ÌâÄ¿ÄÚÈÝ

1£®Ä³Í¬Ñ§Éè¼ÆͼËùʾװÖ÷ֱð½øÐÐ̽¾¿ÊµÑ飨¼Ð³Ö×°ÖÃÒÑÂÔÈ¥£©£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺



ʵÑé Ò©Æ· ÖÆÈ¡ÆøÌå Á¿Æø¹ÜÖеÄÒºÌå
¢ÙCu¡¢Å¨HNO3 H2O
¢ÚCaO¹ÌÌ塢Ũ°±Ë® NH3  
¢Û´Öþ¡¢Ï¡H2SO4 £¨×ãÁ¿£©[ÔÓÖʲ»ÓëÁòËá·´Ó¦]H2H2O
£¨1£©ÒÇÆ÷aµÄÃû³ÆÊǸÉÔï¹Ü£®
£¨2£©¼ì²é¸Ã×°ÖõÄÆøÃÜÐԵķ½·¨ÊÇ£º¹Ø±Õ¿ª¹ØAºÍ·ÖҺ©¶·»îÈû£¬Î¢ÈÈÉÕÆ¿£¬Á¿Æø¹ÜÓÒ¶ËÒºÃæÉý¸ß£¬ËµÃ÷ÆøÃÜÐÔÁ¼ºÃ£®
£¨3£©¸ÃͬѧÈÏΪʵÑé¢Ù¿Éͨ¹ýÊÕ¼¯²âÁ¿NO2ÆøÌåµÄÌå»ý£¬À´Ì½¾¿CuÑùÆ·µÄ´¿¶È£®ÇëÎÊÊÇ·ñ¿ÉÐзñ£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©£»ÊµÑé¢ÙÖÐÉÕÆ¿ÄÚ·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºCu+4H++2NO3-=Cu2++2NO2¡ü+2H2O£®
£¨4£©ÊµÑé¢ÚÖУ¬Á¿Æø¹ÜÖеÄÒºÌå×îºÃÊÇD£º
A£®H2O      B£®±¥ºÍNaHCO3ÈÜÒºC£®±¥ºÍNa2CO3ÈÜÒº      D£®CCl4
£¨5£©±¾ÊµÑéÓ¦¶ÔÁ¿Æø¹Ü¶à´Î¶ÁÊý£¬¶ÁÊýʱӦעÒ⣺
¢Ù»Ö¸´ÖÁÊÒΣ¬¢Ú±£³ÖÁ¿Æø¹ÜÄÚÒºÌåÓë¸ÉÔï¹ÜÄÚÒºÌåÒºÃæÏàƽ£¬¢ÛÊÓÏßÓë°¼ÒºÃæ×îµÍ´¦Ïàƽ£®
£¨6£©ÊµÑé¢Û»ñµÃÒÔÏÂÊý¾Ý£¨ËùÓÐÆøÌåÌå»ý¾ùÒÑ»»Ëã³É±ê×¼×´¿ö£©£®
±àºÅþ£¨º¬ÔÓÖÊ£©ÖÊÁ¿Á¿Æø¹ÜµÚÒ»´Î¶ÁÊýÁ¿Æø¹ÜµÚ¶þ´Î¶ÁÊý
10.5g10.0mL346.5mL
20.5g10.0mL335.0mL
30.5g10.0mL345.5mL
¸ù¾ÝÉÏÊöºÏÀíÊý¾Ý¼ÆËã´ÖMgÖÐMgµÄÖÊÁ¿·ÖÊý72%£®

·ÖÎö £¨1£©ÒÇÆ÷aΪ¸ÉÔï¹Ü£»
£¨2£©ÀûÓÃÆøÌåµÄÈÈÕÍÀäËõ£¬¹Ø±Õ¿ª¹ØAºÍ·ÖҺ©¶·»îÈû£¬Î¢ÈÈÉÕÆ¿£¬Á¿Æø¹ÜÒºÃæÉý¸ß£¬ËµÃ÷ÆøÃÜÐÔÁ¼ºÃ£»
£¨3£©¶þÑõ»¯µªÓëË®·´Ó¦£¬µ¼Ö²ⶨ¶þÑõ»¯µªµÄÌå»ý²»×¼£»CuÓëŨÏõËá·´Ó¦Éú³ÉÏõËáÍ­¡¢¶þÑõ»¯µªÓëË®£»
£¨4£©Á¿Æø¹ÜÖÐÒºÌåµÄÑ¡Ôñ±ê×¼ÊÇ£ººÍ¸ÃÆøÌå²»·´Ó¦£¬²»Èܽâ¸ÃÆøÌ壻
£¨5£©¶ÔÁ¿Æø¹Ü¶ÁÊýʱ£¬Ê×ÏȵÈʵÑé×°Öûָ´µ½ÊÒÎÂÔÙ½øÐÐÏÂÒ»²½²Ù×÷£¬È»ºóµ÷½ÚÁ¿Æø¹Üʹ×óÓÒÒºÃæÏàƽ£¬×îºó¶ÁÊýʱÊÓÏßÓë°¼ÒºÃæ×îµÍ´¦Ïàƽ£»
£¨6£©ÊµÑé¢ÛÖУ¬µÚ2´ÎÓëµÚ1¡¢3´ÎʵÑé²âµÃÇâÆøµÄÌå»ýÏà²î½ÏµÍ£¬Ó¦ÉáÈ¥µÚ2´ÎÊý¾Ý£¬¸ù¾ÝÁ¿Æø¹ÜµÄµÚ¶þ´Î¶ÁÊý-µÚÒ»´Î¶ÁÊý=Éú³ÉÇâÆøµÄÌå»ý£¬ÀûÓõÚ1¡¢3´ÎÊý¾ÝÇó³öÉú³ÉÇâÆøµÄƽ¾ùÖµ£¬ÔÙ¸ù¾Ý·½³ÌʽÇó³ö´ÖþÖÐMgµÄÖÊÁ¿£¬½ø¶ø¼ÆËã³öMgµÄÖÊÁ¿·ÖÊý£®

½â´ð ½â£º£¨1£©ÒÇÆ÷aΪ¸ÉÔï¹Ü£¬¹Ê´ð°¸Îª£º¸ÉÔï¹Ü£»
£¨2£©¹Ø±Õ¿ª¹ØAºÍ·ÖҺ©¶·»îÈû£¬Î¢ÈÈÉÕÆ¿£¬Á¿Æø¹ÜÓÒ¶ËÒºÃæÉý¸ß£¬ËµÃ÷ÆøÃÜÐÔÁ¼ºÃ£¬
¹Ê´ð°¸Îª£ºÎ¢ÈÈÉÕÆ¿£¬Á¿Æø¹ÜÓÒ¶ËÒºÃæÉý¸ß£»
£¨3£©¶þÑõ»¯µªÓëË®·´Ó¦£¬µ¼Ö²ⶨ¶þÑõ»¯µªµÄÌå»ý²»×¼£¬¹Ê·½°¸²»¿ÉÐУ»CuÓëŨÏõËá·´Ó¦Éú³ÉÏõËáÍ­¡¢¶þÑõ»¯µªÓëË®£¬·´Ó¦Àë×Ó·½³ÌʽΪ£ºCu+4H++2NO3-=Cu2++2NO2¡ü+2H2O£¬
¹Ê´ð°¸Îª£º·ñ£»Cu+4H++2NO3-=Cu2++2NO2¡ü+2H2O£»
£¨4£©ÊµÑé¢ÚÖа±Æø¼«Ò×ÈÜÓÚË®£¬ËùÒÔÁ¿Æø¹ÜÖÐÒºÌå²»ÄÜÑ¡ÓÃË®¼°Ë®ÈÜÒº£¬Ö»ÄÜÑ¡ºÍ°±Æø²»·´Ó¦µÄËÄÂÈ»¯Ì¼£¬¹Ê´ð°¸Îª£ºD£»
£¨5£©¸ù¾ÝPV=nRT£¬Îª±£Ö¤²â³öÀ´µÄÆøÌåÌå»ýÊǵ±Ê±´óÆøѹϵÄÌå»ý£¬ÔÚ¶ÁÊýʱӦעÒ⣺¢Ù½«ÊµÑé×°Öûָ´µ½ÊÒΣ¬¢ÚʹÁ¿Æø¹ÜÁ½¶ËÒºÃæÏàƽ£¬Ïàƽ˵Ã÷Á½±ßÒºÃæÉϵÄѹǿһÑù£¬ÕâÑù²â³öÀ´µÄÆøÌåÌå»ý²ÅÊǵ±Ê±´óÆøѹϵÄÌå»ý£¬¢ÛÊÓÏßÓë°¼ÒºÃæ×îµÍ´¦Ïàƽ£¬¸©ÊÓ»òÑöÊÓ»áÔì³ÉÎó²î£»
¹Ê´ð°¸Îª£º±£³ÖÁ¿Æø¹ÜÄÚÒºÌåÓë¸ÉÔï¹ÜÄÚÒºÌåÒºÃæÏàƽ£»
£¨6£©ÊµÑé¢ÛÖУ¬µÚ2´ÎÓëµÚ1¡¢3´ÎʵÑé²âµÃÇâÆøµÄÌå»ýÏà²î½ÏµÍ£¬Ó¦ÉáÈ¥µÚ2´ÎÊý¾Ý£¬Éú³ÉÇâÆøÌå»ýΪ$\frac{£¨346.5-10£©mL+£¨345.5-10£©mL}{2}$=336mL=0.336L£¬
ÉèÉú³É0.346LÇâÆø£¬ÐèMgµÄÖÊÁ¿Îªx£¬Ôò£º
Mg+H2SO4¨TMgSO4+H2¡ü
24g           22.4L
x            0.336L
x=$\frac{24g¡Á0.336L}{22.4L}$=0.36g
ËùÒÔþµÄÖÊÁ¿·ÖÊýΪ $\frac{0.36g}{0.5g}$¡Á100%=72%
¹Ê´ð°¸Îª£º72%£®

µãÆÀ ±¾Ì⿼²éÎïÖʺ¬Á¿µÄ²â¶¨£¬Àí½âÔ­ÀíÊǽâÌâ¹Ø¼ü£¬ÊǶÔѧÉú×ÛºÏÄÜÁ¦µÄ¿¼²é£¬ÐèҪѧÉú¾ß±¸ÔúʵµÄ»ù´¡£¬ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø