ÌâÄ¿ÄÚÈÝ

12£®É½Î÷ÑôȪо°¿óÔø·¢ÉúÍß˹±¬Õ¨Ê¹ʣ¬Ôì³É3ÈËËÀÍö¶àÈËÊÜÉË£®Íß˹´æÔÚÓÚú²ã¼°ÃºÖÜΧµÄÑÒʯ²ãÖУ¬ÊÇú¿ó¾®ÏÂÓк¦ÆøÌåµÄ×ܳƣ¬Êǵ¼ÖÂú¿óʹʵġ°ÒþÐÎɱÊÖ¡±£®
£¨1£©Íß˹ÊôÓÚ»ìºÏÎï £¨Ìî¡°´¿¾»Îï»ò»ìºÏÎ£©£»
£¨2£©Ãº¿ó¾®ÀïµÄÍß˹±¬Õ¨ÒªÂú×ãÈý¸öÌõ¼þ£ºÍß˹Ũ¶ÈÒª´ïµ½±¬Õ¨¼«ÏÞ¡¢×ã¹»µÄÑõÆøºÍÃ÷»ð£»¿ØÖÆºÃÆäÖеÄÌõ¼þ¶¼¿ÉÒÔ±ÜÃⱬըʹʵķ¢Éú£¬¸ù¾ÝÄãËùѧ֪ʶÊÔ¾ÙÒ»±ÜÃâÍß˹±¬Õ¨Ê¹ʷ¢ÉúµÄÓÐЧ´ëÊ©£º½ûÖ¹ÑÌ»ð£®

·ÖÎö £¨1£©¸ù¾ÝÎïÖʵķÖÀà·ÖÎö£»
£¨2£©¸ù¾ÝÍß˹±¬Õ¨µÄÌõ¼þ¡¢±ÜÃâÍß˹±¬Õ¨µÄÓÐЧ´ëÊ©·ÖÎö£®

½â´ð ½â£º£¨1£©¸ù¾ÝÍß˹µÄ´æÔÚ»·¾³£¬ÊÇú¿ó¾®ÏÂÓк¦ÆøÌåµÄ×ܳƿÉÒÔÅжϳö£¬ËüÓ¦¸ÃÊÇ»ìºÏÎ¹Ê±¾Ìâ´ð°¸Îª£º»ìºÏÎ
£¨2£©Íß˹µÄÖ÷Òª³É·ÖÊǼ×Í飬µ±Íß˹Óë¿ÕÆø»ìºÏ´ïµ½¼×ÍéµÄ±¬Õ¨¼«ÏÞʱ£¬ÓöÃ÷»ð»á·¢Éú±¬Õ¨£»Òò´Ë±ÜÃâÍß˹±¬Õ¨£¬Ó¦¸Ã½ûÖ¹Ñ̻𣬾­³£Í¨·ç»»Æø£¬ÒÔ·ÀÖ¹°²È«Ê¹ʵķ¢Éú£®¹Ê´ð°¸Îª£ºÑõÆø£»Ã÷»ð£»½ûÖ¹ÑÌ»ð£»

µãÆÀ ±¾Ì⿼²éÁËÎïÖʵķÖÀ࣬Ò×ȼÒ×±¬Îï·½ÃæµÄ֪ʶ£¬Ñ§»áÓöµ½Í»·¢ÎÊÌâµÄʱºòµÄ×ÔÎÒ±£»¤£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
2£®»¯Ñ§ÐËȤС×é²â¶¨Ä³Ê¯»ÒʯÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý
¢ñ¡¢ËûÃǵķ½·¨ÊÇ£º
£¨1£©È¡3.0gÑùÆ··ÅÈëÉÕ±­£»
£¨2£©¼ÓÈë×ãÁ¿µÄÏ¡ÑÎËáÖÁÇ¡ºÃ²»ÔÙ²úÉúÆøÌåʱ£¬²â¶¨·´Ó¦ºóÉú³ÉµÄCO2ÖÊÁ¿£»
£¨3£©¸ù¾ÝCO2µÄÖÊÁ¿Çó³öÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿¼°ÖÊÁ¿·ÖÊý£®
¢ò¡¢Îª²â¶¨CO2µÄÖÊÁ¿£¬ËûÃÇÏë³öÁËÒÔÏ·½°¸£¬Çë°ïÖúËûÃÇÍê³ÉÉè¼Æ·½°¸£º
£¨1£©Ñ¡ÓÃÈçͼ¼×£¨Ìî¡°¼×¡±»ò¡°ÒÒ¡±£©×°ÖÿɲâÁ¿²úÉúµÄCO2µÄÌå»ý£¬ÔÙÀûÓôËʱCO2µÄÃܶȣ¬¿É¼ÆËãCO2µÄÖÊÁ¿£®Í¼ÖÐÆ¿ÄÚË®ÃæÉÏÓͲãµÄ×÷ÓÃÊÇ·ÀÖ¹¶þÑõ»¯Ì¼ÈÜÓÚË®£®
£¨2£©ËûÃÇÑ¡Óø÷½°¸Ê±²âÁ¿µÄÆøÌåÔÚ³£ÎÂÏÂÌå»ýΪ440mL£®ÒÑÖª³£ÎÂÏÂCO2ÆøÌåµÄÃܶÈΪ2.0g/L£¬ÔòÉÏÊö·´Ó¦·Å³öÆøÌåµÄÖÊÁ¿Îª0.88g£®
Çë¸ù¾Ý»¯Ñ§·´Ó¦·½³Ìʽ¼ÆËã¸Ãʯ»ÒʯÑùÆ·ÖÐCaCO3µÄÖÊÁ¿·ÖÊý£®
¢ó¡¢ÊµÑéºóСÃ÷Ìá³öÓò»Í¬µÄ·½·¨²â¶¨Ä³Ê¯»ÒʯÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£º
£¨1£©È¡mgÑùÆ··ÅÈëÉÕ±­£»
£¨2£©¼ÓÈë×ãÁ¿µÄÏ¡ÑÎËáÖÁ²»ÔÙ²úÉúÆøÌåʱ£¬¹ýÂË£¬Ï´µÓ¡¢¸ÉÔïºó³ÆÁ¿Ê£Óà¹ÌÌåÖÊÁ¿Îªng£»
£¨3£©¸ù¾ÝÊ£Óà¹ÌÌåµÄÖÊÁ¿Çó³öÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ$\frac{m-n}{m}$¡Á100%£®£¨ÓÃm¡¢nµÄ´úÊýʽ±íʾ£©
17£®Ä³ÐËȤС×éµÄͬѧ·¢ÏÖ£¬ÉϸöÔÂ×öʵÑéÓõÄÇâÑõ»¯ÄÆÈÜÒºÍü¼ÇÁËÈûÉÏÆ¿Èû£®ÕâÆ¿ÈÜÒºÓÐûÓбäÖÊÄØ£¿Í¬Ñ§ÃÇÏëһ̽¾¿¾¹£¬ÇëÄãºÍËûÃÇÒ»Æð²ÎÓ룮
¡¾²éÔÄ×ÊÁÏ¡¿ÂÈ»¯±µÈÜÒºÏÔÖÐÐÔ£¬ÇÒÂÈ»¯±µºÍ̼ËáÄÆÄÜ·´Ó¦Éú³É̼Ëá±µ°×É«³ÁµíºÍÂÈ»¯ÄÆ£®
¡¾Ìá³ö²ÂÏë¡¿¼ÙÉèÒ»£º¸ÃÈÜҺûÓбäÖÊ£»
¼ÙÉè¶þ£º¸ÃÈÜÒº²¿·Ö±äÖÊ£»
¼ÙÉèÈý£º¸ÃÈÜҺȫ²¿±äÖÊ£®
¡¾ÊµÑéÑéÖ¤¡¿
ʵÑé²½ÖèʵÑéÏÖÏóʵÑé½áÂÛ
¢ÙÈ¡ÉÙÁ¿¸ÃÈÜÒºÓÚÊÔ¹ÜÖУ¬ÏòÆäÖеμӹýÁ¿µÄÂÈ»¯±µÈÜÒº£»Éú³É°×É«³Áµí¼ÙÉèÒ»²»³ÉÁ¢
¢Ú¾²Öã¬ÏòÉϲãÇåÒºÖеμӷÓ̪ÈÜÒº£®ÈÜÒº±äºì¼ÙÉè¶þ³ÉÁ¢
¡¾ÌÖÂÛ½»Á÷¡¿
£¨1£©²½Öè¢ÙµÎ¼Ó¹ýÁ¿µÄÂÈ»¯±µÈÜÒºµÄÄ¿µÄÊdzý¾¡ÈÜÒºÖеÄ̼ËáÄÆ£¬±ÜÃâ¶Ô¼ìÑéÇâÑõ»¯ÄÆÔì³É¸ÉÈÅ£»
£¨2£©ÓÐͬѧÌá³öÓÃÇâÑõ»¯±µÈÜÒº´úÌæÂÈ»¯±µÈÜÒº×öͬÑùµÄʵÑ飬ҲÄÜ¿´µ½ÏàͬµÄÏÖÏ󣬵óöÏàͬµÄ½áÂÛ£®ÄãͬÒâËûµÄ¹ÛµãÂð£¿ÎªÊ²Ã´£¿²»Í¬Ò⣬ÒòΪ¼ÓÈëÇâÑõ»¯±µÈÜÒºÒýÈëÁËOH-£¬¶Ô¼ìÑéÇâÑõ»¯ÄÆ»áÔì³É¸ÉÈÅ£®
¡¾·´Ë¼¡¿ÇâÑõ»¯ÄƳ¨¿Ú·ÅÖñäÖʵÄÔ­ÒòÊÇ2NaOH+CO2¨TNa2CO3+H2O£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£¬Òò´ËÇâÑõ»¯ÄÆÓ¦ÃÜ·â±£´æ£®
¡¾¶¨Á¿Ñо¿¡¿Ïà¶ÔÔ­×ÓÖÊÁ¿£ºC-12  O-16   Na-23  Ca-40   Cl-35.5
´Ó¸ÃÆ¿ÒѲ¿·Ö±äÖÊΪ̼ËáÄÆµÄÇâÑõ»¯ÄÆÈÜÒºÖÐÈ¡³ö100.0gÈÜÒº£¬ÏòÆäÖмÓÈë100.0gÂÈ»¯¸ÆÈÜÒº£¬ÍêÈ«·´Ó¦ºó¹ýÂË£¬²âµÃÂËÒºµÄÖÊÁ¿Îª190.0g£®
£¨¼ÆËã½á¹û¾«È·µ½Ð¡Êýµãºóһλ£»Óйػ¯Ñ§·½³Ìʽ£ºNa2CO3+CaCl2=CaCO3+2NaCl£©
£¨1£©·´Ó¦Éú³ÉµÄ³ÁµíÖÊÁ¿ÊÇ10g£®
£¨2£©ËùÈ¡µÄÈÜÒºÖк¬Ì¼ËáÄÆµÄÖÊÁ¿ÊǶàÉÙ¿Ë£¿
£¨ 3£©·´Ó¦ºóÂËÒºÖÐÂÈ»¯ÄÆÈÜÖʵÄÖÊÁ¿·ÖÊý£¿

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

¾«Ó¢¼Ò½ÌÍø