ÌâÄ¿ÄÚÈÝ

¸ßһн̲ÄÖÐÓÐÒ»¸öÑÝʾʵÑ飺ÓÃÍÑÖ¬ÃÞ°üסԼ0£®2g¹ýÑõ»¯ÄÆ·ÛÄ©£¬ÖÃÓÚʯÃÞÍøÉÏ£¬ÍùÍÑÖ¬ÃÞÉϵÎË®£¬¿É¹Û²ìµ½ÍÑÖ¬ÃÞ¾çÁÒȼÉÕÆðÀ´¡£

£¨1£©ÓÉʵÑéÏÖÏóËùµÃ³öµÄÓйØNa2O2ºÍH2O·´Ó¦µÄ½áÂÛÊÇ£ºa£®ÓÐÑõÆøÉú³É£»b£®________¡£Na2O2ºÍH2O·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º________¡£

£¨2£©Ä³Ñо¿ÐÔѧϰС×éÄâÓÃͼ10×°ÖýøÐÐʵÑ飬ÒÔÖ¤Ã÷ÉÏÊö½áÂÛ¡£

¢ÙÓÃÒÔÑéÖ¤½áÂÛaµÄʵÑé·½·¨ÊÇ£º________¡£

¢ÚÓÃÒÔÑéÖ¤½áÂÛbµÄʵÑé·½·¨¼°ÏÖÏóÊÇ£º________

£¨3£©ÊµÑ飨2£©ÍùÊÔ¹ÜÖмÓË®ÖÁ¹ÌÌåÍêÈ«ÈܽâÇÒ²»ÔÙÓÐÆøÅÝÉú³Éºó£¬È¡³öÊԹܣ¬ÍùÊÔ¹ÜÖеÎÈë·Ó̪ÊÔÒº£¬·¢ÏÖÈÜÒº±äºì£¬Õñµ´ºó£¬ºìÉ«ÏûÍÊ¡£ÎªÌ½¾¿´ËÏÖÏ󣬸ÃС×éͬѧ´Ó²éÔÄÓйØ×ÊÁÏÖеÃÖª£ºNa2O2ÓëH2O·´Ó¦¿ÉÉú³ÉH2O2£¬H2O2¾ßÓÐÇ¿Ñõ»¯ÐÔºÍƯ°×ÐÔ¡£ÇëÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑ飬֤Ã÷Na2O2ºÍ×ãÁ¿H2O³ä·Ö·´Ó¦ºóµÄÈÜÒºÖÐÓÐH2O2´æÔÚ£¨Ö»ÒªÇóÁгöʵÑéËùÓõÄÊÔ¼Á¼°¹Û²ìµ½µÄÏÖÏ󣩡£

ÊÔ¼Á£º________£»ÏÖÏó£º________¡£

 

´ð°¸£º
½âÎö£º

£¨1£©¸Ã·´Ó¦ÊÇ·ÅÈÈ·´Ó¦  2Na2O2£«2H2O¨T¨T4NaOH£«O2¡ü

£¨2£©¢Ù½«´øÓлðÐǵÄľÌõ¿¿½üµ¼Æø¹Ü¿Úp´¦£¬Ä¾Ìõ¸´È¼

¢Ú½«µ¼Æø¹Üq·ÅÈëË®ÖУ¬·´Ó¦¹ý³ÌÖйܿڴ¦ÓÐÆøÅÝð³ö

£¨3£©Na2SÈÜÒº  ÈÜÒº±ä»ë×Ç£¨»òÊÇ£ºÓÐÉ«²¼Ìõ£»²¼ÌõÍÊÉ«µÈ£©

 


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¸ßһн̲ÄÖÐÓÐÒ»¸öÑÝʾʵÑ飺ÓÃÍÑÖ¬ÃÞ°üסԼ0.2g¹ýÑõ»¯ÄÆ·ÛÄ©£¬ÖÃÓÚʯÃÞÍøÉÏ£¬ÍùÍÑÖ¬ÃÞÉϵÎË®£¬¿É¹Û²ìµ½ÍÑÖ¬ÃÞ¾çÁÒȼÉÕÆðÀ´£®

¡¡¡¡£¨1£©ÓÉʵÑéÏÖÏóËùµÃ³öµÄÓйغͷ´Ó¦µÄ½áÂÛÊÇ£ºa£®ÓÐÑõÆøÉú³É£»b£®________

____________________________________£®

¡¡¡¡ºÍ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ____________________£®

¡¡¡¡£¨2£©Ä³Ñо¿ÐÔѧϰС×éÄâÓÃÏÂͼÖеÄ×°ÖýøÐÐʵÑ飬ÒÔÖ¤Ã÷ÉÏÊö½áÂÛ£®

¡¡¡¡¢ÙÓÃÒÔÑéÖ¤½áÂÛaµÄʵÑé·½·¨ÊÇ£º

¡¡¡¡________________________________________________________________________£®

¡¡¡¡¢ÚÓÃÒÔÑéÖ¤½áÂÛbµÄʵÑé·½·¨¼°ÏÖÏóÊÇ£º

¡¡¡¡________________________________________________________________________£®

¡¡¡¡£¨3£©ÊµÑ飨2£©ÍùÊÔ¹ÜÖмÓË®ÖÁ¹ÌÌåÍêÈ«ÈܽâÇÒ²»ÔÙÓÐÆøÅÝÉú³Éºó£¬È¡³öÊԹܣ¬ÍùÊÔ¹ÜÖеÎÈë·Ó̪ÊÔÒº£¬·¢ÏÖÈÜÒº±äºì£¬Õñµ´ºó£¬ºìÉ«ÏûÍÊ£®ÎªÌ½¾¿´ËÏÖÏ󣬸ÃС×éͬѧ´Ó²éÔÄÓйØ×ÊÁÏÖеÃÖª£ºÓë·´Ó¦¿ÉÉú³É£¬¾ßÓÐÇ¿Ñõ»¯ÐÔºÍƯ°×ÐÔ£®ÇëÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑ飬֤Ã÷ºÍ×ãÁ¿³ä·Ö·´Ó¦ºóµÄÈÜÒºÖÐÓдæÔÚ£®£¨Ö»ÒªÇóÁгöʵÑéËùÓõÄÊÔ¼Á¼°¹Û²ìµ½µÄÏÖÏó£©

¡¡¡¡ÊÔ¼Á£º____________________________________________________£®

¡¡¡¡ÏÖÏó£º____________________________________________________£®

 

¸ßһн̲ÄÖÐÓÐÒ»ÑÝʾʵÑ飬ÓÃÍÑÖ¬ÃÞ°üסԼ0.2g¹ýÑõ»¯ÄÆ·ÛÄ©£¬ÖÃÓÚʯÃÞÍøÉÏ£¬ÍùÍÑÖ¬ÃÞÉϵÎË®£¬¹Û²ìµ½ÍÑÖ¬ÃÞ¾çÁÒȼÉÕÆðÀ´¡£

£¨1£©ÓÉÉÏÊöʵÑéÏÖÏóËùµÃ³öµÄÓйعýÑõ»¯ÄƸúË®·´Ó¦µÄ½áÂÛÊÇ£º

µÚÒ»£¬ÓÐÑõÆøÉú³É£¬µÚ¶þ£¬                ¡£Na2O2¸úË®·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                         £¬

ÆäÖл¹Ô­¼ÁÊÇ            £¬Ñõ»¯¼ÁÊÇ            ¡£

£¨2£©Ä³Ñо¿ÐÔѧϰС×éÄâÓÃAͼËùʾװÖã¨ÆøÃÜÐÔÁ¼ºÃ£©½øÐÐʵÑ飬ÒÔÖ¤Ã÷ÉÏÊö½áÂÛ¡£ÓÃÒÔÑéÖ¤µÚÒ»Ìõ½áÂÛµÄʵÑé·½·¨ÊÇ£º                                                                     

ÓÃÒÔÑéÖ¤µÚ¶þÌõ½áÂÛµÄʵÑé·½·¨ÊÇ£º                                                  

£¨3£©ÊµÑ飨2£©ÍùÊÔ¹ÜÖмÓË®ÖÁ¹ÌÌåÍêÈ«ÈܽâÇÒ²»ÔÙÓÐÆøÅÝÉú³Éºó£¬È¡³öÊԹܣ¬ÍùÊÔ¹ÜÖеÎÈë·Ó̪ÊÔÒº£¬·¢ÏÖÈÜÒºÏȱäºìºóÍÊÉ«¡£ÎªÌ½¾¿ÆäÔ­Òò£¬¸ÃС×éͬѧ´Ó²éÔÄÓйØ×ÊÁÏÖеÃÖª£ºNa2O2ÓëË®·´Ó¦¿ÉÉú³ÉH2O2£¬H2O2¾ßÓÐÇ¿Ñõ»¯ÐÔºÍƯ°×ÐÔ¡£ÇëÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑ飬ÑéÖ¤Na2O2¸ú×ãÁ¿Ë®³ä·Ö·´Ó¦ºóµÄÈÜÒºÖÐÓÐH2O2´æÔÚ¡££¨Ö»ÒªÇóд³öʵÑéËùÓõÄÊÔ¼Á¼°¹Û²ìµ½µÄÏÖÏó£©

ÊÔ¼Á£º                                                   ¡£

ÏÖÏ󣺠                                                                   ¡£

£¨4£©¸ÃС×éͬѧÌá³öÓö¨Á¿µÄ·½·¨Ì½¾¿Na2O2¸úË®·´Ó¦ºóµÄÈÜÒºÖзñº¬ÓÐH2O2£¬ÆäʵÑé·½·¨Îª£º³ÆÈ¡2.6g Na2O2¹ÌÌ壬ʹ֮Óë×ãÁ¿µÄË®·´Ó¦£¬²âÁ¿²úÉúO2µÄÌå»ý²¢ÓëÀíÂÛÖµ±È½Ï£¬¼´¿ÉµÃ³ö½áÂÛ¡£


¢Ù²âÁ¿ÆøÌåÌå»ýʱ£¬±ØÐë´ýÊԹܺÍÁ¿Í²ÄÚµÄÆøÌ嶼ÀäÈ´ÖÁÊÒÎÂʱ½øÐУ¬Ó¦Ñ¡ÓÃÉÏͼװÖÃÖеģ¨ºöÂÔµ¼¹ÜÔÚÁ¿Í²ÖÐËùÕ¼µÄÌå»ý£©         £¨ÌîÐòºÅ£©£¬ÀíÓÉÊÇ                                              ¡£

¢ÚÈôÔÚ±ê×¼×´¿öϲâÁ¿ÆøÌåµÄÌå»ý£¬Ó¦Ñ¡ÓõÄÁ¿Í²µÄ´óС¹æ¸ñΪ          

£¨Ñ¡Ìî¡°100mL¡±¡°200mL¡±¡°500mL¡±»ò¡°1000mL¡±£©¡£

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

¾«Ó¢¼Ò½ÌÍø