ÌâÄ¿ÄÚÈÝ

ÉèNAΪ°¢·üÙ¤µÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®1 mol [Cu(NH3)4]2+ Öк¬ÓЦҼüµÄÊýĿΪ12NA

B£®0.1molÌú·ÛÓë×ãÁ¿Ë®ÕôÆø·´Ó¦Éú³ÉµÄH2·Ö×ÓÊýĿΪ0.1NA

C£®·Ö×ÓÊýĿΪ0.1NAµÄN2ºÍNH3»ìºÏÆøÌ壬ԭ×Ӽ京ÓеĹ²Óõç×Ó¶ÔÊýĿΪ0.3NA

D£®ÓöèÐԵ缫µç½âCuSO4ÈÜÒºÒ»¶Îʱ¼äºó£¬Èô¼ÓÈë0.05molµÄCu2(OH)2CO3¹ÌÌåÇ¡ºÃÄÜʹÈÜÒº»Ö¸´µ½Ô­À´µÄŨ¶È£¬Ôò¸Ãµç½â¹ý³ÌÖÐ×ªÒÆµç×ÓµÄÊýĿΪ0.2NA

C

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£ºA£®µ¥¼ü¶¼ÊǦҼü¹¹³ÉµÄ£¬1 mol [Cu(NH3)4]2+ Öк¬ÓÐ4mol°±Æø£¬ÁíÍ⵪ԭ×ÓÓëÍ­Àë×ÓÖ®¼ä»¹ÓÐÅäλ½¡£¬Òò´Ë¦Ò¼üµÄÊýĿΪ16NA£¬A´íÎó£»B£®¸ù¾Ý·½³Ìʽ3Fe£«4H2OFe3O4£«4H2¡ü¿ÉÖª0.1molÌú·ÛÓë×ãÁ¿Ë®ÕôÆø·´Ó¦Éú³ÉµÄH2·Ö×ÓÊýĿΪ0.133NA£¬B´íÎó£»C£®µªÆøºÍ°±Æø·Ö×ÓÖоùº¬ÓÐ3¶Ôµç×Ó¶Ô£¬Ôò·Ö×ÓÊýĿΪ0.1NAµÄN2ºÍNH3»ìºÏÆøÌ壬ԭ×Ӽ京ÓеĹ²Óõç×Ó¶ÔÊýĿΪ0.3NA£¬CÕýÈ·£»D£®ÓöèÐԵ缫µç½âCuSO4ÈÜÒºÒ»¶Îʱ¼äºó£¬Èô¼ÓÈë0.05molµÄCu2(OH)2CO3¹ÌÌåÇ¡ºÃÄÜʹÈÜÒº»Ö¸´µ½Ô­À´µÄŨ¶È£¬Õâ˵Ã÷·´Ó¦ÖÐÒõ¼«²úÉúÁË0.1molÍ­ºÍ0.05molÇâÆø£¬Ôò¸Ãµç½â¹ý³ÌÖÐ×ªÒÆµç×ÓµÄÊýĿΪ0.3NA£¬D´íÎ󣬴ð°¸Ñ¡C¡£

¿¼µã£º¿¼²é°¢·ü¼ÓµÂÂÞ³£ÊýµÄ¼ÆËã

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

(16·Ö)Á¬¶þÑÇÁòËáÄÆ(Na2S2O4)£¬ÓֳƱ£ÏÕ·Û£¬ÊÇÓ¡Ë¢¹¤ÒµÖÐÖØÒªµÄ»¹Ô­¼Á¡£Ä³¿ÎÌâС×é½øÐÐÈçÏÂʵÑ飺

I£®¡¾²éÔÄ×ÊÁÏ¡¿

£¨1£©Á¬¶þÑÇÁòËáÄÆ(Na2S2O4)ÊÇÒ»ÖÖ°×É«·ÛÄ©£¬Ò×ÈÜÓÚË®£¬ÄÑÈÜÓÚÒÒ´¼¡£

£¨2£©2Na2S2O4£«4HCl= 4NaCl£«S¡ý£«3SO2¡ü£«2H2O£»Na2S2O3£«2HCl= 2NaCl£«S¡ý£«SO2¡ü£«H2O¡£

II£®¡¾ÖƱ¸·½·¨¡¿

75¡æÊ±½«¼×ËáÄÆºÍ´¿¼î¼ÓÈëÒÒ´¼Ë®ÈÜÒºÖУ¬Í¨ÈëSO2½øÐз´Ó¦£¬Íê³ÉÆä·´Ó¦·½³Ìʽ£º

ÀäÈ´ÖÁ40¡«50¡æ£¬¹ýÂË£¬Óà ϴµÓ£¬¸ÉÔïÖÆµÃNa2S2O4¡£

III£®¡¾Na2S2O4µÄÐÔÖÊ¡¿

£¨1£©Na2S2O4ÈÜÒºÔÚ¿ÕÆøÖÐÒ×±»Ñõ»¯£®¿ÎÌâС×é²â¶¨0.050mol/L

Na2S2O4ÈÜÒºÔÚ¿ÕÆøÖÐpH±ä»¯ÈçͼËùʾ£º0¡«t1¶ÎÖ÷ÒªÏÈÉú³ÉHSO3£­£¬¸ù¾ÝpH±ä»¯Í¼£¬HSO3£­µÄµçÀë³Ì¶È Ë®½â³Ì¶È(Ìî¡°£¼¡±»ò¡°£¾¡±)¡£

¿ÎÌâС×éÍÆ²âNa2S2O4ÈÜÒºÔÚ¿ÕÆøÖÐÒ×±»Ñõ»¯£¬0¡«t1¶Î·¢ÉúÀë×Ó·´Ó¦·½³ÌʽΪ ¡£t3ʱÈÜÒºÖÐÖ÷ÒªÒõÀë×Ó·ûºÅÊÇ ¡£

£¨2£©¸ô¾ø¿ÕÆø¼ÓÈÈNa2S2O4¹ÌÌåÍêÈ«·Ö½â£¬µÃµ½¹ÌÌå²úÎïNa2SO3¡¢Na2S2O3ºÍÆøÌåΪ (Ìѧʽ)£®

ÇëÄãÉè¼ÆÊµÑéÑéÖ¤²úÎïÓÐNa2S2O3´æÔÚ£¬Íê³ÉϱíÖÐÄÚÈÝ¡£

(¹©Ñ¡ÔñµÄÊÔ¼Á£ºÏ¡ÑÎËᡢϡÏõËá¡¢BaCl2ÈÜÒº¡¢KMnO4ÈÜÒº)

ʵÑé²½Öè(²»ÒªÇóд³ö¾ßÌå²Ù×÷¹ý³Ì)

Ô¤ÆÚµÄʵÑéÏÖÏóºÍ½áÂÛ

(14·Ö)Á×ÊǵؿÇÖк¬Á¿½ÏΪ·á¸»µÄ·Ç½ðÊôÔªËØ£¬Ö÷ÒªÒÔÄÑÈÜÓÚË®µÄÁ×ËáÑÎÈçCa3(PO4)2µÈÐÎʽ´æÔÚ¡£ËüµÄµ¥Öʺͻ¯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Óá£

£¨1£©°×Á×(P4)¿ÉÓÉCa3(PO4)2¡¢½¹Ì¿ºÍSiO2ÔÚµç¯ÖиßÎÂ(¡«1550¡æ)ÏÂͨ¹ýÏÂÃæÈý¸ö·´Ó¦¹²È۵õ½¡£

¢Ù4Ca3(PO4)2(s)+10C(s)£½12CaO(s)+2P4(s)+10CO2(g£©¦¤H1=£«Q1kJ¡¤mol-1

¢ÚCaO(s)+SiO2(s)£½CaSiO3(s£© ¦¤H2=£­Q2 kJ¡¤mol-1

¢ÛCO2 (g)£«C(s)£½2CO(g) ¦¤H3=£«Q3kJ¡¤mol-1

ÒÑÖª£ºCaSiO3µÄÈÛµã(1546¡æ)±ÈSiO2µÍ¡£

a¡¢Ð´³öÓÉÁ×Ëá¸Æ¿óÖÆÈ¡°×Á××ܵķ´Ó¦·½³Ìʽ_____________________________¡£

b¡¢ÉÏÊö·´Ó¦ÖÐSiO2ÆðºÎ×÷Óã¿______________________________________¡£

£¨2£©°×Á×ÔÚÈȵÄŨÇâÑõ»¯¼ØÈÜÒºÖÐ᪻¯µÃµ½Ò»ÖÖ´ÎÁ×ËáÑÎ(KH2PO2)ºÍÒ»ÖÖÆøÌå (д»¯Ñ§Ê½)¡£

£¨3£©Á×µÄÖØÒª»¯ºÏÎïNaH2PO4¿Éͨ¹ýH3PO4ÓëNaOHÈÜÒº·´Ó¦»ñµÃ¡£¹¤ÒµÉÏΪÁËʹ·´Ó¦µÄÖ÷Òª²úÎïÊÇNaH2PO4£¬Í¨³£½«pH¿ØÖÆÔÚ Ö®¼ä(ÒÑÖªÁ×ËáµÄ¸÷¼¶µçÀë³£ÊýΪ£ºK1= 7.1¡Á103 K2 = 6.3¡Á108 K3 =4.2¡Á1013 lg7.1¡Ö0.9 lg6.3¡Ö0.8 lg¡Ö0.6) ¡£Na2HPO4 ÈÜÒºÏÔ¼îÐÔ£¬ÈôÏòÈÜÒºÖмÓÈë×ãÁ¿µÄCaCl2 ÈÜÒº£¬ÈÜÒºÔòÏÔËáÐÔ£¬ÆäÔ­ÒòÊÇ (ÓÃÀë×Ó·½³Ìʽ±íʾ)¡£

£¨4£©°×Á×Öж¾ºó¿ÉÓÃCuSO4ÈÜÒº½â¶¾£¬½â¶¾Ô­Àí¿ÉÓÃÏÂÁл¯Ñ§·½³Ìʽ±íʾ£º

11P 4+60CuSO4+96H2O£½20Cu3P+24H3PO4+60H2SO4

60molCuSO4ÄÜÑõ»¯°×Á×µÄÎïÖʵÄÁ¿ÊÇ ¡£

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

¾«Ó¢¼Ò½ÌÍø