ÌâÄ¿ÄÚÈÝ

ijζÈÏ£¬ÔÚÒ»ºãÈÝÈÝÆ÷ÖнøÐÐÈçÏ·´Ó¦ N2+3H22NH3£¬ÏÂÁÐÇé¿öÒ»¶¨ÄÜ˵Ã÷·´Ó¦ ÒѴﵽƽºâµÄÊÇ£¨ £©

¢ÙÈÝÆ÷ÄÚѹǿ²»Ëæʱ¼ä¶ø±ä»¯

¢Úµ¥Î»Ê±¼äÄÚ£¬ÓÐ 3molH2 ·´Ó¦£¬Í¬Ê±ÓÐ 2molNH3 Éú³É

¢ÛÆøÌåµÄÃܶȲ»Ëæʱ¼ä¶ø±ä»¯

¢Üµ¥Î»Ê±¼äÄÚ£¬ÓÐ 1molN2 Éú³É£¬Í¬Ê±ÓÐ 2molNH3 Éú³É

¢ÝÓà N2¡¢ H2¡¢NH3 ±íʾµÄ¸Ã·´Ó¦µÄ»¯Ñ§·´Ó¦ËÙÂÊÖ®±ÈΪ 1¡Ã3¡Ã2

¢ÞÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿²»Ëæʱ¼ä¶ø±ä»¯

A£®¢Ù¢Ü¢Þ B£®¢Ù¢Ú¢Û C£® ¢Ú¢Û¢Ý D£®¢Ù¢Ú¢Þ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ijʵÑéС×éÄâÓÃËá¼îÖк͵ζ¨·¨²â¶¨Ê³´×µÄ×ÜËáÁ¿£¨g/100mL£©£¬ÏÖÑûÇëÄã²ÎÓ뱾ʵÑé²¢»Ø´ðÏà¹ØÎÊÌâ¡££¨ÓйØʵÑéҩƷΪ£ºÊÐÊÛʳÓð״×ÑùÆ·500mL¡¢0.1000mol/LNaOH±ê×¼ÈÜÒº¡¢ÕôÁóË®¡¢0.1%¼×»ù³ÈÈÜÒº¡¢0.1%·Ó̪ÈÜÒº¡¢0.1%ʯÈïÈÜÒº£©¡£

ʵÑé²½Ö裺

A£®Óõζ¨¹ÜÁ¿È¡10mLÊÐÊÛ°×´×ÑùÆ·£¬ÖÃÓÚ100mLÈÝÁ¿Æ¿ÖУ¬¼ÓÕôÁóË®£¨Öó·Ð³ýÈ¥CO2²¢Ñ¸ËÙÀäÈ´£©Ï¡ÊÍÖÁ¿Ì¶ÈÏߣ¬Ò¡Ôȼ´µÃ´ý²âʳ´×ÈÜÒº¡£

B£®ÓÃËáʽµÎ¶¨¹ÜÈ¡´ý²âʳ´×ÈÜÒº20.00mLÓÚ׶ÐÎÆ¿ÖУ¬²¢µÎ¼Ó·Óָ̪ʾ¼Á´ýÓá£

C£®¼îʽµÎ¶¨¹ÜÊ¢×°±ê×¼NaOHÈÜÒº£¬¾²Öú󣬶ÁÈ¡Êý¾Ý£¬¼Ç¼ΪNaOH±ê×¼ÈÜÒºÌå»ýµÄ³õ¶ÁÊý¡£

D£®µÎ¶¨£¬²¢¼Ç¼NaOHµÄÖÕ¶ÁÊý¡£Öظ´µÎ¶¨2-3´Î¡£

E£®ÊµÑéÊý¾Ý¼Ç¼

1

2

3

4

V£¨ÑùÆ·£©/mL

20.00

20.00

20.00

20.00

V£¨NaOH£©/mL£¨³õ¶ÁÊý£©

0.00

0.200

0.10

0.00

V£¨NaOH£©/mL£¨ÖÕ¶ÁÊý£©

14.98

15.20

15.12

15.95

V£¨NaOH£©/mL£¨ÏûºÄ£©

14.98

15.00

15.02

15.95

£¨1£©Ôòc£¨ÑùÆ·£©/moL•L-1=_________£»ÑùÆ·×ÜËáÁ¿=_________g/100mL¡£

£¨2£©ÅжÏD²½Öè²Ù×÷ʱÈÜÒºµ½´ïµÎ¶¨ÖÕµãµÄ·½·¨ÊÇ________________¡£

£¨3£©ÈôÉÏÊöB²½Öè²Ù×÷֮ǰ£¬ÏÈÓôý²âÒºÈóϴ׶ÐÎÆ¿£¬Ôò¶ÔµÎ¶¨½á¹û²úÉúµÄÓ°ÏìÊÇ_________(Ìî¡°ÎÞÓ°Ï족¡¢¡°Æ«´ó¡±»ò¡°Æ«Ð¡¡±£¬ÏÂͬ)£»ÈôÔÚD²½ÖèÓüîʽµÎ¶¨¹Üʱ¿ªÊ¼Ã»ÓÐÆøÅÝ£¬ºó²úÉúÆøÅÝ£¬Ôò¶ÔµÎ¶¨½á¹û²úÉúµÄÓ°ÏìÊÇ___________£»ÈôD²½ÖèµÎ¶¨Ç°Æ½ÊÓ¶ÁÊý£¬µÎ¶¨ÖÕµãʱ¸©ÊÓ¶ÁÊý£¬Ôò¶ÔµÎ¶¨½á¹û²úÉúµÄÓ°ÏìÊÇ__________¡£

úÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬ÈËÃǽ«ÀûÓÃúÖÆÈ¡µÄˮúÆø¡¢½¹Ì¿¡¢¼×Ãѵȹ㷺ÓÃÓÚ¹¤Å©ÒµÉú²úÖС£(1)ÒÑÖª:

¢ÙC(s) £« H2O(g) = CO(g)£«H2(g) ¦¤H£½£«131.3 kJ¡¤mol£­1

¢ÚCO2(g) £« H2(g) = CO(g) £« H2O(g) ¦¤H£½+41.3 kJ¡¤mol£­1

Ôò̼ÓëË®ÕôÆø·´Ó¦Éú³É¶þÑõ»¯Ì¼ºÍÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽΪ

¸Ã·´Ó¦ÔÚ (Ìî¡°¸ßΡ±¡¢¡°µÍΡ±»ò¡°ÈκÎζȡ±)ÏÂÓÐÀûÓÚÕýÏò×Ô·¢½øÐС£

(2)ÓÐÈËÀûÓÃÌ¿»¹Ô­·¨´¦ÀíµªÑõ»¯Î·¢Éú·´Ó¦C(s)+2NO(g)N2(g)+CO2(g)¡£ÏòijÃܱÕÈÝÆ÷ÖмÓÈëÒ»¶¨Á¿µÄ»îÐÔÌ¿ºÍNO£¬ÔÚT1¡æʱ£¬²»Í¬Ê±¼ä²âµÃ¸÷ÎïÖʵÄŨ¶ÈÈçϱíËùʾ£º

ʱ¼ä(min)

Ũ¶È(mol¡¤L-1)

0

10

20

30

40

50

(3)Ñо¿±íÃ÷£º·´Ó¦CO(g)+H2O(g)H2(g)+CO2(g)ƽºâ³£ÊýËæζȵı仯ÈçϱíËùʾ£º

ζÈ/¡æ

400

500

800

ƽºâ³£ÊýK

9.94

9

1

Èô·´Ó¦ÔÚ500¡æʱ½øÐУ¬ÉèÆðʼʱCOºÍH2OµÄŨ¶È¾ùΪ0.020 mol¡¤L-1£¬ÔÚ¸ÃÌõ¼þÏ´ﵽƽºâʱ£¬COµÄת»¯ÂÊΪ

(4)ÓÃCO×öȼÁϵç³Øµç½âCuSO4ÈÜÒº¡¢FeCl3ºÍFeCl2»ìºÏÒºµÄʾÒâͼÈçͼ1Ëùʾ£¬ÆäÖÐA¡¢B¡¢D¾ùΪʯīµç¼«£¬CΪͭµç¼«¡£¹¤×÷Ò»¶Îʱ¼äºó£¬¶Ï¿ªK£¬´ËʱA¡¢BÁ½¼«ÉϲúÉúµÄÆøÌåÌå»ýÏàͬ¡£

ͼ1ÖÐA¼«²úÉúµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ

¢Úµç¼«ÎªC¡¢DµÄ×°ÖÃÈÜÒºÖнðÊôÑôÀë×ÓµÄÎïÖʵÄÁ¿ÓëתÒƵç×ÓµÄÎïÖʵÄÁ¿±ä»¯¹ØϵÈçͼ2Ëùʾ£¬ÔòͼÖТÛÏß±íʾµÄÊÇ (ÌîÀë×Ó·ûºÅ)µÄ±ä»¯£»·´Ó¦½áÊøºó£¬ÒªÊ¹±û×°ÖÃÖнðÊôÑôÀë×ÓÇ¡ºÃÍêÈ«³Áµí£¬ÐèÒª mL 5.0 mol¡¤L£­1NaOHÈÜÒº¡£

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

¾«Ó¢¼Ò½ÌÍø