ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿îѹ¤³§TiCl4Ñ̳¾Öк¬ÓдóÁ¿µÄScCl3¡¢MgCl2¼°SiO2С¿ÅÁ£µÈÎïÖÊ£¬Ä³Ñо¿ËùÀûÓÃÉÏÊöÑ̳¾»ØÊÕSc2O3£¬²¢ÖƱ¸îÑ°×·Û£¨TiO2£©£¬Æ乤ÒÕÁ÷³ÌÈçͼËùʾ£º

£¨1£©ÔÚ¿ÕÆøÖÐ×ÆÉÕ²ÝËáîÖ¼´¿ÉµÃµ½Ñõ»¯îÖ£¨Sc2O3£©£¬Æ仯ѧ·´Ó¦·½³ÌʽΪ_____¡£

£¨2£©¡°Ë®ÏࡱµÄÖ÷ÒªÀë×ÓÓÐTiOCl42£­¡¢H+¡¢Cl£­¼°Mg2+£¬Ð´³ö¼ÓÈë´óÁ¿µÄË®²¢¼ÓÈÈ·¢ÉúµÄÏà¹ØÀë×Ó·´Ó¦·½³Ìʽ_____¡£

£¨3£©Ëá½þ¹ý³ÌÖУ¬ÉÔ¹ýÁ¿µÄÑÎËáµÄ×÷ÓóýÈܽâÎüÊÕÑ̳¾Í⣬ÁíÍ⻹ÓеÄ×÷ÓÃÊÇ_______¡£

£¨4£©îѵÄÒ±Á¶Ð·¨Êǽ£Çŵç½â·¨£¨Èçͼ£©¡£ÒÔº¬ÉÙÁ¿CaCl2µÄCaOÈÛÈÚÎï×÷Ϊ½éÖÊ£¬µç½âʱ¡£ÔÚÒõ¼«Éú³ÉµÄCa½øÒ»²½»¹Ô­TiO2µÃîÑ¡£ÀûÓÃÖÐѧËùѧ֪ʶ¿ÉÒÔÔ¤²âCaCl2µÄ×÷Óðüº¬ÔöÇ¿µ¼µçÐÔ¼°______¡£

¡¾´ð°¸¡¿3O2+2Sc2(C2O4)32Sc2O3+12CO2 TiOCl42-+(1+x)H2OTiO2¡¤xH2O+2H++4Cl- ΪÁË·ÀÖ¹TiOCl42-¡¢Mg2+¡¢Se3+Ë®½â ½µµÍCaOµÄÈ۵㣬½ÚÔ¼ÄÜÁ¿

¡¾½âÎö¡¿

(1)ÔÚ¿ÕÆøÖÐ×ÆÉÕ²ÝËáîÖ¼´¿ÉµÃµ½Ñõ»¯îÖ(Sc2O3)£¬Í¬Ê±²ÝËá¸ùÀë×Ó±»Ñõ»¯Éú³É¶þÑõ»¯Ì¼£¬·´Ó¦µÄ»¯Ñ§·´Ó¦·½³ÌʽΪ3O2+2Sc2(C2O4)32Sc2O3+12CO2£»

(2)¸ù¾ÝÁ÷³Ìͼ£¬îѹ¤³§TiCl4Ñ̳¾Öк¬ÓдóÁ¿µÄScCl3¡¢MgCl2¼°SiO2С¿ÅÁ£µÈÎïÖÊ£¬Ëá½þºó£¬¹ýÂ˳ýÈ¥Á˶þÑõ»¯¹è£¬ÔÙÓÃÓлúÈܼÁÝÍÈ¡£¬¡°Ë®ÏࡱÖÐÖ÷ÒªÀë×ÓÓÐTiOCl42-¡¢H+¡¢Cl-¼°Mg2+£¬ÔÚË®ÏàÖмÓÈë´óÁ¿µÄË®²¢¼ÓÈÈ´Ù½øTiOCl42-Ë®½âÉú³ÉTiO2¡¤xH2O£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪTiOCl42-+(1+x)H2OTiO2¡¤xH2O+2H++4Cl-£»

(3) Ëá½þ¹ý³ÌÖУ¬ÉÔ¹ýÁ¿µÄÑÎËáµÄ×÷ÓóýÈܽâÎüÊÕÑ̳¾Í⣬ÁíÍ⻹ÓеÄ×÷ÓÃÊÇΪÁË·ÀÖ¹TiOCl42-¡¢Mg2+¡¢Se3+Ë®½â£»

(4)µç½âÖÊΪ»ìÓÐÂÈ»¯¸ÆµÄÈÛÈÚµÄÑõ»¯¸Æ£¬¼ÓÈëÂÈ»¯¸Æ¿ÉÒÔÔöÇ¿µç½âÖʵĵ¼µçÐÔ£¬Í¬Ê±¿ÉÒÔ½µµÍCaOµÄÈ۵㣬½ÚÔ¼ÄÜÁ¿¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ä³»¯Ñ§Ð¡×éͬѧÓÃÏÂÁÐ×°ÖúÍÊÔ¼Á½øÐÐʵÑ飬̽¾¿O2ÓëKIÈÜÒº·¢Éú·´Ó¦µÄÌõ¼þ¡£

¹©Ñ¡ÊÔ¼Á£º30%H2O2ÈÜÒº¡¢0.1mol/L H2SO4ÈÜÒº¡¢MnO2¹ÌÌå¡¢KMnO4¹ÌÌå

£¨1£©Ð¡×éͬѧÉè¼Æ¼×¡¢ÒÒ¡¢±ûÈý×éʵÑ飬¼Ç¼ÈçÏÂ

²Ù×÷

ÏÖÏó

¼×

ÏòIµÄ׶ÐÎÆ¿ÖмÓÈë___£¬ÏòIµÄ____ÖмÓÈë30%H2O2ÈÜÒº£¬Á¬½ÓI¡¢¢ó£¬´ò¿ª»îÈû

IÖвúÉúÎÞÉ«ÆøÌå²¢°éËæ´óÁ¿°×Îí£»¢óÖÐÓÐÆøÅÝð³ö£¬ÈÜҺѸËÙ±äÀ¶

ÒÒ

Ïò¢òÖмÓÈëKMnO4¹ÌÌ壬Á¬½Ó¢ò¡¢¢ó£¬µãȼ¾Æ¾«µÆ

¢óÖÐÓÐÆøÅÝð³ö£¬ÈÜÒº²»±äÀ¶

±û

Ïò¢òÖмÓÈëKMnO4¹ÌÌ壬¢óÖмÓÈëÊÊÁ¿0.1mol/LH2SO4ÈÜÒº£¬Á¬½Ó¢ò¡¢¢ó£¬µãȼ¾Æ¾«µÆ

¢óÖÐÓÐÆøÅÝð³ö£¬ÈÜÒº±äÀ¶

£¨2£©±ûʵÑéÖÐO2ÓëKIÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ____¡£

£¨3£©¶Ô±ÈÒÒ¡¢±ûʵÑé¿ÉÖª£¬O2ÓëKIÈÜÒº·¢Éú·´Ó¦µÄÊÊÒËÌõ¼þÊÇ___¡£Îª½øÒ»²½Ì½¾¿¸ÃÌõ¼þ¶Ô·´Ó¦ËÙÂʵÄÓ°Ï죬¿É²ÉÈ¡µÄʵÑé´ëÊ©ÊÇ___¡£

£¨4£©Óɼס¢ÒÒ¡¢±ûÈýʵÑéÍƲ⣬¼×ʵÑé¿ÉÄÜÊÇIÖеİ×ÎíʹÈÜÒº±äÀ¶¡£Ñ§Éú½«IÖвúÉúµÄÆøÌåÖ±½ÓͨÈëÏÂÁÐ_____(Ìî×Öĸ)ÈÜÒº£¬Ö¤Ã÷ÁË°×ÎíÖк¬ÓÐH2O2¡£

A£®ËáÐÔKMnO4 B£®FeC12 C£®Na2S D£®Æ·ºì

¡¾ÌâÄ¿¡¿(1)ÒÑÖªKsp[Cu(OH)2]£½2.2¡Á10£­20£¬Ksp[Fe(OH)3]£½2.6¡Á10£­39¡£³£ÎÂÏ£¬Ä³ËáÐÔCuCl2ÈÜÒºÖк¬ÓÐÉÙÁ¿µÄFeCl3£¬ÎªÁ˵õ½´¿¾»µÄCuCl2¡¤2H2O¾§Ì壬Ӧ¼ÓÈë___________(ÌîÑõ»¯ÎïµÄ»¯Ñ§Ê½)£¬µ÷½ÚÈÜÒºµÄpH£½4£¬Ê¹ÈÜÒºÖеÄFe3£«×ª»¯ÎªFe(OH)3³Áµí£¬´ËʱÈÜÒºÖеÄc(Fe3£«)£½________¡£¹ýÂ˺󣬽«ËùµÃÂËÒºµÍÎÂÕô·¢¡¢Å¨Ëõ½á¾§£¬¿ÉµÃµ½CuCl2¡¤2H2O¾§Ìå¡£

(2)ij̼Ëظֹø¯ÄÚË®¹¸µÄÖ÷Òª³É·ÖÊÇ̼Ëá¸Æ¡¢ÁòËá¸Æ¡¢ÇâÑõ»¯Ã¾¡¢ÌúÐâ¡¢¶þÑõ»¯¹èµÈ¡£Ë®¹¸Ð輰ʱÇåÏ´³ýÈ¥¡£ÇåÏ´Á÷³ÌÈçÏ£º

¢ñ.¼ÓÈëNaOHºÍNa2CO3»ìºÏÒº£¬¼ÓÈÈ£¬½þÅÝÊýСʱ£»

¢ò.·Å³öÏ´µÓ·ÏÒº£¬ÇåË®³åÏ´¹ø¯£¬¼ÓÈëÏ¡ÑÎËáºÍÉÙÁ¿NaFÈÜÒº£¬½þÅÝ£»

¢ó.ÏòÏ´µÓÒºÖмÓÈëNa2SO3ÈÜÒº£»

¢ô.ÇåÏ´´ï±ê£¬ÓÃNaNO2ÈÜÒº¶Û»¯¹ø¯¡£

¢ÙÓÃÏ¡ÑÎËáÈܽâ̼Ëá¸ÆµÄÀë×Ó·½³ÌʽÊÇ_____________________________¡£

¢ÚÒÑÖª£º25 ¡æʱÓйØÎïÖʵÄÈܶȻý

ÎïÖÊ

CaCO3

CaSO4

Mg(OH)2

MgCO3

Ksp

2.8¡Á10£­9

9.1¡Á10£­6

1.8¡Á10£­11

6.8¡Á10£­6

¸ù¾ÝÊý¾Ý£¬½áºÏ»¯Ñ§Æ½ºâÔ­Àí½âÊÍÇåÏ´CaSO4µÄ¹ý³Ì________________¡££¨ÓÃÈܽâƽºâ±í´ïʽºÍ±ØÒªµÄÎÄ×ÖÐðÊö¼ÓÒÔ˵Ã÷£©£»ÔÚ²½Öè¢ñ½þÅݹý³ÌÖл¹»á·¢Éú·´Ó¦MgCO3(s)£«2OH£­(aq)Mg(OH)2(s)£«CO32-(aq)£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK£½________(±£ÁôÁ½Î»ÓÐЧÊý×Ö)¡£

¢Û²½Öè¢óÖУ¬¼ÓÈëNa2SO3ÈÜÒºµÄÄ¿µÄÊÇ_______________________________¡£

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

¾«Ó¢¼Ò½ÌÍø