ÌâÄ¿ÄÚÈÝ

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

¾«Ó¢¼Ò½ÌÍø

ÒÑÖª£ºCu2O+2H+¨TCu+Cu2++H2O
£¨1£©ÊµÑé²Ù×÷IµÄÃû³ÆΪ £»ÔÚ¿ÕÆøÖÐ×ÆÉÕ¹ÌÌå»ìºÏÎïDʱ£¬Óõ½¶àÖÖ¹èËáÑÎÖʵÄÒÇÆ÷£¬³ý²£Á§°ô¡¢¾Æ¾«µÆ¡¢ÄàÈý½ÇÍ⣬»¹ÓР£¨ÌîÒÇÆ÷Ãû³Æ£©£®
£¨2£©ÂËÒºAÖÐÌúÔªËصĴæÔÚÐÎʽΪ  £¨ÌîÀë×Ó·ûºÅ£©£¬Éú³É¸ÃÀë×ÓµÄÀë×Ó·½³ÌʽΪ £¬¼ìÑéÂËÒºAÖдæÔÚ¸ÃÀë×ÓµÄÊÔ¼ÁΪ £¨ÌîÊÔ¼ÁÃû³Æ£©£®
£¨3£©½ðÊôEÓë¹ÌÌåF·¢ÉúµÄijһ·´Ó¦¿ÉÓÃÓÚº¸½Ó¸Ö¹ì£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ £®
£¨4£©³£ÎÂÏ£¬µÈpHµÄNaAlO2ºÍNaOHÁ½·ÝÈÜÒºÖУ¬ÓÉË®µçÀë³öµÄc£¨OH-£©Ç°ÕßΪºóÕßµÄ108±¶£¬ÔòÁ½ÖÖÈÜÒºµÄpH= £®
£¨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+Ũ¶È¼õС
d£®´ÖÍ­¾«Á¶Ê±Í¨¹ýµÄµçÁ¿ÓëÒõ¼«Îö³öÍ­µÄÁ¿ÎÞÈ·¶¨¹Øϵ
¢Ú´ÓŨÁòËᡢŨÏõËá¡¢ÕôÁóË®ÖÐÑ¡ÓúÏÊʵÄÊÔ¼Á£¬²â¶¨´ÖÍ­ÑùÆ·ÖнðÊôÍ­µÄÖÊÁ¿·ÖÊý£¬Éæ¼°µÄÖ÷Òª²½ÖèΪ£º³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÑùÆ·¡ú ¡ú¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡ú³ÆÁ¿Ê£Óà¹ÌÌåÍ­µÄÖÊÁ¿£®£¨ÌîȱÉٵIJÙ×÷²½Ö裬²»±ØÃèÊö²Ù×÷¹ý³ÌµÄϸ½Ú£©
²é¿´±¾Ìâ½âÎöÐèÒªµÇ¼
²é¿´½âÎöÈçºÎ»ñÈ¡Óŵ㣿ÆÕͨÓû§£º2¸öÓŵ㡣
ÈçºÎÉêÇëVIPÓû§£¿VIPÓû§£ºÇëÖ±½ÓµÇ¼¼´¿É²é¿´¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2013?ÁÙÒÊһģ£©¹¤ÒµÉÏÓÃij¿óÔü£¨º¬ÓÐCu2O¡¢Al2O3¡¢Fe2O3¡¢SiO2£©ÌáÈ¡Í­µÄ²Ù×÷Á÷³ÌÈçÏ£º

ÒÑÖª£ºCu2O+2H+¨TCu+Cu2++H2O
£¨1£©ÊµÑé²Ù×÷IµÄÃû³ÆΪ
¹ýÂË
¹ýÂË
£»ÔÚ¿ÕÆøÖÐ×ÆÉÕ¹ÌÌå»ìºÏÎïDʱ£¬Óõ½¶àÖÖ¹èËáÑÎÖʵÄÒÇÆ÷£¬³ý²£Á§°ô¡¢¾Æ¾«µÆ¡¢ÄàÈý½ÇÍ⣬»¹ÓÐ
ÛáÛö
ÛáÛö
£¨ÌîÒÇÆ÷Ãû³Æ£©£®
£¨2£©ÂËÒºAÖÐÌúÔªËصĴæÔÚÐÎʽΪ
Fe2+
Fe2+
 £¨ÌîÀë×Ó·ûºÅ£©£¬Éú³É¸ÃÀë×ÓµÄÀë×Ó·½³ÌʽΪ
2Fe3++Cu¨T2Fe2++Cu2+
2Fe3++Cu¨T2Fe2++Cu2+
£¬¼ìÑéÂËÒºAÖдæÔÚ¸ÃÀë×ÓµÄÊÔ¼ÁΪ
KSCNÈÜÒº
KSCNÈÜÒº
£¨ÌîÊÔ¼ÁÃû³Æ£©£®
£¨3£©½ðÊôEÓë¹ÌÌåF·¢ÉúµÄijһ·´Ó¦¿ÉÓÃÓÚº¸½Ó¸Ö¹ì£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2 Al+Fe2O3
 ¸ßΠ
.
 
Al2O3+2Fe
2 Al+Fe2O3
 ¸ßΠ
.
 
Al2O3+2Fe
£®
£¨4£©³£ÎÂÏ£¬µÈpHµÄNaAlO2ºÍNaOHÁ½·ÝÈÜÒºÖУ¬ÓÉË®µçÀë³öµÄc£¨OH-£©Ç°ÕßΪºóÕßµÄ108±¶£¬ÔòÁ½ÖÖÈÜÒºµÄpH=
11
11
£®
£¨5£©¢ÙÀûÓõç½â·¨½øÐдÖÍ­¾«Á¶Ê±£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ
b£®c
b£®c
£¨Ìî´úºÅ£©£®
a£®µçÄÜÈ«²¿×ª»¯Îª»¯Ñ§ÄÜ
b£®´ÖÍ­½ÓµçÔ´Õý¼«£¬·¢ÉúÑõ»¯·´Ó¦
c£®¾«Í­×÷Òõ¼«£¬µç½âºóµç½âÒºÖÐCu2+Ũ¶È¼õС
d£®´ÖÍ­¾«Á¶Ê±Í¨¹ýµÄµçÁ¿ÓëÒõ¼«Îö³öÍ­µÄÁ¿ÎÞÈ·¶¨¹Øϵ
¢Ú´ÓŨÁòËᡢŨÏõËá¡¢ÕôÁóË®ÖÐÑ¡ÓúÏÊʵÄÊÔ¼Á£¬²â¶¨´ÖÍ­ÑùÆ·ÖнðÊôÍ­µÄÖÊÁ¿·ÖÊý£¬Éæ¼°µÄÖ÷Òª²½ÖèΪ£º³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÑùÆ·¡ú
½«Å¨ÁòËáÓÃÕôÁóˮϡÊÍ£¬½«ÑùÆ·ÓëÏ¡ÁòËá³ä·Ö·´Ó¦ºó
½«Å¨ÁòËáÓÃÕôÁóˮϡÊÍ£¬½«ÑùÆ·ÓëÏ¡ÁòËá³ä·Ö·´Ó¦ºó
¡ú¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡ú³ÆÁ¿Ê£Óà¹ÌÌåÍ­µÄÖÊÁ¿£®£¨ÌîȱÉٵIJÙ×÷²½Ö裬²»±ØÃèÊö²Ù×÷¹ý³ÌµÄϸ½Ú£©
£¨2013?·ą́Çø¶þÄ££©¹¤ÒµÉÏÓÃij¿óÔü£¨º¬ÓÐCu2O¡¢Al2O3¡¢Fe2O3¡¢SiO2£©ÌáÈ¡Í­µÄ²Ù×÷Á÷³ÌÈçÏ£º
ÒÑÖª£ºCu2O+2H+=Cu+Cu2++H2O
³ÁµíÎï Cu£¨OH£©2 Al£¨OH£©3 Fe£¨OH£©3 Fe£¨OH£©2
¿ªÊ¼³ÁµípH 5.4 4.0 1.1 5.8
³ÁµíÍêÈ«pH 6.7 5.2 3.2 8.8
£¨1£©¹ÌÌå»ìºÏÎïAÖеijɷÖÊÇ
SiO2¡¢Cu
SiO2¡¢Cu
£®
£¨2£©·´Ó¦¢ñÍê³Éºó£¬ÌúÔªËصĴæÔÚÐÎʽΪ
Fe2+
Fe2+
£®£¨ÌîÀë×Ó·ûºÅ£©
Çëд³öÉú³É¸ÃÀë×ÓµÄÀë×Ó·½³Ìʽ
2Fe3++Cu=Cu2++2Fe2+
2Fe3++Cu=Cu2++2Fe2+
£®
£¨3£©x¡¢y¶ÔÓ¦µÄÊýÖµ·¶Î§·Ö±ðÊÇ
3.2¡ÜpH£¼4.0
3.2¡ÜpH£¼4.0
¡¢
5.2¡ÜpH£¼5.4
5.2¡ÜpH£¼5.4
£®
£¨4£©µç½â·¨»ñÈ¡Cuʱ£¬Òõ¼«·´Ó¦Ê½Îª
Cu2++2e-=Cu
Cu2++2e-=Cu
£¬Ñô¼«·´Ó¦Ê½Îª
2Cl--2e-=Cl2¡ü
2Cl--2e-=Cl2¡ü
£®
£¨5£©ÏÂÁйØÓÚNaClOµ÷pHµÄ˵·¨ÕýÈ·µÄÊÇ
b
b
£®
a£®¼ÓÈëNaClO¿ÉʹÈÜÒºµÄpH½µµÍ
b£®NaClOÄܵ÷½ÚpHµÄÖ÷ÒªÔ­ÒòÊÇÓÉÓÚ·¢Éú·´Ó¦ClO-+H+?HClO£¬ClO-ÏûºÄH+£¬´Ó¶ø´ïµ½µ÷½ÚpHµÄÄ¿µÄ
c£®NaClOÄܵ÷½ÚpHµÄÖ÷ÒªÔ­ÒòÊÇÓÉÓÚNaClOË®½âClO-+H2O?HClO+OH-£¬OH-ÏûºÄH+£¬´Ó¶ø´ïµ½µ÷½ÚpHµÄÄ¿µÄ
£¨6£©ÓÃNaClOµ÷pH£¬Éú³É³ÁµíBµÄͬʱÉú³ÉÒ»ÖÖ¾ßÓÐƯ°××÷ÓõÄÎïÖÊ£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ
5ClO-+2Fe2++5H2O=2Fe£¨OH£©3+Cl-+4HClO
5ClO-+2Fe2++5H2O=2Fe£¨OH£©3+Cl-+4HClO
£®

£¨1£©¸ßεç½â¼¼ÊõÄܸßЧʵÏÖCO2(g) + H2O(g) =CO(g) + H2(g) +O2(g)? £¬¹¤×÷Ô­ÀíʾÒâͼÈçÏ£º

¢Ùµç¼«b·¢Éú??????? £¨Ìî¡°Ñõ»¯¡±»ò¡°»¹Ô­¡±£©·´Ó¦¡£

¢ÚCO2Ôڵ缫a·ÅµçµÄ·´Ó¦Ê½ÊÇ?????????????????????????????? ¡£

£¨2£©¹¤ÒµÉÏÓÃij¿óÔü£¨º¬ÓÐCu2O¡¢Al2O3¡¢Fe2O3¡¢SiO2£©ÌáÈ¡Í­µÄ²Ù×÷Á÷³ÌÈçÏ£º

ÒÑÖª£º Cu2O + 2H+? = Cu + Cu2+ + H2O

³ÁµíÎï

Cu(OH)2

Al(OH)3

Fe(OH)3

Fe(OH)2

¿ªÊ¼³ÁµípH

5.4

4.0

1.1

5.8

³ÁµíÍêÈ«pH

6.7

5.2

3.2

8.8

 

¢Ù¹ÌÌå»ìºÏÎïAÖеijɷÖÊÇ???????????? ¡£

¢Ú·´Ó¦¢ñÍê³Éºó£¬ÌúÔªËصĴæÔÚÐÎʽΪ??????????? ¡££¨ÌîÀë×Ó·ûºÅ£©

Çëд³öÉú³É¸ÃÀë×ÓµÄÀë×Ó·½³Ìʽ??????????????????????????????????????? ¡£

¢ÛxµÄÊýÖµ·¶Î§ÊÇ3.2¡ÜpH£¼4.0£¬y¶ÔÓ¦µÄÊýÖµ·¶Î§ÊÇ????????????? ¡£

¢ÜÏÂÁйØÓÚNaClOµ÷pHµÄ˵·¨ÕýÈ·µÄÊÇ???????? £¨ÌîÐòºÅ£©¡£

a£®¼ÓÈëNaClO¿ÉʹÈÜÒºµÄpH½µµÍ

b£®NaClOÄܵ÷½ÚpHµÄÖ÷ÒªÔ­ÒòÊÇÓÉÓÚ·¢Éú·´Ó¦ClO£­+ H+HClO£¬ClO£­ÏûºÄH+£¬´Ó¶ø´ïµ½µ÷½ÚpHµÄÄ¿µÄ

c£®NaClOÄܵ÷½ÚpHµÄÖ÷ÒªÔ­ÒòÊÇÓÉÓÚNaClOË®½âClO£­+ H2OHClO+OH£­£¬OH£­ÏûºÄH+ £¬´Ó¶ø´ïµ½µ÷½ÚpHµÄÄ¿µÄ

¢ÝʵÑéÊÒÅäÖÆÖÊÁ¿·ÖÊýΪ20.0%µÄCuSO4ÈÜÒº£¬ÅäÖƸÃÈÜÒºËùÐèµÄCuSO4¡¤5H2OÓëH2OµÄÖÊÁ¿Ö®±ÈΪ????????? ¡£

 

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

¾«Ó¢¼Ò½ÌÍø