ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿º¬ÓÐ0.01molFeCl3µÄÂÈ»¯Ìú±¥ºÍÈÜÒºÒò¾ÃÖñäµÃ»ë×Ç£¬½«ËùµÃ·Öɢϵ´ÓÈçͼËùʾװÖÃµÄ A ÇøÁ÷ÏòBÇø£¬ÆäÖÐCÇøÊDz»¶Ï¸ü»»ÖеÄÕôÁóË®¡£ÒÑÖªNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

A.ʵÑéÊÒÖƱ¸ Fe(OH)3½ºÌåµÄ·´Ó¦Îª£ºFeCl3+3H2OFe(OH)3(½ºÌå)+3HCl

B.ÂËÖ½ÉϲÐÁôµÄºìºÖÉ«ÎïÖÊΪFe(OH)3¹ÌÌå¿ÅÁ£

C.ÔÚBÇøµÄÉîºìºÖÉ«·ÖɢϵΪFe(OH)3½ºÌå

D.½øÈëCÇøµÄH+µÄÊýĿΪ0.03NA

¡¾´ð°¸¡¿D

¡¾½âÎö¡¿

A£®±¥ºÍFeCl3ÔÚ·ÐË®ÖÐË®½â¿ÉÒÔÖƱ¸½ºÌ壬»¯Ñ§·½³ÌʽΪFeCl3+3H2OFe(OH)3(½ºÌå)+3HCl £¬ÕýÈ·£¬A²»Ñ¡£»

B£®ÂËÖ½ÉϲãµÄ·ÖɢϵÖÐÐü¸¡¿ÅÁ£Ö±¾¶Í¨³£´óÓÚ10-7 mʱ£¬Îª×ÇÒº£¬²»ÄÜ͸¹ýÂËÖ½£¬Òò´ËÂËÖ½ÉϵĺìºÖÉ«¹ÌÌåΪFe(OH)3¹ÌÌå¿ÅÁ££¬ÕýÈ·£¬B²»Ñ¡£»

C£®½ºÌåµÄÖ±¾¶ÔÚ10-9~10- 7mÖ®¼ä£¬¿ÉÒÔ͸¹ýÂËÖ½£¬µ«²»ÄÜ͸¹ý°ë͸Ĥ£¬Òò´ËÔÚÂËÖ½ºÍ°ë͸Ĥ֮¼äµÄB²ã·ÖɢϵΪ½ºÌ壬ÕýÈ·£¬C²»Ñ¡£»

D£®ÈôFe3£«ÍêÈ«Ë®½â£¬Cl£­È«²¿½øÈëCÇø£¬¸ù¾ÝµçºÉÊغ㣬Ôò½øÈëCÇøµÄH£«µÄÊýĿӦΪ0.03NA¡£µ«ÊÇFe3£«²»Ò»¶¨ÍêÈ«Ë®½â£¬Cl£­Ò²²»¿ÉÄÜͨ¹ýÉøÎöÍêÈ«½øÈëCÇø£¬´ËÍâFe(OH)3½ºÌåÁ£×Óͨ¹ýÎü¸½´øÕýµçºÉµÄÀë×ÓÈçH+¶ø´øÓÐÕýµçºÉ£¬Òò´Ë½øÈëCÇøµÄH£«µÄÊýĿСÓÚ0.03NA£¬´íÎó£¬DÑ¡¡£

´ð°¸Ñ¡D¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿¼×´¼ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£ÀûÓúϳÉÆø(Ö÷Òª³É·ÖΪCO¡¢CO2ºÍH2)ÔÚ´ß»¯¼ÁµÄ×÷ÓÃϺϳɼ״¼£¬¿ÉÄÜ·¢ÉúµÄ·´Ó¦ÈçÏ£º

i CO2(g)+ 3H2(g) CH3OH(g)+ H2O(g) H1=Q kJ¡¤mol-1

ii. CO2(g)+ H2(g) CO(g)+H2O(g) H2=+41 kJ¡¤mol-1

iii. CO(g)+2H2(g) CH3OH(g) H3= 99 kJ¡¤mol-1

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Q=_________

£¨2£©Í¼ÖÐÄÜÕýÈ··´Ó³Æ½ºâ³£ÊýK3(·´Ó¦iiiµÄƽºâ³£Êý)Ëæζȱ仯¹ØϵµÄÇúÏßΪ__£¨Ìî×Öĸ£©

£¨3£©ÈçͼΪµ¥Î»Ê±¼äÄÚCO2+H2¡¢CO+ H2¡¢CO/CO2+H2Èý¸öÌõ¼þÏÂÉú³É¼×´¼µÄÎïÖʵÄÁ¿Å¨¶ÈÓëζȵĹØϵ(Èý¸öÌõ¼þÏÂͨÈëµÄCO¡¢CO2ºÍH2µÄÎïÖʵÄÁ¿Å¨¶ÈÏàͬ)¡£490Kʱ£¬¸ù¾ÝÇúÏßa¡¢cÅжϺϳɼ״¼Ê±Ö÷Òª·¢ÉúµÄ·´Ó¦Îª________(Ìî¡°i¡±»ò¡°iii¡±)£»ÓÉÇúÏßa¿ÉÖª£¬¼×´¼µÄÁ¿ÏÈÔö´óºó¼õС£¬ÆäÔ­ÒòÊÇ_________________________________¡£

£¨4£©ÈçͼÊÇÒÔNaOHÈÜҺΪµç½âÖÊÈÜÒºµÄ¼×´¼È¼Áϵç³Ø:µç¼«aµÄ·´Ó¦Ê½Îª____________£¬Èô¸ôĤΪÑôÀë×Ó½»»»Ä¤£¬ÔòÿתÒÆ6molµç×Ó£¬ÈÜÒºÖÐÓÐ_______mol Na+Ïò____________(Ìî¡°Õý¼«Çø¡±»ò¡°¸º¼«Çø")Òƶ¯¡£

£¨5£©CO2¾­´ß»¯¼ÓÇâÒ²¿ÉÒÔÉú³ÉµÍ̼Ìþ£¬Ö÷ÒªÓÐÁ½¸ö¾ºÕù·´Ó¦£º

·´Ó¦I:CO2(g)+4H2(g) CH4(g)+2H2O(g)

·´Ó¦II :2CO2(g)+6H2(g) C2H4(g)+4H2O(g)

ÔÚ1LºãÈÝÃܱÕÈÝÆ÷ÖгäÈë2molCO2ºÍ4molH2²âµÃƽºâʱÓйØÎïÖʵÄÎïÖʵÄÁ¿Ëæζȱ仯ÈçͼËùʾ¡£T1¡æʱ,CO2µÄת»¯ÂÊΪ______¡£T1¡æʱ£¬·´Ó¦IµÄƽºâ³£ÊýK=_________(±£ÁôÈýλÓÐЧÊý×Ö)¡£

¡¾ÌâÄ¿¡¿Èâ¹ðËá()ÊÇÖƱ¸¸Ð¹âÊ÷Ö¬µÄÖØÒªÔ­ÁÏ£¬Ä³Èâ¹ðËá´Ö²úÆ·Öк¬Óб½¼×Ëá¼°¾Û±½ÒÒÏ©£¬¸÷ÎïÖÊÐÔÖÊÈç±í£º

Ãû³Æ

Ïà¶Ô·Ö×ÓÖÊÁ¿

ÈÛµã(¡æ)

·Ðµã(¡æ)

Ë®ÖÐÈܽâ¶È(25¡æ)

±½¼×È©

106

-26

179.62

΢ÈÜ

¾Û±½ÒÒÏ©

104n

83.1~105

240.6

ÄÑÈÜ

Èâ¹ðËá

148

135

300

΢ÈÜ(ÈÈË®ÖÐÒ×ÈÜ)

ʵÑéÊÒÌá´¿Èâ¹ðËáµÄ²½Öè¼°×°ÖÃÈçÏÂ(²¿·Ö×°ÖÃδ»­³ö)£¬ÊԻشðÏà¹ØÎÊÌ⣺
2g´Ö²úÆ·ºÍ30mLÈÈË®µÄ»ìºÏÎïÂËÒº³ÆÖØ

(1)×°ÖÃAÖг¤²£Á§µ¼¹ÜµÄ×÷ÓÃÊÇ_________£¬²½Öè¢Ùʹ±½¼×È© ËæË®ÕôÆûÀ뿪ĸҺ£¬ÉÏÊö×°ÖÃÖÐÁ½´¦ÐèÒª¼ÓÈȵÄÒÇÆ÷ÊÇ____________(ÓÃ×Öĸ A¡¢B¡¢C¡¢D»Ø´ð)¡£

(2)ÒÇÆ÷XµÄÃû³ÆÊÇ_______£¬¸Ã×°ÖÃÖÐÀäˮӦ´Ó___________¿Ú(Ìîa»òb)ͨÈë¡£

(3)²½Öè¢ÚÖУ¬10%NaOHÈÜÒºµÄ×÷ÓÃÊÇ___________£¬ÒÔ±ã¹ýÂ˳ýÈ¥¾Û±½ÒÒÏ©ÔÓÖÊ¡£

(4)²½Öè¢ÜÖУ¬Ö¤Ã÷Ï´µÓ¸É¾»µÄ×î¼Ñ·½·¨ÊÇ________£¬Èô²úÆ·Öл¹»ìÓÐÉÙÁ¿NaCl£¬½øÒ»²½Ìá´¿»ñµÃÈâ¹ðËᾧÌå·½·¨Îª_________________¡£

(5)Èô±¾ÊµÑéÈâ¹ðËá´Ö²úÆ·ÖÐÓи÷ÖÖÔÓÖÊ50%£¬¼Ó¼îÈܽâʱËðʧÈâ¹ðËá10%£¬½áÊøʱ³ÆÖصõ½²úÆ·0.6g£¬Èô²»¼Æ²Ù×÷Ëðʧ£¬Ôò¼ÓÑÎËá·´Ó¦µÄ²úÂÊԼΪ_____£¥(½á¹û¾«È·ÖÁ0.1%)¡£

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

¾«Ó¢¼Ò½ÌÍø