ÌâÄ¿ÄÚÈÝ


½«0.1 molµÄþºÍÂÁµÄ»ìºÏÎïÈÜÓÚ50 mL 4 mol¡¤L-1 H2SO4ÈÜÒºÖУ¬È»ºóÔٵμÓ

2 mol¡¤L-1µÄNaOHÈÜÒº¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÈôÔڵμÓNaOHÈÜÒºµÄ¹ý³ÌÖУ¬³ÁµíÖÊÁ¿mËæ¼ÓÈëNaOHÈÜÒºµÄÌå»ýVµÄ±ä»¯ÈçͼËùʾ¡£µ±V1£½80 mLʱ£¬¼ÆËã½ðÊô·ÛÄ©ÖÐþµÄÎïÖʵÄÁ¿¼°V2µÄÌå»ý¡£

(2)ÈôÔڵμÓNaOHÈÜÒºµÄ¹ý³ÌÖУ¬ÓûʹMg2£«¡¢Al3£«¸ÕºÃ³ÁµíÍêÈ«£¬¼ÆËãµÎÈëNaOHÈÜÒºµÄÌå»ý¡£


 (1) n(Mg)=0.06mol £¨2·Ö£©;   V2=220mL£¨2·Ö£©

(2)V(NaOH)=200mL£¨2·Ö£©


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

  ¹èÊÇÖØÒªµÄ°ëµ¼Ìå²ÄÁÏ£¬¹¹³ÉÁËÏÖ´úµç×Ó¹¤ÒµµÄ»ù´¡¡£»Ø´ðÏÂÁÐÎÊÌâ:

(1)»ù̬SiÔ­×ÓÖУ¬µç×ÓÕ¼¾ÝµÄÄÜÁ¿×î¸ßµÄµç×Ó²ãµÄ·ûºÅΪ        £¬¸Ãµç×Ó²ã¾ßÓеÄÔ­×Ó¹ìµÀÊýΪ       ¡£

(2)¹èÖ÷ÒªÒÔ¹èËáÑΡ¢        µÈ»¯ºÏÎïµÄÐÎʽ´æÔÚÓڵؿÇÖС£

(3)µ¥ÖÊ¹è´æÔÚÓë½ð¸Õʯ½á¹¹ÀàËÆµÄ¾§Ì壬ÆäÖÐÔ­×ÓÓëÔ­×ÓÖ®¼äÒÔ             Ïà½áºÏ£¬Æä¾§°ûƽ¾ùÓµÓÐ8¸ö¹èÔ­×Ó£¬ÆäÖÐÔÚÃæÐÄλÖù±Ï×         ¸öÔ­×Ó¡£

(4)µ¥Öʹè¿Éͨ¹ý¼×¹èÍé(SiH4)·Ö½â·´Ó¦À´ÖƱ¸¡£¹¤ÒµÉϲÉÓÃMg2SiºÍNH4ClÔÚÒº°±½éÖÊÖз´Ó¦ÖƵÃSiH4£¬Í¬Ê±²úÉúÒ»ÖÖ¿ÉʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                       ¡£

(5)̼ºÍ¹èµÄÓйػ¯Ñ§¼ü¼üÄÜÈçÏÂËùʾ£¬¼òÒª·ÖÎöºÍ½âÊÍÏÂÁÐÓйØÊÂʵ£º

¢Ù¹èÓë̼ͬ×壬ҲÓÐϵÁÐÇ⻯Îµ«¹èÍéÔÚÖÖÀàºÍÊýÁ¿É϶¼Ô¶²»ÈçÍéÌþ¶à£¬Ô­ÒòÊÇ                 ¡£

¢ÚSiH4µÄÎȶ¨ÐÔСÓÚCH4£¬¸üÒ×Éú³ÉÑõ»¯ÎԭÒòÊÇ                                        ¡£

   (6)ÔÚ¹èËáÑÎÖУ¬ËÄÃæÌå(ÈçÏÂͼ(a))ͨ¹ý¹²Óö¥½ÇÑõÀë×Ó¿ÉÐγɵº×´¡¢Á´×´¡¢²ã×´¡¢¹Ç¼ÜÍø×´ËÄ´óÀà½á¹¹ÐÍʽ¡£Í¼(b)ΪһÖÖÎÞÏÞ³¤µ¥Á´½á¹¹µÄ¶à¹èËá¸ù£»ÆäÖÐSiÓëOµÄÔ­×ÓÊýÖ®±ÈΪ           »¯Ñ§Ê½Îª               ¡£

 


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

¾«Ó¢¼Ò½ÌÍø