1¡¢Ñ¡ÔñÌâ ÏòËữ¹ýµÄMnSO4ÈÜÒºÖеμÓ(NH4)2S2O8(¹ý¶þÁòËáï§)ÈÜÒº»á·¢Éú·´Ó¦£ºMn2£«£«S2O82-£«H2O¨D¡úMnO4-£«H£«£«SO42-¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ(¡¡¡¡)
A£®¿ÉÒÔÀûÓø÷´Ó¦¼ìÑéMn2£«
B£®Ñõ»¯ÐԱȽϣºS2O82-£¾MnO4-
C£®MnSO4ÈÜÒº¿ÉÒÔʹÓÃÑÎËáËữ
D£®ÈôÓÐ0.1 molÑõ»¯²úÎïÉú³É£¬ÔòתÒƵç×Ó0.5 mol
²Î¿¼´ð°¸£ºC
±¾Ìâ½âÎö£ºÓÉÓÚS2O82-µÄÑõ»¯ÐÔ´óÓÚMnO4-µÄ£¬µ±ÓÃÑÎËáËữʱCl£»á±»S2O82-Ñõ»¯ÎªCl2¡£
±¾ÌâÄѶȣºÒ»°ã
2¡¢ÊµÑéÌâ £¨15·Ö£©ÎªÁË̽¾¿H2O2¡¢H2SO3ºÍBr2Ñõ»¯ÐÔµÄÏà¶ÔÇ¿Èõ£¬Éè¼ÆÈçÏÂʵÑ飨¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥£©¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÇÆ÷AµÄÃû³Æ_________£¬Æä×÷ÓÃÊÇ___________¡£
£¨2£©ÓÃÒÇÆ÷BµÎ¼ÓÒºÌå²¢²»ÐèÒª´ò¿ª²£Á§Èûc£¬ÔÒòÊÇ____________________________¡£
£¨3£©ÊµÑé¼Ç¼ÈçÏ£¨Ç벹ȫ¿Õ°×£©£º
²½Öè
| ʵÑé²Ù×÷
| ʵÑéÏÖÏó
| ʵÑé½áÂÛ
|
¢ñ
| ´ò¿ª»îÈûa£¬ÖðµÎ¼ÓÈëH2SO3ÈÜÒºÖÁ¹ýÁ¿
| ________________
| __________________________
|
¢ò
| Ïò²½Öè¢ñËùµÃÈÜÒºÖÐÖðµÎ¼ÓÈëH2O2ÈÜÒº
| ¸Õ¿ªÊ¼ÈÜÒºÑÕÉ«ÎÞÃ÷ÏԱ仯£¬¼ÌÐøµÎ¼Ó£¬ÈÜÒº±äΪ³È»ÆÉ«
| __________________________
|
£¨4£©²½Öè¢òÖУ¬¿ªÊ¼Ê±ÑÕÉ«ÎÞÃ÷ÏԱ仯µÄÔÒòÊÇ£¨Ð´³öÒ»Ìõ£©_______________________£¬
²½Öè¢ñÖз´Ó¦µÄÀë×Ó·½³Ìʽ_________________________________________________£¬
²½Öè¢òÖÐÖ÷Òª·´Ó¦µÄÀë×Ó·½³Ìʽ_____________________________________________¡£
²Î¿¼´ð°¸£º£¨1£©ÇòÐÎÀäÄý¹Ü£¨»òÀäÄý¹Ü£© »ØÁ÷ä壨»òÒÔÃâäå»Ó·¢µÈ£© £¨2£©µÎҺ©¶·ºÍÈý¾±Æ¿ÆøѹÏàͨ
£¨3£©¢ñ£º³È»ÆÉ«ÍÊÈ¥ Br2µÄÑõ»¯ÐÔ´óÓÚH2SO3 ¢ò£ºH2O2µÄÑõ»¯ÐÔ´óÓÚBr2
£¨4£©²½Öè1µÄH2SO3ÓйýÁ¿£¬H2O2ÏȺÍH2SO3·´Ó¦£¨H2O2Ũ¶ÈС»òBr£ÓëH2O2·´Ó¦ÂýµÈ¶¼¿É£©
H2SO3+Br2+H2O=4H++SO42-+2Br- H2O2+2Br-£«2H+£½Br2£«2H2O
±¾Ìâ½âÎö£º
ÊÔÌâ½âÎö£º£¨1£©ÒÇÆ÷AµÄÃû³ÆÊÇÇòÐÎÀäÄã¹Ü£¬Æä×÷ÓÃÊÇʹäåÕôÆûÀäÄý»ØÁ÷¡££¨2£©ÓÃÒÇÆ÷BµÎ¼ÓÒºÌå²¢²»ÐèÒª´ò¿ª²£Á§Èûc£¬ÔÒòÊǸÃÒÇÆ÷ÓëÈý¾±Æ¿Ïàͬ£¬ÆøѹÏàµÈ£¬ÒºÌåÄÜ˳ÀûÁ÷³ö¡££¨3£©¢ñ£º´ò¿ª»îÈûa£¬ÖðµÎ¼ÓÈëH2SO3ÈÜÒºÖÁ¹ýÁ¿£¬H2SO3±»äåË®Ñõ»¯£¬ÈÜÒºµÄ³È»ÆÉ«ÍÊÈ¥£¬ËµÃ÷äåµ¥ÖʵÄÑõ»¯ÐÔÇ¿ÓÚÑÇÁòË᣻¢ò£ºÏò²½Öè¢ñËùµÃÈÜÒºÖÐÖðµÎ¼ÓÈëH2O2ÈÜÒº£¬ÈÜÒºÓÖ±äΪ³È»ÆÉ«£¬ËµÃ÷¹ýÑõ»¯Ç⽫äåÀë×ÓÑõ»¯Îªäåµ¥ÖÊ£¬ÔòH2O2µÄÑõ»¯ÐÔ´óÓÚBr2£»£¨4£©²½Öè¢òÖУ¬¿ªÊ¼Ê±ÑÕÉ«ÎÞÃ÷ÏԱ仯¿ÉÄÜÊÇÒòΪ²½Öè1µÄH2SO3ÓйýÁ¿£¬H2O2ÏȺÍH2SO3·´Ó¦£¬»òH2O2Ũ¶ÈС»òBr£ÓëH2O2·´Ó¦ÂýµÈ£»²½Öè¢ñÖУ¬ÑÇÁòËáÓëäåµ¥ÖÊ·´Ó¦£¬Àë×Ó·½³ÌʽΪ£ºH2SO3+Br2+H2O=4H++SO42-+2Br-£»²½Öè¢òÖÐÖ÷Òª·¢Éú¹ýÑõ»¯ÇâºÍäåÀë×ӵķ´Ó¦£¬Àë×Ó·½³ÌʽΪ£º H2O2+2Br-£«2H+£½Br2£«2H2O¡£
¿¼µã£ºÊµÑé·½°¸µÄÉè¼Æ
±¾ÌâÄѶȣºÀ§ÄÑ
3¡¢Ñ¡ÔñÌâ ÔÚŨÑÎËáÖÐH3AsO3ÓëSnCl2·´Ó¦µÄÀë×Ó·½³Ì3SnCl2+12Cl-+2H3AsO3+6H+=2As+3SnCl62-+6M£®¹ØÓڸ÷´Ó¦µÄ˵·¨ÖÐÕýÈ·µÄ×éºÏÊÇ£¨¡¡¡¡£©¢ÙÑõ»¯¼ÁÊÇH3AsO3£»¢Ú»¹ÔÐÔ£ºCl-£¾As£»¢ÛÿÉú³É0.1mol?As£¬·´Ó¦ÖÐתÒƵĵç×ÓµÄÎïÖʵÄÁ¿Îª0.3?mol£®¢ÜMΪOH-£®
A£®¢Ù¢Û
B£®¢Ù¢Ú¢Ü
C£®¢Ú¢Û¢Ü
D£®Ö»ÓТÙ
²Î¿¼´ð°¸£ºÓ¦Îª3SnCl2+12Cl-+2H3AsO3+6H+=2As+3SnCl62-+6M£®
¶ÔÓÚ·´Ó¦3SnCl2+12Cl-+2H3AsO3+6H+=2As+3SnCl62-+6M£®
¢Ù·´Ó¦ÖÐÖ»ÓÐAsÔªËØ»¯ºÏ¼Û½µµÍ£¬ÓÉH3AsO3ÖÐ+3¼Û½µµÍΪ0¼Û£¬ËùÒÔH3AsO3ÊÇÑõ»¯¼Á£¬¹Ê¢ÙÕýÈ·£»
¢Ú·´Ó¦ÖÐÖ»ÓÐSnÔªËØ»¯ºÏ¼ÛÉý¸ß£¬ÓÉSnCl2ÖÐ+2¼ÛÉý¸ßΪSnCl62+ÖÐ+4¼Û£¬ËùÒÔSnCl2ÊÇ»¹Ô¼Á£¬·´Ó¦ÖÐÖ»ÓÐAsÔªËØ»¯ºÏ¼Û½µµÍ£¬ÓÉH3AsO3ÖÐ+3¼Û½µµÍΪ0¼Û£¬ËùÒÔH3AsO3ÊÇÑõ»¯¼Á£¬ËùÒÔ»¹Ô²úÎïΪAs£¬»¹ÔÐÔSnCl2£¾As£»·´Ó¦ÖÐÂÈÔªËØ»¯ºÏ¼ÛΪ·¢Éú±ä»¯£¬¹Ê¢Ú´íÎó£»
¢Û·´Ó¦ÖÐÖ»ÓÐAsÔªËØ»¯ºÏ¼Û½µµÍ£¬±»»¹Ô£¬ÓÉH3AsO3ÖÐ+3¼Û½µµÍΪ0¼Û£¬Ã¿Éú³É0.1mol?As£¬·´Ó¦ÖÐתÒƵĵç×ÓµÄÎïÖʵÄÁ¿Îª0.1mol¡Á3=0.3?mol£¬¹Ê¢ÛÕýÈ·£»
¢Ü¸ù¾ÝÔªËØÊغã¿ÉÖª£¬MӦΪH2O£¬¹Ê¢Ü´íÎó£®
¹Ê¢Ù¢ÛÕýÈ·£®
¹ÊÑ¡£ºA£®
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣº¼òµ¥
4¡¢Ñ¡ÔñÌâ ´ÓʯӢɰÖÆÈ¡²¢»ñµÃ¸ß´¿¹èµÄÖ÷Òª»¯Ñ§·´Ó¦ÈçÏ£º
¢ÙSiO2£«2CSi(´Ö)£«2CO¡ü
¢ÚSi(´Ö)£«2Cl2SiCl4
¢ÛSiCl4£«2H2Si(´¿)£«4HCl
¹ØÓÚÉÏÊö·´Ó¦µÄ·ÖÎö²»ÕýÈ·µÄÊÇ
[? ]
A£®¢Ù¡¢¢ÛÊÇÖû»·´Ó¦£¬¢ÚÊÇ»¯ºÏ·´Ó¦
B£®¸ßÎÂÏ£¬½¹Ì¿ÓëÇâÆøµÄ»¹ÔÐÔ¾ùÇ¿ÓÚ¹è
C£®ÈÎÒ»·´Ó¦ÖУ¬Ã¿ÏûºÄ»òÉú³É28?g¹è£¬¾ùתÒÆ4?molµç×Ó
D£®¸ßÎÂϽ«Ê¯Ó¢É°¡¢½¹Ì¿¡¢ÂÈÆø¡¢ÇâÆø°´Ò»¶¨±ÈÀý»ìºÏ¿ÉµÃ¸ß´¿¹è
²Î¿¼´ð°¸£ºD
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣºÒ»°ã
5¡¢Ñ¡ÔñÌâ ÒÑÖª£º¢Ù2FeCl3+2KI=2FeCl2+2KCl+I2¢Ú2FeCl2+Cl2=2FeCl3£¬ÔòÏÂÁÐ΢Á£»¹ÔÄÜÁ¦ÓÉ´óµ½Ð¡µÄ˳ÐòÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Fe2+£¾Cl-£¾I-
B£®I-£¾Fe2+£¾Cl-
C£®I-£¾Cl-£¾Fe2+
D£®Cl-£¾I-£¾Fe2+
²Î¿¼´ð°¸£º¢Ù2FeCl3+2KI=2FeCl2+2KCl+I2ÖУ¬µâÔªËØ»¯ºÏ¼ÛÉý¸ß£¬ÌúÔªËØ»¯ºÏ¼Û½µµÍ£¬ËùÒÔÂÈ»¯Ìú×÷Ñõ»¯¼Á£¬µâ»¯¼Ø×÷»¹Ô¼Á£¬»¹Ô²úÎïÊÇÂÈ»¯ÑÇÌú£¬Ôò»¹ÔÐÔI-£¾Fe2+£¬
¢Ú2FeCl2+Cl2=2FeCl3ÖУ¬ÂÈÔªËØ»¯ºÏ¼Û½µµÍ£¬ÌúÔªËØ»¯ºÏ¼ÛÉý¸ß£¬ËùÒÔÂÈÆøÊÇÑõ»¯¼Á£¬ÂÈ»¯ÑÇÌúÊÇ»¹Ô¼Á£¬ÂÈ»¯ÌúÊÇÑõ»¯²úÎïºÍ»¹Ô²úÎËùÒÔ»¹ÔÐÔFe2+£¾Cl-£¬
ÔòÕâÈýÖÖÀë×ӵĻ¹ÔÐÔ´óС˳ÐòÊÇ£ºI-£¾Fe2+£¾Cl-£¬
¹ÊÑ¡B£®
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣºÒ»°ã