¡¾ÌâÄ¿¡¿Á×¼°Æ仯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓÐÖØÒªµÄ×÷Óá£
(1)PÔªËØÓа×Áס¢ºìÁס¢ºÚÁ×ÈýÖÖ³£¼ûµÄµ¥ÖÊ¡£
¢ÙÏÖ´ú»¯Ñ§ÖУ¬³£ÀûÓÃ_______ÉϵÄÌØÕ÷Æ×ÏßÀ´¼ø¶¨ÔªËØ¡£
¢Ú°×Á×(P4)Ò×ÈÜÓÚCS2£¬ÄÑÈÜÓÚË®£¬ÔÒòÊÇ_________¡£
¢ÛºÚÁ×ÊÇÒ»ÖÖºÚÉ«ÓнðÊô¹âÔóµÄ¾§Ì壬ÊÇÒ»ÖÖ±Èʯīϩ¸üÓÅÐãµÄÐÂÐͲÄÁÏ¡£°×Áס¢ºìÁ׶¼ÊÇ·Ö×Ó¾§Ì壬ºÚÁ×¾§Ìå¾ßÓÐÓëʯīÏàÀàËƵIJã×´½á¹¹£¬ÈçÏÂͼËùʾ¡£ÏÂÁÐÓйغÚÁ×¾§ÌåµÄ˵·¨²»ÕýÈ·µÄÊÇ_______¡£
A£®ºÚÁ×¾§ÌåÖÐÁ×Ô×ÓÔÓ»¯·½Ê½Îªsp3ÔÓ»¯
B£®ºÚÁ×¾§ÌåÖвãÓë²ãÖ®¼äµÄ×÷ÓÃÁ¦ÊÇ·Ö×Ó¼ä×÷ÓÃÁ¦
C£®ºÚÁ×¾§ÌåµÄÿһ²ãÖÐÁ×Ô×Ó¶¼ÔÚͬһƽÃæÉÏ
D£®PÔªËØÈýÖÖ³£¼ûµÄµ¥ÖÊÖУ¬ºÚÁ×µÄÈ۷еã×î¸ß
(2)Á×»¯ÅðÊÇÒ»ÖÖÄÍĥͿÁÏ£¬Ëü¿ÉÓÃ×÷½ðÊôµÄ±íÃæ±£»¤²ã¡£Á×»¯Åð¿ÉÓÉÈýä廯Á׺ÍÈýä廯ÅðÓÚ¸ßÎÂÏÂÔÚÇâÆøÖз´Ó¦ºÏ³É¡£Á×»¯Åð¾§ÌåµÄ¾§°ûÈçÏÂͼËùʾ£¬ÆäÖÐʵÐÄÇòΪÁ×Ô×Ó¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÁ×»¯ÅðµÄ»¯Ñ§Ê½Îª_________£¬¸Ã¾§ÌåµÄ¾§ÌåÀàÐÍÊÇ___________¡£
¢ÚÈýä廯Á×ÖÐäåÔªËØ»ù̬Ô×ӵĵç×ÓÅŲ¼Ê½Îª_________£¬Èýä廯Åð·Ö×ӵĿռ乹ÐÍÊÇ________£¬ºÏ³ÉÁ×»¯ÅðµÄ»¯Ñ§·½³ÌʽΪ£º___________¡£
¢ÛÔÚÒ»¸ö¾§°ûÖÐÁ×Ô×Ó¿Õ¼ä¶Ñ»ý·½Ê½Îª________£¬Á×Ô×ÓµÄÅäλÊýΪ________£¬¸Ã½á¹¹ÖÐÓÐÒ»¸öÅäλ¼ü£¬Ìṩ¿Õ¹ìµÀµÄÔ×ÓÊÇ_______¡£ÒÑÖª¾§ÌåÖÐBÓëPÔ×ÓµÄ×î½ü¾àÀëΪa nm£¬Ôò¸Ã¾§ÌåµÄÃܶȵıí´ïʽΪ(²»Ð軯¼ò)________g/cm3¡£
¡¾´ð°¸¡¿Ô×Ó¹âÆ× P4¡¢CS2ÊǷǼ«ÐÔ·Ö×Ó£¬H2OÊǼ«ÐÔ·Ö×Ó£¬¸ù¾ÝÏàËÆÏàÈÜÔÀí£¬P4ÄÑÈÜÓÚË® C BP Ô×Ó¾§Ìå 1s22s22p63s23p63d104s24p5»ò[Ar] 3d104s24p5 ƽÃæÈý½ÇÐÎ PBr3£«BBr3£«3H2BP£«6HBr ÃæÐÄÁ¢·½×îÃܶѻý 4 B
¡¾½âÎö¡¿
ÏÖ´ú»¯Ñ§ÖУ¬³£ÀûÓÃÔ×Ó¹âÆ×ÉϵÄÌØÕ÷Æ×ÏßÀ´¼ø¶¨ÔªËØ£»·Ö×ÓµÄÈܽâÐÔ£¬Í¨³£·ûºÏÏàËÆÏàÈÜÔÀí£»ºÚÁ׵ľ§Ìå½á¹¹ÀàËÆÓÚʯī£¬µ«ÓÉÓÚÁ×Ô×Ó×îÍâ²ãµç×ÓÊýΪ5£¬ËùÒÔ·¢Éúsp3ÔÓ»¯£¬ÓÉÓڳɼüµç×ÓÊܹµç×Ó¶ÔµÄÅųâ×÷Óã¬ËùÒÔ²ãÄÚÁ×Ô×Ó²»ÔÚͬһƽÃæÉÏ£»ºÚÁ×ÓбðÓÚºìÁ׺Ͱ×Á×£¬ÓÉÓÚ²ãÄÚÔ×Ó¼äÐγɹ²¼Û¼ü£¬ËùÒÔ²ãÄÚÓ¦ÐγÉÔ×Ó¾§Ìå¡£ÔÚÈýä廯ÅðÖУ¬BÔ×ÓµÄ×îÍâ²ãµç×ÓÈ«²¿²ÎÓë³É¼ü£¬ËùÒÔ·Ö×Ó³ÊƽÃæ½á¹¹£»ÔÚÁ×»¯ÅðÖУ¬ÓÉͼÖпÉÒÔ¿´³ö£¬Ã¿¸öBÔ×ÓÐγÉ4¸ö¹²¼Û¼ü£¬¶øBµÄ¼Ûµç×ÓÖ»ÓÐ3¸ö£¬ËùÒÔÿ¸öPÔ×ÓÌṩ1¶Ô¹Â¶Ôµç×ÓÓëBÔ×Ó(Ìṩ¿Õ¹ìµÀ)ÐγÉÅäλ¼ü£¬¹¹³ÉÕýËÄÃæÌå½á¹¹£»ÔÚBP¾§°ûÖУ¬Ã¿¸öBÔ×ÓÓë2¸öPÔ×Ó¹¹³É¼Ð½ÇΪ109¡ã28¡äµÄµÈÑüÈý½ÇÐΣ¬ÀûÓÃÓàÏÒ¶¨Àí£¬¿É½¨Á¢¾§°û±ß³¤ÓëB-P¼ü³¤¼äµÄ¶¨Á¿¹Øϵ£¬´Ó¶øÀûÓÃB-P¼ü³¤Çó³ö¾§°ûµÄ±ß³¤£¬×îÖÕ½¨Á¢ÃܶÈÓë±ß³¤µÄµÈÁ¿¹Øϵʽ¡£
(1)¢ÙÏÖ´ú»¯Ñ§ÖУ¬¼ø¶¨Ô×Ó£¬³£Ê¹ÓÃÔ×Ó¹âÆ×£¬ÀûÓùâÆ×ÉϵÄÌØÕ÷Æ×ÏßÀ´¼ø¶¨ÔªËØ¡£
¢Ú°×Á×(P4)µÄ½á¹¹ÎªÕýËÄÃæÌ壬Ϊ·Ç¼«ÐÔ·Ö×Ó£¬Ò×ÈÜÓڷǼ«ÐÔ·Ö×ÓCS2£¬ÄÑÈÜÓÚ¼«ÐÔ·Ö×ÓË®£¬ÔÒòÊÇP4¡¢CS2ÊǷǼ«ÐÔ·Ö×Ó£¬H2OÊǼ«ÐÔ·Ö×Ó£¬¸ù¾ÝÏàËÆÏàÈÜÔÀí£¬P4ÄÑÈÜÓÚË®¡£
¢ÛA£®ºÚÁ×¾§ÌåÖÐÁ×Ô×ÓÓëÖÜΧ3¸öPÔ×ÓÐγɹ²¼Û¼ü£¬Ã¿¸öPÔ×Ó»¹ÓÐ1¶Ô¹Â¶Ôµç×Ó£¬ËùÒÔÔÓ»¯·½Ê½Îªsp3ÔÓ»¯£¬AÕýÈ·£»
B£®ºÚÁ×¾§ÌåÖвãÓë²ãÖ®¼ä²»Ðγɹ²¼Û¼ü£¬½öͨ¹ý·¶µÂ»ªÁ¦½áºÏ£¬ËùÒÔ²ã¼ä×÷ÓÃÁ¦ÊÇ·Ö×Ó¼ä×÷ÓÃÁ¦£¬BÕýÈ·£»
C£®ºÚÁ×¾§ÌåµÄÿһ²ãÖÐÁ×Ô×Ó·¢Éúsp3ÔÓ»¯£¬ÓÉÓڹµç×Ó¶ÔµÄÅųâ×÷Óã¬PÔ×Ó²»ÔÚÒ»¸öƽÃæÉÏ£¬C²»ÕýÈ·£»
D£®PÔªËØÈýÖÖ³£¼ûµÄµ¥ÖÊÖУ¬°×Á׺ͺìÁ׶¼ÐγɷÖ×Ó¾§Ì壬¶øºÚÁײãÄÚÁ×Ô×Ó¼äÐγɹ²¼Û¼ü£¬Ï൱ÓÚÔ×Ó¾§Ì壬ËùÒÔºÚÁ×µÄÈ۷еã×î¸ß£¬DÕýÈ·£»
¹ÊÑ¡C¡£
´ð°¸Îª£ºÔ×Ó¹âÆ×£»P4¡¢CS2ÊǷǼ«ÐÔ·Ö×Ó£¬H2OÊǼ«ÐÔ·Ö×Ó£¬¸ù¾ÝÏàËÆÏàÈÜÔÀí£¬P4ÄÑÈÜÓÚË®£»C£»
(2)¢ÙÔÚÁ×»¯Åð¾§ÌåÖУ¬ÀûÓþù̯·¨£¬¿ÉÇó³öº¬PÔ×Ó¸öÊýΪ=4£¬º¬BÔ×Ó¸öÊýΪ4£¬¶þÕßÔ×Ó¸öÊý±ÈΪ1:1£¬ËùÒÔÁ×»¯ÅðµÄ»¯Ñ§Ê½ÎªBP£¬¸Ã¾§ÌåÄÚÔ×Ó¼äÈ«²¿Ðγɹ²¼Û¼ü£¬Ã»ÓÐÆäËü×÷ÓÃÁ¦£¬ËùÒÔ¾§ÌåÀàÐÍÊÇÔ×Ó¾§Ìå¡£
¢ÚÈýä廯Á×ÖÐäåÔªËØ»ù̬Ô×ÓºËÍâµç×ÓÅŲ¼Îª2¡¢8¡¢18¡¢7£¬µç×ÓÅŲ¼Ê½Îª1s22s22p63s23p63d104s24p5»ò[Ar] 3d104s24p5£¬Èýä廯Åð·Ö×ÓÖУ¬BÔ×Ó·¢Éúsp2ÔÓ»¯£¬²»´æÔڹµç×Ó¶Ô£¬ËùÒԿռ乹ÐÍÊÇƽÃæÈý½ÇÐΣ»Èýä廯Á׺ÍÈýä廯ÅðÓÚ¸ßÎÂÏÂÔÚÇâÆøÖз´Ó¦Éú³ÉBPºÍHBr£¬ºÏ³ÉÁ×»¯ÅðµÄ»¯Ñ§·½³ÌʽΪ£ºPBr3£«BBr3£«3H2BP£«6HBr¡£
¢ÛÔÚÒ»¸ö¾§°ûÖУ¬Á×Ô×ӵĵþºÏ·½Ê½ÎªÈý²ãһѻ·£¬ËùÒÔÁ×Ô×Ó¿Õ¼ä¶Ñ»ý·½Ê½ÎªÃæÐÄÁ¢·½×îÃܶѻý£¬Ã¿¸öÁ×Ô×ÓÖÜΧ¾àÀë×î½üµÄÅðÔ×ÓÓÐ4¸ö£¬ËùÒÔÁ×Ô×ÓµÄÅäλÊýΪ4£¬¸Ã½á¹¹ÖÐÓÐÒ»¸öÅäλ¼ü£¬Ìṩ¿Õ¹ìµÀµÄÔ×ÓÊÇB¡£ÒÑÖª¾§ÌåÖÐBÓëPÔ×ÓµÄ×î½ü¾àÀëΪa nm£¬ÎÒÃÇÈ¡Ò»¸öÒÔBΪÖÐÐĵÄÕýËÄÃæÌ壬Ȼºó´ÓÆäÖÐÈ¡Ò»¸öµÈÑüÈý½ÇÐΣ¬´ËʱÁ½ÑüµÄ¼Ð½ÇΪ109¡ã28¡ä£¬ÀûÓÃÓàÏÒ¶¨Àí¿ÉµÃ£¬()2=a2+a2-2aacos109¡ã28¡ä£¬²é±íÖª£¬cos109¡ã28¡ä= -£¬´Ó¶øÇó³öx=nm£¬Ôò¸Ã¾§ÌåµÄÃܶȵıí´ïʽΪ=g/cm3¡£
´ð°¸Îª£ºBP£»Ô×Ó¾§Ì壻1s22s22p63s23p63d104s24p5»ò[Ar] 3d104s24p5£»Æ½ÃæÈý½ÇÐΣ»PBr3£«BBr3£«3H2BP£«6HBr£»ÃæÐÄÁ¢·½×îÃܶѻý£»4£»B£»¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿³£¼ûµÄµç×ÓÆøÌåÓÐBCl3¡¢N2O¡¢SiH4¼°SiHCl3µÈ¡£»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÆøÌåB2H6ÓëÂÈÆø»ìºÏ¿ÉÉú³ÉÆø̬BCl3£¬Ã¿Éú³É1.0 g BCl3·Å³ö5.9 kJµÄÈÈÁ¿£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ______¡£
(2)25¡æʱ·´Ó¦ S(s) +H2(g)H2S(g) Kp1=6.0¡Á105
Si(s)+2H2(g)SiH4(g) Kp2=7.8¡Á106
Ôò·´Ó¦ Si(s)+2H2S(g)SiH4(g)+2S(s) Kp=_____(KpΪÒÔ·Öѹ±íʾµÄƽºâ³£Êý£¬ÏÂͬ)¡£
(3)¶à¾§¹èÖÆÖз¢ÉúһϵÁз´Ó¦£º
(I)4SiHCl3(g)Si(s)+3SiCl4(g)+2H2(g) ¡÷H1=a kJ /mol
(II)SiCl4(g)+H2(g)SiHCl3(g)+HCl(g) ¡÷H2=b kJ/mo1
(III)SiCl2(g)+H2(g)Si(s)+2HCl(g) ¡÷H3=c kJ/mo1
¢Ù·´Ó¦SiHCl3 (g)SiCl2 (g) +HCl (g) ¡÷H=____kJ/mo1 (Óú¬a¡¢b¡¢cµÄ´úÊýʽ±íʾ)
¢Ú·´Ó¦(I)(II)(III)µÄKpÓëζȵĹØϵÈçÏÂͼ
ÊôÓÚÎüÈÈ·´Ó¦µÄÊÇ_________(ÌîI¡¢II»òIII)£»Í¼ÖÐMµã·Öѹ¼äÂú×ã¹Øϵ£ºp(SiCl4)=______(ÓÃÏà¹ØÎïÖʵķÖѹp±íʾ)¡£
(4)Ò»ÖÖÖÆÈ¡N2OµÄ·½·¨Îª O2NNH2(aq)¡úN2O(g)+H2O(1)£¬¸Ã·´Ó¦µÄÀú³ÌÈçÏ£º
(I)O2NNH2(aq)O2NNH-(aq) +H+(aq) (¿ìËÙƽºâ)
(II)O2NNH-(aq) N2O(g) +OH-(aq) (Âý)
(III)H+(aq)+OH-(aq) H2O(1) (¿ì)
¢Ù»î»¯ÄÜ×î´óµÄ·´Ó¦²½ÖèÊÇ_________(ÌîI¡¢II»òIII)¡£
¢ÚÒÑÖª·´Ó¦(I)µÄËÙÂÊ·½³Ìv(Õý)=k1c(O2NNH2)£¬v(Äæ)=k -1 c(O2NNH-)c(H+)£¬(k1¡¢k -1·Ö±ðΪÕýÄæ·´Ó¦ËÙÂʳ£Êý£¬·´Ó¦(I)´ïµ½Æ½ºâʱ£¬Æ½ºâ³£ÊýK=__________(ÓÃk1¡¢k -1±íʾ)¡£
¢ÛÒÑÖª×Ü·´Ó¦ËÙÂÊ·½³ÌΪv=K £¬·´Ó¦(II)µÄv(Õý)=k2c(O2NNH-)£¬ÔòK=________ (ÓÃk1¡¢k -1¡¢k2¡¢k3±íʾ)
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ò»ÖÖÈçͼËùʾµÄÈýÊÒ΢ÉúÎïȼÁϵç³ØÓÃÀ´´¦Àí·ÏË®(·ÏË®ÖиßŨ¶ÈÓлúÎïÓÃC6H12O6±íʾ)£¬ÏÂÁÐÐðÊö²»ÕýÈ·µÄÊÇ
A.aµç¼«·´Ó¦Ê½Îª£ºC6H12O6+6H2O-24e-£½6CO2¡ü+24H£«
B.bµç¼«¸½½üµÄÈÜÒºpHÔö´ó
C.ζÈÔ½¸ß£¬´¦Àí·ÏË®µÄЧÂÊÔ½¸ß
D.Èô·´Ó¦ÖÐתÒƵĵç×ÓÊýΪNA£¬ÔòÉú³É±ê×¼×´¿öÏÂN2µÄÌå»ýΪ2.24L
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¶ÔH2O2·Ö½â·´Ó¦£¬Cu2+Ò²ÓÐÒ»¶¨µÄ´ß»¯×÷Óá£Îª±È½ÏFe3+ºÍCu2+¶ÔH2O2·Ö½âµÄ´ß»¯Ð§¹û£¬Ä³»¯Ñ§Ñо¿Ð¡×éµÄͬѧÃÇ·Ö±ðÉè¼ÆÁËÈçͼ¼×¡¢ÒÒËùʾµÄʵÑé×°Öá£Çë»Ø´ðÏà¹ØÎÊÌ⣺
(1)¶¨ÐÔÈçͼ¼×¿Éͨ¹ý¹Û²ì_________£¬¶¨ÐԱȽϵóö½áÂÛ¡£ÓÐͬѧÌá³ö½«FeCl3¸ÄΪFe2(SO4)3¸üΪºÏÀí£¬ÆäÀíÓÉÊÇ_______¡£
(2)¶¨Á¿ÓÃͼÒÒËùʾװÖÃ×ö¶ÔÕÕÊÔÑ飬ʵÑéʱ¾ùÒÔÉú³É40 mLÆøÌåΪ׼£¬ÆäËû¿ÉÄÜÓ°ÏìʵÑéµÄÒòËؾùÒѺöÂÔ¡£ÊµÑéÖÐÐèÒª²âÁ¿µÄÊý¾ÝÊÇ_______¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Îø(Se)¡¢íÚ(Te)µÄµ¥Öʺͻ¯ºÏÎïÔÚ»¯¹¤Éú²úµÈ·½Ãæ¾ßÓÐÖØÒªÓ¦Óá£
(1)ÏÂÁйØÓÚÎø¡¢íÚ¼°Æ仯ºÏÎïµÄÐðÊö²»ÕýÈ·µÄÊÇ_______¡£
A£®Te λÓÚÖÜÆÚ±íµÄµÚÎåÖÜÆÚ ¢ö A ×å
B£®Se µÄÑõ»¯Îïͨ³£ÓÐSeO2ºÍSeO3
C£®H2TeO4µÄËáÐÔ±ÈH2SO4µÄËáÐÔÇ¿
D£®ÈÈÎȶ¨ÐÔH2Se±ÈH2SÈõ£¬µ«±ÈHBrÇ¿
(2)25¡æ ʱ£¬ÎøËáµÄµçÀëH2SeO4=H+ +£»H+ + Ka2 = 1¡Á10-3£¬Ôò0.1 mol¡¤L-1NaHSeO4ÈÜÒºµÄpHԼΪ________£»NaHSeO4ÈÜÒºÖеÄÎïÁÏÊغã±í´ïʽΪ____________¡£
(3)TeO2΢ÈÜÓÚË®£¬Ò×ÈÜÓÚ½ÏŨµÄÇ¿ËáºÍÇ¿¼î¡£¹¤ÒµÉϳ£ÓÃÍÑô¼«Äà(Ö÷Òªº¬TeO2£¬»¹º¬ÓÐÉÙÁ¿Ag¡¢Au)ΪÔÁÏÖƱ¸µ¥ÖÊíÚ£¬Æ乤ÒÕÁ÷³ÌÈçͼ£º
¢ÙÍÑô¼«ÄàÔÚ¼î½þÇ°Ðèºæ¸É¡¢ÑгɷÛÄ©£¬Ä¿µÄÊÇ____________£»
¢Ú¡°¼î½þ¡±Ê±TeO2·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________£»
¢Û¡°³ÁíÚ¡±Ê±¿ØÖÆÈÜÒºµÄ pH Ϊ 4.5~5.0£¬Éú³ÉTeO2³Áµí£¬ËáÐÔ²»ÄܹýÇ¿µÄÔÒòÊÇ__________£»
¢Ü¡°»¹Ô¡±µÃµ½¹Ì̬íÚΪ´ÖíÚ£¬¶Ô´ÖíÚ½øÐÐÏ´µÓ£¬ÅжÏÏ´µÓ¸É¾»µÄʵÑé²Ù×÷ºÍÏÖÏóÊÇ________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁжÔһЩʵÑéÊÂʵºÍÀíÂÛ½âÊÍÕýÈ·µÄÊÇ
Ñ¡Ïî | ʵÑéÊÂʵ | ÀíÂÛ½âÊÍ |
A | HClÆøÌåÈÜÓÚË®£¬Äܵ¼µç | HClΪÀë×Ó»¯ºÏÎï |
B | HBrµÄËáÐÔÇ¿ÓÚHClµÄËáÐÔ | BrµÄ·Ç½ðÊôÐÔ±ÈClÇ¿ |
C | ÔÚÈÛÈÚ״̬ÏÂÄܹ»µ¼µç | Öк¬ÓÐÀë×Ó¼ü |
D | HFµÄ·Ðµã¸ßÓÚHCl | FµÄ·Ç½ðÊôÐÔ±ÈClÇ¿ |
A.AB.BC.CD.D
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÓÐÎåÖÖÔªËØX¡¢Y¡¢Z¡¢Q¡¢T¡£XÔªËØΪÖ÷×åÔªËØ£¬XÔ×ÓµÄM²ãpÄܼ¶ÉÏÓÐÁ½¸öδ³É¶Ôµç×ÓÇÒÎÞ¿Õ¹ìµÀ£»YÔ×ӵļ۵ç×ÓÅŲ¼Îª3d64s2£»ZÔ×ÓµÄLµç×Ó²ãµÄpÄܼ¶ÉÏÓÐÒ»¸ö¿Õ¹ìµÀ£»QÔ×ÓµÄLµç×Ó²ãµÄPÄܼ¶ÉÏÖ»ÓÐÒ»¶Ô³É¶Ôµç×Ó£»TÔ×ÓµÄMµç×Ó²ãÉÏp¹ìµÀ°ë³äÂú¡£ÏÂÁÐÐðÊö²»ÕýÈ·µÄÊÇ
A.ÔªËØYºÍQ¿ÉÐγɻ¯ºÏÎïY2Q3
B.XÓëTµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎǰÕßµÄËáÐԱȺóÕßÇ¿
C.XºÍQ½áºÏÉú³ÉµÄ»¯ºÏÎïΪÀë×Ó»¯ºÏÎï
D.ZQ2ÊǼ«ÐÔ¼ü¹¹³ÉµÄ·Ç¼«ÐÔ·Ö×Ó
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Í¨³£Çé¿öÏ£¬NCl3ÊÇÒ»ÖÖÓÍ×´ÒºÌ壬Æä·Ö×ӿռ乹ÐÍÓëNH3ÏàËÆ£¬ÏÂÁжÔNCl3ºÍNH3µÄÓйØÐðÊöÕýÈ·µÄÊÇ( )
A. ·Ö×ÓÖÐN¡ªCl¼ü¼ü³¤ÓëCCl4·Ö×ÓÖÐC¡ªCl¼ü¼ü³¤ÏàµÈ
B. NCl3·Ö×ÓÊǷǼ«ÐÔ·Ö×Ó
C. NBr3±ÈNCl3Ò×»Ó·¢
D. ÔÚ°±Ë®ÖУ¬´ó²¿·ÖNH3ÓëH2OÒÔÇâ¼ü(Óá°¡¡±±íʾ)½áºÏÐγÉNH3¡¤H2O·Ö×Ó£¬ÔòNH3¡¤H2OµÄ½á¹¹Ê½Îª
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÔÚÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬½øÐÐÈçÏ»¯Ñ§·´Ó¦£ºCO2(g)£«H2(g) CO(g)£«H2O(g)£¬Æ仯ѧƽºâ³£ÊýKºÍζÈtµÄ¹ØϵÈçϱíËùʾ£º
t ¡æ | 700 | 800 | 830 | 1 000 | 1 200 |
K | 0.6 | 0.9 | 1.0 | 1.7 | 2.6 |
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪK£½_________________________________¡£
£¨2£©¸Ã·´Ó¦Îª________·´Ó¦(Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±)¡£
£¨3£©Ä³Î¶ÈÏ£¬¸÷ÎïÖʵÄƽºâŨ¶È·ûºÏÏÂʽ£º3c(CO2)¡¤c(H2)£½5c(CO)¡¤c(H2O)£¬ÊÔÅжϴËʱµÄζÈΪ______¡£
£¨4£©Èô830 ¡æʱ£¬ÏòÈÝÆ÷ÖгäÈë1 mol CO¡¢5 mol H2O£¬·´Ó¦´ïµ½Æ½ºâºó£¬Æ仯ѧƽºâ³£ÊýK______1.0(Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)¡£
£¨5£©830 ¡æʱ£¬ÈÝÆ÷Öеķ´Ó¦ÒѴﵽƽºâ¡£ÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬À©´óÈÝÆ÷µÄÌå»ý¡£Æ½ºâ____Òƶ¯(Ìî¡°ÏòÕý·´Ó¦·½Ïò¡±¡°ÏòÄæ·´Ó¦·½Ïò¡±»ò¡°²»¡±)¡£
£¨6£©Èô1 200 ¡æʱ£¬ÔÚijʱ¿ÌƽºâÌåϵÖÐCO2¡¢H2¡¢CO¡¢H2OµÄŨ¶È·Ö±ðΪ2 mol¡¤L£1¡¢2 mol¡¤L£1¡¢4 mol¡¤L£1¡¢4 mol¡¤L£1£¬Ôò´ËʱÉÏÊö·´Ó¦µÄƽºâÒƶ¯·½ÏòΪ__________(Ìî¡°Õý·´Ó¦·½Ïò¡±¡°Äæ·´Ó¦·½Ïò¡±»ò¡°²»Òƶ¯¡±)¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com