Ô­×ÓÐòÊýÓÉСµ½´óÅÅÁеÄËÄÖÖ¶ÌÖÜÆÚÔªËØX¡¢Y¡¢Z¡¢W£¬ËÄÖÖÔªËصÄÔ­×ÓÐòÊýÖ®ºÍΪ32£¬ÔÚÖÜÆÚ±íÖÐXÊÇÔ­×Ӱ뾶×îСµÄÔªËØ£¬Y¡¢Z×óÓÒÏàÁÚ£¬Z¡¢WλÓÚͬÖ÷×å¡£
£¨1£©XÔªËØÊÇ        £¨ÌîÃû³Æ£© ,WÔÚÖÜÆÚ±íÖеÄλÖà                          ¡£
£¨2£©XÓëY Ðγɻ¯ºÏÎïµÄµç×ÓʽΪ        £¬XÓëW×é³ÉµÄ»¯ºÏÎïÖдæÔÚ       ¼ü£¨Ìî¡°Àë×Ó¡±¡°¹²¼Û¡±£©¡£
£¨3£©¢Ùд³öʵÑéÊÒÖƱ¸YX3µÄ»¯Ñ§·½³Ìʽ£º                                        ¡£
¢Ú¹¤ÒµÉÏÒ²¿ÉÒÔÑ¡ÔñºÏÊʵÄÌõ¼þ½øÐÐYX3µÄºÏ³É£¬ÈôÒÑÖªÔÚ¸ÃÌõ¼þÏÂÿÉú³É2molYX3ÆøÌåʱ·Å³ö
92.4kJµÄÈÈÁ¿£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                                           ¡£
£¨4£©ÓÉX¡¢Y¡¢Z¡¢WËÄÖÖÔªËØ×é³ÉµÄÒ»ÖÖÀë×Ó»¯ºÏÎïA¡£ÒÑÖª1mol AÄÜÓë×ãÁ¿NaOHŨÈÜÒº·´Ó¦Éú³É±ê×¼×´¿öÏÂ44.8LÆøÌå¡£ÔòAµÄÃû³ÆÊÇ          ¡£

£¨1£©H  µÚÈýÖÜÆÚµÚ¢öA×å
£¨2£© ¹²¼Û
£¨3£©¢Ù2NH4Cl + Ca£¨OH£©2CaCl2 + 2NH3¡ü+ 2H2O
¢ÚN2(g)+3H2(g)2NH3(g)   ¡÷H=-92.4KJ/mol
£¨4£©ÁòËá炙òÑÇÁòËáï§

½âÎöÊÔÌâ·ÖÎö£ºÔ­×ÓÐòÊýÓÉСµ½´óÅÅÁеÄËÄÖÖ¶ÌÖÜÆÚÔªËØX¡¢Y¡¢Z¡¢W£¬ÔÚÖÜÆÚ±íÖÐXÊÇÔ­×Ӱ뾶×îСµÄÔªËØ£¬ÔòXΪHÔªËØ£»Z¡¢WλÓÚͬÖ÷×壬ÉèZµÄÔ­×ÓÐòÊýΪx£¬ÔòWµÄÔ­×ÓÐòÊýΪx+8£¬Y¡¢Z×óÓÒÏàÁÚ£¬YµÄÔ­×ÓÐòÊýΪx-1£¬ÓÉËÄÖÖÔªËصÄÔ­×ÓÐòÊýÖ®ºÍΪ32£¬Ôò1+£¨x-1£©+x+£¨x+8£©£½32£¬½âµÃx£½8£¬¼´YΪNÔªËØ£¬ZΪOÔªËØ£¬WΪSÔªËØ¡£
£¨1£©ÓÉÉÏÊö·ÖÎö¿ÉÖª£¬XÔªËØÊÇHÔªËØ£¬WÊÇÁòÔªËØ£¬Î»ÓÚÔªËØÖÜÆÚ±íµÄµÚÈýÖÜÆÚµÚ¢öA×壬¹Ê´ð°¸Îª£ºH£»µÚÈýÖÜÆÚµÚ¢öA×å¡£
£¨2£©XÓëY ÐγɵĻ¯ºÏÎïÊÇ°±Æø£¬º¬Óм«ÐÔ¼üµÄ¹²¼Û»¯ºÏÎµç×ÓʽΪ¡£XÓëW×é³ÉµÄ»¯ºÏÎïÊÇH2S£¬·Ö×ÓÖÐH¡¢SÖ®¼äÒÔ¹²Óõç×Ó¶ÔÐγɻ¯ºÏÎÔòÒÔ¹²¼Û¼ü½áºÏ£¬¹Ê´ð°¸Îª£º¹²¼Û£»¡£
£¨3£©¢ÙʵÑéÊÒÖƱ¸°±ÆøÊÇÓÃÊìʯ»ÒºÍÂÈ»¯ï§¼ÓÈÈ£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2NH4Cl + Ca£¨OH£©2CaCl2 + 2NH3¡ü+ 2H2O¡£
¢ÚÿÉú³É2mol°±ÆøÆøÌåʱ·Å³ö92.4kJµÄÈÈÁ¿£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪN2(g)+3H2(g)2NH3(g)   ¡÷H£½£­92.4KJ/mol¡£
£¨4£©X¡¢Y¡¢Z¡¢WËÄÖÖÔªËØ×é³ÉµÄÒ»ÖÖÀë×Ó»¯ºÏÎïÖеÄÑôÀë×ÓΪNH4+£¬ÒõÀë×ÓΪÁòËá¸ùÀë×Ó¡¢ÁòËáÇâ¸ùÀë×Ó¡¢ÑÇÁòËá¸ùÀë×Ó¡¢ÑÇÁòËáÇâ¸ùÀë×Ó£¬µ«1mol AÄÜÓë×ãÁ¿NaOHŨÈÜÒº·´Ó¦Éú³É±ê×¼×´¿öÏÂ44.8LÆøÌ壬Ôò1molAÖк¬ÓÐ2molNH4+£¬ÔòAΪÁòËá炙òÑÇÁòËá李£
¿¼µã£º¿¼²é½á¹¹ÓëλÖùØϵ¡¢ÔªËØ»¯ºÏÎïµÄÐÔÖÊ¡¢³£Óû¯Ñ§ÓÃÓïµÈ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨7·Ö£©¾ö¶¨ÎïÖÊÐÔÖʵÄÖØÒªÒòËØÊÇÎïÖʽṹ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Í­Êǹý¶ÉÔªËØ¡£»¯ºÏÎïÖУ¬Í­³£³ÊÏÖ+1¼Û»ò+2¼Û¡£ÓÒͼΪijͭÑõ»¯ÎᄃÌå½á¹¹µ¥Ôª£¬¸ÃÑõ»¯ÎïµÄ»¯Ñ§Ê½Îª                  ¡£
£¨2£©µÚÈýÖÜÆÚ²¿·ÖÔªËØ·ú»¯ÎïµÄÈÛµã¼ûÏÂ±í£º

·ú»¯Îï
NaF
MgF2
SiF4
ÈÛµã/K
1266
1534
183
½âÊÍMgF2ÓëSiF4ÈÛµã²îÒìµÄÔ­Òò                                         ¡£
£¨3£©AºÍBΪµÚÈýÖÜÆÚÔªËØ£¬ÆäÔ­×ӵIJ¿·ÖµçÀëÄÜÈçϱíËùʾ£º
µçÀëÄÜ/kJ¡¤mol£­1
I1
I2
I3
I4
A
578
1817
2745
11578
B
738
1451
7733
10540
ÔòAµÄµç¸ºÐÔ       BµÄµç¸ºÐÔ£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£
£¨4£©Ñо¿ÎïÖÊ´ÅÐÔ±íÃ÷£º½ðÊôÑôÀë×Óº¬Î´³É¶Ôµç×ÓÔ½¶à£¬Ôò´ÅÐÔÔ½´ó£¬´Å¼Ç¼ÐÔÄÜÔ½ºÃ¡£Àë×ÓÐÍÑõ»¯ÎïV2O5¡¢CrO2¡¢Fe3O4ÖУ¬¸üÊʺÏ×÷¼Òô´ø´Å·ÛÔ­ÁϵÄÊÇ__________£¨Ìѧʽ£©¡£
£¨5£©ÊµÑéÊÒ¼ìÑéNi2+¿ÉÓö¡¶þͪë¿ÓëÖ®×÷ÓÃÉú³ÉÐȺìÉ«ÅäºÏÎï³Áµí¡£
¢ÚÔÚÅäºÏÎïÖÐÓû¯Ñ§¼üºÍÇâ¼ü±ê³öδ»­³öµÄ×÷ÓÃÁ¦£¨ÄøµÄÅäλÊýΪ4£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©ÔªËصĵÚÒ»µçÀëÄÜAl¡¡¡¡¡¡Si£¨Ìî¡°>¡±»ò¡°<¡±£©¡£
£¨2£©»ù̬Mn2+µÄºËÍâµç×ÓÅŲ¼Ê½Îª¡¡¡¡¡¡¡¡¡£
£¨3£©¹èÍ飨SinH2n+2£©µÄ·ÐµãÓëÆäÏà¶Ô·Ö×ÓÖÊÁ¿µÄ±ä»¯¹ØϵÈçͼËùʾ,³ÊÏÖÕâÖֱ仯¹ØϵµÄÔ­ÒòÊÇ                                ¡£

£¨4£©ÅðÉ°ÊǺ¬½á¾§Ë®µÄËÄÅðËáÄÆ£¬ÆäÒõÀë×ÓXm-£¨º¬B¡¢O¡¢HÈýÖÖÔªËØ£©µÄÇò¹÷Ä£ÐÍÈçͼËùʾ£º

¢ÙÔÚXm-ÖÐ,ÅðÔ­×Ó¹ìµÀµÄÔÓ»¯ÀàÐÍÓС¡¡¡¡¡£»Åäλ¼ü´æÔÚÓÚ¡¡¡¡Ô­×ÓÖ®¼ä£¨ÌîÔ­×ÓµÄÊý×Ö±êºÅ£©£»m=¡¡¡¡¡¡£¨ÌîÊý×Ö£©¡£
¢ÚÅðÉ°¾§ÌåÓÉNa+¡¢Xm-ºÍH2O¹¹³É£¬ËüÃÇÖ®¼ä´æÔÚµÄ×÷ÓÃÁ¦ÓС¡¡¡¡¡£¨ÌîÐòºÅ£©¡£
A¡¢Àë×Ó¼ü   B¡¢¹²¼Û¼ü   C¡¢½ðÊô¼ü  D¡¢·¶µÂ»ªÁ¦¡¡E¡¢Çâ¼ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨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

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ñо¿ÎïÖʵÄ΢¹Û½á¹¹£¬ÓÐÖúÓÚÈËÃÇÀí½âÎïÖʱ仯µÄ±¾ÖÊ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)C¡¢Si¡¢NÔªËصĵ縺ÐÔÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ                          £¬
C60ºÍ½ð¸Õʯ¶¼ÊÇ̼µÄͬËØÒìÐÎÌ壬¶þÕßÏà±È£¬ÈÛµã¸ßµÄÊÇ       £¬Ô­ÒòÊÇ                           ¡£
(2)A¡¢B¾ùΪ¶ÌÖÜÆÚ½ðÊôÔªËØ£¬ÒÀ¾Ý±íÖÐÊý¾Ý£¬Ð´³öBµÄ»ù̬ԭ×ӵĵç×ÓÅŲ¼Ê½£º               ¡£

µçÀëÄÜ/(kJ¡¤mol£­1)
I1
I2
I3
I4
A
932
1 821
15 390
21 771
B
738
1 451
7 733
10 540
(3)¹ý¶É½ðÊôÀë×ÓÓëË®·Ö×ÓÐγɵÄÅäºÏÎïÊÇ·ñÓÐÑÕÉ«£¬ÓëÆäd¹ìµÀµç×ÓÅŲ¼Óйء£Ò»°ãµØ£¬d0»òd10ÅŲ¼ÎÞÑÕÉ«£¬d1¡«d9ÅŲ¼ÓÐÑÕÉ«¡£Èç[Co(H2O)6]2£«ÏÔ·ÛºìÉ«¡£¾Ý´ËÅжϣº[Mn(H2O)6]2£«       (Ìî¡°ÎÞ¡±»ò¡°ÓС±)ÑÕÉ«¡£
(4)ÀûÓÃCO¿ÉÒԺϳɻ¯¹¤Ô­ÁÏCOCl2¡¢ÅäºÏÎïFe(CO)5µÈ¡£
¢ÙCOCl2·Ö×ӵĽṹʽΪ£¬Ã¿¸öCOCl2·Ö×ÓÄÚº¬ÓР      ¸ö¦Ò¼ü£¬       ¸ö¦Ð¼ü£¬ÆäÖÐÐÄÔ­×Ó²ÉÈ¡       ÔÓ»¯¹ìµÀ·½Ê½¡£
¢ÚFe(CO)5ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·Ö½â·´Ó¦£ºFe(CO)5(s)=Fe(s)£«5CO(g)¡£·´Ó¦¹ý³ÌÖУ¬¶ÏÁѵĻ¯Ñ§¼üÖ»ÓÐÅäλ¼ü£¬ÔòÐγɵĻ¯Ñ§¼üÀàÐÍÊÇ       ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

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

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ϱíΪ²¿·Ö¶ÌÖÜÆÚÔªËØ»¯ºÏ¼Û¼°ÏàÓ¦Ô­×Ӱ뾶µÄÊý¾Ý£º

ÔªËØÐÔÖÊ
ÔªËرàºÅ
A
B
C
D
E
F
G
H
Ô­×Ӱ뾶(nm)
0.102
0.110
0.117
0.074
0.075
0.071
0.099
0.077
×î¸ß»¯ºÏ¼Û
£«6
£«5
£«4
 
£«5
 
£«7
£«4
×îµÍ»¯ºÏ¼Û
£­2
£­3
£­4
£­2
£­3
£­1
£­1
£­4
 
ÒÑÖª£º
¢ÙAÓëD¿ÉÐγɻ¯ºÏÎïAD2¡¢AD3£»
¢ÚEÓëD¿ÉÐγɶàÖÖ»¯ºÏÎÆäÖÐED¡¢ED2Êdz£¼ûµÄ»¯ºÏÎC¿ÉÓÃÓÚÖƹâµç³Ø¡£
(1)EÔÚÖÜÆÚ±íÖÐλÖÃÊÇ               £»
(2)CºÍHµÄÆø̬Ç⻯ÎïµÄÎȶ¨ÐÔÇ¿Èõ¹ØϵΪ             (Ó÷Ö×Óʽ±íʾ)£»
(3)·Ö×Ó×é³ÉΪADG2µÄÎïÖÊÔÚË®ÖлáÇ¿ÁÒË®½â£¬²úÉúʹƷºìÈÜÒºÍÊÉ«µÄÎÞÉ«ÆøÌåºÍÒ»ÖÖÇ¿Ëá¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                 ¡£
(4)¹¤ÒµÉÏ¿ÉÓô¿¼îÈÜÒº´¦ÀíEDºÍED2£¬¸Ã·´Ó¦ÈçÏ£º
ED£«ED2£«Na2CO3=2       £«CO2
ºáÏßÉÏijÑεĻ¯Ñ§Ê½Ó¦Îª       ¡£
(5)ÔÚÒ»ÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦2AD2£«D22AD3 ¦¤H£½£­47 kJ/mol£¬ÔÚÉÏÊöƽºâÌåϵÖмÓÈë18D2£¬µ±Æ½ºâ·¢ÉúÒƶ¯ºó£¬AD2ÖÐ18DµÄ°Ù·Öº¬Á¿       (Ìî¡°Ôö¼Ó¡±¡°¼õÉÙ¡±»ò¡°²»±ä¡±)ÆäÔ­ÒòΪ                                    ¡£
(6)ÇëÉè¼ÆÒ»¸öʵÑé·½°¸£¬Ê¹Í­ºÍÏ¡µÄH2AD4ÈÜÒº·´Ó¦£¬µÃµ½À¶É«ÈÜÒººÍÇâÆø¡£»æ³ö¸ÃʵÑé·½°¸×°ÖÃͼ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÎåÖÖ¶ÌÖÜÆÚÔªËØA¡¢B¡¢C¡¢D¡¢EµÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó¡£AºÍCͬ×壬BºÍDͬ×壬CÀë×ÓºÍBÀë×Ó¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹¡£AºÍB¡¢D¡¢E¾ùÄÜÐγɹ²¼Û»¯ºÏÎï¡£AºÍBÐγɵĻ¯ºÏÎïÔÚË®ÖгʼîÐÔ£¬CºÍEÐγɵĻ¯ºÏÎïÔÚË®ÖгÊÖÐÐÔ¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÎåÖÖÔªËØÖУ¬Ô­×Ӱ뾶×î´óµÄÊÇ         £¬·Ç½ðÊôÐÔ×îÇ¿µÄÊÇ        (ÌîÔªËØ·ûºÅ)¡£
£¨2£©ÓÉAºÍB¡¢D¡¢EËùÐγɵĹ²¼ÛÐÍ»¯ºÏÎïÖУ¬ÈÈÎȶ¨ÐÔ×î²îµÄÊÇ         £¨Óû¯Ñ§Ê½±íʾ£©¡£
£¨3£©AºÍEÐγɵĻ¯ºÏÎïÓëAºÍBÐγɵĻ¯ºÏÎï·´Ó¦£¬²úÎïµÄ»¯Ñ§Ê½Îª        ,ÆäÖдæÔڵĻ¯Ñ§¼üÀàÐÍΪ            ¡£
£¨4£©µ¥ÖÊDÔÚ³ä×ãµÄµ¥ÖÊEÖÐȼÉÕ£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                    £»
£¨5£©µ¥ÖÊEÓëË®·´Ó¦µÄÀë×Ó·½³Ìʽ                                         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ijԪËصĺ˵çºÉÊýÊǵç×Ó²ãÊýµÄ5±¶,ÆäÖÊ×ÓÊýÊÇ×îÍâ²ãµç×ÓÊýµÄ3±¶,¸ÃÔªËصÄÔ­×ӽṹʾÒâͼΪ¡¡¡¡¡¡¡¡¡£ 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸