ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ä³»¯Ñ§Ñо¿ÐÔѧϰС×éÐèÅäÖƺ¬ÓÐNH¡¢Cl£­¡¢K£«¡¢SO42-µÄÖ²ÎïÅàÑøÒº500 mL£¬ÇÒÒªÇó¸ÃÅàÑøÒºÖÐc(Cl£­)£½c(K£«)£½c(SO42¡ª)£½0.4 mol¡¤L-1¡£ÊµÑéÊÒÌṩµÄÒ©Æ·ÓУºNH4Cl¡¢KCl¡¢(NH4)2SO4¡¢K2SO4ºÍÕôÁóË®£»ÌṩµÄʵÑéÒÇÆ÷ÓУº¢ÙÒ©³×¡¡¢ÚÍÐÅÌÌìƽ¡¡¢ÛÉÕ±­¡¡¢Ü½ºÍ·µÎ¹Ü¡¡¢ÝÁ¿Í²¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¸ÃÖ²ÎïÅàÑøÒºÖУ¬NHµÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________mol¡¤L-1£»

£¨2£©¸ÃÑо¿Ð¡×éÅäÖƸÃÖ²ÎïÅàÑøҺʱ£¬»¹ÐëÓõ½µÄ²£Á§ÊµÑéÒÇÆ÷ÓÐ__________¡¢__________£»

£¨3£©¼×ͬѧÓÃNH4Cl¡¢(NH4)2SO4ºÍK2SO4ÈýÖÖÎïÖʽøÐÐÅäÖÆ£¬ÔòÐèÁòËá淋ÄÖÊÁ¿Îª__________g£»

£¨4£©ÈôÅäÖƸÃÅàÑøÒºµÄÆäËû²Ù×÷¾ùÕýÈ·£¬ÔòÏÂÁдíÎó²Ù×÷½«Ê¹ËùÅäÖÆÈÜÒºµÄŨ¶ÈÆ«µÍµÄÊÇ__________¡£

a£®½«ÈÜҺתÒÆÖÁÈÝÁ¿Æ¿ºó£¬Î´Ï´µÓÉÕ±­ºÍ²£Á§°ô

b£®½«ÉÕ±­ÄÚµÄÈÜÒºÏòÈÝÁ¿Æ¿ÖÐתÒÆʱ£¬ÈÝÁ¿Æ¿Öл¹ÓÐÉÙÁ¿µÄË®

c£®½«ÉÕ±­ÄÚµÄÈÜÒºÏòÈÝÁ¿Æ¿ÖÐתÒÆʱ£¬Òò²Ù×÷²»µ±Ê¹²¿·ÖÈÜÒº½¦³öÈÝÁ¿Æ¿

d£®ÓýºÍ·µÎ¹ÜÏòÈÝÁ¿Æ¿ÖмÓˮʱ£¬¸©ÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß

e£®Ò¡ÔȺó·¢ÏÖÒºÃæµÍÓÚÈÝÁ¿Æ¿¿Ì¶ÈÏß

¡¾´ð°¸¡¿0.8 500mLÈÝÁ¿Æ¿ ²£Á§°ô 13.2 ac

¡¾½âÎö¡¿

£¨1£©¸ù¾ÝÑôÀë×ÓËù´øÕýµçºÉ×ÜÊýµÈÓÚÒõÀë×ÓËù´ø¸ºµçºÉ×ÜÊý¼ÆËãNHµÄÎïÖʵÄÁ¿Å¨¶È£»£¨2£©¸ù¾ÝÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº²Ù×÷²½Öè·ÖÎöÒÇÆ÷£»(3)ÒªÇó¸ÃÅàÑøÒºÖÐc(K£«)=0.4mol/L£¬ÔòÁòËá¼ØµÄŨ¶ÈÊÇ0.2mol/L £¬c(SO42-)£½0.4 mol¡¤L-1£¬¸ù¾ÝÁòÔªËØÊغ㣬ÐèÒªÁòËá淋ÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.4mol/L-0.2 mol/L=0.2 mol/L£¬¸ù¾Ým=¼ÆËãÁòËá淋ÄÖÊÁ¿¡££¨4£©¸ù¾Ý·ÖÎöÎó²î¡£

£¨1£©ÒÀ¾ÝÈÜÒºÖеçºÉÊغã¹æÂÉ¿ÉÖª: c(Cl-)+2c(SO42-)=c(K+)+C(NH4+)£¬0.4mol/L+2¡Á0.4mol/L=0.4mol/L+c(NH4+)£¬µÃC(NH4+)=0.8mol/L£»£¨2£©¸ù¾ÝÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº²Ù×÷²½Ö裬ÅäÖÆ500 mLÈÜÒº£¬ÐèÒªÔÚ500 mLÈÝÁ¿Æ¿Öж¨ÈÝ£¬ËùÒÔ»¹ÐëÓõ½µÄ²£Á§ÊµÑéÒÇÆ÷ÓÐ500mLÈÝÁ¿Æ¿;¹ÌÌåÈÜÖʵÄÈܽâʱҪÓò£Á§°ô½Á°è¡¢ÏòÈÝÁ¿Æ¿ÖÐתÒÆÈÜҺʱҪÓò£Á§°ôÒýÁ÷£»(3)ÒªÇó¸ÃÅàÑøÒºÖÐc(K£«)=0.4mol/L£¬ÔòÁòËá¼ØµÄŨ¶ÈÊÇ0.2mol/L £¬c(SO42-)£½0.4 mol¡¤L-1£¬¸ù¾ÝÁòÔªËØÊغ㣬ÐèÒªÁòËá淋ÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.4mol/L-0.2 mol/L=0.2 mol/L¡£ÁòËá淋ÄÖÊÁ¿=0.2 mol¡¤L-113.2g¡££¨4£©a£®½«ÈÜҺתÒÆÖÁÈÝÁ¿Æ¿ºó£¬Î´Ï´µÓÉÕ±­ºÍ²£Á§°ô£¬ÈÜÖÊÎïÖʵÄÁ¿Æ«Ð¡£¬ËùµÃÈÜҺŨ¶ÈÆ«µÍ£¬¹ÊÑ¡a£» b£®½«ÉÕ±­ÄÚµÄÈÜÒºÏòÈÝÁ¿Æ¿ÖÐתÒÆʱ£¬ÈÝÁ¿Æ¿Öл¹ÓÐÉÙÁ¿µÄË®£¬¶ÔŨ¶ÈÎÞÓ°Ï죬¹Ê²»Ñ¡b£» c£®½«ÉÕ±­ÄÚµÄÈÜÒºÏòÈÝÁ¿Æ¿ÖÐתÒÆʱ£¬Òò²Ù×÷²»µ±Ê¹²¿·ÖÈÜÒº½¦³öÈÝÁ¿Æ¿£¬ÈÜÖÊÎïÖʵÄÁ¿Æ«Ð¡£¬ËùµÃÈÜҺŨ¶ÈÆ«µÍ£¬¹ÊÑ¡c£» d£®ÓýºÍ·µÎ¹ÜÏòÈÝÁ¿Æ¿ÖмÓˮʱ£¬¸©ÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏߣ¬ÈÜÒºÌå»ýƫС£¬ËùµÃÈÜҺŨ¶ÈÆ«¸ß£¬¹Ê²»Ñ¡d£» e£®Ò¡ÔȺó·¢ÏÖÒºÃæµÍÓÚÈÝÁ¿Æ¿¿Ì¶ÈÏߣ¬¶ÔŨ¶ÈÎÞÓ°Ï죬¹Ê²»Ñ¡e¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ïØÊÇÒ»ÖÖµÍÈÛµã¸ß·ÐµãµÄÏ¡ÓнðÊô£¬ÓС°µç×Ó¹¤Òµ¼¹Áº¡±µÄÃÀÓþ£¬±»¹ã·ºÓ¦Óõ½¹âµç×Ó¹¤ÒµºÍ΢²¨Í¨ÐŹ¤Òµ£»ïØ£¨Ga£©ÓëÂÁλÓÚͬһÖ÷×壬½ðÊôïصÄÈÛµãÊÇ29.8¡æ£¬·ÐµãÊÇ2403¡æ¡£

£¨1£©¹¤ÒµÉÏÀûÓÃGa(l)ÓëNH3(g)ÔÚ1000¡æ¸ßÎÂϺϳɰ뵼Ìå¹ÌÌå²ÄÁϵª»¯ïØ£¨GaN£©£¬Í¬Ê±Éú³ÉÇâÆø£¬Ã¿Éú³Élmol H2ʱ·Å³ö10.3 kJÈÈÁ¿¡£Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ__________________¡£

£¨2£©ÔÚÃܱÕÈÝÆ÷ÖУ¬³äÈëÒ»¶¨Á¿µÄGaÓëNH3·¢Éú·´Ó¦£¬ÊµÑé²âµÃ·´Ó¦Æ½ºâÌåϵÖÐNH3µÄÌå»ý·ÖÊýÓëѹǿPºÍζÈTµÄ¹ØϵÇúÏßÈçͼ1Ëùʾ¡£

¢Ù ͼ1ÖÐAµãºÍCµã»¯Ñ§Æ½ºâ³£ÊýµÄ´óС¹ØϵÊÇ£ºKA_____KC£¬£¨Ìî¡°£¼¡±¡¢¡°£½¡±»ò¡°£¾¡±£©£¬ÀíÓÉÊÇ__________________________________________________¡£

¢Ú ¸Ã·´Ó¦ÔÚT1ºÍP6Ìõ¼þÏÂÖÁ3minʱ´ïµ½Æ½ºâ£¬´Ëʱ¸Ä±äÌõ¼þ²¢ÓÚDµã´¦ÖØдﵽƽºâ£¬H2µÄŨ¶ÈË淴Ӧʱ¼äµÄ±ä»¯Ç÷ÊÆÈçͼ2Ëùʾ£¨3¡«4 minµÄŨ¶È±ä»¯Î´±íʾ³öÀ´£©£¬Ôò¸Ä±äµÄÌõ¼þΪ__________£¨½ö¸Ä±äζȻòѹǿÖеÄÒ»ÖÖ£©¡£

£¨3£©ÈôÓÃѹǿƽºâ³£ÊýKp±íʾ£¬´ËʱBµã¶ÔÓ¦µÄKp£½__________£¨Óú¬P6µÄʽ×Ó±íʾ£©£¨KpΪѹǿƽºâ³£Êý£¬ÓÃƽºâ·Öѹ´úÌæƽºâŨ¶È¼ÆË㣬ÆøÌåƽºâ·Öѹ=×Üѹ¡ÁÆøÌåÌå»ý·ÖÊý£©

£¨4£©µç½â¾«Á¶·¨Ìá´¿ïصľßÌåÔ­ÀíÈçÏ£ºÒÔ´ÖïØ£¨º¬Zn¡¢Fe¡¢CuÔÓÖÊ£©ÎªÑô¼«£¬ÒԸߴ¿ïØΪÒõ¼«£¬ÒÔNaOHÈÜҺΪµç½âÖÊ¡£ÔÚµçÁ÷×÷ÓÃÏÂʹ´ÖïØÈܽâ½øÈëµç½âÖÊÈÜÒº£¬²¢Í¨¹ýijÖÖÀë×ÓǨÒƼ¼Êõµ½´ïÒõ¼«²¢ÔÚÒõ¼«·ÅµçÎö³ö¸ß´¿ïØ¡£

¢ÙÒÑÖªÀë×ÓÑõ»¯ÐÔ˳ÐòΪ£ºZn2£«£¼Ga3£«£¼Fe2£«£¼Cu2£«¡£µç½â¾«Á¶ïØʱÑô¼«ÄàµÄ³É·ÖÊÇ_________¡£

¢ÚÑô¼«ÈܽâÉú³ÉµÄGa3£«ÓëNaOHÈÜÒº·´Ó¦Éú³ÉGaO2-£¬Ð´³öGaO2-ÔÚÒõ¼«·ÅµçµÄµç¼«·´Ó¦Ê½ÊÇ_____________¡£

¡¾ÌâÄ¿¡¿¾Ûµª»¯Áò(SN)xºÍK3C60ÊÇÁ½ÖÖ²»Í¬ÀàÐ͵ij¬µ¼²ÄÁÏ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÔÚ»ù̬KÔ­×ÓÖУ¬ÄÜÁ¿×îµÍµÄ¿Õ¹ìµÀµÄ·ûºÅÊÇ_____________¡£

(2)S¡¢N¡¢KÈýÖÖÔªËصĵÚÒ»µçÀëÄÜÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ____________¡£

(3)(SN)x¾ßÓÐÀàËÆ»ÆÍ­µÄ½ðÊô¹âÔóºÍµ¼µçÐÔ£¬Æä½á¹¹ÈçÏÂͼ£º

ÒÔS2Cl2ΪԭÁÏ¿ÉÖÆÈ¡(SN)x£ºS2Cl2¡ú¡­S4N4 S2N2 (SN)x

¢Ù(SN)xÖÐNÔ­×ÓµÄÔÓ»¯·½Ê½ÊÇ_________£»(SN)xµÄ¾§ÌåÀàÐÍÊÇ______£»

¢ÚS2Cl2µÄ½á¹¹Ê½Îª________£»

¢ÛAgÔªËØλÓÚÖÜÆÚ±íµÚ5ÖÜÆÚ¡¢IB×壬»ù̬AgÔ­×ӵļ۵ç×ÓÅŲ¼Ê½Îª__________¡£

(4)K3C60ÊÇÓÉ×ãÇòÏ©(C60)Óë½ðÊô¼Ø·´Ó¦Éú³ÉµÄÑΡ£

¢ÙÔÚK3C60¾§°ûÖУ¬C603-¶Ñ»ý·½Ê½ÎªÃæÐÄÁ¢·½½á¹¹£¬Ã¿¸ö¾§°ûÖÐÐγÉ4¸ö°ËÃæÌå¿Õ϶ºÍ8¸öËÄÃæÌå¿Õ϶£¬K+Ìî³äÔÚ¿Õ϶ÖС£¾§°ûÖб»K+Õ¼¾ÝµÄ¿Õ϶°Ù·Ö±ÈΪ___________¡£

¢ÚC60Óë½ð¸Õʯ»¥ÎªÍ¬ËØÒìÐÎÐÝ£¬±È½ÏÁ½ÕßµÄÈ۷е㲢˵Ã÷ÀíÓÉ____________¡£

¢ÛC60µÄ½á¹¹ÊÇÒ»ÖÖ¶àÃæÌ壬Èçͼ¡£¶àÃæÌåµÄ¶¥µãÊý¡¢ÃæÊýºÍÀâ±ßÊýµÄ¹Øϵ×ñÑ­Å·À­¶¨ÂÉ£º¶¥µãÊý+ÃæÊý-Àâ±ßÊý=2¡£C60·Ö×ÓÖÐËùº¬µÄÎå±ßÐκÍÁù±ßÐεĸöÊý·Ö±ðΪ_____¡¢______¡£

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

¾«Ó¢¼Ò½ÌÍø