ÌâÄ¿ÄÚÈÝ

ºãÎÂϵļ×ÒÒÁ½ÃܱÕÈÝÆ÷£¬¼×ÈÝÆ÷Ìå»ý²»±ä£¬ÒÒÈÝÆ÷´øÓÐÀíÏë»îÈû£¬Ìå»ý¿É±ä£¬Á½ÈÝÆ÷µÄÆðʼ״̬ÍêÈ«Ïàͬ£¬¶¼³äÓÐCÆøÌ壮·¢Éú·´Ó¦C(g)A(g)£«B(g)£¬Ò»¶Îʱ¼äºó£¬ÈÝÆ÷Öеķ´Ó¦¾ù´ïµ½Æ½ºâ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

[¡¡¡¡]

A£®

ƽ¾ù·´Ó¦ËÙÂÊ£ºÒÒ£¾¼×

B£®

µ½´ïƽºâʱCµÄÎïÖʵÄÁ¿£ºÒÒ£¾¼×

C£®

µ½´ïƽºâʱCµÄת»¯ÂÊ£ºÒÒ£¾¼×

D£®

µ½´ïƽºâʱAµÄÎïÖʵÄÁ¿£º¼×£¾ÒÒ

´ð°¸£ºC
½âÎö£º

¼×ºãÈÝ£¬Ëæ×Å·´Ó¦µÄ½øÐУ¬Ñ¹Ç¿»áÔö´ó£¬Æ½¾ù·´Ó¦ËÙÂÊ´ó£¬Ïà±È֮ϣ¬×ª»¯ÂÊС£¬´ïµ½Æ½ºâʱ£¬CµÄÎïÖʵÄÁ¿´ó£®


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2013?ºÓÎ÷ÇøÒ»Ä££©ë£¨N2H4£©ºÍ°±ÊǵªµÄÁ½ÖÖ³£¼û»¯ºÏÎÔÚ¿ÆÑ§¼¼ÊõºÍÉú²úÖÐÓй㷺ӦÓã®Çë°´ÒªÇ󻨴ðÏÂÁÐÎÊÌ⣺
£¨1£©N2H4ÖÐNÔ­×ÓºËÍâ×îÍâ²ã´ïµ½8µç×ÓÎȶ¨½á¹¹£®Ð´³öN2H4µÄ½á¹¹Ê½£º
£®
£¨2£©ÊµÑéÊÒÓÃÁ½ÖÖ¹ÌÌåÖÆÈ¡NH3µÄ·´Ó¦»¯Ñ§·½³ÌʽΪ
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2H2O+2NH3¡ü
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2H2O+2NH3¡ü
£®
£¨3£©NH3ÓëNaClO·´Ó¦¿ÉµÃµ½ë£¨N2H4£©£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2NH3+NaClO=N2H4+NaCl+H2O
2NH3+NaClO=N2H4+NaCl+H2O
£®
£¨4£©ëÂÒ»¿ÕÆøÈ¼ÁÏµç³ØÊÇÒ»ÖÖ¼îÐÔ»·±£µç³Ø£¬¸Ãµç³Ø·Åµçʱ£¬¸º¼«µÄ·´Ó¦Ê½Îª
N2H4+4OH--4e-=N2+4H2O
N2H4+4OH--4e-=N2+4H2O
£®
£¨5£©¹¤ÒµÉú²úÄòËØµÄÔ­ÀíÊÇÒÔNH3ºÍCO2ΪԭÁϺϳÉÄòËØ[CO£¨NH2£©2]·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2NH3£¨g£©+CO2£¨g£©?CO£¨NH2£©2£¨l£©+H2O£¨l£©£¬¸Ã·´Ó¦µÄƽºâ³£ÊýºÍζȹØÏµÈçÏ£º
T/¡æ 165 175 185 195
K 111.9 74.1 50.6 34.8
¢Ùìʱä¡÷H
£¼
£¼
0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®
¢ÚÔÚÒ»¶¨Î¶ȺÍѹǿÏ£¬ÈôÔ­ÁÏÆøÖеÄNH3ºÍCO2µÄÎïÖʵÄÁ¿Ö®±È£¨°±Ì¼±È£©
n(NH3)
n(CO2)
=x
£¬ÈçͼÊǰ±Ì¼±È£¨x£©ÓëCO2ƽºâת»¯ÂÊ£¨¦Á£©µÄ¹ØÏµ£®¦ÁËæ×ÅxÔö´ó¶øÔö´óµÄÔ­ÒòÊÇ
c£¨NH3£©Ôö´ó£¬Æ½ºâÕýÏòÒÆ¶¯
c£¨NH3£©Ôö´ó£¬Æ½ºâÕýÏòÒÆ¶¯
£®Í¼ÖÐAµã´¦£¬NH3µÄƽºâת»¯ÂÊΪ
42%
42%
£®
£¨6£©ÔÚºãκãÈÝÃܱÕÈÝÆ÷Öа´Õռס¢ÒÒ¡¢±ûÈýÖÖ·½Ê½·Ö±ðͶÁÏ£¬·¢Éú·´Ó¦£ºN2£¨g£©+3H2£¨g£©?2NH3£¨g£©£¬²âµÃ¼×ÈÝÆ÷ÖÐH2µÄƽºâת»¯ÂÊΪ40%£®
n£¨N2£© n£¨H2£© n£¨NH3£©
¼× 1mol 3mol 0mol
ÒÒ 0.5mol 1.5mol 1mol
±û 0mol 0mol 4mol
¢ÙÅжÏÒÒÈÝÆ÷Öз´Ó¦½øÐеķ½ÏòÊÇ
ÄæÏò
ÄæÏò
£¨Ìî¡°ÕýÏò»ò¡°ÄæÏò¡±£©Òƶ¯£®
¢Ú´ïƽºâʱ£¬¼×¡¢ÒÒ¡¢±ûÈýÈÝÆ÷ÖÐNH3µÄÌå»ý·ÖÊý´óС˳ÐòΪ
¼×=ÒÒ=±û
¼×=ÒÒ=±û
£®
£¨2011?Ö£ÖݶþÄ££©ÀûÓÃËùѧ»¯Ñ§·´Ó¦Ô­Àí£¬½â¾öÒÔÏÂÎÊÌ⣺
£¨1£©Ä³ÈÜÒºº¬Á½ÖÖÏàͬÎïÖʵÄÁ¿µÄÈÜÖÊ£¬ÇÒÆäÖÐÖ»´æÔÚOHÒ»¡¢H+¡¢N
H
+
4
¡¢ClÒ»ËÄÖÖÀë×Ó£¬ÇÒc£¨N
H
+
4
£©£¾c£¨Cl-£©£¾c£¨OH-£©£¾c£¨H+£©£¬ÔòÕâÁ½ÖÖÈÜÖÊÊÇ
NH4ClºÍNH3?H2O
NH4ClºÍNH3?H2O
£®
£¨2£©0.1mol?L-1µÄ°±Ë®Óë0.05mol?L-1µÄÏ¡ÁòËáµÈÌå»ý»ìºÏ£¬ÓÃÀë×Ó·½³Ìʽ±íʾ»ìºÏºóÈÜÒºµÄËá¼îÐÔ£º
NH4++H2ONH3?H2O+H+
NH4++H2ONH3?H2O+H+
£®
£¨3£©ÒÑÖª£ºKsp£¨RX£©=1.8¡Á10-10£¬Ksp£¨RY£©=1.5¡Á10-16£¬Ksp£¨R2Z£©=2.0¡Á10-12£¬ÔòÄÑÈÜÑÎRX¡¢RYºÍR2ZµÄ±¥ºÍÈÜÒºÖУ¬R+Ũ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
R2Z£¾RX£¾RY
R2Z£¾RX£¾RY
£®
£¨4£©ÒÔʯīµç¼«µç½â100mL 0.1mol?L-1CuSO4ÈÜÒº£®ÈôÑô¼«ÉϲúÉúÆøÌåµÄÎïÖʵÄÁ¿Îª0.01mol£¬ÔòÒõ¼«ÉÏÎö³öCuµÄÖÊÁ¿Îª
0.64
0.64
g£®
£¨5£©Ïò20mLÁòËáºÍÑÎËáµÄ»ìºÏÒºÖÐÖðµÎ¼ÓÈëpH=13µÄBa£¨OH£©2ÈÜÒº£¬Éú³ÉBaSO4µÄÁ¿ÈçͼËùʾ£¬BµãÈÜÒºµÄpH=7£¨¼ÙÉèÌå»ý¿ÉÒÔÖ±½ÓÏà¼Ó£©£¬Ôòc£¨HCl£©=
0.2
0.2
mol?L-1£®
£¨6£©ÔÚζȡ¢ÈÝ»ýÏàͬµÄ3¸öÃܱÕÈÝÆ÷ÖУ¬°´Ï±íͶÈë·´Ó¦Î·¢Éú·´Ó¦£¨H2£¨g£©+
I2£¨g£©?2HI£¨g£©¡÷H=-14.9kJ?mol-1£©£¬ÔÚºãΡ¢ºãÈÝÌõ¼þÏ£¬²âµÃ·´Ó¦´ï
µ½Æ½ºâʱµÄÊý¾ÝÈçÏÂ±í£º
ÈÝÆ÷ ¼× ÒÒ ±û
·´Ó¦ÎïͶÈëÁ¿ 1mol H2¡¢1mol I2 2 mol HI 4 mol HI
HIµÄŨ¶È£¨mol?L-1£© C1 C2 C3
·´Ó¦µÄÄÜÁ¿±ä»¯ ·Å³öakJ ÎüÊÕbkJ ÎüÊÕckJ
·´Ó¦Îïת»¯ÂÊ a1 a2 a3
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
AC
AC
£®
A£®¦Á1+¦Á2=1   B£®2¦Á2=¦Á3 C£®a+b=14.9     D£®c1=c2=c3£®

ÇâÔªËØÓëµªÔªËØ¿É×é³É¶àÖÖ΢Á££¬ÈçNH3¡¢NH4+¡¢N2H4µÈ¡£
¢ñ£®£¨1£©·ÖÎö³£¼ûµÄH2OÓëH2O2¡¢CH4ÓëC2H6µÄ·Ö×ӽṹ£¬Ð´³öµÄN2H4µç×Óʽ______¡£
£¨2£©Ä³ÑÎN2H5ClÓëNH4ClÀàËÆ£¬ÊÇ¿ÉÈÜÓÚË®µÄÀë×Ó»¯ºÏÎÆäÈÜÒºÒòË®½â¶ø³ÊÈõËáÐÔ¡£N2H5ClÈÜÒºÏÔËáÐÔÔ­Òò£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©                           ¡£
£¨3£©ÓÐÆßÖÖÎïÖÊ£ºNH3¡¢Mn2O3¡¢ZnCl2¡¢MnO2¡¢NH4Cl¡¢ZnºÍH2O£¬ÊÇп¡ªÃÌµç³ØÖÐÑõ»¯»¹Ô­·´Ó¦µÄijЩ·´Ó¦ÎNH4ClΪÆäÖÐÖ®Ò»£©ºÍijЩÉú³ÉÎNH3ΪÆäÖÐÖ®Ò»£©¡£Ð´³öÉÏÊö»¯Ñ§·´Ó¦·½³Ìʽ£º_____________________________________________________¡£
¢ò£®ÔÚºãÎÂÌõ¼þÏ£¬ÆðʼʱÈÝ»ý¾ùΪ5 LµÄ¼×¡¢ÒÒÁ½ÃܱÕÈÝÆ÷ÖУ¨¼×ΪºãÈÝÈÝÆ÷¡¢ ÒÒΪºãѹÈÝÆ÷£©£¬¾ù½øÐз´Ó¦£ºN2+3H22NH3£¬ÓйØÊý¾Ý¼°Æ½ºâ×´Ì¬ÌØµã¼ûÏÂ±í¡£

ÈÝÆ÷

ÆðʼͶÈë

´ïƽºâʱ

¼×

2 mol N2

3 mol H2

1.5 mol NH3

ͬÖÖÎïÖʵÄÌå»ý·ÖÊýÏàͬ

ÒÒ

a mol N2

b mol H2

1.2 mo l NH3

£¨4£©ÏÂÁÐÄܱíÃ÷ÈÝÆ÷ÒÒÒ»¶¨´ïƽºâ״̬µÄÊÇ____________(Ìî×Öĸ)
A£®ÈÝÆ÷ÄڵĻìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»Ôٱ仯
B£®ÈÝÆ÷ÄڵĵªÔªËصÄÖÊÁ¿²»Ôٱ仯
C£®ÇâÆøµÄÉú³ÉËÙÂÊÓë°±ÆøµÄÏûºÄËÙÂÊÖ®±ÈΪ2¡Ã3
D£®ÐγÉ1 mol N¡ÔN¼üµÄͬʱÐγÉ6 mol N¡ªH¼ü
£¨5£©¼×ÈÝÆ÷ÖеªÆøµÄת»¯ÂÊΪ         ¡£
£¨6£©Æðʼʱ£¬ÈÝÆ÷ÒÒÊÇÈÝÆ÷¼×ѹǿµÄ_________±¶¡£

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

¾«Ó¢¼Ò½ÌÍø