ÌâÄ¿ÄÚÈÝ

¹¤ÒµÉÏÒÔÂÁÍÁ¿ó£¨Ö÷Òª³É·ÖÊÇA12O3£¬º¬ÔÓÖÊFe2O3ºÍSiO2£©ÎªÔ­ÁÏÉú²úÂÁ£¬ÆäÉú²úÁ÷³ÌÈçÏ£º

¾«Ó¢¼Ò½ÌÍø

Çë»Ø´ð£º
£¨1£©¹¤ÒµÉÏAl2O3 Ò±Á¶AlËù²ÉÓõķ½·¨ÊÇ______£¬»¯Ñ§·½³Ìʽ______
£¨2£©¼ÓÈëÊÔ¼Á¼×ºó·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÓÐÁ½¸ö£¬Ò»ÊÇ______£¬¶þÊÇ______£®
£¨3£©ÔÚÈÜÒºBÖмÓÈë¹ýÁ¿ÑÎËáÈÜÒºµÄÄ¿µÄÊÇ______£®
£¨4£©¼ÓÈëÊÔ¼ÁÒÒºó·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ______£®
£¨5£©ÂÁÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£®
SiO2ºÍÑÎËá²»·´Ó¦£¬Fe2O3ºÍÇâÑõ»¯ÄƲ»·´Ó¦¶øÑõ»¯ÂÁÄÜ·´Ó¦£¬¸Ã¹¤ÒÕÁ÷³ÌÔ­ÀíΪ£ºÂÁÍÁ¿ó¼ÓÈëÇâÑõ»¯ÄÆ£¬ÂËÒºCÖк¬ÓÐÆ«ÂÁËá¸ùÀë×Ó¡¢¹èËá¸ùÀë×Ó£¬A³ÁµíΪÑõ»¯Ìú£¬BÈÜÒºÖмӹýÁ¿ÑÎËᣬ³ýÈ¥¹èËá¸ùÀë×Ó£¬½«Æ«ÂÁËá¸ùÀë×Óת»¯ÎªÂÁÀë×Ó£¬½øÈëÂËÒºC£¬³ÁµíDΪ¹èËᣬÊÔ¼ÁÒÒΪ°±Ë®£¬½«ÂÁÀë×Óת»¯ÎªÇâÑõ»¯ÂÁ³Áµí£¬¼ÓÈÈ·Ö½âµÃÑõ»¯ÂÁ£¬
£¨1£©¹¤ÒµÉÏÀûÓõç½âÈÛÈÚµÄAl2O3 Éú³ÉAlÓëÑõÆø£¬ÒÔ´ËÒ±Á¶½ðÊôAl£¬·´Ó¦·½³ÌʽΪ2Al2O3£¨ÈÛÈÚ£©
   µç½â   
.
±ù¾§Ê¯
4Al+3O2¡ü£¬
¹Ê´ð°¸Îª£ºµç½âÈÛÈÚµÄAl2O3£¬2Al2O3£¨ÈÛÈÚ£©
   µç½â   
.
±ù¾§Ê¯
4Al+3O2¡ü£»
£¨2£©¶þÑõ»¯¹èÓëÇâÑõ»¯ÄÆ·´Ó¦Éú³É¹èËáÄÆÓëË®£¬·´Ó¦Àë×Ó·½³ÌʽΪ£ºSiO2+2OH-¨TSiO32-+H2O£¬
Ñõ»¯ÂÁÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³ÉÆ«ÂÁËáÄÆÓëË®£¬·´Ó¦Àë×Ó·½³ÌʽΪ£ºAl2O3+2OH-¨T2AlO2-+H2O£¬
¹Ê´ð°¸Îª£ºSiO2+2OH-¨TSiO32-+H2O£¬Al2O3+2OH-¨T2AlO2-+H2O£»
£¨3£©Óɹ¤ÒÕÁ÷³Ì¿ÉÖª£¬BÈÜÒºÖмӹýÁ¿ÑÎËᣬ³ýÈ¥¹èËá¸ùÀë×Ó£¬½«Æ«ÂÁËá¸ùÀë×Óת»¯ÎªÂÁÀë×Ó£¬½øÈëÂËÒº£»
¹Ê´ð°¸Îª£º³ýÈ¥¹èËá¸ùÀë×Ó£¬½«Æ«ÂÁËá¸ùÀë×Óת»¯ÎªÂÁÀë×Ó£»
£¨4£©°±Ë®ÓëÂÁÀë×Ó·´Ó¦Éú³ÉÇâÑõ»¯ÂÁ³Áµí£¬·´Ó¦Àë×Ó·½³ÌʽΪ£ºAl3++3NH3?H2O¨TAl£¨OH£©3¡ý+3NH4+£¬
¹Ê´ð°¸Îª£ºAl3++3NH3?H2O¨TAl£¨OH£©3¡ý+3NH4+£»
£¨5£©ÂÁÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³ÉÆ«ÂÁËáÄÆÓëÇâÆø£¬·´Ó¦Àë×Ó·½³ÌʽΪ£º2Al+2OH-+2H2O¨T2AlO2-+3H2¡ü£¬
¹Ê´ð°¸Îª£º2Al+2OH-+2H2O¨T2AlO2-+3H2¡ü£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¾«Ó¢¼Ò½ÌÍøÄ¿Ç°Á÷ͨµÄµÚÎåÌ×ÈËÃñ±ÒÓ²±Ò²ÄÁÏ·Ö±ðΪ£º1ÔªÓ²±ÒΪͭо¶ÆÄøºÏ½ð£¬5½ÇÓ²±ÒΪͭо¶ÆÍ­ºÏ½ð£¬1½ÇÓ²±ÒΪӲÂÁºÏ½ð£®Çë»Ø´ð£º£¨ÌâÄ¿Öеİٷֺ¬Á¿¾ùΪÖÊÁ¿·ÖÊý£©
£¨1£©¸ÖÊǺ¬Ì¼Á¿Îª0.03%¡«2%µÄ
 
£¨ÌîÎïÖÊÀà±ðÃû³Æ£©£®
£¨2£©ÈçͼËùʾµÄ×°ÖÃÖУ¬¼×¡¢ÒÒ¡¢±ûÈý¸öÉÕ±­·Ö±ðÊ¢·Å×ãÁ¿µÄÈÜÒº£¬µç¼«¼°ÈÜÒºÈçϱíËùʾ£º
µç¼« a b c d e f
ʯī ʯī Í­ ¸Ö ¸Ö Í­
ÈÜÒº NaClÈÜÒº CuSO4ÈÜÒº CuSO4ÈÜÒº
ͨµçºóµç¼«aÉϿɲúÉúÄÜʹʪÈóµÄµâ»¯¼Øµí·ÛÊÔÖ½±äÀ¶µÄÆøÌ壮
¢ÙÉÏÊö×°ÖÃÖÐM¼«ÎªÖ±Á÷µçÔ´µÄ
 
¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬µç¼«bÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½Îª
 
£®
¢ÚÈôÔÚ¸Ö±ÒÉ϶ÆÍ­£¬Ó¦Ñ¡Ôñ
 
ÉÕ±­£¨Ìî¡°ÒÒ¡±»ò¡°±û¡±£©£¬µ±µç¼«aÉÏÉú³É±ê×¼×´¿öÏÂÆøÌå2240mLʱ£¬ÀíÂÛÉÏ¿ÉÔڵ缫
 
£¨Ìî×Öĸ£©É϶ÆÍ­
 
g£®
£¨3£©¹¤ÒµÉÏÒÔÂÁÍÁ¿ó£¨Ö÷Òª³É·ÖAl2O3?nH2O£¬º¬ÉÙÁ¿µÄÑõ»¯ÌúºÍʯӢµÈÔÓÖÊ£©ÎªÔ­ÁÏÉú²úÂÁ£®Îª³ýÈ¥ÂÁÍÁ¿óÑùÆ·ÖÐÑõ»¯ÌúºÍʯӢÔÓÖÊ£¬Òª½«·ÛË顢ɸѡºóµÄÂÁÍÁ¿óÑùÆ·ÈܽâÔÚ×ãÁ¿µÄÇâÑõ»¯ÄÆÈÜÒºÖд¦Àí£¬Çëд³öÓйط´Ó¦µÄÀë×Ó·½³Ìʽ
 
£®
£¨4£©ÒÑÖªÓ²ÂÁÖк¬Cu£º2.2%¡«5%¡¢Mg£º0.2%¡«3%¡¢Mn£º0.3%¡«1.5%¡¢Si£º0.5%£¬ÆäÓàÊÇAl£®1½ÇÓ²±Òµ¥Ã¶ÖÊÁ¿Îª2.20¿Ë£¬ÈôÏëÖÆµÃ1½ÇÓ²±Ò1°ÙÍòö£¬ÀíÂÛÉÏÖÁÉÙÐèÒªº¬Al2O3 90%µÄÂÁÍÁ¿óÔ¼
 
¶Ö£¨Ð¡Êýµãºó±£ÁôһλÊý×Ö£©£®

ĿǰÁ÷ͨµÄµÚÎåÌ×ÈËÃñ±ÒÓ²±Ò²ÄÁÏ·Ö±ðΪ£º1ÔªÓ²±ÒΪͭо¶ÆÄøºÏ½ð£¬5½ÇÓ²±ÒΪͭо¶ÆÍ­ºÏ½ð£¬1½ÇÓ²±ÒΪӲÂÁºÏ½ð¡£Çë»Ø´ð£º£¨ÌâÄ¿Öеİٷֺ¬Á¿¾ùΪÖÊÁ¿·ÖÊý£©

(1)¸ÖÊǺ¬Ì¼Á¿Îª0.03%¡«2%µÄ___________£¨ÌîÎïÖÊÀà±ðÃû³Æ£©¡£

(2)ÏÂͼËùʾµÄ×°ÖÃÖУ¬¼×¡¢ÒÒ¡¢±ûÈý¸öÉÕ±­·Ö±ðÊ¢·Å×ãÁ¿µÄÈÜÒº£¬µç¼«¼°ÈÜÒºÈçϱíËùʾ£º

µç¼«

a

b

c

d

e

f

ʯī

ʯī

Í­

¸Ö

¸Ö

Í­

ÈÜÒº

NaClÈÜÒº

CuSO4ÈÜÒº

CuSO4ÈÜÒº

ͨµçºóµç¼«aÉϿɲúÉúÄÜʹʪÈóµÄµâ»¯¼Øµí·ÛÊÔÖ½±äÀ¶µÄÆøÌå¡£

¢ÙÉÏÊö×°ÖÃÖÐM¼«ÎªÖ±Á÷µçÔ´µÄ_________¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬µç¼«bÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½Îª__________________________________________¡£

¢ÚÈôÔÚ¸Ö±ÒÉ϶ÆÍ­£¬Ó¦Ñ¡Ôñ_______ÉÕ±­£¨Ìî¡°ÒÒ¡±»ò¡°±û¡±£©£¬µ±µç¼«aÉÏÉú³É±ê×¼×´¿öÏÂÆøÌå2240 mLʱ£¬ÀíÂÛÉÏ¿ÉÔڵ缫_______£¨Ìî×Öĸ£©É϶ÆÍ­_______g¡£

(3)¹¤ÒµÉÏÒÔÂÁÍÁ¿ó£¨Ö÷Òª³É·ÖAl2O3¡¤nH2O£¬º¬ÉÙÁ¿µÄÑõ»¯ÌúºÍʯӢµÈÔÓÖÊ£©ÎªÔ­ÁÏÉú²úÂÁ¡£Îª³ýÈ¥ÂÁÍÁ¿óÑùÆ·ÖÐÑõ»¯ÌúºÍʯӢÔÓÖÊ£¬Òª½«·ÛË顢ɸѡºóµÄÂÁÍÁ¿óÑùÆ·ÈܽâÔÚ×ãÁ¿µÄÇâÑõ»¯ÄÆÈÜÒºÖд¦Àí£¬Çëд³öÓйط´Ó¦µÄÀë×Ó·½³Ìʽ__________________________¡£

(4)ÒÑÖªÓ²ÂÁÖк¬Cu£º2.2%¡«5%¡¢Mg£º0.2%¡«3%¡¢Mn£º0.3%¡«1.5%¡¢Si£º0.5%£¬ÆäÓàÊÇAl¡£1½ÇÓ²±Òµ¥Ã¶ÖÊÁ¿Îª2.20¿Ë£¬ÈôÏëÖÆµÃ1½ÇÓ²±Ò1°ÙÍòö£¬ÀíÂÛÉÏÖÁÉÙÐèÒªº¬Al2O3 90%µÄÂÁÍÁ¿óÔ¼_______¶Ö£¨Ð¡Êýµãºó±£ÁôһλÊý×Ö£©¡£

 

ĿǰÁ÷ͨµÄµÚÎåÌ×ÈËÃñ±ÒÓ²±Ò²ÄÁÏ·Ö±ðΪ£º1ÔªÓ²±ÒΪͭо¶ÆÄøºÏ½ð£¬5½ÇÓ²±ÒΪͭо¶ÆÍ­ºÏ½ð£¬1½ÇÓ²±ÒΪӲÂÁºÏ½ð¡£Çë»Ø´ð£º£¨ÌâÄ¿Öеİٷֺ¬Á¿¾ùΪÖÊÁ¿·ÖÊý£©

(1)¸ÖÊǺ¬Ì¼Á¿Îª0.03%¡«2%µÄ___________£¨ÌîÎïÖÊÀà±ðÃû³Æ£©¡£

(2)ÏÂͼËùʾµÄ×°ÖÃÖУ¬¼×¡¢ÒÒ¡¢±ûÈý¸öÉÕ±­·Ö±ðÊ¢·Å×ãÁ¿µÄÈÜÒº£¬µç¼«¼°ÈÜÒºÈçϱíËùʾ£º

µç¼«

a[À´Ô´:ZXXK]

b

c

d

e

f[À´Ô´:ѧ+¿Æ+ÍøZ+X+X+K]

ʯī

ʯī

Í­

¸Ö

¸Ö

Í­

ÈÜÒº

NaClÈÜÒº

CuSO4ÈÜÒº

CuSO4ÈÜÒº

ͨµçºóµç¼«aÉϿɲúÉúÄÜʹʪÈóµÄµâ»¯¼Øµí·ÛÊÔÖ½±äÀ¶µÄÆøÌå¡£

¢ÙÉÏÊö×°ÖÃÖÐM¼«ÎªÖ±Á÷µçÔ´µÄ_________¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬µç¼«bÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½Îª__________________________________________¡£

¢ÚÈôÔÚ¸Ö±ÒÉ϶ÆÍ­£¬Ó¦Ñ¡Ôñ_______ÉÕ±­£¨Ìî¡°ÒÒ¡±»ò¡°±û¡±£©£¬µ±µç¼«aÉÏÉú³É±ê×¼×´¿öÏÂÆøÌå2240 mLʱ£¬ÀíÂÛÉÏ¿ÉÔڵ缫_______£¨Ìî×Öĸ£©É϶ÆÍ­_______g¡£

(3)¹¤ÒµÉÏÒÔÂÁÍÁ¿ó£¨Ö÷Òª³É·ÖAl2O3¡¤nH2O£¬º¬ÉÙÁ¿µÄÑõ»¯ÌúºÍʯӢµÈÔÓÖÊ£©ÎªÔ­ÁÏÉú²úÂÁ¡£Îª³ýÈ¥ÂÁÍÁ¿óÑùÆ·ÖÐÑõ»¯ÌúºÍʯӢÔÓÖÊ£¬Òª½«·ÛË顢ɸѡºóµÄÂÁÍÁ¿óÑùÆ·ÈܽâÔÚ×ãÁ¿µÄÇâÑõ»¯ÄÆÈÜÒºÖд¦Àí£¬Çëд³öÓйط´Ó¦µÄÀë×Ó·½³Ìʽ__________________________¡£

(4)ÒÑÖªÓ²ÂÁÖк¬Cu£º2.2%¡«5%¡¢Mg£º0.2%¡«3%¡¢Mn£º0.3%¡«1.5%¡¢Si£º0.5%£¬ÆäÓàÊÇAl¡£1½ÇÓ²±Òµ¥Ã¶ÖÊÁ¿Îª2.20¿Ë£¬ÈôÏëÖÆµÃ1½ÇÓ²±Ò1°ÙÍòö£¬ÀíÂÛÉÏÖÁÉÙÐèÒªº¬Al2O3 90%µÄÂÁÍÁ¿óÔ¼_______¶Ö£¨Ð¡Êýµãºó±£ÁôһλÊý×Ö£©¡£

 

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

¾«Ó¢¼Ò½ÌÍø