ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Í­¼°Æ仯ºÏÎïÔÚÉú²úÉú»îÖÐÓй㷺ӦÓÃ

(1)¹¤ÒµÉÏÒÔ»ÆÍ­¿ó(CuFeS2)ΪԭÁÏ£¬²ÉÓûð·¨ÈÛÁ¶¹¤ÒÕÉú²úÍ­µÄÖмä¹ý³Ì»á·¢Éú·´Ó¦:2Cu2O+Cu2S6Cu+SO2¡ü¸Ã·´Ó¦µÄÑõ»¯¼ÁÊÇ________£¬ÑéÖ¤·´Ó¦²úÉúµÄÆøÌåÊÇSO2µÄ·½·¨ÊÇ___________¡£

(2)½«ÉÙÁ¿Í­Ë¿·ÅÈëÊÊÁ¿µÄÏ¡ÁòËáÖУ¬Î¶ȿØÖÆÔÚ50¡æ£¬¼ÓÈëH2O2£¬·´Ó¦Ò»¶Îʱ¼äºó£¬Éýε½60¡æ£¬ÔÙ·´Ó¦Ò»¶Îʱ¼äºó¿ÉÖƵÃÁòËáÍ­£¬Î¶ȿØÖÆÔÚ50¡æ¡«60¡æµÄÁ½¸öÔ­Òò³ýÁ˼ӿ췴ӦËÙÂÊÍ⣬»¹ÓÐ___________¡£ÔÚCuSO4ÈÜÒºÖмÓÈëÒ»¶¨Á¿µÄNa2SO3ºÍNaClÈÜÒº¼ÓÈÈ£¬Éú³ÉCuC1³Áµí£¬Ð´³öÉú³ÉCuClµÄÀë×Ó·½³Ìʽ______________¡£

(3)¼îʽ̼ËáÍ­ÔÚÓлú´ß»¯¼Á¡¢ÑÌ»ðÖÆÔìºÍÑÕÁÏ¡¢Å©Ò©Éú²úÖÐÓй㷺µÄÓ¦Óá£Ä³¹¤³§ÒÔ»ÔÍ­¿ó(Ö÷Òª³É·ÖΪCu2S£¬º¬ÉÙÁ¿Fe2O3¡¢SiO2µÈÔÓÖÊ)ΪԭÁÏÖƱ¸¼îʽ̼ËáÍ­µÄÁ÷³ÌÈçÏÂ:

¢Ù»ÔÍ­¿óÔÚ½þÈ¡Ç°Òª¸»¼¯Óë·ÛË飬·ÛËéµÄºÃ´¦ÊÇ___________¡£

¢Ú½þÈ¡¹ý³ÌÖеõ½Ò»ÖÖµ¥ÖÊ£¬Ð´³ö½þȡʱÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³Ìʽ___________¡£

¢Ûд³ö¡°³ÁÃÌ¡±·´Ó¦Ê±Àë×Ó·½³Ìʽ___________¡£

¢Ü¡°³ýÌú¡±ÕâÒ»²½·´Ó¦ÔÚ25¡æ½øÐУ¬¼ÓÈ백ˮµ÷½ÚÈÜÒºpHΪ4ºó£¬ÈÜÒºÖÐÍ­Àë×Ó×î´óŨ¶È²»³¬¹ý_________mol/L¡£(ÒÑÖªKsp[Cu(OH)2]=2.2¡Á10-20)

¡¾´ð°¸¡¿ Cu2SºÍCu2O ½«ÆøÌåͨÈëÆ·ºìÈÜÒº£¬ÈÜÒºÍÊÉ«£¬¼ÓÈȻָ´Ô­É« ͬʱ·ÀÖ¹H2O2·Ö½â 2Cu2++2Cl-+SO32-+H2O 2CuCl¡ý+ SO42-+2H+ ¿ÉÌá¸ß½þÈ¡ËÙÂʺͽþÈ¡ÂÊ Cu2S+2MnO2 +4H2SO4= 2CuSO4+S+2MnSO4+4H2O Mn2++HCO3-+NH3 = MnCO3¡ý+NH4+ 2.2

¡¾½âÎö¡¿£¨1£©2Cu2O+Cu2S6Cu+SO2¡ü£¬¸Ã·´Ó¦ÖÐÍ­µÄ»¯ºÏ¼ÛÓÉ+1¼Û±äΪ0¼Û£¬Í­ÔªËØÔڸ÷´Ó¦Öеõç×Ó»¯ºÏ¼Û½µµÍ£¬ËùÒԸ÷´Ó¦ÖеÄÑõ»¯¼ÁÊÇCu2O£¬Cu2S£»SO2¾ßÓÐƯ°×ÐÔ£¬ÄÜʹƷºìÈÜÒºÍÊÉ«£¬¼ÓÈȻָ´Ô­É«£¬ËùÒÔ½«ÆøÌåͨÈëÆ·ºìÈÜÒºÈÜÒºÍÊÉ«£¬¼ÓÈȻָ´Ô­É«£»

£¨2£©ÉÙÁ¿Í­Ë¿·ÅÈëÊÊÁ¿µÄÏ¡ÁòËáÖУ¬Í­Ë¿ÓëÏ¡ÁòËá²»·´Ó¦£¬µ«¼ÓÈëH2O2ºóÓÉÓÚË«ÑõË®¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ÔÚËáÐÔÌõ¼þÏ¿ÉÒÔ°ÑÍ­Ñõ»¯³É¶þ¼ÛÍ­Àë×Ó£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºCu+2H++H2O2=Cu2++2H2O£»ÒòΪH2O2ÔڽϸßζÈʱÈÝÒ׷ֽ⣬ËùÒÔζȿØÖÆÔÚ50¡æ-60¡æ£¬¿ÉÒÔ·ÀÖ¹H2O2·Ö½â£»ÔÚCuSO4ÈÜÒºÖмÓÈëÒ»¶¨Á¿µÄNa2SO3ºÍNaClÈÜÒº¼ÓÈÈ£¬Éú³ÉCuCl³Áµí£¬Í­ÔªËصĻ¯ºÏ¼Û½µµÍ£¬ÔòSO32-±»Cu2+Ñõ»¯ÎªSO42-£¬·´Ó¦Îï³ýCu2+¡¢Cl-¡¢SO32-Í⣬»¹ÓÐH2O£¬²úÎïÓÐH+£¬¸ù¾ÝµÃʧµç×ÓÊغãºÍÔ­×ÓÊغãÅäƽµÃ£º2Cu2++2Cl-+SO32-+H2O 2CuCl¡ý+ SO42-+2H+¹Ê´ð°¸Îª£ºÍ¬Ê±·ÀÖ¹H2O2·Ö½â£»2Cu2++2Cl-+SO32-+H2O 2CuCl¡ý+ SO42-+2H+£»

£¨3£©¢ÙËá½þʱ£¬Í¨¹ý·ÛËé¿óʯ»òÕßÉý¸ßζȻòÕß½øÐнÁ°è»ò¶¼Ôö¼ÓËáµÄŨ¶È£¬¶¼¿ÉÒÔÌá¸ß½þÈ¡ËÙÂÊ£¬¹Ê´ð°¸Îª£ºÌá¸ß½þÈ¡ËÙÂÊ£»
¢Ú·´Ó¦ÎïÊǶþÑõ»¯ÃÌ¡¢Áò»¯Í­ºÍÁòËᣬÉú³ÉÎïÊÇS¡¢ÁòËáÍ­¡¢ÁòËáÃÌ£¬·´Ó¦µÄ·½³ÌʽΪ2MnO2+Cu2S+4H2SO4=S¡ý+2CuSO4+2MnSO4+4H2O£¬¹Ê´ð°¸Îª£º2MnO2+Cu2S+4H2SO4=S¡ý+2CuSO4+2MnSO4+4H2O£»
¢Û¡°³ÁÃÌ¡±£¨Mn2+£©¹ý³Ì£¬Ö÷ÒªÊÇʹMn2+Éú³É³ÁµíMnCO3£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪMn2++HCO3-+NH3 = MnCO3¡ý+NH4+ £¬¹Ê´ð°¸Îª£ºMn2++HCO3-+NH3 = MnCO3¡ý+NH4+£»

¢Ü¸ù¾ÝpHΪ4¿ÉÒÔÖªµÀc(OH-)=10-10mol/L£¬ÔÙ¸ù¾ÝKsp[Cu(OH)2]ÖªµÀc(Cu2+)=Ksp/[c2(OH-)]=mol/L=2.2mol/L£»ËùÒÔ´ð°¸Îª2.2¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿2017Äê11ÔÂ16ÈÕÐÂÎű¨µÀ¡°°¢Ë¹Àû¿µÏø´­¡±ÐÂÒ©»ñÅúÉÏÊУ¬Ïø´­ÂʽµµÍ½ü51%£¬ÓлúÎïZ¿ÉÓÃÓÚÖÎÁÆÏø´­¡¢ÏµÍ³ÐÔºì°ßÀÇ´¯µÈ¡£ÆäºÏ³É·ÏßÈçÏÂͼËùʾ

(1)»¯ºÏÎïXµÄÓÐ_______ÖÖ»¯Ñ§»·¾³²»Í¬µÄÇâÔ­×Ó¡£

(2)ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_______.

A.XÊÇ·¼Ï㻯ºÏÎï B.Ni´ß»¯ÏÂYÄÜÓë5molH2¼Ó³É

C.ZÄÜ·¢Éú¼Ó³É¡¢È¡´ú¼°ÏûÈ¥·´Ó¦ D. lmolZ×î¶à¿ÉÓë5molNaOH·´Ó¦

(3)YÓë¹ýÁ¿µÄäåË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________¡£

(4)X¿ÉÒÔÓÉ_______(дÃû³Æ)ºÍM() ·Ö×Ó¼äÍÑË®¶øµÃ£»Ò»¶¨Ìõ¼þÏ£¬M·¢Éú1¸ö¡ªOHµÄÏûÈ¥·´Ó¦µÃµ½Îȶ¨»¯ºÏÎïN (·Ö×ÓʽΪC6H8O2)£¬ÔòNµÄ½á¹¹¼òʽΪ_____(¼ºÖªÏ©´¼Ê½²»Îȶ¨£¬»á·¢Éú·Ö×ÓÖØÅÅ£¬ÀýÈç: )¡£

(5)YÒ²¿ÉÒÔÓë»··ú±ûÍé)·¢ÉúÀàËÆ·´Ó¦¢ÙµÄ·´Ó¦£¬ÆäÉú³ÉÎïµÄ½á¹¹¼òʽΪ_________(дһÖÖ)£»YµÄͬ·ÖÒì¹¹ÌåºÜ¶àÖÖ£¬ÆäÖÐÓб½»·¡¢±½»·ÉÏÓжþ¸öÈ¡´ú»ù(ÇÒ·ÓôÇ»ùµÄλÖúÍÊýÄ¿¶¼²»±ä)¡¢ÊôÓÚõ¥µÄͬ·ÖÒì¹¹ÌåÓÐ_____ÖÖ¡£

(6)¿ÉÓɺÍÓлúÎïH·Ö×Ó¼äÍÑÈ¥Ò»·Ö×ÓË®¶øµÃ£¬ÔòHµÄÃû³ÆΪ________£»Í¨³£²ÉÓÃΪԭÁÏºÏ³É £¬Çë²ÎÕÕÌâĿ·ÏßͼÉè¼Æ¸ÃºÏ³É·Ïß(ÎÞ»úÊÔ¼ÁÈÎÑ¡) _______¡£(ºÏ³É·Ïß³£Óõıíʾ·½Ê½Îª: )

¡¾ÌâÄ¿¡¿Ç°ËÄÖÜÆÚÔªËØA¡¢B¡¢C¡¢D¡¢E¡¢F£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐAºÍBͬÖÜÆÚ£¬¹Ì̬µÄAB2ÄÜÉý»ª£»CºÍEÔ­×Ó¶¼ÓÐÒ»¸öδ³É¶Ôµç×Ó£¬C+±ÈE-ÉÙÒ»¸öµç×Ӳ㣬EÔ­×ӵõ½Ò»¸öµç×Óºó3p¹ìµÀÈ«³äÂú£»D×î¸ß¼ÛÑõ»¯ÎïÖÐDµÄÖÊÁ¿·ÖÊýΪ40%£¬ÇÒºËÄÚÖÊ×ÓÊýµÈÓÚÖÐ×ÓÊý£»FΪºìÉ«µ¥ÖÊ£¬ÓÐF+ºÍF2+Á½ÖÖÀë×Ó¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÔªËص縺ÐÔ£ºD____E (Ì¡¢£¼»ò£½)£»

£¨2£©B¡¢Cµ¥ÖÊÈÛµãB_____C(Ì¡¢£¼»ò£½)£»

£¨3£©AE4ÖÐAÔ­×ÓÔÓ»¯¹ìµÀ·½Ê½Îª£º________ÔÓ»¯£»Æä¹Ì̬¾§ÌåÀàÐÍΪ_____________£»

£¨4£©Ç⻯ÎïµÄ·Ðµã£ºB±ÈD¸ßµÄÔ­Òò______________£»

£¨5£©FµÄºËÍâµç×ÓÅŲ¼Ê½Îª____________________________£»ÏòFµÄÁòËáÑÎÖÐÖðµÎ¼ÓÈ백ˮÏȲúÉú³Áµí£¬ºó³ÁµíÈܽâΪÉîÀ¶É«ÈÜÒº£¬¼ÓÈëÒÒ´¼»áÎö³öÉîÀ¶É«¾§Ì壬¸Ã¾§ÌåµÄ»¯Ñ§Ê½Îª_______£¬ÆäÖйØÓڸþ§ÌåÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ_____________________¡£

A£®¼ÓÈëÒÒ´¼µÄÄ¿µÄÊǽµµÍÈܼÁµÄ¼«ÐÔ£¬´Ùʹ¾§ÌåÎö³ö

B£®FÓëNH3Ö®¼äµÄ»¯Ñ§¼üΪÀë×Ó¼ü

C£®¸ÃÅäºÏÎᄃÌåÖУ¬NÊÇÅäλԭ×Ó£¬NH3ΪÈý½Ç׶ÐÍ

D£®ÅäÀë×ÓÄÚNÔ­×ÓÅÅÁгÉΪƽÃæÕý·½ÐΣ¬ÔòÆäÖÐFÀë×ÓÊÇsp3ÔÓ»¯

E£®Ïò¸Ã¾§ÌåµÄË®ÈÜÒºÖмÓÈëŨBaCl2ÈÜÒºÓа×É«³ÁµíÉú³É

£¨6£©ÔªËØXµÄij¼Û̬ÒõÀë×ÓXn-ÖÐËùÓеç×ÓÕýºÃ³äÂúKºÍLµç×Ӳ㣬CnX¾§ÌåµÄ×îС½á¹¹µ¥ÔªÎªÁ¢·½Ì壬½á¹¹ÈçͼËùʾ¡£¸Ã¾§ÌåÖÐÿ¸öXn-±»________¸öµÈ¾àÀëµÄC+Àë×Ó°üΧ¡£ÒÑÖª¸Ã¾§ÌåµÄÃܶÈΪ¦Ñg£®cm-3£¬°¢·ü¼ÓµÂÂÞ³£ÊýΪNA£¬CnXµÄĦ¶ûÖÊÁ¿ÎªM g/mol£¬C+ºÍXn-¼äµÄ×î¶Ì¾àÀëÊÇ_____________nm¡££¨Áгö¼ÆËãʽ¼´¿É£©

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

¾«Ó¢¼Ò½ÌÍø