ÌâÄ¿ÄÚÈÝ

13¡¢Ä³ÕòÓÐ×ùÁòË᳧£¬É豸¼òª£¬¼¼Êõ³Â¾É£¬¸Ã³§Ã¿ÌìÅÅ·Å´óÁ¿º¬SO2µÄ·ÏÆøºÍº¬ÁòËáµÄËáÐÔ·ÏË®£®µ±µØµÄÆäËû¹¤³§ºÍ¾ÓÃñ¾ùÓÃú̿×÷ȼÁÏ£®Ö»ÒªÏÂÓê¾ÍÏÂËáÓ꣬¶Ô¸ÃÕò»·¾³Ôì³É¼«´óÆÆ»µ£®
£¨1£©¾ÙÒ»Àý˵Ã÷ËáÓê¶Ô»·¾³Ôì³ÉµÄΣº¦£º
ËữÍÁÈÀ£¨»òÎÛȾˮÌå¡¢¸¯Ê´½¨Öþ¼°ÎÄÎï¹Å¼£¡¢¼ÓËÙ½ðÊôÖÆÆ·µÄÐâÊ´µÈ£©
£®
£¨2£©¸ÃÕòijÖÐѧ»·±£Ð¡×éÌá³öÁËÖÎÀíËáÓêµÄÏÂÁдëÊ©£¬ÄãÈÏΪÆäÖв»Í×µÄÊÇ
A
£¨Ñ¡Ìî×Öĸ£©£®
A£®½«ÁòË᳧°áÀë¸ÃÕò                      B£®½¨Òé»·±£²¿ÃÅÏÞÁîÕû¸Ä
C£®½«ÁòË᳧ÅųöµÄ·ÏÆøÖеÄSO2´¦ÀíºóÅÅ·Å   D£®¹¤³§ºÍ¾ÓÃñ¸ÄÓýÏÇå½àµÄȼÁÏ
£¨3£©»·±£Ð¡×éÔÚʵÑéÊÒÓÃ15mLNaOHÏ¡ÈÜÒººÍËáÐÔ·ÏË®½øÐÐÖкͷ´Ó¦ÊµÑ飬»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÍê³É¸ÃʵÑé±ØÐëʹÓõÄÒÇÆ÷ÓÐ
Á¿Í²¡¢ÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü
£®
¢ÚΪÅжÏÇâÑõ»¯ÄÆÇ¡ºÃ±»Öкͣ¬ÔڵμÓËáÐÔ·Ïˮǰ£¬ÐèÏòÇâÑõ»¯ÄÆÈÜÒºÖеμÓ1¡«2µÎ·Ó̪ÊÔÒº£¬µ±Ç¡ºÃÖкÍʱ£¬¹Û²ìµ½µÄÏÖÏóÊÇ
ÈÜÒºÓɺìÉ«±äΪÎÞÉ«
£®
¢Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2NaOH+H2SO4¨TNa2SO4+2H2O
£®
·ÖÎö£º£¨1£©¸ù¾ÝËáÓêµÄ³É·ÖÒÔ¼°ËáµÄ»¯Ñ§ÐÔÖʽøÐзÖÎö£¬
£¨2£©¸ù¾ÝËáÓêµÄ¶¨Òå½øÐзÖÎö£¬
£¨3£©¢Ù¸ù¾ÝÖкͷ´Ó¦²Ù×÷ʱÐèÒªµÄÒÇÆ÷½øÐзÖÎö£¬
¢Ú¸ù¾ÝËá¼îָʾ¼ÁÔÚËá¼îÑÎÖеÄÏÔÉ«½øÐзÖÎö£¬
¢Û¸ù¾ÝÖкͷ´Ó¦µÄÔ­Àí½øÐзÖÎö£®
½â´ð£º½â£º£¨1£©ËáÓêÖк¬ÓдóÁ¿µÄÑÇÁòËá»òÁòËᣬ¶øËá»áÓë½ðÊô·´Ó¦£¬»áÓ뽨ÖþÎïÖеÄ̼ËáÑη´Ó¦£¬»áʹÍÁÈÀËữ£¬¹Ê´ð°¸Îª£ºËữÍÁÈÀ£¨»òÎÛȾˮÌå¡¢¸¯Ê´½¨Öþ¼°ÎÄÎï¹Å¼£¡¢¼ÓËÙ½ðÊôÖÆÆ·µÄÐâÊ´µÈ£©£¬
£¨2£©A¡¢½«ÁòË᳧°áÀë¸ÃÕò£¬Ö»Êǽ«ÎÛȾÎïתÒÆ£¬Ã»ÓдӸù±¾ÉÏÏû³ýÎÛȾÎ¹ÊAÕýÈ·£¬
B¡¢ÏÞÁîÕû¸Ä£¬¿ÉÒÔʹÅųöµÄÆøÌå´ï±ê£¬¿ÉÒÔ¼õÉÙÎÛȾ£¬¹ÊB´íÎó£¬
C¡¢´¦Àí¶þÑõ»¯Áò£¬¼õÉÙÁ˶þÑõ»¯ÁòµÄÅÅ·Å£¬¹ÊC´íÎó£¬
D¡¢¼õÉÙÁË»¯Ê¯È¼ÁϵÄʹÓ㬼õÉÙÁËËáÐÔÆøÌåµÄÅÅ·Å£¬¹ÊD´íÎó£¬
¹ÊÑ¡A£®
£¨3£©¢ÙÐèÒªÓÐÁ¿È¡ÒºÌåµÄ×°ÖãºÁ¿Í²£¬ÐèÒªÓÐÊ¢·ÅÇâÑõ»¯ÄÆ×÷Öкͷ´Ó¦µÄ×°ÖãºÉÕ±­£¬ÐèÒªÓнÁ°è×°Ö㺲£Á§°ô£¬ÐèÒªÓÐУÕý15mlÇâÑõ»¯ÄƵÄ×°Ö㺽ºÍ·µÎ¹Ü£¬
¹Ê´ð°¸Îª£ºÁ¿Í²¡¢ÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü
¢Ú·Ó̪Óö¼î»á±äºìÉ«£¬ÓöËáºÍÑζ¼²»±äÉ«£¬Ö»ÒªºìÉ«ÍÊÈ¥£¬ÈÜÒºÖоͲ»´æÔÚÇâÑõ»¯ÄÆÁË£¬¾ÍÖкÍÍêÈ«ÁË£¬¹Ê´ð°¸Îª£ºÈÜÒºÓɺìÉ«±äΪÎÞÉ«
¢ÛÇâÑõ»¯ÄƺÍÁòËá·¢ÉúÖкͷ´Ó¦Éú³ÉÁòËáÄƺÍË®£¬¹Ê´ð°¸Îª£º2NaOH+H2SO4¨TNa2SO4+2H2O
µãÆÀ£ºÔÚ½â´ËÀàÌâʱ£¬Ê×ÏÈÒª´ÓÖÚ¶àµÄÐðÊöÖÐÕÒ³öÓÐÓõÄ֪ʶ£¬´ËÀàÌâÒ»°ã²»ÄÑ£¬Ó¦ÓõĶ¼ÊÇƽʱµÄ֪ʶ£¬Æðµã¸ß£¬µ«ÊÇÂäµãµÍ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ú¡¢Ê¯ÓͺÍÌìÈ»ÆøÊÇÖØÒªµÄ¡¢²»¿ÉÔÙÉúÄÜÔ´£¬ËüÃÇȼÉÕ²úÉú¹ý¶àµÄ¶þÑõ»¯Ì¼µÈÎïÖÊ»á¶Ô»·¾³Ôì³ÉÑÏÖصÄÓ°Ï죮ÎÒ¹úÕþ¸®¸ß¶ÈÖØÊÓ¡°½ÚÄܼõÅÅ¡±£¬ÎÒ¹ú2007Äê¡°ÊÀ½ç»·¾³ÈÕ¡±µÄÖ÷ÌâºÍ±êʶ£¨Èçͼ1£©£®¾«Ó¢¼Ò½ÌÍø
£¨1£©ÏÂÁдëÊ©·ûºÏÎÒ¹ú¡°ÊÀ½ç»·¾³ÈÕ¡±Ö÷ÌâµÄÊÇ
 
£®£¨Ìî×Öĸ£©
A£®ÏÞÖÆ·¢Õ¹¸ßºÄÄܵIJúÒµ
B£®ÍƹãʹÓá°ÒÒ´¼ÆûÓÍ¡±£¬¼õÉÙÓк¦ÆøÌåÅÅ·Å
C£®ÎªÁ˼õÉÙ·ÏÆø¶Ô±¾µØÇø¿ÕÆøµÄÎÛȾ£¬»¯¹¤³§¼Ó¸ßÑÌ´Ñ
£¨2£©ÌìÈ»ÆøµÄÖ÷Òª³É·ÖÊǼ×Í飬ÆäÍêȫȼÉյĻ¯Ñ§·½³ÌʽΪ
 
ÔÚµÚÒ»´Îͨ¹ý¹ÜµÀÊäËÍÌìÈ»Æøµ½Óû§Ê±£¬±ØÐëÏȽ«¹ÜµÀÄÚ×¢ÂúµªÆø£®Ìî³äµªÆøµÄÄ¿µÄÊÇ
 
 
£®
£¨3£©Í¼2ÖеÄÐÅÏ¢¸æ½ëÈËÀࣺ¶þÑõ»¯Ì¼µÈÆøÌå¹ý¶àµÄÅÅ·Å£¬Ê¹ÎÂÊÒЧӦ¼Ó¾ç£¬µ¼ÖÂÈ«Çò±äů£®Îª¼õÉÙ¶þÑõ»¯Ì¼¹ý¶àµÄÅŷŶԻ·¾³µÄÓ°Ï죬ÄãÄÜ×öµÄÒ»¼þÊ£º
 
£®
£¨4£©Îª½â¾öÄÜÔ´¶ÌȱºÍ»·¾³ÎÛȾµÄÎÊÌ⣬ĿǰÓдý¿ª·¢¡¢ÀûÓõÄÐÂÄÜÔ´ÓÐ
 
£®£¨Ð´Ò»ÖÖ£©
£¨5£©Ä³ÕòÓÐ×ùÁòË᳧£¬É豸¼òª£¬¼¼Êõ³Â¾É£¬¸Ã³§Ã¿ÌìÅÅ·Å´óÁ¿º¬SO2µÄ·ÏÆøºÍº¬H2SO4µÄËáÐÔ·ÏË®£®µ±µØµÄÆäËû¹¤³§ºÍ¾ÓÃñ¾ùÓÃú̿×÷ȼÁÏ£®Ö»ÒªÏÂÓê¾ÍÏÂËáÓ꣬¶Ô¸ÃÕò»·¾³Ôì³É¼«´óÆÆ»µ£®
¢Ù¾ÙÒ»Àý˵Ã÷ËáÓê¶Ô»·¾³Ôì³ÉµÄΣº¦£º
 
£®
¢Ú¸ÃÕòijÖÐѧ»·±£Ð¡×éÌá³öÁËÖÎÀíËáÓêµÄÏÂÁдëÊ©£¬ÄãÈÏΪÆäÖв»Í×µÄÊÇ
 
£®
A£®½«ÁòË᳧°áÀë¸ÃÕò                       B£®½¨Òé»·±£²¿ÃÅÏÞÁîÕû¸Ä
C£®½«ÁòË᳧ÅųöµÄ·ÏÆøÖеÄSO2´¦ÀíºóÅÅ·Å   D£®¹¤³§ºÍ¾ÓÃñ¸ÄÓýÏÇå½àµÄȼÁÏ
¢Û¿ÉÓÃÊìʯ»ÒÀ´´¦ÀíÁòË᳧ÅųöµÄËáÐÔ·ÏË®£¬´¦ÀíÔ­ÀíµÄ»¯Ñ§·½³ÌʽÊÇ
 
£®
£¨2007?³É¶¼£©Ä³ÕòÓÐ×ùÁòË᳧£¬É豸¼òª£¬¼¼Êõ³Â¾É£¬¸Ã³§Ã¿ÌìÅÅ·Å´óÁ¿º¬SO2µÄ·ÏÆøºÍº¬H2SO4µÄËáÐÔ·ÏË®£®µ±µØµÄÆäËû¹¤³§ºÍ¾ÓÃñ¾ùÓÃú̿×÷ȼÁÏ£®Ö»ÒªÏÂÓê¾ÍÏÂËáÓ꣬¶Ô¸ÃÕò»·¾³Ôì³É¼«´óÆÆ»µ£®
£¨1£©·ÖÎö¸ÃÕòÏÂËáÓêµÄÔ­Òò£º
ÅŷŶþÑõ»¯Áò·ÏÆø¡¢ÓÃú×÷ȼÁÏ
ÅŷŶþÑõ»¯Áò·ÏÆø¡¢ÓÃú×÷ȼÁÏ

£¨2£©¾ÙÒ»Àý˵Ã÷ËáÓê¶Ô»·¾³Ôì³ÉµÄΣº¦£º
ËữÍÁÈÀ¡¢¸¯Ê´¡¢ÆÆ»µÉ­ÁÖÖ²Îï¡¢¸¯Ê´½¨ÖþÎï
ËữÍÁÈÀ¡¢¸¯Ê´¡¢ÆÆ»µÉ­ÁÖÖ²Îï¡¢¸¯Ê´½¨ÖþÎï

£¨3£©¸ÃÕòijÖÐѧ»·±£Ð¡×éÌá³öÁËÖÎÀíËáÓêµÄÏÂÁдëÊ©£¬ÄãÈÏΪÆäÖв»Í×µÄÊÇ
A
A

A£®½«ÁòË᳧°áÀë¸ÃÕò                       B£®½¨Òé»·±£²¿ÃÅÏÞÁîÕû¸Ä
C£®½«ÁòË᳧ÅųöµÄ·ÏÆøÖеÄSO2´¦ÀíºóÅÅ·Å    D£®¹¤³§ºÍ¾ÓÃñ¸ÄÓýÏÇå½àµÄȼÁÏ
£¨4£©¿ÉÓÃÊìʯ»ÒÀ´´¦ÀíÁòË᳧ÅųöµÄËáÐÔ·ÏË®£¬´¦ÀíÔ­ÀíµÄ»¯Ñ§·½³ÌʽÊÇ
Ca£¨OH£©2+H2SO4=CaSO4+2H2O
Ca£¨OH£©2+H2SO4=CaSO4+2H2O

£¨5£©Å¨ÁòËáŪµ½ÊÖÉϺóÓ¦Á¢¼´ÓÃĨ²¼²ÁÈ¥£¬ÔÙÓÃË®³åÏ´£¬È»ºóÍ¿ÉÏ̼ËáÇâÄÆ£®ÈôÊÇÏ¡ÁòËáŪµ½ÊÖÉÏ£¬
ÐèÒª
ÐèÒª
£¨Ìî¡°ÐèÒª¡±»ò¡°²»ÐèÒª¡±£©ÕâÑù×ö£¬ÀíÓÉÊÇ
Ï¡ÁòËáÖеÄË®Õô·¢ºó»á±ä³ÉŨÁòËá
Ï¡ÁòËáÖеÄË®Õô·¢ºó»á±ä³ÉŨÁòËá
£®
ijÕòÓÐ×ùÁòË᳧£¬É豸¼òª£¬¼¼Êõ³Â¾É£¬¸Ã³§Ã¿ÌìÅÅ·Å´óÁ¿º¬SO2µÄ·ÏÆøºÍº¬H2SO4µÄËáÐÔ·ÏË®£®µ±µØµÄÆäËû¹¤³§ºÍ¾ÓÃñ¾ùÓÃú̿×÷ȼÁÏ£®Ö»ÒªÏÂÓê¾ÍÏÂËáÓ꣬¶Ô¸ÃÕò»·¾³Ôì³É¼«´óÆÆ»µ£®
£¨1£©·ÖÎö¸ÃÕòÏÂËáÓêµÄÔ­Òò£º
ÅÅ·ÅSO2·ÏÆø¡¢ÓÃú×÷ȼÁÏ
ÅÅ·ÅSO2·ÏÆø¡¢ÓÃú×÷ȼÁÏ
£»
£¨2£©¾ÙÒ»Àý˵Ã÷ËáÓê¶Ô»·¾³Ôì³ÉµÄΣº¦£º
ËữÍÁÈÀ£¨»ò¸¯Ê´¡¢ÆÆ»µÉ­ÁÖÖ²Îï¡¢¸¯Ê´½¨ÖþÎ
ËữÍÁÈÀ£¨»ò¸¯Ê´¡¢ÆÆ»µÉ­ÁÖÖ²Îï¡¢¸¯Ê´½¨ÖþÎ
£»
£¨3£©¶þÑõ»¯ÁòÊÇ´óÆøÎÛȾÎÈÜÓÚË®ËùµÃÈÜÒºµÄpH
СÓÚ
СÓÚ
7£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£»¶þÑõ»¯ÁòÆøÌå¿ÉÓÃÇâÑõ»¯ÄÆÈÜÒºÀ´ÎüÊÕ£¬Çëд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
SO2+2NaOH¨TNa2SO3+H2O
SO2+2NaOH¨TNa2SO3+H2O
£»
£¨4£©¸ÃÕòijÖÐѧ»·±£Ð¡×éÌá³öÁËÖÎÀíËáÓêµÄÏÂÁдëÊ©£¬ÄãÈÏΪÆäÖв»Í×µÄÊÇ
A
A

A£®½«ÁòË᳧°áÀë¸ÃÕò
B£®½¨Òé»·±£²¿ÃÅÏÂÁîÕû¸Ä
C£®½«ÁòË᳧ÅųöµÄ·ÏÆøÖеÄSO2´¦ÀíºóÅÅ·Å
D£®¹¤³§ºÍ¾ÓÃñ¸ÄÓýÏÇå½àµÄȼÁÏ
£¨5£©¿ÉÓÃÊìʯ»ÒÀ´´¦ÀíÁòË᳧ÅųöµÄËáÐÔ·ÏË®£¬´¦ÀíÔ­ÀíµÄ»¯Ñ§·½³ÌʽÊÇ
Ca£¨OH£©2+H2SO4¨TCaSO4+2H2O
Ca£¨OH£©2+H2SO4¨TCaSO4+2H2O
£»
£¨6£©Å¨ÁòËáŪµ½ÊÖÉϺóÓ¦Á¢¼´ÓÃË®³åÏ´£¬È»ºóÍ¿ÉÏ̼ËáÇâÄÆ£®ÈôÊÇÏ¡ÁòËáŪµ½ÊÖÉÏ£¬
ÐèÒª
ÐèÒª
£¨Ìî¡°ÐèÒª¡±»ò¡°²»ÐèÒª¡±£©ÕâÑù×ö£¬ÀíÓÉÊÇ
Ï¡ÁòËáÖеÄË®Õô·¢ºó»á±ä³ÉŨÁòËá
Ï¡ÁòËáÖеÄË®Õô·¢ºó»á±ä³ÉŨÁòËá
£®

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

¾«Ó¢¼Ò½ÌÍø