ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Åð¼°Æ仯ºÏÎïÔÚ¹¤ÒµÉÏÓÐÐí¶àÓÃ;¡£ÒÔÌúÅð¿ó£¨Ö÷Òª³É·ÖΪMg2B2O5¡¤H2OºÍFe3O4£¬»¹ÓÐÉÙÁ¿Fe2O3¡¢FeO¡¢CaO¡¢Al2O3ºÍSiO2µÈ)ΪԭÁÏÖƱ¸ÅðËá(H3BO3)µÄ¹¤ÒÕÁ÷³ÌÈçͼËùʾ£º

ÒÑÖª£ºÅðËáΪ·Ûĩ״¾§Ì壬Ò×ÈÜÓÚË®£¬¼ÓÈȵ½Ò»¶¨Î¶ȿɷֽâΪÎÞË®Îï¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ð´³öMg2B2O5¡¤H2OÓëÁòËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ___¡£ÎªÌá¸ß½þ³öËÙÂÊ£¬³ýÊʵ±Ôö¼ÓÁòËáŨ¶ÈŨ¶ÈÍ⣬»¹¿É²ÉÈ¡µÄ´ëÊ©ÓУ¨Ð´Á½Ìõ£©£º___¡£

£¨2£©¡°½þÔü¡±Öл¹Ê£ÓàµÄÎïÖÊÊÇ£º___£¨Ð´»¯Ñ§Ê½£©¡£

£¨3£©¡°¾»»¯³ýÔÓ¡±ÐèÏȼÓH2O2ÈÜÒº£¬×÷ÓÃÊÇ___¡£È»ºóÔÙµ÷½ÚÈÜÒºµÄpHԼΪ5£¬Ä¿µÄÊÇ___¡£

£¨4£©¡°´ÖÅðËᡱÖеÄÖ÷ÒªÔÓÖÊÊÇ___£¨ÌîÃû³Æ£©¡£

£¨5£©µ¥ÖÊÅð¿ÉÓÃÓÚÉú³É¾ßÓÐÓÅÁ¼¿¹³å»÷ÐÔÄÜÅð¸Ö¡£ÒÔÅðËáºÍ½ðÊôþΪԭÁÏ¿ÉÖƱ¸µ¥ÖÊÅð£¬Óû¯Ñ§·½³Ìʽ±íʾÖƱ¸¹ý³Ì___¡£

¡¾´ð°¸¡¿Mg2B2O5¡¤H2O+2H2SO42MgSO4+2H3BO3 ¼õСÌúÅð¿ó·ÛÁ£¾¶¡¢Ìá¸ß·´Ó¦ÎÂ¶È SiO2ºÍCaSO4 ½«Fe2+Ñõ»¯ÎªFe3+ ʹAl3+ÓëFe3+ÐγÉÇâÑõ»¯Îï¶ø³ýÈ¥ £¨ÆßË®£©ÁòËáþ 2H3BO3B2O3+3H2O¡¢B2O3+3Mg3MgO+2B

¡¾½âÎö¡¿

ÒÔÌúÅð¿ó(Ö÷Òª³É·ÖΪMg2B2O5¡¤H2OºÍFe3O4£¬»¹ÓÐÉÙÁ¿Fe2O3¡¢FeO¡¢CaO¡¢Al2O3ºÍSiO2µÈ)ΪԭÁÏÖƱ¸ÅðËá(H3BO3)£¬ÓÉÁ÷³Ì¿ÉÖª£¬¼ÓÁòËáÈܽ⣬Fe3O4¡¢SiO2²»ÈÜ£¬CaOת»¯ÎªÎ¢ÈÜÓÚË®µÄCaSO4£¬Fe3O4¾ßÓдÅÐÔ£¬¿ÉÒÔ²ÉÓÃÎïÀí·½·¨·ÖÀ룬ÂËÔü1µÄ³É·ÖΪSiO2ºÍCaSO4£»

¡°¾»»¯³ýÔÓ¡±ÐèÏȼÓH2O2ÈÜÒº£¬½«Fe 2+ת»¯ÎªFe 3+£¬µ÷½ÚÈÜÒºµÄpHԼΪ5£¬Ê¹Fe3£«¡¢Al3£«¾ùת»¯Îª³Áµí£¬ÔòÂËÔüΪAl(OH)3¡¢Fe(OH)3£¬È»ºóÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢¹ýÂË·ÖÀë³öH3BO3¡£

(1)Mg2B2O5¡¤H2OÓëÁòËá·´Ó¦Éú³ÉÁòËáþºÍÅðËᣬ»¯Ñ§·½³Ìʽ Mg2B2O5¡¤H2O+2H2SO42MgSO4+2H3BO3£»

ΪÌá¸ß½þ³öËÙÂÊ£¬³ýÊʵ±Ôö¼ÓÁòËáŨ¶ÈŨ¶ÈÍ⣬»¹¿É²ÉÈ¡µÄ´ëÊ©ÓУº¼õСÌúÅð¿ó·ÛÁ£¾¶¡¢Ìá¸ß·´Ó¦Î¶ȣ»

(2)ÀûÓÃFe3O4µÄ´ÅÐÔ£¬¿É½«Æä´Ó¡°½þÔü¡±ÖзÖÀ룬¡°½þÔü¡±Öл¹Ê£ÓàµÄÎïÖÊÊÇSiO2¡¢CaSO4£»

(3)¡°¾»»¯³ýÔÓ¡±ÐèÏȼÓH2O2ÈÜÒº£¬×÷ÓÃÊǽ«Fe2+Ñõ»¯ÎªFe3+ ¡£È»ºóÔÙµ÷½ÚÈÜÒºµÄpHԼΪ5£¬Ä¿µÄÊÇʹAl3+ÓëFe3+ÐγÉÇâÑõ»¯Îï¶ø³ýÈ¥£»

(4)¸ù¾Ý·½³ÌʽMg2B2O5¡¤H2O+2H2SO42MgSO4+2H3BO3£¬·´Ó¦Öл¹Éú³ÉÁòËáþ£¬¡°´ÖÅðËᡱÖеÄÖ÷ÒªÔÓÖÊÊÇ(ÆßË®)ÁòËáþ£»

(5)ÅðËáºÍ½ðÊôþΪԭÁÏ¿ÉÖƱ¸µ¥ÖÊÅð£¬ÅðËáÊÜÈȷֽ⣬Éú³ÉÈýÑõ»¯¶þÅð£¬Ã¾½«Åð»¹Ô­£¬»¯Ñ§·½³Ìʽ£º2H3BO3B2O3+3H2O¡¢B2O3+3Mg3MgO+2B¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÒÑÖª X¡¢Y¡¢Z¡¢Q¡¢R¡¢EÁùÖÖÇ°ËÄÖÜÆÚÔªËØÖУ¬Ô­×ÓÐòÊýX£¼Y£¼Z£¼Q£¼R£¼E£¬Æä½á¹¹»òÐÔÖÊÐÅÏ¢Èç±í£º

ÔªËØ

½á¹¹»òÐÔÖÊÐÅÏ¢

X

Ô­×ÓµÄL²ãÉÏsµç×ÓÊýµÈÓÚpµç×ÓÊý

Y

Ô­×ÓºËÍâµÄL²ãÓÐ3¸öδ³É¶Ôµç×Ó

Z

µØ¿ÇÖк¬Á¿×î¶àµÄÔªËØ

Q

µ¥Öʳ£Î³£Ñ¹ÏÂÊÇÆøÌ壬ԭ×ÓµÄM²ãÉÏÓÐ1¸öδ³É¶ÔµÄpµç×Ó

R

ºËµçºÉÊýÊÇYÓëQµÄºËµçºÉÊýÖ®ºÍ

E

NÄܲãÉÏÖ»ÓÐÒ»¸öµç×Ó£¬K¡¢L¡¢M²ã¾ùÅÅÂúµç×Ó

Çë¸ù¾ÝÐÅÏ¢»Ø´ðÓйØÎÊÌ⣺

(1)д³öÔªËØYµÄÔ­×ÓºËÍâ¼Ûµç×ÓÅŲ¼Í¼£º___________¡£XµÄÒ»ÖÖÇ⻯ÎïÏà¶Ô·Ö×ÓÖÊÁ¿Îª26£¬Æä·Ö×ÓÖеĦҼüÓë¦Ð¼üµÄ¼üÊýÖ®±ÈΪ______¡£

(2)X¡¢Y¡¢ZÈýÖÖÔªËصĵÚÒ»µçÀëÄÜÓɸߵ½µÍµÄÅÅÁÐΪ(дԪËØ·ûºÅ)_____¡£

(3)X¡¢ZÔªËØ·Ö±ðÓëÇâÔªËØÐγɵÄ×î¼òµ¥»¯ºÏÎïÖУ¬·Ðµã½Ï¸ßµÄΪ(д»¯Ñ§Ê½)______£¬Ô­ÒòÊÇ________¡£

(4)XZÓëY2ÊôÓڵȵç×ÓÌ壬д³ö»¯ºÏÎïXZµÄ½á¹¹Ê½£º_____¡£

(5)RµÄÒ»ÖÖÅäºÏÎïµÄ»¯Ñ§Ê½ÎªRCl36H2O¡£ÒÑÖª0.01mol RCl36H2OÔÚË®ÈÜÒºÖÐÓùýÁ¿ÏõËáÒøÈÜÒº´¦Àí£¬²úÉú0.02mol AgCl³Áµí¡£´ËÅäºÏÎï×î¿ÉÄÜÊÇ____¡£

A£®[R(H2O)6]Cl3 B£®[R(H2O)5Cl]Cl2H2O

C£®[R(H2O)4Cl2]Cl2H2O D£®[R(H2O)3Cl3]3H2O

(6)Ïòº¬ÉÙÁ¿ESO4µÄË®ÈÜÒºÖÐÖðµÎµÎÈ백ˮ£¬Éú³ÉÀ¶É«³Áµí£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º__£¬¼ÌÐøµÎ¼Ó°±Ë®ÖÁ¹ýÁ¿£¬³ÁµíÈܽ⣬µÃµ½ÉîÀ¶É«ÈÜÒº£¬Ð´³ö·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º__¡£

¡¾ÌâÄ¿¡¿Ä³Ð£¿Æѧ»î¶¯Ð¡×é¶Ô½Ì²ÄÖС°²â¶¨¿ÕÆøÀïÑõÆøº¬Á¿¡±µÄʵÑ飨Èçͼ1£©½øÐÐÁ˴󵨸Ľø£¬Éè¼ÆÈçͼ2£¨Ñ¡ÓÃÈÝ»ýΪ40mLµÄÊÔ¹Ü×÷Ϊ·´Ó¦ÈÝÆ÷ºÍÈó»¬ÐԺܺõÄ×¢ÉäÆ÷×é×°£©ÊµÑé·½°¸½øÐС£ÇëÄã¶Ô±È·ÖÎöͼ1¡¢Í¼2ʵÑ飬»Ø´ðÏÂÁÐÓйØÎÊÌ⣺

I.Çë½áºÏͼ1»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ö¸³öʵÑéÖÐA¡¢BÒÇÆ÷µÄÃû³Æ£ºA___£»B___¡£

£¨2£©Ð´³öºìÁ×ȼÉÕ·´Ó¦µÄ±í´ïʽ___¡£

£¨3£©ÊµÑéÖеãȼºìÁ׺ó¹Û²ìµ½___£¬´ýºìÁ×ϨÃð²¢ÀäÈ´ÖÁÊÒκ󣬴ò¿ªµ¯»É¼Ð£¬¹Û²ìµ½___¡£

£¨4£©ÓÉ´ËʵÑé¿ÉÍÆ֪ʣÓàÖ÷ÒªÆøÌåµÄ»¯Ñ§ÐÔÖÊÊÇ___(´ðÒ»Ìõ¼´¿É)¡£

II.¸ÄÓÃͼ2ʵÑé·½°¸½øÐУ¬ÊµÑéµÄ²Ù×÷²½ÖèÈçÏ£º¢Ù½«ºìÁ××°ÈëÊÔ¹ÜÖУ¬½«30mLµÄ×¢ÉäÆ÷»îÈûÖÃÓÚ10mL¿Ì¶È´¦£¬²¢°´Í¼2ÖÐËùʾµÄÁ¬½Ó·½Ê½¹Ì¶¨ºÃ£¬ÔÙ½«µ¯»É¼Ð¼Ð½ôÏðƤ¹Ü¢Úµãȼ¾Æ¾«µÆ¢Û³·È¥¾Æ¾«µÆ£¬´ýÊÔ¹ÜÀäÈ´ºóËÉ¿ªµ¯»É¼Ð¢Ü¶ÁȡעÉäÆ÷»îÈûµÄÊý¾Ý¡£

£¨5£©Óëͼ1±È½Ï£¬Í¼2×°ÖõÄÓŵãÊÇ___(´ðÒ»µã¼´¿É)¡£

£¨6£©ÈôÓÃͼ2×°ÖòâµÃµÄÑõÆøÌå»ý·ÖÊýСÓÚ£¬Ôò¿ÉÄܵÄÔ­ÒòÊÇ___(дһÌõ¼´¿É)¡£

£¨7£©Í¼2ʵÑéÖÐ×¢ÉäÆ÷»îÈû½«´Ó10mL¿Ì¶È´¦ÂýÂýÇ°ÒƵ½Ô¼Îª___mL¿Ì¶È´¦²ÅÍ£Ö¹(ºöÂÔµ¼¹ÜÆøÌåÌå»ý)¡£

£¨8£©Èôͼ2×°ÖÃÖиÄÓÃÈÝ»ýΪ80mLÊÔ¹ÜÇÒ²»Ê¹Óüе¯»É¼Ð£¬ÆäËü²Ù×÷¶¼²»±ä£¬ÎªÈ·±£ÊµÑé³É¹¦£¬Ôò¼ÓÈÈÇ°×¢ÉäÆ÷»îÈûÇ°ÑØÖÁÉÙÓ¦µ÷Õûµ½___mL¿Ì¶È´¦(ÌîÕûÊý)¡£

¡¾ÌâÄ¿¡¿ÓÐÒ»º¬NaCl¡¢Na2CO3¡¤10H2OºÍNaHCO3µÄ»ìºÏÎijͬѧÉè¼ÆÈçͼËùʾµÄʵÑé×°Öã¬Í¨¹ý²âÁ¿·´Ó¦²úÉúµÄCO2ºÍH2OµÄÖÊÁ¿£¬À´È·¶¨¸Ã»ìºÏÎïÖи÷×é·ÖµÄÖÊÁ¿·ÖÊý¡£

(1)ʵÑé²½Ö裺

¢Ù°´Í¼(¼Ð³ÖÒÇÆ÷δ»­³ö)×é×°ºÃʵÑé×°Öúó£¬Ê×ÏȽøÐеIJÙ×÷ÊÇ__________¡£

¢Ú³ÆÈ¡ÑùÆ·£¬²¢½«Æä·ÅÈëÓ²Öʲ£Á§¹ÜÖУ¬³ÆÁ¿×°Å¨ÁòËáµÄÏ´ÆøÆ¿CµÄÖÊÁ¿ºÍ×°¼îʯ»ÒµÄUÐιÜDµÄÖÊÁ¿¡£

¢Û´ò¿ª»îÈûK1¡¢K2£¬¹Ø±ÕK3£¬»º»º¹ÄÈë¿ÕÆøÊý·ÖÖÓ£¬ÆäÄ¿µÄÊÇ________¡£

¢Ü¹Ø±Õ»îÈûK1¡¢K2£¬´ò¿ªK3£¬µãȼ¾Æ¾«µÆ¼ÓÈÈÖÁ²»ÔÙ²úÉúÆøÌå¡£×°ÖÃBÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________¡¢________¡£

¢Ý´ò¿ª»îÈûK1£¬»º»º¹ÄÈë¿ÕÆøÊý·ÖÖÓ£¬È»ºó²ðÏÂ×°Öã¬ÔٴγÆÁ¿Ï´ÆøÆ¿CµÄÖÊÁ¿ºÍUÐιÜDµÄÖÊÁ¿¡£

(2)¹ØÓÚ¸ÃʵÑé·½°¸£¬Çë»Ø´ðÏÂÁÐÎÊÌâ¡£

¢ÙÈô¼ÓÈÈ·´Ó¦ºó²»¹ÄÈë¿ÕÆø£¬¶Ô²â¶¨½á¹ûµÄÓ°ÏìÊÇ_______________¡£

¢ÚE´¦¸ÉÔï¹ÜÖÐÊ¢·ÅµÄÒ©Æ·ÊǼîʯ»Ò£¬Æä×÷ÓÃÊÇ_____________£¬Èç¹ûʵÑéÖÐûÓиÃ×°Öã¬Ôò»áµ¼Ö²âÁ¿½á¹ûNaHCO3µÄÖÊÁ¿_____________(Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£

¢ÛÈôÑùÆ·ÖÊÁ¿Îªw g£¬·´Ó¦ºóC¡¢D×°ÖÃÔö¼ÓµÄÖÊÁ¿·Ö±ðΪm1g¡¢m2g£¬Ôò»ìºÏÎïÖÐNa2CO3¡¤10H2OµÄÖÊÁ¿·ÖÊýΪ________(Óú¬w¡¢m1¡¢m2µÄ´úÊýʽ±íʾ)¡£

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

¾«Ó¢¼Ò½ÌÍø