ÀûÓû¯Ñ§Ô­Àí¿ÉÒÔ¶Ô¹¤³§ÅŷŵķÏË®¡¢·ÏÔüµÈ½øÐÐÓÐЧ¼ì²â£®Ä³¹¤³§¶ÔÖƸõ¹¤ÒµÎÛÄàÖÐCr£¨¢ó£©»ØÊÕÓëÔÙÀûÓù¤ÒÕÈçÏ£¨ÁòËá½þÒºÖнðÊôÀë×ÓÖ÷ÒªÊÇCr3+£¬Æä´ÎÊÇFe3+£¬Fe2+£¬Al3+£¬Ca2+£¬Mg2+£©

³£ÎÂϲ¿·ÖÑôÀë×ÓµÄÇâÑõ»¯ÎïÐγɳÁµíʱÈÜÒºµÄpH¼ûÏÂ±í£º
ÑôÀë×ÓFe3+Fe2+Mg2+Al3+Ca2+Cr3+
¿ªÊ¼³ÁµíʱµÄpH1.97.09.64.29.7-
³ÁµíÍêȫʱµÄpH3.29.011.18.011.79.0£¨£¾9.0Èܽ⣩
£¨1£©Ëá½þʱ£¬ÎªÁËÌá¸ß½þÈ¡ÂʿɲÉÈ¡µÄ´ëÊ©ÊÇ
 
£¨ÖÁÉÙÒ»Ìõ£©
£¨2£©µ÷pH=4.0ÊÇΪÁ˳ýÈ¥
 
£¨ÌîFe3+£¬Al3+£¬Ca2+£¬Mg2+£©
£¨3£©ÄÆÀë×Ó½»»»Ê÷Ö¬µÄÔ­ÀíΪMn++n NaR¡úMRn+nNa+£¬±»½»»»µÄÔÓÖÊÀë×ÓÊÇ
 
£¨ÌîFe3+£¬Al3+£¬Ca2+£¬Mg2+£©
£¨4£©ÊÔÅäƽÏÂÁÐÑõ»¯»¹Ô­·´Ó¦·½³Ìʽ£º
 
Na2Cr2O7+
 
SO2+
 
H2O=
 
Cr£¨OH£©£¨H2O£©5SO4+
 
Na2SO4£®
¿¼µã£º½ðÊôµÄ»ØÊÕÓë»·¾³¡¢×ÊÔ´±£»¤,Ñõ»¯»¹Ô­·´Ó¦·½³ÌʽµÄÅäƽ,ÄÑÈܵç½âÖʵÄÈܽâƽºâ¼°³Áµíת»¯µÄ±¾ÖÊ
רÌ⣺µçÀëƽºâÓëÈÜÒºµÄpHרÌâ
·ÖÎö£º£¨1£©Ëá½þʱ£¬ÎªÁËÌá¸ß½þÈ¡ÂʿɲÉÈ¡µÄ´ëÊ©ÊÇÑÓ³¤½þȡʱ¼ä¡¢¼Ó¿ìÈܽâËٶȵȴëÊ©£»ÁòËá½þÈ¡ÒºÖеĽðÊôÀë×ÓÖ÷ÒªÊÇCr3+£¬Æä´ÎÊÇFe3+¡¢Al3+¡¢Ca2+ºÍMg2+£¬Ëá½þÊÇÈܽâÎïÖÊΪÁËÌá¸ß½þÈ¡ÂÊ£¬¿ÉÒÔÉý¸ßζÈÔö´óÎïÖÊÈܽâ¶È£¬Ôö´ó½Ó´¥Ãæ»ýÔö´ó·´Ó¦ËÙÂÊ£¬»ò¼Ó¿ì½Á°èËٶȵȣ»
£¨2£©ÁòËá½þÈ¡ÒºÖеĽðÊôÀë×ÓÖ÷ÒªÊÇCr3+£¬Æä´ÎÊÇFe3+¡¢Al3+¡¢Ca2+ºÍMg2+£¬¼ÓÈëNaOHÈÜҺʹÈÜÒºµ÷½ÚpH=4£¬Fe3+ת»¯Îª³Áµí³ýÈ¥£»
£¨3£©ÄÆÀë×Ó½»»»Ê÷Ö¬½»»»µÄÀë×ÓÊǸÆÀë×ÓºÍþÀë×Ó£»
£¨4£©ÒÀ¾ÝÁ÷³ÌͼÖеÄת»¯¹ØϵºÍ²úÎ½áºÏ¶þÑõ»¯ÁòµÄ»¹Ô­ÐÔ£¬ÀûÓÃÑõ»¯»¹Ô­·´Ó¦Ô­ÀíÅжϣ®
½â´ð£º ½â£º£¨1£©Ëá½þʱ£¬ÎªÁËÌá¸ß½þÈ¡ÂʿɲÉÈ¡µÄ´ëÊ©ÊÇ£ºÔö¼Ó½þȡʱ¼ä¡¢²»¶Ï½Á°è»ìºÏÎï¡¢ÂËÔü¶à´Î½þÈ¡µÈ£¬
¹Ê´ð°¸Îª£ºÉý¸ßζȣ¬½Á°è£¬¹ýÂ˺óÔÙÏòÂËÔüÖмÓÈëH2SO4 £¨¶à´Î½þÈ¡£©£¬Êʵ±ÑÓ³¤½þȡʱ¼ä£»
£¨2£©ÁòËá½þÈ¡ÒºÖеĽðÊôÀë×ÓÖ÷ÒªÊÇCr3+£¬Æä´ÎÊÇFe3+¡¢Al3+¡¢Ca2+ºÍMg2+£¬¼ÓÈëNaOHÈÜҺʹÈÜÒº³Ê¼îÐÔ£¬ÈÜÒºPH=4£¬Fe3+¡¢Al3+ת»¯Îª³Áµí³ýÈ¥£»
¹Ê´ð°¸Îª£ºFe3+£»
£¨3£©ÄÆÀë×Ó½»»»Ê÷Ö¬½»»»µÄÀë×ÓÊǸÆÀë×ÓºÍþÀë×Ó£¬ÒòΪÔÚ´Ë֮ǰ£¬Fe3+±»³ýÈ¥£¬Al3+ת»¯ÎªÆ«ÂÁËá¸ùµÄÐÎʽ£¬¹Ê´ð°¸Îª£ºCa2+¡¢Mg2+£»
£¨4£©¶þÑõ»¯Áò¾ßÓл¹Ô­ÐÔ£¬±»ÂËÒº¢òÖÐͨ¹ýÀë×Ó½»»»ºóµÄÈÜÒºÖÐNa2CrO4Ñõ»¯ÎªÁòËᣬNa2CrO4Ñõ±»»¹Ô­ÎªCrOH£¨H2O£©5SO4£¬Ë®ÈÜÒºÖÐÉú³ÉÁòËá·´Ó¦Éú³ÉÁòËáÄÆ£¬ÒÀ¾ÝÔ­×ÓÊغã·ÖÎöÊéдÅäƽ£»Na2CrO4+3SO2+11H2O=2CrOH£¨H2O£©5SO4¡ý+Na2SO4£¬
¹Ê´ð°¸Îª£º1£»3£»11£»2£»1£®
µãÆÀ£º±¾Ì⿼²éÁËÀë×Ó·½³Ìʽ¡¢»¯Ñ§·½³ÌʽµÄÊéд¡¢ÎïÖʵķÖÀëµÈ֪ʶµã£¬ÄѶȽϴó£¬×¢Òâ»áÔËÓÃÈÜÒºµÄpHÖµ¶ÔÈÜÒºÖеÄÀë×Ó½øÐзÖÀ룬³ýÔÓµÄÔ­ÔòÊÇ£º³ýÈ¥ÔÓÖÊÇÒ²»Òý½øеÄÔÓÖÊ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÄÜÕýÈ·±íʾÏÂÁз´Ó¦µÄÀë×Ó·½³ÌʽµÄÊÇ£¨¡¡¡¡£©
A¡¢ÓÃÍ­×öµç¼«µçCuSO4½âÈÜÒº£º2Cu2++2H2O
 Í¨µç 
.
 
2Cu+O2¡ü+4H+
B¡¢ÔÚNH4Fe£¨SO4£©2ÈÜÒºÖУ¬µÎ¼ÓÉÙÁ¿Ba£¨OH£©2ÈÜÒº£º2NH4++SO42-+Ba2++2OH-¨T2NH3?H2O+BaSO4¡ý
C¡¢ÔÚNa2CO3ÈÜÒºÖÐÖðµÎµÎÈëÏ¡ÑÎËáÈÜÒº£¬·´Ó¦¿ªÊ¼½×¶Î£ºCO32-+H+¨THCO3-
D¡¢ÔÚNa2S2O3ÈÜÒºÖмÓÈëÏ¡ÁòË᣺2S2O32-+4H+¨TSO42-+3S¡ý+2H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÒÀ¾ÝÏà¹ØʵÑéµÃ³öµÄ½áÂÛÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢ÏòijÈÜÒºÖмÓÈëÏ¡ÑÎËᣬ²úÉúÄÜʹʯ»ÒË®±ä»ë×ǵÄÆøÌ壬¸ÃÈÜÒºÒ»¶¨º¬ÓÐCO32-
B¡¢ÏòijÈÜÒºÖеμÓÂÈË®ºó£¬µÎ¼ÓKSCNÈÜÒº£¬ÈÜÒº³ÊºìÉ«£¬¸ÃÈÜÒºÖÐÒ»¶¨º¬ÓÐFe2+
C¡¢ÏòijÈÜÒºÖмÓÈëÂÈ»¯±µÈÜÒº£¬²úÉú²»ÈÜÓÚÏ¡ÑÎËá°×É«³Áµí£¬¸ÃÈÜÒºÒ»¶¨º¬ÓÐSO42-
D¡¢ÍùijÈÜÒºÖмÓÈëNaOH£¬Î¢ÈÈ£¬²úÉúÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壬¸ÃÈÜÒºÖÐÒ»¶¨º¬ÓÐNH4+

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÐÅϢʱ´ú²úÉúµÄ´óÁ¿µç×ÓÀ¬»ø¶Ô»·¾³Ôì³ÉÁ˼«´óµÄÍþв£®Ä³¡°±ä·ÏΪ±¦¡±Ñ§Éú̽¾¿Ð¡×齫һÅú·ÏÆúµÄÏß·°å¼òµ¥´¦Àíºó£¬µÃµ½º¬70%Cu¡¢25%Al¡¢4%Fe¼°ÉÙÁ¿Au¡¢PtµÈ½ðÊôµÄ»ìºÏÎ²¢Éè¼Æ³öÈçÏÂÖƱ¸ÁòËáÍ­¾§ÌåµÄ·Ïߣº

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©µÚ¢Ù²½CuÓëËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ
 
£»µÃµ½ÂËÔü1µÄÖ÷Òª³É·ÖΪ
 
£®
£¨2£©µÚ¢Ú²½¼ÓÈëH2O2µÄ×÷ÓÃÊÇ
 
£¬Ê¹ÓÃH2O2µÄÓŵãÊÇ
 
£»µ÷½ÚpHµÄÄ¿µÄÊÇʹ
 
Éú³É³Áµí£®
£¨3£©Óõڢ۲½ËùµÃCuSO4?5H2OÖƱ¸ÎÞË®CuSO4µÄ·½·¨ÊÇ
 
£®
£¨4£©ÓÉÂËÔü2ÖÆÈ¡Al2£¨SO4£©3?18H2O£¬Ì½¾¿Ð¡×éÉè¼ÆÁËÈýÖÖ·½°¸£º

ÉÏÊöÈýÖÖ·½°¸ÖУ¬
 
·½°¸²»¿ÉÐУ»ÈýÖÖ·½°¸Öоù²ÉÓÃÁËÕô·¢£¬½øÐиòÙ×÷ʱעÒâ
 
ʱֹͣ¼ÓÈÈ£®
£¨5£©Ì½¾¿Ð¡×éÓõζ¨·¨²â¶¨CuSO4?5H2O£¨Mr=250£©º¬Á¿£®È¡a g ÊÔÑùÅä³É100mLÈÜÒº£¬Ã¿´ÎÈ¡20.00mL£¬Ïû³ý¸ÉÈÅÀë×Óºó£¬ÓÃc mol/L EDTA£¨H2Y2-£©±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ƽ¾ùÏûºÄEDTAÈÜÒºb mL£®µÎ¶¨·´Ó¦ÈçÏ£ºCu2++H2Y2-¨TCuY2-+2H+£®£¨EDTA£ºÒÒ¶þ°·ËÄÒÒËá°×É«½á¾§·Ûĩ״¹ÌÌ壬³ÊËáÐÔ£®£©
ÅäÖÆEDTA£¨H2Y2-£©±ê×¼ÈÜҺʱ£¬ÏÂÁÐÒÇÆ÷²»±ØÒªÓõ½µÄÓÐ
 
£¨ÓñàºÅ±íʾ£©£®
¢Ùµç×ÓÌìƽ  ¢ÚÉÕ±­   ¢ÛÁ¿Í²   ¢Ü²£Á§°ô   ¢ÝÈÝÁ¿Æ¿   ¢Þ½ºÍ·µÎ¹Ü   ¢ßÒÆÒº¹Ü
д³ö¼ÆËãCuSO4?5H2OÖÊÁ¿·ÖÊýµÄ±í´ïʽ¦Ø=
 
£»ÏÂÁвÙ×÷»áµ¼ÖÂCuSO4?5H2Oº¬Á¿²â¶¨½á¹ûÆ«¸ßµÄÊÇ
 
£®
a£®Î´¸ÉÔï׶ÐÎÆ¿
b£®µÎ¶¨ÖÕµãʱµÎ¶¨¹Ü¼â×ìÖвúÉúÆøÅÝ
c£®Î´³ý¾²¿ÉÓëEDTA·´Ó¦µÄ¸ÉÈÅÀë×Ó£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÐËȤС×éµÄѧÉú¸ù¾Ý»îÆýðÊôMgÓëCO2·¢Éú·´Ó¦£¬ÍƲâ»îÆýðÊôÄÆÒ²ÄÜÓëCO2·¢Éú·´Ó¦£¬Òò´ËʵÑéС×éÓÃÏÂÁÐ×°ÖýøÐС°ÄÆÓë¶þÑõ»¯Ì¼·´Ó¦¡±µÄʵÑé̽¾¿£¨Î²Æø´¦Àí×°ÖÃÒÑÂÔÈ¥£©£®
ÒÑÖª£º³£ÎÂÏ£¬COÄÜʹһЩ»¯ºÏÎïÖеĽðÊôÀë×Ó»¹Ô­£®
ÀýÈ磺PdCl2+CO+H2O¨TPd¡ý+CO2+2HCl
·´Ó¦Éú³ÉºÚÉ«µÄ½ðÊôîÙ£¬´Ë·´Ó¦Ò²¿ÉÓÃÀ´¼ì²â΢Á¿COµÄ´æÔÚ£®

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Í¨³£ÊµÑéÊÒÖÆÈ¡CO2ÆøÌåµÄÀë×Ó·½³ÌʽÊÇ
 
£¬ÎªÁËʹÖÆÆø×°ÖÃÄÜ¡°Ë濪ËæÓã¬Ëæ¹ØËæÍ£¡±£¬ÉÏͼA´¦Ó¦Ñ¡ÓõÄ×°ÖÃÊÇ
 
£¨Ìîд¡°¢ñ¡±¡¢¡°¢ò¡±»ò¡°¢ó¡±£©£®ÈôÒªÖÆÈ¡¸ÉÔï¡¢´¿¾»µÄCO2ÆøÌ壬װÖÃBÖÐӦʢ·ÅµÄÊÔ¼ÁÊÇ
 
ÈÜÒº£¬×°ÖÃCÖÐӦʢ·ÅµÄÊÔ¼ÁÊÇ
 
£®
£¨2£©¹Û²ìʵÑé×°ÖÃͼ¿ÉÖªNaÓëCO2·´Ó¦µÄÌõ¼þÊÇ
 
£¬¼ì²é×°ÖõÄÆøÃÜÐÔÍêºÃ²¢×°ÈëÒ©Æ·ºó£¬ÔÚµãȼ¾Æ¾«ÅçµÆÇ°£¬´ò¿ª¢óÖеÄֹˮ¼Ð£®´ý×°ÖÃ
 
£¨Ìîд×Öĸ£©ÖгöÏÖ
 
ÏÖÏóʱ£¬ÔÙµãȼ¾Æ¾«ÅçµÆ£¬Õâ²½²Ù×÷µÄÄ¿µÄÊÇ
 
£®
£¨3£©¼ÙÉèCO2ÆøÌåΪ×ãÁ¿£¬ÔÚʵÑé¹ý³ÌÖзֱð²úÉúÒÔÏ¢١¢¢ÚÁ½ÖÖ²»Í¬Çé¿ö£¬Çë·ÖÎö²¢»Ø´ðÎÊÌ⣺
¢ÙÈô×°ÖÃFÖÐÈÜÒºÎÞÃ÷ÏԱ仯£¬×°ÖÃDÖÐÉú³ÉÁ½ÖÖ¹ÌÌåÎïÖÊ£¬ÔòÄÆÓë¶þÑõ»¯Ì¼·´Ó¦µÄÉú³ÉÎïÊÇNa2CO3ºÍ
 
£®
¢ÚÈô×°ÖÃFÖÐÓкÚÉ«³ÁµíÉú³É£¬×°ÖÃDÖÐÖ»Éú³ÉÒ»ÖÖ¹ÌÌåÎïÖÊ£¬ÔòÄÆÓë¶þÑõ»¯Ì¼·´Ó¦µÄÉú³ÉÎïÊÇNa2CO3ºÍ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ij»ìºÏÎïµÄË®ÈÜÒºÖУ¬Ö»¿ÉÄܺ¬ÓÐÒÔÏÂÀë×ÓÖеÄÈô¸ÉÖÖ£ºK+¡¢Ca2+¡¢NH4+¡¢Cl-¡¢CO32-ºÍSO42-£®ÏÖÿ´ÎÈ¡100.00mL½øÐÐʵÑ飺
¢ÙµÚÒ»·Ý¼ÓÈëAgNO3ÈÜÒºÓгÁµí²úÉú£»
¢ÚµÚ¶þ·Ý¼ÓÈë×ãÁ¿NaOHºó¼ÓÈÈ£¬ÊÕ¼¯µ½ÆøÌå0.896L£¨±ê×¼×´¿öÏ£©
¢ÛµÚÈý·Ý¼ÓÈë×ãÁ¿BaCl2ÈÜÒººóµÃ¸ÉÔï³Áµí6.27g£¬³Áµí¾­×ãÁ¿ÑÎËáÏ´µÓ£¬¸ÉÔïºóÊ£Óà2.33g£®
Çë»Ø´ð£º£¨1£©c£¨CO32-£©=
 
£®
£¨2£©K+ÊÇ·ñ´æÔÚ£¿
 
£»Èô´æÔÚ£¬Å¨¶È·¶Î§ÊÇ £¨Èô²»´æÔÚ£¬Ôò²»±Ø»Ø´ðµÚ2ÎÊ£©
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ij»¯Ñ§ÐËȤС×éÀûÓÃÒÔϸ÷×°ÖÃÁ¬½Ó³ÉÒ»ÕûÌ××°Öã¬Ì½¾¿ÂÈÆøÓë°±ÆøÖ®¼äµÄ·´Ó¦£®ÆäÖÐDΪ´¿¾»¸ÉÔïµÄÂÈÆøÓë´¿¾»¸ÉÔï°±Æø·´Ó¦µÄ×°Öã®

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Á¬½ÓºÃ×°Öú󣬱ØÐè½øÐеÄÒ»²½ÊµÑé²Ù×÷ÊÇ
 
£®
£¨2£©×°ÖÃEµÄ×÷ÓÃÊÇ
 
£¬Ï𽺹ÜkµÄ×÷ÓÃÊÇ
 
£®
£¨3£©´Ó×°ÖÃDµÄG´¦ÒݳöµÄβÆøÖпÉÄܺ¬ÓлÆÂÌÉ«µÄÓж¾ÆøÌ壬´¦Àí·½·¨ÊÇ
 
£®
£¨4£©×°ÖÃFÖÐÊÔ¹ÜÄÚ·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£®
£¨5£©½ÓÈëD×°ÖõÄÁ½¸ùµ¼¹Ü×ó±ß½Ï³¤¡¢Óұ߽϶̣¬Ä¿µÄÊÇ
 
£®
£¨6£©ÕûÌ××°ÖôÓ×óÏòÓÒµÄÁ¬½Ó˳ÐòÊÇ£¨j£©½Ó
 
½Ó£¨f£©£¨g£©½Ó
 
½Ó
 
½Ó£¨a£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ϊ²âÊÔNaHCO3ºÍNaCl»ìºÏÎïÖÐNaHCO3µÄº¬Á¿£¬Ä³¿ÎÍâ»î¶¯Ð¡×éÌá³öÏÂÃæÁ½ÖÖ·½°¸²¢½øÐÐÁËʵÑ飨ÒÔÏÂÊý¾ÝΪ¶à´ÎƽÐÐʵÑé²â¶¨½á¹ûµÄƽ¾ùÖµ£©£º
·½°¸Ò»£º½«ag»ìºÏÎï¼ÓÈÈÒ»¶Îʱ¼ä£¬²âµÃÉú³É¸ÉÔïÆøÌåµÄÌå»ýΪ560mL£¨±ê×¼×´¿ö£©£»
·½°¸¶þ£º½«3ag»ìºÏÎïÍêÈ«ÈܽâÓÚ¹ýÁ¿Ï¡ÁòËáÖУ¬½«·´Ó¦ºóµÃµ½µÄÆøÌåÓüîʯ»Ò³ä·ÖÎüÊÕ£¬²âµÃ¼îʯ»ÒÔöÖØ6.93g£®
£¨1£©Ð´³öNaHCO3ÓëÏ¡ÁòËá·´Ó¦µÄÀë×Ó·½³Ìʽ
 
£®
£¨2£©ÔÚ·½°¸¶þÖУ¬
 
£¨Ìî¡°ÄÜ¡°»ò¡°²»ÄÜ¡°£©Ñ¡ÔñŨÑÎËᣬÀíÓÉÊÇ
 
£®
£¨3£©¸ù¾Ý·½°¸Ò»ºÍ·½°¸¶þ²â¶¨µÄ½á¹û¼ÆË㣬»ìºÏÎïÖÐNaHCO3µÄÖÊÁ¿·ÖÊýÒÀ´ÎΪ
 
ºÍ
 
£®
£¨4£©ÈôÅųýʵÑéÒÇÆ÷µÄÓ°ÏìÒòËØ£¬ÊÔ¶ÔÉÏÊöÁ½ÖÖ·½°¸²â¶¨½á¹ûµÄ׼ȷÐÔ×ö³öÅжϺͷÖÎö£®
¢Ù·½°¸Ò»
 
£¨Ì׼ȷ¡±¡°²»×¼È·¡±»ò¡°²»Ò»¶¨×¼È·¡±£©£¬ÀíÓÉÊÇ
 
£»
¢Ú·½°¸¶þ
 
£¨Ì׼ȷ¡±¡°²»×¼È·¡±»ò¡°²»Ò»¶¨×¼È·¡±£©£¬ÀíÓÉÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯ºÏÎïAX3ºÍµ¥ÖÊX2ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦¿ÉÉú³É»¯ºÏÎïAX5£®»Ø´ðÏÂÁÐÎÊÌ⣺
·´Ó¦AX3£¨g£©+X2£¨g£©?AX5£¨g£©ÔÚÈÝ»ýΪ10LµÄÃܱÕÈÝÆ÷ÖнøÐУ®ÆðʼʱAX3ºÍX2¾ùΪ0.2mol£®·´Ó¦ÔÚ²»Í¬Ìõ¼þϽøÐУ¬·´Ó¦Ìåϵ×ÜѹǿËæʱ¼äµÄ±ä»¯ÈçͼËùʾ£®
¢ÙÁÐʽ¼ÆËãʵÑéa´Ó·´Ó¦¿ªÊ¼ÖÁ´ïµ½Æ½ºâʱµÄ·´Ó¦ËÙÂÊ v£¨AX3£©=
 
£®
¢ÚͼÖÐ3×éʵÑé´Ó·´Ó¦¿ªÊ¼ÖÁ´ïµ½Æ½ºâʱµÄ·´Ó¦ËÙÂÊv£¨AX5£©ÓÉСµ½´óµÄ´ÎÐòΪ
 
£¨ÌîʵÑéÐòºÅ£©£»
¢ÛÓëʵÑéaÏà±È£¬ÆäËûÁ½×é¸Ä±äµÄʵÑéÌõ¼þ¼°ÅжÏÒÀ¾ÝÊÇ£ºb
 
£»c£®
 
£®
¢ÜÓÃp0±íʾ¿ªÊ¼Ê±×Üѹǿ£¬p±íʾƽºâʱ×Üѹǿ£¬¦Á±íʾAX3µÄƽºâת»¯ÂÊ£¬Ôò¦ÁµÄ±í´ïʽΪ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸