A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ»¯ºÏÎÆäÖÐA¡¢B¡¢C¡¢D¡¢E¾ùÓɶÌÖÜÆÚÔªËØ×é³É£¬ÑæÉ«·´Ó¦¾ù³Ê»ÆÉ«£¬B¡¢C¡¢E¾ùÓÉÈýÖÖÔªËØ×é³É£¬B¡¢CµÄ×é³ÉÔªËØÏàͬ£¬ÇÒCµÄĦ¶ûÖÊÁ¿±ÈB´ó80g/molÇë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¹ÌÌ廯ºÏÎïAΪdz»ÆÉ«·ÛÄ©£¬¸Ã»¯ºÏÎïÖк¬ÓеĻ¯Ñ§¼üΪ
 
£¨ÌîÐòºÅ£©£®
A£®Àë×Ó¼ü    B£®¼«ÐÔ¹²¼Û¼ü    C£®·Ç¼«ÐÔ¹²¼Û¼ü    D£®Çâ¼ü
£¨2£©±íΪBÓëFʵÑéµÄ²¿·ÖÄÚÈÝ£º
  ÐòºÅÖ÷ҪʵÑé²½Ö輰ʵÑéÏÖÏó
¢ÙÔÚº¬ÓÐBµÄÈÜÒºÖУ¬¼ÓÈëÏ¡H2S04£¬²úÉúdz»ÆÉ«»ë×ǺÍÎÞÉ«Óд̼¤ÐÔÆøζµÄÆøÌ壮
¢Ú20 ml·ÐË®ÖеμÓFµÄ±¥ºÍÈÜÒº1¡«2ml£¬ËùµÃÒºÌå³ÊºìºÖÉ«
¢Û½«ÊµÑé¢ÚµÃµ½µÄºìºÖÉ«ÒºÌå¼ÓÈÈÕô·¢¡¢×ÆÉÕ£¬×îÖյõ½ºì×ØÉ«¹ÌÌå
д³öBÓëÏ¡H2S04·´Ó¦µÄÀë×Ó·½³Ìʽ£º
 
£»Ð´³ö¢ÚÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
 
£®
£¨3£©ÏÖÓÉ6ÖÖÁ£×ÓMn2+¡¢MnO4-¡¢H+¡¢H20¡¢X2Y82-£¨CÖк¬ÓеÄÒõÀë×Ó£©¡¢XY42-×é³ÉÒ»¸öÀë×Ó·½³Ìʽ£¬ÒÑÖªMn2+Ϊ»¹Ô­¼Á£¬µÃµ½1mol MnO4-ÐèÑõ»¯¼ÁµÄÎïÖʵÄÁ¿Îª
 
£®
£¨4£©»¯ºÏÎïDºÍE¿ÉÒÔÏ໥ת»¯£º£¬ÈôÓÐDºÍE?XH20µÄ»ìºÏÎï13.04g£¬¼ÓÈȵ½ÍêÈ«·´Ó¦ºó£¬ÆøÌå²úÎïͨ¹ýŨH2SO4ÔöÖØ3.42g£¬Ê£ÓàÆøÌåͨ¹ý¼îʯ»ÒÔöÖØ2.20g£¬Ôò»ìºÏÎïÖÐDµÄÖÊÁ¿Îª
 
£¬E?XH20µÄ»¯Ñ§Ê½Îª
 
£®
¿¼µã£ºÎÞ»úÎïµÄÍƶÏ
רÌ⣺ÍƶÏÌâ
·ÖÎö£º£¨1£©A¡¢B¡¢C¡¢D¡¢E¾ùÓɶÌÖÜÆÚÔªËØ×é³É£¬ÑæÉ«·´Ó¦¾ù³Ê»ÆÉ«£¬¾ùº¬ÓÐNaÔªËØ£¬¹ÌÌ廯ºÏÎïAΪdz»ÆÉ«·ÛÄ©£¬ÔòÊǹýÑõ»¯ÄÆ£»
£¨2£©¸ù¾Ý¢Ú¢Û¿ÉÖªFÊÇÂÈ»¯ÌúÈÜÒº£®¸ù¾Ý¢Ù¿ÉÖªBÊÇNaS2O3£»
£¨3£©Mn2+±»Ñõ»¯Éú³ÉMnO4-£¬»¯ºÏ¼ÛÉý¸ß5¸öµ¥Î»£®Ñõ»¯¼ÁÊÇX2Y82Ò»£¬Æ仹ԭ²úÎïÊÇSO42-£¬1molÑõ»¯¼ÁµÃµ½2molµç×Ó£¬¸ù¾Ýµç×ӵĵÃʧÊغã¼ÆËãÐèÒªÑõ»¯¼ÁµÄÎïÖʵģ»
£¨4£©¸ù¾Ýת»¯¿ÉÖªDÊÇ̼ËáÇâÄÆ£¬EΪ̼ËáÄÆ£¬Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿¾ÍÊÇË®µÄÖÊÁ¿£¬¼îʯ»ÒÔö¼ÓµÄÖÊÁ¿ÊÇCO2µÄÖÊÁ¿£¬¼ÆËãË®¡¢¶þÑõ»¯Ì¼µÄÎïÖʵÄÁ¿£¬¸ù¾Ý¶þÑõ»¯Ì¼¼ÆËã̼ËáÇâÄƵÄÎïÖʵÄÁ¿£¬½ø¶ø¼ÆËãE?XH2OµÄÖÊÁ¿£¬½áºÏÉú³ÉË®µÄÎïÖʵÄÁ¿¼ÆËãE?XH2OÖнᾧˮµÄÎïÖʵÄÁ¿£¬ÔÙ¼ÆËãE?XH2OÖÐ̼ËáÄƵÄÎïÖʵÄÁ¿£¬½ø¶øÈ·¶¨»¯Ñ§Ê½£®
½â´ð£º ½â£º£¨1£©A¡¢B¡¢C¡¢D¡¢E¾ùÓɶÌÖÜÆÚÔªËØ×é³É£¬ÑæÉ«·´Ó¦¾ù³Ê»ÆÉ«£¬¾ùº¬ÓÐNaÔªËØ£¬¹ÌÌ廯ºÏÎïAΪdz»ÆÉ«·ÛÄ©£¬ÔòÊǹýÑõ»¯ÄÆ£¬¹ýÑõ»¯ÄÆÖк¬ÓÐÀë×Ó¼üºÍ·Ç¼«ÐÔ¼ü£¬¹Ê´ð°¸Îª£ºAC£»
£¨2£©¸ù¾Ý¢Ú¢Û¿ÉÖªFÊÇÂÈ»¯ÌúÈÜÒº£®¸ù¾Ý¢Ù¿ÉÖªBÊÇNaS2O3£¬BºÍÁòËá·´Ó¦µÄ·½³ÌʽΪ£ºS2O32-+2H+=S¡ý+SO2¡ü+H2O£¬¢ÚÊôÓÚÇâÑõ»¯Ìú½ºÌåµÄÖƱ¸£¬·½³ÌʽΪ£ºFeCl3+3H2O
  ¡÷  
.
 
Fe£¨OH£©3£¨½ºÌ壩+3HCl£¬
¹Ê´ð°¸Îª£ºS2O32-+2H+=S¡ý+SO2¡ü+H2O£»FeCl3+3H2O
  ¡÷  
.
 
Fe£¨OH£©3£¨½ºÌ壩+3HCl£»
£¨3£©Mn2+±»Ñõ»¯Éú³ÉMnO4-£¬»¯ºÏ¼ÛÉý¸ß5¸öµ¥Î»£®Ñõ»¯¼ÁÊÇX2Y82Ò»£¬Æ仹ԭ²úÎïÊÇSO42-£¬1molÑõ»¯¼ÁµÃµ½2molµç×Ó£¬ËùÒÔ¸ù¾Ýµç×ӵĵÃʧÊغã¿ÉÖª£¬ÐèÒªÑõ»¯¼ÁµÄÎïÖʵÄÁ¿ÊÇ
1mol¡Á5
2
=2.5mol£¬¹Ê´ð°¸Îª£º2.5mol£»
£¨4£©¸ù¾Ýת»¯¿ÉÖªDÊÇ̼ËáÇâÄÆ£¬EΪ̼ËáÄÆ£¬Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿¾ÍÊÇË®µÄÖÊÁ¿£¬ÎïÖʵÄÁ¿ÊÇ
3.42g
18g/mol
=0.19mol£®¼îʯ»ÒÔö¼ÓµÄÖÊÁ¿ÊÇCO2µÄÖÊÁ¿£¬ËùÒÔCO2µÄÎïÖʵÄÁ¿ÊÇ
2.2g
44g/mol
=0.05mol£¬Ôò£º
2NaHCO3
  ¡÷  
.
 
Na2CO3+CO2¡ü+H2O
0.1mol        0.05mol  0.05mol
Òò´ËD£¨NaHCO3£©µÄÎïÖʵÄÁ¿ÊÇ0.1mol£¬ÖÊÁ¿ÊÇ0.1mol¡Á84g/mol=8.4g£®
ËùÒÔNa2CO3?XH2OµÄÖÊÁ¿ÊÇ13.04g-8.4g=4.64g£¬Òò´ËNa2CO3?XH2OÖнᾧˮµÄÎïÖʵÄÁ¿ÊÇ0.19mol-0.05mol=0.14mol£¬ÖÊÁ¿ÊÇ0.14mol¡Á18g/mol=2.52g£®ËùÒÔNa2CO3?XH2OÖÐNa2CO3µÄÖÊÁ¿ÊÇ4.64g-2.52g=2.12g£¬ÆäÎïÖʵÄÁ¿ÊÇ
2.12g
106g/mol
=0.02mol£¬¹Ên£¨Na2CO3£©£ºn£¨H2O£©=0.02mol£º0.14mol=1£º7£¬¼´E?XH20µÄ»¯Ñ§Ê½ÎªNa2C03?7H2O£¬
¹Ê´ð°¸Îª£º8.4g£»Na2C03?7H2O£®
µãÆÀ£º±¾Ì⿼²éÎÞ»úÎïÍƶϡ¢»¯Ñ§¼ü¡¢·½³ÌʽÊéд¡¢Ñõ»¯»¹Ô­·´Ó¦¡¢»ìºÏÎï¼ÆËãµÈ£¬ÄѶÈÖеȣ¬ÐèҪѧÉú¾ß±¸ÔúʵµÄ»ù´¡£¬ÊǶԻù´¡ÖªÊ¶µÄ×ÛºÏÓ¦Óã®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÈçͼËùʾװÖÃÊÇ»¯Ñ§ÊµÑéÖг£¼ûµÄÒÇÆ÷£¬Ëü³ýÓÃÓÚÏ´ÆøÍ⣬»¹ÓÐÆäËûÓÃ;£º
£¨1£©Ò½ÔºÀï¸ø²¡ÈËÊäÑõʱ£¬ÍùÍùÔÚÑõÆø¸ÖÆ¿Ó벡È˺ôÎüÃæ¾ßÖ®¼ä°²×°Ê¢ÓÐË®µÄ¸Ã×°Öã¬ÓÃÓÚ¹Û²ìÆøÅݲúÉúµÄÇé¿ö£¬ÒÔ±ãµ÷½Ú¹©ÑõËÙÂÊ£¬´ËʱÑõÆøÓ¦´Ó
 
£¨Ìî±êºÅ£¬ÏÂͬ£©¹Ü¿Úµ¼È룮
£¨2£©µ±ÓôË×°ÖÃÊÕ¼¯NOʱ£¬Ó¦²ÉÈ¡µÄÖ÷Òª²Ù×÷²½ÖèÊÇ£º
¢Ù
 
£»
¢Ú
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐʵÑé·½°¸ÖУ¬ÄܴﵽʵÑéÄ¿µÄÊÇ£¨¡¡¡¡£©
Ñ¡ÏîʵÑéÄ¿µÄʵÑé·½°¸
A¼ìÑéCH3CH2BrÔÚNaOHÈÜÒºÖÐÊÇ·ñ·¢ÉúË®½â½«CH3CH2BrÓëNaOHÈÜÒº¹²ÈÈ£®ÀäÈ´ºó£¬È¡³öÉϲãË®ÈÜÒº£¬¼ÓÈëAgNO3ÈÜÒº£¬¹Û²ìÊÇ·ñ²úÉúµ­»ÆÉ«³Áµí
B¼ìÑéFe£¨NO3£©2¾§ÌåÊÇ·ñÒÑÑõ»¯±äÖʽ«Fe£¨NO3£©2ÑùÆ·ÈÜÓÚÏ¡H2SO4ºó£¬µÎ¼ÓKSCNÈÜÒº£¬¹Û²ìÈÜÒºÊÇ·ñ±äºì
CÑéÖ¤Ñõ»¯ÐÔ£ºFe3+£¾I2½«KIºÍFeCl3ÈÜÒºÔÚÊÔ¹ÜÖлìºÏºó£¬Õñµ´£¬¾²Ö㮿ɹ۲ìÈÜÒºÑÕÉ«±ä»¯
DÑéÖ¤Fe£¨OH£©3µÄÈܽâ¶ÈСÓÚMg£¨OH£©2½«FeCl3ÈÜÒº¼ÓÈëMg£¨OH£©2Ðü×ÇÒºÖУ¬Õñµ´£¬¿É¹Û²ìµ½³ÁµíÓÉ°×É«±äΪºìºÖÉ«
A¡¢AB¡¢BC¡¢CD¡¢D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÌúË¿¡¢Í­Æ¬ÔÚÂÈÆøÖÐȼÉÕµÄÑ̵ÄÑÕÉ«·Ö±ðΪ£¨¡¡¡¡£©
A¡¢×ػơ¢×غì
B¡¢×ػơ¢²Ô°×
C¡¢×غ졢²Ô°×
D¡¢×غ졢×Ø»Æ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚNH4ClÓëNH3?H2O×é³ÉµÄ»ìºÏÒºÖУ¬ÏÂÁйØϵһ¶¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢c£¨NH4+£©£¾c£¨Cl-£©
B¡¢c£¨H+£©=c£¨OH-£©
C¡¢c£¨NH4+£©+c£¨NH3?H2O£©£¾c£¨Cl-£©
D¡¢c£¨NH4+£©£¾c£¨Cl-£©+c£¨OH-£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ñõ»¯ÂÁÓëÑÎËá·´Ó¦
 
£»Ñõ»¯ÂÁÓëÇâÑõ»¯ÄÆ·´Ó¦
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ͨ¹ý¶ÔʵÑéÏÖÏóµÄ¹Û²ì¡¢·ÖÎöÍÆÀíµÃ³öÕýÈ·µÄ½áÂÛÊÇ»¯Ñ§Ñ§Ï°µÄ·½·¨Ö®Ò»£®¶ÔÏÂÁÐʵÑéÊÂʵµÄ½âÊͲ»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
 ÏÖÏó½âÊÍ»ò½áÂÛ
ASO2ʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍËÉ«SO2±íÏÖ»¹Ô­ÐÔ
BŨHNO3ÔÚ¹âÕÕÌõ¼þϱä»ÆŨHNO3²»Îȶ¨£¬Éú³ÉÓÐÉ«²úÎïÄÜÈÜÓÚŨÏõËá
CijÈÜÒºÖмÓÈëŨNaOHÈÜÒº¼ÓÈÈ£¬·Å³öÆøÌåÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶¸ÃÈÜÒºÖÐÒ»¶¨º¬ÓÐNH4+
DÂÁƬ·ÅÈëŨÁòËáÖУ¬ÎÞÃ÷ÏԱ仯˵Ã÷ÂÁÓëÀäµÄŨÁòËá²»·¢Éú»¯Ñ§·´Ó¦
A¡¢AB¡¢BC¡¢CD¡¢D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ºÍÏÂÁÐÀë×Ó·´Ó¦·½³ÌʽÏà¶ÔÓ¦µÄ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢Zn2++2OH-=Zn£¨OH£©2¡ý     ZnCO3+2NaOH=Zn£¨OH£©2¡ý+Na2CO3
B¡¢Ba2++SO42-=BaSO4¡ý    Ba£¨OH£©2+H2SO4=BaSO4¡ý+2H2O
C¡¢Ag++Cl-=AgCl¡ý    AgNO3+NaCl=AgCl¡ý+NaNO3
D¡¢Cu+2Ag+=Cu2++2Ag¡ý   Cu+2AgCl=2Ag+CuCl2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÓлúÎïÊôÓÚÌþµÄÊÇ£¨¡¡¡¡£©
A¡¢ÒÒÏ©B¡¢ÒÒ´¼C¡¢ÒÒËáD¡¢Ïõ»ù±½

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸