ÌâÄ¿ÄÚÈÝ

ϱíÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬±íÖÐ×Öĸ·Ö±ð´ú±íijһ»¯Ñ§ÔªËØ£¬Çë»Ø´ðÓйØÎÊÌ⣮
£¨1£©dÔªËØµÄ»ù̬ԭ×ÓÍâΧµç×Ó£¨¼Ûµç×Ó£©ÅŲ¼Îª
3d104s1
3d104s1
£¬kµÄÔªËØ·ûºÅΪ
As
As
£®
£¨2£©eÊÇÖÜÆÚ±íÖеÄ
¢ø
¢ø
×å£¬ÔªËØÔ­×ӽṹʾÒâͼΪ
£®
£¨3£©aµÄÒõÀë×Ó°ë¾¶
´óÓÚ
´óÓÚ
bµÄÑôÀë×Ó°ë¾¶£¨Ìî´óÓÚ¡¢µÈÓÚ¡¢Ð¡ÓÚ£©£®
£¨4£©ÔÚÔªËØÖÜÆÚ±íÖÐÓÐÒ»¸ö¶Ô½ÇÏß¹æÔò£¬ÆäÖаüÀ¨b¡¢cµÄ»¯ºÏÎïµÄÐÔÖÊÊ®·ÖÏàËÆ£®ÔòbµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï·Ö±ðÓëÑÎËáºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ£¨Òª±í´ï³ö¾ßÌåµÄÔªËØ·ûºÅ£¬ÏÂͬ£©
Be£¨OH£©2+2H+¨TBe2++2H2O
Be£¨OH£©2+2H+¨TBe2++2H2O
£»
Be£¨OH£©2+2OH-¨TBeO2-+2H2O
Be£¨OH£©2+2OH-¨TBeO2-+2H2O
£®
£¨5£©Èôaµ¥ÖʵÄȼÉÕÈÈΪ285.8kJ/mol£¬Ð´³ö±íʾÆäȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ
H2£¨g£©+
1
2
O2£¨g£©=H2O£¨l£©£»¡÷H=-285.8kJ/mol
H2£¨g£©+
1
2
O2£¨g£©=H2O£¨l£©£»¡÷H=-285.8kJ/mol
£®
£¨6£©ÔªËØÖÜÆÚ±íÖеÚ6ÖÜÆÚÖеÄïçÏµÔªËØ¹²ÓÐ
15
15
ÖÖ£¬ËüÃÇÔ­×ӵĵç×Ó²ã½á¹¹ºÍÐÔÖÊÊ®·ÖÏàËÆ£®
·ÖÎö£ºÓÉÔªËØÔÚÖÜÆÚ±íÖеÄλÖÿÉÖªaΪHÔªËØ£¬bΪBeÔªËØ£¬cΪAlÔªËØ£¬dΪCuÔªËØ£¬eΪFeÔªËØ£¬fΪPbÔªËØ£¬gΪHeÔªËØ£¬hΪArÔªËØ£¬iΪKrÔªËØ£¬jΪFÔªËØ£¬kΪAsÔªËØ£¬½áºÏÔªËØ¶ÔÓ¦µÄµ¥ÖÊ¡¢»¯ºÏÎïµÄÐÔÖÊÒÔ¼°ÔªËØÖÜÆÚÂÉ֪ʶºÍÌâĿҪÇó¿É½â´ð¸ÃÌ⣮
½â´ð£º½â£º£¨1£©dΪCuÔªËØ£¬Î»ÓÚÖÜÆÚ±íµÚ4ÖÜÆÚ¢ñB×壬»ù̬ԭ×ÓÍâΧµç×Ó£¨¼Ûµç×Ó£©ÅŲ¼Îª3d104s1£¬kΪAsÔªËØ£¬
¹Ê´ð°¸Îª£º3d104s1£»As£»
£¨2£©eΪFeÔªËØ£¬Î»ÓÚÖÜÆÚ±íµÚ4ÖÜÆÚµÚ¢ø×壬ԭ×ӽṹʾÒâͼΪ£¬
¹Ê´ð°¸Îª£º¢ø£»£»
£¨3£©aΪHÔªËØ£¬bΪBeÔªËØ£¬aµÄÒõÀë×ÓÓëbµÄÑôÀë×ÓºËÍâµç×ÓÅŲ¼Ïàͬ£¬Óɺ˵çºÉÊýÔ½´ó°ë¾¶Ô½Ð¡¿ÉÖªaµÄÒõÀë×Ó°ë¾¶´óÓÚbµÄÑôÀë×Ó°ë¾¶£¬
¹Ê´ð°¸Îª£º´óÓÚ£»
£¨4£©bΪBeÔªËØ£¬cΪAlÔªËØ£¬b¡¢cµÄ»¯ºÏÎïµÄÐÔÖÊÊ®·ÖÏàËÆ£®ÔòbµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÓ¦¾ßÓÐÁ½ÐÔ£¬·Ö±ðÓëÑÎËáºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪBe£¨OH£©2+2H+¨TBe2++2H2O¡¢Be£¨OH£©2+2OH-¨TBeO2-+2H2O£¬
¹Ê´ð°¸Îª£ºBe£¨OH£©2+2H+¨TBe2++2H2O£» Be£¨OH£©2+2OH-¨TBeO2-+2H2O£»
£¨5£©aΪHÔªËØ£¬µ¥ÖʵÄȼÉÕÈÈΪ285.8kJ/mol£¬Ôò±íʾÆäȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪH2£¨g£©+
1
2
O2£¨g£©¨TH2O£¨l£©£»¡÷H=-285.8kJ/mol£¬
¹Ê´ð°¸Îª£ºH2£¨g£©+
1
2
O2£¨g£©¨TH2O£¨l£©£»¡÷H=-285.8kJ/mol£»
£¨6£©ÔªËØÖÜÆÚ±íÖеÚ6ÖÜÆÚÖеÄïçÏµÔªËØ¹²ÓÐ 15ÖÖ£¬Ô­×ÓÐòÊý½éÓÚ57¡«71Ö®¼ä£¬
¹Ê´ð°¸Îª£º15£®
µãÆÀ£º±¾Ìâ×ۺϿ¼²éÁËÔªËØÖÜÆÚ±í£¬ÖÜÆÚÂÉ£¬ºËÍâµç×ÓÅŲ¼µÈ֪ʶ£¬×ÛºÏÐÔÇ¿£¬Îª¸ß¿¼³£¼ûÌâÐÍ£¬ÄѶȲ»´ó£¬¿¼ÉúÔÚÆ½Ê±×¢Òâ¼ÓÇ¿»ýÀÛ£¬ÊìÁ·ÕÆÎÕ»ù´¡ÖªÊ¶£¬²é©²¹È±£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ϱíÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£®±íÖÐËùÁеÄ×Öĸ·Ö±ð´ú±íÒ»ÖÖ»¯Ñ§ÔªËØ£®
A B C D E
F G H
I J K
L
M N
O
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©IµÄÔªËØ·ûºÅΪ
Sc
Sc
£¬KµÄÔªËØÃû³ÆÎª
ïØ
ïØ
£®
£¨2£©Ð´³ö»ù̬ʱJÔªËØÔ­×ӵĵç×ÓÅŲ¼Ê½
1s22s22p63s23p63d64s2
1s22s22p63s23p63d64s2
£¬NÔªËØÔ­×ӵļò»¯µç×ÓÅŲ¼Ê½
[Xe]4f145d106s26p5
[Xe]4f145d106s26p5
£®
£¨3£©ÏÂÁжԱÈÕýÈ·µÄÊÇ
cd
cd
£®
a£®Ô­×Ó°ë¾¶H£¾G£¾B£¾A£»          b£®µÚÒ»µçÀëÄÜE£¾D£¾C£¾B£»
c£®µç¸ºÐÔA£¾H£¾G£¾K£»            d£®×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïËáÐÔB£¾A£¾H£¾G£»
£¨4£©ÏÂÁйØÓÚÔªËØÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÒÔ¼°ÔªËØÔ­×ÓµÄÍâΧµç×ÓÅŲ¼ÌصãµÄÓйØÐðÊö²»ÕýÈ·µÄÊÇ
bc
bc
£®
a£®LλÓÚÔªËØÖÜÆÚ±íÖеÚÎåÖÜÆÚ¡¢¢ñA×壬ÊôÓÚsÇøÔªËØ£»
b£®OλÓÚÔªËØÖÜÆÚ±íÖÐµÚÆßÖÜÆÚ¡¢¢øB×壬ÊôÓÚdÇøÔªËØ£»
c£®MµÄÍâΧµç×ÓÅŲ¼Ê½Îª6s1£¬ÊôÓÚdsÇøÔªËØ£»
d£®HËùÔÚ×åµÄÍâΧµç×ÓÅŲ¼Ê½Îªns2np2£¬ÊôÓÚpÇøÔªËØ£»
£¨5£©¶ÌÖÜÆÚÖÐijÖÖÔªËØµÄ×î¸ß¼ÛÑõ»¯Îï1molÓë12molNaOHÇ¡ºÃÍêÈ«·´Ó¦Éú³ÉÕýÑΣ¬¸ÃÔªËØÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃΪ
Èý
Èý
ÖÜÆÚ
¢õA
¢õA
×壮

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

¾«Ó¢¼Ò½ÌÍø