ÌâÄ¿ÄÚÈÝ

ij»¯Ñ§ÐËȤС×éÔÚѧϰ¡°ÁòµÄ»¯ºÏÎïµÄijЩÐÔÖÊ¡±ÖУ¬½øÐÐÁËÈçÏÂʵÑ飺

¡¾ÊµÑéÒ»¡¿Ì½¾¿SO2µÄÐÔÖÊ£¬

°´ÏÂͼËùʾװÖýøÐÐʵÑé¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©×°ÖÃAÖÐÊ¢·ÅÑÇÁòËáÄƵÄÒÇÆ÷Ãû³ÆÊÇ             £¬ÆäÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                    £»

£¨2£©ÊµÑé¹ý³ÌÖУ¬×°ÖÃB¡¢CÖз¢ÉúµÄÏÖÏó·Ö±ðÊÇ                  ¡¢               £¬ÕâЩÏÖÏó·Ö±ð˵Ã÷SO2¾ßÓеÄÐÔÖÊÊÇ         ºÍ          £»×°ÖÃBÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                          £»

£¨3£©×°ÖÃDµÄÄ¿µÄÊÇ̽¾¿SO2ÓëÆ·ºì×÷ÓõĿÉÄæÐÔ£¬Çëд³öʵÑé²Ù×÷¼°ÏÖÏó                   

______________________________________________£»

£¨4£©¸Ã×°ÖÃÓÐÒ»Ã÷ÏÔȱÏÝ£¬ÇëÖ¸Ã÷_______________________£¬²¢ÔÚ¿òͼÄÚ»­³ö²¹³ä×°Öã¨×¢Ã÷Ò©Æ·£©¡£

¡¾ÊµÑé¶þ¡¿ÑéÖ¤×ãÁ¿Ð¿ÓëŨÁòËá·´Ó¦²úÉúµÄÆøÌå³É·ÖÊǶþÑõ»¯ÁòºÍÇâÆø¡£°´ÏÂͼװÖýøÐÐʵÑ飨пÓëŨÁòËá¹²ÈÈʱ²úÉúµÄÆøÌåΪX£¬ÇÒ¸Ã×°ÖÃÂÔÈ¥£©¡£ÊԻشð£º

£¨5£©AÖмÓÈëµÄÊÔ¼Á¿ÉÄÜÊÇ____________£¬×÷ÓÃÊÇ_______________£»

BÖмÓÈëµÄÊÔ¼Á¿ÉÄÜÊÇ____________£¬×÷ÓÃÊÇ_______________£»

EÖмÓÈëµÄÊÔ¼Á¿ÉÄÜÊÇ____________£¬×÷ÓÃÊÇ_______________¡£

£¨6£©¿ÉÒÔÖ¤Ã÷ÆøÌåXÖк¬ÓÐÇâÆøµÄʵÑéÏÖÏóÊÇ£ºCÖУº____________£¬DÖУº__________£»

 

¡¾´ð°¸¡¿

£¨20·Ö£©¡¾ÊµÑéÒ»¡¿(1)ÕôÁóÉÕÆ¿  Na2SO3+H2SO4(Ũ)£½Na2SO4+SO2¡ü

(2)ÈÜÒºÓÉ×ϺìÉ«±äΪÎÞÉ«£¬ÎÞÉ«ÈÜÒº³öÏÖ»ÆÉ«»ë×Ç  »¹Ô­ÐÔºÍÑõ»¯ÐÔ

5SO2+2MnO4-+2H2O£½5SO42£­ +2Mn2++4H+

(3)Æ·ºìÈÜÒºÍÊÉ«ºó£¬¹Ø±Õ·ÖҺ©¶·»îÈû£¬µãȼ£Ä´¦¾Æ¾«µÆ¼ÓÈÈ£¬ÈÜÒº»Ö¸´ºìÉ«

(4)ȱÉÙβÆøÎüÊÕ×°ÖÃ

¡¾ÊµÑé¶þ¡¿

£¨5£©Æ·ºìÈÜÒº£»     ¼ìÑ飻

Ũ£»    ÎüÊÕË®ÕôÆø£»

¼îʯ»Ò£»       ·ÀÉÏ¿ÕÆøÖÐË®ÕôÆø½øÈëDÖÐ

£¨6£©CÖкÚÉ«·ÛÄ©£¨£©±ä³ÉºìÉ«£»DÖа×É«·ÛÄ©±ä³ÉÀ¶É«£»

¡¾½âÎö¡¿(4 ·Ö)£¨1£©0.04   (2)26.7

ÊÔÌâ·ÖÎö£º¡¾ÊµÑéÒ»¡¿£¨1£©¸ù¾ÝÒÇÆ÷µÄ½á¹¹¿ÉÖª£¬×°ÖÃAÖÐÊ¢·ÅÑÇÁòËáÄƵÄÒÇÆ÷Ãû³ÆÊÇÕôÁóÉÕÆ¿£¬ÆäÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇNa2SO3+H2SO4(Ũ)£½Na2SO4+SO2¡ü¡£

£¨2£©SO2¾ßÓл¹Ô­ÐÔ£¬Äܱ»ËáÐÔ¸ßÃÌËá¼ØÈÜÒºÑõ»¯£¬ËùÒÔBÖÐÏÖÏóÊÇÈÜÒºÓÉ×ϺìÉ«±äΪÎÞÉ«£¬·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ5SO2+2MnO4-+2H2O£½5SO42£­ +2Mn2++4H+¡£SO2Ò²¾ßÓÐÑõ»¯ÐÔ£¬ÄÜ°ÑÁò»¯ÄÆÑõ»¯Éú³Éµ¥ÖÊS³Áµí£¬ËùÒÔCÖÐÏÖÏóÊÇÎÞÉ«ÈÜÒº³öÏÖ»ÆÉ«»ë×Ç¡£

£¨3£©SO2µÄƯ°×ÊDz»Îȶ¨µÄ£¬¼ÓÈÈÓÖÄܻӷ¢Ô­À´µÄÑÕÉ«£¬ËùÒÔҪ̽¾¿SO2ÓëÆ·ºì×÷ÓõĿÉÄæÐÔ£¬ÊµÑé²Ù×÷¼°ÏÖÏóÊÇÆ·ºìÈÜÒºÍÊÉ«ºó£¬¹Ø±Õ·ÖҺ©¶·»îÈû£¬µãȼ£Ä´¦¾Æ¾«µÆ¼ÓÈÈ£¬ÈÜÒº»Ö¸´ºìÉ«¡£

£¨4£©SO2ÊÇ´óÆøÎÛȾÎȱÉÙβÆøÎüÊÕ×°Öá£

¡¾ÊµÑé¶þ¡¿£¨5£©¼ìÑéSO2µÄÊÔ¼ÁÊÇÆ·ºìÈÜÒº£¬ÔòAÖеÄÊÔ¼ÁÊÇÆ·ºìÈÜÒº£¬ÓÃÀ´¼ìÑéSO2¡£ÓÉÓÚÆøÌåÖк¬ÓÐË®ÕôÆø£¬»á¸ÉÈÅÇâÆøµÄ¼ìÑ飬Òò´ËBÖеÄÊÔ¼ÁŨÁòËᣬÓÃÀ´ÎüÊÕË®ÕôÆø¡£ÁíÍâ¿ÕÆøÖÐÒ²º¬ÓÐË®ÕôÆø£¬Ê£ÓàEÖÐÓ¦¸ÃÊ¢·Å¼îʯ»Ò£¬ÓÃÀ´·ÀÉÏ¿ÕÆøÖÐË®ÕôÆø½øÈëDÖС£

£¨6£©ÇâÆø»¹Ô­Ñõ»¯Í­Éú³ÉÍ­ºÍË®ÕôÆø£¬ËùÒÔ¿ÉÒÔÖ¤Ã÷ÆøÌåXÖк¬ÓÐÇâÆøµÄʵÑéÏÖÏóÊÇ£ºCÖкÚÉ«·ÛÄ©£¨£©±ä³ÉºìÉ«£»DÖа×É«·ÛÄ©±ä³ÉÀ¶É«¡£

¿¼µã£º¿¼²éÒÇÆ÷µÄʶ±ð£¬»ù±¾ÊµÑé²Ù×÷¡¢SO2µÄÖƱ¸¡¢¼ìÑé¡¢ÐÔÖÊÒÔ¼°Æ¯°×ÐÔ̽¾¿

µãÆÀ£º¸ÃÌâÊǸ߿¼Öеij£¼ûÌâÐÍ£¬ÄѶȴó£¬×ÛºÏÐÔÇ¿£¬¶ÔѧÉúµÄÒªÇó¸ß¡£ÊÔÌâÔÚ×¢ÖضԻù´¡ÖªÊ¶¹®¹ÌºÍѵÁ·µÄͬʱ£¬²àÖضÔѧÉúÄÜÁ¦µÄÅàÑøºÍ½âÌâ·½·¨µÄÖ¸µ¼ÓëѵÁ·£¬ÓÐÀûÓÚÅàÑøѧÉú¹æ·¶ÑϽ÷µÄʵÑéÉè¼ÆÄÜÁ¦ÒÔ¼°ÆÀ¼ÛÄÜÁ¦¡£¸ÃÀàÊÔÌâÖ÷ÒªÊÇÒÔ³£¼ûÒÇÆ÷µÄÑ¡Óá¢ÊµÑé»ù±¾²Ù×÷ΪÖÐÐÄ£¬Í¨¹ýÊÇʲô¡¢ÎªÊ²Ã´ºÍÔõÑù×öÖص㿼²éʵÑé»ù±¾²Ù×÷µÄ¹æ·¶ÐÔºÍ׼ȷÐÔ¼°Áé»îÔËÓÃ֪ʶ½â¾öʵ¼ÊÎÊÌâµÄÄÜÁ¦¡£¸ÃÀàÊÔÌâ×ÛºÏÐÔÇ¿£¬ÀíÂÛºÍʵ¼ùµÄÁªÏµ½ôÃÜ£¬ÓеĻ¹ÌṩһЩеÄÐÅÏ¢£¬Õâ¾ÍÒªÇóѧÉú±ØÐëÈÏÕ桢ϸÖµÄÉóÌ⣬ÁªÏµËùѧ¹ýµÄ֪ʶºÍ¼¼ÄÜ£¬½øÐÐ֪ʶµÄÀà±È¡¢Ç¨ÒÆ¡¢ÖØ×飬ȫÃæϸÖµÄ˼¿¼²ÅÄܵóöÕýÈ·µÄ½áÂÛ¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨9·Ö£©Ä³»¯Ñ§ÐËȤС×éÔÚѧϰÁ˵ç½âÖʵÄÓйØ֪ʶºó£¬ÀÏʦÌá³öÁËÒ»¸ö¡°µç½âÖÊË®ÈÜÒºµ¼µçÄÜÁ¦µÄÇ¿ÈõµÄÓ°ÏìÒòËØ¡±µÄ¿ÎÌâ¡£ÈøÃС×éÀ´Íê³É¿ÎÌâµÄÑо¿£¬ÒÑÖªHClÔÚË®ÖÐÄÜÍêÈ«µçÀë¡£

£¨1£©¸ÃС×éͬѧÊ×ÏÈÓÃÃܶÈΪ1.049 g¡¤cm-3µÄÒÒËᣨCH3COOH£©ºÍÖÊÁ¿·ÖÊýΪ36.5%£¬ÃܶÈΪ1.18 g¡¤cm-3µÄŨÑÎËá·Ö±ðÅäÖÆ1mol/L CH3COOHÈÜÒººÍ1 mol/L HClÈÜÒº¸÷500 mL£¬ÔòËûÃÇÓ¦¸ÃÁ¿È¡ÒÒËáºÍŨÑÎËáµÄÌå»ý·Ö±ðΪ         ºÍ          £»

£¨2£©Ä³Í¬Ñ§ÔÚÅäÖÆ500 mL 1 mol/LµÄCH3COOHÈÜҺʱ£¬ÏòÈÝÁ¿Æ¿ÖÐתÒÆÈÜҺʱ²Ù×÷ÈçͼËùʾ£¬Í¼ÖеĴíÎóÓР                                         £»

£¨3£©ËûÃÇÊ×ÏÈÓÃÏÂͼËùʾװÖòâ1 mol/LµÄCH3COOHÈÜÒººÍ1mol/L HClÈÜÒºµÄµ¼µçÄÜÁ¦£¬½ÓͨµçÔ´ºó·¢ÏÖÓëHClÈÜÒºÏàÁ¬µÄµÆÅݽÏÁÁ£¬ÓÐͬѧÈÏΪÒÒËáÊÇÈõµç½âÖÊ£¬Äã        £¨ÌͬÒ⡱»ò¡°²»Í¬Ò⡱£©ËûµÄ¹Ûµã£¬Í¨¹ý¸ÃÏÖÏóÄãµÃµ½µÄ½áÂÛÊÇ                                       

£¨4£©ËûÃÇÓÖÓÃÉÏÊö×°ÖòâÎïÖʵÄÁ¿Å¨¶ÈÏàͬµÄCuSO4ÈÜÒººÍNaOHÈÜÒºµÄµ¼µçÄÜÁ¦£¬½ÓͨµçÔ´ºó·¢ÏÖÓëCuSO4ÈÜÒºÏàÁ¬µÄµÆÅݽÏÁÁ£¬ÓÐͬѧÈÏΪNaOHÊÇÈõµç½âÖÊ£¬Äã           £¨ÌͬÒ⡱»ò¡°²»Í¬Ò⡱£©ËûµÄ¹Ûµã£¬Í¨¹ý¸ÃÏÖÏóÄãµÃµ½µÄ½áÂÛÊÇ                                       £»

£¨5£©Í¨¹ýÉÏÊö̽¾¿ÊµÑ飬ÄãÄܵõ½µÄ½áÂÛÊÇ                                       

 

£¨9·Ö£©Ä³»¯Ñ§ÐËȤС×éÔÚѧϰÁ˵ç½âÖʵÄÓйØ֪ʶºó£¬ÀÏʦÌá³öÁËÒ»¸ö¡°µç½âÖÊË®ÈÜÒºµ¼µçÄÜÁ¦µÄÇ¿ÈõµÄÓ°ÏìÒòËØ¡±µÄ¿ÎÌâ¡£ÈøÃС×éÀ´Íê³É¿ÎÌâµÄÑо¿£¬ÒÑÖªHClÔÚË®ÖÐÄÜÍêÈ«µçÀë¡£

£¨1£©¸ÃС×éͬѧÊ×ÏÈÓÃÃܶÈΪ1.049 g¡¤cm-3µÄÒÒËᣨCH3COOH£©ºÍÖÊÁ¿·ÖÊýΪ36.5%£¬ÃܶÈΪ1.18 g¡¤cm-3µÄŨÑÎËá·Ö±ðÅäÖÆ1mol/L CH3COOHÈÜÒººÍ1 mol/L HClÈÜÒº¸÷500 mL£¬ÔòËûÃÇÓ¦¸ÃÁ¿È¡ÒÒËáºÍŨÑÎËáµÄÌå»ý·Ö±ðΪ         ºÍ          £»

£¨2£©Ä³Í¬Ñ§ÔÚÅäÖÆ500 mL 1 mol/LµÄCH3COOHÈÜҺʱ£¬ÏòÈÝÁ¿Æ¿ÖÐתÒÆÈÜҺʱ²Ù×÷ÈçͼËùʾ£¬Í¼ÖеĴíÎóÓР                                         £»

£¨3£©ËûÃÇÊ×ÏÈÓÃÏÂͼËùʾװÖòâ1 mol/LµÄCH3COOHÈÜÒººÍ1mol/L HClÈÜÒºµÄµ¼µçÄÜÁ¦£¬½ÓͨµçÔ´ºó·¢ÏÖÓëHClÈÜÒºÏàÁ¬µÄµÆÅݽÏÁÁ£¬ÓÐͬѧÈÏΪÒÒËáÊÇÈõµç½âÖÊ£¬Äã        £¨ÌͬÒ⡱»ò¡°²»Í¬Ò⡱£©ËûµÄ¹Ûµã£¬Í¨¹ý¸ÃÏÖÏóÄãµÃµ½µÄ½áÂÛÊÇ                                       

£¨4£©ËûÃÇÓÖÓÃÉÏÊö×°ÖòâÎïÖʵÄÁ¿Å¨¶ÈÏàͬµÄCuSO4ÈÜÒººÍNaOHÈÜÒºµÄµ¼µçÄÜÁ¦£¬½ÓͨµçÔ´ºó·¢ÏÖÓëCuSO4ÈÜÒºÏàÁ¬µÄµÆÅݽÏÁÁ£¬ÓÐͬѧÈÏΪNaOHÊÇÈõµç½âÖÊ£¬Äã           £¨ÌͬÒ⡱»ò¡°²»Í¬Ò⡱£©ËûµÄ¹Ûµã£¬Í¨¹ý¸ÃÏÖÏóÄãµÃµ½µÄ½áÂÛÊÇ                                       £»

£¨5£©Í¨¹ýÉÏÊö̽¾¿ÊµÑ飬ÄãÄܵõ½µÄ½áÂÛÊÇ                                       

 

£¨9·Ö£©Ä³»¯Ñ§ÐËȤС×éÔÚѧϰÁ˵ç½âÖʵÄÓйØ֪ʶºó£¬ÀÏʦÌá³öÁËÒ»¸ö¡°µç½âÖÊË®ÈÜÒºµ¼µçÄÜÁ¦µÄÇ¿ÈõµÄÓ°ÏìÒòËØ¡±µÄ¿ÎÌâ¡£ÈøÃС×éÀ´Íê³É¿ÎÌâµÄÑо¿£¬ÒÑÖªHClÔÚË®ÖÐÄÜÍêÈ«µçÀë¡£
£¨1£©¸ÃС×éͬѧÊ×ÏÈÓÃÃܶÈΪ1.049 g¡¤cm-3µÄÒÒËᣨCH3COOH£©ºÍÖÊÁ¿·ÖÊýΪ36.5%£¬ÃܶÈΪ1.18 g¡¤cm-3µÄŨÑÎËá·Ö±ðÅäÖÆ1 mol/L CH3COOHÈÜÒººÍ1 mol/L HClÈÜÒº¸÷500 mL£¬ÔòËûÃÇÓ¦¸ÃÁ¿È¡ÒÒËáºÍŨÑÎËáµÄÌå»ý·Ö±ðΪ         ºÍ          £»
£¨2£©Ä³Í¬Ñ§ÔÚÅäÖÆ500 mL 1 mol/LµÄCH3COOHÈÜҺʱ£¬ÏòÈÝÁ¿Æ¿ÖÐתÒÆÈÜҺʱ²Ù×÷ÈçͼËùʾ£¬Í¼ÖеĴíÎóÓР                                         £»

£¨3£©ËûÃÇÊ×ÏÈÓÃÏÂͼËùʾװÖòâ1 mol/LµÄCH3COOHÈÜÒººÍ1 mol/L HClÈÜÒºµÄµ¼µçÄÜÁ¦£¬½ÓͨµçÔ´ºó·¢ÏÖÓëHClÈÜÒºÏàÁ¬µÄµÆÅݽÏÁÁ£¬ÓÐͬѧÈÏΪÒÒËáÊÇÈõµç½âÖÊ£¬Äã       £¨ÌͬÒ⡱»ò¡°²»Í¬Ò⡱£©ËûµÄ¹Ûµã£¬Í¨¹ý¸ÃÏÖÏóÄãµÃµ½µÄ½áÂÛÊÇ                                       

£¨4£©ËûÃÇÓÖÓÃÉÏÊö×°ÖòâÎïÖʵÄÁ¿Å¨¶ÈÏàͬµÄCuSO4ÈÜÒººÍNaOHÈÜÒºµÄµ¼µçÄÜÁ¦£¬½ÓͨµçÔ´ºó·¢ÏÖÓëCuSO4ÈÜÒºÏàÁ¬µÄµÆÅݽÏÁÁ£¬ÓÐͬѧÈÏΪNaOHÊÇÈõµç½âÖÊ£¬Äã           £¨ÌͬÒ⡱»ò¡°²»Í¬Ò⡱£©ËûµÄ¹Ûµã£¬Í¨¹ý¸ÃÏÖÏóÄãµÃµ½µÄ½áÂÛÊÇ                                       £»
£¨5£©Í¨¹ýÉÏÊö̽¾¿ÊµÑ飬ÄãÄܵõ½µÄ½áÂÛÊÇ                                       

£¨9·Ö£©Ä³»¯Ñ§ÐËȤС×éÔÚѧϰÁ˵ç½âÖʵÄÓйØ֪ʶºó£¬ÀÏʦÌá³öÁËÒ»¸ö¡°µç½âÖÊË®ÈÜÒºµ¼µçÄÜÁ¦µÄÇ¿ÈõµÄÓ°ÏìÒòËØ¡±µÄ¿ÎÌâ¡£ÈøÃС×éÀ´Íê³É¿ÎÌâµÄÑо¿£¬ÒÑÖªHClÔÚË®ÖÐÄÜÍêÈ«µçÀë¡£

£¨1£©¸ÃС×éͬѧÊ×ÏÈÓÃÃܶÈΪ1.049 g¡¤cm-3µÄÒÒËᣨCH3COOH£©ºÍÖÊÁ¿·ÖÊýΪ36.5%£¬ÃܶÈΪ1.18 g¡¤cm-3µÄŨÑÎËá·Ö±ðÅäÖÆ1 mol/L CH3COOHÈÜÒººÍ1 mol/L HClÈÜÒº¸÷500 mL£¬ÔòËûÃÇÓ¦¸ÃÁ¿È¡ÒÒËáºÍŨÑÎËáµÄÌå»ý·Ö±ðΪ          ºÍ           £»

£¨2£©Ä³Í¬Ñ§ÔÚÅäÖÆ500 mL 1 mol/LµÄCH3COOHÈÜҺʱ£¬ÏòÈÝÁ¿Æ¿ÖÐתÒÆÈÜҺʱ²Ù×÷ÈçͼËùʾ£¬Í¼ÖеĴíÎóÓР                                          £»

£¨3£©ËûÃÇÊ×ÏÈÓÃÏÂͼËùʾװÖòâ1 mol/LµÄCH3COOHÈÜÒººÍ1 mol/L HClÈÜÒºµÄµ¼µçÄÜÁ¦£¬½ÓͨµçÔ´ºó·¢ÏÖÓëHClÈÜÒºÏàÁ¬µÄµÆÅݽÏÁÁ£¬ÓÐͬѧÈÏΪÒÒËáÊÇÈõµç½âÖÊ£¬Äã        £¨ÌͬÒ⡱»ò¡°²»Í¬Ò⡱£©ËûµÄ¹Ûµã£¬Í¨¹ý¸ÃÏÖÏóÄãµÃµ½µÄ½áÂÛÊÇ                                       

£¨4£©ËûÃÇÓÖÓÃÉÏÊö×°ÖòâÎïÖʵÄÁ¿Å¨¶ÈÏàͬµÄCuSO4ÈÜÒººÍNaOHÈÜÒºµÄµ¼µçÄÜÁ¦£¬½ÓͨµçÔ´ºó·¢ÏÖÓëCuSO4ÈÜÒºÏàÁ¬µÄµÆÅݽÏÁÁ£¬ÓÐͬѧÈÏΪNaOHÊÇÈõµç½âÖÊ£¬Äã            £¨ÌͬÒ⡱»ò¡°²»Í¬Ò⡱£©ËûµÄ¹Ûµã£¬Í¨¹ý¸ÃÏÖÏóÄãµÃµ½µÄ½áÂÛÊÇ                                        £»

£¨5£©Í¨¹ýÉÏÊö̽¾¿ÊµÑ飬ÄãÄܵõ½µÄ½áÂÛÊÇ                                       

 

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

¾«Ó¢¼Ò½ÌÍø