ÌâÄ¿ÄÚÈÝ

19£®ÈçͼËùʾΪʵÑéÊÒÖг£¼ûµÄÆøÌåÖÆ±¸¡¢¾»»¯¡¢ÊÕ¼¯ºÍÐÔÖÊʵÑéµÄ²¿·ÖÒÇÆ÷£®ÊÔ¸ù¾ÝÌâĿҪÇ󣬻شðÏÂÁÐÎÊÌ⣺

£¨1£©ÈôÒÔʯ»ÒʯºÍÏ¡ÑÎËáΪԭÁÏ£¬ÔÚʵÑéÊÒÖÐÖÆ±¸²¢ÊÕ¼¯Ò»Æ¿¸ÉÔï´¿¾»µÄ¶þÑõ»¯Ì¼ÆøÌ壮
¢ÙËùÑ¡ÒÇÆ÷µÄÁ¬½Ó˳ÐòΪAGEF £¨ÌîдÒÇÆ÷ÐòºÅ×Öĸ£©£®
¢ÚÇëд³ö×°ÖÃAÖÐËù·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽCaCO3+2HC1=CaCl2+H2O+CO2¡ü£®
¢Û¼ìÑé¼¯ÆøÆ¿ÖжþÑõ»¯Ì¼ÆøÌåÒÑÊÕ¼¯ÂúµÄ·½·¨ÊÇÄÃÒ»¸ùȼ×ŵÄľÌõ·Åµ½¼¯ÆøÆ¿¿Ú£¬Ä¾ÌõÁ¢¼´Ï¨Ãð£¬ËµÃ÷¶þÑõ»¯Ì¼ÆøÌåÒѼ¯Âú£®
£¨2£©ÉúÌúºÍ¸Ö¶¼ÊÇÌúµÄÖØÒªºÏ½ð£®Ä³»¯Ñ§Ñо¿Ð¡×éµÄͬѧҪ֤Ã÷ijÉúÌúÑùÆ·Öк¬ÓÐÌ¼ÔªËØ£¬²¢²â¶¨Æäº¬Ì¼Á¿£¬ÒÇÆ÷µÄÁ¬½Ó˳ÐòΪ£ºA¡úE¡úC¡úD1¡úB¡úD2£®ÆäÖÐÒÇÆ÷AÖеÄҩƷΪ˫ÑõË®ºÍ¶þÑõ»¯ÃÌ£®ÏÖÈ¡10g¸ÃÑùÆ·ÔÚ´¿ÑõÖÐȼÉÕ£®Çë¸ù¾ÝÌâĿҪÇ󣬷ÖÎöÓйØÊý¾Ý£¬»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÇëд³ö×°ÖÃAÖÐËù·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ2H2O2$\frac{\underline{\;MnO_2\;}}{\;}$2H2O+O2¡ü£®
¢ÚÆäÖÐ×°ÖÃEµÄ×÷ÓÃÊÇÎüÊÕË®ÕôÆø£®
¢Ûµ±ÒÇÆ÷D1ÖгöÏÖ³ÎÇåʯ»ÒË®±ä»ë×ǵÄÏÖÏó£¬Ôò¿ÉÖ¤Ã÷ÉúÌúÑùÆ·Öк¬ÓÐÌ¼ÔªËØ£®
¢Üµ±ÕûÌ××°ÖÃÄڵĻ¯Ñ§·´Ó¦¾ùÍêÈ«½øÐк󣬾­²âÁ¿ÒÇÆ÷D1ÔöÖØm1g£¬ÒÇÆ÷BÔöÖØm2g£¬ÒÇÆ÷D2ÖÊÁ¿Ã»Óб仯£¬Ôò10g¸ÃÊÔÑùÖк¬Ì¼ÔªËصÄÖÊÁ¿Îª$\frac{3£¨m_{1}+m_{2}£©}{11}$g£¨¼ÆËã½á¹û¿ÉΪ·ÖÊýÐÎʽ£©£®

·ÖÎö £¨1£©ÊµÑéÊÒͨ³£ÓôóÀíʯ»òʯ»ÒʯºÍÏ¡ÑÎËá·´Ó¦ÖÆÈ¡¶þÑõ»¯Ì¼£¬·´Ó¦²»ÐèÒª¼ÓÈÈ£¬´óÀíʯºÍʯ»ÒʯµÄÖ÷Òª³É·ÖÊÇ̼Ëá¸Æ£¬ÄܺÍÏ¡ÑÎËá·´Ó¦Éú³ÉÂÈ»¯¸Æ¡¢Ë®ºÍ¶þÑõ»¯Ì¼£» 
¶þÑõ»¯Ì¼²»ÄÜȼÉÕ£¬²»Ö§³ÖȼÉÕ£»
£¨2£©Í¨³£Çé¿öÏ£¬¹ýÑõ»¯ÇâÔÚ¶þÑõ»¯Ã̵Ĵ߻¯×÷ÓÃÏ£¬·Ö½âÉú³ÉË®ºÍÑõÆø£»   
ŨÁòËá¾ßÓÐÎüË®ÐÔ£¬¿ÉÒÔÓÃÀ´¸ÉÔïÇâÆø¡¢ÑõÆø¡¢¶þÑõ»¯Ì¼µÈÆøÌ壻
¶þÑõ»¯Ì¼ÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×Ç£»
¸ù¾Ý×°ÖÃÖÊÁ¿±ä»¯µÄÇé¿ö¿ÉÒÔ¼ÆËã¸ÃÊÔÑùÖк¬Ì¼ÔªËصÄÖÊÁ¿£®

½â´ð ½â£º£¨1£©¢ÙËùÑ¡ÒÇÆ÷µÄÁ¬½Ó˳ÐòΪ£ºÍ¨¹ýA×°ÖÃÖÆÈ¡µÄ¶þÑõ»¯Ì¼µ¼ÈëG×°ÖÃÖгýÈ¥ÂÈ»¯ÇâÆøÌ壬ÔÙͨ¹ýE×°ÖóýȥˮÕôÆø£¬×îºóͨ¹ýF×°ÖÃÀûÓÃÏòÉÏÅÅ¿ÕÆø·¨ÊÕ¼¯µ½¸ÉÔïµÄ¶þÑõ»¯Ì¼ÆøÌ壻
¢Ú×°ÖÃAÖÐËù·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCaCO3+2HC1=CaCl2+H2O+CO2¡ü£»
¢Û¼ìÑé¼¯ÆøÆ¿ÖжþÑõ»¯Ì¼ÆøÌåÒÑÊÕ¼¯ÂúµÄ·½·¨ÊÇ£ºÄÃÒ»¸ùȼ×ŵÄľÌõ·Åµ½¼¯ÆøÆ¿¿Ú£¬Ä¾ÌõÁ¢¼´Ï¨Ãð£¬ËµÃ÷¶þÑõ»¯Ì¼ÆøÌåÒѼ¯Âú£®
¹ÊÌAGEF£»CaCO3+2HC1=CaCl2+H2O+CO2¡ü£»ÄÃÒ»¸ùȼ×ŵÄľÌõ·Åµ½¼¯ÆøÆ¿¿Ú£¬Ä¾ÌõÁ¢¼´Ï¨Ãð£¬ËµÃ÷¶þÑõ»¯Ì¼ÆøÌåÒѼ¯Âú£®
£¨2£©¢Ù×°ÖÃAÖÐËù·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2H2O2$\frac{\underline{\;MnO_2\;}}{\;}$2H2O+O2¡ü£»
¢ÚÆäÖÐ×°ÖÃEµÄ×÷ÓÃÊÇÎüÊÕË®ÕôÆø£»
¢Ûµ±ÒÇÆ÷D1ÖгöÏÖ³ÎÇåʯ»ÒË®±ä»ë×ǵÄÏÖÏó£¬Ôò¿ÉÖ¤Ã÷ÉúÌúÑùÆ·Öк¬ÓÐÌ¼ÔªËØ£»
¢Ü¾­²âÁ¿ÒÇÆ÷D1ÔöÖØm1g£¬ÒÇÆ÷BÔöÖØm2g£¬ÒÇÆ÷D2ÖÊÁ¿Ã»Óб仯£¬ËµÃ÷Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª£ºm1g+m2g£¬
Ôò10g¸ÃÊÔÑùÖк¬Ì¼ÔªËصÄÖÊÁ¿Îª£º£¨m1g+m2g£©¡Á$\frac{12}{44}$¡Á100%=$\frac{3£¨m_{1}+m_{2}£©}{11}$£¬
¹ÊÌ2H2O2$\frac{\underline{\;MnO_2\;}}{\;}$2H2O+O2¡ü£»ÎüÊÕË®ÕôÆø£»³ÎÇåʯ»ÒË®±ä»ë×Ç£»$\frac{3£¨m_{1}+m_{2}£©}{11}$£®

µãÆÀ ʵÑéÏÖÏóÊÇÎïÖÊÖ®¼äÏ໥×÷ÓõÄÍâÔÚ±íÏÖ£¬Òò´ËҪѧ»áÉè¼ÆÊµÑé¡¢¹Û²ìʵÑé¡¢·ÖÎöʵÑ飬Ϊ½ÒʾÎïÖÊÖ®¼äÏ໥×÷ÓõÄʵÖʵ춨»ù´¡£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø