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

(Ìáʾ£ºBaCl2+Na2SO4====BaSO4¡ý+2NaCl)

(1)ÍêÈ«·´Ó¦ºóÉú³ÉBaSO4³Áµí______________g¡£

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

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


±¾Ì⿼²é½áºÏͼÏñ½øÐл¯Ñ§·½³Ìʽ¼ÆËãµÄÄÜÁ¦¡£

(1)ÓÉÌâÖÐͼʾ¿ÉÖª£¬ÍêÈ«·´Ó¦ºóÉú³ÉBaSO4³ÁµíµÄÖÊÁ¿Îª23.3 g¡£

(2)¸ù¾ÝBaSO4µÄÖÊÁ¿¿ÉÇó³öÁòËáÄƵÄÖÊÁ¿£¬¸ù¾ÝÁòËáÄÆÈÜÒºµÄÈÜÖʵÄÖÊÁ¿·ÖÊý£¬¿ÉÇóµÃÁòËáÄÆÈÜÒºµÄÖÊÁ¿¡£

(3)Ç¡ºÃÍêÈ«·´Ó¦ºó£¬ÈÜÒºÖÐÈÜÖÊÊÇÂÈ»¯ÄÆ£¬¸ù¾Ý³ÁµíµÄÖÊÁ¿£¬¿ÉÇóµÃ·´Ó¦Éú³ÉµÄÂÈ»¯ÄƵÄÖÊÁ¿£¬¼ÓÉÏÔ­¹ÌÌåÖÐÂÈ»¯ÄƵÄÖÊÁ¿¼´Îª·´Ó¦ºóÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿£¬½øÒ»²½¿ÉÇóµÃÇ¡ºÃÍêÈ«·´Ó¦Ê±ËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý¡£

´ð°¸£º(1)23.3¡¡(2)142

(3)½â£ºÉèBaCl2µÄÖÊÁ¿Îªx£¬·´Ó¦Éú³ÉµÄNaClµÄÖÊÁ¿Îªy¡£

BaCl2+Na2SO4====BaSO4¡ý+2NaCl

208          233      117

x               23.3 g  y

==

x=20.8 g¡¡y=11.7 g

Ç¡ºÃÍêÈ«·´Ó¦Ê±£¬ÈÜÒºÖÐNaClµÄÖÊÁ¿Îª

11.7 g+(32.5 g-20.8 g)=23.4 g

NaClÈÜÒºÈÜÖʵÄÖÊÁ¿·ÖÊýΪ¡Á100%=9.3%

´ð£º(ÂÔ)


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔÚÈ˵ÄθҺÖÐ,ÑÎËá(HCl)µÄÖÊÁ¿·ÖÊý´óÔ¼ÊÇ0.45%¡«0.6%¡£Ëü¾ßÓÐÒÔÏÂÉúÀí¹¦ÄÜ:¢Ù´Ù½øθµ°°×øµÄ´ß»¯×÷ÓÃ,ʹµ°°×ÖÊË®½â¶ø±»ÈËÌåÎüÊÕ;¢ÚʹÌÇÀàÎïÖʽøÒ»²½Ë®½â;¢Ûɱ¾ú¡£

Çë½áºÏÉÏÊöÄÚÈݻشðÏÂÁÐÎÊÌâ:

(1)Èç¹ûÓÃAl(OH)3µÄ»ìºÏ¼ÁÖÎÁÆθËá¹ý¶à,Ôò·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ¡¡____¡£

(2)ÈËÀàʳÓÃÊÊÁ¿µÄNaClÓÐÀûÓÚ¡¡¡¡¡¡¡¡¡£

A.´Ù½øÑÎËá²úÉú¡¡¡¡¡¡¡¡  B.°ïÖúÏû»¯ºÍÔö½øʳÓû

C.ÖÎÁÆθËá¹ý¶à              D.ÖÎÁÆȱµâÐÔ¼××´ÏÙÖ×´ó

(3)ÌÇÀà¡¢¡¡¡¡¡¡¡¡¡¢ÓÍÖ¬¡¢Î¬ÉúËØ¡¢ÎÞ»úÑκÍË®ÊÇÈËÌåµÄÁù´óÓªÑøÎïÖÊ¡£ÔÚÕâÁù´óÓªÑøÎïÖÊÖÐ,²»¾­¹ýÏû»¯,¾Í¿ÉÒÔÔÚÏû»¯µÀÄÚÖ±½Ó±»ÈËÌåÎüÊÕµÄÓС¡_____¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ij¹¤ÒµÔ°ÇøÐÂÄ⽨һÁòË᳧£¬ÔÚÉú²úÁòËáµÄ¹ý³ÌÖлá²úÉúSO2£¬ÈçSO2Ö±½ÓÏò´óÆøÅÅ·Å£¬ÔòÒ×ÐγÉËáÓ꣬ËáÓêµÄpH<________£»Îª·ÀÖ¹SO2ÎÛȾ´óÆø£¬ÓÐÒÔÏ·½°¸ÄãÈÏΪ¿ÉÐеÄÊÇ________(ÌîÐòºÅ)¡£

¢Ù½¨Ôì¸ßÑÌ´Ñ£¬½«·ÏÆøÏò¸ß¿ÕÅÅ·Å

¢ÚÓ÷ÏNaOHÈÜÒºÎüÊÕSO2

¢ÛÓÃÊìʯ»ÒÎüÊÕSO2

¢ÜÓ÷ÏÁòËáÈÜÒºÎüÊÕSO2

ÇëÈÎÒâд³ö¿ÉÄÜ·¢ÉúµÄÒ»¸ö»¯Ñ§·½³Ìʽ£º_______________________________

_______________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÈçͼËùʾ£¬XÈÜÒºÓëAgNO3¡¢Na2CO3¡¢Na2SO4ÈýÖÖÈÜÒº·¢Éú·´Ó¦¾ùÉú³É°×É«³Áµí¡£ÔòX¿ÉÄÜÊÇÏÂÁÐÄÄÖÖÎïÖʵÄÈÜÒº(¡¡¡¡)

A.HNO3»òKNO3¡¡¡¡¡¡¡¡             B.HCl»òH2SO4

C.BaCl2»òCaCl2                       D.NaOH»òCa(OH)2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


º£Ë®ºÍ¿ÕÆøÊÇÖØÒªµÄ×ÔÈ»×ÊÔ´£¬´Óº£Ë®ºÍ¿ÕÆøÖзÖÀë³ö¸÷ÖֳɷֿɸüºÃµÄ·þÎñÈËÀà¡£

(1)ijС×éÉè¼Æ½«¿ÕÆøÒÀ´Î¾­¹ý¼×¡¢ÒÒ¡¢±ûÈý¸ö×°Öã¬×îÖÕ·ÖÀë³ö½Ï´¿µÄN2¡£

¼×ÖпÉÄÜ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________£»ÒÒÖÐŨÁòËáµÄ×÷ÓÃÊÇ________________£»±ûÖпÉÒÔ¼ÓÈëµÄÊÔ¼ÁΪ__________¡£

(2)º£Ë®Öи»º¬Ã¾ÔªËØ£¬ÏÂͼÊÇ´Óº£Ë®ÖÐÌáȡþµÄÁ÷³Ìͼ¡£Á÷³ÌͼÖÐA¡¢BµÄ»¯Ñ§Ê½ÒÀ´ÎΪ_____________¡¢______________¡£

(3)º£Ë®¾­¹ýÈÕɹºó£¬µÃµ½´ÖÑκÍʳÑεı¥ºÍÈÜÒº¡£ÏòһƿʳÑα¥ºÍÈÜÒºÖмÓÈëÒ»¿é¹æÔòµÄʳÑξ§Ì壬±£³ÖζȺÍË®µÄÖÊÁ¿¶¼²»±ä£¬Ò»¶Îʱ¼äºó£¬Ê³Ñξ§ÌåµÄÖÊÁ¿______(Ìî¡°²»±ä¡±»ò¡°±ä»¯¡±£¬ÏÂͬ)£¬Ê³Ñξ§ÌåµÄÐÎ×´________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁмÒÍ¥ÖеÄ×ö·¨²»ÄÜ·ÀÖ¹²Ëµ¶ÉúÐâµÄÊÇ(¡¡¡¡)

A.²Ëµ¶Ê¹ÓÃÍêºó£¬ÓÃˮϴ¾»£¬²¢²Á¸É

B.¾ÃÖò»ÓÃʱ²Ëµ¶±íÃæÍ¿Ò»²ãÖ²ÎïÓÍ

C.°Ñ²Ëµ¶´æ·ÅÔÚ³±ÊªµÄµØ·½

D.Óò»Ðâ¸Ö²Ëµ¶´úÌæÆÕͨ²Ëµ¶

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÊÂʵ²»ÄÜ˵Ã÷Zn±ÈAg»îÆõÄÊÇ(¡¡¡¡)

A.ZnÄÜÓëAgNO3ÈÜÒº·´Ó¦Öû»³öAg

B.ZnÓëÏ¡ÁòËá·´Ó¦£¬AgÔò²»ÄÜ

C.×ÔÈ»½çÖÐûÓÐÒÔµ¥ÖÊÐÎʽ´æÔÚµÄZn£¬¶øÓÐÒÔµ¥ÖÊÐÎʽ´æÔÚµÄAg

D.ZnµÄÈÛµãΪ420¡æ£¬AgµÄÈÛµãΪ962¡æ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐʵÀý²»ÊôÓÚÖкͷ´Ó¦µÄÊÇ(¡¡¡¡)

A.ÍÁÈÀËữºó¼ÓÈëÊìʯ»Ò¸ÄÁ¼

B.θËá·ÖÃÚ¹ý¶àµÄ²¡ÈË×ñÒ½Öö·þÓú¬ÓÐÇâÑõ»¯ÂÁµÄÒ©ÎïÒÔÖк͹ý¶àθËá

C.Îó涣ҧÈ˵ÄƤ·ô·ÖÃÚ³öÒÏËᣬÈç¹ûÍ¿º¬¼îÐÔÎïÖʵÄÒ©Ë®¾Í¿É¼õÇáÍ´Ñ÷

D.½ðÊô±íÃæÐâÊ´ºó£¬¿ÉÓÃÏ¡ÑÎËá½øÐÐÇåÏ´

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


½«Ò»¶¨ÖÊÁ¿µÄBa(OH)2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏÇ¡ºÃÍêÈ«·´Ó¦£¬Ïò·´Ó¦ºóµÄ»ìºÏÎïÖмÓÈëÏ¡ÑÎËᣬ²úÉúÆøÌåµÄÌå»ýÓë¼ÓÈëÏ¡ÑÎËáµÄÌå»ýµÄ¹ØϵÈçÓÒÏÂͼËùʾ£¬ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨    £©

A£®N µãʱ£¬ËùµÃÈÜÒºµÄpH=7

      B£®Qµãʱ£¬ËùµÃÈÜÒºÖеÄÈÜÖÊÖ»º¬ÓÐBaCl2

      C£®OÖÁP¶Î·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪNaOH+HCl=NaCl+H2O

      D£®PÖÁQ¶Î·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪBa(OH)2+2HCl=BaCl2+2H2O

MµÄÖÊÁ¿/ g

SµÄÖÊÁ¿/ g

M2SµÄÖÊÁ¿/ g

¼×

6.0

2.5

7.5

ÒÒ

7.0

1.5

7.5

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸