ÌâÄ¿ÄÚÈÝ

£¨12·Ö£©Ä³Ñ§ÉúÉè¼ÆÈçÏÂʵÑé×°ÖÃÀûÓÃÂÈÆøÓëÏûʯ»Ò·´Ó¦ÖÆÈ¡ÉÙÁ¿Æ¯°×·Û£¨ÕâÊÇÒ»¸ö·ÅÈÈ·´Ó¦£©£¬¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©AÒÇÆ÷µÄÃû³ÆÊÇ¡¡            ¡¡£¬ËùÊ¢ÊÔ¼ÁÊÇ¡¡¡¡                      ¡¡¡¡¡£

£¨2£©Æ¯°×·Û½«ÔÚUÐ͹ÜÖвúÉú£¬Æ仯ѧ·´Ó¦·½³ÌʽÊÇ                        ¡¡¡¡¡¡¡£

£¨3£©ÓÐͬѧ½¨ÒéÔÚÁ½¸öÏðƤÌ×¹ÜÄڵIJ£Á§¹Ü¿ÚÓ¦¾¡Á¿½ô¿¿£¬Ô­ÒòÊÇ¡¡¡¡                 ¡£

£¨4£©´ËʵÑé½á¹ûËùµÃCa(ClO)2²úÂÊÌ«µÍ¡£¾­·ÖÎö²¢²éÔÄ×ÊÁÏ·¢ÏÖÖ÷ÒªÔ­ÒòÊÇÔÚUÐ͹ÜÖдæÔÚÁ½¸ö¸±·´Ó¦£º¢ÙζȽϸßʱÂÈÆøÓëÏûʯ»Ò·´Ó¦Éú³ÉÁËCa(ClO3)2£¬´Ë¸±·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£¬Îª±ÜÃâ´Ë¸±·´Ó¦µÄ·¢Éú£¬¿É²ÉÈ¡µÄ´ëÊ©ÊÇ             ¡£

¢Úд³öÁíÒ»¸ö¸±·´Ó¦µÄ»¯Ñ§·½³Ìʽ                            ¡£Îª±ÜÃâ´Ë¸±·´Ó¦·¢Éú£¬¿É²ÉÈ¡µÄ´ëÊ©ÊÇ                                                ¡£

£¨5£©ÔÚ¿ÕÆøÖУ¬Æ¯°×·Û»áʧЧ£¬Ô­ÒòÊÇ£¨Ó÷½³Ìʽ±íʾ£©                                ¡£

 

¡¾´ð°¸¡¿

£¨1£©·ÖҺ©¶·£¨1·Ö£©¡¡Å¨ÑÎËᣨ1·Ö£©

£¨2£©Ca(OH)2+Cl2£½CaCl2£«Ca(ClO)2£«H2O£¨2·Ö£©

£¨3£©·ÀÖ¹Cl2½«ÏðƤ¹ÜÑõ»¯£¨1·Ö£©

£¨4£©6Cl2£«6Ca(OH)2£½Ca(ClO3)2£«5CaCl2£«6H2O£¨2·Ö£©

½«UÐιÜÖÃÓÚÀäË®ÖÐÀäÈ´£¨1·Ö£©¡¡¡¡

ÁíÒ»¸±·´Ó¦2HCl£«Ca(OH)2£½CaCl2£«2H2O£¨1·Ö£©¡¡¡¡½«B¡¢CÖ®¼äÁ¬½ÓÒ»¸öÊ¢Óб¥ºÍʳÑÎË®µÄÏ´ÆøÆ¿£¨1·Ö£©

£¨5£©Ca(ClO)2+CO2+ H2O£½CaCO3£«2 HClO  2HClO=2HCl+O2¡ü£¨2·Ö£©

¡¾½âÎö¡¿

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ijѧÉúÉè¼ÆÁËÈçͼËùʾµÄ×°ÖýøÐÐʵÑ飬¸÷×°ÖÃÖÐËù×°ÊÔ¼Á¡¢ÊµÑé²Ù×÷¼°ÊµÑéÏÖÏóÈçÏ£º
¢ñ£®×°ÖÃÖÐËù×°µÄÊÔ¼ÁÊÇ£º¢ÙAƿװÎÞË®ÒÒ´¼£¬ÄÚ·ÅÎÞË®ÑÎX£»¢ÚB¸ÉÔï¹ÜÖÐ×°Éúʯ»Ò£»¢ÛCºÍDÖж¼×°Å¨ÁòË᣻¢ÜEÆ¿ÖÐ×°ÈëÊÔ¼ÁY
¢ò£®ÊµÑé²Ù×÷¼°ÏÖÏóÊÇ£ºÓÃˮԡ¼ÓÈÈAÆ¿£»½«DÖÐŨÁòËỺ»ºµÎÈëEÖÐÓëÊÔ¼ÁY×÷Óã»·¢ÏÖCÖе¼¹ÜÓдóÁ¿ÆøÅݷųö£»AÆ¿ÄÚXÖð½¥±äÉ«£¬´ÓB¹Ü»Ó·¢³öµÄÆøÌå¿Éµãȼ£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©EÆ¿ÖÐËù×°µÄÊÔ¼ÁYÊÇ
c
c
£¨Ìîд±àºÅ£©
a£®±¥ºÍʳÑÎË®¡¡b£®MnO2ºÍNaClµÄ»ìºÏÎï¡¡c£®Å¨ÑÎËá
£¨2£©CÖÐŨH2SO4ËùÆðµÄ×÷ÓÃÊÇ
¸ÉÔïHClÆøÌå
¸ÉÔïHClÆøÌå
£¬DÆ¿ÖÐŨH2SO4ËùÆðµÄ×÷ÓÃÊÇ
ÎüÊÕŨÑÎËáÖлìÓеÄË®·Ö£¬Å¨H2SO4ÈÜÓÚË®·ÅÈÈ£¬ÓÐÀûÓÚHClÆøÌåÒݳö
ÎüÊÕŨÑÎËáÖлìÓеÄË®·Ö£¬Å¨H2SO4ÈÜÓÚË®·ÅÈÈ£¬ÓÐÀûÓÚHClÆøÌåÒݳö
£®
£¨3£©AÆ¿Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
CH3CH2OH+HCl
¡÷
CH3CH2Cl+H2O
CH3CH2OH+HCl
¡÷
CH3CH2Cl+H2O
£¬·´Ó¦ÀàÐÍÊÇ
È¡´ú·´Ó¦
È¡´ú·´Ó¦
£¬ËùÉú³ÉµÄ
ÂÈÒÒÍé
ÂÈÒÒÍé
£¨Ð´Ãû³Æ£©ÔÚB³ö¿Ú´¦µãȼ£®
£¨4£©ÎÞË®ÑÎXÒËÑ¡ÓÃ
ÎÞË®CuSO4
ÎÞË®CuSO4
£¬ËüÄÜÆðָʾ¼Á×÷ÓõÄÔ­ÒòÊÇ
ʵÑé¹ý³ÌÖй۲쵽ÎÞË®CuSO4ÓÉ°×±äÀ¶£¬ËµÃ÷·´Ó¦ÖÐÓÐË®Éú³É£¬ÓëCuSO4½áºÏÉú³ÉCuSO4?5H2O
ʵÑé¹ý³ÌÖй۲쵽ÎÞË®CuSO4ÓÉ°×±äÀ¶£¬ËµÃ÷·´Ó¦ÖÐÓÐË®Éú³É£¬ÓëCuSO4½áºÏÉú³ÉCuSO4?5H2O
£®
£¨5£©´ËʵÑéÄÜÖ¤Ã÷ÒÒ´¼·Ö×ÓÖк¬ÓÐÑõÔ­×ÓµÄÀíÓÉÊÇ
ÎÞË®CuSO4±äÀ¶Ö¤Ã÷ÁË·´Ó¦ÖÐÒ»¶¨ÓÐË®Éú³É£¬Ë®ÖеÄÑõÔªËز»ÄÜÀ´×ÔÓÚHCl£¬¹ÊÖ»ÄÜÓÉÒÒ´¼Ìṩ
ÎÞË®CuSO4±äÀ¶Ö¤Ã÷ÁË·´Ó¦ÖÐÒ»¶¨ÓÐË®Éú³É£¬Ë®ÖеÄÑõÔªËز»ÄÜÀ´×ÔÓÚHCl£¬¹ÊÖ»ÄÜÓÉÒÒ´¼Ìṩ
£®
£¨6£©Èç¹û½«×°ÖÃÖеÄCÆ¿È¥µô£¬ÊµÑéÄ¿µÄÊÇ·ñÄܹ»´ïµ½£¿
²»ÄÜ
²»ÄÜ
£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©£¬ÒòΪ
HCl»Ó·¢Ê±´ø³öË®ÕôÆø£¬Èô²»³ýÈ¥£¬ÎÞ·¨Åж¨Ê¹ÎÞË®CuSO4±äÀ¶µÄË®ÊÇ·ñÀ´×ÔÓÚÒÒ´¼
HCl»Ó·¢Ê±´ø³öË®ÕôÆø£¬Èô²»³ýÈ¥£¬ÎÞ·¨Åж¨Ê¹ÎÞË®CuSO4±äÀ¶µÄË®ÊÇ·ñÀ´×ÔÓÚÒÒ´¼
£®
£¨2009?º£¿ÚÄ£Ä⣩ijѧϰС×é¶ÔÈ˽̰æ½Ì²ÄʵÑé¡°ÔÚ200mLÉÕ±­ÖзÅÈë20gÕáÌÇ£¨C12H22O11£©£¬¼ÓÈëÊÊÁ¿Ë®£¬½Á°è¾ùÔÈ£¬È»ºóÔÙ¼ÓÈë15mLÖÊÁ¿·ÖÊýΪ98%ŨÁòËᣬѸËÙ½Á°è¡±½øÐÐÈçÏÂ̽¾¿£®
£¨1£©¹Û²ìÏÖÏó£ºÕáÌÇÏȱä»Æ£¬ÔÙÖð½¥±äºÚ£¬Ìå»ýÅòÕÍ£¬ÐγÉÊèËɶà¿×µÄº£Ãà×´ºÚÉ«ÎïÖÊ£¬Í¬Ê±Îŵ½´Ì¼¤ÐÔÆøζ£®°´Ñ¹´ËºÚÉ«ÎïÖÊʱ£¬¸Ð¾õ½ÏÓ²£¬·ÅÔÚË®ÖгÊƯ¸¡×´Ì¬£®
ͬѧÃÇÓÉÉÏÊöÏÖÏóÍƲâ³öÏÂÁнáÂÛ£º
¢ÙŨÁòËá¾ßÓÐÇ¿Ñõ»¯ÐÔ    ¢ÚŨÁòËá¾ßÓÐÎüË®ÐÔ     ¢ÛŨÁòËá¾ßÓÐÍÑË®ÐÔ
¢ÜŨÁòËá¾ßÓÐËáÐÔ        ¢ÝºÚÉ«ÎïÖʾßÓÐÇ¿Îü¸½ÐÔ
ÆäÖÐÒÀ¾Ý²»³ä·ÖµÄÊÇ
¢Ú¢Ü
¢Ú¢Ü
£¨ÌîÐòºÅ£©£®
£¨2£©ÎªÁËÑéÖ¤ÕáÌÇÓëŨÁòËá·´Ó¦Éú³ÉµÄÆø̬²úÎͬѧÃÇÉè¼ÆÁËÈçÏÂ×°Öãº

ÊԻشðÏÂÁÐÎÊÌ⣺
¢Ùͼ1µÄAÖÐ×îºÃÑ¡ÓÃÏÂÁÐ×°ÖÃ
¢ò
¢ò
£¨Ìî±àºÅ£©£®
¢Úͼ1µÄ B×°ÖÃËù×°ÊÔ¼ÁÊÇ
Æ·ºìÈÜÒº
Æ·ºìÈÜÒº
£»D×°ÖÃÖÐÊÔ¼ÁµÄ×÷ÓÃÊÇ
¼ìÑéSO2ÊÇ·ñ±»³ý¾¡
¼ìÑéSO2ÊÇ·ñ±»³ý¾¡
£»E×°ÖÃÖз¢ÉúµÄÏÖÏóÊÇ
ÈÜÒº³öÏÖ°×É«»ë×Ç
ÈÜÒº³öÏÖ°×É«»ë×Ç
£®
¢Ûͼ1µÄA×°ÖÃÖÐʹÕáÌÇÏȱäºÚµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ
C12H11O11£¨ÕáÌÇ£©
ŨÁòËá
12C+11H2O
C12H11O11£¨ÕáÌÇ£©
ŨÁòËá
12C+11H2O
£»ºóÌå»ýÅòÕ͵Ļ¯Ñ§·½³ÌʽΪ£º
2H2SO4£¨Å¨£©+C
  ¡÷  
.
 
CO2¡ü+2SO2¡ü+2H2O
2H2SO4£¨Å¨£©+C
  ¡÷  
.
 
CO2¡ü+2SO2¡ü+2H2O
£®
¢ÜijѧÉú°´Í¼2½øÐÐʵÑéʱ£¬·¢ÏÖDÆ¿Æ·ºì²»ÍÊÉ«£¬E×°ÖÃÖÐÓÐÆøÌåÒݳö£¬F×°ÖÃÖÐËáÐÔ¸ßÃÌËá¼ØÈÜÒºÑÕÉ«±ädz£¬ÍƲâF×°ÖÃÖÐËáÐÔ¸ßÃÌËá¼ØÈÜÒºÑÕÉ«±ädzµÄÔ­Òò
COÆøÌåÄÜ»¹Ô­ÈȵÄËáÐÔ¸ßÃÌËá¼Ø
COÆøÌåÄÜ»¹Ô­ÈȵÄËáÐÔ¸ßÃÌËá¼Ø
Æä·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
5CO+6H++2MnO4
  ¡÷  
.
 
5CO2¡ü+2Mn2++3H2O
5CO+6H++2MnO4
  ¡÷  
.
 
5CO2¡ü+2Mn2++3H2O
£®
ijѧÉúÉè¼ÆÁËÈçͼËùʾµÄ×°ÖýøÐÐʵÑ飬¸÷×°ÖÃÖÐËù×°ÊÔ¼Á¡¢ÊµÑé²Ù×÷¼°ÊµÑéÏÖÏóÈçÏ£º
¢ñ.×°ÖÃÖÐËù×°µÄÊÔ¼ÁÊÇ£º¢ÙAƿװÎÞË®ÒÒ´¼£¬ÄÚ·ÅÎÞË®ÑÎX£»¢ÚB¸ÉÔï¹ÜÖÐ×°Éúʯ»Ò£»¢ÛCºÍDÖж¼×°Å¨ÁòË᣻¢ÜEÆ¿ÖÐ×°ÈëÊÔ¼ÁY
¢ò.ʵÑé²Ù×÷¼°ÏÖÏóÊÇ£ºÓÃˮԡ¼ÓÈÈAÆ¿£»½«DÖÐŨÁòËỺ»ºµÎÈëEÖÐÓëÊÔ¼ÁY×÷Óã»·¢ÏÖCÖе¼¹ÜÓдóÁ¿ÆøÅݷųö£»AÆ¿ÄÚXÖð½¥±äÉ«£¬´ÓB¹Ü»Ó·¢³öµÄÆøÌå¿Éµãȼ¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)EÆ¿ÖÐËù×°µÄÊÔ¼ÁYÊÇ________(Ìîд±àºÅ)
a£®±¥ºÍʳÑÎË®¡¡b£®MnO2ºÍNaClµÄ»ìºÏÎï¡¡c£®Å¨ÑÎËá
(2)CÖÐŨH2SO4ËùÆðµÄ×÷ÓÃÊÇ_______________________£¬DÆ¿ÖÐŨH2SO4ËùÆðµÄ×÷ÓÃÊÇ________________________________¡£
(3)AÆ¿Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ______________________£¬·´Ó¦ÀàÐÍÊÇ__________£¬ËùÉú³ÉµÄ________________(дÃû³Æ)ÔÚB³ö¿Ú´¦µãȼ¡£
(4)ÎÞË®ÑÎXÒËÑ¡ÓÃ____________£¬ËüÄÜÆðָʾ¼Á×÷ÓõÄÔ­ÒòÊÇ______________________¡£
(5)´ËʵÑéÄÜÖ¤Ã÷ÒÒ´¼·Ö×ÓÖк¬ÓÐÑõÔ­×ÓµÄÀíÓÉÊÇ________________________¡£
(6)Èç¹û½«×°ÖÃÖеÄCÆ¿È¥µô£¬ÊµÑéÄ¿µÄÊÇ·ñÄܹ»´ïµ½£¿_____(Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)£¬ÒòΪ___________________________¡£

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

¾«Ó¢¼Ò½ÌÍø