ÌâÄ¿ÄÚÈÝ

17£®¶þÑõ»¯ÁòÊÇÔì³É´óÆøÎÛȾµÄÖ÷ÒªÓк¦ÆøÌåÖ®Ò»£¬Ä³ºÏ×÷ѧϰС×éµÄͬѧÄâ²â¶¨Ä³µØ´óÆøÖÐSO2µÄº¬Á¿£¬ÊµÑé²½Öè¼°×°ÖÃÈçÏ£º
²½Öè¢Ù£º³ÆÈ¡a g I2£¬£¨¼ÓÉÙÁ¿KI°ïÖúÈܽ⣩£¬Åä³É500mL 1¡Á10-2mol•L-1 I2ÈÜÒº£»
²½Öè¢Ú£ºÈ¡²½Öè¢ÙÖÐÈÜÒº10mL£¬Ï¡ÊÍÖÁ100mL£»
²½Öè¢Û£ºÁ¿È¡²½Öè¢ÚÖÐÈÜÒº5.00mLÓÚÊÔ¹Ü2ÖУ¬¼ÓÈë2¡«3µÎµí·ÛÈÜÒº£»
²½Öè¢Ü£ºÈçͼ³éÆøN´ÎÖÁÈÜÒºÀ¶É«¸ÕºÃÍÊÈ¥£®£¨ÈÜÒºÖеÄI2±»SO2»¹Ô­ÎªI-£©
£¨1£©²½Öè¢ÙÖÐÅäÖÆ500mL 1¡Á10-2mol•L-1 I2ÈÜҺʱ£¬³ýÓõ½ÌìÆ½¡¢Ò©³×¡¢ÉÕ±­¡¢Á¿Í²¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜÍ⣬»¹Óõ½µÄ²£Á§ÒÇÆ÷ÓÐ1000mLÈÝÁ¿Æ¿£»Ðè³ÆÁ¿µ¥ÖʵâµÄÖÊÁ¿aΪ1.27g£¨±£ÁôСÊýµãºóÁ½Î»£©£®
£¨2£©SO2ÓëµâË®·´Ó¦µÄÀë×Ó·½³ÌʽΪI2+SO2+2H2O=4H++2I-+SO42-£®
£¨3£©ÊµÑéʱ£¬×¢ÉäÆ÷ÿ´Î»º»º³éÆø100mL¿ÕÆø£¬Ê¹¿ÕÆøÍ¨¹ýI2-µí·ÛÈÜÒº£®
¢ÙÈô³éÆøÌ«¿ì£¬½«µ¼Ö½á¹ûƫС£¨Ñ¡Ì¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±¡¢¡°²»±ä¡±£©£®
¢ÚÈôij´ÎʵÑéʱ³éÆø¹²80´Î£¬ÊÔ¼ÆËã¸ÃµØ¿ÕÆøÖÐSO2µÄº¬Á¿£¨mg/L£©£¨Ð´³ö¼ÆËã¹ý³Ì£¬Ã»Óйý³Ì²»¸ø·Ö£©£®

·ÖÎö £¨1£©¸ù¾ÝʵÑé²Ù×÷²½Öè¼°¸÷ÒÇÆ÷µÄ×÷ÓÃѡȡÒÇÆ÷£»¸ù¾Ýn=cV¼ÆËãµâµÄÎïÖʵÄÁ¿£¬¸ù¾Ým=nM¼ÆËãµâµÄÖÊÁ¿£»
£¨2£©¶þÑõ»¯Áò¾ßÓл¹Ô­ÐÔ£¬ÄÜÓëµâË®·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³ÉÁòËáºÍÇâµâË᣻
£¨3£©¢ÙÈô³éÆøÌ«¿ì£¬µ¼Ö¶þÑõ»¯ÁòûÓÐÍêÈ«ºÍµâ·´Ó¦£»
¢ÚÉè¸ÃµØÇø¿ÕÆøÖÐSO2Ũ¶ÈΪc£¬Ç¡ºÃÍêÈ«·´Ó¦Ê±³éÆø80´Î£¬ÔòSO2µÄÁ¿Îª80¡Á100mL¡Á10-3L/mL¡Ácmg/L£¬½áºÏ·´Ó¦¼ÆË㣮

½â´ð ½â£º£¨1£©ÊµÑé²Ù×÷µÄ²½Ö裺¼ÆËã¡¢³ÆÁ¿¡¢ÈÜ½â¡¢ÒÆÒº¡¢Ï´µÓÒÆÒº¡¢¶¨ÈÝ¡¢Ò¡ÔȵȲÙ×÷£¬¸ÃʵÑéÖÐÐèÒªÓÃÌìÆ½³ÆÁ¿¡¢ÓÃÒ©³×ȡҩƷ£¬ÉÕ±­ÈܽâÒ©Æ·£¬ÐèÒª²£Á§°ô½Á°èºÍÒýÁ÷£¬ÐèÒª1000mLÈÝÁ¿Æ¿ÅäÖÆÈÜÒº£¬ÐèÒª½ºÍ·µÎ¹Ü¶¨ÈÝ£»1L 5¡Á10-3mol•L-1ÈÜÒºÖꬵâµÄÎïÖʵÄÁ¿Îª£º1L¡Á5¡Á10-3mol•L-1=5¡Á10-3mol£¬ÖÊÁ¿Îª£º5¡Á10-3mol¡Á254g/mol=1.27g£¬
¹Ê´ð°¸Îª£º1000mLÈÝÁ¿Æ¿£»1.27g£»
£¨2£©SO2ͨÈëäåË®ÖУ¬¶þÑõ»¯Áò±»Ñõ»¯¼Áµâµ¥ÖÊÑõ»¯ÎªÁòËᣬ·´Ó¦µÄ»¯Ñ§·½³ÌʽSO2+I2+2H2O=H2SO4+2HI£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºSO2+I2+2H2O¨T4H++SO42-+2I-£»
¹Ê´ð°¸Îª£ºI2+SO2+2H2O=4H++2I-+SO42-£»
£¨3£©¢ÙÈô³éÆøËٶȹý¿ìʱ£¬¶þÑõ»¯Áò²»Äܱ»ÍêÈ«ÎüÊÕ£¬ÔòËù²âµÃµÄSO2Ũ¶È½«±Èʵ¼Êº¬Á¿Æ«Ð¡£¬
¹Ê´ð°¸Îª£ºÆ«Ð¡£»
¢ÚÉè¸ÃµØÇø¿ÕÆøÖÐSO2Ũ¶ÈΪc£¬Ç¡ºÃÍêÈ«·´Ó¦Ê±³éÆø80´Î£¬ÔòSO2µÄÁ¿Îª80¡Á100mL¡Á10-3L/mL¡Ácmg/L=8cmg£¬
SO2¡«¡«¡«¡«¡«¡«¡«¡«¡«¡«¡«I2
64 g¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡1 mol
8c¡Á10-3 g¡¡¡¡5¡Á10-3 L¡Á5¡Á10-4 mol•L-1
½âµÃc=$\frac{64g¡Á5¡Á10{\;}^{-3}L¡Á5¡Á10{\;}^{-4}mol/L}{1mol¡Á8¡Á10{\;}^{-3}g}$=0.02mg/L£¬
´ð£º¸ÃµØ¿ÕÆøÖÐSO2µÄº¬Á¿Îª0.02mg/L£®

µãÆÀ ±¾Ì⿼²éÎïÖʺ¬Á¿µÄ²â¶¨£¬Îª¸ßƵ¿¼µã£¬°ÑÎÕ·¢ÉúµÄ»¯Ñ§·´Ó¦Îª½â´ðµÄ¹Ø¼ü£¬×¢ÒâÀûÓùØÏµÊ½¼ÆËã¼ò»¯¼ÆË㣬²àÖØ·ÖÎöÓëÓ¦ÓᢼÆËãÄÜÁ¦µÄ×ۺϿ¼²é£¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
7£®ÇâÄÜÊÇÒ»ÖֽྻµÄ¿ÉÔÙÉúÄÜÔ´£¬ÖƱ¸ºÍ´¢´æÇâÆøÊÇÇâÄÜ¿ª·¢µÄÁ½¸ö¹Ø¼ü»·½Ú£®
¢ñÇâÆøµÄÖÆÈ¡
£¨1£©Ë®ÊÇÖÆÈ¡ÇâÆøµÄ³£¼ûÔ­ÁÏ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇAB£¨ÌîÐòºÅ£©£®
A£®H3O+µÄ¿Õ¼ä¹¹ÐÍΪÈý½Ç×¶ÐÎ
B£®Ë®µÄ·Ðµã±ÈÁò»¯Çâ¸ß
C£®±ù¾§ÌåÖУ¬1molË®·Ö×Ó¿ÉÐγÉ4molÇâ¼ü
£¨2£©¿ÆÑÐÈËÔ±Ñо¿³öÒÔîÑËáïÈΪµç¼«µÄ¹â»¯Ñ§µç³Ø£¬ÓÃ×ÏÍâÏßÕÕÉäîÑËáïȵ缫£¬Ê¹Ë®·Ö½â²úÉúÇâÆø£®ÒÑÖªîÑËáïȾ§°û½á¹¹Èçͼ1£¬ÔòÆä»¯Ñ§Ê½ÎªSrTiO3£®

¢òÇâÆøµÄ´æ´¢
£¨3£©Ti£¨BH4£©2ÊÇÒ»ÖÖ´¢Çâ²ÄÁÏ£®
¢ÙTiÔ­×ÓÔÚ»ù̬ʱµÄºËÍâµç×ÓÅŲ¼Ê½ÊÇ1s22s22p63s23p63d24s2»ò[Ar]3d24s2£®
¢ÚTi£¨BH4£©2¿ÉÓÉTiCl4ºÍLiBH4·´Ó¦ÖƵã¬TiCl4 ÈÛµã-25.0¡æ£¬·Ðµã136.94¡æ£¬³£ÎÂÏÂÊÇÎÞɫҺÌ壬ÔòTiCl4¾§ÌåÀàÐÍΪ·Ö×Ó¾§Ì壮
£¨4£©×î½üÄáºÕ³ÏȽø¿ÆÑ§Ñо¿ÖÐÐĽèÖúADFÈí¼þ¶ÔÒ»ÖÖÐÂÐÍ»·Ï©Àà´¢Çâ²ÄÁÏ£¨C16S8£©½øÐÐÑо¿£¬´ÓÀíÂ۽ǶÈÖ¤Ã÷ÕâÖÖ·Ö×ÓÖеÄÔ­×Ó¶¼´¦ÓÚÍ¬Ò»Æ½ÃæÉÏ£¨½á¹¹Èçͼ2Ëùʾ£©£¬Ã¿¸öÆ½ÃæÉÏÏÂÁ½²à×î¶à¿É´¢´æ10¸öH2·Ö×Ó£®
¢ÙÔªËØµç¸ºÐÔ´óС¹ØÏµÊÇ£ºC£¼S£¨Ìî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±£©£®
¢Ú·Ö×ÓÖÐCÔ­×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪsp2£®
¢ÛÓйؼü³¤Êý¾ÝÈçÏ£º
C-SC¨TSC16S8ÖÐ̼Áò¼ü
¼ü³¤/pm181155176
´Ó±íÖÐÊý¾Ý¿ÉÒÔ¿´³ö£¬C16S8ÖÐ̼Áò¼ü¼ü³¤½éÓÚC-SÓëC¨TSÖ®¼ä£¬Ô­Òò¿ÉÄÜÊÇ£º·Ö×ÓÖеÄCÓëSÔ­×ÓÖ®¼äÓЦмü»ò·Ö×ÓÖеÄ̼Áò¼ü¾ßÓÐÒ»¶¨³Ì¶ÈµÄË«¼üÐÔÖÊ£®
¢ÜC16S8ÓëH2΢Á£¼äµÄ×÷ÓÃÁ¦ÊÇ·¶µÂ»ªÁ¦£®

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

¾«Ó¢¼Ò½ÌÍø