±»ÓþΪ¸Ä±äδÀ´ÊÀ½çµÄÊ®´óпƼ¼Ö®Ò»µÄȼÁϵç³Ø¾ßÓÐÎÞÎÛȾ¡¢ÎÞÔëÒô¡¢¸ßЧÂʵÄÌص㣮ÓÒͼΪÇâÑõȼÁϵç³ØµÄ½á¹¹Ê¾Òâͼ£¬µç½âÖÊÈÜҺΪKOHÈÜÒº£¬µç¼«²ÄÁÏΪÊèËɶà¿×ʯī°ô£®µ±ÑõÆøºÍÇâÆø·Ö±ðÁ¬Ðø²»¶ÏµØ´ÓÕý¡¢¸ºÁ½¼«Í¨ÈëȼÁϵç³Øʱ£¬±ã¿ÉÔڱպϻØ·Öв»¶ÏµØ²úÉúµçÁ÷£®ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£© д³öÇâÑõȼÁϵç³Ø¹¤×÷ʱÕý¼«µç¼«·´Ó¦·½³Ìʽ£º
___________                            ¡£
£¨2£©Èç¹û¸ÃÇâÑõȼÁϵç³ØÿתÒÆ0.1molµç×Ó£¬ÏûºÄ±ê×¼×´¿öÏÂ_______LÑõÆø¡£    
£¨3£© Èô½«´ËȼÁϵç³Ø¸Ä½øΪֱ½ÓÒÔ¼×ÍéºÍÑõÆøΪԭÁϽøÐй¤×÷ʱ£¬¸º¼«·´Ó¦Ê½Îª_______¡£ µç³Ø×ÜÀë×Ó·´Ó¦·½³ÌʽΪ___________________________¡£

£¨1£©2H2O£«O2£«4e£­=4OH£­ £¨2·Ö£©
£¨2£©0.56L£¨2·Ö£©
£¨3£©CH4£«10OH£­£­8e£­=CO32-£«7H2O £¨2·Ö£©
CH4£«2O2£«2OH£­=CO32-£«3H2O£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£ºÒòΪÇâÑõȼÁϵç³ØµÄµç½âÖÊÈÜҺΪ¼îÐÔ£¬ËùÒÔÕý¼«·´Ó¦Ê½Îª2H2O£«O2£«4e£­=4OH£­£¬ÇÒÿתÒÆ1molµç×Ó£¬ÏûºÄ±ê×¼×´¿öÏÂÑõÆø5.6L£¬ËùÒÔתÒÆ0.1molµç×Óʱ£¬ÏûºÄ±ê×¼×´¿öÏÂÑõÆø0.56L,µ±ÒÔ¼×ÍéΪȼÁÏʱ£¬¼×ÍéΪ¸º¼«·´Ó¦Îµç¼«·´Ó¦Ê½ÎªCH4£«10OH£­£­8e£­=CO32-£«7H2O£¬Õý¼«·´Ó¦Ê½Îª2O2+4H2O+8e-=8OH-,Á½¸öµç¼«·´Ó¦Ê½ºÏ²¢Îªµç³ØµÄ×Ü·´Ó¦CH4£«2O2£«2OH£­=CO32-£«3H2O¡£
¿¼µã£ºÔ­µç³Ø·´Ó¦·½³ÌʽµÄÊéд¼°Ïà¹Ø¼ÆËã¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨14·Ö£©¹¤ÒµÉÏÓÃij¿óÔü£¨º¬ÓÐCu2O¡¢Al2O3¡¢Fe2O3¡¢SiO2£©ÌáÈ¡Í­µÄ²Ù×÷Á÷³ÌÈçÏ£º

ÒÑÖª£ºCu2O£«2H£«=Cu£«Cu2£«£«H2O
£¨1£©ÊµÑé²Ù×÷¢ñµÄÃû³ÆΪ________£»ÔÚ¿ÕÆøÖÐ×ÆÉÕ¹ÌÌå»ìºÏÎïDʱ£¬Óõ½¶àÖÖ¹èËáÑÎÖʵÄÒÇÆ÷£¬³ý²£Á§°ô¡¢¾Æ¾«µÆ¡¢ÄàÈý½ÇÍ⣬»¹ÓÐ________£¨ÌîÒÇÆ÷Ãû³Æ£©¡£
£¨2£©ÂËÒºAÖÐÌúÔªËصĴæÔÚÐÎʽΪ________£¨ÌîÀë×Ó·ûºÅ£©£¬Éú³É¸ÃÀë×ÓµÄÀë×Ó·½³ÌʽΪ_______________________________________________
________£¬¼ìÑéÂËÒºAÖдæÔÚ¸ÃÀë×ÓµÄÊÔ¼ÁΪ________£¨ÌîÊÔ¼ÁÃû³Æ£©¡£
£¨3£©½ðÊôµ¥ÖÊEÓë¹ÌÌå»ìºÏÎïF·¢ÉúµÄijһ·´Ó¦¿ÉÓÃÓÚº¸½Ó¸Ö¹ì£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________________________________________________¡£
£¨4£©³£ÎÂÏ£¬µÈpHµÄNaAlO2ºÍNaOHÁ½·ÝÈÜÒºÖУ¬ÓÉË®µçÀë³öµÄc£¨OH£­£©Ç°ÕßΪºóÕßµÄ108±¶¡£ÔòÁ½ÖÖÈÜÒºµÄpH£½________¡£
£¨5£©¢ÙÀûÓõç½â·¨½øÐдÖÍ­¾«Á¶Ê±£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ________£¨Ìî´úºÅ£©¡£
a£®µçÄÜÈ«²¿×ª»¯Îª»¯Ñ§ÄÜ
b£®´ÖÍ­½ÓµçÔ´Õý¼«£¬·¢ÉúÑõ»¯·´Ó¦
c£®¾«Í­×÷Òõ¼«£¬µç½âºóµç½âÒºÖÐCu2£«Å¨¶È¼õС
d£®´ÖÍ­¾«Á¶Ê±Í¨¹ýµÄµçÁ¿ÓëÒõ¼«Îö³öÍ­µÄÁ¿ÎÞÈ·¶¨¹Øϵ
¢Ú´ÓŨÁòËᡢŨÏõËá¡¢ÕôÁóË®ÖÐÑ¡ÓúÏÊʵÄÊÔ¼Á£¬²â¶¨´ÖÍ­ÑùÆ·ÖнðÊôÍ­µÄÖÊÁ¿·ÖÊý£¬Éæ¼°µÄÖ÷Òª²½Ö裺³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÑùÆ·¡ú______________¡ú¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡ú³ÆÁ¿Ê£Óà¹ÌÌåÍ­µÄÖÊÁ¿¡££¨ÌîȱÉٵIJÙ×÷²½Ö裬²»±ØÃèÊö²Ù×÷¹ý³ÌµÄϸ½Ú£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÂÁÉú²ú²úÒµÁ´ÓÉÂÁÍÁ¿ó¿ª²É¡¢Ñõ»¯ÂÁÖÆÈ¡¡¢ÂÁµÄÒ±Á¶ºÍÂÁ²Ä¼Ó¹¤µÈ»·½Ú¹¹³É¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¹¤ÒµÉϲÉÓõç½âÑõ»¯ÂÁºÍ±ù¾§Ê¯(Na3AlF6)ÈÛÈÚÌåµÄ·½·¨Ò±Á¶µÃµ½½ðÊôÂÁ£º

2Al2O 4Al£«3O2¡ü
¼ÓÈë±ù¾§Ê¯µÄ×÷Óãº________________________________________________¡£
£¨2£©ÉÏÊö¹¤ÒÕËùµÃÂÁ²ÄÖÐÍùÍùº¬ÓÐÉÙÁ¿FeºÍSiµÈÔÓÖÊ£¬¿ÉÓõç½â·½·¨½øÒ»²½Ìá´¿£¬¸Ãµç½â³ØÖÐÑô¼«µÄµç¼«·´Ó¦Ê½Îª________________£¬ÏÂÁпÉ×÷Òõ¼«²ÄÁϵÄÊÇ__________¡£
A£®ÂÁ²Ä        B£®Ê¯Ä«        C£®Ç¦°å        D£®´¿ÂÁ
£¨3£©Ñô¼«Ñõ»¯ÄÜʹ½ðÊô±íÃæÉú³ÉÖÂÃܵÄÑõ»¯Ä¤¡£ÒÔÏ¡ÁòËáΪµç½âÒº£¬ÂÁÑô¼«·¢ÉúµÄµç¼«·´Ó¦Ê½Îª_____________________________________________________________¡£
£¨4£©ÔÚÂÁÑô¼«Ñõ»¯¹ý³ÌÖУ¬ÐèÒª²»¶ÏµØµ÷Õûµçѹ£¬ÀíÓÉÊÇ_________________¡£
£¨5£©ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ__________________¡£
A£®Ñô¼«Ñõ»¯ÊÇÓ¦ÓÃÔ­µç³ØÔ­Àí½øÐнðÊô²ÄÁϱíÃæ´¦ÀíµÄ¼¼Êõ
B£®ÂÁµÄÑô¼«Ñõ»¯¿ÉÔöÇ¿ÂÁ±íÃæµÄ¾øÔµÐÔÄÜ
C£®ÂÁµÄÑô¼«Ñõ»¯¿ÉÌá¸ß½ðÊôÂÁ¼°ÆäºÏ½ðµÄÄ͸¯Ê´ÐÔ£¬µ«ÄÍÄ¥ÐÔϽµ
D£®ÂÁµÄÑô¼«Ñõ»¯Ä¤¸»Óжà¿×ÐÔ£¬¾ßÓкÜÇ¿µÄÎü¸½ÐÔÄÜ£¬ÄÜÎü¸½È¾Á϶ø³Ê¸÷ÖÖÑÕÉ«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

²¿·ÖÄÑÈÜÎïµÄÑÕÉ«ºÍ³£ÎÂϵÄKspÈçϱíËùʾ£º

 
Cu(OH)2
CuOH
CuCl
Cu2O
ÑÕÉ«
À¶É«
»ÆÉ«
°×É«
שºìÉ«
Ksp(25 ¡æ)
1£®6¡Á10£­19
1£®0¡Á10£­14
1£®2¡Á10£­6
¡ª

ijÑо¿ÐÔѧϰС×é¶Ôµç½âʳÑÎË®½øÐÐÁËÈçÏÂ̽¾¿£º
ʵÑé¢ñ×°ÖÃÈçͼËùʾ£¬½ÓͨµçÔ´ºó£¬·¢ÏÖa¡¢bµç¼«ÉϾùÓÐÆøÅݲúÉú¡£
£¨1£©µç½â¹ý³ÌÖеÄ×ÜÀë×Ó·´Ó¦·½³ÌʽΪ_________________________________________¡£
£¨2£©ÎªÁËÈ·¶¨µçÔ´µÄÕý¡¢¸º¼«£¬ÏÂÁвÙ×÷Ò»¶¨ÐÐÖ®ÓÐЧµÄÊÇ¡¡¡¡   ¡¡¡¡¡£
A£®¹Û²ìÁ½¼«²úÉúÆøÌåµÄÑÕÉ«
B£®ÍùUÐιÜÁ½¶Ë·Ö±ðµÎÈëÊýµÎ·Ó̪ÊÔÒº
C£®ÓÃȼ×ŵÄľÌõ¿¿½üUÐιܿÚ
D£®ÔÚUÐιܿÚÖÃÒ»ÕÅʪÈóµÄµí·Û­KIÊÔÖ½
ʵÑé¢ò°ÑÉÏÊöµç½â×°ÖõÄʯī°ô»»³ÉÍ­°ô£¬ÓÃÖ±Á÷µçÔ´½øÐеç½â£¬×°ÖÃÈçͼËùʾ¡£

¹Û²ìµ½µÄÏÖÏóÈçÏÂËùʾ£º
¢Ù¿ªÊ¼ÎÞÃ÷ÏÔÏÖÏó£¬ËæºóÒºÃæÒÔϵÄÍ­°ô±íÃæÖ𽥱䰵£»
¢Ú5 minºó£¬b¼«¸½½ü¿ªÊ¼³öÏÖ°×É«³Áµí£¬²¢Öð½¥Ôö¶à£¬ÇÒÏòa¼«À©É¢£»
¢Û10 minºó£¬×î¿¿½üa¼«µÄ°×É«³Áµí¿ªÊ¼±ä³ÉºìÉ«£»
¢Ü12 minºó£¬b¼«¸½½üµÄ°×É«³Áµí¿ªÊ¼±ä³É»ÆÉ«£¬È»ºóÖð½¥±ä³É³È»ÆÉ«£»
¢Ýa¼«Ò»Ö±ÓдóÁ¿ÆøÅݲúÉú£»
¢ÞÍ£Ö¹µç½â£¬½«UÐιÜÖÐÐü×ÇÒº¾²ÖÃÒ»¶Îʱ¼äºó£¬ÉϲãÈÜÒº³ÊÎÞÉ«£¬Ã»ÓгöÏÖÀ¶É«£¬Ï²ã³ÁµíÈ«²¿ÏÔשºìÉ«¡£
£¨3£© a¼«·¢ÉúµÄµç¼«·´Ó¦·½³ÌʽΪ________________________________________________________¡£
£¨4£© µç½â5 minºó£¬b¼«·¢ÉúµÄµç¼«·´Ó¦·½³ÌʽΪ___________________________________________¡£
£¨5£©12 minºó£¬b¼«¸½½ü³öÏֵijȻÆÉ«³ÁµíµÄ³É·ÖÊÇ¡¡¡¡¡¡¡¡¡¡£¬Ô­ÒòÊÇ___________________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ðîµç³ØÊÇÒ»ÖÖ¿ÉÒÔ·´¸´³äµç¡¢·ÅµçµÄ×°Öá£ÓÐÒ»ÖÖÐîµç³ØÔÚ³äµçºÍ·Åµçʱ·¢ÉúµÄ·´Ó¦ÎªNiO2+Fe+2H2OFe(OH)2+Ni(OH)2
£¨1£©¸ÃÐîµç³Ø·Åµçʱ£¬·¢Éú»¹Ô­·´Ó¦µÄÎïÖÊÊÇ       £¨Ìî×Öĸ£¬ÏÂͬ£©¡£
A£®NiO2             B£®Fe             C£®Fe(OH)2            D£®Ni(OH)2
£¨2£©ÏÂÁÐÓйظõç³ØµÄ˵·¨ÖÐÕýÈ·µÄÊÇ        
A£®·Åµçʱµç½âÖÊÈÜÒºÏÔÇ¿ËáÐÔ
B£®·Åµçʱ5.6g FeÈ«²¿×ª»¯ÎªFe(OH)2ʱ£¬Íâµç·ÖÐͨ¹ýÁË0.2 molµç×Ó
C£®³äµçʱÑô¼«·´Ó¦ÎªNi(OH)2+2OH£­?2e£­==NiO2+2H2O
D£®³äµçʱÒõ¼«¸½½üÈÜÒºµÄ¼îÐÔ±£³Ö²»±ä
£¨3£©ÓôËÐîµç³Øµç½âº¬ÓÐ0.01 mol CuSO4ºÍ0.01 mol NaClµÄ»ìºÏÈÜÒº100 mL£¬µç½â³ØµÄµç¼«¾ùΪ¶èÐԵ缫¡£µ±ÈÜÒºÖеÄCu2+ È«²¿×ª»¯³ÉCuʱ£¬Ñô¼«²úÉúµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ         £»½«µç½âºóµÄÈÜÒº¼ÓˮϡÊÍÖÁ1L£¬´ËʱÈÜÒºµÄpH=      ¡£
£¨4£©ÓôËÐîµç³Ø½øÐеç½â£¬ÇÒµç½â³ØµÄµç¼«¾ùΪͭµç¼«£¬µç½âÖÊÈÜҺΪŨ¼îÒºÓëNaClÈÜÒºµÄ»ìºÏÒº£¬µç½âÒ»¶Îʱ¼äºó£¬Í¬Ñ§ÃǾªÆæµØ·¢ÏÖ£¬Ñô¼«¸½½ü²»ÊÇÉú³ÉÀ¶É«³Áµí£¬¶øÊÇÉú³ÉºìÉ«³Áµí£¬²éÔÄ×ÊÁϵÃÖª¸ÃºìÉ«³ÁµíÊÇCu2O¡£Ð´³ö¸ÃÑô¼«Éϵĵ缫·´Ó¦Ê½£º                                           ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

µç½âÔ­ÀíÔÚ»¯Ñ§¹¤ÒµÖÐÓй㷺ӦÓá£ÓÒͼ±íʾһ¸öµç½â³Ø£¬×°Óеç½âÒºa£»X¡¢YÊÇÁ½¿éµç¼«°å£¬Í¨¹ýµ¼ÏßÓëÖ±Á÷µçÔ´ÏàÁ¬¡£Çë»Ø´ðÒÔÏÂÎÊÌ⣺

£¨1£©ÈôX¡¢Y¶¼ÊǶèÐԵ缫£¬aÊDZ¥ºÍNaClÈÜÒº£¬ÊµÑ鿪ʼʱ£¬Í¬Ê±ÔÚÁ½±ß¸÷µÎÈ뼸µÎ·Ó̪ÊÔÒº£¬Ôò
¢Ùµç½â³ØÖÐX¼«Éϵĵ缫·´Ó¦Ê½Îª                                  ¡£ÔÚX¼«¸½½ü¹Û²ìµ½µÄÏÖÏóÊÇ                                         ¡£
¢ÚYµç¼«Éϵĵ缫·´Ó¦Ê½Îª                                       £¬¼ìÑé¸Ãµç¼«·´Ó¦²úÎïµÄ·½·¨ÊÇ                                                                                 ¡£
£¨2£©ÈçÒªÓõç½â·½·¨¾«Á¶´ÖÍ­£¬µç½âÒºaÑ¡ÓÃCuSO4ÈÜÒº£¬Ôò
¢ÙXµç¼«µÄ²ÄÁÏÊÇ            £¬µç¼«·´Ó¦Ê½ÊÇ                            ¡£
¢ÚYµç¼«µÄ²ÄÁÏÊÇ            £¬µç¼«·´Ó¦Ê½ÊÇ                            ¡£
£¨ËµÃ÷£ºÔÓÖÊ·¢ÉúµÄµç¼«·´Ó¦²»±Øд³ö£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

°´ÒªÇóÌî¿Õ£º

A                    B
£¨1£©ÔÚAͼÖУ¬Ï¡ÁòËáΪµç½âÖÊÈÜÒº£¬Óõ¼ÏßÁ¬½Óºó£¬Í­Æ¬µç¼«·´Ó¦Ê½        ¡£
£¨2£©ÔÚBͼÖÐÍâ½ÓÖ±Á÷µçÔ´£¬ÈôÒªÔÚa¼«¶ÆÍ­£¬¼ÓÒÔ±ØÒªµÄÁ¬½Óºó£¬¸Ã×°ÖýР       £¬b¼«µç¼«·´Ó¦Ê½                                    ¡£
£¨3£©ÔÚBͼÖÐÍâ½ÓÖ±Á÷µçÔ´£¬Èôµç¼«Îª¶èÐԵ缫£¬µç½âÖÊÈÜÒºÊÇCuSO4ÈÜÒº£¨×ãÁ¿£©£¬µç½â×Ü·´Ó¦Àë×Ó·½³ÌʽΪ                        £¬Òõ¼«ÔöÖØ3.2 g£¬ÔòÑô¼«ÉϷųöµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýÊÇ_____L£¬¼ÓÈëÒ»¶¨Á¿µÄ     ºó£¨Ìѧʽ£©£¬ÈÜÒºÄָܻ´ÖÁÓëµç½âÇ°ÍêÈ«Ò»Ö¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨l£©ÂÁÓëijЩ½ðÊôÑõ»¯ÎïÔÚ¸ßÎÂϵķ´Ó¦³ÆΪÂÁÈÈ·´Ó¦£¬¿ÉÓÃÓÚÒ±Á¶¸ßÈÛµã½ðÊô¡£
ÒÑÖª£º4Al£¨s£©£«3O2£¨g£©=2Al2O3£¨s£© £½-2830kJ¡¤mol-1
 =+230kJ¡¤mol-1
 =-390kJ¡¤mol-1
ÂÁÓëÑõ»¯Ìú·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ                  ¡£
£¨2£©ÈçÏÂͼËùʾ£¬¸÷ÈÝÆ÷ÖÐÊ¢Óк£Ë®£¬ÌúÔÚÆäÖб»¸¯Ê´Ê±ÓÉ¿ìµ½ÂýµÄ˳ÐòÊÇ           £»
¢Ú×°ÖÃÖÐCuµç¼«Éϵĵ缫·´Ó¦Ê½Îª                                          ¡£

£¨3£©·°£¨V£©¼°Æ仯ºÏÎï¹ã·ºÓ¦ÓÃÓÚÐÂÄÜÔ´ÁìÓò¡£È«·°ÒºÁ÷´¢Äܵç³ØÊÇÀûÓò»Í¬¼Û̬Àë×Ó¶ÔµÄÑõ»¯»¹Ô­·´Ó¦À´ÊµÏÖ»¯Ñ§Äܺ͵çÄÜÏ໥ת»¯µÄ×°Öã¬ÆäÔ­ÀíÈçͼËùʾ¡£

¢Ùµ±×ó²ÛÈÜÒºÖð½¥ÓɻƱäÀ¶£¬Æäµç¼«·´Ó¦Ê½Îª                                       ¡£
¢Ú³äµç¹ý³ÌÖУ¬ÓÒ²ÛÈÜÒºÑÕÉ«Öð½¥ÓÉ      É«±äΪ     É«¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ

£¨10·Ö£©Ä³»¯Ñ§ÐËȤС×éµÄͬѧΪÁË̽¾¿ÂÁµç¼«ÔÚµç³ØÖеÄ×÷ÓÃ,Éè¼Æ²¢½øÐÐÁËÒÔÏÂһϵÁÐʵÑé,ʵÑé½á¹û¼Ç¼ÈçÏÂ:

񅧏
µç¼«²ÄÁÏ
µç½âÖÊÈÜÒº
µçÁ÷¼ÆÖ¸Õëƫת·½Ïò
1
Mg¡¢Al
Ï¡ÑÎËá
Æ«ÏòAl
2
Al¡¢Cu
Ï¡ÑÎËá
Æ«ÏòCu
3
Al¡¢C(ʯī)
Ï¡ÑÎËá
Æ«Ïòʯī
4
Mg¡¢Al
ÇâÑõ»¯ÄÆÈÜÒº
Æ«ÏòMg
5
Al¡¢Zn
ŨÏõËá
Æ«ÏòAl
ÊÔ¸ù¾Ý±íÖеÄʵÑéÏÖÏó»Ø´ðÏÂÁÐÎÊÌâ:
£¨1£©ÊµÑé1ºÍ2ÖÐAlËù×÷µÄµç¼«(Õý¼«»ò¸º¼«)         (Ìî¡°Ïàͬ¡±»ò¡°²»Ïàͬ¡±)¡£
£¨2£©ÊµÑé3Öиº¼«·´Ó¦Ê½£º¡¡               £»×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ£º ¡¡                           ¡£
£¨3£©ÊµÑé4ÖÐÂÁ×÷¡¡¡¡¡¡¡¡¼«£¬µç¼«·´Ó¦Ê½£º¡¡                           ¡£
£¨4£©½âÊÍʵÑé5ÖеçÁ÷¼ÆÖ¸ÕëÆ«ÏòÂÁµÄÔ­Òò                            ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸