ÌâÄ¿ÄÚÈÝ

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

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¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
º£ÑóÊÇÉúÃüµÄÒ¡Àº¡¢×ÊÔ´µÄ±¦¿â¡£ÖйúҪʵʩº£ÑóÇ¿¹úÕ½ÂÔ£¬ÊµÏÖÓɺ£Ñó´ó¹úÏòº£ÑóÇ¿¹úÂõ½øµÄÃÎÏë¡£º£Ñó¾­¼ÃÒѾ­³ÉΪÀ­¶¯ÎÒ¹ú¹úÃñ¾­¼Ã·¢Õ¹µÄÖØÒªÒýÇ棬º£Ë®µÄ×ۺϿª·¢¡¢ÀûÓÃÊǺ£Ñ󾭼õÄÒ»²¿·Ö£¬º£Ë®ÖпÉÌáÈ¡¶àÖÖ»¯¹¤Ô­ÁÏ£¬ÏÂÃæÊǹ¤ÒµÉ϶Ժ£Ë®µÄ¼¸Ïî×ÛºÏÀûÓõÄʾÒâͼ¡£ÆäÁ÷³ÌÈçÏÂͼËùʾ£º

£¨1£©Ð´³ö¢Ù¡¢¢Ú·´Ó¦µÄÀë×Ó·½³Ìʽ£º
¢Ù______________________£¬¢Ú______________________¡£
£¨2£©¹¤ÒµÉÏÀûÓõç½â±¥ºÍʳÑÎË®²úÉúµÄÇâÆøºÍÂÈÆøÖÆÈ¡ÑÎËᣬΪÁËÌåÏÖÂÌÉ«»¯Ñ§ÀíÄʹÂÈÆø³ä·Ö·´Ó¦£¬²ÉÈ¡½«ÂÈÆøÔÚÇâÆøÖÐȼÉյİ취£¬¿É±ÜÃâÂÈÆøȼÉÕ²»ÍêÈ«ÎÛȾ¿ÕÆø£¬Çëд³öÂÈÆøÔÚÇâÆøÖÐȼÉÕµÄʵÑéÏÖÏó£º______________________¡£
£¨3£©´ÖÑÎÖк¬ÓÐCa£²£«¡¢Mg£²£«¡¢SO4£²£­µÈÔÓÖÊ£¬¾«Öƺó¿ÉµÃµ½±¥ºÍNaC1ÈÜÒº¡£ÏÖÓÐÏÂÁгýÔÓÊÔ¼Á£ºA£®ÑÎËá   B£®ÇâÑõ»¯±µÈÜÒº   C£®Ì¼ËáÄÆÈÜÒº¡£¾«ÖÆʱ¼ÓÈë¹ýÁ¿³ýÔÓÊÔ¼ÁµÄÕýȷ˳ÐòÊÇ______________¡££¨ÌîÐòºÅ£©
£¨4£©½ðÊôþÔÚ¿ÕÆøÖÐȼÉÕʱ£¬³ýÉú³ÉMgOÍ⣬»¹ÓÐÉÙÁ¿Mg3N2Éú³É¡£°ÑµÈÎïÖʵÄÁ¿µÄ½ðÊôþ·Ö±ð·ÅÔÚ£ºA£®´¿ÑõÆø£¨O2£©ÖУ»B£®¶þÑõ»¯Ì¼ÆøÌåÖУ»C£®¿ÕÆøÖС£ÍêȫȼÉպ󣬵õ½µÄ¹ÌÌåÎïÖʵÄÖÊÁ¿ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ______________¡££¨ÌîÐòºÅ£©
£¨5£©½«µç½â±¥ºÍNaClÈÜÒºÉú³ÉµÄÂÈÆøͨÈëÇâÑõ»¯ÄÆÈÜÒºÖпÉÒԵõ½NaClO¡£Ä³»¯Ñ§ÐËȤС×é̽¾¿NaClOÓëÄòËØCO(NH2)2µÄ·´Ó¦²úÎͨ¹ýʵÑé·¢ÏÖ²úÎï³ýijÖÖÑÎÍ⣬ÆäÓà²úÎﶼÊÇÄܲÎÓë´óÆøÑ­»·µÄÎïÖÊ£¬Ôò¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________¡£
º£Ë®ÖÐÖ÷ÒªÀë×ӵĺ¬Á¿ÈçÏ£º
³É·Ö
º¬Á¿/(mg/L)
³É·Ö
º¬Á¿/(mg/L)
Cl-
18980
Ca2+
400
Na+
10560
HCO3-
142
SO42-
2560
Mg2+
1272
 
£¨1£©³£ÎÂÏ£¬º£Ë®µÄpHÔÚ7.5~8.6Ö®¼ä£¬ÆäÔ­ÒòÊÇ(ÓÃÀë×Ó·½³Ìʽ±íʾ)______________________________¡£
£¨2£©µçÉøÎö·¨µ­»¯º£Ë®Ê¾ÒâͼÈçͼËùʾ£¬ÆäÖÐÒõ(Ñô)Àë×Ó½»»»Ä¤½öÔÊÐíÒõ(Ñô)Àë×Óͨ¹ý¡£Òõ¼«ÉϲúÉúÇâÆø£¬Òõ¼«¸½½ü»¹²úÉúÉÙÁ¿°×É«³Áµí£¬Æä³É·ÖÓÐ________ºÍCaCO3£¬Éú³ÉCaCO3µÄÀë×Ó·½³ÌʽÊÇ_______________¡£

£¨3£©Óú£Ë®¿ÉͬʱÉú²úÂÈ»¯ÄƺͽðÊôþ»òþµÄ»¯ºÏÎÆäÁ÷³ÌÈçÏÂͼËùʾ£º

¢ÙÔÚʵÑéÊÒÖÐÓÉ´ÖÑΡ°Öؽᾧ¡±Öƾ«ÑεIJÙ×÷°üÀ¨Èܽ⡢¹ýÂË¡¢Õô·¢¡­Ï´µÓµÈ²½Ö裻ÓйØÆäÖС°Õô·¢¡±²½ÖèµÄÐðÊöÕýÈ·µÄÊÇ_____¡£
a£®Õô·¢µÄÄ¿µÄÊǵõ½Èȱ¥ºÍÈÜÒº
b£®Õô·¢µÄÄ¿µÄÊÇÎö³ö¾§Ìå
c£®Ó¦ÓÃÓàÈÈÕô¸ÉÈÜÒº
d£®Ó¦Õô·¢ÖÁÓн϶ྦྷÌåÎö³öʱΪֹ
¢ÚÓÉMgCl2ÈÜÒºµÃµ½MgCl2?6H2O¾§Ìåʱ£¬Ò²ÐèÒªÕô·¢£¬Õô·¢µÄÄ¿µÄÊǵõ½Èȱ¥ºÍÈÜÒº£¬ÅжÏÈÜÒºÒѱ¥ºÍµÄÏÖÏóÊÇ_________________________¡£
£¨4£©25¡æʱ£¬±¥ºÍMg(OH)2ÈÜÒºµÄŨ¶ÈΪ5¡Á10-4 mol£¯L¡£
¢Ù±¥ºÍMg(OH)2ÈÜÒºÖеμӷÓ̪£¬ÏÖÏóÊÇ___________________________¡£
¢ÚijѧϰС×é²âº£Ë®ÖÐMg2+º¬Á¿(mg/L)µÄ·½·¨ÊÇ£ºÈ¡Ò»¶¨Ìå»ýµÄº£Ë®£¬¼ÓÈë×ãÁ¿_________£¬ÔÙ¼ÓÈë×ãÁ¿NaOH£¬½«Mg2+תΪMg(OH)2¡£25¡æ£¬¸Ã·½·¨²âµÃµÄMg2+º¬Á¿Óë±íÖÐ1272mg/LµÄ¡°ÕæÖµ¡±±È£¬Ïà¶ÔÎó²îԼΪ______£¥[±£Áô2λСÊý£¬º£Ë®Öб¥ºÍMg(OH)2ÈÜÒºµÄÃܶȶ¼ÒÔl g/cm3¼Æ]¡£

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

¾«Ó¢¼Ò½ÌÍø