ÌâÄ¿ÄÚÈÝ


ŨH2SO4ºÍľ̿ÔÚ¼ÓÈÈʱ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ2H2SO4£¨Å¨£©+CCO2¡ü+2H2O+2SO2¡üÇë´ÓÏÂͼÖÐÑ¡ÓÃËùÐèµÄÒÇÆ÷£¨¿ÉÖØ¸´Ñ¡Óã©×é³ÉÒ»Ì×½øÐи÷´Ó¦²¢¼ì³ö·´Ó¦²úÎïµÄ×°Öá£ÏÖÌṩŨH2SO4¡¢Ä¾Ì¿ºÍËáÐÔKMnO4ÈÜÒº£¬ÆäËû¹Ì¡¢ÒºÊÔ¼Á×ÔÑ¡¡££¨Á¬½ÓºÍ¹Ì¶¨ÒÇÆ÷ÓõIJ£Á§¹Ü¡¢½º¹Ü¡¢Ìú¼Ð¡¢Ìú¼Ų̈¼°¼ÓÈÈ×°ÖõȾùÂÔÈ¥£©

½«ËùÑ¡µÄÒÇÆ÷°´Á¬½Ó˳ÐòÓÉÉÏÖÁÏÂÒÀ´ÎÌîÈëÏÂ±í£¬²¢Ð´³ö¸ÃÒÇÆ÷ÖÐÓ¦¼ÓÊÔ¼ÁµÄÃû³Æ¼°Æä×÷Óá£

Ñ¡ÓõÄÒÇÆ÷£¨Ìî×Öĸ£©

¼ÓÈëµÄÊÔ¼Á

×÷ÓÃ


¡¾´ð°¸¡¿

Ñ¡ÓõÄÒÇÆ÷£¨Ìî×Öĸ£©

¼ÓÈëµÄÊÔ¼Á

×÷ÓÃ

C

ŨH2SO4ºÍľ̿

·´Ó¦Æ÷£¨»ò·¢ÉúÆøÌ壩

B

ÎÞË®CuSO4

¼ì³öH2O

A

Æ·ºìÈÜÒº

¼ì³öSO2

A

ËáÐÔKMnO4ÈÜÒº

ÎüÊÕÓàϵÄSO2

A

³ÎÇåʯ»ÒË®

¼ì³öCO2

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

µ¥¾§¹èÊÇÐÅÏ¢²úÒµÖÐÖØÒªµÄ»ù´¡²ÄÁÏ¡£Í¨³£ÓÃ̼ÔÚ¸ßÎÂÏ»¹Ô­¶þÑõ»¯¹èÖÆµÃ´Ö¹è(º¬Ìú¡¢ÂÁ¡¢Åð¡¢Á×µÈÔÓÖÊ)£¬´Ö¹èÓëÂÈÆø·´Ó¦Éú³ÉËÄÂÈ»¯¹è£¨·´Ó¦Î¶È450¡«500 ¡æ£©£¬ËÄÂÈ»¯¹è¾­Ìá´¿ºóÓÃÇâÆø»¹Ô­¿ÉµÃ¸ß´¿¹è¡£ÒÔÏÂÊÇʵÑéÊÒÖÆ±¸ËÄÂÈ»¯¹èµÄ×°ÖÃʾÒâͼ¡£

Ïà¹ØÐÅÏ¢ÈçÏ£º

a.ËÄÂÈ»¯¹èÓöË®¼«Ò×Ë®½â£»

b.Åð¡¢ÂÁ¡¢Ìú¡¢Á×ÔÚ¸ßÎÂϾùÄÜÓëÂÈÆøÖ±½Ó·´Ó¦Éú³ÉÏàÓ¦µÄÂÈ»¯Î

c.ÓйØÎïÖʵÄÎïÀí³£Êý¼ûÏÂ±í£º

ÎïÖÊ

SiCl4

BCl3

AlCl3

FeCl3

PCl챐5

·Ðµã/¡æ

57.7

12.8

¡ª

315

¡ª

ÈÛµã/¡æ

-70.0

-107.2

¡ª

¡ª

¡ª

Éý»ªÎ¶È/¡æ

¡ª

¡ª

180

300

162

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)д³ö×°ÖÃAÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ____________¡£

(2)×°ÖÃAÖÐg¹ÜµÄ×÷ÓÃÊÇ____________£»×°ÖÃCÖеÄÊÔ¼ÁÊÇ;×°ÖÃEÖеÄhÆ¿ÐèÒªÀäÈ´µÄÀíÓÉÊÇ____________________________________¡£

(3)×°ÖÃEÖÐhÆ¿ÊÕ¼¯µ½µÄ´Ö²úÎï¿Éͨ¹ý¾«Áó£¨ÀàËÆ¶à´ÎÕôÁ󣩵õ½¸ß´¿¶ÈËÄÂÈ»¯¹è£¬¾«ÁóºóµÄ²ÐÁôÎïÖУ¬³ýÌúÔªËØÍâ¿ÉÄÜ»¹º¬ÓеÄÔÓÖÊÔªËØÊÇ____________£¨ÌîÐ´ÔªËØ·ûºÅ£©¡£

(4)ΪÁË·ÖÎö²ÐÁôÎïÖÐÌúÔªËØµÄº¬Á¿£¬ÏȽ«²ÐÁôÎïÔ¤´¦Àí£¬Ê¹ÌúÔªËØ»¹Ô­³ÉFe2+,ÔÙÓÃKMnO4±ê×¼ÈÜÒºÔÚËáÐÔÌõ¼þϽøÐÐÑõ»¯»¹Ô­µÎ¶¨£¬·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ£º5Fe2++MnO-4+8H+====5Fe3++Mn2++4H2O

¢ÙµÎ¶¨Ç°ÊÇ·ñÒªµÎ¼Óָʾ¼Á£¿________£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©£¬Çë˵Ã÷ÀíÓÉ__________________¡£

¢Úijͬѧ³ÆÈ¡5.000 g²ÐÁôÎ¾­Ô¤´¦ÀíºóÔÚÈÝÁ¿Æ¿ÖÐÅäÖÆ³É100 mLÈÜÒº£¬ÒÆÈ¡25.00 mLÊÔÑùÈÜÒº£¬ÓÃ1.000¡Á10-2 mol¡¤L-1 KMnO4±ê×¼ÈÜÒºµÎ¶¨¡£´ïµ½µÎ¶¨ÖÕµãʱ£¬ÏûºÄ±ê×¼ÈÜÒº20.00 mL£¬Ôò²ÐÁôÎïÖÐÌúÔªËØµÄÖÊÁ¿·ÖÊýÊÇ________________¡£


Èý²ÝËáºÏÌúËá¼Ø¾§ÌåK3£ÛFe(C2O4)3£Ý¡¤3H2O¿ÉÓÃÓÚÉãÓ°ºÍÀ¶É«Ó¡Ë¢¡£Ä³Ð¡×齫ÎÞË®Èý²ÝËáºÏÌúËá¼ØÔÚÒ»¶¨Ìõ¼þϼÓÈȷֽ⣬¶ÔËùµÃÆøÌå²úÎïºÍ¹ÌÌå²úÎï½øÐÐʵÑéºÍ̽¾¿¡£ÇëÀûÓÃʵÑéÊÒ³£ÓÃÒÇÆ÷¡¢ÓÃÆ·ºÍÒÔÏÂÏÞÑ¡ÊÔ¼ÁÍê³ÉÑéÖ¤ºÍ̽¾¿¹ý³Ì¡£

ÏÞÑ¡ÊÔ¼Á£ºÅ¨ÁòËá¡¢1.0 mol¡¤L-1 HNO3¡¢1.0 mol¡¤L-1ÑÎËá¡¢1.0 mol¡¤L-1 NaOH¡¢3% H2O2¡¢0.1 mol¡¤L-1 KI¡¢0.1 mol¡¤L-1 CuSO4¡¢20% KSCN¡¢³ÎÇåʯ»ÒË®¡¢Ñõ»¯Í­¡¢ÕôÁóË®¡£

£¨1£©½«ÆøÌå²úÎïÒÀ´Îͨ¹ý³ÎÇåʯ»ÒË®£¨A£©¡¢Å¨ÁòËá¡¢×ÆÈÈÑõ»¯Í­£¨B£©¡¢³ÎÇåʯ»ÒË®£¨C£©£¬¹Û²ìµ½A¡¢CÖгÎÇåʯ»ÒË®¶¼±ä»ë×Ç£¬BÖÐÓкìÉ«¹ÌÌåÉú³É£¬ÔòÆøÌå²úÎïÊÇ______________¡£

£¨2£©¸ÃС×éͬѧ²éÔÄ×ÊÁϺóÍÆÖª£¬¹ÌÌå²úÎïÖУ¬ÌúÔªËØ²»¿ÉÄÜÒÔÈý¼ÛÐÎʽ´æÔÚ£¬¶øÑÎÖ»ÓÐK2CO3¡£ÑéÖ¤¹ÌÌå²úÎïÖмØÔªËØ´æÔڵķ½·¨ÊÇ________________________£¬ÏÖÏóÊÇ_______________________________________________________________________________¡£

£¨3£©¹ÌÌå²úÎïÖÐÌúÔªËØ´æÔÚÐÎʽµÄ̽¾¿¡£

¢ÙÌá³öºÏÀí¼ÙÉè

¼ÙÉè1£º_____________£»¼ÙÉè2£º_____________£»¼ÙÉè3£º_____________¡£

¢ÚÉè¼ÆÊµÑé·½°¸Ö¤Ã÷ÄãµÄ¼ÙÉ裨²»ÒªÔÚ´ðÌ⿨ÉÏ×÷´ð£©

¢ÛʵÑé¹ý³Ì

¸ù¾Ý¢ÚÖз½°¸½øÐÐʵÑé¡£ÔÚ´ðÌ⿨Éϰ´Ï±íµÄ¸ñʽд³öʵÑé²½Öè¡¢Ô¤ÆÚÏÖÏóÓë½áÂÛ¡£

ʵÑé²½Öè

Ô¤ÆÚÏÖÏóÓë½áÂÛ

²½Öè1£º

²½Öè2£º

²½Öè3£º

¡­¡­


ijÑо¿ÐÔѧϰС×é½èÖúA¡«DµÄÒÇÆ÷×°ÖÃÍê³ÉÓйØÊµÑé¡£

£ÛʵÑéÒ»£Ý ÊÕ¼¯NOÆøÌå¡£

£¨1£©ÓÃ×°ÖÃAÊÕ¼¯NOÆøÌ壬ÕýÈ·µÄ²Ù×÷ÊÇ_____________£¨ÌîÐòºÅ£©¡£

a.´Ó¢Ù¿Ú½øÆø£¬ÓÃÅÅË®·¨¼¯Æø                       b.´Ó¢Ù¿Ú½øÆø£¬ÓÃÅÅÆø·¨¼¯Æø

c.´Ó¢Ú¿Ú½øÆø£¬ÓÃÅÅË®·¨¼¯Æø                       d.´Ó¢Ú¿Ú½øÆø£¬ÓÃÅÅÆø·¨¼¯Æø

£ÛʵÑé¶þ£ÝΪÁË̽¾¿¶ÆÐ¿±¡Ìú°åÉÏпµÄÖÊÁ¿·ÖÊýw(Zn)ºÍ¶Æ²ãºñ¶È£¬²éѯµÃ֪пÒ×ÈÜÓÚÇ¿¼î£ºZn+2NaOH====Na2ZnO2+H2¡ü¡£¾Ý´Ë£¬½ØÈ¡Ãæ»ýΪSµÄË«Ãæ¶ÆÐ¿±¡Ìú°åÊÔÑù£¬¼ôËé¡¢³ÆµÃÖÊÁ¿Îªm1 g¡£ÓùÌÌåÉÕ¼îºÍË®×÷ÊÔ¼Á£¬Äâ³öÏÂÁÐʵÑé·½°¸²¢½øÐÐÏà¹ØÊµÑé¡£

·½°¸¼×£ºÍ¨¹ý²âÁ¿ÊÔÑùÓë¼î·´Ó¦Éú³ÉµÄÇâÆøÌå»ýÀ´ÊµÏÖ̽¾¿Ä¿±ê¡£

£¨2£©Ñ¡ÓÃBºÍ_____________£¨ÌîÒÇÆ÷±êºÅ£©Á½¸ö×°ÖýøÐÐʵÑé¡£

£¨3£©²âµÃ³ä·Ö·´Ó¦ºóÉú³ÉÇâÆøµÄÌå»ýΪV L£¨±ê×¼×´¿ö£©£¬w(Zn)=___________¡£

£¨4£©¼ÆËã¶Æ²ãºñ¶È£¬»¹Ðè¼ìË÷µÄÒ»¸öÎïÀíÁ¿ÊÇ___________¡£

£¨5£©Èô×°ÖÃBÖеĺãѹ·ÖҺ©¶·¸ÄΪÆÕͨ·ÖҺ©¶·£¬²âÁ¿½á¹û½«_________£¨Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©¡£

·½°¸ÒÒ£ºÍ¨¹ý³ÆÁ¿ÊÔÑùÓë¼î·´Ó¦Ç°ºóµÄÖÊÁ¿ÊµÏÖ̽¾¿Ä¿±ê¡£Ñ¡ÓÃÒÇÆ÷C×öʵÑ飬ÊÔÑù¾­³ä·Ö·´Ó¦£¬Â˳ö²»ÈÜÎϴµÓ¡¢ºæ¸É£¬³ÆµÃÆäÖÊÁ¿Îªm2 g¡£

£¨6£©w(Zn)=___________¡£

·½°¸±û£ºÍ¨¹ý³ÆÁ¿ÊÔÑùÓë¼î·´Ó¦Ç°ºóÒÇÆ÷¡¢ÊÔÑùºÍÊÔ¼ÁµÄ×ÜÖÊÁ¿£¨Æä²îÖµ¼´ÎªH2µÄÖÊÁ¿£©ÊµÏÖ̽¾¿Ä¿±ê¡£ÊµÑéͬÑùʹÓÃÒÇÆ÷C¡£

£¨7£©´ÓʵÑéÎó²î½Ç¶È·ÖÎö£¬·½°¸±û___________·½°¸ÒÒ£¨Ìî¡°ÓÅÓÚ¡±¡°ÁÓÓÚ¡±»ò¡°µÈͬÓÚ¡±£©¡£

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

¾«Ó¢¼Ò½ÌÍø