ÌâÄ¿ÄÚÈÝ

A¡¢B¡¢C¡¢X¾ùΪÖÐѧ³£¼ûµÄ´¿¾»ÎËüÃÇÖ®¼äÓÐÈçÏÂת»¯¹ØÏµ£¨¸±²úÎïÒÑÂÔÈ¥£©¡£

ÊԻشð£º
£¨1£©ÈôXÊÇÇ¿Ñõ»¯ÐÔµ¥ÖÊ£¬ÔòA²»¿ÉÄÜÊÇ           ¡£
a¡¢S         b¡¢N2          c¡¢Na            d¡¢Mg            e¡¢Al
£¨2£©ÈôXÊǽðÊôµ¥ÖÊ£¬ÏòCµÄË®ÈÜÒºÖеμÓAgNO3ÈÜÒº£¬²úÉú²»ÈÜÓÚÏ¡HNO3­µÄ°×É«³Áµí£¬
ÔòBµÄ»¯Ñ§Ê½Îª                  £»CÈÜÒºÔÚÖü´æÊ±Ó¦¼ÓÈëÉÙÁ¿X£¬ÀíÓÉÊÇ£¨ÓñØÒªµÄÎÄ×ÖºÍÀë×Ó·½³Ìʽ±íʾ£©                                                 £¬¼ìÑé´ËCÈÜÒºÖнðÊôÔªËØ¼Û̬µÄ²Ù×÷·½·¨ÊÇ                                             
_____________________________________________________________________________¡£
£¨3£©ÈôA¡¢B¡¢CΪº¬½ðÊôÔªËØµÄÎÞ»ú»¯ºÏÎXΪǿµç½âÖÊ£¬AÈÜÒºÓëCÈÜÒº·´Ó¦Éú³ÉB£¬ÔòBµÄ»¯Ñ§Ê½Îª        £¬XµÄ»¯Ñ§Ê½¿ÉÄÜΪ£¨Ð´³ö²»Í¬ÀàÎïÖÊ£©      »ò      £¬·´Ó¦¢ÙµÄÀë×Ó·½³ÌʽΪ                             »ò                           ¡£

£¨¹²15·Ö£©£¨1£©de £»(2·Ö)
£¨2£©FeC13£»(1·Ö)   2Fe3++Fe=3Fe2£«¡¢·ÀÖ¹Fe2£«±»Ñõ»¯£»(3·Ö)
ÓÃÊÔ¹ÜÈ¡ÉÙÁ¿CÈÜÒº£¬µÎ¼ÓKSCNÈÜÒº£¬ÎÞÑÕÉ«±ä»¯£¬ÔٵμÓÂÈË®£¨»òÏõËᣩ£¬ÈÜÒº³ÊѪºìÉ«£¬Ö¤Ã÷Ô­ÈÜÒºÖÐÓÐFe2+´æÔÚ¡£(2·Ö)
£¨3£©A1£¨OH£©3£»(1·Ö)   NaOH£¨»òKOH£©£»(1·Ö)   HC1£¨»òÆäËûÇ¿Ëᣩ£»(1·Ö)
A13++3OH-=A1£¨OH£©3¡ý £»(2·Ö)
A1O2-+H++H2O=A1£¨OH£©3¡ý(2·Ö)

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
A¡¢B¡¢C¡¢X¾ùΪÖÐѧ³£¼ûµÄ´¿¾»ÎËüÃÇÖ®¼äÓÐÈçÏÂת»¯¹ØÏµ£¨¸±²úÎïÒÑÂÔÈ¥£©

ÊԻشð£º
£¨1£©ÈôXÊÇÇ¿Ñõ»¯ÐÔµ¥ÖÊ£¬ÔòA ²»¿ÉÄÜÊÇ
de
de
£®
a£®S£»   b£®N2£»c£®Na£»  d£®Mg£»  e£®Al
£¨2£©Èô A¡¢B¡¢CΪÌþµÄº¬ÑõÑÜÉúÎXΪ·Ç½ðÊôµ¥ÖÊ£¬AÓëC·´Ó¦Éú³ÉD£¬DµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª88£¬ÔòCµÄ¹ÙÄÜÍÅΪ
ôÈ»ù
ôÈ»ù
£¬·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽΪ
2CH3CH2OH+O2
´ß»¯¼Á
¡÷
2CH3CHO+2H2O
2CH3CH2OH+O2
´ß»¯¼Á
¡÷
2CH3CHO+2H2O
£®
£¨3£©ÈôA¡¢B¡¢CΪº¬ÓнðÊôÔªËØµÄÎÞ»ú»¯ºÏÎXΪǿµç½âÖÊ£¬A ÈÜÒºÓëCÈÜÒº·´Ó¦¿ÉÉú³ÉB£¬ÔòBµÄ»¯Ñ§Ê½¿ÉÄÜΪ
Al£¨OH£©3
Al£¨OH£©3
£¬XµÄ»¯Ñ§Ê½¿ÉÄÜΪ £¨Ð´³ö²»Í¬ÀàÎïÖÊ£©
NaOH£¨»òKOH£©
NaOH£¨»òKOH£©
»ò
HCl
HCl
£¬·´Ó¦¢ÙµÄÀë×Ó·½³ÌʽΪ
Al3++3OH-=Al£¨OH£©3¡ý
Al3++3OH-=Al£¨OH£©3¡ý
»ò
AlO2-+H++H2O=Al£¨OH£©3¡ý
AlO2-+H++H2O=Al£¨OH£©3¡ý
£®
£¨4£©Èô X ÊǽðÊôµ¥ÖÊ£¬ÏòCµÄË®ÈÜÒºÖеμÓAgNO3ÈÜÒº£¬²úÉú²»ÈÜÓÚÏ¡HNO3µÄ°×É«³Áµí£¬ÔòAµÄµç×ÓʽΪ
£»CÈÜÒºÔÚÖü´æÊ±Ó¦¼ÓÈëÉÙÁ¿X£¬ÀíÓÉÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©
2Fe2++Fe=3Fe3+
2Fe2++Fe=3Fe3+
£¬¼ìÑé´ËCÈÜÒºÖнðÊôÔªËØ¼Û̬µÄ²Ù×÷·½·¨ÊÇ
ÓÃÊÔ¹ÜÈ¡ÉÙÁ¿CÈÜÒº£¬µÎ¼ÓKSCNÈÜÒº£¬ÎÞÑÕÉ«±ä»¯£¬ÔٵμÓÂÈË®£¨»òÏõËᣩ£¬ÈÜÒº³ÊºìÉ«£¬Ö¤Ã÷Ô­ÈÜÒºÖÐÓÐFe2+´æÔÚ
ÓÃÊÔ¹ÜÈ¡ÉÙÁ¿CÈÜÒº£¬µÎ¼ÓKSCNÈÜÒº£¬ÎÞÑÕÉ«±ä»¯£¬ÔٵμÓÂÈË®£¨»òÏõËᣩ£¬ÈÜÒº³ÊºìÉ«£¬Ö¤Ã÷Ô­ÈÜÒºÖÐÓÐFe2+´æÔÚ
£®
A¡¢B¡¢C¡¢X¾ùΪ³£¼ûµÄ´¿¾»ÎËüÃÇÖ®¼äÓÐÈçÏÂת»¯¹ØÏµ£¨¸±²úÆ·ÒÑÂÔÈ¥£©£®

ÊԻشð£º
£¨1£©ÈôXÊÇÇ¿Ñõ»¯ÐÔµ¥ÖÊ£¬ÔòA²»¿ÉÄÜÊÇ
de
de
£®
a£®S         b£®N2         c£®Na        d£®Mg        e£®Al
£¨2£©ÈôXÊǽðÊôµ¥ÖÊ£¬ÏòCµÄË®ÈÜÒºÖеÎÈëAgNO3ÈÜÒº£¬²úÉú²»ÈÜÓÚÏ¡HNO3µÄ°×É«³Áµí£¬ÔòBµÄ»¯Ñ§Ê½Îª
FeCl3
FeCl3
£»CÈÜÒºÔÚÖü´æÊ±Ó¦¼ÓÈëÉÙÁ¿X£¬ÀíÓÉÊÇ£¨ÓñØÒªµÄÎÄ×ÖºÍÀë×Ó·½³Ìʽ±íʾ£©
2Fe3++Fe¨T3Fe2+ ±£Ö¤ÈÜÒºÖÐFe2+ΪÖ÷
2Fe3++Fe¨T3Fe2+ ±£Ö¤ÈÜÒºÖÐFe2+ΪÖ÷
£®¼ìÑé´ËCÈÜÒºÖнðÊôÔªËØ¼Û̬µÄ²Ù×÷·½·¨ÊÇ
ÏòÈÜÒºÖеμÓÉÙÁ¿KSCNÈÜÒº£¬ÎÞÃ÷ÏÔÏÖÏó£¬Í¨ÈëÉÙÁ¿Cl2ÈÜÒº³ÊѪºìÉ«
ÏòÈÜÒºÖеμÓÉÙÁ¿KSCNÈÜÒº£¬ÎÞÃ÷ÏÔÏÖÏó£¬Í¨ÈëÉÙÁ¿Cl2ÈÜÒº³ÊѪºìÉ«
£®
£¨3£©ÈôA¡¢B¡¢CΪº¬Ä³½ðÊôÔªËØµÄÎÞ»ú»¯ºÏÎXΪǿµç½âÖÊ£¬AÈÜÒºÓëCÈÜÒº·´Ó¦Éú³ÉB£¬ÔòBµÄ»¯Ñ§Ê½Îª
Al£¨OH£©3
Al£¨OH£©3
£¬XµÄ»¯Ñ§Ê½¿ÉÄÜΪ£¨Ð´³ö²»Í¬ÀàÎïÖÊ£©
HCl
HCl
»ò
NaOH
NaOH
£®·´Ó¦¢ÙµÄÀë×Ó·½³ÌʽΪ
AlO2-+H++H2O¨TAl£¨OH£©3¡ý»òAl3++3OH-¨TAl£¨OH£©3¡ý
AlO2-+H++H2O¨TAl£¨OH£©3¡ý»òAl3++3OH-¨TAl£¨OH£©3¡ý
£®

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

¾«Ó¢¼Ò½ÌÍø