ÌâÄ¿ÄÚÈÝ

2£®Í¼ÊÇÒ»¸ö»¯Ñ§¹ý³ÌµÄʾÒâͼ£®
£¨1£©Í¼ÖÐÒÒ³ØÊǵç½â³Ø  ×°Öã®
£¨2£©C£¨Pt£©µç¼«µÄÃû³ÆÊÇÑô¼«£®
£¨3£©Ð´³öͨÈëCH3OHµÄµç¼«µÄµç¼«·´Ó¦Ê½ÊÇCH3OH+8OH-6e-¨TCO32-+6H2O£®
£¨4£©ÒÒ³ØÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ4Ag++2H2O$\frac{\underline{\;µç½â\;}}{\;}$4Ag+O2¡ü+4H+£®
£¨5£©µ±ÒÒ³ØÖÐB£¨Ag£©¼«µÄÖÊÁ¿Ôö¼Ó5.40gʱ£¬¼×³ØÖÐÀíÂÛÉÏÏûºÄO2280mL£¨±ê×¼×´¿öÏ£©£»´Ëʱ±û³ØÄ³µç¼«Îö³ö1.6gij½ðÊô£¬Ôò±ûÖеÄijÑÎÈÜÒº¿ÉÄÜÊÇBD
A£®MgSO4       B£®CuSO4       C£®NaCl       D£®CuCl2£®

·ÖÎö ¼×Öиº¼«Îª¼×´¼£¬Õý¼«ÎªÑõÆø£¬ÎªÔ­µç³Ø£¬¸º¼«·´Ó¦ÎªCH3OH+8OH-6e-¨TCO32-+6H2O£¬ÒÒΪµç½â³Ø£¬AÓëÔ­µç³ØµÄÕý¼«ÏàÁ¬£¬ÔòAΪÑô¼«£¬·¢Éú4AgNO3+2H2O$\frac{\underline{\;µç½â\;}}{\;}$4Ag+O2¡ü+4HNO3£¬ÒÒÖÐB¼«Îö³öAg£¬Òõ¼«Éú³ÉÑõÆø£¬ÀûÓõç×ÓÊØºã¿ÉÖª£¬O2¡«4Ag¡«4HNO3£¬±û³ØÄ³µç¼«Îö³ö1.60gij½ðÊô£¬Ñõ»¯ÐÔ±ÈÇâÀë×ÓÇ¿µÄ½ðÊôÀë×Ó¾ù¿ÉÄÜ£¬ÒÔ´ËÀ´½â´ð£®

½â´ð ½â£º£¨1£©ÓÉͼ¿ÉÖª£¬¼×ΪȼÁÏµç³Ø£¬ÔòΪԭµç³Ø£¬¼×ΪÒҵĵçÔ´£¬ÔòÒÒΪµç½â³Ø£¬¹Ê´ð°¸Îª£ºµç½â³Ø£»
£¨2£©DÓëÔ­µç³ØµÄ¸º¼«ÏàÁ¬£¬ËùÒÔDÊǵç½â³ØµÄÒõ¼«£¬ÔòC£¨Pt£©µç¼«µÄÃû³ÆÊÇÑô¼«£¬¹Ê´ð°¸Îª£ºÑô¼«£»
£¨3£©Í¨ÈëCH3OHµÄµç¼«Îª¸º¼«£¬Ê§È¥µç×Ó£¬Ôòµç¼«·´Ó¦Ê½ÎªCH3OH+8OH-6e-¨TCO32-+6H2O£¬¹Ê´ð°¸Îª£ºCH3OH+8OH-6e-¨TCO32-+6H2O£»
£¨4£©ÒÒΪµç½â³Ø£¬AÓëÔ­µç³ØµÄÕý¼«ÏàÁ¬£¬ÔòAΪÑô¼«£¬ÈÜÒºÖÐÒøÀë×Ó¡¢ÇâÑõ¸ùÀë×ӷŵ磬·¢Éú4AgNO3+2H2O $\frac{\underline{\;µç½â\;}}{\;}$4Ag+O2¡ü+4HNO3£¬Àë×Ó·´Ó¦·½³ÌʽΪ£º4Ag++2H2O$\frac{\underline{\;µç½â\;}}{\;}$4Ag+O2¡ü+4H+£»
¹Ê´ð°¸Îª£º4Ag++2H2O$\frac{\underline{\;µç½â\;}}{\;}$4Ag+O2¡ü+4H+£»
£¨5£©Óɵç×ÓÊØºã¿ÉÖª£¬O2¡«4Ag¡«4HNO3£¬n£¨O2£©=$\frac{5.4g}{108g/mol}$¡Á$\frac{1}{4}$=0.0125mol£¬±ê¿öϵÄÌå»ýΪ0.0125mol¡Á22.4L/mol=0.28L=280mL£¬±û³ØÄ³µç¼«Îö³ö1.60gij½ðÊô£¬Ñõ»¯ÐÔ±ÈÇâÀë×ÓÇ¿µÄ½ðÊôÀë×Ó¾ù¿ÉÄÜ£¬ÔòBD·ûºÏ£¬
¹Ê´ð°¸Îª£º280£»BD£®

µãÆÀ ±¾Ì⿼²éÔ­µç³ØºÍµç½â³Ø£¬Ã÷È··¢ÉúµÄµç¼«·´Ó¦¡¢Àë×ӵķŵç˳Ðò¼´¿É½â´ð£¬×¢Òâµç×ÓÊØºãÔÚ¼ÆËãÖеÄÓ¦Óã¬ÌâÄ¿ÄѶÈÖеȣ¬×¢ÖØÁË»ù´¡ÖªÊ¶µÄ¿¼²é£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
13£®½¹ÑÇÁòËáÄÆ£¨Na2S2O5£©Êdz£ÓõÄʳƷ¿¹Ñõ»¯¼ÁÖ®Ò»£®Ä³Ñо¿Ð¡×é½øÐÐÈçÏÂʵÑ飺
ʵÑéÒ»  ½¹ÑÇÁòËáÄÆµÄÖÆÈ¡
²ÉÓÃͼ1×°Öã¨ÊµÑéǰÒѳý¾¡×°ÖÃÄÚµÄ¿ÕÆø£©ÖÆÈ¡Na2S2O5£®×°ÖâòÖÐÓÐNa2S2O5¾§ÌåÎö³ö£¬·¢ÉúµÄ·´Ó¦Îª£º
Na2SO3+SO2¨TNa2S2O5£®

£¨1£©×°ÖÃIÖвúÉúÆøÌåµÄ»¯Ñ§·½³ÌʽΪNa2SO3+H2SO4=Na2SO4+SO2¡ü+H2O£®
£¨2£©Òª´Ó×°ÖâòÖлñµÃÒÑÎö³öµÄ¾§Ì壬¿É²ÉÈ¡µÄ·ÖÀë·½·¨ÊǹýÂË£®
£¨3£©×°ÖâóÓÃÓÚ´¦ÀíÎ²Æø£¬¿ÉÑ¡ÓõĺÏÀí×°Ö㨼гÖÒÇÆ÷ÒÑÂÔÈ¥£©Îªd£¨ÌîÐòºÅ£©£®
ʵÑé¶þ    ½¹ÑÇÁòËáÄÆµÄÐÔÖÊ        Na2S2O5ÈÜÓÚË®¼´Éú³ÉNaHSO3£®
£¨4£©ÒÑÖªNaHSO3ÈÜÒºÖÐHSO3- µÄµçÀë³Ì¶È´óÓÚË®½â³Ì¶È£¬ÔòÆäÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪÈçͼ2c£¨Na+£©£¾c£¨HSO3-£©£¾c£¨H+£©£¾c£¨SO32-£©£¾c£¨OH-£©£®
£¨5£©¼ìÑéNa2S2O5¾§ÌåÔÚ¿ÕÆøÖÐÒѱ»Ñõ»¯µÄʵÑé·½°¸ÊÇÈ¡ÉÙÁ¿Na2S2O5¾§ÌåÓÚÊÔ¹ÜÖУ¬¼ÓÈëÊÊÁ¿Ë®Èܽ⣬µÎ¼ÓÑÎËᣬÕñµ´£¬ÔٵμÓÂÈ»¯±µÈÜÒº£¬Óа×É«³ÁµíÉú³É£®
ʵÑéÈý ÆÏÌѾÆÖп¹Ñõ»¯¼Á²ÐÁôÁ¿µÄ²â¶¨
£¨6£©ÆÏÌѾƳ£ÓÃNa2S2O5×÷¿¹Ñõ»¯¼Á£®²â¶¨Ä³ÆÏÌѾÆÖп¹Ñõ»¯¼ÁµÄ²ÐÁôÁ¿£¨ÒÔÓÎÀëSO2¼ÆË㣩µÄ·½°¸ÈçÏ£º

£¨ÒÑÖª£ºSO2+I2+H2O¨TH2SO4+2HI£©
¢Ù°´ÉÏÊö·½°¸ÊµÑ飬ÏûºÄ±ê×¼I2ÈÜÒº25.00mL£¬¸Ã´ÎʵÑé²âµÃÑùÆ·Öп¹Ñõ»¯¼ÁµÄ²ÐÁôÁ¿£¨ÒÔÓÎÀëSO2¼ÆË㣩Ϊ0.16g•L-1£®
¢ÚÔÚÉÏÊöʵÑé¹ý³ÌÖУ¬ÈôÓв¿·ÖHI±»¿ÕÆøÑõ»¯£¬Ôò²âµÃ½á¹ûÆ«µÍ£¨Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»±ä¡±£©£®

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

¾«Ó¢¼Ò½ÌÍø