ÌâÄ¿ÄÚÈÝ

£¨8·Ö£©ÎªÌ½¾¿Ëá¼îÑÎÖ®¼äµÄ·´Ó¦£¬Ð¡ºìͬѧ×öÁËÒÔÏÂʵÑ飺ÔÚÏ¡ÁòËáÖмÓÈëÁËÒ»¶¨Á¿µÄÏõËá±µÈÜÒº¡£Çë»Ø´ð£º
£¨1£©¹Û²ìµ½µÄʵÑéÏÖÏóÊÇ              £¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                    £¬·ÖÎöʵÑéÏÖÏó£¬ÈÜÒºÖмõÉÙµÄÀë×ÓÊÇ            £¨Ð´Àë×Ó·ûºÅ£©¡£
£¨2£©Ð¡ºìͬѧÔÚÏ¡ÁòËáÖмÓÈëÏõËá±µÈÜҺʱ£¬ÏõËá±µÈÜÒº¿ÉÄܹýÁ¿£¬¼ìÑéÈÜÒºÖк¬ÓÐÉÙÁ¿
ÏõËá±µµÄ·½·¨ÊÇ                                                    ¡£
£¨3£©ÔÚ98gÖÊÁ¿·ÖÊýΪ10%µÄÏ¡ÁòËáÖмÓÈë100gÏõËá±µÈÜÒº£¬Ç¡ºÃÍêÈ«·´Ó¦£¬Ëù¼ÓÏõËá±µÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýÊÇ           £¨¼ÆËã½á¹û¾«È·µ½0£®1%£©¡£
£¨4£©ÈôÒªÅäÖÆ98gÖÊÁ¿·ÖÊýΪ10%µÄÏ¡ÁòËᣬÐèÒªÖÊÁ¿·ÖÊýΪ98%µÄŨÁòËᣨÃܶÈΪ
1£®84g/cm3£©       mL£¨¼ÆËã½á¹û¾«È·µ½0£®1£©¡£ÔÚʵÑéÊÒÓÃŨÁòËáÅäÖÆÏ¡ÁòËáµÄÖ÷Òª²½ÖèÓУº¼ÆËã¡¢            ¡¢»ìÔÈ¡¢ÀäÈ´ÖÁÊÒÎÂ×°Æ¿£¬²¢ÌùÉϱêÇ©¡£
 £¨1£©²úÉú°×É«³Áµí£¬H2SO4 + Ba(NO3)2 = BaSO4¡ý+2HNO3 ,  Ba2+ SO42-
£¨2£©ÔÚÏ¡ÁòËáÈÜÒºÖмÓÈëÏõËá±µÈÜÒººó£¬¾²Öã¬È¡ÉϲãÇåÒºÓÚÊÔ¹ÜÖУ¬µÎÈëÏ¡ÁòËᣬÈôÓа×É«³Áµí£¬ÈÜÒºÖк¬ÓÐÉÙÁ¿ÏõËá±µ£¨ºÏÀí·½·¨¾ù¿É£©£¨3£©26£®1%£¨4£©5£®4 ¡£Á¿È¡
 
ÊÔÌâ·ÖÎö£º£¨1£©¹Û²ìµ½µÄʵÑéÏÖÏóÊDzúÉú°×É«³Áµí£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪH2SO4 + Ba(NO3)2 = BaSO4¡ý+2HNO3·ÖÎöʵÑéÏÖÏó£¬ÓÉÓÚÉú³ÉÁòËá±µ³Áµí£¬¹ÊÈÜÒºÖмõÉÙµÄÀë×ÓÊÇBa2+ SO42-£¨2£©Ð¡ºìͬѧÔÚÏ¡ÁòËáÖмÓÈëÏõËá±µÈÜҺʱ£¬ÏõËá±µÈÜÒº¿ÉÄܹýÁ¿£¬¼ìÑéÈÜÒºÖк¬ÓÐÉÙÁ¿ÏõËá±µµÄ·½·¨ÊÇÔÚÏ¡ÁòËáÈÜÒºÖмÓÈëÏõËá±µÈÜÒººó£¬¾²Öã¬È¡ÉϲãÇåÒºÓÚÊÔ¹ÜÖУ¬µÎÈëÏ¡ÁòËᣬÈôÓа×É«³Áµí£¬ÈÜÒºÖк¬ÓÐÉÙÁ¿ÏõËá±µ£¨ºÏÀí·½·¨¾ù¿É£©£¨3£©ÔÚ98gÖÊÁ¿·ÖÊýΪ10%µÄÏ¡ÁòËáÖмÓÈë100gÏõËá±µÈÜÒº£¬Ç¡ºÃÍêÈ«·´Ó¦£¬ÉèËù¼ÓÏõËá±µÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪX
H2SO4 + Ba(NO3)2 = BaSO4¡ý+2HNO3
261
98g¡Á10%    X¡Á100g
98:261="98g¡Á10%" £ºX¡Á100g
X=26£®1%
£¨4£©ÉèÅäÖÆ98gÖÊÁ¿·ÖÊýΪ10%µÄÏ¡ÁòËᣬÐèÒªÖÊÁ¿·ÖÊýΪ98%µÄŨÁòËáµÄÌå»ýΪX
¸ù¾ÝÈÜҺϡÊÍÇ°ºóÈÜÖʵÄÖÊÁ¿²»±äµÃ£º
98g¡Á10%=1£®84g/cm3¡ÁX¡Á98%
X=5£®4ml  
ÔÚʵÑéÊÒÓÃŨÁòËáÅäÖÆÏ¡ÁòËáµÄÖ÷Òª²½ÖèÓУº¼ÆËã¡¢Á¿È¡¡¢»ìÔÈ¡¢ÀäÈ´ÖÁÊÒÎÂ×°Æ¿£¬²¢ÌùÉϱêÇ©¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø