ÈÜÒº³£ÓÃÓÚ¸¯Ê´Ó¡Ë¢µç·°å,·´Ó¦µÄÀë×Ó·½³ÌʽΪ                    ¡£

(2)Ïòº¬ÓÐ1 mol ¡¢2 mol µÄ¸¯Ê´Í­°åºóµÄ»ìºÏÒºÖмÓÈëÌú·Ûn mol,ÊÔ·ÖÎöÈÜÒºÖÐÑôÀë×ӵĸ÷ÖÖÇé¿öÌîÈëϱí:

nµÄÈ¡Öµ·¶Î§

ÈÜÒºÖÐÑôÀë×Ó

n<1

n<2

n

 (3)¸¯Ê´Í­°åºóµÄ»ìºÏÒºÖÐ,Èô¡¢ºÍµÄŨ¶È¾ùΪ0.10 Çë²ÎÕÕϱí¸ø³öµÄÊý¾ÝºÍÒ©Æ·,¼òÊö³ýÈ¥ÈÜÒºÖк͵ÄʵÑé²½Öè:                              ¡£

ÇâÑõ»¯Î↑ʼ³ÁµíʱµÄpH

ÇâÑõ»¯Îï³ÁµíÍêȫʱµÄpH

Fe3+

1.9

3.2

Fe2+

7.0

9.0

Cu2+

4.7

6.7

ÌṩµÄÒ©Æ·: ¡¢Å¨ÁòËá¡¢NaOHÈÜÒº¡¢CuO¡¢Cu

===

(2)

n<1

¡¢¡¢

n<2

¡¢

n

 

(3)¢ÙͨÈë×ãÁ¿ÂÈÆø½«Ñõ»¯³É;¢Ú¼ÓÈëCuOµ÷½ÚÈÜÒºpHÖÁ3.2 ~4.7;¢Û¹ýÂË¡²³ýÈ¥

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄêÖØÇìÊÐÎåÇø¸ßÈýµÚÒ»´ÎѧÉúѧҵµ÷Ñгé²âÀí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

ÌúÔªËؼ°Æ仯ºÏÎïÓëÈËÀàµÄÉú²úÉú»îϢϢÏà¹Ø£¬ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©µç×Ó¹¤Òµ³£ÓÃ30£¥µÄFeCl3ÈÜÒº¸¯Ê´·óÔÚ¾øÔµ°åÉϵÄÍ­²­£¬ÖÆÔìÓ¡Ë¢µç·°å£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ                        ¡£

£¨2£©ÒÑÖª£ºFe(s)+O2(g)FeO(s)  ¡÷H=£­272 kJ¡¤mol£­1

C(s)+O2(g)CO2(g)    ¡÷H=£­393.5 kJ¡¤mol£­1

2C(s)+O2(g)2CO(g)   ¡÷H=£­221 kJ¡¤mol£­1

Ôò¸ß¯Á¶Ìú¹ý³ÌÖР  FeO(s)+CO(g)Fe(s)+CO2(g)    ¡÷H=               ¡£

£¨3£©Ìúºì£¨Fe2O3£©ÊÇÒ»ÖÖºìÉ«ÑÕÁÏ¡£½«Ò»¶¨Á¿µÄÌúºìÈÜÓÚ160mL 5 mol¡¤L£­1ÑÎËáÖУ¬ÔÙ¼ÓÈë×ãÁ¿Ìú·Û£¬´ý·´Ó¦½áÊø¹²ÊÕ¼¯µ½ÆøÌå2.24L£¨±ê×¼×´¿ö£©£¬¾­¼ì²âÈÜÒºÖÐÎÞFe3£«£¬Ôò²Î¼Ó·´Ó¦µÄÌú·ÛµÄÖÊÁ¿Îª          ¡£

£¨4£©ÒÔH2¡¢O2¡¢ÈÛÈÚÑÎNa2CO3×é³ÉȼÁϵç³Ø£¬²ÉÓõç½â·¨ÖƱ¸Fe(OH)2£¬×°ÖÃÈçÏÂͼËùʾ£¬ÆäÖÐP¶ËͨÈëCO2¡£

¢ÙʯīIµç¼«Éϵĵ缫·´Ó¦Ê½Îª                          ¡£

¢ÚͨµçÒ»¶Îʱ¼äºó£¬ÓҲಣÁ§¹ÜÖвúÉú´óÁ¿µÄ°×É«³Áµí£¬Çҽϳ¤Ê±¼ä²»±äÉ«¡£ÔòÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ   £¨ÌîÐòºÅ£©¡£

A£®X¡¢YÁ½¶Ë¶¼±ØÐëÓÃÌú×÷µç¼«

B£®¿ÉÒÔÓÃNaOHÈÜÒº×÷Ϊµç½âÒº

C£®Òõ¼«·¢ÉúµÄ·´Ó¦ÊÇ£º2H2O£« 2e£­= H2¡ü+ 2OH£­

D£®°×É«³ÁµíÖ»ÄÜÔÚÑô¼«ÉϲúÉú

¢ÛÈô½«ËùµÃFe(OH)2³Áµí±©Â¶ÔÚ¿ÕÆøÖУ¬ÆäÑÕÉ«±ä»¯Îª                                        £¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                      ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012ÄêËս̰æ¸ßÖл¯Ñ§Ñ¡ÐÞ2 4.1 ²ÄÁϵļӹ¤´¦ÀíÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

½ðÊô¼°Æ仯ºÏÎïµÄÓ¦Óúܹ㷺£¬ÏÖ¾Ù¼¸ÖÖ³£¼û½ðÊô¼°Æ仯ºÏÎïµÄÓÃ;£¬»Ø´ðÓйØÎÊÌ⣺

(1)FeCl3ÈÜÒº³£ÓÃÓÚ¸¯Ê´Ó¡Ë¢µç·ͭ°å£¬ÆäÀë×Ó·½³ÌʽΪ£º_______________________________¡£

(2)»ÆÍ­¿óÈÛÁ¶ºóµÃµ½µÄ´ÖÍ­Öк¬ÉÙÁ¿Fe¡¢Ag¡¢AuµÈ½ðÊôÔÓÖÊ£¬Ðè½øÒ»²½²ÉÓõç½â·¨¾«ÖÆ¡£Çë¼òÊö´ÖÍ­µç½âµÃµ½¾«Í­µÄÔ­Àí£º_________________________________________¡£

(3)ij¿ÆÑÐÈËÔ±·¢ÏÖÁÓÖʲ»Ðâ¸ÖÔÚËáÖи¯Ê´»ºÂý£¬µ«ÔÚijЩÑÎÈÜÒºÖи¯Ê´ÏÖÏóÃ÷ÏÔ¡£Çë´ÓϱíÌṩµÄÒ©Æ·ÖÐÑ¡ÔñÁ½ÖÖ(Ë®¿ÉÈÎÑ¡)£¬Éè¼Æ×î¼ÑʵÑ飬ÑéÖ¤ÁÓÖʲ»Ðâ¸ÖÒ×±»¸¯Ê´¡£

ŨÁòËᡢϡÁòËá¡¢CuO¡¢NaOHÈÜÒº

Óйط´Ó¦µÄ»¯Ñ§·½³Ìʽ____________________________________________________¡£

ÁÓÖʲ»Ðâ¸Ö¸¯Ê´µÄʵÑéÏÖÏó________________________________________________¡£

(4)ºÏ½ðÔÚÉú»îºÍÉú²úÉÏÓÃ;ԽÀ´Ô½¹ã·º¡£ÇëÓÃÏà¹ØµÄ֪ʶ½âÊÍÔÚº½¿ÕÒµÉÏÂÁºÏ½ð±ÈÌúºÏ½ðÓÃ;¸ü¹ã·ºµÄÔ­Òò¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

½ðÊô¼°Æ仯ºÏÎïµÄÓ¦Óúܹ㷺£¬ÏÖ¾Ù¼¸ÖÖ³£¼û½ðÊô¼°Æ仯ºÏÎïµÄÓÃ;£¬»Ø´ðÓйØÎÊÌ⣺
(1)FeCl3ÈÜÒº³£ÓÃÓÚ¸¯Ê´Ó¡Ë¢µç·ͭ°å£¬ÆäÀë×Ó·½³ÌʽΪ£º________¡£
(2)»ÆÍ­¿óÈÛÁ¶ºóµÃµ½µÄ´ÖÍ­Öк¬ÉÙÁ¿Fe¡¢Ag¡¢AuµÈ½ðÊôÔÓÖÊ£¬Ðè½øÒ»²½²ÉÓõç½â·¨¾«ÖÆ¡£Çë¼òÊö´ÖÍ­µç½âµÃµ½¾«Í­µÄÔ­Àí£º________¡£
(3)ij¿ÆÑÐÈËÔ±·¢ÏÖÁÓÖʲ»Ðâ¸ÖÔÚËáÖи¯Ê´»ºÂý£¬µ«ÔÚijЩÑÎÈÜÒºÖи¯Ê´ÏÖÏóÃ÷ÏÔ¡£Çë´ÓϱíÌṩµÄÒ©Æ·ÖÐÑ¡ÔñÁ½ÖÖ(Ë®¿ÉÈÎÑ¡)£¬Éè¼Æ×î¼ÑʵÑ飬ÑéÖ¤ÁÓÖʲ»Ðâ¸ÖÒ×±»¸¯Ê´¡£
ŨÁòËᡢϡÁòËá¡¢CuO¡¢NaOHÈÜÒº
Óйط´Ó¦µÄ»¯Ñ§·½³Ìʽ________¡£
ÁÓÖʲ»Ðâ¸Ö¸¯Ê´µÄʵÑéÏÖÏó________¡£
(4)ºÏ½ðÔÚÉú»îºÍÉú²úÉÏÓÃ;ԽÀ´Ô½¹ã·º¡£ÇëÓÃÏà¹ØµÄ֪ʶ½âÊÍÔÚº½¿ÕÒµÉÏÂÁºÏ½ð±ÈÌúºÏ½ðÓÃ;¸ü¹ã·ºµÄÔ­Òò¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸