ÌâÄ¿ÄÚÈÝ

£¨12·Ö£©Ä³Ñ§Ï°Ð¡×éΪ֤Ã÷²¢¹Û²ìÍ­ÓëÏ¡HNO3·´Ó¦µÄ²úÎïÊÇNO£¬Éè¼ÆÁËÈçÏÂͼËùʾµÄʵÑé×°Öá£ÇëÄã¸ù¾ÝËûÃǵÄ˼·£¬»Ø´ðÓйصÄÎÊÌâ¡£

£¨Ò»£©ÊµÑéÒÇÆ÷£º´óÊԹܡ¢²£Á§µ¼¹Ü¡¢ÏðƤÈû¡¢ÉÕ±­¡¢ÃÞ»¨¡¢×¢ÉäÆ÷¡£
£¨¶þ£©ÊµÑéÒ©Æ·£ºÍ­Ë¿¡¢Ï¡ÏõËᡢ̼Ëá¸Æ¿ÅÁ£¡¢ÉÕ¼îÈÜÒº¡£
£¨Èý£©ÊµÑéÔ­Àí£º
Í­ÓëÏ¡ÏõËá·´Ó¦µÄÀë×Ó·´Ó¦·½³Ìʽ            ¢Ù                     ¡£
£¨ËÄ£©ÊµÑé²½Ö裺
1¡¢°´ÓÒͼËùʾÁ¬½ÓºÃ×°Ö㬼ìÑé×°ÖõĠ    ¢Ú     £»
2¡¢ÏòÊÔ¹ÜÖмÓÈëÒ»¶¨Á¿µÄ¹ÌÌåÒ©Æ·£¨Ìѧʽ£©    ¢Û  £¬È»ºóÏòÊÔ¹ÜÖе¹Èë¹ýÁ¿µÄÏ¡ÏõËᣬ²¢Ñ¸ËÙÈû½ô´øÍ­Ë¿ºÍµ¼¹ÜµÄÏðƤÈû£»
3¡¢ÈÃÊÔ¹ÜÖеķ´Ó¦½øÐÐÒ»¶Îʱ¼äºó£¬ÓÃÕºÓÐNaOHÈÜÒºµÄÃÞ»¨ÍÅ·âסµ¼¹Ü¿Ú£»
4¡¢½«Í­Ë¿ÏòÏÂÒƶ¯²åÈëÊÔ¹ÜÒºÌåÖУ¬Ê¹Ö®ÓëÏ¡ÏõËá·´Ó¦£»
5¡¢°Ñ×¢ÉäÆ÷µÄÕë¿×²åÈëÊԹܿڵÄÏðƤÈûÖУ¬»ºÂýÏòÊÔ¹ÜÄÚÍÆÈë¿ÕÆø¡£
£¨Î壩ʵÑéÌÖÂÛ£º
1¡¢ÊµÑé²½Öè¢ÚµÄÄ¿µÄÊÇ£¨Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ£¬½áºÏÎÄ×Ö˵Ã÷£©
                                       ¢Ü                                £»
2¡¢ÊµÑé²½Öè¢ÝµÄÄ¿µÄÊÇ£¨Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¬½áºÏÎÄ×Ö˵Ã÷£©
                                     ¢Ý                                  ¡£
£¨Áù£©ÊµÑéÆÀ¼Û£º
¸Ã×°ÖõÄÓŵãÊÇ£¨ÈÎдһ¸ö¼´¿É£©                  ¢Þ                  £»
ÓÐͬѧÌá³ö£ºÔö¼ÓÓÒͼËùʾװÖã¬ÔÚ²½Öè¢ÚÍê³Éºó£¬µ±ÓÐÃ÷ÏÔÏÖÏóÔÙ³·È¥¸Ã×°Ö㬲¢¼ÌÐø²½Öè¢ÛµÄ¡°ÓÃÕºÓÐNaOHÈÜÒºµÄÃÞ»¨ÍÅ·âסµ¼¹Ü¿Ú¡±¡£ÇëÆÀ¼ÛËûµÄ×ö·¨    ¢ß   ¡£
¢Ù 3Cu + 8H+ + 2NO3£­ = 3Cu2+ + 2NO¡ü+ 4H2O£¨2·Ö£©  ¢ÚÆøÃÜÐÔ£¨1·Ö£© ¢ÛCaCO3£¨1·Ö£©
¢Ü CaCO3 + 2H+ = Ca2+ + CO2¡ü+ H2O £¨2·Ö£©£¬ÀûÓÃCO2ÅųöÊÔ¹ÜÖеĿÕÆø£¬·ÀÖ¹O2ÓëNO·´Ó¦£¨1·Ö£©
¢Ý 2NO + O2 = 2NO2£¨1·Ö£©,ÆøÌåÓÉÎÞÉ«±äºì×ØÉ«Ö¤Ã÷CuÓëÏ¡ÏõËá·´Ó¦Éú³ÉNO£¨1·Ö£©
¢ÞʹÓÿɳ鶯µÄÍ­Ë¿ÄÜËæʱ¿ØÖÆ·´Ó¦µÄ¿ªÊ¼ºÍ½áÊø£¬½ÚÔ¼Ò©Æ·ÓÃÁ¿£¬¼õÉÙÎÛȾÆøÌåµÄ²úÉú¡££¨»òʹÓÃÕºÓÐNaOHÈÜÒºµÄÃÞ»¨ÍÅ·âסµ¼¹Ü¿Ú£¬¿É·ÀÖ¹NOºÍNO2ÆøÌåÒݳöÎÛȾ´óÆø¡££©£¨2·Ö£©
¢ßͨ¹ýʯ»ÒË®±ä»ë×ǵÄÏÖÏó׼ȷÅжÏÊÔ¹ÜÄÚÎÞ¿ÕÆøºó£¬²Å½øÐв½Öè¢Ü£¬Ê¹µÃ³öµÄ½áÂÛ¸ü¿Æѧ¡££¨1·Ö£©
¢ÙÏõËáÊÇÑõ»¯ÐÔËᣬÄܺÍÍ­·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬·½³ÌʽΪ3Cu + 8H+ + 2NO3£­ = 3Cu2+ + 2NO¡ü+ 4H2O¡£
¢ÚÖÆÆø×°ÖÃÁ¬½ÓºÃºó£¬Ê×ÏȼìÑé×°ÖõÄÆøÃÜÐÔ¡£
¢ÛÓÉÓÚ×°ÖÃÖк¬ÓпÕÆø£¬»áÑõ»¯Éú³ÉµÄNO£¬´Ó¶ø¸ÉÈÅʵÑ飬ËùÒÔÐèÒªÏÈÅž¡×°ÖÃÖеĿÕÆø¡£¿ÉÒÔÀûÓÃÏõËáºÍ̼Ëá¸Æ·´Ó¦Éú³ÉµÄCO2À´ÊµÏÖ£¬¼´¼ÓÈëµÄ¹ÌÌåÒ©Æ·ÊÇ̼Ëá¸Æ£¬·´Ó¦µÄ·½³ÌʽΪCaCO3 + 2H+ = Ca2+ + CO2¡ü+ H2O¡£
¢ÝÓÉÓÚNO¼«Ò×±»Ñõ»¯Éú³Éºì×ØÉ«NO2£¬ËùÒÔÄ¿µÄÊÇÖ¤Ã÷CuÓëÏ¡ÏõËá·´Ó¦Éú³ÉNOµÄ£¬·½³ÌʽΪ2NO + O2 = 2NO2¡£
¢ÞÆÀ¼ÛʵÑé¿ÉÒÔ´Ó²Ù×÷µÄÄÑÒ׳̶ȡ¢Î²ÆøµÄ´¦Àí¡¢ÊµÑéÏÖÏóµÈ¿¼ÂÇ¡£¸ù¾Ý×°Öü°ÊµÑé¹ý³Ì¿ÉÖª£¬ÓŵãÊÇʹÓÿɳ鶯µÄÍ­Ë¿ÄÜËæʱ¿ØÖÆ·´Ó¦µÄ¿ªÊ¼ºÍ½áÊø£¬½ÚÔ¼Ò©Æ·ÓÃÁ¿£¬¼õÉÙÎÛȾÆøÌåµÄ²úÉú¡££¨»òʹÓÃÕºÓÐNaOHÈÜÒºµÄÃÞ»¨ÍÅ·âסµ¼¹Ü¿Ú£¬¿É·ÀÖ¹NOºÍNO2ÆøÌåÒݳöÎÛȾ´óÆø¡££©
¢ßÓÉÓÚÔÚÅÅ¿ÕÆøʱ£¬ÎÞ·¨ÅжÏÊÇ·ñÍêÈ«Åž¡£¬ËùÒÔͨ¹ýʯ»ÒË®±ä»ë×ǵÄÏÖÏó£¬À´×¼È·ÅжÏÊÔ¹ÜÄÚÎÞ¿ÕÆøºó£¬²Å½øÐв½Öè¢Ü£¬Ê¹µÃ³öµÄ½áÂÛ¸ü¿Æѧ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ijУ»¯Ñ§Ñо¿ÐÔѧϰС×éÓûÉè¼ÆʵÑéÑéÖ¤Al¡¢Fe¡¢CuµÄ½ðÊô»î¶¯ÐÔ£¬ËûÃÇÌá³öÁËÒÔÏÂÁ½ÖÖ·½°¸¡£ÇëÄã°ïÖúËûÃÇÍê³ÉÓйØʵÑéÏîÄ¿£º
£¨1£©ÓÃÈýÖÖ½ðÊôÓëÑÎËá·´Ó¦µÄÏÖÏóÀ´Åжϣ¬ÊµÑéÖгýÑ¡Ôñ´óСÏàͬµÄ½ðÊôƬÍ⣬»¹ÐèÒª¿ØÖÆ
_____________ ¡¢_____________ Ïàͬ£»ÈôÑ¡ÓÃÒ»ÖÖÑÎÈÜÒºÑéÖ¤ÈýÖÖ½ðÊôµÄ»îÆÃÐÔ£¬¸ÃÊÔ¼ÁΪ                ¡£
£¨2£©¢ÙÈôÓÃFe¡¢Cu×÷µç¼«Éè¼Æ³ÉÔ­µç³Ø£¬ÒÔÈ·¶¨Fe¡¢CuµÄ»î¶¯ÐÔ¡£ÊÔÔÚÏÂÃæ·½¿òÖл­³öÔ­µç³Ø×°ÖÃͼ£¬±ê³öÔ­µç³ØµÄµç¼«²ÄÁϺ͵ç½âÖÊÈÜÒº¡£

ÉÏÊö×°ÖÃÖÐÕý¼«·´Ó¦Ê½Îª                                                ¡£
¢ÚijС×éͬѧ²ÉÓÃAl¡¢Fe×÷Ϊµç¼«£¬Ö»ÓÃÒ»¸öÔ­µç³ØÖ¤Ã÷ÈýÖÖ½ðÊôµÄ»î¶¯ÐÔ£¬Ôòµç½âÖÊÈÜÒº×îºÃÑ¡Óà         ¡£
A£®0.5 mol¡¤L-1ÂÈ»¯ÑÇÌúÈÜÒºB£®0.5 mol¡¤L-1ÂÈ»¯Í­ÈÜÒº
C£®0.5 mol¡¤L-1ÑÎËáD£®0.5 mol¡¤L-1ÂÈ»¯ÑÇÌúºÍ0.5 mol¡¤L-1ÂÈ»¯Í­»ìºÏÈÜÒº
£¨3£©ÀûÓÃÌṩµÄÊÔ¼ÁºÍÓÃÆ·£º0.1 mol¡¤L-1ÑÎËáÈÜÒº¡¢0.1 mol¡¤L-1´×ËáÈÜÒº¡¢0.5 mol¡¤L-1ÂÈ»¯ÂÁÈÜÒº¡¢0.5 mol¡¤L-1ÇâÑõ»¯ÄÆÈÜÒº¡¢PHÊÔÖ½¡£
ÇëÉè¼Æ×î¼òµ¥ÊµÑéÖ¤Ã÷ÇâÑõ»¯ÂÁΪÈõ¼î                                    
£¨12·Ö£©Ä³Ñо¿ÐÔѧϰС×é¶Ô¹ýÁ¿Ì¿·ÛÓëÑõ»¯Ìú·´Ó¦µÄÆøÌå²úÎï³É·Ö½øÐÐÑо¿¡£
(1)Ìá³ö¼ÙÉ裺
¢Ù¸Ã·´Ó¦µÄÆøÌå²úÎïÊÇCO2£»
¢Ú¸Ã·´Ó¦µÄÆøÌå²úÎïÊÇCO£»
¢Û¸Ã·´Ó¦µÄÆøÌå²úÎïÊÇ              ¡£
(2)¼Æ·½°¸ ÈçͼËùʾ£¬½«Ò»¶¨Á¿µÄÑõ»¯ÌúÔÚ¸ô¾ø¿ÕÆøµÄÌõ¼þÏÂÓë¹ýÁ¿Ì¿·ÛÍêÈ«·´Ó¦£¬²â¶¨²Î¼Ó·´Ó¦µÄ̼ԪËØÓëÑõÔªËصÄÖÊÁ¿±È¡£

(3)²éÔÄ×ÊÁÏ£º
µªÆø²»Óë̼¡¢Ñõ»¯Ìú·¢Éú·´Ó¦¡£ÊµÑéÊÒ¿ÉÒÔÓÃÂÈ»¯ï§±¥ºÍÈÜÒººÍÑÇÏõËáÄÆ£¨NaNO2£©±¥ºÍÈÜÒº»ìºÏ¼ÓÈÈ·´Ó¦ÖƵõªÆø¡£Çëд³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º                            ¡£
(4)ʵÑé²½Ö裺
¢Ù°´ÉÏͼÁ¬½Ó×°Ö㬲¢¼ì²é×°ÖõÄÆøÃÜÐÔ£¬³ÆÈ¡3.20gÑõ»¯Ìú¡¢2.00g̼·Û»ìºÏ¾ùÔÈ£¬·ÅÈë48.48gµÄÓ²Öʲ£Á§¹ÜÖУ»
¢Ú¼ÓÈÈÇ°£¬ÏÈͨһ¶Îʱ¼ä´¿¾»¸ÉÔïµÄµªÆø£»
¢ÛֹͣͨÈëN2ºó£¬¼Ð½ôµ¯»É¼Ð£¬¼ÓÈÈÒ»¶Îʱ¼ä£¬³ÎÇåʯ»ÒË®£¨×ãÁ¿£©±ä»ë×Ç£»
¢Ü´ý·´Ó¦½áÊø£¬ÔÙ»º»ºÍ¨ÈëÒ»¶Îʱ¼äµÄµªÆø¡£ÀäÈ´ÖÁÊÒΣ¬³ÆµÃÓ²Öʲ£Á§¹ÜºÍ¹ÌÌå×ÜÖÊÁ¿Îª52.24g£»
¢Ý¹ýÂ˳öʯ»ÒË®ÖеijÁµí£¬Ï´µÓ¡¢ºæ¸Éºó³ÆµÃÖÊÁ¿Îª2.00g¡£
²½Öè¢Ú¡¢¢ÜÖж¼·Ö±ðͨÈëN2£¬Æä×÷Ó÷ֱðΪ                                  ¡£
(5)Êý¾Ý´¦Àí£º
ÊÔ¸ù¾ÝʵÑéÊý¾Ý·ÖÎö£¬Ð´³ö¸ÃʵÑéÖÐÑõ»¯ÌúÓë̼·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
                                    ¡£
(6)ʵÑéÓÅ»¯£º
ѧϰС×éÓÐͬѧÈÏΪӦ¶ÔʵÑé×°ÖýøÒ»²½ÍêÉÆ¡£
¢Ù¼×ͬѧÈÏΪ£ºÓ¦½«³ÎÇåʯ»ÒË®»»³ÉBa(OH)2ÈÜÒº£¬ÆäÀíÓÉÊÇ                  ¡£
¢Ú´Ó»·¾³±£»¤µÄ½Ç¶È£¬ÇëÄãÔÙÌá³öÒ»¸öÓÅ»¯·½°¸½«´ËʵÑé×°ÖýøÒ»²½ÍêÉÆ£º
                                                           ¡£
22¡¢Ä³Ñо¿Ð¡×鰴ͼʾ·½Ïò½«A×°ÖÃÓëÏÂÃæËùʾװÖÃÏàÁ¬£¬Éè¼ÆʵÑ飬Ñо¿ÌúÓëŨÁòËáµÄ·´Ó¦¡£ÇëÄã°ïÖúÍê³ÉÏÂÁÐʵÑ鱨¸æ£¬²¢»Ø´ðÓйØÎÊÌâ¡£


ʵÑéÏÖÏó
½áÂÛ¼°½âÊÍ
¢Ù½«Ìú¶¤£¨×ãÁ¿£©¼ÓÈëÉÕÆ¿ÖУ¬ÔÙÏòÆäÖеμӺ¬amolH2SO4µÄŨÁòËáÈÜÒº
δ¼ûÃ÷ÏÔÏÖÏó
Ô­Òò£º
                   ¡£
¢ÚµãȼA¡¢G´¦¾Æ¾«µÆ
Ƭ¿Ìºó£¬ÉÕÆ¿ÖÐÈÜÒºµÄÑÕÉ«·¢Éú±ä»¯£¬²¢ÓÐÆøÅݲúÉú¡£CÖÐÈÜÒºÑÕÉ«      £¬EÖÐÈÜҺδ¼ûÃ÷ÏԱ仯¡£Ò»¶Îʱ¼äºó£¬ºÚÉ«µÄCuO±ä³ÉÁ˺ìÉ«£¬´ËʱE×°ÖÃÈÜÒºÑÕÉ«ÈÔδÓÐÑÕÉ«±ä»¯
ÉÕÆ¿Öз¢ÉúµÄËùÓз´Ó¦
µÄ»¯Ñ§·½³Ìʽ£º
                         
                         
˵Ã÷²úÉúµÄSO2ÆøÌåÍêÈ«±»DÖÐÈÜÒºÎüÊÕ
ÌîÍêÉÏÊöʵÑ鱨¸æ²¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Á¬½ÓºÃ×°Öúó£¬Ê¢·ÅÒ©Æ·Ç°£¬Ê×ÏÈÒª½øÐР                     ²Ù×÷¡£
£¨2£©×°ÖÃDµÄ×÷ÓÃÊÇ                         £¬ÈôÏÖÔÚʵÑéÊÒûÓÐËáÐÔ¸ßÃÌËá¼ØÈÜÒº£¬ÔòD×°ÖÃÖеÄÈÜÒº¿ÉÓà         £¨Ìî±àºÅ£©´úÌæ
A¡¢ÕôÁóË®       B¡¢Å¨ÁòËá       C¡¢Å¨NaOHÈÜÒº     D¡¢±¥ºÍNaHSO3ÈÜÒº
£¨3£©×°ÖÃFÖеÄÈÜÒºÊÇ               £¬×°ÖÃHµÄ×÷ÓÃÊÇ£º                                    
£¨4£©ÎªÁ˲ⶨ²úÉúµÄSO2µÄÁ¿£¬ÊµÑéºóÏòÉÕÆ¿ÖмÓ×ãÁ¿µÄÑÎËáÖÁÎÞÆøÅݲúÉú£¬ÔÙ¼Ó×ãÁ¿µÄBaCl2ÈÜÒºÖÁ³ÁµíÍêÈ«£¬È»ºó½øÐР          ¡¢             ¡¢¸ÉÔï¡¢³ÆÁ¿¡£ÎªÅжϳÁµíÍêÈ«£¬Ó¦¸Ã½øÐеIJÙ×÷ÊÇ£º                                                  ¡£
£¨5£©Èô£¨4£©ËùµÃ³ÁµíµÄÖÊÁ¿Îªmg£¬Ôò·´Ó¦²úÉúµÄSO2ÔÚ±ê×¼×´¿öµÄÌå»ýΪ£º
                           L£¨Áгö¼ÆËãʽ¼´¿É£©

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

¾«Ó¢¼Ò½ÌÍø