ÌâÄ¿ÄÚÈÝ
£¨1£©¹ØÓÚ¼×ͼʵÑéÏÖÏóÔ¤²âÕýÈ·µÄÊÇ
A£®ÊµÑé¢ñ£ºÕñµ´ºó¾²Öã¬ÈÜÒº²»Ôٷֲ㣬ÇÒ±£³ÖÎÞɫ͸Ã÷
B£®ÊµÑé¢ò£ºÌúƬ×îÖÕÍêÈ«Èܽ⣬ÇÒ¸ßÃÌËá¼ØÈÜÒº±äÎÞÉ«
C£®ÊµÑé¢ó£ºÎ¢ÈÈÏ¡HNO3Ƭ¿Ì£¬ÈÜÒºÖÐÓÐÆøÅݲúÉú£¬¹ã¿ÚÆ¿ÄÚʼÖÕ±£³ÖÎÞÉ«
D£®ÊµÑé¢ô£ºµ±ÈÜÒºÖÁºìºÖÉ«£¬Í£Ö¹¼ÓÈÈ£¬ÈùâÊøͨ¹ýÌåϵʱ¿É²úÉú¶¡´ï¶ûÏÖÏó
£¨2£©ÁòËáÍÔÚÍ¿ÁϹ¤Òµ¡¢Ó¡È¾¹¤Òµ¡¢Å©ÒµµÈ·½ÃæÓÐÖØÒªÓ¦ÓÃ
¢ÙʵÑéÊÒÅäÖÆ480ml 0.1mol/LµÄCuSO4ÈÜÒº£¬ÐèÓÃÍÐÅÌÌìƽ³ÆÈ¡µ¨·¯
¢ÚijͬѧÏëÒªÔÚʵÑéÊÒÖÐÖƱ¸ÁòËáÍ£¬²éѯ×ÊÁϵÃÖª£ºÍм·ÅÈëÏ¡ÁòËáÖв»Èܽ⣬ÈôÔÚÏ¡ÁòËáÖмÓÈëH2O2£¬Íм¿ÉÖð½¥Èܽ⣮ΪÑéÖ¤¸ÃʵÑ飬ʵÑéС×éÉè¼ÆʵÑé×°ÖÃÈçͼÒÒ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
D
D
A£®ÊµÑé¢ñ£ºÕñµ´ºó¾²Öã¬ÈÜÒº²»Ôٷֲ㣬ÇÒ±£³ÖÎÞɫ͸Ã÷
B£®ÊµÑé¢ò£ºÌúƬ×îÖÕÍêÈ«Èܽ⣬ÇÒ¸ßÃÌËá¼ØÈÜÒº±äÎÞÉ«
C£®ÊµÑé¢ó£ºÎ¢ÈÈÏ¡HNO3Ƭ¿Ì£¬ÈÜÒºÖÐÓÐÆøÅݲúÉú£¬¹ã¿ÚÆ¿ÄÚʼÖÕ±£³ÖÎÞÉ«
D£®ÊµÑé¢ô£ºµ±ÈÜÒºÖÁºìºÖÉ«£¬Í£Ö¹¼ÓÈÈ£¬ÈùâÊøͨ¹ýÌåϵʱ¿É²úÉú¶¡´ï¶ûÏÖÏó
£¨2£©ÁòËáÍÔÚÍ¿ÁϹ¤Òµ¡¢Ó¡È¾¹¤Òµ¡¢Å©ÒµµÈ·½ÃæÓÐÖØÒªÓ¦ÓÃ
¢ÙʵÑéÊÒÅäÖÆ480ml 0.1mol/LµÄCuSO4ÈÜÒº£¬ÐèÓÃÍÐÅÌÌìƽ³ÆÈ¡µ¨·¯
12.5g
12.5g
g£¬¢ÚijͬѧÏëÒªÔÚʵÑéÊÒÖÐÖƱ¸ÁòËáÍ£¬²éѯ×ÊÁϵÃÖª£ºÍм·ÅÈëÏ¡ÁòËáÖв»Èܽ⣬ÈôÔÚÏ¡ÁòËáÖмÓÈëH2O2£¬Íм¿ÉÖð½¥Èܽ⣮ΪÑéÖ¤¸ÃʵÑ飬ʵÑéС×éÉè¼ÆʵÑé×°ÖÃÈçͼÒÒ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
Cu+H2O2+H2SO4=CuSO4+2H2O
Cu+H2O2+H2SO4=CuSO4+2H2O
£¬Èô½«H2O2ºÍÏ¡ÁòËá¼ÓÈëÉÕÆ¿ÖеÄ˳Ðòµßµ¹£¬ÊµÑéµÃµ½µÄ½áÂÛÊÇÍм·ÅÈëH2O2Öв»·¢Éú·´Ó¦£¬ÈôÔÚH2O2ÖмÓÈëÏ¡ÁòËᣬÍм¿ÉÖð½¥Èܽâ
Íм·ÅÈëH2O2Öв»·¢Éú·´Ó¦£¬ÈôÔÚH2O2ÖмÓÈëÏ¡ÁòËᣬÍм¿ÉÖð½¥Èܽâ
£®·ÖÎö£º£¨1£©A¡¢×°ÖÃͼÖб½ÄÑÈÜÓÚË®ºÍË®·Ö²ã£»
B¡¢³£ÎÂÏÂÌúÔÚŨÁòËáÖз¢Éú¶Û»¯ÏÖÏó£»
C¡¢ÍºÍÏ¡ÏõËá·¢ÉúÉú³ÉÒ»Ñõ»¯µªÆøÌåÔÚ¹ã¿ÚÆ¿ÖÐÓöµ½¿ÕÆøÖеÄÑõÆø·´Ó¦Éú³Éºì×ØÉ«ÆøÌå¶þÑõ»¯µª£»
D¡¢±¥ºÍÂÈ»¯ÌúÈÜÒº¼ÓÈȵ½ºìºÖÉ«ºóÖƵÄÇâÑõ»¯Ìú½ºÌ壻
£¨2£©¢Ù¸ù¾Ýn=cv¼ÆËãÈÜÖÊCuSO4µÄÎïÖʵÄÁ¿£¬ÀûÓÃCuSO4?5H2OµÄÎïÖʵÄÁ¿µÈÓÚCuSO4µÄÎïÖʵÄÁ¿£¬¸ù¾Ým=nM¼ÆËãCuSO4?5H2OµÄÖÊÁ¿£»
¢ÚË«ÑõË®µÄÑõ»¯ÐÔ±ÈÏ¡ÁòËáÇ¿µÄ¶à£¬¿ÉÒÔ½«ÍÑõ»¯£¬Ñõ»¯ºóµÄͺÍÏ¡ÁòËá·´Ó¦£»
B¡¢³£ÎÂÏÂÌúÔÚŨÁòËáÖз¢Éú¶Û»¯ÏÖÏó£»
C¡¢ÍºÍÏ¡ÏõËá·¢ÉúÉú³ÉÒ»Ñõ»¯µªÆøÌåÔÚ¹ã¿ÚÆ¿ÖÐÓöµ½¿ÕÆøÖеÄÑõÆø·´Ó¦Éú³Éºì×ØÉ«ÆøÌå¶þÑõ»¯µª£»
D¡¢±¥ºÍÂÈ»¯ÌúÈÜÒº¼ÓÈȵ½ºìºÖÉ«ºóÖƵÄÇâÑõ»¯Ìú½ºÌ壻
£¨2£©¢Ù¸ù¾Ýn=cv¼ÆËãÈÜÖÊCuSO4µÄÎïÖʵÄÁ¿£¬ÀûÓÃCuSO4?5H2OµÄÎïÖʵÄÁ¿µÈÓÚCuSO4µÄÎïÖʵÄÁ¿£¬¸ù¾Ým=nM¼ÆËãCuSO4?5H2OµÄÖÊÁ¿£»
¢ÚË«ÑõË®µÄÑõ»¯ÐÔ±ÈÏ¡ÁòËáÇ¿µÄ¶à£¬¿ÉÒÔ½«ÍÑõ»¯£¬Ñõ»¯ºóµÄͺÍÏ¡ÁòËá·´Ó¦£»
½â´ð£º½â£º£¨1£©A¡¢×°ÖÃͼÖб½ÄÑÈÜÓÚË®ºÍË®·Ö²ã£¬±½²ã×غìÉ«£»¹ÊA´íÎó£»
B¡¢³£ÎÂÏÂÌúÔÚŨÁòËáÖз¢Éú¶Û»¯ÏÖÏó£»×èÖ¹·´Ó¦½øÐУ¬ÌúƬ²»ÄÜÈܽ⣬¸ßÃÌËá¼ØÈÜÒº²»±äÉ«£¬¹ÊB´íÎó£»
C¡¢ÍºÍÏ¡ÏõËá·¢ÉúÉú³ÉÒ»Ñõ»¯µªÆøÌåÔÚ¹ã¿ÚÆ¿ÖÐÓöµ½¿ÕÆøÖеÄÑõÆø·´Ó¦Éú³Éºì×ØÉ«ÆøÌå¶þÑõ»¯µª£»¹ÊC´íÎó£»
D¡¢±¥ºÍÂÈ»¯ÌúÈÜÒº¼ÓÈȵ½ºìºÖÉ«ºóÖƵÄÇâÑõ»¯Ìú½ºÌ壻½ºÌå¾ßÓж¡´ï¶ûÏÖÏ󣬹âÏßͨ¹ý³öÏÖÒ»Ìõ¹âÁÁµÄͨ·£¬¹ÊDÕýÈ·£»
¹Ê´ð°¸Îª£ºD£»
£¨2£©¢ÙÅäÖÆÈÜÒºµÄÌå»ýΪ480ml£¬¶øÈÝÁ¿Æ¿µÄ¹æ¸ñûÓÐ480ml£¬Ö»ÄÜÑ¡ÓÃ500mlÈÝÁ¿Æ¿£¬CuSO4µÄÎïÖʵÄÁ¿n=cV=0.5L¡Á0.1mol?L-1=0.05mol£¬CuSO4?5H2OµÄÎïÖʵÄÁ¿µÈÓÚCuSO4µÄÎïÖʵÄÁ¿£¬ËùÒÔCuSO4?5H2OµÄÖÊÁ¿0.05mol¡Á250g/mol=12.5g£¬¹Ê´ð°¸Îª£º12.5£»
¢ÚÒòΪ˫ÑõË®ÔÚËáÐÔÈÜÒºÖÐÏÈ°ÑÍÑõ»¯³ÉÑõ»¯Í£¬µ±È»ÕâÊÇÒ»¸ö΢ÈõµÄ·´Ó¦£¬ÐγÉÒ»¸öƽºâ£¬µ«ÊÇÐγɵÄÑõ»¯ÍÂíÉϾͻᱻϡÁòËáÈܽ⣬ƽºâ±»´òÆÆ£¬·´Ó¦³¯Õý·½Ïò½øÐУ¬¹Ê¶øÖð½¥Èܽ⣬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCu+H2O2+H2SO4=CuSO4+2H2O£»¹Ê´ð°¸Îª£ºCu+H2O2+H2SO4=CuSO4+2H2O£»
Èô½«H2O2ºÍÏ¡ÁòËá¼ÓÈëÉÕÆ¿ÖеÄ˳Ðòµßµ¹£¬ÊµÑéµÃµ½µÄ½áÂÛÊÇ£ºÍм·ÅÈëH2O2Öв»·¢Éú·´Ó¦£¬ÈôÔÚH2O2ÖмÓÈëÏ¡ÁòËᣬÍм¿ÉÖð½¥Èܽ⣻
¹Ê´ð°¸Îª£ºÍм·ÅÈëH2O2Öв»·¢Éú·´Ó¦£¬ÈôÔÚH2O2ÖмÓÈëÏ¡ÁòËᣬÍм¿ÉÖð½¥Èܽ⣻
B¡¢³£ÎÂÏÂÌúÔÚŨÁòËáÖз¢Éú¶Û»¯ÏÖÏó£»×èÖ¹·´Ó¦½øÐУ¬ÌúƬ²»ÄÜÈܽ⣬¸ßÃÌËá¼ØÈÜÒº²»±äÉ«£¬¹ÊB´íÎó£»
C¡¢ÍºÍÏ¡ÏõËá·¢ÉúÉú³ÉÒ»Ñõ»¯µªÆøÌåÔÚ¹ã¿ÚÆ¿ÖÐÓöµ½¿ÕÆøÖеÄÑõÆø·´Ó¦Éú³Éºì×ØÉ«ÆøÌå¶þÑõ»¯µª£»¹ÊC´íÎó£»
D¡¢±¥ºÍÂÈ»¯ÌúÈÜÒº¼ÓÈȵ½ºìºÖÉ«ºóÖƵÄÇâÑõ»¯Ìú½ºÌ壻½ºÌå¾ßÓж¡´ï¶ûÏÖÏ󣬹âÏßͨ¹ý³öÏÖÒ»Ìõ¹âÁÁµÄͨ·£¬¹ÊDÕýÈ·£»
¹Ê´ð°¸Îª£ºD£»
£¨2£©¢ÙÅäÖÆÈÜÒºµÄÌå»ýΪ480ml£¬¶øÈÝÁ¿Æ¿µÄ¹æ¸ñûÓÐ480ml£¬Ö»ÄÜÑ¡ÓÃ500mlÈÝÁ¿Æ¿£¬CuSO4µÄÎïÖʵÄÁ¿n=cV=0.5L¡Á0.1mol?L-1=0.05mol£¬CuSO4?5H2OµÄÎïÖʵÄÁ¿µÈÓÚCuSO4µÄÎïÖʵÄÁ¿£¬ËùÒÔCuSO4?5H2OµÄÖÊÁ¿0.05mol¡Á250g/mol=12.5g£¬¹Ê´ð°¸Îª£º12.5£»
¢ÚÒòΪ˫ÑõË®ÔÚËáÐÔÈÜÒºÖÐÏÈ°ÑÍÑõ»¯³ÉÑõ»¯Í£¬µ±È»ÕâÊÇÒ»¸ö΢ÈõµÄ·´Ó¦£¬ÐγÉÒ»¸öƽºâ£¬µ«ÊÇÐγɵÄÑõ»¯ÍÂíÉϾͻᱻϡÁòËáÈܽ⣬ƽºâ±»´òÆÆ£¬·´Ó¦³¯Õý·½Ïò½øÐУ¬¹Ê¶øÖð½¥Èܽ⣬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCu+H2O2+H2SO4=CuSO4+2H2O£»¹Ê´ð°¸Îª£ºCu+H2O2+H2SO4=CuSO4+2H2O£»
Èô½«H2O2ºÍÏ¡ÁòËá¼ÓÈëÉÕÆ¿ÖеÄ˳Ðòµßµ¹£¬ÊµÑéµÃµ½µÄ½áÂÛÊÇ£ºÍм·ÅÈëH2O2Öв»·¢Éú·´Ó¦£¬ÈôÔÚH2O2ÖмÓÈëÏ¡ÁòËᣬÍм¿ÉÖð½¥Èܽ⣻
¹Ê´ð°¸Îª£ºÍм·ÅÈëH2O2Öв»·¢Éú·´Ó¦£¬ÈôÔÚH2O2ÖмÓÈëÏ¡ÁòËᣬÍм¿ÉÖð½¥Èܽ⣻
µãÆÀ£º±¾Ì⿼²éÁ˳£¼ûÎïÖʵÄÖÆÈ¡ÔÀí£¬×°Ö÷ÖÎö£¬ÊµÑé·½·¨µÄÉè¼ÆÅжϣ¬×¢ÒâʵÑéµÄ»ù±¾²Ù×÷·½·¨ºÍ×¢ÒâÊÂÏ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿