¡¾ÌâÄ¿¡¿½«º¬ÓÐC¡¢H¡¢OµÄÓлúÎï3.24 g£¬×°ÈëÔªËØ·ÖÎö×°Öã¬Í¨Èë×ãÁ¿µÄO2ʹËüÍêȫȼÉÕ£¬½«Éú³ÉµÄÆøÌåÒÀ´Îͨ¹ýÂÈ»¯¸Æ¸ÉÔï¹ÜAºÍ¼îʯ»Ò¸ÉÔï¹ÜB¡£²âµÃA¹ÜÖÊÁ¿Ôö¼ÓÁË2.16g£¬B¹ÜÔö¼ÓÁË9.24g¡£ÒÑÖª¸ÃÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª108¡£

(1)ȼÉÕ´Ë»¯ºÏÎï3.24g£¬ÐëÏûºÄÑõÆøµÄÖÊÁ¿ÊÇ______________________£»

(2)¸Ã»¯ºÏÎïµÄ·Ö×ÓʽΪ____________£»

(3)¸Ã»¯ºÏÎï1·Ö×ÓÖдæÔÚ1¸ö±½»·£¬ÇÒÔں˴Ź²ÕñÇâÆ×ͼÖÐÓÐ4×éÎüÊÕ·å¡£ÊÔд³öËü¿ÉÄܵĽṹ¼òʽ_______________¡¢_______________¡£

¡¾´ð°¸¡¿8.16g C7H8O

¡¾½âÎö¡¿

(1)A¹ÜÖÊÁ¿Ôö¼ÓÁË2.16gΪÉú³ÉË®µÄÖÊÁ¿£¬B¹ÜÔö¼ÓÁË9.24gΪÉú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿£¬¸ù¾ÝÖÊÁ¿Êغã¼ÆËãÏûºÄÑõÆøµÄÖÊÁ¿£»

(2)¼ÆËãÓлúÎˮ¡¢¶þÑõ»¯Ì¼µÄÎïÖʵÄÁ¿£¬¸ù¾ÝÔ­×ÓÊغã¼ÆËãÓлúÎï·Ö×ÓÖÐC¡¢HÔ­×ÓÊýÄ¿£¬½áºÏÏà¶Ô·Ö×ÓÖÊÁ¿È·¶¨OÔ­×ÓÊýÄ¿£»

(3)ÓлúÎﺬÓб½»·£¬ºË´Å¹²ÕñÇâÆ×´æÔÚËĸö·å£¬ËµÃ÷º¬ÓÐ4ÖÖHÔ­×Ó£¬½áºÏ·Ö×ÓʽÊéд¿ÉÄܵĽṹ¡£

(1)ÊÔ¹ÜAÖÊÁ¿Ôö¼Ó2.16gΪÉú³ÉË®µÄÖÊÁ¿£¬B¹Ü¼îʯ»ÒÎüCO2Ôö¼Ó9.24g£¬ÓÉÖÊÁ¿Êغ㶨ÂÉ£¬¿ÉÖªÏûºÄÑõÆøµÄÖÊÁ¿m(O2)=2.16g+9.24g-3.24g=8.16g£»

(2)ÊÔ¹ÜAÖÊÁ¿Ôö¼Ó2.16gΪÉú³ÉË®µÄÖÊÁ¿£¬ÔòË®µÄÎïÖʵÄÁ¿n(H2O)=2.16g¡Â18g/mol=0.12mol£¬¼îʯ»Ò¸ÉÔï¹ÜBÎüÊÕÔöÖØCO2µÄÖÊÁ¿Îª9.24g£¬¶þÑõ»¯Ì¼µÄÎïÖʵÄÁ¿n(CO2)=9.24g¡Â44g/mol=0.21mol£¬¸ÃÓлúÎïÖÊÁ¿Îª3.24g£¬Ïà¶Ô·Ö×ÓÖÊÁ¿ÊÇ108£¬ËùÒÔÓлúÎïµÄÎïÖʵÄÁ¿Îªn(ÓлúÎï)=3.24g¡Â108g/mol=0.03mol£¬ËùÒÔÓлúÎï·Ö×ÓÖУ¬N(C)==7£¬ N(H)==8£¬ N(O)==1£¬Òò´Ë¸ÃÓлúÎïµÄ·Ö×ÓʽΪ£ºC7H8O£»

(3)Èô¸ÃÓлúÎﺬÓб½»·£¬ÇҺ˴Ź²ÕñÇâÆ×´æÔÚËĸö·å£¬¿ÉÄܵĽṹ¼òʽΪ£º¡¢¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¸ù¾ÝÏÂÁл¯ºÏÎ¢ÙHCl ¢ÚNaOH ¢ÛNH4Cl ¢ÜCH3COONa ¢ÝCH3COOH ¢ÞNH3H2O¢ßNa2CO3£¬Çë¸ù¾Ý×ÖĸÐòºÅÌáʾµÄÎïÖʻشðÏÂÁÐÎÊÌâ¡£

(1)¢ÛÈÜÒºÖÐÀë×ÓŨ¶È´óС˳ÐòΪ______________________¡£ÓÃÀë×Ó·½³Ìʽ±íʾ¢ßÈÜÒºÏÔ¼îÐÔµÄÔ­Òò_______________________________¡£

(2)³£ÎÂÏ£¬pH=11µÄ¢ÜµÄÈÜÒºÖУ¬ÓÉË®µçÀë³öÀ´µÄc(OH-)=____________¡£ÒÑÖª³£ÎÂÏ¢ݺ͢޵ĵçÀë³£Êý¾ùΪ1.7¡Á10-5 mol¡¤L-1£¬Ôò·´Ó¦£ºCH3COOH+NH3H2OCH3COO-+NH4++H2OµÄƽºâ³£ÊýΪ______________¡£

(3)³£ÎÂÏ£¬¹ØÓÚpHÖµÏàͬµÄ¢ÙºÍ¢ÝÁ½ÖÖÈÜÒº£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ________¡£

A.Á½ÖÖÈÜÒºÖÐË®µÄµçÀë³Ì¶ÈÏàͬ B.c(CH3COO-)=c(Cl-)

C.c(CH3COOH)>c(HCl) D.ÓëµÈŨ¶ÈµÄÇâÑõ»¯ÄÆÈÜÒº·´Ó¦£¬´×ËáÏûºÄµÄÌå»ýÉÙ

(4)³£ÎÂÏ£¬½«0.10 mol/LµÄ¢ÙÈÜÒººÍ0.30 mol/L£»¢ÚÈÜÒºµÈÌå»ý»ìºÏ£¬³ä·Ö·´Ó¦ºó»Ö¸´ÖÁ³£Î£¬ÈÜÒºµÄpH=________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ï®ÔÚÓлúºÏ³É¡¢µç³ØµÈÁìÓòÖÐÓÐÖØÒªµÄ×÷Óá£

I. µÄÖƱ¸ºÍÓ¦ÓÃÈçÏÂͼËùʾ¡£

(1)ï®ÔªËØÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃ_______________________¡£

(2)д³öAµÄµç×Óʽ___________________________¡£

(3)ÊÇÓлúºÏ³ÉÖг£ÓõĻ¹Ô­¼Á£¬ÊÔд³ö·´Ó¦¢ÛµÄ»¯Ñ§·½³Ìʽ_________________¡£

II.Á×ËáÑÇÌúï®ÊÇÐÂÐÍï®Àë×Óµç³ØµÄÊ×Ñ¡µç¼«²ÄÁÏ£¬ÊÇÒÔÌú°ôΪÑô¼«£¬Ê¯Ä«ÎªÒõ¼«£¬µç½âÁ×Ëá¶þÇâ李¢ÂÈ»¯ï®»ìºÏÈÜÒº£¬Îö³öÁ×ËáÑÇÌú﮳Áµí£¬ÔÚ800¡æ×óÓÒ¡¢¶èÐÔÆøÌå·ÕΧÖÐìÑÉÕÖƵá£ÔÚï®Àë×Óµç³ØÖУ¬ÐèÒªÒ»ÖÖÓлú¾ÛºÏÎï×÷ΪÕý¸º¼«Ö®¼äï®Àë×ÓǨÒƵĽéÖÊ£¬¸ÃÓлú¾ÛºÏÎïµÄµ¥ÌåÖ®Ò»(ÓÃM±íʾ)µÄ½á¹¹¼òʽÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(4)ÖƱ¸Á×ËáÑÇÌúï®±ØÐëÔÚ¶èÐÔÆøÌå·ÕΧÖнøÐУ¬ÆäÔ­ÒòÊÇ_______________¡£

(5)Ñô¼«Éú³ÉÁ×ËáÑÇÌú﮵ĵ缫·´Ó¦Ê½Îª___________________¡£

(6)д³öMÓë×ãÁ¿ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ_____________________¡£

(7)¸Ãµç³Ø³äµçʱÑô¼«µÄÁ×ËáÑÇÌúï®Éú³ÉÁ×ËáÌú£¬Ôò·ÅµçʱÕý¼«µÄµç¼«·´Ó¦Ê½Îª___________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³Ñ§ÉúÓûÓÃÒÑÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÑÎËáÀ´²â¶¨Î´ÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÇâÑõ»¯ÄÆÈÜÒº¡£

£¨1£©ÈôµÎ¶¨¿ªÊ¼ºÍ½áÊøʱ£¬ËáʽµÎ¶¨¹ÜÖеÄÒºÃæÈçͼËùʾ£ºÔòËùÓÃÑÎËáµÄÌå»ýΪ__mL¡£µÎ¶¨ÖÕµãʱµÄÏÖÏóÊÇ£ºµ±µÎÈë×îºóÒ»µÎÑÎËáʱ£¬×¶ÐÎÆ¿ÖÐÈÜÒºÑÕÉ«ÓÉ___£¬ÇÒ30s²»»Ö¸´Ô­É«¡£

£¨2£©Ä³Ñ§Éú¸ù¾ÝÈý´ÎʵÑé·Ö±ð¼Ç¼ÓйØÊý¾ÝÈçÏÂ±í£º

µÎ¶¨´ÎÊý

´ý²âÇâÑõ»¯ÄÆ

ÈÜÒºµÄÌå»ý/mL

0.1000mol¡¤L£­1ÑÎËáµÄÌå»ý/mL

µÎ¶¨Ç°¿Ì¶È

µÎ¶¨ºó¿Ì¶È

ÈÜÒºÌå»ý

µÚÒ»´Î

25.00

0.00

26.11

26.11

µÚ¶þ´Î

25.00

1.56

30.30

28.74

µÚÈý´Î

25.00

0.22

26.31

26.09

ÇëÑ¡ÓÃÆäÖкÏÀíÊý¾ÝËã³ö¸ÃÇâÑõ»¯ÄÆÈÜÒºÎïÖʵÄÁ¿Å¨¶È(¼ÆËã½á¹û±£Áô4λÓÐЧÊý×Ö)£ºc(NaOH)=__mol¡¤L£­1¡£

£¨3£©ÓÉÓÚ´íÎó²Ù×÷£¬Ê¹µÃÉÏÊöËù²âÇâÑõ»¯ÄÆÈÜÒºµÄŨ¶ÈÆ«¸ßµÄÊÇ__(Ìî×Öĸ)¡£

A.µÎ¶¨Ç°µÎ¶¨¹ÜÖÐÓÐÆøÅÝ£¬µÎ¶¨ºóÏûʧ B.¼îʽµÎ¶¨¹ÜÁ¿È¡NaOHÈÜҺʱ£¬Î´½øÐÐÈóÏ´²Ù×÷

C.µÎ¶¨Ê±´ïµ½µÎ¶¨ÖÕµãʱ¸©ÊÓ¶ÁÊý D.׶ÐÎÆ¿È¡ÓÃNaOH´ý²âҺǰ¼ÓÉÙÁ¿Ë®Ï´µÓ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¸ù¾ÝÏÂÁÐʵÑé²Ù×÷ºÍÏÖÏóËùµÃ³öµÄ½áÂÛÕýÈ·µÄÊÇ

Ñ¡Ïî

ʵÑé²Ù×÷ºÍÏÖÏó

½á ÂÛ

A

ÏòÒ»¶¨Å¨¶ÈµÄNa2SiO3 ÈÜÒºÖÐͨÈëÊÊÁ¿CO2 ÆøÌ壬 ³öÏÖ°×É«³Áµí¡£

H2SiO3 µÄËáÐÔ±ÈH2CO3µÄËáÐÔÇ¿

B

½«ÉÙÁ¿Fe(NO3)2¼ÓË®Èܽâºó£¬µÎ¼ÓÏ¡ÁòËáËữ£¬ÔٵμÓKSCNÈÜÒº£¬ÈÜÒº±ä³ÉѪºìÉ«

Fe(NO3)2ÒѱäÖÊ

C

ÊÒÎÂÏ£¬²âµÃ£º0.1mol¡¤L-1 Na2SO3ÈÜÒºµÄpHԼΪ10£»0.1mol¡¤L-1 NaHSO3ÈÜÒºµÄpHԼΪ5¡£

HSO3£­ ½áºÏH+ µÄÄÜÁ¦±ÈSO32£­µÄÇ¿

D

·Ö±ðÏò25mLÀäË®ºÍ25mL·ÐË®ÖеÎÈë6µÎFeCl3 ±¥ºÍÈÜÒº£¬Ç°ÕßΪ»ÆÉ«£¬ºóÕßΪºìºÖÉ«¡£

ζÈÉý¸ß£¬Fe3+µÄË®½â³Ì¶ÈÔö´ó

A.AB.BC.CD.D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿³£ÎÂÏ£¬ÏÂÁÐÈÜÒºÖи÷Àë×ÓŨ¶È¹ØϵÕýÈ·µÄÊÇ

A.pH£½12µÄ°±Ë®ÓëpH£½2µÄÑÎËáµÈÌå»ý»ìºÏ£ºc(Cl£­)>c(NH4+)>c(H£«)>c(OH£­)

B.Ũ¶ÈΪ0.1mol¡¤L£­1µÄ̼ËáÄÆÈÜÒº£ºc(Na£«)£½c(CO32£­)£«c(HCO3£­)£«c(H2CO3)

C.0.1mol¡¤L£­1µÄNaHSÈÜÒºÖÐÀë×ÓŨ¶È¹Øϵ£ºc(OH£­)=c(H+)£­c(S2£­)+c(H2S)

D.´×ËáÈÜÒºÓëNaOHÈÜÒº»ìºÏºó£¬ËùµÃÈÜÒº³ÊÖÐÐÔ£ºc(Na£«)>c(CH3COO£­)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿PVAcÊÇÒ»ÖÖ¾ßÓÐÈÈËÜÐÔµÄÊ÷Ö¬£¬¿ÉºÏ³ÉÖØÒª¸ß·Ö×Ó²ÄÁÏM£¬ºÏ³É·ÏßÈçÏ£º

¼ºÖª£ºR¡¢R¨@¡¢R¨@¨@ΪHÔ­×Ó»òÌþ»ù

I. R'CHO+ R"CH2CHO

II. RCHO+

£¨1£©±ê×¼×´¿öÏ£¬4.48LÆø̬ÌþAµÄÖÊÁ¿ÊÇ5.2g£¬ÔòAµÄ½á¹¹¼òʽΪ___________________¡£

£¨2£©¼ºÖªA¡úBΪ¼Ó³É·´Ó¦£¬ÔòXµÄ½á¹¹¼òʽΪ_______£»BÖйÙÄÜÍŵÄÃû³ÆÊÇ_________¡£

£¨3£©·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽΪ______________________¡£

£¨4£©EÄÜʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«£¬·´Ó¦¢ÚµÄ·´Ó¦ÊÔ¼ÁºÍÌõ¼þÊÇ_______________________¡£

£¨5£©·´Ó¦¢ÛµÄ»¯Ñ§·½³ÌʽΪ____________________________¡£

£¨6£©ÔÚE¡úF¡úG¡úHµÄת»¯¹ý³ÌÖУ¬ÒÒ¶þ´¼µÄ×÷ÓÃÊÇ__________________________¡£

£¨7£©¼ºÖªMµÄÁ´½ÚÖгý±½»·Í⣬»¹º¬ÓÐÁùÔª»·×´½á¹¹£¬ÔòMµÄ½á¹¹¼òʽΪ_________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Èý²ÝËáºÏÌúËá¼Ø¾§Ì壨K3[Fe(C2O4)3]¡¤xH2O£©ÊÇÒ»ÖÖ¹âÃô²ÄÁÏ£¬ÔÚ110¡æ¿ÉÍêȫʧȥ½á¾§Ë®¡£Îª²â¶¨¸Ã¾§ÌåÖÐÌúµÄº¬Á¿£¬Ä³ÊµÑéС×é×öÁËÈçÏÂʵÑ飺

Ìúº¬Á¿µÄ²â¶¨

²½ÖèÒ»£º³ÆÁ¿5.00 gÈý²ÝËáºÏÌúËá¼Ø¾§Ì壬ÅäÖƳÉ250 mLÈÜÒº¡£

²½Öè¶þ£ºÈ¡ËùÅäÈÜÒº25.00 mLÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÏ¡H2SO4Ëữ£¬µÎ¼ÓKMnO4ÈÜÒºÖÁ²ÝËá¸ùÇ¡ºÃÈ«²¿Ñõ»¯³É¶þÑõ»¯Ì¼£¬Í¬Ê±MnO4£­±»»¹Ô­³ÉMn2+¡£Ïò·´Ó¦ºóµÄÈÜÒºÖмÓÈëһС³×п·Û£¬¼ÓÈÈÖÁ»ÆÉ«¸ÕºÃÏûʧ£¬¹ýÂË£¬Ï´µÓ£¬½«¹ýÂ˼°Ï´µÓËùµÃÈÜÒºÊÕ¼¯µ½×¶ÐÎÆ¿ÖУ¬´ËʱÈÜÒºÈÔ³ÊËáÐÔ¡£

²½ÖèÈý£ºÓÃ0.010 mol/L KMnO4ÈÜÒºµÎ¶¨²½Öè¶þËùµÃÈÜÒºÖÁÖյ㣬ÏûºÄKMnO4ÈÜÒº20.02 mL£¬µÎ¶¨ÖÐMnO4£­±»»¹Ô­³ÉMn2+¡£Öظ´²½Öè¶þ¡¢²½ÖèÈý²Ù×÷£¬µÎ¶¨ÏûºÄ0.010 mol/L KMnO4ÈÜÒº·Ö±ðΪ19.98 mLºÍ20.00 mL¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©µÎ¶¨¹ý³ÌÖУ¬¸ßÃÌËá¼ØÈÜҺӦʢװÔÚ__________µÎ¶¨¹ÜÖУ¨Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±£©¡£

£¨2£©ÓÃÀë×Ó·½³Ìʽ±íʾ²½Öè¶þÖÐÉæ¼°µ½µÄÏà¹Ø»¯Ñ§·´Ó¦£º________________£» Zn + 2Fe3+ = 2Fe2+ + Zn2+¡£

£¨3£©²½ÖèÈýÖеζ¨ÖÕµãµÄÅж¨£º____________________¡£

£¨4£©ÔÚ²½Öè¶þÖУ¬Èô¼ÓÈëµÄKMnO4ÈÜÒºµÄÁ¿²»¹»£¬Ôò²âµÃµÄÌúº¬Á¿____________¡£ÔÚ²½ÖèÈýÖУ¬ÈôµÎ¶¨Ç°ÑöÊÓ¶ÁÊý£¬µÎ¶¨ºó¸©ÊÓ¶ÁÊý£¬Ôò²âµÃµÄÌúº¬Á¿__________¡££¨Ñ¡Ìî¡°Æ«µÍ¡±¡¢¡°Æ«¸ß¡±¡¢¡°²»±ä¡±£©

£¨5£©ÊµÑé²âµÃ¸Ã¾§ÌåÖÐÌúµÄÖÊÁ¿·ÖÊýΪ__________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¶¾ÖØʯµÄÖ÷Òª³É·ÖBaCO3(º¬Ca2£«¡¢Mg2£«¡¢Fe3£«µÈÔÓÖÊ)£¬ÊµÑéÊÒÀûÓö¾ÖØʯÖƱ¸BaCl2¡¤2H2OµÄÁ÷³ÌÈçÏ£º

£¨1£©¶¾ÖØʯÓÃÑÎËá½þÈ¡Ç°Ðè³ä·ÖÑÐÄ¥£¬Ä¿µÄÊÇ________¡£ÊµÑéÊÒÓÃ37%µÄÑÎËáÅäÖÆ15%µÄÑÎËᣬ³ýÁ¿Í²Í⻹ÐèʹÓÃÏÂÁÐÒÇÆ÷ÖеÄ________¡£

a£®ÉÕ±­¡¡b£®500mLÈÝÁ¿Æ¿¡¡c£®²£Á§°ô¡¡d£®µÎ¶¨¹Ü

£¨2£©ÒÑÖª²»Í¬ÔÓÖÊÀë×Ó¿ªÊ¼³ÁµíºÍ³ÁµíÍêÈ«µÄpHÈçÏ£º

¼ÓÈëNH3¡¤H2Oµ÷½ÚpH=8¿É³ýÈ¥_______ (ÌîÀë×Ó·ûºÅ)£¬´Ëʱ£¬ÈÜÒºÖиÃÀë×ÓµÄŨ¶ÈΪ_______mol¡¤L£­1¡£¼ÓÈëNaOHµ÷pH=12.5£¬ÈÜÒºÄÚÊ£ÓàµÄÑôÀë×ÓÖÐ_______ÍêÈ«³Áµí£¬_____________ (ÌîÀë×Ó·ûºÅ)²¿·Ö³Áµí¡£¼ÓÈëH2C2O4ʱӦ±ÜÃâ¹ýÁ¿£¬Ô­ÒòÊÇ___________¡££¨ÒÑÖª£ºKsp(BaC2O4)=1.6¡Á10£­7,Ksp(CaC2O4)=2.3¡Á10£­9, Ksp£ÛFe(OH)3£Ý =2.6¡Á10£­39£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸