ijÁòË᳧·ÏË®Öк¬ÓÐÉÙÁ¿ÁòËᣬΪ´ï±êÅÅ·Å£¬¼¼ÊõԱСÕŶԷÏË®ÖÐÁòËáµÄº¬Á¿½øÐмì²â£®
£¨1£©ÅäÖÆÈÜÒº£º
ÓûÅäÖÆÈÜÖÊÖÊÁ¿·ÖÊýΪ4%µÄNaOHÈÜÒº100g£¬ÐèÒªNaOH¹ÌÌå¡¡¡¡g£¬Ë®¡¡¡¡ml£¨Ë®µÄÃܶÈΪ1g/cm3£©£»
£¨2£©¼ì²â·ÖÎö£º
È¡·ÏË®ÑùÆ·98g£¬ÏòÆäÖÐÖðµÎ¼ÓÈëNaOHÈÜÒºÖÁÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ÏûºÄ4%µÄNaOHÈÜÒº20g£®£¨¼ÙÉè·ÏË®ÖÐÆäËü³É·Ö¾ù²»ºÍNaOH·´Ó¦£»·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2NaOH+H2SO4═Na2SO4+2H2O£©
ÊÔ¼ÆËã·ÏË®ÖÐÁòËáµÄÖÊÁ¿·ÖÊý£¨Ð´³ö¼ÆËã¹ý³Ì£©£®
½â£º£¨1£©ÈÜÖÊÖÊÁ¿=ÈÜÒºÖÊÁ¿¡ÁÈÜÖÊÖÊÁ¿·ÖÊý=100g¡Á4%=4g£¬ËùÒÔÈܼÁÖÊÁ¿Îª£º100g﹣4g=96g£¬¹ÊÐèË®µÄÌå»ýΪ=96cm3£¬£¬1cm3=1ml£¬¹ÊÐèË®96ml£®
¹Ê´ð°¸Îª£º4£»96£»
£¨2£©ÇâÑõ»¯ÄƵÄÖÊÁ¿=20g¡Á4%=0.8g
ÉèÁòËáµÄÖÊÁ¿Îªx
H2SO4+2NaOH═Na2SO4+2H2O
98 80
x 0.8g
x=0.98g
·ÏË®ÖÐÁòËáµÄÖÊÁ¿·ÖÊý=¡Á100%=1.01%
´ð£º·ÏË®ÖÐÁòËáµÄÖÊÁ¿·ÖÊýΪ1.01%£®
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ijÁòË᳧·ÏË®Öк¬ÓÐÉÙÁ¿ÁòËᣬΪ´ï±êÅÅ·Å£¬¼¼ÊõԱСÕŶԷÏË®ÖÐÁòËáµÄº¬Á¿½øÐмì²â£®
£¨1£©ÅäÖÆÈÜÒº£º
ÓûÅäÖÆÈÜÖÊÖÊÁ¿·ÖÊýΪ4%µÄNaOHÈÜÒº100g£¬ÐèÒªNaOH¹ÌÌå¡¡¡¡g£¬Ë®¡¡¡¡ml£¨Ë®µÄÃܶÈΪ1g/cm3£©£»
£¨2£©¼ì²â·ÖÎö£º
È¡·ÏË®ÑùÆ·98g£¬ÏòÆäÖÐÖðµÎ¼ÓÈëNaOHÈÜÒºÖÁÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ÏûºÄ4%µÄNaOHÈÜÒº20g£®£¨¼ÙÉè·ÏË®ÖÐÆäËü³É·Ö¾ù²»ºÍNaOH·´Ó¦£»·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2NaOH+H2SO4═Na2SO4+2H2O£©
ÊÔ¼ÆËã·ÏË®ÖÐÁòËáµÄÖÊÁ¿·ÖÊý£¨Ð´³ö¼ÆËã¹ý³Ì£©£®
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2013Äê½ËÕÊ¡ËÕÖÝÊÐÖп¼»¯Ñ§Ä£ÄâÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com