ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿ÏÖÓк¬Ì¼ËáÄƵÄʳÑÎÑùÆ·£¬Îª²â¶¨ÑùÆ·ÖÐÂÈ»¯ÄƵĺ¬Á¿£¬¿ÎÍâ»î¶¯Ð¡×éµÄͬѧ³ÆÈ¡10gʳÑÎÑùÆ·ÓÚÉÕ±ÖУ¬²¢½«40gÏ¡ÑÎËáƽ¾ù·Ö³ÉËĴμÓÈëÉÕ±ÖУ¬ÊµÑéÊý¾Ý¼ûÏÂ±í£º
ʵÑéÐòºÅ | ¼ÓÈëÏ¡ÑÎ ËáµÄÖÊÁ¿/g | ·´Ó¦ºóÉÕ±ÖРʣÓàÎïÖʵÄÖÊÁ¿£¯g |
µÚÒ»´Î | 10 | 19.56 |
µÚ¶þ´Î | 10 | 29.12 |
µÚÈý´Î | 10 | 38.9 |
µÚËÄ´Î | 10 | 48.9 |
(1)Çó²úÉúCO2ÆøÌåµÄ×ÜÖÊÁ¿¡£
(2)ÇóÑùÆ·ÖÐÂÈ»¯ÄƵĴ¿¶È¡£
(3)ÇóʹÓõÄÏ¡ÑÎËáµÄÈÜÖÊÖÊÁ¿·ÖÊý¡£
¡¾´ð°¸¡¿½â£º£¨1£©²úÉú¶þÑõ»¯Ì¼ÆøÌåµÄÖÊÁ¿Îª1.1g
£¨2£©Éè10gÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿ÎªX
Na2 CO3+2HCl===2NaCl+ H2O + CO2¡ü
106 44
X 1.1g
106/x=44/1.1g
X=2.65g
ËùÒÔÑùÆ·ÖÐNaClµÄ´¿¶ÈΪ(10g-2.65g)/10g¡Á100%=73.5%
(3)Éè10gÏ¡ÑÎËáÖд¿ÑÎËáµÄÖÊÁ¿Îªy
Na2 CO3+2HCl===2NaCl+ H2O + CO2¡ü
73 44
Y 0.44g
73/y=44/0.44g
Y=0.73g
ËùÒÔʹÓÃÏ¡ÑÎËáµÄÖÊÁ¿·ÖÊýΪ0.73g/10g¡Á100%=7.3%
´ð£ºÂÔ
¡¾½âÎö¡¿ÓÉÖÊÁ¿Êغ㶨ÂÉÖª£¬(1)Çó²úÉúCO2ÆøÌåµÄ×ÜÖÊÁ¿¡Ã10g£«40g£48.9g£½1.1g. (2)Éè10gÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿ÎªX
Na2 CO3+2HCl===2NaCl+ H2O + CO2¡ü
106 44
X 1.1g
106/x=44/1.1g X=2.65g
ËùÒÔÑùÆ·ÖÐNaClµÄ´¿¶ÈΪ(10g-2.65g)/10g¡Á100%=73.5%
(3)Éè10gÏ¡ÑÎËáÖд¿ÑÎËáµÄÖÊÁ¿Îªy
Na2 CO3+2HCl===2NaCl+ H2O + CO2¡ü
73 44
Y 0.44g
73/y=44/0.44g
Y=0.73g
ËùÒÔʹÓÃÏ¡ÑÎËáµÄÖÊÁ¿·ÖÊýΪ0.73g/10g¡Á100%=7.3%
´ð¡Ã(1)²úÉúCO2ÆøÌåµÄ×ÜÖÊÁ¿ÊÇ1.1g. ¢Æ ÑùÆ·ÖÐNaClµÄ´¿¶ÈΪ73.5%.¢ÇʹÓÃÏ¡ÑÎËáµÄÖÊÁ¿·ÖÊýΪ7.3%.
µã¾¦¡Ã±¾ÌâÖ÷Òª¿¼²éÖÊÁ¿Êغ㶨ÂɵÄÓ¦ÓÃÒÔ¼°¸ù¾Ý»¯Ñ§·½³Ìʽ½øÐмÆËã¡£
¡¾ÌâÄ¿¡¿Ð¡Ã÷ºÍСÑÞÔÚʵÑéÊÒÅäÖÆʯ»Òˮʱ£¬Ð¡Ã÷ͬѧ½«Á½Ò©³×Êìʯ»ÒÑùÆ··ÅÈëСÉÕ±ÖУ¬ÏòÆäÖмÓÈëÒ»¶¨Á¿µÄÕôÁóË®£¬³ä·Ö½Á°èºó·¢ÏÖÉÕ±µ×²¿ÈÔÓв»ÈÜÐԵĹÌÌ壬ÓÚÊÇËûÈÏΪÊìʯ»ÒÑùÆ·ÒѾ±äÖÊ¡£Ð¡ÑÞͬѧ²»Í¬ÒâËûµÄÅжϣ¬ÄãÈÏΪСÑÞͬѧµÄÀíÓÉÊÇ ____________________¡£Õë¶Ô¸ÃÊìʯ»ÒÑùÆ·µÄ³É·Ö£¬Í¬Ñ§ÃÇÕ¹¿ªÁËÌÖÂÛ£¬ÇëÄã²ÎÓëÌÖÂÛ²¢Íê³ÉʵÑ鱨¸æ.
¡¾Ìá³öÎÊÌâ¡¿Êìʯ»ÒÑùÆ·µÄÖ÷Òª³É·ÖÊÇʲô?
¡¾×÷³ö²ÂÏë¡¿¼×ͬѧ²ÂÏëÊÇCa(OH)2£»ÒÒͬѧ²ÂÏëÊÇCaCO3;ÄãÈÏΪÊÇ ________________¡£
¡¾ÊµÑéÓë½áÂÛ¡¿ÇëÄãͨ¹ýʵÑéÑéÖ¤ÄãµÄ²ÂÏë:
ʵÑé²½Öè | ʵÑéÏÖÏó | ʵÑé½áÂÛ |
È¡ÉÙÁ¿µÄÊìʯ»ÒÑùÆ·ÓÚÉÕ±ÖУ¬¼ÓÊÊÁ¿µÄË®³ä·ÖÈܽâºó¹ýÂË¡£ ÏòÂËÒºÖмÓÈëÊÊÁ¿_____ÈÜÒº; ÏòÂ˳ö¹ÌÌåÖмÓÈëÊÊÁ¿ _____. | _________£» _____________¡£ | ²ÂÏëÕýÈ· |
¡¾ÍØÕ¹ÓëÓ¦Óá¿Ð¡×éͬѧ·´Ë¼ÁËÊìʯ»Ò±äÖʵÄÔÒò£¬ÈÏʶµ½ÊµÑéÊÒÓ¦ _____ ±£´æÊìʯ»Ò.