ÌâÄ¿ÄÚÈÝ

ijУ³õÈý£¨5£©°àµÄѧÉú¶Ô×æ¹úÓƾõÄÎÄ»¯ÓÐŨºñµÄÐËȤ£¬ËûÃǶԱʡ¢Ä«¡¢Ñâ¡¢Ö½¡°ÎÄ·¿Ëı¦¡±½øÐпÆѧ̽¾¿£®Í¨¹ýµ÷²éÑо¿ÊÕ¼¯µ½ÒÔÏÂÐÅÏ¢£º
¢ÙÎÒÊ¡µÄÎĸÛÕòµô³öµÄ±ÊÎÅÃûÈ«¹ú£¬±»ÓþΪ¡°±Ê¶¼¡±£®
¢ÚÄ«¿ÉÓÉËÉÖ¦²»ÍêȫȼÉÕÉú³ÉµÄÑÌìƼӹ¤ÖƳɣ¬ÎÒ¹úµÄ¹ú»­Ö÷ÒªÊÇÓÃÄ«»­µÄ£»
¢ÛÖÆÑą̂µÄºñÁÏÈ¡×ÔÓÚ×ÔÈ»½çµÄ´óÀíʯ£»
¢ÜÔìÖ½ÓõÄÖ½½¬ÊÇÀûÓÃľ²Ä¼Ó¹¤µÃµ½µÄ£¬Ä¾²ÄÖ÷Òª³É·ÖÊÇÏËάËØ£®
ÇëÄãÀûÓÃÒÑѧ֪ʶ£¬¼ÌÐøÍê³ÉÏÂÁÐ̽¾¿£º
£¨1£©Í¨¹ýʵÑéÑéÖ¤Ñą̂µÄºñÁÏΪ´óÀíʯ£¬ÐèÒªµÄÊÔ¼ÁÊÇ
Ï¡ÑÎËᡢʯ»ÒË®
Ï¡ÑÎËᡢʯ»ÒË®
£¬·´Ó¦»¯Ñ§·½³ÌʽΪ
CaCO3+2HCl=CaCl2+H2O+CO2¡ü£¬Ca£¨OH£©2+CO2=CaCO3¡ý+H2O
CaCO3+2HCl=CaCl2+H2O+CO2¡ü£¬Ca£¨OH£©2+CO2=CaCO3¡ý+H2O
£®
£¨2£©ÎªÁËÖ¤Ã÷Ö½ÖÐÏËάËغ¬ÓÐ̼ԪËØ£¬¿É²ÉÓõķ½·¨ÊÇ
È¡ÉÙÁ¿Ö½½¬ºæ¸É¡¢µãȼ£¬½«Õ´ÓгÎÇåʯ»ÒË®µÄÉÕ±­ÕÖÔÚ»ðÑæÉÏ·½£¬Ê¯»ÒË®±ä»ë×Ç
È¡ÉÙÁ¿Ö½½¬ºæ¸É¡¢µãȼ£¬½«Õ´ÓгÎÇåʯ»ÒË®µÄÉÕ±­ÕÖÔÚ»ðÑæÉÏ·½£¬Ê¯»ÒË®±ä»ë×Ç
£®
£¨3£©Öйú»­¡°ÇåÃ÷ÉϺÓͼ¡±¾­¾Ã²»ÍÊÉ«µÄÔ­ÒòÊÇ
Ì¿ºÚµÄ»¯Ñ§ÐÔÖÊÎȶ¨£¬³£ÎÂϲ»ÓëÆäËüÎïÖÊ·´Ó¦
Ì¿ºÚµÄ»¯Ñ§ÐÔÖÊÎȶ¨£¬³£ÎÂϲ»ÓëÆäËüÎïÖÊ·´Ó¦
£®
£¨4£©ÇëÄã¸ù¾ÝÐÅÏ¢¢ÜÌáÒ»Ìõ»·±£½¨Òé
Ö½ÕÅË«ÃæÓ㨻ò·ÏÖ½Òª»ØÊÕÀûÓõȣ©
Ö½ÕÅË«ÃæÓ㨻ò·ÏÖ½Òª»ØÊÕÀûÓõȣ©
£®
·ÖÎö£º£¨1£©´óÀíʯÊôÓÚ̼ËáÑΣ¬ËùÒÔÀûÓÃÆäÓöËá»á·Å³ö¶þÑõ»¯Ì¼µÄÐÔÖʼø¶¨¼´¿É£®
£¨2£©¿ÉÒÔ¸ù¾ÝÎïÖʵÄÐÔÖÊ·½Ãæ½øÐзÖÎö¡¢Åжϣ¬´Ó¶øµÃ³öÕýÈ·µÄ½áÂÛ£®
£¨3£©Ä«Ö­µÄÖ÷Òª×é³ÉÔªËØÊÇ̼£¬Ôò̼ÔÚ³£ÎÂϵĻ¯Ñ§ÐÔÖÊÎȶ¨£¬²»Ò×·¢Éú·´Ó¦Ê¹µÃ×Ö»­²»±äÉ«£®
£¨4£©ÈËÀàÔÚÏûºÄ¸÷ÖÖÄÜÔ´µÄͬʱ£¬²»¿É±ÜÃâµÄ¶Ô»·¾³Ôì³ÉÓ°Ï죬½ÚÄܼõÅÅÊǵ±½ñÊÀ½çµÄÖ÷Ì⣬½ÚÄܼõÅÅÒª´ÓÎÒÃÇÉí±ßµÄСÊÂ×öÆð£®
½â´ð£º½â£º£¨1£©È¡ÑùµÎ¼ÓÑÎËᣬÈç¹û·´Ó¦ºóÉú³ÉÆøÌåÇÒ¸ÃÆøÌåÄÜʹʯ»ÒË®±ä»ë×Ç£¬¼´¿ÉÖ¤Ã÷¸ÃÎïÖÊÊÇ´óÀíʯ£®Æä·´Ó¦·½³ÌʽΪ£ºCaCO3+2HCl=CaCl2+H2O+CO2¡ü£¬Ca£¨OH£©2+CO2=CaCO3¡ý+H2O£®
¹Ê´ð°¸Îª£ºÏ¡ÑÎËᣬʯ»ÒË®£»CaCO3+2HCl=CaCl2+H2O+CO2¡ü£¬Ca£¨OH£©2+CO2=CaCO3¡ý+H2O£»
£¨2£©Ö¤Ã÷Ö½½¬ÖеÄÏËάËغ¬ÓÐ̼ԪËØ£¬¿ÉÒÔ²ÉÓõķ½·¨ÊÇ£ºÈ¡ÉÙÁ¿Ö½½¬ºæ¸É¡¢µãȼ£¬½«Õ´ÓгÎÇåʯ»ÒË®µÄÉÕ±­ÕÖÔÚ»ðÑæÉÏ·½£¬Ê¯»ÒË®±ä»ë×Ç£®»òÔÚ»ðÑæÉÏ·½ÕÖÒ»¸ö²£Á§Æ¬£¬Èô³öÏÖºÚÉ«Ñ̳¾£¬Ö¤Ã÷º¬ÓÐ̼ԪËØ£®
¹Ê´ð°¸Îª£ºÈ¡ÉÙÁ¿Ö½½¬ºæ¸É¡¢µãȼ£¬½«Õ´ÓгÎÇåʯ»ÒË®µÄÉÕ±­ÕÖÔÚ»ðÑæÉÏ·½£¬Ê¯»ÒË®±ä»ë×Ç£®
£¨3£©Ì¼ÊÇÄ«µÄÖ÷Òª×é³ÉÔªËØ£¬Ì¼ÔÚ³£ÎÂϵĻ¯Ñ§ÐÔÖÊÎȶ¨ÊÇ×Ö»­³¤ÆÚ²»±äÉ«µÄÖ÷ÒªÔ­Òò£®
¹Ê´ð°¸Îª£ºÌ¿ºÚµÄ»¯Ñ§ÐÔÖÊÎȶ¨£¬³£ÎÂϲ»ÓëÆäËüÎïÖÊ·´Ó¦£»
£¨4£©ÔÚƽʱÉú»îÖÐÈç¹ûÄܹ»°ÑÖ½ÕÅË«ÃæÓ㨻ò·ÏÖ½Òª»ØÊÕÀûÓõȣ©£¬½ÚÔ¼ÁËÄÜÔ´£¬·ûºÏ½ÚÄܼõÅŵÄÀíÄ
¹Ê´ð°¸Îª£ºÖ½ÕÅË«ÃæÓ㨻ò·ÏÖ½Òª»ØÊÕÀûÓõȣ©£®
µãÆÀ£º±¾Ì⿼²éµÄÊÇ̼ÔÚ³£ÎÂÏ»¯Ñ§ÐÔÖÊÎȶ¨£¬Ö½È¼ÉÕÉú³ÉµÄÎïÖʵÄÐÔÖÊ£¬Ö»ÓÐÀí½â¡¢ÕÆÎÕÏà¹ØµÄ֪ʶ²ÅÄܶÔÎÊÌâ×ö³öÕýÈ·µÄÅжϣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¾«Ó¢¼Ò½ÌÍøÏÂÃæÊÇijУ³õÈýÈý°à»¯Ñ§ÐËȤС×é¾Í¿ÕÆøÖÐÑõÆøµÄº¬Á¿½øÐеÄʵÑé̽¾¿£º
£¨1£©¸Õ¿ªÊ¼´ó¼ÒÈÏΪ£ºÒª±£Ö¤Ñ¡ÔñµÄÒ©Æ·Ò×ÓëÑõÆø·´Ó¦£¬µ«ÓÖ²»Äܸú¿ÕÆøÖеÄÆäËü³É·Ö·´Ó¦£¬¶øÇÒÖ»Óеõ½µÄÉú³ÉÎïΪ¹ÌÌå²Å²»»áÓ°Ïì²â¶¨½á¹û£®ÄÇôÈç¹ûÈÃÄã´ÓÒÔÏÂһЩÎïÖÊÖÐÑ¡Ôñ³öijÎïÖÊÀ´×ö´ËʵÑ飬ÄãÑ¡£¨Ìî±àºÅ
 
£©£¬
A£®À¯Öò        B£®ºìÁ×        C£®Áò·Û         D¡¢Ä¾Ì¿       E¡¢Ìú·Û
£¨2£©²¿·Öͬѧֱ½Ó°´¿Î±¾ÖеÄʵÑé×°Öã¨Èçͼ£©ºÍÒªÇó½øÐÐʵÑ飬Ëù¹Û²ìµ½µÄÏÖÏóÊÇ£º
 
£®
ÆäÖÐÁù¸öʵÑéС×éËùµÃµ½µÄÊý¾ÝÈçÏÂ±í£º
£¨×¢£º¼¯ÆøÆ¿ÈÝ»ýΪ100mL£©
×é    ±ð 1 2 3 4 5 6
½øÈ뼯ÆøÆ¿ÖÐË®µÄÌå»ý£¨mL£© 20 21 19 20 22 18
ͨ¹ý¶ÔʵÑé½á¹ûµÄ½»Á÷·ÖÎö£¬´ó¶àÊýͬѧ¶¼ÑéÖ¤³öÑõÆøÔ¼Õ¼¿ÕÆøÌå»ýµÄ
 
£®²¢Íƶϳö¼¯ÆøÆ¿ÖÐÊ£ÓàÆøÌåµÄÐÔÖÊÊÇ
 
£®
£¨3£©²¿·ÖͬѧÔÚѧÍêËá¡¢¼î¡¢ÑεÄÐÔÖʺó£¬ÈÏΪֻҪ¶ÔÉÏÊöʵÑé×÷Êʵ±¸Ä½ø£¬¼´Ê¹ÓÃľ̿¡¢Áò·ÛÀ´½øÐÐÉÏÊöʵÑ飬ÕÕÑù¿ÉÒԵõ½ÏàÓ¦½áÂÛ£®ËûÃǵÄ×ö·¨ÊÇÊÂÏÈÔÚ¼¯ÆøÆ¿ÄÚ×¢ÈëÉÙÁ¿
 
ÈÜÒºÀ´ÎüÊÕ
 
ÆøÌ壬Çëд³öÎüÊÕÆøÌåµÄ·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
 £¨ÈÎÑ¡Ò»ÖÖ˵Ã÷¼´¿É£©£®

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

¾«Ó¢¼Ò½ÌÍø