ÌâÄ¿ÄÚÈÝ

È¥Ä궬ÌìÖÁ½ñÄê´ºÌ죬ÎÒ¹úÎ÷ÄϵØÇøÔâÓö´ó·¶Î§³ÖÐø¸Éºµ£®È«¹úÈËÃñÍŽáÒ»Ö£¬¹²Í¬¿¹ºµ£®
£¨1£©ÓÐЩ´åׯ´òÉȡÓõØÏÂË®£®¼ìÑéµØÏÂË®ÊÇӲˮ»¹ÊÇÈíË®£¬¿ÉÓõÄÎïÖÊÊÇ________£®
£¨2£©ÓÐЩ´åÃñÈ¡»ë×ǵĿÓË®×÷Éú»îÓÃË®£®ÓÐͬѧÀûÓÃËùѧµÄ֪ʶ½«»ë×ǵĿÓË®ÓÃͼËùʾµÄ¼òÒ×¾»Ë®Æ÷½øÐо»»¯£¬ÆäÖÐСÂÑʯ¡¢Ê¯Ó¢É³µÄ×÷ÓÃÊÇ________£®
£¨3£©Èç¹ûµØÏÂˮӲ¶È´ó£¬»òÕß¿ÓË®Öв¡Ô­Î¢ÉúÎï¹ý¶à£¬¶¼¿ÉÒÔ²ÉÈ¡________·½·¨£¬À´½µµÍÓ²¶ÈºÍɱÃð²¡Ô­Î¢ÉúÎ
£¨4£©Éí´¦ºµÇøÖ®ÍâµÄÎÒÃÇÄÜ×öЩʲô£¿________¡¢________£¨ÁоÙÁ½ÖÖ×ö·¨£©

½â£º£¨1£©Ó²Ë®Óë·ÊÔíË®»ìºÏ²úÉú´óÁ¿µÄ¸¡Ôü£¬ÈíË®Óë·ÊÔíË®»ìºÏ²úÉú½Ï¶àµÄÅÝÄ­£¬¼ìÑéµØÏÂË®ÊÇӲˮ»¹ÊÇÈíË®£¬¿ÉÓõÄÎïÖÊÊÇ·ÊÔíË®£¬¹ÊÌ·ÊÔíË®£»
£¨2£©ÔÚ¼òÒ×¾»Ë®Æ÷ÖУ¬Ð¡ÂÑʯºÍʯӢɰÄÜ×èÖ¹²»ÈÜÐÔ¹ÌÌå¿ÅÁ£Í¨¹ý£¬Æðµ½Á˹ýÂ˵Ä×÷Ó㬹ÊÌ¹ýÂË£»
£¨3£©Ó²Ë®Öк¬Óн϶àµÄ¿ÉÈÜÐÔ¸Æþ»¯ºÏÎ¼ÓÈȺóÄÜ·Ö½âÉú³É²»ÈÜÐÔ¸Æþ»¯ºÏÎ¼ÓÈÈ¿ÉÒÔɱËÀϸ¾ú΢ÉúÎ¹ÊÈç¹ûµØÏÂˮӲ¶È´ó£¬»òÕß¿ÓË®Öв¡Ô­Î¢ÉúÎï¹ý¶à£¬¶¼¿ÉÒÔ²ÉÈ¡¼ÓÈÈÖó·Ð·½·¨£¬¹ÊÌ¼ÓÈÈÖó·Ð£»
£¨4£©Éí´¦ºµÇøÖ®ÍâµÄÎÒÃÇÐèÒªÔÚƽʱµÄÉú»îÖÐ×¢Òâ½ÚÔ¼ÓÃË®£¬¿ÉÒÔ´ÓËæÊֹرÕË®ÁúÍ·¡¢ÌÔÃ×Ë®½½»¨¡¢Ï´ÒÂË®³å²ÞËùµÈ·½Ãæ½øÐÐÃèÊö£¬¹ÊÌËæÊֹرÕË®ÁúÍ·£¬ÌÔÃ×Ë®½½»¨£®
·ÖÎö£º¸ù¾ÝÒÑÓеÄ֪ʶ½øÐзÖÎö£¬¼ìÑéӲˮºÍÈíˮʹÓõÄÊÇ·ÊÔíË®£¬ÔÚ¼òÒ×¾»Ë®Æ÷ÖУ¬Ð¡ÂÑʯºÍʯӢɰÄÜ×èÖ¹²»ÈÜÐÔ¹ÌÌå¿ÅÁ£Í¨¹ý£¬Æðµ½Á˹ýÂ˵Ä×÷Óã¬Ó²Ë®Öк¬Óн϶àµÄ¿ÉÈÜÐÔ¸Æþ»¯ºÏÎ¼ÓÈȺóÄÜ·Ö½âÉú³É²»ÈÜÐÔ¸Æþ»¯ºÏÎ¼ÓÈÈ¿ÉÒÔɱËÀϸ¾ú΢ÉúÎÔÚƽʱµÄÉú»îÖÐҪעÒâ½ÚÔ¼ÓÃË®£®
µãÆÀ£º±¾Ì⿼²éÁ˾»Ë®µÄ֪ʶ£¬Íê³É´ËÌ⣬¿ÉÒÔÒÀ¾ÝÒÑÓеÄ֪ʶ½áºÏÌâ¸ÉÐðÊö½øÐУ®
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»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