ÌâÄ¿ÄÚÈÝ

£¨14·Ö£©Áò»¯ÎïÔÚ×ÔÈ»½çÖеIJ¿·ÖÑ­»·¹ØϵÈçÏ¡£

£¨1£©H2SÔÚ¿ÕÆøÖпÉÒÔȼÉÕ¡£
ÒÑÖª£º 2H2S(g) + O2(g) 2S(s) + 2H2O(g)  ¦¤H= £­442.38 kJ/mol    ¢Ù
S(s) + O2(g)  SO2(g)             ¦¤H=£­297.04 kJ/mol    ¢Ú
H2S(g)ÓëO2(g)·´Ó¦²úÉúSO2(g)ºÍH2O(g)µÄÈÈ»¯Ñ§·½³ÌʽÊÇ     ¡£
£¨2£©SO2ÊÇ´óÆøÎÛȾÎº£Ë®¾ßÓÐÁ¼ºÃµÄÎüÊÕSO2µÄÄÜÁ¦£¬Æä¹ý³ÌÈçÏ¡£
¢Ù SO2ÈÜÓÚº£Ë®Éú³ÉH2SO3£¬H2SO3×îÖÕ»áµçÀë³öSO32¡ª£¬ÆäµçÀë·½³ÌʽÊÇ     ¡£
¢Ú SO32¡ª¿ÉÒÔ±»º£Ë®ÖеÄÈܽâÑõÑõ»¯ÎªSO42¡ª¡£º£Ë®µÄpH»á     £¨Ìî¡°Éý¸ß¡± ¡¢¡°²»±ä¡±»ò¡°½µµÍ¡±£©¡£
¢Û Ϊµ÷Õûº£Ë®µÄpH£¬¿É¼ÓÈëÐÂÏʵĺ£Ë®£¬Ê¹ÆäÖеÄHCO3¡ª²ÎÓë·´Ó¦£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ     ¡£
¢Ü ÔÚÉÏÊö·´Ó¦µÄͬʱÐèÒª´óÁ¿¹ÄÈë¿ÕÆø£¬ÆäÔ­ÒòÊÇ     ¡£
£¨3£©×ÔÈ»½çµØ±í²ãÔ­ÉúÍ­µÄÁò»¯Îï¾­Ñõ»¯¡¢ÁÜÂË×÷Óúó±ä³ÉCuSO4ÈÜÒº£¬ÏòµØÏÂÉî²ãÉø͸£¬Óöµ½ÄÑÈܵÄZnS£¬ÂýÂýת±äΪͭÀ¶£¨CuS£©£¬Óû¯Ñ§ÓÃÓï±íʾÓÉZnSת±äΪCuSµÄ¹ý³Ì£º     ¡£
£¨4£©SO2ºÍO2ÔÚH2SO4ÈÜÒºÖпÉÒÔ¹¹³ÉÔ­µç³Ø£¬Æ为¼«·´Ó¦Ê½ÊÇ     ¡£
£¨14·Ö£¬Ã¿¿Õ2·Ö£©
£¨1£©2H2S(g)+3O2(g)  2SO2(g)+2H2O(g) ¦¤H= £­1036.46 kJ/mol £¨·½³ÌʽºÍ¼ÆËã¸÷1·Ö£©
£¨2£©¢Ù H2SO3HSO3¡ª+ H+  HSO3¡ª SO3 2¡ª+ H+   £¨Ã¿¸ö1·Ö£©
     ¢Ú ½µµÍ
¢Û HCO3 ¡ª+ H+  CO2 ¡ü+ H2O£¨ÎïÖÊ1·Ö£¬Åäƽ1·Ö£©
¢ÜÌá¸ßÍÑÁòº£Ë®µÄÈܽâÑõ£¬½«SO32-Ñõ»¯³ÉΪSO42-£¬ÓÐÀûÓÚƽºâH2SO3HSO3¡ª+ H+  HSO3¡ª SO3 2¡ª+ H+ ÕýÏòÒƶ¯£¨1·Ö£©£¬Ìá¸ß¶þÑõ»¯ÁòµÄת»¯ÂÊ£¬Í¬Ê±Æðµ½¼Ó¿ì·´Ó¦ËÙÂʵÄ×÷Óã¨1·Ö£©¡£
£¨3£©

£¨4£©¸º¼«£ºSO2 - 2e¡ª+ 2H2O  SO4 2¡ª+ 4H+£¨ÎïÖÊ1·Ö£¬Åäƽ1·Ö£©

ÊÔÌâ·ÖÎö£º£¨1£©¸ù¾Ý¸Ç˹¶¨ÂɵÃÄ¿±ê·½³Ìʽ=¢Ù+2¡Á¢Ú£¬ËùÒÔH2S(g)ÓëO2(g)·´Ó¦²úÉúSO2(g)ºÍH2O(g)µÄÈÈ»¯Ñ§·½³ÌʽÊÇ2H2S(g)+3O2(g) 2SO2(g)+2H2O(g) ¦¤H= £­1036.46 kJ/mol 
£¨2£©¢ÙÑÇÁòËáÊÇÈõËᣬ·Ö²½µçÀëµÃSO3 2¡ª£¬µçÀë·½³ÌʽÊÇH2SO3HSO3¡ª+ H+  HSO3¡ª SO3 2¡ª+ H+   
¢ÚSO32¡ª¿ÉÒÔ±»º£Ë®ÖеÄÈܽâÑõÑõ»¯ÎªSO42¡ª£¬ÁòËáÊÇÇ¿ËᣬÈÜÒºÖеÄÇâÀë×ÓŨ¶ÈÔö´ó£¬pH½µµÍ£»
¢ÛÐÂÏʵĺ£Ë®ÏÔËáÐÔ£¬ÓëHCO3¡ª·´Ó¦Éú³É¶þÑõ»¯Ì¼ºÍË®£¬Àë×Ó·½³ÌʽΪHCO3 ¡ª+ H+  CO2 ¡ü+ H2O
¢Üº£Ë®¾ßÓÐÁ¼ºÃµÄÎüÊÕSO2µÄÄÜÁ¦£¬´óÁ¿¹ÄÈë¿ÕÆø£¬¿ÉÒÔÌá¸ßº£Ë®Öеĺ¬ÑõÁ¿£¬½«SO32-Ñõ»¯³ÉΪSO42-£¬ÓÐÀûÓÚƽºâH2SO3HSO3¡ª+ H+  HSO3¡ª SO3 2¡ª+ H+ ÕýÏòÒƶ¯£¬Ìá¸ß¶þÑõ»¯ÁòµÄת»¯ÂÊ£¬Í¬Ê±Æðµ½¼Ó¿ì·´Ó¦ËÙÂʵÄ×÷Óã»
£¨3£©CuS±ÈZnS¸üÄÑÈÜ£¬Áò»¯Ð¿´æÔÚZnS(s) Zn2+(aq)+S2£­(aq)ƽºâ£¬µ±Óöµ½ÁòËáÍ­ÈÜҺʱ£¬S2£­ÓëCu2+½áºÏÉú³ÉCuS³Áµí£¬Ê¹ÈܽâƽºâÕýÏòÒƶ¯£¬×îÖÕZnSÈ«²¿×ª»¯ÎªCuS£¬Óû¯Ñ§ÓÃÓï±íʾΪ

£¨4£©¸º¼«·¢ÉúÑõ»¯·´Ó¦£¬ÔªËØ»¯ºÏ¼ÛÉý¸ß£¬ËùÒÔÊǶþÑõ»¯ÁòÔÚ¸º¼«·¢Éú·´Ó¦£¬Éú³ÉÁòËá¸ùÀë×Ó£¬µç¼«·´Ó¦Ê½ÎªSO2 - 2e¡ª+ 2H2O  SO4 2¡ª+ 4H+
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨14·Ö£©Ñо¿Òø¼°Æ仯ºÏÎï¾ßÓÐÖØÒªÒâÒå¡£
£¨l£©ÒÑÖª£º
Ag2O£¨s£©+2HC1£¨g£©   2AgC1£¨s£©+H2O£¨1£© ¡÷H1=£­324£®4 kJ¡¤mo1£­1
2Ag£¨s£©+1/2O2£¨g£©   Ag2O£¨s£©            ¡÷H2=£­30£®6 kJ¡¤mo1£­1
H2£¨g£©+C12£¨g£©    2HC1£¨g£©           ¡÷H3=£­184£®4 kJ£®¡¤mo1£­1
2H2£¨9£©+O2£¨g£©   2H2O£¨1£©           ¡÷H4=£­571£®2 l¡¤mo1£­1
д³öÂÈÆøÓëÒøÉú³É¹ÌÌåÂÈ»¯ÒøµÄÈÈ»¯Ñ§·½³Ìʽ________¡£
£¨2£©ÃÀÀöµÄÒøÊγ£ÓÃFe(NO3)3ÈÜҺʴ¿Ì£¬Ð´³öFe3+ÓëAg·´Ó¦µÄÀë×Ó·½³Ìʽ___        _£»ÒªÅж¨Fe(NO3)3ÈÜÒºÖÐNO3¡ªÊÇ·ñÔÚÒøÊÎÊ´¿ÌÖз¢Éú·´Ó¦£¬¿ÉÈ¡                   µÄÏõËáÈÜÒº£¬È»ºó¸ù¾ÝÆäÊÇ·ñÓëAg·¢Éú·´Ó¦À´Åж¨¡£
£¨3£©Òøп¼îÐÔµç³ØµÄµç½âÖÊÈÜҺΪKOHÈÜÒº£¬·Åµçʱ£¬Õý¼«Ag2O2ת»¯ÎªAg£¬¸º¼«Znת»¯ÎªZn(OH)2£¬ÔòÕý¼«·´Ó¦Ê½Îª           £¬¸º¼«¸½½üÈÜÒºµÄpH ___  £¨Ìî¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±£©¡£
£¨4£©µç½â·¨¾«Á¶Òøʱ£¬´ÖÒøÓ¦ÓëÖ±Á÷µçÔ´µÄ       ¼«ÏàÁ¬£¬µ±ÓÃAgNO3ºÍHNO3»ìºÏÈÜÒº×öµç½âÖÊÈÜҺʱ£¬·¢ÏÖÒõ¼«ÓÐÉÙÁ¿ºì×ØÉ«ÆøÌ壬Ôò²úÉú¸ÃÏÖÏóµÄµç¼«·´Ó¦Ê½Îª____¡£

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

¾«Ó¢¼Ò½ÌÍø