ÌâÄ¿ÄÚÈÝ

£¨Ã¿¿Õ1·Ö£¬¹²10·Ö£©¸ù¾ÝÒÑѧ֪ʶ£¬ÇëÄã»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©º¬ÓÐ8¸öÖÊ×Ó£¬10¸öÖÐ×ÓµÄÔ­×ӵĻ¯Ñ§·ûºÅ__________¡£
£¨2£©×îÍâ²ãµç×ÓÅŲ¼Îª4s24p1µÄÔ­×ӵĺ˵çºÉÊýΪ__________¡£
£¨3£©Ä³ÔªËر»¿Æѧ¼Ò³Æ֮ΪÈËÌå΢Á¿ÔªËØÖеġ°·À°©Ö®Íõ¡±£¬ÆäÔ­×ÓµÄÍâΧµç×ÓÅŲ¼ÊÇ4s24p4£¬¸ÃÔªËصÄÃû³ÆÊÇ_________¡£
£¨4£©¸ù¾ÝVSEPRÄ£ÐÍ£¬H3O+µÄ·Ö×ÓÁ¢Ìå½á¹¹Îª£º        £¬SO2µÄÁ¢Ìå½á¹¹Îª£º          ¡£
£¨5£©ÖÜÆÚ±íÖÐ×î»îÆõķǽðÊôÔªËØÔ­×ӵĹìµÀ±íʾʽΪ__________    ¡£
( 6 ) ÈýÂÈ»¯Ìú³£ÎÂÏÂΪ¹ÌÌ壬ÈÛµã282¡ãC£¬·Ðµã315¡ã£¬ÔÚ300¡ãCÒÔÉÏÒ×Éý»ª¡£Ò×ÈÜÓÚË®£¬Ò²Ò×ÈÜÓÚÒÒÃÑ¡¢±ûͪµÈÓлúÈܼÁ¡£¾Ý´ËÅжÏÈýÂÈ»¯Ìú¾§ÌåΪ________¡£
( 7 ) ijԪËغËÍâÓÐÈý¸öµç×Ӳ㣬×îÍâ²ãµç×ÓÊýÊǺËÍâµç×Ó×ÜÊýµÄ1/6£¬Ð´³ö¸ÃÔªËØÔ­×ӵĵç×ÓÅŲ¼Ê½ÊÇ__________¡£
( 8 )д³ö¸õÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃ__________£¬ËüλÓÚ__________Çø¡£
£¨1£© £¨1·Ö£©  £¨2£©31 £¨1·Ö£© £¨3£©Îø £¨1·Ö£© £¨4£©Èý½Ç׶ÐΣ¨1·Ö£©¡¢VÐÎ £¨1·Ö£©
£¨5£©£¨1·Ö£©£¨6£© ·Ö×Ó¾§Ìå £¨1·Ö£© £¨7£© 1s22s22p63s2»ò[Ne]3s2 £¨1·Ö£©
£¨8£© µÚËÄÖÜÆÚ¢öB×壨1·Ö£©¡¡d£¨1·Ö£©

ÊÔÌâ·ÖÎö£º£¨1£©ÔÚ±íʾԭ×Ó×é³ÉʱԪËØ·ûºÅµÄ×óϽDZíʾÖÊ×ÓÊý£¬×óÉϽDZíʾÖÊÁ¿Êý£¬Ôòº¬ÓÐ8¸öÖÊ×Ó£¬10¸öÖÐ×ÓµÄÔ­×ӵĻ¯Ñ§·ûºÅ¡£
£¨2£©×îÍâ²ãµç×ÓÅŲ¼Îª4s24p1µÄÔ­×ӵĺ˵çºÉÊýΪ2+8+18+3£½31¡£
£¨3£©Ä³ÔªËر»¿Æѧ¼Ò³Æ֮ΪÈËÌå΢Á¿ÔªËØÖеġ°·À°©Ö®Íõ¡±£¬ÆäÔ­×ÓµÄÍâΧµç×ÓÅŲ¼ÊÇ4s24p4£¬ÔòÆäÔ­×ÓÐòÊýÊÇ2+8+18+6£½34£¬ËùÒÔ¸ÃÔªËصÄÃû³ÆÊÇÎø¡£
£¨4£©H3O+Öм۲ãµç×Ó¶ÔÊýÊÇ4£¬ÆäÖÐÑõÔ­×Óº¬ÓÐ1¶Ô¹Â¶Ôµç×Ó£¬ËùÒÔ·Ö×ÓÁ¢Ìå½á¹¹ÎªÈý½Ç׶ÐΣ»SO2·Ö×Ó¼Û²ãµç×Ó¶ÔÊýÊÇ3£¬ÆäÖÐÁòÔ­×Óº¬ÓÐ1¶Ô¹Â¶Ôµç×Ó£¬ËùÒÔÁ¢Ìå½á¹¹ÎªVÐΡ£
£¨5£©ÖÜÆÚ±íÖÐ×î»îÆõķǽðÊôÔªËØÊÇFÔªËØ£¬ÔòÆäÔ­×ӵĹìµÀ±íʾʽΪ¡£
£¨6£©ÈýÂÈ»¯Ìú³£ÎÂÏÂΪ¹ÌÌ壬ÈÛµã282¡ãC£¬·Ðµã315¡ã£¬ÔÚ300¡ãCÒÔÉÏÒ×Éý»ª¡£Ò×ÈÜÓÚË®£¬Ò²Ò×ÈÜÓÚÒÒÃÑ¡¢±ûͪµÈÓлúÈܼÁ¡£¾Ý´ËÅжÏÈýÂÈ»¯Ìú¾§ÌåΪ·Ö×Ó¾§Ìå¡£
£¨7£©Ä³ÔªËغËÍâÓÐÈý¸öµç×Ӳ㣬×îÍâ²ãµç×ÓÊýÊǺËÍâµç×Ó×ÜÊýµÄ1/6£¬ÔòÉè×îÍâ²ãµç×ÓÊýÊÇx£¬ËùÒÔ£¨10+x£©¡Â6£½x£¬½âµÃx£½2£¬Òò´Ë¸ÃÔªËØÔ­×ӵĵç×ÓÅŲ¼Ê½ÊÇ1s22s22p63s2»ò[Ne]3s2¡£
£¨8£©¸õÔªËØÔÚÖÜÆÚ±íÖеÄλÖõÚËÄÖÜÆÚ¢öB×å¡£ÇøµÄÃû³ÆÀ´×ÔÓÚ°´ÕÕ¹¹ÔìÔ­Àí×îºóͨÈëµç×ӵĹìµÀÃû³Æ£¬ËùÒÔ¸õÔªËØλÓÚdÇø¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¡¾ÎïÖʽṹÓëÐÔÖÊ¡¿£¨15·Ö£©
¸ù¾ÝÔªËØÖÜÆÚ±íÖеÚËÄÖÜÆÚÔªËصÄÏà¹Ø֪ʶ£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©µÚËÄÖÜÆÚÔªËصĻù̬ԭ×ӵĵç×ÓÅŲ¼ÖÐ4s¹ìµÀÉÏÖ»ÓÐ1¸öµç×ÓµÄÔªËØÓÐ_______ÖÖ£»
д³öCu+µÄºËÍâµç×ÓÅŲ¼Ê½_________¡£
£¨2£©°´µç×ÓÅŲ¼£¬¿É½«ÖÜÆÚ±íÀïµÄÔªËØ»®·Ö³ÉÎå¸öÇøÓò£¬µÚËÄÖÜÆÚÔªËØÖÐÊôÓÚsÇøµÄÔªËØÓÐ___________ÖÖ£¬ÊôÓÚdÇøµÄÔªËØÓÐ________ÖÖ¡£
£¨3£©CaO¾§°ûÈçͼËùʾ£¬CaO¾§ÌåÖÐCa2£«µÄÅäλÊýΪ________£»CaOµÄÑæÉ«·´Ó¦Îª×©ºìÉ«£¬Ðí¶à½ðÊô»òËüÃǵĻ¯ºÏÎﶼ¿ÉÒÔ·¢ÉúÑæÉ«·´Ó¦£¬ÆäÔ­ÒòÊÇ________¡£

£¨4£©Óɵþµª»¯¼Ø(KN3)ÈÈ·Ö½â¿ÉµÃ´¿£¬ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ________£¨ÌîÑ¡Ïî×Öĸ£©¡£
A£®NaN3ÓëKN3½á¹¹ÀàËÆ£¬Ç°Õß¾§¸ñÄܽÏС
B£®¾§Ìå¼ØµÄ¾§°û½á¹¹ÈçÓÒͼËùʾ£¬Ã¿¸ö¾§°ûÖзÖ̯2¸ö¼ØÔ­×Ó
C£®µªµÄµÚÒ»µçÀëÄÜ´óÓÚÑõ
D£®µªÆø³£ÎÂϺÜÎȶ¨£¬ÊÇÒòΪµªµÄµç¸ºÐÔС
£¨5£©¶þÑõ»¯îÑ(TiO2)Êdz£Óõġ¢¾ßÓнϸߴ߻¯»îÐÔºÍÎȶ¨ÐԵĹâ´ß»¯¼Á¡£O2ÔÚÆä´ß»¯×÷ÓÃÏ£¬¿É½«CN£­Ñõ»¯³ÉCNO£­¡£CN£­µÄµç×ÓʽΪ________£¬CNO£­µÄÖÐÐÄÔ­×ÓµÄÔÓ»¯·½Ê½Îª____________¡£
£¨6£©ÔÚCrCl3ÈÜÒºÖУ¬Ò»¶¨Ìõ¼þÏ´æÔÚ×é³ÉΪ£¨nºÍx¾ùΪÕýÕûÊý£©µÄÅäÀë×Ó£¬½«Æäͨ¹ýÇâÀë×Ó½»»»Ê÷Ö¬(R-H)£¬¿É·¢ÉúÀë×Ó½»»»·´Ó¦£º
½«º¬0.0015molµÄÈÜÒº£¬ÓëR-HÍêÈ«½»»»ºó£¬ÖкÍÉú³ÉµÄÐèŨ¶ÈΪ
0.1200 mol/LNaOHÈÜÒº25.00 mL£¬Ôò¸ÃÅäÀë×ӵĻ¯Ñ§Ê½Îª____________¡£
[»¯Ñ§Ò»Ñ¡ÐÞ3£ºÎïÖʽṹÓëÐÔÖÊ]£¨15·Ö£©
£¨1£©Ô­×ÓÐòÊýСÓÚ36µÄX¡¢Y¡¢Z¡¢WËÄÖÖÔªËØ£¬ÆäÖÐXÊÇÔªËØÖÜÆÚ±íÔ­×Ӱ뾶×îСµÄÔªËØ£¬YÔ­×Ó»ù̬ʱ×îÍâ²ãµç×ÓÊýÊÇÆäÄÚ²ãµç×ÓÊýµÄ2±¶£¬ZÔ­×Ó»ù̬ʱ2pÔ­×Ó¹ìµÀÉÏÓÐ3¸öδ   ³É¶ÔµÄµç×Ó£¬wµÄÔ­×ÓÐòÊýΪ29¡£»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙY2X2·Ö×ÓÖÐYÔ­×Ó¹ìµÀµÄÔÓ»¯ÀàÐÍΪ_________£¬1molY2X2º¬ÓмüµÄÊýĿΪ____    __________________.
¢Ú»¯ºÏÎïZX3µÄ·Ðµã±È»¯Ì¨ÎïYX4µÄ¸ß£¬ÆäÖ÷ÒªÔ­ÒòÊÇ_____________¡£
¢ÛÔªËØYµÄÒ»ÖÖÑõ»¯ÎïÓëÔªËØzµÄÒ»ÖÖÑõ»¯ÎﻥΪµÈµç×ÓÌ壬ԪËØzµÄÕâÖÖÑõ»¯ÎïµÄ·Ö×ÓʽÊÇ_____________¡£
£¨2£©ÌúÔªËØÄÜÐγɶàÖÖÅäºÏÎÈ磺Fe(CO)x
¢Ù»ù̬Fe3£«µÄM²ãµç×ÓÅŲ¼Ê½Îª___________________________¡£
¢ÚÅäºÏÎïFe(CO)xµÄÖÐÐÄÔ­×Ó¼Ûµç×ÓÊýÓëÅäÌåÌṩµç×ÓÊýÖ®ºÍΪ18£®Ôòx=_________¡£³£ÎÂϳÊҺ̬£¬ÈÛµãΪ-20.5¡æ£¬·ÐµãΪ103¡æ£¬Ò×ÈÜÓڷǼ«ÐÔÈܼÁ£¬¾Ý´Ë¿ÉÅжÏFe(CO)x¾§ÌåÊôÓÚ
£¨ÌÌåÀàÐÍ£©£º
£¨3£©OºÍNaÐγɵÄÒ»ÖÖÖ»º¬ÓÐÀë×Ó¼üµÄÀë×Ó»¯ºÏÎïµÄ¾§°û½á¹¹Èçͼ£¬¾àÒ»¸öÒõÀë×ÓÖÜΧ×î½üµÄËùÓÐÑôÀë×ÓΪ¶¥µã¹¹³ÉµÄ¼¸ºÎÌåΪ_________¡£ÒÑÖª¸Ã¾§°ûµÄÃܶÈΪ g/cm£¬°¢·ü¼ÓµÂÂÞ³£ÊýΪ£¬Ç󾧰û±ß³¤a=________cm¡££¨Óú¬µÄ¼ÆËãʽ±íʾ£©

£¨4£©ÏÂÁÐÓйصÄ˵·¨ÕýÈ·µÄÊÇ_____________¡£
A£®µÚÒ»µçÀëÄÜ´óС£ºS>P>Si
B£®µç¸ºÐÔ˳Ðò£ºC<N<O<F
C£®ÒòΪ¾§¸ñÄÜCa0±ÈKCl¸ß£¬ËùÒÔKCl±ÈCaOÈÛµãµÍ
D£®SO2ÓëCO2µÄ»¯Ñ§ÐÔÖÊÀàËÆ£¬·Ö×ӽṹҲ¶¼³ÊÖ±ÏßÐÍ£¬ÏàͬÌõ¼þÏÂSO2µÄÈܽâ¶È¸ü´ó
E.·Ö×Ó¾§ÌåÖУ¬¹²¼Û¼ü¼üÄÜÔ½´ó£¬¸Ã·Ö×Ó¾§ÌåµÄÈ۷еãÔ½¸ß
¡¾Ñ¡ÐÞ3ÎïÖʽṹÓëÐÔÖÊ¡¿£¨15·Ö£©
VIA×åµÄÑõ¡¢Áò¡¢Îø(Se)¡¢íÚ(Te)µÈÔªËØÔÚ»¯ºÏÎïÖг£±íÏÖ³ö¶àÖÖÑõ»¯Ì¬£¬º¬VIA×åÔªËصĻ¯Ì¨ÎïÔÚÑо¿ºÍÉú²úÖÐÓÐÐí¶àÖØÒªÓÃ;¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Sµ¥Öʵij£¼ûÐÎʽΪS8£¬Æä»·×´½á¹¹ÈçÏÂͼËùʾ£¬SÔ­×Ó²ÉÓõĹìµÀÔÓ»¯·½Ê½ÊÇ      £»
 
£¨2£©Ô­×ӵĵÚÒ»µçÀëÄÜÊÇÖ¸Æø̬µçÖÐÐÔ»ù̬ԭ×Óʧȥһ¸öµç×Óת»¯ÎªÆø̬»ù̬ÕýÀë
×ÓËùÐèÒªµÄ×îµÍÄÜÁ¿£¬O¡¢S¡¢SeÔ­×ӵĵÚÒ»µçÀëÄÜÓÉ´óµ½Ð¡µÄ˳ÐòΪ               £»
£¨3£©SeÔ­×ÓÐòÊýΪ       £¬ÆäºËÍâM²ãµç×ÓµÄÅŲ¼Ê½Îª                    £»
£¨4£©H2SeµÄËáÐÔ±ÈH2S         £¨Ìî¡°Ç¿¡±»ò¡°Èõ¡±£©¡£Æø̬SeO3·Ö×ÓµÄÁ¢Ìå¹¹ÐÍ
Ϊ                 £¬SO32-Àë×ÓµÄÁ¢Ìå¹¹ÐÍΪ                   £»
£¨5£©H2SeO3µÄK1ºÍK2·Ö±ðΪ2.7x l0-3ºÍ2.5x l0-8£¬H2SeO4µÚÒ»²½¼¸ºõÍêÈ«µçÀ룬
K2Ϊ1.2X10-2£¬Çë¸ù¾Ý½á¹¹ÓëÐÔÖʵĹØϵ½âÊÍ£º
¢ÙH2SeO3ºÍH2SeO4µÚÒ»²½µçÀë³Ì¶È´óÓÚµÚ¶þ²½µçÀëµÄÔ­Òò£º                          
                                                   £»
¢Ú H2SeO4±È H2SeO3ËáÐÔÇ¿µÄÔ­Òò£º                                                  
                                                                              
£¨6£©ZnSÔÚÓ«¹âÌå¡¢¹âµ¼Ìå²ÄÁÏ¡¢Í¿ÁÏ¡¢ÑÕÁϵÈÐÐÒµÖÐÓ¦Óù㷺¡£Á¢·½ZnS¾§Ìå½á¹¹ÈçÏÂͼËùʾ£¬Æ侧°û±ß³¤Îª540.0 pm£®ÃܶÈΪ                    £¨ÁÐʽ²¢¼ÆË㣩£¬aλÖÃS2-Àë×ÓÓëbλÖÃZn2+Àë×ÓÖ®¼äµÄ¾àÀëΪ               pm£¨ÁÐʾ±íʾ£©

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

¾«Ó¢¼Ò½ÌÍø