ÌâÄ¿ÄÚÈÝ

13£®Ä³Í¬Ñ§¶ÔÉúÌúµÄ×é³É½øÐÐÑо¿£®³ÆÈ¡ÉúÌúÑùÆ·40g£¬°Ñ500gÏ¡ÁòËá·Ö5´Î¼ÓÈëÑùÆ·ÖУ¬²âµÃʵÑéÊý¾Ý¼ûÏÂ±í£º£¨ÉúÌúÖеÄÔÓÖʲ»ÈÜÓÚË®£¬Ò²²»ºÍÏ¡ÁòËá·´Ó¦£©
ʵÑéÐòºÅ¼ÓÈëÏ¡ÁòËáµÄÖÊÁ¿/gÊ£Óà¹ÌÌåµÄÖÊÁ¿/g
µÚ1´Î10028.8
µÚ2´Î10017.6
µÚ3´Î1006.4
µÚ4´Î100m
µÚ5´Î1001.2
£¨1£©µÚ4´ÎÊ£Óà¹ÌÌåµÄÖÊÁ¿m=1.2g
£¨2£©¸ÃÉúÌúÑùÆ·ÖÐÌúµÄÖÊÁ¿·ÖÊýΪ97%£®
£¨3£©40gµÄÉúÌúÑùÆ·ÍêÈ«·´Ó¦ºóÉú³ÉÇâÆøµÄÖÊÁ¿ÊǶàÉÙ£¿£¨Çëд³ö¼ÆËãµÄ¹ý³Ì£©
£¨4£©ÔÚͼÖл­³ö40gÑùÆ·ÖвúÉúÆøÌåµÄÖÊÁ¿Óë¼ÓÈëÏ¡ÁòËáµÄÖÊÁ¿¹ØÏµ£®

·ÖÎö ÓɵÚ1¡¢2¡¢3´Î·´Ó¦¿ÉÖª£¬100gÏ¡ÁòËáÇ¡ºÃºÍ11.2gÌú·´Ó¦£¬ÓɵÚ4¡¢5´Î·´Ó¦¿ÉÖª£¬µÚ4´Î·´Ó¦µÄÌúµÄÖÊÁ¿ÊÇ5.2g£¬¼´m=6.4-5.2=1.2£»
ÌúºÍÏ¡ÁòËá·´Ó¦Éú³ÉÁòËáÑÇÌúºÍÇâÆø£¬¸ù¾Ý·´Ó¦µÄ»¯Ñ§·½³ÌʽºÍÌṩµÄÊý¾Ý¿ÉÒÔ¼ÆËãÌúµÄÖÊÁ¿ºÍÉú³ÉÇâÆøµÄÖÊÁ¿£¬½øÒ»²½¿ÉÒÔ¼ÆËãÉúÌúÖÐÌúµÄÖÊÁ¿·ÖÊý£®

½â´ð ½â£º£¨1£©µÚ4´Î·´Ó¦ÖУ¬Ìú²»×㣬¼´µÚ4´Î·´Ó¦ÖÐÌúÒѾ­ÍêÈ«·´Ó¦£¬Òò´ËµÚ4´ÎÊ£Óà¹ÌÌåµÄÖÊÁ¿m=1.2g£®
¹ÊÌ1.2£®
£¨2£©¸ÃÉúÌúÑùÆ·ÖÐÌúµÄÖÊÁ¿Îª£º40g-1.2g=38.8g£¬ÌúµÄÖÊÁ¿·ÖÊýΪ£º$\frac{38.8g}{40g}$¡Á100%=97%£¬
´ð£º¸ÃÉúÌúÑùÆ·ÖÐÌúµÄÖÊÁ¿·ÖÊýΪ97%£®
£¨3£©ÉèÉú³ÉÇâÆøµÄÖÊÁ¿Îªx£¬
Fe+H2SO4¨TFeSO4+H2¡ü£¬
56                          2
38.8g                     x
$\frac{56}{38.8g}$=$\frac{2}{x}$£¬
x=1.4g£¬
´ð£º40gµÄÉúÌúÑùÆ·ÍêÈ«·´Ó¦ºóÉú³ÉÇâÆøµÄÖÊÁ¿ÊÇ1.4g£®
£¨4£©100gÏ¡ÁòËáÇ¡ºÃºÍ11.2gÌú·´Ó¦£¬Éè38.8gÌúÇ¡ºÃºÍÖÊÁ¿ÎªyµÄÏ¡ÁòËá·´Ó¦£¬
$\frac{100}{11.2g}$=$\frac{y}{38.8g}$£¬
y=346.4g£¬
40gÑùÆ·ÖвúÉúÆøÌåµÄÖÊÁ¿Óë¼ÓÈëÏ¡ÁòËáµÄÖÊÁ¿¹ØÏµÈçÏÂͼËùʾ£º

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éѧÉúÔËÓüÙÉè·¨ºÍ»¯Ñ§·½³Ìʽ½øÐмÆËãºÍÍÆ¶ÏµÄÄÜÁ¦£¬¼ÆËãʱҪעÒâ¹æ·¶ÐÔºÍ׼ȷÐÔ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
4£®Ð¡ºÎ½øÐÐÑÎËáºÍÇâÑõ»¯ÄÆÖкͷ´Ó¦µÄʵÑ飬ÔÚÏòÉÕ±­ÖеÄÇâÑõ»¯ÄÆÈÜÒºµÎ¼ÓÁËÒ»¶¨Á¿µÄÏ¡ÑÎËáºó£¬²Å·¢ÏÖÍü¼ÇÁ˼ÓÈëָʾ¼Á£®ÎªÁËÅжϸ÷´Ó¦ËùµÃÈÜÒºµÄËá¼îÐÔ£¬Ëû½øÐÐÁËÈçÏÂ̽¾¿£º
¡¾Ìá³öÎÊÌ⡿СºÎҪ̽¾¿µÄÎÊÌâÊÇÑÎËáºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦ËùµÃÈÜÒºµÄËá¼îÐÔ£®
¡¾²ÂÏëÓë¼ÙÉè¡¿ËùµÃÈÜÒº¿ÉÄÜÏÔ¼îÐÔ£¬Ò²¿ÉÄÜÏÔËáÐÔ£¬»¹¿ÉÄÜÏÔÖÐÐÔ£®
ÈôÈÜÒºÏÔ¼îÐÔ£¬ÔòÈÜÒºÖÐʹÆäÏÔ¼îÐÔµÄÀë×ÓÊÇOH- £¨ÌîдÀë×Ó·ûºÅ£©£®Îª±ÜÃâ¼îÐÔÈÜÒºÎÛȾ»·¾³£¬Òª¶ÔËùµÃÈÜÒº½øÐгÁµí´¦Àí£¬Ð¡ºÎÏòÈÜÒºÖмÓÈëÂÈ»¯Ã¾ÖÁ²»ÔÚ²úÉú³Áµí£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽMgCl2+2NaOH=Mg£¨OH£©2¡ý+2NaCl£®
¡¾ÊµÑéÓë·ÖÎö¡¿Ð¡ºÎ´ÓÉÕ±­ÖÐÈ¡Á˲¿·Ö·´Ó¦ºóµÄÈÜÒºÖÃÓÚÒ»ÊÔ¹ÜÖУ¬²¢ÏòÊÔ¹ÜÖеμÓÁ˼¸µÎÎÞÉ«·Ó̪ÊÔÒº£¬Õðµ´£¬¹Û²ìµ½ÊÔÒº²»±äÉ«£¬ÓÚÊÇËûµÃ³öÁËÈÜÒºÏÔÖÐÐԵĽáÂÛ£®Ð¡ÕÅÈÏΪСºÎµÃ³öµÄ½áÂÛ²»ÕýÈ·£¬ÄãÈÏΪСÕŵÄÀíÓÉÊÇÑÎËá¹ýÁ¿Ê±£¬ÊÔÒºÒ²²»±äÉ«£®ÎªÁ˽øÒ»²½È·¶¨ÈÜÒºµÄËá¼îÐÔ£¬Ð¡ÕÅÈ¡ÊÔ¹ÜÖеÄÈÜÒº×öÒÔϼ¸¸öʵÑ飬ÄãÈÏΪ¿ÉÐеÄÊÇBD£®£¨ÈÜÓÚË®µÄ¶þÑõ»¯Ì¼ºöÂÔ²»¼Æ£©£®
A£®È¡Ñù£¬µÎ¼ÓÁòËáÄÆÈÜÒº£¬¹Û²ìÏÖÏó
B£®È¡Ñù£¬¼ÓÈëпÁ££®¹Û²ìÏÖÏó
C£®È¡Ñù£¬Í¨Èë¶þÑõ»¯Ì¼£¬¹Û²ìÏÖÏó
D£®È¡Ñù£¬µÎ¼ÓÇâÑõ»¯ÄÆÈÜÒº£¬¹Û²ìÏÖÏó
¡¾½»Á÷ÍØÕ¹¡¿µÚ¶þÌ죬СÃ÷Ò²×öͬÑùµÄʵÑ飮µ±ËûÏòÇâÑõ»¯ÄÆÈÜÒº¼ÓÏ¡ÑÎËáʱ£¬·¢ÏÖÈÜÒºÓÐÆøÅÝð³ö£¬ÇëÄãд³ö²úÉúÆøÅݵĻ¯Ñ§·½³ÌʽNa2CO3+2HCl=2NaCl+H2O+CO2¡ü£®

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

¾«Ó¢¼Ò½ÌÍø