ÌâÄ¿ÄÚÈÝ

3£®Ä³¹«Ë¾Éú²ú³öµÄ´¿¼î²úÆ·Öо­¼ì²âÖ»º¬ÓÐÂÈ»¯ÄÆÔÓÖÊ£®Îª²â¶¨²úÆ·ÖÐ̼ËáÄÆµÄÖÊÁ¿·ÖÊý£¬20¡æÊ±£¬³ÆÈ¡¸Ã²úÆ·ÑùÆ·26.5g£¬¼ÓÈ뵽ʢÓÐÒ»¶¨ÖÊÁ¿Ï¡ÑÎËáµÄÉÕ±­ÖУ¬Ì¼ËáÄÆÓëÏ¡ÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¬ÆøÌåÍêÈ«Òݳö£¬µÃµ½²»±¥ºÍNaClÈÜÒº£»·´Ó¦¹ý³ÌÓþ«ÃÜÒÇÆ÷²âµÃÉÕ±­ÄÚ»ìºÏÎïµÄÖÊÁ¿£¨m£©Ó뷴Ӧʱ¼ä£¨t£©¹ØÏµÈçͼËùʾ£®
£¨1£©Éú³ÉCO2µÄÖÊÁ¿Îª8.8g£»
£¨2£©ÇóÇ¡ºÃÍêÈ«·´Ó¦ºóËùµÃµ½²»±¥ºÍNaClÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®

·ÖÎö ÂÈ»¯ÄƲ»ÄܺÍÏ¡ÑÎËá·´Ó¦£¬Ì¼ËáÄÆºÍÏ¡ÑÎËá·´Ó¦Éú³ÉÂÈ»¯ÄÆ¡¢Ë®ºÍ¶þÑõ»¯Ì¼£¬·´Ó¦Ç°ºóµÄÖÊÁ¿²î¼´Îª·´Ó¦Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿£¬¸ù¾Ý¶þÑõ»¯Ì¼ÖÊÁ¿¿ÉÒÔ¼ÆËã̼ËáÄÆµÄÖÊÁ¿ºÍ·´Ó¦Éú³ÉÂÈ»¯ÄƵÄÖÊÁ¿£¬½øÒ»²½¿ÉÒÔ¼ÆËãÇ¡ºÃÍêÈ«·´Ó¦ºóËùµÃµ½²»±¥ºÍNaClÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®

½â´ð ½â£º£¨1£©Éú³ÉCO2µÄÖÊÁ¿Îª£º172.5g-163.7g=8.8g£¬
¹ÊÌ8.8g£®
£¨2£©Éè̼ËáÄÆÖÊÁ¿Îªx£¬Éú³ÉÂÈ»¯ÄÆÖÊÁ¿Îªy£¬
Na2CO3+2HCl¨T2NaCl+H2O+CO2¡ü£¬
106                    117             44
x                         y              8.8g
$\frac{106}{x}$=$\frac{117}{y}$=$\frac{44}{8.8g}$£¬
x=21.2g£¬y=23.4g£¬
Ç¡ºÃÍêÈ«·´Ó¦ºóËùµÃµ½²»±¥ºÍNaClÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ£º$\frac{26.5g-21.2g+23.4g}{163.7g}$¡Á100%=17.5%£¬
´ð£ºÇ¡ºÃÍêÈ«·´Ó¦ºóËùµÃµ½²»±¥ºÍNaClÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýÊÇ17.5%£®

µãÆÀ ²îÁ¿·¨ÔÚ¼ÆËãÖеÄÓ¦Óúܹ㷺£¬½â´ðµÄ¹Ø¼üÊÇÒª·ÖÎö³öÎïÖʵÄÖÊÁ¿²îÓëÒªÇóµÄδ֪ÊýÖ®¼äµÄ¹ØÏµ£¬ÔÙ¸ù¾Ý¾ßÌåµÄÊý¾ÝÇó½â£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
18£®½ðÊôÄÆ¼°Æä²¿·Ö»¯ºÏÎïµÄ»¯Ñ§ÐÔÖʱȽϻîÆÃ£¬±£´æ²»µ±½ÏÒ×±äÖÊ£¬ÇëÄã¸ù¾ÝËùѧ֪ʶ»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Èçͼ1ÊǽðÊôÄÆÔÚÔªËØÖÜÆÚ±íÖеÄÐÅÏ¢£¬ÏÂÁйØÓÚÄÆÔªËØµÄ˵·¨²»ÕýÈ·µÄÊÇC
A£®ÄÆÊôÓÚ½ðÊôÔªËØ         B£®ÄƵĺ˵çºÉÊýΪ11
C£®Ïà¶ÔÔ­×ÓÖÊÁ¿Îª22.99g   D£®Ô­×ÓºËÍâÓÐ11¸öµç×Ó
£¨2£©¹ýÑõ»¯ÄÆ£¨Na2O2£©Óöµ½Ë®»á·¢Éú·´Ó¦£º2Na2O2+2H2O=4NaOH+x¡ü£¬Çëд³öxµÄ»¯Ñ§Ê½O2£®
£¨3£©ÎªÁËÑé֤ʵÑéÊÒ±£´æµÄNaOHÊÇ·ñ±äÖÊ£¬Ä³Í¬Ñ§Éè¼ÆÁËÈçͼ2ËùʾʵÑé·½°¸½øÐÐ̽¾¿£¬Çë¾Ýͼ»Ø´ð£º
¢ÙʵÑéÖвⶨÈÜÒºpHµÄ·½·¨ÊÇÓýྻ¸ÉÔïµÄ²£Á§°ôպȡÈÜÒº£¬µÎÔÚpHÊÔÖ½ÉÏ£¬°Ñ±äÉ«µÄpHÊÔÖ½Óë±ê×¼±ÈÉ«¿¨¶ÔÕÕ£¬¼´¿ÉµÃÈÜÒºµÄpH£®
¢ÚÇâÑõ»¯ÄƱäÖʵÄÔ­ÒòÊÇ2NaOH+CO2¨TNa2CO3+H2O £¨Óû¯Ñ§·½³Ìʽ±íʾ£©£¬¸ù¾ÝʵÑéÏÖÏó£¬ÄãÈÏΪ¸ÃÇâÑõ»¯ÄƹÌÌåÒѱäÖÊ£¨Ìî¡°ÒÑ¡±»ò¡°Î´¡±£©£®
¢ÛʵÑéÖмÓÈë¹ýÁ¿ÂÈ»¯±µÈÜÒºµÄÄ¿µÄÊÇʹ̼ËáÄÆÍêÈ«·´Ó¦£®
¢ÜÈô²âµÃÈÜÒºEµÄpH£¾7£¬ÔòÑùÆ·AµÄ³É·ÖÊÇÇâÑõ»¯ÄƺÍ̼ËáÄÆ£®
£¨4£©È¡ÒѱäÖʵÄÇâÑõ»¯ÄÆÑùÆ·ÅäÖÆ³É100gÈÜÒº£¬ÏòÆäÖмÓÈëÈÜÖÊÖÊÁ¿·ÖÊýΪ7.3%µÄÏ¡ÑÎËá100g£¬ÍêÈ«·´Ó¦ºóµÃµ½ÖÐÐÔÈÜÒº197.8g£®
ÊÔ¼ÆËãËùÈ¡ÑùÆ·Öк¬ÔÓÖÊ̼ËáÄÆµÄÖÊÁ¿£®

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

¾«Ó¢¼Ò½ÌÍø