ÌâÄ¿ÄÚÈÝ

WO3¿ÉÓÃÓÚÖƱ¸µç×èÔª¼þ¡¢µç×ÓÉäÏßÆÁµÈ¡£Æ乤ҵÉú²úÁ÷³ÌÈçÏ£º

£¨1£©²Ù×÷XµÄÄ¿µÄÊÇΪÁË»ñµÃ´¿¾»µÄÖÙÎÙËá茶§Ì壬¸Ã²Ù×÷°üÀ¨£º½«ÓÃÑÎËáÖкͺóµÄÈÜÒº     ¡¢ÀäÈ´½á¾§¡¢     ¡¢µÍκæ¸É¡£
£¨2£©Êµ¼Ê¹¤ÒµÉú²úÖУ¬´ÖÖÙÎÙËá茶§Ì壨º¬ÉÙÁ¿NH4Cl¾§Ì壩¿É²»¾­Ìá´¿¾ÍÖ±½Ó×ÆÉÕ£¬ÆäÔ­ÒòÊÇ     ¡£
£¨3£©ÒÑÖª£ºÖÙÎÙËá茶§Ìå[x(NH4)2O¡¤yWO3¡¤zH2O]ÊÜÈÈ·Ö½âµÄ»¯Ñ§·½³ÌʽÈçÏ£º
x(NH4)2O¡¤yWO3¡¤zH2O¡úWO3 +NH3¡ü+H2O¡ü(δÅäƽ)¡£
ijͬѧΪ²â¶¨ÖÙÎÙËá茶§ÌåµÄ×é³É£¬½øÐÐÈçÏÂʵÑ飺
¢Ù׼ȷ³ÆÈ¡16.21gÑùÆ·£¬ÑÐϸ×ÆÉÕ£»
¢Ú½«²úÉúµÄÆøÌåͨÈë×°Óмîʯ»Ò¸ÉÔï¹Ü£¬³ä·ÖÎüÊճƵøÉÔï¹ÜÔöÖØ1.44g£»
¢Û³ÆÁ¿ÀäÈ´ºóµÄ¹ÌÌåÖÊÁ¿Îª13.92g¡£
ͨ¹ý¼ÆËãÈ·¶¨´ËÖÙÎÙËá茶§ÌåµÄ»¯Ñ§Ê½£¨Ð´³ö¼ÆËã¹ý³Ì£©¡£

£¨1£©¼ÓÈÈŨËõ ¹ýÂË¡¢Ï´µÓ
£¨2£©ÔÓÖÊNH4Cl±»×ÆÉÕ·Ö½âºóÈ«²¿×ª»¯ÎªÆøÌå
£¨3£©m(NH3)=16.21g-13.92g-1.44g=0.85g
n(NH4+)=n(NH3)=" 0.85g" ¡Â17g¡¤mol¡ª1=0.05mol 
ÑùÆ·Öнᾧˮ£ºn(H2O)="1.44" g ¡Â18 g¡¤mol¡ª1-0.05mol¡Â2="0.055" mol
n(WO3)=13.92g¡Â232 g¡¤mol¡ª1=0.06mol
x£ºy£ºz="0.025" mol£º0.06mol£º0.055 mol=5£º12£º11
¹ÊÖÙÎÙËá茶§ÌåµÄ»¯Ñ§Ê½Îª5(NH4)2O¡¤12WO3¡¤11H2O¡£

½âÎöÊÔÌâ·ÖÎö£º£¨1£©½«ÈÜÒºÕô·¢Å¨Ëõºó²ÅÄÜÀäÈ´½á¾§£¬¾§ÌåÓëÈÜÒº·Ö¿ª²ÉÓùýÂ˵ķ½·¨£¬²¢ÒªÏ´µÓ³ýÈ¥¾§Ìå±íÃæÔÓÖÊ£»£¨2£©NH4ClÊÜÈÈ·Ö½âΪHClºÍNH3£¬´Ó¹ÌÌåÖзÖÀ룻
£¨3£©ÎïÖÊ×é³ÉµÄ¼ÆËã
¸ù¾Ý¢Ú²½¸ÉÔï¹ÜÔöÖØΪˮµÄÁ¿
ÑùÆ·Öнᾧˮ£ºn(H2O)="1.44" g ¡Â18 g¡¤mol¡ª1-0.05mol¡Â2="0.055" mol
¸ù¾ÝÖÊÁ¿Êغã¼ÆËã²úÉú°±ÆøµÄÁ¿£¬¼ÆËãNH4+µÄÁ¿
m(NH3)=16.21g-13.92g-1.44g=0.85g
n(NH4+)=n(NH3)=" 0.85g" ¡Â17g¡¤mol¡ª1="0.05mol"
¸ù¾Ý¢Û²½µÃµ½¹ÌÌåΪWO3£¬n(WO3)=13.92g¡Â232 g¡¤mol¡ª1=0.06mol
ËùÒÔÓÐx£ºy£ºz="0.025" mol£º0.06mol£º0.055 mol=5£º12£º11
¹ÊÖÙÎÙËá茶§ÌåµÄ»¯Ñ§Ê½Îª5(NH4)2O¡¤12WO3¡¤11H2O¡£
¿¼µã£º¿¼²é¹¤ÒµÁ÷³ÌÔ­Àí¼°ÎïÖÊ×é³É¼ÆËãÓйØÎÊÌâ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÒÑ֪ij¡°84Ïû¶¾Òº¡±Æ¿Ì岿·Ö±êÇ©ÈçͼËùʾ£¬¸Ã¡°84Ïû¶¾Òº¡±Í¨³£Ï¡ÊÍ100±¶(Ìå»ýÖ®±È)ºóʹÓá£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)¸Ã¡°84Ïû¶¾Òº¡±µÄÎïÖʵÄÁ¿Å¨¶ÈԼΪ       mol¡¤L£­1¡£
(2)ijͬѧȡ100 mL¸Ã¡°84Ïû¶¾Òº¡±£¬Ï¡ÊͺóÓÃÓÚÏû¶¾£¬Ï¡ÊͺóµÄÈÜÒºÖÐc(Na£«)£½        mol¡¤L£­1¡£
(3)¸Ãͬѧ²ÎÔĸá°84Ïû¶¾Òº¡±µÄÅä·½£¬ÓûÓÃNaClO¹ÌÌåÅäÖÆ480 mLº¬NaClOÖÊÁ¿·ÖÊýΪ25%µÄÏû¶¾Òº¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ       (ÌîÐòºÅ)¡£

A£®ÈçͼËùʾµÄÒÇÆ÷ÖУ¬ÓÐÈýÖÖÊDz»ÐèÒªµÄ£¬»¹ÐèÒªÒ»ÖÖ²£Á§ÒÇÆ÷ 
B£®ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴ¾»ºó£¬Ó¦ºæ¸Éºó²ÅÄÜÓÃÓÚÈÜÒºÅäÖÆ 
C£®ÅäÖƹý³ÌÖУ¬Î´ÓÃÕôÁóˮϴµÓÉÕ±­ºÍ²£Á§°ô¿ÉÄܵ¼Ö½á¹ûÆ«µÍ 
D£®ÐèÒª³ÆÁ¿NaClO¹ÌÌåµÄÖÊÁ¿Îª143.0 g 

(4)¡°84Ïû¶¾Òº¡±ÓëÏ¡ÁòËá»ìºÏʹÓÿÉÔöÇ¿Ïû¶¾ÄÜÁ¦£¬Ä³Ïû¶¾Ð¡×éÈËÔ±ÓÃ98%(ÃܶÈΪ1.84 g¡¤cm£­3)µÄŨÁòËáÅäÖÆ2 000 mL 2.3 mol¡¤L£­1µÄÏ¡ÁòËáÓÃÓÚÔöÇ¿¡°84Ïû¶¾Òº¡±µÄÏû¶¾ÄÜÁ¦¡£
¢ÙËùÅäÖƵÄÏ¡ÁòËáÖУ¬H£«µÄÎïÖʵÄÁ¿Å¨¶ÈΪ       mol¡¤L£­1¡£
¢ÚÐèÓÃŨÁòËáµÄÌå»ýΪ       mL¡£

ÄÆÊÇ»îÆõļî½ðÊôÔªËØ£¬ÄƼ°Æ仯ºÏÎïÔÚÉú²úºÍÉú»îÖÐÓй㷺µÄÓ¦Óá£
Íê³ÉÏÂÁмÆË㣺
(1)µþµª»¯ÄÆ(NaN3)ÊÜײ»÷ÍêÈ«·Ö½â²úÉúÄƺ͵ªÆø£¬¹Ê¿ÉÓ¦ÓÃÓÚÆû³µ°²È«ÆøÄÒ¡£Èô²úÉú40.32 L(±ê×¼×´¿öÏÂ)µªÆø£¬ÖÁÉÙÐèÒªµþµª»¯ÄÆ________g¡£
(2)ÄÆ­¼ØºÏ½ð¿ÉÔں˷´Ó¦¶ÑÖÐÓÃ×÷ÈȽ»»»Òº¡£5.05 gÄÆ­¼ØºÏ½ðÈÜÓÚ200 mLË®Éú³É0.075 molÇâÆø¡£
¢Ù¼ÆËãÈÜÒºÖÐÇâÑõ¸ùÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶È(ºöÂÔÈÜÒºÌå»ý±ä»¯)¡£
_____________________________________________________________
¢Ú¼ÆË㲢ȷ¶¨¸ÃÄÆ­¼ØºÏ½ðµÄ»¯Ñ§Ê½¡£
_____________________________________________________________
(3)ÇâÑõ»¯ÄÆÈÜÒº´¦ÀíÂÁÍÁ¿ó²¢¹ýÂË£¬µÃµ½º¬ÂÁËáÄƵÄÈÜÒº¡£Ïò¸ÃÈÜÒºÖÐͨÈë¶þÑõ»¯Ì¼£¬ÓÐÏÂÁз´Ó¦£º
2NaAlO2£«3H2O£«CO2=2Al(OH)3¡ý£«Na2CO3
ÒÑ֪ͨÈë¶þÑõ»¯Ì¼336 L(±ê×¼×´¿öÏÂ)£¬Éú³É24 mol Al(OH)3ºÍ15 mol Na2CO3£¬ÈôͨÈëÈÜÒºµÄ¶þÑõ»¯Ì¼Îª112 L(±ê×¼×´¿öÏÂ)£¬¼ÆËãÉú³ÉµÄAl(OH)3ºÍNa2CO3µÄÎïÖʵÄÁ¿Ö®±È¡£
_________________________________________________________________
(4)³£ÎÂÏ£¬³ÆÈ¡²»Í¬ÇâÑõ»¯ÄÆÑùÆ·ÈÜÓÚË®£¬¼ÓÑÎËáÖкÍÖÁpH£½7£¬È»ºó½«ÈÜÒºÕô¸ÉµÃÂÈ»¯Äƾ§Ì壬Õô¸É¹ý³ÌÖвúÆ·ÎÞËðʧ¡£

 
ÇâÑõ»¯ÄÆÖÊÁ¿(g)
ÂÈ»¯ÄÆÖÊÁ¿(g)
¢Ù
2.40
3.51
¢Ú
2.32
2.34
¢Û
3.48
3.51
 
ÉÏÊöʵÑé¢Ù¢Ú¢ÛËùÓÃÇâÑõ»¯Äƾù²»º¬ÔÓÖÊ£¬ÇÒʵÑéÊý¾Ý¿É¿¿¡£Í¨¹ý¼ÆË㣬·ÖÎöºÍ±È½ÏÉϱí3×éÊý¾Ý£¬¸ø³ö½áÂÛ¡£

µªÊǵØÇòÉϺ¬Á¿·á¸»µÄÒ»ÖÖÔªËØ£¬°±¡¢ëÂ(N2H4)ºÍµþµªËᶼÊǵªÔªËصÄÖØÒªÇ⻯ÎÔÚ¹¤Å©ÒµÉú²ú¡¢Éú»îÖÐÓÐ×ÅÖØ´ó×÷Óá£
(1)ºÏ³É°±Éú²ú¼¼ÊõµÄ´´Á¢¿ª±ÙÁËÈ˹¤¹ÌµªµÄ;¾¶£¬¶Ô»¯Ñ§¹¤Òµ¼¼ÊõÒ²²úÉúÁËÖØÒªÓ°Ïì¡£
¢ÙÔڹ̶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬½øÐÐÈçÏ»¯Ñ§·´Ó¦£ºN2(g)£«3H2(g)2NH3(g)¡¡¦¤H£¼0£¬Æäƽºâ³£ÊýKÓëζÈTµÄ¹ØϵÈçÏÂ±í¡£

T/K
298
398
498
ƽºâ³£ÊýK
4.1¡Á106
K1
K2
 
Ôò¸Ã·´Ó¦µÄƽºâ³£ÊýµÄ±í´ïʽΪ________£»ÅжÏK1________K2(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±)¡£
¢ÚÏÂÁи÷ÏîÄÜ˵Ã÷¸Ã·´Ó¦ÒѴﵽƽºâ״̬µÄÊÇ________(Ìî×Öĸ)¡£
a£®ÈÝÆ÷ÄÚN2¡¢H2¡¢NH3µÄŨ¶ÈÖ®±ÈΪ1¡Ã3¡Ã2
b£®v(N2)Õý£½3v(H2)Äæ
c£®ÈÝÆ÷ÄÚѹǿ±£³Ö²»±ä
d£®»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
¢ÛÒ»¶¨Î¶ÈÏ£¬ÔÚ1 LÃܱÕÈÝÆ÷ÖгäÈë1 mol N2ºÍ3 mol H2²¢·¢ÉúÉÏÊö·´Ó¦¡£ÈôÈÝÆ÷ÈÝ»ýºã¶¨£¬10 min´ïµ½Æ½ºâʱ£¬ÆøÌåµÄ×ÜÎïÖʵÄÁ¿ÎªÔ­À´µÄ£¬ÔòN2µÄת»¯ÂÊΪ________£¬ÒÔNH3µÄŨ¶È±ä»¯±íʾ¸Ã¹ý³ÌµÄ·´Ó¦ËÙÂÊΪ________¡£
(2)ë¿ÉÓÃÓÚ»ð¼ýȼÁÏ¡¢ÖÆÒ©Ô­Áϵȡ£
¢ÙÔÚ»ð¼ýÍƽøÆ÷ÖÐ×°ÓÐëÂ(N2H4)ºÍҺ̬H2O2£¬ÒÑÖª0.4 molҺ̬N2H4ºÍ×ãÁ¿ÒºÌ¬H2O2·´Ó¦£¬Éú³ÉÆø̬N2ºÍÆø̬H2O£¬·Å³ö256.6 kJµÄÈÈÁ¿¡£¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ________________________________________________________________________¡£
¢ÚÒ»ÖÖëÂȼÁϵç³ØµÄ¹¤×÷Ô­ÀíÈçͼËùʾ¡£¸Ãµç³Ø¹¤×÷ʱ¸º¼«µÄµç¼«·´Ó¦Ê½Îª_____________________________________¡£

¢Û¼ÓÈÈÌõ¼þÏÂÓÃҺ̬ëÂ(N2H4)»¹Ô­ÐÂÖÆCu(OH)2À´ÖƱ¸ÄÉÃ×¼¶Cu2O£¬Í¬Ê±·Å³öN2¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________________________¡£
ëÂÓëÑÇÏõËá(HNO2)·´Ó¦¿ÉÉú³ÉµþµªËᣬ8.6 gµþµªËáÍêÈ«·Ö½â    ¿É·Å³ö6.72 LµªÆø(±ê×¼×´¿öÏÂ)£¬ÔòµþµªËáµÄ·Ö×ÓʽΪ________¡£

ÈçͼΪʵÑéÊÒijŨÑÎËáÊÔ¼ÁÆ¿±êÇ©ÉϵÄÓйØÊý¾Ý£¬ÊÔ¸ù¾Ý±êÇ©ÉϵÄÓйØÊý¾Ý»Ø´ðÏÂÁÐÎÊÌ⣺

ÑÎËá
·Ö×Óʽ£ºHCl
Ïà¶Ô·Ö×ÓÖÊÁ¿£º36.5,Ãܶȣº1.19 g¡¤cm£­3
HClµÄÖÊÁ¿·ÖÊý£º36.5%
 
£¨1£©¸ÃŨÑÎËáÖÐHClµÄÎïÖʵÄÁ¿Å¨¶ÈΪ__        ____mol¡¤L£­1¡£
£¨2£©È¡ÓÃÈÎÒâÌå»ýµÄ¸ÃÑÎËáÈÜҺʱ£¬ÏÂÁÐÎïÀíÁ¿Öв»ËæËùÈ¡Ìå»ýµÄ¶àÉÙ¶ø±ä»¯µÄÊÇ________¡£
A£®ÈÜÒºÖÐHClµÄÎïÖʵÄÁ¿    B£®ÈÜÒºµÄŨ¶È
C£®ÈÜÒºÖÐCl£­µÄÊýÄ¿         D£®ÈÜÒºµÄÃܶÈ
£¨3£©Ä³Ñ§ÉúÓûÓÃÉÏÊöŨÑÎËáºÍÕôÁóË®ÅäÖÆ500 mLÎïÖʵÄÁ¿Å¨¶ÈΪ0.400 mol¡¤L£­1µÄÏ¡ÑÎËá¡£ 
¢Ù¸ÃѧÉúÐèÒªÁ¿È¡___    _____mLÉÏÊöŨÑÎËá½øÐÐÅäÖÆ¡£
¢ÚÔÚÅäÖƹý³ÌÖУ¬ÏÂÁÐʵÑé²Ù×÷¶ÔËùÅäÖƵÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈÓкÎÓ°Ï죿(ÔÚÀ¨ºÅÄÚÌîA±íʾ¡°Æ«´ó¡±¡¢ÌîB±íʾ¡°Æ«Ð¡¡±¡¢ÌîC±íʾ¡°ÎÞÓ°Ï족)¡£
a£®ÓÃÁ¿Í²Á¿È¡Å¨ÑÎËáʱ¸©Êӹ۲찼ҺÃæ (  )
b£®¶¨Èݺó¾­Õñµ´¡¢Ò¡ÔÈ¡¢¾²Ö㬷¢ÏÖÒºÃæϽµ£¬ÔÙ¼ÓÊÊÁ¿µÄÕôÁóË® (  )
£¨4£©¢Ù¼ÙÉè¸Ãͬѧ³É¹¦ÅäÖÆÁË0.400 mol¡¤L£­1µÄÑÎËᣬËûÓÖÓøÃÑÎËáÖкͺ¬0.4 g NaOHµÄNaOHÈÜÒº£¬Ôò¸ÃͬѧÐèÈ¡________mLÑÎËá¡££¨¾«È·µ½Ð¡Êýµãºóһ룩
¢Ú¼ÙÉè¸ÃͬѧÓÃÐÂÅäÖƵÄÑÎËáÖкͺ¬0.4 g NaOHµÄNaOHÈÜÒº£¬·¢ÏֱȢÙÖÐËùÇóÌå»ýƫС£¬Ôò¿ÉÄܵÄÔ­ÒòÊÇ________¡£
A£®Å¨ÑÎËá»Ó·¢£¬Å¨¶È²»×ã
B£®ÅäÖÆÈÜҺʱ£¬Î´Ï´µÓÉÕ±­
C£®ÅäÖÆÈÜҺʱ£¬¸©ÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß
D£®¼Óˮʱ³¬¹ý¿Ì¶ÈÏߣ¬ÓýºÍ·µÎ¹ÜÎü³ö

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø