ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿îѼ°Æ仯ºÏÎïÔÚ»¯¹¤¡¢Ò½Ò©¡¢²ÄÁϵÈÁìÓòÓÐ׏㷺µÄÓ¦Óá£

(1)»ù̬îÑÔ­×ӵļ۵ç×ÓÅŲ¼Ê½Îª_____________£¬ÓëîÑͬÖÜÆÚµÄÔªËØÖУ¬»ù̬ԭ×ÓµÄδ³É¶Ôµç×ÓÊýÓëîÑÏàͬµÄÓÐ____________ÖÖ¡£

(2)îѱȸÖÇá¡¢±ÈÂÁÓ²£¬ÊÇÒ»ÖÖÐÂÐ˵Ľṹ²ÄÁÏ£¬îѵÄÓ²¶È±ÈÂÁ´óµÄÔ­ÒòÊÇ_________¡£

(3)ÔÚŨµÄTiCl3µÄÑÎËáÈÜÒºÖмÓÈëÒÒÃÑ£¬²¢Í¨ÈëHClÖÁ±¥ºÍ£¬¿ÉµÃµ½ÅäλÊýΪ6¡¢×é³ÉΪTiCl3¡¤6H2OµÄÂÌÉ«¾§Ì壬¸Ã¾§ÌåÖÐÁ½ÖÖÅäÌåµÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º5£¬Ôò¸ÃÅäºÏÀë×ӵĻ¯Ñ§Ê½Îª___________¡£

(4)°ë¼ÐÐĽṹ´ß»¯¼ÁMÄÜ´ß»¯ÒÒÏ©¡¢±ûÏ©¡¢±½ÒÒÏ©µÄ¾ÛºÏ£¬Æä½á¹¹ÈçÏÂͼËùʾ¡£

¢Ù×é³ÉMµÄÔªËØÖУ¬µç¸ºÐÔ×î´óµÄÊÇ_________(ÌîÃû³Æ)¡£

¢ÚMÖÐ̼ԭ×ÓµÄÔÓ»¯·½Ê½Îª____________¡£

¢ÛMÖв»º¬________(Ìî´úºÅ)¡£

a£®¦Ð¼ü b£®¦Ò¼ü c£®Àë×Ó¼ü d£®Åäλ¼ü

(5)½ðºìʯ(TiO2)ÊǺ¬îѵÄÖ÷Òª¿óÎïÖ®Ò»¡£Æ侧°û½á¹¹(¾§°ûÖÐÏàͬλÖõÄÔ­×ÓÏàͬ)ÈçͼËùʾ¡£

¢ÙA¡¢B¡¢C¡¢D 4ÖÖ΢Á££¬ÆäÖÐÑõÔ­×ÓÊÇ________(Ìî´úºÅ)¡£

¢ÚÈôA¡¢B¡¢CµÄÔ­×Ó×ø±ê·Ö±ðΪA(0£¬0£¬0)¡¢B(0.69a£¬0.69a£¬c)¡¢C(a£¬a£¬c)£¬ÔòDµÄÔ­×Ó×ø±êΪD(0.19a£¬____£¬___)£»îÑÑõ¼üµÄ¼ü³¤d=______(ÓôúÊýʽ±íʾ)¡£

¡¾´ð°¸¡¿ 3d24s2 3 TiÔ­×ӵļ۵ç×ÓÊý±ÈAl¶à£¬½ðÊô¼ü¸üÇ¿ [TiCl(H2O)5]2+ Ñõ sp2 sp3 c BD 0.81a 0.5c

¡¾½âÎö¡¿ÊÔÌâ·ÖÎö£º(1)îÑÔ­×ÓºËÍâÓÐ22¸öµç×Ó£¬¸ù¾ÝºËÍâµç×ÓÅŲ¼¹æÂÉд»ù̬îÑÔ­×ӵļ۵ç×ÓÅŲ¼Ê½£»»ù̬ԭ×ÓµÄδ³É¶Ôµç×ÓÊýΪ2£»µÚËÄÖÜÆÚÖÐδ³É¶Ôµç×ÓÊýΪ2µÄÔªËØÓÐGe¡¢Se¡¢Ni£¬ÓÐ3ÖÖ£»(2) TiÔ­×ӵļ۵ç×ÓÊýÊÇ4¡¢ÂÁÔ­×ӵļ۵ç×ÓÊýÊÇ3£»(3).ÅäλÊýΪ6£¬Á½ÖÖÅäÌåµÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º5£¬ËùÒÔÅäÌåÖÐÓÐ1¸öÂÈÔ­×Ó¡¢5¸öË®·Ö×Ó£»(4) ¢Ù×é³ÉMµÄÔªËØÓÐTi¡¢C¡¢H¡¢O¡¢Cl£¬·Ç½ðÊôÐÔԽǿµç¸ºÐÔÔ½´ó£»¢ÚMÖÐÓÐË«¼ü̼ºÍµ¥¼ü̼ԭ×ÓÁ½ÖÖ£»¢Ûµ¥¼üΪ¦Ò¼ü¡¢Ë«¼üÖÐ1¸öÊǦҼü¡¢1¸öÊǦмü£¬¸ù¾ÝMµÄ½á¹¹Í¼£¬»¹ÓÐÅäλ¼ü£»(5) ¢Ù¸ù¾Ý¾ù̯ԭÔò£¬¾§°ûÖй²ÓÐÔ­×Ó£¬¾§°ûÖÐÏàͬλÖõÄÔ­×ÓÏàͬ£¬¸ù¾ÝîÑÑõÔ­×Ó±ÈÊÇ1:2·ÖÎö£»¢Ú¸ù¾Ý¾§°û½á¹¹·ÖÎöDÔ­×Ó×ø±ê£»¸ù¾Ýͼʾ£¬

½âÎö£º(1)îÑÔ­×ÓºËÍâÓÐ22¸öµç×Ó£¬»ù̬îÑÔ­×ӵļ۵ç×ÓÅŲ¼Ê½Îª3d24s2£»»ù̬îÑÔ­×ÓµÄδ³É¶Ôµç×ÓÊýΪ2£»µÚËÄÖÜÆÚÖÐδ³É¶Ôµç×ÓÊýΪ2µÄÔªËØ»¹ÓÐ3ÖÖ£¬·Ö±ðÊÇGe¡¢Se¡¢Ni£» (2) TiÔ­×ӵļ۵ç×ÓÊýÊÇ4¡¢ÂÁÔ­×ӵļ۵ç×ÓÊýÊÇ3£¬TiÔ­×ӵļ۵ç×ÓÊý±ÈAl¶à£¬½ðÊô¼ü¸üÇ¿£¬ËùÒÔîѵÄÓ²¶È±ÈÂÁ´ó£»(3).ÅäλÊýΪ6£¬Á½ÖÖÅäÌåµÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º5£¬ËùÒÔÅäÌåÖÐÓÐ1¸öÂÈÔ­×Ó¡¢5¸öË®·Ö×Ó£¬ËùÒÔ¸ÃÅäºÏÀë×ӵĻ¯Ñ§Ê½Îª[TiCl(H2O)5]2+£»(4) ¢Ù×é³ÉMµÄÔªËØÓÐTi¡¢C¡¢H¡¢O¡¢Cl£¬ÆäÖÐOµÄ·Ç½ðÊôÐÔ×îÇ¿£¬·Ç½ðÊôÐÔԽǿµç¸ºÐÔÔ½´ó£¬ËùÒԵ縺ÐÔ×î´óµÄÊÇÑõ£»¢ÚMÖÐÓÐË«¼ü̼ºÍµ¥¼ü̼ԭ×ÓÁ½ÖÖ£¬ËùÒÔMÖÐ̼ԭ×ÓµÄÔÓ»¯·½Ê½Îªsp2¡¢ sp3£»¢Ûµ¥¼üΪ¦Ò¼ü¡¢Ë«¼üÖÐ1¸öÊǦҼü¡¢1¸öÊǦмü£¬¸ù¾ÝMµÄ½á¹¹Í¼£¬»¹ÓÐÅäλ¼ü£¬Ã»ÓÐÀë×Ó¼ü£¬¹ÊÑ¡c£»(5) ¢Ù¸ù¾Ý¾ù̯ԭÔò£¬¾§°ûÖй²ÓÐÔ­×Ó£¬¾§°ûÖÐÏàͬλÖõÄÔ­×ÓÏàͬ£¬¸ù¾ÝîÑÑõÔ­×Ó±ÈÊÇ1:2£¬¿ÉÖªÑõÔ­×ÓÊÇBD£»¢Ú¸ù¾Ý¾§°û½á¹¹£¬ÈôA¡¢B¡¢CµÄÔ­×Ó×ø±ê·Ö±ðΪA(0£¬0£¬0)¡¢B(0.69a£¬0.69a£¬c)¡¢C(a£¬a£¬c)£¬ÔòDÔ­×Ó×ø±êÊÇ(0.19a£¬0.81a£¬0.5c)£»¸ù¾Ýͼʾ£¬£¬Ôòd= ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿»ÆÌú¿óʯµÄÖ÷Òª³É·ÖΪFeS2ºÍÉÙÁ¿FeS(¼ÙÉèÆäËüÔÓÖÊÖв»º¬Ìú¡¢ÁòÔªËØ£¬ÇÒ¸ßÎÂϲ»·¢Éú»¯Ñ§±ä»¯)£¬ËüÊÇÎÒ¹ú´ó¶àÊýÁòË᳧ÖÆÈ¡ÁòËáµÄÖ÷ÒªÔ­ÁÏ¡£Ä³»¯Ñ§ÐËȤС×é¶Ô¸Ã»ÆÌú¿óʯ½øÐÐÈçÏÂʵÑé̽¾¿¡£½«m1g¸Ã»ÆÌú¿óʯµÄÑùÆ··ÅÈëÈçͼװÖÃ(¼Ð³ÖºÍ¼ÓÈÈ×°ÖÃÂÔ)µÄʯӢ¹ÜÖУ¬´Óa´¦²»¶ÏµØ»º»ºÍ¨Èë¿ÕÆø£¬¸ßÎÂ×ÆÉÕ»ÆÌú¿óÑùÆ·ÖÁ·´Ó¦ÍêÈ«¡£Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ4FeS2+11O2=2Fe2O3+8SO2£¬4FeS+7O2=2Fe2O3+4SO2

£¨ÊµÑéÒ»£©²â¶¨ÁòÔªËصĺ¬Á¿

·´Ó¦½áÊøºó£¬½«ÒÒÆ¿ÖеÄÈÜÒº½øÐÐÈçÏ´¦Àí£º

(1)¹ÄÈë¿ÕÆøµÄ×÷ÓÃÊÇ_____________________¡£

(2)·´Ó¦½áÊøºó£¬¸øÒÒÆ¿ÈÜÒºÖмÓÈë×ãÁ¿H2O2ÈÜÒºµÄÄ¿µÄÊÇ___________(Óû¯Ñ§·½³Ìʽ±íʾ)¡£H2O2¿ÉÒÔ¿´³ÉÊÇÒ»ÖÖºÜÈõµÄËᣬд³öÆäÖ÷ÒªµÄµçÀë·½³ÌʽΪ____________________¡£

(3)¸Ã»ÆÌúɰʯÖÐÁòÔªËصÄÖÊÁ¿·ÖÊýΪ____________________(Áгö±í´ïʽ¼´¿É)¡£

£¨ÊµÑé¶þ£©²â¶¨ÌúÔªËصĺ¬Á¿

¢ÙÓÃ×ãÁ¿Ï¡ÁòËáÈܽâʯӢ¹ÜÖеĹÌÌå²ÐÔü¢Ú¼Ó»¹Ô­¼ÁʹÈÜÒºÖеÄFe3+Ç¡ºÃÍêȫת»¯ÎªFe2+ºó£¬¹ýÂË¡¢Ï´µÓ ¢Û½«¹ýÂËҺϡÊÍÖÁ250mL

¢ÜÈ¡25.00mLÏ¡ÊÍÒº£¬ÓÃ0.100mol¡¤L-1µÄËáÐÔKMnO4ÈÜÒºµÎ¶¨

(4)²½Öè¢ÚÖУ¬ÈôÓÃÌú·Û×÷»¹Ô­¼Á£¬ÔòËù²âµÃµÄÌúÔªËصĺ¬Á¿__________(Ìî¡°Æ«´ó¡±¡° ƫС¡± »ò¡°ÎÞÓ°Ï족)¡£

(5)Çëд³ö²½Öè¢ÚÖÐÏ´µÓµÄ·½·¨____________________¡£

(6)ijͬѧһ¹²½øÐÐÁËËĴεζ¨ÊµÑ飬ʵÑé½á¹û¼Ç¼ÈçÏ£º

µÚÒ»´Î

µÚ¶þ´Î

µÚÈý´Î

µÚËÄ´Î

ÏûºÄKMnO4ÈÜÒºÌå»ý/ml

25.00

25.03

20.00

24.97

¸ù¾ÝËù¸øÊý¾Ý£¬¼ÆËã¸ÃÏ¡ÊÍÒºÖÐFe2+µÄÎïÖʵÄÁ¿Å¨¶Èc(Fe2+)=__________¡£

¡¾ÌâÄ¿¡¿¹¤ÒµÉÏÀûÓ÷ÏÄø´ß»¯¼Á(Ö÷Òª³É·ÖΪNi£¬»¹º¬ÓÐÒ»¶¨Á¿µÄZn¡¢Fe¡¢SiO2¡¢CaOµÈ)ÖƱ¸²ÝËáÄø¾§ÌåµÄÁ÷³ÌÈçÏ£º

(1)Çëд³öÒ»ÖÖÄÜÌá¸ß¡°Ëá½þ¡±ËÙÂʵĴëÊ©£º________________________£»ÂËÔüIµÄ³É·ÖÊÇ____________(Ìѧʽ)¡£

(2)³ýÌúʱ£¬¿ØÖƲ»Í¬µÄÌõ¼þ¿ÉÒԵõ½²»Í¬µÄÂËÔüII¡£ÒÑÖªÂËÔüIIµÄ³É·ÖÓëζȡ¢pHµÄ¹ØϵÈçͼËùʾ£º

¢ÙÈô¿ØÖÆζÈ40¡æ¡¢pH=8£¬ÔòÂËÔüIIµÄÖ÷Òª³É·ÖΪ_________________________(Ìѧʽ)¡£

¢ÚÈô¿ØÖÆζÈ80¡æ¡¢pH=2£¬¿ÉµÃµ½»ÆÌú·¯ÄÆ[Na2Fe6(SO4)4(OH)12](ͼÖÐÒõÓ°²¿·Ö)£¬Ð´³öÉú³É»ÆÌú·¯ÄƵÄÀë×Ó·½³Ìʽ£º___________________________________________¡£

(3)ÒÑÖª³ýÌúºóËùµÃ100 mLÈÜÒºÖÐc(Ca2+)=0.01mol¡¤L-1£¬¼ÓÈë100 mL NH4FÈÜÒº£¬Ê¹Ca2+Ç¡ºÃ³ÁµíÍêÈ«¼´ÈÜÒºÖÐc(Ca2+)=1¡Á10-5 mol¡¤L-1£¬ÔòËù¼Óc(NH4F)=_________mol¡¤L-1¡£[ÒÑÖªKsp(CaF2)=5.29¡Á10-9]

(4)¼ÓÈëÓлúÝÍÈ¡¼ÁµÄ×÷ÓÃÊÇ________________________¡£

(5)ij»¯Ñ§¶ÆÄøÊÔ¼ÁµÄ»¯Ñ§Ê½ÎªMxNi(SO4)y(MΪ+1¼ÛÑôÀë×Ó£¬NiΪ+2¼Û£¬x¡¢y¾ùΪÕýÕûÊý)¡£Îª²â¶¨¸Ã¶ÆÄøÊÔ¼ÁµÄ×é³É£¬½øÐÐÈçÏÂʵÑ飺

I£®³ÆÁ¿28.7g¶ÆÄøÊÔ¼Á£¬ÅäÖÆ100 mLÈÜÒºA£»

¢ò£®×¼È·Á¿È¡10.00 mLÈÜÒºA£¬ÓÃ0.40 mol¡¤L-1µÄEDTA(Na2H2Y)±ê×¼ÈÜÒºµÎ¶¨ÆäÖеÄNi2+(Àë×Ó·½³ÌʽΪNi2++H2Y2-=NiY2-+2H+)£¬ÏûºÄEDTA±ê×¼ÈÜÒº25.00mL£»

¢ó£®ÁíÈ¡10.00 mLÈÜÒºA£¬¼ÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬µÃµ½°×É«³Áµí4.66g¡£

¢ÙÅäÖÆ100 mL¶ÆÄøÊÔ¼Áʱ£¬ÐèÒªµÄÒÇÆ÷³ýÒ©³×¡¢ÍÐÅÌÌìƽ¡¢²£Á§°ô¡¢ÉÕ±­¡¢Á¿Í²¡¢½ºÍ·µÎ¹ÜÍ⣬»¹ÐèÒª________________________¡£

¢Ú¸Ã¶ÆÄøÊÔ¼ÁµÄ»¯Ñ§Ê½Îª________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø