»¯ºÏÎïA¡¢BÊÇÖÐѧ³£¼ûµÄÎïÖÊ£¬ÆäÒõÑôÀë×Ó¿É´ÓϱíÖÐÑ¡Ôñ

ÑôÀë×Ó

K£« ¡¢Na£« ¡¢Fe2£« ¡¢Ba2£«¡¢NH

ÒõÀë×Ó

OH£­¡¢NO¡¢I£­¡¢HCO¡¢AlO¡¢HSO

(1)ÈôA¡¢BµÄË®ÈÜÒº¾ùΪÎÞÉ«£¬BµÄË®ÈÜÒº³Ê¼îÐÔ£¬ÇÒ»ìºÏºóÖ»²úÉú²»ÈÜÓÚÏ¡ÏõËáµÄ°×É«³Áµí¼°ÄÜʹºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌå¡£

¢ÙBµÄ»¯Ñ§Ê½Îª_________________________________________________________¡£

¢ÚA¡¢BÈÜÒº»ìºÏºó¼ÓÈȳÊÖÐÐÔ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ

________________________________________________________________________¡£

(2)ÈôAµÄË®ÈÜÒº³ÊdzÂÌÉ«£¬BµÄË®ÈÜÒºÎÞÉ«ÇÒÆäÑæÉ«·´Ó¦Îª»ÆÉ«¡£ÏòAµÄË®ÈÜÒºÖмÓÈëÏ¡ÑÎËáÎÞÃ÷ÏÔÏÖÏó£¬ÔÙ¼ÓÈëBºóÈÜÒº±ä»Æ£¬µ«A¡¢BµÄË®ÈÜÒº»ìºÏÒàÎÞÃ÷ÏԱ仯¡£Ôò

¢ÙAΪ___________________________________________________________¡£

¢Ú¾­·ÖÎöÉÏÊö¹ý³ÌÖÐÈÜÒº±ä»ÆµÄÔ­Òò¿ÉÄÜÓÐÁ½ÖÖ£º

¢ñ.________________________________________________________________________¡£

¢ò.________________________________________________________________________¡£

¢ÛÇëÓÃÒ»¼òÒ×·½·¨Ö¤Ã÷ÉÏÊöÈÜÒº±ä»ÆµÄÔ­Òò__________________________________¡£

¢ÜÀûÓÃÈÜÒº±ä»ÆÔ­Àí£¬½«ÆäÉè¼Æ³ÉÔ­µç³Ø£¬Èôµç×ÓÓÉaÁ÷Ïòb£¬Ôòb¼«µÄµç¼«·´Ó¦Ê½Îª

________________________________________________________________________

________________________________________________________________________¡£


½âÎö£º(1)¸ù¾ÝA¡¢BµÄÐÔÖʺͷ´Ó¦ÏÖÏó¿ÉÖª£¬AΪNH4HSO4£¬BΪBa(OH)2£¬¶þÕß1¡Ã1»ìºÏ·´Ó¦£¬¼ÓÈȺóÈÜÒºÏÔÖÐÐÔ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºH£«£«SO£«NH£«Ba2£«£«2OH£­BaSO4¡ý£«NH3¡ü£«2H2O¡£

(2)ÈôAµÄË®ÈÜÒº³ÊdzÂÌÉ«£¬ËµÃ÷AΪÑÇÌúÑΣ»BµÄË®ÈÜÒºÎÞÉ«ÇÒÆäÑæÉ«·´Ó¦Îª»ÆÉ«£¬BΪÄÆÑΡ£ÏòAµÄË®ÈÜÒºÖмÓÈëÏ¡ÑÎËáÎÞÃ÷ÏÔÏÖÏó£¬ÔÙ¼ÓÈëBºóÈÜÒº±ä»Æ£¬ËµÃ÷BΪÏõËáÄÆ£»ÓÉA¡¢BµÄË®ÈÜÒº»ìºÏÒàÎÞÃ÷ÏԱ仯֪£¬AÖ»ÄÜÊÇFeI2¡£

´ð°¸£º(1)¢ÙBa(OH)2¡¡¢ÚH£«£«SO£«NH£«Ba2£«£«2OH£­BaSO4¡ý£«NH3¡ü£«2H2O[À´Ô´:ѧ&¿Æ&Íø]

(2)¢ÙFeI2¡¡¢Ú¢ñ.½öÓÐI£­±»Ñõ»¯³ÉI2ʹÈÜÒº³Ê»ÆÉ«

¢ò.I£­¡¢Fe2£«¾ù±»Ñõ»¯Ê¹ÈÜÒº³Ê»ÆÉ«

¢ÛÈ¡ÉÙÁ¿±ä»ÆÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó¼¸µÎKSCNÈÜÒº£¬Èô±äºìÔò¢òºÏÀí(ÆäËûºÏÀí´ð°¸Òà¿É)

¢ÜNO£«4H£«£«3e£­===NO¡ü£«2H2O


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


±ê×¼×´¿öÏ£¬½«a L SO2ºÍCl2×é³ÉµÄ»ìºÏÆøÌåͨÈë100 mL 0.1 mol¡¤L£­1Fe2(SO4)3ÈÜÒºÖУ¬³ä·Ö·´Ó¦ºó£¬ÈÜÒºµÄ×Ø»ÆÉ«±ädz¡£Ïò·´Ó¦ºóµÄÈÜÒºÖмÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬½«ËùµÃ³Áµí¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïºó³ÆÖØ£¬ÆäÖÊÁ¿Îª11.65 g¡£ÔòÏÂÁйØÓڸùý³ÌµÄÍƶϲ»ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®ËùµÃ³ÁµíΪ0.05 molµÄBaSO4

B£®»ìºÏÆøÌåÖÐSO2µÄÌå»ýΪ0.448 L

C£®a L»ìºÏÆøÌåµÄÎïÖʵÄÁ¿Îª0.04 mol

D£®aµÄÈ¡Öµ·¶Î§Îª0.672<a<0.896

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁжÔÂÈÆøÐÔÖʵÄÐðÊö²»ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®»ÆÂÌÉ«                  B£®ÄÜÈÜÓÚË®

C£®Ò×Òº»¯                  D£®²»Ö§³ÖȼÉÕ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


(1)ÓÎÓ¾³ØË®Öеĺ¬ÂÈÆøÁ¿Ó¦¸Ã¿ØÖÆÔÚ0.5 mg¡¤L£­1ÖÁ1.0 mg¡¤L£­1Ö®¼ä£¬Ð´³öÂÈÆøÈÜÓÚË®µÄ»¯Ñ§·½³Ìʽ

________________________________________________________________________¡£

ÈÜÓÚË®ºó²úÉúµÄ________¿Éɱ¾úÏû¶¾¡£

(2)ÏÂͼÏÔʾһÐÇÆÚÖÐÿÌì19ʱʱӾ³ØË®ÖеÄÂÈÆøº¬Á¿£¬Äļ¸ÌìʹÓÃÓ¾³Ø²»°²È«________________________________________________________________________¡£

(3)ÄãÈÏΪÄļ¸ÌìµÄÌìÆø»ðÈÈ¡¢Ñô¹âÇ¿ÁÒ________£¬Ëµ³öÒ»ÖÖÀíÓÉÊÇ______________¡£

(4)Èô°ÑһƬ×ÏÉ«µÄ»¨°ê·ÅÈëÂÈË®ÖУ¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ____________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÒ´¼ÓëÖظõËá¸ùÀë×ÓÔÚËáÐÔÈÜÒºÖÐÄÜ·¢ÉúÈçÏ·´Ó¦£ºC2H5OH£«Cr2O£«H£«¨D¡úCO2¡ü£«Cr3£«£«H2O£¬µ±Õâ¸ö·½³ÌʽÅäƽºó£¬H£«µÄ»¯Ñ§¼ÆÁ¿ÊýΪ(¡¡¡¡)

A£®10                               B£®12

C£®14                               D£®16

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¼×¡¢ÒÒ¡¢±ûÊÇÈýÖÖ²»º¬ÏàͬÀë×ӵĿÉÈÜÐÔÇ¿µç½âÖÊ¡£ËüÃÇËùº¬Àë×ÓÈçϱíËùʾ£º

ÑôÀë×Ó

NH¡¢Na£«¡¢Mg2£«

ÒõÀë×Ó

OH£­¡¢NO¡¢SO

È¡µÈÖÊÁ¿µÄÈýÖÖ»¯ºÏÎïÅäÖÆÏàͬÌå»ýµÄÈÜÒº£¬ÆäÈÜÖÊÎïÖʵÄÁ¿Å¨¶È£ºc(¼×)>c(ÒÒ)>c(±û)£¬ÔòÒÒÎïÖÊ¿ÉÄÜÊÇ(¡¡¡¡)

¢ÙMgSO4¡¡¢ÚNaOH¡¡¢Û(NH4)2SO4¡¡¢ÜMg(NO3)2¡¡¢ÝNH4NO3

A£®¢Ù¢Ú                             B£®¢Û¢Ü

C£®¢Û¢Ý                             D£®¢Ù¢Ý

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¸ßÌúËá¼Ø(K2FeO4)ÊÇÒ»ÖÖÐÂÐÍ¡¢¸ßЧ¡¢¶à¹¦ÄÜÂÌÉ«Ë®´¦Àí¼Á£¬±ÈCl2¡¢O2¡¢ClO2¡¢KMnO4Ñõ»¯ÐÔ¸üÇ¿£¬ÎÞ¶þ´ÎÎÛȾ£¬¹¤ÒµÉÏÊÇÏÈÖƵøßÌúËáÄÆ£¬È»ºóÔÚµÍÎÂÏ£¬Ïò¸ßÌúËáÄÆÈÜÒºÖмÓÈëKOHÖÁ±¥ºÍ£¬Ê¹¸ßÌúËá¼ØÎö³ö¡£

(1)¸É·¨ÖƱ¸¸ßÌúËá¼ØµÄÖ÷Òª·´Ó¦Îª£º2FeSO4 £« 6Na2O2===2Na2FeO4 £« 2Na2O £« 2Na2SO4 £« O2¡ü

¢Ù¸Ã·´Ó¦ÖеÄÑõ»¯¼ÁÊÇ________£¬»¹Ô­¼ÁÊÇ________£¬Ã¿Éú³É1 mol Na2FeO4תÒÆ________molµç×Ó¡£

¢Ú¼òҪ˵Ã÷K2FeO4×÷Ϊˮ´¦Àí¼ÁʱËùÆðµÄ×÷ÓÃ__________________________________

________________________________________________________________________¡£

(2)ʪ·¨ÖƱ¸¸ßÌúËá¼Ø(K2FeO4)µÄ·´Ó¦ÌåϵÖÐÓÐÁùÖÖÁ£×Ó£ºFe(OH)3¡¢ClO£­¡¢OH£­¡¢FeO¡¢Cl£­¡¢H2O¡£

¢Ùд³ö²¢Åäƽʪ·¨ÖƸßÌúËá¼Ø·´Ó¦µÄÀë×Ó·½³Ìʽ£º____________________________

________________________________________________________________________¡£

¢ÚÿÉú³É1 mol FeO תÒÆ________molµç×Ó£¬Èô·´Ó¦¹ý³ÌÖÐתÒÆÁË0.3 molµç×Ó£¬Ôò»¹Ô­²úÎïµÄÎïÖʵÄÁ¿Îª________mol¡£

¢ÛµÍÎÂÏ£¬ÔÚ¸ßÌúËáÄÆÈÜÒºÖмÓÈëKOHÖÁ±¥ºÍ¿ÉÎö³ö¸ßÌúËá¼Ø(K2FeO4)£¬ËµÃ÷ʲôÎÊÌâ________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


½«Ê¢ÓÐNH4HCO3·ÛÄ©µÄСÉÕ±­·ÅÈëÊ¢ÓÐÉÙÁ¿´×ËáµÄ´óÉÕ±­ÖС£È»ºóÏòСÉÕ±­ÖмÓÈëÑÎËᣬ·´Ó¦¾çÁÒ£¬´×ËáÖð½¥Äý¹Ì¡£Óɴ˿ɼû(¡¡¡¡)

A£® NH4HCO3ºÍÑÎËáµÄ·´Ó¦ÊÇ·ÅÈÈ·´Ó¦

B£®¸Ã·´Ó¦ÖУ¬ÈÈÄÜת»¯Îª²úÎïÄÚ²¿µÄÄÜÁ¿

C£®·´Ó¦ÎïµÄ×ÜÄÜÁ¿¸ßÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿

D£®·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºNH4HCO3£«HCl===NH4Cl£«CO2¡ü£«H2O¡¡¦¤H£½£«Q

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁйØÓÚÓ¦ÓÃÌ×¹ÜʵÑé×°Ö㨲¿·Ö×°ÖÃδ»­³ö£©½øÐеÄʵÑ飬ÐðÊö´íÎóµÄ£¨    £©

A£®ÀûÓü××°ÖÿÉÒÔÖÆÈ¡ÉÙÁ¿H2

B£®ÀûÓÃÒÒ×°ÖÿÉÒÔÑéÖ¤Na2O2ÓëË®·´Ó¦¼ÈÉú³ÉÑõÆø£¬ÓַųöÈÈÁ¿

C£®ÀûÓñû×°ÖÃÑéÖ¤KHCO3ºÍK2CO3µÄÈÈÎȶ¨ÐÔ£¬XÖÐÓ¦·ÅµÄÎïÖÊÊÇK2CO3

D£®ÀûÓö¡×°ÖÃÖÆÈ¡SO2£¬²¢¼ìÑéÆ仹ԭÐÔ£¬Ð¡ÊÔ¹ÜÖеÄÊÔ¼Á¿ÉΪËáÐÔKMnO4ÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸