ÌâÄ¿ÄÚÈÝ

(10)ÒÑÖªA¡¢B¡¢C¡¢DΪԭ×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÔªËØ,ÆäÖÐA¡¢DͬÖ÷×å.AÔªËØµÄµ¥ÖÊÊÇÃܶÈ×îСµÄÆøÌå,BÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ2±¶,ÇÒÆä×îÍâ²ãµç×ÓÊý±ÈCÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÉÙ2¡£

ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©Ð´³öBµÄÔªËØ·ûºÅ£º          ¡£

£¨2£©CÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃΪ                    ¡£

£¨3£©Ð´³öBA4µÄµç×Óʽ£º                

£¨4£©Óõç×Óʽ±íʾÀë×Ó»¯ºÏÎïD2CµÄÐγɹý³Ì¡£

                                                          

£¨5£©Ð´³öD2C2ÓëA2C·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

                                                     

 

(1£©C    (2)µÚ¶þÖÜ¢öA×å    £¨3£©

(4)  

£¨5£©2 Na2O2+2H2O=4NaOH+O2¡ü

½âÎö:ÃܶÈ×îСµÄÆøÌåÊÇÇâÆø£¬AÊÇÇâÔªËØ¡£ÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ2±¶ÊÇÌ¼ÔªËØ£¬BÊÇ̼£¬CµÄ×îÍâ²ãµç×ÓÊýÊÇ6¡£A¡¢DͬÖ÷×åA¡¢B¡¢C¡¢DΪԭ×ÓÐòÊýÒÀ´ÎÔö´ó£¬¹ÊCÊÇÑõÔªËØ£¬DÊÇÄÆÔªËØ¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø