ïÈ(Sr)ÊÇÈËÌå±ØÐèµÄ΢Á¿ÔªËØ£¬Æäµ¥Öʺͻ¯ºÏÎïµÄ»¯Ñ§ÐÔÖÊÓë¸Æ¡¢±µµÄÏàËÆ¡£ÊµÑéÊÒÓú¬Ì¼ËáïȵķÏÔü(º¬SrCO3 38.40%£¬SrO12.62%£¬CaCO3 38.27%£¬BaCO3 2.54%£¬ÆäËü²»ÈÜÓÚÏõËáµÄÔÓÖÊ8.17%)ÖƱ¸ÏõËáïÈ´ÖÆ·µÄ²¿·ÖʵÑé¹ý³ÌÈçÏ£º
£¨1£©ÊÐÊÛŨÏõËáµÄÖÊÁ¿·ÖÊýΪ65%£¬ÃܶÈΪ1.4g/cm3£¬ÒªÅäÖÆ30%Ï¡ÏõËá500mL£¬»¹ÐèÒª²éÔĵÄÊý¾ÝÊÇ £¬ÈôÅäÖƹý³ÌÖв»Ê¹ÓÃÌìƽ£¬Ôò±ØÐëÒª¼ÆËãµÄÊý¾ÝÊÇ £¬±ØÐëҪʹÓõÄÒÇÆ÷ÊÇ ¡£
ÒÑÖªÁ½ÖÖÑεÄÈܽâ¶È(g/100 gË®)Èçϱí
ζÈ/¡æÎïÖÊ |
0 |
20 |
30 |
45 |
60 |
80 |
100 |
Sr(NO3)2 |
28.2 |
40.7 |
47 |
47.2 |
48.3 |
49.2 |
50.7 |
Ca(NO3)2¡¤4H2O |
102 |
129 |
152 |
230 |
300 |
358 |
408 |
£¨2£©ÓɽþÈ¡ºóµÃµ½µÄ»ìºÏÎïÖƱ¸ÏõËáïÈ´ÖÆ·µÄʵÑé²½ÖèÒÀ´ÎΪ£º¹ýÂË¡¢ ¡¢ ¡¢Ï´µÓ£¬¸ÉÔï¡£
ÒÑÖª£¬ÏõËá¸ÆÄÜÈÜÓÚÓлúÈܼÁAÖС£Ê½Á¿£ºSr(NO3)2¨C212¡¢Ba(NO3)2¨C261¡¢Ca(NO3)2¨C164
£¨3£©ÖƵõÄÏõËáïÈ´ÖÆ·Öк¬ÉÙÁ¿Ca(NO3)2¡¢Ba(NO3)2µÈÔÓÖÊ¡£²â¶¨ÏõËáïÈ´¿¶ÈµÄʵÑéÈçÏ£º³ÆÈ¡5.39gÏõËáïÈÑùÆ·£¬¼ÓÈë×ãÁ¿µÄÓлúÈܼÁA£¬¾¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïºó£¬Ê£Óà¹ÌÌå5.26g£¬½«´Ë¹ÌÌåÅä³É250 mLµÄÈÜÒº£¬È¡³ö25.00 mL£¬µ÷½ÚpHΪ7£¬¼ÓÈëָʾ¼Á£¬ÓÃŨ¶ÈΪ0.107mol/LµÄ̼ËáÄÆÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ̼ËáÄÆÈÜÒº22.98mL¡£
µÎ¶¨¹ý³ÌµÄ·´Ó¦£ºSr2£«£«CO32£¡ú SrCO3¡ý¡¡¡¡Ba2£«£«CO32£¡ú BaCO3¡ý
¢ÙµÎ¶¨Ñ¡ÓõÄָʾ¼ÁΪ £¬µÎ¶¨ÖÕµã¹Û²ìµ½µÄÏÖÏóΪ ¡£
¢Ú¸ÃÏõËáïÈ´ÖÆ·ÖУ¬ÏõËáïȵÄÖÊÁ¿·ÖÊýΪ £¨Ð¡Êýµãºó±£ÁôÁ½Î»£©¡£ÈôµÎ¶¨Ç°ÑùÆ·ÖÐCa(NO3)2ûÓгý¾¡£¬Ëù²â¶¨µÄÏõËáïÈ´¿¶È½«»á (Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°²»±ä¡±)¡£
£¨1£©30%Ï¡ÏõËáµÄÃÜ¶È Å¨ÏõËáºÍÕôÁóË®µÄÌå»ý Á¿Í²¡¢ÉÕ±¡¢²£Á§°ô
£¨2£©Õô·¢½á¾§ ³ÃÈȹýÂË
£¨3£©¢Ù·Ó̪ ÈÜÒº±äΪºìÉ«ÇÒ30s²»±äÉ« ¢Ú0.95£¨212x+261y=5.26 x+y=0.0246£© Æ«¸ß
¡¾½âÎö¡¿
ÊÔÌâ·ÖÎö£º£¨1£©¸ù¾ÝÈÜÖÊÖÊÁ¿ÏàµÈ£¬»¹ÐèÒª²éÔĵÄÊý¾ÝÊÇ30%Ï¡ÏõËáµÄÃܶȣ¬ÈôÅäÖƹý³ÌÖв»Ê¹ÓÃÌìƽ£¬Ôò±ØÐëÒª¼ÆËãµÄÊý¾ÝÊÇŨÏõËáºÍÕôÁóË®µÄÌå»ý£¬±ØÐëҪʹÓõÄÒÇÆ÷ÊÇÁ¿Í²¡¢ÉÕ±¡¢²£Á§°ô¡£
£¨2£©ÓɱíÖеÄÊý¾Ý¿ÉÒÔ¿´³ö£¬Sr(NO3)2µÄÈܽâ¶ÈËæ×ÅζȵÄÉý¸ß±ä»¯²»´ó£¬¶øÏõËá¸ÆµÄÈܽâ¶ÈËæζȱ仯½Ï´ó£¬Òò´Ë¿Éͨ¹ýÓɽþÈ¡ºóµÃµ½µÄ»ìºÏÎïÖƱ¸ÏõËáïÈ´ÖÆ·µÄʵÑé²½ÖèÒÀ´ÎΪ£ºÕô·¢½á¾§£¬³ÃÈȹýÂË£¬Ï´µÓ£¬¸ÉÔïµÃµ½¡£
£¨3£©¢Ù·ÖÎöµÎ¶¨¹ý³Ì¿ÉÖª£¬ÑùÆ·ÈÜÒºÎÞÉ«£¬µÎÈë̼ËáÄƳÁµíÍêÈ«¿ÉÒÔµÎÈë·Ó̪ÊÔҺָʾÖյ㣬µÎÈë×îºóÒ»µÎÈÜÒº³ÊºìÉ«£¬°ë·ÖÖÓÄÚ²¿ÍÊÉ«£¬¹Ê´ð°¸Îª£º·Ó̪£»ÈÜÒºÓÉÎÞ±äΪdzºìÉ«30ÃëÄÚ²»ÍËÉ«£»¢ÚÈôµÎ¶¨Ç°ÑùÆ·ÖÐCa£¨NO3£©2ûÓгý¾¡£¬¶àÏûºÄ±ê×¼ÈÜҺ̼ËáÄÆ£¬ÒÀ¾ÝµÎ¶¨Îó²î·ÖÎö£¬c£¨´ý²âÒº£©=£¬ÏûºÄ±ê×¼Òº¶à£¬Ëù²â¶¨µÄÏõËáïÈ´¿¶È»áÆ«¸ß£¬¹Ê´ð°¸Îª£ºÆ«¸ß¡£ÒÑÖª£¬ÏõËá¸ÆÄÜÈÜÓÚÓлúÈܼÁAÖУ¬ÖÊÁ¿Îª5.39-5.26=0.03g¡£ºÍ̼Ëá¸ù·´Ó¦µÄÖ»ÓбµÀë×ÓºÍïÈÀë×Ó£¬Å¨¶ÈΪ0.107mol/LµÄ̼ËáÄÆÈÜÒº22.98mL£¬ÎïÖʵÄÁ¿Îª0.00246mol£¬ÉèÏõËá±µºÍÏõËáïȵÄÎïÖʵÄÁ¿ÎªyºÍx£¬½¨Á¢·½³ÌʽΪ212x+261y=5.26
x+y=0.0246£¬ÔÙÓÉÏõËáïȵÄÖÊÁ¿ÓÉ×ÜÖÊÁ¿5.39¼õÈ¥ÏõËá¸ÆºÍÏõËá±µµÄÖÊÁ¿£¬ÇóµÃÖÊÁ¿·ÖÊýΪ0.95¡£
¿¼µã£º±¾Ì⿼²éÁËÎïÖÊÖƱ¸ÊµÑé·½°¸µÄ·ÖÎöÅжϣ¬·ÖÀë»ìºÏÎïµÄÁ÷³Ì£¬»¯Ñ§·´Ó¦ËÙÂʵÄÓ°ÏìÒòËØ·ÖÎö£¬µÎ¶¨ÊµÑéµÄ²½ÖèºÍָʾ¼ÁÑ¡Ôñ£¬Îó²î·ÖÎöÓ¦Óá£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com