ÌâÄ¿ÄÚÈÝ

15£®ÈçͼÊǸߵÈÉúÎïϸ°ûÓÐÑõºôÎü¹ý³ÌµÄͼ½â£¬¾Ýͼ»Ø´ð£º
£¨1£©Í¼ÖТÛËù´ú±íµÄÎïÖÊÊÇCO2£®
£¨2£©¢Ü¡¢¢Ý¡¢¢Þ´ú±íÄÜÁ¿£¬ÆäÖÐ×î¶àµÄ´úºÅÊÇ¢Þ£®
£¨3£©ÓÐÑõºôÎüµÄ·´Ó¦Ê½£ºC6H12O6+6H2O+6O2$\stackrel{ø}{¡ú}$6CO2+12H2O+ÄÜÁ¿£®
£¨4£©ÈËÌ弡ϸ°ûÄÚÈç¹ûȱÑõ£¬C6H12O6Ñõ»¯µÄ×îÖÕ²úÎïÊÇÈéËᣮ
£¨5£©Öü²ØÊ±¼ä¾ÃÁËµÄÆ»¹ûÄÚ²¿µÄϸ°û½øÐкôÎüʱ£¬ÏûºÄÒ»¶¨Á¿µÄÆÏÌÑÌÇ¿ÉÒÔ²úÉúAĦ¶ûµÄCO2£¬ÆäÖ²ÎïÌåҶƬÔÚÕý³£Éú³¤Ê±ÏûºÄͬÑùÊýÁ¿µÄÆÏÌÑÌǿɲúÉúCO23AĦ¶û£®

·ÖÎö 1¡¢ÓÐÑõºôÎüµÄ¹ý³Ì£º
1¡¢C6H12O6$\stackrel{ø}{¡ú}$2±ûͪËá+4[H]+ÄÜÁ¿        £¨Ï¸°ûÖÊ»ùÖÊ£©
2¡¢2±ûͪËá+6H2O$\stackrel{ø}{¡ú}$6CO2+20[H]+ÄÜÁ¿     £¨ÏßÁ£Ìå»ùÖÊ£©
3¡¢24[H]+6O2$\stackrel{ø}{¡ú}$12H2O+ÄÜÁ¿             £¨ÏßÁ£ÌåÄÚĤ£©
2¡¢¸ù¾ÝÌâÒâºÍͼʾ·ÖÎö¿ÉÖª£ºÍ¼ÖТٴú±íC3H4O3£¨±ûͪËᣩ£¬¢Ú´ú±íH2O£¨Ë®£©£¬¢Û´ú±íCO2£¨¶þÑõ»¯Ì¼£©£¬¢Ü¡¢¢Ý´ú±íÉÙÁ¿ÄÜÁ¿£¬¢Þ´ú±í´óÁ¿ÄÜÁ¿£®

½â´ð ½â£º£¨1£©Í¼ÖТÛËù´ú±íµÄÎïÖÊÊÇCO2£®
£¨2£©¢Ü¡¢¢Ý¡¢¢Þ´ú±íÄÜÁ¿£¬ÆäÖТܴú±íÓÐÑõºôÎüµÚÒ»½×¶ÎËùÊͷŵÄÉÙÁ¿ÄÜÁ¿¡¢¢Ý´ú±íÓÐÑõºôÎüµÚ¶þ½×¶ÎËùÊͷŵÄÉÙÁ¿ÄÜÁ¿£¬¢Þ´ú±íÓÐÑõºôÎüµÚÈý½×¶ÎËùÊͷŵĴóÁ¿ÄÜÁ¿£¬¹ÊÄÜÁ¿×î¶àµÄ´úºÅÊÇ¢Þ£®
£¨3£©ÓÐÑõºôÎüµÄ·´Ó¦Ê½£ºC6H12O6+6H2O+6O2$\stackrel{ø}{¡ú}$6CO2+12H2O+ÄÜÁ¿£®
£¨4£©ÈËÌ弡ϸ°ûÄÚÈç¹ûȱÑõ£¬C6H12O6Ñõ»¯µÄ×îÖÕ²úÎïÊÇÈéËᣮ
£¨5£©Öü²ØÊ±¼ä¾ÃÁËµÄÆ»¹ûÄÚ²¿µÄϸ°û½øÐÐÎÞÑõºôÎüʱ£¬ÏûºÄÒ»¶¨Á¿µÄÆÏÌÑÌÇ¿ÉÒÔ²úÉúAĦ¶ûµÄCO2£¬¹Ê¿ÉµÃÆäÖ²ÎïÌåҶƬ½øÐÐÓÐÑõºôÎüÏûºÄͬÑùÊýÁ¿µÄÆÏÌÑÌǿɲúÉúCO2£º$\frac{A}{2}$¡Á6=3AĦ¶û£®
¹Ê´ð°¸Îª£º
£¨1£©CO2 
£¨2£©¢Þ
£¨3£©C6H12O6+6H2O+6O2$\stackrel{ø}{¡ú}$6CO2+12H2O+ÄÜÁ¿
£¨4£©ÈéËá   
£¨5£©3A

µãÆÀ ±¾ÌâÒÔͼÐÎÎªÔØÌ壬¿¼²éѧÉú¶Ôϸ°ûºôÎüÏà¹ØÖªÊ¶µÄÀí½âºÍ·ÖÎöÄÜÁ¦£®ºôÎü×÷Óùý³ÌÊÇ¿¼²éµÄÖØµãºÍÄѵ㣬¿ÉÒÔͨ¹ýÁ÷³Ìͼ·ÖÎö£¬±í¸ñ±È½Ï£¬µäÐÍÁ·Ï°·ÖÎöÇ¿»¯Ñ§ÉúµÄÀí½â£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø