ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿´Ó¶¬ÇàÖÐÌáÈ¡³öµÄÓлúÎïA¿ÉÓúϺϳɿ¹½á³¦Ñ×Ò©ÎïY¼°ÆäËû»¯Ñ§Æ·£¬ºÏ³É·ÏßÈçͼ£º

¸ù¾ÝÉÏÊöÐÅÏ¢»Ø´ð£º

(1)Çëд³öYÖк¬Ñõ¹ÙÄÜÍŵÄÃû³Æ_________________________¡£

(2)д³ö·´Ó¦¢ÛµÄ·´Ó¦ÀàÐÍ£º________________________________¡£

(3)д³ö·´Ó¦¢ÙµÄ»¯Ñ§·½³Ìʽ£º_______________________________________¡£

(4)AµÄͬ·ÖÒì¹¹ÌåIºÍJÊÇÖØÒªµÄÒ½Ò©ÖмäÌ壬ÔÚŨÁòËáµÄ×÷ÓÃÏÂIºÍJ·Ö±ðÉú³ÉºÍ£¬¼ø±ðIºÍJµÄÊÔ¼ÁΪ____________________¡£

(5)GµÄͬ·ÖÒì¹¹ÌåÖУ¬Âú×ãÏÂÁÐÌõ¼þµÄÓÐ_____________ÖÖ¡£

¢ÙÄÜ·¢ÉúÒø¾µ·´Ó¦ ¢ÚÄÜÓëÂÈ»¯ÌúÈÜÒº·¢ÉúÏÔÉ«·´Ó¦

ÆäÖк˴Ź²ÕñÇâÆ×ÏÔʾËÄÖÖ²»Í¬ÀàÐ͵ÄÎüÊÕ·å¡£ÇÒÆä·åÃæ»ýÖ®±ÈΪ1:2:2:1µÄ½á¹¹¼ò ʽΪ____(дһÖÖ)¡£

(6)AµÄÁíÒ»ÖÖͬ·ÖÒì¹¹ÌåKÓÃÓںϳɸ߷Ö×Ó²ÄÁÏM()£¬K ¿ÉÓÉL()ÖƵá£Çëд³öÒÔLΪԭÁÏÖƵÃM µÄºÏ³É·ÏßÁ÷³Ìͼ( ÎÞ»úÊÔ¼ÁÈÎÓÃ)¡£Á÷³ÌͼʾÀýÈçÏ£º___________________________

¡¾´ð°¸¡¿ ôÈ»ù¡¢£¨·Ó£©ôÇ»ù È¡´ú·´Ó¦ FeCl3ÈÜÒº»òäåË® 9 »ò»ò HOOCCH2ClNaOOCCH2OHHOOCCH2OH

¡¾½âÎö¡¿±¾ÌâÖ÷Òª¿¼²éÓлúÎïµÄ½á¹¹ÓëÐÔÖÊ¡£

(1)YÖк¬Ñõ¹ÙÄÜÍŵÄÃû³Æ£ºôÈ»ù¡¢ôÇ»ù¡£

(2)·´Ó¦¢ÛÊÇÏõ»¯·´Ó¦£¬·´Ó¦ÀàÐÍ£ºÈ¡´ú·´Ó¦¡£

(3)BÊǼ״¼£¬·´Ó¦¢Ù·¢Éúõ¥ÔÚ¼îÐÔÌõ¼þϵÄË®½â·´Ó¦£¬·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º¡£

(4)Iº¬Óд¼ôÇ»ù£¬Jº¬ÓзÓôÇ»ù£¬ÀûÓ÷ÓÓë´¼ÐÔÖʵIJî±ð¿ÉÒÔ¼ø±ðIºÍJ£¬¼ø±ðÊÔ¼ÁΪFeCl3ÈÜÒº»òäåË®¡£

(5)GÊÇÁÚôÇ»ù±½¼×Ëá¡£¢Ù±íÃ÷¸ÃGµÄͬ·ÖÒì¹¹Ì庬ÓÐÈ©»ù£»¢Ú±íÃ÷¸ÃGµÄͬ·ÖÒì¹¹Ì庬ÓзÓôÇ»ù¡£¶ÔÓÚ¸ÃGµÄͬ·ÖÒì¹¹Ì壬µ±±½»·²àÁ´Îª¡ªOH¡¢¡ªOH¡¢¡ªCHOʱ£¬ÓÐ6Öֽṹ£¬µ±±½»·²àÁ´Îª¡ªOH¡¢¡ªOOCHʱ£¬ÓÐ3Öֽṹ£¬¹²9Öֽṹ¡£ÆäÖк˴Ź²ÕñÇâÆ×ÏÔʾËÄÖÖ²»Í¬ÀàÐ͵ÄÎüÊշ壬ÇÒÆä·åÃæ»ýÖ®±ÈΪ1:2:2:1µÄ½á¹¹¼òʽΪ»ò»ò¡£

(6)ÒÔLΪԭÁÏÖƵÃMµÄºÏ³É·ÏßÁ÷³Ìͼ£ºHOOCCH2ClNaOOCCH2OHHOOCCH2OH¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÈýÂÈÑõÁ×(POCl2)ÊÇÖØÒªµÄ»ù´¡»¯¹¤Ô­ÁÏ£¬¹ã·ºÓÃÓÚÖÆÒ©¡¢È¾»¯¡£ËܽºÖú¼ÁµÈÐÐÒµ¡£Ä³ÐËȤС×éÄ£ÄâPCl3Ö±½ÓÑõ»¯·¨ÖƱ¸POCl3£¬ÊµÑé×°ÖÃÉè¼ÆÈçÏ£º

ÓйØÎïÖʵIJ¿·ÖÐÔÖÊÈçÏÂ±í£º

ÈÛµã/¡æ

·Ðµã/¡æ

ÆäËû

PCl3

-112

75.5

ÓöË®Éú³ÉH3PO3ºÍHCl£¬ÓöO2Éú³ÉPOCl3

POCl3

2

105.3

ÓöË®Éú³ÉH3PO4ºÍHCl£¬ÄÜÈÜÓÚPCl3

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)×°ÖÃAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________________¡£

(2)B×°ÖõÄ×÷Óóý¹Û²ìO2µÄÁ÷ËÙÖ®Íâ¡£»¹ÓÐ__________________________¡£

(3)C×°ÖÿØÖÆ·´Ó¦ÔÚ60¡æ¡«65¡æ½øÐУ¬ÆäÖ÷ҪĿµÄÊÇ_______________________¡£

(4)ͨ¹ý·ð¶û¹þµÂ·¨¿ÉÒԲⶨÈýÂÈÑõÁײúÆ·ÖÐClÔªËغ¬Á¿£¬ÊµÑé²½ÖèÈçÏ£º

¢ñ.È¡xg ²úÆ·ÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë×ãÁ¿NaOH ÈÜÒº£¬´ýÍêÈ«·´Ó¦ºó¼ÓÏ¡ÏõËáÖÁËáÐÔ¡£

¢ò.Ïò׶ÐÎÆ¿ÖмÓÈë0.1000mol/L µÄAgNO3ÈÝÒº40.00mL£¬Ê¹Cl-ÍêÈ«³Áµí¡£

¢ó.ÏòÆäÖмÓÈë2mLÏõ»ù±½£¬ÓÃÁ¦Ò¡¶¯£¬Ê¹³Áµí±íÃæ±»ÓлúÎ︲¸Ç¡£

¢ô.¼ÓÈëָʾ¼Á£¬ÓÃcmol/LNH4SCN ÈÜÒºµÎ¶¨¹ýÁ¿Ag+ÖÁÖյ㣬¼ÇÏÂËùÓà Ìå»ýVmL¡£

ÒÑÖª£ºKsp(AgCl)=3.2¡Á10-10£¬Ksp(AgSCN)=2¡Á10-12

¢ÙµÎ¶¨Ñ¡ÓõÄָʾ¼ÁÊÇ________________(Ìî±êºÅ)¡£

a.FeCl2 b.NH4Fe(SO4)2 c.µí·Û d.¼×»ù³È

¢ÚClÔªËصÄÖÊÁ¿°Ù·Öº¬Á¿Îª(ÁгöËãʽ)____________________¡£

¢Û²½Öè¢ó¼ÓÈëÏõ»ù±½µÄÄ¿µÄÊÇ_________________£¬ÈçÎ޴˲Ù×÷£¬Ëù²âClÔªËغ¬Á¿½«»á____________Ìî¡°Æ«´ó¡±¡° ƫС¡±»ò¡°²»±ä¡±)¡£

¡¾ÌâÄ¿¡¿¿Æѧ¼ÒÌá³ö¡°ÂÌÉ«×ÔÓÉ¡±¹¹Ï룬°Ñ¿ÕÆøÖеÄCO2ת»¯Îª¿ÉÔÙÉúÄÜÔ´¼×´¼£¨CH3OH£©£®¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCO£¨g£©+2H2£¨g£©¡úCH3OH£¨g£©+H2O£¨g£©¡÷H
£¨1£©ÈôÔÚÒ»¸ö¶¨Î¶¨ÈݵÄÈÝÆ÷Öз¢ÉúÉÏÊö·´Ó¦£¬·´Ó¦´ïµ½Æ½ºâµÄ±êÖ¾ÊÇ £®
A.CO2ºÍCH3OHµÄŨ¶ÈÏàµÈ
B.H2µÄ°Ù·Öº¬Á¿±£³Ö²»±ä
C.ÈÝÆ÷ÄÚѹǿ±£³Ö²»±ä
D.3vÕý£¨H2£©=vÄ棨H2O£©
E.ÈÝÆ÷ÖлìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
£¨2£©Èô½«CO2ºÍH2µÄ»ìºÏÆøÌå·Ö³ÉÎåµÈ·Ý£¬½«ËüÃÇ·Ö±ð³äÈëζȲ»Í¬¡¢ÈÝ»ýÏàͬµÄºãÈÝÈÝÆ÷Öз¢ÉúÉÏÊö·´Ó¦£®·´Ó¦Ïàͬʱ¼äºó£¬²âµÃ¼×´¼µÄÌå»ý·ÖÊý £¨CH3OH£©Ó뷴ӦζÈTµÄ¹ØϵÈçͼ1£¬ÔòÉÏÊöCO2ת»¯Îª¼×´¼µÄ·´Ó¦µÄ¡÷H0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±¡¢¡°=¡±£©
£¨3£©ÉÏÊö·´Ó¦ÔÚʵ¼ÊÉú²úÖвÉÓõÄζÈÊÇ300¡æ£¬ÆäÄ¿µÄÊÇ £®
£¨4£©300¡æʱ£¬½«CO2ÓëH2°´1£º3µÄÌå»ý±È³äÈëijÃܱÕÈÝÆ÷ÖУ¬CO2µÄƽºâת»¯ÂÊ£¨a£©ÓëÌåϵ×Üѹǿ£¨P£©µÄ¹ØϵÈçͼ2Ëùʾ£®»Ø´ðÎÊÌ⣺ ¢ÙÈôÆäËûÌõ¼þ²»±ä£¬½«AµãµÄÌå»ýѹËõÖÁÔ­À´µÄÒ»°ë£¬Ò»¶Îʱ¼äºó·´Ó¦Ôٴδﵽƽºâ£¬ÓëԭƽºâÏà±È½ÏÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ £®
A.CO2µÄŨ¶È¼õС
B.Õý·´Ó¦ËÙÂÊÔö´ó£¬Äæ·´Ó¦ËÙÂʼõС
C.CO2ºÍH2µÄÌå»ý±ÈÈÔÈ»ÊÇ1£º3
D.H2µÄÌå»ý·ÖÊý¼õС¢ÚBµãƽºâ»ìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿Îª¡¡22.7g/mol¡¡£¨±£ÁôһλСÊý£©£®

¡¾ÌâÄ¿¡¿CO¡¢CO2ÊÇ»¯Ê¯È¼ÁÏȼÉÕµÄÖ÷Òª²úÎï¡£

(1)½«º¬0.02mol CO2ºÍ0.01 mol COµÄ»ìºÏÆøÌåͨÈëÓÐ×ãÁ¿Na2O2¹ÌÌåµÄÃܱÕÈÝÆ÷ÖУ¬Í¬Ê±²»¶ÏµØÓõç»ð»¨µãȼ£¬³ä·Ö·´Ó¦ºó£¬¹ÌÌåÖÊÁ¿Ôö¼Ó_____g¡£

(2)ÒÑÖª£º2CO(g)+O2(g)==2CO2(g) ¡÷H=-566.0kJ/mol£¬¼üÄÜEo-o=499.0kJ/mol¡£Ôò·´Ó¦:CO(g)+O2(g)CO2(g)+O(g)µÄ¡÷H=_________kJ/mol¡£

(3)ÔÚijÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦:2CO2(g)2CO(g)+O2(g)£¬1molCO2ÔÚ²»Í¬Î¶ÈϵÄƽºâ·Ö½âÁ¿ÈçͼËùʾ¡£

¢ÙºãκãÈÝÌõ¼þÏÂ,Äܱíʾ¸Ã¿ÉÄæ·´Ó¦´ïµ½Æ½ºâ״̬µÄÓÐ___(Ìî×Öĸ)¡£

A.COµÄÌå»ý·ÖÊý±£³Ö²»±ä B.ÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä

C.ÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿±£³Ö²»±ä

D.µ¥Î»Ê±¼äÄÚ£¬ÏûºÄCOµÄŨ¶ÈµÈÓÚÉú³ÉCO2µÄŨ¶È

¢Ú·ÖÎöÉÏͼ£¬Èô1500¡æʱ·´Ó¦´ïµ½Æ½ºâ״̬£¬ÇÒÈÝÆ÷Ìå»ýΪ1L£¬Ôò´Ëʱ·´Ó¦µÄƽºâ³£ÊýK=___(¼ÆËã½á¹û±£Áô1λСÊý)¡£

¢ÛÏò2LµÄºãÈÝÃܱÕÈÝÆ÷ÖгäÈë2molCO2(g)£¬·¢Éú·´Ó¦: 2CO2(g)2CO(g)+O2(g)£¬²âµÃζÈΪT¡æʱ£¬ÈÝÆ÷ÄÚO2µÄÎïÖʵÄÁ¿Å¨¶ÈËæʱ¼äµÄ±ä»¯ÈçÇúÏßIIËùʾ¡£Í¼ÖÐÇúÏßI±íʾÏà¶ÔÓÚÇúÏßII½ö¸Ä±äÒ»ÖÖ·´Ó¦Ìõ¼þºó£¬c(O2)Ëæʱ¼äµÄ±ä»¯£¬Ôò¸Ä±äµÄÌõ¼þÊÇ____£»a¡¢bÁ½µãÓÃCOŨ¶È±ä»¯±íʾµÄ¾»·´Ó¦ËÙÂʹØϵΪva(CO)____(Ìî¡°>¡±¡°<¡±»ò¡°=¡±)vb(CO)¡£

¡¾ÌâÄ¿¡¿»¯Ê¯È¼ÁÏ¿ª²É¡¢¼Ó¹¤¹ý³Ì²úÉúµÄH2S·ÏÆø¿ÉÒÔͨ¹ý¶àÖÖ·½·¨½øÐÐÖÎÀí£®¿ÉÒÔÖÆÈ¡ÇâÆø£¬Í¬Ê±»ØÊÕÁòµ¥ÖÊ£¬¼ÈÁ®¼ÛÓÖ»·±£¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÒÑÖª£º2H2(g)+O2(g)=2H2O(g) ¡÷H1

S(s)+ O2(g)=SO2(g) ¡÷H2

2S(s)S2(g) ¡÷H3

2H2S(g)+SO2(g)=3S(s)+2H2O(g) ¡÷H4

Ôò·´Ó¦2H2S(g)2H2(g)+ S2(g)µÄ¡÷H=_____

(2)¹¤ÒµÉϲÉÖݸßÎÂÈÈ·Ö½âHzSµÄ·½·¨ÖÆÈ¡H2£¬ÔÚĤ·´Ó¦Æ÷ÖзÖÀë³öH2¡£

ÔÚºãÈÝÃܱÕÈÝÆ÷ÖУ¬×èH2SµÄÆðʼŨ¶È¾ùΪ0.009 mol/L¿ØÖƲ»Í¬Î¶ȽøÐÐH2S·Ö½â£º £¬ÊµÑé¹ý³ÌÖвâµÃH2SµÄת»¯ÂÊÈçͼËùʾ¡£ÇúÏßa±íʾH2SµÄƽºâת»¯ÂÊÓëζȵĹØϵ£¬ÇúÏßb±íʾ²»Í¬Î¶ÈF·´Ó¦¾­¹ýÏàͬʱ¼äʱH2SµÄת»¯ÂÊ¡£

¢ÙÔÚ935¡æʱ£¬¸Ã·´Ó¦¾­¹ýt s H2SµÄת»¯ÂÊ´ïµ½PµãÊýÖµ£¬ÔòÔÚtsÄÚÓÃH2Ũ¶È±ä»¯±íʾµÄƽ¾ù·´Ó¦ËÙÂÊv(H2)=________¡£

¢ÚζÈÉý¸ß£¬Æ½ºâÏò____·½ÏòÒƶ¯£¨Ìî¡°Õý·´Ó¦¡±¡°Äæ·´Ó¦¡±£©£¬Æ½ºâ³£Êý____£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©¡£985¡æʱ¸Ã·´Ó¦µÄƽºâ³£ÊýΪ________¡£

¢ÛËæ×ÅH2S·Ö½âζȵÄÉý¸ß£¬ÇúÏßbÖð½¥ÏòÇúÏßa¿¿½ü£¬ÆäÔ­ÒòÊÇ___________¡£

(3)µç½â·¨ÖÎÀíÁò»¯ÇâÊÇÏÈÓÃFeCl3ÈÜÒºÎüÊÕº¬H2SµÄ¹¤Òµ·ÏÆø£¬ËùµÃÈÜÒºÓöèÐԵ缫µç½â£¬Ñô¼«ÇøËùµÃÈÜҺѭ»·ÀûÓá£

¢Ù½øÈëµç½â³ØµÄÈÜÒºÖУ¬ÈÜÖÊÊÇ____¡£

¢ÚÑô¼«µÄµç¼«·´Ó¦Ê½Îª________________¡£

¢Ûµç½â×Ü·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ____________¡£

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

¾«Ó¢¼Ò½ÌÍø