ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÈçͼÊÇÓÉ4¸ö̼ԭ×Ó½áºÏ³ÉµÄ6ÖÖÓлúÎÇâÔ­×ÓûÓл­³ö£©¡£

£¨1£©Ð´³öÓлúÎï(a)µÄϵͳÃüÃû·¨µÄÃû³Æ_____¡£

£¨2£©ÓлúÎï(e)ÓÐÒ»ÖÖͬÀà±ðµÄͬ·ÖÒì¹¹Ì壬ÊÔд³öÆä½á¹¹¼òʽ_____¡£

£¨3£©ÉÏÊöÓлúÎïÖÐÓë(c)»¥ÎªÍ¬·ÖÒì¹¹ÌåµÄÊÇ_____£¨ÌîÐòºÅ£©¡£

£¨4£©Ð´Ò»ÖÖÓë(b)»¥ÎªÍ¬ÏµÎïÇÒËùº¬Ì¼Ô­×ÓÊý×îÉÙµÄÓлúÎïµÄ½á¹¹¼òʽ_____¡£

£¨5£©ÉÏÊöÓлúÎïÖв»ÄÜÓëäåË®·´Ó¦Ê¹ÆäÍÊÉ«µÄÓÐ_____£¨ÌîÐòºÅ£©¡£

£¨6£©Ð´³öÓлúÎï(d)·´Ó¦Éú³É¸ß·Ö×Ó»¯ºÏÎïµÄ»¯Ñ§·½³Ìʽ_____¡£

¡¾´ð°¸¡¿2-¼×»ù±ûÍé CH3-C¡ÔC-CH3 bf CH2=CH2 af nCH2=CH-CH=CH2

¡¾½âÎö¡¿

£¨a£©Îª2-¼×»ù±ûÍ飬£¨b£©Îª2-¼×»ù±ûÏ©£¬£¨c£©Îª2-¶¡Ï©£¬£¨d£©Îª1,3-¶¡¶þÏ©£¬£¨e£©Îª1-¶¡È²£¬£¨f£©Îª»·¶¡Í飬¾Ý´ËÀ´·ÖÎö¸÷СÌâ¼´¿É¡£

£¨1£©ÓлúÎa£©ÊÇÒ»¸öÓÐÖ§Á´µÄ±ûÍ飬Òò´ËΪ2-¼×»ù±ûÍ飻

£¨2£©eΪ1-¶¡È²£¬Í¬Àà±ðµÄͬ·ÖÒì¹¹ÌåҲΪȲ£¬¼´2-¶¡È²£¬Æä½á¹¹¼òʽΪ£»

£¨3£©¿ÉÒÔ´Ó²»±¥ºÍ¶ÈµÄ½Ç¶ÈÀ´¿¼ÂÇ£¬£¨c£©ÖнöÓÐ1¸ö²»±¥ºÍ¶È£¬¶ø£¨b£©£¨f£©Ò²¸÷Ö»ÓÐ1¸ö²»±¥ºÍ¶È£»

£¨4£©£¨b£©Îªµ¥Ï©Ìþ£¬×î¼òµ¥µÄµ¥Ï©ÌþΪÒÒÏ©£¬Æä½á¹¹¼òʽΪ£»

£¨5£©£¨a£©ºÍ£¨f£©ÎÞ²»±¥ºÍ¼ü£¬Òò´Ë²»ÄÜÓëäåË®·¢Éú·´Ó¦£»

£¨6£©d¿ÉÒÔ·¢Éú¼Ó¾Û·´Ó¦µÃµ½¸ß·Ö×Ó»¯ºÏÎÆä·´Ó¦·½³ÌʽΪnCH2=CH-CH=CH2¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Í¨¹ý³Áµí£­Ñõ»¯·¨´¦Àíº¬¸õ·ÏË®£¬¼õÉÙ·ÏÒºÅŷŶԻ·¾³µÄÎÛȾ£¬Í¬Ê±»ØÊÕK2Cr2O7¡£ÊµÑéÊÒ¶Ôº¬¸õ·ÏÒº(º¬ÓÐCr3+¡¢Fe3+¡¢K+¡¢SO42£­¡¢NO3£­ºÍÉÙÁ¿Cr2O72£­)»ØÊÕÓëÔÙÀûÓù¤ÒÕÈçÏ£º

ÒÑÖª£º¢ÙCr(OH)3 + OH£­ = CrO2£­ + 2H2O£» ¢Ú2CrO2£­ + 3H2O2 + 2OH£­ = 2CrO42£­ + 4H2O£»

¢ÛH2O2ÔÚËáÐÔÌõ¼þϾßÓл¹Ô­ÐÔ£¬Äܽ«+6¼ÛCr»¹Ô­Îª+3¼ÛCr¡£

(1)ÈçͼÊÇÓÃKOH¹ÌÌåÅäÖÆ250mL 6 mol¡¤L£­1 KOHÈÜÒºµÄ¹ý³ÌʾÒâͼ¡£

¢ÙÇëÄã¹Û²ìͼʾÅжϣ¬ÆäÖв»ÕýÈ·µÄ²Ù×÷ÓÐ(ÌîÐòºÅ)________£»

¢ÚÆäÖÐÅäÖÆ250 mLÈÜÒºµÄÌå»ýÈÝÆ÷ÊÇ(ÌîÃû³Æ)_________________£»

¢ÛÈç¹ûÓÃͼʾµÄ²Ù×÷ÅäÖÆÈÜÒº£¬ËùÅäÖƵÄÈÜҺŨ¶È½«________(Ìî¡°Æ«´ó¡±»ò¡°Æ«Ð¡¡±)¡£

(2)ÂËÒº¢ñËữǰ£¬½øÐмÓÈȵÄÄ¿µÄÊÇ_________________¡£±ùÔ¡¡¢¹ýÂ˺ó£¬Ó¦ÓÃÉÙÁ¿ÀäˮϴµÓK2Cr2O7£¬ÆäÄ¿µÄÊÇ_______¡£

(3)³ÆÈ¡²úÆ·ÖظõËá¼ØÊÔÑù2.000gÅä³É250mLÈÜÒº£¬È¡³ö25.00mLÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë10mL 2mol¡¤L£­1H2SO4ºÍ×ãÁ¿µâ»¯ÄÆ(¸õµÄ»¹Ô­²úÎïΪCr3+)£¬·ÅÓÚ°µ´¦5min£¬È»ºó¼ÓÈë100mLË®£¬¼ÓÈë3mL µí·Ûָʾ¼Á£¬ÓÃ0.1200 mol¡¤L£­1Na2S2O3±ê×¼ÈÜÒºµÎ¶¨(I2+2S2O32£­=2I£­+S4O62£­)¡£

¢Ùд³öÖظõËá¼ØÑõ»¯µâ»¯ÄƵÄÀë×Ó·½³Ìʽ_________¡£

¢ÚµÎ¶¨ÖÕµãµÄÏÖÏóΪ_________¡£

¢ÛÈôʵÑéÖй²ÓÃÈ¥Na2S2O3±ê×¼ÈÜÒº30.00mL£¬ËùµÃ²úÆ·ÖеÄÖظõËá¼ØµÄ´¿¶ÈΪ_________(ÉèÕû¸ö¹ý³ÌÖÐÆäËüÔÓÖʲ»²ÎÓë·´Ó¦)¡£

¢ÜÈôµÎ¶¨¹ÜÔÚʹÓÃǰδÓÃNa2S2O3±ê×¼ÈÜÒºÈóÏ´£¬²âµÃµÄÖظõËá¼ØµÄ´¿¶È½«_____________(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°²»±ä¡±)¡£

¡¾ÌâÄ¿¡¿¢ñ.ÒûÓÃË®Öк¬ÓÐÉé»áµ¼ÖÂÉéÖж¾£¬Ë®ÌåÖÐÈܽâµÄÉéÖ÷ÒªÒÔAs(¢ó)ÑÇÉéËáÑκÍAs(¢õ)ÉéËáÑÎÐÎʽ´æÔÚ¡£

(1)ÉéÓëÁ×ΪͬһÖ÷×åÔªËØ£¬ÉéµÄÔ­×ÓÐòÊýΪ________________¡£

(2)¸ù¾ÝÔªËØÖÜÆÚÂÉ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ__________________¡£

a.ËáÐÔ£º H2SO4£¾H3PO4£¾H3AsO4

b.Ô­×Ӱ뾶£ºS£¾P£¾As

c.·Ç½ðÊôÐÔ£ºS£¾P£¾As

(3)¹ØÓÚµØÏÂË®ÖÐÉéµÄÀ´Ô´ÓжàÖÖ¼ÙÉ裬ÆäÖÐÒ»ÖÖÈÏΪÊǸ»º¬ÉéµÄ»ÆÌú¿ó(FeS2)±»Ñõ»¯ÎªFe(OH)3£¬Í¬Ê±Éú³ÉSO42-£¬µ¼ÖÂÉéÍÑÀë¿óÌå½øÈëµØÏÂË®¡£FeS2±»O2Ñõ»¯µÄÀë×Ó·½³ÌʽΪ______¡£

¢ò.(4)A¡¢B¡¢C¡¢XÊÇÖÐѧ»¯Ñ§³£¼ûµÄÎïÖÊ£¬A¡¢B¡¢C¾ùÓɶÌÖÜÆÚÔªËØ×é³É£¬×ª»¯¹ØϵÈçͼ£º

ÈôA¡¢B¡¢CÖоùº¬Í¬Ò»ÖÖ³£¼û½ðÊôÔªËØ£¬¸ÃÔªËØÔÚCÖÐÒÔÒõÀë×ÓÐÎʽ´æÔÚ£¬½«A¡¢CµÄË®ÈÜÒº»ìºÏ¿ÉµÃBµÄ°×É«½º×´³Áµí¡£

¢ÙAÖк¬ÓеĽðÊôÔªËØΪ________(дԪËØÃû³Æ)£¬ËüÔÚÖÜÆÚ±íÖеÄλÖÃΪ______¡£

¢Ú¸Ã½ðÊôÔªËصĵ¥ÖÊÓëijÑõ»¯ÎïÔÚ¸ßÎÂÏ·´Ó¦£¬³£ÓÃÓÚº¸½ÓÌú¹ì¼°¶¨Ïò±¬ÆÆ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________¡£

¢ó.Ìú¼°Æ仯ºÏÎïÔÚÉú»îÖÐÓй㷺ӦÓá£

(5)Fe3+»ù̬ºËÍâµç×ÓÅŲ¼Ê½Îª______¡£

(6)µª»¯Ìú¾§ÌåµÄ¾§°û½á¹¹ÈçͼËùʾ¡£¸Ã¾§ÌåÖÐÌú¡¢µªµÄ΢Á£¸öÊýÖ®±ÈΪ______¡£

(7)ijÌúµÄ»¯ºÏÎï½á¹¹¼òʽÈçͼËùʾ£º

¢ÙÉÏÊö»¯ºÏÎïÖÐËùº¬ÓеķǽðÊôÔªËصĵ縺ÐÔÓÉ´óµ½Ð¡µÄ˳ÐòΪ______(ÓÃÔªËØ·ûºÅ±íʾ)¡£

¢ÚÉÏÊö»¯ºÏÎïÖеªÔ­×ÓµÄÔÓ»¯·½Ê½Îª______¡£

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

¾«Ó¢¼Ò½ÌÍø