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
£¨2£©¸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
£¨3£©»·±£Ð¡×éÔÚʵÑéÊÒÓÃ15mLNaOHÏ¡ÈÜÒººÍËáÐÔ·ÏË®½øÐÐÖкͷ´Ó¦ÊµÑ飬»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÍê³É¸ÃʵÑé±ØÐëʹÓõÄÒÇÆ÷ÓÐ
¢ÚΪÅжÏÇâÑõ»¯ÄÆÇ¡ºÃ±»Öкͣ¬ÔڵμÓËáÐÔ·Ïˮǰ£¬ÐèÏòÇâÑõ»¯ÄÆÈÜÒºÖеμÓ1¡«2µÎ·Ó̪ÊÔÒº£¬µ±Ç¡ºÃÖкÍʱ£¬¹Û²ìµ½µÄÏÖÏóÊÇ
¢Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
£¨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£®
£¨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
£¨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¿