ÏÂÁÐʵÑé²Ù×÷¡¢ÏÖÏóÓë½áÂÛ¶ÔÓ¦¹ØÏµÕýÈ·µÄÊÇ£¨¡¡¡¡£©
Ñ¡Ïî ʵÑé²Ù×÷ ʵÑéÏÖÏóÓë½áÂÛ
A ½«SO2ͨÈëËáÐÔKMnO4ÈÜÒºÖÐ ËáÐÔKMnO4ÈÜÒºÍÊÉ«£¬Ö¤Ã÷SO2¾ßÓÐÆ¯°×ÐÔ
B ÏòÑùÆ·ÈÜÒºÖÐÏȵμӹýÁ¿µÄÏ¡ÑÎËᣬÔٵμÓBaCl2ÈÜÒº µÎ¼ÓÏ¡ÑÎËáÎÞÏÖÏ󣬵μÓBaCl2ºó³öÏÖ°×É«³Áµí£®ËµÃ÷ÑùÆ·ÈÜÒºÖÐÒ»¶¨º¬ÓÐSO42-
C ijÈÜÒºÖмÓÈë×ãÁ¿Å¨NaOHÈÜÒº£¬¼ÓÈÈ£¬ÔÚÊԹܿڷÅһƬʪÈóµÄºìɫʯÈïÊÔÖ½ ʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬Ôò¸ÃÑùÆ·Öк¬ÓÐNH4+
D ijÈÜÒºÖмÓÈëÑÎËᣬ²¢½«²úÉúµÄÆøÌåͨÈë³ÎÇåʯ»ÒË® ³ÎÇåʯ»ÒË®±ä»ë×Ç£¬ËµÃ÷¸ÃÈÜÒºÖÐÒ»¶¨º¬ÓÐCO32-
A¡¢AB¡¢BC¡¢CD¡¢D
¿¼µã£º»¯Ñ§ÊµÑé·½°¸µÄÆÀ¼Û
רÌ⣺ʵÑéÆÀ¼ÛÌâ
·ÖÎö£ºA£®¶þÑõ»¯ÁòÓëËáÐÔ¸ßÃÌËá¼Ø·¢ÉúÑõ»¯»¹Ô­·´Ó¦£»
B£®µÎ¼ÓÏ¡ÑÎËáÎÞÏÖÏ󣬵μÓBaCl2ºó³öÏÖ°×É«³Áµí£¬¿ÉÅųýCl-ºÍSO32-µÄÓ°Ï죻
C£®ÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌåÖ»ÓÐNH3£»
D£®¿ÉÄÜΪSO32-»òHCO3-µÈÀë×Ó£®
½â´ð£º ½â£ºA£®¶þÑõ»¯ÁòÓëËáÐÔ¸ßÃÌËá¼Ø·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬±íÏÖ³ö¶þÑõ»¯ÁòµÄ»¹Ô­ÐÔ£¬¶ø²»ÊÇÆ¯°×ÐÔ£¬¹ÊA´íÎó£»
B£®µÎ¼ÓÏ¡ÑÎËáÎÞÏÖÏ󣬵μÓBaCl2ºó³öÏÖ°×É«³Áµí£¬¿ÉÅųýCl-ºÍSO32-µÄÓ°Ï죬¿É˵Ã÷ÑùÆ·ÈÜÒºÖÐÒ»¶¨º¬ÓÐSO42-£¬¹ÊBÕýÈ·£»
C£®ÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌåÖ»ÓÐNH3£¬ÄÜÖ¤Ã÷¸ÃÑùÆ·Öк¬ÓÐNH4+£¬¹ÊCÕýÈ·£»
D£®ÄÜÓëËá·´Ó¦Éú³Éʹ³ÎÇåʯ»ÒË®±ä»ë×ÇµÄÆøÌåµÄ¿ÉÄÜΪSO32-»òHCO3-µÈÀë×Ó£¬²»Ò»¶¨ÎªCO32-£¬¹ÊD´íÎó£®
¹ÊÑ¡BC£®
µãÆÀ£º±¾Ì⿼²é»¯Ñ§ÊµÑé·½°¸µÄÆÀ¼Û£¬Éæ¼°ÎïÖʵÄÐÔÖÊ̽¾¿¡¢Àë×Ó¼ìÑéµÈ£¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦¡¢ÊµÑéÄÜÁ¦ºÍÆÀ¼ÛÄÜÁ¦µÄ¿¼²é£¬Îª¸ß¿¼³£¼ûÌâÐÍ£¬×¢Òâ°ÑÎÕÎïÖʵÄÐÔÖÊÒÔ¼°ÊµÑéµÄÑÏÃÜÐԺͿÉÐÐÐÔµÄÆÀ¼Û£¬ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÀë×Ó·½³ÌʽµÄÊéдÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢ÂÈ»¯ÑÇÌúÓëÂÈË®µÄ·´Ó¦£º2Fe2++Cl2=2Fe3++2Cl-
B¡¢ÊµÑéÊÒÓôóÀíʯºÍÏ¡ÑÎËáÖÆÈ¡CO2£º2H++CO32-=CO2¡ü+H2O
C¡¢ÏòAlCl3ÈÜÒºÖмÓÈë¹ýÁ¿µÄNaOHÈÜÒº£ºAl3++3OH-=Al£¨OH£©3¡ý
D¡¢ÌúºÍÈý¼ÛÌú·´Ó¦£ºFe+Fe3+=2Fe2+

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

HAÈÜÒºÖдæÔÚÏÂÁÐÆ½ºâ£ºHA£¨ºìÉ«£©?H++A-£¨»ÆÉ«£©£¬Í¨ÈëSO2¿É¹Û²ìµ½£¨¡¡¡¡£©
A¡¢»ÆÉ«±äÉîB¡¢»ÆÉ«±ädz
C¡¢ºìÉ«±äÉîD¡¢ÍÊΪÎÞÉ«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

µÈÎïÖÊÁ¿Å¨¶ÈµÄHCNÈÜÒºÓëNaCNÈÜÒºµÈÌå»ý»ìºÏ£¬»ìºÏÒºÏÔ¼îÐÔ£¬ÏÂÁÐ˵·¨´íÎóµÄ£¨¡¡¡¡£©
A¡¢c£¨CN-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©
B¡¢c£¨Na+£©£¾c£¨CN-£©£¾c£¨OH-£©£¾c£¨H+£©
C¡¢c£¨Na+£©+c£¨H+£©=c£¨CN-£©+c£¨OH-£©
D¡¢2c£¨OH-£©+c£¨CN-£©=2c£¨H+£©+c£¨HCN£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔªËØAµÄÑôÀë×ÓaAm+ÓëÔªËØBµÄÒõÀë×ÓbBn-¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£®ÒÔϹØÓÚA¡¢BÔªËØÐÔÖʵıȽÏÖУ¬´íÎóµÄÊÇ£¨¡¡¡¡£©
A¡¢Ô­×ÓÐòÊý£ºA£¾B
B¡¢Ô­×Ó°ë¾¶£ºA£¼B
C¡¢Àë×Ó°ë¾¶£ºaAm+£¼bBn-
D¡¢ÔªËØËùÔÚµÄÖÜÆÚÊý£ºA£¾B

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

³£ÎÂʱ£¬ÈôË®ÈÜÒºÖÐÓÉË®µçÀë²úÉúµÄc£¨OH-£©Îª1¡Á10-11£¬Âú×ã´ËÌõ¼þµÄÈÜÒºÖÐÒ»¶¨¿ÉÒÔ´óÁ¿¹²´æµÄÀë×Ó×éÊÇ£¨¡¡¡¡£©
A¡¢Al3+¡¢Na+¡¢Cl-¡¢N
O
-
3
B¡¢K+¡¢N
O
-
3
¡¢N
H
+
4
¡¢S
O
2-
4
C¡¢Na+¡¢K+¡¢Al
O
-
2
¡¢Cl-
D¡¢Na+¡¢K+¡¢N
O
-
3
¡¢Cl-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÓÃÎïÖʵÄÁ¿¶¼ÊÇ0.1molµÄCH3COOHºÍCH3COONaÅä³É1L»ìºÏÈÜÒº£¬ÒÑÖªÆäÖÐc£¨CH3COO-£©´óÓÚc£¨Na+£©£¬¶Ô¸Ã»ìºÏÈÜÒºµÄÏÂÁÐÅжÏÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢c£¨H+£©£¾c£¨OH-£©
B¡¢c£¨CH3COO-£©+c£¨CH3COOH£©=0.2 mol?L-1
C¡¢c£¨CH3COOH£©£¾c£¨CH3COO-£©
D¡¢c£¨CH3COO-£©+c£¨OH-£©=0.1¡¡mol?L-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ͼÖÐAΪµ­»ÆÉ«µÄ¹ÌÌ壬BÎªÆøÌ壬CΪÈÜÒº£¬E¡¢FΪ°×É«³Áµí£®ÆäÖУ¬Óв¿·ÖÎïÖÊÊ¡ÂÔ£®

£¨1£©Ð´³ö»¯Ñ§Ê½£ºA
 
£¬B
 
£¬C
 
£¬E
 
£¬F
 
£®
£¨2£©Ð´³öB¡úAµÄ»¯Ñ§·´Ó¦·½³Ìʽ
 
£®
£¨3£©Ð´³öC¡úFµÄÀë×Ó·´Ó¦·½³Ìʽ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐʵÑéÖУ¬ËùѡװÖò»ºÏÀíµÄÊÇ£¨¡¡¡¡£©
A¡¢·ÖÀëNa2CO3ÈÜÒººÍCCl4£¬Ñ¡¢Ü
B¡¢ÓÃCC14ÌáÈ¡µâË®Öеĵ⣬ѡ¢Û
C¡¢ÓÃFeC12ÈÜÒºÎüÊÕC12£¬Ñ¡¢Ý
D¡¢´ÖÑÎÌá´¿£¬Ñ¡¢ÙºÍ¢Ú

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸