ÌâÄ¿ÄÚÈÝ
ÏÂͼÊÇXXXÅÆ¸ß¸ÆÆ¬µÄ±êÇ©£¬Çë»Ø´ð£º| XXXÅÆ¸ß¸ÆÆ¬ ³É·Ö£ºÃ¿Æ¬º¬Ì¼Ëá¸Æ1.5¿Ë £¨Ï൱ÓÚº¬¸Æ0.6¿Ë£© ¹æ¸ñ£º¹²30Ƭ£¨1£©±êÇ©ÖУ¬¡°Ï൱ÓÚº¬¸Æ0.6¿Ë¡±£¬ÊÇÈçºÎ¼ÆËã³öµÄ£¿Çëд³öÄãµÄ¼ÆËã·½·¨£® £¨2£©ÎªÈ·¶¨±êÇ©µÄÕæÊµÐÔ£¬Ð¡Î÷ͬѧȡÁËһƬ¸ÆÆ¬£¬µ·Ëé·ÅÈëÊ¢ÓÐ×ãÁ¿Ï¡ÑÎËáµÄÉÕ±ÖУ¬³ä·Ö·´Ó¦ºó£¬²âµÃ²úÉúµÄ¶þÑõ»¯Ì¼Ìå»ýΪVÉý£¬ÔòVΪ¶àÉÙʱ£¬±êÇ©ÕæÊµ£®£¨¼ÆËã½á¹û¾«È·µ½Ð¡ÊýµãºóÁ½Î»£© £¨Ìáʾ£º¢Ù¸ÆÆ¬ÖÐÆäËû³É·Ö²»ÓëÑÎËá·´Ó¦£»¢ÚʵÑéÌõ¼þÏ£¬¶þÑõ»¯Ì¼ÃܶÈΪ1.96¿Ë/Éý£®£© ·ÖÎö£º£¨1£©ÀûÓû¯Ñ§Ê½¼ÆËã̼Ëá¸ÆÖиƵÄÖÊÁ¿£¬¼´ÀûÓÃ̼Ëá¸ÆµÄÖÊÁ¿³ËÒÔ̼Ëá¸ÆÖиƵÄÖÊÁ¿·ÖÊý£® £¨2£©ÀûÓû¯Ñ§·½³Ìʽ¼ÆËã½â¾ö£¬ÀûÓÃһƬ¸ÆÆ¬ÖеÄ̼Ëá¸ÆµÄ·´Ó¦ÎïÖÊÁ¿Çó¶þÑõ»¯Ì¼Éú³ÉÎïµÄÖÊÁ¿¼´¿É£¬ÆäÖаüº¬ÀûÓÃÃܶȼÆËãÌå»ýÓëÖÊÁ¿µÄת»¯£® ½â´ð£º½â£º£¨1£©1.5g¡Á
£¨2£©CaCO3 +2HCl=CaCl2+H2O+CO2¡ü 100 44 1.5g 1.96g/L¡ÁV
V=0.34L ´ð£ºVÊÇ0.34Lʱ£¬±êÇ©ÕæÊµ£® µãÆÀ£º´ËÌâÊÇ»¯Ñ§Ê½¼°»¯Ñ§·½³Ìʽ¼ÆË㣬ÊǶԳ£¹æ¼ÆËãµÄÁ·Ï°Óë¹®¹Ì£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÏÂͼÊÇXXXÅÆ¸ß¸ÆÆ¬µÄ±êÇ©£¬Çë»Ø´ð£º
£¨2£©ÎªÈ·¶¨±êÇ©µÄÕæÊµÐÔ£¬Ð¡Î÷ͬѧȡÁËһƬ¸ÆÆ¬£¬µ·Ëé·ÅÈëÊ¢ÓÐ×ãÁ¿Ï¡ÑÎËáµÄÉÕ±ÖУ¬³ä·Ö·´Ó¦ºó£¬²âµÃ²úÉúµÄ¶þÑõ»¯Ì¼Ìå»ýΪVÉý£¬ÔòVΪ¶àÉÙʱ£¬±êÇ©ÕæÊµ£®£¨¼ÆËã½á¹û¾«È·µ½Ð¡ÊýµãºóÁ½Î»£© £¨Ìáʾ£º¢Ù¸ÆÆ¬ÖÐÆäËû³É·Ö²»ÓëÑÎËá·´Ó¦£»¢ÚʵÑéÌõ¼þÏ£¬¶þÑõ»¯Ì¼ÃܶÈΪ1.96¿Ë/Éý£®£©
ÏÂͼÊÇXXXÅÆ¸ß¸ÆÆ¬µÄ±êÇ©£¬Çë»Ø´ð£º
£¨2£©ÎªÈ·¶¨±êÇ©µÄÕæÊµÐÔ£¬Ð¡Î÷ͬѧȡÁËһƬ¸ÆÆ¬£¬µ·Ëé·ÅÈëÊ¢ÓÐ×ãÁ¿Ï¡ÑÎËáµÄÉÕ±ÖУ¬³ä·Ö·´Ó¦ºó£¬²âµÃ²úÉúµÄ¶þÑõ»¯Ì¼Ìå»ýΪVÉý£¬ÔòVΪ¶àÉÙʱ£¬±êÇ©ÕæÊµ£®£¨¼ÆËã½á¹û¾«È·µ½Ð¡ÊýµãºóÁ½Î»£© £¨Ìáʾ£º¢Ù¸ÆÆ¬ÖÐÆäËû³É·Ö²»ÓëÑÎËá·´Ó¦£»¢ÚʵÑéÌõ¼þÏ£¬¶þÑõ»¯Ì¼ÃܶÈΪ1.96¿Ë/Éý£®£© |