ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÅäÖÆÈÜÖÊÖÊÁ¿·ÖÊýΪµÄµâ¾ÆËùÐèµâºÍ¾Æ¾«µÄÖÊÁ¿¡££¨¾Æ¾«µÄÃܶÈÊÇ£©

¢Ù¼ÆË㣺ÆäÖеâ__________ g£¬¾Æ¾«__________ g£¬¾Æ¾«µÄÌå»ýΪ__________ mL¡£

¢Ú³ÆÁ¿£ºÓÃ__________³ÆÁ¿ËùÐèµâµÄÖÊÁ¿£¬µ¹ÈëÉÕ±­ÖС£

¢ÛÈܽ⣺ÓÃ__________ºÁÉý£¨ÇëÌîд»ò£©Á¿Í²È¡ËùÐè¾Æ¾«µÄÌå»ý£¬µ¹ÈëÊ¢ÓеâµÄÉÕ±­ÀÓÃ__________½Á°è£¬Ê¹µâÈܽ⡣ÕâÑùµÃµ½µÄÈÜÒº¼´ÎªÈÜÖʵÄÖÊÁ¿·ÖÊýΪµÄµâ¾Æ¡£

¢Ü×°Æ¿²¢Ìù±êÇ©£º°ÑÅäºÃµÄÈÜҺװÈëÊÔ¼ÁÆ¿ÖУ¬¸ÇºÃÆ¿Èû²¢ÌùÉϱêÇ©¡£±êÇ©ÉÏÌîдµÄÄÚÈÝÊÇ__________ºÍ__________£¬·Åµ½ÊÔ¼Á¹ñÖС£

¡¾´ð°¸¡¿0.4 19.6 24.5 ÍÐÅÌÌìƽ 50 ²£Á§°ô µâ¾Æ 2%

¡¾½âÎö¡¿

¢ÙÈÜÖÊÖÊÁ¿=ÈÜÒºÖÊÁ¿¡ÁÈÜÖʵÄÖÊÁ¿·ÖÊý£¬Ò½ÔºÒªÅäÖÆ20¿ËÖÊÁ¿·ÖÊýΪ2%µÄµâ¾ÆÓÃÀ´Ïû¶¾£¬ÐèµâµÄÖÊÁ¿=20g¡Á2%=0.4g£»ÈܼÁÖÊÁ¿=ÈÜÒºÖÊÁ¿-ÈÜÖÊÖÊÁ¿£¬ÔòËùÐè¾Æ¾«µÄÖÊÁ¿=20g-0.4g=19.6g£»¾Æ¾«µÄÃܶÈΪ0.8g/cm3£¬¸ù¾ÝV= = =24.5cm3=24.5mL

¢Ú³ÆÁ¿£ºÓÃÍÐÅÌÌìƽ³ÆÁ¿ËùÐèµâµÄÖÊÁ¿£¬µ¹ÈëÉÕ±­ÖС£
¢ÛÈܽ⣺ÓÃ50ºÁÉýÁ¿Í²È¡ËùÐè¾Æ¾«µÄÌå»ý£¬µ¹ÈëÊ¢ÓеâµÄÉÕ±­ÀÓà ²£Á§°ô½Á°è£¬Ê¹µâÈܽ⡣ÕâÑùµÃµ½µÄÈÜÒº¼´Îª20gÈÜÖʵÄÖÊÁ¿·ÖÊýΪ2%µÄµâ¾Æ¡£
¢Ü×°Æ¿²¢Ìù±êÇ©£º°ÑÅäºÃµÄÈÜҺװÈëÊÔ¼ÁÆ¿ÖУ¬¸ÇºÃÆ¿Èû²¢ÌùÉϱêÇ©¡£±êÇ©ÉÏÌîдµÄÄÚÈÝÊÇ µâ¾ÆºÍ 2%£¬·Åµ½ÊÔ¼Á¹ñÖС£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø