ÌâÄ¿ÄÚÈÝ

(13·Ö)ºìשÊÇÓÃÕ³ÍÁ¸ßÎÂÉÕ½á¶ø³É£¬ÒòÆäÑÕÉ«³ÊºìÉ«»òרºìÉ«¶øµÃÃû,³£ÓÃ×÷½¨Öþ²ÄÁÏ¡£
(1)¸ù¾ÝºìשµÄÑÕÉ«¡¢²Â²âÆä¿ÉÄܺ¬ÓеijɷÝÊÇ£¨Ð´»¯Ñ§Ê½£©                     
(2)ΪÁ˼ìÑéÄãµÄ²Â²âÊÇ·ñÕýÈ·£¬ÇëÉè¼ÆÒ»¸öʵÑé·½°¸£¬¼òÒªµØÐ´³ö¸÷²½²Ù×÷¹ý³ÌÒÔ¼°×îºóµÄʵÑé·ÖÎöºÍ½áÂÛ£¬Éæ¼°»¯Ñ§·´Ó¦µÄд³ö»¯Ñ§·½³Ìʽ(ʵÑéÓÃÆ·ÈÎÈ¡£¬¸ÃʵÑé·Ö¼¸²½×Ô¼º¾ö¶¨£¬²»±ØÐ´ÊµÑé×°ÖÃÖеÄÒÇÆ÷°²×°)
ʵÑé²½Ö裺¢ÙÓÃÌú´¸ÇÃËéºìש£¬È¡Ð¡¿éÄ¥³É·ÛÄ©¡£
¢Ú
¢Û
©«
©«
ʵÑé·ÖÎö¼°½áÂÛ£º                                                            
                                                             
(3)ºì×©Ôø¶ÔÎÒ¹ú½¨ÉèÆðµ½ÖØÒª×÷Ó㬾ÍĿǰÎÒ¹úʵ¼ÊÇé¿ö¿´£¬ÄãÊÇÈÏΪӦ¼ÌÐøÊ¹ÓûòÕß
ÊǽûֹʹÓúìש£¿            £¬ÀíÓÉÊÇ                                     
                                                                     ¡£

£¨13·Ö£©
£¨1£©£¨2·Ö£©Fe2O3
£¨2£©²½Öè¼°»¯Ñ§·½³Ìʽ£¨5·Ö£©
¢Ú½«ºìש·ÛÄ©·ÅÈëÉÕ±­£¬¼ÓÈëÊÊÁ¿Ï¡ÑÎËᣬ½Á°è£ºFe2O3 + 6HCl = 2FeCl3 + 3H2O
¢Û½«»ìºÏÒºµ¹Èë¹ýÂËÆ÷¹ýÂË
¢ÜÔÚÂËÒºÖеÎÈëKSCNÈÜÒº£ºFeCl3 + 3KSCN = Fe(SCN)3 + 3KCl
·ÖÎö¼°½áÂÛ(4·Ö),ÈôÈÜÒº³ÊѪºìÉ«£¬Ö¤Ã÷ºìשº¬ÓÐFe2O3£¬ÈôÈÜÒº²»³ÊѪºìÉ«£¬ËµÃ÷ºìש²»º¬Fe2O3¡£
£¨3£©£¨2·Ö£©Ó¦½ûֹʹÓá£ÒòΪ»áÏûºÄ´óÁ¿µÄ¹úÍÁ×ÊÔ´¡£

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

(13·Ö)ºìשÊÇÓÃÕ³ÍÁ¸ßÎÂÉÕ½á¶ø³É£¬ÒòÆäÑÕÉ«³ÊºìÉ«»òרºìÉ«¶øµÃÃû,³£ÓÃ×÷½¨Öþ²ÄÁÏ¡£

(1)¸ù¾ÝºìשµÄÑÕÉ«¡¢²Â²âÆä¿ÉÄܺ¬ÓеijɷÝÊÇ£¨Ð´»¯Ñ§Ê½£©                     

(2)ΪÁ˼ìÑéÄãµÄ²Â²âÊÇ·ñÕýÈ·£¬ÇëÉè¼ÆÒ»¸öʵÑé·½°¸£¬¼òÒªµØÐ´³ö¸÷²½²Ù×÷¹ý³ÌÒÔ¼°×îºóµÄʵÑé·ÖÎöºÍ½áÂÛ£¬Éæ¼°»¯Ñ§·´Ó¦µÄд³ö»¯Ñ§·½³Ìʽ(ʵÑéÓÃÆ·ÈÎÈ¡£¬¸ÃʵÑé·Ö¼¸²½×Ô¼º¾ö¶¨£¬²»±ØÐ´ÊµÑé×°ÖÃÖеÄÒÇÆ÷°²×°)

ʵÑé²½Ö裺¢Ù ÓÃÌú´¸ÇÃËéºìש£¬È¡Ð¡¿éÄ¥³É·ÛÄ©¡£

            ¢Ú

            ¢Û

            ©«

            ©«

ʵÑé·ÖÎö¼°½áÂÛ£º                                                            

                                                                            

(3)ºì×©Ôø¶ÔÎÒ¹ú½¨ÉèÆðµ½ÖØÒª×÷Ó㬾ÍĿǰÎÒ¹úʵ¼ÊÇé¿ö¿´£¬ÄãÊÇÈÏΪӦ¼ÌÐøÊ¹ÓûòÕß

ÊǽûֹʹÓúìש£¿            £¬ÀíÓÉÊÇ                                      

                                                                     ¡£

 

(13·Ö)ºìשÊÇÓÃÕ³ÍÁ¸ßÎÂÉÕ½á¶ø³É£¬ÒòÆäÑÕÉ«³ÊºìÉ«»òרºìÉ«¶øµÃÃû,³£ÓÃ×÷½¨Öþ²ÄÁÏ¡£

(1)¸ù¾ÝºìשµÄÑÕÉ«¡¢²Â²âÆä¿ÉÄܺ¬ÓеijɷÝÊÇ£¨Ð´»¯Ñ§Ê½£©                     

(2)ΪÁ˼ìÑéÄãµÄ²Â²âÊÇ·ñÕýÈ·£¬ÇëÉè¼ÆÒ»¸öʵÑé·½°¸£¬¼òÒªµØÐ´³ö¸÷²½²Ù×÷¹ý³ÌÒÔ¼°×îºóµÄʵÑé·ÖÎöºÍ½áÂÛ£¬Éæ¼°»¯Ñ§·´Ó¦µÄд³ö»¯Ñ§·½³Ìʽ(ʵÑéÓÃÆ·ÈÎÈ¡£¬¸ÃʵÑé·Ö¼¸²½×Ô¼º¾ö¶¨£¬²»±ØÐ´ÊµÑé×°ÖÃÖеÄÒÇÆ÷°²×°)

ʵÑé²½Ö裺¢Ù ÓÃÌú´¸ÇÃËéºìש£¬È¡Ð¡¿éÄ¥³É·ÛÄ©¡£

            ¢Ú

            ¢Û

            ©«

            ©«

ʵÑé·ÖÎö¼°½áÂÛ£º                                                            

                                                                             

(3)ºì×©Ôø¶ÔÎÒ¹ú½¨ÉèÆðµ½ÖØÒª×÷Ó㬾ÍĿǰÎÒ¹úʵ¼ÊÇé¿ö¿´£¬ÄãÊÇÈÏΪӦ¼ÌÐøÊ¹ÓûòÕß

ÊǽûֹʹÓúìש£¿             £¬ÀíÓÉÊÇ                                      

                                                                      ¡£

 

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

¾«Ó¢¼Ò½ÌÍø