ÌâÄ¿ÄÚÈÝ

ÓÐ6Æ¿°×É«¹ÌÌåÊÔ¼Á£¬·Ö±ðÊÇÂÈ»¯±µ¡¢ÇâÑõ»¯ÄÆ¡¢ÁòËáÄÆ¡¢ÁòËá李¢ÎÞË®ÁòËáÍ­¡¢Ì¼ËáÄÆ¡£ÏÖÖ»ÌṩÕôÁóË®£¬Í¨¹ýÏÂÃæµÄʵÑé²½Öè¼´¿É¼ø±ðËüÃÇ¡£ÇëÌîдÏÂÁпհס£

£¨1£©¸÷È¡ÊÊÁ¿¹ÌÌåÊÔ¼Á·Ö±ð¼ÓÈë6Ö§ÊÔ¹ÜÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®£¬Õñµ´ÊԹܣ¬¹Û²ìµ½µÄÏÖÏóÊÇ                                                            ¡£

£¨2£©·Ö±ðȡδ¼ì³öµÄÈÜÒº£¬ÍùÆäÖмÓÈëÉÏÊöÒѼì³öµÄÈÜÒº£¬¹Û²ìµ½µÄÏÖÏóºÍÏàÓ¦µÄÀë×Ó·½³Ìʽ·Ö±ðÊÇ¡£±»¼ì³öµÄÎïÖʵĻ¯Ñ§Ê½£¨·Ö×Óʽ£©ÊÇ                                  ¡£

£¨3£©¼ø±ðÓàÏÂδ¼ì³öÎïÖʵķ½·¨ºÍ¹Û²ìµ½µÄÏÖÏóÊÇ                                  ¡£

£¨1£©6ÖÖ¹ÌÌåÈ«²¿Èܽ⣬5Ö§ÊÔ¹ÜÖеõ½ÎÞÉ«ÈÜÒº£¬1Ö§ÊÔ¹ÜÖеõ½À¶É«ÈÜÒº£¬¸ÃÈÜÒºÊÇCuSO4ÈÜÒº

£¨2£©Ba2++BaSO4¡ý(°×É«³Áµí£©£¬Cu2++2OH-Cu(OH)2¡ý(À¶É«³Áµí£©£¬Cu2++ CuCO3¡ý(À¶É«³Áµí£©BaCl2

(3)(i£©·Ö±ðÈ¡£¨2£©ÖÐÄܲúÉúÀ¶É«³ÁµíµÄÁ½ÖÖÈÜÒº£¬ÏòÆäÖмÓÈëBaCl2ÈÜÒº£¬Óа×É«³ÁµíÉú³ÉµÄÊÇNa2CO3ÈÜÒº£¬ÎÞ³ÁµíÉú³ÉµÄÊÇNaOHÈÜÒº£»

£¨ii)·Ö±ðÈ¡ÉÙÁ¿Î´¼ì³öµÄÈÜÒº£¬ÍùÆäÖмÓÈëNaOHÈÜÒº£¬ÎÞÃ÷ÏÔÏÖÏóµÄÊÇNa2SO4ÈÜÒº£¬Óд̼¤ÐÔÆøζÆøÌå²úÉúµÄÊÇ(NH4)2SO4ÈÜÒº


½âÎö:

±¾ÌâÊ×ÏÈͨ¹ý¹ÌÌåÈÜÓÚË®ºó£¬³ÊÀ¶É«ÈÜÒºµÄÊÇCuSO4,¼ø±ð³öCuSO4£»È»ºóͨ¹ýCuSO4¼ø±ð³öBaCl2,ÔÙͨ¹ýBaCl2¼ø±ð³öNaOHÈÜÒº£¬×îºóͨ¹ýNaOHÈÜÒº¼ø±ð¿ªÊ£ÓàµÄÎïÖÊ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÓÐ6Æ¿°×É«¹ÌÌåÊÔ¼Á£¬·Ö±ðÊÇÂÈ»¯±µ¡¢ÇâÑõ»¯ÄÆ¡¢ÁòËáÄÆ¡¢ÁòËá李¢ÎÞË®ÁòËáÍ­¡¢Ì¼ËáÄÆ£¬ÏÖÖ»ÌṩÕôÁóË®£¬Í¨¹ýÏÂÃæµÄʵÑé²½Öè¼´¿ÉÇ©±ðËüÃÇ£®ÇëÌîдÏÂÁпհףº
£¨1£©¸÷È¡ÊÊÁ¿¹ÌÌåÊÔ¼Á·Ö±ð¼ÓÈë6Ö§ÊÔºÍÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®£¬Õñµ´ÊԹܣ¬¹Û²ìµ½µÄÏÖÏóÊÇ
6ÖÖ¹ÌÌåÈ«²¿Èܽ⣬5Ö§ÊÔ¹ÜÖеõ½ÎÞÉ«ÈÜÒº£¬1Ö§ÊÔ¹ÜÖеõ½À¶É«ÈÜÒº
6ÖÖ¹ÌÌåÈ«²¿Èܽ⣬5Ö§ÊÔ¹ÜÖеõ½ÎÞÉ«ÈÜÒº£¬1Ö§ÊÔ¹ÜÖеõ½À¶É«ÈÜÒº
±»¼ì³öµÄÎïÖʵĻ¯Ñ§Ê½£¨·Ö×Óʽ£©ÊÇ
CuSO4
CuSO4

£¨2£©·Ö±ðȡδ¼ì³öµÄÈÜÒº£¬ÍùÆäÖмÓÈëÉÏÊöÒѼì³öµÄÈÜÒº£¬¹Û²ìµ½µÄÏÖÏóºÍÏàÓ¦µÄÀë×Ó·½³Ìʽ·Ö±ðΪ£º
1Ö§ÊÔ¹ÜÖÐÓа×É«³ÁµíÉú³É£º
Ba2++SO42-=BaSO4¡ý
Ba2++SO42-=BaSO4¡ý
£®
2Ö§ÊÔ¹ÜÖÐÓÐÀ¶É«³ÁÉú³É£º
Cu2++2OH-=Cu£¨OH£©2
Cu2++2OH-=Cu£¨OH£©2
¡¢
2Cu2++2CO32-+H2O=Cu2£¨OH£©2CO3¡ý+CO2¡ü
2Cu2++2CO32-+H2O=Cu2£¨OH£©2CO3¡ý+CO2¡ü
£®
±»¼ì³öµÄÎïÖʵĻ¯Ñ§Ê½£¨·Ö×Óʽ£©ÊÇ
BaCl2
BaCl2

£¨3£©¼ø±ðÓàÏÂδ¼ì³öÎïÖʵķ½·¨ºÍ¹Û²ìµ½µÄÏÖÏó
£¨i£©·Ö±ðÈ¡£¨2£©ÖÐÄܲúÉúÀ¶É«³ÁµíµÄÁ½ÖÖÈÜÒº£¬ÏòÆäÖмÓÈëBaCl2ÈÜÒº£¬Óа×É«³ÁµíÉú³ÉµÄÊÇ
Na2CO3ÈÜÒº
Na2CO3ÈÜÒº
£¬ÎÞ³ÁµíÉú³ÉµÄÊÇ
NaOHÈÜÒº
NaOHÈÜÒº
£®
£¨ii£©·Ö±ðÈ¡ÉÙÁ¿Î´¼ì³öµÄÈÜÒº£¬ÍùÆäÖмÓÈëNaOHÈÜÒº£¬ÎÞÃ÷ÏÔÏÖÏóµÄÊÇ
Na2SO4ÈÜÒº
Na2SO4ÈÜÒº
£¬Óд̼¤ÐÔÆøζÆøÌå²úÉúµÄÊÇ
£¨NH4£©2SO4ÈÜÒº
£¨NH4£©2SO4ÈÜÒº
£®

£¨14·Ö£©ÓÐ6Æ¿°×É«¹ÌÌåÊÔ¼Á£¬·Ö±ðÊÇÂÈ»¯±µ¡¢ÇâÑõ»¯ÄÆ¡¢ÁòËáÄÆ¡¢ÁòËá李¢ÎÞË®ÁòËáÍ­¡¢Ì¼ËáÄÆ£¬ÏÖÖ»ÌṩÕôÁóË®£¬Í¨¹ýÏÂÃæµÄʵÑé²½Öè¼´¿ÉÇ©±ðËüÃÇ¡£ÇëÌîдÏÂÁпհףº
£¨1£©¸÷È¡ÊÊÁ¿¹ÌÌåÊÔ¼Á·Ö±ð¼ÓÈë6Ö§ÊÔºÍÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®£¬Õñµ´ÊԹܣ¬¹Û²ìµ½µÄÏÖÏóÊÇ                                                                        
±»¼ì³öµÄÎïÖʵĻ¯Ñ§Ê½£¨·Ö×Óʽ£©ÊÇ              
£¨2£©·Ö±ðȡδ¼ì³öµÄÈÜÒº£¬ÍùÆäÖмÓÈëÉÏÊöÒѼì³öµÄÈÜÒº£¬¹Û²ìµ½µÄÏÖÏóºÍÏàÓ¦µÄÀë×Ó·½³Ìʽ·Ö±ðΪ£º
1Ö§ÊÔ¹ÜÖÐÓа×É«³ÁµíÉú³É£º ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡   ¡£
2Ö§ÊÔ¹ÜÖÐÓÐÀ¶É«³ÁÉú³É£º                        ¡¢________________________¡£
±»¼ì³öµÄÎïÖʵĻ¯Ñ§Ê½£¨·Ö×Óʽ£©ÊÇ              
£¨3£©¼ø±ðÓàÏÂδ¼ì³öÎïÖʵķ½·¨ºÍ¹Û²ìµ½µÄÏÖÏó
£¨i£©·Ö±ðÈ¡£¨2£©ÖÐÄܲúÉúÀ¶É«³ÁµíµÄÁ½ÖÖÈÜÒº£¬ÏòÆäÖмÓÈëBaCl2ÈÜÒº£¬Óа×É«³ÁµíÉú³ÉµÄÊÇ__________________£¬ÎÞ³ÁµíÉú³ÉµÄÊÇ_________________¡£
£¨ii£©·Ö±ðÈ¡ÉÙÁ¿Î´¼ì³öµÄÈÜÒº£¬ÍùÆäÖмÓÈëNaOHÈÜÒº£¬ÎÞÃ÷ÏÔÏÖÏóµÄÊÇ_____________£¬Óд̼¤ÐÔÆøζÆøÌå²úÉúµÄÊÇ___________________¡£

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

¾«Ó¢¼Ò½ÌÍø