ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿Ë®ÖеÄÈܽâÑõ£¨DO£©µÄ¶àÉÙÊǺâÁ¿Ë®ÌåË®ÖʵÄÖØÒªÖ¸±ê¡£Ä³»¯Ñ§Ð¡×é²â¶¨Ä³ºÓÁ÷ÖÐÑõµÄº¬Á¿£¬¾²éÔÄÓйØ×ÊÁÏÁ˽⵽ÈܽâÑõ²â¶¨¿ÉÓá°µâÁ¿·¨¡±£¬
¢ñ£®ÓÃÒÑ׼ȷ³ÆÁ¿µÄÁò´úÁòËáÄÆ£¨Na2S2O3£©¹ÌÌåÅäÖÆÒ»¶¨Ìå»ýµÄcmol/L±ê×¼ÈÜÒº£»
¢ò£®ÓÃË®ÑùÆ¿È¡ºÓÁ÷ÖÐË®Ñùv1mL²¢Á¢¼´ÒÀ´ÎÐò×¢Èë1.0mLMnCl2ÈÜÒººÍ1.0mL¼îÐÔKIÈÜÒº£¬Èû½ôÆ¿Èû£¨Æ¿ÄÚ²»×¼ÓÐÆøÅÝ£©£¬·´¸´Õðµ´ºó¾²ÖÃÔ¼1Сʱ£»
¢ó£®ÏòË®ÑùÆ¿ÖмÓÈë1.0mLÁòËáÈÜÒº£¬Èû½ôÆ¿Èû£¬Õñµ´Ë®ÑùÆ¿ÖÁ³ÁµíÈ«²¿Èܽ⣬´ËʱÈÜÒº±äΪ»ÆÉ«£» ¢ô£®½«Ë®ÑùÆ¿ÄÚÈÜҺȫÁ¿µ¹Èë׶ÐÎÆ¿ÖУ¬ÓÃÁò´úÁòËáÄƱê×¼ÈÜÒºµÎ¶¨£»
V£®´ýÊÔÒº³Êµ»ÆÉ«ºó£¬¼Ó1mLµí·ÛÈÜÒº£¬¼ÌÐøµÎ¶¨µ½Öյ㲢¼Ç¼ÏûºÄµÄÁò´úÁòËáÄÆÈÜÒºÌå»ýΪv2¡£
ÒÑÖª£ºI2 +2Na2S2O3 =2NaI+Na2S4O6
£¨1£©Ôڵ樻·½ÚÖÐʹÓõÄÒÇÆ÷Óеζ¨¹Ü¼Ð¡¢Ìú¼Ų̈¡¢ÉÕ±¡¢×¶ÐÎÆ¿ºÍ________________________¡£
£¨2£©ÔÚ²½Öè¢òÖУ¬Ë®ÑùÖгöÏÖÁËMnMnO3³Áµí£¬Àë×Ó·½³ÌʽΪ4Mn2++O2+8OH£2MnMnO3¡ý+4H2O¡£
£¨3£©²½Öè¢óÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ _______________________________________________________________¡£
£¨4£©µÎ¶¨Ê±£¬ÈÜÒºÓÉ__________É«µ½______________É«£¬ÇÒ°ë·ÖÖÓÄÚÑÕÉ«²»Ôٱ仯¼´´ïµ½µÎ¶¨Öյ㡣
£¨5£©ºÓË®ÖеÄÈܽâÑõΪ_____________________________mg/L¡£
£¨6£©µ±ºÓË®Öк¬Óн϶àNO3£Ê±£¬²â¶¨½á¹û»á±Èʵ¼ÊÖµ________(ÌîÆ«¸ß¡¢Æ«µÍ»ò²»±ä)
¡¾´ð°¸¡¿¼îʽµÎ¶¨¹Ü MnMnO3+2I£+6H+I2+2Mn2++3H2O À¶ ÎÞ Æ«¸ß
¡¾½âÎö¡¿
(1)µÎ¶¨»·½ÚÖÐʹÓõÄÒÇÆ÷Óеζ¨¹Ü¼Ð¡¢Ìú¼Ų̈¡¢ÉÕ±¡¢×¶ÐÎÆ¿ºÍµÎ¶¨¹Ü£¬µÎ¶¨¹Ü×°Na2S2O3ÈÜÒº£¬Na2S2O3ÏÔ¼îÐÔ£»
(3)¢òÖÐÓÐMnMnO3³ÁµíÉú³ÉͬʱÓеâÀë×ÓÊ£Ó࣬¼ÓÈë1.0mLÁòËáÈÜÒº£¬Èû½ôÆ¿Èû£¬Õñµ´Ë®ÑùÆ¿£¬³ÁµíÈ«²¿Èܽ⣬´ËʱÈÜÒº±äΪ»ÆÉ«£¬ËµÃ÷²úÉúI2£¬¸ù¾ÝÑõ»¯»¹Ô·´Ó¦·½³Ìʽ·ÖÎö£»
(4)´ý²âÒºÖÐÓÐI2£¬Óõí·ÛÈÜÒº×öָʾ¼Á£¬ÈÜҺΪÀ¶É«£¬ÖÕµãʱΪÎÞÉ«£»
(5)¸ù¾Ý¹Øϵʽ£¬½øÐж¨Á¿·ÖÎö£»
(6)º¬Óн϶àNO3£Ê±£¬ÔÚËáÐÔÌõ¼þÏ£¬ÐγÉÏõËᣬ¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬»áÑõ»¯Na2S2O3¡£
(1)µÎ¶¨»·½ÚÖÐʹÓõÄÒÇÆ÷Óеζ¨¹Ü¼Ð¡¢Ìú¼Ų̈¡¢ÉÕ±¡¢×¶ÐÎÆ¿ºÍµÎ¶¨¹Ü£¬µÎ¶¨¹Ü×°Na2S2O3ÈÜÒº£¬Na2S2O3ÏÔ¼îÐÔ£¬ÓüîʽµÎ¶¨¹Ü£¬¹Ê´ð°¸Îª£º¼îʽµÎ¶¨¹Ü£»
(3)¢òÖÐÓÐMnMnO3³ÁµíÉú³ÉͬʱÓеâÀë×ÓÊ£Ó࣬¼ÓÈë1.0mLÁòËáÈÜÒº£¬Èû½ôÆ¿Èû£¬Õñµ´Ë®ÑùÆ¿£¬³ÁµíÈ«²¿Èܽ⣬´ËʱÈÜÒº±äΪ»ÆÉ«£¬ËµÃ÷²úÉúI2£¬¼´µâ»¯ºÏ¼ÛÉý¸ß£¬MnµÄ»¯ºÏ¼Û»á½µµÍ£¬Àë×Ó·½³ÌʽΪMnMnO3+2I£+6H+I2+2Mn2++3H2O£¬¹Ê´ð°¸Îª£ºMnMnO3+2I£+6H+I2+2Mn2++3H2O£»
(4)´ý²âÒºÖÐÓÐI2£¬Óõí·ÛÈÜÒº×öָʾ¼Á£¬ÈÜҺΪÀ¶É«£¬ÖÕµãʱΪÎÞÉ«£¬¹Ê´ð°¸Îª£ºÀ¶£»ÎÞ£»
(5)ÓÉ4Mn2++O2+8OH£2MnMnO3¡ý+4H2O£¬MnMnO3+2I£+6H+I2+2Mn2++3H2O£¬I2 +2Na2S2O3 =2NaI+Na2S4O6£¬¿ÉÖª¹Øϵʽ£¬¼´£¬ÔòÑõÆøµÄÎïÖʵÄÁ¿x=£¬v1mLË®ÑùÖÐÈܽâÑõ=£¬¹Ê´ð°¸Îª£º£»
(6)º¬Óн϶àNO3£Ê±£¬ÔÚËáÐÔÌõ¼þÏ£¬ÐγÉÏõËᣬ¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬»áÑõ»¯Na2S2O3£¬¼´Na2S2O3ÓÃÁ¿Ôö¼Ó£¬½á¹ûÆ«¸ß£¬¹Ê´ð°¸Îª£ºÆ«¸ß¡£
¡¾ÌâÄ¿¡¿ÀûÓ÷ϾÉпÌúƤÖƱ¸´ÅÐÔFe3O4½ºÌåÁ£×Ó¼°¸±²úÎïZnO¡£ÖƱ¸Á÷³ÌͼÈçÏ£º
ÒÑÖª£ºZn¼°»¯ºÏÎïµÄÐÔÖÊÓëAl¼°»¯ºÏÎïµÄÐÔÖÊÏàËÆ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÃNaOHÈÜÒº´¦Àí·Ï¾ÉпÌúƤµÄ×÷ÓÃÓÐ_________________________________________¡£
A£®È¥³ýÓÍÎÛ | B£®Èܽâ¶Æп²ã | C£®È¥³ýÌúÐâ | D£®¶Û»¯ |
£¨2£©µ÷½ÚÈÜÒºAµÄpH¿É²úÉúZn(OH)2³Áµí£¬ÎªÖƵÃZnO£¬ºóÐø²Ù×÷²½ÖèÊÇ______________¡£
£¨3£©ÓÉÈÜÒºBÖƵÃFe3O4½ºÌåÁ£×ӵĹý³ÌÖУ¬Ðë³ÖÐøͨÈëN2£¬ÔÒòÊÇ___________________¡£
£¨4£©Fe3O4½ºÌåÁ£×ÓÄÜ·ñÓüõѹ¹ýÂË·¢ÊµÏÖ¹ÌÒº·ÖÀ룿____________£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©£¬ÀíÓÉÊÇ_________________________________________________¡£
£¨5£©ÓÃÖظõËá¼Ø·¨£¨Ò»ÖÖÑõ»¯»¹ÔµÎ¶¨·¨£©¿É²â¶¨²úÎïFe3O4ÖеĶþ¼ÛÌúº¬Á¿¡£ÈôÐèÅäÖÆŨ¶ÈΪ0.01000 mol¡¤L£1µÄK2Cr2O7±ê×¼ÈÜÒº250 mL£¬Ó¦×¼È·³ÆÈ¡_______g K2Cr2O7(±£Áô4λÓÐЧÊý×Ö£¬ÒÑÖªM(K2Cr2O7)£½294.0 g¡¤mol£1)¡£
ÅäÖƸñê×¼ÈÜҺʱ£¬ÏÂÁÐÒÇÆ÷Öв»±ØÒªÓõ½µÄÓÐ_____________¡££¨ÓñàºÅ±íʾ£©¡£
¢Ùµç×ÓÌìƽ ¢ÚÉÕ± ¢ÛÁ¿Í² ¢Ü²£Á§°ô ¢ÝÈÝÁ¿Æ¿ ¢Þ½ºÍ·µÎ¹Ü ¢ßÒÆÒº¹Ü
£¨6£©µÎ¶¨²Ù×÷ÖУ¬Èç¹ûµÎ¶¨Ç°×°ÓÐK2Cr2O7±ê×¼ÈÜÒºµÄµÎ¶¨¹Ü¼â×첿·ÖÓÐÆøÅÝ£¬¶øµÎ¶¨½áÊøºóÆøÅÝÏûʧ£¬Ôò²â¶¨½á¹û½«__________£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±£©¡£