ÌâÄ¿ÄÚÈÝ

£¨11·Ö£©µç½â·¨ÖÆÉÕ¼îµÄÔ­ÁÏÊDZ¥ºÍʳÑÎË®£¬ÓÉÓÚ´ÖÑÎÖк¬ÓÐÄàɳºÍCa2+¡¢Mg2+¡¢Fe3+¡¢SO42-ÔÓÖÊ£¬²»·ûºÏµç½âÒªÇó£¬Òò´Ë±ØÐë¾­¹ý¾«ÖÆ¡£Ä³Ð£ÊµÑéС×龫ÖÆ´ÖÑÎË®µÄʵÑé¹ý³ÌÈçÏ£º

Çë»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©²Ù×÷aµÄÃû³ÆÊÇ____________¡£
£¨2£©ÔÚ¢ò²½ÖУ¬¼ÓÈë¹ýÁ¿ÊÔ¼Á¼×ºó£¬Éú³ÉÁËÁ½ÖÖ´óÁ¿µÄ³Áµí£¬ÔòÊÔ¼Á¼×Ϊ________ÈÜÒº¡£ÊÔ¼ÁÒÒΪ         £¬¹ÌÌåFΪ                       ¡£
£¨3£©ÔÚµÚ¢õ²½ÖУ¬¼ÓÈëÊÔ¼Á¶¡Ö±µ½ÈÜÒºÎÞÃ÷ÏԱ仯ʱ£¬Ð´³ö´Ë¹ý³ÌµÄ»¯Ñ§·½³Ìʽ                                     ¡¢                                 _  ¡£
£¨1£©¹ýÂË£¨1·Ö£©£¨2£©NaOH¡¢BaCl2¡¢CaCO3ºÍBaCO3£¨Ã¿¿Õ2·Ö£©£¨3£©NaOH + HCl =" NaCl" + H2O       Na2CO3 + 2HCl =" 2NaCl" + H2O + CO2¡ü£¨Ã¿¿Õ2·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÁòËṤҵÉú²úÓ¦¿¼ÂÇ×ۺϾ­¼ÃЧÒæÎÊÌâ¡£
£¨1£©Èô´ÓÏÂÁÐËĸö³ÇÊÐÖÐÑ¡ÔñÒ»´¦Ð½¨Ò»×ùÁòË᳧£¬ÄãÈÏΪ³§Ö·ÒËÑ¡ÔÚ           µÄ½¼Çø£¨ÌîÑ¡ÏîµÄ±êºÅ£©
A£®Óзḻ»ÆÌú¿ó×ÊÔ´µÄ³ÇÊÐB£®·ç¹âÐãÀöµÄÂÃÓγÇÊÐ
C£®ÏûºÄÁòËáÉõ¶àµÄ¹¤Òµ³ÇÊÐD£®ÈË¿Ú³íÃܵÄÎÄ»¯¡¢ÉÌÒµÖÐÐijÇÊÐ
£¨2£©¾Ý²âË㣬½Ó´¥·¨ÖÆÁòËá¹ý³ÌÖУ¬Èô·´Ó¦Èȶ¼Î´±»ÀûÓã¬ÔòÿÉú²ú1t 98%ÁòËáÐèÏûºÄ3.6¡Á105kJÄÜÁ¿¡£Çëͨ¹ý¼ÆËãÅжÏ,Èô·´Ó¦:SO2£¨g£©+1/2O2£¨g£© ?SO3£¨g£©;¡÷H=£­98£®3kJ¡¤mol£­1£»·Å³öµÄÈÈÁ¿ÄÜÔÚÉú²ú¹ý³ÌÖеõ½³ä·ÖÀûÓÃ,ÔòÿÉú²ú1t98%ÁòËáÖ»ÐèÍâ½çÌṩ£¨»ò¿ÉÏòÍâ½çÊä³ö£©          Ç§½¹ÄÜÁ¿£»
£¨3£©CuFeS2ÊÇ»ÆÌú¿óµÄÁíÒ»³É·Ö£¬ìÑÉÕʱ£¬CuFeS2ת»¯ÎªCuO¡¢Fe2O3ºÍSO2£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                              ¡£
£¨4£©ÓÉÁòË᳧·ÐÌÚ¯ÅųöµÄ¿óÔüÖк¬ÓÐFe2O3¡¢CuO¡¢CuSO4£¨ÓÉCuOÓëSO3ÔÚ·ÐÌÚ¯Öл¯ºÏ¶ø³É£©£¬ÆäÖÐÁòËáÍ­µÄÖÊÁ¿·ÖÊýËæ·ÐÌگζȲ»Í¬¶ø±ä»¯£¨¼ûÏÂ±í£©
·ÐÌگζÈ/¡æ
600
620
640
660
¿óÔüÖÐCuSO4µÄÖÊÁ¿·ÖÊý/%
9.3
9.2
9.0
8.4
ÒÑÖªCuSO4ÔÚµÍÓÚ660¡æʱ²»»á·Ö½â£¬Çë¼òÒª·ÖÎöÉϱíÖÐCuSO4µÄÖÊÁ¿·ÖÊýËæζÈÉý¸ß¶ø½µµÍµÄÔ­Òò                                         ¡£
£¨10·Ö£©MnO2ºÍпÊÇÖÆÔì¸Éµç³ØµÄÖ÷ÒªÔ­ÁÏ¡£Ä³µØÓÐÈíÃÌ¿óºÍÉÁп¿óÁ½×ù¿óɽ£¬ËüÃǵÄÖ÷Òª³É·ÝΪ£º
ÈíÃÌ¿ó£ºMnO2º¬Á¿¡Ý65%       Al2O3º¬Á¿Îª4%
ÉÁп¿ó£ºZnSº¬Á¿¡Ý80%   FeS¡¢ CuS¡¢ CdSº¬Á¿¸÷Ϊ2%
µç½â·¨Éú²úMnO2´«Í³µÄ¹¤ÒÕÖ÷ÒªÁ÷³ÌΪ£ºÈíÃÌ¿ó¼Óú»¹Ô­±ºÉÕ£»ÓÃÁòËá½þ³ö±ºÉÕÁÏ£»½þ³öÒº¾­¾»»¯ºóÔÙ½øÐеç½â£¬MnO2ÔÚµç½â³ØµÄÑô¼«Îö³ö¡£
µç½âпµÄ´«Í³Éú²ú¹¤ÒÕΪ£ºÉÁп¿ó¸ßÎÂÑõ»¯ÍÑÁòÔÙÓÃÈÈ»¹Ô­·¨»¹Ô­µÃ´Öп£º
2ZnS£«O22ZnO£«2SO2  2C£«O22CO  ZnO£«COZn(g)£«CO2
½«ÓÃÈÈ»¹Ô­·¨ÖƵõĴÖпÈÜÓÚÁòËᣬÔÙµç½âZnSO4ÈÜÒº¿ÉÉú²ú´¿¶ÈΪ99.95%µÄп¡£
ÏÖÔÚÉú²úMnO2ºÍпµÄй¤ÒÕÖ÷ÒªÊÇͨ¹ýµç½â»ñµÃMnO2ºÍп£¬¸±²úÆ·ÊÇÁò¡¢½ðÊôÍ­ºÍïÓ¡£¼ò»¯Á÷³Ì¸ÜͼÈçÏ£º

ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ZnµÄÔ­×ÓÐòÊýΪ30£¬ËüÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊÇ           £»ÈíÃÌ¿ó¡¢ÉÁп¿ó·ÛδÓëÁòËáÈÜÒº¹²ÈÈʱÎö³öÁòµÄ·´Ó¦ÎªÑõ»¯£­»¹Ô­·´Ó¦£¬ÀýÈ磺MnO2£«ZnS£«2H2SO4£½MnSO4£«ZnSO4£«S¡ý£«2H2O£¬¾Ý´Ëд³öMnO2ÔÚËáÐÔÈÜÒºÖзֱðFeS·¢ÉúÑõ»¯£­»¹Ô­·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                  £»ÈíÃÌ¿óÖÐAl2O3ÈÜÓÚÁòËáµÄÀë×Ó·½³Ìʽ                         £»ÓÉÂËÔü¼×»ØÊÕÁò»ÆµÄʵÑé·½·¨ÊÇ                              £»
£¨2£©ÓÃÀë×Ó·½³Ìʽ±íʾ½þ³öÒºAÓëÊÊÁ¿Zn·Û×÷Óõõ½ÂËÒºBÓëÂËÔüÒҵĹý³Ì     ______________________________________________________________________¡£
£¨3£©ÂËÔü±ûµÄ»¯Ñ§³É·ÖÊÇ              £»
£¨4£©ÓÃÌúºÍ²¬µç¼«µç½âMnSO4ºÍZnSO4µÄ»ìºÏÈÜÒº¿ÉÒԵõ½ZnºÍMnO2£¬µç½âʱ£¬Ìú×ö       ¼«£¬Ìú¼«·¢ÉúµÄµç¼«·´Ó¦Îª                      ¡£

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

¾«Ó¢¼Ò½ÌÍø