ÌâÄ¿ÄÚÈÝ
22¡¢¾Ý±¨µÀ£¬´Ó2005Äê1ÔÂ1ÈÕÆ𣬹óÖÝÊ¡¸÷µØ»·±£²¿ÃŽ«¶ÔʵÑéÊÒÀàÎÛȾʵʩÑϸñµÄ»·¾³¼à¹Ü£®ÓÉÓÚѧУ»¯Ñ§ÊµÑéÊÒÒªÅŷųɷָ´ÔÓµÄÎÛȾÎËùÒÔÒ²±»ÁÐΪ»·¾³¼à¹Ü·¶Î§£®Ä³Ð£»¯Ñ§ÐËȤС×éµÄͬѧÔÚ¼×ʵÑéÊÒÖнøÐÐÁËÑõÆø·Ö±ðÓë̼¡¢Áò¡¢Ìú·´Ó¦µÄʵÑéºó£¬ÎªÁ˽â¸ÃʵÑé²úÉúµÄÆøÌå¶Ô¿ÕÆø³É·ÖÔì³ÉµÄÓ°Ï죬½Ó×ÅÉè¼ÆÁËÈçÏÂʵÑé×°ÖýøÐÐʵÑ飨ͼÖжà¿×ÇòÅݵÄ×÷ÓÃÊÇ£ºÔö´óÆøÌåÓëÈÜÒºµÄ½Ó´¥Ãæ»ý£¬Ê¹·´Ó¦³ä·Ö½øÐУ©£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÊµÑéÔÚûÓÐÊÜÎÛȾµÄÒÒʵÑéÊÒÖнøÐУ¬È¡³öÊÊÁ¿¼×ʵÑéÊÒÖпÕÆøÑùÆ·µÄ·½·¨ÊÇ
£¨2£©½«È¡µÃµÄ¿ÕÆøÑùÆ·°´ÉÏͼËùʾµÄ×°ÖýøÐÐʵÑé²â¶¨£º×°ÖÃAµÄ×÷ÓÃÊÇ
£¨3£©ÈôͨÈë¿ÕÆøÑùÆ·100mL£¬ÊµÑé½áÊøºó£¬Á¿Í²ÖÐÒºÌåµÄÌå»ýΪ99mL£¨µ¼¹ÜÄÚÒºÌåºöÂÔ²»¼Æ£©£¬ËµÃ÷×°ÖÃ
£¨4£©Çëд³öÄãÔÚ×ö»¯Ñ§ÊµÑéʱ£¬¼õÉÙʵÑé¶Ô»·¾³ÎÛȾµÄÒ»ÖÖ×ö·¨£º
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÊµÑéÔÚûÓÐÊÜÎÛȾµÄÒÒʵÑéÊÒÖнøÐУ¬È¡³öÊÊÁ¿¼×ʵÑéÊÒÖпÕÆøÑùÆ·µÄ·½·¨ÊÇ
ÓÃ×¢ÉäÆ÷³éÈ¡
£®£¨2£©½«È¡µÃµÄ¿ÕÆøÑùÆ·°´ÉÏͼËùʾµÄ×°ÖýøÐÐʵÑé²â¶¨£º×°ÖÃAµÄ×÷ÓÃÊÇ
ÎüÊÕÆøÌåÑùÆ·ÖеĶþÑõ»¯Ì¼ºÍ¶þÑõ»¯Áò
£¬ÆäÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2NaOH+CO2=Na2CO3+H2O£»2NaOH+SO2=Na2SO3+H2O
×°ÖÃBÖеÄÏÖÏóÊÇ
Ë®ÑØ×ŵ¼¹Ü½øÈËCÖÐ
£»×°ÖÃCµÄ×÷ÓÃÊÇ
²âÁ¿¶þÑõ»¯Ì¼ºÍ¶þÑõ»¯Áò±»ÎüÊÕºóÊ£ÓàÆøÌåµÄ´ÖÂÔÌå»ý
£®£¨3£©ÈôͨÈë¿ÕÆøÑùÆ·100mL£¬ÊµÑé½áÊøºó£¬Á¿Í²ÖÐÒºÌåµÄÌå»ýΪ99mL£¨µ¼¹ÜÄÚÒºÌåºöÂÔ²»¼Æ£©£¬ËµÃ÷×°ÖÃ
A
ÖÐÎüÊÕÁË1mLÆøÌ壮ÈôÕâЩÆøÌå´óÁ¿Åŷŵ½¿ÕÆøÖУ¬Òò²úÉú
ËáÓêºÍÎÂÊÒЧӦ
£¬»á¶Ô»·¾³Ôì³É²»ÀûÓ°Ï죮£¨4£©Çëд³öÄãÔÚ×ö»¯Ñ§ÊµÑéʱ£¬¼õÉÙʵÑé¶Ô»·¾³ÎÛȾµÄÒ»ÖÖ×ö·¨£º
¿ØÖÆʵÑéÒ©Æ·ÓÃÁ¿
£®·ÖÎö£º£¨1£©¸ù¾ÝÈ¡ÆøÌå·½·¨£»£¨2£©ÓÉÓÚÇâÑõ»¯ÄÆÈÜÒºÊôÓÚ¼îÄÜÓëËáÐÔÆøÌå·´Ó¦À´¿¼ÂDZ¾Ì⣻¸ù¾ÝÆøÌåÅųöµÄÒºÌåÌå»ýÀ´²âÆøÌåµÄÌå»ý£»£¨3£©¸ù¾ÝʵÑéËù²úÉúµÄÆøÌåÀ´¿¼ÂǶԻ·¾³µÄÓ°Ï죻£¨4£©¼õÉÙʵÑé¶Ô»·¾³ÎÛȾҪ´Ó½ÚÔ¼Ò©Æ·¿¼ÂÇ£®
½â´ð£º½â£º£¨1£©È¡³öÊÊÁ¿¼×ʵÑéÊÒÖпÕÆøÑùÆ·µÄ·½·¨ÓУº½«¼¯ÆøƿװÂúË®ÓÃë²£Á§Æ¬¸ÇºÃ£¬½øÈë¼×ʵÑéÊÒ½«Ë®µ¹µô£¬ÔÙ¸ÇÉÏë²£Á§Æ¬¼´¿É£¬Ò²¿ÉÒÔÓÃ×¢ÉäÆ÷³éÈ¡£»
£¨2£©ÓÉÓÚÔÚ¼×ʵÑéÊÒÖнøÐÐÁËÑõÆø·Ö±ðÓë̼¡¢Áò¡¢Ìú·´Ó¦µÄʵÑ飬ÄÜÉú³É¶þÑõ»¯Ì¼¡¢¶þÑõ»¯ÁòÆøÌ壬²¢ÇÒÕâÁ½ÖÖÆøÌ嶼ÊôÓÚËáÐÔÆøÌ壬ÄÜÓë¼î·´Ó¦£¬ÓÖÒòΪÇâÑõ»¯ÄÆÈÜÒºÊôÓÚ¼îÄÜÓëËáÐÔÆøÌå·´Ó¦£¬ËùÒÔ×°ÖÃAµÄ×÷ÓÃÊÇÎüÊÕÆøÌåÑùÆ·ÖеĶþÑõ»¯Ì¼ºÍ¶þÑõ»¯Áò£»³ýÈ¥ËáÐÔÆøÌåºóÊ£ÓàµÄÆøÌå»á½øÈëB×°Öã¬Ê¹ÆäѹǿÔö´ó£¬°ÑÒºÌåѹÈëCÖУ»½øÈëCÖÐË®µÄÌå»ýÊǶàÉÙ£¬ÄÇôÔÚB ÖеÄÆøÌåÌå»ý¾ÍÊǶàÉÙ£»
£¨3£©ÈôͨÈë¿ÕÆøÑùÆ·100mL£¬ÊµÑé½áÊøºó£¬Á¿Í²ÖÐÒºÌåµÄÌå»ýΪ99mL£¬ËùÒÔ±»ÎüÊÕµÄËáÐÔÆøÌåÊÇ1mL£¬ÒòΪËáÐÔÆøÌåÊǶþÑõ»¯Ì¼¡¢¶þÑõ»¯Áò£¬¶þÑõ»¯Ì¼¶àÁË»áÔì³ÉÎÂÊÒЧӦ£¬¶þÑõ»¯ÁòÔö¶à»áÔì³ÉËáÓꣻ
£¨4£©ÓÉÓÚͬѧÃÇÔÚʵÑéÊÒÖл¹²»ÄÜ´¦ÀíÎÛȾÐÔÆøÌ壬ËùÒÔÖ»ÄÜÔÚ×ö»¯Ñ§ÊµÑéʱ£¬¼õÉÙʵÑéÒ©Æ·£¬¼õÉÙ¶Ô»·¾³ÎÛȾµÄÆøÌåÅÅ·Å£®¹Ê´ð°¸Îª£º£¨1£©ÓÃ×¢ÉäÆ÷³éÈ¡£¨»òÓü¯ÆøƿװÂúË®ºóÔÚÊÒÄÚ½«Ë®µ¹³ö£¬²Á¸É£¬Æ¿¿ÚÏòÉÏ£¬¸ÇÉϲ£Á§Æ¬¡¢Ê¹ÓÃÆøÄÒÊÕ¼¯µÈ£©
£¨2£©ÎüÊÕÆøÌåÑùÆ·ÖеĶþÑõ»¯Ì¼ºÍ¶þÑõ»¯Áò£»2NaOH+CO2=Na2CO3+H2O£»2NaOH+SO2=Na2SO3+H2O£»Ë®ÑØ×ŵ¼¹Ü½øÈËCÖУ¨»òˮλϽµ£©£»²âÁ¿¶þÑõ»¯Ì¼ºÍ¶þÑõ»¯Áò±»ÎüÊÕºóÊ£ÓàÆøÌåµÄ´ÖÂÔÌå»ý
£¨3£©A£»ËáÓêºÍÎÂÊÒЧӦ
£¨4£©¿ØÖÆʵÑéÒ©Æ·ÓÃÁ¿£¨»ò¶ÔÎÛȾÎï½øÐд¦ÀíºóÔÙÅÅ·Å¡¢²»ÂÒµ¹·ÏÒººÍÂÒÈÓ·ÏÎïµÈ£©
£¨2£©ÓÉÓÚÔÚ¼×ʵÑéÊÒÖнøÐÐÁËÑõÆø·Ö±ðÓë̼¡¢Áò¡¢Ìú·´Ó¦µÄʵÑ飬ÄÜÉú³É¶þÑõ»¯Ì¼¡¢¶þÑõ»¯ÁòÆøÌ壬²¢ÇÒÕâÁ½ÖÖÆøÌ嶼ÊôÓÚËáÐÔÆøÌ壬ÄÜÓë¼î·´Ó¦£¬ÓÖÒòΪÇâÑõ»¯ÄÆÈÜÒºÊôÓÚ¼îÄÜÓëËáÐÔÆøÌå·´Ó¦£¬ËùÒÔ×°ÖÃAµÄ×÷ÓÃÊÇÎüÊÕÆøÌåÑùÆ·ÖеĶþÑõ»¯Ì¼ºÍ¶þÑõ»¯Áò£»³ýÈ¥ËáÐÔÆøÌåºóÊ£ÓàµÄÆøÌå»á½øÈëB×°Öã¬Ê¹ÆäѹǿÔö´ó£¬°ÑÒºÌåѹÈëCÖУ»½øÈëCÖÐË®µÄÌå»ýÊǶàÉÙ£¬ÄÇôÔÚB ÖеÄÆøÌåÌå»ý¾ÍÊǶàÉÙ£»
£¨3£©ÈôͨÈë¿ÕÆøÑùÆ·100mL£¬ÊµÑé½áÊøºó£¬Á¿Í²ÖÐÒºÌåµÄÌå»ýΪ99mL£¬ËùÒÔ±»ÎüÊÕµÄËáÐÔÆøÌåÊÇ1mL£¬ÒòΪËáÐÔÆøÌåÊǶþÑõ»¯Ì¼¡¢¶þÑõ»¯Áò£¬¶þÑõ»¯Ì¼¶àÁË»áÔì³ÉÎÂÊÒЧӦ£¬¶þÑõ»¯ÁòÔö¶à»áÔì³ÉËáÓꣻ
£¨4£©ÓÉÓÚͬѧÃÇÔÚʵÑéÊÒÖл¹²»ÄÜ´¦ÀíÎÛȾÐÔÆøÌ壬ËùÒÔÖ»ÄÜÔÚ×ö»¯Ñ§ÊµÑéʱ£¬¼õÉÙʵÑéÒ©Æ·£¬¼õÉÙ¶Ô»·¾³ÎÛȾµÄÆøÌåÅÅ·Å£®¹Ê´ð°¸Îª£º£¨1£©ÓÃ×¢ÉäÆ÷³éÈ¡£¨»òÓü¯ÆøƿװÂúË®ºóÔÚÊÒÄÚ½«Ë®µ¹³ö£¬²Á¸É£¬Æ¿¿ÚÏòÉÏ£¬¸ÇÉϲ£Á§Æ¬¡¢Ê¹ÓÃÆøÄÒÊÕ¼¯µÈ£©
£¨2£©ÎüÊÕÆøÌåÑùÆ·ÖеĶþÑõ»¯Ì¼ºÍ¶þÑõ»¯Áò£»2NaOH+CO2=Na2CO3+H2O£»2NaOH+SO2=Na2SO3+H2O£»Ë®ÑØ×ŵ¼¹Ü½øÈËCÖУ¨»òˮλϽµ£©£»²âÁ¿¶þÑõ»¯Ì¼ºÍ¶þÑõ»¯Áò±»ÎüÊÕºóÊ£ÓàÆøÌåµÄ´ÖÂÔÌå»ý
£¨3£©A£»ËáÓêºÍÎÂÊÒЧӦ
£¨4£©¿ØÖÆʵÑéÒ©Æ·ÓÃÁ¿£¨»ò¶ÔÎÛȾÎï½øÐд¦ÀíºóÔÙÅÅ·Å¡¢²»ÂÒµ¹·ÏÒººÍÂÒÈÓ·ÏÎïµÈ£©
µãÆÀ£ºÕÆÎղɼ¯Ä³Öֵط½µÄ¿ÕÆø·½·¨£ºÅÅË®·¨¡¢³éÆø·¨£»Á˽âͨ¹ýÅÅË®·¨²âÆøÌåµÄÌå»ýµÄ·½·¨£¬²¢ÄÜÁ˽â¿ÕÆøÎÛȾÇé¿ö£¬ÈçºÎ×ö²ÅÄܼõÉÙ¿ÕÆøÎÛȾ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿