¸ßÌúËáÄÆ(Na2FeO4)ÊÇÒ»ÖÖÐÂÐÍÏû¶¾¼Á£¬¹¤ÒµÉÏÖƱ¸¸ßÌúËáÄÆÓÐÏÂÁÐËÄÖÖ·½·¨£º

a£®2Fe(OH)3£«3NaClO£«4NaOH===2Na2FeO4£«3NaCl£«5H2O£»

b£®2FeSO4£«6Na2O2===2Na2FeO4£«2Na2O£«2Na2SO4£«O2¡ü£»

c£®Fe2O3£«3Na2O2===2Na2FeO4£«Na2O£»

d£®Fe(NO3)3£«NaOH£«Cl2¨D¡úNa2FeO4£«NaNO3£«NaCl£«H2O¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÏÂÁÐÅжÏÕýÈ·µÄÊÇ________(ÌîÐòºÅ)¡£

A£®·½·¨a¡¢b¡¢c¶¼¿ÉÔÚË®ÈÜÒºÖнøÐÐ

B£®¸ù¾Ýa¡¢b¿ÉÖªNaClO¡¢Na2O2µÄÑõ»¯ÐÔ¾ùÇ¿ÓÚNa2FeO4µÄ

C£®FeSO4Ö»Óл¹Ô­ÐÔ£¬Ã»ÓÐÑõ»¯ÐÔ

D£®ÄÜÓÃKSCNÈÜÒº¼ìÑébµÄ²úÎïÖÐÊÇ·ñº¬ÓÐFeSO4

(2)¶ÔÓÚ·½·¨cÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________(ÌîÐòºÅ)¡£

A£®Na2O2¼ÈÊÇÑõ»¯¼ÁÓÖÊÇ»¹Ô­¼Á

B£®»¹Ô­²úÎïÖ»ÓÐNa2O

C£®3 mol Na2O2·¢Éú·´Ó¦£¬ÓÐ6 molµç×ÓתÒÆ

D£®ÔÚNa2FeO4ÖÐFeΪ£«4¼Û£¬¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ÄÜÏû¶¾É±¾ú

(3)¶ÔÓÚ·½·¨d£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ________¡£

¢Úд³öNa2FeO4ÓëH2O·´Ó¦µÄÀë×Ó·½³Ìʽ£º____________________________£¬

ÇëÄãÍƲâNa2FeO4³ýÁËÄÜÏû¶¾É±¾úÍ⣬ÁíÒ»¸öÓÃ;ÊÇ____________________¡£


½âÎö¡¡(1)Ñ¡ÏîA£¬·½·¨b¡¢cÖж¼ÓÐNa2O2²ÎÓë·´Ó¦£¬ÈôÔÚË®ÈÜÒºÖнøÐУ¬Na2O2ÓëH2O·´Ó¦£¬´Ó¶øÎÞ·¨µÃµ½Na2FeO4¡£Ñ¡ÏîB£¬·´Ó¦ÖÐNaClO¡¢Na2O2¾ù»ñµÃµç×Ó£¬ÎªÑõ»¯¼Á¡£Na2FeO4ΪÑõ»¯²úÎËùÒÔNaClO¡¢Na2O2µÄÑõ»¯ÐÔ¾ùÇ¿ÓÚNa2FeO4µÄÑõ»¯ÐÔ¡£Ñ¡ÏîC£¬FeSO4ÖеÄFeÔªËØΪ£«2¼Û£¬ÔÚ·´Ó¦ÖпÉʧȥµç×Ó£¬Ò²¿ÉµÃµ½µç×Ó£¬ËùÒÔFeSO4¼ÈÓÐÑõ»¯ÐÔ£¬Ò²Óл¹Ô­ÐÔ¡£Ñ¡ÏîD£¬FeSO4²»ÄÜÓëKSCNÈÜÒº·´Ó¦ÉúºìÉ«ÈÜÒº¡£

(2)Ñ¡ÏîA£¬¸ù¾Ý·´Ó¦£ºFe2O3£«3Na2O2===2Na2FeO4£«Na2O£¬Na2O2ÔÚ·´Ó¦ÖÐÖ»»ñµÃµç×Ó£¬Ö»×÷Ñõ»¯¼Á¡£Ñ¡ÏîB£¬Na2O2¶ÔÓ¦µÄ²úÎï°üº¬Na2FeO4¡¢Na2O£¬¶þÕ߶¼ÊÇ»¹Ô­²úÎѡÏîC,3 mol Na2O2·¢Éú·´Ó¦£¬ÓÐ6 molµç×ÓתÒÆ£¬ÕýÈ·¡£Ñ¡ÏîD£¬Na2FeO4ÖÐFeÔªËصĻ¯ºÏ¼ÛΪ£«6¼Û¡£

(3)¢ÙÑõ»¯¼ÁΪCl2£¬Æ仯ѧ¼ÆÁ¿ÊýΪ3£»»¹Ô­¼ÁΪFe(NO3)3£¬Æ仯ѧ¼ÆÁ¿ÊýΪ2£¬¶þÕßµÄÎïÖʵÄÁ¿Ö®±ÈΪ3¡Ã2¡£¢Ú¸ù¾ÝÌâÄ¿ÐÅÏ¢£¬·´Ó¦ÎïΪNa2FeO4¡¢H2O£¬Éú³ÉÎïΪFe(OH)3¡¢O2ºÍNaOH£¬ÔòÓУºNa2FeO4£«H2O¨D¡úFe(OH)3(½ºÌå)£«NaOH£«O2¡ü£¬ÅäƽµÄ·´Ó¦»¯Ñ§·½³ÌʽΪ4Na2FeO4£«10H2O===4Fe(OH)3(½ºÌå)£«8NaOH£«3O2¡ü£¬ÆäÀë×Ó·´Ó¦·½³ÌʽΪ4FeO£«10H2O===4Fe(OH)3(½ºÌå)£«8OH£­£«3O2¡ü¡£Na2FeO4¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ÄÜÏû¶¾£¬Éú³ÉµÄFe(OH)3½ºÌå¾ßÓнϴóµÄ±íÃæ»ý£¬ÄÜÎü¸½Ë®ÖеÄÐü¸¡ÎÆðµ½¾»Ë®µÄ×÷Óá£

´ð°¸¡¡(1)B¡¡(2)C

(3)¢Ù3¡Ã2¡¡¢Ú4FeO£«10H2O===4Fe(OH)3(½ºÌå)£«8OH£­£«3O2¡ü¡¡Éú³ÉµÄÇâÑõ»¯Ìú½ºÌ壬¾ßÓнϴóµÄ±íÃæ»ý£¬ÄÜÎü¸½ÔÓÖʶø´ïµ½¾»Ë®µÄÄ¿µÄ


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ijͬѧ³ÆÈ¡9 gµí·ÛÈÜÓÚË®£¬²â¶¨µí·ÛµÄË®½â°Ù·ÖÂÊ¡£Æä³ÌÐòÈçÏ£º

µí·ÛÈÜÒº»ìºÏÎ

¼ÓÈëCÈÜÒº¨D¡úשºìÉ«³ÁµíD

(1)¸÷²½¼ÓÈëµÄÊÔ¼ÁΪ£º

A________£¬B__________£¬C__________¡£

(2)¼ÓÈëAÈÜÒº¶ø²»¼ÓÈëBÈÜÒºÊÇ·ñ¿ÉÒÔ________(Ìî¡°¿ÉÒÔ¡±»ò¡°²»¿ÉÒÔ¡±)£¬ÆäÀíÓÉÊÇ________________________________________________________________________

________________________________________________________________________¡£

(3)µ±Îö³ö1.44 gשºìÉ«³Áµí£¬µí·ÛµÄË®½âÂÊÊÇ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÓйػ¯Ñ§ÓÃÓïÖÐ×îÄÜÌåÏÖµªÔ­×ÓºËÍâµç×ÓÔ˶¯×´Ì¬µÄÊÇ                    (¡¡¡¡)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


G(ÒìÎìËᱡºÉ´¼õ¥)ÊÇÒ»ÖÖÖÎÁÆÐÄÔಡµÄÒ©Îï¡£ÆäºÏ³ÉÏß·ÈçÏ£º

(1)AµÄÃû³ÆΪ________¡£

(2)GÖк¬Ñõ¹ÙÄÜÍÅÃû³ÆΪ________¡£

(3)DµÄ·Ö×ÓÖк¬ÓÐ________ÖÖ²»Í¬»¯Ñ§»·¾³µÄÇâÔ­×Ó¡£

(4)EÓëÐÂÖƵÄÇâÑõ»¯Í­·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________

__________________________________________________________¡£

(5)д³ö·ûºÏÏÂÁÐÌõ¼þµÄAµÄËùÓÐͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º____________¡£

a£®·Ö×ÓÖÐÓÐ6¸ö̼ԭ×ÓÔÚÒ»ÌõÖ±ÏßÉÏ£»

b£®·Ö×ÓÖк¬ÓСªOH¡£

(6)ÕýÎìÈ©¿ÉÓÃ×÷ÏãÁÏ¡¢Ï𽺴ٽø¼ÁµÈ£¬Ð´³öÒÔÒÒ´¼ÎªÔ­ÁÏÖƱ¸CH3(CH2)3CHOµÄºÏ³É·ÏßÁ÷³Ìͼ(ÎÞ»úÊÔ¼ÁÈÎÑ¡)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔÚÒ»¸öÑõ»¯»¹Ô­·´Ó¦ÌåϵÖз´Ó¦Îï¡¢Éú³ÉÎï¹²ÁùÖÖÁ£×Ó£¬Fe3£«¡¢NO¡¢Fe2£«¡¢NH¡¢H£«¡¢H2O£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ                                      (¡¡¡¡)¡£

A£®¸Ã·´Ó¦ËµÃ÷Fe(NO3)2ÈÜÒº²»Ò˼ÓËáËữ

B£®¸Ã·´Ó¦ÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁÎïÖʵÄÁ¿Ö®±ÈΪ8¡Ã1

C£®ÈôÓÐ1 mol NO·¢ÉúÑõ»¯·´Ó¦£¬×ªÒƵç×Ó5 mol

D£®Èô½«¸Ã·´Ó¦Éè¼Æ³ÉÔ­µç³Ø£¬Ôò¸º¼«·´Ó¦ÎªFe3£«£«e£­===Fe2£«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐŨ¶È¹ØϵÕýÈ·µÄÊÇ(¡¡¡¡)

A£®ÂÈË®ÖУºc(Cl2)£½2[c(ClO£­)£«c(Cl£­)£«c(HClO)]

B. ÂÈË®ÖУºc(Cl£­)>c(H£«)>c(OH£­)>c(ClO£­)

C£®µÈÌå»ýµÈŨ¶ÈµÄÇâÑõ»¯ÄÆÓë´×Ëá»ìºÏ£ºc(Na£«)£½c(CH3COO£­)

D£®Na2CO3ÈÜÒºÖУºc(Na£«)>c(CO)>c(OH£­)>c(HCO)>c(H£«)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¹¤ÒµÉÏÖƱ¸BaCl2µÄ¹¤ÒÕÁ÷³ÌͼÈçÏ£º

ijÑо¿Ð¡×éÔÚʵÑéÊÒÓÃÖؾ§Ê¯(Ö÷Òª³É·ÖBaSO4)¶Ô¹¤Òµ¹ý³Ì½øÐÐÄ£ÄâʵÑé¡£²é±íµÃ£º

BaSO4(s)£«4C(s)4CO(g)£«BaS(s)

¦¤H1£½£«571.2 kJ¡¤mol£­1¡¡¢Ù

BaSO4(s)£«2C(s)2CO2(g)£«BaS(s)

¦¤H2£½£«226.2 kJ¡¤mol£­1¡¡¢Ú

(1)ÆøÌåÓùýÁ¿NaOHÈÜÒºÎüÊÕ£¬µÃµ½Áò»¯ÄÆ¡£ÔòNa2SË®½âµÄÀë×Ó·½³ÌʽΪ

________________________________________________________________________¡£

(2)ÏòBaCl2ÈÜÒºÖмÓÈëAgNO3ºÍKBr£¬µ±Á½ÖÖ³Áµí¹²´æʱ£¬£½________¡£

[Ksp(AgBr)£½5.4¡Á10£­13£¬Ksp(AgCl)£½2.0¡Á10£­10]

(3)·´Ó¦£ºC(s)£«CO2(g)2CO(g)µÄ¦¤H£½______kJ¡¤mol£­1¡£

(4)ʵ¼ÊÉú²úÖбØÐë¼ÓÈë¹ýÁ¿µÄÌ¿£¬Í¬Ê±»¹ÒªÍ¨Èë¿ÕÆø£¬ÆäÄ¿µÄÊÇ

________________________________________________________________________£¬

________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


²éÔÄ×ÊÁÏ·¢ÏÖ£¬½ðÊôÄƲ»½öÄܸúÑõÆøºÍË®·´Ó¦£¬»¹Äܸú¶àÖÖÆäËüÎïÖÊ·¢Éú·´Ó¦£¬ÆäÖаüÀ¨Óë¾Æ¾«ÔÚ³£ÎÂÏ·´Ó¦¡£ÒªÑо¿½ðÊôÄƸú¾Æ¾«·´Ó¦µÄÐÔÖÊÒÔ¼°ËüÓëË®·´Ó¦µÄÒìͬ µã£¬ÏÂÁеÄÑо¿·½·¨ÖÐûÓÐÓõ½µÄÊÇ                                                                        £¨    £©

    A£®ÊµÑé·¨           B£®¹Û²ì·¨           C£®·ÖÀà·¨           D£®±È½Ï·¨

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÓÃCH4´ß»¯»¹Ô­NOx¿ÉÒÔÏû³ýµªÑõ»¯ºÏÎïµÄÎÛȾ¡£ÀýÈ磺

¢ÙCH4£¨g£©+4NO2£¨g£©=4NO£¨g£©+CO2£¨g£©+2H2O£¨g£©£»¡÷H=£­574kJ/mol

¢ÚCH4£¨g£©+4NO£¨g£©=2N2£¨g£©+CO2£¨g£©+2H2O£¨g£©£»¡÷H=£­1160kJ/mol

ÏÂÁÐ˵·¨ÖдíÎóµÄÊÇ£¨   £©

A£®µÈÎïÖʵÄÁ¿µÄCH4ÔÚ·´Ó¦¢Ù¡¢¢ÚÖÐתÒƵç×ÓÊýÏàͬ

B£®ÓÉ·´Ó¦¢Ù¿ÉÍÆÖª£ºCH4£¨g£©+4NO2£¨g£©=4NO£¨g£©+CO2£¨g£©+2H2O£¨l£©£»¡÷H>£­574kJ/mol

C£®4NO2£¨g£©+2N2£¨g£©=8NO£¨g£©£»¡÷H=+586kJ/mol

D£®ÈôÓñê×¼×´¿öÏÂ4.48L CH4°ÑNO2»¹Ô­ÎªN2£¬Õû¸ö¹ý³ÌÖÐתÒƵĵç×Ó×ÜÊýΪ1.6NA

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸