ÌâÄ¿ÄÚÈÝ
ÒÑÖª£ºAΪº¬½ðÊôÀë×ӵĵ»ÆÉ«¹ÌÌ廯̨ÎE¡¢XΪ¿ÕÆøÖг£¼ûÆøÌ壬A¡¢B¡¢C¡¢Dº¬ÓÐÏàͬµÄ½ðÊôÀë×Ó£¬Æäת»¯¹ØÏµÈçÏÂͼ£¨²¿·Ö²úÎïÒÑÂÔÈ¥£©¡£
![]()
Çë»Ø´ðÏÂÁÐÎÊÌâ
£¨1£©ÕâÖÖ½ðÊôÀë×ÓµÄÀë×ӽṹʾÒâͼΪ_____________;
£¨2£©XµÄµç×Óʽ_______________;
£¨3£©BÖÐËùº¬»¯Ñ§¼üµÄÀàÐÍÊÇ_____________;
³£Î³£Ñ¹Ï£¬7.8gAÓë×ãÁ¿µÄË®³ä·Ö·´Ó¦·Å³öÈÈÁ¿a kJ£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
_________________________________________________________________.
£¨4£©¢ÙCÒ²¿Éת»¯ÎªB£¬Ð´³ö¸Ãת»¯µÄ»¯Ñ§·½³Ìʽ_______________________________;
¢ÚBÓëD·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______________________________________.
£¨5£©½«Ò»¶¨Á¿µÄÆøÌåXͨÈë2LBµÄÈÜÒºÖУ¬ÏòËùµÃÈÜÒºÖбßÖðµÎ¼ÓÈëÏ¡ÑÎËá±ßÕñµ´ÖÁ¹ýÁ¿£¬²úÉúµÄÆøÌåÓëÑÎËáÎïÖʵÄÁ¿µÄ¹ØÏµÈçͼ£¨ºöÂÔÆøÌåµÄÈܽâºÍHClµÄ»Ó·¢£©¡£
![]()
Çë»Ø´ð£ºaµãÈÜÒºÖÐËùº¬ÈÜÖʵĻ¯Ñ§Ê½Îª__________ £¬a£bÖ®¼äµÄ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_________________________¡£
£¨1£©Na+
(1·Ö)
£¨2£©
(1·Ö)
£¨3£©£¨¼«ÐÔ£©¹²¼Û¼ü¡¢Àë×Ó¼ü£¨2·Ö£©2Na2O2(s)+2H2O(l)=4NaOH(aq)+ O2(g)¡ü¡÷H=-20akJ/mol(2·Ö)
£¨4£©¢ÙCa(OH)2+Na2CO3=CaCO3¡ý+2NaOH»ò Ba(OH)2+Na2CO3=BaCO3¡ý+2NaOH
(2·Ö)
¢ÚOH-+HCO3-=CO32-+ H2O(2·Ö)
£¨5£©Na2CO3¡¢NaCl (2·Ö) CO32- + H+ = HCO3-(2·Ö)
ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ
| A£® | B£® | C£® | D£® |
| ͨµçÒ»¶Îʱ¼äºó£¬½Á°è¾ùÔÈ£¬ÈÜÒºµÄpHÔö´ó | ¼×µç¼«Éϵĵ缫·´Ó¦Îª£º 2Cl£ -2e£ = Cl2¡ü | Ptµç¼«Éϵĵ缫·´Ó¦Îª£ºO2£«2H2O£«4e£==4OH£ | ×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º 2Fe3++Cu=Cu2++ 2Fe2+ |
|
|