóÏ´µÓºó²ÐÁôÓÐÉÙÁ¿µÄË®________________£»

¢Ú¶¨ÈÝʱ£¬ÑöÊÓÈÝÁ¿Æ¿µÄ¿Ì¶ÈÏß________________£»

¢ÛתÒÆÈÜҺʱ£¬²£Á§°ôµÄ϶˿¿ÔÚÈÝÁ¿Æ¿µÄ¿Ì¶ÈÏßÒÔÉϵÄÄÚ±Ú________________¡£

¡¾ÌâÄ¿¡¿ÑÇÏõõ£ÂÈ(ClNO)ÊÇÓлúºÏ³ÉÖеÄÖØÒªÊÔ¼Á£¬¹¤ÒµÉÏ¿ÉÓÃNOÓëCl2ºÏ³É£¬»Ø´ðÏÂÁÐÎÊÌ⣺

(1)¡ª¶¨Ìõ¼þÏ£¬µªÑõ»¯ÎïÓëÐü¸¡ÔÚ´óÆøÖеĺ£ÑÎÁ£×ÓÏ໥×÷ÓÃʱ»áÉú³ÉÑÇÏõõ£ÂÈ£¬Éæ¼°ÈÈ»¯Ñ§·½³ÌʽºÍƽºâ³£ÊýÈçÏÂ±í£º

ÐòºÅ

ÈÈ»¯Ñ§·½³Ìʽ

ƽºâ³£Êý

¢Ù

2NO2(g) +NaCl(s) NaNO3(s) +ClNO(g) ¡÷H1

K1

¢Ú

4NO2(g) +2NaCl(s) 2NaNO3(s)+2NO(g)+Cl2(g) ¡÷H2

k2

¢Û

2NO(g)+Cl2(g) 2ClNO(g) ¡÷H3

K3

K3=_______(ÓÃK1¡¢K2±íʾ)¡£

(2)25¡æʱ£¬ÏòÌå»ýΪ2LÇÒ´øÆøѹ¼ÆµÄºãÈÝÃܱÕÈÝÆ÷ÖÐͨÈë0.08molNOºÍ0.04molCl2·¢Éú·´Ó¦£º2NO(g)+Cl2(g) 2ClNO(g) ¡÷H3¡£

¢Ù ÏÂÁÐÃèÊöÄÜ˵Ã÷¸Ã·´Ó¦ÒѴﵽƽºâ״̬µÄÊÇ_____£¨ÌîÐòºÅ£©

a.vÕý(Cl2)=2vÄæ(NO) b.ÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä

c.ÈÝÆ÷ÄÚÆøÌåѹǿ±£³Ö²»±ä d.ÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿±£³Ö²»±ä

¢ÚÈô·´Ó¦ÆðʼºÍƽºâʱζÈÏàͬ£¬²âµÃ·´Ó¦¹ý³ÌÖÐѹǿ£¨P)Ëæʱ¼ä£¨t)µÄ±ä»¯ÈçͼIÇúÏßaËùʾ£¬Ôò¡÷H3______0(Ìî¡° >¡±¡¢¡° < ¡±»ò¡°²»È·¶¨"£©£»ÈôÆäËûÌõ¼þÏàͬ£¬½ö¸Ä±äijһÌõ¼þʱ£¬²âµÃÆäѹǿ(P)Ëæʱ¼ä£¨t)µÄ±ä»¯ÈçͼIÇúÏßbËùʾ£¬„t¸Ä±äµÄÌõ¼þÊÇ_________¡£

¢ÛͼIIÊǼס¢ÒÒͬѧÃè»æÉÏÊö·´Ó¦Æ½ºâ³£ÊýµÄ¶ÔÊýÖµ£¨lgK)Óëζȵı仯¹Øϵ£¬ÆäÖÐÕýÈ·µÄÇúÏßÊÇ____(Ìî¡°¼×¡±»ò¡°ÒÒ¡±£©£»mֵΪ_______¡£

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

¾«Ó¢¼Ò½ÌÍø