ÌâÄ¿ÄÚÈÝ

19£®Ò˲ýÊÐÄϽò¹ØÒ»´øº¬ÓзḻµÄʯ»Òʯ×ÊÔ´£¬ÎÒÃÇÖªµÀ£¬¸ßÎÂìÑÉÕʯ»Òʯ£¨Ö÷Òª³É·ÝÊÇ̼Ëá¸Æ£©¿ÉÖÆµÃÉúʯ»Ò£¨¼´Ñõ»¯¸Æ£©ºÍ¶þÑõ»¯Ì¼£®ÎªÁË·ÖÎö²úÆ·µÄ´¿¶È£¬Ð¡Ë´Í¬Ñ§È¡Ê¯»ÒʯÑùÆ·22g£¬³ä·ÖìÑÉÕºó³ÆµÃÊ£Óà¹ÌÌåµÄÖÊÁ¿Îª13.2g£¨Ìáʾ£º¼Ù¶¨ÔÓÖʲ»·´Ó¦£©£®¼ÆË㣺
£¨1£©Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿ÊǶàÉÙ¿Ë£¿ÕâЩ¶þÑõ»¯Ì¼ÔÚ±ê×¼×´¿öϵÄÌå»ýÊǶàÉÙÉý£¿£¨±ê×¼×´¿öÏÂCO2µÄÃܶÈ1.977g/L£©
£¨2£©ÖƵÃÉúʯ»ÒµÄÖÊÁ¿ÊǶàÉÙ¿Ë£¿Æä²úÆ·µÄ´¿¶ÈÊǶàÉÙ£¿

·ÖÎö ̼Ëá¸ÆÔÚ¸ßÎÂÌõ¼þÏ·ֽâÉú³ÉÑõ»¯¸ÆºÍ¶þÑõ»¯Ì¼£¬¸ù¾Ý·´Ó¦µÄ»¯Ñ§·½³Ìʽ¼°ÆäÌṩµÄÊý¾Ý¿ÉÒÔ½øÐÐÏà¹Ø·½ÃæµÄ¼ÆË㣮

½â´ð ½â£º£¨1£©¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉ¿ÉÖª£¬Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª£º22g-13.2g=8.8g£¬
¶þÑõ»¯Ì¼µÄÌå»ýΪ£º8.8g¡Â1.977g/L=4.5L£¬
£¨2£©ÉèÉú³ÉÉúʯ»ÒµÄÖÊÁ¿Îªx£¬
CaCO3$\frac{\underline{\;¸ßÎÂ\;}}{\;}$CaO+CO2¡ü
                      56   44
                      x    8.8g
$\frac{56}{x}$=$\frac{44}{8.8g}$£¬
x=11.2g£¬
ÔòËùµÃ²úÆ·µÄ´¿¶ÈΪ£º$\frac{11.2g}{13.2g}$¡Á100%=84.8%£¬
´ð£ºËùµÃ²úÆ·µÄ´¿¶ÈΪ84.8%£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éѧÉúÔËÓüÙÉè·¨ºÍ»¯Ñ§·½³Ìʽ½øÐмÆËãºÍÍÆ¶ÏµÄÄÜÁ¦£¬¼ÆËãʱҪעÒâ¹æ·¶ÐÔºÍ׼ȷÐÔ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø