ÌâÄ¿ÄÚÈÝ

1£®Ë®ÊÇÈËÀàÉú»îÖв»¿ÉȱÉÙµÄÎïÖÊ£®

£¨1£©ÏÂÁÐÈÕ³£Éú»îºÍÉú²úÖг£¼ûµÄË®£¬ÊôÓÚ´¿¾»ÎïµÄÊÇB£®
A£®¿óȪˮ  B£®ÕôÁóË®  C£®ºÓË®  D£®ÓêË®
£¨2£©Èçͼ1ËùʾµÄ¼òÒ×¾»»¯Ë®µÄ×°Öã¬ÔËÓÃÁËÈçϾ»Ë®·½·¨ÖеÄBD  £¨ÌîÐòºÅ£©£®
A¡¢Ïû¶¾  B¡¢¹ýÂË  C¡¢ÕôÁó  D¡¢Îü¸½
£¨3£©Ä³Í¬Ñ§Éè¼ÆÁ˵ç½âË®¼òÒ××°Öã¬ÆäÖÐA¡¢Bµç¼«ÓɽðÊôÇú±ðÕëÖÆ³É£®Í¨µçÒ»¶Îʱ¼äºóµÄÏÖÏóÈçͼ2Ëùʾ£¬ÔòÓëµçÔ´¸º¼«ÏàÁ¬µÄÊÇBµç¼«[ÌîA»òB]£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2H2O$\frac{\underline{\;ͨµç\;}}{\;}$2H2¡ü+O2¡ü£®
£¨4£©×ÔÀ´Ë®Öк¬ÓÐÉÙÁ¿Ca£¨HCO3£©2µÈ¿ÉÈÜÐÔ»¯ºÏÎÉÕˮʱCa£¨HCO3£©2·¢Éú·Ö½â·´Ó¦£¬Éú³ÉÄÑÈÜÐÔµÄ̼Ëá¸Æ¡¢Ë®ºÍ¶þÑõ»¯Ì¼£®Çëд³öCa£¨HCO3£©2ÊÜÈÈ·Ö½âµÄ»¯Ñ§·½³ÌʽCa£¨HCO3£©2$\frac{\underline{\;\;¡÷\;\;}}{\;}$CaCO3¡ý+H2O+CO2¡ü£»Èô½à²ÞÁéµÄÖ÷Òª³É·ÖÊÇÑÎËᣬÔòÓýà²ÞÁé³ýË®¹¸µÄ»¯Ñ§·½³ÌʽΪCaCO3+2HCl=CaCl2+CO2¡ü+H2O£®
£¨5£©¸ù¾ÝÉÏÊöµç½âˮʵÑéÏÖÏó£¬ÈôÒÑ֪ˮµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª18£¬±ê×¼×´¿öÏÂÇâÆøÃܶȣº0.089g/L£¬ÑõÆøÃܶȣº1.429g/L£®Ôò¼ÆËãÒ»¸öË®·Ö×ÓÖÐÑõÔ­×Ó¸öÊýµÄÊýѧ±í´ïʽΪ18¡Á$\frac{1.429g/L¡Á1L}{0.089g/L¡Á2L+1.429g/L¡Á1L}$¡Â16£®

·ÖÎö ÎïÖÊ·ÖΪ»ìºÏÎïºÍ´¿¾»Î»ìºÏÎïÊÇÓÉÁ½ÖÖ»òÁ½ÖÖÒÔÉϵÄÎïÖÊ×é³É£»´¿¾»ÎïÊÇÓÉÒ»ÖÖÎïÖÊ×é³É£®Ë®µÄ¾»»¯·½·¨ÓУºÏû¶¾¡¢¹ýÂË¡¢ÕôÁó¡¢Îü¸½µÈ£»µç½âË®µÄʵÑéÖУ¬ÊÇË®ÔÚͨµçµÄÌõ¼þÏÂÉú³ÉÇâÆøºÍÑõÆø£¬ÆäÖÐÊÇÕýÑõ¸ºÇ⣮Ca£¨HCO3£©2ÊÜÈÈ·Ö½âÉú³É̼Ëá¸ÆºÍË®ºÍ¶þÑõ»¯Ì¼£»Ì¼Ëá¸ÆºÍÑÎËá·´Ó¦Éú³ÉÂÈ»¯¸ÆºÍË®ºÍ¶þÑõ»¯Ì¼£¬Å䯽¼´¿É£®ÓйØÔ­×Ó¸öÊýµÄ¼ÆËãÒªÕýÈ·£®

½â´ð ½â£º£¨1£©»ìºÏÎïÊÇÓÉÁ½ÖÖ»òÁ½ÖÖÒÔÉϵÄÎïÖÊ×é³É£¬´¿¾»ÎïÊÇÓÉÒ»ÖÖÎïÖÊ×é³É£¬ÕôÁóË®ÊôÓÚ´¿¾»Î¹Ê´ð°¸Îª£ºB£®
£¨2£©ÈçÓÒͼËùʾµÄ¼òÒ×¾»»¯Ë®µÄ×°Öã¬Ð¡ÂÑʯºÍʯӢɰÆð¹ýÂË×÷Ó㬻îÐÔÌ¿ÆðÎü¸½×÷Ó㬹ʴð°¸Îª£ºBD
£¨3£©Ë®ÔÚͨµçµÄÌõ¼þÏÂÉú³ÉÇâÆøºÍÑõÆø£¬Å䯽¼´¿É£¬ÆäÖÐÊÇÕýÑõ¸ºÇ⣬Òò´ËÓëµçÔ´¸º¼«ÏàÁ¬µÄÊÇÌå»ý±È½Ï´óµÄÇâÆø£¬¼´B£®¹Ê´ð°¸Îª£ºB£¬2H2O$\frac{\underline{\;ͨµç\;}}{\;}$2H2¡ü+O2¡ü
£¨4£©Ca£¨HCO3£©2ÊÜÈÈ·Ö½âÉú³É̼Ëá¸ÆºÍË®ºÍ¶þÑõ»¯Ì¼£»Ì¼Ëá¸ÆºÍÑÎËá·´Ó¦Éú³ÉÂÈ»¯¸ÆºÍË®ºÍ¶þÑõ»¯Ì¼£¬Å䯽¼´¿É£®¹Ê´ð°¸Îª£ºCa£¨HCO3£©2$\frac{\underline{\;\;¡÷\;\;}}{\;}$ CaCO3¡ý+H2O+CO2¡ü£»CaCO3+2HCl=CaCl2+CO2¡ü+H2O£®
£¨5£©Ïȸù¾ÝÃܶȹ«Ê½µÄ±äÐι«Ê½£ºm=¦Ñv£¬ÇóÑõÔªËØÔÚÒ»¸öË®·Ö×ÓÖÐÕ¼µÄÖÊÁ¿ÊÇ£º
18¡Á$\frac{1.429g/L¡Á1L}{0.089g/L¡Á2L+1.429g/L¡Á1L}$£»ÔÙ¸ù¾ÝÇóÔ­×Ó¸öÊýµÄ¹«Ê½£ºÔ­×Ó¸öÊý=$\frac{ÖÊÁ¿}{Ïà¶ÔÔ­×ÓÖÊÁ¿}$£¬¿ÉÒÔ¼ÆËãÑõÔ­×ӵĸöÊý£®
Ϊ£º18¡Á$\frac{1.429g/L¡Á1L}{0.089g/L¡Á2L+1.429g/L¡Á1L}$¡Â16
´ð°¸£º
£¨1£©B£»
£¨2£©BD£»
£¨3£©B£¬2H2O$\frac{\underline{\;ͨµç\;}}{\;}$2H2¡ü+O2¡ü
£¨4£©Ca£¨HCO3£©2$\frac{\underline{\;\;¡÷\;\;}}{\;}$ CaCO3¡ý+H2O+CO2¡ü£»CaCO3+2HCl=CaCl2+CO2¡ü+H2O£®
£¨5£©18¡Á$\frac{1.429g/L¡Á1L}{0.089g/L¡Á2L+1.429g/L¡Á1L}$¡Â16

µãÆÀ ±¾¿¼µã¿¼²éÁËÓйØË®µÄ¾»»¯¡¢µç½âË®µÄʵÑéºÍÓйصļÆË㣬»¹¿¼²éÁË»¯Ñ§·½³ÌʽµÄÊéд£¬×ÛºÏÐԱȽÏÇ¿£¬Ò»¶¨Òª¼ÓÇ¿¼ÇÒ䣬×ÛºÏÕÆÎÕ£®Êéд·½³Ìʽʱ£¬Òª×¢ÒâÅ䯽£¬ÓйصļÆËãҪ׼ȷ£®´ËÀ࿼µãÖ÷Òª³öÏÖÔÚÌî¿ÕÌâºÍʵÑéÌâÖУ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
12£®¡°¶Ô±ÈʵÑ顱ÊÇ»¯Ñ§Ñ§Ï°ÖÐÐÐÖ®ÓÐЧµÄ˼ά·½·¨£®Ä³»¯Ñ§Ñ§Ï°Ð¡×éµÄͬѧÔÚѧÍêÁËÏà¹ØµÄ»¯Ñ§ÖªÊ¶ºó£¬×ß½øÊµÑéÊÒ×öÁËÈçÏÂʵÑ飬ÇëÄã²ÎÓë²¢»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Í¨¹ýÊÔÑéA£¬¿ÉÒÔ˵Ã÷ȼÉÕµÄÌõ¼þÖ®Ò»ÊÇζȴﵽ¿ÉȼÎïµÄ×Å»ðµã£¬ÊµÑéÖÐʹÓÃͭƬ£¬ÊÇÀûÓÃÁËÍ­µÄµ¼ÈÈÐÔ£¨ÌîÒ»ÌõÎïÀíÐÔÖÊ£©£®
£¨2£©¶ÔÓÚʵÑéB£¬Ò»¶Îʱ¼ä¹Û²ìµ½ÊԹܢÙÖеÄÌú¶¤Ã÷ÏÔÐâÊ´£¬Óɴ˵óö£ºÌúÉúÐâµÄÖ÷ÒªÌõ¼þÊÇÌúÓëË®ºÍ¿ÕÆøÖ±½Ó½Ó´¥£®Óû³ýÈ¥ÌúÐâ¿ÉÓÃÑÎËáÏ´È¥£¬ÌúÖÆÆ·³ýÐâʱ²»ÄÜ£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©³¤Ê±¼ä½þÔÚËáÈÜÒºÖУ®
£¨3£©ÊµÑéCÊÇÀûÓÃÌå»ýÏàͬ²¢³äÂúCO2µÄÈíËÜÁÏÆ¿¡¢µÈÁ¿µÄË®£¨Æ¿¢Ù£©ºÍNaOHÈÜÒº£¨Æ¿¢Ú£©½øÐÐʵÑ飬¸ù¾ÝËÜÁÏÆ¿±ä±ñµÄ³Ì¶ÈÖ¤Ã÷CO2ºÍNaOHÈÜÒºÖеÄÈÜÖÊȷʵ·¢ÉúÁË·´Ó¦£¬ÕâÒ»·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2NaOH+CO2¨TNa2CO3+H2O£®
£¨4£©ÊµÑéDµÄÉÕ±­¢ÚÖгÊÏÖµÄÏÖÏóÄÜ˵Ã÷·Ö×ÓÊDz»¶ÏÔ˶¯µÄ£®µ±ÉÕ±­¢ÙÖÐÒºÌåʱŨ°±Ë®Ê±£¬ÉÕ±­¢ÚÖеķÓ̪ÈÜÒºÓÉÎÞÉ«±äΪºìÉ«£»µ±ÉÕ±­¢ÙÖÐÒºÌå»»³ÉŨÑÎËᣬÇÒÉÕ±­¢ÚÖÐÒºÌå»»³ÉµÎÓзÓ̪NaOHÈÜҺʱ£¬Ò»¶Îʱ¼äºó£¬ÈÜÒºÑÕÉ«µÄ±ä»¯ÊǺìÉ«Öð½¥ÍÊÈ¥±äΪÎÞÉ«£®ÆäÖÐÑÎËáÓëNaOH·´Ó¦µÄ»¯Ñ§·½³ÌʽΪHCl+NaOH¨TNaCl+H2O£¬ÊôÓÚ¸´·Ö½â·´Ó¦£¨Ìî·´Ó¦ÀàÐÍ£©£®

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

¾«Ó¢¼Ò½ÌÍø