ÌâÄ¿ÄÚÈÝ

£¨8·Ö£©ÎªÁËÑо¿Íâ½çÌõ¼þ¶Ô¹ýÑõ»¯Çâ·Ö½âËÙÂʵÄÓ°Ï죬ijͬѧ×öÁËÒÔÏÂʵÑé¡£
ʵÑé±àºÅ
ʵÑé²Ù×÷
ʵÑéÏÖÏó
 
¢Ù
·Ö±ðÔÚÊÔ¹ÜA¡¢BÖмÓÈë 5 mL 5£¥(ÈÜҺŨ¶È) H2O2ÈÜÒº£¬¸÷µÎÈë2 µÎÏàͬŨ¶ÈµÄCuSO4ÈÜÒº¡£´ýÊÔ¹ÜÖоùÓÐÊÊÁ¿ÆøÅݳöÏÖʱ£¬½«ÊÔ¹ÜA·ÅÈëÊ¢ÓÐ5¡æ×óÓÒÀäË®µÄÉÕ±­ÖнþÅÝ£»½«ÊÔ¹ÜB·ÅÈëÊ¢ÓÐ40¡æ×óÓÒÈÈË®µÄÉÕ±­ÖнþÅÝ¡£
 
ÊÔ¹ÜAÖв»ÔÙ²úÉúÆøÅÝ£»
ÊÔ¹ÜBÖвúÉúµÄÆøÅÝÁ¿Ôö´ó¡£
¢Ú
ÁíÈ¡Á½Ö§ÊԹֱܷð¼ÓÈë5mL 5£¥H2O2ÈÜÒººÍ5 mL 10£¥H2O2ÈÜÒº
ÊÔ¹ÜA¡¢BÖоùδÃ÷ÏÔ¼ûµ½ÓÐÆøÅݲúÉú¡£
£¨1£©¹ýÑõ»¯Çâ·Ö½âµÄ»¯Ñ§·½³ÌʽΪ                                ¡£
£¨2£©ÊµÑé¢ÙµÄÄ¿µÄÊÇ                                           ¡£
ʵÑéÖеμÓCuSO4ÈÜÒºµÄÄ¿µÄÊÇ                                  ¡£
£¨3£©ÊµÑé¢Úδ¹Û²ìµ½Ô¤ÆÚµÄʵÑéÏÖÏó£¬ÎªÁË°ïÖú¸Ãͬѧ´ïµ½ÊµÑéÄ¿µÄ£¬ÄãÉè¼ÆµÄʵÑé·½°¸ÊÇ                                                     ¡££¨ÓÃʵÑéÖÐËùÌṩµÄ¼¸ÖÖÊÔ¼Á£©¡£
£¨4£©¶ÔÓÚH2O2·Ö½â·´Ó¦£¬Fe2(SO4)3ÈÜÒºÒ²ÓÐÒ»¶¨µÄ´ß»¯×÷Óá£Îª±È½ÏFe2(SO4)3ºÍCuSO4ÈÜÒº¶ÔH2O2·Ö½âµÄ´ß»¯Ð§¹û¡£Ä³»¯Ñ§Ñо¿Ð¡×éµÄͬѧ·Ö±ðÉè¼ÆÁËÈçͼ¼×¡¢ÒÒËùʾµÄʵÑé¡£Çë»Ø´ðÏà¹ØÎÊÌ⣺

ÈçͼËùʾ£¬Í¬Ñ§ÃÇ·Ö±ð´Ó¶¨ÐԺͶ¨Á¿½Ç¶È½øÐÐÁ˱Ƚϡ£
¢Ù¶¨ÐÔ·ÖÎö£ºÈçͼ¼×¿Éͨ¹ý¹Û²ì                    £¬¶¨ÐԱȽϵóö½áÂÛ¡£
¢Ú¶¨Á¿·ÖÎö£ºÓÃͼÒÒËùʾװÖÃ×ö¶ÔÕÕÊÔÑ飬ʵÑéʱ¾ùÒÔÉú³É40mLÆøÌåΪ׼£¬ÆäËü¿ÉÄÜÓ°ÏìʵÑéµÄÒòËؾùÒѺöÂÔ¡£ÊµÑéÖÐÐèÒª²âÁ¿µÄÊý¾ÝÊÇ               ¡£
£¨5£©Í¨¹ý¶ÔÉÏÊöʵÑé¹ý³ÌµÄ·ÖÎö£¬ÔÚʵÑéÉè¼Æʱ£¬Òª¿¼ÂÇ_________·½·¨µÄÓ¦Óá£
£¨1£©2H2O2  MnO2  2H2O + O2¡ü   
£¨2£©Ñо¿Î¶ȶÔH2O2·Ö½âËÙÂʵÄÓ°Ïì
¼Ó¿ìH2O2·Ö½âËÙÂÊ£¬Ê¹ÊµÑéÏÖÏóÒ×Óڹ۲젠
£¨3£©½«Á½Ö§ÊÔ¹Üͬʱ·ÅÈëÊ¢ÓÐÏàͬζÈÈÈË®µÄÉÕ±­ÖУ¬»òÏòÁ½Ö§ÊÔ¹ÜÖÐͬʱµÎÈëÏà
ͬµÎÊý¡¢Å¨¶ÈÒ»ÑùµÄÁòËáÍ­ÈÜÒº£¬²úÉúÆøÅݵÄËÙÂÊ¿ìµÄÊÇ10£¥H2O2ÈÜÒº£¬·´Ö®
ÊÇ5£¥H2O2ÈÜÒº¡££¨2·Ö£©
£¨4£©¢ÙÈÜÒºÖÐÆøÅݲúÉúµÄËÙÂÊ  ¢ÚÊÕ¼¯40mLÆøÌåËùÐèµÄʱ¼ä
£¨5£©¿ØÖƱäÁ¿£¨»ò¶Ô±È£©

ÊÔÌâ·ÖÎö£º£¨1£©¹ýÑõ»¯Çâ·Ö½âµÄ»¯Ñ§·½³ÌʽΪ2H2O2  MnO2  2H2O + O2¡ü¡£
£¨2£©ÊµÑé¢ÙÖпØÖƵÄÊÇ·´Ó¦µÄζȣ¬ÆäÄ¿µÄÊÇÑо¿Î¶ȶÔH2O2·Ö½âËÙÂʵÄÓ°Ïì¡£
ʵÑéÖеμÓCuSO4ÈÜÒºµÄÄ¿µÄÊǼӿìH2O2·Ö½âËÙÂÊ£¬Ê¹ÊµÑéÏÖÏóÒ×Óڹ۲졣
£¨3£©ÊµÑé¢Úδ¹Û²ìµ½Ô¤ÆÚµÄʵÑéÏÖÏó£¬ÎªÁ˴ﵽʵÑéÄ¿µÄ£¬¿ÉÒÔÕâÑù×ö£º½«Á½Ö§ÊÔ¹Üͬʱ·ÅÈëÊ¢ÓÐÏàͬζÈÈÈË®µÄÉÕ±­ÖУ¬»òÏòÁ½Ö§ÊÔ¹ÜÖÐͬʱµÎÈëÏàͬµÎÊý¡¢Å¨¶ÈÒ»ÑùµÄÁòËáÍ­ÈÜÒº£¬²úÉúÆøÅݵÄËÙÂÊ¿ìµÄÊÇ10£¥H2O2ÈÜÒº£¬·´Ö®ÊÇ5£¥H2O2ÈÜÒº¡£
£¨4£©¢Ù¶¨ÐÔ·ÖÎö£ºÍ¨¹ý¹Û²ì¼×ͼÖÐÈÜÒºÖÐÆøÅݲúÉúµÄËÙÂÊ£¬¶¨ÐԱȽϵóö½áÂÛ¡£
¢Ú¶¨Á¿·ÖÎö£ºÓÃͼÒÒËùʾװÖÃ×ö¶ÔÕÕÊÔÑ飬ʵÑéʱ¾ùÒÔÉú³É40mLÆøÌåΪ׼£¬ÆäËü¿ÉÄÜÓ°ÏìʵÑéµÄÒòËؾùÒѺöÂÔ¡£ÊµÑéÖÐÐèÒª²âÁ¿µÄÊý¾ÝÊÇÊÕ¼¯40mLÆøÌåËùÐèµÄʱ¼ä¡£
£¨5£©Í¨¹ý¶ÔÉÏÊöʵÑé¹ý³ÌµÄ·ÖÎö£¬ÔÚʵÑéÉè¼Æʱ£¬Òª¿¼ÂÇ¿ØÖƱäÁ¿£¨»ò¶Ô±È£©·½·¨µÄÓ¦Óá£
µãÆÀ£ºÊéд»¯Ñ§·½³ÌʽҪ×ñÑ­¿Í¹ÛÊÂʵºÍÖÊÁ¿Êغ㶨ÂÉÁ½¸öÔ­Ôò£¬×¢Ò⻯ѧʽҪÕýÈ·£¬²»ÒªÍü¼Ç·´Ó¦Ìõ¼þ¡¢ÆøÌå»òÕß³Áµí·ûºÅ¡£
¿ØÖƱäÁ¿·¨ÊÇʵÑé̽¾¿µÄ³£Ó÷½·¨£¬ÔÚʵÑéÉè¼ÆºÍ̽¾¿¹ý³ÌÖУ¬ÍùÍùͨ¹ýºÏÀíµÄ¿ØÖÆij¸ö±äÁ¿£¬Í¨¹ý²»Í¬µÄʵÑéÏÖÏóÀ´·ÖÎöʵÑé½áÂÛ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø