ÌâÄ¿ÄÚÈÝ

A¡¢B¡¢C¡¢D¡¢EÊǺ˵çºÉÊýÒÀ´ÎÔö´óµÄÎåÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬AÔªËصÄÔ­×ÓºËÄÚÖ»ÓÐ1¸öÖÊ×Ó£¬BÔªËصÄÔ­×Ӱ뾶ÊÇÆäËùÔÚÖ÷×åÖÐ×îСµÄ£¬BµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½ÎªHBO3£»CÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊý±È´ÎÍâ²ã¶à4¸ö£»CµÄÒõÀë×ÓÓëDµÄÑôÀë×Ó¾ßÓÐÏàͬµÄµç×ÓÅŲ¼£¬Á½ÔªËØ¿ÉÐγɻ¯ºÏÎïD2C£»C¡¢EͬÖ÷×壮
£¨1£©DÔÚÖÜÆÚ±íÖеÄλÖà         £»BµÄÔ­×ÓºËÍâµç×ÓÅŲ¼Ê¾Òâͼ            £»
£¨2£©EÔªËØÐγÉ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½Îª               £»
£¨3£©ÔªËØC¡¢D¡¢EÐγɵÄÔ­×Ӱ뾶´óС¹ØϵÊÇ                £¨ÓÃÔªËØ·ûºÅ±íʾ£©¡£
£¨4£©C¡¢D¿ÉÐγɻ¯ºÏÎïD2C2£¬D2C2º¬ÓеĻ¯Ñ§¼üÊÇ                      £»
£¨5£©A¡¢CÁ½ÖÖÔªËØÐγɵÄÔ­×Ó¸öÊýÖ®±ÈΪ1:1µÄ»¯ºÏÎïµç×Óʽ                  £»
£¨6£©BµÄÇ⻯ÎïÓëBµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï·´Ó¦µÄÀë×Ó·½³Ìʽ                 ¡£

£¨1£©µÚÈýÖÜÆÚµÚ¢ñA×å   £¨2·Ö£©       £¨2·Ö£©
£¨2£©H2SO4                £¨2·Ö£©
£¨3£©Na  > S > O         (2·Ö£©
£¨4£©Àë×Ó¼ü¡¢¹²¼Û¼ü £¨2·Ö£©
£¨5£©      £¨2·Ö£©
£¨6£©NH3 + H+ = NH4+   £¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©AÔªËصÄÔ­×ÓºËÄÚÖ»ÓÐ1¸öÖÊ×Ó£¬AÊÇHÔªËØ£»BµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½ÎªHBO3£¬BΪµÚÎåÖ÷×åÔªËØ£¬ÇÒÔ­×Ӱ뾶ÔÚͬ×åÖÐ×îС£¬ËùÒÔBÊÇNÔªËØ£»CÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊý±È´ÎÍâ²ã¶à4¸ö£¬ËµÃ÷CÖ»ÄÜÓÐ2²ãµç×Ó£¬ËùÒÔCÊÇOÔªËØ£»CµÄÒõÀë×ÓÓëDµÄÑôÀë×Ó¾ßÓÐÏàͬµÄµç×ÓÅŲ¼£¬Á½ÔªËØ¿ÉÐγɻ¯ºÏÎïD2C£¬DΪ+1¼ÛÔªËØ£¬ËùÒÔDÊÇNa£»C¡¢EͬÖ÷×壬ÔòEÊÇSÔªËØ£®ËùÒÔNaÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊǵÚÈýÖÜÆÚµÚ¢ñA×壻NµÄÔ­×ÓºËÍâµç×ÓÅŲ¼Ê¾Òâͼ£»
£¨2£©SÔªËØÐγÉ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½ÎªH2SO4£»
£¨3£©¸ù¾ÝÔ­×Ӱ뾶µÄ´óС¹æÂÉ£¬ÔªËØO¡¢Na¡¢SÐγɵÄÔ­×Ӱ뾶´óС¹ØϵÊÇNa  > S > O£»
£¨4£©¹ýÑõ»¯ÄÆÖмÈÓÐÀë×Ó¼ü£¬ÓÖÓй²¼Û¼ü£»
£¨5£©A¡¢CÁ½ÖÖÔªËØÐγɵÄÔ­×Ó¸öÊýÖ®±ÈΪ1:1µÄ»¯ºÏÎïΪ¹ýÑõ»¯Ç⣬Êǹ²¼Û»¯ºÏÎµç×ÓʽΪ£»
£¨6£©BµÄÇ⻯ÎïÊÇ°±Æø£¬ÓëBµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÏõËá·´Ó¦Éú³ÉÏõËá泥¬Àë×Ó·½³ÌʽΪNH3 + H+ = NH4+¡£
¿¼µã£º¿¼²éÔªËØÍƶϣ¬ÔªËØÔÚÔªËØÖÜÆÚ±íÖÐλÖõÄÅжϣ¬µç×Óʽ¡¢Ô­×ӽṹʾÒâͼ¡¢Àë×Ó·½³Ìʽ¡¢»¯Ñ§Ê½µÄÊéд£¬»¯Ñ§¼üµÄÅжϣ¬Ô­×Ӱ뾶µÄ±È½Ï

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨15·Ö£©ÒÑÖªA¡¢B¡¢C¡¢D¡¢E¶¼ÊÇÖÜÆÚ±íÖÐÇ°ËÄÖÜÆÚµÄÔªËØ£¬ËüÃǵĺ˵çºÉÊýÒÀ´ÎÔö¼Ó¡£Ïà¹ØÐÅÏ¢ÈçϱíËùʾ£¬¸ù¾ÝÍƶϻشðÏÂÁÐÎÊÌ⣺£¨´ðÌâʱA¡¢B¡¢C¡¢D¡¢EÓÃËù¶ÔÓ¦µÄÔªËØ·ûºÅ±íʾ£©

A
AµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎﻯѧʽΪH2AO3
B
BÔªËصĵÚÒ»µçÀëÄܱÈͬÖÜÆÚÏàÁÚÁ½¸öÔªËض¼´ó
C
CÔ­×ÓÔÚͬÖÜÆÚÔ­×ÓÖа뾶×î´ó£¨Ï¡ÓÐÆøÌå³ýÍ⣩£¬Æäµ¥ÖÊÑæɫΪ»ÆÉ«
D
ZµÄ»ù̬ԭ×Ó×îÍâ²ãµç×ÓÅŲ¼Ê½Îª3s23p2
E
EÓëCλÓÚ²»Í¬ÖÜÆÚ£¬EÔ­×ÓºËÍâ×îÍâ²ãµç×ÓÊýÓëCÏàͬ£¬ÆäÓà¸÷²ãµç×Ó¾ù³äÂú
£¨1£©CÔÚÖÜÆÚ±íÖÐλÓÚµÚ¡¡¡¡¡¡ÖÜÆÚµÚ¡¡¡¡×壬E»ù̬ԭ×ÓºËÍâµç×ÓÅŲ¼Ê½ÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨2£©A¡¢B¡¢DÈýÖÖÔªËص縺ÐÔÓÉ´óµ½Ð¡ÅÅÁÐ˳ÐòΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡£¬ÆäÖÐAµÄ×î¸ß¼ÛÂÈ»¯Îï¹¹³É¾§ÌåµÄ΢Á£¼ä×÷ÓÃÁ¦Îª¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨3£©AºÍBµÄ×î¼òµ¥Ç⻯ÎïÖнÏÎȶ¨µÄÊÇ¡¡¡¡¡¡¡¡£¨Ìѧʽ£©¡£BµÄ×î¼òµ¥Ç⻯ÎïºÍEµÄºÚÉ«Ñõ»¯Îï¹ÌÌåÔÚ¼ÓÈÈʱ¿É·´Ó¦£¬Ð´³öÆä·´Ó¦·½³Ìʽ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨4)EµÄµ¥Öʺ͹ýÑõ»¯ÇâÔÚÏ¡ÁòËáÖпɷ´Ó¦£¬ÓÐÈ˽«Õâ¸ö·´Ó¦Éè¼Æ³ÉÔ­µç³Ø£¬Çëд³öÕý¼«·´Ó¦·½³Ìʽ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨5£©ÃºÈ¼ÉÕ²úÉúµÄÑÌÆøÖÐÓÐBµÄÑõ»¯Î»áÒýÆðÑÏÖصĻ·¾³ÎÊÌ⣬Òò´Ë£¬³£ÓÃAH4´ß»¯»¹Ô­ÒÔÏû³ýÎÛȾ£¬ÒÑÖª£º 
¢Ù AH4(g)+2 BO2£¨g)£½ B2(g)+AO2(g)+2H2O (g)  ¡÷H1£½£­867kJ£¯mol
¢Ú 2BO2(g) ?B2O4(g)  ¡÷H2=£­56.9 kJ£¯mol
д³öAH4ºÍB2O4·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡

£¨14·Ö£©Ï±íΪԪËØÖÜÆÚ±íµÄÒ»²¿·Ö£º

×å
ÖÜÆÚ
 
 
 
1
¢Ù
 
 
 
 
 
 
 
2
 
 
 
 
 
¢Ú
 
 
3
¢Û
 
 
¢Ü
 
¢Ý
¢Þ
 
¢ñ£®Óû¯Ñ§ÓÃÓï»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÔªËØ¢ÜÔÚÖÜÆÚ±íÖеÄλÖ㺡¡¡¡¡¡¡¡                                           ¡¡¡¡£»
£¨2£©¢Ú¢Û¢ÝµÄÔ­×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ¡¡¡¡¡¡¡¡                                   ¡¡¡¡£»
£¨3£©¢Ü¢Ý¢ÞµÄÆø̬Ç⻯ÎïµÄÎȶ¨ÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòÊÇ¡¡¡¡¡¡                       ¡¡£»
£¨4£©¢Ù¢Ú¢Û¢ÞÖеÄijЩԪËØ¿ÉÐγɼȺ¬Àë×Ó¼üÓÖº¬¼«ÐÔ¹²¼Û¼üµÄ»¯ºÏÎд³öÆäÖÐÁ½ÖÖ»¯
ºÏÎïµÄµç×Óʽ£º¡¡¡¡¡¡                                 ¡¡¡¡¡¡¡£
¢ò£®ÓÉÉÏÊö²¿·ÖÔªËØ×é³ÉµÄÎïÖʼ䣬ÔÚÒ»¶¨Ìõ¼þÏ£¬¿ÉÒÔ·¢ÉúÏÂͼÖеı仯£¬ÆäÖÐAÊÇÒ»
ÖÖµ­»ÆÉ«¹ÌÌå¡£Ôò£º

£¨1£©Ð´³ö¹ÌÌåAÓëÒºÌåX·´Ó¦µÄÀë×Ó·½³Ìʽ£º¡¡¡¡¡¡¡¡¡¡¡¡      ¡¡¡¡¡¡¡¡£»
£¨2£©ÆøÌåYÊÇÒ»ÖÖ´óÆøÎÛȾÎֱ½ÓÅÅ·Å»áÐγÉËáÓê¡£¿ÉÓÃÈÜÒºBÎüÊÕ£¬µ±BÓëYÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã1ÇÒÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ËùµÃÈÜÒºDµÄÈÜÖÊΪ¡¡¡¡ ¡¡¡¡£¨Ìѧʽ£©£»ÒÑÖªÈÜÒºDÏÔËáÐÔ£¬ÔòDÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ                           ¡¡£»
£¨3£©ÔÚ100 mL 18 mol/LµÄFŨÈÜÒºÖмÓÈë¹ýÁ¿Í­Æ¬£¬¼ÓÈÈʹ֮³ä·Ö·´Ó¦£¬²úÉúÆøÌåµÄÌå»ý£¨±ê¿öÏ£©¿ÉÄÜΪ£º¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
A£®40.32 L    B£®30.24 L     C£®20.16 L     D£®13.44 L

ϱíÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö, Õë¶Ô±íÖеĢ١«¢áÖÖÔªËØ,ÌîдÏÂÁпհ×:

   Ö÷×å
ÖÜÆÚ
¢ñA
¢òA
¢óA
¢ôA
¢õA
¢öA
¢÷A
0
  2
 
 
 
¢Ù
¢Ú
¢Û
 
 
  3
¢Ü
 
¢Ý
 
 
¢Þ
¢ß
¢à
  4
¢á
 
 
 
 
 
 
 
 
£¨1£©ÔÚÕâЩԪËØÖÐ,»¯Ñ§ÐÔÖÊ×î²»»îÆõÄÊÇ:        (Ìî¾ßÌ廯ѧÓÃÓï,ÏÂͬ)¡£
£¨2£©ÔÚ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÖУ¬ËáÐÔ×îÇ¿µÄ»¯ºÏÎïµÄ»¯Ñ§Ê½ÊÇ              £¬¼îÐÔ×îÇ¿µÄ»¯ºÏÎïµÄµç×ÓʽÊÇ£º               ¡£
£¨3£©×î¸ß¼ÛÑõ»¯ÎïÊÇÁ½ÐÔÑõ»¯ÎïµÄÔªËØÊÇ                £»Ð´³öËüµÄÑõ»¯ÎïÓëÇâÑõ»¯ÄÆ·´Ó¦µÄÀë×Ó·½³Ìʽ                                                ¡£
£¨4£©¢ÚÇ⻯ÎïÓë¢ÛµÄµ¥ÖÊÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                    ¡£
£¨5£© ¢Ú¿ÉÒÔÐγɶàÖÖÑõ»¯ÎÆäÖÐÒ»ÖÖÊǺì×ØÉ«ÆøÌ壬ÊÔÓ÷½³Ìʽ˵Ã÷¸ÃÆøÌå²»Ò˲ÉÓÃÅÅË®·¨ÊÕ¼¯µÄÔ­Òò                                  ¡£
£¨6£© Óýṹʽ±íʾԪËØ¢ÙÓë¢ÛÐγɵĻ¯ºÏÎï            £¬¸Ã»¯ºÏÎïÔÚ¹ÌÌåʱË׳Ơ       £¬ÊôÓÚ       ¾§Ì壬ָ³öËüµÄÒ»ÖÖÓÃ;               ¡£

ϱíΪԪËØÖÜÆÚ±íµÄÒ»²¿·Ö£º

    ×å
ÖÜÆÚ
 
 
 
1
¢Ù
 
 
 
 
 
 
 
2
 
 
 
 
 
¢Ú
 
 
3
¢Û
 
 
¢Ü
 
¢Ý
¢Þ
 
 
¢ñ£®Çë²ÎÕÕÔªËآ٣­¢ÞÔÚ±íÖеÄλÖã¬Óû¯Ñ§ÓÃÓï»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÔªËØ¢ÚµÄÀë×ӽṹʾÒâͼ______________¡£
£¨2£©¢Ú¡¢¢Û¡¢¢ÝµÄÀë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ_________________________¡£
£¨3£©ÔªËØ¢ÜÓë¢ÞÐγɻ¯ºÏÎïµÄµç×ÓʽÊÇ_________________________¡£
¢ò£®ÓÉÉÏÊö²¿·ÖÔªËØ×é³ÉµÄÎïÖʼ䣬ÔÚÒ»¶¨Ìõ¼þÏ£¬¿ÉÒÔ·¢ÉúÏÂͼËùʾµÄ±ä»¯£¬ÆäÖÐAÊÇÒ»ÖÖµ­»ÆÉ«¹ÌÌå¡£Çë»Ø´ð£º

£¨4£©Ð´³ö¹ÌÌåAÓëÒºÌåX·´Ó¦µÄÀë×Ó·½³Ìʽ                                     ¡£
£¨5£©ÆøÌåYÊÇÒ»ÖÖ´óÆøÎÛȾÎֱ½ÓÅÅ·Å»áÐγÉËáÓê¡£¿ÉÓÃÈÜÒºBÎüÊÕ£¬µ±BÓëYÎïÖʵÄÁ¿Ö®±ÈΪ1:1ÇÒÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ËùµÃÈÜÒºD¡£ÒÑÖªÈÜÒºDÏÔËáÐÔ£¬ÔòDÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ                              ¡£
£¨6£©ÔÚ500¡æ£¬101kPaʱ£¬ÆøÌåCÓëÆøÌåY·´Ó¦Éú³É0£®2molÆøÌåEʱ£¬·Å³öakJÈÈÁ¿£¬Ð´³ö¸ÃÌõ¼þÏ·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                                                        ¡£
£¨7£©ÈôÆøÌåCÓëYÔÚºãÈݾøÈȵÄÌõ¼þÏ·´Ó¦£¬ÏÂÁÐ˵·¨ÄÜÅжϴﵽƽºâ״̬µÄÊÇ           ¡£
A£®Î¶Ȳ»±ä   B£®ÆøÌå×Üѹǿ²»±ä   C£®»ìºÏÆøÌåµÄÃܶȲ»±ä    D£®»ìºÏÆøÌåµÄƽ¾ù·Ö×ÓÁ¿²»±ä

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

¾«Ó¢¼Ò½ÌÍø