ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿K3Fe(A2B4)3¡¤3H2O(M)ÊÇÖƱ¸¸ºÔØÐÍ»îÐÔÌú´ß»¯ÖƵÄÖ÷ÒªÔ­ÁÏ£¬Ò²ÊÇһЩÓлú·´Ó¦´ß»¯¼Á£¬Òò¶ø¾ßÓй¤ÒµÉú²ú¼ÛÖµ¡£A¡¢B¾ùΪ³£¼û¶ÌÖÜÆڷǽðÊôÔªËØ¡£Ä³Ñо¿Ð¡×齫´¿¾»µÄ»¯ºÏÎïMÔÚÒ»¶¨Ìõ¼þϼÓÈȷֽ⣬¶ÔËùµÃÆøÌå²úÎïÓйÌÌå²úÎïµÄ×é³É½øÐÐ̽¾¿¡£

£¨1£©¾­ÊµÑé·ÖÎö£¬ËùµÃÆøÌå²úÎïÓɼס¢ÒÒºÍË®ÕôÆø×é³É¡£¼×¡¢ÒÒ¾ùÖ»º¬A¡¢BÁ½ÖÖÔªËØ£¬¼×ÄÜʹ³ÎÇåµÄʯ»ÒË®½»»ë×Ç£¬ÒÒ³£ÓÃÓÚ¹¤ÒµÁ¶Ìú¡£

¢Ùд³ö¼×µÄµç×Óʽ:_________¡£

¢Úд³ö¹¤ÒµÁ¶ÌúµÄ»¯Ñ§·´Ó¦·½³Ìʽ:_______________¡£

£¨2£©¸ÃÑо¿Ð¡×é²éÔÄ×ÊÁϺóÍÆÖª£¬¹ÌÌå²úÎïÖв»´æÔÚ+3¼ÛÌú£¬ÑÎÒ²Ö»ÓÐK2AB3¡£Ð¡×éͬѧ¶Ô¹ÌÌå²úÎï½øÐнøÐÐÈçͼËùʾ¶¨Á¿·ÖÎö¡£

¢Ù²Ù×÷IµÄ¾ßÌå²Ù×÷Ϊ____________£»¡°¹ÌÌåÔöÖØ¡±ËµÃ÷¹ÌÌå²úÎïÖж¨º¬ÓеÄÎïÖʵĻ¯Ñ§Ê½Îª_____________¡£

¢Úд³öÓëËáÐÔKMnO4ÈÜÒº·´¸®µÄ½ðÊôÑôÀë×ÓµÄÀë×ӽṹʾÒâͼ:_____________£»ÓÃËáÐÔKMnO4ÈÜҺȥµÎ¶¨¸ÃÈÜҺʱ£¬ÖÕµãÑÕÉ«±ä»¯Îª______________¡£

¢ÛÒÑÖªÔÚËáÖÝÌõ¼þÏÂKMnO4µÄ»¹Ô­²úÎïΪMn2+£¬ÒÔÉÏʵÑéÊý¾Ý·ÖÎö¼ÆËã¹ÌÌå²úÎïÖи÷ÎïÖʼ°ËüÃÇÖ®¼äµÄÎïÖʵÄÁ¿Ö®±ÈΪ_____________¡£

£¨3£©Ð´³ö¸´ÑÎM¼ÓÈÈ·Ö½âµÄ»¯Ñ§·½³Ìʽ:_______________¡£

¡¾´ð°¸¡¿ 3CO+Fe2O32Fe+3CO2 ¼ÓË®Èܽ⣬¹ýÂË£¨Î´¼ÓË®Èܽ⣬²»¸ø·Ö£© Fe ÎÞÉ«±äΪdzºìÉ« n(K2CO3)©Un(Fe)©Un(FeO)=3©U1©U1 2K3Fe(C2O4)3¡¤3H2O3K2CO3+Fe +FeO +5CO2¡ü+4CO¡ü+6H2O

¡¾½âÎö¡¿A¡¢B¾ùΪ³£¼û¶ÌÖÜÆڷǽðÊôÔªËØ£¬¼×¡¢ÒÒ¾ùÖ»º¬A¡¢BÁ½ÖÖÔªËØ£¬¼×ÄÜʹ³ÎÇåµÄʯ»ÒË®±ä»ë×Ç£¬ÍƳöA¡¢BΪ̼¡¢ÑõÔªËØ¡£¸ù¾ÝÌâÒ⣬K3Fe(C2O4)3¡¤3H2O ·Ö½â£¬Ö»ÓÐÒ»ÖÖÑÎÉú³É£¨K2CO3£©£¬Í¬Ê±ÓÐCO2ºÍCO²úÉú¡££¨1£©¢Ù¼×ÊǶþÑõ»¯Ì¼£¬µç×ÓʽΪ: £»¢Ú¹¤ÒµÁ¶ÌúµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ: 3CO+Fe2O32Fe+3CO2£»£¨2£©¢Ù½«¹ÌÌå²úÎï¼ÓË®Èܽ⣬¹ýÂ˺óËùµÃ¹ÌÌåÓëÁòËáÍ­ÈÜÒº·´Ó¦£¬¹ÌÌåÖÊÁ¿ÔöÖØ0.04g£¬ËµÃ÷ÓÐÌúµ¥ÖÊ´æÔÚ£¬»¯Ñ§Ê½ÎªFe£»¢ÚÓëËáÐÔKMnO4ÈÜÒº·´Ó¦µÄ½ðÊôÑôÀë×ÓÑÇÌúÀë×ÓµÄÀë×ӽṹʾÒâͼ: £»ÓÃËáÐÔKMnO4ÈÜҺȥµÎ¶¨¸ÃÈÜҺʱ£¬ÖÕµãÑÕÉ«±ä»¯ÎªÎÞÉ«±äΪdzºìÉ«£»¢Û£»½«µÈÁ¿¹ÌÌå²úÎïÓëÏ¡ÁòËá·´Ó¦ºóÓÃËáÐÔKMnO4ÈÜÒºµÎ¶¨£¬¿ÉÇó³ö£¬ÆäÖÐ0.01molÀ´×ÔÓÚÌúµ¥ÖÊ£¬ËµÃ÷¹ÌÌåÖдæÔÚFeO£»¡£ÔÙ£¬ËùÒÔn(K2CO3)©Un(Fe)©Un(FeO)=3©U1©U1£»£¨3£©¸ù¾ÝÉÏÊö¼ÆË㣬¸´ÑÎM¼ÓÈÈ·Ö½âµÄ»¯Ñ§·½³Ìʽ: 2K3Fe(C2O4)3¡¤3H2O3K2CO3+Fe +FeO +5CO2¡ü+4CO¡ü+6H2O¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿º£ÑóÖ²ÎïÈ纣´ø¡¢º£ÔåÖк¬ÓзḻµÄµâÔªËØ£¬Ö÷ÒªÒԵ⻯ÎïÐÎʽ´æÔÚ¡£ÓÐÒ»»¯Ñ§¿ÎÍâС×éÓú£´øΪԭÁÏÖÆÈ¡ÉÙÁ¿µâµ¥ÖÊ£¬ËûÃǽ«º£´ø×ÆÉճɻң¬ÓÃË®½þÅÝÒ»¶Îʱ¼ä(ÒÔÈõ⻯Îï³ä·ÖÈܽâÔÚË®ÖÐ)£¬µÃµ½º£´ø»ÒÐü×ÇÒº£¬È»ºó°´ÒÔÏÂʵÑéÁ÷³ÌÌáÈ¡µ¥Öʵ⣺

£¨1£©×ÆÉÕº£´øʱ£¬³ýÐèÒªÈý½Å¼Ü¡¢ÄàÈý½ÇÍ⣬»¹ÐèÒªÓõ½µÄʵÑéÒÇÆ÷ÊÇ___________¡£

A.ÉÕ±­ B.±íÃæÃó C.ÛáÛö D.¾Æ¾«µÆ

£¨2£©Ö¸³öÌáÈ¡µâµÄ¹ý³ÌÖÐÓйصÄʵÑé²Ù×÷Ãû³Æ£º¢Ù___________£¬¢Û___________¡£

£¨3£©²Ù×÷¢ÛÖÐËùÓõÄÓлúÊÔ¼ÁÊÇCCl4£¬¼òÊöÑ¡ÔñÆäÀíÓÉ_________________________________¡£

£¨4£©²Ù×÷¹ý³Ì¢Û¿ÉÒÔ·Ö½âΪÈçϼ¸²½£º

A. °ÑÊ¢ÓÐÈÜÒºµÄ·ÖҺ©¶··ÅÔÚÌú¼Ų̈µÄÌúȦÖУ»

B. °Ñ50 mLµâË®ºÍ15 mL CCl4¼ÓÈë·ÖҺ©¶·ÖУ¬²¢¸ÇºÃ²£Á§Èû£»

C. ¼ìÑé·ÖҺ©¶·»îÈûºÍÉÏ¿Ú²£Á§ÈûÊÇ·ñ©Һ£»

D. µ¹×ªÂ©¶·Õñµ´£¬²¢²»Ê±Ðý¿ª»îÈû·ÅÆø£¬×îºó¹Ø±Õ»îÈû£¬°Ñ·ÖҺ©¶··ÅÕý£»

E. Ðý¿ª»îÈû£¬ÓÃÉÕ±­½ÓÊÕÈÜÒº£»

F. ´Ó·ÖҺ©¶·ÉÏ¿Úµ¹³öÉϲãË®ÈÜÒº£»

G. ½«·ÖҺ©¶·ÉϿڵIJ£Á§Èû´ò¿ª»òʹ²£Á§ÈûÉϵݼ²Û»òС¿×¶Ô×¼·ÖҺ©¶·¿ÚÉϵÄС¿×£»

H. ¾²Öᢷֲ㡣

a.ÕýÈ·²Ù×÷²½ÖèµÄ˳ÐòÊÇ(Ìîд¸÷²Ù×÷µÄ±àºÅ×Öĸ)£º______¡ú______¡ú_____¡úA¡ú____¡úG¡úE¡úF¡£_____

b.ÉÏÊöG²½²Ù×÷µÄÄ¿µÄÊÇ£º_________________________________£»

c.×îºóµâµÄÓлúÈÜÒºÊÇͨ¹ý___________»ñµÃ(Ì©¶·ÉÏ¿Ú¡±»ò¡°Â©¶·Ï¿ڡ±)¡£

£¨5£©ÇëÉè¼ÆÒ»ÖÖ¼ìÑéÌáÈ¡µâºóµÄË®ÈÜÒºÖÐÊÇ·ñ»¹º¬Óе¥ÖʵâµÄ¼òµ¥·½·¨£º___________________¡£

£¨6£©´Óº¬µâµÄÓлúÈÜÒºÖÐÌáÈ¡µâºÍ»ØÊÕÓлúÈÜÒº£¬»¹ÐèÒª¾­¹ýÕôÁ󣬹۲ìÏÂͼËùʾʵÑé×°Öã¬Ö¸³öÆäËùÓдíÎóÖ®´¦__________________________________________¡£

£¨7£©½øÐÐÉÏÊöÕôÁó²Ù×÷ʱ£¬Ê¹ÓÃˮԡµÄÓŵãÊÇ_______________________________________£¬×îºó¾§Ì¬µâÔÚ__________Àï¾Û¼¯¡£

¡¾ÌâÄ¿¡¿²¬£¨Pt£©¼°Æ仯ºÏÎïÓÃ;¹ã·º¡£

£¨1£©ÔÚÔªËØÖÜÆÚ±íÖУ¬²¬ÔªËØÓëÌúÔªËØͬ×壬Ôò²¬ÔªËØλÓÚ ______¡£

A£®sÇø B£®pÇø C£®dÇø D£®ds Çø E£®fÇø

£¨2£©¶þÂȶþßÁऺϲ¬ÊÇÓÉPt2+ ¡¢Cl-ºÍßÁऽáºÏÐγɵIJ¬ÅäºÏÎÓÐ˳ʽºÍ·´Ê½Á½ÖÖͬ ·ÖÒì¹¹Ìå¡£¿ÆѧÑо¿±íÃ÷£¬·´Ê½·Ö×ÓºÍ˳ʽ·Ö×ÓÒ»Ñù¾ßÓп¹°©»îÐÔ¡£

¢ÙCl£­µÄÍâΧµç×ÓÅŲ¼Ê½Îª ______¡£

¢ÚßÁषÖ×ÓÊÇ´óÌå»ýƽÃæÅäÌ壬Æä½á¹¹¼òʽÈçÓÒͼËùʾ£¬¸ÃÅäÌåµÄÅäλԭ×ÓÊÇ_____¡£ßÁषÖ×ÓÖУ¬Ì¼¡¢µªÔ­×ӵĹìµÀÔÓ»¯·½Ê½·Ö±ðÊÇ ___¡¢____£¬¸÷ÔªËصĵ縺ÐÔÓÉ´óµ½Ð¡µÄ˳ÐòΪ_______¡£

¢Û¶þÂȶþßÁऺϲ¬·Ö×ÓÖдæÔÚµÄ΢Á£¼ä×÷ÓÃÁ¦ÓÐ _____£¨ÌîÐòºÅ£©¡£

a£®Àë×Ó¼ü b£®Åäλ¼ü c£®½ðÊô¼ü d£®·Ç¼«ÐÔ¼ü e£®Çâ¼ü

¢Ü¶þÂȶþßÁऺϲ¬·Ö×ÓÖУ¬Pt2+µÄÅäλÊýÊÇ4£¬µ«ÊÇÆä¹ìµÀÔÓ»¯·½Ê½²¢²»ÊÇsp3¡£¼òÊöÀíÓÉ£º ________¡£

¢Ý·´Ê½¶þÂȶþßÁऺϲ¬·Ö×ӽṹÈçͼËùʾ£¬¸Ã·Ö×ÓÊÇ _____·Ö×Ó£¨Ñ¡Ìî¡°¼«ÐÔ¡±¡¢¡°·Ç¼«ÐÔ¡±£©¡£

£¨3£©½ðÊô²¬Á¢·½¾§°ûÑØ x¡¢y»òzÖáµÄͶӰͼÈçÓÒͼËùʾ¡£Èô½ðÊô²¬µÄÃܶÈΪd g¡¤cm£­3£¬Ôò¾§°û²ÎÊýa£½_____nm£¨Áгö¼ÆËãʽ¼´¿É£©¡£

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

¾«Ó¢¼Ò½ÌÍø