ÌâÄ¿ÄÚÈÝ

7£®Ä³¿óɽµÄʯ»ÒʯÑùÆ·Öк¬ÓжþÑõ»¯¹èÔÓÖÊ£¨¶þÑõ»¯¹èÊÇÒ»ÖּȲ»ÈÜÓÚˮҲ²»ÓëÑÎËá·´Ó¦ÇÒÄ͸ßεĹÌÌ壩£®ÃûʦѧУµÄͬѧÃÇÏë²â¶¨¸ÃÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£¬ËûÃDzÉÈ¡ÁËÒ»¿éʯ»ÒʯÑùÆ·£¬½«ÆäÇÃËéºó£¬³Æ³ö6g·ÅÈëÉÕ±­ÄÚ£¨ÉÕ±­ÖÊÁ¿Îª20g£©£®È»ºó¼ÓÈë50gµÄÏ¡ÑÎËᣬÓò£Á§±­½Á°èÖÁ²»ÔÙ²úÉúÆøÅÝΪֹ£®·´Ó¦ËùÐèʱ¼ä£¨t£©ºÍÉÕ±­¼°ÆäËùÊ¢ÎïÖÊ×ÜÖØÁ¿£¨m£©µÄ¹ØÏµÈçͼËùʾ£®ÊԻشð£º
£¨1£©½«Ê¯»ÒʯÑùÆ·ÇÃËéµÄÖ÷ҪĿµÄÊǼӿìºÍÏ¡ÑÎËá·´Ó¦µÄËÙÂÊ£®
£¨2£©ÊµÑé½áÊøÊ±£¬¹²·Å³ö¶þÑõ»¯Ì¼2.2¿Ë£®
£¨3£©¸Ãʯ»ÒʯÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿£¨½á¹û±£ÁôһλСÊý£©

·ÖÎö ·´Ó¦Îï½Ó´¥Ô½³ä·Ö£¬·´Ó¦ËÙÂÊÔ½¿ì£»
·´Ó¦Ç°ºóµÄÖÊÁ¿²î¼´Îª·´Ó¦Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿£¬¸ù¾Ý¶þÑõ»¯Ì¼µÄÖÊÁ¿¿ÉÒÔ¼ÆËã̼Ëá¸ÆµÄÖÊÁ¿£¬½øÒ»²½¿ÉÒÔ¼ÆËã¸Ãʯ»ÒʯÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£®

½â´ð ½â£º£¨1£©½«Ê¯»ÒʯÑùÆ·ÇÃËéµÄÖ÷ҪĿµÄÊÇÔö´ó̼Ëá¸ÆºÍÏ¡ÑÎËáµÄ½Ó´¥Ãæ»ý£¬¼Ó¿ì·´Ó¦ËÙÂÊ£®
¹ÊÌ¼Ó¿ìºÍÏ¡ÑÎËá·´Ó¦µÄËÙÂÊ£®
£¨2£©ÊµÑé½áÊøÊ±£¬¹²·Å³ö¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª£º6g+20g+50g-73.8g=2.2g£¬
¹ÊÌ2.2£®
£¨3£©Éè̼Ëá¸ÆµÄÖÊÁ¿Îªx£¬
CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü£¬
 100                  44
  x                  2.2g
$\frac{100}{x}$=$\frac{44}{2.2g}$£¬
x=5g£¬
¸Ãʯ»ÒʯÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ£º$\frac{5g}{6g}$¡Á100%=83.3%£¬
´ð£º¸Ãʯ»ÒʯÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýÊÇ83.3%£®

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

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø