³ÆÈ¡NaClºÍBaCl2µÄ¹ÌÌå»ìºÏÎï32.5 g£¬¼ÓÈë100 gÕôÁóË®£¬ÍêÈ«ÈܽâºóÏò¸Ã»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈëÖÊÁ¿·ÖÊýΪ10£¥µÄNa2SO4ÈÜÒº£¬·´Ó¦Éú³ÉBaSO4³ÁµíµÄÖÊÁ¿ÓëËù¼ÓÈëµÄNa2SO4ÈÜÒºµÄÖÊÁ¿¹ØϵÈçÏÂͼËùʾ£®ÊԻشðÏÂÁÐÎÊÌ⣺

(Ìáʾ£ºBaCl2£«Na2SO4£½BaSO4¡ý£«2NaCl)

(1)ÍêÈ«·´Ó¦ºóÉú³ÉBaSO4³Áµí________g£®

(2)Ç¡ºÃÍêÈ«·´Ó¦Ê±ÏûºÄNa2SO4ÈÜÒºµÄÖÊÁ¿ÊǶàÉÙ¿Ë£¿

(3)Ç¡ºÃÍêÈ«·´Ó¦Ê±ËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿(¾«È·µ½0.1£¥)

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?ÇìÑô£©³ÆÈ¡NaClºÍBaCl2µÄ¹ÌÌå»ìºÏÎï32¡¢5g£¬¼ÓÈë100gÕôÁóË®£¬ÍêÈ«ÈܽâºóÏò¸Ã»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈëÖÊÁ¿·ÖÊýΪ10%µÄNa2SO4ÈÜÒº£¬·´Ó¦Éú³ÉBaSO4³ÁµíµÄÖÊÁ¿ÓëËù¼ÓÈëµÄNa2SO4ÈÜÒºµÄÖÊÁ¿¹ØϵÈçͼËùʾ£®ÊԻشðÏÂÁÐÎÊÌ⣺£¨Ìáʾ£ºBaCl2+Na2SO4¨TBaSO4¡ý+2NaCl£©
£¨1£©ÍêÈ«·´Ó¦ºóÉú³ÉBaSO4³Áµí
23.3
23.3
g£®
£¨2£©Ç¡ºÃÍêÈ«·´Ó¦Ê±ÏûºÄNa2SO4ÈÜÒºµÄÖÊÁ¿ÊǶàÉÙ¿Ë£¿
£¨3£©Ç¡ºÃÍêÈ«·´Ó¦Ê±ËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿£¨¾«È·µ½0.1%£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?ºþ±±Ä£Ä⣩³ÆÈ¡NaClºÍBaCl2µÄ¹ÌÌå»ìºÏÎï49.3g£¬¼ÓÈë167gÕôÁóË®£¬ÍêÈ«ÈܽâºóÏò¸Ã»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈë142gµÄNa2SO4ÈÜÒº£¬Ç¡ºÃ·´Ó¦Éú³ÉBaSO4³ÁµíµÄÖÊÁ¿Îª23.3g£®£¨Ìáʾ£ºBaCl2+Na2SO4=BaSO4¡ý+2NaCl£©£®¼ÆËãÇ¡ºÃÍêÈ«·´Ó¦Ê±ËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2013Äê³õÖбÏÒµÉýѧ¿¼ÊÔ£¨¸ÊËàÇìÑô¾í£©»¯Ñ§£¨½âÎö°æ£© ÌâÐÍ£º¼ÆËãÌâ

³ÆÈ¡NaClºÍBaCl2µÄ¹ÌÌå»ìºÏÎï32¡¢5g£¬¼ÓÈë100gÕôÁóË®£¬ÍêÈ«ÈܽâºóÏò¸Ã»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈëÖÊÁ¿·ÖÊýΪ10%µÄNa2SO4ÈÜÒº£¬·´Ó¦Éú³ÉBaSO4³ÁµíµÄÖÊÁ¿ÓëËù¼ÓÈëµÄNa2SO4ÈÜÒºµÄÖÊÁ¿¹ØϵÈçͼËùʾ£®ÊԻشðÏÂÁÐÎÊÌ⣺£¨Ìáʾ£ºBaCl2+Na2SO4¨TBaSO4¡ý+2NaCl£©

£¨1£©ÍêÈ«·´Ó¦ºóÉú³ÉBaSO4³Áµí          g£®

£¨2£©Ç¡ºÃÍêÈ«·´Ó¦Ê±ÏûºÄNa2SO4ÈÜÒºµÄÖÊÁ¿ÊǶàÉÙ¿Ë£¿

£¨3£©Ç¡ºÃÍêÈ«·´Ó¦Ê±ËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿£¨¾«È·µ½0.1%£©

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2013½ì½­ËÕÑγdzõ¼¶ÖÐѧ¾ÅÄ꼶µÚ¶þѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º¼ÆËãÌâ

³ÆÈ¡NaClºÍBaCl2µÄ¹ÌÌå»ìºÏÎï32.5g£¬¼ÓÈë100gÕôÁóË®£¬ÍêÈ«ÈܽâºóÏò¸Ã»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈëÖÊÁ¿·ÖÊýΪ10%µÄNa2SO4ÈÜÒº£¬·´Ó¦Éú³ÉBaSO4³ÁµíµÄÖÊÁ¿ÓëËù¼ÓÈëµÄNa2SO4ÈÜÒºµÄÖÊÁ¿¹ØϵÈçÏÂͼËùʾ¡£ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©ÍêÈ«·´Ó¦ºóÉú³ÉBaSO4³Áµí     g¡£

£¨2£©Ç¡ºÃÍêÈ«·´Ó¦Ê±ÏûºÄNa2SO4ÈÜÒºµÄÖÊÁ¿ÊÇ      g¡£

£¨3£©Ç¡ºÃÍêÈ«·´Ó¦Ê±ËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿£¨¾«È·µ½0.1%£©       

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

³ÆÈ¡NaClºÍBaCl2µÄ¹ÌÌå»ìºÏÎï32.5g£¬¼ÓÈë100gÕôÁóË®£¬ÍêÈ«ÈܽâºóÏò¸Ã»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈëÖÊÁ¿·ÖÊýΪ10%µÄNa2SO4ÈÜÒº£¬·´Ó¦Éú³ÉBaSO4³ÁµíµÄÖÊÁ¿ÓëËù¼ÓÈëµÄNa2SO4ÈÜÒºµÄÖÊÁ¿¹ØϵÈçÏÂͼËùʾ¡£ÊԻشðÏÂÁÐÎÊÌ⣺

£¨Ìáʾ£ºBaCl2+Na2SO4=BaSO4¡ý+2NaCl£©

£¨1£©ÍêÈ«·´Ó¦ºóÉú³ÉBaSO4³Áµí      g¡££¨1·Ö£©

£¨2£©Ç¡ºÃÍêÈ«·´Ó¦Ê±ÏûºÄNa2SO4ÈÜÒºµÄÖÊÁ¿ÊǶàÉÙ¿Ë£¿£¨2·Ö£©

£¨3£©Ç¡ºÃÍêÈ«·´Ó¦Ê±ËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿£¨¾«È·µ½0.1%£©£¨3·Ö£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸