ÌâÄ¿ÄÚÈÝ

δ³ÉÄêÈ˺ÍÀÏÄêÈËÊÇÐèÒª²¹¸ÆµÄÖ÷ÒªÈËȺ£¬Ä¿Ç°ÊÐÃæÉÏÓÐÐí¶à²¹¸ÆµÄ¸ÆƬ³öÊÛ£¬ÈçͼÊǸÇÖиǸ߸ÆƬµÄ²¿·Ö˵Ã÷£¬Ä³¿ÎÍâÐËȤС×éÓû¼ìÑé˵Ã÷µÄÕæᣬ½øÐÐÁËÈëϵÄʵÑ飺
¢ÙÈ¡1ƬƬ¼ÁÑÐËé
¢Ú½«Æä¼ÓÈë×ãÁ¿Ï¡ÑÎËáÖÐ
¢ÛÍêÈ«·´Ó¦ºóÊÕ¼¯µ½0.55g¶þÑõ»¯Ì¼
ÊÔ·ÖÎö£º
£¨1£©¸ß¸ÆƬÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£¿
£¨2£©Í¨¹ý¼ÆËãÅжϴ˸ÆƬÖиƺ¬Á¿Óë±ê×¢ÊÇ·ñÊôʵ£®£¨¼Ù¶¨¸ÆƬÖÐÆäËû³É·Ý²»ÓëÑÎËá·´Ó¦£¬1g=1000mg£©
¿¼µã£º¸ù¾Ý»¯Ñ§·´Ó¦·½³ÌʽµÄ¼ÆËã,»¯ºÏÎïÖÐijԪËصÄÖÊÁ¿¼ÆËã
רÌ⣺Óйػ¯Ñ§·½³ÌʽµÄ¼ÆËã
·ÖÎö£º£¨1£©¸ù¾Ý̼Ëá¸ÆÓëÑÎËá·´Ó¦µÄ·½³Ìʽ£¬½«¶þÑõ»¯Ì¼µÄÖÊÁ¿´úÈë¼ÆËã¿ÉÅжÏ̼Ëá¸ÆµÄÖÊÁ¿£»
£¨2£©Ì¼Ëá¸ÆµÄÖÊÁ¿³ËÒÔ¸ÆÔÚ̼Ëá¸ÆÖÐËùÕ¼ÓеÄÖÊÁ¿·ÖÊý£¬Çó³ö¸ÆÔªËصÄÖÊÁ¿Óë²¹¸Æ¼ÁµÄ˵Ã÷±È½ÏÅж¨ÊÇ·ñÊôʵ£®
½â´ð£º½â£º
£¨1£©ÒòÑÎËá×ãÁ¿£¬Ôò̼Ëá¸ÆÍêÈ«·´Ó¦£»
Éè¸ÆƬÖÐ̼Ëá¸ÆµÄÖÊÁ¿Îªx
CaCO3+2HCl¨TCaCl2+CO2¡ü+H2O
100               44
x                0.55g
100
44
=
x
0.55g

x¨T1.25g
¸ß¸ÆƬÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý=
1.25g
2.5g
¡Á100%
=50%
£¨2£©Ôò1.25g̼Ëá¸ÆÖк¬¸ÆÔªËصÄÖÊÁ¿Îª£º1.25g¡Á
40
40+12+16¡Á3
¡Á100%
=0.5g
0.5g¡Á1000mg/g=500mg
Óë˵Ã÷ÖÐÿƬº¬¸Æ500mg¡¢Ò»Ö£¬Ôò˵Ã÷Êôʵ£¬
´ð°¸£º
£¨1£©¸ß¸ÆƬÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý50%
£¨2£©´Ë¸ÆƬÖиƺ¬Á¿Óë±ê×¢Êôʵ£®
µãÆÀ£º±¾Ì⿼²éÀûÓû¯Ñ§Ê½ºÍ»¯Ñ§·´Ó¦·½³ÌʽµÄ¼ÆË㣬ѧÉúÃ÷È·³£»¯Ñ§Ê½¼ÆËãµÄ·½·¨¡¢»¯Ñ§·´Ó¦ÖкÎÖÖÎïÖʵÄÖÊÁ¿¿É´úÈë·½³Ìʽ¼ÆËãÊǽâ´ðµÄ¹Ø¼ü£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø