ÌâÄ¿ÄÚÈÝ

ÏÂÁÐÎïÖʶ¼¿ÉÓÃÓÚË®Öʵľ»»¯»òɱ¾úÏû¶¾¡£

£¨1£©Ã÷·¯ºÍ»îÐÔ̼£ºÏòË®ÖмÓÈëÃ÷·¯½Á°èÈܽ⣬¾²ÖÃÒ»¶Îʱ¼äºó£¬½øÐР        £¨Ìî²Ù×÷Ãû³Æ£©£¬´Ó¶ø³ýÈ¥¹ÌÌåС¿ÅÁ££»ÔÙÏòËùµÃµÄ³ÎÇåÒºÖмÓÈë»îÐÔÌ¿£¬ÀûÓÃÆä

            ³ýȥˮÖеÄÑÕÉ«ºÍÒìζ¡£

£¨2£©³ôÑõ(»¯Ñ§Ê½O3)£ºÓÐÇ¿Ñõ»¯ÐÔ£¬¿ÉÓÃÓÚÓÎÓ¾³Ø¡¢Éú»îÓÃË®¡¢ÎÛË®µÄɱ¾úºÍÏû¶¾¡£³ôÑõºÍÑõÆø×é³ÉÔªËØÏàͬ¶ø»¯Ñ§ÐÔÖʲ»Í¬µÄÔ­ÒòÊÇ                         ¡£

£¨3£©Æ¯°×·Û£º¿ÉÓÃÓÚË®µÄɱ¾úÏû¶¾£¬ÆäÓÐЧ³É·ÖÊÇ´ÎÂÈËá¸Æ[»¯Ñ§Ê½ÎªCa(ClO)2]¡£´ÎÂÈËá¸Æ¿É·¢ÉúÈçÏ·´Ó¦£ºCa(ClO)2 + X + H2O£½CaCO3¡ý+2HClO£¬ÔòXµÄ»¯Ñ§Ê½Îª                 ¡£

£¨4£©ClO2£º×ÔÀ´Ë®³§³£ÓÃClO2Ïû¶¾¡£ÖÆÈ¡ClO2µÄ·´Ó¦Æä΢¹Û¹ý³ÌÈçͼËùʾ£¨ÆäÖÐ ±íʾÂÈÔ­×Ó£¬ ±íʾÄÆÔ­×Ó£¬ ±íʾÑõÔ­×Ó£©¡£

Çëд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ                      ¡£

£¨1£©¹ýÂË   Îü¸½ÐÔ   £¨2£©·Ö×Ó¹¹³É²»Í¬ £¨3£©CO2  

£¨4£©Cl2 + 2NaClO2 = 2ClO2 + 2NaCl

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
Ë®ÊÇÈËÀàÉú»îÖв»¿ÉȱÉÙµÄÎïÖÊ£®
£¨1£©È¥Ä궬ÌìÖÁ½ñÄê´ºÌ죬ÎÒ¹úÎ÷ÄϵØÇøÔâÓö´ó·¶Î§³ÖÐø¸Éºµ£®ÓÐЩ´åׯ´òÉȡÓõØÏÂË®£®¼ìÑéµØÏÂË®ÊÇӲˮ»¹ÊÇÈíË®£¬¿ÉÓõÄÎïÖÊÊÇ
·ÊÔíË®
·ÊÔíË®
£¬²â¶¨µØÏÂË®µÄËá¼î¶È¿ÉÓÃ
pHÊÔÖ½
pHÊÔÖ½
£®
£¨2£©ÓÐЩ´åÃñÈ¡»ë×ǵĿÓË®×÷Éú»îÓÃË®£®ÓÐͬѧÀûÓÃËùѧµÄ֪ʶÓÃÈçͼ1ËùʾµÄ¼òÒ×¾»Ë®Æ÷½øÐо»»¯£¬ÆäÖÐСÂÑʯ¡¢Ê¯Ó¢É³µÄ×÷ÓÃÊÇ
¹ýÂË
¹ýÂË
£»»îÐÔÌ¿µÄ×÷ÓÃÊÇ
Îü¸½
Îü¸½
£®
£¨3£©Èç¹ûµØÏÂˮӲ¶È´ó£¬»òÕß¿ÓË®Öв¡Ô­Î¢ÉúÎï¹ý¶à£¬¶¼¿ÉÒÔ²ÉÈ¡
Öó·Ð
Öó·Ð
·½·¨£¬À´½µµÍÓ²¶ÈºÍɱÃð²¡Ô­Î¢ÉúÎ
£¨4£©ÏÂÁÐÎïÖʶ¼¿ÉÓÃÓÚË®Öʵľ»»¯»òɱ¾úÏû¶¾£®
¢Ù³ôÑõ£¨»¯Ñ§Ê½O3£©£ºÓÐÇ¿Ñõ»¯ÐÔ£¬¿ÉÓÃÓÚÓÎÓ¾³Ø¡¢Éú»îÓÃË®¡¢ÎÛË®µÄɱ¾úºÍÏû¶¾£®³ôÑõºÍÑõÆø×é³ÉÔªËØÏàͬ¶ø»¯Ñ§ÐÔÖʲ»Í¬µÄÔ­ÒòÊÇ
·Ö×Ó¹¹³É²»Í¬
·Ö×Ó¹¹³É²»Í¬
£®
¢ÚClO2£º×ÔÀ´Ë®³§³£ÓÃClO2Ïû¶¾£®ÖÆÈ¡ClO2µÄ·´Ó¦Æä΢¹Û¹ý³ÌÈçͼ2Ëùʾ£¨ÆäÖР±íʾÂÈÔ­×Ó£¬ ±íʾÄÆÔ­×Ó£¬±íʾÑõÔ­×Ó£©£®Çëд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ
Cl2+2NaClO2¨T2ClO2+2NaCl
Cl2+2NaClO2¨T2ClO2+2NaCl
£®
Ë®ÊÇÈËÀàÉú»îÖв»¿ÉȱÉÙµÄÎïÖÊ£®

£¨1£©È¥Ä궬ÌìÖÁ½ñÄê´ºÌ죬ÎÒ¹úÎ÷ÄϵØÇøÔâÓö´ó·¶Î§³ÖÐø¸Éºµ£®ÓÐЩ´åׯ´òÉȡÓõØÏÂË®£®¼ìÑéµØÏÂË®ÊÇӲˮ»¹ÊÇÈíË®£¬¿ÉÓõÄÎïÖÊÊÇ
·ÊÔíË®
·ÊÔíË®
£¬²â¶¨µØÏÂË®µÄËá¼î¶È¿ÉÓÃ
pHÊÔÖ½»òpH¼Æ
pHÊÔÖ½»òpH¼Æ
£®
£¨2£©ÓÐЩ´åÃñÈ¡»ë×ǵĿÓË®×÷Éú»îÓÃË®£®ÓÐͬѧÀûÓÃËùѧµÄ֪ʶ»ë×ǵĿÓË®ÓÃͼËùʾµÄ¼òÒ×¾»Ë®Æ÷½øÐо»»¯£¬ÆäÖÐСÂÑʯ¡¢Ê¯Ó¢É³µÄ×÷ÓÃÊÇ
¹ýÂË
¹ýÂË
£»»îÐÔÌ¿µÄ×÷ÓÃÊÇ
Îü¸½
Îü¸½
£®
£¨3£©Èç¹ûµØÏÂˮӲ¶È´ó£¬»òÕß¿ÓË®Öв¡Ô­Î¢ÉúÎï¹ý¶à£¬¶¼¿ÉÒÔ²ÉÈ¡
Öó·Ð
Öó·Ð
·½·¨£¬À´½µµÍÓ²¶ÈºÍɱÃð²¡Ô­Î¢ÉúÎ
£¨4£©ÏÂÁÐÎïÖʶ¼¿ÉÓÃÓÚË®Öʵľ»»¯»òɱ¾úÏû¶¾£®
¢Ù³ôÑõ£¨»¯Ñ§Ê½O3£©£ºÓÐÇ¿Ñõ»¯ÐÔ£¬¿ÉÓÃÓÚÓÎÓ¾³Ø¡¢Éú»îÓÃË®¡¢ÎÛË®µÄɱ¾úºÍÏû¶¾£®³ôÑõºÍÑõÆø×é³ÉÔªËØÏàͬ¶ø»¯Ñ§ÐÔÖʲ»Í¬µÄÔ­ÒòÊÇ
·Ö×Ó²»Í¬
·Ö×Ó²»Í¬
£®
¢ÚƯ°×·Û£º¿ÉÓÃÓÚË®µÄɱ¾úÏû¶¾£¬ÆäÓÐЧ³É·ÖÊÇ´ÎÂÈËá¸Æ[»¯Ñ§Ê½ÎªCa£¨ClO£©2]£®´ÎÂÈËá¸Æ¿É·¢ÉúÈçÏ·´Ó¦£ºCa£¨ClO£©2+X+H2O=CaCO3+2HClO£¬ÔòXµÄ»¯Ñ§Ê½Îª
CO2
CO2
£®
¢ÛClO2£º×ÔÀ´Ë®³§³£ÓÃClO2Ïû¶¾£®ÖÆÈ¡ClO2µÄ·´Ó¦Æä΢¹Û¹ý³ÌÈçͼËùʾ£¨ÆäÖР±íʾÂÈÔ­×Ó£¬ ±íʾÄÆÔ­×Ó£¬ ±íʾÑõÔ­×Ó£©£®Çëд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ
Cl2+2NaClO2¨T2ClO2+2NaCl
Cl2+2NaClO2¨T2ClO2+2NaCl
£®
£¨5£©²úÎïÖÐÓÐË®Éú³ÉµÄ·´Ó¦ºÜ¶à£¨ÈçͼËùʾ£©£¬Çë°´ÒªÇóд³öÒ»¸ö»¯Ñ§·½³Ìʽ£®
¢Ù²úÎïÖÐÓÐË®Éú³ÉµÄ»¯ºÏ·´Ó¦£º
2H2+O2
 µãȼ 
.
 
2H2O
2H2+O2
 µãȼ 
.
 
2H2O
£»
¢Ú²úÎïÖÐÓÐË®Éú³ÉµÄ¸´·Ö½â·´Ó¦£º
NaOH+HCl¨TNaCl+H2O
NaOH+HCl¨TNaCl+H2O
£»
¢Û²úÎïÖÐÓÐË®Éú³ÉµÄÆäËû·´Ó¦£º
CO2+Ca£¨OH£©2¨TCaCO3¡ý+H2O
CO2+Ca£¨OH£©2¨TCaCO3¡ý+H2O
£®

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

¾«Ó¢¼Ò½ÌÍø