A¡¢B¡¢C¡¢D¡¢E¾ùΪÖÐѧ»¯Ñ§µÄ³£¼ûµ¥ÖÊ»ò»¯ºÏÎËüÃÇÖ®¼äµÄ·´Ó¦¹ØϵÈçͼËùʾ£º
£¨1£©ÈôAÊǶÌÖÜÆÚÖÐÔ­×Ӱ뾶×î´óµÄÔªËع¹³ÉµÄµ¥ÖÊ£¬E¼È¿ÉÈÜÓÚÑÎËáÓÖ¿ÉÈÜÓÚNaOHÈÜÒº£¬EÈÜÓÚNaOHÈÜÒºµÄÀë×Ó·½³ÌʽÊÇ______£»¹¤ÒµÉÏÒ±Á¶AµÄ»¯Ñ§·´Ó¦·½³ÌʽÊÇ______£®
£¨2£©ÈôCÊǼȺ¬Óм«ÐÔ¼üÓÖº¬ÓзǼ«ÐÔ¼üµÄËÄÔ­×Ó·Ö×Ó£¬ÔòʵÑéÊÒÖÆÈ¡CµÄ»¯Ñ§·½³ÌʽÊÇ______£»1mol CÍêȫȼÉÕÉú³ÉҺ̬ˮʱ·ÅÈÈ1300kJ£¬ÔòCÍêȫȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽÊÇ______£®AÓëÑÎBµÄÈÜÒº·´Ó¦Ê±Ö»Éú³ÉÆøÌåC¡¢Ì¼Ëá¸Æ³ÁµíºÍË®£¬ÔòBµÄ»¯Ñ§Ê½ÊÇ______£®

½â£º£¨1£©AÊǶÌÖÜÆÚÖÐÔ­×Ӱ뾶×î´óµÄÔªËع¹³ÉµÄµ¥ÖÊ£¬Ó¦ÎªNa£¬E¼È¿ÉÈÜÓÚÑÎËáÓÖ¿ÉÈÜÓÚNaOHÈÜÒº£¬Ó¦ÎªAl£¨OH£©3£¬
Al£¨OH£©3ÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪAl£¨OH£©3+OH-=AlO2-+2H2O£¬¹¤ÒµÒ±Á¶NaÓõç½âÈÛÈÚµÄNaClµÄ·½·¨£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2NaCl£¨ÈÛÈÚ£© 2 Na+Cl2¡ü£¬
¹Ê´ð°¸Îª£ºAl£¨OH£©3+OH-=AlO2-+2H2O£»2NaCl£¨ÈÛÈÚ£© 2 Na+Cl2¡ü£»
£¨2£©¼Èº¬Óм«ÐÔ¼üÓÖº¬ÓзǼ«ÐÔ¼üµÄËÄÔ­×Ó·Ö×ÓÓÐC2H2»òH2O2£¬ÆäÖÐÄÜȼÉÕµÄÊÇC2H2£¬ÔòAΪCaC2£¬ÓëË®·´Ó¦Éú³ÉC2H2ºÍCa£¨OH£©2£¬ÓëBµÄÈÜÒº·´Ó¦Éú³ÉC2H2¡¢Ë®ºÍ̼Ëá¸Æ£¬
ÔòBΪCa£¨HCO3£©2£¬
1molC2H2ÍêȫȼÉÕÉú³ÉҺ̬ˮʱ·ÅÈÈ1300kJ£¬Ôò2molC2H2ÍêȫȼÉշųö2600kJµÄÈÈÁ¿£¬
ËùÒÔÍêȫȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ2C2H2£¨g£©+5O2£¨g£©=4CO2£¨g£©+2H2O£¨l£©¡÷H=-2600 kJ?mol-1£¬
¹Ê´ð°¸Îª£ºCaC2+2H2O¡úC2H2¡ü+Ca£¨OH£©2£»2C2H2£¨g£©+5O2£¨g£©=4CO2£¨g£©+2H2O£¨l£©¡÷H=-2600 kJ?mol-1£»Ca£¨HCO3£©2£®
·ÖÎö£º£¨1£©AÊǶÌÖÜÆÚÖÐÔ­×Ӱ뾶×î´óµÄÔªËع¹³ÉµÄµ¥ÖÊ£¬Ó¦ÎªNa£¬E¼È¿ÉÈÜÓÚÑÎËáÓÖ¿ÉÈÜÓÚNaOHÈÜÒº£¬Ó¦ÎªAl£¨OH£©3£¬¹¤ÒµÒ±Á¶NaÓõç½âÈÛÈÚµÄNaClµÄ·½·¨£»
£¨2£©¼Èº¬Óм«ÐÔ¼üÓÖº¬ÓзǼ«ÐÔ¼üµÄËÄÔ­×Ó·Ö×ÓÓÐC2H2»òH2O2£¬ÆäÖÐÄÜȼÉÕµÄÊÇC2H2£¬ÔòAΪCaC2£¬ÓëË®·´Ó¦Éú³ÉC2H2ºÍCa£¨OH£©2£¬ÓëBµÄÈÜÒº·´Ó¦Éú³ÉC2H2¡¢Ë®ºÍ̼Ëá¸Æ£¬
ÔòBΪCa£¨HCO3£©2£¬½áºÏÎïÖʵĻ¯Ñ§ÐÔÖʺÍÌâÄ¿ÒªÇó½â´ð¸ÃÌ⣮
µãÆÀ£º±¾Ì⿼²éÎÞ»úÎïµÄÍƶϣ¬ÌâÄ¿ÄѶÈÖеȣ¬±¾Ìâ×¢Òâ¼Èº¬Óм«ÐÔ¼üÓÖº¬ÓзǼ«ÐÔ¼üµÄËÄÔ­×Ó·Ö×ÓÓÐC2H2»òH2O2£¬ÆäÖÐÄÜȼÉÕµÄÊÇC2H2£¬×¢ÒâÈÈ»¯Ñ§·½³ÌʽµÄÊéд£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2012?Íò°²ÏØÄ£Ä⣩ÈçͼÖРA¡¢B¡¢C¡¢D¡¢E¾ùΪÓлú»¯ºÏÎÒÑÖª£ºCÄܸúNaHCO3·¢Éú·´Ó¦£¬CºÍDµÄÏà¶Ô·Ö×ÓÖÊÁ¿ÏàµÈ£¬ÇÒEΪÎÞÖ§Á´µÄ»¯ºÏÎ

¸ù¾ÝÉÏͼ»Ø´ðÎÊÌ⣺
£¨1£©ÒÑÖªEµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª102£¬ÆäÖÐ̼¡¢ÇâÁ½ÖÖÔªËصÄÖÊÁ¿·ÖÊý·Ö±ðΪ58.8%¡¢9.8%£¬ÆäÓàΪÑõ£¬ÔòEµÄ·Ö×ÓʽΪ
C5H10O2
C5H10O2
£»C·Ö×ÓÖеĹÙÄÜÍÅÃû³ÆÊÇ
ôÈ»ù
ôÈ»ù
£»»¯ºÏÎïB²»ÄÜ·¢ÉúµÄ·´Ó¦ÊÇ
e
e
£¨Ìî×ÖĸÐòºÅ£©£º
a£®¼Ó³É·´Ó¦  b£®È¡´ú·´Ó¦  c£®ÏûÈ¥·´Ó¦ d£®õ¥»¯·´Ó¦  e£®Ë®½â·´Ó¦  f£® Öû»·´Ó¦
£¨2£©·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ
CH3COOH+CH3CH2CH2OH
ŨÁòËá
¡÷
CH3COOCH2CH2CH3+H2O
CH3COOH+CH3CH2CH2OH
ŨÁòËá
¡÷
CH3COOCH2CH2CH3+H2O
£®
£¨3£©·´Ó¦¢ÚʵÑéÖмÓÈȵÄÄ¿µÄÊÇ£º
¢ñ
¼Ó¿ì·´Ó¦ËÙÂÊ
¼Ó¿ì·´Ó¦ËÙÂÊ
£»¢ò
¼°Ê±½«²úÎïÒÒËá±ûõ¥Õô³ö£¬ÒÔÀûÓÚƽºâÏòÉú³ÉÒÒËá±ûõ¥µÄ·½ÏòÒƶ¯
¼°Ê±½«²úÎïÒÒËá±ûõ¥Õô³ö£¬ÒÔÀûÓÚƽºâÏòÉú³ÉÒÒËá±ûõ¥µÄ·½ÏòÒƶ¯
£®
£¨4£©AµÄ½á¹¹¼òʽÊÇ
£®
£¨5£©Í¬Ê±·ûºÏÏÂÁÐÈý¸öÌõ¼þµÄBµÄͬ·ÖÒì¹¹ÌåµÄÊýÄ¿ÓÐËĸö£®
¢ñ£®º¬Óмä¶þÈ¡´ú±½»·½á¹¹¢ò£®ÊôÓÚ·Ç·¼ÏãËáõ¥¢ó£®Óë FeCl3 ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£®Ð´³öÆäÖÐÈÎÒâÒ»¸öͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ͼÖРA¡¢B¡¢C¡¢D¡¢E¾ùΪÓлú»¯ºÏÎÒÑÖª£ºCÄܸúNaHCO3·¢Éú·´Ó¦£¬CºÍDµÄÏà¶Ô·Ö×ÓÖÊÁ¿ÏàµÈ£¬ÇÒEΪÎÞÖ§Á´µÄ»¯ºÏÎ

¸ù¾Ýͼ»Ø´ðÎÊÌ⣺
£¨1£©C·Ö×ÓÖеĹÙÄÜÍÅÃû³ÆÊÇ£º
ôÈ»ù
ôÈ»ù
£»
ÏÂÁз´Ó¦ÖУ¬»¯ºÏÎïB²»ÄÜ·¢ÉúµÄ·´Ó¦ÊÇ
e
e
£¨Ìî×ÖĸÐòºÅ£©£º
a¡¢¼Ó³É·´Ó¦  b¡¢È¡´ú·´Ó¦  c¡¢ÏûÈ¥·´Ó¦    d¡¢õ¥»¯·´Ó¦  e¡¢Ë®½â·´Ó¦  f¡¢Öû»·´Ó¦
£¨2£©·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ
CH3COOH+CH3CH2CH2OH
ŨÁòËá
¼ÓÈÈ
CH3COOCH2CH2CH3+H2O
CH3COOH+CH3CH2CH2OH
ŨÁòËá
¼ÓÈÈ
CH3COOCH2CH2CH3+H2O
£®
£¨3£©AµÄ½á¹¹¼òʽÊÇ
£®
£¨4£©Í¬Ê±·ûºÏÏÂÁÐÈý¸öÌõ¼þµÄBµÄͬ·ÖÒì¹¹ÌåµÄÊýÄ¿ÓÐ
4
4
¸ö£®
¢ñ£®º¬Óмä¶þÈ¡´ú±½»·½á¹¹£»¢ò£®ÊôÓÚ·Ç·¼ÏãËáõ¥£»¢ó£®Óë FeCl3 ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£®
д³öÆäÖÐÈÎÒâÒ»¸öͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ
д³öËÄÕßÖ®Ò»¼´¿É
д³öËÄÕßÖ®Ò»¼´¿É

£¨5£©³£ÎÂÏ£¬½«CÈÜÒººÍNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬Á½ÖÖÈÜÒºµÄŨ¶ÈºÍ»ìºÏºóËùµÃÈÜÒºpHÈçÏÂ±í£º
ʵÑé±àºÅ CÎïÖʵÄÁ¿Å¨¶È£¨mol?L-1£© NaOHÎïÖʵÄÁ¿Å¨¶È£¨mol?L-1£© »ìºÏÈÜÒºµÄpH
m 0.1 0.1 pH=9
n 0.2 0.1 pH£¼7
´Óm×éÇé¿ö·ÖÎö£¬ËùµÃ»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨OH-£©=
10-5
10-5
mol?L-1£®
n×é»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
c£¨CH3COO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©
c£¨CH3COO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÈçͼËùʾ£¬A¡¢B¡¢C¡¢D¡¢E¾ùΪÓлú»¯ºÏÎÒÑÖª£ºCÄÜÓëNaHCO3ÈÜÒº·´Ó¦£¬CºÍDµÄÏà¶Ô·Ö×ÓÖÊÁ¿ÏàµÈ£¬ÇÒD¾­´ß»¯Ñõ»¯ºóµÄ²úÎï²»ÄÜ·¢ÉúÒø¾µ·´Ó¦£®

¸ù¾Ý¿òͼÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©C·Ö×ÓÖйÙÄÜÍŵÄÃû³ÆÊÇ
ôÈ»ù
ôÈ»ù
£»»¯ºÏÎïB²»ÄÜ·¢ÉúµÄ·´Ó¦ÊÇ
E
E
£¨ÌîÐòºÅ£©£®
A£®¼Ó³É·´Ó¦    B£®È¡´ú·´Ó¦    C£®ÏûÈ¥·´Ó¦    D£®õ¥»¯·´Ó¦    E£®Ë®½â·´Ó¦
£¨2£©Ð´³öEµÄ½á¹¹¼òʽ£º
CH3COOCH£¨CH3£©2
CH3COOCH£¨CH3£©2
£®
£¨3£©Ð´³ö·´Ó¦¢ÙµÄ»¯Ñ§·½³Ìʽ£º
£»Ð´³öBÎïÖÊ·¢ÉúËõ¾Û·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
£®
£¨4£©Í¬Ê±·ûºÏÏÂÁÐÈý¸öÌõ¼þµÄBµÄͬ·ÖÒì¹¹ÌåÓжàÖÖ£®
a£®±½»·ÉÏÓÐÁ½¸öÈ¡´ú»ùÇÒ±½»·ÉϵÄÒ»ÂÈ´úÎïÓÐ2ÖÖ£»
b£®ÄÜÓëFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£»
c£®ÄÜ·¢ÉúË®½â·´Ó¦£®¢Ù1mol BµÄͬ·ÖÒì¹¹ÌåÄÜÓë3mol NaOH·´Ó¦£¬ÊÔд³öÆä½á¹¹¼òʽ£º
£®
¢ÚÆäÖÐÓÐÒ»ÖÖͬ·ÖÒì¹¹ÌåµÄºË´Å¹²ÕñÇâÆ×ÓÐ6×é·å£¬ÇÒ·åÃæ»ýÖ®±ÈΪ1£º2£º2£º2£º2£º1£¬Ôò¸Ãͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽΪ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2008?º£ÄÏ£©A¡¢B¡¢C¡¢D¡¢E¾ùΪÓлú»¯ºÏÎËüÃÇÖ®¼äµÄ¹ØϵÈçͼËùʾ£¨Ìáʾ£ºRCH=CHR¡äÔÚËáÐÔ¸ßÃÌËá¼ØÈÜÒºÖз´Ó¦Éú³ÉRCOOHºÍR¡äCOOH£¬ÆäÖÐRºÍR¡äΪÍé»ù£©£®

£¨1£©Ö±Á´»¯ºÏÎïAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Ð¡ÓÚ90£¬A·Ö×ÓÖÐ̼¡¢ÇâÔªËصÄ×ÜÖÊÁ¿·ÖÊýΪ0.814£¬ÆäÓàΪÑõÔªËØ£¬ÔòAµÄ·Ö×ÓʽΪ
C5H10O
C5H10O
£®
£¨2£©ÒÑÖªBÓëNaHCO3ÈÜÒºÍêÈ«·´Ó¦£¬ÆäÎïÖʵÄÁ¿Ö®±ÈΪ1£º2£¬ÔòÔÚŨÁòËáµÄ´ß»¯Ï£¬BÓë×ãÁ¿µÄC2H5OH·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
HOOC-CH2-COOH+2C2H5OH
ŨÁòËá
¡÷
C2H5OOC-CH2-COOC2H5+2H2O
HOOC-CH2-COOH+2C2H5OH
ŨÁòËá
¡÷
C2H5OOC-CH2-COOC2H5+2H2O
£¬·´Ó¦ÀàÐÍΪ
õ¥»¯·´Ó¦
õ¥»¯·´Ó¦
£®
£¨3£©A¿ÉÒÔÓë½ðÊôÄÆ×÷Ó÷ųöÇâÆø£¬ÄÜʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«£¬ÔòAµÄ½á¹¹¼òʽÊÇ
HO-CH2-CH2-CH=CH-CH3
HO-CH2-CH2-CH=CH-CH3
£®
£¨4£©DµÄͬ·ÖÒì¹¹ÌåÖУ¬ÄÜÓëNaHCO3ÈÜÒº·´Ó¦·Å³öCO2µÄÓÐ
2
2
£®ÖÖ£¬ÆäÏàÓ¦µÄ½á¹¹¼òʽÊÇ
CH3CH2CH2COOH¡¢CH3CH£¨CH3£©COOH
CH3CH2CH2COOH¡¢CH3CH£¨CH3£©COOH
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡¢B¡¢C¡¢D¡¢E¾ùº¬ÓÐͬһÖÖ¶ÌÖÜÆÚÔªËØ£¬ÆäÖÐAÊǵ¥ÖÊ£¬B³£ÎÂÏÂÊÇÆø̬Ç⻯ÎC¡¢DÊÇÑõ»¯ÎEÊÇDºÍË®·´Ó¦µÄ²úÎϱíËùÁи÷×éÎïÖÊÖУ¬ÎïÖÊÖ®¼äͨ¹ýÒ»²½·´Ó¦¾ÍÄÜʵÏÖÈçͼËùʾת»¯µÄÊÇ£¨¡¡¡¡£©
 ÎïÖʱàºÅ  ÎïÖÊת»¯¹Øϵ  A  D  E
 ¢Ù  ¾«Ó¢¼Ò½ÌÍø  Si  SiO2  H2SiO3
 ¢Ú  N2  NO2  HNO3
 ¢Û  S  SO3  H2SO4
 ¢Ü  Na  Na2O2  NaOH
A¡¢¢Ú¢ÛB¡¢¢Ú¢Ü
C¡¢¢Ù¢Û¢ÜD¡¢¢Ù¢Ú¢Û¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸