¡¾»¯Ñ§¡ª¡ªÑ¡ÐÞ3£ºÎïÖʽṹÓëÐÔÖÊ¡¿(15·Ö)
±×åÔªËصĵ¥Öʺͻ¯ºÏÎïºÜ¶à£¬ÎÒÃÇ¿ÉÒÔÀûÓÃËùѧÎïÖʽṹÓëÐÔÖʵÄÏà¹Ø֪ʶȥÈÏʶºÍÀí½âËüÃÇ¡£
£¨1£©Â±×åÔªËØλÓÚÔªËØÖÜÆÚ±íµÄ_________Çø£»äåµÄ¼Ûµç×ÓÅŲ¼Ê½Îª____________________¡£
£¨2£©ÔÚÒ»¶¨Å¨¶ÈµÄÈÜÒºÖУ¬Çâ·úËáÊÇÒÔ¶þ·Ö×ÓµÞºÏ(HF)2ÐÎʽ´æÔڵġ£Ê¹Çâ·úËá·Ö×ӵ޺ϵÄ×÷ÓÃÁ¦ÊÇ________¡£
£¨3£©Çë¸ù¾ÝϱíÌṩµÄµÚÒ»µçÀëÄÜÊý¾ÝÅжϣ¬×îÓпÉÄÜÉú³É½ÏÎȶ¨µÄµ¥ºËÑôÀë×ӵıËØÔ­×ÓÊÇ_________¡£

 
·ú
ÂÈ
äå
µâ
µÚÒ»µçÀëÄÜ
(kJ/mol)
1681
1251
1140
1008
£¨4£©ÒÑÖªµâËá(HIO3)ºÍ¸ßµâËá(H5IO6)µÄ½á¹¹·Ö±ðÈçͼI¡¢IIËùʾ£º

Çë±È½Ï¶þÕßËáÐÔÇ¿Èõ£ºHIO3_____ H5IO6(Ìî¡°£¾¡±¡¢ ¡°£¼¡±»ò¡°£½¡±)¡£
£¨5£©ÒÑÖªClO2£­ Îª½ÇÐÍ£¬ÖÐÐÄÂÈÔ­×ÓÖÜΧÓÐËĶԼ۲ãµç×Ó¡£ClO2£­ ÖÐÐÄÂÈÔ­×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪ___________£¬Ð´³öÒ»¸öClO2£­ µÄµÈµç×ÓÌå__________¡£
£¨6£©ÏÂͼΪµâ¾§Ì徧°û½á¹¹¡£ÓйØ˵·¨ÖÐÕýÈ·µÄÊÇ_____________¡£

A£®µâ·Ö×ÓµÄÅÅÁÐÓÐ2ÖÖ²»Í¬µÄÈ¡Ïò£¬2ÖÖÈ¡Ïò²»Í¬µÄµâ·Ö×ÓÒÔ4ÅäλÊý½»ÌæÅäλÐγɲã½á¹¹
B£®Óþù̯·¨¿É֪ƽ¾ùÿ¸ö¾§°ûÖÐÓÐ4¸öµâÔ­×Ó
C£®µâ¾§ÌåΪÎÞÏÞÑÓÉìµÄ¿Õ¼ä½á¹¹£¬ÊÇÔ­×Ó¾§Ìå
D£®µâ¾§ÌåÖдæÔÚµÄÏ໥×÷ÓÃÓзǼ«ÐÔ¼üºÍ·¶µÂ»ªÁ¦
£¨7£©ÒÑÖªCaF2¾§Ìå(¼ûͼ)µÄÃܶÈΪ¦Ñg/cm3£¬NAΪ°¢·ü¼ÓµÂÂÞ³£Êý£¬ÀâÉÏÏàÁÚµÄÁ½¸öCa2£«µÄºË¼ä¾àΪa cm£¬ÔòCaF2µÄÏà¶Ô·Ö×ÓÖÊÁ¿¿ÉÒÔ±íʾΪ___________¡£

£¨1£©P (1·Ö)4S24P5(1·Ö)    £¨2£© Çâ¼ü(1·Ö)
£¨3£©µâ(2·Ö)       £¨4£©£¾(2·Ö)
£¨5£©sp3 (2·Ö) Cl2O (»òOF2µÈºÏÀí´ð°¸) (2·Ö)
£¨6£©AD(2·Ö)
£¨7£©a3¦ÑNA/£´(2·Ö)

½âÎöÊÔÌâ·ÖÎö£º¢Å±×åÔªËØλÓÚÖÜÆÚ±íµÚ17ÁÐ(PÇø)£¬äå(Ö÷×åÔªËØ)Ô­×ӵļ۵ç×ÓÅŲ¼Îª4S24P5£»
¢ÆÇâ·úËá·Ö×Ó¼äͨ¹ýÇâ¼üµÞºÏ³É(HF)2¡£
¢ÇÓÉÓÚµâµÄµÚÒ»µçÀëÄÜÔÚ±ËØÔ­×ÓÖÐÏà¶Ô½ÏС£¬×îÓпÉÄÜÉú³É½ÏÎȶ¨µÄµ¥ºËÑôÀë×Ó¡£
¢ÈHIO3µÄ·ÇôÇ»ùÑõÔ­×ÓÊý£¨2£©¶à£¬ËáÐÔ½ÏH5IO6 [·ÇôÇ»ùÑõÔ­×ÓÊýΪ1]Ç¿¡£
¢ÉÓÉÓÚClO2£­ µÄÖÐÐÄÂÈÔ­×ÓÖÜΧÓÐËĶԼ۲ãµç×Ó£¬ÆäÖÐÐÄÂÈÔ­×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪsp3£»¸ù¾Ý¡°Ôö¼õÕÒÁÚ¡¢Í¬×廥»»¡±µÄÔ­Ôò¿ÉÈ·¶¨ClO2£­ µÄµÈµç×ÓÌåΪCl2O»òOF2¡£
¢Êµâ¾§Ì徧°û½á¹¹ÊôÓÚÃæÐÄÁ¢·½¾§°û£¬Ã¿¸ö¾§°ûÖк¬ÓÐ8¡Á£«6¡Á£½4¸öµâ·Ö×Ó£¬µ«µâÔ­×ÓÊýΪ8£¬BÏî´íÎ󣻵⾧ÌåÊôÓÚ·Ö×Ó¾§Ì壬CÏî´íÎó¡£
¢ËÔÚCaF2¾§°ûÖк¬ÓÐCa2£«£º8¡Á£«6¡Á£½4¸ö£¬º¬ÓÐF£­£º8¸ö£¬Ï൱ÓÚÓÐ4¸öCaF2£¬Ôò¦Ñ£½£¬»¯¼òµÃM£½)a3¦ÑNA/£´¡£
¿¼µã£º±¾Ì⿼²éÎïÖʽṹÓëÐÔÖÊ(ÔªËصķÖÇø¡¢¼Ûµç×ÓÅŲ¼Ê½¡¢Çâ¼ü¡¢µÚÒ»µçÀëÄÜ¡¢º¬ÑõËáµÄËáÐÔ´óС¡¢¾§°ûµÄÏà¹Ø¼ÆËãÓëÅжÏ)¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÔªËØÖÜÆÚ±íÓëÔªËØÖÜÆÚÂÉÔÚѧϰ¡¢Ñо¿ºÍÉú²úʵ¼ùÖÐÓкÜÖØÒªµÄ×÷Óá£Ï±íÁгöÁË¢Ù¡«¢á¾ÅÖÖÔªËØÔÚÖÜÆÚ±íÖеÄλÖá£

×å
ÖÜÆÚ
¢ñA
¢òA
¢óA
¢ôA
¢õA
¢öA
¢÷A
0
2
 
 
 
¢Ý
 
¢Þ
 
 
3
¢Ù
¢Û
¢Ü
 
 
 
¢ß
¢á
4
¢Ú
 
 
 
 
 
¢à
 
Çë»Ø´ð£º£¨1£©Ðγɻ¯ºÏÎïÖÖÀà×î¶àµÄÔªËØÊÇ          (ÌîÔªËØ·ûºÅ)¡£
£¨2£©ÔÚ¢Ù¡¢¢Ú¡¢¢ÛÈýÖÖÔªËصÄÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÖУ¬¼îÐÔ×îÇ¿µÄÊÇ         (Ìѧʽ)¡£ 
£¨3£©¢Ù¡¢¢Ú¡¢¢ÛÈýÖÖÔªËØ°´Àë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòÒÀ´ÎΪ               (ÌîÀë×Ó·ûºÅ)¡£
£¨4£©¢ÞÔªËØÐγɵľßÓÐÇ¿Ñõ»¯ÐÔµÄÇ⻯Îï½á¹¹Ê½ÊÇ              £¬ ¸ÃÔªËØÁíÒ»ÖÖÇ⻯ÎïÔÚ³£ÎÂÏÂÓë¢Ú·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                     ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨12·Ö£©¡¾Ñ¡×öÌâ¡¿±¾Ìâ°üÀ¨A¡¢BÁ½Ð¡Ì⣬ÇëÑ¡¶¨ÆäÖÐһСÌ⣬²¢ÔÚÏàÓ¦µÄ´ðÌâÇøÓòÄÚ×÷´ð¡£Èô¶à×ö£¬Ôò°´AСÌâÆÀ·Ö¡£
A£®[ÎïÖʽṹÓëÐÔÖÊ]
ϱíΪԪËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬ÆäÖеÄ×Öĸ´ú±íÏàÓ¦µÄÔªËØ¡£

a
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
b
c
d
 
 
 
e
 
 
 
 
 
 
 
 
 
 
f
 
g
 
 
 
 
 
 
 
 
 
 
 
h
I
 
 
 
 
 
 
 
 
£¨1£©ÔªËØhµÄËļÛÑôÀë×ÓµÄÍâΧµç×ÓÅŲ¼Ê½Îª     ¡£
£¨2£©ÔªËØc¡¢d¡¢e¡¢fµÄµÚÒ»µçÀëÄÜ£¨I1£©ÓÉСµ½´óµÄ˳ÐòΪ     ¡££¨ÓÃÏàÓ¦ÔªËصÄÔªËØ·ûºÅ±íʾ£©
£¨3£©±íÖÐËùÁеÄÔªËØÖ®¼ä¿ÉÒÔÐγɶàÖÖÎÞ»ú»¯ºÏÎïºÍÓлú»¯ºÏÎÔòËüÃÇÐγɵĻ¯ºÏÎïÖ®Ò»¡ª¡ªÁÚ¼×»ù±½¼×È©µÄ·Ö×ÓÖÐ̼ԭ×Ó¹ìµÀµÄÔÓ»¯ÀàÐÍΪ     ¡£
1 mol ±½¼×È©·Ö×ÓÖк¬ÓЦҼüµÄÊýĿΪ     ¡£
£¨4£©ÔªËØdÓëeÐγɵĻ¯ºÏÎï³£ÓÃÓÚÖÆ×÷     ²ÄÁÏ£¬ÆäÔ­ÒòÊÇ     ¡£
£¨5£©±íÖÐÓйØÔªËØÐγɵÄÒ»ÖÖÀë×Ӻ͵¥ÖÊd3»¥ÎªµÈµç×ÓÌ壬Ôò¸ÃÀë×ӵĻ¯Ñ§Ê½Îª     ¡£
£¨6£©ÔªËØIµÄºÏ½ð¿ÉÓÃÀ´´¢´æaµÄµ¥ÖÊ£¬¸ÃºÏ½ðµÄ¾§°û½á¹¹ÈçͼËùʾ£¬Ôò´ËºÏ½ðµÄ»¯Ñ§Ê½Îª     ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÔªËØ£¬ÆäÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐBÓëCͬÖÜÆÚ£¬DÓëEºÍFͬÖÜÆÚ£¬AÓëDͬÖ÷×壬CÓëFͬÖ÷×壬CÔªËصÄÔ­×Ó×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄÈý±¶£¬DÊÇËùÔÚÖÜÆÚÔ­×Ӱ뾶×î´óµÄÖ÷×åÔªËØ¡£ÓÖÖªÁùÖÖÔªËØËùÐγɵij£¼ûµ¥ÖÊÔÚ³£Î³£Ñ¹ÏÂÓÐÈýÖÖÊÇÆøÌ壬ÈýÖÖÊǹÌÌå¡£
Çë»Ø´ðÏÂÁÐÎÊÌâ¡£
(1)ÔªËØDÔÚÖÜÆÚ±íÖеÄλÖà             ¡£
(2)C¡¢D¡¢FÈýÖÖÔªËØÐγɵļòµ¥Àë×ӵİ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ(ÓÃÀë×Ó·ûºÅ±íʾ)        ¡£
(3)ÈôEÊǷǽðÊôÔªËØ£¬Æäµ¥ÖÊÔÚµç×Ó¹¤ÒµÖÐÓÐÖØÒªÓ¦Óã¬Çëд³öÆäÑõ»¯ÎïÈÜÓÚÇ¿¼îÈÜÒºµÄÀë×Ó·½³Ìʽ£º                                  ¡£
(4)ÈôEÊǽðÊôÔªËØ£¬Æäµ¥ÖÊÓëÑõ»¯Ìú·´Ó¦³£ÓÃÓÚº¸½Ó¸Ö¹ì£¬Çëд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º    ¡£Ò±Á¶½ðÊôEʱ£¬ÓÃʯī×öµç¼«¡£EÔÚ(Ìî¡°Òõ¡±»ò¡°Ñô¡±)      ¼«µÃµ½£¬µç½â¹ý³ÌÖУ¬Ñô¼«Ê¯Ä«ÐèÒª²»¶Ï²¹³ä£¬½áºÏµç¼«·´Ó¦ËµÃ÷ÆäÔ­ÒòÊÇ                 ¡£
(5)ÓÉA¡¢B¡¢CÈýÖÖÔªËØÒÔÔ­×Ó¸öÊý±È4£º2£º3Ðγɻ¯ºÏÎïX£¬XÖÐËùº¬»¯Ñ§¼üÀàÐÍÓР     ¡£ÍÁÈÀÖк¬ÓÐXÖеÄÑôÀë×ÓÔÚÏõ»¯Ï¸¾úµÄ´ß»¯×÷ÓÃϱ»ÑõÆøÑõ»¯ÎªÆäÒõÀë×Ó£¬Ð´³öÆäÀë×Ó·½³Ìʽ£º   ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨14·Ö£©A¡¢B¡¢C¡¢D 4ÖÖÔªËØ£¬AÔªËØËù´¦µÄÖÜÆÚÊý¡¢Ö÷×åÐòÊý¡¢Ô­×ÓÐòÊý¾ùÏàµÈ£»BµÄÔ­×Ӱ뾶ÊÇÆäËùÔÚÖ÷×åÖÐ×îСµÄ£¬BµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½ÎªHBO3£»CÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊý±È´ÎÍâ²ãÉÙ2¸ö£»CµÄÒõÀë×ÓÓëDµÄÑôÀë×Ó¾ßÓÐÏàͬµÄµç×ÓÅŲ¼£¬Á½ÔªËØ¿ÉÐγɻ¯ºÏÎïD2C£®
£¨1£©BÔªËصÄÃû³Æ___      _____£»BÔÚÖÜÆÚ±íÖеÄλÖõڠ ______ÖÜÆÚ£¬µÚ________×壻
£¨2£©A¡¢BÐγɵĻ¯ºÏÎïµÄµç×Óʽ____                       ____£»
£¨3£©CµÄÔªËØ·ûºÅ________£¬CµÄ×î¸ß¼ÛÑõ»¯ÎïµÄ»¯Ñ§Ê½____              ____£»
£¨4£© DµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïµÄ»¯Ñ§Ê½____     ____£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(15·Ö)
A¡¢B¡¢D¡¢E¡¢FΪ¶ÌÖÜÆÚÔªËØ£¬·Ç½ðÊôÔªËØA×îÍâ²ãµç×ÓÊýÓëÆäÖÜÆÚÊýÏàͬ£¬BµÄ×îÍâ²ãµç×ÓÊýÊÇÆäËùÔÚÖÜÆÚÊýµÄ2±¶¡£B ÔÚDÖгä·ÖȼÉÕÄÜÉú³ÉÆä×î¸ß¼Û»¯ºÏÎïBD2¡£E£«ÓëD2£­¾ßÓÐÏàͬµÄµç×ÓÊý¡£AÔÚFÖÐȼÉÕ£¬²úÎïÈÜÓÚË®µÃµ½Ò»ÖÖÇ¿Ëá¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ           £¬Ð´³öÒ»ÖÖ¹¤ÒµÖƱ¸µ¥ÖÊFµÄÀë×Ó·½³Ìʽ             ¡£
£¨2£©B¡¢D¡¢E×é³ÉµÄÒ»ÖÖÑÎÖУ¬EµÄÖÊÁ¿·ÖÊýΪ43%£¬ÆäË×ÃûΪ              £¬ÆäË®ÈÜÒºÓëFµ¥ÖÊ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                                  £»ÔÚ²úÎïÖмÓÈëÉÙÁ¿KI£¬·´Ó¦ºó¼ÓÈËCC14²¢Õñµ´£¬ÓÐ »ú²ãÏÔ       É«¡£
£¨3£©ÓÉÕâЩԪËØ×é³ÉµÄÎïÖÊ£¬Æä×é³ÉºÍ½á¹¹ÐÅÏ¢ÈçÏÂ±í£º

ÎïÖÊ
 
×é³ÉºÍ½á¹¹ÐÅÏ¢
 
a
 
º¬ÓÐAµÄ¶þÔªÀë×Ó»¯ºÏÎï
 
b
 
º¬ÓзǼ«ÐÔ¹²¼Û¼üµÄ¶þÔªÀë×Ó»¯ºÏÎÇÒÔ­×ÓÊýÖ®±ÈΪ1:1
 
c
 
»¯Ñ§×é³ÉΪBDF2
 
d
 
Ö»´æÔÚÒ»ÖÖÀàÐÍ×÷ÓÃÁ¦Çҿɵ¼µçµÄµ¥Öʾ§Ìå
 
aµÄ»¯Ñ§Ê½Îª     £»bµÄ»¯Ñ§Ê½Îª              £»£»cµÄµç×ÓʽΪ                £»
dµÄ¾§ÌåÀàÐÍÊÇ                 ¡£
£¨4£©ÓÉAºÍB¡¢DÔªËØ×é³ÉµÄÁ½ÖÖ¶þÔª»¯ºÏÎïÐγÉÒ»ÀàÐÂÄÜÔ´ÎïÖÊ¡£Ò»ÖÖ»¯ºÏÎï·Ö×Óͨ¹ý    ¼ü¹¹³É¾ßÓпÕÇ»µÄ¹ÌÌ壻ÁíÒ»ÖÖ»¯ºÏÎï(ÕÓÆøµÄÖ÷Òª³É·Ö)·Ö×Ó½øÈë¸Ã¿ÕÇ»£¬Æä·Ö×ӵĿռä½á¹¹Îª      ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(9·Ö)ÔªËØÖÜÆÚÂÉÊÇÖ¸µ¼ÎÒÃÇѧϰԪËؼ°Æ仯ºÏÎï֪ʶµÄÖØÒª¹¤¾ß¡£ÒÑÖªÑõ×åÔªËØ£¨²»º¬Po£©µÄ²¿·Ö֪ʶÈçϱíËùʾ£¬ 

ÔªËØ
8O
16S
34Se
52Te
µ¥ÖÊÈ۵㣨¡æ£©
-218.4
113
 
450
µ¥Öʷе㣨¡æ£©
-183
444.6
685
1390
ÔªËØÖ÷Òª»¯ºÏ¼Û
-2
-2,+4,+6
-2,+4,+6
 
Ô­×Ӱ뾶
Öð½¥Ôö´ó
µ¥ÖÊÓëH2·´Ó¦Çé¿ö
µãȼʱÒ×»¯ºÏ
¼ÓÈÈ»¯ºÏ
¼ÓÈÈÄÑ»¯ºÏ
²»ÄÜÖ±½Ó»¯ºÏ
Çë×ܽá±íÖÐ֪ʶ¹æÂÉ£¬²¢ÒÀ¾ÝÔªËØÖÜÆÚÂɻشðÏÂÁÐÎÊÌ⣺
£¨1£©µ¥ÖÊÎøµÄÈ۵㷶Χ¿ÉÄÜÊÇ________________¡£
£¨2£©ÔªËØíÚµÄÖ÷Òª»¯ºÏ¼Û¿ÉÄÜÓÐ________________¡£
£¨3£©Áò¡¢Îø¡¢íÚµÄÇ⻯ÎïË®ÈÜÒºµÄËáÐÔÓÉÇ¿ÖÁÈõµÄ˳ÐòÊÇ________________(Óû¯Ñ§Ê½±íʾ)¡£
£¨4£©ÇâÎøËáÓнÏÇ¿µÄ__________£¨Ìî¡°Ñõ»¯ÐÔ¡±»ò¡°»¹Ô­ÐÔ¡±£©£¬Òò´Ë¶ÖÃÔÚ¿ÕÆøÖг¤ÆÚ±£´æÒ×±äÖÊ£¬Æä¿ÉÄÜ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________________________________¡£
£¨5£©ÏÂͼËùʾΪÑõ×åÔªËص¥ÖÊÓëH2·´Ó¦¹ý³ÌÖеÄÄÜÁ¿±ä»¯Ê¾Òâͼ£¬ÆäÖÐa¡¢b¡¢c¡¢d·Ö±ð±íʾÑõ×åÔªËØÖÐijһԪËصĵ¥ÖÊ£¬×Ý×ø±êΪÏàͬÎïÖʵÄÁ¿µÄµ¥ÖÊÓëH2·´Ó¦¹ý³ÌÖеÄÄÜÁ¿±ä»¯£¨ÄÜÁ¿±ä»¯£¾0±íʾ·ÅÈÈ£¬ÄÜÁ¿±ä»¯£¼0±íʾÎüÈÈ£©¡£Ôò£ºb´ú±í___________  _____£¬ d´ú±í                 (¾ùдµ¥ÖÊÃû³Æ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨16·Ö£©µÈµç×ÓÔ­ÀíµÄ»ù±¾¹ÛµãÊÇ£ºÔ­×ÓÊýÏàͬÇÒ¼Ûµç×Ó×ÜÊýÏàµÈµÄ·Ö×Ó»òÀë×Ó¾ßÓÐÏàͬµÄ»¯Ñ§¼üÀàÐͺͿռ乹ÐÍ£¬»¥³ÆΪµÈµç×ÓÌå¡£µÈµç×ÓÌåµÄ½á¹¹ÏàËÆ£¬ÎïÀíÐÔÖÊÏà½ü¡£È磺N2¡¢COÓëC22£­¡¢CN£­ÎªµÈµç×ÓÌå¡£
£¨1£©ÒÑÖªCaC2ΪÀë×Ó»¯ºÏÎÔòCaC2µÄµç×ÓʽΪ                ¡£
£¨2£©¾Û±ûÏ©ëæË׳ÆÈËÔìÑò룬ÓɱûÏ©ëæ·Ö×ÓCH2=CH¡ªCN¾­¾ÛºÏ·´Ó¦Éú³É£»ÔòCH2=CH¡ªCNÖÐCÔ­×ÓµÄÔÓ»¯·½Ê½Îª         £»·Ö×ÓÖЦҼüºÍ¦Ð¼üÊýÖ®±ÈΪ          ¡£
£¨3£©CO³£Óë¹ý¶É½ðÊôÔ­×ÓMÐγÉÅäºÏÎïM(CO)n £¬ÆäÖÐÂú×ãÖÐÐÄÔ­×Ó¼Ûµç×ÓÊýÓëÅäλÌåÌṩµç×Ó×ÜÊýÖ®ºÍΪ18£¬ÈôMΪFe£¬Ôòn=       ¡£
£¨4£©COÓëN2µÄ½á¹¹ÏàËÆ£¬·Ö×ÓÖк¬Óй²¼ÛÈý¼ü£¬¿É±íʾΪC¡ÔO £»Ï±íÊÇÁ½ÕߵļüÄÜÊý¾Ý£¨µ¥Î»£ºkJ¡¤mol£­1£©

 
C£­O
C=O
C¡ÔO
CO
357.7
798.9
1071.9
 
N£­N
N=N
N¡ÔN
N2
154.8
418.4
941.7
COÓëN2Öл¯Ñ§ÐÔÖʽϻîÆõÄÊÇ       £»½áºÏÊý¾Ý˵Ã÷Ô­Òò                                 ¡£
£¨5£©Fe3+£¬Fe2+£¬Co3+£¬Co2+¶¼ÄÜÓëCN-ÐγÉÅäºÏÎï¡£ÁòËáÑÇÌúÈÜÒºÖмÓÈë¹ýÁ¿KCNÈÜÒº£¬¿ÉÎö³ö»ÆÉ«¾§ÌåK4[Fe(CN)6]£»ÈôÔÚÉÏÊöÈÜÒºÖÐÔÙͨÈëÂÈÆøºó£¬¿ÉÎö³öÉîºìÉ«¾§ÌåK3[Fe(CN)6]£»ÔÚK4[Fe(CN)6]ºÍK3[Fe(CN)6]¾§ÌåÖж¼²»´æÔÚµÄ΢Á£¼ä×÷ÓÃÁ¦ÊÇ                  ¡££¨Ìî±êºÅ£©
¡¡A£®Àë×Ó¼ü¡¡¡¡ B£®¹²¼Û¼ü¡¡¡¡ C£®½ðÊô¼ü¡¡¡¡D£®Åäλ¼ü¡¡¡¡ E£®·¶µÂ»ªÁ¦
£¨6£©Ð´³öÓëNO3£­»¥ÎªµÈµç×ÓÌåµÄ·Ö×Ó            £¨Ð´³öÒ»ÖÖ£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(12·Ö)¡¾»¯Ñ§¡ª¡ªÎïÖʽṹÓëÐÔÖÊ¡¿
£¨1£©¹ý¶É½ðÊôÔªËØÌúÄÜÐγɶàÖÖÅäºÏÎÈ磺[Fe(H2NCONH2)6](NO3)3[ÈýÏõËáÁùÄòËغÏÌú(¢ó)ºÍFe(CO)xµÈ¡£
¢Ù»ù̬Fe3+µÄM²ãµç×ÓÅŲ¼Ê½Îª               £»
¢ÚÄòËØ(H2NCONH2)·Ö×ÓÖÐCÔ­×ÓµÄÔÓ»¯·½Ê½ÊÇ             £»
¢ÛÅäºÏÎïFe(CO)xµÄÖÐÐÄÔ­×Ó¼Ûµç×ÓÊýÓëÅäÌåÌṩµç×ÓÊýÖ®ºÍΪ18£¬Ôòx=        ¡£
Fe(CO)x³£ÎÂϳÊҺ̬£¬ÈÛµãΪ-20.5¡æ£¬·ÐµãΪ103¡æ£¬Ò×ÈÜÓڷǼ«ÐÔÈܼÁ£¬¾Ý´Ë¿ÉÅжÏFe(CO)x¾§ÌåÊôÓÚ         (ÌÌåÀàÐÍ)¡£
£¨2£©ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ         (Ìî×ÖĸÐòºÅ)¡£

A£®µÚÒ»µçÀëÄÜ´óС£ºS>P>Si
B£®µç¸ºÐÔ˳Ðò£ºC<N<O<F
C£®ÒòΪ¾§¸ñÄÜCaO±ÈKCl¸ß£¬ËùÒÔKClµÄÈÛµã±ÈCaOÈÛµãµÍ
D£®·Ö×Ó¾§ÌåÖУ¬¹²¼Û¼ü¼üÄÜÔ½´ó£¬¸Ã·Ö×Ó¾§ÌåµÄÈ۷еãÔ½¸ß
£¨3£©OºÍNaµÄÒ»ÖÖÖ»º¬ÓÐÀë×Ó¼üµÄ»¯ºÏÎïµÄ¾§°û½á¹¹Èçͼ£¬¾àÒ»¸öÒõÀë×ÓÖÜΧ×î½üµÄËùÓÐÑôÀë×ÓΪ¶¥µã¹¹³ÉµÄ¼¸ºÎÌåΪ         ¡£ÒÑÖª¸Ã¾§°ûµÄÃܶÈΪg¡¤cm-3,°¢·ü¼ÓµÂÂÞ³£ÊýΪNA£¬Ç󾧰û±ß³¤a=   cm¡£(Óú¬¡¢NAµÄ¼ÆËãʽ±íʾ)

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸