¡¾ÌâÄ¿¡¿Á×´æÔÚÓÚÈËÌåËùÓÐϸ°ûÖУ¬ÊÇά³Ö¹Ç÷ÀºÍÑÀ³ÝµÄ±ØÒªÎïÖÊ£¬Ò²ÊÇÖØÒªµÄ·Ç½ðÊôÔªËØ¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)»ù̬Á×Ô×ӵļ۵ç×ÓÅŲ¼Ê½Îª________________¡£
(2)NºÍPͬÖ÷×åÇÒÏàÁÚ,PF3ºÍNH3¶¼ÄÜÓëÐí¶à¹ý¶É½ðÊôÐγÉÅäºÏÎï,µ«NF3È´²»ÄÜÓë¹ý¶É½ðÊôÐγÉÅäºÏÎï,ÆäÔÒòÊÇ____________________________________________¡£
(3)Á×ËáΪÈýÔªËá,Æä½á¹¹Ê½Îª¡£PO43-µÄ¿Õ¼ä¹¹ÐÍΪ____________£»Ð´³öÒ»ÖÖÓëPO43-»¥ÎªµÈµç×ÓÌåÇÒÊôÓڷǼ«ÐÔ·Ö×ӵĻ¯Ñ§Ê½£º____________________¡£
(4)½«Á×Ëá¼ÓÈÈ¿É·¢Éú·Ö×Ó¼äÍÑË®Éú³É½¹Á×Ëá(H4P2O7)¡¢ÈýÁ×ËáÒÔ¼°¸ß¾ÛÁ×Ëᣬ½¹Á×ËáµÄËáÐÔÇ¿ÓÚÁ×ËáµÄÔÒòÊÇ______________________________________________________¡£
(5)Á×»¯Åð(BP)ÊÇÊܵ½¸ß¶È¹Ø×¢µÄÄÍĥͿÁÏ¡£BPÖÐBºÍPÔ×Ó¾ùÐγɹ²¼Û¼ü£¬ÆäÖдæÔÚÅäλ¼ü,Åäλ¼üÖÐÌṩ¹Âµç×Ó¶ÔµÄÊÇ____________(ÌîÔªËØ·ûºÅ)Ô×Ó;Á×»¯ÅðµÄ¾§ÌåÀàÐÍÊÇ____________£¬ÆäÖÐBÔ×ÓµÄÔÓ»¯·½Ê½ÊÇ____________ÔÓ»¯,1molBPÖк¬ÓÐ____________molB-P¼ü¡£
(6)Cu3PµÄ¾§°û½á¹¹ÈçͼËùʾ,P3-µÄÅäλÊýΪ____________,Cu+µÄ°ë¾¶Îªapm,P3-µÄ°ë¾¶Îªbpm,°¢·ü¼ÓµÂÂÞ³£ÊýµÄֵΪNA,ÔòCuP¾§ÌåµÄÃܶÈΪ____________g¡¤cm-3(Óú¬a¡¢b¡¢NAµÄ´úÊýʽ±íʾ)¡£
¡¾´ð°¸¡¿ 3s23p3 FÔ×ӵ縺ÐÔÇ¿£¬ÎüÒýNÔ×ӵĵç×Ó,ʹÆäÄÑÒÔ¸ø³öµç×Ó¶ÔÐγÉÅäλ¼ü ÕýËÄÃæÌåÐÎ CCl4(»òSiF4) ½¹Á×ËáÖзÇôÇ»ùÑõµÄÊýÄ¿±ÈÁ×ËáÖзÇôÇ»ùÑõµÄÊýÄ¿¶à P Ô×Ó¾§Ìå sp3 4 6 g/ cm3
¡¾½âÎö¡¿(1)Á×Ϊ15ºÅÔªËØ£¬»ù̬Á×Ô×ÓºËÍâÓÐ15¸öµç×Ó£¬×îÍâ²ãµç×ÓÊýΪ5£¬¹Ê»ù̬Á×Ô×ӵļ۵ç×ÓÅŲ¼Ê½Îª3s23p3£»(2)NºÍPͬÖ÷×åÇÒÏàÁÚ,PF3ºÍNH3¶¼ÄÜÓëÐí¶à¹ý¶É½ðÊôÐγÉÅäºÏÎï,µ«NF3È´²»ÄÜÓë¹ý¶É½ðÊôÐγÉÅäºÏÎï,ÆäÔÒòÊÇFÔ×ӵ縺ÐÔÇ¿£¬ÎüÒýNÔ×ӵĵç×Ó,ʹÆäÄÑÒÔ¸ø³öµç×Ó¶ÔÐγÉÅäλ¼ü£»(3) PO43-µÄ¿Õ¼ä¹¹ÐÍΪÕýËÄÃæÌåÐΣ»µÈµç×ÓÌåÊÇÖ¸¼Ûµç×ÓÊýºÍÔ×ÓÊý£¨ÇâµÈÇáÔ×Ó²»¼ÆÔÚÄÚ£©ÏàͬµÄ·Ö×Ó¡¢Àë×Ó»òÔ×ÓÍÅ£¬ÓëPO43-»¥ÎªµÈµç×ÓÌåÇÒÊôÓڷǼ«ÐÔ·Ö×ӵĻ¯Ñ§Ê½ÓÐCCl4»òSiF4£»(4)½«Á×Ëá¼ÓÈÈ¿É·¢Éú·Ö×Ó¼äÍÑË®Éú³É½¹Á×Ëá(H4P2O7)¡¢ÈýÁ×ËáÒÔ¼°¸ß¾ÛÁ×Ëᣬ½¹Á×ËáµÄËáÐÔÇ¿ÓÚÁ×ËáµÄÔÒòÊǽ¹Á×ËáÖзÇôÇ»ùÑõµÄÊýÄ¿±ÈÁ×ËáÖзÇôÇ»ùÑõµÄÊýÄ¿¶à£»(5) BPÖÐBºÍPÔ×Ó¾ùÐγɹ²¼Û¼ü£¬ÆäÖдæÔÚÅäλ¼ü,Åäλ¼üÖÐÌṩ¹Âµç×Ó¶ÔµÄÊÇPÔ×Ó; £¨3£©ÓÉÓÚBPÊôÓÚÄÍĥͿÁÏ£¬ËùÒÔÐγɵľ§ÌåÓ¦¸ÃÊÇÔ×Ó¾§Ìå¡£BÔ×ÓÐγÉ4¸ö¹²¼Û¼ü£¬ËùÒÔÿÉú³É1molBP£¬¹²ÐγÉ4molB-P¼ü£¬ÆäÖÐBÔ×ÓµÄÔÓ»¯·½Ê½ÊÇsp3ÔÓ»¯£»(6)¸ù¾ÝCu3PµÄ¾§°û½á¹¹¿ÉÖª£¬P3-ÔÚ¾§°ûµÄ¶¥µã£¬¹ÊÅäλÊýΪ6£»¸ù¾ÝÔ×Ó¾ù̯·¨¼ÆË㣬ÿ¸ö¾§°ûÖÐÓиöP3-£¬12¸öCu+£¬¾§°û±ß³¤Îª2£¨a+b£©pm,Ìå»ýΪ8£¨a+b£©3pm3=8£¨a+b£©3cm3£¬ÔòÃܶÈΪ=g/ cm3¡£
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¿¡¿¶ÔÓÚ¿ÉÄæ·´Ó¦mA(g)+nB(g)pC(g)+qD(g)+Q (Q>0)£¬ÇÒm+n>p+q£¬Ê¹Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯µÄÌõ¼þÊÇ( )
A.½µÎ¡¢½µÑ¹B.ÉýΡ¢½µÑ¹C.½µÎ¡¢ÔöѹD.ÉýΡ¢Ôöѹ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÈÜÒººÍ½ºÌå±¾ÖÊÇø±ðÊÇ£¨ £©
A.΢Á£Ö±¾¶µÄ´óСB.¹Û²ìÍâ¹ÛC.µçÓ¾D.¶¡´ï¶ûЧӦ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÑÖªÍÔÚ³£ÎÂÏÂÄܱ»Å¨HNO3Èܽ⣬·´Ó¦Îª:
Cu£«4HNO3===Cu(NO3)2£«2NO2¡üÊ®2H2O¡£
£¨1£©ÓÃË«ÏßÇÅ·¨±ê³öµç×ÓתÒƵķ½ÏòºÍÊýÄ¿________________________ ¡£
£¨2£©ÉÏÊö·´Ó¦ÖУ¬Ñõ»¯¼ÁÊÇ______£¬Ñõ»¯²úÎïÊÇ________£¬»¹Ô¼ÁÓ뻹ԲúÎïµÄÎïÖʵÄÁ¿Ö®±ÈΪ________¡£
£¨3£©ÈôÓÐ1mol Cu±»Ñõ»¯£¬ÔòתÒƵç×ÓµÄÊýĿΪ________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁз´Ó¦ÊôÓÚ·ÅÈÈ·´Ó¦µÄÊÇ£¨ £©
A.ÄÆÓëË®µÄ·´Ó¦B.̼Ëá¸Æ·Ö½â·´Ó¦C.Ì¿ºÍË®ÕôÆø·´Ó¦D.Ë®½á±ù
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿½«Ò»¶¨ÖÊÁ¿µÄп¡¢ÂÁ»ìºÏÎïÓë×ãÁ¿µÄÏ¡ÁòËá·´Ó¦£¬Éú³É2.8 L(±ê¿ö)ÇâÆø£¬Ô»ìºÏÎïµÄÖÊÁ¿¿ÉÄÜÊÇ(¡¡¡¡)
A. 2 g B. 1 g C. 8 g D. 10 g
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿»¯ºÏÎïMÊÇÖØÒªµÄÓлúºÏ³ÉÖмäÌå,HΪ¸ß·Ö×Ó»¯ºÏÎï,ÆäºÏ³É·ÏßÈçͼËùʾ£º
ÒÑÖª£º¢Ù(RΪÌþ»ù);
¢Ú2R-CH2CHO¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)JµÄÃû³ÆΪ____________¡£EÖÐËùº¬¹ÙÄÜÍŵÄÃû³ÆΪ____________¡£HµÄ½á¹¹¼òʽΪ____________¡£
(2)C¡úDÉæ¼°µÄ·´Ó¦ÀàÐÍÓÐ____________________________¡£
(3)A¡úB+FµÄ»¯Ñ§·½³ÌʽΪ___________________________________¡£
(4)DÓëÒø°±ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________________________________________¡£
(5)·ûºÏÏÂÁÐÌõ¼þµÄMµÄͬ·ÖÒì¹¹ÌåÓÐ____________ÖÖ(²»¿¼ÂÇÁ¢ÌåÒì¹¹)¡£
¢Ù¹ÙÄÜÍÅÖÖÀàºÍÊýÄ¿ÓëMÏàͬ
¢Ú·Ö×ÓÖк¬ÓÐ1¸ö-CH3ºÍ1¸ö-CH2CH2-
¢Û²»º¬-CH2CH2CH2-
(6)Çë½áºÏËùѧ֪ʶºÍÉÏÊöÐÅÏ¢£¬Ð´³öÒÔ±½¼×È©ºÍÒ»ÂÈÒÒÍéΪÔÁÏ(ÎÞ»úÊÔ¼ÁÈÎÑ¡)£¬ÖƱ¸ÜлùÒÒÈ©()µÄºÏ³É·Ïߣº_________________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÎïÖÊÖÐÊôÓÚµç½âÖÊ£¬µ«ÔÚ¸ø¶¨Ìõ¼þϲ»Äܵ¼µçµÄÊÇ
A. Һ̬ÂÈ»¯Çâ B. ÕáÌÇ C. ÂÁ D. Ï¡ÏõËá
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿(1)0.5 mol Na2CO3Öк¬ÓÐ___________¸öNa+¡£
(2)ÖÊÁ¿¶¼ÊÇ50 gµÄ HCl¡¢NH3¡¢CO2¡¢O2ËÄÖÖÆøÌ壬ÔÚÏàͬζȺÍÏàͬѹǿÌõ¼þÏ£¬Ìå»ý×î´óµÄÊÇ____________¡£
(3)ÅäÖÆ90 mL 0.1 mol/L CuSO4ÈÜÒº£¬ÐèÒªµ¨·¯________g¡£
(4)Ñõ»¯»¹Ô·´Ó¦3S+6KOH=2K2S+K2SO3+3H2OÖУ¬Ñõ»¯¼ÁÓ뻹ԼÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ_____________£¬Èô·´Ó¦ÖÐÏûºÄÁË0.6molS£¬Ôò·´Ó¦ÖÐתÒƵĵç×ÓΪ________mol¡£
(5)ÏÖÓÐÏÂÁÐ10ÖÖÎïÖÊ£º¢ÙH2O¡¢¢ÚMg¡¢¢ÛCH3COOH¡¢¢ÜNaOH¡¢¢ÝCuSO4¡¤5H2O¡¢¢Þµâ¾Æ¡¢¢ßC2H5OH¡¢¢àÑÎËᣬ(½«ÐòºÅÌîÔÚÏàÓ¦µÄ¿Õ¸ñÄÚ)ÆäÖУ¬ÊôÓÚÇ¿µç½âÖʵÄÊÇ_____________¡£
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