ÌâÄ¿ÄÚÈÝ


ÒÒÏ©ÊÇʯÓÍÁÑ½âÆøµÄÖ÷Òª³É·Ö£¬¡£Çë»Ø´ðÏÂÁÐÎÊÌâ¡£

(1)ÒÒÏ©µÄµç×Óʽ____________£¬½á¹¹¼òʽ____________¡£

(2)¼ø±ð¼×ÍéºÍÒÒÏ©µÄÊÔ¼ÁÊÇ______(ÌîÐòºÅ)¡£

A£®Ï¡ÁòËá B£®äåµÄËÄÂÈ»¯Ì¼ÈÜÒº

C£®Ë® D£®ËáÐÔ¸ßÃÌËá¼ØÈÜÒº

(3)ÏÂÁÐÎïÖÊÖУ¬¿ÉÒÔͨ¹ýÒÒÏ©¼Ó³É·´Ó¦µÃµ½µÄÊÇ______(ÌîÐòºÅ)¡£

A£®CH3CH3 B£®CH3CHCl2

C£®CH3CH2OH D£®CH3CH2Br

(4)ÒÑÖª 2CH3CHO£«O2 2CH3COOH¡£ÈôÒÔÒÒϩΪÖ÷ÒªÔ­ÁϺϳÉÒÒËᣬÆäºÏ³É·ÏßÈçÏÂͼËùʾ¡£


·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽΪ____________________________________¡£

¹¤ÒµÉÏÒÔÒÒϩΪԭÁÏ¿ÉÒÔÉú²úÒ»ÖÖÖØÒªµÄºÏ³ÉÓлú¸ß·Ö×Ó»¯ºÏÎÆä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________________________________£¬·´Ó¦ÀàÐÍÊÇ__________________¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¹¤ÒµÉÏ¿ÉÓÃÈíÃÌ¿ó£¨Ö÷Òª³É·ÖÊÇMnO2£©ºÍ»ÆÌú¿ó£¨Ö÷Òª³É·ÖÊÇFeS2£©ÎªÖ÷ÒªÔ­ÁÏÖÆ±¸¸ßÐÔÄÜ´ÅÐÔ²ÄÁÏ̼ËáÃÌ£¨MnCO3£©¡£Æä¹¤ÒµÁ÷³ÌÈçÏ£º

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÎªÁËÌá¸ßÈܽþ¹¤ÐòÖÐÔ­ÁϵĽþ³öÂÊ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÓУ¨Ð´Ò»Ìõ£©              ¡£

£¨2£©³ýÌú¹¤ÐòÖУ¬ÔÚ¼ÓÈëʯ»Òµ÷½ÚÈÜÒºµÄpHǰ£¬¼ÓÈëÊÊÁ¿µÄÈíÃÌ¿ó£¬Æä×÷ÓÃÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©                                                 ¡£

£¨3£©¾»»¯¹¤ÐòµÄÄ¿µÄÊdzýÈ¥ÈÜÒºÖеÄCu2+¡¢Ca2+µÈÔÓÖÊ¡£Èô²âµÃÂËÒºÖÐc(F-)=0.01 mol•L-1£¬ÂËÒºÖвÐÁôµÄc(Ca2+)=               ¡²ÒÑÖª£ºKsp(CaF2)=1.46¡Á10-10¡³£¬

    £¨4£©³ÁÃ̹¤ÐòÖУ¬298K¡¢c(Mn2+)Ϊ1.05 mol•L-1ʱ£¬ÊµÑé²âµÃMnCO3µÄ²úÂÊÓëÈÜÒºpH¡¢·´Ó¦Ê±¼äµÄ¹ØÏµÈçÓÒͼËùʾ¡£¸ù¾ÝͼÖÐÐÅÏ¢µÃ³öµÄ½áÂÛÊÇ                                     ¡£

    £¨5£©³ÁÃ̹¤ÐòÖÐÓÐCO2Éú³É£¬ÔòÉú³ÉMnCO3µÄÀë×Ó·½³ÌʽÊÇ___________           ¡£

£¨6£©´Ó³ÁÃ̹¤ÐòÖеõ½´¿¾»MnCO3µÄ²Ù×÷·½·¨ÊÇ   

                       £¬³ÁÃ̹¤ÐòÖÐÔõÑùÖ¤Ã÷MnCO3ÒѾ­Ï´µÓ¸É¾»                                         

                                       £¬¸±²úÆ·AµÄ»¯Ñ§Ê½ÊÇ              ¡£

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

¾«Ó¢¼Ò½ÌÍø