ÌâÄ¿ÄÚÈÝ

¹ý̼ËáÄÆ£¨2Na2CO3¡¤3H2O2£©ÊÇÒ»ÖÖ¶àÓÃ;µÄÐÂÐÍÏû¶¾¼Á£¬¿ÉÓÐЧɱÃð¡°¼×ÐÍ    H1N1Á÷¸Ð¡±²¡¶¾¡£ÒÑÖª¹ý̼ËáÄÆÊÇÒ»ÖÖ¿ÉÈÜÓÚË®µÄ°×ɫϸС¿ÅÁ£×´·ÛÄ©£¬50¡æ¿É·Ö½â£¬Æä3£¥µÄË®ÈÜÒºµÄpHԼΪ10.5£¬¹ý̼ËáÄƾßÓÐNa2C03ºÍH202µÄË«ÖØÐÔÖÊ¡£

   £¨1£©ÎªÌ½¾¿¹ý̼ËáÄƵÄÐÔÖÊ£¬Ä³Í¬Ñ§ÓÃÊÔ¹ÜÈ¡ÊÊÁ¿¹ý̼ËáÄÆÈÜÒº£¬µÎ¼Ó·Ó̪ÊÔÒº¡£¿ªÊ¼¿ÉÄܹ۲쵽µÄÏÖÏóÊÇ              £¬²úÉú´ËÏÖÏóµÄÔ­ÒòÊÇ             £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£¬Î¢ÈȲ¢Õñµ´ÊԹܺóÓÖ¿ÉÄܹ۲쵽             ÏÖÏó¡£

   £¨2£©ÒÑÖª¹ý̼ËáÄÆÓöÏ¡ÁòËá¿É²úÉúÁ½ÖÖÆøÌ塣ij»¯Ñ§¿ÎÍâ»î¶¯Ð¡×éÀûÓÃÒÔÏÂ×°ÖÃÍê³ÉÁËϵÁÐʵÑé¡£  

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ù¼×ͬѧÓÃ×°ÖÃIÑéÖ¤²úÉúµÄÁ½ÖÖÆøÌ壬BÖÐÊ¢ÓÐ×ãÁ¿µÄBa£¨OH£©2ÈÜÒº£¬Èô¹Û²ìµ½µÄÏÖÏóÊÇ               £¬ÔòÖ¤Ã÷ÓР              ÆøÌåÉú³É£»¼òÊöÑéÖ¤ÁíÒ»ÖÖÆøÌåµÄ·½·¨                                                   £»

¢ÚÒÒ²àѧ°ÑÉÏÊö×°ÖÃ×éºÏ£¬ÓÃÓڲⶨ2Na2CO3¡¤3H2OÑùÆ·ÖÐNa2CO3µÄº¬Á¿¡£°´ÆøÁ÷´Ó×óµ½Óҵķ½Ïò£¬×°ÖÃI¡¢II¡¢IIIµÄÁ¬½Ó˳ÐòÊÇ                       £¨Ìî×°ÖÃÐòºÅ£©£»  B¡¢EÖÐÓ¦·Ö±ðÊ¢·Å               ¡¢            £»×°ÖÃIIIÖÐͨ¿ÕÆøµÄÄ¿µÄÊÇ                                  ¡£

 

¡¾´ð°¸¡¿

£¨1£©ÈÜÒº±äºì£¨1·Ö£©CO2-3+H2OHCO-3+OH-£¨2·Ö£©ÈÜÒºÍÊÉ«£¨1·Ö£©

   £¨2£©¢Ù²úÉú°×É«³Áµí£¨1·Ö£©CO2£¨1·Ö£©

È¡´øÓлðÐǵÄľÌõ¿¿½üBÆ¿µÄµ¼¹Ü³ö¿Ú£¬ÈôľÌõ¸´È¼ËµÃ÷ÓÐÑõÆøÉú³É£¨2·Ö£©

¢ÙIII¡úI¡úII£¨2·Ö£©Å¨H2SO4£¨1·Ö£©NaOHÈÜÒº£¨1·Ö£©

·ÀÖ¹¿ÕÆøÖеĶþÑõ»¯Ì¼¡¢Ë®ÕôÆøµÈÆøÌå½øÈëCÖУ¨1·Ö£©

½«×°ÖÃIÖеĶþÑõ»¯Ì¼È«²¿ÅÅÈëCÖУ¨2·Ö£©

¡¾½âÎö¡¿ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2010?Öî³ÇÊÐÄ£Ä⣩¹ý̼ËáÄÆ£¨2Na2CO3?3H2O2£©ÊÇÒ»ÖÖÐÂÐ͸ßЧ¹ÌÌåƯ°×ɱ¾ú¼Á£¬Ëü¾ßÓÐÎÞ³ô¡¢ÎÞ¶¾¡¢ÎÞÎÛȾµÄÌص㣬±»´óÁ¿Ó¦ÓÃÓÚÏ´µÓ¡¢Ó¡È¾¡¢·ÄÖ¯¡¢ÔìÖ½¡¢Ò½Ò©ÎÀÉúµÈÁìÓòÖУ¬ÒÀÍд¿¼î³§ÖƱ¸¹ý̼ËáÄƿɽµµÍÉú²ú³É±¾£¬ÆäÉú²úÁ÷³ÌÈçÏ£®

ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©¹ý̼ËáÄÆÈÜÓÚË®ºóËùµÃÈÜÒºÒ»¶¨ÏÔ
¼îÐÔ
¼îÐÔ
£¨¡°ËáÐÔ¡±¡¢¡°¼îÐÔ¡±»ò¡°ÖÐÐÔ¡±£©£®
£¨2£©²Ù×÷¢ÙÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐ
ÆÕͨ©¶·¡¢ÉÕ±­¡¢²£Á§°ô
ÆÕͨ©¶·¡¢ÉÕ±­¡¢²£Á§°ô
£¨ÌîдÒÇÆ÷Ãû³Æ£©£®
£¨3£©¹¤Òµ´¿¼îÖг£º¬ÓÐÉÙÁ¿NaCl£¬Ä³Ð£»¯Ñ§¿ÎÍâ»î¶¯Ð¡×éÉè¼ÆÈçͼËùʾװÖ㬲ⶨ¹¤Òµ´¿¼îÖÐNa2CO3µÄº¬Á¿£®

¢ÙÒª¼ìÑ鹤ҵ´¿¼îÖÐÔÓÖʵĴæÔÚ£¬×îºÃÑ¡ÓÃÏÂÁÐÊÔ¼ÁÖеÄ
BD
BD
£¨Ñ¡ÌîÐòºÅ£©£®
A£®ÇâÑõ»¯±µÈÜÒº     B£®Ï¡ÏõËá    C£®ÁòÇ軯¼ØÈÜÒº    D£®ÏõËáÒøÈÜÒº
¢Ú¼ìÑé×°ÖÃBÆøÃÜÐԵķ½·¨ÊÇ£ºÈû½ô´ø³¤¾±Â©¶·µÄÈý¿×Ïð½ºÈû£¬¼Ð½ôµ¯»É¼Ðºó£¬´Ó©¶·×¢ÈëÒ»¶¨Á¿µÄË®£¬Ê¹Â©¶·ÄÚµÄË®Ãæ¸ßÓÚÆ¿ÄÚµÄË®Ã棬ֹͣ¼ÓË®ºó£¬Èô
©¶·ÖÐÓëÊÔ¼ÁÆ¿ÖеÄÒºÃæ²î±£³Ö²»Ôٱ仯»ò©¶·ÖеÄÒºÃæ²»ÔÙϽµ
©¶·ÖÐÓëÊÔ¼ÁÆ¿ÖеÄÒºÃæ²î±£³Ö²»Ôٱ仯»ò©¶·ÖеÄÒºÃæ²»ÔÙϽµ
£¬ËµÃ÷×°Öò»Â©Æø£®
¢Û×°ÖÃAµÄ×÷ÓÃÊÇ
³ýÈ¥¿ÕÆøÖÐCO2£¬·ÀÖ¹Ó°Ïì²âÁ¿½á¹û
³ýÈ¥¿ÕÆøÖÐCO2£¬·ÀÖ¹Ó°Ïì²âÁ¿½á¹û
£®×°ÖÃCÖеÄÊÔ¼ÁΪ
ŨÁòËá
ŨÁòËá
£®
¢ÜijͬѧÈÏΪÔÚD×°ÖúóÓ¦ÔÙÁ¬½ÓE×°Öã¨×°ÓÐÊʵ±ÊÔ¼Á£©£¬ÄãÈÏΪÊÇ·ñ±ØÒª£¿
±ØÒª
±ØÒª
£¨Ñ¡Ìî¡°±ØÒª¡±»ò¡°²»±ØÒª¡±£©£¬ÅжϵÄÀíÓÉÊÇ
ÒòΪװÖÃE»áÎüÊÕ¿ÕÆøÖеĶþÑõ»¯Ì¼ºÍË®ÕôÆø£¬Ó°Ïì²âÁ¿½á¹û£®
ÒòΪװÖÃE»áÎüÊÕ¿ÕÆøÖеĶþÑõ»¯Ì¼ºÍË®ÕôÆø£¬Ó°Ïì²âÁ¿½á¹û£®
£®
£¨2012?Õã½­Ä£Ä⣩¹ý̼ËáÄÆ£¨2Na2CO3?3H2O2£©ÊÇÒ»ÖÖ¼¯Ï´µÓ¡¢Æ¯°×¡¢É±¾úÓÚÒ»ÌåµÄÑõϵƯ°×¼Á£®Ä³ÐËȤС×éÖƱ¸¹ý̼ËáÄƵÄʵÑé·½°¸ºÍ×°ÖÃʾÒâͼÈçÏ£º

ÒÑÖª£ºÖ÷·´Ó¦    2Na2CO3 £¨aq£©+3H2O2 £¨aq£©  2Na2CO3?3H2O2 £¨s£©
¡÷H£¼0
¸±·´Ó¦    2H2O2=2H2O+O2¡ü
µÎ¶¨·´Ó¦  6KMnO4+5£¨2Na2CO3?3H2O2£©+19H2SO4=3K2SO4+6MnSO4+10Na2SO4+10CO2¡ü+15O2¡ü+34H2O
50¡ãCʱ   2Na2CO3?3H2O2 £¨s£© ¿ªÊ¼·Ö½â
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Í¼ÖÐÖ§¹ÜµÄ×÷ÓÃÊÇ
ƽºâѹǿ
ƽºâѹǿ
£®
£¨2£©²½Öè¢ÙµÄ¹Ø¼üÊÇ¿ØÖÆζȣ¬Æä´ëÊ©ÓÐ
Àäˮԡ
Àäˮԡ
¡¢
´ÅÁ¦½Á°è
´ÅÁ¦½Á°è
ºÍ
»ºÂýµÎ¼ÓH2O2ÈÜÒº
»ºÂýµÎ¼ÓH2O2ÈÜÒº
£®
£¨3£©ÔÚÂËÒºXÖмÓÈëÊÊÁ¿NaCl¹ÌÌå»òÎÞË®ÒÒ´¼£¬¾ù¿ÉÎö³ö¹ý̼ËáÄÆ£¬Ô­ÒòÊÇ
½µµÍ²úÆ·µÄÈܽâ¶È£¨ÑÎÎö×÷Óûò´¼Îö×÷Óã©
½µµÍ²úÆ·µÄÈܽâ¶È£¨ÑÎÎö×÷Óûò´¼Îö×÷Óã©
£®
£¨4£©²½Öè¢ÛÖÐÑ¡ÓÃÎÞË®ÒÒ´¼Ï´µÓ²úÆ·µÄÄ¿µÄÊÇ
ϴȥˮ·Ý£¬ÀûÓÚ¸ÉÔï
ϴȥˮ·Ý£¬ÀûÓÚ¸ÉÔï
£®
£¨5£©ÏÂÁÐÎïÖÊÖУ¬»áÒýÆð¹ý̼ËáÄÆ·Ö½âµÄÓÐ
AB
AB
£®
A£®Fe2O3 B£®CuO
C£®Na2SiO3 D£®MgSO4
£¨6£©×¼È·³ÆÈ¡0.2000g ¹ý̼ËáÄÆÓÚ250mL ×¶ÐÎÆ¿ÖУ¬¼Ó50mL ÕôÁóË®Èܽ⣬ÔÙ¼Ó50mL 2.0mol?L-1 H2SO4£¬ÓÃ2.000¡Á10-2 mol?L-1 KMnO4 ±ê×¼ÈÜÒºµÎ¶¨ÖÁÖÕµãʱÏûºÄ30.00mL£¬Ôò²úÆ·ÖÐH2O2µÄÖÊÁ¿·ÖÊýΪ
0.2550
0.2550
£®

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

¾«Ó¢¼Ò½ÌÍø